Page 147 - Essentials of applied mathematics for scientists and engineers

book Mobk070 March 22, 2007 11:7

SOLUTIONS WITH LAPLACE TRANSFORMS 137
The solution for W(x, t) is then obtained by superposition:

∞
2
2
−n π t
W(x, t) = K n e sin(nπx)
n=0
and using the orthogonality principle
1 1
x 1

2
2
e −λ (x − 1) sin(nπx)dx = K n sin (nπx)dx = K n
6 2
x=0 n=0
so
1
∞
2
2
−λ x 2 −n π t
W(x, t) = e (x − 1) sin(nπx)dx e sin(nπx)
3
n=1
x=0

x 1 x 2 2
∞
2
2
P(x,λ, t) = (1 − x ) + (x − 1) sin(nπx)dx sin(nπx) e −n π t e −λ
6 x=0 3
n=1
and
x
! "
2
P(x,λ, t − λ) = (1 − x )e −λ
6
∞ 1
2
2
2
2
x 2 −n π t n π λ−λ
+ (x − 1) sin(nπx)dx sin(nπx) e e
x=0 3
n=1
1
∞
∂ 2 2 x 2 −n π t (n π −1)λ
2
2
2
2
P(x,λ, t − λ) = n π (1 − x )sin(nπx)dxe e
∂t 3
n=1 x=0
According to Duhamel’s theorem, the solution for u(x, t)isthen
1 t
x
∞
2
2
2 2 2 −n π (t−λ) −λ
u(x, t) = (1 − x )n π sin(nπx)dx sin(nπx) e dλ
3
n=1
x=0 λ=0
1
∞ 2 2
2
2
n π x 2 −t −n π t
= (1 − x )sin(nπx)dx [e − e ]sin(nπx)
2
2
n π − 1 3
n=1
x=0
Example 8.9. Reconsider Example 8.6 in which u t = u xx on the half space, with
u(x, 0) = 0
u(0, t) = f (t)