Page 591 - Advanced_Engineering_Mathematics o'neil

16.2 Wave Motion on an Interval 571

2
2
∂ y ∂ y
= c 2 for 0 < x < L,t > 0,
∂t 2 ∂x 2
y(0,t) = y(L,t) = 0for t ≥ 0,
y(x,0) = 0,
and
∂y
(x,0) = g(x) for 0 < x < L.
∂t
Again, let y(x,t) = X(x)T (t). The problem for X is the same as before,

X + λX = 0; X(0) = X(L) = 0
2
2
with eigenvalues λ n = n π /L and eigenfunctions sin(nπx/L). The problem for T , however, is
2
different, The differential equation is still
n π c
2 2 2
2
T + λc T = T + T = 0
L 2
but now the zero initial displacement gives us y(x,0) = X(x)T (0) = 0, so T (0) =0. Solutions of
this problem for T (t) have the form

nπct
T n (t) = c n sin .
L
Now we have functions
nπx nπct

y n (x,t) = X n (x)T n (t) = c n sin sin
L L
that satisfy the wave equation, the boundary conditions, and the initial condition y(x,0) = 0. To
satisfy the initial velocity condition, we will generally (depending on g) need a superposition
∞

nπx nπct
y(x,t) = c n sin sin .
L L
n=1
We must choose the c n ’s to satisfy
∂y
nπc nπx
∞
= c n sin = g(x).
∂t L L
t=0 n=1
Then
∞
L
nπx
g(x) = c n sin ,
nπc L
n=1
This is the Fourier sine expansion of L g(x). Therefore choose the coefficients
nπc
2 L L nπξ
c n = g(ξ)sin dξ,
L nπc 0 L
or
2 L nπξ
c n = g(ξ)sin dξ.
nπc 0 L
With this choice of the coefficients the solution is
∞
2
1 L nπx nπct
y(x,t) = g(ξ)sin(nπξ/L)dξ sin sin . (16.6)
πc n 0 L L
n=1

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October 14, 2010 15:23 THM/NEIL Page-571 27410_16_ch16_p563-610