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590 CHAPTER 16 The Wave Equation

with roots ±s/c. The general solution for Y(s, x) is
Y(x,s) = k 1 e sx/c + k 2 e −sx/c .

We need two initial conditions to solve for k 1 and k 2 . First, transform y(0,t) = 0 to obtain
Y(0,s) = 0. Then
Y(0,s) = k 1 + k 2 = 0
so k 1 =−k 2 . This means that Y(x,s) has the form

Y(x,s) = k sinh(sx/c),
in which k is an arbitrary constant. To determine k, transform the other initial condition
E(∂y/∂t)(L,t) = f (t) to obtain EY (L,s) = F(s).

Apply this to Y(x,s) = k sinh(sx/c) to obtain
s
Ek cosh(sL/c) = F(s),
c
or
c 1
k = F(s) .
E s cosh(sL/c)
We now have
c sinh(sx/c)
Y(x,s) = F(s) .
E s cosh(sL/c)
The solution to the original problem is
−1
y(x,t) = L [Y(x,s)](t).
Because of the generality of the problem, f (t) is unspecified and we cannot proceed beyond this
point. However, there are special cases of interest in which we can complete the solution. We
will consider two such cases.
Case 1 Suppose f (t) = K, constant.
Now
K
F(s) = LK =
s
so
cK sinh(sx/c)
Y(x,s) = .
E s cosh(sL/c)
2
We can take the inverse transform of this expression by making use of the geometric series:
∞
1
n
= (−1) ξ n
1 + ξ
n=0
for |ξ| < 1. Now write
sinh(sx/c) e sx/c − e −sx/c
=
cosh(sL/c) e sL/c + e −sL/c
e
e
e sx/c −sL/c − e −sx/c −sL/c
=
e
1 + e −sL/c −sL/c
e −(L−x)s/c − e −(L+x)s/c
=
1 + e −2sL/c
1
−(L−x)s/c −(L+x)s/c
= e − e
1 + e −2sL/c
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October 14, 2010 15:23 THM/NEIL Page-590 27410_16_ch16_p563-610