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It is known that Laplace equation in polar coordinates has the following view:
1 1
u rr + u r + u θθ = 0. (10.39)
r r 2

The boundary conditions are imposed on the unit circle r = 1, and so, by (10.38), take
the form
u(1, θ) = h(θ). (10.40)

Keep in mind that, in order to be single-valued functions of x, y, the solution u(r, θ) and
its boundary values h(θ) must both be 2π periodic functions of the angular coordinate:

u(r, θ + 2π) = u(r, θ), h(θ + 2π) = h(θ). (10.41)

Polar separation of variables is based on the following product

u(r, θ) = v(r)w(θ), (10.42)

that assumes that the solution u(r, θ) is a product of functions of the individual polar
variables. Substituting (10.42) into the polar form (10.39) of Laplace’s equation, we find
1 1
00
00
0
v (r)w(θ) + v (r)w(θ) + v(r)w (θ) = 0.
r r 2
We now separate variables by moving all the terms involving r onto one side of the equation
and all the terms involving θ onto the other. This is accomplished by first multiplying the
2
equation by r /v(r)w(θ), and then moving the last term to the right side:
00
2 00
0
r v (r) + rv (r) w (θ)
= − = λ.
v(r) w(θ)
As in the rectangular case, a function of r can equal a function of θ if and only if both are
equal to a common separation constant, which we call λ. The partial differential equation
thus splits into a pair of ordinary differential equations

2 00
00
0
r v (r) + rv − λv = 0, w + λw = 0, (10.43)
that will prescribe the separable solution (10.42). Observe that both have the form of
eigenfunction equations in which the separation constant λ plays the role of the eigenvalue,
and we are only interested in nonzero solutions or eigenfunctions.
We have already solved the eigenvalue problem for w(θ) in previous classes . According
to (10.41), w(θ + 2π) = w(θ) must be a 2π periodic function. Therefore, our periodic
boundary value problem has the nonzero eigenfunctions

1, sin nθ, cos θ for n = 1, 2, . . . (10.44)

2
corresponding to the eigenvalues (separation constants) λ = n , where n = 0, 1, 2, . . . .
Fixing the value of λ, the remaining ordinary differential equation

2 00
0
2
r v (r) + rv − n v = 0 (10.45)
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