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Last edited by hbghlyj 2022-11-10 21:04Schwarz and Poisson formulas - planetmath
Schwarz integral formula - Wikipedia
Deriving the Poisson Integral Formula from the Cauchy Integral Formula
complex.pdf(page 83 bottom, 84 top)
Lemma 12.20. If $u$ is harmonic on $B(0, r)$ for $r>1$ then for all $w \in B(0,1)$ we have
\[
u(w)=\frac{1}{2 \pi} \int_0^{2 \pi}u\left(e^{i \theta}\right) \frac{1-|w|^2}{\left|e^{i \theta}-w\right|^2} d \theta=\frac{1}{2 \pi} \int_0^{2 \pi} u\left(e^{i \theta}\right) \Re\left(\frac{e^{i \theta}+w}{e^{i \theta}-w}\right) d \theta .
\]
Proof. Take, as before, $f(z)$ holomorphic with $\Re(f)=u$ on $B(0, r)$. Then letting $\gamma$ be a parametrization of the positively oriented unit circle we have
\[
f(w)=\frac{1}{2 \pi i} \int_\gamma \frac{f(z) d z}{z-w}-\frac{1}{2 \pi i} \int_\gamma \frac{f(z) d z}{z-\bar{w}^{-1}}
\]
where the first term is $f(w)$ by the integral formula and the second term is zero because $f(z) /\left(z-\bar{w}^{-1}\right)$ is holomorphic inside all of $B(0,1)$. Gathering the terms, this becomes
\[
f(w)=\frac{1}{2 \pi} \int_\gamma f(z) \frac{1-|w|^2}{|z-w|^2} \frac{d z}{i z}=\frac{1}{2 \pi} \int_0^{2 \pi} f\left(e^{i \theta}\right) \frac{1-|w|^2}{\left|e^{i \theta}-w\right|^2} d \theta .
\]
The advantage of this last form is that the real and imaginary parts are now easy to extract, and we see that
\[
u(w)=\int_0^{2 \pi} u\left(e^{i \theta}\right) \frac{1-|w|^2}{\left|e^{i \theta}-w\right|^2} d \theta .
\]
Finally for the second integral expression note that if $|z|=1$ then
\[
\frac{z+w}{z-w}=\frac{(z+w)(\bar{z}-\bar{w})}{|z-w|^2}=\frac{1-|w|^2+(\bar{z} w-z \bar{w})}{|z-w|^2} .
\]
from which one readily sees the real part agrees with the corresponding factor in our first expression. |
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