Page 139 - Essentials of applied mathematics for scientists and engineers

book Mobk070 March 22, 2007 11:7

SOLUTIONS WITH LAPLACE TRANSFORMS 129
which yields

x = 0
y cos y + H sin y = 0

The solution for the second term of U is
√
lim (s − iy)(1 + H)sinh( s ς)e s τ
√ √ √
s → iy s s cosh( s ) + H sinh s )

or
s τ
P(ς, s )e
Q (ς, s )

s =−y 2
where

√ √ √
Q = s s cosh s + H sinh s

√ √ √ 1 √ 1 √ H √

Q = s cosh s + H sinh s + s √ cosh s + sinh s + √ cosh s
2 s 2 2 s
√
s (1 + H) √ s √

Q = cosh s + sinh s
2 2
√ √
s (1 + H) s s √

Q = − cosh s
2 2H
H(H + 1) + y 2
2
Q (s =−y ) = iy cos(y)
2H
while
2
2
P(s =−y ) = (1 + H)i sin(yς)e −y τ
2
2
−(1 + H)sin(y n ς)e −y τ −2H(H + 1) sin(y n ς)e −y τ
=
u n (ς, τ) = 2
H(H+1)+y 2 y n cos(y n ) [H(H + 1) + y ]y n cos(y n )
2H

2(H + 1) sin ς y n −y τ
2
= e n
H(H + 1) + y 2 sin y n
The solution is therefore
∞
2(H + 1)
2
sin ς y n −y τ
u(ς, τ) = e n
H(H + 1) + y 2 sin y n
n=1