384 Chapter 6 Laplace Transform
The only solution occurs whenξ=0, for otherwise both terms of this equa-
tion have the same sign.
After settingξ=0, we find that Eq. (1) reduces to
tan(η)=1
ηγ,for which there is an infinite number of solutions. We shall number the posi-
tive solutionsη 1 ,η 2 ,.... Now, we have found that the roots ofp(s)arer 0 = 0
andrk=(iηk)^2 =−η^2 k. The computation of the inverse transform follows.
Part a. (r 0 =0.) The limit ofsU(x,s)asstendstozeroiseasilyfoundtobe
- Thus this root contributes 1·e^0 t=1tou(x,t).
Part b. (rk=−ηk^2 .) First, we compute
p′(s)=cosh(√
s)
+
√
sγsinh(√
s)
+
1
2
√
s( 1 +γ)sinh(√
s)
+
1
2 γscosh(√
s)
.
Using the fact that cosh(√rk)+√rkγsinh(√rk)=0, we may reduce the fore-
going to
p′(rk)=−^1
2 γ(
1 +γ+η^2 kγ^2)
cos(ηk).Hence the contribution tou(x,t)ofrkis
q(rk)
p′(rk)exp(rkt)=− 2 γcos(ηkx)−ηkγsin(ηkx)
( 1 +γ+η^2 kγ^2 )cos(ηk)exp(
−η^2 kt)
.
Part c. The construction of the final solution is left to the reader. We note
that an attempt to solve this problem by separation of variables would find
difficulties, for the eigenfunctions arenotorthogonal.
Example 2.
Sometimes one is interested not in the complete solution of a problem, but
only in part of it. For example, in the problem of heat conduction in a semi-
infinite solid with time-varying boundary conditions, we may seek that part of
the solution that persists after a long time. (This may or may not be a steady-
state solution.) Any initial condition that is bounded inxgives rise only to
transient temperatures; these being of no interest, we assume a zero initial con-
dition. Thus, the problem to be studied is
∂^2 u
∂x^2 =∂u
∂t,^0 <x<∞,^0 <t,
u( 0 ,t)=f(t), 0 <t,
u(x, 0 )= 0 , 0 <x<∞.