Page 92 - Essentials of applied mathematics for scientists and engineers

book Mobk070 March 22, 2007 11:7

82 ESSENTIALS OF APPLIED MATHEMATICS FOR SCIENTISTS AND ENGINEERS
must be zero. The boundary condition at ζ = 1is

Z ζ =−C 1 (γ L/r 1 )sin(γ L/r 1 ) = 0, or γ L/r 1 = nπ (5.39)

The boundary condition at η = 1 requires

C 3 [J (β) + BJ 0 (β)] = 0or

0
(5.40)
BJ 0 (β) = β J 1 (β)
which is the transcendental equation for finding β m . Also note that

2
2
λ = γ + β 2 m (5.41)
n
By superposition we write the final form of the solution as
∞ ∞
2
2
−(γ +β )τ
U(τ, η, ς) = K nm e n m J 0 (β m η)cos(nπς) (5.42)
n=0 m=0
K nm is found using the orthogonality properties of J 0 (β m η) and cos(nπζ) after using the initial
condition.

1 1 1 1

2
2
rJ 0 (β m η)dη cos(nπς)dς = K nm rJ (β m η)dη cos (nπς)dς (5.43)
0
r=0 ς=−1 r=0 ς=−1
Example 5.5 (Heat transfer in a sphere). Consider heat transfer in a solid sphere whose
surface temperature is a function of θ, the angle measured downward from the z-axis (see Fig.
1.3, Chapter 1). The problem is steady and there is no heat source.

∂ 2 1 ∂ ∂u
r (ru) + sin θ = 0
∂r 2 sin θ ∂θ ∂θ
u(r = 1) = f (θ) (5.44)

u is bounded
Substituting x = cos θ,

∂ 2 ∂ ∂u
2
r (ru) + (1 − x ) = 0 (5.45)
∂r 2 ∂x ∂x
We separate variables by assuming u = R(r)X(x). Substitute into the equation, divide by RX
and find
2
r [(1 − x )X ]

2
(r) =− =±λ (5.46)
R X