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648 CHAPTER 18 The Potential Equation

18.4 Poisson’s Integral Formula

We will derive an integral formula due to Poisson for the Dirichlet problem for a disk. Suppose
the disk is centered at the origin and has a radius of 1, and that u(1,θ) = f (θ). By equation
(18.3), the solution with R = 1is
∞
π
1 n

u(r,θ) = 1 + 2 r cos(n(ξ − θ)) f (ξ)dξ.
2π −π n=1
The quantity

∞
1 n
P(r,ζ) = 1 + 2 r cos(nζ)
2π
n=1
is called the Poisson kernel. In terms of this kernel function, the solution is
π
u(r,θ) = P(r,ξ − θ) f (ξ)dξ.
−π
We will evaluate the sum in the Poisson kernel, yielding Poisson’s integral formula for the
solution.
iζ
Let z be a complex number. In polar form, z = re with r < 1 inside the unit disk and ζ an
argument of z. By Euler’s formula,
n
n
n inζ
n
z =r e =r cos(nζ) + ir sin(nζ).
n
This enables us to recognize r cos(nζ), which appears in the Poisson kernel, as the real part of
n
z and write

∞ ∞

n
1 + 2 r cos(nζ) = Re 1 + 2 z n .
n=1 n=1
Now suppose |z|=r < 1. Then this sum is just a geometric series:
∞
z
n
z = .
1 − z
n=1
Combining these observations, we have

∞ ∞
n n z
1 + 2 r cos(nθ) = Re 1 + 2 z = Re 1 + 2
1 − z
n=1 n=1
iζ
1 + z 1 +re
= Re = Re .
1 − z 1 −re iζ
To extract this real part, compute
1 +re iζ 1 +re iζ 1 −re −iζ
=
1 −re iζ 1 −re iζ 1 −re −iζ
iζ
2
1 −r +r(e − e −iζ )
=
iζ
1 +r −r(e + e −iζ )
2
2
1 −r + 2ir sin(ζ)
= .
1 +r − 2r cos(ζ)
2
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October 14, 2010 15:27 THM/NEIL Page-648 27410_18_ch18_p641-666