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576 CHAPTER 16 The Wave Equation

This will be simplified if we choose ψ so that
ψ (x) + Ax = 0.

Integrate twice to get
x 3
ψ(x) =−A + Cx + D,
6
with C and D constants of integration. To find C and D look at the boundary conditions. First,

y(0,t) = Y(0,t) + ψ(0) = 0.
This will be just y(0,t) = Y(0,t) if we make
ψ(0) = 0.

This requires that we choose D = 0. Next,
L 3
y(L,t) = Y(L,t) + ψ(L) = Y(L,t) − A + CL = 0.
6
This reduces to y(L,t) = Y(L,t) if we choose C so that

L 3
ψ(L) =−A + CL = 0.
6
2
With C = AL /6, we have
1
2
2
ψ(x) = Ax(L − x ).
6
With this ψ,
Y(0,t) = Y(L,t) = 0.
Next relate the initial conditions for y to initial conditions for Y.First,
1
2
2
Y(x,0) = y(x,0) − ψ(x) =−ψ(x) = Ax(x − L ),
6
then
∂Y ∂y
(x,0) = (x,0) = 1.
∂t ∂t
Now we have an initial-boundary value problem for Y:
2
∂ Y ∂Y 2
= for 0 < x < L,t > 0,
∂t 2 ∂x 2
Y(0,t) = Y(L,t) = 0for t ≥ 0,
and
1 ∂Y
2
2
Y(x,0) = Ax(x − L ), (x,0) = 1for 0 < x < L.
6 ∂t
We know the solution of this problem. By equations (16.5) and (16.6),
2
L 1
∞
2
2
Y(x,t) = Aξ(ξ − L )sin(nπξ/L)dξ sin(nπx/L)cos(nπt/L)
L 0 6
n=1
∞
2
1 L
+ sin(nπξ/L)dξ sin(nπx/L)sin(nπt/L)
π n 0
n=1
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October 14, 2010 15:23 THM/NEIL Page-576 27410_16_ch16_p563-610