Advanced Engineering Mathematics 10th Edition , ERWIN KREYSZIG Solution Manual by Ace Test Banks

Advanced Engineering Mathematics 10th Edition By ERWIN KREYSZIG

Email: Richard@qwconsultancy.com

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Part A. ORDINARY DIFFERENTIAL EQUATIONS (ODEs) CHAPTER 1

First-Order ODEs

Major Changes There is more material on modeling in the text as well as in the problem set. Some additions on population dynamics appear in Sec. 1.5. Team Projects, CAS Projects, and CAS Experiments are included in most problem sets. SECTION 1.1. Basic Concepts. Modeling, page 2 Purpose. To give the students a first impression of what an ODE is and what we mean by solving it. The role of initial conditions should be emphasized since, in most cases, solving an engineering problem of a physical nature usually means finding the solution of an initial value problem (IVP). Further points to stress and illustrate by examples are: The fact that a general solution represents a family of curves. The distinction between an arbitrary constant, which in this chapter will always be denoted by c, and a fixed constant (usually of a physical or geometric nature and given in most cases). The examples of the text illustrate the following. Example 1: the verification of a solution Examples 2 and 3: ODEs that can actually be solved by calculus with Example 2 giving an impression of exponential growth (Malthus!) and decay (radioactivity and further applications in later sections) Example 4: the straightforward solution of an IVP Example 5: a very detailed solution in all steps of a physical IVP involving a physical constant k Background Material. For the whole chapter we need integration formulas and techniques from calculus, which the student should review. General Comments on Text This section should be covered relatively rapidly to get quickly to the actual solution methods in the next sections. Equations (1)–(3) are just examples, not for solution, but the student will see that solutions of (1) and (2) can be found by calculus. Instead of (3), one could perhaps take a third-order linear ODE with constant coefficients or an Euler–Cauchy equation, both not of great interest. The present (3) is included to have a nonlinear ODE (a concept that will be mentioned later when we actually need it); it is not too difficult to verify that a solution is y⫽ with arbitrary constants a, b, c, d.

ax ⫹ b cx ⫹ d 1

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Problem Set 1.1 will help the student with the tasks of Solving y r ⫽ f (x) by calculus Finding particular solutions from given general solutions Setting up an ODE for a given function as solution, e.g., y ⫽ ex Gaining a first experience in modeling, by doing one or two problems Gaining a first impression of the importance of ODEs without wasting time on matters that can be done much faster, once systematic methods are available. Comment on “General Solution” and “Singular Solution” Usage of the term “general solution” is not uniform in the literature. Some books use the term to mean a solution that includes all solutions, that is, both the particular and the singular ones. We do not adopt this definition for two reasons. First, it is frequently quite difficult to prove that a formula includes all solutions; hence, this definition of a general solution is rather useless in practice. Second, linear differential equations (satisfying rather general conditions on the coefficients) have no singular solutions (as mentioned in the text), so that for these equations a general solution as defined does include all solutions. For the latter reason, some books use the term “general solution” for linear equations only; but this seems very unfortunate.

SOLUTIONS TO PROBLEM SET 1.1, page 8 2. y ⫽ eⴚx >2 ⫹ c 2

4. y ⫽ ceⴚ1.5x 6. y ⫽ a cos x ⫹ b sin x 8. y ⫽ ⫺

1 0.23

eⴚ0.2x ⫹ c1x 2 ⫹ c2x ⫹ c3

10. y ⫽ peⴚ2.5x

12. y 2 ⫺ 4x 2 ⫽ 12 14. y ⫽ 4 ⫺ 4 sin2 x 16. Substitution of y ⫽ cx ⫺ c2 into the ODE gives y r 2 ⫺ xy r ⫹ y ⫽ c2 ⫺ xc ⫹ (cx ⫺ c2) ⫽ 0. Similarly, y ⫽ 14 x 2,

y r ⫽ 12 x,

18. eⴚ3.6k ⫽ 12 , k ⫽ 0.19254, ˛

thus

(a) eⴚk ⫽ 0.825,

1 2 1 1 2 4 x ⫺ x (2 x) ⫹ 4 x ⫽ 0.

(b) 3.012 # 10ⴚ31.

20. k follows from e18,000k ⫽ 12, k ⫽ (ln 12)>18,000 ⫽ ⫺0.000039. Answer: e35,000k ⫽ 0.26y0. Since the decay is exponential, 36,000 ⫽ 2 # 18,000 would give (y0>2)>2 ⫽ 0.25y0.

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SECTION 1.2. Geometric Meaning of y r ⴝ f (x, y ). Direction Fields, Euler’s Method, page 9 Purpose. To give the student a feel for the nature of ODEs and the general behavior of fields of solutions. This amounts to a conceptual clarification before entering into formal manipulations of solution methods, the latter being restricted to relatively small—albeit important—classes of ODEs. This approach is becoming increasingly important, especially because of the graphical power of computer software. It is the analog of conceptual studies of the derivative and integral in calculus as opposed to formal techniques of differentiation and integration. Comment on Order of Sections This section could equally well be presented later in Chap. 1, perhaps after one or two formal methods of solution have been studied. Euler’s method has been included for essentially two reasons, namely, as an early eye opener to the possibility of numerically obtaining approximate values of solutions by step-by-step computations and, secondly, to enhance the student’s conceptual geometric understanding of the nature of an ODE. Furthermore, the inaccuracy of the method will motivate the development of much more accurate methods by practically the same basic principle (in Sec. 21.1). Problem Set 1.2 will help the student with the tasks of: Drawing direction fields and approximate solution curves Handling your CAS in selecting appropriate windows for specific tasks A first look at the important Verhulst equation (Prob. 4) Bell-shaped curves as solutions of a simple ODE Outflow from a vessel (analytically discussed in the next section) Discussing a few types of motion for given velocity (Parachutist, etc.) Comparing approximate solutions for different step size SOLUTIONS TO PROBLEM SET 1.2, page 11 2. Ellipses x 2 ⫹ 14 y 2 ⫽ c. If your CAS does not give you what you expected, change the given point. 4. Verhulst equation, to be discussed as a population model in Sec. 1.5. The given points correspond to constant solutions [(0, 0) and (0, 2)] , an increasing solution through (0, 1), and a decreasing solution through (0, 3). 6. Solution y(x) ⫽ ⫺arctan [1>(x ⫹ c)] , not needed for doing the problem. 2 8. ODE of the bell-shaped curves y ⫽ ceⴚx . 10. ODE of the outflow from a vessel, to be discussed in Sec. 1.3. 12. y ⫽ 21t ⫹ 1, not needed to do the problem. 14. y(x) ⫽ sin (x ⫹ 14 p), not needed to do the problem. 16. (a) Your PC may give you fields of varying quality, depending on the choice of the region graphed, and good choices are often obtained only after some trial and error. Enlarging generally gives more details. Subregions where ƒ y r ƒ is large are usually critical and often tend to give nonsense.

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(b) 2x ⫹ 18yy r ⫽ 0. Your CAS will produce the direction field well, even at points of the x-axis where the tangents of solution curves are vertical. (c) y 2 ⫹ x 2 ⫽ c (not needed for doing the problem). (d) y ⫽ ceⴚx>2 by remembering calculus. 18. y ⫽ ex. The computed value for x ⫽ 0.1 shows that its error has decreased by about a factor 10. This is typical for this “first-order method” (Euler’s method), as will be seen in Sec. 21.1. xn

y(x n)

Error

Error in Prob. 17

0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.09 0.10

1.010000 1.020100 1.030301 1.040604 1.051010 1.061520 1.072135 1.082857 1.093685 1.104622

1.010050 1.020201 1.030455 1.040811 1.051271 1.061837 1.072508 1.083287 1.094174 1.105171

0.000050 0.000101 0.000154 0.000207 0.000261 0.000317 0.000373 0.000430 0.000489 0.000549

0.005171

20. The error is first negative, then positive, and finally decreases as the solution (which is decreasing for all positive x) approaches the limit 0. The computed values are: xn

Error

0 0.2 0.4 0.6 0.8 1.0 1.2 1.4 1.6 1.8 2.0

1.0000 1.0000 0.9984 0.9729 0.8502 0.5541 0.2471 0.1205 0.0647 0.0373 0.0227

0 0.0000 ⫺0.0083 ⫺0.0453 ⫺0.0972 ⫺0.0541 0.0396 0.0363 0.0223 0.0130 0.0076

SECTION 1.3. Separable ODEs. Modeling, page 12 Purpose. To familiarize the student with the first “big” method of solving ODEs, the separation of variables, and an extension of it, the reduction to separable form by a transformation of the ODE, namely, by introducing a new unknown function. The section includes standard applications that lead to separable ODEs, namely, 1–3. Three simple2 separable ODEs with solutions involving tan x, an exponential function, eⴚx (bell-shaped curves) 4. The ODE of the exponential function, having various applications, such as in radiocarbon dating

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5. A mixing problem for a single tank 6. Newton’s law of cooling 7. Torricelli’s law of outflow In reducing to separability we consider 8. The transformation u ⫽ y>x, giving perhaps the most important reducible class of ODEs Ince’s classical book [A11] contains many further reductions as well as a systematic theory of reduction for certain classes of ODEs. Comment on Problem 5 From the implicit solution we can get two explicit solutions y ⫽ ⫹ 2c ⫺ (6x)2 representing semi-ellipses in the upper half-plane, and y ⫽ ⫺2c ⫺ (6x)2 representing semi-ellipses in the lower half-plane. [Similarly, we can get two explicit solutions x(y) representing semi-ellipses in the left and right half-planes, respectively.] On the x-axis, the tangents to the ellipses are vertical, so that y r (x) does not exist. Similarly for x r (y) on the y-axis. This also illustrates that it is natural to consider solutions of ODEs on open rather than on closed intervals. Comment on Separability An analytic function f (x, y) in a domain D of the xy-plane can be factored in D, f (x, y) ⫽ g(x)h(y), if and only if in D, fxy f ⫽ fx fy [D. Scott, American Math. Monthly 92 (1985), 422–423]. Simple cases are easy to decide, but this may save time in cases of more complicated ODEs, some of which may perhaps be of practical interest. You may perhaps ask your students to derive such a criterion. Comments on Application Each of those examples can be modified in various ways, for example, by changing the application or by taking another form of the tank, so that each example characterizes a whole class of applications. The many ODEs in the problem set, much more than one would ordinarily be willing and have the time to consider, should serve to convince the student of the practical importance of ODEs; so these are ODEs to choose from, depending on the students’ interest and background. Comment on Footnote 3 Newton conceived his method of fluxions (calculus) in 1665–1666, at the age of 22. Philosophiae Naturalis Principia Mathematica was his most influential work. Leibniz invented calculus independently in 1675 and introduced notations that were essential to the rapid development in this field. His first publication on differential calculus appeared in 1684.

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SOLUTIONS TO PROBLEM SET 1.3, page 18 2. y 3 dy ⫽ ⫺x 3 dx, 14 y 4 ⫽ ⫺14 x 4 ⫹ ~ c , so that multiplication by 4 gives the answer y 4 ⫹ x 4 ⫽ c. These are curves that lie between a circle and a square, outside the circle and inside the square that touch the circle at the points of intersection with the axes. The figure shows a quarter of such a curve for c ⫽ 1. 1

1 t

Sec. 1.3.

Prob. 2. Quarter of the solution curve

4. Separation, integration, and taking exponents gives dy>y ⫽ p cot 2px dx,

ln ƒ y ƒ ⫽ 12 ln ƒ sin 2px ƒ ⫹ c,

and y ⫽ c1sin 2px. 6. Separation of variables, integration, and taking the reciprocal gives dy y

⫽ e2xⴚ1 dx,

1 ⫺ ⫽ 12 e2xⴚ1 ⫹ ~ c y

y⫽

2 c ⫺ e2xⴚ1

8. From the ODE and the suggested transformation we obtain y r ⫽ v r ⫺ 4 ⫽ v2,

hence

v r ⫽ v2 ⫹ 4.

Separation of variables and integration gives dv v2 ⫹ 4

⫽ dx

and

v 1 ⫽x⫹~ c. 2 arctan 2

This implies v ⫽ 2 tan (2x ⫹ c) and gives the answer y ⫽ v ⫺ 4x ⫽ 2 tan (2x ⫹ c) ⫺ 4x. 10. From the transformation and the ODE we have yr ⫽ urx ⫹ u ⫽ 1 ⫹

y ⫽ 1 ⫹ u, x

hence

u r x ⫽ 1.

Separation of variables, integration, and again using the transformation gives du ⫽ dx>x, u ⫽ ln x ⫹ c,

y ⫽ ux ⫽ x (ln x ⫹ c).

12. Separation of variables and integration gives dy 1 ⫹ 4y 2

⫽ dx

and

1 ~ 2 arctan 2y ⫽ x ⫹ c .

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Hence arctan 2y ⫽ 2x ⫹ c. Solving for y gives the general solution y ⫽ 12 tan (2x ⫹ c) and c ⫽ ⫺2 from the initial condition. 14.

2 dr ⫽ ⫺2t dt, ln r ⫽ ⫺t 2 ⫹ ~ c . The general solution is r ⫽ ceⴚt and c ⫽ r0 from r the initial condition.

16. From the transformation and the ODE we have y⫽v⫺x⫹2

y r ⫽ v r ⫺ 1 ⫽ v2.

and

Hence v r ⫽ v2 ⫹ 1. Separation of variables and integration gives dv v ⫹1 2

⫽ dx

and

arctan v ⫽ x ⫹ c

hence v ⫽ tan (x ⫹ c).

From this and the transformation we obtain y ⫽ v ⫺ x ⫹ 2 ⫽ 2 ⫺ x ⫹ tan (x ⫹ c). From the initial condition we get y(0) ⫽ 2 ⫹ tan c ⫽ 0 and c ⫽ 0, so that the answer is y ⫽ tan x ⫺ x ⫹ 2. 18. On the left, integrate g from y0 to y. On the right, integrate f (x) over x from x 0 to x. In Prob. 12,

冮

w dw ⫽

冮 (⫺4t) dt. 2

20. Let k B and k D be the constants of proportionality for the birth rate and death rate, respectively. Then y r ⫽ k By ⫺ k Dy, where y(t) is the population at time t. By separating variables, integrating, and taking exponents, dy>y ⫽ (k B ⫺ k D) dt,

ln y ⫽ (k B ⫺ k D)t ⫹ c*,

y ⫽ ce(kBⴚkD)t.

22. The acceleration is a ⫽ 9 ⴢ 106 meters>sec2, and the distance traveled is 5.5 meters. This is obtained as follows. Since s(0) ⫽ 0 (i.e., we count time from the instant the particle enters the accelerator), we have for a motion of constant acceleration s(t) ⫽ a

(A)

t2 ⫹ bt 2

and the velocity is v(t) ⫽ s r (t) ⫽ at ⫹ b. From the given data we thus obtain v(0) ⫽ b ⫽ 103 and v(10ⴚ3) ⫽ 10ⴚ3a ⫹ 103 ⫽ 104 so that a ⫽ 103(104 ⫺ 103) ⫽ 107 ⫺ 106 ⫽ 9 ⴢ 106. Finally, with this a and that b, from (A) we get s(10ⴚ3) ⫽ 9 # 106 #

10ⴚ6 ⫹ 103 # 10ⴚ3 ⫽ 5.5 [m] . 2

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24. Let y(t) be the amount of salt in the tank at time t. Then each gallon contains y>400 lb of salt. 2¢t gal of water run in during a short time ¢t, and ⫺¢y ⫽ 2¢t (y>400) ⫽ ¢t y>200 is the loss of salt during ¢t. Thus ¢y>¢t ⫽ ⫺y>200, y r ⫽ ⫺0.005y, y(t) ⫽100eⴚ0.005t. Answer: y(60) ⫽ 100eⴚ0.3 ⫽ 74 [lb]. 26. The model is y r ⫽ ⫺Ay ln y with A ⬎ 0. Constant solutions are obtained from y r ⫽ 0 when y ⫽ 0 and 1. Between 0 and 1 the right side is positive (since ln y ⬍ 0), so that the solutions grow. For y ⬎ 1 we have ln y ⬎ 0; hence the right side is negative, so that the solutions decrease with increasing t. It follows that y ⫽ 1 is stable. The general solution is obtained by separation of variables, integration, and two subsequent exponentiations: dy>(y ln y) ⫽ ⫺A dt,

ln (ln y) ⫽ ⫺At ⫹ c*,

ln y ⫽ ceⴚAt,

y ⫽ exp (ceⴚAt).

28. This follows from the inquality 1>26 ⫽ 0.016 ⬎ 0.010 ⬎ 1>27 ⫽ 0.0078.

30. Acceleration y s ⫽ 7t. Hence y r ⫽ 7t 2>2, y ⫽ 7t 3>6, y r (10) ⫽ 350 (initial speed of further flight ⫽ end speed upon return from peak), y(10) ⫽ 7000>6 ⫽ 1167 (height reached after the 10 sec). At the peak, v ⫽ 0, s ⫽ 0, say; thus for the further flight (measured from the peak), s(t) ⫽ (g>2)t 2 ⫽ 4.9t 2, v(t) ⫽ 9.8t ⫽ 350 (see before). This gives the further flight time to the peak t ⫽ t 1 ⫽ 350>9.8 ⫽ 35.7 and the further height s(t 1) ⫽ 4.9t 12 ⫽ 6245, approximately. Answer: 1167 ⫹ 6245 ⫽ 7412 [m]. 32. W ⫽ mg in Fig. 15 is the weight (the force of attraction acting on the body). Its component parallel to the surface in mg sin a, and N ⫽ mg cos a. Hence the friction is 0.2mg cos a, and it acts against the direction of motion. From this and Newton’s second law, noting that the acceleration is dv>dt (v the velocity), we obtain m

dv ⫽ mg sin a ⫺ 0.2mg cos a dt ⫽ m # 9.80(0.500 ⫺ 0.2 # 0.866) ⫽ 3.203 m.

The mass m drops out, and two integrations give v ⫽ 3.203t

and

s ⫽ 3.203

t2 . 2

Since the slide is 10 meters long, the last equation with s ⫽ 10 gives the time t ⫽ 12 # 10>3.203 ⫽ 2.50. From this we obtain the answer v ⫽ 3.203 # 2.50 ⫽ 8.01 [meters>sec] . 34. TEAM PROJECT. (a) Note that at the origin, x>y ⫽ 0>0, so that y r is undefined at the origin. (b) (xy) r ⫽ y ⫹ xy r ⫽ 0, y r ⫽ ⫺y>x. (c) y ⫽ cx. Here the student should learn that c must not appear in the ODE. y>x ⫽ cy r >x ⫺ y>x 2 ⫽ 0, y r ⫽ y>x.

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(d) The right sides ⫺x>y and y>x are the slopes of the curves. Orthogonality is important and will be discussed further in Sec. 1.6. (e) No. 36. Team Project. B now depends on h, namely, by the Pythagorean theorem, B(h) ⫽ pr 2 ⫽ p(R 2 ⫺ (R ⫺ h)2) ⫽ p(2Rh ⫺ h2). Hence you can use the ODE h r ⫽ ⫺26.56(A>B)1h in the text, with constant A as before and the new B. The latter makes further calculations different from those in Example 5. From the given outlet size A ⫽ 5 cm2 and B(h) we obtain dh 5 ⫽ ⫺26.56 # 1h. dt p(2Rh ⫺ h2) Now 26.56 # 5> p ⫽ 42.27, so that separation of variables gives (2Rh1>2 ⫺ h3>2) dh ⫽ ⫺42.27 dt. By integration, 4 3>2 ⫺ 25 h5>2 ⫽ ⫺42.27t ⫹ c. 3 Rh

From this and the initial conditions h(0) ⫽ R we obtain 4 5>2 ⫺ 25 R 5>2 ⫽ 0.9333R5>2 ⫽ c. 3R

Hence the particular solution (in implict form) is 4 3>2 ⫺ 25 h5>2 ⫽ ⫺42.27t ⫹ 0.9333R 5>2. 3 Rh

The tank is empty (h ⫽ 0) for t such that 0 ⫽ ⫺42.27t ⫹ 0.9333R5>2;

hence

t⫽

0.9333 5>2 R ⫽ 0.0221R5>2. 42.27

For R ⫽ 1 m ⫽ 100 cm this gives t ⫽ 0.0221 # 1005>2 ⫽ 2210 [sec] ⫽ 37 [min] . The tank has water level R>2 for t in the particular solution such that 4 R 3>2 2 R5>2 R 3>2 ⫺ ⫽ 0.9333R 5>2 ⫺ 42.27t. 3 2 5 25>2 The left side equals 0.4007R5>2. This gives t⫽

0.4007 ⫺ 0.9333 5>2 R ⫽ 0.01260R 5>2. ⫺42.27

For R ⫽ 100 this yields t ⫽ 1260 sec ⫽ 21 min. This is slightly more than half the time needed to empty the tank. This seems physically reasonable because if the water level is R> 2, this means that 11>16 of the total water volume has flown out, and 5>16 is left—take into account that the velocity decreases monotone according to Torricelli’s law.

R=h r h

Problem Set 1.3. Tank in Team Project 36

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SECTION 1.4. Exact ODEs. Integrating Factors, page 20 Purpose. This is the second “big” method in this chapter, after separation of variables, and also applies to equations that are not separable. The criterion (5) is basic. Simpler cases are solved by inspection, more involved cases by integration, as explained in the text. Comment on Condition (5) Condition (5) is equivalent to (6 s ) in Sec. 10.2, which is equivalent to (6) in the case of two variables x, y. Simple connectedness of D follows from our assumptions in Sec. 1.4. Hence the differential form is exact by Theorem 3, Sec. 10.2, part (b) and part (a), in that order. Method of Integrating Factors This greatly increases the usefulness of solving exact equations. It is important in itself as well as in connection with linear ODEs in the next section. Problem Set 1.4 will help the student gain skill needed in finding integrating factors. Although the method has somewhat the flavor of tricks, Theorems 1 and 2 show that at least in some cases one can proceed systematically—and one of them is precisely the case needed in the next section for linear ODEs. In Example 2, exactness is seen from 0 cos y sinh x ⫹ 1 ⫽ ⫺sin y sinh x. 0y 0 (⫺sin y cosh x) ⫽ ⫺sin y sinh x. 0x In Example 3, separation of variables gives dy dx ⫽ , y x

y ⫽ cx.

SOLUTIONS TO PROBLEM SET 1.4, page 26 2. Exact, x 4 ⫹ y 4 ⫽ c. Note that an ODE f (x) dx ⫹ g(y) dy ⫽ 0 is always exact. 4. Exact. The test gives 3e3u ⫽ 3e3u. By integration, u⫽

冮 e dr ⫽ re ⫹ c(u). 3u

Hence u u ⫽ 3re3u ⫹ c r ⫽ 3re3u,

c r ⫽ 0,

c ⫽ const

6. The new ODE is 3(y ⫹ 1)2xⴚ4 dx ⫺ 2(y ⫹ 1)x ⴚ3 dy ⫽ 0. It is exact, M y ⫽ Nx ⫽ 6(y ⫹ 1)x ⴚ4.

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The general solution is (y ⫹ 1) 2x ⴚ3 ⫽ c. 8. Exact; the test gives ⫺ex sin y on both sides. Integrate M with respect to x: u ⫽ ex cos y ⫹ k(y).

u y ⫽ ⫺ex sin y ⫹ k r .

Differentiate:

Equate this to N ⫽ ⫺ex sin y. Hence k r ⫽ 0, k ⫽ const. Answer: ex cos y ⫽ c. 10. y cos (x ⫹ y) dx ⫹ [y cos (x ⫹ y) ⫹ sin (x ⫹ y)] dy ⫽ 0 is exact because [y cos (x ⫹ y)] y ⫽ cos (x ⫹ y) ⫺ y sin (x ⫹ y) ⫽ [y cos (x ⫹ y) ⫹ sin (x ⫹ y)] x. By inspection or systematically, y sin (x ⫹ y) ⫽ c. 2

12. (2xyex )y ⫽ 2xex ⫽ (ex )x shows exactness. By integration, 2

yex ⫽ c. y(0) ⫽ 2 gives c ⫽ 2. Answer: y ⫽ 2eⴚx . 2

14. The integrating factor gives the exact ODE (a ⫹ 1) x ay bⴙ1 dx ⫹ (b ⫹ 1) x aⴙ1y b dy ⫽ d(x aⴙ1y bⴙ1) ⫽ 0. The general solution is x aⴙ1y bⴙ1 ⫽ c and c ⫽ 1 from the initial condition. 16. Team Project. (a) ey cosh x ⫽ c (b) R* ⫽ tan y, F ⫽ 1>cos y. Separation: dy>cos2 y ⫽ ⫺(1 ⫹ 2x) dx,

tan y ⫽ ⫺x ⫺ x 2 ⫹ c.

x 2 ⫺ y 2 ⫽ cx;

divide by x. (d) Separation is simplest. y ⫽ cx ⴚ3>4. R ⫽ ⫺9>(4x), F(x) ⫽ x ⴚ9>4, x 3y 4 ⫽ c. R* ⫽ 3>y, F*(y) ⫽ y 3. 18. CAS Project.

(a) Theorem 1 does not apply. Theorem 2 gives

1 dF ⫺1 2 ⫽ 2 (0 ⫹ 2y sin x) ⫽ ⫺ , y F dy y sin x

冮

2 1 F ⫽ exp ⫺ dy ⫽ 2 . y y

The exact ODE is y ⴚ2 dy ⫺ sin x dx ⫽ 0, as one could have seen by inspection—any equation of the form f (x) dx ⫹ g(y) dy ⫽ 0

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is exact! We now obtain

冮

u ⫽ ⫺sin x dx ⫽ cos x ⫹ k(y) u y ⫽ k r (y) ⫽

⫽ sin x dx,

1 ⫺ ⫽ ⫺cos x ⫹ c, y

y 1 u ⫽ cos x ⫺ ⫽ c. y

1 k⫽⫺ , y

(b) Yes, y r ⫽ y 2 sin x,

dy y

y⫽

1 . cos x ⫹ 苲 c ˛

(c) The vertical asymptotes that some CAS programs draw disturb the graph. From the solution in (b) the student should conclude that for each initial condition y(x 0) ⫽ y0 with y0 ⫽ 0 there is a unique particular solution because from (b), 1 ⫺ y0 cos x 0 苲 . c ⫽ y0 (d) y ⬅ 0. SECTION 1.5. Linear ODEs. Bernoulli Equation. Population Dynamics, page 27 Purpose. Linear ODEs are of great practical importance, as Problem Set 1.5 illustrates (and even more so are second-order linear ODEs in Chap. 2). We show that the homogeneous ODE of the first order is easily separated and the nonhomogeneous ODE is solved, once and for all, in the form of an integral (4) by the method of integrating factors. Of course, in simpler cases one does not need (4), as our examples illustrate. Comment on Notation We write y r ⫹ p(x)y ⫽ r(x). p(x) seems standard, r(x) suggests “right side.” The notation y r ⫹ p(x)y ⫽ q(x) used in some calculus books (which are not concerned with higher order ODEs) would be shortsighted here because later, in Chap. 2, we turn to second-order ODEs y s ⫹ p(x)y r ⫹ q(x)y ⫽ r(x), where we need q(x) on the left, thus in a quite different role (and on the right we would have to choose another letter different from that used in the first-order case). Comment on Content Bernoulli’s equation appears occasionally in practice, so the student should remember how to handle it. A special Bernoulli equation, the Verhulst equation, plays a central role in population dynamics of humans, animals, plants, and so on, and we give a short introduction to this interesting field, along with one reference in the text.

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Riccati and Clairaut equations are less important than Bernoulli’s, so we have put them in the problem set; they will not be needed in our further work. Input and output have become common terms in various contexts, so we thought this a good place to mention them. Problems 15–20 express properties that make linearity important, notably in obtaining new solutions from given ones. The counterparts of these properties will, of course, reappear in Chap. 2. Comment on Footnote 7 Eight members of the Bernoulli family became known as mathematicians; for more details, see p. 220 in Ref. [GenRef 2] listed in App. 1. Examples in the Text. The examples in the text concern the following. Example 1 illustrates the use of the integral formula (4) for the linear ODE (1). Example 2 deals with the RL-circuit for which the underlying physics is rather simple and straightforward and the solution exhibits exponential approach to a constant value (48>11 A). Several particular solutions are shown in Fig. 19. Example 3 on hormone level is an input–output problem, eventually giving a periodic steady-state solution, after an exponential term has decreased to zero, theoretically as t : ⬁ , practically after a very short time, as shown in Fig. 20. Example 4 concerns the logistic or Verhulst ODE, perhaps the practically most important case of a Bernoulli ODE. The Bernoulli ODE is reduced to a linear ODE by setting u ⫽ y1ⴚa (a ⫽ 1), giving (10). Example 5 concerns population dynamics, based on Malthus’s and Verhulst’s ODEs, both of which are autonomous. This concept is defined in connection with (13) and will be of central interest in the theory and application of systems of ODEs in Chap. 4, in particular, in Sec. 4.5 when we shall discuss the Lotka–Volterra population model. Problem Set 1.5 stikes a balance between formal problems (3–13) for linear ODEs, experimentation (Prob. 14), some basic theory (15–21), formal problems (22–28) for nonlinear ODEs, a project (29) on transformation, two ODEs of lesser importance (Clairaut and Riccati ODEs in Team Project 30, showing singular solutions), and, finally, a variety of modeling problems (31–40) taken from various fields. SOLUTIONS TO PROBLEM SET 1.5, page 34 4. The standard form (1) is y r ⫺ 2y ⫽ ⫺4x, so that (4) gives

冮

y ⫽ e2x c eⴚ2x (⫺4x) dx ⫹ c d ⫽ ce2x ⫹ 2x ⫹ 1. 6. From (4) with p ⫽ 2, h ⫽ 2x, r ⫽ 4 cos 2x we obtain

冮

y ⫽ eⴚ2x c e2x 4 cos 2x dx ⫹ c d ⫽ eⴚ2x [e2x(cos 2x ⫹ sin 2x ⫹ c] . It is perhaps worthwhile mentioning that integrals of this type can more easily be evaluated by undetermined coefficients. Also, the student should verify the result by differentiation, even if it was obtained by a CAS. From the initial condition we obtain y (14 p) ⫽ ce⫺p>2 ⫹ 0 ⫹ 1 ⫽ 3;

hence

The answer can be written y ⫽ 2ep>2ⴚ2x ⫹ cos 2x ⫹ sin 2x.

c ⫽ 2ep>2.

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8. In (4) we have p ⫽ tan x, h ⫽ ⫺ln (cos x), eh ⫽ 1>cos x, so that (4) gives y ⫽ (cos x) c

冮 cos x e cos x

⫺0.01x

dx ⫹ c d ⫽ [⫺100 e⫺0.01x ⫹ c] cos x.

The initial condition gives y(0) ⫽ ⫺100 ⫹ c ⫽ 0; hence c ⫽ 100. The particular solution is y ⫽ 100 (1 ⫺ e⫺0.01x ) cos x. The factor 0.01, which we include in the exponent, has the effect that the graph of y shows a long transition period. Indeed, it takes x ⫽ 460 to let the exponential function eⴚ0.01x decrease to 0.01. Choose the x-interval of the graph accordingly. 10. The standard form (1) is yr ⫹

y⫽

cos x

cos2 x

Hence h ⫽ 3 tan x, and (4) gives the general solution y ⫽ eⴚ3 tan x c

e3 tan x

冮 cos x dx ⫹ c d . 2

To evaluate the integral, observe that the integrand is of the form 1 3 (3 tan x)

r e3 tan x;

that is, 1 3 tan x ). 3 (e

Hence the integral has the value 13 e3 tan x. This gives the general solution y ⫽ eⴚ3 tan x [ 13 e3 tan x ⫹ c] ⫽ 13 ⫹ ceⴚ3 tan x. The initial condition gives from this y(14p) ⫽ 13 ⫹ ce⫺3 ⫽ 43 ; ˛

hence

c ⫽ e3.

The answer is y ⫽ 13 ⫹ e3ⴚ3 tan x. 12. y ⫽ cx ⴚ4 ⫹ x 4 is the general solution. The initial condition gives c ⫽ 1. 14. CAS Experiment (a) y ⫽ x sin (1>x) ⫹ cx. c ⫽ 0 if y(2> p) ⫽ 2> p. y is undefined at x ⫽ 0, the point at which the “waves” of sin (1>x) accumulate; the factor x makes them smaller and smaller. Experiment with various x-intervals. (b) y ⫽ x n3sin (1>x) ⫹ c4. y(2> p) ⫽ (2> p)n. n need not be an integer. Try n ⫽ 12. Try n ⫽ ⫺1 and see how the “waves” near 0 become larger and larger. 16. Substitution gives the identity 0 ⫽ 0. These problems are of importance because they show why linear ODEs are preferable over nonlinear ones in the modeling process. Thus one favors a linear ODE over a nonlinear one if the model is a faithful mathematical representation of the problem. Furthermore, these problems illustrate the difference between homogeneous and nonhomogeneous ODEs.

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18. We obtain ( y1 ⫺ y2 ) r ⫹ p( y1 ⫺ y2) ⫽ y r1 ⫺ y r2 ⫹ py1 ⫺ py2 ⫽ ( y r1 ⫹ py1) ⫺ (y r2 ⫹ py2) ⫽r⫺r ⫽ 0. 20. The sum satisfies the ODE with r1 ⫹ r2 on the right. This is important as the key to the method of developing the right side into a series, then finding the solutions corresponding to single terms, and finally, adding these solutions to get a solution of the given ODE. For instance, this method is used in connection with Fourier series, as we shall see in Sec. 11.5. 22. Bernoulli equation. First solution method: Transformation to linear form. Set y ⫽ 1>u. Then y r ⫹ y ⫽ ⫺u r >u 2 ⫹ 1>u ⫽ 1>u 2. Multiplication by ⫺u 2 gives the linear ODE in standard form u r ⫺ u ⫽ ⫺1.

u ⫽ cex ⫹ 1.

General solution

Hence the given ODE has the general solution y ⫽ 1>(cex ⫹ 1). From this and the initial condition y(0) ⫽ ⫺13, we obtain y(0) ⫽ 1>(c ⫹ 1) ⫽ ⫺13,

c ⫽ ⫺4,

y ⫽ 1>(1 ⫺ 4ex).

Answer:

Second solution method: Separation of variables and use of partial fractions. dy 1 1 ⫽a ⫺ b dy ⫽ dx. y y(y ⫺ 1) y⫺1 Integration gives ln ƒ y ⫺ 1 ƒ ⫺ ln ƒ y ƒ ⫽ ln `

y⫺1 ` ⫽ x ⫹ c*. y

Taking exponents on both sides, we obtain y⫺1 1 ⫽1⫺ ⫽苲 c ex, y y

1 ⫽1⫺苲 c ex, y

y⫽

1 . 1 ⫹ cex

We now continue as before. 24. u ⫽ y2, yy r ⫹ y2 ⫽ ⫺x, 12 u r ⫹ u ⫽ ⫺x, u r ⫹ 2u ⫽ ⫺2x; hence u ⫽ eⴚ2x c ⫺

冮 e 2x dx ⫹ c d ⫽ ⫺ x ⫹ ce 2x

1 2

ⴚ2x

y ⫽ 1u

26. This ODE can simply be solved by separating variables, cot y dy ⫽ dx>1x ⫺ 12,

ln ƒ sin y ƒ ⫽ ln ƒ x ⫺ 1 ƒ ⫹ 苲 c

hence y ⫽ arcsin [ĉ1x ⫺ 12]

x ⫽ 1 ⫹ c sin y

with c ⫽ ⫺1 from the initial condition. As an alternative, we can regard it as an ODE for the unknown function x ⫽ x(y) and solve it by (4) with x and y interchanged.

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28. Using the given transformation y 2 ⫽ z, we obtain the linear ODE 1 z r ⫹ a1 ⫺ b z ⫽ xex, x which we can solve by (4) with z instead of y, z ⫽ xeⴚx a

冮 x e xe dx ⫹ cb ⫽ xe 1 e ⫹ c2 ⫽ cxe ⫹ xe . 1

ⴚx 1 2x 2

ⴚx

1 2

From this we obtain y ⫽ 1z. 30. Team Project. (a) y ⫽ Y ⫹ v reduces the Riccati equation to a Bernoulli equation by removing the term h(x). The second transformation, v ⫽ 1>u, is the usual one for transforming a Bernoulli equation with y 2 on the right into a linear ODE. Substitute y ⫽ Y ⫹ 1>u into the Riccati equation to get Y r ⫺ u r >u 2 ⫹ p1Y ⫹ 1>u2 ⫽ g (Y 2 ⫹ 2Y>u ⫹ 1>u 22 ⫹ h. ˛

Since Y is a solution, Y r ⫹ pY ⫽ gY 2 ⫹ h. There remains ⫺u r >u 2 ⫹ p>u ⫽ g ( 2Y>u ⫹ 1>u 2). ˛

Multiplication by ⫺u 2 gives u r ⫺ pu ⫽ ⫺g(2Yu ⫹ 1). Reshuffle terms to get u r ⫹ (2Yg ⫺ p)u ⫽ ⫺g, the linear ODE as claimed. (b) Substitute y ⫽ Y ⫽ x to get 1 ⫺ 2x 4 ⫺ x ⫽ ⫺x 4 ⫺ x 4 ⫺ x ⫹ 1, which is true. Now substitute y ⫽ x ⫹ 1>u. This gives 1 ⫺ u r >u 2 ⫺ (2x 3 ⫹ 1)(x ⫹ 1>u) ⫽ ⫺x 2(x 2 ⫹ 2x>u ⫹ 1>u 2) ⫺ x 4 ⫺ x ⫹ 1.

Most of the terms cancel on both sides. There remains ⫺u r >u 2 ⫺ 1>u ⫽ ⫺x 2>u 2. Multiplication by ⫺u 2 finally gives u r ⫹ u ⫽ x 2. The general solution is u ⫽ ceⴚx ⫹ x 2 ⫺ 2x ⫹ 2 and y ⫽ x ⫹ 1>u. Of course, instead performing this calculation we could have used the general formula in (a), in which 2Yg ⫺ p ⫽ 2x(⫺x 2) ⫹ 2x 3 ⫹ 1 ⫽ 1

and

⫺g ⫽ ⫹x 2.

(c) By differentiation, 2y r y s ⫺ y r ⫺ xy s ⫹ y r ⫽ 0, y s (2y r ⫺ x) ⫽ 0. (A) y s ⫽ 0, y ⫽ cx ⫹ a. By substitution, c2 ⫺ xc ⫹ cx ⫹ a ⫽ 0, a ⫽ ⫺c2, y ⫽ cx ⫺ c2, a family of straight lines. (B) y r ⫽ x>2, y ⫽ x 2>4 ⫹ c*. By substitution into the given ODE, x 2>4 ⫺ x 2>2 ⫹ x 2>4 ⫹ c* ⫽ 0, c* ⫽ 0, y ⫽ x 2>4, the envelope of the family; see Fig. 6 in Problem Set 1.1. 32. k 1(T ⫺ Ta) follows from Newton’s law of cooling. k 2(T ⫺ Tw) models the effect of heating or cooling. T ⬎ Tw calls for cooling; hence k 2(T ⫺ Tw) should be negative in this case; this is true, since k 2 is assumed to be negative in this formula. Similarly for heating, when heat should be added, so that the temperature increases. The given model is of the form T r ⫽ kT ⫹ K ⫹ k 1C cos (p>12)t.

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This can be seen by collecting terms and introducing suitable constants, k ⫽ k 1 ⫹ k 2 (because there are two terms involving T ), and K ⫽ ⫺k 1A ⫺ k 2Tw ⫹ P. The general solution is T ⫽ cekt ⫺ K>k ⫹ L(⫺k cos (pt>12) ⫹ (p>12) sin (pt>12)),

where L ⫽ k 1C>(k 2 ⫹ p2>144). The first term solves the homogeneous ODE T r ⫽ kT and decreases to zero. The second term results from the constants A (in Ta), Tw, and P. The third term is sinusoidal, of period 24 hours, and time-delayed against the outside temperature, as is physically understandable. 34. y r ⫽ ky(1 ⫺ y) ⫽ f (y), where k ⬎ 0 and y is the proportion of infected persons. Equilibrium solutions are y ⫽ 0 and y ⫽ 1. The first, y ⫽ 0, is unstable because f (y) ⬎ 0 if 0 ⬍ y ⬍ 1 but f (y) ⬍ 0 for negative y. The solution y ⫽ 1 is stable because f (y) ⬎ 0 if 0 ⬍ y ⬍ 1 and f (y) ⬍ 0 if y ⬎ 1. The general solution is y⫽

1 1 ⫹ ceⴚkt

It approaches 1 as t : ⬁ . This means that eventually everybody in the population will be infected. 36. The model is y r ⫽ Ay ⫺ By 2 ⫺ Hy ⫽ Ky ⫺ By 2 ⫽ y(K ⫺ By) where K ⫽ A ⫺ H. Hence the general solution is given by (12) in Example 4 with A replaced by K ⫽ A ⫺ H. The equilibrium solutions are obtained from y r ⫽ 0; hence they are y1 ⫽ 0 and y2 ⫽ K>B. The population y2 remains unchanged under harvesting, and the fraction Hy2 of it can be harvested indefinitely—hence the name. 38. For the first 3 years you have the solution y1 ⫽ 4>(5 ⫺ 3eⴚ0.8t) from Prob. 36. The idea now is that, by continuity, the value y1(3) at the end of the first period is the initial value for the solution y2 during the next period. That is, y2(3) ⫽ y1(3) ⫽ 4>(5 ⫺ 3eⴚ2.4). Now y2 is the solution of y r ⫽ y ⫺ y 2 (no fishing!). Because of the initial condition this gives y2 ⫽ 4>(4 ⫹ e3ⴚt ⫺ 3e0.6ⴚt). Check the continuity at t ⫽ 3 by calculating y2(3) ⫽ 4>(4 ⫹ e0 ⫺ 3eⴚ2.4). Similarly, for t from 6 to 9 you obtain y3 ⫽ 4>(5 ⫺ e4.8ⴚ0.8t ⫹ e1.8ⴚ0.8t ⫺ 3eⴚ0.6ⴚ0.8t). This is a period of fishing. Check the continuity at t ⫽ 6: y3(6) ⫽ 4>(5 ⫺ e0 ⫹ eⴚ3 ⫺ 3eⴚ5.4). This agrees with y2(6) ⫽ 4>(4 ⫹ eⴚ3 ⫺ 3eⴚ5.4).

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40. Let y denote the amount of fresh air measured in cubic feet. Then the model is obtained from the balance equation “Inflow minus Outflow equals the rate of change”; that is, y r ⫽ 600 ⫺

600 y ⫽ 600 ⫺ 0.03y. 20,000

The general solution of this linear ODE is y ⫽ ceⴚ0.03t ⫹ 20,000. The initial condition is y(0) ⫽ 0 (initially no fresh air) and gives y(0) ⫽ c ⫹ 20,000 ⫽ 0;

hence

c ⫽ ⫺20,000.

The particular solution of our problem is y ⫽ 20,000(1 ⫺ eⴚ0.03t). This equals 90% if t is such that eⴚ0.03t ⫽ 0.1 thus if t ⫽ (ln 0.1)>(⫺0.03) ⫽ 77 [min] . SECTION 1.6. Orthogonal Trajectories. Optional, page 36 Purpose. To show that families of curves F (x, y, c) ⫽ 0 can be described by ODEs y r ⫽ f (x, y) and the switch to ~ y r ⫽ ⫺1>f (x, ~ y ) produces as general solution the orthogonal trajectories. This is a nice application that may also help the student to gain more selfconfidence, skill, and a deeper understanding of the nature of ODEs. We leave this section optional, for reasons of time. This will cause no gap. The reason ODEs can be applied in this fashion results from the fact that general solutions of ODEs involve an arbitrary constant that serves as the parameter of this oneparameter family of curves determined by the given ODE, and then another general solution similarly determines the one-parameter family of the orthogonal trajectories. Curves and their orthogonal trajectories play a role in several physical applications (e.g., in connection with electrostatic fields, fluid flows, and so on). Problem Set 1.6 should help the student to obtain skill in representing families of curves (Probs. 1–3), finding trajectories (4–10), and understanding some basic physical and geometric applications of trajectories (11–16). This will also involve the Cauchy–Riemann equations, which are basic in complex analysis. SOLUTIONS TO PROBLEM SET 1.6, page 38 2. (x ⫺ c)2 ⫹ ( y ⫺ c3)2 ⫺ r 2 ⫽ 0 gives a circle of radius r with center (x 0, y0) ⫽ (c, c3) on the cubic parabola. Since this center has distance r ⫽ 2c2 ⫹ c6, we have r 2 ⫽ c2 ⫹ c6. 4. y r ⫽ 2x, ~ y r ⫽ ⫺1>(2x), ~ y ⫽ ⫺ 1 ln x ⫹ ~ c . Note that these curves and their OTs are 2

congruent. This is typical of ODEs y r ⫽ f (x) with f not depending on y.

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6. Differentiating the given formula, we obtain xy r ⫹ y ⫽ 0.

y yr ⫽ ⫺ . x

Thus

This is the differential equation of the given hyperbolas. Hence the differential equation of the orthogonal trajectories is x ~ y r ⫽ ~. y Separation of variables and integration gives 1 ~2 ~ y dy ⫽ x dx, y ⫽ 1x2 ⫹ ~ c. 2

Answer: The hyperbolas x 2 ⫺ ~ y 2 ⫽ c* are the orthogonal trajectories of the given hyperbolas. 8. Squaring the given formula, differentiating, and solving algebraically for y r, we obtain y 2 ⫺ x ⫽ c,

2yy r ⫽ 1,

yr ⫽

1 . 2y

This is the differential equation of the given curves. Hence the differential equation of the orthogonal trajectories is ~ y r ⫽ ⫺2y~. By separation of variables and integration we obtain ln ƒ ~ y ƒ ⫽ ⫺2x ⫹ ~ c. Taking exponents gives the answer ~ y ⫽ c*eⴚ2x . 10. x 2 ⫹ y 2 ⫺ 2cy ⫽ 0. Solve algebraically for 2c: x 2 ⫹ y2 x2 ⫽ ⫹ y ⫽ 2c. y y Differentiation gives x 2y r 2x ⫺ 2 ⫹ y r ⫽ 0. y y By algebra, x2 2x y r a⫺ 2 ⫹ 1b ⫽ ⫺ . y y ˛

Solve for y r : yr ⫽ ⫺

2x y

冒

y2 ⫺ x 2 y

b⫽

⫺2xy y ⫺ x2 2

This is the ODE of the given family. Hence the ODE of the trajectories is ~ y2 ⫺ x2 y 1 ~ x ~ yr ⫽ ⫽ a ⫺ ~ b. ~ y 2xy 2 x

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To solve this equation, set u ⫽ ~ y >x. Then 1 1 ~ y r ⫽ xu r ⫹ u ⫽ au ⫺ b. u 2 Subtract u on both sides to get xu r ⫽ ⫺

u2 ⫹ 1 . 2u

Now separate variables, integrate, and take exponents, obtaining c2 dx 2u du ⫽⫺ , ln (u 2 ⫹ 1) ⫽ ⫺ln ƒ x ƒ ⫹ c1, u2 ⫹ 1 ⫽ . x x u2 ⫹ 1 Write u ⫽ ~ y >x and multiply by x 2 on both sides of the last equation. This gives ~ y 2 ⫹ x 2 ⫽ c2x.

The answer is (x ⫺ c3)2 ⫹ ~ y 2 ⫽ c32. Note that the given circles all have their centers on the y-axis and pass through the origin. The result shows that their orthogonal trajectories are circles, too, with centers on the x-axis and passing through the origin. 12. Setting y ⫽ 0 gives from x 2 ⫹ (y ⫺ c) 2 ⫽ 1 ⫹ c2 the equation x 2 ⫹ c2 ⫽ 1 ⫹ c2; hence x ⫽ ⫺1 and x ⫽ 1, which verifies that those circles all pass through ⫺1 and 1, each of them simultaneously through both points. Subtracting c2 on both sides of the given equation, we obtain x 2 ⫹ y 2 ⫺ 2cy ⫽ 1,

x 2 ⫹ y 2 ⫺ 1 ⫽ 2cy,

x2 ⫺ 1 ⫹ y ⫽ 2c. y

Emphasize to your class that the ODE for the given curves must always be free of c. Having accomplished this, we can now differentiate. This gives 2x x2 ⫺ 1 ⫺a ⫺ 1b y r ⫽ 0. y y2 This is the ODE of the given curves. Replacing y r with ⫺1>y~ r and y with ~ y , we obtain the ODE of the trajectories:

冒

2x x2 ⫺ 1 ⫺ a ⫺ 1b ~ ~ y y2

(⫺y~ r ) ⫽ 0.

Multiplying this by ~ y r , we get 2xy~ r x2 ⫺ 1 ⫹ ⫺ 1 ⫽ 0. ~ ~ y y2 Multiplying this by ~ y 2>x 2, we obtain ~ 2y~~ yr y2 y2 1 d ~ 1 ⫹1⫺ 2⫺ 2⫽ a b ⫹ 1 ⫺ 2 ⫽ 0. x dx x x x x

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By integration, ~ y2 x

⫹x⫹

1 ⫽ 2c*. x

Thus,

~ y 2 ⫹ x 2 ⫹ 1 ⫽ 2c*x.

We see that these are the circles ~ y 2 ⫹ (x ⫺ c*)2 ⫽ c*2 ⫺ 1 dashed in Fig. 25, as claimed. 14. By differentiation, 2x

2yy r

⫹ 2

⫽ 0,

yr ⫽ ⫺

Hence the ODE of the orthogonal trajectories is a 2y~ ~ By separation, yr ⫽ 2 . b x

2x>a 2

b 2x . ⫽ ⫺ 2y>b 2 a 2y dy~ a 2 dx . ⫽ ~ y b2 x

Integration and taking exponents gives ln ƒ ~ yƒ ⫽

a2 b

ln ƒ x ƒ ⫹ c**,

2 2 ~ y ⫽ c*x a >b .

This shows that the ratio a >b has substantial influence on the form of the trajectories. For a 2 ⫽ b 2 the given curves are circles, and we obtain straight lines as trajectories. a 2>b 2 ⫽ 2 gives quadratic parabolas. For higher integer values of a 2>b 2 we obtain parabolas of higher order. Intuitively, the “flatter” the ellipses are, the more rapidly the trajectories must increase to have orthogonality. Note that our discussion also covers families of parabolas; simply interchange the roles of the curves and their trajectories. Note further that, in the light of the present answer, our example in the text turns out to be typical. 16. y ⫽ 兰 f (x) dx ⫹ c. Since c is just an additive constant, the statement about the curves follows; these curves are obtained from any one of them by translation in the y-direction. Similarly for the OTs, whose ODE is ~ y r ⫽ ⫺1>f (x) with the function on the right independent of ~ y. 2

SECTION 1.7. Existence and Uniqueness of Solutions for Initial Value Problems, page 38 Purpose. To give the student at least some impression of the theory that would occupy a central position in a more theoretical course on a higher level. Short Courses. This section can be omitted. Comment on Iteration Methods Iteration methods were used rather early in history, but it was Picard who made them popular. Proofs of the theorems in this section (given in books of higher level, e.g., [A11]) are based on the Picard iteration (see CAS Project 6). Iterations are well suited for the computer because of their modest storage demand and usually short programs in which the same loop or loops are used many times, with different data. Because integration is generally not difficult for a CAS, Picard’s method has gained some popularity during the past few decades.

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Example 1 is simple, involving only y ⫽ tan x, and is typical inasmuch as it illustrates that the actual interval of existence is much larger than the interval guaranteed by Existence Theorem 1. Example 2 shows that IVPs violating uniqueness can be constructed relatively easily. Lipschitz and Hölder conditions play a basic role in the theory of PDEs on a level substantially higher than that of our Chap.12. SOLUTIONS TO PROBLEM SET 1.7, page 42 2. The initial condition is given at the point x ⫽ 2. The coefficient of y r is 0 at that point, so from the ODE we already see that something is likely to go wrong. Separating variables, integrating, and taking exponents gives dy 2 dx ⫽ , y x⫺2

ln ƒ y ƒ ⫽ 2 ln ƒ x ⫺ 2 ƒ ⫹ c*,

y ⫽ c(x ⫺ 2)2.

This last expression is the general solution. It shows that y(2) ⫽ 0 for any c. Hence the initial condition y(1) ⫽ 1 cannot be satisfied. This does not contradict the theorems because we first have to write the ODE in standard form: y r ⫽ f (x, y) ⫽

2y . x⫺2

This shows that f is not defined when x ⫽ 2 (to which the initial condition refers). 4. For k ⫽ 0 we still get no solution, violating the existence as in Prob. 2. For k ⫽ 0 we obtain infinitely many solutions, because c remains unspecified. Thus in this case the uniqueness is violated. Neither of the two theorems is violated in either case. x2 x3 . . . x nⴙ1 6. CAS Project. (b) yn ⫽ ⫹ ⫹ ⫹ , y ⫽ ex ⫺ x ⫺ 1 2! 3! (n ⫹ 1)! (c) y0 ⫽ 1, y1 ⫽ 1 ⫹ 2x, y2 ⫽ 1 ⫹ 2x ⫹ 4x 2 ⫹ y(x) ⫽

8x 3 . . . , 3

1 ⫽ 1 ⫹ 2x ⫹ 4x 2 ⫹ 8x 3 ⫹ . . . 1 ⫺ 2x

(d) y ⫽ (x ⫺ 1)2, y ⫽ 0. It approximates y ⫽ 0. General solution y ⫽ (x ⫹ c)2. (e) y r ⫽ y would be a good candidate to begin with. Perhaps you write the initial choice as y0 ⫹ a; then a ⫽ 0 corresponds to the choice in the text, and you see how the expressions in a are involved in the approximations. The conjecture is true for any choice of a constant (or even of a continuous function of x). It was mentioned in footnote 10 that Picard used his iteration for proving his existence and uniqueness theorems. Since the integrations involved in the method can be handled on the computer quite efficiently, the method has gained in importance in numerics. 8. The student should get an understanding of the “intermediate” position of a Lipschitz condition between continuity and (partial) differentiability. The student should also realize that the linear equation is basically simpler than the nonlinear one. The calculation is straightforward because we have f (x, y) ⫽ r(x) ⫺ p(x)y

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and this implies that f (x, y2) ⫺ f (x, y1) ⫽ ⫺p(x)(y2 ⫺ y1).

(A)

This becomes a Lipschitz condition if we note that the continuity of p(x) for ƒ x ⫺ x 0 ƒ ⬉ a implies that p(x) is bounded, say ƒ p(x) ƒ ⬉ M for all these x. Taking absolute values on both sides of (A) now gives ƒ f (x, y2) ⫺ f (x, y1) ƒ ⬉ M ƒ y2 ⫺ y1 ƒ . 10. By separation and integration, dy 2x ⫺ 1 ⫽ 2 dx, y x ⫺x

ln ƒ y ƒ ⫽ ln ƒ x 2 ⫺ x ƒ ⫹ c*.

Taking exponents gives the general solution y ⫽ c(x 2 ⫺ x). From this we can see the answers: No solution if y(0) ⫽ k ⫽ 0 or y(1) ⫽ k ⫽ 0. A unique solution if y(x 0) equals any y0 and x 0 ⫽ 0 or x 0 ⫽ 1. Infinitely many solutions if y(0) ⫽ 0 or y(1) ⫽ 0. This does not contradict the theorems because f (x, y) ⫽

2x ⫺ 1 x2 ⫺ x

is not defined when x ⫽ 0 or 1. SOLUTIONS TO CHAP. 1 REVIEW QUESTIONS AND PROBLEMS, page 43 12. y ⫽ tanh (x ⫹ c). Note that the solution curves are congruent. 14. y ⫽ x 2 ⫹ cx. The figure also shows the solution curves through (⫺1, 1) [thus, y(⫺1) ⫽ 1], (1, 0.1), (1, 1), and (1, 2). y 2

y(x)

–2

–1

–2

Problem 14. Direction field of xy r ⫽ y ⫹ x 2 ˛

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16. Solution y ⫽ 1>(1 ⫹ 4eⴚx). Computations: xn

Error

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0

0.2000 0.2160 0.2329 0.2509 0.2696 0.2893 0.3098 0.3312 0.3534 0.3762 0.3997

0 0.0005 0.0010 0.0015 0.0021 0.0026 0.0032 0.0037 0.0041 0.0046 0.0048

0.2

0.4

y 0.4

0.35

0.3

0.25

0.2 0

0.6

0.8

Problem 16. Solution curve and computed values

18. y ⫽ ce0.4x ⫺ 25 cos x ⫺ 10 sin x. 20. This Bernoulli equation (a Verhulst equation if b ⬍ 0) can be reduced to linear form, as shown in Example 4 of Sec. 1.5 (except for the notation). The general solution is (see (12) in Sec. 1.5) y⫽

1 . ⫺ b>a

ⴚax

22. The general solution of this linear differential equation is obtained as explained in Sec. 1.6,

冮

y ⫽ eⴚ2x a e2x eⴚ2x dx ⫹ cb ⫽ 1x ⫹ c2eⴚ2x 2

From this and the initial condition y(0) ⫽ ⫺4.3 we have c ⫽ ⫺4.3. Answer: y ⫽ (x ⫺ 4.3)eⴚ2x . 2

24. To solve this Bernoulli equation we set u ⫽ y ⴚ2. Then y ⫽ u ⴚ1>2, y r ⫽ ⫺12u ⴚ3>2 u r . Substitution into the given ODE gives ⫺12 u ⴚ3>2 u r ⫹ 12 u ⴚ1>2 ⫽ u ⴚ3>2. We now multiply by ⫺2u 3>2, obtaining u r ⫺ u ⫽ ⫺2.

General solution:

u ⫽ cex ⫹ 2.

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Hence y ⫽ u ⫺1>2 ⫽

1 2cex ⫹ 2

From this and the initial condition y(0) ⫽ 13 we get c ⫽ 7. Answer: y⫽

1 22 ⫹ 7ex

26. Theorem 1 in Sec. 1.4 gives the integrating factor F ⫽ 1>x 2. We thus obtain the exact equation 1 1 sinh y dy ⫺ 2 cosh y dx ⫽ 0. x x By inspection or systematically by integration (as explained in Sec. 1.4), we obtain 1 d a cosh yb ⫽ 0; x

thus,

1 cosh y ⫽ c. x

From this and the initial condition we get 13 # 1 ⫽ c. Answer: cosh y ⫽ 13 x. 28. We proceed as in Sec. 1.3. The time rate of change y r ⫽ dy>dt equals the inflow of salt minus the outflow per minute. y r ⫽ 20 ⫺

20 y. 500

The initial condition is y(0) ⫽ 80. This gives the particular solution y ⫽ 500 ⫺ 420e⫺0.04t. The limiting value is 500 lb; 95% are 475 lb, so that we get the condition 500 ⫺ 420eⴚ0.04t ⫽ 475, from which we can determine t ⫽ 25 ln

420 ⫽ 70.5 [min]; 25

so it will take a little over an hour. 30. By Newton’s law of cooling, since the surrounding temperature is 100°C and the initial temperature of the metal is T(0) ⫽ 20, we first obtain T(t) ⫽ 100 ⫺ 80ekt. k can be determined from the condition that T(1) ⫽ 51.5; that is, T(1) ⫽ 100 ⫺ 80ek ⫽ 51.5, so that k ⫽ ln (48.5>80) ⫽ ⫺0.500. With this value of k we can now find the time at which the metal has the temperature 99.9°C. 99.9 ⫽ 100 ⫺ 80eⴚ0.5t,

0.1 ⫽ 80eⴚ0.5t,

t⫽

ln 800 ⫽ 13.4. 0.5

Answer: The temperature of the metal has practically reached that of the boiling water after 13.4 min.

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CHAPTER 2

Second-Order Linear ODEs

Major Changes Among linear ODEs those of second order are by far the most important ones from the viewpoint of applications, and from a theoretical standpoint they illustrate the theory of linear ODEs of any order (except for the role of the Wronskian). For these reasons we consider linear ODEs of third and higher order in a relatively short separate chapter, Chap. 3. Section 2.2 combines all three cases of the roots of the characteristic equation governing homogeneous linear ODEs with constant coefficients. (In some of the previous editions the complex case was discussed in a separate section, which seems of no great advantage to the student.) Section 2.3 is a short introduction to differential operators. Modeling begins in Sec. 2.4 with the mass–spring system, which is now derived more simply than before and in a better logical order. After a discussion of the Euler–Cauchy equation and its application to electric fields between concentric spheres in Sec. 2.6, we discuss in Sec. 2.7 the existence and uniqueness of the solution of IVPs involving the homogeneous linear ODE of second order. This is the end of discussing homogeneous ODEs. It is followed in Sec. 2.7 by the method of undetermined coefficients for nonhomogeneous ODEs, which is basic in applications since it is simpler than the general method (variation of parameters, Sec. 2.10) and covers many, if not most of the standard engineering applications. Modeling of forced mechanical oscillations is discussed in Sec. 2.8, and electric RLC-circuits in Sec. 2.9. Note that we have placed the RL-circuit, governed by a firstorder ODE into Sec. 1.5, which the student may perhaps wish to review. This was a request by various users of the book, as a stepping stone that may lessen difficulties and simplify the derivation of the model from physics. SECTION 2.1. Homogeneous Linear ODEs of Second-Order, page 46 Purpose. To extend the basic concepts from first-order to second-order ODEs and to present the basic properties of linear ODEs. Comment on the Standard Form (1) The form (1), with 1 as the coefficient of y s , is practical, because if one starts from f (x)y s g(x)y r h(x)y r苲(x), one usually considers the equation in an interval I in which f (x) is nowhere zero, so that in I one can divide by f (x) and obtain an equation of the form (1). Points at which f (x) 0 require a special study, which we present in Chap. 5. Main Content, Important Concepts Linear and nonlinear ODEs Homogeneous linear ODEs (to be discussed in Secs. 2.1–2.6) Superposition principle for homogeneous ODEs General solution, basis, linear independence Initial value problem (2), (4), particular solution Reduction to first order (text and Probs. 3–10) 26

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Comment on the Three ODEs after (2) These are for illustration, not for solution, but should a student ask, answers are that the first will be solved by methods in Secs. 2.7 and 2.10, the second is a Bessel equation (Sec. 5.5) and the third has the solutions 1c1x c2 with any c1 and c2. Comment on Footnote 1 In 1760, Lagrange gave the first methodical treatment of the calculus of variations. The book mentioned in the footnote includes all major contributions of others in the field and made him the founder of analytical mechanics. Examples in the Text. The examples show the following. Example 1 shows the superposition of solutions of the homogeneous linear ODE. Examples 2 and 3 are counter-examples to the superposition for a nonhomogeneous linear ODE and a nonlinear ODE. Example 4 is an initial value problem, suggesting the concepts of a general solution, a particular solution, and a basis. Examples 5 and 6 give further illustrations of those concepts. Example 7 shows the reduction of order of y s p(x)y r q(x)y 0 using a known solution y1, followed by the derivation of a general formula for a second solution y2 y1

冮y e 1

2 1

ⴚ 兰p dx

dx.

Hence solving the ODE for finding a second solution is reduced to two integrations, and the student should understand that this is a simpler task. Comment on Terminology p and q are called the coefficients of (1) and (2). The function r on the right is not called a coefficient, to avoid the misunderstanding that r must be constant when we talk about an ODE with constant coefficients.

SOLUTIONS TO PROBLEM SET 2.1, page 53 2. y s

dy r dy r dy dz z dx dy dx dy

4. z y r , 2xz r 3z. Separation of variables and integration gives dz 3 dx, z 2x

ln ƒ z ƒ 32 ln ƒ x ƒ 苲 c,

Integrating once more, we have y

冮 z dx c x 1

5>2

c2.

z cx 3>2.

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6. The formula in the text was derived under the assumption that the ODE is in standard form; in the present case, ys

2 y r y 0. x

Hence p 2>x, so that eⴚ兰p dx x ⴚ2. It follows from (9) in the text that U

# 1

cos x

1 cos2 x

The integral of U is tan x; we need no constants of integration because we merely want to obtain a particular solution. The answer is y2 y1 tan x

sin x . x

8. z r 1 z 2, dz>(1 z 2) dx, arctan z x c1, z tan (x c1), y ln ƒ cos (x c1) ƒ c2 This is an obvious use of problems from Chap. 1 in setting up problems for this section. The only difficulty may be an unpleasant additional integration. dy dz 1 10. z , z a1 y b z 2 0, divide by z, separate variables, and integrate: dx dy 1 dz a1 b dy, z y

ln ƒ z ƒ y ln ƒ y ƒ 苲 c.

Take exponentials, separate again, and integrate: dy c z eⴚy, y dx

冮 ye dy cx c .

yey dy c dx,

Evaluation of the integral gives the answer (y 1)ey c1x c2. 12. z r (1 z2) 1>2, (1 z2)ⴚ1>2 dz dx, arcsinh z x c1. From this we have z sinh (x c1), y cosh (x c1) c2. From the boundary conditions y(1) 0, y( 1) 0 we get cosh (1 c1) c2 0 cosh ( 1 c1) c2. Hence c1 0 and then c2 cosh 1. The answer is (see the figure) y cosh x cosh 1. y –1

– 0.5

0.5

– 0.54

Section 2.1. Problem 12

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14. y s 1>y r , y s y r 1,

dz 2 z 1. Integration with respect to y gives dy z3 y c. 3

Hence z 3 3y 苲 c and thus dy z (3y 苲 c )1>3. dt By separation of variables, dy dt. (3y 苲 c )1>3 By integration, 苲 1 苲 2>3 t 苲 c. 2 (3y c ) Hence (3y 苲 c )2>3 2t c2 and thus 3y 苲 c 1 (2t c2)3>2. The answer is y 13 (2t c2)3>2 c1. 16. y (2.2 0.8x)eⴚ0.3x 18. y 4.3x 3.8x ln x SECTION 2.2. Homogeneous Linear ODEs with Constant Coefficients, page 53 Purpose. To show that homogeneous linear ODEs with constant coefficients can be solved by algebra, namely, by solving the quadratic characteristics equation (3). The roots may be: (Case I)

Real distinct roots

(Case II)

A real double root (“Critical case”)

(Case III)

Complex conjugate roots

In Case III the roots are conjugate because the coefficients of the ODE, and thus of (3), are real, a fact the student should remember. To help poorer students, we have shifted the derivation of the real form of the solutions in Case III to the end of the section, but the verification of these real solutions is done immediately when they are introduced. This will also help to a better understanding.

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The student should become aware of the fact that Case III includes both undamped (harmonic) oscillations (if c 0) and damped oscillations. Also it should be emphasized that in the transition from the complex to the real form of the solutions we use the superposition principle. Furthermore, one should emphasize the general importance of the Euler formula (11), which we shall use on various occasions. Examples in the Text. The examples show the following. Examples 1 and 2 concern Case I, the case of distinct real roots. In this case, as well as in the other two cases, an initial value problem requires the solution of a system of two linear equations in two unknowns, whose values are determined by the two initial conditions. A typical solution in Case I is shown in Fig. 30. Examples 3 and 4 concern Case II, the case of a real double root, which is the limiting case between Cases I and III. Figure 31 shows a typical solution, having a real root at x 1.5, which is the solution of 3 2x 0, where 3 2x is a factor in the solution of the IVP in Example 4. Example 5 concerns Case III, in which one obtains solutions (9), representing oscillations. These may be damped as in Fig. 32, or of increasing maximum amplitude if a 0, or of constant maximum amplitude if a 0, as in Example 6, giving a harmonic oscillations. Comment on How to Avoid Working in Complex The average engineering student will profit from working a little with complex numbers. However, if one has reasons for avoiding complex numbers here, one may apply the method of eliminating the first derivative from the equation, that is, substituting y = uv and determining v so that the equation for u does not contain u r . For v this gives 2v r av 0.

A solution is

v e ax>2.

With this v, the equation for u takes the form u s (b 14 a 2)u 0 and can be solved by remembering from calculus that cos vx and sin vx reproduce under two differentiations, multiplied by v2. This gives (9), where v 2b 14 a 2. Of course, the present approach can be used to handle all three cases. In particular, u s 0 in Case II gives u c1 c2x at once. SOLUTIONS TO PROBLEM SET 2.2, page 59 2. y c1 cos 6x c2 sin 6x 4. y eⴚ2x (c1 cos px c2 sin px) 6. y (c1 c2x)e1.6x 8. y e ⴚx>2 (c1 cos (13x) c2 sin (13x)) 10. y eⴚ1.2x (c1 cos (1.4px) c2 sin (1.4px)) 12. y c1eⴚ5x c2eⴚ4x 2 14. y (c1 c2x)eⴚk x

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16. y s 1.7y r 11.18y 0 18. y s 4p2y 0 20. y s 6.2y r 14.02y 0 22. A general solution is y(x) eⴚ2x(c1 cos px c2 sin px). The first initial condition gives y(12) e ⴚ1(0 c2) 1, hence c2 e. The derivative is y r (x) eⴚ2x( 2c1 cos px 2c2 sin px c1p sin px c2p cos px). From this and the second initial condition we obtain y r (12) eⴚ1( 2c2 c1p) eⴚ1( 2e c1p) 2 c1eⴚ1p 2. Hence c1 0. This gives the answer y eⴚ2xⴙ1 sin px. Notice that it depends on the initial condition whether both solutions of a basis appear in the particular solution or just one; this is worthwhile pointing out to the students. 24. A general solution is y c1eⴚx>2 c2e3x>2. The first initial condition gives y( 2) c1e c2eⴚ3 e. The derivative is y r 12 c1eⴚx>2 32 c2e3x>2. From this and the second initial condition we have y r ( 2) 12 c1e 32 c2eⴚ3 e>2. Division by e and by 12, respectively, gives c1 c2eⴚ4 1 c1 3c2eⴚ4 1. The solution of this system is c2 0, c1 1. Hence the answer is y eⴚx>2.

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26. A general solution is y c1ekx c2eⴚkx. From this and the first initial condition we have c1 c2 1. The derivative is y r c1kekx c2keⴚkx. From this and the second initial condition we obtain c1 c2 1>k. The solution of this system is c1

k 1 , 2k

k 1 . 2k

Hence the answer (the particular solution of the IVP) is y [(k 1)ekx (k 1)eⴚkx ]>(2k). 28. A general solution is y c1eⴚx>4 c2ex>2. From this and the first initial condition we have y(0) c1 c2 0.2. The derivative is y r 14 c1eⴚx>4 12 c2ex>2. From this and the second initial condition we have y r (0) 14 c1 12 c2 0.325. The solution of this system is c1 0.3, c2 0.5. Hence the particular solution satisfying the initial condition is y 0.3eⴚx>4 0.5ex>2. 30. A general solution is y (c1 c2x)e5x>3.

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This is a case of a double root of the characteristic equation. The first initial condition yields y(0) c1 3.3. By differentiation, y r [c2 53 (c1 c2x)]e5x>3. From this and the second initial condition we obtain y r (0) c2 53 c1 c2 5.5 10. Hence c2 4.5, so that the solution of the IVP is y (3.3 4.5x)e5x>3. 32. Independent if a 0 34. Dependent since ln (x 3) 3 ln x 36. If one of the functions is identically zero, the set is linearly dependent because c1 f1(x) c2 # 0 0 holds with any c2 0 (and c1 0). The intervals given in the problems are just a reminder that linear independence or independence always refers to some interval. In the present case we could choose as the interval the real axis, or the positive half-axis if a logarithm is involved. 38. Team Project. (a) We obtain (l l1)(l l2) l2 (l1 l2)l l1l2 l2 al b 0. Comparison of coefficients gives a (l1 l2), b l1l2. (b) y s ay r 0. (i) y c1eⴚax c2e0x c1eⴚax c2. (ii) z r az 0, where z y r , z ceⴚax and the second term comes in by integration: y

冮z dx 苲c e 1

ⴚax

苲 c 2.

(d) e(k m)x and ekx satisfy y s (2k m)y r k(k m)y 0, by the coefficient formulas in part (a). By the superposition principle, another solution is e(k m)x ekx . m We now let m : 0. This becomes 0>0, and by l’Hôpital’s rule (differentiation of numerator and denominator separately with respect to m, not x!) we obtain xekx>l xekx. The ODE becomes y s 2ky r k 2y 0. The characteristic equation is l2 2kl k 2 (l k) 2 0 and has a double root. Since a 2k, we get k a>2, as expected.

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SECTION 2.3. Differential Operators. Optional, page 60 Purpose. To take a short look at the operational calculus of second-order differential operators with constant coefficients. This parallels and confirms our discussion of ODEs with constant coefficients. A discussion of the case of variable coefficients would exceed the level and the area of interest of the book, sidetrack the attention of the student, and give no substantial additional insights that might be helpful to our further work. SOLUTIONS TO PROBLEM SET 2.3, page 61 2. The first function gives 6x 3 9x 2 9x 9x 2 3x 3. The second function gives 9e3x 9e3x 0. The third function gives 4 sin 4x 4 cos 4x 3 cos 4x 3 sin 4x 7 cos 4x sin 4x. 4. For the first function, (D 6I)(6 6 cos 6x 36x 6 sin 6x) 36 sin x 36 36 cos 6x 36 36 cos 6x 216x 36 sin 6x 72 72 cos 6x 216x. For the second function, (D 6I )(eⴚ6x 6xeⴚ6x 6xeⴚ6x) 6eⴚ6x 6eⴚ6x 0. 6. (D 2.8I ) (D 1.2I ), y c1eⴚ2.8x c2eⴚ1.2x 8. (D 13iI )(D 13iI ), y c1 cos 13x c2 sin 13x 10. (D 2.4I )2, y (c1 c2x)eⴚ2.4x 12. [D (1.5 0.5i)I][D (1.5 0.5i)I], y e ⴚ1.5x(c1 cos 12 x c2 sin 12 x) 14. y is a solution, as follows from the superposition principle in Sec. 2.1 since the ODE is homogeneous and linear. In applying l’Hopital’s rule, regard y as a function of ␮, the variable that approaches the limit, whereas l is fixed. Differentiation of the numerator with respect to ␮ gives xe␮x 0 and differentiation of the denominator gives 1. The limit of this is xelx. SECTION 2.4. Modeling of Free Oscillations of a Mass–Spring System, page 62 Purpose. To present a main application of second-order constant-coefficient ODEs my s cy r ky 0 resulting as models of motions of a mass m on an elastic spring of modulus k ( 0) under linear damping c ( 0) applying Newton’s second law and Hooke’s law. These are free motions (no driving force). Forced motions follow in Sec. 2.8.

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This system should be regarded as a basic building block of more complicated systems, a prototype of a vibrating system that shows the essential features of more sophisticated systems as they occur in various forms and for various purposes in engineering. The quantitative agreement between experiments of the physical system and its mathematical model is surprising. Indeed, the student should not miss performing experiments if there is an opportunity, as I had as a student of Prof. Blaess, the inventor of a (now obscure) graphical method for solving ODEs. Main Content, Important Concepts Restoring force ky, damping force cy r , force of inertia my s

No damping, harmonic oscillations (4), natural frequency v0>(2p) Overdamping, critical damping, nonoscillatory motions (7), (8) Underdamping, damped oscillations (10)

In the text, the derivation of the model has been simplified by clarifying the role of the force F0, which has no effect on the motion. The model, like many others, is obtained from Newton’s second law. We discuss the undamped case c 0 and the damped case c 0 separately because the types of motion are basically different, as follows. The undamped case c 0 gives a harmonic motion (4) for an infinite time interval (practically: for a long time). The damped case c 0 gives a damped motion, which is either oscillatory or, if c is large enough, is a nonoscillatory approach to zero. Hence it is interesting that the formal distinction of Cases I–III mechanically corresponds to quite different types of motion. No damping (c 0) means no loss of the energy corresponding to the initial displacement and initial velocity. Make sure that the student understands the physics behind (4*), that shows the phase shift. Examples in the Text. The examples illustrate the following. Example 1 discusses the undamped case c 0. Example 2 compares the three cases, the three types of motion, graphically shown in Fig. 40, namely, Case I giving a rapid approach to zero, Case II looking almost the same, also showing a rapid and monotone approach to zero, and, finally, Case III a damped oscillation, of a frequency smaller than that of the harmonic oscillation when c 0. Problem Set 2.4. Problems 1–6 enhance the physical understanding and insight into the basic properties of the undamped model. Team Project 10 shows that the model in the text is in fact the prototype of various physical systems governed by the same mathematical formulas. Problems 11–19 play a role for the damped case similar to that of Probs. 1–9 for the undamped model. CAS Project 20 shows the “continuity” in the transitions between Cases I–III, that is also illustrated in Fig. 47. SOLUTIONS TO PROBLEM SET 2.4, page 69 2. W 20 and s0 2 gives k W>s0 10 by Hooke’s law. Thus f

2k>m 2k>(W>g) 210>(20>980) v0 3.52 [Hz]. 2p 2p 2p 2p

From this we get the period 1>f 0.284 [sec].

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4. No because the frequency depends only on k>m, not on initial conditions. 6. By Hooke’s law, F1 k 1 8 stretches spring S1 by 8, and F2 k2 12 stretches spring S2 by 12. Hence the unknown k of the combination of the springs stretches S1 by k>k 1 k>8 and S2 by k>k 2 k>12. And k is such that the sum of these stretches equals 1, because k is the force that corresponds to the stretch 1 of the combination. Thus k k 1, k1 k2

1 1 1 . k1 k2 k

Answer: k 4.8.

8. my s p # 0.32yg, where p # 0.32y is the volume of water displaced when the buoy is depressed y meters from its equilibrium position, and g 9800 nt is the weight of water per cubic meter. Thus y s v02y 0, where v02 p # 0.32g>m and the period is 2p>v0 2; hence m p # 0.32g>v02 0.32g> p 280.7 W mg 280.7 # 9.80 2750.86 [nt]

(about 620 lb).

10. Team Project. (a) By Prob. 7 the frequency is g 1 1 9.80 0.498, 2p B L 2p B 1 so it takes about 2 sec to complete 1 cycle. Answer: It ticks about 30 times per minute. (b) W ks0 8. Now s0 1 because the system has its equilibrium position 1 cm below the horizontal line. Also, m W>g, so that v

W>s0 k 2g 2980 31.3, B m B W>g

and we get the general solution y A cos 31.3t B sin 31.3t. The initial conditions give y(0) A 0 and y r (0) 31.3B 10. Hence B 0.319 and the answer is y 0.319 sin 31.3t [cm]. (c) u(t) 0.5235 cos 3.7t 0.0943 sin 3.7t [rad] 12. y 0 gives c1 c2eⴚ2bt, which has at most one solution because the exponential function is monotone. 14. Case (II) of (5) with c 24mk 24 # 500 # 4500 3000 [kg/sec], where 500 kg is the mass per wheel. 16. 2p>v* since Eq. (10) and y r 0 give tan (v*t d) a>v*; tan is periodic with period p>v*. 18. If an extremum is at t 0, the next one is at t 1 t 0 p>v*, by Prob. 16. Since the cosine and sine in (10) have period 2p>v*, the amplitude ratio is exp( at 0)>exp( at 1) exp( a(t 0 t 1)) exp(ap>v*).

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The natural logarithm is ap>v*, and maxima alternate with minima. Hence ¢ 2pa>v* follows. For the ODE, ¢ 2p # 1>(12 24 ⴢ 5 22) p. 20. CAS Project. (a) The three cases appear, along with their typical solution curves, regardless of the numeric values of k>m, y(0), etc. (b) The first step is to see that Case II corresponds to c 2. Then we can choose other values of c by experimentation. In Fig. 47 the values of c (omitted on purpose; the student should choose!) are 0 and 0.1 for the oscillating curves, 1, 1.5, 2, 3 for the others (from below to above). (c) This addresses a general issue arising in various problems involving heating, cooling, mixing, electrical vibrations, and the like. One is generally surprised how quickly certain states are reached whereas the theoretical time is infinite. (d) General solution y(t) eⴚct>2(A cos v*t B sin v*t), where v* 12 24 c2. The first initial condition y(0) 1 gives A 1. For the second initial condition we need the derivative (we can set A = 1) c c y r (t) eⴚct>2 a cos v*t B sin v*t v* sin v*t v*B cos v*tb . 2 2 From this we obtain y r (0) c>2 v*B 0, B c>(2v*) c> 24 c2. Hence the particular solution (with c still arbitrary, 0 c 2) is y(t) eⴚct>2 acos v*t

c 24 c2

sin v*tb.

It derivative is, since the cosine terms drop out, y r (t) eⴚct>2( sin v*t) a

2 24 c

1 24 c2 b 2 224 c 2

eⴚct>2 sin v*t.

The tangent of the y-curve is horizontal when y r 0, for the first positive time when v*t p, thus t t2 p>v* 2p> 24 c 2. Now the y-curve oscillates between eⴚct>2, and (11) is satisfied if eⴚct>2 does not exceed 0.001. Thus ct 2 ln 1000, and t t 2 gives the best c satisfying (11). Hence c

2 ln 1000 t2

(ln 1000)2

(4 – c2).

The solution of this is c 1.821, approximately. For this c we get by substitution v* 0.4141, t 2 7.587, and the particular solution y(t) eⴚ0.9103t(cos 0.4139t 2.199 sin 0.4139t) The graph shows a positive maximum near 15, a negative minimum near 23, a positive maximum near 30, and another negative minimum at 38.

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(e) The main difference is that Case II gives y (1 t)eⴚt which is negative for t 1. The experiments with the curves are as before in this project. SECTION 2.5. Euler–Cauchy Equations, page 71 Purpose. Algebraic solution of the Euler–Cauchy equation, which appears in certain applications (see our Example 4) and which we shall need again in Sec. 5.4 as the simplest equation to which the Frobenius method applies. We have three cases; this is similar to the situation for constant-coefficient equations, to which the Euler–Cauchy equation can be transformed (Team Project 20(d)); however, this fact is of theoretical rather than of practical interest. Comment on Footnote 4 Euler worked in St. Petersburg 1727–1741 and 1766–1783 and in Berlin 1741–1766. He investigated Euler’s constant (See. 5.6) first in 1734, used Euler’s formula (Secs. 2.2, 13.5, 13.6) beginning in 1740, introduced integrating factors (Sec. 1.4) in 1764, and studied conformal mappings (Chap. 17) starting in 1770. His main influence on the development of mathematics and mathematical physics resulted from his textbooks, in particular from his famous Introductio in analysin infinitorum (1748), in which he also introduced many of the modern notations (for trigonometric functions, etc.). Euler was the central figure of the mathematical activity of the 18th century. His Collected Works are still incomplete, although some seventy volumes have already been published. Cauchy worked in Paris, except during 1830–1838, when he was in Turin and Prague. In his two fundamental works, Cours d ’Analyse (1821) and Résumé des leçons données à l’École royale polytechnique (vol. 1, 1823), he introduced more rigorous methods in calculus, based on an exactly defined limit concept; this also includes his convergence principle (Sec. 15.1). Cauchy also was the first to give existence proofs in ODEs. He initiated complex analysis; we discuss his main contributions to this field in Secs. 13.4, 14.2–14.4, and 15.2, His famous integral theorem (Sec. 14.2) was published in 1825 and his paper on complex power series and their radius of convergence (Sec. 15.2), in 1831. Examples in the Text. The examples illustrate the following. Examples 1–3 and Fig. 48 illustrate Cases I–III, respectively. In particular, Example 3 shows the derivation of real solutions from complex ones. Example 4 shows the occurrence of an Euler–Cauchy equation in connection with the electric potential field between concentric spheres kept at different constant potentials. Here, the student may wish to find a solution formula for arbitrary r1, r2 and potentials v1, v2. SOLUTIONS TO PROBLEM SET 2.5, page 73 2. y c1x 5 c2x ⴚ4 4. y c1 c2>x 6. y c1x 0.5 c2x ⴚ0.2 8. y (c1 c2 ln x)x 2 10. y x(c1 cos (2 ln x) c2 sin (2 ln x))

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12. y 1.2x 2 0.8x 3 14. y c1 cos (3 ln x) c2 sin (3 ln x) is a general solution, and from the initial conditions we obtain the answer y 56 sin (3 ln x) because y(1) c1 cos 0 c2 sin 0 c1 0 and 3 3 y r (x) c1( sin (3 ln x) # x c2 cos (3 ln x) # x c1 # 0 c2 # 3 2.5, so that c2 56. 16. A general solution is y(x) (c1 c2 ln x)x 2, so that y(1) c1 p. The derivative is y r (x)

c2 2 x (c1 c2 ln x) # 2x, x

so that y r (1) c2 2c1 c2 2p 2p , hence c2 4p. This gives the answer y ( p 4p ln x)x 2. 18. The auxiliary equation is 9 (m (m 1) 13 m 19) 9 (m2 23 m 19) 9 (m 13)2 0 has the double root 31. Hence a general solution is y(x) (c1 c2 ln x) x 1>3. ˛

By the first initial condition, y(1) c1 1. Differentiation gives y r (x)

c2 1>3 x (c1 c2 ln x) # 13 xⴚ2>3. x

The second initial condition thus gives y r (1) c2 c1 # 13 0.

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Hence c2 13. This yields the answer (the solution of the initial value problem) y (1 13 ln x) x 1>3. 20. Team Project. (a) The student should realize that the present steps are the same as in the general derivation of the method in Sec. 2.1. An advantage of such specific derivations may be that the student gets a somewhat better understanding of the method and feels more comfortable with it. Of course, once a general formula is available, there is no objection to applying it to specific cases, but often a direct derivation may be simpler. In that respect the present situation resembles, for instance, that of the integral solution formula for first-order linear ODEs in Sec. 1.5. (b) The Euler–Cauchy equation to start from is x 2y s (1 2m s)xy r m(m s)y 0 where m (1 a)>2, the exponent of the one solution we first have in the critical case. For s : 0 the ODE becomes x 2y s (1 2m)xy r m 2y 0. Here 1 2m 1 (1 a) a, and m 2 (1 a) 2>4, so that this is the Euler–Cauchy equation in the critical case. Now the ODE is homogeneous and linear; hence another solution is Y (x mⴙs x m)>s. L’Hôpital’s rule, applied to Y as a function of s (not x, because the limit process is with respect to s, not x), gives (x mⴙs ln ƒ x ƒ )>1 : x m ln ƒ x ƒ

as s : 0.

This is the expected result. (c) This is less work than perhaps expected, an exercise in the technique of differentiation (also necessary in other cases). We have y x m ln x, and with (ln x) r 1>x we get y r mx mⴚ1 ln ƒ x ƒ x mⴚ1 y s m(m 1)x mⴚ2 ln ƒ x ƒ mx mⴚ2 (m 1)x mⴚ2. Since x m x (1 a)>2 is a solution, in the substitution into the ODE the ln-terms drop out. Two terms from y s and one from y r remain and give x 2(mx mⴚ2 (m 1)x mⴚ2) ax m x m(2m 1 a) 0 because 2m 1 a. # # (d) t ln x, dt>dx 1>x, y r y t r y >x, where the dot denotes the derivative with respect to t. By another differentiation, y s (y >x) r y >x 2 y >( x 2).

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Substitution of y r and y s into (1) gives the constant-coefficient ODE

y y ay by y (a 1)y by 0. The corresponding characteristic equation has the roots l 12 (1 a) 214(1 a)2 b. With these l, solutions are elt (et)l (eln ƒ x ƒ)l xl. (e) telt (ln ƒ x ƒ )el ln ƒ x ƒ (ln ƒ x ƒ )(eln ƒ x ƒ)l xl ln ƒ x ƒ . SECTION 2.6. Existence and Uniqueness of Solutions. Wronskian, page 74 Purpose. To explain the theory of existence of solutions of ODEs with variable coefficients in standard form (that is, with y s as the first term, not, say, f (x)y s ) y s p(x)y r q(x)y 0

(1)

and of their uniqueness if initial conditions y(x0) K0,

(2)

y r (x0) K1

are imposed. Of course, no such theory was needed in the last sections on ODEs for which we were able to write all solutions explicitly. Main Content. The theorems show the following. Theorem 1 shows that the continuity of the coefficients suffices for the existence and uniqueness of a solution of the initial values problem (1), (2). Theorem 2 gives a criterion for linear dependence and independence involving the Wronskian. Simple basic applications are shown in Examples 1 and 2. Theorem 3 on the existence of a general solution follows from Theorems 1 and 2 by the trick of using two special initial value problems; this idea is worth remembering. For Theorem 4 see below. Comment on Wronskian For n 2, where linear independence and dependence can be seen immediately, the Wronskian serves primarily as a tool in our proofs; the practical value of the independence criterion will appear for higher n in Chap. 3. Comment on General Solution Theorem 4 shows that linear ODEs (actually, of any order) have no singular solutions. This also justifies the term “general solution,” on which we commented earlier. We did not pay much attention to singular solutions, which sometimes occur in geometry as envelopes of one-parameter families of straight lines or curves. Altogether, this provides a general theory that is useful in practice. SOLUTIONS TO PROBLEM SET 2.6 page 79

冟冟

冟

e4x eⴚ1.5x 2.5x 1 4x ⴚ1.5x e 4e 1.5e 4 x 1>x 2 4. W x 1 1>x 2 2. W

冟

1 5.5e2.5x 1.5

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6. We use the abbreviations c cos vx, s sin vx. Then W

冟

eⴚxc eⴚx( c v s)

冟

eⴚxs c e 2x ⴚx e ( s v c) c v s

冟

s veⴚ2x. s v c

8. Equation (6*) saves much work and avoids sources of errors. We obtain y2 r a y b (tan (ln x)) r 1

1 # 1, cos2 (ln x) x

Multiplication by y 12 x 2k cos2 (ln x) gives W x 2k 1. 10. x 2y s (m 1 m 2 1)xy r m 1m 2y 0. For the Wronskian we obtain from (6*) x m1 W a m2 b r x 2m2 (m 1 m 2)x m1ⴙm2ⴚ1. x From the initial conditions we obtain the particular solution y 2x m1 4x m2. 12. x 2y s 3xy r 4y 0. The Wronskian is W

冟

x2 2x

冟

x 2 ln x x 3. 2x ln x x

The solution of the initial value problem is y (4 2 ln x)x 2. 14. y1 eⴚkx cos px, y2 eⴚkx sin px. By (6*), W (y2>y1) r y 12 (tan px) r eⴚ2kx cos2 px peⴚ2kx.

The characteristics equation is (l k)2 p2 0. This gives the ODE y s 2ky r 1k2 p22y 0. From the initial conditions we obtain the particular solution y eⴚkx(cos px sin px). 16. Team Project. (a) c1e x c2e ⴚx c*1 cosh x c*2 sinh x. Expressing cosh and sinh in terms of exponential functions [see (17) in App. 3.1], we have 1 * * x 1 * * ⴚx; 2 (c1 c2 )e 2 (c1 c2 )e

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hence c1 12(c*1 c*2 ), c2 12(c*1 c*2 ). The student should become aware that for second-order ODEs there are several possibilities for choosing a basis and making up a general solution. For this reason we say “a general solution,” whereas for first-order ODEs we said “the general solution.” (b) If two solutions are 0 at the same point x 0, their Wronskian is 0 at x 0, so that these solutions are linearly dependent by Theorem 2. (c) The derivatives would be 0 at that point. Hence so would be their Wronskian W, implying linear dependence. (d) y2>y1 is constant in the case of linear dependence; hence the derivative of this quotient is 0, whereas in the case of linear independence this is not the case. This makes it likely that such a formula should exist. (e) The first two derivatives of y1 and y2 are continuous at x 0 (the only x at which something could happen). Hence these functions qualify as solutions of a secondorder ODE. y1 and y2 are linearly dependent for x 0 as well as for x 0 because, in each of these two intervals, one of the functions is identically 0. On 1 x 1 they are linearly independent because c1y1 c2y2 0 gives c1 0 when x 0, and c2 0 when x 0. The Wronskian is W y1 y 2r y2 y 1r e

0 # 3x 2 x 3 # 0 f 0 x 3 # 0 0 # 3x 2

if e

x 0 . x 0

The Euler–Cauchy equation satisfied by these functions has the auxiliary equation (m 3)m m(m 1) 2m 0. Hence the ODE is xy s 2y r 0. Indeed, xy s1 2y1r x # 6x 2 # 3x2 0 if x 0, and 0 0 for x 0. Similarly for y2. Now comes the point. In the present case the standard form, as we use it in all our present theorems, is ys

2 yr 0 x

and shows that p(x) is not continuous at 0, as required in Theorem 2. Thus there is no contradiction. This illustrates why the continuity assumption for the two coefficients is quite important. (f) According to the hint given in the enunciation, the first step is to write the ODE (1) for y1 and then again for y2. That is, y1s py 1r qy1 0 y s2 py r2 qy2 0

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where p and q are variable. The hint then suggests eliminating q from these two ODEs. Multiply the first equation by y2, the second by y1, and add: (y1 y s2 y s1 y2) p(y1 y2r y1r y2) W r pW 0 where the expression for W r results from the fact that y1r y 2r appears twice and drops out. Now solve this by separating variables or as a homogeneous linear ODE. In Prob. 6 we have p 2, hence W ceⴚ2x by integration from 0 to x, where c y1(0)y2r (0) y2(0)y1r (0) 1 # v 0 # ( 1) v. SECTION 2.7. Nonhomogeneous ODEs, page 79 Purpose. We show that for getting a general solution y of a nonhomogeneous linear ODE we must find a general solution yh of the corresponding homogeneous ODE and then—this is our new task—any particular solution yp of the nonhomogeneous ODE, y yh yp. Main Content, Important Concepts General solution, particular solution Continuity of p, q, r suffices for existence and uniqueness. A general solution exists and includes all solutions. Comment on Methods for Particular Solutions The method of undetermined coefficients is simpler than that of variation of parameters (Sec. 2.10), as is mentioned in the text, and it is sufficient for many applications, of which Secs. 2.8 and 2.9 show standard examples. Comment on General Solution Theorem 2 shows that the situation with respect to general solutions is practically the same for homogeneous and nonhomogeneous linear ODEs. Comment on Table 2.1 It is clear that the table could be extended by the inclusion of products of polynomials times cosine or sine and other cases of limited practical value. Also, a 0 in the last pair of lines gives the previous two lines, which are listed separately because of their practical importance. General Comments on Text. Determination of Constants. For a good understanding, it is important to realize that a general solution of a nonhomogeneous linear ODE contains two kinds of constants, namely, (I) the constants in Table 2.1, which depend on the right side of the ODE, but not on the initial conditions, and must be determined first, (II) the two arbitrary constants in a general solution yh of the homogeneous ODE (and thus in a general solution y yh yp of the given ODE) and must be determined after the constants in 1I2 have been determined, and are to be determined by using the initial conditions. Examples 1–3 illustrate this important fact, which, for weaker students, sometimes causes difficulties in understanding.

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Examples in the Text. Examples 1–3 are very similar, illustrating Rules (a), (b), (c). In particular, the student should realize that Example 3 on the sum rule is not more difficult than the other two examples. The only additional idea in the case, say, r r1 r 2 on the right, is to split yp into a sum, yp yp1 yp2, whose coefficients, according to Table 2.1, can be determined for yp1 and yp2 separately. Hence we have to determine two sets of constants from two systems of algebraic equations, each of which is not larger than it would be had we only yp1 or only yp2 on the right side of the given ODE. Problem Set 2.7 This problem set begins with the determination of general solutions of nonhomogeneous linear ODEs, Probs. 1–6 with a single term on the right, Probs. 7–9 with a sum of two terms each, and Prob. 10 showing a simple extension of the method of undetermined coefficients beyond functions r shown in Table 2.1. Problems 11–18 concern IVPs for nonhomogeneous linear ODEs with one term on the right (Probs. 11–14 and 17) and with two terms on the right (Probs. 15, 16, and 18). CAS Project 19 should make the student aware that, depending on the initial conditions and on the kind of the homogeneous ODE, the solution y yh yp may approach yp as: x : , or may contain an increasing ƒ yh ƒ , or may be of the form y yp with yh absent. Team Project 20 is an invitation to explore more general functions on the right, and to what Euler–Cauchy equation the present method can be extended.

SOLUTIONS TO PROBLEM SET 2.7, page 84 2. y c1eⴚ3.2x c2eⴚ1.8x 0.00999 cos x 0.0105 sin x. This is a typical solution of a forced oscillation problem in the overdamped case. The general solution of the homogeneous ODE dies out, practically after some short time (theoretically never), and the transient solution goes over into a harmonic oscillation whose frequency is equal to that of the driving force (or electromotive force). Note that the input (the driving force) is a cosine, whereas the output (the response) is a cosine and sine; this means a phase shift. It is due to the presence of a y r -term, mechanically a linear damping force, as we shall see in the next section. 2 4. y c1e3x c2eⴚ3x cos px. Observe that the ODE has no y r -term, 1 (p>3)2 so we have no phase shift, and the output is a pure cosine term, just like the input. Compare this with Prob. 2. 6. A general solution of the homogeneous ODE is yh eⴚx>2(c1 cos px c2 sin px). We see that the function on the right side of the ODE is a solution of the homogeneous ODE. Hence we have to apply the Modification Rule, starting from yp xeⴚx>2(K cos px M sin px). Substitution gives K 1>(2p); M 0. Hence the answer is y yh yp e ⴚx>2 ac1 cos px c2 sin px

x cos pxb. 2p

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Note that the output involves cosine, whereas the input involves sine; and, although we have a y r -term, the output is a single term. Compare this with Probs. 2 and 4, which differ from the present situation. 1 x sin 3x. An important point is that the 8. y A cos 3x B sin 3x 18 cos x 18 Modification Rule applies to the second term on the right. Hence the best way seems to split yp additively, yp yp1 yp2, where yp1 K 1 cos x M 1 sin x,

yp2 K 2x cos 3x M 2x sin 3x.

In Prob. 9 the situation is similar. 10. 2x sin x is not listed in the table because it is of minor practical importance. However, by looking at its derivatives, we see that yp Kx cos x Mx sin x N cos x P sin x should be general enough. Indeed, by substitution and collecting cosine and sine terms separately we obtain (1)

(2K 2Mx 2P 2M) cos x 0

(2)

( 2Kx 2M 2N 2K) sin x 2x sin x.

In (1) we must have 2Mx 0; hence M 0 and then P K. In (2) we must have 2Kx 2x; hence K 1, so that P 1 and from (2), finally, 2N 2K 0, hence N 1. Answer: y (c1 c2x)eⴚx (1 – x) cos x sin x. 12. The Modification Rule is needed. The answer is y 1.8 cos 2x sin 2x 3x cos 2x. 14. The characteristic equation of the homogeneous ODE has a double root 2. The function on the right is such that the Modification Rule does not apply. A general solution of the nonhomogeneous ODE is y (c1 c2x)eⴚ2x 14 eⴚ2x sin 2x. From this and the initial conditions we obtain the answer y (1 x)eⴚ2x 14 eⴚ2x sin 2x. 16. yh c1e2x c2, yp C1xe2x C2eⴚ2x by the Modification Rule for a simple root. y e2x 32 3xe2x 12 eⴚ2x. 18. The Basic Rule and the Sum Rule are needed. We obtain yh eⴚx(A cos 3x B sin 3x) y eⴚx cos 3x 0.4 cos x 1.8 sin x 6 cos 3x sin 3x.

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20. Team Project. (b) Perhaps the simplest way is to take a specific ODE, e.g., x 2y s 6xy r 6y r(x) and then experiment by taking various r(x) to find the form of choice functions. The simplest case is a single power of x. However, almost all the functions that work as r(x) in the case of an ODE with constant coefficients can also be used here. SECTION 2.8. Modeling: Forced Oscillations. Resonance, page 85 Purpose. To extend Sec. 2.4 from free to forced vibrations by adding an input (a driving force, here assumed to be sinusoidal). Mathematically, we go from a homogeneous to a nonhomogeneous ODE which we solve by undetermined coefficients. New Features Undamped Forced Oscillations. Resonance (Fig. 55) Resonance appears if the physical system is (theoretically) undamped (in practice, if it has small enough damping that the damping effect can be neglected), and if the input frequency is exactly equal to the natural frequency of the system. Then a solution (11)

y At sin v0t

has a factor t which makes it increase to (Fig. 55). The approach to resonance as v : v0 ( 1k>m2 is also characterized by the resonance factor in Fig. 54. Undamped Forced Oscillations. Beats (Fig. 56) Beats occur if the input frequently is approximately equal to the natural frequency of the physical system. Then (12)

y K (cos vt cos v0t)

with K F0>[m(v02 v2)]. Damped Forced Oscillations. For these the transient solution approaches the steady-state solution as t : , practically after some time which may often be rather short. If c 0, there is no more true resonance, but the maximum amplitude (16) may still be large, as Fig. 57 illustrates. Also, there is a phase lag ␩, discontinuous and equal to 0 or p when c 0, and continuous and monotone when c 0 (Fig. 58). Problem Set 2.8 Problems 3–7 concern damped systems. Hence a general solution of such a physical system is that of a homogeneous linear ODE and approaches 0 as t : , so that solving these problems amounts to determining a particular solution of the corresponding ODE. Problems 8–15 amount to finding a general solution of the nonhomogeneous ODE. Problems 16–20 are IVPs for nonhomogeneous linear ODEs. Problem 17 is of the kind that will occur in connection with partial sums of Fourier series in Chap. 11. Problem 18 is a typical example illustrating the rapidity of approach to the steady-state solution. Beats are considered in Probs. 21, 22, and 25 with v 0.9, whereas for v farther away from v0 1 (corresponding to the natural frequency of the physical system) the form of vibrations cannot be guessed immediately (see Fig. 60).

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Practical resonance is considered experimentally in Team Experiment 23. Continuity Conditions Nonhomogeneous linear ODEs with a driving force acting for some finite interval of time only will require the idea of continuity conditions of y and y r at the instant of time when the driving force becomes identically zero. This makes such problems more involved, as Prob. 24 illustrates, and motivates the application of an “operational method,” such as the Laplace transform (Chap. 6). SOLUTIONS TO PROBLEM SET 2.8, page 91 2. Problems 2, 8, 18. Note that the damping and restoring terms must have positive coefficients, and that Prob. 12 shows resonance; hence it is not a candidate. 4. yp cos 4t 0.6 sin 4t 1 1 1 6. yp 10 cos t 90 cos 3t 15 sin t 45 sin 3t ⴚt 8. y e (A cos 1.5t B sin 1.5t) 0.6 cos 1.5t 0.2 sin 1.5t 10. y A cos 4t B sin 4t 7t sin 4t 12. y eⴚt(A cos 2t B sin 2t) 2 sin t. Note that, whereas a single term on the right side of the ODE will usually produce two terms in the solution (the response), the present problem shows that sometimes the opposite will also occur. 14. y A cos t B sin t eⴚt(cos t 2 sin t). Note that this does not give resonance, but, on the contrary, yp : 0, which is understandable because the driving force on the right approaches 0 as t approaches . 16. A general solution is y A cos 5t B sin 5t sin t. From the initial condition we obtain A 1 and B 0. Hence the answer is y cos 5t sin t. Note that, whereas in general both solutions of a basis occur in the solution of an IVP, here we have only one of them. Of course, this could be changed by changing the initial conditions. 18. y eⴚ4t cos t 26.8 sin 0.5t 6.4 cos 0.5t. At t 1.2 the exponential term has decreased to less than 1% of its original value. This marks the end of the transition from a practical point of view. t 1.8 is the time when that term has become less than 1>10 of a percent in absolute value. 20. A general solution is y A cos 15 t B sin 15 t (cos pt sin pt)>(p2 5). Using the initial conditions, we obtain the answer y

1 p acos 15 t sin 15 t cos pt sin ptb. p 5 15 2

22. y 100 cos 4.9t 98 cos 5t. (a) changes the coefficient of cos 5t (b) changes the amplitude

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24. If 0

p, then a particular solution yp K 0 K 1t K 2t 2

gives y sp 2k 2 and 1

y sp yp K0 2K2 K1t K2t2 1

t2;

thus, K2

1 2

K1 0,

K0 1 2K2 1

Hence a general solution is y A cos t B sin t 1

2 2

t2.

From this and the first initial conditions, y(0) A 1

A a1

The derivative is y r A sin t B cos t

and gives y r (0) B 0. Hence the solution is (I)

y(t) (1 2> p2)(1 cos t) t 2> p2

if 0

and if t p, then (II)

y y2 A2 cos t B2 sin t

with A2 and B2 to be determined from the continuity conditions y(p) y2(p),

y r (p) y 2r (p).

So we need from (I) and (II) y(p) 2(1 2> p2) 1 1 4> p2 y2(p) A2 and y r (t) (1 2> p2) sin t 2t> p2

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and from this and (II), y r (p) 2> p B cos p B2. This gives the solution y (1 4> p2) cos t (2> p) sin t

t p.

(1 2> p2)(1 cos t) t 2> p2

if 0

Answer: y e

(1 4> p ) cos t (2> p) sin t 2

if t p

The function in the second line gives a harmonic oscillation because we disregarded damping. SECTION 2.9. Modeling: Electric Circuits, page 93 Purpose. To discuss the current in an RLC-circuit with sinusoidal input E 0 sin vt, as follows. Modeling the RLC-circuit Solving the model (1) for the current I(t) Discussion of a typical IVP Discussion of a electrical–mechanical analogy Modeling the RLC-circuit. The student should first review the special case of an RL-circuit in Example 2 of Sec. 1.5, which is modeled by a first-order ODE, using Kirchhoff’s KVL. The present addition of a capacitor is very simple in terms of setting up the model, resulting in a second-order ODE. Proceed stepwise in this way: Write the voltage drop across the capacitor in the form (1>C)Q (rather than CQ) is a standard convention to obtain generally more convenient numbers. Solve the model (1) by the method of undetermined coefficients (see Sec. 2.7). ATTENTION! The right side in (1) is E 0v cos vt, because of differentiation. In solving, two quantities of practical importance are introduced, namely, the reactance (3)

S vL 1>(vC)

and the impedance (also called the apparent resistance) 2R 2 C 2. (Its complex analog, the complex impedance Z R iS, is mentioned in the answer to Prob. 20.) Example 1 shows a typical IVP with Ih rapidly going to 0, as illustrated in Fig. 62, so that the transient current rapidly approaches a harmonic steady-state current. Table 2.2 shows a strictly quantitative electrical–mechanical analogy, which is used in transducers, as explained in the text.

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SOLUTIONS TO PROBLEM SET 2.9, page 98 2. This is another special case of a circuit that leads to an ODE of first order, RI r I>C E r vE 0 cos vt. Integration by parts gives the solution I(t) eⴚt>(RC) c

冮

v E0 t>(RC) e cos vt dt c d R

ce ⴚt>(RC) ceⴚt>(RC)

vE0C 1 (vRC)2

(cos vt vRC sin vt)

vE 0C 21 (vRC)2

sin (vt d),

where tan d 1>(vRC). The first term decreases steadily as t increases, and the last term represents the steady-state current, which is sinusoidal. The graph of I(t) is similar to that in Fig. 62. 4. The integral that occurs can be evaluated by integration by parts, as is shown (with other notations) in standard calculus texts. From (4) in Sec. 1.5 we obtain I eⴚRt>L c

E 0 Rt>L e sin vt dt c d L

ceⴚRt>L ce ⴚRt>L

冮

E0 R v2L2 2

(R sin vt vL cos vt)

E0 2R v L 2

2 2

sin (vt d),

6. E 2t 2, E r 4t, 0.5I s 200I 4t, I r (0) 0 follows from

d arctan

I s 400I 8t,

I(0) 0 is given.

LI r (0) Q(0)>C E(0) 0. Answer: I 0.02(t 0.05 sin 20t). 8. E r 1000 cos 2t, 0.5I s 4I r 10I 1000 cos 2t, so that the steady-state solution is I 62.5 (cos 2t sin 2t) A. 10. The ODE is I s 2I r 20I 157 # 3.

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The steady-state solution is I 33 cos 3t 18 sin 3t. Note that if you let C decrease, the sine term in the solution will become smaller and smaller, compared with the cosine term. 12. The ODE is 0.1I s 0.2I r 0.5I 220 # 314 cos 314t. Its characteristic equation is 0.1[(l 1) 2 4] 0. Hence a general solution of the homogenous ODE is eⴚt(A cos 2t B sin 2t). The transient solution (rounded to 4 decimals) is I eⴚt(A cos 2t B sin 2t ) 7.0064 cos 314t 0.0446 sin 314t A. 14. Write l1 a b and l2 a b, as in the text before Example 1. Here a R>(2L) 0, and b can be real or imaginary. If b is real, then b R>(2L) 0 (and l2 0, of course). If b is imaginary, because R2 4L>C R 2. Hence l1 then Ih(t) represents a damped oscillation, which certainly goes to zero as t : . 16. The ODE is 0.2I s 8I r 80I 1000 cos 10t. A general solution is I (c 1 c2t)eⴚ20t 6 cos 10t 8 sin 10t. The initial conditions are I(0) 0, Q(0) 0, which because of (1 r ), that is, LI r (0) RI102

Q(0) E(0) 0, C

leads to I r (0) 0. This gives I(0) c1 6 0, I r (0) 20c1 c2 80 0,

c1 6 c2 200.

Hence the answer is I (6 200t)eⴚ20t 6 cos 10t 8 sin 10t.

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18. The characteristic equation of the homogenous ODE is 0.2(l 8)(l 10) 0. The initial conditions are I(0) 0 as given, I r (0) E(0)>L 820 by formula 1 s in the text and Q(0) 0. Also, E r 8200 sin 10t. The ODE is I s 8I r 80I 8200 sin 10t. The answer is I 160 eⴚ8t 205eⴚ10t 45 cos 10t 5 sin 10t. 20. 苲 I p Keivt, 苲 I pr ivKeivt, 苲 I ps v2Keivt . Substitution gives a v2L ivR

1 b Keivt E 0veivt. C

Divide this by veivt on both sides and solve the resulting equation algebraically for K, obtaining (A)

E0 E0 S iR 1 b iR avL vC

where S is the reactance given by (3). To make the denominator real, multiply the numerator and the denominator of the last expression by S iR. This gives K

E 0(S iR) S 2 R2

The real part of Keivt is (Re K)(Re eivt) (Im K)(Im eivt)

E0S S R 2

cos vt

E0 S R2 2

E0R S R2 2

sin vt

(S cos vt R sin vt),

in agreement with (2) and (4). We mention that (A) can be written K

E0 iZ

where Z R iS R i avL

1 b vC

is called the complex impedance. Note that its absolute value ƒ Z ƒ 2R2 S2 is the impedance, as defined in the text.

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SECTION 2.10. Solution by Variation of Parameters, page 99 Purpose. To discuss the general method for particular solutions, which applies in any case but may often lead to difficulties in integration (which we, by and large, have avoided in our problems, as the subsequent solutions show). Comments The ODE must be in standard form, with 1 as the coefficient of y s —students tend to forget that. Here we do need the Wronskian, in contrast with Sec. 2.6 where we could get away without it. SOLUTIONS TO PROBLEM SET 2.10, page 102 2. y1 cos 3x,

y2 sin 3x,

W 3, r csc 3x. Hence in (2),

冮 W dx 3 冮 sin 3x dx 3 y2r

1 sin 3x

1 3x dx ln ƒ sin 3x ƒ . 冮 W dx 13 冮 cos sin 3x 9 y1r

Answer: y A cos 3x B sin 3x 4. y1 e2x cos x,

y2 e2x sin x,

冮W

y2r y1r

冮冮

1 x cos 3x (sin 3x) ln ƒ sin 3x ƒ 3 9

W e4x. Hence in (2),

(e2x sin x)e2x>sin x

dx x

(e2x cos x)e2x>sin x

dx ln ƒ sin x ƒ .

e4x

Answer: y 3A cos x B sin x x cos x (sin x) ln ƒ sin x ƒ 4e2x. 6. y1 eⴚ3x, y2 xeⴚ3x, W e6x. Hence in (2),

冮W

y2r y1r

冮冮

(xeⴚ3x)16eⴚ3x>(x 2 1) e

ⴚ6x

(eⴚ3x)16eⴚ3x>(x 2 1) e

ⴚ6x

冮 x 16x 1 dx 8 ln (x 1)

冮 x 16 1 dx 16 arctan x.

Answer: y (c1 c2x)eⴚ3x 8[ ln (x2 1) 2x arctan x] eⴚ3x.

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8. y c1 cos 2x c2 sin 2x 18 cosh 2x 10. y1 eⴚx cos x, y2 eⴚx sin x, W eⴚ2x. Hence in (2),

冮W

y2r y1r

冮冮

(eⴚx sin x) 4eⴚx>cos3 x

(eⴚx cos x) 4eⴚx>cos3 x

dx 4 tan x.

eⴚ2x

2 cos2 x

This gives the particular solution eⴚx c (cos x)

2 2 ⴚx 2 4 sin x 4(sin x) tan x e a b d cos x cos2 x

eⴚx[ 2(cos 2x)>cos x]. Answer: y eⴚx[A cos x B sin x 2(cos 2x)>cos x]. 12. The right side suggests the following choice of a basis of solutions: y1 cosh x,

y2 sinh x.

Then W 1, and

冮

yp cosh x (sinh x)>cosh x dx sinh x (cosh x)>cosh x dx (cosh x) ln ƒ cosh x ƒ x sinh x. 14. TEAM PROJECT. (a) y1 eⴚ3x, y2 eⴚx, W 2eⴚ4x, r 65 cos 2x. From (2), yp eⴚ3x

冮

ⴚ3x eⴚx65 cos 2x 65 cos 2x ⴚx e dx e dx 2eⴚ4x 2eⴚ4x

冮

65 a eⴚ3x e3x cos 2x dx eⴚx ex cos 2x dxb 2

ⴚ3x 1 3x ⴚx 1 x 65 2 ( e 13 e (3 cos 2x 2 sin 2x) e 5 e (cos 2x 2 sin 2x))

cos 2x 8 sin 2x. Answer: y c1eⴚ3x c2eⴚx cos 2x 8 sin 2x. This was much more work than that for undetermined coefficients.

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(b) We can treat x 2 on the right by undetermined coefficients, obtaining the contribution x 2 4x 6 to the solution. We could treat it by the other method, but we would have to evaluate additional integrals of an exponential function times a power of x. We treat the other part, 35x 3>2ex, by the method of this section, calling the resulting function yp1. We need y1 ex, y2 xex, W e2x. From this and (2), xex

冮 e 35x e dx xe 冮 e 35x e dx 35 a e 冮 x dx xe 冮 x dxb 4e x .

yp1 ex

3>2 x

5>2

3>2

x 7>2

Complete answer: y (c1 c2x)ex 4exx 7>2 x 2 4x 6. (c) If the right side is a power of x, say, r r0x k, then substitution of yp Cx k gives x2y s axy r by (k(k 1) ak b)Cxk r0 xk. This can be solved for C. To explore further possibilities, one may work “backwards”; that is, assume a solution, substitute it on the left, and see what from one gets as a right side. SOLUTIONS TO CHAPTER 2 REVIEW QUESTIONS AND PROBLEMS, page 102 8. y c1eⴚ4x c2e3x 10. y eⴚ0.1x(A cos 0.4x B sin 0.4x) 12. y (c1 c2x)eⴚ2px 14. y c1x 3 c2x ⴚ3 16. y eⴚx(A cos x B sin x) eⴚx cos 2x 18. y r z, y s (dz>dy)z by the chain rule, yz dz>dy 2z 2, dz>z 2dy>y, ln ƒ z ƒ 2 ln ƒ y ƒ c*, z c1y2 y r , dy>y2 c1 dx, 1>y c1x c2; hence y 1>(c苲1x 苲 c2). Also, y 0 is a solution. 20. Obtain the particular solution by undetermined coefficients. Answer: y 3ex 5e2x 3 cos x sin x 22. The auxiliary equation is m(m 1) 15m 49 (m 7) 2 0. Hence a general solution is y (c1 c2 ln ƒ x ƒ )xⴚ7. From the initial conditions, c1 2, c2 3.

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24. I c1eⴚ1999.87t c2eⴚ0.125008t A 26. E r 220 ⴢ 314 cos 314t, I eⴚ50t(A cos 150t B sin 150t) 0.847001 sin 314t 1.985219 cos 314t A 28. The equation is 0.125y s 1.125y cos t 4 sin t; thus, y s 9y 8 cos t 32 sin t. The solution satisfying the initial conditions is y cos 3t 43 sin 3t cos t 4 sin t, as obtained by the method of undetermined coefficients. The last two terms result from the driving force. In the first two terms, v 0 2k>m 3. This shows that resonance would occur if the driving force had the frequency v(2p) 3>(2p). 30. C*(v) is given by (16), Sec. 2.8. The maximum is obtained by equating the derivative to zero; this gives (15*) in Sec. 2.8, which for our numerical values becomes 16 2(24 v2), so that v 4. Equation (16) in Sec. 2.8 then gives the maximum amplitude C*(vmax)

2 # 1 # 10 424 # 12 # 24 16

0.5590.

To check this result, we determine the general solution, using the method of undetermined coefficients, finding y(t) eⴚ2t(A cos 2 15t B sin 215t) 0.25 cos 4t 0.5 sin 4t, and confirm the result by calculating the amplitude 20.252 0.52 0.5590.

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CHAPTER 3

Higher Order Linear ODEs

Chapters 1 and 2 demonstrate and illustrate that first- and second-order ODEs are, by far, the most important ones in usual engineering and physical applications, and the linear ODEs of orders 1 and 2 foreshadow much of the general theory. This implies that the material of the present chapter is less important than that of the previous chapters. For this reason the present chapter was created from the previous edition by rearranging and extending material previously contained in Chap. 2. This rearrangement parallels the arrangement in Chap. 2, in order to facilitate comparisons. Root Finding For higher order ODEs you may need Newton’s method or some other method from Sec. 19.2 (which is independent of other sections in numerics) in work on a calculator or with your CAS (which may give you a root-finding method directly). Linear Algebra The typical student may have taken an elementary linear algebra course simultaneously with a course on calculus and will know much more than is needed in Chaps. 2 and 3. Thus Chaps. 7 and 8 need not be taken before Chap. 3. In particular, although the Wronskian becomes useful in Chap. 3 (whereas for n ⫽ 2 one hardly needs it), a very modest knowledge of determinants will suffice. (For n ⫽ 2 and 3, determinants are treated in a reference section, Sec. 7.6.) SECTION 3.1. Homogeneous Linear ODEs, page 105 Purpose. Extension of the basic concepts and theory in Secs. 2.1 and 2.6 to homogeneous linear ODEs of any order n. This shows that practically all the essential facts carry over without change. Linear independence, now more involved than for n ⫽ 2, causes the Wronskian to become indispensable (whereas for n ⫽ 2 it played a marginal role). Main Content, Important Concepts Superposition principle for the homogeneous ODE (2) (Theorem 1) General solution (3), basis, particular solution Linear independence and dependence of functions (see (4) and Examples 1 and 2) Linear combination General solution of (2) with continuous coefficients exists. Existence and uniqueness of solution of initial value problem (2), (5) (Theorem 2, Example 4) Linear independence of solutions, Wronskian (6) (Theorem 3, Example 5) Existence of general solution (Theorem 4) General solution includes all solutions of (2) (Theorem 5) Comment on Order of Material In Chap. 2 we first gained practical experience and skill and presented the theory of the homogeneous linear ODE at the end of the discussion, in Sec. 2.6. In this chapter, with all the experience gained on second-order ODEs, it is more logical to present the whole 58

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theory at the beginning and the solution methods (for linear ODEs with constant coefficients) afterward. Similarly, the same logic applies to the nonhomogeneous linear ODE, for which Sec. 3.3 contains the theory as well as the solution methods. Problem Set 3.1 Problems 1–6 should emphasize the importance of the Wronskian for higher-order ODEs. Team Project 7 is another demonstration of the importance of linearity in connection with ODEs. Problems 8–16 invite students to deepen their understanding of linear independence and dependence from a practical point of view, with Probs. 8–15 involving special functions and Prob. 16 concerning the basic general theorems. SOLUTIONS TO PROBLEM SET 3.1, page 111 2. Problems 1–6 should give the student a first impression of the changes in the transition from n ⫽ 2 to general n. 8. Functions y1 ⫽ 0, y2, Á , yn are always linearly dependent because (4) holds with any k 1 ⫽ 0 and the other k j’s all 0. 10. Linearly independent. First proof. Divide (4) by e2x to obtain c1 ⫹ c2 x ⫹ c3 x 2 ⫽ 0. Set x ⫽ 0 to get c1 ⫽ 0. Etc. Second proof. The functions are solutions of the ODE y t ⫺ 6y s ⫹ 12y r ⫺ 8y ⫽ 0. Their Wronskian equals 2e6x. 12. Linearly dependent because cos 2x ⫽ cos2 x ⫺ sin2 x. This problem is typical of the use of functional relations for proofs of linear dependence; for instance, ln x 2 ⫽ 2 ln x, etc. 14. Linearly dependent because cos2 x ⫹ sin2 x ⫽ 1 16. Team Project. (a) (1) No. If y1 ⬅ 0, then (4) holds with any k 1 ⫽ 0 and the other k j all zero. (2) Yes. If S were linearly dependent on I, then (4) would hold with a k j ⫽ 0 on I, hence also on J, contradicting the assumption. (3) Not necessarily. For instance, x 2 and x ƒ x ƒ are linearly dependent on the interval 0 ⬍ x ⬍ 1, but linearly independent on ⫺1 ⬍ x ⬍ 1. (4) Not necessarily. See the answer to (3). (5) Yes. See the answer to (2). (6) Yes. By assumption, k1 y1 ⫹ Á ⫹ kp yp ⫽ 0 with k 1, Á , k p not all zero (this refers to the functions in S), and for T we can add the further functions with coefficients all zero; then the condition for linear dependence of T is satisfied. (b) We can use the Wronskian for testing linear independence only if we know that the given functions are solutions of a homogeneous linear ODE with continuous coefficients. Other means of testing are the use of functional relations, e.g., ln x 2 ⫽ 2 ln x or trigonometric identities, or the evaluation of the given functions at several values of x, to see whether we can discover proportionality.

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SECTION 3.2. Homogeneous Linear ODEs with Constant Coefficients, page 111 Purpose. Extension of the algebraic solution method for constant-coefficient ODEs from n ⫽ 2 (Sec. 2.2) to any n, and discussion of the increased number of possible cases: Real different roots (Theorems 1 and 2, Example 1) Complex simple roots (Example 2) Real multiple roots ((7), Example 3) Complex multiple roots (11) Combinations of the preceding four basic cases Comment on Numerics In practical cases, one may have to use Newton’s method or another method for computing (approximate values of) roots in Sec. 19.2. Problem Set 3.2 Problems 1–13 concern general solutions and IVPs. Reduction of order when solutions are known (for instance, are obtained by inspection) was considered for second-order ODEs in Sec. 2.1 and is extended to third-order ODEs in the present Project 14 as well as in the open-ended CAS Experiment 15, where the difficulty seems to be in finding ODEs simple enough so that they can be successfully handled as requested. SOLUTIONS TO PROBLEM SET 3.2, page 116 2. From the characteristic equation l4 ⫹ 2l2 ⫹ 1 ⫽ (l2 ⫹ 1)2 ⫽ 0 we conclude that the corresponding real general solution is y ⫽ (c1 ⫹ c2x) cos x ⫹ (c3 ⫹ c4x) sin x. 4. From the characteristic equation l3 ⫺ l2 ⫺ l ⫹ 1 ⫽ (l ⫹ 1)(l ⫺ 1)2 we conclude that a general solution is y ⫽ c1eⴚx ⫹ (c2 ⫹ c3x)ex. 6. The characteristic equation is l5 ⫹ 8l3 ⫹ 16l ⫽ l(l ⫹ 4)2. Hence a general solution is y ⫽ c1 ⫹ (c2 ⫹ c3 x) cos 2x ⫹ (c4 ⫹ c5 x) sin 2x. 8. y ⫽ 10e

ⴚ4x

(cos 1.5x ⫺ sin 1.5x) ⫹ 0.05e0.5x

5 3 3 10. y ⫽ ex(16 cos x ⫺ 16 sin x) ⫹ eⴚx(16 cos x ⫺ 23 16 sin x)

12. y ⫽ 1 ⫹ 3eⴚ2x ⫺ eⴚx 14. Project. (a) Divide the characteristic equation by l ⫺ l1 if el1x is known. (b) The idea is the same as in Sec. 2.1.

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(c) We first produce the standard form because this is the form under which the equation for z was derived. Division by x 3 gives yt ⫺

3 6 6 1 y s ⫹ a 2 ⫺ 1b y r ⫺ a 3 ⫺ b y ⫽ 0. x x x x

With y1 ⫽ x, y 1r ⫽ 1, y s1 ⫽ 0, and the coefficients p1 and p2 from the standard equation, we obtain 3 3 6 xz s ⫹ c 3 ⫹ a⫺ b x d z r ⫹ c 2 a⫺ b # 1 ⫹ a 2 ⫺ 1b x d z ⫽ 0. x x x Simplification gives xz s ⫹ a⫺

6 6 ⫹ ⫺ xb z ⫽ x(z s ⫺ z) ⫽ 0. x x

Hence z ⫽ c1ex ⫹ 苲 c 2eⴚx. By integration we get the answer

冮

y2 ⫽ x z dx ⫽ (c1ex ⫹ c2eⴚx ⫹ c3)x.

SECTION 3.3. Nonhomogeneous Linear ODEs, page 116 Purpose. To show that the transition from n ⫽ 2 (Sec. 2.7) to general n introduces no new ideas but generalizes all results and practical aspects in a straightforward fashion; this refers to existence, uniqueness, and the need for a particular solution yp to get a general solution in the form y ⫽ yh ⫹ yp. Comment on Elastic Beams This is an important standard example of a fourth-order ODE that needs no lengthy preparations and explanations. Vibrating beams follow in Problem Set 12.3. This leads to PDEs, since time t comes in as an additional variable. Comment on Variation of Parameters This method looks perhaps more complicated than it is; also the integrals may often be difficult to evaluate, and handling the higher order determinants that occur may require some more skill than the average student will have at this time. Thus it may be good to discuss this matter only rather briefly. Problem Set 3.3 concerns: Determination of general solutions (Probs. 1–7) and solving IVPs (Probs. 8–13), with yp obtained by the method of undetermined coefficients, where Prob. 6 shows an ODE reducible to second order. CAS Experiment 14 invites students to explore the extent to which the method of undetermined coefficients can be used, and Writing Report 15 should establish this method as a special case of variation of parameters for the ODE indicated.

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SOLUTIONS TO PROBLEM SET 3.3, page 122 2. y ⫽ c1eⴚ2x ⫹ c2eⴚx ⫹ c3ex ⫹ 2x 3 ⫺ 3x 2 ⫹ 15x ⫺ 8 4. The characteristic equation is

l3 ⫹ 3l2 ⫺ 5l ⫺ 39 ⫽ (l2 ⫹ 6l ⫹ 13)(l ⫺ 3) ⫽ 3(l ⫹ 3)2 ⫹ 44(l ⫺ 3) ⫽ 0.

Hence a general solution of the homogeneous ODE is yh ⫽ eⴚ3x(c1 cos 2x ⫹ c2 sin 2x) ⫹ c3e3x. Using the method of undetermined coefficients, substitute yp ⫽ a cos x ⫹ b sin x, obtaining yp ⫽ 7 cos x ⫹ sin x. 6. Set y r ⫽ z to obtain z s ⫹ 4z ⫽ sin x. By undetermined coefficients, z ⫽ c1 cos 2x ⫹ c2 sin 2x ⫹ 13 sin x. By integration, y⫽

冮 z dx ⫽ c cos 2x ⫹ c sin 2x ⫺ 3 cos x ⫹ c . 1

8. A general solution of the homogeneous ODE is obtained from the characteristic equation (l2 ⫺ 1)(l2 ⫺ 4) ⫽ l4 ⫺ 5l2 ⫹ 4 ⫽ 0 in the form yh ⫽ c1ex ⫹ c2eⴚx ⫹ c3e2x ⫹ c4eⴚ2x, A particular solution of the nonhomogeneous ODE is obtained by undetermined coefficients, namely, eⴚ3x>4. From this and the initial conditions we obtain the answer y ⫽ 14 ex ⫹ 32 eⴚx ⫺ eⴚ2x ⫹ 14 eⴚ3x

10. y1 ⫽ x, y2 ⫽ x ln x, y3 ⫽ x (ln x)2,

2 Answer: x 2 ⫹ x ln x ⫹ 11 2 x (ln x)

12. The characteristic equation of the homogeneous ODE l3 ⫺ 2l2 ⫺ 9l ⫹ 18 ⫽ 0 has the roots ⫺3, 2, and 3. Hence a general solution of the homogeneous ODE is yh ⫽ c1eⴚ3x ⫹ c2e2x ⫹ c3e3x. This also shows that the function on the right is a solution of the homogeneous ODE. Hence the Modification Rule applies, and the particular solution obtained is yp ⫽ ⫺0.2xe2x. A general solution of the nonhomogeneous ODE is y ⫽ yh ⫹ yp ⫽ c1eⴚ3x ⫹ c2e2x ⫹ c3e3x ⫺ 0.2xe2x. From this and the initial conditions we obtain y ⫽ (4.5 ⫺ 0.2x)e2x.

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14. CAS EXPERIMENT. The first equation has as a general solution 8 y ⫽ (c1 ⫹ c2 x ⫹ c3 x 2)ex ⫹ 105 exx 7>2,

so in cases such as this, one could try yp ⫽ x 1>2(a0 ⫹ a1x ⫹ a2 x 2 ⫹ a3 x 3)ex. However, the equation alone does not show much, so another idea is needed. One could modify the right side systematically and see how the solution changes. The solution of the second suggested equation shows that the equation is not accessible by undetermined coefficients; its solution is 7 3 y ⫽ c1x ⴚ1 ⫹ c2 x ⫹ c3 x 2 ⫹ 18 x 3 ln x ⫺ 32 x .

And one could perhaps modify this equation, too, in an attempt to obtain a form of solution that might be suitable for undetermined coefficients.

SOLUTIONS TO CHAPTER 3 REVIEW QUESTIONS AND PROBLEMS, page 122 6. The characteristic equation is l4 ⫺ 3l2 ⫺ 4 ⫽ (l2 ⫺ 1)(l2 ⫹ 4) ⫽ 0. Hence a general solution is y ⫽ c1eⴚ2x ⫹ c2e2x ⫹ c3 cos x ⫹ c4 sin x. 8. yp is obtained by the method of undetermined coefficients. A general solution of the nonhomogeneous ODE is y ⫽ c1ex ⫹ c2eⴚx ⫹ c3e4x ⫺ 5e2x. 10. The auxiliary equation is m(m ⫺ 1)(m ⫺ 2) ⫹ 3m(m ⫺ 1) ⫺ 2m ⫽ m(m 2 ⫺ 3) ⫽ 0. Hence a general solution is y ⫽ c1 ⫹ c2x13 ⫹ c3eⴚ13. 12. The characteristic equation is l3 ⫺ l ⫽ l(l ⫹ 1)(l ⫺ 1) ⫽ 0. A particular solution is obtained by undetermined coefficients. We thus obtain as a general solution of the nonhomogeneous ODE y ⫽ c1 ⫹ c2 cosh x ⫹ c3 sinh x ⫺ 125 36 cosh 0.8x. 14. The characteristic equation is quadratic in l2, the roots being ⫾2 and ⫾3. A particular solution of the nonhomogeneous ODE is obtained by undetermined coefficients, namely, eⴚx>2. Hence a general solution of the nonhomogeneous ODE is y ⫽ c1eⴚ2x ⫹ c2e2x ⫹ c3eⴚ3x ⫹ c4e3x ⫹ 12 eⴚx.

16. The characteristic polynomial is l3 ⫺ l2 ⫺ l ⫹ 1 ⫽ (l ⫹ 1)(l ⫺ 1)2.

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Hence a general solution is y ⫽ c1eⴚx ⫹ c2ex ⫹ c3 xex. Answer: y ⫽ sinh x 18. The characteristic equation is quadratic in l2, the roots being ⫺5, ⫺1, 1, 5. The right side requires a quadratic polynomial whose coefficients can be determined by substitution. Finally, the initial conditions are used to determine the four arbitrary constants in the general solution of the nonhomogeneous ODE thus obtained. The answer is y ⫽ 5eⴚx ⫹ eⴚ5x ⫹ 6.16 ⫹ 4x ⫹ 2x 2. 20. Two of four possible terms resulting from the homogeneous ODE are not visible in the answer. The student should recognize that all or some or none of the solutions of a basis of the homogeneous ODE may be present in the final answer; this will depend on the initial conditions. l3 ⫹ 3l2 ⫹ 3l ⫹ 1 ⫽ (l ⫹ 1)3; hence a general solution of the homogeneous equation is yh ⫽ (c1 ⫹ c2x ⫹ c3x 2)eⴚx. By the method of undetermined coefficients and from the initial conditions we get the answer y ⫽ (1 ⫹ x 2) eⴚx ⫺ 2 cos x ⫺ 2 sin x.

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CHAPTER 4 Systems of ODEs. Phase Plane. Qualitative Methods Major Changes This chapter was completely rewritten in the eighth edition, on the basis of suggestions by instructors who have taught from it and my own recent experience. The main reason for rewriting was the increasing emphasis on linear algebra in our standard curricula, so that we can expect that students taking material from Chap. 4 have at least some working knowledge of 2 ⫻ 2 matrices. Accordingly, Chap. 4 makes modest use of 2 ⫻ 2 matrices. n ⫻ n matrices are mentioned only in passing and are immediately followed by illustrative examples of systems of two ODEs in two unknown functions, involving 2 ⫻ 2 matrices only. Section 4.2 and the beginning of Sec. 4.3 are intended to give the student the impression that, for first-order systems, one can develop a theory that is conceptually and structurally similar to that in Chap. 2 for a single ODE. Hence if the instructor feels that the class may be disturbed by n ⫻ n matrices, omission of the latter and explanation of the material in terms of two ODEs in two unknown functions will entail no disadvantage and will leave no gaps of understanding or skill. To be completely on the safe side, Sec. 4.0 is included for reference, so that the student will have no need to search through Chap. 7 or 8 for a concept or fact needed in Chap. 4. Basic throughout Chap. 4 is the eigenvalue problem (for 2 ⫻ 2 matrices), consisting first of the determination of the eigenvalues l1, l2 (not necessarily numerically distinct) as solutions of the characteristic equation, that is, the quadratic equation

a11 ⫺ l

a12

a21

a22 ⫺ l

2 ⫽ (a11 ⫺ l)(a22 ⫺ l) ⫺ a12a21 ⫽ l2 ⫺ (a11 ⫹ a22)l ⫹ a11a22 ⫺ a12a21 ⫽ 0,

and then an eigenvector corresponding to l1 with components x 1, x 2 from (a11 ⫺ l1)x 1 ⫹ a12x 2 ⫽ 0 and an eigenvector corresponding to l2 from (a11 ⫺ l2)x 1 ⫹ a12x 2 ⫽ 0. It may be useful to emphasize early that eigenvectors are determined only up to a nonzero factor and that, in the present context, normalization (to obtain unit vectors) is hardly of any advantage. If there are students in the class who have not seen eigenvalues before (although the elementary theory of these problems does occur in every up-to-date introductory text on beginning linear algebra), they should not have difficulties in readily grasping the meaning of these problems and their role in this chapter, simply because of the numerous examples and applications in Sec. 4.3 and in later sections. Section 4.5 includes three famous applications, namely, the pendulum and van der Pol equations and the Lotka–Volterra predator–prey population model. 65

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SECTION 4.0. For Reference: Basics of Matrices and Vectors, page 124 Purpose. This section is for reference and review only, the material being restricted to what is actually needed in this chapter, to make it self-contained. Main Content Matrices, vectors Algebraic matrix operations Differentiation of vectors Eigenvalue problems for 2 ⫻ 2 matrices Important Concepts and Facts Matrix, column vector and row vector, multiplication Linear independence Eigenvalue, eigenvector, characteristic equation Some Details in Content Most of the material is explained in terms of 2 ⫻ 2 matrices, which play the major role in Chap. 4; indeed, n ⫻ n matrices for general n occur only briefly in Sec. 4.2 and at the beginning in Sec. 4.3. Hence the demand of linear algebra on the student in Chap. 4 will be very modest, and Sec. 4.0 is written accordingly. In particular, eigenvalue problems lead to quadratic equations only, so that nothing needs to be said about difficulties encountered with 3 ⫻ 3 or larger matrices. Example 1. Although the later sections include many eigenvalue problems, the complete solution of such a problem (the determination of the eigenvalues and corresponding eigenvectors) is given in Sec. 4.0. Emphasize to your students that the eigenvalues of a given square matrix are uniquely determined (and some of them can very well be 0), whereas eigenvectors must not be zero vectors and are determined only up to a nonzero multiplicative constant. SECTION 4.1. Systems of ODEs as Models in Engineering Applications, page 130 Purpose. In this section the student will gain a first impression of the importance of systems of ODEs in physics and engineering and will learn why they occur and why they lead to eigenvalue problems. Main Content Mixing problem Electrical network Conversion of single equations to system (Theorem 1) The possibility of switching back and forth between systems and single ODEs is practically quite important because, depending on the situation, the system or the single ODE will be the better source for obtaining the information sought in a specific case. Background Material. Secs. 2.4, 2.8. Short Courses. Take a quick look at Sec. 4.1, skip Sec. 4.2 and the beginning of Sec. 4.3, and proceed directly to solution methods in terms of the examples in Sec. 4.3. Some Details on Content Example 1 extends the physical system in Sec. 1.3, consisting of a single tank, to a system of two tanks. The principle of modeling remains the same. The problem leads to a typical

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eigenvalue problem, and the solutions show typical exponential increases and decreases to a constant value. Problem Set 4.1 The mixing problems (Probs. 1–6) should lead to an understanding of the physical parameters involved (tank size, flow rate, amount of fertilizer), similarly in the networks in Probs. 7–9. Problems 10–13 show conversions from ODEs to first-order systems. Problem 14 shows the principle of extending the physical system in Sec. 2.5 to a system of more than one mass and spring, with (11) probably best understood by looking at Fig. 81. SOLUTIONS TO PROBLEM SET 4.1, page 136 2. The system is y1r ⫽ 0.02y2 ⫺ 0.01y1 y r2 ⫽ 0.01y1 ⫺ 0.02y2 where 0.01 appears because we divide by the content of the tank T1, which is twice the old value. In proper order, the system becomes y 1r ⫽ ⫺0.01y1 ⫹ 0.02y2 y r2 ⫽ 0.01y1 ⫺ 0.02y2. As a single vector equation,

d. 0.01 ⫺0.02 A has the eigenvalues l1 ⫽ 0 and l2 ⫽ ⫺0.03 and corresponding eigenvectors y r ⫽ Ay,

A⫽ c

where

x (1) ⫽ c

1 0.5

⫺0.01

x (2) ⫽ c

1 ⫺1

0.02

respectively. The corresponding general solution is y ⫽ c1x (1) ⫹ c2x (2)eⴚ0.03t. From the initial values, y(0) ⫽ c1 c

1 0.5

d ⫹ c2 c

1 ⫺1

d⫽c

0 150

In components this is c1 ⫹ c2 ⫽ 0, 0.5c1 ⫺ c2 ⫽ 150. Hence c1 ⫽ 100, c2 ⫽ ⫺100. This gives the solution y ⫽ 100 c

1 0.5

d ⫺ 100 c

1 ⫺1

d e ⴚ0.03t.

In components, y1 ⫽ 100(1 ⫺ e ⴚ0.03t) y2 ⫽ 100(12 ⫹ eⴚ0.03t). 4. With a⫽

Flow rate Tank size

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we can write the system that models the process in the following form: y1r ⫽ ay2 ⫺ ay1 y r2 ⫽ ay1 ⫺ ay2, ordered as needed for the proper vector form y1r ⫽ ⫺ay1 ⫹ ay2 y r2 ⫽ ay1 ⫺ ay2. In vector form, y r ⫽ Ay,

A⫽ c

where

⫺a

The characteristic equation is (l ⫹ a)2 ⫺ a 2 ⫽ l2 ⫹ 2al ⫽ 0. Hence the eigenvalues are 0 and ⫺2a. Corresponding eigenvectors are

c d

and

1 ⫺1

respectively. The corresponding “general solution” is y ⫽ c1 c

1 1

d ⫹ c2 c

1 ⫺1

d eⴚ2at.

This result is interesting. It shows that the solution depends only on the ratio a, not on the tank size or the flow rate alone. Furthermore, the larger a is, the more rapidly y1 and y2 approach their limit. The term “general solution” is in quotation marks because this term has not yet been defined formally, although it is clear what is meant. 6. The matrix of the system is ⫺0.02

0.02

A ⫽ D 0.02

⫺0.04

0 0.02 T . ⫺0.02

0.02

The characteristic polynomial is l3 ⫹ 0.08l2 ⫹ 0.0012l ⫽ l(l ⫹ 0.02)(l ⫹ 0.06). This gives the eigenvalues and corresponding eigenvectors 1 l1 ⫽ 0, x (1) ⫽ D 1 T ,

1 l2 ⫽ ⫺0.02,

x (2) ⫽ D 0 T ,

1 l3 ⫽ ⫺0.06,

⫺1

Hence a “general solution” is 1

y ⫽ c1 D 1 T ⫹ c2 D 0 T e ⴚ0.02t ⫹ c3 D ⫺2 T e ⴚ0.06t. 1

⫺1

x (3) ⫽ D ⫺2 T .

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We use quotation marks since the concept of a general solution has not yet been defined formally, although it is clear what is meant. 8. The first ODE remains as before. The second ODE is obviously changed to I r2 ⫽ 0.4I 1r ⫺ 0.54I 2. Substitution of the first ODE into the new second one, as in the text, gives I 2r ⫽ ⫺1.6I1 ⫹ 1.06I2 ⫹ 4.8. Hence the matrix of the new system is A⫽ c

⫺4

⫺1.6

1.06

Its eigenvalues are ⫺1.5 and ⫺1.44. The corresponding eigenvectors are x (1) ⫽ [1 0.625]T and x (2) ⫽ [1 0.64]T, respectively. The corresponding general solution is y ⫽ c1x (1)e ⴚ1.5t ⫹ c2x (2)e ⴚ1.44t. 10. The system is y 1r ⫽ y2 y r2 ⫽ ⫺2y1 ⫺ 3y2. The matrix has the eigenvalues ⫺1 and ⫺2 and corresponding eigenvectors [1 and [1 ⫺2]T, respectively. From this

⫺1]T

y ⫽ y1 ⫽ c1e ⴚ t ⫹ c2e ⴚ2t and the second equation gives the derivative y2 ⫽ y r . 12. The system is y r1 ⫽ y2 y r2 ⫽ y3 y r3 ⫽ 2y1 ⫹ y2 ⫺ 2y3 The eigenvalues of its matrix are 1, ⫺1, ⫺2. The eigenvectors are [1 1 1]T, [1 ⫺1 1]T, [1 ⫺2 4]T, respectively. The corresponding general solution is y ⫽ c1 [1

1]Tet ⫹ c2 [1

⫺1

1]Te ⴚt ⫹ c3 [1

⫺2

4]Te ⴚ2t.

14. TEAM PROJECT. (a) From Sec. 2.5 we know that the undamped motions of a mass on an elastic spring are governed by my s ⫹ ky ⫽ 0 or my s ⫽ ⫺ky where y ⫽ y(t) is the displacement of the mass. By the same arguments, for the two masses on the two springs in Fig. 81 we obtain the linear homogeneous system (11)

m 1y s1 ⫽ ⫺k 1y1 ⫹ k 2( y2 ⫺ y1) m 2y s2 ⫽ ⫺k 2( y2 ⫺ y1)

for the unknown displacements y1 ⫽ y1(t) of the first mass m 1 and y2 ⫽ y2(t) of the second mass m 2. The forces acting on the first mass give the first equation, and the

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forces acting on the second mass give the second ODE. Now m 1 ⫽ m 2 ⫽ 1, k 1 ⫽ 3, and k 2 ⫽ 2 in Fig. 81 so that by ordering (11) we obtain y1s ⫽ ⫺5y1 ⫹ 2y2 y s2 ⫽ 2y1 ⫺ 2y2 or, written as a single vector equation, ys ⫽ c

y s1 y s2

d⫽ c

⫺5

⫺2

d c d. y1 y2

(b) As for a single equation, we try an exponential function of t, y ⫽ xevt.

y s ⫽ v2xevt ⫽ Axevt.

Then

Then, writing v2 ⫽ l and dividing by evt, we get Ax ⫽ lx. Eigenvalues and eigenvectors are l1 ⫽ ⫺1,

x (1) ⫽ c

1 2

and

l2 ⫽ ⫺6,

x (2) ⫽ c

2 ⫺1

Since v ⫽ ⫾ 2l and 2⫺1 ⫽ i and 2⫺6 ⫽ i 26, we get y ⫽ x (1)(c1eit ⫹ c2e ⴚit ) ⫹ x (2)(c3ei26t ⫹ c4e ⴚi26t ) or, by (7) in Sec. 2.3, y ⫽ a1x (1) cos t ⫹ b1x (1) sin t ⫹ a2x (2) cos 26t ⫹ b2x (2) sin 26t where a1 ⫽ c1 ⫹ c2, b1 ⫽ i(c1 ⫺ c2), a2 ⫽ c3 ⫹ c4, b2 ⫽ i(c3 ⫺ c4). These four arbitrary constants can be specified by four initial conditions. In components, this solution is y1 ⫽ a1 cos t ⫹ b1 sin t ⫹ 2a2 cos 26t ⫹ 2b2 sin 26t y2 ⫽ 2a1 cos t ⫹ 2b1 sin t ⫺ a2 cos 26t ⫺ b2 sin 26t. (c) The first two terms in y1 and y2 give a slow harmonic motion, and the last two a fast harmonic motion. The slow motion occurs if, at some instant, both masses are moving downward or both upward. For instance, if a1 ⫽ 1 and the three other constants are zero, we get y1 ⫽ cos t, y2 ⫽ 2 cos t; this is an example of such a motion. The fast motion occurs if, at each instant, the two masses are moving in opposite directions, so that one of the two springs is extended, whereas the other is simultaneously compressed. For instance, if a2 ⫽ 1 and the other constants are zero, we have y1 ⫽ 2 cos 26t, y2 ⫽ ⫺cos 26t; this is a fast motion of the indicated type. Depending on the initial conditions, one or the other motion will occur or a superposition of both. SECTION 4.2 Basic Theory of Systems of ODEs. Wronskian, page 137 Purpose. This survey of some basic concepts and facts on nonlinear and linear systems is intended to give the student an impression of the conceptual and structural similarity of the theory of systems to that of single ODEs.

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Content, Important Concepts Standard form (1) of first-order systems, nonlinear or linear. The point is that each equation contains only one derivative, which appears on the left. For instance, Kirchhoff’s Voltage Law (KVL) for electric systems gives a system of ODEs that can be transformed into the form (1) by algebra and differentiation. Form of corresponding initial value problems (1), (2). If an ODE is converted to a system, using Theorem 1 of Sec. 4.1, a corresponding IVP written as in Chaps. 2 and 3 converts to the form (1), (2) and conversely. Existence and Uniqueness Theorem 1 for solutions of IVPs (1), (2). Note that the (sufficient) conditions correspond to those for single ODEs. Standard form and notations (3) for linear systems of ODEs. These notations agree with the usual notations for matrices involving double subscripts. Homogeneous and nonhomogeneous linear systems of ODEs. Basis, general solution (5), Wronskian (7), which is the determinant of the fundamental matrix Y (see (6)) such that a general solution can be written (8) y ⫽ Yc, c the vector with the arbitrary constants as components. Background Material. Sec. 2.6 contains the analogous theory for single equations. See also Sec. 1.7. Short Courses. This section may be skipped, as mentioned before.

SECTION 4.3. Constant-Coefficient Systems. Phase Plane Method, page 140 Purpose. Typical examples show the student the rich variety of pattern of solution curves (trajectories) near critical points in the phase plane, along with the process of actually solving homogeneous linear systems. This will also prepare the student for a good understanding of the systematic discussion of critical points in the phase plane in Sec. 4.4. Main Content Solution method for homogeneous linear systems Examples illustrating types of critical points Solution when no basis of eigenvectors is available (Example 6) Important Concepts and Facts Trajectories as solution curves in the phase plane Phase plane as a means for the simultaneous (qualitative) discussion of a large number of solutions Basis of solutions obtained from basis of eigenvectors Background Material. Short review of eigenvalue problems from Sec. 4.0, if needed. Short Courses. Omit Example 6. Some Details on Content In addition to developing skill in solving homogeneous linear systems, the student is supposed to become aware that it is the kind of eigenvalues that determines the type of critical point. The examples show important cases. (A systematic discussion of all cases follows in the next section.) Example 1. Two negative eigenvalues give a node. Example 2. A real double eigenvalue gives a node. Example 3. Real eigenvalues of opposite sign give a saddle point.

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Example 4. Pure imaginary eigenvalues give a center, and working in complex is avoided by a standard trick, which can also be useful in other contexts. Example 5. Genuinely complex eigenvalues give a spiral point. Some work in complex can be avoided, if desired, by differentiation and elimination. The first ODE is y2 ⫽ y r1 ⫹ y1.

(a)

By differentiation and from the second ODE as well as from (a), y s1 ⫽ ⫺y r1 ⫹ y r2 ⫽ ⫺y r1 ⫺ y1 ⫺ (y r1 ⫹ y1) ⫽ ⫺2y r1 ⫺ 2y1. Complex solutions e(ⴚ1⫾i)t give the real solution y1 ⫽ e ⴚt(A cos t ⫹ B sin t). From this and (a) there follows the expression for y2 given in the text. Example 6 shows that the present method can be extended to include cases when A does not provide a basis of eigenvectors, but then becomes substantially more involved. In this way the student will recognize the importance of bases of eigenvectors, which also play a role in many other contexts. A further illustration of the situation is given in Probs. 8 and 16.

SOLUTIONS TO PROBLEM SET 4.3, page 147 2. The eigenvalues are 3 and 9. Eigenvectors are [3 The corresponding general solution is

⫺1]T and [3

1]T, respectively.

y1 ⫽ 3c1e3t ⫹ 3c2e9t y2 ⫽⫺c1e3t ⫹ c2e9t. 4. The matrix has the double eigenvalue ⫺6. An eigenvector is [1 vector u needed is obtained from (A ⫹ 6I) u ⫽ c

We can take u ⫽ 30,

⫺2

du⫽ c

1 ⫺1

⫺1]T. Hence the

⫺124. With this we obtain as a general solution y1 ⫽ c1e ⴚ6t ⫹ c2te ⴚ6t y2 ⫽ ⫺c1e ⴚ6t ⫺ c2(t ⫹ 12)e ⴚ6t.

6. The eigenvalues are complex, 2 ⫹ 2i and 2 ⫺ 2i. Corresponding complex eigenvectors are [1 ⫺i]T and [1 i]T, respectively. Hence a complex general solution is y1 ⫽ c1e(2⫹2i)t ⫹ c2e(2–2i)t y2 ⫽ ⫺ic1e(2⫹2i)t ⫹ ic2e(2–2i)t. From this and the Euler formula we obtain a real general solution y1 ⫽ e2t3(c1 ⫹ c2) cos 2t ⫹ i(c1 ⫺ c2) sin 2t4 ⫽ e2t (A cos 2t ⫹ B sin 2t), y2 ⫽ e2t3(⫺ic1 ⫹ ic2) cos 2t ⫹ i(⫺ic1 ⫺ ic2) sin 2t4 ⫽ e2t(⫺B cos 2t ⫹ A sin 2t) where A ⫽ c1 ⫹ c2 and B ⫽ i(c1 ⫺ c2).

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8. The eigenvalue of algebraic multiplicity 2 is 9. An eigenvector is [1 is no basis of eigenvectors. A first solution is y (1) ⫽ c

1 ⫺1

⫺1]T. There

d e9t.

A second linearly independent solution is (see Example 6 in the text) y (2) ⫽ c

1 ⫺1

d te9t ⫹ c d e9t u1 u2

with 3u 1 u 24T determined from (A ⫺ 9I) u ⫽ c

⫺1

dc d ⫽ c u1

⫺1

Thus u 1 ⫹ u 2 ⫽ ⫺1. We can take u 1 ⫽ 0, u 2 ⫽ ⫺1. This gives the general solution y ⫽ c1y (1) ⫹ c2y (2) ⫽ (c1 ⫹ c2t) c

1 ⫺1

d e9t ⫹ c2 c

0 ⫺1

d e9t.

10. The eigenvalues are 4 and ⫺3. The corresponding general solution is y1 ⫽ c1e4t ⫹ c23 ⴚ3t y2 ⫽ c1e4t ⫺ 52 c2e ⴚ3t. From this and the initial conditions we obtain y1(0) ⫽ c1 ⫹ c2 ⫽ 0 y2(0) ⫽ c1 ⫺ 52 c2 ⫽ 7. Hence the solution of the IVP is y1 ⫽ 2e4t ⫺ 2e ⴚ3t y2 ⫽ 2e4t ⫹ 5e ⴚ3t. 12. The eigenvalues are 0 and 2. The corresponding general solution is y1 ⫽ c1 ⫹ c2e2t y2 ⫽ ⫺13 c1 ⫹ 13 c2e2t. From this and the initial values we obtain y1(0) ⫽

c1 ⫹ c2 ⫽ 12

y2(0) ⫽ ⫺13 c1 ⫹ 13 c2 ⫽

Hence the solution of the IVP is y1 ⫽ 3 ⫹ 9e2t y2 ⫽ ⫺1 ⫹ 3e2t. 14. The eigenvalues are ⫺1 ⫹ i and ⫺1 ⫺ i. The corresponding real general solution is y1 ⫽ e ⴚt (c1 cos t ⫹ c2 sin t) y2 ⫽ e ⴚt (⫺c2 cos t ⫹ c1 sin t).

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From this and the initial conditions we obtain the solutions of the IVP y1 ⫽ eⴚt cos t y2 ⫽ eⴚt sin t. 16. The system is y 1r ⫽ 8y1 ⫺ y2 y r2 ⫽ y1 ⫹ 10y2.

(a) (b)

10(a) ⫹ (b) gives 10y r1 ⫹ y r2 ⫽ 81y1; hence y r2 ⫽ ⫺10y r1 ⫹ 81y1.

(c)

Differentiating (a) and using (c) gives 0 ⫽ y s1 ⫺ 8y r1 ⫹ y r2 ⫽ y s1 ⫺ 8y r1 ⫺ 10y r1 ⫹ 81y1 ⫽ y s1 ⫺ 18y r1 ⫹ 81y1. A general solution is y1 ⫽ (c1 ⫹ c2t)e9t. From this and (a) we obtain y2 ⫽ ⫺y r1 ⫹ 8y1 ⫽ 3⫺c2 ⫺ 9(c1 ⫹ c2t) ⫹ 8(c1 ⫹ c2t)4e9t ⫽ (⫺c2 ⫺ c1 ⫺ c2t)e9t, in agreement with the answer to Prob. 8. 18. The restriction of the inflow from outside to pure water is necessary to obtain a homogeneous system. The principle involved in setting up the model is Time rate of change ⫽ Inflow ⫺ Outflow. For Tank T1 this is (see Fig. 88) y r1 ⫽ a12 ⴢ 0 ⫹

4 16 y2 b ⫺ y1. 200 200

For Tank T2 it is y r2 ⫽

16 4 ⫹ 12 y1 ⫺ y2. 200 200

Performing the divisions and ordering terms, we have y r1 ⫽ ⫺0.08y1 ⫹ 0.02y2 y r2 ⫽ 0.08y1 ⫺ 0.08y2. The eigenvalues of the matrix of this system are ⫺0.04 and ⫺0.12. Eigenvectors are [1 2]T and [1 ⫺2]T, respectively. The corresponding general solution is y ⫽ c1 c

1 2

d e ⴚ0.04t ⫹ c2 c

1 ⫺2

d e ⴚ0.12t.

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The initial conditions are y1(0) ⫽ 100, y2(0) ⫽ 200. This gives c1 ⫽ 100, c2 ⫽ 0. In components the answer is y1 ⫽ 100e ⴚ0.04t y2 ⫽ 200e ⴚ0.04t. Both functions approach zero as t : ⬁, a reasonable result because pure water flows in and mixture flows out.

SECTION 4.4. Criteria for Critical Points. Stability, page 148 Purpose. Systematic discussion of critical points in the phase plane from the standpoints of both geometrical shapes of trajectories and stability. Main Content Table 4.1 for the types of critical points Table 4.1 for the stability behavior Stability chart (Fig. 92), giving Tables 4.1 and 4.2 graphically Important Concepts Node, saddle point, center, spiral point Stable and attractive, stable, unstable Background Material. Sec. 2.4 (needed in Example 2). Short Courses. Since all these types of critical points already occurred in the previous section, one may perhaps present just a short discussion of stability. Some Details on Content The types of critical points in Sec. 4.3 now recur, and the discussion shows that they exhaust all possibilities. With the examples of Sec. 4.3 fresh in mind, the student will acquire a deeper understanding by discussing the stability chart and by reconsidering those examples from the viewpoint of stability. This gives the instructor an opportunity to emphasize that the general importance of stability in engineering can hardly be overestimated. Example 2, relating to the familiar free vibrations in Sec. 2.4, gives a good illustration of stability behavior, namely, depending on c, attractive stability, stability (and instability if one includes “negative damping,” with c ⬍ 0, as it will recur in the next section in connection with the famous van der Pol equation). Problem Set 4.4 Problems 1–10 give straightforward applications of the criteria in this section. Problems 11–12 are related to previous material, for instance, to oscillations of a mass on a spring. For Prob. 13 let the student reconsider the discussion of (10)–(13) in Sec. 4.3 as discussed there and then, in addition, from the new viewpoint of stability. Problems 15–17 concern the basic concept of perturbation, which, in practice, may be due to inaccuracies of measurement. The purpose of Probs. 18–20 is fairly obvious.

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SOLUTIONS TO PROBLEM SET 4.4, page 151 2. p ⫽ ⫺7, q ⫽ 12 ⬎ 0, ¢ ⫽ 49 ⫺ 48 ⬎ 0, stable and attractive improper node, y1 ⫽ c1e⫺4t, y2 ⫽ c2eⴚ3t 4. p ⫽ 0, q ⫽ ⫺9, saddle point, always unstable. A general solution is y1 ⫽ c1eⴚ3t ⫹ c2e3t y2 ⫽ ⫺5c1eⴚ3t ⫹ c2e3t. 6. p ⫽ ⫺12, q ⫽ 27, ¢ ⫽ 144 ⫺108 ⬎ 0, stable and arrtactive node. A general solution is y1 ⫽ c1eⴚ3t ⫹ c2eⴚ9t y2 ⫽ ⫺3c1eⴚ3t ⫹ 3c2eⴚ9t. 8. p ⫽ ⫺3,

q ⫽ ⫺10, saddle point, always unstable. A general solution is y1 ⫽ c1eⴚ5t ⫹ 4c2e2t y2 ⫽ ⫺c1eⴚ5t ⫹ 3c2e2t.

10. p ⫽ ⫺2, q ⫽ 5, ¢ ⫽ 4 ⫺ 20 ⬍ 0, stable and attractive spiral. The components of a general solution are y1 ⫽ c1e(ⴚ1⫹2i)t ⫹ c2e(ⴚ1ⴚ2i)t ⫽ eⴚt((c1 ⫹ c2) cos 2t ⫹ i(c1 ⫺ c2) sin 2t) ⫽ eⴚt(A cos 2t ⫹ B sin 2t), y2 ⫽ (⫺1 ⫹ 2i)c1e(ⴚ1⫹2i)t ⫹ (⫺1 ⫺ 2i)c2e(ⴚ1ⴚ2i)t ⫽ eⴚt((⫺c1 ⫺ c2 ⫹ 2ic1 ⫺ 2ic2) cos 2t ⫹ i(2ic1 ⫹ 2ic2 ⫺ c1 ⫹ c2) sin 2t) ⫽ eⴚt((⫺A ⫹ 2B) cos 2t ⫺ (2A ⫹ B) sin 2t) where A ⫽ c1 ⫹ c2 and B ⫽ i(c1 ⫺ c2). 12. y ⫽ A cos 13t ⫹ B sin 13t. The trajectories are the ellipses 1 2 2 9 y 1 ⫹ y 2 ⫽ const.

This is obtained as in Example 4 in Sec. 4.3. 14. y r1 ⫽ ⫺dy1>dt, y r2 ⫽ ⫺dy2>dt, reversal of the direction of motion; to get the usual form, we have to multiply the transformed system by ⫺1, which amounts to multiplying the matrix by ⫺1, changing p into ⫺p, but leaving q and ¢ unchanged. In the example, we get an unstable node. 16. At a center, p ⫽ a11 ⫹ a22 ⫽ 0, q ⫽ det A ⬎ 0, hence ¢ ⬍ 0. Under the change, p changes into a11 ⫹ k ⫹ a22 ⫹ k ⫽ 2k ⫽ 0; q remains positive because (a11 ⫹ k)(a22 ⫹ k) ⫺ a12a21 ⫽ q ⫹ k 2 ⬎ 0. Finally, ¢ remains unchanged because (p ⫹ 2k)2 ⫺ 4(q ⫹ k 2) ⫽ (2k)2 ⫺ 4(q ⫹ k 2) ⫽ ⫺4q ⬍ 0. Hence we obtain a spiral point, which is unstable if k ⬎ 0 and stable and attractive if k ⬍ 0.

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We can reason more simply as follows. For a center, the eigenvalues are pure imaginary (to have closed trajectories). An eigenvalue l of A gives an eigenvalue l ⫹ k of A, causing a damped oscillation (when k ⬍ 0) or an increasing one (when k ⬎ 0), thus a spiral.

SECTION 4.5 Qualitative Methods for Nonlinear Systems, page 152 Purpose. As a most important step, in this section we extend phase plane methods from linear to nonlinear systems and nonlinear ODEs. The particular importance of phase plane methods for nonlinear ODEs and systems results from the difficulty (or impossibility) of solving them, explicitly as explained in the text. Main Content Critical points of nonlinear systems Their discussion by linearization Transformation of single autonomous ODEs Applications of linearization and transformation techniques Important Concepts and Facts Linearized system (3), condition for applicability Linearization of pendulum equations Self-sustained oscillations, van der Pol equation Short Courses. Linearization at different critical points seems the main issue that the student is supposed to understand and handle in practice. Examples 1 and 2 may help students to gain skill in that technique. The other material can be skipped without loss of continuity. Some Details on Content This section is very important, because from it the student should learn not only techniques (linearization, etc.) but also the fact that phase plane methods are particularly powerful and important in application to systems or single ODEs that cannot be solved explicitly. The student should also recognize that it is quite surprising how much information these methods can give. This is demonstrated by the pendulum equation (Examples 1 and 2) for a relatively simple system, and by the famous van der Pol equation for a single ODE, which has become a prototype for self-sustained oscillations of electrical systems of various kinds. We also discuss the famous Lotka–Volterra predator–prey model. For the Rayleigh and Duffing equations, see the problem set. Problem Set 4.5 Problems 1–3 concern the undamped pendulum and limit cycles and their deformation in the famous van der Pol equation. Problems 4–8 are formal and concern linearization of systems in the case of several limit points, usually of a different type. Problems 9–13 concern linearization of single nonlinear second-order ODEs, which are first converted to systems. Team Project 14 discusses further self-sustained oscillations.

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SOLUTIONS TO PROBLEM SET 4.5, page 159 2. A limit cycle is approached by trajectories (from inside and outside). No such approach takes place for a closed trajectory. 4. Writing the system in the form y r1 ⫽ y1(4 ⫺ y1) y r2 ⫽ y2 we see that the critical points are (0, 0) and (4, 0). For (0, 0) the linearized system is y r1 ⫽ 4y1 y r2 ⫽ y2. The matrix is

Hence p ⫽ 5, q ⫽ 4, ¢ ⫽ 25 ⫺ 16 ⫽ 9. This shows that the critical point at (0, 0) is an unstable node. For (4, 0) the translation to the origin is y1 ⫽ 4 ⫹ 苲 y1,

y2 ⫽ 苲 y2.

This gives the transformed system 苲 y 1r ⫽ (4 ⫹ 苲 y1)(⫺y苲1) 苲 y 2r ⫽ 苲 y2 and the corresponding linearized system 苲苲1 y 1r ⫽ ⫺4y 苲 y r2 ⫽ 苲 y 2. Its matrix is

⫺4

Hence 苲 p ⫽ ⫺3, 苲 q ⫽ ⫺4. Thus the critical point at (4, 0) is a saddle point, which is always unstable. 6. At (0, 0), y 1r ⫽ y2, y r2 ⫽ ⫺y1, p ⫽ 0, q ⫽ 1, ¢ ⫽ ⫺4, center. The other critical point y 2, is at (⫺1, 0). We set y1 ⫽ ⫺1 ⫹ 苲 y1, y2 ⫽ 苲 y 2. Then ⫺y1 – y 21 ⬇ 苲 y1. Hence 苲 y 1r ⫽ 苲苲苲 y 2r ⫽ y 1. This gives a saddle point. 8. The system may be written y 1r ⫽ y2(1 ⫺ y2) y r2 ⫽ y1(1 ⫺ y1). From this we see immediatly that there are four critical points, at (0, 0), (0, 1), (1, 0), (1, 1). At (0, 0) the linearized system is y 1r ⫽ y2 y r2 ⫽ y1.

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The matrix is

Hence p ⫽ 0, q ⫽ ⫺1, so that we have a saddle point. y1, y2 ⫽ 1 ⫹ 苲 y 2. This gives the transformed At (0, 1) the transformation is y1 ⫽ 苲 system 苲 y r1 ⫽ (1 ⫹ 苲 y 2)(⫺y苲2) 苲苲 y r2 ⫽ y 1(1 ⫺ 苲 y 1). Linearization gives 苲 y 1r ⫽ ⫺y苲2

with matrix

⫺1

苲 y 2r ⫽ 苲 y1 1 0 苲 for which p苲 ⫽ 0, q苲 ⫽ 1, ¢ ⫽ ⫺4, and we have a center. y1, y2 ⫽ 苲 y 2. The transformed system is At (1, 0) the transformation is y1 ⫽ 1 ⫹ 苲苲 y 1r ⫽ 苲 y 2(1 ⫺ 苲 y 2) 苲苲 y r2 ⫽ (1 ⫹ y1)(⫺y苲1). Its linearization is 苲 y 1r ⫽

苲 y2

with matrix

苲 y 2r ⫽ ⫺y苲1 ⫺1 0 苲 p ⫽ 0, 苲 q ⫽ 1, ¢ ⫽ ⫺4, so that we get another center. for which 苲 At (1, 1) the transformation is y1 ⫽ 1 ⫹ 苲 y1, y2 ⫽ 1 ⫹ 苲 y2. The transformed system is 苲2) 苲 y 2)(⫺y y 1r ⫽ (1 ⫹ 苲苲苲1). y 2r ⫽ (1 ⫹ 苲 y1)(⫺y Linearization gives 苲 y 1r ⫽ ⫺y苲2 苲 y 2r ⫽ ⫺y苲1

with matrix

⫺1

q ⫽ ⫺1, and we have another saddle point. for which p苲 ⫽ 0, 苲 10. y s ⫹ y ⫺ y 3 ⫽ 0 written as a system is y1r ⫽ y2 y 2r ⫽ ⫺y1 ⫹ y 31. Now ⫺y1 ⫹ y 31 ⫽ y1(⫺1 ⫹ y 21) ⫽ 0 shows that there are three critical points, at (y1, y2) ⫽ (0, 0), (⫺1, 0), and (1, 0). The linearized system at (0, 0) is y 1r ⫽ y2 y 2r ⫽ ⫺y1.

Matrix:

⫺1

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From the matrix we see that p ⫽ a11 ⫹ a22 ⫽ 0, q ⫽ 1. Hence (0, 0) is a center (see Sec. 3.4). For the next critical point we have to linearize at (⫺1, 0) by setting y1 ⫽ ⫺1 ⫹ 苲 y1,

y2 ⫽ 苲 y 2.

Then y1(⫺1 ⫹ y 21 ⫽ (⫺1 ⫹ 苲 y1)3⫺1 ⫹ (⫺1 ⫹ 苲 y1)24 2 ⫽ (⫺1 ⫹ 苲 y1)3⫺2y苲1 ⫹ 苲 y 14 ⬇ 2y苲1. Hence the linearized system is 苲 y 1r ⫽ 苲 y2 苲 y 2r ⫽ 2y苲1.

Matrix:

Hence q ⫽ det A ⫽ ⫺2 ⬍ 0, that is, the critical point at (⫺1, 0) is a saddle point. Similarly, to linearize at (1, 0), set y1 ⫽ 1 ⫹ 苲 y1,

y2 ⫽ 苲 y 2.

Then ⫺y1 ⫹ y 31 ⬇ 2y苲1 and we obtain another saddle point, as just before. 12. y 1r ⫽ y2, y 2r ⫽ y1(⫺9 ⫺ y1). (0, 0) is a critical point. The linearized system at (0, 0) is y 1r ⫽

with matrix

y r2 ⫽ ⫺9y1

⫺9

for which p ⫽ 0 and q ⫽ 9 ⬎ 0, so that we have a center. A second critical point is at (⫺9, 0). The transformation is y ⫽ ⫺9 ⫹ 苲 y , y ⫽苲 y . 1

This gives the transformed system 苲 y 1r ⫽ 苲 y2 苲 y r ⫽ (⫺9 ⫹ 苲 y1)(⫺y苲1). 2

Its linearization is 苲 y 1r ⫽ 苲 y2 苲 y r2 ⫽ 9y苲1

with matrix

for which 苲 q ⫽ ⫺9 ⬍ 0, so that we have a saddle point. 14. TEAM PROJECT. (a) Unstable node if ␮ ⭌ 2, unstable spiral point if 2 ⬎ ␮ ⬎ 0, center if ␮ ⫽ 0, stable and attractive spiral point if 0 ⬎ ␮ ⬎ ⫺2, stable and attractive node if ␮ ⬉ ⫺2. (c) As a system we obtain (A) (B)

y r1 ⫽ y2 y r2 ⫽ ⫺(v20y1 ⫹ by 31).

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The product of the left side of (A) and the right side of (B) equals the product of the right side of (A) and the left side of (B): y r2y2 ⫽ ⫺(v20y1 ⫹ by 31)y r1. Integration on both sides and multiplication by 2 gives y 22 ⫹ v20 y 21 ⫹ 12 by 41 ⫽ const. For positive b these curves are closed because then by 3 is a proper restoring term, adding to the restoring due to the y-term. If b is negative, the term by 3 has the opposite effect, and this explains why then some of the trajectories are no longer closed but extend to infinity in the phase plane. For generalized van der Pol equations, see e.g., K. Klotter and E. Kreyzig, On a class of self-sustained oscillations. J. Appl. Math. 27(1960), 568–574. SECTION 4.6. Nonhomogeneous Linear Systems of ODEs, page 160 Purpose. We now turn from homogeneous linear systems considered so far to solution methods for nonhomogeneous systems. Main Content Method of undetermined coefficients Modification for special right sides Method of variation of parameters Short Courses. Select just one of the preceding methods. Some Details on Content In addition to understanding the solution methods as such, the student should observe the conceptual and technical similarities to the handling of nonhomogeneous linear ODEs in Secs. 2.7–2.10 and understand the reason for this, namely, that systems can be converted to single equations and conversely. For instance, in connection with Example 1 in this section, one may point to the Modification Rule in Sec. 2.7, or, if time permits, establish an even more definite relation by differentiation and elimination of y2, y1s ⫽ ⫺3y 1r ⫹ y 2r ⫹ 12eⴚ2t ⫽ ⫺3y 1r ⫹ ( y1 ⫺ 3y2 ⫹ 2eⴚ2t) ⫹ 12eⴚ2t ⫽ ⫺3y 1r ⫹ y1 ⫺ 3( y 1r ⫹ 3y1 ⫹ 6eⴚ2t) ⫹ 14eⴚ2t ⫽ ⫺6y r1 ⫺ 8y1 ⫺ 4eⴚ2t, solving this for y1 and then getting y2 from the solution. Problem Set 4.6 Problems 2–7 concern general solutions of nonhomogeneous systems, where the particular solution needed is obtained by the method of undetermined coefficients. Problems 8 and 9 are general questions on the method of undetermined coefficients. Problems 10–15 concern initial value problems; here it is important to emphasize that the initial conditions are to be used only after a general solution of the nonhomogeneous system has been obtained. Typical applications to nonhomogeneous systems are shown in Probs. 17–20.

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SOLUTIONS TO PROBLEM SET 4.6, page 163 2. The matrix of the system has the eigenvalues 2 and ⫺2. Eigenvectors are [1 1]T and [1 ⫺3]T, respectively. Hence a general solution of the homogeneous system is y (h) ⫽ c1 c

1 1

d e2t ⫹ c2 c

1 ⫺3

d eⴚ2t.

We determine y (p) by the method of undetermined coefficients, starting from y (p) ⫽ c

a1 cos t ⫹ b1 sin t a2 cos t ⫹ b2 sin t

Substituting this and its derivative into the given nonhomogeneous system, we obtain, in terms of components, ⫺a1 sin t ⫹ b1 cos t ⫽ (a1 ⫹ a2) cos t ⫹ (b1 ⫹ b2) sin t ⫹ 10 cos t ⫺a2 sin t ⫹ b2 cos t ⫽ (3a1 ⫺ a2) cos t ⫹ (3b1 ⫺ b2) sin t ⫺ 10 sin t. By equating the coefficients of the cosine and of the sine in the first of these two equations we obtain b1 ⫽ a1 ⫹ a2 ⫹ 10,

⫺a1 ⫽ b1 ⫹ b2.

Similarly from the second equation, b2 ⫽ 3a1 ⫺ a2,

⫺a2 ⫽ 3b1 ⫺ b2 ⫺ 10.

The solution is a1 ⫽ ⫺2, b1 ⫽ 4, a2 ⫽ ⫺4, b2 ⫽ ⫺2. This gives the answer y1 ⫽ c1e2t ⫹ c2eⴚ2t ⫺ 2 cos t ⫹ 4 sin t y2 ⫽ c1e2t ⫺ 3c2eⴚ2t ⫺ 4 cos t ⫺ 2 sin t. 4. From the characteristic equation, l1 ⫽ 2,

x (1) ⫽ c

4 1

l2 ⫽ ⫺4,

x (2) ⫽ c

1 1

y (p) can be obtained by the method of undetermined coefficients, starting from y (p) ⫽ a cosh t ⫹ b sinh t. Substitution gives y (p) r ⫽ a sinh t ⫹ b cosh t 4 ⫺8 2 0 ⫽ c d (a cosh t ⫹ b sinh t) ⫹ c d cosh t ⫹ c d sinh t. 2 ⫺6 1 2 Comparing sinh terms and cosh terms (componentwise), we get from this a1 ⫽ 4b1 ⫺ 8b2 a2 ⫽ 2b1 ⫺ 6b2 ⫹ 2 b1 ⫽ 4a1 ⫺ 8a2 ⫹ 2 b2 ⫽ 2a1 ⫺ 6a2 ⫹ 1

f sinh terms f cosh terms.

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To solve this, one can substitute the first two equations into the last two, solve for b1 ⫽ 2, b2 ⫽ 1, and then get from the first two equations a1 ⫽ a2 ⫽ 0. This gives the general solution y1 ⫽ 4c1e2t ⫹ c2eⴚ4t ⫹ 2 sinh t,

y2 ⫽ c1e2t ⫹ c2eⴚ4t ⫹ sinh t.

6. From the characteristic equation we obtain l1 ⫽ 4,

x (1) ⫽ c

1 1

x (2) ⫽ c

l2 ⫽ ⫺4,

1 ⫺1

By the method of undetermined coefficients, we set y (p) ⫽ u ⫹ vt ⫹ wt 2. By substitution, y (p) r ⫽ v ⫹ 2wt ⫽ c

d (u ⫹ vt ⫹ wt 2) ⫹ c d ⫹ c 0

⫺16

d t 2.

We now compare componentwise the constant terms, linear terms, and quadratic terms: v1 ⫽ 4u 2 v2 ⫽ 4u 1 ⫹ 2

f constant terms

2w1 ⫽ 4v2 2w2 ⫽ 4v1

f linear terms

0 ⫽ 4w2 0 ⫽ 4w1 ⫺ 16

f quadratic terms.

We then obtain, in this order, w1 ⫽ 4,

w2 ⫽ 0, v1 ⫽ 0,

v2 ⫽ 2,

u 1 ⫽ 14 (v2 ⫺ 2) ⫽ 0,

u 2 ⫽ 0.

The corresponding general solution is y ⫽ c1x (1)e4t ⫹ c2x (2)eⴚ4t ⫹ vt ⫹ wt 2; in components, y1 ⫽ c1e4t ⫹ c2eⴚ4t ⫹ 4t 2, 10. y ⫽ c1x (1) et ⫹ c2x (2)e2t, x (1) ⫽ c

y2 ⫽ c1e4t ⫺ c2eⴚ4t ⫹ 2t.

d , x (2) ⫽ c

d . Now et on the right is a

⫺1 ⫺5 solution of the homogeneous system. Hence, to find y (p), we have to proceed as in Example 1, setting y (p) ⫽ utet ⫹ vet. Substitution gives y (p) r ⫽ u(t ⫹ 1)et ⫹ vet ⫽ c

⫺3

⫺4

d (utet ⫹ vet) ⫹ c

5 ⫺6

d et.

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Equating the terms in et (componentwise) gives u 1 ⫹ v1 ⫽ ⫺3v1 ⫺ 4v2 ⫹ 5 u 2 ⫹ v2 ⫽ 5v1 ⫹ 6v2 ⫺ 6 and the terms in tet give u 1 ⫽ ⫺3u 1 ⫺ 4u 2 u 2 ⫽ 5u 1 ⫹ 6u 2. Hence u 1 ⫽ 1, u 2 ⫽ ⫺1, v1 ⫽ 1, v2 ⫽ 0. This gives the general solution y1 ⫽ c1et ⫹ 4c2e2t ⫹ tet ⫹ et,

y2 ⫽ ⫺c1et ⫺ 5c2e2t ⫺ tet.

From the initial conditions we obtain c1 ⫽ ⫺2, c2 ⫽ 5. Answer: y1 ⫽ ⫺2et ⫹ 20e2t ⫹ tet ⫹ et,

y2 ⫽ 2et ⫺ 25e2t ⫺ tet.

12. A general solution of the homogeneous system is x (1) ⫽ c

y (h) ⫽ c1x (1)e3t ⫹ c2x (2)eⴚt,

2 1

x (2) ⫽ c

2 ⫺1

Answer: y1 ⫽ 2eⴚt ⫹ t 2,

y2 ⫽ ⫺eⴚt ⫺ t.

14. The matrix of the homogeneous system

⫺1

has the eigenvalues 2i and ⫺2i and eigenvectors [2 Hence a complex general solution is y (h) ⫽ c1 c

2 i

d e2it ⫹ c2 c

i]T and [2 2

⫺i

⫺i]T, respectively.

d eⴚ2it.

By Euler’s formula this becomes, in components, y 1(h) ⫽ (2c1 ⫹ 2c2) cos 2t ⫹ i(2c1 ⫺ 2c2) sin 2t y 2(h) ⫽ (ic1 ⫺ ic2) cos 2t ⫹ (⫺c1 ⫺ c2) sin 2t. Setting A ⫽ c1 ⫹ c2 and B ⫽ i(c1 ⫺ c2), we can write y 1(h) ⫽ 2A cos 2t ⫹ 2B sin 2t y 2(h) ⫽ B cos 2t ⫺ A sin 2t. Before we can consider the initial conditions, we must determine a particular solution y (p) of the given system. We do this by the method of undetermined coefficients, setting y 1(p) ⫽ a1et ⫹ b1eⴚt y 2(p) ⫽ a2et ⫹ b2eⴚt.

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Differentiation and substitution gives a1et ⫺ b1eⴚt ⫽ 4a2et ⫹ 4b2eⴚt ⫹ 5et a2et ⫺ b 2eⴚt ⫽ ⫺a1et ⫺ b1eⴚt ⫺ 20eⴚt. Equating the coefficients of et on both sides, we get a1 ⫽ 4a2 ⫹ 5,

a2 ⫽ ⫺a1,

hence

a1 ⫽ 1,

a2 ⫽ ⫺1.

Equating the coefficients of e⫺t, we similarly obtain ⫺b1 ⫽ 4b2,

⫺b2 ⫽ ⫺b1 ⫺ 20,

hence

b1 ⫽ ⫺16,

b2 ⫽ 4.

Hence a general solution of the given nonhomogeneous system is y1 ⫽ 2A cos 2t ⫹ 2B sin 2t ⫹ et ⫺ 16eⴚt y2 ⫽ B cos 2t ⫺ A sin 2t ⫺ et ⫹ 4eⴚt. From this and the initial conditions we obtain y1(0) ⫽ 2A ⫹ 1 ⫺16 ⫽ 1,

y2(0) ⫽ B ⫺ 1 ⫹ 4 ⫽ 0.

The solution is A ⫽ 8, B ⫽ ⫺3. This gives the answer (the solution of the initial value problem) y1 ⫽ 16 cos 2t ⫺ 6 sin 2t ⫹ et ⫺ 16eⴚt y2 ⫽ ⫺3 cos 2t ⫺ 8 sin 2t ⫺ et ⫹ 4eⴚt. 18. The equations are I1r ⫽ ⫺2I1 ⫹ 2I2 ⫹ 440 sin t

(a) and

冮

8I2 ⫹ 2 I2 dt ⫹ 2(I2 ⫺ I1) ⫽ 0. Thus

冮

I2 ⫽ ⫺0.25 I2 dt ⫹ 0.25(I1 ⫺ I2), which, upon differentiation and insertion of I1r from (a) and simplification, gives I r2 ⫽ ⫺0.4I1 ⫹ 0.2I2 ⫹ 88 sin t.

(b)

The general solution of the homogeneous system is as in Prob. 17, and the method of undetermined coefficients gives, as a particular solution, ⫺ 20. A ⫽ c

⫺3

1.25

⫺1

J⫽⫺

1 352 1 616 cos t ⫹ c c d d sin t. 3 44 3 132

5 500 1 125 1 ⴚ0.5t 125 e ⫺ eⴚ3.5t ⫹ c d c d c d. 3 2 21 ⫺2 7 1

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SOLUTIONS TO CHAPTER 4 REVIEW QUESTIONS AND PROBLEMS, page 164 12. The matrix

has the eigenvalues 5 and 1 with eigenvectors [1 a general solution is y ⫽ c1 c

1 0

0]T and [0

1]T, respectively. Hence

d e5t ⫹ c2 c d et. 0 1

Since p ⫽ 6, q ⫽ 5, ¢ ⫽ 36 ⫺ 20⬎0, the critical point is an unstable node. 14. Eigenvalues ⫺1, 6. Eigenvectors [1 ⫺1]T, [4 3]T. The corresponding general solution is y1 ⫽ c1e ⴚt ⫹ 4c2e6t,

y2 ⫽ ⫺c1e ⴚt ⫹ 3c2e6t.

The critical point at (0, 0) is a saddle point, which is always unstable. 16. Eigenvalues ⫺4i and 4i. y1 ⫽ c1 cos 4t ⫹ c2 sin 4t, y2 ⫽ ⫺c2 cos 4t ⫹ c1 sin 4t. Center, which is always stable. 18. Saddle point; see Table 4.1 (b) in Sec. 4.4. 20. y1 ⫽ 2c1et ⫹ c2e ⴚ2t ⫹ 2tet ⫹ et, y2 ⫽ ⫺c1et ⫺ 2c2e ⴚ2t ⫺ tet 22. The matrix

has the eigenvalues 3 and ⫺1 with eigenvectors [1 2]T and [1 ⫺2]T, respectively. A particular solution of the nonhomogeneous system is obtained by the method of undetermined coefficients. We set y 1(p) ⫽ a1 cos t ⫹ b1 sin t,

y 2(p) ⫽ a2 cos t ⫹ b2 sin t.

Differentiation and substitution gives the following two equations, where c ⫽ cos t and s ⫽ sin t. (1) (2)

⫺a1s ⫹ b1c ⫽ a1c ⫹ b1s ⫹ a2c ⫹ b2s ⫹ s ⫺a2s ⫹ b2c ⫽ 4a1c ⫹ 4b1s ⫹ a2c ⫹ b2s.

From the cosine terms and the sine terms in (1) we obtain b1 ⫽ a1 ⫹ a2,

⫺a1 ⫽ b1 ⫹ b2 ⫹ 1.

Similarly from (2), b 2 ⫽ 4a1 ⫹ a2,

⫺a2 ⫽ 4b1 ⫹ b2.

Hence a1 ⫽ ⫺0.3, a2 ⫽ 0.4, b1 ⫽ 0.1, b2 ⫽ ⫺0.8. This gives the answer y1 ⫽ c1e3t ⫹ c2e ⴚt ⫺ 0.3 cos t ⫹ 0.1 sin t y2 ⫽ 2c1e3t ⫺ 2c2e ⴚt ⫹ 0.4 cos t ⫺ 0.8 sin t.

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24. The balance equations are 16 6 y1r ⫽ ⫺200 y1 ⫹ 100 y2 16 16 y 2r ⫽ 200 y1 ⫺ 100 y2.

Note that the denominators differ. Note further that the outflow to the right must be included in the balance equation for T2. The matrix is

⫺0.08

0.06

0.08

⫺0.16

It has the eigenvalues ⫺0.2 and ⫺0.04 with eigenvectors [1 ⫺2]T and [1.5 1]T, respectively. The initial condition is y1(0) ⫽ 160, y2(0) ⫽ 0. This gives the answer y1 ⫽ 40eⴚ0.2t ⫹ 120eⴚ0.04t y2 ⫽ ⫺80eⴚ0.2t ⫹ 80eⴚ0.04t. 26. The ODEs of the system are I1r ⫺ I2r ⫹ 5I1 ⫽ 0 I2 ⫺ I1 ⫹ 1.25I2r ⫽ 0, written in the usual form Ir ⫽ c

A⫺B

⫺A

where A ⫽ R>L ⫽ 0.8 and B ⫽ 1>(RC ) ⫽ 5. A general solution is I ⫽ c1 c

1 ⫺4

d eⴚt ⫹ c2 c

c1 ⫽ ⫺13

The initial conditions give satisfying the initial conditions is

4 ⫺1

d eⴚ4t.

and c2 ⫽ 13. Hence the particular solution

I1 ⫽ ⫺13eⴚt ⫹ 43eⴚ4t I2 ⫽ 43eⴚt ⫺ 13eⴚ4t. 28. cos y2 ⫽ 0 when y2 ⫽ (2n ⫹ 1)p>2, where n is any integer. This gives the location of the critical points, which lie on the y2-axis y1 ⫽ 0 in the phase plane. For (0, 12p) the transformation is y2 ⫽ 12p ⫹ 苲 y 2. Now 苲2. cos y2 ⫽ cos (12p ⫹ 苲 y 2) ⫽ ⫺sin 苲 y 2 ⬇ ⫺y Hence the linearized system is 苲 y 1r ⫽ ⫺y苲2 苲 y 2r ⫽ 3y苲1. For its matrix

⫺1

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we obtain p苲 ⫽ 0, q苲 ⫽ 3 ⬎ 0, so that this is a center. Similarly, by periodicity, the critical points at (4n ⫹ 1)p>2 are centers. 苲 For (0, ⫺12p) the transformation is y2 ⫽ ⫺12p ⫹ 苲 y 2. From 苲 cos y ⫽ cos (⫺1p ⫹ y ) ⫽ sin 苲 y ⬇苲 y 2

we obtain the linearized system 苲 y 1r ⫽ 苲 y2 苲苲 y 2r ⫽ 3y 1 with matrix

for which q苲 ⫽ ⫺3 ⬍ 0, so that this point and the points with y2 ⫽ (4n ⫺ 1)p>2 on the y2-axis are saddle points. 30. Critical points at (0, 0) and (0, ⫺1). The linearized systems are 苲 y 1r ⫽ 2y2 y 1r ⫽ ⫺2y苲2 and 苲 yr2 ⫽ ⫺8y1 y 2r ⫽ ⫺8y苲1 where y ⫽ 苲 y and y ⫽ ⫺1 ⫹ 苲 y . At (0, 0) the system has a center and at (0, ⫺1) 1

a saddle point.

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CHAPTER 5

Series Solutions of ODEs. Special Functions

Changes of Text Extensive changes have been made in this chapter. Section 5.1 has been rewritten and the material on the theory of the power series method (the previous Sec. 5.2) has been incorporated into it. The material on the Sturm–Liouville problem, orthogonal functions, and orthogonal eigenfunction expansions (the previous Secs. 5.7 and 5.8) has been moved to Chap. 11 where it is a natural extension of the discussions about Fourier series. The overview of some techniques required for higher special functions and the frequent need for a CAS in exploring them remains. SECTION 5.1. Power Series Method, page 167 Purpose. A simple introduction to the technique of the power series method in terms of simple examples whose solution the student knows very well. Of course, one should emphasize repeatedly that for simple ODEs, such as that in Example 2, one does not need the present method, whereas Example 3 is a first case in which we do need it. Main Content, Important Concepts Power series (1) and (2) Basic examples known from calculus (Example 1, geometric series) ODE (4) [see also (12)] to be solved by inserting (2), (3), (5) Special Legendre equation (Example 3) Partial sum, remainder, convergence Convergence, convergence interval Radius of convergence (Example 4) Theorem 1: Existence of power series solutions Operations on power series Problem Set 5.1 For reviewing power series in more detail the student should use a calculus book, preferably his or her own. To mention Airy functions in Prob. 10 will do no harm and will give the student the impression that much research has been done on special functions and their power series, most of which are known to the usual CAS. This impression if further deepened and expanded in CAS Probs. 18–19. A figure such as 106 should be familiar to the student from calculus. SOLUTIONS TO PROBLEM SET 5.1, page 174 2. 1 4. ⬁ 6. y ⫽ a0 (1 ⫹ x) The purpose of Probs. 6–9 is to give the student a simple introduction to the technique of the power series method in terms of simple examples whose solutions the student can easily obtain otherwise. 8. y ⫽ a3 x 3 ⫺ k>3 89

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10. We obtain 1 4 1 1 4 1 5 a 0(1 ⫺ 16 x 3 ⫺ 24 x ⫺ 120 x 5 ⫹ Á ) ⫹ a1(x ⫹ 12 x 2 ⫹ 16 x 3 ⫺ 24 x ⫺ 30 x ⫺ Á ).

This solution can be expressed in terms of Airy functions, y ⫽ ex>2 (a0 Ai(14 ⫺ x) ⫹ a1Bi(14 ⫺ x)); but this representation is rather complicated, so that for numeric purposes it will be practical to use the power series (a partial sum with sufficiently many terms) directly. 12. y ⫽ a1x ⫹ a0(1 ⫺ x 2 ⫺ 13 x 4 ⫺ 15 x 6 ⫺ 17 x 8 ⫺ Á ). [This is a particular case of Legendre’s equation (n ⫽ 1), which we consider in Sec. 5.2.] 14. y ⫽ (a0 ⫹ a1x) a1 ⫹ x 2 ⫹

2 x4 x6 ⫹ ⫹ Á b ⫽ (a0 ⫹ a1x)ex 2! 3!

32 4 128 5 3 16. s ⫽ 54 ⫺ 4x ⫹ 8x 2 ⫺ 32 s(0.2) ⫽ 0.69900 3 x ⫹ 3 x ⫺ 15 x , 15 35 3 63 5 18. s ⫽ 8 x ⫺ 4 x ⫹ 8 x . This is the Legendre polynomial P5, a solution of this Legendre equation with parameter n ⫽ 5, to be discussed in Sec. 5.2. The initial conditions were chosen accordingly, so that a second linearly independent solution (a Legendre function) does not appear in the answer. s(0.5) ⫽ 0.089844.

SECTION 5.2. Legendre’s Equation. Legendre Polynomials Pn (x ), page 175 Purpose. This section on Legendre’s equation, one of the most important ODEs, and its solution is more than just an exercise on the power series method. It should give the student a feel for the usefulness of power series in exploring properties of special functions and for the wealth of relations between functions of a one-parameter family (with parameter n). Legendre’s equation occurs again in Secs. 11.5 and 11.6. Comment on Literature and History For literature on Legendre’s equation and its solutions, see Refs. [GenRef 1] and [GenRef10]. Legendre’s work on the subject appeared in 1785 and Rodrigues’s contribution (see Prob. 12), in 1816. Problem Set 5.2 Problems 1–9 on Legendre polynomials and functions are straightforward illustrations of the simpler examples of these functions, also showing how a CAS can be used to discover properties. In particular, Prob. 9 is of general interest in connection with any coefficient recursion similar to (3). The problem set also provides a good place to discuss the idea of a generating function and illustrate it in terms of Legendre polynomials. Problems 11–15 discusses a small portion of particularly important basic formulas selected from a large set that can be found in reference books as well as in classical monographies (see [GenRef1] in App. 1). SOLUTIONS TO PROBLEM SET 5.2, page 179 6. We know that at the endpoints of the interval ⫺1 ⬉ x ⬉ 1 all the Legendre polynomials have the values ⫾1. It is interesting that in between they are strictly less than 1 in absolute value (P0 excluded). Furthermore, absolute values between 12 and

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1 are taken only near the endpoints, so that in an interval, say ⫺0.8 ⬉ x ⬉ 0.8, they are less than 21 in absolute value (P0, P1, P2 excluded). 10. Team Project. (a) Following the hint, we obtain (1 ⫺ 2xu ⫹ u 2)ⴚ1>2 ⫽ 1 ⫹ 12 (2xu ⫺ u 2)

(A)

⫹ Á ⫹

1 # 3 Á (2n ⫺ 1) (2xu ⫺ u 2)n ⫹ Á 2 # 4 Á (2n)

and for the general term on the right, (2xu ⫺ u 2)m ⫽ (2x)mu m ⫺ m(2x) mⴚ1u m⫹1 m(m ⫺ 1) ⫹ (2x)mⴚ2u m⫹2 ⫹ Á . 2!

(B)

Now u n occurs in the first term of the expansion (B) of (2xu ⫺ u 2)n, in the second term of the expansion (B) of (2xu ⫺ u 2)nⴚ1, and so on. From (A) and (B) we see that the coefficients of u n in those terms are 1 # 3 Á (2n ⫺ 1) (2x)n ⫽ an x n 2 # 4 Á (2n) ⫺

[see (8)],

1 # 3 Á (2n ⫺ 3) 2n n ⫺ 1 (n ⫺ 1)(2x)nⴚ2 ⫽ ⫺ anxnⴚ2 ⫽ anⴚ2xnⴚ2 2 # 4 Á (2n ⫺ 2) 2n ⫺ 1 4

and so on. This proves the assertion. (b) Set u ⫽ r1>r2 and x ⫽ cos u. (c) Use the formula for the sum of the geometric series and set x ⫽ 1 and x ⫽ ⫺1. Then set x ⫽ 0 and use (1 ⫹ u 2)ⴚ1>2 ⫽ a a

⫺12 m

b u2m

12. We have n n (x 2 ⫺ 1)n ⫽ a (⫺1)m a b (x 2)nⴚm. m m⫽0

Differentiating n times, we can express the product of occurring factors (2n ⫺ 2m)(2n ⫺ 2m ⫺ 1) Á as a quotient of factorials and get M (2n ⫺ 2m)! nⴚ2m dn n! 2 n [(x ⫺ 1) ] ⫽ (⫺1)m x a n m!(n ⫺ m)! (n ⫺ 2m)! dx m⫽0

with M as in (11). Divide by n!2n. Then the left side equals the right side in Rodrigues’s formula, and the right side equals the right side of (11). 14. Abbreviate 1 ⫺ 2xu ⫹ u 2 ⫽ U. Differentiation of (13) with respect to u gives ⬁

1 ⫺ U ⴚ3>2(⫺2x ⫹ 2u) ⫽ a nPn(x)u nⴚ1. 2 n⫽0

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Multiply this equation by U and represent U⫺1>2 by (13): ⬁

⬁

(x ⫺ u) a Pn(x)u n ⫽ (1 ⫺ 2xu ⫹ u 2) a nPn(x)u nⴚ1. n⫽0

n⫽0

In this equation, u has the coefficients xPn(x) ⫺ Pnⴚ1(x) ⫽ (n ⫹ 1)Pn⫹1(x) ⫺ 2nxPn(x) ⫹ (n ⫺ 1)Pnⴚ1(x). Simplifying gives the asserted Bonnet recursion. SECTION 5.3. Extended Power Series Method: Frobenius Method, page 180 Purpose. To introduce the student to the Frobenius method (an extension of the power series method), which is important for ODEs with coefficients that have singularities, notably Bessel’s equation, so that the power series method can no longer handle them. This extended method requires more patience and care. Main Content, Important Concepts Theorem 1 characterizes the ODEs for which the Frobenius method can be used and what form of solutions can be expected. Indicial Equation (4): its role in determining the kind of series solutions to be expected. Theorem 2: Extension of Theorem 1, three cases of roots of the indicial equations and corresponding forms of solution, illustrative Examples 1–3 of these forms. Regular and singular points Short Courses. Take a quick look at those bases in Frobenius’s theorem, see how it fits with the Euler–Cauchy equation, and omit everything else. Comment on “Regular Singular” and “Irregular Singular” These terms are used in some books and papers, but there is hardly any need for confusing the student by using them, simply because we cannot do (and don’t do) anything about “irregular singular points.” A simple use of “regular” and “singular” (as in complex analysis, where holomorphic functions are also known as “regular analytic functions”) may thus be the best terminology. Comment on Footnote 5 Gauss was born in Braunschweig (Brunswick) in 1777. At the age of 16, in 1793 he discovered the method of least squares (Secs. 20.5, 25.9). From 1795 to 1798 he studied at Göttingen. In 1799 he obtained his doctor’s degree at Helmstedt. In 1801 he published his first masterpiece, Disquisitiones arithmeticae (Arithmetical Investigations, begun in 1795), thereby initiating modern number theory. In 1801 he became generally known when his calculations enabled astronomers (Zach, Olbers) to rediscover the planet Ceres, which had been discovered in 1801 by Piazzi at Palermo but had been visible only very briefly. He became the director of the Göttingen observatory in 1807 and remained there until his death. In 1809 he published his famous Theoria motus corporum coelestium in sectionibus conicis solem ambientium (Theory of the Heavenly Bodies Moving About the Sun in Conic Sections; Dover Publications, 1963), resulting from his further work in astronomy. In 1814 he developed his method of numeric integration (Sec. 19.5). His Disquisitiones generales circa superficies curvas (General Investigations Regarding Curved Surfaces, 1828) represents the foundation of the differential geometry of surfaces and contributes to conformal mapping (Sec. 17.1). His clear conception of the complex plane dates back to his thesis, whereas his first publication on this topic was not before 1831. This is typical: Gauss left many of his

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most outstanding results (non-Euclidean geometry, elliptic functions, etc.) unpublished. His paper on the hypergeometric series published in 1812 is the first systematic investigation into the convergence of a series. This series, generalizing the geometric series, allows a study of many special functions from a common point of view. Problem Set 5.3 Problems 2–13: Only in simpler cases will it be possible to recognize the series as one of a known function; this task is included to make students aware that even an unfamiliar series may be an expansion of an elementary function. Gauss’s hypergeometric ODE (15), series (16), and function F(a, b, c; x) play a central role in special functions, simply because they include an incredible number of familiar elementary and higher special functions. Perhaps it is worth mentioning that (15) may be regarded as a “natural” extension of the Euler–Cauchy equation.

SOLUTIONS TO PROBLEM SET 5.3, page 186 2. y1 ⫽ x ⫹ 2, y2 ⫽ 1>(x ⫹ 2). Check: Set x ⫹ 2 ⫽ z, to get an Euler–Cauchy equation. x x2 x3 ⫹ ⫺ ⫹ ⫺ Á b. From this and (10) we obtain 4. y1 ⫽ x a1 ⫺ 1!2! 2!3! 3!4! 7 3 35 y2 ⫽ ⫺y1 ln x ⫹ 1 ⫺ 34 x 2 ⫹ 36 x ⫺ 1728 x4 ⫹ ⫺ Á .

In the present case, k and A1 in (10) are at first arbitrary, and our y2 corresponds to the choice k ⫽ 1 and A1 ⫽ 0. Choosing A1 ⫽ 0, we obtain the above expression for y2 plus A1y1. 6. b0 ⫽ 0, c0 ⫽ ⫺2, r(r ⫺ 1) ⫺ 2 ⫽ (r ⫺ 2)(r ⫹ 1), r1 ⫽ 2, r2 ⫽ ⫺1, 9 4 13 6 y1 ⫽ x 2 (1 ⫺ 12 x 2 ⫹ 56 x ⫺ 336 x ⫹ Á)

y2 ⫽ 8. y1 ⫽ 1 ⫹

(2 # 4)

⫹ 2

y2 ⫽ y1 ln x ⫺

9 1 7 (12 ⫺ 6x 2 ⫹ x 4 ⫺ x 6 ⫹ Á ). x 2 4

⫹ 2

x6 (2 # 4 # 6) 2

⫹ Á,

x2 3x 4 11x 6 ⫺ # ⫺ ⫺ Á 4 8 16 64 # 6 # 36

10. y1 ⫽ x ⴚ1 cos 2x, y2 ⫽ x ⴚ1 sin 2x 12. b0 ⫽ 6, c0 ⫽ 6, r1 ⫽ ⫺2, r2 ⫽ ⫺3; the series are y1 ⫽

1 2 2 2 4 4 1 sin 2x ⫺ ⫹ x ⫺ x ⫹⫺Á ⫽ x2 3 15 315 2 x3

y2 ⫽

1 2 2 4 3 cos 2x ⫺ ⫹ x⫺ x ⫹⫺ Á ⫽ . x3 x x3 3 45

14. Team Project. (b) In (11b) of Sec. 5.1, an⫹1 an hence R ⫽ 1.

⫽

(a ⫹ n)(b ⫹ n) (n ⫹ 1)(c ⫹ n)

: 1,

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(c) In the third and fourth lines, arctan x ⫽ x ⫺ 13 x 3 ⫹ 15 x 5 ⫺ 17 x 7 ⫹ ⫺ Á 1 1 #3 1#3 #5 arcsin x ⫽ x ⫹ # x 3 ⫹ # # x 5 ⫹ # # # x 7 ⫹ Á 2 3 2 4 5 2 4 6 7

(| x | ⬍ 1) (| x | ⬍ 1).

(d) The roots can be read off from (15), brought to the form (1 r ) by multiplying it by x and dividing by 1 ⫺ x; then b0 ⫽ c in (4) and c0 ⫽ 0. 16. y ⫽ A(1 ⫹ 4x) ⫹ B>xF(⫺12, ⫺32, 32, x) 18. t 2 ⫺ 3t ⫹ 2 ⫽ (t ⫺ 1)(t ⫺ 2) ⫽ 0. Hence the transformation is x ⫽ t ⫺ 1. It gives the ODE 4(x 2 ⫺ x)y s ⫺ 2y r ⫹ y ⫽ 0. To obtain the standard form of the hypergeometric equation, multiply this ODE by ⫺14 . It is clear that the factor 4 must be absorbed, but don’t forget the factor ⫺1; otherwise your values for a, b, c will not be correct. The result is x(1 ⫺ x)y s ⫹ 12 y r ⫺ 14 y ⫽ 0. Hence ab ⫽ 14, b ⫽

1 1 ,a⫹b⫹1⫽a⫹ ⫹ 1 ⫽ 0, a ⫽ ⫺12, b ⫽ ⫺12, c ⫽ 12. 4a 4a

This gives y1 ⫽ F(⫺12, ⫺12, 12; t ⫺ 1). In y2 we have a ⫺ c ⫹ 1 ⫽ ⫺12 ⫺ 12 ⫹ 1 ⫽ 0; hence y2 terminates after the first term, and, since 1 ⫺ c ⫽ 12, y2 ⫽ x 1>2 # 1 ⫽ 2t ⫺ 1. 20. y ⫽ c1F(⫺1, 13, 13; t ⫹ 1) ⫹ c2(t ⫹ 1) 2>3 F(⫺13, 1, 53; t ⫹ 1) SECTION 5.4. Bessel’s Equation. Bessel Functions J␯(x ), page 187 Purpose. To derive the Bessel functions of the first kind J␯ and J⫺␯ by the Frobenius method. (This is a major application of that method.) To show that these functions constitute a basis if ␯ is not an integer but are linearly dependent for integer ␯ ⫽ n (so that we must look later, in Sec. 5.5, for a second linearly independent solution). To show that various ODEs can be reduced to Bessel’s equation (see Problem Set 5.4). Main Content, Important Concepts Derivation just mentioned Linear independence of J␯ and Jⴚ␯ if ␯ is not an integer Linear dependence of J␯ and Jⴚ␯ if ␯ ⫽ n ⫽ 1, 2, Á Gamma function as a tool Short Courses. No derivation of any of the series. Discussion of J0 and J1 (which are similar to cosine and sine). Mention Theorem 2.

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Comment on Special Functions Since various institutions no longer find time to offer a course in special functions, Bessel functions may give another opportunity (together with Sec. 5.2) for getting at least some feel for the flavor of the theory of special functions, which will continue to be of significance to the engineer and physicist. For this reason we have added some material on basic relations for Bessel functions in this section. Problem Set 5.4 Problems 2–10 concern a few of the large number of ODEs reducible to Bessel’s equation. The latter contains a single parameter, and Prob. 5 shows how a second parameter can be introduced. CAS Experiment 12 and Prob. 14 give an impression of the accuracy of the asymptotic formula (14). Elimination of the first derivative from an ODE (Probs. 16–18) is a standard process used for various practical and theoretical purposes. Problems 19–25 show that (21) is a backbone of the whole theory; in particular, it can be used to obtain Bessel’s equation (1). SOLUTIONS TO PROBLEM SET 5.4, page 195 2. c1J2>7(x) ⫹ c2Jⴚ2>7(x) 4. c1J1>3(eⴚx) ⫹ c2 Jⴚ1>3(eⴚx) 6. 1x(c1J1>2 (1x) ⫹ c2 Jⴚ1>2(1x)) ⫽ x 1>4(c苲1 sin 1x ⫹ 苲 c 2 cos 1x) 8. c1J1(2x ⫹ 1), Jⴚ1 ⫽ ⫺J1 10. x ␯(c1J␯ (x ␯) ⫹ c2 Jⴚ␯ (x ␯)), ␯ ⫽ 0, ⫾1, ⫾2, Á 12. CAS Experiment. (b) x 0 ⫽ 1, x 1 ⫽ 2.5, x 2 ⫽ 20, approximately. It increases with n (c) (14) is exact. (d) It oscillates. (e) Formula (24b) with ␯ ⫽ 0. 14. This problem should give the student a feel for the applicability and accuracy of an asymptotic formula. A corresponding formula for y␯ is included in the next problem set. Zeros of J0(x) Approximation (14) Exact Value 2.35619 5.49779 8.63938 11.78097

2.40483 5.52008 8.65373 11.79153

Error 0.04864 0.02229 0.01435 0.01056

Zeros of J1(x) Approximation (14) Exact Value 3.92699 7.06858 10.21018 13.35177

3.83171 7.01559 10.17347 13.32369

Error ⫺0.09528 ⫺0.05299 ⫺0.03671 ⫺0.02808

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18. For ␯ ⫽ ⫾12 the ODE (27) in Prob. 17 becomes u s ⫹ u ⫽ 0. Hence for y we obtain the general solution y ⫽ x ⴚ1>2u ⫽ x ⴚ1>2(A cos x ⫹ B sin x). We can now obtain A and B by comparing with the first term in (20). Using ⌫(32) ⫽ 12 ⌫(12) ⫽ 12 1p (see (23)) we obtain, for ␯ ⫽ 12, the first term x 1>2>(21>2 # 12 ⌫(12)) ⫽ 22x> p.

This gives (22a) because the series of sin x starts with the power x. For ␯ ⫽ ⫺12 the first term is x ⴚ1>2>(21>2 # 12 ⌫(12)) ⫽ 22> px. This gives (22b). 20. (21b) with ␯ ⫺ 1 instead of ␯ is (x ⴚ␯⫹1J␯ⴚ1) r ⫽ ⫺x ⴚ␯⫹1J␯ . Now use (21a) (x ␯J␯) r ⫽ x ␯J␯ⴚ1; solve it for J␯ⴚ1 to obtain J␯ⴚ1 ⫽ x ⴚ␯(x ␯J␯) r ⫽ x ⴚ␯(␯x ␯ⴚ1J␯ ⫹ x ␯J r␯) ⫽ ␯x ⴚ1J␯ ⫹ J r␯. Substitute this into the previous equation on the left. Then perform the indicated differentiation: (x ⴚ␯⫹1(␯xⴚ1J␯ ⫹ J r␯)) r ⫽ (␯xⴚ␯J␯ ⫹ xⴚ␯⫹1J r␯) r ⫽ ⫺␯2xⴚ␯ⴚ1J␯ ⫹␯x ⴚ␯J r␯ ⫹ (⫺␯ ⫹1)x ⴚ␯J r␯ ⫹ x ⴚ␯⫹1J s␯ . Equating this to the right side of the first equation and dividing by x ⴚ␯⫹1 gives J s␯ ⫹

1 ␯2 J r␯ ⫺ 2 J␯ ⫽ ⫺J␯. x x

Taking the term on the right to the left (with a plus sign) and multiplying by x 2 gives (1). 22. Integrate (21b) and (21d). 24. We obtain

冮 x J (x) dx ⫽ 冮 x (x J (x)) dx ⫽ 冮 x (⫺x J (x)) r dx ⫽ ⫺x x J (x) ⫹ 冮 2xx J (x) dx ⫽ ⫺x J (x) ⫹ 2 冮 x J (x) dx ⫽ ⫺x J (x) ⫹ 2 冮 (⫺x J ) r dx ⴚ1

ⴚ3

2 ⴚ3

ⴚ1

ⴚ3

ⴚ2

⫽ ⫺x ⴚ1J3(x) ⫺ 2x ⴚ2J2(x) ⫹ c

(trivial) ((21b) with ␯ ⫽ 3) (by parts) (simplify) ((21b) with ␯ ⫽ 2) (trivial).

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SECTION 5.5. Bessel Functions Y␯(x). General Solution, page 196 Purpose. Derivation of a second independent solution, which is still missing in the case of ␯ ⫽ n ⫽ 0, 1, Á . Main Content Detailed derivation of Y0(x) Cursory derivation of Yn(x) for any n General solution (9) valid for all ␯, integer or not Short Courses. Omit this section. Comment on Hankel Functions and Modified Bessel Functions These are included for completeness, but will not be needed in our further work. SOLUTIONS TO PROBLEM SET 5.5, page 200 2. x ⴚ2 (c1J2(x) ⫹ c2Y2(x)). Here Jⴚ2 could not be used, because of linear dependence. 4. Substitute y ⫽ ux 1>2 and its derivatives into the given equation and multiply the resulting equation by x 3>2 to get x 2u s ⫹ xu r ⫹ A x 3 ⫺ 14 B u ⫽ 0.

Now introduce z as given in the problem statement to get the answer y ⫽ 1x 3c1J1>3 A 23x 3>2 B ⫹ c2 J⫺1>3 A 23x 3>2 B4 .

6. c1J0(12 1x) ⫹ c2Y0(12 1x) 8. 1x (c1J1>6 (13 kx 3) ⫹ c2Y1>6 (13 kx 3)). Jⴚ1>6 could be used instead of Y1>6. 10. CAS Experiment. (a) Y0 and Y1, similarly as for J0 and J1. (b) Accuracy is best for Y0. x n increases with n; actual values will depend on the scales used for graphing. (c), (d) Y0

By (11)

Exact

By (11)

Exact

By (11)

Exact

1 2 3 4 5 6 7 8 9 10

0.785 3.9270 7.0686 10.210 13.352 16.493 19.635 22.777 25.918 29.060

0.894 3.958 7.086 10.222 13.361 16.501 19.641 22.782 25.923 29.064

2.356 5.498 8.639 11.781 14.923 18.064 21.206 24.347 27.489 30.631

2.197 5.430 8.596 11.749 14.897 18.043 21.188 24.332 27.475 30.618

3.927 7.0686 10.210 13.352 16.493 19.635 22.777 25.918 29.060 32.201

3.384 6.794 10.023 13.210 16.379 19.539 22.694 25.846 28.995 32.143

These values show that the accuracy increases with x (for fixed n), as expected. For a fixed m (number of zero) it decreases with increasing n (order of Yn). 12. Set x ⫽ is in (1), Sec. 5.4, to get the present ODE (12) in terms of s. 14. For x ⫽ 0 all the terms of the series are real and positive.

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SOLUTIONS TO CHAPTER 5 REVIEW QUESTIONS AND PROBLEMS, page 200 12. ex, ex ln x 14. (x ⫹ 1)3>4, (x ⫹ 1)1>4; Euler–Cauchy equation with independent variable x ⫹ 1 16. b0 ⫽ 0, c0 ⫽ ⫺2, r(r ⫺ 1) ⫺ 2 ⫽ 0, r1 ⫽ 2, r2 ⫽ ⫺1. The series are 9 4 13 6 y1 ⫽ x 2(1 ⫺ 12 x 2 ⫹ 56 x ⫺ 336 x ⫹ Á ),

9 1 7 y2 ⫽ x (12 ⫺ 6x 2 ⫺ x 4 ⫺ x 6 ⫹ Á ). 2 4 18. x ⴚ2 cos x 2, x ⴚ2 sin x 2 x2 x4 x6 20. y1 ⫽ 1 ⫹ 2 ⫹ ⫹ ⫹ Á 2 (2 # 4)2 (2 # 4 # 6) 2 y2 ⫽ y1 ln x ⫺

3x 4 11x 6 x2 ⫺ # ⫺ ⫺ Á 4 8 16 64 # 6 # 36

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CHAPTER 6

Laplace Transforms

In the ninth edition, this chapter underwent major changes, which have been retained and further extended. This concerns a more natural order of the material and changing emphasis placed on the various topics. This has made the chapter more teachable and simpler, a goal reached by placing Dirac’s delta function in a separate section, discussing partial fractions earlier, and in terms of practical problems rather than in terms of impractical and lengthy general formulas. Also, to have a better flow of ideas, convolution and nonhomogeneous linear ODEs now appear earlier, whereas differentiation and integration of transforms (not of functions!) are now presented near the end of the chapter and with less emphasis. SECTION 6.1. Laplace Transform. Linearity. First Shifting Theorem (s-Shifting), page 204 Purpose. To explain the basic concepts, to present a short list of basic transforms, and to show how these are derived from the definition. Main Content, Important Concepts Transform, inverse transform, linearity First shifting theorem Table 6.1 Existence and its practical significance Comment on Table 6.1 After working for a while in this chapter, the student should be able to memorize these transforms. Further transforms in Sec. 6.9 are derived as we go along, many of them from Table 6.1. Problem Set 6.1 The problem set addresses the two main tasks in connection with Laplace transform, namely, to find transforms of given functions (Probs. 1–16 and 33–36), and to find functions for given transforms (that is, to find inverse transforms, Probs. 25–32 and 37–45). This includes the task of finding formulas for graphically given functions, and, as techniques, integration, reduction by partial fractions, the use of Table 6.1, and s-shifting (in Probs. 33–45). Problems 17–24 include a modest amount of theory compatible with the level of this book. SOLUTIONS TO PROBLEM SET 6.1, page 210 2. a 2>s ⫺ 2ab>s 2 ⫹ 2b 2>s 3 1 1 1 1 s 4. l(cos2 vt) ⫽ l a ⫹ cos 2vtb ⫽ ⫹ # 2 2 2 2s 2 s ⫹ 4v2 1 1 1 4 1 3t (e ⫺ eⴚ5t); transform a ⫺ b⫽ 2 2 s⫺3 s⫹5 (s ⫹ 1)2 ⫺ 16 This can be checked by the first shifting theorem.

6. eⴚt sinh 4t ⫽

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8. 1.5 sin (3t ⫺ p>2) ⫽ 1.5 (sin 3t cos p>2 ⫺ cos 3t sin p>2) ⫽ ⫺1.5 cos 3t; transform ⫺1.5s>(s 2 ⫹ 9) k 10. s (1 ⫺ eⴚcs) 1 ⫺ eⴚs 1 12. ⫺ 2s 2 s se b

14. k

冮e

ⴚst

k dt ⫽ s (eⴚas ⫺ eⴚbs)

16.

冮

eⴚstt dt ⫹

冮 e (2 ⫺ t) dt. Integration by parts gives ⴚst

eⴚstt 1 1 ⫺ s ` ⫹ s 0 ⫽⫺ ⫽

冮

eⴚst dt ⫺

eⴚst(2 ⫺ t) 2 1 ` ⫺ s s 1

冮e

ⴚst

eⴚs 1 eⴚs 1 ⫺ 2 (eⴚs ⫺ 1) ⫹ ⫹ 2 (eⴚ2s ⫺ eⴚs) s s s s

(1 ⫺ eⴚs)2

(⫺eⴚs ⫹ 1 ⫹ eⴚ2s ⫺ eⴚs) ⫽ 2

18. Use Prob. 10 with k ⫽ 1 and c ⫽ 2 to obtain 1 1 eⴚ2s l( f1) ⫽ l(1) ⫺ l( f ) ⫽ s ⫺ s (1 ⫺ eⴚ2s) ⫽ s . 2

20. No matter how large we choose M and k, we have et ⬎ Mekt for all t greater than some t 0 because t 2 ⬎ ln M ⫹ kt for all sufficiently large t (and fixed positive M and k). 22. Let st ⫽ t, t ⫽ t>s, dt ⫽ dt>s. Then ⬁

l(1> 1t) ⫽

冮e t

ⴚst ⴚ1>2

⫽

冮

⬁

ⴚ1>2

t eⴚt a s b

1 s dt

⬁

⫽ s ⴚ1>2

冮e t

ⴚt ⴚ1>2

⫽ s ⴚ1>2⌫(12) ⫽ 1p>s. 24. Let f ⫽ lⴚ1(F ), g ⫽ lⴚ1(G). Since the transform is linear, we obtain aF ⫹ bG ⫽ al( f ) ⫹ bl(g) ⫽ l(af ⫹ bg). Now apply lⴚ1 on both sides to get the desired result, lⴚ1(aF ⫹ bG) ⫽ lⴚ1l(af ⫹ bg) ⫽ af ⫹ bg ⫽ alⴚ1(F ) ⫹ blⴚ1(G). Note that we have proved much more than just the claim, namely, the following. Theorem. If a linear transformation has an inverse, the inverse is linear.

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26. The inverse transform is 5t ⴚ5t 5 cosh 5t ⫹ 15 sinh 5t ⫽ 13 ⫹ 12 . 5 e 5 e

28. In terms of partial fractions, the given function is 1 1 1 a ⫺ b. 13 ⫹ 12 s ⫺ 13 s ⫹ 12 Hence the inverse transform is 1 (et13 ⫺ eⴚt12). 13 ⫹ 12 30. The inverse transform is 4 cosh 4t ⫹ 8 sinh 4t ⫽ 6e4t ⫺ 2eⴚ4t. 1 (eⴚat ⫺ eⴚbt). If a ⫽ b, then teⴚbt by the shifting theorem (or b⫺a from the first result by l’Hôpital’s rule, taking derivatives with respect to a).

32. If a ⫽ b, then

34.

k(s ⫹ a) (s ⫹ a)2 ⫹ v2

1 1 s⫺1 s⫹1 36. l a (et ⫺ eⴚt) cos tb ⫽ a ⫺ b 2 (s ⫹ 1)2 ⫹ 1 2 (s ⫺ 1)2 ⫹ 1 ⫽ ⫽

1 2s 2 ⫺ 4 a b 2 s4 ⫹ 4 s2 ⫺ 2 s4 ⫹ 4

38. 3t 2eⴚt 40. 2et sinh 2t ⫽ e3t ⫺ eⴚt 42. (a0 ⫹ a1t ⫹ 12 a2t 2) eⴚt 44. eⴚkt(a cos pt ⫹ b sin pt)

SECTION 6.2. Transforms of Derivatives and Integrals. ODEs, page 211 Purpose. To get a first impression of how the Laplace transform solves ODEs and initial value problems, the task for which it is designed. Main Content, Important Concepts (1) l( f r ) ⫽ sl( f ) ⫺ f (0) Extension of (1) to higher derivatives [(2), (3)] Solution of an ODE, subsidiary equation Transform of the integral of a function (Theorem 3) Transfer function (6) Shifted data problems (Example 6)

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Comment on ODEs The last of the three steps of solution is the hardest, but we shall derive many general properties of the Laplace transform (collected in Sec. 6.8) that will help, along with formulas in Table 6.1 and those in Sec. 6.9, so that we can proceed to ODEs for which the present method is superior to the classical one. Problem Set 6.2 Problems 1–15 concern IVPs, including those with shifted data (12–15). Problems 16–22 show how to obtain new transforms from known ones by differentiation, the basic formulas being (1) and (2). Conversely, Probs. 23–29 show how to obtain new inverse transforms by integration (Theorem 3). Project 30 extends Theorem 1 to the case when a discontinuity (finite jump) occurs. SOLUTIONS TO PROBLEM SET 6.2, page 216 2. The subsidiary equation is sY ⫺ 1.5 ⫹ 2Y ⫽ 0. Hence (s ⫹ 2)Y ⫽ 1.5,

Y ⫽ 1.5>(s ⫹ 2),

y ⫽ 1.5eⴚ2t.

4. We obtain (s 2 ⫹ 9)Y ⫽

10 . s⫹1

The solution of this subsidiary equation is Y⫽

10 1 s⫺1 . ⫽ ⫺ 2 s⫹1 (s ⫹ 1)(s 2 ⫹ 9) s ⫹9

This gives the solution y ⫽ eⴚt ⫺ cos 3t ⫹ 13 sin 3t. 6. The subsidiary equation is (s 2 ⫺ 6s ⫹ 5)Y ⫽ 3.2s ⫹ 6.2 ⫺ 6 # 3.2 ⫹ 29s>(s 2 ⫹ 4). The solution Y is Y⫽

(3.2s ⫺ 13)(s 2 ⫹ 4) ⫹ 29s (s ⫺ 1)(s ⫺ 5)(s 2 ⫹ 4)

In terms of partial fractions, this becomes Y⫽

1 2 0.2s ⫺ 2.4 # 2 . ⫹ ⫹ s⫺1 s⫺5 s2 ⫹ 4

The inverse transform of this gives the solution y ⫽ et ⫹ 2e5t ⫹ 0.2 cos 2t ⫺ 2.4 sin 2t. 8. The subsidiary equation is (s ⫺ 2)2Y ⫽ 8.1s ⫹ 3.9 ⫺ 4 # 8.1. Its solution is Y⫽

8.1s ⫺ 28.5 (s ⫺ 2)2

⫽

8.1(s ⫺ 2) ⫹ 8.1 # 2 ⫺ 28.5 (s ⫺ 2)2

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This can be written Y⫽

8.1 12.3 ⫺ . s⫺2 (s ⫺ 2)2

The inverse transform (the solution of the IVP) is y ⫽ 8.1e2t ⫺ 12.3te2t. It has the form expected in the case of a double root of the characteristic equation. 10. (s 2 ⫹ 0.04)Y ⫽ ⫺25s ⫹ 0.04>s 3. The solution is Y⫽⫺

25 1 ⫹ 3. s s

The inverse transform of Y (the solution of the IVP) is y ⫽ ⫺25 ⫹ 0.5t 2. Note that the initial values are such that the general solution of the homogeneous ODE yh ⫽ c1 cos 0.2t ⫹ c2 sin 0.2t does not contribute to the solution of the IVP. t ⫹ 4, so that the “shifted problem” is 12. t ⫽ ~ ~y s ⫺ 2y~ r ⫺ 3y~ ⫽ 0, ~y (0) ⫽ ⫺3,

~y r (0) ⫽ ⫺17.

Hence the corresponding subsidiary equation is ~ (s 2 ⫺ 2s ⫺ 3)Y ⫽ ⫺3s ⫺ 17 ⫹ (⫺2)(⫺3). Its solution is ~ Y⫽

⫺3s ⫺ 11 2 5 ⫽ ⫺ . (s ⫹ 1)(s ⫺ 3) s⫹1 s⫺3

Inversion gives ~y ⫽ 2eⴚt~ ⫺ 5e3t~ . This is the solution of the “shifted problem,” and the solution of the given problem is y ⫽ 2eⴚ(tⴚ4) ⫺ 5e3(tⴚ4). t ⫹ 2, so that the “shifted equation” is 14. t ⫽ ~ ~y s ⫹ 2y~ r ⫹ 5y~ ⫽ 50t~ . The corresponding subsidiary equation is ~ 3(s ⫹ 1)2 ⫹ 44Y ⫽ ⫺4s ⫹ 14 ⫺ 2 # 4 ⫹ 50>s 2. Its solution is 4 4 ~ 10 Y⫽ 2 ⫺ ⫹ . s s (s ⫹ 1)2 ⫹ 4 The inverse transform is ~y ⫽ 10t~ ⫺ 4 ⫹ 2eⴚt~ sin 2t~. Hence the given problem has the solution y ⫽ 10(t ⫺ 2) ⫺ 4 ⫹ 2eⴚ(tⴚ2) sin 2(t ⫺ 2).

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16. f ⫽ t cos 4t, f r ⫽ cos 4t ⫺ 4t sin 4t, f s ⫽ ⫺8 sin 4t ⫺ 16t cos 4t. Hence l( f s ) ⫽

⫺32 s ⫹ 16 2

⫺ 16l( f ) ⫽ s 2l( f ) ⫺ s # 0 ⫺ 1

and thus (s 2 ⫹ 16)l( f ) ⫽

⫺32 s 2 ⫹ 16

⫹1⫽

s 2 ⫺ 16 s 2 ⫹ 16

Hence the answer is s 2 ⫺ 16

l( f ) ⫽

(s 2 ⫹ 16) 2

18. f ⫽ cos2 2t, f r ⫽ ⫺4 cos 2t sin 2t ⫽ ⫺2 sin 4t. Hence l( f r ) ⫽

⫺8 s ⫹ 16 2

⫽ sl( f ) ⫺ 1.

Solving for sl( f ) gives sl( f ) ⫽

s2 ⫹ 8 s 2 ⫹ 16

Division by s gives the answer s2 ⫹ 8

l( f ) ⫽

s(s 2 ⫹ 16)

20. f ⫽ sin4 t, f r ⫽ 4 sin3 t cos t, f s ⫽ 12 sin2 t cos2 t ⫺ 4 sin4 t. From this and (2), it follows that l( f s ) ⫽ 12l(sin2 t cos2 t) ⫺ 4l( f ) ⫽ s 2l( f ). Collecting terms and using Prob. 19 with v ⫽ 2 and sin 2a ⫽ 2 sin a cos a, we obtain (s 2 ⫹ 4)l( f ) ⫽ 3l(sin2 2t) ⫽

3 #8 s(s ⫹ 16) 2

Answer l( f ) ⫽

24 s(s ⫹ 4)(s 2 ⫹ 16) 2

22. Project. We derive (a). We have f (0) ⫽ 0 and f r (t) ⫽ cos vt ⫺ vt sin vt, f s(t) ⫽ ⫺2v sin vt ⫺ v2f (t).

f r (0) ⫽ 1

By (2), l( f s ) ⫽ ⫺2v

v s ⫹v 2

⫺ v2l( f ) ⫽ s 2l( f ) ⫺ 1.

Collecting l( f )-terms, we obtain l( f )(s 2 ⫹ v2) ⫽

⫺2v2 s ⫹v 2

Division by s 2 ⫹ v2 on both sides gives (a).

⫹1⫽

s 2 ⫺ v2 s 2 ⫹ v2

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In (b) on the right we get from (a) l(sin vt ⫺ vt cos vt) ⫽

v s2 ⫹ v

⫺v 2

s 2 ⫺ v2 (s 2 ⫹ v2)2

Taking the common denominator and simplifying the numerator, v(s 2 ⫹ v2) ⫺ v(s 2 ⫺ v2) ⫽ 2v3 we get (b). (c) is shown in Example 1. (d) is derived the same way as (b), with ⫹ instead of ⫺, so that the numerator is v(s 2 ⫹ v2) ⫹ v(s 2 ⫺ v2) ⫽ 2vs 2, which gives (d). (e) is similar to (a). We have f (0) ⫽ 0 and obtain f r(t) ⫽ cosh at ⫹ at sinh at, f s(t) ⫽ 2a sinh at ⫹ a 2f (t).

f r(0) ⫽ 1

By (2) we obtain l( f s ) ⫽

2a 2 s ⫺a 2

⫹ a 2l( f ) ⫽ s 2l( f ) ⫺ 1.

Hence l( f )(s 2 ⫺ a 2) ⫽

2a 2 s2 ⫺ a

⫹1⫽ 2

s2 ⫹ a2 s2 ⫺ a2

Division by s 2 ⫺ a 2 gives (e). (f) follows similarly. We have f (0) ⫽ 0 and, furthermore, f r(t) ⫽ sinh at ⫹ at cosh at, f r(0) ⫽ 0 f s(t) ⫽ 2a cosh at ⫹ a 2f (t) s l( f s(t)) ⫽ 2a 2 ⫹ a 2l( f ) ⫽ s 2l( f ) 2 s ⫺a 2as l( f )(s 2 ⫺ a 2) ⫽ 2 . s ⫺ a2 Division by s 2 ⫺ a 2 gives formula (f). 24. We start from lⴚ1 a

1 b ⫽ e2pt s ⫺ 2p

and integrate twice. The first integration gives 1 2pt (e ⫺ 1). 2p Multiplication by 20 and another integration from 0 to t gives the answer 20t 1 10t 20 2pt 2pt ⫺ 1) ⫺ ⫽ ⫺ 1) ⫺ . 2 (e 2 (e 2p p 4p 5p

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26. The inverse of 1>(s 2 ⫺ 1) is sinh t. A first integration from 0 to t gives cosh t ⫺ 1. Another integration from 0 to t yields sinh t ⫺ t. This can be confirmed as follows. 1 1 1 la 4 b⫽la 2 ⫺ 2 b ⫽ sinh t ⫺ t. 2 s ⫺s s ⫺1 s 28. The transform of 3s ⫹ 4 s2 ⫹ k 2 is 3 cos kt ⫹

4 sin kt. A first integration from 0 to t gives k 3 4 sin kt ⫺ 2 (cos kt ⫺ 1). k k

Another integration from 0 to t gives the answer ⫺

3 4 1 (cos kt ⫺ 1) ⫺ 2 a sin kt ⫺ tb . k2 k k

30. Project. (a) Theorems 1 and 2 are crucial in solving ODEs, whereas Theorem 3 serves as a tool for obtaining new transforms, that is, as one of various tools for this purpose. (b) In the integration by parts shown in the proof of Theorem 1 we now have to integrate from 0 to a and then from a to ⬁ , thus obtaining f (a ⫺ 0)eⴚas from the upper limit of integration of the first integral and ⫺f (a ⫹ 0)eⴚas from the lower limit of integration of the second integral. (c) Direct integration of the defining integral formulas gives l( f ) ⫽ (1 ⫺ eⴚ(s⫹1))>(s ⫹ 1) and l( f r ) ⫽ (eⴚ(s⫹1) ⫺ 1)>(s ⫹ 1). Formula (1*) confirms this by straightforward calculation and simplification since f (0) ⫽ 1, a ⫽ 1, and f (a ⫹ 0) ⫺ f (a ⫺ 0) ⫽ 0 ⫺ eⴚ1. SECTION 6.3. Unit Step Function (Heaviside Function). Second Shifting Theorem (t-Shifting), page 217 Purpose 1. To introduce the unit step function u(t ⫺ a), which, together with Dirac’s delta (Sec. 6.4), greatly increases the usefulness of the Laplace transform. 2. To find the transform of ~ f (t) ⫽ 0 if f ⬍ a

and

~ f (t) ⫽ f (t ⫺ a) if t ⬎ a

where the transform of f (t) (0 ⬍ t ⬍ ⬁) is known (Theorem 1, called Second Shifting Theorem).

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Main Content, Important Concepts Unit step function (1), its transform (2) Second shifting theorem (Theorem 1) Comment on the Unit Step Function Problem Set 6.3 shows that u(t ⫺ a) is the basic function for representing discontinuous functions. Problem Set 6.3 Problems 2–11 concern the role of the unit step function in t-shifting in the second shifting theorem. Problems 12–17 show the application of that theorem in finding inverse transforms. IVPs, some with discontinuous inputs (right sides of the ODEs) are solved in Probs. 18–27; this is a typical task using the Laplace transform. Models of electric circuits follow in Probs. 28–40; these have a discontinuous EMF (electromotive force); the most general of these circuits are RLC-circuits (Probs. 38–40). SOLUTIONS TO PROBLEM SET 6.3, page 223 2. t(1 ⫺ u(t ⫺ 2)) ⫽ t ⫺ 3(t ⫺ 2) ⫹ 24u(t ⫺ 2). Hence the transform is 1 eⴚ2s 2eⴚ2s . 2 ⫺ 2 ⫺ s s s

4. (cos 4t)(1 ⫺ u(t ⫺ p)) ⫽ cos 4t ⫺ (cos 4(t ⫺ p))u(t ⫺ p). Hence the transform is s s ⫹ 16 2

(1 ⫺ eⴚps).

6. sin (t ⫺ 2)p ⫽ sin (t ⫺ 4)p ⫽ sin pt because of periodicity. The representation in terms of unit step functions is (sin pt)(u(t ⫺ 2) ⫺ u(t ⫺ 4)) and thus gives the transform

p s ⫹ p2 2

(eⴚ2s ⫺ eⴚ4s).

8. The given function is represented by

t 2(u(t ⫺ 1) ⫺ u(t ⫺ 2)) ⫽ 3(t ⫺ 1)2 ⫹ 2(t ⫺ 1) ⫹ 14u(t ⫺ 1) ⫺ 3(t ⫺ 2)2 ⫹ 4(t ⫺ 2) ⫹ 44u(t ⫺ 2)

and thus has the transform 2 2 1 2 4 4 a 3 ⫹ 2 ⫹ b eⴚs ⫺ a 3 ⫹ 2 ⫹ b eⴚ2s s s s s s s 10. 12 (et ⫺ eⴚt)(1 ⫺ u(t ⫺ 2)) ⫽ 12 (et ⫺ eⴚt) ⫺ 12 (e(tⴚ2)⫹2 ⫺ e(ⴚt⫹2)ⴚ2)u(t ⫺ 2). Hence the transform is 1 # 1 1 1 1 e2 eⴚ2 ⫺ # ⫺ a ⫺ b eⴚ2s. 2 s⫺1 2 s⫹1 2 s⫺1 s⫹1

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Alternatively, by using the addition formula (22) in App. 3.1 we obtain the transform in the form cosh 2 s sinh 2 ⴚ2s ⫺a 2 ⫹ 2 be . s ⫺1 s ⫺1 s ⫺1 1

12. s ⴚ3 has the inverse t 2>2, hence (s ⫺ 1)ⴚ3 has the inverse ett 2>2 (first shifting), and eⴚ3s>(s ⫺ 1)3 has the inverse 12 etⴚ3(t ⫺ 3)2u(t ⫺ 3) (second shifting). 14. 4u(t ⫺ 2) ⫺ 8u(t ⫺ 5) 16. (sinh (2t ⫺ 2))u(t ⫺ 1) ⫺ (sinh (2t ⫺ 6))u(t ⫺ 3) 18. (9s 2 ⫺ 6s ⫹ 1)Y ⫽ 27s ⫹ 9 ⫺ 18 ⫽ 27s ⫺ 9. Hence Y⫽

27s ⫺ 9 9s ⫺ 6s ⫹ 1 2

⫽

3s ⫺ 1 (s ⫺ 13) 2

⫽

3 s ⫺ 13

Hence the answer is y ⫽ 3et>3. 19 190 288 s⫺5⫹ . Division by 3 ⫹ 12 12 s s 2 ⫹10s ⫹ 24 and expansion in terms of partial fractions gives

20. (s 2 ⫹ 10s ⫹ 24)Y ⫽ (s ⫹ 4)(s ⫹ 6)Y ⫽

Y⫽

12 5 12 ⫺ 2⫹ . s s3 s

Hence the answer is y ⫽ 6t 2 ⫺ 5t ⫹ 19 12 . Note that it does not contain a contribution from the general solution of the homogeneous ODE. 22. In terms of unit step functions, the function on the right is r(t) ⫽ 4t31 ⫺ u(t ⫺ 1)4 ⫹ 8u(t ⫺ 1) ⫽ 4t ⫺ 34(t ⫺ 1) ⫺ 44u(t ⫺ 1). Answer: y⫽ •

4eⴚt ⫺ eⴚ2t ⫹ 2t ⫺ 3

if 0 ⬍ t ⬍ 1

(4 ⫺ 8e)eⴚt ⫹ (3e2 ⫺ 1)eⴚ2t ⫹ 4

t ⬎ 1.

24. The subsidiary equation is 1 eⴚs (s 2 ⫹ 3s ⫹ 2)Y ⫽ s ⫺ s . It has the solution Y⫽

1 ⫺ eⴚs 1 1 1 ⫽a ⫹ ⫺ b (1 ⫺ eⴚs). 2s 2(s ⫹ 2) s⫹1 s(s ⫹ 3s ⫹ 2) 2

This gives the answer y ⫽ 12 ⫹ 12 eⴚ2t ⫺ eⴚt ⫺ (12 ⫹ 12 eⴚ2(tⴚ1) ⫺ eⴚ(tⴚ1))u(t ⫺ 1); that is, y⫽ •

1 1 ⴚ2t ⫺ eⴚt 2 ⫹ 2e

if 0 ⬍ t ⬍ 1

1 ⴚ2t (1 ⫺ e2) ⫺ eⴚt(1 ⫺ e) 2e

t ⬎ 1.

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26. t ⫽ p ⫹ ~ t , ~y s ⫹ 2y~ r ⫹ 5y~ ⫽ r~, r~ ⫽ 103⫺1 ⫹ u(t~ ⫺ p)4 sin ~ t , ~y (0) ⫽ 1, ~y r (0) ⫽ ~ ⴚp ⫺2 ⫹ 2e . The solution in terms of t is ~y (t~) ⫽ eⴚtⴚp sin 2t~ ⫹ cos ~ t ⫺ 2 sin ~ t ~ ~ ~ ⴚt ⫹p ⫹ u(t ⫺ p)3e (⫺cos 2t ⫹ 1 sin 2t~) ⫺ cos ~ t ⫹ 2 sin ~ t 4. 2

In terms of t, y(t) ⫽ eⴚt sin 2t ⫺ cos t ⫹ 2 sin t ⫹ u(t ⫺ 2p)3eⴚt⫹2p(⫺cos 2t ⫹ 12 sin 2t) ⫹ cos t ⫺ 2 sin t4. 28. i r ⫹ 1000i ⫽ 40 sin tu(t ⫺ p). The subsidiary equation is sI ⫹ 1000I ⫽ ⫺40

eⴚps s2 ⫹ 1

Its solution is I⫽

⫺40eⴚps ⫺40eⴚps 1 s ⫺ 1000 a ⫺ b. ⫽ 1,000,001 s ⫹ 1000 s⫹1 (s 2 ⫹ 1)(s ⫹ 1000)

The inverse transform is i ⫽ u(t ⫺ p) c ⫺

40,000 40 (cos t ⫹ e ⴚ1000(tⴚp)) ⫹ sin t d ; 1,000,001 1,000,001

hence i ⫽ 0 if t ⬍ p, and i ⬇ 0.04 sin t if t ⬎ p. 30. (0.5i r ⫹ 10i ⫽ 200t(1 ⫺ u(t ⫺ 2)). The subsidiary equation is (0.5s ⫹ 10)I(s) ⫽ 200(1 ⫺ eⴚ2s(1 ⫹ 2s))>s 2. Its solution is I ⫽ 400(⫺1 ⫹ eⴚ2s ⫹ 2seⴚ2s)>(s 2(s ⫹ 20)). The solution of the problem (the current in the circuit) is i ⫽ eⴚ20t ⫹ 20t ⫺ 1 ⫹ u(t ⫺ 2)3⫺20t ⫹ 1 ⫹ 39eⴚ20(tⴚ2)4. 32. We obtain I ⫽ 14 # 105 hence

seⴚ4sⴚ12 2 # 106 6 # 105 ⴚ4sⴚ12 ⫽a ⫺ be (s ⫹ 10)(s ⫹ 3) s ⫹ 10 s⫹3

i ⫽ u(t ⫺ 4) 32 # 106eⴚ10(tⴚ4)ⴚ12 ⫺ 6 # 105eⴚ3(tⴚ4)ⴚ124. t

34. 10i ⫹ 100

冮 i(t) dt ⫽ 100(u(t ⫺ 0.5) ⫺ u(t ⫺ 0.6)). Divide by 10 and take the 0

transform, using Theorem 3 in Sec. 6.2, 10 10 I ⫹ s I ⫽ s (eⴚ0.5s ⫺ eⴚ0.6s). Solving for I ⫽ l(i) gives I⫽

10 (eⴚ0.5s ⫺ eⴚ0.6s). s ⫹ 10

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The inverse transform is i(t) ⫽ 10(e ⴚ10(tⴚ0.5) u(t ⫺ 0.5) ⫺ e ⴚ10(tⴚ0.6) u(t ⫺ 0.6)). Hence i(t) ⫽ 0 i(t) ⫽ 10e

ⴚ10(tⴚ0.5)

t ⬍ 0.5

0.5 ⬍ t ⬍ 0.6

t ⬎ 0.6.

i(t) ⫽ 10(eⴚ10(tⴚ0.5) ⫺ eⴚ10(tⴚ0.6)) ⫽ 10eⴚ10t(e5 ⫺ e6) ⫽ ⫺2550eⴚ10t

Jumps occur at t ⫽ 0.5 (upward) and at t ⫽ 0.6 (downward) because the right side has those jumps and the term involving the integral (representing the charge on the capacitor) cannot change abruptly; hence the first term, Ri(t), must jump by the amounts of the jumps on the right, which have size 100, and since R ⫽ 10, the current has jumps of size 10. 36. i s ⫹ 4i ⫽ 200(1 ⫺ t 2)(1 ⫺ u(t ⫺ 1)). Observing that t 2 ⫽ (t ⫺ 1)2 ⫹ 2(t ⫺ 1) ⫹ 1, we obtain the subsidiary equation 1 2 2 2 (s 2 ⫹ 4)I ⫽ 200 a ⫺ 3 b ⫹ 200eⴚs a 2 ⫹ 3 b . s s s s Its solution is I ⫽ 200

s 2 ⫺ 2 ⫹ eⴚs(2s ⫹ 2) s 3(s 2 ⫹ 4)

4 3s 3 ⫽ 25 a ⫺ 3 ⫺ 2 b s s s ⫹4 ⫹ 25eⴚs a⫺

1 4 4 s⫺4 ⫹ 2⫹ 3⫹ 2 b. s s s s ⫹4

Its inverse transform is i ⫽ 75 ⫺ 50t 2 ⫺ 75 cos 2t ⫹ u(t ⫺ 1)3⫺75 ⫹ 50t 2 ⫹ 25 cos (2t ⫺ 2) ⫺ 50 sin (2t ⫺ 2)4. t

38. i r ⫹ 4i ⫹ 20

冮 i(t) dt ⫽ 34e (1 ⫺ u(t ⫺ 4)). The subsidiary equation is ⴚt

20 34 as ⫹ 4 ⫹ s b I ⫽ s ⫹ 1 (1 ⫺ eⴚ4sⴚ4). Its solution is I⫽

34s (s ⫹ 1)(s ⫹ 4s ⫹ 20) 2

(1 ⫺ eⴚ4sⴚ4).

The inverse transform is i ⫽ ⫺2eⴚt ⫹ eⴚ2t(2 cos 4t ⫹ 9 sin 4t) ⫹ u(t ⫺ 4)32eⴚt ⫺ e2(2ⴚt)(9 sin 4(t ⫺ 4) ⫹ 2 cos 4(t ⫺ 4))4.

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40. The model t

i r ⫹ 2i ⫹ 10

冮 i(t) dt ⫽ 255(1 ⫺ u(t ⫺ 2p)) sin t. 0

This gives the subsidiary equation c (s ⫹ 2) ⫹

10 255 (1 ⫺ eⴚ2ps). dI ⫽ 2 s s ⫹1

Its solution is I⫽

255s(1 ⫺ eⴚ2ps) (s 2 ⫹ 2s ⫹ 10)(s 2 ⫹ 1)

⫽a

27s ⫹ 6 s ⫹1 2

⫺

27(s ⫹ 1) ⫹ 33 (s ⫹ 1)2 ⫹ 9

b (1 ⫺ eⴚ2ps).

The inverse transform (the solution of the problem) is i ⫽ 27 cos t ⫹ 6 sin t ⫺ eⴚt(27 cos 3t ⫹ 11 sin 3t) ⫹ u(t ⫺ 2p)3eⴚ(tⴚ2p)(27 cos 3t ⫹ 11 sin 3t) ⫺ (27 cos t ⫹ 6 sin t)4. SECTION 6.4. Short Impulses. Dirac’s Delta Function. Partial Fractions, page 225 Purpose. Modeling of short impulses by Dirac’s delta function (also known as unit impulse function). The text includes a remark that this is not a function in the usual sense of calculus but a “generalized function” or “distribution.” Details cannot be discussed on the level of this book; they can be found in books on functional analysis or on PDEs. See, e.g., L. Schwartz, Mathematics for the Physical Sciences, Paris: Hermann, 1966. The French mathematician LAURENT SCHWARTZ (1915–2002) created and popularized the theory of distributions. See also footnote 2. Main Content Definition of Dirac’s delta (3) Sifting property (4) Transform of delta (5) Application to mass–spring systems and electric networks More on partial fractions (Example 4) For the beginning of the discussion of partial fractions in the present context, see Sec. 6.2. Problem Set 6.4 CAS Project 1 involves exploring the nature of solutions with the damping or spring constant being changed in the presence of one or two Dirac delta functions on the right. CAS Experiment 2 models a wave of constant area acting for shorter and shorter times. Problems 3–12 model vibrating systems with driving forces consisting of unit step functions, Dirac’s delta functions, and other functions. Problems 14–15 concern rectifiers, sawtooth waves, and staircase functions.

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SOLUTIONS TO PROBLEM SET 6.4, page 230 2. CAS Experiment. Students should become aware that careful observation of graphs may lead to discoveries or to more information about conjectures that they may want to prove or disprove. The curves branch from the solution of the homogeneous ODE at the instant at which the impulse is applied, which by choosing, say a ⫽ 1, 2, 3, Á , gives an interesting joint graph. 4. The subsidiary equation is (s 2 ⫹ 16)Y ⫽ 2s ⫹ 4eⴚ3ps where 2s comes from y(0). The solution is Y⫽

2s ⫹ 4eⴚ3ps s 2 ⫹ 16

The inverse transform (the solution of the IVP) is y ⫽ 2 cos 4t ⫹ u(t ⫺ 3p) sin 4t. 2 1 0

8 t

⫺1

⫺2

Section 6.4. Problem 4

6. The subsidiary equation is (s 2 ⫹ 4s ⫹ 5)Y ⫽ 3 ⫹ eⴚs where 3 comes from y r (0). The solution is Y⫽

3 ⫹ eⴚs (s ⫹ 2)2 ⫹ 1

The inverse transform (the solution of the initial value problem) is y ⫽ 3eⴚ2t sin t ⫹ u(t ⫺ 1)eⴚ2(tⴚ1) sin (t ⫺ 1). 8. The subsidiary equation is (s 2 ⫹ 3s ⫹ 2)Y ⫽ s ⫺ 1 ⫹ 3 ⫹

10 s ⫹1 2

⫹ 10eⴚs.

In terms of partial fractions, its solution is Y⫽

⫺2 6 3s ⫺ 1 1 1 ⫹ ⫺ 2 ⫹ 10 a ⫺ b eⴚs. s ⫹1 s⫹2 s⫹1 s⫹1 s⫹2

Its inverse transform is y ⫽ ⫺2eⴚ2t ⫹ 6eⴚt ⫺ 3 cos t ⫹ sin t ⫹ 10u(t ⫺ 1)3eⴚt⫹1 ⫺ eⴚ2(tⴚ1)4.

Without the d-term, the solution is ⫺3 cos t ⫹ sin t ⫺ 2eⴚ2t ⫹ 6eⴚt and approaches a harmonic oscillation fairly soon. With the d-term the first half-wave has a maximum amplitude of about 5, but from about t ⫽ 8 or 10 on its graph practically coincides

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with the graph of that harmonic oscillation (whose maximum amplitude is 110). This is physically understandable, since the system has damping that eventually consumes the additional energy due to the d-term. 10. The subsidiary equation is (s 2 ⫹ 5s ⫹ 6)Y ⫽ eⴚps>2 ⫺

s s ⫹1 2

eⴚps.

Its solution is Y⫽a

0.1(s ⫹ 1) ⴚps 1 1 0.4 0.3 ⫺ b eⴚps>2 ⫺ a⫺ ⫹ ⫹ be . s⫹2 s⫹3 s⫹2 s⫹3 s2 ⫹ 1

The inverse transform of Y is y ⫽ u(t ⫺ 12 p) 3eⴚ2t⫹p ⫺ eⴚ3t⫹3p>24 ⫺ 0.1u(t ⫺ p) 3⫺4eⴚ2t⫹2p ⫹ 3eⴚ3t⫹3p ⫺ cos t ⫺ sin t4. This solution is zero from 0 to 12 p and then increases rapidly. Its first negative halfwave has a smaller maximum amplitude (about 0.1) than the continuation as a harmonic oscillation with maximum amplitude of about 0.15. 0.15 0.1 0.05 0

⫺0.05

⫺0.1

Section 6.4. Problem 10

12. The subsidiary equation is (s 2 ⫹ 2s ⫹ 5)Y ⫽ 1 ⫺ 2s ⫹

25 s2

⫺ 100eⴚps.

Its solution is Y⫽

⫺s 2 ⫹ 2s 3 ⫺ 25 ⫹ 100s 2eⴚps s 2((s ⫹ 1)2 ⫹ 4)

Its inverse transform (the solution of the IVP) is y ⫽ 5t ⫺ 2 ⫺ 50u(t ⫺ p)eⴚt⫹p sin 2t. This is essentially a straight line, sharply deformed between p and about 8. 40 30 20 10 0 ⫺10

6 t

Section 6.4. Problem 12

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14. TEAM PROJECT. (a) If f (t) is piecewise continuous on an interval of length p, then its Laplace transform exists, and we can write the integral from zero to infinity as the series of integrals over successive periods: ⬁

l( f ) ⫽

冮 e f (t) dt ⫽ 冮 e f dt ⫹ 冮 e f dt ⫹ 冮 e f dt ⫹ Á . ⴚst

ⴚst

If we substitute t ⫽ t ⫹ p in the second integral, t ⫽ t ⫹ 2p in the third integral, Á , t ⫽ t ⫹ (n ⫺ 1)p in the nth integral, Á , then the new limits in every integral are 0 and p. Since f (t ⫹ p) ⫽ f (t),

f (t ⫹ 2p) ⫽ f (t),

etc., we thus obtain l( f ) ⫽

冮

eⴚstf (t) dt ⫹

冮

冮e

eⴚs(t⫹p)f (t) dt ⫹

ⴚs(t⫹2p)

f (t) dt ⫹ Á .

The factors that do not depend on t can be taken out from under the integral signs; this gives l( f ) ⫽ 31 ⫹ e

ⴚsp

⫹e

⫹ Á4

ⴚ2sp

冮 e f (t) dt. ⴚst

The series in brackets 3 Á 4 is a geometric series whose sum is 1>(1 ⫺ eⴚps). The theorem now follows. (b) From (11) we obtain l( f ) ⫽

p>v

1 1⫺e

ⴚ2ps>v

冮 e

ⴚst

sin vt dt.

Using 1 ⫺ eⴚ2ps>v ⫽ (1 ⫹ eⴚps>v)(1 ⫺ eⴚps>v) and integrating by parts or noting that the integral is the imaginary part of the integral p>v

冮 e 0

(ⴚs⫹iv)t

p>v

1 dt ⫽ e(ⴚs⫹iv)t ` ⫺s ⫹ iv 0

⫽

⫺s ⫺ iv (⫺eⴚsp>v ⫺ 1) s 2 ⫹ v2

we obtain the result. (c) From (11) we obtain the following equation by using sin vt from 0 to p>v and ⫺sin vt from p>v to 2p>v: v

1 ⫹ eps>v

s 2 ⫹ v2 eps>v ⫺ 1

⫽ ⫽

eⴚps>2v ⫹ eps>2v

s 2 ⫹ v2 eps>2v ⫺ eⴚps>2v cosh (ps>2v)

v s ⫹v 2

ps>2v)

2 sinh (

This gives the result. (d) The sawtooth wave has the representation k f (t) ⫽ p t

if 0 ⬍ t ⬍ p,

f (t ⫹ p) ⫽ f (t).

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Integration by parts gives

冮

t eⴚstt dt ⫽ ⫺ s eⴚst `

p 0

1 ⫹ s

冮e

ⴚst

p 1 ⫽ ⫺ eⴚsp ⫺ 2 (eⴚsp ⫺ 1) s s and thus from (11) we obtain the result l( f ) ⫽

k keⴚps ⫺ ps 2 s(1 ⫺ eⴚps)

(s ⬎ 0).

SECTION 6.5. Convolution. Integral Equations, page 232 Purpose. To find the inverse h(t) of a product H(s) ⫽ F(s)G(s) of transforms whose inverses are known. Main Content, Important Concepts Convolution f * g, its properties Convolution theorem Application to ODEs and integral equations Comment on Occurrence In an ODE the transform R(s) of the right side r(t) is known from Step 1. Solving the subsidiary equation algebraically for Y(s) causes the transform R(s) to be multiplied by the reciprocal of the factor of Y(s) on the left (the transfer function Q(s); see Sec. 6.2). This calls for the convolution theorem, unless one sees some other way or shortcut. Very Short Courses. This section can be omitted. Problem Set 6.5 This set concerns integrations needed to obtain convolutions (Probs. 1–7), the use of the latter in solving a certain class of integrations (Probs. 8–14), the effect of varying a parameter in an integral equation (CAS Experiment 15), a problem (Prob. 16) on general properties of convolution. Problems 17–26 show how to obtain inverse transforms by evaluating integrals that define convolutions.

SOLUTIONS TO PROBLEM SET 6.5, page 237 t

2. 1 * sin vt ⫽

cos vt t

冮 sin vt dt ⫽ ⫺ v ` ⫽ 1 ⫺ vcos vt 0

4. We obtain

冮

1 cos vt cos (vt ⫺ vt) dt ⫽ 2 ⫽

1 c t cos vt ⫹ 2

冮 3cos vt ⫹ cos (2vt ⫺ vt)4 dv 0

sin vt ⫺ sin (⫺vt) 2v

d ⫽

1 1 sin vt. t cos vt ⫹ 2 2v

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6. eat * ebt ⫽

冮

eateb(tⴚt) dt ⫽ ebt

冮e

(aⴚb)t

dt ⫽

eat ⫺ ebt a⫺b

8. The integral equation can be written y(t) ⫹ 4y(t) * t ⫽ 2t. By the convolution theorem it has the transform Y ⫹ 4Y>s 2 ⫽ 2>s 2. The solution is Y ⫽ 2>(s 2 ⫹ 4). Its inverse transform is y ⫽ sin 2t. 10. In terms of convolution the given integral equation is y(t) ⫺ y(t) * sin 2t ⫽ sin 2t. By the convolution theorem its transform is Y ⫺ 2Y>(s 2 ⫹ 4) ⫽ 2>(s 2 ⫹ 4). Multiplication by s 2 ⫹ 4 gives Y(s 2 ⫹ 2) ⫽ 2;

thus

Y⫽

2 s ⫹2 2

The inverse transform (the solution of the integral equation) is y ⫽ 12 sin (t 12). 12. The integral equation can be written y(t) ⫹ y(t) * cosh t ⫽ t ⫹ et. This implies, by the convolution theorem, that its transform is Y⫹

s 1 1 Y⫽ 2⫹ . s⫺1 s2 ⫺ 1 s

The solution is Y⫽

1 s2 ⫺ 1 1 1 1 b⫽ 2⫹ . a ⫹ s⫺1 s s ⫹ s ⫺ 1 s2 s 2

Hence its inverse transform gives the answer y(t) ⫽ t ⫹ 1. This result can easily be checked by substitution into the given equation and integration. 14. Y a1 ⫺

1 2 1 b ⫽ ⫺ 3 , hence s s2 s Y⫽

The answer is y ⫽ 1 ⫹ cosh t.

2s 2 ⫺ 1 s 1 ⫽ ⫹ 2 . 2 s s(s ⫺ 1) s ⫺1

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16. Team Project. (a) Setting t ⫺ t ⫽ p, we have t ⫽ t ⫺ p, dt ⫽ ⫺dp, and p runs from t to 0; thus f*g⫽

冮

f (t)g(t ⫺ t) dt ⫽

冮 g( p) f (t ⫺ p)(⫺dp) t

0 t

⫽

冮 g( p) f (t ⫺ p) dp ⫽ g * f. 0

(b) Interchanging the order of integration and noting that we integrate over the shaded triangle in the figure, we obtain ( f * g) * v ⫽ v * ( f * g) ⫽

⫽

tⴚp

冮 v( p) 冮 f (t)g(t ⫺ p ⫺ t) dt dp 冮

tⴚt

f (t)

冮 g(t ⫺ t ⫺ p)v( p) dp dt 0

⫽ f * (g * v).

t p=t– =t–p

Section 6.5. Team Project 16(b)

(d) Let t ⬎ k. Then ( fk * f )(t) ⫽

冮 1k f (t ⫺ t) dt ⫽ f (t ⫺ ~t ) for some ~t between 0 0

and k. Now let k : 0. Then ~ t : 0 and fk(t ⫺ ~ t ) : d(t), so that the formula follows. 2 (e) s Y ⫺ sy(0) ⫺ y r (0) ⫹ v2Y ⫽ l(r) has the solution Y⫽

y r (0) 1 v s v a b l(r) ⫹ y(0) 2 2 ⫹ 2 v s 2 ⫹ v2 v s ⫹v s ⫹ v2

etc. t

18. e * e

⫽

冮a e

at a(tⴚt)

dt ⫽ e

冮 dt ⫽ te 0

20. 9 * eⴚ3t ⫽ 9

冮e 0

ⴚ3t

dt ⫽ 3 ⫺ 3eⴚ3t

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22. u(t ⫺ a) * e2t ⫽ t⬍a

冮

e2(tⴚt) dt ⫽ e2t

冮e

ⴚ2t

dt ⫽

1 2(tⴚa) (e ⫺ 1) if t ⬎ a and 0 if 2

24. 48 (sin t) * (sin 5t) ⫽ 48

冮 sin t sin 5(t ⫺ t) dt. Using formula (11) in App. 3.1, convert 0

the product in the integrand to a sum and integrate, obtaining t

冮 3⫺cos (5t ⫺ 4t) ⫹ cos (t ⫺ 5t ⫹ 5t)4 dt 0

⫽ 24[⫺ 14 sin (⫺5t ⫹ 4t) ⫹ 16 sin (6t ⫺ 5t)] `

t 0

⫽ 3(sin t ⫺ sin 5t) ⫹ 2(sin t ⫹ sin 5t) ⫽ 10 sin t ⫺ 2 sin 5t.

SECTION 6.6. Differentiation and Integration of Transforms. ODEs with Variable Coefficients, page 238 Purpose. To show that, roughly, differentiation and integration of transforms (not of functions, as before!) correspond to multiplication and division, respectively, of functions by t, with application to the derivation of further transforms and to the solution of Laguerre’s differential equation. Comment on Application to Variable-Coefficient Equations This possibility is rather limited; our Example 3 is perhaps the best elementary example of practical interest. Very Short Courses. This section can be omitted. Problem Set 6.6 Problems 2–11 concern single or twofold applications of differentiation with respect to s, as the only method wanted for solving these problems, whereas in Probs. 14–20 the student has first to select a suitable one among several possible methods suggested. Problem 13 is an invitation to study Laguerre polynomials in somewhat more detail, in particular, to compare the locations of the extrema depending on the parameter n.

SOLUTIONS TO PROBLEM SET 6.6, page 241 4 24s r 2. F(s) ⫽ ⫺3 a 2 b ⫽ 2 s ⫺ 16 (s ⫺ 16)2 4. We have l(eⴚt cos t) ⫽

s⫹1 (s ⫹ 1) ⫹ 1 2

⫽

s⫹1 s ⫹ 2s ⫹ 2 2

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Differentiation and simplification gives the answer s⫹1

F r (s) ⫽ ⫺ a 2 s ⫹ 2s ⫹ 2

s ⫹ 2s ⫹ 2 ⫺ (s ⫹ 1)(2s ⫹ 2) r b ⫽⫺ 2

(s 2 ⫹ 2s ⫹ 2)2

⫽

s 2 ⫹ 2s (s 2 ⫹ 2s ⫹ 2)2

6. We need two differentiations. We can drop the two minus signs. Starting from the transform of sin 3t, we obtain 3 ⫺6s s r b ⫽a 2 a 2 b s ⫹9 (s ⫹ 9)2 ⫺6(s 2 ⫹ 9)2 ⫹ 6s # 2(s 2 ⫹ 9) # 2s ⫽ (s 2 ⫹ 9)4 ⫺6s 2 ⫺ 54 ⫹ 24s 2 ⫽ (s 2 ⫹ 9)3 18(s 2 ⫺ 3) ⫽ 2 . (s ⫹ 9)3 8. ⫺ a

2(s ⫹ k)

1 (s ⫹ k) ⫹ 1 2

10. l(t nekt) ⫽

b ⫽

((s ⫹ k)2 ⫹ 1)2

can be obtained from l(ekt) ⫽

(s ⫺ k) differentiations,

n⫹1

(n)

1 a b s⫺k

(nⴚ1)

⫺1 ⫽a b (s ⫺ k)2

⫽ Á ⫽

1 by n subsequent s⫺k

(⫺1)nn! (s ⫺ k)n⫹1

and multiplication by (⫺1)n (to take care of the minus sign in (1) in each of the n steps), or much more simply, by the first shifting theorem, starting from l(t n) ⫽ n!>s n⫹1. 12. CAS Project. Students should become aware that usually there are several possibilities for calculations, and they should not rush into numeric work before carefully selecting formulas. (b) Use the usual rule for differentiating a product n times. Some of the Laguerre polynomials are l 2 ⫽ 1 ⫺ 2t ⫹ 12 t 2 l 3 ⫽ 1 ⫺ 3t ⫹ 32 t 2 ⫺ 16 t 3 1 4 l 4 ⫽ 1 ⫺ 4t ⫹ 3t 2 ⫺ 23 t 3 ⫺ 24 t 5 3 5 4 1 5 2 l 5 ⫽ 1 ⫺ 5t ⫹ 5t ⫺ 3 t ⫺ 24 t ⫺ 120 t . 14. By differentiation we have ⫺8s 4 r . b ⫽ 2 a 2 s ⫹ 16 (s ⫹ 16)2

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Hence the answer is 18 t sin 4t. By integration we see that

冮

⬁

2 ds~ ⫽ 2 2 2 ~ s ⫹ 16 s (s ⫹ 16)

has the inverse transform 18 sin 4t and gives the same answer. By convolution, t

(cos 4t) * (14 sin 4t) ⫽

冮 cos 4t sin (4t ⫺ 4t) dt 0

and gives the same answer. 16. By (6),

冮

⬁

f 2s~ ⫹ 6 1 ds~ ⫽ 2 ⫽ la b. 2 2 ~ ~ t (s ⫹ 6s ⫹ 10) s ⫹ 6s ⫹ 10

The inverse transform of the integral is eⴚ3t sin t. Answer: teⴚ3t sin t. s r 18. aarccot p b ⫽ ⫺

20. ln

1> p 2

1 ⫹ apb s

⫽

⫺p s ⫹ p2 2

s⫹a ⫽ ln (s ⫹ a) ⫺ ln (s ⫹ b) ⫽ ⫺ s⫹b

shows that the answer is (sin pt)>t.

冮

⬁

ds ⫹ s⫹a

⬁

冮 sds⫹ b . This shows that the s

answer is (⫺eⴚat ⫹ eⴚbt)>t.

SECTION 6.7. Systems of ODEs, page 242 Purpose. This section explains the application of the Laplace transform to systems of ODEs in terms of three typical examples: a mixing problem, an electrical network, and a system of several (two) masses on elastic springs. Note that Examples 1 and 2 in the text concern first-order systems, whereas the system in Example 3 is of second order. Problem Set 6.7 Team Project 1 concerns Examples 1 and 2 of Sec. 4.1, namely, a comparison of the usual method and the present transform method. Problems 2–10 involve IVPs for first-order systems, Probs. 11–13 second-order systems, and Probs. 14–15 systems of three ODEs.

SOLUTIONS TO PROBLEM SET 6.7, page 246 2. The subsidiary equations are sY1 ⫹ Y2 ⫽ 1 Y1 ⫹ sY2 ⫽

2s s ⫹1 2

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The solution is Y1 ⫽

s s ⫹1 2

Y2 ⫽

1 s ⫹1 2

The inverse transform is y1 ⫽ cos t, y2 ⫽ sin t. 4. The subsidiary equations are sY1 ⫽ 4Y2 ⫺

s ⫹ 16 36 sY2 ⫽ 3 ⫺ 3Y1 ⫺ 2 . s ⫹ 16 2

The solution is 4

Y1 ⫽

s ⫹ 16 2

Y2 ⫽

3s s ⫹ 16 2

The inverse transform is y1 ⫽ sin 4t, y2 ⫽ 3 cos 4t. 6. The subsidiary equations are sY1 ⫺ 1 ⫽ 5Y1 ⫹ Y2 sY2 ⫹ 3 ⫽ Y1 ⫹ 5Y2. The solution is Y1 ⫽ Y2 ⫽

s⫺8 (s ⫺ 5) ⫺ 1 2

⫺3s ⫹ 16 (s ⫺ 5)2 ⫺ 1

⫽ ⫽

(s ⫺ 5) ⫺ 3 (s ⫺ 5)2 ⫺ 1 ⫺3(s ⫺ 5) ⫹ 1 (s ⫺ 5)2 ⫺ 1

The inverse transform is y1 ⫽ e5t cosh t ⫺ 3e5t sinh t ⫽ 2e4t ⫺ e6t y2 ⫽ ⫺3e5t cosh t ⫹ e5t sinh t ⫽ ⫺2e4t ⫺ e6t. 8. The subsidiary equations are sY1 ⫽ 4 ⫺ 2Y1 ⫹ 3Y2 sY2 ⫽ 3 ⫹ 4Y1 ⫺ Y2. The solution is Y1 ⫽

4s ⫹ 13 3 1 ⫽ ⫹ s⫺2 s⫹5 s ⫹ 3s ⫺ 10

Y2 ⫽

3s ⫹ 22 4 1 ⫽ ⫺ . s⫺2 s⫹5 s 2 ⫹ 3s ⫺ 10

The inverse transform is y1 ⫽ 3e2t ⫹ eⴚ5t,

y2 ⫽ 4e2t ⫺ eⴚ5t.

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10. The subsidiary equations are sY1 ⫽ 1 ⫺ Y2 sY2 ⫽ ⫺Y1 ⫹

2s s2 ⫹ 1

⫺

2eⴚ2pss s2 ⫹ 1

The solution is Y1 ⫽ ⫽

s(s 2 ⫺ 1 ⫹ 2eⴚ2ps) s4 ⫺ 1 s s2 ⫹ 1

⫹

2seⴚ2ps s4 ⫺ 1 1

⫽ Y2 ⫽ ⫽

s 2 2 s ⫹ eⴚ2ps a ⫹ ⫺ 2 b s⫺1 s⫹1 s2 ⫹ 1 s ⫹1 s 2 ⫺ 1 ⫺ 2s 2eⴚ2ps s4 ⫺ 1 1 1 1 1 1 ⫺ b⫺ 2 ⫹ eⴚ2ps a a b. 2 s⫹1 s⫺1 s2 ⫹ 1 s ⫹1

The inverse transform is y1 ⫽ cos t ⫹ u(t ⫺ 2p) 3⫺cos t ⫹ 12 (eⴚt⫹2p ⫹ etⴚ2p)4 y2 ⫽ sin t ⫹ u(t ⫺ 2p) 3⫺sin t ⫹ 12 (eⴚt⫹2p ⫺ etⴚ2p)4. Thus y1 ⫽ cos t and y2 ⫽ sin t if 0 ⬍ t ⬍ 2p, y1 ⫽ cosh (t ⫺ 2p), and y2 ⫽ ⫺sinh (t ⫺ 2p) if t ⬎ 2p. 12. The subsidiary equations are s 2Y1 ⫽ s ⫺ 2Y1 ⫹ 2Y2 s 2Y2 ⫽ 3s ⫹ 2Y1 ⫺ 5Y2. The solution is Y1 ⫽ Y2 ⫽

s(s 2 ⫹ 11) s 4 ⫹ 7s 2 ⫹ 6 s(3s 2 ⫹ 8) s ⫹ 7s ⫹ 6 4

⫽ ⫽

2s s2 ⫹ 1 s s ⫹1 2

⫺ ⫹

s s2 ⫹ 6 2s s ⫹6 2

Hence the inverse transform is y1 ⫽ 2 cos t ⫺ cos t 16,

y2 ⫽ cos t ⫹ 2 cos t16.

14. The subsidiary equations are 4sY1 ⫺ 8 ⫹ sY2 ⫺ 2sY3 ⫽ 0 ⫺2sY1 ⫹ 4

1 ⫹ sY3 ⫽ s 16 2sY2 ⫺ 4sY3 ⫽ ⫺ 2 . s

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The solution is 1 1 Y1 ⫽ 2 a ⫹ 3 b , s s

Y2 ⫽

2 , s2

Y3 ⫽

1 4 2 ⫹ 3. s s

The inverse transform is y1 ⫽ 2 ⫹ t 2,

y2 ⫽ 2t,

y3 ⫽ t ⫹ 2t 2.

16. The subsidiary equations are s 2Y1 ⫺ s ⫺ 1 ⫽ ⫺8Y1 ⫹ 4Y2 ⫹ s 2Y2 ⫺ s ⫹ 1 ⫽ ⫺8Y2 ⫹ 4Y1 ⫺

11 s ⫹1 2

s ⫹1 2

The solution, in terms of partial fractions, is Y1 ⫽ Y2 ⫽

s s ⫹4 2

⫹ ⫺

1 s ⫹1 2

The inverse transform is y1 ⫽ cos 2t ⫹ sin t y2 ⫽ cos 2t ⫺ sin t. 18. The new salt contents are y1 ⫽ 100 ⫺ 62.5eⴚ0.24t ⫺ 37.5eⴚ0.08t y2 ⫽ 100 ⫹ 125eⴚ0.24t ⫺ 75eⴚ0.08t. Setting 2t ⫽ t gives the old solution, except for notation. 20. For 0 ⬉ t ⬉ 2p the solution is as in Prob. 19; for i 1 we have i 1 ⫽ ⫺26eⴚ2t ⫺ 16eⴚ8t ⫹ 42 cos t ⫹ 15 sin t. For t ⬎ 2p we have to add to this further terms whose form is determined by this solution and the second shifting theorem, u(t ⫺ 2p)326eⴚ2t⫹4p ⫹ 16eⴚ8t⫹16p ⫺ 42 cos t ⫺ 15 sin t4. The cosine and sine terms cancel, so that i 1 ⫽ ⫺26(1 ⫺ e4p)eⴚ2t ⫺ 16(1 ⫺ e16p)eⴚ8t

t ⬎ 2p.

Similarly, for i 2 we obtain i2 ⫽ •

⫺26eⴚ2t ⫹ 8eⴚ8t ⫹ 18 cos t ⫹ 12 sin t ⫺26(1 ⫺ e4p)eⴚ2t ⫹ 8(1 ⫺ e16p)eⴚ8t.

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SOLUTIONS TO CHAPTER 6 REVIEW QUESTIONS AND PROBLEMS, page 251 12. Use s-shifting to give the answer s⫹1 (s ⫹ 1) ⫹ 16 2

⫺

8 (s ⫹ 1)2 ⫹ 16

14. We have 16t 2 ⫽ 16(t ⫺ 14)2 ⫹ 8(t ⫺ 14) ⫹ 1. The transform is 32 8 1 . 3 ⫹ 2 ⫹ s s s This times eⴚs>4 gives the answer (the transform of the given function). 16. u(t ⫺ 2p) sin t ⫽ u(t ⫺ 2p) sin (t ⫺ 2p). Hence the transform is eⴚ2ps>(s 2 ⫹ 1).

18. The transform of sin vt is v>(s 2 ⫹ v2), and that of cos vt is s>(s 2 ⫹ v2). Hence, by convolution, the given function (use (11) in Appendix 3.1 in integration) (sin vt) * (cos vt) ⫽ 12 t sin vt has the transform vs (s ⫹ v2)2 2

20. 1.25(e4t ⫺ eⴚ2t) ⫽ 2.5et sinh 3t 1 16

1 2

# 18

. Hence 18 eⴚt>2 sin 12 t. s2 ⫹ s ⫹ 2 (s ⫹ 12)2 ⫹ 14 24. The given transform suggests the differentiation 22.

⫽ 1

s r s 2 ⫹ 6.25 ⫺ s # 2s ⫽ ⫺ s 2 ⫺ 6.25 a 2 b ⫽ s ⫹ 6.25 (s 2 ⫹ 6.25)2 (s 2 ⫹ 6.25)2 and shows that the answer is t cos 2.5t. 26.

2s ⫺ 10 s

⫽

2 s

⫺

10 s3

. This shows that the inverse transform is

u(t ⫺ 5) 32(t ⫺ 5) ⫺ 5(t ⫺ 5)24 ⫽ u(t ⫺ 5) 3⫺5t 2 ⫹ 52t ⫺ 1354. 28.

3s s ⫺ 2s ⫹ 2 2

⫽

3(s ⫺ 1) ⫹ 3 (s ⫺ 1)2 ⫹ 1

. Hence the inverse transform is 3et(cos t ⫹ sin t).

30. y ⫽ ⫺cos 4t ⫹ u(t ⫺ p) sin 4t. To see the impact of 4d(t ⫺ p), graph both the solution and the term ⫺cos 4t, perhaps in a short t-interval with midpoint p.

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32. y ⫽ cos 2t ⫹ 123u(t ⫺ p) ⫺ u(t ⫺ 2p)4 sin 2t. The curve has cusps at t ⫽ p and 2p (abrupt changes of the tangent direction). 34. y1 ⫽ 12u(t ⫺ p) sin 2t, y2 ⫽ u(t ⫺ p) cos 2t. Hence y1 is continuous at p, whereas y2 has an upward jump of 1 at that point. 36. y1 ⫽ ⫺6e4t ⫹ 2, y2 ⫽ ⫺3e4t ⫺ 1. Note that, from the given system, we see that y2r ⫽ 12 y1r , so that y2 ⫽ 12 y1 ⫹ const. ⫺0.2 1 ⫺0.4 0.5 ⫺0.6 0

3.5

4.5

⫺0.8

5.5

6.5

t ⫺0.5

⫺1 3

3.05

3.1

3.15 t

3.2

3.25

3.3

⫺1

Problem 32

Problem 30 ⫺4 ⫺4.2 ⫺4.4 ⫺4.6 ⫺4.8 ⫺5 0

0.01

0.02 t

0.03

0.04

Problem 36

38. ⫺k 1y1 is the force in the first spring when extended by an amount y1; it tends to pull m 1 to the left, hence the minus. Similarly, a positive y2 ⫺ y1 causes the spring force k 2(y2 ⫺ y1) in the second spring, pulling m 1 to the right. The spring force ⫺k 2(y2 ⫺ y1) in the second equation pulls m 2 to the left. Finally, a y2 ⬎ 0 causes a compression of the third spring and a spring force ⫺k 3y2 in the negative direction, pushing m 2 to the left. 40. We now have masses m 1, m 2, m 3 in a row, connected to each other by two springs and connected to the two walls at the ends by the other two springs. The two springs pulling or pushing at each of the three masses give two terms in each of the three ODEs m 1 y1s ⫽ ⫺k 1y1 ⫹ k 2( y2 ⫺ y1) m 2 y2s ⫽ ⫺k 2( y2 ⫺ y1) ⫹ k 3( y3 ⫺ y2) m 3 y3s ⫽ ⫺k 3( y3 ⫺ y2) ⫺ k 4 y3. 42. q ⫽ •

1 ⫺ 12 (eⴚt ⫹ cos t ⫹ sin t)

if 0 ⬍ t ⬍ p

1 ⴚp ⫺ 3) cos t ⫺ (eⴚp ⫹ 1) sin t4 2 3(e

ⴚt

44. i 1 ⫽ 2(1 ⫺ e ),

ⴚt

i 2 ⫽ 2e .

t ⬎ p.

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Part B LINEAR ALGEBRA. VECTOR CALCULUS Part B consists of Chap. 7

Linear Algebra: Matrices, Vectors, Determinants. Linear Systems

Chap. 8

Linear Algebra: Matrix Eigenvalue Problems

Chap. 9

Vector Differential Calculus. Grad, Div, Curl

Chap. 10 Vector Integral Calculus. Integral Theorems Hence we have retained the previous subdivision of Part B into four chapters. Chapter 9 is self-contained and completely independent of Chaps. 7 and 8. Thus, Part B consists of two large independent units, namely, Linear Algebra (Chaps. 7, 8) and Vector Calculus (Chaps. 9, 10). Chapter 10 depends on Chap. 9, mainly because of the occurrence of div and curl (defined in Chap. 9) in the Gauss and Stokes theorems in Chap. 10.

CHAPTER 7 Linear Algebra: Matrices, Vectors, Determinants. Linear Systems Changes The order of the material in this chapter and its subdivision into sections has been retained, but various local changes have been made to increase the usefulness of this chapter for applications. SECTION 7.1. Matrices, Vectors: Addition and Scalar Multiplication, page 257 Purpose. Explanation of the basic concepts. Explanation of the two basic matrix operations. The latter derive their importance from their use in defining vector spaces, a fact that should perhaps not be mentioned at this early stage. Its systematic discussion follows in Sec. 7.4, where it will fit nicely into the flow of thoughts and ideas. No Prerequisites. Although most of the students may have some working knowledge on the simplest parts of linear algebra, we make no prerequisites. Main Content, Important Concepts Matrix, square matrix, main diagonal Double subscript notation Row vector, column vector, transposition Equality of matrices Matrix addition Scalar multiplication (multiplication of a matrix by a scalar)

126

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Comments on Important Facts One should emphasize that vectors are always included as special cases of matrices and that those two operations have properties [formulas (3), (4)] similar to those of operations for numbers, which is a great practical advantage. Content and Significance of the Examples Example 1 of the text gives a first impression of the main application of matrices, that is, to linear systems, whose significance and systematic discussion will be explained later, beginning in Sec. 7.3. Example 2 gives a simple application showing the usefulness of matrix addition. Example 3 elaborates on equality of matrices. Examples 4 and 5 concern the two basic algebraic operations of addition and scalar multiplication. Purpose and Structure of Problem Set The questions in Probs. 1–7 should help the student to reflect on the basic concepts in this section. Problems 8–16 should help the student in gaining technical skill. Problems 17–20 show applications in connection with forces in mechanics and with electrical networks. SOLUTIONS TO PROBLEM SET 7.1, page 261 2. 100, 810, 960, 0 4. 4, 0, 1; a11, a22; 4, ⫺1 6. B ⫽ (1>1.609344)A; see inside of the front cover. 8. They illustrate the commutativity (3a), the role (3c) of the zero matrix in addition, and an example of subtraction. Calculation gives 0

D 32

26T ,

⫺6

⫺6.4

⫺16.0

same, B, D⫺23.2

⫺19.8

⫺19.4T .

⫺5.0

1.6

11.8

⫺14

10. They illustrate (4c) and (4b), respectively. Calculation gives 0

D72

60T ,

same, D 55

44T ,

⫺22

⫺36

same.

⫺22

12. The first two expressions illustrate (3b) and (3a). The other two expressions illustrate the role of the zero matrix. Namely, the third expression is simply 4D, whereas the last expression is undefined since 0C is the 3 ⫻ 2 zero matrix and cannot be added to A because A is of size 3 ⫻ 3. Hence the last expression is undefined. Calculation gives 1

8T ,

⫺2

same,

4D,

undefined.

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14. This illustrates proportionality: The second expression equals ⫺4 times the first. Furthermore, it shows that the sum of nonzero vectors can be the zero vector; in mechanics, if these are forces, their resultant is the zero force, indicating that these forces are in equilibrium, their joint action is zero. Calculation gives ⫺20

D 30T , D⫺120T , ⫺10

D0T .

undefined,

16. This illustrates commutativity and the role of zero. Calculation gives ⫺59.55

0 D135T ,

same,

undefined, D⫺253.80T .

119.10

18. From Prob. 17 we have 4.5 u ⫹ v ⫹ w ⫹ p ⫽ 0, p ⫽ ⫺(u ⫹ v ⫹ w) ⫽ D 27.0T . ⫺9.0 20. TEAM PROJECT. (b) The nodal incidence matrices are ⫺1

⫺1

D 1

⫺1

and

⫺1

3 3

2 1

2 4

3 2

3 5

SECTION 7.2. Matrix Multiplication, page 263 Purpose. Matrix multiplication, the third and last algebraic operation, is defined and discussed, with emphasis on its “unusual” properties; this also includes its representation by inner products of row and column vectors.

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The motivation of this multiplication is given by formulas (6)–(8) in connection with linear transformations. Main Content, Important Facts Definition of matrix multiplication (“rows times columns”) Properties of matrix multiplication Matrix products in terms of inner products of vectors Linear transformations motivating the definition of matrix multiplication AB ⫽ BA in general, so the order of factors is important AB ⫽ 0 does not imply A ⫽ 0 or B ⫽ 0 or BA ⫽ 0 (AB)T ⫽ BTAT Short Courses. Products in terms of row and column vectors and the discussion of linear transformations could be omitted. Comment on Notation For transposition, T seems preferable over a prime, which is often used in the literature, but will be needed to indicate differentiation in Chap. 9. Comments on Content and Significance of the Examples in the Text Matrix multiplication is shown for the three possible cases, namely, for products of two matrices (in Example 1), for a matrix times a vector (the most important case in the next sections, in Example 2), and for products of row and column vectors (in Example 3). Most important, matrix multiplication is not commutative (Example 4). Example 5 shows how a matrix product can be expressed in terms of row and column vectors. Example 6 illustrates the computation of matrix products on parallel processors. The operation of transposition (Example 7) transforms, as a special case, row vectors into column vectors and conversely. It is also used in the definition of the very important symmetric and skew-symmetric matrices (Example 8). Further special square matrices are triangular matrices (Example 9), diagonal and scalar matrices, including unit matrices of various sizes n (Example 10). Examples 11–13 show some typical applications. Formula (10d) for the transposition of a product should be memorized. In motivating matrix multiplication by linear transformations, one may also illustrate the geometric significance of noncommutativity by combining a rotation with a stretch in the x-direction in both orders and show that a circle transforms into an ellipse with main axes in the direction of the coordinate axes or rotated, respectively.

SOLUTIONS TO PROBLEM SET 7.2, page 270 2. The 3 ⫻ 3 zero matrix, as follows directly from the definitions. 4. n(n ⫺ 1) ⫹ l, where l is the number of different main diagonal entries (which are all 0), hence 13 when n ⫽ 4. 6. Triangular are U1 ⫹ U2, U1U2, hence U12, which, by transposition, implies the same for L 1 ⫹ L 2, L 1L 2, and L12.

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8. c

d, c

d , etc. Problems 7 and 8 should serve as eye 0 0 b 0 ⫺a ⫺a openers, to see what can happen under multiplication. 10. The entry ckj of (AB)T is cjk of AB, which is Row j of A times Column k of B. On the right, ckj is Row k of BT, hence Column k of B, times Column j of AT, hence Row j of A. 29

⫺4

⫺6

12. D 8

12T ,

D⫺4

12T ,

D⫺6

0T ,

d, c

same

The student should observe, and may prove, that the first expression, AAT, is symmetric for any square matrix. Also, BBT, BTB, and B2 are equal for any square matrix that is symmetric. The 3 ⫻ 3 matrices are important in applications of physics (mechanics, for instance, and geometry). The matrices in this problem set are simple, so that one needs no technology. A has a nice spectrum, ⫺3 of multiplicity 1 and 5 of multiplicity 2. It is a main purpose of Probs. 11–20 that they make the student aware of the various laws that we discussed in the text, to acquire familiarity with them and really to understand what they mean, including traps to be avoided. 10

14. D 0

18T ,

D 0

6T ,

⫺9

⫺5

16. D 3 4

18. [1],

10 same, D⫺2T 27

⫺1T , undefined,

D⫺8T , [0

⫺8 2]

0 3

⫺6

D 1

⫺2

0T ,

⫺1

0 [8

⫺4

⫺9],

D⫺8T 2

20. [7], [17]; these are quadratic forms in the components of the vectors b and a, respectively, to be discussed in the next chapter (in Sec. 8.4). The other two matrix 12] and (and vector) products are [⫺3 ⫺24

⫺10

Equality of the two rows occurs just by chance since C Tb ⫽ c

5 5

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24. M ⫽ AB ⫺ BA must be the 2 ⫻ 2 zero matrix. It has the form M⫽ c ⫽ c

a11

a12

a21

a22

d⫺c

a11

a12

a21

a22

2a11 ⫹ 3a12 ⫺ 2a11 ⫺ 3a21

3a11 ⫹ 4a12 ⫺ 2a12 ⫺ 3a22

2a21 ⫹ 3a22 ⫺ 3a11 ⫺ 4a21

3a21 ⫹ 4a22 ⫺ 3a12 ⫺ 4a22

a21 ⫽ a12 from m 11 ⫽ 0 (also from m 22 ⫽ 0). a22 ⫽ a11 ⫹ 23a12 from m 12 ⫽ 0 (also from m 21 ⫽ 0). Answer: A⫽ c

a11

a12

a11 ⫹ 23 a12

26. The transition probabilities can be given in a matrix From N A⫽ c

From T

0.8

0.5

0.2

0.5

To N To T

and multiplication of [1 0]T by A, A2, A3 gives [0.8 [0.722 0.278]T. 28. The matrix of the transition probabilities is A⫽ c

0.9

0.002

0.1

0.998

0.2]T, [0.74 0.26]T, and

The initial state is [1200 98,800]T. Hence multiplication by A gives the further states (rounded) [1278 98,722]T, [1347 98,653]T, [1410 98,590]T, indicating that a substantial increase is likely. 30. Team Project. (b) Use induction on n. True if n ⫽ 1. Take the formula in the problem as the induction hypothesis, multiply by A, and simplify the entries in the product by the addition formulas for the cosine and sine to get An⫹1. (c) These formulas follow directly from the definition of matrix multiplication. (d) A scalar matrix would correspond to a stretch or contraction by the same factor in all directions. (e) Rotations about the x 1-, x 2-, x 3-axes through u, ␸, c, respectively.

SECTION 7.3. Linear Systems of Equations. Gauss Elimination, page 272 Purpose. This section centers around the Gauss elimination for solving linear systems of m equations in n unknowns x 1, Á , x n, its practical use as well as its mathematical justification (leaving the—more demanding—general existence theory to the next sections).

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Main Content, Important Concepts Nonhomogeneous, homogeneous, coefficient matrix, augmented matrix Gauss elimination in the case of the existence of I. a unique solution (Example 2) II. infinitely many solutions (Example 3) III. no solutions (Example 4) Pivoting Elementary row operations, echelon form Background Material.

All one needs here is the multiplication of a matrix and a vector.

Comments on Content The student should become aware of the following facts: 1. Linear systems of equations provide a major application of matrix algebra and justification of the definitions of its concepts. 2. The Gauss elimination (with pivoting) gives meaningful results in each of the Cases I⫺III. 3. This method is a systematic elimination that does not look for unsystematic “shortcuts” (depending on the size of the numbers involved and still advocated in some older precomputer-age books). Algorithms for programs of Gauss’s and related methods are discussed in Sec. 20.1, which is independent of the rest of Chap. 20 and can thus be taken up along with the present section in case of time and interest. Comments on Examples in the Text Example 1 and Fig. 158 show geometric interpretations of linear systems in 2 and 3 unknowns. Example 2 on an electrical network of Ohm’s resistors shows Gauss elimination with pivoting and back substitution in the case of a unique solution. Theorem 1 is central; it proves that the three kinds of elementary row operations leave solution sets unchanged, thus justifying Gauss elimination. Examples 2, 3, and 4 illustrate the Gauss elimination for the three possible cases that a linear system has a unique solution, or infinitely many solutions, or no solution, respectively. The section closes with a few comments on the row-echelon form, that is, on the form into which Gauss elimination transforms linear systems. Comments on Problems Problems 1–14 on Gauss elimination give further illustrations of those three cases. Problem 15 concerns equivalence (its general definition), which is of general mathematical interest, involving reflexivity, symmetry, and transitivity. Electrical networks of Ohm’s resistors (no inductances or capacitances) are discussed as linear systems in Probs.17–20 and some further models in Probs. 21–23. Project 24 presents the idea of representing matrix operations (as discussed before) in terms of standard matrices, called elementary matrices. Such representations are helpful, for instance, in designing algorithms, whereas computations themselves generally proceed directly.

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SOLUTIONS TO PROBLEM SET 7.3, page 280 2. x ⫽ 0.4, y ⫽ 1.2 4. No solution 6. x ⫽ 2y ⫹ 1 ⫽ 2t ⫹ 1, y ⫽ t arbitrary, z ⫽ 4 8. No solution 10. No solution 12. w ⫽ x ⫺ 2y ⫽ t 1 ⫺ 2t 2, x ⫽ t 1 arbitrary, y ⫽ t 2 arbitrary, z ⫽ 3 14. w ⫽ 0, x ⫽ 3z ⫽ 3t, y ⫽ 2z ⫹ 1 ⫽ 2t ⫹ 1, z ⫽ t arbitrary 18. Currents at the lower node: ⫺I1 ⫹ I2 ⫹ I3 ⫽ 0 (minus because I1 flows out). Voltage in the left circuit: 4I1 ⫹ 12I2 ⫽ 12 ⫹ 24 and in the right circuit 12I2 ⫺ 8I3 ⫽ 24 (minus because I3 flows against the arrow of E 2). Hence the augmented matrix of the system is ⫺1

D 4

36T .

⫺8

The solution is I1 ⫽ 27 11 A,

I2 ⫽ 24 11 A,

3 I3 ⫽ 11 A.

22. P1 ⫽ 6, P2 ⫽ 10, D1 ⫽ S1 ⫽ 18, D2 ⫽ S2 ⫽ 26 24. PROJECT. (a) B and C are different. For instance, it makes a difference whether we first multiply a row and then interchange, and then do these operations in reverse order.

B⫽E

a11

a12

a31

a32

a21 ⫺ 5a11

a22 ⫺ 5a12

8a41

8a42

a12

a31 ⫺ 5a11

a32 ⫺ 5a12

a21

a22

8a41

8a42

C⫽E

a11

(b) Premultiplying A by E makes E operate on rows of A. The assertions then follow almost immediately from the definition of matrix multiplication. SECTION 7.4. Linear Independence. Rank of a Matrix. Vector Space, page 282 Purpose. This section introduces some theory centered around linear independence and rank, in preparation for the discussion of the existence and uniqueness problem for linear systems of equations (Sec. 7.7).

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Main Content, Important Concepts Linear independence Real vector space Rn, dimension, basis Rank defined in terms of row vectors Rank in terms of column vectors Invariance of rank under elementary row operations Short Courses. For the further discussion in the next sections, it suffices to define linear independence and rank. Comments on Rank and Vector Spaces Of the three possible equivalent definitions of rank, (i) By row vectors (our definition), (ii) By column vectors (our Theorem 3), (iii) By submatrices with nonzero determinant (Sec. 7.7), the first seems to be most practical in our context. Introducing vector spaces here, rather than in Sec. 7.1, we have the advantage that the student immediately sees an application (row and column spaces). Vector spaces in full generality follow in Sec. 7.9. Comments on Text and Problem Set Examples 1–5 concern the same three vectors, in Examples 2–5 as row vectors of a matrix. Theorem 1 states that rank is invariant under row reduction. Example 3 determines rank by Gauss reduction to echelon form. Since we defined rank in terms of row vectors, rank in terms of column vectors becomes a theorem (Theorem 3). Theorem 4 results from Theorems 2 and 3, as indicated. The text then continues with the definition of vector space in general, of vector space Rn, and of row space and column space of a matrix A, both of which have the same dimension, equal to rank A. This is immediately illustrated in Probs. 1–10 of the problem set. Problems 12–16 give a further discussion of rank. Problems 17–25 concern linear independence and dependence. Sets of vectors that form, or do not form, vector spaces follow in Probs. 27–35. The discussion of vector space will be continued and extended in the last section (Sec. 7.9) of this chapter, which we leave optional since we shall not make further use of it.

SOLUTIONS TO PROBLEM SET 7.4, page 287 2. The matrix is symmetric. Row reduction gives

a ⫺ b 2>a

Hence for general a ⫽ 0 and a 2 ⫽ b 2 the rank is 2. A basis is {[a [0, a 2 ⫺ b 2]}.

b],

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4. The matrix is symmetric. Row reduction gives 6

⫺4

6T .

Hence a basis of the row space is {[3 ⫺2 0], [0 1 3], [0 0 1]}. The rank is 3. Note that this matrix has the “nice” eigenvalues 6 and 3 ⫾ 129. 6. The matrix is skew-symmetric. Row reduction gives ⫺1

⫺4

D 0

as can be seen without calculation. Hence the rank of the matrix is 2. Bases are {[1 0 4], [0 1 0]} and the same vectors transposed (as column vectors). 8. The matrix has rank 4. Row reduction gives as a basis of the row space: [1 2

8], [0

21],

5],

1].

Row reduction of the tranpose gives a basis of the column space of the given matrix. 10. The matrix is symmetric. Row reduction gives 5

⫺2

⫺4

0 V. 2

0], [0

Hence the rank is 3, and a basis is [5

⫺2

0],

⫺1].

12. AB and its transpose (AB)T ⫽ BTAT have the same rank. 14. This follows directly from Theorem 3. 16. A proof is given in Ref. [B3], vol 1, p. 12. (See App. 1 of the text.) 18. Yes. We mention that these are the row vectors of the 4 ⫻ 4 Hilbert matrix; see the Index of the textbook. 20. No. It is remarkable that A ⫽ [ajk] with ajk ⫽ j ⫹ k ⫺ 1 has rank 2 for any size n of the matrix. 22. No. Quite generally, if one of the vectors v(1), Á , v(m) is 0, then (1) holds with any c1 ⫽ 0 and c2, Á , cm all zero. 24. No, by Theorem 4. 26. Three steps; the first two vectors remain; these form a linearly independent subset of the given set. 28. No, if k ⫽ 0; yes, if k ⫽ 0, dimension 2, basis [1 0 0], [0 1 ⫺3].

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30. Yes, dimension 2, basis [0 Á 0 1 0], [0 Á 0 1]. 32. Yes, dimension 1. The two given equations form a linear system with coefficient matrix

⫺2

The solution is x 1 ⫽ 5t 1, x 2 ⫽ ⫺4t 1, x 3 ⫽ ⫺23t 1 with arbitrary t 1. Hence a basis is [5 ⫺4 ⫺23]. 34. No

SECTION 7.5. Solutions of Linear Systems: Existence, Uniqueness, page 288 Purpose. The student should see that the totality of solutions (including the existence and uniqueness) can be characterized in terms of the ranks of the coefficient matrix and the augmented matrix. Main Content, Important Concepts Augmented matrix Necessary and sufficient conditions for the existence of solutions Implications for homogeneous systems rank A ⫹ nullity A ⫽ n Background Material. Short Courses. examples.

Rank (Sec. 7.4)

Brief discussion of the first two theorems, illustrated by some simple

Comments on Content This section should make the student aware of the great importance of rank. It may be good to have students memorize the condition 苲 rank A ⫽ rank A for the existence of solutions. Students familiar with ODEs may be reminded of the analog of Theorem 4 (see Sec. 2.7). This section may also provide a good opportunity to point to the roles of existence and uniqueness problems throughout mathematics (and to the distinction between the two).

SECTION 7.7. Determinants. Cramer’s Rule, page 293 For second- and third-order determinants see the reference Sec. 7.6. Main Content of This Section nth-order determinants General properties of determinants Rank in terms of determinants (Theorem 3) Cramer’s rule for solving linear systems by determinants (Theorem 4)

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General Comments on Determinants Our definition of a determinant seems more practical than that in terms of permutations (because it immediately gives those general properties), at the expense of the proof that our definition is unambiguous (see the proof in App. 4). General properties are given for order n, from which they can be easily seen for n ⫽ 3 when needed. The importance of determinants has decreased with time, but determinants will remain in eigenvalue problems (characteristic determinants), ODEs (Wronskians!), integration and transformations (Jacobians!), and other areas of practical interest. Comments on Examples Examples 1–3 show expansions of determinants in the simplest cases. Example 4 illustrates the role of triangular matrices in the present context. The theorems show properties of determinants, in particular the relation to rank (Theorem 3) and Cramer’s rule for n equations in n unknowns. Note that the cases n ⫽ 2 and n ⫽ 3 were considered in Sec. 7.6. Comments on Problems Problems 1–15 illustrate general properties of determinants and their evaluation. Problems 17–19 compare the (impractical) determination of rank by determinants and by row reduction. Team Project 20 concerns some applications of linear systems to analytic geometry all using the vanishing of determinants. SOLUTIONS TO PROBLEM SET 7.7, page 300 8. ⫺7.87 10. 1 12. a 3 ⫹ b 3 ⫹ c3 ⫺ 3abc 14. 216. Note that

2#2

⫺2

2 ⫽ 18 # 12 ⫽ 216.

16. det A n ⫽ (⫺1)nⴚ1(n ⫺ 1). True for n ⫽ 2, a 2-simplex on R 1, that is, a segment (an interval), because det A 2 ⫽ (⫺1)2ⴚ1(2 ⫺ 1) ⫽ ⫺1. Assume true for n as just given. Consider A n⫹1. To get the first row with all entries 0, except for the first entry, subtract from Row 1 the expression 1 (Row 2 ⫹ . . . ⫹ Row (n ⫹ 1)). n⫺1 The first component of the new row is ⫺n>(n ⫺ 1), whereas the other components are all 0. Develop det A n⫹1 by this new first row and notice that you can then apply the above induction hypothesis, det A n⫹1 ⫽ ⫺ as had to be shown.

n (⫺1)nⴚ1(n ⫺ 1) ⫽ (⫺1)nn, n⫺1

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18. 3 because interchange of Rows 1 and 2 and row reduction gives 4

⫺6T .

20. Team Project. (a) Use row operation (subtraction of rows) on D to transform the last column of D into the form [0 0 1]T and then develop D ⫽ 0 by this column. (b) For a plane the equation is ax ⫹ by ⫹ cz ⫹ d # 1 ⫽ 0, so that we get the determinantal equation

6 ⫽ 0.

The plane is 3x ⫹ 4y ⫺ 2z ⫽ 5. (c) For a circle the equation is a(x 2 ⫹ y 2) ⫹ bx ⫹ cy ⫹ d # 1 ⫽ 0, so that we get

x 2 ⫹ y2

x 21 ⫹ y 21

x 22 ⫹ y 22

x 23 ⫹ y 23

6 ⫽ 0.

The circle is x 2 ⫹ y 2 ⫺ 4x ⫺ 2y ⫽ 20. (d) For a sphere the equation is a(x 2 ⫹ y 2 ⫹ z 2) ⫹ bx ⫹ cy ⫹ dz ⫹ e # 1 ⫽ 0, so that we obtain x 2 ⫹ y2 ⫹ z2

x 21 ⫹ y 21 ⫹ z 21

7 x 22 ⫹ y 22 ⫹ z 22

1 7 ⫽ 0.

x 23 ⫹ y 23 ⫹ z 23

x 24 ⫹ y 24 ⫹ z 24

The sphere through the given points is x 2 ⫹ y 2 ⫹ (z ⫺ 1)2 ⫽ 16. (e) For a general conic section the equation is ax 2 ⫹ bxy ⫹ cy 2 ⫹ dx ⫹ ey ⫹ f # 1 ⫽ 0,

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so that we get

x 21

x 1y1

y 21

x 22

x 2y2

y 22

x 23

x 3y3

y 23

x 24

x 4y4

y 24

x 25

x 5y5

y 25

8 ⫽ 0.

22. In Cramer’s rule we have D⫽2

⫺4

2 ⫽ 24 5 2 ⫺24 ⫺4 2 ⫽ ⫺48 D1 ⫽ 2 0 2 2 ⫺24 2 ⫽ 120. D2 ⫽ 2 5 0 Hence x ⫽ D1>D ⫽ ⫺48>24 ⫽ ⫺2, and y ⫽ D2>D ⫽ 120>24 ⫽ 5. 24. D ⫽ ⫺60, D1 ⫽ ⫺60, D2 ⫽ 180, and D3 ⫽ ⫺240. Hence x ⫽ 1, y ⫽ ⫺3, and z ⫽ 4. SECTION 7.8. Inverse of a Matrix. Gauss–Jordan Elimination, page 301 Purpose. To familiarize the student with the concept of the inverse Aⴚ1 of a square matrix A, its conditions for existence, and its computation. Main Content, Important Concepts AAⴚ1 ⫽ Aⴚ1A ⫽ I Nonsingular and singular matrices Existence of Aⴚ1 and rank Gauss–Jordan elimination (AC)ⴚ1 ⫽ C ⴚ1Aⴚ1 Cancellation laws (Theorem 3) det (AB) ⫽ det (BA) ⫽ det A det B Short Courses. Theorem 1 without proof, Gauss–Jordan elimination, formulas (4*) and (7). Comments on Content Although in this chapter we are not concerned with operations count (Chap. 20), it would make no sense to first mislead the student by using Gauss–Jordan for solving Ax ⫽ b and then later, in numerics, correct the false impression by explaining why Gauss elimination is better because back substitution needs fewer operations than the diagonalization of a triangular matrix. Thus Gauss–Jordan should be applied only when Aⴚ1 is wanted.

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The “unusual” properties of matrix multiplication, briefly mentioned in Sec. 7.2 can now be explored systematically by the use of rank and inverse. Formula (4*) is worth memorizing. SOLUTIONS TO PROBLEM SET 7.8, page 308 2. c

cos 2u

⫺sin 2u

sin 2u

cos 2u

Note that the given matrix corresponds to a rotation through an angle 2u. If 2u is replaced by ⫺2u (rotation in the opposite sense), this gives the inverse, which corresponds to a rotation through ⫺2u. 4. The entries of the inverse are the same as for a diagonal matrix, but their position on the other diagonal is different. The inverse is 0

0.4

D 0

⫺2.5

0 T.

6. Note that, due to the special form of the given matrix, the 2 ⫻ 2 minor in the right lower corner of the inverse has the form of the inverse of a 2 ⫻ 2 matrix; the inverse is ⫺14

D 0

⫺13T .

⫺3

8. The matrix is singular. It is interesting that this is not the case for the 2 ⫻ 2 matrix

10. The inverse equals the transpose. This is the defining property of orthogonal matrices to be discussed in Sec. 8.3. 12. I ⫽ (A2)ⴚ1A2. Multiply this by Aⴚ1 from the right on both sides of the equation. This gives Aⴚ1 ⫽ (A2)ⴚ1A. Do the same operation once more to get the formula to be proved. 14. I ⫽ I T ⫽ (Aⴚ1A)T ⫽ AT(Aⴚ1)T. Now multiply the first and last expression by (AT)ⴚ1 from the left, obtaining (AT)ⴚ1 ⫽ (Aⴚ1)T. 16. Rotation through 2u. The inverse represents the rotation through ⫺2u. Replacement of 2u by ⫺2u in the matrix gives the inverse. 18. Multiplication by A from the right interchanges Rows 1 and 2 of A. The inverse of this is the interchange that gives back the original matrix. Hence the inverse of the given matrix should equal the matrix itself, as is the case. 20. Straightforward calculation, particularly simple because of the zeros. And instructive because we now see distinctly why the inverse has zeros at the same positions as the given matrix does.

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SECTION 7.9. Vector Spaces, Inner Product Spaces, Linear Transformations, Optional, page 309 Purpose. In this optional section we extend our earlier discussion of vector spaces Rn and C n, define inner product spaces, and explain the role of matrices in linear transformations of R n into R m. Main Content, Important Concepts Real vector space, complex vector space Linear independence, dimension, basis Inner product space Linear transformation of Rn into Rm Background Material. Vector spaces R n (Sec. 7.4) Comments on Content The student is supposed to see and comprehend how concrete models (Rn and C n, the inner product for vectors) lead to abstract concepts, defined by axioms resulting from basic properties of those models. Because of the level and general objective of this chapter, we have to restrict our discussion to the illustration and explanation of the abstract concepts in terms of some simple typical examples. Most essential from the viewpoint of matrices is our discussion of linear transformations, which, in a more theoretically oriented course of a higher level, would occupy a more prominent position. Comment on Footnote 4 Hilbert’s work was fundamental to various areas in mathematics; roughly speaking, he worked on number theory 1893–1898, foundations of geometry 1898–1902, integral equations 1902–1912, physics 1910–1922, and logic and foundations of mathematics 1922–1930. Closest to our interests here is the development in integral equations, as follows. In 1870 Carl Neumann (Sec. 5.6) had the idea of solving the Dirichlet problem for the Laplace equation by converting it to an integral equation. This created general interest in integral equations. In 1896 Vito Volterra (1860–1940) developed a general theory of these equations, followed by Erik Ivar Fredholm (1866–1927) in 1900–1903 (whose papers caused great excitement), and Hilbert since 1902. This gave the impetus to the development of inner product and Hilbert spaces and operators defined on them. These spaces and operators and their spectral theory have found basic applications in quantum mechanics since 1927. Hilbert’s great interest in mathematical physics is documented by Ref. [GenRef3], a classic full of ideas that are of importance to the mathematical work of the engineer. For more details, see G. Birkhoff and E. Kreyszig. The establishment of functional analysis. Historia Mathematica 11 (1984), pp. 258–321. Further Comments on Content and Comments on Problems It is important to understand that matrices form vector spaces (Example 1, etc.) and so do polynomials up to a fixed degree n (Example 2). An inner product space is obtained from a vector space V by defining an inner product on V ⫻ V. In addition to the usual dot product notation the student should perhaps also become aware of the standard notation (a, b), used, e.g., in functional analysis and applications. The last part of the section concerns the role of matrices in connection with linear transformations.

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The fundamental importance of (3)–(5) (Cauchy–Schwarz, triangle inequalities, parallelogram equality) will not appear on our level, but should perhaps be mentioned in passing, along with the verifications required in Probs. 23–25. The problems center around vector spaces, supplementing Problem Set 7.4, linear spaces and inverse matrices, as well as norm and inner product (to be substantially extended in numerics in Chap. 20). SOLUTIONS TO PROBLEM SET 7.9, page 318 2. Take the difference of the given representation and another representation v ⫽ k 1a (1) ⫹ . . . ⫹ k na (n), obtaining v ⫺ v ⫽ a (cj ⫺ k j) a(j) ⫽ 0. Hence cj ⫺ k j ⫽ 0 because of the linear independence of the basis vectors. This implies k j ⫽ cj for j ⫽ 1, Á , n, the uniqueness. 4. Yes, dimension 3 because of the skew symmetry, at most three entries of the nine entries of such a matrix can be different (and not zero). A basis is 0

D⫺1

0T ,

D 0

0T ,

1T .

⫺1

6. Yes. Dimension 2. Basis cos 2x, sin 2x. Note that these functions are solutions of the ODE y s ⫹ 4y ⫽ 0. To mention this connection with vector spaces would not have added much to our discussion in Chap. 2. Similarly for the next problem (Prob. 7). 8. No, because det (A 1 ⫹ A 2) ⫽ det A 1 ⫹ det A 2 in general. 10. Yes, dimension 4, basis 3

D 0

0T ,

1T ,

0T .

⫺5

12. The inverse transformation is obtained by calculating the inverse matrix. This gives x 1 ⫽ ⫺15 y1 ⫹ 25 y2 x 2 ⫽ 45 y1 ⫺ 35 y2. 14. x 1 ⫽

4y1 ⫺ 2y2 ⫹ 2y3

x 2 ⫽ ⫺2y1 ⫺ 4y2 ⫹ 4y3 x 3 ⫽ ⫺4y1 ⫹ 2y2 ⫹ 8y3 16. 214 ⫹ 19 ⫹ 14 ⫹ 19 ⫽ 212 ⫹ 29 ⫽ 213 18 18. 9 20. 1 22. Yes. Vectors [v1 v2 v3]T orthogonal to the given vector must satisfy 2v1 ⫹ v3 ⫽ 0. 1 0]T, [1 0 ⫺2]T. A basis of this two-dimensional vector space is [0

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24. We obtain 1 ƒ (a, b) ƒ ⫽ ƒ aT b ƒ ⫽ ƒ 12 # 23 ⫹ 13 (⫺23 ) ⫺ 12 # 13 ƒ ⫽ ƒ ⫺18 ƒ ⫽ 0.056

# 1 ⫽ 0.850. ⬉ 储 a 储储 b 储 ⫽ 213 18 SOLUTIONS TO CHAPTER 7 REVIEW QUESTIONS AND PROBLEMS, page 318 12. A, ⫺B 14. The first expression is the 1 ⫻ 1 matrix [⫺1], which we can identify with its only entry, ⫺1; similarly in Prob. 15. The second expression is the 3 ⫻ 3 matrix: 14

⫺6

0T .

⫺35

⫺15

16. Note that the symmetric A has a symmetric inverse: ⫺3.2

2.2

D 2.2

⫺1.2

⫺2.8

1.8

⫺2.8 1.8T . ⫺2.2

The matrix B is singular. 18. Both matrices are equal; calculation gives 22.92

⫺14.72

19.08

D⫺14.72

9.52

⫺12.28T .

19.08

⫺12.28

15.92

Symmetry follows by noting that (A2)T ⫽ (AT)2 ⫽ A2. 20. Because of the symmetry of A and the skew symmetry of B, this expression equals ⫺4

2A # 2B ⫽ 4AB ⫽ D⫺72

⫺28T.

⫺52

⫺8

⫺28

22. x ⫽ 1, y ⫽ t arbitrary, z ⫽ 3t ⫹ 2 3 24. x ⫽ 13 y ⫽ t 1 arbitrary, z ⫽ t 2 arbitrary 2 t 1 ⫺ 2 t 2 ⫹ 2, 26. No solution 28. x ⫽ 14 , y ⫽ ⫺12 , z ⫽ 32 30. Ranks 1, 1. The first row of the matrices equals ⫺3 times the second row. 32. Ranks 1, 2, so that there exists no solution. 34. I1 ⫽ 12 A, I2 ⫽ 4 A, I3 ⫽ 16 A

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CHAPTER 8

Linear Algebra: Matrix Eigenvalue Problems

Prerequisite for this chapter is some familiarity with the notion of a matrix and with the two algebraic operations for matrices. Otherwise the chapter is independent of Chap. 7, so that it can be used for teaching eigenvalue problems and their applications, without first going through the material in Chap. 7. SECTION 8.1. The Matrix Eigenvalue Problem. Determining Eigenvalues and Eigenvectors, page 323 Purpose. To familiarize the student with the determination of eigenvalues and eigenvectors of real matrices and to give a first impression of what one can expect (multiple eigenvalues, complex eigenvalues, etc.). Main Content, Important Concepts Eigenvalue, eigenvector Determination of eigenvalues from the characteristic equation Determination of eigenvectors Algebraic and geometric multiplicity, defect Comments on Content To maintain undivided attention on the basic concepts and techniques, all the examples in this section are formal, and typical applications are put into a separate section (Sec. 8.2). The distinction between the algebraic and geometric multiplicity is mentioned in this early section, and the idea of a basis of eigenvectors (“eigenbasis”) could perhaps be mentioned briefly in class, whereas a thorough discussion of this in a later section (Sec. 8.4) will profit from the increased experience with eigenvalue problems, which the student will have gained at that later time. The possibility of normalizing any eigenvector is mentioned in connection with Theorem 2, but this will be of greater interest to us only in connection with orthonormal or unitary systems (Secs. 8.4 and 8.5). In our present work we find eigenvalues first and are then left with the much simpler task of determining corresponding eigenvectors. Numeric work (Secs. 20.62–20.9) may proceed in the opposite order, but to mention this here would perhaps just confuse the student. Further Comments on Examples and Theorems The simple examples should give the student a first impression of what to expect. In particular, Example 4 shows that eigenvalue problems lead to work in complex, even if the matrices are real. This is an important point to emphasize. Theorem 1 shows that for matrices, in contrast to differential equations, eigenvalue problems involve no existence questions since the existence of an eigenvalue is always guaranteed. Theorems 2 and 3 concern the notion of eigenspace and the invariance of eigenvalues under transposition. Comments on Problems Problems 1–16 involve straightforward calculations to gain skill and an understanding of the concepts involved. Sidetracking attention by solving cubic or higher-order equations is avoided. 144

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Problems 17–20 illustrate simple applications to analytic geometry. Actual applications of eigenvalue problems follow in the next section, as has been mentioned before. Problems 21–25 illustrate some important simple facts.

SOLUTIONS TO PROBLEM SET 8.1, page 329 2. This zero matrix, like any square zero matrix, has the eigenvalue 0. The algebraic multiplicity and geometric multiplicity are both equal to 2, and we can choose the basis [1 0]T, [0 1]T. 4. Eigenvalues are 5 and 0 and eigenvectors [1 2]T and [⫺2 1]T, respectively. The matrix is symmetric, and for such a matrix it is typical that the eigenvalues are real and the eigenvectors orthogonal. Also, make the students aware of the fact that 0 can very well be an eigenvalue. 6. Eigenvalues are 1 and 3 and eigenvectors [1 0]T and [1 1]T, respectively. Note that for such a triangular matrix, the main diagonal entries are still the eigenvalues (as for a diagonal matrix; cf. Prob. 1), but the eigenvectors are no longer orthogonal. 8. The characteristic equation is (a ⫺ l)2 ⫹ b 2 ⫽ 0.

Solutions are

l ⫽ a ⫾ ib.

Eigenvectors are obtained from (a ⫺ l)x 1 ⫹ bx 2 ⫽ ⫿ibx 1 ⫹ bx 2 ⫽ 0. Hence we can take x 1 ⫽ 1 and x 2 ⫽ ⫾i. Note that b has dropped out, and the eigenvectors are the same as in Example 4 of the text. Here the student should see and understand that eigenvalue problems for real matrices may lead to complex eigenvalues. 10. The characteristic equation is (cos u ⫺ l)2 ⫹ sin2 u ⫽ 0. Solutions (eigenvalues) are l ⫽ cos u ⫾ i sin u. Eigenvectors are obtained from (l ⫺ cos u)x 1 ⫹ (sin u)x 2 ⫽ (sin u)(⫾ix 1 ⫹ x 2) ⫽ 0, say, x 1 ⫽ 1, x 2 ⫽ ⫿i. Note that this matrix represents a rotation through an angle u, and this linear transformation preserves no real direction in the x 1x 2-plane, as would be the case if the eigenvectors were positive real. This explains why these vectors must be complex. 12. 3, [1

0]T; 4, [5

0]T; 1, [7,

⫺4 2]T

14. Develop the characteristic determinant by the second row, obtaining (12 ⫺ l)[(2 ⫺ l)(4 ⫺ l) ⫹ 1] ⫽ (12 ⫺ l)(l ⫺ 3)2. Eigenvectors for the eigenvalues 12 and 3 are [0 respectively, and we get no basis for R 3.

0]T and [⫺1

16. The indicated division of the characteristic polynomial gives (l4 ⫺ 22l2 ⫹ 24l ⫹ 45)>(l ⫺ 3)2 ⫽ l2 ⫹ 6l ⫹ 5.

1]T,

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The eigenvalues and eigenvectors are

18. c

l1 ⫽ 3,

l2 ⫽ ⫺1

⫺1

l3 ⫽ ⫺5,

[⫺11

(a) 1, c

(b) ⫺1, c

1]T 5

1]T.

d . (a) Any point of the x 1-axis is mapped onto 0 ⫺1 0 1 itself. (b) Any point (0, x 2) on the x 2-axis is mapped onto (0, ⫺x 2), so that [0 x 2]T is an eigenvector corresponding to l ⫽ ⫺1. 1

1]T with a defect of 1

1 2

20. D 12

1 2

0T . The eigenvalue 1 with eigenvectors D0T and D1T (which span the plane

1 0 0 0 1 x 2 ⫽ x 1) indicates that every point in the plane x 2 ⫽ x 1 is mapped onto itself. The other eigenvalue 0 with eigenvector [1 ⫺1 0]T indicates that any point on the line x 2 ⫽ ⫺x 1, x 3 ⫽ 0 (which is perpendicular to the plane x 2 ⫽ x 1) is mapped onto the origin. The student should perhaps make a sketch to see what is going on geometrically. 24. By Theorem 1 in Sec. 7.8 the inverse exists if and only if det A ⫽ 0. On the other hand, from the product representation D(l) ⫽ det (A ⫺ lI) ⫽ (⫺1)n(l ⫺ l1)(l ⫺ l2) Á (l ⫺ ln) of the characteristic polynomial we obtain det A ⫽ (⫺1)n(⫺l1)(⫺l2) Á (⫺ln) ⫽ l1l2 Á ln. Hence Aⴚ1 exists if and only if 0 is not an eigenvalue of A. Furthermore, let l ⫽ 0 be an eigenvalue of A. Then Ax ⫽ lx. ⴚ1

Multiply this by A

from the left: Aⴚ1Ax ⫽ lAⴚ1x.

Now divide by l: 1 x ⫽ Aⴚ1x. l SECTION 8.2. Some Applications of Eigenvalue Problems, page 329 Purpose. Matrix eigenvalue problems are of greatest importance in physics, engineering, geometry, etc., and the applications in this section and in the problem set are supposed to give the student at least some impression of this fact. Main Content Applications of eigenvalue problems in Elasticity theory (Example 1) Probability theory (Example 2) Biology (Example 3) Mechanical vibrations (Example 4)

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Short Courses. Of course, this section can be omitted, for reasons of time, or one or two of the examples can be considered quite briefly. Comments on Content The examples in this section have been selected from the viewpoint of modest prerequisites, so that not too much time will be needed to set the scene. Example 4 illustrates why real matrices can have complex eigenvalues (as mentioned before, in Sec. 8.1), and why these eigenvalues are physically meaningful. (For students familiar with systems of ODEs, one can easily pick further examples from Chap. 4.) Comments on Problems Problems 1–12 are similar to the applications shown in the examples of the text. Problems 13–15 show an interesting application of eigenvalue problems to production, typical of various other applications of eigenvalue theory in economics included in various textbooks in economic theory.

SOLUTIONS TO PROBLEM SET 8.2, page 333 2. Eigenvalues and eigenvectors are 1.6, [1 ⫺1]T and 2.4, [1 1]T. These vectors are orthogonal, as is typical of a symmetric matrix. Directions are ⫺45° and 45°, respectively. 4. Extension factors 9 ⫹ 215 ⫽ 13.47 and 9 ⫺ 215 ⫽ 4.53 in the directions given by [1 2 ⫹ 15]T and [1 2 ⫺ 15]T (76.7° and ⫺13.3°, respectively). 6. 2, [1

1]T;

1 2 , [1

⫺1]T; directions 45° and ⫺45°, respectively.

8. [1 1 1]T, as could also be seen without calculation because A has row sums equal to 1, which would not be the case in general. 10. Growth rate 2. The other two eigenvalves are not needed; they could be determined by dividing the characteristic polynomial by l ⫺ 2; they are ⫺1 ⫾ 10.6. The sum of all eigenvalues is the trace, hence 0. 12. Growth rate 1.3748. The other eigenvalues are complex or negative and are not needed. The sum of all eigenvalues equals the trace, that is, 0, except for a roundoff error. This 4 ⫻ 4 Leslie matrix corresponds to a classification of the population into four classes. 14. A has the same eigenvalues as AT, and AT has row sums 1, so that it has the eigenvalue 1 with eigenvector x ⫽ [1 Á 1]T. Leontief is a leader in the development and application of quantitative methods in empirical economical research, using genuine data from the economy of the United States to provide, in addition to the “closed model” of Prob. 13 (where the producers consume the whole production), “open models” of various situations of production and consumption, including import, export, taxes, capital gains and losses, etc. See W. W. Leontief, The Structure of the American Economy 1919–1939 (Oxford: Oxford University Press, 1951). H. B. Cheney and P. G. Clark, Interindustry Economics (New York: Wiley, 1959). 16. This follows by comparing the coefficient of lnⴚ1 in the development of the characteristic determinant D (l) with that obtained from the product representation.

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18. The first statement follows from Ax ⫽ lx,

(kA)x ⫽ k(Ax) ⫽ k(lx) ⫽ (kl)x,

the second by induction and multiplication of Akxj ⫽ lkj xj by A from the left. 20. det (L ⫺ lI) ⫽ ⫺l3 ⫹ l 12l 21l ⫹ l 13l 21l 32 ⫽ 0. Hence l ⫽ 0. If all three eigenvalues are real, at least one is positive since trace L ⫽ 0. The only other possibility is l1 ⫽ a ⫹ ib, l2 ⫽ a ⫺ ib, l3 real (except for the numbering of the eigenvalues). Then l3 ⬎ 0 because l1l2l3 ⫽ (a 2 ⫹ b 2)l3 ⫽ det L ⫽ l 13l 21l 32 ⬎ 0. SECTION 8.3. Symmetric, Skew-Symmetric, and Orthogonal Matrices, page 334 Purpose. To introduce the student to the three most important classes of real square matrices and their general properties and eigenvalue theory. Main Content, Important Concepts The eigenvalues of a symmetric matrix are real. The eigenvalues of a skew-symmetric matrix are pure imaginary or zero. The eigenvalues of an orthogonal matrix have absolute value 1. Further properties of orthogonal matrices. Comments on Content The student should memorize the preceding three statements on the locations of eigenvalues as well as the basic properties of orthogonal matrices (orthonormality of row vectors and of column vectors, invariance of inner product, determinant equal to 1 or ⫺1). Furthermore, it may be good to emphasize that, since the eigenvalues of an orthogonal matrix may be complex, so may be the eigenvectors. Similarly for skew-symmetric matrices. Both cases are simultaneously illustrated by A⫽ c

0 ⫺1

1 0

c d 1

with eigenvectors

and

1 ⫺i

corresponding to the eigenvalues i and ⫺i, respectively. Further Comments on the Three Classes of Matrices in This Section Reality of eigenvalues is a main reason for the importance of symmetric matrices—many quantities in physics, such as mass, energy, etc., are real. Formula (4) brings in skew-symmetric matrices in a rather natural fashion. Theorem 3 explains the importance of orthogonal matrices. Typical examples of the spectra of the matrices considered in this section are illustrated by Probs. 1–10––most importantly, Probs. 4 and 8. Problems 13–20 should help the student gain a deeper understanding of the concepts and properties of the three classes of matrices considered in this section. SOLUTIONS TO PROBLEM SET 8.3, page 338 2. Eigenvalues a ⫾ ib. Symmetric if b ⫽ 0; then the eigenvalues are real. Skew-symmetric if a ⫽ 0; then the eigenvalues are pure imaginary (or zero). Orthogonal if a 2 ⫹ b 2 ⫽ 1; then the eigenvalues have an absolute value of 1.

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4. The characteristic equation is

cos u ⫺ l

⫺sin u

sin u

cos u ⫺ l

2 ⫽ l2 ⫺ (2 cos u) l ⫹ 1 ⫽ 0.

Hence the eigenvalues are l ⫽ cos u ⫾ i sin u. If u ⫽ 0 (the identity transformation) we have l ⫽ 1 with multiplicity 2. If u ⫽ 0, we obtain the eigenvectors from x 2 ⫽ ⫿ix 1,

[1 ⫿i],

say,

which are complex; indeed, no (real) direction is preserved under a rotation. 6. a ⫹ 2k, [1 1 1]T; a ⫺ k, [1 0 ⫺1]T, [1 ⫺1 0]T; symmetric (for real a and k) 8. Orthogonal, a rotation about the x 1-axis through an angle u. Eigenvalues 1 and cos u ⫾ i sin u. Compare with Prob. 4. 10. Orthogonal; eigenvalues 1 and 7>18 ⫾ (5>11>18)i, all of absolute value 1. 12. CAS Experiment (a) AT ⫽ Aⴚ1, BT ⫽ Bⴚ1, (AB)T ⫽ BTAT ⫽ Bⴚ1Aⴚ1 ⫽ (AB)ⴚ1. Also (Aⴚ1)T ⫽ (AT)ⴚ1 ⫽ (Aⴚ1)ⴚ1. In terms of rotations it means that the composite of rotations and the inverse of a rotation are rotations. (b) The inverse is

cos u

sin u

⫺sin u

cos u

(c) To a rotation of about 36.87°. No limit. For a student unfamiliar with complex numbers this may require some thought. (d) Limit 0, approach along some spiral. (e) The matrix is obtained by using familiar values of cosine and sine, A⫽ c

13>2

⫺12

1 2

13>2

16. Let Ax ⫽ lx (x ⫽ 0), Ay ⫽ ␮y (y ⫽ 0). Then lx T ⫽ (Ax)T ⫽ x TAT ⫽ x TA. Thus lx Ty ⫽ x TAy ⫽ x T␮y ⫽ ␮x Ty. Hence if l ⫽ ␮, then x Ty ⫽ 0, which proves orthogonality. 18. det A ⫽ det (AT) ⫽ det (⫺A) ⫽ (⫺1)n det A ⫽ ⫺det A ⫽ 0 if n is odd. Hence the answer is no. For even n ⫽ 2, 4, Á we have

⫺1

etc,

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20. Yes, for instance, 1 2

13>2

D 13>2

⫺12

0T .

SECTION 8.4. Eigenbases. Diagonalization. Quadratic Forms, page 339 Purpose. This section exhibits the role of bases of eigenvectors (“eigenbases”) in connection with linear transformations and contains theorems of great practical importance in connection with eigenvalue problems. Main Content, Important Concepts Bases of eigenvectors (Theorems 1, 2) Similar matrices have the same spectrum (Theorem 3) Diagonalization of matrices (Theorem 4) Principal axes transformation of forms (Theorem 5) Short Courses. Complete omission of this section or restriction to a short look at Theorems 1 and 5. Comments on Content Theorem 1 on similar matrices has various applications in the design of numeric methods (Chap. 20), which often use subsequent similarity transformations to tridiagonalize or (nearly) diagonalize matrices on the way to approximations of eigenvalues and eigenvectors. The matrix X of eigenvectors [see (5)] also occurs quite frequently in that context. Theorem 2 is another result of fundamental importance in many applications, for instance, in those methods for numerically determining eigenvalues and eigenvectors. Its proof is substantially more difficult than the proofs given in this chapter. The theorems in this section give sufficient conditions for the existence of eigenbases (⫽ bases of eigenvectors), namely, the almost trivial Theorem 1 as well as the very important Theorem 2, exhibiting another basic property of symmetric matrices. This is followed in Theorems 3 and 4 by similarity of matrices and its application to diagonalization. The second part of the section concerns the principal axes transformation of quadratic forms and its application to conic sections. The extension of these ideas and results to complex matrices and forms follows in the next section, the last one of this chapter. SOLUTIONS TO PROBLEM SET 8.4, page 345 2.

Â ⫽ c

⫺29

⫺42

l ⫽ ⫺1, y ⫽ c

5 7

x ⫽ Py ⫽ c

0 1

Similarly, for the second eigenvalue we obtain l ⫽ 1,

y⫽ c

2 3

x ⫽ Py ⫽ c

⫺1 ⫺1

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l ⫽ ⫺1,

Â ⫽ D 6

10T ; l ⫽ 3,

⫺8

⫺14

l ⫽ 2,

151

y ⫽ [26 y ⫽ [0

⫺16]T,

0]T,

y ⫽ [⫺2

⫺2]T

x ⫽ [0 1

0]T

x ⫽ [⫺1

1]T,

x ⫽ [4

⫺1]T

Project. (a) This follows immediately from the product representation of the characteristic polynomial of A. n

(b) C ⫽ AB, c11 ⫽ a a1lbl1, c22 ⫽ a a2lbl2, etc. Now take the sum of these n sums. l⫽1

l⫽1

Furthermore, trace BA is the sum of n

~ c 11 ⫽ a b1mam1, Á , ~ cnn ⫽ a bnmamn, m⫽1

m⫽1

involving the same n terms as those in the double sum of trace AB. (c) By multiplications from the right and from the left we readily obtain ~ A ⫽ P 2ÂP ⫺2. (d) Interchange the corresponding eigenvectors (columns) in the matrix X in (5). 10. The eigenvalues of A are 1 and ⫺1. A matrix of corresponding eigenvectors is x⫽ c

Its inverse is x ⴚ1 ⫽ c

⫺1

Hence diagonalization gives D ⫽ Xⴚ1AX ⫽ c

⫺1

Compare with Prob. 2, where the same matrix X underwent another similarity transformation. 12. A has the eigenvalues 10 and ⫺5. A matrix of eigenvectors is X⫽ c

⫺1

Its inverse is Xⴚ1 ⫽ c

1>150

11>150

13>150

⫺7>150

Hence diagonalization gives D ⫽ Xⴚ1AX ⫽ c

⫺5

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14. A has the eigenvalues ⫺2, 4, 1. A matrix of eigenvectors is ⫺1

X ⫽ D1

1T .

Its inverse is

ⴚ1

⫽ D⫺2

⫺2

⫺1 3T . ⫺2

Hence diagonalization gives

D⫽X

⫺2

AX ⫽ D 0

0T .

ⴚ1

16. A has the eigenvalues ⫺4, 2, 0. A matrix of corresponding eigenvectors is ⫺1

X ⫽ D0

1T .

Its inverse is

ⴚ1

1 2

0T .

⫺12

1 2

⫽D

Diagonalization thus gives

D⫽X

⫺4

Ax ⫽ D 0

0T .

ⴚ1

18. The symmetric coefficient matrix is C⫽ c

⫺3

It has the eigenvalues 5 and ⫺5. Hence the transformed form is 5y 21 ⫺ 5y 22 ⫽ 10;

thus,

y 21 ⫺ y 22 ⫽ 2.

This is a hyperbola. The matrix X whose columns are normalized eigenvectors of C gives the relation between y and x in the form x⫽ c

2> 15

1> 15

⫺2> 15

d y.

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20. The symmetric coefficient matrix is C⫽ c

Its eigenvalues are 0 and 10. Hence the transformed form is 10y 22 ⫽ 10. This represents a pair of parallel straight lines 10y 22 ⫽ 10,

y2 ⫽ ⫾1.

thus

The matrix X whose columns are normalized eigenvectors of C gives the relation between y and x in the form x⫽

1 1 c 110 ⫺3

3 1

d y.

22. The symmetric coefficient matrix is C⫽ c

Its eigenvalues are 1 and 16. Hence the transformed form is y 21 ⫹ 16y 22 ⫽ 16. This represents an ellipse. The matrix whose columns are normalized eigenvectors of C gives the relation between y and x in the form x⫽

1 ⫺2 c 15 1

1 2

d y.

24. Transform Q (x) by (9) to the canonical form (10). Since the inverse transform y ⫽ X⫺1x of (9) exists, there is a one-to-one correspondence between all x ⫽ 0 and y ⫽ 0. Hence the values of Q (x) for x ⫽ 0 coincide with the values of (10) on the right. But the latter are obviously controlled by the signs of the eigenvalues in the three ways stated in the theorem. This completes the proof.

SECTION 8.5. Complex Matrices and Forms. Optional, page 346 Purpose. This section is devoted to the three most important classes of complex matrices and corresponding forms and eigenvalue theory. Main Content, Important Concepts Hermitian and skew-Hermitian matrices Unitary matrices, unitary systems Location of eigenvalues (Fig. 163) Quadratic forms, their symmetric coefficient matrix Hermitian and skew-Hermitian forms

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Background Material. Section 8.3, which the present section generalizes. The prerequisites on complex numbers are very modest, so that students will need hardly any extra help in that respect. Short Courses. This section can be omitted. The importance of these matrices results from quantum mechanics as well as from mathematics itself (e.g., from unitary transformations, product representations of nonsingular matrices A ⫽ UH, U unitary, H Hermitian, etc.). The determinant of a unitary matrix (see Theorem 4) may be complex. For example, the matrix A⫽

1⫹i 1 c 12 0

0 1

is unitary and has det A ⫽ i. Comments on Problems Complex matrices appear in quantum mechanics; see Prob. 7, etc. Problems 13–20 give an impression of calculations for complex matrices. Normal matrices, defined in Prob.18, play an important role in a more extended theory of complex matrices.

SOLUTIONS TO PROBLEM SET 8.5, page 351 2. Skew-Hermitian. Eigenvalues and eigenvectors are ⫺i, [⫺1 ⫹ i 2]T and 2i, [2

1 ⫹ i]T.

4. Skew-Hermitian, as well as unitary, eigenvalues i and ⫺i, eigenvectors [1 1]T and [1 ⫺1]T, respectively. 6. Hermitian. Eigenvalues and eigenvectors are ⫺4, [i

⫺1 ⫺ i 1]T

0, [1 0 4, [i

i]T

1⫹i

1]T.

8. Eigenvectors are as follows. (Multiplication by a complex constant may change them drastically!) For A

[1 ⫺ 3i 5]T,

For B

[2 ⫹ i

For C

1]T,

[1 ⫺ 3i [2 ⫹ i

i]T, [1

⫺2]T

⫺5i]T

⫺1]T.

10. The matrix is skew-Hermitian. Calculation gives x TAx ⫽ [⫺2i

8][⫺18 ⫹ 24i

⫺6 ⫹ 4i]T ⫽ 68i.

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12. The matrix is Hermitian. We obtain the real value [1

⫺i

i]A[1 i

⫺i]T ⫽ [1 ⫺i

i][⫺4i

4 ⫺ 2i]T ⫽ 4.

14. (BA)T ⫽ (BA)T ⫽ A T B T ⫽ A(⫺B) ⫽ ⫺AB. For the matrices in Example 2. AB ⫽ c

1 ⫹ 19i

5 ⫹ 3i

⫺23 ⫹ 10i

⫺1

16. The inverse of a product UV of unitary matrices is (UV)ⴚ1 ⫽ Vⴚ1U ⴚ1 ⫽ V T U T ⫽ (UV )T. This proves that UV is unitary. We show that the inverse Aⴚ1 ⫽ B of a unitary matrix A is unitary. We obtain Bⴚ1 ⫽ (Aⴚ1)ⴚ1 ⫽ (A T)ⴚ1 ⫽ (Aⴚ1)T ⫽ B T, as had to be shown. 18. A TA ⫽ A2 ⫽ AA T if A is Hermitian, A TA ⫽ ⫺A2 ⫽ A(⫺A) ⫽ AA T if A is skewHermitian, A TA ⫽ A⫺1A ⫽ I ⫽ AA⫺1 ⫽ AA T if A is unitary. 20. For instance,

is not normal. A normal matrix that is not Hermitian, skew-Hermitian, or unitary is obtained if we take a unitary matrix and multiply it by 2 or some other real factor different from ⫾1.

SOLUTIONS TO CHAPTER 8 REVIEW QUESTIONS AND PROBLEMS, page 352 12. The eigenvalues are ⫺1 and 1. Corresponding eigenvectors are [2 3]T and [1 2]T, respectively. Note that this basis is not orthogonal. 14. One of the eigenvalues is 9. Its algebraic and geometric multiplicities are 2. Corresponding linearly independent eigenvectors are [1 0 ⫺2]T and [0 1 2]T. The other eigenvalue is 4.5. A corresponding eigenvector is [2 ⫺2 1]T. 16. The eigenvalues of A are ⫺5 and 25. The similar matrix, having the same eigenvalues, is Â ⫽ P ⫺1AP ⫽

1 ⫺1 c 6 2

2 ⫺1

100

d⫽c

18. A has the eigenvalues ⫺2, ⫺1, 2. The inverse of P is 1

⫺8

P ⴚ1 ⫽ D0

⫺3T .

⫺5

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The similar matrix Â, having the same eigenvalues, is 1

⫺8

Â ⫽ D0

⫺4

⫺26

⫺3T D 0

⫺1

⫺7

⫺35

⫺259

345

⫺27T .

⫺1

⫺7

6T ⫽ D 11

20. Eigenvalues are 65 and 520. A matrix of eigenvectors is X⫽ c

⫺8

Note that these vectors are orthogonal. Its inverse is Xⴚ1 ⫽

1 8 c 65 1

1 ⫺8

Diagonalization gives 1 ⫺8 c 56 1

1 ⫺8

520

⫺4160

d⫽c

520

22. The symmetric coefficient matrix is C⫽ c

⫺3

Its eigenvalues are 8 and 18; they are both positive real. The transformed form is 8y 21 ⫹ 18y 22 ⫽ 36. This is the canonical form; there is no y1y2-term. It represents an ellipse. The matrix X whose columns are normalized eigenvectors of C gives the relationship between y and x in the form x⫽

1 3 c 110 1

1 ⫺3

d y.

24. The symmetric coefficient matrix is C⫽ c

⫺5

Its eigenvalues are 13 and ⫺13. The transformed form is 13y 21 ⫺ 13y 22 ⫽ 13(y1 ⫹ y2)(y1 ⫺ y2) ⫽ 0. It represents two perpendicular straight lines through the origin. The matrix X whose columns are normalized eigenvectors of C gives the relationship between x and y in the form x⫽

1 3 c 113 2

⫺3 2

d y.

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CHAPTER 9

Vector Differential Calculus. Grad, Div, Curl

This chapter is independent of the previous two chapters (7 and 8). Formulas for grad, div, and curl in curvilinear coordinates are placed for reference in App. A3.4. SECTION 9.1. Vectors in 2-Space and 3-Space, page 354 Purpose. We introduce vectors in 3-space given geometrically by (families of parallel) directed segments or algebraically by ordered triples of real numbers, and we define addition of vectors and scalar multiplication (multiplication of vectors by numbers). Main Content, Important Concepts Vector, norm (length), unit vector, components Addition of vectors, scalar multiplication Vector space R3, linear independence, basis Comments on Content Our discussions in the whole chapter will be independent of Chaps. 7 and 8, and there will be no more need for writing vectors as columns and for distinguishing between row and column vectors. Our notation a ⫽ [a1, a2, a3] is compatible with that in Chap. 7. Engineers seem to like both notations a ⫽ [a1, a2, a3] ⫽ a1i ⫹ a2 j ⫹ a3k, preferring the first for “short” components and the second in the case of longer expressions. The student is supposed to understand that the whole vector algebra (and vector calculus) has resulted from applications, with concepts that are practical, that is, they are “made to measure” for standard needs and situations; thus, in this section, the two algebraic operations resulted from forces (forming resultants and changing magnitudes of forces); similarly in the next sections. The restriction to three dimensions (as opposed to n dimensions in the previous two chapters) allows us to “visualize” concepts, relations, and results and to give geometrical explanations and interpretations. On a higher level, the equivalence of the geometric and the algebraic approach (Theorem 1) would require a consideration of how the various triples of numbers for the various choices of coordinate systems must be related (in terms of coordinate transformations) for a vector to have a norm and direction independent of the choice of coordinate systems. Teaching experience makes it advisable to cover the material in this first section rather slowly and to assign relatively many problems, so that the student gets a feel for vectors in R3 (and R2) and the interrelation between algebraic and geometric aspects. Comments on Problems Problems 1–10 illustrate components and length. Operations on vectors (addition, scalar multiplication) in Probs. 11–20 are followed by applications to forces and velocities in Probs. 31–37. This includes questions on equilibrium and relative velocity.

157

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SOLUTIONS TO PROBLEM SET 9.1, page 360 2. v ⫽ [1, 1, ⫺1]; ƒ v ƒ ⫽ 13; ƒ u ƒ ⫽ [1> 13, 1> 13, ⫺1> 13] 4. v ⫽ [⫺2, ⫺8, ⫺4]. A line segment in space, of length 14 ⫹ 64 ⫹ 16 ⫽ 184, with the origin as midpoint. The unit vector is u ⫽ [⫺1> 121, ⫺4> 121, ⫺2> 121]. 6. Q: (4, 2, 13), ƒ v ƒ ⫽ 1189 8. Position vector of Q: (13.1, 0.8, ⫺2.0), ƒ v ƒ ⫽ 1176.25 10. Q: (0, 0, 0), ƒ v ƒ ⫽ 118 12. [4, 7, 8]. This illustrates (4b). Problem 13 illustrates (4a). 14. [15, ⫺3, 0] ⫽ 3[5, ⫺1, 0] 16. 9[32 ⫺ 53 , 1 ⫹ 13 , ⫺83 ] ⫽ [⫺32 , 12, ⫺24] 18. [0, 26, 0], [0, ⫺26, 0] 22. 0, equilibrium 24. [1, 1, 0] 26. v ⫽ ⫺(p ⫹ q ⫹ u) ⫽ [⫺4, ⫺9, 3] 28. [1> 12, 1> 12, 0]. Unit vectors will play a role in fixing (determining) directions. 30. v ⫽ [⫺1, ⫺1, v3] with arbitrary v3 32. 2 ⬉ ƒ p ⫹ q ƒ ⬉ 10; nothing about direction. Application: suitable lengths of the portions of an arm. 34. vB ⫺ vA ⫽ [⫺450> 12, 450> 12] ⫺ [⫺550> 12, ⫺550> 12] ⫽ [100> 12, 1000> 12] 36. Choose a coordinate system whose axes contain the mirrors. Let u ⫽ [u 1, u 2] be incident. Then the first reflection gives, say, v ⫽ [u 1, ⫺u 2], and the second w ⫽ [⫺u 1, ⫺u 2] ⫽ ⫺u. The reflected ray is parallel to the incoming ray, with the direction reversed. 38. Team Project. (a) The idea is to write the position vector of the point of intersection P in two ways and then to compare them, using that a and b are linearly independent vectors. Thus l(a ⫹ b) ⫽ a ⫹ ␮(b ⫺ a). l ⫽ 1 ⫺ ␮ are the coefficients of a and l ⫽ ␮ those of b. Together, l ⫽ ␮ ⫽ 12, expressing bisection. (b) The idea is similar to that in part (a). It gives l(a ⫹ b) ⫽ 12 a ⫹ ␮ 12 (b ⫺ a). l ⫽ 12 ⫺ 12 ␮ from a and l ⫽ 12 ␮ from b, resulting in l ⫽ 14 , thus giving a ratio A 34 B : A 14 B ⫽ 3:1. (c) Partition the parallelogram into four congruent parallelograms. Part (a) gives 1:1 for a small parallelogram, hence 1: (1 ⫹ 2) for the large parallelogram. (d) v (P) ⫽ 12 a ⫹ l(b ⫺ 12 a) ⫽ 12 b ⫹ ␮(a ⫺ 12 b) has the solution l ⫽ ␮ ⫽ 13 , which gives by substitution v (P) ⫽ 13 (a ⫹ b) and shows that the third median OQ passes through P and OP equals 23 of ƒ v (Q) ƒ ⫽ 12 ƒ a ⫹ b ƒ , dividing OQ in the ratio 2:1, too. (e) In the figure in the problem set, a ⫹ b ⫹ c ⫹ d ⫽ 0; hence c ⫹ d ⫽ ⫺(a ⫹ b). Also, AB ⫽ 12 (a ⫹ b), CD ⫽ 12 (c ⫹ d) ⫽ ⫺12 (a ⫹ b), and for DC we get ⫹12 (a ⫹ b),

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which shows that one pair of sides is parallel and of the same length. Similarly for the other pair. (f) Let a, b, c be edge vectors with a common initial point (see the figure below). Then the four (space) diagonals have the midpoints AG: BH: EC: DF:

1 2 (a ⫹ b ⫹ c) a ⫹ 12 (b ⫹ c ⫺ a) c ⫹ 12 (a ⫹ b ⫺ c) b ⫹ 12 (a ⫹ c ⫺ b),

and these four position vectors are equal. H

D G

b F

C a B

Section 9.1. Parallelepiped in Team Project 38(f)

(g) Let v1, Á , vn be the vectors. Their angle is a ⫽ 2p>n. The interior angle at each vertex is b ⫽ p ⫺ (2p>n). Put v2 at the terminal point of v1, then v3 at the terminal point of v2, etc. Then the figure thus obtained is an n-sided regular polygon, because the angle between two sides equals p ⫺ a ⫽ b. Hence v1 ⫹ v2 ⫹ Á ⫹ vn ⫽ 0. (Of course, for even n the truth of the statement is immediately obvious.)

SECTION 9.2. Inner Product (Dot Product), page 361 Purpose. We define, explain, and apply a first kind of product of vectors, the dot product a • b, whose value is a scalar. Main Content, Important Concepts Definition (1) Dot product in terms of components Orthogonality Length and angle between vectors in terms of dot products Cauchy–Schwarz and triangle inequalities Comment on Dot Product This product is motivated by work done by a force (Example 2), by the calculation of components of forces (Example 3), and by geometric applications such as those given in Examples 5 and 6. “Inner product” is more modern than “dot product” and is also used in more general settings (see Sec. 7.9).

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Comments on Text Figure 178 shows geometrically why the inner product can be positive or negative or— this is the most important case—zero, in which the vectors are called orthogonal. This includes the case of two zero vectors in the definition, in which case the angle is no longer defined. Equations (6)–(8) concern relationships that also extend to more abstract setting, where they turn out to be of basic importance; see [GenRef7]. Examples 2–6 in the text show some simple applications of inner products in mechanics and geometry that motivate these products. Further applications will appear as we proceed. Comments on Problems Problems 1–10 illustrate the various laws for inner products. Problems 11–16 include a modest amount of theory. Problems 17–40 add further applications of inner products in mechanics and geometry, including a generalization of the concept of component in Probs. 36–40 that is quite useful. SOLUTIONS TO PROBLEM SET 9.2, page 367 2. ⫺132,

660

4. ƒ [5, ⫺3, 13] ƒ ⫽ 1203 ⫽ 14.25, ƒ a ƒ ⫹ ƒ b ƒ ⫽ 135 ⫹ 180 ⫽ 14.86 ⬎ 14.25. This illustrates the very important triangle inequality (7). 6. 0 by (8). The left side of (8) is the sum of the squares of the diagonals; the right side equals the sum of the squares of the four sides of the parallelogram. 8. 65 • 44 ⫽ 2860 10. 68, ⫺24. Note that there is no formula according to which the two given expressions could be equal. 12. u • (v ⫺ w) ⫽ 0 implies nothing if u ⫽ 0 and implies orthogonality of u and v ⫺ w if u ⫽ 0. 16. ƒ a ⫹ b ƒ 2 ⫽ (a ⫹ b) • (a ⫹ b) ⫽ ƒ a ƒ 2 ⫹ 2 ƒ a ƒ ƒ b ƒ ⫹ ƒ b ƒ 2 ⫽ ( ƒ a ƒ ⫹ ƒ b ƒ ) 18. 0; the vectors p and v ⫽ [6, 7, 5] are orthogonal. 20. [6, ⫺3, ⫺3] • [2, ⫺1, ⫺1] ⫽ 18 ⫽ ƒ p ƒ ƒ v ƒ because p has the direction of AB ⫽ v ⫽ [2, ⫺1, ⫺1]. 22. arccos (5> 114 ⴢ 5) ⫽ arccos 0.9449 ⫽ 0.3335 ⫽ 19.1° 24. a ⫹ c ⫽ [2, 1, 2],

b ⫹ c ⫽ [4, 2, 3], hence g ⫽ arccos (16> 19 ⴢ 29) ⫽ 0.1389 ⫽ 7.96°.

26. ƒ c ƒ 2 ⫽ ƒ a ⫺ b ƒ 2 ⫽ (a ⫺ b) • (a ⫺ b) ⫽ ƒ a ƒ 2 ⫹ ƒ b ƒ 2 ⫺ 2 ƒ a ƒ ƒ b ƒ cos g 28. gA ⫽ arccos (AB, AC) ⫽ arccos (3>(3 ⴢ 13)) ⫽ 54.74°, gB ⫽ arccos (BA, BC) ⫽ 35.26°, g3 ⫽ arccos (CA, CB) ⫽ 90° 30. The distance of P: 3x ⫹ y ⫹ z ⫽ 9 from the origin is 9> ƒ a ƒ ⫽ 9> 111 ⫽ 2.714 (Hesse’s normal form). The plane parallel to P through A is 3x ⫹ y ⫹ z ⫽ 3 ⴢ 1 ⫹ 1 ⴢ 0 ⫹ 1 ⴢ 2 ⫽ 5.

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Its distance from the origin is 5> 111 ⫽ 1.508. This gives the answer 2.714 ⫺ 1.508 ⫽ 1.206. 32. Necessary and sufficient is the orthogonality of the normal vectors [3, 0, 1] and [8, ⫺1, c]. Hence [3, 0, 1] • [8, ⫺1, c] ⫽ 0,

c ⫽ ⫺24.

34. Let the mirrors correspond to the coordinate planes. If the ray [v1, v2, v3] first hits the yz-plane, then the xz-plane, and then the xy-plane, it will be reflected to [⫺v1, v2, v3], [⫺v1, ⫺v2, v3], [⫺v1, ⫺v2, ⫺v3]; hence the angle is 180°, the reflected ray will be parallel to the incident ray but will have the opposite direction. Corner reflectors have been used in connection with missiles; their aperture changes if the axis of the missile deviates from the tangent direction of the path. See E. Kreyszig, On the theory of corner reflectors with unequal faces. Ohio State University: Antenna Lab Report 601>19. 36. 6> 114 38. ⫺34> 117 ⫽ ⫺2117. Note that the vectors have exactly opposite directions; this is a case in which the component will have a minus sign. Also ƒ a ƒ > ƒ b ƒ ⫽ 168>17 gives the factor 2. 40. Nothing because ƒ b ƒ appears in the numerator a • b as well as in the denominator. SECTION 9.3. Vector Product (Cross Product), page 368 Purpose. We define and explain a second kind of product of vectors, the cross product a ⴛ b, which is a vector perpendicular to both given vectors (or the zero vector in some cases). Main Content, Important Concepts Definition of cross product, its components (2), (2**) Right- and left-handed coordinate systems Properties (anticommutative, not associative) Scalar triple product Prerequisites. Elementary use of second- and third-order determinants (see Sec. 7.6) Comment on Motivations Cross products were suggested by the observation that, in certain applications, one associates with two given vectors a third vector perpendicular to the given vectors (illustrations in Examples 4–6). Scalar triple products can be motivated by volumes and linear independence (Theorem 2 and Example 6). Comments on Problems Problems 1–10 should help in obtaining an intuitive understanding of the cross product and give further motivation of this concept by applications. Problems 11–23 compare various products, with emphasis on those of three factors. Team Project 24 concerns standard formulas needed in working with dot and cross products and their combination. Problems 25–35 show some further applications in mechanics and geometry, to emphasize further that the definitions of these products are motivated by applications.

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SOLUTIONS TO PROBLEM SET 9.3, page 374 2. a ⴛ (b ⫺ c) ⫽ 0, thus b ⫽ c or b ⫺ c has the direction of a or ⫺a. 4. The cross product is i

v ⫽ a ⴛ b ⫽ 43

2 4 ⫽ (8 ⫺ 0)i ⫹ (2 ⫺ 6)j ⫹ (0 ⫺ 4)k

2 ⫽ [8, ⫺4, ⫺4].

Its length is ƒ v ƒ ⫽ 196, which equals the right side of (12), 229 ⴢ 5 ⫺ 72 ⫽ 1145 ⫺ 49. Using the definition of the length of a vector product and the given hint, we obtain (12) by taking the square roots of ƒ a ⴛ b ƒ 2 ⫽ ƒ a ƒ 2 ƒ b ƒ 2 sin2 g ⫽ ƒ a ƒ 2 ƒ b ƒ 2(1 ⫺ cos2 g) ⫽ (a • a)(b • b) ⫺ (a • b)2. 6. Instead of vd you now have 2vd ⫽ 2 ƒ w ⴛ r ƒ , hence ƒ v ƒ doubles. 8. We obtain i

v ⫽ 4 10> 12

10> 12

k 0 4 ⫽ [⫺1012, 1012, ⫺1012] ⫺2

so that the speed is 1600. 12. [150, ⫺195, ⫺315],

[⫺150, 195, 315], ⫺75, ⫺75

14. 0 because of anticommutativity 16. The first expression gives [⫺4, ⫺6, ⫺14] • [5, ⫺1, 3] ⫽ ⫺56. The second expression looks totally different but, of course, gives the same value: [⫺3, 2, 0] • [10, ⫺13, ⫺21] ⫽ ⫺56. 18. [⫺7, 14, 0]. The student should note and understand why both product vectors lie in the plane of a and b, why neither of them is zero, and why they are the same. This should become clear by drawing little sketches of the factors and products. 20. Formula (14) shows that the two expressions are equal, namely, equal to [91, 70, 0]. The intermediate calculation of the second expression is 21[1, 4, ⫺2] ⫺ (⫺14)[5, ⫺1, 3] ⫽ [21, 84, ⫺42] ⫹ [70, ⫺14, 42]. 22. 28,

24. Team Project. To prove (13), we choose a right-handed Cartesian coordinate system such that the x-axis has the direction of d and the xy-plane contains c. Then the vectors in (13) are of the form b ⫽ [b1, b2, b3],

c ⫽ [c1, c2, 0],

d ⫽ [d1, 0, 0].

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Hence by (2**), i

c ⴛ d ⫽ 4 c1

0 4 ⫽ ⫺c2d1k,

b ⴛ (c ⴛ d) ⫽ 4 b1

b3 4 .

⫺c2d1

The “determinant” on the right equals [⫺b2c2d1,

b1c2d1,

0]. Also,

(b • d)c ⫺ (b • c)d ⫽ b1d1[c1, c2, 0] ⫺ (b1c1 ⫹ b2c2)[d1, 0, 0] ⫽ [⫺b2c2d1,

b1d1c2,

0].

This proves (13) for our special coordinate system. Now the length and direction of a vector and a vector product, and the value of an inner product, are independent of the choice of the coordinates. Furthermore, the representation of b ⴛ (c ⴛ d) in terms of i, j, k will be the same for right-handed and left-handed systems, because of the double cross multiplication. Hence, (13) holds in any Cartesian coordinate system, and the proof is complete. Equation (14) follows from (13) with b replaced by a ⴛ b. To prove (15), we note that a • [b ⴛ (c ⴛ d)] equals (a

b [c ⴛ d]) ⫽ (a ⴛ b) • (c ⴛ d)

by the definition of the triple product, as well as (a • c)(b • d) ⫺ (a • d)(b • c) by (13) (take the dot product by a). The last formula, (16), follows from familiar rules of interchanging the rows of a determinant. 26. m ⫽ [2, 3, 2] ⴛ [1, 0, 3] ⫽ [9, ⫺4, ⫺3]; m ⫽ 1106 ⫽ 10.3 28. The midpoints are M 1: (3.5, 0, 0)

Midpoint of AB,

M 2: (6.5, 0.5, 0)

Midpoint of BC,

M 3: (6, 2.5, 0)

Midpoint of CD,

M 4: (3, 2, 0)

Midpoint of DA.

The cross product of adjacent sides of Q is M 1M 4 ⴛ M 1M 2 ⫽ [0.5, 2, 0] ⴛ [3, 0.5, 0] ⫽ [0, 0, ⫺6.25]. Its length 6.25 is the area of Q. 30. A normal vector is N ⫽ AB ⴛ AC ⫽ [3, 0, ⫺2.25] ⴛ [⫺1, 6, 3.75] ⫽ [13.5, ⫺9, 18]. Hence the plane is represented by N • r ⫽ 13.5x ⫺ 9y ⫹ 18z ⫽ c with c obtained by substituting the coordinates of C (or of A or B) c ⫽ ⫺9 ⴢ 8 ⫹ 18 ⴢ 4 ⫽ 0.

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32. 10 34. Edge vectors are [3 ⫺ 1, 7 ⫺ 3, 12 ⫺ 6], [8 ⫺ 1, 8 ⫺ 3, 9 ⫺ 6], [2 ⫺ 1, 2 ⫺ 3, 8 ⫺ 6]. The mixed triple product of these vectors is ⫺90 (or +90). This gives the answer 15.

SECTION 9.4. Vector and Scalar Functions and Their Fields. Vector Calculus: Derivatives, page 375 Purpose. To get started on vector differential calculus, we discuss vector functions and their continuity and differentiability. Differentiation of scalar and vector functions will be needed throughout the rest of the chapter for developing the differential geometry of curves with application to mechanics (Sec. 9.5) and the three operators, gradient, with application to directional derivatives (Sec. 9.7), divergence (Sec. 9.8), and curl (Sec. 9.3). The form of these operators in curvilinear coordinates is given in App. A3.4. Main Content, Important Concepts Vector and scalar functions and fields Continuity, derivative of vector functions (9), (10) Differentiation of dot, cross, and triple products, (11)–(13) Partial derivatives Comment on Content This parallels calculus of functions of one variable and can be surveyed quickly. Further Comments on Text A vector field (or scalar field) may be given along a straight line, along a curve (Fig. 195) or a surface (Fig. 196) or in a three-dimensional region of space. In practice, these are the most important cases for the engineer. Important applications are scalar fields in space (Example 1), velocity fields of rotations (Example 2), and the gravitational field of masses (Example 3). Convergence, continuity, and differentiability of vector functions are defined in connection with (4), (8), and (9), and these concepts relating to vector functions can be expressed in terms of components. In particular, formula (10) states that a vector function can be differentiated componentwise. Formulas (11)–(13) are immediate consequences of familiar differentiation rules. An extension of this to partial differentiation is illustrated in Example 5. Comments on Problems Although there is practically not much difference in working in the plane and in space, we begin in Probs. 1–8 with the former case, where visualization and graphing (sketching) is much simpler. Extension to space follows in Probs. 9–14. Those first problems concern scalar fields, which are simpler than vector fields, which may technically be regarded as triples of (coordinate-dependent!) scalar fields (which conceptually they are not!).

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The set ends with a few problems (22–25) in differential calculus. Here the student should consult and review material from his or her calculus text. SOLUTIONS TO PROBLEM SET 9.4, page 380 2. Hyperbolas T ⫽ xy ⫽ const with the coordinate axes as asymptotes. 4. Straight lines through the origin (planes through the z-axis) y>x ⫽ const. 6. x>(x 2 ⫹ y 2) ⫽ c, hence x ⫽ c(x 2 ⫹ y 2). Division by c (⫽ 0) gives x 2 2 c⫽x ⫹y ,

thus

1 1 ax ⫺ b ⫹ y 2 ⫽ 2 . 2c 4c

These are circles with center at 1>(2c) on the y-axis and radius 1>(2 ƒ c ƒ ), so that they all pass through the origin. 8. CAS Project. A CAS can graphically handle these more complicated functions, whereas the paper-and-pencil method is relatively limited. This is the point of this project. Note that all these functions occur in connection with Laplace’s equation, so that they are real or imaginary parts of complex analytic functions. 10. Ellipsoids of revolution. The ellipsoid 9x 2 ⫹ 9y 2 ⫹ z 2 ⫽ c2 intersects the axes at c>3, c>3, and c, respectively. 12. Congruent circular cones z ⫽ 2x 2 ⫹ y 2 ⫹ c with apex at z ⫽ c on the z-axis. 14. Congruent parabolic cylinders with vertical generators and the xz-plane as plane of symmetry. 16. This could be the velocity field of a counterclockwise rotation about the origin. Indeed, at a point (x, y) the vector v is perpendicular to the segment from the origin to (x, y). Also, ƒ v ƒ ⫽ 2x 2 ⫹ y 2, that is, the speed is proportional to the distance of the point from the origin (the axis of rotation in space), as it should be for such a rotation. 18. v has radial direction away from the origin. 20. Clockwise rotation; compare with Prob. 16. 22. r r ⫽ [⫺6 sin 2t, 6 cos 2t, 4]. The second derivative is r s ⫽ [⫺12 cos 2t, ⫺12 sin 2t, 0]. This problem has to do with a helix, as we shall see in the next section. 24. y1x ⫽ [ex cos y, ex sin y], v1y ⫽ [⫺ex sin y, ex cos y]. Similarly, for the second given function, v2 x ⫽ [⫺sin x cosh y, ⫺cos x sinh y] and v2y ⫽ [cos x sinh y, ⫺sin x cosh y]. SECTION 9.5. Curves. Arc Length. Curvature. Torsion, page 381 Purpose. Discussion of space curves as an application of vector functions of one variable, the use of curves as paths in mechanics (and as paths of integration of line integrals in Chapter 10). Role of parametric representations, interpretation of derivatives in mechanics, completion of the discussion of the foundations of differential–geometric curve theory.

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Main Content, Important Concepts Parametric representation (1) Orientation of a curve Circle, ellipse, straight line, helix Tangent vector (7), unit tangent vector (8), tangent (9) Length (10), arc length (11) Arc length as parameter [cf. (14)] Velocity, acceleration (16)–(19) Centripetal acceleration, Coriolis acceleration Curvature, torsion, Frenet formulas (Prob. 50) Short Courses. This section can be omitted. Comments on Text This long section gives an overview of the differential geometry of curves in space, as needed in mechanics, where velocity and tangential and normal acceleration are basic; see (17), (18), and (18*). The discussion begins with parametric representations (Examples 1–4), tangents (Example 5), and arc length (for the helix in Example 6). Then the section turns to mechanics, discussing centripetal and centrifugal forces (Example 7) and Coriolis acceleration appearing in the superposition of rotations, as for the motion of missiles (Example 8 and Fig. 211). We finally discuss curvature ␬ and torision t and related concepts shown in Fig. 212; since this is of minor interest to the engineer, we leave this last part of the section optional. The culmination of this are the Frenet formulas (Probs. 54 and 55), which imply that ␬(s) and t(s), if sufficiently differentiable, determine a curve uniquely, except for its position in space. Comments on Problems These follow the train of thoughts in the text, discussing first parametric representations in detail (Probs. 1–23). Here, Prob. 23 shows a list of classical curves the engineer may need from time to time. Problems 24–28 concern the representations of tangents. Problems 29–32 involve only integrals that are simple, which is generally not the case in connection with lengths of curves. Problems 35–46 concern mechanics. The remaining Probs. 47–55 correspond to the optional parts of the text regarding curvature and torsion. SOLUTIONS TO PROBLEM SET 9.5, page 390 2. Straight line through (a, b, c) in the direction of the vector [1, 3, ⫺5]. 4. Circle of radius 5 and center (2, ⫺1) in the plane x ⫽ ⫺2, which is parallel to the yz-plane. 6. This is an ellipse with center (a, b) and semi-axes 3 and 2, oriented clockwise because of the minus sign. Because of the factor p the whole curve is obtained if we let t vary from 0 to 2p> p ⫽ 2. 8. x 2 ⫺ y 2 ⫽ cosh2 t ⫺ sinh2 t ⫽ 1 gives a hyperbola in the plane z ⫽ 2.

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10. Hyperbola xz ⫽ 1 in the plane y ⫽ 2. 12. The yz-plane is x ⫽ 0. The center (4, 0) has the distance 5 from (0, 3). Hence a representation is r ⫽ [0, 4 ⫹ 5 cos t, 0 ⫹ 5 sin t]. 14. A vector from (1, 1, 1) to (4, 0, 2) is b ⫽ [3, ⫺1, 1]. Hence a representation is r ⫽ [1, 1, 1] ⫹ bt ⫽ [1 ⫹ 3t, 1 ⫺ t, 1 ⫹ t]. 16. Ellipse r ⫽ [cos t, sin t, sin t]. Since the plane makes an angle of 45° with the xy-plane, the semi-axes of the ellipse are 12 (in the y-direction) and 1 (in the x-direction); indeed, the apex at (0, 1, 1) has distance 12 from the origin. 18. r ⫽ [5 cos t, 5 sin t, 2t]. 20. This linear system of two equations in three unknowns has the solution r ⫽ [75 ⫺ t, 45 ⫹ t, t] where z ⫽ t remains arbitrary. Hence this may be regarded as a parametric representation of the straight line of intersection of the two planes given by the two equations. Obviously, this line is determined by a point through which it passes and a direction, given by a vector v. As a point we can choose the intersection of the line with the plane z ⫽ 0 (the xy-plane), for which the given equations, with z ⫽ 0, yield 2x ⫺ y ⫽ 2,

x ⫹ 2y ⫽ 3

and have the solution x ⫽ 75 , y ⫽ 45 . Hence a ⫽ [75 , 45 , 0] is a point on the line of intersection, call it L. The direction of the latter is given by the vector product for the two normal vectors of the given planes, that is i

v ⫽ 42

⫺1

k 3 4 ⫽ [⫺5, 5, 5]. ⫺1

Hence a parametric representation of the straight line of intersection of the two planes is r ⫽ a ⫹ t v ⫽ a ⫹ [⫺5t, 5t, 5t] ⫽ [75 ⫺ 5t, 45 ⫹ 5t, 5t]. 24. P corresponds to t ⫽ 2; indeed, r(2) ⫽ [2, 2, 1]. Differentation gives r r (t) ⫽ [1, t, 0]

and at P,

r r (2) ⫽ [1, 2, 0].

The unit tangent vector in the direction of r r (t) is u r (t) ⫽ (1 ⫹ t 2)ⴚ1>2[1, t, 0]

and at P,

u r (2) ⫽ [1> 15, 2> 15].

This gives the representation of the tangent of C at P in the form q (w) ⫽ r (2) ⫹ wr r (2) ⫽ [2 ⫹ w, 2 ⫹ 2w, 0]. The tangent at P has a positive slope, as expected. 26. Differentiation gives a tangent vector r r (t) ⫽ [⫺sin t, cos t, 9].

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P corresponds to the parametric value t ⫽ 2p. The value of r r at P is [0, 1, 9]. A representation of the tangent at P is q (w) ⫽ [1 ⫹ 0, 0 ⫹ w, 18p ⫹ 9w]. 28. A tangent vector is r r (t) ⫽ [1, 2t, 3t 2]. The corresponding unit tangent vector is u ⫽ (1 ⫹ 4t 2 ⫹ 9t 4)ⴚ1>2[1, 2t, 3t 2]. P corresponds to t ⫽ 1. Hence at P we have r r ⫽ [1, 2, 3] and u ⫽ 14ⴚ1>2[1, 2, 3]. A representation of the tangent at P is q (w) ⫽ [1 ⫹ w, 1 ⫹ 2w, 1 ⫹ 3w]. 30. The initial and terminal point of the arc correspond to t ⫽ 0 and t ⫽ 2p. Differentiation gives a tangent vector r r ⫽ [⫺4 sin t, 4 cos t, 5]. The integrand needed is 2r r • r r ⫽ 141. Hence the length is l ⫽ 2p 141. 32. r r ⫽ [⫺3a cos2 t sin t, 3a sin2 t cos t]. Taking the dot product and applying trigonometric simplification gives r r • r r ⫽ 9a 2 cos4 t sin2 t ⫹ 9a 2 sin4 t cos2 t ⫽ 9a 2 cos2 t sin2 t ⫽

9a 2 2 sin 2t. 4

From this we obtain as the length in the first quadrant l⫽

3 a 2

p>2

冮 sin 2t dt ⫽ ⫺ 3a4 (cos p ⫺ cos 0) ⫽ 3a2 .

Answer: 6a

34. We obtain ds 2 ⫽ dx 2 ⫹ dy 2 ⫽ (dr cos u ⫺ r sin u du)2 ⫹ (dr sin u ⫹ r cos u du)2 ⫽ dr2 ⫹ r2 du2 ⫽ (r r 2 ⫹ r2) du2. For the cardioid. r2 ⫹ r r 2 ⫽ a 2(1 ⫺ cos u)2 ⫹ a 2 sin2 u ⫽ 2a 2(1 ⫺ cos u) ⫽ 4a 2 sin2 12 u

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so that 2p

l ⫽ 2a

冮 sin 12 u du ⫽ 8a. 0

36. v ⫽ r r ⫽ [8, 6, 0], ƒ v ƒ ⫽ 10, a ⫽ 0, a nonaccelerated motion (uniform motion, motion of constant speed) 38. v ⫽ r r ⫽ [⫺sin t, 2 cos t, 0], ƒ v ƒ ⫽ (sin2 t ⫹ 4 cos2 t)1>2, a ⫽ [⫺cos t, ⫺2 sin t, 0]. Hence the tangential acceleration is atan ⫽

⫺3 sin t cos t sin2 t ⫹ 4 cos2 t

[⫺sin t,

2 cos t,

and has the magnitude ƒ atan ƒ , where ƒ atan ƒ 2 ⫽

9 sin2 t cos2 t sin2 t ⫹ 4 cos2 t

40. The velocity is v ⫽ [⫺2 sin t ⫺ 2 sin 2t, 2 cos t ⫺ 2 cos 2t]. From this we obtain the square of the speed ƒ v ƒ 2 ⫽ v • v ⫽ (⫺2 sin t ⫺ 2 sin 2t)2 ⫹ (2 cos t ⫺ 2 cos 2t)2. Performing the squares and simplifying gives ƒ v ƒ 2 ⫽ 8(1 ⫹ sin t sin 2t ⫺ cos t cos 2t) ⫽ 8(1 ⫺ cos 3t) ⫽ 16 sin2

3t . 2

Hence ƒ v ƒ ⫽ 4 sin

3t . 2

a ⫽ [⫺2 cos t ⫺ 4 cos 2t, ⫺2 sin t ⫹ 4 sin 2t]. We use (18*). By straightforward simplification (four terms cancel), a • v ⫽ 12(cos t sin 2t ⫹ sin t cos 2t) ⫽ 12 sin 3t. Hence (18*) gives atan ⫽

12 sin 3t 16 sin2 (3t>2)

anorm ⫽ a ⫺ atan. 42. The velocity vector is v ⫽ [c cos t ⫺ ct sin t, c sin t ⫹ ct cos t, c]. Hence the square of the speed is ƒ v ƒ 2 ⫽ c2(t 2 ⫹ 2).

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Another differentiation gives the acceleration a ⫽ [⫺2c sin t ⫺ ct cos t, 2c cos t ⫺ ct sin t, 0]. The tangential acceleration is atan ⫽

ct t ⫹2 2

[cos t ⫺ t sin t, sin t ⫹ t cos t, 1]

and the normal acceleration is anorm ⫽ a ⫺ atan. This is a spiral on a cone. 44. R ⫽ 3.85 # 108 m, ƒ v ƒ ⫽ 2pR>(2.36 # 106) ⫽ 1025 [m>sec], ƒ v ƒ ⫽ vR, ƒ a ƒ ⫽ v2> R ⫽ ƒ v ƒ 2>R ⫽ 0.0027 [m>sec 2], which is only 2.8 # 10⫺4 g, where g is the acceleration due to gravity at the Earth’s surface. 46. R ⫽ 3960 ⫹ 450 ⫽ 4410 [mi], 2pR ⫽ 100 ƒ v ƒ , ƒ v ƒ ⫽ 277.1 mi>min, g ⫽ ƒ a ƒ ⫽ v2R ⫽ ƒ v ƒ 2>R ⫽ 17.41 [mi>min2] ⫽ 25.53 [ft>sec2] ⫽ 7.78 [m>sec2].

Here we used ƒ v ƒ ⫽ vR. 48. We denote derivatives with respect to t by primes. In (22), u⫽

dr dt ⫽ rr , ds ds

dt 1 ⫽ ⫽ (r r • r r )ⴚ1>2. ds sr

[See (12).]

Thus in (22), 2

du dt d 2t d 2t ⫽ r s a b ⫹ r r 2 ⫽ r s (r r • r r )ⴚ1 ⫹ r r 2 ds ds ds ds where d 2t d dt dt 1 ⫽ a b ⫽ ⫺ (r r • r r )ⴚ3>2 2(r s • r r )(r r • r r )ⴚ1>2 dt ds ds 2 ds 2 ⫽ ⫺(r s • r r )(r r • r r )ⴚ2. Hence du ⫽ r s (r r • r r )ⴚ1 ⫺ r r (r s • r r )(r r • r r )ⴚ2 ds du du • ⫽ (r s • r s )(r r • r r )ⴚ2 ⫺ 2(r s • r r )2(r r • r r )ⴚ3 ⫹ (r r • r r )ⴚ3(r s • r r )2 ds ds ⫽ (r s • r s )(r r • r r )ⴚ2 ⫺ (r s • r r )2(r r • r r )ⴚ3. Taking square roots, we get (22*). 50. t ⫽ ⫺p • (u ⴛ p) r ⫽ ⫺p • (u r ⴛ p ⫹ u ⴛ p r ) ⫽ 0 ⫺ (p u p r ) ⫽ ⫹(u p p r ). Now u ⫽ r r , p ⫽ (1>␬)r s ; hence p r ⫽ (1>␬)r t ⫹ (1>␬) r r s . Inserting this into the triple product (the determinant), we can simplify the determinant by familiar rules and let the last term in p r disappear. Pulling out 1>␬ from both p and p r , we obtain the second formula in (23**). 52. From r (t) ⫽ [a cos t, a sin t, r r ⫽ [⫺a sin t,

ct] we obtain

a cos t, c],

r r • r r ⫽ a 2 ⫹ c2 ⫽ K 2.

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Hence, by integration, s ⫽ Kt. Consequently, t ⫽ s>K. This gives the indicated representation of the helix with arc length s as parameter. Denoting derivatives with respect to s also by primes, we obtain r(s) ⫽ Ba cos u(s) ⫽ r r (s) ⫽ B⫺ r s (s) ⫽ B⫺

s , K

a sin

a s sin , K K

a s cos , K K2

⫺

b r (s) ⫽ B

a s sin , K K2

a K2

s c sin , K K

c s , 2 cos K K

t (s) ⫽ ⫺p (s) • b r (s) ⫽

⫽

a a 2 ⫹ c2

⫺sin ⫺

c 2

⫽

s , K

s c cos , K K

c s , 2 sin K K

K 2 ⫽ a 2 ⫹ c2 c R K

1 s r s (s) ⫽ B⫺cos , ␬ (s) K

b(s) ⫽ u(s) ⫻ p(s) ⫽ B

cs R, K

a s cos , K K

␬(s) ⫽ ƒ r s ƒ ⫽ 2r s • r s ⫽ p (s) ⫽

s , K

0R c

a ⫹ c2 2

a R K

Positive c gives a right-handed helix and positive torsion; negative c gives a lefthanded helix and negative torsion. 54. p r ⫽ (1>␬)u r implies the first formula, u r ⫽ ␬p. The third Frenet formula was given in the text before (23). To obtain the second Frenet formula, use p r ⫽ (b ⴛ u) r ⫽ b r ⴛ u ⫹ b ⴛ u r ⫽ ⫺tp ⴛ u ⫹ b ⴛ ␬p ⫽ ⫹tb ⫺ ␬u. In differential geometry (see [GenRef8] in App. 1) it is shown that the whole differential–geometric theory of curves can be obtained from the Frenet formulas, whose solution shows that the natural equations ␬ ⫽ ␬(s), t ⫽ t(s) determine a curve uniquely, except for its position in space.

SECTION 9.6. Calculus Review: Functions of Several Variables. Optional, page 392 Purpose. To give students a handy reference and some help on material known from calculus that they will need in their further work.

SECTION 9.7. Gradient of a Scalar Field. Directional Derivative, page 395 Purpose. To discuss gradients and their role in connection with directional derivatives, surface normals, and the generation of vector fields from scalar fields (potentials).

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Main Content, Important Concepts Gradient, nabla operator Directional derivative, maximum increase, surface normal Vector fields as gradients of potentials Laplace’s equation Comments on Content This is probably the first section in which one should no longer rely on knowledge from calculus, although relatively elementary calculus books usually include a passage on gradients. Potentials are important; they will occur at a number of places in our further work. Further Comments on Text Figure 215 illustrates the directional derivatives geometrically. Note that s can be positive, zero, or negative. Theorem 1 is needed because the gradient in (1) involves coordinates. Figure 216 illustrates a major geometric application of the gradient. The notion of potential is basic, and Theorem 3 states one of the most important examples. Coulomb’s law (12) is of the same form as Newton’s law of graviation in (8); thus the two are governed by the same theory. Comments on Problems Problems 1–17 require specific calculations and show some general foumulas for the gradient and the Laplacian. Problems 18–23 and 43–45 concern vector fields and their potentials. Problems 30–35 show applications to curve and surface theory. Directional derivatives are considered in Probs. 36–42. Hence the problem set reflects the many-sided aspects of the gradient and its applications.

SOLUTIONS TO PROBLEM SET 9.7, page 402 2. v ⫽ grad f ⫽ [18x, 8y] 4. v ⫽ grad f ⫽ [2(x ⫺ 4), 2( y ⫹ 6)] 6. v ⫽ grad f ⫽ (x 2 ⫹ y 2)ⴚ2[4xy 2, ⫺4yx 2] 8. Applying the product rule to each component of ⵜ( fg) and collecting terms, the formula follows, [( fg)x, ( fg)y, ( fg)z] ⫽ [ fxg, fyg, fz g] ⫹ [ fgx, fgy, fgz]. 10. Apply the product rule twice to each of the three terms of ⵜ2, obtaining ( fg)xx ⫽ fxxg ⫹ 2fxgx ⫹ fgxx and so on, and reorder and collect terms into three sums that make up the right side of the formula. 12. v ⫽ ⵜf ⫽ c ⫺

d , v(1, 1) ⫽ [0, ⫺12], so that the gradient at (x ⫹ y ) (x ⫹ y 2)2 (1, 1) is pointing in the negative y-direction. 14. v ⫽ grad f ⫽ ⫺(x 2 ⫹ y 2 ⫹ z 2) ⴚ3>2[x, y, z]. Its value at P is [⫺0.0015, 0, ⫺0.0020]. x 2 ⫺ y2 2

2 2

,⫺

2xy

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16. v ⫽ ⵜf ⫽ [50x, 18y, 32z] ⫽ ⫺k[x, y, z] has solutions precisely for the points on the three principal axes of the ellipsoids, that is, for P on the coordinate axes. 18. v ⫽ ⵜf ⫽ [2x ⫺ 6, ⫺2y]. The value at P is [⫺8, ⫺10]. The curves f ⫽ const are hyperbolas with asymptotes y ⫽ ⫾(x ⫺ 3). 20. v ⫽ ⵜf ⫽ c 1 ⫹

2x 2

2xy

d . At (1, 1) this equals [1, ⫺0.5]. x ⫹y (x ⫹ y ) (x ⫹ y 2)2 We mention that this is the real part of the complex analytic function z ⫹ 1>z, where z ⫽ x ⫹ iy (we write z since z is used as coordinate in space), giving the flow around the circle x 2 ⫹ y 2 ⫽ 1, that is, a cylinder in space with axis intersecting the xy-plane at the origin. This flow and extensions of it will be discussed in the chapter on complex analysis and potential theory (in Sec. 18.4). 1

⫺ 2

2 2

,⫺

22. The x-component of v ⫽ ⵜf ⫽ [ex cos y, ⫺ex sin y] must be zero; thus y ⫽ ⫾(2n ⫹ 1)p>2. Then sin y ⫽ ⫾1. We must have sin y ⫽ ⫺1 to obtain ⫺ex sin y ⬎ 0 (upward flow), hence y ⫽ ⫺12 p ⫹ 2np. 24. ⫺ⵜT ⫽ [⫺6x, 4y]. At P this gives [⫺15.0, 7.2]. 26. ⫺ⵜT ⫽ [⫺2x, ⫺2y, ⫺8z] at P is [⫺4, 2, ⫺16]. 28. ⫺ⵜz ⫽ [⫺2x, ⫺18y], ⵜz(P) ⫽ [⫺8, ⫺18]. Hence a vector in the direction of steepest ascent is [⫺1, ⫺2.25]. 30. ⵜf ⫽ [8x, 18y],

ⵜf ( p) ⫽ [16, 12114]

32. [a, b, c]. Planes have constant normal direction. 34. ⵜf ⫽ [4x 3, 4y 3, 4z 3], ⵜf ( p) ⫽ [32, 4, 256]. The intersection of this surface with planes parallel to the coordinate planes are curves each of which is between a circle and a square of portions of four tangents to that circle whose center is the origin of the plane of the circle. 36. ⵜf ⫽ [4x, 4y]. From (5*) we thus obtain the answer [⫺1, ⫺3] • [12, 12]> 110 ⫽ ⫺48> 110. 38. Da f ⫽ [1, 1, 3] • [4, 4, ⫺2]> 111 ⫽ 2> 111 40. (ⵜf ) • a> ƒ a ƒ ⫽ (x 2 ⫹ y 2)ⴚ1[2x, 2y] • [1, ⫺1] ⫽ (x 2 ⫹ y 2]ⴚ1(2x ⫺ 2y)> 12 at (3, 0) equals 12>3. 42. 0 without calculation because on the axes an ellipsoid f ⫽ const has a tangent plane perpendicular to the axis, whereas a lies in that plane at the x-axis, so that ⵜf and a are perpendicular to each other. 44. f ⫽ yex ⫹ 13 z 3

SECTION 9.8. Divergence of a Vector Field, page 402 Purpose. To explain the divergence (the second of the three concepts grad, div, curl) and its physical meaning in fluid flows. Main Content, Important Concepts Divergence of a vector field Continuity equations (5), (6) Incompressibility condition (7)

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Comment on Content The interpretation of the divergence in Example 2 depends essentially on our assumption that there are no sources or sinks in the box. From our calculations it becomes plausible that, in the case of sources or sinks, the divergence may be related to the net flow across the boundary surfaces of the box. To confirm this and to make it precise we need integrals; we shall do this in Sec. 10.8 (in connection with Gauss’s divergence theorem). Moving div and curl to Chap. 10? Experimentation has shown that this would perhaps not be a good idea, simply because it would combine two substantial difficulties, that of understanding div and curl themselves and that of understanding the nature and role of the two basic integral theorems by Gauss and Stokes, in which div and curl play the key role. Comments on Problems Project 9 concerns some standard formulas useful in working with the divergence. CAS Experiment should help the student in gaining an intuitive understanding of the divergence. Formula (3) is basic, as the problems should further emphasize.

SOLUTIONS TO PROBLEM SET 9.8, page 405 2. ⫺xz sin xyz ⫹ xy cos xyz; p 4. 0. Hence this field is solenoidal, regardless of the special form of v1, v2, v3. 6. 0. Hence the vector field is solenoidal. The potential of this vector function is that of a point source (or point mass) at the origin. 0v3 8. div v ⫽ 2 ⫹ . Of course, there are many ways of satisfying the conditions. For 0z instance, (a) v3 ⫽ 0, (b) v3 ⫽ ⫺z ⫺ 13 z 3. The point of the problem is that the student gets used to the definition of the divergence and recognizes that div v can have different values and also the sign can differ in different regions of space. 10. (a) Parallel flow. (b) Outflow on the left and right, no flow across the other sides; hence div v ⬎ 0. (c) Outflow left and right, inflow from above and below, balance perhaps zero; by calculation, div v ⫽ 0. Etc. dy dz dx 12. v ⫽ v1i ⫹ v2 j ⫹ v3k ⫽ i⫹ j⫹ k ⫽ xi. Hence div v ⫽ 1, and dt dt dt dx ⫽ x, dt

dy dt

⫽ 0,

dz ⫽ 0. dt

By integration, x ⫽ c1et, y ⫽ c2, z ⫽ c3, and r ⫽ xi ⫹ yj ⫹ zk. Hence r(0) ⫽ c1i ⫹ c2 j ⫹ c3k

and

r(1) ⫽ c1ei ⫹ c2 j ⫹ c3k.

This shows that the cube in Prob. 9 is now transformed into the rectangular parallelepiped bounded by x ⫽ 0, x ⫽ e, y ⫽ 0, y ⫽ 1, z ⫽ 0, z ⫽ 1, whose volume is e. 14. No. u ⫽ v concerns components, whereas div u and div v are sums of contributions from all three components.

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16. The gradient is ⵜf ⫽ exyz [ yz, zx, xy]. Hence (3) gives div (ⵜf ) ⫽ exyz( y 2z 2 ⫹ z 2x 2 ⫹ x 2y 2). The work in the direct calculation is practically the same. From Probs. 15–20 the student should understand that relation (3) is quite natural. 18. The gradient is ⵜf ⫽ [⫺x(x 2 ⫹ y 2)ⴚ1>2, ⫺y(x 2 ⫹ y 2)ⴚ1>2, 1]. Application of the definition of the divergence now gives ⵜ2f ⫽ ⫺(x 2 ⫹ y 2)ⴚ1>2 ⫹ x 2(x 2 ⫹ y 2)ⴚ3>2 ⫺ (x 2 ⫹ y 2)ⴚ1>2 ⫹ y 2(x 2 ⫹ y 2)ⴚ3>2 which simplifies to ⫺(x 2 ⫹ y 2)ⴚ1>2. 20. The gradient is ⵜf ⫽ [2e2x cosh 2y, 2e2x sinh 2y], so that for the Laplacian we obtain div (ⵜf ) ⫽ 4e2x cosh 2y ⫹ 4e2x cosh 2y ⫽ 8e2x cosh 2y, ~ ~ whereas for f ⫽ e2x cos 2y we have ⵜ2f ⫽ 0. SECTION 9.9. Curl of a Vector Field, page 406 Purpose. We introduce the curl of a vector field (the last of the three concepts grad, div, curl) and interpret it in connection with rotations (Example 2 and Theorem 1). A main application of the curl follows in Sec. 10.9 in Stokes’s integral theorem. Experience has shown that it is generally didactically preferable to defer Stokes’s theorem to a later section and first to give the student a feel for the curl independent of an integral theorem. Main Content Definition of the curl (1) Curl and rotations (Theorem 1) Gradient fields are irrotational (Theorem 2) Irrotational fields, conservative fields Comments on Text The curl is suggested by rotations; see Theorem 1. We have now reached the point at which we can state basic relations among the three operators grad, div, curl (Theorem 2). Since Definition 1 involves coordinates, we have to prove that curl v is a vector; see Theorem 3. Comments on Problems Calculations (Probs. 4–8) are followed by typical applications in fluid mechanics (Probs. 9–13). As in the previous two sections, we finally present general formulas, this time for div and curl, and request some corresponding calculations (Probs. 14–20).

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SOLUTIONS TO PROBLEM SET 9.9, page 408 2. (a) Nothing, in general. (b) curl v is parallel to the x-axis or 0. 4. curl v ⫽ [0, 0, 5 ⫺ 4y] 6. curl v ⫽ 0. Recall from Theorem 3 in Sec. 9.7 with r0 ⫽ 0 and x 2 ⫹ y 2 ⫹ z 2 ⫽ r 2 that the present vector field is a gradient field, so that we must have curl v ⫽ 0. 2 2 2 8. curl v ⫽ [⫺2yeⴚy , ⫺2zeⴚz , ⫺2xeⴚx ] 10. curl v ⫽ [0, 0, ⫺cos x csc2 x], div v ⫽ sec x tan x, compressible. Streamlines are obtained as follows. By the definition of the velocity vector and its present given form, v ⫽ [x r , y r , z r ] ⫽ [sec x, csc x, 0]. Equating the first components gives xr ⫽

dx ⫽ sec x, dt

cos x dx ⫽ dt.

By integration, sin x ⫽ t ⫹ c1. Hence x ⫽ arcsin (t ⫹ c1). From this and the second components, y r ⫽ cosec x ⫽

1 . 1 ⫹ c1

By integration, y ⫽ ln (t ⫹ c1) ⫹ c2. Equating the third components and integrating, we finally have z ⫽ c3. 12. curl v ⫽ [0, 0, 2]. Also div v ⫽ 0, incompressible. Streamlines are helices obtained as follows. As in Prob. 10 we first have v ⫽ [x r , y r , z r ] ⫽ [⫺y, x, p]. In components, x r ⫽ ⫺y,

y r ⫽ x,

z r ⫽ p.

From the first two components, by differentiation and substitution, x s ⫽ ⫺y r ⫽ ⫺x. A general solution is x ⫽ a cos t ⫹ b sin t. From this and the first components, y ⫽ ⫺x r ⫽ a sin t ⫺ b cos t. From the third component, z ⫽ pt ⫹ c3.

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The helix obtained lies on the cylinder of radius 2a 2 ⫹ b 2 and axis the z-axis. Indeed, x 2 ⫹ y 2 ⫽ a 2 cos2 t ⫹ 2ab cos t sin t ⫹ b 2 sin2 t ⫹ a 2 sin2 t ⫺ 2ab cos t sin t ⫹ b 2 sin2 t ⫽ a 2 ⫹ b 2. 14. Project. Parts (b) and (d) are basic. They follow from the definitions by direct calculation. Part (a) follows by decomposing each component accordingly. (c) In the first component in (1) we now have f v3 instead of v3, etc. Product differentiation gives ( f v3)y ⫽ fyv3 ⫹ f # (v3)y. Similarly for the other five terms in the components. fyv3 and the corresponding five terms give (grad f ) ⴛ v and the other six terms f # (v3)y, etc. give f curl v. (d) For twice continuously differentiable f the mixed second derivatives are equal, so that the result follows from ⵜf ⫽ fxi ⫹ fy j ⫹ fzk and (1), which gives curl (ⵜf ) ⫽ [( fz)y ⫺ ( fy)z]i ⫹ [( fx)z ⫺ ( fz)x] j ⫹ [( fy)x ⫺ ( fx)y]k. (e) Write out and compare the 12 terms on either side. 16. [xy ⫺ zx, yz ⫺ xy, zx ⫺ yz]. Confirmation by (c) in Project 14: (ⵜg) ⴛ v ⫽ [1, 1, 1] ⴛ [ yz, zx, xy] ⫽ [xy ⫺ zx,

yz ⫺ xy,

zx ⫺ yz], g curl v ⫽ 0

because v is a gradient field, namely, v ⫽ grad f. 18. div (u ⴛ v) ⫽ div [xyz ⫺ x 2z, xyz ⫺ y 2x, xyz ⫺ z 2y] ⫽ ( yz ⫺ 2xz) ⫹ (xz ⫺ 2yx) ⫹ (xy ⫺ 2zy) ⫽ ⫺yz ⫺ zx ⫺ xy. Confirmation by (e) in Project 14: div (u ⴛ v) ⫽ v • curl u ⫺ u • curl v ⫽ [ yz, zx, xy] • [⫺1, ⫺1, ⫺1] ⫺ u • curl (ⵜf ) ⫽ ⫺yz ⫺ zx ⫺ xy ⫺ 0. 20. div ([2xyz ⫹ y 2z ⫹ yz 2, x 2z ⫹ 2xyz ⫹ xz 2, x 2y ⫹ xy 2 ⫹ 2xyz]) ⫽ 2(yz ⫹ zx ⫹ xy). Confirmation. By Problem Set 9.7, ⵜ( fg) ⫽ f ⵜg ⫹ gⵜf. From this and Problem Set 9.8, div (ⵜ( fg)) ⫽ div ( f ⵜg) ⫹ div (gⵜf ) ⫽ f ⵜ2g ⫹ gⵜ2f ⫹ 2ⵜf • ⵜg ⫽ 0 ⫹ 0 ⫹ 2[yz, xz, xy] • [1, 1, 1] ⫽ 2( yz ⫹ zx ⫹ xy). SOLUTIONS TO CHAPTER 9 REVIEW QUESTIONS AND PROBLEMS, page 409 12. [0, 0, 50], [2, ⫺19, ⫺5], [⫺2, 19, 5], 0 14. ⫺1250, ⫺1250, ⫺1250, undefined 16. [4> 165, 7> 165, 0], [3> 135, ⫺1> 135, 5> 135]; 5> 135 is the projection of a in the direction of b. Similarly, 5> 165 is the projection of b in the direction of a.

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18. 1110 ⫽ 10.49 ⬍ 165 ⫹ 135 ⫽ 13.98 illustrates the triangle inequality (7) in Sec. 9.2. 20. If u ⴛ v ⫽ 0. Always. 22. a ⫹ b ⫹ c ⫹ v should have no x-component; thus 4 ⫹ 3 ⫺ 6 ⫹ v1 ⫽ 0, v1 ⫽ ⫺1, v2 and v3 arbitrary. 24. We are looking for the normal vectors [4, ⫺1, 3] and [1, 2, 4] of the planes. We obtain g ⫽ arccos

14 ⫽ 0.928 ⫽ 53.19°. 126121

26. The condition is v • w> ƒ w ƒ ⫽ w • v> ƒ v ƒ . The answer is ƒ v ƒ ⫽ ƒ w ƒ or v and w are orthogonal, so that the numerators are zero and the size of the denominators does not matter. 28. The moment m ⫽ ƒ m ƒ ⫽ ƒ r ⴛ p ƒ of a force p about a point Q is zero if p ⫽ 0 or p is acting in a straight line through Q, which makes p and r parallel (or exactly opposite or r ⫽ 0). 30. This is a helix. P corresponds to t ⫽ p>4. By differentiation, r r (t) ⫽ [⫺3 sin t, 3 cos t, 4]. At P the velocity is v ⫽ r r ⫽ [⫺3> 12, 3> 12, 4]. The speed is ƒ v ƒ ⫽ 5. The acceleration vector is r s (t) ⫽ [⫺3 cos t, ⫺3 sin t, 0]. It is parallel to the xy-plane. Its absolute value, the acceleration, is constant, just as the speed, namely, ƒ r s (t) ƒ ⫽ 3. 32. grad f ⫽ [ y, x ⫺ z, ⫺y] at P is [7, 2, ⫺7]. The value of f at P is 2 # 7 ⫺ 0 ⫽ 14. Hence the value of f grad f at P is [98, 28, ⫺98]. 34. [⫺2, ⫺4, ⫺2], [2y, 6z, 2x] 36. 4y 2 ⫹ 12z 2 ⫹ 8x 2 ⫹ 2xz; 192 [ y, x ⫺ z, ⫺y] • [2y, 2z, 4x ⫹ z] (grad f ) • v 38. Dv f ⫽ has at P the value ⫽ ƒvƒ 24y 2 ⫹ 4z 2 ⫹ (4x ⫹ z)2 [1, ⫺1, ⫺1] • [2, 4, 6] 14 ⫹ 16 ⫹ 36

⫽

⫺8 . 156

40. 0 since v appears in two rows (componentwise) of this scalar triple product.

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SECTION 10.1. Line Integrals, page 413 Purpose. To explain line integrals in space and in the plane conceptually and technically with regard to their evaluation by using the representation of the path of integration. Main Content, Important Concepts Line integral (3), (3 r ), its evaluation Its motivation by work done by a force (“work integral”) General properties (5) Dependence on path (Theorem 2) Background Material. Parametric representation of curves (Sec. 9.5); a couple of review problems may be useful. Comments on Content The integral (3) is more practical than (8) (more direct in view of subsequent material), and work done by a force motivates it sufficiently well. Independence of path will be settled in the next section. Further Comments on Text and Problem Set 10.1 Examples 1 and 2 show that the evaluation of line integrals in the plane and in space is conceptually the same, making the difference just a technical one. Example 3 illustrates a main motivation of (3), which we take as a definition of line integral. Theorem 1 and Project 12 in the problem set concern direction preserving and reversing transformations of line integrals, as they are needed in various applications. Kinetic energy appears in connection with the work integral; see Example 4. The basic fact of path dependence is emphasized in Theorem 2 and again in Project 12. Problems 2–11 and 15–20 concern the evaluation of integrals (3) and (8), respectively. SOLUTIONS TO PROBLEM SET 10.1, page 418 2. r ⫽ [t, 4t 2], so that the integrand is F (r (t)) • r r ⫽ [16t 4, ⫺t 2] • [1, 8t] ⫽ 16t 4 ⫺ 8t 3.

Integration gives 16t 5>5 ⫺ 8t 4>4. The limits of integration are 0 and 1, so that the answer is 65. In the next problem (Prob. 3) we integrate the same F over a shorter path, but the value of the integral will be larger. This is not unusual. 4. For instance, we may take r ⫽ [2 ⫺ 2t, 2t]

(0 ⬉ t ⬉ 1).

Then we obtain the integrand F(r (t)) • r r ⫽ [(2 ⫺ 2t) 2t, (2 ⫺ 2t)2(2t)2] • [⫺2, 2] ⫽ ⫺4t(2 ⫺ 2t) ⫹ 8t 2(2 ⫺ 2t)2. 4 Integration gives ⫺ 15 .

179

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6. Helix on the cylinder x 2 ⫹ z 2 ⫽ 4, F (C ) ⫽ [2 cos t ⫺ t, t ⫺ 2 sin t, 2 sin t ⫺ 2 cos t]. Answer: ⫺8p ⫹ 2p2 8. C looks similar to the curve in Fig. 210, Sec. 9.5. The integrand is F (r (t)) • r r (t) ⫽ [et, cosh t 2, sinh t 3] • [1, 2t, 3t 2] ⫽ et ⫹ 2t cosh t 2 ⫹ 3t 2 sinh t 3. Integration from t ⫽ 0 to t ⫽ 12 gives the answer e1>2 ⫹ sinh 14 ⫹ cosh 18 ⫺ 2. 10. Here we integrate around a triangle in space. For the three sides and corresponding integrals we obtain 1

r1 ⫽ [t, t, 0],

r1r ⫽ [1, 1, 0],

F (r1 (t)) ⫽ [t, 0, 2t],

冮 t dt ⫽ 12 0

r2 ⫽ [1, 1, t],

r2r ⫽ [0, 0, 1],

F (r2 (t)) ⫽ [1, ⫺t, 2],

r3 ⫽ [1 ⫺ t,

1 ⫺ t, 1 ⫺ t],

r3r ⫽ [⫺1, ⫺1, ⫺1],

冮 2 dt ⫽ 2 0

F(r3(t)) ⫽ [1 ⫺ t, ⫺1 ⫹ t,

冮 (⫺2 ⫹ 2t) dt ⫽ ⫺2 ⫹ 1.

2 ⫺ 2t],

Hence the answer is 32. 12. Project. (a) r ⫽ [cos t, sin t], r r ⫽ [⫺sin t, cos t]. From F ⫽ [ xy, obtain F(r(t)) ⫽ [cos t sin t, ⫺sin2 t]. Hence the integral is ⫺2

p>2

⫺y 2] we

冮 cos t sin t dt ⫽ ⫺ 23 . 2

Setting t ⫽ p , we have r ⫽ [cos p 2, 2

sin p 2] and

F(r(p)) ⫽ [cos p 2 sin p 2, Now r r ⫽ [⫺2p sin p 2,

冮

2p>2

(b) r ⫽ [t, t n],

⫺sin2 p 2].

2p cos p 2], so that the integral is

2 (⫺2p cos p 2 sin2 p 2 ⫺ 2p cos p 2 sin2 p 2) dp ⫽ ⫺ . 3

F(r(t)) ⫽ [t n⫹1,

⫺t 2n],

r r ⫽ [1, nt nⴚ1]. The integral is

冮 (t 0

n⫹1

⫺ nt 3nⴚ1) dt ⫽

1 1 ⫺ . n⫹2 3

(c) The limit is ⫺ 13. The first portion of the path gives 0, since y ⫽ 0. The second portion is r2 ⫽ [1, t], so that F(r2(t)) ⫽ [t, ⫺t 2], r ⫽ [0, 1]. Hence the integrand is ⫺t 2, which upon integration gives ⫺ 13. 2 r ⫽ [t, 43 t], 0 ⬉ t ⬉ 3, L ⫽ 5, ƒ F ƒ ⫽ 2t 4 ⫹ 16 9 t . The derivative is

14.

1 4 16 2 ⴚ1>2 (4t 3 ⫹ 32 2 [t ⫹ 9 t ] 9 t).

The expression in parentheses (. . .) has the root t ⫽ 0, but no further real roots. Hence the maximum of ƒ F ƒ is taken at (3, 4), so that we obtain the bound L ƒ F ƒ ⬉ 5281 ⫹ 16 ⫽ 5297 ⬍ 50.

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Calculation gives r r ⫽ [1, 43 ],

181

F(r(t)) ⫽ [t 2, 43 t]. The integral is

冮 A t ⫹ 169 t B dt ⫽ 9 ⫹ 8 ⫽ 17. 3

The result is typical, that is, the point of the ML-inequality is generally not to obtain a very sharp upper bound for integrals, but to show that an integral remains bounded in some limit process. Also, instead of L, you may sometimes have to be satisfied with using an upper bound for L if L itself is complicated. 16. 3t ⫹ cosh t ⫹ 5 sinh t integrated from 0 to 1 gives 3 7 2 ⫹ sinh 1 ⫹ 5 cosh 1 ⫺ 5 ⫽ sinh 1 ⫹ 5 cosh 1 ⫺ 2 .

The projection of this space curve into the xy-plane is a cosh curve. Its projection into the xz-plane is a sinh curve. From these projections one can conclude the general shape of C. 18. F(r(t)) ⫽ [sin t, cos t, 0] integrated from 0 to p>4 gives c⫺

1 1 ⫹ 1, , 0d. 12 12

20. C is an exponentially increasing curve in the plane x ⫽ y from (0, 0, 0) to (5, 5, e5). The representation of C gives F(r(t)) ⫽ [tet, tet, t 4]. Integration from 0 to 5 gives [4e5 ⫹ 1, 4e5 ⫹ 1, 625]. SECTION 10.2. Path Independence of Line Integrals, page 419 Purpose. Independence of path is a basic issue on line integrals, and we discuss it here in full. Main Content, Important Concepts Definition of independence of path Relation to gradient (Theorem 1), potential theory Integration around closed curves Work, conservative systems Relation to exactness of differential forms Comment on Content We see that our text pursues three ideas by relating path independence to (i) gradients (potentials Theorem 1), (ii) closed paths (Theorem 2), and (iii) exactness of the form under the integral sign (Theorem 3*). The complete proof of the latter needs Stokes’s theorem, so here we leave a small gap to be easily filled in Sec. 10.9. It would not be a good idea to delay introducing the important concept of path independence until Stokes’s theorem is available. Simple connectedness of domains is further emphasized in Example 4.

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The determination of a potential is shown in Example 2. Problem 1, writing a report on the concepts and relations in this section, should help to understand the various ideas. The remaining problems shed light on path dependence and independence from several angles, with experimentation and computer application. SOLUTIONS TO PROBLEM SET 10.2, page 425 2. No. The origin is a boundary point of this “degenerate annulus,” which therefore is not simply connected (but doubly connected). 4. The exactness test for path independence gives (F2 ) x ⫽ 8xe4y ⫽ (F1 ) y. We find F ⫽ grad f, where f (x, y) ⫽ x 2e4y so that for the integral we have f (6, 1) ⫺ f (4, 0) ⫽ 36e4 ⫺ 16e0 ⫽ 1949. 2 6. For the dx-term and the dy-term, the exactness test of path independence gives 2xyex ⫹ 2 y 2 ⫹ z 2 ⫽ 2xyex ⫹ y 2 ⫹ z 2, etc. We find f ⫽ 12 exp (x 2 ⫹ y 2 ⫹ z 2). Evaluation at the limits gives 1 2 2 (e ⫺ 1).

8. The test for path independence gives ⫺z sin yz on both sides in the first equation, and so on. We find f ⫽ x cos yz and from this and the limits of integration f (3, p, 3) ⫺ f (5, 3, p) ⫽ 3 cos 3p ⫺ 5 cos 3p ⫽ 2. 10. Project. (a) 2y 2 ⫽ x 2 from (6*). (b) r ⫽ [t, bt], 0 ⬉ t ⬉ 1, represents the first part of the path. By integration we obtain b>4 ⫹ b 2>2. On the second part, r ⫽ [1, t], b ⬉ t ⬉ 1. Integration gives 2(1 ⫺ b 3)>3. Equating the derivative of the sum of the two expressions to zero gives b ⫽ 1> 12 ⫽ 0.70711. The corresponding maximum value of I is 1>(6 12) ⫹ 23 ⫽ 0.78452. (c) The first part is y ⫽ x>c or r ⫽ [t, t>c], 0 ⬉ t ⬉ c. The integral over this portion is c3>4 ⫹ c>2. For the second portion r ⫽ [t, 1], c ⬉ t ⬉ 1, the integral is (1 ⫺ c3)>3. For c ⫽ 1 we get I ⫽ 0.75, the same as in (b) for b ⫽ 1. This is the maximum value of I for the present paths through (c, 1) because the derivative of I with respect to c is positive for 0 ⬉ c ⬉ 1. 12. CAS Experiment. The circle passes through (0, 0) and (1, 1) if its center P lies on the line y ⫽ 1 ⫺ x, so that P is (a, 1 ⫺ a), a arbitrary. Then the radius is r ⫽ 2a 2 ⫹ (1 ⫺ a)2. One would perhaps except a value near a ⫽ 12, for which the circle is smallest. The experiment gives for the circle r ⫽ [a ⫹ r cos t, 1 ⫺ a ⫹ r sin t]

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a minimal value Imin ⫽ 50.85137100, approximately, of the integral when a ⫽ 0.4556, approximately. 14. Dependent; indeed, we obtain sinh xz ⫹ zx cosh xz ⫽ ⫺sinh xz ⫺ zx cosh xz. 16. Independent, e ⫽ ey, etc., xey ⫺ yez, aeb ⫺ bec 18. Dependent because for y

F ⫽ [yz cos xy, xz cos xy, ⫺2 sin xy] we obtain curl F ⫽ [⫺3x cos xy, 3y cos xy, 0] ⫽ 0. 20. The point of the problem is to make the student think of the nature of (6) and (6 r ). The constructions are trivial. Start from an F satisfying (6), for instance, F ⫽ [1, 1, 1] and, to violate the third equation (6 r ), add to F1 ⫽ 1 a function of y, e.g., y. SECTION 10.3. Calculus Review: Double Integrals. Optional, page 426 Purpose. We need double integrals (and line integrals) in the next section and review them here for completeness, suggesting that the student go on to the next section. Content Definition, evaluation, and properties of double integrals Some standard applications Change of variables, Jacobians (6), (7) Polar coordinates (8) Historical Comment The two ways of evaluating double integrals explained in the text give the same result. For continuous functions this was known at least to Cauchy. Some calculus books call this Fubini’s theorem, after the Italian mathematician GUIDO FUBINI (1879–1943; 1939–1943 professor at New York University), who in 1907 proved the result for arbitrary Lebesgue-integrable functions (published in Atti Accad. Naz. Lincei, Rend., 161, 608–614). Comments on Text and Problems We can present here only an absolute minimum of what we shall need, and students should be encouraged to supplement this by material from their calculus books, if needed. Centers of gravity of other domains (cross sections appearing in engineering design) can be found in engineering handbooks. SOLUTIONS TO PROBLEM SET 10.3, page 432 2. Integration over y gives 19x 3>3. Integration over x now gives the answer 76>3. 4. This order of integration is less practical since it requires splitting the integral into two parts 0

冮冮

(x 2 ⫹ y 2) dy dx ⫹

ⴚ3 ⴚx

冮冮 (x ⫹ y ) dy dx. 2

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This complication occurs quite often. By integration over y, we get from the first part 0

冮 c x (3 ⫹ x) ⫹ 273 ⫹ x3 d dx ⫽ 27. 2

ⴚ3

The other part gives 27, too. Answer: 54 2

冮 (cosh 2y ⫺ cosh y) dy ⫽ 21 sinh 4 ⫼ sinh 2 0

8. After the integration over x, we have

冮

p>4

10.

1ⴚx2

冮冮冮 0

1ⴚx2

dy dz dx ⫽

冮冮 0

1 1 cos3 y sin y dy ⫽ . 3 16

1ⴚx2

8 15

(1 ⫺ x 2) dz dx ⫽

12. x ⫽ b>2 for reasons of symmetry. Since the given R and its left half (the triangle with vertices (0, 0), (b>2, 0), (b>2, h)) have the same y, we can consider that half, for which M ⫽ 14 bh. We obtain 4 y⫽ bh

b>2

2hx>b

冮冮

4 y dy dx ⫽ bh

⫽

冮

b>2

1 2hx a b dx 2 b

4 # 1 2h # 1 b h a b a b ⫽ . bh 2 b 3 2 3

Note that y is the same value as in the next problem (Prob. 13), for obvious reasons. 14. The total mass is p

M⫽

冮冮 r dr du ⫽ p2 (r ⫺ r ). 2 2

2 1

We thus obtain x ⫽ 0 by symmetry and y⫽

冮 (r sin u) r dr du

p(r 22 ⫺ r 21) r

⫽ ⫽

2 (r 22 ⫺ r 21)

ⴢ2ⴢ

4(r 32 ⫺ r 31) 3p(r 22 ⫺ r 21)

1 3 (r 2 ⫺ r 31) 3

With r1 ⫽ 0 and r2 ⫽ r this gives the answer to Prob. 15. 16. x ⫽ y ⫽ 4r>3p from Prob. 15 without calculation.

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18. We obtain b>2

冮冮

Ix ⫽

b>2

1 2hx a b dx ⫹ 3 3

⫽

冮冮

y 2 dy dx ⫹

冮

⫽

2hx>b

2hⴚ2hx>b

y 2 dy dx

b>2 0

冮

1 x c 2h a1 ⫺ b d dx 3 b

b>2

1 1 1 bh3 ⫹ bh3 ⫽ bh3. 24 24 12

Each of these two halves of R contributes half to the moment Ix about the x-axis. Hence we could have simplified our calculation and saved half the work. Of course, this does not hold for Iy. We obtain b>2

Iy ⫽

冮冮 0

⫽

冮

x 2 dy dx ⫹

冮冮

2hⴚ2hx>b

x 2 dy dx

b>2 0

b>2

⫽

2hx>b

x2 a

2hx b dx ⫹ b

冮 x a2h ⫺ 2hxb b dx 2

b>2

1 3 11 3 7 3 b h⫹ b h⫽ b h. 32 96 48

20. We denote the right half of R by R1 ´ R2, where R1 is the rectangular part and R2 the triangular. The moment of interia Ix1 of R1 with respect to the x-axis is b>2

Ix1 ⫽

冮冮 0

y 2 dy dx ⫽

冮

b>2

1 h3 dx ⫽ bh3. 3 6

Similarly for the triangle R2 we obtain a>2

Ix2 ⫽

⫽

冮冮

y 2 dy dx

b>2

a>2

1 h3(2x ⫺ a)3 dx 3 (b ⫺ a)3

冮

b>2

⫽

h(2xⴚa)>(bⴚa)

1 3 h (a ⫺ b). 24

Together, 1 h3 Ix ⫽ (3b ⫹ a) 2 24

and

Ix ⫽

Iy is the same as in Prob. 19; that is, Iy ⫽

h(a 4 ⫺ b 4) . 48(a ⫺ b)

1 3 h (3b ⫹ a). 12

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This can be derived as follows, where we integrate first over x and then over y, which is simpler than integrating in the opposite order, where we would have to add the two contributions, one over the square and the other over the triangle, which would be somewhat cumbersome. Solving the equation for the right boundary y⫽

h (a ⫺ 2x) a⫺b

for x, we have x⫽

1 (ah ⫺ (a ⫺ b)y) 2h

and thus 1 Iy ⫽ 2

冮冮 0

(ahⴚ(aⴚb)y)>2h

x 2 dx dy

⫽

冮 24h1 (ah ⫺ (a ⫺ b)y) dy 3

⫽

h 3 h(a 4 ⫺ b 4) (a ⫹ a 2b ⫹ ab 2 ⫹ b 3) ⫽ . 96 96(a ⫺ b)

Now we multiply by 2, because we considered only the right half of the profile. SECTION 10.4. Green’s Theorem in the Plane, page 433 Purpose. To state, prove, and apply Green’s theorem in the plane, relating line and double integrals. Comment on the Role of Green’s Theorem in the Plane This theorem is a special case of each of the two “big” integral theorems in this chapter, Gauss’s and Stokes’s theorems (Secs. 10.7, 10.9), but we need it as the essential tool in the proof of Stokes’s theorem. The present theorem must not be confused with Green’s first and second theorems in Sec. 10.8. Comments on Text and Problems Equation (1) is the basic formula in this section and later on in applications of Theorem 1. Eq. (1 r ) shows its vectorial form, and other forms of Green’s theorem are included in Project 12 of the problem set. The proof of Theorem 1 proceeds componentwise, (2) and (3) relating to F1 when F2 ⫽ 0. For F2 when F1 ⫽ 0 the calculation is similar, with a change of the direction of an integration as noted in the proof. Cancelation of integrals over subdivisions, as it occurs here (see Fig. 238), is a standard idea that will occur quite often in the sequel. The important Examples 2–4 need no further comments. The resulting integral formula (9) has applications in theory and practice. A basic formula for harmonic functions (solutions of Laplace’s equation whose second partial derivatives are continuous) is considered in Probs. 18–20. All these formulas and calculational problems emphasize the great importance of the present theorem.

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SOLUTIONS TO PROBLEM SET 10.4, page 438 2. Integrate (F2)x ⫺ (F1)y ⫽ 2 ⫺ 12y over y from ⫺2 to 2 and the result over x from ⫺2 to 2, obtaining (8 ⫺ 0) # 4 ⫽ 32. 4. We obtain 1

冮冮 2x sinh 2y dy dx 1

冮 (x cosh 2x ⫺ x cosh 2x ) dx

⫽

x 1 1 ⫽ c sinh 2x ⫺ cosh 2x ⫺ sinh 2x 2 d 2 4 4 0 1 1 1 1 sinh 2 ⫺ cosh 2 ⫺ sinh 2 ⫹ 2 4 4 4 1 1 1 ⫽ ⫹ sinh 2 ⫺ cosh 2. 4 4 4 ⫽

冮冮 (⫺cosh x ⫺ sinh y) dy dx. Evaluation over y gives 3

冮 (cosh x ⫺ 2x cosh x ⫺ cosh 3x) dx 1

1 (14 sinh 3 ⫹ sinh 9). 3 8. F ⫽ grad (eⴚx cos y), so that the integral around a closed curve is zero. Also the integrand in (1) on the left is identically zero. 10. This is a portion of a circular ring (annulus) bounded by the circles of radii 1 and 2 centered at the origin, in the first quadrant bounded by y ⫽ x and the y-axis. The integrand is ⫺1>y 2 ⫺ 2x 2y. We use polar coordinates, obtaining ⫽ 2(cosh 3 ⫺ cosh 1) ⫹ sinh 1 ⫺

p>2

冮冮 a⫺ r sin1 u ⫺ 2r cos u sin ub r dr du p>4

⫽

p>2

冮 c ⫺ sinln 2u ⫺ 25 (32 ⫺ 1) cos u sin u d du p>4

⫽ (ln 2) acot

p 62 p p ⫺ cot b ⫹ acos3 ⫺ cos2 b 2 4 15 2 4 31 ⫽ ⫺ln 2 ⫺ 1512 ⫽ ⫺2.155. p

12. Project. We obtain div F in (11) from (1) if we take F ⫽ [F2, ⫺F1]. Taking n ⫽ [ y r , ⫺x r ] as in Example 4, we get from (1) the right side in (11), (F ⴢ n) ds ⫽ aF2

dy ds

⫹ F1

dx b ds ⫽ F2 dy ⫹ F1 dx. ds

Formula (12) follows from the explanation of (1 r ).

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Furthermore, div F ⫽ 7 ⫺ 3 ⫽ 4 times the area of the disk of radius 2 gives 16p. For the line integral in (11) we need s r ⫽ c 2 cos , 2

s 2 sin d , 2

s r r ⫽ c ⫺sin , 2

s cos d , 2

n ⫽ [y r ,

⫺x r ]

where s varies from 0 to 4p. This gives

冯

F • n ds ⫽

冯

(7xy r ⫹ 3yx r ) ds ⫽

冮 a14 cos 2s ⫺ 6 sin 2s b ds ⫽ 16p. 2

In (12) we have curl F ⫽ 0 and F • r r ⫽ ⫺14 cos

s s s s sin ⫺ 6 cos sin ⫽ ⫺10 sin s 2 2 2 2

which gives zero upon integration from 0 to 4p. 14. ⵜ2w ⫽ 2y ⫹ 2x, so that we obtain 1

冮冮 0

21ⴚx2

(2y ⫹ 2x) dy dx

0 1

冮 (1 ⫺ x ⫹ 2x 21 ⫺ x ) dx

⫽

⫽ cx ⫺

1 1 3 2 x ⫺ (1 ⫺ x 2)3>2 d ` 3 3 0

2 2 4 ⫹ ⫽ . 3 3 3

⫽

Confirmation in polar coordinates. We have ⵜ2w ⫽ 2r cos u ⫹ 2r sin u, so that p>2

冮冮 2r (cos u ⫹ sin u) r dr du ⫽

冮

p>2

2 (cos u ⫹ sin u) du 3 1

2 ⫽ (sin u ⫺ cos u) ` 3 0 ⫽

2 4 (1 ⫹ 1) ⫽ . 3 3

16. ⵜ2w ⫽ 2 ⫹ 2 ⫽ 4. Answer: 4 ⴢ 4p. Confirmation: For C, use s s r ⫽ c 2 cos , 2 sin d . 2 2 The corresponding unit tangent vector is s s # r ⫽ c ⫺sin , cos d . 2 2

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The outer unit normal vector is s s n ⫽ c cos , sin d . 2 2 Furthermore, s s grad w ⫽ [2x, 2y] ⫽ c 2 ⴢ 2 cos , 2 ⴢ 2 sin d , 2 2 so that (grad w) • n ⫽ 4 cos2

s s ⫹ 4 sin2 ⫽ 4 2 2

and 4p

冮 4 ds ⫽ 16p. s⫽0

18. Set F1 ⫽ ⫺wwy and F2 ⫽ wwx in Green’s theorem, where subscripts x and y denote partial derivatives. Then (F2)x ⫺ (F1)y ⫽ w 2x ⫹ w 2y because ⵜ 2w ⫽ 0, and F1 dx ⫹ F2 dy ⫽ (⫺wwyx r ⫹ wwzy r ) ds ⫽ w(grad w) • ( y r i ⫺ x r j) ds ⫽ w(grad w) • n ds ⫽ w

0w ds 0n

where primes denote derivatives with respect to s. 20. The integrand on the left is ƒ grad w ƒ 2 ⫽ 4(x 2 ⫹ y 2). Integration over y gives 4x 2y ⫹ 43 y 3. We have to integrate over y from 0 to y ⫽ 1 ⫺ x. These limits give in the previous formula 4x 2(1 ⫺ x) ⫹ 43 (1 ⫺ x)3. Integration over x now gives 4 3 4 (1 ⫺ x) x ⫺ x4 ⫺ ⴢ . 3 3 4 4

Inserting the limits 0 and 1 we finally have the answer 23 . SECTION 10.5. Surfaces for Surface Integrals, page 439 Purpose. The section heading indicates that we are dealing with a tool in surface integrals, and we concentrate our discussion accordingly. Main Content, Important Concepts Parametric surface representation (2) (see also Fig. 241) Tangent plane Surface normal vector N, unit surface normal vector n Short Courses. Discuss (2) and (4) and a simple example.

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Comments on Text and Problems The student should realize and understand that the present parametric representations are the two-dimensional analog of parametric curve representations. Examples 1–3 and Probs. 1–8 and 14–19 concern some standard surfaces of interest in applications. We shall need only a few of these surfaces, but these problems should help students grasp the idea of a parametric representation and see the relation to representations (1). Moreover, it may be good to collect surfaces of practical interest in one place for possible reference.

SOLUTIONS TO PROBLEM SET 10.5, page 442 2. Circles, straight lines through the origin. A normal vector is i N ⫽ 4 cos v ⫺u sin v

sin v

0 4 ⫽ [0,

u cos v

u] ⫽ uk.

At the origin this normal vector is the zero vector, so that (4) is violated at (0, 0). This can also be seen from the fact that all the lines v ⫽ const pass through the origin, and the curves u ⫽ const (the circles) shrink to a point at the origin. This is a consequence of the choice of the representation, not of the geometric shape of the present surface (in contrast with the cone, where the apex has a similar property, but for geometric reasons). 4. The parameter curves u ⫽ const are ellipses, namely, the intersections of the cylinder with planes z ⫽ const; a and b are their semi-axes. The curves v ⫽ const are the generating straight lines of the cylinder, which are perpendicular to the xy-plane. A normal vector is [⫺b cos v, ⫺a sin v, 0]. Note that these normal vectors are parallel to the xy-plane, which is geometrically obvious. For b ⫽ a we obtain the representation in Example 1 of the text. Note further that the normal vectors are independent of u; they are parallel along each generator v ⫽ const, which is also geometrically obvious. 6. z ⫽ arctan ( y>x), helices (hence the name!), horizontal straight lines. This surface is similar to a spiral staircase, without steps (as in the Guggenheim Museum in New York). A normal vector is [sin v,

⫺cos v,

u].

8. z ⫽ x 2>a 2 ⫺ y 2>b 2, hyperbolas, parabolas; a normal vector is [⫺2bu 2 cosh v,

2au 2 sinh v,

abu].

10. ru is tangent to the curves v ⫽ const, and rv is tangent to the curves u ⫽ const, and ru • rv ⫽ 0 if and only if we have orthogonality, as follows directly from the definition of the inner product in Sec. 9.2. 12. N(0, 0) ⫽ 0 in Prob. 2 (polar coordinates); see the answer to Prob. 2. N ⫽ 0 in Prob. 3 at the apex of the cone, where no tangent plane and hence no normal exists. In Probs. 5 (paraboloid) and 7 (ellipsoid) the situation is similar to that in the case of polar coordinates. In Prob. 8 the origin is a saddle point. In each of these cases, one can find a representation for which N(0, 0) ⫽ 0; for Prob. 5 this is shown in Prob. 11. See also Example 4 in the text for the sphere.

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14. z ⫽ 6 ⫺ 2x ⫺ 1.5y, hence r(u, v) ⫽ [u, v,

6 ⫺ 2u ⫺ 1.5v].

A normal vector is i

N ⫽ 41

⫺2 4 ⫽ [2,

⫺1.5

1.5,

1].

More simply, [4, 3, 2] by applying grad to the given representation. Or [4, 3, 2] by remembering that a plane can be represented by N • [x, y, z] ⫽ c. 16. Generalizing a representation of a sphere in the text suggests r(u, v) ⫽ [cos v cos u,

cos v sin u,

3 sin v]

because then x 2 ⫹ y 2 ⫽ cos2 v (cos2 u ⫹ sin2 u) ⫽ cos2 v and 19 z 2 ⫽ sin2 v, so that the sum of the two expressions is cos2 v ⫹ sin2 v ⫽ 1, as it should be. 18. x 2 ⫹ 4y 2 suggests choosing x ⫽ 2u cos v, y ⫽ u sin v because then x 2 ⫹ 4y 2 ⫽ 4u 2(cos2 v ⫹ sin2 v) ⫽ 4u 2, hence z ⫽ 2u. Together, r(u, v) ⫽ [2u cos v,

u sin v,

2u]

and N ⫽ ru ⴛ rv ⫽ [⫺2u cos v,

⫺4u sin v,

2u].

20. Project. (a) ru(P) and rv(P) span T(P). r* varies over T(P). The vanishing of the scalar triple product implies that r* ⫺ r(P) lies in the tangent plane T(P). (b) Geometrically, the vanishing of the dot product means that r* ⫺ r(P) must be perpendicular to ⵜg, which is a normal vector of S at P. (c) Geometrically, fx (P) and fy (P) span T(P), so that for any choice of x*, y* the point (x*, y*, z*) lies in T(P). Also, x* ⫽ x, y* ⫽ y gives z* ⫽ z, so that T(P) passes through P, as it should. SECTION 10.6. Surface Integrals, page 443 Purpose. We define and discuss surface integrals with and without taking into account surface orientations. Main Content Surface integrals (3) ⬅ (4) ⬅ (5) Change of orientation (Theorem 1) Integrals (6) without regard to orientation; also (11) Comments on Content The right side of (3) shows that we need only N but not the corresponding unit vector n. An orientation results automatically from the choice of a surface representation, which determines ru and rv and thus N. The existence of nonorientable surfaces is interesting but is not needed in our further work.

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Further Comments and Suggestions Emphasize to your students that the integral of (3) is a scalar, and the orientation results from N, that is, from the choice of parameters u, v. Note further that the three terms in (5) give three double integrals over regions in the coordinate planes. (3) and (5) are compared in Example 1. Example 2 is of a similar character. Example 3 illustrates the effect of interchanging parameters, resulting in a factor minus, in the result. Orientability is extended beyond our need, but the student should see this excursion into topology in Fig. 248 and perhaps also in Probs. 17 and 18. Formula (8) and Examples 4 and 5 (sphere and doughnut) are typical applications of (6), in which orientation no longer appears. Moments of inertia of surfaces appear in engineering construction work from time to time. A simple special case of constant distance (as well as of constant mass) is shown in Example 6. The text concludes with a look at representations z ⫽ f (x, y). Problems 1–11 concern integrals (3) and Probs. 12–16, more briefly, integrals (6). Problems 17 and 18 are worth noting, also for historical reasons. Further applications are shown in Probs. 19–25. Problem 26 is important; it defines the first fundamental form of a surface and shows that this form determines the metric on that surface, as was first shown in full generality by Gauss. SOLUTIONS TO PROBLEM SET 10.6, page 450 2. z ⫽ 1 ⫺ x ⫺ y, r ⫽ [u, v, 1 ⫺ u ⫺ v], hence F (r) • N ⫽ [ev, eu, 1] • ([1, 0, ⫺1] ⴛ [0, 1, ⫺1]) ⫽ [ev, eu, 1] • [1, 1, 1] ⫽ 1 ⫹ eu ⫹ ev. Integration gives the answer 2e ⫺ 1. 4. Quarter of a circular cylinder of radius 5 and height 2 in the first octant with the z-axis as axis. A parametric representation is r ⫽ [5 cos u,

5 sin u, v],

0 ⬉ u ⬉ 12 p,

0 ⬉ v ⬉ 2.

From this we obtain N ⫽ [5 cos u, 5 sin u, 0] F • N ⫽ 5e5 sin u cos u ⫺ 5ev sin u. Integration over u from 0 to 21 p gives e5 ⫺ 1 ⫺ 5ev. Integration of this over v from 0 to 2 gives the answer 2(e5 ⫺ 1) ⫺ 5(e2 ⫺ 1) ⫽ 2e5 ⫺ 5e2 ⫹ 3 ⫽ 262.88. 6. r ⫽ [u, v, u ⫹ v2]. This is a parabolic cylinder, the parameter curves being parabolas and straight line generators of the cylinder. For the normal vector we obtain N ⫽ [⫺1, ⫺2v, 1].

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F on the surface is F(r(u, v)) ⫽ [cosh v, 0, sinh u]. This gives the integrand F • N ⫽ ⫺cosh v ⫹ sinh u. Integration over v from 0 to u gives ⫺sinh u ⫹ u sinh u. Integration of this over u from 0 to 1 gives the answer [⫺cosh u ⫹ u cosh u ⫺ sinh u] ƒ 10 ⫽ 1 ⫺ sinh 1 ⫽ ⫺0.1752. 8. We may choose 2 ⬉ u ⬉ 5, 0 ⬉ v ⬉ p>2.

r ⫽ [u, cos v, sin v], The integrand is

F • N ⫽ [tan (u cos v), u, cos v] • [0, ⫺cos v, ⫺sin v]. Integration gives p>2

冮冮 (⫺u cos v ⫺ cos v sin v) dv du ⫽ ⫺12. 10. Portion of a circular cone with the z-axis as axis. A parametric representation is r ⫽ [u cos v, u sin v,

(0 ⬉ u ⬉ 2, 0 ⬉ v ⬉ p).

4u]

From this, N ⫽ [⫺4u cos v,

⫺4u sin v,

u],

F ⫽ [u 2 sin2 v,

u 2 cos2 v, 256u 4].

The integrand is F • n ⫽ ⫺4u 3 sin2 v cos v ⫺ 4u 3 cos2 v sin v ⫹ 256u 5. Integration over u from 0 to 2 gives ⫺16 sin2 v cos v ⫺ 16 cos2 v sin v ⫹

8192 . 3

Integration of this over v from 0 to p gives p

8192 8192 16 3 16 32 cos3 v ⫹ vd ⫽ ⫺ ⫹ p ⫽ 8567.98. c ⫺ sin v ⫹ 3 3 3 3 3 0 12. r ⫽ [u, v,

1 ⫺ u ⫺ v], G(r) ⫽ cos u ⫹ sin v, ƒ N ƒ ⫽ 13. Integration gives

1ⴚu

冮冮

(cos u ⫹ sin v)13 dv du ⫽ (2 ⫺ cos 1 ⫺ sin 1)13 ⫽ 1.0708.

14. r ⫽ [cos v cos u, vector is

sin v],

0 ⬉ u ⬉ p,

0 ⬉ v ⬉ p>2. A normal

cos2 v sin u,

cos v sin v]

and

cos v sin u,

r ⫽ [cos2 v cos u,

ƒ N ƒ ⫽ cos v.

On S, G ⫽ a cos v cos u ⫹ b cos v sin u ⫹ c sin v. The integrand is this expression times cos v. Integration over u from 0 to p gives 0 ⫹ 2b cos2 v ⫹ pc sin v cos v.

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Integration over v from 0 to p>2 gives the answer 1 2

16. r ⫽ [u cos v, u sin v,

p(b ⫹ c)

u 2], hence u sin v G(r) ⫽ arctan u cos v ⫽ v.

Furthermore, ƒ N ƒ ⫽ 24u 4 ⫹ u 2, so that the integrand is G(r) ƒ N(r) ƒ ⫽ vu24u 2 ⫹ 1. Integration gives 3

冮冮 1

p>2

vu24u 2 ⫹ 1 dv du ⫽

p2 8

冮 u24u ⫹ 1 du 2

⫽

ⴢ

1 2 ⴢ (4u 2 ⫹ 1)3>2 ` 8 3 1

1 2 3>2 ⫽ p (37 ⫺ 53>2) ⫽ 22.00. 96 18. Möbius reported that Gauss had shown him the “double ring,” most likely in connection with counting the intertwinements of two curves, physically related to the electromagnetic field near two wires. Both Möbius and Listing independently published the invention of the Möbius strip in 1858, three years after Gauss had died. It is possible that they got the idea from Gauss, who usually published full theories, rather than isolated results. Listing was a student of Gauss and the author of the earliest book on topology, which he (and Riemann and others) called Analysis situs. 22. B is represented by y ⫽ 0, z ⫽ h>2. The square of the distance of a point (x, y, z) on S from B is (x ⫺ x)2 ⫹ (y ⫺ 0)2 ⫹ (z ⫺ 12 h)2 ⫽ y 2 ⫹ (z ⫺ 12 h)2. S can be represented by r ⫽ [cos u, sin u, v], where 0 ⬉ u ⬉ 2p, 0 ⬉ v ⬉ h. Hence the moment of inertia is (with dA ⫽ r du dz ⫽ 1 # du dv) h

I⫽

冮冮 0

⫽

冮

1 c sin u ⫹ av ⫺ hb d du dv 2 2

a p ⫹ av ⫺

⫽ hp ⫹

h3 ⴢ 2p 12

⫽ hp a1 ⫹

h2 b. 6

1 hb ⴢ 2p b dv 2

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24. Proof for a lamina S of density s. Choose coordinates so that B is the z-axis and ␬ is the line x ⫽ k in the xz-plane. Then I␬ ⫽

冮冮 [(x ⫺ k) ⫹ y ]s dA ⫽ 冮冮 (x ⫺ 2kx ⫹ k ⫹ y )s dA 2

⫽

冮冮 (x ⫹ y )s dA ⫺ 2k 冮冮 xs dA ⫹ k 冮冮 s dA 2

⫽ IB ⫺ 2k # 0 ⫹ k 2M, the second integral being zero because it is the first moment of the mass about a line through the center of gravity. For a mass distributed in a region in space the idea of proof is the same. 26. Team Project. (a) Use dr ⫽ ru du ⫹ rv dv. This gives (13) and (14) because dr • dr ⫽ ru • ru du 2 ⫹ 2ru • rv du dv ⫹ ru • rv dv2. (b) E, F, G appear if you express everything in terms of dot products. In the numerator, a • b ⫽ (ru g r ⫹ rv h r ) • (ru p r ⫹ rv q r ) ⫽ Eg r p r ⫹ F (g r q r ⫹ h r p r ) ⫹ Gh r q r and similarly in the denominator. (c) This follows by Lagrange’s identity (Problem Set 9.3), ƒ ru ⴛ rv ƒ 2 ⫽ (ru ⴛ rv) • (ru ⴛ rv) ⫽ (ru • ru)(rv • rv) ⫺ (ru • rv)2 ⫽ EG ⫺ F 2. (d) r ⫽ [u cos v, u sin v], ru ⫽ [cos v, sin v], ru • ru ⫽ cos2 v ⫹ sin2 v ⫽1, etc. (e) By straightforward calculation, E ⫽ (a ⫹ b cos v)2, and F ⫽ 0 (the coordinate curves on the torus are orthogonal!), and G ⫽ b 2. Hence, as expected, 2EG ⫺ F 2 ⫽ b(a ⫹ b cos v). SECTION 10.7. Triple Integrals. Divergence Theorem of Gauss, page 452 Purpose, Content Proof and application of the first “big” integral theorem in this chapter, Gauss’s theorem, preceded by a short discussion of triple integrals (probably known to most students from calculus). Comment on Proof The proof is simple: 1. Cut (2) into three components. Take the third, (5). 2. On the left, integrate

(8)

冮冮冮 0z dz dx dy over z to get 0F3

冮冮[F (upper surface) ⫺ F (lower surface)] dx dy 3

integrated over the projection R of the region in the xy-plane (Fig. 252).

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3. Show that the right side of (5) equals (8). Since the third component of n is cos g, the right side is

冮冮F cos g dA ⫽ 冮冮F dx dy 3

冮冮F (upper) dx dy ⫺ 冮冮F (lower) dx dy,

⫽

where minus comes from cos g ⬍ 0 in Fig. 252, lower surface. This is the proof. Everything else is (necessary) accessory. Comments on Problems Problems 1–8 concern triple integrals and Probs. 9–18 the divergence theorem itself. Masses in space lead to triple integrals over the region of mass distribution; an application to some standard regions is given in Probs. 19–25.

SOLUTIONS TO PROBLEM SET 10.7, page 457 c

冮冮冮 0

xyz dx dy dz ⫽

1 2

冮冮 0

a 2 yz dy dz ⫽

1 4

冮 a b dz ⫽ 81 a b c 2 2

2 2 2

4. Integration over x from 0 to 3 ⫺ y ⫺ z gives eⴚyⴚz ⫺ eⴚ3. Integration over y from 0 to 3 – z then gives eⴚz ⫺ 4eⴚ3 ⫹ zeⴚ3. Integration over z from 0 to 3 finally gives the answer 1 ⫺ 8.5eⴚ3. Note that the region T in space is bounded by portions of the coordinate planes and of the plane x ⫹ y ⫹ z ⫽ 3, from which one obtains the limits of integration. 6. We can represent the region T: x 2 ⫹ z 2 ⬉ 16, ƒ y ƒ ⬉ 4, a portion of a solid cylinder, by [r cos u, v, r sin u],

0 ⬉ r ⬉ 4,

0 ⬉ u ⬉ 2,

Then the volume element is dV ⫽ r dr du dv and the density is s ⫽ r 4(cos2 u sin2 u)v2. Integration over r from 0 to 4 gives 2048 (cos2 u sin2 u)v2. 3

⫺4 ⬉ v ⬉ 4.

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Integration of this over u from 0 to 2p gives 512 2 v p. 3 Integration of this over v from ⫺4 to 4 finally gives the answer 65,536 9

p ⫽ 22,876.4.

8. From (3) in Sec. 10.5 with variable r instead of constant a we have x ⫽ r cos v cos u,

y ⫽ r cos v sin u,

z ⫽ r sin v.

Hence x ⫹ y ⫽ r cos v. The volume element is 2

dV ⫽ r 2 cos v dr du dv. The intervals of integration are 0 ⬉ r ⬉ a, 0 ⬉ u ⬉ 2p, 0 ⬉ v ⬉ 12 p. The integrand is r 4 cos3 v. Integration over r, u, and v gives a 5>5, 2p, and 23. The product of these is the answer 4pa 5>15. 10. The outer normal vectors of the faces x ⫽ ⫾1 are n ⫽ [⫾1, 0, 0]. This gives F • n ⫽ ⫾1, integrated over the two faces ⫾1 ⴢ 6 ⴢ 2 ⫽ ⫾12, and ⫺12 ⫹ 12 ⫽ 0. No contribution from the faces y ⫽ ⫾3 since n ⫽ [0, ⫾1, 0], so that F • n ⫽ [x 2, 0, z 2] • [0, ⫾1, 0] ⫽ 0. The face z ⫽ 0 gives F • n ⫽ [x 2, 0, 0] • [0, 0, ⫺1] ⫽ 0. The face z ⫽ 2 gives F • n ⫽ [x 2, 0, 4] • [0, 0, 1] ⫽ 4 times the area 2 ⴢ 6 ⫽ 12, hence 48, as expected. This calculation is so simple only because of the simplicity of the surface and the region. Let the student make a sketch so that he/she understands the form of the outer normals. 12. div F ⫽ 3x 2 ⫹ 3y 2 ⫹ 3z 2 ⫽ 3r 2, dV ⫽ r 2 cos v dr du dv. Intervals of integration 0 ⬉ r ⬉ 5, 0 ⬉ u ⬉ 2p, 0 ⬉ v ⬉ 12p. The integrand is 3r 4 cos v. Integration over r, v, and u gives successively 1875 cos v,

1875,

3750p.

Note that S consists of a hemisphere and a disk! 14. div F ⫽ ⫺sin z. Integration over z gives cos 2 ⫺ 1. Multiplication by the crosssectional area gives the answer 9p(cos 2 ⫺ 1) ⫽ ⫺40.04. 16. div F ⫽ ⫺sinh x. Integration over z from 0 to 1 ⫺ x ⫺ y gives (1 ⫺ x ⫺ y) sinh x. Integration of this over y from 0 to 1 ⫺ x gives 12 (sinh x) (1 ⫺ x)2. Integration of this over x from 0 to 1 gives the answer cosh 1 ⫺ 32 ⫽ 0.0431. 18. div F ⫽ y ⫹ z ⫹ x. A parametrization of the cone is [r cos v, r sin v, u]. The volume element is r dr du dv. Integration over r from 0 to 2u gives 8 3 3 (sin v ⫹ cos v) ⫹ 2u .

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Integration over v from 0 to 2p gives 0 ⫹ 4pu 3. Integration of this over u from 0 to 2 gives the answer 16p. 20. From (3) in Sec. 10.5 with variable r instead of constant a we have x ⫽ r cos v cos u,

y ⫽ r cos v sin v,

z ⫽ r sin v.

Hence x 2 ⫹ v2 ⫽ r 2 cos2 v. The volume element is dV ⫽ r 2 cos v dr du dv. The intervals of integration are 0 ⬉ r ⬉ a, 0 ⬉ u ⬉ 2p, ⫺12 p ⬉ v ⬉ 12p. The integrand is r 4 cos3 v. Writing the integrals as a product of three integrals, integration over r, u, and v gives a 5>5, 2p, and 43, respectively. The product of these is the answer 8pa 5>15. 22. r 2 ⫽ y 2 ⫹ z 2. Integration over r from 0 to 1x gives x 2>4. Integration of this over x from 0 to h gives h3>12. Answer: h3p>6. 24. ph510 ⫽ ph3>6 gives h ⫽ 15>3. For h ⬎ 15>3 the moment Ix is larger for the cone because the mass of the cone is spread out farther than that of the paraboloid when x ⬎ 1.

SECTION 10.8. Further Applications of the Divergence Theorem, page 458 Purpose. To represent the divergence free of coordinates and to show that it measures the source intensity (Example 1); to use Gauss’s theorem for deriving the heat equation governing heat flow in a region (Example 2); to obtain basic properties of harmonic functions. Main Content, Important Concepts Total flow (1) out of a region Divergence as the limit of a surface integral; see (2) Heat equation (5) (to be discussed further in Chap. 12) Properties of harmonic functions (Theorems 1–3) Green’s formulas (8), (9) Short Courses. This section can be omitted. Comments on (2) Equation (2) is sometimes used as a definition of the divergence, giving independence of the choice of coordinates immediately. Also, Gauss’s theorem follows more readily from this definition, but, since its proof is simple (see Sec. 10.7. in this Manual), that saving is marginal. Also, it seems that our Example 2 in Sec. 9.8 motivates the divergence at least as well (and without integrals) as (2) in the present section does for a beginner. General Comments on Text and Problems Team Project 12 shows how the ideas of the text in Theorems 1–3 can be further extended and supplemented. These ideas are basic in the theory of partial differential equations and their applications in physics, where the latter have helped in the discovery of mathematical theorems on our present level as well as in more general and more abstract theories, using functional analysis and measure theory.

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SOLUTIONS TO PROBLEM SET 10.8, page 462 2. A representation of the cylinder is r ⫽ [2 cos u, 2 sin u], which also equals the normal vector N, as is seen by geometry or by calculating ru ⴛ rv. Here v ⫽ z varies from 0 to h and u from 0 to 2p. It follows that n ⫽ [cos u, sin u]. Hence the directional derivative is 0f 0n

⫽ ⵜf • n ⫽ [⫺8 cos u sin u ⫺ 8 sin u cos u, 0] • [cos u, sin u] ⫽ ⫺16 cos2 u sin u.

The integral over u from 0 to 2p or from ⫺p to p is zero since the integrand is odd. Integration of this over v ⫽ z from 0 to h gives zero, as had to be shown. 4. ⵜ2g ⫽ 4, grad f • grad g ⫽ [1, 0, 0] • [0, 2y, 2z] ⫽ 0. Integration of 4x over the box gives 12. Also, f 0g>0n for the six surfaces gives (x ⫽ 0)

0[⫺1,

(x ⫽ 1)

1[1,

0] • [0,

2y, 2z],

integral 0

( y ⫽ 0)

x[0,

⫺1, 0] • [0,

integral 0

( y ⫽ 2)

x[0,

0] • [0,

(z ⫽ 0)

x[0,

⫺1] • [0, 2y,

(z ⫽ 3)

x[0,

1] • [0,

0] • [0,

2y,

2z], 2z],

2z],

2y,

0], 6],

integral 0

integral of 4x gives 2 3 ⫽ 6 integral 0 integral of 6x gives 3 2 ⫽ 6.

6. The volume integral of 12x 2y 2 ⫺ 2y 4 is 43 ⫺ 25. The surface integral of x 2n • [0,

4y 3,

0] ⫺ y 4n • [2x, 0,

0] ⫽ n • [2xy 4,

4y 3,

is ⫺25 (x ⫽ 1) and 43 ( y ⫽ 1) and 0 for the other faces. 8. z ⫽ hr>a (make a sketch). For the disk S1: z ⫽ h, 0 ⬉ r ⬉ a,

冮冮

z dx dy ⫽

冮冮

hr dr du ⫽ 2ph

a 2 ⫽ pha . 2

For the conical portion, whose normal vector has a negative z-component, hence the minus sign in the formula, 2p

冮冮 z dx dy ⫽ ⫺ 冮冮 hra r dr du ⫽ ⫺2p ha a3 ⫽ ⫺p 23 ha . 2

Together pha 2>3. 10. F ⫽ [x, y, z], div F ⫽ 3. In (2), Sec. 10.7, we obtain F • n ⫽ ƒ F ƒ ƒ n ƒ cos ␾ ⫽ 2x 2 ⫹ y 2 ⫹ z 2 cos ␾ ⫽ r cos ␾.

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12. Team Project. (a) Put f ⫽ g in (8). (b) Use (a). (c) Use (9). (d) h ⫽ f ⫺ g is harmonic and 0h>0n ⫽ 0 on S. Thus h ⫽ const in T by (b). (e) Use div (grad f ) ⫽ ⵜ2f and (2). SECTION 10.9. Stokes’s Theorem, page 463 Purpose. To prove, explain, and apply Stokes’s theorem, relating line integrals over closed curves and surface integrals. Main Content Formula (2) ⬅ (2*) Further interpretation of the curl (see also Sec. 9.9) Path independence of line integrals (leftover from Sec. 10.2) Comment on Orientation Since the choice of right-handed or left-handed coordinates is essential to the curl (Sec. 9.9), surface orientation becomes essential here (Fig. 253). Comment on Proof The proof is simple: 1. Cut (2*) into components. Take the first, (3). 2. Using N1 and N3, cast the left side of (3) into the form (7). 3. Transform the right side of (3) by Green’s theorem in the plane into a double integral and show equality with the integral obtained on the left. Further Comments on Text and Problems Examples 1 and 3 show typical calculations in connection with Stoke’s theorem. In connection with Example 2, emphasize that Green’s theorem in the plane (a special case of Stokes’s theorem) was needed in the proof of the general case of Stokes’s theorem, as is mentioned in Example 2 (but was perhaps missed by the student). Basic applications of Stokes’s theorem to fluid flow and to work done in displacements around closed curves are illustrated in Examples 4 and 5. Problems 1–10 concern direct integration of surface integrals and Probs. 13–20 concern the evaluation of such integrals with Stokes’s theorem as a tool for evaluating line integrals around closed curves. SOLUTIONS TO PROBLEM SET 10.9, page 468 2. The rectangle S is represented by 0 ⬉ x ⬉ 4, 0 ⬉ y ⬉ 1, z ⫽ 2. The curl is curl F ⫽ [⫺3 cosh z, ⫺1, 13 cos y]. A normal vector of S is N ⫽ [0, 0, 1]. Hence (curl F) • N ⫽ 13 cos y.

Integration over x from 0 to 4 and over y from 0 to p>2 gives the answer ⫾52. 4. The curl is curl F ⫽ [0, 2z, ⫺2x] ⫽ [0, 2xy, ⫺2x].

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A normal vector of S is N ⫽ [⫺y, ⫺x, 1]. Hence the integrand is (curl F) • N ⫽ 0 ⫺ 2x 2y ⫺ 2x. Integration over x from 0 to 1 and over y from 0 to 4 gives the answer ⫿28>3. 6. The curl is curl F ⫽ [0, 0, ⫺3x 2 ⫺ 3y 2]. A normal vector of S is N ⫽ [0, 0, 1]. Hence the integrand is (curl F) • N ⫽ ⫺3x 2 ⫺ 3y 2 ⫽ ⫺3r 2 where x ⫽ r cos u, y ⫽ r sin u are polar coordinates. The integral is ⫺3

冮冮

(x ⫹ y ) dx dy ⫽ ⫺3

冮冮

3 4 3 2 r ⴢ r dr du ⫽ ⫺ r ⴢ 2p ` ⫽ ⫺ p. 4 2 0

Hence the answer is ⫿ 32 p. 8. This is a portion of a cone. A representation is r ⫽ [u cos v, u sin v, u]

(0 ⬉ u ⬉ h, 0 ⬉ v ⬉ p).

The integrand is (curl F) • N ⫽ [2y, 2z, 2x] • (⫺x, ⫺y, 2x 2 ⫹ y 2 ] ⫽ [2u sin v, 2u, 2u cos v] • [⫺u cos v, ⫺u sin v, u] ⫽ ⫺2u 2(cos v sin v ⫹ sin v ⫺ cos v). Integration over u from 0 to h and v from 0 to p (since y ⭌ 0) gives ⫺

2h3 4h3 (0 ⫹ 2 ⫺ 0) ⫽ ⫺ . 3 3

Hence the answer is ⫿4h3>3. 10. r ⫽ [cos u, sin u], F • r r ⫽ [sin3 u, ⫺cos3 u] • [⫺sin u, cos u] ⫽ ⫺sin4 u ⫺ cos4 u. Integration over u from 0 to 2p gives ⫺3p>4 ⫺ 3p>4. 14. The curl is curl F ⫽ [3y 2, 3z 2, 3x 2]. The circle is the boundary curve of a circular disk of radius 3 in the yz-plane with normal [1, 0, 0]. Hence (curl F) • N ⫽ 3y 2. To integrate, we introduce polar coordinates defined by r ⫽ [2, u cos v, u sin v]

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and obtain, using dy dz ⫽ u du dv, 3

冮冮 3u (cos v) u du dv ⫽ 2434 p. 2

16. The curl is curl F ⫽ [0, ⫺ex, ⫺ey]. A normal vector is [0, 0, 1], as in Prob. 15. Hence the surface integral is (sketch the triangle!) 1

冮冮 ⫺e dy dx ⫽ ⫺e ⫹ 2. y

18. S bounded by C can be represented by r ⫽ [v, 2 cos u, 2 sin u]

(0 ⬉ v ⬉ h, 0 ⬉ u ⬉ p).

A normal vector with a nonnegative z-component is N ⫽ ru ⴛ rv ⫽ [0, 2 cos u, 2 sin u]. The curl is curl F ⫽ [⫺2, 0, 1]. The inner product is (curl F) • N ⫽ 2 sin u. Integration over u from 0 to p (since z ⭌ 0) and over v gives h

冮冮 2 sin u du dv ⫽ 4h. 20. A representation of the portion of the cylinder bounded by C is r ⫽ [v, 2 cos u, 2 sin u],

(0 ⬉ u ⬉ p>2, 0 ⬉ v ⬉ p).

A normal vector of the cylinder is N ⫽ ru ⴛ rv ⫽ [0, 2 cos u, 2 sin u]. The curl is curl F ⫽ [0, 0, ⫺sin v] and (curl F) • N ⫽ ⫺2 sin u sin v.

Integration over u from 0 to p>2 and over v from 0 to p gives ⫺4. SOLUTIONS TO CHAPTER 10 REVIEW QUESTIONS AND PROBLEMS, page 469 12. Exact. By integration we find that F ⫽ grad f,

where

f ⫽ sin xy ⫹ ez.

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Substituting the limits of integration gives f ( 12 , p, 1) ⫺ f (p, 1, 0) ⫽ sin 12 p ⫺ sin p ⫹ e1 ⫺ e0 ⫽ e. 14. Not exact. Use Green’s theorem. We obtain curl F ⫽ [0, 0, 3x 2 ⫹ 3y 2],

N ⫽ [0, 0, 1].

Hence (curl F) • N ⫽ 3x 2 ⫹ 3y 2. Using polar coordinates, we integrate 2p

p. 冮冮 3r ⴢ r dr du ⫽ 1875 2 2

The answer is ⫾1875p>2. 16. Not exact. We obtain r r ⫽ [⫺2 sin t,

2 cos t,

and F(r) ⫽ [4 cos2 t,

4 sin2 t,

8 sin2 t cos t].

The inner product is F • r r ⫽ ⫺8 cos2 t sin t ⫹ 32 sin2 t cos t. Integration gives 8 32 3 3 p 16 3 cos t ⫹ 3 sin t ƒ 0 ⫽ ⫺ 3 .

18. Use Stokes’s theorem. r ⫽ [u, v, u], N ⫽ [⫺1, 0, 1]. Furthermore, curl F ⫽ [0, ⫺p cos px,

⫺p(sin px ⫹ cos py)]

⫽ [0, ⫺p cos pu,

⫺p(sin pu ⫹ cos pv)].

The inner product is (curl F(r)) • N ⫽ ⫺p(sin pu ⫹ cos pv). Integration over u from 0 to 1, gives ⫺p cos pv ⫺ 2. Integration of this over v from 0 to 2 gives ⫺4. Answer: ⫿4. 20. Exact, F ⫽ grad (exz ⫹ cosh 2y). Integration from (⫺1, ⫺1, 1) to (1, 1, 1) gives e ⫹ cosh 2 ⫺ (eⴚ1 ⫹ cosh 2) ⫽ 2 sinh 1. 22. M ⫽ pa 4>4, x ⫽ 0 (why?), y ⫽

1 M

冮冮 (r sin t)r ⴢ r dr dt ⫽ 58ap 2

24. x ⫽ 0 by symmetry. Furthermore, 1

M⫽

冮 (1 ⫺ x ) dx ⫽ 85 4

⫺1

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and 1 y⫽ ⫽ M

冮冮

1ⴚx4

y dy dx ⫽

⫺1 0

冮 165 (1 ⫺ x ) dx ⫽ 49. 4 2

⫺1

26. k drops out in both integrands x f>M and y f>M. 28. div F ⫽ 3, V ⫽ 43 pabc. Answer: 4pabc 30. Direct integration. We have r ⫽ [2 cos u cos v,

2 cos u sin v,

(0 ⬉ u ⬉ 12 p,

sin u]

0 ⬉ v ⬉ 2p).

From this, ru ⫽ [⫺2 sin u cos v,

⫺2 sin u sin v,

cos u]

rv ⫽ [⫺2 cos u sin v,

2 cos u cos v,

N ⫽ [⫺2 cos u cos v,

⫺2 cos u sin v,

⫺4 cos u sin u].

The inner product is F • N ⫽ (⫺2 cos 2u)(cos v ⫹ sin v) ⫺ 4

cos u sin u.

Integration of cos v ⫹ sin v over v from 0 to 2p gives 0. Integration of ⫺4 cos u sin u over u from 0 to p>2 gives ⫺2. Integration of this constant over v from 0 to 2p gives ⫺4p (or ⫹ 4p if we change the orientation by interchanging u and v). 32. By direct integration. We can represent the paraboloid in the form r ⫽ [u cos v, u sin v, u 2]. A normal vector is N ⫽ ru ⴛ rv ⫽ [⫺2u 2 cos v, ⫺u 2 sin v, u]. On the paraboloid, F ⫽ [u 2 sin2 v, u 2 cos2 v, u 4]. The inner product is F • N ⫽ ⫺2u 4 cos v sin2 v ⫺ 2u 4 cos2 v sin v ⫹ u 5. Integration over v from 0 to 2p gives 0 ⫹ 0 ⫹ 2pu 5. Integration of this over u from 0 to 3 (note that z ⫽ u 2 varies from 0 to 9) gives 2p36>6 ⫽ 243p. 34. By Gauss’s theorem, T can be represented by r ⫽ [r cos u, r sin u,

v],

where

0 ⬉ r ⬉ 1, 0 ⬉ u ⬉ 2p,

0 ⬉ v ⬉ 5.

The divergence of F is div F ⫽ 1 ⫹ x ⫹ 1 ⫽ 2 ⫹ x ⫽ 2 ⫹ r cos u. The integrand is (2 ⫹ r cos u)r. Integration over r from 0 to 1 gives 1 ⫹ 13 cos u. Integration of this over u from 0 to 2p gives 2p. Integration of this over v from 0 to 5 gives the answer 10p. Note that this is the integral of the term 2 in div F, whereas r cos u gives 0.

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Part C FOURIER ANALYSIS. PARTIAL DIFFERENTIAL EQUATIONS (PDEs) CHAPTER 11

Fourier Analysis

SECTION 11.1. Fourier Series, page 474 Purpose. To derive the Euler formulas (6) for the coefficients of a Fourier series (5) of a given function of period 2p, using as the key property the orthogonality of the trigonometric system. Main Content, Important Concepts Periodic function Trigonometric system, its orthogonality (Theorem 1) Fourier series (5) with Fourier coefficients (6) Representation by a Fourier series (Theorem 2) Comment on Notation If we write a0>2 instead of a0 in (1), we must do the same in (6.0) and see that (6.0) then becomes (6a) with n ⫽ 0. This is merely a small notational convenience (but may be a source of confusion to poorer students). Comment on Fourier Series Whereas their theory is quite involved, practical applications are simple, once the student has become used to evaluating integrals in (6) that depend on n. Figure 260 should help students understand why and how a series of continuous terms can have a discontinuous sum. Comment on the History of Fourier Series Fourier series were already used in special problems by Daniel Bernoulli (1700–1782) in 1748 (vibrating string, Sec. 12.3) and Euler (Sec. 2.5) in 1754. Fourier’s book of 1822 became the source of many mathematical methods in classical mathematical physics. Furthermore, the surprising fact that Fourier series, whose terms are continuous functions, may represent discontinuous functions led to a reflection on, and generalization of, the concept of a function in general. Hence the book is a landmark in both pure and applied mathematics. [That surprising fact also led to a controversy between Euler and D. Bernoulli over the question of whether the two types of solution of the vibrating string problem (Secs. 12.3 and 12.4) are identical; for details, see E. T. Bell, The Development of Mathematics, New York: McGraw-Hill, 1940, p. 482.] A mathematical theory of Fourier series was started by Peter Gustav Lejeune Dirichlet (1805–1859) of Berlin in 1829. The concept of the Riemann integral also resulted from work on Fourier series. Later on, these series became the model case in the theory of orthogonal functions (Sec. 5.7). An English translation of Fourier’s book was published by Dover Publications in 1955. Further Comments on Text and Problems Figure 260, showing the rectangular periodic wave and partial sums of its Fourier series, is typical and should give the student an intuitive feel for convergence of Fourier series, notably of their behavior near discontinuities. The latter are of great practical importance, as will appear as we proceed. 205

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The essential property of the trigonometric system is its orthogonality. Other orthogonal systems will follow later in connection with families of solutions of linear ODEs (Legendre polynomials, Bessel functions, etc.). The integrals needed for Fourier coefficients are those of calculus, but their dependence on n will create a new situation, and the student will need some time and particular attention to become familiar with them. Problem 25 suggests that students look critically at the speed of convergence, which will depend on the power of 1>n in the Fourier coefficients, which in turn depends on continuity of f (x) to be developed. The figures in Probs. 16–21 should help students to get familiar with “piecewise given” functions, as they will occur in practical work on potential problems, heat flow, and so on. SOLUTION OF PROBLEM SET 11.1, page 482 2. 2p>n, 2p>n, k, k, k>n, k>n 4. f (x ⫹ p) ⫽ f (x) implies f (ax ⫹ p) ⫽ f (a[x ⫹ ( p>a)]) ⫽ f (ax) or g [ x ⫹ ( p>a)] ⫽ g (x), where g (x) ⫽ f (ax). Thus g (x) has the period p>a. This proves the first statement. The other statement follows by setting a ⫽ 1>b. p 4 1 1 ⫺ acos x ⫹ cos 3x ⫹ cos 5x ⫹ Á b 12. 2 p 9 25 14. 13 p2 ⫺ 4(cos x ⫺ 14 cos 2x ⫹ 19 cos 3x ⫺ ⫹ Á ) 2 2 2 1 1 1 sin x ⫹ sin 2x ⫺ sin 3x ⫺ sin 4x ⫹ sin 5x ⫹ cos 6x ⫹ Á 16. p 2 9p 4 25p 6 1 2 1 1 ⫹ asin x ⫹ sin 3x ⫹ sin 5x ⫹ Á b. f (x) ⫺ 12 is odd. 18. 2 p 3 5 1 1 1 1 (2 ⫹ 5p) sin 5x ⫹ Á d ⫺ sin 2x 20. c (2 ⫹ p) sin x ⫹ (⫺2 ⫹ 3p) sin 3x ⫹ p 9 25 2 1 1 ⫺ sin 4x ⫺ sin 6x ⫺ Á 4 6 22. CAS Experiment. Experimental approach to Fourier series. This should help the student obtain a feel for the kind of series to expect in practice, and for the kind and quality of convergence, depending on continuity properties of the sum of the series. (a) The 2p-periodic function f (x) ⫽ x (⫺p ⬍ x ⬍ p) has discontinuities at ⫾p. The instructor will notice the Gibbs phenomenon (see Sec. 11.7) at the points of discontinuity. (b) f (x) ⫽ 1 ⫹ x> p if ⫺p ⬍ x ⬍ 0 and 1 ⫺ x> p if 0 ⬍ x ⬍ p, is continuous throughout, and the accuracy is much better than in (a). (c) f (x) ⫽ p2 ⫺ x 2 has about the same continuity as (b), and the approximation is good. The coefficients in (a) involve 1>n, whereas those in (b) and (c) involve 1>n 2. This is typical. See also CAS Experiment 25. 24. CAS Experiment. The student should recognize the importance of the interval in connection with orthogonality, which is the basic concept in the derivation of the Euler formulas. For instance, for sin 3x sin 4x the integral equals sin a ⫺ 17 sin 7a, and the graph suggests orthogonality for a ⫽ p, as expected.

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1 0.5

–2␲

␲

–␲

2␲

–0.5 –1

Section 11.1. Integral in Problem 24 as a function of a

SECTION 11.2. Arbitrary Period. Even and Odd Functions, Half-Range Expansions, page 483 Purpose. The three topics considered in this section are listed in the title. The three main points are as follows. The transition from period 2p to period 2L amounts to a linear transformation in x. From (5), (6) in Sec. 11.1 it produces (5), (6) in this section. For even functions the Fourier series reduces to a cosine series (hence a series without sine terms) (5*) with coefficients (6*). For odd functions the Fourier series reduces to a sine series (5**) with coefficients (6**). For period 2p the corresponding simpler formulas are separately listed in the Summary. Typical illustrations of all this are shown in Examples 1–5. The third and last topic is half-range expansions, typically illustrated in Example 6 and Fig. 272. This will be applied to physical problems of vibrations and heat conduction in the next chapter. In the problem set we start with some general questions on even and odd functions (Probs. 1–7), followed by Fourier series developments for various periods (Probs. 8–17) and by some general problems (Probs. 18–22). Finally, half-range expansions are needed in Probs. 23–30, cosine series as well as sine series in each case, that is, functions f (x) given on an interval 0 ⬉ x ⬉ L are to be 苲 represented on ⫺L ⬉ x ⬉ L as a cosine series as well as a sine series, of a function f (x) of period 2L, obtained by extending the given function from 0 ⬉ x ⬉ L to ⫺L ⬉ x ⬉ L as an even or odd function, respectively. SOLUTIONS TO PROBLEM SET 11.2, page 490 2. Even, even, neither, odd, even 4. Odd for sums and for products of an odd number 2k ⫹ 1 of factors, f (⫺x) ⫽ f1(⫺x) Á f2k⫹1(⫺x) ⫽ (⫺1)2k⫹1 f1(x) Á f2k⫹1(x) ⫽ ⫺f (x). Even for products of an even number of factors. 6. Odd. This is important in connection with the integrand in the Euler formulas for the Fourier coefficients. It implies the simplification of the Fourier series of an odd function to a Fourier sine series and of the Fourier series of an even function to a Fourier cosine series. 8. Even, L ⫽ 1, full-wave rectification of g (x) ⫽ x (⫺1 ⬍ x ⬍ 1), p ⫽ 2. Cf. Prob. 12 in Sec. 11.1. The Fourier series is 1 4 1 1 ⫺ 2 acos px ⫹ cos 3px ⫹ cos 5px ⫹ Á b. 2 9 25 p

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10. Odd, L ⫽ 4. Cf. also Prob. 21 in Sec. 11.1. The series is 8

asin

px 4

1 px 1 3px sin ⫹ sin ⫹ Áb 2 2 3 4

⫹

12. Even. Use the answer to Prob. 11, multiply by ⫺1, add 1, replace x by x>2. The series is 2 4 px 1 1 3px 1 5px ⫹ 2 acos ⫺ cos px ⫹ cos ⫺ cos ⫹⫺Áb p 3 2 4 9 2 16 2 14. Even, L ⫽ 12, full-wave rectification of a cosine current. The series is 2

⫹

16. Odd, L ⫽ 1,

cos 2px ⫺

1 1 cos 4px ⫹ cos 6px ⫺ ⫹ Á b . 3ⴢ5 5ⴢ7

f (x) ⫽ ⫺x 2 if ⫺1 ⬍ x ⬍ 0,

3 a(

p 1ⴢ3

f (x) ⫽ x 2 if 0 ⬍ x ⬍ 1, series

1 1 (9p2 ⫺ 4) sin 3px ⫹ (25p2 ⫺ 4) sin 5px ⫹ Á b 27 125 1 1 1 ⫺ asin 2px ⫹ sin 4px ⫹ sin 6px ⫹ Á b p 2 3

p2 ⫺ 4) sin px ⫹

V0 18. bn ⫽ 0, a0 ⫽ p , an ⫽ 100V0

冮

1>200

cos 100pt cos 100npt dt

ⴚ1>200

⫽ 50V0

冮

1>200

cos 100(n ⫹ 1)pt dt ⫹ 50V0

ⴚ1>200

冮

1>200

cos 100(n ⫺ 1)pt dt ;

ⴚ1>200

hence the series is V0

⫹ ⫹

V0 2

cos 100pt

2V0

1 1 1 cos 200pt ⫺ cos 400pt ⫹ cos 600pt ⫺ ⫹ Á b 1ⴢ3 3ⴢ5 5ⴢ7

20. Set x ⫽ ⫺1. Then f (⫺1) ⫽ 1 ⫽

1 4 1 1 ⫺ 2 a⫺1 ⫺ ⫺ ⫺ Á b, etc. p 3 4 9

22. Multiply by ⫺1 and then add 1: ⫺ 24. L ⫽ 4, (a) (b)

1 4 1 1 ⫺ 2 acos px ⫹ cos 3px ⫹ cos 5px ⫹ Á b ⫹ 1 2 9 25 p

1 2 px 1 3px 1 5px ⫺ acos ⫺ cos ⫹ cos ⫺⫹Áb 2 p 4 3 4 5 4 2

asin

1 3px 1 5px 1 3px sin ⫹ sin ⫺ sin 4 2 3 4 5 4 3 2 9px 1 7px 1 1 5px 1 11px ⫹ sin ⫹ sin ⫺ sin ⫹ sin ⫹ Áb 7 4 9 4 5 2 11 4 ⫺ sin

⫹

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26. L ⫽ p, (a)

209

3p 2 1 1 1 1 ⫺ acos x ⫹ cos 2x ⫹ cos 3x ⫹ cos 5x ⫹ cos 6x 8 p 2 9 25 18 ⫹

(b) a1 ⫹

1 1 1 1 cos 7x ⫹ cos 9x ⫹ cos 10x ⫹ cos 11x ⫹ Á b 49 81 50 121

b sin x ⫺

1 1 2 1 sin 2x ⫹ a ⫺ b sin 3x ⫺ sin 4x 2 3 9p 4

1 2 1 1 2 ⫹a ⫹ b sin 5x ⫺ sin 6x ⫹ a ⫺ b sin 7x ⫹ Á 5 25p 6 7 49p The student should be invited to find the two functions that the sum of the series represents. This can be done by graphing sin x ⫺ 12 sin 2x ⫹ 13 sin 3x ⫹ ⫺ Á ⫽ x>2 and then (2> p)(sin x ⫺ 3ⴚ2 sin 3x ⫹ ⫺ Á ) ⫽ f2, where f2 (x) ⫽ b

x>2

p>2 ⫺ x>2

if ⫺p>2 ⬍ x ⬍ p>2 if

p>2 ⬍ x ⬍ 3p>2.

The first of these functions is discountinuous, the coefficients being proportional to 1>n, whereas f2 is continuous, its Fourier coefficients being proportional to 1>n 2, so that they go to zero much faster than the others. 28. (a) (b)

L 4L px 1 3px 1 5px ⫺ 2 acos ⫹ cos ⫹ cos ⫹ Áb 2 L 9 L 25 L p 2L

asin

px L

⫺

1 2px 1 3px 1 4px sin ⫹ sin ⫺ sin ⫹⫺ Áb 2 L 3 L 4 L

30. Shift by p. This cosine series is obtained. The sine series needs to be multiplied by ⫺1. The result is that the coefficients of odd multiples of p remain the same, whereas the coefficients of even multiples of p are multiplied by ⫺1.

SECTION 11.3. Forced Oscillations, page 492 Purpose. To show that mechanical or electrical systems with periodic but nonsinusoidal input may respond predominantly to one of the infinitely many terms in the Fourier series of the input, giving an unexpected output; see Fig. 277, where the output frequency is essentially five times that of the input. Further Comments on Text and Problems Example 1 in the text is sufficient to show the general idea and is typical of the problems. Problems 2 and 20 are particularly important. In fact, students should decrease the damping term in (4), letting it approach zero. This will give an additional better understanding of the present situation.

SOLUTIONS TO PROBLEM SET 11.3, page 494 2. For k ⫽ 49 (and c ⫽ 0.05 as before) the amplitudes are C1 ⫽ 0.0265, C3 ⫽ 0.0035, C5 ⫽ 0.0021, C7 ⫽ 0.0742, C9 ⫽ 0.0005, C11 ⫽ 0.0001. An increase of the damping constant c ( ⬎ 0) increases Dn for all n, hence it decreases all amplitudes Cn.

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4. r r (t) is given by the sine series in Example 1 with k ⫽ ⫺1. The new Cn is n times the old. Hence C5 is now so large that the output is practically a cosine vibration having five times the input frequency. 6. y ⫽ C1 cos vt ⫹ C2 sin vt ⫹

1 v ⫺a 2

sin at ⫹

1 v ⫺ b2 2

sin bt

8. The given r (t) is r (t) ⫽

p 4

cos t if ⫺

⬍t⬍

and ⫺

p 4

cos t if

p 2

⬍t⬍

3p . 2

The corresponding Fourier series, a Fourier cosine series, is r (t) ⫽

1 1 1 1 ⫹ # cos 2t ⫺ # cos 4t ⫹ # cos 6t ⫺ ⫹ Á . 2 1 3 3 5 5 7

Substituting this into the above ODE and solving it gives the answer y ⫽ C1 cos vt ⫹ C2 sin vt ⫹ ⫺

1 3 # 5(v2 ⫺ 16)

10. y ⫽ C1 cos vt ⫹ C2 sin vt ⫹ ⫺

1 3 # 5(v2 ⫺ 16)

1 2v

cos 4t ⫺

1 2

⫹

1 1 # 3 (v2 ⫺ 4)

cos 2t

cos 4t ⫹ ⫺ Á . ⫺

1 1 # 3 (v2 ⫺ 4) 1

5 # 7(v2 ⫺ 36)

cos 2t

cos 6t ⫺ Á

14. The Fourier series is a Fourier sine series, as given and derived in Example 1 of Sec. 11.1 (k ⫽ 1) with coefficients bn ⫽ 4>(np)

(n odd).

Hence the ODE must be solved with the right side rn (t) ⫽ (4>np) sin nt

(n odd).

The steady-state solution of this ODE is ⴥ

y ⫽ a (An cos nt ⫹ Bn sin nt) n⫽1 n odd

where An ⫽ ⫺4nc>(npDn),

Bn ⫽ (4 ⫺ 4n 2)>(npDn)

with Dn ⫽ (1 ⫺ n 2)2 ⫹ n 2c2 as in Prob. 13. So we have b1 ⫽ 4> p, b2 ⫽ 0, b3 ⫽ 4>(3p), etc. and no cosine terms in the Fourier series on the right side of the ODE. In the solution the damping constant appears with the cosine terms (and in Dn), causing a phase shift, which is zero if c ⫽ 0. Also, increasing c increases Dn, hence it decreases the amplitudes; this is physically understandable. 16. For the right side we have the Fourier sine series 4

asin t ⫺

1 1 sin 3t ⫹ sin 5t ⫺ ⫹ Á b 9 25

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with the coefficients bn ⫽ 4>(n 2p) if n ⫽ 1, 5, 9, Á , and bn ⫽ ⫺4>(n 2p) if n ⫽ 3, 7, 11, Á . Substitution of this series into the ODE gives y ⫽ A1 cos t ⫹ B1 sin t ⫹ A3 cos 3t ⫹ B3 sin 3t ⫹ Á with coefficients

An ⫽ ⫺ncbn>Dn,

Bn ⫽ (1 ⫺ n 2)bn>Dn,

Dn ⫽ (1 ⫺ n 2)2 ⫹ n 2c2.

The damping constant c appears in the cosine terms, causing a phase shift, which is zero if c ⫽ 0. Also, c increases Dn, hence it decreases the amplitudes, which is physically understandable. 18. The ODE in Probs. 17–19 is the same, except for the changing right sides, whose Fourier series we use term-by-term, as in the text. The solution of the ODE is of the general form ⴥ

I ⫽ A0 ⫹ a 1An cos nt ⫹ Bn sin nt) n⫽1

with coefficients obtained by substitution An ⫽ ⫺

n 2 ⫺ 10 an, Dn

Bn ⫽

10n an , Dn ⫽ (n 2 ⫺ 1022 ⫹ 100n 2, Dn

in particular, A0 ⫽ a0>10, the ODE being I s ⫹ 10I r ⫹ 10I ⫽ an cos nt. For the present problem we have the Fourier series 100 ⫹ 100p ⫺

800

acos t ⫹

1 1 cos 3t ⫹ cos 5t ⫹ Á b. 9 25

Hence A0 ⫽ 10 ⫹ 10p and all the other An and Bn with n even are zero. The formula for the an is ⫺800>(pn 2) where n is odd. Numerically evaluating the terms, we obtain the solution (the current in the RLC-circuit) I ⫽ 41.416 ⫺ 12.662 cos t ⫺ 14.069 sin t ⫺ 0.031 cos 3t ⫺ 0.942 sin 3t ⫹ 0.056 cos 5t ⫺ 0.187 sin 5t ⫹ Á . 20. Cn ⫽ 2A2n ⫹ B 2n ⫽ 4>(n 2p 1Dn), Dn ⫽ (n 2 ⫺ k)2 ⫹ n 2c2 with An and Bn obtained as solutions of (k ⫺ n 2) An ⫹

ncBn ⫽ 4>n 2p

⫺ncAn ⫹ (k ⫺ n 2)Bn ⫽ 0. SECTION 11.4. Approximation by Trigonometric Polynomials, page 495 Purpose. We show how to find “best” approximations of a given function by trigonometric polynomials of a given degree N. Important Concepts Trigonometric polynomial Square error, its minimum (6) Bessel’s inequality, Parseval’s identity Short Courses. This section can be omitted.

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Comment on Quality of Approximation This quality can be measured in many ways. Particularly important are (i) the absolute value of the maximum deviation over a given interval, and (ii) the mean square error considered here. See Ref. [GenRef7] in App. 1. Further Comments The ideas in this section play a basic role in more advanced applied and abstract courses. CAS Experiment 10 will give the student a feel for the size of the error of the present approximation and its size as a function of the number of terms considered, that is, for the rapidity of its decrease. For other types of interpolation and approximation and for numeric work, see Secs. 19.3, 19.4, and 20.5. SOLUTIONS OF PROBLEM SET 11.4, page 498 2. an ⫽ 0 since f (x) ⫽ x is odd. Calculation of the bn gives the approximating trigonometric polynomial F ⫽ 2 asin x ⫺

N⫹1

(⫺1) 1 sin 2x ⫹ Á ⫹ 2 N

sin Nxb .

From this, the minimum square error is obtained as shown in Example 1 of the text; note that the present function and that in Example 1 differ just by an additive constant p. 4. F ⫽

p2 3

⫺ 4 acos x ⫺

N⫹1

(⫺1) 1 1 cos 2x ⫹ cos 3x ⫺ ⫹ Á ⫹ 4 9 N2

cos Nxb ;

E* ⫽ 4.14, 1.00, 0.38, 0.18, 0.10 6. The function in Prob. 3 is continuous, the function in Prob. 5 is not; indeed, it is the derivative of the function in Prob. 3, and differentiation produces a factor n in each term. 8. The approximating trigonometric polynomial of minimum square error is F⫽

⫺

p 1#3

cos 2x ⫹

1 1 cos 4x ⫹ # cos 6x ⫹ Á b , 3 #5 5 7

From this we obtain E* ⫽ 0.5951, 0.0292, 0.0292, 0.0066, 0.0066, etc. These values are small; indeed, a graph shows that the four terms given are such that this partial sum of F* approximates f (x) rather accurately. 1

–␲

␲

Section 11.4. Problem 8

10. CAS Experiment. Factors are the continuity or discontinuity and the speed with which the coefficients go to zero, 1>n, 1>n 2.

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For f (x) given on ⫺p ⬍ x ⬍ p some data are as follows ( f, decrease of the coefficients, continuity or not, smallest N such that E* ⬍ 0.1). f ⫽ x, 1>n, discontinuous, N ⫽ 126 f ⫽ x 2, 1>n 2, continuous, N ⫽ 5 f ⫽ x 3, 1>n 2, discontinuous, E* ⫽ 6.105 for N ⫽ 200 f ⫽ x 4, 1>n, continuous, N ⫽ 40 f ⫽ x 6, 1>n, continuous. For N ⫽ 200 we still have E* ⫽ 0.1769. For f in Prob. 9 we have 1>n 2, continuity, N ⫽ 2. The functions f (x) ⫽ c

(x ⫹ 12 p)2k ⫺ (12 p)2k

⫺p ⬍ x ⬍ 0

⫺(⫺x ⫺ 12p)2k ⫹ (12p)2k

0⬍x⬍p

with k ⫽ 1, 2, 3, 4 have coefficients proportional to 1>n 2 and E* ⬍ 0.1 when N ⭌ 1 (k ⫽ 1), 5 (k ⫽ 2), 11 (k ⫽ 3), 21 (k ⫽ 4). These data indicate that the whole situation is more complex than one would at first assume. So the student may need your help and guidance. 14. The Fourier series is cos2 x ⫽ 12 ⫹ 12 cos 2x and Parseval’s identity (8) gives 1 1 1 ⫹ ⫽ 2 4 p

冮 cos x dx. 4

ⴚp

Section 11.5. Sturm–Liouville Problems. Orthogonal Functions, page 498 Purpose. Discussion of eigenvalue problems for ordinary second-order ODEs (1) under boundary conditions (2). Main Content, Important Concepts Sturm–Liouville equations, Sturm–Liouville problem Reality of eigenvalues Orthogonality of eigenfunctions Orthogonality of Legendre polynomials and Bessel functions Short Courses. Omit this section. Comment on Importance This theory owes its significance to two factors. On the one hand, boundary value problems involving practically important ODEs (Legendre’s, Bessel’s, etc.) can be cast into Sturm– Liouville form, so that here we have a general theory with several important particular cases. On the other hand, the theory gives important general results on the spectral theory of those problems.

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Comment on Existence of Eigenvalues This theory is difficult. Quite generally, in problems where we can have infinitely many eigenvalues, the existence problem becomes nontrivial, in contrast with matrix eigenvalue problems (Chap. 8), where existence is trivial, a consequence of the fact that a polynomial equation f (x) ⫽ 0 ( f not constant) has at least one solution and at most n numerically different ones (where n is the degree of the polynomial). Move of This Theory from Chap. 5 to Chap. 11 Sturm–Liouville theory is motivated to a large extent by the use of orthogonality in connection with Fourier series and by generalizations from y s ⫹ ly ⫽ 0 (in the separation of the vibrating string PDE, the wave PDE) to more general linear ODEs (Legendre and Bessel above all). The Sturm–Liouville material is now close to one of its main applications in PDEs (Chap. 12). More importantly, orthogonality seems more complicated to grasp than other theories and needs digestion time and more maturity than the average student probably has in Part A on ODEs. At least, this is my observation resulting from teaching these matters many times to engineers, physicists, and mathematicians, representing groups of various interests and maturity in applied mathematics. SOLUTIONS TO PROBLEM SET 11.5, page 503 2. If ym is a solution of (1), so is z m because (1) is linear and homogeneous; here, l ⫽ lm is the eigenvalue corresponding to ym. Also, multiplying (2) with y ⫽ ym by c, we see that z m also satisfies the boundary conditions. This proves the assertion. 4. a ⫽ ⫺p, b ⫽ p, c ⫽ p, k ⫽ 0. Problems 2 to 6 are most useful in applications; they concern situations that appear rather frequently. 6. Perform the differentiations in (1), divide by p, and compare; that is, py s ⫹ p r y r ⫹ (q ⫹ lr)y ⫽ 0,

pr q r y s ⫹ p y r ⫹ a p ⫹ l p b y ⫽ 0.

Hence f ⫽ p r >p, p ⫽ exp ( 兰 f dx), q>p ⫽ g, q ⫽ gp, r>p ⫽ h, r ⫺ hp. A reason for performing this transformation may be the discovery of the weight function needed for determining the orthogonality. We see that

冮

r (x) ⫽ h (x)p (x) ⫽ h (x) exp a f (x) dxb . This problem shows that a Sturm–Liouville equation is rather general; more precisely, it is equivalent to a general second-order homogeneous linear ODE whose coefficient of the unknown function y (x) contains a parameter l, the coefficient being of the form Q ⫹ lR. 8. l ⫽ (mp>L)2, m ⫽ 1, 2, Á ; ym ⫽ sin (mpx>L) 10. We need y ⫽ A cos kx ⫹ B sin kx y r ⫽ ⫺Ak sin kx ⫹ Bk cos kx. From the boundary conditions we obtain y (0) ⫽ A ⫽ y (1) ⫽ A cos k ⫹ B sin k y r (0) ⫽ Bk ⫽ y r (1) ⫽ ⫺Ak sin k ⫹ Bk cos k.

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Ordering gives (1 ⫺ cos k)A ⫺ (sin k)B ⫽ 0 (k sin k) A ⫹ k (1 ⫺ cos k) B ⫽ 0. By eliminating A and then requiring B ⫽ 0 (to have y ⫽ 0, an eigenfunction) or simply by noting that for this homogeneous system to have a nontrivial solution A, B, the determinant of its coefficients must be zero; that is, k (1 ⫺ cos k)2 ⫹ k sin2 k ⫽ k (2 ⫺ 2 cos k) ⫽ 0; hence cos k ⫽ 1, k ⫽ 2mp, so that the eigenvalues and eigenfunctions are lm ⫽ (2mp)2, y0 ⫽ 1,

m ⫽ 0, 1, Á ;

ym (x) ⫽ cos (2m p x), sin (2mpx),

m ⫽ 1, 2, Á .

12. A general solution y ⫽ ex (A cos kx ⫹ B sin kx), k ⫽ 1l, of this ODE with constant coefficients is obtained as usual. The Sturm–Liouville form of the ODE is obtained by using the formulas in Prob. 6, (eⴚ2xy r ) r ⫹ eⴚ2x (k 2 ⫹ 1)y ⫽ 0. From this and the boundary conditions we expect the eigenfunctions to be orthogonal on 0 ⬉ x ⬉ 1 with respect to the weight function eⴚ2x. Now, from that general solution and y (0) ⫽ A ⫽ 0, we see that we are left with y ⫽ ex sin kx. From the second boundary condition y (1) ⫽ 0 we now obtain y (1) ⫽ e sin k ⫽ 0,

k ⫽ mp,

m ⫽ 1, 2, Á .

Hence the eigenvalues and eigenfunctions are lm ⫽ (mp)2,

ym ⫽ ex sin mpx.

The orthogonality is as expected (because ex cancels). 14. Team Project. (a) We integrate over x from ⫺1 to 1, hence over u defined by x ⫽ cos u from p to 0. Using (1 ⫺ x 2)ⴚ1>2 dx ⫽ ⫺du, we thus obtain 1

冮 cos (m arc cos x) cos (n arc cos x)(1 ⫺ x )

2 ⴚ1>2

ⴚ1

冮

1 ⫽ cos mu cos nu du ⫽ (cos (m ⫹ n)u ⫹ cos (m ⫺ n)u) du, 2 0 0 which is zero for integer m ⫽ n. (b) Following the hint, we calculate 兰 eⴚxx kL n dx ⫽ 0 for k ⬍ n:

冮

⬁

ⴚx

1 x L n(x) dx ⫽ n! k

冮

⬁

d k n ⴚx x n (x e ) dx ⫽ ⫺ dx n! 0 k

⫽ Á ⫽ (⫺1)k

k n!

ⴥ

冮

⬁ kⴚ1

d n ⴚx nⴚ1 (x e ) dx dx

nⴚk

冮 dxd 0

nⴚ1

nⴚk (x

n ⴚx

e ) dx ⫽ 0.

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SECTION 11.6. Orthogonal Series. Generalized Fourier Series, page 504 Purpose. To show how families (sequences) of orthogonal functions, as they arise in eigenvalue problems and elsewhere, are used in series for representing other functions, and to show how orthogonality becomes crucial in simplifying the determination of the coefficients of such a series by integration. Main Content, Important Concepts Standard notation (ym, yn) Orthogonal expansion (1), eigenfunction expansion Fourier constants (2) Fourier–Legendre series (Example 1) Fourier–Bessel series (Example 2) Orthogonality of Bessel functions (Theorem 1) Mean square convergence Completeness, also called totality (Theorem 2) Comments on Text Formula (2) for the Fourier constants (the coefficients of an orthogonal series) generalizes (6) in Sec. 11.2 for the Fourier coefficients of a Fourier series. Note that for Fourier–Bessel series (9) with coefficients (10) you obtain infinitely many orthogonal families, each consisting of infinitely many Bessel functions. In many applications of these series (and other orthogonal series), it turns out that one needs only a relatively small number of terms for obtaining a reasonable accuracy. Completeness of orthogonal families of functions guarantees that the set of given functions to be developed is sufficiently large to be of practical (and theoretical!) interest. In practical work, numeric methods may be needed for obtaining values of Fourier constants. SOLUTIONS TO PROBLEM SET 11.6, page 509 2. 23 P2 (x) ⫹ 2P1 (x) ⫹ 43 P0 (x). This is probably most simply obtained by the method of undetermined coefficients, beginning with the highest power, x 2 and P2 (x). The point of these problems is to make the student aware that these developments look totally different from the usual expansions in terms of powers of x. 8 4. P0(x), P1(x), 13 P0(x) ⫹ 23 P2(x), 35 P1(x) ⫹ 25 P3(x), 15 P0(x) ⫹ 47 P2(x) ⫹ 35 P4(x) 4m 6. The series may contain all even powers, not just powers x . 8. m 0 ⫽ 5. The size of m 0, that is, the rapidity of convergence seems to depend on the variability of f (x). A discontinuous derivative (e.g., as for ƒ sin x ƒ occurring in connection with rectifiers) makes it virtually impossible to reach the goal. Let alone when f (x) itself is discontinuous. In the present case the series is f (x) ⫽ 0.95493P1(x) ⫺ 1.15824P3 (x) ⫹ 0.21929P5 (x) ⫺ ⫹ Á . Rounding seems to have considerable influence in Prob. 8–13. 10. f (x) ⫽ 0.7468P0 (x) ⫺ 0.4460P2 (x) ⫹ 0.0739P4 (x) ⫺ Á , m 0 ⫽ 4 12. f (x) ⫽ 0.6116P0 (x) ⫺ 0.7032P2 (x) ⫹ 0.0999P4 (x) ⫹ Á , m 0 ⫽ 4. Compare with Prob. 13!

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217 ⴥ

14. Team Project. (b) A Maclaurin series f (t) ⫽ a ant n has the coefficients an ⫽ n⫽0 f (n) (0)>n! We thus obtain f (n) (0) ⫽

2 2 d n txⴚt2>2 dn (e )` ⫽ ex >2 n (eⴚ(xⴚt) >2) ` . n dt dt t⫽0 t⫽0

If we set x ⫺ t ⫽ z, this becomes f (n) (0) ⫽ ex >2(⫺1)n 2

n 2 d n ⴚz 2>2 n x 2>2 d (e ) ` ⫽ (⫺1) e (eⴚx >2) ⫽ Hen (x). n n dz dx z⫽x

(d) We write eⴚx >2 ⫽ v, v(n) ⫽ d nv>dx n, etc., and use (21). By integrations by parts, for n ⬎ m, 2

冮

⬁

vHem Hen dx ⫽ (⫺1)n

ⴚ⬁

冮

⬁

Hemv(n) dx ⫽ (⫺1)nⴚ1

ⴚ⬁

冮 He r v m

(nⴚ1)

ⴚ⬁

⬁

⫽ (⫺1)nⴚ1m

冮 He

(nⴚ1) dx ⫽ mⴚ1v

ⴚ⬁ ⬁

⫽ (⫺1)nⴚmm!

冮 He v 0

(nⴚm)

dx ⫽ 0.

ⴚ⬁

(e) nHen ⫽ nxHenⴚ1 ⫺ nHenr ⴚ1 from (22) with n ⫺ 1 instead of n. In this equation, the first term on the right equals xHenr by (21). The last terms equals ⫺Hens, as follows by differentiation of (21).

SECTION 11.7. Fourier Integral, page 510 Purpose. Beginning in this section, we show how ideas from Fourier series can be extended to nonperiodic functions defined on the real line, leading to integrals instead of series. Main Content, Important Concepts Fourier integral (5) Existence Theorem 1 Fourier cosine integral, Fourier sine integral, (10)–(13) Application to integration Short Courses. This section can be omitted. Comments on Text and Problems The simplest example on Fourier series can also serve here as an introduction (Example 1 and Fig. 280) to motivate the present extension to Fourier integrals as well as the result in Theorem 1, which we must leave without proof (a reference is given in the text); so the situation is somewhat similar to that on Fourier series near the beginning of the chapter.

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It is interesting that we shall now be able to prove and understand the Gibbs phenomenon in terms of the sine integral and Dirichlet’s discontinuous factor in Example 2. See, in particular, Fig. 283. Fourier cosine and Fourier sine integral in (10) and (11) are analogs of Fourier cosine and Fourier sine series. Example 3 shows a basic application. The evaluation of integrals by the present method is shown in Probs. 1–6 for the Fourier integral itself and in Probs. 7–12 and 16–20 for Fourier cosine and sine integrals, respectively. CAS Experiments 13 and 15 should help the student in gaining additional insight beyond the present formalism. Project 14 is a first step into transform theory similar to that of the Laplace transform in Chap. 6, the latter being of much greater importance to the engineer. SOLUTIONS TO PROBLEM SET 11.7, page 517 2. Use (11) and f (x) ⫽ (p>2) sin x

(0 ⬉ x ⬉ p) to get, with the help of (11) in App. 3.1, p

B (w) ⫽

冮 sin v sin wv dv ⫽ 1sin⫺pww . 2

4. Use (10). Also use (11) in App. A3.1. Take f (x) ⫽ 12 p cos x. Then A (w) ⫽

冮

p>2

p ⴚ p> 2 2

6. Take f (x) ⫽ peⴚx cos x give

cos v cos wv dv ⫽ 2

1 ⫺ w2

(x ⬎ 0). Then (11) in this section and (11) in App. A3.1 ⬁

冮 pe cos v sin wv dv

2 B (w) ⫽ p

ⴚv

⬁

⫽

cos 12 pw

⬁

冮 e sin (w ⫹ 1)v dv ⫹ 冮 e sin (w ⫺ 1)v dv. ⴚv

ⴚv

Integrate by parts, obtaining B (w) ⫽

w⫹1

w⫺1 2w 3 ⫹ ⫽ . 1 ⫹ (w ⫹ 1)2 1 ⫹ (w ⫺ 1)2 w4 ⫹ 4

Now use the first formula in (11) to obtain the result. 8. A ⫽

冮 v cos wv dv ⫽ p2w asin w ⫺ w2 sin w ⫹ w2 cos wb, so that the answer is 2

⬁

2 2 cos xw b sin w ⫹ cos w d dw. c a1 ⫺ 冮 p w w w 2

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Although many students will do the actual integration by their CAS, problems of the present type have the merit of illustrating the ideas of integral representations and transforms, a rather deep and versatile creation, and the techniques involved, such as the proper choice of integration variables and integration limits. Moreover, graphics will help in understanding the transformation process and its properties, for instance, with the help of Prob. 18 or similar experiments. 10.

ⴥ

sin aw ⫺ aw cos aw cos xw dw 冮 p w 4

12. A ⫽

冮 e cos wv dv ⫽ p a 2

ⴚv

1 ⫺ eⴚa (cos wa ⫺ w sin wa) 1 ⫹ w2

b, so that the integral

representation is 2

冮

ⴥ

1 ⫺ eⴚa (cos wa ⫺ w sin wa) 1 ⫹ w2

p 0

cos xw dw.

14. Project. (a) Formula (a1): Setting wa ⫽ p, we have from (10) f (ax) ⫽

冮

⬁

A (w) cos axw dw ⫽

冮 A a a b cos xp a . p

If we again write w instead of p, we obtain (a1). Formula (a2): From (11) with f (v) replaced by vf (v) we have B*(w) ⫽

ⴥ

dA 冮 vf (v) sin wv dv ⫽ ⫺dw

p 0

where the last equality follows from (10). Formula (a3) follows by differentiating (10) twice with respect to w, ⬁

d 2A 2 f * (v) cos wv dv, 2 ⫽ ⫺ p 0 dw

冮

f * (v) ⫽ v2f (v).

(b) In Prob. 7 we have 2 A ⫽ p wⴚ1 sin w. Hence by differentiating twice we obtain 2 As ⫽ p (2wⴚ3 sin w ⫺ 2 wⴚ2 cos w ⫺ w ⴚ1 sin w). By (a3) we now get the result, as before, x 2f (x) ⫽

⬁

冮 a⫺ w2 ⫹ w1 b sin w ⫹ w2 cos w d cos xw dw.

p 0 c 2

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dA 2 cos w sin w ⫽⫺ a ⫺ b. dw p w w2

This agrees with the result obtained by using (11). Note well that here we are dealing with a relation between the two Fourier transforms under consideration. (d) The derivation of the following formulas is similar to that of (a1)–(a3).

冮

⬁

1 w f (bx) ⫽ B a b sin xw dw b 0 b

(d1)

(b ⬎ 0)

⬁

xf (x) ⫽

(d2)

冮 C*(w) cos xw dw,

C* (w) ⫽

x f (x) ⫽ 2

(d3)

冮

dB , dw

⬁

D* (w) ⫽ ⫺

D* (w) sin xw dw,

B as in (11)

d 2B dw 2

16. From (11) we obtain B (w) ⫽

2 v sin wv dv ⫽ (sin aw ⫺ aw cos aw) 冮 p pw 2

so that the answer is f (x) ⫽

ⴥ

冮 sin aw ⫺waw cos aw sin xw dw. 2

p 0

18. B (w) ⫽

2 w(1 ⫹ cos pw) gives the integral representation cos v sin wv dv ⫽ 冮 p p w ⫺1 2

⬁

pw) sin xw dw. 冮 w(1w⫹ cos ⫺1

p 0

Note that the Fourier sine series of the odd periodic extension of f (x) of period 2p has the Fourier coefficients bn ⫽ (2> p)n(1 ⫹ cos np)>(n 2 ⫺ 1). Compare this with B (w). 20.

冮

⬁

p 0

w ⫺ (w cos w ⫹ sin w)>e sin xw dw 1 ⫹ w2

SECTION 11.8. Fourier Cosine and Sine Transforms, page 518 Purpose. Fourier cosine and sine transforms are obtained immediately from Fourier cosine and sine integrals, respectively, and we investigate some of their properties. Content Fourier cosine and sine transforms Transforms of derivatives (8), (9)

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Comment on Purpose of Transforms Just as the Laplace transform (Chap. 6), these transforms are designed for solving differential equations. We show this for PDEs in Sec. 12.7. Short Courses. This section can be omitted.

SOLUTIONS TO PROBLEM SET 11.8, page 522 2. f (x) ⫽

2 Bp

⬁

1 if 0 ⬍ x ⬍ 1

冮 f (w) cos wx dw ⫽ e ⫺1 if 1⬍ x ⬍ 2. c

If you got your answer by a CAS (e.g., by Maple) in a somewhat unusual form, plot it to see that it is correct. 4. This is standard integral of calculus. If you want to do it by your CAS, you may have trouble for general a (⬎0); however, you should still be able to see what the limit is and your CAS should be able to do evaluation for any fixed number a. For instance, this is the situation for Maple. 6. We have f s(x) ⫽ 2 ⫽ g(x) if 0 ⬍ x ⬍ 1, By integration, gc (w) ⫽ 2

f s(x) ⫽ 0 if x ⬎ 1. 2 sin w . Bp w

Hence (5a) with f r (0) ⫽ 0 would give ⫺2

2 sin w , B p w3

just one of the three terms shown in the answer to Prob. 5. This should show the student the importance of the continuity assumptions in the present and similar cases. 8. The defining integral (1a) has no limit, ⬁

冮 k cos wx dx ⫽ k lim sinwwx x :⬁

(w fixed!).

Similarly for (2a). 10. Use f s(x) ⫽ a 2f (x) to obtain from (5b) fs( f s(x)) ⫽ a 2 fs( f (x)) ⫽ ⫺w 2 fs ( f (x)) ⫹

2 w, Bp

hence by collecting terms (a 2 ⫹ w 2)fs( f (x)) ⫽

2 w Bp

and so on. 12. fs(xeⴚx >2) ⫽ ⫺fs((eⴚx >2) r ) ⫽ wfc(eⴚx >2) ⫽ weⴚw >2 from formula 4 in Table I (see Sec. 11.10 of text). 2

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14. Formula 4 in Table II (see Sec. 11.10 of text) with a ⫽ 12 gives fs(x ⴚ1>2) ⫽

1 p 1 1 w ⴚ1>2 ⌫ a b sin ⫽ w ⴚ1>2 ⌫ a b . p 4 Bp 2 2 1 2

On the other hand, by formula 2 in Table II, fs(x ⴚ1>2) ⫽ w ⴚ1>2. Comparison proves ⌫(12 ) ⫽ 1p. SECTION 11.9 Fourier Transform. Discrete and Fast Fourier Transforms, page 522 Purpose. Derivation of the Fourier transform from the complex form of the Fourier integral; explanation of its physical meaning and its basic propertites. Main Content, Important Concepts Complex Fourier integral (4) Fourier transform (6), its inverse (7) Spectral representation, spectral density Transforms of derivatives (9), (10) Convolution f * g Comments on Content The complex Fourier integral is relatively easily obtained from the real Fourier integral in Sec. 11.7, and the definition of the Fourier transform is then immediate. Note that convolution f * g differs from that in Chapter 6, and so does the formula (12) in the convolution theorem (we now have a factor 12p). Short Courses. This section can be omitted. SOLUTIONS TO PROBLEM SET 11.9, page 533 2. This involves a transformation of exponential functions into a sine, as mentioned in Prob. 1. Integration of the defining integral gives 1

1 ⫺i e(2ⴚw)ix dx ⫽ (e(2ⴚw)i ⫺ e ⴚ(2ⴚw)i) 12p ⴚ1 12p (2 ⫺ w)

冮

⫽ ⫽

⫺i 2i sin (2 ⫺ w) 12p (2 ⫺ w) 2 sin (w ⫺ 2) . w⫺2

4. By integration of the defining integral we obtain

冮

1 1 # e(kⴚiw)x e(kⴚiw)x dx ⫽ ` 12p ⴚ ⬁ 12p k ⫺ iw ⴚ ⬁ ⫽

1 1 # . 12p k ⫺ iw

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6. 12> p(1 ⫹ w 2), as obtained by integration and simplification, namely

冮

⬁

1 ex(1ⴚiw) dx ⫹ ex(ⴚ1ⴚiw) dx d c 12p ⴚ⬁ 0 ⫽

1 1 1 a ⫹ b 12p 1 ⫺ iw 1 ⫹ iw

⫽

1 2 2 1 ⫽ . B p 1 ⫹ w2 12p 1 ⫹ w 2

8. By integration by parts we obtain 1 12p

冮 xe

ⴚxⴚiwx

dx ⫽

ⴚ1

ⴚ(1⫹iw)x

0 0 1 xe 1 ⴚ(1⫹iw)x a ` ⫹ e dxb 12p ⫺(1 ⫹ iw) ⴚ1 1 ⫹ iw ⴚ1

冮

⫽

1 e1⫹iw e⫺(1⫹iw)x a⫺(⫺1) ⫹ ` b 12p ⫺(1 ⫹ iw) ⫺(1 ⫹ iw)(1 ⫹ iw) ⴚ1

⫽

1 ⫺e1⫹iw 1 ⫺ e1⫹iw a ⫺ b (1 ⫹ iw)2 12p 1 ⫹ iw

⫽ ⫽

1 12p(⫺w ⫹ i)2 1 12p(⫺w ⫹ i)2

(1 ⫹ e1⫹iw(⫺1 ⫺ i(⫺w ⫹ i))) (1 ⫹ iwe1⫹iw).

Problems 2 to 11 should help the student get a feel for integrating complex exponential functions and for their transformation into cosine and sine, as needed in this context. Here, it is taken for granted that complex exponential functions can be handled in the same fashion as real ones, which will be justified in Part D on complex analysis. The problems show that the technicalities are rather formidable for someone who faces these exponential functions for the first time. This is so for relatively simple f (x), and since a CAS will give all the results without difficulty, it would make little sense to deal with more complicated f (x), which would involve increased technical difficulties but no new ideas. Furthermore, the present problem illustrates the following situation. Maple gave a result quite different and much more complicated than that obtained by the usual formal integration. In such a case, consider the difference between the result obtained and that expected and try to see whether you can reduce this difference to zero, thereby confirming what you had expected. Problem 9 is of a similar type, that is, the difference between the result expected and that obtained can be reduced to zero. 10. By integration by parts, 1

1 xeⴚiwx 1 1 xe ⴚiwxdx ⫽ a ` ⫺ eⴚiwx dx b 12p ⴚ1 12p ⫺iw ⴚ1 ⫺iw ⴚ1

冮

eiw 1 1 eⴚiw ⫹ ⫺ a eⴚiwx ` b 12p ⫺iw (⫺iw)2 ⫺iw ⴚ1 1

⫽

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⫽ ⫽ ⫽

1 2 cos w 1 a ⫹ 2 (e ⴚiw ⫺ eiw )b ⫺iw 22p w 2 Bp

i cos w i sin w ⫺ b w w2

2 # i (w cos w ⫺ sin w). B p w2

12. Let f (x) ⫽ xe ⴚx (x ⬎ 0) and g(x) ⫽ e ⴚx (x ⬎ 0). Then f r ⫽ g ⫺ f and by (9) iw f( f ) ⫽ f( f r ) ⫽ f( g) ⫺ f( f ). From this and formula 5 in Table III (see Sec. 11.10 of text) with a ⫽ 1, (1 ⫹ iw) f( f ) ⫽ f( g) ⫽

1 . 12p (1 ⫹ iw)

Now divide by 1 ⫹ iw. 14. Formula 8 with ⫺b instead of b and b instead of c is i eib(aⴚw) ⫺ eⴚib(aⴚw) . a⫺w 12p Multiply numerator and denominator by ⫺1, use sin (⫺a) ⫽ ⫺sin a and the formula for the sine Prob. 1. This gives i 2i sin b(a ⫺ w) 2 sin b(w ⫺ a) i eib(aⴚw) ⫺ eⴚib(aⴚw) ⫽ ⫽ . Bp w⫺a w⫺a w⫺a 12p 12p 16. Team Project. (a) Use t ⫽ x ⫺ a as a new variable of integration. (b) Use c ⫽ 3b. Then (a) gives e2ibw f( f (x)) ⫽

eibw ⫺ e⫺ibw 2i sin bw 2 # sin bw . ⫽ ⫽ Bp w iw12p iw12p

(c) Replace w with w ⫺ a. This gives a new factor eiax. (d) We see that fˆ(w) in formula 7 is obtained from fˆ(w) in formula 1 by replacing w with w ⫺ a. Hence by (c), f (x) in formula 1 times eiax should given f (x) in formula 7, which is true. Similarly for formulas 2 and 8.

20.

1 1 E 4 1

⫺1

⫺i

⫺1

⫺i

⫺1

22. c 1

1 2

⫺12

f1 ⫹ f2 f1 ⫺ f2

14i

⫺4 ⫹ 8i

⫺1

⫺6

⫺4 ⫺ 8i

U E

U⫽E U 9

d⫽c d f1 f2

24. The powers of w are 1

1⫺i 12

⫺i

⫺1 ⫺ i 12

⫺1

⫺1 ⫹ i 12

1⫹i . 12

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This is the second row of F. The third row is 1

⫺i

⫺1

⫺i

⫺1 i,

and so on. SOLUTIONS TO CHAPTER 11 REVIEW QUESTIONS AND PROBLEMS, page 537 12. f (x) ⫺ 1 is an odd function. 14. The even function feven(x) ⫽ b

⫺x>2 x>2

if ⫺1 ⬍ x ⬍ 0 if

0⬍x⬍1

and the odd function fodd(x) ⫽ x>2

(⫺1 ⬍ x ⬍ 1),

respectively. 16.

p 2

⫺

4 1 1 acos x ⫹ cos 3x ⫹ cos 5x ⫹ Á b 11 9 25

18. The function and series in Example 1 of Sec. 11.1. 20. p2 ⫺ 12 (cos x ⫺ 14 cos 2x ⫹ 19 cos 3x ⫺⫹ Á ) 22. y ⫽ C1 cos vt ⫹ C2 sin vt ⫹ ⫹

1 25

ⴢ

cos 5t

p2 ⫺ 25

p 2v

2 ⫺

cos t 1 cos 3t ⫹ ⴢ 2 2 p p ⫺1 9 p ⫺9 4

⫹Áb

24. By a factor k 2, as can be seen directly from the formula in Sec. 11.4. 26. 22> p(w ⫹ sin w ⫺ 2w cos w)>w 2 28. k[(ibw ⫹ 1)e ⴚibw ⫺ (iaw ⫹ 1)e ⴚiaw]>(w 2 12p) 30. 1>((2 ⫹ iw)22p)

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CHAPTER 12

Partial Differential Equations (PDEs)

SECTION 12.1 Basic Concepts of PDEs, page 540 Purpose. To familiarize the student with the following: Concept of solution, verification of solutions Superposition principle for homogeneous linear PDEs PDEs solvable by methods for ODEs SOLUTIONS TO PROBLEM SET 12.1, page 542 2. c ⫽ 1, u tt ⫽ 2 ⫽ u xx. Problems 2–13 should give the student a first impression of what kind of solutions to expect, and of the great variety of solutions compared with those of ODEs. It should be emphasized that although the wave and the heat equations look so similar, their solutions are basically different. It could be mentioned that the boundary and initial conditions are basically different, too. Of course, this will be seen in great detail in later sections, so one should perhaps be cautious not to overload students with such details before they have seen a problem being solved. 4. Any c and k 6. u t ⫽ ⫺u ⫽ u xx, c ⫽ 1 8. u t ⫽ ⫺9u ⫽ ⫺c2v2u, c ⫽ 3>v 14. Team Project. (a) Denoting derivatives with respect to the entire argument x ⫹ ct and x ⫺ ct, respectively, by a prime, we obtain by differentiating twice u xx ⫽ v s ⫹ w s ,

u tt ⫽ v sc2 ⫹ w sc2

and from this the desired result. (c) The student should realize that u ⫽ 1> 2x 2 ⫹ y 2 is not a solution of Laplace’s equation in two variables, but satisfies the remarkable Poisson equation shown under (b). 16. Integrate twice with respect to y, u y ⫽ c1(x),

u ⫽ c1(x)y ⫹ c2 (x)

With the “constants” of integration c1(x) and c2(x) arbitrary. Problems 16–25 will help the student get used to the notations in this chapter; in particular, y will now occur as an independent variable. Second-order PDEs in this set will also help review the solution methods in Chap. 2, which will play a role in separating variables. 18. u ⫽ c1(x)e0.4y ⫹ c2(x)eⴚ0.4y 20. The characteristic equation is 2l2 ⫹ 9l ⫹ 4 ⫽ 2(l ⫹ 4)(l ⫹ 12 ) ⫽ 0. Hence a general solution of the homogeneous PDE is u h(x, y) ⫽ c1( y)eⴚ4x ⫹ c2( y)eⴚ0.5x. 226

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A particular solution u p of the nonhomogeneous PDE is obtained by the method of undetermined coefficients, u p(x, y) ⫽ 3 cos x ⫺ sin x.

22. Set u x ⫽ v to get vy ⫽ v, vy>v ⫽ 1, v ⫽ c(x)ey, and u⫽

冮 v dx ⫽ c (x)e ⫹ c ( y). 1

24. By the given PDE and the chain rule, yz x ⫺ xz y ⫽ y(z rrx ⫹ z uux) ⫺ x(z rry ⫹ z uuy) ⫽ 0.

(A)

Differentiate r ⫽ x 2 ⫹ y 2 by parts and divide by 2r, 2

(B)

rx ⫽ x>r,

ry ⫽ y>r.

Now z r has in (A) the coefficients (use (B)) yrx ⫺ xry ⫽ yx>r ⫺ xy>r ⫽ 0 so that (A) reduces to z u ⫽ 0. That is, z(r, u) depends only on r, not on the angle u, as for a sphere, a circular cylinder, and so on. SECTION 12.3. Solution by Separating Variables. Use of Fourier Series, page 545 Purpose. This first section in which we solve a “big” problem has several purposes: 1. To familiarize the student with the wave equation and with the typical initial and boundary conditions that physically meaningful solutions must satisfy. 2. To explain and apply the important method of separation of variables, by which the PDE is reduced to two ODEs. 3. To show how Fourier series help to get the final answer, thus seeing the reward of our great and long effort in Chap. 11. 4. To discuss the eigenfunctions of the problem, the basic building blocks of the solution, which lead to a deeper understanding of the whole problem. Steps of Solution 1. Setting u ⫽ F(x)G(t) gives two ODEs for F and G. 2. The boundary conditions lead to sine and cosine solutions of the ODEs. 3. A series of those solutions with coefficients determined from the Fourier series of the initial conditions gives the final answer. SOLUTIONS TO PROBLEM SET 12.3, page 551 2. If Assumption 3 is violated, the string can move in various planes. For large displacements and> or angles the PDE would no longer be linear and have constant coefficients. Lack of elasticity entails loss of mechanical energy by conversion into heat (damping). If homogeneity were dropped, it is hard to see what would happen; one would first have to be specific and state in what way homogeneity is changed and perhaps support theoretical results by physical experiments. 6. k(cos pt sin px ⫺ 12 cos 2pt sin 2px)

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a4 cos pt sin px ⫺

3 4 cos 2pt sin 2px ⫹ cos 3pt sin 3px 2 27

3 cos 4pt sin 4px ⫹⫺ Á b 16 8 1 1 1 10. 2 a cos 2pt sin 2px ⫺ cos 6pt sin 6px ⫹ cos 10pt sin 10px ⫺ Á b 4 36 100 p ⫺

There are more graphically posed problems than in previous editions, so that CASusing students will have to make at least some additional effort in solving these problems. 12.

p2 ⫺

acos pt sin px ⫹

1 1 cos 3pt sin 3px ⫺ cos 5pt sin 5px 9 25

1 cos 7pt sin 7px ⫹ Á b 49 ⬁

14. u ⫽ a Bn* sin nt sin nx, n⫽1

Bn* ⫽

0.04

pn 3

sin

np 2

16. Fn ⫽ sin (npx>L), Gn ⫽ an cos (cn 2p2t>L2) 18. For the string the frequency of the nth mode is proportional to n, whereas for the beam it is proportional to n 2. 20. F(0) ⫽ A ⫹ C ⫽ 0, C ⫽ ⫺A, F r(0) ⫽ b(B ⫹ D) ⫽ 0, D ⫽ ⫺B. Then F(x) ⫽ A(cos bx ⫺ cosh bx) ⫹ B(sin bx ⫺ sinh bx) F s (L) ⫽ b2[⫺A(cos bL ⫹ cosh bL) ⫺ B(sin bL ⫹ sinh bL)] ⫽ 0 F t (L) ⫽ b3[A(sin bL ⫺ sinh bL) ⫺ B(cos bL ⫹ cosh bL)] ⫽ 0. The determinant (cos bL ⫹ cosh bL)2 ⫹ sin2 bL ⫺ sinh2 bL of this system in the unknowns A and B must be zero, and from this the result follows. From (23) we have cos bL ⫽

⫺1 ⬇0 cosh bL

because cosh bL is very large. This gives the approximate solutions bL ⬇ 12 p, 32 p, 52 p, Á (more exactly, 1.875, 4.694, 7.855, Á ). SECTION 12.4. D’Alembert’s Solution of the Wave Equation. Characteristics, page 553 Purpose. To show a simpler method of solving the wave equation, which, unfortunately, is not so universal as separation of variables. Comment on Order of Sections Section 12.12 on the solution of the wave equation by the Laplace transform may be studied directly after this section. We have placed that material at the end of this chapter because some students may not have studied Chap. 6 on the Laplace transform, which is not a prerequisite for Chap. 12. Comment on Footnote 1 D’Alembert’s Traité de dynamique appeared in 1743 and his solution of the vibrating string problem in 1747; the latter makes him, together with Daniel Bernoulli

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(1700–1782), the founder of the theory of PDEs. In 1754 d’Alembert became Secretary of the French Academy of Science and as such the most influential man of science in France.

SOLUTIONS TO PROBLEM SET 12.4, page 556 2. u(0, t) ⫽ 12[ f (ct) ⫹ f (⫺ct)] ⫽ 0, f (⫺ct) ⫽ ⫺f (ct), so that f is odd. Also u(L, t) ⫽ 12[ f (ct ⫹ L) ⫹ f (⫺ct ⫹ L)] ⫽ 0 hence f (ct ⫹ L) ⫽ ⫺f (⫺ct ⫹ L) ⫽ f (ct ⫺ L). This proves the periodicity. 4. (1>(2p))(np>2) # 80.83 ⫽ 20.21n 10. Hyperbolic, wave equation. Characteristic equation y r 2 ⫺ 16 ⫽ ( y r ⫹ 4)( y r ⫺ 4) ⫽ 0. New variables are v ⫽ ␾ ⫽ y ⫹ 4x,

w ⫽ ° ⫽ y ⫺ 4x.

By the chain rule, u x ⫽ 4u v ⫺ 4u w u xx ⫽ 16u vv ⫺ 16u vw ⫺ 16u wv ⫹ 16u ww and ⫺16u yy ⫽ ⫺16u vv ⫺ 16u vw ⫺ 16u wv ⫺ 16u ww. Assuming u vw ⫽ u wv, as usual, we have u vw ⫽ 0, solvable by two integrations, as shown in the text. 12. Parabolic. Characteristic equation y r 2 ⫹ 2y r ⫹ 1 ⫽ ( y r ⫹ 1)2 ⫽ 0. New variables v ⫽ £ ⫽ x, w ⫽ ° ⫽ x ⫹ y. By the chain rule, ux ⫽ uv ⫹ uw u xx ⫽ u vv ⫹ 2u vw ⫹ u ww u xy ⫽ u vw ⫹ u ww u yy ⫽ u ww. Substitution of this into the PDE gives the expected normal form u vv ⫽ 0. 14. Hyperbolic. New variables x ⫽ v and xy ⫽ w. The latter is obtained from ⫺xy r ⫺ y ⫽ 0,

yr 1 y ⫽ ⫺x,

ln ƒ y ƒ ⫽ ⫺ln ƒ x ƒ ⫹ c.

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By the chain rule we obtain, in these new variables from the given PDE by cancellation of ⫺yu yy against a term in xu xy and division of the remaining PDE by x, the PDE u w ⫹ xu vw ⫽ 0.

(The normal form is u vw ⫽ ⫺u w>x ⫽ ⫺u w>v.) We set u w ⫽ z and obtain c(w) z⫽ v .

1 z v ⫽ ⫺ v z,

By integration with respect to w we obtain the solution 1 1 u ⫽ v f1(w) ⫹ f2(v) ⫽ x f1(xy) ⫹ f2(x). Note that the solution of the next problem (Problem 15) is obtained by interchanging x and y in the present problem. 16. Elliptic. The characteristic equation is y r 2 ⫺ 2y r ⫹ 10 ⫽ 3y r ⫺ (1 ⫺ 3i)43y r ⫺ (1 ⫹ 3i)4 ⫽ 0. Complex solutions are £ ⫽ y ⫺ (1 ⫺ 3i) x ⫽ const,

° ⫽ y ⫺ (1 ⫹ 3i) x ⫽ const.

This gives the solutions of the PDE: u ⫽ f1( y ⫺ (1 ⫺ 3i)x) ⫹ f2( y ⫺ (1 ⫹ 3i)x). Since the PDE is linear and homogeneous, real solutions are the real and the imaginary parts of u. 18. Parabolic. Characteristic equation y r 2 ⫹ 6y r ⫹ 9 ⫽ ( y r ⫹ 3)2. New variables v ⫽ £ ⫽ x, w ⫽ ° ⫽ y ⫽ 3x. By the chain rule, u x ⫽ u v ⫹ 3u w u xx ⫽ u vv ⫹ 6u vw ⫹ 9u ww u xy ⫽ u vw ⫹ 3u ww u yy ⫽ u ww. Substitution into the PDE gives the expected normal form u vv ⫹ 6u vw ⫹ 9u ww ⫺6u vw ⫺ 18u ww ⫹ 9u ww ⫽ u vv ⫽ 0. Solution u ⫽ f1(v) ⫹ f2(w) ⫽ f1(x) ⫹ f2( y ⫹ 3x) where f1 and f2 are any twice differentiable functions of the respective variables. 20. The Tricomi equation is elliptic in the upper half-plane and hyperbolic in the lower, because of the coefficient y. u ⫽ F(x)G( y) gives yF s G ⫽ ⫺FG s , and k ⫽ 1 gives Airy’s equation.

Fs Gs ⫽⫺ ⫽ ⫺k F yG

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SECTION 12.6. Heat Equation: Solution by Fourier Series. Steady Two-Dimensional Heat Problems. Dirichlet Problem, page 558 Purpose. This section has two purposes: 1. To solve a typical heat problem by steps similar to those for the wave equation, pointing to the two main differences: only one initial condition (instead of two) and u t (instead of u tt), resulting in exponential functions in t (instead of cosine and sine in the wave equation). 2. Solution of Laplace’s equation (which can be interpreted as a time-independent heat equation in two dimensions). Comments on Content Additional points to emphasize are More rapid decay with increasing n, Difference in time evolution in Figs. 295 and 291, Zero can be an eigenvalue (see Example 4), Three standard types of boundary value problems, Analogy of electrostatic and (steady-state) heat problems. Problem Set 12.6 includes additional heat problems and types of boundary conditions.

SOLUTIONS TO PROBLEM SET 12.6, page 566 2. (c2p2>L2)20 ⫽ ln 2, c2 ⫽ 0.0035L2 6. u ⫽ ⫹

2 asin 0.1

pxeⴚ0.01752p t ⫺ 2

2 1 sin 0.3pxeⴚ0.01752(3p) t 9

2 1 sin 0.5pxeⴚ0.01752(5p) t ⫺ ⫹ Á b 25

8. u I ⫽ U1 ⫹ (U2 ⫺ U1)x>L. This is the solution of (1) with 0u>0t ⫽ 0 satisfying the boundary conditions. 10. u(x, 0) ⫽ f (x) ⫽ 100, U1 ⫽ 100, U2 ⫽ 0, u I ⫽ 100 ⫺ 10x. Hence 2 Bn ⫽ 10

冮 [100 ⫺ (100 ⫺ 10x)] sin n10px dx 0

2 ⫽ 10

冮 10x sin n10px dx 0

200 ⫽ ⫺ np cos np ⫽

(⫺1)n⫹1 # 63.66. n

This gives the solution ⬁

u(x, t) ⫽ 100 ⫺ 10x ⫹ 63.66 a n⫽1

(⫺1)n⫹1 npx ⴚ1.752(np>10)2t sin e . n 10

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For x ⫽ 5 this becomes

u(5, t) ⫽ 50 ⫹ 63.663eⴚ1.729t ⫺ 13 eⴚ1.556t ⫹ 15 eⴚ4.323t ⫺ ⫹ Á 4.

Obviously, the sum of the first few terms is a good approximation of the true value at any t ⬎ 0. We find: t u(5, t) 12. u ⫽

p 2

⫺

acos x eⴚt ⫹

1 99

2 94

3 88

10 61

50 . 50

1 1 cos 3x eⴚ9t ⫹ cos 5x eⴚ25t ⫹ Á b 9 25

14. u(x, t) ⫽ cos 2x eⴚ4t 16. c2vxx ⫽ vt, v(0, t) ⫽ 0, v(p, t) ⫽ 0, v(x, 0) ⫽ f (x) ⫹ Hx(x ⫺ p)>(2c2), so that, as in (9) and (10), u(x, t) ⫽ ⫺

Hx(x ⫺ p) 2c

⬁

⫹ a Bn sin nx eⴚc n t 2

n⫽1

where 2

Bn ⫽

冮 af (x) ⫹ Hx(x2c⫺ p) b sin nx dx. 2

18. u ⫽

⬁ (2n ⫹ 1)py 1 (2n ⫹ 1)px sinh sin a p n⫽1 (2n ⫹ 1) sinh 2(2n ⫹ 1)p 20 20

440

20. CAS Project. (a) u ⫽ 80 (sin px sinh py)>sinh 2p (b) u y(x, 0, t) ⫽ 0, u y(x, 2, t) ⫽ 0, u ⫽ 0 22. u ⫽ u I ⫹ u II, where 1 (2n ⫺ 1)px sinh 3(2n ⫺ 1)py>244 sin a p n⫽1 2n ⫺ 1 24 sinh (2n ⫺ 1)p

uI ⫽

4U1

u II ⫽

4U0

⬁

a 2n ⫺ 1 p n⫽1

sin

(2n ⫺ 1)px sinh 3(2n ⫺ 1)p(1 ⫺ y>24)4 . 24 sinh (2n ⫺ 1)p

24. u ⫽ F(x)G( y), F ⫽ A cos px ⫹ B sin px, u x(0, y) ⫽ F r (0)G( y) ⫽ 0, B ⫽ 0, G ⫽ C cosh py ⫹ D sinh py, u y(x, b) ⫽ F(x)G r (b) ⫽ 0, C ⫽ cosh pb, D ⫽ ⫺sinh pb, G ⫽ cosh ( pb ⫺ py). For u ⫽ cos px cosh p(b ⫺ y) we get u x(a, y) ⫹ hu(a, y) ⫽ (⫺p sin pa ⫹ h cosh pa) cosh p(b ⫺ y) ⫽ 0. Hence p must satisfy tan ap ⫽ h>p, which has infinitely many positive real solutions p ⫽ g1, g2, Á , as you can illustrate by a simple sketch. Answer: u n ⫽ cos gn x cosh gn(b ⫺ y), where g ⫽ gn satisfies g tan ga ⫽ h. To determine coefficients of series of u n’s from a boundary condition at the lower side is difficult because that would not be a Fourier series, the gn’s being only approximately regularly spaced. See [C3], pp. 114–119, 167.

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SECTION 12.7. Heat Equation: Modeling Very Long Bars. Solution by Fourier Integrals and Transforms, page 568 Purpose. Whereas we solved the problem of a finite bar in the last section by using Fourier series, we show that for an infinite bar (practically, a long insulated wire) we can use the Fourier integral for the same purpose. Figure 299 shows the time evolution for a “rectangular” initial temperature (100°C between x ⫽ ⫺1 and ⫹1, zero elsewhere), giving bell-shaped curves as for the density of the normal distribution. We also show typical applications of the Fourier transform and the Fourier sine transform to the heat equation. Short Courses. This section can be omitted. SOLUTIONS TO PROBLEM SET 12.7, page 574 2. From (8) and (6) we obtain 1 A⫽p

冮 cos pv dv ⫽ 2 sinp ap ,

B⫽0

ⴚa

and thus

2 u⫽p

ⴥ

冮 sinpap cos px e

ⴚc2p2t

dp.

4. A ⫽

⬁

2 e cos pv dv ⫽ , B ⫽ 0. Hence 冮 p( p ⫹ 1) p 1

ƒvƒ

ⴚ⬁

u⫽

ⴥ

cos px e 冮 p ⫹1 p 2

ⴚc2p2t

dp.

6. A ⫽ 0, B ⫽

冮 v sin pv dv ⫽ p2 # sin p ⫺p p cos p , etc. 2

ⴚ1

8. Simple straightforward integration. 10. See (36) in App. A3.1. 14. u(x, t) ⫽

1⫺x 1⫹x U0 , where t ⬎ 0 erf ⫹ erf 2 c 2c1t 2c1t d

SECTION 12.8. Modeling: Membrane, Two-Dimensional Wave Equation, page 575 Purpose. A careful derivation of the two-dimensional wave equation governing the motions of a drumhead, from physical assumptions (the analog of the modeling in Sec. 12.2). SECTION 12.9. Rectangular Membrane. Double Fourier Series, page 577 Purpose. To solve the two-dimensional wave equation in a rectangle 0 ⬉ x ⬉ a, 0 ⬉ y ⬉ b (“rectangular membrane”) by separation of variables and double Fourier series.

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Comment on Content New features as compared with the one-dimensional case (Sec. 12.3) are as follows: 1. We have to separate twice, first by u ⫽ F(x, y)G(t), then the Helmholtz equation for F by F ⫽ H(x)Q( y). 2. We get a double sequence of infinitely many eigenvalues lmn and eigenfunctions u mn; see (9), (10). 3. We need double Fourier series (easily obtainable from the usual Fourier series) to get a solution that also satisfies the initial conditions.

SOLUTIONS TO PROBLEM SET 12.9, page 584 2. Modeling is the art of recognizing and neglecting minor factors and circumstances, and formulating major factors so that they become mathematically accessible, leading to a model that can be solved. No assumption in any model can be satisfied exactly; in particular, in Assumption 2 the tension will change during the motion. 4. Bmn ⫽ 16>(mnp2) if m, n odd, 0 otherwise 6. Bmn ⫽ (⫺1)m⫹18>(mnp2) if n odd, Bmn ⫽ 0 if n is even. Note that this follows from the answer to Prob. 5 by interchanging m and n. 8. Bmn ⫽ 64a 2b 2>(m 3n 3p6) if m and n are odd, Bmn ⫽ 0 otherwise 10. The program will give you 85 ⫽ 5 # 17 ⫽ 22 ⫹ 92 ⫽ 62 ⫹ 72 145 ⫽ 5 # 29 ⫽ 11 ⫹ 122 ⫽ 82 ⫹ 92 185 ⫽ 5 # 37 ⫽ 42 ⫹ 132 ⫽ 82 ⫹ 112 221 ⫽ 13 # 17 ⫽ 52 ⫹ 142 ⫽ 102 ⫹ 112 377 ⫽ 13 # 29 ⫽ 42 ⫹ 192 ⫽ 112 ⫹ 162 493 ⫽ 17 # 29 ⫽ 32 ⫹ 222 ⫽ 132 ⫹ 182 etc. 12. 0.01 cos 12t sin x sin y 18. A ⫽ ab, b ⫽ A>a, so that from (9) with m ⫽ n ⫽ 1 by differentiating with respect to a and equating the derivative to zero, we obtain l211 r 1 1 r 1 a2 r ⫺2 2a a 2 2 b ⫽ a 2 ⫹ 2 b ⫽ a 2 ⫹ 2 b ⫽ 3 ⫹ 2 ⫽ 0; c p a b a A a A hence a 4 ⫽ A2, a 2 ⫽ A, b ⫽ A>a ⫽ a.

SECTION 12.10. Laplacian in Polar Coordinates. Circular Membrane. Fourier–Bessel Series, page 585 Purpose. Detailed derivation of the transformation of the Laplacian into polar coordinates. Derivation of the function that models vibrations of a circular membrane. Comment on Content The transformation is a typical case of a task often required in applications. It is done by two applications of the chain rule.

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In solving the wave equation we concentrate on the simpler case of radially symmetric vibrations, that is, vibrations independent of the angle. (For eigenfunctions depending on the angle, see Probs. 22–25.) We do three steps: 1. u ⫽ W(r)G(t) gives for W Bessel’s equation with ␯ ⫽ 0, hence solutions W(r) ⫽ J0(kr). 2. We satisfy the boundary condition W(R) ⫽ 0 by choosing suitable values of k. 3. A Fourier–Bessel series (18) helps to get the solution (17) of the entire problem. Short Courses. This section can be omitted. SOLUTIONS TO PROBLEMS SET 12.10, page 591 2. If u ⫽ u(r) and we set u r ⫽ v, then ⵜ2u ⫽ u s ⫹ u r >r ⫽ v r ⫹ v>r ⫽ 0. Hence v r >v ⫽ ⫺1>r, ln v ⫽ ⫺ln r ⫹ ~ c ⫽ ln (c1>r), v ⫽ c1>r. By integration, u ⫽ c1 ln r ⫹ c2. 4. Team Project. (a) r 2 cos 2u ⫽ r 2(cos2 u ⫺ sin2 u) ⫽ x 2 ⫺ y 2, r 2 sin 2u ⫽ 2xy, etc. (c) u ⫽

400

ar sin u ⫹

1 3 1 r sin 3u ⫹ r 5 sin 5u ⫹ Á b 3 5

(d) The form of the series results as in (b), and the formulas for the coefficients follow from ⬁

nⴚ1

u r (R, u) ⫽ a nR

(An cos nu ⫹ Bn sin nu) ⫽ f (u).

n⫽1

(f) u ⫽ ⫺(r ⫹ 4>r)(sin u)>3 by separating variables 6. 300 r cos u ⫹ 100 r 3 cos 3u 8.

r sin u ⫹

1 2 2 3 1 r sin 2u ⫺ r sin 3u ⫺ r 4 sin 4u ⫹ Á 2 9p 4

Except for the presence of the variable r, this is just another important application of Fourier series, and we concentrate on a few simple practically important types of boundary values. Of course, earlier problems on Fourier series can now be modified by introducing the powers of r and considered from the present point of view. 10. To get u ⫽ 0 on the x-axis, the idea is to extend the given potential from 0 ⬍ u ⬍ p skew-symmetrically to the whole boundary circle r ⫽ 1; that is, u(1, u) ⫽ b

110u(p ⫺ u)

if 0 ⬍ u ⬍ p

110u(p ⫹ u)

if ⫺p ⬍ u ⬍ 0.

(given)

Then you obtain (valid in the whole disk and thus in the semidisk) u(r, u) ⫽

880

ar sin u ⫹

1 3 1 5 Áb. 3 r sin 3u ⫹ 3 r sin 5u ⫹ 3 5

12. CAS Project. (b) Error 0.04863 (m ⫽ 1), 0.02229, 0.01435, 0.01056, 0.00835, 0.00691, 0.00589, 0.00513, 0.00454, 0.00408 (m ⫽ 10) (c) The approximation of the partial sums is poorest for r ⫽ 0.

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(d) The radii of the nodal circles are u 2: a1>a2 ⫽ 0.43565 u 3: a1>a3 ⫽ 0.27789 a2>a3 ⫽ 0.63788 u 4: a1>a4 ⫽ 0.2039 a2>a4 ⫽ 0.4681 a3>a4 ⫽ 0.7339

Comparison 0.435>0.500 ⫽ 0.87 0.278>0.333 ⫽ 0.83 0.638>0.667 ⫽ 0.96 0.204>0.250 ⫽ 0.82 0.468>0.500 ⫽ 0.94 0.734>0.750 ⫽ 0.98.

We see that the larger radii are better approximations of the values of the nodes of the string than the smaller ones. The smallest quotient does not seem to improve (to get closer to 1); on the contrary, e.g., for u 6 it is 0.80. The other ratios seem to approach 1 and so does the sum of all of them divided by m ⫺ 1.

14. lm>(2p) ⫽ cam>(2pR) increases with decreasing R. 16. The reason is that f (0) ⫽ 1. The partial sums equal 1.10801

0.96823

1.01371

0.99272

1.00436

the last value having 3-digit accuracy. Musically the values indicate substantial contributions of overtones to the overall sound. 18. Differentiation brings in a factor 1>lm ⫽ R>(cam). 22. On Notation. n is standard for Legendre polynomials and for Bessel functions of integer order. Hence we needed another letter for numbering the zeros of J1, J2, Á , and we took m. Hence, for example, the positive zeros of J2 are numbered a21, a22, a23, Á . (In the 9th Edition we used the probably less advantageous opposite order a12, a22, a32, Á .) For consistency, we should have numbered the positive zeros of J0 by a01, a02, a03, Á , but this would make formulas unnecessarily clumsy, and we wrote a1, a2, Á , in particular since the “problem” occurred only at the very end, in the last problems of Sec. 12.10. SECTION 12.11. Laplace’s Equation in Cylindrical and Spherical Coordinates. Potential, page 593 Purpose. 1. Transformation of the Laplacian into cylindrical coordinates (which is trivial because of Sec. 12.9) and spherical coordinates; some remarks on areas in which Laplace’s equation is basic. 2. Separation of the Laplace equation in spherical coordinates and application to a typical boundary value problem. For simplicity we consider a boundary value problem for a sphere with boundary values depending only on ␾. We do three steps: 1. u ⫽ G(r)H(␾) and separation gives for H Legendre’s equation. 2. Continuity requirements restrict H to Legendre polynomials. 3. A Fourier–Legendre series (18) helps to get the solution (17) of the interior problem. Similarly for the exterior problem, whose solution is (20). Short Courses. Omit the derivation of the Laplacian in cylindrical and spherical coordinates.

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SOLUTIONS TO PROBLEM SET 12.11, page 598 4. By (11 r ) in Sec. 5.3 we have (cf. Fig. 312) u 1 ⫽ A1r cos ␾ ⫽ 0

␾ ⫽ 12 p.

This is the xy-plane. Similarly, u 2 ⫽ A2

r2 2

(3 cos2 ␾ ⫺ 1) ⫽ 0

cos ␾ ⫽

1 13

cos ␾ ⫽ 0

and

and u 3 ⫽ A3

r3 (5 cos3 ␾ ⫺ 3 cos ␾) ⫽ 0 2

3 . B5

10. By (5), ⵜ2u ⫽ u s ⫹ u r >r ⫽ 0. Separation and integration gives u s >u r ⫽ ⫺1>r,

ln ƒ u r ƒ ⫽ ⫺ln ƒ r ƒ ⫹ c1.

Taking exponents and integrating again gives u r ⫽ c>r 12. u ⫽ ⫺80 ln r>(ln 2) ⫹ 300

u ⫽ c ln r ⫹ k.

and

14. v ⫽ F(r)G(t), F s ⫹ k 2F ⫽ 0, G ⫹ c2k 2G ⫽ 0, Fn ⫽ sin (npr>R), Gn ⫽ Bn exp (⫺c2n 2p2t>R 2), Bn ⫽ ~ 2n ⫹ 1 16. f (w) ⫽ w, An ⫽ 2

2 R

冮 rf (r) sin npR r dr ⴚ0

冮 wP (w) dw. Since w ⫽ P (w) and the P (w) are orthogonal n

ⴚ1

on the interval ⫺1 ⬉ w ⬉ 1, we obtain A1 ⫽ 1, An ⫽ 0 (n ⫽ 0, 2, 3, Á ). Answer: u ⫽ r cos ␾. Of course, this is at once seen by integration. 18. By definition, P2 (cos ␾) ⫽ 32 cos2 ␾ ⫺ 12. Hence 1 ⫺ cos2 ␾ ⫽ ⫺ 23 P2 (cos ␾) ⫹ 23 and u ⫽ ⫺23 r 2P2 (cos ␾) ⫹ 23 ⫽ r 2(⫺cos2 ␾) ⫹ 13) ⫹ 23. 20. u ⫽ 4r 3P3 (cos ␾) ⫺ 2r 2P2 (cos ␾) ⫹ rP1 (cos ␾) ⫺ 2 22. In Prob. 16, f (␾) ⫽ cos ␾; hence u int ⫽ r cos ␾,

u ext ⫽ r ⴚ2 cos ␾.

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In Prob. 19, f (␾) ⫽ cos 2␾; hence f (␾) ⫽ 2 cos2 ␾ ⫺ 1, so that 2 cos2 ␾ ⫺ 1 ⫽ 43 P2(cos ␾) ⫺ 13 and thus u int ⫽ 43 r 2P2 (cos ␾) ⫺ 13 and u ext ⫽

4 1 P2 (cos ␾) ⫺ . 3r 3r 3

24. Team Project. (a) The two drops over a portion of the cable of length ¢x are ⫺Ri¢x and ⫺L(0i>0t)¢x, respectively. Their sum equals the difference u x⫹ ¢x ⫺ u x. Divide by ¢x and let ¢x : 0. (c) To get the first PDE, differentiate the first transmission line equation with respect to x and use the second equation to replace i x and i xt: ⫺u xx ⫽ Ri x ⫹ Li xt ⫽ R(⫺Gu ⫺ Cu t) ⫹ L(⫺Gu t ⫺ Cu tt). Now collect terms. Similarly for the second PDE. 1 (d) Set ⫽ c2. Then u t ⫽ c2u xx, the heat equation. By (9), (10), Sec. 12.6, RC u⫽

4U0

asin

px ⴚl t l

2 1

⫹

1 3px ⴚl32t Á sin e ⫹ b, 3 l

l2n ⫽

n 2p2 . l nRC

(e) u ⫽ U0 cos (pt>(l1LC sin (px>l) SECTION 12.12. Solution of PDEs by Laplace Transforms, page 600 Purpose. For students familiar with Chap. 6 we show that the Laplace transform also applies to certain PDEs. In such an application the subsidiary equation will generally be an ODE. Short Courses. This section can be omitted. SOLUTIONS TO PROBLEM SET 12.12, page 602 4. w ⫽ w(x, t), W ⫽ l{w(x, t)} ⫽ W(x, s). The subsidiary equation is 0W 0W x ⫹ xl{wt(x, t)} ⫽ ⫹ x(sW ⫺ w(x, 0)) ⫽ xl(1) ⫽ and w(x, 0) ⫽ 1. 0x 0x s By simplification, 0W x ⫹ xsW ⫽ x ⫹ . 0x s By integration of this first-order ODE with respect to x we obtain W ⫽ c(s)eⴚsx >2 ⫹ 2

1 1 . 2 ⫹ s s

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For x ⫽ 0 we have w(0, t) ⫽ 1 and W(0, s) ⫽ l {w(0, t)} ⫽ l {1} ⫽

1 1 1 ⫽ c(s) ⫹ 2 ⫹ . s s s

Hence c(s) ⫽ ⫺1>s 2, so that 2 1 1 1 W ⫽ ⫺ 2 eⴚsx >2 ⫹ 2 ⫹ . s s s

The inverse Laplace transform of this solution of the subsidiary equation is w(x, t) ⫽ ⫺(t ⫺ 12 x 2) u(t ⫺ 12 x 2) ⫹ t ⫹ 1 ⫽ c 1

t⫹1

if t ⬍ 12 x 2

⫹1

if t ⬎ 12 x 2.

6. w(x, t) ⫽ t ⫹ 1 ⫺ (t ⫺ x ) u(t ⫺ x ) ⫽ b 2

t⫹1

if t ⬉ x 2

x2 ⫹ 1

if t ⭌ x 2

as obtained from W(x, s) ⫽

1⫹s s

⫹ c(s)eⴚsx

with c(s) ⫽ ⫺1>s 2 as obtained from w(0, t) ⫽ 1, W(0, s) ⫽ 1>s. 8. W ⫽ l{w}, Wxx ⫽ (100s 2 ⫹ 100s ⫹ 25)W ⫽ (10s ⫹ 5)2W. The solution of this ODE is W ⫽ c1(s)eⴚ(10s⫹5)x ⫹ c2(s)e(10s⫹5)x with c2(s) ⫽ 0, so that the solution is bounded. c1(s) follows from W(0, s) ⫽ l{w(0, t)} ⫽ l{sin t} ⫽

1 s ⫹1 2

⫽ c1(s).

Hence W⫽

1 s ⫹1 2

eⴚ(10s⫹5)x.

The inverse Laplace transform (the solution of our problem) is w ⫽ lⴚ1{W} ⫽ eⴚ5xu(t ⫺ 10x) sin (t ⫺ 10x), a traveling wave decaying with x. Here u is the unit step function (the Heaviside function). 10. From W ⫽ F(s)eⴚ(x>c)1s and the convolution theorem we have w ⫽ f * lⴚ1{eⴚk1s},

x k ⫽ c.

From this and formula 39 in Sec. 6.9 we get, as asserted, t

w⫽

冮 f (t ⫺ t) 22kpt e 3

ⴚk2>(4t)

dt.

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12. W0(x, s) ⫽ s ⴚ1eⴚ1sx>c, l{u(t)} ⫽ 1>s, and since w(x, 0) ⫽ 0, W(x, s) ⫽ F(s)sW0 (x, s) ⫽ F(s)[sW0 (x, s) ⫺ w(x, 0)] 0w0 ⫽ F(s) l e f. 0t Now apply the convolution theorem. SOLUTIONS TO CHAPTER 12 REVIEW QUESTIONS AND PROBLEMS, page 603 16. u ⫽ A( y) cos 5x ⫹ B( y) sin 5x 18. u ⫽ g( y)(1 ⫺ eⴚx) ⫹ f ( y) 20. Parabolic, y r 2 ⫺ 6y r ⫹ 9 ⫽ ( y r ⫺ 3)2, v ⫽ x, z ⫽ y ⫺ 3x, u ⫽ xf1( y ⫺ 3x) ⫹ f2( y ⫺ 3x) 22. cos 8t sin 4x 1 4 1 24. u ⫽ acos 2t sin x ⫺ cos 6t sin 3x ⫹ cos 10t sin 5x ⫺ ⫹ Á b p 9 25 400 px ⴚ0.001143t 1 3px ⴚ0.01029t 26. u ⫽ 2 asin e ⫺ sin e ⫹⫺Áb 100 9 100 p 28. u ⫽ p2 ⫺ 12(cos x eⴚt ⫺ 14 cos 2x eⴚ4t ⫹ 19 cos 3x eⴚ9t ⫺ ⫹ Á ) 32 1 1 30. u ⫽ p ⫺ cos 6x eⴚ36t ⫹ Á b a cos 2x eⴚ4t ⫹ p 4 36 32. u ⫽

⬁

2 a a 3 3 m ⫽1 n⫽1 m n m,n odd

sin mx sin ny eⴚc (m ⫹n )t 2

34. l11>(2p) ⫽ cp(11 ⫹ 1)>(2p) ⫽ 1> 12 36. Area pR 2>2 ⫽ 1, R ⫽ 12> p, and

ck 11>(2p) ⫽ k 11>(2p) ⫽ a11>(2pR) ⫽ 3.832>(2p 12> p) ⫽ 3.832> 18p.

38. u ⫽

(u 0 ⫺ u 1)r0r1 (r1 ⫺ r0)r

⫹

u 1r1 ⫺ u 0r0 r1 ⫺ r0 , where r is the distance from the center of the

spheres 40. f (␾) ⫽ 4 cos3 ␾. Now, by (11 r ), Sec. 5.3, cos3 ␾ ⫽ 25 P3(cos ␾) ⫹ 35 P1(cos ␾). Answer: u ⫽ 85 r 3P3(cos ␾) ⫹ 12 5 rP1(cos ␾).

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Part D. COMPLEX ANALYSIS Major Changes In the eighth edition, conformal mapping was distributed over several sections in the first chapter on complex analysis. It has now been given greater emphasis by consolidation of that material in a separate chapter (Chap. 17), which can be used independently of a CAS (just as any other chapter) or in part supported by the graphic capabilities of a CAS. Thus in this respect one has complete freedom. Recent teaching experience has shown that the present arrangement seems to be preferable over that of the 8th edition.

CHAPTER 13 Complex Numbers and Functions. Complex Differentiation SECTION 13.1. Complex Numbers and Their Geometric Representation, page 608 Purpose. To discuss the algebraic operations for complex numbers and the representation of complex numbers as points in the plane. Main Content, Important Concepts Complex number, real part, imaginary part, imaginary unit The four algebraic operations in complex Complex plane, real axis, imaginary axis Complex conjugates Two Suggestions on Content 1. Of course, at the expense of a small conceptual concession, one can also start immediately from (4), (5), z ⫽ x ⫹ iy,

i 2 ⫽ ⫺1

and go on from there. 2. If students have some knowledge of complex numbers, the practical division rule (7) and perhaps (8) and (9) may be the only items to be recalled in this section. (But I personally would do more in any case.) SOLUTIONS TO PROBLEM SET 13.1, page 612 2. Note that z ⫽ 1 ⫹ i and iz ⫽ i ⫺ 1 ⫽ ⫺1 ⫹ i lie on the bisecting lines of the first and second quadrants. 4. ⫺12 ⫺ 14i, ⫺10 ⫺ 6i, ⫺29 ⫹ 54i, 3 ⫺ 2i 6. z 1z 2 ⫽ 0 if and only if Re (z 1z 2) ⫽ x 2 x 1 ⫺ y2 y1 ⫽ 0

and

Im (z 1z 2) ⫽ y2 x 1 ⫹ x 2 y1 ⫽ 0.

Let z 2 ⫽ 0, so that x 22 ⫹ y 22 ⫽ 0. Now x 22 ⫹ y 22 is the coefficient determinant of our homogeneous system of equations in the “unknowns” x 1 and y1, so that this system has only the trivial solution; hence z 1 ⫽ 0. 241

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8. 7 ⫹ 24i, 7 ⫺ 24i 10. 1>z 22 ⫽ (3 ⫹ 4i)>25, 12. We obtain

3 Re (1>z 22) ⫽ 25 ,

1>Re (z 22) ⫽ 13

z1 (⫺2 ⫺ 11i)(2 ⫹ i) ⫺15 ⫹ 20i ⫽ ⫽ ⫺3 ⫹ 4i, z2 ⫽ 5 5 and z2 (2 ⫺ i)(⫺2 ⫺ 11i) ⫺15 ⫺ 20i 1 ⫽ ⫽ . z1 ⫽ 125 125 ⫺3 ⫹ 4i 14. ⫺3 ⫺ 4i y 16. ⫺ 2 , x ⫹ y2

⫺

2xy (x ⫹ y 2)2 2

18. (1 ⫹ i)4 ⫽ ⫺4, (1 ⫹ i)16 ⫽ 256, Re [(1 ⫹ i)16z 2] ⫽ 256(x 2 ⫺ y 2) 20. Im (1>z 2 ⫽ Im (z 2>(zz )2) ⫽ 2xy>(x 2 ⫹ y 2)2 SECTION 13.2. Polar Form of Complex Numbers. Powers and Roots, page 613 Purpose. To give the student a firm grasp of the polar form, including the principal value Arg z, and its application in multiplication and division. Main Content, Important Concepts Absolute value ƒ z ƒ , argument u, principal value Arg u Triangle inequality (6) Multiplication and division in polar form nth root, nth roots of unity (16) SOLUTIONS TO PROBLEM SET 13.2, page 618 2. 132 (cos 34 p ⫹ i sin 34 p) 4. 5 (cos p ⫹ i sin p) 6. Simplification shows that the quotient equals ⫺2. Answer: 2(cos p ⫹ i sin p). 8. Division shows that the given quotient equals 3 ⫹ 2i. Hence the polar form is 232 ⫹ 22 (cos arctan 23 ⫹ i sin arctan 23). 10. p, ⫺p ⫹ arctan (51) ⫽ ⫺2.944, 2.944 12. ⫺3p>4 14. p ⫺ 0.0997 ⫽ 3.0419, ⫺p ⫹ 0.0997 ⫽ ⫺3.0419 16. 3 ⫹ 127i 18. ⫺5 ⫹ 5i 20. Team Project. (a) Use (15). (b) Use (10) in App. 3.1 in the form cos 12 u ⫽ 212 (1 ⫹ cos u),

sin 12 u ⫽ 212 (1 ⫺ cos u),

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multiply them by 1r, 1r cos 12 u ⫽ 212 (r ⫹ r cos u),

1r sin 12 u ⫽ 212 (r ⫺ r cos u),

use r cos u ⫽ x, and finally choose the sign of the Im 1z in such a way that sign [(Re 1z) (Im 1z)] ⫽ sign y. (c) ⫾17(1 ⫺ i), ⫾(4 ⫺ 5i), ⫾(2 ⫹ 13i) 22. The three values are 3 2 5 (cos 13 u ⫹ i sin 13 u) 3 2 5 (cos (13 u ⫹ 23 p) ⫹ i sin (13 u ⫹ 23 p)) 3 2 5 (cos (13 u ⫹ 43 p) ⫹ i sin (13 u ⫹ 43 p))

where u ⫽ arctan 43 . 24. ⫾(1 ⫾ i) kp kp 26. cos ⫹ i sin , k ⫽ 0, 1, Á , 7; that is, ⫾1, ⫾i, ⫾(1 ⫾ i)>2. 4 4 28. 3 ⫹ 2i, 3 ⫺ 4i 30. z 2 ⫽ ⫾18i, z ⫽ ⫾(3 ⫾ 3i). Each of the two factors with real coefficients is obtained if you take one solution and its complex conjugate: (z ⫺ 3 ⫺ 3i) (z ⫺ 3 ⫹ 3i) ⫽ z 2 ⫺ 6z ⫹ 18 (z ⫹ 3 ⫹ 3i)(z ⫹ 3 ⫺ 3i) ⫽ z 2 ⫹ 6z ⫹ 18. The product of the two factors is (z 2 ⫹ 18)2 ⫺ (6z)2 ⫽ z 4 ⫹ 36z 2 ⫹ 324 ⫺ 36z 2 ⫽ z 4 ⫹ 324. 32. ƒ z 1 ⫹ z 2 ƒ ⫽ ƒ 1 ⫹ 5i ƒ ⫽ 126 ⫽ 5.10 ⬍ ƒ z 1 ƒ ⫹ ƒ z 2 ƒ ⫽ 110 ⫹ 120 ⫽ 7.63 34. ƒ z ƒ ⫽ 2x 2 ⫹ y 2 ⭌ 2x 2 ⫽ ƒ x ƒ , etc. SECTION 13.3. Derivative. Analytic Function, page 619 Purpose. To define (complex) analytic functions—the class of functions complex analysis is concerned with—and the concepts needed for that definition, in particular, derivatives. This is preceded by a collection of a few standard concepts on sets in the complex plane that we shall need from time to time in the chapters on complex analysis. Main Content, Important Concepts Unit circle, unit disk, open and closed disks Domain, region Complex function Limit, continuity Derivative Analytic function Comment on Content The most important concept in this section is that of an analytic function. The other concepts resemble those of real calculus. The most important new idea is connected with the limit: the approach in infinitely many possible directions. This yields the negative result in Example 4 and—much more importantly—the Cauchy–Riemann equations in the next section.

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SOLUTIONS TO PROBLEM SET 13.3, page 624 2. Open unit disk without the origin z ⫽ 0. An old-fashioned term for this is deleted neighborhood; we shall not use this expression. 4. Open horizontal infinite strip bounded by the parallel straight lines y ⫽ ⫺p and p. 6. We obtain 1 x Re a b ⫽ 2 ⬍ 1, z x ⫹ y2

x ⬍ x 2 ⫹ y 2,

1 1 2 2 4 ⬍ (x ⫺ 2 ) ⫹ y ,

hence 1 2 1 4 ⬍ ƒz ⫺ 2ƒ ,

1 1 2 ⬍ ƒz ⫺ 2ƒ.

This is the open exterior of the circle with center 12 on the x-axis and radius 12 ; hence this circle passes through the origin z ⫽ 0. 8. Geometrically, we need hardly any calculation. The left side of the inequality is the distance d1 of z from ⫺i, and the right side is the distance d2 of z from i. These distances are equal if and only if z lies on the x-axis. Sketch it. Strict inequality d1 ⬎ d2 means that z is farther away from ⫺i than from i; this is the case if and only if z lies in the upper half-plane. Analytically, we have ƒ x ⫹ iy ⫹ i ƒ 2 ⭌ ƒ x ⫹ iy ⫺ i ƒ 2, that is, x 2 ⫹ ( y ⫹ 1)2 ⭌ x 2 ⫹ ( y ⫺ 1)2 and thus y 2 ⫹ 2y ⫹ 1 ⭌ y 2 ⫺ 2y ⫹ 1, so that 4y ⭌ 0 and y ⭌ 0, the upper half-plane. 10. We obtain u(x, y) ⫽ Re f (z) ⫽ 5 (x 2 ⫺ y 2) ⫺ 12x ⫹ 3 and u(4, ⫺3) ⫽ ⫺10. Similarly, v(x, y) ⫽ Im f (z) ⫽ 10xy ⫺ 12y ⫹ 2 and v(4, ⫺3) ⫽ ⫺82. 12. We obtain u (x, y) ⫽ Re f (z) ⫽

x 2 ⫹ y2 ⫺ 4 (x ⫹ 2)2 ⫹ y 2

and u (0, 8) ⫽ 15 17 .

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Furthermore, v(x, y) ⫽ Im f (z) ⫽

4y (x ⫹ 2)2 ⫹ y 2

and 8 v(0, 8) ⫽ 17 .

14. Yes, because (r 2 cos 2u)>r ⫽ r cos 2u : 0 ⫽ f (0)

as r : 0.

16. No, since (r 2 sin 2u)>r 2 ⫽ sin 2u depends on u, on the direction of approach to 0, so that it has no limit, by definition. 18. Since the differentiation rules in complex are the same as in calculus, the purpose of Probs. 18–23 is to help the student in getting accustomed to handling complex expressions. At present, by the quotient rule, f r(z) ⫽

z ⫹ i ⫺ (z ⫺ i) # 1 (z ⫹ i)

⫽

2i (z ⫹ i)2

Thus, f r(i) ⫽ 2i>(2i)2 ⫽ ⫺i>2. 20. 0 since the quotient is constant, equal to ⫺i>2. 22. f r (z) ⫽ 3 (iz 3 ⫹ 3z 2)2(3iz 2 ⫹ 6z), f (2i) ⫽ 0 because the last factor of f r(z) is zero at 2i. 24. Team Project. (a) Use Re f (z) ⫽ [ f (z) ⫹ f (z)]>2, Im f (z) ⫽ [ f (z) ⫺ f (z)]>2i. (b) Assume that lim z:z0 f (z) ⫽ l 1, lim z:z0 f (z) ⫽ l 2, l 1 ⫽ l 2. For every P ⬎ 0 there are d1 ⬎ 0 and d2 ⬎ 0 such that ƒ f (z) ⫺ l j ƒ ⬍ P

when

0 ⬍ ƒ z ⫺ z 0 ƒ ⬍ dj,

j ⫽ 1, 2.

Hence for P ⫽ ƒ l 1 ⫺ l 2 ƒ >2 and 0 ⬍ ƒ z ⫺ z 0 ƒ ⬍ d, where d ⬉ d1, d ⬉ d2, we have ƒ l 1 ⫺ l 2 ƒ ⫽ ƒ [ f (z) ⫺ l 2] ⫺ [ f (z) ⫺ l 1] ƒ ⬉ ƒ f (z) ⫺ l 2 ƒ ⫹ ƒ f (z) ⫺ l 1 ƒ ⬍ 2P ⫽ ƒ l 1 ⫺ l 2 ƒ . (c) By continuity, for any P ⬎ 0 there is a d ⬎ 0 such that ƒ f (z) ⫺ f (a) ƒ ⬍ P when ƒ z ⫺ a ƒ ⬍ d. Now ƒ z n ⫺ a ƒ ⬍ d for all sufficiently large n since lim z n ⫽ a. Thus ƒ f (z n) ⫺ f (a) ƒ ⬍ P for these n. (d) The proof is as in calculus. We write f (z) ⫺ f (z 0) ⫺ f r(z 0) ⫽ h. z ⫺ z0 Then, from the definition of a limit, it follows that for any given P ⬎ 0 there is a d ⬎ 0 such that ƒ h ƒ ⬍ P when ƒ z ⫺ z 0 ƒ ⬍ d. From this and the triangle inequality, ƒ f (z) ⫺ f (z 0) ƒ ⫽ ƒ z ⫺ z 0 ƒ ƒ f r (z 0) ⫹ h ƒ ⬉ ƒ z ⫺ z 0 ƒ ƒ f r (z 0) ƒ ⫹ ƒ z ⫺ z 0 ƒ P, which approaches 0 as ƒ z ⫺ z 0 ƒ : 0.

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(e) The quotient in (4) is ¢x>¢z, which is 0 if ¢x ⫽ 0 but 1 if ¢y ⫽ 0, so that it has no limit as ¢z : 0. (f)

f (z ⫹ ¢z) ⫺ f (z) ¢z

⫽

(z ⫹ ¢z)( z ⫹ ¢z) ⫺ zz ¢z

⫽z

¢z ¢z

⫹ z ⫹ ¢z .

When z ⫽ 0 the expression on the right approaches zero as ¢z : 0. When z ⫽ 0 and ¢z ⫽ ¢x, then ¢z ⫽ ¢x and that expression approaches z ⫹ z. When z ⫽ 0 and ¢z ⫽ i ¢y, then ¢z ⫽ ⫺i ¢y and that expression approaches ⫺z ⫹ z. This proves the statement. SECTION 13.4. Cauchy–Riemann Equations. Laplace’s Equation, page 625 Purpose. To derive and explain the most important equations in this chapter, the Cauchy–Riemann equations, a system of two PDEs, which constitute the basic criterion for analyticity. Main Content, Important Concepts Cauchy–Riemann equations (1) These equations as a criterion for analyticity (Theorems 1 and 2) Derivative in terms of partial derivatives, (4), (5) Relation of analytic functions to Laplace’s equation Harmonic function, harmonic conjugate Comment on Content (4), (5), and Example 3 will be needed occasionally. The relation to Laplace’s equation is basic, as mentioned in the text. SOLUTIONS TO PROBLEM SET 13.4, page 629 2. No, f (z) ⫽ u ⫹ iv ⫽ 0 ⫹ i ƒ z ƒ 2, u ⫽ 0, v ⫽ ƒ z ƒ 2. 4. No, u x ⫽ ex cos y ⫽ vy ⫽ ⫺ex cos y. 6. Yes, when z ⫽ 0, ⫾1, ⫾i. 8. No, u ⫽ Arg 2pz, v ⫽ 0. 10. Yes, when z ⫽ 0. 12. No. 14. Yes, f (z) ⫽ 12 z 2 ⫹ c, c a real constant. Indeed using the Cauchy–Riemann equations, we obtain v ⫽ xy vy ⫽ x ⫽ u x u ⫽ 12 x 2 ⫹ k ( y) u y ⫽ k r ( y) ⫽ ⫺vx ⫽ ⫺y hence k ( y) ⫽ ⫺12 y 2 ⫹ c so that u ⫽ 12 x 2 ⫺ 12 y 2 ⫹ c.

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16. Yes, f (z) ⫽ sin x cosh y ⫹ i cos x sinh y ⫹ ic because, using the Cauchy–Riemann equations, we obtain u ⫽ sin x cosh y u x ⫽ cos x cosh y ⫽ vy v ⫽ cos x sinh y ⫹ h (x) vx ⫽ ⫺sin x sinh y ⫹ h r (x) ⫽ ⫺u y ⫽ ⫺sin x sinh y hence h r (x) ⫽ 0, h ⫽ c ⫽ const (real), so that we obtain the expression given at the beginning, which is sin z ⫹ ic, as we shall see in Sec. 13.6. 18. We obtain f (z) ⫽ z 3 ⫹ ic ⫽ x 3 ⫺ 3xy 2 ⫹ i(3x 2y ⫺ y 3 ⫹ c) because, using the Cauchy–Riemann equations, we have u ⫽ x 3 ⫺ 3xy 2 u x ⫽ 3x 2 ⫺ 3y 2 ⫽ vy v ⫽ 3x 2y ⫺ y 3 ⫹ h (x) vx ⫽ 6xy ⫹ h r (x) ⫽ ⫺u y ⫽ 6xy so that h r (x) ⫽ 0, h ⫽ c ⫽ const (real). 22. By differentiation we obtain ⵜ2u ⫽ (⫺a 2 ⫹ 4)u ⫽ 0, so a ⫽ ⫾2. Hence we can take u ⫽ cos 2x cosh 2y and obtain, by using the first Cauchy–Riemann equation, u x ⫽ ⫺2 sin 2x cosh 2y ⫽ vy and by integration v ⫽ ⫺sin 2x sinh 2y ⫹ h (x). The second Cauchy–Riemann equation now gives vx ⫽ ⫺2 cos 2x sinh 2y ⫹ h r (x) ⫽ ⫺u y ⫽ ⫺2 cos 2x sinh 2y. Hence h r ⫽ 0 and h (x) ⫽ const, and the answer is v ⫽ ⫺sin 2x sinh 2y ⫹ c which will give the imaginary part of cos 2z; see Sec. 13.6. 24. ⵜ2u ⫽ a 2 cosh ax cos y ⫺ cosh ax cos y ⫽ 0, hence a ⫽ 1. From this and the Cauchy–Riemann equations we obtain, by differentiation, u ⫽ cosh x cos y u x ⫽ sinh x cos y ⫽ vy v ⫽ sinh x sin y ⫹ h(x)

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vx ⫽ cosh x sin y ⫹ h r(x) ⫽ ⫺u y ⫽ cosh x sin y, so that h r (x) ⫽ 0, h (x) ⫽ c ⫽ const (real), and v ⫽ sinh x sin y ⫹ c, the imaginary part of cosh z ⫹ ic, as we shall see in Sec. 13.6. 30. Team Project. (a) u ⫽ const, u x ⫽ u y ⫽ 0, vx ⫽ vy ⫽ 0 by (1), v ⫽ const, and f ⫽ u ⫹ iv ⫽ const. (b) Same idea as in (a). (c) f r ⫽ u x ⫹ ivx ⫽ 0 by (4). Hence vy ⫽ 0, u y ⫽ 0 by (1), f ⫽ u ⫹ iv ⫽ const. SECTION 13.5. Exponential Function, page 630 Purpose. Sections 13.5–13.7 are devoted to the most important elementary functions in complex, which generalize the corresponding real functions, and we emphasize properties that are not apparent in real. Basic Properties of the Exponential Function Derivative and functional relation as in real Euler formula, polar form of z Periodicity with 2pi, fundamental region ez ⫽ 0 for all z SOLUTIONS TO PROBLEM SET 13.5, page 632 2. e3(cos 4 ⫹ i sin 4) ⫽ ⫺13.129 ⫺ 15.201, 6. e

11pi>2

⫽ eⴚpi>2 ⫽ ⫺i

8. 2 r exp (i(u ⫹ 2kp)>n), 10. e 12.

20.086

(cos 1.8 ⫺ i sin 1.8) ⫽ 1.822(⫺0.227 ⫺ 0.974i) ⫽ ⫺0.414 ⫺ 1.774i

4. e

0.6

pi>4

ⴚ3pi>4

e 1

; e

k ⫽ 0, 1, Á ,

ⴚpi>4

n ⫺ 1, r ⫽ ƒ z ƒ , u ⫽ Arg z

3pi>4

ei arctan (y>(1⫺x))

2(x ⫺ 1) ⫹ y 14. eⴚpx cos py, ⫺eⴚpx sin py x y y b acos 2 ⫺ i sin 2 b 16. e1>z ⫽ exp a 2 2 2 x ⫹y x ⫹y x ⫹ y2 2

18. Team Project. (a) e1>z is analytic for all z ⫽ 0. ez is not analytic for any z. The last function is analytic if and only if k ⫽ 1. (b) (i) ex sin y ⫽ 0, sin y ⫽ 0. Answer: On the horizontal lines y ⫽ ⫾np, n ⫽ 0, 1, Á . (ii) eⴚx ⬍ 1, x ⬎ 0 (the right half-plane). (iii) ez ⫽ exⴚiy ⫽ ex(cos y ⫺ i sin y) ⫽ ex(cos y ⫹ i sin y) ⫽ ez. Answer: All z. (d) f r ⫽ u x ⫹ ivx ⫽ f ⫽ u ⫹ iv, hence u x ⫽ u, vx ⫽ v. By integration, u ⫽ c1( y)ex,

v ⫽ c2( y)ex.

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By the first Cauchy–Riemann equation, u x ⫽ vy ⫽ c2r ex,

thus

c1 ⫽ c2r

( r ⫽ d>dy).

thus

c1r ⫽ ⫺c2.

By the second Cauchy–Riemann equation, u y ⫽ c1r ex ⫽ ⫺vx ⫽ ⫺c2ex,

Differentiating the last equation with respect to y, we get c1s ⫽ ⫺c2r ⫽ ⫺c1,

hence

c1 ⫽ a cos y ⫹ b sin y.

Now for y ⫽ 0 we must have u (x, 0) ⫽ c1 (0)ex ⫽ ex,

c1 (0) ⫽ 1,

v (x, 0) ⫽ c2 (0)e ⫽ 0,

c2 (0) ⫽ 0.

a ⫽ 1,

Also, b ⫽ c1r (0) ⫽ ⫺c2(0) ⫽ 0. Together c1( y) ⫽ cos y. From this, c2( y) ⫽ ⫺c1r ( y) ⫽ sin y. This gives f (z) ⫽ e (cos y ⫹ i sin y). 20. z ⫽ ln 5 ⫹ (arctan 34 ⫹ 2np)i x

21. z ⫽ ln 2 ⫹ (1 ⫹ 2n)pi SECTION 13.6. Trigonometric and Hyperbolic Functions, Euler’s Formula, page 633 Purpose. Discussion of basic properties of trigonometric and hyperbolic functions, with emphasis on the relations between these two classes of functions as well as between them and the exponential function; here we see, on an elementary level, that investigation of special functions in complex can add substantially to their understanding. Main Content Definitions of cos z and sin z (1) Euler’s formula in complex (5) Definitions of cosh z and sinh z (11) Relations between trigonometric and hyperbolic functions Real and imaginary parts (6) and Prob. 3 SOLUTIONS TO PROBLEM SET 13.6, page 636 2. The right side is cosh z 1 cosh z 2 ⫹ sinh z 1 sinh z 2 ⫽ 14 (ez1 ⫹ eⴚz1) (ez2 ⫹ eⴚz2) ⫹ 14 (ez1 ⫺ eⴚz1) (ez2 ⫺ eⴚz2). If we multiply out, then, because of the minus signs, the products ez1eⴚz2 and eⴚz1ez2 cancel in pairs. There remains, as asserted, 2 # 14 (ez1⫹z2 ⫹ eⴚz1ⴚz2) ⫽ cosh (z 1 ⫹ z 2). Similarly for the other formula. 6. i sinh p ⫽ 267.74i

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8. cosh p ⫽ 11.59, and cosh pi ⫽ 12 (epi ⫹ eⴚpi) ⫽ 12 (⫺1 ⫺ 1) ⫽ ⫺1 10. ⫺6.548 ⫺ 7.619i,

⫺6.581 ⫺ 7.582i

12. We obtain cos 12 pi ⫽ cosh 12 p ⫽ 2.509 and cos ( 12p ⫺ 12pi) ⫽ cos 12 p cos 12 pi ⫺ sin 12 p sin 12 pi ⫽ 0 ⫺ 1 # i sinh 12 p ⫽ ⫺2.301i. 14. From (7a) we obtain ƒ cos z ƒ 2 ⫽ cos2 x ⫹ sinh2 y ⫽ cos2 x ⫹ cosh2 y ⫺ 1. Hence ƒ cos z ƒ 2 ⭌ sinh2 y from the first equality, and ƒ cos z ƒ 2 ⬉ cosh2 y from the second equality because cos2 x ⫺ 1 ⬉ 0. Now take square roots. The inequality for ƒ sin z ƒ is obtained similarly. 16. cos x sinh y ⫽ 0, x ⫽ 12 p ⫾ 2np, cosh y ⫽ 100, cosh y ⬇ 12 ey for large y, ey ⬇ 200, y ⬇ 5.29832 (agrees to 4D with the solution of cosh y ⫽ 100). Answer: z ⫽ 12 p ⫾ 2np ⫾ 5.29832i. 18. (a) cosh x cos y ⫽ ⫺1, (b) sinh x sin y ⫽ 0. From (b) we have x ⫽ 0 or y ⫽ ⫾np. Then y ⫽ (2n ⫹ 1)p and x ⫽ 0 from (a). Answer: z ⫽ (2n ⫹ 1)pi. 20. We obtain tan z ⫽

sin z cos z sin z ⫽ . ƒ cos z ƒ 2 cos z

Hence the denominator is (use cosh2 y ⫽ sinh2 y ⫹ 1 to simplify) cos2 x cosh2 y ⫹ sin2 x sinh2 y ⫽ cos2 x ⫹ sinh2 y. Insert sin z and cos z into the numerator and multiply out. Then for the real part of the product we get sin x cos x cosh2 y ⫺ cos x sin x sinh2 y ⫽ sin x cos x and for the imaginary part, using sin2 x ⫹ cos2 x ⫽ 1, we get sin2 x cosh y sinh y ⫹ cos2 x cosh y sinh y ⫽ cosh y sinh y. SECTION 13.7. Logarithm. General Power, Principal Value, page 636 Purpose. Discussion of the complex logarithm, which extends the real logarithm ln x (defined for positive x) to an infinitely many-valued relation (3) defined for all z ⫽ 0; definition of general powers z c. Comment on Notation ln z is also denoted by log z, but for the engineer, who also needs logarithms log x of base 10, the notation ln is more practical; this notation is widely used in mathematics.

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Important Formulas Real and imaginary parts (1) Relation of the principal value to the other values (3) Relations between ln and the exponential function (4) Functional relation in complex (5) Derivative (6) General power (8)

SOLUTIONS TO PROBLEM SET 13.7, page 640 6. ln 412 ⫹ pi>4 ⫽ 1.733 ⫹ 0.785i 8. ln 12 ⫾ pi>4 ⫽ 0.347 ⫾ 0.785i 10. 2.708 ⫾ 3.135i. This also is a reminder that the principal value Arg z has a jump along the negative part of the real axis. 12. ln e ⫽ 1 ⫾ 2npi, n ⫽ 0, 1, Á 14. ln 7 ⫹ (1 ⫾ 2n)pi, n ⫽ 0, 1, Á 16. ln 5 ⫹ (arctan 34 ⫾ 2np)i ⫽ 1.609 ⫹ (0.644 ⫾ 2np)i, n ⫽ 0, 1, Á 18. z ⫽ eⴚpi>2 ⫽ ⫺i 20. z ⫽ ee ⴚ pi ⫽ eee ⴚ pi ⫽ ⫺ee ⫽ ⫺15.154 22. Using the definition, we obtain (2i)2i ⫽ e2i Ln 2i ⫽ e2i (ln ƒ2iƒ⫹pi>2) ⫽ eⴚpe2i ln 2 ⫽ eⴚpei ln 4 ⫽ eⴚp(cos (ln 4) ⫹ i sin (ln 4)) ⫽ 0.0079 ⫹ 0.0425i 24. e

(1⫹i)Ln(1ⴚi)

⫽e

(1⫹i)(ln12⫺ pi>4)

⫽ exp (ln 12 ⫺ pi>4 ⫹ i ln 12 ⫹ p>4)

⫽ 12ep>4(cos (⫺14 p ⫹ ln 12) ⫹ i sin (⫺14 p ⫹ ln 12)) ⫽ 2.8079 ⫺ 1.3179i.

Note that this is the complex conjugate of the answer to Prob. 23. 26. i i>2 ⫽ ei>2 ln i ⫽ e(i>2)pi>2 ⫽ eⴚp>4 ⫽ 0.456 28. We obtain exp (13 Ln (3 ⫹ 4i) ⫽ exp ( 13 (ln 5 ⫹ i arctan 43)) 3 ⫽2 5 [cos (13 arctan 43) ⫹ i sin (13 arctan 43)]

⫽ 1.6289 ⫹ 0.5202i. 30. Team Project. (a) w ⫽ arccos z, z ⫽ cos w ⫽ 12 (eiw ⫺ eⴚiw). Multiply by 2eiw to get a quadratic equation in eiw, e2iw ⫺ 2zeiw ⫹ 1 ⫽ 0.

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A solution is eiw ⫽ z ⫹ 2z 2 ⫺ 1, and by taking logarithms we get the given formula arccos z ⫽ w ⫽ ⫺i ln (z ⫹ 2z 2 ⫺ 1). (b) Similarly, z ⫽ sin w ⫽

1 iw (e ⫺ eⴚiw), 2i

2izeiw ⫽ e2iw ⫺ 1, e2iw ⫺ 2izeiw ⫺ 1 ⫽ 0, eiw ⫽ iz ⫹ 2⫺z 2 ⫹ 1. Now take logarithms, etc. (c) cosh w ⫽ 12 (ew ⫹ eⴚw) ⫽ z, (ew)2 ⫺ 2zew ⫹ 1 ⫽ 0, ew ⫽ z ⫹ 2z 2 ⫺ 1. Take logarithms. (d) z ⫽ sinh w ⫽ 12 (ew ⫺ eⴚw), 2zew ⫽ e2w ⫺ 1, ew ⫽ z ⫹ 2z 2 ⫹ 1. Take logarithms. sin w eiw ⫺ eⴚiw e2iw ⫺ 1 (e) z ⫽ tan w ⫽ ⫽ ⫺i iw , ⴚiw ⫽ ⫺i 2iw cos w e ⫹e e ⫹1 e2iw ⫽

i⫺z , i⫹z

w⫽

i⫺z i i⫹z 1 ln ⫽ ln 2i i ⫹ z 2 i⫺z

(f) This is similar to (e). SOLUTIONS TO CHAPTER 13 REVIEW QUESTIONS AND PROBLEMS, page 641 2. You obtain the complex conjugate. The absolute value. 12. (1 ⫺ i)10 ⫽ [(1 ⫺ i)2]5 ⫽ (⫺2i)5 ⫽ ⫺32i 14. ⫾(1 ⫹ i)> 12 16. i, ⫺i 18. 12.04e⫾0.08314i 20. ei arctan (0.8>0.6) ⫽ e0.9273i 22. ⫾4(1 ⫺ i) 13 2p 2p 1 24. 1, cos ⫾ i sin ⫽ ⫾ 3 3 2 2 26. f (z) ⫽ ⫺1>z 28. f (z) ⫽ cos 3z 30. f (z) ⫽ sin 2z 32. ln 1 ⫹ i Arg (0.6 ⫹ 0.8i) ⫽ 0 ⫹ i arctan (0.8>0.6) ⫽ 0.927i 34. We obtain sinh (1 ⫹ pi) ⫽ sinh 1 cos p ⫹ i cosh 1 sin p ⫽ ⫺sinh 1 ⫽ ⫺1.175, and, totally different from this, sin (1 ⫹ pi) ⫽ sin 1 cosh p ⫹ i cos 1 sinh p ⫽ 9.754 ⫹ 6.240i.

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CHAPTER 14

Complex Integration

Comment The introduction to the chapter mentions two reasons for the importance of complex integration. Another practical reason is the extensive use of complex integral representations in the higher theory of special functions; see, for instance, Ref. [GenRef10] listed in App. 1. SECTION 14.1. Line Integral in the Complex Plane, page 643 Purpose. To discuss the definition, existence, and general properties of complex line integrals. Complex integration is rich in methods, some of them very elegant. In this section we discuss the first two methods, integration by the use of path and (under suitable assumptions given in Theorem 1!) by indefinite integration. Main Content, Important Concepts Definition of the complex line integral Existence Basic properties Indefinite integration (Theorem 1) Integration by the use of path (Theorem 2) Integral of 1>z around the unit circle (basic!) ML-inequality (13) (needed often in our work) Comment on Content Indefinite integration will be justified in Sec. 14.2, after we have obtained Cauchy’s integral theorem. We discuss this method here for two reasons: (i) to get going a little faster and, more importantly, (ii) to answer the students’ natural question suggested by calculus, that is, whether the method works and under what condition—that it does not work unconditionally can be seen from Example 7! SOLUTIONS TO PROBLEM SET 14.1, page 651 2. Straight segment from (3, 1) sloping downward to (6, ⫺2). 4. Parabola y ⫽ 1 ⫺ x 2 (⫺1 ⬉ x ⬉ 1), opening downward. 6. Circle of radius 1 and center (1, 1), oriented clockwise. 8. Quartercircle from (5, 0) to (0, ⫺5) in the fourth quadrant, center 0. 10. Ellipse with coordinate axes as axes, semiaxes of lengths 2 (in x-direction) and 1. 12. First along the x-axis and then along the parallel to the y-axis gives C1:

z 1 ⫽ t 1 (0 ⬉ t 1 ⬉ 2),

C2:

z 2 ⫽ 2 ⫹ it 2,

(0 ⬉ t 2 ⬉ 1).

First along the y-axis and then parallel to the x-axis gives a second path consisting of C3 and C4, C3:

z 3 ⫽ it 3 (0 ⬉ t 3 ⬉ 1),

C4:

z 4 ⫽ t4 ⫹ i

(0 ⬉ t 4 ⬉ 2).

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14. z(t) ⫽ eⴚit (0 ⬉ t ⬉ 2p) Emphasize here, and elsewhere, the advantage of parametric representations that they represent the entire curve, whereas y ⫽ y (x) gives only the upper half, and y r(x) : ⬁ as x : ⫾1, i.e., vertical tangents. Also, essential for our present purpose is the fact that a parametric representation gives an orientation of the curve. 16. z(t) ⫽ 3 cos t ⫹ 2i sin t (0 ⬉ t ⬉ 2p) 18. Hyperbola z(t) ⫽ t ⫹ i>t

(1 ⬉ t ⬉ 5)

20. Ellipse z(t) ⫽ 2 ⫹ 15 cos t ⫹ (⫺1 ⫹ 2 sin t)i

(0 ⬉ t ⬉ 2p)

22. We have Re z ⫽ t

z r ⫽ 1 ⫹ (t ⫺ 1)i

and

so that

冮

Re z dz ⫽

冮 t (1 ⫹ (t ⫺ 1)i) dt ⫽ 4 ⫹ i. 14 3

Compare with Prob. 21. Both paths have the same endpoints, so that the comparison illustrates path dependence, as expected because the integrand is not analytic. 24. First method,

冮

cos 2z dz ⫽ 12 sin 2z `

pi ⴚpi

⫽ sin 2pi ⫽ i sinh 2p ⫽ 267.7i

26. By linearity we can integrate the two terms separately. The integral of z is zero by Theorem 1. The integral of 1>z equals 2pi; see Example 5 in the text. 28. ⫺5 # 2p ⫺ 0 by Example 6. 1

30. Re z 2 ⫽ x 2 ⫺ y 2. (1) Upward, z(t) ⫽ it, z(t) ⫽ i,

冮 ⫺t i dt ⫽ ⫺ 13 i 2

(2) To the right, z(t) ⫽ t ⫹ i, z(t) ⫽ 1,

冮 (t ⫺ 1) dt ⫽ ⫺ 13 ⫺ 1 2

(3) Down, z (t) ⫽ 1 ⫹ it, t goes from 1 to 0, z (t) ⫽ i,

冮 (1 ⫺ t )i dt ⫽ i a⫺1 ⫹ 13 b 2

冮

1 # (4) To the left, z (t) ⫽ t, t goes from 1 to 0, z(t) ⫽ 1, t 2 dt ⫽ ⫺ . 3 1 Answer: ⫺1 ⫺ i 34. Team Experiment. (b) (i) 12.8i, (ii) 12(e2⫹4i ⫺ 1) (c) The integral of Re z equals 21 p2 ⫺ 2ai. The integral of z equals 12p2. The integral of Re (z 2) equals p3>3 ⫺ pa 2>2 ⫺ 2api. The integral of z 2 equals p3>3. (d) The integrals of the four functions in (c) have, for the present paths, the values 1 2 2 api, 0, (4a ⫺ 2)i>3, and ⫺2i>3, respectively. Parts (c) and (d) may also help in motivating further discussions on path independence and the principle of deformation of path.

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SECTION 14.2. Cauchy’s Integral Theorem, page 652 Purpose. To discuss and prove the most important theorem in this chapter, Cauchy’s integral theorem, which is basic by itself and has various basic consequences to be discussed in the remaining sections of the chapter. Main Content, Important Concepts Simply connected domain Cauchy’s integral theorem, Cauchy’s proof (Goursat’s proof in App. 4) Independence of path Principle of deformation of path Existence of indefinite integral Extension of Cauchy’s theorem to multiply connected domains

SOLUTIONS TO PROBLEM SET 14.2, page 659 2. (a) z ⫽ 0 outside C. (b) z ⫽ 0, ⫾4i outside C. 4. No, it would contradict the deformation principle. # 6. (a) z(t) ⫽ (1 ⫹ i)t (0 ⬉ t ⬉ 1), z(t) ⫽ 1 ⫹ i, hence

冮

ez dz ⫽

冮e

(1 ⫹ i) dt ⫽ e(1⫹i)t ƒ 0 ⫽ e1⫹i ⫺ 1.

(1⫹i)t

In (b) we can choose

z 1(t 1) ⫽ t 1 (0 ⬉ t 1 ⬉ 1), z 1 ⫽ 1 z 2(t 2) ⫽ 1 ⫹ it 2 (0 ⬉ t 2 ⬉ 1),

z2 ⫽ i

and obtain the corresponding integrals 1

冮 e dt ⫹ 冮 e t1

1⫹it2

i dt 2 ⫽ e ⫺ 1 ⫹ e1⫹i ⫺ e

⫽ e1⫹i ⫺ 1. 2z ⫹ 3i 4 2 ⫽ ⫺ . From this, the principle of 2 1 z ⫹4 z ⫺ i>2 z ⫹ i>2 deformation of path, and (3) we obtain the answer

8. Team Experiment. (b) (i)

4 ⴢ 2pi ⫺ 2 ⴢ 2pi ⫽ 4pi. (ii) Similarly, 1

z⫹1 2 2 ⫽ ⫹ . 2 z z⫹2 z ⫹ 2z

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Now z ⫽ ⫺2 lies outside the unit circle. Hence the answer is 12 ⴢ 2pi ⫽ pi. (c) The integral of z, Im z, z 2, Re z 2, Im z 2 equals 12, a>6, 13, 13 ⫺ a 2>30 ⫺ ia>6, a>6 ⫺ ia 2>30, respectively. Note that the integral of Re z 2 plus i times the integral of Im z 2 must equal 13 . Of course, the student should feel free to experiment with any functions whatsoever. 10. 0, yes. The points 14 z ⫽ (2n ⫹ 1)p>2, thus z ⫽ (4n ⫹ 2)p, at which cos 14 z ⫽ 0, lie outside the unit circle. 2p

12.

冮 e

ⴚ3it

ⴚ

ieit dt ⫽ ⫺12 e 2it ƒ 0 ⫽ 0, no

14.

冮 e ie dt ⫽ 0 it

16.

1> p 1 1 ⫽ . Answer p 2pi ⫽ 2i by deformation and (11) in Sec. 14.1. pz ⫺ 1 z ⫺ 1> p 1 4

⫽ , so that the integral equals 14 ⴢ 2pi ⫽ 12 pi by the principle of z ⫺ 34 4z ⫺ 3 deformation of path. The theorem does not apply. 20. Sketch C to see what is going on. Ln (1 ⫺ z) is not analytic at z ⫽ 1 and on the portion x ⬎ 1 of the real axis. Since this lies outside the contour, the integral is 0 by Cauchy’s theorem, which applies. 18.

22.

冮 x dx ⫽ 0, z(t) ⫽ e

(0 ⬉ t ⬉ p); hence the integral over the semicircle is

ⴚ1

冮 (cos t)ie dt ⫽ i 冮 (e ⫹ e )e dt ⫽ i [0 ⫹ p] ⫽ pi. 1 2

ⴚit

1 2

24. We obtain 1

1 2 2 ⫺ ⫽ . z⫺1 z2 ⫺ 1 z⫹1 For the first fraction, integration around the left loop gives 0 and around the right loop ( 12 )2pi ⫽ pi. For the second fraction, with the minus sign, clockwise integration around the left loop gives (⫺1)(⫺12)2pi ⫽ pi and for the integration around the right loop 0. Answer 2pi. 26. 0 because the points ⫾4npi at which sinh 12 z ⫽ 0 all lie outside the contour of integration, so that Cauchy’s theorem applies. 28. 0 because z ⫽ ⫾p, ⫾2p, Á as well as ⫾2 and ⫾2i lie outside the contour C. 30. The integrand equals 1 1 2z 1 1 ⫹ ⫹ . ⫹ 2⫽ z z ⫹4 z z ⫹ 2i z ⫺ 2i 2

The integrals of the three terms on the right equal ⫺2pi, ⫺2pi, 0, respectively, so that the answer is ⫺4pi, the minus sign arising because we integrate clockwise.

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SECTION 14.3. Cauchy’s Integral Formula, page 660 Purpose. To prove, discuss, and apply Cauchy’s integral formula, the second major consequence of Cauchy’s integral theorem (the first being the justification of indefinite integration). Comment on Examples The student has to find out how to write the integrand as a product f (z) times 1>(z ⫺ z 0), and the examples (particularly Example 3) and problems are designed to give help in that technique. SOLUTIONS TO PROBLEM SET 14.3, page 663 2. The integrand is not analytic at ⫾1. z ⫽ 1 lies inside the contour, z ⫽ ⫺1 outside. Hence, by Cauchy’s formula, the integral has the value 2pi ⴢ

z2 1 ` ⫽ 2pi ⴢ ⫽ pi. z ⫹ 1 z⫽1 2

4. z ⫽ ⫺1 lies inside C, z ⫽ 1 outside. Hence Cauchy’s formula gives for z2 z2 1 ⴢ ⫽ 2 z ⫺1 z⫺1 z⫹1

2pi ⴢ

the value

z2 1 ` ⫽ 2pi ⴢ a⫺ b ⫽ ⫺pi. z ⫺ 1 z⫽ⴚ1 2

Note that by the deformation principle this must be the same value as in Prob. 1. 6. The given function e2z> p e2 z ⫽ pz ⫺ i z ⫺ i> p is not analytic at z ⫽ i> p inside the unit circle. By Cauchy’s formula the integral equals 2pi 2i>p 2 2 ⫽ 2i acos p ⫹ i sin p b ⫽ ⫺1.189 ⫹ 1.608i. pe 8. From z 2 sin z 4z ⫺ 1

(z 2 sin z)>4

⫽

z ⫺ 14

we obtain by Cauchy’s formula 1 2pi ⴢ 14 z 2 sin z ƒ z⫽1/4 ⫽ 32 pi sin 14 ⫽ 0.02429i.

10. Team Project. (a) Eq. (2) is

冯

z3 ⫺ 6

i 3 dz ⫽ a b ⫺ 6d c 2 z ⫺ 12 i ⫽ a⫺

冯 z⫺ i⫹ 冯 dz

i ⫺ 6b 2pi ⫹ 8

1 2

z 3 ⫺ (12 i) z ⫺ 12 i

冯 az ⫹ 12 iz ⫺ 14 b dz ⫽ 14 p ⫺ 12pi 2

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because the last integral is zero by Cauchy’s integral theorem. The result agrees with that in Example 2, except for a factor 2. (b) Using (12) in App. A3.1, we obtain (2) in the form

冯 z⫺ p sin z 1 2

dz ⫽ (sin 12 p)

冯 z⫺ p 冯 dz

1 2

⫽ 2pi ⫹

冯

⫹

sin z ⫺ sin 12 p z ⫺ 12 p

2 sin (12 z ⫺ 14 p) cos (12 z ⫹ 14 p) z ⫺ 12 p

dz.

As r in Fig. 357 approaches 0, the integrand approaches 0. Indeed, the expression 2 sin (21 z ⫺ 14 p) z ⫺ 12 p approaches 1 (calculus!), whereas the cosine factor approaches cos 12 p, which is 0. 12. The integrand is z z . ⫽ z 2 ⫹ 4z ⫹ 3 (z ⫹ 1)(z ⫹ 3) z ⫽ ⫺1 lies inside the contour, z ⫽ ⫺3 outside. Accordingly, Cauchy’s integral formula gives 2pi

z ⫺1 ` ⫽ 2pi ⫽ ⫺pi. z ⫹ 3 z⫽ⴚ1 2

14. z ⫽ 0 lies inside the contour. The solutions of ez ⫺ 2i ⫽ 0 lie outside because ez ⫽ 2i, z ⫽ ln 2i ⫾ 2npi and ƒ ln 2i ƒ ⬎ ln 2 ⬎ 0.6. We thus obtain the answer 2pi

ez 2pi 2 ⫽ ⫽ ⫺0.8p ⫹ 0.4pi ⫽ ⫺2.51 ⫹ 1.26i. z e ⫺ 2i z⫽0 1 ⫺ 2i

16. 2pi tan i ⫽ ⫺2p tanh 1 ⫽ ⫺4.785 18. 4z 2 ⫺ 8iz ⫽ 4z(z ⫺ 2i) ⫽ 0 at z ⫽ 2i is in the “annulus” in the figure and z ⫽ 0 is not. Hence

冯 4z ⫺ 8iz dz ⫽ 2pi ⴢ 14 ⴢ z ` sin z

sin z

⫽ z⫽2i

pi 4

sinh 2 ⫽ 2.849i

20. For z 1 ⫽ z 2 use Example 6 in Sec. 14.1 with m ⫽ ⫺2. For z 1 ⫽ z 2 use partial fractions 1 1 1 ⫽ ⫺ . (z ⫺ z 1)(z ⫺ z 2) (z 1 ⫺ z 2)(z ⫺ z 1) (z 1 ⫺ z 2)(z ⫺ z 2)

SECTION 14.4. Derivatives of Analytic Functions, page 664 Purpose. To discuss and apply the most important consequence of Cauchy’s integral formula, the theorem on the existence and form of the derivatives of an analytic function.

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Main Content Formulas for the derivatives of an analytic function (1) Cauchy’s inequality Liouville’s theorem Morera’s theorem (inverse of Cauchy’s theorem) Comments on Content Technically the application of the formulas for derivatives in integration is practically the same as that in the last section. The basic importance of (1) in giving the existence of all derivatives of an analytic function is emphasized in the text. SOLUTIONS TO PROBLEM SET 14.4, page 667 2. We obtain z6

z6 ⫽ . (2z ⫺ 1)6 26(z ⫺ 12)6 Hence the solution is 2pi 1 2pi 1 3pi ⴢ 6 ⴢ (z 6)(5) ` ⴢ 6 ⴢ 6!z ` . ⫽ ⫽ 5! 2 5! 2 32 z⫽1>2 z⫽1>2 4. We have to differentiate twice, obtaining 2pi z (e cos z) s ` 2! z⫽ p>4

⫽ piez(cos z ⫺ sin z) r

⫽ piez(cos z ⫺ sin z ⫺ sin z ⫺ cos z) ⫽ ⫺2ez sin z.

Evaluation at z ⫽ p>4 gives ⫺22ep>4 ⫽ ⫺3.102. 6. z ⫽ 12 lies within the contour, z ⫽ 2i lies outside. Accordingly, differentiating once, we obtain 2pi c

1 (z ⫺ 2i)2

r ⴚ3 d ⫽ 2pi(⫺2)(z ⫺ 2i) ƒ z⫽i>2 ⫽ ⫺4pi(⫺3i>2)ⴚ3 ⫽⫺

32 p. 27

8. We have to differentiate twice, so that (1 s ) gives 2pi 3 (z ⫹ sin z) s ⫽ pi(6z ⫺ sin z) ƒ z⫽i 2! ⫽ pi(6i ⫺ sin i) ⫽ ⫺6p ⫹ p sinh 1 ⫽ ⫺15.158.

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10. We have to differentiate once and obtain from (1 r ) 2pi a

4z 3 ⫺ 6 r 6 ⫽ 2pi a8z ⫹ 2 b ` b ` z z z⫽1⫹i z⫽1⫹i ⫽ 2pi(8 ⫹ 8i ⫹ 6>2i) ⫽ 2pi(8 ⫹ 5i) ⫽ p(⫺10 ⫹ 16i) ⫽ ⫺31.42 ⫹ 50.27i.

12. We have to differentiate once and obtain from (1), with the minus sign because we integrate clockwise, ⫺2pi a

exp (z 2) r 1 ⫽ ⫺2pi a2 exp (z 2) ⫺ 2 exp (z 2)b ` b ` z z z⫽2i z⫽2i ⫽ ⫺2pieⴚ4 (2 ⫹ 14 ) ⫽ ⫺92 peⴚ4i ⫽ ⫺0.2589i.

14. Differentiating once, we obtain from (1), since z ⫽ ⫺1 lies inside the contour and z ⫽ 2 outside, 2pi a

Ln (z ⫹ 3) Ln (z ⫹ 3) r 1 b ` ⫺ ⫽ 2pi a b` (z ⫺ 2)2 z⫺2 (z ⫹ 3)(z ⫺ 2) z⫽ⴚ1 z⫽ⴚ1 ⫽ 2pi a

Ln 2 1 ⫺ b ⫺6 9

⫽ ⫺1.531i. 16. z ⫽ 0 lies outside the “annulus” bounded by the two circles of C. The point z ⫽ 2i lies inside the larger circle but outside the smaller. Hence the integral equals (differentiate once) 2pi a

e2z e4i 2e2z 2e4i 1 ⫺ 2b` ⫺ b ⫽ pe4i a2 ⫹ ib . ⫽ 2pi a z 2i ⫺4 2 z z⫽2i

18. 0 when n is a negative integer or 0, by Cauchy’s integral theorem. Also 0 when n ⫽ 1, 3, 5, Á because then we have in (1) an even number of differentiations, which reproduces sinh z, which is 0 at z ⫽ 0. When n ⫽ 2, 4, Á , We have to differentiate an odd number of times, producing cosh z, which is 1 at z ⫽ 0, so that by (1) the answer is 2pi>(n ⫺ 1)! 20. Team Project. (a) If no such z existed, we would have ƒ f (z) ƒ ⬉ M for every ƒ z ƒ , which means that the entire function f (z) would be bounded, hence a constant by Liouville’s theorem. cnⴚ1 c0 ⫹ Á ⫹ n b, cn ⫽ 0, n ⬎ 0. (b) Let f (z) ⫽ c0 ⫹ c1z ⫹ Á ⫹ cn z n ⫽ z n acn ⫹ z z Set ƒ z ƒ ⫽ r. Then ƒ f (z) ƒ ⬎ r n a ƒ cn ƒ ⫺

ƒ cnⴚ1 ƒ ƒ c0 ƒ ⫺Á⫺ nb r r

and ƒ f (z) ƒ ⬎ 12 r n ƒ cn ƒ for sufficiently large r. From this the result follows.

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(c) ƒ ez ƒ ⬎ M for real z ⫽ x with x ⬎ R ⫽ ln M. On the other hand, ƒ ez ƒ ⫽ 1 for any pure imaginary z ⫽ iy because ƒ eiy ƒ ⫽ 1 for any real y (Sec. 13.5). (d) If f (z) ⫽ 0 for all z, then g ⫽ 1>f would be analytic for all z. Hence by (a) there would be values of z exterior to every circle ƒ z ƒ ⫽ R at which, say, ƒ g (z) ƒ ⬎ 1 and thus ƒ f (z) ƒ ⬍ 1. This contradicts (b). Hence f (z) ⫽ 0 for all z cannot hold. SOLUTIONS TO CHAPTER 14 REVIEW QUESTIONS AND PROBLEMS, page 668 18. Not in general. 20. The ML-inequality gives such a bound. 22. The integral of z is zero. ƒ z ƒ is not analytic, so we must use a representation z (t) ⫽ eⴚit (0 ⬉ t ⬉ 2p) of the path, obtaining

冯

ƒ z ƒ dz ⫽

24. z(t) ⫽ t ⫹ it 3,

冮 1 ⴢ (⫺ie ) dt ⫽ e ƒ ⴚit

ⴚit 2p t⫽0

⫽ 0.

(0 ⬉ t ⬉ 3), z ⫽ 1 ⫹ 3it 2,

Re z (t) ⫽ t, hence

冮 t(1 ⫹ 3it ) dt ⫽ 4.5 ⫹ 60.75i . 2

26. z 2 ⫹ z 2 ⫽ 2(x 2 ⫺ y 2). From 0 to 2 we have z ⫽ x ⫽ t (0 ⬉ t ⬉ 2), dz ⫽ dt, and thus

冮

2 2

2t dt ⫽

2 3 2 16 t ` ⫽ . 3 0 3

From 2 to 2 ⫹ i we have z ⫽ 2 ⫹ it (0 ⬉ t ⬉ 2), dz ⫽ i dt, and thus 2

冮 2(2 ⫺ t )i dt ⫽ 323 i . 2

2pi 28. 2pi(Ln z) r ƒ z⫽2i ⫽ z ` ⫽p z⫽2i (1⫹i)p

30. ⫺cos z ƒ 0

⫽ cosh p ⫹ 1

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CHAPTER 15

Power Series, Taylor Series

Power series and, in particular, Taylor series, play a much more fundamental role in complex analysis than they do in calculus. The student may do well to review what has been presented about power series in calculus but should become aware that many new ideas appear in complex, mainly owing to the use of complex integration.

SECTION 15.1. Sequences, Series, Convergence Tests, page 671 Purpose. The beginnings on sequences and series in complex is similar to that in calculus (differences between real and complex appear only later). Hence this section can almost be regarded as a review from calculus plus a presentation of convergence tests for later use. Main Content, Important Concepts Sequences, series, convergence, divergence Comparison test (Theorem 5) Ratio test (Theorem 8) Root test (Theorem 10)

SOLUTIONS TO PROBLEM SET 15.1, page 679 2. Convergent to 0, hence bounded, since n! grows faster than ƒ 3 ⫹ 4i ƒ n ⫽ 5n. Convergence implies that there are no limit points other than the limit itself. Note that z n is the nth term of the Maclaurin series of e3⫹4i. 4. Unbounded, hence divergent, ƒ z n ƒ ⫽ 5n>2 6. Unbounded since cos npi ⫽ cosh np ⬎ n hence divergent 8. Divergent; all terms have absolute value 1, and a graph suggests that every point on the unit circle is a limit point. 10. Bounded, divergent, ⫾1> 12 ⫾ i, 0, 1, ⫺2 12. For any P ⬎ 0 there is an N ⫽ N(P) such that ƒ z n ⫺ l ƒ ⬍ P>3,

ƒ z n* ⫺ l* ƒ ⬍ P>3.

Hence by the triangle inequality, for all n ⬎ N we have * ⫺ l*) ƒ ƒ (z n ⫹ z n*) ⫺ (l ⫹ l*) ƒ ⫽ ƒ (z n ⫺ l) ⫹ (z n ⬉ ƒ z n ⫺ l ƒ ⫹ ƒ z n* ⫺ l* ƒ ⬉ P>3 ⫹ P>3 ⬍ P. This proof is typical of many similar ones. Much less important is termwise multiplication of sequences, but a similar theorem holds true for this case. Namely, under the assumptions just made on the convergence of the two sequences, it follows that the sequence z 1z *1 , z 2z *2 , Á is convergent and has the limit ll*. The proof is more complicated, as follows. 262

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The two sequences are bounded, ƒ z n ƒ ⬍ K, ƒ z n* ƒ ⬍ K. Since they converge, for an P ⬎ 0 there is an N ⫽ N(P) (such that ƒ z n ⫺ l ƒ ⬍ P>(3K), ƒ z n* ⫺ l* ƒ ⬍ P>(3 ƒ l ƒ ) (l ⫽ 0; the case l ⫽ 0 is rather trivial), hence ƒ z n z n* ⫺ ll* ƒ ⫽ ƒ (z n ⫺ l )z n* ⫹ (z n* ⫺ l*)l ƒ * ƒ ⫹ ƒ zn * ⫺ l* ƒ ƒ l ƒ ⬉ ƒ zn ⫺ l ƒ ƒ zn ⬍ P>3 ⫹ P>3 ⬍ P (n ⬎ N ). 16. Convergent. In connection with Maclaurin series in Sec. 15.4 the sum will turn out to be e20⫹30i, of absolute value 0.5 # 109. 18. Divergent 20. Divergent because `

n⫹i 1 1 ⫹ i>n 3 ` ⫽ ` ` ⬎ 3n 2 ⫹ 2i 3n 1 ⫹ 2i>(3n 2) n

and the harmonic series diverges. 22. Divergent because 1> 1n ⬎ 1>n for n ⫽ 2, 3, Á and the harmonic series diverges. 24. Divergent by the ratio test because `

(3i)n⫹1(n ⫹ 1)!>(n ⫹ 1)n⫹1 z n⫹1 ` ⫽ ` ` (3i)nn!>n n zn nn

⫽ 3(n ⫹ 1)

(n ⫹ 1)n⫹1 n

⫽3a

n b n⫹1

⫽

n⫹1 a n b 3 1 a1 ⫹ n b

3 e ⬎ 1.

26. It is essential that, from some n on, the test ratio does not become greater than a fixed q ⬍ 1 instead of coming arbitrarily close to 1, as is the case, for instance, for the harmonic series, which diverges. 30. The form of the estimate of Rn suggests we use the fact that the ratio test is a comparison test based on the geometric series. This gives z n⫹2 z n⫹3 Rn ⫽ z n⫹1 ⫹ z n⫹2 ⫹ Á ⫽ z n⫹1 a1 ⫹ z ⫹z ⫹ Áb, n⫹1

z n⫹2 `z ` ⬉ q, n⫹1

n⫹1

z n⫹3 z n⫹3 z n⫹2 `z ` ⫽ `z ` ⬉ q 2, n⫹1 n⫹2 z n⫹1

ƒ Rn ƒ ⬉ ƒ z n⫹1 ƒ (1 ⫹ q ⫹ q 2 ⫹ Á ) ⫽

ƒ z n⫹1 ƒ 1⫺q

etc., .

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For the given series we obtain the test ratio 1 n⫹1⫹i n n (n ⫹ 1)2 ⫹ 1 ` ⴢ ` ⫽ 2 n⫹1 n⫹i 2(n ⫹ 1) B n 2 ⫹ 1 ⫽

1 n 4 ⫹ 2n 3 ⫹ 2n 2 1 ⬍ ; 4 2 B n ⫹ 2n 3 ⫹ 2n 2 ⫹ 2n ⫹ 1 2

from this with q ⫽ 12 we have ƒ Rn ƒ ⬉

ƒ z n⫹1 ƒ 1⫺q

⫽

ƒn ⫹ 1 ⫹ iƒ 2n(n ⫹ 1)

⫽

2(n ⫹ 1)2 ⫹ 1 2n(n ⫹ 1)

⬍ 0.05.

Hence n ⫽ 5 (by computation), and s⫽

31 661 ⫹ i ⫽ 0.96875 ⫹ 0.688542i. 32 960

Exact to 6 digits is 1 ⫹ 0.693147i. SECTION 15.2. Power Series, page 680 Purpose. To discuss the convergence behavior of power series, which will be basic to our further work (and which is simpler than that of series having arbitrary complex functions as terms). Comment. Most complex power series appearing in practical work and applications have real coefficients because most of the complex functions of practical interest are obtained from calculus by replacing the real variable x with the complex variable z ⫽ x ⫹ iy, retaining the real coefficients. Accordingly, in the problem set we consider primarily power series with real coefficients, also because complex coefficients would neither provide additional difficulties nor contribute new ideas. Proof of the Assertions in Example 6 苲苲苲苲 R ⫽ 1>L follows from R ⫽ 1> l by noting that in the case of convergence, L ⫽ l (the 苲 only limit point). l exists by the Bolzano–Weierstrass theorem, assuming boundedness n n of {2 ƒ an ƒ }. Otherwise, 2 ƒ an ƒ ⬎ K for infinitely many n and any given K. Fix z ⫽ z 0 and take K ⫽ 1> ƒ z ⫺ z 0 ƒ to get n

2 ƒ an(z ⫺ z 0)n ƒ ⬎ K ƒ z ⫺ z 0 ƒ ⫽ 1 and divergence for every z ⫽ z 0 by Theorem 9, Sec. 15.1. Now, by the definition of a limit point, for a given P ⬎ 0 we have, for infinitely many n, n 苲苲 l ⫺ P ⬍ 2ƒ an ƒ ⬍ l ⫹ P;

hence for all z ⫽ z 0 and those n, (*)

n 苲苲 (l ⫺ P) ƒ z ⫺ z 0 ƒ ⬍ 2 ƒ an(z ⫺ z 0)n ƒ ⬍ (l ⫹ P) ƒ z ⫺ z 0 ƒ .

The right inequality holds even for all n ⬎ N (N sufficiently large), by the definition of a greatest limit point.

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n 苲 Let l ⫽ 0 . Since 2 ƒ an ƒ ⭌ 0, we then have convergence to 0. Fix any z ⫽ z 1 ⫽ z 0. n Then for P ⫽ 1>(2 ƒ z 1 ⫺ z 0 ƒ ) ⬎ 0 there is an N such that 2 ƒ an ƒ ⬍ P for all n ⬎ N; hence

ƒ an(z 1 ⫺ z 0)n ƒ ⬎ Pn ƒ z 1 ⫺ z 0 ƒ n ⫽

1 , 2n

and convergence for all z 1 follows by the comparison test. 苲苲 Let l ⬎ 0. We establish 1>l as the radius of convergence of (1) by proving 苲 convergence of the series (1) if ƒ z ⫺ z 0 ƒ ⬍ 1>l , 苲 divergence of the series (1) if ƒ z ⫺ z 0 ƒ ⬎ 1>l . 苲苲 Let ƒ z ⫺ z 0 ƒ ⬍ 1>l . Then, say, ƒ z ⫺ z 0 ƒ l ⫽ 1 ⫺ b ⬍ 1. With this and P ⫽ b>(2 ƒ z ⫺ z 0 ƒ ) ⬎ 0 in (*), for all n ⬎ N, n 苲 2 ƒ an(z ⫺ z 0)n ƒ ⬍ l ƒ z ⫺ z 0 ƒ ⫹ P ƒ z ⫺ z 0 ƒ ⫽ 1 ⫺ b ⫹ 12 b ⬍ 1. Convergence now follows from Theorem 9, Sec. 15.1. 苲苲 Let ƒ z ⫺ z 0 ƒ ⬎ 1>l . Then ƒ z ⫺ z 0 ƒ l ⫽ 1 ⫹ c ⬎ 1. With this and P ⫽ c>(2 ƒ z ⫺ z 0 ƒ ) ⬎ 0 in (*), for infinitely many n, n 苲 2 ƒ an(z ⫺ z 0)n ƒ ⬎ l ƒ z ⫺ z 0 ƒ ⫺ P ƒ z ⫺ z 0 ƒ ⫽ 1 ⫹ c ⫺ 12 c ⬎ 1, and divergence follows.

SOLUTIONS TO PROBLEM SET 15.2, page 684 ⴥ 1 4. a n (z ⫺ 4 ⫹ 3pi)n 6 n⫽0 6. Center ⫺1, radius of convergence 14 because in (6)

n⫹1 ⫽

1 . 4

8. The center is pi. In (6) we obtain n n>n!

(n ⫹ 1)n⫹1>(n ⫹ 1)!

⫽

n n(n ⫹ 1) (n ⫹ 1)n⫹1

⫽

1 1 ⫽ n⫹1 n 1 n a n b a1 ⫹ n b

1 . e

10. Center 2i, radius of convergence ⬁ because n

1>n n n⫹1 1 ⫽ (n ⫹ 1) a b ⫽ (n ⫹ 1) a1 ⫹ b 1>(n ⫹ 1)n⫹1 n n

⬁.

12. Center 0, radius of convergence 8 14. Center 0, radius of convergence ⬁ because, by (6), 1>(2n(n!)2) 1>(22n⫹2(n ⫹ 1)2(n!)2)

⫽ 4(n ⫹ 1)2

⬁.

From Sec. 5.5 you see that this is the complex analog of the Maclaurin series of the Bessel function J0 (x), so that is the Maclaurin series of J0 (z) for complex z, as will follow in Sec. 15.4.

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16. Center 0. From (6) we obtain the radius of convergence 2>27 because (3n)!>(2n(n!)3) (n ⫹ 1)3 ⫽ 2 (3n ⫹ 3)!>(2n⫹1((n ⫹ 1)!)3) (3n ⫹ 3)(3n ⫹ 2)(3n ⫹ 1)

2 . 27

18. Center 0, radius of convergence ⬁ . We mention that this is the series (36) of the error function in App. 3.1, extended to a complex variable z. Formula (6) gives (2n ⫹ 3)(n ⫹ 1)! (2n ⫹ 3)(n ⫹ 1) ⫽ (2n ⫹ 1)n! 2n ⫹ 1

⬁.

20. Team Project. (a) The faster the coefficients go to zero, the larger ƒ an>an⫹1 ƒ becomes. (b) (i) Nothing. (ii) R is multiplied by 1>k. (iii) The new series has radius of convergence 1>R. (c) In Example 6 we took the first term of one series, then the first term of the other, and so on alternately. We could have taken, for instance, the first three terms of one series, then the first five terms of the other, then again three terms and five terms, and so on; or we could have mixed three or more series term by term. (d) No, because ƒ 30 ⫹ 10i ƒ ⬎ ƒ 31 ⫺ 6i ƒ . SECTION 15.3. Functions Given by Power Series, page 685 Purpose. To show what operations on power series are mathematically justified and to prove the basic fact that power series represent analytic functions. Main Content Termwise addition, subtraction, and multiplication of power series Termwise differentiation and integration (Theorems 3, 4) Analytic functions and derivatives (Theorem 5) Comment on Content That a power series is the Taylor series of its sum will be shown in the next section. SOLUTIONS TO PROBLEM SET 15.3, page 689 6. ƒ z>(2p) ƒ 2 ⬍ 1 by integrating the geometric series. Thus ƒ z ƒ ⬍ R ⫽ 2p. 8. 15 , where 1>(n(n ⫹ 1)) can be produced by two integrations of the geometric series. 10. The binomial coefficient n(n ⫺ 1) Á (n ⫺ k ⫹ 1) n a b⫽ k! k consists of the fixed k!, which has no effect on R, and factors n(n ⫺ 1) Á (n ⫺ k ⫹ 1) as obtained by differentiation. Since g(z>2)n has R ⫽ 2, the answer is 2. 12. ⬁, because 2n(2n ⫺ 1) results from differentiation, and for the coefficients without these factors we have in the Cauchy–Hadamard formula n

n⫹1 1>n n ⫽a b (n ⫹ 1) 1>(n ⫹ 1)n⫹1 n

⬁

as n : ⬁.

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14. 1, by applying Theorem 3 to gz n⫹m. 16. This follows from f (z) ⫽ a0 ⫹ a1z ⫹ a2 z 2 ⫹ a3 z 3 ⫹ Á ⫽ f (⫺z) ⫽ a0 ⫺ a1z ⫹ a2 z 2 ⫺ a3 z 3 ⫹ ⫺ Á . 18. This is a useful formula for binomial coefficients. It follows from p

p q (1 ⫹ z)p(1 ⫹ z)q ⫽ a a b z n a a b z m n m n⫽0 m⫽0 p⫹q

⫽ (1 ⫹ z)p⫹q ⫽ a a

p⫹q

r⫽0

b zr

by equating the coefficients of z on both sides. To get z nz m ⫽ z r on the left, we must have n ⫹ m ⫽ r; thus m ⫽ r ⫺ n, and this gives the formula in the problem. 20. Team Project. (a) Division of the recursion relation by an gives r

an⫹1 anⴚ1 an ⫽ 1 ⫹ an . Take the limit on both sides, denoting it by L: L⫽1⫹

1 . L

Thus L2 ⫺ L ⫺ 1 ⫽ 0, L ⫽ (1 ⫹ 15)>2 ⫽ 1.618, an approximate value reached after just ten terms. (b) The list is 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233. In the recursion, an is the number of pairs of rabbits present and anⴚ1 is the number of pairs of offsprings from the pairs of rabbits present at the end of the preceding month. (c) Using the hint, we calculate ⴥ

ⴥ

n⫽0

(1 ⫺ z ⫺ z 2) a anz n ⫽ a (an ⫺ anⴚ1 ⫺ anⴚ2)z n ⫽ 1 where aⴚ1 ⫽ aⴚ2 ⫽ 0, and Theorem 2 gives a0 ⫽ 1, a1 ⫺ a0 ⫽ 0, an ⫺ anⴚ1 ⫺ anⴚ2 ⫽ 0 for n ⫽ 2, 3, Á . The converse follows from the uniqueness of a power series representation (see Theorem 2). SECTION 15.4. Taylor and Maclaurin Series, page 690 Purpose. To derive and explain Taylor series, which include those for real functions known from calculus as special cases. Main Content Taylor series (1), integral formula (2) for the coefficients Singularity, radius of convergence Maclaurin series for ez, cos z, sin z, cosh z, sinh z, Ln (1 ⫹ z) Theorem 2 connecting Taylor series to the last section

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Comment The series just mentioned, with z ⫽ x, are familiar from calculus. SOLUTIONS TO PROBLEM SET 15.4, page 697 ⴥ

4. (2 ⫹ z) a z 2n ⫽ 2 ⫹ z ⫹ 2z 2 ⫹ z 3 ⫹ 2z 4 ⫹ Á ,

R⫽1

n⫽0

ⴥ 1 ⫽ a (⫺3i)nz n ⫽ 1 ⫺ 3iz ⫺ 9z 2 ⫹ 27iz 3 ⫹ Á , 1 ⫺ (⫺3iz) n⫽0

2 6 1 8. 12 ⫺ 12 cos 2z ⫽ z 2 ⫺ 13 z 4 ⫹ 45 z ⫺ 315 z8 ⫹ ⫺ Á ,

R ⫽ 13

R⫽⬁

10. The series is f⫽z⫹

2 3 22 23 z ⫹ z5 ⫹ z7 ⫹ Á , 1ⴢ3 1ⴢ3ⴢ5 1ⴢ3ⴢ5ⴢ7

R ⫽ ⬁.

It can be obtained in several ways. (a) Integrate the Maclaurin series2 of the integrand termwise and form the Cauchy product with the series of ez . (b) f satisfies the differential equation f r ⫽ 2z f ⫹ 1. Use this, its derivatives f s ⫽ 2( f ⫹ z f r ), etc., f (0) ⫽ 0, f r (0) ⫽ 1, etc., and the coefficient formulas in (1). (c) Substitute ⴥ

ⴥ

n⫽0

f ⫽ a anz n and f r ⫽ a nanz nⴚ1 into the differential equation and compare coefficients; that is, apply the power series method (Sec. 5.1). 12. z ⫺

z9 z 13 z5 ⫹ ⫺ ⫹⫺Á; R⫽ ⬁ 2!5 4!9 6!13

14. z ⫺

z5 z7 z3 ⫹ ⫺ ⫹⫺Á; R⫽ ⬁ 3!3 5!5 7!7

16. First of all, since sin (w ⫹ 2p) ⫽ sin w and sin (p ⫺ w) ⫽ sin w, we obtain all values of sin w by letting w vary in a suitable vertical strip of width p, for example, in the strip ⫺p>2 ⬉ u ⬉ p>2. Now since sin a

p 2

⫺ iyb ⫽ sin a

p 2

⫹ iyb ⫽ cosh y

and sin a⫺

p 2

⫺ iyb ⫽ sin a⫺

p 2

⫹ iyb ⫽ ⫺cosh y,

we have to exclude a part of the boundary of that strip, so we exclude the boundary in the lower half-plane. To solve our problem we have to show that the value of the series lies in that strip. This follows from ƒ z ƒ ⬍ 1 and ` Re az ⫹

3 3 3 1 z 1 z 1 ƒzƒ ⫹ Á b ` ⬉ `z ⫹ ⫹ Á ` ⬉ ƒzƒ ⫹ ⫹Á 2 3 2 3 2 3

⫽ arcsin ƒ z ƒ ⬍

p 2

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18. We obtain from the sum of the geometric series ⴥ 1 ⫽ ⫺i a i n(z ⫺ i)n i(1 ⫺ i(z ⫺ i)) n⫽0

⫽ ⫺i ⫺ i ⴢ i(z ⫺ i) ⫺ i(⫺1)(z ⫺ i)2 ⫺ Á ,

R ⫽ 1.

20. We obtain 1 1 ⫹ cos 2z 2 2 1 1 ⫽ ⫺ cos (2z ⫺ p) 2 2

cos2 z ⫽

⫽

1 4 1 42 1 az ⫺ p b ⫹ ⫺ Á d , c az ⫺ p b ⫺ 2 2! 2 4! 2

1 1 1 (z ⫺ pi)2 ⫹ (z ⫺ pi)4 ⫹ (z ⫺ pi)6 ⫹ Á , 2! 4! 6! 24. We obtain 22. 1 ⫹

R ⫽ ⬁.

R⫽⬁

ez(zⴚ2) ⫽ e(zⴚ1⫹1)(zⴚ1ⴚ1) ⫽ e(zⴚ1) ⴚ1 2

1 ⴥ (z ⫺ 1) ⫽e a n!

n⫽0

1 1 1 ⫽ e a1 ⫹ (z ⫺ 1)2 ⫹ 2! (z ⫺ 1)4 ⫹ 3! (z ⫺ 1)6 ⫹ Á b ,

R ⫽ ⬁.

SECTION 15.5. Uniform Convergence. Optional, page 698 Purpose. To explain the concept of uniform convergence. To show that power series have the advantage that they converge uniformly (exact formulation in Theorem 1). To discuss properties of general uniformly convergent series. Main Content Uniform convergence of power series (Theorem 1) Continuous sum (Theorem 2) Termwise integration (Theorem 3) and differentiation (Theorem 4) Weierstrass test for uniform convergence (Theorem 5) The test in Theorem 5 is very simple, conceptually and technically in its application. SOLUTIONS TO PROBLEM SET 15.5, page 704 2. R ⫽ 7, uniform convergence for ƒ z ƒ ⬉ 7 ⫺ d, d ⬎ 0 4. R ⫽ ⬁, uniform convergence on any bounded set 6. ƒ tanh n 2 ƒ ⫽ 1. Convergence for ƒ z 2 ƒ ⬍ 12 . Uniform convergence for ƒ z ƒ ⫽ 1> 12 ⫺ d, d⬎0 8. R ⫽ 1> 13, uniform convergence for ƒ z ⫺ 1 ƒ ⬉ 1> 13 ⫺ d, d ⬎ 0 10. The Maclaurin series of cosh z converges for all z. Use Theorem 1.

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12. ƒ z n ƒ ⬉ 1, 1>(n 3 cosh n ƒ z ƒ ) ⬉ 1>n 3 ⬍ 1>n 2 (n ⭌ 2) and g 1>n 2 converges. Use the Weierstrass M-test. zn 1 1 14. ` 2n ` ⬉ n ⬉ n and g 2ⴚn converges. Use the Weierstrass M-test. ƒzƒ ƒzƒ ⫹ 1 2 16. ƒ tanh ƒ z ƒ ƒ ⬉ 1 (see App. 3.1), 1>(n(n ⫹ 1)) ⬍ 1>n 2 and g 1>n 2 converges. Use the Weierstrass M-test. 18. Team Project. (a) Convergence follows from the comparison test (Sec. 15.1). Let * be the remainders of (1) and (5), respectively. Since (5) converges, Rn (z) and Rn * ⬍ P for all n ⬎ N (P). Since for given P ⬎ 0 we can find an N (P) such that Rn * and therefore ƒ fm(z) ƒ ⬉ M m for all z in the region G, we also have ƒ Rn(z) ƒ ⬉ Rn ƒ Rn(z) ƒ ⬍ P for all n ⬎ N (P) and all z in the region G. This proves that the convergence of (1) in G is uniform. (b) Since f0r ⫹ f1r ⫹ Á converges uniformly, we may integrate term by term, and the resulting series has the sum F(z), the integral of the sum of that series. Therefore, the latter sum must be F r(z). (c) The converse is not true. (d) Noting that this is a geometric series in powers of q ⫽ (1 ⫹ z 2)ⴚ1, we have q ⫽ ƒ 1 ⫹ z 2 ƒ ⴚ1 ⬍ 1, 1 ⬍ ƒ 1 ⫹ z 2 ƒ 2 ⫽ (1 ⫹ x 2 ⫺ y 2)2 ⫹ 4x 2y 2, the exterior of a lemniscate. The series converges also at z ⫽ 0. (e) We obtain (add and subtract 1) ⴥ

ⴥ 1 2 ⫽ x a⫺1 ⫹ b a 2 m 2 m m⫽1 (1 ⫹ x ) m⫽0 (1 ⫹ x )

x2 a

⫽ ⫺x 2 ⫹ 1⫺

⫽ ⫺x 2 ⫹ 1 ⫹ x 2 ⫽ 1.

1 1 ⫹ x2

20. We obtain 2 ƒ Bn ƒ ⫽ ` L

冮 f (x) sin npL x dx ` ⬍ L2 ML 0

where M is such that ƒ f (x) ƒ ⬍ M on the interval of integration. Thus ƒ Bn ƒ ⬍ K (⫽ 2M). Now when t ⭌ t 0 ⬎ 0, ƒ u n ƒ ⫽ ` Bn sin

2 npx ⴚln2t e ` ⬍ Keⴚlnt0 L

because ` sin

npx ` ⬉ 1 and the exponential function decreases in a monotone fashion L as t increases. From this, ` Consider

2 0u n ` ⫽ ƒ ⫺l2nu n ƒ ⫽ l2n ƒ u n ƒ ⬍ l2nKe⫺lnt0 0t

ⴥ

a lnKe n⫽1

2 ⴚln t0

when t ⭌ t 0.

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Since ln ⫽

271

cnp , for the test ratio we have L 2

2 l2n⫹1K exp (⫺ln⫹1 t 0) n⫹1 cp ⫽a b exp c ⫺(2n ⫹ 1) a b t0 d 2 2 n L lnK exp (⫺lnt 0)

as n : ⬁, and the series converges. From this and the Weierstrass test it follows that 0u n 0u g , etc. converges uniformly and, by Theorem 4, has the sum 0t 0t

SOLUTIONS TO CHAPTER 15 REVIEW QUESTIONS AND PROBLEMS, page 706 12. R ⫽ 14 14. R ⫽ ⬁ 16. R ⫽ 1, ⫺ln (1 ⫺ z) 18. R ⫽ ⬁, sin pz ⴚ1 z b 20. R ⫽ 5, a1 ⫺ 3 ⫹ 4i 1 ⴥ 22. a n(n ⫺ 1)z nⴚ2, R ⫽ 1 2 n⫽2 ⴥ

24. a (⫺pz)n,

R ⫽ 1> p

n⫽0

26. [(z ⫺ i) ⫹ i]4 ⫽ 1 ⫺ 4i(z ⫺ i) ⫺ 6(z ⫺ i)2 ⫹ 4i(z ⫺ i)3 ⫹ (z ⫺ i)4, a binomial expansion readily obtainable from Taylor’s theorem. 28. We obtain 1 ⫽ z ⫺ 2i ⫹ 2i

1 z ⫺ 2i 2i a1 ⫹ b 2i (⫺1)n

ⴥ

⫽ a

n⫹1 n⫽0 (2i)

(z ⫺ 2i)n

⫽ ⫺ 12 i ⫹ 14 (z ⫺ 2i) ⫹ 18 i(z ⫺ 2i)2 ⫺ Á , 30. e

zⴚpi⫹pi

ⴥ

⫽⫺a n⫽0

(z ⫺ pi)n n!

R ⫽ ⬁ ; here we used eⴚpi ⫽ ⫺1.

R ⫽ 2.

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CHAPTER 16

Laurent Series. Residue Integration

This is another powerful and elegant integration method that has no analog in calculus. It uses Laurent series (roughly, series of positive and negative powers of z), more precisely, it uses just a single term of such a series (the term in 1>(z ⫺ z 0), whose coefficient is called the residue of the sum of the series that converges near z 0). SECTION 16.1. Laurent Series, page 708 Purpose. To define Laurent series, to investigate their convergence in an annulus (a ring, in contrast to Taylor series, which converge in a disk), to discuss examples. Major Content, Important Concepts Laurent series Convergence (Theorem 1) Principal part of a Laurent series Techniques of development (Examples 1–5) SOLUTIONS TO PROBLEM SET 16.1, page 714 ⴥ

(⫺1)n ⴚ2nⴚ2 1 1 1 ⴚ10 z ⫽ z ⴚ2 ⫺ z ⴚ4 ⫹ z ⴚ6 ⫺ z ⴚ8 ⫹ z ⫺⫹Á, n! 2 6 24 n⫽0

2. a

0 ⬍ ƒzƒ ⬍ ⬁ ⴥ

(⫺1)np2n⫹1 2nⴚ1 1 1 z ⫽ pz ⴚ1 ⫺ p3z ⫹ p5z 3 ⫺ ⫹ Á , (2n ⫹ 1)! 6 120 n⫽0

4. a

ⴥ 22n⫹1 2nⴚ1 8 32 3 6. a z ⫽ 2z ⴚ1 ⫹ z ⫹ z ⫹ Á, (2n ⫹ 1)! 6 120 n⫽0

0 ⬍ ƒzƒ ⬍ ⬁

8. We obtain m ⴥ 1 ⴥ z 1 z 1 n ⫽ e a m! a z z2 1 ⫺ z z 2 m⫽0 n⫽0

⫽

n 1 ⴥ 1 n ≤z 2 a ¢ a z n⫽0 m⫽0 m!

⫽

1 2 5 8 65 2 ⫹ ⫹ z⫹ z ⫹ Á, 2 ⫹ z 2 3 24 z

0 ⬍ ƒ z ƒ ⬍ R ⫽ 1.

To see how the second line is obtained from the first, write the first few terms of both Maclaurin series one below the other and take n ⫽ 0, then n ⫽ 1, n ⫽ 2, and so on until the form of the second line becomes clear to you. 10. We may proceed as follows: 6 z 2 ⫺ 3i [3 ⫹ (z ⫺ 3)]2 ⫺ 3i 9 ⫺ 3i ⫹ ⫹1 ⫽ ⫽ (z ⫺ 3)2 (z ⫺ 3)2 (z ⫺ 3)2 z⫺3 The next problem (Prob. 11) is of a similar type. 272

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12. Using the binomial series, we obtain 1 z (z ⫺ i) 2

⫽

[i ⫹ (z ⫺ i)]2 (z ⫺ i) nⴚ1 ⴥ ⫺2 (z ⫺ i) ⫽ a a b i nⴚ2 n n⫽0

⫽ ⫺(z ⫺ i)ⴚ1 ⫺ 2i ⫹ 3 (z ⫺ i) ⫹ 4i (z ⫺ i)2 ⫺ Á , convergent for 0 ⬍ ƒ z ⫺ i ƒ ⬍ 1. 14. Using the Maclaurin series of the exponential function, we obtain eaz eab⫹a (zⴚb) ⫽ z⫺b z⫺b ⴥ

an (z ⫺ b)nⴚ1 n! n⫽0

⫽ eab a

⫽ eab((z ⫺ b)ⴚ1 ⫹ a ⫹ 12 a 2 (z ⫺ b) ⫹ 16 a 3 (z ⫺ b)2 ⫹ Á ) convergent for 0 ⬍ ƒ z ⫺ b ƒ ⬍ ⬁. If a ⫽ 0, the series reduces to its first term, which is the principal part. 16. Successive differentiation and use of cos 14 p ⫽ sin 14 p ⫽ 1> 12 gives 1 D 12

1 az ⫺

p 3 4

⫹

p 2

az ⫺

⫺

1 ⫹ 3!

2! az ⫺

d ⫽

c sin az ⫺

z⫺

p 4

⫹ Á T,

R ⫽ ⬁.

This can also be obtained from sin z ⫽ sin c az ⫺

p 4

b⫹

p 4

1 22

p 4

b ⫹ cos az ⫺

p 4

and substitution of the Taylor series with center 41 p on the right. ⴥ

ⴥ

ⴚⴥ

18. Team Project. (a) Let a an (z ⫺ z 0)n and a cn (z ⫺ z 0)n be two Laurent series of the same function f (z) in the same annulus. We multiply both series by (z ⫺ z 0)ⴚkⴚ1 and integrate along a circle with center at z 0 in the interior of the annulus. Since the series converge uniformly, we may integrate term by term. This yields 2piak ⫽ 2pick. Thus, ak ⫽ ck for all k ⫽ 0, ⫾1, Á . (b) No, because tan (1>z) is singular at 1>z ⫽ ⫾p>2, ⫾3p>2, Á , hence at z ⫽⫾2> p, ⫾23 p, Á , which accumulate at 0. (c) These series are obtained by termwise integration of the integrand. The second function is Si (z)>z 3, where Si(z) is the sine integral [see (40) in App. A3.1]. Answer: z z2 1 1 ⫹ ⫹ ⫹ ⫹ Á, z 2!2 3!3 4!4 z2 1 1 ⫺ ⫹ ⫺⫹Á. z2 3!3 5!5

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20. We obtain ⴥ 1 ⫽ a (⫺1)n (z ⫺ 1)n 1 ⫹ (z ⫺ 1) n⫽0

⫽ 1 ⫺ (z ⫺ 1) ⫹ (z ⫺ 1)2 ⫺ ⫹ Á ,

0 ⬍ ƒ z ⫺ 1 ƒ ⬍ 1,

and ⴥ

1 (z ⫺ 1) a1 ⫹

1 b z⫺1

⫽ a (⫺1)n (z ⫺ 1)ⴚnⴚ1 n⫽0

⫽ (z ⫺ 1)ⴚ1 ⫺ (z ⫺ 1)ⴚ2 ⫹ (z ⫺ 1)3 ⫺ ⫹ Á convergent for ƒ z ⫺ 1 ƒ ⬎ 1. ⴥ ⫺2 (z ⫺ i)n ⫽ a b n⫹2 , ƒ z ⫺ i ƒ ⬍ 1, a i z [i ⫹ (z ⫺ i)]2 n⫽0 n ⴥ ⫺2 1 1 in ⫽ ⫽ a b , ƒz ⫺ iƒ ⬎ 1 a 2 n (z ⫺ i)n⫹2 z2 i n⫽0 2 (z ⫺ i) a1 ⫹ b z⫺i ⴥ a n 24. a (z ⫺ 1)nⴚ4, an ⫽ sinh 1 (n even), an ⫽ cosh 1 (n odd), ƒ z ⫺ 1 ƒ ⬎ 0 n!

22.

⫽ 2

n⫽0

SECTION 16.2. Singularities and Zeros. Infinity, page 715 Purpose. Singularities just appeared in connection with the convergence of Taylor and Laurent series in the last sections, and since we now have the instrument for their classification and discussion (i.e., Laurent series), this seems the right time for doing so. We also consider zeros, whose discussion is somewhat related. Main Content, Important Concepts Principal part of a Laurent series convergent near a singularity Pole, behavior (Theorem 1) Isolated essential singularity, behavior (Theorem 2) Zeros are isolated (Theorem 3) Relation between poles and zeros (Theorem 4) Point ⬁, extended complex plane, behavior at ⬁ Riemann sphere SOLUTIONS TO PROBLEM SET 16.2, page 719 2. z 4 ⫺ 81 has simple zeros at ⫾3 and ⫾3i. Hence the given function has third-order zeros at these points. 4. The zeros of tan z are at 0, ⫾p, ⫾2p, Á . Hence the answer is 0, ⫾p>2, ⫾p, Á , second order. 6. (12 ⫾ n) pi, fourth order, because cosh z ⫽ 12 (ez ⫹ eⴚz) ⫽ 0; and this implies

hence

e2z ⫽ ⫺1

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2z ⫽ Ln (⫺1) ⫾ 2npi ⫽ (1 ⫾ 2n)pi so that z has the values given at the beginning. Using the familiar location of the zeros of the cosine, we can also proceed as follows. From Sec. 13.5 we have cosh z ⫽ cos iz and cos iz ⫽ 0 for iz ⫽ ⫾(2n ⫹ 1) p>2,

hence

z ⫽ ⫿(2n ⫹ 1) pi>2, in agreement with the previous values. 8. 12 p ⫾ 2np, n ⫽ 0, 1, Á , third order 10. ⫾18, third order, (⫾1 ⫾ i) 1np, simple, because 2

ez ⫽ 1,

z 2 ⫽ Ln 1 ⫾ 2npi ⫽ ⫾2npi;

hence z ⫽ 1⫾2npi ⫽ ⫾

1⫾i 12np. 12

12. Team Project. (a) f (z) ⫽ (z ⫺ z 0)ng(z) gives f r(z) ⫽ n(z ⫺ z 0)nⴚ1g(z) ⫹ (z ⫺ z 0)ng r(z) which implies the assertion because g (z 0) ⫽ 0. (b) f (z) as in (a) implies 1>f (z) ⫽ (z ⫺ z 0)ⴚnh(z), where h(z) ⫽ 1>g(z) is analytic at z 0 because g(z 0) ⫽ 0. (c) f (z) ⫺ k ⫽ 0 at those points. Apply Theorem 3. (d) f1(z) ⫺ f2(z) is analytic in D and zero at each z n. Hence its zeros are not isolated because that sequence converges. Thus it must be constant, since otherwise it would contradict Theorem 3. And that constant must be zero because it is zero at those points. Thus f1(z) and f2(z) are identical in D. 14. Third-order pole at z ⫽ i, essential singularity at z ⫽ ⬁ . 16. Simple poles at z ⫽ (2n ⫹ 1)>2, n ⫽ 0, ⫾1, ⫾2, Á because cos pz has simple zeros at these z. Essential singularity at z ⫽ ⬁. 18. Essential singularity at z ⫽ 1, third-order pole at z ⫽ ⬁ . 20. Expressing cos z and sin z in terms of exponential functions, we have 1 iz 1 1 (e ⫹ eⴚiz) ⫺ (eiz ⫺ eⴚiz) ⫽ ((1 ⫹ i) eiz ⫹ (1 ⫺ i) eⴚiz) ⫽ 0. 2 2i 2 Hence e2iz ⫽ ⫺

1⫺i ⫽i 1⫹i

and 2iz ⫽ Ln i ⫾ 2npi. This gives simple poles at z ⫽ 14 p ⫾ np, n ⫽ 0, 1, Á . Note that these are the points of intersection of the curves of cos x and sin x. Since they have different tangent directions at these points, the corresponding zeros must be simple, giving simple poles. Our analysis has shown that these are the only points at which the complex functions cos z and sin z are equal, so that their difference gives those zeros, but no further zeros, hence no further poles.

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22. 1>(z ⫺ p) alone would give a simple pole at z ⫽ ⬁. In the given function the factor sin z has a simple zero at z ⫽ p, and the Taylor series with center z ⫽ p is 1 1 (z ⫺ p)2 ⫺ (z ⫺ p)4 ⫹ ⫺ Á . 3! 5!

⫺1 ⫹

Answer: Essential singularity at ⬁. The given function is entire. 24. For ƒ z ƒ small enough we have ƒ 1 ⫹ z ƒ ⬎ 1> 12, ƒ 1 ⫺ z ƒ ⬎ 1> 12; hence ƒ 1 ⫺ z 2 ƒ ⫽ ƒ 1 ⫹ z ƒ ƒ 1 ⫺ z ƒ ⬎ 12 and 1 1 1 1 ` 3 ⫺ ` ⫽ 3 ` 1 ⫺ z2 ` ⬎ 2ƒzƒ3 z ƒzƒ z

⬁

ƒ z ƒ : 0.

This motivates the idea of the proof. To prove the theorem, let f (z) have a pole of mth order at some point z ⫽ z 0. Then f (z) ⫽ ⫽

bm (z ⫺ z 0)

⫹

m c

(z ⫺ z 0)

bmⴚ1 (z ⫺ z 0)mⴚ1

1⫹

bmⴚ1 bm

⫹Á

(z ⫺ z 0) ⫹ Á d ,

bm ⫽ 0.

For given M ⬎ 0, no matter how large, we can find a d ⬎ 0 so small that ƒ bm ƒ m

⬎ 2M

and

` c1 ⫹

bmⴚ1 bm

(z ⫺ z 0) ⫹ Á d ` ⬎

1 2

for all ƒ z ⫺ z 0 ƒ ⬍ d. Then ƒ f (z) ƒ ⬎

ƒ bm ƒ 1 dm 2

⬎ M.

Hence ƒ f (z) ƒ : ⬁ as z : z 0. SECTION 16.3. Residue Integration Method, page 719 Purpose. To explain and apply this most elegant integration method. Main Content, Important Concepts Formulas for the residues at poles (3)–(5) Residue theorem (several singularities inside the contour) Comment The extension from the case of a single singularity to several singularities (residue theorem) is immediate. SOLUTIONS TO PROBLEM SET 16.3, page 725 4. 0 at 0 because the Laurent series of (cos z)>z 4 has only even powers. 6. ⫺1 at (2n ⫹ 1)p>2, n ⫽ 0, ⫾1, ⫾2, Á ; note that tan z is periodic with p, so that the residues must be equal.

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8. ⫾14 p at ⫿1 10. z 2 ⫺ iz ⫹ 2 ⫽ (z ⫺ 2i)(z ⫹ i) shows that the given function has two simple poles at 2i and ⫺i with residues z4 16 16 ⫽ ` ⫽ ⫺ i at z ⫽ 2i z ⫹ i z⫽2i 3i 3 and z4 1 1 ` ⫽ ⫽ i. z ⫺ 2i z⫽ⴚi ⫺3i 3 This calculation used (3), and (4) gives the same results, of course, with about the same amount of work. 12. 1 at z ⫽ 1; see the similar Example 7 in the text. 14. z 2 ⫺ 4z ⫹ 5 ⫽ (z ⫹ 1)(z ⫺ 5). Simple poles ⫺1, residue (⫺1 ⫺ 23)>(⫺2 ⫺ 4) ⫽ 4, and at 5, residue (5 ⫺ 23)>(10 ⫺ 4) ⫽ ⫺3, by (4). Both poles lie inside the contour. Answer: 2pi(4 ⫺ 3) ⫽ 2pi. 16. Essential singularity at 0, residue 1 obtained from the Laurent series e1>z ⫽ 1 ⫹

1 1 ⫹ ⫹ Á. z 2!z 2

Answer: 2pi 18. z 4 ⫺ 2z 3 ⫽ z 3(z ⫺ 2). Third-order pole at 0, residue 1 z⫹1 s 1 ⫺3 6 3 r a b ` ⫽ a ⫽ ⫽⫺ . 2b ` 2! z ⫺ 2 2 2(⫺8) 8 (z ⫺ 2) z⫽0 z⫽0 Simple pole at 2, residue z⫹1 3 ` ⫽ . 4z 3 ⫺ 6z 2 z⫽2 8 Both poles lie inside the contour. Hence the answer is 0. 20. (z 2 ⫹ 1)3 ⫽ (z ⫹ i)3 (z ⫺ i)3. Third-order poles at ⫾i with residues [(z ⫹ i)ⴚ3] s ƒ z⫽i ⫽ [⫺3(z ⫹ i)ⴚ4] r ƒ z⫽i ⫽ 12(z ⫹ i)ⴚ5 ƒ z⫽i ⫽ 12(2i)ⴚ5 ⫽ ⫺3i>8 and [(z ⫺ i)ⴚ3] s ƒ z⫽ⴚi ⫽ [⫺3(z ⫺ i)ⴚ4] r ƒ z⫽ⴚi ⫽ 12(z ⫺ i)ⴚ5 ƒ z⫽ⴚi ⫽ 12(⫺2i)ⴚ5 ⫽ 3i>8. Answer: 0 22. Simple poles at ⫾12. By (4) the residues are obtained from z 2 sin z 1 ⫽ z sin z. 8z 8

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The answer is 2pi [18 ⴢ 12 ⴢ sin 12 ⫹ 18 (⫺12 ) sin (⫺12)] ⫽ 14 pi sin 12. 24. sin 4z ⫽ 0 at 0, ⫾p>4 inside C, ⫾p>2, Á outside C. This gives three simple poles at ⫺p>4, 0, p>4 to be taken into account, with residues exp (⫺z 2)

Res

sin 4z

z⫽z0

⫽

exp (⫺z 20)

1 p2 1 1 ⫺p2 ⫽ ⫺ exp a⫺ b , , ⫺ exp a b, 4 cos 4z 0 4 16 4 4 16

respectively, and by the residue theorem the answer 2pi p2 a1 ⫺ 2 exp a⫺ bb ⫽ ⫺0.1245i. 4 16 SECTION 16.4. Residue Integration of Real Integrals, page 725 Purpose 1. To show that certain classes of real integrals over finite or infinite intervals of integration can also be evaluated by residue integration. Comment on Content Since residue integration requires a closed path, one must have methods for producing such a path. We see that for the finite intervals in the text, this is done by (2), perhaps preceded by a translation and change of scale if another interval is given. (This is not shown in the text.) In the case of an infinite interval, we start from a finite one, close it by some curve in complex (here, a semicircle; Fig. 374), blow it up, and make assumptions on the integrand such that we can prove (once and for all) that the value of the integral over the complex curve added goes to zero. Purpose 2. Extension of the second of the two methods just mentioned to integrals of practical interest in connection with Fourier integral representations (Sec. 11.7) and to discuss the case of singularities on the real axis. Main Content Integrals involving cos and sin (1), their transformation (2) Improper integral (4), Cauchy principal value Fourier integrals Poles on the real axis (Theorem 1), Cauchy principal value SOLUTIONS TO PROBLEM SET 16.4, page 733 2. Note that in each of the present problems we are dealing with a whole class of integrals obtainable by replacing numerical coefficients with parameters a, b, etc., as follows. The present integral is of the form

冮

du 1 ⫽ a ⫹ b cos u 2 ⫽ ⫽

冮 a ⫹ dub cos u 0

1 2

冯 iz [a ⫹ b(z ⫹ 1>z)]

1 ib

冯 z ⫹ 2az>b ⫹ 1

1 2

(a ⬎ b ⬎ 0).

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The zeros of the denominator, z1 ⫽ ⫺

a ⫹ K, b

z2 ⫽ ⫺

a ⫺ K, b

K2 ⫽

a2 ⫺1 b2

give poles at z1 (inside the unit circle C because a ⬎ b ⬎ 0) and z 2 (outside C) with the residue at z1 given by Res

z⫽z1

1 1 1 ⫽ ` ⫽ z ⫹ 2az>b ⫹ 1 z ⫺ z 2 z1 2K 2

and the integral from 0 to p equals 1

2pi #

2ibK

⫽

2a ⫺ b 2 2

and 2p>bK if we integrate from 0 to 2p. In our case a ⫽ p, b ⫽ 3 and the answer is p 2p2 ⫺ 9. Note that for a ⫽ b the integrand is infinite at u ⫽ p, and the value of the integral approaches ⬁ as a : b ⬎ 0 from above. 4. The integral equals 1 i

1 ⫹ 2(z ⫹ z ⴚ1) 1 ⴚ1 dz ⫽ ⫺4i z[17 ⫺ 4(z ⫹ z )] C

冯

2z 2 ⫹ z ⫹ 2 z(z ⫺ z 1)(z ⫺ z 2) dz

where z 1 ⫽ 14 (inside the unit circle) and z 2 ⫽ 4 (outside) give simple poles. The residue at z ⫽ 0 is 2>(z 1z 2) ⫽ 2, and at z ⫽ 14 it is 2 1 16 ⫹ 4 ⫹ 2

( 14) (14 ⫺ 4)

⫽⫺

38 . 15

This gives the answer 38 4p 2pi a2 ⫺ b⫽ . ⫺4i 15 15 6. The integral equals ⫺14 (z ⫺ 1>z)2

冯 i[5z ⫺ 2(z ⫹ 1)] C

dz ⫽

⫺14

(z ⫺ 1) dz. 冯 z [z ⫺ 5z>2 ⫹ 1] ⫺2i 2

From (5*), Sec. 16.3, we obtain for the integrand of the last integral (without the factor in front of it) at z ⫽ 0 (second-order pole) the residue (z 2 ⫺ 1)2 r ⫽5 c 2 d z ⫺ 5z>2 ⫹ 1 z⫽0 2 by straightforward differentiation. Also, z 2 ⫺ 5z>2 ⫹ 1 ⫽ (z ⫺ 2)(z ⫺ 12) and for the simple pole at z ⫽ 12 (inside the unit circle) we get the residue 9

16 3 (z 2 ⫺ 1)2 ` ⫽ ⫽⫺ . 2 z 2(z ⫺ 2) z⫽1>2 ( 14)(⫺ 32 )

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In front of the integral, (⫺14)>(⫺2i) ⫽ ⫺i>8. Together,

2pi(⫺i>8)(52 ⫺ 32) ⫽ p>4.

Note that the answers to Probs. 5 and 6 are different, although the denominators are equal and the numerators are similar and periodic; of course, this is not enough. We would obtain the same result in Prob. 6 as in Prob. 5 if we replaced the two cosines in Prob. 5 by two corresponding sines in Prob. 6. Furthermore, by adding the answers to Probs. 5 and 6 we obtain 2p>3, which is the value of the integral 2p

冮 5 ⫺ du4 cos u 0

and can be used as a check because cos2 u ⫹ sin2 u ⫽ 1. 8. The integral is of the form 2p

dz 冮 a ⫺ dub sin u ⫽ 冯 iz[a ⫺ (b>2i)(z ⫺ 1>z)] C

⫽

dz 冯 z ⫺ (2ai>b)z ⫺ 1 ⫺b 2

⫽

⫺2 b

冯 (z ⫺ z dz)(z ⫺ z ) C

(a ⬎ b ⬎ 0)

where the zeros of the denominator (the poles of the integrand) are z1 ⫽

ai ⫹ iK, b

z2 ⫽

ai ⫺ iK, b

K2 ⫽

a2 ⫺1 b2

and from (3), Sec. 16.3, we obtain the residue of the pole inside the unit circle 1 1 1 ⫽ ⫽⫺ . z2 ⫺ z1 2iK z⫽z2 (z ⫺ z 1)(z ⫺ z 2)

Res

Hence the integral equals 2pi(⫺2>b)(⫺1>(2Ki)) ⫽ 2p>(bK) ⫽ 2p> 2a 2 ⫺ b 2.

In our case, a ⫽ 8, b ⫽ 2 gives the answer 2p> 160 ⫽ p> 115. Note that the present formula p> 2a 2 ⫺ b 2 also occurred in the solution to Prob. 2. The reason is that ⫺sin u ⫽ cos u*, where u* ⫽ u ⫹ 12 p and the integrand is periodic with period 2p. 10. Third-order pole at z ⫽ i (and at z ⫽ ⫺i in the lower half-plane) with residue 1 1 s ⫽ 6 . c 3 d 2 (z ⫹ i) z⫽i (2i)5

Answer: 2pi ⴢ 6>(2i)5 ⫽ 3p>8. 12. Second-order pole at z1 ⫽ 1 ⫹ 2i in the upper half-plane (and at z 2 ⫽ 1 ⫺ 2i in the lower) with residue c

1 ⫺2 1 r ⫽ ⫺2 . 2 d 3 ⫽ 3 ⫽ 32i (z ⫺ 1 ⫹ 2i) z⫽z1 (z1 ⫺ z 2) (4i)

Answer: 2pi(1>32i) ⫽ p>16.

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14. Denote the integrand by f (x). The complex function f (z) has simple poles at z1 ⫽ epi>4 and z 2 ⫽ e3pi>4 in the upper half-plane (and two further ones in the lower half-plane) with residues 1⫹i i ⫽⫺ 2 12(⫺1 ⫹ i) 18

and

1⫺i i ⫽⫺ 212(1 ⫹ i) 18

respectively, as obtained from (4) in Sec. 16.3. Hence the answer is (⫺i> 12)2pi ⫽ p 12. 16. Second-order poles at z1 ⫽ i and z 2 ⫽ ⫺i (in the lower half-plane). By (5) Sec. 15.3, we get the residue c

e2iz r ⫽ e2iz [2i(z ⫹ i) ⫺ 2] ` ⫽ eⴚ2 (⫺6) ⫽ ⫺3eⴚ2i . 2d 4 (z ⫹ i) z⫽i (z ⫹ i)3 (2i)3 z⫽i

Multiplying the imaginary part ⫺3eⴚ2>4 by ⫺2p gives the answer 3peⴚ2>2 ⫽ 0.6378.

18. z 4 ⫹ 5z 2 ⫹ 4 ⫽ (z 2 ⫹ 1)(z 2 ⫹ 4) shows that we have two simple poles at z1 ⫽ i and z 2 ⫽ 2i in the upper half-plane (and two at ⫺i and ⫺2i in the lower). By (4), Sec. 16.3, the sum of the residues of e4iz>(z 4 ⫹ 5z 2 ⫹ 4) at z1 and z 2 is eⴚ4 eⴚ8 i i ⴚ8 ⫹ ⫽ ⫺ eⴚ4 ⫹ e . 6 12 4i ⫹ 10i 4(2i)3 ⫹ 20i 3

From the first formula in (10) we thus obtain the answer

p(2eⴚ4 ⫺ eⴚ8)>6. 20. Simple poles at 2, ⫺1 ⫹ i13 (and ⫺1 ⫺ i 13 in the lower half-plane, not needed). Residues ⫺16, (1 ⫹ i13)>12, respectively. Answer:

pi a⫺ b ⫹ 1 6

2 13 pi(1 ⫹ i 13) ⫽ ⫺ p. 12 6

22. Equivalently to sequences, you may use a parameter p, obtaining ⴚp

ⴥ

冮 x ⫺ ix 冮 x dx⫺ ix ⫽ p ⫺ 2 arctan p. ⴚⴥ

⫹

Now let p : 0. 24. The integrand is p>q, where p ⫽ 1 and q(x) ⫽ ⫺x 4 ⫹ 3x 2 ⫺ 4; hence q(z) ⫽ z 4 ⫹ 3z 2 ⫺ 4. In (4) in Sec. 16.3 we need the derivative q r(z) ⫽ 4z 3 ⫹ 6z. Simple poles are at 2i, ⫺1, 1 1 and 1 (and ⫺2i, not needed), with residues i>20, ⫺ 10 , and 10 , respectively. Hence the value of the integral is 2pi ⴢ

i 1 1 p ⫹ pi a⫺ ⫹ b⫽⫺ . 20 10 10 10

A graph of the integrand makes it plausible that the integral should have a negative value. 26. Simple poles at z 0 ⫽ ⫾1, i (and ⫺i outside, in the lower half-plane), with residue z 20>(4z 30) ⫽ 1>(4z 0), as obtained from (4) in Sec. 16.3. Hence the answer is i 1 1 1 2pi a⫺ b ⫹ pi a⫺ ⫹ b ⫽ p. 4 4 4 2

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28. Team Project. (b) The integral of eⴚz around C is zero. Writing it as the sum of four integrals over the four segments of C, we have 2

冮

eⴚx dx ⫹ ieⴚa 2

ⴚa

冮

ⴚa

ey ⴚ2ayi dy ⫹ eb 2

冮 e

ⴚx2ⴚ2ibx

⫹ ieⴚa

冮e

y2⫹2ayi

dy ⫽ 0.

Let a : ⬁. Then the terms having the factor eⴚa approach zero. Taking the real part of the third integral, we thus obtain 2

⫺

冮

ⴚⴥ

eⴚx cos 2bx dx ⫽ 2 2

ⴥ

冮

ⴥ

eⴚx cos 2bx dx ⫽ eⴚb 2

冮 e

ⴚx2

dx ⫽ eⴚb 1p. 2

ⴚⴥ

2 Answer: 12 eⴚb 1

(c) Use the fact that the integrands are odd. SOLUTIONS TO CHAPTER 16 REVIEW QUESTIONS AND PROBLEMS, page 733 12. Essential singularity at z ⫽ 0; this point lies inside the contour. The residue z ⫽ 2 can be seen from the Laurent series e2>z ⫽ 1 ⫹

22 2 ⫹ ⫹ Á, z 2!z 2

which converges for ƒ z ƒ ⬎ 0. This gives the answer 4pi. 14. Simple poles at ⫾2i. In Prob. 13, both were inside C. Here, only z ⫽ 2i lies inside C, and the answer is 5z 3 ⫺40i ⫽ ⫺20pi. ⫽ 2pi ⴢ 2 2 ⴢ 2i z⫽2i z ⫹ 4

2pi Res

16. z 3 ⫺ 9z ⫽ z(z ⫹ 3)(z ⫺ 3). Simple poles at ⫺3, 0, 3 inside C with residues 15z ⫹ 9 3z 2 ⫺ 9

⫽ ⫺2, ⫺1, 3,

respectively. Answer: 0. 18. Simple poles at z ⫽ 0 inside C and ⫾14 , ⫾12 , Á all outside C. The residue 14 at z ⫽ 0 is obtained from the Laurent series cot 4z ⫽

1 ⫹Á 4z

or from the formula 1 ` ⫽ . (sin 4z) r z⫽0 4 cos 4z

The answer is pi>2.

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20. The integral equals (z ⫺ 1>z)>(2i)

冯 iz[3 ⫹ (z ⫹ 1>z)] 1 2

dz ⫽ ⫺

z2 ⫺ 1

冯 z[z ⫹ 6z ⫹ 1] dz. C

At the simple pole at z ⫽ 0 the residue is ⫺1. At the simple pole at ⫺3 ⫹ 18 (inside the unit circle) the residue is (⫺3 ⫹ 18)2 ⫺ 1 3(⫺3 ⫹ 18)2 ⫹ 12(⫺3 ⫹ 18) ⫹ 1

⫽ 1.

Answer: 0. Simpler: integrate from ⫺p to p and note that the integrand is odd. 22. 1 ⫹ 4z 4 ⫽ 0 gives simple poles of f (z) ⫽ 1>(1 ⫹ 4z 4) at ⫾12 ⫹ 12 i in the upper halfplane and at ⫾12 ⫺ 12 i in the lower half-plane. The residues of the first two poles are (use the formula for the residue of a simple pole) 1 1 1 ⫺ i. 3 ` 1 1 ⫽ ⫿ 8 8 16z ⫾2⫹2 i The sum of these two residues is ⫺i>4. Answer: 2pi(⫺ 14 i) ⫽ 12 p. 24. x 2 ⫺ 4ix ⫽ x(x ⫺ 4i). Simple poles of 1>(z 2 ⫺ 4iz) at 0 and 4i. The residues are Res

1 1 i i ⫽ ⫽ ,⫺ . 2z ⫺ 4i 4 4 z 2 ⫺ 4iz

This gives the answer

pi ⴢ 14 i ⫹ 2pi(⫺ 14 i) ⫽ ⫺ 14 p ⫹ 12 p ⫽ 14 p.

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CHAPTER 17

Conformal Mapping

This chapter was new in the ninth edition. SECTION 17.1. Geometry of Analytic Functions: Conformal Mapping, page 737 Purpose. To show conformality (preservation of angles in size and sense) of the mapping by an analytic function w ⫽ f (z); exceptional are points with f r(z) ⫽ 0. Main Content, Important Concepts Concept of mapping. Surjective, injective, bijective Conformal mapping (Theorem 1) Magnification, Jacobian Examples. Joukowski airfoil Comment on the Proof of Theorem 1 The crucial point is to show that w ⫽ f (z) rotates all straight lines (hence all tangents) passing through a point z 0 through the same angle a ⫽ arg f r(z 0), but this follows from (3). This in a nutshell is the proof, once the stage has been set. SOLUTIONS TO PROBLEM SET 17.1, page 741 6. See Fig. 379. We have u ⫽ x 2 ⫺ y 2,

v ⫽ 2xy.

For x ⫽ c, u ⫽ c ⫺ v >(4c ), while for y ⫽ k, u ⫽ u 2>(4k 2) ⫺ k 2. These are families of parabolas. For x ⫽ 1, 2, 3, 4 2

u ⫽ x 2 ⫺ v2>(4x 2) ⫽ 1 ⫺ v2>4, 4 ⫺ v2>16,

and for y ⫽ 1, 2, 3, 4,

u ⫽ v2>(4y 2) ⫺ y 2 ⫽ v2>4 ⫺ 1,

v2>16 ⫺ 4,

9 ⫺ v2>36,

16 ⫺ v2>64

v2>36 ⫺ 9,

v2>64 ⫺ 16.

8. w ⫽ 1>z so ƒ w ƒ ⫽ 1> ƒ z ƒ ƒ w ƒ ⫽ 3, 2, 1, 12 , 13 and Arg w ⫽ ⫺Arg z

Arg w ⫽ 0, ⫿ p>4, ⫿ p>2, ⫿3p>4, ⫿ p.

We see that the unit circle is mapped onto itself, but only 1 and ⫺1 are mapped onto themselves. eiu is mapped onto eⴚiu, the complex conjugate, for example, i onto ⫺i. 10. CAS Experiment. Orthogonality is a consequence of conformality because in the w-plane, u ⫽ const and v ⫽ const are orthogonal. We obtain (a) u ⫽ x 4 ⫺ 6x 2y 2 ⫹ y 4, v ⫽ 4x 3y ⫺ 4xy 3 (b) u ⫽ x>(x 2 ⫹ y 2), v ⫽ ⫺y>(x 2 ⫹ y 2) 2 2 2 (c) u ⫽ (x ⫺ y )>(x ⫹ y 2)2, v ⫽ ⫺2xy>(x 2 ⫹ y 2)2 (d) u ⫽ 2x>((1 ⫺ y)2 ⫹ x 2), v ⫽ (1 ⫺ x 2 ⫺ y 2)>((1 ⫺ y 2)2 ⫹ x 2). 284

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12. 1 ⬍ ƒ w ƒ ⬍ 27, 0 ⬍ Arg w ⬍ 3p>2 14. On the line x ⫽ 1 we have z ⫽ 1 ⫹ iy, w ⫽ u ⫹ iv ⫽ 1>z ⫽ (1 ⫺ iy)>(1 ⫹ y 2), so that v ⫽ ⫺y(1 ⫹ y 2) and, furthermore, u⫽

1 1⫹y

, 2

u (1 ⫺ u) ⫽

1 1⫹y

a1 ⫺ 2

1 1⫹y

b⫽ 2

y2 (1 ⫹ y 2)2

⫽ v2.

Having obtained a relation between u and v, we have solved the problem. We now have u 2 ⫺ u ⫹ v2 ⫽ 0, (u ⫺ 12)2 ⫹ v2 ⫽ 14 and by taking roots ƒ w ⫺ 12 ƒ ⫽ 12 , a circle through 0 and 1 whose interior is the image of x ⬎ 1. Note the relation to Prob. 15. 16. ƒ w ƒ ⬎ 2, Im w ⬍ 0. This is the exterior of the circle ƒ w ƒ ⫽ 2 in the lower half-plane Im w ⬍ 0. 18. Annulus 1>e ⬉ ƒ w ƒ ⬉ e2 cut along the negative real axis. 20. The rectangle ⫺ln 2 ⬉ u ⬉ 0, 0 ⬉ v ⬍ p>2 22. 2z ⫺ 2z ⴚ3 ⫽ 0, z 4 ⫽ 1, hence ⫾1, ⫾i 24. The derivative (5z 4 ⫺ 80) exp (z 5 ⫺ 80z) is zero at z ⫽ ⫾2 and ⫾2i since the exponential function has no zeros. 26. p cos pz ⫽ 0 at z ⫽ (2n ⫹ 1)>2, n ⫽ 0, ⫾1, ⫾2, Á 28. By the Taylor series, since the first few derivatives vanish at z 0, f (z) ⫽ f (z 0) ⫹ (z ⫺ z 0)kg(z),

g(z 0) ⫽ 0, since f (k)(z 0) ⫽ 0.

Hence arg [ f (z) ⫺ f (z 0)] ⫽ k arg (z ⫺ z 0) ⫹ arg g(z). Now the angle a from the x-axis to the tangent of a smooth curve C at z 0 is a ⫽ lim arg (z ⫺ z 0)

(z : z 0 along C).

Similarly for the angle b of the tangent to the image at f (z 0): b ⫽ lim arg [ f (z) ⫺ f (z 0)] ⫽ ka ⫹ lim arg g(z) ⫽ ka ⫹ arg g(z 0). Note that g ⫽ arg g(z 0) is defined since g(z 0) ⫽ 0. Hence the angle b2 ⫺ b1 between the tangents of the images of two curves at f (z 0) is b2 ⫺ b1 ⫽ ka2 ⫹ g ⫺ (ka1 ⫹ g) ⫽ k(a2 ⫺ a1), as asserted. 30. M ⫽ 3 ƒ z ƒ 2 ⫽ 1 on the circle ƒ z ƒ ⫽ 1> 13, J ⫽ 9 ƒ z ƒ 4 3 32. M ⫽ ƒ w r ƒ ⫽ 2> ƒ z ƒ 3 ⫽ 1 on ƒ z ƒ ⫽ 2 2 ⫽ 1.26 2 34. M ⫽ ƒ w r ƒ ⫽ 1> ƒ z ⫺ 1 ƒ ⫽ 1 when ƒ z ⫺ 1 ƒ 2 ⫽ (x ⫺ 1)2 ⫹ y 2 ⫽ 1, the circle with center z ⫽ 1 and radius 1, which passes through the origin. SECTION 17.2. Linear Fractional Transformations (Möbius Transformations), page 742 Purpose. Systematic discussion of linear fractional transformations (Möbius transformations), which owe their importance to a number of interesting properties shown in this section and the next one.

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Main Content Definition (1) Special cases (3), Example 1 Images of circles and straight lines (Theorem 1) One-to-one mapping of the extended complex plane Inverse mapping (4) Fixed points SOLUTIONS TO PROBLEM SET 17.2, page 745 2. This follows by direct calculation and simplification. 4. z ⫽ 1>w ⫽ u>(u 2 ⫹ v2) ⫺ iv>(u 2 ⫹ v2). x ⫽ c ⫽ 1>k ⫽ u>(u 2 ⫹ v2) or u 2 ⫹ v2 ⫽ ku from which (u ⫺ k>2)2 ⫹ v2 ⫽ (k>2)2. This is the circle with center w ⫽ u ⫽ 1>(2k) and radius 1>(2k). Note that all these circles pass through the origin (see Fig. 387). For instance, for x ⫽ k ⫽ 12 you get (u ⫺ 1)2 ⫹ v2 ⫽ 1 shown in the figure. 6. w ⫽ z ⫽ x ⫺ iy ⫽ z ⫽ x ⫹ iy, y ⫽ 0 (the x-axis), w ⫽ 1>z ⫽ z, z 2 ⫽ 1, z ⫽ ⫾1. The center of a rotation is a fixed point. 8. We obtain (z ⫹ i)w ⫽ z ⫺ i,

z(w ⫺ 1) ⫽ ⫺iw ⫺ 1

so that z⫽

⫺iw ⫺ i . w⫺1

10. We obtain w(⫺12 iz ⫺ 1) ⫽ z ⫺ 12 i, hence z⫽

⫺2w ⫹ i . iw ⫹ 2

Mappings of this kind will occur in the next section in connection with mapping disks onto disks. 12. None. A translation has no fixed points in the finite plane. 14. No fixed point when a ⫽ 1, that is, a translation has no fixed points in the finite plane; z ⫽ ⫺b>(a ⫺ 1) when a ⫽ 1, for instance, z ⫽ 0 when b ⫽ 0, which is a rotation about the origin and, for ƒ a ƒ ⫽ 1, combined with a uniform dilatation or contraction. 16. We obtain z(z ⫹ ai) ⫽ aiz ⫺ 1,

z 2 ⫽ ⫺1

so that z ⫽ ⫾i. Note that for a ⫽ 1 the denominator is zero when z ⫽ ⫺i, and so is the numerator. 18. We have k(z ⫺ 1)(z ⫹ 1) ⫽ k(z 2 ⫺ 1)

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and, by comparing with (5), c ⫽ k,

a ⫺ d ⫽ 0,

b⫽k

so that w⫽

az ⫹ b , bz ⫹ a

including the identity mapping w ⫽ z when b ⫽ 0. 20. We obtain a fixed-point equation (5) without solutions if and only if we choose c ⫽ 0, a ⫺ d ⫽ 0, and b ⫽ 0. This gives a translation w ⫽ (az ⫹ b)>a ⫽ z ⫹ b>a

(a ⫽ 0)

as expected. SECTION 17.3. Special Linear Fractional Transformations, page 746 Purpose. Continued discussion to show that linear fractional transformations map “standard domains” conformally onto each other. Main Content Determination by three points and their images Mappings of standard domains (disks, half-planes) Angular regions SOLUTIONS TO PROBLEM SET 17.3, page 750 2. The inverse is z⫽

w⫹i u ⫹ iv ⫹ i (u ⫹ iv ⫹ i)(1 ⫺ v ⫺ iu) ⫽ ⫽ . iw ⫹ 1 1 ⫺ v ⫹ iu (1 ⫺ v)2 ⫹ u 2

Multiplying out the numerator, a number of terms drop out, and the real part of the numerator is 2u. This gives Re z ⫽ x in the form x⫽

2u (1 ⫺ v)2 ⫹ u 2

This yields the circles 2

1 1 (v ⫺ 1)2 ⫹ au ⫺ b ⫽ 2 c c as claimed. 4. Problem 18 in Problem Set 17.2 applies since the present mapping has the fixed points ⫾1. The answer to Prob. 18 is w(z) ⫽

az ⫹ b . bz ⫹ a

Since z ⫽ 0 maps onto w ⫽ ⫺i, we have w(0) ⫽ b>a ⫽ ⫺i, so that b ⫽ ⫺ai and division by a finally gives the mapping in Example 1, that is, w(z) ⫽

z⫺i az ⫺ ai ⫽ . ⫺aiz ⫹ a ⫺iz ⫹ 1

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The point of this problem is to illustrate that sometimes one need not use the general formula (2), but can proceed directly. However, these cases are not too frequent. 6. The inverse is z⫽

4w ⫺ 1 . ⫺2w ⫹ 1

The fixed points z ⫽ ⫺34 ⫾ 14 117 are obtained as solutions of 2z 2 ⫺ (1 ⫺ 4)z ⫺ 1 ⫽ 0. 8. w ⫽ 1>(z ⫹ 1) almost without calculation 10. Formula (2) gives w ⫹ 1 ⫺⬁ z ⫺2i 2z ⴢ ⫽ ⴢ ⫽ . w⫺⬁ 1 z ⫺ i ⫺i z⫺i Replacing infinity on the left as indicated in the text, we get w ⫹ 1 on the left, so that the answer is w⫽

2z z⫹i ⫺1⫽ . z⫺i z⫺i

Caution! In setting up further problems by starting from the result, which is quite easy, one should check how complicated the solution, starting from (2), will be; this may often involve substantial work until one reaches the final form. 12. By (2), w ⫹ 1 ⫺⬁ z 4i ⴢ ⫽ ⴢ ; w⫺⬁ 1 z ⫹ 2i 2i hence, after getting rid of infinity, w⫹1⫽

2z , z ⫹ 2i

w⫽

2z z ⫺ 2i ⫺1⫽ . z ⫹ 2i z ⫹ 2i

Second solution. 0 maps onto ⫺1, hence d ⫽ ⫺b. Next, 2i maps onto 0, hence 2ia ⫹ b ⫽ 0. Finally, ⫺2i maps onto ⬁, hence ⫺2ic ⫹ d ⫽ 0. Choosing a ⫽ 1 gives b ⫽ ⫺2i, d ⫽ 2i, c ⫽ 1 and confirms our previous result. 14. w ⫽ iz ⫹ 1 ⫹ i, a rotation, combined with a translation 16. w ⫽ (2z ⫹ 3)>(3z ⫹ 2) 18. The requirement is that w⫽u⫽

ax ⫹ b cx ⫹ d

must come out real for all real x. Hence the four coefficients must be real, except possibly for a common complex factor. 20. w ⫽ ⫺(z 2 ⫹ i)>(iz 2 ⫹ 1) SECTION 17.4. Conformal Mapping by Other Functions, page 750 Purpose. So far we have discussed mapping properties of z n, ez, and linear fractional transformations. We now add to this a discussion of trigonometric and hyperbolic functions.

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SOLUTIONS TO PROBLEM SET 17.4, page 754 2. Ray Arg w ⫽ k from 0 to ⬁ 4. 1 ⬍ ƒ w ƒ ⬍ e, 12 ⬍ Arg w ⬍ 1; see Fig. 385 in Sec. 17.1. 6. The portion of 1 ⬍ ƒ w ƒ ⬍ ⬁ in the first quadrant 10. If x ⫽ c, then sin z ⫽ u ⫹ iv ⫽ sin x cosh y ⫹ i cos x sinh y in Sec. 13.6 gives u ⫽ sin c cosh y, v ⫽ cos c sinh y. Hence if c ⫽ 0, then u ⫽ 0, so that the y-axis maps onto the v-axis. If c ⫽ 12 p, then v ⫽ 0, u ⭌ 1. If c ⫽ ⫺12 p, then v ⫽ 0, u ⬉ ⫺1. If c ⫽ 0, ⫾12 p, ⫾p, Á , then we obtain hyperbolas u2 sin2 c

⫺

v2 cos2 c

⫽ 1.

12. The region in the upper half-plane bounded by portions of the u-axis, the ellipse u 2>cosh2 1 ⫹ v2>sinh2 1 ⫽ 1 and the hyperbola u 2 ⫺ v2 ⫽ 12 . Indeed, for x ⫽ ⫾p>4 we get (see the formula for sin z in the Sec. 13.6) sin (⫾14 p ⫹ iy) ⫽ sin (⫾14 p) cosh y ⫹ i cos (⫾14 p) sinh y ⫽ ⫾(1> 12) cosh y ⫹ i(1> 12) sinh y and from this 1 ⫽ cosh2 y ⫺ sinh2 y ⫽ 2u 2 ⫺ 2v2,

u 2 ⫺ v2 ⫽ 12 .

thus

For y ⫽ 0 we get v ⫽ 0 (a portion of the u-axis). For y ⫽ 1 we get u ⫽ sin x cosh 1,

v ⫽ cos x sinh 1;

hence 1 ⫽ sin2 x ⫹ cos2 x ⫽

u2 cosh2 1

⫹

v2 sinh2 1

14. The region in the right half-plane bounded by the v-axis and the hyperbola 4u 2 ⫺ 4 2 3 v ⫽ 1 because sin z ⫽ u ⫹ iv ⫽ sin x cosh y ⫹ i cos x sinh y reduces to sin iy ⫽ i sinh y

when x ⫽ 0; thus u ⫽ 0 (the v-axis) is the left boundary of that region. For x ⫽ p>6 we obtain sin (p>6 ⫹ iy) ⫽ sin (p>6) cosh y ⫹ i cos (p>6) sinh y; thus u ⫽ 12 cosh y,

v ⫽ 12 13 sinh y

and we obtain the right boundary curve of that region from 1 ⫽ cosh2 y ⫺ sinh2 y ⫽ 4u 2 ⫺ 43 v2, as asserted. 16. 0, ⫾i>2, ⫾i, Á

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18. w ⫽ cos z ⫽ cos x cosh y ⫺ i sin x sinh y (Sec. 13.6) gives u ⫽ cos x cosh k,

v ⫽ ⫺sin x sinh k.

For k ⫽ 0 we thus obtain u2 cosh2 k

⫹

v2 sinh2 k

⫽ cos2 x ⫹ sin2 x ⫽ 1.

For k ⫽ 0 we have u ⫽ cos x, hence ⫺1 ⬉ u ⬉ 1,

v ⫽ 0.

20. The upper boundary maps onto the ellipse u2 cosh2 1

⫹

v2 sinh2 1

⫽1

and the lower boundary onto the ellipse u2

cosh 2

sinh2 12

⫹ 21

⫽ 1.

Since 0 ⬍ x ⬍ 2p, we get the entire ellipses as boundaries of the image of the given domain, which therefore is an elliptical ring. Now the vertical boundaries x ⫽ 0 and x ⫽ 2p map onto the same segment sinh 12 ⬉ u ⬉ sinh 1 of the u-axis because for x ⫽ 0 and x ⫽ 2p we have v ⫽ 0.

u ⫽ cosh y,

Answer: Elliptical annulus between those two ellipses and cut along that segment. See the figure.

Section 17.4. Problem 20

22. The lower boundary segment maps onto cos 1 ⬉ u ⬉ 1 (v ⫽ 0). The left and right boundary segments map onto portions of the hyperbola u2 cos2 1

⫺

v2 sin2 1

⫽ 1.

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The upper boundary segment maps onto a portion of the ellipse u2 cosh2 1

⫹

v2 sinh2 1

⫽ 1.

24. ln 2 ⬉ u ⬉ ln 3, p>4 ⬉ v ⬉ p>2 SECTION 17.5. Riemann Surfaces. Optional, page 754 Purpose. To introduce the idea and some of the simplest examples of Riemann surfaces, on which multivalued relations become single-valued, that is, functions in the usual sense. Short Courses. This section may be omitted. SOLUTIONS TO PROBLEM SET 17.5, page 756 2. By the hint we have w ⫽ 1r1eiu1>2 1r2eiu2>2 ⫽ 1r1r2 ei(u1⫹u2)>2. If we move from A in the first sheet (see the figure), we get into the second sheet at B (dashed curve) and get back to A after two loops around the branch point 1 (of first order). Similarly for a loop around z ⫽ 2 (without encircling z ⫽ 1); this curve is not shown in the figure. If we move from C and back to C as shown, we do not cross the cut, we stay in the same sheet, and we increase u1 and u2 by 2p each. Hence (u1 ⫹ u2)>2 is increased by 2p, and we have completed one loop in the w-plane. This makes it plausible that two sheets will be sufficient for the present w and that the cut along which the two sheets are joined crosswise is properly chosen. It now seems plausible that the Riemann surface of w ⫽ 1(z ⫺ a1)(z ⫺ a2) Á (z ⫺ ak) with a1, Á , ak all different, has two sheets and these points as branch points, and needs k>2 branch cuts when k is even, and (k ⫹ 1)>2 branch cuts when k is odd; one of the latter cuts extends to infinity.

C A

␪2

␪1 1

Section 17.5. Problem 2

4. ⫺1 ⫺ 2i, two sheets 6. i>3, infinitely many sheets

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8. The Riemann surface of the first of these two functions has a branch point at z ⫽ 0 and two sheets. The second expression, 2ez, has no branch points because ez is never zero (see Sec. 13.5); it has two sheets, but these are nowhere connected with each other. Hence 2ez represents two different functions, namely, ez>2 and ⫺ez>2. 10. Branch points at ⫾1 and ⫾2, as shown in the figure, together with the cuts. If we pass a single cut, we get into the other sheet. If we cross two cuts, we are back in the sheet in which it started. The figure shows one path (A) that encircles two branch points and stays entirely in one sheet. The path from B and back to B also encloses two branch points, and since it crosses two cuts, part of it is in one sheet and part of it is in the other. A discussion in terms of coordinates as in Prob. 2 would be similar to the previous one. Various other paths can be drawn and discussed in the figure. B

–2

–1

Section 17.5. Problem 10

SOLUTIONS TO CHAPTER 17 REVIEW QUESTIONS AND PROBLEMS, page 756 12. 1> p ⬍ ƒ w ƒ ⬍ p, v ⬎ 0 14. y ⫽ 0 maps onto the nonnegative real axis u ⫽ x 2, v ⫽ 0. The other boundary y ⫽ 2 gives u ⫽ x 2 ⫺ 4, v ⫽ 4x. Elimination of x gives the parabola u ⫽ v2>16 ⫺ 4,

with apex at u ⫽ ⫺4 and opening to the right. Answer: The domain to the right of that parabola except for the nonnegative u-axis. 1 2 16. u ⫽ 16 v ⫺ 4. See Example 1 and Fig. 379 in Sec. 17.1. y ⫽ ⫺2 and 2 are mapped onto the same parabola. The formulas in Example 1 of Sec. 17.1 show that this is true for any straight lines x ⫽ ⫾c (see the previous Prob. 15) and y ⫽ ⫾k since these formulas contain c2 ⫽ (⫺c)(⫺c) and k 2, respectively. 1 18. z ⫽ reiu, w ⫽ r eⴚiu; hence the first quadrant of the interior of the unit disk in the z-plane maps onto the exterior of ƒ w ƒ ⫽ 1 in the fourth quadrant of the w-plane. 20. ⫺p>4 ⬉ arg w ⬉ 0 22. We have w⫽

1 1 ⫺ iy ⫽ 1 ⫹ iy 1 ⫹ y2

and thus obtain u 2 ⫹ v2 ⫽

1 ⫹ y2 (1 ⫹ y )

2 2

⫽

1 1 ⫹ y2

⫽u

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that is, (u ⫺ 12)2 ⫹ v2 ⫽ 14 , the circle with center z ⫽ 12 and radius 12 , thus passing through w ⫽ 0 and 1. In the text it is shown that geometrical arguments may often replace calculations. In the present problem, that is as follows. Since z ⫽ 1 ⫹ iy lies outside ƒ z ƒ ⫽ 1, is a straight line, extends to infinity, and maps z ⫽ 1 onto w ⫽ 1, its image is a circle, passing through 0 and 1. Since this image lies inside ƒ w ƒ ⫽ 1, its center must lie on the u-axis, otherwise it would intersect ƒ w ƒ ⫽ 1 at some point different from 1; hence that center lies at u ⫽ 12 . 24. w ⫽ 1>z by inspection. This illustrates that, in many cases, one hardly needs the general formula (2) in Sec. 17.3. Sometimes, one can also use points and their images, one after another, and determine the coefficents of the mapping function stepwise. 26. Translation w ⫽ z ⫹ 2i 28. w ⫽ 1 ⫹ iz, a translation combined with a rotation, first rotating, then translating 30. z ⫽ ⫾212, ⫾2 12i 32. ⫾i 34. z ⫽ ⫾1 36. To map a quarter-disk onto a disk, we have to quadruple angles; thus w ⫽ z 4 maps the given quarter-disk onto the unit disk ƒ w ƒ ⫽ 1, and the exterior is obtained by taking the reciprocal. Answer: w ⫽ 1>z 4 38. w ⫽ 2>z maps ƒ z ƒ ⫽ 1 onto ƒ w ƒ ⫽ 2 and the interior onto the exterior. Hence w ⫹ 2 ⫽ 2>z will give the answer 2 2⫺z w ⫽ ⫺2 ⫹ z ⫽ z . 40. z 2>4 maps the semidisk onto the unit disk; 4>z 2 maps its interior onto the exterior; 4p>z 2 maps this onto the exterior of the circle with center 0 and radius p, so that the answer is w ⫽ p ⫹ 4p>z 2.

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CHAPTER 18

Complex Analysis and Potential Theory

This is perhaps the most important justification for teaching complex analysis to engineers, and it also provides for nice applications of conformal mapping. SECTION 18.1. Electrostatic Fields, page 759 Purpose. To show how complex analysis can be used to discuss and solve two-dimensional electrostatic problems and to demonstrate the usefulness of complex potential, a major concept in this chapter. Main Content, Important Concepts Equipotential lines Complex potential (2) Combination of potentials by superposition The main reason for using complex methods in two-dimensional potential problems is the possibility of using the complex potential, whose real and imaginary parts both have a physical meaning, as explained in the text. This fact should be emphasized in teaching from this chapter. SOLUTIONS TO PROBLEM SET 18.1, page 762 2. We have £ ⫽ a ln r ⫹ b (see Example 2 of the text) and obtain from this and the boundary conditions £(1) ⫽ a ln 1 ⫹ b ⫽ b ⫽ 400 £(2) ⫽ a ln 2 ⫹ b ⫽ a ln 2 ⫹ 400 ⫽ 0. Hence a ⫽ ⫺400>ln 2. Together this gives the answer £(r) ⫽ 400 a1 ⫺

1 ln rb . ln 2

F(z) ⫽ 400 a1 ⫺

1 Ln zb . ln 2

The complex potential is

4. Less because the logarithmic curve is concave; it is not a straight line; see Fig. 397, where the distance between equipotential lines (circles), such as U ⫽ 100, 200, 300, 400, becomes smaller and smaller near the inner circle. To calculate the actual value, we have to take £(r) ⫽ a ln r ⫹ b at r ⫽ 2 and r ⫽ 6 and use the boundary values (the given potentials), that is, £(2) ⫽ a ln 2 ⫹ b ⫽ 300 £(6) ⫽ a ln 6 ⫹ b ⫽ 100 and solve for a and b, obtaining (rounded) a ⫽ ⫺182.05, 294

b ⫽ 426.19.

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Inserting these values, we obtain, for the potential at r ⫽ 4, the value £(4) ⫽ 173.8, substantially less than 200. This contrasts with the equidistance of the equipotential curves (surfaces) in the case of parallel plates, which are straight lines (planes in space). 6. £ is expected to be linear because of the boundary (two parallel straight lines). From the boundary conditions and by inspection. £(x, y) ⫽ 220(y ⫺ x)>k. This is the real part of the complex potential F(z) ⫽ ⫺220(1 ⫹ i)z>k. The real potential can also be obtained systematically, starting from £(x, y) ⫽ ax ⫹ by ⫹ c. By the first boundary condition, for y ⫽ x this is zero: ax ⫹ by ⫹ c ⫽ 0.

(1)

By the second boundary condition, for y ⫽ x ⫹ k this equals 220: ax ⫹ b(x ⫹ k) ⫹ c ⫽ 220.

(2) Subtract (1) from (2) to get

bk ⫽ 220,

b ⫽ 220>k.

Substitute this into (1) to get ax ⫹

220 220 x ⫹ c ⫽ aa ⫹ b x ⫹ c ⫽ 0. k k

Since this is an identity in x, the coefficients must vanish, a ⫽ ⫺220>k and c ⫽ 0. This gives the above result. 8. CAS Experiment. (a) x 2 ⫺ y 2 ⫽ c, xy ⫽ k (b) xy ⫽ c, x 2 ⫺ y 2 ⫽ k; the rotation caused by the multiplication by i leads to the interchange of the roles of the two families of curves. (c) x>(x 2 ⫹ y 2) ⫽ c gives (x ⫺ 1>(2c))2 ⫹ y 2 ⫽ 1>(4c2). Also, ⫺y>(x 2 ⫹ y 2) ⫽ k gives the circles x 2 ⫹ (y ⫹ 1>(2k))2 ⫽ 1>(4k 2). All circles of both families pass through the origin. (d) Another interchange of the families, compared to (c), (y ⫺ 1>(2c))2 ⫹ x 2 ⫽ 1>(4c2), (x ⫺ 1>(2k))2 ⫹ y 2 ⫽ 1>(4k 2). iz ⫹ 1 ⫹ i # 10. w ⫽ w ⫽ 0 gives z ⫽ z 0 ⫽ ⫺1 ⫹ i, Arg z 0 ⫽ 3p>4. Hence at z 0 the z⫹1⫹i

potential is (1> p)3p>4 ⫽ 34 . By considering the three given points and their images we see that the potential on the unit circle in the w-plane is 0 for the quarter-circle in the first quadrant and 1 for the other portion of the circle. This corresponds to a conductor consisting of two portions of a cylinder separated by small slits of insulation at w ⫽ 1 and w ⫽ i, where the potential jumps.

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12. £ ⫽ 500 ⫺ 100xy 14. Proceed as in Prob. 13 and note that Fig. 402 can be obtained from Fig. 401 in a rather obvious fashion.

SECTION 18.2. Use of Conformal Mapping. Modeling, page 763 Purpose. To show how conformal mapping helps in solving potential problems by mapping given domains onto simpler ones or onto domains for which the solution of the problem (subject to the transformed boundary conditions) is known. The theoretical basis of this application of conformal mapping is given by Theorem 1, characterizing the behavior of harmonic functions under conformal mapping. Problem 10 gives a hint on possibilities of generalizing potential problems for which the solution is known or can be easily obtained. The idea extends to more sophisticated situations. SOLUTIONS TO PROBLEM SET 18.2, page 766 2. It avoids the use of the existence of a conjugate harmonic. For some details on this concept, see Ref. [GenRef6], Sec. 193 on harmonic and subharmonic functions. 4. Figure 386 in Sec. 17.1 shows D, which is a semi-infinite horizontal strip, and D*, which is the upper half of the unit circular disk. Since u ⫽ ex cos y and v ⫽ ex sin y, the potential in D is £(x, y) ⫽ £*(u (x, y), v(x, y)) ⫽ 4u(x, y)v(x, y) ⫽ 4e2x cos y sin y ⫽ 2e2x sin 2y. This is a harmonic function. Its boundary values are 0 on the horizontal lines y ⫽ 0 and y ⫽ p and 2 sin 2y on the vertical segment x ⫽ 0 of D. 6. u2 ⫺ v2 ⫽ e2x (cos2 y ⫺ sin2 y) ⫽ e2x cos 2y ⫽ £(x, y) ⫽ Re e2z is harmonic, ⵜ2£ ⫽ 0. 8. £ ⫽ 2u(x, y)v (x, y) ⫽ 2 sin x cos x cosh y sinh y ⫽ 12 sin 2x sinh 2y, which is 0 if x ⫽ 0, 12 p or y ⫽ 0, and 12 sin 2x sinh 2 if y ⫽ 1. 10. We map 0 : r0, 2c : ⫺r0, obtaining from (2), with b ⫽ z 0, the conditions: r0 ⫽

⫺z 0 ⫺1

⫽ z 0,

⫺r0 ⫽

2c ⫺ z 0 2z 0c ⫺ 1

⫽

2c ⫺ r0 2r0c ⫺ 1

hence r0 ⫽

1 a1 ⫺ 21 ⫺ 4c2 b . 2c

r0 is real for positive c ⬉ 12 . Continue as in Example 1 of the text. Note that with increasing c (⬍ 12 ) the image (an annulus) becomes slimmer and slimmer. 12. ⫾i are fixed points, and straight lines are mapped onto circles or straight lines. From this the assertion follows.

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Alternatively, it also follows very simply by setting x ⫽ 0 and calculating ƒ w ƒ . 14. The potential on the y-axis is 1 ⫹ iy 1 ⫺ iy 6000 1 ⫹ 2iy ⫺ y 2 ⫽ Arg p 1 ⫹ y2 6000 2y ⫽ arctan p 1 ⫺ y2

£⫽

6000

Arg

and its derivative 1200>(p (1 ⫹ y 2)) has a maximum at y ⫽ 0. 16. Apply w ⫽ z 4. 18. By Theorem 1 in Sec. 17.2. z⫺a V0 V0 20. £ ⫽ p (⫺Arg (z ⫺ a) ⫹ Arg (z ⫹ a)), F ⫽ p i Ln z ⫹ a SECTION 18.3. Heat Problems, page 767 Purpose. To show that previous examples and new ones can be interpreted as potential problems in time-independent heat flow. Comment on Interpretation Change Boundary conditions of importance in one interpretation may be of no interest in another; this is about the only handicap in a change of interpretation. In other words, one should emphasize that, whereas the unifying underlying theory remains the same, problems of interest will change from field to field of application. This can be seen most distinctly by comparing the problem sets in this chapter.

SOLUTIONS TO PROBLEM SET 18.3, page 769 2. This is an approximate model of a long slender metal plate with insulated faces and edges kept at the indicated temperatures. By inspection we find that the temperature distribution is T(x, y) ⫽ 10 ⫹ 7.5(y ⫺ x). This is the real part of the complex potential F(z) ⫽ 10 ⫺ 7.5(1 ⫹ i)z. A systematic derivation is as follows. The boundary and boundary values suggest that T(x, y) is linear in x and y, T(x, y) ⫽ ax ⫹ by ⫹ c. From the boundary conditions, (1)

T(x, x ⫺ 4) ⫽ ax ⫹ b(x ⫺ 4) ⫹ c ⫽ ⫺20.

(2)

T(x, x ⫹ 4) ⫽ ax ⫹ b(x ⫹ 4) ⫹ c ⫽ 40.

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By addition, 2ax ⫹ 2bx ⫹ 2c ⫽ 20. Since this is an identity in x, we must have a ⫽ ⫺b and c ⫽ 10. From this and (1), ⫺bx ⫹ bx ⫺ 4b ⫹ 10 ⫽ ⫺20. Hence b ⫽ ⫺7.5. This agrees with our result obtained by inspection. F(z) ⫽ 200 ⫹ 10iz 2 4. T(x, y) ⫽ 200 ⫺ 20xy, y 150 150i 6. p arctan x ⫽ Re a⫺ p Ln zb y 8. Arg z ⫽ arctan x ⫽ Re (⫺i Ln z) is a basic building block when the boundary values have jumps. The reason is that it is a harmonic function, so that we may use it for the present purpose, and that it is discontinuous at 0, that is, it is 0 for positive x and p for negative x. Accordingly, we obtain a jump j at x ⫽ 0 by using j> p. And we obtain such a jump at x ⫽ a, as in the figure, by using Ln (z ⫺ a) instead of Ln z. To obtain an arbitrary jump with arbitrary values on both sides, we just have to include a few constants: in the figure we add T1 and obtain the right size of the jump by taking (T2 ⫺ T1)> p. Together we thus have in the figure T(x, y) ⫽ T1 ⫹

T2 ⫺ T1

Ln (z ⫺ a),

where the minus sign results from i # i. With this in mind, it is not difficult to produce arbitrary discontinuous boundary potentials by going along the real axis from right to left. Note that the solution of Prob. 7 is based on the same principle, with p>2 instead of p because we have a quadrant instead of a half-plane. 10. This generalizes Prob. 9. We obtain, using the idea explained in the solution to Prob. 8, T(x, y) ⫽ T1 ⫹

T2 ⫺ T1

Arg (z ⫺ b) ⫹

T3 ⫺ T2

Arg (z ⫺ a)

i i ⫽ Re aT1 ⫺ p (T2 ⫺ T1) Ln (z ⫺ b) ⫺ p (T3 ⫺ T2) Ln (z ⫺ a)b . 12. The temperature distribution is T(x, y) ⫽

Arg

cosh z ⫺ 1 . cosh z ⫹ 1

The boundary in the figure is mapped onto the u-axis; cosh x, 0 ⬉ x ⬍ ⬁, gives u ⭌ 1; cosh (x ⫹ pi) ⫽ ⫺cosh x, 0 ⬉ x ⬍ ⬁, gives u ⬍ ⫺1; the vertical segment (x ⫽ 0) is mapped onto ⫺1 ⬍ u ⬍ 1. Now use the boundary values in Prob. 11. 14. (400> p) Arg z. This is similar to Example 3. 16. The angle is 60° ⫽ p>3. Answer: T(x, y) ⫽ 20 ⫹

480 # 3

Arg z ⫽ Re a20 ⫺

1440i

Ln zb .

18. 100 ⫺ (260> p) Arg z ⫽ Re (100 ⫹ (260i> p) Ln z) 20. The lines of heat flow are perpendicular to the isotherms, and heat flows from higher to lower temperatures. Accordingly, heat flows from the portion of higher temperature

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of the unit circle ƒ Z ƒ ⫽ 1 to that kept at a lower temperature, along the circular arcs that intersect the isotherms at right angles. Of course, as temperatures on the boundary, we must choose values that are physically possible, for example, 10°C and 300°C. SECTION 18.4. Fluid Flow, page 771 Purpose. To give an introduction to complex analysis in potential problems of fluid flow. These two-dimensional flows are given by their velocity vector field, and our presentation in the text begins with an explanation of handling this field by complex methods. It is interesting that we use complex potentials as before, but whereas in electrostatics the real part (the real potential) is of central interest, here it is the imaginary part of the complex potential which gives the streamlines of the flow. Important Concepts Stream function °, streamlines ° ⫽ const Velocity potential £ , equipotential lines £ ⫽ const Complex potential F ⫽ £ ⫹ i° Velocity V ⫽ F r(z) Circulation (6), vorticity, rotation (9) Irrotational, incompressible Flow around a cylinder (Example 2, Team Project 16) SOLUTIONS TO PROBLEM SET 18.4, page 776 2. On circles x 2 ⫹ y 2 ⫽ const. No. 4. ƒ F r(eiu) ƒ 2 ⫽ (1 ⫺ cos 2u)2 ⫹ sin2 2u ⫽ 2 ⫺ 2 cos 2u, which is 0 for u ⫽ 0 and 4 for u ⫽ p>2, and the maximum speed is the square root of this. 6. Flow against a horizontal wall (the x-axis) 8. F ⫽ (1 ⫺ i)Kz> 12 because V ⫽ F r ⫽ (1 ⫹ i)K> 12 y and ƒV ƒ ⫽ K. 10. F(z) ⫽ iz 2 ⫽ i (x 2 ⫺ y 2) ⫺ 2xy gives the streamlines x 2 ⫺ y 2 ⫽ const.

The equipotential lines are

xy ⫽ const. The velocity vector is V ⫽ F r ⫽ ⫺2iz ⫽ ⫺2y ⫺ 2ix.

Section 18.4. Problem 10

See the figure. 12. w ⫽ f (z) ⫽ iz 2>K, F*(w) ⫽ ⫺iKw, F(z) ⫽ F*(w) ⫽ ⫺iKiz 2>K ⫽ z 2 14. F(z) ⫽ iz 3 ⫽ i(x 3 ⫹ 3ix 2y ⫺ 3xy 2 ⫺ iy 3) ⫽ ⫺3x 2y ⫹ y 3 ⫹ i(x 3 ⫺ 3xy 2) gives the streamlines x (x 2 ⫺ 3y 2) ⫽ const. This includes the three straight-line asymptotes x ⫽ 0 and y ⫽ ⫾x> 13 (which make 60° angles with one another, dividing the plane into six angular regions of angle 60° each),

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and we could interpret the flow as a flow in such a region. This is similar to the case F(z) ⫽ z 2, where we had four angular regions of 90° opening each (the four quadrants of the plane) and the streamlines were hyperbolas. In the present case the streamlines look similar but they are “squeezed” a little so that each stays within its region, whose two boundary lines it has for asymptotes. The velocity vector is V ⫽ ⫺6xy ⫹ 3i(y 2 ⫺ x 2) so that V2 ⫽ 0 on y ⫽ x and y ⫽ ⫺x. See the figure. y

Section 18.4. Problem 14

16. F(z) ⫽ z 2 ⫹ 1>z 2, ° ⫽ (r 2 ⫺ 1>r 2) sin 2u ⫽ 0 if r ⫽ 1 (the cylinder wall) or u ⫽ 0, ⫾p>2, p. The unit circle and the axes are streamlines. z ⫽ ⫾1, ⫾i are stagnation points. For large ƒ z ƒ the flow is “parallel” to the x-axis and also to the y-axis. For smaller ƒ z ƒ ⭌ 1 it is a flow in the first quadrant around a quarter of ƒ z ƒ ⬉ 1. Similarly in the other quadrants. 18. w ⫽ arccosh z implies z ⫽ x ⫹ iy ⫽ cosh w ⫽ cos iw ⫽ sin (iw ⫹ 12 p). Along with an interchange of the roles of the z- and w-planes, this reduces the present problem to the consideration of the sine function in Sec. 17.4 (compare with Fig. 391). We now have the hyperbolas x2 sin2 c

⫺

y2 cos2 c

⫽1

as streamlines, where c is different from the zeros of sine and cosine, and as limiting cases the y-axis and the two portions of the aperture.

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20. Team Project. (b) F(z) ⫽ ⫺

301

iK iK K ln z ⫽ ⫺ ln ƒ z ƒ ⫹ arg z. 2p 2p 2p

Hence the streamlines are circles K ln ƒ z ƒ ⫽ const, 2p

ƒ z ƒ ⫽ const.

thus

The formula also shows the increase of the potential K arg z 2p

£(x, y) ⫽

under an increase of arg z by 2p, as asserted in (b). 1 1 ln (z ⫹ a) (source). F2(z) ⫽ ⫺ ln (z ⫺ a) (sink). The minus (d) F1(z) ⫽ 2p 2p sign has the consequence that the flow is directed radially inward toward the sink because the velocity vector V is V ⫽ F r (z) ⫽ ⫺ ⫽⫺

1 # 1 # 1 1 ⫽⫺ 2p z ⫺ a 2p x ⫺ iy ⫺ a 1 # x ⫺ a ⫹ iy . 2p (x ⫺ a)2 ⫹ y 2

For instance, at z ⫽ a ⫹ i (above the sink), V⫽⫺

i , 2p

which is directed vertically downward, that is, in the direction of the sink at a. (e) The addition gives iK 1 F(z) ⫽ z ⫹ z ⫺ 2p ln z ⫽x⫹

x x ⫹y 2

⫹ i ay ⫺

⫹

K 2p

y x ⫹y 2

arg z

⫺

K 2p

ln 2x 2 ⫹ y 2 b .

Hence the streamlines are °(x, y) ⫽ Im F(z) ⫽ y ⫺

y x ⫹y 2

⫺

K 2p

ln 2x 2 ⫹ y 2 ⫽ const.

In both flows that we have added, ƒ z ƒ ⫽ 1 is a streamline; hence the same is true for the flow obtained by the addition. Depending on the magnitude of K, we may distinguish among three types of flow having two or one or no stagnation points on the cylinder wall. The speed is ƒV ƒ ⫽ ƒ F r(z) ƒ ⫽ ƒ F r(z) ƒ ⫽ ` a1 ⫺

1 iK `. b⫺ z2 2pz

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We first note that ƒV ƒ : 1 as ƒ z ƒ : ⬁; actually, V : 1, that is, for points at a great distance from the cylinder the flow is nearly parallel and uniform. The stagnation points are the solutions of the equation V ⫽ 0, that is, z2 ⫺

(A)

iK z ⫺ 1 ⫽ 0. 2p

We obtain z⫽

iK 4p

⫾

⫺K 2 . 2 ⫹ 1 B 16p

If K ⫽ 0 (no circulation), then z ⫽ ⫾1, as in Example 2. As K increases from 0 to 4p, the stagnation points move from z ⫽ ⫾1 up on the unit circle until they unite at z ⫽ i. The value K ⫽ 4p corresponds to a double root of the equation (A). If K ⬎ 4p, the roots of (A) become imaginary, so that one of the stagnation points lies on the imaginary axis in the field of flow while the other one lies inside the cylinder, thus losing its physical meaning. SECTION 18.5. Poisson’s Integral Formula for Potentials, page 777 Purpose. To represent the potential in a standard region (a disk ƒ z ƒ ⬉ R) as an integral (5) over the boundary values; to derive from (5) a series (7) that gives the potential and for ƒ z ƒ ⫽ R is the Fourier series of the boundary values. So here we see another important application of Fourier series, much less obvious than that of vibrational problems, where one can “see” the cosine and sine terms of the series. Comment on Footnote 2 Poisson’s discovery (1812) that Laplace’s equation holds only outside the masses (or charges) resulted in the Poisson equation (Sec. 12.1). The publication on the Poisson distribution (Sec. 24.7) appeared in 1837. SOLUTIONS TO PROBLEM SET 18.5, page 781 6. £(r, u) ⫽ 5 ⫺ r 2 cos 2u 8. £(r, u) ⫽ 3r sin u ⫺ r 3 sin 3u, as follows from sin3 u ⫽ 34 sin u ⫺ 14 sin 3u. 10. cos3 2u ⫽ 34 cos 2u ⫹ 14 cos 6u. Hence the answer is £(r, u) ⫽ 12r 2 cos 2u ⫹ 4r 6 cos 6u. 12. This is a sine series plus the constant term k>2 because £(1, u) ⫺ k>2 is odd. Answer: £(r, u) ⫽

k 2k 1 1 ⫹ ar sin u ⫹ r 3 sin 3u ⫹ r 5 sin 5u ⫹ Á b . 2 p 3 5

14. The given boundary potential is an even function. Answer: £(r, u) ⫽

4 1 1 1 5 ⫺ 2 ar cos u ⫹ r 3 cos 3u ⫹ r cos 5u ⫹ Á b . p 2 9 25

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16. The given boundary potential is odd. Answer: £(r, u) ⫽ ⫺2(r sin u ⫹ 12 r 2 sin 2u ⫹ 13 r 3 sin 3u ⫹ Á ). 18. The boundary potential is neither even nor odd, and we obtain a series of both cosine and sine terms, corresponding to the representation ƒ u ƒ >2 ⫹ u>2 of the boundary potential. Answer: £(r, u) ⫽

1 3 1 5 r cos 3u ⫹ r cos 5u ⫹ Á b 4 p 9 25 1 1 ⫹ r sin u ⫺ r 2 sin 2u ⫹ r 3 sin 3u ⫺ ⫹ Á . 2 3

⫺

ar cos u ⫹

20. Team Project. (a) r ⫽ 0 in (5) gives £(0) ⫽

1 2p

冮 £(R, a) da. Note that the 0

interval of integration has length 2p, not 2pR. (b) ⵜ2u ⫽ 0, u ⫽ g(r)h(u), gs h ⫹ gr gs r 2 g ⫹ r g ⫽ n 2,

1 1 gr h ⫹ 2 ghs ⫽ 0; hence by separating variables r r

hs ⫽ ⫺n 2, h

h ⫽ an cos nu ⫹ bn sin nu.

Also, A solution is r n>R n.

r 2g s ⫹ rg r ⫺ n 2g ⫽ 0.

r ° ⫽ °(0) ⫹ a a b (⫺bn cos nu ⫹ an sin nu). R n⫽1 (d) From the series for £ and ° we obtain by addition ⴥ r F(z) ⫽ a0 ⫹ i°(0) ⫹ a a b [(an ⫺ ibn) cos nu ⫹ i(an ⫺ ibn) sin nu] R n⫽1 n

ⴥ r ⫽ a0 ⫹ i°(0) ⫹ a a b (an ⫺ ibn)einu , R n⫽1 n

1 an ⫺ ibn ⫽ p

冮 £(R, a)e 0

ⴚina

da.

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Using z ⫽ reiu, we have the power series ⴥ

F(z) ⫽ a0 ⫹ i°(0) ⫹ a

an ⫺ ibn Rn

n⫽1

z n.

SECTION 18.6. General Properties of Harmonic Functions. Uniqueness Theorem for the Dirchlet Problem, page 781 Purpose. We derive general properties of analytic functions and from them corresponding properties of harmonic functions. Main Content, Important Properties Mean value of analytic functions over circles (Theorem 1) Mean value of harmonic functions over circles, over disks (Theorem 2) Maximum modulus theorem for analytic functions (Theorem 3) Maximum principle for harmonic functions (Theorem 4) Uniqueness theorem for the Dirichlet problem (Theorem 5) Comment on Notation Recall that we introduced F to reserve f for conformal mappings (beginning in Sec. 18.2), and we continue to use F also in this last section of Chap. 18. SOLUTIONS TO PROBLEM SET 18.6, page 784 2. Use (2), obtaining 1 F (2) ⫽ 32 ⫽ 2p 2 ⫽ 2p

冮 2(⫺2 ⫹ e ) da ia 4

冮 ((⫺2) ⫹ Á ) da 4

2 ((⫺2)4 # 2p ⫹ 0 ⫹ ⫹ Á ) 2p ⫽ 32. ⫽

4. Note first that F(z) ⫽ (z ⫺ 1)ⴚ2 is analytic in the disk ƒ z ⫺ z 0 ƒ ⫽ ƒ z ⫹ 1 ƒ ⬉ 1. From (2) and the binomial series we obtain F(z 0) ⫽ F(⫺1) ⫽

⫽

1 2p

1 1 ⫽ 4 2p

冮 (⫺2 ⫹ e ) da ia ⴚ2

冮 [(⫺2) ⫹ Á ] da ⫽ 41 . ⴚ2

6. The first statement in Theorem 2 is obtained by setting r ⫽ 0 in Poisson’s integral formula and noting that we can choose any point as center 0 and radius R of the circle of integration, as long as this circle remains in the (simply connected) domain in which £ is harmonic.

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8. The value £(3, 8) ⫽ 9 ⫺ 64 ⫽ ⫺55 is obtained from the integral 1

£(3, 8) ⫽ p

冮冮 [(3 ⫹ r cos a) ⫺ (8 ⫹ r sin a) ]r dr da 2

1 ⫽p

⫽

冮冮 (9 ⫺ 64)r ⫹ (6 cos a ⫺ 16 sin a)r ⫹ (cos a ⫺ sin a)r ] dr da 2

p c 1

⫺55 #

1 # 1 2 ⫹ 0 ⫹ (p ⫺ p) # d ⫽ ⫺55. 2 4

12. Team Project. (a) (i) Polar coordinates show that ƒ F(z) ƒ ⫽ ƒ z ƒ 2 assumes its maximum 41 and its minimum 5 at the boundary points 5 ⫹ 4i and 1 ⫹ 2i, respectively, and at no interior point. (ii) At z ⫽ ⫾i we obtain the maximum ƒ F(z) ƒ ⫽ [sin2 x ⫹ sinh2 y]1>2 ⫽ sinh 1 ⫽ 1.1752 and at z ⫽ 0 the minimum 0. (iii) Use the fact that ƒ ez ƒ ⫽ ex is monotone. (b) No, F(z) is not analytic. (c) ƒ sin z ƒ 2 ⫽ sin2 x ⫹ sinh2 y (Sec. 13.6) shows that, for instance, in the disk ƒ z ⫺ p>2 ƒ ⫽ 1 the maximum 21 ⫹ sinh2 1 ⫽ 1.5431 is taken at z ⫽ p>2 ⫾ i, not at p>2. (d) Since the interior B of C is simply connected, Theorem 3 applies. By Theorem 3, the maximum of ƒ F(z) ƒ is assumed on C. We shall obtain a contradiction from the assumption that F(z) has no zeros inside C. Then, by the last statement in Theorem 3, it would follow that ƒ F(z) ƒ also has its minimum on C. But on C the absolute value ƒ F(z) ƒ is constant by assumption, so that the minimum equals the maximum, implying that ƒ F(z) ƒ is constant in D and, by Example 3 in Sec. 12.4, F(z) ⫽ const in D. This contradicts the assumption that F(z) is not constant in D, hence our assumption that F(z) has no zeros inside C must be false, and the proof is complete. z ⫽ 1, Max ⫽ e 14. ƒ F(z) ƒ ⫽ exp (x 2 ⫺ y 2), 16. Since ƒ z ƒ ⬉ 1, the triangle inequality yields ƒ az ⫹ b ƒ ⬉ ƒ a ƒ ⫹ ƒ b ƒ . The maximum lies on ƒ z ƒ ⫽ 1. Write a ⫽ ƒ a ƒ eia, z ⫽ eiu, b ⫽ ƒ b ƒ eib. Choose u ⫽ b ⫺ a. Then ƒ az ⫹ b ƒ ⫽ ` ƒ a ƒ eia⫹i(b⫺a) ⫹ ƒ b ƒ eib ` ⫽ ( ƒ a ƒ ⫹ ƒ b ƒ ) ƒ eib ƒ ⫽ ƒ a ƒ ⫹ ƒ b ƒ . Answer: ƒ a ƒ ⫹ ƒ b ƒ , taken at z ⫽ ei(b⫺a). Check by substitution: az ⫹ b ⫽ ƒ a ƒ eiaei(bⴚa) ⫹ ƒ b ƒ eib ⫽ ( ƒ a ƒ ⫹ ƒ b ƒ )eib and ƒ eib ƒ ⫽ 1. 18. ex ⬉ eb with equality only at x ⫽ b. Also, sin y ⬉ 1 with equality only at y ⫽ p>2; and (b, p>2) lies on the boundary. 20. £ ⫽ exp (x 2 ⫺ y 2) cos 2xy, R: ƒ z ƒ ⬉ 1, x ⭌ 0, y ⭌ 0. Yes, (u 1, v1) ⫽ (1, 0) is the image of (x 1, y1) ⫽ (1, 0); this is typical. (u 1, v1) is found by noting that on the boundary (semicircle), £* ⫽ eu cos (21 ⫺ u 2) increases monotone with u. Similarly for R.

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SOLUTIONS TO CHAPTER 18 REVIEW QUESTIONS AND PROBLEMS, page 785 2. Complex integration, conformal mapping, higher special functions and their series, potential theory. 4. Electrostatics, corresponding Dirichlet boundary value problems. Heat conduction that is stationary (time-independent), so that the heat equation reduces to the Laplace equation; Dirichlet and mixed boundary value problems. Incompressible flow problems, such as fluid flows around cylinders, without and with circulation, airflow around profiles, for instance, Joukowski flow. Certain problems admit interpretations in more than one area of physics, but one should note the following. Whereas in electrostatics, the main interest concentrates on equipotential lines, in fluid flow one is mainly interested in the form of streamlines along which particles of the flow are moving. 6. It is of importance that both the real part and the imaginary part of the complex potential have a physical meaning. This gives uniformity of approach to applications in different areas, regardless of whether the potential or its harmonic conjugate is more important from the viewpoint of physics. Also, in applications of methods of complex analysis, the complex potential is the more natural and often simpler instrument in solving problems. 8. The idea is to map the domain in which a solution is wanted onto a domain for which the solution is known or is more easily obtainable. The method works because under a conformal mapping a harmonic function goes over into a harmonic function. 10. Note that the transition from the Cauchy formula, which is complex, to the Poisson formula, which is real, is accomplished by the addition of an integral that has the value zero. 12. In this case, the general form of the potential is (derivation in Sec. 18.1) £ ⫽ a ln r ⫹ b. From the boundary conditions, U1 ⫽ 0 ⫹ b ⫽ 200,

b ⫽ 200

U2 ⫽ a ln 10 ⫹ 200 ⫽ 2000. Hence a ⫽ 1800>ln 10 ⫽ 781.7. The answer is £ ⫽ a ln r ⫹ b ⫽ Re (a Ln z ⫹ b) ⫽

1800 ln r ⫹ 200. ln 10

14. Since the logarithmic curve is convex, the value in Prob. 12 is larger than that in the present problem, which involves a linear function. 16. Arg z ⫽ const 18. £ ⫽ 1000 ⫺ (1200> p) Arg z 20. Circular arcs, as follows from orthogonality. 22. y(x ⫹ 1) ⫽ const, flow within the corner formed by the straight lines y ⫽ 0 and x ⫽ ⫺1. The streamlines are hyperbolas. 24. We obtain the harmonic function £(x, y) ⫽ £* (u(x, y), v (x, y)) ⫽ £*(x 2 ⫺ y 2, 2xy) ⫽ exp (x 2 ⫺ y 2) sin 2xy.

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PART E Numeric Analysis The subdivision into three chapters has been retained. All three chapters have been updated in the light of computer requirements and developments. A list of suppliers of software (with addresses, etc.) can be found at the beginning of Part E of the book and another list at the beginning of Part G.

CHAPTER 19

Numerics in General

SECTION 19.1. Introduction, page 790 Purpose. To familiarize the student with some facts of numerical work in general, regardless of the kind of problem or the choice of method. Main Content, Important Concepts Floating-point representation of numbers, overflow, underflow Roundoff Concept of algorithm Stability Errors in numerics, their propagation, error estimates Loss of significant digits Short Courses. Mention the roundoff rule and the definitions of error and relative error.

SOLUTIONS TO PROBLEM SET 19.1, page 796 2. ⫺0.7644 ⴢ 102, 0.6010 ⴢ 106, ⫺0.1000 # 10ⴚ4 Numerics needs practical experience. Experience cannot be taught but it must be gained. However, the present problem set should serve as an eye opener, illustrating various aspects and some (not all!) unexpected facts occurring in numerics. 4. (a ⫹ b) ⫹ c ⫽ a ⫹ (b ⫹ c) in general, because of rounding, so that the order matters. Take small terms first. Example: (0.0004 ⫹ 0.0004) ⫹ 1.000 ⫽ 1.001

(4S)

but (1.000 ⫹ 0.0004) ⫹ 0.0004 ⫽ 1.000 ⫹ 0.0004 ⫽ 1.000

(4S).

6. 61.1 ⫺ 7.5 ⴢ 15.5 ⫹ 11.2 ⴢ 3.94 ⫹ 2.80 ⫽ 61.1 ⫺ 116 ⫹ 44.1 ⫹ 2.80 ⫽ ⫺8.0 ((x ⫺ 7.5)x ⫹ 11.2)x ⫹ 2.80 ⫽ (⫺3.56 # 3.94 ⫹ 11.2)3.94 ⫹ 2.80 ⫽ (⫺14.0 ⫹ 11.2)3.94 ⫹ 2.80 ⫽ ⫺11.0 ⫹ 2.80 ⫽ ⫺8.20 Exact: ⫺8.336016 307

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8. x ⫽ 20 ⫾ 1398 ⫽ 20.00 ⫾ 19.95, so that (4) gives 39.95 and 0.05, and (5) gives 39.95 and 2.000>39.95 ⫽ 0.05006, in error by less than one unit of the last digit. The 10S-value is 0.05006265674. 10. The roots are x ⫽ k ⫾ 1a. A change from a ⫽ 0 to, say, a ⫽ 10 ⴚ4 , produces a change of 1a ⫽ 100 in the roots. This is typical of the very large sensitivity of roots of polynomials that occurs quite often. 14. (a) 14.00000 ⫺ 2.00000 ⫽ 0, whereas (b) gives 0.000001>4.00000 ⫽ 0.25 ⴢ 10ⴚ6. 16. (a) 1.00000 ⫺ 0.999800 ⫽ 0.000200. (b) We obtain 1 ⫺ cos x ⫽ 1 ⫺ (cos2 12 x ⫺ sin2 12 x) ⫽ 2 sin2 12 x (see (7) in App. A3.1). The corresponding 6S-computation is 2 sin2 0.01 ⫽ 2 ⴢ 0.009999832 ⫽ 2 ⴢ 0.0000999966 ⫽ 0.000199993, which is 6S-exact. 18. (a) (1 ⫺ cos 0.005)>sin 0.005 ⫽ (1 ⫺ 0.999988)>0.00499998 ⫽ 0.000050>0.00499998 ⫽ 0.00240001 (b) Taking the numerator as in Prob. 16 and using (7) in App. A3.1 also for the present denominator, we obtain 2 sin2 12 x>(2 sin 12 x cos 12 x) ⫽ tan 12 x,

tan 0.0025 ⫽ 0.00250001.

20. Correct answer is 4.53999 # 10ⴚ5. (a) Forty terms give a 6S-sum that does not change any more by adding further terms, the sum being 0.00802204, the error being 4.539993 ⴢ 10ⴚ5 ⫺ 0.00802204 ⫽ ⫺7.9766401 ⫻ 10ⴚ3, so that the relative error is 7.9766401 ⫻ 10ⴚ3>4.539993 ⴢ 10ⴚ5 ⫽ 175.69719. The partial sum used is eⴚ10 ⬇ 1 ⫺ 10 ⫹ 50 ⫹ ⫺ Á ⫹ 1.225617 ⴢ 10ⴚ8. (b) Taking 27 terms and then taking the reciprocal gives 1>22026.3 ⫽ 4.54003 ⫻ 10ⴚ5 with error 4.53999 ⴢ 10ⴚ5 ⫺ 4.54003 ⫻ 10ⴚ5 ⫽ ⫺4.0 ⫻ 10ⴚ10 and relative error 4.0 ⫻ 10ⴚ10>4.53999 ⴢ 10ⴚ5 ⫽ 8.810592 ⫻ 10ⴚ6. The work is less, but the result is useful, in contrast to (a). 22. Check backward: (0.10011)2 ⫽ 1 ⴢ 2ⴚ1 ⫹ 0 ⴢ 2ⴚ2 ⫹ 0 ⴢ 2ⴚ3 ⫹ 1 ⴢ 2ⴚ4 ⫹ 1 ⴢ 2ⴚ5 ⫽ 0.59375. 24. A binary machine number is a (finite!) sum of terms each of which has a finite decimal representation. The converse is not true. Take 0.1, for example. Prove that a number a is a binary machine number if and only if a ⫽ m>2n, where m and n are integers. Indeed, if its binary representation is finite, you can convert it to this form by taking the common denominator. And conversely. 26. (a) In ⫽ 1.718, 1.000, 0.7180, 0.5640, 0.4620, 0.4080, 0.2700, 0.8280, and ⫺3.906. (b) Later, because the roundoff errors become smaller with increasing k. 28. No, because am (the computer number for 1>m) is zero from some m on. This is typical. Indeed, more generally, since convergence of a series of numbers implies that the terms must approach zero, the corresponding computer numbers must be zero from some m on, which means convergence.

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SECTION 19.2. Solution of Equations by Iteration, page 798 Purpose. Discussion of the most important methods for solving equation f (x) ⫽ 0, a very important task in practice. Main Content, Important Concepts Solution of f (x) ⫽ 0, by iteration (3) x n⫹1 ⫽ g (x n) Condition sufficient for convergence (Theorem 1) Newton (–Raphson) method (5) Speed of convergence, order Secant, bisection, false position methods Comments on Content Fixed-point iteration gives the opportunity to discuss the idea of a fixed point, which is also of basic significance in modern theoretical work (existence and uniqueness of solutions of differential, integral, and other functional equations). The less important method of bisection and method of false position are included in the problem set. SOLUTIONS TO PROBLEM SET 19.2, page 807 2. The sequences corresponding to the initial values x 0 ⫽ 1, 0.5, 2 are (rounded) 1, 0, 1, 0, 1, Á 0.50000, 0.87500, 0.33008, 0.96404, 0.10405, 0.99887, Á 2, ⫺7, 344, ⫺0.40708 ⴢ 108 Á 4. g ⫽ cosec x, x 0 ⫽ 1, x 1 ⫽ 1.8840, Á , x 17 ⫽ 1.11416 (exact to 6S) 6. x ⫽ g (x) ⫽ 15 (x 2 ⫹ 1.01 ⫹ 1.88>x). Starting from x 0 ⫽ 1 gives the sequence 1, 0.778, 0.806347, 0.79840, 0.800447, 0.799881, 0.800032, 0.799991, 0.800002, 0.799999, 0.800000 the last value being exact. 4 8. x ⫽ 2 x ⫹ 0.12; 1, 1.02874, Á 1.03717 (6S exact, 7 steps) 10. x ⫽ 1>cosh x; 1, 0.64805, 0.82140, 0.73706, Á approaches 0.76501 (5S-exact) in a nonmonotone fashion. 12. Roots r1 ⫽ 1.56155 (6S-value), r2 ⫽ ⫺1 (exact), r3 ⫽ ⫺2.56155 (6S-value). (1) r1, 12 steps, (2) r1, about 25 steps, (3) r2, about 15 steps, (4) r3, about 10 steps, (5) r2, but alternating between ⫺0.9916 Á and ⫺1.008 Á (Maple program), and (6) r1, 3 steps. This is Newton’s method. Choosing x 0 ⫽ 2.0, you will obtain divergence in Cases (3) and (5) but convergence to the same limit in the other cases. 14. x n⫹1 ⫽ x n ⫺ (x 3n ⫺ 7)>(3x 2n) ⫽ (2x n ⫹ 7>x 2n)>3 16. Convergence to the same limit, perhaps more steps needed, and convergence may take place in a nonmonotone fashion. The reason becomes clear from the graph of f. 18. (a) 0.906180 (6S-exact, 4 steps, x 0 ⫽ 1). (b) 13 25 ⫹ 27 170 ⫽ 0.906179846 20. 2, 1.537902, 1.557099, 1.557146, 1.557146 22. 2, 2.452, 2.473; 39.02°C

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24. (a) ALGORITHM REGULA FALSI ( f, a 0, b0, P, N), Method of False Position. This algorithm computes the solution c of f (x) ⫽ 0 ( f continuous) within the tolerance P. INPUT: Continuous function f, initial interval [a0, b0], tolerance P, maximum number of iterations N. OUTPUT: A solution c (within the tolerance P), or a message of failure. For n ⫽ 0, 1, Á , N ⫺ 1 do: Compute c ⫽

an f (bn) ⫺ bn f (an) f (bn) ⫺ f (an)

If f (c) ⫽ 0 then OUTPUT c Stop. [Procedure completed] Else if f (an)f (bn) ⬍ 0 then set an⫹1 ⫽ an and bn⫹1 ⫽ c. Else set an⫹1 ⫽ c and bn⫹1 ⫽ bn. If ƒ an⫹1 ⫺ bn⫹1 ƒ ⬍ P ƒ c ƒ then OUTPUT c. Stop. [Procedure completed] End OUTPUT [aN, bN] and a message “Failure”. Stop. [Unsuccessful completion; N iterations did not give an interval of length not exceeding the tolerance.] End REGULA FALSI (b) 1.18921, 26. 1, 28. 1,

0.64171,

1.55715

0.7, 0.577094, 0.534162, 0.531426, 0.531391, 0.531391 2, 0.765035, 0.742299, 0.739103, 0.739085 (6S-exact)

SECTION 19.3. Interpolation, page 808 Purpose. To discuss methods for interpolating (or extrapolating) given data (x 0, f0), Á , (x n, fn), all x j different, arbitrarily or equally spaced, by polynomials of degree not exceeding n. Main Content, Important Concepts Lagrange interpolation (4) (arbitrary spacing) Error estimate (5) Newton’s divided difference formula (10) (arbitrary spacing) Newton’s difference formulas (14), (18) (equal spacing) Short Courses. Lagrange’s formula briefly, Newton’s forward difference formula (14). Comment on Content For given data, the interpolation polynomial pn (x) is unique, regardless of the method by which it is derived. Hence the error estimate (5) is generally valid (provided f is n ⫹ 1 times continuously differentiable). SOLUTIONS TO PROBLEM SET 19.3, page 819 2. This parallels Example 3. From (5) with n ⫽ 1 we obtain P1 (9.3) ⫽ (x ⫺ 9)(x ⫺ 9.5)

f s(t) 2

x⫽9.3

⫽ 0.3(⫺0.2)(⫺12 t ⴚ2) ⫽ 0.03t ⴚ2

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where 9 ⬉ t ⬉ 9.5. Now t ⴚ2 is a monotone function of t. Hence, its extrema occur at 9.0 and 9.5. We thus obtain 0.03>9.52 ⫽ 0.0003 ⬉ a ⫺ a苲 ⬉ 0.03>9.02 ⫽ 0.0004. This gives the answer (add 苲 a ⫽ 2.2297 on both sides) 2.2300 ⬉ a ⬉ ln 9.3 ⫽ 2.2301. The exact 4S-value is 2.2300. 4. From (5) we obtain P2(9.2) ⫽ (x ⫺ 9)(x ⫺ 9.5)(x ⫺ 11) ⫺ ⫽

0.036 t3

(ln t) t 3!

x⫽9.2

We now have to find the extrema of the right side in the interval 9 ⬉ t ⬉ 11. Usually, one would have to equate the next derivative to zero and solve the resulting equation by the Newton–Raphson method or by some other root-finding method. Presently, the right side is monotone; hence, its extreme values occur at the end of the interval 9 ⬉ t ⬉ 11. This gives 0.000027 ⬉ P2(9.2) ⫽ a ⫺ a苲 ⬉ 0.000050 and by adding 苲 a ⫽ 2.2192 from Example 2, 2.2192 ⬉ a ⬉ 2.2193. 5S-exact is 2.2192. 6. From L 0(x) ⫽ x 2 ⫺ 20.5x ⫹ 104.5 1 (⫺x 2 ⫹ 20x ⫺ 99) L 1(x) ⫽ 0.75 1 L 2(x) ⫽ (x 2 ⫺ 18.5x ⫹ 85.5) 3 (see Example 2) and the 5S-values of the logarithm in the text we obtain p2(x) ⫽ ⫺0.005233x 2 ⫹ 0.205017x ⫹ 0.775950. This gives the values and errors 2.2407,

error 0

2.3028,

error ⫺0.0002

2.3517,

error ⫺0.0003

2.4416,

error 0.0007

2.4826,

error 0.0024.

It illustrates that in extrapolation one may usually get less accurate values than one does in interpolation. p2(x) would change if we took more accurate values of the logarithm. Small changes in initial values can produce large differences in final values. Examples in which the difference in accuracy between interpolation and extrapolation is larger can easily be constructed.

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8. Fast decrease for decreasing x smaller than 9, rapid increase for increasing x larger than 11. 6 4 2 0 ⫺2

10 x

⫺4

Problem 8. Function (x ⫺ 9)(x ⫺ 9.5)(x ⫺ 11)

10. From (5) we obtain P2(0.75) ⫽ (x ⫺ 0.25)(x ⫺ 0.5)(x ⫺ 1)

f t(t) 6

x⫽0.75

⫽ ⫺0.005208f t(t) where, by differentiation, f t(t) ⫽

2 ⫺4 (1 ⫺ 2t 2)eⴚt . 1p

Another differentiation shows that f t is monotone on the interval 0.25 ⬉ t ⬉ 1 because f iv(t) ⫽

2 8t (3 ⫺ 2t 2)eⴚt ⫽ 0 1p

on that interval. Hence the extrema of f t occur at the ends of the interval, so that we obtain f t(0.25) ⫽ ⫺1.8550,

f t(1) ⫽ 0.83021

and thus ⫺0.00432 ⬉ a ⫺ 苲 a ⬉ 0.00966 and by adding 苲 a ⬉ 0.70929 (see Prob. 9) 0.70497 ⬉ a ⬉ 0.71895. Exact: 0.71116 (5S). 12. The difference table is xj

f (x j)

1.0

0.94608

1.5

1.32468

1st Diff.

2nd Diff.

3rd Diff

0.37860 ⫺0.09787 0.28073 2.0

1.60541

2.5

1.77852

⫺0.10762 0.17311

⫺0.00975

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The interpolating values and errors are p1(1.25) ⫽ f (1.0) ⫹ 0.5 # 0.37860 ⫽ 1.13538 (P ⫽ 0.011017) 0.5(⫺0.5) # (⫺0.09787) ⫽ 1.14761 (P ⫽ ⫺0.00116) p2(1.25) ⫽ p1(1.25) ⫹ 2 0.5(⫺0.5)(⫺1.5) # (⫺0.00975) ⫽ 1.14700 (P ⫽ ⫺0.00055) p3(1.25) ⫽ p2(1.25) ⫹ 6 Note the decrease of the error. 14. p2(x) ⫽ 1.0000 ⫺ 0.0112r ⫹ 0.0008r (r ⫺ 1)>2 ⫽ x 2 ⫺ 2.580x ⫹ 2.580, r ⫽ (x ⫺ 1)>0.02; 0.9943, 0.9835, 0.9735, exact to 4S 16. From (10) and the difference table j

f (x)

0.25

0.27633

0.5

0.5250

f [x j, x j⫹1]

f [x j, x j⫹1, x j⫹2]

0.97667 ⫺0.44304 0.64640 2

1.0

0.84270

we obtain p2(0.75) ⫽ 0.2763 ⫹ (0.75 ⫺ 0.25) # 0.9767 ⫹ (0.75 ⫺ 0.25)(0.75 ⫺ 0.5) # (⫺0.4430) ⫽ 0.7093, in agreement with Prob. 9, except for a roundoff error. 18. With the change in j the difference table is j

fj ⫽ cosh x j

⫺3

0.5

1.127626

⫺2

0.6

1.185465

ⵜfj

ⵜ2fj

ⵜ3fj

0.057839 0.011865 0.069704

⫺1

0.7

1.255169

0.8

1.337435

0.000697 0.012562

0.082266

From this and (18) we obtain x ⫺ 0.8 0.1 (x ⫺ 0.8)(x ⫺ 0.7)

p3(x) ⫽ 1.337435 ⫹ 0.082266 # ⫹ 0.0012562 # ⫹ 0.000697 #

0.01 # 2! (x ⫺ 0.8)(x ⫺ 0.7)(x ⫺ 0.6) 0.001 # 3!

and with x ⫽ 0.56 this becomes 1.337435 ⫹ 0.082266(⫺2.4) ⫹ 0.012562(⫺2.4)(⫺1.4)>2 ⫹ 0.000697(⫺2.4)(⫺1.4)(⫺0.4)>6 ⫽ 1.160945. This agrees with Example 5. The correct last digit is 1 (instead of 5 here or 4 in Example 5).

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20. Team Project. (a) For p1(x) we need x ⫺ x1 L 0 ⫽ x ⫺ x ⫽ 19 ⫺ 12x, 0

x ⫺ x0 L 1 ⫽ x ⫺ x ⫽ ⫺18 ⫹ 2x

p1(x) ⫽ 2.19722(19 ⫺ 2x) ⫹ 2.25129(⫺18 ⫹ 2x) ⫽ 1.22396 ⫹ 0.10814x p1(9.2) ⫽ 2.21885. Exact: 2.21920, error 0.00035. For p2 we need 2 L 0 ⫽ 104.5 ⫺ 41 2 x ⫹ x 4 2 L 1 ⫽ ⫺132 ⫹ 80 3 x ⫺ 3x 1 2 L 2 ⫽ 28.5 ⫺ 37 6 x ⫹ 3x .

This gives (with 10S-values for the logarithm) p2(x) ⫽ 0.779466 ⫹ 0.204323x ⫺ 0.0051994x 2, hence p2(9.2) ⫽ 2.21916, error 0.00004. The error estimate is p2(9.2) ⫺ p1(9.2) ⫽ 0.00031. (b) Extrapolation gives a much larger error. The difference table is 0.2

0.9980

0.4

0.9686

⫺0.0294 ⫺0.1243 0.6

0.8443 ⫺0.3085

0.8

0.5358 ⫺0.5358

1.0

⫺0.0949 ⫺0.1842 ⫺0.2273

0.0000

The differences not shown are not needed. Taking x ⫽ 0.6, 0.8, 1.0 gives the best result. Newton’s formula (14) with r ⫽ 0.1>0.2 ⫽ 0.5 gives 0.8443 ⫹ 0.5 # (⫺0.3085) ⫹

0.5 # (⫺0.5) 2

# (⫺0.2273) ⫽ 0.7185, P ⫽ ⫺0.0004.

Similarly, by taking x ⫽ 0.4, 0.6, 0.8 we obtain 0.9686 ⫹ 1.5 # (⫺0.1243) ⫹

1.5 # 0.5 # (⫺0.1842) ⫽ 0.7131, 2

P ⫽ 0.0050.

Taking x ⫽ 0.2, 0.4, 0.6, we extrapolate and get a much poorer result: 0.9980 ⫹ 2.5 # (⫺0.0294) ⫹

2.5 # 1.5 # (⫺0.0949) ⫽ 0.7466, 2

P ⫽ ⫺0.0285.

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SECTION 19.4. Spline Interpolation, page 820 Purpose. Interpolation of data (x 0, f0), Á , (x n, fn) by a (cubic) spline, that is, a twice continuously differentiable function g(x) which, in each of the intervals between adjacent nodes, is given by a polynomial of third degree at most, on [x 0, x 1] by q0(x),

on [x 1, x 2] by q1(x)

Á,

on [x nⴚ1, x n] by qnⴚ1(x).

Short Courses. This section may be omitted. Comments on Content Higher order polynomials tend to oscillate between nodes—the polynomial P10(x) in Fig. 434 is typical—and splines were introduced to avoid that phenomenon. This motivates their application. It is stated in the text that splines also help lay the foundation of CAD (computeraided design). If we impose the additional condition (3) with given k 0 and k n (tangent direction of the spline at the beginning and at the end of the total interval considered), then for given data the cubic spline is unique. SOLUTIONS TO PROBLEM SET 19.4, page 826 2. Writing f (x j) ⫽ fj, f (x j⫹1) ⫽ fj⫹1, x ⫺ x j ⫽ F, and x ⫺ x j⫹1 ⫽ G, we get (6) in the form qj(x) ⫽ fjc2j G2(1 ⫹ 2cjF) ⫹ fj⫹1c2j F 2(1 ⫺ 2cjG) ⫹ k jc2j FG2 ⫹ k j⫹1c2j F 2G. If x ⫽ x j, then F ⫽ 0, so that because cj ⫽ 1>(x j⫹1 ⫺ x j), qj(x j) ⫽ fjc2j (x j ⫺ x j⫹1)2 ⫽ fj. Similarly, if x ⫽ x j⫹1, then G ⫽ 0 and qj(x j⫹1) ⫽ fj⫹1c2j (x j⫹1 ⫺ x j)2 ⫽ fj⫹1. This verifies (4). By differentiating (6) we obtain qrj (x) ⫽ fjc2j [2G (1 ⫹ 2cjF) ⫹ 2cjG2] ⫹ fj⫹1c2j [2F (1 ⫺ 2cjG) ⫺ 2cjF 2] ⫹ k jc2j [G2 ⫹ 2FC] ⫹ k j⫹1c2j [2FG ⫹ F 2]. If x ⫽ x j, then F ⫽ 0, and in the first line the expression in the brackets [ Á ] reduces to x j ⫺ x j⫹1 2G(1 ⫹ cjG) ⫽ 2(x j ⫺ x j⫹1) a1 ⫹ x ⫺ x b ⫽ 0. j⫹1

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In the second and fourth lines we obtain zero. There remains qrj (x j) ⫽ k jc2j (x j⫹1 ⫺ x j)2 ⫽ k j. Similarly, if x ⫽ x j⫹1, then G ⫽ 0 and qjr (x j⫹1) ⫽ fj⫹1c2j [2(x j⫹1 ⫺ x j) ⫺ 2cj(x j⫹1 ⫺ x j)2] ⫹ k j⫹1c2j (x j⫹1 ⫺ x j)2 ⫽ 0 ⫹ k j⫹1 # 1 ⫽ k j⫹1. This verifies (5). 4. This derivation is simple and straightforward. 8. This derivation is simple and straightforward. p2(x) ⫽ x 2[ f (x) ⫺ p2(x)] r ⫽ 4x 3 ⫺ 2x ⫽ 0 gives the points of maximum deviation x ⫽ ⫾1> 12 and by inserting this, the maximum deviation itself, ƒ f (1> 12) ⫺ p2(1> 12) ƒ ⫽ ƒ 14 ⫺ 12 ƒ ⫽ 14. For the spline g(x) we get, taking x ⭌ 0, [ f (x) ⫺ g(x)] r ⫽ 4x 3 ⫹ 2x ⫺ 6x 2 ⫽ 0. A solution is x ⫽ 12 . The corresponding maximum deviation is 1 1 f (12) ⫺ g(12) ⫽ 16 ⫺ (⫺ 14 ⫹ 2 # 18) ⫽ 16 ,

which is merely 25% of the previous value. 10. We obtain q0 ⫽ ⫺ 34 (x ⫹ 2)2 ⫹ 34 (x ⫹ 2)3 3 3 2 ⫽ 3 ⫹ 6x ⫹ 15 4 x ⫹ 4x

q1 ⫽ 34 (x ⫹ 1) ⫹ 32 (x ⫹ 1)2 ⫺ 54 (x ⫹ 1)3 ⫽ 1 ⫺ 94 x 2 ⫺ 54 x 3 q2 ⫽ 1 ⫺ 94 x 2 ⫹ 54 x 3 q3 ⫽ ⫺34 (x ⫺ 1) ⫹ 32 (x ⫺ 1)2 ⫺ 34 (x ⫺ 1)3 3 3 2 ⫽ 3 ⫺ 6x ⫹ 15 4 x ⫺ 4x .

12. n ⫽ 3, h ⫽ 2, so that (14) is k 0 ⫹ 4k 1 ⫹ k 2

⫽ 32 ( f2 ⫺ f0) ⫽ 60

k 1 ⫹ 4k 2 ⫹ k 3 ⫽ 32 ( f3 ⫺ f1) ⫽ 48. Since k 0 ⫽ 0 and k 3 ⫽ ⫺12, the solution is k 1 ⫽ 12, k 2 ⫽ 12. In (13) with j ⫽ 0 we have a00 ⫽ f0 ⫽ 1, a01 ⫽ k 0 ⫽ 0, a02 ⫽ 34 (9 ⫺ 1) ⫺ 12 (12 ⫹ 0) ⫽ 0 a03 ⫽ 28 (1 ⫺ 9) ⫹ 14 (12 ⫹ 0) ⫽ 1. y 50 40 30 20 10 0

Section 19.4. Spline in Prob. 12

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From this and, similarly, from (13) with j ⫽ 1 and j ⫽ 2 we get the spline g(x) consisting of the three polynomials (see the figure) q0(x) ⫽ 1 ⫹ x 3

(0 ⬉ x ⬉ 2)

q1(x) ⫽ 9 ⫹ 12(x ⫺ 2) ⫹ 6(x ⫺ 2) ⫺ 2(x ⫺ 2) ⫽ 25 – 36x ⫹ 18x ⫺ 2x 3 (2 ⬉ x ⬉ 4) 2

q2(x) ⫽ 41 ⫹ 12(x ⫺ 4) ⫺ 6(x ⫺ 4)2 ⫽ ⫺103 ⫹ 60x ⫺ 6x 2

(4 ⬉ x ⬉ 6).

14. q0(x) ⫽ 2 ⫹ x , q1(x) ⫽ 3 ⫹ 3(x ⫺ 1) ⫹ 3(x ⫺ 1) ⫺ (x ⫺ 1) , q2(x) ⫽ 8 ⫹ 6(x ⫺ 2) ⫺ 2(x ⫺ 2)3. Note that this is not a natural spline because g s (3) ⫽ ⫺12 ⫽ 0. 16. We obtain 3

q0 ⫽ 2 ⫹ x 2 ⫺ x3 q1 ⫽ ⫺ 2 ⫺ 8 (x ⫺ 2) ⫺ 5(x ⫺ 2)2 ⫹ 5(x ⫺ 2)3 q2 ⫽ 2 ⫹ 32(x ⫺ 4) ⫹ 25(x ⫺ 4)2 ⫺ 11(x ⫺ 4)3. The data of Prob. 16 are obtained from those of Prob. 15 by subtracting 2 from the f-values, leaving k 0 and k 3 as they were. Hence, to obtain the answer to Prob. 16, subtract 2 from each of the three polynomials in the answer to Prob. 15. 18. The purpose of the experiment is to see that the advantage of the spline over the polynomial increases drastically with m. The greatest deviation of pm occurs at the ends. Formulas for the pm as a CAS will give them are p4(x) ⫽ 1 ⫺ 54 x 2 ⫹ 14 x 4 7 4 1 6 2 p6(x) ⫽ 1 ⫺ 49 36 x ⫹ 18 x ⫺ 36 x 91 4 5 6 1 2 8 p8(x) ⫽ 1 ⫺ 205 144 x ⫹ 192 x ⫺ 96 x ⫹ 576 x

and so on. Note that the polynomials of degrees 8 and more have their greatest deviation from y ⫽ 0 (their maximum or minimum) away from 0 near the endpoints of the interval ⫺m ⬉ x ⬉ m, where the splines deviate least from 0 in absolute value. 20. Team Project. (b) x(t) ⫽ 12 t ⫹ 52 t 2 ⫺ 2t 3, y(t) ⫽ 12 t ⫹ (14 13 ⫺ 1)t 2 ⫹ (12 ⫺ 14 13)t 3 (c) x(t) ⫽ t ⫹ 2t 2 ⫺ 2t 3, y(t) ⫽ t ⫹ (12 13 ⫺ 2)t 2 ⫹ (1 ⫺ 12 13)t 3 Note that the tangents in (b) and (c) are not parallel. SECTION 19.5. Numeric Integration and Differentiation, page 827 Purpose. Evaluation of integrals of empirical functions, functions not integrable by elementary methods, etc. Main Content, Important Concepts Simpson’s rule (7) (most important), error (8), (10) Trapezoidal rule (2), error (4), (5) Gaussian integration Degree of precision of an integration formula Adaptive integration with Simpson’s rule (Example 6) Numerical differentiation Short Courses. Discuss and apply Simpson’s rule.

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Comments on Content The range of numerical integration includes empirical functions, as measured or recorded in experiments, functions that cannot be integrated by the usual methods, or functions that can be integrated by those methods but lead to expressions whose computational evaluation would be more complicated than direct numerical integration of the integral itself. Simpson’s rule approximates the integrand by quadratic parabolas. Approximations by higher order polynomials are possible, but lead to formulas that are generally less practical. Numerical differentiation can sometimes be avoided by changing the mathematical model of the problem. SOLUTIONS TO PROBLEM SET 19.5, page 839 2. Let Aj and Bj be lower and upper bounds for f in the jth subinterval. Then A ⬉ J ⬉ B, where A ⫽ hSAj and B ⫽ hSBj. In Example 1 the maximum and minimum of the integrand in a subinterval occur at the ends. Hence we obtain A and B by choosing the left and right endpoints, respectively, of each subinterval instead of the midpoints as in (1). Answer: 0.714 ⬉ J ⬉ 0.778, rounded to 3S. In Prob. 1, the answer is larger than the 6S-value because the integrand is convex for x ⬍ 1> 12, and concave only for the shorter interval 1> 12 ⬍ x ⬍ 1. 4. h ⫽1, J1 ⫽ 0.5; h ⫽ 0.5, J0.5 ⫽ 0.28125, P0.5 ⫽ 13 (0.28125 ⫺ 0.5) ⫽ ⫺0.07292 (actual error ⫺0.08125); h ⫽ 0.25, J0.25 ⫽ 0.22070, P0.25 ⫽ 13 (0.22070 ⫺ 0.28125) ⫽ ⫺0.02018 (actual error ⫺0.02070). The agreement is very good. The same is true in Prob. 5, where we integrate a trigonometric function (instead of a single power of x). 6. h ƒ 12 P0 ⫹ P1 ⫹ Á ⫹ Pn⫺1 ⫹ 12 Pn ƒ ⬉ [(b ⫺ a)>n]nu ⫽ (b ⫺ a)u. This is similar to the corresponding proof for Simpson’s rule given in the text. 8. 0.693150. Exact to 6S: ln 2 ⫽ 0.693147; hence Simpson’s rule here gives a 5S-exact value. 10. 0.07392816. Exact to 7S: 0.07392811 ⫽ ⫺12 (exp (⫺0.42) ⫺ 1) 12. 0.785398514 (9S-exact 0.785398164) 14. C ⫽ ⫺0.54>90 in (9), ⫺0.000694 ⬉ P ⬉ ⫺0.000094 (actual error ⫺0.000292). In (10), 1 P0.5 ⬇ 15 (0.864956 ⫺ 0.868951) ⫽ ⫺0.000266.

Note that the absolute value of this is less than that of the actual error, and we must carefully distinguish between bounds and approximate values. 16. 0.94508, 0.94583, 5S-exact: 0.94608. An application of (5) gives the 5S-exact value 0.94583 ⫹ 0.00025 ⫽ 0.94608. 18. From (10) and Prob. 17 we obtain 1 0.94608693 ⫹ 15 (0.94608693 ⫺ 0.946145) ⫽ 0.946083

which is exact to 6S, the error being 7 units of the 8th decimal. 20. We obtain 5

1 1 1 1 25 a4 a sin a j ⫺ b ⫹ 2 a sin k 2 ⫹ sin b ⫽ 0.545941. 24 j⫽1 4 8 16 16 k⫽1 The exact 6S-value is 0.545962.

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22. Since the cosine is even, we can take terms together,

冮

p>2

cos x dx ⫽

⬇

p>2

1 2

冮

c 2A1 cos

ⴚp>2

cos x dx ⫽

冮 cos 12 pt dt

p 4

ⴚ1

1 1 pt 1 ⫹ 2A2 cos pt 2 ⫹ A3 ⴢ 1 d ⫽ 1.0000000552 2 2

where t 1 ⫽ 0.9061798459, t 2 ⫽ 0.5384693101, t 3 ⫽ 0. Note the high accuracy obtained with very few computational operations. 24. x ⫽ (t ⫹ 1)>1.6 gives x ⫽ 0 when t ⫽ ⫺1 and x ⫽ 1.25 when t ⫽ 1. Also, dx ⫽ dt>1.6. The computation gives 0.545963, the 6S-value of the Fresnel integral S(1.25) is 0.545962. 26. Team Project. The factor 24 ⫽ 16 comes in because we have replaced h by 14 h, giving 1 2 for h2 now (14 h)2 ⫽ 16 h . In the next step (with h>8) the error P43 has the factor 1 6 1>(2 ⫺ 1) ⫽ 63 , etc. For f (x) ⫽ eⴚx the table of J and P values is J11 ⫽ 1.135335 P21 ⫽ ⫺0.066596 J21 ⫽ 0.935547

J22 ⫽ 0.868951 P31 ⫽ ⫺0.017648

J31 ⫽ 0.882604

P32 ⫽ ⫺0.000266 J32 ⫽ 0.864956

J33 ⫽ 0.864690

Note that J33 is exact to 4D. For f (x) ⫽ 14 px 4 cos 14 px the table is J11 ⫽ 0 P21 ⫽ 0.185120 J21 ⫽ 0.555360

J22 ⫽ 0.74048 P31 ⫽ 0.168597

J31 ⫽ 1.06115

P32 ⫽ 0.032618 J32 ⫽ 1.22975

P41 ⫽ 0.049140 J41 ⫽ 1.20857

J33 ⫽ 1.26237 P42 ⫽ 0.001864

J42 ⫽ 1.25771

P43 ⫽ ⫺0.00004 J43 ⫽ 1.25957

J44 ⫽ 1.25953

Note that J44 is exact to 5D. 28. 0.240, which is not exact. It can be shown that the error term of the present formula is h3f (4)(␰)>12, whereas that of (15) is h4f (5)(␰)>30, where x 2 ⫺ h ⬍ ␰ ⬍ x 2 ⫹ h. In our case this gives the exact value 0.240 ⫹ 0.016 ⫽ 0.256 and 0.256 ⫹ 0 ⫽ 0.256, respectively. 30. Differentiating (14) in Sec. 19.3 with respect to r and using dr ⫽ dx>h we get df (x) dr

⫽ hf r(x) ⬇ ¢f0 ⫹

2r ⫺ 1 2 3r 2 ⫺ 6r ⫹ 2 3 ¢ f0 ⫹ ¢ f0 ⫹ Á . 2! 3!

Now x ⫽ x 0 gives r ⫽ (x ⫺ x 0)>h ⫽ 0 and the desired formula follows.

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SOLUTIONS TO CHAPTER 19 REVIEW QUESTIONS AND PROBLEMS, page 841 16. ⫺0.46903 ⴢ 102, 0.31740 ⴢ 10ⴚ3, 0.77143 ⴢ 101, ⫺0.29667 ⴢ 103 18. 0.38755>5.29731 ⫽ 0.073160. The 4S-computation gives 0.3876>(5.682 ⫺ 0.3842) ⫽ 0.3876>5.298 ⫽ 0.07316. The 3S-computation resulting from this is 0.388>(5.68 ⫺ 0.384) ⫽ 0.388>5.30 ⫽ 0.0732. The resulting 2S-computation is 0.39>(5.7 ⫺ 0.38) ⫽ 0.39>5.3 ⫽ 0.074. For 1S we obtain 0.4>(6 ⫺ 0.4) ⫽ 0.4>6 ⫽ 0.07. See Prob. 17 for a case in which rounding has a much larger effect. 20. ⫺3.145 ⬉ d ⬉ ⫺3.035 a 2a a ⫺ 苲 a a2 ⫺ 苲 a2 a⫹ 苲 a a⫺苲 22. Pr (a苲2) ⫽ ⴢ ⬇ ⴢ ⫽ 2Pr (a苲 ) ⫽ 2 a a a a a 24. x 1 ⫽ 50 ⫹ 7151 ⫽ 99.990, x 2 ⫽ 0.01, x 2 ⫽ 1>99.990 ⫽ 0.010001 26. 0.5, 0.924207, 0.829106, 0.824146, 0.824132, 0.824132. Answer: ⫾0.824132 28. 0.406 (3S-exact 0.411) 30. q0(x) ⫽ x 3, q1(x) ⫽ 1 ⫹ 3(x ⫺ 1) ⫹ 3(x ⫺ 1)2 ⫺ (x ⫺ 1)3, q2(x) ⫽ 6 ⫹ 6(x ⫺ 2) ⫺ 2 (x ⫺ 2)3; p ⫽ ⫺83 x ⫹ 92 x 2 ⫺ 56 x 3 32. 0.90450 (5S-exact 0.90452) 34. (a) (⫺0 ⫹ 0.43)>0.4 ⫽ 0.16, (b) (⫺0.12 ⫹ 0.33)>0.2 ⫽ 0.13, exact 0.12

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CHAPTER 20

Numeric Linear Algebra

SECTION 20.1. Linear Systems: Gauss Elimination, page 844 Purpose. To explain the Gauss elimination, which is a solution method for linear systems of equations by systematic elimination (reduction to triangular form). Main Content, Important Concepts Gauss elimination, back substitution Pivot equation, pivot, choice of pivot Operations count, order [e.g., O(n 3)] Comments on Content This section is independent of Chap. 7 on matrices (in particular, independent of Sec. 7.3, where the Gauss elimination is also considered). Gauss’s method and its variants (Sec. 20.2) are the most important solution methods for those systems (with matrices that do not have too many zeros). The Gauss–Jordan method (Sec. 20.2) is less practical because it requires more operations than the Gauss elimination. Cramer’s rule (Sec. 7.7) would be totally impractical in numeric work, even for systems of modest size.

SOLUTIONS TO PROBLEM SET 20.1, page 851 2. x 1 ⫽ t 1 arbitrary, 4. x 1 ⫽ 0,

x 2 ⫽ (25>42)t 1

x 2 ⫽ ⫺3

6. x 1 ⫽ (30.60 ⫹ 15.48x 2)>25.38, x 2 arbitrary. Remember from Sec. 7.1 that one also writes x 2 ⫽ t 1 ⫽ first arbitrary unknown and thus x 1 ⫽ (30.60 ⫹ 15.48t 1)>25.38, x 2 ⫽ t 1 (arbitrary). 5

8. D0

⫺4

⫺3T

0 0 0 5 Hence the system has no solution. 10. Gauss reduction to triangular form gives 4

⫺4

1 2

0T .

Back substitution gives x 1 ⫽ ⫺58 t 1,

x 2 ⫽ 18 t 1,

x 3 ⫽ t 1 arbitrary.

Note that rank A ⫽ 2. 321

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12. Gauss reduction to triangular form gives 5

⫺4

⫺3T .

⫺7

⫺118

⫺65 47

⫺39 47

⫺221 47 T .

This shows that the system has no solution. 14. Gauss reduction to triangular form gives ⫺47 D

Back substitution gives the solution x 1 ⫽ t 1 arbitrary,

x 2 ⫽ 3t 1 ⫺ 5,

x 3 ⫽ ⫺5t 1 ⫹ 14.

16. Gauss reduction to triangular form gives 1.6

⫺0.8

2.4

⫺4.8

3.6

⫺39

3.6

2.4

10.2

2.93333

⫺7.33333

3.2

Back substitution now gives x 1 ⫽ 1.5,

x 2 ⫽ ⫺3.5, x 3 ⫽ 4.5,

x ⫽ ⫺2.5.

18. Team Project. (a) (i) a ⫽ 1 to make D ⫽ a ⫺ 1 ⫽ 0; (ii) a ⫽ 1, b ⫽ 3; (iii) a ⫽ 1, b ⫽ 3. (b) x 1 ⫽ 12 (3x 3 ⫺ 1), x 2 ⫽ 12 (⫺5x 3 ⫹ 7), x 3 arbitrary is the solution of the system. The second system has no solution. (c) det A ⫽ 0 can change to det A ⫽ 0 because of roundoff. (d) (1 ⫺ 1>P)x 2 ⫽ 2 ⫺ 1>P eventually becomes x 2>P ⬇ 1>P, x 2 ⫽ 1, x 1 ⫽ (1 ⫺ x 2)>P ⬇ 0. The exact solution is x 1 ⫽ 1>(1 ⫺ P), x 2 ⫽ (1 ⫺ 2P)>(1 ⫺ P). We obtain it if we take x 1 ⫹ x 2 ⫽ 2 as the pivot equation. (e) The exact solution is x 1 ⫽ 1, x 2 ⫽ ⫺4. The 3-digit calculation gives x 2 ⫽ ⫺4.5, x1 ⫽ 1.27 without pivoting and x 2 ⫽ ⫺6, x 1 ⫽ 2.08 with pivoting. This shows that 3S is simply not enough. The 4-digit calculation give x 2 ⫽ ⫺4.095, x 1 ⫽ 1.051 without pivoting and the exact result x 2 ⫽ ⫺4, x 1 ⫽ 1 with pivoting. SECTION 20.2. Linear Systems: LU-Factorization, Matrix Inversion, page 852 Purpose. To discuss Doolittle’s, Crout’s, and Cholesky’s methods, three methods for solving linear systems that are based on the idea of writing the coefficient matrix as a product of two triangular matrices (“LU-factorization”). Furthermore, we discuss matrix inversion by the Gauss–Jordan elimination.

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Main Content, Important Concepts Doolittle’s and Crout’s methods for arbitrary square matrices Cholesky’s method for positive definite symmetric matrices Numerical matrix inversion Short Courses. Doolittle’s method and the Gauss–Jordan elimination. Comment on Content L suggests “lower triangular” and U “upper triangular.” For Doolittle’s method, these are the same as the matrix of the multipliers and of the triangular system in the Gauss elimination. The point is that in the present methods, one solves one equation at a time, no systems. SOLUTIONS TO PROBLEM SET 20.2, page 857 2. c

⫺18.5

x 1 ⫽ ⫺4

1.5

x1 ⫽

4. D⫺1

0T D0

3T ,

x2 ⫽

1 2

x 2 ⫽ 10

1 0 0 ⫺92 x 3 ⫽ ⫺4 6. Team Project. (a) The formula for the entries of L ⫽ [l jk] and U ⫽ [u jk] are l j1 ⫽ aj1 u 1k ⫽

j ⫽ 1, Á , n

a1k

k ⫽ 2, Á , n

l 11 kⴚ1

l jk ⫽ ajk ⫺ a l jsu sk

j ⫽ k, Á , n; k ⭌ 2

s⫽1 jⴚ1

u jk ⫽

1 aajk ⫺ a l jsu sk b l jj s⫽1

k ⫽ j ⫹ 1, Á , n; j ⭌ 2.

(b) The factorization and the solutions are 3

D18

⫺6

⫺54

0T D0

⫺3

23 15

1 ⫺12 T , y ⫽ D15 T,

1 ⫺15 4 x ⫽ D 15 T.

2 5

⫺4

D⫺4

0T D0

12T (Doolittle)

4 3

2 5

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⫺4

D⫺4

0T D0

4 3 T (Crout)

⫺4

D⫺4

0T D0

(Cholesky).

(d) For fixing the notation, for n ⫽ 4 we have a1

b2 A ⫽ LU ⫽ E 0

U⫽E

U E

where a1 ⫽ a1,

aj ⫽ aj ⫺ bjgj⫺l,

g1 ⫽ c1>a1,

gj ⫽ cj>aj,

j ⫽ 2, Á , n j ⫽ 2, Á , n ⫺ 1

(e) If A is symmetric. 2

x1 ⫽

8. D3

0T D0

8T ,

x2 ⫽

x 3 ⫽ ⫺4

10. D0

0 T D0

1 2

1 2 23

12. E

2 16.

1 D1 9 2

⫺2 2 1

U E

x 1 ⫽ ⫺18 x2 ⫽

3 4

x3 ⫽

⫽

⫽ ⫺2

⫽

⫽ 14

1 2T ⫽ ⫺2

1 T 1 A . Hence A is orthogonal. 9 3

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1000

150

⫺300

31.25

⫺50T

18. D 150

325

⫺300 ⫺50 100 Note that det A ⫽ 0.000016, det A⫺1 ⫽ 1>det A ⫽ 62,500. Furthermore, the condition number of A (Sec 20.4) will be 0.25 ⴢ 1450 ⫽ 362.5, showing that A is ill-conditioned. 20. det A ⫽ 0 as given, but rounding makes det A ⫽ 0 and may completely change the situation with respect to existence of solutions of linear systems, a point to be watched for when using a CAS. In the present case we get (a) ⫺0.00000035, (b) ⫺0.00001998, (c) ⫺0.00028189, (d) 0.002012, (e) 0.0038. SECTION 20.3 Linear Systems: Solution by Iteration, page 858 Purpose. To familiarize the student with the idea of solving linear systems by iteration, to explain in what situations that is practical, and to discuss the most important method (Gauss–Seidel iteration) and its convergence. Main Content, Important Concepts Distinction between direct and indirect methods Gauss–Seidel iteration, its convergence, its range of applicability Matrix norms Jacobi iteration Short Courses. Gauss–Seidel iteration only. Comments on Content The Jacobi iteration appeals by its simplicity but is of limited practical value. A word on the frequently occurring sparse matrices may be good. For instance, we have about 99.5% zeros in solving the Laplace equation in two dimensions by using a 1000 ⫻ 1000 grid and the usual five-point pattern (Sec. 21.4). SOLUTIONS TO PROBLEM SET 20.3, page 863 2. The eigenvalues of I ⫺ A are 0.5, 0.5, ⫺1. Here A is 12 times the coefficient matrix of the given system; thus, 0

⫺12

I ⫺ A ⫽ D⫺12

⫺12 T .

⫺12

4. The exact solution is x ⫽ [3 ⫺9 6]T is reached at Step 8 rather quickly owing to the fact that the spectral radius of C is 0.125, hence rather small; here 0

1 4

C ⫽ D0

1 16

1 4 T.

1 64

1 16

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6. Row 1 becomes Row 3, Row 2 becomes Row 1, and Row 3 becomes Row 2. The exact solution is x 1 ⫽ 0.5, x 2 ⫽ ⫺2.5, x 3 ⫽ 4.0. The choice of x0 is of much less influence than one would expect. We obtain for x0 ⫽ [0 0 0]T 1, [0, ⫺1.75000, 3.89286] 2, [0.350000, ⫺2.45715, 3.99388] 3, [0.491430, ⫺2.49755, 3.99966] 4, [0.499510, ⫺2.49987, 3.99998] 5, [0.499974, ⫺2.50000, 4.00000] 6, [0.500000, ⫺2.50000, 4.00000] For x0 ⫽ [1 1

1, [⫺0.200000, ⫺1.88333, 3.91190] 2, [0.376666, ⫺2.46477, 3.99497] 3, [0.492954, ⫺2.49798, 3.99971] 4, [0.499596, ⫺2.49988, 3.99998] 5, [0.499976, ⫺2.50000, 4.00000] 6, [0.500000, ⫺2.50000, 4.00000] for x0 ⫽ [10 10

10]

1, [⫺2, ⫺3.08333, 4.08333] 2, [0.616666, ⫺2.53333, 4.00476] 3, [0.506666, ⫺2.50190, 4.00027] 4, [0.500380, ⫺2.50010, 4.00001] 5, [0.500020, ⫺2.50000, 4.00000] 6, [0.500000, ⫺2.50000, 4.00000] and for x0 ⫽ [100 100

100]T 1, [⫺20, ⫺15.0833, 5.79761] 2, [3.01666, ⫺3.21905, 4.10271] 3, [0.643810, ⫺2.54108, 4.00587] 4, [0.508216, ⫺2.50235, 4.00034] 5, [0.500470, ⫺2.50013, 4.00001] 6, [0.500026, ⫺2.50000, 4.00000]

2 The spectral radius of C (and its only nonzero eigenvalue) is 35 ⫽ 0.057. 8. The exact solution is 2, 0, 1. Step 10 gives 32.00144 ⫺0.00221311 0.9997794T. The spectral radius (23)3>2 ⫽ 0.544331 of C is relatively large. 10. Interchange the first equation and the last equation. Then the exact solution ⫺2.5, 2, 4.5 is reached at Step 11, the spectral radius of

⫺14

⫺18

C ⫽ D0

1 24

5 ⫺16 T

1 5

1 10

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being 1> 215 ⫽ 0.258199. (The eigenvalues are complex conjugates and the third eigenvalue is 0, as always for the present C.) 12. In (a) we obtain C ⫽ ⫺(I ⫹ L)ⴚ1U

1 ⫽ ⫺ D⫺0.1 ⫺0.09

1 ⫺0.1

0.1

0T D0

0.1T

⫺0.100

⫽ D0

0.010

⫺0.090T

0.009

0.019

and 储 C 储 ⫽ 0.2 ⬍ 1 by (11), which implies convergence by (8). In (b) we have 1

D10

1T ⫽ (I ⫹ L) ⫹ U

1 1

⫽ D10

0T ⫹ DD0

1T .

From this we compute 1

C ⫽ ⫺(I ⫹ L)ⴚ1U ⫽ ⫺ D⫺10

0T D0

⫺10

⫽ ⫺ D0

⫺10

⫺99T

980

⫺1

⫽ D0

⫺99

⫺10 99T . ⫺980

Developing the characteristic determinant of C by its first column, we obtain ⫺l ⫹ 10 ⫺l 2 ⫺99

99 ⫺l ⫺ 980

2 ⫽ ⫺l (l2 ⫹ 970l ⫹ 1)

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which shows that one of the eigenvalues is greater than 1 in absolute value, so that we have divergence. In fact, l ⫽ ⫺0, 0.001, and ⫺970, approximately. 1

5.5

2.5625

14. D1T , D⫺10.75T , D⫺7.75

3.3125

8.5

2.94531

3.03906

T , D⫺9.21875T , D⫺8.84375T , D⫺9.02734T .

5.5625

6.3125

5.94531

6.03906

Step 5 of the Gauss–Seidel iteration gives the better result 32.99969 1

⫺1.8125

Exact: 33

⫺9

⫺2.51691

⫺2.53287

⫺9.00015 5.999964T. ⫺2.29583

⫺2.63594

64T.

16. D1T , D 2.58333T , D 2.81875T , D 2.14931T , D 2.07710T , D 1.96657T 1

1.7

3.95

4.33667

4.60875

4.51353

Step 5 of the Gauss–Seidel iteration gives the more accurate result 3⫺2.49475 1.99981

4.495804T.

Exact: 3⫺2.5

4.54T.

18. 2151 ⫽ 12.29, 13 (column sum norm), 11 (row sum norm) 20. 218k 2 ⫽ 4.24 ƒ k ƒ , 4 ƒ k ƒ , 4 ƒ k ƒ SECTION 20.4. Linear Systems: Ill-Conditioning, Norms, page 864 Purpose. To discuss ill-conditioning quantitatively in terms of norms, leading to the condition number and its role in judging the effect of inaccuracies on solutions. Main Content, Important Concepts Ill-conditioning, well-conditioning Symptoms of ill-conditioning Residual Vector norms Matrix norms Condition number Bounds for effect of inaccuracies of coefficients on solutions Comment on Content Reference [E9] in App. 1 gives some help when Aⴚ1, needed in ␬(A), is unknown (as is the case in practice). SOLUTIONS TO PROBLEM SET 20.4, page 871 2. 13, 9, 8, [12 ⫺18 1]. Note that the l 2-norm of a vector with integer components will generally not be an integer. 4. 4k ⫹ k 2 ⫹ k 3, 2k 6 ⫹ k 4 ⫹ 16k 2, k 3, [1>k 4>k 2 1] 6. 1, 1, 1. The given vector is a unit vector. 8. Square the double inequality to be proved, that is, 储 x 储2⬁ ⬉ 储 x 储22 ⬉ 储 x 储21

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and write this inequality in terms of components, obtaining 2 n 2 ⬉ a xj ⬉ a a ƒ xj ƒ b . n

(max ƒ x j ƒ ) j

j⫽1

Now perform the square on the right. 10. 储A储1 ⫽ 6.3, 储Aⴚ1 储1 ⫽ 220 51 ⫽ 4.3137, so that ␬ ⫽ 27.176. ⴚ1 12. ␬ ⫽ 储A储1 ⴢ 储A 储1 ⫽ 13 ⴢ 13 ⫽ 169. The matrix is ill-conditioned. # 5100 14. ␬ ⫽ 51 50 4999 ⫽ 1.0406. For the unit matrix we would have ␬ ⫽ 1, and the problem shows the effect of small off-diagonal entries, as they may occur in numeric diagonalization. 16. A is the 4 ⫻ 4 Hilbert matrix times 21. Its inverse is

Aⴚ1 ⫽

⫺120

240

⫺140

1 ⫺120 E 21 240

1200

⫺2700

1680

⫺2700

6480

⫺4200

⫺140

1680

⫺4200

2800

This gives the condition number (in both norms, since A and Aⴚ1 are symmetric) ␬ ⫽ 43.75 ⴢ

13,620 21

⫽ 28,375.000

showing that A is very ill-conditioned. 18. The product of the two matrices equals AB ⫽ c

4.7

10.8

2.0

7.2

Hence for the l 1-norm (column “sum” norm) we have (12) in the form 18.0 ⬍ 5.0 ⴢ 6.3 ⫽ 31.5 and for the l ⬁-norm (row “sum” norm) 15.5 ⬍ 4.0 ⴢ 6.6 ⫽ 26.4. 20. The systems have the solutions [1

1]T

and (4S-values) [0.8455 1.273]T. Hence a change of 0.2% in b has produced changes of 16 and 27% in the components of the solution. The matrix is ill-conditioned; the condition number is (4S-value) ␬ (A) ⫽ 4.7 ⴢ 42.73 ⫽ 200.8. 22. By (12), 1 ⫽ 储 I 储 ⫽ 储 AAⴚ1储 ⬉ 储 A 储 ⴢ 储 Aⴚ1储 ⫽ ␬ (A). For the Frobenius norm, 1n ⫽ 储 I 储 ⬉ ␬ (A).

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24. Team Project. (a) Formula (18a) is obtained from max ƒ x j ƒ ⬉ S ƒ x k ƒ ⫽ 储 x 储1 ⬉ n max ƒ x j ƒ ⫽ n储 x 储⬁. Equation (18b) follows from (18a) by division by n. (b) To get the first inequality in (19a), consider the square of both sides and then take square roots on both sides. The second inequality in (19a) follows by means of the Cauchy–Schwarz inequality and a little trick worth remembering, a ƒ xj ƒ ⫽ a 1 ⴢ ƒ xj ƒ ⬉

2 a ƒ x j ƒ ⫽ 1n 储x储2 .

To get (19b), divide (19a) by 1n. (c) Let x ⫽ 0. Set x ⫽ 储 x 储y. Then 储 y 储 ⫽ 储 x 储>储 x 储 ⫽ 1. Also, Ax ⫽ A (储 x 储y) ⫽ 储 x 储 Ay since 储 x 储 is a number. Hence 储Ax储>储 x 储 ⫽ 储 Ay 储, and in (9), instead of taking the maximum over all x ⫽ 0, since 储 y 储 ⫽ 1 we only take the maximum over all y of norm 1. Write x for y to get (10) from this. (d) These “axioms of a norm” follow from (3), which are the axioms of a vector norm. SECTION 20.5. Least Squares Method, page 872 Purpose. To explain Gauss’s least squares method of “best fit” of straight lines to given data (x 0, y0), Á , (x n, yn) and its extension to best fit of quadratic polynomials, etc. Main Content, Important Concepts Least squares method Normal equations (4) for straight lines Normal equations (8) for quadratic polynomials Short Courses. Discuss the linear case only. Comment. Normal equations are often ill-conditioned, so that results may be sensitive to roundoff. For another (theoretically much more complicated) method, see Ref. [E5], p. 201. SOLUTIONS TO PROBLEM SET 20.5, page 875 2. 2.676 ⫺ 1.216x. Note the change of the slope. 4. Hook’s law F ⫽ ks gives the spring modulus k ⫽ F>s; for the present data, s (F) ⫽ 0.03349 ⫹ 0.3139F,

hence

k ⫽ 1>0.3139 ⫽ 3.186.

6. U ⫽ ⫺5.20 ⫹ 53.4i. Estimate: R ⫽ 53.4⍀. Note that the line does not pass through the origin, as it should. This is typical. 8. 2.955 ⫺ 1.159x ⫹ 0.932x 2 10. y ⫽ 2.29 ⫺ 0.433t ⫹ 0.105t 2 12. b0n ⫹ b1 Sx j ⫹ b2 Sx 2j ⫹ b3 Sx 3j ⫽ Syj b0 Sx j ⫹ b1 Sx 2j ⫹ b2 Sx 3j ⫹ b3 Sx 4j ⫽ Sx jyj b0 Sx 2j ⫹ b1 Sx 3j ⫹ b2 Sx 4j ⫹ b3 Sx 5j ⫽ Sx 2j yj b0 Sx 3j ⫹ b1 Sx 4j ⫹ b2 Sx 5j ⫹ b3 Sx 6j ⫽ Sx 3j yj

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14. Team Project. (a) We substitute Fm(x) into the integral and perform the square. This gives 储 f ⫺ Fm 储2 ⫽

冮

f 2 dx ⫺ 2 a aj j⫽0

冮

fyj dx ⫹ a a ajak j⫽0 k⫽0

冮 y y dx. j k

This is a quadratic function in the coefficients. We take the partial derivative with respect to any one of them, call it al, and equate this derivative to zero. This gives b

0⫺2

冮 fy dx ⫹ 2 a a 冮 y y dx ⫽ 0. l

j⫽0

j l

Dividing by 2 and taking the first integral to the right gives the system of normal equations, with l ⫽ 0, Á , m. (b) In the case of a polynomial we have

冮

yjyl dx ⫽

冮x

j⫹l

which can be readily integrated. In particular, if a ⫽ 0 and b ⫽ 1, integration from 0 to 1 gives 1>( j ⫹ l ⫹ 1), and we obtain the Hilbert matrix as the coefficient matrix. SECTION 20.6. Matrix Eigenvalue Problems: Introduction, page 876 Purpose. This section is a collection of concepts and a handful of theorems on matrix eigenvalues and eigenvectors that are frequently needed in numerics; some of them will be discussed in the remaining sections of the chapter and others can be found in more advanced or more specialized books listed in Part E of App. 1. The section frees both the instructor and the student from the task of locating these matters in Chaps. 7 and 8, which contain much more material and should be consulted only if problems on one or another matter are wanted (depending on the background of the student) or if a proof might be of interest. SECTION 20.7. Inclusion of Matrix Eigenvalues, page 879 Purpose. To discuss theorems that give approximate values and error bounds of eigenvalues of general (square) matrices (Theorems 1, 2, 4, Example 2) and of special matrices (Theorem 6). Main Content, Important Concepts Gerschgorin’s theorem (Theorem 1) Sharpened Gerschgorin’s theorem (Theorem 2) Gerschgorin’s theorem improved by similarity (Example 2) Strict diagonal dominance (Theorem 3) Schur’s inequality (Theorem 4), normal matrices Perron’s theorem (Theorem 5) Collatz’s theorem (Theorem 6) Short Courses. Discuss Theorems 1 and 6.

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Comments on Content It is important to emphasize that one must always make sure whether or not a thoerem applies to a given matrix. Some theorems apply to any real or complex square matrices whatsoever, whereas others are restricted to certain classes of matrices. The exciting Gerschgorin theorem was one of the early theorems on numerics for eigenvalues; it appeared in the Bull. Acad. Sciences de l’URSS (Classe mathém, 7-e série, Leningrad, 1931, p. 749), and shortly thereafter in the German Zeitschrift für angewandte Mathematik und Mechanik. SOLUTIONS TO PROBLEM SET 20.7, page 884 2. 5, 8, 9; radii 2 ⴢ 10ⴚ2. Estimates of this kind can be useful when a matrix has been diagonalized numerically and some very small nonzero off-diagonal entries are left. 4. 1, 4, 12; radii 1, 3, 4. Since the matrix is symmetric, this gives the two overlapping intervals 0 ⬉ l ⬉ 2 and 1 ⬉ l ⬉ 7 and the separate interval 8 ⬉ l ⬉ 16. The actual spectrum is (4S-values) {0.8786, 3.047, 13.07}, again illustrating Theorem 2; cf. Example 1. 6. 10, 6, 3; radii 0.3, 0.1, 0.2; actually (4S) 10.01, 6.000, 2.994 8. T with t 11 ⫽ t 22 ⫽ 1, t 33 ⫽ 34 gives 1

0.1

T ⴚ1AT ⫽ D0

0 T D 0.1

1 34

⫺0.2

0.1

⫺6.8

⫽ D 0.1

0 T

⫺0.2 34

⫺0.2

0 T D0

Note that the disk with center 3 is still disjoint from that with center 10. Indeed, 10 ⫺ 6.9 ⫽ 3.1

3 ⫹ 0.2 34 ⫽ 3.00588.

whereas

The next integer, 35, is already too large and leads to overlapping, 3 ⫹ 0.2 35 ⫽ 3.057 ⬎ 10 ⫺ 7.1 ⫽ 2.9. 10. An example is A⫽ c

The eigenvalues are ⫺1 and 1, so that the entire spectrum lies on the circle. A similar looking 3 ⫻ 3 matrix or 4 ⫻ 4 matrix, etc., can be constructed with some or all of its eigenvalues on the circle. 12. 1181 ⫽ 13.45 14. 1145.1 ⫽ 12.05 16. 183 ⫽ 9.110

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18. Proofs follow readily from the definitions of these classes of matrices. This normality is of interest, since normal matrices have important properties; in particular, they have an orthonormal set of n eigenvectors, where n is the size of the matrix. See [B3], vol. 1, pp. 268–274. 20. This is a continuity proof. Let S ⫽ D1 ´ D2 ´ Á ´ Dp without restriction, where Dj is the Gerschgorin disk with center ajj. We write A ⫽ B ⫹ C, where B ⫽ diag (ajj) is the diagonal matrix with the main diagonal of A as its diagonal. We now consider A t ⫽ B ⫹ tC

for 0 ⬉ t ⬉ 1.

Then A 0 ⫽ B and A 1 ⫽ A. Now by algebra, the roots of the characteristic polynomial ft(l) of A t (that is, the eigenvalues of A t) depend continuously on the coefficients of ft(l), which in turn depend continuously on t. For t ⫽ 0 the eigenvalues are a11, Á , ann. If we let t increase continuously from 0 to 1, the eigenvalues move continuously and, by Theorem 1, for each t, lie in the Gerschgorin disks with centers ajj and radii trj

rj ⫽ a ƒ ajk ƒ .

where

k⫽j

Since at the end, S is disjoint from the other disks, the assertion follows. SECTION 20.8. Power Method for Eigenvalues, page 885 Purpose. Explanation of the power method for determining approximations and error bounds for eigenvalues of real symmetric matrices. Main Content, Important Concepts Iteration process of the power method Rayleigh quotient (the approximate value) Improvement of convergence by a spectral shift Scaling (for eigenvectors) Short Courses. Omit spectral shift. Comments on Content The method is simple but converges slowly, in general. Symmetry of the matrix is essential to the validity of the error bound (2). The method as such can be applied to more general matrices. SOLUTIONS TO PROBLEM SET 20.8, page 887 2. The Rayleigh quotients for the first five steps are (4S-values) q ⫽ 0, 6, 7.846, 7.990, 7.999. These approximate the larger eigenvalue 8, the other one being ⫺2, which is included in the first two of the five resulting inclusion intervals. [q ⫺ d, q ⫹ d] ⫽ [⫺4, 4] [2, 10] [6.615, 9.077] [7.678, 8.302] [7.921, 8.078]

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where d ⫽ 4, 4, 1.231, 0.312, 0.078. Note that the first interval includes only ⫺2. 4. Computation gives (4S-values) q

⫾d

q⫺d

q⫹d

Error

1.333

2.175

⫺0.841

3.508

5.867

6.574

1.742

4.832

8.316

0.626

7.158

0.474

6.684

7.632

0.042

7.197

0.119

7.078

7.317

0.003

7.1998

0.02983

7.170

7.230

0.0002

Note that the error is much less than the error bound. This is typical. Since the bound is optimal, it simply means that the present case is not the worst possible. 6. q ⫽ 12.333, 12.962, 12.998; ƒ P ƒ ⬉ 2.944, 0.614, 0.142. The spectrum is {⫺1, 3, 13}. 8. q ⫽ 10.5000, 11.1303, 11.1831; (4S-values)

ƒ P ƒ ⬉ 2.9580, 1.3689, 0.9637. The spectrum is

s ⫽ {⫺7.956, 0.4426, 4.283, 11.232}. 10. Let z 1, Á , z n be orthonormal eigenvectors corresponding to the eigenvalues l1, Á , ln, respectively. Then for any initial vector x0 we have (summations from 1 to n) x0 ⫽ S cjz j x1 ⫽ S cjljz j xs ⫽ S cjlsj z j xs⫹1 ⫽ S cjls⫹1 zj j and, using the last step, for the Rayleigh quotient q⫽

x s⫹1Txs x sTxs

⫽

S c2j l2s⫹1 j S c2j l2s j

⬇ lm

where lm is of maximum absolute value, and the quality of the approximation increases with increasing number of steps s. Here we have to assume that cm ⫽ 0, that is, that not just by chance we picked an x0 orthogonal to the eigenvector z m of lm. The chance that this happens is practically zero, but should it occur, then rounding will bring in a component in the direction of z m and one should eventually expect good approximations, although perhaps only after a great number of steps. 12. CAS Experiment. (a) We obtain 16, 41.2, 34.64, 32.888, 32.317, 32.116, 32.043, 32.0158, 32.0059, 32.0022, etc. The spectrum is {32, 12, 8}. Eigenvectors are [3 [3 ⫺2 1]T.

⫺7]T, [1

⫺1]T,

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(b) For instance, for A ⫺ 10I we get q ⫽ 6,

22.5,

22.166,

22.007,

22.00137,

etc. Whereas for A the ratio of eigenvalues is 32>12, for A ⫺ 10I we have the spectrum {22, 2, ⫺2}, hence the ratio 22>2. This explains the improvement of convergence. (d) The eigenvalues are l ⫽ ⫾1. Corresponding eigenvectors are z1 ⫽ B

2 1

z2 ⫽ B

1 ⫺2

and we have chosen x0 as 3

x0 ⫽ B

R ⫽ z 1 ⫹ z 2,

⫺1

so that x1 ⫽ z 1 ⫺ z 2 ⫽ [1

3]T

x2 ⫽ z 1 ⫹ z 2 ⫽ [3

⫺1]T

etc. From this, x 0Tx1 ⫽ 0 and for the error bound we get x1Tx1

d⫽ R

⫺ q2 ⫽

x 0Tx0

10 ⫺ 0 ⫽ 1, B 10

and similarly in all further steps. This illustrates that our error bound is best possible in general. (e) A in (a) provides an example, q

41.2

34.64

32.888

32.317

5.5

1.45

0.557

0.225

etc. Also A ⫺ 10I dose, q

22.5

22.166

22.007

22.00137

3.0

0.072

0.023

0.00032

Further examples can easily be found. For instance, the matrix ⫺2

⫺1

has the spectrum {⫺5, 2}, but we obtain, with x0 ⫽ 31

etc.

14T,

⫺7

⫺4.3208

⫺5.2969

0.37736

0.2006

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SECTION 20.9. Tridiagonalization and QR-Factorization, page 888 Purpose. Explanation of an optimal method for determining the whole spectrum of a real symmetric matrix by first reducing the matrix to a tridiagonal matrix with the same spectrum and then applying the QR-method, an iteration in which each step consists of a factorization (5) and a multiplication. Comment on Content The n ⫺ 2 Householder steps (n ⫻ n the size of the matrix) correspond to similarity transformations; hence the spectrum is preserved. The same holds for QR. But we can perform any number of QR steps, depending on the desired accuracy. SOLUTIONS TO PROBLEM SET 20.9, page 896 2. We have to do n ⫺ 2 ⫽ 1 Householder step. We obtain S ⫽ 12 and 1 2 22 ⫹ 12

v1 ⫽ [0

1>(12 22 ⫹ 12)]T.

This gives 1

P1 ⫽ D0

⫺1> 12

1 ⫺ 1>(2 ⫹ 12)

From this and the given matrix we obtain the tridiagonal matrix 0 B ⫽ P1AP1 ⫽ D⫺12 0

⫺12

0T . ⫺1

Hence ⫺1 is an eigenvalue. The other eigenvalues, ⫺1 and 2, are now obtained by solving the remaining quadratic characteristic equation. 4. We have to do two Householder steps because n ⫺ 2 ⫽ 2. Step1. We obtain S1 ⫽ 4.2426 and v1 ⫽ [0

0.98560

0.11957

0.11957]T.

This gives 1

⫺0.94281

⫺0.23570

0.97140

⫺0.02860

⫺0.23570

⫺0.02860

0.97140

P1 ⫽ E

From this and the given matrix follows 5 ⫺4.2426

A1 ⫽ E

⫺4.2426

⫺1

3.5

1.5

⫺1

1.5

3.5

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Step 2.

337

We obtain S2 ⫽ 1.4142 and v2 ⫽ [0

0.38268

⫺0.92388]T.

0.70711

⫺0.70711

This gives

P2 ⫽ E

From this and the given matrix we obtain the tridiagonal matrix 5 B ⫽ A2 ⫽ E

⫺4.2426

⫺4.2426 0

0 ⫺1.4142

6 ⫺1.4142

0 0

6. The matrices B in the first three steps are 1.2925 ⫺0.2942

D⫺0.2942

0.8402

0.1260T

0.1260

0.2072

1.3976 ⫺0.1697

D⫺0.1697

0.7608

0.0303T

0.0303

0.1817

1.4289 ⫺0.0886

D⫺0.0886

0.7310

0.0074T .

0.0074

0.1801

The eigenvalues are 1.44, 0.72, 0.18. The calculations were done with 8S and then rounded in the results to 4D. We see that the approximations of the eigenvalues are somewhat better than can be expected by looking at the size of the off-diagonal entries. Similarly for Probs. 7–9. 8. The matrices B in the first three steps are 14.2004

⫺0.0444

D⫺0.0444

⫺6.3046

0.0668T

0.0668

2.1042

14.2005

⫺0.0197

D⫺0.0197

⫺6.3052

0.0223T

0.0223

2.1047

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14.2005

⫺0.0088

D⫺0.0088

⫺6.3052

0.0074T

0.0074

2.1048

⫺6.3052,

2.1048.

The eigenvalues are (5S) 14.200,

Note that the off-diagonal entries of the matrix B are small and that the approximations of the eigenvalues are more exact than in the other problems. SOLUTIONS TO CHAPTER 20 REVIEW QUESTIONS AND PROBLEMS, page 896 14. [4 16. [3

t 1 12 t 1] T 3t 1 ⫹ 2 t 1]T

⫺10

18. D⫺2.9924

3.1859

2.9578T

⫺5.6669

5.9641

5.9710

48 ⫺8 ⫺6 1 20. D⫺8 39 1T 226 ⫺6 1 29 22. Reorder to get convergence. Equation 1 becomes 2 2 becomes 3 3 becomes 1. The iteration gives ⫺1.5067

⫺2.0193

⫺1.9992

D 8.1753T , D 7.9925T ,

D 8.0003T

⫺1.0846

⫺1.0001

⫺0.9967

The solution is x 1 ⫽ ⫺2, x 2 ⫽ 8, x 3 ⫽ ⫺1 24. 12.0, 171.50 ⫽ 8.4558, 8.1 26. 1, 1, 1 28. 152 30. 9.1 32. (0.3 ⫹ 4.3 ⫹ 2.8) (10 ⫹ 10 ⫹ 10) ⫽ 222 34. 1.3 ⫹ 0.7x 36. Centers 2.0, 4.4, 2.8; radii 3.4, 2.1, 4.6, respectively. The eigenvalues are (4S-values) 0.05313, 4.573 ⫾ 2.514i.

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38. Centers 5, 6, 8; radii 2, 1, 1, respectively. Eigenvalues (4S-values) 4.186, 6.471, 8.343. Note that the first disk includes the second and touches the third, the point x ⫽ 7 being common to all three disks. The third disk includes one eigenvalue and so does the second disk. The first disk includes the first two eigenvalues. 40. Computation gives n

76.67

92.46

95.81

96.46

28.96

14.62

6.58

2.80

The spectrum is {2.635, 40.77, 96.60} (4S-values).

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CHAPTER 21

Numerics for ODEs and PDEs

SECTION 21.1. Methods for First-Order ODEs, page 901 Purpose. To explain three numerical methods for solving initial value problems y r ⫽ f (x, y), y(x 0) ⫽ y0 by stepwise computing approximations to the solution at x 1 ⫽ x 0 ⫹ h, x 2 ⫽ x 0 ⫹ 2h, etc. Main Content, Important Concepts Euler’s method (3) Automatic variable step size selection Improved Euler method (8), Table 21.1 Classical Runge–Kutta method (Table 21.3) Error and step size control Runge–Kutta–Fehlberg method Backward Euler’s method Stiff ODEs Comments on Content Euler’s method is good for explaining the principle but is too crude to be of practical value. The improved Euler method is a simple case of a predictor–corrector method. The classical Runge–Kutta method is of order h4 and is of great practical importance. Principles for a good choice of h are important in any method. f in the equation must be such that the problem has a unique solution (see Sec. 1.7). SOLUTIONS TO PROBLEM SET 21.1, page 910 2. y ⫽ sin 12 px. Since the values obtained give y9 ⫽ 1.01170 ⬎ 1, y10 comes out complex and is meaningless. xn

Error

0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0

0.15708 0.31221 0.46144 0.60079 0.72636 0.83433 0.92092 0.98214 1.01170 —

⫺0.00065 ⫺0.00319 ⫺0.00745 ⫺0.01301 ⫺0.01926 ⫺0.02531 ⫺0.02991 ⫺0.03109 ⫺0.02401 —

4. This is a special Riccati equation. Set y ⫹ x ⫽ u, then u r ⫽ u 2 ⫹ 1 and u r >(u 2 ⫹ 1) ⫽ 1.

340

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By integration, arctan u ⫽ x ⫹ c and u ⫽ tan (x ⫹ c) ⫽ y ⫹ x so that y ⫽ tan (x ⫹ c) ⫺ x and c ⫽ 0 from the initial condition. Hence y ⫽ tan x ⫺ x. The calculation is xn

y(x n)

Error

0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0

0.000000 0.001000 0.005040 0.014345 0.031513 0.059764 0.103293 0.167821 0.261488 0.396394

0.000335 0.002710 0.009336 0.022793 0.046302 0.084137 0.142288 0.289639 0.360158 0.557408

0.000335 0.001710 0.004296 0.008448 0.014789 0.024373 0.038996 0.061818 0.098670 0.161014

Although the ODE is similar to that in Prob. 3, the error is greater by about a factor of 10. This is understandable because tan x becomes infinite as x : 12 p. 6. y ⫽ tan 2x. Note that the error is first negative and then positive and rapidly increasing, due to the behavior of the tangent. xn

Error ⫻ 105

0.05 0.10 0.15 0.20 0.25 0.30 0.35 0.40 0.45 0.50

0.10050 0.20304 0.30981 0.42341 0.54702 0.68490 0.84295 1.02989 1.25930 1.55379

⫺17 ⫺33 ⫺48 ⫺62 ⫺72 ⫺76 ⫺66 ⫺25 ⫹86 362

8. The solution is y ⫽ 1>(1 ⫹ 4eⴚx). The 10S-computation, rounded to 6D, gives xn

y(x n)

Error y(x n) ⫺ yn

0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0

0.216467 0.233895 0.252274 0.271587 0.291802 0.312876 0.334754 0.357366 0.380633 0.404462

0.216481 0.233922 0.252317 0.271645 0.291875 0.312965 0.334858 0.357486 0.380767 0.404610

0.000014 0.000028 0.000043 0.000058 0.000073 0.000089 0.000104 0.000119 0.000134 0.000148

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10. We obtain the following with h ⫽ 0.1 in Prob. 7 and now with h ⫽ 0.05 and see that the error now decreases to about 14 of its former value, as expected by halfing h in a second-order method. x

h ⫽ 0.1

h ⫽ 0.05

0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0

0.00003 0.00006 0.00009 0.00016 0.00029 0.00054 0.00106 0.00220 0.00488 0.01187

0.00001 0.00001 0.00001 0.00003 0.00005 0.00011 0.00023 0.00051 0.00119 0.00305

12. The solution is y ⫽ 1>(1 ⫹ 4eⴚx). The computation gives xn

Error y(x n) ⫺ yn

0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0

0.2164806848 0.2339223328 0.2523167036 0.2716446148 0.2918751118 0.3129648704 0.3348578890 0.3574855192 0.3807668746 0.4046096381

0.043 ⴢ 10ⴚ8 0.085 ⴢ 10ⴚ8 0.128 ⴢ 10ⴚ7 0.170 ⴢ 10ⴚ7 0.209 ⴢ 10ⴚ7 0.249 ⴢ 10ⴚ7 0.284 ⴢ 10ⴚ7 0.318 ⴢ 10ⴚ7 0.346 ⴢ 10ⴚ7 0.370 ⴢ 10ⴚ7

To apply (10), we need the calculation of five steps with h ⫽ 0.2. We obtain (the third column would not be needed) xn

Error y(x n) ⫺ yn

0.2 0.4 0.6 0.8 1.0

0.2339222103 0.2716443697 0.3129645099 0.3574850569 0.4046090919

1.310 ⴢ 10ⴚ7 2.621 ⴢ 10ⴚ7 3.854 ⴢ 10ⴚ7 4.941 ⴢ 10ⴚ7 5.832 ⴢ 10ⴚ7

Hence the required error estimate is (0.4046096381 ⫺ 0.4046090919)>15 ⫽ 0.364 ⴢ 10ⴚ7. The actual error is 0.370 ⴢ 10ⴚ7; the estimate is much closer to the actual error than one can expect in general.

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14. We obtain the solution y ⫽ ex⫺1>x, and the computation gives xn

Error ⫻ 109

1.1 1.2 1.3 1.4 1.5 1.6 1.7 1.8 1.9 2.0

1.0047008 1.0178355 1.0383528 1.0655889 1.0991473 1.1388240 1.1845602 1.2364116 1.2945277 1.3591406

68 113 144 168 188 206 224 241 260 281

It is interesting that, from x ⫽ 1.2 on, the error is almost a linear function. 16. The solution is y ⫽ 3 cos x ⫺ 2 cos2 x. The computation gives xn

Error: y(x n) ⫺ yn

Error in Prob. 15

0.2 0.4 0.6 0.8 1.0

1.019137566 1.066471079 1.113634713 1.119282901 1.036951866

0.1173 ⴢ 10ⴚ5 0.5194 ⴢ 10ⴚ5 0.14378 ⴢ 10ⴚ4 0.36749 ⴢ 10ⴚ4 0.101888 ⴢ 10ⴚ3

0.74 ⴢ 10ⴚ7 0.328 ⴢ 10ⴚ6 0.904 ⴢ 10ⴚ6 0.228 ⴢ 10ⴚ5 0.614 ⴢ 10ⴚ5

Note that the ratio of the errors is about the same for all x n, about 24, as can be expected in doubling the step size in the case of a fourth-order method. 18. From y r ⫽ x ⫹ y and the given formula we get, with h ⫽ 0.2, k 1 ⫽ 0.2(x n ⫹ yn) k 2 ⫽ 0.2[x n ⫹ 0.1 ⫹ yn ⫹ 0.1(x n ⫹ yn)] ⫽ 0.2[1.1(x n ⫹ yn) ⫹ 0.1] k*3 ⫽ 0.2[x n ⫹ 0.2 ⫹ yn ⫺ 0.2(x n ⫹ yn) ⫹ 0.4[1.1(x n ⫹ yn) ⫹ 0.1]] ⫽ 0.2[1.24(x n ⫹ yn) ⫹ 0.24] and from this yn⫹1 ⫽ yn ⫹ 16 [1.328(x n ⫹ yn) ⫹ 0.128]. The computed values are xn

Error

0.0 0.2 0.4 0.6 0.8 1.0

0.000000 0.021333 0.091655 0.221808 0.425035 0.717509

0.000000 0.000067 0.000165 0.000312 0.000505 0.000771

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20. CAS Experiment. (b) The computation gives xn

Error Estimate (10) ⴢ 109

Error ⴢ 109

0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0

1.2003346725 1.4027100374 1.6093362546 1.8227932298 2.0463025124 2.2841368531 2.5422884689 2.8296387346 3.1601585865 3.5574085377

3.0 3.7 5.9 9.4 13.1 14.4 ⫹5.0 ⫺38.8 ⫺191.1 ⫺699.9

⫺0.4 ⫺1.9 ⫺5.0 ⫺11.0 ⫺22.5 ⫺44.7 ⫺88.4 ⫺177.6 ⫺369.0 ⫺813.0

SECTION 21.2. Multistep Methods, page 911 Purpose. To explain the idea of a multistep method in terms of the practically important Adams–Moulton method, a predictor–corrector method that in each computation uses four preceding values. Main Content, Important Concepts Adams–Bashforth method (5) Adams–Moulton method (7) Self-starting and not self-starting Numerical stability, fair comparison Short Courses. This section may be omitted. SOLUTIONS TO PROBLEM SET 21.2, page 915 2. The solution is y ⫽ exp (x 2). The computation gives xn

Error 䡠 106

0.4 0.5 0.6 0.7 0.8 0.9 1.0

1.173518 1.284044 1.433365 1.632375 1.896573 2.248047 2.718486

⫺7 ⫺19 ⫺35 ⫺59 ⫺92 ⫺139 ⫺205

4. The comparison shows that, in the present case, RK is somewhat better. The comparison is fair since we have four evaluations per step for RK but only two for AM. The errors are x

0.4

0.6

0.8

1.0

AM RK

⫺0.7 ⴢ 10ⴚ5 0.1 ⴢ 10ⴚ5

⫺3.5 ⴢ 10ⴚ5 0.8 ⴢ 10ⴚ5

⫺9.2 ⴢ 10ⴚ5 4.0 ⴢ 10ⴚ5

⫺20.5 ⴢ 10ⴚ5 17.5 ⴢ 10ⴚ5

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6. The exact solution is y ⫽ tan x ⫹ x ⫹ 1. The Adams–Moulton calculation gives Starting yn

xn 0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0

Predicted y*n

Corrected yn

Exact Values

Error ⴢ 106 of yn

1.822798 2.046315 2.284161 2.542332 2.829714 3.160288 3.557626

1.000000 1.200335 1.402710 1.609336 1.822793 2.046302 2.284137 2.542288 2.829639 3.160158 3.557408

0 0.08 0.16 0.2 ⫺4.9 ⫺12.4 ⫺24.2 ⫺43.5 ⫺75.6 ⫺130 ⫺218

1.000000 1.200335 1.402710 1.609336 1.822715 2.046197 2.283978 2.542027 2.829171 3.159247 3.555451

8. The solution is y ⫽ 12 tanh 2x. Computation gives xn

Exact

Error ⴢ 106

Exact

Error ⴢ 106

0.1 0.2 0.3 0.4 0.5

0.098686 0.189971 0.268519 0.332007 0.380726

0.098688 0.189974 0.268525 0.332018 0.380797

1 3 6 11 71

0.6 0.7 0.8 0.9 1.0

0.416701 0.442532 0.460706 0.473306 0.481949

0.416827 0.442676 0.460834 0.473403 0.482014

127 144 128 97 65

Error ⴢ 109

1.2 1.4 1.6 1.8 2.0

3.07246 3.15595 3.24962 3.35261 3.46410

⫺22 ⫺42 ⫺59 ⫺781 ⫺1273

2.2 2.4 2.6 2.8 3.0

3.58330 3.70945 3.84188 3.97995 4.12311

⫺1564 ⫺1691 ⫺1696 ⫺1615 ⫺1480

10. We obtain

14. y ⫽ exp (x 2). Some of the values and errors are xn

yn (h ⫽ 0.05)

Error ⴢ 106

yn (h ⫽ 0.1)

Error ⴢ 106

0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0

1.010050 1.040817 1.094188 1.173535 1.284064 1.433388 1.632404 1.896612 2.248105 2.718579

⫺6 ⫺14 ⫺24 ⫺38 ⫺58 ⫺87 ⫺131 ⫺197 ⫺297

1.01005 1.040811 1.094224 1.173623 1.284219 1.433636 1.632782 1.897175 2.248931 2.719785

⫺50 ⫺112 ⫺194 ⫺307 ⫺466 ⫺694 ⫺1023 ⫺1503

The errors differ by a factor of 4 to 5, approximately.

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SECTION 21.3. Methods for Systems and Higher Order ODEs, page 915 Purpose. Extension of the methods in Sec. 21.1 to first-order systems and to higher order ODEs. Content Euler’s method for systems (5) Classical Runge–Kutta method extended to systems (6) Runge–Kutta–Nyström method (7) Short Courses. Discuss merely Runge–Kutta (6), which shows that this “vectorial extension” of the method is conceptually quite simple.

SOLUTIONS TO PROBLEM SET 21.3, page 922 2. The exact solution is (see Fig. 86 in Sec. 4.3) y1 ⫽ 4eⴚx sin x y2 ⫽ 4eⴚx cos x. Computation gives the following values and errors: x

Error

0 0.2 0.4 0.6 0.8 1.0

0 0.8 1.28 1.504 1.5360 1.3488

0 ⫺0.149 ⫺0.236 ⫺0.264 ⫺0.2467 ⫺0.19664

4 3.2 2.40 1.664 1.0304 0.51712

0 0.01 0.07 0.148 0.2218 0.27794

We see that the values are much too inaccurate to be of any practical value, and we recall that we briefly mentioned this method only for illustrating that the extension from a single ODE to a system is quite straightforward. 4. The solution y1 ⫽ eⴚ2x ⫹ eⴚ4x, y2 ⫽ eⴚ2x ⫺ eⴚ4x. The computation is xn

y1,n

Error

y2,n

Error

0.1 0.2 0.3 0.4 0.5

1.4 1.00 0.728 0.5392 0.4054

0.0890 0.1196 0.1220 0.1120 0.0978

0.2 0.28 0.296 0.2800 0.2499

⫺0.0516 ⫺0.0590 ⫺0.0484 ⫺0.0326 ⫺0.0174

The figure shows (for x ⫽ 0, Á , 1) that these values give a qualitatively correct impression, although they are rather inaccurate. Note that the error of y1 is not monotone.

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1.5

0.5

0.2

0.4

0.6

0.8

Section 21.3. Solution curves and computed values in Problem 4

6. Solution y1 ⫽ 2ex, y2 ⫽ 2eⴚx (see also Example 3 in Sec. 4.3). The computation is

y1,n

Error

y2,n

Error

0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0

2.2 2.42 2.662 2.9282 3.2210 3.5431 3.8974 4.2872 4.7159 5.1875

0.0104 0.0228 0.0378 0.0554 0.0764 0.1011 0.1302 0.1638 0.2033 0.2491

1.8 1.62 1.458 1.3122 1.1810 1.0629 0.95659 0.86093 0.77484 0.69736

0.0097 0.0175 0.0236 0.0284 0.0321 0.0347 0.03659 0.03773 0.03830 0.03840

The figure illustrates that the error of y1 is monotone increasing and is positive (the points lie below that curve), and similarly for y2.

4 3 2 y2

1 0

0.2

0.4

0.6

0.8

Section 21.3. Solution curves and computed values (the dots) in Problem 6

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8. The errors are smaller by a factors 103 to 105. Computation gives x

Error ⴢ 105

0 0.2 0.4 0.6 0.8 1.0

0 0.650667 1.044190 1.239570 1.289335 1.238233

0 ⫺4.0 ⫺5.0 ⫺4.1 ⫺2.0 ⫹0.6

4 3.2096 2.469541 1.811705 1.252075 0.794934

0 4.3 8.1 11.0 12.7 13.1

10. The errors are smaller by a factor 103, approximately. The computation gives xn

y1,n

Error ⴢ 105

y2,n

Error ⴢ 105

0 0.1 0.2 0.3 0.4 0.5

2 1.489133 1.119760 0.850119 0.651328 0.503301

0 ⫺8 ⫺11 ⫺11 ⫺10 ⫺9

0 0.148333 0.220888 0.247515 0.247342 0.232469

0 8 10 10 9 7

12. Division by x gives y s ⫹ y r >x ⫹ y ⫽ 0. The system is y1r ⫽ y2, y2r ⫽ y s ⫽ ⫺y2>x ⫺ y1. Because of the factor 1>x we have to choose x 0 ⫽ 0. The computation gives, with the initial values taken from Ref. [GenRef1] in App. 1: xn

J0(xn)

J 0⬘(xn)

106 䡠 Error of J0(xn)

1 1.5 2 2.5 3 3.5

0.765198 0.511903 0.224008 ⫺0.048289 ⫺0.260055 ⫺0.380298

⫺0.440051 ⫺0.558002 ⫺0.576897 ⫺0.497386 ⫺0.339446 ⫺0.137795

0 ⫺76 ⫺117 ⫺95 ⫹3 170

Note that, although the step is very large, the values are relatively accurate. Also, recall that J0r ⫽ ⫺J1. 0.8 0.6 0.4

0.2 0

1.5

2.5

3.5

–0.2 –0.4

J'0

–0.6

Section 21.3. Solution curves in Problem 12. Computed values lie practically on the curves.

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14. We obtain k 1 ⫽ 0.1x nyn k 2 ⫽ k 3 ⫽ 0.1(x n ⫹ 0.1)(yn ⫹ 0.1ynr ⫹ 0.05k 1) k 4 ⫽ 0.1(x n ⫹ 0.2)(yn ⫹ 0.2ynr ⫹ 0.2k 2) and the following numeric values, whose accuracy is about the same as that in Example 2, but the work was much less.

ynr

y (x) Exact (8D)

108 ⴢ Error of yn

0.0 0.2 0.4 0.6 0.8 1.0

0.35502806 0.30370304 0.25474212 0.20979975 0.16984600 0.13529219

⫺0.25881940 ⫺0.25240464 ⫺0.23583070 ⫺0.21279172 ⫺0.18641134 ⫺0.15914608

0.35502806 0.30370315 0.25474235 0.20980006 0.16984632 0.13529242

0 11 23 31 32 23

SECTION 21.4. Methods for Elliptic PDEs, page 922 Purpose. To explain numerical methods for the Dirichlet problem involving the Laplace equation, the typical representative of elliptic PDEs. Main Content, Important Concepts Elliptic, parabolic, hyperbolic equations Dirichlet, Neumann, mixed problems Difference analogs (7), (8) of Poisson’s and Laplace’s equations Coefficient scheme (9) Liebmann’s method of solution (identical with Gauss–Seidel, Sec. 20.3) Peaceman–Rachford’s ADI method (15) Short Courses. Omit the ADI method. Comments on Content Neumann’s problem and the mixed problem follow in the next section, including the modification in the case of irregular boundaries. The distinction between the three kinds of PDEs (elliptic, parabolic, hyperbolic) is not merely a formal matter because the solutions of the three types behave differently in principle, and the boundary and initial conditions are different; this necessitates different numerical methods, as we shall see.

SOLUTIONS TO PROBLEM SET 21.4, page 930 4. u 11 ⫽ 92.86, u 21 ⫽ 90.18, u 12 ⫽ 81.25, u 22 ⫽ 75.00, u 23 ⫽ 47.32, u 31 ⫽ u 11, etc., by symmetry

u 13 ⫽ 57.14,

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6. Gauss gives the values of the exact solution u(x, y) ⫽ x 3 ⫺ 3xy 2. Gauss–Seidel needs about 10 steps for producing 5S-values: n

u 11

u 21

u 12

u 22

1 2 3 4 5 6 7 8 9 10

50.25 19.031 3.2578 ⫺0.6855 ⫺1.6714 ⫺1.9178 ⫺1.9795 ⫺1.9949 ⫺1.9987 ⫺1.9997

44.062 12.516 4.6289 2.6572 2.1643 2.0411 2.0103 2.0026 2.0006 2.0002

31.062 ⫺0.484 ⫺8.371 ⫺10.343 ⫺10.836 ⫺10.959 ⫺10.990 ⫺10.997 ⫺10.999 ⫺11.000

5.031 ⫺10.742 ⫺14.686 ⫺15.671 ⫺15.918 ⫺15.979 ⫺15.995 ⫺15.999 ⫺16.000 ⫺16.000

It is interesting that it takes only 4 or 5 steps to turn the values away from the starting values to values that are already relatively close to the respective limits. 8. 165, 165, 165, 165 by Gauss. The Gauss–Seidel computation gives n

u11

u21

u12

u22

1 2 3 4 5 6 7 8

132.50 152.81 161.95 164.24 164.81 164.95 164.99 165.00

140.63 158.91 163.48 164.62 164.90 164.98 165.00 165.00

152.81 161.95 164.24 164.81 164.95 164.99 165.00 165.00

10. The values of the exact solution of the Laplace equation are u (1, 1) ⫽ ⫺4,

u (2, 1) ⫽ u (1, 2) ⫽ ⫺7,

u (2, 2) ⫽ ⫺64.

Gauss gives ⫺2, ⫺5, ⫺5, ⫺62. Corresponding errors are ⫺2, ⫺2, ⫺2, ⫺2. Gauss– Seidel needs about 10 steps for producing 5S-values: n

u11

u21

u12

u22

1 2 3 4 5 6 7 8 9 10

50.50 24.8125 4.7031 ⫺0.3242 ⫺1.5811 ⫺1.8953 ⫺1.9738 ⫺1.9935 ⫺1.9984 ⫺1.9996

48.625 8.406 ⫺1.648 ⫺4.162 ⫺4.791 ⫺4.948 ⫺4.987 ⫺4.997 ⫺4.999 ⫺5.000

⫺35.188 ⫺55.297 ⫺60.324 ⫺61.581 ⫺61.895 ⫺61.974 ⫺61.993 ⫺61.998 ⫺62.000 ⫺62.000

12. This shows the importance of good starting values; it then does not take long until the approximations come close to the solution. A rule of thumb is to take a rough

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estimate of the average of the boundary values at the points that enter the linear system. By starting from 0 we obtain [0.09472 0.10149

0.31799

0.32138]

(Step 3)

[0.10741 0.10783

0.32433

0.32455]

(Step 5).

14. All the isotherms must begin and end at a corner. The diagonals are isotherms u ⫽ 25, because of the data obtained and for reasons of symmetry. Hence we obtain a qualitative picture as the figure shows. In Prob. 7 the situation is similar. 0°C

25°C 31.25°C

50°C

18.75°C 0°C

Section 21.4. Problem 14

16. Step 1. First come rows j ⫽ 1, j ⫽ 2; for these, (14a) is j ⫽ 1, j ⫽ 2,

i ⫽ 1.

u 01 ⫺ 4u 11 ⫹ u 21 ⫽ ⫺u 10 ⫺ u 12

i ⫽ 2.

u 11 ⫺ 4u 21 ⫹ u 31 ⫽ ⫺u 20 ⫺ u 22

i ⫽ 1.

u 02 ⫺ 4u 12 ⫹ u 22 ⫽ ⫺u 11 ⫺ u 13

i ⫽ 2. u 12 ⫺ 4u 22 ⫹ u 32 ⫽ ⫺u 21 ⫺ u 22. Six of the boundary values are zero, and the two on the upper edge are u 13 ⫽ u 23 ⫽ 13>2 ⫽ 0.866025. Also, on the right we substitute the starting values 0. With this, our four equations become ⫺4u 11 ⫹ u 21 ⫽ 0 u 11 ⫺ 4u 21 ⫽ 0 ⫺4u 12 ⫹ u 22 ⫽ ⫺0.866025 u 12 ⫺ 4u 22 ⫽ ⫺0.866025. From the first two equations, u 11 ⫽ 0,

u 21 ⫽ 0

and from the other two equations, u 12 ⫽ 0.288675,

u 22 ⫽ 0.288675.

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Step 1. Now come columns; for these, (14b) is i ⫽ 1, i ⫽ 2,

j ⫽ 1.

u 10 ⫺ 4u 11 ⫹ u 12 ⫽ ⫺u 01 ⫺ u 21

j ⫽ 2.

u 11 ⫺ 4u 12 ⫹ u 13 ⫽ ⫺u 02 ⫺ u 22

j ⫽ 1.

u 20 ⫺ 4u 21 ⫹ u 22 ⫽ ⫺u 11 ⫺ u 31

j ⫽ 2. u 21 ⫺ 4u 22 ⫹ u 23 ⫽ ⫺u 12 ⫺ u 32. With the boundary values and the previous solution on the right, this becomes ⫺4u 11 ⫹ u 12 ⫽ 0 u 11 ⫺ 4u 12 ⫽ ⫺0.866025 ⫺ 0.288675 ⫺4u 21 ⫹ u 22 ⫽ 0 u 21 ⫺ 4u 22 ⫽ ⫺0.866025 ⫺ 0.288675. The solution is u 11 ⫽ 0.07698 u 21 ⫽ 0.07698 u 12 ⫽ 0.30792 u 22 ⫽ 0.30792. Step 2. Rows. We can use the previous equations, changing only the right sides: ⫺4u 11 ⫹ u 21 ⫽ ⫺0.30792 u 11 ⫺ 4u 21 ⫽ ⫺0.30792 ⫺4u 12 ⫹ u 22 ⫽ ⫺0.866025 ⫺ 0.07698 u 12 ⫺ 4u 22 ⫽ ⫺0.866025 ⫺ 0.07698 Solution: u 11 ⫽ u 21 ⫽ 0.102640,

u 12 ⫽ u 22 ⫽ 0.314335.

Step 2. Columns. The equations with the new right sides are ⫺4u 11 ⫹ u 12 ⫽ ⫺0.102640 u 11 ⫺ 4u 12 ⫽ ⫺0.866025 ⫺ 0.314335 ⫺4u 21 ⫹ u 22 ⫽ ⫺0.102640 u 21 ⫺ 4u 22 ⫽ ⫺0.866025 ⫺ 0.314335. Final result (solution of these equations): u 11 ⫽ 0.106061 u 21 ⫽ 0.106061 u 12 ⫽ 0.321605 u 22 ⫽ 0.321605. Exact 3D values: u 11 ⫽ u 21 ⫽ 0.108,

u 12 ⫽ u 22 ⫽ 0.325.

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18. CAS Project. (b) The solution of the linear system (rounded to integers), with the values arranged as the points in the xy-plane, is 160

170

157

110

138

145

125

138

145

125

160

170

157

110

Twenty steps gave accuracies of 3S–5S, with slight variations, between the components of the output vector. SECTION 21.5. Neumann and Mixed Problems. Irregular Boundary, page 931 Purpose. Continuing our discussion of elliptic PDEs, we explain the ideas needed for handling Neumann and mixed problems and the modifications required when the domain is no longer a rectangle. Main Content, Important Concepts Mixed problem for a Poisson equation (Example 1) Modified stencil (6) (notation in Fig. 460) Comments on Content Neumann’s problem can be handled as explained in Example 1 on the mixed problem. In all the cases of an elliptic PDE we need only one boundary condition at each point (given u or given u n). SOLUTIONS TO PROBLEM SET 21.5, page 935 2. The exact solution of the Poisson equation is u ⫽ x 2y 2. The approximate solution results from Au ⫽ b, where ⫺4

⫺4

Y, 0

⫺4

⫺20

⫺4

⫺103

A⫽I

b⫽I

where the six equations correspond to P11, P21, P31, P12, P22, P32, in our usual order. The components of b are of the form a ⫺ c with a resulting from 2(x 2 ⫹ y 2) and c from the boundary values; thus, 4 ⫺ 0 ⫽ 4, 10 ⫺ 0 ⫽ 10, 20 ⫺ 12 ⫽ 8, 10 ⫺ 9 ⫽ 1, 16 ⫺ 36 ⫽ ⫺20, 26 ⫺ 81 ⫺ 48 ⫽ ⫺103. The solution of this system agrees with the values obtained at the Pjk from the exact solution, u 11 ⫽ 1, u 21 ⫽ u 12 ⫽ 4, u 22 ⫽ 16, and u 31 ⫽ 9, u 32 ⫽ 36 on the boundary. u 41 ⫽ u 21 ⫹ 12 and u 42 ⫽ u 22 ⫹ 48 produced entries 2 in A and ⫺12 and ⫺48 in b.

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4. 0 ⫽ u 01,x ⫽

1 (u 11 ⫺ u ⴚ1,1) gives u ⴚ1,1 ⫽ u 11. Similarly, u 41 ⫽ u 21 ⫹ 3 from the 2h

condition on the right edge, so that the equations are ⫺4u 01 ⫹ 2u 11

⫽1

u 01 ⫺ 4u 11 ⫹ u 21

⫽ ⫺0.25 ⫹ 0.75 ⫽ 0.5

u 11 ⫺ 4u 21 ⫹ u 31 ⫽ ⫺1 2u 21 ⫺ 4u 31 ⫽ ⫺2.25 ⫺ 1.25 ⫺ 3 ⫽ ⫺6.5. u 01 ⫽ ⫺0.25, u 11 ⫽ 0, u 21 ⫽ 0.75, u 31 ⫽ 2; this agrees with the values of the exact solution u(x, y) ⫽ x 2 ⫺ y 2 of the problem. 6. Exact solution u ⫽ 9y sin 13 px. Linear system Au ⫽ b, where ⫺4

⫺4

Y, 1

2a b⫽I 2a

⫺4

3a ⫹ c

⫺4

3a ⫹ c

A⫽I

a ⫽ ⫺8.54733, c ⫽ ⫺2243 ⫽ ⫺15.5885. The solution of this system is (exact values of u in parentheses) u 11 ⫽ u 21 ⫽ 8.46365 (exact 92 23 ⫽ 7.79423) u 12 ⫽ u 22 ⫽ 16.8436 (exact 923 ⫽ 15.5885) u 13 ⫽ u 23 ⫽ 24.9726 (exact 27 2 13 ⫽ 23.3827). 14. Let v denote the unknown boundary potential. Then v occurs in Au ⫽ b, where ⫺4

⫺4

2 3

⫺4

A⫽E

⫺v

b⫽E

⫺v ⫺83 v

v [5 10 10 16]T. From this and 5v>19 19 ⫽ 220 (the potential at P11) we have v ⫽ 836 V as the constant boundary potential on the indicated portion of the boundary. 16. Two equations are as usual: The solution of this linear system is u ⫽

⫺4u 11 ⫹ u 21 ⫹ u 12 ⫺ u 11 ⫺ 4u 21

2⫽2

⫺ 0.5 ⫽ 2

where the right side is due to the fact that we are dealing with the Poisson equation. The third equation results from (6) with a ⫽ p ⫽ q ⫽ 1 and b ⫽ 12 . We get 2 c

u 22 2

⫹

u 1.5>2 3 4

u 02 u 11 2 ⫹ ⫹ 3 ⫺ 1 u 12 d ⫽ 2. 2 2 2 3

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The first two terms are zero and u 02 ⫽ ⫺2; these are given boundary values. There remains 2 3 u 11 ⫺ 3u 12 ⫽ 1 ⫹ 1 ⫽ 2.

Our three equations for the three unknowns have the solution u 11 ⫽ ⫺1.5,

u 21 ⫽ ⫺1,

u 12 ⫽ ⫺1.

SECTION 21.6. Methods for Parabolic PDEs, page 936 Purpose. To show the numerical solution of the heat equation, the prototype of a parabolic equation, on the region given by 0 ⬉ x ⬉ 1, t ⭌ 0, subject to one initial condition (initial temperature) and one boundary condition on each of the two vertical boundaries. Content Direct method based on (5), convergence condition (6) Crank–Nicolson method based on (8) Special case (9) of (8) Comment on Content Condition (6) restricts the size of time steps too much, a disadvantage that the Crank– Nicolson method avoids.

SOLUTIONS TO PROBLEM SET 21.6, page 941 4. CAS Experiment. u(0, t) ⫽ u(1, t) ⫽ 0, u(0.2, t) ⫽ u(0.8, t), u(0.4, t) ⫽ u(0.6, t), where x ⫽ 0.2

x ⫽ 0.4

t⫽0

0.587785

0.951057

t ⫽ 0.04

0.393432 0.399274 0.396065

0.636586 0.646039 0.640846

t ⫽ 0.08

0.263342 0.271221 0.266879

0.426096 0.438844 0.431818

t ⫽ 0.12

0.176267 0.184236 0.179830

0.285206 0.298100 0.290970

t ⫽ 0.16

0.117984 0.125149 0.121174

0.190901 0.202495 0.196063

t ⫽ 0.2

0.078972 0.085012 0.081650

0.127779 0.137552 0.132112

Explicit CN Exact (6D)

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6. Note that h ⫽ 0.2 and k ⫽ 0.01 gives r ⫽ 0.25. The computation gives t

x ⫽ 0.2

x ⫽ 0.4

x ⫽ 0.6

x ⫽ 0.8

0.00 0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08

0.2 0.2 0.1875 0.171875 0.156250 0.141602 0.128174 0.115967 0.104904

0.4 0.35 0.3125 0.281250 0.253906 0.229492 0.207520 0.187683 0.169754

0.2 0.2 0.1875 0.171875 0.156250 0.141602 0.128174 0.115967 0.104904

8. u(x, 0) ⫽ u(1 ⫺ x, 0) and the boundary conditions imply u(x, t) ⫽ u(1 ⫺ x, t) for all t. The calculation gives (0, 0.2, 0.35, 0.35, 0.2, 0) (0, 0.1875, 0.3125, 0.3125, 0.1875, 0) (0, 0.171875, 0.28125, 0.28125, 0.171875, 0) (0, 0.156250, 0.253906, 0.253906, 0.156250, 0) (0, 0.141602, 0.229492, 0.229492, 0.141602, 0) 10. The boundary condition on the left is that the normal derivative is zero. We have, by (5), u 0, j⫹1 ⫽ (1 ⫺ 2r)u 0j ⫹ r (u 1j ⫹ u ⴚ1j). Now, by the central difference formula (see hint) for the normal derivative (partial derivative with respect to x) we get 0u 0j 1 (u 1j ⫺ u ⴚ1j). 0⫽ ⫽ 0x 2h Hence, the previous formula becomes u 0, j⫹1 ⫽ (1 ⫺ 2r)u 0j ⫹ 2ru 1j. We have r ⫽ 0.25 so u 0, j⫹1 ⫽ 12 (u 0j ⫹ u 1j). The underlying idea is quite similar to that in Sec. 21.5. The computation gives t

x⫽0

x ⫽ 0.2

x ⫽ 0.4

x ⫽ 0.6

x ⫽ 0.8

x⫽1

0 0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.09 0.10 0.11 0.12

0 0 0 0 0 0 0.001 0.004 0.010 0.019 0.029 0.039 0.046

0 0 0 0 0 0.002 0.007 0.016 0.027 0.039 0.048 0.054 0.054

0 0 0 0 0.008 0.025 0.049 0.073 0.091 0.097 0.089 0.068 0.041

0 0 0 0.031 0.085 0.144 0.187 0.201 0.178 0.122 0.048 ⫺0.028 ⫺0.085

0 0 0.125 0.279 0.397 0.437 0.379 0.236 0.043 ⫺0.150 ⫺0.295 ⫺0.352 ⫺0.308

0 0.5 0.866 1 0.866 0.5 0 ⫺0.5 ⫺0.866 ⫺1 ⫺0.866 ⫺0.5 0

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12. We obtain (10S-computation, rounded to 4D) t

x ⫽ 0.2

x ⫽ 0.4

x ⫽ 0.6

x ⫽ 0.8

0.04

0.1636

0.2545

0.1636

0.08

0.1074

0.1752

0.1074

0.12

0.0735

0.1187

0.0735

0.16

0.0498

0.0807

0.0498

0.20

0.0339

0.0548

0.0339

Exact: 0.20

0.0331

0.0535

0.0331

14. We need the matrix

A⫽

⫺1

0 .

⫺1

3S-values computed by Crank–Nicolson for x ⫽ 0.1 (and 0.9), 0.2 (and 0.8), 0.3 (and 0.7), 0.4 (and 0.6), 0.5 and t ⫽ 0.01, 0.02, Á , 0.05 are 0.0754

0.141

0.190

0.220

0.230

0.0669

0.126

0.172

0.201

0.210

0.0600

0.114

0.156

0.182

0.192

0.0541

0.103

0.141

0.166

0.174

0.0489

0.093

0.128

0.150

0.158

0.10275

0.14115

5S-values for t ⫽ 0.04 are 0.054112

0.16565

0.17407.

The corresponding values in Prob. 15 are 0.10182

0.16727.

Exact 5S-values computed by (9) and (10) in Sec. 12.5 (two nonzero terms suffice) are 0.053946

0.10245

0.14074

0.16519

We see that the present values are better than those in Prob. 15.

0.17358.

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0.2

0.4

0.6

0.8

Section 21.6. u (x, t ) for constant t ⫽ 0.01, • • • , 0.05 as polygons with the Crank–Nicolson values as vertices in Problem 14

SECTION 21.7. Method for Hyperbolic PDEs, page 942 Purpose. Explanation of the numerical solution of the wave equation, the prototype of a hyperbolic PDE, on a region of the same type as in the last section, subject to initial and boundary conditions that guarantee the uniqueness of the solution. Comments on Content We now have two initial conditions (given initial displacement and given initial velocity), in contrast to the heat equation in the last section, where we had only one initial condition. The computation by (6) is simple. Formula (8) gives the values of the first time-step in terms of the initial data. SOLUTIONS TO PROBLEM SET 21.7, page 944 2. Computation, rounded to 3D, gives t

x ⫽ 0.2

x ⫽ 0.4

x ⫽ 0.6

x ⫽ 0.8

0 0.2 0.4 0.6 0.8 1.0 1.2 1.4 1.6 1.8 2.0

0.032 0.048 0.056 0.016 ⫺0.072 ⫺0.128 ⫺0.072 0.016 0.056 0.048 0.032

0.096 0.088 0.064 ⫺0.016 ⫺0.112 ⫺0.144 ⫺0.112 ⫺0.016 0.064 0.088 0.096

0.144 0.112 0.016 ⫺0.064 ⫺0.088 ⫺0.096 ⫺0.088 ⫺0.064 0.016 0.112 0.144

0.128 0.072 ⫺0.016 ⫺0.056 ⫺0.048 ⫺0.032 ⫺0.048 ⫺0.056 ⫺0.016 0.072 0.128

4. Since u(x, 0) ⫽ f (x), the derivation is immediate. Formula (9) results if the integral equals 2kgi. 6. Exact solution u(x, t) ⫽ (x ⫹ t)2. The values obtained in the computation are those of the exact solution. u 11, u 21, u 31, u 41 are obtained from (8) and the initial conditions u i0 ⫽ (0.2i)2,

gi ⫽ 2 (0.2i),

i ⫽ 0, Á , 5.

In connection with the left boundary condition we can use the central difference formula 1 (u 1, j ⫺ u ⴚ1, j) ⬇ u x (0, jk) ⫽ 2jk 2h to obtain u ⴚ1, j and then (8) to compute u 01 and (6) to compute u 0, j⫹1.

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8. From (12), Sec. 12.4, with c ⫽ 1 we get the exact solution 1 u(x, t) ⫽ 2

冮

x⫹ct

sin ps ds ⫽

xⴚct

1 [cos p(x ⫺ ct) ⫺ cos p(x ⫹ ct)]. 2p

From (8) we have kgi ⫽ 0.1gi ⫽ 0.1sin 0.1pi. Because of the symmetry with respect to x ⫽ 0.5 we need to list only the following values (with the exact values in parentheses): t

x ⫽ 0.1

x ⫽ 0.2

x ⫽ 0.3

x ⫽ 0.4

x ⫽ 0.5

0.0 0.1

0 0.030902 (0.030396) 0.058779 (0.057816) 0.080902 (0.079577) 0.095106 (0.093549)

0 0.058779 (0.057816) 0.111803 (0.109973) 0.153884 (0.151365) 0.180902 (0.177941)

0 0.080902 (0.079577) 0.153884 (0.151365) 0.211803 (0.208337) 0.248990 (0.244914)

0 0.095106 (0.093549) 0.180902 (0.177941) 0.248990 (0.244914) 0.292705 (0.287914)

0 0.100000 (0.098363) 0.190211 (0.187098) 0.261803 (0.257518) 0.307768 (0.302731)

0.2 0.3 0.4

10. By (13), Sec. 12.4, with c ⫽ 1 the left side of (6) is u i, j⫹1 ⫽ u(ih, (j ⫹ 1)h) ⫽ 12[ f (ih ⫹ ( j ⫹ 1)h) ⫹ f (ih ⫺ ( j ⫹ 1)h)]

(A)

and the right side is the sum of the six terms u iⴚ1, j ⫽ 12 [ f ((i ⫺ 1)h ⫹ jh) ⫹ f ((i ⫺ 1)h ⫺ jh)], u i⫹1, j ⫽ 12 [ f (i ⫹ 1)h ⫹ jh) ⫹ f ((i ⫹ 1)h ⫺ jh)], ⫺u i, jⴚ1 ⫽ ⫺12[ f (ih ⫹ ( j ⫺ 1)h) ⫹ f (ih ⫺ ( j ⫺ 1)h)]. Four of these six terms cancel in pairs, and the remaining expression equals the right side of (A). SOLUTIONS TO CHAPTER 21 REVIEW QUESTIONS AND PROBLEMS, page 945 18. y ⫽ ex. Computed values are xn

y(x n)

Error ⴢ 106

Error in Prob. 17

0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.09 0.10

1.010000 1.020100 1.030301 1.040604 1.051010 1.061520 1.072135 1.082857 1.093685 1.104622

1.010050 1.020201 1.030455 1.040811 1.051271 1.061837 1.072508 1.083287 1.094174 1.105171

50 101 154 207 261 312 373 430 489 549

0.005171

We see that the error of the last value has decreased by a factor of 10, approximately, due to the smaller step.

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20. y ⫽ 2eⴚx ⫹ x 2 ⫹ 1. xn

Error ⴢ 104

0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0

2.8205 2.6790 2.5738 2.5033 2.4662 2.4612 2.4871 2.5428 2.6276 2.7404

⫺8 ⫺15 ⫺22 ⫺27 ⫺32 ⫺36 ⫺39 ⫺42 ⫺44 ⫺46

22. (a) 0.1, 0.2034, 0.3109, 0.4217, 0.5348, 0.6494, 0.7649, 0.8806, error 0.0044, 0.0122, Á , 0.0527. (b) 0.2055, 0.4276, 0.6587, 0.8924, error 0.0101, 0.0221, 0.0322, 0.0409. (c) 0.4352, 0.9074, error 0.0145, 0.0258 24. Solution y ⫽ tan x ⫺ x ⫹ 4. xn

Error ⴢ 106

0.8 1.0

4.22969 4.55686

⫺52 ⫹548

The starting values were obtained by classical Runge–Kutta. 26. Exact solution 4y 21 ⫹ y 22 ⫽ 16 (ellipse). The computation gives xn

y1,n

y2,n

0 0.2 0.4 0.6 0.8 1.0 1.2 1.4 1.6 1.8 2.0

2 2 1.68 1.04 0.1312 ⫺0.9440 ⫺2.0402 ⫺2.9853 ⫺3.6041 ⫺3.7451 ⫺3.3095

0 ⫺1.6 ⫺3.2 ⫺4.544 ⫺5.376 ⫺5.4810 ⫺4.7258 ⫺3.0936 ⫺0.7053 2.1779 5.1740

28. The exact solution is y1 ⫽ ⫺6e9t ⫹ 3e3t, y2 ⫽ ⫺2e9t ⫺ e3t. The computation gives xn

y1,n

(y1,n)

y2,n

(y2,n)

0 0.05 0.10 0.15

⫺3.00000 ⫺5.92338 ⫺10.70492 ⫺18.43227

0 ⫺0.00099 ⫺0.00312 ⫺0.00734

⫺3.00000 ⫺4.29813 ⫺6.26802 ⫺9.28071

0 ⫺0.00033 ⫺0.00104 ⫺0.00245

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30. From the 3D-values given below we see that at each point x ⬎ 0 the temperature oscillates with a phase lag and a maximum amplitude that decreases with decreasing x. t

x⫽0

x ⫽ 0.2

x ⫽ 0.4

x ⫽ 0.6

x ⫽ 0.8

0 0.02 0.04 0.06 0.08 0.10 0.12 0.14 0.16 0.18 0.20 0.22 0.24

0 0 0 0 0 0 0 0 0 0 0 0 0

0 0 0 0 0 0.031 0.054 0.086 0.095 0.094 0.068 0.034 ⫺0.009

0 0 0 0 0.063 0.108 0.172 0.189 0.188 0.135 0.067 ⫺0.019 ⫺0.086

0 0 0 0.125 0.217 0.313 0.325 0.289 0.176 0.041 ⫺0.105 ⫺0.206 ⫺0.252

0 0 0.250 0.433 0.563 0.541 0.406 0.162 ⫺0.105 ⫺0.345 ⫺0.479 ⫺0.485 ⫺0.353

32. u (P21) ⫽ 500, u (P22) ⫽ 200, u ⫽ 100 at all other gridpoints 34. u (P11) ⫽ u (P22) ⫽ u (P33) ⫽ 35, u (P21) ⫽ u (P32) ⫽ 20, u (P31) ⫽ 10, u (P12) ⫽ u (P23) ⫽ 50, u (P13) ⫽ 60

x ⫽ 1.0 0 0.5 0.866 1 0.866 0.5 0 ⫺0.5 ⫺0.866 ⫺1 ⫺0.866 ⫺0.5 0

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Part F. Optimization. Graphs CHAPTER 22 Unconstrained Optimization. Linear Programming SECTION 22.1. Basic Concepts. Unconstrained Optimization: Method of Steepest Descent, page 951 Purpose. To explain the concepts needed throughout this chapter. To discuss Cauchy’s method of steepest descent or gradient method, a popular method of unconstrained optimization. Main Content, Important Concepts Objective function Control variables Constraints, unconstrained optimization Cauchy’s method SOLUTIONS TO PROBLEM SET 22.1, page 953 2. The level curves f (x) ⫽ const are concentric circles. Hence ⫺ⵜf (x) points from any x ⫽ x0 to the center 0, which should be reached in one step. Indeed, ⵜf (x) ⫽ (2x 1, 2x 2), z(t) ⫽ (x1, x 2) ⫺ t(2x1, 2x 2) ⫽ (x1(1 ⫺ 2t), x 2(1 ⫺ 2t)) so that g(t) ⫽ f (z (t)) ⫽ (1 ⫺ 2t)2 (x 21 ⫹ x 22) and the derivative is g r(t) ⫽ ⫺4(1 ⫺ 2t) (x 21 ⫹ x 22). Thus g r(t) ⫽ 0 gives t ⫽ 12 , and z(t) ⫽ x ⫺ 12 (2x1, 2x2) ⫽ 0, so that the center, the location of the minimum, is reached after one step. 4. f (x) ⫽ (x 1 ⫺ 2.5)2 ⫹ 0.5 (x 2 ⫺ 3.0)2 ⫹ 14.2. The computation gives

362

Step

f (x)

1 2 3 4 5

2.3333 2.5556 2.4815 2.5062 2.4979

3.3333 3.1111 3.0370 3.0125 3.0041

14.2833 14.2093 14.2010 14.2001 14.2000

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6. The computation gives for Steps 1–5 x1

f (x)

2.66667 ⫺1.77778 ⫺4.74074 3.16049 8.42798

⫺1.33333 ⫺3.55556 2.37037 6.32099 ⫺4.21399

5.3334 ⫺9.4815 16.8560 ⫺29.9662 53.2732

For the other x0, the values x 1 and x 2 are interchanged and f (x) is multiplied by ⫺1. 8. f (x) ⫽ x 21 ⫺ x 2 gives z(t) ⫽ x ⫺ t(2x1, ⫺1) ⫽ ((1 ⫺ 2t)x 1, x 2 ⫹ t), hence g(t) ⫽ (1 ⫺ 2t)2x 21 ⫺ x 2 ⫺ t, g r(t) ⫽ ⫺4 (1 ⫺ 2t)x 21 ⫺ 1 ⫽ 0. From this, 1 ⫺ 2t ⫽ ⫺

1 4x 21

t⫽

1 1 ⫹ 2. 2 8x 1

For this t, z(t) ⫽ A ⫺

1 1 1 , x2 ⫹ ⫹ 2 B . 4x 1 2 8x 1

From this, with x 1 ⫽ 1, x 2 ⫽ 1, we get successively z (1) ⫽ A ⫺14 , z (2) ⫽ A 1,

z (3) ⫽ A ⫺14 ,

1 ⫹ 12 ⫹ 18 B

1 ⫹ 2 ⴢ 12 ⫹ 18 ⫹ 2 B

1 ⫹ 3 ⴢ 12 ⫹ 2 ⴢ 18 ⫹ 2 B

etc.

The student should sketch this to see that it is reasonable. The process continues indefinitely, as had to be expected, with x 1 ⫽ ⫺14 , 1, Á alternatingly, and x 2 and ƒ f (x) ƒ monotone increasing. SECTION 22.2. Linear Programming, page 954 Purpose. To discuss the basic ideas of linear programming in terms of very simple examples involving two variables, so that the situation can be handled graphically and the solution can be found geometrically. To prepare conceptually for the case of three or more variables x1, Á , x n. Main Content, Important Concepts Linear programming problem Its normal form. Slack variables Feasible solution, basic feasible solution Optimal solution

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Comments on Content Whereas the function to be maximized (or minimized) by Cauchy’s method was arbitrary (differentiable), but we had no constraints, we now simply have a linear objective function, but constraints, so that calculus no longer helps. No systematic method of solution is discussed in this section; these follow in the next sections.

SOLUTIONS TO PROBLEM SET 22.2, page 957 2. Ordinarily a vertex of a region is the intersection of only two straight lines given by inequalities taken with the equality sign. Here, (5, 4) is the intersection of three such lines. This may merit special attention in some cases, as we discuss in Sec. 22.4. 6. The first inequality could be dropped from the problem because it does not restrict the region determined by the other inequalities. Note that that region is unbounded (stretches to infinity). This would cause a problem in maximizing an objective function with positive coefficients. 8. Nonnegativity is an immediate consequence of the definition of a slack variable. We need as many slack variables as we have inequalities that we want to convert into equations, with each one giving one of the constraints. 10. No. For instance, f ⫽ 5x 1 ⫹ 2x 2 gives maximum profit f ⫽ 12 for every point on the segment AB because AB has the same slope as f ⫽ const does. 12. Location not unique; f ⫽ 225 on the segment from (3, 4) to (0, 10). 14. f (56 , 76) ⫽ 100 3 16. f (0, 5) ⫽ 10 18. The function to be maximized is f ⫽ 11x1 ⫹ 15x 2 where x1 and x 2 are the number of bottles of N and S produced. Constraints arise from the production (the availability) of drugs A, B, C, namely, 2x 1 ⫹ x 2 ⬉ 1400

(Drug A)

x 1 ⫹ x 2 ⬉ 800

(Drug B)

x 1 ⫹ 3x 2 ⬉ 1800

(Drug C).

The values of f at the corners of the feasibility region are f (0, 0) ⫽ 0 and f (0, 600) ⫽ 9000 f (300, 500) ⫽ 10,800 f (600, 200) ⫽ 9600 f (700, 0) ⫽ 7700. The maximum profit of $10,800 is achieved by producing 300 bottles of N and 500 bottles of S.

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20. x 1 ⫽ Number of days of operation of Kiln I, x 2 ⫽ Number of days of operation of Kiln II. Objective function f ⫽ 400x 1 ⫹ 600x 2. Constraints: 3000x1 ⫹ 2000x 2 ⭌ 18,000

(Gray bricks)

2000x1 ⫹ 5000x 2 ⭌ 34,000

(Red bricks)

300x1 ⫹ 1500x 2 ⭌

(Glazed bricks).

9000

fmin ⫽ f (2, 6) ⫽ 4400, as can be seen from a sketch of the region in the x 1x 2-plane resulting from the constraints in the first quadrant. Operate Kiln I two days and Kiln II six days in filling that order. Note that the region determined by the constraints in the first quadrant of the x 1x 2-plane is unbounded, which causes no difficulty because we minimize (not maximize) the objective function. 22. x 1 units of A and x 2 units of B cost f ⫽ 1.8x 1 ⫹ 2.1x 2. Constraints are 15x1 ⫹ 30x 2 ⭌ 150

(Protein)

600x1 ⫹ 500x 2 ⭌ 3900

(Calories).

From a sketch of the region we see that fmin ⫽ f (4, 3) ⫽ 13.50. Hence the minimumcost diet consists of 4 units A and 3 units B. SECTION 22.3. Simplex Method, page 958 Purpose. To discuss the standard method of linear programming for systematically finding an optimal solution by a finite sequence of transformations of matrices. Main Content, Important Concepts Normal form of the problem Initial simplex table (initial augmented matrix) Pivoting, further simplex tables (augmented matrices) Comment on Concepts and Method The given form of the problem involves inequalities. By introducing slack variables we convert the problem to the normal form. This is a linear system of equations. The initial simplex table is its augmented matrix. It is transformed by first selecting the column of a pivot and then the row of that pivot. The rules for this are entirely different from those for pivoting in connection with solving a linear system of equations. The selection of a pivot is followed by a process of elimination by row operations similar to that in the Gauss–Jordan method (Sec. 7.8). This is the first step, leading to another simplex table (another augmented matrix). The next step is done by the same rules, and so on. The process comes to an end when the first row of the simplex table obtained contains no more negative entries. From this final simplex table one can read the optimal solution of the problem because the first row corresponds to the objective function f (x) to be maximized (or minimized). SOLUTIONS TO PROBLEM SET 22.3, page 961 2. Matrices with Rows 2 and 3 and Columns 4 and 5 interchanged. Pivots remain the same as before, with 5 now in Column 2 and Row 2 and 7.2 in T1 now in Column 3 and Row 3. The maximum is 840 and is achieved at (10, 5), as before.

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4. The matrices and pivot selections are 1

⫺1

T0 ⫽ D0

550T

650

550>3 ⬎ 650>5, pivot 5 1

⫺15

1 5

130

Row 1 ⫹ 15 Row 3

T1 ⫽ D0

8 5

⫺35

160T

Row 2 ⫺ 35 Row 3

650

160> 85 ⬍ 650>4, pivot 85 1

1 8

150

Row 1 ⫹ 18 Row 2

T2 ⫽ D0

8 5

⫺35

160T

⫺52

5 2

250

Row 3 ⫺ 52 Row 2

fmax ⫽ 150 at x 1 ⫽ 250>5 ⫽ 50, x 2 ⫽ 160>(85) ⫽ 100. 6. From the given data we have the augmented matrix (the initial simplex table) 1

⫺1

T0 ⫽ D0

1200T .

1600

The pivot is 4 since 1600>4 ⬍ 1200>2. The indicated calculations give 1

⫺12

1 4

400

Row 1 ⫹ 14 Row 3

T1 ⫽ D0

⫺12

400T

Row 2 ⫺ 24 Row 3

1600

Row 3.

The pivot is 2 in Row 2. The indicated calculations give 1

1 4

1 8

500

Row 1 ⫹ 14 Row 2

T2 ⫽ D0

⫺12

400T

Row 2

⫺1

3 2

1200

Row 3 ⫺ 22 Row 2.

This shows that the solution is fa

1200 400 , b ⫽ 500. 4 2

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8. The matrices and pivot selections are 1

⫺90

⫺50

T0 ⫽ E

pivot 3 in row 4 1

⫺20

720

8 3

⫺13

2 3

⫺13

780

⫺4

2 3

⫺13

⫺32

3 2

T1 ⫽ E

pivot 23 in row 3

T2 ⫽ E

2 fmax ⫽ 780 at x 1 ⫽ 21 3 ⫽ 7, x 2 ⫽ 2> 3 ⫽ 3. 10. Remember that we are looking for a minimum rather than for a maximum! The matrices and pivot selections are

⫺4

T0 ⫽ E

60 20 4 ⫽ 15 ⬍ 1 ⫽ 20, pivot 4

⫺23 2

15 2

⫺52

⫺150

Row 2

5 4

⫺54

⫺14

Row 3 ⫺ 14 Row 2

Row 4

T1 ⫽ E

60 30 5 ⬎ 3 , pivot 3

Row 1 ⫺ 10 4 Row 2

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⫺33 2

⫺52

⫺225

Row 1 ⫺ 15 6 Row 4

⫺13

⫺53

Row 2 ⫺ 53 Row 4

25 12

⫺14

5 12

35 2

5 Row 3 ⫹ 12 Row 4

T2 ⫽ E

Row 4

30 fmin ⫽ ⫺225 at x 1 ⫽ 0, x 2 ⫽ 10 4 ⫽ 2.5, x 3 ⫽ 3 ⫽ 10. 12. From the given data we obtain the augmented matrix

⫺2

⫺3

⫺1

4.8

9.9

⫺1

0.2

T0 ⫽ E

The pivot is 10. The indicated calculations give 0

⫺3

⫺45

1 5

1.98

2 Row 1 ⫹ 10 Row 3

9 10

1 ⫺10

3.81

1 Row 2 ⫺ 10 Row 3

9.9

Row 3

⫺1

0.2

T1 ⫽ E

Row 4

The next pivot is 1 in Row 4 and Column 3. The indicated calculations give 0

⫺19 5

1 5

2.58

Row 1 ⫹ 3 Row 4

19 10

1 ⫺10

⫺1

3.61

Row 2 ⫺ Row 4

9.9

Row 3

⫺1

0.2

T2 ⫽ E

Row 4

The last pivot needed is 19 10 in Row 2 and Column 4. We obtain 1

9.8

T3 ⫽ E 0

19 10

1 ⫺10

⫺1

3.61

⫺10 19

20 19

10 19

1 ⫺19

9 19

2.1

Row 1 ⫹ 2 Row 2

Hence a solution of our problem is fa

8 2.1 3.61 , , b ⫽ f (0.8, 2.1, 1.9) ⫽ 9.8. 10 1 19>10

Row 2 Row 3 ⫺ 10 19 Row 2 Row 4 ⫹ 10 19 Row 2

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Actually, all solutions are 2.5 ⫺ 0.5x 1,

(x1,

2.3 ⫺ 0.5x 1)

where x1 is arbitrary, satisfying 0 ⬉ x1 ⬉ 0.8, thus giving a straight segment with endpoints (0, 2.5, 2.3) and (0.8, 2.1, 1.9), where x1 ⫽ 0.8 results from solving the system of three equations of the constraints taken with equality signs. The reason for the nonuniqueness is that the plane f (x 1, x 2, x 3) ⫽ 9.8 contains an edge of the region to which x 1, x 2, x 3 are restricted, whereas in general it will have just a single point (a vertex) in common with that region. 14. From the given data we obtain the augmented matrix 1

⫺2

⫺3

T0 ⫽ D0

105T .

126

The pivot is 5 in Row 2. The indicated calculations give 1

⫺95

2 5

T1 ⫽ D0

105T

21 5

⫺35

Row 1 ⫹ 25 Row 2

Row 3 ⫺ 35 Row 2.

105 21 Since 63> (21 5 ) ⬍ 3 , the pivot is 5 in Row 3. The indicated calculations give

1 7

3 7

9 Row 1 ⫹ 21 Row 3

T2 ⫽ D0

10 7

⫺57

60T

Row w ⫺ 57 Row 3.

21 5

⫺35

Hence the result is fmax ⫽ f a

60 63 , b ⫽ f (12, 15) ⫽ 69. 5 21>5

SECTION 22.4. Simplex Method: Difficulties, page 962 Purpose. To explain ways of overcoming difficulties that may arise in applying the simplex method. Main Content, Important Concepts Degenerate feasible solution Artificial variable (for overcoming difficulties in starting) Short Courses. Omit this section because these difficulties occur only quite infrequently in practice.

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SOLUTIONS TO PROBLEM SET 22.4, page 968 2. In the second step of Prob. 1 we had a choice of pivot, and, in the present problem, owing to the rule of choice, we took the other pivot. The result remains the same. 1

⫺7

⫺14

0 T0 ⫽ E 0

⫺14

Row 1 ⫹ 7 Row 2

Row 2

⫺7

Row 3 ⫺ 7 Row 2

Row 4

Row 1 ⫹ Row 3

Row 2

⫺7

Row 3

1 2

1 ⫺14

T1 ⫽ E

T2 ⫽ E

1 Row 4 ⫺ 14 Row 3

This gives x 1 ⫽ 6, x 2 ⫽ 3, and x 3 ⫽ x 4 ⫽ x 5 ⫽ 0, f (6, 3) ⫽ 84. 4. The calculation is as follows. 1

⫺300

⫺500

T0 ⫽ E

⫺350

150

4500

⫺1

Row 2 ⫺ Row 3

Row 3

⫺2

Row 4 ⫺ 2 Row 3

⫺200

175

4500

Row 1 ⫹ 175 Row 4

⫺72

Row 2 ⫺ 72 Row 4

⫺12

Row 3 ⫺ 12 Row 4

⫺2

Row 4

T1 ⫽ E

T2 ⫽ E 0 0

Row 1 ⫹ 150 Row 3

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z ⫽ 4500 is the same as in the step before. But we shall now be able to reach the maximum f (10, 5) ⫽ 5500 in the final step. 1

100 3

175 3

5500

⫺72

Row 2

⫺13

2 3

Row 3 ⫺ 13 Row 2

1 3

⫺16

Row 4 ⫹ 13 Row 2

T3 ⫽ E

Row 1 ⫹ 100 3 Row 2

10 30 We see that x 1 ⫽ 20 2 ⫽ 10, x 2 ⫽ 2 ⫽ 5, x 3 ⫽ 0, x 4 ⫽ 6 ⫽ 5, x 5 ⫽ 0, z ⫽ 5500. Problem 5 shows that the extra step (which gave no increase of z ⫽ f (x)) could have been avoided if we had chosen 4 (instead of 2) as the first pivot. 12 24 6. The maximum f (12 11 , 11 , 0) ⫽ 11 is obtained as follows.

⫺1

T0 ⫽ D0

12T

⫺37

⫺67

12 7

Row 1 ⫹ 17 Row 3

T1 ⫽ D0

22 7

44 7

24 7 T

Row 2 ⫺ 57 Row 3

Row 3

3 22

24 11

3 Row 1 ⫹ 22 Row 2

T2 ⫽ D0

22 7

44 7

24 7 T

Row 2

⫺7

⫺14 11

84 11

Row 3 ⫺ 28 22 Row 2

12 24 From T2 we see that x 1 ⫽ 22 11 , x 2 ⫽ 11 , x 3 ⫽ 0, z ⫽ 11 . 8. We wish to maximize the modified objective function 苲 f ⫽ ⫺4x 1 ⫹ x 2. We introduce an artificial variable x 6 is defined by

x 3 ⫽ ⫺2 ⫹ x 1 ⫹ x 2 ⫹ x 6. A corresponding objective function is ⬇ 苲 f ⫽ f ⫺ Mx 6 ⫽ (⫺4 ⫹ M)x1 ⫹ (1 ⫹ M)x 2 ⫺ Mx 3 ⫺ 2M. The corresponding matrix is 4⫺M

⫺1 ⫺ M

⫺2M

⫺1

⫺2

T0 ⫽ E

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⫺5

⫺8

0 T1 ⫽ E 0

⫺1

Row 2

⫺2

Row 3 ⫹ 2 Row 2

⫺1

Row 4 ⫺ 5 Row 2

⫺3

Row 1 ⫹ 1 Row 3

⫺35

⫺15

Row 2 ⫺ 15 Row 3

⫺2

Row 3

23 5

1 5

T2 ⫽ E

Row 1 ⫺ (⫺4 ⫹ M) Row 2

Row 4 ⫹ 15 Row 3 ⬇ We see that x 1 ⫽ 1, x 2 ⫽ 1, x 3 ⫽ 0, x 4 ⫽ 0, x 5 ⫽ 41, f ⫽ ⫺3 0

SOLUTIONS TO CHAPTER 22 REVIEW QUESTIONS AND PROBLEMS, page 968 4. Replace ⫺f by ⫹f. 10. Nine steps give the solution [⫺1

2]T to 6S. Steps 1–5 give

⫺1.01462 ⫺0.888521 ⫺1.00054 ⫺0.995857 ⫺1.00002

3.77670 2.07432 2.06602 2.00276 2.00245

⫺9.8414 ⫺12.8826 ⫺12.9956 ⫺12.9998 ⫺13.0000

12. The values obtained are x1

⫺1.04366 ⫺0.758212 ⫺1.01056 ⫺0.941538 ⫺1.00255

0.231924 1.51642 1.57250 1.88308 1.89664

Gradients (times a scalar) are obtained by calculating differences of subsequent values. Orthogonality follows from the fact that we change direction when we are tangent to a level curve and then proceed perpendicular to it. Such a zigzag curve is certainly not a shortest curve, and there should be better ways; this is the idea of more refined methods, considered, e.g., in [E25], listed in App. 1. 18. The augmented matrix of the given data is 1

⫺1

T0 ⫽ E

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The pivot is 2 in Row 3 and Column 2. The calculation gives 1

⫺12

1 2

Row 1 ⫹ 12 Row 3

3 2

⫺12

Row 2 ⫺ 12 Row 3

Row 3

Row 4.

T1 ⫽ E 0 0

The next pivot is 32 in Row 2. The calculation gives 1

1 3

20 3

Row 1 ⫹ 13 Row 2

0 T2 ⫽ E 0

3 2

⫺12

Row 2

⫺23

4 3

Row 3 ⫺ 23 Row 2

⫺23

1 3

2 3

Row 4 ⫺ 23 Row 2.

U 20

We see from the last matrix that for the maximum we have fa

20>3 5 10 10 20 , b⫽fa , b⫽ . 2 3>2 3 3 3

20. The matrix of the given data is 1

⫺60

⫺30

T0 ⫽ D0

1800T

200

6300

The pivot is 200. The calculation gives 1

⫺24

3 10

T1 ⫽ D0

⫺15

540T

Row 2 ⫺ 15 Row 3

200

6300

Row 3

2250

Row 1 ⫹ 24 36 Row 2

540T

Row 2

6000

Row 3 ⫺ 20 36 Row 2

1890

60 Row 1 ⫹ 200 Row 3

The next pivot is 36. The calculation gives 1

2 3

1 6

T2 ⫽ D0

⫺15

200

⫺59

10 9

Hence the solution is fa

6000 540 , b ⫽ f (30, 15) ⫽ 2250. 200 36

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CHAPTER 23

Graphs. Combinatorial Optimization

SECTION 23.1. Graphs and Digraphs, page 970 Purpose. To explain the concepts of a graph and a digraph (directed graph) and related concepts, as well as their computer representations. Main Content, Important Concepts Graph, vertices, edges Incidence of a vertex v with an edge, degree of v Digraph Adjacency matrix Incidence matrix Vertex incidence list, edge incidence list Comment on Content Graphs and digraphs have become more and more important, due to an increase of supply and demand—a supply of more and more powerful methods of handling graphs and digraphs, and a demand for those methods in more and more problems and fields of application. Our chapter, devoted to the modern central area of combinatorial optimization, will give us a chance to get a feel for the usefulness of graphs and digraphs in general.

SOLUTIONS TO PROBLEM SET 23.1, page 974 4.

8. G1

10. E

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12. G0

14.

375

16. Join v1 to v2, Á , vn, then v2 to v3, Á , vn, then v3 to v4, Á , vn, etc.; then take the sum 1 2 Á (n 1) 12 n(n 1), the number of edges you have used in that process.

18.

Vertex

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Edge e4

20. Vertex 1 2 3 4

Incident Edges e1, e2, e3, e4 e1 e2, e3 e4

SECTION 23.2. Shortest Path Problems. Complexity, page 975 Purpose. To explain a method (by Moore) of determining a shortest path from a given vertex s to a given vertex t in a graph, all of whose edges have length 1. Main Content, Important Concepts Moore’s algorithm (Table 23.1) BFS (Breadth First Search), DFS (Depth First Search) Complexity of an algorithm Efficient, polynomially bounded Comment on Content The basic idea of Moore’s algorithm is quite simple. A few related ideas and problems are illustrated in the problem set.

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SOLUTIONS TO PROBLEM SET 23.2, page 979 2. The length of a shortest path is 5. No uniqueness. 4. There are 3 shortest paths, of length 4 each: s

(A)

(B)

(C)

Which one we obtain in backtracking depends on the numbering (not labeling!) of the vertices and on the backtracking rule. For the rule in Example 1 and the numbering shown in the following figure we get (B). 1 0 s 1 4 t 3

3 4

2 2 6

5 1 3 10

8 3 7 2

4 12

9 3

If we change the rule and let the computer look for largest (instead of smallest) numbers, we get (A). 6. n 1. If it had more, a vertex would appear more often than once and the corresponding cycle could be omitted. One edge. 8. This is true for l 0 since then v s. Let it be true for an l 1. Then l(vlⴚ1) l 1 for the predecessor vlⴚ1 of v on a shortest path s : v. We claim that when vlⴚ1 gets labeled, v is still unlabeled (so that we shall have l(v) l as wanted). Indeed, if v were labeled, it would have a label less than l, hence distance less than l by Prob. 5, contradicting that v has distance l. 10.

12. No 14. Let T: s : s be a shortest postman trail and v any vertex. Since T includes each edge, T visits v. Let T1: s : v be the portion of T from s to the first visit of v and T2: v : s the other portion of T. Then the trail v : v consisting of T2 followed by T1 has the same length as T and solves the postman problem. 16. 1 2 3 4 5 3 1, 1 3 4 5 3 2 1, and these two trails traversed in the opposite sense 20. From m to 10m, 2.5m, m 4.6

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SECTION 23.3. Bellman’s Principle. Dijkstra’s Algorithm, page 980 Purpose. This section extends the previous one to graphs whose edges have any (positive) length and explains a popular corresponding algorithm (by Dijkstra). Main Content, Important Concepts Bellman’s optimality principle, Bellman’s equations Dijkstra’s algorithm (Table 23.2) Comment on Content Throughout this chapter, one should emphasize that algorithms are needed because most practical problems are so large that solution by inspection would fail, even if one were satisfied with approximately optimal solutions. SOLUTIONS TO PROBLEM SET 23.3, page 983 2. Let j be the vertex that gave k its present label L k, namely, L j l jk. After this label had been assigned, j did not change its label, since it was then removed from ᐀ᏸ. Next, find the vertex which gave j its permanent label, etc. This backward search traces a path from 1 to k, who length is exactly L k. 4. Dijkstra’s algorithm gives 苲 15, L 苲 2, L 苲 10, L 苲 6 1. L 1 0, L 2 3 4 5 2. L 3 2 苲 min {15, 2 l } 15 3. L 2 32 苲 min {10, 2 l } 10 L 4 34 苲 min {6, 2 l } 5 L 5 35 2. L 5 5 苲 min {15, 5 l } 15 3. L 2 52 苲 min {10, 5 l } 9 L 4 54 2. L 4 9 苲 min {15, 9 l } 14 3. L 2 42 2. L 2 14 The answer is (1, 3), (2, 4), (3, 5), (4, 5); L 2 14, L 3 2, L 4 9, L 5 5 6. The algorithm gives 苲 2, L 苲 6, L 苲 8, L 苲 1. L 1 0, L 2 3 4 5 2. L 2 2, k 2 苲 min {6, 2 l } 5 3. L 3 23 苲 min {8, 2 l } 8 L 4 24 苲 min { , 2 } L 5 2. L 3 5, k 3 苲 min {8, 5 l } 8 3. L 4 34 苲 min { , 5 } L 5 2. L 4 8, k 4 苲 min { , 8 l } 28 3. L 5 45 2. L 5 28, k 5,

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so that the answer is (1, 2), (1, 4), (2, 3), (4, 5); L 2 2, L 3 5, L 4 8, L 5 28 8. Dijkstra’s algorithm gives 苲苲苲苲苲 1. L 1 0, L 2 8, L 3 10, L 4 , L 5 5, L 6 2. L 5 5 苲 3. L 2 min {8, 5 l 52} 7 苲 min {10, 5 l } 10 L 3 53 苲 min { , 5 l } 10 L 4 54 苲 min { , 5 l } 7 L 6 56 2. L 2 7 苲 3. L 3 min {10, 7 l 23} 9 苲 min {10, 7 l } 10 L 4 24 苲 L min {7, 7 l 6 26} 7 2. L 6 7 苲 3. L 3 min {9, 7 l 63} 9 苲 min {10, 7 l } 8 L 4 64 2. L 4 8 苲 3. L 3 min {9, 8 l 43} 9 2. L 3 9 The answer is (1, 5), (2, 3), (2, 5), (4, 6), (5, 6); L 2 7, L 3 9, L 4 8, L 5 5, L6 7 10. CAS PROBLEM. Dijkstra’s Algorithm. Write a program and apply it to Probs. 4–6. SECTION 23.4. Shortest Spanning Trees: Greedy Algorithm, page 984 Purpose. After the discussion of shortest paths between two given vertices, this section is devoted to the construction of a tree in a given graph that is spanning (contains all vertices of the graph) and is of minimum length. Main Content, Important Concepts Tree Cycle Kruskal’s greedy algorithm (Table 23.3) Comment on Content Figure 491 illustrates that Kruskal’s algorithm does not necessarily give a tree during each intermediate step, in contrast to another algorithm to be discussed in the next section. SOLUTIONS TO PROBLEM SET 23.4, page 987 2w 2w6 ì w 2. 4 3 L 30 ê 2w1 5w1 Note that trees, just like general graphs, can be sketched in different ways. 2 3 ì 4. 2 w 1 w 4 L 12 ê 2w 1 5

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2w 6w5 ì w 6. 7 8 1 L 40 ê ì 2w1 2 ê ê w 2w1 2 4 3 8. Order the edges in descending order of length and delete them in this order, retaining an edge only if it would lead to the omission of a vertex or to a disconnected graph. 10. Order the edges in descending order of length and choose them in this order, rejecting when a cycle would arise. 14. Let P1: u : v and P2: u : v be different. Let e (w, x) be in P1 but not in P2. Then P1 without e together with P2 is a connected graph. Hence it contains a path P3: w : x. Hence P3 together with e is a cycle in T, a contradiction. 16. Extend an edge e into a path by adding edges to its ends if such edges exist. A new edge attached at the end of the path introduces a new vertex, or closes a cycle, which contradicts our assumption. This extension terminates on both sides of e, yielding two vertices of degree 1. 18. True for n 2. Assume truth for all trees with fewer than n vertices. Let T be a tree with n 2 vertices, and (u, v) an edge of T. Then T without (u, v) contains no path u : v, by Prob. 14. Hence this graph is disconnected. Let G1, G2 be its connected components, having n 1 and n 2 vertices, hence n 1 1 and n 2 1 edges, respectively, by the induction hypothesis, so that G has n 1 1 n 2 1 1 n 1 edges. 20. If G is a tree, it has no cycles, and has n 1 edges by Prob. 18. Conversely, let G have no cycles and n 1 edges. Then G has 2 vertices of degree 1 by Prob. 16. Now prove connectedness by induction. True when n 2. Assume true for n k 1. Let G with k vertices have no cycles and k 1 edges. Omit a vertex v and its incident edge e, apply the induction hypothesis and add e and v back on.

SECTION 23.5. Shortest Spanning Trees: Prim’s Algorithm, page 988 Purpose. To explain another algorithm (by Prim) for constructing a shortest spanning tree in a given graph whose edges have arbitrary (positive) lengths. Comments on Content In contrast to Kruskal’s greedy algorithm (Sec. 23.4), Prim’s algorithm gives a tree at each intermediate step. The problem set illustrates a few concepts that can be included into the present cycle of ideas.

SOLUTIONS TO PROBLEM SET 23.5, page 990 2. In Step 2 we first select a smallest l 1j for the n 1 vertices outside U; these are n 2 comparisons. Step 3 then requires n 2 updatings (pairwise comparisons). In the next round we have n 3 comparisons in Step 2 and n 3 updatings in Step 3, and so on, until we finally end up with 1 comparison and 1 updating. The sum of all these numbers is (n 2)(n 1) O(n2). 4. An algorithm for minimum spanning trees must examine each entry of the distance matrix at least once, because an entry not looked upon might have been one that should have been included in a shortest spanning tree. Hence, examining the relevant given information is already O(n 2) work.

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6. The algorithm gives

Vertex

Initial Label

2 3 4 5

l 12 6 l 13 1 l 15 15

Relabeling (I)

(II)

(III)

l 34 10 l 25 9

l 54 2

l 32 3 l 34 10 l 15 15

We see that we got L 15.

(1, 3), (3, 2), (2, 5), (5, 4);

The tree has the length L 15. To visualize the effect of the algorithm, use the graph (the figure) and for each step circle U and then go along the “circle” and look for the shortest edge that crosses it. 8. The algorithm gives Relabeling

Vertex

Initial Label

(I)

(II)

(III)

(IV)

2 3 4 5 6 7 8

l 12 3 l 18 8

l 23 4 l 26 10 l 27 7 l 28 7

l 34 3 l 35 5 l 36 2 l 37 6 l 28 7

l 64 1 l 35 5

l 35 5

l 37 6 l 28 7

(V)

(VI)

l 37 6 l 28 7

l 28 7

We see that we got (1, 2), (2, 3), (3, 6), (6, 4), (3, 5), (3, 7), (2, 8). The length is L 28. 10. We obtain, in this order, the tree (1, 2), (2, 8), (8, 7), (8, 6), (6, 5), (2, 4), (4, 3). The length is 40. The tree is the same as before. 12. The algorithm proceeds as follows.

Vertex

Initial Label

2 3 4 5 6

l 12 20 l 15 8 l 16 30

Relabeling (I)

(II)

(III)

l 12 20 l 53 6 l 54 12

l 32 4

l 34 2

l 16 30

Hence we got successively (1, 5), (5, 3), (3, 4), (3, 2), (2, 6);

L 30.

(IV)

l 16 30

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In Prob. 2 of Sec. 23.4 we got the same edges, but in the order (3, 4), (2, 3), (3, 5), (1, 5), (2, 6). 14. TEAM PROJECT. (a) P(1) 16, P(2) 22, P(3) 12 (b) From the graph we have P(1) 16 d(1, 5) max {6, 4, 2, 16, 14} P(2) 22 d(2, 5) max {6, 10, 8, 22, 8} P(3) 12 d(3, 6) d(3, 5) max {4, 10, 6, 12, 12} P(4) 18 d(4, 5) max {2, 8, 6, 18, 16} P(5) 24 d(5, 6) max {16, 22, 12, 18, 24} P(6) 24 d(6, 5) max {14, 8, 12, 16, 24} From this and the definitions it follows that d(G) 24, r (G) 12, and the center is {3}. Note that the 30 values are the entries of a symmetric 6 6 distance matrix with zero diagonal entries. Similarly for (c). (c) For the spanning tree in Example 1 of the text we obtain P(1) max {2, 4, 12, 18, 5} 18 d(1, 5) P(2) max {2, 6, 14, 20, 7} 20 d(2, 5) P(3) max {4, 6, 8, 14, 1} 14 d(3, 5) P(4) max {12, 14, 8, 6, 9} 14 d(4, 1) P(5) max {18, 20, 14, 6, 15} 20 d(5, 6) P(6) max {5, 7, 1, 9, 15} 15 d(6, 5) so that diameter and radius are d(T ) max {P(1), Á , P(6)} 20 P(2) P(5) r(T ) min {P(1), Á , P(6)} 14 P(3) P(4) and the center is {3, 4}. It is interesting to compare this with the values for the given graph, namely, 苲 P(1) max {2, 4, 12, 14, 5} 14 d(1, 5) 苲 P(2) max {2, 6, 11, 16, 7} 16 d(2, 5) 苲 P(3) max {4, 6, 8, 10, 1} 10 d(3, 5) 苲 P(4) max {12, 11, 8, 6, 9} 12 d(4, 1) 苲 P(5) max {14, 16, 10, 6, 9} 16 d(5, 2) 苲 P(6) max {5, 7, 1, 9, 9} 9 d(6, 5) d(6, 4) so that diameter and radius are d(G) 16 P(2) r(G) 9 P(6) and the center is {6}. (e) Let T * be obtained from T by deleting all endpoints ( vertices of degree 1) together with the edges to which they belong. Since, for fixed u, max d(u, v) occurs only when v is an endpoint, P(u) is one less in T * than it is in T. Hence the vertices of minimum

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eccentricity in T are the same as those in T *. Thus T has the same center as T *. Delete the endpoints of T * to get a tree T ** whose center is the same as that of T, etc. The process terminates when only one vertex or two adjacent vertices are left. (f) Choose a vertex u and find a farthest v1. From v1 find a farthest v2. Find w such that d(w, v1) is as close as possible to being equal to 12 d(v1, v2).

SECTION 23.6. Flows in Networks, page 991 Purpose. After shortest paths and spanning trees we discuss in this section a third class of practically important problems, the optimization of flows in networks. Main Content, Important Concepts Network, source, target (sink) Edge condition, vertex condition Path in a digraph, forward edge, backward edge Flow augmenting path Cut set, Theorems 1 and 2 Augmenting path theorem for flows (Theorem 3) Max-flow min-cut theorem Comment on Content An algorithm for determining flow augmenting paths follows in the next section.

SOLUTIONS TO PROBLEM SET 23.6, page 997 2. T {4, 5, 6, 7}, cap (S, T) 7 10 17 4. T {3, 4, 5, 6, 7}, cap (S, T ) 7 8 15 6. T {2, 4, 6}, cap (S, T ) 20 10 4 3 13 50

8, 4

s 1

8, 5

2 6, 1

4, 2

10,

7 t

2, 1

5 4

6, 5

Fig. 499.

4, 2

Problem 2, 4, 5

8. T {3, 6}, f 11 3 14 12. Flow augmenting paths are P1: 1 2 4 6, ¢f 1 P2: 1 3 5 6, ¢f 1 P3: 1 2 3 5 6, ¢f 1 P4: 1 2 3 4 5 6, ¢f 1, etc.

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14. Flow augmenting paths are P1: 1 2 4 5, ¢f 2 P2 : 1 2 5, ¢f 2 P3: 1 2 3 5, ¢f 3 P4: 1 3 5, ¢f 5, etc. 16. The maximum flow is f 14. It can be realized by f12 8, f13 6, f24 8, f43 4, f35 10, f45 4. 18. The maximum flow is f 4. It is realized by f12 2, f13 2, f24 1, f23 1, f35 1, f34 2, f45 0, f46 3, f56 1. f is unique, but the way in which it is achieved is not, in general. In the present case we can change f45 from 0 to 1, f46 from 3 to 2, f56 from 1 to 2. 20. For instance, change f35 to 0, f32 to 5, f25 to 8.

SECTION 23.7. Maximum Flow: Ford–Fulkerson Algorithm, page 998 Purpose. To discuss an algorithm (by Ford and Fulkerson) for systematically increasing a flow in a network (e.g., the zero flow) by constructing flow augmenting paths until the maximum flow is reached. Main Content, Important Concepts Forward edge, backward edge Ford–Fulkerson algorithm (Table 23.8) Scanning of a labeled vertex Comment on Content Note that this is the first section in which we are dealing with digraphs.

SOLUTIONS TO PROBLEM SET 23.7, page 1000 2. Not more work than in Example 1. Steps 1–7 are similar to those in the example and give the flow augmenting path P1: 1 2 3 6, which augments the flow from 0 to 11. In determining a second flow augmenting path we scan 1, labeling 2 and 4 and getting ¢ 2 9, ¢ 4 10. In scanning 2, that is, trying to label 3 and 5, we cannot label 3 because cij c23 fij f23 11, and we cannot label 5 because f52 0. In scanning 4 (i.e., labeling 5) we get ¢ 5 7. In scanning 5 we cannot label 3 because f35 0, and we further get ¢ 6 3. Hence a flow augmenting path is P2: 1 4 5 6 and ¢ t 3. Together we get the maximum flow 11 3 14 because no further flow augmenting paths can be found. The result agrees with that in Example 1.

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4. At each vertex, the inflow and the outflow are increased by the same amount. 6. Scanning the vertices in the order of their numbers, we get a flow augmenting path with ¢ t 1 and then P1: 1 2 4 6 P2: 1 3 4 6 with ¢ t 1, but no further flow augmenting path. Since the initial flow was 2, this gives the total flow f 4. 8. The given flow equals 9. We first get the flow augmenting path P1: 1 2 5

with

¢ t 2,

with

¢ t 5,

then the flow augmenting path P2: 1 3 5 and finally the flow augmenting path P3: 1 2 3 5

with

¢ t 1.

The maximum flow is 9 2 5 1 17. 10. Start from the zero flow. If it is not maximum, there is an augmenting path by which we can augment the flow by an amount that is an integer, since the capacities are integers, etc. 14. The forward edges of the set are used to capacity; otherwise one would have been able to label their other ends. Similarly for the backward edges of the set, which carry no flow. 苲 16. Let G have k edge-disjoint paths s : t, and let f be a maximum flow in G. Define 苲苲 on those paths a flow f by f (e) 1 on each of their edges. Then f k f since f is maximum. Now let G* be obtained from G by deleting edges that carry no portion 苲苲 of f . Then, since each edge has capacity 1, there exist f edge-disjoint paths in G*, 苲苲 hence also in G, and f k. Together, f k. 18. Since (S, T ) is a cut set, there is no directed path s : t in G with the edges of (S, T ) deleted. Since all edges have capacity 1, we thus obtain cap (S, T) q. Now let E 0 be a set of q edges whose deletion destroys all directed paths s : t, and let G0 denote G without these q edges. Let V0 be the set of all those vertices v in G0 for which there is a directed path s : v. Let V1 be the set of the other vertices in G. Then (V0, V1) is a cut set since s 僆 V0 and t 僆 V1. This cut set contains none of the edges of G0, by the definition of V0. Hence all the edges of (V0, V1) are in E 0, which has q edges. Now (S, T ) is a minimum cut set, and all the edges have capacity 1. Thus, cap (S, T ) cap (V0, V1) q. Together, cap (S, T ) q. 20. f 7

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SECTION 23.8. Bipartite Graphs. Assignment Problems, page 1001 Purpose. As the last class of problems, in this section we explain assignment problems (of workers to jobs, goods to storage spaces, etc.), so that the vertex set V of the graph consists of two subsets S and T and vertices in S are assigned (related by edges) to vertices in T. Main Content, Important Concepts Bipartite graph G (V, E) (S, T; E) Matching, maximum cardinality matching Exposed vertex Alternating path, augmenting path Matching algorithm (Table 23.9) Comment on Content A few additional problems on graphs, related to the present circle of ideas as well as of a more general nature, are contained in the problem set. SOLUTIONS TO PROBLEM SET 23.8, page 1005 2. Yes. S {1, 4}, T {2, 3} 4. No. For instance, if you put 1 into S, then 2, 3, 4 should go into T, but this contradicts since 2, 3 cannot both be in T, they are adjacent (endpoints of the same edge). 6. No, as for a triangle, septangle, etc., whereas square, hexagon, Á are bipartite. 8. Yes; a graph is not bipartite if it has a nonbipartite subgraph. 10. 5 1 ⴚ 4 3 ⴚ 6 2 12. 1 4 ⴚ 3 6 ⴚ 7 8 14. The path 5 1ⴚ4 3ⴚ6 2 is augmenting and gives 5ⴚ1 4ⴚ3 6ⴚ2 and (5, 1), (4, 3), (6, 2) is a maximum cardinality matching. 16. n 1n 2 18. One might perhaps mention that the particular significance of K 5 and K 3,3 results from Kuratowski’s theorem, stating that a graph G is planar if and only if G contains no subdivision of K 5 or K 3,3 (that is, G contains no subgraph obtained from K 5 or K 3,3 by subdividing the edges of these graphs by introducing new vertices on them). 20. Period

x1 x2 x3

y4 y1 —

y3 y4 y2

y1 y3 y4

— y2 y3

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22. 4 24. No; K 5 is not planar. 26. max d(u) n. Let u1, Á , u n and v1, Á , vn denote the vertices of S and T, respectively. Color edges (u 1, v1), Á , (u 1, vn) by colors 1, Á , n, respectively, then edges (u 2, v1), Á , (u 2, vn) by colors 2, Á , n, 1, respectively, etc., cyclicly permuted. SOLUTIONS TO CHAPTER 23 REVIEW QUESTIONS AND PROBLEMS, page 1006 12.

To vertex 1

2 D0

From vertex 1

14.

16.

18. 4 20. 2 w 2 ì 1w4 L 10 ê 2w1 3w5 22. The maximum flow is f 7. 24. 1 2 ⴚ 3 5

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Part G. PROBABILITY, STATISTICS CHAPTER 24

Data Analysis. Probability Theory

SECTION 24.1. Data Representation. Average. Spread, page 1011 Purpose. To discuss standard graphical representations of data in statistics. To introduce concepts that characterize the average size of the data values and their spread (their variability). Main Content, Important Concepts Stem-and-leaf plot Histogram Boxplot Absolute frequency, relative frequency Cumulative relative frequency Outliers Mean Variance, standard deviation Median, quartiles, interquartile range Comment on Content The graphical representations of data to be discussed in this section have become standard in connection with statistical methods. Average size and variability give the two most important general characterizations of data. Relative frequency will motivate probability as its theoretical counterpart. This is a main reason for presenting this material here before the beginning of our discussion of probability in this chapter. Randomness is not mentioned in this section because the introduction of samples (random samples) as a concept can wait until Chap. 25 when we shall need them in connection with statistical methods. The connection with this section will then be immediate and will provide no difficulty or duplication. SOLUTIONS TO PROBLEM SET 24.1, page 1015 2. qL ⫽ 2, qM ⫽ 5, qU ⫽ 6 4. qL ⫽ 69.7, qM ⫽ 70.5, qU ⫽ 71.2 6. qL ⫽ qM ⫽ 15, qU ⫽ 15.5 8. qL ⫽ 85, qM ⫽ 87, qU ⫽ 89 10. qL ⫽ ⫺0.51, qM ⫽ ⫺0.18, qU ⫽ 0.25 12. x ⫽ 4.3, s ⫽ 2.541, IQR ⫽ 4 14. x ⫽ 70.487, s ⫽ 1.0467, IQR ⫽ 71.2 ⫺ 69.7 ⫽ 1.5 16. x ⫽ 86.3, s ⫽ 3.63, IQR ⫽ 89 ⫺ 85 ⫽ 4 18. Outliers lie 89 ⫹ 1.5 ⴢ IQR ⫽ 95 or higher or 85 ⫺ 6 ⫽ 79 or lower. Hence the only outlier is 76. Its omission increases the mean from 86.3 to 86.8 and decreases the standard deviation (of course!), very substantially from 3.63 to 2.77. 387

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20. x min ⬉ x j ⬉ x max. Now sum over j from 1 to n. Then divide by n to get x min ⬉ x ⬉ x max. SECTION 24.2. Experiments, Outcomes, Events, page 1015 Purpose. To introduce basic concepts needed throughout Chaps. 24 and 25. Main Content, Important Concepts Experiment Sample space S, outcomes, events Union, intersection, complements of events Mutually exclusive events Representation of sets by Venn diagrams Comment on Content To make the chapter self-contained, we explain the modest amount of set-theoretical concepts needed in the next sections, although most students will be familiar with these matters.

SOLUTIONS TO PROBLEM SET 24.2, page 1017 2. 4 outcomes: HH, HT, TH, TT where H ⫽ Head, T ⫽ Tail 4. This is an example of a “waiting time problem” (and so is the next problem). We wait for the first Six. The sample space is infinite, the outcomes are (S ⫽ Six, S c ⫽ No Six) S, S cS,

S cS cS,

S cS cS cS,

Á.

6. The space of ordered triples of nonnegative numbers 8. 10 outcomes, by choosing persons 1, 2 1, 3 1, 4 1, 5 2, 3 2, 4 2, 5 3, 4 3, 5 4, 5 Section 24.4 will help us to get the answer without listing cases. This is important because in other cases the number of outcomes can be very much larger. 10. (5, 5, 5) 僆 A 傽 B, hence the answer is no. Note that A 傽B ⫽ {(3, 6, 6), (4, 5, 6), (4, 6, 5), (5, 4, 6), (5, 5, 5), (5, 6, 4), (6, 3, 6), (6, 4, 5), (6, 5, 4), (6, 6, 3)}. Note further that the answer to the same question in the case of 2 dice is yes; see Prob. 11.

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12. The subsets are ⭋, {a}, {b}, {c}, {a, b}, {a, c}, {b, c}, S. 14. We have A ⫽ {(R, R), (R, L), (L, R)} B ⫽ {(L, R), (R, L), (L, L)} C ⫽ {(R, R)} D ⫽ {(L, L)}. Hence A 傽 B ⫽ {(R, L), (L, R)} ⫽ ⭋ C傽D⫽⭋ so that the answer is no, yes. 18. Obviously, A 債 B implies A 傽 B ⫽ A. Conversely, if A 傽 B ⫽ A, then every element of A must also be in B, by the definition of intersection; hence A 債 B. 20. For instance, for the first formula we can proceed as follows (see the figure). On the right, A 傼 B: All except 3 A 傼 C: All except 5 and the intersection of these two is Right side: All except 3 and 5. On the left, A⫽1傼2傼6傼7 B傽C⫽4傼7 and the union of these two gives the same as on the right. Similarly for the other formula. C 3 4

2 7

1 6

Section 24.2. Problem 20

SECTION 24.3. Probability, page 1018 Purpose. To introduce 1. Laplace’s elementary probability concept based on equally likely outcomes, 2. The general probability concept defined axiomatically.

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Main Content, Important Concepts Definition 1 of probability Definition 2 of probability Motivation of the axioms of probability by relative frequency Complementation rule, addition rules Conditional probability Multiplication rule, independent events Sampling with and without replacement Comments on Content Whereas Laplace’s definition of probability takes care of some applications and some statistical methods (for instance, nonparametric methods in Sec. 25.8), the major part of applications and theory will be based on the axiomatic definition of probability, which should thus receive the main emphasis in this section. Sampling with and without replacement will be discussed in detail in Sec. 24.7. SOLUTIONS TO PROBLEM SET 24.3, page 1024 1 2. 12 36 ⫽ 3 4. (a) (i) 0.1 ⴢ 0.92 ⴢ 3 ⫹ 0.12 ⴢ 0.9 ⴢ 3 ⫹ 0.13 ⫽ 27.1%, as can be seen by noting that

E ⫽ {DNN,

NDN,

NND,

DDN,

DND, NDD,

DDD}.

(ii) The complement of the answer in Prob. 3 is 1 ⫺ 0.729 ⫽ 27.1%. 10 89 10 9 90 10 9 8 (b) (i) 100 䡠 90 99 䡠 98 䡠 3 ⫹ 100 䡠 99 䡠 98 䡠 3 ⫹ 100 䡠 99 䡠 98 ⫽ 27.35% (ii) 1 ⫺ 0.7265 ⫽ 27.35% 19 20 10 6. Increase. The probabilities are ( 20 30 )( 29 ) for RR, ( 30 )( 29 ) for RL and LR. For RR the probability decreases, whereas the other two probabilities increase. 8 Answer: 780 870 ⫽ 0.89655 ⬎ 9 ⫽ 0.88889

8. 0.94 ⬇ 65.6% 10. We list the outcomes that favor the event whose probability we want to determine, and after each outcome the corresponding probability (F ⫽ female, M ⫽ male): FF MFF FMF MMFF MFMF FMMF

1 4 1 8 1 8 1 16 1 16 1 16

1 This gives the answer 16 . Note that the result does not depend on the number n (⭌ 6) of cards, but only on the ratio F>M.

12. P 4 ⫽ 0.99 gives P ⫽ 0.99749 as the probability that a single switch does not fail during a given time interval, and the answer is the complement of this, namely, 0.25%.

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14. Drawing without replacement from the (hypothetically infinite) production process that is going on. The probabilities are (a) 0.982 ⫽ 96.04% (b) 2 ⴢ 0.98 ⴢ 0.02 ⫽ 3.92% (c) 0.022 ⫽ 0.04% and the sum is 1. 16. We have P(A) ⫽ 24 ⫽ 12,

P(B) ⫽ 12,

P(C) ⫽ 12

and P(A 傽 B) ⫽ 14,

P(B 傽 C) ⫽ 14,

P(C 傽 A) ⫽ 14,

but P(A 傽 B 傽 C) ⫽ 0 because there is no chip numbered 111; here P(A 傽 B 傽 C) ⫽ P(A)P(B)P(C) ⫽ 18. 18. Using Theorem 4, first for the intersection of A 傽 B and C and then for the intersection of A and B, we obtain the desired extension, P(A 傽 B 傽 A) ⫽ P((A 傽 B) 傽 C) ⫽ P(A 傽 B)P(C ƒ A 傽 B) ⫽ P(A)P(B ƒ A)P(C ƒ A 傽 B). SECTION 24.4. Permutations and Combinations, page 1024 Purpose. To discuss permutations and combinations as tools necessary for systematic counting in experiments with a large number of outcomes. Main Content Theorems 1–3 contain the main properties of permutations and combinations we must know. Formulas (5)–(14) contain the main properties of factorials and binomial coefficients we need in practice. Comment on Content The student should become aware of the surprisingly large size of the numbers involved in (1)–(4), even for relatively modest numbers n of given elements, a fact that would make attempts to list cases a very impractical matter. SOLUTIONS TO PROBLEM SET 24.4, page 1028 2. The 5!>3! ⫽ 120 6 ⫽ 20 permutations are ⴢ

ⴢ

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5 The a b ⫽ 10 combinations without repetition are obtained from the previous list 2 by regarding the two pairs consisting of the same two letters (in opposite orders) 5⫹2⫺1 6 as equal. The a b ⫽ a b ⫽ 15 combinations with repetitions consist of 2 2 the 10 combinations just mentioned plus the 5 combinations aa

uu.

4. 2!3!5!>10! ⬇ 0.04% (by Theorem 1) 50 6. a b ⫽ 230,300 samples, by Theorem 3 4 52 8. a b ⫽ 635,013,559,600 13 100 10. There are a b samples of 10 from 100; hence the probability of picking a particular 10 100 97 one is 1^ a b . Now the number of samples containing the 3 male mice is a b 10 7 because these are obtained by picking the 3 male mice and then 7 female mice from 97 97, which can be done in a b ways. Hence the answer is 7 a

97 100 2 ⫽ 0.074%. b^a b⫽ 2695 7 10

12. This concerns combinations. We obtain 9 1 3 2 1 䡠䡠 ⫽ ⫽ 1^a b , 9 8 7 84 3 which is very small, about 1%, and 9 6 5 4 5 6 䡠䡠 ⫽ ⫽ a b^a b 9 8 7 21 3 3 which is about 25%. 14. Team Project. (a) There are n choices for the first thing and we terminate with the kth thing, for which we have n ⫺ k ⫹ 1 choices. (b) The theorem holds when k ⫽ 1. Assuming that it holds for any fixed positive k, n⫹k b . From the we show that the number of combinations of (k ⫹ 1)th order is a k⫹1 n⫹k⫺1 b combinations of (k ⫹ 1)th order assumption it follows that there are a k of n elements whose first element is 1 (this is the number of combinations of kth order n⫹k⫺2 b combinations of (k ⫹ 1)th of n elements with repetitions). Then there are a k order whose first element is 2 (this is the number of combinations of kth order of the n ⫺ 1 elements 2, 3, Á , n, the combinations no longer containing 1 because the n⫹k⫺3 b combinations containing 1 have just been taken care of). Then there are a k combinations of (k ⫹ 1)th order of the n ⫺ 2 elements 3, 4, Á , n whose first element is 3, etc., and, by (13),

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nⴚ1 n⫹k⫺1 n⫹k⫺2 k k⫹s n⫹k b⫹a b⫹Á⫹a b⫽ a a b⫽a b. k k k k k⫹1 s⫽0

(d) a kb nⴚk is obtained by picking k of the n factors (a ⫹ b)(a ⫹ b) Á (a ⫹ b)

(n factors)

and choosing a from each of k factors (and b from the remaining n ⫺ k factors); by n Theorem 3, this can be done in a b ways. k (e) Apply the binomial theorem to (1 ⫹ b)p(1 ⫹ b)q ⫽ (1 ⫹ b)p⫹q. b r has the coefficient a

r p⫹q p q b on the right and a a b a b on the left. r k r⫺k k⫽0

SECTION 24.5. Random Variables. Probability Distributions, page 1029 Purpose. To introduce the concepts of discrete and continuous random variables and their distributions (to be followed up by the most important special distributions in Secs. 24.7 and 24.8). Main Content, Important Concepts Random variable X, distribution function F(x) Discrete random variable, its probability function Continuous random variable, its density Comments on Content The definitions in this section are general, but the student should not be scared because the number of distributions one needs in practice is small, as we shall see. Discrete random variables occur in experiments in which we count, continuous random variables in experiments in which we measure. For both kinds of random variables X the definition of the distribution function F(x) is the same, namely, F(x) ⫽ P(X ⬉ x), so that it permits a uniform treatment of all X. For discrete X the function F(x) is piecewise constant; for continuous X it is continuous. For obtaining an impression of the distribution of X the probability function or density is more useful than F(x). SOLUTIONS TO PROBLEM SET 24.5, page 1034 2. Using (10), we have k

冮

x 2 dx ⫽

53k ⫽ 1, 3

k⫽

3 . 125

Hence F(x) ⫽ 0 (x ⬉ 0), F(x) ⫽ x 3>125 (0 ⬍ x ⬉ 5), F(x) ⫽ 1 (x ⬎ 5). Of course, the point of Probs. 1 and 2 is to make the student aware of the conceptual difference of the two kinds of probability distributions, roughly, discrete coming up if we count, and continuous if we measure.

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4. c ⫽ 1.9, 苲 c ⫽ 1.8 6. The point of this problem is to tell the student to be careful with respect to ⬍ and ⬉ as well as with ⬎ and ⭌, where the distinction does not matter in the continuous case but does matter for a discrete distribution. The answers are 1313 %, 5313 %, 3313 %, 0, 6623 %, 8623 %, 3313 %, 8623 %. 8. f (x) ⫽ F r(x) gives f (x) ⫽ 0 if x ⬍ 0, f (x) ⫽ 3eⴚ3x if x ⬎ 0. This illustrates that a density may have discontinuities, whereas the distribution function of a continuous distribution must be continuous. These first few problems should help the student to recognize the conceptual distinction between probability and density as well as the fact, that in many cases, the distribution function is given by different formulas over different intervals and that the same holds for the density. Furthermore, F(x) ⫽ 0.9 gives 0.1 ⫽ eⴚ3x; hence x ⫽ 13 ln 10 ⫽ 0.7675. 10. To have the area under the density curve equal to 1, we must have k ⫽ 5. If A denotes “Defective” and B ⫽ Ac, we have P(B) ⫽ 5

冮

120.09

dx ⫽ 0.9,

P(A) ⫽ 0.1,

119.91

so that about 50 of the 500 axles will be defective. This can also be seen without calculation. 12. By differentiation, f (x) ⫽ 0.4x

if 2 ⬍ x ⬍ 3,

f (x) ⫽ 0

otherwise.

Furthermore, P(2.5 ⬍ X ⬉ 5) ⫽ F(5) ⫺ F(2.5) ⫽ 1 ⫺ 0.45 ⫽ 0.55, that is, 55%. 14. The outcomes and their probabilities are (A: Six. B ⫽ Ac) P(A) ⫽ 16 P(BA) ⫽ 56 ⴢ 16 P(BBA) ⫽ 56 ⴢ 56 ⴢ 16 . etc. Hence the event X ⫽ x: First Six in rolling x times has the probability xⴚ1

1 5 f (x) ⫽ a b 6 6

x ⫽ 1, 2, Á .

We can now verify (6) by applying the sum formula of the geometric series: ⴥ

a f (x) ⫽ x⫽1

Here x ⫺ 1 ⫽ y.

xⴚ1

1 ⴥ 5 a a6b 6 x⫽1

⫺6 1 ⴥ 5 a b ⫽ ⫽ 1. a 6 y⫽0 6 1 ⫺ 56 y

⫽

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SECTION 24.6. Mean and Variance of a Distribution, page 1035 Purpose. To introduce the two most important parameters of a distribution, the mean ␮ of X (also called expectation of X), which measures the central location of the values of X, and the variance s2 of X, which measures the spread of those values. Main Content, Important Concepts Mean ␮ given by (1) Variance s2 given by (2), standard deviation s Standardized random variable (6) Short Courses. Mention definitions of mean and variance and go on to the special distributions in the next two sections. Comments on Content Important practical applications follow in Secs. 24.7, 24.8, and later. The transformation theorem (Theorem 2) will be basic in Sec. 24.8 and will have various applications in Chap. 25. Moments (8) and (9) will play no great role in our further work, but would be more important in a more theoretical approach on a higher level. We shall use them in Sec. 25.2. SOLUTIONS TO PROBLEM SET 24.6, page 1038 2. ␮ ⫽ 3.5, as follows by symmetry, without calculation. The variance is s2 ⫽ 16 (2 ⴢ 2.52 ⫹ 2 ⴢ 1.52 ⫹ 2 ⴢ 0.52) ⫽ 2.916667. 4. Mean 0 and variance 1, by Theorem 2 6. k ⫽ 34, to make the area under the density curve equal to 1; see Example 5 in Sec. 24.5. Furthermore ␮ ⫽ 0 because of symmetry. The variance is s2 ⫽

3 4

冮 x (1 ⫺ x ) dx ⫽ 15 . 2

ⴚ1

Note that the area of this symmetric distribution between ⫺s and s equals about 63%, as integration shows. (For the normal distribution in Sec. 24.8 the area under the density curve between ␮ ⫺ s and ␮ ⫹ s is not much different, namely 68%, although the density extends to ⫾⬁. 8. The probability function (12) j,

j ⫽ 1, 2, Á

satisfies (6) in Sec. 24.5: ⴥ

ⴚ1

1 j 1 ⴥ 1 j a a 2b ⫽ 2 a a2b j⫽1 j⫽1

⫽

1 ⴥ 1 k 1 1 ⫽1 a a2b ⫽ 2 2 k⫽0 1 ⫺ 12

Now, by again using the geometric series, ⴥ 1 j ax ⫽1⫺x

j⫽0

( j ⫺ 1 ⫽ k).

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we have ⴥ

a jx

jⴚ1

⫽

j⫽1

1 (1 ⫺ x)2

and when x ⫽ 12 , ⴥ

1 jⴚ1

⫽

a j(2) j⫽1

1 (1 ⫺ 12)2

⫽4

and by multiplying by 12 ⴥ 1 j ␮ ⫽ a j a b ⫽ 2. 2 j⫽1

That is, on the average, the first Head will appear in the second trail. 10. 20.8% because for nondefective bolt we obtain k

冮

1⫹0.06

0.06

f (x) dx ⫽ 750 ⴢ 2

1ⴚ0.06

冮 (0.1 ⫹ z)(0.1 ⫺ z) dz ⫽ 0.792 0

where x ⫺ 1 ⫽ z. 12. About 70 14.

1 2

冮 x dx ⫽ 13 2

⫺1

16. We are asking for the sale x such that F(x) ⫽ 0.95. Integration of f (x) gives F(x) ⫽ 3x 2 ⫺ 2x 3

if 0 ⬉ x ⬍ 1.

From this we get the solution 0.865, meaning that, with a probability of 95% , the sale will not exceed 8650 gallons (because we measure here in ten thousands of gallons). Thus P(X ⬉ 0.865) ⫽ 0.95 and the complementary event that the sale will exceed 8650 gallons thus has a 5% chance, P(X ⬎ 0.865) ⫽ 0.05 and then the tank will be empty if it has a capacity of 8650 gallons. 18. Writing k ⫽ 0.001 (and in fact for any positive k) we obtain, by integrating by parts, ⴥ

␮⫽k

冮 xe 0

ⴚkx

dx ⫽

1 . k

Hence in our case the answer is 1000 hours. 20. Team Project. (a) E(X ⫺ ␮) ⫽ E(X ) ⫺ ␮E(1) ⫽ ␮ ⫺ ␮ ⫽ 0. Furthermore, s2 ⫽ E([X ⫺ ␮]2) ⫽ E(X 2 ⫺ 2␮X ⫹ ␮2) ⫽ E(X 2) ⫺ 2␮E(X ) ⫹ ␮2E(1) where E(X) ⫽ ␮ and E(1) ⫽ 1, so that the result follows. The formula obtained has various practical and theoretical applications.

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(b) g(X) ⫽ X and the definition of expectation gives the defining formula for the mean. Similarly for (11). For E(1) we get the sum of all possible values or the integral of the density taken over the x-axis, and in both cases the value is 1 because of (6) and (10) in Sec. 24.5. (c) E(X k) ⫽ (b k⫹1 ⫺ a k⫹1)>[(b ⫺ a)(k ⫹ 1)] by straightforward integration. (d) Set x ⫺ ␮ ⫽ 1, write t instead of t, set t ⫽ ⫺t, and use f (␮ ⫺ t) ⫽ f (␮ ⫹ t). Then ⴥ

E([X ⫺ ␮] ) ⫽ 3

冮 t f (␮ ⫹ t) dt ⫽ 冮 t ( f (␮ ⫹ t) dt ⫹ 冮 t f (␮ ⫹ t) dt 3

ⴚⴥ

ⴥ

⫽

ⴥ

冮 (⫺t) f (␮ ⫺ t)(⫺dt) ⫹ 冮 t f (␮ ⫹ t) dt ⫽ 0. 3

ⴥ

(e) ␮ ⫽ 2, s ⫽ 2, g ⫽ 4>2 ⫽ 12 (g) f (1) ⫽ 12 , f (⫺4) ⫽ 13 , f (5) ⫽ 16 . But for distributions of interest in applications, the skewness will serve its purpose. 2

3>2

SECTION 24.7. Binomial, Poisson, and Hypergeometric Distributions, page 1039 Purpose. To introduce the three most important discrete distributions and to illustrate them by typical applications. Main Content, Important Concepts Binomial distribution (2)–(4) Poisson distribution (5), (6) Hypergeometric distribution (8)–(10) Short Courses. Discuss the binomial and hypergeometric distributions in terms of Examples 1 and 4. Comments on Content The “symmetric case” p ⫽ q ⫽ 12 of the binomial distribution with probability function (2*) is of particular interest. Formulas (3) and (4) will be needed from time to time. The approximation of the binomial distribution by the normal distribution will be discussed in the next section.

SOLUTIONS TO PROBLEM SET 24.7, page 1044 4. For drawing with replacement, we have 4 6 x 14 4ⴚx f (x) ⫽ a b a b a b 20 x 20 the values being 0.2401 and summing up to 1.0.

0.416

0.2646

0.0756

0.0081

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For drawing without replacement we have 6 14 20 f (x) ⫽ a b a b^a b x 4⫺x 4 with the values 0.2066

0.4508

0.2817

0.0578

0.0031.

The difference is (note the distribution of signs) 0.0335 ⫺0.0392

⫺0.0171

0.0178

0.0050.

6. f (x) ⫽ a

100 b 0.04x0.96100ⴚx ⬇ 4xeⴚ4>x!. Values 0.018, 0.073, 0.147, 0.195, 0.195, x 0.156. Sum 0.784. This leaves 21.6% for the remaining x-values. 8. Let X be the number of calls per minute. By assumption the average number of calls per minute is 300>60 ⫽ 5. Hence X has a Poisson distribution with mean ␮ ⫽ 5. By assumption, the board is overtaxed if X ⬎ 10. From Table A6. App. 5, we see that the complementary event X ⬉ 10 has probability P(X ⬉ 10) ⫽ 0.9863. Hence the answer is 1.4%. 10. p ⫽ 0.02, q ⫽ 0.98, hence 0.9815 ⫽ 74% 12. (a) ␮ ⫽ 6 defects per 300 m, f (x) ⫽ 6xeⴚ6>x!, (b) f (0) ⫽ eⴚ6 ⫽ 0.00248 ⬇ 0.25% 14. Let X be the number of customers per minute. The average number is 120>60 ⫽ 2 per minute. Hence X has a Poisson distribution with mean 2. Waiting occurs if X ⬎ 4. The probability of the complement is P(X ⬉ 4) ⫽ 0.9473 (see Table 6). Hence the answer is 1 ⫺ 0.9473 ⫽ 5 14 %. 16. Team Project. (a) In each differentiation we get a factor x j by the chain rule, so that G(k)(t) ⫽ a x kj etxjf (x j). j

If we now set t ⫽ 0, the exponential function becomes 1 and we are left with the definition of E(X k). Similarly for a continuous random variable. (d) By differentiation, G r(t) ⫽ n( pet ⫹ q)nⴚ1pet, G s(t) ⫽ n(n ⫺ 1)( pet ⫹ q)nⴚ2( pet)2 ⫹ n( pet ⫹ q)nⴚ1pet This gives, since p ⫹ q ⫽ 1, E(X 2) ⫽ G s(0) ⫽ n(n ⫺ 1)p 2 ⫹ np. From this we finally obtain the desired result, s2 ⫽ E(X 2) ⫺ ␮2 ⫽ n(n ⫺ 1)p 2 ⫹ np ⫺ n 2p 2 ⫽ npq. (e) G(t) gives G(0) ⫽ 1 and furthermore, G r(t) ⫽ eⴚ␮ exp [␮et] ␮et ⫽ ␮etG(t) G s(t) ⫽ ␮et[G(t) ⫹ G r (t)] E(X 2) ⫽ G s(0) ⫽ ␮ ⫹ ␮2 s2 ⫽ E(X 2) ⫺ ␮2 ⫽ ␮.

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(f) By definition. ␮ ⫽ a xf (x) ⫽

M N⫺M 1 xa b a b N a x n⫺x a b n

(summation over x from 0 to n). Now M xM(M ⫺ 1) Á (M ⫺ x ⫹ 1) xa b⫽ x! x M(M ⫺ 1) Á (M ⫺ x ⫹ 1) ⫽ (x ⫺ 1)! ⫽Ma

M⫺1 b. x⫺1

Thus ␮⫽

M M⫺1 N⫺M a b a b. N a x⫺1 n⫺x a b n

Now (14), Sec. 24.4, is p q p⫹q a a k b ar ⫺ kb ⫽ a r b (summation over k from 0 to r). With p ⫽ M ⫺ 1, k ⫽ x ⫺ 1, q ⫽ N ⫺ M, r ⫺ k ⫽ n ⫺ x we have p ⫹ q ⫽ N ⫺ 1, r ⫽ k ⫹ n ⫺ x ⫽ n ⫺ 1 and the formula gives ␮⫽

M N⫺1 M a b⫽n n. N n⫺1 a b n

18. Let X be the number of defective screws in a simple of size n. The process will be halted if X ⭌ 1. The manufacturer wants n to be such that P(X ⭌ 1) ⬇ 0.95 when p ⫽ 0.1, thus P(X ⫽ 0) ⫽ q n ⫽ 0.9n ⬇ 0.05, and n ln 0.9 ⬇ ln 0.05, n ⫽ 28.4. Answer: n ⫽ 28 or 29. It is perhaps worthwhile to point to the fact that the situation is typical; in the case of a discrete distribution, one will in general not be able to fulfill percentage requirements exactly. SECTION 24.8. Normal Distribution, page 1045 Purpose. To discuss Gauss’s normal distribution, the practically and theoretically most important distribution. Use of the normal tables in App. 5. Main Content, Important Concepts Normal distribution, its density (1) and distribution function (2) Distribution function £(z). Tables A7, A8 in App. 5 De Moivre–Laplace limit theorem Short Course. Emphasis on the use of Tables A7 and A8 in terms of some of the given examples and problems.

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Comments on Content Although the normal tables become superfluous when a CAS (Maple, Mathematica. etc.) is used, their discussion may be advisable for a better understanding of the distribution and its numerical values. Applications of the De Moivre–Laplace limit theorem follow in Chap. 25. Bernoulli’s law of large numbers is included in the problem set.

SOLUTIONS TO PROBLEM SET 24.8, page 1050 2. From Table A7 in Appendix 5 we get P(X ⬉ 112.5) ⫽ F(112.5) ⫽ £ a

112.5 ⫺ 105 b ⫽ £(1.5) ⫽ 0.9332 5

P(X ⬎ 100) ⫽ 1 ⫺ F(100) ⫽ 1 ⫺ £ a

100 ⫺ 105 b ⫽ 1 ⫺ £(⫺1) 5

⫽ 1 ⫺ (1 ⫺ £(1)) ⫽ 0.8413, P(110.5 ⬍ X ⬍ 111.25) ⫽ £ a

111.25 ⫺ 105 110.5 ⫺ 105 b ⫺ £a b 5 5

⫽ £(1.25) ⫺ £(1.1) ⫽ 0.8944 ⫺ 0.8643 ⫽ 0.0301. 4. We have P(X ⬉ c) ⫽ £ a

c ⫺ 3.6 b ⫽ 0.5, 0.1

Thus, from Table A8 in Appendix 5, c ⫺ 3.6 ⫽ 0, 0.1

c ⫽ 3.6,

This could be seen without calculation. Next, P(X ⬎ c) ⫽ 1 ⫺ £ a

c ⫺ 3.6 b ⫽ 0.1, 0.1

£a

c ⫺ 3.6 b ⫽ 0.9. 0.1

From the table, c ⫺ 3.6 ⫽ 1.282, 0.1

c ⫽ 3.7282.

Finally, P(⫺c ⬍ X ⫺ 3.6 ⬉ c) ⫽ £ a c ⫽ 3.291, 0.1

c ⫺c b ⫺ £ a b ⫽ 99.9%, 0.1 0.1 c ⫽ 0.3291.

6. Smaller. This should help the student in qualitative thinking and an understanding of standard deviation and variance.

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8. We get the maximum load c from the condition P(X ⬉ c) ⫽ £ a

c ⫺ 1500 b ⫽ 5%. 50

By Table A8 in Appendix 5. c ⫺ 1500 ⫽ ⫺1.645. 50

c ⫽ 1418 kg.

10. Applying the De Moivre–Laplace theorem. We get 10

P⫽ aa x⫽0

⬇ £a

1000 b 0.01x0.991000ⴚx x

10 ⫺ 10 ⫹ 0.5 0 ⫺ 10 ⫺ 0.5 b ⫺ £a b ⫽ 0.564. 19.9 19.9

The value of the binomial distribution in the first line is 0.583; the relative error of the approximation is about 314 %. 12. The complementary event has the probability 15,000 ⫺ 12,000 P(X ⬉ 15,000) ⫽ £ a b ⫽ £(1.5) ⫽ 0.9332. 2000 Hence the answer is 6.7%. 14. Team Project. (c) Let e denote the exponential function in (1). Then (s12pf ) s ⫽ a⫺

x⫺␮ s2

eb ⫽ a⫺

x⫺␮ 2 ⫹ a b b e ⫽ 0, (x ⫺ ␮)2 ⫽ s2, 2 2 s s 1

hence x ⫽ ␮ ⫾ s. (d) Proceeding as suggested, we obtain £ 2(ⴥ) ⫽

冮

ⴥ

冮

ⴥ

2 2 1 1 eⴚu >2 du eⴚv >2 dv 12p ⴚⴥ 12p ⴚⴥ

1 ⫽ 2p

ⴥ

冮冮 e

ⴚu2>2 ⴚv2>2

ⴚⴥ ⴚⴥ

1 du dv ⫽ 2p

ⴥ

冮冮e

ⴚr 2>2

r dr du.

The integral over u equals 2p, which cancels the factor in front, and the integral over r equals 1, which proves the desired result. (e) Writing b instead of s in (1) and using (x ⫺ ␮)>b ⫽ u and dx ⫽ bdu, we obtain s2 ⫽

ⴥ

1 1 x⫺␮ 2 (x ⫺ ␮)2 exp c ⫺ a b d dx 2 b 12p b ⴚⴥ

冮

ⴥ ⴥ b2 2 1 2 2 ⴚu2>2 ⫽ b u e b du ⫽ u 2eⴚu >2 du 12p b ⴚⴥ 12p ⴚⴥ

冮

⫽

ⴥ

冮 (⫺u)(⫺ue 12p

冮

ⴚu2>2

) du

ⴚⴥ

⫽ b2 c

ⴥ

冮

ⴥ

2 2 1 1 (⫺u)eⴚu >2 ` ⫹ eⴚu >2 du d ⫽ b2(0 ⫹ 1) ⫽ b2. 12p 12p ⴚⴥ ⴚⴥ

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(f) We have X P a ` n ⫺ p ` ⬍ Pb ⫽ P([p ⫺ P]n ⬍ X ⬍ [p ⫹ P]n) and apply (11) with a ⫽ ( p ⫺ P)n, b ⫽ ( p ⫹ P)n. Then, since np cancels, b ⫽ (Pn ⫹ 0.5)> 1npq,

a ⫽ ⫺b,

and a : ⫺ⴥ, b : ⴥ as n : ⴥ. Hence the above probability approaches £(ⴥ) ⫺ £(⫺ⴥ) ⫽ 1 ⫺ 0 ⫽ 1. (g) Set x* ⫽ c1x ⫹ c2. Then (x ⫺ ␮)>s ⫽ (x* ⫺ ␮*)>s* and F(x*) ⫽ P(X* ⬉ x*) ⫽ P(X ⬉ x) ⫽ £((x ⫺ ␮)>s) ⫽ £((x* ⫺ ␮*)>s*). SECTION 24.9. Distributions of Several Random Variables, page 1051 Purpose. To discuss distributions of two-dimensional random variables, with an extension to n-dimensional random variables near the end of the section. Main Content, Important Concepts Discrete two-dimensional random variables and distributions Continuous two-dimensional random variables and distributions Marginal distributions Independent random variables Addition of means and variances Short Courses. Omit this section. (Use the addition theorems for means and variances in Chap. 25 without proof.) Comments on Content The addition theorems (Theorems 1 and 3) resulting from the present material will be needed in Chap. 25; this is the main reason for the inclusion of this section. Note well that the addition theorem for variances holds for independent random variables only. In contrast, the addition of means is true without that condition. SOLUTIONS TO PROBLEM SET 24.9, page 1059 1 2. The answers are 0 and 32 . Since the density is constant in that triangle, these results can be seen from a sketch of the triangle and the regions determined by the inequalities x ⬎ 4, y ⬎ 4 and x ⬉ 1, y ⬉ 1, respectively, without any integrations. 1 4. We have to integrate f (x, y) ⫽ 32 over y from 0 to 8 ⫺ x, where this upper integration limit follows from x ⫹ y ⫽ 8. This gives the density of the desired marginal distribution in the form

f1(x) ⫽

冮

8ⴚx

1 1 1 dy ⫽ ⫺ x 32 4 32

if 0 ⬉ x ⬉ 8 and

0 otherwise.

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6. By Theorem 1 the mean is 10,000 䡠 10 ⫽ 100 kg. By Theorem 3, assuming independence (which is reasonable), we find the variance 10,000 䡠 (0.05)2 ⫽ 25, hence the standard deviation 5 grams. Note that the mean is multiplied by n ⫽ 10,000, whereas the standard deviation is multiplied only by 1n ⫽ 100. 8. From the given distributions we obtain f1(x) ⫽ 25

if 0.98 ⬍ x ⬍ 1.02 and

f2( y) ⫽ 25 if 1.00 ⬍ y ⬍ 1.04

and

0 otherwise, 0 otherwise.

A pin fits that hole if X ⬍ 1, and P(X ⬍ 1) ⫽ 50%. 10. No. Whereas for the mean it is not essential that the trials are not independent and one still obtains from ␮ ⫽ M>N (single trial) the result ␮ ⫽ nM>N (n trials) via Theorem 1, one cannot use Theorem 3 here; indeed, the variance s2 ⫽ M(N ⫺ M)>N 2 (single trial) does not lead to (10), Sec. 24.7. 12. By Theorem 1 the mean is 105 lb. By Theorem 3, assuming independence, we get the variance 0.04 ⫹ 0.25, hence the standard deviation 10.29 ⫽ 0.539 lb. 14. Independent. f1(x) ⫽ 2eⴚ2x if x ⬎ 0, f2( y) ⫽ 2eⴚ2y if y ⬎ 0, 1.83% 16. (X, Y ) takes a value in A, B, C, or D (see the figure) with probability F(b1, b2), a value in A or C with probability F(a1, b2), a value in C or D with probability F(b1, a2), a value in C with probability F(a1, a2), hence a value in B with probability given by the right side of (2). Y = b2 A

Y = a2

X = a1

X = b1

18. x 2 ⫹ y 2 ⫽ 1 implies y ⫽ ⫾21 ⫺ x 2, x ⫽ ⫾21 ⫺ y 2; this gives the limits of integration in the integrals for the marginal densities. k times the area p must equal 1, hence k ⫽ 1> p. The marginal distributions have the densities f1(x) ⫽

冮 f (x, y) dy ⫽ p 冮

21ⴚx2

冮

21ⴚy2

ⴚ21ⴚx2

2 dy ⫽ p 21 ⫺ x 2 if ⫺1 ⬉ x ⬉ 1

and f2(x) ⫽

1 f (x, y) dx ⫽ p

冮

ⴚ21ⴚy

2 dx ⫽ p 21 ⫺ y 2 if ⫺1 ⬉ y ⬉ 1.

20. In the continuous case, (18) is obtained from (17) by differentiation, and (17) is obtained from (18) by integration. In the discrete case the proof results from the following theorem. Two random variables X and Y are independent if and only if the events of the form a1 ⬍ X ⬉ b1 and a2 ⬍ Y ⬉ b2 are independent. This theorem can be proved as follows. From (2), Sec. 24.5, we have P(a1 ⬍ X ⬉ b1)P(a2 ⬍ Y ⬉ b2) ⫽ [F1(b1) ⫺ F1(a1)][F2(b2) ⫺ F2(a2)].

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In the case of independence of the variables X and Y we conclude from (17) that the expression on the right equals F(b1, b2) ⫺ F(a1, b2) ⫺ F(b1, a2) ⫹ F(a1, a2). Hence, by (2), P(a1 ⬍ X ⬉ b1)P(a2 ⬍ Y ⬉ b2) ⫽ P(a1 ⬍ X ⬉ b1, a2 ⬍ Y ⬉ b2). This means independence of a1 ⬍ X ⬉ b1 and a2 ⬍ Y ⬉ b2; see (14), Sec. 24.3. Conversely, suppose that the events are independent for any a1, b1, a2, b2. Then P(a1 ⬍ X ⬉ b1)P(a2 ⬍ Y ⬉ b2) ⫽ P(a1 ⬍ X ⬉ b1, a2 ⬍ Y ⬉ b2). Let a1 : ⫺ⴥ, a2 : ⫺ⴥ and set b1 ⫽ x, b2 ⫽ y. This yields (17), that is, X and Y are independent. SOLUTIONS TO CHAPTER. 24 REVIEW QUESTIONS AND PROBLEMS, page 1060 8. Mathematica, Maple, etc. will give you values of densities and distribution functions for given x (and conversely) directly, without first applying a transformation to the standardized normal distribution. 12. Q L ⫽ 13.2, Q M ⫽ 13.3, Q U ⫽ 13.5 14. x ⫽ 13.14, s ⫽ 0.602, s 2 ⫽ 0.363 16. HHH, HHT, HTH, THH, HTT, THT, TTH, TTT 18. Obviously, A 債 B implies A 傽 B ⫽ A. Conversely, if A 傽 B ⫽ A, then every element of A must also be in B, by the definition of intersection; hence A 債 B. 12 9 3 9 20. (a) a b ⫽ 220, (b) a b ⫽ 84, (c) a b a b ⫽ 108. 3 3 1 2 9 3 3 (d) a b a b ⫽ 27, (e) a b ⫽ 1. Note that the sum of (b) through (e) is 220. 1 2 3 22. k ⫽ 1.1565, 26.9% 24. 17.29, 10.71, 19.152.

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CHAPTER 25 Mathematical Statistics SECTION 25.1. Introduction. Random Sampling, page 1063 Purpose. To explain the role of (random) samples from populations. Main Content, Important Concepts Population Sample Random numbers, random number generator Sample mean: x; see (1) Sample variance s 2; see (2) Comments on Content Sample mean and sample variance are the two most important parameters of a sample. x measures the central location of the sample values and s 2 their spread (their variability). Small s 2 may indicate high quality of production, high accuracy of measurement, etc. Note well that x and s 2 will generally vary from sample to sample taken from the same population, whose mean ␮ and variance s2 are unique, of course. This is an important conceptual distinction that should be mentioned explicitly to the students.

SECTION 25.2. Point Estimation of Parameters, page 1065 Purpose. As a first statistical task we discuss methods for obtaining approximate values of unknown population parameters from samples; this is called estimation of parameters. Main Content, Important Concepts Point estimate, interval estimate Method of moments Maximum likelihood method

SOLUTIONS TO PROBLEM SET 25.2, page 1067 2. Since s2 is known and constant, 0 ln />0␮ ⫽ 0 in Example 1 gives ␮ˆ ⫽ x as before and 0 ln />0s ⫽ 0. Thus, Prob. 2 is a counterpart to Prob. 1; one of the parameters has disappeared, and for the other we obtain the same estimate as before. 4. / ⫽ 1(b ⫺ a)n is maximum if b ⫺ a is as small as possible, that is, a equal to the smallest sample value and b equal to the largest. 6. The likelihood function is (we can drop the binomial factors) / ⫽ p k1 (1 ⫺ p)nⴚk1 Á p km (1 ⫺ p)nⴚkm ⫽ p k1 ⫹

Á ⫹k

(1 ⫺ p)nmⴚ(k1 ⫹

Á ⫹k ) m

. 405

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The logarithm is ln / ⫽ (k 1 ⫹ Á ⫹ k m) ln p ⫹ [nm ⫺ (k 1 ⫹ Á ⫹ k m)] ln (1 ⫺ p). Equating the derivative with respect to p to zero, we get, with a factor ⫺1 from the chain rule, 1 (k 1 ⫹ Á ⫹ k m) p ⫽ [nm ⫺ (k 1 ⫹ Á ⫹ k m)]

1 . 1⫺p

Multiplication by p (1 ⫺ p) gives (k 1 ⫹ Á ⫹ k m)(1 ⫺ p) ⫽ [nm ⫺ (k 1 ⫹ Á ⫹ k m)]p. By simplification, 8. We obtain / ⫽ f (x 1) f (x 2) Á f (x n) ⫽ p (1 ⫺ p)x1 ⴚ1p (1 ⫺ p)x2 ⴚ1 Á p (1 ⫺ p)xn ⴚ1 ⫽ p n (1 ⫺ p)x1 ⫹

Á ⫹x ⴚn n

The logarithm is n

ln / ⫽ n ln p ⫹ ¢ a x m ⫺ n≤ ln (1 ⫺ p). m⫽1

Differentiation with respect to p gives, with a factor ⫺1 from the chain rule, n

0 ln / n x ⫺n ⫽p ⫺ a m . 0p 1⫺p m⫽1 Equating this derivative to zero gives n

n (1 ⫺ p̂) ⫽ p̂ ¢ a x m ⫺ n≤ . m⫽1

Thus p̂ ⫽ 1>x. 2 , by Prob. 8. 10. p̂ ⫽ 2>(7 ⫹ 6) ⫽ 13 12. ␮ ⫽ 1>u, ␮ˆ ⫽ x 14. û ⫽ 1>x ⫽ 2. F (x) ⫽ 1 ⫺ eⴚ2x if x ⭌ 0 and 0 otherwise. A graph shows that the step 苲 function F (x) (the sample distribution function) approximates F (x) reasonably well. (For goodness of fit, see Sec. 25.7.) SECTION 25.3. Confidence Intervals, page 1068 Purpose. To obtain interval estimates (“confidence intervals”) for unknown population parameters for the normal distribution and other distributions. Main Content, Important Concepts Confidence interval for ␮ if s2 is known Confidence interval for ␮ if s2 is unknown

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t-distribution, its occurrence (Theorem 2) Confidence interval for s2 Chi-square distribution, its occurrence (Theorem 3) Distribution of a sum of independent normal random variables Central limit theorem Comments on Content The present methods are designed for the normal distribution, but the central limit theorem permits their extension to other distributions, provided we can use sufficiently large samples. The theorems giving the theory underlying the present methods also serve as the theoretical basis of tests in the next section. Hence these theorems are of basic importance. We see that, although our task is the development of methods for the normal distribution, other distributions (t and chi-square) appear in the mathematical foundation of those methods. SOLUTIONS TO PROBLEM SET 25.3, page 1077 2. CONF0.95 {46.47 ⬉ ␮ ⬉ 52.87}; longer 2.576>1.960; cf. Table 25.1 in Sec. 25.3. 4. CONF0.95 {74.26 ⬉ ␮ ⬉ 75.36} 6. Cf. Example 2. n ⫽ 166. The ordinate in Fig. 526 is L>s ⫽ 2>5 ⫽ 0.4, and the figure gives about n ⫽ 165, actually slighly more. The point of the problem is to let the student reflect on the dependence of the confidence interval on the sample size and the standard deviation, to arrive at an intuitive understanding of the role of the variance. 8. n ⫽ 24,000, x ⫽ 12,012, p̂ ⫽ x>n ⫽ 0.5005. Now the random variable X ⫽ Number of heads in 24,000 trials is approximately normal with mean 24,000 p and variance 24,000 p (1 ⫺ p). Estimators are 24,000 p̂ ⫽ 12,012

and

24,000 p̂ (1 ⫺ p̂) ⫽ 5999.99.

For the standardized normal random variable we get, from Table A8 in App. 5 (or from a CAS) and £(c) ⫽ 0.995, the value c ⫽ 2.576 ⫽

c* ⫺ 12,012 26000

and c* ⫺ 12,012 ⫽ 2.57626000 ⫽ 199.5 so that CONF0.99 {11,813 ⬉ ␮ ⬉ 12,212} and by division by n, CONF0.99 {0.492 ⬉ p ⬉ 0.509}. 10. n ⫺ 1 ⫽ 4; F (c) ⫽ 0.995 gives c ⫽ 4.60. From the sample we compute x ⫽ 661.2,

s 2 ⫽ 22.70.

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Hence k ⫽ 9.8 in Table 25.2. Step 4. This gives the confidence interval CONF0.99 {651.4 ⬉ ␮ ⬉ 671.0}. 14. n ⫺ 1 ⫽ 7 degrees of freedom, F (c1) ⫽ 0.025, c1 ⫽ 1.69, F (c2) ⫽ 0.975, c2 ⫽ 16.01 from Table A10. From the sample, x ⫽ 17.7625,

(n ⫺ 1)s 2 ⫽ 7s 2 ⫽ 0.73875.

Hence k 1 ⫽ 0.437, k 2 ⫽ 0.046. The answer is CONF0.95 {0.046 ⬉ s2 ⬉ 0.437}. 16. n ⫺ 1 ⫽ 9 degrees of freedom, F (c1) ⫽ 0.025, c1 ⫽ 2.70, F (c2) ⫽ 0.975, c2 ⫽ 19.02 from Table A10. From the sample, x ⫽ 253.5,

9s 2 ⫽ 54.5.

Hence k 1 ⫽ 54.5>2.70 ⫽ 20.19, k 2 ⫽ 54.5>19.02 ⫽ 2.87. From Table 25.3 we thus obtain the (rather long!) confidence interval CONF0.95 {2.9 ⬉ s2 ⬉ 20.2}. 18. A normal distribution with mean 3 ⴢ 14 ⫺ 8 ⫽ 34 and variance 9 ⴢ 2 ⫹ 5 ⫽ 23 20. By Theorem 1, the load Z is normal with mean 40N and variance 4N, where N is the number of bags. Now P (Z ⬉ 2000) ⫽ £ a

2000 ⫺ 40N 22N

b ⫽ 0.95

gives the condition 2000 ⫺ 40N

⭌ 1.645 22N by Table A8. The answer is N ⫽ 49 (since N must be an integer). SECTION 25.4. Testing of Hypotheses. Decisions, page 1077 Purpose. Our third big task is testing of hypotheses. This section contains the basic ideas and the corresponding mathematical formalism. Applications to further tasks of testing follow in Secs. 25.5–25.7. Main Content, Important Concepts Hypothesis (null hypothesis) Alternative (alternative hypothesis), one- and two-sided Type I error (probability a ⫽ significance level) Type II error (probability b; 1 ⫺ b ⫽ power of a test) Test for ␮ with known s2 (Example 2) Test for ␮ with unknown s2 (Example 3) Test for s2 (Example 4) Comparison of means (Example 5) Comparison of variances (Example 6)

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Comment on Content Special testing procedures based on the present ideas have been developed for controlling the quality of production processes (Sec. 25.5), for assessing the quality of produced goods (Sec. 25.6), for determining whether some function F (x) is the unknown distribution function of some population (Sec. 25.7), and for situations in which the distribution of a population need not be known in order to perform a test (Sec. 25.8). SOLUTIONS TO PROBLEM SET 25.4, page 1086 4. If the hypothesis p ⫽ 0.5 is true, X ⫽ Number of heads in 4040 trials is approximately normal with ␮ ⫽ 2020, s2 ⫽ 1010 (Sec. 24.8). Hence P (X ⬉ c) ⫽ £ ([c ⫺ 2020]> 21010) ⫽ 0.95, c ⫽ 2072 ⬎ 2048, do not reject the hypothesis. 9 6. Left-sided test, s2>n ⫽ 20 ⫽ 0.45. From Table A8 in App. 5 we obtain P (X ⬉ c)␮⫽60.0 ⫽ £ a

c ⫺ 60.0 20.45

b ⫽ 0.05.

Hence c ⫽ 60.0 ⫺ 1.64520.45 ⫽ 58.9 ⬎ x and we reject the hypothesis. 8. We obtain h (57.0) ⫽ P (X ⬉ c)␮⫽57 ⫽ £ a

58.9 ⫺ 57.0 20.45

b ⫽ £ (2.83) ⫽ 99.77%.

12. Hypothesis ␮0 ⫽ 35,000, alternative ␮ ⬎ 35,000. Using the given data and Table A9, we obtain t ⫽ (37,000 ⫺ 35,000)>(5000> 225) ⫽ 2.00 ⬎ c ⫽ 1.71. Hence we reject the hypothesis and assert that the manufacturer’s claim is justified. 14. Hypothesis H0: not better. Alternative H1: better. Under H0 the random variable X ⫽ Number of cases cured in 400 cases is approximately normal with mean ␮ ⫽ np ⫽ 300 and variance s2 ⫽ npq ⫽ 75. From Table A8 and a ⫽ 5% we get (c ⫺ 300)> 275 ⫽ 1.645,

c ⫽ 300 ⫹ 1.645275 ⫽ 314.

Since the observed value 310 is not greater than c, we do not reject the hypothesis. This indicates that the results obtained so far do not establish the superiority. 16. We test the hypothesis s20 ⫽ 25 against the alternative that s2 ⬍ 25. As in Example 4, we now get Y ⫽ (n ⫺ 1)S 2>s02 ⫽ 27S 2>25 ⫽ 1.08S 2. From Table A10 with 27 degrees of freedom and the condition P (Y ⬍ c) ⫽ a ⫽ 5%

we get

c ⫽ 16.2.

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Since y ⫽ 1.08s 2 ⫽ 1.08 ⴢ 3.52 ⫽ 13.23 ⬍ c and the test is left-sided, we reject the hypothesis and assert that it will be less expensive to replace all the batteries simultaneously. 18. We test the hypothesis s12 ⫽ s22 against the alternative s12 ⬎ s22. We proceed as in Example 6. By computation, v0 ⫽ s12>s22 ⫽ 350>61.9 ⫽ 5.65. For a ⫽ 5% and (5, 6) degrees of freedom. Table A11 gives 4.39. Since 5.65 is greater, we reject the hypothesis and assert that the variance of the first population is greater than that of the second. 20. In this two-sided test, we use (11), obtaining t0 ⫽

12 ⴢ 18 ⴢ 28 #

10 ⫺ 14

⫽ ⫺3.58. 211 ⴢ 9 ⫹ 17 ⴢ 9 From the t-table with n 1 ⫹ n 2 ⫺ 2 ⫽ 28 degrees of freedom we obtain c2 ⫽ 2.05 corresponding to 97 12 % and ⫺2.05 for 2 12 %, by symmetry. Since ⫺3.58 ⬍ ⫺2.05, we reject the hypothesis and assert that the population means are different. B

SECTION 25.5. Quality Control, page 1091 Purpose. Quality control is a testing procedure performed every hour (or every half hour, etc.) in an ongoing process of production in order to see whether the process is running properly (“is under control,” is producing items satisfying the specifications) or not (“is out of control”), in which case the process is halted in order to search for the trouble and remove it. These tests may concern the mean, variance, range, etc. Main Content Control chart for the mean Control chart for the variance Comment on Content Control charts have also been developed for the range, the number of defectives, the number of defects per unit, for attributes, etc. (see the problem set). SOLUTIONS TO PROBLEM SET 25.5, page 1091 2. 1 ⫾ 3 ⴢ 0.02> 14 ⫽ 1 ⫾ 0.03 4. Decrease by a factor 12 ⫽ 1.41. By a factor 2.58>1.96 ⫽ 1.32. Hence the two operations have almost the same effect. 6. LCL ⫽ 3.5, UCL ⫽ 6.5 8. The sample range tends to increase with increasing n, whereas s remains unchanged. 10. The random variable Z ⫽ Number of defectives in a sample of size n has the variance npq. Hence X ⫽ Z>n has the variance s2 ⫽ npq>n 2 ⫽ 0.04 # 0.96>100 ⫽ 0.000384. This gives UCL ⫽ 0.04 ⫹ 3s ⫽ 0.0988. From the given values we see that the process is out of control. 14. LCL ⫽ n␮0 ⫺ 2.58s1n, UCL ⫽ n␮0 ⫹ 2.58s1n, as follows from Theorem 1 in Sec. 25.3.

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SECTION 25.6. Acceptance Sampling, page 1092 Purpose. This is a test for the quality of a produced lot designed to meet the interests of both the producer and the consumer of the lot, as expressed in the terms listed below. Main Content, Important Concepts Sampling plan, acceptance number, fraction defective Operating characteristic curve (OC curve) Acceptable quality level (AQL) Rejectable quality level (RQL) Rectification Average outgoing quality limit (AOQL) Comments on Content Basically, acceptance sampling first leads to the hypergeometric distribution, which, however, can be approximated by the simpler Poisson distribution and simple formulas resulting from it, or in other cases by the binomial distribution, which can in turn be approximated by the normal distribution. Typical cases are included in the problem set. SOLUTIONS TO PROBLEM SET 25.6, page 1095 2. We expect a decrease of values because of the exponential function in (3), which involves n. The probabilities are 0.9098 (down from 0.9825), 0.7358, 0.0404. 4. P (A; u) ⫽ eⴚ20u (1 ⫹ 20u) from (3). From Fig. 540 we find a and b. For u ⫽ 2% we obtain P (A; 0.02) ⫽ 93.8%, hence a ⫽ 6.2%. Also b ⫽ P (A; 0.15) ⫽ 20%, which is very poor. 1

0.8

0.6

0.4

0.2

0.1

0.2

Section 25.6. Problem 4

6. AOQ(u) ⫽ ueⴚ25u (1 ⫹ 25u), AOQ r ⫽ 0, umax ⫽ 0.0647, AOQ(umax ) ⫽ 0.0336 8. From the definition of the hypergeometric distribution we now obtain P (A; u) ⫽ a

20u 0

20 ⫺ 20u 3

b^a

20 3

b⫽

(20 ⫺ 20u)(19 ⫺ 20u)(18 ⫺ 20u) 6840

This gives P (A; 0.1) ⫽ 0.72 (instead of 0.81 in Example 1) and P (A; 0.2) ⫽ 0.49 (instead of 0.63), a decrease in both cases, as had to be expected.

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10. The approximation is u0 (1 ⫺ u)2 and is fairly accurate, as the following values show: ␪

Exact (2D)

Approximate

0.0 0.2 0.4 0.6 0.8 1.0

1.00 0.63 0.35 0.15 0.03 0.00

1.00 0.64 0.36 0.16 0.04 0.00

12. For u ⫽ 0.05 we should get P (A; u) ⫽ 0.98. (Figure 540 illustrates this, for different values.) Since n ⫽ 100, we get np ⫽ 5 and the variance npq ⫽ 5 # 0.95 ⫽ 4.75. Using the normal approximation of the binomial distribution, we thus obtain, with c to be determined, c 100 c ⫺ 5 ⫹ 0.5 0 ⫺ 5 ⫺ 0.5 b 0.05x0.95100ⴚx ⬇ £ a b ⫺ £a b ⫽ 0.98. aa x 24.75 24.75 x⫽0 £a

c ⫺ 5 ⫹ 0.5

b ⫽ £a

24.75 From this and Table A8 we get (by interpolation)

⫺5.5 24.75

b ⫹ 0.98 ⫽ 0.9859.

c ⫽ 4.5 ⫹ 2.21424.75 ⫽ 9.325. The answer is that we should choose 9 or 10 as c. 14. P (A; u) ⫽ eⴚ20u (1 ⫹ 20u). [uP (A; u)] r ⫽ 0 gives u ⫽ u0 ⫽ 0.0809, u0P (A; u0) ⫽ 0.0420. SECTION 25.7. Goodness of Fit. X2-Test, page 1096 Purpose. The x2-test is a test for a whole unknown distribution function, as opposed to the previous tests for unknown parameters in known types of distributions. Main Content Chi-square (x2) test Test of normality Comments on Content The present method includes many practical problems, some of which are illustrated in the problem set. Recall that the chi-square distribution also occurred in connection with confidence intervals and in our basic section on testing (Sec. 25.4). SOLUTIONS TO PROBLEM SET 25.7, page 1099 49 2. x02 ⫽ 49 20 ⫹ 60 ⫽ 3.27 ⬍ c ⫽ 3.84 (1 degree of freedom, s ⫽ 5%), which supports the claim. 4. The hypothesis that the coin is fair is accepted, in contrast with Prob. 3, because we now obtain

x02 ⫽

(4 ⫺ 5)2 5

⫹

(6 ⫺ 5)2 5

⫽ 0.4 ⬍ 3.84.

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If we suspect the opposite, we should try to confirm our suspicion by taking larger samples. The situation is typical. 6. x02 ⫽ 2.33 ⬍ c ⫽ 11.07 8. K ⫽ 2 classes (dull, sharp). Expected values 10 dull, 390 sharp; 1 degree of freedom; hence 49 x02 ⫽ 49 10 ⫹ 390 ⫽ 5.03 ⬎ 3.84.

Reject the claim. Here, two things are interesting. First, 16 dull bits (an excess of 60% over the expected value!) would not have been sufficient to reject the claim at 49 the 5% level. Second, 49>10 contributes much more to x02 than 390 does; in other applications the situation will often be qualitatively similar. 10. Team Project. n ⫽ 3 ⴢ 77 ⫽ 231. 2 (a) aj ⫽ 231 20 ⫽ 11.55, K ⫽ 20. ␹0 ⫽ 24.32 ⬍ c ⫽ 30.14 (a ⫽ 5%, 19 degrees of freedom). Accept the hypothesis. (b) x 02 ⫽ 13.10 ⬎ c ⫽ 3.84 (a ⫽ 5%, 1 degree of freedom). Reject the hypothesis. (c) x 02 ⫽ 14.7 ⬎ c ⫽ 3.84 (a ⫽ 5%, 1 degree of freedom). Reject the hypothesis. 12. The maximum likelihood estimates for the two parameters are x ⫽ 59.87, ŝ ⫽ 1.504. K ⫺ 1 ⫺ 2 ⫽ 2 degrees of freedom. From Table A10 we get the critical value 9.21 ⬎ ␹02 ⫽ 6.22. Accept the hypothesis that the population from which the sample was taken is normally distributed. ␹20 is obtained as follows. x 58.5 59.5 60.5 61.5 ⬁

x⫺x s

£a

⫺0.91 ⫺0.25 0.42 1.08

x⫺x s b

Expected

Observed

Terms in (1)

14.31 17.51 20.50 15.68 11.00

14 17 27 8 13

0.01 0.01 2.06 3.78 0.36

0.181 0.402 0.662 0.860 1.000

␹02 ⫽ 6.22 Slightly different results due to rounding are possible. 14. ␮ˆ ⫽ x ⫽ 23 50 ⫽ 0.46. 1 degree of freedom, since we estimated ␮. We thus obtain ␹02 ⫽

(33 ⫺ 31.56)2 31.56

⫹

(11 ⫺ 14.52)2 14.52

⫹

(6 ⫺ 3.34)2 3.34

⫽ 3.04 ⬍ 3.84.

Hence we accept the hypothesis. SECTION 25.8. Nonparametric Tests, page 1100 Purpose. To introduce the student to the ideas of nonparametric tests in terms of two typical examples selected from a wide variety of tests in that field. Main Content Median, a test for it Trend, a test for it Comment on Content Both tasks have not yet been considered in the previous sections. Another approach to trend follows in the next section.

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SOLUTIONS TO PROBLEM SET 25.8, page 1102

2. 12 6 ⫹ 6 ⴢ A 12 B 6 ⫹ 15 ⴢ A 12 B 6 ⫽ 34% is the probability of at most 2 negative values if ␮苲 ⫽ 0. which we do not reject. 4. Under the hypothesis that no adjustment is needed, shorter and longer pipes are equally probable. We drop the 4 pipes of exact length from the sample. Then the probability that under the hypothesis one gets 3 or fewer longer pipes among 18 pipes is, since np ⫽ 9 and s2 ⫽ npq ⫽ 4.5, 1

3⫺9⫹2 0⫺9⫺2 18 1 18 b a b ⬇ £a b ⫺ £a b 2 x 24.5 24.5 x⫽0 3

⫽ £ (⫺2.6) ⫺ £ (⫺4.5) ⬇ 0.0047 and we reject the hypothesis and assert that the process needs adjustment. 6. Under the hypothesis the probability of obtaining at most 3 negative differences (80 ⫺ 85, 90 ⫺ 95, 60 ⫺ 75) is 15 15 15 1 15 a b c 1 ⫹ a b ⫹ a b ⫹ a b d ⫽ 1.76%. 2 1 2 3 We reject the hypothesis and assert that B is better. 8. We drop 0 from the sample. Let X ⫽ Number of positive values. Under the hypothesis we get the probability 9 1 9 P (X ⫽ 9) ⫽ a b a b ⫽ 0.2%. 9 2 Accordingly, we reject the hypothesis that there is no difference between A and B and assert that the observed difference is significant. 10. Let X ⫽ Number of higher values among 9 values. If the hypothesis is true, a value greater than 50°C is as probable as a value less than 50°C, and thus has probability 1 2 . Hence, under the hypothesis the probability of getting at most higher value is P ⫽ A 12 B 9 ⫹ 9 ⴢ A 12 B 9 ⫽ 2% ⬍ 11%.

Hence we reject the hypothesis and assert that the setting is too low. 12. n ⫽ 5 values, with 2 transpositions, namely, 101.1 before 100.4 and 100.8, so that from Table A12 we obtain P (T ⬉ 2) ⫽ 0.117 and we do not reject the hypothesis. 14. We order by increasing x. Then we have 10 transpositions: 418 ⬎ 301, 352, 395, 375, 388 395 ⬎ 375, 388 465 ⬎ 455 521 ⬎ 455, 490.

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Hypothesis no trend, alternative positive trend, P (T ⬉ 10) ⫽ 1.4% by Table A12 in App. 5. Reject the hypothesis. SECTION 25.9. Regression. Fitting Straight Lines. Correlation, page 1103 Purpose. This is a short introduction to regression analysis, restricted to linear regression, and to correlation analysis; the latter is presented without proofs. Main Content, Important Concepts Distinction between regression and correlation Gauss’s least squares method Sample regression line, sample regression coefficient Population regression coefficient, a confidence interval for it Sample covariance sxy Sample correlation coefficient r Population covariance sXY Population correlation coefficient r Independence of X and Y implies r ⫽ 0 (“uncorrelatedness”) Two-dimensional normal distribution If (X, Y ) is normal, r ⫽ 0 implies independence of X and Y. Test for r ⫽ 0 SOLUTIONS TO PROBLEM SET 25.9, page 1111 2. y ⫽ 2.745 ⫺ 0.454x 4. n ⫽ 9, Sx i ⫽ 183, x ⫽ 20.33, Syi ⫽ 440, y ⫽ 48.89. Sx iyi ⫽ 7701. (n ⫺ 1)s 2x ⫽ 8s 2x ⫽ 944. From (11) we thus obtain k1 ⫽

9 # 7701 ⫺ 183 # 440 ⫽ ⫺1.32. 9 # 944

This gives the answer y ⫺ 48.89 ⫽ ⫺1.32 (x ⫺ 20.33), hence y ⫽ 75.72 ⫺ 1.32x. 6. n ⫽ 5, x i ⫽ 17.8, x ⫽ 3.56, yi ⫽ 565, y ⫽ 113, Sx iyi ⫽ 2399, (n ⫺ 1)sx2 ⫽ 4sx2 ⫽ 11.252. From (11) we thus obtain k1 ⫽

5 # 2399 ⫺ 17.8 # 565 ⫽ 34.447. 5 # 11.252

This gives the answer y ⫺ 113 ⫽ 34.447 (x ⫺ 3.56), hence y ⫽ ⫺9.632 ⫹ 34.447x. 8. Equation (5) is y ⫺ 10 ⫽ 1.98 (x ⫺ 5); The spring modulus is 1>1.98.

thus

y ⫽ 0.1 ⫹ 1.98x.

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10. y ⫽ ⫺120.5 ⫹ 9.15x, y (35) ⫽ 199.75. The negative constant ⫺120.5 simply indicates that our linear interpolation by the least squares principle is meaningful only over a relatively short interval where we can approximate the actual function y (x) by a linear function. 12. c ⫽ 4.30 from Table A9 (t-distribution) with n ⫺ 2 ⫽ 2 degrees of freedom, corresponding to 9712 %. From the sample we compute n ⫽ 4,

x ⫽ 1.75

y ⫽ 1.95

and, by (11) (see also (7), (8), (9a)) k1 ⫽

S(x j ⫺ x)(yj ⫺ y) S(x j ⫺ x)2

⫽ ⫺0.4542,

as expected from the answer to Prob. 2. From (15), q0 ⫽ 0.1314

and

K ⫽ 0.2131.

Hence the confidence interval is CONF0.95 {⫺0.6673 ⬉ ␬1 ⬉ ⫺0.2411}. 14. n ⫽ 9, c ⫽ 2.36, x ⫽ 20.33, y ⫽ 48.89, k 1 ⫽ ⫺1.3197, q0 ⫽ 77.15, k ⫽ 0.2550, CONF0.95 {⫺1.57 ⬉ ␬1 ⬉ ⫺1.06}. SOLUTIONS TO CHAPTER 25 REVIEW QUESTIONS AND PROBLEMS, page 1111 14. x ⫽ 20.325, s 2 ⫽ 4.551, s ⫽ 2.133 16. k ⫽ 1.96 ⴢ 5> 2500 (see Table 25.1 in Sec. 25.3). CONF0.95 {21.56 ⬉ ␮ ⬉ 22.44} 18. n ⫺ 1 ⫽ 3 degrees of freedom, F (c1) ⫽ 0.025, c1 ⫽ 0.22, F (c2) ⫽ 0.975, c2 ⫽ 9.35 from Table A10 in Appendix 5; hence k 1 ⫽ 0.7>0.22 ⫽ 3.18, k 2 ⫽ 0.7>9.35 ⫽ 0.07 20. x ⫽ 376.3, y ⫽ 335.3, sx2 ⫽ 1009.3, sy2 ⫽ 869.3, t 0 ⫽ 1.64 ⬍ c ⫽ 2.13 (a ⫽ 5%, 4 degrees of freedom): do not reject the hypothesis. 22. Because the sample size n is finite. 24. We drop the two rods of exact length. Then we have a sample of 18. Under the hypothesis that no adjustment is needed longer rods and shorter rods have the same probability 12 . Hence the probability of getting three or fewer shorter rods is

A 12 B 18 (1 ⫹ 18 ⫹ 153 ⫹ 816) ⫽ 0.0038

and we reject the hypothesis and accept the alternative.