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584 CHAPTER 16 The Wave Equation

EXAMPLE 16.10
We will use the Fourier transform to solve the wave equation on the real-line subject to the initial
conditions g(x) = 0 and

cos(x) for −π/2 ≤ x ≤ π/2
f (x) =
0 for |x| >π/2.
With zero initial velocity, we can use the solution (16.11). We need the Fourier transform of the
initial position function:
∞

ˆ
f (ω) = f (ξ)e −iωξ dξ
−∞

2
π/2 2cos(πω/2)/(1 − ω ) for ω = 1
= cos(ξ)e −iωξ dξ =
π/2 for ω = 1.
−π/2
f is continuous because
ˆ
2cos(πω/2) π
lim = .
ω→1 1 − ω 2 2
The solution is
∞
1 2cos(πω/2)
y(x,t) = Re cos(ωct)e iωx dω .
π 1 − ω 2
−∞
The integral is complex because the Fourier transform is a complex quantity. The real part is the
solution of the problem. In this case, we can extract the real part by writing
e iωx = cos(ωx) + i sin(ωx),
obtaining

1 ∞ 2cos(πω/2)
y(x,t) = cos(ωx)cos(ωct)dω.
π 1 − ω 2
−∞

SECTION 16.3 PROBLEMS

In each of Problems 1 through 6, solve the wave equa- sin(x) for −π ≤ x ≤ π
3. c = 4, f (x) = 0, and g(x) =
tion on the real line for the given initial position f 0 for |x| >π
and initial velocity g, first by separation of variables
and the Fourier integral, and then by using the Fourier 2 −|x| for −2 ≤ x ≤ 2
4. c = 1, g(x) = 0, and f (x) =
transform. The same solution should result from both 0 for |x| > 2
methods.
e −2x for x ≥ 1
5. c = 3, f (x) = 0, and g(x) =
0 for x < 1
1. c = 12, f (x) = e −5|x| , g(x) = 0
6. c = 2, f (x) = 0, and
⎧
2. c = 8, g(x) = 0, ⎪1 for 0 ≤ x ≤ 2
⎨

8 − x for 0 ≤ x ≤ 8 g(x) = −1for −2 ≤ x < 0
and f (x) = ⎪
0 for x < 0and for x > 8. ⎩ 0 for x > 2and for x < −2
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October 14, 2010 15:23 THM/NEIL Page-584 27410_16_ch16_p563-610