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We find the equation
2
2
α
α
2
2
ρ (α(α − 1) + α − n ) = ρ [α − n ] = 0.
If n =6= 0 we thus have two independent solutions (as should be)
n
R n (ρ) = Cρ −n + Dρ .
The term with the negative power of ρ diverges as ρ goes to zero. This is not accept-
able for a physical quantity (like the temperature). We keep the regular solution,

n
R n (ρ) = ρ .
For n = 0 we find only one solution, but it is not very hard to show (e.g., by substi-
tution) that the general solution is

R 0 (ρ) = C 0 + D 0 ln ρ.

We reject the logarithm since it diverges at ρ = 0. In summary, we have
∞
X
A 0 n
u(ρ, φ) = + ρ (A n cos nφ + B n sin nφ).
2
n=1
The one remaining boundary condition can now be used to determine the coefficients
A n and B n ,

(
∞
X 100, if 0 < φ < π,
A 0 n
U(c, φ) = + c (A n cos nφ + B n sin nφ) =
2 0, if π < φ < 2π.
n=1
We find
Z π
1
A 0 = 100dφ = 1000,
π 0
Z π
1 100 π
n
c A n = 100 cos nφdφ = sin nφ = 0,
π 0 πn 0
(
Z π
1 100 π 200/(πn), if n is odd,
n
c B n = 100 sin nφdφ = − cos nφ =
π 0 πn 0 0, if n is even.
In summary
200 X sin nφ
ρ n
u(ρ, φ) = 50 + ( ) . (10.55)
π c n
n is odd
We clearly see the dependence of u on the pure number r/c, rather than ρ.

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