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Application of Laplace Transform
.

where we have used the following properties (for z = x + iy)

sinh z = cos y sinh x + i sin y cosh x,

cosh z = cos y cosh x + i sin y sinh x

to obtain the last equality.
Therefore,

∞ (∫ x )
∑ ∑ 2 2
Res = 2 e −n π t f(u) sin πnudu sin πnx.
0
n=1

Similarly, the inverse of the second integral is given by

∞ (∫ 1 )
∑ 2 2
2 e −n π t f(u) sin πnudu sin πnx.
x
n=1

Finally,

∞ (∫ 1 )
∑ 2 2
u(x, t) = 2 e −n π t f(u) sin πnu du sin πnx.
n=1 0

The same result is obtained when we solve this problem by the separation-
. . . . of-variables method.

One-Dimensional Wave Equation

The wave motion of a string initially lying on the x-axis with one end at the origin
can be described by the equation

2
2
∂ y ∂ y
= a 2 , x > 0, t > 0.
∂t 2 ∂x 2
The displacement is only in the vertical direction and is given by y(x, t) at
√
position x and time t. The constant a is given by a = T/ρ, where T is the
tension on the string and ρ its mass per unit length. The same equation happens
to describe the longitudinal vibrations in a horizontal beam, where y(x, t)
.
represents the longitudinal displacement of a cross section at x and time t.

Example 5.24, Solve

2
2
∂ y ∂ y
= a 2 , x > 0, t > 0,
∂t 2 ∂x 2
for
+
1. y(x, 0 ) = 0, x > 0,
. . . . .

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