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16.5 Laplace Transform Techniques 593

2

h(t)
1

(0,1)
0
0 0.2 0.4 0.6 0.8 1
x t
(4L/c, 0) (8L/c, 0)
–1

–2

FIGURE 16.11 Profiles of the bar at different FIGURE 16.12 Square wave of height 1 and
times with f (t) = Iδ(t). period 4L/c.

1
L e −αs = H(t − α)
s
the analysis of case 1, with this adjustment, yields
∞
cI
1
Y(x,s) = (−1) n e −(((2n+1)L−x)/c)s
E s
n=0
∞
cI
1
− (−1) n e −(((2n+1)L+x)/c)s .
E s
n=0
Invert this term by term to obtain
∞
cI
n (2n + 1)L − x
y(x,t) = (−1) H t −
E c
n=0
∞
cI
n (2n + 1)L + x
− (−1) H t − .
E c
n=0
Figure 16.11 shows a graph of this solution for c = 2 and I = E = L = 1.
The graph shows the profile at times t =0.4,0.7,0.9 (with maximum point 2 achieved farther
to the right as t increases, and t = 1.3,1.8, with minimum −2 achieved farther to the right as t
increases.
As in case 1, this case also has a simple form if we focus on the right end of the bar. In this
case, the Laplace transform of the solution is

cI 1
Y(L,s) = tanh(sL/c).
E s
This also has a simple inverse and we obtain y(L,s) = h(t), where h is the square wave of
Figure 16.12, with a height of 1 and a period of 4L/c.

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October 14, 2010 15:23 THM/NEIL Page-593 27410_16_ch16_p563-610