%%%%		-*- 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:   Maximum Principle and Uniqueness of Solutions
%%%%

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\title{Maximum Principle and Uniqueness of Solutions}
\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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% - Keep it simple, no one is interested in your street address.

\date{ % (optional)
Fall
%Spring
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\the\year}

\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]
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% 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 explore:
\begin{itemize}
\item the Maximum Principle for solutions to the heat equation and its
  justification,
\item the dependence of solutions to the heat equation on the initial
  and boundary conditions, and
\item the uniqueness of solutions to the heat equation and its justification.
\end{itemize}

\end{frame}
%%

\begin{frame}\frametitle{The Maximum Principle}

\begin{theorem}%[The Maximum Principle]
Consider the initial boundary value problem
\begin{align*}
u_{t}	&= k \, u_{xx}	,
\quad 0 < x < L	,	\quad t > 0	\\
u(0,t)	&= a(t)
\quad\mbox{and}\quad
u(L,t) = b(t)	,
\quad t > 0	\\
u(x,0)	&= f(x)	,
\quad 0 \le x \le L
\end{align*}
where the diffusion constant $k > 0$, and the
functions $a(t)$, $b(t)$, and $f(x)$ are $\mathcal{C}^{2}$ (twice
continuously differentiable) on their respective intervals.
Let $T > 0$ be a fixed time and let
\[
A = \max_{0 \le t \le T} \{ a(t) \}	,
\quad
B = \max_{0 \le t \le T} \{ b(t) \}	,
\quad\mbox{and}\quad
F = \max_{0 \le x \le L} \{ f(x) \}	.
\]
If $M = \max \{ A, B, F \}$ and if $u(x,t)$ is any $\mathcal{C}^{2}$
solution of the initial boundary value problem, then $u(x,t) \le M$
for all $0 \le x \le L$ and $0 \le t \le T$.
\end{theorem}

\end{frame}
%%

\begin{frame}\frametitle{Illustration}

\begin{center}
\resizebox{0.80\textwidth}{!}{\includegraphics{maxregion}}
\end{center}

\end{frame}
%%

\begin{frame}\frametitle{Example}

Consider the IBVP:
\begin{align*}
u_{t} &= 9 u_{xx}
\mbox{ for $0 < x < 3$ and $t > 0$}	\\
u(0,t) &= 0 = u(3,t)	\\
u(x,0) &= 6 \sin \frac{\pi x}{3} + 2 \sin \pi x
\end{align*}
Find an upper bound for the solution.

\end{frame}
%%

\begin{frame}\frametitle{Solution (1 of 3)}

\begin{itemize}
\item
According to the Maximum Principle, the maximum occurs either where
$x=0$, $x=3$, or $t = 0$.
\item
Either the upper bound is $u(0,t) = u(3,t) = 0$ or the maximum occurs
where
\begin{align*}
u(x,0) = f(x) &= 6 \sin \frac{\pi x}{3} + 2 \sin \pi x	.	\\
\uncover<2->{%
f'(x) &= 2 \pi (\cos \frac{\pi x}{3} + \cos \pi x)	\\
	&=
2 \pi \left( \cos \left( \frac{2 \pi x}{3} - \frac{\pi x}{3} \right) +
\cos \left( \frac{2 \pi x}{3} + \frac{\pi x}{3} \right) \right)	\\
	&=
4 \pi \cos \frac{2 \pi x}{3} \cos \frac{\pi x}{3}
}
\end{align*}
\end{itemize}

\end{frame}
%%

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

\[
f'(x) = 4 \pi \cos \frac{2 \pi x}{3} \cos \frac{\pi x}{3} = 0
\]
has critical numbers $x = 3/4$, $x = 9/4$ (both maxima) and $x = 3/2$
(minimum) in the
interval $[0,3]$.
\begin{center}
\resizebox{0.80\textwidth}{!}{\includegraphics{example01}}
\end{center}

\end{frame}
%%

\begin{frame}\frametitle{Solution (3 of 3)}

\[
u(x,t) \le f(3/4) = f(9/4) = 4 \sqrt{2}
\]
\begin{center}
\resizebox{0.80\textwidth}{!}{\includegraphics{example01surf}}
\end{center}

\end{frame}
%%

\begin{frame}\frametitle{Minimum Principle}

\begin{corollary}
Consider the initial boundary value problem
\begin{align*}
u_{t}	&= k u_{xx}	,
\quad 0 < x < L	,	\quad t > 0	\\
u(0,t)	&= a(t)
\quad\mbox{and}\quad
u(L,t) = b(t)	,
\quad t > 0	\\
u(x,0)	&= f(x)	,
\quad 0 \le x \le L
\end{align*}
where the diffusion constant $k > 0$, and the
functions $a(t)$, $b(t)$, and $f(x)$ are $\mathcal{C}^{2}$
on their respective intervals.
Let $T > 0$ be a fixed time and let
\[
\alpha = \min_{0 \le t \le T} \{ a(t) \}	,
\quad
\beta = \min_{0 \le t \le T} \{ b(t) \}	,
\quad\mbox{and}\quad
\gamma = \min_{0 \le x \le L} \{ f(x) \}	.
\]
If $\mu = \min \{ \alpha, \beta, \gamma \}$ and if $u(x,t)$ is any
$\mathcal{C}^{2}$ solution of the initial boundary value problem, then
$u(x,t) \ge \mu$ for all $0 \le x \le L$ and $0 \le t \le T$.
\end{corollary}

\end{frame}
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\begin{frame}\frametitle{Continuous Dependence on BC and IC}
\fontsize{10pt}{12}\selectfont

\begin{theorem}
Consider the two initial boundary value problems
\[
\begin{array}{lcr}
\begin{array}{rcl}
u_{t} & = & k u_{xx}	\\
u(0,t) & = & a_{1}(t)	\\
u(L,t) & = & b_{1}(t)	\\
u(x,0) & = & f_{1}(x)
\end{array}
{} & \qquad & {}
\begin{array}{rcl}
v_{t} & = & k v_{xx}	\\
v(0,t) & = & a_{2}(t)	\\
v(L,t) & = & b_{2}(t)	\\
v(x,0) & = & f_{2}(x)
\end{array}
\end{array}
\]
defined for $0 \le x \le L$ and $t \ge 0$.
Let $T > 0$ and suppose there exists $\epsilon \ge 0$ such that
\begin{align*}
\left| f_{1}(x) - f_{2}(x) \right|
	&\le \epsilon \mbox{ for $0 \le x \le L$,} \\
\left| a_{1}(t) - a_{2}(t) \right|
	&\le \epsilon \mbox{ for $0 \le t \le T$, and}	\\
\left| b_{1}(t) - b_{2}(t) \right|
	&\le \epsilon \mbox{ for $0 \le t \le T$.}
\end{align*}
If $u(x,t)$ and $v(x,t)$ are $\mathcal{C}^{2}$ solutions respectively
to the two initial boundary value problems, then for all $0 \le x \le
L$ and $0 \le t \le T$,
\[
\left| u(x,t) - v(x,t) \right| \le \epsilon	.
\]
\end{theorem}

\end{frame}
%%

\begin{frame}\frametitle{Proof}

\begin{itemize}
\item Let $U(x,t) = u(x,t) - v(x,t)$, then
\[
U_{t} = u_{t} - v_{t} = k u_{xx} - k v_{xx} = k U_{xx}	.
\]
\item For $x = 0$, $|U(0,t)| = |a_{1}(t) - a_{2}(t)| \le \epsilon$ by
  assumption and thus
\[
-\epsilon \le U(0,t) \le \epsilon
\mbox{ for $0 \le t \le T$.}
\]
Likewise $-\epsilon \le U(L,t) \le \epsilon$ for $0 \le t \le T$.
\item For $t = 0$, $|U(x,0)| = |f_{1}(x)-f_{2}(x)| \le \epsilon$ by
  assumption and thus
\[
-\epsilon \le U(x,0) \le \epsilon
\mbox{ for $0 \le t \le L$.}
\]
\item Applying the Maximum and the Minimum
Principles yields
$-\epsilon \le U(x,t) \le \epsilon$ or
\[
\left| u(x,t) - v(x,t) \right| \le \epsilon
\]
for $0 \le x \le L$ and $0 \le t \le T$.
\end{itemize}

\end{frame}
%%

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

\begin{corollary}%[Uniqueness]
Consider the initial boundary value problem
\begin{align*}
u_{t}	&= k u_{xx} + g(x,t)	,
\quad 0 < x < L	,	\quad t > 0	\\
u(0,t)	&= a(t)
\quad\mbox{and}\quad
u(L,t) = b(t)	,
\quad 0 < x < L	\\
u(x,0)	&= f(x)	,
\quad 0 \le x \le L
\end{align*}
where the diffusion constant $k > 0$, the function $g(x,t)$ is
continuous, and the
functions $a(t)$, $b(t)$, and $f(x)$ are $\mathcal{C}^{2}$
on their respective intervals.
If there exists a $\mathcal{C}^{2}$ solution to this initial boundary
value problem then it is unique.
\end{corollary}

\end{frame}
%%

\begin{frame}\frametitle{Proof}

\begin{itemize}
\item For the purposes of contradiction,
suppose there are two solutions $u(x,t)$ and $v(x,t)$.
\item Define $U(x,t) = u(x,t) - v(x,t)$, then
\begin{align*}
U_{t} &= u_{t} - v_{t}	\\
	&= k u_{xx} + g(x,t) - (k v_{xx} + g(x,t))	\\
	&= k (u_{xx} - v_{xx})	\\
U_{t} &= k U_{xx}
\mbox{ for $0 < x < L$ and $t > 0$.}
\end{align*}
\item $U(0,t) = u(0,t)-v(0,t) = a(t)-a(t) = 0$ and $U(L,t) = 0$ and
  $U(x,0) = 0$
\item By the Maximum and Minimum Principles $U(x,t) = 0$ for all $0 \le
  x \le L$ and $t \ge 0$ which implies $u(x,t) = v(x,t)$.
\end{itemize}

\end{frame}
%%

\begin{frame}\frametitle{Homework}

\begin{itemize}
\item Read Section $4.3$
\item Exercises: 21, 22, 23
\end{itemize}

\end{frame}
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