348 Chapter 5 Higher Dimensions and Other Coordinates
Figure 15 Graphs of the solution of the example problem, withu(φ, 0 )positive
in the north and negative in the south. The functionu(φ,t)is shown as a function
ofφin the range 0 toπfor times chosen so that the dimensionless timekt/R^2
takes the values 0, 0.01, 0.1, and 1; for convenience,T 0 =100.
problem is
1
ρ^2∂
∂ρ(
ρ^2 ∂∂ρu)
+ρ (^2) sin^1 (φ)∂φ∂
(
sin(φ)∂∂φu)
=c^12 ∂(^2) u
∂t^2 ,
0 <ρ<a, 0 <φ<π, 0 <t,
u(a,φ,t)= 0 , 0 <φ<π, 0 <t,
u(ρ, φ, 0 )=f(ρ, φ), 0 <ρ<a, 0 <φ<π,
∂u
∂t
(ρ, φ, 0 )=g(ρ, φ), 0 <ρ<a, 0 <φ<π.
(12)
As usual, we require in addition thatube bounded asρ→0andasφ→ 0
andφ→π.
First, we seek solutions in the product formu(ρ, φ,t)=R(ρ)(φ)T(t).
Insertinguin this form into the partial differential equation (12) and manip-
ulating, we find
1
ρ^2(
(ρ^2 R′)′
R +(sin(φ)′)′
sin(φ))
=T
′′
c^2 T. (13)Both sides of this equation must have the same and constant value, say,−λ^2.
Thus, we must have
(ρ^2 R′)′
R+(sin(φ)′)′
sin(φ)=−λ^2 ρ^2.