Cmjv03i02p0233

J. Comp. & Math. Sci. Vol.3 (2), 233-236 (2012)

Solution of Telegraph and Radio Equations Involving A-Function of One Variable S. S. SHRIVASTAVA and PINKEY SIKARWAR Department of Mathematics Govt. P. G. College, Shahdol, M. P., India.

(Received on : April 3, 2012) ABSTRACT The aim of this paper is to first we evaluate an integral involving A-function of one variable and then we make its applications to solve two boundary value problems on (i) radio equation (ii) telegraph equation under certain conditions.

Keywords: Integral, A-function, radio equation, boundary value problems, telegraph equation, variable.

(iii) x ≠ 0 and parameters aj, αj, bk and βk (j = 1 to p and k = 1 to q) are all complex.

1. INTRODUCTION The A-function of one variable is defined by Gautam1 and we will represent here in the following manner: A

((ap, αp))

m, n

[x|

p, q

((bq, βq))

]=

where i = √(− 1) and

1 2π i L

∫ θ(s) xs ds

(1)

where ୮ ୯ k = Im (∑ଵ α୨ ∑ଵ β୨ )

n

m

Π Γ(aj + sαj) Π Γ(1 – bj – sβj)

(i)

m

j=1

j=1

θ (s) =

(2) p

q

Π Γ(1 – aj – sαj) Π Γ(bj + sβj) j=m+1

The integral in the right hand side of is convergent if (i) x ≠ 0, k = 0, h > 0, |arg(ux)| < πh/2 (ii) x > 0, k = 0 = h, (ν − σω) < − 1

p

n

q

h = Re ( Σ αj – Σ αj + Σ βj – Σ βj ) j=1

୮

j=m+1

j=1

୯

u = ∏ଵ α୨ ஑ౠ ∏ଵ β୨ ஒౠ

j=n+1

(ii) m, n, p and q are non-negative numbers in which m ≤ p, n ≤ q.

p

q

1

1

(3)

j=m+1

ν = Re ( Σ aj – Σ bj ) − (p - q)/2 ,

Journal of Computer and Mathematical Sciences Vol. 3, Issue 2, 30 April, 2012 Pages (131-247)

(4)

234

S. S. Shrivastava, et al., J. Comp. & Math. Sci. Vol.3 (2), 233-236 (2012) q

From Gradshteyn2,1, following modified form of the integral is used in the present investigation:

p

w = Re ( Σ βj – Σ αj ) 1

and

1

s = σ + it is on path L when |t| →∞.

L sin ½ nπ Γ(ω)

L

∫ {sin (πx/L)}ω – 1 sin (nπx/L) dx = 0

2

ω–1

(5) Γ{½ (ω ± n + 1)}

where n is any integer and Re (ω ) > 0.

2. INTEGRAL The integral involving the A-function of one variable to be evaluated is: L

∫ 0

(a , α ) (sin (πx/L))ω – 1 sin (nπx/L) Am, l [z (sin (πx/L)) λ | j j 1, p ] dx (bj, βj)1, q p, q

= 2 1 – ω L sin ½ nπ A

m + 1, l

[z 2 – λ |(ω, λ), (aj, αj)1, p

p + 1, q + 2

],

(bj, βj)1, q, (1/2 + ω/2 ± n/2, λ/2)

(6)

provided that λ ≥ 0, Re (ω) > 0 and |arg(uz)| < πh/2, where u and h are given in section 1.

Proof of (6) Replace the A-function by its equivalent contour integral as given in (1), change the order of integration, evaluate the inner integral with the help of (6) and finally interpret it with (1), to get (6).

3. APPLICATION TO TELEGRAPH EQUATION In this section, we consider a problem on telegraph equation in under certain boundary conditions. The telegraph equation is given by

∂2e ∂e = RC ∂x2 ∂t

(7)

where e is the potential, R is resistance and C is capacitance per unit length of the cable respectively.

If we take the following boundary conditions e (0, t) = 0, e (L, t) = 0

(8)

and initial condition e (x, 0) = f (x),

(9)

then the solution of (7) is given by3:

Journal of Computer and Mathematical Sciences Vol. 3, Issue 2, 30 April, 2012 Pages (131-247)

235

S. S. Shrivastava, et al., J. Comp. & Math. Sci. Vol.3 (2), 233-236 (2012) ∞

L

e (x, t) = Σ

Bn sin (nπx/L) exp [– (nπ/RCL)2t]

n=1

Bn = (2/L) ∫

f(x) sin (nπx/L) dx,

(11)

0

(10)

Now we shall consider the problem of determine e (x, t), where

where n is any integer and e (x, 0) = f(x) = sin (πx/L)ω – 1 A m, l

[z (sin (πx/L)) λ |

p, q

(aj, αj)1, p ]. (bj, βj)1, q

(12)

Solution of the Problem: Combining (11) and (12) and making the use of integral (6), we derive m + 1, l

Bn = 2 2 – ω sin ½ nπ A

[z 2 – λ |

p + 1, q + 2

(ω, λ), (aj, αj)1, p (bj, βj)1, q, (1/2 + ω/2 ± n/2, λ/2)

],

(13)

Putting the value of Bn from (13) in (10), we get the following required solution of the problem ∞

e (x, t) = 2 2 – ω Σ

n=1

×

sin ½ nπ sin (nπx/L) exp [– (nπ/RCL)2t] × A m + 1, l

[z 2 – λ |

p + 1, q + 2

(ω, λ), (aj, αj)1, p (bj, βj)1, q, (1/2 + ω/2 ± n/2, λ/2)

],

(14)

provided that λ ≥ 0, Re (ω) > 0 and |arg(uz)| < πh/2, where u and h are given in section 1.

4. APPLICATION TO RADIO EQUATION In this section, we consider a problem on radio equation in under certain boundary conditions. The radio equation is given by ∂2e ∂2e 2 = LC ∂x ∂t2

C is capacitance per unit length of the cable respectively.

Since the ends are suddenly grounded, the boundary conditions are given by e (0, t) = 0, e (l, t) = 0 (16) and initial condition

(15)

where e is the potential, L is inductance and

e (x, 0) = f (x), ∂e/∂t = 0, then the solution of (15) is given by3.

Journal of Computer and Mathematical Sciences Vol. 3, Issue 2, 30 April, 2012 Pages (131-247)

(17)

236

S. S. Shrivastava, et al., J. Comp. & Math. Sci. Vol.3 (2), 233-236 (2012) ∞

e (x, t) = Σ sin (nπx/l) [Cn cos (nπ/l√LC)t + Dn sin (nπ/l√LC)t]

(18)

n=1

where n is any integer and l

Cn = (2/l) ∫

f(x) sin (nπx/l) dx,

(19)

0

and

Dn = 0.

(20)

Now we shall consider the problem of determine e (x, t), where m, k

e (x, 0) = f(x) = sin (πx/l)ω – 1 Ap, q [z (sin (πx/l)) λ |

(aj, αj)1, p ]. (bj, βj)1, q

(21)

Solution of the Problem Combining (19) and (21) and making the use of integral (6), we derive Cn = 2 2 – ω sin ½ nπ Am + 1, k

[z 2 – λ | (ω, λ), (aj, αj)1, p

p + 1, q + 2

],

(bj, βj)1, q, (1/2 + ω/2 ± n/2, λ/2)

(22)

Putting the value of Cn and Dn from (22) and (20) respectively in (18), we get the following required solution of the problem ∞

e (x, t) = 2 2 – ω Σ

n=1

×

sin ½ nπ sin (nπx/l) cos (nπ/l√LC)t × (ω, λ), (aj, αj)1, p m + 1, k [z 2 – λ | ], p + 1, q + 2 (bj, βj)1, q, (1/2 + ω/2 ± n/2, λ/2)

A

provided that λ ≥ 0, Re (ω) > 0 and |arg(uz)| < πh/2, where u and h are given in section 1.

REFERENCES 1. Gautam, G. P. and Goyel, A. N.: Ind. J. Pure and Appl. Math., 12, 1094-1105 (1981).

(23)

2. Gradshteyn, I. S. and Ryzhik, I. M.: Tables of Integrals, Series and Products , Academic Press, Inc. New York, (1980).

3. Nath Vishwa: Engineering Mathematics, Satya Prakashan, New Delhi, (1988).

Journal of Computer and Mathematical Sciences Vol. 3, Issue 2, 30 April, 2012 Pages (131-247)