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16.4 Wave Motion in a Semi-Infinite Medium 585

16.4 Wave Motion in a Semi-Infinite Medium

We will solve the wave equation on a half-line 0 ≤ x < ∞. The problem is
2
2
∂ y ∂ y
= c 2 for 0 ≤ x < ∞,t > 0,
∂t 2 ∂x 2
y(0,t) = 0for t ≥ 0
and
∂y
y(x,0) = f (x), (x,0) = g(x) for 0 ≤ x < ∞.
∂t
We will seek a bounded solution, considering first the case that g(x) = 0. Separate variables
by putting y(x,t) = X(x)T (t) to obtain
2
X + λX = 0 and T + λc T = 0.

Unlike the problem on the entire line, we have a boundary condition at the fixed left end. This
means that y(0,t) = X(0)T (t) = 0, so X(0) = 0, and the problem for X is

X + λX = 0; X(0) = 0.
Upon considering cases as we have done before, we find the eigenvalues λ≥0 and eigenfunctions

X ω = b ω sin(ωx).
Because the motion is assumed to begin from rest,
∂y

(x,0) = X(x)T (0) = 0,
∂t
so T (0) = 0. The problem for T is

2
2

T + c ω T = 0; T (0) = 0

with solutions that are constant multiples of cos(ωct). For each ω ≥ 0, we now have a function
y ω (x,t) = b ω sin(ωx)cos(ωct)
that satisfies the wave equation, the boundary condition and the initial condition that the
string starts from rest. To satisfy the condition y(x,0) = f (x), we generally need the
superposition
∞

y(x,t) = b ω sin(ωx)cos(ωct).
0
Then
∞

y(x,0) = b ω sin(ωx) = f (x),
0
so we must choose
2 ∞
b ω = f (ξ)sin(ωξ)dξ,
π 0
which is the coefficient in the Fourier sine integral representation of f on [0,∞). This solves the
problem when g(x)=0. A similar analysis allows us to write an integral solution when f (x)=0
and the string has an initial velocity of (∂y/∂t)(x,0) = g(x).

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October 14, 2010 15:23 THM/NEIL Page-585 27410_16_ch16_p563-610