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Problem of the Week #70 - September 30th, 2013

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Chris L T521

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Jan 26, 2012
995
Here's this week's problem.

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Problem: Let $u$ be a harmonic function in the unit disk that is continuous on its closure. Deduce Poisson's integral formula
\[u(z_0)=\frac{1}{2\pi}\int_0^{2\pi}\frac{1-|z_0|^2}{|e^{i\theta}-z_0|^2}u(e^{i\theta})\,d\theta\quad\text{for }|z_0|<1\]
from the special case $z_0=0$ (the mean value theorem). Show that if $z_0=re^{i\varphi}$, then
\[\frac{1-|z_0|^2}{|e^{i\theta}-z_0|^2}=\frac{1-r^2}{1-2r\cos(\theta-\varphi)+r^2}=P_r(\theta-\varphi)\]
(which is know as the Poisson kernel.)

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Hint:
Set $u_0(z)=u(T(z))$ where
\[T(z)=\frac{z-z_0}{1-\overline{z_0}z}.\]
Prove that $u_0$ is harmonic. Then apply the mean value theorem to $u_0$, and make a change of variables in the integral.


Remember to read the POTW submission guidelines to find out how to submit your answers!
 
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Chris L T521

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Jan 26, 2012
995
No one answered this week's question. You can find my solution below.

If $u(z)$ is harmonic in the closed disc $|z|\leq 1$, then\[z=S(T(z))=\frac{T(z)+z_0}{1+\overline{z_0}T(z)}\]
maps $|T(z)|\leq 1$ onto $|z|\leq 1$ with $T(z)=0$ corresponding to $z=z_0$. Thus, $u(S(T(z)))$ is harmonic in $|T(z)|\leq 1$ and by the mean value property we have
\[u(z_0) = \frac{1}{2\pi}\int_{|T(z)|= 1}u(S(T(z)))\,d\arg\,T(z)\] where $\theta=\arg T(z)$. Now,
\[z=\frac{T(z)+z_0}{1+\overline{z_0}T(z)}\implies T(z)=\frac{z-z_0}{1-\overline{z_0}z}.\]
It follows that
\[\begin{aligned}\,d\arg\,T(z) &= -i\frac{d T(z)}{T(z)}\\ &= -i\frac{(1-\overline{z_0}z)+\overline{z_0}(z-z_0)}{(1-\overline{z_0}z)^2}\cdot\frac{1-\overline{z_0}z}{z-z_0}\,dz\\ &=-i\left(\frac{1}{z-z_0}+\frac{\overline{z_0}}{1-\overline{z_0}z}\right)\,dz\\ &= \left(\frac{z}{z-z_0}+\frac{\overline{z_0}z}{1-\overline{z_0}z}\right)\cdot\left(-i\frac{\,dz}{z}\right)\\ &= \left(\frac{z}{z-z_0}+\frac{\overline{z_0}z}{1-\overline{z_0}z}\right)\,d\theta.\end{aligned}\]
If $z\overline{z}=1$, then
\[\begin{aligned}\frac{z}{z-z_0}+\frac{\overline{z_0}z}{1-\overline{z_0}z} &= \frac{z}{z-z_0}+\frac{\overline{z_0}z}{z\overline{z}-\overline{z_0}z}\\ &=\frac{z}{z-z_0}+\frac{\overline{z_0}}{\overline{z}-\overline{z_0}}\\ &= \frac{z(\overline{z}-\overline{z_0})+\overline{z_0}(z-z_0)}{|z-z_0|^2}\\ &= \frac{1-|z_0|^2}{|z-z_0|^2}\end{aligned}\]
Therefore,
\[\begin{aligned}u(z_0) &= \frac{1}{2\pi}\int_{|T(z)|=1}u(S(T(z)))\,d\arg T(z)\\ &= \frac{1}{2\pi}\int_{|T(z)|=1}\left(\frac{z}{z-z_0}+\frac{\overline{z_0}z}{1-\overline{z_0}z}\right)u(z)\,d\theta\\ &= \frac{1}{2\pi}\int_{|T(z)|=1}\frac{1-|z_0|^2}{|z-z_0|^2}u(z)\,d\theta.\end{aligned}\]
Letting $z=e^{i\theta}$, we get Poisson's integration formula
\[u(z_0)=\frac{1}{2\pi}\int_0^{2\pi}\frac{1-|z_0|^2}{|e^{i\theta}-z_0|^2}u(e^{i\theta})\,d\theta.\]
If $z_0=re^{i\varphi}$, then it follows that
\[\begin{aligned}\frac{1-|z_0|^2}{|e^{i\theta}-z_0|^2} &= \frac{1-r^2|e^{i\varphi}|^2}{|e^{i\theta}-e^{i\varphi}|^2}\\ &=\frac{1-r^2}{|(\cos\theta-r\cos\varphi)-i(\sin\theta-r\sin\varphi)|^2}\\ &= \frac{1-r^2}{(\cos\theta-r\cos\varphi)^2+(\sin\theta-r\sin\varphi)^2}\\ &= \frac{1-r^2}{(\cos^2\theta+\sin^2\theta)-2r(\sin\theta\sin\varphi+\cos\theta\cos\varphi)+r^2(\sin^2\varphi+\cos^2\varphi)}\\ &=\frac{1-r^2}{r^2-2r\cos(\theta-\varphi)+1}\\ &= P_r(\theta-\varphi). \hspace{.5in}\blacksquare\end{aligned}\]
 
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