%%%%		-*- Mode: LaTeX -*-
%%%% Author:    J. Robert Buchanan
%%%% Address:   Department of Mathematics
%%%%		Millersville University
%%%%            P.O. Box 1002
%%%%            Millersville, PA 17551-0302
%%%% Phone:     717-871-7305
%%%% FAX:	717-871-7948
%%%% Email:     Robert.Buchanan@millersville.edu
%%%%
%%%% Date:      06/30/2022
%%%%
%%%% Purpose:   Poisson's Formula
%%%%

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\title{Poisson's Formula}
\subtitle{MATH 467 \textit{Partial Differential Equations}} % (optional)


\author % (optional, use only with lots of authors)
{J~Robert~Buchanan}
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\institute[Millersville University of Pennsylvania]
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{
  Department of Mathematics
}
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\subject{Talks}
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\begin{document}

\begin{frame}
  \titlepage
\end{frame}

%\begin{frame}
%  \frametitle{Outline}
%  \tableofcontents
  % You might wish to add the option [pausesections]
%\end{frame}


% Since this a solution template for a generic talk, very little can
% be said about how it should be structured. However, the talk length
% of between 15min and 45min and the theme suggest that you stick to
% the following rules:  

% - Exactly two or three sections (other than the summary).
% - At *most* three subsections per section.
% - Talk about 30s to 2min per frame. So there should be between about
%   15 and 30 frames, all told.

%\section{Introduction}

\begin{frame}\frametitle{Objectives}

In this lesson we will learn about
\begin{itemize}
\item a closed form solution to Laplace's equation on a disk,
\item properties of harmonic functions.
\end{itemize}

\end{frame}
%%

\begin{frame}\frametitle{Background}

Consider Laplace's equation (in polar coordinates) on a disk:
\begin{align*}
u_{rr} + \frac{1}{r} u_{r} + \frac{1}{r^{2}} u_{\theta\theta} &= 0
\quad\mbox{for $0 < r < a$ and $-\infty < \theta < \infty$}	\\
u(a,\theta) &= f(\theta)
\end{align*}
The solution can be expressed as
\[
u(r,\theta) = \frac{\alpha_{0}}{2} + \sum_{n=1}^{\infty} \left(
  \frac{r}{a} \right)^{n} \left[ \alpha_{n} \cos (n \theta) +
  \beta_{n} \sin (n \theta) \right]
\]
where
\begin{align*}
\alpha_{0} &= \frac{1}{\pi} \int_{-\pi}^{\pi} f(\theta) \, d\theta
  \\
\alpha_{n} &= \frac{1}{\pi} \int_{-\pi}^{\pi} f(\theta) \cos (n
                  \theta) \, d\theta	\\
\beta_{n} &= \frac{1}{\pi} \int_{-\pi}^{\pi} f(\theta) \sin (n
                  \theta) \, d\theta	.
\end{align*}

\end{frame}
%%

\begin{frame}\frametitle{Closed Form Solution (1 of 3)}
\fontsize{10pt}{12}\selectfont

We can express the Fourier series solution as
\begin{align*}
u(r,\theta)
	= {} & \frac{1}{2 \pi} \int_{-\pi}^{\pi} f(t) \, dt +
\sum_{n=1}^{\infty} \left( \frac{r}{a} \right)^{n} \left[
\frac{\cos (n \theta)}{\pi} \int_{-\pi}^{\pi} f(t) \cos (n t) \, dt
              \right]	\\
	& + \sum_{n=1}^{\infty} \left( \frac{r}{a} \right)^{n} \left[
\frac{\sin (n \theta)}{\pi} \int_{-\pi}^{\pi} f(t) \sin (n t) \, dt
\right]	\\
	= {} & \frac{1}{2 \pi} \int_{-\pi}^{\pi} f(t) \, dt	\\
	& + \sum_{n=1}^{\infty} \left( \frac{r}{a} \right)^{n} \left[
\frac{1}{2 \pi} \int_{-\pi}^{\pi} f(t) [ \cos (n (\theta +t)) + \cos (n
              (\theta -t)) ] \, dt \right]	\\
	& + \sum_{n=1}^{\infty} \left( \frac{r}{a} \right)^{n} \left[
\frac{1}{2 \pi} \int_{-\pi}^{\pi} f(t) [ \cos (n (\theta - t)) - \cos
              (n (\theta + t)) ] \, dt
\right]
\end{align*}

\end{frame}
%%

\begin{frame}\frametitle{Closed Form Solution (2 of 3)}

\begin{align*}
u(r,\theta)
	&=
\frac{1}{2 \pi} \int_{-\pi}^{\pi} f(t) \, dt +
\sum_{n=1}^{\infty} \left( \frac{r}{a} \right)^{n}
\frac{1}{\pi} \int_{-\pi}^{\pi} f(t) \cos (n (\theta - t)) \, dt	\\
	&=
\frac{1}{\pi} \int_{-\pi}^{\pi} \frac{1}{2} f(t) \, dt +
\frac{1}{\pi} \int_{-\pi}^{\pi} \sum_{n=1}^{\infty} \left( \frac{r}{a} \right)^{n}
f(t) \cos (n (\theta - t)) \, dt	\\
	&=
\frac{1}{\pi} \int_{-\pi}^{\pi} f(t) \left[ \frac{1}{2} +
\sum_{n=1}^{\infty} \left( \frac{r}{a} \right)^{n}
\cos (n (\theta - t)) \right] \, dt 	\\
	&=
\frac{1}{\pi} \int_{-\pi}^{\pi} f(t) \left[ \frac{1}{2} +
\sum_{n=1}^{\infty} \left( \frac{r}{a} \right)^{n}
\Re{e^{i n(\theta-t)}} \right] \, dt 	\\
	&=
\frac{1}{\pi} \int_{-\pi}^{\pi} f(t) \left[ \frac{1}{2} +
\Re{\sum_{n=1}^{\infty} \left( \frac{r e^{i (\theta - t)}}{a}
              \right)^{n}} \right] \, dt
\end{align*}

\end{frame}
%%

\begin{frame}\frametitle{Closed Form Solution (3 of 3)}
\fontsize{10pt}{12}\selectfont

\begin{align*}
u(r,\theta)
	&=
\frac{1}{\pi} \int_{-\pi}^{\pi} f(t) \left[ -\frac{1}{2} +
\Re{\sum_{n=0}^{\infty} \left( \frac{r e^{i (\theta - t)}}{a}
              \right)^{n}} \right] \, dt	\\
	&=
\frac{1}{\pi} \int_{-\pi}^{\pi} f(t) \left[ -\frac{1}{2} +
              \Re{\frac{1}{1 - \frac{r e^{i (\theta - t)}}{a}}} \right] \, dt	\\
	&=
\frac{1}{\pi} \int_{-\pi}^{\pi} f(t) \left[ -\frac{1}{2} +
              \Re{\frac{1}{1 - \frac{r}{a} \cos (\theta - t) - i
              \frac{r}{a} \sin (\theta - t)}} \right] \, dt	\\
	&=
\frac{1}{\pi} \int_{-\pi}^{\pi} f(t) \left[ -\frac{1}{2} +
\frac{1 - \frac{r}{a} \cos (\theta - t)}{1 + \frac{r^{2}}{a^{2}} -
              \frac{2 r}{a} \cos (\theta - t)} \right] \, dt	\\
u(r,\theta)	&=
\frac{1}{\pi} \int_{-\pi}^{\pi} \frac{f(t) \left( \frac{1}{2} -
              \frac{r^{2}}{2 a^{2}} \right)}{1 + \frac{r^{2}}{a^{2}} -
              \frac{2 r}{a} \cos (\theta - t)} \, dt
\end{align*}
This is known as \textbf{Poisson's integral formula}.

\end{frame}
%%

\begin{frame}\frametitle{Main Result}

\begin{theorem}
If $u(r,\theta)$ is a solution to Laplace's equation in polar
coordinates on the disk:
\begin{align*}
\Delta u &= 0
\quad\mbox{for $0 < r < a$ and $-\infty < \theta < \infty$}	\\
u(a,\theta) &= f(\theta)
\end{align*}
then
\[
u(r,\theta) = \frac{1}{2\pi} \int_{-\pi}^{\pi} \frac{f(t) (a^{2} -
  r^{2})}{a^{2} - 2 a r \cos (t - \theta) + r^{2}} \, dt	.
\]
\end{theorem}

\end{frame}
%%

\begin{frame}\frametitle{Mean Value at the Origin}

\begin{corollary}
If $u(r,\theta)$ is a solution to Laplace's equation in polar
coordinates on the disk:
\begin{align*}
\Delta u &= 0
\quad\mbox{for $0 < r < a$ and $-\infty < \theta < \infty$}	\\
u(a,\theta) &= f(\theta)
\end{align*}
then
\[
u(0,\theta) = \frac{1}{2\pi} \int_{-\pi}^{\pi} f(t) \, dt	.
\]
\end{corollary}

\end{frame}
%%

\begin{frame}\frametitle{Proof}

\begin{align*}
u(r,\theta)
	&=
\frac{1}{2\pi} \int_{-\pi}^{\pi} \frac{f(t) (a^{2} -
  r^{2})}{a^{2} - 2 a r \cos (t - \theta) + r^{2}} \, dt	\\
u(0,\theta)
	&=
\frac{1}{2\pi} \int_{-\pi}^{\pi} f(t) \, dt
\end{align*}

\end{frame}
%%

\begin{frame}\frametitle{Mean Value Result}

\begin{theorem}
Let $u(x, y)$ be a harmonic function on $\Omega$, an
open, connected subset of $\mathbb{R}^{2}$, let $(x_{0}, y_{0}) \in
\Omega$, and let $a > 0$ be chosen so that
\[
D_{a}(x_{0}, y_{0}) =
\{ (x,y) \, | \, (x-x_{0})^{2} + (y-y_{0})^{2} < a^{2} \}
\subset \Omega
\]
and
\[
C_{a}(x_{0}, y_{0}) =
\{ (x,y) \, | \, (x-x_{0})^{2} + (y-y_{0})^{2} = a^{2} \}
\subset \Omega	,
\]
then
\[
u(x_{0}, y_{0}) =
\frac{1}{2 \pi} \int_{-\pi}^{\pi}
u(x_{0} + a \cos \theta, y_{0} + a \sin \theta) \, d\theta	.
\]
That is $u(x_{0}, y_{0})$ is the mean value of $u$ on $C_{a}(x_{0}, y_{0})$, the
boundary of $D_{a}(x_{0}, y_{0})$.
\end{theorem}

\end{frame}
%%

\begin{frame}\frametitle{Proof (1 of 2)}

\begin{itemize}
\item Let $\xi = x - x_{0}$ and $\eta = y - y_{0}$, then $D_{a}(x=x_{0},
y=y_{0}) = D_{a}(\xi=0, \eta=0)$.
\item Define the function $U(\xi, \eta) = u(x_{0} + \xi, y_{0} +
  \eta)$.
\item Since $u$ is harmonic on $\Omega$ then by the
chain rule
\[
U_{\xi\xi} + U_{\eta\eta} = 0	.
\]
\end{itemize}

\end{frame}
%%

\begin{frame}\frametitle{Proof (2 of 2)}

\begin{itemize}
\item Using Poisson's formula, then for $(\xi, \eta) \in D_{a}(0, 0)$
\begin{align*}
U(\xi, \eta)
	&=
\frac{1}{2 \pi} \int_{-\pi}^{\pi}
\frac{(a^{2} - r^{2}) U(a \cos t, a \sin t)}{a^{2} - 2 a r \cos (t-\theta) + r^{2}} \, dt	\\
	&=
\frac{1}{2 \pi} \int_{-\pi}^{\pi}
\frac{(a^{2}-r^{2}) u(x_{0} + a \cos t, y_{0} + a \sin t)}{a^{2} - 2 a
              r \cos(t-\theta) + r^{2}} \, dt
\end{align*}
\item If $(\xi,\eta) = (0,0)$
\[
U(0, 0) = \frac{1}{2 \pi} \int_{-\pi}^{\pi}
u(x_{0} + a \cos t, y_{0} + a \sin t) \, dt
\]
which is equivalent to
\[
u(x_{0}, y_{0}) = \frac{1}{2 \pi} \int_{-\pi}^{\pi}
u(x_{0} + a \cos t, y_{0} + a \sin t) \, dt	.
\]
\end{itemize}

\end{frame}
%%

\begin{frame}\frametitle{Maximum/Minimum Principle}

\begin{theorem}
Let $\Omega$ be an open, connected, and bounded domain in
$\mathbb{R}^{2}$ with boundary $\partial\Omega$.
Assume that $u(x, y) \in \mathcal{C}^{2}(\Omega) \cap
\mathcal{C}(\overline{\Omega})$ where $\overline{\Omega} =
\Omega \cup \partial\Omega$, then
\begin{itemize}
\item if $u_{xx} + u_{yy} \ge 0$ for all $(x, y) \in \Omega$, then
\[
\max_{(x, y) \in \overline{\Omega}} u(x, y) =
\max_{(x, y) \in \partial\Omega} u(x, y)	,
\]
\item if $u_{xx} + u_{yy} \le 0$ for all $(x, y) \in\Omega$, then
\[
\min_{(x, y) \in \overline{\Omega}} u(x, y) =
\min_{(x, y) \in \partial\Omega} u(x, y)	.
\]
\end{itemize}
\end{theorem}

\end{frame}
%%

\begin{frame}\frametitle{Maximum/Minimum Principle for Laplace's Equation}

\begin{corollary}
Let $\Omega$ be an open, connected, and bounded domain in $\mathbb{R}^{2}$.
Assume that $u(x,y) \in \mathcal{C}^{2}(\Omega) \cap
\mathcal{C}(\overline{\Omega})$ where $\overline{\Omega} = \Omega \cup
\partial\Omega$ and $u(x,y)$ satisfies Laplace's equation on $\Omega$, then
\[
\max_{(x, y) \in \overline{\Omega}} u(x, y) =
\max_{(x, y) \in \partial\Omega} u(x, y)
\]
and
\[
\min_{(x, y) \in \overline{\Omega}} u(x, y) =
\min_{(x, y) \in \partial\Omega} u(x, y)	.
\]
\end{corollary}

\end{frame}
%%

\begin{frame}\frametitle{Uniqueness of Solutions}

\begin{theorem}[Uniqueness]
Suppose that $\Omega$ is an open, connected, and bounded domain in
$\mathbb{R}^{2}$ with $f(x,y) \in \mathcal{C}(\Omega)$ and $\phi(x,y)
\in \mathcal{C}(\partial \Omega)$.
If $u(x,y) \in \mathcal{C}^{2}(\Omega) \cap
\mathcal{C}(\overline{\Omega})$ where $\overline{\Omega} = \Omega \cup
\partial\Omega$ is a solution to the boundary value problem,
\begin{align*}
\triangle u &= f(x,y)
\mbox{ for $(x, y) \in \Omega$}	\\
u(x, y)	&= \phi(x,y)
\mbox{ for $(x, y) \in \partial\Omega$,}
\end{align*}
then this solution is unique.
\end{theorem}

\end{frame}
%%

\begin{frame}\frametitle{Homework}

\begin{itemize}
\item Read Section $6.8$--$6.9$
\item Exercises: 28
\end{itemize}

\end{frame}
%%

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