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The separated solutions are just the same as for a circle except that we do not
throw out the functions r −n and log r, as these functions are C perfectly finite within
the annulus. So the solution is

∞
1 X
n
n
u(r, θ) = (C 0 + D 0 log r) + (C n r + D n r −n ) cos nθ + (A n r + B n r −n ) sin nθ.
2
n=1
The coefficients are determined by setting r = a and r = b.
Example 10.6 The Exterior of a Circle
The Dirichlet problem for the exterior of a circle is

2
2
2
u xx + u yy = 0 for x + y > a ,
2
2
2
u = h(θ) for x + y = a ,
2
2
u bounded as x + y → ∞.
We follow the same reasoning as in the interior case. But now, instead of finiteness
at the origin, we have imposed boundedness at infinity. Therefore, r +n is excluded
and r −n is retained. So we have
∞
1 X
u(r, θ) = A 0 + r −n (A n cos nθ + B n sin nθ).
2
n=1
The boundary condition means

1 X
h(θ) = A 0 + a −n (A n cos nθ + B n sin nθ),
2

so that Z
a n π
A n = h(θ) cos nθdθ
π −π
and Z
a n π
B n = h(θ) sin nθdθ.
π −π

This is the complete solution but it is one of the rare cases when the series can
actually be summed. Comparing it with the interior case, we see that the only
difference between the two sets of formulas is that r and a are replaced by r −1 and
−1
a . Therefore, we get Poisson’s formula with only this alteration. The result can be
written as
Z 2π h(φ) dφ
2
2
u(r, θ) = (r − a ) 2 2
0 a − 2ar cos(θ − φ) + r 2π
for r > a.

These three examples illustrate the technique of separating variables in polar coordinates.

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