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588 CHAPTER 16 The Wave Equation

A Problem on a Half-Line We will solve the boundary value problem
2
2
∂ y 2 ∂ y
= c − A for x > 0,t > 0,
∂t 2 ∂x 2
∂y
y(x,0) = (x,0) = 0,
∂t
and
y(0,t) = 0,
with A as a positive constant. Because the half-line is unbounded to the right, we also impose the
condition that
∂y
lim (x,t) = 0
x→∞ ∂x
for t ≥ 0.
This problem models an infinitely long string lying on the nonnegative x-axis, with its left
end (x = 0) fastened. The string is pulled downward by a force of constant magnitude A.
Apply the Laplace transform L, with respect t, to the wave equation, using the boundary
conditions and the operational rule for taking the transform of derivatives. In this process L, and
∂/∂x interchange because x is independent of t. The wave equation transforms to
2
∂y ∂ Y A
2 2
s Y(x,s) − sy(x,0) − (x,0) = c − .
∂t ∂x 2 s
Substitute the boundary conditions into this and let Y (x,s) denote ∂Y/∂x. This differential

equation becomes
s 2 A

Y (x,s) − Y(x,s) = .
c 2 c s
2
Think of this as a nonhomogeneous constant coefficient second order differential equation, whose
general solution is the general solution of the associated homogeneous equation plus a particular
3
solution of the nonhomogeneous equation. By inspection, Y p =−A/s is a particular solution.
The characteristic equation of the associated homogeneous equation is
s
2
2
λ − = 0
c
with roots ±s/c. The general solution of the associated homogeneous equation is
Y(x,s) = k 1 e sx/c + k 2 e −sx/c .
Therefore, the general solution for Y(x,s) is
A
Y(x,s) = k 1 e sx/c + k 2 e −sx/c − .
s 3
To solve for k 1 and k 2 , we need two pieces of information. Use the boundary conditions. Take the
transform of y(0,t) = 0 to obtain Y(0,s) = 0. Then
A
Y(0,s) = k 1 + k 2 − = 0,
s 3
s 3 − k 1 . Thus far,
so k 2 = A
A −sx/c A

sx/c
Y(x,s) = k 1 e + − k 1 e − .
s 3 s 3
Now use the limit boundary condition. Applying the transform, we have
L[lim(∂/∂x)y(x,t)]= lim Y (x,s) = 0.

x→∞ x→∞

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October 14, 2010 15:23 THM/NEIL Page-588 27410_16_ch16_p563-610