Page 113 - 6099

they are finite is the ”boundary condition” at r = 0. Summing the remaining solutions,
we have
∞
1 X
n
u = A 0 + r (A n cos nθ + B n sin nθ). (10.30)
2
n=1
Finally, we use the inhomogeneous BCs at r = a. Setting r = a in the series above, we
require that
∞
1 X
n
h(θ) = A 0 + a (A n cos nθ + B n sin nθ).
2
n=1
This is precisely the full Fourier series for h(θ), so we know that
Z 2π
1
A n = h(φ) cos nφdφ (10.31)
πa n 0
Z 2π
1
B n = h(φ) sin nφdφ (10.32)
πa n 0
(10.33)

Equations (10.30) to (10.32) constitute the full solution. of our problem.
Now comes an amazing fact. The series (10.30) can be summed explicitly! In fact,
let’s plug (10.31) and (10.32) directly into (10.30) to get
∞
Z 2π dφ X r n Z 2π
u(r, θ) = h(φ) + n h(φ)(cos nφ cos nθ + sin nφ sin nθ)dφ =
0 2π n=1 πa 0
!
2π X dφ
Z ∞
r n
= h(φ) 1 + 2 ( ) cos n(θ − φ) .
a
0 2π
n=1
The term in braces is exactly the series we summed by writing it as a geometric series of
complex numbers; namely,
∞ ∞ i(θ−φ) −i(θ−φ)
X X re re
r n in(θ−φ)
r n −in(θ−φ)
1 + ( ) e + ( ) e = 1 + + =
2 a a − re i(θ−φ) a − re −i(θ−φ)
n=1 n=1
2
a − r 2
= .
2
a − 2at cos(θ − φ) + r 2
Therefore,
2π h(φ) dφ
Z
2
2
u(r, θ) = (a − r ) 2 2 . (10.34)
0 a − 2ar cos(θ − φ) + r 2π
This single formula (10.34), known as Poisson’s formula, replaces the triple of formulas
(10.30)- (10.32). It expresses any harmonic function inside a circle in terms of its boundary
values.
The Poisson formula can be written in a more geometric way as follows. Write x =
(x, y) as a point with polar coordinates (r, θ). We could also think of x as the vector
0
from the origin 0 to the point (x, y). Let x be a point on the boundary.
x : polar coordinates (r, θ)

106