ADVANCED ENGINEERING MATHEMATICS SI EDITION 8TH EDITION BY PETER O NEIL, SOLUTIONS MANUAL by QuentinAxel

INSTRUCTOR'S SOLUTIONS MANUAL TO ACCOMPANY

ADVANCED ENGINEERNG MATHEMATICS 8th EDITION

PETER V. O’NEIL

Contents

First-Order Differential Equations 1.1 Terminology and Separable Equations 1.2 The Linear First-Order Equation 1.3 Exact Equations 1.4 Homogeneous, Bernoulli and Riccati Equations 2 Second-Order Differential Equations 2.1 The Linear Second-Order Equation 2.2 The Constant Coefficient Homogeneous Equation 2.3 Particular Solutions of the Nonhomogeneous Equation 2.4 The Euler Differential Equation 2.5 Series Solutions 3 The Laplace Transform 3.1 Definition and Notation 3.2 Solution of Initial Value Problems 3.3 The Heaviside Function and Shifting Theorems 3.4 Convolution 3.5 Impulses and the Dirac Delta Function 3.6 Systems of Linear Differential Equations iii

1 1 12 19 28 37 37 41 46 53 58 69 69 72 77 86 92 93

CONTENTS 4

Sturm-Liouville Problems and Eigenfunction Expansions 4.1 Eigenvalues and Eigenfunctions and Sturm-Liouville Problems 4.2 Eigenfunction Expansions 4.3 Fourier Series 5 The Heat Equation 5.1 Diffusion Problems on a Bounded Medium 5.2 The Heat Equation With a Forcing Term F (x, t) 5.3 The Heat Equation on the Real Line 5.4 The Heat Equation on a Half-Line 5.5 The Two-Dimensional Heat Equation 6 The Wave Equation 6.1 Wave Motion on a Bounded Interval 6.2 Wave Motion in an Unbounded Medium 6.3 d’Alembert’s Solution and Characteristics 6.4 The Wave Equation With a Forcing Term K(x, t) 6.5 The Wave Equation in Higher Dimensions 7 Laplace’s Equation 7.1 The Dirichlet Problem for a Rectangle 7.2 The Dirichlet Problem for a Disk 7.3 The Poisson Integral Formula 7.4 The Dirichlet Problem for Unbounded Regions 7.5 A Dirichlet Problem in 3 Dimensions 7.6 The Neumann Problem 7.7 Poisson’s Equation 8 Special Functions and Applications 8.1 Legendre Polynomials 8.2 Bessel Functions 8.3 Some Applications of Bessel Functions 9 Transform Methods of Solution 9.1 Laplace Transform Methods 9.2 Fourier Transform Methods 9.3 Fourier Sine and Cosine Transforms 10 Vectors and the Vector Space Rn 10.1 Vectors in the Plane and 3− Space 10.2 The Dot Product 10.3 The Cross Product 10.4 n− Vectors and the Algebraic Structure of Rn 10.5 Orthogonal Sets and Orthogonalization 10.6 Orthogonal Complements and Projections 11 Matrices, Determinants and Linear Systems 11.1 Matrices and Matrix Algebra 11.2. Row Operations and Reduced Matrices 11.3 Solution of Homogeneous Linear Systems 11.4 Nonhomogeneous Systems 11.5 Matrix Inverses 11.6 Determinants 11.7 Cramer’s Rule 11.8 The Matrix Tree Theorem

101 101 107 114 137 137 147 150 153 155 157 157 167 173 190 192 197 197 202 205 205 208 211 217 221 221 235 251 263 263 268 271 275 275 277 278 280 284 287 291 291 295 299 306 313 315 318 320

v 12

Eigenvalues, Diagonalization and Special Matrices 12.1 Eigenvalues and Eigenvectors 12.2 Diagonalization 12.3 Special Matrices and Their Eigenvalues and Eigenvectors 12.4 Quadratic Forms 13 Systems of Linear Differential Equations 13.1 Linear Systems 13.2 Solution of X0 = AX When A Is Constant 13.3 Exponential Matrix Solutions 13.4 Solution of X0 = AX + G for Constant A 13.5 Solution by Diagonalization 14 Nonlinear Systems and Qualitative Analysis 14.1 Nonlinear Systems and Phase Portraits 14.2 Critical Points and Stability 14.3 Almost Linear Systems 14.4 Linearization 15 Vector Differential Calculus 15.1 Vector Functions of One Variable 15.2 Velocity, Acceleration and Curvature 15.3 The Gradient Field 15.4 Divergence and Curl 15.5 Streamlines of a Vector Field 16 Vector Integral Calculus 16.1 Line Integrals 16.2 Green’s Theorem 16.3 Independence of Path and Potential Theory 16.4 Surface Integrals 16.5 Applications of Surface Integrals 16.6 Gauss’s Divergence Theorem 16.7 Stokes’s Theorem 17 Fourier Series 17.1 Fourier Series on [−L, L] 17.2 Sine and Cosine Series 17.3 Integration and Differentiation of Fourier Series 17.4 Properties of Fourier Coefficients 17.5 Phase Angle Form 17.6 Complex Fourier Series 17.7 Filtering of Signals

323 323 327 332 336 339 339 341 348 350 353 359 359 363 364 369 373 373 376 381 385 387 391 391 393 398 405 408 412 414 419 419 423 428 430 432 435 438

CONTENTS 18

Fourier Transforms 18.1 The Fourier Transform 18.2 Fourier sine and Cosine Transforms 19 Complex Numbers and Functions 19.1 Geometry and Arithmetic of Complex Numbers 19.2 Complex Functions 19.3 The Exponential and Trigonometric Functions 19.4 The Complex Logarithm 19.5 Powers 20 Complex Integration 20.1 The Integral of a Complex Function 20.2 Cauchy’s Theorem 20.3 Consequences of Cauchy’s Theorem 21 Series Representations of Functions 21.1 Power Series 21.2 The Laurent Expansion 22 Singularities and the Residue Theorem 22.1 Classification of Singularities 22.2 The Residue Theorem 22.3 Evaluation of Real Integrals 23 Conformal Mappings 23.1 The Idea of a Conformal Mapping 23.2 Construction of Conformal Mappings

441 441 448 451 451 455 461 467 468 473 473 477 479 485 485 492 497 497 499 505 515 515 533

Chapter 1

First-Order Differential Equations 1.1

Terminology and Separable Equations

1. The differential equation is separable because it can be written 3y 2

dy = 4x, dx

or, in differential form, 3y 2 dy = 4x dx. Integrate to obtain y 3 = 2x2 + k. This implicitly defines a general solution, which can be written explicitly as y = (2x2 + k)1/3 , with k an arbitrary constant. 2. Write the differential equation as x

dy = −y, dx

which separates as 1 1 dy = − dx y x if x 6= 0 and y 6= 0. Integrate to get ln |y| = − ln |x| + k. Then ln |xy| = k, so xy = c 1

CHAPTER 1. FIRST-ORDER DIFFERENTIAL EQUATIONS with c constant (c = ek ). y = 0 is a singular solution, satisfying the original differential equation. 3. If cos(y) 6= 0, the differential equation is y sin(x + y) = dx cos(y) sin(x) cos(y) + cos(x) sin(y) = cos(y) = sin(x) + cos(x) tan(y). There is no way to separate the variables in this equation, so the differential equation is not separable. 4. Write the differential equation as ex ey

dy = 3x, dx

which separates in differential form as ey dy = 3xe−x dx. Integrate to get ey = −3e−x (x + 1) + c, with c constant. This implicitly defines a general solution. 5. The differential equation can be written x or

dy = y 2 − y, dx

1 1 dy = dx, y(y − 1) x

and is therefore separable. Separating the variables assumes that y 6= 0 and y 6= 1. We can further write 1 1 1 − dy = dx. y−1 y x Integrate to obtain ln |y − 1| − ln |y| = ln |x| + k. Using properties of the logarithm, this is ln

y−1 = k. xy

1.1. TERMINOLOGY AND SEPARABLE EQUATIONS Then

y−1 = c, xy

with c = ek constant. Solve this for y to obtain the general solution y=

1 . 1 − cx

y = 0 and y = 1 are singular solutions because these satisfy the differential equation, but were excluded in the algebra of separating the variables. 6. The differential equation is not separable. 7. The equation is separable because it can be written in differential form as 1 sin(y) dy = dx. cos(y) x This assumes that x 6= 0 and cos(y) 6= 0. Integrate this equation to obtain − ln | cos(y)| = ln |x| + k. This implicitly defines a general solution. From this we can also write sec(y) = cx with c constant. The algebra of separating the variables required that cos(y) 6= 0. Now cos(y) = 0 if y = (2n+1)π/2, with n any integer. Now y = (2n+1)π/2 also satisfies the original differential equation, so these are singular solutions. 8. The differential equation itself requires that y 6= 0 and x 6= −1. Write the equation as 2y 2 + 1 x dy = y dx x and separate the variables to get 1 1 dy = dx. y(2y 2 + 1) x(x + 1) Use a partial fractions decomposition to write this as 1 2y 1 1 − dy = − dx. y 2y 2 + 1 x x+1 Integrate to obtain ln |y| −

1 ln(1 + 2y 2 ) = ln |x| − ln |x + 1| + c 2

CHAPTER 1. FIRST-ORDER DIFFERENTIAL EQUATIONS with c constant. This implicitly defines a general solution. We can go a step further by writing this equation as ! x y = ln ln p +c x+1 1 + 2y 2 and take the exponential of both sides to get x y p =k , x+1 1 + 2y 2 which also defines a general solution. 9. The differential equation is dy = ex − y + sin(y), dx and this is not separable. It is not possible to separate all terms involving x on one side of the equation and all terms involving y on the other. 10. Substitute sin(x − y) = sin(x) cos(y) − cos(x) sin(y), cos(x + y) = cos(x) cos(y) − sin(x) sin(y), and cos(2x) = cos2 (x) − sin2 (x) into the differential equation to get the separated differential form (cos(y) − sin(y)) dy = (cos(x) − sin(x)) dx. Integrate to obtain the implicitly defined general solution cos(y) + sin(y) = cos(x) + sin(x) + c. 11. If y 6= −1 and x 6= 0, we obtain the separated equation 1 y2 dy = dx. y+1 x To make the integration easier, write this as 1 1 y−1+ dy = dx. 1+y x Integrate to obtain 1 2 y − y + ln |1 + y| = ln |x| + c. 2

1.1. TERMINOLOGY AND SEPARABLE EQUATIONS

This implicitly defines a general solution. The initial condition is y(3e2 ) = 2, so put y = 2 and x = 3e2 to obtain 2 − 2 + ln(3) = ln(3e2 ) + c. Now ln(3e2 ) = ln(3) + ln(e2 ) = ln(3) + 2, so ln(3) = ln(3) + 2 + c. Then c = −2 and the solution of the initial value problem is implicitly defined by 1 2 y − y + ln |1 + y| = ln |x| − 2. 2 12. Integrate 1 dy = 3x2 dx, y+2 assuming that y 6= −2, to obtain ln |2 + y| = x3 + c. This implicitly defines a general solution. To have y(2) = 8, let x = 2 and y = 8 to obtain ln(10) = 8 + c. The solution of the initial value problem is implicitly defined by ln |2 + y| = x3 + ln(10) − 8. We can take this a step further and write 2+y ln = x3 − 8. 10 By taking the exponential of both sides of this equation we obtain the explicit solution 3 y = 10ex −8 − 2. 13. With ln(y x ) = x ln(y), we obtain the separated equation ln(y) dy = 3x dx. y Integrate to obtain (ln(y))2 = 3x2 + c. For y(2) = e3 , we need (ln(e3 ))2 = 3(4) + c,

CHAPTER 1. FIRST-ORDER DIFFERENTIAL EQUATIONS or 9 = 12 + c. Then c = −3 and the solution of the initial value problem is defined by (ln(y))2 = 3x2 − 3. Solve this to obtain the explicit solution √ 2 y = e 3(x −1) if |x| > 1. 2

14. Because ex−y = ex e−y , the variables can be separated to obtain 2

2yey dy = ex dx. Integrate to get 2

ey = ex + c. To satisfy y(4) = −2 we need e4 = e4 + c so c = 0 and the solution of the initial value problem is implicitly defined by 2 ey = ex , which reduces to the simpler equation x = y2 . √ Because y(4) = −2, the explicit solution is y = − x for x > 0. 15. Separate the variables to obtain y cos(3y) dy = 2x dx. Integrate to get 1 1 y sin(3y) + cos(3y) = x2 + c, 3 9 which implicitly defines a general solution. For y(2/3) = π/3, we need 1 4 1π sin(π) + cos(π) = + c. 33 9 9 This reduces to

1 4 = + c, 9 9 so c = −5/9 and the solution of the initial value problem is implicitly defined by 1 1 5 y sin(3y) + cos(3y) = x2 − , 3 9 9 or 3y sin(3y) + cos(3y) = 9x2 − 1. −

1.1. TERMINOLOGY AND SEPARABLE EQUATIONS

16. Let T (t) be the temperature function. By Newton’s law of cooling, T 0 (t) = k(T −60) for some constant k to be determined. This equation is separable and is easily solved to obtain: T (t) = 60 + 30ekt . To determine k, use the fact that T (10) = 88: T (10) = 60 + 30e10k = 88. Then e10k =

14 88 − 60 = , 30 15

1 ln(14/15). 10

Now we know the temperature function completely: T (t) = 60 + 30ekt = 60 + 30 e10k t/10 14 = 60 + 30 . 15

t/10

We want to know T (20), so compute T (20) = 60 + 30

14 15

2 ≈ 86.13

degrees Fahrenheit. To see how long it will take for the object to reach 65 degrees, solve for t in T (t) = 65 = 60 + 30

14 15

t/10 .

Then

14 15

t/10 =

65 − 60 1 = , 30 6

so t ln(14/15) = ln(1/6) = − ln(6). 10 The object reaches 65 degrees at time t=−

10 ln(6) ≈ 259.7 ln(14/15)

minutes.

CHAPTER 1. FIRST-ORDER DIFFERENTIAL EQUATIONS 17. Suppose the thermometer was removed from the house at time t = 0, and let T (t) be the temperature function. Let A be the ambient temperature outside the house (assumed constant). By Newton’s law, T 0 (t) = k(t − A). We are also given that T (0) = 70 and T (5) = 60. Further, fifteen minutes after being removed from the house, the thermometer reads 50.4, so T (15) = 50.4. We want to determine A, the constant outside temperature. From the differential equation for T , 1 dT = kdt. T −A Integrate this, as we have done before, to get T (t) = A + cekt . Now, T (0) = 70 = A + c, so c = 70 − A and T (t) = A + (70 − A)ekt . Now use the other two conditions: T (5) = A + (70 − A)e5k = 15.5 and T (15) = A + (70 − A)e15k = 50.4. From the equation for T (5), solve for e5k to get e5k =

60 − A . 70 − A

Then e15k = e5k

60 − A 70 − A

3 .

Substitute this into the equation T (15) to get 3 60 − A = 50.4 − A. (70 − A) 70 − A Then (60 − A)3 = (50.4 − A)(70 − A)2 . The cubic terms cancel and this reduces to the quadratic equation 10.4A2 − 1156A + 30960 = 0, with roots 45 and (approximately) 66.15385. Clearly the outside temperature must be less than 50, and must therefore equal 45 degree.

1.1. TERMINOLOGY AND SEPARABLE EQUATIONS

18. The amount A(t) of radioactive material at time t is modeled by A0 (t) = kA, A(0) = e3 together with the given half-life of the material, A(ln(2)) =

1 3 e . 2

Solve this (as in the text) to obtain A(t) = Then

t/ ln(2) 1 e3 . 2

3/ ln(2) 1 = 1 tonne. A(3) = e 2 3

19. The problem is like Problem 18, and we find that the amount of Uranium235 at time t is t/(4.5(109 )) 1 , U (t) = 10 2 with t in years. Then 1/4.5 1 ≈ 8.57 kg. U (10 ) = 10 2 9

20. At time t there will be A(t) = 12ekt grams, and A(4) = 12e4k = 9.1. Solve this for k to get 1 9.1 k = ln . 4 12 The half-life of this element is the time t∗ it will take for there to be 6 grams, so ∗ A(t∗ ) = 6 = 12eln(9.1/12)t /4 . Solve this to get t∗ =

4 ln(1/2) ≈ 10.02 minutes. ln(9.1/12)

21. Let

Z ∞ I(x) =

e−t −(x/t) dt.

The integral we want is I(3). Compute Z ∞ 1 −t2 −(x/t)2 I 0 (x) = −2x e dt. 2 t 0

CHAPTER 1. FIRST-ORDER DIFFERENTIAL EQUATIONS Let u = x/t, so t = x/u and x dt = − 2 du. u Then I 0 (x) = −2x

Z 0 2 2 2 −x u e−(x/u) −u 2 du 2 x u ∞

= −2I(x). Then I(x) satisfies the separable differential equation I 0 = −2I, with general solution of the form I(x) = ce−2x . Now observe that √ Z ∞ 2 π e−t dt = I(0) = = c, 2 0 in which we used a standard integral that arises often in statistics. Then √ π −2x e . I(x) = 2 Finally, put x = 3 for the particular integral of interest: √ Z ∞ 2 2 π −6 e . I(3) = e−t −(9/t) dt = 2 0 22. Begin with the logistic equation P 0 (t) = aP (t) − bP (t)2 , in which a and b are positive constants. Then dP = P (a − bP ) dt so

1 dP = dt P (a − bP )

and the variables are separated. To make the integration easier, write this equation as b 1 11 + dP = dt. aP a a − bP Integrate to obtain 1 b ln(P ) − ln(a − bP ) = t + c. a a if P (t) > − and a − bP (t) > 0. Using properties of the logarithm, we can write this equation as P ln = at + k, a − bP

1.1. TERMINOLOGY AND SEPARABLE EQUATIONS

in which k = ac is still constant. Then P = eat+k = ek eat = Keat , a − bP in which K = ek is a positive constant. Now suppose the initial population (say at time zero) is p0 . Then P (0) = p0 and p0 = K. a − bp0 We now have

P p0 = eat . a − bP a − bp0

It is a straightforward algebraic manipulation to solve this for P (t): P (t) =

ap0 eat . a − bp0 + bp0

This is the solution of the logistic equation with P (0) = p0 . Because a − bp0 > 0 by assumption, then bp0 eat < a − bp0 + bpeat , so P (t) <

a ap0 at e = . bp0 eat b

This means that this population function is bounded above. Further, by multiplying the numerator and denominator of P (t) by e(−at, we have ap0

lim P (t) = lim

t→∞

t→∞ (a − bp0 )e−at + bp0

= lim

ap0

t→∞ bp0

a . b

23. With a and b as given, and p0 = 3, 929, 214 (the population in 1790), the logistic population function for the United States is P (t) =

123, 141.5668 e0.03134t . 0.03071576577 + 0.0006242342283e0.03134t

If we attempt an exponential model Q(t) = Aekt , then take A = Q(0) = 3, 929, 214, the population in 1790. To find k, use the fact that Q(10) = 5308483 = 3929214e10k and we can solve for k to get 1 5308483 k= ln ≈ 0.03008667012. 10 3929214

CHAPTER 1. FIRST-ORDER DIFFERENTIAL EQUATIONS

year 1790 1800 1810 1820 1830 1840 1850 1860 1870 1880 1890 1900 1910 1920 1930 1940 1950 1960 1970 1980

population 3,929,213 5,308,483 7,239,881 9,638,453 12,886,020 17,169,453 23,191,876 31,443,321 38,558,371 50,189,209 62,979766 76,212,168 92,228,496 106,021,537 123,202,624 132,164,569 151,325,798 179,323,175 203,302,031 226,547,042

P (t) 3,929,214 5,336,313 7,228,171 9,757,448 13,110,174 17,507,365 23.193,639 30,414,301 39,374,437 50,180,383 62,772,907 76,873,907 91,976,297 107,398,941 122,401,360 136,329,577 148,679,224 150,231,097 167,943,428 174,940,040

percent error 0 0.52 -0.16 1.23 1.90 2.57 0.008 -3.27 2.12 -0.018 -0.33 0.87 -0.27 1.30 -0.65 3.15 -1.75 -11.2 -17.39 -22.78

Q(t) 3,929,214 5,308,483 7,179,158 7,179,158 13,000,754 17,685,992 23,894,292 32,281,888 43,613,774 58,923,484 79,073,491 107,551,857 145,303,703 196,312,254

percent error 0 0 -0.94 0.53 1.75 3.61 3.03 2.67 13.11 17.40 26.40 41.12 57.55 83.16

Table 1.1: Census data for Problem 23

The exponential model, using these two data points (1790 and 1800 populations), is Q(t) = 3929214e0.03008667012t . Table 1.1 uses Q(t) and P (t) to predict later populations from these two initial figures. The logistic model remains quite accurate until about 1960, at which time it loses accuracy quickly. The exponential model becomes quite inaccurate by 1870, after which the error becomes so large that it is not worth computing further. Exponential models do not work well over time with complex populations, such as fish in the ocean or countries throughout the world.

1.2

The Linear First-Order Equation

1. With p(x) = −3/x, and integrating factor is R

e (−3/x) dx = e−3 ln(x) = x−3 for x > 0. Multiply the differential equation by x−3 to get x−3 y 0 − 3x−4 = 2x−1 .

1.2. THE LINEAR FIRST-ORDER EQUATION

or d −3 2 (x y) = . dx x Integrate to get x−3 y = 2 ln(x) + c, with c an arbitrary constant. For x > 0 we have a general solution y = 2x3 ln(x) + cx3 . In the last integration, we can allow x < 0 by replacing ln(x) with ln |x| to derive the solution y = 2x3 ln |x| + cx3 for x 6= 0. R

2. e dx = ex is an integrating factor. Multiply the differential equation by ex to get 1 y 0 ex + yex = (e2x − 1). 2 Then (ex y)0 =

1 2x (e − 1) 2

and an integration gives us ex y =

1 2x 1 e − x + c. 4 2

Then y=

1 x 1 −x e − xe + ce−x 4 2

is a general solution, with c an arbitrary constant. R

3. e 2 dx = e2x is an integrating factor. Multiply the differential equation by e2x : y 0 e2x + 2ye2x = xe2x , or (e2x y)0 = xe2x . Integrate to get e2x y =

1 2x 1 2x xe − e + c. 2 4

giving us the general solution y=

1 1 x − + ce−2x . 2 4

CHAPTER 1. FIRST-ORDER DIFFERENTIAL EQUATIONS 4. For an integrating factor, compute R

e sec(x) dx = eln | sec(x)+tan(x)| = sec(x) + tan(x). Multiply the differential equation by this integrating factor: y 0 (sec(x) + tan(x)) + sec(x)(sec(x) + tan(x))y = y 0 (sec(x) + tan(x)) + (sec(x) tan(x) + sec2 (x))y = ((sec(x) + tan(x))y)0 = cos(x)(sec(x) + tan(x)) = 1 + sin(x). We therefore have ((sec(x) + tan(x))y)0 = 1 + sin(x). Integrate to get y(sec(x) + tan(x)) = x − cos(x) + c. Then y=

x − cos(x) + c . sec(x) + tan(x)

This is a general solution. If we wish, we can also observe that 1 cos(x) = sec(x) + tan(x) 1 + sin(x) to obtain

cos(x) y = (x − cos(x) + c) 1 + sin(x) 2 x cos(x) − cos (x) + c cos(x) = . 1 + sin(x) 5. First determine the integrating factor R

e −2 dx = e−2x . Multiply the differential equation by e−2x to get (e−2x y)0 = −8x2 e−2x . Integrate to get e−2x y =

−8x2 e−2x dx = 4x2 e−2x + 4xe−2x + 2e−2x + c.

This yields the general solution y = 4x2 + 4x + 2 + ce2x .

1.2. THE LINEAR FIRST-ORDER EQUATION

6. e 3 dx = e3x is an integrating factor. Multiply the differential equation by e3x to get (e3x y)0 = 5e5x − 6e3x . Integrate this equation: e3x y = e5x − 2e3x + c. Now we have a general solution y = e2x − 2 + ce−3x . We need y(0) = 2 = 1 − 2 + c, so c = 3. The unique solution of the initial value problem is y = e2x + 3e−3x − 2. 7. x − 2 is an integrating factor for the differential equation because R

e (1/(x−2)) dx = eln(x−2) = x − 2. Multiply the differential equation by x − 2 to get ((x − 2)y)0 = 3x(x − 2). Integrate to get (x − 2)y = x3 − 3x2 + c. This gives us the general solution y=

1 (x3 − 3x2 + c). x−2

Now we need y(3) = 27 − 27 + c = 4, so c = 4 and the solution of the initial value problem is y=

1 (x3 − 3x2 + 4). x−2

8. e (−1) dx = e−x is an integrating factor. Multiply the differential equation by e−x to get: (ye−x )0 = 2e3x . Integrate to get ye−x =

2 3x e + c, 3

CHAPTER 1. FIRST-ORDER DIFFERENTIAL EQUATIONS and we have the general solution y=

2 4x e + cex . 3

We need

2 + c = −3, 3 so c = −11/3 and the initial value problem has the solution y(0) =

2 4x 11 x e − e . 3 3

9. First derive the integrating factor 2

e (2/(x+1)) dx = e2 ln(x+1) = eln((x+1) ) = (x + 1)2 . Multiply the differential equation by (x + 1)2 to obtain 0 (x + 1)2 y = 3(x + 1)2 . Integrate to obtain (x + 1)2 y = (x + 1)3 + c. Then y =x+1+

c . (x + 1)2

Now y(0) = 1 + c = 5 so c = 4 and the initial value problem has the solution y =x+1+

4 . (x + 1)2

10. An integrating factor is 5/9

e (5/9x) dx = e(5/9) ln(x) = eln(x

)

= x5/9 .

Multiply the differential equation by x5/9 to get (yx5/9 )0 = 3x32/9 + x14/9 . Integrate to get yx5/9 = Then y=

27 41/9 9 x + x23/9 + c. 41 23

27 4 9 x + x2 + cx−5/9 . 41 23

Finally, we need 27 9 + − c = 4. 41 23 Then c = −2782/943, so the initial value problem has the solution y(−1) =

23 4 9 2782 −5/9 x + x2 − x . 41 23 943

1.2. THE LINEAR FIRST-ORDER EQUATION

11. Let (x, y) be a point on the curve. The tangent line at (x, y) must pass through (0, 2x2 ), and so has slope y0 =

y − 2x2 . x

This is the linear differential equation y0 −

1 y = −2x. x

An integrating factor is R

e− (1/x) dx = e− ln(x) = eln(1/x) =

1 , x

so multiply the differential equation by 1/x to get 1 0 1 y − 2 y = −2. x x This is

1 y x

0 = −2.

Integrate to get 1 y = −2x + c. x Then y = −2x2 + cx, in which c can be any number. 12. Let A(t) be the number of pounds of salt in the tank at time t ≥ 0. Then dA = rate salt is added − rate salt is removed dt A(t) =6−2 . 50 + t We must solve this subject to the initial condition A(0) = 25. The differential equation is 2 A0 + A = 6, 50 + t which is linear with integrating factor R

e 2/(50+t) dt = e2 ln(50+t) = (50 + t)2 . Multiply the differential equation by (50 + t)2 to get (50 + t)2 A0 + 2(50 + t)A = 6(50 + t)2 .

CHAPTER 1. FIRST-ORDER DIFFERENTIAL EQUATIONS This is

0 (50 + t)2 A = 6(50 + t)2 .

Integrate this equation to get (50 + t)2 A = 2(50 + t)3 + c, which we will write as A(t) = 2(50 + t) +

c . (50 + t)2

We need c so that

c = 25, 2500 so c = 187, 500. The number of pounds of salt in the tank at time t is A(0) = 100 +

A(t) = 2(50 + t) −

187, 500 . (50 + t)2

13. Let A1 (t) and A2 (t) be the number of pounds of salt in tanks 1 and 2, respectively, at time t. Then A01 (t) =

5 5A1 (t) − ; A1 (0) = 20 2 100

and

5A1 (t) 5A2 (t) − ; A2 (0) = 90. 100 150 Solve the linear initial value problem for A1 (t) to get A02 (t) =

A1 (t) = 50 − 30e−t/20 . Substitute this into the differential equation for A2 (t) to get A02 +

1 5 3 A2 = − e−t/20 ; A2 (0) = 90. 30 2 2

Solve this linear problem to obtain A2 (t) = 75 + 90e−t/20 − 75e−t/30 . Tank 2 has its minimum when A02 (t) = 0, and this occurs when 2.5e−t/30 − 4.5e−t/20 = 0. This occurs when et/60 = 9/5, or t = 60 ln(9/5). Then A2 (t)min = A2 (60 ln(9/5)) =

5450 81

pounds.

1.3. EXACT EQUATIONS

1.3

Exact Equations

In these problems it is assumed that the differential equation has the form M (x, y) + N (x, y)y 0 = 0, or, in differential form, M (x, y) dx + N (x, y) dy = 0. 1. With M (x, y) = 2y 2 + yexy and N (x, y) = 4xy + xexy + 2y. Then ∂N ∂M = 4y + exy + xyexy = ∂x ∂y for all (x, y), so the differential equation is exact on the entire plane. A potential function ϕ(x, y) must satisfy ∂ϕ = M (x, y) = 2y 2 + yexy ∂x and

∂ϕ = N (x, y) = 4xy + xexy + 2y. ∂y

Choose one to integrate. If we begin with ∂ϕ/∂x = M , then integrate with respect to x to get ϕ(x, y) = 2xy 2 + exy + α(y), with α(y) the “constant” of integration with respect to x. Then we must have ∂ϕ = 4xy + xexy + α0 (y) = 4xy + xexy + 2y. ∂y This requires that α0 (y) = 2y, so we can choose α(y) = y 2 to obtain the potential function ϕ(x, y) = 2xy 2 + exy + y 2 . The general solution is defined implicitly by the equation 2xy 2 + exy + y 2 = c, , with c an arbitrary constant. 2. ∂M/∂y = 4x = ∂N/∂x for all (x, y), so this equation is exact on the entire plane. For a potential function, we can begin by integrating ∂ϕ = 2x2 + 3y 2 ∂y to get ϕ(x, y) = 2x2 y + y 3 + c(x). Then

∂ϕ = 4xy + 2x = 4xy + c0 (x). ∂x

CHAPTER 1. FIRST-ORDER DIFFERENTIAL EQUATIONS Then c0 (x) = 2x so we can choose c(x) = x2 to obtain the potential function ϕ(x, y) = 2x2 y + y 3 + x2 . The general solution is defined implicitly by 2x2 y + y 3 + x2 = k, with k an arbitrary constant. 3. ∂M/∂y = 4x + 2x2 and ∂N/∂x = 4x, so this equation is not exact (on any rectangle). 4.

∂N ∂M = −2 sin(x + y) + 2x cos(x + y) = , ∂y ∂x for all (x, y), so this equation is exact on the entire plane. Integrate ∂ϕ/∂x = M or ∂ϕ/∂y = N to obtain the potential function ϕ(x, y) = 2x cos(x + y). The general solution is defined implicitly by 2x cos(x + y) = k with k an arbitrary constant.

5. ∂M/∂y = 1 = ∂N/∂x, for x 6= 0, so this equation is exact on the plane except at points (0, y). Integrate ∂ϕ/∂x = M or ∂ϕ/∂y = N to find the potential function ϕ(x, y) = ln |x| + xy + y 3 for x 6= 0. The general solution is defined by an equation ln |x| + xy + y 3 = k. 6. For the equation to be exact, we need ∂M ∂N = αxy α−1 = = −2xy α−1 . ∂y ∂x This will hold if α = −2. With this choice of α, the (exact) equation is 3x2 + xy −2 − x2 y −3 y 0 = 0. Routine integrations produce a potential function ϕ(x, y) = x3 +

x2 . 2y 2

The general solution is defined by the equation x3 +

x2 = k, 2y 2

for y 6= 0.

1.3. EXACT EQUATIONS

7. For this equation to be exact, we need ∂M ∂N = 6xy 2 − 3 = = −3 − 2αxy 2 . ∂y ∂x This will be true if α = −3. By integrating, we find a potential function ϕ(x, y) = x2 y 3 − 3xy − 3y 2 and a general solution is defined implicitly by x2 y 3 − 3xy − 3y 2 = k. 8. We have ∂M ∂N = 2 − 2y sec2 (xy 2 ) − 2xy 3 sec2 (xy 2 ) tan(xy 2 ) = , ∂y ∂x for all (x, y), so this equation is exact over the entire plane. By integrating ∂ϕ/∂x = 2y − y 2 sec2 (xy 2 ) with respect to x, we find that ϕ(x, y) = 2xy − tan(xy 2 ) + c(y). Then ∂ϕ = 2x − 2xy sec2 (xy 2 ) ∂y = 2x − 2xy sec2 (xy 2 ) + c0 (y). Then c0 (y) = 0 and we can choose c(y) = 0 to obtain the potential function ϕ(x, y) = 2xy − tan(xy 2 ). A general solution is defined implicitly by 2xy − tan(xy 2 ) = k. For the solution satisfying y(1) = 2, put x = 1 and y = 2 into this implicitly defined solution to get 4 − tan(4) = k. The solution of the initial value problem is defined implicitly by 2xy − tan(xy 2 ) = 4 − tan(4). 9. Because ∂M/∂y = 12y 2 = ∂N/∂x, this equation is exact for all (x, y). Straightforward integrations yield the potential function ϕ(x, y) = 3xy 4 − x.

CHAPTER 1. FIRST-ORDER DIFFERENTIAL EQUATIONS A general solution is defined implicitly by 3xy 4 − x = k. To satisfy the condition y(1) = 2, we must choose k so that 48 − 1 = k, so k = 47 and the solution of the initial value problem is specified by the equation 3xy 4 − x = 47. In this case we can actually write this solution explicitly with y in terms of x.

10. First, 1 1 y ∂M = ey/x − ey/x − 2 ey/x ∂y x x x ∂N y , = − 2 ey/x = x ∂x so the equation is exact for all (x, y) with x 6= 0. For a potential function, we can begin with ∂ϕ = ey/x ∂y and integrate with respect to y to get ϕ(x, y) = xey/x + c(x). Then we need ∂ϕ y y = 1 + ey/x − ey/x = ey/x − ey/x + c0 (x). ∂x x x This requires that c0 (x) = 1 and we can choose c(x) = x. Then ϕ(x, y) = xey/x + x. The general solution of the differential equation is implicitly defined by xey/x + x = k. To have y(1) = −5, we must choose k so that e−5 + 1 = k. The solution of the initial value problem is given by xey/x + x = 1 + e−5 . This can be solve for y to obtain the explicit solution 1 + e−5 y = x ln x+1 for x + 1 > 0.

1.3. EXACT EQUATIONS

11. First, ∂M ∂N = −2x sin(2y − x) − 2 cos(2y − x) = , ∂y ∂x so the differential equation is exact for all (x, y). For a potential function, integrate ∂ϕ = −2x cos(2y − x) ∂y with respect to y to get ϕ(x, y) = −x sin(2y − x) + c(x). Then we must have ∂ϕ = x cos(2y − x) − sin(2y − x) ∂x = x cos(2y − x) − sin(2y − x) + c0 (x). Then c0 (x) = 0 and we can take c(x) to be any constant. Choosing c(x) = 0 yields ϕ(x, y) = −x sin(2y − x). The general solution is defined implicitly by −x sin(2y − x) = k. To satisfy y(π/12) = π/8, we need −

π sin(π/6) = k, 12

so choose k = −π/24 to obtain the solution defined by −x sin(2y − x) = −

π 24

which of course is the same as x sin(2y − x) = We can also write y=

π . 24

π 1 x + arcsin 2 24x

for x 6= 0. 12.

∂M ∂N = ey = ∂y ∂x so the differential equation is exact. Integrate ∂ϕ = ey ∂x

CHAPTER 1. FIRST-ORDER DIFFERENTIAL EQUATIONS with respect to x to get ϕ(x, y) = xey + c(y). Then

∂ϕ = xey + c0 (y) = xey − 1, ∂y

so c0 (y) = −1 and we can let c(y) = −y. This gives us the potential function ϕ(x, y) = xey − y. The general solution is given by xey − y = k. For y(5) = 0 we need 5−0=k so k = 5 and the solution of the initial value problem is given by xey − y = 5. 13. ϕ + c is also a potential function if ϕ is because ∂(ϕ + c) ∂ϕ = ∂x ∂x and

∂ϕ ∂(ϕ + c) = . ∂y ∂y

The function defined implicitly by ϕ(x, y) = k is the same as that defined by ϕ(x, y) + c = k if k is arbitrary. 14. (a) ∂M ∂N = 1 and = −1 ∂y ∂x so this equation is not exact over any rectangle in the plane. (b) Multiply the differential equation by x−2 to obtain yx−2 − x−1 y 0 = 0. This is exact because

∂M ∗ ∂N ∗ = x−2 = . ∂y ∂x

1.3. EXACT EQUATIONS

This new equation has potential function ϕ(x, y) = −yx−1 and so has general solution defined implicitly by −

y = k. x

This also defines a general solution of the original differential equation. (c) Multiply the differential equation by y −2 to obtain y −1 − xy −2 y 0 = 0. This is exact on any region of the plane not containing y = 0, because ∂M ∗∗ ∂N ∗∗ = −y −2 = . ∂y ∂x The new equation has potential function ϕ(x, y) = xy −1 , so its general solution is defined implicitly by xy −1 = k. It is easy to check that this also defines a solution of the original differential equation. (d) Multiply the differential equation by xy −2 to obtain xy −2 − x2 y −3 y 0 = 0. This is exact (on any region not containing y = 0) because ∂M ∗∗∗ ∂N ∗∗∗ = −2xy −3 = . ∂y ∂x Integrate ∂ϕ/∂x = xy −2 with respect to x to obtain ϕ(x, y) = Then

1 2 −2 x y + c(y). 2

∂ϕ = −x−2 y −3 + c0 (y) = −x2 y −3 , ∂y

so c0 = 0 and we can choose c(y) = 0. Then ϕ(x, y) =

1 2 −2 x y 2

and we can define a general solution of this differential equation as x2 y −2 = k. Here we absorbed the factor of 1/2 into the arbitrary constant c. This again defines a solution of the original differential equation.

CHAPTER 1. FIRST-ORDER DIFFERENTIAL EQUATIONS (e) The original differential equation can be written as the linear equation y0 −

1 y = 0. x

This has integrating factor −1

e −(1/x) dx = e− ln(x) = eln(x

)

= x−1 .

Multiply the differential equation by x−1 to write this equation as (x−1 y)0 = 0, so x−1 y = c implicitly defines the general solution. (f) The methods of (b) through (e) yield the same general solution. For example, in (b) we obtained −yx−1 = c, which we can write as y = −cx. Because c is an arbitrary constant, this general solution can be written y = kx. And in (d) we obtained x2 y −2 = c, and this gives the same solutions as y 2 = cx2 , or y = kx. 15. First, ∂M 3 ∂N = x − y −5/2 and = 2x. ∂y 2 ∂x and these are not equal on any rectangle in the plane. In differential form, the differential equation is (xy + y −3/2 ) dx + x2 dy = 0. Multiply this equation by xa y b to get (xa+1 y b+1 + xa y b−3/2 ) dx + xa+2 y b dy = 0 = M ∗ dx + N ∗ dy. For this to be exact, we need ∂M ∗ 3 a+1 b = (b + 1)x y + (b − xa y b−5/2 ∂y 2 ∂N ∗ = = (a + 2)xa+1 y b . ∂x Divide this equation by xa y b to get 3 (b + 1)x + b − y −5/2 = (a + 2)x. 2 This will hold for all x and y if we let b = 3/2 and then choose a and b so that b + 1 = a + 2. Thus choose a=

1 3 and b = 2 2

1.3. EXACT EQUATIONS

to get the integrating factor µ(x, y) = x1/2 b3/2 . Multiply the original differential equation by this to get (x3/2 y 5/2 + x1/2 ) dx + x5/2 y 1/2 dy = 0. To find a potential function, integrate ∂ϕ = x5/2 y 3/2 ∂y with respect to y to get ϕ(x, y) =

2 5/2 5/2 x y + c(x). 5

Then we need ∂ϕ = x3/2 y 5/2 + c0 (x) = x3/2 y 5/2 + x1/2 . ∂x Therefore c0 (x) = x1/2 , so c(x) = 2x3/2 /3 and ϕ(x) =

2 5/2 5/2 2 3/2 x y + x . 5 3

The general solution of the original differential equation is given implicitly by 2 2 (xy)5/2 + x3/2 = k. 5 3 In this we must have x 6= 0 and y 6= 0 to ensure that the integrating factor µ(x, y) 6= 0. 16. It is routine to verify that the differential equation is not exact. To find an integrating factor, first multiply by xa y b to get (2xa y b+2 − 9xa+1 y b+1 ) dx + (3xa+1 y b+1 − 6xa+2 y b ) dy = 0. For this to be exact, we must have ∂M = 2(b + 2)xa y b+1 − 9(b + 1)xa+1 y b ∂y ∂N = = 3(a + 1)xa y b+1 − 6(a + 2)x+1 y b . ∂x Divide by xa y b and rearrange terms to obtain (2(b + 2) − 3(a + 1))y = (9(b + 1) − 6(a + 2))x. Because x and y are independent, both coefficients must be zero: 2(b + 2) − 3(a + 1) = 0 and 9(b + 1) − 6(a + 2) = 0.

CHAPTER 1. FIRST-ORDER DIFFERENTIAL EQUATIONS Solve these to get a = b = 1, so µ(x, y) = xy is an integrating factor. Multiply the differential equation by xy to obtain, in differential form, (2xy 3 − 9x2 y 2 ) dx + (3x2 y 2 − 6x3 y) dy = 0. This equation is exact. For a potential function, integrate ∂ϕ = 2xy 3 − 9x2 y 2 ∂x with respect to x to get ϕ(x, y) = x2 y 3 − 3x3 y 2 + c(y). Then

∂ϕ = 3x2 y 2 − 6x3 y + c0 (y) = 3x2 y 2 − 6x3 y. ∂y

Then c0 (y) = 0 and we can choose c(y) = 0 got s potential function ϕ(x, y) = x2 y 3 − 3x3 y 2 . A general solution of this equation, and also the original equation, is given by x2 y 3 − 3x3 y 2 = k. This requires that µ(x, y) 6= 0.

1.4

Homogeneous, Bernoulli and Riccati Equations

1. This is a Riccati equation and one solution (by inspection) is S(x) = x. Let y = x + 1/z to obtain 2 1 0 1 1 1 1 2 − 2z = 2 x + − x+ + 1. z x z x z This simplifies to 1 1 z = − 2, x x a linear equation with integrating factor z0 +

e (1/x) dx = eln(x) = x. The differential equation for z can therefore be written 1 (xz)0 = − . x Integrate to get xz = − ln(x) + c,

1.4. HOMOGENEOUS, BERNOULLI AND RICCATI EQUATIONS so z=−

ln(x) c c − ln(x) + = . x x x

for x > 0. Then y =x+

x 1 =x+ z c − ln(x)

for x > 0. 2. This is a Bernoulli equation with α = −4/3. Put v = y 7/3 , so y = v 7/3 . Substitute this into the differential equation to get 3 −4/7 0 7 3/7 14 v v v + = 2. 7 3x 3x This simplifies to the linear differential equation v0 +

7 14 v= 2 3x 3x

which has integrating factor R

e 7/3x dx = e(7/3) ln(x) = eln(x

7/3

)

= x7/3

for x > 0. Multiply the differential equation by x7/3 to get (x7/3 v)0 =

14 1/3 x . 3

Integrate to get vx7/3 =

7 4/3 x + c. 2

Because v = y 7/3 , this gives us 2y 7/3 x7/3 − 7x4/3 = k, in which k = 2c is an arbitrary constant. This equation implicitly defines the general solution. 3. This is a Bernoulli equation with α = 2, so let v = y 1−α = y −1 for y 6= 0 and y = 1/v. Compute y0 =

dy dv 1 = − 2 xv 0 . dv dx v

The differential equation becomes x x 1 − 2 v0 + = 2 . v v v This is v 0 − xv = −x,

CHAPTER 1. FIRST-ORDER DIFFERENTIAL EQUATIONS 2

a linear equation with integrating factor e−x /2 . We can therefore write 2

(e−x /2 v)0 = −xe−x /2 . Integrate to get 2

e−x /2 v = e−x /2 + c, so

v = 1 + ce−x /2 . The original differential equation has the general solution 1 1 , = v 1 + ce−x2 /2

in which c is an arbitrary constant. 4. This equation is homogeneous. With y = ux we obtain u + xu0 = u +

1 . u

Then

du 1 = , dx u a separable equation. In differential form, this is x

u du =

1 dx. x

Integrate to get 1 2 u = ln |x| + c. 2 Then

1 2 x = 2 ln |x| + k, y2

where k = 2c is an arbitrary constant. This implicitly defines the general solution. 5. This differential equation is homogeneous and setting y = ux gives us u + xu0 =

u . 1+u

This is the separable equation x

du u = −u dx 1+u

which, in terms of x and y, is 1 1 1 + du = − dx. u2 u x

1.4. HOMOGENEOUS, BERNOULLI AND RICCATI EQUATIONS

Integrate to get 1 + ln |u| = − ln |x| + c. u With u = y/x this reduces to −x + y ln |y| = cy, with c an arbitrary constant. 6. This is a Riccati equation and one solution (by inspection) is S(x) = 4. After some routine computation we obtain the general solution y =4+

6x3 . c − x3

7. The differential equation is exact, with general solution defined implicitly by xy − x2 − y 2 = c. 8. The differential equation is homogeneous, and y = ux yields the general solution defined by y y sec + tan = cx. x x 9. The differential equation is of Bernoulli type with α = −3/4. The general solution is defined by 5(xy)7/4 + 7x−5/4 = c. 10. The differential equation is homogeneous and y = ux leads to the separable differential equation 1 − u + u2 1 du = dx. x Integrate and set u = y/x to obtain the general solution implicitly defined by 2 2y − x √ arctan √ = ln |x| + c. 3 3x 11. The equation is Bernoulli with α = 2 and the change of variables v = y −1 leads to the general solution y =2+

2 . cx2 − 1

12. The equation is homogeneous and y = ux leads to the general solution defined by 1 x2 = ln |x| + c. 2 y2

CHAPTER 1. FIRST-ORDER DIFFERENTIAL EQUATIONS

13. The differential equation is Riccati and one solution is S(x) = ex . A general solution is given explicitly by y=

2ex . ce2x − 1

14. The equation is Bernoulli with α = 2 and a general solution is given by y=

2 . 3 + cx2

15. For the first part, y ax + by + c a + b(y/x)c/x F =F =f dx + py + r d + p(y/x) + r/x x if and only if c = r = 0. Next, suppose x = X + h and y = Y + k. Then dY a(X + h) + b(Y + k) + c =F dX d(x + h) + p(Y + k) + r aX + bY + c + ah + bk + c =F . dX + pY + r + dh + pk + r This equation is homogeneous exactly when h and k can be chosen so that ah + bk = −c and dh + pk = −r. This 2 × 2 system of algebraic equations has a solution exactly when the determinant of the coefficients is nonzero, and this is the condition that a b = ap − bd 6= 0. d p 16. Comparing this with problem 15, we have a = 0, b = 1, c = −3, d = p = 1 and r = −1. The system to solve for h and k is k = 3, h + k = 1. Then k = 3 and h = −2. Let X = x − 2, Y = y + 3 to obtain dY Y = . dX X +Y This is a homogeneous equation solved in problem 5. The general of the current problem is defined by (y − 3) ln |y − 3| − (x + 2) = c(y − 3), with c an arbitrary constant.

1.4. HOMOGENEOUS, BERNOULLI AND RICCATI EQUATIONS

17. Let X = x − 2, Y = y + 3 to get the homogeneous equation dY 3X − Y = . dX X +Y The general solution of the original equation (in terms of x and y) is defined by 3(x − 2)2 − 2(x − 2)(y + 3) − (y + 3)2 = c, with c an arbitrary constant. 18. Set X = x + 5, Y = y + 1 to obtain the implicitly defined general solution (x + 5)2 + 4(x + 5)(y + 1) − (y + 1)2 = c. 19. Let X = x − 2, Y = y + 1 to obtain the general solution given by (2x + y − 3)2 = c(y − x + 3). 20. Suppose at time t = 0 the dog is at the origin of an x, y− coordinate system, and the person is at (A, 0). The person moves directly upward and at time t is at (A, vt), while the dog is at (x, y) and runs toward the person at a speed 2v. The tangent to the dog’s path joins these two points, and so has slope vt − y y0 = . A−x To find the equation of the dog’s path, we will first eliminate t from this equation. In the time the person has moved vt units upward, the dog has run 2vt units along its path of motion, so Z x

2vt =

1+ 0

dy dξ

2 !1/2 dξ.

Then 1 dy vt = y + (A − x) = dx 2

Z x

Then 2(A − x)y 0 =

Z x

dy dξ

2 !

dy dx

dy dξ

2 ! dξ.

dξ − 2y.

Differentiate this equation to get 2(A − x)y 00 − 2y 0 =

so 2(A − x)y 00 = 1 + (y 0 )2

2 !

1/2

− 2y 0 ,

CHAPTER 1. FIRST-ORDER DIFFERENTIAL EQUATIONS together with the conditions y(0) = y 0 (0) = 0. Now let u = y 0 and rewrite the resulting equation to get 1 1 du = dx. 2 1/2 2(A − x) (1 + u ) This has the general solution p 1 ln(u + 1 + u2 ) = − ln(A − x) + c. 2 Use the condition that y 0 (0) = u(0) = 0 to obtain u+

1 + u2 =

A A−x

1/2 .

In terms of y, we now have y0 +

√ p A 1 + (y 0 )2 = √ ; y(0) = 0. A−x

But p 1 + (y 0 )2 = 2(A − x)y 00 , so

√ A y + 2(A − x)y = √ . A−x 0

Let w = y 0 to get √ 1 A w + w= . 2(A − x) 2(A − x)3/2 0

√ This linear differential equation has integrating factor 1/ A − x, so 0 w A √ . = 2(A − x)2 A−x Integrate this to get √ w=

√ A 1 √ + c A − x. 2 A−x

Use the fact that w(0) = 0 to get √ A 1 1 √ √ w= − √ A − x = y0 . 2 A−x 2 A Integrate this to get √ √ 1 y = − A A − x + √ (A − x)3/2 + c. 3 A

1.4. HOMOGENEOUS, BERNOULLI AND RICCATI EQUATIONS

Because y(0) = 0, √ √ 1 2 y = − A A − x + √ (A − x)3/2 + A. 3 3 A Now the dog catches the person at x = A, so they meet at (A, 2A/3). This is also the point (A, vt), so vt = 2A/3 and they meet at time t=

2A . 3v

21. It is convenient to use polar coordinates to formulate a model for this problem. Put the origin at the submarine at the time of sighting, and the polar axis the line from there to the destroyer at this time (the point (9, 0)). Initially the destroyer should steam at speed 2v directly toward the origin, until it reaches (3, 0). During this time the submarine, moving at speed v, will have moved three units from the point where it was sighted. Let θ = ϕ be the ray (half-line) along which the submarine is moving. Upon reaching (3, 0), the destroyer should execute a search pattern along a path r = f (θ). The object is to choose this path so that the sub and the destroyer both reach (f (ϕ, ϕ) at the same time T after the sighting. From sighting to interception, the destroyer travels a distance Z ϕp (f (θ))2 + (f 0 (θ))2 dθ, 6+ 0

so T =

1 2v

Z ϕp (f (θ))2 + (f 0 (θ))2 dθ . 0

For the submarine, 1 f (ϕ). v Equate these two expressions for T and differentiate with respect to ϕ to get 1p (f (ϕ))2 + (f 0 (ϕ))2 = f 0 (ϕ). 2 Denote the variable as θ and rearrange the last equation to obtain T =

1 f 0 (θ) = ±√ . f (θ) 3 The positive sign here indicates that the destroyer should execute a starboard (left) turn, while the negative sign is for a portside turn. Taking the positive sign, solve for f (θ) to get √

f (θ) = keθ/ 3 .

CHAPTER 1. FIRST-ORDER DIFFERENTIAL EQUATIONS Now f (0) = k = 3, so the path of the destroyer is part of the graph of √

f (θ) = 3seθ/ 3 . After sailing directly to (3, 0), the destroyer should execute √this spiral pattern. A similar conclusion follows if the negative sign of 1/ 3 is used. This shows that the destroyer can carry out a maneuver that will take it directly over the submarine at some time. However, there is no way to solve for the specific time, so it is unknown when this will occur. 22. (a) Observe that each bug follows the same curve of pursuit relative to the center from which it starts. Place a polar coordinate system as suggested √ and determine the pursuit curve for the bug starting at θ = 0, r = a/ 2. At t > 0, the bug will be at (f (θ), θ) and its target is at (f (θ), θ + π/2). Show that dy dy/dθ f 0 (θ) sin(θ) + f (θ) cos(θ) = = 0 . dx dx/dθ f (θ)cos(θ) − f (θ) sin(θ) At the same time, the direction of the tangent must be from the position (f (θ), θ) to the target location (f (θ), θ + π/2), so we also have dy f (θ) sin(θ + π/2) − f (θ) sin(θ) = dx f (θ) cos(θ + π/2) − f (θ) cos(θ) cos(θ) − sin(θ) sin(θ) − cos(θ) = = . − sin(θ) − cos(θ) sin(θ) + cos(θ) Equate these two expressions for dy/dx and rearrange terms to get f 0 (θ) + f (θ) = 0. √ Further, f (0) = a/ 2. This is a separable, and also linear, differential equation, and the initial value problem has the solution a r = f (θ) = √ e−θ . 2 This is the pursuit curve (in polar coordinates). (b) The distance traveled is Z ∞p r2 + (r0 )2 dθ 0

Z ∞ " = 0

Z ∞ =a

a √ e−θ 2

2 #1/2 a −θ + −√ e dθ 2

e−θ dθ = a.

√ (c) Because r = ae−θ / 2 > 0 for all θ, no bug actually catches its quarry. The actual distance between pursuer and quarry is ae−θ .

Chapter 2

Second-Order Differential Equations 2.1

The Linear Second-Order Equation

1. It is a routine exercise in differentiation to show that y1 (x) and y2 (x) are solutions of the homogeneous equation, while yp (x) is a solution of the nonhomogeneous equation. The Wronskian of y1 (x) and y2 (x) is W (x) =

sin(6x) cos(6x) = −6 sin2 (x) − 6 sin2 (x) = −6, 6 cos(6x) −6 sin(6x)

and this is nonzero for all x, so these solutions are linearly independent on the real line. The general solution of the nonhomogeneous differential equation is 1 y = c1 sin(6x) + c2 cos(6x) + (x − 1). 36 For the initial value problem, we need y(0) = c2 −

1 = −5 36

so c2 = −179/36. And y 0 (0) = 2 = 6c1 +

1 36

so c1 = 71/216. The unique solution of the initial value problem is y(x) =

71 179 1 sin(6x) − cos(6x) + (x − 1). 216 36 36

2. The Wronskian of e4x and e−4x is W (x) =

e4x 4e4x

e−4x = −8 6= 0 −4e−4x

CHAPTER 2. SECOND-ORDER DIFFERENTIAL EQUATIONS so these solutions of the associated homogeneous equation are independent. With the particular solution yp (x) of the nonhomogeneous equation, this equation has general solution 1 1 y(x) = c1 e4x + c2 e−4x − x2 − . 4 32 From the initial conditions we obtain y(0) = c1 + c2 −

1 = 12 32

and y 0 (0) = 4c1 − 4c2 = 3. Solve these to obtain c1 = 409/64 and c2 = 361/64 to obtain the solution y(x) =

1 409 4x 361 −4x 1 2 e + e − x − . 64 64 4 32

3. The associated homogeneous equation has solutions e−2x and e−x . Their Wronskian is e−2x e−x = e−3x W (x) = −2x −2e −e−x and this is nonzero for all x. The general solution of the nonhomogeneous differential equation is y(x) = c1 e−2x + c2 e−x +

15 . 2

For the initial value problem, solve y(0) = −3 = c1 + c2 +

15 2

and y 0 (0) = −1 = −2c1 − c2 to get c1 = 23/2, c2 = −22. The initial value problem has solution y(x) =

23 −2x 15 e − 22e−x + . 2 2

4. The associated homogeneous equation has solutions y1 (x) = e3x cos(2x), y2 (x) = e3x sin(2x). The Wronskian of these solutions is W (x) =

e3x cos(2x) cos(2x) − 2e3x sin(2x)

e3x sin(2x) = e6x 6= 0 sin(2x) + 2e3x cos(2x)

2.1. THE LINEAR SECOND-ORDER EQUATION

for all x. The general solution of the nonhomogeneous equation is 1 y(x) = c1 e3x cos(2x) + c2 e3x sin(2x) − ex . 8 To satisfy the initial conditions, it is required that y(0) = −1 = c1 −

1 8

and

1 = 1. 8 Solve these to obtain c1 = −7/8 and c2 = 15/8. The solution of the initial value problem is 3c1 + 2c2 −

15 1 7 y(x) = − e3x cos(2x) + e3x sin(2x) − ex . 8 8 8 5. The associated homogeneous equation has solutions y1 (x) = ex cos(x), y2 (x) = ex sin(x). These have Wronskian W (x) =

ex cos(x) ex sin(x) = e2x 6= 0 ex cos(x) − ex sin(x) ex sin(x) + ex cos(x)

so these solutions are independent. The general solution of the nonhomogeneous differential equation is 5 5 y(x) = c1 ex cos(x) + c2 ex sin(x) − c2 − 5x − . 2 2 We need y(0) = c1 −

5 =6 2

and y 0 (0) = 1 = c1 + c2 − 5. Solve these to get c1 = 17/2 and c2 = −5/2 to get the solution y(x) =

17 x 5 5 5 e cos(x) − ex sin(x) − x2 − 5x − . 2 2 2 2

6. Suppose y1 and y2 are solutions of the homogeneous equation (2.2). Then y100 + py10 + qy1 = 0 and y200 + py20 + qy2 = 0.

CHAPTER 2. SECOND-ORDER DIFFERENTIAL EQUATIONS Multiply the first equation by y2 and the second by −y1 and add the resulting equations to obtain y100 y2 − y200 y1 + p(y10 y2 − y20 y1 ) = 0. We want to relate this equation to the Wronskian of these solutions, which is W = y1 y20 − y2 y10 . Now W 0 = y1 y200 − y2 y100 . Then W 0 + pW = 0. This is a linear first-order differential equation for W . Multiply this equation by the integrating factor R

e p(x) dx to obtain R

W e p(x) dx + pW e p(x) dx = 0, which we can write as

W e p(x) dx

= 0.

Integrate this to obtain R

W e p(x) dx = k, with k constant. Then W (x) = ke−

p(x) dx

This shows that W (x) = 0 for all x (if k = 0), and W (x) 6= 0 for all x (if k 6= 0). Now suppose that y1 and y2 are independent and observe that d y2 y1 y20 − y2 y10 1 = = 2 W (x). dx y1 y12 y1 If k = 0, then W (x) = 0 for all x and the quotient y2 /y1 has zero derivative and so is constant: y2 =c y1 for some constant c. But then y2 (x) = cy1 (x), contradicting the assumption that these solutions are linearly independent. Therefore k 6= 0 and W (x) 6= − for all x, as was to be shown.

2.2. THE CONSTANT COEFFICIENT HOMOGENEOUS EQUATION

7. The Wronskian of x2 and x3 is W (x) =

x2 x3 = x4 . 2x 3x2

Then W (0) = 0, while W (x) 6= 0 if x 6= 0. This is impossible if x2 and x3 are solutions of equation (2.2) for some functions p(x) and q(x). We conclude that these functions are not solutions of equation (2.2). 8. It is routine to verify that y1 (x) and y2 (x) are solutions of the differential equation. Compute x x2 = x2 . W (x) = 1 2x Then W (0) = 0 but W (x) > 0 if x 6= 0. However, to write the differential equation in the standard form of equation (2.2), we must divide by x2 to obtain 2 2 y 00 − y 0 + 2 y = 0. x x This is undefined at x = 0, which is in the interval −1 << 1, so the theorem does not apply. 9. If y2 (x) and y2 (x) both have a relative extremum (max or min) at some x0 within (a, b), then y 0 (x0 ) = y20 (x0 ) = 0. But then the Wronskian of these functions vanishes at 0, and these solutions must be independent. 10. By assumption, ϕ(x) is the unique solution of the initial value problem y 00 + py 0 + qy = 0; y(x0 ) = 0. But the function that is identically zero on I is also a solution of this initial value problem. Therefore these solutions are the same, and ϕ(x) = 0 for all x in I. 11. If y1 (x0 ) = y2 (x0 ) = 0, then the Wronskian of y1 (x) and y2 (x) is zero at x0 , and these two functions must be linearly dependent.

2.2

The Constant Coefficient Homogeneous Equation

1. From the differential equation we read the characteristic equation λ2 − λ − 6 = 0, which has roots −2 and 3. The general solution is y(x) = c1 e−2x + c2 e3x .

CHAPTER 2. SECOND-ORDER DIFFERENTIAL EQUATIONS 2. The characteristic equation is λ2 − 2λ + 10 = 0 with roots 1 ± 3i. We can write a general solution y(x) = c1 ex cos(3x) + c2 ex sin(3x). 3. The characteristic equation is λ2 + 6λ + 9 = 0 with repeated roots −3, −3. Then y(x) = c1 e−3x + c2 xe−3x is a general solution. 4. The characteristic equation is λ2 − 3λ = 0 with roots 0, 3, and y(x) = c1 + c2 e3x is a general solution. 5. characteristic equation λ2 + 10λ + 26 = 0, with roots −5 ± i; general solution y(x) = c1 e−5x cos(x) + c2 e−5x sin(x). 6. characteristic equation λ2 +6λ−40 = 0, with roots 4, −10; general solution y(x) = c1 e4x + c2 e−10x . √ 7. characteristic equation λ2 +3λ+18 = 0, with roots −3/2±3 7i/2; general solution √ ! √ ! 3 7x 3 7x −3x/2 −3x/2 y(x) = c2 e cos + c2 e sin . 2 2 8. characteristic equation λ2 + 16λ + 64 = 0, with repeated roots −8, −8; general solution y(x) = e−8x (c1 + c2 x). 9. characteristic equation λ2 − 14λ + 49 = 0, with repeated roots 7, 7; general solution y(x) = e7x (c1 + c2 x).

2.2. THE CONSTANT COEFFICIENT HOMOGENEOUS EQUATION

√ 10. characteristic equation λ2 −6λ+7 = 0, with roots 3± 2i; general solution √ √ y(x) = c1 e3x cos( 2x) + c2 e3x sin( 2x). In each of Problems 11–20 the solution is found by finding a general solution of the differential equation and then using the initial conditions to find the particular solution of the initial value problem. 11. The differential equation has characteristic equation λ2 + 3λ = 0, with roots 0, −3. The general solution is y(x) = c1 + c2 e−3x . Choose c1 and c2 to satisfy: y(0) = c1 + c2 = 3, y 0 (0) = −3c2 = 6. Then c2 = −2 and c1 = 5, so the unique solution of the initial value problem is y(x) = 5 − 2e−3x . 12. characteristic equation λ2 + 2λ − 3 = 0, with roots 1, −3; general solution y(x) = c1 ex + c2 e−3x . Solve y(0) = c1 + c2 = 6, y 0 (0) = c1 − 3c2 = −2 to get c1 = 4 and c2 = 2. The solution is y(x) = 4ex + 2e−3x . 13. The initial value problem has the solution y(x) = 0 for all x. This can be seen by inspection or by finding the general solution of the differential equation and then solving for the constants to satisfy the initial conditions. 14. y(x) = e2x (3 − x) 15. characteristic equation λ2 + λ − 12 = 0, with roots 3, −4. The general solution is y(x) = c1 e3x + c2 e−4x . We need y(2) = c1 e6 + c2 e−8 = 2 and y 0 (2) = 3c1 e6 − 4c2 e−8 = −1. Solve these to obtain c1 = e−6 , c2 = e8 .

CHAPTER 2. SECOND-ORDER DIFFERENTIAL EQUATIONS The solution of the initial value problem is y(x) = e−6 e3x + e8 e−4x . This can also be written y(x) = e3(x−2) + e−4(x−2) .

16.

√ y(x) =

√ 6 x √6x e e − e− 6x 4

17. y(x) = ex−1 (29 − 17x) 18. √ √ 8 √ sin( 23)e5x/2 cos( 23x/2) e5 23 √ √ 8 − √ cos( 23)e5x/2 sin( 23x/2) 5 e 23

y(x) =

19.

h √ y(x) = e(x+2)/2 cos( 15(x + 2)/2) √ 5 + √ sin( 15(x + 2)/2) 15

20. y(x) = ae(−1 +

√ √ 5)x/2 + be(−1− 5)x/2 ,

where a=

√ ! √ 9+7 5 √ e−2+ 5 and b = 2 5

! √ √ 7 5−9 √ e−2− 5 2 5

21. (a) The characteristic equation is λ2 − 2αλ + α2 = 0, with α as a repeated root. The general solution is y(x) = (c1 + c2 x)eαx . (b) The characteristic equation is λ2 − 2αλ + (α2 − 2 ) = 0, with roots α + , α − . The general solution is y (x) = c1 e(α+ )x + c2 e(α− )x . We can also write y (x) = c1 e x + c2 e− x eαx . In general, lim y (x) = (c1 + c2 )eαx 6= y(x).

→0

Note, however, that the coefficients in the differential equations in (a) and (b) can be made arbitrarily close by choosing sufficiently small.

2.2. THE CONSTANT COEFFICIENT HOMOGENEOUS EQUATION

22. With a2 = 4b, one solution is y1 (x) = e−ax/2 . Attempt a second solution y2 (x) = u(x)e−ax/2 . Substitute this into the differential equation to get a2 a 00 0 0 u − au + u + a u − u + bu e−ax/2 = 0. 4 2 Because a2 − 4b = 0, this reduces to u00 (x) = 0. Then u(x) = cx + d, with c and d arbitrary constants, and the functions (cx + d)e−ax/2 are also solutions of the differential equation. If we choose c = 1 and d = 0, we obtain y2 (x) = xe−ax/2 as a second solution. Further, this solution is independent from y1 (x), because the Wronskian of these solutions is W (x) =

e−ax/2 −(a/2)e−ax/2

−ax/2

xe−ax/2 = e−ax , − (a/2)xe−ax/2

and this is nonzero. 23. The roots of the characteristic equation are √ √ −a − a2 − 4b −a + a2 − 4b and λ2 = . λ1 = 2 2 Because a2 − 4b < a2 by assumption, λ1 and λ2 are both negative (if a2 − 4b ≥ 0), or complex conjugates (if a2 − 4b < 0). There are three cases. Case 1 - Suppose λ1 and λ2 are real and unequal. Then the general solution is y(x) = c1 eλ1 x + c2 eλ2 x and this has limit zero as x → ∞ because λ1 and λ2 are negative. Case 2 - Suppose λ1 = λ2 . Now the general solution is y(x) = (c1 + c2 x)eλ1 x , and this also has limit zero as x → ∞. Case 3 - Suppose λ1 and λ2 are complex. Now the general solution is h i p p y(x) = c1 cos( 4b − a2 x/2) + c2 sin( 4b − a2 x/2) e−ax/2 , and this has limit zero as x → ∞ because a > 0.

√

If, for example, a = 1 and b = −1, then one solution is e(−1+ 5)x/2 , and this tends to ∞ as x → ∞.

CHAPTER 2. SECOND-ORDER DIFFERENTIAL EQUATIONS

2.3

Particular Solutions of the Nonhomogeneous Equation

1. Two independent solutions of y 00 + y = 0 are y1 (x) = cos(x) and y2 (x) = sin(x), with Wronskian W (x) =

cos(x) − sin(x)

sin(x) = 1. cos(x)

Let f (x) = tan(x) and use equations (2.7) and (2.8). First, Z u1 (x) = −

y2 (x)f (x) =− W (x)

Z tan(x) sin(x) dx

sin2 (x) dx cos(x) Z 1 − cos2 (x) =− dx cos(x) Z Z = cos(x) dx − sec(x) dx Z

=−

= sin(x) − ln | sec(x) + tan(x)|. Next, Z u2 (x) =

y1 (x)f (x) dx = W (x)

Z cos(x) tan(x) dx

Z sin(x) dx = − cos(x).

= The general solution is

y(x) = c1 cos(x) + c2 sin(x) + u1 (x)y1 (x) + u2 (x)y2 (x) = c1 cos(x) + c2 sin(x) − cos(x) ln | sec(x) + tan(x)|.

2. Two independent solutions of the associated homogeneous equation are y1 (x) = e3x and y2 (x) = ex . Their Wronskian is W (x) = −2e4x . Compute 2ex cos(x + 3) dx u1 (x) = − −2e4x Z = e−3x cos(x + 3) dx Z

=−

3 −3x 1 e cos(x + 3) + e−3x sin(x + 3) 10 10

2.3. PARTICULAR SOLUTIONS OF THE NONHOMOGENEOUS EQUATION47 and Z

2e3x cos(x + 3) dx −2e4x

e−x cos(x + 3) dx

u2 (x) = = =

1 −x 1 e cos(x + 3) − e−x cos(x + 3). 2 2

The general solution is y(x) = c1 e3x + c2 ex 3 1 − cos(x + 3) + sin(x + 3) 10 10 1 1 + cos(x + 3) − sin(x + 3). 2 2 More compactly, the general solution is y(x) = c1 e3x + c2 ex +

2 1 cos(x + 3) − sin(x + 3). 5 5

For Problems 3–6, some details of the calculations are omitted. 3. The associated homogeneous equation has independent solutions y1 (x) = cos(3x) and y2 (x) = sin(3x), with Wronskian 3. The general solution is y(x) = c1 cos(3x) + c2 sin(3x) + 4x sin(3x) +

4 cos(3x) ln | cos(3x)|. 3

4. y1 (x) = e3x and y2 (x) = e−x , with W (x) = −4e−2x . With f (x) = 2 sin2 (x) = 1 − cos(2x) we find the general solution y(x) = c1 e3x + c2 e−x −

7 4 1 + cos(2x) + sin(2x). 3 65 65

5. y1 (x) = ex and y2 (x) = e2x , with Wronskian W (x) = e3x . With f (x) = cos(e−x ), we find the general solution y(x) = c1 ex + c2 e2x − e2x cos(e−x ). 6. y1 (x) = e3x and y2 (x) = e2x , with Wronskian W (x) = e−5x . Use the identity 8 sin2 (4x) = 4 cos(8x) − 1 in determining u1 (x) and u2 (x) to write the general solution y(x) = c1 e3x + c2 e2x +

2 58 40 + cos(8x) + sin(8x). 3 1241 1241

CHAPTER 2. SECOND-ORDER DIFFERENTIAL EQUATIONS

In Problems 7–16 the method of undetermined coefficients is used to find a particular solution of the nonhomogeneous equation. Details are included for Problems 7 and 8, and solutions are outlined for the remainder of these problems. 7. The associated homogeneous equation has independent solutions y1 (x) = e2x and e−x . Because 2x2 +5 is a polynomial of degree 2, attempt a second degree polynomial yp (x) = Ax2 + Bx + C for the nonhomogeneous equation. Substitute yp (x) into this nonhomogeneous equation to obtain 2A − (2Ax + B) − 2(Ax2 + Bx + C) = 2x2 + 5. Equating coefficients of like powers of x on the left and right, we have the equations −2A = 2( coefficients of x2 ) −2A − 2B − 0( coefficients of x 2A − 2B − 2C = 5( constant term.) Then A = −1, B = 1 and C = −4. Then yp (x) = −x2 + x − 4 and a general solution of the (nonhomogeneous) equation is y = c1 e2x + c2 e−x − x2 + x − 4. 8. y1 (x) = e3x and y2 (x) = e−2x are independent solutions of the associated homogeneous equation. Because e2x is not a solution of the homogeneous equation, attempt a particular solution yp (x) = Ae2x of the nonhomogeneous equation. Substitute this into the differential equation to get 4A − 2A − 6A = 8, so A = −2 and a general solution is y(x) = c1 e3x + c2 e−2x − 2e2x . 9. y1 (x) = ex cos(3x) and y2 (x) = ex sin(3x) are independent solutions of the associated homogeneous equation. Try a particular solution yp (x) = Ax2 + Bx + C to obtain the general solution y(x) = c1 ex cos(3x) + c2 ex sin(3x) + 2x2 + x − 1.

2.3. PARTICULAR SOLUTIONS OF THE NONHOMOGENEOUS EQUATION49 10. For the associated homogeneous equation, y1 (x) = e2x cos(x) and y2 (x) = e2x sin(x). Try yp (x) = Ae2x to get A = 21 and obtain the general solution y(x) = c1 e2x cos(x) + c2 e2x sin(x) + 21e2x . 11. For the associated homogeneous equation, y1 (x) = e2x and y2 (x) = e4x . Because ex is not a solution of the homogeneous equation, attempt a particular solution of the nonhomogeneous equation of the form yp (x) = Aex . We get A = 1, so a general solution is y(x) = c1 e2x + c2 e4x + ex . 12. y1 (x) = e−3x and y2 (x) = e−3x . Because f (x) = 9 cos(3x) (which is not a solution of the associated homogeneous equation), attempt a particular solution yp (x) = A cos(3x) + B sin(3x). This attempt includes both a sine and cosine term even though f (x) has only a cosine term, because both terms may be needed to find a particular solution. Substitute this into the nonhomogeneous equation to obtain A = 0 and B = 1/2, so a general solution is y(x) = (c1 + c2 x)e−3x +

1 sin(3x). 2

In this case yp (x) contains only a sine term, although f (x) has only the cosine term. 13. y1 (x) = ex and y2 (x) = e2x . Because f (x) = 10 sin(x), attempt yp (x) = A cos(x) + B sin(x). Substitute this into the (nonhomogeneous) equation to find that A = 3 and B = 1. A general solution is y(x) = c1 ex + c2 e2x + 3 cos(x) + sin(x). 14. y1 (x) = 1 and y2 (x) = e−4x . Finding a particular solution yp (x) for this problem requires some care. First, f (x) contains a polynomial term and an exponential term, so we are tempted to try yp (x) as a second degree polynomial Ax2 + Bx + C plus an exponential term De3x to account for the exponential term in the equation. However, note that y1 (x) = 1, a constant solution, is one term of the proposed polynomial part, so multiply this part by x to try yp (x) = Ax3 + Bx2 + Cx + De3x . Substitute this into the nonhomogeneous differential equation to get 6Ax + 2B + 9De3x − 4(3Ax2 + 2Bx + C + 3De3x ) = 8x2 + 2e3x .

CHAPTER 2. SECOND-ORDER DIFFERENTIAL EQUATIONS Matching coefficients of like terms, we conclude that 2B − 4C = 0( from the constant terms) 6A − 8B = 0( from the x terms), −12A = 8( from the x2 terms) −3D = 2. Then D = − 32 , A = − 23 , B = − 21 and C = − 41 . Then 2 1 1 2 yp (x) = − x3 − x2 − x − e3x 3 2 4 3 and a general solution of the nonhomogeneous equation is 1 1 2 2 y(x) = c1 + c2 e−4x − x3 − x2 − x − e3x . 3 2 4 3

15. y1 (x) = e2x cos(3x) and y2 (x) = e2x sin(3x). Try yp x = Ae2x + Be3x . This will work because neither e2x nor e3x is a solution of the associated homogeneous equation. Substitute yp (x) into the differential equation and obtain A = 1/3, B = −1/2. The differential equation has general solution 1 1 y(x) = [c1 cos(3x) + c2 sin(3x)]e2x + e2x − e3x . 3 2 16. y1 (x) = ex and y2 (x) = xex . Try yp (x) = Ax + B + C cos(3x) + D sin(3x). This leads to the general solution y(x) = (c1 + c2 x)ex + 3x + 6 +

3 cos(3x) − 2 sin(3x). 2

In Problems 17–24 the strategy is to first find a general solution of the differential equation, then solve for the constants to find a solution satisfying the initial conditions. Problems 17–22 are well suited to the use of undetermined coefficients, while Problems 23 and 24 can be solved fairly directly using variation of parameters. 17. y1 (x) = e2x and y2 (x) = e−2x . Because e2x is a solution of the associated homogeneous equation, use xe2x in the method of undetermined coefficients, attempting yp (x) = Axe2x + Bx + C.

2.3. PARTICULAR SOLUTIONS OF THE NONHOMOGENEOUS EQUATION51 Substitute this into the nonhomogeneous differential equation to obtain 4Axe2x + 4Axe2x − 4Axe2x − 4Bx − 4C = −7e2x + x. Then A = −7/4, B = −1/4 and C = 0, so the differential equation has the general solution 1 7 y(x) = c1 e2x + c2 e−2x − xe2x − x. 4 4 We need y(0) = c1 + c2 = 1 and y 0 (0) = 2c1 − 2c2 −

7 1 − = 3. 4 4

Then c1 = 7/4 and c2 = −3/4. The initial value problem has the unique solution 7 3 7 1 y(x) = e2x − e−2x − xe2x − x. 4 4 4 4 18. y1 = 1 and y2 (x) = e−4x are independent solutions of the associated homogeneous equation. For yp (x), try yp (x) = Ax + B cos(x) + C sin(x), with the term Ax because 1 is a solution of the homogeneous equation. This leads to the general solution y(x) = c1 + c2 e−4x + 2x − 2 cos(x) + 8 sin(x). Now we need y(0) = c1 + c2 − 2 = 3 and y 0 (0) = −4c2 + 2 + 8 = 2. Then c1 = 3 and c2 = 2, so the initial value problem has the solution y(x) = 3 + 2e−4x + 2x − 2 cos(x) + 8 sin(x). 19. We find the general solution 1 7 y(x) = c1 e−2x + c2 e−6x + e−x + . 5 12 The solution of the initial value problem is y(x) =

3 −2x 19 −6x 1 −x 7 e − e + e + . 8 120 5 12

CHAPTER 2. SECOND-ORDER DIFFERENTIAL EQUATIONS

20. 1 and e3x are independent solutions of the associated homogeneous equation. Attempt a particular solution yp (x) = Ae2x cos(x) + Be2x sin(x) of the nonhomogeneous equation to find the general solution 1 y(x) = c1 + c2 e3x − e2x (cos(x) + 3 sin(x)). 5 The solution of the initial value problem is y(x) =

1 1 + e3x − (cos(x) + 3 sin(x)). 5 5

21. e4x and e−2x are independent solutions of the associated homogeneous equation. The nonhomogeneous equation has general solution y(x) = c1 e4x + c2 e−2x − 2e−x − e2x . The solution of the initial value problem is y(x) = 2e4x + 2e−2x − 2e−x − e2x . 22. The general solution of the differential equation is " √ !# √ ! 3 3 x/2 x + c2 sin x + 1. y(x) = e c1 cos 2 2 It is easier to fit the initial conditions specified at x = 1 if we write this general solution as " ! !# √ √ 3 3 x/2 y(x) = e d1 cos (x − 1) + d2 sin (x − 1) + 1. 2 2 Now

√ 1 1/2 3 1/2 y(1) = e d1 + 1 = 4 and y (1) = e d1 + e d2 = −2. 2 2 √ Solve these to obtain d1 = 3e−1/2 and d2 = −7e−1/2 / 3. The solution of the initial value problem is " ! !# √ √ 3 7 3 y(x) = e(x−1)/2 3 cos (x − 1) − √ sin (x − 1) + 1. 2 2 3 1/2

23. The differential equation has general solution y(x) = c1 ex + c2 e−x − sin2 (x) − 2. The solution of the initial value problem is y(x) = 4e−x − sin2 (x) − 2.

2.4. THE EULER DIFFERENTIAL EQUATION

24. The general solution is y(x) = c1 cos(x) + c2 sin(x) − cos(x) ln | sec(x) + tan(x)|, and the solution of the initial value problem is y(x) = 4 cos(x) + 4 sin(x) − cos(x) ln | sec(x) + tan(x)|.

2.4

The Euler Differential Equation

Details are included with solutions for Problems 1–2, while just the solutions are given for Problems 3–10. These solutions are for x > 0. 1. Read from the differential equation that the characteristic equation is r2 + r − 6 = 0 with roots 2, −3. The general solution is y(x) = c1 x2 + c2 x−3 . 2. The characteristic equation is r2 + 2r + 1 = 0 with repeated root −1, −1. The general solution is y(x) = c1 x−1 + c2 x−1 ln(x). We can also write y(x) =

1 (c1 + c2 ln(x)) x

for x > 0. 3. y(x) = c1 cos(2 ln(x)) + c2 sin(2 ln(x)) 4. y(x) = c1 x2 + c1

1 x2

y(x) = c1 x2 + c1

1 x4

6. y(x) =

1 (c2 cos(3 ln(x)) + c2 sin(3 ln(x))) x2

7. y(x) = c1

1 1 + c2 3 x2 x

CHAPTER 2. SECOND-ORDER DIFFERENTIAL EQUATIONS 8. y(x) = x2 (c1 cos(7 ln(x)) + c2 sin(7 ln(x))) 9. y(x) =

1 (c1 + c2 ln(x)) x12

10. y(x) = c1 x7 + c2 x5 11. The general solution of the differential equation is y(x) = c1 x3 + c2 x−7 . From the initial conditions, we need y(2) = 8c1 + 2−7 c2 = 1 and y 0 (2) = 3c1 22 − 7c2 2−8 = 0. Solve for c1 and c2 to obtain the solution of the initial value problem 7 x 3 3 x −7 y(x) = + . 10 2 10 2 12. The initial value problem has the solution y(x) = −3 + 2x2 . 13. y(x) = x2 (4 − 3 ln(x)) 14. y(x) = −4x−12 (1 + 12 ln(x)) 15. y(x) = 3x6 − 2x4 16. y(x) =

11 2 17 −2 x + x 4 4

17. With Y (t) = y(et ), use the chain rule to get y 0 (x) =

dY dt 1 = Y 0 (t) dt dx x

and then d 1 0 y (x) = Y (t) dx x 1 1 d = − 2 Y 0 (t) + (Y 0 (t)) x x dx 1 1 dY 0 dt = − 2 Y 0 (t) + x x dt dx 1 0 1 1 00 = − 2 Y (t) + Y (t) x xx 1 = 2 (Y 00 (t) − Y 0 (t)). x 00

2.4. THE EULER DIFFERENTIAL EQUATION

Then x2 y 00 (x) = Y 00 (t) − Y 0 (t). Substitute these into Euler’s equation to get Y 00 (t) + (A − 1)Y 0 (t) + BY (t) = 0. This is a constant coefficient second-order homogeneous differential equation for Y (t), which we know how to solve. 18. If x < 0, let t = ln(−x) = ln |x|. We can also write x = −et . Note that 1 1 dt = (−1) = dx −x x just as in the case x > 0. Now let y(x) = y(−et ) = Y (t) and proceed with chain rule differentiations as in the solution of Problem 17. First, y 0 (x) =

dY dt 1 = Y 0 (t) dt dx x

and d 1 0 Y (t) dx x 1 1 dt 00 = − 2 Y 0 (t) + Y (t) x x dx 1 1 = − 2 Y 0 (t) + 2 Y 00 (t). x x

y 00 (x) =

Then x2 y 00 (x) = Y 00 (t) − Y 0 (t) just as we saw with x > 0. Now Euler’s equation transforms to Y 00 + (A − 1)Y 0 + BY = 0. We obtain the solution in all cases by solving this linear constant coefficient second-order equation. Omitting all the details, we obtain the solution of Euler’s equation for negative x by replacing x with |x| in the solution for positive x. For example, suppose we want to solve x2 y 00 + xy 0 + y = 0 for x < 0. Solve this for x > 0 to get y(x) = c1 cos(ln(x)) + c2 sin(ln(x)). The solution for x < 0 is y(x) = c1 cos(ln |x|) + c2 sin(ln |x|).

CHAPTER 2. SECOND-ORDER DIFFERENTIAL EQUATIONS

19. The problem to solve is x2 y 00 − 5dxy 0 + 10y = 0; y(1) = 4, y 0 (1) = −6. We know how to solve this problem. Here is an alternative method, using the transformation x = et , or t = ln(x) for x > 0 (since the initial conditions are specified at x = 1). Euler’s equation transforms to Y 00 − 6Y 0 + 10Y = 0. However, also transform the initial conditions: Y (0) = y(1) = 4, Y 0 (0) = (1)y 0 (1) = −6. This differential equation for Y (t) has general solution Y (t) = c1 e3t cos(t) + c2 e3t sin(t). Now Y (0) = c2 = 4 and Y 0 (0) = 3c1 + c2 = −6, so c2 = −18. The solution of the transformed initial value problem is Y (t) = 4e3t cos(t) − 18e3t sin(t). The original initial value problem therefore has the solution y(x) = 4x3 cos(ln(x)) − 19x3 sin(ln(x)) for x > 0. The new twist here is that the entire initial value problem (including initial conditions) was transformed in terms of t and solved for Y (t), then this solution Y (t) in terms of t was transformed back to the solution y(x) in terms of x. 20. Suppose x2 y 00 + Axy 0 + By = 0 has repeated roots. Then the characteristic equation r2 + (A − 1)r + B = 0 has (1 − A)/2 as a repeated root, and we have only one solution y1 (x) = x(1−A)/2 so far. For another solution, independent from y1 , look for a solution of the form y2 (x) = u(x)y1 (x). Then y20 = u0 y1 + uy10 and y200 = u00 y1 + 2u0 y10 + uy100 .

2.4. THE EULER DIFFERENTIAL EQUATION

Substitute y2 into the differential equation to get x2 (u00 y1 + 2u0 y10 + uy100 ) + Ax(u0 y1 + uy10 ) + Buy1 = 0. Three terms in this equation cancel, because u(x2 y100 + Axy10 + By1 ) = 0 by virtue of y1 being a solution. This leaves x2 u00 y1 + 2x2 u0 y10 + Axu0 y1 = 0. Assuming that x > 0, divide by x to get xu00 y1 + 2xu0 y10 + Au0 y1 = 0. Substitute y1 (x) = x(1−A)/2 into this to obtain 00 (1−A)/2

xu x

+ 2xu

1−A 2

x(−1−A)/2 + Au0 x(1−A)/2 = 0.

Divide this by x(1−A)/2 to get xu00 + (1 − A)u0 + Au0 = 0, and this reduces to xu00 + u0 = 0. Let z = u0 to obtain xz 0 + z = 0, or (xz)0 = 0. Then xz = c, constant, so z = u0 =

c . x

Then u(x) = c ln(x) + d. We only need one second solution, so let c = 1 and d = 0 to get u(x) = ln(x). A second solution, independent from y1 (x), is y2 (x) = y1 (x) ln(x), as given without derivation in the chapter.

CHAPTER 2. SECOND-ORDER DIFFERENTIAL EQUATIONS

2.5

Series Solutions

2.5.1

Power Series Solutions

1. Put y(x) =

P∞

n=0 an x

y 0 − xy = =

into the differential equation to obtain

∞ X n=1 ∞ X

nan xn−1 − nan xn−1 −

n=1

∞ X n=0 ∞ X

an xn+1 an−2 xn−1

n=2

= a1 + (2a2 − a0 )x +

∞ X

(nan − an−2 )xn−1

n=3

= 1 − x. Then a0 is arbitrary, a1 = 1, 2a2 − a0 = −1, and an =

1 an−2 for n = 3, 4, · · · . n

This is the recurrence relation. If we set a0 = c0 + 1, we obtain the coefficients 1 1 1 a2 = c0 , a4 = c0 , a6 = c0 , 2 2·4 2·4·6 and so on. Further, a1 = 1, a3 =

1 1 1 , a5 = , a7 = 3 3·5 3·5·7

and so on. The solution can be written y(x) = 1 +

+ c0

∞ X

1 x2n+1 3 · 5 · · · 2n + 1 n=0 ! ∞ X 1 2n x . 1+ 2 · 4 · · · 2n n=1

2. Write y 0 − x3 y =

∞ X

nan xn−1 −

n=1

∞ X

an xn+3

n=0

= a1 + 2a2 x + 3a3 x2 +

∞ X

(nan − an−4 )xn−1 = 4.

n=4

The recurrence relation is an =

1 an−4 for n = 4, 5, · · · , n

2.5. SERIES SOLUTIONS

with a0 arbitrary, a1 = 4 and a2 = a3 = 0. This yields the solution ∞ X

1 x4n+1 1 · 5 · 9 · · · (4n + 1) n=0 ! ∞ X 1 4n . + a0 1 + x 4 · 8 · 12 · · · 4n n=1

y(x) = 4

3. Write y 0 + (1 − x2 )y =

∞ X

nan xn−1 +

n=1

∞ X

an xn −

n=0

∞ X

an xn+2

n=0

= (a1 + a0 ) + (2a2 + a1 )x +

∞ X

(nan + an−1 − an−3 )xn−1

n=3

= x. The recurrence relation is nan + an−1 − an−3 = 0 for n = 3, 4, · · · . Here a0 is arbitrary, a1 + a0 = 0 and 2a2 + a1 = 1. This gives us the solution 1 7 19 1 y(x) = a0 1 − x + x2 + x3 − x4 + x5 + · · · 2! 3! 4! 5! 1 2 1 3 1 4 11 5 31 6 + x − x + x + x − x + ··· . 2! 3! 4! 5! 6! 4. Begin with y 00 + 2y 0 − xy =

∞ X

n(n − 1)an xn−2 +

n=2

∞ X

2nan xn−1 +

n=1

∞ X

an xn+1

n=0

= (2a2 + 2a1 ) + (3 · 2a3 + 2 · 2a2 + a0 )x ∞ X + (n(n − 1)an + 2(n − 1)an−1 + an−2 )xn−2 = 0. n=1

The recurrence relation is n(n − 1)an + 2(n − 1)an−1 + an−2 = 0 for n = 4, 5, · · · . Further, a0 and a1 are arbitrary, a2 = −A1 and 6a3 + 4a2 + a0 = 0. Taking a0 = 1, a1 = 0, we obtain the solution 1 1 1 1 y1 (x) = 1 − x3 + x4 − x5 + x6 + · · · . 6 12 30 60

CHAPTER 2. SECOND-ORDER DIFFERENTIAL EQUATIONS With a0 = 0, a1 = 1 we get a second, linearly independent solution 2 5 7 y2 (x) = x − x2 + x3 − x4 + x5 + · · · . 3 12 60 5. Write 00

y − xy + y = = 2a2 + a0 +

∞ X

n(n − 1)n x

n−2

n=2 ∞ X

−

∞ X

nan x +

n=1

(n + 2)(n + 1)an+2 xn −

n=0

∞ X

an xn

n=0 ∞ X

nan xn +

n=1

∞ X

an xn = 3.

n=0

Here a0 and a1 are arbitrary and a2 = (3 − a0 )/2. The recurrence relation is n−1 for n = 1, 2, · · · . an+2 = (n + 2)(n + 1) This yields the general solution 3 − a0 2 3 − a0 4 x + x 2 4! 3(3 − a0 ) 6 3 · 5(3 − a0 ) 8 + x + x + ··· . 6! 8!

y(x) = a0 + a1 x +

6. Begin with y 00 + xy 0 + xy =

∞ X

n(n − 1)axn xn−2 +

n=2

= 2a2 +

∞ X

∞ X n=1

nan xn +

∞ X

an xn+1

n=0

(n(n − 1)an + (n − 2)an−2 + an−3 )xn−2 = 0.

n=3

Here a0 and a1 are arbitrary and a2 = 0. The recurrence relation is an = −

(n − 2)an−2 − an−3 for n = 3, 4, · · · . n(n − 1)

With a0 = 1 and a1 = 0, we obtain one solution 2 3 y1 (x) = 1 − x3 + x5 3 2·3·4·5 1 3·5 + x6 − x7 + · · · . 2·3·5·6 2·3·4·5·6·7 With a0 = 0 and a1 = 1, we obtain a second, linearly independent solution 1 3 1 4 x − x 2·3 3·4 3 3·5 + x5 + x6 + · · · . 2·3·4·5 2·3·5·6

y2 (x) = x −

2.5. SERIES SOLUTIONS

7. We have ∞ X

y 00 − x2 y 0 + 2y = −

∞ X

n(n − 1)an xn−2

n=2 ∞ X

nan xn+1 +

n=1

2an xn

n=0

= 2a2 + 2a0 + (6a3 + 2a1 )x ∞ X + (n(n − 1)an − (n − 3)an−3 + 2an−2 )xn−2 = x. n=1

Then a0 and a1 are arbitrary, a2 = −a0 , and 6a3 +2a1 = 1. The recurrence relation is (n − 3)an−3 − 2an−2 an = n(n − 1) for n = 4, 5, · · · . The general solution has the form 1 4 1 5 1 6 2 y(x) = a0 1 − x + x − x − x + · · · 6 10 90 1 3 1 4 1 5 7 6 + a1 x − x + x + z − x + ··· 3 12 30 180 1 1 1 1 7 1 8 + x3 − x5 + x6 + x − x + ··· . 6 6 60 1260 480 Note that a0 = y(0) and a1 = y 0 (0). The third series represents the solution obtained subject to y(0) = y 0 (0) = 0. 8. Using the Maclaurin expansion for cos(x), we have y 0 + xy =

∞ X

nan xn − 1 +

n=1

= a1 +

∞ X

an xn+1

n=0 ∞ X

(2na2n + a2n−2 )x2n−1

n=0

∞ X

((2n + 1)a2n+1 + a2n−1 )x2n

n=1

= cos(x) =

∞ X (−1)n n=0

(2n)!

x2n .

a0 is arbitrary and a1 = 1. The recurrence relation is a2n = −

1 −a2n−1 + (−1)n /((2n)!) a2n−2 and a2n+1 = 2n 2n + 1

CHAPTER 2. SECOND-ORDER DIFFERENTIAL EQUATIONS for n = 1, 2, · · · . This yields the solution 1 2 1 4 2 · 4 · 66 y(x) = a0 1 − x + x − + ··· 2 2·4 x 3 3 13 5 79 7 633 9 + x− x + x − x + x + ··· . 3! 5! 7! 9! 9. We have 00

y + (1 − x)y + 2y =

∞ X

n(n − 1)an xn−2

n=2

∞ X

nan xn−1 −

n=1

∞ X

nan xn + 2

n=1

= (2a2 + a1 + 2a0 ) +

∞ X

an xn

n=0 ∞ X

(n(n − 1)an + (n − 1)an−1 − (n − 4)a2n−2 )xn−2

n=3

= 1 − x2 . Then a0 and a1 are arbitrary, 2a2 + a1 + 2a0 = 1, 6a3 + 2a2 + a1 = 0, and 12a4 + 3a3 = −1. The recurrence relation is an =

−(n − 1)an−1 + an−4 an−2 n(n − 1)

for n = 5, 6, · · · . The general solution is 1 4 1 5 1 3 2 y(x) = a0 1 − x + x − x + x − · · · 3 12 30 1 1 1 1 1 6 1 7 + a1 x − x2 + x2 − x3 − x4 − x + x + ··· . 2 2 6 24 360 2520 Here a0 = y(0) and a1 = y 0 (0). 10. Using the Maclaurin expansion of ex , we have y 00 + xy 0 =

∞ X

n(n − 1)an xn−2 +

n=2

∞ X

nan xn

n=1

= 2a2 +

∞ X

(n(n − 1)an + (n − 2)an−2 )xn−2

n=3

=−

∞ X

1 xn−2 . (n − 2)! n=3

Then a0 and a1 are arbitrary, a2 = 0 and an =

−(n − 2)an−2 − 1/(n − 2)! n(n − 1)

2.5. SERIES SOLUTIONS

for n = 3, 4, · · · . This leads to the solution 1 3 3 5 15 7 105 9 y(x) = a0 + a1 x − x + x − x + x + ··· 3! 5! 7! 9! 1 3 1 4 2 5 3 6 11 7 19 8 + − x − x + x + x − x + x + ··· . 3! 4! 5! 6! 7! 8! Note that a0 = y(0) and a1 = y 0 (0).

2.5.2

Frobenius Solutions

1. Substitute y(x) =

P∞

n=0 cn x

n+r

xy 00 + (1 − x)y 0 + y =

into the differential equation to get

∞ X

(n + r)(n + r − 1)cn xn+r−2

n=0

∞ X

(n + r)cn xn+r−1 −

n=0

∞ X

(n + r)cn xn+r +

n=0

= r2 c0 xr−1 +

∞ X

cn xn+r

n=0

((n + r)2 cn − (n + r − 2)cn−1 )xn+r−1

n=1

= 0. Because c0 is assumed to be nonzero, r must satisfy the indicial equation r2 = 0, so r1 = r2 = 0. One solution has the form y1 (x) =

∞ X

cn xn ,

n=0

while a second solution has the form y2 (x) = y1 (x) ln(x) +

∞ X

c∗n xn .

n=0

For the first solution, choose the coefficients to satisfy c0 = 1 and cn =

n−2 cn−1 for n = 1, 2, · · · . n2

This yields the solution y1 (x) = 1 − x. The second solution is therefore y2 (x) = (1 − x) ln(x) +

∞ X

c∗n xn .

n=0

CHAPTER 2. SECOND-ORDER DIFFERENTIAL EQUATIONS Substitute this into the differential equation to obtain 2 1−x 1−x x − − + (1 − x) − ln(x) + x x2 x ∞ ∞ X X + (1 − x) ln(x) + n(n − 1)c∗n xn−1 + (1 − x) c∗n xn−1 n=2

∞ X

n=1

c∗n xn

n=1

= (−3 + c∗1 ) + (1 + 4c∗2 )x +

∞ X

(n2 c∗n − (n − 2)c∗n−1 )xn−2

n=3

= 0. The coefficients are determined by c∗1 = 3, c∗2 = −1/4, and c∗n =

n−2 for n = 3, 4, · · · . n2

A second solution is y2 (x) = (1 − x) ln(x) + 3x −

∞ X

1 xn . n(n − 1) n=2

2. Omitting some routine details, the indicial equation is r(r − 1) = 0, so r1 = 1 and r2 = 0. There are solutions y1 (x) =

∞ X

cn xn+1 and y2 (x) = ky1 (x) ln(x)

n=0

∞ X

c∗n xn .

n=0

For y1 , the recurrence relation is cn =

2(n + r − 2) cn−1 (n + r)(n + r − 1)

for n = 1, 2, · · · . With r = 1 and c0 = 1, this yields y1 (x) = x, a solution that can be seen by inspection from the differential equation. For the second solution, substitute y2 (x) into the differential equation to get (2c∗0 + k) + 2(c∗2 − k)x ∞ X + (n(n − 1)c∗n − 2(n − 2)c∗n−1 )xn−1 = 0. n=1

2.5. SERIES SOLUTIONS

Choose c∗0 = 1 to obtain k = −2. c∗1 is arbitrary, and we will take c∗1 = 0. Finally, c∗2 = −2 and c∗n =

2(n − 2) ∗ c for n = 3, 4, · · · . n(n − 1) n−1

This yields the second solution y2 (x) = −2x ln(x) + 1 −

∞ X

2n xn . n!(n − 1) n=2

3. The indicial equation is r2 − 4r = 0, so r1 = 4 and r2 = 0. There are solutions of the form y1 (x) =

∞ X

cn x

n+4

and y2 (x) = ky1 (x) ln(x) +

n=0

∞ X

c∗n xn .

n=0

With r = 4 the recurrence relation is n+1 cn−1 for n = 1, 2, · · · . cn = n Then y1 (x) = x4 (1 + 2x + 3x2 + 4x3 + · · · ). Using the geometric series, we can observe that d (1 + x + x2 + x3 + · · · ) dx d 1 x4 = x4 . = dx 1 − x (1 − x)2

y1 (x) = x4

This gives us the second solution y2 (x) =

3 − 4x . (1 − x)2

4. The indicial equation is 4r2 − 9 = 0, with roots r1 = 3/2 and r2 = −3/2. There are solutions y1 (x) =

∞ X

cn xn+3/2 and y2 (x) = ky1 (x) ln(x) +

n=0

∞ X

c∗n xn−3/2 .

n=0

Upon substituting these into the differential equation, we obtain " # ∞ n X (−1) y1 (x) = x3/2 1 + x2n n n!(5 · 7 · 9 · · · (2n + 3)) 2 n=1 and

" y2 (x) = x

−3/2

∞ X

# (−1)n+1 2n 1+ x . 2n+1 n!(3) · · · (2n − 3) n=1

CHAPTER 2. SECOND-ORDER DIFFERENTIAL EQUATIONS 5. The indicial equation is 4r2 − 2r = 0, with roots r1 = 1/2 and r2 = 0. There are solutions of the form ∞ X

y1 (x) =

cn xn+1/2 and y2 (x) =

n=0

∞ X

c∗n xn .

n=0

Substitute these into the differential equation to get # " ∞ X (−1)n n 1/2 x y1 (x) = x 1+ 2n n!(3 · 5 · 7 · · · (2n + 1)) n=1 1 1 2 1 3 1 = x1/2 1 − x + x − x + x4 + · · · 6 120 5040 362880 and y2 (x) = 1 +

∞ X

(−1)n xn n n!(1 · 3 · 5 · · · (2n − 1)) 2 n=1

1 1 3 1 1 x + x4 + · · · . = 1 − x + x2 − 2 24 720 40320 6. The indicial equation is 4r2 − 1 = 0, with roots r1 = 1/2 and r2 = −1/2. There are solutions y1 (x) =

∞ X

cn x

n+1/2

and y2 (x) = ky1 (x) ln(x) +

∞ X

c∗n xn−1/2 .

n=0

After substituting these into the differential equation, we obtain the simple solutions y1 (x) = x1/2 and y2 (x) = x−1/2 . These solutions are consistent with the observation that, upon division by 4, the differential equation is an Euler equation. 7. The indicial equation is r2 − 3r + 2 = 0, with roots r1 = 2 and r2 = 1. There are solutions y(x) =

∞ X n=0

cn xn+2 and

∞ X

c∗n xn−2 .

n=0

Substitute these in turn into the differential equation to obtain the solutions 1 1 1 y1 (x) = x2 + x4 + x6 + x8 + · · · 3! 5! 7! and 1 1 1 y2 (x) = x − x2 + x3 − x4 + x5 − · · · . 2! 3! 4! We can recognize these series as y1 (x) = x sinh(x) and y2 (x) = xe−x .

2.5. SERIES SOLUTIONS

8. The indicial equation is r2 − 2r = 0, with roots r1 = 2, r2 = 0. There are solutions y1 (x) =

∞ X

cn xn+2 and y2 (x) = ky1 (x) ln(x) +

n=0

∞ X

c∗n xn .

n=0

The recurrence relation for the c0n s is cn =

−2 for n = 3, 4, · · · n(n − 2)

and we obtain, with c0 = 1, y1 (x) =

∞ X (−1)n 2n+1 n=0

n(n + 2)

xn+2

2 1 1 1 6 = x2 − x3 + x4 − x5 + x + ··· . 3 6 45 540 For the second solution, substitute y2 (x) into the differential equation to get 2c∗0 − c∗1 +

∞ X (−1)n 2n k n(n − 2)c∗n + c∗n−1 + xn−1 = 0. 2 n((n − 2)!) n=1

Setting c∗0 = 1 for simplicity, we obtain c∗1 = 2, k = −2, c∗2 arbitrary (we take this to be zero), and the recurrence relation 1 (−1)n 2n+1 ∗ ∗ cn = − 2cn−1 + n(n − 2) n((n − 2)!)2 for n = 3, 4, · · · . We obtain the second solution y2 (x) = −2y1 ln(x) + 1 + 2x +

157 6 16 3 25 4 x − x + x − ··· . 9 36 1350

9. The indicial equation is 2r2 = 0, with roots r1 = r2 = 0. There are solutions y1 (x) =

∞ X

cn xn and y2 (x) = y1 (x) ln(x) +

n=0

∞ X

c∗n xn .

n=1

Upon substituting these into the differential equation, we obtain the independent solutions y1 (x) = 1 − x and

y2 (x) = (1 − x) ln

x x−2

− 2.

CHAPTER 2. SECOND-ORDER DIFFERENTIAL EQUATIONS

10. The indicial equation is r2 − 1 = 0, with roots r1 = 1 and r2 = −1. There are solutions y1 (x) =

∞ X n=0

cn xn+1 and y2 (x) = ky1 (x) ln(x) +

∞ X

c∗n xn−1 .

n=0

Substitute each of these into the differential equation to get # " ∞ X (−1)n (1 · 4 · 7 · · · (3n − 2)) 3n x y1 (x) = x 1 + 3n n!(5 · 8 · 11 · · · (3n + 2)) n=1 and

" # ∞ X 1 (−1)n+1 (1 · 2 · 5 · · · (3n − 1)) 3n y2 (x) = 1+ x . x 3n n!(4 · 7 · · · (3n − 2)) n=1

Chapter 3

The Laplace Transform 3.1

Definition and Notation

1. F (s) = 2.

3(s2 − 4) (s2 + 4)2

G(s) =

8 (s + 4)2 + 64

H(s) =

14 7 − 2 s2 s + 49

4. W (s) =

s s2 + 9

5. K(s) = −

−

s s2 + 49

10 3 + 2 (s + 4)2 s +9

6. r(t) =

7 sinh(3t) 3

7. q(t) = cos(8t) 8.

√ √ 5 g(t) = √ sin( 12t) − 4 cos( 8t) 12 1 p(t) = e−42t − t3 e−3t 6 69

CHAPTER 3. THE LAPLACE TRANSFORM

10.

5 f (t) = − t sin(t) 2

11. (a) From the definition, Z R

e−st f (t) dt.

F (s) = L[f ](s) = lim

R→∞

For each R, let N be the largest integer such that (N + 1)T ≤ R to write Z R

e−st f (t) dt =

N Z (n+1)T X n=0

e−st f (t) dt +

Z R

e−st f (t) dt.

(N +1)T

By choosing R sufficiently large, we can make the last integral on the right as small as we like. Further, R → ∞ as N → ∞, so Z ∞ ∞ Z (n+1)T X −st e f (t) dt = e−st f (t) dt. 0

n=0

(b) Use the periodicity of f (t) and the change of variables u = t − nT to write Z (n+1)T Z T −st e f (t) dt = e−s(u+nT ) f (u + nT ) dt nT

= e−snT

Z T

e−su f (u) du,

because f (u + nT ) = f (u). (c) Use the results of (a) and (b) to write ∞ Z (n+1)T X L[f ](s) = e−st f (t) dt =

n=0 ∞ X

e−snT

n=0 "∞ X

Z T f (t) dt 0

e−snT

e−st f (t) dt,

n=0

because the summation is independent of t. (d) For s > 0, 0 < e−st < 1 and we can use the geometric series to obtain ∞ X

e−snt =

∞ X

e−sT

n=0

L[f ](s) =

1 1 − e−sT

Therefore

Z T

1 . 1 − e−sT

e−st f (t) dt.

3.1. DEFINITION AND NOTATION

12. f has period T = 6, and Z 6 Z 3 5 e−st f (t) dx = 5e−st = (1 − e−3s ). s 0 0 Then 5 1 − e−3s s 1 − e−6s 5 1 − e−3s = −3s s (1 − e )(1 + e−3s ) 5 = . s(1 − e−3s )

L[F ](s) =

13. f has period T = π/ω and Z T

e−st f (t) dt =

Z π/ω

Ee−st sin(ωt) dt

Eω (1 + e−πs/ω ) . s2 + ω 2 1 − e−πs/ω

14. f has period T = 25 and, from the graph,   0 for 0 < t ≤ 5, f (t) = 5 for 5 < t ≤ 10,   0 for 10 < t ≤ 25. Now, Z 25

e−st f (t) dt =

Z 10

Then L[f ](s) =

5e−st dt =

5 −5s e (1 − e−5s ). s

5e−5s (1 − e−5s ) . s(1 − e−25s )

15. f has period 6 and f (t) = t/3 for 0 ≤ t < 6. Compute Z T 0

e−st f (t) dt =

Z 6 0

1 −st 1 te dt = 2 (1 − 6se−6s − e−6s ). 3 3s

Then L[f ](s) =

1 1 − 6se−6s − e−6s . 3s2 1 − e−6s

16. f is periodic with period 8, and ( 3t/2 for 0 < t < 2, f (t) = 0 for 2 ≤ t ≤ 8.

CHAPTER 3. THE LAPLACE TRANSFORM Then Z 8 1 e−st f (t) dt 1 − e−8s 0 3 1 − 2e−2s − e−2s . = 2 2s 1 − e−8s

L[f ](s) =

17. Here

( h for 0 < t ≤ a, f (t) = 0 for a < t ≤ 2a,

with period 2a. Then Z 2a Z a h e−st f (t) dt = he−st dt = (1 − e−as ). s 0 0 Then L[f ](s) =

h 1 − e−as . s 1 − e−2as

18. T = 2a and ( ht/a for 0 ≤ t < a, f (t) = −h(t − 2a)/a for a < t ≤ 2a. Then h (1 − e−as )2 as2 1 − e−2as h 1 − e−as = 2 . as 1 + e−as

L[f ](s) =

This can also be written in terms of the hyperbolic tangent function: as h L[f ](s) = 2 tanh . as 2

3.2

Solution of Initial Value Problems

In many of these problems a partial fractions decomposition is used to find the inverse transform of the transform of the solution (hence find the solution). Partial fractions are reviewed in a web module. 1. Transform the differential equation, using the operational formula, to obtain 1 sY (s) − y(0) + 4Y (s) = . s With y(0) = 3, this is 1 sY − 3 + 4Y = . s

3.2. SOLUTION OF INITIAL VALUE PROBLEMS Then Y (s) =

1 1 1 − 3s −3 = . s+4 s s(s + 4)

Decompose this into a sum of simpler fractions: Y (s) =

A B + . s s+4

It these fractions are added, the numerator must equal the numerator of the original fraction, 1 − 3s: A(s + 4) + Bs = 1 − 3s. Then (A + B)s + 4A = 1 − 3s. Matching coefficients of like powers of x, this requires that A + B = −3 and 4A = 1. Then A = 1/4 and B = −13/4. Now Y (s) =

1 1 13 1 − . 4s 4 s+4

Now we immediately read from Table 3.1 that y(t) = L−1 [Y ](t) =

1 13 −4t − e . 4 4

This is the solution of the initial value problem. 2. The transform of the differential equation is sY (s) − 5 − 9Y (s) =

1 , s2

in which we have incorporated the initial condition y(0) = 5. Then Y (s) =

1 + 5s2 . s2 (s − 9)

Use a partial fractions decomposition to write this as Y (s) =

A B C + 2+ . s s s−9

The numerator of the sum of these fractions must equal 1 + 5s2 , so As(s − 9) + B(s − 9) + Cs2 = 1 + 5s2 .

CHAPTER 3. THE LAPLACE TRANSFORM Equate coefficients of like powers of x on both sides of this equation to get A + C = 5, −9A + B = 0, −9B = 1. Then A = −1/81, B = −1/9 and C = 406/81, so Y (s) = −

1 1 1 1 406 1 + − . 81 s 9 s2 81 s − 9

From this we read the inverses to obtain 1 1 406 9t y(t) = − − t + e . 81 9 81 3. Take the transform of the differential equation and insert the initial data and solve for Y (s) to get s 1 . Y (s) = s + 4 s2 + 1 Use a partial fractions decomposition to obtain Y (s) = −

4 1 1 4s + 1 + . 17 s + 4 17 s2 + 1

The inverse of this is the solution: 4 4 1 y(t) = − e−4t + cos(t) + sin(t). 17 17 17 4. Take the transform of the differential equation to get

Then

sY − 1 + 2Y =

1 . s+1

1 Y (s) = s+2

1 +1 s+1

1 . s+1

The solution is y(t) = e−t . 5. The transform of the initial value problem is sY − 4 − 2Y =

1 1 − . s s2

Then 1 1 1 Y (s) = − +4 s − 2 s s2 1 1 1 1 17 1 = − + . 2 s2 4s 4 s−2 The solution is y(t) =

1 1 17 t − + e2t . 2 4 4

3.2. SOLUTION OF INITIAL VALUE PROBLEMS

6. Transform the differential equation, using the operational formula for the second derivative to get s2 Y (s) − sy(0) − y 0 (0) + Y (s) =

1 . s

Inserting the initial conditions, we have s2 Y − 6s + Y =

1 . s

Then 1 Y (s) = 2 s +1

1 + 6s s

1 = +5 s

s 2 s +1

The solution is the inverse of this expression: y(t) = 1 + 5 cos(t). 7. Transform the differential equation to obtain s2 Y − sy(0) − y 0 (0) − 4(sY − y(0)) + 4Y =

s . s2 + 1

Insert the initial conditions to get (s − 2)2 Y =

s + s − 5. s2 + 1

Then Y (s) =

s (s2 + 1)(s − 2)2

s−5 . (s − 2)2

With a some manipulation, write this as Y (s) =

s3 − 5s2 + 2s − 5 . (s2 + 1)(s − 2)2

Expand this in partial fractions: Y (s) =

As + B C D + + . s2 + 1 s − 2 (s − 2)2

To determine the coefficients, we need the numerator in the sum of these fractions to equal the numerator in Y (s): (As + B)(s − 2)2 + C(s − 2)(s2 + 1) + D(s2 + 1) = (A + C)s3 + (−4A + B − 2C + D)s2 + (4A − 4B + C)s + (4B − 2C + D) = s3 − 5s2 + 2s − 5.

CHAPTER 3. THE LAPLACE TRANSFORM Matching coefficients of powers of s, we obtain: A + C = 1, −4A + B − 2C + D = −5, 4A − 4B + C = 2, 4B − 2C + D = −5. Then A=

3 4 22 13 ,B = − ,C = ,D = − . 25 25 25 5

Then Y (s) =

s 4 1 22 1 13 1 3 . − + − 25 s2 + 1 25 s2 + 1 25 s − 2 5 (s − 2)2

These terms are easily inverted to obtain the solution y(t) =

3 4 22 13 cos(t) − sin(t) + e2t − te2t . 25 25 25 5

8. Transform the differential equation to obtain 1 2 Y (s) = 2 s + 9 s3 2 1 2 1 2 s = − + . 3 2 9s 81 s 81 s + 9 The solution is y(t) =

2 2 1 2 t − + cos(3t). 9 81 81

9. Upon transforming the differential equation and inserting the initial conditions, we have 1 1 1 Y (s) = 2 + − 2s + 1 s + 16 s s2 15 1 1 1 1 33 s 1 − − . = + 16 s2 16 s 16 s2 + 16 64 s2 + 16 The solution is y(t) =

1 33 15 (1 + t) − cos(4t) + sin(4t). 16 16 64

10. Transforming the differential equation, we obtain 1 1 Y (s) = − 10 (s − 2)(s − 3) s + 1 29 1 39 1 1 1 = − + . 3 s−2 4 s − 3 12 s + 1 The solution is y(t) =

29 2t 39 3t 1 e − e + e−t . 3 4 12

3.3. THE HEAVISIDE FUNCTION AND SHIFTING THEOREMS

11. Begin with the definition of the Laplace transform and integrate by parts: Z ∞ 0 L[f ](s) = e−st f 0 (t) dt 0 Z ∞ b = e−st f (t) a − −se−st f (t) dt Z ∞0 e−st f (t) dt = −f (0) + s 0

= sF (s) − f (0). 12. Apply the definition of the transform to f 00 (t) and then use equation (3.1) to obtain Z ∞ L[f 00 ](s) = e−st f 00 (t) dt 0 Z ∞ −st ∞ =e − −se−st f 0 (t) dt 0 0

= −f 0 (0) + s(sF (s) − f (0)) = s2 F (s) − sf (0) − f 0 (0).

3.3

The Heaviside Function and Shifting Theorems

In each of Problems 1–15, we will indicate shifting f (t) by a, replacing t with t − a, by writing [f (t)]t→t−a . Similarly, if we replace s with s − a in the transform F (s) of f (t), we will write [F (s)]s→s−a or sometimes L[f (t)]s→s−a . This notation is sometimes useful in applying a shifting theorem or inverse shifting theorem. 1. Apply the shifting theorem: L[(t3 − 3t + 2)e−2t ](s) = L[t3 − 3t + 2]s→s+2 3 2 6 − + . = 4 2 (s + 2) (s + 2) s+2 2. We know that L[t](s) =

1 2 and L[−2](s) = − . 2 s s

CHAPTER 3. THE LAPLACE TRANSFORM Use the shifting theorem to obtain L[te−3t − 2e−3t ](s) = L[te−3t ](s) − 2L[−2e−3t ](s) 2 1 − = s2 s→s+3 s s→s+3 2 1 − = . (s + 3)2 s+3 3. First write f (t) = [1 − H(t − 7)] + H(t − 7) cos(t) = [1 − H(t − 7)] + H(t − 7) cos((t − 7) + 7) = [1 − H(t − 7)] + cos(7)H(t − 7) cos(t − 7) − sin(7)H(t − 7) sin(t − 7). Then L[f ](s) =

1 s 1 1 − e−7s + 2 cos(7)e−7s − 2 sin(7)e−7s . s s +1 s +1

1 s − s2 s→s+4 s2 + 1 s→s+4 1 s+4 = − . 2 (s + 4) (s + 4)2 + 1

L[f ](s) =

5. First, write the function as f (t) = t + (1 − 4t)H(t − 3) = t + (1 − 4(t − 3) + 3)H(t − 3) = t − 11H(t − 3) − 4(t − 3)H(t − 3). Then L[f ](s) =

11 4 1 − e−3s − 2 e−3s . s2 s s

6. First, write f (t) = [2(t − π) + 2π + sin(t − π)][1 − H(t − π)]. Then L[f ](s) =

2 1 2 2π −πs s − 2 − e−πs − e − 2 e−πs . s2 s + 1 s2 s s +1

7. Replace s with s + 1 in the transform of 1 − t2 + sin(t) to get L[f ](s) =

1 2 1 − + . 2 s + 1 (s + 1) (s + 1)2 + 1

3.3. THE HEAVISIDE FUNCTION AND SHIFTING THEOREMS

8. In terms of the Heaviside function, f (t) = t2 + (1 − t − 4t2 )H(t − 2) = t2 + [1 − (t − 2) − 2 − 4((t − 2) + 2)2 ]H(t − 2) = t2 − [17 + 17(t − 2) + 4(t − 2)2 ]H(t − 2). Then L[f ](s) =

2 − s3

17 17 8 + 2 + 3 s s s

e−2s .

9. First, write f (t) = (1 − H(t − 2π)) cos(t) + H(t − 2π)(2 − sin(t)). Then l[f ](s) =

s + 2 s +1

2 s 1 − 2 − 2 s s +1 s +1

e−2πs .

10. First write f (t) = −4(1 − H(t − 1)) + e−t H(t − 3) = −4 + 4H(t − 1) + e−3 e−(t−3) H(t − 3). Then

4 4 e−3 −3s L[f ](s) = − + e−s + e . s s s+1

11. Because L[t cos(t)](s) =

s2 − 9 , (s2 + 9)2

we obtain the transform of te−t cos(t) by replacing s with s + 1: L[te−t cos(t)](s) =

(s + 1)2 − 9 . ((s + 1)2 + 9)2

12. The transform of 1 − cosh(t) is 1 s − 2 , s s −1 then L[et (1 − cosh(t))](s) =

1 s−1 − . s − 1 (s − 1)2 − 1

This can be written L[et (1 − cosh(t))](s) =

1 s−1 − . s − 1 s(s − 2)

CHAPTER 3. THE LAPLACE TRANSFORM

13. First, put f (t) in terms of the Heaviside function as f (t) = (1 − H(t − 16))(t − 2) − H(t − 16) = t − 2 + (1 − t)H(t − 16). Then L[f ](s) =

1 2 − + s2 s

1 1 − s s2

e−16s .

14. In terms of the Heaviside function, f (t) = 1 − cos(2t) − (1 − cos(2t))H(t − 3π). Then

F (s) =

1 s − s s2 + 1

(1 − e−3πs ).

15. Replace s with s + 5 in the transform of t4 + 2t2 + 1 to obtain F (s) =

24 4 1 + + . 5 3 (s + 5) (s + 5) (s + 5)2

16. Notice that

1 . (s + 2)2 + 8 √ √ This is the transform of (1/2 2) sin(2 2t), with s replaced by s + 2. Therefore, √ 1 f (t) = √ e−2t sin(2 2t). 2 2 F (s) =

17. Write F (s) =

1 . (s − 2)2 + 1

This is the transform of sin(t) with s replaced by s − 2. Therefore f (t) = e2t sin(t). 18. −1

−5s e 1 −1 (t) = L 3 s s3 t→t−5 1 = (t − 5)2 H(t − 5). 2

19. Because 3/(s2 + 9) is the transform of sin(3t), then f (t) =

1 sin(3(t − 2))H(t − 2). 3

3.3. THE HEAVISIDE FUNCTION AND SHIFTING THEOREMS

20. Because the transform of 3e−2t is 3/(s + 2), then f (t) = e−2(t−4) H(−4). 21. We recognize that F (s) = so

1 (s + 3)2 − 2

√ 1 f (t) = √ e−3t sinh( 2t). 2

22. Write F (s) = to obtain

s−4 (s − 4)2 − 6

√ f (t) = e4t cosh( 6t).

23. Write F (s) =

(s + 3) − 1 (s + 3)2 − 8

to obtain √ √ 1 f (t) = e−3t cosh(2 2t) − √ e−3t sinh(2 2t). 2 2 24. Put a = 1 and F (s) = 1/(s − 5) in the second shifting theorem. Then f (t) = e5t and 1 −s −1 L e (t) = e5(t−1) H(t − 1). s−5 25. First use a partial fractions decomposition to write 1 s(s2 + 16)

1 1 1 s . − 2 16 s 16 s + 16

From this, f (t) =

1 (1 − cos(4(t − 21)))H(t − 21). 16

26. By the first shifting theorem, Z t L e−2t e2w cos(3w) dw = F (s + 2), 0

where

Z t F (s) = L

e2w cos(3w) dw .

CHAPTER 3. THE LAPLACE TRANSFORM Now d dt

Z t

e2w cos(3w) dw = e2t cos(3t).

By the operational rule for the Laplace transform, applied to the case of a first derivative, we have Z d t 2w e cos(3w) , dw = L e2t cos(3t) (s) dt 0 Z t e2w cos(3w) dw (s) = sF (s). =L

Then F (s) =

1 1 s−2 L[e2t cos(3t)](s) = . s s (s − 2)2 + 9

Therefore Z t −2t 2w L e e cos(3w) dw = 0

s . (s + 2)(s2 + 9)

27. The initial value problem is y 00 + 4y = 3H(t − 4); y(0) = 1, y 0 (0) = 0. This transforms to Y (s) =

s s 3 1 − 2 e−4s + 2 . 4 s s +4 s +4

Invert this to obtain the solution 3 y(t) = cos(2t) + (1 − cos(2(t − 4)))H(t − 4). 4 28. The problem is y 00 − 2y 0 − 3y = 12H(t − 4); y(0) = 1, y 0 (0) = 0. Transform this problem and solve for Y (s) to obtain Y (s) =

1 1 3 1 3 4 −4s + + + − e . 4 s−3 s+1 s−3 s+1 s

Invert this to obtain the solution y(t) =

1 3t (e + 3e−t ) + (e3(t−4) + 3e−(t−4) − 4)H(t − 4). 4

3.3. THE HEAVISIDE FUNCTION AND SHIFTING THEOREMS

29. The problem is y (3) − 8y 0 = 2H(t − 6); y(0) = y 0 (0) = 0. The transform of the problem is 1 1 1 1 s Y (s) = − + + e−6s . 4s 12 s − 2 6 s2 + 2s − 4 Invert this to obtain the solution √ 1 −2(t−6) 1 −(t−6) 1 + e cos( 3(t − 6)) H(t − 6). y(t) = − + e 4 12 6 30. The problem is y 00 + 5y 0 + 6y = −2(1 − H(t − 3)); y(0) = y 0 (0) = 0. Transform this problem to obtain 2 1 1 1 − − H(t − 3). Y (s) = s + 2 3 s + 3 3s The inverse of this is the solution 1 2 2 y(t) = e−2t − e−3t − − e−2(t−3) − e−3(t−3) H(t − 3). 3 3 3 31. The problem is y (3) − y 00 + 4y 0 − 4y = 1 + H(t − 5); y(0) = y 0 (0) = 0, y 00 = 1. Transform this to obtain 1 2 1 3 s 2 1 Y (s) = − + − − (1 − e−5s ). 4s 5 s − 1 20 s2 + 4 5 s2 + 4 Invert this for the solution 1 2 3 1 y(t) = − + et − cos(2t) − sin(2t) 4 5 20 5 1 2 t−5 3 1 − cos(2(t − 5)) − sin(2(t − 5)) H(t − 5). − − + e 4 5 20 5 32. The initial value problem is y 00 − 4y 0 + 4y = t + 2H(t − 3); y(0) = −2, y 0 (0) = 1. Take the transform of the problem to obtain 1 1 9 1 43 1 + 2− − 4s 4s 4s−2 4 (s − 2)2 1 1 1 1 + − + e−3s . 2s 2 s − 2 (s − 2)2

Y (s) =

CHAPTER 3. THE LAPLACE TRANSFORM The solution is 1 1 9 43 + t − e2t − te2t 4 4 4 4 1 1 2(t−3) + − e + (t − 3)e2(t−3) H(t − 3). 2 2

y(t) =

33. The current i(t) is modeled by Li0 + Ri = k(1 − H(t − 5)); i(0) = 0. Transform the problem and solve for I(s) to get k (1 − e−5s ) Ls + R 1 k 1 − (1 − e−5s ). = R s s + R/L

I(s) =

Invert this to obtain the solution for the current: k k i(t) = (1 − e−Rt/L ) − (1 − e−R(t−5)/L )H(t − 5). R R In Problems 34–38, the Heaviside formula is used to compute an inverse transform. One way to use this formula efficiently is to begin with F (s), which has the form p(s) F (s) = . (s − a1 )(s − a2 ) · · · (s − an ) For the first term, cover up the s − a1 factor and evaluate the resulting quotient at a1 to get the coefficient of ea1 t . Then cover up the s − a2 factor and evaluate the resulting quotient at a2 for the coefficient of ea2 t . Continue this through the n simple zeros of the denominator to obtain f (t). 34 . Write F (s) =

s+1 , (s − 1)(s − 2)

so a1 = 1 and a2 = 2. Then 2 t 3 2t e + e −1 1 = −2et + 3e2t .

f (t) =

35. With

s2 (s − 1)(s − 2)(s + 5) we have a1 = 1, a2 = 2 and a3 = −5. Then F (s) =

12 22 2t 52 et + e + e5t (−1)(6) (1)(7) (−6)(−7) 1 4 25 = et + e2t + e5t . 6 7 42

f (t) =

3.3. THE HEAVISIDE FUNCTION AND SHIFTING THEOREMS 36. With F (s) = then f (t) =

s+4 , (s + 3i)(s − 3i)

4 + 3i 3it 4 − 3i −3it e + e . 6i −6i

This is correct, but we can write the inverse in terms of trigonometric functions using Euler’s formula: 4 + 3i 4 − 3i (cos(3t) + i sin(3t)) + (cos(3t) − i sin(3t)) 6i −6i 1 [4 cos(3t) + 3i sin(3t) + 4i sin(3t) − 3 sin(3t) − 4 cos(3t) + 3i cos(3t) + 4i sin(3t) + 3 sin(3t)] = 6i 4 = cos(3t) + sin(3t). 3

f (t) =

37. Here F (s) =

s2 + 2s − 1 . (s − 3)(s − 5)(s + 8)

Then 14 34 5t 47 e3t + e + e−3t (−2)(11) (2)(3) (−11)(−13) 17 47 −3t 7 e . = − e3t + e5t + 11 13 143

f (t) =

38. Now F (s) =

s . (s − 2i)(s + 2i)(s − 3i)(s + 4i)

Using Heaviside’s formula and Euler’s formula, as in Problem 36, we obtain f (t) =

1 1 cos(2t) − cos(3t). 5 5

39. Write (s − aj )

p(s) p(s) = q(s) (q(s) − q(aj )/(s − aj ))

and take the limit as s → aj . Finally, use the fact that lim

s→aj

q(s) − q(aj ) = q 0 (aj ). s − aj

40. We will find A1 . The process for each Aj is the same. First observe that (s − a1 )

p(s) s − a1 s − a1 = A1 + A2 + · · · + An , q(s) s − a2 s − an

CHAPTER 3. THE LAPLACE TRANSFORM in which we use the fact that the zeros of q(s) are simple, so a1 6= aj for j = 2, · · · , n. In the limit as s → a1 , the right side of this equation approaches A1 , because all other terms have limit zero. But in this limit, (s − a1 )p(s)/q(s) is exactly the quotient of p(s) with the polynomial q1 (s) obtained from q(s) by deleting the factor s − a1 . This yields Heaviside’s formula.

3.4

Convolution

1. Let F (s) =

1 s2 + 4

and G(s) =

1 s2 − 4

Then L−1 [F ](t) =

1 1 sin(2t) and L−1 [G](t) = sinh(2t). 2 2

By the convolution theorem, 1 1 sin(2t) ∗ sinh(2t) 2 2 Z 1 t = sin(2(t − τ )) sinh(τ ) dτ 4 0 1 t = [sin(2(t − τ )) cosh(2τ ) + cos(2(t − τ )) sinh(2τ )]0 16 1 = (sinh(2t) − sin(2t)). 16

L−1 [F (s)G(s)](t) =

2. Choose F (s) =

s 1 and G(s) = e−2s . s2 + 16 s

Then L[F (s)G(s)](t) = cos(4t) ∗ H(t − 2) Z t = cos(4(t − τ ))H(τ − 2) dτ (0 0 for t < 2, = Rt cos(4(t − τ )) dτ for t ≥ 2. 2 Upon carrying out the last integration, we have L−1

e−2s 1 (s) = sin(4(t − 2))H(t − 2). 2 s + 16 4

3.4. CONVOLUTION

3. There are two cases. Suppose first that a2 6= b2 . Then 1 1 s (t) = cos(at) ∗ sin(bt) L−1 2 2 2 2 s +a s +b b Z t 1 cos(a(t − τ )) sin(bτ ) dτ = b 0 Z t 1 = [sin((b − a)τ + at) + sin((b + a)τ − at) dτ ] dτ 2b 0 t 1 cos((b − a)τ + at) cos((b + a)τ − at) = − − 2b b−a b+a 0 1 cos(bt) cos(bt) cos(at) cos(at) = − − + + 2b b−a b+a b−a b+a cos(at) − cos(bt) . = (b − a)(b + a) If a2 = b2 , then s 1 1 L−1 2 (t) = cos(at) ∗ sin(at) s + a2 s2 + b2 a Z 1 t = cos(a(t − τ )) sin(aτ ) dτ a 0 Z t 1 = (sin(at) + sin(2aτ − at)) dτ 2a 0 t 1 1 = τ sin(at) − cos(a(2τ − t)) 2a 2a 0 1 t sin(at). = 2a 4. First write

s 1 s = . (s − 3)(s2 + 5) s − 3 s2 + 5

Then L

−1

√ s 3t 5t) (t) = e ∗ cos( (s − 3)(s2 + 5) Z t √ = cos( 5τ )e3(t−τ ) dτ 0 Z t √ = e3t cos( 5τ )e−3τ dτ 0

t √ √ √ 1 −3τ e − 3 cos( 5τ ) + 5 sin( 5τ ) 14 0 √ √ √ 3 3t 3 5 = e − cos( 5t) + sin( 5t). 14 14 14

= e3t

CHAPTER 3. THE LAPLACE TRANSFORM 5. First, −1 L Then −1 L

1 1 1 1 −1 (t) = t sin(at). (t) = 2 (1−cos(at)) and L s(s2 + a2 ) a s2 + a2 a 1 1 (t) = 2 [1 − cos(at)] ∗ sin(at) s(s2 + a2 ) a Z t a [1 − cos(a(t − τ ))] sin(aτ ) dτ = 3 a 0 t 1 1 1 1 = 3 − cos(aτ ) − τ sin(at) + cos(2aτ − at) a a 2 4a 0 1 1 = 4 (1 − cos(at)) − 3 sin(at). a 2a

6. −1

1 1 1 (t) = e5t ∗ t3 s4 s − 5 6 Z t 1 = e5(t−τ ) τ 3 dτ 6 0 Z t 1 = e5t τ 3 e−5τ dτ 6 0 1 1 1 1 1 5t e − t3 − t2 − t− . = 625 30 50 125 625

7. L

−1

1 e−4s (t) = e−2t ∗ H(t − 4) s+2 s Z t = e−2(t−τ ) dτ 4 ( 1 −2(t−4) e if t > 4, = 2 0 if t ≤ 4.

We can therefore write the inverse transform as 1 e−4s 1 L−1 (t) = (1 − e−2(t−4) )H(t − 4). s+2 s 2 8. L

−1

√ 2 1 sin( 5t) 2 √ (t) = t ∗ s2 s2 + 5 5 Z t √ 1 =√ τ 2 sin( 5(t − τ )) dτ 5 0 √ 1 2 2 = t− + cos( 5t). 5 25 25

3.4. CONVOLUTION

9. Take the transform of the initial value problem and solve for Y (s) to get Y (s) =

F (s) 1 1 = − F (s). s2 − 5s + 6 s−3 s−2

By the convolution theorem, y(t) = e3t ∗ f (t) − e2t ∗ f (t). 10. Take the transform of the initial value problem to obtain F (s) s + (s + 6)(s + 4) (s + 6)(s + 4) 1 1 1 1 3 + . − −2 = F (s) 2 s+4 s+6 s+6 s+4

Y (s) =

Then y(t) =

1 1 −4t e ∗ f (t) − e−6t ∗ f (t) + 3e−6t − 2e−4t . 2 2

For Problems 11–16 the solution is given, but the details (similar to those of Problems 9 and 10) are omitted. 11. y(t) =

1 6t 1 e ∗ f (t) − e2t ∗ f (t) + 2e6t − 5e2t 4 4

12. y(t) =

1 5t 1 1 3 e ∗ f (t) − e−t ∗ f (t) + e5t + e−t 6 6 2 2

13. y(t) =

1 1 sin(3t) ∗ f (t) − cos(3t) + sin(3t) 3 3

14. y(t) =

1 4 sinh(kt) ∗ f (t) − 2 cosh(kt) − sinh(kt) k k

15. y(t) =

1 2t 1 1 1 1 4 e ∗ f (t) + e−2t ∗ f (t) − et ∗ f (t) − e2t − e−2t + et 4 12 3 4 12 3

16. √ √ 1 3t 1 −3t 2 √2t 2 −√2t y(t) = e ∗ f (t) − e ∗ f (t) − e ∗ f (t) − e ∗ f (t) 42 42 28 28

CHAPTER 3. THE LAPLACE TRANSFORM

17. The integral equation can be expressed as f (t) = −1 + f (t) ∗ e−3t . Take the transform of this equation to obtain 1 F (s) F (s) = − + . s s+3 Then F (s) = −

s+3 1 1 31 = − . s(s + 2) 2s+2 2s

Invert to obtain the solution f (t) =

1 −2t 3 e − . 2 2

18. The equation is f (t) = −1 + f (t) ∗ sin(t). Take the transform of this equation and solve for F (s) to obtain F (s) = − Then

(s2 + 1) 1 1 = − 2 − 4. s4 s s

1 f (t) = −t − t3 . 6

19. The equation is f (t) = e−t + f (t) ∗ 1. Take the transform and solve for F (s) to get F (s) =

s 1 1 1 1 = + . (s − 1)(s + 1) 2s+1 2s−1

Then f (t) =

1 −t 1 t e + e = cosh(t). 2 2

20. The equation is f (t) = −1 + t − 2f (t) ∗ sin(t). From this we obtain (1 − s)(s2 + 1) s2 (s2 + 3) 1 1 s 2 1 11 2 = − + . − 3 s2 3 s 3 s2 + 3 3 s2 + 3

F (s) =

Invert this to get √ √ √ 1 2 2 3 1 f (t) = t − − cos( 3t) + sin( 3t). 3 3 3 9

3.4. CONVOLUTION

21. The equation is f (t) = 3 + f (t) ∗ cos(2t). From this we obtain F (t) =

3(s2 + 4) 3 3 = + 2 . s(s2 − s + 4) s s −s+4

The inverse of this is √ 2 15 t/2 f (t) = 3 + e sin 5

√ ! 15 t . 2

22. Write the equation as Z t f (t) = cos(t) +

f (τ )e−2(t−τ ) dτ = cos(t) + f (t) ∗ e−2t .

Transform this and solve for F (s) to obtain s(s + 2) (s + 1)(s2 + 1) 3 1 s 1 1 + . + =− 2(s + 1) 2 s2 + 1 2 s2 + 1

F (s) =

Invert this to get the solution f (t) =

1 1 −t 3 e + cos(t) + sin(t). 2 2 2

23. Let F = L[f ] and G = L[g]. Then Z ∞ F (s)G(s) = F (s)

e−sτ g(τ ) dτ.

Now recall that e−sτ F (s) = L[H(t − τ )f (t − τ )](s). Substitute this into the expression for F (s)G(s) to get Z ∞ F (s)G(s) = L[H(t − τ )f (t − τ )](s)g(τ ) dτ. 0

But, from the definition of the Laplace transform, Z ∞ L[H(t − τ )f (t − τ )] = e−st H(t − τ )f (t − τ ) dt. 0

Then Z ∞ Z ∞ F (s)G(s) = 0 Z0 ∞ Z ∞

= 0

e−st H(t − τ )f (t − τ ) dt g(τ ) dτ

e−st g(τ )H(t − τ )f (t − τ ) dτ dt.

CHAPTER 3. THE LAPLACE TRANSFORM But H(t − τ ) = 0 if 0 ≤ t < τ , while H(t − τ ) = 1 if t ≥ τ . Therefore Z ∞Z ∞ e−st g(τ )f (t − τ ) dt dτ. F (s)G(s) = 0

The last integration is over the wedge in the t, τ − plane consisting of points (t, τ ) with 0 ≤ τ ≤ t < ∞. Reverse the order of integration to write Z ∞Z t e−st g(τ )f (t − τ ) dτ dt F (s)G(s) = 0 0 Z t Z ∞ −st = e g(τ )f (t − τ ) dτ dt 0 Z0 ∞ −st = e (f ∗ g)(t) dt 0

= L[f ∗ g](s). This is what we wanted to show.

3.5

Impulses and the Dirac Delta Function

In Problem 1 details of the solution are given. For Problems 2–5, the details are similar and only the solution is given. 1. Transform the initial value problem to obtain (s2 + 5s + 6)Y (s) = 3e−2s − 4e−5s . Using a partial fractions decomposition, this gives us 1 1 1 1 −3s − − e −4 e−5s . Y (s) = 3 s+2 s+3 s+2 s+3 Invert this to obtain the solution h i h i y(t) = 3 e−2(t−2) − e−3(t−2) H(t − 2) − 4 e−2(t−5) − e−3(t−5) H(t − 5). 2. y(t) =

4 2(t−3) e sin(3(t − 3))H(t − 3) 3

3. y(t) = 6(e−2t − e−t + te−t ) 4. y(t) = 3 cos(4t) + 3 sin(4(t − 5π/8))H(t − 5π/8)

3.6. SYSTEMS OF LINEAR DIFFERENTIAL EQUATIONS

5. y(t) = (B + 9)e−2t − (B + 6)e−3t 6. Begin with Z ∞

Z ∞ f (t)δ (t − a) dt =

1 [H(t − a) − H(t − a − )]f (t) dt

Z a+ f (t) dt.

== a

By the mean value theorem for integrals, there is some t between t and t + such that Z a+ 1 f (t) dt = f (t ). a Then

Z ∞ f (t)δ (t − a) dt = f (t ). 0

Now let → 0+. Then a + → a, so t → a also. Because f (t) is assumed to be continuous, f (t ) → f (a). Finally, we have Z ∞ f (t)δ (t − a) dt =

lim

→0+

Z ∞

f (t) lim δ (t − a) dt →0+

Z0 ∞

f (t)δ(t − a)

= 0

= lim f (t ) = f (a). →0+

3.6

Systems of Linear Differential Equations

1. Take the transform of the system: sX − 2sY =

1 , sX − X + Y = 0. s

Then 1 2 4 =− 2 − + , s s 2s − 1 1−s 1 1 2 Y (s) = 2 =− 2 − + . s (2s − 1) s s 2s − 1

X(s) =

s2 (2s − 1)

Apply the inverse transform to get the solution x(t) = −t − 2 + 2et/2 , y(t) = −t − 1 + et/2 .

CHAPTER 3. THE LAPLACE TRANSFORM 2. Take the transform of the system to obtain 2sX + (s − 3)Y = 0, sX + sY = Then

1 . s2

s−3 2 X(s) = − 2 , Y (s) = 2 . s (s + 3) s (s + 3)

Using partial fractions, we can write 1 2 1 21 2 1 + 3− − , 9 s 3 s2 s 9s+3 21 2 1 2 1 Y (s) = − + + . 9 s 3 s2 9s+3

X(s) =

Inverting, we obtain the solution 1 2 2 2 − t + t2 − e−3t , 9 3 2 9 2 2 2 −3t y(t) = − + t + e . 9 3 9

x(t) =

3. After transforming the system, we obtain sX + (2s − 1)Y =

1 , 2sX + Y = 0. s

Then 1 4 16 1 X(s) = − 2 = − + , s (4s − 3) 9s 9(4s − 3) 3s2 2 8 2 =− + . Y (s) = s(4s − 3) 3s 3(4s − 3) Invert these to obtain 1 4 (1 − e3t/4 ) + t, 9 3 2 y(t) = (−1 + e3t/4 ). 3

x(t) =

4. The transformed system is sX + sY − X =

s , sX + 2sY = 0. s2 + 1

Then 2s 1 4s − 2 4 =− 2 + , s3 − 2s2 + s − 2 5 s + 1 5(s − 2) s 1 2s − 1 2 Y (s) = − 3 = − . 2 2 s − 2s + s − 2 5 s + 1 5(s − 2)

X(s) =

3.6. SYSTEMS OF LINEAR DIFFERENTIAL EQUATIONS

The solution is 4 2 4 x(t) = − cos(t) + sin(t) + e2t , 5 5 5 2 1 2 2t y(t) = cos(t) − sin(t) − e . 5 5 5 5. The system transforms to 3sX − Y =

2 sX + sY − Y = 0. s2

Then 2(s − 1) 3 1 1 9 = + + 3− , s2 (3s − 2) 4s 2s2 s 4(3s − 2) 2 3 1 2(3s − 2) Y (s) = − 2 = + −9 s (3s − 2) 2s s2 .

X(s) =

Then 3 1 1 3 + t + t2 − e2t/3 , 4 2 2 4 3 2t/3 3 . y(t) = + t − e 2 2

x(t) =

6. Apply the transform to get sX + 4sY − Y = 0sX + 2Y =

1 . s+1

Then 1 32 5 4s − 1 = + − s(4s2 + s − 3) 3s 21(4s − 3) 7(s + 1) 4 1 1 Y (s) = − 2 =− + . 4s + s − 3 7(4s − 3) 7(s + 1)

X(s) =

Then 1 8 5 + e3t/4 − e−t , 3 21 7 1 3t/4 1 −t y(t) = − e + e . 7 7

x(t) =

7. The transform of the system is sX + 2X − sY = 0, sX + Y + X =

2 . s3

CHAPTER 3. THE LAPLACE TRANSFORM Then s+1 2 1 1 = 2+ 2 − , s2 (s2 + 2s + 2) s s + 2s + 2 s 1 2(s + 2) 1 2 Y (s) = 3 2 =− 2 + 2 + . s (s + 2s + 2) s s + 2s + 2 s3

X(s) =

The solution is x(t) = t + e−t cos(t) − 1 y(t) = −t + e−t sin(t) + t2 . In inverting X(s) and Y (s), the terms involving s2 + 2s + 2 can be treated by using a shifting theorem, expressing these as functions of s + 1. 8. The system transforms to sX + 4X − Y = 0, sX + sY =

1 . s2

Then 1 1 1 1 1 = − − + , s2 (s + 5) 125s 25s2 125(s + 5) 5s2 s+4 1 1 1 4 Y (s) = 3 =− + + + . s (s + 5) 125s 25s2 125(s + 5) 5s3

X(s) =

The solution is 1 1 1 −5t 1 − t− e + t2 125 25 125 10 1 1 −3t 2 2 1 + t+ e + t . y(t) = − 125 25 125 5

x(t) =

9. First, sX + sY + X − Y = 0, sX + 2sY + X =

1 . s

Then 1−s −2 1 1 = − + , s(s + 1)2 s + 1)2 s+1 s 1 1 1 Y (s) = = − . s(s + 1) s s+1

X(s) =

Then x(t) = 1 − e−t (2t + 1), y(t) = 1 − e−t . 10. The transform of the system is 6 sX + 2sY − X = 0, 4sX + 3sY + Y = − . s

3.6. SYSTEMS OF LINEAR DIFFERENTIAL EQUATIONS

Then, X(s) = −

12 6(s − 1) , Y (s) = . 5s2 + 2s + 1 s(5s2 + 2s + 1)

By using a shifting theorem, invert these to get x(t) = −6e−t/5 sin(2t/5), y(t) = −6 + 6e−t/5 (cos(2t/5) + sin(2t/5)) . 11. The system transforms to sX − 2sY + 3X = 0, X − 4sY + 3sZ =

1 1 , X − 2sY + 3sZ = − . s2 s

Then 2 2 1 − − 2, 9s 9(s + 3) 3s 1s+1 −1 1 Y (s) = − = 3 − 2, 3 2 s 2s 2s 2 2 2 16 2 s2 + 3s + 1 =− − + − . Z(s) = − 3 s3 (s + 3) 9s 81s 81(s + 3) 27s2

X(s) =

s+1

s2 (s + 3)

The solution is 2 1 2 + t − e−3t , 9 3 9 1 y(t) = − t(t + 2) 4 16 1 2 2 z(t) = − t − t2 − + e−3t . 27 9 81 81

x(t) =

12. The loop currents in the circuit satisfy 5i01 + 5i1 − 5i02 = 1 − H(t − 4) sin(2(t − 4)), −5i01 + 5i02 + 5i2 = 0. Apply the Laplace transform to these equations and solve for I1 (s) and I2 (s) to get 1 2e−4s s+1 − 2 I1 (s) = 5(2s + 1) s s + 4 1 1 1 2 2 s 9 = − − − 2 + 2 e−4s 5 s 2s + 1 85 2s + 1 s + 4 s + 4 1 2e−4s I2 (s) = 1− 2 2s + 1 s +4 1 2 2 s 8 = + − 2 − 2 e−4s . 5(2s + 1) 85 2s + 1 s + 4 s + 4

CHAPTER 3. THE LAPLACE TRANSFORM Apply the inverse Laplace transform to obtain the solution for the currents: 1 1 −t/2 i1 (t) = 1− e 5 2 2 9 −(t−4)/2 − e − cos(2(t − 4)) + sin(2(t − 4)) H(t − 4), 85 2 1 −t/2 e i2 (t) = 10 i 2 h −(t−4)/2 e − cos(2(t − 4)) − 4 sin(2(t − 4)) H(t − 4). + 85

13. The equations for the loop currents are 20i01 + 10(i1 − i2 ) = E(t) = 5H(t − 5), 30i02 + 10i2 + 10(i2 − i1 ) = 0. Initial conditions are i1 (0) = i2 (0) = 0. Transform the system to obtain 5(30s + 20)e−5s s(600s2 + 700s + 100) 1 1 27 1 = − − e−5s , s 10(s + 1) 5 6s + 1 50e−5s I2 (s) = 2 s(600s + 700s + 100) 1 10 18 1 = + − e−5s . 2s s + 1 5 6s + 1 I1 (s) =

Invert these to obtain the current functions: 1 9 i1 (t) = 1 − e−(t−5) − e−(t−5)/6 H(t − 5), 10 10 1 3 1 + e−(t−5) − e−(t−5)/6 H(t − 5). i2 (t) = 2 10 10 14. Let x1 (t) and x2 (t) be the amounts of salt (in pounds) in tanks 1 and 2, respectively, at time t. Now x01 (t) = rate of change of salt in tank 1 = (rate salt is added ) − ( rate salt is removed). Then x01 (t) =

1 3 5 + x2 − x1 . 3 18 60

3.6. SYSTEMS OF LINEAR DIFFERENTIAL EQUATIONS

Similarly, 5 5 x1 − x2 + 11(H(t − 4) − H(t − 6)). 60 18 Initial conditions are x1 (0) = 11, x2 (0) = 7. Transform this system to obtain 4 (12s + 1)X1 − 2X2 = + 132, s 396 −4s −6s −3X1 + (36s + 10)X2 = (e − e ) + 252. s Then 4752s2 + 1968s + 40 + 792(e−4s − e−6s ) X1 (s) = s(432s2 + 156s + 4) 6 108 99 27 3888 10 = − + +2 + − (e−4s − e−6s ), s 3s + 1 36s + 1 s 3s + 1 36s + 1 3024s2 + 648s + 12 + 396(12s + 1)(e−4s − e−6s ) X2 (s) = s(432s2 + 156s + 4) 3 9 36 99 81 2592 = + + + − − (e−4s − e−6s ). s 3s + 1 36s + 1 s 3s + 1 36s + 1 x02 =

Apply the inverse transform to obtain the solution: x1 (t) = 10 − 2e−t/3 + 3e−t/36 + 2(99 + 9e−(t−4)/3 − 108e−(t−4)/6 )H(t − 4) − 2(99 + 9e−(t−6)/3 − 108e−(t−6)/36 )H(t − 6), x2 (t) = 3 + 3e−t/3 + e−t/36 + (99 − 27e−(t−4)/3 − 72e−(t−4)/36 )H(t − 4) − (99 − 27e−(t−6)/3 − 72e−(t−6)/36 )H(t − 6). 15. Using the notation of the preceding problem, we can write 3 6 x1 + x2 , x01 = − 200 100 4 4 x1 − x2 + 5H(t − 3). x02 = 200 200 Initial conditions are x1 (0) = 10, x2 (0) = 5. Apply the transform to this initial value problem and rearrange terms to obtain (100s + 3)X1 − 3X2 = 1000, −2X1 + (100s + 4)X2 = 500 + 500e−3s . Solve these to get 100000s + 5500 + 1500e−3s 10000s2 + 700s + 6 50 900 300 150 = + + − e−3s , 50s + 3 100s + 1 100s + 1 50s + 3

X1 (s) =

100

CHAPTER 3. THE LAPLACE TRANSFORM and 50000s + 3500 + (50000s + 1500)e−3s 10000s2 + 700s + 6 50 600 150 200 =− + + + e−3s . 50s + 3 100s + 1 50s + 3 100s + 1

X2 (s) =

Apply the inverse transform to obtain the solution: i1 (t) = e−3t/50 + 9e−t/100 + 3(e−(t−3)/100 − e−3(t−3)/50 )H(t − 3), i2 (t) = −e−3t/50 + 6e−t/100 + (3e−3(t−3)/50 + 2e−(t−3)/100 )H(t − 3).

Chapter 4

Sturm-Liouville Problems and Eigenfunction Expansions 4.1

Eigenvalues, Eigenfunctions and Sturm-Liouville Problems

For these problems, an eigenfunction is found for each eigenvalue, and it is understood that nonzero constant multiples of eigenfunctions are also eigenfunctions. 1. The problem is regular on [0, L]. To find the eigenvalues and eigenfunctions, take cases on λ. Case 1. If λ = 0, the differential equation is y 00 = 0, with solutions y = a + bx. Now y(0) = a = 0, so y = bx. But then y 0 (L) = b = 0 also, so this case has only the trivial solution and 0 is not an eigenvalue of this problem. Case 2. If λ is negative, say λ = −α2 with α > 0, then the differential equation is y 00 − α2 y = 0 with general solution y = c1 eαx + c2 e−αx . Now y(0) = c1 + c2 = 0, so c2 = −c1 and y(x) = c1 eαx − e−αx = 2c1 sinh(αx). 101

102CHAPTER 4. STURM-LIOUVILLE PROBLEMS AND EIGENFUNCTION EXPANSIONS From the other boundary condition, y 0 (L) = 2c1 α cosh(αL) = 0. But cosh(αL) > 0 and α > 0, so we must have c1 = 0 and this case also has only the trivial solution. This problem has no positive eigenvalue. Case 3. Suppose λ is positive, write λ = α2 , with α > 0. Now the differential equation is y 00 + α2 y = 0, with general solution y(x) = c1 cos(αx) + c2 sin(αx). Immediately y(0) = c1 = 0, so y(x) = c2 sin(αx). From the other boundary condition, we must have y 0 (L) = c2 α cos(αL) = 0. We need to be able to choose c2 6= 0 to have nontrivial solutions. This requires that we α must be chosen to satisfy cos(αL) = 0. We know that the zeros of the cosine function have the form (2n − 1)π/2 for integer π, so let (2n − 1)π , αL = 2 with n = 1, 2, · · · . Then acceptable values of α are α=

(2n − 1)π . 2L

Because λ = α2 , the eigenvalues of this problem, indexed by n, are λn =

(2n − 1)π 2L

for n = 1, 2, · · · . Corresponding eigenfunctions are (2n − 1)π ϕn (x) = sin x , 2L or any nonzero constant multiple of this function.

4.1. EIGENVALUES, EIGENFUNCTIONS AND STURM-LIOUVILLE PROBLEMS103 2. The problem is regular on [0, L]. The eigenvalues are λ0 = 0, λn = n2 for n = 1, 2, · · · . Corresponding eigenfunctions are ϕn (x) = cos(nπx) for n = 0, 1, 2, · · · . Any nonzero constant multiple of an eigenfunction is also an eigenfunction (for the same eigenvalue). Notice in this example that 0 is an eigenvalue. The number zero can be an eigenvalue, but the trivial function (identically zero) cannot be an eigenfunction. 3. The problem is regular on [0, L]; λn =

2 1 π n− 2 4

is an eigenvalue for n = 1, 2, · · · , with eigenfunctions (2n − 1)π . ϕn (x) = cos 8 4. The problem is periodic on [0, π]. Eigenvalues are λn = 4n2 for n = 0, 1, 2, · · · and eigenfunctions are ϕn (x) = an cos(2nx) + bn sin(2nx) with an and bn constant and not both zero. 5. The problem is periodic on [−3π, 3π]. Eigenvalues are λ0 = 0 and λn =

n2 for n = 1, 2, · · · . 9

Eigenfunctions are ϕn (x) = an cos(nx/3) + bn sin(nx/3) for n = 0, 1, 2, · · · , with an and bn constant and not both zero. 6. The problem is regular on [0, π]. To find the eigenvalues and eigenfunctions, take cases on λ. Case 1. If λ = 0, then y = ax + b for constants a and b. Because y(0) = b = 0, then y = ax. But then y(π) + 2y 0 (π) = 0 = aπ, so a = 0 and this case has only the trivial solution. 0 is not an eigenvalue of this problem.

104CHAPTER 4. STURM-LIOUVILLE PROBLEMS AND EIGENFUNCTION EXPANSIONS Case 2. If λ = −α2 , with α > 0, then y(x) = c1 eαx + c2 e−αx . Now y(0) = c1 + c2 = 0, so c2 = −c1 and y(x) = c1 (eαx − e−αx ). From the boundary condition at π, we must have y(π) + 2y 0 (π) = c1 (eαπ − e−απ ) = 0. But e−απ < eαπ because the exponential function is strictly increasing. Therefore c1 = 0 and this case admits only the trivial solution. The problem has no negative eigenvalue. Case 3. If λ = α2 with α > 0, then y(x) = c1 cos(αx) + c2 sin(αx). Immediately, y(0) = c1 = 0, so y(x) = c2 sin(αx). The second boundary condition is y(π) + 2y 0 (π)0 = c2 sin(απ) + 2c2 α cos(απ). To look for nontrivial solutions, suppose c2 6= 0. Then this equation is sin(απ) = −2α cos(απ) or tan(απ) = −2α. This is a transcendental equation, which cannot be solved by algebraic manipulations. There are infinitely many positive solutions, however, because the graphs of y = tan(απ) and y = −2α intersect infinitely often in the right half-plane. Let the first coordinates of these points of intersection be α1 , α2 , · · · , in increasing order. Then the eigenvalues of this problem are λn = αn2 . Using a numerical approximation program, the first four eigenvalues are approximately λ1 ≈ 0.48705, λ2 ≈ 2.54914, λ2 ≈ 6.56059, λ4 ≈ 12.56423. Corresponding eigenfunctions are p ϕn (x) = sin( λn x).

4.1. EIGENVALUES, EIGENFUNCTIONS AND STURM-LIOUVILLE PROBLEMS105 7. The problem is regular on [0, 1]. The analysis to find eigenvalues and eigenfunctions is similar to that done for Problem 6. Take cases on λ. It is routine to check that λ = 0 or λ < 0 lead only to the trivial solution, so the problem has no negative eigenvalue and zero is not an eigenvalue. If λ = α2 for α > 0, then y(x) = c1 cos(αx) + c2 sin(αx). Using the first boundary condition, y(0) − 2y 0 (0) = 0 = c1 − 2c2 α so c1 = 2αc2 and y(x) = 2αc2 cos(αx) + c2 sin(αx). Now use the boundary condition at 1: y 0 (1) = −2α2 c2 sin(α) + c2 α cos(α) = 0. For a nontrivial solution we need to be able to choose c2 nonzero. This requires that −2α sin(α) + cos(α) = 0, or

1 . 2α Solutions of this equation must be numerically approximated. There are infinitely many positive solutions α1 < α2 < · · · , and the eigenvalues are λj = αj2 . The first four eigenvalues are tan(α) =

λ1 ≈ 0.42676, λ2 ≈ 10.8393, λ3 ≈ 40.4702, λ4 ≈ 89.8227. Corresponding eigenfunctions are p p √ ϕn (x) = 2 λ cos( λn x) + sin( λn x). 8. The problem is regular on [0, 1]. The differential equation has characteristic equation r2 + 2r + (1 + λ) = 0, √ with roots r = −1 ± λi. The general solution of the differential equation is √ √ y(x) = c1 e−x cos( λx) + c2 e−x sin( λx). Because y(0) = 0 = c1 , we have just √ y(x) = c2 e−x sin( λx).

106CHAPTER 4. STURM-LIOUVILLE PROBLEMS AND EIGENFUNCTION EXPANSIONS The boundary condition at 1 requires that √ y(1) = 0 = c2 e−1 sin( λ). This can be satisfied with c2 6= 0 if √ λ = nπ, a positive zero of the sine function, for n = 1, 2, · · · . The eigenvalues are therefore λn = n2 π 2 and the eigenfunctions are ϕn (x) = e−x sin(nπx). 9. The problem is regular on [0, π]. The differential equation can be written y 00 + 2y 0 + λy = 0. and the characteristic equation has roots √ −1 ± 1 − λ. Here it is convenient to take cases on 1 − λ. Case 1. If 1 − λ = 0, then λ = 1 and the general solution is y(x) = c1 e−x + c2 xe−x . Now y(0) = c1 = 0, so y(x) = c2 xe−x . And y(π) = 0 = c2 πe−π = 0 forces c2 = 0, so this case has only the trivial solution and 0 is not an eigenvalue. Case 2. If 1 − λ is positive, say 1 − λ = α2 with α > 0, then y(x) = c1 e(−1+α)x + c2 e(−1−α)x . Now y(0) = c1 + c2 = 0 so c2 = −c1 and y(x) = c1 e(−1+α)x − e(−1−α)x . Then

y(π) = c1 e(−1+α)π − e(−1−α)π .

4.2. EIGENFUNCTION EXPANSIONS

107

If c1 6= 0, this requires that eαπ = e−απ , which is impossible if α > 0. The problem has no negative eigenvalue. Case 3. If 1 − λ is negative, write 1 − λ = −α2 . Now y(x) = c1 e−x cos(αx) + c2 e−x sin(αx). Immediately y(0) = c1 = 0. Next, y(π) = c2 e−π sin(απ) = 0. To have c2 6= 0, we must choose α=

√ λ − 1 = n,

any positive integer. Then λ = 1 + n2 , so the eigenvalues are λn = 1 + n2 for n = 1, 2, · · · . Eigenfunctions are ϕn (x) = e−x sin(nx). 10. The problem is regular on [0, 8]. Details are similar to those of Problem 9 and we find eigenvalues λn = 8 + n2 π 2 for n = 1, 2, · · · , and eigenfunctions ϕn (x) = e3x sin(nπx). 11. If λn = 1 − 1/n, then the eigenvalues are listed in increasing order, and lim λn = 0.

n→∞

This is impossible by Theorem 4.1.

4.2

Eigenfunction Expansions

In Problems 1–5, the weight function is p(x) = 1 (read from the differential equation). In Problem 6 the differential equation must be put into standard Sturm-Liouville form to read the weight function p(x) as the coefficient of λ. In graphing partial sums of eigenfunctions and comparing them to the function, note the differences in the number of terms that must be taken to have the partial sum fit reasonably close to the function. The convergence theorem does not give any information about how fast an eigenfunction expansion converges to the function.

108CHAPTER 4. STURM-LIOUVILLE PROBLEMS AND EIGENFUNCTION EXPANSIONS 1. It is routine to find the eigenfunctions ϕn (x) = sin

nπx 2

for this problem. The expansion has the form ∞ X

cn sin(nπx/2),

n=1

where

R2 0

cn =

(1 − ξ) sin(nπξ/2) dξ . R2 2 sin (nπξ/2) dξ 0

These integrals are Z 2

sin2 (nπξ/2) dξ = 1

and

Z 2 (1 − ξ) sin(nπξ/2) dξ = 0

2(1 + (−1)n ) . nπ

The eigenfunction expansion on [0, 2] is ∞ X 2(1 + (−1)n )

nπ

n=1

sin(nπx/2).

Figure 4.1 shows a graph of f (x) = 1 − x and the fortieth partial sum of this expansion. By the convergence theorem, this expansion converges to 1 − x for 0 < x < 2. Clearly the expansion converges to 0 at both x = 0 and x = 2 because the eigenfunctions vanish there. 2. The problem has eigenfunctions ϕn (x) = sin((2n − 1)x/2). The eigenfunction expansion on [0, π] has the coefficients cn =

8 (−1)n+1 π (2n − 1)2

and the expansion is ∞ X

cn sin((2n − 1)x/2).

n=1

Figure 4.2 shows a graph of the function and the fifth partial sum of this eigenfunction expansion. Unlike Problem 1, this expansion converges very rapidly to the function. By the convergence theorem, it converges to x for 0 < x < π. The graph suggests that it converges to x at the endpoints as well, but this is not given by the theorem.

4.2. EIGENFUNCTION EXPANSIONS

109

Figure 4.1: Comparison of 1 − x and the fortieth partial sum of its eigenfunction expansion on [0, 2].

Figure 4.2: Comparison of x and the fifth partial sum of its eigenfunction expansion on [0, π].

110CHAPTER 4. STURM-LIOUVILLE PROBLEMS AND EIGENFUNCTION EXPANSIONS

Figure 4.3: Comparison of f (x) and the sixtieth partial sum of its eigenfunction expansion on [0, 4].

3. The eigenfunctions are ϕn (x) = cos((2n − 1)πx/8). The coefficients in the expansion of f (x) on [0, 4] are R2 R4 − cos((2n − 1)πx/8) dx + 2 cos((2n − 1)πx/8) dx cn = 0 R4 cos2 ((2n − 1)πx/8) dx 0 i h √ 4 (−1)n+1 + 2(cos(nπ/2) − sin(nπ/2)) . = (2n − 1)π The expansion has the form ∞ X

cn cos((2n − 1)πx/8).

n=1

Figure 4.3 compares f (x) with the sixtieth partial sum of this eigenfunction expansion. The theorem tells us that the expansion converges to f (x) on (0, 2) and on(2, 4), as well at to 0 at x = 0 (average of left and right limits there). 4. The eigenfunctions are ϕ0 (x) = 1, ϕn = cos(nx) for n = 1, 2, · · · .

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111

Figure 4.4: Comparison of f (x) and the tenth partial sum of its eigenfunction expansion in Problem 4.

The coefficients in the eigenfunction expansion of f (x) = sin(2x) on [0, π] are Z 1 π sin(1ξ) dξ = 0, c0 = π 0 Z 2 π c2 = sin(2ξ) cos(2ξ) dξ = 0, π 0 and, for n = 1, 3, 4, 5, · · · , Z 2 π 4 (−1)n − 1 cn = sin(2ξ) cos(nξ) dξ = . π 0 π n2 − 4 The eigenfunction expansion is 4 π

∞ X n=1,n6=2

4 ((−1)n − 1) cos(nx). π n2 − 4

Figure 4.4 compares a graph of f (x) with the tenth partial sum of this eigenfunction expansion. The theorem tells us that this expansion converges to sin(2x) for 0 < x < π. 5. The eigenfunctions are ϕ0 (x) = 1, ϕn (x) = an cos(nx/3) + bn sin(nx/3)

112CHAPTER 4. STURM-LIOUVILLE PROBLEMS AND EIGENFUNCTION EXPANSIONS

Figure 4.5: Comparison of f (x) and the fifth partial sum of its eigenfunction expansion in Problem 5.

for n = 1, 2, · · · . The coefficients in the eigenfunction expansion of x2 on [−3π, 3π] are Z 3π 1 ξ 2 dξ = 3π 2 , 6π −3π 1 2 36 an = ξ cos(nξ/3) dξ = 2 (−1)n , 3π n Z 3π 1 bn = ξ 2 sin(nξ/3) dξ = 0. 3π −3π a0 =

The eigenfunction expansion is 2

3π + 36

∞ X (−1)n n=1

cos(nx/3).

Figure 4.5 shows f (x) and the fifth partial sum of this eigenfunction expansion. By the theorem, the expansion converges to x2 for −3π < x < 3π. 6. The eigenfunctions are ϕn (x) = e−x sin(nπx)

4.2. EIGENFUNCTION EXPANSIONS

113

Figure 4.6: Comparison of f (x) and the eightieth partial sum of its eigenfunction expansion in Problem 6.

for n = 1, 2, · · · . Notice that the differential equation of the problem can written in standard Sturm-Liouville form as 0 e2x y 0 + (1 + λ)e2x y = 0, with the coefficient of λ equal to e2x . Therefore, in the expansion of a function in terms of these eigenfunctions, the weight function p(x) = e2x must be used. With this in mind, the coefficients in this expansion are R1 1/2

e2ξ e−ξ sin(nπξ) dξ

cn = R 1

e2ξ e−2ξ sin2 (nπξ) dξ 0 R1 ξ e sin(nπξ) dξ 1/2 = R1 2 sin (nπξ) dξ 0 1/2

(nπ cos(nπ/2) − sin(nπ/2)) − 2enπ(−1)n . 1 + n2 π 2

Figure 4.6 shows a graph of the function and the eightieth partial sum of this eigenfunction expansion. 7. Recall that the complex conjugate of z = a + ib is z = a − ib. Suppose λ is an eigenvalue of a Sturm-Liouville problem, with eigenfunction ϕ(x). By

114CHAPTER 4. STURM-LIOUVILLE PROBLEMS AND EIGENFUNCTION EXPANSIONS taking the complex conjugate of the Sturm-Liouville differential equation and appropriate boundary conditions, it is routine to check that λ is also an eigenvalue with eigenfunction ϕ(x). If λ is complex and not real, then λ 6= λ, so the eigenfunctions must be orthogonal with respect to the weight function p, and Z b p(x)ϕ(x)ϕ(x) dx = 0. a

Now, ϕ(x)ϕ(x) = |ϕ(x)|2 , so

Z b

p(x)|ϕ(x)|2 | dx = 0.

This is impossible because p(x) > 0 on (a, b) and ϕ(x) is continuous and not identically zero on the interval. This contradiction shows that λ = λ, so λ is real.

4.3

Fourier Series

1. The Fourier series of f (x) = 4 on [−3, 3] has the form ∞ X 1 [an cos(nπx/3) + bn sin(nπx/3)]. a0 + 2 n=1

All that is left is to compute the coefficients. First, because f (x) is an even function, each bn = 0. Compute Z 2 3 a0 = 4 dx = 8 3 −3 and, for n = 1, 2, · · · , an =

2 3

Z 3 4 cos(nπx/3) dξ = 0. 0

With each an = 0 for n = 1, 2, · · · , the Fourier series is 1 a0 = 4. 2 This series consists of a single term, namely the constant term (which seems obvious by hindsight, if not noticed immediately). This one-term series converges to 4 on [−3, 3]. 2. Because f (x) is an odd function on [−1, 1], each an = 0. Compute the Fourier coefficients Z 1 2 (−1)n . bn = 2 −x sin(nπx) dx = nπ 0

4.3. FOURIER SERIES

115

Figure 4.7: Comparison of f (x) and the thirtieth partial sum of the Fourier series in Problem 2.

The Fourier series of −x on [−1, 1] is ∞

2 X (−1)n sin(nπx). π n=1 n This series converges to −x for −1 < x < 1, and to 0 at x = −1 and at x = 1. The latter is consistent with the convergence theorem because lim (−x) + lim (−x) = 1 − 1 = 0.

x→−1+

x→1−

Figure 4.7 is a graph of f (x) and the thirtieth partial sum of the Fourier series. 3. Because cosh(πx) is an even function, each bn = 0 in the Fourier series, which will have the appearance ∞ X 1 a0 + an cos(nπx). 2 n=1

Compute Z 1 a0 = 2

cosh(πx) dx 0

2 sinh(π) π

116CHAPTER 4. STURM-LIOUVILLE PROBLEMS AND EIGENFUNCTION EXPANSIONS

Figure 4.8: Comparison of f (x) and the eighth partial sum of the Fourier series in Problem 3.

and, for n = 1, 2, · · · , Z 1 2 sinh(π) (−1)n an = 2 cosh(πx) cos(nπx) dx = . π 1 + n2 0 The Fourier series is ∞ X 2 sinh(π) (−1)n 1 sinh(π) + cos(nπx). π π 1 + n2 n=1

This series converges to cosh(πx) for −1 ≤ x ≤ 1. Figure 4.7 shows a graph of f (x) and the eighth partial sum of this Fourier series. 4. Omitting the routine integrations, the Fourier series of 1 − |x| on [−2, 2] is ∞ 8 X 1 cos((2n − 1)πx/2). π 2 n=1 (2n − 1)2 This series converges to 1 − |x| for −2 ≤ x ≤ 2. Figure 4.9 shows a graph of f (x) and the fifth partial sum of this Fourier series. 5. The Fourier series of f (x) on [−π, π] is ∞

1 16 X sin((2n − 1)x). π n=1 (2n − 1)2

4.3. FOURIER SERIES

117

Figure 4.9: Comparison of f (x) and the fifth partial sum of the Fourier series in Problem 4.

This series converges to   −4 4   0

for −π < x < 0, for 0 < x < π, for x = 0, −π, π.

Figure 4.10 is a graph of f (x) and the twentieth partial sum of this Fourier series. 6. Because f (x) = sin(2x) is periodic of period π, this function is its own Fourier series on this [−π, π]. 7. The Fourier series of f (x) on [−2, 2] is ∞ 13 X 16 1 n + (−1) sin(nπx/2) . cos(nπx/2) + 3 (nπ)2 nπ n=1 This series converges to ( x2 − x + 3 7

for −2 < x < 2, for x = ±2.

118CHAPTER 4. STURM-LIOUVILLE PROBLEMS AND EIGENFUNCTION EXPANSIONS

Figure 4.10: Comparison of f (x) and the twentieth partial sum of the Fourier series in Problem 5.

At the endpoints, 1 1 (f (−2+) + f (2−)) = (9 + 5) = 7. 2 2 Figure 4.11 shows f (x) and the twentieth partial sum of this Fourier series. 8. The Fourier series of f (x) on [−5, 5] is ∞ X 1 a0 + [an cos(nπx/5) + bn sin(nπx/5)], 2 n=1

where a0 =

1 5

Z 5 f (x) dx = −5

71 , 6

Z 1 5 an = f (x) cos(nπx/5) dx 5 −5 5((−1)n − 1) 50(−1)n = + n2 π 2 n2 π 2

4.3. FOURIER SERIES

119

Figure 4.11: Comparison of f (x) and the twentieth partial sum of the Fourier series in Problem 7.

and Z 1 5 f (x) sin(nπx/5) dx 5 −5 1 5(−1)n − 3 3 (50 − n2 π 2 − 50(−1)n + 26n2 π 2 (−1)n ). = nπ n π

bn =

This series converges to  x    1 + x2 1/2    31/2

for −5 < x < 0, for 0 < x < 5, for x = 0, at x = ±5.

Figure 4.12 shows f (x) and the fiftieth partial sum of this Fourier series on [−5, 5]. 9. The Fourier series of f (x) on [−π, π] is ∞ 3 2X 1 + sin((2n − 1)x). 2 π n−1 2n − 1

120CHAPTER 4. STURM-LIOUVILLE PROBLEMS AND EIGENFUNCTION EXPANSIONS

Figure 4.12: Comparison of f (x) and the fiftieth partial sum of the Fourier series in Problem 8.

This series converges to   1 2   3/2

for −π < x < 0, for 0 < x < π, for x = 0, −π, π,

Figure 4.13 shows the function and the thirtieth partial sum of the Fourier series in Problem 9. 10. The Fourier series of f (x) on [−π, π] is ∞ 2 4 X (−1)n − cos(nx). π π n=1 4n2 − 1

This converges to cos(x/2) − sin(x) for −π < x < π, and to 0 at x = −π and at x = π. Figure 4.14 is a graph of f (x) and the fourth partial sum of the Fourier series. 11. The Fourier series of cos(x) on [−3, 3] is ∞ X (−1)n+1 1 sin(3) + 6 sin(3) cos(nπx/3). 3 π 2 n2 − 9 n=1

4.3. FOURIER SERIES

121

Figure 4.13: Comparison of f (x) and the thirtieth partial sum of the Fourier series in Problem 9.

Figure 4.14: Comparison of f (x) and the fourth partial sum of the Fourier series in Problem 10.

122CHAPTER 4. STURM-LIOUVILLE PROBLEMS AND EIGENFUNCTION EXPANSIONS

Figure 4.15: Comparison of f (x) and the fifth partial sum of the Fourier series in Problem 11.

This converges to cos(x) on [−3, 3]. Figure 4.15 is a graph of f (x) and the fifth partial sum of this Fourier expansion on [−3, 3]. It might seem at first that cos(x) should be its own Fourier expansion, but this problem illustrates the importance of the interval. If you expand cos(x) in a Fourier series on [−π, π], you obtain just cos(x). But this is not the expansion on [−3, 3]. 12. The Fourier expansion of f (x) on [−1, 1] is ∞ 3 X 1 − (−1)n 1 − 2(−1)n − cos(nπx) + sin(nπx) . 4 n=1 n2 π 2 nπ This converges to  1−x    0  1/2    1

for −1 < x < 0, for 0 < x < 1, at x = 0, for x = ±1.

Figure 4.16 is a graph of f (x) and the thirtieth partial sum of this Fourier series.

4.3. FOURIER SERIES

123

Figure 4.16: Comparison of f (x) and the thirtieth partial sum of the Fourier series in Problem 12.

13. The Fourier series of f (x) on [−3, 3] converges to  3/2      2x    −2  0      1/2    2 x

for x = ±3, for −3 < x < −2, for x = −2, for −2 < x < 1, for x = 1, for 1 < x < 3.

14. The Fourier series converges to   1      0 2    1/2    3/2

for x = ±1 and for 1/2 < x < 3/4, for −1 < x < 1/2, for 3/4 < x < 1, for x = 1/2, for x = 3/4.

124CHAPTER 4. STURM-LIOUVILLE PROBLEMS AND EIGENFUNCTION EXPANSIONS 15. The Fourier series converges to  −1 for x = ±4,    3/2 for x = −2,  5/2 for x = 2,    f (x) for all other x in [−4, 4]. 16. Suppose f (x) is an even function on [−L, L]. In the Fourier series for f (x) on [−L, L], the sine terms have coefficients Z 1 L f (x) sin(nπx/L) dx bn = L −L ! Z 0 Z L 1 = f (x) sin(nπx/L) dx + f (x) sin(nπx/L) dx . L −L 0 In the next to last integral in this equation, let x = −ξ. Using the fact that f (x) is an even function, and sin(−nπx/L) = − sin(nπx/L) then Z 0 f (x) sin(nπx/L) dx −L

Z 0 =

f (−ξ) sin(−nπξ/L)(−1) dξ L

Z L =−

f (ξ) sin(nπξ/L) d/xi. 0

Therefore each bn = 0. For the cosine terms, the coefficients are Z 1 L f (x) cos(nπx/L) dx an = L −L ! Z 0 Z L 1 = f (x) cos(nπx/L) dx + f (x) cos(nπx/L) dx . L −L 0 In the integral from −L to 0, make the change of variables x = −ξ, except now note that cos(nπx/L) = cos(−nπx/L) to show that Z 2 L f (x) cos(nπx/L) dx. an = L 0 17. The argument is like that used in Problem 16, except now use the fact that f (−x) = −f (x). 18. Define fe (x) =

(x) + f (−x) 2

4.3. FOURIER SERIES

125

and

f (x) − f (−x) . 2 It is routine to check that fe (x) is an even function and fo (x) is odd on [−L, L]. fo (x) =

19. Suppose f (x) is both even and odd on [−L, L]. Then, for every x in this interval, f (x) = f (−x) = −f (x). But then f (x) = 0, so f (x) is identically zero on the interval. 20. The Fourier cosine series for f (x) = 4 on [0, 3] is just the constant 4, converging to 4 at each point of the interval. The sine expansion of 4 on [0, 3] is ∞ X

Bn sin(nπx/3),

n=1

where Bn =

2 3

Because

Z 3 4 sin(nπx/3) dx = 0

( 2 1 − (−1) = 0 n

8 (1 − (−1)n ). nπ

if n is odd, if n is even,

this sine series can also be written 16 X 1 ∞ sin((2n − 1)πx/3). n =1 π 2n − 1 This sums over just the odd positive integers. The sine expansion converges to ( 4 for 0 < x < 4, 0 for x = 0 and for x = 3. 21. The cosine series is ∞

−

4 X (−1)n cos((2n − 1)πx/2). π n=1 2n − 1

This converges to   1 0   −1

for 0 ≤ x < 1, for x = 1, for 1 < x ≤ 2.

126CHAPTER 4. STURM-LIOUVILLE PROBLEMS AND EIGENFUNCTION EXPANSIONS

Figure 4.17: Comparison of f (x) and the thirtieth partial sum of the Fourier cosine series in Problem 21.

Figure 4.17 compares the function to the thirtieth partial sum of this cosine series. The sine series is ∞

2X (1 + (−1)n − 2 cos(nπ/2)) sin(nπx/2), π n=1 which converges to   1 0   −1

for 0 < x < 1, for x = 0, 1, 2, for 1 < x < 2.

Figure 4.18 shows f (x) and the seventieth partial sum of this sine expansion on [0, 2]. 22. The cosine series is 1 cos(x) + 2

∞ X n=1,n6=2

2n sin(nπ/2) cos(nx/2). π(n2 − 4)

4.3. FOURIER SERIES

127

Figure 4.18: Comparison of f (x) and the seventieth partial sum of the Fourier sine series in Problem 21.

This converges to  0 for 0 ≤ x < π,    −1/2 for x = π,  cos(x) for π < x < 2π,    1 for x = 2π, Graphs of the function and the fifteenth partial sum of this cosine series are shown in Figure 4.19. The sine series is −

∞ X 2 2n sin(x/2) − ((−1)n + cos(nπ/2)) sin(nx/2). 2 − 4)π 3π (n n=3

This converges to  0 for 0 ≤ x < π,    −1/2 for x = π,  cos(x) for π < x < 2π,    0 for x = 2π. Figure 4.20 shows the function and the fortieth partial sum of its sine expansion in Problem 22.

128CHAPTER 4. STURM-LIOUVILLE PROBLEMS AND EIGENFUNCTION EXPANSIONS

Figure 4.19: Comparison of f (x) and the fifteenth partial sum of the cosine series in Problem 22.

Figure 4.20: Comparison of f (x) and the fortieth partial sum of the sine expansion in Problem 22.

4.3. FOURIER SERIES

129

23. The cosine series is ∞

1−

1 8 X cos((2n − 1)πx), 2 π n=1 (2n − 1)2

converging to 2x for 0 ≤ x ≤ 1. The sine series is

∞

−

4 X (−1)n sin(nπx), π n=1 n

converging to 2x for 0 < x < 1 and to 0 for x = 0 and for x = 1. 24. The cosine series is ∞ 4 10 X (−1)n + cos(nπx/2), 3 π 2 n=1 n2

converging to x2 for 0 ≤ x ≤ 2. The sine expansion is ∞ 8 X (−1)n 2(1 − (−1)n ) − + sin(nπx/2). π n=1 n n3 π 2 This converges to x2 for 0 ≤ x < 2 and to 0 at x = 2. 25. The cosine series is −1 − e−1 + 2

∞ X 1 − (−1)n e−1 n=1

1 + n2 π 2

cos(nπx),

converging to e−x for 0 ≤ x ≤ 1. The sine series is 2π

∞ X n=1

n n −1 (1 − (−1) e ) sin(nπx), 1 + n2 π 2

converging to e−x for 0 < x < 1 and to 0 at x = 0 and at x = 1. 26. The cosine expansion of f (x) is ∞ 1 X + An cos(nπx/3), 2 n=1

where An =

−6(1 + (−1)n ) + 4nπ sin(2nπ/3) + 12 cos(2nπ/3) . n2 π 2

130CHAPTER 4. STURM-LIOUVILLE PROBLEMS AND EIGENFUNCTION EXPANSIONS

Figure 4.21: f (x) and the fortieth partial sum of the cosine series in Problem 26.

This cosine series converges to   x 1   2−x

for 0 ≤ x < 2, for x = 2, for 2 < x ≤ 3.

This fortieth partial sum of this cosine expansion of f (x) is shown in Figure 4.21. The sine expansion is

∞ X

Bn sin(nπx/3),

n=1

where Bn =

1 + sin(2nπ/3) − 4nπ cos(2nπ/3) + 2nπ(−1)n . n2 π 2

This sine series converges to  x    1  2−x    0

for 0 ≤ x < 2, for x = 2, for 2 < x < 3, for x = 3.

4.3. FOURIER SERIES

131

Figure 4.22: f (x) and the thirtieth partial sum of the sine series in Problem 26.

Figure 4.22 shows the function and the thirtieth partial sum of its sine expansion on [0, 3]. 27. The cosine expansion is ∞ 1 4X1 − + cos(nπ/5) sin(2nπ/5) cos(nπx/5), 5 π n=1 n

converging to   1      1/2 0    −1/2    −1

for 0 ≤ x < 1, for x = 1, for 1 < x < 3, for x = 3, for 3 < x < 5.

Figure 4.23 shows the function and the sixtieth partial sum of this cosine expansion. The sine series is ∞

4X 1 (1 + (−1)n − 2 cos(nπ/5) cos(2nπ/5)) sin(nπx/5), π n=1 2n

132CHAPTER 4. STURM-LIOUVILLE PROBLEMS AND EIGENFUNCTION EXPANSIONS

Figure 4.23: f (x) and the sixtieth partial sum of the cosine series in Problem 27.

converging to   1      1/2 0   −1/2    −1

for 0 < x < 1, for x = 1, for 1 < x < 3, x = 0, or x = 5, for x = 3, for 3 < x < 5.

Figure 4.24 shows f (x) and the one hundredth partial sum of its sine expansion on [0, 5]. 28. The cosine series is ∞ 5 16 X 1 4 + cos(nπ/4) − 3 sin(nπ/4) cos(nπx/4), 6 π 2 n=1 n2 n π converging to x2 for 0 ≤ x ≤ 1 and to 1 for 1 ≤ x ≤ 4. Figure 4.25 shows the function and the tenth partial sum of this cosine expansion. The sine series is ∞ X 16 64 2(−1)n sin(nπ/2) + 3 3 (cos(nπ/4) − 1) − sin(nπx/4), n2 π 2 n π nπ n=1

4.3. FOURIER SERIES

133

Figure 4.24: Comparison of f (x) and the hundredth partial sum of the sine series in Problem 27.

Figure 4.25: f (x) and the tenth partial sum of the Fourier cosine series in Problem 28.

134CHAPTER 4. STURM-LIOUVILLE PROBLEMS AND EIGENFUNCTION EXPANSIONS

Figure 4.26: f (x) and the eightieth partial sum of the sine series in Problem 28.

and this converges to x2 for 0 ≤ x ≤ 1, to 1 if 1 ≤ x < 4, and to 0 at x = 4. Figure 4.26 shows f (x) and the eightieth partial sum of its sine expansion on [0, 4]. 29. The cosine series is ∞ 24 X 1 4 n n −1 − 2 2(−1) + 2 2 (1 − (−1) ) cos(nπx/3), π n=1 n2 n π converging to 1 − x2 for 0 ≤ x ≤ 2. The sine series is ∞ 2X1 48 1 + 7(−1)n − 2 2 sin(nπx/2). π n=1 n n π This series converges to 1 − x2 for 0 < x < 2 and to 0 at x = 0 and x = 2. 30. It is routine to determine the Fourier sine expansion of sin(x) on [0, π]: ∞ 4X 1 2 − cos(2nx). π π n=1 4n2 − 1

4.3. FOURIER SERIES

135

This converges to sin(x) for 0 ≤ x ≤ π. Put x = π/2 in this expansion to obtain ∞ 2 4 X (−1)n sin(π/2) = 1 = − π π n=1 4n2 − 1 because cos(nπ) = (−1)n . Solve this for the series in question to obtain ∞ X (−1)n

4n2 − 1 n=1

1 π − . 2 4

136CHAPTER 4. STURM-LIOUVILLE PROBLEMS AND EIGENFUNCTION EXPANSIONS

Chapter 5

The Heat Equation 5.1

Diffusion Problems on a Bounded Medium

For the first three problems, separate the variables by letting u(x, t) = X(x)T (t). The boundary conditions u(0, t) = u(L, t) = 0 leads to eigenvalues λn =

n2 π 2 L2

for the separation constant, and corresponding eigenfunctions Xn (x) = sin(nπx/L). Corresponding solutions for T are 2

Tn (t) = e−n π kt/L . Solutions of these problems therefore all have the form u(x, t) =

∞ X

cn sin(nπx/L)e−kn π t/L ,

n=1

in which cn is determined by the initial condition u(x, 0) = f (x) by cn =

2 L

Z L f (ξ) sin(nπξ/L) dξ. 0

Therefore, for these problems, all we need do is evaluate these integrals for the coefficients. 1. With f (x) = x(L − x), cn =

2 L

Z L ξ(L − ξ) sin(nπξ/L) dξ = 0

4L2 (1 − (−1)n ). n3 π 3

137

138

CHAPTER 5. THE HEAT EQUATION The solution is u(x, t) =

∞ X 4L2

n3 π 3 n=1

(1 − (−1)n ) sin(nıx/L)e−kn π t/L .

We can also observe that ( 0 1 − (−1) = 2 n

if n is even, if n is odd,

so the solution can also be written by summing over just the odd positive integers. This can be done by replacing n with 2n − 1 in the summation: ∞

u(x, t) =

2 2 2 1 8L2 X sin((2n − 1)πx/L)e−k(2n−1) π t/L . 3 3 π n=1 (2n − 1)

2. Now k = 4 and u(x, 0) = f (x) = x2 (L − x), so the coefficients are cn =

2 L

Z L

ξ 2 (L − ξ) sin(nπξ/L) dξ = −

4L3 (1 + 2(−1)n ). n3 π 3

The solution is ∞

u(x, t) =

4L3 X π 3 n=1

1 + 2(−1)n n3

sin(nπx/L)e−4n π t/L .

3. Here k = 3 and f (x) = L(1 − cos(2πx/L)). Compute ( Z 8((−1)n −1) 2 L nπ(n2 −4) L(1 − cos(2πξ/L)) sin(nπξ/L) dξ = cn = L 0 0

if n 6= 2, for n = 2,

Because (−1)n − 1 = 0 if n is even, and −2 if n is odd, we actually have cn = −

16L nπ(n2 − 4)

for n = 1, 3, 5, · · · . We can write the solution as ∞

u(x, t) = −

2 2 2 16L X 1 sin((2n−1)πx/L)e−3(2n−1) π t/L . π n=1 (2n − 1)((2n − 1)2 − 4)

Problems 4–7 have insulated boundary conditions, so separation of variables by putting u(x, t) = X(x)T (t) leads to eigenvalues and eigenfunctions λ0 = 1, X1 = 1 and, for n = 1, 2, · · · , λn =

n2 π 2 , Xn (x) = cos(nπx/L). L2

5.1. DIFFUSION PROBLEMS ON A BOUNDED MEDIUM We also find that

139

Tn (t) = e−kn π t/L

as in the case of boundary conditions u(0, t) = u(L, t) = 0. Now the solution has the form ∞ X 2 2 2 1 u(x, t) = c0 + cn cos(nπx/L)e−kn π t/L , 2 n=1 where

2 f (ξ) cos(nπξ/L) dξ. L 4. Now k = 1, f (x) = sin(x) and L = π. The coefficients in the series for the solution are 4 2 c0 = sin(ξ) dξ = π π and, for n = 1, 2, · · · , ( n Z − π2 1+(−1) if n 6= 1, 2 π 2 −1 n sin(ξ) cos(nξ) dξ = cn = π 0 0 if n = 1. cn =

Because 1 + (−1)n = 0 if n is odd, each c2n−1 = 0 and we need sum over only even n. For n even cn = −

4 1 . π n2 − 1

The solution is u(x, t) =

∞ 2 2 4X 1 − cos(2nx)e−4n t . π π n=1 4n2 − 1

Here we let n = 1, 2, · · · in the summation, but replaced n with 2n in the terms of the series to sum over all the even positive integers. 5. Now k = 4, L = 2π and f (x) = x(2π − x)2 . The coefficients are Z 1 2π 4 c0 = ξ(2π − ξ)2 dξ = π 3 π 0 3 and, for n = 1, 2, · · · , 1 cn = π =−

Z 2π

ξ(2π − ξ)2 cos(nξ/2) dξ

16 2 2 (n π − 6(1 − (−1)n )). πn4

The solution is u(x, t) = −

2 3 π 3 ∞ X 16 n2 π 2 − 6(1 − (−1)n ) π n=1

cos(nx/2)e−n t .

140

CHAPTER 5. THE HEAT EQUATION

6. Now L = 3, k = 4 and f (x) = x sin(πx). Compute Z 2 3 2 c0 = ξ sin(πξ) dξ = , 3 0 π Z 18(−1)n 2 3 ξ sin(πξ) cos(nπξ/3) dξ = − cn = 3 0 π(n2 − 9) for n = 1, 2, 4, 5, · · · , and, for n = 3, Z 1 2 3 ξ sin(πξ) cos(πξ) dξ = − . c3 = 3 0 2π The solution is 2 1 1 − cos(πx)e−4π t π 2π ∞ X 18(−1)n+1 −4n2 π2 t/9 e . + π(n2 − 9)

u(x, t) =

n=1,n6=3

7. In this problem L = 6, k = 2 and f (x) = x cos(πx/4). The coefficients in the solution are Z 1 6 8 2 + 3π c0 = , ξ cos(πξ/4) dξ = − 3 0 3 π2 and, for n = 1, 2, · · · , Z 1 6 ξ cos(πξ/4) cos(nπξ/6) dξ cn = 6 0 24(−18 − 8n2 − 27π(−1)n + 12πn2 (−1)n ) = . π 2 (2n − 3)2 (2n + 3)2 The solution is 4 2 + 3π 3 π2 ∞ X 2 2 + cn cos(nπx/6)e−n π t/18 .

u(x, t) = −

n=1

8. The initial-boundary value problem modeling this setting is ut = kuxx for 0 < x < L, t > 0, ux (0, t) = ux (L, t) = 0, u(x, 0) = B. The coefficients in the series solution are Z 2 L c0 = B dξ = 2B L 0

5.1. DIFFUSION PROBLEMS ON A BOUNDED MEDIUM

141

and, for n = 1, 2, · · · , cn =

2 L

Z L B cos(nπξ/L) dξ = 0. 0

The solution is u(x, t) = B. This is consistent with intuition - with no energy loss, the bar maintains a constant temperature. 9. The initial-boundary value problem for the temperature function is ut = uxx for 0 < x < L, t > 0, u(0, t) = ux (L, t) = 0, Bx . u(x, 0) = L Separate variables by putting u(x, t) = X(x)T (t) to obtain X 00 + λX = 0; X(0) = X 0 (L) = 0 and T 00 + λkT = 0. By taking cases on λ, we find the eigenvalues and corresponding eigenfunctions: 2 (2n − 1)π and Xn (x) = sin((2n − 1)πx/2L). λn = 2L Further, 2

Tn (x) = e−k(2n−1) π t/4L . The solution has the form u(x, t) =

∞ X

cn sin((2n − 1)πx/2L)e−k(2n−1) π t/4L .

n=1

The coefficients are Z L

B ξ sin((2n − 1)πξ/2L) dξ 0 L −8B = 2 (−1)n . π (2n − 1)2

cn =

2 L

The solution is ∞

u(x, t) = −

2 2 2 8B X (−1)n sin((2n − 1)πx/2L)e−k(2n−1) π t/4L . π 2 n=1 (2n − 1)2

142

CHAPTER 5. THE HEAT EQUATION

10. The initial-boundary value problem for the temperature function is ut = 9uxx for 0 < x < 2, t > 0, u(0, t) = ux (2, t) = 0, u(x, 0) = x2 . A routine separation of variables leads to eigenvalues and eigenfunctions λn =

n2 π 2 , Xn (x) = sin((2n − 1)πx/4) 4

and solutions for the time-dependent parts as 2

Tn (t) = e−9n π t/4 . The solution has the form u(x, t) =

∞ X

cn sin((2n − 1)πx/4)e−9n π t/4 ,

n=1

where Z 2 cn =

ξ 2 sin((2n − 1)πξ/4) dξ

=−

64 2 + (2n − 1)π(−1)n . π3 (2n − 1)3

11. Make the transformation u(x, t) = eαx+βt v(x, t). Following the discussion of the text, let α = −A/2 = −4/2 = −2 and β = k(B − A2/4) = −2 also, so u(x, t) = e−2x−2t v(x, t) and v is the solution of the problem vt = vxx for 0 < x < π, t > 0, v(0, t) = v(π, t) = 0, v(x, 0) = e2x u(x, 0) = xe2x (π − x). This has the solution v(x, t) =

∞ X

cn sin(nx)e−n t ,

n=1

where 2 cn = π

Z π

ξe2ξ (π − ξ) sin(nξ) dξ

4 24n − 2n3 + 16nπ + 4n3 π − 24ne2π (−1)n 2 3 π(4 + n ) +2e2π n3 (−1)n + 16nπe2π (−1)n + 4n3 πe2π (−1)n .

=−

5.1. DIFFUSION PROBLEMS ON A BOUNDED MEDIUM

143

The solution of the original problem is u(x, t) = e−2x−2t v(x, t). 12. Again following the discussion in the text, we have A = 6, B = 0, k = 1, L = 4 and u(x, 0) = f (x) = 1. Find that α = −3 and β = −9 to let u(x, t) = e−3x−9t v(x, t). The v(x, t) is the solution of the standard problem vt = vxx for 0 < x < 4, t > 0, v(0, t) = v(4, t) = 0, v(x, 0) = e3x f (x) = e3x . This problem has the solution v(x, t) =

∞ X

cn sin(nπx/4)e−n π t/16 ,

n=1

where cn = =

1 2

Z 4

e3ξ sin(nπξ/4) dξ

2nπ (1 − e12 (−1)n ). 144 + n2 π 2

The original problem has the solution u(x, t) = e−3x−9t v(x, t) = e−3x−9t

∞ X

cn sin(nπx/4)e−n π t/16 .

n=1

13. Here we have A = 6, B = 0, k = 1, L = π and u(x, 0) = f (x) = x(π − x). Let α = 3 and β = −9 and let u(x, t) = e3x−9t v(x, t). The v(x, t) satisfies vt = vxx for 0 < x < π, t > 0, v(0, t) = v(π, t) = 0, v(x, 0) = e−3x f (x) = x(π − x)e−3x . The solution of this problem is v(x, t) =

∞ X

cn sin(nx)e−n t ,

n=1

144

CHAPTER 5. THE HEAT EQUATION where 2 cn = π =

Z π

e−3ξ ξ(π − ξ) sin(nξ) dξ

4n (1 − (−1)n e−3π )(3π(n2 + 9) + n2 − 27). π(n2 + 9)3

The original problem has the solution v(x, t) = e3x−9t v(x, t). 14. Because of the nonhomogeneous boundary conditions, we cannot solve this problem by a standard separation of variables. Transform the problem by letting u(x, t) = v(x, t) + ψ(x), where Lψ(x) must be chosen so that we have a problem for v(x, t) that we know how to solve. Substitute u(x, t) into the initial-boundary value problem to obtain vt = 16vxx + 16ψ 00 (x) for 0 < x < 1, t > 0, v(0, t) + ψ(0) = u(0, t) = 2, v(1, t) + ψ(1) = u(x, t) = 5, v(x, 0) + ψ(x) = u(x, 0) = x(1 − x)2 . First, simplify the differential equation for v by setting ψ 00 (x) = 0, so ψ(x) = cx + d. Now, we will have v(0, t) = 0 if ψ(0) = 2, and v(1, t) = 0 if ψ(1) = 5. Therefore choose d = 2 and c = 3, so ψ(x) = 3x + 2. The problem for v is vt = 16vxx for 0 < x < 1, t > 0, v(0, t) = v(1, t) = 0, v(x, 0) = u(x, 0) − ψ(x) = sin(πx) − 3x − 2. This problem has the solution v(x, t) =

∞ X

cn sin(nπx)e−16n π t ,

n=1

where Z 1 (sin(πξ) − 3ξ − 2) sin(nπξ) dξ

cn = 2 0

−1 + 10(−1)n nπ

5.1. DIFFUSION PROBLEMS ON A BOUNDED MEDIUM

145

if n = 2, 3, · · · , and Z 1 −14 + π c1 = 2 (sin(πξ) − 3ξ − 2) sin(πξ) dξ = . π 0 This determines v(x, t), and u(x, t) = v(x, t) + 3x + 2. 15. Let u(x, t) = v(x, t) + ψ(x). To get a standard problem for v(x, t), choose ψ(x) so that ψ 00 = 0 and ψ(0) = T, ψ(L) = 0. Then ψ(x) =

T (L − x) L

The problem for v is vt = kvxx for 0 < x < L, t > 0, v(0, t) = v(L, t) = 0, v(x, 0) = u(x, 0) − ψ(x) = x(L − x)2 −

T (L − x). L

This has the solution v(x, t) =

∞ X

cn sin(nπx/L)e−kn π t/L ,

n=1

where Z L T ξ(1 − ξ)2 − (L − ξ) sin(nπx/L) dξ L 0 2 2 2 = 3 3 −n π T + 4L3 + 2L3 (−1)n . n π

cn =

2 L

16. Here we have k = 4, L = 9 and u(x, 0) = x sin((9 − x)π). Let v(x, t) = ekAt v(x, t) and the problem for v(x, t) is vt = 4vxx for 0 < x < 9, t > 0, v(0, t) = v(9, t) = 0, v(x, 0) = u(x, 0) = x sin((9 − x)π). The solution for v(x, t) is v(x, t) =

∞ X

cn sin(nπx/9)e−4n π t/81 ,

n=1

146

CHAPTER 5. THE HEAT EQUATION where cn =

2 9

=−

Z 9 ξ sin((9 − ξ)π) sin(nπξ/9) dξ 0

324n(1 + (−1)n ) π 2 (n − 9)2 (n + 9)2

for n 6= 9, and 2 c9 = 9

Z 9 ξ sin((9 − ξ)π) sin(πξ) dξ = 0

9 . 2

The solution of the original problem is u(x, t) = ekAt v(x, t). 17. Let u(x, t) = v(x, t) + h(x) and substitute this into the initial-boundary value problem to choose h(x) and obtain a standard problem for v(x, t). We find that x h(x) = T 1 − L and the problem for v(x, t) is vt = 9vxx for 0 < x < L, t > 0, v(0, t) = v(L, t) = 0, x v(x, 0) = −T 1 − . L This has the solution v(x, t) =

∞ X

cn sin(nπx/L)e−9n π t/L ,

n=1

where 2 cn = L

Z L 0

2T ξ sin(nπξ/L) dξ = − . −T 1 − L nπ

Then ∞ 2 2 2 x 2T X 1 − sin(nπx/L)e−9n π t/L . u(x, t) = T 1 − L π n=1 n

18. The diffusion equation in this case is ut = duxx −

dη ux . L

Let u(x, t) = eαx+βt v(x, t) and solve for α and β to obtain a standard problem for v(x, t). We find that α=

dη d3 η 2 and β = − 2 . 2L 4L

5.2. THE HEAT EQUATION WITH A FORCING TERM F (X, T )

147

The problem for v(x, t) is vt = dvxx for 0 < x < L, t > 0, v(0, t) = v(L, t) = 0, kL −dηx/2L v(x, 0) = e−dηx/2L u(x, 0) = 1 − e−η(1−x/L) . e dη The solution for v(x, t) is v(x, t) =

∞ X

cn sin(nπx/L)e−dn π t/L ,

n=1

where Z L

kL −dηξ/2L e 1 − e−η(1−ξ/L) dξ 0 dη 4Lk d − 2 − de−η + 2e−dη/2 = 2 2 d η (d − 2)

cn =

2 L

if d 6= 2. If d = 2, then cn = −

Lk −η (e + 1 + η). η2

This determines v(x, t) and therefore u(x, t).

5.2

The Heat Equation With a Forcing Term F (x, t)

In Problems 1–5, notation of the text is used for Bn (t), bn and Tn (t). Note that the second term in the solution for u(x, t) is the solution to the problem without the forcing. 1. Here k = 4, L = π, f (x) = x(π − x) and F (x, t) = t. We need Z 2t 2 π Bn (t) = t sin(nξ) dξ = (1 − (−1)n ), π 0 nπ Z 2 π 4 bn = f (ξ) sin(nξ) dξ = (1 − (−1)n ), π 0 nπ 3 and Z t Tn (t) =

e−4n (t−τ ) Bn (τ ) dτ + bn e−4n t

2 1 (1 − (−1)n )(−1 + 4n2 t + e−4n t ). 8πn5

148

CHAPTER 5. THE HEAT EQUATION The solution is u(x, t) = +

∞ X

2 1 (1 − (−1)n )(−1 + 4n2 t + e−4n t ) sin(nx) 5 8πn n=1

∞ X 2 4 (1 − (−1)n ) sin(nx)e−4n t . 3 πn n=1

2. Compute 1 Bn (t) = 2

Z 4 ξ sin(t) sin(nπξ/4) dξ = 0

and bn =

1 2

Z 4 sin(nπξ/4) dξ = 0

8 (−1)n+1 sin(t) nπ

2 (1 − (−1)n ) nπ

and Tn (t) = +

∞ X

2 2 128(−1)n (16 cos(t) − n2 π 2 sin(t) − 16e−n π t/16 ) sin(nπx/4) 4 4 nπ(n π + 256) n=1

∞ X 2

nπ n=1

(1 − (−1)n ) sin(nπx/4)e−n π t/16 .

3. First, 2 Bn (t) = 5 =

bn =

2 5

Z 5

Z 5 t cos(ξ) sin(nπξ/5) dξ 0

2t ((−1)n+1 (nπ + 5) + nπ), n2 π 2 − 25

ξ 2 (5 − ξ) sin(nπξ/5) dξ =

500 ((−1)n+1 − 1) n3 π 3

and Tn (t) =

50(1 − cos(5)(−1)n ) 2 2 −n2 π 2 t/25 n π t − 25 + 25e . n3 π 3 (n2 π 2 − 25)

The solution is u(x, t) =

∞ X 50(1 − cos(5)(−1)n ) n=1

n3 π 3 (n2 π 2 − 25)

∞ X 500

n3 π 3 n=1

(n2 π 2 t − 25 + 25e−n π t/25 ) sin(nπx/5) 2

((−1)n+1 − 1) sin(nπx/5)e−n π t/25 .

5.2. THE HEAT EQUATION WITH A FORCING TERM F (X, T )

149

Figure 5.1: Solution surface for Problem 3, without effects of the forcing term included.

Figure 5.2: Solution surface for Problem 3, including effects of the forcing term.

150

CHAPTER 5. THE HEAT EQUATION

Sometimes a graphic can display features of a solution. This is done for Problem 3 (Section 5.2). Figure 5.1 shows part of a surface plot of the solution without the forcing term, and Figure 5.2 shows the solution with the forcing term. In Figure 5.1, the temperature decreases quickly to zero, without the introduction of new energy, while in Figure 5.2 this does not occur. 4. First we need Z 1 Bn (t) =

K sin(nπξ/2) dξ = 0

2K (1 − cos(nπ/2)), nπ

b1 = 1 and bn = 0 for n = 2, 3, · · · , and Tn (t) =

2 2 2K (1 − cos(nπ/2))(1 − e−n π t ). 3 3 n π

The solution is u(x, t) =

∞ X 2K

n3 π 3 n=1

(1 − cos(nπ/2))(1 − e−n π t ) sin(nπx/2) 2

+ sin(πx/2)e−π t . 5. First compute 2 Bn (t) = 3 bn =

2 3

Z 3 ξt sin(nπξ/3) dξ = 0

Z 3

and Tn (t) =

K sin(nπξ/3) dξ = 0

6t (−1)n+1 , nπ

2K (1 − (−1)n ), nπ

2 2 27(−1)n+1 (16n2 π 2 − 9 + 9e−16n π t/9 ). 128n5 π 5

The solution is u(x, t) =

∞ X 27(−1)n+1 n=1

∞ X 2K n=1

5.3

128n5 π 5 nπ

(16n2 π 2 − 9 + 9e−16n π t/9 ) sin(nπx/3)

(1 − (−1)n ) sin(nπx/5)e−16n π t/9 .

The Heat Equation on the Real Line

1. With f (x) = e−|x| , compute Z 1 ∞ −|ξ| 8 1 aω = e cos(ωξ) dξ = π −∞ π 16 + ω 2

5.3. THE HEAT EQUATION ON THE REAL LINE

151

and bω = 0 because f (x) is an even function on the real line. The solution is Z 2 1 8 ∞ cos(ωx)e−ω kt . u(x, t) = 2 π 0 16 + ω The solution can also be written in the form Z ∞ 2 1 e−|ξ| e−(x−ξ) /4kt dξ. u(x, t) = √ 2 πkt −∞ 2. The coefficients in the Fourier integral solution are aω = 0 because f (x) is an odd function, and Z 2 sin(ωπ) 1 π sin(ξ) sin(ωξ) dξ = . bω = π −π π(ω 2 − 1) The solution is u(x, t) =

2 π

Z ∞ 0

2 sin(ωπ) sin(ωx)e−ω kt . 2 ω −1

Alternatively, we can write the solution as Z π 2 1 u(x, t) = √ sin(ξ)e−(x−ξ) /4kt dξ. 2 πkt −π 3. The coefficients are Z 1 1 4 ξ cos(ωξ) dξ = (4ω sin(4ω) + cos(4ω) − 1) aω = π 0 πω 2 and bω =

1 π

Z 4 ξ sin(ωξ) dξ = 0

1 (sin(4ω) − 4ω cos(4ω)). πω 2

The solution is Z ∞ u(x, t) =

(aω cos(ωx) + bω sin(ωx))e−ω kt dω.

We can also write 1 u(x, t) = √ 2 πkt

Z 4

ξe−(x−ξ) /4kt dξ.

4. We need aω =

1 π

Z 1 −1

e−ξ cos(ωξ) dξ =

2 cos(ω) sinh(1) + ω sin(ω) cosh(1) π ω2 + 1

152

CHAPTER 5. THE HEAT EQUATION and 1 bω = π

Z 1

e−ξ sin(ωξ) dξ =

−1

2 ω cos(ω) sinh() − sin(ω) cosh(1)ω 2 + 1. π

The solution is Z ∞

2 u(x, t) = π

(aω cos(ωx) + bω sin(ωx))e−ω kt dω.

The solution can also be written 1 u(x, t) = √ 2 πkt

Z 1

e−ξ e−(x−ξ) /4kt dξ.

−1

In each of Problems 5–8, the solution has the form Z ∞ 2 u(x, t) = [aω cos(ωx) + bω sin(ωx)]e−ω kt 0

and just aω and bω are given 5. aω = 0 because f (x) is an even function, and bω = 6. aω = − and bω =

4(1 − cos(ω)) . πω

2 (2 sin(ω) + 3 sin(3ω) − 7 sin(9ω)) πω

2 (2 cos(ω) + 3 cos(3ω) − 7 cos(9ω) + 2) πω

7. Each bω = 0, while aω = 8. aω =

2 cos(πω/2) . π(1 − ω 2 )

2 (cos(ω) + 2ω sin(ω) − 2 + cos(2ω) + 3ω sin(2ω)), πω 2

and bω =

2 (− sin(ω) + 2ω cos(ω) + sin(2ω) − 3ω cos(2ω)) πω 2

9. Let

Z ∞ F (x) =

e−ζ cos(xζ) dζ.

5.4. THE HEAT EQUATION ON A HALF-LINE

153

By differentiating under the integral sign and then integrating by parts, we obtain Z ∞ 2 F 0 (x) = −ζe−ζ sin(xζ) dζ 0 ∞ Z ∞ 2 1 −ζ 2 1 = ζe sin(xζ) − x e−ζ cos(xζ) dζ 2 2 0 0 Z ∞ 1 −ζ 2 =− x e cos(xζ) dζ 2 0 1 = − xF (x). 2 The linear differential equation 1 F 0 (x) + xF (x) = 0 2 has the general solution 2

F (x) = ke−x /4 , with k an arbitrary constant. However, we also know that Z ∞ 2 1√ F (0) = e−ζ dζ = π, 2 0 an integral that can be found in tables and is widely used in statistics. Therefore Z ∞ 2 1 √ −x2 /4 F (x) = e−ζ cos(xζ) dζ = πe . 2 0 Now let x = α/β to obtain Z ∞ αζ 1 √ −α2 /4β 2 −ζ 2 e cos dζ = πe . β 2 0 Finally, this integral is half the value of the integral of the same function from −∞ to ∞, so Z ∞ √ 2 2 αζ −ζ 2 dζ = πe−α /4β . e cos β −∞ This is equation (5.17).

5.4

The Heat Equation on a Half-Line

In each of Problems 1–4, the initial condition is u(x, 0) = f (x) and the solution has the form Z ∞ 2

bω sin(ωx)e−kω t dω,

u(x, t) = 0

where bω =

2 π

Z ∞ f (ξ) sin(ωξ) dξ. 0

154

CHAPTER 5. THE HEAT EQUATION

1. Compute 2 bω = π

Z ∞

e−αξ sin(ωξ) dξ =

2 ω , π ω 2 + α2

so the solution is 2 π

u(x, t) =

Z ∞

ω 2 + α2

sin(ωx)e−kω t dω.

2. First, bω =

2 π

Z ∞

ξe−αξ sin(ωξ) dξ =

2 αω(α2 + ω 2 )2 , π

so the solution is u(x, t) =

Z ∞

4 π

2 αω sin(ωx)e−kω t dω. (α2 + ω 2 )2

3. The coefficients are 2 bω = π

Z h sin(ωξ) dξ = 0

2 1 − cos(hω) , π ω

and the solution is u(x, t) =

2 π

Z ∞ 0

2 1 − cos(hω) sin(ωx)e−kω t dω. ω

4. The coefficients are bω =

2 π

Z 2 ξ sin(ωξ) dξ = 0

2 sin(2ω) − 2ω cos(2ω) , π ω2

so the solution is 2 u(x, t) = π

Z ∞ 0

sin(2ω) − 2ω cos(2ω) ω2

sin(ωx)e−kω t dω.

In Problems 5–8, the heat equation is to be solved on the half-line x > 0, but the initial condition is now the insulation condition ux (x, 0) = f (x). Now the solution is Z ∞

aω cos(ωx)e−ω kt dω,

u(x, t) = 0

where aω =

2 π

Z ∞ f (ξ) cos(ωξ) dξ. 0

We will just give aω for each problem.

5.5. THE TWO-DIMENSIONAL HEAT EQUATION

155

5. 2 aω = π =

Z 4 ξ(ξ + 1) cos(ωξ) dξ 0

2 (20ω 2 sin(4ω) − ω − 2 sin(4ω) + 9ω cos(4ω)) πω 3

6. 2 aω = π =

Z π

ξ 2 cos(ωξ) dξ

2 (−2 sin(πω) + ω 2 π 2 sin(πω) + 2ωπ cos(πω) πω 3

7. Z 2 9 aω = 4 cos(ωξ) dξ π 5 8(sin(9ω) − sin(5ω)) = πω 8. 2 aω = π =

5.5

Z ∞

e−ξ sin(ξ) cos(ωξ) dξ

4 − 2ω 2 π(2 + 2ω + ω 2 )(2 − 2ω + ω 2 )

The Two-Dimensional Heat Equation

With the condition of zero initial temperature on the sides of the rectangle, and initial temperature u(x, y, 0) = f (x, y), the solution is ∞ X ∞ X

cnm sin(nπx/L) sin(mπy/K)e−αnm kt ,

n=1 n=1

where αnm =

n2 π 2 m2 π 2 + L2 K2

and cnm =

4 LK

Z LZ K f (ξ, η) sin(nπξ/L) sin(mπη/K) dη dξ. 0

In the problems we will give the values of αnm and cnm for the particular initial temperature function.

156

CHAPTER 5. THE HEAT EQUATION

1. Here k = 1, L and K are positive numbers, and f (x, y) = x(L − x)y 2 (K − y). Because f (x, y) is a product of a function of x and a function of y, we have ! Z ! Z L K 4 2 cnm = ξ(L − ξ) sin(nπξ/L) dξ η (K − η) sin(mπη/K) dη LK 0 0 =−

16 3 3 6 n m π (1 − (−1)n )(1 + 2(−1)m )) L2 K 3

and

n2 π 2 m2 π 2 + . L2 K2 2. Now k = 4, L = 2 and K = 3, and αnm =

f (x, y) = x2 (2 − x)(3 − y) sin(y). Now Z π Z π 4 2 ξ (2 − ξ) sin(nπξ/2) dξ (3 − η) sin(η) sin(mη/3) dη π2 0 0 2 3 16L K = 3 3 6 (−1 + (−1)n )(1 + 2(−1)m ). n m π

cnm =

And αnm =

m2 π 2 n2 π 2 + . 4 9

3. Now k = 1 and L = K = π, and Z π Z π 4 cnm = 2 sin(ξ) sin(nξ) dξ η cos(η/2) cos(mη) dη . π 0 0 Now, Z π 0

( π/2 sin(ξ) sin(nξ) dξ = 0

if n = 1, for n = 2, 3, · · · .

Therefore, in the double summation for u(x, y, t), we have only c1m terms and the summation is for m = 1 to ∞. Completing the computation of the integrations with respect to η, we obtain c1,m =

32m(−1)m+1 . (4m2 − 1)2

Further, α1m = 1 + m2 . The solution is u(x, y, t) =

∞ X 32m(−1)m+1 m=1

(4m2 − 1)2

sin(x) sin(my)e−(1+m )t .

Chapter 6

The Wave Equation 6.1

Wave Motion on an Interval

For each of Problems 1–8, the problem involves the wave equation on [0, L], with fixed ends, initial position y(x, 0) = f (x), and initial velocity yt (x, 0) = g(x). The solution is y(x, t) =

∞ X

[an cos(nπct/L) + bn sin(nπct/L)] sin(nπx/L),

n=1

where an =

2 L

Z L

and 2 bn = nπc

f (ξ) sin(nπξ/L) dξ 0

Z L g(ξ) sin(nπξ/L) dξ. 0

1. Here c = 1, L = 2, the initial position is given by f (x) = 0, and the initial velocity is ( 2x for 0 ≤ x ≤ 1, g(x) = 0 for 1 < x ≤ 2. In the general expression for the solution, then, we have an = 0 for n = 1, 2, · · · and bn =

2 nπ

Z 2 g(ξ) sin(nπξ/2) 0

Z 1 2 2ξ sin(nπξ/2) dξ nπ 0 8 = 3 3 [2 sin(nπ/2) − nπ cos(nπ/2)]. n π

157

158

CHAPTER 6. THE WAVE EQUATION The solution is ∞

y(x, t) =

8 X 1 [2 sin(nπ/2) − nπ cos(nπ/2)] sin(nπx/2) sin(nπt/2). π 3 n=1 n3

2. Now c = 3, L = 4, f (x) = 2 sin(πx) and g(x) = 0, so the string is lifted to its initial position and released from rest. Now each bn = 0, and ( Z 2 if n = 4, 1 4 1 sin(πξ) sin(nπξ/4)dξ = an = 2 0 0 for n = 1, 2, 3, 5, 6, · · · . The solution is y(x, t) = 2 sin(πx) cos(3πt). 3. Each an = 0 and bb =

1 nπ

Z 3 ξ(3 − ξ) sin(nπx/3) dξ = 0

54 (1 − (−1)n ). n4 π 4

The solution is ∞ X 54 y(x, t) = (1 − (−1)n ) sin(nπx/3) sin(2nπt/3). 4 π4 n n=1

Because (1 − (−1)n ) is 2 if n is odd, and zero if n is even, we can also write the solution by summing only over the odd positive integers. This is achieved by replacing n with 2n − 1 in the expression being summed, and replacing each (1 − (−1)n ) with 2: y(x, t) =

∞ X

108 sin((2n − 1)πx/3) sin(2(2n − 1)πt/3). (2n − 1)4 π 4 n=1

4. After carrying out the integrations for the coefficients, we get ∞

y(x, t) = sin(x) cos(3t) +

1 4 X sin((2n − 1)x) sin(3(2n − 1)t). 3π n=1 (2n − 1)2

5. The solution is y(x, t) =

∞ √ 24 X (−1)n+1 sin((2n − 1)x/2) cos((2n − 1) 2t). 2 π n=1 (2n − 1)

6. The solution is ∞

y(x, t) =

5 X 1 [5 sin(4nπ/5)+nπ cos(4nπ/5)] sin(nπx/5) sin(2nπt/5). π 3 n=1 n3

6.1. WAVE MOTION ON AN INTERVAL

159

7. The solution is ∞

y(x, t) = −

32 X 1 sin((2n − 1)πx/2) cos((3(2n − 1)πt/2) π 3 n=1 (2n − 1)3 ∞

4 X 1 [cos(nπ/4) − cos(nπ/2)] sin(nπx/2) sin(3nπt/2). π 2 n=1 n2

8. The solution is ∞

y(x, t) = sin(2x) cos(10t) +

2X 1 sin(nx) sin(5nt). 5 n=1 n2

9. Let y(x, t) = Y (x, t) + ψ(x) and substitute into the wave equation ytt = Ytt = 3yxx + 2x = 3Yxx + 3ψ 00 (x) + 2x. Choose ψ(x) so that 3ψ 00 (x) + 2x = 0. This means that 1 ψ(x) = − x3 + cx + d. 9 Now, y(0, t) = Y (0, t) + ψ(0) = 0 so let d = 0 to have ψ(0) = 0. Then Y (0, t) = 0. Next

8 + 2c = 0. 9 We will have Y (2, t) = 0 if c = 4/9. This means that y(2, t) = Y (2, t) + ψ(2) = Y (2, t) −

4 1 1 ψ(x) = − x3 + x = x(4 − x2 ). 9 9 9 The problem for Y (x, t) is Ytt = 3Yxx for 0 < x < 2, t > 0, Y (0, t) = Y (2, t) = 0, 1 Y (x, 0) = y(x, 0) − ψ(x) = x(x2 − 4). 9 The solution for Y (x, t) is Y (x, t) =

∞ X 32 (−1)n

3 n3 π 3 n=1

√ sin(nπx/2) cos(nπ t/2).

The original problem has the solution 1 y(x, t) = Y (x, t) + x(4 − x2 ). 9

160

CHAPTER 6. THE WAVE EQUATION

10. Here L = 4 and c = 3. Let y(x, t) = Y (x, t) + ψ(x) and substitute this into the wave equation to get Ytt = 9(Yxx + ψ 00 (x)) + x2 . Set 9ψ 00 (x) + x2 = 0. By two integrations, ψ(x) = −

1 4 x + cx + d. 108

Now, y(0, t) = Y (0, t) + ψ(0) = 0 will give us Y (0, t) = 0 if ψ(0) = 0. Thus choose d = 0. Next, y(4, t) = Y (4, t) + ψ(4) = 0 will give us Y (4, t) = 0 if ψ(4) = 0. Then c = 16/27, so 1 4 16 x + x. ψ(x) = 108 27 The problem for Y (x, t) is Ytt = 9Yxx for 0 < x < 4, t > 0, Y (0, t) = Y (4, t) = 0, Y (x, 0) = y(x, 0) − ψ(x) = sin(πx) +

x4 16 − x, 108 27

Yt (x, 0) = 0. This has a solution of the form Y (x, t) =

∞ X

an cos(3nπt/4) sin(nπx/4),

n=1

where Z 16 1 4 ξ4 − ξ sin(nπξ/4) dξ sin(πξ) + 2 0 108 27 8 1 = 128n2 − 128n1 (−1)n − 2048 9 n5 π 5 (n2 − 16) +2048(−1)n + 64n4 π 2 (−1)n − 1024n2 π 2 (−1)n ,

an =

if n 6= 4, and 1 ξ4 16 sin(πξ) + − ξ sin(πξ) dξ 2 108 27 3 1 8 + 9π = . 9 π3

a4 =

With these, the solution of the original problem is y(x, t) = Y (x, t) + ψ(x).

6.1. WAVE MOTION ON AN INTERVAL

161

11. Let y(x, t) = Y (x, t) + ψ(x). Substitute this into the wave equation to get ytt = Ytt = yxx = Yxx + ψ 00 (x) − cos(x). This will give us Ytt = Yxx if ψ 00 (x) = cos(x), which means that ψ(x) = − cos(x) + cx + d. Now y(0, t) = 0 = Y (0, t) + ψ(0) = −1 + d. This will give us Y (0, t) = 0 if d = 1. Next, y(2π, t) = 0 = Y (2π, t) − cos(2π) + 2πc + 1. This will give us Y (2π, 0) = 0 if c = 0. Then ψ(x) = − cos(x) + 1. Finally, y(x, 0) = Y (x, 0) − cos(x) + 1 = 0 implies that Y (x, 0) = cos(x) − 1. And yt (x, 0) = Yt (x, 0) = x. The problem for Y (x, t) is: Ytt = Yxx for 0 < x < 2π, t > 0, Y (0, t) = Y (2π, t) = 0, Y (x, 0) = cos(x) − 1, Yt (x, 0) = x. This has a solution of the form Y (x, t) =

∞ X

[an cos(nt/2) + bn sin(nt/2)] sin(nx/2),

n=1

where an = = and 2 bn = π

1 π (

Z 2π (cos(ξ) − 1) sin(nξ/2) dξ 0 16 nπ(n2 −4)

if n is odd,

if n is even,

Z 2π ξ sin(nξ/2) dξ = 0

8 (−1)n+1 . n2

These coefficients determine Y (x, t), and then y(x, t) = Y (x, t)+1−cos(x).

162

CHAPTER 6. THE WAVE EQUATION

12. Let y(x, t) = Y (x, t) + ψ(x) and substitute into the wave equation to get Ytt = 9(Yxx + ψ 00 (x)) + 5x3 . Choose ψ(x) so that 9ψ 00 (x) = −5x3 . This requires that ψ(x) = −

1 5 x + cx + d. 36

To have Y (0, t) = Y (4, t) = 0, we need ψ(0) = ψ(4) = 0. We get ψ(0) = 0 by choosing d = 0. And ψ(4) = −

1 4 4 + 4c = 0 36

requires that c = 64/9. Then ψ(x) = −

1 5 64 x + x. 36 9

Now Y satisfies: Ytt = 9Yxx for 0 < x < 4, t > 0, Y (0, t) = Y (4, t) = 0, Y (x, 0) = f (x) − ψ(x) = 1 − cos(πx) +

1 5 64 x − x, yt (x, 0) = 0. 36 9

This has the solution ∞ X

Y (x, t) =

an cos(3nπt/4) sin(nπx/4),

n=1

where 1 an = 2

Z 4 0

1 5 64 1 − cos(πξ) + ξ − x sin(nπξ/4) dξ. 36 9

We find that a4 =

20 8π 2 − 3 9 π5

while, for n 6= 4, 32

−9n5 π 5 + 30720nπ(−1)n + 9n5 π 5 (−1)n +320n5 π 3 (−1)n − 5120n3 π 3 (−1)n − 1920n3 π(−1)n .

an =

9n6 π 6 (n2 − 16)

The original problem has the solution y(x, t) = Y (x, t) + ψ(x).

6.1. WAVE MOTION ON AN INTERVAL

163

13. Let y(x, t) = Y (x, t) + ψ(x) and substitute this into the wave equation of the problem to choose ψ(x) so that 7ψ 00 (x) + e−x = 0, ψ(0) = ψ(2) = 0. This leads to

1 1 1 ψ(x) = − e−x + (e−2 − 1)x + . 7 14 7

Now Ytt = 7Yxx for 0 < x < 2, t > 0, Y (0, t) = Y (2, t) = 0, Y (x, 0) = −ψ(x), Yt (x, 0) = 5x. This has the solution Y (x, t) =

∞ h i X √ √ an cos(nπ 7t/2) + bn sin(nπ 7t/2) sin(nπx/2), n=1

where Z 2 1 −ξ 1 1 an = e − (e−2 − 1)ξ − sin(nπξ/2) dξ 7 14 7 0 2 (−4 − n2 π 2 e−2 (−1)n + e−1 n2 π 2 (−1)n + 4e−1 (−1)n ), = 7nπ(4 + n2 π 2 ) and 2 √ bn = nπ 7

Z 2 5ξ sin(nπξ/2) dξ = 0

40(−1)n+1 √ . n2 π 2 7

These coefficients determine Y (x, t), and then y(x, t) = Y (x, t) + ψ(x). 14. Let y(x, t) = Y (x, t) + ψ(x). Solve 4ψ 00 + cos(πx) = 0; ψ(0) = ψ(4) = 0 to get ψ(x) =

1 (cos(πx) − 1). 4π 2

Then Ytt = 4Yxx for 0 < x < 4, t > 0, Y (0, t) = Y (4, t) = 0, Y (x, 0) = x(π − x) − ψ(x), Yt (x, 0) = x2 . Then Y (x, t) =

∞ X

[an cos(nπt/2) + bn sin(nπt/2)] sin(nπx/4) dx,

n=1

164

CHAPTER 6. THE WAVE EQUATION where Z 4 1 ξ(4 − ξ) − 2 (cos(πξ) − 1) sin(nπξ/4) dξ 4π 0 8 =− 3 3 2 (128 − 7n2 + 7n2 (−1)n − 128(−1)n ), n π (n − 16)

1 an = 2

and 1 bn = πn =−

Z 4

ξ 2 sin(nπξ/4) dξ

64 (2(1 − (−1n ) + n2 π 2 (−1)n ). n4 π 4

Finally, y(x, t) = Y (x, t) + ψ(x). 15. (a) Substitute y(x, t) = X(x)T (t) into the fourth-order differential equation to get X (4) − λX = 0, T 00 + λa4 λT = 0, with λ the separation constant. Note - by rearranging terms differently, we can reach different separated equations for X and T . For example, we could have kept the a4 factor with the X terms. (b) Consider cases on λ, noting that the boundary conditions are X 00 (0) = X 00 (π) = X (3) (0) = X (3) (π) = 0. Case 1 - Suppose λ = 0. Then X (4) (x) = 0 and four integrations give us X(x) = A + Bx + Cx2 + DX 3 . The boundary conditions force C = D = 0, while A and B are arbitrary. Therefore 0 is an eigenvalue of this problem, with eigenfunctions X0 (x) = A + Bx, with A and B not both zero. In this case solutions for T are T (t) = α + βt. Case 2 - Suppose λ < 0. The notation is simplified if we set λ = −4α4 , with α > 0. The differential equation for X is X (4) + 4α4 X = 0, with characteristic equation r4 + 4α4 = 0. This has roots (1 + i)α, (1 − i)α, (−1 + i)α and (−1 − i)α. In this case solutions are X(x) = eαx (A cos(αx) + B sin(αx)) + e−αx (C cos(αx) + D sin(αx)).

6.1. WAVE MOTION ON AN INTERVAL

165

Apply the boundary conditions to this general solution to obtain: B − D = 0, A − B − C − D = 0, − Aeαπ sin(απ) + Beαπ cos(απ) + Ce−απ sin(απ) − De−απ cos(απ) − Aeαπ (cos(απ) + sin(απ)) + Beαπ (cos(απ) − sin(απ))

= 0,

− Ce−απ (cos(απ) − sin(απ)) − De−απ (cos(απ) + sin(απ)) = 0. This is a 4 × 4 homogeneous system of linear algebraic equations. This system has a nontrivial solution if and only if the determinant of the coefficients is zero: cosh(2απ) − cos(2απ) = 0. But this equation is satisfied only by α = 0, and in this case α > 0. Therefore the system has only the trivial solution A = B = C = D = 0, and this problem has no negative eigenvalue. Case 3 - Suppose λ > 0, say λ = α4 with α > 0. Now X (4) − α4 X = 0, and the characteristic equation has roots α, −α, αi, −αi. The general solution is X(x) = A cos(αx) + B sin(αx) + C cosh(αx) + D sinh(αx). The boundary conditions give us four equations: − A + C = 0, − A cos(απ) − B sin(απ) + C cosh(απ) + D sinh(απ) = 0, − B + D = 0, A sin(απ) + B cos(απ) + C sinh(απ + D cosh(απ) = 0. From the first and third equations, A = C and B = D. This reduces the system to the second and fourth equations in two unknowns: C(cosh(απ) − cos(απ)) + D(cosh(απ) − sin(απ)) = 0, C(sinh(απ) + sin(απ)) + D(cosh(απ) − cos(απ)) = 0. This 2 × 2 system has a nontrivial solution if and only if the determinant of the system is nonzero. This requires that cos(απ) cosh(απ) = 1. It may not be obvious, but this equation has infinitely many positive solutions for α (two are 2.499752670 and 0.000000207171091). If these solutions for α are listed α1 < α2 < · · · , then λn = αn4 is an eigenvalue of the problem. Eigenfunctions then have the form Xn (x) = A cos(αn x) + B sin(αn x) + C cosh(αn x) + D sinh(αn x).

166

CHAPTER 6. THE WAVE EQUATION

16. Separation of variables yields X 00 + λX = 0; X(0) = X(L) = 0 and T 00 + AT 0 + (B + c2 λ)T = 0; T 0 (0) = 0. The problem for X(x) is a familiar one, with eigenvalues and eigenfunctions n2 π 2 , Xn (t) = sin(nπx/L). λn = L2 With these eigenvalues, the characteristic equation for the differential equation for T is r2 + Ar + (B + c2 n2 π 2 /L2 ) = 0 with roots A 1 r=− ± 2 2

A2 − 4

c2 n 2 π 2 B+ . L2

The given condition that A2 L2 < 4(V L2 + c2 n2 π 2 ) ensures that these roots for r are complex. Let rn = 4(BL2 + c2 n2 π 2 ) − A2 L2 . The roots are then r=−

A rn ± i. 2 2L

Then Tn (t) = e−At/2 [an cos(rn t/2L) + bn sin(rn t/2L)]. Now T 0 (0) = 0 gives us −Aan /2 + bn rn /2L = 0, so bn =

AL an . rn

By superposition, u(x, t) = e−At/2

∞ X

AL sin(rn t/2L) sin(nπx/L). an cos(rn t/2L) + rn n=1

To satisfy u(x, 0) = f (x), choose an =

2 L

Z L f (ξ) sin(nπξ/L) dξ. 0

6.2. WAVE MOTION IN AN UNBOUNDED MEDIUM

6.2

167

Wave Motion in an Unbounded Medium

For the wave equation on a the real line, ytt = c2 yxx for − ∞ < x < ∞, t > 0, y(x, 0) = f (x), yt (x, 0) = 0, specifying an initial position but zero initial velocity, a solution can be found very much like the problem for an interval [0, L], with Fourier integrals replacing the Fourier series seen in the bounded interval case. The solution is Z ∞ y(x, t) = [aω cos(ωx) + bω sin(ωx)] cos(ωct) dω, 0

where aω =

1 π

Z ∞

1 π

Z ∞

and bω =

f (ξ) cos(ωξ) dξ −∞

f (ξ) sin(ωξ) dξ. −∞

If y(x, 0) = 0 and yt (x, 0) = g(x) (string released without initial displacement, but with initial velocity g(x)), then the solution is Z ∞ y(x, t) = [αω cos(ωx) + βω sin(ωx)] sin(ωct) dω, 0

where αω =

1 πωc

Z ∞

1 πωc

Z ∞

and βω =

g(ξ) cos(ωξ) dξ −∞

g(ξ) sin(ωξ) dξ. −∞

If the problem has f (x) and g(x) both nonzero, then the solution is the sum of the solution with zero initial velocity, and the solution with no initial displacement. 1. With c = 5, f (x) = e−5|x| and g(x) = 0, compute Z 10 1 ∞ −5|ξ| e cos(ωξ) dξ = aω = π −∞ (25 + ω 2 )π and bω = 0 because f (x) is an even function. The solution is Z 10 ∞ 1 y(x, t) = cos(ωx) cos(12ωt) dω. π 0 25 + ω 2

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2. The coefficients are

and

(8 − ξ) cos(ωξ) dξ = 0

Z 8

1 π

bω =

Z 8

1 π

aω =

(8 − ξ) sin(ωξ) dξ = 0

1 − cos(8ω) πω 2

8ω − sin(8ω) . πω 2

The solution is Z ∞

1 − cos(8ω) cos(ωx) cos(8ωt) dω y(x, t) = πω 2 0 Z ∞ 8ω − sin(8ω) + sin(ωx) cos(8ωt) dω. πω 2 0 3. Compute the coefficients to determine the solution Z ∞ sin(πω) y(x, t) = sin(ωx) sin(4ωt) dω. − 2πω(ω 2 − 1) 0 4. The solution is Z ∞ y(x, t) = 0

2 (2 − cos(2ω)) cos(ωx) cos(ωt) dω. πω 2

5. The solution is Z ∞ y(x, t) =

[αω cos(ωx) + βω sin(ωx)] sin(3ωt) dω, 0

where 1 3πω

Z ∞

1 βω = 3πω

Z ∞

αω =

e−2ξ cos(ωξ) dξ =

1 3e2 πω

2 cos(ω) − ω sin(ω) 4 + ω2

and e−2ξ sin(ωξ) dξ =

1 2 sin(ω) + ω cos(ω) . 3e2 πω 4 + ω2

Z ∞

6. The solution is y(x, t) = 0

1 − cos(2ω) πω 2

sin(ωx) sin(2ωt) dω.

7. The solution for the problem with the given displacement and zero initial velocity is Z ∞ y1 (x, t) = [aω cos(ωx) + bω sin(ωx)] cos(7ωt) dω, 0

6.2. WAVE MOTION IN AN UNBOUNDED MEDIUM

169

where Z 1 5 f (ξ) cos(ωξ) dξ π −1 sin(ω) − 2 sin(2ω) + 3 sin(5ω) = πω

aω =

and Z 1 5 f (ξ) sin(ωξ) dξ bω = π −1 cos(ω) + 2 cos(2ω) − 3 cos(5ω) = . πω The solution for the problem with the given velocity, but zero initial displacement, is Z ∞ y2 (x, t) = [αω cos(ωx) + βω sin(ωx)] sin(7ωt) dω, 0

where αω =

1 7πω

=−

Z 1

e−|ξ| cos(ωξ) dξ

−1

2 (e−1 cos(ω) − ωe−1 sin(ω) − 1) 7π(1 + ω 2 )

and βω =

1 7πω

Z 1

e−|ξ| sin(ωξ) dξ = 0.

−1

The solution of the problem with initial displacement f (x) and initial velocity g(x) is y(x, t) = y1 (x, t) + y2 (x, t). 8. Let y1 (x, t) be the solution of the problem with initial displacement f (x) and zero initial velocity, and y2 (x, t) the solution with no initial displacement, but initial velocity g(x). Then y(x, t) = y1 (x, t) + y2 (x, t). Now,

Z ∞ y1 (x, t) =

√ [aω cos(ωx) + bω sin(ωx)] cos( 7ωt).

where 1 aω = π =

Z 3π sin(ξ) cos(ωξ) dξ −π

4 cos(πω) sin2 (πω) π(ω 2 − 1)

170

CHAPTER 6. THE WAVE EQUATION and 1 bω = π = And

Z ∞ y2 (x, t) =

Z 3π sin(ξ) sin(ωξ) dξ −π

4 sin(πω) cos2 (πω) . π(1 − ω 2 )

√ [αω cos(ωx) + βω sin(ωx)] sin( 7ωt),

where Z ∞ 1 αω = √ g(ξ) cos(ωξ) dξ 7πω −∞ 2 √ (8ω 2 sin(4ω) − sin(4ω) + 4ω cos(4ω)) = 2 πω 7 and Z ∞ 1 βω = √ g(ξ) sin(ωξ) dξ 7πω −∞ −2 (1 + 8ω 2 cos(4ω) − cos(4ω) − 4ω sin(4ω)). =√ 7πω 4 Then y(x, t) = y1 (x, t) + y2 (x, t). 9. Following the notation of Problems 7 and 8, form y1 (x, t) with coefficients 1 aω = π

Z 2 ξ cos(ωξ) dξ = 0 −2

and Z 1 2 bω = ξ sin(ωξ) dξ π −2 2 sin(2ω) − 4 cos(2ω) = . πω 2 And y2 (x, t) has coefficients Z 3 4 ξ 2 cos(ωξ) dξ πω −3 8 = (9ω 2 sin(3ω) − 2 sin(3ω) + 6ω cos(3ω)) πω 4

αω =

and βω = 0.

6.2. WAVE MOTION IN AN UNBOUNDED MEDIUM

171

The solution is Z ∞ y(x, t) =

an cos(ωx) cos(ωt/4) Z 0∞

βω sin(ωx) sin(ωt/4). 0

The the problem on a half-line [0, ∞), there is a boundary condition which we will take to be y(0, t) = 0 along with initial conditions y(x, 0) = f (x), yt (x, 0) = g(x) for x > 0. The solution has the form Z ∞ y(x, t) = [Aω cos(ωct) + Bω sin(ωct)] sin(nωx) dω, 0

where

2 π

Z ∞

2 Bω = π

Z ∞

Aω = and

f (ξ) sin(ωξ) dξ 0

g(ξ) sin(ωξ) dξ. 0

10. With zero initial velocity, we need only compute Z 2 1 ξ(1 − ξ) sin(ωξ) dξ Aω = π 0 2 2 sin(ω) (1 − cos(ω)) − . = π ω3 ω2 The solution is y(x, t) =

2 π

Z ∞ 0

2 sin(ω) (1 − cos(ω)) − sin(ωx) cos(3ωt) dω. ω3 ω2

11. Here Aω = 0 and Z 11 2 2 sin(ωξ) dξ 3πω 4 4(cos(4ω) − cos(11ω)) = . 3πω 2

Bω =

The solution is y(x, t) =

4 3π

Z ∞ 0

cos(4ω) − cos(11ω) sin(ωx) sin(3ωt) dω. ω2

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CHAPTER 6. THE WAVE EQUATION

12. Aω = 0 and 2 Bω = 2πω =

Z 5π/2 cos(ξ) sin(ωξ) dξ π/2

sin(ωπ/2) − sin(5ωπ/2) . πω(ω 2 − 1)

The solution is Z ∞ y(x, t) = 0

sin(ωπ/2) − sin(5ωπ/2) sin(ωx) sin(2ωt) dω. πω(ω 2 − 1)

13. With g(x) = 0, Bω = 0 and 2 Aω = π =−

Z ∞

−2e−ξ sin(ωξ) dξ

4ω . π(1 + ω 2 )

The solution is 4 y(x, t) = − π

Z ∞ 0

ω sin(ωx) cos(6ωt) dω. 1 + ω2

14. Now Aω = 0 and Z 3 2 Bω = ξ 2 (3 − ξ) sin(ωξ) dξ 14πω 0 3 (2 sin(3ω) − 4ω cos(3ω) − 3ω 2 sin(3ω) − 2ω). = 7πω 5 The solution is Z ∞ y(x, t) =

Bω sin(ωx) sin(14ωt) dω. 0

15. Compute Z 2 1 f (ξ) sin(ωξ) dξ π 0 2 sin(ω) = 2 π − ω2

Aω =

and Z 4 2 √ Bω = g(ξ) sin(ωξ) dξ 13πω 0 2 =√ (1 − 2 cos(ω) + cos(4ω)). 13πω 2

6.3. D’ALEMBERT’S SOLUTION AND CHARACTERISTICS

173

The solution is Z ∞ y(x, t) =

√ √ [Aω cos( 13ωt) + Bω sin( 13ωt)] sin(ωx) dω.

16. Compute the coefficients Z 12ω 2 ∞ −3ξ ξe sin(ωξ) dξ = Aω = π 0 π(9 + ω 2 )2 and Bω =

2 5πω

Z 2π cos(ξ) sin(ωξ) dξ = 0

4 sin2 (πω) . (w2 − 1)π

The solution is Z ∞ y(x, t) =

[Aω cos(5ωt) + Bω sin(5ωt)] sin(ωx) dω. 0

6.3

d’Alembert’s Solution and Characteristics

1. The characteristics are the lines x − t = k1 and x + t = k2 , with k1 and k2 arbitrary real numbers. d’Alembert’s solution of the problem is Z 1 1 x+t 2 2 y(x, t) = [(x − t) + (x + t) ] + −ξ dξ 2 2 x−t x+t 1 1 = [x2 − 2xt + t2 + x2 + 2xt + t2 − ξ 2 2 2 x−t = x2 − xt + t2 . 2. characteristics: x − 4t = k1 , x + 4t = k2 ; 1 (x − 4t)2 − 2(x − 4t) + (x + 4t)2 − 2(x + 4t) 2 Z 1 x+4t + (cos(ξ) dξ 8 x−4t 1 = x2 + 16t2 − 2x + cos(x) sin(4t) 4

y(x, t) =

3. characteristics: x − 7t = k1 , x + 7t = k2 ; y(x, t) =

1 49 [cos(π(x − 7t)) + cos(π(x + 7t))] + t − x2 t − t3 2 3

This solution can also be written y(x, t) =

49 1 cos(πx) cos(7πt) + t − x2 t − t3 . 2 3

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CHAPTER 6. THE WAVE EQUATION

4. characteristics: x − 5t = k1 , x + 5t = k2 ; 1 [sin(2(x − 5t)) + sin(2(x + 5t))] + x3 t + 25xt3 2 = sin(2x) cos(10t) + x3 t + 25xt3

y(x, t) =

5. characteristics: x − 14t = k1 , x + 14t = k2 ; 1 x−14t e + ex+14t + xt 2 = ex cosh(14t) + xt.

y(x, t) =

6. characteristics: x − 12t = k1 , x + 12t = k2 ; y(x, t) = x2 + 144t2 − 5x + 3t. 7. characteristics x −

√ √ 3t = k1 , x + 3t = k2 ; √ i 1 h −3|x−√3t| + e−3|x+ 3t| e 2 " √ !# √ ! 1 x − 3t x + 3t + √ sin − sin 2 2 3

y(x, t) =

8. characteristics: x − 9t = k1 , x + 9t = k2 ; 1 (2 − cos(x − 9t) − cos(x + 9t)) 2 1 + e−x+9t (cos(−x + 9t) − sin(−x + 9t)) 36 1 − e−x−9t (cos(x + 9t) + sin(x + 9t)) 36

y(x, t) =

9. The solution with y(x, 0) = f (x) = sin(x) is y(x, t) =

1 (sin(x − t) + sin(x + t)). x

With y(x, 0) = sin(x) + , the solution is y (x, t) =

1 sin(x − t) + + sin(x + t) + = y(x, t) + . 2

10. Let y(x, t) = F (x − ct) + G(x + ct) = (x − ct)2 − (x − ct) + (x + ct) cos(x + ct).

6.3. D’ALEMBERT’S SOLUTION AND CHARACTERISTICS

175

Compute yx (x, t) = 2(x − ct) − 1 + cos(x + ct) − (x + ct) sin(x + xt), yxx (x, t) = 2 − sin(x + ct) − sin(x + ct) − (x + ct) cos(x + ct), yt (x, t) = 2(x − ct)(−c) + c + c cos(x + ct) − c(x + ct) sin(x + ct), and ytt (x, t) = 2c2 − c2 sin(x + ct) − c2 sin(x + ct) − c2 (x + ct) cos(x + ct). Now it is routine to check that ytt = c2 yxx . 11. Now y(x, t) = e−3(x−ct) + sin(4(x + ct) = e−3x e3ct + sin(4(x + ct)). Then yx = −3e−3x e3ct + 4 cos(4(x + ct)), yxx = 9e−3x e3ct − 16 sin(4(c + ct)), yt = 3ce−3x e3ct + 4c cos(4(x + ct)), ytt = 9c2 e−3x e3ct − 16c2 sin(4(x + ct)). It is easy to check that ytt = c2 yxx . In each of Problems 12–17, with c = 1 and g(x) = 0, the forward wave is F (x, t) = 21 f (x − t) and the backward wave is B(x, t) = 12 f (x + t). The solution is y(x, t) = F (x, t) + B(x, t). 12. Figures 6.1–6.7 are graphs of y(x, t) for 1 3 t = 0, , , 2, 3, 4, 6 2 4 respectively. 13. Figures 6.8–6.13 show graphs of y(x, t) for times 3 1 3 7 t = 0, , , , 1, . 2 4 8 2

176

CHAPTER 6. THE WAVE EQUATION

Figure 6.1: y(x, 0) in Problem 12.

Figure 6.2: y(x, 1/2), Problem 12.

Figure 6.3: y(x, 3/4), Problem 12.

6.3. D’ALEMBERT’S SOLUTION AND CHARACTERISTICS

177

Figure 6.4: y(x, 2), Problem 12.

Figure 6.5: y(x, 3), Problem 12.

Figure 6.6: y(x, 4), Problem 12.

178

CHAPTER 6. THE WAVE EQUATION

Figure 6.7: y(x, 6), Problem 12.

Figure 6.8: y(x, 0), Problem 13.

Figure 6.9: y(x, 1/2), Problem 13.

6.3. D’ALEMBERT’S SOLUTION AND CHARACTERISTICS

179

Figure 6.10: y(x, 3/4), Problem 13.

Figure 6.11: y(x, 7/8), Problem 13.

Figure 6.12: y(x, 1), Problem 13.

180

CHAPTER 6. THE WAVE EQUATION

Figure 6.13: y(x, 3/2), Problem 13.

Figure 6.14: y(x, 0), Problem 14.

Figure 6.15: y(x, 1/4), Problem 14.

6.3. D’ALEMBERT’S SOLUTION AND CHARACTERISTICS

181

Figure 6.16: y(x, 1/2), Problem 14.

Figure 6.17: y(x, 3/4), Problem 14.

Figure 6.18: y(x, 1), Problem 14.

182

CHAPTER 6. THE WAVE EQUATION

Figure 6.19: y(x, 5/4), Problem 14.

14. Figures 6.14–6.21 show the graphs of y(x, t) for times 5 3 1 1 3 t = 0, , , , 1, , , 2. 4 2 4 4 2 15. Figures 6.22–6.27 show graphs of y(x, t) for 1 1 3 3 t = 0, , , , 1, . 4 2 4 2 16. Figures 6.28–6.34 show graphs of y(x, t) for 5 7 1 1 3 t = 0, , , , 1, , . 4 2 4 4 4 17. Figures 6.35–6.41 show graphs of y(x, t) for 1 1 3 5/4 7 5 t = 0, , , , , . 4 2 4 , 4 2 18. Derive d’Alembert’s solution as follows. Begin with the fact that any solution of the wave equation on the line must look like y(x, t) = F (x + ct) + G(x − ct), for some twice differentiable functions F and G. To satisfy the initial conditions, we need y(x, 0) = F (x) + G(x) = f (x) and, by chain rule differentiations, yt (x, 0) = −cF 0 (x) + cG0 (x) = g(x).

6.3. D’ALEMBERT’S SOLUTION AND CHARACTERISTICS

183

Figure 6.20: y(x, 3/2), Problem 14.

Figure 6.21: y(x, 2), Problem 14.

Figure 6.22: y(x, 0), Problem 15.

184

CHAPTER 6. THE WAVE EQUATION

Figure 6.23: y(x, 1/4), Problem 15.

Figure 6.24: y(x, 1/2), Problem 15.

Figure 6.25: y(x, 3/4), Problem 15.

6.3. D’ALEMBERT’S SOLUTION AND CHARACTERISTICS

185

Figure 6.26: y(x, 1), Problem 15.

Integrate the last equation to obtain Z 1 x −F (x) + G(x) = g(ξ) dξ − F (0) + G(0). c 0 Add this to the equation for y(x, 0) to obtain Z 1 x 2G(x) = f (x) + g(ξ) dξ − F (0) + G(0). c 0 Solve this for G(x) to obtain 1 1 G(x) = f (x) + 2 2c

Z x 0

1 1 g(ξ) dξ − F (0) + G(0). 2 2

But then, from the equation for y(x, 0), F (x) = f (x) − G(x) Z x 1 1 1 1 = f (x) − g(ξ) dξ + F (0) − G(0). 2 2c 0 2 2 From the last two equations, y(x, t) = F (x − ct) + G(x + ct) Z x−ct 1 1 1 = f (x − ct) − g(ξ) dξ + (F (0) − G(0)) 2 2c 0 2 Z x 1 1 1 + f (x + ct) + g(ξ) dξ − (F (0) − G(0)). 2 2c 0 2 Upon combining the integrals and canceling terms where possible, this yields d’Alembert’s formula. 19. We know that y(x, t) =

1 1 (f (x − ct) + f (x + ct)) + 2 2c

Z x+ct g(ξ) dξ x−ct

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CHAPTER 6. THE WAVE EQUATION

Figure 6.27: y(x, 3/2), Problem 15.

Figure 6.28: y(x, 0), Problem 16.

Figure 6.29: y(x, 1/4), Problem 16.

6.3. D’ALEMBERT’S SOLUTION AND CHARACTERISTICS

187

Figure 6.30: y(x, 1/2), Problem 16.

and ỹ(x, t) =

1 ˜ 1 (f (x − ct) + f˜(x + ct)) + g̃(ξ) dξ. 2 2c

Then y(x, t) − ỹ(x, t) Z x+ct 1 1 (f (x − ct) + f (x + ct)) + g(ξ) dξ 2 2c x−ct 1 Z x+ct 1 ˜ ˜ f (x − ct) + f (x + ct) − − g̃(ξ) dξ 2 2 x−ct 1 1 = (f (x − ct) − f˜(x − ct)) + (f (x + ct) − f˜(x + ct)) 2 2 Z x+ct 1 (g(ξ) − g̃(ξ)) dξ. + 2c x−ct =

188

CHAPTER 6. THE WAVE EQUATION

Figure 6.31: y(x, 3/4), Problem 16.

Figure 6.32: y(x, 1), Problem 16.

Figure 6.33: y(x, 5/4), Problem 16.

6.3. D’ALEMBERT’S SOLUTION AND CHARACTERISTICS

189

Figure 6.34: y(x, 7/4), Problem 16.

Figure 6.35: y(x, 0), Problem 17.

190

CHAPTER 6. THE WAVE EQUATION velocity go (x), the d’Alembert solution is y(x, t) =

1 1 (fo (x − ct) + fo (x + ct)) + 2 2c

Z x+ct g(ξ) dξ. x−ct

This function satisfies the wave equation for all x, and therefore also for all x > 0. Further, if x ≥ 0, y(x, 0) =

1 1 (fo (x) + fo (x)) = (f (x) + f (x)) = f (x). 2 2

And, 1 1 (−cfo0 (x) + cfo0 (x)) + (cgo (x) − (−c)go (x)) 2 2c = go (x) = g(x).

yt (x, 0) =

Therefore y(x, t) is also a solution of the problem on the half-line x ≥ 0.

6.4

The Wave Equation With a Forcing Term K(x, t)

1. Here c = 4, f (x) = x, g(x) = e−x and K(x, t) = x + t. The solution is Z 1 x+4t −ξ 1 ((x − 4t) + (x + 4t)) + e dξ 2 8 x−4t Z Z 1 t x+4t−4T + (X + T ) dX dT 8 0 x−4t+4T 1 −x+4t e − e−x−4t =x+ 8 Z t + (xt − xT + tT − T 2 ) dT

y(x, t) =

1 1 1 = x + e−x e4t − e−4t + t3 + xt2 . 8 6 2 We can also write this solution as 1 1 1 y(x, t) = x + e−x sinh(4t) + t3 + xt2 . 4 6 2 2. We are given c = 2, f (x) = sin(x), g(x) = 2x and K(x, t) = 2xt. The

6.4. THE WAVE EQUATION WITH A FORCING TERM K(X, T )

191

solution is Z 1 1 x+2t 2ξ dξ (sin(x − 2t) + sin(x + 2t)) + 2 4 x−2t Z Z 1 t x+2t−2T (X + T ) dX dT 8 0 x−2t+2T 1 1 = (sin(x − 2t) + sin(x + 2t)) + (x + 2t)2 − (x − 2t)2 2 4 Z 1 t + T (x + 2t − 2T )2 − (x − 2t + 2T )2 dT. 4 0

y(x, t) =

After carrying out the last integration, this expression reduces to y(x, t) =

1 1 (sin(x − 2t) + sin(x + 2t)) + 2xt + xt3 . 2 3

3. The solution is Z x+8t 1 1 (f (x − 8t) + f (x + 8t)) + cos(2ξ) dξ 2 16 x−8t Z t Z x+8t−8T 1 + XT 2 dX dT 16 0 x−8t+8T 1 = x2 + 64t2 − x + (sin(−2x + 16t) + sin(2x + 16t)) 32 Z t + −xT 2 (−t + T ) dT

y(x, t) =

= x2 + 64t2 − x +

1 1 (sin(−2x + 16t) + sin(2x + 16t)) + xt4 . 32 32

4. Z 1 x+4t −ξ 1 (x − 4t)2 + (x + 4t)2 + ξe dξ 2 8 x−4t Z Z 1 t x+4t−4T + X sin(T ) dX dT 8 0 x−4t+4T 1 1 1 = x2 + 16t2 + e−x+4t + xe−x+4t − te−x+4t 8 8 2 1 1 1 − e−x−4t − xe−x+4t − te−x−4t 8 8 2 Z t + x(t − T ) sin(T ) dT

y(x, t) =

1 1 = x2 + 16t2 + (1 + x)e−x+4t − t e−x+4t + e−x−4t 8 2 1 − (1 + x)e−x−4t + xt − x sin(t). 8

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CHAPTER 6. THE WAVE EQUATION

5. 1 1 (cosh(x − 3t) + cosh(x + 3t)) + 2 6 Z t Z x+3t−3T 1 3XT 3 dX dT + 6 0 x−3t+3T

Z x+3t dξ

y(x, t) =

x−3t

Z t 1 (cosh(x − 3t) + cosh(x + 3t)) + t + −3xT 3 (T − t) dT 2 0 1 3 = (cosh(x − 3t) + cosh(x + 3t)) + t + xt5 . 2 20

6. 1 ((1 + x − 7t) + (x + x + 7t)) 2 Z t Z x+7t−7T 1 + (X − cos(T )) dX dT 14 0 x−7t+7T Z t =1+x+ (xt − xT − t cos(T ) + T cos(T )) dT

y(x, t) =

1 = x + xt2 + cos(t). 2

6.5

The Wave Equation in Higher Dimensions

For Problems 1–3 the solution has the form z(x, y, t) =

∞ ∞ X X

anm sin(nπx/L) sin(mπy/K) cos(αnm πct),

n=1 m=1

where

r αnm =

and anm =

4 LK

n2 m2 + L2 K2

Z LZ K f (ξ, η) sin(nπξ/L) sin(mπη/K) dη dξ. 0

1. Because f (x, y) = x2 y is a product of a function of x and a function of y, we can compute the coefficients as a product of integrals: Z 2π Z K 1 ξ 2 sin(nξ/2) dξ η sin(mη/2) dη anm = 2 π 0 0 32(−1)m = 2(1 − (−1)n ) + n2 π 2 (−1)n . 3 mn π Further, αnm =

1 p 2 n + m2 . 2π

6.5. THE WAVE EQUATION IN HIGHER DIMENSIONS

193

Because c = 1, the solution is z(x, y, t) =

∞ X ∞ X 32(−1)m

(2(1 − (−1)n ) 3π mn n=1 m=1 p + n2 π 2 (−1)n sin(nx/2) sin(my/2) cos( n2 + m2 t/2).

2. Now c = 3, L = 1 and K = 4, and f (x, y) = sin(πx) cos(πy/2). The coefficients in the solution are Z 1 Z 4 anm = sin(πξ) sin(nπξ) dξ cos(πη/2) sin(mπη/4) dη. 0

Now

( 0 for n = 2, 3, · · · , sin(πξ) sin(nπξ) dξ = 1/2 for n = 1. 0

Z 1

And ( 0 m cos(πη/2) sin(mπη/4) dη = ) − 4m(−1+(−1) 0 π(m2 −4)

Z 4

if m = 2, if m 6= 2.

This means that n will only assume the value n = 1 and the solution is a single summation over m, with m = 2 excluded. Further, r m2 α1m = 1 + 16 for m = 1, 3, 4, 5, · · · . Then z(x, y, t) =

∞ X

−

m=1,m6=2

4m(−1 + (−1)m ) sin(nπx) sin(mπy/4) cos(3α1m πt). 2π(m2 − 4)

3. We have c = 1, L = K = π and f (x, y) = xey . A routine integration yields Z π Z π 4 anm = 2 ξ sin(nξ) dξ eη sin(mη) dη π 0 0 4(−1)n+1 m π (1 − e (−1)m ). = πn2 (m2 + 1) Further, αnm =

1p 2 n + m2 . π

The solution is z(x, y, t) =

∞ X ∞ X

p anm sin(nx) sin(my) cos(2 n2 + m2 t).

n=1 m=1

194

CHAPTER 6. THE WAVE EQUATION

4. Now suppose that the membrane is initially unmoved and that an initial velocity is given by zt (x, y, 0) = g(x, y). Separate variables in the twodimensional wave equation. Because the boundary conditions are the same as in the zero initial velocity case, we obtain λn =

n2 π 2 , Xn (x) = sin(nπx/L) L2

µm =

m2 π 2 , Ym (y) = sin(mπy/K). K2

and

The problem for T is T 00 + c2

n2 π 2 m2 π 2 + 2 L K2

T = 0,

but now we have T (0) = 0, not T 0 (0) = 0. Then Tnm (t) = cos(αnm πct), where

n2 m2 + 2. 2 L K We therefore are led to attempt a solution αm =

∞ X ∞ X

z(x, y, t) =

bnm sin(nπx/L) sin(mπy/K) cos(αnm πct).

n=1 m=1

We need zt (x, y, 0) =

∞ ∞ X X

αnm πc sin(nπx/L) sin(mπy/K) = g(x, y).

n=1 m=1

This is a Fourier double series expansion of g(x, y), and the coefficients are Z LZ K 4 bnm αnm πc = g(ξ, η) sin(nπξ/L) sin(mπη/K). LK 0 0 Then bnm =

4 LKαnm πc

Z LZ K g(ξ, η) sin(nπξ/L) sin(mπη/K). 0

5. Suppose c = 3, L = K = π, f (x, y) = 0 and g(x, y) = xy. Now αnm =

1p 2 n + m2 . π

6.5. THE WAVE EQUATION IN HIGHER DIMENSIONS

195

Compute 4 (−1)n+m bnm = √ . nm 3 n2 + m2 The solution is ∞ X ∞ X

z(x, y, t) =

p bnm sin(nx) sin(mx) cos(3 n2 + m2 t).

n=1 m=1

√ 6. With L = K = 2π, we have αnm = 21 n2 + m2 . Compute Z 2π Z 2π 4 √ sin(mη) dη bnm = sin(nξ) dξ 1 (2π)(2π) 2π n2 + m2 2π 0 0 1 4 = √ (1 − (−1)n )(1 − (−1)m ). 2 2 2 π n + m nm The solution is z(x, y, t) =

∞ X

p 2 2 sum∞ m=1 bnm sin(nx/2) sin(my/2) cos( n + m t)

n=1

196

CHAPTER 6. THE WAVE EQUATION

Chapter 7

Laplace’s Equation 7.1

The Dirichlet Problem for a Rectangle

1. Substitute u(x, y) = X(x)Y (y) into Laplace’s equation to obtain X 00 + λX = 0; X(0) = X(1) = 0 and Y 00 − λY = 0; Y (π) = 0. Solutions for X are λ = n2 π 2 , Xn (x) = sin(nπx). With these values of λ, the problem for Y (y) has solutions that are constant multiples of sinh(nπ(π − y). To find a solution satisfying the boundary condition u(x, 0) = sin(πx), use a superposition u(x, y) =

∞ X

an sin(nπx) sinh(nπ(π − y)).

n=1

We need u(x, 0) =

∞ X

an sin(πx) sinh(nπ 2 ) = sin(πx).

n=1

We can compute the Fourier coefficients of this sine expansion, or simply observe that we can take an = 0 for n = 2, 3, · · · and a1 =

1 . sinh(π 2 )

The solution is u(x, y) =

1 (sin(πx) sinh(π(π − y))). sinh(π 2 ) 197

198

CHAPTER 7. LAPLACE’S EQUATION

2. Separate variables and use the boundary conditions u(x, 0) = u(x, 2) = 0 to obtain the problem Y 00 + λY = 0; Y (0) = Y (2) = 0, with eigenvalues and eigenfunctions λn =

n2 π 2 , Yn (y) = sin(nπy/2). L2

The problem for X is X 00 −

n2 π 2 X = 0; X(3) = 0. L2

Solutions are constant multiples of sinh(nπ(3 − x)/2). Now attempt a solution of the form u(x, y) =

∞ X

bn sinh(nπ(3 − x)/2) sin(nπy/2).

n=1

To satisfy u(3, y) = 0, we need u(0, y) = y(2 − y) =

∞ X

bn sinh(3nπ/2) sin(nπy/2).

n=1

This is the Fourier sine expansion of y(2 − y) on [0, 2], and the coefficients are Z 2 1 bn = η(2 − η) sin(nπη/2) dη sinh(3nπ/2) 0 16 1 − (−1)n = . sinh(3nπ/2) n3 π 3 The solution is ∞

u(x, y) =

16 X 1 − (−1)n sinh(nπ(3 − x)/2) sin(nπy/2). π 3 n=1 n3 sinh(3nπ/2)

3. After separating the variables and applying the boundary conditions, we find that the solution has the form u(x, y) =

∞ X n=1

sinh(nπy) sin(nπx). sinh(4nπ)

The coefficients must be determined so that u(x, 4) =

∞ X

an sin(nπx) = x cos(πx/2).

n=1

7.1. THE DIRICHLET PROBLEM FOR A RECTANGLE

199

This is a Fourier sine expansion of x cos(πx/2) on [0, 1]. Choose the coefficients as Z 1 32n(−1)n+1 . ξ cos(πξ/2) sin(nπξ) dξ = 2 an = 2 π (4n2 − 1)2 0 These determine the solution. 5. There are nonhomogeneous boundary conditions on two sides of the rectangle, so write u(x, y) = v(x, y) + w(x, y) where ∇2 v = 0; v(0, y) = v(π, y) = v(x, 0) = 0, v(x, π) = x sin(πx) and ∇2 w = 0; w(x, 0) = w(x, π) = w(0, y) = 0, w(w, y) = sin(y). These are defined on 0 < x < 2, 0 < y < π. Solve these problems independently. First, separate variables in the problem for w to find that it has a solution of the form ∞ X sinh(nx) w(x, y) = bn sin(ny) . sinh(2n) n=1 Observe that we can solve this problem for w by taking b1 = 1 and all other bn = 0, so sinh(x) . w(x, y) = sin(y) sinh(2) The problem for v has a solution of the form v(x, y) =

∞ X

an sin(nπx/2)

n=1

We need v(x, π) = x sin(πx) =

∞ X

sinh(nπy/2) . sinh(nπ 2 /2)

an sin(nπx/2).

n=1

This is a Fourier sine expansion of x sin(πx) on [0, 2], so choose Z 2 bn =

ξ sin(nξ) sin(nπξ/2) dξ (0

16n n π 2 ((n2 −4)2 ) ((−1) − 1)

for n = 1, 3, 4, 5, · · · ,

for n = 2.

200

CHAPTER 7. LAPLACE’S EQUATION Then sinh(πy) sinh(π 2 ) ∞ X n

v(x, y) = sin(πx) +

10 π2

n=1,n6=2

(n2 − 4)2

((−1)n − 1) sin(nπx/2)

sinh(nπy/2) . sinh(nπ 2 /2)

6. A routine separation of variables, together with the zero boundary conditions on sides y = 0, y = b and x = 0, lead to a solution of the form u(x, y) =

∞ X

an sin((2n − 1)πy/2)

n=1

sinh((2n − 1)πx/2b) . sinh((2n − 1)πa/2b)

Now we need to choose the coefficients so that u(a, y) = g(y) =

∞ X

an sin((2n − 1)πy/2b).

n=1

Then 2 an = b

Z b g(η) sin((2n − 1)πη/2b) dη. 0

7. Separation of variables and the zero boundary conditions on x = 0, x = a and y = 0 yield a general form of the solution: u(x, y) =

∞ X

an sin((2n − 1)πx/2a)

n=1

sinh((2n − 1)πy/2a) . sinh((2n − 1)πb/2a)

Now we need u(x, 0) = f (x) =

∞ X

an sin((2n − 1)πx/2a).

n=1

This leads us to choose 2 an = a

Z a f (ξ) sin((2n − 1)πξ/2a) dξ. 0

8. The homogeneous conditions on the sides x = 0, x = a and y = 0 lead to the form of the solution: u(x, y) =

∞ sin(nπx/a)

n=1

Then 2

u(x, a) = x(x − a) =

∞ X

sinh(nπy/a) . sinh(nπb/a)

an sin(nπx/a).

n=1

7.1. THE DIRICHLET PROBLEM FOR A RECTANGLE

201

The coefficients in this eigenfunction expansion are Z 2 a ξ(ξ − a)2 sin(nπξ/a) dξ an = a 0 4 = 3 3 (1 + 2nπ(−1)n ). n π 9. Write the solution as u(x, y) = v(x, y) + w(x, y), where v is the solution of the problem ∇2 v = 0, v(x, 0) = v(x, 1) = v(4, y) = 0, v(0, y) = sin(πy), and w is the solution of ∇2 (w) = 0, w(x, 0) = w(x, 1) = w(0, y) = 0, w(4, y) = y(1 − y). These problems are defined on 0 ≤ x ≤ 4, 0 ≤ y ≤ 1. A separation of variables yields a general form of the solution of the problem for v: v(x, y) =

∞ X

an sin(nπy)

n=1

We need v(0, y) = sin(πy) =

sinh(nπ(4 − x)) . sinh(4nπ)

∞ X

an sin(nπy),

n=1

so by observation we can let a1 = 1 and an = 0 for n = 2, 3, · · · . Then v(x, y) = sin(πy)

sinh(π(4 − x)) . sinh(4π)

Another separation of variables leads to a general form of the solution for w: ∞ X w(x, y) = bn sin(nπy) sinh(nπx). n=1

Then w(4, y) = y(1 − y) =

∞ X

bn sinh(4nπ) sin(nπy),

n=1

so Z 1 2 ξ(1 − ξ) sin(nπξ) dξ sinh(4nπ) 0 4(1 − (−1)n ) = 3 3 . n π sinh(4nπ)

bn =

202

CHAPTER 7. LAPLACE’S EQUATION

7.2

The Dirichlet Problem for a Disk

For each of Problems 1–8, a solution u(r, θ) =

∞ n X 1 r a0 + [an cos(nθ) + bn sin(nθ)], 2 R n=1

where an =

1 πRn

Z π

1 πRn

Z π

f (ξ) cos(nξ) dξ −π

for n = 0, 1, 2, · · · and bn =

f (ξ) sin(nξ) dξ −π

for n = 1, 2, · · · . 1. We can see by observation that u(r, θ) = 1 is a solution. This can also be obtained the long way by carrying out the integrations, obtaining a0 = 2 and an = bn = 0 for n = 1, 2, · · · . 2. As with Problem 1, we can obtain the solution by inspection by matching coefficients of the proposed series solution to settle on only one nonzero coefficient, namely a4 . If we carry out the integrations, we get Z π 1 an = 8 cos(4ξ) cos(nξ) dξ = 0 π3n −π for n = 0, 1, 2, 3, 5, 6, · · · , and Z π 1 bn = 8 cos(4ξ) sin(nξ) dξ = 0 π3n −π for n = 1, 2, · · · , while 1 a4 = π34

Z π 8 cos(4ξ) cos(4ξ) dξ = −π

8 . 34

The solution therefore consists of a single term: u(r, θ) = 8 3. Calculate a0 = an =

1 π

Z π

2n π

−π

Z π −π

r 4 3

cos(4θ).

(ξ 2 − ξ) dξ =

2π 2 , 3

(ξ 2 − ξ) cos(nξ) dξ =

4(−1)n n2 2 n

7.2. THE DIRICHLET PROBLEM FOR A DISK and

1 bn = n 2 π

Z π

(ξ 2 − ξ) sin(nξ) dξ =

−π

203

2(−1)n . n2n

The solution is u(r, θ) =

∞ n X π2 (−1)n r [2 cos(nθ) + n sin(nθ)]. +2 3 2 n2 n=1

4. After carrying out the integrations, the solution is u(r, θ) = −

∞ X 1 (−1)n n r n sin(nθ). r sin(θ) + 2 10 n2 − 1 5 n=2

5. The solution is u(r, θ) =

∞ 2 X (−1)n r n sinh(π) + sinh(π)[cos(nθ) + n sin(nθ)]. π π n=1 n2 + 1 4

6. If we write sin2 (θ) = (1 − cos(2θ))/2, we can identify the only nonzero coefficients in the series solution as a0 = 1/2 and a2 = −1. The solution is 1 r u(r, θ) = − cos(2θ). 2 2 7. After the integrations, we obtain the solution ∞ X 4(−1)n+1 r n 1 u(r, θ) = 1 − π 2 + cos(nθ). 3 n2 8 n=1

8. After integrating to find the coefficients, the solution is 2 cosh(2π) − sinh(2π) 4π ∞ 2 X (−1)n r n + [cosh(2π)(8π + 2nπ 2 ) + sinh(2π)(n2 + 4)] cos(nθ) π n=1 (n2 + 4)2 4

u(r, θ) =

∞

2 X (−1)n r n [cosh(2π)(4nπ + n3 π 3 ) − 4n sinh(2π)] sin(nθ). π n=1 (n2 + 4)2 4

9. Let U (r, θ) = u(r cos(θ), r sin(θ). The problem given in rectangular coordinates converts to the following problem in polar coordinates: ∇2 U (r, θ) = 0 for 0 ≤ r < 4, U (4, θ) = 16 cos2 (θ). If we write 16 cos2 (θ) = 8(1 + cos(2θ), we can recognize by inspection that 1 a0 = 8, a2 (42 ) = 8, 2

204

CHAPTER 7. LAPLACE’S EQUATION and all other an = 0. The solution in polar coordinates is r 2 U (r, θ) = 8 + 8 cos(2θ). 4 Because the original problem was posed in rectangular coordinates, convert this to rectangular coordinates by using x = r cos(θ), y = r sin(θ), and the identity cos(2θ) = 2 cos2 (θ) − 1, to obtain 1 u(x, y) = 8 + (x2 − y 2 ). 2

10. In polar coordinates this Dirichlet problem is ∇2 U (r, θ) = 0 for 0 ≤ r < 3, U (3, θ) = 3(cos(θ) − sin(θ)). In the series solution for this problem, identify 3a1 = 3 and 3b1 = 3, with all other coefficients zero. Then U (r, θ) = r(cos(θ) − sin(θ)). In rectangular coordinates, this solution is u(x, y) = x − y. Perhaps by hindsight, this could have been seen with no computation because x − y is harmonic on the entire plane. 11. In polar coordinates this problem is ∇2 U (r, θ) = 0 for 0 ≤ r < 2, U (2, θ) = 4(cos2 (θ) − sin2 (θ)) = 4 cos(2θ). Identify a2 22 = 4, with all other coefficients zero, so U (r, θ) = r2 cos(2θ). In rectangular coordinates the solution is u(x, y) = x2 − y 2 . 12. In polar coordinates the problem is ∇2 U (r, θ) = 0 for 0 ≤ r < 5, U (5, θ) = 25 sin(θ) cos(θ). If we write

25 sin(2θ) 2 we can identify the coefficients in the series solution: a2 (52 ) = 25/2, with all other coefficients zero. The solution is 1 U (r, θ) = r2 sin(2θ). 2 To convert this to rectangular coordinates, use 1 2 r sin(2θ) = r sin(θ)r cos(θ) 2 to obtain u(x, y) = xy. U (5, θ) =

7.3. THE POISSON INTEGRAL FORMULA

7.3

205

The Poisson Integral Formula

In Problems 1–4 the idea is to use the Poisson integral formula to write the requested solution value as an integral which can be approximated by a numerical technique. This assumes the availability of software that will do this. For Problem 1, the Poisson integrals are given for each approximate value calculated. For Problems 2, 3 and 4, we give just the approximate values. 1. With R = 1 and f (θ) = θ, the solution is Z pi 1 − r2 1 ξ dξ. U (r, θ) = 2 2π −π 1 + r − 2r cos(ξ − θ) Then

1 U (1/2, π) = 2π

Z π

3ξ dξ = 0. 5 − 4 cos(ξ − π) −π

Next, U (3/4, π/3) =

1 2π

Z π

7ξ dξ ≈ 0.8826128645. 25 − 14 cos(ξ − π/3) −π

And 1 U (1/5, π/4) = 2π

Z π

24ξ dξ ≈ 0.2465422. −π 26 − 10 cos(ξ − π/4)

2. U (1, π/6) ≈ 0.0033829, U (3, 7π/2) ≈ 0, U (1, π/4) ≈ 0, U (2.5, π/12) ≈ 0.132145. 3. U (4, π) ≈ −16.4654, U (12, π/6) ≈ 0.0694, U (8, π/4) ≈ 1.5281 4. U (5.5, 3π/5) ≈ 0.8230986392, U (4, 2π/7) ≈ 1.609925382, U (1, π) ≈ 3.986586385, U (4, 9π/4) ≈ .461163821

7.4

The Dirichlet Problem for Unbounded Regions

1. If we put f (ξ) = K in equation (7.7), we get the solution Z Ky ∞ 1 u(x, y) = dξ π −∞ y 2 + (ξ − x)2 Z L 1 Ky lim dξ = 2 π L→∞ −L y + (ξ − x)2 K L−x −L − x = lim arctan − arctan π L→∞ y y K π π = + = K. π 2 2

206

CHAPTER 7. LAPLACE’S EQUATION Or, we can avoid this computation by observing that u(x, y) = K is harmonic on the entire plane, and equals K on the real line (the boundary of the upper half-plane). Therefore the solution is u(x, y) = K.

2. By equation (7.7), the solution of this problem for the upper half-plane is Z y ∞ 1 u(x, y) = e−|ξ| dξ. π −∞ y 2 + (ξ − x)2 3. By equation (7.7), the solution is Z ξ y ∞ dξ. u(x, y) = π −∞ y 2 + (ξ − x)2 4. The solution is Z y k 1 u(x, y) = dξ π −k y 2 + (ξ − x)2 ξ−x k 1 = arctan π y −k 1 k−x −k − x = arctan − arctan π y y 1 k−x k+x = arctan + arctan , π y y in which the last line makes use of the fact that the arctangent is an odd function. 5. Suppose u(x, y) is harmonic on the upper half-plane and u(x, 0) = f (x). Then the function v(x, y) defined by v(x, y) = u(x, −y) on the lower halfplane is harmonic, and v(x, 0) = f (x). But we know an integral formula for u(x, y). Therefore the problem for the lower half-plane has the solution Z y ∞ f (ξ) v(x, y) = u(x, −y) = − dξ. π −∞ y 2 + (ξ − x)2 for all x and for y < 0. 6. By the result of Problem 5, the solution of this problem for the lower half-plane is Z f (ξ) y ∞ u(x, y) = − dξ π −∞ y 2 + (ξ − x)2 Z Z −1 y 1 1 y 0 dξ − dξ =− 2 2 2 π −1 y + (ξ − x) π 0 y + (ξ − x)2 1 x x+1 x−1 =− 2 arctan − arctan − arctan . π y y y

7.4. THE DIRICHLET PROBLEM FOR UNBOUNDED REGIONS

207

7. The boundary of the right quarter-plane consists of the nonnegative horizontal and vertical axes. Define two Dirichlet problems, in each of which boundary data is nonzero on just one part of the boundary: Problem 1 ∇2 v = 0 for x > 0, y > 0 and v(x, 0) = f (x), v(0, y) = 0, and Problem 2 ∇2 w = 0 for x > 0, y > 0 and w(x, 0) = 0, w(0, y) = g(y). Both problems can be solved by separation of variables and Fourier integrals, obtaining Z Z ∞ 2 ∞ f (ξ) sin(ωξ) dξ sin(ωx)e−ωy dω v(x, y) = π 0 0 and w(x, y) =

2 π

Z ∞ Z ∞ 0

g(η) sin(ωη) dη sin(ωy)e−ωx dω.

The solution of the original problem is u(x, y) = v(x, y) + w(x, y). 8. Using the result of Problem 7, compute Z ∞ bω = e−3ξ sin(ωξ) dξ = 0

ω . 9 + ω2

The solution is u(x, y) =

2 π

Z ∞ 0

ω sin(ωx)e−ωy dω. 9 + ω2

9. Z 1 Ay 8 dξ u(x, y) = π 4 y 2 + (ξ − x)2 A x−4 x−8 = arctan − arctan π y y 10. The two-dimensional heat equation is ut = k(uxx + uyy ) = k∇2 u. In the steady-state case, ut = 0 and u satisfies Laplace’s equation. Use the result of Problem 7. The solution is Z 2 ∞ u(x, y) = Bω sin(ωy)e−ωx dω, π 0 where Bω =

2 π

Z ∞ 0

e−η sin(ωη) dη =

ω . 1 + ω2

208

CHAPTER 7. LAPLACE’S EQUATION

11. Using the results of Problem 7, the solution is Z ∞ u(x, y) = [bω sin(ωx)e−ωy + Bω sin(ωy)e−ωx ] dω, 0

where Z 2 ∞ ξ sin(ωξ) dξ π 0 2 sin(πω) − 2ωπ cos(πω) = πω 2

bω =

and Z 2 ∞ 2 Bω = η sin(ωη) dη π 0 4(cos(πω) − 1) − 2ω 2 π 2 cos(πω) + 4π sin(πω) = . πω 3

7.5

A Dirichlet Problem in 3 Dimensions

1. Let u(x, y, z) = X(x)Y (y)Z(z) to separate variables, obtaining first that X 00 + λX = 0; X(0) = X(1) = 0 and Y 00 + µY − 0; Y (0) = Y (1) = 0. Then λn = n2 π 2 , Xn (x) = sin(nπx) and µm = m2 π 2 , Ym (y) = sin(mπy). Further, Z 00 − (n2 + m2 )π 2 Z = 0; Z(0) = 0. This leads to functions unm (x, y, z) = cnm sin(nπx) sin(mπy) sinh(αnm πz), √ where αnm = n + m2 . To satisfy the condition u(x, y, 1) = xy, use a superposition u(x, y, z) =

∞ X ∞ X

cnm sin(nπx) sin(mπy) sinh(αnm πz).

n=1 m=1

We must choose the coefficients so that u(x, y, 1) = xy =

∞ X ∞ X

sin(nπx) sin(mπy) sinh(αnm π).

n=1 m=1

7.5. A DIRICHLET PROBLEM IN 3 DIMENSIONS

209

This is a double Fourier series on the square 0 ≤ x ≤ 1, 0 ≤ y ≤ 1, and we know from experience with the heat and wave equations that Z 1 Z 1 4 ξ sinh(nπξ) dξ η sin(mπη) dη sinh(αnm π) 0 0 4(−1)n+m = . nmπ 2 sinh(αnm π)

cnm =

2. Again separate variables, except now exploit the zero boundary conditions on sides z = 0, z = 1, y = 0, y = 2π and x = 0 to obtain Sturm-Liouville problems for Y (y) and Z(z): Y 00 + λY = 0; Y (0) = Y (2π) = 0 and Z 00 + µZ = 0; Z(0) = Z(1) = 0. Then λn =

n2 , Yn (y) = sin(ny/2) 4

and µm = m2 π 2 , Zm (z) = sin(nπz). Further Xnm (x) = sinh(αnm x/2) √ where αnm = n2 + 4m2 π 2 . Therefore use the superposition u(x, y, z) =

∞ ∞ X X

cnm sin(ny/2) sin(mπz) sinh(αnm x/2).

n=1 m=1

For the solution, choose the coefficients so that u(2π, y, z) = z. Thus choose Z Z 1 1 1 2π 2 sin(nη/2) dη 2 sin(mπτ ) dτ sinh(αnm π) π 0 0 4 = (1 − (−1)n )(1 − (−1)m ). sinh(αnm π)

cnm =

3. The solution is the sum of the solutions of the following two problems: ∇2 w = 0, w(0, y, z) = w(1, y, z) = w(x, 0, z) = w(x, 2π, z) = w(x, y, 0) = 0, w(x, y, π) = 1

210

CHAPTER 7. LAPLACE’S EQUATION and ∇2 v = 0, v(0, y, z) = v(1, y, z) = v(x, y, 0) = v(x, y, π) = v(x, 0, z) = 0, v(x, 2π, z) = xz 2 . Each of these problems is solved by a separation of variables. For the first, we obtain ∞ X ∞ p X w(x, yz) = anm sin(nπx) sin(my/2) sinh( 4n2 π 2 + m2 z/2), n=1 m=1

in which Z 1 Z 2π 1 1 √ 2 sin(nπξ) dξ sin(nπη) dη 2 2 2 π sinh( 4n π + m π/2) 0 0 1 1 − (−1)n 1 − (−1)m √ = . nπ mπ sinh( n2 π 2 + m2 π/2)

anm =

For the second problem, obtain v(x, y, z) =

∞ ∞ X X

p bnm sin(nπx) sin(mz) sinh( n2 π 2 + m2 y),

n=1 m=1

in which Z 1 Z π 4 √ ξ sin(nπξ) dξ τ 2 sin(mτ ) dτ ) π 2 sinh( n2 π 2 + m2 2π) 0 0 (−1)n+1 2 − 2(−1)m + m2 π 2 (−1)m 4 √ . = nπ m3 π 2 sinh( n2 π 2 + m2 2π)

bnm =

The original problem has solution u(x, y, z) = w(x, y, z) + v(x, y, z). 4. The solution u(x, y, z) is a sum of the solutions of the following two problems: ∇2 (v) = 0, v(0, y, z) = 0, v(1, y, z) = sin(πy) sin(z), v(x, 0, z) = v(x, 2, z) = 0, v(x, y, 0) = v(x, y, π) = 0 and ∇2 w = 0, w(0, y, z) = w(1, y, z) = 0, w(x, 0, z) = w(x, 2, z) = 0, w(x, y, 0) = x2 (1 − x)y(2 − y), w(x, y, π) = 0.

7.6. THE NEUMANN PROBLEM

211

Solve these problems by separation of variables. The first problem has a solution of the form v(x, y, z) =

∞ X ∞ X

anm sinh(m2 + (mπ/2)2 ) sin(nπy/2) sin(mz),

n=1 m=1

while the second problem has a solution of the form w(x, y, z) =

∞ X ∞ X

p bnm sin(nπx) ∼ (mπy/2) sinh( (nπ)2 + (mπ/2)2 (π−z)).

n=1 m=1

√ The coefficients anm can be obtained by inspection. Observe that a21 π 2 + 1 = 1 and all other anm = 0. The coefficients bnm require integrations: bnm = Z 1 Z 2 2 2 √ ξ (1 − ξ) sin(nπξ) dξ η(2 − η) sin(mπη/2) dη sinh(π 4n2 + m2 /2) 0 0 64 = − 6 (1 + 2(−1)n )(1 − (−1)m ). π ’

7.6

The Neumann Problem

1. First, Z 1 4 cos(πx) dx = 0, 0

so this problem may have a solution. (If this integral were nonzero, we could conclude that the problem has no solution). A separation of variables, making use of the three homogeneous boundary conditions, leads to the problems X 00 + λX = 0; X 0 (0) = X 0 (1) = 0 and Y 00 − λY = 0; Y 0 (1) = 0. These have solutions of the form λn = n2 π 2 , Xn (x) = cos(nπy) and Yn (x) = cosh(nπ(1 − y)). Thus attempt a solution of the Neumann problem of the form u(x, y) = c0 +

∞ X

cn cosh(nπ(1 − y)) cos(nπy).

n=1

212

CHAPTER 7. LAPLACE’S EQUATION The condition uy (0) = 4 cos(πx) requires that ∞ X

−cn nπ sinh(nπ) cos(nπx) = 4 cos(πx).

n=1

This is satisfied if we put cn = 0 for n = 2, 3, 4, · · · , and choose c1 so that −c1 π sinh(1) cos(πx) = 4 cos(πx). Therefore −c1 π sinh(π) = 4, and c1 = −

4 . π sinh(π)

The solution is u(x, y) = c0 −

4 cosh(π(1 − y)) cos(πx). π sinh(π)

Here c0 is an arbitrary constant, so this solution is not unique. 2. First check that

Z π

y−

π dy = 0, 2

so it is worthwhile to look for a solution. From the zero boundary conditions at y = 0 and y = π, separation of variables gives us a solution of the form u(x, y) = c0 +

∞ X

[cn cosh(nx) + dn cosh(1 − x)] cos(ny).

n=1

The boundary conditions on the edges x = 0 and x = 1 give us ∞ X ∂u π (0, y) = y − = −ndn sinh(n) cos(ny) ∂x 2 n=1

and

∞ X ∂u (1, y) = cos(y) = ncn sinh(n) cos(ny). ∂x n=1

The coefficients are given by 1 2 cn = n sinh(n) π ( 0 = 1/ sinh(1) and dn =

−1 n sinh(n)

Z π 0

Z π cos(y) cos(ny) dy 0

if n 6= 1, if n = 1

y−

π 2(1 − (−1)n ) dy = 2 πn2 sinh(n)

7.6. THE NEUMANN PROBLEM

213

for n = 1, 2, · · · . The solution is u(x, y) = c0 + +

1 cosh(x) sinh(1)

∞ X 2((−1)n − 1)

πn2 sinh(n)

n=1

cosh(n(1 − x)) cos(ny).

Rπ 3. A solution may exist because 0 cos(3x) dx = 0. From the zero boundary conditions on edges x = 0 and x = π, separation of variables yields a solution of the form u(x, y) = c0 +

∞ X

[cn cosh(ny) + dn cosh(n(π − y))] cos(nx).

n=1

Now

∞ X ∂u −ndn sinh(nπ) cos(nx) (x, 0) = cos(3x) = ∂y n=1

so d3 = −

1 3 sinh(3π)

and dn = 0 if n 6= 3. Next, the boundary condition at y = π gives us ∞ X ∂u ncn sinh(nπ) cos(nx). (x, π) = 6x − 3π = ∂u n=1

Then Z 1 2 π cn = (6x − 3π) cos(nx) dx n sinh(nπ) π 0 12 1 ((−1)n − 1). = n sinh(nπ) n2 π The solution is u(x, y) = c0 − +

cosh(3(π − y)) cos(3x) 3 sinh(3π)

∞ X 12((−1)n − 1) n=1

n3 π sinh(nπ)

cosh(ny) cos(nx).

4. Let u(x, y) = X(x)Y (y) and use the boundary conditions to obtain the problems X 00 − λX = 0; X 0 (0) = X 0 (π) = 0 and Y 00 + λY = 0.

214

CHAPTER 7. LAPLACE’S EQUATION We find that X0 (x) = 1 and Xn (x) = cos(nx) for n = 1, 2, · · · , while Yn (y) = cn cosh(ny) + dn cosh(n(π − y)) for n = 1, 2, · · · . We therefore seek a solution of the form u(x, y) = c0 +

∞ X

[cn cosh(ny) + dn cosh(n(π − y))] cos(nx).

n=1

At the edge y = π, we have u(x, y) = 0 = c0 +

∞ X

[cn cosh(nπ) + dn ] cos(nx).

n=1

Along the edge y = 0, we have u(x, 0) = f (x) = c0 +

∞ X

[cn + dn cosh(nπ)] cos(nx).

n=1

In order to have a solution, the coefficients must be chosen to satisfy the equations c0 = 0, cn cosh(nπ) + dn = 0, Z 2 π f (ξ) cos(nξ) dξ cn + dn cosh(nπ) = π 0 for n = 1, 2, · · · . For n ≥ 1, the determinant of the coefficients of this system of equations is cosh(nπ) 1 = cosh2 (nπ) − 1 = sinh2 (nπ) 6= 0. 1 cosh(nπ) Therefore there is a unique solution of these algebraic equations for the coefficients cn and dn for n = 1, 2, · · · . The problem therefore has a unique solution (recalling that c0 = 0). 5. With u(x, y) = X(x)Y (y), we obtain: X 00 − λX = 0 and Y 00 + λY = 0; Y (0) = Y (1) = 0. Then Yn (y) = sin(nπy) and Xn (x) = cn cosh(nπx) + dn cosh(nπ(1 − x))

7.6. THE NEUMANN PROBLEM

215

for n = 1, 2, · · · . Look for a solution of the form u(x, y) =

∞ X

[cn cosh(nπx) + dn cosh(nπ(1 − x))] sin(nπy).

n=1

To solve for the constants, use the other two boundary conditions. First, ∞ X ∂u (1, y) = 0 = nπcn sinh(nπ) sin(nπy) ∂x n=1

so we each cn = 0. Next, ∞ X ∂u (0, y) = 3y 2 − 2y = −nπdn sinh(nπ) sin(nπy). ∂x n=1

Then Z 1 −2 (3η 2 − 2η) sin(nπη) dη nπ sinh(nπ) 0 2 [n2 π 2 (−1)n + 6(1 − (−1)n )] = 4 4 n π sinh(nπ)

dn =

for n = 1, 2, · · · . The solution is u(x, y) = ∞ X 2 [n2 π 2 (−1)n + 6(1 − (−1)n )] cosh(nπ(1 − x)) sin(nπy). 4 π 4 sinh(nπ) n n=1 Rπ 6. Because −pi sin(3θ) dθ = 0, this problem may have a solution. Such a solution will have the form ∞ X 1 [an rn cos(nθ) + bn rn sin(nθ)]. u(r, θ) = a0 + 2 n=1 The boundary conditions give us ∂u (R, θ) = sin(3θ) ∂r ∞ X = [nan Rn−1 cos(nθ) + nbn Rn−1 sin(nθ)]. n=1

By inspection, we see that we can choose each an = 0 and bn = 0 for n 6= 3. Further, 3b3 R2 = 1, so b3 =

1 3R2

and the solution is u(r, θ) =

1 R r 3 a0 + sin(nθ). 2 3 R

216

CHAPTER 7. LAPLACE’S EQUATION

Rπ 6. First check that −π cos(2θ) dθ = 0, a necessary condition for a solution to exist. A solution must have the form u(r, θ) =

∞ X 1 a0 + [an cos(nθ) + bn sin(nθ)]. 2 n=1

From the boundary condition at r = R, we have ∂u (R, θ) = cos(2θ) ∂r ∞ X = [nan Rn−1 cos(nθ) + nbn Rn−1 sin(nθ)]. n=1

As in the preceding problem, compare coefficients on both sides of this equation to choose each bn = 0 and an = 0 except for n = 2. Further, 2a2 R = 1. The solution is R r 2 1 cos(2θ). u(r, θ) = a0 + 2 2 R 8. Because

Z ∞

xe−|x| dx = 0

−∞

this problem may have a solution. Using the general solution developed in Section 7.6.3, we can immediately write the solution Z ∞ 1 u(x, y) = ln(y 2 + (ξ − x)2 )ξe−|ξ| dξ + c. 2π −∞ 9. Because

Z ∞

e−|ξ| sin(ξ) dξ = 0,

−∞

a necessary condition for a solution to exist is satisfied. The solution is Z 1 u(x, y) = −∞∞ ln(y 2 + (ξ − x)2 )e−|ξ| sin(ξ) dξ. 2π 10. A solution of the Dirichlet problem for the lowr half-plane was requested in Problem 5, Section 7.4. Use this result and the line of argument used to solve the Neumann problem for the upper half-plane to obtain the solution Z ∞ 1 ln(y 2 + (ξ − x)2 )f (ξ) dξ. u(x, y) = − 2π −∞ 11. Problem 7, Section 7.4 requested a solution of the Dirichlet problem for the right quarter-plane. Using this, we are led to attempt a solution for the Neumann problem for the right quarter-plane of the form Z ∞ u(x, y) = aω cos(ωx)e−ωy dω. 0

7.7. POISSON’S EQUATION Now

217

∂u (x, 0) = ∂y

Z ∞ −ωaω cos(ωx) dω. 0

This tells us to choose aω = −

2 πω

Z ∞ f (ξ) cos(ωξ) dξ. 0

12. Attempt a solution of the form Z ∞ u(x, y) = bω e−ωy sin(ωx) dω. 0

This is harmonic and has been chosen to satisfy u(0, y) = 0. Now Z ∞ ∂u (x, 0) = f (x) = −ωbω sin(ωx) dω, ∂y 0 so choose bω = −

7.7

2 πω

Z ∞ f (ξ) sin(ωξ) dξ. 0

Poisson’s Equation

1. Write u(x, y) = v(x, y) + w(x, y), where v is the solution of the Dirichlet problem ∇2 v = 0 for 0 < x < 1, 0 < y < 1, v(x, 0) = v(x, 1) = 0, v(1, y) = 0, v(0, y) = y and w is the solution of the problem ∇2 w = 0 for 0 < x < 1, 0 < y < 1, w(0, y) = w(1, y) = w(x, 0) = w(x, 1) = 0, w(x, y) = xy for 0 < x < 1, 0 < y < 1. For the first problem, for v(x, y), separate variables to obtain the solution: v(x, y) =

∞ X

an sin(nπ(1 − x)) sin(nπy),

n=1

where an =

2(−1)n+1 . nπ sinh(nπ)

218

CHAPTER 7. LAPLACE’S EQUATION The problem for w(x, y) has the solution w(x, y) =

∞ X ∞ X

knm sin(nπx) sin(mπy),

n=1 m=1

where Z 1 Z 1 −4 ξ sin(nπξ dξ η sin(mπη) dη knm = 2 2 π (n + m2 ) 0 0 4(−1)n+m+1 = 2 . (n + m2 )nmπ 4 2. Split the Poisson problem into two problems, as in Figure 7.6. The first is a Dirichlet problem: ∇2 v = 0 for 0 < x < 1, 0 < y < 2, v(0, y = v(1, y) = v(x, 2) = 0, v(x, 0) = x2 . This has the solution v(x, y) =

∞ X

cn sin(nπx) sinh(nπ(2 − y)),

n=1

where Z 1 2 ξ 2 sin(nπξ) dξ sinh(2nπ) 0 −2 + 2(−1)n − n2 π 2 (−1)n 2 . = sinh(2nπ) n3 π 3

cn =

The second problem is ∇2 w = 0, w(x, y) = 0 on the boundary of the rectangle, w(x, y) = x sin(y) for 0 < x < 1, 0 < y < 2. This problem has the solution w(x, y) =

∞ X ∞ X

knm sin(nπx) sin(mπy/2)

n=1 m=1

where Z 1 Z 2 8 ξ sin(nπξ) dξ η 2 sin(mπη/2) dη π 2 (4n2 + m2 ) 0 0 16 (−1)n+1 mπ sin(2)(−1)m = 2 . 2 2 π (4n + m ) nπ −4 + m2 π 2

knm = −

7.7. POISSON’S EQUATION

219

3. Split the problem into two problems, as we have been doing. However, the first problem (see Figure 7.6) must itself be broken up into two problems, in the first of which v(0, y) = 1 and v(π, y) = 0, and in the second of which v(0, y) = 0 and v(π, y) = 0. Applying straightforward separation of variables to these problems, we obtain v1 (x, y) =

∞ X

2 (1 − (−1)n ) sin(my) sinh(n(π − x)) nπ sinh(nπ) n=1

for the Dirichlet problem with v(0, y) = 1 and v(π, y) = 0. If v(0, y) = 0 and v(π, y) = y, we obtain v2 (x, y) =

∞ X

2 (−1)n+1 sin(ny) sinh(nx). n sinh(nπ) n=1

For the problem for w in Figure 7.6, we have w(x, y) =

∞ X ∞ X

knm sin(nx) sin(my),

n=1 m=1

where knm = Z πZ π Z π 4 − 2 n2 π 2 + m2 π 2 ξ 2 sin(nξ) dξ η 2 sin(mη) dη π 0 0 0 4 (2 − 2(−1)n + n2 π 2 (−1)n )(−2 + 2(−1)m − m2 π 2 (−1)m ). = 2 2 π (n + m2 )

220

CHAPTER 7. LAPLACE’S EQUATION

Chapter 8

Special Functions and Applications 8.1

Legendre Polynomials

For Problems 1–4 and 6, graphs of the function and the sixth partial sum of its Fourier-Legendre expansion on[−1, 1] appear nearly indistinguishable within the scale of the graph. The “most“ functions many terms of this expansion are needed to achieve a good fit between the partial sum and the function. This is seen in Problem 5, where the sixth partial sum is a poor fit to the function, while the fiftieth partial sum is much closer (though still a poor fit). 1. The coefficients are cn =

2n + 1 2

Z 1 sin(πx/2)Pn (x) dx. −1

Carrying out these integrations, we obtain 12 , π2 168(π 2 − 10) 660(π 4 − 112π 2 + 1008) c3 = , c5 = . 4 π π6

c0 = c2 = c4 = 0, c1 =

Figure 1 shows a graph of f (x) and

n=0 cn Pn (x).

2. The coefficients are cn =

2n + 1 2

Z 1

e−x Pn (x) dx.

−1

221

222

CHAPTER 8. SPECIAL FUNCTIONS AND APPLICATIONS

Figure 8.1: Graph of sin(πx/2) and the sixth partial sum of its Fourier-Legendre expansion.

The first six coefficients are 1 3 5 35 (e − e−1 ), c1 = − , c2 = e − e−1 , 2 e 2 2 35 259 −1 259 −1 c3 = e− e , c4 = 162e − e , 2 2 2 3619 26741 −1 c5 = e− e . 2 2

c0 =

Figure 2 is a graph of f (x) = e−x and the sixth partial sum of its FourierLegendre expansion. 3. The coefficients are 2n + 1 cn = 2

Z 1

sin2 (x)Pn (x) dx.

−1

The first six are 1 1 c0 = − sin(1) cos(1) + , c1 = c3 = c5 = 0, 2 2 5 15 15 c2 = − sin(1) cos(1) + − cos2 (1), 8 8 4 585 585 531 c4 = − + cos2 (1) + sin(1) cos(1). 32 16 32

8.1. LEGENDRE POLYNOMIALS

223

Figure 8.2: Graph of e−x and the sixth partial sum of its Fourier-Legendre expansion.

Figure 8.3 shows the function and the sixth partial sum of this FourierLegendre expansion. 4. The coefficients are cn =

2n + 1 2

Z 1 (cos(x) − sin(x))Pn (x) dx. −1

The first six coefficients are c0 = sin(1), c1 = −3 sin(1) + 3 cos(1), c2 = 15 cos(1) − 10 sin(1), c4 = 549 sin(1) − 855 cos(1), c5 = −5940 sin(1) + 9250 cos(1). Figure 8.4 is a graph of the function and the sixth partial sum. 5. The coefficients are 2n + 1 cn = 2

Z 1 f (x)Pn (x) dx. −1

The first six coefficients are c0 = c2 = c4 = 0, c1 =

3 7 11 , c3 = − , c5 = . 2 8 16

224

CHAPTER 8. SPECIAL FUNCTIONS AND APPLICATIONS

Figure 8.3: Graph of sin2 (x) and the sixth partial sum of its Fourier-Legendre expansion.

Figure 8.4: Graph of cos(x) − sin(x) and the sixth partial sum of its FourierLegendre expansion.

8.1. LEGENDRE POLYNOMIALS

225

Figure 8.5: Graph of f (x) and the sixth partial sum of its Fourier-Legendre expansion.

Figure 8.6: Graph of f (x) and the fiftieth partial sum of its Fourier-Legendre expansion.

226

CHAPTER 8. SPECIAL FUNCTIONS AND APPLICATIONS

Figure 8.7: Graph of (x + 1) cos(x) and the sixth partial sum of its FourierLegendre expansion.

Figure 8.5 shows a graph of the function and this partial sum. For this function the sixth partial sum does not fit the function well at all on [−1, 1]. Figure 8.6 shows the fiftieth partial sum, a better fit to the function. 6. The coefficients are 2n + 1 n

Z 1 (x + 1) cos(x)Pn (x) dx. −1

The first six coefficients are c0 = sin(1), c1 = −3 sin(1) + 6 cos(1), c2 = 15 cos(1) − 10 sin(1) c3 = 238 sin(1) − 378 cos(1), c4 = 549 sin(1) − 855 cos(1) c5 = −34969 sin(1) + 54461 cos(1). Figure 8.7 shows a graph of (x + 1) cos(x) and the sixth partial sum of the Fourier-Legendre expansion. 7. For n = 7, [n/2] = [7/2] = 3, so P7 =

3 X

(14 − 2k)! (−1)k 7 xn−2k 2 k!(7 − k)!(7 − 2k)!

k=0

429 7 693 5 315 3 35 x − x + x − x. 16 16 16 16

8.1. LEGENDRE POLYNOMIALS

227

For n = 8, [n/2] = [4] = 4 and 4 X

P8 (x) =

(16 − 2k)! (−1)k 8 x8−2k 2 k!(8 − k)!(8 − 2k)!

k=0

6435 8 3003 6 3465 4 x − x + x 128 32 64 315 2 35 − x + . 33 128 =

For n = 9, [n/2] = [9/2] = 4 and 4 X

P9 (x) =

(18 − 2k)! (−1)k 9 x9−2k 2 k!(9 − k)!(9 − 2k)!

k=0

12155 9 6435 7 9009 5 x − x + x 128 32 64 1155 3 315 − x + x. 32 128 =

For n = 10, [n/2] = [5] = 5 and

P10 (x) =

5 X

(20 − 2k)! (−1)k 10 x10−2k 2 k!(10 − k)!(10 − 2k)

k=0

46189 10 109395 8 45045 6 x − x + x 256 256 128 15015 4 3465 2 63 − x + x − . 128 256 256 =

8. Put n = 0 into the integral formula to obtain 1 π

Z π dθ = 0

1 (π) = 1 = P0 (x). π

Next put n = 1 to get 1 π

Z πZ π p x + x2 − 1 cos(θ) dθ 0

iπ p 1h = xθ + x2 − 1 sin(θ) π 0 1 = (xπ) = x = P1 (x). π

228

CHAPTER 8. SPECIAL FUNCTIONS AND APPLICATIONS With n = 2, we have Z 2 p 1 π x + x2 − 1 cos(θ) dθ π 0 Z p 1 π 2 x + 2 x2 − 1 cos(θ) + (x2 − 1) cos2 (θ) dθ = π 0 Z p 1 π 1 x2 + 2 x2 − 1 cos(θ) + (x2 − 1)(1 + cos(2θ)) dθ = π 0 2 π p 1 2 1 2 2 = x θ + 2 x − 1 sin(θ) + (x − 1)(θ + sin(2θ)/2) π 2 0 1 2 1 2 x π + (x − −1)π = π 2 1 3 = x2 − = P2 (x). 2 2 Finally, wit n = 3, we have Z 3 p 1 π x + x2 − 1 cos(θ) dθ π 0 Z p 1 π 3 = x + 3x2 x2 − 1 cos(θ) − 3x3 cos2 (θ) dθ π 0 Z p p 1 π + −3x cos2 (θ) + x2 x2 − 1 cos3 (θ) − x2 − 1 cos3 (θ) dθ π 0 1 3 3 = x3 π + πx3 − πx π 2 2 5 3 3 = x − x = P3 (x). 2 2

9. Let Qn (x) =

1 π

Z π

n x2 − 1 cos(θ) dθ

for n = 0, 1, 2, · · · . The strategy is to show that Qn (x) satisfies the same recurrence relation (8.7) that the Legendre polynomials do. From Problem 8, we also have that Q0 (x) = P0 (x) and Q1 (x) = P1 (x). Then the recurrence relation will give us Q2 (x) = P2 (x), and then Q4 (x) = P4 (x), and so on. To show that Qn (x) satisfies equation (8.7), first substitute the integral for Qn (x) into this equation and rearrange terms to obtain Z p p 1 π −n(x2 − 1) sin2 (θ) + x2 − 1 cos(θ)[x + x2 − 1 cos(θ)] π 0 n−1 p × x + x2 − 1 cos(θ) dθ.

8.1. LEGENDRE POLYNOMIALS

229

Now integrate this by parts, with n p u = x + x2 − 1 cos(θ) and dv = cos(θ) dθ to obtain 1 π

Z π

n p p x2 − 1 cos(θ) x2 − 1 cos(θ) dθ

n−1 p 1 x + x2 − 1 cos(θ) = n(x2 − 1) sin2 (θ). π Use this in the substitution of Qn (x) into equation (8.7) to show that Qn (x) satisfies this recurrence relation. This shows that Qn (x) = Pn (x). 10. We want to show that Z 1

Pn2 (x) dx =

−1

2 2n + 1

for n = 0, 1, 2, · · · . Let An be the coefficient of xn in Pn (x). We have an explicit formula for An in equation (8.8). For convenience, denote Z 1 pn = Pn2 (x) dx. −1

Consider the polynomial q(x) = Pn (x) −

An xPn−1 (x). An−1

Both Pn (x) and xPn−1 (x) are of degree n, and all terms involving xn in q(x) cancel (by design), so q(x) has degree ≤ n − 1. Further, Pn (x) = q(x) + Then Pn2 (x) = q(x)Pn (x) +

An xPn−1 (x). An−1 An xPn−1 (x)Pn (x). An−1

Integrate this equation, using Theorem 8.3, to get Z 1 An pn = xPn−1 (x)Pn (x) dx. An−1 −1 Now use the recurrence relation to write xPn (x) =

n+1 n Pn+1 (x) + Pn−1 (x). 2n + 1 2n + 1

230

CHAPTER 8. SPECIAL FUNCTIONS AND APPLICATIONS Then xPn (x)Pn−1 (x) =

n+1 n Pn+1 (x)Pn−1 (x) + P 2 (x). 2n + 1 2n + 1 n−1

Finally, we have Z 1 An pn = xPn (x)Pn−1 (x) dx An−1 −1 Z 1 Z 1 n An n+1 2 Pn+1 (x)Pn−1 (x) dx + Pn−1 (x) dx . = An−1 2n + 1 −1 2n + 1 −1 Connecting the first and last expressions in this equation, we have pn =

An n pn−1 . An−1 2n + 1

This is a recurrence relation for the numbers pn . Because we know An , this is 1 · 3 · 5 · · · (2n − 3)(2n − 1) (n − 1)! n pn−1 n! 1 · 3 · 5 · · · (2n − 3) 2n + 1 2n − 1 pn . = 2n + 1

pn =

From this we deduce pn : Z Z 1 n 1 1 2 P0 (x)2 dx = dx = , 3 −1 3 −1 3 3 32 2 p2 = p1 = = , 5 53 5 5 2 p3 = p3 = , 7 7 7 2 p4 = p3 = , 9 9 p1 =

and so on. A straightforward induction now yields pn =

2 . 2n + 1

11. Put x = t = 1/2 into the generating formula for the Legendre polynomials to get n ∞ X 1 1 1 p = Pn . 2 2 3/4 n=0 Then

∞ X 2 1 1 √ = P . n n 2 2 3 n=0

8.1. LEGENDRE POLYNOMIALS Then

∞ X n=0

231

P n+1 n

1 1 =√ . 2 3

12. One way to derive these results is to use the recurrence relation. Another way is to use the formula given in Problem 7: [n/2]

Pn (x) =

X k=0

(2n − 2k)! (−1)k n xn−2k . 2 k!(n − k)!(n − 2k)!

To do this, first observe that 2n + 1 1 = n+ = n, 2 2 so P2n+1 (x) =

n X k=0

(4n + 2 − 2k)! x2n+1−2k . (−1)k 2n+1 2 k!(2n + 1 − k)!(2n + 1 − 2k)!

Each term of this sum involves a positive power of x, so P2n+1 (0) = 0. Next, so P2n (x) =

n X

2n = [n] = n, 2

(4n − 2k)! x2n−2k . (−1)k 2n 2 k!(2n − k)!(2n − 2k)!

k=0

The constant term in this polynomial occurs when k = n, so P2n (0) =

(−1)n (2n)! . 22n (n!)2

13. Apply the law of cosines to the triangle in the diagram to get R2 = r2 + d2 − 2rd cos(θ). Then

r2 R2 r = 1 − 2 cos(θ) + 2 . 2 d d d

Then ϕ(x, y, z) =

1 1d 1 1 . = = q R dR d 1 − 2 r cos(θ) + r2 d

232

CHAPTER 8. SPECIAL FUNCTIONS AND APPLICATIONS For the remainder of the problem, consider two cases on r/d. First, suppose r/d < 1, so r < d. Put x = cos(θ) and t = r/d in the generating function to obtain ∞

r n 1X , Pn (cos(θ)) d n=0 d

ϕ(r) = or ϕ(r) =

∞ X n=0

1 dn+1

Pn (cos(θ))rn .

If r/d > 1, so r > d, now write d2 R2 d cos(θ) + = 1 − 2 . r2 r r2 Then

r 1 . =q 2 R d 1 − 2 cos(θ) + d r2

Again, comparing this with the generating function, we have n ∞ 1X d ϕ(r) = Pn (cos(θ)) , r n=0 r and this is equivalent to ∞

ϕ(r) =

1X n d Pn (cos(θ))e−n . r n=0

14. With f (ϕ) = ϕ2 , the solution is u(ρ, ϕ) =

∞cn

n=0

ρ n R

Pn (cos(ϕ)),

in which cn = =

2n + 1 2 2n + 1 2

Z 1 f (arccos(ξ))Pn (ξ), dξ −1 Z 1

(arccos(ξ))2 Pn (ξ) dξ.

−1

Many of these coefficients can be explicitly computed. For example, 1 2 3 10 π − 2, c1 = − π 2 , c2 = , 2 8 9 7 2 8 11 2 c3 = − π , c4 = , c5 = − π . 128 25 512 c0 =

8.1. LEGENDRE POLYNOMIALS

233

With R = 1, use of the twenty-first partial sum of the solution series yields the following approximations: u(1, π/4) ≈ 0.615257, u(1, π/6) ≈ 0.272881, u(1, π/8) ≈ 0.154213. These can be compared to f (ϕ) as a check on their accuracy. 15. With f (ϕ) = sin(ϕ), we have 2n + 1 cn = 2 and u(ρ, ϕ) =

Z 1 sin(arccos(ξ))Pn (ξ) dξ −1 ∞ X

n=0

ρ n R

Pn (cos(ϕ)).

In computing the coefficients, use can be made of the identity p sin(arccos(ξ)) = 1 − ξ 2 . With R = 1, and using the twenty-first partial sum of the solution, we obtain the approximations u(1, π/4) ≈ 0.707274, u(1, π/6) ≈ 0.500761, u(1, π/8) ≈ 0.382683. 16. With f (ϕ) = ϕ3 , let cn =

2n + 1 2

and u(ρ, ϕ) =

Z 1

∞ X

(arccos(ξ))3 Pn (ξ) dξ

−1

ρ n

n=0

Pn (n, cos(ϕ)).

With R = 1, the twenty-first partial sum of this series yields the following approximate values: u(1, π/4) ≈ 0.476973, u(1, π/6) ≈ 0.143548, u(1, π/8) ≈ 0.060559. 17. With f (ϕ) = 2 − ϕ2 , let cn =

2n + 1 2

and u(ρ, ϕ) =

Z 1

(2 − arccos2 (ξ))Pn (ξ) dξ

−1 ∞ X n=0

ρ n R

Pn (cos(ϕ)).

With R = 1, use the twenty-first partial sum to approximate: u(1, π/4) ≈ 1.384743, u(1, π/6) ≈ 1.725844, u(1, π/8) ≈ 1.845787.

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CHAPTER 8. SPECIAL FUNCTIONS AND APPLICATIONS

18. Now f (ϕ) = ϕ − cos(ϕ). Let cn =

2n + 1 2

and u(ρ, ϕ) =

Z 1 (arccos(ξ) − ξ)Pn (ξ) dξ −1 ∞ X n=0

ρ n R

Pn (cos(ϕ)).

With R = 1, the twenty-first partial sum of this series gives us the following approximate values: u(1, π/4) ≈ 0.077951, u(1, π/6) ≈ −0.342074, u(1, π/8) ≈ −0.531180. 19. In spherical coordinates, the Dirichlet problem to be solved is: 2 cot(ϕ) uρρ + uϕϕ + uϕ = 0, R1 < ρ < R2 , −π/2 ≤ ϕ ≤ π/2, ρ ρ2 u(R1 , ϕ) = T, U (R2 , ϕ) = 0. This problem can be solved by separation of variables. Let u(ρ, ϕ) = F (ρ)Φ(ϕ). This results in:

2 λ F 00 + F 0 − 2 F = 0 ρ ρ

and Φ00 + cot(ϕ)Φ0 + λΦ = 0. The equation for Φ(ϕ) has the bounded solution Φn (ϕ) = Pn (cos(ϕ)), corresponding to an eigenvalue λn = n(n + 1) of Legendre’s equation. For n = 0, 1, 2, · · · , solutions for F (ρ) are Fn (ρ) = an ρn + bn ρ−n−1 . Attempt a superposition u(ρ, ϕ) =

∞ X

(an ρn + bn ρ−n−1 )Pn (cos(ϕ)).

n=0

We require that u(R1 , ϕ) = T =

∞ X

(an R1n + bn R1−n−1 )Pn (cos(ϕ)).

n=0

8.2. BESSEL FUNCTIONS

235

And the condition at ρ = R2 is that r(R2 , ϕ) = 0 =

∞ X

(an R2n + bn R2−n−1 )Pn (cos(ϕ)).

n=0

Recalling that P0 (cos(ϕ)) = 1, these equations are satisfied if we choose the coefficients so that a0 + b0 R1−1 = T, a0 + b0 R2−1 = 0 and, for n = 1, 2, · · · , let an = bb = 0. We should therefore let a0 =

T R1 R2 T R1 and b0 = − . R1 − R2 R1 − R2

The solution is T R1 R2 u(ρ, ϕ) = −1 . R1 − R2 ρ

8.2

Bessel Functions

1. With f (x) = e−x , the expansion in terms of zero-order Bessel functions is ∞ X

cn J0 (jn x),

n=1

where jn is the nth (in increasing order) positive zero of J0 (x) and cn =

2 J12 (jn )

Z 1

ξe−ξ J0 (jn ξ) dξ.

Figure 8.8 shows the function and the tenth partial sum of this expansion, and Figure 8.9 shows the twenty-fifth partial sum. 2. With f (x) = x, let jn be the nth positive zero of J2 (x). The Fourier-Bessel expansion of f (x) on (0, 1) is ∞ X

cn J2 (jn x),

n=0

where cn =

2 J32 (jn )

Z 1

ξ 2 J2 (jn ξ) dξ.

Figure 8.10 shows the function and the fifteenth partial sum of this expansion, and Figure 8.11 has the thirtieth partial sum.

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Figure 8.8: Graph of e−x and the tenth partial sum of its Fourier-Bessel expansion.

Figure 8.9: Graph of e−x and the twenty-fifth partial sum of its Fourier-Legendre expansion.

8.2. BESSEL FUNCTIONS

237

Figure 8.10: Graph of x and the fifteenth partial sum of its Fourier-Legendre expansion.

Figure 8.11: Graph of x and the thirtieth partial sum of its Fourier-Legendre expansion.

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CHAPTER 8. SPECIAL FUNCTIONS AND APPLICATIONS

Figure 8.12: Graph of x2 e−2x and the twentieth partial sum of its FourierLegendre expansion.

3. Let jn be the nth positive zero of J1 (x) and Z 1 2 cn = 2 ξ 3 e−2ξ J1 (jn ξ) dξ. J2 (jn ) 0 The Fourier-Bessel expansion is ∞ X

cn J1 (jn x).

n=1

Figure 8.12 shows the function and the twentieth partial sum of this series, while Figure 8.13 has the fortieth partial sum. 4. Let jn be the nth positive zero of J2 (x) and Z 1 2 cn = 2 ξ 2 cos(2ξ)J2 (jn ξ) dξ. J3 (jn ) 0 The Fourier-Bessel expansion is ∞ X

cn J2 (jn x).

n=0

Figure 8.14 is the tenth partial sum of this expansion, and Figure 8.15 the thirtieth.

8.2. BESSEL FUNCTIONS

239

Figure 8.13: Graph of x2 e−2x and the fortieth partial sum of its FourierLegendre expansion.

Figure 8.14: Graph of x cos(x) and the tenth partial sum of its Fourier-Legendre expansion.

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CHAPTER 8. SPECIAL FUNCTIONS AND APPLICATIONS

Figure 8.15: Graph of x cos(x) and the thirtieth partial sum of its FourierLegendre expansion.

5. Let jn be the nth positive zero of J4 (x), and Z 1 2 cn = 2 ξ sin(3ξ)J4 (jn ξ) dξ. J5 (jn ) 0 The Fourier-Bessel expansion is ∞ X

cn J4 (jn x).

n=1

Figure 8.16 shows the function and the twentieth partial sum of this series, while Figure 8.17 has the fortieth partial sum. 6. Let jn be the nth positive zero of J1 (x). Let Z 1 2 cn = 2 ξ 3 cos(πξ)J1 (jn ξ) dξ. J2 (jn ) 0 The Fourier-Bessel expansion is ∞ X

cn J1 (jn x).

n=1

Figures 8.18 and 8.19 compare graphs of the function with the twentieth and fiftieth partial sums of this expansion, respectively.

8.2. BESSEL FUNCTIONS

241

Figure 8.16: Graph of sin(3x) and the twentieth partial sum of its FourierLegendre expansion.

Figure 8.17: Graph of sin(3x) and the fortieth partial sum of its FourierLegendre expansion.

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CHAPTER 8. SPECIAL FUNCTIONS AND APPLICATIONS

Figure 8.18: Graph of x2 cos(πx) and the twentieth partial sum of its FourierLegendre expansion.

Figure 8.19: Graph of x2 cos(πx) and the fiftieth partial sum of its FourierLegendre expansion.

8.2. BESSEL FUNCTIONS

243

7. A formal proof that r J1/2 (x) =

2 sin(x) πx

can be made by using the Maclaurin expansion of sin(x) on the right side and manipulating the coefficients to obtain the series defining J1/2 (x). We will be less formal here and essentially carry out this type of argument using a few terms of the series so that it is apparent what is happening. Begin with J1/2 (x) =

∞ X n=0

(−1)n 22n+1/2 n!Γ(n + 1/2 + 1)

x2n+1/2

∞

(−1)n xX x2n 2n 2 n=0 2 n!Γ(n + 1/2 + 1) r 1 1 x = − x2 2 Γ(1/2 + 1) 22 Γ(1 + 1/2 + 1) 1 1 1 + 4 − 6 x6 + 8 x8 + · · · . 2 2!Γ(2 + 1/2 + 1) 2 3!Γ(3 + 1/2 + 1) 2 4!Γ(4 + 1/2 + 1) r

Now we need to know some values of the gamma function. First, Z ∞ Γ(1/2) = t−1/2 e−t dt. 0

Letting t = u2 , this is Z ∞ Γ(1/2) = 0

1 −u2 e 2u du u

Z ∞ =2

e−u du =

√ π.

Here we have used the well-known result that Z ∞ 2 1√ e−u du = π, 2 0 which can be derived using double integrals or complex integration, and is widely used in probability and statistics. Then, using the factorial property of the gamma function, we can evaluate Γ(n+1/2+1) for various values of n. In particular, √ √ 1 1 1 3 π 3 π Γ 1+ +1 = 1+ Γ 1+ = = , 2 2 2 2 2 4 √ √ 1 1 1 53 π 3·5 π Γ 2+ +1 = 2+ Γ 2+ = = , 2 2 2 2 4 23

244

CHAPTER 8. SPECIAL FUNCTIONS AND APPLICATIONS √ 1 1 1 3·5·7 π , Γ 3+ +1 = 3+ Γ 3+ = 2 2 2 24 and so on. In general, if n is a positive integer, then √ 1 3 · 5 · · · (2n + 1) π Γ n+ +1 = . 2 2n+1 Now back to the series for J1/2 (x). We can now write r x 2 22 √ − 2 √ x2 J1/2 (x) = 2 π 2 ·3 π 3 2 24 √ x4 − + 4 √ x6 6 2 2!3 · 5 π 2 3!3 · 57̇ π 25 8 √ + 8 x −· . 2 4! · 3 · 5 · 7 · 9 π This simplifies to r x 2 1 1 √ − √ x2 + J1/2 (x) = √ x4 2 π 3 π 2 · 2 · 35̇ π 1 1 √ x6 + 3 √ x8 + · · · − 2 2 3!3 · 5 · 7 π 2 2·3·5·7·9 π r 1 x 1 1 √ 2 − x2 + = x4 2 π 3 2·2·3·5 1 1 − 2 x6 + 3 x8 + · · · . 2 3! · 3 · 5 · 7 2 ·2·3·4·3·5·7·9 Finally, write this as r 1 2 1 x 1 √ 2 1− J1/2 (x) = x + x4 2 π 2·3 2·2·2·3·5 1 1 x6 + 4 x8 · · · − 3 2 3!3 · 5 · 7 2 ·2·3·4·3·5·7·9 r 2 1 3 1 = x− x + x5 πx 2·3 2·2·2·3·5 1 1 − 3 x7 + 4 x9 − · · · 2 3! · 3 · 5 · 7 2 ·2·3·4·3·5·7·9 r ∞ n 2 X (−1) = x2n+1 πx n=0 (2n + 1)! r 2 = sin(x). πx A similar argument shows that r J−1/2 (x) =

2 cos(x). πx

8.2. BESSEL FUNCTIONS

245

8. We want to show that r J3/2 (x) = f (x) =

2 sin(x) − cos(x) . πx x

Using the Maclaurin expansions of sin(x) and cos(x), we obtain r X ∞ 2 sin(x) (−1)n+1 2n − cos(x) = x . πx x (2n + 1)! n=1 We also have J3/2 (x) =

∞ x 3/2 X

k=0

(−1)k x2k . 22k k!Γ(k + 3/2 + 1)

Using the result from Problem 7 that Γ(n + 3/2 + 1) =

1 · 3 · · · (2n + 1) · (2n + 3) , 2n+1

it is routine to verify that these expressions for J3/ (x) and f (x) match up term for term, using the n = k + 1 rule (so the n = 1 term equals the k = 0 term, the n = 2 term equals the k = 1 term, and so on. A similar argument shows that r J−3/2 (x) =

2 πx

cos(x) − sin(x) − x

9. First, recall that J00 (x) = −J1 (x). Then Z α α J1 (αs) ds == −J0 (x) = J0 (0) − J0 (α) = 1 0

because J0 (0) = 1 and J0 (α) = 0 by choice of α. Now change variables by s = αx in the integral to get Z α Z 1 J1 (s) ds = J1 (αx)α dx = 1 0

and this implies that Z 1 J1 (αx) dx = 0

1 . α

10. (a) Let u(x) = J0 (αx). Then u0 (x) = αJ00 (αx) and u00 (x) = α2 J000 (αx) so xu00 + u0 + α2 xu = α2 J000 (αx) + αJ00 (αx) + α2 xJ0 (αx) = α[αxJ000 (αx) + J00 (αx) + αxJ0 (αx)] = 0

246

CHAPTER 8. SPECIAL FUNCTIONS AND APPLICATIONS because J0 (αx) satisfies Bessel’s equation of order ν = 0. Similarly, if v(x) = J0 (βx), then xv 00 + v 0 + β 2 xv = 0. (b) Multiply the equation for u by v and the equation for v by u and subtract the resulting equations to get xuv 00 − xvu00 + uv 0 − vu0 + (β 2 − α2 )xuv = 0. This can be written as (β 2 − α2 )xuv = −[x(uv 0 − vu0 )]. (c) Integrate both sides of the equation derived in part (b) to obtain Z (β 2 − α2 ) xJ0 (αx)J0 (βx) dx = −x(βJ0 (αx)J00 (βx) − αJ0 (βx)J00 (αx)). Upon multiplying through by the −1 coefficient on the right side side of this equation, we obtain Lommel’s integral.

11. By equation (8.23), (xn Jn (x))0 = xn Jn−1 (x). Then

xn Jn−1 (x) dx = xn Jn (x).

In similar fashion, equation (8.24) immediately yields the second integral. 12. These results follow from the integrals given in Problem 11. We know from the first conclusion of Problem 11 that Z xn Jn−1 (x) dx = xn Jn (x). Let s = αx to obtain αn

xn Jn−1 (αx) dx = α2 xn Jn (x).

The second conclusion follows from the second integral formula of Problem 11. 13. Define

Z 1 In,k =

(1 − x2 )k xn+1 Jn (αx) dx.

For part (a), begin with a result from Problem 11: Z sn Jn−1 (s) ds = xs Jn (s).

8.2. BESSEL FUNCTIONS

247

Replacing n with n + 1, we have Z sn+1 Jn (s) ds = sn+1 Jn+1 (s). Then

Z α

sn+1 Jn (s) ds = sn+1 Jn+1 (s)

α 0

= αn+1 Jn+1 (α).

Now let s = αx to get Z 1

αn+1 xn+1 Jn (αx) dx = αn+1 Jn+1 (α).

Then

Z 1

xn+1 Jn (αx) dx =

1 Jn+1 (α). α

But, Z 1 In,0 =

xn+1 Jn (αx) dx.

Therefore

1 Jn+1 (α). α Now use the integral of Problem 12, with n + 1 in place of n, to write 1 n+1 d n+1 x Jn+1 (αx) . x Jn (αx) = dx α In,0 =

Upon substituting this into the definition of In,k , we have Z 1 In,k =

(1 − x2 )k

d dx

1 n+1 x Jn+1 (αx) α

dx.

This completes part (b). For part (c), apply integration by parts to the integral of part (b): Z 1

2 k d

(1 − x )

In,k = 0

1 n+1 x Jn+1 (αx) dx α

1 1 = (1 − x2 )k xn+1 Jn+1 (αx) α 0 Z 1 1 n+1 − x Jn+1 (αx)k(1 − x2 )k−1 (−2x) dx α 0 Z 2k 1 = (1 − x2 )k−1 xn+2 Jn+1 (αx) dx α 0 2k = In+1,k−1 . α

248

CHAPTER 8. SPECIAL FUNCTIONS AND APPLICATIONS This relates In,k to the value of this integral when n is increased by 1 and k is decreased by 1. In particular, if we carry out k repetitions of this operation, eventually increasing n to n + k, and decreasing k to k to 0, we obtain 2k In+1,k−1 α 2k 2(k − 1) In+2,k−2 = α α 2 2 k(k − 1) = In−2,k−2 α2 22 k(k − 1) 2(k − 2) = In+3,k−2 α2 α 23 k(k − 1)(k − 2) = In+3,k−3 α3 2k k! = · · · k In+k,0 . α

In,k =

Because k is a positive integer, we can write k! = Γ(k + 1) in this expression. For part (e), combine the results of parts (a) and (d) to write Z 1 2k Γ(k + 1) Jn+k+1 (α). (1 − x2 )k xn+1 Jn (αx) dx = αk+1 0 For part (f), write this equation as Jn+k+1 (α) =

αk+1 2k Γ(k + 1)

Z 1

(1 − x2 )k xn+1 Jn (αx) dx.

The rest is just notation to provide the appropriate perspective. In the last equation, replace α with x and x with t to get Z 1 xk+1 Jn+k+1 (x) = k tn+1 (1 − t2 )k Jn (xt) dt. 2 Γ(k + 1) 0 Finally, for part (g), let m − n = k + 1 to get Z 1 2xm−n Jm (x) = m−n tn+1 (1 − t2 )m−n−1 Jn (xt) dt. 2 Γ(m − n) 0 In these results, it is not necessary that k be an integer, because k! has been replaced by Γ(k+1), which is defined if k+1 > 0. For the expressions derived in this problem, it is enough to have n + −1, k > −1 and, in part (g),m > n > −1.

8.2. BESSEL FUNCTIONS

249

14. Use a result of part (g) of Problem 13, with n = −1/2 and m > −1/2 to write 1/2 Z 1 2xm+1/2 2 Jm (x) = m+1/2 t1/2 (1 − t2 )m−1/2 cos(xt) dt πxt 2 Γ(m + 1/2) 0 Z 1 xm (1 − t2 )m−1/2 cos(xt) dt. = m−1 2 Γ(m + 1/2) 0 This is the requested result with m used in place of n. 15. Start with the following result from Problem 14: Z 1 xm Mm (x) = m−1 (1 − t2 )m−1/2 cos(xt) dt. 2 Γ(m + 1/2) 0 Make the change of variables t = sin(θ) to get Z π/2 xm Jm (x) = m−1 (cos2 (t))m−1/2 cos(x sin(θ)) dθ 2 Γ(m + 1/2) 0 Z π/2 xm cos2m (θ) cos(x sin(θ)) dθ. = m−1 2 Γ(m + 1/2) 0 16. Let y = xa Jν (bxc ) and compute the derivatives: y 0 = axa−1 Jν (bxc ) + xa bcxc−1 Jν0 (bxc ) and y 00 = a(a − 1)xa−2 Jν (bxc ) + [2axa−1 bcxc−1 + xa bc(c − 1)xc−2 Jν0 (bxc )] + xa b2 c2 x2c−2 Jν00 (bxc ). Substitute these into the differential equation and simplify to obtain c2 xa−2 [(bxc )2 Jν00 + bxc Jν0 (bxc ) + ((bxc )2 − ν 2 )Jν (bxc )] = 0. In Problems 17–24, the strategy is to match the given differential equation to the differential equation of Problem 16 by choosing a, b, c and ν. This makes it possible to write a general solution in terms of Bessel functions 17. The differential equation matches that of Problem 16 if 7 1 . 2a − 1 = − , 2c − 2 = 0, b2 c2 = 1, and a2 − ν 2 c2 = 3 144 Then

1 1 , b = c = 1, and ν = . 3 4 We can write a general solution a=

y(x) = c1 x1/3 J1/4 (x) + c2 x1/3 J−1/4 (x).

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CHAPTER 8. SPECIAL FUNCTIONS AND APPLICATIONS

18. We need 4 −(2a − 1) = 1, 2c − 2 = 2, b2 c2 = 4, and a2 − ν 2 c2 = − . 9 Then a = 0, c = 2, b = 1 and ν =

1 . 3

Then y(x) = c1 J1/3 (x2 ) + c2 J−1/3 (x2 ) is a general solution. 19. Choose a = 3, c = 4, b = 2, ν = 1/2 to get y(x) = c1 x3 J1/2 (2x4 ) + c2 J−1/2 (2x4 ). 20. Let a = −1, c = b = 2, ν = 3/4 to obtain 1 1 y(x) = c1 J3/4 (2x2 ) + c2 J−3/4 (2x2 ). x x 21. Let a = 2, c = 3, b = 1, ν = 2/3 to get y(x) = c1 x2 J2/3 (x3 ) + c2 x2 J−2/3 (x3 ). 22. Choose a = 4, c = 3, b = 2, ν = 3/4 for a general solution y(x) = c1 x4 J3/4 (2x3 ) + c2 x4 J−3/4 (2x3 ). 23. Here we get a = b = 0, so this method produces only the trivial solution. However, if the differential equation is multiplied by x2 , we obtain x2 y 00 + xy 0 −

1 y = 0, 16

which is an Euler equation with general solution y(x) = c1 x1/4 + c2 x−1/4 . 24. With a = −2, c = b = 3 and ν = 1/2, the general solution is y(x) = c1 x−2 J1/2 (3x3 ) + c2 x−2 J−1/2 (3x3 ).

8.3. SOME APPLICATIONS OF BESSEL FUNCTIONS

251

Figure 8.20: First normal mode in Problem 1 at times t = 1/4, 1/2, 3/4.

8.3

Some Applications of Bessel Functions

1. With f (r) = r(1 − r) and g(r) = r2 , the coefficients in the solution are an =

2 2 J1 (jn )

Z 1

Rs2 (1 − RS)J0 (jn s) ds

and bn =

2 R jn c J12 (jn )

Z 1

R2 s3 J0 (jn s) ds.

The solution is z(r, t) =

∞ X

zn (r, t),

n=1

where jn ct jn jn ct + bn sin J0 r . zn (r, t) = an cos R R R Figures 8.20 through 8.23 show graphs of the first four normal modes of the solution times t = 1/2, 1, 2 and 4, for R = 1, c = 2, f (r) = r(1 − r) and g(r) = 0.

252

CHAPTER 8. SPECIAL FUNCTIONS AND APPLICATIONS

Figure 8.21: Second normal mode in Problem 1.

Figure 8.22: Third normal mode in Problem 1.

8.3. SOME APPLICATIONS OF BESSEL FUNCTIONS

253

Figure 8.23: Fourth normal mode in Problem 1.

2. With f (r) = r cos(πr/2) and g(r) = e−r , the coefficients an and bn in the nth normal mode are Z 1 2 an = 2 Rs2 cos(πRs/2)J0 (Rs) ds J1 (jn ) 0 and 2 R bn = jn c J12 (jn )

Z 1

se−Rs J0 (jn s) ds.

With R = 1, c = 2, f (r) = r cos(πr/2 and g(r) = 0, Figures 8.24–8.27 show the first four normal modes for times t = 1/4, 1/2, 3/4. 3. With f (r) = r2 (1 − r) and g(r) = r + r2 , the coefficients are an =

2 J12 (jn )

Z 1

R2 s3 (1 − Rs)J0 (jn s) ds

and bn =

R 2 2 jn c J1 (jn )

Z 1

Rs2 (1 + Rs)J0 (jn s) ds.

With R = 1, c = 2, f (r) = r2 (1 − r) and g(r) = 0, Figures 8.28–8.31 show the first four normal modes for times t = 1/4, 1/2, 3/4.

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CHAPTER 8. SPECIAL FUNCTIONS AND APPLICATIONS

Figure 8.24: First normal mode in Problem 2.

Figure 8.25: Second normal mode in Problem 2.

8.3. SOME APPLICATIONS OF BESSEL FUNCTIONS

255

Figure 8.26: Third normal mode in Problem 2.

Figure 8.27: Fourth partial sum in Problem 2.

256

CHAPTER 8. SPECIAL FUNCTIONS AND APPLICATIONS

Figure 8.28: First normal mode in Problem 3.

Figure 8.29: Second normal mode in Problem 3.

8.3. SOME APPLICATIONS OF BESSEL FUNCTIONS

257

Figure 8.30: Third normal mode in Problem 3.

Figure 8.31: Fourth normal mode in Problem 3.

258

CHAPTER 8. SPECIAL FUNCTIONS AND APPLICATIONS

Figure 8.32: First normal mode in Problem 4.

4. The coefficients in the solution are given by Z 1 2 an = 2 s sin(πs)J0 (jn s) ds. J1 (jn ) 0 Figures 8.32–8.35 show the first four normal modes, with R = 1, c = 2, g(r) = 0 and f (r) = sin(πr). In each of Problems 5–8, the solution has the form u(r, t) =

∞ X

an J0 (jn r)e−2jn t .

n=1

For each, the expression for the coefficients is given. 5. an =

2 J12 (jn )

Z 1 ξ(1 + cos(πξ))J0 (jn ξ) dξ. 0

6. 2 an = 2 J1 (jn ) 7. an =

2 2 J1 (jn )

Z 1

ξ(1 − ξ 2 )J0 (jn ξ) dξ.

Z 1

ξ 2 cos(3πξ/2)J0 (jn ξ).

8.3. SOME APPLICATIONS OF BESSEL FUNCTIONS

259

Figure 8.33: Second normal mode in Problem 4.

Figure 8.34: Third normal mode in Problem 4.

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CHAPTER 8. SPECIAL FUNCTIONS AND APPLICATIONS

Figure 8.35: Fourth normal mode in Problem 4.

8. an =

2 2 J1 (jn )

Z 1

ξ(2ξ 3 − ξ − 1)J0 (jn ξ) dξ.

Figures 8.36–8.39 show graphs of u(r, t0 ) for t0 equal to 1/10, 1/5, 1/7 and 2/5.

8.3. SOME APPLICATIONS OF BESSEL FUNCTIONS

261

Figure 8.36: Temperature profiles at specific times in Problem 5.

Figure 8.37: Temperature profiles in Problem 6.

262

CHAPTER 8. SPECIAL FUNCTIONS AND APPLICATIONS

Figure 8.38: Temperature profiles in Problem 7.

Figure 8.39: Temperature profiles in Problem 8.

Chapter 9

Transform Methods of Solution 9.1

Laplace Transform Methods

1. Apply the Laplace transform with respect to t to the partial differential equation to get K s2 Y (x, s) = c2 Y 00 (x, s) + . s Here primes denote differentiation with respect to x and the initial conditions have been inserted in using the operational rule for the derivatives. Write this equation as s2 K Y 00 − 2 Y = − 2 . c c s Think of this as a second-order differential equation in x, with s carried along as a parameter. The general solution is Y (x, s) = c1 esx/c + c2 e−sx/c +

K . s3

Here c1 and c2 are “constants”, but may involve s, because x is the variable of the differential equation. Now Y (0, s) = [y(0, t)](s) = F (s) = c1 + c2 +

K . s3

We need limx→∞ y(0, t) = 0, so lims→∞ Y (x, s) = 0. Therefore c1 = 0 and K c2 = F (s) − 3 . s We now have K K Y (x, s) = F (s) − 3 e−sx/c + 3 . s s 263

264

CHAPTER 9. TRANSFORM METHODS OF SOLUTION The solution is the inverse Laplace transform of Y (x, s). Recalling the formula for the inverse Laplace transform of e−as F (s), we obtain x K x 2 x 1 2 y(x, t) = f t − − t− H t− + Kt , c 2 c c 2 in which H is the Heaviside function.

2. Apply the Laplace transform with respect to t to the partial differential equation to get 9s2 Y (x, s) + Y 00 (x, s) − 6sY 0 (x, s) = 0. Then Y 00 − 6sY 0 + 9s2 Y = 0. This second-order differential equation has characteristic equation r2 − 6sr + 9s2 = 0, with repeated root r = 3s. Then Y (x, s) = c1 e3sx + c2 xe3sx . Now, L[y(0, t)](s) = Y (0, s) = 0 = c1 , so Y (x, s) = c2 xe3xs . Next, L[y(2, t)](s) = F (s) = 2c2 e6s , so c2 =

1 −6s e F (s). 2

Then

1 1 −6s e F (s)xe3xs = xF (s)e(3x−6)s . 2 2 The solution is obtained by inverting this to get Y (x, s) =

y(x, t) =

1 xf (t − (6 − 3x))H(t − (6 − 3x)). 2

3. From the partial differential equation and the initial conditions, s2 Y (x, s) = c2 Y 00 − Then Y 00 −

A . s2

s2 A = 2, 2 c s

9.1. LAPLACE TRANSFORM METHODS

265

with general solution A . s4

Y (x, s) = c1 esx/c + c2 e−sx/c −

Because limx→∞ y(x, t) = 0, we must have lims→∞ Y (x, s) = 0, so c1 = 0 and A Y (x, s) = c2 e−sx/c − 4 . s Next, y(0, t) = 0, so A Y (0, s) = c2 − 4 s and then A c2 = 4 . s Then A A Y (x, s) = 4 e−sx/c − 4 . s s The solution is the inverse of this, x A 3 x 3 A H t− t− − t . y(x, t) = 6 c c 6 5. Transform the partial differential equation to get s2 Y (x, s) = c2 Y 00 (x, s) − Then Y 00 −

Ax . s2

s2 Ax Y = 2 2. 2 c c s

This has general solution Y (x, s) = c1 esx/c + c2 e−sx/c −

Ax . s4

The condition that limx→∞ y(x, t) = 0 forces lims→∞ Y (x, s) = 0, so c1 = 0 and Ax Y (x, t) = c2 e−sx/c − 4 . s Next, L[y(0, t)](s) = F (s) = Y (0, s) = c2 . Then Y (x, s) = F (s)e−sx/c −

Ax . s4

Invert this for the solution x x 1 y(x, t) = f t − H t− − Axt3 . c c 6

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6. Take the Laplace transform (in t) in the heat equation and use the initial conditions to get s U 00 − U = 0, k with general solution √ √ U (x, s) = c1 e s/kx + c2 e− s/kx . Because limx→∞ u(x, t) = 0, we have lims→∞ U (x, s) = 0, so c1 = 0 and √ U (x, s) = c2 e− s/kx . Next take the transform of u(0, t) = t2 to get U (0, s) =

2 = c2 , s3

2 −√s/kx . e s3 As preparation for using the convolution theorem, think of this as the product 2 1 −√s/kx U (x, s) = 2 e . s s U (x, s) =

By the convolution theorem, the inverse of this is the solution x √ u(x, t) = 2t ∗ erfc . 2 kt 7. Take the transform with respect to t of the heat equation to get sU (x, s) − e−x = kU 00 (x, s), or

1 s U = − e−x . k k The associated homogeneous equation of this nonhomogeneous equation has the general solution √ √ Uh (x, s) = c1 e s/kx + c2 e− s/kx . U 00 −

For a particular solution, use undetermined coefficients, trying Up (x, s) = Ae−x . Substitute this into the nonhomogeneous differential equation to get A−

s 1 A=− , k k

9.1. LAPLACE TRANSFORM METHODS so A=

267

s−k .

Then, √ U (x, s) = Uh (x, s) + Up (x, s) = c1 e

s/kx

√ + c2 e− s/kx +

1 −x e . s−k

Because limx→∞ u(x, t) = 0, choose c1 = 0, so √ U (x, s) = c2 e− s/kx +

1 −x e . s−k

Take the transform of u(0, t) = 0 to get U (0, s) = c2 + Then c2 = − so U (x, s) = −

1 . s−k

1 s−k

1 −√s/kx 1 −x e + e . s−k s−k

Using the convolution theorem, write h √ i u(x, t) = −e−kt ∗ L−1 e− s/kx (t) + ekt e−x . By consulting a table, we obtain h i √ 2 x e−x /4kt . L−1 e−(x/ k)s (t) = √ 3 2 πkt Then,

2 x u(x, t) = −e−kt ∗ √ e−x /4kt + ekt−x . 3 2 πkt

8. Apply the transform (in t) to the heat equation to get s U = 0, k

U 00 − with general solution √ U (x, s) = c1 e

s/kx

√ + c2 e− x/kx .

Now u(x, 0) forces c1 = −c2 and the transformed solution has the form p U (x, s) = c sinh( s/kx).

268

CHAPTER 9. TRANSFORM METHODS OF SOLUTION With u(1, t) = f (t), we get U (1, s) = F (s), so c= Then

F (s) p . sinh( s/k)

p sinh( s/kx) p F (s) U (x, s) = sinh( s/k)

and the solution is u(x, t) = L−1 [U (x, s)](t).

9.2

Fourier Transform Methods

For the first four problems, the solution is given by Z ∞ 2 1 u(x, t) = √ f (ξ)e−(x−ξ) /4kt . 2 πkt −∞ 1. With f (x) = e−4|x| , the solution is Z ∞ 2 1 √ e−4|ξ| e−(x−ξ) /4kt dξ. 2 πkt −∞ 2. The solution is 1 u(x, t) = √ 2 πkt

Z π −π

3. 1 u(x, t) = √ 2 πkt 4. 1 u(x, t) = √ 2 πkt

sin(ξ)e−(x−ξ) /4kt dξ.

Z 4

ξe−(x−ξ) /4kt dξ.

Z 1

e−ξ−(x−ξ) /4kt dξ.

−1

5. Take the Fourier transform of the wave equation with respect to x to get yb00 = 144(iω)2 yb = −144ω 2 yb, or yb00 + 144ω 2 yb = 0. Then yb(ω, t) = c1 cos(12ωt) + c2 sin(12ωt), in which primes denote differentiation with respect to t. Because yt (x, 0) = 0, then c1 = 0 and yb(ω, t) = c1 cos(12ωt).

9.2. FOURIER TRANSFORM METHODS

269

Next, y(x, 0) = f (x), so yb(ω, 0) = fb(ω). Then c1 = fb(ω), so yb(ω, t) = fb(ω) cos(12ωt). It is routine to compute (or use a software routine to find) fb(ω) =

10 . 25 + ω 2

Then yb(ω, t) =

10 cos(144ωt). 25 + ω 2

Finally, use the integral formula for the inverse Fourier transform to obtain Z ∞ 1 10 iωx y(x, t) = Re cos(12ωt)e dω . 2π −∞ 25 + ω 2 Of course y(x, t) is a real quantity, so in the last line we have taken the real part of the integral. If we replace eiωx = cos(ωx) + i sin(ωx) in this integral, we can write more explicitly Z ∞ 1 10 y(x, t) = cos(ωx) cos(12ωt) dω. 2π −∞ 25 + ω 2 6. Apply the Fourier transform (in x) to the initial-boundary value problem to get yb00 + 64ω 2 yb = 0, Z 8 1 − 8iω − e−8ωi yb(ω, 0) = (8 − ξ)e−iωξ dξ = , ω2 0 yb0 (ω, 0) = 0. This problem has the solution yb(ω, t) =

1 − 8ωi − e−8ωi cos(8ωt). ω2

Invert this to obtain

1 y(x, t) = Re 2π

Z ∞

1 − 8ωi − e−8ωi iωx cos(8ωt)e dω . ω2 −∞

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CHAPTER 9. TRANSFORM METHODS OF SOLUTION

7. Apply the Fourier transform to the initial-boundary value problem to get yb00 + 16ω 2 yb = 0, yb(ω, 0) = 0, Z π 2i sin(πω) sin(ξ)e−iωξ dξ = yb0 (ω, 0) = . ω2 − 1 −π The solution of the transformed problem is yb(ω, t) =

2i sin(πω) sin(4ωt). 4ω(ω 2 − 1)

Invert this to obtain the solution Z ∞ 1 i sin(πω) iωx sin(4ωt)e dω . y(x, t) = Re 2π −∞ 2ω(ω 2 − 1) 8. The initial-boundary value problem transforms (in x) to yb00 + ω 2 y = 0, Z 2 2 (2 − |ξ|)e−iωξ dξ = 2 (1 − cos(2ω)), yb(ω, 0) = ω −2 yb0 (ω, 0) = 0. This has solution yb(ω, t) =

2 (1 − cos(2ω)) cos(ωt)eiωx dω. ω2

9. The problem transforms (in x) to yb00 + 9ω 2 yb = 0, yb(ω, 0) = 0, yb0 (ω, 0) = F[e−2x H(x − 1)](ω) =

(2 − iω)e−(2+iω) . 4 + ω2

This problem has the solution yb(ω, t) =

(2 − iω)e−(2+iω) . 3ω(4 + ω 2 )

Then 1 y(x, t) = 2π

Z ∞

(2 − iω)e−(2+iω) iωx sin(3ωt)e dω . 3ω(4 + ω 2 ) −∞

9.3. FOURIER SINE AND COSINE TRANSFORM METHODS

271

10. Transform the problem to get yb00 + 4ω 2 yb = 0, yb(ω, 0) = 0, Z 2 2(1 − cos(2ω)) yb0 (ω, 0) = . g(ξ)e−iωξ dξ = ω −2 This has the solution yb(ω, t) =

1 − cos(2ω) sin(2ωt). ω2

Invert this to get y(x, t) = Re

1 2π

Z ∞

1 − cos(2ω) iωx sin(2ωt)e dω . ω2 −∞

11. This problem was solved by the Fourier transform in the text, so we can use the result to immediately write Z 0 Z 4 −1 1 y dξ + dξ u(x, y) = 2 2 π −4 y 2 + (ξ − x)2 0 y + (ξ − x) y x+4 x−4 =− arctan + arctan . π y y 12. The solution is Z ξ y 1 dξ u(x, y) = π 0 y 2 + (ξ − x)2 1 − ln(x2 + y 2 ) − ln((x − 1)2 + y 2 ) − 2x arctan(x/y) + 2x arctan((x − 1)/y) . 2π

9.3

Fourier Sine and Cosine Transform Methods

1. Take a Fourier sine transform (in x) of the problem to get ybS00 + 9ω 2 ybS = 0, Z 1 2(1 − cos(ω) − ω sin(ω) ybS (ω, 0) = ξ(1 − ξ) sin(ωξ) dξ = , ω3 0 ybS0 (ω, 0) = 0. The solution of the transformed problem is 2(1 − cos(ω)) − ω sin(ω) ybS (ω, t) = cos(3ωt). ω3 Invert this to obtain the solution Z 2 ∞ 2(1 − cos(ω)) − ω sin(ω) y(x, t) = sin(ωx) cos(3ωt) dω. π 0 ω3

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CHAPTER 9. TRANSFORM METHODS OF SOLUTION

2. Apply the sine transform to the problem to get ybS00 + 9ω 2 ybS = 0, ybS (ω, 0) = 0, Z 11 2(cos(4ω) − cos(11ω)) 0 ybS = 2 sin(ωξ) dξ = . ω 4 This problem has the solution ybS (ω, t) =

2(cos(4ω) − cos(11ω)) sin(3ωt). 3ω 2

Invert this to obtain the solution Z ∞ 4 cos(4ω) − cos(11ω) y(x, t) = sin(ωx) sin(3ωt). 3π 0 ω2 3. The sine transformed problem is ybS00 + 4ω 2 ybS = 0, ybS (ω, 0) = 0, Z 5π/2 sin(ωπ/2) − sin(5ωπ/2) . ybS0 (ω, 0) = cos(ξ) sin(ωξ) dξ = ω2 − 1 π/2 This has solution ybS (ω, t) =

sin(ωπ/2) − sin(5ωπ/2) sin(2ωt). 2ω(ω 2 − 1)

Invert this to obtain the solution Z 2 ∞ sin(ωπ/2) − sin(5ωπ/2) y(x, t) = sin(ωx) sin(2ωt). π 0 2ω(ω 2 − 1) 4. Upon taking the sine transform (in x) of the problem, we get ybS00 + 36ω 2 ybS = 0, Z ∞ ybS (ω, 0) = −2e−ξ sin(ωξ) dξ 0 0 ybS (ω, 0) = 0.

This has the solution ybS (ω, t) = −

2ω cos(6ωt). 1 + ω2

Invert this to obtain y(x, t) = −

4 π

Z ∞ 0

ω sin(ωx) cos(6ωt). 1 + ω2

9.3. FOURIER SINE AND COSINE TRANSFORM METHODS

273

5. After applying the sine transform to the initial-boundary value problem, we obtain ybS00 + 196ω 2 ybS = 0, ybS (ω, 0) = 0, Z 3 ybS0 (ω, 0) = ξ 2 (3 − ξ) sin(ωξ) dξ 0

3 = 4 (2 sin(3ω) − 4ω cos(3ω) − 3ω 2 sin(3ω) − 2ω). ω The transformed problem has the solution ybS (ω, t) = 3 (2 sin(3ω) − 4ω cos(3ω) − 3ω 2 sin(3ω) − 2ω) sin(14ωt). 14ω 5 Then y(x, t) =

2 π

Z ∞ 0

ybS (ω, t) sin(ωx) dω.

6. Apply the Fourier sine transform with respect to x to the problem to get u b0S + ω 2 u bS + tb uS = 0, u bS (ω, 0) = FS xe−x =

2ω . (1 + ω 2 )2

This problem (involving a linear first-order differential equation) has the solution 2 2 2ω u bS (ω, t) = e−(ω t−t /2) . (1 + ω 2 )2 Invert this to solve the problem: Z 2 2 4 ∞ ω e−ω t−t /2 sin(ωx) dω. u(x, t) = 2 2 π 0 (1 + ω ) 7. Apply the Fourier cosine transform in x to the problem to get u b0C + (1 + ω 2 )b uC = −f (t); u bC (ω, 0) = 0. This has the solution 2

u bC (ω, t) = e−(1+ω )t = −f (τ ) ∗ e

Z t

f (τ )e(1+ω )τ dτ

0 −(1+ω 2 )t

Invert this to obtain u(x, t) = −

2 π

Z ∞

f (t) ∗ e−(1+ω )t cos(ωx) dω.

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CHAPTER 9. TRANSFORM METHODS OF SOLUTION

Chapter 10

Vectors and the Vector Space Rn 10.1

Vectors in the Plane and 3−Space

1. √ √ 2)i + 3j, F − G = (2 − 2)i − 9j + 10k, √ √ 2F = 4i − 6j + 10k, 3G = 3 2i + 18j − 15k, k F k= 38

F + G = (2 +

2. F + G = i + 4j − 3k, F − G = i − 4j − 3k, √ 2F = 2i − 6k, 3G = 12j, k F k= 10 3. F + G = 3i − k, F − G = i − 10j + k, 2F = 2i − 6k, 3G = 3i + 15j − 3k, k F k=

√ 29

4. √ √ F + G = ( 2 + 8)i + j − 4k, F − G = ( 2 − 9)i + j − 8k, √ 2F = 4i − 6j + 10k, 3G = 24i + 6k, k F k= 39 5. F + G = 3i − j + 3k, F − G = −i + 3j − k, 2F = 2i + 2j + 2k, 3G = 6i − 6j + 6k, k F k=

√ 3

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CHAPTER 10. VECTORS AND THE VECTOR SPACE RN

276

6. The vector F = −5i + j − 2k is in the direction √ from the first point to the second, but this vector has length k F k= 30. The vector 5 √ F 30 has length 5 and extends from the first point to the second. 7.

9 √ (−5i − 4j + 2k) 45

12 √ (10i − 3j − 4k) 125

4 (−4i + 7j + 4k) 9

10. The vector M = i + j − 3k is oriented from (1, 0, 4) to (2, 1, 1). The vector V = (x − 1)i + yj + (z − 4)k is parallel to M exactly when (x, y, z) is on the line through the two given points. Then, for some t, V = tM, so (x − 1)i + yj + (z − 4)k = t(i + j − 3k). Then x − 1 = t, y = t, z − 4 = −3t. Then x = 1 + t, y = t, z = 4 − 3t are parametric equations of the line through (1, 0, 4) and (2, 1, 1). Here t varies over all real values. As a check, notice that these points are on the line given by these parametric equations for t = 0 and t = 1 respectively. 11. x = 3 − 6t, y = 1 − t, z = 0 12. x = 2, y = 1, z = 1 − 3t This line is parallel to the z− axis, and also passes through (2, 1, 0) when t = 0. 13. x = 0, y = 1 − t, z = 3 − 2t 14. x = 1 − 3t, y = −2t, z = −4 + 9t 15. x = 2 − 3t, y = −3 + 9t, z = 6 − 2t

10.2. THE DOT PRODUCT

10.2

277

The Dot Product

In Problems 1–6, F is the first vector given in the problem G is the second, and θ is the angle between F and G. 1. F · G = 2 and cos(θ) =

F·G 2 =√ . k F kk G k 14

F and G are not orthogonal. √ 2. F · G = 8, cos(θ) = 8/ 82 and the vectors are not orthogonal. √ √ 3. F · G = −23, cos(θ) = −23/ 29 41 and the vectors are not orthogonal. √ √ 4. F · G = −63, cos(θ) = −63/ 75 74 and the vectors are not orthogonal. 5. F · G = −18, cos(θ) = −9/10 and the vectors are not orthogonal. 6. F · G = 4, cos(θ) = 2/3 and the vectors are not orthogonal. In Problems 7–12, if the given point is (x0 , y0 , z0 ) and the normal to the proposed plane is N = ai + bj + ck, then the plane through the point and having N as normal vector has equation a(x − x0 ) + b(y − y0 ) + c(z − z0 ) = 0. The constant terms are usually collected to write this equation in the form ax + by + cz = k. 7. The plane has equation 3(x + 1) − (y − 1) + 4(z − 2) = 0, or 3x − y + 4z = 4. 8. x − 2y = −1 9. 4x − 3y + 2z = 25 10. −3x + 2y = 1 11. 7x + 6y − 5z = −26 12. 4x + 3y + z = −6 In each of Problems 13–17, the projection of v onto u is the vector u·v u. k u k2

CHAPTER 10. VECTORS AND THE VECTOR SPACE RN

278 13.

proju v = −

9 u 14

14. −

11 u 30

15. 1 u 62 16. 73 u 101 17. 15 u 53

10.3

The Cross Product

1. i j F × G = −3 6 −1 −2

k 1 = 8i + 2j + 12k 1

and i j G × F = −1 −2 −3 6

k 1 = −8i − 2j − 12k = −F × G 1

2. F × G = i + 12j + 6k = −G × F 3. F × G = −8i − 12j − 5k = −G × F 4. F × G = 112k = −G × F In Problems 5–9, the three given points are used to find two nonparallel vectors in the plane that is sought (assuming that the points are not collinear). The cross product of these vectors is a normal to the plane. We than have a point on the plane and a normal to the plane, so we can find an equation of the plane.

10.3. THE CROSS PRODUCT

279

5. Vectors from the first point to the second and third points are F = 4i − j − 6k and G = i − k. Compute N = F × G = i − 2j + k. Because this cross product is not the zero vector, the given points are not collinear. The plane containing these points has equation ((x + 1)i + (y − 1)j + (z − 6)k) · N = 0. This is x + 1 − 2(y − 1) + z − 6 = 0, or x − 2y + z = 3. 6. The points are not collinear and the plane containing them has equation x + 2y + 6z = 12. 7. 2x − 11y + z = 0 8. 5x + 16y − 2z = −1 9. 29x + 37y − 12z = 30 In Problems 10–12, recall that a plane ax + by + cz = k has normal vector ai + bj + ck, or any nonzero multiple of this vector. 10. N = 8i − j + k, or any nonzero scalar multiple of this vector. 11. N = i − j + 2k 12. n = i − 3j + 2k 13. If two sides of a parallelogram meet at a point, with an angle θ between the sides, then the area of the parallelogram is the product of the lengths of these sides, times the cosine of θ. Now suppose F and G are vectors drawn from a corner of the parallelogram, along two incident sides, and θ the angle between these vectors. Then k F kk G k cos(θ) is the product of the lengths of incident sides, times the angle between them, hence is the area of the parallelogram. But k F kk G k cos(θ) =k F × G k so the magnitude of this cross product is the area of the parallelogram.

CHAPTER 10. VECTORS AND THE VECTOR SPACE RN

280

14. The vector N = G × H is normal to the base of the parallelopiped (box) having sides along F, G, H (drawn as arrows from a corner of the box). From Problem 13, the area of the base of this box is N. Then (|G × F) · F = F · N =k N kk F k cos(θ) is the magnitude of the volume of the box having incident sides F, G, H, because k F k cos(θ) = ± altitude of the box.

10.4

n− Vectors and the Algebraic Structure of Rn

1. Let F =< −2, 1, −1, 4 >. Then O =< 0, 0, 0, 0 > is in S (the scalar multiple of 0 with F). Further, if a and b are real numbers, then aF + bF = (a + b)F, so a sum of scalar multiples of F is again a scalar multiple of F, hence is in S. Finally, a(bF) = (ab)F so a scalar multiple of a vector in S is in S. Therefore S is a subspace of R4 . 2. < 0, 0, 0, 0, 0, 0 > is in S, having zero third and fifth components. Further, if F and G have zero third and fifth components, so does any scalar multiple of these vectors, as well as any sum of scalar multiples of these vectors. Therefore S is a subspace of R6 . 3. S is not a subspace of R5 . The zero vector does not have fourth component 1, and a sum of vectors with fourth component 1 does not have fourth component 1. S fails on scalar multiples as well. 4. S is not a subspace of R8 . The zero vector is in S, but a sum of vectors of length less than 1 need not have length less than 1, and scalar multiples could also fail to have length less than 1. 5. S is not a subspace of R4 . For example, < 1, 1, 1, 0 > and < 0, 1, 1, 1 > are both in S, but their sum is not, having no zero component. 6. The zero vector is in S, and a linear combination of two vectors with equal first and fourth components will have equal first and fourth components. S is a subspace of R5 . In Problems 7–16, keep in mind that a set of vectors is linearly dependent if some linear combination of these vectors, with at least one nonzero coefficient, is equal to the zero vector. And the vectors are linearly independent exactly when the only linear combination of these vectors adding up to the zero vector is the trivial one, with all zero coefficients.

10.4. N − VECTORS AND THE ALGEBRAIC STRUCTURE OF RN

281

7. If a(3i + 2j) + b(i − j) =< 0, 0 >, then 3a + b = 0 and 2a − b = 0. From the second equation, b = 2a, and then the first equation is 5a = 0, so a = 0 and then b = 0. The only linear combination of these vectors that equals the zero vector is the trivial combination, so the vectors are linearly independent. We could also observe that neither vector is a linear combination of the other, which would be the case of these two vectors were linearly dependent. 8. These vectors are linearly dependent because no one of them is a linear combination of the others. It is also easy to check that the only linear combination of these vectors that equals < 0, 0, 0, 0 > is the trivial linear combination (all coefficients zero). 9. These two vectors are linearly independent because neither is a scalar multiple of the other (which would occur for two linearly dependent vectors). 10. These vectors are linearly dependent because 4 < 1, 0, 0, 0 > −6 < 0, 1, 1, 0 > + < −4, 6, 6, 0 >=< 0, 0, 0, 0 > . 11. The vectors are linearly dependent because 2 < 1, 2, −3, 1 > + < 4, 0, 0, 2 > − < 6, 4, −6, 4 >=< 0, 0, 0, 0 > . 12. Suppose that α < 0, 1, 1, 1 > +β < −3, 2, 4, 4 > +γ < −2, 2, 34, 2 > + δ < 1, 1, −6, 2 >=< 0, 0, 0, 0 > . Looking at the components of the sum on the left, we obtain −3β − 2γ + δ = 0, α + 2β + 2γ + δ = 0, α + 4β + 34γ − 6δ = 0, α + 4β + 2γ + 2δ = 0. It is routine to solve these equations and show that α = β = γ = δ = 0. Therefore the vectors are linearly independent.

282

CHAPTER 10. VECTORS AND THE VECTOR SPACE RN

13. The vectors are linearly dependent because 2 < 1, −2 > −2 < 4, 1 > + < 6, 6 >=< 0, 0 > . 14. The vectors are linearly independent (each has a component of 1 where the other two have components of 0). 15. The vectors are linearly independent. 16. The vectors are linearly independent. In each of Problems 17–21 it is routine to show that S is not empty, and that a linear combination of vectors of S is in S, showing that S is a subspace of the appropriate Rn . We will show how to find a basis for S. 17. Every vector in S has the form x < 1, 0, 0, −1 > +y < 0, 1, −1, 0 > . Therefore the two vectors < 1, 0, 0, −1 > and < 0, 1, −1, 0 > span S. These vectors are also linearly independent, and so form a basis for S, which has dimension 2. 18. Every vector in S is of the form x < 1, 0, 2, 0 > +y < 0, 1, 0, 3 > . The independent vectors < 1, 0, 2, 0 > and < 0, 1, 0, 3 > therefore span S and form a basis. S has dimension 2. 19. S consists of all vectors in Rn of the form < x1 , 0, x2 , · · · , xn−1 > . The n − 1 independent vectors < 1, 0, 0, · · · , 0 >, < 0, 0, 1, 0, · · · , 0 >, · · · , < 0, 0, 0, · · · , 0, 1 > span S, so S has dimension n − 1. 20. S is spanned by the independent vectors < 1, 1, 0, 0, 0, 0 >, < 0, 0, 1, 1, 0, 0 >, < 0, 0, 0, 0, 0, 1 > and so has dimension 3. 21. Every vector in S is a scalar multiple of 0, 1, 0, 2, 0, 3, 0 > so this vector forms a basis for S and S has dimension 1.

10.4. N − VECTORS AND THE ALGEBRAIC STRUCTURE OF RN

283

22. The spanning vectors are independent, so they form a basis for the subspace S of R3 that they span. Further, by inspection, < 3, 1, 0 >= 3 < 1, −1, 0 > +2 < 0, 1, 0 > . 23. The spanning vectors are independent and form a basis for the subspace S of R3 that they span. Further, by inspection, < −5, −3, −3 >= −5 < 1, 1, 1 > +2 < 0, 1, 1 > . 24. It is routine to check that the three spanning vectors are independent, hence form a basis for the subspace S of R4 that they span. By inspection, < 4, 5, 9, 4 >= 2 < 2, 1, 1, 0 > +3 < 0, 1, 1, 0 > +4 < 0, 0, 1, 1 > . 25. The spanning vectors are independent and so form a basis for the subspace S of R4 that they span. This subspace has dimension 3. For X to be in S, we need numbers a, b and c so that a < 1, 0, −3, 2 > +b < 1, 0, −1, 1 >=< −4, 0, 10, −5 > . This requires that a + b = −4, −3a + b = 10, and 2a + b = −5. Then a = −3, b = −1, so < −4, 0, 10, −7 >= −3 < 1, 0, −3, 2 > − < 1, 0, −1, 1 > and < −4, 0, 10, −4 > is in S. 26. All that is needed is to show that the dot product of the two vectors is zero: (X − Y) · (X + Y) = X · X + X · Y − Y · X − Y · Y =k X k2 − k Y k2 = 0. 27. Because U is in S, which is spanned by V1 , · · · , Vk , we must have U = c1 V1 + · · · + ck Vk , so V1 , · · · , Vk , U are linearly independent. 28. If U1 , · · · , Uk are linearly independent, then they form a basis for the subspace of Rn spanned by S. If these vectors are not independent, then at least one of the vectors must be a linear combination of the others. Suppose Uk is a linear combination of U1 , · · · , Uk−1 . Then U1 , · · · , Uk−1

CHAPTER 10. VECTORS AND THE VECTOR SPACE RN

284

span S. If these vectors are linearly independent, they form a basis for S. If not, one of these vectors, say Uk−1 is a linear combination of the others. Now U1 , · · · , Uk−2 span S. Continue in this way. If these vectors are linearly independent, they form a basis for S. If not, one must be a linear combination of the others, and then we obtain a set of k − 3 of the original set of vectors that spans S. Continue in this way until a linearly independent set of the original spanning vectors spans S. This set is a basis for S.

29. Suppose the set of vectors is V1 , · · · , Vk , O. Then 0V1 + 0V2 + · · · + 0Vk + 1O = O is a linear combination of the vectors adding up to the zero vector, and with a nonzero coefficient. This makes the original set of vectors (including the zero vector) linearly dependent.

10.5

Orthogonal Sets and Orthogonalization

1. k V1 + · · · + Vk k2 = (V1 + · · · + Vk ) · (V1 + · · · + Vk ) = V1 · (V1 + · · · + Vk ) + V2 · (V1 + · · · + Vk ) + · · · + Vk · (V1 + · · · + Vk ) = V1 · V1 + V2 · V2 + · · · + Vk · Vk =k V1 k2 + k V2 k2 + · · · k Vk k2 , in which we have used the fact that Vi · Vj = 0 if i 6= j.

2. Let Y =X−

k X (X · Vj )Vj . j=1

10.5. ORTHOGONAL SETS AND ORTHOGONALIZATION

285

Compute 0 ≤k Y k2 = Y · Y     k k X X = X − (X · Vj )Vj  · X − (X · Vj )Vj  j=1

j=1

= X · X − 2X ·

k X

(X · Vj )Vj

j=1

    k k X X +  (X · Vj )Vj  ·  (X · Vj )Vj  j=1

j=1

= X · X − 2X

k X

(X · Vj )Vj

j=1

k k X X (X · Vj )(X · Vr )Vj · Vr . j=1 r=1

Because Vj · Vr = 0 if j 6= r, and Vj · Vj = 1, the double sum collapses to just terms in which j = r and we have 0 ≤k X k2 −2

k k X X (X · Vj )2 + (X · Vj )2 j=1

=k X k2 −

j=1

k X

(X · Vj )2 .

j=1

Then k X (X · Vj )2 ≤k X k2 . j=1

3. Reason as in Problem 2, except now use the fact that n X (X · Vj )Vj = X. j=1

In each of Problem 4–11, the given vectors are denoted X1 , · · · , Xk in the given order. 4. Let V1 = X1 and X2 · X1 X1 X1 · X1 18 = X2 + X1 17 =< 52/17, −13/17, 0 > .

V2 = X2 −

CHAPTER 10. VECTORS AND THE VECTOR SPACE RN

286

5. Take V1 = X1 and V2 = X2 +

11 X1 =< 0, 4/5, 2/5, 0 > . 5

6. Let V1 = X1 . Next, 7 V2 = X2 + X1 =< 0, 4/3, 13/6, 29/6 > . 6 Finally, 3 43/2 V3 = X3 − V1 − V2 6 179/6 129 1 V2 = X3 − V1 − 2 179 1 < 0, 7, −11, 3 > . = 179 7. Let V1 = X1 and V2 = X2 + =

5 X1 26

1 < 109, 0, −41, 58 > . 26

Finally, let 331/26 17 X1 − V1 26 651/26 331 17 V2 = X3 − V1 − 26 651 1 = < −962, 0, −1406, 0, 814 > . 651

V3 = X3 −

8. V1 = X1 , 5 V2 = X2 − X1 7 1 = < 0, 0, −1, −19, 40 >, 9 2 17 V3 = X3 + V1 + V2 9 9 1 = < 0, 218, −341, 279, 62 >, 218 and 6 13 435 X1 + V2 − V3 9 3 1179 1 = < 0, 248, 88, −24, −32 > . 393

V4 =

10.6. ORTHOGONAL COMPLEMENTS AND PROJECTIONS

287

9. V1 = X1 , V2 = X2 − =

1 < 21, −8, −60, −31, −18, 0 >, 10

V3 = X3 − =

1 X1 10

163/10 3 X1 − V2 10 269/10

1 < −423, −300, 489, −759, 132, 0 >, 269

and V4 = X4 + =

13/2 4455/269 15 X1 − V2 − V4 10 260/10 4095/269

1 < 337, −145, 250, 29, −9, 0 > . 91

10. V1 = X1 and 3 1 V2 = X2 + X1 = < 0, 0, −3, 3, 0, 0 > . 2 2 11. V1 = X1 and V2 = X2 because X2 and X1 are orthogonal. Finally, V3 = X3 +

10.6

4 4 1 V1 + V2 = < 0, −8, 0, −8, 0, 16 > . 12 2 3

Orthogonal Complements and Projections

1. Let V1 =< 1, −1, 0, 0 > and V2 =< 1, 1, 0, 0 >. These form an orthogonal basis for S. Then u · V1 u · V2 V1 + V2 V1 · V1 V2 · V2 = −4V1 + 2V2 =< −2, 6, 0, 0 >

uS =

and u⊥ = u − uS =< 0, 0, 1, 7 > . The distance between u and S is k u⊥ k=

√ 50.

2. Let V1 =< 1, 0, 0, 2, 0 > and V2 =< −2, 0, 0, 1, 0 > .

CHAPTER 10. VECTORS AND THE VECTOR SPACE RN

288

We find that uS =

2 1 V1 + V2 =< 0, 0, 0, 1, 0 > 5 5

and u⊥ =< 0, −4, −4, 0, 3 > . The distance between u and S is k u⊥ k=

√ 41.

3. Let V1 =< 1, −1, 0, 1, −1 >, V2 =< 1, 0, 0, −1, 0 >, V3 =< 0, −1, 0, 0, 1 > . Then uS =

1 7 V1 + V2 − 3V3 = < 9, −1, 0, 5, −13 > . 2 2

and

1 < −1, −1, 6, −1, −1 > . 2 The distance between u and S is √ k u⊥ k= 10. u⊥ =

4. Let V1 =< 1, −1, 0, 0 > and V2 =< 1, 1, 6, 1 > . Then uS = −3V1 +

31 V2 =< −86/39, 148/39, 62/13, 31/39 > 39

and

1 < 203, 203, −230, −226 > . 309 The distance between u and S is 1 p k u⊥ k= 186, 394. 309 u⊥ =

5. Let V1 =< 1, 0, 1, 0, 1, 0, 0 > and V2 =< 0, 1, 0, 1, 0, 0, 0 > . We find that 1 1 uS = 3V1 + V2 = < 6, 1, 6, 1, 6, 0, 0 > 2 2 and

1 < 10, 1, −4, −1, −6, −6, 8 > . 2 The distance between u and S is 1√ k u⊥ k= 254. 2 u⊥ =

10.6. ORTHOGONAL COMPLEMENTS AND PROJECTIONS

289

6. If every vector in S ⊥ is orthogonal to every vector in S, then, going the other way, every vector in S is orthogonal to every vector in S ⊥ . Then (S ⊥⊥ ) = S. 7. Let v1 , · · · , vk be an orthogonal basis for S, and u1 , · · · ur an orthogonal basis for S ⊥ . Then the vectors u1 , · · · , uk , v1 , · · · vr span Rn because every n− vector is a sum of a vector in S (a linear combination of v1 , · · · , vk ) and a vector in S ⊥ (a linear combination of u1 , · · · , ur ). Further, the set of vectors u1 , · · · , uk , v1 , · · · vr are linearly independent, because no one of them is a linear combination of others in this set. These k + r vectors therefore form a basis for Rn . Because Rn has dimension n, n = k + r. That is, dimension(S) + dimension (S ⊥ ) = dimension(Rn ). 8. Let u =< 1, −1, 3, −3 >. The idea is to use an orthogonal basis for S to produce uS , which is the vector we seek. The given vectors V1 =< 1, 0, 1, 0 > and V2 =< −2, 0, 2, 1 > form an orthogonal basis for S. Then 1 1 uS = 2V1 + V2 = < 16, 0, 20, 1 > . 9 9 This is the vector in S closest to u. 9. Denote V1 =< 2, 1, −1, 0, 0 >, V2 =< −1, 2, 0, 1, 0 >, V3 =< 0, 1, 1, −2, 0 > . These form an orthogonal basis for S. With u =< 4, 3, −3, 4, 7 >, compute 7 4 V1 + V2 − V3 3 3 1 = < 11, 9, −11, 11, 0 > . 3

uS =

CHAPTER 10. VECTORS AND THE VECTOR SPACE RN

290 10. Denote

V1 =< 0, 1, 1, 0, 0, 1 >, V2 =< 0, 0, 3, 0, 0, −3 >, V3 =< 6, 0, 0, −2, 0, 0 > . With u =< 0, 1, 1, −2, −2, 6 >, compute 8 5 1 V1 − V2 + V3 3 6 10 =< 3/5, 8/3, 1/6, −1/5, 0, 31/6 > .

uS =

This is the vector in S closest to u.

Chapter 11

Matrices, Determinants and Linear Systems 11.1

Matrices and Matrix Algebra

1.  14 −2 6 −5 −6 2A − 3B =  10 −26 −43 −8 

2.  19 2  6 −2  −5A + 3B =  −28 38  −27 35 

3. 2

2 + 2x − x2 4 + 2x + 2ex + 2xex

A + 2AB =

12x + (1 − x)(x + ex + 2 cos(x)) −22 − 2x + e2x + 2ex cos(x)

4. −3A − 4B = (18) This is a 1 × 1 matrix, which we can think of as just the single matrix entry, in this case 18. 5. 4A + 8B =

−36 0 68 128 −40 −36

196 20 −8 72

6. A2 − B2 =

−17 6

18 −40 8 27 − = 1 −5 −39 11

10 40

291

292CHAPTER 11. MATRICES, DETERMINANTS AND LINEAR SYSTEMS 7. BA is not defined;   −10 −34 −16 −30 −14 −2 −11 −8 −45 AB =  10 −5 1 15 61 −63 8.

AB =

−16 17

0 12 −32 ; BA = 28 −14 0

9. AB = (115);  3 −18 −6 −42 66 −2 12 4 28 −44     −6 36 12 84 −132 BA =    0 0 0 0 0  4 −24 −8 −56 88 

10.

 48 1 1 −58  −96 2 2 220   ; BA = 76 AB =  −288 −22 −22 −68 50 −16 6 6 184 

152 136

11. AB is not defined; 410 BA = 17

36 −56 253 40

227 −1

12. BA is not defined; AB =

−22 −42

30 −10 −4 45 30 6

13. AB is not defined; BA = −16 −13 −5 14. Neither AB nor BA is defined. 15. BA is not defined; AB =

39 −84 −23 38

21 3

16. AB is not defined; BA = 28

17. AB is 14 × 14, BA is 21 × 21. 18. Neither AB nor BA is defined.

11.1. MATRICES AND MATRIX ALGEBRA

293

19. AB is not defined, BA is 4 × 2. 20. AB is 1 × 3, BA is not defined. 21. AB is not defined, BA is 7 × 6. 22. There are infinitely many examples. Here is one. Let 2 1 2 1 6 A= ,B = ,C = 8 4 −1 1 −1

0 . 1

Then B 6= C, but BA = CA = 23. The adjacency matrix of G is  0 1  A= 1 0 0

1 1 1 1 1

12 6

6 . 3

1 0 0 1 1

 0 1  1 . 1 0

0 1 1 0 1

Compute 

2 7  A3 =  7 4 4

7 8 9 9 9

7 9 8 9 9

4 9 9 6 7

  4 14 17 9   4  9  and A = 17 18 7 6 18

17 34 33 26 26

17 33 34 26 26

18 26 26 25 24

 18 26  26 . 24 25

The number of v1 − −v4 walks of length 3 is (A)314 = 4 and the number of v1 − −v4 walks of length 4 is (A4 )14 = 18. The number of v2 − −v3 walks of length 3 is 9 and the number of v2 − −v4 walks of length 4 is 26. 24. The adjacency matrix of H is  0 1  A= 1 0 1

1 0 1 0 1

1 1 0 1 0

0 0 1 0 1

 1 1  0 . 1 0

Then 

3 2  A2 =  1 2 1

2 3 1 2 1

1 1 3 0 3

2 2 0 2 0

  1 19 18 1   4  3  and A = 11  14 0 3 11

18 19 11 14 11

11 11 20 4 20

14 11 4 12 4

 11 11  20 . 4 20

The number of v1 − −v4 walks of length 4 is (A4 )14 = 14, the number of v2 − −v3 walks of length 2 is (A2 )23 = 1.

294CHAPTER 11. MATRICES, DETERMINANTS AND LINEAR SYSTEMS 25. The adjacency matrix of K is 

0  1  A= 11 01 1  1 1

1 0

1 1

1 0

1 1

0 1

 1 0  .  1 0

Then  4 2  A2 =  3 3 2 and

2 3 2 2 3

3 2 4 3 2

3 2 3 4 2

  10 2 10 3  3   2  , A = 11 11  2 10 3

 42 32  A4 =  41 41 32

32 30 32 32 30

41 32 42 31 32

41 32 41 42 32

10 6 10 10 6

11 10 10 11 10

11 10 11 10 10

 10 6  10  10 6

 32 30  32 . 32 30

The number of v4 − −v5 walks of length 2 is 2 and the number of v2 − −v3 walks of length 3 is 10. The number of v1 − −v2 walks of length 4 is 32 and the number of v4 − −v5 walks of length 4 is 32. 26. (a) The i, i element of A2 is the number of vi − −vi walks of length 2 in the graph. Each such walk has the form vi − −vj − −vi for some j 6= i, hence corresponds to a vertex vj adjacent to vi in the graph. Therefore A2ii counts the number of vertices adjacent to vi in the graph, and this is the degree of vi . (b) The i − −i element of A3 is the number of vi − −vi walks of length 3 in the graph. Any such walk has the form vi −−vj −−vk −−vi with j 6= k and neither j nor k equal to i. These three vertices therefore form the vertices of a triangle in the graph. However, each such triangle is counted twice in the i, i element of A3 because this triangle actually represents two vI −−vi walks of length 3, namely vi − −vj − −vk − −vi and vi − −vk − −vj − −vi . Therefore (A3 )ii = 2(number of triangles in G). 27. Let M be the set of all n × n real matrices. Each such matrix has nm elements in its n rows and m columns (some possibly the same number, but differentiated by location). If we string out the rows, with row 2 following row 1, then row 3 following row 2, and so on, we have a string of nm elements, which we can think of as a vector in Rnm . Each nm − −vector corresponds to an n × m matrix, and each n × m matrix produces a unique nm − − vector. In this way the n × m matrices correspond in a one-to-one fashion with the vectors in Rnm .

11.2. ROW OPERATIONS AND REDUCED MATRICES

11.2

295

Row Operations and Reduced Matrices

1. A is 3×4, so multiply A on the left by the matrix Ω formed by performing this elementary row operation on I3 . Thus form   1 √0 0 Ω = 0 3 0 . 0 0 1 This choice of Ω can be checked by verifying that ΩA = B. 2.

 1 0 Ω= 0 0

0 1 6 0

0 0 1 0

 0 0  0 1

3. In this case more than one elementary operation is performed, so Ω is a product of elementary operations, the right-most factor performing the first operation, then the next to right-most, the second operation, and so on, with the left factor performing the last operation:    √  5 0 0 0 1 0 1 0 13 Ω = 0 1 0 1 0 0 0 1 0 . 0 0 1 0 0 1 0 0 1 If these products are carried out, we get   0 5 √0 1 0 13 0 0 1 4.



1 Ω = −1 0 5.

Ω=

 1 Ω = 0 0 Then

0 1 1

0 1

 0 0 1 1 0 0 0 1 0

  0 1 1 = −1 0 0

 0 0 0 1 1 0

√ 0 √ 15 0 1 3 = 15 0 1 1 3

1 1 0 0  0 1 0 0 1 0

0 0 1

0 1 0

 0 √1 0  3 4 0

 0 0 1 1 0 0 1 0 1

0 1 0

 0 0 . 1



 0 0 √1 3 1 0 . Ω= √ 4 + 31 4

296CHAPTER 11. MATRICES, DETERMINANTS AND LINEAR SYSTEMS 7.

 1 Ω = 0 0  1 Ω = 0 0

  0 1 0 0 1 1 14 1 0 0 0 0 0 1 0

0 0 1 0 1 0

 0 0 0 0 5 1

 1 1 0 0 0 0

0 1 0

If these are multiplied out, we get 

0 Ω = 0 5

1 0 0

0 1 0

   0 1 0 0 0 =  0 0 4 4 14 1 0

0 1 3

 0 1 0 0 1 0

0 0 1

 0 1 . 0

 3 1 . 0

In these and later problems, the delta notation is sometimes useful: ( 1 if i = j, δij = 0 if i 6= j. 9. Let A = [aij ] be an n × m matrix. Now, B = [bij ] and E = [eij ] are obtained, respectively, by interchanging rows s and t of A and In . Then, for i 6= s and i 6= t, bij = aij and eij = δij . If i = s, bsj = atj and esj = δtj . And if i = t, bij = asj and eij = δsj . Now consider the i, j − − element of EA. For i 6= s and i 6= t, (EA)sj =

n X

eik akj = aij = bij .

k=1

For i = s, (EA)sj =

n X

esk akj =

n X

k=1

n X

δtk akj = atj = bsj .

And for i = t, (EA)tj =

k=1

eik akj =

δsk akj = asj = btj

k=1

for j = 1, · · · , m. Therefore EA = B. 10. Let A be n × m. B and E are formed, respectively, by multiplying row s of A and In by α. Then, for i 6= s, bij = aij and eij = δij . For i = s, bsj = αasj and esj = αδsj .

11.2. ROW OPERATIONS AND REDUCED MATRICES

297

Now consider the i, j − −element of EA. For i 6= s, (EA)ij =

n X

eik akj =

k=1

and (EA)sj =

n X

αδik akj = aij = bij

k=1

n X

esk akj =

k=1

n X

αδsk akj = bsj

k=1

for j = 1, 2, · · · , m. This shows that EA = B. 11. Let A be n × m. Now B and E are formed, respectively, from A and In by adding α times row s to row t. For i 6= t, bij = aij and eij = δij , while for i = t, btj = atj + αasj and etj = δtj + αδsj . Now consider the i, j − − element of of EA. For i 6= t, (EA)ij =

n X

eik akj =

k=1

n X

δik akj = aij .

k=1

For i = t, (EA)tj =

n X

etk akj =

k=1

n X

(δtk + αakj )

k=1

= atj + αasj = bsj . This shows that EA = B. In Problems 12–23, keep in mind that a given matrix can be reduced by different sequences of row operations, but the end result must be the same - a matrix has only one reduced form. There may therefore be different matrices Ω1 and Ω2 such that Ω1 A = AR = Ω2 A. We give one such matrix for each problem. 12. A is reduced by simply adding row two to row one:     1 1 0 1 0 5 Ω = 0 1 0 and AR = 0 1 2 . 0 0 1 0 0 0 13. We can reduce A by first subtracting row two from row one of I2 , then multiplying row one of the resulting matrix by 1/3: 1 0 1 −1 1/3 −1/3 I2 = → → = Ω. 0 1 0 1 0 1 Then ΩA = AR =

1 0

0 1

1/3 0

4/3 . 0

298CHAPTER 11. MATRICES, DETERMINANTS AND LINEAR SYSTEMS 14. We can reduce A by first interchanging rows two and four of I4 , then (on the resulting matrix), multiplying row one by −1, then adding row two to row one:     1 0 0 0 1 0 0 0   0 1 0 0  → 0 0 0 1 I4 =  0 0 1 0 0 0 1 0 0 1 0 0 0 0 0 1     −1 0 0 0 −1 0 0 1  0 0 0 1    →  0 0 0 1 = Ω. →  0 0 1 0  0 0 1 0 0 1 0 0 0 1 0 0 Then



1 −4 −1 0 0 0 ΩA =  0 0 0 0 0 0

 0 1  = AR . 0 0

15. A is in reduced form, so A = AR . 16. Starting with I4 , subtract row one from row four, subtract 3 times row one from row three, interchange rows two and three, and then interchange rows one and two:     1 0 0 0 1 0 0 0  0 1 0 0  0 1 0 0    I4 →   0 0 1 0 → −3 0 1 0 −1 0 0 1 −1 0 0 1     −3 0 1 0 1 0 0 0  1 0 0 1 −3 0 0    →  0 1 0 0 →  − 1 0 0 . −1 0 0 1 −1 0 0 1 Then



1 0 ΩA =  0 0

 0 1  = AR . 0 0

For Problems 17–23, just Ω and AR are given. 17.

Ω=

18.

0 1 1 , AR = 1 −2 0

1 . 0

   −8 −2 38 1 1  37 43 −7 , AR = 0 Ω= 270 19 −29 11 0

0 1 0

 0 0 = I3 . 1

11.3. SOLUTION OF HOMOGENEOUS LINEAR SYSTEMS 19.

−1/3 Ω= 0

20.

Ω=

21.

0 1/2

0 1 −4/3 −4/3 , AR = 1 0 0 0 1 1 , AR = 1/2 0



0 1

  0 0 1 1 1 Ω =  4 −4 −8 , AR = 0 4 −4 8 8 0

22.



0 Ω= 0 −1/7 23.



0 3/2

0 1/2

0 1 0

 0 −3/4 0 3  1 0

  1/2 −1 1 0 1  , AR = 0 2/7 −3/7 0

0 0 1 0 1 3  1 0 −6 0 0 −1

11.3

299

0 1 0

 0 0 1

   1 0   0  , AR = 0 0 0 0 1

Solution of Homogeneous Linear Systems

In Problems 1–12, we use the facts that (1) the system AX = O has the same solutions as the reduced system AR X = O, and (2) the solution of the reduced system can be read by inspection from the reduced matrix AR . 1. The coefficient matrix and its reduced form are 1 2 −1 1 1 0 1 −1 A= , AR = . 0 1 −1 1 0 1 −1 1 Because AR has two nonzero rows and m = 4, the solution space has dimension m − 2 = 2, which means that the general solution is in terms of two of the unknowns, which can be given any values. Specifically, from the reduced matrix, x1 = −x3 + x4 x2 = x3 − x4 . All solutions are given by         x1 −x3 + x4 −1 1 x2   x3 − x4  1 −1    = x3   + x4   . X= x3  =   1 0 x3 x4 x4 0 1

300CHAPTER 11. MATRICES, DETERMINANTS AND LINEAR SYSTEMS It looks a little neater to write the general solution     1 −1 −1 1    X = α  1 +β 0  1 0 in which α and β are arbitrary real numbers. The solution space of this system is the subspace of R4 having basis vectors < −1, 1, 1, 0 >, < 1, −1, 0, 1 > . 2. The coefficient matrix is 

−3 A= 0 0 and

 1 AR = 0 0

0 1 0

1 −1 1 1 0 −3

 1 1 0 4 2 1

 0 1/9 11/9 0 2/3 13/3  . 1 −2/3 −1/3

With x4 = α and x5 = β, the general solution is     −11/9 −1/9 −13/3 −2/3      + β  1/3  . 2/3 X = α      0   1  1 0 The solution space is a subspace of R5 having dimension 2 and basis < −1/9, −2/3, 2/3, 1, 0 >, < −11/9, −13/3, 1/3, 0, 1 > . 3. The coefficient matrix and its reduced form are    −2 1 2 1 0 A =  1 −1 0 , AR = 0 1 1 1 0 0 0

 0 0 = I3 . 1

Here A has rank 3, the number of nonzero rows of A, and m− rank (A) = 3 − 3 = 0, so the solution space has dimension zero, consisting just of the trivial solution   0 X = 0 . 0 This is consistent with the reduced system being the system x1 = 0, x2 = 0, x3 = 0.

11.3. SOLUTION OF HOMOGENEOUS LINEAR SYSTEMS

301

4. The coefficient matrix and its reduced form are 4 1 −3 1 1 0 −1/2 0 A= , AR = . 2 0 −1 0 0 1 −1 1 The general solution can be written   1/2  1   X = α  1  + β 0 − 101 . 0 The solution space is the subspace of R4 having basis vectors < 1/2, 1, 1, 0 >, < 0, −1, 0, 1 > . 5. The coefficient matrix is  1 −1 3 −1 4 2 −2 1 1 0  A= 1 0 −2 0 1 0 0 1 1 −1 

and



1 0 AR =  0 0

0 1 0 0

 0 9/4 0 7/4  . 0 5/8  1 −13/8

0 0 1 0

The solution space has dimension 5 − 4 = 1 and the general solution is   −9/4 −7/4    X = α −5/8 .  13/8  1 The single vector < −9/4, −7/4, −5/8, 13/8, 1 > is a basis for the solution space. 6. The coefficient matrix is  6 −1 A = 1 0 1 0 and

 1 AR = 0 0

 1 0 0 0 −1 2  0 1 −2

0 0 1 −1 0 0

 0 −2 0 −12 . 1 −4

302CHAPTER 11. MATRICES, DETERMINANTS AND LINEAR SYSTEMS The general solution is     2 0 12 1        X = α 1 + β  0  . 4 0 1 0 The solution space is the subspace of R5 of dimension 2, having basis < 0, 1, 1, 0, 0 >, < 2, 12, 0, 4, 1 > . 7. The coefficient matrix is  −10 −1 4 1  0 1 −1 3 A=  2 −1 0 0 0 1 0 −1

 1 −1 0 0  1 0 0 1

and

  1 0 0 05/6 5/9 0 1 0 0 2/3 10/9 . AR =  0 0 1 0 8/3 13/9 0 0 0 1 2/3 1/9 From the reduced system read the general solution     −5/9 −5/6 −10/9 −2/3     −13/9 −8/3     X = α  + β  −1/9  .   −2/3  0   1  1 0 The solution space is a subspace of R6 having dimension 2, with basis vectors < −5/6, −2/3, −8/3, −2/3, 1, 0 >, < −5/9, −10/9, −13/9, −1/9, 0, 1 > . 8. The coefficient matrix is  1 0 A= 0 0

0 1 0 0

0 0 0 0

 0 7/6 −5/4 0 −20/3 9/2  . 1 14/3 −11/2 1 −3 2

From the reduced matrix we can write the general solution     −7/6 5/4  20/3  −9/2     −14/3    + β  11/2  . X = α  3   −2       1   0  0 1

11.3. SOLUTION OF HOMOGENEOUS LINEAR SYSTEMS

303

The solution space is the subspace of R6 of dimension 2, having basis vectors < −7/6, 20/3, −14/3, 3, 1, 0 >, < 5/4, −9/2, 11/2, −2, 0, 1 > . 9. There is no x3 in the equations, so we actually have a system of three equations in the four unknowns x1 , x2 , x4 and x5 . The coefficient matrix is   0 1 −3 1 A = 2 −1 1 0 2 −3 0 4 and

1 0 AR = 0 0

0 −5/14 0 1 − 6/7

−11/17

Now x1 , x2 and x4 depend on x5 and 

 5/14 11/7  X = α  6/7  . 1

The solution space is the subspace of R4 of dimension 1, having basis vector < 5/14, 11/7, 6/7, 1 >. 10. The coefficient matrix is  4 −3 0 0 2 0 A= 3 −2 0 2 1 −3 and

 1 0 AR =  0 0

0 1 0 0

0 0 1 0

 1 1 −3 1 −1 −6  0 4 −1 4 0 0

 0 −41/6 −2/3 0 −49/6 −1/3 . 0 −13/6 −7/3 1 23/6 −4/3

The general solution is 

   41/6 2/3  49/6  1/3      13/6     + β 7/3 . X = α −23/6 4/3      1   0  0 1 The solution space is a subspace of R6 of dimension 2 and having basis vectors < 41/6, 49/6, 13/6, −23/6, 1, 0 >, < 2/3, 1/3, 7/3, 4/3, 0, 1 > .

304CHAPTER 11. MATRICES, DETERMINANTS AND LINEAR SYSTEMS 11. The coefficient matrix is  1 −2 0 0 A= 1 0 2 0 and



1 0 AR =  0 0

0 1 0 0

0 0 1 −1 1 −1 1 −2 0 0 −1 2 0 −3 1 0

 1 3  0 0

 0 −1 2 0 0 −1 3/2 −1/2 . 0 0 −2/3 3  1 −1 4/3 0

0 0 1 0

The general solution is       0 −2 1 1/2 −3/2 1        −3   2/3  0        + β −4/3 + γ  0  . 1 X = α       1  0   0         0   1  0 1 0 0 The solution space is the subspace of R7 of dimension 3, with basis vectors < 1, 1, 0, 1, 1, 0, 0 >, < −2, −3/2, 2/3, −4/3, 0, 1, 0 >, < 0, 1/2, −3, 0, 0, 0, 1 > . 12. The coefficient matrix is  2 0 0 0 2 0  A= 0 0 1 0 1 −1 0 1 0 and

 1 0  AR =  0 0 0

0 1 0 0 0

0 0 1 0 0

 0 −4 0 1 1 0 0 −1 1 −1  −4 0 0 0 1  1 0 0 0 0 0 − 1 1 −1 0 0 0 0 1 0

0 0 0 0 1

 −3 7/2 −1/2 −1/2 1/2 −1/2  −2/3 2/3 −1  . −1/6 1/6 −1/2 −3/2 3/2 −1/2

The general solution is 

     3 −7/2 1/2 1/2 −1/2 1/2       2/3 −2/3  1        1/6 −1/6    + γ 1/2 . +β X = α 3/2 −3/2 1/2        1   0   0         0   1   0  0 0 1

11.3. SOLUTION OF HOMOGENEOUS LINEAR SYSTEMS

305

The solution space is the subspace of R8 having basis vectors < 3, 1/2, 2/3, 1/6, 3/2, 1, 0, 0 >, < −7/2, −1/2, −2/3, −1/6, −3/2, 0, 1, 0 >, < 1/2, 1/2, 1, 1/2, 1/2, 0, 0, 1 > . 13. The answer is yes. All that is required is that m − rank(A) > 0, so that the solution space has positive dimension, hence non-zero vectors, which are solutions of the system. As a specific example, consider the system AX = O, where   1 0 3 A = 0 1 −1 . 3 0 9 This is a homogeneous system with three equations in three unknowns. We find that   1 0 3 AR = 0 1 −1 . 0 0 0 Then AR has two nonzero rows, so the solution space of AX = O has dimension 3 − 2 = 1 > 0. In fact, the general solution is   3 X = α 1 . 1 14. Suppose A is n × m. Let the columns of A be C1 , · · · , Cm , written as n × 1 column matrices. Now notice that, if   a1  a2    X= .   ..  am then AX = O is equivalent to a1 C1 + · · · + am Cm = O, the n × 1 zero matrix. With these preliminaries, we can prove the proposition. If the columns of A are linearly dependent, then there are numbersa1 , · · · , am not all zero such that a1 C1 + · · · + am Cm = O, and then X, formed as a column with these coefficients as elements, is a nontrivial solution of AX = O.

306CHAPTER 11. MATRICES, DETERMINANTS AND LINEAR SYSTEMS Conversely, suppose the system has a nontrivial solution   a1  a2    X = X =  . .  ..  am Then a1 C1 + · · · + am Cm = O, and at least one of these coefficients is nonzero, so the columns of A are linearly dependent. 15. Let the rows of A be R1 , · · · , Rn . These are vectors in Rm . Let R be the row space of A, which is the subspace of Rm spanned by the row vectors. Now, X is in the solution space S(A) of the homogeneous system with coefficient matrix A exactly when AX = O. This is true exactly when the dot product Rj · X = O for j = 1, · · · , n, which is true exactly when each row is orthogonal to X. But this is equivalent to X being orthogonal to every linear combination of these rows, hence to every vector in the row space R of A. This makes the solution space of the system the orthogonal complement of the row space: R⊥ = S(A). Because the columns of At are the rows of A, similar reasoning shows that the solution space S(At ) of the system At X = O has the column space C of A as its orthogonal complement: C ⊥ = S(At ).

11.4

Nonhomogeneous Systems

1. The augmented matrix is  .. 3 −2 1 . 6   .  ..  [A..B] =  1 10 −1 . 2   .. −3 −2 1 . 0 

with reduced form  1 .  [A..B]R = 0  0

0 1 0

. 0 .. . 0 .. . 1 ..

 1   . 1/2  4

11.4. NONHOMOGENEOUS SYSTEMS

307

The reduced forms of the matrix of coefficients of the system and the augmented matrix have the same number of nonzero rows, so both A . and [A..B] have the same rank (in this case 3). Therefore this system is consistent. We can read the solution from the reduced form of the augmented matrix:   1 X = 1/2 . 4 In this case the solution is unique because m minus the rank of A is 3 − 3 = 0, so the associated homogeneous system has only the trivial solution. 2. The augmented matrix is  4 −2 ..  [A.B] = 1 0  2 −3 and

 .. .1   .. , 0 −3 .8  .. 0 1 . 16 3

  .. 8  1 0 0 −3 . ..   . . . [A.B]R = 0 1 0 −7/3 . 0    .. 0 0 1 52/9 . −31/3

. Because A and [A..B] have the same rank 3, this system is consistent. Read from the reduced augmented matrix that     3 8  7/3   0     X= −31/3 + α −52/9 . 1 0 3. We have  1 2 −3 0 .  [A..B] = 3 0 −2 0  0 1 0 −1

 .. 0 −1 . 0 .  1 0 .. 1  .. 0 6 . 3

and  0 1 0 0 −1 .  [A..B]R = 0 1 0 −1 0  0 0 1 −3/2 −1/2

. 17/2 .. .. 6 . . 51/4 ..

 9/2   . 3   25/4

308CHAPTER 11. MATRICES, DETERMINANTS AND LINEAR SYSTEMS . A and [A..B] have the same rank 3, so the system has solutions. Read from the reduced augmented matrix that         −17/2 0 1 9/2  −6   0   1   3            1/2 3/2 25/4  + α   + β   + γ −51/4 . X=  0   0   1   0           0   1   0   0  1 0 0 0 4. From the system, we have  ..  2 −3 . 1 .   [A..B] = −1 3 ... 0   .. 1 −4 . 3 

and  1 ..  [A.B]R = 0  0

. 0 .. . 1 .. . 0 ..

 0  . 0  1

The left two columns of this reduced augmented matrix form AR , and this has two nonzero rows, while the reduced form of the augmented matrix has three nonzero rows. Because the rank of A does not equal the rank . of [A..B], this system is inconsistent. In this case the absence of solutions can be seen easily from the reduced augmented matrix, which is the system x1 = 0, x2 = 0, 0x1 + 0x2 = 1. Clearly the third of these equations can have no solution. 5. The augmented matrix is  0 3  1 −3 .. [A.B] =   0 1 

1 −1

0 −4 0

1 −6 0

 .. . 10   . 4 −1 .. 8    .. 0 1 . −9  .. 0 1 . 0 0

11.4. NONHOMOGENEOUS SYSTEMS and its reduced form is 

 .. . −4   . −2 1 .. −4  .  . −7 9/2 .. −38   .. 1 −3/2 3/4 . −11/2

1 0 0 0  0 1 0 0 . [A..B]R =   0 0 1 0  0

309

−2

. Both A and [A..B] have the same rank 4 (number of nonzero rows in their reduced forms), so this system has a solution. From the reduced augmented matrix we read the general solution       −2 2 −4  −1   2   −4           7   −38   + α   + β −9/2 . X= −3/4 3/2 −11/2        0   1   0  1 0 0 6. We have  2 −3  A = 0 3  2 −3 and 

. 0 1 .. . 1 −1 .. . 10 0 ..

 1  0  0

 .. . 11/20   .. . . 1/30   .. 1 −1/10 . −1/10

1 0 0 1/20 ..  [A.B]R = 0 1 0 −9/30  0

Because A has rank 3, the same as the rank of the augmented matrix, this system is consistent. Read the general solution from the reduced augmented matrix:     11/20 −1/20  1/30   9/30     X= −1/10 + α  1/10  . 0 1 7. The augmented matrix is  8 −4 .  [A..B] = 0 1  0 0

. 0 1/2 10 .. . 0 1 −1 .. . 1 −3 2 ..

 1  2  0

310CHAPTER 11. MATRICES, DETERMINANTS AND LINEAR SYSTEMS with reduced form  .. 3/4 . 9/8  . . −1 .. 2   .. 2 . 0

 1 ..  [AR .C] = 0  0

1/2

−3

The system is consistent (same number of zero rows in AR and the reduced augmented matrix), and       −3/4 −1/2 9/8  1   −1   2             X =  0  + α 3  + β  −2  .  0   1   0  1 0 0 8. The augmented matrix is  .. 2 0 −3 .  .. 1 −1 .  . 2 −4 1 ..

 1  1  2

with reduced form  .. 3/4  1 0 0 .   .. 0 1 0 . −1/12 .   .. 0 0 1 . 1/6 

Both the matrix of coefficients and the augmented matrix have rank 3, so the system is consistent. The system has the unique solution   3/4 X = −1/12 . 1/6 This unique solution could have been foreseen from the fact that the number of columns of A minus its rank is 3 − 3 = 0, so the associated homogeneous system has only the trivial solution. 9. The augmented matrix is  0

−3

−1

 . 1 .. 2  . 1 0 .. −4

11.4. NONHOMOGENEOUS SYSTEMS with reduced form  1 1 0

3/14

0 −3/14

311

 . 1 −1/14 .. −29/7 . . 0 1/14 .. 1/7

The matrix of coefficients and its augmented matrix have rank 2, so the system is consistent. The general solution is             −1 −3/14 1 −1 −29/7 1/14 0  0  0 0 1  0               3/14   0  −1/14 0 0  1/7      0                    X=  0 +α  0 +β 1+γ  0 +δ  0 +  0  .   1  0  0 0  0      0          0  1 0 0  0  1 0 0 0 0 0 10. The augmented matrix of coefficients is   .. 3 −2 . −1   .. 4 3 . 4 with reduced form

 1 0

. 0 .. . 1 ..

 5/17 

16/17

The rank of the coefficient matrix equals 2, the same as the rank of the augmented matrix, so the system is consistent. The solution is unique: 5/17 X= . 16/17 11. The augmented matrix is  7 −3 4  2 1 −1  0

 .. 0 . −7  . 4 .. 6   .. −3 . −5

and the reduced augmented matrix is  1 0 0 19/15  0 1 0 −3  0 0 1 −67/13

 .. . 22/15   .. . . −5   .. . −121/15

312CHAPTER 11. MATRICES, DETERMINANTS AND LINEAR SYSTEMS The rank of A and of the augmented matrix is 3, so the system is consistent. The general solution is     −19/15 22/15  3   −5     X= −121/15 + α  67/15  . 1 0 12. The augmented matrix of coefficients is  −6 −4 5   2 −6 1  −6

with reduced form  1  0  0

 .. . 2  .. . −5  .. −11 . 1

 .. 0 . −137/48  . . 0 .. 1/6   .. 1 . 41/24

0 1 0

The coefficient matrix and its augmented matrix have the same rank 3, so the system is consistent. Because the number of unknowns is also 3, this system has a unique solution   −137/48 X =  1/6  . 41/24 13. The augmented matrix is 

..  4 −1 4 .  . 1 1 −5 ..  . −2 1 7 ..

 1  0  4

and its reduced form is  1  0  0

0 1 0

. 0 .. . 0 .. . 1 ..

 16/57  . 99/57  23/57

A and the augmented matrix both have rank 3, which is also the number of unknowns, so the system has a unique solution   16/57 X = 99/57 . 23/57

11.5. MATRIX INVERSES

313

14. The augmented matrix is  −6  1  1

 .. . 0  . −1 .. −5  .. −7 . 0

2 −1 4

and its reduced form is  21/23 1 0 0  0 1 0 −11/23  0 0 1 −171/23

 .. . −15/23  .. . . −25/23  .. . 40/23

A and its augmented matrix have the same rank 3, so the system is consistent. The solution is not unique because m − 3 = 4 − 3 = 1, so the associated homogeneous system as a solution space of dimension 1. The general solution is     −21/23 −15/23  11/23  −25/23    X=  40/23  + α  171/23  . 1 0 15. Write



 α1  α2    X= .   ..  αm

Let the columns of A be C1 , · · · , Cm . Then AX = B if and only if α1 C1 + · · · + αm Cm = B. This means that X can be a solution if and only if X is a linear combination of the columns of A, hence is in the column space of A.

11.5

Matrix Inverses

Problem 1 shows all of the row operations used to reduce the matrix and find the inverse, or show that the matrix is singular. For Problems 2–10, just the inverse is given if the matrix is nonsingular. 1. Reduce  .. −1 2 . . 2 1 ..

1 0

  0 −1 → add 2 times row one to row two  1 0

. 2 .. . 5 ..

1 2

 0 1

314CHAPTER 11. MATRICES, DETERMINANTS AND LINEAR SYSTEMS . 1 −2 .. −1  → multiply row one by −1 . 0 5 .. 2 

→ multiply row two by 1/5

 1 −2 0

→ add 2 times row two to row one

 1

 0 1

.. .

−1

.. .2/5

1/5

. 0 .. −1/5 . 1 .. 2/5

 0  2/5 . 1/5

Because I2 has appeared on the left, the given matrix is nonsingular and the right two columns of this augmented matrix form the inverse matrix: 1 −1 2 −1 . A = 5 2 1 2. The matrix is singular (has no inverse) because we find that 1 1/4 AR = 6= I2 . 0 0 3. A

−1

4. A−1 = 5. A 6.

−1

−2 1

2 5

4 −4 − 1

1 = 12 1 4

1 = 12

3 −2 −3 6

  64 −4 49 1 −8 4 −7 A−1 = 56 0 0 14   −6 11 2 1 3 10 −1 A−1 = 31 1 −7 10

8. The matrix is singular because 

1 AR =  0 0

0 1 0

 3 1 6= I3 . 0

11.6. DETERMINANTS 9.

315



 6 −6 0 1 A−1 = − −3 −9 2  12 3 −3 −2

10. A has no inverse because  1 AR = 0 0 11.

 28/27 14/9  . 0

0 1 0

     −1 −1 8 4 1 −23     1  −9 2 −5 14   2  = 1 −75 X = A−1 B = 2 −5 3   0  11  −9  11  2 3 3 −2 −1 −5 14

12.



5 1  −10 X = A−1 B = 55 5

5 34 6

    5 4 9 1   23 0 = 15 11 17 5 13

13.  11 1  3 = 28 8 14.

15.

11.6

X = A−1 B     12 9 −4 22 1 16 5  5  = 27 7 24 4 8 30

 4 4 1 7 −6 X = A−1 B = 52 1 11

    4 0 −1 1  58  39 −5 = 52 0 13 −66

     0 5 −15 −15 −21 1 1 10   0  =  14  X = A−1 B = − −10 15 25 5 −7 −5 10 0 0

Determinants

In Problems 1–6 we provide a sequence of row and/or column operations leading to a determinant that is easily evaluated. Other sequences of operations can also be used. 1. Add 2 times row two to row one and then −7 times row two to row three to obtain −2 1 7

4 1 0 16 7 16 7 6 3 = 1 6 3 = (−1)3+1 (1) = −22. −42 −17 0 4 0 −42 −17

316CHAPTER 11. MATRICES, DETERMINANTS AND LINEAR SYSTEMS 2. Add 3 times row two to row one, then add row two to row three to obtain 2 −3 14 1 −13 −1

44 7 1 = 14 1 5

0 10 44 1 1 = (−1)2+2 (1) 1 0 6

10 = 254. 6

3. Add column two to column one, then 2 times column two to column three: −4 5 −2 3 2 −2

1 5 6 5 = 1 3 0 −2 6

21 1 14 = (−1)3+2 (−2) 1 0

21 = −14. 14

4. Add 2 times row three to row one and 2 times row three to row two to obtain 2 −5 8 28 −5 0 28 −5 4 3 8 = 30 3 0 = (−1)3+3 = −936. 30 3 13 0 −4 13 0 −4 5. Add 2 times column three to column one and then add column three to column two to obtain 27 17 −2 5 1 12 0 = 1 0 14 7 −7

3 5 27 3 12 0 = (−1)3+3 (−7) = −2, 247. 1 12 0 −7

6. Add column one to column two, then 3 times column two to column three, then 2 times column one to column four to obtain −3 0 −3 3 9 6 1 −1 1 −2 15 6 = 7 8 7 1 1 5 2 3 2 1 −1 3 −1 = (−1)1+1 (−3) 8 3

18 22 5

0 18 22 5

0 8 19 7

8 −1 19 = −3 8 7 3

18 22 5

8 19 . 7

Now the problem of evaluating a 4 × 4 determinant has been reduced to one of a 3 × 3 determinant. In this determinant, add 18 times column one to column two and then 8 times column one to column three to obtain −1 8 3

18 22 5

8 −1 19 = 8 7 3

0 166 59

0 166 83 = (−1) 59 31

83 = −249. 31

Putting the pieces together, −3 3 9 6 1 −2 15 6 = (−3)(−249) = 747. 7 1 1 5 2 1 −1 3

11.6. DETERMINANTS

317

For Problems 7–10, we just give the value of the determinant. 7. −122 8. 293 9. 72 10. −2, 667 11. −15, 698 12. 1, 693 13. 3, 372 14.

a2 1 a b2 = 0 b − a c2 0 c−a

1 a 1 b 1 c

a2 b − a2 c2 − a2 2

1 a a2 1 a a2 = 0 b − a (b − a)(b + a) = (b − a)(c − a) 0 1 b + a 0 c − a (c − a)(c + a) 0 1 c+a = (b − a)(c − a)

1 b+a = (b − a)(c − a)(c − b). 1 c+a

15. Add columns two, three and four to column one, then factor (a + b + c + d) out of column one to obtain a b b c c d d a

c d a b

1 b c d 1 c d a = (a + b + c + d) 1 d a b 1 a b c

d a b c

Now add (−1)row two + row three − row four to row one and factor out b − a + d − c from the new row one to obtain 1 b c 1 c d (a + b + c + d)(b − a + d − c) 1 d a 1 a b

d a . b c

16. Define the function 1 z L(x, y) = 1 x2 1 x3

y y2 = (y2 − y3 )x + (x3 − x2 )y + x2 y3 − x3 y2 . y3

318CHAPTER 11. MATRICES, DETERMINANTS AND LINEAR SYSTEMS Notice that L(x, y) has the form ax+by +c, where a = y2 −y3 , b = x3 −x2 and c = x2 y3 − x3 y2 are given numbers. The graph of L(x, y) = 0 is a straight line in the x, y− plane. Because L(x2 , y2 ) = 0 and L(x3 , y3 ) = 0, both of these points are on the line L(x, y) = 0. Finally, L(x1 , y1 ) = 0 (x1 , y1 ) is also on this line, and this occurs only if 1 x1 1 x2 1 x3

y1 y2 = 0. y3

17. Use induction to prove that, for n = 2, 3, · · · , the determinant of an n × n upper triangular matrix is the product of the main diagonal elements. For n = 2, this is obvious because a11 0

a12 = a11 a22 . a22

Now suppose the statement is true for some n ≥ 2. We want to prove that it is true for n + 1. Let A be an n + 1 × n + 1 upper triangular matrix. Then ai1 = 0 for i = 2, · · · , n + 1, so by expanding by column one, we have |A| = a11 |B|, where B is the n × n upper triangular matrix obtained by deleting row one and column one of A. By the inductive hypothesis, |B| = a22 a33 · · · an+1,n+1 . Then |A| = a11 a22 · · · an+1,n+1 . This completes the proof by induction.

11.7

Cramer’s Rule

1. |A| = 47 6= 0 so Cramer’s rule applies: x1 =

1 5 −4 1 15 11 = − , x2 = 47 −4 1 47 47 8

100 5 . =− −1 47

2. |A| = −3 and the solution is x1 = −

1 3 3 0

1 1 4 = −1, x2 = − 1 3 1

3 = 1. 0

11.7. CRAMER’S RULE

319

3. |A| = 132 and the solution is

x1 =

0 −4 3 66 1 1 −5 5 −1 = − =− , 132 132 2 −4 6 1

x2 =

1 132

8 0 3 114 19 1 −5 −1 = − =− , 132 22 −2 −4 1

and x3 =

8 −4 0 24 2 1 5 −5 = = . 132 11 −2 6 −4

1 132

4. |A| = 108 and the solution is x1 = −

63 7 165 55 −943 9 = − , x2 = − = − , x3 = =− . 108 12 108 36 108 4

5. |A| = −6 and x1 =

5 10 5 , x2 = − , x3 = − . 6 3 6

6. |A| = −130 and x1 =

197 255 1260 42 173 , x2 = , x3 = , x4 = , x5 = . 130 130 130 130 130

7. |A| = 4 and the solution is x1 = −

172 109 43 37 = −86, x2 = − , x3 = − , x4 = . 2 2 2 2

8. |A| = 12 and x1 =

117 63 3 21 , , x3 = , x4 = . 12 12 2 12

9. |A| = 93 and x1 =

33 409 1 116 , x2 = − , x3 = − , x4 = . 93 93 93 93

10. |A| = 42, x1 =

69 162 24 54 , x2 = , x3 = , x4 = − . 21 21 21 21

320CHAPTER 11. MATRICES, DETERMINANTS AND LINEAR SYSTEMS

11.8

The Matrix Tree Theorem

1. The tree matrix for this graph is  2 0 0 2  T= −1 −1  0 −1 −1 0

 −1 0 −1 −1 −1 0   4 −1 −1 . −1 3 −1 −1 −1 3

Evaluate any 4 × 4 cofactor of T to obtain 21 as the number of spanning trees in the labeled graph. 2.



 4 −1 −1 −1 0 −1 −1 3 −1 −1 0 0   −1 −1 2  0 0 0   T= 4 −1 −1 −1 −1 0  0 0 0 −1 2 −1 −1 0 0 −1 −1 3 and each cofactor yields 55 spanning trees in G.

 4 −1 0 −1 −1 −1 −1 2 −1 0 0 0    0 −1 3 −1 −1 0    T=  −1 0 −1 4 −1 −1 −1 0 −1 −1 3 0 −1 0 0 −1 0 2 

and each cofactor gives 61 as the number of spanning trees 4.

 4 −1 −1 0 −1 −1 −1 3 −1 −1 0 0   −1 −1 3 −1 0 0   T=   0 −1 −1 3 −1 0  −1 0 0 −1 3 0 −1 0 0 0 −1 2 

and each cofactor gives 64 for the number of spanning trees. 5.



 3 −1 0 0 −1 −1 −1 3 −1 0 −1 0     0 −1 4 −1 −1 −1   T= 0 −1 2 −1 0  0  −1 −1 −1 −1 4 0 −1 0 −1 0 0 2 and the number of spanning trees is 61.

11.8. THE MATRIX TREE THEOREM

321

6. The tree matrix for Kn is the n × n matrix  n−1 −1 −1 ···  −1 n − 1 −1 ···   −1 −1 n − 1 ··· T=  .. .. ..  . . . ··· −1

−1

···

 −1 −1   −1  . ..  .  n−1

We will compute −1)1+1 M11 , which is the n−1×n−1 determinant formed by deleting row one and column one of T. In M11 , add the last n − 2 rows to row one to obtain the new n − 1 × n − 1 determinant still equal to M11 : 1 1 1 −1 n − 1 −1 −1 n−1 M11 = −1 .. .. .. . . . −1 −1 −1

··· ··· ··· ··· ···

1 −1 −1 .. .

n−1

Subtract column one of this determinant from each other column. This does not change the value of the determinant, and we now have 1 0 0 0 ··· −1 n 0 0 · · · M11 = −1 −1 n 0 · · · .. .. .. .. .. . . . . . −1 0 0 0 · · ·

0 0 0 . .. . 0

This is a lower triangular n − 1 × n − 1 determinant, which is therefore equal to the product of the diagonal elements, which is nn−2 . Therefore the number of spanning trees in Kn is nn−2 .

322CHAPTER 11. MATRICES, DETERMINANTS AND LINEAR SYSTEMS

Chapter 12

Eigenvalues and Diagonalization 12.1

Eigenvalues and Eigenvectors

1. pA (λ) = |λI2 − A| = λ2 − 2λ − 5 √ √ so the eigenvalues of A are 1 + 6 and 1 − 6, with eigenvectors, respectively, √ √ 6 − 6 V1 = , V2 = . 2 1 The Gerschgorin circles are of radius 3 about (1, 0) and radius 2 about (1, 0). 2. The characteristic equation of A is λ2 − 2λ − 8 = 0. Eigenvalues and eigenvectors are 0 6 λ1 = 4, V1 = , λ2 = −2, V2 = . 4 −1 Gerschgorin circles are of radius 1 about (4, 0) and radius 0 about −2. The circle of radius 0 is a “degenerate” circle, consisting of just the center point. 3. The characteristic equation is λ2 + 3λ − 10 = 0 323

324

CHAPTER 12. EIGENVALUES AND DIAGONALIZATION and eigenvalues and eigenvectors are 7 0 λ1 = −5, V1 = , λ2 = 2, V2 = . −1 1 One Gerschgorin circle has radius 1 and center (2, 0), and the other is the degenerate circle of radius 0 about (−5, 0).

4. pA (λ) = λ2 − 10λ + 18, λ1 = 5 +

√ 7, V1 =

√ 2√ 2√ , λ2 = 5 − 7, V2 = 1− 7 1+ 7

Gerschgorin circles have radius 2, center (6, 0), and radius 3, center (4, 0). 5. pA (λ) = λ2 − 3λ + 14, √ −1 + 47i 4 √ √ −1 − 47i λ2 = (3 − 47i)/2, V2 = 4 √ λ1 = (3 + 47i)/2, V1 =

The Gerschgorin circles have radius 6, center (1, 0), and radius 2, center (2, 0). 6. pA (λ) = λ2 , and eigenvalues are λ1 = λ2 = 0. All engenvectors are nonzero scalar multiples of 1 V= . 0 The lone Gerschgorin circle has radius 1 and center the origin. 7. pA (λ) = λ3 − 5λ2 + 6λ.       2 0 0 λ1 = 0, V1 = 1 , λ2 = 2, V2 = 1 , λ3 = 3, V3 = 2 0 0 3 The Gershgorin circle has radius 3, center (0, 0). 8. pA (λ) = (λ + 1)(λ2 − λ − 7)     2√ 0 √ λ1 = 1, V1 = 0 , λ2 = (1 + 29)/2, V2 = 5 + 29 1 0 and

  2√ √ λ3 = (1 − 29)/2, V3 = 5 − 29 0

Gershgorin circles have radius 1, center (−2, 0), and radius 1, center (3, 0).

12.1. EIGENVALUES AND EIGENVECTORS

325

9. pA (λ) = λ2 (λ + 3)     1 1 λ1 = −3, V1 = 0 , λ2 = λ3 = 0, V2 = 0 0 3 All eigenvectors associated with the double eigenvalue 0 are constant multiples of V2 . The Gershgorin circle has radius 2, center (−3, 0). 10. pA (λ) = λ(λ2 + 2)       1 1 0 √ √  , , λ3 = − 2i, V3 =  −1  λ1 = 0, V1 = 1 , λ2 = 2i, V2 =  −1 √ √ 0 2i − 2i The Gershgorin circles have center (0, 0) and radii 1 and 2. 11. pA (λ) = (λ + 14)(λ − 2)2 ,     −16 0 λ1 = −14, V1 =  0  λ2 = λ3 = 2, V2 = 0 . 1 1 All eigenvectors associated with λ2 are constant multiples of V2 . The Gershgorin circles have radius 1, center (−14, 0) and radius 3, center (2, 0). 12. pA (λ) = (λ − 3)(λ2 + λ − 42),       30 0 0 λ1 = 6, V1 =  1  , λ2 = 3, V2 = −2 , λ3 = −7, V3 = 8 5 5 −1 The Gerschgorin circles have radius 9, center (−2, 0) and radius 5, center (1, 0). 13. pA (λ) = λ(λ2 − 8λ + 7),       14 6 0 λ1 = 0, V1 =  7  , λ2 = 1, V2 = 0 , λ3 = 7, V3 = 0 10 5 1 The Gershgorin circles have radius 2, center (1, 0), and radius 5, center (7, 0). 14. pA (λ) = λ2 (λ2 + 2λ − 1),     1 0 2 0    λ1 = λ2 = 0, V1 =  0 , V2 = 1 1 0

326

CHAPTER 12. EIGENVALUES AND DIAGONALIZATION so the repeated eigenvalue 0 has two linearly independent eigenvectors associated with it. For the other two eigenvalues,     1√ 1√ √ √ 1 + 2 1 − 2    λ3 = −1 + 2, V3 =   0  , λ4 = −1 − 2, V4 =  0  0 0 The Gershgorin circles have radius 1, center (−2, 0) and radius 2, center (0, 0).

15. pA (λ) = (λ − 1)(λ − 2)(λ2 + λ − 13),     −2 0 −11 0    λ1 = 1, V1 =   0  , λ2 = 2, V2 = 1 1 0   √  √ 53 − 7 − 53 − 7 √ √     −1 + 53 0 0  , λ4 = −1 − 53 , V4 =   λ3 = , V3 =      0 0 2 2 2 2 The Gershgorin circles have radius 2, center (−4, 0) and radius 1, center (3, 0). 16. pA (λ) = λ2 (λ − 1)(λ − 5),    1 1 0 −4    λ1 = 1, V1 =   0  , λ2 = 5, V2 = 0 0 0   0 0  , λ3 = λ4 = 0, V3 =  1 0 

All eigenvectors associated with the double eigenvalue 0 are constant multiples of just the one eigenvector V3 . The Gershgorin circles have radius 10, center (5, 0), radius 9, center (1, 0), radius 9, center (0, 0). 17. We know that AE = λE. Then A(AE) = A2 E = A(λE) = λAE = λ(λE) = λ2 E. This says that λ2 is an eigenvalue of A2 with eigenvector E. It is not a routine inductive argument to show that λn is an eigenvalue of An with eigenvector E, for any positive integer n.

12.2. DIAGONALIZATION

327

18. The characteristic polynomial of A is pA (λ) = |λIn − A|. The pA (0) = | − A| is the constant term in this polynomial. But | − A| = (−1n )|A| because −A is A with each of the n rows multiplied by −1, and each such multiplication introduces a factor of −1 into the determinant of the resulting matrix. Now, if A is singular, then |A| = 0. But then pA (0) = (−1)n |A| = 0, so 0 is a root of the characteristic polynomial, and is therefore an eigenvalue of A.

12.2

Diagonalization

1. The characteristic equation is λ2 − 3λ + 4 = 0, so the eigenvalues are √ √ 3 − 7i 3 + 7i and λ2 = . λ1 = 2 2 Corresponding eigenvectors are 2√ 2√ and V2 = . V1 = −3 − 7i −3 + 7i The matrix

2√ 2√ −3 + 7i −3 − 7i

diagonalizes A, and P

−1

√ (3 + 7i)/2 AP = 0

0 √ . (3 − 7i)/2

If we wrote the eigenvectors in different order in defining the columns of P, then the columns of P−1 AP would be reversed. 2. The characteristic equation is λ2 − 8λ + 12 = 0 so the eigenvalues are λ1 = 2 and λ2 = 6. Corresponding eigenvectors are −1 3 V1 = and . 1 1

328

CHAPTER 12. EIGENVALUES AND DIAGONALIZATION The matrix

3 1

−1 1

diagonalizes A, and −1

2 AP = 0

0 . 6

3. The characteristic equation is λ2 −2λ+1 = 0, with repeated root 1. Every eigenvector is a scalar multiple of 0 V1 = . 1 Because A does not have two independent eigenvectors, A is not diagonalizable. 4. Eigenvalues and corresponding eigenvalues are 1 3 λ1 = −5, V1 = , λ2 = 9, V2 = . 0 14 The matrix

1 0

3 14

−5 0

diagonalizes A, and A

−1

AP =

0 . 9

5. The eigenvalues and corresponding eigenvectors are       5 0 0 λ1 = 0, V1 = 1 , λ2 = 5, V2 = 1 , λ3 = −2, V3 = −3 . 0 2 0 The matrix



 P=0 5 0  1 1 −3 0 0 2

diagonalizes A and  0 P−1 AP = 0 0

 0 0 5 0 . 0 −2

6. Eigenvalues and corresponding eigenvectors are     0 −2 √ 4√  λ1 = 0, V1 = −3 , λ2 = (3 + 17)/2, V2 =  1 3 + 17

12.2. DIAGONALIZATION and

329

  0 √ 4√  . λ3 = (3 − 17)/2, V3 =  3 − 17

The matrix

 −2 P = −3 1

0 4√ 3 + 17

 0 4√  3 − 17

diagonalizes A, and  0 P−1 AP = 0 0

 0 . (3 + 17)/2 0 √ 0 (3 − 17)/2 0 √

7. Eigenvalues and corresponding eigenvectors are     −3 0 λ1 = 1, V1 = 1 , λ2 = λ3 = −2, V2 =  1  . 0 0 All eigenvectors associated with the repeated eigenvalue −2 are scalar multiples of V2 . Because A does not have three linearly independent eigenvectors, A is not diagonalizable. 8. Eigenvalues and corresponding eigenvectors are       1 0 0 λ1 = 2, V1 = 0 , λ2 = 2 + i, V2 = −i , λ3 = 2 − i, V3 =  i  . 0 1 1 Let

  1 0 0 P = 0 −i i  . 0 1 −1

Then P diagonalizes A and 

2 P−1 AP = 0 0

0 2+i 0

 0 0 . 2−i

9. The characteristic equation is (λ − 1)(λ − 4)(λ2 + 5λ + 5) = 0. √ √ Eigenvalues are λ1 = 1, λ2 = 4, λ3 = (−5+ 5)/2, and λ4 = (−5− 5)/2). Corresponding eigenvectors are     1 0 0 1    V1 =  0 , V2 = 0 , 0 0

330

CHAPTER 12. EIGENVALUES AND DIAGONALIZATION    0 0 √ √ (2 + 3 5)/41 (2 − 3 5)/41    √ √ V3 =   (−1 + 5)/2  , V4 =  (−1 − 5)/2  . 1 1 

Let  1 0 0 1 P= 0 0 0 0 Then P diagonalizes A:  1 0 0 P−1 AP = 0 1 0 0 0 0

0 √ (2 − 3 √5)/41 (−1 + 5)/2 1

 0 √ (2 + 3 √5)/41 . (−1 − 5)/2  1

0 00√ (−5 − 5)/2

√ (−5 + 5)/2

 0 .

10. The characteristic equation is (λ + 2)4 = 0, so A has a repeated eigenvalue −2 of multiplicity 4. We find that A has only three independent eigenvectors,       0 0 0 0 1 0   ,   , and   . 0 0 1 1 0 0 Because A does not have four independent eigenvectors, A is not diagonalizable. 11. Let 

λ1 0   D=0  .. .

0 λ2 0 .. .

0 0 λ3 .. .

··· 0 0 .. .



0 ··· ··· ··· ···

0  0 . ..  .  λn

Then D = P−1 AP, so A = PDP−1 . Then Ak = (PDP−1 )k = (PDP−1 )(PDP−1 · · · (PDP−1 ) = PDk P−1 , with the interior pairings of P and P−1 canceling.

12.2. DIAGONALIZATION

331

Problems 12–15 can be solved using the idea of Problem 11, coupled with the fact that the kth power of a diagonal matrix is the diagonal matrix formed by raising each diagonal element to the kth power. 12. Eigenvalues of A are −1, −6. Make respective eigenvectors the columns of a matrix −3 1 P= . 2 1 Then P−1 = Then P

−1

−1/5 1/5

−1 0 AP = D = . 0 −6

Then 5

2/5 . 3/5

A = PD P

−1

−3111 −4665 . −3110 −4666

13. Eigenvalues of A are −1, −5, and the matrix of respective eigenvectors, 4 0 P= 1 1 diagonalizes A to D=

−1 0 . 0 −5

Now, P−1 =

1/4 −1/4

0 1

so compute A6 = PD6 P−1 =

1 −3906

0 . 15625

p √ 14. Eigenvalues of A are −3 + (10), −3 − 10. Form P from respective eigenvectors: √ √ 1 + 10 1 − 10 P= . 3 3 Then P diagonalizes A to √ −3 + 10 0√ D= . 0 −3 − 10 Then A4 = PD4 P−1 =

493 −694 . −684 949

332

CHAPTER 12. EIGENVALUES AND DIAGONALIZATION

15. Eigenvalues of A are

√ √ 2, − 2. Form √ √ 2 − 2 P= . 1 1

Then P

−1

Let

√ √2/4 = − 2/4

1/2 . 1/2

√ 2 0 √ . B= 0 − 2

Then 6

−1

A = PD P

8 = 0

0 . 8

16. Suppose A2 is an n × n diagonalizable matrix. Then A must have n linearly independent eigenvectors V1 , · · · , Vn , corresponding, respectively, to eigenvalues λ1 , · · · , λn . We know that pA (λj ) = (A2 − λj In )X = O for j = 1, · · · , n. But then p p pA (λj ) = (A − λj In )(A + λj In ) p p = pA ( λj )pA (− λj ) = 0. Then p p pA ( λj ) = 0 or pA (− λj ) = 0. p p Then λj or −λj is an eigenvalue of A with eigenvector Vj . This means that A has n linearly independent eigenvectors, and is therefore diagonalizable.

12.3

Special Matrices and Their Eigenvalues and Eigenvectors

In Problems 1–12, find independent eigenvectors for the given matrix. Normalize these by dividing each eigenvector by its length. These normalized eigenvectors form columns of an orthogonal matrix that diagonalizes the given matrix. It is routine to show that eigenvectors are orthogonal by taking their dot product. We will omit the arithmetic of this verification. 1. We find eigenvectors V1 =

1 −2 , V2 = . 2 1

12.3. SPECIAL MATRICES AND THEIR EIGENVALUES AND EIGENVECTORS333 Divide each by its norm to get unit eigenvectors and form the orthogonal matrix √ √ 1/√5 −2/√ 5 Q= . 2/ 5 1/ 5 This is an orthogonal matrix that diagonalizes A. 2.

10 10 √ √ , V2 = . 7 + 149 7 − 149

V1 =

Normalize these to form columns of the orthogonal matrix   √ 10 √ √ 10 √ 298+14 149 298−14 149 √ . Q =  7+√149 √ √ 7− 149 √ √ 298+14 149

298−14 149

3. Eigenvectors are V1 =

√ √ 1+ 2 1− 2 , V2 = . 1 1

Normalize these to form  Q=

√

2 √1− √

√ 1√

√ 1 √

4+2 2 4+2 2

4. Eigenvalues are

√

2 √1+ √



4−2 2  . 4−2 2

√

√ 17 and − 17, with eigenvectors, respectively, √ √ 17 − 4 − 17 − 4 V1 = , V2 = . 1 1

Normalize these vectors to form an orthogonal matrix Q that diagonalizes A: √ √ 1 17 − 4 − 17 − 4 p Q= . √ 1 1 34 − 8 17 √ √ 5. Eigenvalues of A are 3, 2 − 1 and − 2 − 1, with corresponding eigenvectors       0 √ 1 √1 V1 = 0 , V2 =  2 − 1 V3 = − 2 − 1 . 1 0 0 For an orthogonal matrix that diagonalizes A, let   √ 1√ 0 √ 1 √ 4−2 2 4+2 2 √ √  2−1 − 2−1   Q = 0 √4−2√2 √4+2√2  . √ √   − 2−1 √ 1 √ 2−1√ √ 4−2 2

4+2 2

334

CHAPTER 12. EIGENVALUES AND DIAGONALIZATION

6. Eigenvalues of A are √ 1 √ 1 1, (5 + 5), (5 − 5) 2 2 with eigenvectors, respectively,       0√ 0√ 1 V1 = 0 , V2 = −1 + 5 , V3 = −1 − 5 . 0 2 2 Normalize these eigenvectors and use them as columns of an orthogonal matrix that diagonalizes A:   1 0√ 0√ 0 √−1+ 5 √−1− √5  √  Q= 10−2 5 10+2 5  .  2 2 √ 0 √ √ √ 10−2 5

10+2 5

7. Eigenvalues are

√ √ 1 1 7, (5 + 41), (5 − 41) 2 2 with corresponding eigenvectors     √  √  0 5 − 41 5 + 41  , V3 =  . V1 = 1 , V2 =  0 0 0 4 4 The following orthogonal matrix diagonalizes A:   √ √ √ 5− 41√ 1 √ 5+ 41√ 82+10 41 82−10 41   . Q= 0 0  1 4 4 √ 0 √ √ √ 82+10 41

82−10 41

√ √ 8. Eigenvalues are 0, 1 + 17, 1 − 17, with corresponding eigenvectors     √  √  0 1 + 17 1 − 17 V1 = 0 , V2 =  −4  , V3 =  −4  . 1 6 6 For an orthogonal matrix that diagonalizes A, use the normalized eigenvectors as columns: √ √   √1− 17 0 √1+ 17 √ √ 70+2 17 70−2 17   √ −4 √  . 0 √ −4 √ Q=  70−2 17 70+2 17  √ 6√ 1 √ 6 √ 70−2 17

70+2 17

12.3. SPECIAL MATRICES AND THEIR EIGENVALUES AND EIGENVECTORS335 9. The matrix is not hermitian, skew-hermitian or unitary. Eigenvalues are 2, 2. 10. The matrix is not hermitian, skew-hermitian or unitary. Eigenvalues are −1, −1. √ √ 11. The matrix is skew-hermitian because St = −S. Eigenvalues are 0, 3i, − 3i. 12. The matrix is unitary. Eigenvalues are √ √ √ √ 1 1 1, (1 + i)( 2 + 6i), (−1 − i)(− 2 + 6i). 4 4 13. The matrix is not unitary, hermitian or skew-hermitian. Eigenvalues are 2, i, −i. 14. The matrix is hermitian, because Ht = H. Eigenvalues are √ √ −1 − 41 −1 + 41 , . 1, 2 2 15. Suppose H is hermitian. Then H = Ht . Then HHt = HHt = HH = HH. 16. If H is hermitian, then Ht = H, so the diagonal elements satisfy hjj = hjj , so these diagonal elements must be real. 17. Suppose S is skew-hermitian. Then St = −S, so sjj = −sjj for j = 1, 2, · · · , n. Write sjj = ajj + ibjj . Then sjj = ajj + ibjj = −ajj + ibjj = −ajj + ibjj . But then each ajj = −ajj , so ajj = 0 for j = 1, 2, · · · , n. This makes each diagonal element of S either pure imaginary (if bjj 6= 0) or zero. 18. Suppose U and V are unitary n × n matrices. Then t

U−1 = U and V−1 = V . But then t

(UV)−1 = V−1 U−1 = V U = ((U )(V ))t = UV , so UV satisfies the equation defining a matrix to be unitary.

336

CHAPTER 12. EIGENVALUES AND DIAGONALIZATION

12.4

Quadratic Forms

1. The matrix of the quadratic form is −5 A= 2

2 . 3

√ √ This matrix has eigenvalues −1 + 2 5, −1 − 2 5 and the quadratic form has standard form √ √ (−1 + 2 5)y12 + (−1 − 2 5)y22 . 2. The matrix is

4 −6 −6 1

with eigenvalues

√ √ 1 1 (5 + 153) and (5 − 153). 2 2 The standard form of this quadratic form is √ √ 1 1 (5 + 153)y12 + (5 − 153)y22 . 2 2

3. The matrix is

with eigenvalues 2 ±

4. The matrix is

with eigenvalues 3 ±

−3 2

2 7

√ 29. The standard form is √ √ (2 + 29)y12 + (2 − 29)y22 .

4 −2 −2 1

√ 17)/2. The standard form is √ √ 1 1 (3 + 17)y12 + (3 − 17)y22 . 2 2

5. The matrix is

with eigenvalues 2 ±

0 −3 −3 4

√ 13. The standard form is √ √ (2 + 13)y12 + (2 − 13)y22 .

12.4. QUADRATIC FORMS

337

6. The matrix is

5 2

2 2

with eigenvalues 1, 6. The standard form is y12 + 6y22 . 7. The matrix is

with eigenvalues 1 ±

0 −1 −1 2

√ 2. the standard form is √ √ (1 + 2)y12 + (1 − 2)y22 .

338

CHAPTER 12. EIGENVALUES AND DIAGONALIZATION

Chapter 13

Systems of Linear Differential Equations 13.1

Linear Systems

1. The two given solutions are linearly independent because neither is a constant multiple of the other. Use them as columns of a fundamental matrix 2t −e Ω(t) = e2t

3e6t . e6t

Notice that Ω(0) has a nonzero determinant, which is also an indicator that the columns are independent. Now we have a general solution X(t) = Ω(t)C, in which C is a 2 × 1 matrix of constants. To satisfy the initial condition, we need to choose C so that 0 Ω(0)C = . 4 This requires that

−1 1

3 c1 0 = . 1 c2 4

Then c1 = 3, c2 = 1, so the solution of the initial value problem is X(t) =

2t −e e2t

3e6t e6t

3 −3e2t + 3e6t = . 1 3e2t + e6t

339

340 CHAPTER 13. SYSTEMS OF LINEAR DIFFERENTIAL EQUATIONS 2. The two solutions are independent because Φ1 (0) and Φ2 (0) are independent in R2 . We can form a fundamental matrix 5t e e3t Ω(t) = . 3e5t e3t We now have a general solution X(t) = Ω(t)C, in which C is an arbitrary 2 × 1 matrix of real numbers. To solve the initial value problem, we need −2 1 1 c1 −2 X(0) = = Ω(0)C = = . 1 3 1 c2 1 Then c1 = 3/2 and c2 = −7/2, so the solution of the initial value problem is 5t 1 3e5t −7e3t e e3t 3/2 . X(t) = Ω(t)C = = 3e5t e3t −7/2 2 9e5t − 7e3t 3. Because Φ(0) and Φ(0) are independent in R2 , these solutions are independent. Form the fundamental matrix ! √ √ √ √ (2 + 2 3)e√(1+2 3t) (2 − 2 3)e√(1−2 3)t Ω(t) = . e(1+2 3)t e(1−2 3)t Then X(t) = Ω(t)C is a general solution. To solve the initial value problem, we need 2 Ω(0)C = 2 and we find that we must choose 1√ 1√ c1 = 1 − 3, c2 = 1 + 3. 6 6 4. From the given solutions we can write the fundamental matrix Ω(t) = √ √ √ √ √ √ 15 sin( 15t/2) 15 cos( 15t/2) − sin( 15t/2) 3t/2 − cos( 15t/2) − √ √ e 8 cos( 15t/2) 8 cos( 15t/2) The general solution is Ω(t)C. For the solution of the initial value problem, solve for C so that 0 Ω(0)C = . 7 We find that 7 √ C= 8(1 + 15)

√ 15 . 1

13.2. SOLUTION OF X0 = AX WHEN A IS CONSTANT

341

5. Form the fundamental matrix  t e Ω(t) = et 0

 e−3t 3e−3t  . e−3t

−et 0 et

The general solution is X(t) = Ω(t)C. To solve the initial value problem, we need C such that   1 Ω(0)C = −3 . 5 Solve this equation to get 

 24 C =  14  . −9

13.2

Solution of X0 = AX When A Is Constant

In these problems, different fundamental matrices may be found, depending on the choice of eigenvectors used corresponding to eigenvalues of the coefficient matrix. In each problem, the general solution has the form X = Ω(t)C, where Ω(t) is a fundamental matrix. We will give one choice for Ω(t). 1. Ω(t) =

3t 7e 5e3t

e−4t

2et −3et

e6t e6t

Ω(t) =

1 e2t −1 e2t

 t e Ω(t) = et et

e−t e−t 2e−t

1 2e3t 3e3t Ω(t) =  6 −13 −2e3t 

Ω(t) =

2et et

 e2t 2e2t  e2t  −e−4t 2e−4t  e−4t

e−t e−t

342 CHAPTER 13. SYSTEMS OF LINEAR DIFFERENTIAL EQUATIONS 7. A has eigenvalues and eigenvectors −2i 2 + 2i, 2i 1 , 2 − 2i, . 1 Write

2i 0 2 = +i 1 1 0

to form two independent solutions Φ1 (t) = e2t and

−2 sin(2t) cos(2t)

2 cos(2t) Φ2 (t) = e . sin(2t) 2t

Use these as columns of a fundamental matrix −2 sin(2t) 2 cos(2t) Ω(t) = . cos(2t) sin(2t) √ 8. A has complex eigenvalues, with the eigenvalue 3+ 5i having eigenvector √ √ 2 2 + 5i 5 = . +i 3 3 0 Two independent solutions are √ √ √ 2t 2 cos( 5t) − √ 5 sin( 5t) Φ1 (t) = e 3 cos( 5t) and

√ √ √ √ 5 cos( 5t) −√ 5 sin( 5t) Φ2 (t) = e . 3 sin( 5t) 2t

These form the columns of a fundamental matrix. 9. A has eigenvalues 2, 5, 5, with corresponding eigenvectors     1 −3 0 , −3 . 0 1 All eigenvectors associated with 5 are scalar multiples of this eigenvector. Immediately we can write two independent solutions   1 Φ1 (t) = e2t 0 0

13.2. SOLUTION OF X0 = AX WHEN A IS CONSTANT and

343

  −3 Φ2 (t) = e5t −3 . 1

For a third solution, denote the eigenvector associated with 5 as E and let Φ3 (t) = Ete5t + Ke5t . Substitute this into X0 = AX and use the fact that AE = E to obtain E + 5K = AK. If we let

  a K = b , c

we obtain −3a + 5b + 6c = −3 3b + 9c = −3 −b − 3c = 1. Then a = −2/3, b = −1, c = 0, so K=

−2/3 −1 0

and we obtain the third solution   −3te5t − (2/3)e5t Φ3 (t) =  −3te5t − e5t  . te5t The three solutions obtained are linearly independent and can be used to form the columns of a fundamental matrix. 10. The coefficient matrix has eigenvalues 2, −2 − 3 and we can use the independent eigenvectors to form the fundamental matrix   2t 4e 0 e−3t Ω(t) = 3e2t e−t −3e−3t  . 3e2t e−2t 2e−3t 11. The coefficient matrix has eigenvalues 2+2i and 2−2i, with corresponding eigenvectors 2i −2i , . 1 0 From these form two independent solutions which for the columns of the fundamental matrix −2e2t sin(2t) 2e2t cos(2t) . Ω(t) = e2t cos(2t) e2t sin(2t)

344 CHAPTER 13. SYSTEMS OF LINEAR DIFFERENTIAL EQUATIONS 12. The coefficient matrix has eigenvalues −1 ± 2i, and the eigenvector associated with −1 + 2i is 5 . −1 + 2i Form the fundamental matrix 5 cos(2t) Ω(t) = et − cos(2t) − 2 sin(2t)

5 sin(2t) . 2 cos(2t) − sin(2t)

13. The coefficient matrix has eigenvalues 1 ± i. An eigenvector associated with 1 + i is 2+i . 1 Form the fundamental matrix t 2 cos(t) − sin(t) Ω(t) = e cos(t)

cos(t) + 2 sin(t) . sin(t)

14. Eigenvalues of the coefficient matrix are −1, −1, 2. Eigenvectors are     0 2 1 , 1 . 1 1 The second eigenvector is associated with 2, and every eigenvector associated with −1 is a scalar multiple of the first. Immediately we have two independent solutions   0 Φ1 (t) = 1 e−t 1 and

  2 Φ2 (t) = 1 e2t . 1

Try Φ3 (t) = Ete−t + Ke−t , where E is the eigenvector associated with the eigenvalue −1, and   a K = b c is to be determined. Substitute Φ3 (t) into X0 = AX and use the fact that AE = −E to get E − K = AK.

13.2. SOLUTION OF X0 = AX WHEN A IS CONSTANT

345

Solve the equations for a, b and c to get a = 1, b = 2, c = 0, in which c is arbitrary and has been chosen to be zero. Then       0 1 1 Φ3 (t) = 1 te−t + 2 e−t = t + 2 e−t . 1 0 t These solutions form the columns of a fundamental matrix for the system. 15. The coefficient matrix has eigenvalues −2, −1 + 2i, −1 − 2i. From −2 we obtain the solution   0 Φ(t) = e−2t 0 . 1 An eigenvalue for −1 + 2i is     1 0 1 + i 2 . 3 0 This gives us two more solutions: 

 cos(2t) Φ2 (t) = e−t cos(2t) − 2 sin(2t) 3 cos(2t)

and

 sin(2t) Φ3 (t) = e−t 2 cos(2t) + sin(2t) . 3 sin(2t) 

These solutions form the columns of a fundamental matrix. 16. The coefficient matrix has eigenvalues 2, 2, and all eigenvectors are scalar multiples of 0 E= . 1 One solution is

Φ1 (t) =

0 . e2t

For a second solution, try Φ2 (t) = Ete2t + Ke2t . Substitute this into the differential equation to find that we can take 1/5 K= . 0 This yields a second solution Φ2 (t) =

2t e /5 . te2t

346 CHAPTER 13. SYSTEMS OF LINEAR DIFFERENTIAL EQUATIONS 17. The coefficient matrix has eigenvalues 3, 3, and every eigenvector is a scalar multiple of 1 E= . 0 One solution is 3t e Φ1 (t) = . 0 Attempt a second solution Φ2 (t) = Ete3t + Ke3t . Solve for K to get K=

0 . 1/2

Then Φ2 (t) =

te3t . e3t /2

These solutions form columns of a fundamental matrix. 18. The eigenvalues of A are 1, 1, 1 and all eigenvectors are scalar multiples of   0 E = 0 . 1 One solution is   0 Φ1 (t) =  0  . et For a second solution, let Φ2 (t) = Etet + Ket and obtain   1/4 K =  0 . 0 Then Φ2 (t) =

t e /4 . tet

For a third solution, let Φ3 (t) =

1 2 Et + Ktet + Met . 2

13.2. SOLUTION OF X0 = AX WHEN A IS CONSTANT

347

Substitute this into the system and solve for   1/2 M =  0 . −2 This gives us 1 2 1  1 2t + 4t + 2 . 0 Φ3 (t) = et  1 2 2t − 2 √ √ 19. The coefficient matrix has eigenvalues 4 + 29i and 4 − 29i, with corresponding eigenvectors √ −2 − 29 . +i 3 0 This gives us two independent solutions: √ √ √ 4t −2 cos( 29t) +√ 29 sin( 29t) Φ1 (t) = e 3 cos( 29t) and

√ √ √ − 29 cos( 29t) √ − 2 sin( 29t) . 3 sin( 29t) √ √ 20. The coefficient matrix has eigenvalues −1, −3 + 10, −3 − 10, with corresponding eigenvectors √  √      −4 − √10 −4 + √10 3 V1 = 1 , V2 = 10 + 3 10 , V3 = 10 − 3 10 . 6 4 4 Φ2 (t) = e4t

The general solution is √

√

X(t) = V1 e−t + V2 e(−3+ 10)t + V3 e(−3− 10)t . These solutions can also be used as columns of a fundamental matrix. 21. The coefficient matrix has eigenvalues 1, 1, 3, 0, with corresponding eigenvectors         0 1 3 2 −2 0 2 0        V1 =   , V2 =   , V3 =   , V4 =   . −2 0 2 1 1 0 0 0 We can write a general solution X(t) = V1 et + V2 et + V3 e3t + V4 .

348 CHAPTER 13. SYSTEMS OF LINEAR DIFFERENTIAL EQUATIONS

13.3

Exponential Matrix Solutions

In Problems 1–8, the exponential matrix can be obtained using the Putzer algorithm or a software program.

1. e

1 sin(2t) cos(2t) − 21 sin(2t) 2 = − 52 sin(2t) cos(2t) + 21 sin(2t)

2. eAt =

2 −3t 1 +3 3e 2 −3t 2 − e 3 3

1 1 −3t 3 − 3e 1 −3t + 23 3e

3. eAt = e3t

cos(2t) + sin(2t) 2 sin(2t)

− sin(2t) cos(2t) − sin(2t)

3t/2

! √ √ √ − √27 sin( 7t/2) cos( 7t/2) + 17 sin( 7t/2) √ √ √ √4 sin( 7t/2) cos( 7t/2) − √17 sin( 7t/2) 7

5. eAt = et

cos(2t) 2 sin(2t) − 12 sin(2t) cos(2t)

6. eAt =

4 5t 5 + 5e 2 5t − 5 e + 52

− 25 e5t + 25 1 5t 4 5 + 5e

−t/2

√ √ cos(3 3t/2) + √13 sin(3 3t/2) √ √2 sin(3 3t/2) 3

! √ − √23 sin(3 3t/2) √ √ cos(3 3t/2) − √13 sin(3 3t/2)

13.3. EXPONENTIAL MATRIX SOLUTIONS

349

8. eAt is the 3 × 3 matrix [eij ], where 3 2t 2 1 e + cos(t) − sin(t), 5 5 5 1 1 2 e12 = sin(t) + cos(t) − e2t 5 5 5 1 1 1 e13 = sin(t) − cos(t) + e2t 5 5 5 3 3 2t 4 e21 = cos(t) − e − sin(t) 5 5 5 1 2t 4 3 e22 = e + cos(t) + sin(t) 5 5 5 1 1 7 e23 = sin(t) + cos(t) − e2t 5 5 5 3 3 2t 1 e31 = − cos(t) + e − sin(t) 5 5 5 1 1 2t 3 e32 = cos(t) − e − sin(t) 5 5 5 1 2t 4 2 e33 = e + cos(t) − sin(t) 5 5 5 e11 =

9. First, because D is a diagonal matrix, Dn is a diagonal matrix for any positive integer n, and the diagonal element of Dn is dnj . Then

eDt =

∞ X 1

n! n=0

Dn tn

and the diagonal element of eDt is ∞ X 1

n! n=0

(dj )n tn ,

which is edj t .

10. Notice that Bn = (P−1 AB)n = (P−1 AP)(P−1 AP) · · · (P−1 AP) = P−1 An P.

350 CHAPTER 13. SYSTEMS OF LINEAR DIFFERENTIAL EQUATIONS Then Bt

= =

∞ X 1

n! n=0 ∞ X 1

n! n=0

=P =P

(P−1 AP)n tn P−1 An Ptn ∞ X 1

−1

n=0 −1 At

! n n

A t

11. From the result of Problem 10, eAt = PeDt P−1 , where P−1 AP = D, the diagonal matrix having the eigenvalues λ1 , · · · , λn of A down its diagonal. But from the result of Problem 9, eDt is the diagonal matrix having diagonal elements eλj t .

13.4

Solution of X0 = AX + G for Constant A

1. The coefficient matrix A has the repeated eigenvalue 3, 3, and every eigenvector is a scalar multiple of

1 . −1

One solution of the homogeneous system X0 = AX is e

1 . −1

Using methods from Section 13.2, find a second, independent solution of this homogeneous system to write the fundamental matrix Ω(t) = e3t

1 + 2t −2t

2t . 1 − 2t

For a particular solution of the nonhomogeneous system, first compute Ω−1 (t) = e−3t

1 − 2t −2t . 2t 1 + 2t

13.4. SOLUTION OF X0 = AX + G FOR CONSTANT A

351

Now use variation of parameters to compute a solution of the nonhomogeneous system. For this method, we need Z u(t) = Ω−1 (t)G(t) dt Z 1 − 2t −2t −3et = e3t dt 2t 1 + 2t e3t Z −2t 6te − 3e−2t − 2t −3te−2t − t2 = dt = . −6te−2t + 1 + 2t (3/2)(1 + 2t)e−2t + t + t2 The general solution is X(t) = Ω(t)C + Ω(t)u(t) 2t c1 3t 1 + 2t =e −2t 1 − 2t c2 2t −3te−2t − t2 3t 1 + 2t +e −2t 1 − 2t (3/2)(1 + 2t)e−2t + t + t2 e3t (c1 (1 + 2t) + 2c2 t) + t2 e3t = 3t . e (−2c1 t + c2 (1 − 2t)) + (t − t2 )e3t + 3et /2 2. A has eigenvalues 0, 0, and every eigenvector is a scalar multiple of 2 . 1 The associated homogeneous system has this eigenvector as a constant solution. Apply the method of Section 13.2 to find a second, linearly independent solution and produce the fundamental matrix 2 1 + 2t Ω(t) = . 1 t Use this fundamental matrix and variation of parameters to find the general solution of the nonhomogeneous system 2c1 + c2 (1 + 2t) + t + t2 − 2t3 . X(t) = c1 + c2 t + 2t2 − t3 3. A has eigenvalues 6, 6 and eigenvectors are scalar multiples of 1 . 1 We find the fundamental matrix Ω(t) =

1 1+t 1 t

352 CHAPTER 13. SYSTEMS OF LINEAR DIFFERENTIAL EQUATIONS for the associated homogeneous system. Use this and variation of parameters to find the general solution of the nonhomogeneous system: c + c2 (1 + t) + 2t + t2 − t3 X(t) = e6t 1 . c1 + c2 t + 4t2 − t3 4. A has eigenvalues 2, 2, 2, and there are two linearly independent eigenvectors     1 0 0 , 1 . 0 1 We find the fundamental matrix  1 Ω(t) = e2t 0 0

0 1 1

 0 −4t  1 − 4t

of the associated homogeneous system. A general solution for the nonhomogeneous system is   c1 e2t X(t) =  (c2 − 4c3 t)e2t + 1  . (c2 + c3 (1 − 4t))e2t + 1 5. A has eigenvalues 1, 1, 3, 3. The eigenvalue 3 has two independent eigenvectors     0 0 1 −9  , , 0  2  0 1 and 1 has the eigenvector   0 0  , 0 1 with all other eigenvectors associated with 1 scalar multiples of this one. A fundamental matrix for the homogeneous system X0 = AX is   0 et 0 0  0 −2et et −9e3t  . Ω(t) =  0 0 0 2e3t  et −5tet e3t 0 The nonhomogeneous system has general solution   c2 et  −2c2 et + (c3 − 9c4 )e3t + et  . X(t) =    2c4 e3t t 3t t (c1 − 5c2 t)e + c3 e + (1 + 3t)e

13.5. SOLUTION BY DIAGONALIZATION

353

For Problems 6–9, the idea is to find a general solution for the system, then solve for the constants to obtain the solution satisfying the initial condition. For these problems only the solution of the initial value problem is given. 6.

−1 + e2t −5t + (3 + 5t)e2t

X(t) =

(−1 − 14t)et (3 − 14t)et

  13t − (8 + 12t + 3t2 )e2t  4et + (7 + 2t)e2t X(t) =  −et − e2t

 (6 + 12t + (1/)t2 )e−2t  (2 + 12t + (1/2)t2 )e−2t X(t) =  (3 + 38t + 66t2 + (13/6)t3 )e−2t 

13.5

Solution by Diagonalization

1. The coefficient matrix A is diagonalized by 1 1 P= 1 4 and we find that P−1 = The system for Z is −1 Z0 = 0

1 3

4 −1 . −1 1

1 4 −1 0 0 Z+ . 2 10 cos(t) 3 −1 1

This is the system 0 1 −10 cos(t) −z1 z1 Z (t) = = + . z20 2z2 3 10 cos(t) 0

Solve these two independent differential equations to get −t c1 e − (5/3) cos(t) − (5/3) sin(t) Z(t) = . c2 e2t − (4/3) cos(t) + (2/3) sin(t) Then

c1 e−t + c2 e2t − 3 cos(t) − sin(t) . c1 e−t + 4c2 e2t − 7 cos(t) + sin(t)

X(t) = PZ(t) =

354 CHAPTER 13. SYSTEMS OF LINEAR DIFFERENTIAL EQUATIONS 2. The coefficient matrix has eigenvalues 2, 6 and is diagonalized by 3 1 P= . −1 1 We find that P−1 =

1 4

1 −1 . 1 3

With X = PZ, the transformed system has the solution c1 e2t − 1 − e3t Z(t) = . c2 e6t − 1/3 − et The original system has the general solution 3c1 e2t + c2 e6t − 4e3t − 10/3 X(t) = . −c1 e2t + c2 e6t + 2/3 3. The coefficient matrix has eigenvalues 0, 2 and is diagonalized by 1 1 P= −1 1 which has inverse −1

1 = 2

1 −1 . 1 1

The system for Z(t) has the solution c1 − 2t + e3t Z(t) = . c2 e2t − 1 + 3e3t Then

X(t) =

c1 + 2c2 e2t − 1 − 2t + 4e3t . −c1 + c2 e2t − 1 + 2t + 2e3t

4. A has eigenvalues 1, 7, and we find 1 1 −5 1 5 −1 . P= ,P = −1 1 6 1 1 We obtain Z(t) =

c1 et + (1/15) cos(3t) − (3/15) sin(3t) + 20/3 . c2 e7t + (7/87) cos(3t) − (2/58) sin(3t) − 4/21

Then X(t) =

c1 et + 5c2 e7t + (68/145) cos(3t) − (54/145) sin(3t) + 40/7 . −c1 et + c2 e7t + (2/145) cos(3t) + (24/145) sin(3t) − 48/7

13.5. SOLUTION BY DIAGONALIZATION

355

5. A has eigenvalues 3i, −3i and is diagonalized by 1+i 1−i P= . 3 3 We find that P−1 =

1 6

−3i 1 + i . 3i 1 − i

The transformed problem for Z has the solution d1 e3it + ((2 − i)/6)e2t Z(t) = . d2 e−3it + ((2 + i)/6)e2t If Euler’s formula is used on the complex exponential terms, we obtain the real solution c1 (cos(3t) − sin(3t)) − c2 (cos(3t) + sin(3t)) + e2t X(t) = . 3c1 cos(3t)03c2 sin(3t) + 2e2t 6. The coefficient matrix is A=

1 1

with eigenvalues 0, 2. A is diagonalized by 1 1 P= −1 1 and we find that

1/2 −1/2 . 1/2 1/2

P−1 =

The uncoupled system is 2t 0 0 1/2 −1/2 6e 0 Z = Z+ . 0 2 1/2 1/2 2e2t The initial conditions convert to Z(0) = P

−1

3 X(0) = . 3

Solve for Z to obtain Z(t) =

2 + e2t . 3e2t + 4te2t

The solution of the initial value problem for X is 2 + 4(1 + t)e2t X(t) = PZ(t) = . −2 + 2(1 + t)e2t

356 CHAPTER 13. SYSTEMS OF LINEAR DIFFERENTIAL EQUATIONS 7. The coefficient matrix has eigenvalues 0, 3 and is diagonalized by 2 −1 P= . 1 1 Compute P The uncoupled system is 0 0 Z = 0

−1

1 = 3

1 . 2

1 −1

0 1/2 Z+ 3 −1/3

1/3 2t . 2/3 5

The initial condition is Z(0) = P Then

Z(t) =

−1

X(0) =

25/3 . 11/3

(1/3)t2 + (5/3)t + 25/3 . (127/27)e3t + (2/9)t − 28/27

The solution of the initial value problem for X is −(127/27)e3t + (2/3)t2 + (28/9)t + 478/27 X(t) = PZ(t) = . (127/27)e3t + (1/3)t2 + (17/9)t + 197/27 8. The coefficient matrix has eigenvalues i, −i and is diagonalized by 5 5 P= . 2−i 2+i We find that P

−1

1 = 10

1 − 2i 5i . 1 + 2i −5i

With X = PZ, the uncoupled system is i 0 (1 − 2i)/10 i/2 5 sin(t) Z0 = Z+ , 0 −i (1 + 2i)/10 −i/2 0 with initial conditions Z(0) = P−1 X(0) =

1 + i/2 . 1 − i/2

Solve these uncoupled equations for z1 and z2 and then obtain 10 cos(t) − (5/2)t sin(t) − 5t cos(t) X(t) = PZ(t) = . 5 cos(t) + (5/2) sin(t) − (5/2)t sin(t)

13.5. SOLUTION BY DIAGONALIZATION

357

9. The coefficient matrix has eigenvalues 1, 1, −3, but there are two independent eigenvectors associated with the repeated eigenvalue 1 and A is diagonalized by   1 −1 1 P = 1 0 3 . 0 1 1 We find that



 3 −2 3 P−1 =  1 −1 2  . −1 1 −1

With X = PZ we obtain the uncoupled system      −3e−3t 3 =2 3 1 0 0 −1 2  t . Z0 = 0 1 0  Z +  1 0 −1 1 −1 0 0 −3 The initial conditions are 

 11 Z(0) = P−1 X(0) =  6  . −4 Solve this uncoupled system to obtain   (5/2)et − (8/3)e−3t + 3te−3t + (8/9) + (4/3)t X(t) = PZ(t) =  (27/4)et − (113/12)e−3t + 9te−3t + (5/3) + 3t  . (17/4)et − (113/36)e−3t + 3te−3t + (8/9) + (4/3)t 10. The coefficient matrix has eigenvalues 1, 2, 2 and is diagonalized by   1 1 1 P = 1 1 0 1 0 1 and we find that

  −1 1 1 0 −1 . P−1 =  1 1 −1 0

The uncoupled system is      1 0 0 −1 1 1 − 0 −1  t  . Z0 = 0 2 0 Z +  1 0 0 2 1 −1 0 2et The initial conditions are Z(0) = P−1 X(0).

358 CHAPTER 13. SYSTEMS OF LINEAR DIFFERENTIAL EQUATIONS Solve this initial value problem for Z(t) and then obtain   −(1/4)e2t + (2 + 2t)et − (3/4) − (1/2)t . e2t + (2 + 2t)et − 1 − t X(t) = PZ(t) =  (5/4)e2t + 2tet − (3/4) − (1/2)t

Chapter 14

Nonlinear Systems and Qualitative Analysis 14.1

Nonlinear Systems and Phase Portraits

1. The coefficient matrix is A=

3 −5 , 5 −7

with eigenvalues −2, −2. Every eigenvector is a scalar multiple of 1 . 1 The origin is an improper nodal sink. 2. The coefficient matrix has eigenvalues −3, 4 and the origin is a saddle point. For Problems 3 − −16, only the eigenvalues of the coefficient matrix, and the classification of the origin, are given. As typical cases, phase portraits are drawn for the systems of Problems 3, 5, 6, 7 and 11. Phase portraits are included for the systems of Problems 3, 4, 5 and 11. 3. eigenvalues 2i, −2i; center 4. 2, 3, nodal source 5. 4 + 5i, 4 − 5i, spiral point 6. −3, −5, nodal sink 7. 3, 3, and the coefficient matrix does not have two independent eigenvectors; improper node 359

360CHAPTER 14. NONLINEAR SYSTEMS AND QUALITATIVE ANALYSIS

Figure 14.1: Center of Problem 3.

Figure 14.2: Nodal source of Problem 4.

14.1. NONLINEAR SYSTEMS AND PHASE PORTRAITS

361

Figure 14.3: Spiral source of Problem 5. √ √ 31i, − 31i, center √ √ 9. eigenvalues −2 + 3i, −2 − 3i, spiral sink

10. −13, −13, and the coefficient matrix does not have two independent eigenvectors; improper node √ √ 11. 5, − 5, saddle point √ √ 12. 3 + 5i, 3 − 5i, spiral source √ √ 13. −3 + 7, −3 − 7, both eigenvalues negative, nodal sink 14. 11, 11, and the coefficient matrix does not have two independent eigenvectors, improper node √ √ 15. 2 + 3, 2 − 3, nodal source √ √ 16. 13i, − 13i, center 17. (a) First, write dx 1 = x dt t as

1 1 dx = dt. x t

Integrate to get ln(x) = ln(t) + c

362CHAPTER 14. NONLINEAR SYSTEMS AND QUALITATIVE ANALYSIS

Figure 14.4: Saddle point of Problem 11.

for x > 0, t > 0. Then x = ct with c constant. Put this into the second equation to get 1 y 0 = ct − y. t Write this as 1 y 0 + y = ct, t or ty 0 + y = ct2 . Then (ty)0 = ct2 . Integrate to get ty =

c 3 t + d. 3

c 2 d t + . 3 t

Then

(b) Suppose x(t0 ) = 1 and y(t0 ) = 0. Then it is routine to solve for c and d from part (a) to obtain x(t) =

1 1 2 1 t, y(t) = t − t20 . t0 3t0 3t

(c) In part (b), we have trajectories through (1, 0) at any time t0 6= 0. However, these translations are not trajectories of each other.

14.2. CRITICAL POINTS AND STABILITY

14.2

363

Critical Points and Stability

In Problems 1–16, the stability type of the origin is given, based on information in Problems 1–16 of Section 14.1. 1. The origin is an improper node that is both stable and asymptotically stable 2. The origin is an unstable saddle point 3. stable but not asymptotically stable center 4. unstable nodal source 5. unstable spiral source 6. stable and asymptotically stable nodal sink 7. unstable improper node 8. stable but not asymptotically stable center 9. stable and asymptotically stable spiral sink 10. stable and asymptotically stable improper node 11. unstable saddle point 12. unstable spiral point 13. stable and asymptotically stable nodal sink 14. unstable improper node 15. unstable nodal source 16. stable but not asymptotically stable center √ √ 17. If = 0, the eigenvalues are 5i, − 5i and the origin is a center, which is stable but not asymptotically stable. If > 0, the eigenvalues are 1p 1 p 1 + ( − 2)2 − 24, ( − 2)2 − 24. 2 2 2 √ These have positive real part. If 0 <√ < 2(1 + 6), then the origin is an unstable spiral point. √ If > 2(1 + 6), the origin is an unstable saddle point. If = 2(1 + 6), the origin is an unstable improper node.

364CHAPTER 14. NONLINEAR SYSTEMS AND QUALITATIVE ANALYSIS 18. If = 0, the eigenvalues are −3, −3 and every eigenvector is a scalar multiple of 1 . −1 In this case the origin is a stable and asymptotically stable improper node. If > 0, the eigenvalues are − 6 1p − 6 1p + ( + 10)2 − 100, − ( + 10)2 − 100. 2 2 2 2 These eigenvalues are real and distinct. Further, if 0 < < 9/8, these eigenvalues are both negative, so the origin is a stable and asymptotically stable nodal sink. If > 9/8, then the eigenvalues are of opposite sign and the origin is an unstable saddle point. If = 9/8, then the origin is not an isolated singularity.

14.3

Almost Linear Systems

In Problem 1, the details of showing that the system is almost linear are included. Problems 2–10 omit this demonstration and concentrate on analyzing the critical point (0, 0). 1. To show that the system is almost linear, consider lim (x,y)→(0,0)

x2 p

x2 + y 2

r2 cos2 (θ) r→0 r

= lim

= lim r cos2 (θ) = 0. r→0

The origin is a critical point and the matrix of coefficients of the linear part is 1 −1 , 1 2 which has eigenvalues √ √ 1 1 (3 + 3i) and (3 − 3)i. 2 2 The origin is an unstable spiral point of the linear part, hence also of the given system. 2. The matrix of coefficients of the linear part is 1 −1 1 1 with eigenvalues 1 + i, 1 − i. The origin is an unstable spiral source.

14.3. ALMOST LINEAR SYSTEMS

365

3. The linear part has matrix A(0,0) =

−2 1

2 4

√ √ with eigenvalues 1+ 11, 1− 11. These are of opposite sign, so the origin is an unstable saddle point. 4. The matrix of the linear part is

−2 −3 1 4

√ with eigenvalues 1 ± 6, which are of opposite sign. The origin is an unstable saddle point. 5. The linear part has matrix

3 12 , −1 −3

√ √ with eigenvalues 3i, − 3i. The origin is a center for the linear part of the system, so the nonlinear system could have a center or a spiral point there. 6. The linear part has matrix 2 −4 , 1 1 √ with eigenvalues (3 ± 15i)/2. The origin is a spiral point and is unstable because the real part of the eigenvalues is positive. 7. The linear part has matrix

−3 −4 . 1 1

This matrix has eigenvalues −1, −1, and all eigenvectors are scalar multiples of −2 . 1 The origin is a stable improper nodal sink. 8. The linear part has matrix

−3 −4 −1 1

√ with eigenvalues −1 ± 2 2. Both eigenvalues are positive, so the origin is an unstable nodal source.

366CHAPTER 14. NONLINEAR SYSTEMS AND QUALITATIVE ANALYSIS 9. The linear part has matrix

−2 −1 −4 1

with eigenvalues 2, −3, so the origin is an unstable saddle point. 10. The linear part has matrix

1 −2

0 1

with eigenvalues 1, 1. All eigenvectors associated with this eigenvalue are scalar multiples of 0 , 1 so the origin is an unstable (positive eigenvalue) improper node. 11. Refer to these as systems I and II, in the order given. (a) For each system the linear part has coefficient matrix 0 1 A= −1 0 with eigenvalues i, −i. Therefore the origin is a center for each system. (b) For The first system, use polar coordinates, with x = r cos(θ), y = r sin(θ) and p x2 + y 2 = r. Then p −x x2 + y 2 −r2 cos(θ) p lim = lim r→0 r (x,y)→(0,0) x2 + y 2 = lim −r cos(θ) = 0, r→0

independent of θ. And, similarly p −y x2 + y 2 p = lim −r sin(θ) = 0. lim r→0 (x,y)→(0,0) x2 + y 2 Therefore the first system is almost linear. The argument is essentially the same for the second system. (c) With r2 = x2 + y 2 , we have rr0 = xx0 + yy 0 ,

14.3. ALMOST LINEAR SYSTEMS

367

where primes denote differentiation with respect to t. Now insert the expressions for x0 and y 0 from the differential equations of system I to obtain p p rr0 = x(y − x x2 + y 2 ) + y(−x − y x2 + y 2 ) p = −(x2 + y 2 ) x2 + y 2 = −r3 . Then

dr = −r2 for system I. dt Similarly, if we insert the expressions for x0 and y 0 from system II, we obtain p p rr0 = x(y + x x2 + y 2 ) + y(−x + y x2 + y 2 ) p = (x2 + y 2 ) x2 + y 2 r0 =

= r3 . Then

dr = r2 for system II. dt

(d) For system I, dr = −r2 . dt This is the separable equation 1 − 2 dr = dt. r This shows that, for system I, r0 (t) < 0, so the distance between the point and the origin is decreasing with time. Now integrate the differential equation for r(t) to get 1 = t + c. r To satisfy the initial condition r(t0 ) = r0 , we need 1 = t0 + c, r0 so c= Then r(t) =

1 − t0 . r0

1 t − t0 + 1/r0

368CHAPTER 14. NONLINEAR SYSTEMS AND QUALITATIVE ANALYSIS for system I. Then r(t) → 0 as t → ∞. The first system is asymptotically stable at the origin. (e) By an entirely analogous derivation, for system II, r0 = r2 > 0 so r(t) is increasing with time for system II. Solve this separable differential equation subject to r(t0 ) = r0 to get r(t) =

1 . 1/r0 + t0 − t

Then for system II, r(t) → ∞ as t → t0 + 1/r0 from the left. We conclude from this that the second system is unstable at the origin. Parts (d) and (e) show that, in the case of a center at the origin, behavior of the the linear part of an almost linear system at the origin does not provide definitive information about the stability of the origin for the nonlinear system. 12. The linear part of the system has coefficient matrix α −1 A= . 1 −α The eigenvalues are

√ √ α2 − 1 and − α2 − 1. There are the following cases:

1. If 0 < |α < 1, the eigenvalues are pure imaginary. In this case the origin is a center of the linear part and may be a center or spiral point of the almost linear system. 2. If |α| > 1, then the eigenvalues are real and of opposite sign, so the origin is an unstable saddle point of the linear part, hence also of the nonlinear system. 3. If α = ±1, then A is a singular matrix and the theory developed for almost linear systems does not apply. The constants h and k will influence quantitative aspects of the phase portrait of the system, but not the qualitative behavior. 13. Using results from Problem 11, we have rr0 = xx0 + yy 0 = x(y + x(x2 + y 2 )) + y(−x + y(x2 + y 2 )) = (x2 + y 2 )(x2 + y 2 ) = r4 .

14.4. LINEARIZATION

369

Then dr = r3 . dt This is separable 1 dr = dt. r3 Integrate to get 1 − r−2 = t + c. 2 Then 1 , r(t) = √ k − 2 t where k = 2c is an arbitrary constant which is determined by specifying a point that the trajectory is to pass through at some positive time. If < 0, then 1

r(t) = p

k + 2| |t

→0

as t → ∞. In this case trajectories approach the origin as t → ∞ and the nonlinear system is asymptotically stable. However, something different happens if > 0. Suppose r(0) = ρ, so a trajectory starts at a point ρ units from the origin at time zero. Then k = 1/ρ2 and 1 r(t) = p . (1/ρ)2 − 2 t Now, as t starts at zero and increases toward 1/(2 ρ2 ), r(t) → ∞, so there is a time close to which the point is arbitrarily far from the origin. This makes the origin unstable.

14.4

Linearization

1. For critical points other than the origin, solve x − y + x2 = 0, x + 2y = 0 to get (−3/2, 3/4). We find that A(−3/2,3/4) =

−2 −1 , 1 2

√ √ which has eigenvalues 3 and − 3. This critical point is an unstable saddle point of the linear part, and therefore of the given system.

370CHAPTER 14. NONLINEAR SYSTEMS AND QUALITATIVE ANALYSIS 2. The critical points are ! √ −1 − 17 ,4 . 2

! √ −1 + 17 ,4 , 2 We have A((−1+√17)/2,4) =

√ 17 −7

−1 √ , 2 − 17

with eigenvalues q q √ √ 1 − 25 − 2 17, 1 + 25 − 2 17. These are of opposite sign, being approximately equal to −3.093 · · · and 5.093 · · · . This critical point is an unstable saddle point. Next,

√ −1 − 17 √ , −7 2 + 17 p p √ √ with eigenvalues 1 − 25 + 2 17 and 1 + 25 + 2 17. These are also of opposite sign, so this critical point is also an unstable saddle point. A((−1−√17)/2,4) =

3. The critical point other than the origin is (−5, −5). We find that A(−5,−5) =

−2 2 . 1 −6

√ √ This has eigenvalues −4 + 6, −4 − 6, which are unequal and both negative. This critical point is a stable and asymptotically stable nodal sink of the nonlinear system. 4. The only critical point other than the origin is (−20, 5). We find that A(−20,5) =

−2 −13 , 1 4

having eigenvalues 1 + 2i, 1 − 2i. These are complex with positive real part, so (−20, 5) is an unstable spiral source. 5. The system has one critical point other than the origin, (−1/2, 1/8). Calculate 3 12 A(−1/2,1/8) = , −1/4 −3 √ √ with eigenvalues are 6, − 6, so this critical point is an unstable saddle point.

14.4. LINEARIZATION

371

6. The system has two critical points other than the origin, (1, −2) and (−2/3, 2/9). First, −4 −1 A(1,−2) = 3 1 √ and the eigenvalues are (−3± 13)/2. These have opposite sign, so (1, −2) is an unstable saddle point. A(−2/3,2/9) =

8/3 −6 , −1/2 1

√ with eigenvalues (11 ± 97)/2. These are both positive, so (−2/3, 2/9) is an unstable nodal source. 7. Aside from the origin, the system has critical points (1/2, −1/2). We have −2 −3 A(1/2,−1/2) = 1 1 √ with eigenvalues (−1 ± 3i)/2, so (1/2, −1/2) is a spiral point, stable and asymptotically stable because the real part of the eigenvalues is negative. 8. The critical points other than the origin are r √ ! r √ ! 7 21 7 21 ,− and − , . 3 4 3 4 First, A√ (

√ = 7/3,− 21/4)

−3 −4 −5/2 −4/3

√ with eigenvalues (−13 ± 385)/6, both positive, so this critical point is an unstable nodal source. Next, A(−√7/3,√21/4) =

−3 −4 . 9/2 −4/3

√ This has eigenvalues (−13 ± 623i)/6. This critical point is a stable and asymptotically stable spiral point. 9. The critical point other than the origin is (−3/8, −3/2). We find that −2 −2 A(−3/8,−3/2) = −4 1 √ with eigenvalues (−1 ± 23i)/2, so (−3/8, −3/2) is a stable and asymptotically stable spiral point.

372CHAPTER 14. NONLINEAR SYSTEMS AND QUALITATIVE ANALYSIS 10. The system has two critical points other than the origin: √ ! √ ! √ 2+ 2 √ 2− 2 3 + 2 2, , 3 − 2 2, . 2 2 First, A(3+2√2,(2+√2)/2) =

√ 1√ −4 − 2√ 2 (−1 + 2 2)/2 11 + 7 2

with eigenvalues 7 1 6+ √ − 2 2

q q √ √ 7 1 √ 128 2 + 90, 6 + 128 2 + 190. + 2 2

These are approximately 1.3188 · · · and 20.5806 · · · , both positive, so this critical point is an unstable proper nodal source, A(3−2√2,(2−√2)/2) =

√ 1√ −4 + 2√ 2 (−1 − 2 2)/2 11 − 7 2

with eigenvalues 1 7 7− √ − 2 2

q q √ √ 1 7 190 − 128 2, 6 − √ + 190 − 128 2. 2 2

These are approximately −0.4481 · · · and 2.5486 · · · , having opposite signs, so this critical point is an unstable saddle point.

Chapter 15

Vector Differential Calculus 15.1

Vector Functions of One Variable

In Problems 1 and 2 the details of the differentiation are carried out both ways. For Problems 3–8 just the derivative is given. 1. First use the product rule: (f (t)F(t))0 = f 0 (t)F(t) + f (t)F0 (t) = (−12 sin(3t))F(t) + 4 cos(3t)(6tj + 2k) = −12 sin(3t)i + (24t cos(3t) − 36t2 sin(3t))j + (8 cos(3t) − 24t sin(3t))k. If we first carry out the product, we have f (t)F(t) = 4 cos(3t)i + 12t2 cos(3t)j + 8t cos(3t)k, so (f (t)F(t))0 = −12 sin(3t)i + (24t cos(3t) − 36t2 sin(3t))j + (8 cos(3t) − 24t sin(3t))k. 2. (F(t) · G(t)) = (i − 6tk) · (i + cos(t)k) + (ti − 3t2 k) · (− sin(t)k) = 1 − 6t cos(t) + 3t2 sin(t) 373

374

CHAPTER 15. VECTOR DIFFERENTIAL CALCULUS

3. Apply the product rule for cross products: i j 0 (F(t) × G(t))0 = 1 1 − cos(t)

i k 0 + t 0 t

j 1 sin(t)

k 4 1

= −tj − cos(t)k + (1 − 4 sin(t))i − tj + t sin(t)k = (1 − 4 sin(t))i − 2tj − (cos(t) − t sin(t))k. 4. (F(t) × G(t))0 = (3t2 − 2t sinh(t) − t2 cosh(t))i − 2tj − (sinh(t) + t cosh(t))k 5. (f (t)F(t))0 = (1 − 8t2 )i + (6t2 cosh(t) − (1 − 2t3 ) sinh(t))j + (et − 6t2 et − 2t3 et )k 6. (F(t) · G(t))0 = sin(t) + t cos(t) + 4 + 5t4 7. (F(t) × G(t))0 = tet (2 + t)(j − k) 8. (F(t) · G(t))0 = −16 cos2 (t) + 16 sin2 (t) 9. F(t) = sin(t)i + cos(t)j + 45tk is a position vector for the curve, and F0 (t) = cos(t)i − sin(t)j + 45k is a tangent vector. The distance function along C is Z t Z t√ √ s(t) = k F0 (τ ) k dτ = 2026 dτ = 2026t. 0

√ Then t = s/ 2026 and we can write a position vector in terms of s: s s 45s G(s) = F(t(s)) = sin √ i + cos √ j+ √ k. 2026 2026 2026 Now we can write a tangent vector in terms of s: 1 s s G0 (s) = √ cos √ i − sin √ j + 45k . 2026 2026 2026

15.1. VECTOR FUNCTIONS OF ONE VARIABLE

375

10. F(t) = t3 (i + j + k) is a position vector, and a tangent vector is F0 (t) = 3t2 (i + j + k). A distance function along C is given by Z t Z t √ s(t) = k F0 (ξ) k dξ = 3 3(t3 + 1). −1

−1

Then t=

s √ −1 3

Let

G(s) = F(t(s)) =

1/3 .

s √ − 1 (i + j + k). 3

This gives us the unit tangent vector 1 G0 (s) = √ (i + j + k). 3 11. F = t2 (2i + 3j + 4k) is a position vector for the curve, and F0 (t) = 2t(2i + 3j + 4k) is a tangent vector. The distance function along the curve is given by Z t √ Z t √ 0 s(t) = k F (ξ) k dξ = 2 29 ξ dξ = 29(t2 − 1). 1

Then t= Let

q √ 1 + s/ 29.

G(s) = F(t(s)) =

s √ + 1 (2i + 3j + 4k). 29

Then

1 G0 (s) = √ (2i + 3j + k) 29 is a unit tangent vector. 12. Because F(t) × F0 (t) = O, we must have k F(t) kk F0 (t) k sin(θ) = 0 where θ is the angle between these two vectors. Then at least one of F(t) and F0 (t) is the zero vector, or the angle between these vectors is zero (so the vectors are parallel). Consider cases. If F(t) = O then the particle simply remains at the origin for all time.

376

CHAPTER 15. VECTOR DIFFERENTIAL CALCULUS If F0 (t) = O, then F(t) is a constant vector and the particle does not move. In the case that F(t) and F0 (t) are parallel, then the position and tangent vectors are parallel, so the velocity vector is directed along the path of motion and the motion is in a straight line. We could also argue as follows in the last case. If the position and tangent vectors are parallel, then for some number c, F0 (t) = cF(t). Then x0 (t)i + y 0 (t)j + z 0 (t)k = c(x(t)i + y(t)j + z(t)k). Then x0 (t) = cx(t), y 0 (t) = cy(t), z 0 (t) = cz(t). Then x = x0 ect , y = y0 ect , z = z0 ect , where F(0) = x0 i + y0 j + z0 k. But these are parametric equations of a straight line.

15.2

Velocity, Acceleration and Curvature

In Problems 1–10, we can compute v = F0 (t), a(t) = F00 (t), v(t) =k v(t) k by straightforward calculations. Next, T(t) =

1 1 v(t) = F0 (t). 0 v(t) k F (t) k

Tangential and normal components of the acceleration can be obtained as q dv aT = and aN = k a k2 −a2T . dt The unit normal is N(t) =

1 (a(t) − aT T(t)). aN

In this way it is not necessary to compute s(t) and write vectors in terms of s, which is often awkward. We can also compute N(t) =

1 dT . k dT/dt k dt

Curvature is often easily computed as κ=

aN . v2

15.2. VELOCITY, ACCELERATION AND CURVATURE

377

We can also compute curvature as κ=

k T0 (t) k . k F0 (t) k

κ can also be obtained from a formula requested in Problem 13: κ=

k F0 (t) × F00 (t) k . k F0 (t) k3

1. The velocity is v(t) = F0 (t) = 3i + 2tk and the speed is v(t) =k v(t) k=

p 9 + 4t2 .

The acceleration is a(t) = F00 (t) = 2k. A unit tangent is 1 (3i + 2tk). T(t) = √ 9 + 4t2 The curvature is κ=

k T0 (t) k 6 = . k F0 (t) k (9 + 4t2 )3/2

Finally, aT = and aN =

dv 4t =√ dt 9 + 4t2

q 6 k a k2 −a2T = √ . 9 + 4t2

2. v(t) = (sin(t) + t cos(t))i + (cos(t) − t sin(t))j, a(t) = (2 cos(t) − t sin(t))i − (2 sin(t) + t cos(t))j, 1 T(t) = √ v, 1 + t2 p v(t) = 1 + t2 , t 2 + t2 aT = √ , aN = , 1 + t2 1 + t2 κ=

2 + t2 (+t2 )3/2

378

CHAPTER 15. VECTOR DIFFERENTIAL CALCULUS

3. v(t) = 2i − 2j + k, v = 3, 1 (2i − 2j + k), 3 aT = aN = κ = 0

4. v(t) = et (sin(t) + cos(t))i + et (cos(t) − sin(t))k, v(t) =

√ t 2e ,

a(t) = 2et (cos(t)i − sin(t)k), 1 T(t) = √ ((sin(t) + cos(t))i + (cos(t) − sin(t))k), 2 √ 1 aT = 2et = aN , κ = √ e−t 2 5. v(t) = −3e−t (i + j − 2k), a(t) = 3e−t (i + j − 2k), √ 1 v(t) = 3 6e−t , T(t) = √ (−i − j + 2k), 6 √ −t aT = −3 6e , aN = 0, κ = 0 6. v(t) = −α sin(t)i + βj + α cos(t)k, v(t) =

p α2 + β 2 ,

a(t) = −α cos(t)i − α sin(t)k, 1

T(t) = p

α2 + β 2

(−α sin(t)i + βj + α cos(t)k),

aT = 0, aN = α, κ =

α α2 + β 2

7. p v(t) = 2 cosh(t)j − 2 sinh(t)k, v(t) = 2 cosh(2t), a(t) = 2 sinh(t)j − 2 cosh(t)k, 1

T(t) = p

cosh(t)

(cosh(t)j − sinh(t)k),

2 sinh(2t) 2 a(t) = p , aN = p , cosh(2t) cosh(2t) κ=

1 2(cosh(2t))3/2

Here we have used the identity cosh(2t) = cosh2 (t) + sinh2 (t).

15.2. VELOCITY, ACCELERATION AND CURVATURE 8.

379

1 1 (i − j + 2k), a(t) = − 2 (i − j + 2k), t t √ 1 6 v(t) = , T(t) = √ (i − j + 2k), t 6 √ 6 aT = − 2 , aN = 0, κ = 0 t

v(t) =

9. v(t) = 2t(αi + βj + γk), a(t) = 2(αi + βj + γk) p v(t) = 2|t| α2 + β 2 + γ 2 , T(t) = p

1 α2 + β 2 + γ 2

(αi + βj + γk)

aN = 0, κ = 0, and p aN = 2(sgn(t)) α2 + β 2 + γ 2 , where

( 1 sgn(t) = −1

if t > 0, if t < 0.

10. v(t) = (3 cos(t) − 3t sin(t))j − (3 sin(t) + 3t cos(t))k, p v(t) = 3 1 + t2 a(t) = (−6 sin(t) − 3t cos(t))j − (6 cos(t) − 3t sin(t))k, 1 T(t) = √ ((cos(t) − t sin(t))j − (sin(t) + t cos(t))k, 1 + t2 (3t2 + 6)2 3t , aN = √ , aT = √ 1 + t2 1 + t2 κ=

(3t2 + 6)2 9(1 + t2 )3/2

11. The position vector for a straight line has the form F(t) = (a + bt)i + (c + dt)j + (p + ht)k. The tangent vector is the constant vector T(t) = bi + dj + hk. Then T0 (t) = O, so κ = 0.

380

CHAPTER 15. VECTOR DIFFERENTIAL CALCULUS Conversely, suppose C is a smooth curve having zero curvature. Then κ =k T0 (s) k=k F00 (s) k= 0. If we write F(s) = f (s)i + g(s)j + h(s)k, this means that f 00 (s) = g 00 (s) = h00 (s) = 0. But then f (s) = a + bs, g(s) = c + ds, h(s) = p + hs for some constants a, b, c, d, p, h. This makes F(s) the position vector of a straight line.

12. We may suppose that C is a circle of radius r about the origin in the x, y−plane (translations and rotations will not affect the curvature). A position vector for C in polar coordinates has the form F(t) = r cos(t)i + r sin(t)j with 0 ≤ t ≤ 2π. Then F0 (t) == r sin(t)i + r cos(t)j is a tangent vector, which has length r. A unit tangent is T = − sin(t)i + cos(t)j. Then T0 (t) = − cos(t)i − sin(t)j. Then κ= 13. First write T(t) =

1 k T0 (t) k = . k F0 (t) k r

1 1 0 f 0 (t) = F (t). k F0 (t) k v(t)

This enables us to write F0 = vT. Now, F00 (t) is the acceleration a(t), and T × T = O, so vT × F00 = vT(aT T + aN N) = vaT T × T + vaN T × N = vaN T × N = v(v 2 κ)T × N. But T and N are orthogonal unit vectors, so k T × N k= 1.

15.3. THE GRADIENT FIELD

381

Then k F0 × F00 k= v 3 κ. Finally, v =k F0 k so 3

κ=

15.3

k F0 (t) × F00 (t) k . k F0 (t) k

The Gradient Field

1. ∇ϕ(x, y, z) =

∂ ∂ ∂ (xyz)i + (xyz)j + (xyz)k = yzi + xzj + xyk, ∂x ∂y ∂z ∇ϕ(1, 1, 1) = i + j + k

The maximum value of Du (1, 1, 1) is k ∇ϕ(1, 1, 1) k= √ The minimum value is − 3.

√ 3.

2. ∇ϕ(x, y, z) = (2xy − z cos(xz))i + x2 j − x cos(xz)k, √ √ ! 2π 2 i+j− k. ∇ϕ(1, −1, π/4) = −2 − 8 2 The maximum value of Du (1, −1, π/4) is q

√ 11/2 + π 2 /32 + π/ 2.

The minimum value is the negative of this maximum value. 3. ∇ϕ(x, y, z) = (2y + ez )i + 2xj + xez k, ∇ϕ(−2, 1, 6) = (2 + e6 )i − 4j − 2e6 k. The maximum value of Du (−2, 1, 6) is p 20 + 4e6 + 5e12 and the minimum value is the negative of this.

382

CHAPTER 15. VECTOR DIFFERENTIAL CALCULUS

4. ∇ϕ(x, y, z) = −yz sin(xyz)i − xz sin(xyz)j − xy sin(xyz)k, ∇ϕ(−1, 1, π/2) =

π π i − j − k. 2 2

The maximum value of Du (−1, 1, π/2) is r 1+

π2 . 4

The minimum value is the negative of this radical. 5. ∇ϕ(x, y, z) = 2y sinh(2xy)i + 2x sinh(2xy)j − cosh(z)k, ∇ϕ(0, 1, 1) = − cosh(1)k. The maximum value of Du (0, 1, 1) is cosh(1). The minimum value is − cosh(1). 6. 1 [xi + yj + zk] ∇ϕ(x, y, z) = p x2 + y 2 + z 2 1 ∇ϕ(2, 2, 2) = √ (i + j + k). 3 maxDu =k ∇ϕ(2, 2, 2) k= 1, and the minimum is −1. 7. Du ϕ(x, y, z) = ∇ϕ(x, y, z) · u 1 = ((8y 2 − z)i + 16xyj − xk) · √ (i + j + k) 3 1 2 = √ (8y − z + 16xy − x) 3 8. Du ϕ(x, y, z) 1 = (− sin(x − y)i + sin(x − y)jez k) · √ (i − j + 2k) 6 1 z = √ (−2 sin(x − y) + 2e ) 6

15.3. THE GRADIENT FIELD

383

9. Du (x, y, z) 1 = (2xyz 3 i + x2 z 3 j + 3x2 yz 2 k) · √ (2j + k) 5 1 = √ (2x2 z 3 + 3x2 yz 2 ) 5 10. Du (x, y, z) 1 = ((z + y)i + (z + x)j + (y + x)k) · √ (i − 4k) 17 1 = √ (z − 3y − 4x) 17 11. Let ϕ(x, y, z) = x2 + y 2 + z 2 so the level surface is given by ϕ(x, y, z) = 4. The gradient provides a normal vector N(x, y, z) = ∇ϕ(x, y, z) = 2xi + 2yj + 2zk. Then

√ √ N(1, 1, 2) = 2i + 2j + 2 2k √ is normal to the surface at (1, 1, 2). The tangent plane at this point has the equation √ √ 2(x − 1) + 2(y − 1) + 2 2(z − 2) = 0,

or x+y+

√ 2z = 4.

√ The normal line at (1, 1, 2) has parametric equations √ x = y = 1 + 2t, z = 2(1 + 2t) for all real t. 12. Write ϕ(x, y, z) = x2 + y − z, so the level surface is given by ϕ(x, y, z) = 0. A normal vector at (−1, 1, 2) is given by N(−1, 1, 2) = ∇ϕ(−1, 1, 2) = −2i + j − k. The tangent plane at (−1, 1, 2) has the equation −2x + y − z = 1. The normal line at (−1, 1, 2) is given by x = −1 − 2t, t = 1 + t, z = 2 − t in which t takes on all real values.

384

CHAPTER 15. VECTOR DIFFERENTIAL CALCULUS

13. Let ϕ(x, y, z) = x2 − y 2 − z 2 . The normal vector at (1, 1, 0) is N(1, 1, 0) = ∇ϕ(1, 1, 0) = 2i − 2j. The tangent plane at (1, 1, 0) has equation 2x − 2y = 0 or x = y. The normal line at (1, 1, 0) has parametric equations x = 1 + 2t, y = 1 − 2t, z = 0. 14. Let ϕ(x, y, z) = x2 − y 2 + z 2 . A normal vector at (1, 1, 0) is given by N = 2i − 2j. The tangent plane at (1, 1, 0) has equation y = x. The normal line at this point has parametric equations x = 1 + 2t, y = 1 − 2t, z = 0. 15. A normal vector is given by N = ∇(2x − cos(xyz))

= 2i. (1,π,1)

The tangent plane has equation x = 1 and the normal line at the point has parametric equations x = 1 + 2t, y = π, z = 1. 16. A normal vector at the point is N = ∇(3x4 + 3y 4 + 6z 4 )

= 12(i + j + 2k). (1,1,1)

The tangent plane at (1, 1, 1) has equation x + y + 2z = 4 and the normal line has parametric equations x = 1 + 12t, y = 1 + 12t, z = 1 + 24t. 17. Because ∇ϕ(x, y, z) = i + k, the normal to the surface ϕ(x, y, z) = c is the constant vector N(x, y, z) = i + k. The surface must therefore be the plane x + z = c.

15.4. DIVERGENCE AND CURL

15.4

385

Divergence and Curl

1. ∇·F=

∂ ∂ ∂ (x) + (y) + (2z) = 4, ∂x ∂y ∂z

i ∇ × F = ∂/∂x x

j ∂/∂y y

k ∂/∂z = 0i + 0j + 0k = O, 2z

∇ · (∇ × F) = 0. 2. ∇ · F = xz cosh(xyz), ∇ × F = −xy cosh(xyz)i + yz cosh(xyz)k ∂ ∂ (−xy cosh(xyz)) + (yz cosh(xyz)) ∇ · (∇ × F) = ∂x ∂z = cosh(xyz)(y − y) + sinh(xyz)(−xy 2 z + xy 2 z) = 0. 3. ∇ · F = 2y + xey + 2, ∇ × F = (ey − 2x)k, ∇ · (∇ × F) =

∂ y (e − 2x) = 0. ∂x

4. ∇ · F = xzexyz + 3z 2 , ∇ × F = −xyexyz i + (yexyz − 1)k, ∇ · (∇ × F) = −yexyz − xy 2 exyz + yexyz + xy 2 exyz = 0. 5. ∇ · F = cosh(x) + xz sinh(xyz) − 1, ∇ × F = (−1 − xy sinh(xyz))i − j + yz sinh(xyz)k, ∇ · (∇ × F) = (−y + y) sinh(xyz) + ((−xy 2 z + xy 2 z) cosh(xyz) = 0. 6. ∇ · F = cosh(x − z) + 2 + 1 = 3 + cosh(x − z), ∇ × F = −2yi − cosh(x − z)j, ∇ · (∇ × F) =

∂ ∂ (2 y) + (− cosh(x − z)) = 0. ∂x ∂y

386

CHAPTER 15. VECTOR DIFFERENTIAL CALCULUS

7. ∇ϕ = i − j + 4zk, i ∇ × (∇ϕ) = ∂/∂x 1

j ∂/∂y −1

k ∂/∂z = 0. 4z

8. ∇ϕ = (18yz + ex )i + 18xzj + 18xyk, ∇ × (∇ϕ) = (18x − 18x)i + (18y − 18y)j + (18z − 18z)k = O. 9. ∇ϕ = −6x2 yz 2 i − 2x3 z 2 j − 4x3 yzk, ∇ × (∇ϕ) = (−4x3 z + 4x3 z)i + (−12x2 yzi + 12x2 yz)j + (6x2 z 2 − 6x2 z 2 )k = O. 10. ∇ϕ = z cos(xz)i + x cos(xz)k,

∇ × (∇ϕ) =

i j ∂/∂x ∂/∂y z cos(xz) 0

k ∂/∂z x cos(xz)

= 0i + (cos(xz) − xz sin(xz) − cos(xz) + xz sin(xz))j + 0k = O. 11. ∇ϕ = (cos(x + y + z) − x sin(x + y + z))i − x sin(x + y + z)j − x sin(x + y + z)k,

∇ × (∇ϕ) = (−x cos(x + y + z) + x cos(x + y + z))i + (− sin(x + y + z) − x cos(x + y + z) + sin(x + y + z) + x cos(x + y + z))j + (− sin(x + y + z) − x cos(x + y + z) + sin(x + y + z) + x cos(x + y + z))k = O. 12. ∇ϕ = ex+y+z (i + j + k), ∇ × (∇ϕ) = (ex+y+z − ex+h+z )(i + j + k) = O.

15.5. STREAMLINES OF A VECTOR FIELD

387

13. Let F = f i + gj + hk. Then ∇ · (ϕF) = ∇ · (ϕf i + ϕgj + ϕhk) ∂ ∂ ∂ = (ϕf ) + (ϕg) + (ϕh) ∂x ∂y ∂z = ϕx f + ϕy g + ϕz h + ϕ(fx + gy + hz ) = ∇ϕ · F + ϕ(∇ · F). Next, i j k ∇ × (ϕF) = ∂/∂x ∂/∂y ∂/∂z ϕf ϕg ϕh ∂ ∂ = (ϕh) − (ϕg) i ∂y ∂z ∂ ∂ + (ϕf ) − (ϕh) j ∂z ∂x ∂ ∂ + (ϕg) − (ϕf k ∂x ∂y ∂ϕ ∂ϕ ∂ϕ ∂ϕ ∂ϕ ∂ϕ = h− g i+ f− h j+ g− f k ∂y ∂z ∂z ∂x ∂x ∂y ∂h ∂g ∂f ∂h ∂g ∂f +ϕ − i+ − j+ − k ∂y ∂z ∂z ∂x ∂x ∂y = ∇ϕ × F + ϕ(∇ × F).

15.5

Streamlines of a Vector Field

1. The streamlines satisfy dy dz dx = − 2 = . y z Integrate dx = −(1/y 2 ) dy to obtain x=

1 + c1 . y

Next integrate dx = (1/z) dz to get x = ln |z| + c2 . Using x as the parameter, we can write equations of the streamline for this vector field: 1 x = x, y = , z = ex−c2 . x − c1

388

CHAPTER 15. VECTOR DIFFERENTIAL CALCULUS For the streamline through (2, 1, 1), let x = 2. Then 1 and 1 = e2−c2 . 2 − c1

Thenc1 = 1 and c2 = 2, so this streamline has parametric equations x = x, y =

1 , z = ex−2 . x−1

2. Streamlines satisfy 1 dx = − dy = dz 2 and two integrations give us y = −2x + c1 , z = x + c2 . For the streamline through (0, 1, 1), choose c1 = c2 = 1. 3. We have x dx =

dy dz = . ex −1

Integrate xex dx = dy to obtain y = xex − ex + c1 . Integrate x dx = −dz to get x2 = −2z + c2 . Using x as parameter, streamlines are given by y = xex − ex + c1 , z =

1 (c2 − x2 ). 2

For the streamline through (2, 0, 4), we need e2 + c1 = 0 and 4 =

1 (c2 − 4). 2

Then c1 = −e2 and c2 = 12, so this streamline has parametric equations x = x, y = xex − ex − e2 , z =

1 (12 − x2 ). 2

4. Streamlines satisfy dy dx = , dz = 0. cos(y) sin(x) Integrate sin(x) dx = cos(y) dy and dz = 0 to get − cos(x) + c1 = sin(y) and z = c2 . To pass through (π/2, 0, −4) we need c1 = 0 and c2 = −4. This streamline, with x as parameter, is x = x, y = arcsin(− cos(x)), z = −4.

15.5. STREAMLINES OF A VECTOR FIELD

389

5. Streamlines satisfy dz dy =− . z 2e cos(y) This is the separable equation cos(y) dy = −2ez dz. Integrate this to get sin(y) = c2 − 2ez . We also have x √= c1 . For the streamline through 3, π/4, 0), we need c1 = 3 and c2 = 2 + 2/2. With y as parameter, this streamline is given by ! √ 1 2 x = 3, y = y, z = ln + 1 − sin(y) . 4 2 6. Streamlines satisfy dx dy dz =− = 3. 3x2 y z Integrate the equations 1 1 1 3 dx = − dy and dy = − 3 dz 2 x y y z to obtain

1 1 = −3 ln |y| + c1 and 2 ln |y| + c2 = 2 . x z For the streamline passing through (2, 1, 6), we need c1 = 1/2 and c2 = 1/36. With y as parameter, this streamline has parametric equations x=

2 6 , y = y, z = p . 1 + 6 ln(y) 1 + 72 ln(y)

7. Circular streamlines about the origin in the x, y − −plane can be written as x2 + y 2 = r2 , so x dx + y dy = 0. Then

dy dx = − , dz = 0. y x

A vector field having these streamlines is F(x, y) =

1 1 i − j. x y

390

CHAPTER 15. VECTOR DIFFERENTIAL CALCULUS

Chapter 16

Vector Integral Calculus 16.1

Line Integrals

1. On C, x = t, y = t and z = t3 , so Z Z 1 x dx − dy + z dz = (t(1) − (1) + t3 (3t2 )) dt C

Z 1 =

(t − 1 + 3t5 ) dt = 0

2. Z

Z 1

−4x dx + y dy − yz dz = C

(−4(−t2 )(−2t)02 − 0) dt

Z 1 =

−8t3 dt = −2

3. Z 2

Z (x + y) ds = C

p (2t 1 + 1 + 4t2 dt

√ p 2 1 26 2 2t 2 + 4t2 dt = (2 + 4t2 )3/2 = 6 3 0 0

Z 2 =

4. Parametric equations of C are x = t, y = 1 + t, z = 1 − 2t for 0 ≤ t ≤ 1. Then Z C

x2 z ds =

Z 1

√ t2 (1 − 2t) 6 dt

√ Z 1 2 1 = 6 (t − 2t3 ) dt = − √ . 6 0 391

392

CHAPTER 16. VECTOR INTEGRAL CALCULUS

5. Z 3

Z F · dR = C

(cos(t)i + t2 j + tk) · (i − 2tj + 0k) dt

Z 3 =

(cos(t) − 2t3 ) dt = sin(3) −

81 2

√ √ 28 6 4xy ds = 4t 6 dt = 3 C 1 Z 2

7. Parametrize C as x = 2 cos(t), y = 2 sin(t), z = 0 for 0 ≤ t ≤ 2π. Then Z Z 2π (2 cos(t)i + 2 sin(t)j) · (−2 sin(t)i + 2 cos(t)j) dt F · dR = 0

Z 2π =

(−4 cos(t) sin(t) + 4 cos(t) sin(t)) dt = 0. 0

8. Parametrize C by x = 1, y = t, z = t2 for 0 ≤ t ≤ 2 to get Z Z 2 p yz ds = t(t2 ) 1 + 4t2 dt C

Z 2 = 0

p √ 1 1 + 4t2 dt = (391 17 + 1). 120

The last integration can be carried out by an integration by parts. 9. Z 9

Z −xyz dz = C

√ −z z dz

4 9 2 422 = − z 5/2 = − 5 5 4

10.

Z 3

Z xz dy = C

t(−4t2 ) dt = −80

11. Parametrize the line segment as x = y = z = 1 + 3t for 0 ≤ t ≤ 1. The work done is Z Z 1 F · dR = ((1 + 3t)2 − 2(1 + 3t)2 + 1 + 3t)(3) dt C

1 (1 + 3t)3 27 (1 + 3t)2 = − =− . 2 3 2 0

16.2. GREEN’S THEOREM

393

12. Parametrize the shape and location of the wire by x = y = z = t for 0 ≤ t ≤ 3. The mass M is √ Z Z 3 √ 27 3 M= . δ(x, y, z) ds = 3t 3 dt = 2 C 0 Because the density function and the position of the wire are symmetric in the first octant, we will have x = y = z, so we need only compute Z 3 2 √ xδ(x, y, z) ds 27 3 0 Z 3 √ 2 = √ t(3t) 3 dt = 2. 27 3 0

The centroid is (2, 2, 2). 13. Take F(x) = f (x)i and R(t) = tj for a ≤ t ≤ b. The graph of the curve is defined by this position vector is the interval [a, b], and Z b

Z F · dR = C

16.2

f (x) dx. a

Green’s Theorem

1. The work done by F is ∂ ∂ (x) − (xy) dA work = xy dx + x dy = ∂y C Ω ∂x Z 1 Z 6x Z 4 Z 8−2x = (1 − x) dy dx + (1 − x) dy dx I

Z 1

Z 4 6x(1 − x) dx +

(8 − 2x)(1 − x) dx = −8 1

2. I F · dR

work = IC

(ex − y + x cosh(x)) dx + (y 3/2 + x) dy ZZ ∂ x ∂ 3/2 (y + x) − (e − y + x cosh(x)) dA = ∂x ∂y ZZ D = 2 dA = 2(area of D) = 2(62 )π = 72π

394

CHAPTER 16. VECTOR INTEGRAL CALCULUS

3. I

(− cosh(4x4 ) + xy) dx + (e−y + x) dy ZZ ∂ ∂ −y 4 = (e + x) − (− cosh(4x ) + xy) dA ∂y D ∂x ZZ Z 3Z 7 = (1 − x) dy dx (1 − x) dA =

work =

Z 3 6(1 − x) dx = −12

= 1

4. I

ZZ F · dR =

ZZ D

∂ ∂ (−x) − (2y) dA ∂x ∂y

(−3) dA = −3(area of D)

= D

= −3(16π) = −48π 5. I C

˙ FdR ZZ

∂ ∂ 2 (−2xy) − (x ) dA ∂y D ∂x ZZ Z 6 Z (22−2y)/5 = (−2y) dA = −2y dx dy

Z 6 (3y − 18)

= 1

(y+4)/5

2y dy = −40 5

6. Z

ZZ F · dR =

ZZ D =

∂ ∂ (x − y) − (x + y) dA ∂x ∂y

0 dA = 0 D

ZZ ∂ 2 F · dR = (8xy ) dA = 8y 2 dA. C D ∂x D To evaluate this integral, change to polar coordinates x = r cos(θ), y = r sin(θ), with 0 ≤ θ ≤ 2π and 0 ≤ r ≤ 4. We get ZZ Z 2π Z 4 8y 2 dA = 8r2 sin2 (θ)r dr dθ D

Z 2π = 0

sin2 (θ) dθ

Z 4

8r3 dr = 512π.

16.2. GREEN’S THEOREM

395

8. I

ZZ F · dR =

ZZ D =

∂ 2 ∂ (cos(2y) − e3y + 4x) − (x − y) dA ∂x ∂y

5 dA = 125 D

9. I

ZZ F · dR =

ZZ D =

∂ x ∂ x (−e sin(y)) − (e cos(y)) dA ∂x ∂y

(−ex sin(y) + ex sin(y)) dA = 0

10. ∂ 2 ∂ 2 (−xy ) − (x y) dA F · dR = ∂y C D ∂x ZZ Z π/2 Z 2 = (−y 2 − x2 ) dA = (−r2 )r dr dθ

π 2

Z 2

−r3 dr = −2π

11. I C

∂ ∂ (xy 2 − ecos(y) ) − (xy) dA ∂y D ∂x ZZ Z 3 Z 5−5x/3 = (y 2 − x) dA = (y 2 − x) dy dx D 0 0 Z 3 Z 3 1 5x 5x = 5− dx − x 5− ], dx 3 3 0 3 0 95 = 4

˙ = FdR

12. (a) By Green’s theorem with F = −yi, I ZZ ZZ ∂ ∂ −y dx = (0) − (−y) dA = dA = area of D. ∂y C D ∂x D (b) Now apply Green’s theorem with F = xj to get I ZZ ZZ ∂ ∂ F · dR = (x) − (0) dA = dA = area of D. ∂y C D ∂x D (c) Add the results of parts (a) and (b).

396

CHAPTER 16. VECTOR INTEGRAL CALCULUS

13. By Green’s theorem, I ZZ ∂ ∂ ∂ ∂u ∂ ∂u − dx + dy = − − dA ∂y ∂x ∂x ∂y ∂y C D ∂x ZZ 2 ∂ u ∂2u + 2 dA. = 2 ∂y D ∂x 14. Assume that C is the join of two curves in two ways. First, C has an upper piece, the graph of y = p(x), and a lower piece, the graph of y = q(x), for a ≤ x ≤ b. Then D consists of the points (x, y) with a ≤ x ≤ b, q(x) ≤ y ≤ p(x). And C also has a left piece, the graph of x = α(y), and a right piece, the graph of x = β(y), for c ≤ y ≤ d. Then D also consists of the points (x, y) such that c ≤ y ≤ d, α(y) ≤ x ≤ β(y). Now use both of these descriptions of D as follows. Using the second (looking at D from left to right), I Z d Z c g(x, y) dy = g(β(y), y) dy + g(α(y), y) dy. C

Note that, on the right part of C, we take y varying from d to c to maintain a counterclockwise orientation around C. Further, Z d Z β(y) ZZ ∂g ∂g dA = dx dy c α(y) ∂y D ∂x Z d = (g(β(y), y) − g(α(y), y)) dy. c

Then

∂ dA. ∂x C D This is “half” of the conclusion of Green’s theorem. For the rest, use the first description of C. Now, looking from the lower to the upper piece of C, and keeping in mind the counterclockwise orientation on C, we have I Z a Z b f (x, y) dx = f (x, p(x)) dx + f (x, q(x)) dx g(x, y) dy =

Z b =−

(f (x, p(x)) − f (x, q(x))) dx a

and ZZ

∂f dA = D ∂y

Z b Z p(x) a

q(x)

∂f dA ∂y

Z b (f (x, p(x)) − f (x, q(x))) dx.

= a

16.2. GREEN’S THEOREM

397

Then I

ZZ f (x, y) dx = −

∂f dA. D ∂y

Upon adding these two results, we have I

ZZ f (x, y) dx + g(x, y) dy =

∂f ∂g − ∂x ∂y

dA.

15. If C does not enclose the origin, then Green’s theorem applies and I C

F · dR ZZ

= ZZ D =

∂ x −y 2 − 2y − + x dA x2 + y 2 ∂y x2 + y 2

∂ ∂x

0 dA = 0. D

If C does enclose the origin, let K be a circle about the origin of sufficiently small radius r that K is in the region enclosed by C. Then, using the extended Green’s theorem and polar coordinates, we have I

I F · dR =

F · dR K Z 2π

−r sin(θ) 2 2 + r cos (θ) (−r sin(θ)) dθ r2 0 Z 2π r cos(θ) + − 2r sin(θ) (r cos(θ)) dθ r2 0 Z 2π = (1 − r2 cos2 (θ) sin(θ) − 2r2 sin(θ) cos(θ)) dθ =

=θ+

2π r3 cos2 (θ) − r2 sin2 (θ) = 2π. 3 0

16. If C does not enclose the origin, then I

ZZ F · dR =

∂ ∂x

x − y2 x2 + y 2

−

∂ ∂y

−y + 3x x2 + y 2

dA = 0,

because all terms in the integrand cancel. If C does enclose the origin, let K be a circle about the origin and entirely in the region enclosed by C.

398

CHAPTER 16. VECTOR INTEGRAL CALCULUS Then I

I F · dR =

F · dR K

Z 2π

−r sin(θ) + 3r cos(θ) (−r sin(θ)) dθ = r2 0 Z 2π r cos(θ) + − r sin(θ) (r cos(θ)) dθ r2 0 Z 2π = (1 − 4r2 sin(θ) cos(θ)) dθ = 2π. 0

17. By a calculation like those of Problems 15 and 16, obtain I F · dR = 0 C

if C does not enclose the origin, and I I F · dR = F · dR = 0 C

if C does enclose the origin, and K is a circle about the origin entirely in the region enclosed by C. 18. If C does not enclose the origin, then by Green’s theorem we get I F · dR = 0. C

If C encloses the origin, let K be a circle of radius r about the origin and entirely in the region enclosed by C. Using polar coordinates for K, a straightforward calculation yields I I F · dR = F · dR = 0. C

19. By arguing as in the last four problems, obtain I F · dR = 0 C

whether or not C encloses the origin.

16.3

Independence of Path and Potential Theory

1. First observe that

∂ 3 ∂ (y ) = (3xy 2 − 4) ∂y ∂x

16.3. INDEPENDENCE OF PATH AND POTENTIAL THEORY

399

on the entire plane, so this vector function is conservative. To find a potential function, we can begin with ∂ϕ = y3 ∂x and integrate with respect to x to get ϕ(x, y) = xy 3 + k(y). Then we must have ∂ϕ = 3xy 2 + k 0 (y) = 3xy 2 − 4 ∂y to conclude that k 0 (y) = −4, so we can choose k(y) = −4y. Then ϕ(x, y) = xy 3 − 4y is a potential function for F. 2. First, ∂ ∂ (6y + yexy ) = 6 + exy + xyexy = (6x + xexy ) ∂y ∂x for all (x, y). Then F has a potential function defined for all (x, y). To find a potential function, we can begin with ∂ϕ = 6y + yexy ∂x and integrate with respect to x to get ϕ(x, y) = 6xy + exy + k(y). Then we must have ∂ϕ = 6y + xexy + k 0 (y) = 6y + yexy . ∂y Then k 0 (y) = 0 and we can choose k(y) to be any constant, say k(y) = 0 for convenience, to get ϕ(x, y) = 6xy + exy . 3. F is conservative over the entire plane because ∂ ∂ (16x) = (2 − y 2 ) = 0 ∂y ∂x for all (x, y). To find a potential function, we can begin with ∂ϕ = 16x ∂x

400

CHAPTER 16. VECTOR INTEGRAL CALCULUS and integrate with respect to y to get ϕ(x, y) = 8x2 + k(y). Then

∂ϕ = k 0 (y) = 2 − y 2 ∂y

so k(y) = 2y − y 3 /3 and 1 ϕ(x, y) = 8x2 + 2y − y 3 3 is a potential function. 4. First, ∂ ∂ (2xy cos(x2 )) = 2x cos(x2 ) = (sin(x2 )) ∂y ∂x for all (x, y). This vector field is conservative. For a potential function, we can begin with ∂ϕ = sin(x2 ) ∂y to conclude that ϕ(x, y) = y sin(x2 ) + k(x). Then we need ∂ϕ = 2x cos(x2 ) = 2xy cos(x2 ) + k 0 (x). ∂x Then k 0 = 0 and we can choose k(y) = 0 to get the potential function ϕ(x, y) = y sin(x2 ). 5. First, if (x, y) 6= (0, 0), then ∂ 2x 4xy ∂ 2y = − = . ∂y x2 + y 2 (x2 + y 2 )2 ∂y x2 + y 2 Then F is conservative on the plane with the origin removed. For a potential function, we can begin with 2x ∂ϕ = 2 ∂x x + y2 and integrate with respect to x to get ϕ(x, y) = ln(x2 + y 2 ) + k(y). Then we need

∂ϕ 2y 2y = 2 = 2 + k 0 (y). 2 ∂y x +y x + y2 We can choose k(y) = 0 to obtain the potential function ϕ(x, y) = ln(x2 + y 2 ).

16.3. INDEPENDENCE OF PATH AND POTENTIAL THEORY

401

6. In this simple case, we can see a potential function by inspection: ϕ(x, y, z) = x2 − y 2 + z 2 for all (x, y, z). 7. By inspection, ϕ(x, y, z) = x − 2y + z is a potential function for F, for all (x, y, z). 8. A routine calculation shows that ∇ × F = O for all (x, y, z), so F is conservative. A potential function must satisfy ∂ϕ ∂ϕ ∂ϕ = yz cos(x), = z sin(x) + 1, = y sin(x). ∂x ∂y ∂z Integrate the first of these equations with respect to x to get ϕ(x, y) = yz sin(x) + k(y, z). Next, we need ∂k ∂ϕ = z sin(x) + 1 = z sin(x) + . ∂y ∂y Then

∂k =1 ∂y

Integrate this with respect to y to get k(y, z) = y + c(z). So far we have ϕ(x, y, z) = yz sin(x) + y + c(z). Finally, using the last equation, we have ∂ϕ = y sin(x) + c0 (z). ∂z Then c0 (z) = 0, so c(z) can be any constant. For convenience, choose c(z) = 0. Then ϕ(x, y, z) = yz sin(x) + y is a potential function for F. 9. We find that ∇ × F 6= O, so this vector field is not conservative. 10. ∇ × F 6= O, so F is not conservative. In Problems 11–20 we provide a potential function to use in evaluating the line integral, but do not include the details of finding this potential function.

402

CHAPTER 16. VECTOR INTEGRAL CALCULUS

11. By integrating, we find a potential function ϕ(x, y) = x3 (y 2 − 4y). Then

Z F · dR = ϕ(2, 3) − ϕ(1, 1) = −24 − 3 = −27. C

12. ϕ(x, y) = ex cos(y) and Z e2 F · dR = ϕ(2, π/4) − ϕ(0, 0) = √ − 1. 2 C 13. In any region not containing points of the y− axis, we can use the potential function ϕ(x, y) = x2 y − ln |y|. If C does not cross the x− axis, then Z F · dR = ϕ(2, 2) − ϕ(1, 3) C

= 8 − ln(2) − 3 + ln(3) = 5 + ln(3/2). 14. With ϕ(x, y) = x + 3y 2 − cos(y), we have Z F · dR = ϕ(1, 3) − ϕ(0, 0) = 29 − cos(3). C

15. ϕ(x, y) = x3 y 2 − 6xy 3 , so Z F · dR = ϕ(1, 1) − ϕ(0, 0) = −5. C

16. ϕ(x, y, z) = xy cos(xz), so Z F · dR = ϕ(1, 1, 7) − ϕ(1, 0, π) = cos(7). C

17. ϕ(x, y, z) = x − 3y 3 z, so Z F · dR = ϕ(0, 3, 5) − ϕ(1, 1, 1) = −403. C

18. ϕ(x, y, z) = −8xy 2 − 4zy, so Z F · dR = ϕ(1, 3, 2) − ϕ(−2, 1, 1) = −108. C

16.3. INDEPENDENCE OF PATH AND POTENTIAL THEORY

403

19. ϕ(x, y, z) = 2x3 eyz , so Z F · dR = ϕ(2, 1, −1) − ϕ(0, 0, 0) = 2e−2 . C

20. ϕ(x, y, z) = xy − 2x2 z + z 3 , so Z F · dR = ϕ(3, 1, 4) − ϕ(1, 1, 1) = −5. C

21. Let C be a smooth path of motion having position vector R(t) = x(t)i + y(t)j + z(t)k. Let L(t) be the sum of the potential and kinetic energies. Then m k R0 (t) k −ϕ((xt), y(t), z(t)) 2 m = R0 (t) · R0 (t) − ϕ(x(t), y(t), z(t)). 2

L(t) =

Then m ∂ϕ 0 ∂ϕ 0 ∂ϕ 0 (2R00 (t) · R0 (t)) − x (t) − y (t) − z (t) 2 ∂x ∂y ∂z = (mR00 (t) · R0 (t) − ∇ϕ · R0 (t) = (mR00 (t) − ∇ϕ) · R0 (t).

L0 (t) =

Now, ∇ϕ is the force acting on the particle, so by Newton’s second law, mR00 = ∇ϕ. Therefore L0 (t) = 0. 22. We want to show that, in the plane, a potential function exists for F(x, y) = f (x, y)i + g(x, y)j if ∂f ∂g = . ∂x ∂y We will use this condition to construct a potential function for F (on the relevant region D). First observe that, if K is a closed path in D, then by Green’s theorem, I ZZ ∂g ∂f F · dR = − dA = 0, ∂y K M ∂x R where M is the region enclosed by K. This means that C F · dR is independent of path in D. Fix a point P0 : (a, b) in D. Then, for any P in D, define Z F · dR

ϕ(x, y) = C

404

CHAPTER 16. VECTOR INTEGRAL CALCULUS in which C is any smooth path in D from P0 to P . We claim that ∇ϕ = F. To show this, we will first show that ∂ϕ = f (x, y). ∂x Choose ∆x small enough that (x + ∆x, y) is also in D. Now ϕ(x + ∆x, y) − ϕ(x, y) Z (x+∆x,y) Z (x,y) = F · dR − F · dR P0

Z (x,y)

Z (x+∆x,y)

(x,y)

F · dR

F · dR −

F · dR +

Z (x,y)

Z (x+∆x,y) F · dR

= (x,y)

Z (x+∆x,y) =

f (ξ, η) dξ + g(ξ, η) dη. (x,y)

The last line integral is over the horizontal segment from (x, y) to (x + ∆x, y), which can be parametrized by ξ = x + t∆x, η = y for 0 ≤ t ≤ 1. On this segment, dξ = (∆x) dt and dη = 0, because y is constant on this segment. Then Z 1 ϕ(x + ∆x, y) − ϕ(x, y) = ∆x f (x + t∆x, y) dt. 0

Then ϕ(x + ∆x, y) − ϕ(x, y) = ∆x

Z 1 f (x + t∆x, y) dt. 0

By the mean value theorem for integrals, there is some t0 in (0, 1) such that Z 1

f (x + t∆x, y) dt = f (x + t0 ∆x, y). 0

Then

ϕ(x + ∆x, y) − ϕ(x, y) = f (x + t0 ∆x, y). ∆x In the limit as ∆x → 0, we must have t0 → 0, so x + t0 ∆x → x and f (x + t0 ∆x, y) → f (x, y). Then ∂ϕ ϕ(x + ∆x, y) − ϕ(x, y) = lim ∂x ∆x→0 ∆x = lim f (x + t0 ∆x, y) = f (x, y). ∆x→0

16.4. SURFACE INTEGRALS

405

A similar argument, using a vertical segment from (x, y) to (x, y + ∆y), shows that ∂ϕ = g(x, y). ∂y

16.4

Surface Integrals

1. On the surface, z = 10 − x − 4y, so p √ dσ = 1 + (∂z/∂x)2 + (∂z/∂y)2 dA = 3 2 dA. Then ZZ

√ 3 2x dA

ZZ x dσ = Σ

√ Z √ Z 5/2 Z 10−4y 3 2 5/2 =3 2 (10 − 4y)2 dy x dx dy = 2 0 0 0 √ 5/2 √ 2 3 (10 − 4y) = 125 2. = 8 0

2. On the surface, z = x, so dσ =

1 + 12 + 02 dA =

√ 2 dA

and ZZ

y dσ = Σ

√ 2 2y dA

D √ √ Z 2Z 4 2 128 2 = 2 . y dx dy = 3 0 0

3. On Σ, dσ =

p 1 + 4x2 + 4y 2 dA,

and D is the annulus 2 ≤ x2 + y 2 ≤ 7. Then, using polar coordinates, ZZ Σ

Z 2π Z √7 p dσ = r 1 + 4r2 r dr dθ √ 0

√7 1 π 2 3/2 = 2π (1 + 4r ) = (293/2 − 27). √ 12 6 2

4. On the surface, z = (25 − 4x − 8y)/10, so dσ =

3 1 + (2/5)2 + (4/5)2 dA = √ dA. 5

406

CHAPTER 16. VECTOR INTEGRAL CALCULUS Then ZZ

3 (x + y) √ dA 5 D Z 1Z x 3 (x + y) dy dx =√ 5 0 0 Z 1 3 3 2 3√ =√ 5. x dx = 2 5 0 2

(x + y) dσ = Σ

5. On the surface, z 2 = x2 + y 2 , so 2z

∂z ∂z = 2x and 2z = 2y. ∂x ∂y

Then x ∂z y ∂z = and = . ∂x z ∂y z Then r dσ =

x 2 z

y 2 z

dA =

√ 2 dA.

Then ZZ

√ p 2 x2 + y 2 dA

ZZ z dσ = Σ

√ Z π/2 Z 4 2 28π √ = 2 2. r dr dθ = 3 0 2 6. On Σ, dσ =

√ 3 dA and z = x + y, so ZZ √ ZZ xy(x + y) dA xyz dσ = 3 D

√ Z 1Z 1 2 1 = 3 (x y + xy 2 ) dy dx = √ . 3 0 0 √ 7. On the surface, dσ = 1 + 4x2 dA, so Z ZZ p y dσ = y 1 + 4x2 dA Σ

Z 2Z 3 p Z 9 2p = y 1 + 4x2 dy dx = 1 + 4x2 dx 2 0 0 0 √ √ 9 = ln(4 + 17 + 4 17). 8

16.4. SURFACE INTEGRALS

407

p 8. On the surface, dσ = 1 + 4(x2 + y 2 ) dA, so ZZ ZZ p x2 dσ = x2 1 + 4(x2 + y 2 ) dA Σ

Z 2π Z 2 =

p r2 cos2 (θ) 1 + 4r2 r dr dθ

Z 2π =

cos2 (θ) dθ

Z 2

p 1 + 4r2 dr.

For the θ−integration, use the identity 1 (1 + cos(2θ)) 2

cos2 (θ) =

and for the r−integration, use the substitution u = 1 + 4r2 . These give us ZZ

Z 17 Z 1 1 2π (1 + cos(2θ) dθ (u3/2 − u1/2 ) du x dσ = 2 32 0 1 Σ √ π = (782 17 + 2). 240 2

9. On Σ, dσ =

√ 3 dA and z = x − y, so ZZ ZZ √ z dσ = 3(x − y) dA Σ

√ Z 1Z 5 √ (x − y) dy dx = −10 3. = 3 0

10. On Σ, dσ =

p 1 + 4y 2 dA and z = 1 + y 2 , so ZZ ZZ p xyz dσ = xy(1 + y 2 ) 1 + 4y 2 dA Σ

Z 1Z 1 = 0

1 = 2

p xy(1 + y 2 ) 1 + 4y 2 dy dx

Z 1

p y(1 + y 2 ) 1 + 4y 2 dy.

For this integral, let u = 1 + 4y 2 to get ZZ xyz dσ = Σ

1 1 2 32

Z 5 1

(u3/2 + 3u1/2 ) du =

1 16

√ 3 5 5− 5

408

CHAPTER 16. VECTOR INTEGRAL CALCULUS

16.5

Applications of Surface Integrals

1. The triangular shell is on the plane 6x + 2y + 3z = 6, which is the plane through the three given points. The projection of Σ onto the x, y−plane is the set D of points (x, y) such that 0 ≤ y ≤ 3 − 2x. On Σ, 2 z = 2 − y − 2x. 3 Then

r dσ =

4 + 4 dA 9

and ZZ m=

(xz + 1) dσ ZZ Σ 2 = x 2 − y − 2x + 1 dA 3 D Z 1 Z 3−3x 7 49 2 = . x 2 − y − 2x + 1 dy dx = 3 0 0 3 12

The first coordinate of the center of mass is Z 12 x= x(xz + 1) dσ 49 Σ ZZ 12 7 2 = x x 2 − y − 2x + 1 dy dx 49 3 3 D Z 1 Z 3−3x 4 2 = x x 2 − y − 2x + 1 dx 7 0 0 3 Z 1 12 24 2 12 4 12 = . − x + x + x dx = 7 7 7 35 0 The second coordinate is ZZ 12 y= y(xz + 1) dσ 49 Σ Z Z 4 1 3−3x 2 = y y 2 − y − 2x + 1 dy dx 7 0 0 3 Z 1 24 18 36 12 18 = − x − x2 + x3 − x4 + dx 7 7 7 7 7 0 33 = . 35 And, without all the details, the third coordinate is ZZ 12 24 z= z(xz + 1) dσ = . 49 35 Σ

16.5. APPLICATIONS OF SURFACE INTEGRALS

409

2. On Σ, 3 dA. z The part of the sphere above√the plane z = 1 projects onto the x, y−−plane onto the disk D of radius 2 2 about the origin. The mass of the shell is ZZ ZZ 3 1 p m= dA. K dA = 3K 2 + y2 ) z 9 − (x Σ D dσ =

1 + (x/z)2 + (y/z)2 dA =

Use polar coordinates: Z 2π Z 2√2

r dr dθ 9 − r2 0 0 i2√2 h p 2 = 6πK − 9 − r = 12Kπ.

m = 3K

√

By symmetry, x = y = 0. Finally, ZZ 1 z= Kz dσ m ZZ Σ 1 3 dA = Kz m z D 1 3K (area of D) = (24Kπ) = 2. = m m The center of mass is (0, 0, 2). 3. On the surface, dσ =

1 + (x/z)2 + (y/z)2 dA =

√ 2 dA.

Then √ ZZ K dσ = K 2

ZZ mass = Σ

Z 2π Z 3 D

√ r dr dθ = 9πK 2.

By symmetry, x = y = 0, and ZZ 1 z= z dσ m Σ √ √ Z Z 2K 2π 3 2 18 2Kπ = r dr dθ = = 2. m m 0 0 The center of mass is (0, 0, 2). 4. On the surface, dσ =

1 + 4(x2 + y 2 ) dA

410

CHAPTER 16. VECTOR INTEGRAL CALCULUS and Σ projects onto the x, y − −plane to the quarter annulus D consisting of points (x, y) with x ≥ 0, y ≥ 0 and 1 ≤ x2 + y 2 ≤ 9. The mass is ZZ

xy p dσ 1 + 4(x2 + y 2 ) Σ ZZ Z π/2 Z 3 = r3 sin(θ) cos(θ) dr dθ xy dA = 0 D 0 π/2 1 43 1 2 sin (θ) r = 2 4 1 0

= 10. Because the region is symmetric about the line y = x, and δ(x, y, z) = δ(r, θ) = √

1 1 + 4r2

is independent of θ, we must have x = y. Compute ZZ 1 x= xδ(x, y, z) dσ m Σ ZZ Z π/2 Z 3 1 1 121 2 = x y dA = . r4 cos2 (θ) sin(θ) dr dθ = 10 10 75 D 0 1 Finally, ZZ 1 zδ(x, y, z) dσ = (16 − x2 − y 2 )xy dA 10 Σ D Z π/2 Z 3 1 331 = (16 − r2 )r3 sin(θ) cos(θ) dr dθ = . 10 0 48 1

1 z= m

The center of mass is (121/75, 121/75, 331/48). 5. By symmetry of the surface and the density function, x = y = 0. Further, p dσ = 1 + 4x2 + 4y 2 dA. Then ZZ

1 + 4x2 + 4y 2 dσ

m= Σ

ZZ =

(1 − 4x2 + 4y 2 ) dA

D Z 2π Z √6

= 0

(1 + 4r2 )r dr dθ

= 2π(39) = 78π.

16.5. APPLICATIONS OF SURFACE INTEGRALS

411

Finally, ZZ 1 zδ(x, y, z) dσ m Σ ZZ 1 = (6 − x2 − y 2 )(1 + 4x2 + 4y 2 ) dA m D Z √6 Z 1 2π (6 − r2 )(1 + 4r2 )r dr dθ = m 0 0 162π 27 = = . m 13

6. By symmetry, x = y = 0. On the surface, dσ =

1 + (x/z)2 + (y/z)2 dA =

1 dA. z

The mass is ZZ m=

K Kz dσ = Kπ Σ

ZZ z(1/z) dA = D

1 1 area of D = . π 4

7. A unit normal to the plane x + 2y + z = 8 is 1 n = √ (i + 2j + k). 6 Then

1 F · n = √ (x + 2y − z). 6 On Σ, z = 8 − x − 2y, so F · n = 2x + 4y − 8. √ √ Further, dσ = 1 + 4 + 1 dA = 6 dA, so the flux of F across the surface is ZZ ZZ Z 4 Z 8−2y 128 F·n dσ = (2x+4y −8) dA = (2x+4y −8) dx dy = . 3 Σ D 0 0 8. The sphere x2 + y 2 + z 2 = 4 has unit normal n=

1 (xi + yj + zk). 2

Then F·n=

1 2 (x z − yz). 2

Further, dσ =

p 1 + (−x/z)2 + (−y/z)2 dA.

412

CHAPTER 16. VECTOR INTEGRAL CALCULUS Therefore the flux of the force across Σ is ZZ ZZ F · n dσ = (x2 − y) dA. Σ

Use polar coordinates on D, which is the disk 0 ≤ r ≤ Then the flux is Z 2π Z √3 0

16.6

(r2 cos2 (θ) − r sin(θ))r dr dθ =

9 4

Z 2π

√

3, 0 ≤ θ ≤ 2π.

cos2 (θ) dθ =

9π . 4

Gauss’s Divergence Theorem

1. ∇ · F = 1, so ZZ

ZZZ F · n dσ =

∇ · F dV

= volume of V =

256π 4 3 π4 = . 3 3

2. ∇ · F = 4 − 6 = −2, so ZZZ ∇ · F dV = −2(volume of M ) = −2π(22 )(2) = −16π. M

3. ∇ · F = 0, so ZZZ ∇ · F dV = 0. M

4. ∇ · F = 3(x2 + y 2 + z 2 ), so ZZZ ZZZ ∇ · F dV = 3 (x2 + y 2 + z 2 ) dV. M

Using spherical coordinates, we have Z 2π Z π Z 1

ZZZ ∇ · F dV =

Z 2π =

3ρ4 sin(ϕ) dρ dϕ dθ

Z π

sin(ϕ) dϕ 0 0 3 12π = (2π)(2) = . 5 5

Z 1

dθ

3ρ4 dρ

16.6. GAUSS’S DIVERGENCE THEOREM

413

5. With ∇ · F = 4, compute ZZZ 8π ∇ · F dV = 4(volume of )V = . 3 M

6. ∇ · F = 2 + x, so Z 4 ZZZ Z 3Z 2Z 4 (2 + x) dx = 96. (2 + x) dx dy dz = 6 ∇·F= 0

7. ∇ · F = 2(x + y + z), so, using cylindrical coordinates, we have ZZZ Z 2π Z √2 Z √2 ∇ · F dV = 2 (r cos(θ) + r sin(θ) + z)r dz dr dθ. 0

Do these integrations in turn. First, Z √2 √ 1 (r2 (cos(θ)+sin(θ))+rz) dz = r2 (cos(θ)+sin(θ))( 2−r)+ r(2−r2 ). 2 r Next, Z √2 √ 1 1 1 r2 (cos(θ) + sin(θ))( 2 − r) + r(2 − r2 ) dr = (cos(θ)+sin(θ))+ , 2 3 2 0 and finally, Z 2π 0

Therefore

1 1 (cos(θ) + sin(θ)) + 3 2

dθ = π.

ZZZ ∇ · F dV = 2π. M

8. ∇ · F = 1 + 2x, so compute ZZZ ZZZ ∇ · F dV = (1 + 2x) dV M

Z 2 =

ZZ dz

(1 + 2x) dA, D

where D is the plane region given in polar coordinates by 0 ≤ r ≤ 0 ≤ θ ≤ 2π. Now compute ZZZ ZZ (1 + 2x) dV = 2 (1 + 2r cos(θ))r dr dθ M

√

# √ Z 4 2 2π = 2 2π + cos(θ) dθ = 4π. 3 0 "

414

CHAPTER 16. VECTOR INTEGRAL CALCULUS

9. With the given conditions on F, Σ and M , we have ZZ ZZZ (∇ × F) · n dσ = ∇ · (∇ × F) dV = 0 Σ

because ∇ · (∇ × F) = 0. 10. Apply the divergence theorem to get ZZ ZZZ 1 1 F · n dσ = (∇ · R) dV 3 3 Σ M ZZZ 1 3 dV = volume of M. = 3 M

16.7

Stokes’s Theorem

1. The surface is a function of θ and ϕ, and (θ, ϕ) varies over its parameter domain 0 ≤ θ ≤ 2π, 0 ≤ ϕ ≤ π. This is a rectangle in the θ, ϕ− plane, with lower side L1 , upper right side L2 , top L3 and left side L4 . For orientation, imagine (θ, ϕ) moves around this rectangle counterclockwise, starting along L1 from the origin. We want to know what each side maps to on the surface. On L1 , the point (θ, 0) moves from (0, 0) to (2π, 0) as θ increases from 0 to 2π. The image point Σ(θ, 0) = (R cos(θ), R sin(θ), 0) moves from (R, 0, 0) along the circle x2 + y 2 = R2 in the plane z = 0, all the way around to end at (R, 0, 0). Then the point (2π, ϕ) moves up along L2 as ϕ increases from 0 to π. Image points of points (2π, ϕ) on L2 are Σ(2π, ϕ) = (R cos(ϕ), 0, R sin(ϕ)) which is a half-circle x2 + z 2 = R2 in the y = 0 plane, starting at (R, 0, 0) and ending at (−R, 0, 0). From (2π, π), (θ, ϕ) now moves left along L3 . The points are (θ, π), but θ varies from 2π to 0 to maintain counterclockwise orientation on the rectangle. The image points of L3 are Σ(θ, π) = (−R cos(θ), −R sin(θ), 0) asθ varies from2π to 0. The image of L3 on the surface consists of the points Σ(θ, π) = (−R cos(θ), −R sin(θ), 0),

16.7. STOKES’S THEOREM

415

and this point moves along the half-circle x2 + y 2 = R 2 from (−R, 0, 0) to (−R, 0, 0) in the z = 0 plane. Finally, on L4 , θ = 0 and ϕ varies from π to 0. Image points are Σ(0, ϕ) = (R cos(ϕ), 0, R sin(ϕ)) from (−R, 0, 0) to (R, 0, 0). Now trace out the image point on the surface as (θ, ϕ) moves over all four sides of the rectangle. This curve on the graph of the surface is the boundary of Σ. 2. The boundary curve C of the surface can be parametrized by x = 2 cos(t), y = 2 sin(t), z = 0 for 0 ≤ t ≤ 2π. On C, R(t) = 2 cos(t)i + 2 sin(t)j + 0k. Then F · dR = (−16 cos2 (t) sin2 (t) − 16 cos2 (t) sin2 (t)) dt = −32 cos2 (t) sin2 (t). Then

Z 2π

I F dR = C

Now evaluate

−32 cos2 (t) sin2 (t) dt = −8π.

RR Σ

(∇ × F) · n dσ. First, ∇ × F = −(x2 + y 2 )k.

Further, we can use the gradient to find a normal to Σ: ∇(x2 + y 2 + z 2 ) = 2(xi + yj + zk). This gives us the normal vector n= Then dσ =

1 (xi + yj + zk). 2

p 2 1 + (x/z)2 + (y/z)2 dA = dA. z

Then ZZ

ZZ (∇ × F) · n dσ = − Σ

(x2 + y 2 ) dA

D Z 2π Z 2

=− 0

r3 dr dθ = −8π.

416

CHAPTER 16. VECTOR INTEGRAL CALCULUS

In Problems 3–8, one side of Stokes’s theorem is computed in detail, with the choice being determined by which side appears to be the easiest computation. 3. The boundary curve C of the surface is the top of the parabolic bowl. This is the circle of radius 3 about (0, 0, 9). Parametrize C by x = 3 cos(t), y = 3 sin(t), z = 9 for 0 ≤ t ≤ 2π. On C, F(t) = 9 cos(t) sin(t)i + 27 sin(t)j + 27 cos(t)k. Further, dR = (−3 sin(t)i + 3 cos(t)j) dt. Then F · dR = (−27 cos(t) sin2 (t) + 81 cos(t) sin(t)) dt. A routine integration gives Z 2π

I F · dR = C

Evaluation of

(−27 cos(r) sin2 (t) + 81 cos(t) sin(t)) dt = 0.

RR Σ

(∇ × F) dσ involves considerably more labor.

4. Compute ∇ × F = i + j + k. A unit normal to Σ is 1 (xi + yj + zk) n= p 2 x + y2 + z2 This is

1 n= √ p (xi + yj − 2k). 2 x2 + y 2

Further, s dσ =

√ x2 y2 + dA = 2 dA. x2 + y 2 x2 + y 2

Then ZZ

ZZ (∇ · F) dσ = Σ

x+y−z p dA x2 + y 2 D

ZZ = D

x+y p −1 x2 + y 2

! dA.

p Here we used the fact that z = x2 + y 2 on σ. This integral is easily evaluated using polar coordinates to get ZZ (∇ × F) dσ = −16π. Σ

16.7. STOKES’S THEOREM

417

For the line integral around the boundary C of Σ, parametrize C by x = 4 sin(t), y = 4 cos(t), z = 4 for 0 ≤ t ≤ 2π. This orientation is determined by n. Notice that the normal in this problem has negative k component, unlike the normal in the preceding problem, where the surface was a parabolic bow. Parametrize C by Z 2π

I F · dR = C

(−16 cos(t) sin(t) − 16 cos2 (t)) dt = −16π.

5. The boundary curve of Σ is the circle x2 + y 2 = 6 in the x, y− plane. Parametrize C by √ √ x = 6 cos(t), y = 6 sin(t), z = 0 for 0 ≤ t ≤ 2π. Then

Z 2π

I F · dR = C

6 cos2 (t)6 sin2 (t) dt = 0.

6. The boundary C of the surface must be parametrized by three smooth curves. This is not difficult, but it is tedious. This leads us to try the surface integral side of Stokes’s theorem. First, ∇ × F = (x − y)i − yj − xk and

Finally, dσ = ZZ

1 n = √ (2i + 4j + k). 21 √ 21 dA, so ZZ (∇ × F) dσ = Σ

(x − 6y) dA D Z 2 Z 4−2y

(x − 6y) dx dy = −

= 0

7. The circulation is

H C

32 . 3

F · dR. Take Σ to be the disk 0 ≤ x2 + y 2 ≤ 1

with boundary C parametrized by x = cos(t), y = sin(t), z = 0 for 0 ≤ t ≤ 2π. The proper unit normal to Σ (a disk in the x, y− plane) is n = k. Now, ∇ × F = −azj + (2xy + 1)k.

418

CHAPTER 16. VECTOR INTEGRAL CALCULUS Then (∇ × F) · n = 2xy + 1. Further, dσ = dA, so I ZZ F · dR = (∇ × F) · n dσ C

Z 2π Z 1

ZZ =

(2xy + 1) dA =

(2r3 cos(θ) sin(θ) + r) dr dθ = π.

8. First, ∇ × F = −i − j − k. A normal to the surface is n = i + 4j + k so we have the unit normal 1 n = √ (i + 4j + k). 18 C is the boundary of the part of the plane x + 4y + z = 12 lying in the first octant, and consists of three line segments: the line from (0, 0, 12) to (12, 0, 0), then from (12, 0, 0) to (0, 3, 0), and then from (0, 3, 0) to (0, 0, 12). We can think of this portion of the plane in the first octant as having equation z = 12 − x − 4y with (x, y) varying over the triangle D bounded by the segment [0, 12] on the x− axis, the segment [0, 3] on the y− axis, and the line x + 4y = 12. D has area 1 area of D = (12)(3) = 18. 2 By Stokes’s theorem the circulation is I ZZ F · dT ds = (∇ × F) · n dσ. C

Now

6 (∇ × F) · n = − √ . 18

And dσ =k N k dx dy =

√ 18.

Then ZZ

6 − √ dx dy 18 D = −6(area of D) = −6(18) = −108.

(∇ × F) · n dσ = D

Chapter 17

Fourier Series 17.1

Fourier Series on [−L, L]

1. The Fourier coefficients are Z 1 3 4 dx = 8, a0 = 3 −3 Z 1 3 an = 4 cos(nπξ/3) dξ = 0, 3 −3 and bn =

1 3

Z 3 4 sin(nπξ/3) dξ = 0. −3

The Fourier series of 4 on [−3, 3] is just 1 a0 2 or 4, as we might expect. This converges to 4 on [−3, 3]. 2. The Fourier coefficients an are zero for n = 0, 1, 2, · · · because f is an odd function on [−1, 1]. Next, Z 1 2 (−1)n . bn = f (ξ) sin(nπξ) dξ = nπ −1 The Fourier series of f (x) on [−1, 1] is ∞

2X1 (−1)n sin(nπx). π n=1 n This series converges to ( −x for −1 < x < 1, 0 for x = ±1. 419

420

CHAPTER 17. FOURIER SERIES

3. Because cosh(πx) is an even function on [−1, 1], each bn = 0. Compute Z 1 2 a0 = cosh(πξ) dξ = sinh(π) π −1 and, for n = 1, 2, · · · , Z 1 2 sinh(π) (−1)n an = cosh(πξ) cos(nπξ) dξ = . π n2 + 1 −1 The Fourier series is ∞ X 1 2 sinh(π) (−1)n sinh(π) + cos(nπx). π π n2 + 1 n=1

This converges to cosh(πx) for −1 ≤ x ≤ 1. For Problems 4–10, we give just the Fourier series and analyze its convergence. 4. The series of f (x) on [−2, 2] is ∞

8 X 1 cos((2n − 1)πx/2). 2 π n=1 (2n − 1)2 This converges to 1 − |x| for −2 ≤ x ≤ 2. 5. The series of f (x) on [−π, π] is ∞

16 X 1 sin((2n − 1)x), π n=1 2n − 1 converging to   −4 4   0

for −π < x < 0, for 0 < x < π, for x = π and for x = −π.

6. Because sin(2x) is odd and periodic of period π, then this is itself the Fourier expansion of sin(2x) on [−π, π]. 7. The Fourier series of f (x) on [−2, 2] is ∞ 13 X 16 4 + (−1)n cos(nπx/2) + sin(nπx/2) . 3 (nπ)2 nπ n=1 This converges to ( f (x) 7

for −2 < x < 2, for x = 2 and for x = −2.

17.1. FOURIER SERIES ON [−L, L]

421

Convergence at the endpoints is determined by 1 1 (f (−2+) + f (2−)) = (9 + 5) = 7. 2 2 8. The Fourier series of f (x) on [−5, 5] is ∞ X 1 a0 + [an cos(nπx/5) + bn sin(nπx/5)] , 2 n=1

where 1 a0 = 5

Z 5 f (ξ) dξ = −5

71 6

and, for n = 1, 2, · · · , an =

1 5

Z 5 f (ξ) cos(nπξ/5) dξ = −5

55(−1)n − 5 (nπ)2

and bn =

1 5

Z 5 f (ξ) sin(nπξ/5) dξ = −5

n2 π 2 − 50 + (50 − 21n2 π 2 )(−1)n . n3 π 3

This Fourier series converges to  −x for −5 < x < 0,    1 + x2 for 0 < x < 5,  1/2 at x = 0,    31/2 for x = 5 and for x = −5. 9. The Fourier expansion of f (x) on [−π, π] is ∞ 3 2X 1 + sin((2n − 1)x) 2 π n=1 2n − 1

converging to   for −π < x < 0, 1 2 for 0 < x < π,   3/2 at x = 0, x = π and at x = −π. 10. The Fourier series of f (x) on [−π, π] is ∞ 4 X (−1)n 2 − sin(x) − cos(nx), π π n=1 4n2 − 1

converging to f (x) for −π < x < π and to 0 at x = ±π.

422

CHAPTER 17. FOURIER SERIES

11. The Fourier series is ∞ X 1 (−1)n+1 sin(3) + 6 sin(3) cos(nπx/3), 3 n2 π 2 − 9 n=1

converging to cos(x) on [−3, 3]. 12. The series is ∞ X 1 − 2(−1)n 1 1 − (−1)n cos(nπx) + sin(3) − sin(nπx) , 3 n2 π 2 nπ n=1 converging to  1−x    0  1/2    1

for −1 < x < 0, for 0 < x < 1, for x = 0, for x = ±1.

13. The Fourier series converges to  3/2 for x = ±3,      2x for −3 < x < −2,    −2 for x = −2, 0 for −2 < x < 1,      1/2 for x = 1,    2 x for 1 < x < 3. 14.  (1 − 2π)/2 for x = ±π,     3/2 for x = 1,  2x − 2 for −π < x < 1,    3 for 1 < x < π 15.  2  (2 + π )/2 for x = ±π,  x2 for −π < x < 0, 1 for x = 0,    2 for 0 < x < π 16.   (cos(2) + sin(2))/2 for x = ±2,  cos(x) for −2 < x < 0, 1/2 for x = 0,    sin(x) for 0 < x < 2

17.2. SINE AND COSINE SERIES 17.

  −1 0   1

18.

  1      0 2    1/2    3/2

423

for −4 < x < 0, for x = ±4 and for x = 0, for 0 < x < 4

for x = ±1 and for 1/2 < x < 3/4, for −1 < x < 1/2, for 3/4 < x < 1, for x = 1/2, for x = 3/4

19.  −4    3/2 5/2    f (x)

17.2

for x = ±4, for x = −2, for x = 2, for all other x in [−4, 4]

Sine and Cosine Series

1. The cosine series is just 4, a single term. The sine series is ∞

16 X 1 sin((2n − 1)πx/3), π n=1 2n − 1 converging to 0 for x = 0 or x = 3 and to 4 for 0 < x < 3. 2. The cosine series is ∞

−

4 X (−1)n cos((2n − 1)πx/2), π n=1 2n − 1

converging to   1 0   −1

for 0 ≤ x < 1, for x = 1, for 1 < x ≤ 2.

The sine series is ∞

2X1 (1 + (−1)n − 2 cos(nπ/2)) sin(nπx/2), π n=1 n converging to   1 0   −1

for 0 < x < 1, for x = 0, 1, 2, for 1 < x < 2.

424

CHAPTER 17. FOURIER SERIES

3. The cosine series is 1 cos(x) + 2

∞ X n=1,n6=2

2n sin(nπ/2) cos(nx/2) π(n2 − 4)

converging to  0 for 0 < x < π,    −1/2 for x = π, cos(x) for π < x < 2π,    1 for x = 2π. The sine series is ∞ X 2n 2 sin(x/2) − ((−1)n + cos(nπ/2)) sin(nx/2), − 2 − 4) 3π (n n=3

converging to  0 for 0 < x < π,    −1/2 for x = π,  cos(x) for π < x < 2π,    0 for x = 2π. 4. The cosine series is ∞

1−

8 X 1 cos((2n − 1)πx), 2 π n=1 (2n − 1)2

converging to 2x for 0 ≤ x ≤ 1. The sine series is ∞

−

4 X (−1)n sin(nπx), π n=1 n

converging to 2x for 0 ≤ x < 1 and to 0 at x = 1. 5. The cosine series is 4 16 (−1)n + cos(nπx/2), 3 π 2 n2 converging to x2 for 0 ≤ x ≤ 2. The sine expansion is −

∞ 8 X (−1)n 2(1 − (−1)n ) + sin(nπx/2), π n=1 n n3 π 2

converging to x2 for 0 ≤ x < 2 and to 0 at x = 2.

17.2. SINE AND COSINE SERIES

425

6. The cosine series is −1 −

∞ X 1 1 − (−1)n e−1 cos(nπx), +2 e 1 + n2 π 2 n=1

converging to e−x for 0 ≤ x ≤ 1. The sine series is 2π

∞ X n=1

n n −1 (1 − (−1) e ) sin(nπx), 1 + n2 π 2

converging to e−x for 0 < x < 1 and to 0 at x = 0 and at x = 1. 7. The cosine expansion is ∞ 1 X 1 + [−6(1 + (−1)n ) + 12 cos(2nπ/3) + 4nπ sin(2nπ/3)] cos(nπx/3), 2 n=1 n2 π 2

converging to   x 1   2−x

for 0 ≤ x < 2, for x = 2, for 2 < x ≤ 3.

The sine series is ∞ X n=1

1 n2 π 2

[12 sin(2nπ/3) − 4nπ cos(2nπ/3) + 12nπ(−1)n ] sin(nπx/3),

converging to  x    1  2−x    0

for 0 ≤ x < 2, for x = 1, for 2 < x < 3, for x = 3.

8. The cosine series is ∞

1 1X1 − + cos(nπ/5) sin(2nπ/5) cos(nπx/5), 5 π n=1 n converging to   1      1/2 0    −1/2    −1

for 0 ≤ x < 1, for x = 1, for 1 < x < 3, for x = 3, for 3 < x < 5.

426

CHAPTER 17. FOURIER SERIES The sine series is ∞

4X 1 (1 + (−1)n − 2 cos(nπ/5) cos(2nπ/5)) sin(nπx/5), π n=1 2n converging to   1      1/2 0   −1/2    −1

for 0 < x < 1, for x = 1, for 1 < x < 3 and for x = 0 and x = 5, for x = 3, for 3 < x < 5.

9. The cosine expansion is ∞ 4 5 16 X 1 + 2 cos(nπ/4) − sin(nπ/4) cos(nπx/4), 6 π n=1 n2 n3 π converging to x2 for 0 ≤ x < 1 and to 1 for 1 < x ≤ 4. The sine series is ∞ X 16 61 2 sin(nπ/4) + 3 3 (cos(nπ/4) − 1) − nπ sin(nπx/4), n2 π 2 n π (−1)n n=1 converging to x2 for 0 ≤ x < 1, to 1 if 1 < x < 4, and to 0 at x = 4. 10. The cosine series is −1 −

∞ 24 X 1 4 n n (1 − (−1) ) cos(nπx/2), 2(−1) + π 2 n=1 n2 n2 π 2

converging to 1 − x2 for 0 ≤ x ≤ 2. The sine series is ∞ 2X1 48 n n 1 + 7(−1) − 2 2 (−1) sin(nπx/2), π n=1 n n π converging to 1 − x2 for 0 < x < 2 and t 0 at x = 0 and at x = 2. 11. The Fourier cosine expansion of sin(x) on [0, π] is ∞ 2 4X 1 − cos(2nx). 2 π π n=1 4n − 1

This converges to sin(x) for 0 ≤ x ≤ π. Put x = π/2 into this series to obtain ∞ 4X 1 2 cos(π/2) = 0 = − cos(nπ). π π n=1 4n2 − 1

17.2. SINE AND COSINE SERIES

427

Upon putting cos(nπ) = (−1)n we obtain ∞ X π 2 1 π (−1)n = − 1 = − . 2−1 4n 4 π 2 4 n=1 12. Expand Go (x) in a Fourier series on [−L, L]. This series will be ∞ X 1 a0 + [an cos(nπx/L) + bn sin(nπx/L)], 2 n=1

in which an and bn are the Fourier coefficients for Go (x) on this interval. Now example these coefficients. First, Z 1 L a0 = Go (x) dx = 0 L −L because Go (x) is an odd function on [−L, L]. For n = 1, 2, · · · , Go (x) cos(nπx/L) is also odd (product of an even and an odd function), so Z 1 L an = Go (x) cos(nπx/L) dx = 0. L −L Therefore the Fourier expansion of Go (x) contains only sine terms. Further, Z 1 L bn = Go (x) sin(nπx/L) dx L −L Z 2 L Go (x) sin(nπx/L) dx = L 0 Z 2 L = g(x) sin(nπx/L) dx, L 0 because Go (x) sin(nπx/L) is an even function (product of two odd functions), and Go (x) = g(x) for 0 ≤ x ≤ L. This is the Fourier sine coefficient of g(x) on [0, L], and the Fourier series of Go (x) on [−L, L] yields the Fourier sine expansion of g(x) on [0, L]. 13. Use an argument similar to that made for Problem 12, except now the Fourier coefficients of Ge (x) satisfy Z −L L an = Ge (x) cos(nπx/L) dx L −L Z 2 L = g(x) cos(nπx/L) dx L 0 and

Z 1 L Ge (x) sin(nπx/L) dx = 0 L −L because Ge (x) cos(nπx/L) is even and Ge (x) sin(nπx/L) is odd, and Ge (x) = g(x) for 0 ≤ x ≤ L. bn =

428

CHAPTER 17. FOURIER SERIES

17.3

Integration and Differentiation of Fourier Series

1. The Fourier expansion of f (x) on [−π, π] is ∞ X 1 (−1)n (−1)n − 1 π+ cos(nx) + sin(nx) . 4 n2 π n n=1 Because f is continuous and piecewise smooth on [−π, π], this series converges to f (x) for −π < x < π. f (x) satisfies the conditions of the theorem on term by term integration, so this series can be integrated term by term to obtain Z x π f (ξ) dξ = (x + π) 4 −π ∞ X 1 (−1)n 1 n ((−1) − 1) sin(nx) + cos(nx) − 2 . + n3 π n2 n n=1 2. The function is continuous and piecewise smooth on [−1, 1]. Further, f 00 (x) exists on (−1, 1) and f (−1) = f (1) and f 00 (x) exists on (−1, 0) and (0, 1). The Fourier series of f (x) on [−1, 1] is ∞ 1 4 X 1 − cos((2n − 1)πx). 2 π 2 n=1 (2n − 1)2

Termwise differentiation yields the series ∞

4X 1 sin((2n − 1)πx). π n=1 2n − 1 It is routine to check that this is the Fourier expansion, on [−1, 1], of ( −1 for −1 < x < 0, f 0 (x) = 1 for 0 < x < 1. 3. The Fourier expansion of f (x) on [−π, π] is 1−

∞ X 1 (−1)n cos(x) − 2 cos(nx), 2 n2 − 1 n=2

converging to x sin(x) for −π ≤ x ≤ π. The function is continuous on [−π, π], and f 0 (x) is piecewise continuous. Further, f (−π) = f (π) and f 00 (x) exists on (−π, π). We can differentiate the Fourier series term by term to obtain, for −π < x < π, f 0 (x) = sin(x) + x cos(x) ∞ X 1 n(−1)n = cos(x) + 2 cos(nx). 22 n2 − 1 n=2

17.3. INTEGRATION AND DIFFERENTIATION OF FOURIER SERIES429 4. The Fourier expansion of x2 on [−3, 3] is ∞

36 X (−1)n cos(nπx/3). π 2 n=1 n2

This converges to x2 for −3 ≤ x ≤ 3. Further, f (3) = f (−3). f is continuous and piecewise smooth, and f 00 exists on (−3, 3). Term by term differentiation yields ∞

2x = −

12 X 1 sin(nx) π n=1 n

for −3 < x < 3. It is routine to verify this by expanding 2x in a Fourier series on [−3, 3]. 5. Let the Fourier coefficients of f on [−L, L]be an and bn , as usual. Let the Fourier coefficients of f 0 (x) be An , Bn . Notice that A0 =

2 L

Z L

f 0 (ξ) dξ = f (L) − f (L ) = 0

−L

because f (L) = f (−L). Now we will develop some inequalities aimed at showing uniform convergence of the Fourier series of f (x) on [−L, L]. Begin with an integration by parts: Z 1 L 0 f (ξ) cos(nπξ/L) dξ An = L −L =

1 n1 L [f (x) cos(nπx/L)]−L + L πL

Z L f (ξ) sin(nπξ/L) dξ. −L

Now, f (L) cos(nπ) − f (−L) cos(−nπ) = 0 for integer n, again because f (L) = f (L ). Then nπ 1 An = L L

Z L f (ξ) cos(nπξ/L) dξ = −L

nπ an L

for n = 1, 2, · · · . A similar integration by parts gives us Bn = −

nπ an . L

Observe that 0≤

|An | −

1 n

= A2n −

2 1 |An | + 2 n n

430

CHAPTER 17. FOURIER SERIES and, similarly, 0 ≤ Bn2 −

2 1 |Bn | + 2 . n n

Add these two inequalities to get 2 2 (|An | + |Bn |) ≤ A2n + Bn2 + 2 . n n Multiply this by 1/2 to obtain 1 1 1 (|An | + |Bn |) ≤ (A2n + Bn2 ) + 2 . n 2 n On the left, insert |An | =

nπ|an | nπ|an | and |Bn | = L L

to obtain |an | + |bn | ≤ Now,

∞ X

L 2 L 1 (A + Bn2 ) + . 2π n π n2

A2n and

n=1

∞ X

Bn2

n+1

both converge, by Bessel’s inequality. Therefore, by the comparison test for nonnegative series, we conclude that ∞ X

(|an | + |bn |)

n=1

converges. Finally, observe that, on [−L, L], |an cos(nπx/L)| + bn sin(nπx/L)| ≤ |an | + |bn |. By what is often known as the Weierstrass M - test (in this case with Mn = |an | + |bn | the Fourier series of f (x) on [−L, L] converges uniformly on this interval.

17.4

Properties of Fourier Coefficients

1. The argument is like that for sine series, but is notationally a little messier because of the additional constant term in the cosine expansion. Let N X 1 SN = A0 + An cos(nπx/L). 2 n=1

17.4. PROPERTIES OF FOURIER COEFFICIENTS

431

Now Z L 0≤

(g(x) − SN (x)) dx 0

Z L

2 (g(x))2 dx − 2g(x)SN (x) + SN (x) dx

= 0

! N X 1 g(x) (g(x)) dx − 2 = A0 + An cos(nπx/L) dx 2 0 0 n=1 ! ! Z L N N X X 1 1 A0 + A0 + An cos(nπx/2) An cos(nπx/L) dx + 2 2 0 n=1 n=1 Z L Z L = (g(x))2 dx − A0 g(x) dx Z L

Z L

−2

N Z L X n=1

! N N N X 1 2 XX A0 An cos(nπx/L) dx An Am cos(nπx/L) cos(mπx/L) + A + 4 0 n=1 m=1 n=1

Z L + 0

Z L =

g(x)An cos(nπx/L) dx

(g(x))2 dx −

N X L 2 A0 − L A2n 2 n=1

N L 2 XL 2 A0 + A 4 2 n n=1 Z L N LX 2 L A . = (g(x))2 − A20 − 4 2 n=1 n 0

Here we have used the fact that Z L cos(nπx/L) dx = 0 0

and that Z L 0

( L/2 cos(nπx/L) cos(mπx/L) dx = 0

if n = m, if n 6= m,

Upon rearranging terms in the first and last lines (which are connected by an inequality), we have Z N 2 L 1 2 X 2 A0 + An ≤ (g(x))2 dx. 2 L 0 n=1

432

CHAPTER 17. FOURIER SERIES

2. Use Bessel’s inequality to conclude that N X

A2n and

n=1

∞ X

Bn2

n=1

converge. Then lim A2n = lim Bn2 = 0.

n→∞

Then lim An = lim Bn = 0.

n→∞

Upon inserting the integral expressions for An and Bn , we obtain Riemann’s lemma. Problems 3 and 4 obtained by adapting the argument of the text to the notation cosine expansions on [0, L] and Fourier series on [−L, L], similar to the solution of Problem 1.

17.5

Phase Angle Form

1. (αf + βg)(x + p) = αf (x + p) + βg(x + p) = αf (x) + βg(x) = (αf + βg)(x). 2. g(x + p/α) = f (α(x + p/α)) = f (αx + p) = f (αx) = g(x) and h(x + αp) = f ((x + αp)/α) = f (x/α + p) = f (x/α) = h(x). 3. f (x + p + h) − f (x + p) h→0 h f (x + h) − f (x) = lim = f 0 (x). h→0 h

f 0 (x + p) = lim

4. Expand f (x) in a Fourier series on [0, 2] to get ∞

1−

2X1 sin(nπx). π n=1 n

One way to obtain this expansion using the standard formulas for the Fourier coefficients on an interval [−L, L] is write f (x), which has period 2, on the interval [−1, 1]: ( 2 + x for −1 ≤ x < 0, f (x) = x for 0 < x < 1.

17.5. PHASE ANGLE FORM

433

One way to write the phase angle form is to use the identity π sin(nπx) = cos nπx − 2 to obtain

∞

1−

2X1 π . cos nπx − π n=1 n 2

The amplitude spectrum points are (0, 1) and (nπ, −1/nπ) for n = 1, 2, · · · . 5. The Fourier series is ∞ 1 2X 1 + sin((2n − 1)πx), 2 π n=1 2n − 1

with phase angle form ∞

π 2X 1 cos (2n − 1)πx − . π n=1 2n − 1 2

Points of the amplitude spectrum are (0, 1), (nπ, 1/((2n − 1)π)) for n = 1, 2, · · · . 6. The Fourier series is 16 +

∞ 48 X 1 π cos(nπx/2) − sin(nπx/2) π 2 n=1 n2 n

and the phase angle form is ∞ √ nπx 48 X 1 + n2 π 2 16 + 2 cos + arctan(nπ) . π n=1 n2 2 Points of the amplitude spectrum are (0, 16),

! √ nπ 24 1 + n2 π 2 , . 2 n2 π 2

7. The Fourier series is ∞ 19 X 2 + αn cos(nπx/2) + βn sin(nπx/2), 8 n2 π 2 n=1

where αn = nπ sin(3nπ/2) + cos(3nπ/2)

434

CHAPTER 17. FOURIER SERIES and βn = sin(3nπ/2) −

nπ − nπ cos(3nπ/2). 2

The phase angle form is ∞

nπx 19 1 X + 2 dn cos + δN , 8 π n=1 2 where dn =

8 + 5n2 π 2 − 12nπ sin(3nπ/2) + 4(n2 π 2 − 2) cos(3nπ/2)

and

δn = arctan

nπ/2 + nπ cos(3nπ/2) − sin(3nπ/2) nπ sin(3nπ/2) + cos(3nπ/2) − 1

8. The Fourier series is ∞

n 8X sin(2nπx) 2 π n=1 4n − 1 and the phase angle form is ∞

n π 8X cos 2nπx − . π n=1 4n2 − 1 2 9. We can write

( x for 0 ≤ x < 1, f (x) = x − 2 for 1 < x < 2

and f (x + 2)f (x), so f has period 2. The Fourier series is ∞

2 X (−1)n+1 sin(nπx). π n=1 n The phase angle form is ∞

2X1 π cos nπx + (−1)n+1 . π n=1 n 2 10. We can write

( −kx for −1 ≤ x < 0, f (x) = kx for 0 ≤ x < 1,

with f (x + 2) = f (x). The Fourier series is ∞ k X 2k + ((−1)n − 1) cos(nπx). 2 n=1 n2 π 2

17.6. COMPLEX FOURIER SERIES

435

The phase angle form is ∞ k 2k X + 2 cos((2n − 1)πx − π). 2 π n=1

11. We can write

  1 for 0 ≤ x < 1, f (x) = 2 for 1 < x < 3,   1 for 3 < x < 4

with f (x + 4) = f (x). The Fourier series is ∞ 3 X (−1)n + cos((2n − 1)πx/2). 2 n=1 2n − 1

The phase angle form is ∞ 2X πx π 3 + cos (2n − 1) + (1 − (−1)n ) . 2 π n=1 2 2

12. Write

( k f (x) = 0

for 0 ≤ x < 1, for 1 < x < 2

with f (x + 2) = f (x). The Fourier series is ∞ k 2k X 1 + sin((2n − 1)πx). 2 π n=1 2n − 1

The phase angle form is ∞ k 2k X 1 π + cos (2n − 1)πx − . 2 π n=1 2n − 1 2

17.6

Complex Fourier Series

1. Compute 1 d0 = 3 and dn =

1 3

Z 3

Z 3 2t dt = 3 0

2xe2nπix/3 dx =

3 i. nπ

The complex Fourier series of f (x) is 3+

3i π

∞ X n=−∞,n6=0

1/n e

2nπix/3

436

CHAPTER 17. FOURIER SERIES converging to ( 3 for x = 0 or x = 3, 2x for 0 < x < 3. Points of the frequency spectrum are 2nπ 3 (0, 3), , , 3 nπ in which n is a nonzero integer.

2. The complex Fourier series of f is 4 + 3

∞ X

n=−∞,n6=0

2i 2 − n2 π 2 nπ

enπix .

This converges to ( 2 x2

for x = 0 and for x = 2, for 0 < x < 2.

Points of the frequency spectrum are r (0, 4/3), nπ, 2

1 1 + 2 2 4 4 n π n π

! .

3. The complex Fourier series of the function is 1 3 − 4 2π

∞ X n=−∞,n6=0

1 (sin(nπ/2) + (cos(nπ/2) − 1)i)enπix/2 , n

converging to   1/2 0   1

for x = 0, 1 or 4, for 0 < x < 1, for 1 < x < 4.

Points of the frequency spectrum are q nπ 1 2 2 (0, 3/4), , sin (nπ/2) + (cos(nπ/2) − 1) . 2 2nπ 4. The complex series is 1 3i − − 2 π

∞ X n=−∞,n6=0

1 nπix/3 e , n

17.6. COMPLEX FOURIER SERIES

437

converging to ( −2 for x = 0 and for x = 6, 1 − x for 0 < x < 6. Points of the frequency spectrum are (0, 1/2),

nπ 3 , 2 nπ

5. The complex Fourier series is 1 3i + 2 π

e(n−1)πix/2 ,

n=−∞,n6=0

converging to   1/2 for x = 0, 2, 4, −1 for 0 < x < 2,   2 for 2 < x < 4. Points of the frequency spectrum are (0, 1/2),

nπ 3 , 2 (2n − 1)π

6. The complex Fourier series is ∞ X

1 − e−5 2nπix/5 e , 5 + 2nπi n=−∞ converging to

2nπ 1 − e−5 p 2 2 , √ 25 + 4n π . 3 29

7. The complex Fourier series is 2 1 − 2 2 π

∞ X n=−∞,n6=0

1 e(2n−1)πix , (2n − 1)2

converging to f (x) for 0 ≤ x ≤ 2. Points of the frequency spectrum are 2 1 (0, 1/2), nπ, 2 . π (2n − 1)2

438

CHAPTER 17. FOURIER SERIES

17.7

Filtering of Signals

1. The complex Fourier coefficients of f are d0 = 0 and, for n 6= 0, Z 0 Z 2 i 1 −nπit/2 −nπit/2 −e dt + e dt = dn = ((−1)n − 1). 4 −2 nπ 0 The complex Fourier series is ∞ X n=−∞,n6=0

i ((−1)n − 1)enπit/2 . nπ

After some routine calculation using Euler’s formula, we obtain the series ∞

4X 1 sin((2n − 1)πt/2). π n=1 2n − 1 The N th partial sum is N

SN (t) =

4X 1 sin((2n − 1)πt/2). π n=1 2n − 1

The N th Cesáro sum is formed by inserting factors 1 − |n|/N : N 4X 2n − 1 1 σN (t) = 1− sin((2n − 1)πt/2). π n=1 N 2n − 1 2. The N th partial sum of the Fourier series of f is SN (t) =

N 13 X (an cos(nπt/2) + bn sin(nπt/2)), + 8 n=1

where an = −

2 (−nπ sin(nπ/2) + cos(nπ/2)(−4 − n2 π 2 ) + 4(−1)n ) n3 π 3

and bn =

2 n3 π 3

(sin(nπ/2)(4 − n2 π 2 ) + nπ cos(nπ/2) + 4(−1)n ).

σN (t) is obtained by putting a factor of 1 − |n|/N into SN (t). 3. We obtain SN (t) = and σN (t) =

N X n=1

N X 2

nπ n=1

1−

(cos(nπ/2) − (−1)n ) sin(nπt)

n 2 (cos(nπ/2) − (−1)n ) sin(nπt). N nπ

17.7. FILTERING OF SIGNALS

439

4. The N th partial sums are SN (t) = +

1 sin(3) 6 N X −1 n2 π 2 − 9 n=1

[3 sin(3)(−1)n cos(nπt/3) + nπ(−1 + (−1)n ) cos(3) sin(nπt/3)]

σN (t) is obtained by putting a factor of 1 − n/N ) into the summation of SN (t). 5. We find that N 5 − 6(−1)n 17 X 1 − (−1)n + sin(nπt) cos(nπt) + SN (t) = 4 n2 π 2 nπ n=1 and σN (t) =

N 17 X 5 − 6(−1)n n 1 − (−1)n + cos(nπt) + sin(nπt) 1− 4 N n2 π 2 nπ n=1

6. The partial sum of the Fourier series is SN (t) =

N X 2

nπ n=1

(1 − (−1)n ) sin(nπt/2).

The Cesáro, Hamming and Gaussian filtered partial sums are, respectively, σN (t) =

N X 2 n=1

HN (t) =

N X 2

nπ n=1

nπ

1−

n (1 − (−1)n ) sin(nπt/2), N

(0.54 + 0.46 cos(nπ/N ))(1 − (−1)n ) sin(nπt/2),

GN (t) =

N X

e−n π /N (1 − (−1)n ) sin(nπt/2).

n=1

Here α = 1 has been used in the Gauss filter. 7. The partial sums are SN (t) = 1 +

σN (t) = 1 +

N X 2

nπ n=1

N X 2 n=1

nπ

(1 − 3(−1)n ) sin(nπt/2),

1−

n (1 − 3(−1)n ) sin(nπt/2), N

440

CHAPTER 17. FOURIER SERIES HN (t) = 1 +

2 (0.54 + 0.46 cos(nπ/N ))(1 − 3(−1)n ) sin(nπt/2), nπ

GN (t) = 1 +

N X 2

nπ n=1

e−n π /N (1 − 3(−1)n ) sin(nπt/2).

Chapter 18

Fourier Transforms 18.1

The Fourier Transform

Z 0 fb(ω) =

e−iωx dx +

−1

Z 1

e−iωx dx =

2i (cos(ω) − 1). ω

The amplitude spectrum is the graph of |fb(ω)| =

2 (cos(ω) − 1) . ω

2. Write f (x) = sin(x)[H(x + k) − H(x − k)] with H the Heaviside function. Now use the modulation theorem to write i 2 sin(k(ω + 1)) 2 sin(k(ω − 1)) fb(ω) = − 2 ω+1 ω−1 sin(k(ω + 1)) sin(k(ω − 1)) = − i. ω+1 ω−1 The amplitude spectrum is a graph of sin(k(ω + 1)) sin(k(ω − 1)) − . |fb(ω)| = ω+1 ω−1 3. Write f (x) = 5[H(x + 4 − 7) − H(x + 4 + 7)] to obtain fb(ω) = 5e−7iω

2 sin(4ω) ω

10 −7iω e sin(4ω). ω

441

442

CHAPTER 18. FOURIER TRANSFORMS The amplitude spectrum is a graph of |fb(ω)| =

10 sin(4ω) . ω

4. Use time shifting to write r π −ω2 /12 −5iω b f (ω) = 5 e e . 3 The amplitude spectrum is a graph of r π −ω2 /12 b e . |f (ω)| = 5 3 5. Z ∞ fb(ω) =

e−x/4 e−iωx dx

e−(iω+1/4)x ∞ 4e−(iω+1/4)k = . = −(iω + 1/4) k 1 + 4iω The amplitude spectrum is a graph of 4e−k/4 |fb(ω)| = √ . 1 + 16ω 2 6.

2 fb(ω) = 3 (k 2 ω 2 sin(kω) + 2kω cos(kω) − 2 sin(kω)). ω The amplitude spectrum is a graph of |fb(ω)|.

7. fb(ω) = πe−|ω| . The amplitude spectrum is a graph of this function, which is nonnegative and hence equals its own magnitude. 8. Write f (x) = 3e−6 H(x − 2)e−3(x−2) to obtain fb(ω) = 3e−6

e−2iω 3 + iω

We can also write

1 e−2(3+iω) . 3 + iω The amplitude spectrum is a graph of fb(ω) =

3e−6 |fb(ω)| = √ . 9 + ω2

18.1. THE FOURIER TRANSFORM 9.

443

24 e2iω . 16 + ω 2 The amplitude spectrum is a graph of fb(ω) =

24 . 16 + ω 2

|fb(ω)| = 10. First write

f (x) = e−6 H(x − 3)e−2(x−3) to obtain e fb(ω) =

−3(2+iω)

2 + iω The amplitude spectrum is a graph of

e−6 |fb(ω)| = √ . 4 + ω2 11.

r f (x) = 18

12. Write fb(ω) =

2 −4ix −8x2 e e π

e−4(ω−5i) 3 + (ω − 5)i

to get f (x) = e

5ix b−1

e−4iω = e5ix H(x − 4)e−3(x−4) . 3 + iω

13. Write fb(ω) =

e2(ω−3)i 5 + (ω − 3)i

to obtain f (x) = e

3ix b−1

e2iω 5 + iω

= e3ix H(x + 2)e−5(x+2) = H(x + 2)e−(10+(5−3i)x) . 14. Write

10 sin(3ω) 10 sin((ω + π)) fb(ω) = =− . ω+π ω+π

Then f (x) = −5e−πix fb−1

2 sin(3ω) ω

= e−πix [H(x + 3) − H(x − 3)].

444

CHAPTER 18. FOURIER TRANSFORMS

15. Write fb(ω) =

1 + iω 2 2 = − . (3 + iω)(2 + iω) 3 + iω 2 + iω

Then f (x) = (2e−3x − e−2x )H(x). 16. fb−1

1 = H(x)e−x ∗ H(x)e−2x (1 + iω)(2 + iω) Z ∞ = H(ξ)e−ξ H(x − ξ)e−2(x−ξ) dξ −∞ Z x = H(x)e−2x eξ dξ = H(x)e−2x (ex − 1) 0

= H(x)(e−x − e−2x ). 17. fb−1

1 (1 + iω)2

= H(x)e−x ∗ H(x)e−x Z ∞ = H(ξ)e−ξ H(x − ξ)e−(x−ξ) dξ −∞ Z x = H(x)e−x dξ = H(x)xe−x . 0

18. fb−1

1 sin(3ω) = [H(x + 3) − H(x − 3)] ∗ H(x)e−2x (2 + iω)ω 2 Z 1 ∞ = (H(x + 3) − H(x − 3))H(x − ξ)e−2(x−ξ) dξ 2 −∞ Z x Z x 1 −2x −2ξ 2ξ H(x + 3) e dξ − H(x − 3) e dξ = e 2 −3 3 1 1 = (1 − e−2(x+3) )H(x + 3) − (1 − e−2(x−3) H(x − 3). 4 4

19. Compute Z ∞ −∞

|f (x)|2 dx =

1 2π

Z ∞

1 fb(ω)fb(ω) dω = 2π −∞

Z ∞

|fb(ω)|2 dω.

−∞

20. One way to compute the energy is to start with fb[H(x)e−2x ](ω) =

1 2 + iω

18.1. THE FOURIER TRANSFORM

445

and use Parseval’s theorem (Problem 19) to get Z ∞ Z ∞ 2 1 1 |f (x)|2 dx = dω 2π −∞ 2 + iω −∞ Z ∞ 1 1 dω = 2π −∞ 4 + ω 2 ω ∞ 1 = arctan 4π 2 −∞ 1 = . 4 We could also proceed directly in this example: Z ∞ Z ∞ 1 (H(x)e−2x )2 dx = e−4x dx = . 4 −∞ 0 Here the direct computation is easier, but sometimes it is useful to be aware of the use of Parseval’s theorem. 21. Begin with F

Z 1 3 −iωx 1 (H(x + 3) − H(x − 3)) (ω) = e dx 2 2 −3 =

e3iω − e−3iω . 2iω

Use the symmetry property of the transform to get sin(3x) F (ω) = π[H(−ω + 3) − H(−ω − 3)] x = π[H(ω + 3) − H(ω − 3)]. Now use Parseval’s identity to write 2 Z 3 Z ∞ 1 sin(3x) dx = π 2 dω = 3π. x 2π −3 −∞ 22. Let yb(ω) = fb[y(x)](ω) and transform the differential equation to obtain yb(−ω 2 + 6iω + 5) = fb[δ(t − 3)](ω) = e−3iω . Then yb(ω) =

e−3iω −ω 2 + 6iω + 5

e−3iω (1 + iω)(5 + iω) 1 e−3iω e3 iω = − . 4 1 + iω 5 + iω =

446

CHAPTER 18. FOURIER TRANSFORMS Invert this to obtain the solution 1 y(t) = (H(x − 3)e−(x−3) − H(x − 3)e−5(x−3) ). 4

23. Z 5 fbwin (ω) =

x2 e−iωx dx

−5

2 = 3 (25ω 2 sin(5ω) + 10ω cos(5ω) − 2 sin(5ω)). ω Because w(x) = 1 and the support of g is [−5, 5], then tC = 0. For the RMS bandwidth of the window function, we have R 5 2 !1/2 x dx 10 −5 =√ . wRMS = 2 R5 3 dx −5 24. Compute Z 4π fbwin (ω) = =

cos(ax)e−iωx dx

−4π

2 (ω sin(4πω) cos(4aπ) − a cos(4πω) sin(4aπ)). ω 2 − a2

Because w(x) is constant on on [−4π, 4π], tC = 0. We also have R 4π 2 !1/2 x dx 8π −4π wRMS = 2 =√ . R 4π 3 dx −4π 25. Compute Z 1 fbwin (ω) =

e−x e−iωx dx =

1 (1 − e−4(1+iω) ) 1 + iω

1 (1 − e−4 (cos(4ω) − i sin(4ω))(1 − iω) 1 + ω2 1 − e−4 cos(4ω) + e−4 sin(4ω) = 1 + ω2 −4 e sin(4ω) + (e−4 cos(4ω) − 1)ω +i . 1 + ω2 =

We also have

x dx tC = R0 4 =2 dx 0 and R4 wRMS = 2

(x − 2)2 dx R4 dx 0

!1/2

4 =√ . 3

18.1. THE FOURIER TRANSFORM

447

26. Z 1 fbwin (ω) =

ex sin(πx)e−iωx dx

−1

Z 1 =

sin(πx)e(1−iω)x dx

−1

π[2 sinh(1)(1 + π 2 ) − 2ω 2 cos(ω) + cosh(1)ω sin(ω)] (1 + (π + ω)2 )(1 + (π − ω)2 ) π[sinh(1)(2ω 2 sin(ω) − (2 + 2π 2 ) sin(ω)) + cosh(1)ω cos(ω)] . +i (1 + (π + ω)2 )(1 + (π − ω)2 ) =

Finally, compute tC = 0 and R1

x2 dx −1 R1 dx −1

wwin = 2

!1/2

2 =√ . 3

27. Z 2 fbwin (ω) =

(x + 2)2 e−iωx dx

−2

4 ((4ω 2 − 1) sin(2ω) + 2ω cos(2ω)) ω3 8i + 2 (2ω cos(2ω) − sin(2ω)). ω =

With w(x) = 1 and support [−2, 2], we have tC = 0. Finally, R2

x2 dx −2 R2 dx −2

wRMS = 2

!1/2

4 =√ . 3

28. We have Z 5π fb(ω) =

e−iωx dx

3π

1 5πiω (e − e3πiω ) iω 2eπiω e4πiω − e−4πiω =− ω 2i =−

= −2eπiω sin(4πω) = −2 cos(πω) sin(4πω) − 2i sin(πω) sin(4πω). Finally, R 5π

x dx tC = R3π5π = 4π dx 3π

448

CHAPTER 18. FOURIER TRANSFORMS and R 5π wRMS = 2

18.2

3π

(x − 4π)2 dx R 5π dx 3π

!1/2

2π =√ . 3

Fourier Sine and Cosine Transforms

In these problems the integrations are straightforward and details are omitted. 1. Z ∞

1 , 1 + ω2 0 Z ∞ ω e−x sin(ωx) dx = fbS (ω) = 1 + ω2 0

fbC (ω) =

e−x cos(ωx) dx =

2. a2 − ω 2 , (a2 + ω 2 )2 2aω fbS (ω) = 2 (a + ω 2 )2

fbC (ω) =

3. 1 sin(K(ω + 1)) sin(K(ω − 1)) + for ω 6= ±1, fbC (ω) = 2 ω+1 ω−1 1 K + sin(2K) fbC (1) = fbC (−1) = 2 2 ω 1 cos((ω + 1)K) cos((ω − 1)K) − + for ω 6= ±1, ω2 − 1 2 ω+1 ω−1 1 1 fbS (1) = (1 − cos(2K)), fbS (−1) = − (1 − cos(2K)) 4 4

fbS (ω) =

4. 1 fbC (ω) = (2 sin(Kω) − sin(2Kω)), ω 1 fbS (ω) = (1 − 2 cos(Kω) + cos(2Kω)) ω 5. 1 1 1 fbC (ω) = + , 2 1 + (ω + 1)2 1 + (ω − 1)2 1 ω+1 ω−1 fbS (ω) = + 2 1 + (ω + 1)2 1 + (ω − 1)2

18.2. FOURIER SINE AND COSINE TRANSFORMS

449

6. fbC (ω) =

1 (cosh(2K) cos(2Kω) − cosh(K) cos(Kω)) 1 + ω2

1 (ω sinh(2K) sin(2Kω) − ω sinh(K) sin(Kω)) 1 + ω2 1 (cosh(2K) cos(2Kω) − cosh(K) sin(Kω)) fbS (ω) = 1 + ω2 1 + (−ω sinh(2K) cos(2Kω) + ω sinh(K) cos(kω)) 1 + ω2 +

7. Suppose, for each positive number L, f (4) (x) is piecewise continuous on [0, L], f (3) (x) is continuous, and, as x → ∞, f (j) (x) → 0 for j = 1, 2, 3. Then we can integrate by parts four times to obtain Z ∞ FS [f (4) (x)](ω) = f (4) (x) sin(ωx) dx 0 h i∞ = f (3) (x) sin(ωx) − ωf 00 (x) cos(ωx) − ω 2 f 0 (x) sin(ωx) + ω 3 f (x) cos(ωx) 0 Z ∞ + ω4 f (x) sin(ωx) dx 0

= ω 4 FS (ω) − ω 3 f (0) + ωf 00 (0). 8. Assuming the same conditions as in Problem 7, four integrations by parts give us Z ∞ FS [f (4) (x)](ω) = f (4) (x) cos(ωx) dx 0 h i∞ (3) = f (x) cos(ωx) + ωf 00 (x) sin(ωx) − ω 2 f 0 (x) cos(ωx) − ω 3 f (x) sin(ωx) 0 Z ∞ + ω4 f (x) cos(ωx) dx 0

= ω 4 {C (ω) + ω 2 f 0 (0) − d(3) (0).

450

CHAPTER 18. FOURIER TRANSFORMS

Chapter 19

Complex Numbers and Functions 19.1

Geometry and Arithmetic of Complex Numbers

1. (3 − 4i)(6 + 2i) = (18 + 8) + (−24 + 6)i = 26 − 18i 2. i(6 − 2i) + |1 − i| = 6i + 2 + 3.

√ 2

2 + i 4 + 7i 1 + 18i 2+i = = 4 − 7i 4 − 7i 4 + 7i 65 (−1 + 5i)(16 − 2i) 1 (2 + i) − (3 − 4i) = = (−3 + 41i) (5 − i)(3 + i) (16 + 2i)(16 − 2i) 130

5. (17 − 6i)(−3 − 12i) = (17 − 6i)(−3 + 12i) = 4 + 228i 6.

3i 3 =√ −4 + 8i 80

7. i3 − 4i2 + 2 = −i + 4 + 2 = 6 − i 8. (3 + i)3 = 27 + 3(32 ) + 3(3)i2 + i2 = 18 + 26i 451

452

CHAPTER 19. COMPLEX NUMBERS AND FUNCTIONS

−6 + 2i 1 − 8i

2 (−6 + 2i)(1 + 8i) = (1 − 8i)(1 + 8i) 1 (−22 − 46i)2 = = (−1632 + 2024i) 2 65 4225

10. (−1 − 8i)(2i)(4 − i) = (−3 − 8i)(2 + 8i) = 58 − 40i In each of Problems 11–16, n denotes an arbitrary integer. 11. |3i| = 3, arg(3i) = 12.

π + 2nπ 2

√ 3π + 2nπ | − 2 + 2i| = 2 2, arg(−2 + 2i) = 4

13. | − 3 + 2i| =

√ 13, arg(−3 + 2i) = − arctan(2/3) + (2n + 1)π

14. |8 + i| =

√ 65, arg(8 + i) = arctan(1/8) + 2nπ

15. | − 4| = 4, arg(−4) = (2n + 1)π 16. |3 + 4i| = 5, arg(3 + 4i) = − arctan(4/3) + 2nπ √ 17. Because | − 2 + 2i| = 2 2 and 3π/34 is an argument, the polar form of −2 + 2i is √ −2 + 2i = 2 2e3iπ/4 . Here we did not add the customary 2nπ to the argument because, first, we need only one argument to write the polar form, and second, e2nπi = 1 for any integer n. 18. | − 7i| = 7 and an argument of −7i is 3π/2 (or −π/2 if you prefer), so the polar form is −7i = 7e3πi/2 . We could also write −7i = 7e−πi/2 . 19. |5 − 2i| =

√ 29 and an argument of 5 − 2i is − arctan(2/5), so √ 5 − 2i = 29e− arctan(2/5)i .

19.1. GEOMETRY AND ARITHMETIC OF COMPLEX NUMBERS

453

√ 20. | − 4 − i| = 17 and an argument of −4 − i is π + arctan(π/4) (look at the line from the origin to (−4, −1) in the third quadrant), so √ −4 − i = 29e(π+arctan(π/4))i . 21. 8+i= 22. −12 + 3i =

√

65earctan(1/8)i .

√ 153e− arctan(1/8)i .

23. Because i2 = −1, we have i4n = (i2 )2n = ((−1)2 )n = 1, i4n+1 = i4n i = i, i4n+2 = i4n i2 = i2 = −1, i4n+3 = i4n i3 = i2 i = −i. 24. Because (a + bi)2 = a2 − b2 + 2abi, we have Re((a + bi)2 ) = a2 − b2 and Im((a + bi)2 ) = 2ab. 25. Suppose first that z, w, u form vertices of a triangle, labeled in clockwise order around the triangle. The sides of the triangle are vectors represented by the complex numbers w−z, u−w, and z −u. This triangle is equilateral if and only if the sides have the same length, or |w − z| = |u − w| = |z − u| and each of the vector sides can be rotated by θ = 2π/3 radians clockwise to coincide with another side. This occurs exactly when u − w = (w − z)e−2πi/3 and z − u = (u − w)e−2πi/3 . Dividing these equations, we have u−w w−z = . z−u u−w Then (u − w)(u − w) = (w − z)(z − u). Then u2 − 2uw + w2 = wz + zu − uw − z 2 . Then z 2 + w2 + u2 = zw + zu + wu.

454

CHAPTER 19. COMPLEX NUMBERS AND FUNCTIONS

26. Let z = x + iy. Then z 2 = (z)2 if and only if (x + iy)2 = (x − iy)2 if and only if x2 − y 2 + 2ixy = x2 − y 2 − 2ixy if and only if 2ixy = −2ixy if and only if 2ixy = 0. Now, 2ixy = 0 can happen in two ways. If x = 0, then x + iy = iy is pure imaginary. If y = 0, then x + iy = x is real. 27. Suppose first that |z| = 1. Then p p |z| = x2 + (−y)2 = x2 + y 2 = |z| = 1 also. Then z−w z−w = 1 − zw zz − zw |z − w| = = 1. |z||z − w| If |w| = 1, then z−w z−w = 1 − zw ww − zw 1 z−w = =1 |w| z − w because |z − w| = |w − z| = |w − z|. 28. Compute |z + w|2 + |z − w|2 = (z + w)(z + w)(z − w) = zz + zw + wz + ww + zz + zw − wz − zw = 2zz + 2ww = 2 |z|2 + |w|2 . 29. M consists of all x + iy with y < 7. This is the half-plane lying below the horizontal line y = 7. The boundary points are all points x + 7i on the “edge” of M . M is open because it does no contain any of its boundary points (all points of M are interior points). 30. S consists of all points outside the circle of radius 2 about the origin. The boundary points of S are the points |z| = 2 on the circumference of the circle. None of these boundary points are in S. Every point of S is an interior point, and S is open.

19.2. COMPLEX FUNCTIONS

455

31. U consists of all points in the vertical strip between the vertical lines x = 1 and x = 3, including points on the line x = 3, but none of the points on the line x = 1. The boundary points of U are the points 1 + iy and 3 + iy on these lines. U is not closed because there are boundary points of U that do not belong to U . U is not open because U contains some of its boundary points (so not every point of U is an interior point). 32. V consists of all points inside the rectangle having vertices (2, 1), (3, 1), (2, −1) and (3, −1), and the points 3 + iy for −1 < y < 1 on the right side of this rectangle. The boundary points are all the points on the four sides of the rectangle. Some boundary points (on the right side) are in V , and the other boundary points (on the other three sides) are not. Therefore V is not closed, because it does not contain all of its boundary points. V is not open, because it does contain some of its boundary points. 33. W consists of all x+iy with x > y 2 . These are the points “enclosed” by the parabola x = y 2 , which opens to the right from the origin. The boundary √ points are the points on the parabola, which are the points x + i x for x ≥ 0. W does not contain any of its boundary points, and is open. W is not closed. 34. One way to envision R is to think of certain points on vertical segments of length 1. Begin with n = 1. The points 1+

1 i m

are the points 1 + i, 1 + (1/2)i, 1 + (1/3)i, · · · , going down the line x = 1 and approaching arbitrarily lose to (but not reaching) 1 on the horizontal axis. When n = 2, we have points 1 1 + i, 2 m moving down from x = 1/2 and approaching arbitrarily close to x = 1/2 as m is chosen larger. And so on - as n increases, these points remain on parallel horizontal lines, but move down vertical segments closer to the imaginary axis. The boundary points are all points (1/m)i and all points 1/n for positive integer n and m. None of these boundary points belong to R, so R is not closed. But R is open, because every point of R is an interior point.

19.2

Complex Functions

1. f (z) = z − i = x + iy − i = x + (y − 1)i

456

CHAPTER 19. COMPLEX NUMBERS AND FUNCTIONS so u(x, y) = x and v(x, y) = y − 1. The Cauchy-Riemann equations for this function are ∂u ∂v =1= ∂x ∂y and

∂u ∂v =0=− . ∂y ∂x

Because u and v are continuous with continuous first partial derivatives, and the Cauchy-Riemann equations are satisfied for all z, f (z) is differentiable for all z. 2. f (z) = z 2 − iz = x2 − y 2 + 2ixy − ix + y = x2 − y 2 + y + (2xy − x)i. Let u(x, y) = x2 − y 2 + y and v(x, y) = 2xy − x. Then ∂v ∂u = 2x = ∂x ∂y and

∂u ∂v = −2y + 1 = ∂x ∂y

so the Cauchy-Riemann equations are satisfied at every point. Because u and v are continuous with continuous partial derivatives at all points, f (z) is differentiable for all z. p 3. f (z) = |x + iy| = x2 + y 2 , so p u(x, y) = x2 + y 2 and v(x, y) = 0. If z 6= 0, then ∂u x y ∂u =p =p , 2 2 2 ∂x ∂y x +y x + y2 and

∂v ∂v = = 0. ∂x ∂y

The Cauchy-Riemann equations are not satisfied at any nonzero z. To check what happens at z = 0, compute ∂u u(h, 0) − u(0, 0) (0, 0) = lim h→0 ∂x h √ h2 = lim h→0 h |h| = lim . h→0 h

19.2. COMPLEX FUNCTIONS

457

This limit does not exist, because ( 1 |h| = h −1

if h > 0, if h < 0.

Similarly, (∂u/∂y)(0, 0) does not exist. Therefore the Cauchy-Riemann equations are not satisfied at any point, including the origin, and f (z) is not differential for any z. 4. f (z) is defined for all nonzero z. For z 6= 0, f (z) =

2z + 1 1 =2+ . z z

Then 1 x + iy x − iy =2+ 2 x + y2 y x2 − 2 i. =2+ 2 x + y2 x + y2

f (z) = 2 +

Then u(x, y) = 2 +

x2 x2 + y 2

y and v(x, y) = − 2 . x + y2

For z 6= 0 compute the partial derivatives ∂u y 2 − x2 ∂u −2xy = 2 , = 2 , 2 2 ∂x (x + y ) ∂y (x + y 2 )2 and

∂v 2xy ∂v y 2 − x2 = 2 , = . ∂x (x + y 2 )2 ∂y (x2 + y 2 )2

The Cauchy-Riemann equations hold for all nonzero z. Because u, v and its partial derivatives are continuous for all nonzero z, f (z) is differentiable for all z 6= 0. 5. f (z) = i|z|2 = (x2 + y 2 )i, so u(x, y) = 0 and v(x, y) = x2 + y 2 . Then

and

∂u ∂u = =0 ∂x ∂y ∂v ∂v = 2x, = 2y. ∂x ∂y

458

CHAPTER 19. COMPLEX NUMBERS AND FUNCTIONS The Cauchy-Riemann equations are satisfied only at z = 0, so f (z) is certainly not differentiable at any nonzero z. To check at z = 0, fall back on the definition of the derivative: z(h) − f (0) i|h|2 ihh = lim = lim = lim ih = 0. h→0 h→0 h h→0 h h→0 h lim

Therefore f 0 (0) = 0. 0 is the only point at which this function is differentiable. 6. f (z) = x + iy + y = (x + y) + yi so let u(x, y) = x + y and v(x, y) = y. The Cauchy-Riemann equations fail to hold at any point, because ∂v ∂u = 1 but = 0. ∂y ∂x 7. First, y x + iy =1+ i x x for x 6= 0. This function is defined for all z except for points on the imaginary axis. For x 6= 0, we can let f (z) =

u(x, y) = 1, v(x, y) = Now

and

y . x

∂u ∂u = =0 ∂x ∂y ∂v y ∂v 1 = − 2, = . ∂x x ∂y x

These are not satisfied at any z at which the function is defined. Therefore f (z) is not differentiable at any point at which it is defined. 8. Write f (z) = (x + iy)3 − 8(x + iy) + 2 = x3 − 3xy 2 − 8x + 2 + (3x2 y − y 3 − 8y)i. Let u(x, y) = x3 − 3xy 2 − 8x + 2 and v(x, y) = 3x2 y − y 3 − 8y. Then

∂u ∂v = 3x2 − 3y 2 − 8 = ∂x ∂y

19.2. COMPLEX FUNCTIONS

459

and ∂u ∂v = −6xy = − . ∂y ∂x The Cauchy-Riemann equations are satisfied at every z. Further, u, v and their first partial derivatives are continuous for all (x, y), so f (z) is differentiable for all z. 9. First, f (z) = (z)2 = (x − iy)2 = x2 − y 2 − 2xyi, so let u(x, y) = x2 − y 2 and v(x, y) = −2xy. Then ∂v ∂u = 2x but = −2x, ∂x ∂y while ∂v ∂u = −2y = . ∂x ∂y The Cauchy-Riemann equations hold only a z = 0, so this is the only point at which f (z) might have a derivative. To check this, look at f (h) − f (0) (h)2 = lim = lim h→0 h→0 h h→0 h lim

h h=0 h

because h/h has magnitude 1 and h → 0 if h → 0. Therefore f 0 (0) = 0, and 0 is the only point at which the function has a derivative. 10. If z = x + iy, then f (z) = iz + |z| =

p x2 + y 2 − y + xi.

Let u(x, y) =

p x2 + y 2 − y and v(x, y) = x.

Then ∂u x ∂u y =p , = −1 + p , 2 2 2 ∂x x + y ∂y x + y2 while ∂v ∂v = 1, = 0. ∂x ∂y Using these, it is routine to check that f (z) is not differentiable at any z.

460

CHAPTER 19. COMPLEX NUMBERS AND FUNCTIONS

11. For z 6= 0, write 1 1 = −4x − 4iy + z x + iy x − iy = −4x − 4yi + 2 x + y2

f (z) = −4z +

for (x, y) 6= (0, 0). Let x u(x, y) = −4x + 2 and v(x, y) = x + y2 Then

−4y −

y x2 + y 2

∂u y 2 − x2 ∂u −2xy , , = −4 + 2 = 2 ∂x (x + y 2 )2 ∂y (x + y 2 )2

and

∂v 2xy ∂v y 2 − x2 = 2 , = −4 + . ∂x (x + y 2 )2 ∂y (x2 + y 2 )2 The Cauchy-Riemann equations are satisfied at each nonzero z. Because u, v and the partial derivatives are continuous for (x, y) 6= (0, 0), f (z) is differentiable for all nonzero z.

12. First, x + (y − 1)i z−i = z+i x + (y + 1)i x2 + y 2 − 1 − 2xi = . x2 + (y + 1)2

f (z) =

Let u(x, y) =

2x x2 + y 2 − 1 and v(x, y) = − 2 . 2 2 x + (y + 1) x + (y + 1)2

for all (x, y) with x2 + (y + 1)2 6= 0. The partial derivatives are ∂u 4x(y + 1) = 2 , ∂x x + (y + 1)2 2(y + 1)2 − 2x2 ∂u = 2 , ∂y x + (y + 1)2 ∂v 2x2 − 2(y + 1)2 ∂v 4x(y + 1) = 2 , = 2 . 2 ∂x x + (y + 1) ∂y x + (y + 1)2 For (x, y) with x2 + (y − 1)2 6= 0, the Cauchy-Riemann equations are satisfied. Further, u(x, y) and v(x, y) are continuous with continuous partial derivatives. Therefore f (z) is differentiable at all z with z 6= i. 13. Let zn = xn + iyn and z0 = x0 + iy0 . Write f (z) = u(x, y) + iv(x, y). Because u and v are continuous at (x0 , y0 ), then f (zn ) = u(xn , yn ) + iv(xn , yn ) → u(x0 , y0 ) + iv(x0 , y0 ) = f (z0 ).

19.3. THE EXPONENTIAL AND TRIGONOMETRIC FUNCTIONS

19.3

461

The Exponential and Trigonometric Functions

1. ei = e0+i = e0 (cos(1) + i sin(1)) = cos(1) + i sin(1) 2. There are several ways to proceed. One is to use sin(x + iy) = sin(x) cosh(y) + i cos(x) sinh(y) to get sin(1 − 4i) = sin(1) cosh(4) − i cos(1) sinh(4). We could also have begun with the definition of sin(z) and used Euler’s formula. 3. Use the fact that cos(x + iy) = cos(x) cosh(y) − i sin(x) sinh(y) to get cos(3 + 2i) = cos(3) cosh(2) − i sin(3) sinh(2). 4. From their definitions, the trigonometric and hyperbolic functions are related by sin(z) = −i sinh(iz) and cos(z) = cosh(iz). Then −i sinh(−3) sin(3i) = cos(3i) cosh(−3) i sinh(3) = . cosh(3) = i tanh(3)

tan(3i) =

5. e5+2i = e3 e2i = e3 cos(2) + ie3 sin(2). 6. cos(1 − πi/4) sin(1 − πi/4) cos(1) cosh(π/4) + i sin(1) sinh(π/4) = . sin(1) cosh(π/4) − i cos(1) sinh(π/4)

cot(1 − πi/4) =

462

CHAPTER 19. COMPLEX NUMBERS AND FUNCTIONS To identify the real and imaginary parts of this quotient, multiply the numerator and denominator by the conjugate of the denominator to obtain cot(1 − iπ/4) =

(cos(1) cosh(π/4) + i sin(1) sinh(π/4))(sin(1) cosh(π/4) + i cos(1) sinh(π/4)) sin2 (1) cosh2 (π/4) + cos2 (1) sinh2 (π/4)

sin(1)cos(1)(cosh2 (π/4) − sinh2 (π/4)) + i sinh(π/4) cosh(π/4)(sin2 + cos2 (1)) sin2 (1) cosh2 (π/4) + cos2 (1) sinh2 (π/4) sin(1) cos(1) + i sinh(π/4) cosh(π/4) = . sin2 (1) cosh2 (π/4) + cos2 (1) sinh2 (π/4) =

7. 1 (1 − cos(2(1 + i))) 2 1 = [1 − cos(2) cosh(2) + i sin(2) sinh(2)]. 2

sin2 (1 + i) =

8. cos(2 − i) − sin(2 − i) = cos(2) cosh(1) + i sin(2) sinh(1) − sin(2) cosh(1) − i cos(2) sinh(1) = cosh(1)[cos(2) − sin(2)] − i sinh(1)[cos(2) − sin(2)] 9. eiπ/2 = cos(π/2) + i sin(π/2) = i 10. sin(ei ) = sin(cos(1) + i sin(1)) = sin(cos(1)) cosh(sin(1))i cos(cos(1)) sinh(sin(1)) 11. Begin with 2

ez = ex −y +2ixy = ex −y [cos(2xy) + i sin(2xy)]. Then 2

u(x, y) = ex −y cos(2xy) and v(x, y) = ex −y sin(2xy).

19.3. THE EXPONENTIAL AND TRIGONOMETRIC FUNCTIONS

463

Now compute 2 2 ∂u = ex −y [2x cos(2xy) − 2y sin(2xy)], ∂x 2 2 ∂u = ex −y [−2y cos(2xy) − 2x sin(2xy)], ∂y 2 2 ∂v = ex −y [2x sin(2xy) + 2y sin(2xy)], ∂x 2 2 ∂v = ex −y [−2y sin(2x) + 2x cos(2xy)]. ∂y

Then u and v satisfy the Cauchy-Riemann equations for all (x, y). 12. Begin by writing 1 x y 1 = = 2 − 2 i. z x + iy x + y2 x + y2 Then e

1/z

x/(x2 +y 2 )

y y − i sin cos x2 + y 2 x2 + y 2

= u(x, y) + iv(x, y). Take the partial derivatives: 2 2 ∂u ex/(x +y ) y y 2 2 = 2 (y − x ) cos + 2xy sin , ∂x (x + y 2 )2 x2 + y 2 x2 + y 2 2 2 ∂u ex/(x +y ) y y 2 2 = 2 −2xy cos − (x − y ) sin , ∂y (x + y 2 )2 x2 + y 2 x2 + y 2 2 2 ex/(x +y ) y y ∂v 2 2 = 2 (x − y ) sin + 2xy cos , ∂x (x + y 2 )2 x2 + y 2 x2 + y 2 2 2 ex/(x +y ) y y ∂v 2 2 = 2 2xy cos − (x − y ) cos . ∂y (x + y 2 )2 x2 + y 2 x2 + y 2 It is easy to verify that u and v satisfy the Cauchy-Riemann equations for all (x, y). 13. f (z) = zez = (x + iy)ex (cos(y) + i sin(y)) = xex cos(y) − yex sin(y) + (yex cos(y) + xex sin(y))i = u(x, y) + iv(x, y). Then

∂u ∂v = ex [cos(y) + x cos(y) − y sin(y)] = ∂x ∂y

464

CHAPTER 19. COMPLEX NUMBERS AND FUNCTIONS and

∂u ∂v = ex [−x sin(y) − sin(y) − y cos(y)] = − . ∂y ∂x

The Cauchy-Riemann equations are satisfied for all z. 14. f (z) = cos2 (z) = =

1 (1 − cos(2z)) 2

1 1 − (cos(2x) cosh(2y) − i sin(2x) sinh(2y)). 2 2

Then u(x, y) =

1 1 1 − cos(2x) cosh(2y), v(x, y) = sin(2x) sinh(2y). 2 2 2

Now, ∂u = sin(2x) cosh(2y), ∂x ∂u = − cos(2x) sinh(2y), ∂y ∂v = cos(2x) sinh(2y), ∂x ∂v = sin(2) cosh(2y). ∂y The Cauchy-Riemann equations are satisfied at every (x, y). 15. Suppose ez = 2i. With z = x + iy, then ex cos(y) + iex sin(y) = 2i. Then ex cos(y) = 0 and ex sin(y) = 2. Because ex 6= 0, cos(y) = 0, so y=

(2n + 1)π 2

in which n can be any integer. Now we have 2n + 1 ex sin π = 2. 2 Now ex > 0 for real x, so sin((2n + 1)π/2) > 0. But ( 1 if n is even, 2n + 1 sin π = 2 −1 if n is odd.

19.3. THE EXPONENTIAL AND TRIGONOMETRIC FUNCTIONS

465

Therefore n must be even, say n = 2m. Now we have y=

4m + 1 π. 2

Now we are left with ex = 2, so x = ln(2). All the solutions of ez = 2i are ln(2) +

4m + 1 π 2

with m any integer. 16. For the first identity, write sin(z) cos(w) + cos(z) sin(w) 1 iz = (e − e−iz )(eiw + e−iw ) + (eiz + e−iz )(eiw − e−iw ) 4i 1 i(z+w) = e − e−i(z+w) 2i = sin(z + w). The second identity is proved by a similar manipulation. 17. Use the polar form of the given equation. If z = reiθ , the equation is ez = er eiθ = −2. Because θ is real, |eiθ | = 1, so |ez | = er = | − 2| = 2. Then r = ln(2). Next, we must also have eiθ = −1 = cos(θ) + i sin(θ). Then sin(θ) = 0, so θ = nπ, in which (so far) n can be any integer. But cos(θ) = −1 means that n must be odd, so θ = (2m + 1)π in which m can be an y integer. Then z = ln(2) + (2m + 1)π, with m any integer, are all the solutions for z. 18. We want all z such that sin(z) = i. We need, for z = x + iy, sin(x) cosh(y) + i cos(x) sinh(y) = i. Then sin(x) cosh(y) = 0 and cos(x) sinh(y) = 1.

466

CHAPTER 19. COMPLEX NUMBERS AND FUNCTIONS Because cosh(y) > 0 for y real, we must have sin 9x) = 0, so x = nπ, with n as yet any integer. The second equation now becomes cos(nπ) sinh(y) = 1. Then sinh(y) =

1 = (−1)n . cos(nπ)

Then ey − e−y = 2(−1)n . Write this equation as e2y − 2(−1)n ey − 1 = 0 and consider cases. First, if n is even, then (−1)n = 1 and we have a quadratic equation for ey : e2y − 2ey − 1 = 0 with roots ey = 1 ±

√

2. √ Because 1 − 2 < 0 and ey > 0, discard this root and set √ ey = 1 + 2. Then y = ln(1 +

√

2) and solutions for z are z = 2mπ + ln(1 +

√ 2)i,

with m any integer. In the case that n is odd, write n = 2m + 1. Now (−1)n = −1 and we have e2y + 2ey − 1 = 0, with roots −1 ± so in this case

√

2. Again, we can only use the positive root −1 + y = ln(−1 +

√ 2,

√ 2)

and solutions for z are z = (2m + 1)π + ln(−1 +

√ 2)i,

with m any integer.

19.4. THE COMPLEX LOGARITHM

19.4

467

The Complex Logarithm

1. In polar form, z = −4i = 4e3nπi/2 so log(−4i) = ln(4) +

√ 2. First, 2 − 2i = 2 2e7πi/4 , so √ log(2 − 2i) = ln(2 2) +

+ 2nπ i.

7π + 2nπ i. 4

3. −5 = 5eπi , so log(−5) = ln(5) + (2n + 1)πi. √ arctan(5)i , so 4. 1 + 5i = 26e log(1 + 5i) = 5. −9 + 2i =

√

1 ln(26) + (arctan(5) + 2nπ)i. 2

85e(arctan(−2/9)+π)i , so

log(−9 + 2i) =

1 ln(85) + (− arctan(2/9) + (2n + 1)π)i. 2

6. 5 is its own polar form (argument zero), so log(5) = ln(5) + 2nπi. 7. Note that log(zw), log(z) and log(w) all have infinitely many different values, so we cannot expect to write the complex logarithm of the product as the sum of the logarithms of the factors. What we can show is that every value of log(zw) is the sum of a value of log(z) and a value of log(w). Suppose that z and w are nonzero. Let θz be any argument of z, and θw any argument of w. Then z = |z|e(θz +2nπ)i and w = |w|e(θw +2mπ)i . Then zw = |z||w|e(θz +θw +2kπ)i , while log(z) + log(w) = ln(|z|) + ln(|w|) + (θz + θw + 2(n + m)π)i. This means that for any choice of n and m, we can choose k = n + m to obtain a value of log(zw) that is equal to log(z) + log(w). 8. The argument is nearly identical to that used in Problem 7, except now z |z| (θz −θw +2(n−m)π)i = e . w |w|

468

CHAPTER 19. COMPLEX NUMBERS AND FUNCTIONS

19.5

Powers

In these problems, n denotes an arbitrary integer. 1. i1+i = e(1+i) log(i) = e(1+i)(π/2+2nπ)i π i h π = e−(π/2+2nπ) cos + 2nπ + i sin + 2nπ 2 2 = ie−(π/2+2nπ) . 2. √

(1 + i)2i = e2i log(1+i) = e2i(ln( 2)+i(π/4+2nπ)) = e−(π/2+4nπ) [cos(ln(2)) + i sin(ln(2))]. 3. ii = ei log(i) = ei(i(π/2+2nπ)) = e−π/2+2nπ . This is consistent with Problem 1, because i1+i = iii . 4. (1 + i)2−i = e(2−i) log(1+i) √

= e(2−i)(ln( 2)+i(π/4+2nπ)) π h π √ √ i + 4nπ − ln( 2) + i sin + 4nπ − ln( 2) = eln(2)+π/2+2nπ cos 2 √ 2 √ = 2π/4+2nπ [sin(ln( 2)) + i cos(ln( 2))]. 5. (−1 + i)−3i = e−3i log(−1+i) √

= e−3i ln( 2)+i(3π/4+2nπ) √ √ = e9π/4+6nπ [cos(3 ln( 2)) + i sin(3 ln( 2))]. 6. (1 − i)1/3 =

√

2e−i(π/4+2nπ)

1/3

= 21/6 e−i(π/12+2nπ) . We obtain the three distinct cube roots by choosing n = 0, 1, 2. Other choices of n repeat those obtained for these values.

19.5. POWERS

469

7. 1/4 i1/4 = ei(π/2+2nπ) = ei(π/8+nπ/2) , the the four fourth roots obtained for n = 0, 1, 2, 3. Other choices of n repeat these roots. 8. 161/4 = (16e2nπi )1/4 = 2enπi/2 = 2[cos(nπ/2) + i sin(nπ/2)] with the distinct fourth roots obtained by using n = 0, 1, 2, 3. 9. (−4)2−i = e(2−i) log(−4) = e(2−i)(ln(4)+i(π+2nπ)) = e2 ln(4)+π+2nπ [cos(ln(4)) − i sin(ln(4))]. 10. 6−2−3i = e(−2−3i) log(6) = e(−2−3i)(ln(6)+2nπi) = e−2 ln(6)+6nπ e−(3 ln(6)+4nπ)i 1 6nπ = e [cos(3 ln(6)) − i sin(3 ln(6))]. 36 11. 1/4 (−16)1/4 = 16ei(π+2nπ) = 2ei(π/4+nπ/2) π nπ i h π nπ + + i sin + . = 2 cos 4 2 4 2 We obtain the four fourth roots by taking n = 0, 1, 2, 3. These fourth roots are √ √ √ √ 2(1 + i), 2(−1 + i), 2(−1 − i), 2(1 − i). 12. First compute 1+i (1 + i)(1 + i) = = i. 1−i (1 − i)(1 + i) We therefore want i1/3 . These roots are 1/3 i1/3 = ei(π/2+2nπ) = ei(π/6+2nπ/3) π 2nπ π 2nπ = cos + + i sin + . 6 3 6 3

470

CHAPTER 19. COMPLEX NUMBERS AND FUNCTIONS The three roots are obtained for n = 0, 1, 2 and are 1 √ 1 √ ( 3 + i), (− 3 + i), −i. 2 2

13. These are the sixth roots of unity: 1/6 = enπi/3 11/6 = e2nπi = cos(nπ/3) + i sin(nπ/3). These sixth roots are obtained for n = 0, 1, 2, 3, 4, 5, and are √ √ √ √ 1 1 1 1 1, (1 + 3i), (−1 + 3i), −1, (−1 − 3i), (1 − 3i). 2 2 2 2 14. (7i)3i = e3i log(7i) = e3i(ln(7)+i(π/2+2nπ)) = e−2(π/2+2nπ) [cos(3 ln(7)) + i sin(3 ln(7))]. 15. Let ω be any nth root of 1 different from 1. The numbers ω j , for j = 0, 1, · · · , n − 1 are distinct, hence are all of the nth roots of 1. It is therefore enough to show that n−1 X

ω j = 0.

j=0

But this is a finite geometric series, whose sum is known: n−1 X

ωj =

j=0

1 − ωn =0 1−ω

because ω = 1. The conclusion can alsoP be proved as follows. Let ω1 , · · · , ωn be the nth n roots of unity. Let S = j=1 ωj . Now, one of the nth roots of unity is 1, but the other n − 1 roots are different from 1. Pick one root that does not equal 1, say, possibly by relabeling, ω1 6= 1. The numbers ω1 ω1 , ω1 ω2 , · · · , ω1 ωn are also nth roots of unity and are distinct, so this list includes all the nth roots of unity. The sum of these numbers is therefore S: S=

n X

ω1 ωj = ω1 S.

j=1

But then S(1 − ω1 ) = 0. Because ω1 6= 1, then S = 0.

19.5. POWERS

471

16. First recall that, if a 6= 1, then n−1 X

aj =

j=0

1 − an . 1−a

Further, replacing a with −a, n−1 X

n−1 X

j=0

(−a)j =

(−1)j aj =

1 − (−1)n an . 1+a

Apply this result with a = e2π/n , so an = e2πi = 1, to obtain ( n−1 n X 0 if n is even, 1 − (−1) (−1)j e2πij/n = = 2πi/n 2πi/n 1 + e 2/(1 + e ) if n is odd. j=0

472

CHAPTER 19. COMPLEX NUMBERS AND FUNCTIONS

Chapter 20

Complex Integration 20.1

The Integral of a Complex Function

1. In this problem f (z) = 1 is differentiable for all z and we can write an antiderivative F (z) = z. The curve has initial point γ(1) = 1 − i and terminal point γ(3) = 9 − 3i, so Z f (x) dz = F (9 − 3i) − F (1 − i) = 9 − 3i − (1 − i) = 8 − 2i. γ

We can also evaluate the integral by using the parametric equations of the curve. On γ, z = γ(t) = t2 − it, so dz = 2t − i and Z Z f (z) dz = dz γ

Z 3 (2t − i) dt

= 1

= t2 − it

= (9 − 3i) − (1 − i) 1

= 8 − 2i. 2. f (z) = z 2 − iz is differentiable for all z, with antiderivative F (z) =

1 3 i 2 z − z . 3 2

Because the curve extends from 2 to 2i, then Z 8 4 f (z) dz = F (2i) − F (1) = − + i. 3 3 γ If we want to use a parametrization of the curve, we can use polar coor473

474

CHAPTER 20. COMPLEX INTEGRATION dinates and write γ(t) = 2eit for 0 ≤ t ≤ π/2. Then Z Z π/2 f (z) dz = (4e2it − 2ieit )(2ieit dt γ

Z π/2 =

(8ie3it + 4e2it ) dt

π/2 8 3it e − 2ie2it 3 0 8 4 = − + i. 3 3

−

3. f (z) = Re(z) is not differentiable, so there is no antiderivative. There are many ways to parametrize the curve. One is by setting γ(t) = 1 + (1 + i)t for 0 ≤ t ≤ 1. On γ, f (z) = 1 + t and dz = (1 + i) dt, so Z Z 1 f (z) dz = (1 + t)(1 + i) dt γ

Z 1 =

(1 + i + (1 + i)t) dt 0

= (1 + i)t + =

1+i 2 1 t 2 0

3 (1 + i). 2

4. Parametrize γ(t) = 4eit for π/2 ≤ t ≤ 3π/2. Then Z Z 3π/2 1 1 dz = 4ieit dt = πi. it z 4e γ π/2 5. F (z) = (z − 1)2 /2 is an antiderivative of f (z), which is differentiable for all z, so Z 13 f (z) dz = F (1 − 4i) − F (2i) = − + 2i. 2 γ 6. F (z) = iz 3 /3 is an antiderivative of f (z), which is differentiable for all z, so Z 1 f (z) dz = F (3 + i) − F (1 + 2i) = (−28 + 29i). 3 γ 7. f (z) is differentiable for all z and has antiderivative F (z) = − cos(2z)/2, so Z f (z) dz = F (−4i) − F (−i) γ

1 1 = − (cos(−8i) − cos(−2i)) = − [cosh(8) − cosh(2)]. 2 2

20.1. THE INTEGRAL OF A COMPLEX FUNCTION

475

8. f (z) is differentiable for all z and has antiderivative F (z) = z + z 3 /3, so Z f (z) dz = F (3i) − F (−3i) = −12i. γ

9. f (z) is differentiable for all z and has antiderivative F (z) = −i sin(z), so Z f (z) dz = F (2 + i) − F (0) = −i sin(2 + i) γ

= −i[sin(2) cosh(1) + i cos(2) sinh(1)] = − cos(2) sinh(1) − i sin(2) cosh(1). 10. f (z) has no antiderivative, so parametrize the curve, say by γ(t) = −4 + (4 + i)t for 0 ≤ t ≤ 1. On the curve, z(t) = −4 + (4 + i)t and dz = (4 + i) dt and |z|2 = |4(t − 1) + it|2 = 16(t − 1)2 + t2 Then Z

Z 1

|z| dz = γ

(16(t − 1)2 + t2 )(4 + i) dt

4+i [16(t − 1)3 + t3 ] dt 3 17 = (4 + i). 3 =

11. Use the antiderivative F (z) = (z − i)4 /4 to get Z f (z) dz = F (2 − 4i) − F (0) = 10 + 210i. γ

12. Use the antiderivative F (z) = −ieiz to get Z f (z) dz = F (−4 − i) − F (−2) γ

= −i e1−4i − e−2i = −e sin(4) + sin(2) + [cos(2) − e cos(1)]i. 13. f (z) has no antiderivative because this function is not differentiable. Parametrize the curve by γ(t) = (−4 + 3i)t for 0 ≤ t ≤ 1. Then Z Z 1 iz dz = −i(4t − 3ti)(−4 + 3i) dt γ 0 3 25 = (−4 + 3i) − 2i = i. 2 2

476

CHAPTER 20. COMPLEX INTEGRATION

14. f (z) = Im(z) is not differentiable, so parametrize γ(t) = 4eit for 0 ≤ t ≤ 2π. Then Z Z 2π Im(z) dz = 4 sin(t)4ieit dt γ

Z 2π =

16(− sin2 (t) + i cos(t) sin(t)) dt = −16π.

0 2

15. f (z) = |z| has no antiderivative, so write γ(t) = (1 + i)t − i for 0 ≤ t ≤ 1 to get Z Z 1 2 (t2 + (t − 1)2 )(1 + i) dt = (1 + i). |z|2 dz = 3 0 γ 16. Begin with Z

cos(z 2 ) dz ≤ 8πM,

where 8π is the length of the circle γ, and M is a bound for cos(z 2 ) on γ, so | cos(z 2 )| ≤ M for z on γ. There are of course many possible choices for M . Indeed, if we find a bound, then any larger number is also a bound. To find one such number, let z = x + iy so z 2 = x2 − y 2 + 2ixy and | cos(z 2 )| = | cos(x2 − y 2 + 2ixy)| = | cos(x2 − y 2 ) cosh(2xy) − i sin(x2 − y 2 ) sinh(2xy)| ≤ cosh(2xy) + | sinh(2xy)| = e2xy . in which we have used the facts that the real hyperbolic cosine is positive, and that the real sine and cosine functions are bounded by 1. Now, for points on γ, x = 4 cos(t) and y = 4 sin(t) for 0 ≤ t ≤ 2π, so e2xy = e2(4 cos(t))(4 sin(t)) = e32 sin(t) cos(t) = e16 sin(2t) ≤ e16 . We can choose M = e16 to obtain Z cos(z 2 ) dz ≤ 8πe16 . γ

This is a massively huge bound for this integral, and is not claimed to be an approximation of the integral in any sense. But this is what we get with the quick and crude bounds made along the way. With more effort we may get a smaller bound. However, in some applications, it is enough to know that a certain term is bounded, and the size of the bound may not matter.

20.2. CAUCHY’S THEOREM 17. The length of γ is

477

√ 5. Now we need a number M such that 1 ≤ M on γ. 1+z

Notice that the point on γ closest to z = −1 is 2 + i, so for z on the curve, √ |z + i| = |z − (−1)| ≥ |2 + i + i| = 10. Then

1 1 1 = ≤√ . 1+z |1 + z| 10 √ We can therefore use M = 1/ 10 to get the bound √ Z 1 5 1 dz ≤ √ = √ . 10 2 γ 1+z

20.2

Cauchy’s Theorem

1. sin(z) is differentiable for all z, hence on an open set containing the curve and all points enclosed by the curve. By Cauchy’s theorem, I sin(z) dz = 0. γ

2. The circle encloses i, at which f (z) is not defined (hence not differentiable), so Cauchy’s theorem does not apply. Evaluate the integral by parametrizing γ(t) = 2i + 2eit for 0 ≤ t ≤ 2π. Then I Z 2π 2i 2i + 6eit it dz = 3ie dt 3eit γ z−i 0 Z 2π = (−2 + 6ieit ) dt = −4π. 0

3. γ encloses 2i, at which f (z) is not defined. Parametrize γ(t) = 2i + 2eit for 0 ≤ t ≤ 2π. Then I

1 dt = (z − 2i)3 γ

Z 2π 0

i = 4

1 2ieit dt (2eit )3

Z 2π

e−2it dt = 0.

This integral happens to be zero, but we could not conclude this from Cauchy’s theorem, which does not apply here.

478

CHAPTER 20. COMPLEX INTEGRATION

4. Because z 2 sin(z) is differentiable everywhere, and in particular on the curve and throughout the points it encloses, then Cauchy’s theorem applies and I z 2 sin(z) dz = 0. γ

5. f (z) = z is not differentiable. Write γ(t) = eit for 0 ≤ t ≤ 2π. Then I Z 2π z dz = e−it ieit dt = 2πi. γ

6. f (z) = 1/z is differentiable except at the origin, which is enclosed by the curve. Parametrize γ(t) = 5eit for 0 ≤ t ≤ 2π. Then I Z 2π Z 2π 1 1 dz = 5ieit dt −it z 5e γ 0 0 Z 2π = ie2it dt = 0. 0

7. Because f (z) = zez is differentiable on the curve and throughout the region it encloses, then by Cauchy’s theorem, I zez dz = 0. γ

8. A polynomial is differentiable on the entire complex plane, so I (z 2 − 4z + i) dz = 0. γ

9. f (z) = |z|2 is not differentiable at any point other than 0, Cauchy’s theorem does not apply. Write γ(t) = 7eit for 0 ≤ t ≤ 2π to obtain I Z 2π |z|2 dz = 49(7ieit ) dt = 0. γ

10. f (z) = sin(1/z) is not differentiable at 0, which is not on or enclosed by γ, so Cauchy’s theorem applies and I sin(1/z) dz = 0. γ

11. f (z) = Re(z) is not differentiable, so write γ(t) = 2eit for 0 ≤ t ≤ 2π. Then I Z 2π Re(z) dz = 2 cos(t)(2ieit ) dt γ

Z 2π =

[4i cos2 (t) − 4 cos(t) sin(t)] dt = 4πi.

20.3. CONSEQUENCES OF CAUCHY’S THEOREM

479

12. f (z) = z 2 is differentiable for ll z, so by Cauchy’s theorem, I z 2 dz = 0 γ

for any simple closed path in the plane. We can therefore concentrate on the integral of just Im(z) around the square. Let S1 be the left side, S2 the lower side, S3 the right side, and S4 the top side, and orient the rectangle counterclockwise. We can parametrize each side: S1 : γ1 (t) = −2it, S2 : γ2 (t) = 2t − 2i, S3 : γ3 (t) = 2 − 2i(1 − t), S4 : γ4 (t) = 2(1 − t). For each side, t varies from 0 to 1. Then I Z 1 Im(z) dz = (−2t)(−2i) dt γ

Z 1 +

Z 1

Z 1 (−2)(1 − t)(2i) dt +

(−2)2 dt + 0

0 dt 0

= 2i − 4 − 2i = −4. In total, then, I

(z 2 + Im(z)) dz = −4.

20.3

Consequences of Cauchy’s Theorem

1. Because 2i is the center of the circle γ, we can apply Cauchy’s integral formula with f (z) = z 4 to obtain I z4 dz = 2πif (2i) = 2πi(2i)4 = 32πi. z − 2i γ 2. By Cauchy’s integral formula with f (z) = sin(z 2 ), I sin(z 2 ) dz = 2πif (5) = 2πi sin(25). γ z−5 3. Use Cauchy’s integral formula, with f (z) = z 2 − 4z + i, to obtain I 2 z − 4z + i dz = 2πif (1 − 2i) γ z − 1 + 2i = 2πi[(1 − 2i)2 − 4(1 − 2i) + i] = 2πi(−8 + 7i) = −14π − 16πi.

480

CHAPTER 20. COMPLEX INTEGRATION

4. Apply the Cauchy integral formula for derivatives with n = 1 and f (z) = 2z 3 to obtain I 2z 3 dz = 2πif 0 (2) = 48πi. 2 γ (z − 2) 5. We can use the Cauchy integral formula for derivatives with n = 1 and f (z) = iez : I iez dz = 2πif 0 (2 − i) 2 γ (z − 2 + i) = 2πi(ie2−i ) = −2πe2 [cos(1) − i sin(1)]. 6. Apply Cauchy’s formula for derivatives with n = 2 and f (z) = cos(z − i) to obtain I cos(z − i) 2πi 00 dz = f (−2i) 3 2 γ (z + 2i) = −πi cos(−3i) = −πi cosh(3). 7. With f (z) = z sin(3z) and n = 2, Cauchy’s formula for derivatives gives us I 2πi 00 z sin(3z) dz = f (−4) 3 (z + 4) 2 γ = πi[6 cos(12) − 36 sin(12)]. 8. γ is not a closed curve, so parametrize γ(t) = 1 − t − it for 0 ≤ t ≤ 1. On the curve, p f (γ(t)) = 2iγ(t)|γ(t)| = 2i[(1 − t) + it] 1 − 2t + 2t2 so Z 1

Z 2iz|z| dz = γ

p 2i[(1 − t) + it] 1 − 2t + 2t2 dt

Z 1p Z 1 p 2 =2 1 − 2t + 2t dt + 2i (2t − 1) 1 − 2t + 2t2 dt 0 0 ! √ √ 2+1 . = 1 + 24 ln √ 2−1 These definite integrals can be evaluated using Z p −1 + 4t p 1 − 2t + 2t2 dt = 1 − 2t + 4t2 8 √ 7 2 4 1 + sinh−1 √ t− 32 4 7

20.3. CONSEQUENCES OF CAUCHY’S THEOREM and

481

p 1 (2t − 1) 1 − 2t + 2t2 dt = (1 − 2t + 2t2 )3/2 . 3

9. d −(2 + i) sin(z 4 ) dz = −2πi(2 + i) (sin(z 4 )) 2 (z + 4) dz z=−4 γ 3 4 = 2πi(1 − 2i) 4z cos(z ) z=−4

= −512π(1 − 2i) cos(256). 10. γ is not a closed curve. An antiderivative of (z − i)2 is (z − i)3 /3, so Z −i 1 (z − i)2 dz = (z − i)3 3 i γ 1 8 = (−2i)3 = i. 3 3 11. Parametrize γ(t) = 3 − t + (1 − 6t)i for 0 ≤ t ≤ 1. Then Z Z 1 Re(z + 4) dz = (7 − t)(−1 − 6i) dt γ

= (−1 − 6i)

13 13 = − − 39i. 2 2

12. 3z 2 cosh(z) d dz = 2πi (3z 2 cosh(z)) 2 dz z=−2i γ (z + 2i) 2 = 2πi 6 cosh(z) − 3z sinh(z) z=−2i

= 2πi[−12i cosh(2i) + 12 sinh(2i)] = 24π[cos(2) − sin(2)]. 13. First evaluate

ez dz γ z

by Cauchy’s integral formula to obtain Z z e dz = 2πiez = 2πi. z=0 γ z Now evaluate this integral by parametrizing γ(t) = eit for 0 ≤ t ≤ 2π: I z Z 2π cos(t)+i sin(t) e e dz = ieit dt z eit 0 γ Z 2π Z 2π =i ecos(t) cos(sin(t)) dt − ecos(t) sin(sin(t)) dt 0

= 2πi.

482

CHAPTER 20. COMPLEX INTEGRATION By equating the real parts of both sides of this equation, and then the imaginary parts, we obtain Z 2π ecos(t) cos(sin(t)) dt = 2π 0

and

Z 2π

ecos(t) sin(sin(t)) dt = 0.

The first integral is not obvious. The second could be done without complex analysis by observing that the integral from 0 to π is the negative of the integral from π to 2π. 14. First write f (z) =

z − 4i z − 4i = . z 3 + 4z z(z − 2i)(z + 2i)

Now let γ1 , γ2 and γ3 be nonintersecting circles in the region bounded by γ and having centers, respectively, 0, 2i, −2i. By the extended deformation theorem, I 3 I X f (z) dz = f (z) dz. γ

n=1

γj

Consider each integral on the right. On and in the interior of γ1 , think of f (z) =

(z − 4i)/(z − 2i)(z + 2i) . z

Apply the Cauchy integral formula to obtain I z − 4i f (z) dz = 2πi (z − 2i)(z + 2i) z=0 γ1 −4i = 2πi = 2π. (−2i)(2i) For the integral over γ2 , we can apply Cauchy’s integral formula to I I (z − 4i)/z(z + 2i) f (z) dz = z − 2i γ2 γ2 −2i π = 2πi =− . (2i)(4i) 2 And, for the integral over γ3 , think of f (z) =

(z − 4i)/z(z − 2i) z + 2i

and apply Cauchy’s integral formula in this region to obtain I −6i 3π f (z) dz = 2πi = . (−2i)(−4i) 2 γ3

20.3. CONSEQUENCES OF CAUCHY’S THEOREM

483

Finally, we have I f (z) dz = 2π − γ

π 3π − = 0. 2 2

484

CHAPTER 20. COMPLEX INTEGRATION

Chapter 21

Series Representations of Functions 21.1

Power Series

In each of Problems 1–6, the strategy is to take the limit of the magnitude of the ratio of successive terms of the series. The series converges when this limit (if it exists) is less than 1. 1. Take the limit of the magnitude of successive terms: (n + 2)/2n+1 1n+2 |z + 3i| = (z + 3i) n (n + 1)/2 2n+1 1 → |z + 3i|. 2 The series converges (absolutely) if 1 |z + 3i| < 1 2 or |z + 3i| < 2. The power series has radius of convergence 2 and open disk of convergence |z + 3i| < 2, the open disk of radius 2 about the center −3i. 2. 1/(2n + 3)2 (z − i)n+1 = 1/(2n + 1)2 (z − i)n

2n + 1 2n + 3

|z − i|2

→ |(z − i)2 |. if |z − i| < 1. This power series has radius of convergence 1, and open disk of convergence |z − i| < 1, the open disk of radius 1 about the center i. 485

486

CHAPTER 21. SERIES REPRESENTATIONS OF FUNCTIONS

3. (n + 1)n+1 /(n + 2)n+1 (z − 1 + 3i) nn /(n + 1)n (n + 1)2n+1 = n |z − 1 + 3i| n (n + 2)n+1 n n+1 n+1 n+1 = |z − 1 + 3i| n n+2 n n 1 1 + 1/n 1 + 1/n = 1+ |z − 1 + 3i| n 1 + 2/n 1 + 2/n and the limit of this quantity is less than 1 if |z − 1 + 3i| < 1. The radius of convergence is 1 and the disk of convergence is the open disk of radius 1 centered at 1 − 3i. In this limit, we have used (several times) the fact that x n lim 1 + = ex . n→∞ n 4. (2i/(5 + i))n+1 |z + 3 − 4i| (2i/(5 + i)n )n 2i → (z + 3 − 4i) 5+i 2 = √ |z + 3 − 4i|, 26 and this limit is less than 1 when

√ 26 . |z + 3 − 4i| < 2 √ This power series has radius of convergence 26/2, and the open disk of √ convergence is the disk of radius 26/2 centered at −3 + 4i. 5.

in+1 /2n+2 1 (z + 8i) → |z + 8i| in /2n 2 This ratio has limit < 1 if |z + 8i| < 2. The power series has radius of convergence 2 and the open disk of convergence is the open disk of radius 2 centered at −8i.

6. n + 2√ (1 − i)n+1 /(n + 3) (z − 3) = 2|z − 3| n (1 − i) /(n + 2) n+1 √ → 2|z − 3|

21.1. POWER SERIES

487

√ and this limit is less √ than 1 if |z − 3| < 1/ 2. This power series has radius of convergence 1/ 2 and the open disk of convergence is the open disk of √ radius 1/ 2 centered at 3. 7. No. The power series has center 2i. If the series converges at 0, it must also converge at the point i that is closer to the center 2i than 0 is. 8. The center of this power series is 4 − 2i. Now, 1 + i is closer to 4 − 2i than i is, so if the series converges at i, it must also converge at 1 + i. In each of Problems 9–14, we attempt to use known series to derive the requested series. 9. Assuming that we know the series for cos(z): cos(z) =

∞ X (−1)n

(2n)! n=0

z 2n .

This converges for all z. Replace z with 2z to obtain cos(2z) =

∞ X (−1)n 22n n=0

(2n)!

z 2n .

In writing these series, be careful with factorials. For example, in general (2n)! 6= 2n!. 10. We know that ez =

∞ X 1

n! n=0

for all z. Then e−z =

∞ X 1

n! n=0

(−z)n =

∞ X (−1)n n=0

zn.

The expansion of e−z about −3i will have powers of z + 3i, so write e−z = e−z−3i+3i = e3i e−(z+3i) ∞ X (−1)n 3i =e (z + 3i)n , n! n=0 converging for all z. 11. This is just a rearrangement of the given polynomial into powers of z−2+i. This can be done algebraically, or we can write the Maclaurin series of this polynomial about 2 − i. This series will be f (z) = z 2 − 3z + i = c0 + c1 (z − 2 + i) + z2 (z − 2 + i)2 ,

488

CHAPTER 21. SERIES REPRESENTATIONS OF FUNCTIONS where c0 = f (2 − i) = −3, c1 = f 0 (2 − i) = 1 − 2i and c2 =

1 00 f (2 − i) = 1. 2

The expansion of f (z) about 2 − i is z 2 − 3z + i = −3 + (1 − 2i)(z − 2 + i) + (z − 2 + i)2 . 12. Use the series for ez and sin(z): ∞ X (−1)n 2n+1 z and sin(z) = z e = n! (2n + 1)! n=0 n=0 z

∞ X 1

to write ez − i sin(z) =

∞ X 1 n=0

zn − i

∞ X (−1)n 2n+1 z . (2n + 1)! n=0

13. Like Problem 11, this can be done as an algebraic rearrangement of terms in f (z) = (z − 9)2 , or as a power series about 1 + i, which will be in powers of z − i − i. Using the latter approach, compute the coefficients c0 = f (1 + i) = 63 − 16i, c1 = f 0 (1 + i) = −16 + 2i and c2 =

1 00 f (1 + i) = 1. 2

Then (z − 9)2 = 63 − 16i + (−16 + 2i)(z − 1 − i) + (z − 1 − i)2 . 14. Replace z with z + i in the Maclaurin expansion of sin(z) to obtain ∞ X (−1)n (z + i)2n+1 . (2n + 1)! n=0

This converges for all z. 15. We know that f (0) = 1, f 0 (0) = i, and f 00 (z) = 2f (z) + 1. Compute f 00 (0) = 2f (0) + 1 = 3, f (3) (0) = 2f 00 (0) = 2i, f (4) (0) = 2f 00 (0) = 6, f (5) (0) = 2f (3) (0) = 4i.

21.1. POWER SERIES

489

Now use Taylor’s formula for the coefficients (in this case, about 0, 1 (n) f (0) n! to write the first six terms of the expansion: cn =

3 2i 6 4i 1 + iz + z 2 + z 3 + z 4 + z 5 . 2 3! 4! 5! In this problem it is not difficult to write the entire Maclaurin expansion, because an inductive argument shows that f (2n) (0) = 2n + 2n−1 and f (2n+1) (0) = 2n i. 16. With f (z) = sin2 (z), we will need f 0 (z) = 2 sin(z) cos(z) = sin(2z), f 00 (z) = 2 cos(2z), f (3) (z) = −4 sin(2z), f (4) (z) = −8 cos(2z), f (5) (z) = 16 sin(2z), f (6) (z) = 32 cos(2z). (a) Using these derivatives, we can find the first seven terms of the power series expansion of f (z) about 0: 2 1 sin2 (z) = z 2 − z 4 + z 6 + · · · . 3 45 (b) Multiply

1 1 5 1 1 5 z − z3 + z + ··· z − z3 + z + ··· 6 120 6 120 1 1 1 1 1 = z2 − + z4 + + + z6 + · · · 6 6 120 36 120 2 6 1 z + ··· , = z2 − z4 = 3 45

sin2 (z) =

(c) Use the definition of sin(z) to write 2 1 sin2 (z) = − eiz − e−iz 4 1 = − e2iz + e−2iz − 2 4 (2iz)2 (2iz)7 1 1 1 + 2iz + + ··· + + ··· = − 2 4 2! 7! 1 (2iz)2 (2iz)7 − 1 − 2iz + + ··· − + ··· 4 2! 7! 1 1 4 8 = − 1 − 4z 2 + z 4 − z 6 + · · · 2 4 3 45 1 2 = z2 − z4 + z6 + · · · 3 45

490

CHAPTER 21. SERIES REPRESENTATIONS OF FUNCTIONS (d) We can also write 1 1 − cos(2z) 2 2 1 1 (2z)2 (2z)4 (2z)6 == − 1− + − + ··· 2 2 2! 4! 6! 1 1 2 4 = − 1 − z2 + z4 − z6 + · · · 2 2 3 45 1 2 = z2 − z4 + z6 + · · · 3 45

sin2 (z) =

17. Let z be a complex number and consider the integral 1 2πi

zn ezw dw. n+1 γ n!w

Here γ is the unit circle about the origin, oriented counterclockwise as usual. Expand ezw in its Maclaurin series and parametrize γ(t) = eit for 0 ≤ t ≤ π to write zn 1 ezw dw = n+1 2πi γ n!w

∞

z n X (zw)k dw n+1 k! γ n!w k=0 I X ∞ 1 z n+k wk−n−1 = dw 2πi γ n!k! k=0 Z 2π X i(k−n−1)t ∞ 1 z n+k = n!k!ieit dt 2πi 0 e k=0 Z 2π n+k ∞ X 1 z = ei(k−n)t dt. 2π 0 n!k!

k=0

Now, Z 2π e

i(k−n)t

( 0 dt = 2π

if k 6= n, if k = n.

We therefore have 1 2πi

(z n )2 zn ezw dw = . n+1 (n!)2 γ n!w

21.1. POWER SERIES

491

Finally, we can write ∞ zn 1 2n X 1 z = ezw dw 2 n+1 (n!) 2πi n!w n=0 n=0 Z 2π X ∞ it 1 zn = eze eit dt i(n+1)t 2πi 0 n=0 n!e # Z 2π " X ∞ 1 (ze−it )n zeit = e dt 2π 0 n! n=0 Z 2π −it it 1 = eze eze dt 2π 0 Z 2π it −it 1 = ez(e +e ) dt 2π 0 Z 2π 1 e2z cos(t) dt. = 2π 0 ∞ X

18. f (z) has a zero of order 3 at 0 because z 3 has a zero of order 3 there and cos(0) 6= 0. 19. f (z) has a zero of order 4 at 0 because z 2 has a zero of order 2 at 0 and sin2 (z) also has a zero of 2 there (because sin(z) has a zero of order 1 at 0). 20. f (z) = (z − π/2)2 cos(z) has a zero of order 3 at π/2. 21. f (z) has a zero of order 3 at 3π/2 because cos(z) has a simple zero there. 22. f (z) has a zero of order 4 at 0 because cos(z) has a simple zero at π/2. 23. f (z) is not defined at z = 0, so we cannot really speak of it having a zero there. However, notice something interesting. Using the Maclaurin expansion of sin(z), with z 4 in place of z, and divided by z 2 , we can write ∞ X 1 (−1)n 8n+2 4 sin(z ) = z . z2 (2n + 1)! n=0

This power series converges for all z, and has the value 0 at 0. We can therefore extend f (z) by giving it the value 0 at z = 0, and obtain a differentiable function. This extended function has a zero of order 2 at 0. 24. As in Problem 23, f (z) is not defined at π. However, (z − π)5 has a fifth order zero at π, and sin2 (z) has a second order zero at π, and if we extend f (z) by giving it the value 0 at π, this extended function has a third order zero at π.

492

CHAPTER 21. SERIES REPRESENTATIONS OF FUNCTIONS

5. Compute the kth derivative of f (z) at z0 using each series, obtaining f k (z0 ) = =

∞ X n=0 ∞ X

an (n)(n − 1) · · · (n − k + 1)(z − z0 )n−k bn (n)(n − 1) · · · (n − k + 1)(z − z0 )n−k .

n=0

Then f k (z0 ) = k!ak = k!bk , so for each k = 0, 1, 2, · · · , ak =

21.2

1 (k) f (z0 ) = bk . k!

The Laurent Expansion

Problems 1–10 are solved using manipulations of known series, such as geometric series and power series for exponential and trigonometric functions. It is sometimes best, in seeking an expansion about z0 , to focus on getting an expression involving powers of z − z0 , using algebraic manipulations, or sometimes integration and differentiation. In particular, it is useful to know the geometric series ∞ X 1 = rn 1 − r n=0

and

∞ X 1 (−1)n rn , = 1 + r n=0

for |r| < 1. 1. We want an expansion in powers of z − i. To this end, begin with 2z 1 1 = + . 2 1+z z−i z+i The first term is already an expansion in powers of z − i (having only one term). For the second term, write 1 1 1 = = z+i 2i + (z − i) 2i 1 − z−i 2i n ∞ 1 X z−i = (−)n 2i n=0 2i =

∞ X (−1)n

(2i)n+1 n=0

(z − i)n .

21.2. THE LAURENT EXPANSION

493

This expansion is valid for 1 z−i = |z − i| < 1, 2i 2 or |z − i| < 2. The Laurent expansion of f (z) about i is therefore ∞ X 1 (−1)n (z − i)n . + z − i n=0 (2i)n+1

This represents f (z) in the annulus 0 < |z − i| < 2. 2. For z 6= 0, write ∞ sin(z) 1 X (−1)n 2n+1 = 2 z z2 z n=0 (2n + 1)!

∞ X (−1)n 2n−1 z . (2n + 1)! n=0

This series represents sin(z)/z 2 in the annulus 0 < |z < ∞, which is the plane with the origin removed. 3. If z 6= 0, then " # ∞ X 1 − cos(2z) 1 (−1)n 2n = 2 1− (2z) z2 z (2n)! n=0 =

∞ X (−1)n+1 4n n=1

(2n)!

z 2n−2 .

4. Write 2n ∞ X i (−1)n i 2 z cos =z z (2n)! z n=0 2

∞ X

1 2−2n z , (2n)! n=0

in which we have used the fact that i2n = (−1)n . This expansion represents the function in the annulus 0 < |z| < ∞. 5. The denominator is already in terms of z − 1, so concentrate on the numerator: z2 ((z − 1) + 1)2 1 + 2(z − 1) + (z − 1)2 = =− 1−z 1−z z−1 1 =− − 2 − (z − 1). z−1

494

CHAPTER 21. SERIES REPRESENTATIONS OF FUNCTIONS This represents the function for 0 < |z − 1| < ∞, the complex plane with 1 removed.

6. As with Problem 5, algebraic manipulation will be enough here. Write z2 + 1 1 z2 + 1 1 1 + [(z − 1/2) + 1/2]2 = = 2z − 1 2 z − 1/2 2 z − 1/2 5 1 1 1 1 = + + z− 8 z − 1/2 2 2 2 for 0 < z − 12 < ∞. 7. Use the exponential series to obtain ∞ ∞ 1 z2 1 X 1 2n X 1 2n−2 e z = z , = z2 z 2 n=0 n! n! n=0

for 0 < |z| < ∞. 8. Use the Maclaurin expansion of the sine function to get ∞ ∞ 1 1 X (−1)n 2n+1 X (−1)n 2n−1 sin(4z) = z = z , z z n=0 (2n + 1)! (2n + 1)! n=0

for 0 < |z| < ∞. 9. The denominator is already a power of z − i, so we can write 2i + (z − i) 2i z+i = =1+ z−i z−i z−i for 0 < |z − i| < ∞. 10. Use the Maclaurin expansion of sinh(z) (or write sinh(z) = (ez − e−z )/2 and use the exponential series) to obtain X 2n+1 X ∞ ∞ 1 1 1 1 sinh = = z −6n−3 3 3 z (2n + 1)! z (2n + 1)! n=0 n=0 for 0 < |z| < ∞. 11. By Cauchy’s integral formula, for any z enclosed by Γ1 , I 1 f (w) dw. f (z) = 2πi Γ1 w − z Because Γ2 does not enclose z, I 1 f (w) dw = 0 2πi Γ2 w − z

21.2. THE LAURENT EXPANSION

495

by Cauchy’s theorem. The factor of 1/2πi was included in the last equation so we can add these two integrals to get I I 1 f (w) f (w) f (z) = dw + dw . 2πi Γ2 w − z Γ1 w − z Orientation on both curves is counterclockwise. In this sum of integrals, L1 and L2 are traversed in both directions, so the integrals over these segments are zero. The integrals in square brackets therefore give us the integrals over γ1 and γ2 , but counterclockwise on γ1 and clockwise on γ2 . Reversing this orientation on γ1 so that all integrals are over counterclockwise curves, we have I I f (w) f (w) 1 dw − dw . f (z) = 2πi γ2 w − z γ1 w − z Now manipulate the 1/(w − z) factor in each integral to derive the result we want. For the integral over γ2 , write 1 1 = w−z w − z0 − (z − z0 ) 1 1 = w − z0 1 − (z − z0 )/(w − w0 ) n ∞ X 1 z − z0 = w − z0 n=0 w − z0 =

∞ X

1 (z − z0 )n . n+1 (w − z ) 0 n=0

This geometric expansion is valid because, for w on γ2 , w − z0 < 1. z − z0 For the integral over γ1 , use the fact that, for w on this curve, w − z0 < 1. z − z0 Now we have 1 1 = w−z w − z0 − (z − z0 ) −1 1 = z − z0 1 − (w − z0 )/(z − z0 ) ∞ 1 X w − z0 =− z − z0 n=0 z − z0 =−

∞ X

(w − z0 )n

n=0

1 . (z − z0 )n+1

496

CHAPTER 21. SERIES REPRESENTATIONS OF FUNCTIONS Substitute these expressions into the sum of integrals representing f (z) and interchange the integrals with the summation to obtain ! I ∞ X 1 f (w) dw (z − z0 )n f (z) = 2πi γ2 n=0 (w − z0 )n+1 ! n+1 I ∞ X 1 1 n + f (w)(w − z0 ) dw 2πi γ1 n=0 z − z0 I ∞ X f (w) 1 dw (z − z0 )n = n+1 2πi (w − z ) 0 γ2 n=0 I ∞ X 1 1 n f (w)(w − z0 ) dw + . n+1 2πi (z − z 0) γ1 n=0 Finally, use the deformation theorem to replace these integrals over γ1 and γ2 with integrals over Γ, which is any simple closed path in the annulus and enclosing z0 . This gives us f (z) =

∞ X

cn (z − z0 )n ,

n=−∞

with the integral expressions given for the coefficients cn .

Chapter 22

Singularities and the Residue Theorem 22.1

Singularities

1. cos(z)/z has one singularity, a double pole at z = 0. 2. f (z) has a double pole at −i and a simple pole at 1. 3. e1/z (z + 2i) has an essential singularity at 0. 4. Note that sin(z − π) sin(z) =− z−π z−π so lim

z→π

sin(z) sin(z − π) = − lim z→π z−π z−π sin(z) = − lim = 1 6= 0. z→0 z

Because this limit is nonzero, the function has a removable singularity at z = π. 5. The function has a double pole at 1 and simple poles at i and −i. 6. The function has a double pole at −1. 7. Write

z−i z−i 1 = = , z2 + 1 (z + i)(z − i) z+i

so the function has a simple pole at −i. 497

498

CHAPTER 22. SINGULARITIES AND THE RESIDUE THEOREM

8. We need to know if sinh(z) has any zeros other than z = 0. For this, solve 1 z e − e−z = 0. 2 For this to be true, we must have sinh(z) =

e2z = 1 so 2z = 2nπi, or z = nπi, with n any integer. These are simple zeros of sinh(z) because cosh(nπi) 6= 0 for any integer n. For n = 0, observe that 0 is a simple zero of both sin(z) and sinh(z), so 0 is a removable singularity of the function. 9. The denominator has simple zeros at 1, −1, i, −i and these are simple poles of the function because the numerator does not vanish at any of these numbers. 10. tan(z) = sin(z)/ cos(z) has simple poles at the zeros of cos(z), which are simple and occur at points z = (2n + 1)π/2, in which n any integer. 11. sec(z) = 1/ cos(z) has simple poles at the zeros of cos(z), which are the simple zeros (2n + 1)π/2 with n any integer. 12. e1/z(z+1) has essential singularities at 0 and −1. One way to see this is to write 1 1 1 = − z+1 z z+1 so e1/z(z+1) = e1/z e−1/(z+1) . 13. Suppose f is differentiable at z0 and f (z0 ) 6= 0, while g has a pole of order m at z0 . We want to show that the product f g has a pole of order m at z0 . Because g has a pole of order m at z0 , the Laurent expansion in some annulus about z0 has the form g(z) =

∞ X k + cn (z − z0 )n . (z − z0 )m n=−m+1

with k 6= 0. Then (z − z0 )m g(z) = k +

∞ X

cn (z − z0 )n+m .

n=−m+1

If we denote the power series on the right as h(z), then (z − z0 )m g(z) = h(z),

22.2. THE RESIDUE THEOREM

499

where h(z0 ) = k 6= 0. Further, in some annulus about z0 , f (z)g(z) =

f (z)h(z) . (z − z0 )m

Because f (z0 )h(z0 ) 6= 0, f (z)g(z) has a pole of order m at z0 .

22.2

The Residue Theorem

1. The function has simple poles at 1 and −2i, both enclosed by γ. Keep in mind that only singularities enclosed by the curve are relevant in evaluating the integral by the residue theorem. Compute d 1 + z2 z→1 dz z + 2i (z + 2i)(2z) − (1 + z 2 ) = lim z→1 (z + 2i)2 4i , = −3 + 4i

Res(f, 1) = lim

and

1 + z2 −3 = . z→−2i (z − 1)2 −3 + 4i

Res(f, −2i) = lim Then

4i 3 1 + z2 dz = 2πi − = 2πi. 2 −3 + 4i −3 + 4i γ (z − 1) (z + 2i)

2. γ encloses i, which is a double pole and the only singularity of the function. Then I d 2z dz = 2πiRes(f, i) = lim (2z) = 4πi. 2 z→i dz γ (z − i) 3. The only singularity of ez /z is a simple pole at 0, and this is not enclosed by γ, so I z e dz = 0 γ z by Cauchy’s theorem. 4. f (z) has singularities at 2i and −2i. These are both simple poles, and both are enclosed by γ. Then I cos(z) dz = 2πiRes(f, 2i) + 2πiRes(f, −2i) 4 + z2 γ = 2πi

cos(2i) cos(−2i) + 2πi 4i −4i

= 0.

500

CHAPTER 22. SINGULARITIES AND THE RESIDUE THEOREM

√ √ 5. The function has simple poles at 6i and − 6i, both enclosed by γ. Then I h √ √ i z+i dz = 2πi Res(6, 6i) + Res(f, − 6i) 2 γ z +6 "√ # √ 6+1 6−1 √ + √ = 2πi = 2πi. 2 6 2 6 6. f (z) has a simple pole at −1/2. This is the only singularity, and it is enclosed by γ. Then I 1 1 z−i dz = 2πiRes(f, 01/2) = − +i . 2 2 γ 2z + 1 7. z/ sinh2 (z) has a simple pole at 0 and double poles at nπi, for every nonzero integer n. The only singularity enclosed by γ is 0, so I f (z) dz = 2πiRes(f, 0). γ

Compute this residue as Res(f, 0) = lim zf (z) = lim z→0

= lim

z→0

z2 sinh2 (z)

z→0 z 2 + 1 z 4 + · · · 6

= lim

z→0 1 + 1 z 2 + · · · 6

Then

= 1.

z dz = 2πi. 2 γ sinh (z)

8. cos(z)/zez has only one singularity, a simple pole at 0, and this is enclosed by γ. Therefore I cos(z) dz = 2πiRes(f, 0) z γ ze cos(z) = 2πi. z→0 ez

= 2πi lim

9. f (z) has simple poles at i, 3i and −3i. Only the pole at −3i is enclosed by the curve, so I iz dz = 2πiRes(f, −3i) 2 γ (z + 9)(z − i) iz 1 πi = 2πi lim = 2πi − =− . z→−3i (z − 3i)(z − i) 8 4

22.2. THE RESIDUE THEOREM

501

10. e3/z has an essential singularity at 0 and no other singularities. 0 also lies in the region bounded by the curve. We need the residue of f (z) at 0. Here we do not have a formula, but we can look at the Laurent expansion about 0: n ∞ X 2 3 1 . e3/z = 2 n! z n=0 The coefficient of 1/z in this expansion is zero, so the residue is zero and I 2 e3/z dz = 0. γ

11. f (z) has only one singularity, a simple pole at −4i, and this is outside the region bounded by the curve. By Cauchy’s theorem, I 8z − 4i + 1 dz = 0. z + 4i γ 12. f (z) has a simple pole at 1 − 2i, which is enclosed by γ, so I z2 dz = 2πiRes(f, 1 − 2i) γ z − 1 + 2i = 2πi((−1 + 2i)2 ) = 2π(4 − 3i). 13. The singularities of coth(z) = cosh(z)/ sinh(h) are the zeros of sinh(z). This means that coth(z) has simple poles at nπi, with n any integer. Only the simple pole at 0 is enclosed by the curve, so I cosh(0) coth(z) dz = Res(f, 0) = 2πi = 2πi. cosh(0) γ 14. f (z) has simple poles at 2, 2e2πi/3 and 2e4πi/3 . Only 2 is enclosed by γ, so I (1 − z)2 dz = 2πiRes(f, 2) 3 γ z −8 (1 − z)2 = 2πi lim 2 z→2 3z 1 πi = 2πi = . 12 6 15. 0 and 4i are simple poles of f (z) and both are enclosed by γ, so I e2z dz = 2πi [Res(f, 0) + Res(f, 4i)] γ z(z − 4i) e4i 1 = 2πi − + 4i 4i π = [cos(8) − 1 + i sin(8)]. 2

502

CHAPTER 22. SINGULARITIES AND THE RESIDUE THEOREM

16. The function has a double pole at 1 and γ is assumed to be a closed path that enclosed 1, so I z2 dz = 2πiRes(f, 1) 2 γ (z − 1) d 2 = 2πi lim (z ) = 4πi. z→1 dz 17. z0 is a zero of order 2 of h(z), but g(z0 ) 6= 0. We want to show that Res(g/h, z0 ) =

2g 0 (z0 ) 2 g(z0 )h(3) (z0 ) . − h00 (z0 ) 3 (h00 (z0 ))2

To do this, first write h(z) = (z − z0 )2 ϕ(z), with ϕ(z0 ) 6= 0. Then d g(z) (z − z0 )2 z→z0 dz h(z) d g(z) = lim (z − z0 )2 ϕ(z) z→z0 dz (z − z0 )2 ϕ(z) d g(z) = lim z→z0 dz ϕ(z) ϕ(z0 )g 0 (z0 ) − ϕ0 (z0 )g(z0 ) = . (ϕ(z0 ))2

Res(g/h, z0 ) = lim

Now, h0 (z) = 2(z − z0 )ϕ(z) + (z − z0 )2 ϕ0 (z). , h00 (z) = 2ϕ(z) + 4(z − z0 )ϕ0 (z) + (z − z0 )2 ϕ00 (z), and h(3) (z0 ) = 6ϕ0 (z) + 6(z − z0 )ϕ00 (z) + (z − z0 )2 ϕ(3) (z). Then

1 00 1 h (z0 ) and ϕ0 (z0 ) = h(3) (z0 ). 2 6 Substituting these into the expression for the residue, we have ϕ(z0 ) =

Res(g/h, z0 ) =

2g 0 (z0 ) 2 g(z0 )h(3) (z0 ) − . h00 (z0 ) 3 (h00 (z0 ))2

18. Suppose first that f has a zero of order k at z0 in G. Consider the residue of f 0 /f at z0 . First, write f (z) = (z − z0 )k g(z),

22.2. THE RESIDUE THEOREM

503

for some function g(z) that is differentiable on an open disk about z0 , and g(z0 ) 6= 0. Then f 0 (z) k(z − z0 )k−1 g(z) + (z − z0 )k g 0 (z) = f (z) (z − z0 )k g(z) g 0 (z) k + . = z − z0 g(z) Now g 0 /g is differentiable at z0 so the last equation implies that f 0 /f has a simple pole at z0 , and Res(f 0 /f, z0 ) = k. Now see what happens if f has a pole of order m at z1 . In some annulus about z1 , ∞ X f (z) = dn (z − z1 )n n=−m

with d−m 6= 0. Then (z − z1 )m f (z) =

∞ X

dn (z − z1 )n =

n=−m

∞ X

dn−m (z − z1 )n = h(z),

n=0

with h(z) differentiable at z1 and h(z1 ) = d−m 6= 0. Then f (z) =

1 h(z). (z − z1 )m

Now, in some disk about z1 , f 0 (z) −m(z − z1 )−m−1 h(z) + (z − z1 )−m h0 (z) = f (z) (z − z1 )−m h(z) 0 −m h (z) = + . z − z1 h(z) This implies that Res(f 0 /f, z1 ) = −m. Therefore, the sum of the residues of f 0 /f at poles of f enclosed by γ in G counts each zero of f enclosed by γ, according to its multiplicity, and each pole of f enclosed by γ, according to the negative of its multiplicity. If Z is the number of zeros of f enclosed by γ, counting multiplicities, and P the number of poles of f enclosed by γ, also counting multiplicities, then by the residue theorem, I 0 f (z) dz = 2πi(Z − P ). γ f (z)

504

CHAPTER 22. SINGULARITIES AND THE RESIDUE THEOREM

19. By the residue theorem, with g(z) = z/(2 + z 2 ), I h √ i √ z 2i) + Res(g, − 2i) dz = 2πi Res(g, 2 γ 2+z "√ √ # 2i − 2i √ = 2πi √ + 2 2i −2 2i = 2πi. To use the argument principle, write g(z) =

f 0 (z) 1 2z , = f (z) 2 2 + z2

with f (z) = 2 + z 2 . Then f 0 /f = 2g. Now f (z) has two simple zeros enclosed by γ, and no poles, so Z = 2, P = 0, and I I z 1 2z dz = dz 2 2 + z 2 1 + z2 γ γ 1 = (2πi)(Z − P ) = 2πi. 2 20. Let g(z) = tan(z). Then g has simple poles enclosed by γ at ±π/2. By the residue theorem, I tan(z) dz = 2πi[Res(g, π/2) + Res(g, −π/2)] γ sin(π/2) sin(−π/2) = 2πi + − sin(π/2) − sin(−π/2) = 2πi(−1 + (−1)] = −4πi. To use the argument principle, write g(z) = −f 0 (z)/f (z) with f (z) = cos(z). Now f (z) has no poles, and simple zeros at ±π/2 enclosed by γ. Then Z = 2 and P = 0, so I I 0 f (z) g(z) dz = − = −2πi(Z − P ) = −4πi. γ γ f (z) √ 21. g(z) = (z + 1)/(z 2 + 2z + 4) has simple poles at −1 ± 3i enclosed by γ. By the residue theorem I h √ √ i z+1 dz = 2πi Res(g, −1 − 3) + Res(g, −1 + 3) 2 γ z + 2z + 4 " # √ √ 1 − 1 − 3i −1 + 3i + 1 √ √ = 2πi + 2(−1 − 3i) + 2 2(−1 + 3i) + 2 1 1 = 2πi + = 2πi. 2 2

22.3. EVALUATION OF REAL INTEGRALS

505

To use the argument principle, write z+1 z 2 + 2z + 4

1 f 0 (z) 2 f (z)

where f (z) = z 2 + 2z + 4. f (z) has z = 2 zeros enclosed by γ and no poles (P = 0), so I 2z + 2 dz = πi(Z − P ) = 2πi. 2 + 2z + 4 z γ 22. Because p(z) has exactly n simple zeros enclosed by γ, then p0 (z)/p(z) has simple poles at z1 , · · · , zn and p0 (zj ) = 1. p0 (zj )

Res(p0 /p, zj ) =

By the residue theorem, I 0 n X p (z) dz = 2πi Res(p0 /p, zj ) = 2nπi. γ p(z) j=1 To use the argument principle, note that p(z) has exactly n simple zeros enclosed by γ, and a polynomial has no poles, so I 0 p (z) dz = 2πi(n − 0) = 2nπi. γ p(z)

22.3

Evaluation of Real Integrals

1. With z = eiθ , 1 cos(θ) = 2

1 z+ z

and dθ =

1 dz iz

so Z 2π 0

1 dθ = 2 − cos(θ)

z 2 − 4z + 1

= 2i

The integrand has simple poles at z1 = 2 − is enclosed by γ, and Res(f, 2 − Then

Z 2π 0

√

3) =

dz 1 γ 2 − 2 (z + 1/z) iz

2(2 −

dz.

√ √ 3 and z2 = 2 + 3. Only z1

1 1 √ =− √ . 3) − 4 2 3

1 −1 2π dθ = 2i(2πi) √ = √ . 2 − cos(θ) 2 3 3

506

CHAPTER 22. SINGULARITIES AND THE RESIDUE THEOREM

2. f (z) = 1/(z 4 + 1) has two simple poles in the upper half-plane: 1 1 z1 = √ (1 + i) and z2 = √ (−1 + i). 2 2 We need the residues of f (z) at these points: 1 1 and Res(f, z2 ) = − z2 . 4z12 4

Res(f, z1 ) = Then

Z ∞

1 2πi π dx = − (z1 + z2 ) = √ . 4+1 x 4 2 −∞

3. f (z) =√1/(1 + z 6 ) has simple √ poles in the upper half-plane at z1 = i, z2 = ( 3 + i)/2, and z3 = (− 3 + i)/2. The residues of f (z) at these poles are 1 1 Res(f, zj ) = 5 = − zj , 6zj 6 so

Z ∞

1 1 2π dx = 2πi (z + z + z ) = . 1 2 3 6 1 + x 6 3 −∞

Then

Z ∞ 0

1 π dx = . 1 + x6 3

4. Z 2π 0

1 1 dz 1 iz 1 + (z − 1/z) γ 2 I 1 =2 dz. 2 + 12iz + 1 z γ

1 dθ = 6 + sin(θ)

√ √ The integrand has simple poles at z1 = (−6+ 37)i and z2 = (−6− 37)i. Of these, only z1 is enclosed by γ. Further, Res(f, (−6 +

√ 37)i) =

1 1 = √ . 2z1 + 12i 2 37i

Then, recalling the factor of 2 in front of the previously written integral, we have Z 2π 1 1 2π dθ = 2πi √ =√ . 6 + sin(θ) 37i 37 0 5. Let f (z) =

ze2iz . z 4 + 16

22.3. EVALUATION OF REAL INTEGRALS

507

√ f has simple √ poles in the upper half-plane at z1 = (1 + i)/ 2 and z2 = (−1 + i)/ 2. Compute the residues: √

√

e2 2i(−1+i) e2 2 (−1 − i) Res(f, z1 ) = and Res(f, z2 ) = 16i −16i to obtain √

" x sin(2x) dx = Im 2πi 4 −∞ x + 16

Z ∞

e−2 2 8

√

e2 2i − e−2 2i 2i

√

√ πe−2 2 = sin(2 2). 4 2 6. f (z) = 1/(z√ − 2z + 6) has one simple pole in the upper half-plane, and it is z1 = 1 + 5i. The residue there is

Res(f, 1 + so

√ 1 5i) = √ 2 5i

Z ∞

−∞

x2 − 2x + 6

π dx = √ . 5

7. First use the identity cos2 (x) = to write

Z ∞

1 (1 + cos(2x)) 2

1 cos2 (x) dx = 2 2 2 −∞ (x + 4)

Z ∞

1 + cos(2x) dx. 2 2 −∞ (x + 4)

Let f (z) =

1 + e2iz . (z 2 + 4)2

Then f has a pole of order 2 in the upper half-plane at 2i. Compute d 1 + e2iz 1 + 5e−1 Res(f, 2i) = lim = . 2 z→2i dx (z + 2i) 32i Then Z ∞

cos2 (x) 1 1 + 5e−1 dx = Re 2πi 2 2 2 32i −∞ (x + 4) π = (1 + 5e−4 ). 32

508

CHAPTER 22. SINGULARITIES AND THE RESIDUE THEOREM

8. Complex methods will work with this integral, but it is easier to observe that, with the change of variables θ = 2π − ϕ, Z 2π Z π sin(θ) sin(ϕ) dθ = − dϕ. 2 + sin(θ) 2 + sin(ϕ) π 0 Then

Z 2π 0 2

sin(θ) dθ = 0. 2 + sin(θ)

9. Let f (z) = z /(z + 4) . The only singularity of f in the upper half-plane is 2i, which is a double pole. Compute d z2 i =− . Res(f, 2i) = lim 2 z→2i dz (z + 2i) 8 Then

Z ∞

x2 i π dx = 2πi − = . 2 + 4)2 (x 8 4 −∞

10. Let f (z) =

eiβx (z 2 + α2 )2

The only singularity of f in the upper half-plane is a double pole at αi. We need d eiαz Res(f, αi) = lim z→αi dz (z + αi)2 =−

(αβ + 1)e−αβ i. 4α3

Then Z ∞

cos(βx) (αβ + 1)e−αβ dx = 2πi − i 2 2 2 4α3 −∞ (x + α ) =

(αβ + 1)eαβ π . 2α3

11. Let

eiαz . z2 + 1 The only singularity f has in the upper half-plane is a simple pole at i. Compute e−α Res(f, i) = . 2i Then −α Z ∞ cos(αx) e dx = 2πi = πe−α . 2 2i −∞ x + 1 f (z) =

22.3. EVALUATION OF REAL INTEGRALS

509

12. Let

z 2 eiαz . (z 2 + β 2 )2 This function has a double pole at βi in the upper half-plane. The residue there is f (z) =

d z 2 eiαz z→βi dz (z + βi)2

Res(f, βi) = lim

= lim (2zeiαz (z + βi)−2 + iz 2 αz(z + βi)−2 − 2z 2 eiαz (z + βi)−3 ) z→βi

= 2βi(2βi)−2 + iα(−β)2 e−αβ (2βi)−2 − 2(−β 2 )e−αβ (2βi)−3 =

e−αβ i(αβ − 1). 4β

Then Z ∞

−αβ e x2 cos(αx) dx = 2πi i(αβ − 1 2 2 2 4β −∞ (x + β ) π −αβ = e (1 − αβ). αβ

13. Begin with Z 2π

1 dθ 2 cos2 (θ) + β 2 sin2 (θ) α 0 I 1 1 = dz 2 (z + 1/z)2 /4 − β 2 (z − 1/z)2 /4 iz α γ I 4 z = dz. i γ (α2 − β 2 )z 4 + 2(α2 + β 2 )z 2 + (α2 − β 2 )

Singularities of the integrand satisfy z2 =

β−α β−α or z 2 = . β+α β+α

Because α and β are positive, β−α β+α < 1 and > 1. β+α β−α The simple poles enclosed by the unit circle are the square roots z1 and z2 of (β − α)/(β + α). The residue of the integrand at each of these poles can be computed using Corollary 22.1. Omitting the arithmetic, we obtain Res(f, zj ) = for j = 1, 2. Then Z 2π 0

1 8αβ

2π

dθ = (2πi) = . i 8αβ αβ α2 cos2 (θ) + β 2 sin2 (θ)

510

CHAPTER 22. SINGULARITIES AND THE RESIDUE THEOREM

14. Let g(θ) =

1 . α + sin2 (θ)

Write Z π/2

Z π/2 g(θ) dθ =

Z π g(θ) dθ +

g(θ) dθ π/2

Z 2π

Z 3π/2

g(θ) dθ.

g(θ) dθ +

+ π

3π/2

In the second, third and fourth integrals on the right, put, respectively, θ = π − u, θ = π + u, θ = 2π − u. This leads to Z π/2

Z 1 2π 1 dθ 4 0 α + sin2 (θ) I 1 1 1 = dz 4 γ α − (z − 1/z)2 /4 iz I z =i dz. 4 − (2 + 4α)z 2 + 1 z γ

g(θ) dθ = 0

The integrand has two simple poles bounded by the unit circle, and satisfying p zj2 = (1 + 2α) − 2 α(α + 1). Omitting the arithmetic, the residues at each of these poles is the same, given by −1 Res(f, zj ) = p . 8 α(α + 1) Then Z π/2 0

" # −2 1 dθ = i(2πi) p α + sin2 (θ) 8 α(α + 1) π = p . 2 α(α + 1)

15. Let Γ be the given rectangular path. The four sides are: Γ1 : z = t, −R ≤ t ≤ R, Γ2 : z = R + it, 0 ≤ t ≤ β, Γ3 : z = t + iβ, −R ≤ t ≤ R, Γ4 : z = −R + it, 0 ≤ t ≤ β.

22.3. EVALUATION OF REAL INTEGRALS

511

These are, respectively, the lower side, right side, top side and left side of the rectangle. In carrying out the integrations, limits of integration must be consistent with counterclockwise orientation of Γ. 2

Because e−z is differentiable for all z, then by Cauchy’s theorem, I

e−z dz = 0 =

4 Z X

e−z dz.

Γj

j=1

Evaluate each of the four integrals in the sum on the right as follows. Z

Z R

−z 2

dz = −R

Γ1

−z 2

Z β

e−(r +2Rti−t ) i dt

dz =

Γ2

e−t dt,

= ie−R

Z β

et [cos(2Rt) − i sin(2Rt)] dt,

Z e

−z 2

Z R

e−(t +2βti−β ) dt

dz = −R

Γ3

= e−β

Z R

e−t [cos(2βt) − i sin(2βt)] dt,

−R

and Z e Γ4

−z 2

Z 0 dz =

e−(R −2Rti−t ) i dt

= ie−R

Z β

[et [− cos(2βt) − i sin(2βt)] dt.

0 2

Now let R → ∞. The terms having a factor of e−R go to zero in the limit, and upon adding these integrals over the sides of the rectangle, we obtain, using x as the variable of integration on the line, Z ∞ Z ∞ 2 2 e−x dx − e−β [cos(2βx) − i sin(2βx)] dx = 0. −∞

−∞

Now, e−x sin(2βx) is an odd function on the real line, so Z ∞ 2 e−x sin(2βx) dx = 0. −∞

We are therefore lift with Z ∞ Z ∞ 2 2 eβ cos(2βx) dx = e−x dx. −∞

−∞

512

CHAPTER 22. SINGULARITIES AND THE RESIDUE THEOREM Finally, use the known result that Z ∞ √ 2 e−x dx = π −∞

to conclude that Z ∞

e−x cos(2βx) dx =

√ −β 2 πe .

−∞ 2

Finally, because e−x cos(2βx) is an even function on the real line, then Z ∞ e

−x2

√ π −β 2 cos(2βx) dx = e . 2

16. Denote the given closed curve as Γ. By Cauchy’s theorem, I 2 eiz dz = 0. Γ

Γ consists of the segment Γ1 on the x− axis, the circular arc Γ2 , and the segment Γ3 from the end of this arc to the origin. On Γ1 , z = x and Z

iz 2

Z R dz =

Γ1

[cos(x2 ) + i sin(x2 )] dx.

On Γ2 , z = Reiθ and Z e

iz 2

Z π/4 dz =

Γ2

eiR e2iθ dθ.

On Γ3 , z = reiπ/4 and Z e Γ3

iz 2

Z 0 dz =

e−r eiπ/4 dr,

with the limits of integration set to maintain counterclockwise orientation around Γ. Now we want to take the limit as R → ∞. The integral over Γ3 has limit 2 zero because of the factor e−r . The integral over Γ1 only has R in its upper limit of integration and becomes an integral over the half-line. The integral over Γ2 is not as obvious. Make the change of variable u = 2θ to get Z Z 2 1 π/2 iR2 cos(u)−R2 sin(u) e iReiu/2 du. eiz dz = 2 0 Γ2

22.3. EVALUATION OF REAL INTEGRALS

513

Then Z

eiz dz ≤

Γ2

R 2

Z π/2

eiR cos(u) eiu/2 e−R sin(u) du

Z R π/2 −R2 sin(u) = e du 2 0 R π ≤ 2 2πR2 π = →0 4R as R → ∞. In the limit, we therefore obtain Z ∞ Z ∞ 2 2 2 iπ/4 [cos(x ) + i sin(x )] dx − e e−x dx = 0. 0

Because

√

Z ∞ e

−x2

dx =

and

π , 2

1 eiπ/4 = √ (1 + i), 2

we finally have Z ∞

Z ∞

cos(x ) dx + i 0

√ π sin(x ) dx = √ (1 + i). 2 2 2

From the real and imaginary parts of both sides of this equation, we have r Z ∞ Z ∞ 1 π cos(x2 ) dx = sin(x2 ) dx = . 2 2 0 0 17. First observe that, because the integrand is an even function, Z ∞ Z 1 ∞ x sin(αx) x sin(αx) dx = dx. x4 + β 4 2 −∞ x4 + β 4 0 Now, f (z) =

zeiαz z4 + β4

has simple poles in the upper half-plane at z1 = βeiπ/4 and z2 = βe3πi/4 . Compute the residues of f at these poles: iαz eiαβzk ze = . Res(f, zk ) = 4z 3 z=zk 4zk2 In particular, Res(f, z1 ) =

3iπ/4 1 iαβeiπ/4 1 e and Res(f, z2 ) = eiαβe . 4β 2 i −4β 2 i

514

CHAPTER 22. SINGULARITIES AND THE RESIDUE THEOREM Then Z ∞ 0

√ x sin(αx) 1 2πi iαβ(1+i)/√2 iαβ(−1+i)/ 2 1 dx = e − e Im x4 + β 4 2 4β 2 i √ −αβ/ 2 αβ πe sin √ . = 2 2β 2

18. First write Z 2π 0

1 dθ = (α + β cos(θ))2

Z π 0

Z 2π 1 1 dθ+ dθ. (α + β cos(θ))2 (α + β cos(θ))2 π

In the last integral on the right, put θ = 2π − u to show that the two integrals on the right are equal. Therefore Z π Z 1 1 2π 1 dθ = dθ 2 2 0 (α + β cos(θ))2 0 (α + β cos(θ)) I 1 1 1 = dz 2 2 γ (α + β(z + 1/z)/2) iz I 2 z dz. = 2 i γ βz + 2αz + β)2 The function f (z) =

z (βz 2 + 2αz + β)2

has double poles at the zeros of βz 2 + 2αz + β, which are p p −α + α2 − β 2 −α − α2 − β 2 z1 = and z2 = . β β Because z2 lies outside the unit disk, we need only the residue at z1 : d z Res(f, z1 ) = lim z→z1 dz β 2 (z − z1 )2 1 αβ 2 = 2 β 4(α2 − β 2 )3/2 α = . 2 4(α − β 2 )3/2 Then Z π 0

2 α 1 dθ = (2πi) (α + β cos(θ))2 i 4(α2 − β 2 )3/2 πα = 2 . (α − β 2 )3/2

Chapter 23

Conformal Mappings and Applications 23.1

Conformal Mappings

For Problems 1–3, the image of the given rectangle is given as a graph for each part of the problem. 1. (a) The rectangle defined by 0 ≤ x ≤ π, 0 ≤ y ≤ π maps to the sector 1 ≤ r ≤ eπ , 0 ≤ θ ≤ π. See Figure 23.1. (b) This rectangle maps to the sector π π 1 ≤ r ≤ e, − ≤ θ ≤ . e 2 2 See Figure 23.2. (c) The rectangle maps to the sector 1 ≤ r ≤ e, 0 ≤ θ ≤

π . 4

See Figure 23.3. (d) The rectangle maps to the sector e ≤ r ≤ e2 , 0 ≤ θ ≤ π. See Figure 23.4. (e) The rectangle maps to the sector π π 1 ≤ r ≤ e2 , − ≤ θ ≤ . e 2 2 See Figure 23.5. 515

516

CHAPTER 23. CONFORMAL MAPPINGS AND APPLICATIONS

Figure 23.1: Image of the rectangle 0 ≤ x ≤ π, 0 ≤ y ≤ π under w = ez .

Figure 23.2: Image of −1 ≤ x ≤ 1, −π/2 ≤ y ≤ π/2.

23.1. CONFORMAL MAPPINGS

517

Figure 23.3: Image of the rectangle 0 ≤ x ≤ 1, 0 ≤ y ≤ π/4.

Figure 23.4: Image of the rectangle 1 ≤ x ≤ 2, 0 ≤ y ≤ π.

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CHAPTER 23. CONFORMAL MAPPINGS AND APPLICATIONS

Figure 23.5: Image of the rectangle −1 ≤ x ≤ 2, −π/2 ≤ y ≤ π/2.

2. To determine the image of a rectangular area with sides parallel to the axes, we need to know the images of horizontal and vertical lines under this mapping. First write w = cos(z) = cos(x + iy) = u + iv = cos(x) cosh(y) − i sin(x) sinh(y), so u = cos(x) cosh(y) and v = − sin(x) sinh(y). This means that points x, y) map to points (cos(x) cosh(y), − sin(x) sinh(y)) in the w−plane. Consider the images of vertical and horizontal lines, beginning with a vertical line x = a. If cos(a) and sin(a) are nonzero, then u2 v2 − = 1. cos2 (a) sin2 (a) This is the image of one branch of a hyperbola in the w−plane, because cosh(y) > 0, forcing u > 0 if cos(a) > 0 and u < 0 if cos(a) < 0. Next consider the case that sin(a) = 0 or cos(a) = 0. If sin(a) = 0, then a = nπ, with n an integer, and cos(a) = (−1)n . If n is even, say n = 2m, then cos(a) = 1. Because cosh(y) ≥ 1, the vertical line x = 2mπ maps to the interval u ≥ 1, v = 0 on the real line in the

23.1. CONFORMAL MAPPINGS

519

Figure 23.6: Image of the rectangle 0 ≤ x ≤ 1, 1 ≤ y ≤ 2 under w = cos(z).

Figure 23.7: Image of the rectangle π/2 ≤ x ≤ π, 1 ≤ y ≤ 3.

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CHAPTER 23. CONFORMAL MAPPINGS AND APPLICATIONS

Figure 23.8: Image of the rectangle 0 ≤ x ≤ π, −π/2 ≤ y ≤ π.

w−plane. If n is odd, say n = 2m + 1, then cos(a) = −1 and a vertical line x = (2m + 1)π maps to the interval u ≤ −1, v = 0 in the w−plane. For the case cos(a) = 0, suppose a = (2n + 1)π/2, with n any integer. The image of a point on the line x = a is (0, − sin((2n + 1)π/2) sinh(y), varying over the entire imaginary axis in the w−plane as y varies over all real values. This takes care of images of horizontal lines. Now look at a horizontal line y = b. The image of a point on this line is a point (u, v) with u = cos(x) cosh(b), y = − sin(x) sinh(b). If b 6= 0, these points lie on the ellipse u2 v2 + = 1. cosh2 (b) sinh2 (b) The image of a horizontal x = b 6= 0 is an ellipse. If b = 0, then y = 0 is the real line. Now w = u = cos(x), and the image of the real line is the interval −1 ≤ u ≤ 1 on the real line in the w−plane. For the images of the rectangles of parts (a) through (e), see Figures 23.6– 23.10, respectively. 3. For the images of the given rectangles of parts (a) through (e), see Figures 23.11–23.15, respectively.

23.1. CONFORMAL MAPPINGS

521

Figure 23.9: Image of the rectangle π ≤ x ≤ 2π, 1 ≤ y ≤ 2.

Figure 23.10: Image of the rectangle 0 ≤ x ≤ π/2, 0 ≤ y ≤ 1.

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CHAPTER 23. CONFORMAL MAPPINGS AND APPLICATIONS

Figure 23.11: Image of the rectangle 0 ≤ x ≤ π/2, 0 ≤ y ≤ π/2.

Figure 23.12: Image of the rectangle π/4 ≤ x ≤ π, 0 ≤ y ≤ π/2.

23.1. CONFORMAL MAPPINGS

523

Figure 23.13: Image of the rectangle 0 ≤ x ≤ 1, 0 ≤ y ≤ π/6.

Figure 23.14: Image of the rectangle π/2 ≤ x ≤ 3π, 0 ≤ y ≤ π/2.

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CHAPTER 23. CONFORMAL MAPPINGS AND APPLICATIONS

Figure 23.15: Image of the rectangle 0 ≤ x ≤ π/2, 0 ≤ y ≤ 1.

Figure 23.16: Image of the rectangle 1 ≤ x ≤ 2, 1 ≤ y ≤ 2.

23.1. CONFORMAL MAPPINGS

525

4. Write z = reiθ so w = z 2 = r2 e2iθ . If r varies from 0 to ∞, so does r2 . And as θ varies from π/4 to 5π/4, 2θ varies from π/2 to 5π/2. This is an interval of length 2π. Therefore the image of the sector π/4 ≤ θ ≤ 5π/4 is the entire plane. 5. The analysis proceeds like that of Problem 4. Let z = reiθ . If π/6 ≤ θ ≤ π/3, then π/2 ≤ 3θ ≤ π, so image points under this mapping lie in the second quadrant. It is routine to verify that this mapping is onto the second quadrant. 6. Let u = reiθ . Then w = u + iv =

1 2

1 reiθ + e−iθ . e

Using Euler’s formula, we obtain 1 1 1 1 u= r+ cos(θ) and v = r− sin(θ). 2 r 2 r It is routine to check that 2 2 u v + 1 =1 1 2 (r + 1/r) 2 (r − 1/r) assuming that r 6= 1. This is an ellipse in the w−plane. Because r + 1/r > r − 1/r, this ellipse has foci at (±c, 0), where " 2 2 # 1 1 1 2 r+ c = − r− = 1. 4 r r This means that a circle of radius r 6= 1 maps to an ellipse with foci (±1, 0) in the w−plane. If r = 1, then v = 0 and u = 2 cos(θ), so the image of the unit circle about the origin in the x, y−plane is the interval [−2, 2] on the real axis in the w−plane. 7. Using some of the analysis from Problem 6, a half-line θ = k maps to points u + iv with 1 1 1 1 u= r+ cos(k), v = r− sin(k). 2 r 2 r If sin(k) 6= 0 and cos(k) 6= 0, then a little algebraic manipulation shows that v2 u2 − = 1. cos2 (k) sin2 (k) This is the equation of a hyperbola with foci (±c, 0), where c2 = cos2 (k) + sin(k) = 1.

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CHAPTER 23. CONFORMAL MAPPINGS AND APPLICATIONS

Figure 23.17: Image of the rectangle in Problem 8(a) with α = 2.

Finally, the cases cos(k) = 0 and sin(k) = 0 must be considered separately. If cos(k) = 0, then k = (2n + 1)π/2. Now u = 0 and −∞ < v < ∞, so the image is the imaginary axis in the x−plane. If sin(k) = 0, then k = nπ with n an integer. Now v = 0 and 1 1 r+ (−1)n . u= 2 r This is the half-interval u ≥ 1 on the real axis in the w−plane if n is even, and u ≤ −1 if n is odd. 8. (a) First let w = cos(z). The analysis of Problem 2 tells us the images of vertical and horizontal lines. Figure 23.16 shows the image of D for α = 2. Different choices of α result in different images. (b) For w = sin(z), use some of the analysis of Problem 3. Figure 23.17 shows the image of D for α = 2. 9. Write w = 2z 2 = 2(x + iy)2 = 2(x2 − y 2 ) + 4ixy = u + iv. Then vertical line x = 0 maps to u = −2y 2 , v = 0, so the image is the negative u− axis. Other vertical lines x = a 6= 0 map onto parabolas u = 2a2 −

v2 . 2a2

23.1. CONFORMAL MAPPINGS

527

Figure 23.18: Image of the rectangle in Problem 8b with α = 2.

The horizontal line y = 0 maps to u = 2y 2 ≥ 0, the positive u−axis. Other horizontal lines y = b 6= 0 map onto the parabolas u=

v2 − 2b2 . 8b2

Figure 23.18 shows the image of the rectangle 0 ≤ x ≤ 3/2, −3/2 ≤ y ≤ 3/2. 10. Let w = ez = ex+iy for all real x and for 0 ≤ y ≤ 2π. If we write w = ex cos(y) + iex sin(y), then y varies over an entire period of sin(y) and cos(y), so every point in the w−plane, except the origin, is the image of a point in the z−plane under this mapping. 11. If Re(z) = −4, then (z + z)/2 = −4, so z + z = −8. Now, w = 2i/z, so z = 2i/w, so 2i 2i z+z = − = −8. w w Multiply this by ww and rearrange terms to obtain 8ww − 2i(w − w) = 0.

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CHAPTER 23. CONFORMAL MAPPINGS AND APPLICATIONS

Figure 23.19: Image of the rectangle in Problem 9 with α = 2.

Now put w = u + iv to get 2(u2 + v 2 ) + v = 0, or

1 u + v+ 4 2

2 =

1 . 4

This is the equation of a circle of radius 1/2 centered at (0, −1/4) in the w− plane. This is the image of the line x = −4 under the given mapping. 12. If w = 2iz − 4, then z=

w+i . 2i

Now Re(z) = (z + z)/2 becomes w+4 w+4 − = 10. 2i 2i Let w = u + iv to obtain w−w = Im(w) = v = 10. 2i This is a horizontal line in the w− plane.

23.1. CONFORMAL MAPPINGS

529

13. From the mapping, solve for z: z=

−1 . w+i

Substitute this into the given line to obtain −1 1 1 −1 1 1 − + + = 1. 2 w+i w−i 2i w + i w − i Multiply this equation by 2i(w + i)(w − i) and rearrange terms, putting w = u + iv to obtain 4(u2 + v 2 ) + 7v + u = 3. Complete the square to write this as 2 2 7 1 1 + v+ = . u+ 8 8 32 √ This is a circle of radius 1/2 2 and center (−1/8, −7/8), and is the image of the given line. 14. Solve the mapping for z to obtain |z| = 4 =

−w − 1 + i . 2w − 1

Then |w + 1 − i| = 4|2w − 1|. Put w = u + iv to get (u + 1)2 + (v − 1)2 = 16(2u − 1)2 + 64v 2 . Rearrange terms to write this as 2 2 1 208 11 + v+ = . u− 21 63 3969 p This is the equation of a circle of radius 208/3969 and center (11/21, −1/63). 15. Invert the mapping to obtain z=

5 + iw . 2−w

Then

5 − v + iu z − z = 2Re(z) = 2Re 2 − u − iv 2((5 − v)(2 − u) − uv) = (u − 2)2 + v 2 20 − 4v − 10u = . (u − 2)2 + v 2

530

CHAPTER 23. CONFORMAL MAPPINGS AND APPLICATIONS Next, 1 (2 − u)u + (5 − v)v . (z − z) = Im(z) = 2i (u − 2)2 + v 2 Substitute these into the equation of the given line and clear fractions to obtain 2 19 377 (u − 1)2 + v + = . 4 16 √ This is the equation of a circle with radius 377/4 and having center (1, −19/4).

16. From the mapping, obtain z=

−2 . w − 3i − 1

Substitute this into |z − i| = 1 to get −2 − i = 1. w − 3i − 1 Then |w − 3i − 1| = | − 2 − iw − 3 + i|. Let w = u + iv in this expression to obtain (u − 1)2 + (v − 3)2 = (u − 1)2 + (v − 5)2 . From this, v = 4. The map is a translation followed by an inversion that sends the given circle to the vertical line v = 4. 17. Substitute the given values into equation (23.1) to obtain (1 − w)(1 + 2i)(−1)(3 − z) = (1 − z)(1)(1 + i)(1 − i − w). Solve for w to obtain w=

(1 + 4i)z − (3 + 8i) . (2 + 3i)z − (4 + 7i)

18. Substitute the given values into equation (23.1) and solve for w to obtain w=

(1 + i)z − (2 + 2i) . (3 − i)z − 2

19. Here w3 = ∞ so use equation (23.2), obtaining (1 + i − w)(1 − 2i)(4 − z) = (1 − z)(−2 + 2i)(4 − 2i). Then w=

(33 + i)z − (48 + 16i) . 5(z − 4)

23.1. CONFORMAL MAPPINGS

531

20. Using equation (23.2), we obtain w=

(9 − 7i) − (21 + 27i) . 13(z + 1)

21. Using equation (23.1), we find that w=

(3 + 22i) + 4 − 75i . (2 + 3i) − (21 − 4i)

22. Let f be a conformal mapping from the z−plane to the w−plane. Let g be a conformal mapping from the w−plane to the Z−plane. We want to show that g ◦ f is conformal. Certainly g ◦ f is differentiable. Let C1 and C2 be paths in D intersecting at P at an angle θ (the angle between their tangents at P ). Because f is conformal, f (C1 ) and f (C2 ) are paths in the w−plane intersecting at angle θ at f (P ). Because g is conformal, g(f (C1 )) and g(f (C2 )) are paths in the Z−plane intersecting at the same angle θ. Therefore g ◦ f preserves angles. Further, g ◦ f preserves orientation. If θ is the angle between C1 and C2 , going counterclockwise, then this sense of orientation is preserved by f , then by g, and is therefor also preserved by g ◦ f . Therefore g ◦ f is conformal. 23. If we require that a conformal mapping be differentiable, then immediately T (z) = z is disqualified. But it is also easy to see directly that this mapping reverses orientation. For example, let C1 be the nonnegative real axis and C2 the nonnegative imaginary axis. The sense of rotation from C1 to C2 is counterclockwise. But T maps C1 to itself and C2 to the negative imaginary axis, reversing the orientation to clockwise. Therefore T is not conformal. 24. Suppose T is a bilinear transformation that is not the identity map and not a translation. Write az + b T (z) = . cz + d If z is a fixed point of T , then T (z) = z =

az + b . cz + d

Then cz 2 + (d − a)z − b = 0. If c 6= 0, this is a quadratic equation with two roots, so T has two fixed points (if these roots are distinct) and one fixed point (if the roots are repeated). This argument fails if c = 0, in which case T is a translation T (z) =

b a z+ d d

532

CHAPTER 23. CONFORMAL MAPPINGS AND APPLICATIONS if b 6= 0. If T cannot be a translation, then b = 0 and then T (z) = (a/d)z = kz. If a = d this is the identity mapping T (z) = z. If a 6= d, then T is a magnification/rotation, which has only 0 as a fixed point.

25. Let

az + b . cz + d By the argument of Problem 24, if T is not a translation or the identity mapping, then T can have at most two fixed points. Therefore, if T has three fixed points, then T is either a translation or the identity map. But a translation has no fixed point, so in this case T must be the identity map. T (z) =

26. First, observe that, if az + b cz + d is a bilinear map, then ad−bc 6= 0, which guarantees that T has an inverse. This inverse is −dw + b z = T −1 (w) = cw − a T (z) =

which is also bilinear. The composition T −1 ◦ T is the identity map I of the z−plane to itself, I(z) = z. Now suppose T (zj ) = S(zj ) for j = 1, 2, 3. Then (S −1 ◦ T )(zj ) = S −1 (T (zj )) = S −1 (S(zj )) = I(zj ) = zj . Then S −1 ◦ T has three fixed points and is therefore the identity map: S −1 ◦ T = I. Then S = S ◦ I = S ◦ (S −1 ◦ T ) = (S −1 ◦ S) ◦ T = I ◦ T = T. 27. Given z2 , z3 , z4 , let P be the unique bilinear transformation that maps z2 → 1, z3 → 0, z4 → ∞. Then by definition of the cross ratio, P (z1 ) = [z1 , z2 , z3 , z4 ]. Now let T be any bilinear transformation. Then [T (z1 ), T (z2 ), T (z3 ), T (z4 )] = R(T (z1 )),

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533

where R is the unique bilinear transformation that maps T (z2 ) = 1, T (z3 ) = 0, T (z4 ) = ∞. Then R ◦ T = P . Then [T (z1 ), T (z2 ), T (z3 ), T (z4 )] = R(T (z1 )) = P (z1 ) = [z1 , z2 , z3 , z4 ]. 28. Let w = T (z) = 1 −

z3 − z4 z − z2 . z3 − z2 z − z4

A routine calculation shows that w2 = T (z2 ) = 1, w3 = T (z3 ) = 0, and T (z4 ) = ∞. Because three points and their images uniquely determine a bilinear transformation, T is the unique bilinear transformation mapping z2 → 1, z3 → 0 and z4 → ∞. Therefore [z1 , z2 , z3 , z4 ] = T (z4 ). 29. In the definition of cross ratio, w2 , w3 and w4 all lie on an (extended) line, the real axis. Because bilinear transformations map lines/circles to lines/circles, then [z1 , z2 , z3 , z4 ] is real exactly when z1 , z2 , z3 , z4 ] all lie on the same line or circle.

23.2

Construction of Conformal Mappings

If a conformal mapping is requested between two domains, there will in general be many different possible mappings. In each of the solutions below, one mapping is found, but other approaches may yield other suitable mappings. 1. Both domains are open disks, having radii 3 and 6, and different centers, 0 and 1 − i, respectively. We can map |z| < 3 ont |w − 1 + i| < 6 by using a scaling factor of 2 and a translation to superimpose the center of the initial domain onto the center of the image domain. Thus compose z → 2z → 2z + 1 − i. One mapping that does what we want is w = 2z + 1 − i. Now |w − 1 + i| = 2|z| = 2(3) = 6 if |z| = 3.

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CHAPTER 23. CONFORMAL MAPPINGS AND APPLICATIONS

2. First perform an inversion (to map the interior of the unit disk to the exterior), w1 = 1/z. Then expand by a factor of 18 by w2 = 18w1 (18 because (1/3)(18) = 6, the radius of the target disk, then, then translate to the new center w3 = w2 + 1 − i. Putting these together, we have: w=

(1 − i)z + 18 18 +1−i= . z z

3. We will need an inversion at some stage because we are mapping the interior of a disk to the exterior of another disk. First translate by w1 = z + 2i, so the image disk in the w−plane has the origin as its center. Next invert by 1 w2 = . z + 2i Next scale by a factor of 2 to match the radii of the bounding circles: w3 = 2w2 =

2 . z + 2i

Finally, translate to have center 2: w4 = w3 + 3 =

2 3z + 2 + 6i +3= . z + 2i z + 2i

4. A mapping of the half-plane Re(z) > 1 onto the half-plane Im(z) > −1 can be achieved by first rotating counterclockwise by π/2 (by w1 = iz), and then shifting down two units by w2 = w1 − 2i. This is all done by the mapping w = iz − 2i = (z − 2)i. 5. We can map the line Re(z) = 0 to the circle by |w| = 4 by a bilinear transformation. The domain Re(z) < 0 consists of numbers to the left of the imaginary axis, which is the boundary of this domain. Choose three points on this axis, ordered upward so the region Re(z) < 0 is on the left as we walk up the line. Next choose three points on the image circle |w| = 4, counterclockwise so the interior of the circle is on the left as we walk counterclockwise around the circle. Convenient choices are z1 = −i, z2 = 0, z3 = i and w1 = −4i, w2 = 4, w3 = −4i. Using equation (23.1), we find the bilinear transformation mapping zj → wj : 1+z w = T (z) = 4 . 1−z As a check, z = −1, which has negative real part, maps to 0, interior to the disk |w| < 4. Thus T maps Re(z) < 1 to |w| < 4, rather than to the exterior |w| > 4.

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535

6. The domain Im(z) > −4 consists of all x + iy lying above the horizontal line y = −4. The boundary of this domain is this line. We want to map this domain onto |w − i| > 2, the exterior of the circle of radius 2 and center i in the w−plane. The circle |w − i| = 2 is the boundary of this domain. Choose three points on the line y = −4 in the z−plane, ordered from left to right so the domain is on the left. Choose three points on the circle in the w−plane, clockwise so as we walk around this circle, the exterior is on the left. Convenient choices are z1 = −1 − 4i, z2 = −4i, z3 = 1 − 4i and w1 = 3i, w2 = 2 + i, w3 = −i. The bilinear transformation mapping zj → wj (use equation (23.1)) is w=

(−2 + i)z − (3 + 10i) . z + 3i

7. Because the boundary of the wedge in the w−plane is not a circle or line, a bilinear transformation will not work here. However, wedges suggest polar representations. Let z = reiθ for 0 < θ < π. These are points in the upper half-plane. Let w = z 1/3 = r1/3 eiθ/3 = ρeiϕ . Here ρ > 0 and 0 ≤ ϕ ≤ π/3. This mapping is conformal because dw 1 = z −2/3 6= 0 dz 3 for z in the upper half-plane, and the mapping takes the open upper halfplane onto the open wedge 0 < θ < π/2. 8. Let z = x + iy = reiθ with y > 0. Then arg(z) = θ is unique (restricted to 0 ≤ θ < 2π). Further, w = ln(r) + iθ. Because r can be any positive number, ln(r) varies over all the real numbers. Further, Im(z) = θ in (0, π). Therefore log(z) is in the strip 0 < Im(z) < π. To show that this mapping is onto, choose any w = u + iv in this strip. Let z = rew . Then Im(z) = eu sin(v) > 0 and log(z) = u + iv = w. The mapping is therefore onto. Finally, 1 d (log(z)) = 6= 0, dz z so the mapping is conformal.

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9. The solution of this problem requires some familiarity with the gamma and beta functions. To show that f maps the upper half-plane onto the given rectangle, we will evaluate the function at −1, 0, 1 and ∞ and then show that these are the vertices of that rectangle. First, it is obvious that f (0) = 0. Next, Z 1 f (1) = 2i

(ξ 2 − 1)−1/2 ξ −1/2 dξ

Z 1 = 2i 0

Z 1 =2

(1 − ξ 2 )−1/2 −1/2 ξ dξ i

(1 − ξ 2 )−1/2 ξ −1/2 dξ.

Let ξ = u1/2 to obtain Z 1 f (1) =

(1 − u)−1/2 u−3/4 du

= B(1/4, 1/2) =

Γ(1/4)Γ(1/2) = c, Γ(3/4)

in which B(x, y) is the beta function and Γ(x) is the gamma function. Next, write Z 1 f (−1) = 2i (ξ 2 − 1)−1/2 ξ −1/2 dξ. 0

Let ξ = −u to obtain Z 1 f (−1) = 2i

(1 − u2 )−1/2 u−1/2 du

= iB(1/4, 1/2) = i

Γ(1/4)Γ(1/2) = ic. Γ(3/4)

Finally, Z ∞ f (∞) = 2i

(ξ + 1)−1/2 (ξ − 1)−1/2 ξ −1/2 dξ

Z 1 = 2i

(ξ + 1)−1/2 (ξ − 1)−1/2 ξ −1/2 dξ

Z 0∞ + 2i

(ξ + 1)−1/2 (ξ − 1)−1/2 ξ −1/2 dξ.

The first integral in the last line of the last equation is B(1/4, 1/2). In

23.2. CONSTRUCTION OF CONFORMAL MAPPINGS

537

the second integral, put ξ = 1/u to get Z 0 f (∞) = c + 2i 1

Z 1 = c + 2i

1+u u

−1/2

1−u u

−1/2

1/2

1 u2

(1 − u2 )−1/2 u−1/2 du = (1 + i)u.

Solutions - Problems in Statistics Section 1 Measures of Center and Variation 1. (a) The mean is x=

−4 − 6 + 2.5 + 3 + 8 + 5 − 3 − 4 + 8 + 3 + 2.2 = 1.3364. 11

For the median, write the list in nondecreasing order: −6, −4, −4, −3, 2.2, 2.5, 3, 3, 5, 8, 8. With an odd number N = 11 of numbers in the list, the median is the sixth number, 2.5. Here, 6 is obtained as N +1 12 = = 6. 2 2 There are ﬁve numbers of the list to the left of 2.5, and ﬁve numbers of the set to the right. The standard deviation s is

239.44 = 4.8933. 10

(b) The mean is x=

1+1+1−1+2+3−1+4+2 = 1.3333. 9

Arranged in nondecreasing order, the sequence is −1, −1, 1, 1, 1, 2, 2, 3, 4. With N = 9 numbers, the median is the ﬁfth number from the left. This is 1 (not just any 1, but the third of three ones, from the left). There are four numbers of the set to the left of this 1, and four numbers to the right. The standard deviation is

22 = 1.6583. 8

3 − 4 + 2 + 1.5 − 4 − 4 + 2 + 1 + 7 = 0.5. 9

In nondecreasing order, the list is −4, −4, −4, 1, 1.5, 2, 2, 3, 7. 1

The median is 1.5. The standard deviation is

115 = 3.7914. 8

(d) The mean is 9.3 + 9.5 + 9.7 + 10 + 8.4 + 8.7 + 8.8 + 8.8 + 4.1 = 8.5889. 9 As a nondecreasing list, we have x=

4.1, 8.4, 8.7, 8.8, 8.8, 9.3, 9.5, 9.7, 10. The median is 8.8. This is the ﬁfth number from the left in the list, and is the second 8.8 from the left. The standard deviation is

24, 849 = 1.7624. 8

(e) The mean is −16 − 14 − 10 + 0 + 0 + 1 + 1 + 3 + 5 + 7 = −2.3. 10 The data is already listed in nondecreasing order. the median is the average of the ﬁfth and sixth numbers in this list. This median is x=

0+1 1 = . 2 2 In this case that the number of data points is even, the median need not be a number in the set. The standard deviation is

584.1 = 8.0561. 9

2. Making use of the frequency table, compute the mean as 4(−12) + 2(−9.7) + 6(−8) + 4(−7.6) + 12(−5.1) + 3(4) = −6.2903. 31 The data is given in nondecreasing order. The number of entries is an odd number, 31, so the median is the sixteenth number from the left, −7.6 (the right-most −7.6 in the list). This is obtained by using the frequencies to count up to the sixteenth number from the left. x=

The standard deviation is s=

512.73 = 4.1341. 30 2

2 3 4 5 6 7

3 4 5 6 7 8

4 5 6 7 8 9

5 6 7 8 9 10

6 7 8 9 10 11

7 8 9 10 11 12

Table 1: Sums of pairs of dice, Problem 1, Section 2.

3. The mean is x=

4(−3) + 2(−1) + 6(0) + 4(1) + 12(3) + 3(4) = 1.2258. 31

The median is the sixteenth number from the left, or 1 (the last 1 to the right in the ordered list). The standard deviation is s=

151.42 = 2.2466. 30

Section 2 Random Variables and Probability Distributions 1. If we roll two dice, there are thirty-six possible outcomes. The sums of the numbers that can come up on the two dice are listed in Table 1. For example, if o is the outcome that one die comes up 2 and the other 3, then the sum of the dice is 5, so X(o) = 5. The table gives all of the values that X(o) can take on, over all outcomes o of the experiment. Each value is listed as often as it occurs as a value of X. For example, 4 occurs three times, because X(o) = 4 for three diﬀerent outcomes (namely (2, 2), (1, 3) and (3, 1)). Deﬁne a probability distribution P on X by by letting P (x) be the probability of x, for each value x that X can assume. For example, since 2 occurs once out of 36 entries in this table, assign to this value of X the probability P (2) =

1 . 36

Similarly, 11 occurs twice, so give the value 11 of X the probability P (11) =

2 . 36

Since 3 occurs twice in the table, P (3) = 2/36, and so on. Calculating P (n) for each n in the table, we obtain 2 1 , P (3) = P (11) = , 36 36 4 3 , P (5) = P (9) = , P (4) = P (10) = 36 36 6 5 , P (7) = . P (6) = P (8) = 36 36

P (2) = P (12) =

Notice that

x P (x) = 1, as required for a probability function.

The mean of X is μ= xP (x) x

2 3 4 5 6 1 +3 +4 +5 +6 +7 36 36 36 36 36 36 4 3 2 1 5 +9 + 10 + 11 + 12 +8 36 36 36 36 36

= 7. This is interpreted to mean that, on average, we expect to come up with a seven if we roll two dice. This is a reasonable expectation in view of the fact that there are more ways to roll 7 than any other sum with two dice. The standard deviation is

σ=

(x − 7)2 P (x). x

To compute this, ﬁrst compute (x − 7)2 P (x) x

1 2 3 + (3 − 7)2 + (4 − 7)2 36 36 36 4 5 6 + (6 − 7)2 + (7 − 7)2 + (5 − 7)2 36 36 36 5 4 3 + (9 − 7)2 + (10 − 7)2 + (8 − 7)2 36 36 36 2 1 + (11 − 7)2 + (12 − 7)2 36 36 = (2 − 7)2

= 5.8333. Then σ=

√

5.8333 = 2.4152. 4

2. Flip four coins, with sixteen possible outcomes. If o is an outcome, X(o) can have only two values, namely 1 if two, three, or four tails are in o, or 3 otherwise (one tail or no tails in o). There are ﬁve outcomes with one tail or no tail, and eleven with two or more tails, so 5 11 and P (3) = . 16 16

P (1) = The mean is μ=

xP (x) = 1

11 16

5 16

26 = 1.625. 16

For the standard deviation of X, compute (x − μ)2 P (x) x

= (1 − 1.625)2

11 16

+ (3 − 1.625)2

5 16

= 0.85938. Then σ=

√

0.85938 = 0.92703.

3. We have X(1) = 0, X(2) = X(3) = X(5) = X(7) = X(11) = X(13) = X(17) = X(19) = 1, X(4) = X(6) = X(9) = X(10) = X(14) = X(15) = 2, X(8) = X(12) = X(18) = X(20) = 3, X(16) = 4. The values assumed by X are 0, 1, 2, 3, 4. From the list of values, we get P (0) =

8 6 4 1 1 , P (1) = , P (2) = , P (3) = , P (4) = . 20 20 20 20 20

These are the probabilities of the values of the random variable X. The mean of X is μ=

xP (x)

8 6 1 +1 +2 20 20 20 1 4 +4 +3 20 20

= 1.8. 5

(1,1) (2,1) (3,1) (4,1) (5,1) (6,1)

(1,2) (2,2) (3,2) (4,2) (5,2) (6,2)

(1,3) (2,3) (3,3) (4,3) (5,3) (6,3)

(1,4) (2,4) (3,4) (4,4) (5,4) (6,4)

(1,5) (2,5) (3,5) (4,5) (5,5) (6,5)

(1,6) (2,6) (3,6) (4,6) (5,6) (6,6)

Table 2: Outcomes of rolling two dice, Problem 4, Section 2.

For the standard deviation of X, ﬁrst compute (x − μ)2 P (x) x

1 8 2 + (1 − 1.8) = (0 − 1.8) 20 20 6 4 + (3 − 1.8)2 + (2 − 1.8)2 20 20 1 2 = 0.9600. + (4 − 1.8) 20 2

Then σ=

(0.9600) = 0.9798.

4. The outcomes of two rolls of the dice are displayed in Table 2. Then X(n, n) = n for n = 1, 2, 3, 4, 5, 6, X(1, 2) = X(2, 1) = X(2, 4) = X(4, 2) = X(3, 6) = X(6, 3) = 2, X(1, 3) = X(3, 1) = X(2, 6) = X(6, 2) = 3, X(1, 4) = X(4, 1) = 4, X(1, 5) = X(5, 1) = 5, X(1, 6) = X(6, 1) = 6, X(2, 3) = X(3, 2) = X(4, 6) = X(6, 4) = 3/2, X(2, 5) = X(5, 2) = 5/2, X(3, 4) = X(4, 3) = 4/3, X(5, 3) = X(3, 5) = 5/3, X(4, 5) = X(5, 4) = 5/4, X(5, 6) = X(6, 5) = 6/5. Using this list to compute probabilities, we ﬁnd that 7 5 1 , P (2) = , P (3) = , 36 36 36 3 3 3 P (4) = , P (5) = , P (6) = . 36 36 36 P (1) =

The mean of X is μ= P (x) x

1 7 5 3 3 3 +2 +3 +4 +5 +6 36 36 36 36 36 36 3 4 5 2 4 2 5 2 5 2 6 2 + + + + + + 2 36 2 36 3 36 3 36 4 36 5 36 =1

= 2.6916. Using μ, we can compute the standard deviation of X: (x − μ)2 P (x) x

1 7 5 + (2 − 2.6916)2 + (3 − 2.6916)2 36 36 36 3 3 3 + (5 − 2.6916)2 + (6 − 2.6916)2 + (4 − 2.6916)2 36 36 36 4 2 2 + (3/2 − 2.6916)2 + (5/2 − 2.6916)2 + (4/3 − 2.6916)2 36 36 36 2 2 21 + (5/4 − 2.6916)2 + (6/5 − 2.6916)2 + (5/3 − 2.6916)2 36 36 36 = (1 − 2.6916)2

= 2.2442. Then σ=

√

2.2442 = 1.4981.

5. Draw cards from a (ﬁfty-two card) deck. There are 52 C2 ways to do this, disregarding order. If o is an outcome in which both cards are numbered, then X(o) equals the sum of the numbers on the cards. If exactly one of the cards is a face card or ace, then X(o) = 11, and if both cards are chosen from the face cards or aces, then X(o) = 12. Therefore the values of X(o) for all possible outcomes are 4, 5, · · · , 20. By a routine but tedious counting of the ways the numbered cards can take on various possible totals, we obtain 16 22 32 6 , P (5) = , P (6) = , P (7) = , 1326 1326 1326 1326 48 54 640 38 , P (9) = , P (10) = , P (11) = , P (8) = 1326 1326 1326 1326 64 54 48 190 , P (13) = , P (14) = , P (15) = , P (12) = 1326 1326 1326 1326 32 22 16 38 , P (17) = , P (18) = , P (19) = , P (16) = 1326 1326 1326 1326 6 P (20) = . 1326 P (4) =

Using these, compute the mean of X: μ= xP (x) = 11.566 x

and the standard deviation of X: √ σ= (x − μ)2 P (x) = 5.4289 = 2.33. x

6. X takes on values 1, 2, π. In particular, X(1) = X(5) = X(7) = X(11) = X(13) = X(17) = X(19) = X(23) = X(25) = X(29) = π, X(2) = X(4) = · · · = X(even integer) = 1, X(3) = X(9) = X(15) = X(21) = X(27) = 2. These enable us to write the probability distribution P (1) =

1 5 1 10 1 15 = , P (2) = = , P (π) = = . 30 2 30 6 30 3

The mean of X is μ=1

1 1 1 5 π +2 +π = + , 2 6 3 6 3

and this is approximately 1.8805. The standard deviation of X is the square root of (x − μ)2 P (x) x

= (1 − 1.8805)2

1 1 1 + (2 − 1.8805)2 + (π − 1.8805)2 , 2 6 3

which is approximately 0.92014. Then √ σ = 92014 = 0.95924.

Section 3 The Binomial and Poisson Distributions

1. (a) P (2) =

8 (0.43)2 (1 − 0.43)6 = 0.17756. 2

(b) P (3) = (c) P (3) =

4 (0.7)3 (1 − 0.7) = 0.4116. 3 6 (0.5)3 (1 − 0.5)3 = 0.3125. 3

(d) P (2) + P (3) + P (4) + P (5) 10 10 (0.6)2 (0.4)8 + = (0.6)3 (0.4)7 2 3 10 10 4 6 + (0.6) (0.4) + (0.6)5 (0.4)5 4 5 = 0.36521. (e) P (7) + P (8) =

8 8 (0.4)8 (0.4)7 (0.6) + 8 7

= 0.0085197. (f) P (2) + P (3) + P (4) 10 10 10 (0.58)3 (0.42)7 + (0.58)4 (0.42)6 = (0.58)2 (0.42)8 + 3 4 2 = 0.19908. (g) 10 10 3 7 (0.35)7 (0.65)3 P (3) + P (7) = (0.35) (0.65) + 7 3 = 0.27342. (h) P (1) + P (3) + P (5) 7 7 7 (0.24)3 (0.76)4 + (0.24)5 (0.76)2 = (0.24)(0.76)6 + 3 5 1 = 0.49481.

2. (a) The probability of getting exactly one question right is 6 P (1) = (0.25)(0.75)5 = 0.35596. 1 (b) The probability of getting exactly two questions right is 6 P (2) = (0.25)2 (0.75)4 = 0.29663. 2 (c) The probability of getting four or ﬁve right is 6 6 P (4) + P (5) = (0.25)4 (0.75)2 + (0.25)5 (0.75) = 0.037354. 4 5 (d) The probability of getting all six questions right is 6 P (6) = (0.25)6 = 0.00024414. 6 3. Now (referring to Problem 2) the number of choices of choices for each question is three instead of four. This changes the probability of guessing the right answer on any particular question from 1/4 to 1/3, leading to the following probabilities. (a) The probability of getting exactly one question right is 6 P (1) = (1/3)(2/3)5 = 0.26337. 1 (b) The probability of getting exactly two questions right is 6 P (2) = (1/3)2 (2/3)4 = 0.32922. 2 (c) The probability of getting four or ﬁve right is 6 6 P (4) + P (5) = (1/3)4 (2/3)2 + (1/3)5 (2/3) = 0.098765. 4 5 (d) The probability of getting all six right is 6 P (6) = (1/3)6 = 0.0013717. 6

4. Now, again referring to the setting of Problem 2, assume that there are ten questions, with four choices. The probability of guessing the right answer now is 0.25 on any question. (a) The probability of getting exactly one question right is 10 P (1) = (0.25)(0.75)9 = 0.18771. 1 (b) The probability of getting exactly two questions right is 10 P (2) = (0.25)2 (0.75)8 = 0.25157. 2 (c) The probability of getting four or ﬁve questions right is P (4) + P (5) 10 10 = (0.25)4 (0.75)6 + (0.25)5 (0.75)5 = 0.20440. 4 5 (d) The probability of getting all ten questions right is 10 P (10) = (0.25)10 = 9.5367(10−7 ), 10 and this is for all practical purposes zero (e) The probability of getting ﬁve, six or seven questions right is P (5) + P (6) + P (7) 10 10 10 5 5 6 4 (0.25) (0.75) + (0.25)7 (0.75)3 = (0.25) (0.75) + 6 7 5 = 7.7711(10−2 ). 5. We want the probability that two or fewer starters are defective. With N = 100, p = 0.032 (probability of producing a defective starter), compute P (≤ 2) = P (0) + P (1) + P (2) 100 100 0 100 (0.032)(1 − 0.032)99 = (0.032) (1 − 0.032) + 1 0 100 + (0.032)2 (1 − 0.032)98 = 0.37585. 2 6. With N = 1000 and p = 0.07, compute 1000 P (92) = (0.07)92 (1 − 0.07)908 = 0.0014477. 92 11

7. Let μ = 295/320 = 0.922 (to three places). (a) P (3) =

(0.922)3 e−0.922 = 0.052. 3!

(b) P (2) + P (3) + P (4) + P (5) (0.922)3 e−0.922 (0.922)2 e−0.922 + 2! 3! (0.922)5 e−0.922 (0.922)4 e−0.922 + + 4! 5! = 0.23519. =

8. (a) P (6) =

(0.9)6 e−0.9 = 3.0009(10−4 ). 6!

P (10) =

(0.85)10 e−0.85 = 2.3189(10−8 ) 10!

(b)

(0.92)4 e−0.92 = 0.011896. 4!

(0.87)8 e−0.87 = 3.4103(10−6 ). 8!

(e) P (1) + P (2) + P (3) + P (4) + P (5) (0.94)2 e−0.94 (0.94)3 e−0.94 (0.94)e−0.94 + + 1! 2! 3! (0.94)4 e−0.94 (0.94)5 e−0.94 + + 4! 5! = 0.60894. =

(f) P (3) + P (4) + P (5) (0.64)4 e−0.64 (0.64)5 e−0.64 (0.64)3 e−0.64 + + 3! 4! 5! = 0.027196.

(g) P (3) + P (8) = (0.75)8 e−0.75 (0.75)3 e−0.75 + 3! 8! = 0.033214.

(h) P (1) + P (3) + P (10) (0.97)3 e−0.97 (0.97)10 e−0.97 (0.97)e−0.97 + + 1! 3! 10! = 0.42537.

9. Put μ = 476/500 = 0.952. The probability that any region was hit exactly three times is (0.952)3 e−0.952 P (3) = = 0.055502. 3! The probability that any region was hit from two to six times is P (2) + P (3) + P (4) + P (5) + P (6) (0.952)3 e−0.952 (0.952)4 e−0.952 (0.952)2 e−0.952 + + 2! 3! 4! (0.952)5 e−0.952 (0.952)6 e−0.952 + + 5! 6! = 0.24653. =

10. Take μ = 997850/1000000 = 0.998. We want P (3) =

(0.998)3 e−0.998 = 0.061068. 3!

Section 4 Normally Distributed Data and Bell Curves 1. Roll a die ninety times (or, equivalently, roll ninety dice). We are interested in the outcome that a 6 comes up (by this, we mean a six on at least one die). (a) With N = 90, the number of rolls, and p = 1/6, the probability of having any particular face come up on one roll, the mean is 1 μ = N p = 90 = 15. 6 13

The standard deviation is σ=

N p(1 − p) =

1 5 1 . 90 =5 6 6 2

This is approximately 3.5355. (b) Since 2σ 2 = 25, the bell curve for this experiment is the graph of the exponential function √

2 2 2 1 1 e(x−μ) /2σ = √ e−(x−15) /25 . 5 π 2πσ

60+1/2 30−1/2

e−(x−15) /25 dx ≈ (2.0549)(10−5 ).

This is nearly zero. It is extremely unlikely to get this many sixes in thirty to sixty rolls. (d) The probability that a six comes up between 2 and 80 times is 161/2

1 P (2 ≤ x ≤ 80) = √ 5π

3/2

e−(x−15) /25 dx ≈ 0.99993.

It is very likely to obtain a six at least once between two and eighty rolls. (e) The probability that a six comes up at least thirty-ﬁve times is 1 P (x ≥ 35) = √ 5π

∞

69/2

e−(x−15) /25 dx ≈ 1.7396(10−8 ).

This event is very unlikely. (f) The probability that we get fewer than twenty sixes (that is, nineteen or fewer) is 1 P (x ≤ 19) = √ 5π

39/2 −∞

e−(x−15) /25 dx ≈ 0.89845.

(g) The probability that exactly forty-ﬁve heads came up is 1 P (x = 45) = √ 5π

91/2 89/2

e−(x−15) /25 dx ≈ 3.2795(10−17 ).

2. (a) Assuming an honest coin, the mean is μ = 100(1/2) = 50 and the standard deviation is σ = 100(1/2)(1/2) = 5. 14

(b) The bell curve for this experiment is the graph of the exponential function 2 1 √ e−(x−50) /50 . 5 2π (c) 81/2

1 P (20 ≤ x ≤ 40) = √ 5 2π

39/2

e−(x−50) /50 dx ≈ 0.028717.

(d) 101/2

1 P (10 ≤ x ≤ 50) = √ 5 2π

59/2

(e) 1 P (x ≥ 30) = √ 5 2π

∞

e−(x−50) /50 dx ≈ 0.53983.

50/2

e−(x−50) /50 dx ≈ 0.99998.

(f) Fewer than sixty means ﬁfty-nine or fewer occurrences. This probability is 119/2 2 1 e−(x−50) /50 dx ≈ 0.97128. P (x ≤ 59) = √ 5 2π −∞ (g) 1 P (x = 55) = √ 5 2π

111/2 109/2

e−(x−50) /50 dx ≈ 0.048394.

3. (a) Take p = 0.52, so q = 1 − p = 0.48. The mean is μ = (350)(0.52) = 182. The standard deviation is σ = (350)(0.52)(0.48) = 9.3467. (b) Compute 2σ 2 = 174.72. A bell curve for this experiment is the graph of the exponential function 2 1 √ e−(x−182) /174.72 . 9.3467 2π

1 √ 9.3467 2π

∞ 439/2

e−(x−182) /174.72 dx ≈ 3.0087(10−5 ).

(d) P (x ≥ 150) =

∞

1 √ 9.3467 2π

299/2

e−(x−182) /174.72 dx ≈ 0.99974.

(e) P (120 ≤ x ≤ 250) =

1 √ 9.3467 2π

501/2 239/2

e−(x−182) /174.72 dx ≈ 1.

The integral does not equal 1, but is approximately 1 within the accuracy of the integral approximation. For all practical purposes, between 120 and 250 girls, inclusive, is a virtual certainty. (f) P (x = 180) =

1 √ 9.3467 2π

361/2 359/2

e−(x−182) /174.72 dx ≈ 0.041698.

4. The mean is μ = (1000)(0.58) = 580 and σ = (1000)(0.58)(0.42) = 15.608. Compute 2σ 2 = 487.2. The bell curve for this experiment is the graph of the exponential function 2 1 √ e−(x−580) /487.2 . (15.608) 2π

Now we can compute some probabilities. (a) P (x ≥ 400) =

∞

1 √ (15.608) 2π

799/2

e−(x−580) /487.2 dx ≈ 0.99998.

(b) P (400 ≤ x ≤ 600) =

1201/2

1 √ (15.608) 2π

799/2

e−(x−580) /487.2 dx ≈ 0.99547.

1 √ (15.608) 2π

901/2 −∞

e−(x−580) /487.2 dx ≈ 5.3300((10)−17 ).

(d) P (x = 520) =

1 √ (15.608) 2π

1041/2 1039/2

e−(x−580) /487.2 dx ≈ 1.5831(10−5 ).

5. Take μ = 750 and σ = 65. The 2σ 2 = 8450 and the bell curve is the graph of the exponential 2 1 √ e−(x−750) /8450 . (65) 2π The probability that a typical passenger rides between two hundred ﬁfty and six hundred miles per month is P (250 ≤ x ≤ 600) =

1201/2

1 √ (65) 2π

e−(x−750) /8450 dx

499/2

≈ 0.010724. By contrast, P (600 ≤ x ≤ 900) =

1801/2

1 √ (65) 2π

e−(x−750) /8450 dx

1199/2

≈ 0.97941. It is very likely that the average passenger rides between six hundred and nine hundred miles per month, but unlikel that this passenger rides between 250 and 600 miles. 6. We have μ = 940 and σ = 76. Then 2σ 2 = 11552 and the appropriate exponential function for this distribution is 2 1 √ e−(x−940) /11552 . (76) 2π

Then P (750 ≤ x ≤ 1000) =

2001/2

1 √ (76) 2π

1499/2

e−(x−940) /11552 dx

≈ 0.7809. And P (300 ≤ x ≤ 500) =

1 √ (76) 2π

1001/2 599/2 −9

≈ 3.6713(10

e−(x−940) /11552 dx

7. Multiply these percentages by 100 to convert .247 to 24.7 and .021 to 2.1. Now μ = 24.7 and σ = 2.1, so 2σ 2 = 8.82 and the appropriate exponential function for this problem is 2 1 √ e−(x−24.7) /8.82 . (2.1) 2π

Now we can calculate 1 √ (2.1) 2π = 0.53935.

P (24.5 ≤ x ≤ 27.0) =

55/2 24

e−(x−24.7) /8.82 dx

This is the probability that a batter hit between .245 and .270, as these averages are usually written in baseball statistics reports. Next, P (x ≥ 30) =

1 √ (2.1) 2π

∞ 59/2

e−(x−24.7) /8.82 dx

= 0.01135. The reason this does not appear consistent with what we actually see in the American league is the unrealistically small standard deviation assumed in this problem. 8. We have N = 2000 and p = 0.42 (probability of a head on a single ﬂip). Then μ = N p = 2000(0.42) = 840 and σ=

(2000)(0.42)(0.58) = 22.073.

(a) We want to go one standard deviation to the right and left of the mean. Compute 840 − 22.073 = 817.93 so take a = 817, since numbers of heads must be an integer. And 840 + 22.073 = 862.07, so take b = 863. (b) Now we want to take points two standard deviations to the left and right of the mean. These are 840 − 2(22.073) = 795.85 and 840 + 2(22.073) = 884.15. Thus choose a = 795 and b = 885. (c) For a 0.99 probability of seeing between a and b heads, choose a three standard deviations to the left of the mean, and b three standard deviations to the right. First compute 840 − 3(22.073) = 773.78 and 840 + 32(22.073) = 906.22. Thus choose a = 773 and b = 907. 18

x 6.2 6.2 5.2 7 5.6 6.4

x 5.8 5 6 5.2 6.4 5.8

x 6.8 5.6 6.8 5.6 6 7

x 6.2 7 5.4 5.4 6.4 6

x 4.2 5.6 6 5.6 6.4 6.4

Table 3: Sample means in Problem 1, Section 5.

9. The probability of a die coming up even is 0.5. Now μ = 550(0.5) = 275 and σ = 11.726. (a) For a probability of 0.68 of seeing between a and b even rolls, go one standard deviation to the left and right of the mean. Compute 275 − 11.726 = 263.27 and 275 + 11.726 = 286.73. Thus choose a = 263 and b = 287. (b) For a probability of 0.95 of seeing between a and b even tosses, go two standard deviations to each side of the mean. Compute 275 − 2(11.726) = 251.55 and 275 + 2(11.726) = 298.45. Thus choose a = 251 and b = 299. (c) Now go three standard deviations to the left and right of the mean. Compute 275 − 3(11.726) = 239.82 and 275 + 3(11.726) = 310.18. Now choose a = 239 and b = 311. Section 5 Sampling Distributions and the Central Limit Theorem 1. (a) Table 3 gives the sample means corresponding to the samples in Table 10 of the chapter. The mean of the sample means is 179.2/30, or 5.97. The mean of the entire population (the 150 numbers individual of all the samples) is 896/150, or 5.97. (b) We ﬁnd that the standard deviation of the population is √ 25871 , σ= 75 19

x 4.857 5.0 4.143 4.429 4.286 2.857

x 5.286 2.857 4.143 3.714 4.0 3.571

x 4.714 4.142 5.286 2.714 1.714 4.429

x 3.423 2.857 4.143 4.143 3.286 4.714

x 4.857 3.857 5.286 5.286 4.571 3.143

Table 4: Sample means in Problem 2, Section 5.

which is approximately 2.14459. The standard deviation of the sample means is σmeans = 0.662885. √ Since the sample size is n = 5, compute σ/ 5 = 0.95909, differing from σmeans by about 0.296. The approximation in this example is not as good as we might want in practice because the sample size of five is very small. 2. (a) The sample means are given in Table 4, with entries corresponding to the samples given in Table 11 of the chapter. The mean of the sample means is 119.90/30, which is 3.9966. The mean of the population is 851/210, which is 4.053. The difference between these two is due to rounding in computing the sample means. (b) For the entire population, we find that √ 2 31001 , σ= 105 while the standard deviation of the sample means is σmeans = 0.8565047426. With samples of size 7, compute √ 1 2 31001 σ √ = √ = 1.2676. 105 7 7 Again, the approximation would improve with larger samples. 3. Take μ = 670 as the average score and σ = 105. Now let 105 μmeans = 670 and σmeans = √ . 30 Then 2(σmeans )2 = 735. Using these numbers, define the exponential function 2 30 √ e(x−670) /735 . 2π(105) 20

(a) The probability that all thirty students score in the 700 − 750 range is 30 2π(105) ≈ 0.0588.

P (700 ≤ x ≤ 750) = √

750 700

e(x−670) /735 dx

This event is not very likely. (b) The probability that all thirty recruits score at or above average is 30 2π(105) ≈ 0.5.

P (x ≥ 670) = √

900 670

e(x−670) /735 dx

2 = 3150 2σmeans

so the probability that seven of the candidates score 800 or higher is √ 900 2 7 √ e(x−670) /3150 dx P (800 ≤ x ≤ 900) = (105) 2π 800 ≈ 0.00052701. This event is extremely unlikely, which makes sense in view of the fact that the average is 670 and seven is nearly one third of the recruiting class. 4. Since six recruits are chosen, let n = 6. This assumes that recruits do not share pants. Take μ = 31 and σ = 3.2. Then μmeans = 31 and 2 2σmeans =

1 (3.2)2 = 3.4133. 3

(a) The probability that the mean length of the sample is at least twentynine inches is approximately √ ∞ 2 6 √ e−(x−31) /3.4133 dx P (x ≥ 29) = (3.2) 2π 29 ≈ 0.9371. Notice that three sample standard deviations to the left of the mean is nearly 27.

(b) √ 6 √ (3.2) 2π ≈ 0.87421.

P (29 ≤ x ≤ 33) =

e−(x−31) /3.4133 dx

P (x ≤ 32) =

−∞

e−(x−31) /3.4133 dx

5. The braces can handle an average of μ = 5000 pounds per square inch, with a standard deviation of σ = 800. To sample √ twenty braces, choose n = 20. Then μmean = 5000 and σmeans = 800/ 20. Compute 2 2σmeans =

1 (640000) = 64000. 10

(a) The probability that the mean pressure tolerance is at least 4, 800 pounds per square inch is √ ∞ 2 20 √ P (x ≥ 4800) = e−(x−5000) /64000 dx 800 2π 4800 ≈ 0.86822. (b) √ 20 √ 800 2π ≈ 0.66516.

P (4700 ≤ x ≤ 5100) =

5100 4700

e−(x−5000) /64000 dx

P (x ≥ 4500) =

∞ 4500

e−(x−5000) /64000 dx

6. Now μ = 17 and √ σ = 1.8. Sample size is n = 30. Then μmeans = 17 and σmeans = 1.8/ 30. We need 2 2σmeans =

2 (1.8)2 = 0.216. 30

(a) Compute √

30 √ (1.8) 2π ≈ 0.4572.

P (16.8 ≤ x ≤ 17.2) =

17.2 16.8

e−(x−17) /0.216 dx

(b) √

30 √ (1.8) 2π ≈ 0.81934.

P (x ≥ 16.7) =

∞ 16.7

e−(x−17) /0.216 dx

30 √ (1.8) 2π ≈ 0.2714.

P (x ≥ 17.2) =

∞ 17.2

e−(x−17) /0.216 dx

Section 6 Conﬁdence Intervals and Estimating Population Proportion 1. We have n = 1200, and the best estimate for p is p̃ = 0.02. Then q̃ = 0.98. (a) For a 99 percent conﬁdence level, zα/2 = 2.575. Compute = (2.575)

(0.02)(0.98) = 0.0104. 1200

The conﬁdence interval is 0.02 − 0.01 < p < 0.02 + 0.01 which is the interval 0.01 < p < 0.03. (b) With a 95 percent conﬁdence level, zα/2 = 1.96 and = (1.96)

(0.02)(0.98) = 0.008. 1200

Now the conﬁdence interval is 0.02 − 0.008 < p < 0.02 + 0.008, or 0.012 < p < 0.028. 23

(0.02)(0.98) = 0.0065. 1200

The confidence interval is 0.02 − 0.0065 < p < 0.02 + 0.0065. This is the interval 0.0135 < p < 0.0265. (d) Replicate parts (a) through (c), with 1200 reduced to 800. Replace 1200 by 800 in the preceding calculations to obtain the following new confidence intervals. For a 99 percent confidence interval, (0.02)(0.98) = 0.0127, = (2.575) 800 so the new confidence interval is 0.0073 < p < 0.0327. For a 95 percent confidence interval, now = 0.0097 and the confidence interval is 0.0103 < p < 0.0297. For a 90 percent confidence interval, we get = 0.0081 for a confidence interval of 0.0119 < p < 0.0281. As should not be surprising, decreasing the sample size increases the width of the respective confidence intervals. 2. Now n = 750 and p̃ = 0.22. (a) For a 99 percent confidence interval, compute (0.22)(0.78) = 0.039 = (2.575) 750 for a confidence interval of 0.22 − 0.039 < p < 0.22 + 0.039. This is the interval 0.181 < p < 0.259.

(b) For a 95 percent confidence interval, now (0.22)(0.78) = 0.029 = (1.96) 800 and the confidence interval is 0.191 < p < 0.249. 3. Take n = 200 and p̃ = 87/200 = 0.435. (a) For a 95 percent confidence interval, compute (0.435)(0.565) = (1.96) = 0.069 200 for a confidence interval of 0.366 < p < 0.504. (b) For an 85 percent confidence interval, compute (0.435)(0.565) = (1.44) = 0.05. 200 The confidence interval is 0.22 − 0.02965 < p < 0.22 + 0.02965. This is the interval 0.19036 < p < 0.24964. 4. Here n = 100 and p̃ = 0.07. (a) For a 99 percent confidence interval, compute (0.07)(0.93) = 0.066. = (2.575) 100 The interval is 0.004 < p < 0.136. (b) For a 90 percent confidence interval, use (0.07)(0.93) = 0.042. = (1.645) 100 This confidence interval is 0.028 < p < 0.112. 25

5. (a) For a 95 percent conﬁdence level, choose zα/2 = 1.96. We want = 0.02, so (zα/2 )2 p̃q̃ (1.96)2 = p̃q̃. n= 2 (0.02)2 We have no information with which to estimate p̃, so take p̃q̃ = 0.25 to obtain (1.96)2 (0.25) = 2401. n= (0.02)2 The survey should include this number of people. (b) If the executive is willing to allow a ﬁve percent maximum error, now choose = 0.05 and compute n=

(1.96)2 (0.25) = 384.16. (0.05)2

Since the number of people must be an integer, choose n = 384. This relaxation in error tolerance significantly reduces the number of people that must be surveyed. (c) Now suppose we use the previous year’s figure to estimate p̃ = 0.37. Let = 0.02. For a 95 percent confidence level. Now compute n=

(1.96)2 (0.37)(0.68) = 2238.69. (0.02)2

The number of people that should be surveyed now is 2239. (d) If we use more current information to estimate p̃ = 0.32, then the calculation in part (c) is adjusted to n=

(1.96)2 (0.32)(0.68) = 2089.83, (0.02)2

suggesting that 2090 people be surveyed. With a lower population, we do not need to interview as many people. 6. (a) Let = 0.03. For a 99 percent confidence level in the survey’s results, compute (2.575)2 (0.25) = 1841.8, n= (0.03)2 so the official should survey 1842 people. (b) If 95 percent confidence level is good enough, compute n=

(1.96)2 (0.25) = 1067.211. (0.03)2

Now survey 1067 people.

(2.575)2 (0.12)(0.88) = 1750.5. (0.02)2

Here the oﬃcial should survey 1751 people. 7. (a) An estimate of the number of people is provided by n=

(1.44)2 (0.25) = 105.8, (0.07)2

so survey 106 people. (b) Now adjust the estimate to (1.44)2 (0.25) = 36. (0.12)2

This suggests that 36 people should be surveyed. A much higher error tolerance allows a survey involving fewer people, as we would expect intuitively. Here, in the absence of more information, we used p̃q̃ = 0.25. (c) Estimate

1.44 0.07

2 (0.06)(0.94) = 23.868.

Now we can get by with surveying 24 people. Section 7 Estimating Population Mean and the Student t - Distribution

1. (a) For a 99 percent conﬁdence level, compute 1.04 σ = zα/2 √ = (2.575) √ = 0.14315. n 350 Then 71.857 = 72 − 0.143 = x − < μ < x + = 72 + 0.143 = 72.143. (b) Estimate n=

σzα/2

(1.04)(2.575) 0.2

Thus choose n = 180.

2 = 179.29.

2. (a) for a 95 percent conﬁdence level, estimate the maximum error in the population mean as 0.7 = (1.96) √ = 0.194. 50 Then 1.19 − 0.194 < μM 119 + 0.194 so 11881 < μ < 119.194. (b) Estimate

2 (0.7)(2.575) = 81.225. n= 0.2 Choose n = 82 as a good estimate for the sample size.

3. (a) Compute

= (1.96)

2.4 10

= 0.47.

Then estimate 106 − 0.47 < μ < 106 + 0.47. Then 105.53 < μ < 106.47. (b) Estimate

2 (2.4)(2.575) = 3819.2. 0.1 Then 3820 is a good sample size to use.

4. Compute

0.05 s t = tα/2 √ = (2.009) √ = 0.0142. n 50 Here the given critical value is for a 95 percent conﬁdence interval. The conﬁdence interval is 0.8 − 0.0142 < μ < 0.8 + 0.0142. This is the interval 0.7858 < μ < 0.8142.

5. Estimate

0.7 t = (1.992) √ = 0.161. 75 The conﬁdence interval is 3 − 0.161 < μ < 3 + 0.161. This is the interval 2.839 < μ < 3.161. 28

6. Estimate

0.01 = 0.0014. t = (1.971) √ 200

The conﬁdence interval is 7 − 0.0014 < μ < 7 + 0.0014. This is 6.9986 < μ < 7.0014. Section 8 Correlation and Regression 1. From the data in Table 18, compute 10

xi = 52.38,

i=1 10

x2i = 420.88,

i=1 10

yi2 = 631.9,

i=1 2

yi = 73.05,

i=1

= 2743.7,

i=1

= 5336.3,

i=1

x = 5.238, y = 7.305,

xi yi = 502.50.

i=1

Then 10(502.50) − (52.38)(73.05) = 0.99895. c= 10(420.88) − 2743.7 10(631.9) − 5336.3 There is therefore a signiﬁcant linear correlation between the xi and the yi pairs. Next, compute b=

10(502.50) − (52.38)(73.05) = 0.81813 10(420.88) − 2743.7

and a = y − bx = 3.0196. The regression line has the equation y = 3.0196 + 0.81813x,

Now compute

631.9 − (3.0196)(73.05) − (0.81813)(502.50) 8 = 0.1612.

Sr =

With n = 10 and a 95 percent conﬁdence level, tα/2 = 2.306. If x = 0.4, the regression line equation returns a value of y = 3.0196 + 0.81813(0.4) = 3.3469. For this value of x, compute

r = (2.306)(0.1612) 1 +

10(0.4 − 5.238)2 1 + 10 10(420.88) − 2743.7

= 0.41722. The conﬁdence interval for this value of y (for x = 0.4) is 3.3469 − 0.41722 < y < 3.3469 + 0.41722, or 2.9297 < y < 3.7641. If x = 6.2, the predicted value of y is y = 3.0196 + 0.81813(6.2) = 8.092. Now

r = (2.306)(0.1612) 1 +

10(6.2 − 5.238)2 1 + 10 10(420.88) − 2743.7

= 0.39099. Now the conﬁdence interval is 7.701 < y < 8.4830. If x = 15.1, the predicted value of y is y = 3.0196 + 0.81813(15.1) = 15.373. Compute

r = (2.306)(0.1612) 1 +

10(15.373 − 5.238)2 1 + 10 10(420.88) − 2743.7

= 0.49888. The conﬁdence interval is 14.874 < y < 15.872. 30

2. From the data in Table 19, compute 10

xi = 47.3,

i=1 10

x2i = 441.43,

i=1 10

yi = −103.99,

i=1 10 i=1 2

yi2 = 2947.8151,

= 2237.3,

i=1

= 10813.92,

i=1

x = 4.73, y = −10.399,

xi yi = −1128.866.

i=1

Then 10(−1128.866) − (47.3)(−103.99) c= 10(441.43) − (47.3) 10(2947.8151) − 10813.92 = −0.99931. There is a signiﬁcant linear correlation between the xi s and the yi s. Next, compute b=

10(−1128.866) − (47.3)(−103.99) = −2.9262 10(441.43) − 2237.3

and a = −10.399 − (−2.9262)(4.73) = 3.4419. The regression line has the equation y = 3.4419 − 2.9262x. Now compute

2947.8151 − 3.4419(−103.99) + 2.9262(−1128.866) 8 = 0.55347.

Sr =

With n = 10 and a 95 percent conﬁdence level, tα/2 = 2.306. If x = 5.2, the regression line equation returns a predicted value of y = 3.4419 − 2.9262(5.2) = −11.774. For this value of x, compute

r = (2.306)(0.55347) 1 + = 1.3392. 31

10(5.2 − 4.73)2 1 + 10 10(441.43) − 2237.3

The conﬁdence interval for this value of y (for x = 0.4) is −13.113 < y < −10.435. If x = 8.6, the predicted value of y is y = 3.4419 − 2.9262(8.6) = −21.723. Now

10(8.6 − 4.73)2 1 + 10 10(441.43) − 2237.3

r = (2.306)(0.55347) 1 + = 1.3798. Now the conﬁdence interval is

−23.103 < y < −20.343. If x = 17.8, the predicted value of y is y = 3.4419 − 2.9262(17.8) = −48.644. Compute

1 10(17.8 − 4.73)2 + 10 10(441.43) − 2237.3

r = (2.306)(0.55347) 1 + = 1.7522. The conﬁdence interval is

−50.396 < y < −46.892. 3. From the data in Table 20, compute 10

xi = 147.11,

i=1 10

x2i = 3200.2,

yi2 = 676.79,

i=1 2

yi = 12.25,

i=1

i=1 10

= 21642,

i=1

= 150.06,

i=1

x = 14.711, y = 1.225,

xi yi = 215.02.

i=1

Then 10(215.02) − (147.11)(12.25) = 0.042039. c= 10(3200.2) − 21641 10(676.79) − 150.06 There is no signiﬁcant linear correlation between the xi s and the yi s. 32

4. From the data in Table 21, compute 10

xi = 200.4,

i=1 10

x2i = 5701.38,

i=1 10

yi = 1000.3,

i=1 10

yi2 = 130147.33,

i=1 2

= 40160.16,

i=1

= 1000600,

i=1

x = 20.04, y = 100.03,

xi yi = 27156.78.

i=1

Then c=

10(27156.78) − (20.04)(1000.3) = 3.5321. 10(130147.33) − 1000600

10(5701.38) − 40160.16

There is no signiﬁcant linear correlation between the xi s and the yi s. For the remainder of these problems, we omit some routine details of the arithmetic. 5. From the data in Table 22, compute 10

xi = 127.0,

i=1 10

x2i = 2904.4,

i=1 10

yi2 = 1945.0,

i=1 2

yi = 97.42,

i=1

= 16129,

i=1

= 9490.7,

i=1

x = 12.7, y = 9.742,

xi yi = 2369.6.

i=1

Compute c=

10(2369.2) − (127)(97.42) = 0.99845. 10(2904.4) − 16129 10(1945) − 9490.7

This indicates a signiﬁcant linear correlation between the data sets. Next compute b = 0.87678 and a = −1.3931 33

to obtain the equation of the regression line y = −1.3931 + 0.87678x. Next, compute Sr = 0.66229. If x = 0.6, the regression line predicts that y = −0.86703. We get r = 1.6823, so the confidence interval is −2.5493 < y < 0.81527. If x = 3, the regression line predicts that y = −1.2372. Now r = 1.6540 and the confidence interval is −0.4168 < y < 2.8912. If x = 39, we get y = 32.801. Now r = 1.9532 and the confidence interval is 30.848 < y < 34.754. Finally, if x = 42.1, we get y = 35.519. Now r = 2.0314 and the confidence interval is 33.4883 < y < 37.55. 6. From the data in Table 23, 10

xi = 51.8,

i=1 10

x2i = 279.82,

i=1 10

yi = −32.9,

i=1 10 i=1 2

yi2 = 1608.15,

= 2683.24,

i=1

= 1082.41,

i=1

x = 5.18, y = −3.29,

xi yi = −171.34.

i=1

Compute 10(−171.34) − (51.8)(−32.9) c= 10(279.82) − 2683.24 10(1608.15) − 1082.41 = 0.00699. There is no signiﬁcant linear correlation for this data set. 34

7. Note that now there are 15 data points. Other than this, we proceed in the same way as in Problems 1 - 6, but with a diﬀerent value for tα/2 . We obtain 15

xi = 277.5,

i=1 15

x2i = 7315.9,

i=1 15

yi = 398.65,

i=1 15

yi2 = 16115,

i=1 2

= 77006,

i=1

= 158920,

i=1

x = 18.5, y = 26.577,

xi yi = 10843.

i=1

Compute c = 0.99919, so there is a significant linear correlation between the xi points and the yi points. Further, compute b = 1.5892 and a = −2.8232 so the regression line has the equation y = −2.8322 + 1.5892x. Next we find that Sr = 0.82149. If x = −4, the regression line gives y = −9.18. With n = 15 in this problem, a table of t - distribution critical values gives tα/2 = 2.160, from which we compute r = 2.1588. The confidence interval is −11.339 < y < −7.0212. If x = 11.1, we get y = 14.817 and r = 1.9794 for a confidence interval of 12.838 < y < 16.796. If x = 30.1, we get y = 45.012 and r = 2.0122 for a confidence interval of 43 < y < 47.014. If x = 66.7, we get y = 103.18 and r = 2.7656 for a confidence interval of 100.41 < y < 105.95. 35

8. There are n = 15 data points. Compute 15

xi = 127.5,

i=1 15

x2i = 2095.17,

i=1 15

yi = −213.83,

i=1 15

yi2 = 7214.7485,

i=1

= 16256.25,

i=1

= 45723.2689

i=1

x = 8.5, y = −14.2553,

xi yi = −3868.95.

i=1

Compute c=

15(−3868.95) − (127.5)(−213.83) 15(7315.9) − 16256.25 15(7214.7485) − 45723.2689

= −0.40257. There is no signiﬁcant correlation between the xi s and the yi s. 9. We will need the following computations: xi = 674, yi = 533, x2i = 46506, yi2 = 29073, 2 2 xi = 454276, yi = 284089 x = 67.4, y = 53.3, xi yi = 36755. (a) Compute c=

10(36755) − (674)(533) 10(46506) − 454276 10(29073) − 284089

= 0.98172. Therefore there is a signiﬁcant linear correlation between the two data sets. (b) Compute b=

10(36755) − (674)(533) = 0.7704 10(46506) − 454275

and a = 53.3 − (0.7704)(67.4) = 1.375. 36

The equation of the regression line is y = 1.375 + 0.7704x. (c) Compute

c2 = (0.98172)2 = 0.96377.

We conclude that about 96.4 percent of the team’s wins are explained by this player’s presence on the court. (d) Suppose this player plays in 85 games. Compute from the regression line equation that y(85) = 1.375 + 0.7704(85) = 66.859, giving us the estimate that the team will win 67 of these games. For the conﬁdence interval of this prediction, compute 29073 − (1.375)(533) − (0.7704)(36755) Sr = 8 = 1.7347. Then

r = 2.306(1.7347) 1 +

10(85 − 67.4)2 1 + 10 10(46506) − 454276

= 4.7115. The conﬁdence interval is 67 − 4.7115 < y < 67 + 4.7115 or, since the number of games must be an integer, 62 < y < 72. 10. (a) First compute the numbers we will need: xi = 72600, yi = 2255, x2i = 586680000, yi2 = 563855, 2 2 xi = 5270760000, yi = 5085025 x = 7260, y = 225.5, xi yi = 18180500. Now compute c=

10(18180500) − (72600)(2255) 10(586680000) − 5270760000 10(563855) − 5085025

= 0.99605. 37

There is a signiﬁcant linear correlation between the xi s and the yi s. (b) Compute b=

10(18180500) − (72600)(2255) = 0.030354. 10(586680000) − 5270760000

And a = 225.5 − (0.030354)(7260) = 5.1300. The regression line has the equation y = 5.1300 + 0.030354x. (c) From the regression line, estimate that the number of patients, out of 17, 000, having this side effect is y = 5.1300 + (0.030354)(17000) = 521.15. Set the number of patients at 522. (d) Suppose now we consider a population of 17, 500 patients. From the regression line, the number with this side effect is estimated to be 5.1300 + 0.030354(17500) = 536.325. or, in integer terms, 537. For a 95 percent confidence interval, first compute 563855 − 5.13(2255) − 0.030345(18180500) Sr = 8 = 8.6572. Then r = (2.306)(8.6572)

10(17500 − 7260)2 1 + 10 10(586680000) − 5270760000

= 33.757. The conﬁdence interval is 537 − 33.757 < y < 537 + 33.757, or 503 < y < 571.

11. As usual, ﬁrst we must compute some numbers we will need: xi = 151, yi = 58, 2 xi = 2453, yi2 = 358, 2 2 xi = 22801, yi = 3364 x = 15.1, y = 5.8, xi yi = 934. Now we can compute c=

10(934) − (151)(58) = 0.95235. 10(2453) − 22801 10(358) − 3364

There is a signiﬁcant linear correlation between the number of visits and the number of thefts. (b) Compute b=

10(934) − (151)(58) = 0.33661 10(2453) − 22801

and a = 5.8 − (0.33661)(15.1) = 0.71719. The regression line has the equation y = 0.71719 + 0.33661x. (c) Since

c2 = (0.95235)2 = 0.90697,

it is estimated that this person is responsible for about 90 percent of the major jewelry thefts in the city each year. (d) Compute y(25) = 0.71719 + 0.33661(25) = 9.1324. The regression line suggests that, with twenty-five visits, this person will be involved in 9 major thefts. For the confidence interval, first compute 358 − (0.71719)(58) − (0.33661)(934) = 0.50155. Sr = 8 Then

r = (2.306)(0.50155)

= 1.4943. 39

10(25 − 15.1)2 1 + 10 10(2453) − 22801

The conﬁdence interval is 9 − 1.4943 < y < 9 + 1.4943 or, in terms of integer numbers of thefts, 7 < y < 11. 12. Begin with some computations: xi = 4950, yi = 43246, x2i = 2490950, yi2 = 190193552, 2 2 xi = 24502500, yi = 1870216516 x = 495, y = 4324.6, xi yi = 21765805. (a) Compute c=

10(21765805) − (4950)(43246) 10(2490950) − 24502500 10(190193552) − 1870216516

= 0.99926. There is a signiﬁcant linear correlation between the additive and children who are cavity free. (b) Compute b=

10(21765805) − (4950)(43246) = 8.8215 10(2490950) − 24502500

and a = 4324.6 − (8.8215)(495) = −42.043. The regression line has the equation y = −42.043 + 8.8215x. (c) Compute

c2 = (0.99926)2 = 0.99852.

This estimates that 99.8 percent of the children who were cavity free have this circumstance explained by the chemical additive to the water supply. (d) First, y(650) = −42.043 + 8.8215(650) = 5691.9. The estimate is that adding 650 pounds of the chemical will result in 5692 children being cavity free. 40

To determine a 95 percent conﬁdence interval for this estimate, ﬁrst compute 190193552 − (−42.043)(43246) − (8.8215)(21765805) Sr = 8 = 24.225. Then r = (2.305)(24.225)

10(650 − 495)2 1 + 10 10(2490950) − 24502500

= 72.628. The 95 percent conﬁdence interval is 5691.9 − 72.628 < y < 5691.9 + 72.628, or 5619 < y < 5765.

Solutions - Problems in Probability Section 1 Events, Sample Spaces and Probability 1. If three coins are flipped, the outcomes are HT T, HT H, HHT, HHH, T T T, T T H, T HT, T HH. There are eight outcomes. We can also count these outcomes as follows. An outcomes has the form (x, y, z), where each of x, y and z is an H or a T , independently. There are two possibilities for x, and for each, two for y for four possibilities from the first two flips. There are also two possibilities for z, with each of these four possibilities from the first two flips, making 8 outcomes in all. (a) There are three outcomes with exactly two heads, so Pr(exactly two heads) =

3 . 8

(b) For at least two heads, we could have exactly two heads (three outcomes) or all three heads (one outcome), for four ways this can happen. Then 1 4 Pr(at least two heads) = = . 8 2 2. With two dice being rolled, there are 36 outcomes (6 possibilities for the ﬁrst die, and, for each of these, 6 for the second die). (a) Suppose both dice come up even. Then each may come up 2, 4 or 6, independent of the other. The outcomes with both dice even are: (2, 2), (2, 4), (2, 6), (4, 2), (4, 4), (4, 6), (6, 2), (6, 4), (6, 6). There are 9 outcomes with both dice even, so Pr(both dice come up even) =

9 1 = . 36 4

We can also count the outcomes in this case as follows. An outcome has the form (a, b). Here, a can be 2, 4 or 6. With each of these, b can be 2, 4 or 6, for 3 · 3 = 9 outcomes. (b) The outcomes with one die even and the other odd are (1, 2), (1, 4), (1, 6), (3, 2), (3, 4), (3, 6), (5, 2), (5, 4), (5, 6), (2, 1), (4, 1), (6, 1), (2, 3), (4, 3), (6, 3), (2, 5), (4, 5), (6, 5). There are 18 outcomes with one die odd and the other even. We could also count these as follows. If the ﬁrst die is odd, then the second is even, 1

and there are three possibilities for each die, for 9 outcomes. But there are also 9 outcomes with the ﬁrst even and the second odd, yielding 18 outcomes. Then Pr(one die is even, the other is odd) =

1 18 = . 36 2

(c) The dice total exactly 6 in the following ways: (1, 5), (2, 4), (3, 3), (4, 2), (5, 1). Then Pr(the two dice total six) =

5 . 36

(d) Two dice cannot total 14, so this is not an outcome of the experiment, and there is no probability that the dice total 14. In some contexts, it is also reasonable to assign a probability of zero to an outcome that cannot happen. 3. Three dice are rolled. With six possibilities independently for each die (the number that comes up on one die does not inﬂuence what comes up on another), there are 6 · 6 · 6 = 216 outcomes. (a) The dice can come up the same in six ways, namely: (1, 1, 1), (2, 2, 2), (3, 3, 3), (4, 4, 4), (5, 5, 5), (6, 6, 6). Then Pr(all dice come up the same) =

6 . 216

(b) The outcomes with the dice totalling 15 are (6, 6, 3), (6, 5, 4), (6, 4, 5), (6, 3, 6), (5, 6, 4), (5, 5, 3), (5, 4, 6), (4, 6, 5), (4, 5, 6), (3, 6, 6). These have been displayed in a way to suggest the systematic way they were listed. In counting objects by enumerating them, it is helpful to have a scheme in mind to avoid duplications or omissions. In this case, we started with outcomes having 6 on the ﬁrst die, then those having 5, and so on. There are 10 outcomes with the dice totaling 15. Then Pr(the dice total )15 =

10 . 216

(c) Suppose the ﬁrst die comes up 1 and the third comes up 4. This leaves six outcomes, since the second die can come up 1, 2, 3, 4, 5 or 6. Therefore Pr(ﬁrst die 1, second die 4) =

6 216

(d) Suppose the first two dice come up even. There are now three ways the first die can come up, and, independently, three ways the second die can come up, and, independent of these two tosses, six ways the third die can come up. This is a total of 3 · 3 · 6, or 54 outcomes. Then Pr(first two dice come up even) =

1 54 = . 216 4

4. Two cards are drawn from a standard 52 card deck. Before considering probabilities of various outcomes, we need to be very clear on how this draw is made. First, we will assume that one card is drawn, then a second card. This is called drawing without replacement. The effect of this is that there are 52 cards available on the first draw, but only 51 on the second. If we had drawn with replacement, both draws would be from 52 cards. Second, we will assume that the order of the cards makes a difference. Thus, after the draw, we think of one card as the first card, and the other card as the second card. With these agreements, think of an outcome of the draw of two cards as a pair (a, b), where a is the first card drawn (52 choices) and b the second (51 choices). The total number of two-card hands in this case is 52 · 51, for 2652 outcomes. (a) If the first card is a face card, there are 16 possibilities for this draw, then 51 for the second, so Pr(first card a face card) =

16 4 16(51) = = . 52(51) 52 13

(b) If both cards are red, then (since half of the cards are red), there are 26 possibilities for the ﬁrst card, then 25 for the second, so Pr(both cards are red) =

26(25) 25 = . 52(51) 102

(c) Suppose one care is a club and the other a diamond. There are 13 clubs and 13 diamonds, so there can be 13(13) draws with a diamond, then a club, and the same number with a club, then a diamond. Then Pr(one diamond, one club) =

13 2(13)(13) = . 52(51) 108

(d) Suppose both cards are aces. There are four aces, so there are four possibilities for the first card, then three for the second, for 12 outcomes. Then 3 3 12 = = . Pr(both aces) = 52(51) 13(51) 663 This is approximately 0.0045, explaining why it is not good to bet on two aces in two draws. 5. Flip two coins and then roll a die. We can record an outcome as (x, y, z), with x and y can independently be H or T and z can be any integer 1, 2, · · · , 6. There are 2 possibilities for each coin and 6 for the dice, yielding 2(2)(6) = 24 outcomes. (a) Suppose one head, one tail and an even number come up. If the head comes up first, then the other coin is T and there are three possibilities for the die (2, 4 or 6). If the tail comes up first, there are also three possibilities for the two coins and the die. There are therefore 6 outcomes with one head, one tail and an even number on the die. Then Pr(one H, one T , and an even number) =

1 6 = . 24 4

(b) There are two outcomes with both coins heads and the die a 1 or 4. These outcomes are (H, H, 1) and (H, H, 4). Then Pr(both heads, 1 or 4 on the die) =

1 2 = . 24 12

(c) Suppose at least one tail comes up, with a 4, 5 or 6 on the die. Now the outcomes all have the form (T, H, x), (H, T, x) or (T, T, x) in which x can be 4, 5 or 6. This yields 9 outcomes. Therefore Pr(at least one tail, and 4, 5 or 6) =

3 9 = . 24 8

6. Two dice are rolled and a coin is ﬂipped. Each outcome can be recorded as a triple (x, y, z), where each of x and y is, independently, an integer from 1 through 6, and z is an H or T . There are 6(6)(2) = 72 outcomes. (a) For the dice to total at least 9, then can total 9, 10, 11 or 12. List the number of ways this can happen. total 9 (3, 6, x), (4, 5, x), (5, 4, x), (6, 3, x), total 10 (4, 6, x), (5, 5, x), (6, 4, x), total 11 (5, 6, x), (6, 5, x), total 12 (6, 6, x). 4

Since x is H or T in each, there are 8 + 6 + 4 + 2 = 20 outcomes of this experiment. Then Pr(dice total at least 9) =

5 20 = . 72 18

(b) Suppose the dice total at least 9 and the coin comes up heads. Now (from (a)) there are 10 outcomes, so Pr(dice total at least 9, coin is H) =

5 10 = . 72 36

(c) Now suppose the coin is a tail and both dice come up the same. Now there are six outcomes, namely (1, 1, T ), (2, 2, T ), · · · , (6, 6, T ) so Pr(tail, both dice the same) =

1 6 = . 72 12

(d) Suppose the dice both come up even and the coin is a tail. Now the outcomes have the appearance (x, y, T ), where x and y can independently be 2, 4 or 6. There are 9 outcomes of this form, so Pr(tail, both dice even) =

1 9 = . 72 8

Section 2 Four Counting Principles 1. The fact that these are letters of the alphabet that we are arranging in order is irrelevant. The issue is that there are nine distinct objects. The number of arrangements is 9!, which is 362, 880. 2. There are 26 letters in the English slphabet. The problem is one of determining the number of ways of choosing 17 objects from 26 objects, with order taken into account. This is 26! 26! = (26 − 17)! 9! = (10)(11)(12) · · · (23)(24)(25) = 156, 000.

26 P17 =

3. Since any of the nine integers can be used in any of the nine places of the ID number, there are 9 ways the ﬁrst digit can be chosen, 9 ways the second digit can be chosen, and so on. The total number of codes is 99 , or 3.87420489(10)8 . 5

4. The first plan (all different symbols in the different places) allows for 5! = 120 passwords. The second plan (choosing with replacement) allows for 55 = 3, 125 passwords. 5. These 7 symbols have 7! = 5, 040 permutations or arrangements. If a is fixed as the first symbol, then there are six symbols to choose in any order for the other six places, and the number of choices is 6! = 720. If a is fixed as the first symbol, and g as the fifth, then we have five symbols left to choose in any order for the remaining five places. The number of ways of doing this is 5! = 120. 6. We want to find n so that n! ≥ 20, 000. Clearly there are infinitely many such integers n, but we would like the smallest possible n. With a little experimentation, we find that 8! = 40, 320, more than enough, while 7! = 5040 is too small, so choose n = 8. For n! ≥ 100, 000, try some values of n. We find that 10! = 3, 628, 800, more than enough, while 9! = 362, 880 is too small. The most efficient choice in this case is n = 10. 7. There are 12! = 479, 001, 600 outcomes. 8. There are 10 letters from a through l, inclusive. If three of the places are fixed (it does not matter which three), there are seven letters available to put in any order into the remaining seven places. There are 7! = 5, 040 ways to do this. 9. We want to pick 7 objects from 25, taking order into account. The number of ways to do this is 25 P7 =

25! = (19)(20)(21)(22)(23)(24)(25) = 2, 422, 728, 000. 18!

10. The number of ballots is 16! = 12(13)(14)(15)(16) = 524, 160. 16 P5 = 11! 11. Because order is important, the number of possibilities is 22 P6 =

22! = 53, 721, 360. 16!

12. (a) The number of choices is 20 P3 =

20! = 6, 840. 17!

(b) If the list begins with 4, there are only two numbers to choose from the remaining nineteen numbers. There are 19 P 2 =

19! = (18)(19) = 342 17! 6

ways to do this. This result does not depend on which number is fixed in the first position. The percentage of choices beginning with 4 is 100(342/6, 840), or 5 percent. This is reasonable from a common sense point of view, since, with 20 number to choose from, we would expect 5 percent to begin with any particular one of the numbers. (c) This question is really the same as the question in (b), since it does not matter which number is fixed. The answer is that 5 percent of the choices ends in 9. (d) If two places are fixed at 3 and 15, then there are eighteen numbers left, from which we want to choose one. There are 18 such choices. 13. Without order (and, we assume, without replacement), the number of ten card hands is 52! = 15, 820, 024, 220. 52 C10 = 10!42! 14. Disregarding order, the number of nine man lineups that can be formed from a seventeen person roster is 17! = 24, 310. 8!9! If the order makes a difference, then the number is 17 C9 =

17 P9 =

17! = 8!17 C9 = 8, 821, 612, 800. 9!

15. The number of combinations is 20 C4 =

16. The number is 40 C12 =

20! = 4, 845. 4!16!

40! = 5, 586, 853, 480. 12!28!

17. The number of outcomes of flipping five coins is 25 = 32. (a) The number of ways of getting exactly two heads from the five flips is 5 C2 = 5!/(2!3!) = 10. The probability of getting exactly two heads (or exactly two tails) is Pr(exactly two heads) =

5 10 = . 32 16

(b) We get at least two heads if we get exactly two, or exactly three, or exactly four, or exactly ﬁve heads. The sum of the number of ways of doing each of these is 5 C2 +5 C3 +5 C4 +5 C5 = 10 + 10 + 5 + 1 = 26.

Therefore Pr(at least two heads) =

26 13 = . 32 16

18. If four dice are rolled, the number of outcomes is 64 = 1, 296. (a) We need the number of ways the four dice can come up with exactly two of the dice showing 4. We get exactly two dice showing 4 by choosing any two of the four dice to come up 4, and allowing the other two dice to come up any of 1, 2, 3, 5, 6, ﬁve possibilities for the other two. The number of ways this can happen, by the multiplication principle, is (5)(5)4 C2 = 25

4! 4! = 25 = 150. 2!2! 2!2!

The probability of getting exactly two four’s is Pr(exactly two four’s) =

23 150 = . 1296 216

(b) The number of ways of getting exactly three four’s is 5(4 C1 ) = 20, so Pr(exactly three four’s) =

20 5 = . 1296 324

(c) We get at least two four’s if we get exactly two four’s, exactly three four’s, or all four tosses coming up 4. There are 150 ways of getting exactly two, and 20 ways of getting exactly three four’s. Clearly in four tosses there is one way of getting four four’s. Therefore Pr(at least two four’s) =

19 150 + 20 + 1 = . 1296 144

(d) To total 22, the dice could come up 6, 6, 6, 4, in any order (four ways), or 6, 6, 5, 5 in any order (six ways). Therefore Pr(dice total 22) =

5 10 = . 1296 648

19. The number of ways of drawing two cards out of 52 cards, without regard to order, is 52 C2 = 1, 326. (a) We get two kings if we happen to get two of the four kings, and there are 4 C2 = 6 ways of doing this. Then Pr(two kings are drawn) =

1 6 = . 1326 221

(b) The aces and face cards constitute 16 of the 52 cards. If none of the two cards is drawn from these sixteen cards, then the two cards are drawn from the remaining 36 cards. Disregarding order, there are 36 C2 = 630 ways to do this. Therefore Pr(no ace or face card) =

105 630 = . 1326 221

20. The number of ways of choosing four letters from 26 letters, with order, is 26 P4 =

26! 26! = = (23)(24)(25)(26) = 14, 950. (26 − 4)! 22!

(a) If the ﬁrst letter is set at q, then we are left to choose, with order, three letters from the remaining 25 letters. There are 25 P3 = 2, 300 ways to do this, so 2 2300 = . Pr(ﬁrst letter is q) = 14950 13 (b) Suppose a and b are two of the letters. How many ways can this occur? Imagine a string of four boxes. Pick any two, and put a and b in these boxes. There are 4 P2 = 12 ways to do this. For the other two boxes, put (keeping track of the order) any two of the remaining 24 letters. There are 24 P2 ways to do this. The number of ordered strings of four letters with a and b two of the letters, is (4 P2 )(24 P2 ) =

4! 24! = 564. 2! 22!

Then Pr(a and b were chosen) =

7 560 = . 14950 187

(c) The probability of abdz occurring is 1/14950, since this string can occur in exactly one way. 21. The number of ways of choosing, without order, three of the eight bowling balls is 8 C3 = 56. (a) For none of the balls to be defective, they had to come from the six nondefective ones. There are 6 C3 = 20 ways to do this. Then Pr(none defective) =

5 20 = . 56 14

(b) There are two ways to take one defective ball. The other two would have to be taken from the six nondefective ones, which can be done in 6 C2 = 15 ways. Then Pr(exactly one defective ball) =

15 30 = . 56 28

(c) In choosing three bowling balls, there are 3 ways of picking the two defective ones. Then there are six ways of choosing the third ball as nondefective. Therefore Pr(both defective balls are chosen) =

6(3) 9 = . 56 28

22. This problem involves tossing dice, except now each die has only four faces, numbered 1, 2, 3, 4. The number of outcomes is 47 = 16, 384, since each of the seven tosses has four possibilities. (a) There is one way all seven dice can come up 3, so Pr(all come up 3) =

1 . 16384

(b) There are 7 C5 = 21 ways of picking ﬁve of the dice and imagining they come up 1, and imagining the other two come up 4. Therefore Pr(ﬁve ones, two fours) =

21 . 16384

(c) The only ways to roll a total of 26 are to roll (in any order), 4, 4, 4, 4, 4, 4, 2 and there are seven ways to do this (seven possible locations for the 2), or to roll 4, 4, 4, 4, 4, 3, 3 and there are 7 C2 = 21 ways to do this. Therefore Pr(total 26) =

7 7 + 21 = . 16384 4096

(d) The sum is at least 26 if the sum is 26, 27, or, the largest possible, 28. We already know from (c) that there are 28 ways to total exactly 26. To roll a 27, we must get (in any order), 4, 4, 4, 4, 4, 4, 3 and there are 7 ways to do this. To roll 28, we must get all four’s, and there is one way to do this. Therefore Pr(total at least 26) =

9 28 + 7 + 1 = . 16384 4096

23. Taking order into account, there are 20 P5 =

20! = (16)(17)(18)(19)(20) = 1, 860, 480 15!

ways to choose 5 of the balls. (a) There is only one way to choose the balls numbered 1, 2, 3, 4, 5 in this order. the probability is Pr(select 1, 2, 3, 4, 5 in this order) = 10

1 . 1860480

(b) We want the probability that ball number 3 was drawn (somewhere in the five drawings). We therefore need the number of ordered choices of five of the twenty numbers, that include the number 3. We can think of this as choosing, in order, four of the nineteen numbers 1, 2, 4, 5, · · · , 18, 19, 20, and then inserting 3 in any of the positions from the first through fifth numbers of the drawing. This will result in all ordered sequences of length five from the twenty numbers, and containing the number 3 in some position. There are therefore 5(19 P4 ) = 465, 120 such sequences. Therefore Pr(selecting a 3) =

465120 1 = . 1860480 4

(c) We must count all the drawings (sequences) that contain at least one even number. This could mean the sequence contains one, two, three, four or ﬁve even numbers. This is a complicated counting problem if approached directly. It is easier to count the sequences that have no even number, hence are formed just from the ten even integers from 1 through 20. If this number is N0 , then the number X of sequences having an even number is X = 1, 860, 480 − N0 . Now N0 is easy to compute, since this is the number of ordered ﬁve-term sequences of the ten odd numbers. Thus N0 =10 P5 = 30, 240. Then X = 1860480 - 30240 = 1830240. Then

1830240 . 1860480 This is approximately 0.984, so it is very likely that an even numbered ball was drawn. Pr(an even number was drawn) =

24. We need to be clear what an outcome of this experiment looks like. An outcome can be written as a string abc, with each letter representing (in some way) a chosen drawer out of the nine available drawers. The number of outcomes, without regard to order, is 9 C3 = 84. (a) A person gets at least 1,000 dollars by picking exactly one drawer with the thousand dollar bill and two without. There are 2(7 C2 ) ways to do this. Or, we could pick both drawers with the thousand dollar bill and one of the others. There are 7 ways to do this. The number of ways of getting at least a thousand dollars is therefore 42 + 7, and Pr(getting at least one thousand dollars) =

7 49 = . 84 12

(b) A person cannot end up with less than one dollar. This is not an outcome in the sample space. We may also assign a probability of zero to this proposed outcome. 11

(c) The payoff is 1.50 exactly when the person chooses three drawers, each containing fifty cents. The number of ways this can happen is 7 C3 = 35, so 5 35 = . Pr(1.50 payoff) = 84 12 Notice that the sum of the probabilities of (a) and (c) is 1. This is because, in choosing three drawers, the person must either choose all drawers having fifty cents, or must choose at least one of the drawers having a thousand dollar bill. 25. The number of five-card hands (disregarding the order of the deal) is 52 C5 = 2, 598, 960. (a) The number of hands containing exactly one jack and exactly one king is 4(4)(44 C3 ) = 211, 904. This is because there are four ways of getting one jack, four ways of getting one king, and then we choose (without order) three cards from the remaining 44 cards. Thus Pr(exactly one jack and exactly one king) =

211904 . 2598960

This is approximately 0.082, so this event is quite likely. (b) The hand will contain at least two aces if it has exactly two aces, exactly three aces, or exactly four aces. The number of such hands is (4 C2 )(48 C3 ) + (4 C3 )(48 C2 ) + (4 C4 )(48 C1 ) = 108, 336. For the ﬁrst term, choose two aces out of four, then three cards from the remaining forty-eight cards, and similarly for the other two terms for drawing three aces or drawing four aces. Then Pr(draw at least two aces) =

108336 . 2598960

This is approximately 0.042. As we might expect, a hand with at least two aces is not very likely. 26. There are 101 numbers to choose from, and 101 C2 = 668, 324, 943, 343, 021, 950, 370

ways to do this (without regard to order). (a) There are twenty-one numbers in the 80 to 100 range, inclusive. The number of ways of choosing twenty of these (without order) is 21 C20 = 21. Therefore Pr(all numbers are larger than 79) = 12

21 . 668324943343021950370

This is very close to zero. In the real world no sane person would bet on drawing all the twenty numbers from the eighty to one hundred range. (b) There are 100 C19 = 132, 341, 52, 939, 212, 267, 400

ways to choose a five: choose one five, then nineteen numbers from the remaining one hundred numbers. Therefore Pr(choose a five) =

132341572939212267400 . 668324943343021950370

This is approximately 0.190. Since twenty numbers is nearly 1/5 the 101 numbers, it is not surprising that the probability of getting any particular number is close to 1/5. Section 3 Complementary Events

1. There are many ways seven dice can come up with at least two fours (call this event E). We can count these, but it may be easier to look at the complementary event E C , which is that fewer than two dice come up 4. This is the event that exactly one die comes up 4, or none of them do. If no 4 comes up, then each of the seven dice has ﬁve possible numbers showing, for 57 possibilities. If exactly one die comes up with a four, then the other dice have ﬁve possible numbers showing, and this can occur in 7(56 ) ways. This means that E C has 57 + 7(56 ) = 187, 500 outcomes. Then, since the total number of outcomes of seven rolls is 67 = 279, 939, we have 187500 . Pr(E C ) = 279936 The probability we are interested in is Pr(E) = Pr(at least one 4) = 1 −

187500 . 279936

This is approximately 0.330. 2. In fourteen coin tosses, there are 214 = 16, 384 outcomes. Now let E be the event that at least three of the fourteen coins come up heads. E has many outcomes in it. It may be easier to deal with E C , which is the event that fewer than three of the fourteen coins come up heads. E C consists of the events: no head comes up, one head comes up, or two heads come up. 13

No heads can come up in one way (all the coins are tails). Exactly one head can occur in fourteen ways (one of the fourteen coins is a head). Exactly two heads can occur in 14 C2 = 91 ways (any two of the fourteen coins come up heads). Then Pr(fewer than three heads) =

106 1 + 14 + 91 = . 16384 16384

Then

106 . 16384 This is approximately 0.994, suggesting that it is very likely in fourteen ﬂips to see at least three heads. Pr(at least three heads) = 1 −

3. Draw 5 cards from the deck. Without regard to the order of the draw, there are 52 C5 = 2, 598, 960 hands that can occur. Consider the event E that at least one card is a face card, or is numbered 4 or higher. E will take some work to enumerate. Thus look at the smaller complementary event E C , which is the event that the hand has no ace, jack, king or queen, and every card numbers 2 or 3. This leaves just two’s and three’s to form the hand. With these eight cards, the number of possible hands is 8 C5 = 56. Then 56 . Pr(E C ) = 2598960 Then 56 Pr(E) = 1 − , 2598960 which is approximately 0.999. 4. Two coins are ﬂipped and dive dice are rolled. The number of outcomes is 22 65 = 31, 104. We want the probability of the event E that two heads and at least one four come up. E is a large event in the sense of containing many outcomes. Look at the perhaps simpler event E C , which is the event that two heads did not come up, or at least one 4 did not come up. (Note: the negation of a proposition A and B is (notA) or (notB).) This happens if two heads do not come up, and the dice can be anything (3(65 ) ways), or if there are two heads and no die comes up four (this happens in 55 ways). Therefore Pr(E ) =

3(65 ) + 55 26453 = . 31104 31104

This is about 0.85. Then Pr(E) = 1 − approximately 0.15.

26453 , 31104

5. Let E be the event that at least one of the four numbers selected from 1, 2, · · · , 55 is greater than 4. To compute Pr(E) we must count the outcomes in E, and this is tedious. Look at the apparently simpler event E C , which is the event that all four numbers chosen are less than or equal to 4, hence must be chosen from the numbers 1, 2, 3, 4. Since we are choosing four numbers, there is only one way to do this (without regard to order) from these four numbers. Since the total number of ways of choosing four numbers from fifty-five numbers is 55 C4 , then 1 1 . Pr(E C ) = = 341055 55 C4 Then 1 , Pr(E) = 1 − 341055 which is nearly 1, as we might expect intuitively. 6. Suppose N ≥ 2. We want the probability that at least two have the same birthday. Let p be this probability of this event E. We would guess that this depends on N , and that p increases toward 1 as N increases. If N is fairly large, there is a huge number of possibilities to consider in computing p. It might be easier to look at the complementary event E C that no two people have the same birthday. If q is the probability of this, then p = 1 − q. Suppose the year has 365 days. Make a list of N numbers by listing each person’s birthday (as a numbered day from 1 to 365). The total number of sequences that can be formed in this way is (365)(365) · · · (365) = (365)N , since each person could have any of the 365 days as a birthday. Now count the number of sequences of birthdays in which no two people have the same birthday. In this case, the first person has 365 possible birthdays, but the second has 364, the third 363, and so on. The number of sequences with no duplicated birthdays is (365)(364) · · · (365 − N + 1). Therefore

(365)(364) · · · (365 − N + 1) . (365)N We can compute this for any N to obtain q, hence p. q=

Try computing p for some small values of N . With N = 5 we get p ≈ 0.024. With N = 15, we get p ≈ 0.25. With n = 25, we get p = 0.57. This is a surprising result. With only twenty-ﬁve people, there is a better than even chance that at least two will have the same birthday! For N ≥ 60, the probability exceeds 0.99, making it highly likely that a group of sixty or more people will have at least two people with the same birthday. 15

Section 4 Conditional Probability 1. (a) The experiment has four outcomes. Let E be the event that the ﬁrst coin comes up heads. Then E = {HH, HT }, and Pr(E) =

1 . 2

(b) Let U be the event that at least one coin came up heads. Then U = {HH, HT, T H} and E ∩ U = {HH, HT }. The conditional probability of E knowing U is Pr(E|U ) =

number of outcomes inE ∩ U 2 = . number of outcomes in U 3

Alternatively, we could compute Pr(E|U ) =

2/4 2 Pr(E ∩ U ) = = . Pr(U ) 3/4 3

2. (a) Let E be the event that exactly three of the four coins come up heads. Then 1 4 Pr(E) = 4 = . 2 4 (b) We must be clear on what we mean by: if we know that one came up tails. Consider two cases. Suppose ﬁrst that we interpret this to mean that we know that exactly one coin came up tails. Let this event be U . We know that exactly three coins came up heads, so Pr(E|U ) = 1. On the other hand, suppose we interpret the condition of the question to mean that we know at least one coin came up tails. Let U be this event. Then U contains all outcomes except HHHH. Further, E ∩ U = E, so now Pr(E) 1/4 4 Pr(E ∩ U ) = = = . Pr(E|U ) = Pr(U ) Pr(U ) 15/16 15 3. Flip four coins. There are 24 = 16 outcomes. (a) Let E be the event that at least three come up tails. Then E has ﬁve outcomes, namely T T T H, T T HT, T HT T, HT T T, T T T T . Then Pr(E) =

5 . 16

(b) We want the probability that at least three come up tails, having seen two come up tails. This means we know that there were at least two tails. Let U be the event that at leat two tails come up. Then U consists of the outcomes HHT T, HT HT, T HHT, T HT H, T T HH, HT T H, T T T H, T T HT, T HT T, HT T T, T T T T. Then Pr(U ) = 11/16. Further, E ∩ U = E, so Pr(E|U ) =

Pr(E) 5 5 Pr(E ∩ U ) = = = . Pr(U ) Pr(U ) 16 11

4. There are 52 C6 = 20, 358, 520 six card hands from a ﬁfty-two card deck (disregarding order). (a) Let E be the event that exactly two face cards (jack, queen, king) were dealt. In this case, two cards were chosen from the 12 face cards, and the remaining four cards from the other 40 cards. Thus E contains (12 C2 )(40 C4 ) = 6, 031, 740 outcomes. Then Pr(E) =

6031740 , 20358520

which is about 0.296. (b) Let U be the event that a kind was dealt. We want the conditional probability that exactly two face cards were dealt, knowing that a king was dealt. This can occur if one, two, three or four kings were dealt, along with the other cards. Count: hands with one king: 4(48 C5 ) hands with two kings: 6(48 C4 ) hands with three kings: 4(48 C3 ) hands with four kings: 48 C2 . This is the number of outcomes in U , and equals 8, 087, 008. Now we need to consider E ∩ U , which consists of all hands with exactly two face cards and knowing that a king was dealt. In this case, one of the two face cards is a king and the other face card can be another king, a jack or a queen. There are 4(11)(40 C4 ) = 4, 021, 160 outcomes in this event. Then Pr(E ∩ U ) Pr(E|U ) = Pr(U ) 4021160 number of outcomes in E ∩ U = , = number of outcomes in U 8087008 which is approximately 0.497. 17

5. Roll dice four times. First we want the probability that the dice total exactly 19. The number of outcomes of the four rolls is 64 = 1, 296. Now we want the number of outcomes in the event E that the dice total 19. Here are the ways the dice can total 19: 6, 6, 6, 1 ( four ways), 6, 6, 5, 2 (twelve ways) 6, 6, 4, 3 ( twelve ways), 6, 5, 5, 3 ( twelve ways) 6, 5, 4, 4 ( twelve ways), 5, 5, 5, 4 ( four ways). E has 56 outcomes in it, and Pr(E) =

56 , 1296

which is about 0.043. Now suppose we want the probability that the dice total exactly 19, given that one die came up 1 (let this event be U ). Consistent with this, we could have one, two, three or four dice come up 1. We need to count the number of outcomes in U . But U consists of all outcomes except those in which no die came up 1. There are 54 = 625 such outcomes. Therefore U has 65 − 54 = 1296 − 625 = 671 outcomes. Further E ∩ U consists of outcomes totalling 19 and having at least one 1, and there are only four such outcomes. Therefore Pr(E|U ) =

4 , 671

which is approximately 0.059. 6. Roll two dice. We want the probability that they sum to at least 9, knowing that one die came up even. Two dice can sum to at least 9 if they sum to 9, 10, 11 or 12. This can happen in the following ways: sum to 9: 6, 3 or 5, 4 (two ways) sum to 10: 5, 5 or 6, 4 (three ways) sum to 11: 6, 5 (two ways) sum to 12: 6, 6 (one way.) Let E be the event that the dice sum to at least 9, and U the event that one die came up even. We interpret this to mean that one or both dice can come up even. Then E ∩ U consists of eight outcomes. We must count the outcomes in U . These consist of exactly one even: eighteen such outcomes, or, both even: nine outcomes. 18

Then

number of outcomes in E ∩ U 8 = , number of outcomes in U 17 which is approximately 0.47. Pr(E|U ) =

7. Deal a five card hand without regard to order. We want the probability that the hand has four aces, knowing that a four of spades was dealt. Let E be the event that the hand has four aces, and U the event that a four of spades was dealt. Now E ∩ U consists of all outcomes in which there are four aces and a four of spades. Disregarding order, there is one five card hand having four aces and a four of spades, hence one outcome in E ∩ U . We need to count the outcomes in U . If one card is a four of spades, the other four can be any four of the remaining fifty-one cards. There are therefore 51 C4 = 249, 900

unordered hands having the four of spades. Then Pr(E|U ) =

1 , 249, 900

which is very nearly zero. 8. Let E be the event that seven coin flips will produce at least five heads, and U the event that four flips come up heads. Then E ∩ U is the event that five, six or seven heads come up, and U is the event that four, five, six or seven heads come up. The number of outcomes in E ∩ U is 7 C5 + (7 C6 ) + (7 C7 ), which equals 29. The number of outcomes in U is 29 + (7 C4 ), or 64. Then Pr(E|U ) =

29 . 64

This is approximately 0.453. 9. Toss four dice. We want the probability that all four come up an odd number. This probability is Pr(all odd in four tosses) =

34 1 . = 4 6 16

Now let E be the event that all four dice come up odd and let U be the event that one die came up 1 and another, 5. We want Pr(E|U ). The number of outcomes in U is 2(4 C2 )(6)(6), or 432. Next, E ∩U consists of all outcomes in which it is know that one die comes up 1 and one comes

up 5, and all four come up odd. this means we roll 1, 5, odd, and odd, and there are 2(4 C2 )(3)(3) = 108 ways to do this. Then Pr(E|U ) =

108 1 = . 432 4

Finally, suppose we know that the second die came up 6. Now the outcome that all four dice came up odd is not in the conditional sample space, and may be thought of as having probability 0.

Section 5 Independent Events 1. Flip four coins. E is the event that exactly one coin comes up heads, and U is the event that at least three come up tails. Now, E ∩ U consists of the outcomes in which one coin comes up heads and the others are all tails. This can happen in four ways, so Pr(E ∩ U ) =

1 4 = . 24 4

But Pr(E) = 1/4 and Pr(U ) = 5/16. Since Pr(E ∩ U ) = Pr(E)Pr(U ), then E and U are not independent. This is intuitively apparent, since the knowledge that exactly one coin comes up heads inﬂuences the probability that at least three come up tails, by removing one possible way this can happen. 2. Draw two cards from a deck. E is the event that both cards are aces. U is the event that one card is a diamond and the other, a spade. Assume that the order of the draw does not count. Now Pr(E) = And Pr(U ) =

4 C2 52 C2

1 6 = . 1326 221

13 (13)(13) 169 = . = 1326 102 52 C2

Finally, E ∩ U consists of all outcomes in which the ace od diamonds and the ace of spaces are drawn. This can happen in one way, so Pr(E ∩ U ) =

1 = Pr(E)Pr(U ). 1326

E and U are not independent.

3. Two dice are rolled and E is the event that they total more than 11. U is the event that at least one die comes up even. E can happen only one way (they both come up 6), so Pr(E) =

1 1 . = 62 36

And U consists of all outcomes with exactly one die coming up even (eighteen ways this can happen) and all outcomes with both even (nine ways). Therefore 3 27 = . Pr(U ) = 36 4 Finally, E ∩U consists of all outcomes with at least one even, and totalling more than 11. This can happen in one way (both dice 6), so Pr(E ∩ U ) =

1 . 36

Since Pr(E ∩ U ) = Pr(E)Pr(U ), these events are not independent. 4. E is the event that at least three of the six children are girls, and U is the event that at least two are girls. In the absence of information, assume that the probability of having a boy is 1/2, and that of a girl, also 1/2. The number of outcomes in E is 6 C3 +6 C4 +6 C5 +6 C6 = 42.

Then

21 42 . = 26 32 The number of outcomes in U is 42 +6 C2 , or 57. Then Pr(E) =

Pr(U ) =

57 57 = . 26 64

Here E ∩ U = E, so Pr(E ∩ U ) =

21 = Pr(E)Pr(U ). 32

E and U are not independent. 5. Flip two coins and roll two dice. E is the event that at least one head comes up. U is the event that at least one 6 comes up. E ∩ U is the event that at least one coin comes up heads and at least one die comes up 6. The number of outcomes in E is (3)(36), which is 108. Then Pr(E) = 21

3 108 = . 22 62 4

The number of outcomes in U is 4(11), so 11 44 = . 4(36) 36

Pr(U ) =

Finally, E ∩ U consists of all outcomes with at least one head and at least one 6. There are 3(11) = 33 outcomes in this event. Therefore Pr(E ∩ U ) =

33 11 = . 4(36) 48

Now observe that Pr(E)Pr(U ) =

3 11 11 = = Pr(E ∩ U ), 4 36 48

so E and U are independent. 6. Pick two cards from a 52 card deck, with replacement. E is the event that the ﬁrst card drawn was a king. U is the event that the second card drawn was an ace. E ∩ U is the event that the ﬁrst card drawn was a king and the second card, an ace. E can happen in 4(52) = 208 ways. Then 1 4(52) . = 522 13

Pr(E) =

Note that the number of hands is 522 because we are drawing with replacement. U can happen in 4(52) = 208 ways, so Pr(U ) =

1 . 13

E ∩ U can occur in 4(4) = 16 ways, so Pr(E ∩ U ) =

1 16 = Pr(E)Pr(U ), = 522 169

so E and U are independent. It is interesting to recalculate these results, if the cards are chosen without replacement. 7. Deal two cards from a 52 card deck, without replacement. E is the event that the ﬁrst card drawn was a jack of diamonds. U is the event that the second card was a club or spade. E ∩ U is the event that the ﬁrst card was a jack of diamonds and the second card a club or spade. E can happen in 51 ways, since the deal is without replacement. then Pr(E) =

1 51 = . 52(51) 52

We must count the outcomes in U . If the first card drawn was a club or spade, there are 26 choices for this card, and then 25 for the second, for (26)(25) outcomes. If the first card drawn was not a club or spade, then there are 26 choices for the first card and 26 for the second as well for 262 outcomes. Then U has 26(25 + 26), or (26)(51) outcomes. Then Pr(U ) =

1 26(51) = . 52)(51) 2

Finally, E ∩ U has 26 outcomes in it, so Pr(E ∩ U ) =

1 26 = . 52(51) 102

Now compute Pr(E)Pr(U ) =

1 1 1 = , 52 2 104

so E and U are not independent. 8. Flip four coins. E is the event that the ﬁrst coin is a head. U is the event that the last coin is a tail. Then E ∩ U is the event that the ﬁrst coin is a head and the last coin, a tail. E can happen in 23 = 8 ways, so Pr(E) =

1 23 = . 24 2

U can also happen in 8 ways, so Pr(U ) =

1 . 2

These make sense since on any ﬂip there is a 1/2 probability of getting a head or tail. E ∩ U can happen in four ways, so Pr(E ∩ U ) =

1 4 = . 16 4

Now observe that Pr(E)Pr(U ) = Pr(E ∩ U ) so these events are independent. 9. There are 24 = 16 outcomes of four ﬂips. The following have at least two heads: HHHH, HHHT, HHT H, HT HH, T HHH, HHT T, HT HT, HT T H, T T HH, T HT H, T HHT.

The probability of a head is 0.4, so the probability of a tail must be 0.6. Then Pr(at least two heads) = (0.4)4 + 3(0.4)3 (0.6) + 6(0.4)2 (0.6)2 = 0.4864. The probability of getting exactly two heads is Pr(exactly two heads) = 6(0.4)2 (0.6)2 , which is approximately 0.3456. 10. The probability of rolling a 5 in three tosses may be thought of as zero, since the outcome is impossible with three tosses of dice that can come up only 1, 4 and 6. 11. Let r be the probability of drawing a red, g the probability of a green, and b the probability of a blue marble. Since any drawn marble must be blue, green or red, then r + b + g = 1. But b = 2g = 3r, so 11 2 r + 3r + r = r=1 3 2 so

3 2 3 6 2 ,g = = , and b = . 11 2 11 11 11 Now we know the probability of picking any particular color of a marble out of the jar. Imagine picking three marbles. We want the probability of picking exactly two red ones. The possible draws with exactly two red marbles are rrb, rbr, brr, rrg, rgr, grr. r=

Then Pr(exactly two red marbles in three draws) 2 2 6 3 2 108 2 +3 = . =3 11 11 11 11 1331 This is approximately 0.081.

Section 6 Tree Diagrams

50 5 1/2 10

1/2 1/2 1/4

1/2

1/4

10 1/4

1/2

1/4

1/2

10 1/2 1/2 1200

Figure 1: Tree diagram for Problem 1, Section 6.

1. The tree diagram in Figure 1 shows the outcomes. We get the probabilities by multiplying numbers on edges of paths. the outcomes are payoﬀs of 0, 5, 10, 50 and 1200 dollars. From the edges of the tree, read: 1 1 1 = , Pr(0) = 4 2 8 1 1 1 = , Pr(5) = 4 2 8 1 1 1 Pr(10) = 4 = , 4 2 2 1 1 1 = , Pr(50) = 4 2 8 1 1 1 = . Pr(1200) = 4 2 8 As a check, notice that these probabilities sum to 1.

diamonds

stocks cash

1/4

key

whistle

1/2

1/4 1/2 0

1/3 1/3 1/3

1/3

1/3 confederate currency 1/3 stocks 1/3 1/3 perfume 1/3 mansion

Figure 2: Tree diagram for Problem 2, Section 6.

2. From the tree diagram (Figure 2), read: 2 1 7 1 1 = +2 , 3 2 3 18 1 1 1 1 = , Pr(diamonds) = 3 2 4 24 1 1 1 1 Pr(cash) = = , 3 2 4 24 1 1 1 1 = , Pr(whistle) = 3 2 4 24 2 1 1 Pr(Confederate currency) = = , 3 9 2 1 1 Pr(perfume) = = , 3 9 2 1 1 Pr(mansion) = = , 3 9 2 1 1 11 1 1 + . = Pr(stocks) = 3 2 4 3 72 Pr(receive nothing) =

3. From the tree diagram in Figure 3, obtain 1 1 1 = , Pr(Ford) = 2 6 3 9 1 1 1 Pr(Chevrolet) = , = 6 3 18 1 1 1 Pr(VW) = = , 6 2 12 1 1 1 Pr(Porsche) = 2 = , 6 2 6 1 1 1 Pr(Lamborghini) = = , 6 2 12 1 1 1 Pr(tricycle) = = , 6 2 12 1 1 1 = , Pr(Mercedes) = 2 6 3 9 1 1 1 = , Pr(Honda) = 6 3 18 1 1 1 Pr(tank) = = , 6 2 12 1 1 1 Pr(bicycle) = 3 = , 6 4 8 1 1 1 Pr(Stanley Steamer) = = . 6 4 24 4. We do not need a tree diagram for this problem, but we have drawn one in Figure 4 to visualize the possibilities. Of course, we do not want to put 700 dots for the seats in the jumbo jet, so these are just indicated. Because only half of the seats on the jet have odd numbers, then 1 1 1 = . Pr(odd numbered jet seat) = 350 3 700 6

Ford

Chevrolet VW 1/3

Ford

1/3

Porsche

1/2

1/3

1/2

Lamborghini 1/6

1/6

1/2 1/2

tricycle

1/6

Mercedes 1/6

1/3 Mercedes 1/3

1/6 1/3

1/6 1/2

bicycle 1/4

1/2

bicycle

Civic tank

Porsche

1/4 bicycle 1/4 1/4

Stanley Steamer

Figure 3: Tree diagram for Problem 3, Section 6.

1 seat Piper Cub 8 seats 1/3

1/8 each 1/3 company jet 700 seats

1/3

1/700 each

Jumbo Jet

Figure 4: Tree diagram for Problem 4, Section 6.

years experience ≥ 10 5 to 10 years exp. 1 to 5 years exp. < 1 year experience

per cent of time products defective 1 3 5.7 11.3

per cent as a decimal 0.01 0.03 0.057 0.113

Table 1: Information for Problem 1, Section 7.

5. From the tree diagram in Figure 5, we read the following probabilities: 1 1 1 = , Pr(0) = 5 5 4 4 1 1 1 Pr(20) = 2 = , 5 4 10 1 1 1 Pr(nickel) = = , 5 4 20 1 1 1 Pr(chair) = = , 5 4 20 1 1 1 = , Pr(500) = 4 5 4 5 1 1 1 Pr(1000) = = , 5 4 20 1 1 1 Pr(1500) = 4 = , 5 4 5 1 1 1 = . Pr(lion) = 2 5 4 10

Section 7 Bayes’s Theorem 1. The company hired people in four groupings. The following table displays the given information. The table below displays these groupings by work experience, and, for each, the percentage this group constitutes of the total work force (written as a decimal), and a reiteration of the percentage (as a decimal) of products that are defective. By writing these percentages as decimals, they can be thought of as probabilities. For example, if 45 percent of the workforce is in one category, then the probability that a worker is in that category is taken to be 0.45.

0 0 0

1/4 1/4

chair

1/4 1/4

house 1

0 0 1/5

1000

1/4

house 2

1/4 .05

1/5 500 1/5

500 1/4

1/4

1/5

500

1/4

house 3 1/4

500

1/5 1500

1/4 1/4

1500

house 4

house 5 1/4

1/4

1500

1/4

1500

1/4

1/4 1/4

lion

Figure 5: Tree diagram for Problem 5, Section 6.

group ≥ 10 5 to 10 years exp. 1 to 5 years exp. < 1 year experience

per cent of work force (Ej ) 0.45 0.37 0.07 0.11

per cent of time products defective 0.01 0.03 0.057 0.113

Table 2: Percentages of workforce and products defective, by group (Problem 1, Section 7).

For each category or grouping, we want the probability that a worker is in this grouping, knowing that an item is defective. Let U be the event that an item is defective, and Ej the event that a worker is in grouping j (column two of Table 2). We will use the equation Pr(Ej |U ) =

for j = 1, 2, 3, 4. The numbers Pr(U |Ej ) are read from column three of the last table. Now compute Pr(E1 |U ) = (0.45)(0.01) (0.45)(0.01) + (0.37)(0.03) + (0.07)(0.057) + (0.11)(0.113) = 0.141. This is the probability that a worker was in the ten or more years experience group, knowing that a product was defective. For j = 2, 3, 4, the denominator in the Bayes formula is the same and we need only adjust the numerator (which is one term of the denominator). Compute Pr(E2 |U ) = (0.37)(0.03) (0.45)(0.01) + (0.37)(0.03) + (0.07)(0.057) + (0.11)(0.113) = 0.347, Pr(E3 |U ) = (0.07)(0.057 (0.45)(0.01) + (0.37)(0.03) + (0.07)(0.057) + (0.11)(0.113) = 0.125 and Pr(E4 |U ) = (0.11)(0.0113) (0.45)(0.01) + (0.37)(0.03) + (0.07)(0.057) + (0.11)(0.113) = 0.039. 2. We can display the information given as follows: Let U be the probability that a keyboard is defective (column three). Let E1 and E2 be the probabilities that a keyboard is from a particular 33

city Los Angeles Detroit

per cent of keyboards manufactured 0.37 0.63

defective 0.02 0.053

Table 3: Information for Problem 2, Section 7. group adult men adult women boys girls

survive < 1 year 619 471 155 190

survive ≥ 2 years 257 320 104 152

total group pop. 876 791 155 190

Table 4: Information for Problem 3, Section 7.

city (column two). We want the probability that a keyboard is from a particular city, knowing that it is defective. Compute Pr(manufactured in Los Angeles| the keyboard is defective) = Pr(E1 |U ) (0.37)(0.02) (0.37)(0.02) + (0.63)(0.053) = 0.181. =

Finally, Pr(manufactured in Detroit| the keyboard is defective) = Pr(E2 |U ) (0.63)(0.053) (0.37)(0.02) + (0.63)(0.053) = 0.819. =

3. Display the information of the problem: (a) Suppose a patient is selected. We want the probability that this patient is an adult man, if we are told that he or she survived at least two more years. Thus we want to compute Pr(adult man | survived ≥ 2 years). The total number of patients is 876 + 791 + 155 + 190 = 2012. The probability of choosing an adult man is therefore (to the best of our information) approximately 876/2012. Arguing similarly for the other probabilities in

region Lower Amazon Sub-Saharan Africa Newark, New Jersay

per cent of total 0.62 0.25 0.13

fatal 0.14 0.06 0.005

Table 5: Information for Problem 4, Section 7.

the general Bayes formula, we ﬁrst obtain Pr(adult man was chosen | survived ≥ 2 years) (190/2012)(35/190) (876/2012)(257/876) + (791/2012)(320/791) + (155/2012)(104/155) + (190/2012)(152/190) = 0.309. (b) Now compute Pr(girl was chosen | survived < 1 year) (190/2012)(38/190) (190/2012)(38/190) + (155/2012)(51/155) + (791/2012)(471/791) + (876/2012)(619/876) = 0.032. 4. In decimal form, the information is For each of these three locations, we want the probability that the herb came from that location, knowing that it was fatal. Let Ej be the event that the herb came from location j, ordered down column one of the table. Then Pr(came from Lower Amazon | knowing fatal ) = Pr(E1 |U ) Pr(E1 )Pr(U |E1 ) Pr(E1 )Pr(U |E1 ) + Pr(E2 )Pr(U |E2 ) + Pr(E3 )Pr(U |E3 ) (0.62)(0.14) = (0.62)(0.14) + (0.25)(0.06) + (0.13)(0.005) = 0.847. =

Similarly, Pr(E2 |U ) (0.25)(0.06) (0.62)(0.14) + (0.25)(0.06) + (0.13)(0.005) = 0.146, =

and

Pr(E3 |U ) (0.13)(0.005) (0.62)(0.14) + (0.25)(0.06) + (0.13)(0.005) = 0.0063. =

5. Let Ej be the event that the gun is from city j in the table. Let U be the event that the gun explodes. The denominator in the general Bayes formula for this problem is (0.15)(0.02) + (0.07)(0.01) + (0.21)(0.06) + (0.05)(0.02) + (0.09)(0.03) + (0.44)(0.09) = 0.0594. Then (0.15)(0.02) = 0.0505, 0.0594 (0.07)(0.01) = 0.0118, Pr(E2 |U ) = 0.0594 (0.21)(0.06) = 0.2125, Pr(E3 |U ) = 0.0594 (0.04)(0.02) Pr(E4 |U ) = = 0.0135, 0.0594 (0.09)(0.03) = 0.0455, Pr(E5 |U ) = 0.0594 (0.44)(0.09) Pr(E6 |U ) = = 0.666. 0.0594

Pr(E1 |U ) =

Section 8 Expected Value

1. First compute the probability that at least three heads come up in seven ﬂips. It is easier to work with the complementary event that fewer than three heads come up, which occurs if no heads, exactly one head, or exactly two heads come up. The number of ways this can happen is 7 C2 +7 C1 +7 C0 = 29.

Since there are 27 = 128 outcomes, then Pr(fewer than three heads) =

29 128

and

29 99 = . 128 128 The player wins ﬁve dollars if there are at least three heads. Therefore the player’s expected value is Pr(at least three heads) = 1 −

29 99 (5) − (9), 128 128 which works out to about one dollar, eighty three cents. On average the play expects to win this amount each time the game is played. 2. Toss five dice. The player wins if at least three dice come up 5 or 6. First we need the probability of this event. To get at least three dice to come up 5 or 6, we can have exactly three, exactly four, or exactly five dice come up 5 or 6. for exactly three, imagine picking three of the five dice (5 C3 = 10 ways) to come up 5 or 6 (eight ways to arrange three dice in fives and sixes). The other two dice can have any other number, for four possibilities each. There are therefore (10)(8)(4)(4) = 1280 ways of getting exactly three fives or sixes. Now for exactly four or exactly five fives or sixes. Pick any four of the dice (5 C4 = 5 ways to do this). Thee are sixteen ways to pick four of the five dice to be fives or sixes. The remaining die can have any of the other four numbers. Therefore there are (16)(5)(4) = 320 ways of getting exactly four fives or sixes. Finally, for exactly five fives or sixes, each die is a five or six. The number of ways of doing this is 5 C5 +5 C4 +5 C3 +5 C2 +5 C1 +5 C0 = 32.

The number of ways of getting at least three dice to come up with ﬁves or sixes is 1280 + 320 + 32, which is 1632. Therefore Pr(at least three come up 5 or 6) = The expected value to the player is therefore 17 17 (4), (50) − 1 − 81 81 and this is approximately thirty-three cents.

17 1632 . = 65 81

3. Deal six cards, without regard to order. There are 52 C6 = 20, 358, 520 hands (assuming a 52 card deck). The number of hands with no aces is 48 C6 = 12, 271, 512. Therefore Pr(no ace) =

12271512 . 20358520

which is approximately 0.603. Then Pr(at least one ace) = 1 −

12271512 , 20358520

which is approximately 0.397. For the player, the expected value is approximately (0.397)(45) − (0.603)(30), which is approximately −0.225. The player can expect on average to lose between 22 and 23 cents per game. 4. Flip nine coins. There are 29 = 512 outcomes. We need the probability that seven or more come up tails. This happens if exactly seven, or exactly eight, or exactly nine coins come up tails. This occurs in 9 C7 +9 C8 +9 C9 = 46

ways. Then 46 , 512 which is about 0.09. The expected value for the player is 46 46 (50) − 1 − (3), 512 512 Pr(at least seven tails) =

which is about 1.7617. The player can on average expect to win about one dollar, seventy-six cents per game. 5. There are 20 C3 = 1140 ways to draw three marbles from the jar, disregarding the order of the draw. We need the probability that at least two are even-numbered. This happens if exactly two or all three have even numbers. For exactly two even, we can pick two of the ten even numbers in 10 C2 ways, and with each choice, any of the ten odd numbers, for 10(10 C2 ) ways of picking exactly two even-numbered marbles. We can pick exactly three even in 10 C3 ways. Therefore the number of ways of picking two or three even numbered marbles is (10)(10 C2 ) +10 C3 = 570. Therefore Pr(two or three even numbered marbles come up) = 38

570 1 = . 1140 2

The expected value to the player is 1 1 −7 = −2. 3 2 2 On average, the player can expect to lose two dollars per game. 6. If 16 dice are rolled, there are 616 = 2, 821, 109, 907, 456 outcomes. We need the probability that at least five come up six. At least five means exactly five, or exactly six, · · · , exactly sixteen sixes. It is easier to look at the complementary event that fewer than five sixes come up. This occurs with no sixes, or exactly one, two, three or four sixes. There are 516 ways to have no sixes (each die comes up 1, 2, 3, 4 or 5). There are (16 C2 )(514 ) ways to get exactly two sixes (pick two dice and imagine them sixes, and leave the other fourteen dice with no sixes). Similarly, there are (16 C3 )513 ways to get exactly three sixes, and (16 C4 )512 ways to get exactly four sixes. Then 516 + (16 C2 )514 + (16 C3 )513 + (16 C4 )512 616 16 14 5 + 120(5 ) + 560(513 ) + 1820(512 ) = . 616 This is approximately 0.714. This means that Pr(< 5 sixes) =

Pr(≥ 5 sixes) = 1 − Pr(< 5 sixes), and this is approximately 1 − 0.714, or 0.286. The expected value for the player is (0.286)(70)(1) − (0.714)(4), which is approximately 17.164. The player can expect to average a win of about seventeen dollars, sixteen cents per game. 7. We need the probability that, in choosing four of the hats (without regard to order), we draw at least three red hats (meaning exactly two or exactly three red hats). If we draw all three red hats, then we must draw one blue hat, and there are four ways to do this. If we draw exactly two red hats, there are three ways to do this with the three red hats available, and then there are 4 C2 ways of picking the other two hats from the four blue hats. This means that there are 3(6) = 18 ways of picking exactly two red hats. Therefore there are 22 ways of drawing at least two red hats, and Pr(≥ 2 red hats) =

22 . 35

The payoﬀ for the player is 22 22 (10) − 1 − (5), 35 35 which is about 4.42. The player can expect on average to win four dollars and forty two cents per game. 39

8. Fix six coins and roll four dice. We need the probability that four heads or a total of at least 24 on the dice are rolled. We interpret four heads to mean at least four, not exactly four (if we have five heads, then we have four of them). Thus we need Pr(≥ 4 heads or ≥ 24 total on the dice). There are 26 64 outcomes of six coin flips and four dice rolls. Next, there are 22 ways to flip at least four heads (6 C4 = 15 ways of getting exactly four, 6 ways of getting exactly five heads, and 1 way of getting all six heads). With each of these, the dice can come out any 64 ways, for (22)(64 ) possible ways to flip at least four heads. There is only one way for four dice to total 24, so there are 26 outcomes of six coin flips and four dice rolls, with the dice totalling 24. Then Pr(≥ 4 heads or ≥ 24 total on the dice) =

(22)(64 ) + 26 , 26 64

and this is approximately 0.345. The expected value of the player is therefore approximately (0.345)(25) − (1 − 0.345)(3), which is approximately 6.66. The player can expect on average to win six dollars and sixty-six cents for each game.