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Application of Laplace Transform
.

Solution. By transforming the equation we obtain

2
d Y
2
− s Y = 0,
dx 2
whose solution is given by

Y (x, s) = c cosh sx + c sinh sx.
2
1
Then 0 = Y (0, s) = c and Y (x, s) = c sinh sx. Moreover,
1
2
a
Y (l, s) = = c sinh sl
2
s
and c = a/s sinh sl. Thus,
2

a sinh sx
Y (x, s) = .
s sinh sl

This function has simple poles at s = nπi/l, n = 0, ±1, ±2, . . .
n

a sinh sx x cosh sx ax
Res(0) = lim se ts = a lim = .
s→0 s sinh sl s→0 l cosh sl l

For n = ±1, ±2, . . . ,

( )
( ) ( ) ts πni
πni πni e sinh sx s −
Res = a lim s − = a lim l ×
l s→ πni l s sinh sl s→ πni sinh sl
l l
ts
e sinh sx a e πnit/l sinh πnix a πnx
n πnit/l
× lim = · l = (−1) e sin .
πni s l cosh πni nπi/l πn l
s→
l
Therefore,
∞
∑ ax ∑ a πnx
y(x, t) = Res = + (−1) n e πnit/l sin =
l πn l
n=−∞
n̸=0
∞
ax 2a ∑ (−1) n πnx πnt
= + sin cos ,
l π n l l
n=1

. . . . by the complex inversion formula.
For a more detailed study of operational calculus you can take advantage of
the references.

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