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16.3 Wave Motion in an Infinite Medium 581

∞

y(x,t) = [a ω cos(ωx) + b ω sin(ωx)]cos(ωct)dω
0
1 ∞ ∞
= f (ξ)cos(ωξ)dξ cos(ωx)
π 0 −∞
∞

+ f (ξ)sin(ωξ)dξ sin(ωx) cos(ωct)dω
−∞
1 ∞ ∞
= [cos(ωξ)cos(ωx) + sin(ωξ)sin(ωx)] f (ξ)cos(ωct)dω dξ.
π
−∞ 0
Upon applying a trigonometric identity to the term in square brackets, we have
1 ∞ ∞
y(x,t) = cos(ω(ξ − x)) f (ξ)cos(ωct)dω dξ. (16.8)
π −∞ 0
We will use this form of the solution when we solve this problem using the Fourier transform.

Zero Initial Displacement We will solve
2
2
∂ y 2 ∂ y
= c for −∞ < x < ∞,t > 0
∂t 2 ∂x 2
and
∂y
y(x,0) = 0, (x,0) = g(x) for −∞ < x < ∞.
∂t
Letting y(x,t) = X(x)T (t), the analysis proceeds exactly as in the case of zero initial velocity,
except now we find that T ω (t) = sin(ωct) because T (0) = 0 instead of T (0) = 0. For ω ≥ 0, we

have functions
y ω (x,t) =[a ω cos(ωx) + b ω sin(ωx)]sin(ωct),
which satisfy the wave equation and the condition y(x,0) = 0. To satisfy (∂y/∂t)(x,0) = g(x),
attempt a superposition
∞

y(x,t) = [a ω cos(ωx) + b ω sin(ωx)]sin(ωct)dω. (16.9)
0
Compute
∂y ∞
(x,t) = [a ω cos(ωx) + b ω sin(ωx)]ωc cos(ωct)dω.
∂t 0
We must choose the coefficients so that
∂y ∞
(x,0) = ωc[a ω cos(ωx) + b ω sin(ωx)]dω = g(x).
∂t 0
We can do this by choosing ωca ω and ωcb ω as the Fourier coefficients in the integral expansion
of g. Thus, let
1 ∞ 1 ∞
a ω = g(ξ)cos(ωξ)dξ and b ω = g(ξ)sin(ωξ)dξ.
πcω −∞ πcω −∞
With these choices, equation (16.9) is the solution of the initial-boundary value problem.

EXAMPLE 16.9
Suppose the initial displacement is zero and the initial velocity is given by

e x for 0 ≤ x ≤ 1
g(x) =
0 for x < 0 and for x > 1.
Compute the coefficients

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October 14, 2010 15:23 THM/NEIL Page-581 27410_16_ch16_p563-610