%%%%		-*- 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:      04/12/2016
%%%%
%%%% Purpose:   D'Alembert's Solution to the Wave Equation
%%%%

\documentclass[10pt]{beamer}

\mode<presentation>
{
  %\usetheme{Warsaw}
  \usetheme{default}
  %\usetheme{CambridgeUS}
  % or ...

  %\setbeamercovered{transparent}
  % or whatever (possibly just delete it)
}


\usepackage[english]{babel}
% or whatever

\usepackage[latin1]{inputenc}
% or whatever

\usepackage{times}
\usepackage[T1]{fontenc}
% Or whatever. Note that the encoding and the font should match. If T1
% does not look nice, try deleting the line with the fontenc.


\title{D'Alembert's Solution to the Wave Equation}
\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}
% This is only inserted into the PDF information catalog. Can be left
% out. 



% If you have a file called "university-logo-filename.xxx", where xxx
% is a graphic format that can be processed by latex or pdflatex,
% resp., then you can add a logo as follows:

% \pgfdeclareimage[height=0.5cm]{university-logo}{university-logo-filename}
% \logo{\pgfuseimage{university-logo}}



% Delete this, if you do not want the table of contents to pop up at
% the beginning of each subsection:
\AtBeginSubsection[]
{
  \begin{frame}<beamer>
    \frametitle{Outline}
    \tableofcontents[currentsection,currentsubsection]
  \end{frame}
}


% If you wish to uncover everything in a step-wise fashion, uncomment
% the following command: 

%\beamerdefaultoverlayspecification{<+->}


\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 a change of variable technique which simplifies the wave
  equation,
\item d'Alembert's solution to the wave equation which avoids the
  summing of a Fourier series solution.
\end{itemize}

\end{frame}
%%

\begin{frame}\frametitle{Wave Equation}

Consider the initial value problem for the unbounded,
homogeneous one-dimensional
wave equation
\begin{align*}
u_{tt} &= c^{2} u_{xx}
\mbox{ for $-\infty < x < \infty$ and $t > 0$}	\\
u(x,0) &= f(x)	\\
u_{t}(x,0) &= g(x)	.
\end{align*}

\uncover<2->{%
Rewrite the PDE by making the change of variables
\begin{align*}
\xi &= x + c \, t	\\
\eta &= x - c \, t	.
\end{align*}
}

\end{frame}
%%

\begin{frame}\frametitle{Change of Variables}

First partial derivatives:
\begin{align*}
u_{x}
	&=
u_{\xi} \xi_{x} + u_{\eta} \eta_{x} = u_{\xi} + u_{\eta}	\\
u_{t}
	&=
u_{\xi} \xi_{t} + u_{\eta} \eta_{t} = c (u_{\xi}  - u_{\eta})
\end{align*}
Second partial derivatives:
\begin{align*}
u_{xx}
	&=
u_{\xi\xi} \xi_{x} + u_{\xi\eta} \eta_{x} + u_{\eta\xi} \xi_{x} +
u_{\eta\eta} \eta_{x}
	=
u_{\xi\xi} + 2 u_{\xi\eta} + u_{\eta\eta}	\\
u_{tt}
	&=
c (u_{\xi\xi} \xi_{t} + u_{\xi\eta} \eta_{t}) -
c (u_{\eta\xi} \xi_{t} + u_{\eta\eta}\eta_{t})
	=
c^{2} (u_{\xi\xi} - 2 u_{\xi\eta} + u_{\eta\eta})
\end{align*}

\end{frame}
%%

\begin{frame}\frametitle{Substitution into Wave Equation}

\begin{align*}
u_{tt} &= c^{2} u_{xx}	\\
c^{2} (u_{\xi\xi} - 2 u_{\xi\eta} + u_{\eta\eta})
	&=
c^{2} (u_{\xi\xi} + 2 u_{\xi\eta} + u_{\eta\eta})	\\
\uncover<2->{%
u_{\xi\eta} &= 0
}
\end{align*}
\uncover<3->{%
Integrate both sides of the equation.
\begin{align*}
u_{\xi} &= \phi(\xi)	\\
u(\xi,\eta) &= \int \phi(\xi) \, d\xi	\\
	&= \Phi(\xi) + \Psi(\eta)
\end{align*}
Functions $\Phi$ and $\Psi$ are arbitrary smooth functions.
}

\end{frame}
%%

\begin{frame}\frametitle{Return to the Original Variables}

\begin{align*}
u(\xi,\eta)
	&= \Phi(\xi) + \Psi(\eta)	\\
u(x,t)
	&= \Phi(x + c \, t) + \Psi(x - c \, t)
\end{align*}
This is referred to as \textbf{d'Alembert's general solution to the
  wave equation}.

\vspace*{0.10in}

\uncover<2->{%
\textbf{Question}: can $\Phi$ and $\Psi$ be chosen to satisfy the
initial conditions?
}

\end{frame}
%%

\begin{frame}\frametitle{Plucked String (1 of 2)}

\begin{align*}
u(x,t) &= \Phi(x + c \, t) + \Psi(x - c \, t)	\\
u(x, 0) &= \Phi(x) + \Psi(x) = f(x)	\\
u_{t}(x, 0) &= c \, \Phi'(x) - c \, \Psi'(x) = 0	\\
0 &= \Phi'(x) - \Psi'(x)
\end{align*}
Integrate the last equation.
\[
\Phi(x) = \Psi(x) + K
\]
where $K$ is an arbitrary constant.

\vspace*{0.10in}

\uncover<2->{%
Substituting into the equation for the initial displacement produces
\begin{align*}
2 \Psi(x) + K &= f(x)	\\
\Psi(x) &= \frac{1}{2} \left( f(x) - K \right)	\\
\Phi(x) &= \frac{1}{2} \left( f(x) + K \right)	.
\end{align*}
}

\end{frame}
%%

\begin{frame}\frametitle{Plucked String (2 of 2)}

Consequently if $f$ is twice differentiable, then
\[
u(x,t) = \frac{1}{2} \left[ f(x+c \, t) + f(x - c \, t) \right]
\]
solves the initial value problem describing the plucked string.

\end{frame}
%%

\begin{frame}\frametitle{Struck String (1 of 2)}

\begin{align*}
u(x,t) &= \Phi(x + c \, t) + \Psi(x - c \, t)	\\
u(x, 0) &= \Phi(x) + \Psi(x) = 0	\\
u_{t}(x, 0) &= c \, \Phi'(x) - c \, \Psi'(x) = g(x)
\end{align*}
Differentiating the 2nd equation reveals $\Phi'(x) = -\Psi'(x)$

\vspace*{0.10in}

\uncover<2->{%
Substituting into the 3rd equation produces
\begin{align*}
2 c \, \Phi'(x) &= g(x)	\\
\Phi(x) &= \frac{1}{2 c} \int_{0}^{x} g(s) \, ds + K	\\
\Psi(x) &= -\frac{1}{2 c} \int_{0}^{x} g(s) \, ds - K	.
\end{align*}
}

\end{frame}
%%

\begin{frame}\frametitle{Struck String (2 of 2)}

Consequently if $g$ is continuously differentiable, then
\begin{align*}
u(x,t)
	&=
\frac{1}{2 c} \left[ \int_{0}^{x + c \, t} g(s) \, ds -
\int_{0}^{x - c \, t} g(s) \, ds \right]	\\
	&=
\frac{1}{2 c} \left[ \int_{0}^{x + c \, t} g(s) \, ds +
\int_{x - c \, t}^{0} g(s) \, ds \right]	\\
	&=
\frac{1}{2 c} \int_{x - c \, t}^{x + c \, t} g(s) \, ds
\end{align*}
solves the initial value problem describing the struck string.

\end{frame}
%%

\begin{frame}\frametitle{Nonzero Displacement and Velocity}

By the Principle of
Superposition, the general solution is
\[
u(x, t) = \frac{1}{2} \left[ f(x - c \, t) + f(x + c \, t) \right] +
\frac{1}{2 c} \int_{x - c \, t}^{x + c \, t} g(s) \, ds	.
\]

\end{frame}
%%

\begin{frame}\frametitle{Example: Plucked String}

Determine the solution to the initial value problem:
\begin{align*}
u_{tt} &= c^{2} u_{xx}
\mbox{ for $-\infty < x < \infty$ and $t > 0$}	\\
u(x,0) &= f(x) = \left\{ \begin{array}{rl}
2 & \mbox{if $-1 < x < 1$}	\\
0 & \mbox{otherwise}
\end{array} \right.	\\
u_{t}(x,0) &= 0
\end{align*}

\end{frame}
%%

\begin{frame}\frametitle{Solution (1 of 7)}

Using d'Alembert's solution
\[
u(x,t) = \frac{1}{2} \left[ f(x + c \, t) + f(x - c \, t) \right]	.
\]
\textbf{Note}:
\begin{itemize}
\item Along lines where $x + c \, t$ is constant the term $f(x + c \,
  t)$ is constant.
\item Likewise along lines where $x - c \, t$ is constant the term $f(x - c \,
  t)$ is constant.
\item These lines are called \textbf{characteristics}.
\end{itemize}

\end{frame}
%%

\begin{frame}\frametitle{Solution (2 of 7)}

\begin{align*}
f(x) &= \left\{ \begin{array}{rl}
2 & \mbox{if $-1 < x < 1$}	\\
0 & \mbox{otherwise}
\end{array} \right.	\\
f(x + c \, t) &= \left\{ \begin{array}{rl}
2 & \mbox{if $-1 < x + c \, t < 1$}	\\
0 & \mbox{otherwise}
\end{array} \right.	\\
\frac{1}{2} f(x + c \, t)
	&=
\left\{ \begin{array}{rl}
1 & \mbox{if $-1 - c \, t < x < 1 - c \, t$}	\\
0 & \mbox{otherwise}
\end{array} \right.	\\
\uncover<2->{%
\frac{1}{2} f(x - c \, t)
	&=
\left\{ \begin{array}{rl}
1 & \mbox{if $-1 + c \, t < x < 1 + c \, t$}	\\
0 & \mbox{otherwise}
\end{array} \right.
}
\end{align*}
\uncover<3->{%
\textbf{Remark}: the characteristics where $x + c \, t = \pm 1$ and $x
- c \, t = \pm 1$ help determine the solution.
}

\end{frame}
%%

\begin{frame}\frametitle{Solution (3 of 7)}

\begin{center}
\resizebox{1.00\textwidth}{!}{\includegraphics{chars}}
\end{center}

\end{frame}
%%

\begin{frame}\frametitle{Solution (4 of 7)}

\begin{description}
\item[Region 1:] $\{ (x, t) \, | \, x + c \, t < -1 \}$
\item[Region 2:] $\{ (x, t) \, | \, -1 < x - c \, t \mbox{ and } x + c \, t < 1 \}$
\item[Region 3:] $\{ (x, t) \, | \, 1 < x - c \, t \}$
\item[Region 4:] $\{ (x, t) \, | \, 1 < x + c \, t \mbox{ and }  -1 < x - c \, t < 1 \}$
\item[Region 5:] $\{ (x, t) \, | \, 1 < x + c \, t \mbox{ and } x - c \, t < -1 \}$
\item[Region 6:] $\{ (x, t) \, | \, -1 < x + c \, t < 1 \mbox{ and } x - c \, t < -1 \}$
\end{description}

\end{frame}
%%

\begin{frame}\frametitle{Solution (5 of 7)}

\begin{align*}
u(x, t)
	&=
\frac{1}{2} f(x + c \, t) + \frac{1}{2} f(x - c \, t)	\\
	&=
\left\{ \begin{array}{rl}
0 & \mbox{if $x + c \, t < -1$}	\\
2 & \mbox{if $-1 < x - c \, t < 1$ and $-1 < x + c \, t < 1$}	\\
0 & \mbox{if $1 < x - c \, t$}	\\
1 & \mbox{if $1 < x + c \, t$ and $-1 < x - c \, t < 1$}	\\
0 & \mbox{if $1 < x + c \, t$ and $x - c \, t < -1$}	\\
1 & \mbox{if $-1 < x + c \, t < 1$ and $x - c \, t < -1$}
\end{array} \right.
\end{align*}

\end{frame}
%%

\begin{frame}\frametitle{Solution (6 of 7)}

\begin{center}
\resizebox{1.00\textwidth}{!}{\includegraphics{dens}}
\end{center}

\end{frame}
%%

\begin{frame}\frametitle{Solution (7 of 7)}

\begin{center}
\resizebox{1.00\textwidth}{!}{\includegraphics{snaps01}}
\end{center}

\end{frame}
%%

\begin{frame}\frametitle{Example: Struck String}

Determine the solution to the initial value problem:
\begin{align*}
u_{tt} &= c^{2} u_{xx}
\mbox{ for $-\infty < x < \infty$ and $t > 0$}	\\
u(x,0) &= 0	\\
u_{t}(x,0) &= g(x) = \left\{ \begin{array}{rl}
1 & \mbox{if $-1 < x < 1$}	\\
0 & \mbox{otherwise}
\end{array} \right.
\end{align*}

\end{frame}
%%

\begin{frame}\frametitle{Solution (1 of 6)}

Define the function
\[
G(z) = \int_{0}^{z} g(w) \, dw = 
\left\{ \begin{array}{rl}
-1 & \mbox{if $z < -1$}	\\
z & \mbox{if $-1 \le  z \le 1$}	\\
1 & \mbox{if $z > 1$.}
\end{array} \right.
\]
then
\[
u(x,t) = \frac{1}{2 c} \left[ G(x + c \, t) - G(x - c \, t) \right]	.
\]
As in the previous example, the characteristics $x + c \, t = \pm 1$
and $x - c \, t = \pm 1$ divide the $xt$-plane into six regions.

\end{frame}
%%

\begin{frame}\frametitle{Solution (2 of 6)}

\begin{center}
\resizebox{1.00\textwidth}{!}{\includegraphics{chars}}
\end{center}

\end{frame}
%%

\begin{frame}\frametitle{Solution (3 of 6)}

\begin{description}
\item[Region 1:] $\{ (x, t) \, | \, x + c \, t < -1 \}$
\item[Region 2:] $\{ (x, t) \, | \, -1 < x - c \, t \mbox{ and } x + c \, t < 1 \}$
\item[Region 3:] $\{ (x, t) \, | \, 1 < x - c \, t \}$
\item[Region 4:] $\{ (x, t) \, | \, 1 < x + c \, t \mbox{ and }  -1 < x - c \, t < 1 \}$
\item[Region 5:] $\{ (x, t) \, | \, 1 < x + c \, t \mbox{ and } x - c \, t < -1 \}$
\item[Region 6:] $\{ (x, t) \, | \, -1 < x + c \, t < 1 \mbox{ and } x - c \, t < -1 \}$
\end{description}

\end{frame}
%%

\begin{frame}\frametitle{Solution (4 of 6)}

\begin{align*}
u(x, t)
	&=
\frac{1}{2 c} G(x + c \, t) - \frac{1}{2 c} G(x - c \, t)	\\
	&=
\frac{1}{2 c} \left\{ \begin{array}{rl}
0 & \mbox{if $x + c \, t < -1$}	\\
2 c \, t & \mbox{if $-1 < x - c \, t$ and $x + c \, t < 1$}	\\
0 & \mbox{if $1 < x - c \, t$}	\\
1 - x + c \, t & \mbox{if $1 < x + c \, t$ and $-1 < x - c \, t < 1$}	\\
2 & \mbox{if $1 < x + c \, t$ and $x - c \, t < -1$}	\\
1 + x + c \, t & \mbox{if $x - c \, t < -1$ and $-1 < x + c \, t < 1$}
\end{array} \right.
\end{align*}

\end{frame}
%%

\begin{frame}\frametitle{Solution (5 of 6)}

\begin{center}
\resizebox{1.00\textwidth}{!}{\includegraphics{example04dens}}
\end{center}

\end{frame}
%%

\begin{frame}\frametitle{Solution (6 of 6)}

\begin{center}
\resizebox{1.00\textwidth}{!}{\includegraphics{example04surf}}
\end{center}

\end{frame}
%%

\begin{frame}\frametitle{Domain of Dependence (1 of 2)}

In general the solution to the initial value problem:
\begin{align*}
u_{tt} &= c^{2} u_{xx}
\mbox{ for $-\infty < x < \infty$ and $t > 0$}	\\
u(x,0) &= f(x)	\\
u_{t}(x,0) &= g(x)
\end{align*}
can be expressed as
\[
u(x,t) = \frac{1}{2} \left[ f(x + c \, t) + f(x - c \, t) \right] +
\frac{1}{2c} \int_{x-c \, t}^{x + c \, t} g(s) \, ds	.
\]
At the point $(x_{0}, t_{0})$ then
\[
u(x_{0},t_{0}) = \frac{1}{2} \left[ f(x_{0} + c \, t_{0}) + f(x_{0} - c \, t_{0}) \right] +
\frac{1}{2c} \int_{x_{0}-c \, t_{0}}^{x_{0} + c \, t_{0}} g(s) \, ds	.
\]

\end{frame}
%%

\begin{frame}\frametitle{Domain of Dependence (2 of 2)}

\[
u(x_{0},t_{0}) = \frac{1}{2} \left[ f(x_{0} + c \, t_{0}) + f(x_{0} - c \, t_{0}) \right] +
\frac{1}{2c} \int_{x_{0}-c \, t_{0}}^{x_{0} + c \, t_{0}} g(s) \, ds	.
\]

\textbf{Remarks}:
\begin{itemize}
\item $u(x_{0},t_{0})$ depends only on the values of $f(x_{0} \pm c \,
  t)$ and $g(s)$ for $x_{0} - c \, t_{0} \le s \le x_{0} + c \,
  t_{0}$.
\item The interval $[x_{0} - c \, t_{0}, x_{0} + c \, t_{0}]$ is
  called the \textbf{domain of dependence}.
\end{itemize}

\end{frame}
%%

\begin{frame}\frametitle{Domain of Dependence Illustrated}

\begin{center}
\resizebox{1.00\textwidth}{!}{\includegraphics{domdep}}
\end{center}

\end{frame}
%%

\begin{frame}\frametitle{Domain of Influence}

The point $(x_{0}, t_{0})$ influences the solution $u(x,t)$ for $t \ge
t_{0}$ at all points between the characteristics passing through
$(x_{0}, t_{0})$.
\begin{align*}
\frac{t - t_{0}}{x - x_{0}} &= \frac{\pm 1}{c}	\\
c (t - t_{0}) &= \pm (x - x_{0})	\\
\pm x_{0} + c (t - t_{0}) &= \pm x
\end{align*}

\end{frame}
%%

\begin{frame}\frametitle{Domain of Influence Illustrated}

\begin{center}
\resizebox{1.00\textwidth}{!}{\includegraphics{dominf}}
\end{center}

\end{frame}
%%

\begin{frame}\frametitle{Finite Length String}

D'Alembert's solution to the wave equation can be adapted to the wave
equation with $0 < x < L$.
\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*}

\end{frame}
%%

\begin{frame}\frametitle{Case: Plucked String}

\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) &= 0
\end{align*}
We have used separation of variables and Fourier series to determine
\begin{align*}
u(x,t)
	&=
\sum_{n=1}^{\infty} a_{n} \cos \frac{c \, n \, \pi \, t}{L}
\sin \frac{n \, \pi \, x}{L}	\\
\uncover<2->{%
	&=
\frac{1}{2} \left[
\sum_{n=1}^{\infty} a_{n} \sin \frac{n \pi (x + c \, t)}{L} +
\sum_{n=1}^{\infty} a_{n} \sin \frac{n \pi (x - c \, t)}{L}
\right]	\\
	&=
\frac{1}{2} \left[ f(x + c \, t) + f(x - c \, t) \right]	,
}
\end{align*}
\uncover<2->{%
where $f(x)$ is the odd, $2L$-periodic extension of the initial displacement.
}

\end{frame}
%%

\begin{frame}\frametitle{Case: Struck String}

\begin{align*}
u_{tt} &= c^{2} u_{xx}
\mbox{ for $0 < x < L$ and $t > 0$}	\\
u(0,t) &= u(L, t) = 0	\\
u(L,t) &= 0	\\
u(x,0) &= 0	\\
u_{t}(x,0) &= g(x)
\end{align*}
We have used separation of variables and Fourier series to determine
\begin{align*}
u(x,t)
	&=
\sum_{n=1}^{\infty} b_{n} \sin \frac{c \, n \, \pi \, t}{L}
\sin \frac{n \, \pi \, x}{L}	\\
\uncover<2->{%
	&=
\frac{1}{2} \sum_{n=1}^{\infty}
\left[
b_{n} \cos \frac{n \pi (x - c \, t)}{L} - b_{n} \cos \frac{n \pi (x + c \, t)}{L}
\right]	.
}
\end{align*}

\end{frame}
%%

\begin{frame}\frametitle{Integrating Term by Term}

\begin{align*}
u(x,t)
	&=
\frac{1}{2} \sum_{n=1}^{\infty}
\left[
b_{n} \cos \frac{n \pi (x - c \, t)}{L} - b_{n} \cos \frac{n \pi (x + c \, t)}{L}
\right]	\\
	&=
\frac{1}{2} \sum_{n=1}^{\infty} b_{n} \frac{n \pi}{L}
\int_{x - c \, t}^{x + c \, t} \sin \frac{n \pi s}{L} \, ds	\\
	&=
\frac{1}{2} \int_{x - c \, t}^{x + c \, t} \left(
\sum_{n=1}^{\infty} \left[ b_{n} \frac{n\pi}{L} \right]
\sin \frac{n \pi s}{L} \right) \, ds	\\
	&=
\frac{1}{2 c} \int_{x - c \, t}^{x + c \, t} \left(
\sum_{n=1}^{\infty} \left[ b_{n} \frac{c n \pi}{L} \right]
\sin \frac{n \pi s}{L} \right) \, ds	\\
	&=
\frac{1}{2 c} \int_{x - c \, t}^{x + c \, t} g(s) \, ds
\end{align*}
where $g(x)$ is the odd, $2L$-periodic extension of the initial velocity.

\end{frame}
%%

\begin{frame}\frametitle{Example}

Find the solution to the initial boundary value problem
\begin{align*}
u_{tt}	&= 4 u_{xx}
\mbox{ for $0 < x < 1$ and $t > 0$}	\\
u(0, t)	&= u(L, t) = 0	\\
u(x, 0)	&= 0	\\
u_{t}(x, 0)	&= \left\{ \begin{array}{rl}
0	& \mbox{if $x < 1/4$}	\\
1	& \mbox{if $1/4 \le x \le 3/4$}	\\
0	& \mbox{if $3/4 < x < 1$.}
\end{array} \right.
\end{align*}
Let $g(x)$ be the odd, 2-periodic extension of $u_{t}(x,0)$.

\end{frame}
%%

\begin{frame}\frametitle{Solution (1 of 6)}

Let $g_{o}(x)$ be the odd, 2-periodic extension of $u_{t}(x,0)$.
\[
g_{o}(x)
	=
\left\{ \begin{array}{rl}
0	& \mbox{if $0 < x < 1/4$}	\\
1	& \mbox{if $1/4 < x < 3/4$}	\\
0	& \mbox{if $3/4 < x < 5/4$}	\\
-1	& \mbox{if $5/4 < x < 7/4$}	\\
0	& \mbox{if $7/4 < x < 2$}
\end{array} \right.
\]

\end{frame}
%%

\begin{frame}\frametitle{Solution (2 of 6)}

\begin{center}
\resizebox{1.00\textwidth}{!}{\includegraphics{godd}}
\end{center}

\end{frame}
%%

\begin{frame}\frametitle{Solution (3 of 6)}

Define the function $\displaystyle G(x) = \int_{0}^{x} g_{o}(s) \, ds$.
\uncover<2->{%
\begin{align*}
G(x)
	&=
\left\{ \begin{array}{rl}
\int_{0}^{x} 0 \, ds & \mbox{if $x < 1/4$}	\\
\int_{1/4}^{x} 1 \, ds & \mbox{if $1/4 \le x \le 3/4$}	\\
\int_{1/4}^{3/4} 1 \, ds & \mbox{if $3/4 < x < 5/4$}	\\
\int_{1/4}^{3/4} 1 \, ds + \int_{5/4}^{x} (-1) \, ds & \mbox{if $5/4 < x < 7/4$}	\\
\int_{1/4}^{3/4} 1 \, ds + \int_{5/4}^{7/4} (-1) \, ds & \mbox{if $7/4 < x < 2$}
\end{array} \right.	\\
	&=
\left\{ \begin{array}{rl}
0 & \mbox{if $x < 1/4$}	\\
x - 1/4 & \mbox{if $1/4 \le x \le 3/4$}	\\
1/2 & \mbox{if $3/4 < x < 5/4$}	\\
-x + 7/4 & \mbox{if $5/4 < x < 7/4$}	\\
0 & \mbox{if $7/4 < x < 2$}
\end{array} \right.
\end{align*}
}

\end{frame}
%%

\begin{frame}\frametitle{Solution (4 of 6)}

\begin{align*}
G(x+2t)
	&=
\left\{ \begin{array}{rl}
0 & \mbox{if $x+2t < 1/4$}	\\
x+2t - 1/4 & \mbox{if $1/4 \le x+2t \le 3/4$}	\\
1/2 & \mbox{if $3/4 < x+2t < 5/4$}	\\
-x-2t + 7/4 & \mbox{if $5/4 < x+2t < 7/4$}	\\
0 & \mbox{if $7/4 < x+2t < 2$}
\end{array} \right.	\\
G(x-2t)
	&=
\left\{ \begin{array}{rl}
0 & \mbox{if $x-2t < 1/4$}	\\
x-2t - 1/4 & \mbox{if $1/4 \le x-2t \le 3/4$}	\\
1/2 & \mbox{if $3/4 < x-2t < 5/4$}	\\
-x+2t + 7/4 & \mbox{if $5/4 < x-2t < 7/4$}	\\
0 & \mbox{if $7/4 < x-2t < 2$}
\end{array} \right.
\end{align*}

\end{frame}
%%

\begin{frame}\frametitle{Solution (5 of 6)}

Using d'Alembert's solution to the wave equation, then
\begin{align*}
u(x,t)
	&=
\frac{1}{2c} \left[ G((x+c \, t) \pmod{2}) - G((x-c \, t) \pmod{2}) \right]	\\
	&=
\frac{1}{4} \left[ G((x+2t) \pmod{2}) - G((x-2t) \pmod{2}) \right]	.
\end{align*}

\end{frame}
%%

\begin{frame}\frametitle{Solution (6 of 6)}

\begin{center}
\resizebox{0.85\textwidth}{!}{\includegraphics{example05surf}}
\end{center}

\end{frame}
%%

\begin{frame}\frametitle{Homework}

\begin{itemize}
\item Read Sections $5.2$ and $5.3$
\item Exercises: 6--10
\end{itemize}

\end{frame}
%%

%%%
\end{document}

%%%%
%%%% EOF
%%%%
