276 Chapter 4 The Potential Equation
The Dirichlet problem on a disk can now be stated as
1
r∂
∂r(
r∂v
∂r)
+
1
r^2∂^2 v
∂θ^2 =^0 ,^0 ≤r<c, (1)
v(c,θ)=f(θ ), (2)
v(r,θ+ 2 π)=v(r,θ), 0 <r<c, (3)
v(r,θ)bounded asr→ 0 +. (4)By assumingv(r,θ)=R(r)Q(θ )we can separate variables. The potential equa-
tion becomes
1
r
(
rR′)′
Q+r^12 RQ′′= 0.Separation is effected by dividing through byRQ/r^2 :
r(rR′(r))′
R(r)+Q
′′(θ )
Q(θ )= 0.
As usual, both terms must be constant. We know that ifQ′′/Qis a posi-
tive constant, thenQwill be exponential, not periodic. Therefore we choose
Q′′/Q=−λ^2 and obtain this singular eigenvalue problem:
Q′′+λ^2 Q= 0 , (5)
Q(θ+ 2 π)=Q(θ ). (6)The accompanying equation forR(r)is
r(
rR′)′
−λ^2 R= 0 , (7)
R(r)bounded asr→ 0 +. (8)The general solution of Eq. (5) (ifλ>0) is
Q(θ )=Acos(λθ )+Bsin(λθ ).This function is periodic for allλ, but the period is 2πonly ifλis an inte-
ger. Thus we haveλn=n,n= 1 , 2 ,....Inaddition,ifλ=0, we have a peri-
odic solution that is any constant. Thus, the solution of the singular eigenvalue
problem of Eqs. (5) and (6) is
λ 0 = 0 , Q 0 (θ )= 1 ,
λn=n, Qn(θ )=Acos(nθ)+Bsin(nθ), n= 1 , 2 , 3 ,....The novelty here is that we have two eigenfunctions for each eigenvaluen=
1 , 2 ,....