34.4. LAPLACE EQUATION 687

on the disc of radius R if on the boundary of this disk,

u(R,θ) = f (θ)

where f (0) = f (2π) . This last condition is necessary because θ = 0 and θ = 2π corre-spond to the same point on the boundary of this disk. Note how everything is in terms of thevariables r,θ and that in terms of these variables, the circular disk is actually a rectangle.

Use the method of separation of variables. Look for a solution to the equation which isof the form R(r)Θ(θ) .

r2R′′ (r)Θ(θ)+ rR′ (r)Θ(θ)+R(r)Θ′′ (θ) = 0

So divide by RΘ. This leads to

r2 R′′

R+ r

R′

R+

Θ′′

Θ= 0

HenceΘ′′

Θ=−λ = r2 R′′

R+ r

R′

Rfor some constant λ . First consider Θ. You must have Θ(0) = Θ(2π) . Also

Θ′′+λΘ = 0

Multiply both sides by Θ and integrate. This leads to

Θ′ (θ)Θ(θ) |2π

0 −∫ 2π

0

(Θ′)2 dθ +λ

∫ 2π

0Θ

2dθ = 0

The boundary terms disappear because you must also have Θ′ (2π) = Θ′ (0). Therefore,to have a solution, it is necessary that λ ≥ 0. If λ = 0, you need to have Θ′ = 0 andso Θ(θ) = C a constant. Otherwise, you need λ = µ2,µ > 0. Then the solution to theequation is

C1 cos µθ +C2 sin µθ

and you need to have Θ(0) = Θ(2π). Therefore, it is required that µ2π is an integermultiple of 2π so µ = n for n an integer. Thus the eigenvalues are the nonnegative integersand you get

Θn (θ) = (an cos(nθ)+bn sin(nθ)) , n = 0,1,2, · · ·

It follows that for each of these n,

r2R′′n + rR′n−n2Rn = 0

This is an Euler equation and you look for solutions in the form R(r) = rα . Then tofind α, you insert this into the equation.

r2α (α−1)rα−2 + rαrα−1−n2rα = 0

and so you get the indicial equation

α (α−1)+α−n2 = (α−n)(α +n) = 0