%%%%		-*- 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:      02/14/2025
%%%%
%%%% Purpose:   Wave Equation With Homogeneous Boundary Conditions
%%%%

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\title{Wave Equation With Homogeneous Boundary Conditions}
\subtitle{\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]
% (optional, but mostly needed)
{
  Department of Mathematics
}
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Fall
%Spring
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\the\year}

\subject{Talks}
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  \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
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% of between 15min and 45min and the theme suggest that you stick to
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%   15 and 30 frames, all told.

%\section{Introduction}

\begin{frame}\frametitle{Objectives}

In this lesson we will learn:
\begin{itemize}
\item how to solve the wave equation with homogeneous Dirichlet
  boundary conditions using separation of variables,
\item how to solve the wave equation with homogeneous Neumann
  boundary conditions using separation of variables,
\item terms for describing the components of the solution to the wave equation.
\end{itemize}

\end{frame}
%%

\begin{frame}\frametitle{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*}
Since the PDE is linear and homogeneous and the boundary conditions
are homogeneous and of Dirichlet type, the method of separation of
variables and the Principle of Superposition apply.

\end{frame}
%%

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

Assume a product solution of the form $u(x, t) = X(x) T(t)$,
differentiate and substitute into the wave equation.
\begin{align*}
u_{tt}
	&= c^{2} u_{xx}	\\
X(x) T''(t)
	&= c^{2} X''(x) T(t)	\\
\frac{T''(t)}{c^{2} T(t)}
	&= \frac{X''(x)}{X(x)}
	= -\lambda
\end{align*}
where $\lambda$ is a constant.

\vspace*{0.10in}

This implies the boundary value problem for $X(x)$.
\begin{align*}
X''(x) + \lambda X(x)
	&= 0	\\
X(0)
	&= 0	\\
X(L)
	&= 0
\end{align*}

\end{frame}
%%

\begin{frame}\frametitle{Eigenvalues and Eigenfunctions}

The only non-trivial eigenfunctions are
\[
X_{n}(x)
	= \sin \frac{n \pi x}{L}
\]
corresponding to the eigenvalues $\displaystyle \lambda_{n} =
\frac{n^{2} \pi^{2}}{L^{2}}$ for $n \in \mathbb{N}$.

\vspace*{0.10in}

With these eigenvalues, the implied ODE for function $T_{n}(t)$ has
the form
\[
T_{n}''(t) + \frac{c^{2} n^{2} \pi^{2}}{L^{2}} T_{n}(t)
	= 0
\]
and consequently solution
\[
T_{n}(t)
	= a_{n} \cos \frac{c n \pi t}{L} + b_{n} \sin \frac{c n \pi t}{L}	.
\]

\end{frame}
%%

\begin{frame}\frametitle{Product Solutions}

Functions of the form
\[
u_{n}(x, t)
	= X_{n}(x) T_{n}(t)
	= \left( a_{n} \cos \frac{c n \pi t}{L} + b_{n} \sin \frac{c n
            \pi t}{L} \right) \sin \frac{n \pi x}{L}
\]
for $n \in \mathbb{N}$
solve the wave equation and satisfy the homogeneous Dirichlet boundary
conditions.
These solutions are called \textbf{fundamental solutions}.

\vspace*{0.10in}

By the Principle of Superposition a sum of fundamental solutions will
also solve the wave equation and satisfy the homogeneous Dirichlet boundary
conditions.
\[
u(x, t)
	= \sum_{n=1}^{N} \left( a_{n} \cos \frac{c n \pi t}{L} + b_{n}
          \sin \frac{c n \pi t}{L} \right) \sin \frac{n \pi x}{L}
\]

\end{frame}
%%

\begin{frame}\frametitle{Fundamental Solutions}

\[
u_{n}(x, t)
	= \left( a_{n} \cos \frac{c n \pi t}{L} + b_{n} \sin \frac{c n
            \pi t}{L} \right) \sin \frac{n \pi x}{L}
\]
\begin{itemize}
\item This solution is known as the \textbf{$n$th harmonic}.
\item Solution is periodic in $t$ with period $\displaystyle \frac{2
    L}{c n}$.
\item The number of oscillations per $2 \pi$ units of time is called
  the \textbf{natural frequency} and is $\displaystyle \frac{c n
    \pi}{L}$.
\item The number of oscillations per unit time is called the
  \textbf{frequency} and is $\displaystyle \frac{c n}{2 L}$.
\item The \textbf{wavelength} of the solution is $\displaystyle
  \frac{2 L}{n}$.
\item The intensity of the solution is given by the \textbf{amplitude}
  $\sqrt{a_{n}^{2} + b_{n}^{2}}$.
\end{itemize}

\end{frame}
%%

\begin{frame}\frametitle{First Harmonic}

\[
u_{1}(x, t)
	= \left( a_{1} \cos \frac{c \pi t}{L} + b_{1} \sin \frac{c \pi
            t}{L} \right) \sin \frac{\pi x}{L}
\]
\begin{itemize}
\item This solution is known as the \textbf{first harmonic} or the
  \textbf{fundamental mode}.
\item The number of oscillations per $2 \pi$ units of time is called
  the \textbf{fundamental frequency} and is $\displaystyle \frac{c \pi}{L}$.
\item First harmonic is periodic in $t$ with period $\displaystyle \frac{2
    L}{c}$.
\item The $n$th harmonic has a frequency which is $n$ times the
  fundamental frequency.
\end{itemize}


\end{frame}
%%

\begin{frame}\frametitle{Initial Displacement}

We will assume a Fourier series solution to the IBVP.
\begin{align*}
u(x, t)
	&= \sum_{n=1}^{\infty}
\left( a_{n} \cos \frac{c n \pi t}{L} + b_{n} \sin \frac{c n \pi t}{L}
\right)
\sin \frac{n \pi x}{L}	\\
u(x, 0)
	&= \sum_{n=1}^{\infty} a_{n} \sin \frac{n \pi x}{L}
	= f(x)
\end{align*}
Suppose $f(x)$ can be extended to $\mathbb{R}$ as a $2L$--periodic,
odd function, then $f(0) = f(L) = f(-L)$ and
\[
a_{n}
	= \frac{2}{L} \int_{0}^{L} f(x) \sin \frac{n \pi x}{L} \, dx	,
\]
for $n \in \mathbb{N}$.

\end{frame}
%%

\begin{frame}\frametitle{Initial Velocity}

If $u(x, t)$ can be differentiated term by term, then
\begin{align*}
u_{t}(x, t)
	&= \sum_{n=1}^{\infty}
\frac{c n \pi}{L} \left( -a_{n} \sin \frac{c n \pi t}{L} + b_{n} \cos \frac{c n \pi t}{L}
\right)
\sin \frac{n \pi x}{L}	\\
u_{t}(x, 0)
	&= \sum_{n=1}^{\infty}
\frac{c n \pi}{L} b_{n} \sin \frac{n \pi x}{L} = g(x)
\end{align*}
Again, if $g(x)$ can be extended to $\mathbb{R}$ as a $2L$-periodic,
odd function then
\begin{align*}
\frac{c n \pi}{L} b_{n}
	&= \frac{2}{L} \int_{0}^{L} g(x) \sin \frac{n \pi x}{L} \, dx	\\
b_{n}
	&= \frac{2}{c n \pi} \int_{0}^{L} g(x) \sin \frac{n \pi x}{L} \, dx
\end{align*}
for $n \in \mathbb{N}$.

\end{frame}
%%

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

Find the solution to the following IBVP.
\begin{align*}
u_{tt}
	&= u_{xx}
\mbox{ for $0 < x < 10$ and $t > 0$}	\\
u(0, t)
	&= u(10, t) = 0	\\
u(x, 0)
	&= \left\{ \begin{array}{rl}
2x/5 & \mbox{if $0 \le x \le 5/2$,}	\\
1 & \mbox{if $5/2 < x <15/2$,}	\\
4 - 2x/5 & \mbox{if $15/2 \le x \le 10$}
\end{array} \right.	\\
u_{t}(x, 0)
	&= 0
\end{align*}

\end{frame}
%%

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

The formal solution can be expressed as
\[
u(x, t)
	= \sum_{n=1}^{\infty} \left( a_{n} \cos \frac{n \pi t}{10} +
  b_{n} \sin \frac{n \pi t}{10} \right) \sin \frac{n \pi x}{10}	.
\]
Since $u_{t}(x, 0) = 0$ then $b_{n} = 0$ for all $n \in \mathbb{N}$.
\begin{align*}
a_{n}
	&= \frac{2}{10} \int_{0}^{10} u(x, 0) \sin \frac{n \pi x}{10} \, dx	\\
	&= \frac{8}{n^{2} \pi^{2}} \left( \sin \frac{n \pi}{4} + \sin \frac{3 n
              \pi}{4} \right)	\\
\uncover<2->{%
u(x, t) &= \frac{8}{\pi^{2}} \sum_{n=1}^{\infty}
\frac{\sin \frac{n \pi}{4} + \sin \frac{3 n \pi}{4}}{n^{2}}
\cos \frac{n \pi t}{10} \sin \frac{n \pi x}{10}
}
\end{align*}

\end{frame}
%%

\begin{frame}\frametitle{Solution (2 of 2)}
\fontsize{9pt}{10.8}\selectfont

Let $f(x)$ be the odd, 20--periodic extension of $u(x, 0)$, then
\begin{align*}
f(x)
	&= \frac{8}{\pi^{2}} \sum_{n=1}^{\infty}
\frac{\sin \frac{n \pi}{4} + \sin \frac{3 n \pi}{4}}{n^{2}}
\sin \frac{n \pi x}{10}	\\
\uncover<2->{%
f(x+t)
	&= \frac{8}{\pi^{2}} \sum_{n=1}^{\infty}
\frac{\sin \frac{n \pi}{4} + \sin \frac{3 n \pi}{4}}{n^{2}}
\sin \frac{n \pi (x+t)}{10}	\\
	&= \frac{8}{\pi^{2}} \sum_{n=1}^{\infty}
\frac{\sin \frac{n \pi}{4} + \sin \frac{3 n \pi}{4}}{n^{2}}
\left[ \sin \frac{n \pi x}{10} \cos \frac{n \pi t}{10} + \cos
   \frac{n \pi x}{10} \sin \frac{n \pi t}{10} \right]	\\
\uncover<3->{%
f(x-t)
	&= \frac{8}{\pi^{2}} \sum_{n=1}^{\infty}
\frac{\sin \frac{n \pi}{4} + \sin \frac{3 n \pi}{4}}{n^{2}}
\sin \frac{n \pi (x-t)}{10}	\\
\uncover<4->{%
	&= \frac{8}{\pi^{2}} \sum_{n=1}^{\infty}
\frac{\sin \frac{n \pi}{4} + \sin \frac{3 n \pi}{4}}{n^{2}}
\left[ \sin \frac{n \pi x}{10} \cos \frac{n \pi t}{10} - \cos
   \frac{n \pi x}{10} \sin \frac{n \pi t}{10} \right]	\\
\uncover<5->{%
f(x+t) + f(x-t)
	&= \frac{16}{\pi^{2}} \sum_{n=1}^{\infty}
\frac{\sin \frac{n \pi}{4} + \sin \frac{3 n \pi}{4}}{n^{2}}
\cos \frac{n \pi t}{10} \sin \frac{n \pi x}{10} = 2 u(x, t)	\\
u(x, t)
	&= \frac{1}{2} \left( f(x+t) + f(x-t) \right)
}
}
}
}
\end{align*}

\end{frame}
%%

\begin{frame}\frametitle{20--Periodic Extension of Initial Displacement}

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

\end{frame}
%%

\begin{frame}\frametitle{Shifting Initial Displacement Horizontally}

\begin{center}
\resizebox{1.00\textwidth}{!}{\includegraphics{fshift3}}
\end{center}

\end{frame}
%%

\begin{frame}\frametitle{Adding Shifts of Initial Displacement}

\begin{center}
\resizebox{1.00\textwidth}{!}{\includegraphics{fsumshift3}}
\end{center}

\end{frame}
%%

\begin{frame}\frametitle{Check by Differentiation}

Consider the following IBVP.
\begin{align*}
u_{tt}
	&= u_{xx}
\mbox{ for $0 < x < 10$ and $t > 0$}	\\
u(0, t)
	&= u(10, t) = 0	\\
u(x, 0)
	&= \left\{ \begin{array}{rl}
2x/5 & \mbox{if $0 \le x \le 5/2$,}	\\
1 & \mbox{if $5/2 < x <15/2$,}	\\
4 - 2x/5 & \mbox{if $15/2 \le x \le 10$}
\end{array} \right.	\\
u_{t}(x, 0)
	&= 0
\end{align*}
Show by direct differentiation that
\[
u(x, t)
	= \frac{1}{2} \left( f(x+t) + f(x-t) \right)
\]
solves the IBVP when $f$ is the odd, 20--periodic extension of $u(x, 0)$.

\end{frame}
%%

\begin{frame}\frametitle{Solution}

\begin{align*}
u(x, t)
	&= \frac{1}{2} \left( f(x+t) + f(x-t) \right)	\\
u_{xx}
	&= \frac{1}{2} \left( f''(x+t) + f''(x-t) \right)	\\
u_{tt}
	&= \frac{1}{2} \left( f''(x+t) + f''(x-t) \right)	\\
\uncover<2->{%
u_{tt}
	&= u_{xx}
}
\end{align*}

\uncover<3->{%
Initial displacement:
\[
u(x, 0)
	= \frac{1}{2} \left( f(x) + f(x) \right) = f(x)
\]
\uncover<4->{%
Initial velocity:
\[
u_{t}(x, 0)
	= \frac{1}{2} \left( f'(x) - f'(x) \right) = 0
\]
\uncover<5->{%
Dirichlet boundary conditions:
\begin{align*}
u(0, t)
	&= \frac{1}{2} (f(t) + f(-t)) = 0 \mbox{ ($f$ is odd)}	\\
u(10, t)
	&= \frac{1}{2} (f(10+t) + f(10-t)) = 0 \mbox{ ($f$ is
           odd, 20-periodic)}
\end{align*}
}
}
}

\end{frame}
%%

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

Find the solution to the following IBVP.
\begin{align*}
u_{tt}
	&= u_{xx}
\mbox{ for $0 < x < 10$ and $t > 0$}	\\
u(0, t)
	&= u(10, t)
	= 0	\\
u(x, 0)
	&= 0	\\
u_{t}(x, 0)
	&= \left\{ \begin{array}{rl}
x/5 & \mbox{if $0 \le x \le 5$,}	\\
2 - x/5 & \mbox{if $5 < x \le 10$}
\end{array} \right.
\end{align*}

\end{frame}
%%

\begin{frame}\frametitle{Solution (1 of 4)}

The formal solution can be expressed as
\[
u(x, t)
	= \sum_{n=1}^{\infty} \left( a_{n} \cos \frac{n \pi t}{10} +
  b_{n} \sin \frac{n \pi t}{10} \right) \sin \frac{n \pi x}{10}	.
\]
Since $u(x, 0) = 0$ then $a_{n} = 0$ for all $n \in \mathbb{N}$.
\begin{align*}
b_{n}
	&= \frac{2}{n \pi} \int_{0}^{10} u_{t}(x, 0) \sin \frac{n \pi x}{10} \, dx	\\
	&= \frac{80}{n^{3} \pi^{3}} \sin \frac{n \pi}{2}	\\
\uncover<2->{%
u(x, t)
	&= \frac{80}{\pi^{3}} \sum_{n=1}^{\infty}
\frac{\sin \frac{n \pi}{2}}{n^{3}}
\sin \frac{n \pi t}{10} \sin \frac{n \pi x}{10}
}
\end{align*}

\end{frame}
%%

\begin{frame}\frametitle{Solution (2 of 4)}

Let $g(x)$ be the odd, 20--periodic extension of $u_{t}(x, 0)$, then
\[
g(x)
	= \frac{8}{\pi^{2}} \sum_{n=1}^{\infty}
\frac{\sin \frac{n \pi}{2}}{n^{2}}
\sin \frac{n \pi x}{10}	.
\]
Define $\displaystyle G(x) = \int_{0}^{x} g(s) \, ds$ and integrate
the Fourier series term by term.
\begin{align*}
G(x)
	&= \frac{8}{\pi^{2}} \sum_{n=1}^{\infty}
\frac{\sin \frac{n \pi}{2}}{n^{2}}
\int_{0}^{x} \sin \frac{n \pi s}{10} \, ds	\\
	&= \frac{8}{\pi^{2}} \sum_{n=1}^{\infty}
\frac{\sin \frac{n \pi}{2}}{n^{2}}
\left[ -\frac{10}{n \pi} \cos \frac{n \pi s}{10} \right]_{s=0}^{s=x}	\\
	&= \frac{80}{\pi^{3}} \sum_{n=1}^{\infty}
\frac{\sin \frac{n \pi}{2}}{n^{3}}
\left( 1 - \cos \frac{n \pi x}{10} \right)	\\
	&= \frac{80}{\pi^{3}} \sum_{n=1}^{\infty}
\frac{\sin \frac{n \pi}{2}}{n^{3}} -
\frac{80}{\pi^{3}} \sum_{n=1}^{\infty}
\frac{\sin \frac{n \pi}{2}}{n^{3}} \cos \frac{n \pi x}{10}
\end{align*}

\end{frame}
%%

\begin{frame}\frametitle{Solution (3 of 4)}
\fontsize{9pt}{10.8}\selectfont

\begin{align*}
G(x+t)
	&= \int_{0}^{x+t} g(s) \, ds	\\
	&= \frac{80}{\pi^{3}} \sum_{n=1}^{\infty}
\frac{\sin \frac{n \pi}{2}}{n^{3}} -
\frac{80}{\pi^{3}} \sum_{n=1}^{\infty} \frac{\sin \frac{n \pi}{2}}{n^{3}}
\cos \frac{n \pi (x+t)}{10}	\\
	&= \frac{80}{\pi^{3}} \sum_{n=1}^{\infty}
\frac{\sin \frac{n \pi}{2}}{n^{3}} -
\frac{80}{\pi^{3}} \sum_{n=1}^{\infty} \frac{\sin \frac{n \pi}{2}}{n^{3}}
\left[ \cos \frac{n \pi x}{10} \cos \frac{n \pi t}{10} -
\sin \frac{n \pi x}{10} \sin \frac{n \pi t}{10} \right]	\\
G(x-t)
	&= \int_{0}^{x-t} g(s) \, ds	\\
	&= \frac{80}{\pi^{3}} \sum_{n=1}^{\infty}
\frac{\sin \frac{n \pi}{2}}{n^{3}} -
\frac{80}{\pi^{3}} \sum_{n=1}^{\infty} \frac{\sin \frac{n \pi}{2}}{n^{3}}
\cos \frac{n \pi (x-t)}{10}	\\
	&= \frac{80}{\pi^{3}} \sum_{n=1}^{\infty}
\frac{\sin \frac{n \pi}{2}}{n^{3}} -
\frac{80}{\pi^{3}} \sum_{n=1}^{\infty} \frac{\sin \frac{n \pi}{2}}{n^{3}}
\left[ \cos \frac{n \pi x}{10} \cos \frac{n \pi t}{10} +
\sin \frac{n \pi x}{10} \sin \frac{n \pi t}{10} \right]
\end{align*}

\end{frame}
%%

\begin{frame}\frametitle{Solution (4 of 4)}
\fontsize{9pt}{10.8}\selectfont

\begin{align*}
G(x+t)
	&= \frac{80}{\pi^{3}} \sum_{n=1}^{\infty}
\frac{\sin \frac{n \pi}{2}}{n^{3}} -
\frac{80}{\pi^{3}} \sum_{n=1}^{\infty} \frac{\sin \frac{n \pi}{2}}{n^{3}}
\left[ \cos \frac{n \pi x}{10} \cos \frac{n \pi t}{10} -
\sin \frac{n \pi x}{10} \sin \frac{n \pi t}{10} \right]	\\
G(x-t)
	&= \frac{80}{\pi^{3}} \sum_{n=1}^{\infty}
\frac{\sin \frac{n \pi}{2}}{n^{3}} -
\frac{80}{\pi^{3}} \sum_{n=1}^{\infty} \frac{\sin \frac{n \pi}{2}}{n^{3}}
\left[ \cos \frac{n \pi x}{10} \cos \frac{n \pi t}{10} +
\sin \frac{n \pi x}{10} \sin \frac{n \pi t}{10} \right]
\end{align*}
\uncover<2->{%
Subtract the two equations.
\begin{align*}
G(x+t) - G(x-t)
	&= \frac{80}{\pi^{3}} \sum_{n=1}^{\infty}
\frac{\sin \frac{n \pi}{2}}{n^{3}}
\left( \sin \frac{n \pi x}{10} \sin \frac{n \pi t}{10} + \sin \frac{n
   \pi x}{10} \sin \frac{n \pi t}{10} \right)	\\
\int_{0}^{x+t} g(s) \, ds - \int_{0}^{x-t} g(s) \, dx
	&= \frac{160}{\pi^{3}} \sum_{n=1}^{\infty}
\frac{\sin \frac{n \pi}{2}}{n^{3}} \sin \frac{n \pi t}{10} \sin
              \frac{n \pi x}{10}	\\
\int_{x-t}^{x+t} g(s) \, ds
	&= \frac{160}{\pi^{3}} \sum_{n=1}^{\infty}
\frac{\sin \frac{n \pi}{2}}{n^{3}} \sin \frac{n \pi t}{10} \sin
              \frac{n \pi x}{10}	\\
\frac{1}{2} \int_{x-t}^{x+t} g(s) \, ds
	&=
\frac{80}{\pi^{3}} \sum_{n=1}^{\infty}
\frac{\sin \frac{n \pi}{2}}{n^{3}} \sin \frac{n \pi t}{10} \sin
              \frac{n \pi x}{10} = u(x, t)
\end{align*}
}

\end{frame}
%%

\begin{frame}\frametitle{Illustration}

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

\end{frame}
%%

\begin{frame}\frametitle{Check by Differentiation}

Consider the following IBVP.
\begin{align*}
u_{tt}
	&= u_{xx}
\mbox{ for $0 < x < 10$ and $t > 0$}	\\
u(0, t)
	&= u(10, t)
	= 0	\\
u(x, 0)
	&= 0	\\
u_{t}(x, 0)
	&= \left\{ \begin{array}{rl}
x/5 & \mbox{if $0 \le x \le 5$,}	\\
2 - x/5 & \mbox{if $5 < x \le 10$}
\end{array} \right.
\end{align*}
Show by direct differentiation that
\[
u(x, t)
	= \frac{1}{2} \int_{x-t}^{x+t} g(s) \, ds
\]
solves the IBVP when $g$ is the odd, 20--periodic extension of $u_{t}(x, 0)$.

\end{frame}
%%

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

\begin{align*}
u(x, t)
	&= \frac{1}{2} \int_{x-t}^{x+t} g(s) \, ds	\\
u_{t}(x, t)
	&= \frac{1}{2} \left( g(x+t) + g(x-t) \right)	\\
u_{tt}(x, t)
	&= \frac{1}{2} \left( g'(x+t) - g'(x-t) \right)	\\
u_{x}(x, t)
	&= \frac{1}{2} \left( g(x+t) - g(x-t) \right)	\\
u_{xx}(x, t)
	&= \frac{1}{2} \left( g'(x+t) - g'(x-t) \right)	\\
u_{tt}
	&= u_{xx}
\end{align*}

\end{frame}
%%

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

\begin{itemize}
\item Displacement when $t = 0$,
\[
u(x, 0)
	= \frac{1}{2} \int_{x}^{x} g(s) \, ds = 0
\]
if $g$ is continuous at $x$.
\item Velocity when $t = 0$,
\[
u_{t}(x, 0)
	= \frac{1}{2} \left( g(x) + g(x) \right) = g(x)	.
\]
\item Boundary condition at $x = 0$,
\[
u(0, t)
	= \frac{1}{2} \int_{-t}^{t} g(s) \, ds = 0
\]
since $g(x)$ is odd.
\item Boundary condition at $x = 10$,
\[
u(10, t)
	= \frac{1}{2} \int_{10-t}^{10+t} g(s) \, ds = 0
\]
since $g(x)$ is odd, 20-periodic.
\end{itemize}

\end{frame}
%%

\begin{frame}\frametitle{Combination}

Suppose $u(x, t)$ and $v(x, t)$ solve the respective IBVPs for $0 < x <
L$ and $t > 0$:
\begin{columns}
\begin{column}{0.50\textwidth}
\begin{align*}
u_{tt}
	&= c^{2} u_{xx}	\\
u(0, t)
	&= u(L, t)
	= 0	\\
u(x, 0)
	&= f(x)	\\
u_{t}(x, 0)
	&= 0
\end{align*}
\end{column}
%
\begin{column}{0.50\textwidth}
\begin{align*}
v_{tt}
	&= c^{2} v_{xx}	\\
v(0, t)
	&= v(L, t)
	= 0	\\
v(x, 0)
	&= 0	\\
v_{t}(x, 0)
	&= g(x)
\end{align*}
\end{column}
\end{columns}

\vspace*{0.10in}

\textbf{Question}: what IBVP would $w(x, t) = u(x, t) + v(x, t)$ solve?

\uncover<2->{%
\begin{align*}
w_{tt}
	&= c^{2} w_{xx}
\mbox{ for $0 < x < L$ and $t > 0$}	\\
w(0, t)
	&= w(L, t)
	= 0	\\
w(x, 0)
	&= f(x)	\\
w_{t}(x, 0)
	&= g(x)
\end{align*}
}

\end{frame}
%%

\begin{frame}\frametitle{Example}

Find the solution to the IBVP:
\begin{align*}
u_{tt}
	&= u_{xx}
\mbox{ for $0 < x < 1$ and $t > 0$}	\\
u(0, t)
	&= u(1, t) = 0	\\
u(x, 0)
	&= \sin (2 \pi x)	\\
u_{t}(x, 0)
	&= \sin (\pi x)
\end{align*}

\end{frame}
%%

\begin{frame}\frametitle{Solution}

Let $f(x) = \sin (2 \pi x)$ and $g(x) = \sin (\pi x)$ which both odd
functions and $2$--periodic.
\begin{align*}
u(x, t)
	&= \frac{1}{2} \left( \sin (2 \pi (x+t)) + \sin (2 \pi (x-t)) \right) +
\frac{1}{2} \int_{x-t}^{x+t} \sin (\pi s) \, ds	\\
	&= \sin (2 \pi x) \cos (2 \pi t) + \frac{1}{\pi} \sin (\pi x) \sin (\pi t)
\end{align*}

\end{frame}
%%

\begin{frame}\frametitle{Graph}

\begin{center}
\resizebox{1.00\textwidth}{!}{\includegraphics{example04}}
\end{center}

\end{frame}
%%

%\begin{frame}\frametitle{Homework}

%\begin{itemize}
%\item Read Section $5.1$
%\item Exercises: 1--5
%\end{itemize}

%\end{frame}
%%

%%%
\end{document}

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%%%% EOF
%%%%
