Page 87 - Essentials of applied mathematics for scientists and engineers

book Mobk070 March 22, 2007 11:7

SOLUTIONS USING FOURIER SERIES AND INTEGRALS 77
form

2
2
2
2 2
U mn (τ, ξ, η) = K nm e −(n π +m π r )τ sin(nπξ)sin(mπη) (5.13)
for all m, n = 1, 2, 3, 4, 5,...
Superposition now tells us that
∞ ∞
2
2
2 2
2
−(n π +m π r )τ
K nm e sin(nπξ)sin(mπ) (5.14)
n=1 m=1
Using the initial condition
∞ ∞

g(ξ, η) = K nm sin(nπξ)sin(mπη) (5.15)
n=1 m=1
We have a double Fourier series, and since both sin(nπξ) and sin(mπη) are members of
orthogonal sequences we can multiply both sides by sin(nπξ)sin(mπη)dξdη and integrate over
the domains.
1 1

g(ξ, η)sin(nπξ)sin(mπη)dξdη
ξ=0 η=0
1
1
2
2
= K nm sin (nπξ)dξ sin (mπη)dη
η=0
ξ=0
K nm
= (5.16)
4
Our solution is

1 1
∞ ∞
2
2 2
2
2
−(n π +m π r )τ
4 g(ξ, η)sin(nπξ)sin(mπη)dξdη e sin(nπξ)sin(mπη) (5.17)
n=1 m=1
ξ=0 η=0
Example 5.2 (A convection boundary condition). Reconsider the problem defined by (2.1)
in Chapter 2, but with different boundary and initial conditions,
u(t, 0) = u 0 = u(0, x) (5.18)
ku x (t, L) − h[u 1 − u(t, L)] = 0 (5.19)

The physical problem is a slab with conductivity k initially at a temperature u 0 suddenly exposed
at x = L to a fluid at temperature u 1 through a heat transfer coefficient h while the x = 0 face
is maintained at u 0 .