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The other boundary condition involves both R and Φ.
As usual we consider the cases λ > 0, λ < 0 and λ = 0 separately. Consider
the Φ equation first, since this has the most restrictive explicit boundary conditions
(10.54).
2
λ = −α < 0 We have to solve

00
2
Φ = α Φ,
which has as a solution
Φ(φ) = A cos αφ + B sin αφ.

Applying the boundary conditions, we get

A = A cos(2απ) + B sin(2απ),

Bα = −Aα sin(2απ) + Bα cos(2απ).

If we eliminate one of the coefficients from the equation, we get

2
A = A cos(2απ) − A sin(2απ) /(1 − cos(2απ))
which leads to
2
2
sin(2απ) = −(1 − cos(2απ)) ,
which in turn shows
2 cos(2απ) = 2,

and thus we only have a non-zero solution for α = n, an integer. We have found

2
λ n = n , Φ n (φ) = A n cos nφ + B n sin nφ.
λ = 0 We have
00
Φ = 0.
This implies that
Φ = Aφ + B.
The boundary conditions are satisfied for A = 0,

Φ 0 (φ) = B n .

λ > 0 The solution (hyperbolic sines and cosines) cannot satisfy the boundary con-
ditions.
Now let me look at the solution of the R equation for each of the two cases (they
can be treated as one),

0
00
2
2
ρ R (ρ) + ρR (ρ) − n R(ρ) = 0.
Let us attempt a power-series solution
α
R(ρ) = ρ .

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