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has the form of a second order Euler equation for the radial component v(r). Its solutions
k
are obtained by the substitution v(r) = r . We discover that this is a solution if and only
if
2
2
k − n = 0, and hence k = ±n.
Therefore, for n 6= 0, we find two linearly independent solutions
n
v 1 (r) = r , v 2 (r) = r −n , n = 1, 2, . . . . (10.46)

If n = 0, there is an additional logarithmic solution
v 1 (r) = 1, v 2 (r) = ln r, n = 0. (10.47)

Combining (10.44) and (10.46), (10.47), we produce a complete list of separable polar
coordinate solutions to the Laplace equation:

n
n
1, r cos nθ, r sin nθ, (10.48)
ln r, r −n cos nθ, r −n sin nθ n = 1, 2, . . . . (10.49)

Now, the solutions (10.48) are continuous (in fact analytic) at the origin, whereas the
solutions (10.49) have singularities as r = 0. The latter are not relevant since we require
the solution u to remain bounded and smooth - even at the center of the disk. Thus, we
should only use the former to form a candidate series solution

∞
X
a 0 n n
u(r, θ) = + (a n r cos nθ + b n r sin nθ) (10.50)
2
n=1
to the Dirichlet boundary value problem. The coefficients a n , b n will be prescribed by the
boundary conditions (10.40). Substituting r = 1, we find
∞
X
a 0 n
u(1, θ) = + (a n r cos nθ + b n sin nθ) = h(θ).
2
n=1
We recognize this as a standard Fourier series for the 2π periodic function h(θ). Therefore,

π π
Z Z
1 1
a n = h(θ) cos nθdθ, b n = h(θ) sin nθdθ (10.51)
π π
−π −π
are precisely its Fourier coefficients.
Example 7.1 Consider the Dirichlet boundary value problem on the unit disk with

u(1, θ) = θ for − π < θ < π.

The boundary data can be interpreted as a wire in the shape of a single turn of a spiral
helix sitting over the unit circle, with a jump discontinuity, of magnitude 2π, at (−1; 0).
Consider the function h(θ) = θ at (−π; π). Its Fourier coefficients

2
a 0 = a n = 0, b n = (−1) n+1 .
n

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