Math 411: Honours Complex Variables - University of Alberta

98 CHAPTER 14. HARMONIC FUNCTIONS
Definition. Let r > 0. The Poisson kernel for Br(0) is defined as
for z ∈ Br(0) and ζ ∈ ∂Br(0).
Pr(ζ,z) := r2 −|z| 2
2π|ζ −z| 2
Lemma 14.1. Let D ⊂ C be open, let r > 0 be such that Br[0] ⊂ D, and let
f: D → C be holomorphic. Then we have
for z ∈ Br(0).
f(z) = 1
2π
� 2π
0
f(re iθ ) r2 −|z| 2
|reiθ dθ =
−z| 2
� 2π
f(re
0
iθ )Pr(re iθ ,z)dθ.
Proof. Fix z ∈ Br(0), and define g(w) := f(w)
r2 , which is holomorphic for w in
−wz
Br+ǫ(0) for some ǫ > 0. The Cauchy Integral Formula then yields
f(z)
r2 � � 2π
1 g(ζ) 1 g(re
= g(z) = dζ =
−|z| 2 2πi ζ −z 2πi
iθ )ireiθ reiθ dθ
−z
= 1
2π
so that
∂Br(0)
0
� 2π
f(re
0
iθ )reiθ (r2 −reiθz)(re iθ � 2π
1 f(re
dθ=
−z) 2π 0
iθ )
(re−iθ −z)(reiθ � 2π
1
dθ=
−z) 2π 0
f(z) = 1
� 2π
2π 0
f(re iθ ) r2 −|z| 2
|reiθ dθ.
−z| 2
Remark. If we apply Lemma 14.1 to the function f = 1 we see for all z ∈ Br(0) that
� 2π
0
Pr(re iθ ,z)dθ = 1.
Theorem 14.2 (Poisson’s Integral Formula). Let r > 0, and let u: Br[0] → R be
continuous such that u|Br(0) is harmonic. Then
holds for all z ∈ Br(0).
u(z) =
� 2π
u(re
0
iθ )Pr(re iθ ,z)dθ
Proof. Suppose first that u extends to BR(0) for some R > r as a harmonic function.
Then u has a harmonic conjugate v on BR(0), so that f := u + iv is holomorphic.
By Lemma 14.1, we have, for z ∈ Br(0), that
=
u(z)+iv(z) = f(z)
� 2π
f(re
0
iθ )Pr(re iθ � 2π
,z)dθ =
0
u(re iθ )Pr(re iθ � 2π
,z)dθ+i v(re
0
iθ )Pr(re iθ ,z)dθ,
f(reiθ )
|reiθ dθ,
−z| 2