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580 CHAPTER 16 The Wave Equation

We can consolidate cases 1 and 3 by allowing λ = 0incase3.
2
2
The equation for T is T + c ω T = 0 with solutions of the form

T ω (t) = a cos(ωct) + b sin(ωct).
But
∂y
(x,0) = X(x)T (0) = X(x)ωcb = 0,

∂t
and this is satisfied if we choose b = 0. This leaves T ω (t) as a constant multiple of cos(ωct).
Thus far, for every ω ≥ 0, we have functions
y ω (x,t) = X ω (x)T ω (t) =[a ω cos(ωx) + b ω cos(ωx)]cos(ωct),

which satisfy the wave equation and the initial condition (∂y/∂t)(x,0) = 0 for all x. We need to
satisfy the initial condition y(x,0) = f (x). For the similar problem on a bounded interval, we
attempted a superposition ∞ y n (x,t). Now the eigenvalues fill out the entire nonnegative real
n=1
∞
line, so the superposition has the form y ω (x,t)dω. Thus, attempt a solution
0
∞

y(x,t) = [a ω cos(ωx) + b ω sin(ωx)]cos(ωct)dω. (16.7)
0
The initial condition requires that
∞

y(x,0) = [a ω cos(ωx) + b ω sin(ωx)]dω = f (x).
0
This is a Fourier integral expansion of f (x) on the real line, so choose a ω and b ω as the Fourier
integral coefficients of f :
1 ∞
a ω = f (ξ)cos(ωξ)dξ
π
−∞
and
1 ∞
b ω = f (ξ)sin(ωξ)dξ.
π
−∞
With this choice of coefficients, equation (16.7) is the solution.

EXAMPLE 16.8
We will solve the problem
2
2
∂ y ∂ y
= c 2 for −∞ < x < ∞,t > 0,
∂t 2 ∂x 2
∂y
y(x,0) = e −|x| , (x,0) = 0for −∞ < x < ∞.
∂t
Compute the Fourier coefficients of the initial position function. First,
1 ∞ 2
a ω = e −|ξ| cos(ωξ)dξ = .
π π(1 + ω )
2
−∞
Because e −|ξ| sin(ωξ) is an odd function of ξ, b ω = 0. The solution is
2 ∞ 1
y(x,t) = cos(ωx)cos(ωct)dω.
π 0 1 + ω 2
The solution (16.7) is sometimes seen in a different form. If we insert the integrals for the
coefficients into the solution, we have

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October 14, 2010 15:23 THM/NEIL Page-580 27410_16_ch16_p563-610