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582 CHAPTER 16 The Wave Equation

1 ∞ 1 1
ξ
a ω = g(ξ)cos(ωξ)dξ = e cos(ωξ)dξ
πcω πcω
−∞ 0
1 e cos(ω) + eω sin(ω) − 1
=
πcω 1 + ω 2
and
1 ∞ 1 1
ξ
b ω = g(ξ)sin(ωξ)dξ = e sin(ωξ)dξ
πcω πcω 0
−∞
1 eω cos(ω) − e sin(ω) − ω
=− .
πcω 1 + ω 2
The solution is
∞
1 e cos(ω) + eω sin(ω) − 1

y(x,t) = cos(ωx)sin(ωct)dω
πcω 1 + ω 2
0
∞
1 eω cos(ω) − e sin(ω) − ω

− sin(ωx)sin(ωct)dω.
πcω 1 + ω 2
0
As with motion on a bounded interval, the general initial-boundary value problem is solved
by adding the solution with no initial velocity to the solution with no initial displacement.
Solution by Fourier Transform
We will illustrate the use of the Fourier transform to solve the initial-boundary value prob-
lem for the wave equation on the real line. Let the initial displacement be f (x) and the initial
velocity, g(x).
Begin by taking the Fourier transform of the wave equation. This transform must be taken
with respect to x, since −∞ < x < ∞, the appropriate range of values for this transform.
Throughout this transform process, t is carried along as a symbol. Applying the transform, we
have
∂ y ∂ y
2 2
2
F (ω) = c F (ω). (16.10)
∂t 2 ∂x 2
On the right side of equation (16.10), we must transform a second derivative with respect to x,
the variable of the function being transformed. Use the operational rule for the Fourier transform
to write
∂ y
2
2
2
F (ω) = (iω) ˆy(ω,t) =−ω ˆy(ω,t).
∂x 2
For the left side of equation (16.10), the derivative with respect to t passes through the transform
in the x-variable:
2 ∞ 2
∂ y ∂ y −iωx
F (ω) = (x,t)e dx
∂t 2 −∞ ∂t 2
∂ 2 ∞ −iωx ∂ 2
ˆ
= y(x,t)e dx = f (ω,t).
∂t 2 −∞ ∂t 2
The transformed wave equation is
∂ 2
2
2
ˆ y(ω,t) =−c ω ˆy(x,t).
∂t 2

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October 14, 2010 15:23 THM/NEIL Page-582 27410_16_ch16_p563-610