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16.5 Laplace Transform Techniques 587

The general solution is

ˆ y S (ω,t) = a ω cos(ωct) + b ω sin(ωct).
Since
a ω =ˆy S (ω,0) = F S [y(x,0)](ω) = F S [0](ω) = 0

and
∂ ˆy S
(ω,0) = ωcb ω =ˆg S (ω)
∂t
then
1
b ω = ˆ g S (ω).
ωc
This gives us
1
ˆ y S (ω,t) = ˆ g S (ω)sin(ωct),
ωc
which is the sine transform of the solution. We obtain the solution by inverting:
2 ∞ 1
y(x,t) = ˆ g S (ω)sin(ωx)sin(ωct)dω.
π 0 ωc
If we have a wave equation on the half-line with zero initial velocity and initial position given
by f , then we can proceed as we have just done, but using the Fourier cosine transform instead
of the sine transform. This is because the information given now fits within the framework of
the operational rule for the cosine transform. As usual, a problem with initial displacement and
velocity can be solved as the sum of the solution with zero initial velocity and the solution with
zero initial position.

SECTION 16.4 PROBLEMS

In each of Problems 1 through 5, solve the problem for 3. c = 2, f (x) = 0, and
wave equation on the half-line for the given c, initial posi-
cos(x) for π/2 ≤ x ≤ 5π/2
tion f and initial velocity g, first by separation of variables,
g(x) =
then by using an appropriate Fourier transform. 0 for 0 ≤ x <π/2and for x > 5π/2
−x
4. c = 6, f (x) =−2e ,and g(x) = 0

x(1 − x) for 0 ≤ x ≤ 1

1. c = 3, g(x) = 0, and f (x) = 2
0 for x > 1 x (3 − x) for 0 ≤ x ≤ 3
5. c=14, f (x)=0, and g(x)=
⎧ 0 for x > 3
⎪0 for 0 ≤ x ≤ 4
⎨
2. c = 3, f (x) = 0, and g(x) = 2 for 4 < x ≤ 11
⎪
0 for x > 11
⎩
16.5 Laplace Transform Techniques
The Laplace transform is well suited to solving certain problems involving wave motion, both on
closed intervals and on the half-line. We will illustrate this by solving one problem on x > 0 and
another on a closed interval.

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October 14, 2010 15:23 THM/NEIL Page-587 27410_16_ch16_p563-610