%%%%		-*- 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:      11/01/2022
%%%%
%%%% Purpose:   Energy Integral and Uniqueness of Solutions
%%%%

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\title{Energy Integral and Uniqueness of Solutions}
\subtitle{MATH 467 \textit{Partial Differential Equations}} % (optional)


\author % (optional, use only with lots of authors)
{J~Robert~Buchanan}
% - Use the \inst{?} command only if the authors have different
%   affiliation.

\institute[Millersville University of Pennsylvania]
% (optional, but mostly needed)
{
  Department of Mathematics
}
% - Use the \inst command only if there are several affiliations.
% - Keep it simple, no one is interested in your street address.

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

\subject{Talks}
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% out. 



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% Delete this, if you do not want the table of contents to pop up at
% the beginning of each subsection:
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{
  \begin{frame}<beamer>
    \frametitle{Outline}
    \tableofcontents[currentsection,currentsubsection]
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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:
\begin{itemize}
\item how to express the total energy in a vibrating string as an
  integral,
\item how to use the energy integral to establish the uniqueness of
  solutions to the wave equation.
\end{itemize}

\end{frame}
%%

\begin{frame}\frametitle{Initial Boundary Value Problem}

Consider the initial boundary value problem for a string of length
$L>0$.
\begin{align*}
u_{tt} &= c^{2} u_{xx}
\mbox{ for $0 < x < L$ and $t > 0$}	\\
u(0,t) &= u(L, t) = 0	\\
u(x,0) &= f(x)	\\
u_{t}(x,0) &= g(x)
\end{align*}

During the derivation of the wave equation we set $c^{2} = T_{0}/\rho$
where $\rho$ is the mass density of the string and $T_{0}$ is the
tension in the string.

\end{frame}
%%

\begin{frame}\frametitle{Kinetic Energy}

\textbf{Kinetic energy} for a point mass is defined as $m v^{2}/2$
where $v$ is velocity.
For the distributed mass of the string, kinetic energy is
\[
K(t) = \frac{1}{2} \int_{0}^{L} \rho (u_{t}(x,t))^{2} \, dx	.
\]

\end{frame}
%%

\begin{frame}\frametitle{Potential Energy}
\fontsize{10pt}{12}\selectfont

\begin{itemize}
\item While at rest, a portion of the string of length $\Delta x$ is
  under the tension force $T_{0}$.
\item When the segment of the string is displaced by $u(x, t)$, its
  length is approximately $ds = \sqrt{1 + (u_{x}(x, t))^{2}} \Delta
  x$.
\item Assuming the displacement is small, then
\[
ds
	= \sqrt{1 + (u_{x}(x, t))^{2}} \Delta x
	\approx \left( 1 + \frac{1}{2} (u_{x}(x, t))^{2} \right) \Delta x	.
\]
\item The amount by which the length of the segment of the string is
  changed is
\[
\left( 1 + \frac{1}{2} (u_{x}(x, t))^{2} \right) \Delta x - \Delta x
	=
\frac{1}{2} (u_{x}(x, t))^{2} \Delta x	.
\]
\item The \textbf{potential energy} in the string due to displacement $u(x,
  t)$ is the work done against force $T_{0}$ to stretch the string
\[
P(t) = \frac{1}{2} \int_{0}^{L} T_{0} (u_{x}(x,t))^{2} \, dx	.
\]
\end{itemize}

\end{frame}
%%

\begin{frame}\frametitle{Total Energy}

The \textbf{total energy} is the sum of the kinetic and potential
energies.
\[
E(t) = \frac{1}{2} \int_{0}^{L} \left[ \rho (u_{t}(x,t))^{2} + T_{0}
  (u_{x}(x,t))^{2} \right] \, dx
\]

\end{frame}
%%

\begin{frame}\frametitle{Conservation of Energy (1 of 2)}

Differentiate the total energy with respect to time.
\begin{align*}
E'(t)
	&= \frac{1}{2} \int_{0}^{L}
\left[ 2 \rho u_{t}(x,t) u_{tt}(x,t) + 2 T_{0} u_{x}(x,t) u_{xt}(x,t) \right] \,
              dx	\\
	&= \int_{0}^{L}
\left[ c^{2} \rho u_{t}(x,t) u_{xx}(x,t) + T_{0} u_{x}(x,t) u_{xt}(x,t) \right] \,
              dx	\\
	&= T_{0} \int_{0}^{L}
\left[ u_{t}(x,t) u_{xx}(x,t) + u_{x}(x,t) u_{xt}(x,t) \right] \,
              dx	\\
	&= T_{0} \int_{0}^{L}
\frac{d}{dx} \left[ u_{t}(x,t) u_{x}(x,t) \right] \,
              dx	\\
	&= T_{0} \left[ u_{t}(x,t) u_{x}(x,t) \right]_{x=0}^{x=L}	\\
	&= T_{0} \left[ u_{t}(L,t) u_{x}(L,t) - u_{t}(0,t) u_{x}(0,t) \right]
\end{align*}

\end{frame}
%%

\begin{frame}\frametitle{Conservation of Energy (2 of 2)}

\[
E'(t) = T_{0} \left[ u_{t}(L,t) u_{x}(L,t) - u_{t}(0,t) u_{x}(0,t)
\right]
\]
Since $u(0,t) = u(L,t) = 0$ for all $t \ge 0$ then $u_{t}(0,t) =
u_{t}(L,t) = 0$.
Consequently $E'(t) = 0$ and total energy is conserved.

\vspace*{0.10in}

\uncover<2->{%
The total energy for any $t \ge 0$ is the initial energy present in
the string.
\begin{align*}
E(t) &= E(0) =
\frac{1}{2} \int_{0}^{L} \left[ \rho (u_{t}(x,0))^{2} + T_{0}
  (u_{x}(x,0))^{2} \right] \, dx	\\
	&=
\frac{1}{2} \int_{0}^{L} \left[ \rho (g(x))^{2} + T_{0}
  (f'(x))^{2} \right] \, dx
\end{align*}
}

\end{frame}
%%

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

\begin{theorem}
The solution to the initial boundary value problem:
\begin{align*}
u_{tt} &= c^{2} u_{xx}
\mbox{ for $0 < x < L$ and $t > 0$}	\\
u(0,t) &= u(L, t) = 0	\\
u(x,0) &= f(x)	\\
u_{t}(x,0) &= g(x)
\end{align*}
is unique.
\end{theorem}

\end{frame}
%%

\begin{frame}\frametitle{Proof}

\begin{itemize}
\item Suppose there exist two solutions $u_{1}(x,t)$ and $u_{2}(x,t)$
  and define $v(x,t) = u_{1}(x,t) - u_{2}(x,t)$.
\item Function $v(x,t)$ solves the following initial boundary value
  problem.
\begin{align*}
v_{tt} &= c^{2} v_{xx}
\mbox{ for $0 < x < L$ and $t > 0$}	\\
v(0,t) &= v(L, t) = 0	\\
v(x,0) &= 0	\\
v_{t}(x,0) &= 0
\end{align*}
\item The total energy of solution $v(x,t)$ is 0 which implies
  $v_{t}(x,t) = 0$ and $v_{x}(x,t) = 0$ which implies $v(x,t)$ is
  constant and therefore $v(x,t) = 0$.
\end{itemize}

\end{frame}
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\begin{frame}\frametitle{Homework}

\begin{itemize}
\item Read Section $5.5$
\item Exercises: 24--26
\end{itemize}

\end{frame}
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