Page 136 - Essentials of applied mathematics for scientists and engineers

book Mobk070 March 22, 2007 11:7

126 ESSENTIALS OF APPLIED MATHEMATICS FOR SCIENTISTS AND ENGINEERS
If f (τ) = 1, F(s ) = 1/s , a particular solution, V ,is
√
sinh s ς
V = √
s sinh s

where
−1
v = L V (s )

Now,
√
√
√ √ (ς s ) 3 (ς s ) 5
sinh s ς ς s + 3! + 5! + ...
√ = √ √ √
sinh s s + ( s ) 3 + ( s ) 5 + ...
3! 5!
√ √
and so there is a simple pole of Ve s τ at s = 0. Also, since when sinh s = 0, sinhς s not
√
2
2
necessarily zero, there are simple poles at sinh s = 0or s =−n π . The residue at the pole
s = 0is
lim s τ
sV (s )e = ς
s → 0
2
2
and since V (s ) e s τ has the form P(s )/Q(s ) the residue of the pole at −n π is
√ 2 2
2
2
P(ς, −n π ) 2 2 sinh ς se −n π τ sin(nπς) 2 2
e −n π τ = √ = 2 e −n π τ
2
2
Q (−n π ) s cosh √ s + sinh √ s nπ cos(nπ)

2 s =−n π 2
2
The solution for v(ζ, τ)isthen
∞ n
2(−1)
2
2
−n π τ
v(ς, τ) = ς + e sin(nπς)
nπ
n=1
The solution for the general case as originally stated with u(1,τ) = f (τ) is obtained by
first differentiating the equation for v(ζ, τ) and then noting the following:
√
sinh ς s
U(ς, s ) = sF(s ) √
s sinh s
and

L f (τ) = sF(s ) − f (τ = 0)

so that

U(ς, s ) = f (τ = 0)V (ς, s ) + L f (s ) V (ς, s )