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586 CHAPTER 16 The Wave Equation

EXAMPLE 16.11
We will solve the problem
2
2
∂ y ∂ y
= 16 for 0 ≤ x < ∞,t > 0,
∂t 2 ∂x 2
y(0,t) = 0for t ≥ 0
∂y
(x,0) = 0for0 ≤ x < ∞
∂t
and

sin(πx) for 0 ≤ x ≤ 4
y(x,0) = f (x) =
0 for x > 4.
To write the solution, we need only compute the coefficients
2 ∞
b ω = f (ξ)sin(ωξ)dξ
π 0
2
2 4 2cos (ω) − 1
= sin(πξ)sin(ωξ)dξ = 8sin(ω)cos(ω) .
π 0 ω − π 2
2
The solution is
∞ 2cos (ω) − 1
2
y(x,t) = 8sin(ω)cos(ω) sin(ωx)cos(4ωt)dω.
2
ω − π 2
0
16.4.1 Solution by Fourier Sine or Cosine Transform
We will illustrate the use of a Fourier transform to solve the problem on a half-line. Having
considered the case of zero initial velocity for the separation of variables solution, we will assume
here an initial velocity g(x) but zero initial position f (x) = 0. 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) = 0, (x,0) = g(x) for 0 ≤ x < ∞.
∂t
On the half-line, we can try a Fourier sine or a Fourier cosine transform of y(x,t), thinking of
x as the variable and t as a parameter. The choice depends on the operational rules for these
transforms. The cosine transform requires that we know something about the derivative (with
respect to x)of y at x = 0, while the sine transform requires information about the function itself
at x =0. Since we are given that y(0,t)=0, this is the kind of information we have, so choose the
sine transform. Apply F S to the wave equation and use the fact that the derivative with respect to
t goes through the transform to get
2
∂ 2 ∂ y
2 2 2 2 2 2
ˆ y S = c F S =−c ω ˆy S (ω,t) + ωc y(0,t) =−c ω ˆy S (ω,t).
∂t 2 ∂x 2
This is a differential equation for ˆy S (ω,t) with t as the variable:
2
∂ ˆy S 2 2
+ c ω ˆy S (ω,t) = 0.
∂t 2

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October 14, 2010 15:23 THM/NEIL Page-586 27410_16_ch16_p563-610