6.4 More Difficult Examples 387
At the pointss=0,s=±iπ,s=±( 2 n− 1 )iπ,n= 2 , 3 ,...,U(x,s)becomes
undefined. The computation of the parts of the inverse transform correspond-
ing to the points other than±iπis easily carried out. However, at these two
troublesome points, our usual procedure will not work. Instead of expecting a
partial-fraction decomposition containing
A− 1
s+iπ+ A^1
s−iπand other terms of the same sort, we must seek terms like
A− 1 (s+iπ)+B− 1
(s+iπ)^2 +A 1 (s−iπ)+B 1
(s−iπ)^2 ,whose contribution to the inverse transform ofUwould be
A− 1 e−iπt+B− 1 te−iπt+A 1 eiπt+B 1 teiπt.One can computeA 1 andB 1 , for example, by noting that
B 1 =slim→iπ[
(s−iπ)^2 U(x,s)]
,
A 1 =slim→iπ{
(s−iπ)[
U(x,s)−(s−B^1 iπ) 2]}
and similarly forA− 1 andB− 1. The limit forB 1 is not too difficult. For example,
B 1 =slim→iπ{ π
s^2 (s+iπ)cosh(^12 s)−cosh(
s(^12 −x))
(
cosh(^12 s))
/(s−iπ)}
= π
−π^2 ( 2 iπ)cosh(^12 iπ)−cosh(
iπ(^12 −x))
1
2 sinh(1
2 iπ)
=−π^12 cos(
π(
1
2 −x))
=−π^12 sin(πx).The limit forA 1 is rather more complicated but may be computed by
L’Hôpital’s rule. Nevertheless, becauseB− 1 =B 1 ,wealreadyseethatu(x,t)
contains the term
B 1 teiπt+B− 1 te−iπt=−π^2 t 2 sin(πx)cos(πt),whose amplitude increases with time. This, of course, is the expected reso-
nance phenomenon.