1540470959-Boundary_Value_Problems_and_Partial_Differential_Equations__Powers

346 Chapter 5 Higher Dimensions and Other Coordinates

The general solution of the partial differential equation that is bounded in the
region 0<ρ<c,0<φ<πis thus the linear combination

u(ρ, φ)=

∑∞

n= 0

bnρnPn

(

cos(φ)

)

. (5)

Atρ=c, the boundary condition becomes

u(c,φ)=

∑∞

n= 0

bncnPn

(

cos(φ)

)

=f(φ), 0 <φ<π. (6)

The coefficientsbnare then found to be

bn=

2 n+ 1

2 cn

∫π

0

f(φ)Pn

(

cos(φ)

)

sin(φ)dφ. (7)

B. Heat Equation on a Spherical Shell

The temperature on a spherical shell satisfies the three-dimensional heat equa-
tion. If initially there is no dependence onθ, then there will never be such de-
pendence. Furthermore, if the shell is thin (thickness much less than average
radiusR), we may also assume that temperature does not vary in the radial
direction. The heat equation then becomes one-dimensional:

1

sin(φ)

∂

∂φ

(

sin(φ)∂u

∂φ

)

=R

2

k

∂u

∂t

, 0 <φ<π, 0 <t, (8)

u(φ, 0 )=f(φ), 0 <φ<π. (9)

Naturally, we require boundedness ofuatφ=0andφ=π.
The assumption of a product form for the solution,u(φ,t)=(φ)T(t),
leads to the conclusion that

(sin(φ)′(φ))′

sin((φ))

=R

(^2) T′(t)
kT(t)
=−μ^2.
Thus, we have the eigenvalue problem
(
sin(φ)′

)′

+μ^2 sin(φ)= 0 , 0 <φ<π,

( 0 ) and (π)bounded.

The solution of this problem was found in Section 5.9 to beμ^2 =n(n+ 1 )and

n(φ)=Pn

(

cos(φ)

)

, n= 0 , 1 , 2 ,....

Obviously, the other factor in a product solution must be

Tn(t)=exp

(

−n(n+ 1 )kt/R^2

)

.