%%%%		-*- 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:      03/31/2025
%%%%
%%%% Purpose:   Solving Hyperbolic PDEs Using Green's Functions
%%%%

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\title{Solving Hyperbolic PDEs Using Green's Functions}
\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]
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{
  Department of Mathematics
}
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Fall
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\subject{Talks}
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\begin{frame}
  \titlepage
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%\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:
\begin{itemize}
\item develop Green's functions for hyperbolic partial differential
  equations, and
\item use Green's functions for solving the nonhomogeneous wave equation.
\end{itemize}

\end{frame}
%%

\begin{frame}\frametitle{d'Alembert's Solution}

Consider 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)
\mbox{ for $-\infty < x < \infty$}	\\
u_{t}(x, 0)
	&= g(x)
\mbox{ for $-\infty < x < \infty$,}
\end{align*}
whose solution can be expressed as
\[
u(x, t)
	= \frac{1}{2} [ f(x - c t) + f(x + c t) ] +
\frac{1}{2 c} \int_{x - c t}^{x + c t} g(\xi) \, d\xi	.
\]

\end{frame}
%%

\begin{frame}\frametitle{Green's Function}

\[
\frac{1}{2} [ f(x - c t) + f(x + c t) ]
	= \frac{1}{2} \int_{-\infty}^{\infty} f(\xi) [ \delta(\xi - (x - c
        t)) + \delta(\xi - (x + c t)) ] \, d\xi
\]
The Heaviside function allows the second term of d'Alembert's
solution (the ``struck string'') to be written in the form of an
integral over all the real numbers.
\[
\frac{1}{2 c} \int_{x - c t}^{x + c t} g(\xi) \, d\xi
	= \frac{1}{2 c} \int_{-\infty}^{\infty} g(\xi) [ H(\xi - (x - c t)) - H(\xi - (x
+ c t)) ] \, d\xi
\]
Define the function
\[
%\label{eq:green:wave:waveGreen}
G(x, t; \xi, \tau)
	= \frac{1}{2 c} H(c (t-\tau) - |x - \xi|)
\]
then d'Alembert's solution can be written as
\[
u(x, t)
	= \int_{-\infty}^{\infty} \left[ f(\xi) G_{\tau}(x, t; \xi, 0) -
g(\xi) G(x, t; \xi, 0) \right] \, d\xi	.
\]

\end{frame}
%%

\begin{frame}\frametitle{Nonhomogeneous Wave Equation}

%\label{eq:green:wave:infinitestring}
\begin{align*}
u_{tt} - c^{2} u_{xx}
	&= F(x, t)
\mbox{ for $-\infty < x < \infty$ and $t > 0$}	\\
u(x, 0)
	&= f(x)
\mbox{ for $-\infty < x < \infty$}	\\
u_{t}(x, 0)
	&= g(x)
\mbox{ for $-\infty < x < \infty$}
\end{align*}
Replace function $v$ in the adjoint equation with the Green's function.
\[
\iint_{R} L[u] G \, dA
	= \oint_{\partial R} [ c^{2} (u G_{\xi} - u_{\xi} G)
        \mathbf{i} + (u_{\tau} G - u G_{\tau}) \mathbf{j} ] \cdot \mathbf{N} \, ds +
\iint_{R} u L^{\ast}[G] \, dA	.
\]
The solution can be written as
%\label{eq:green:wave:unboundedstringsoln}
\begin{align*}
u(x, t)
	= {} & \int_{-\infty}^{\infty} \left[ f(\xi) G_{\tau}(x, t; \xi, 0) - g(\xi) G(x,
t; \xi, 0) \right] \, d\xi	\\
	&+ \int_{0}^{t} \int_{-\infty}^{\infty} F(\xi, \tau) G(x, t;
   \xi, \tau) \, d\xi \, d\tau	.
\end{align*}

\end{frame}
%%

\begin{frame}\frametitle{Semi-Infinite Wave Equation}

%\label{eq:green:wave:semiinfiniteIBVP}
\begin{align*}
u_{tt} - c^{2} u_{xx}
	&= F(x, t)
\mbox{ for $x > 0$ and $t > 0$}	\\
u(x, 0)
	&= f(x)
\mbox{ for $x > 0$}	\\
u_{t}(x, 0)
	&= g(x)
\mbox{ for $x > 0$}	\\
u(0, t)
	&= h(t)
\mbox{ for $t > 0$.}
\end{align*}
The Green's function for the infinite string placed a source at $x$.
For the semi-infinite wave equation suppose in addition to the source
at $x$ a source opposite in sign is placed at $-x$.
Define the function
\[
%\label{eq:green:wave:semiinfiniteGreen}
G(x, t; \xi, \tau)
	= \frac{H(c (t-\tau) - |x - \xi|) - H(c (t-\tau) - |x + \xi|)}{2 c}	.
\]

\end{frame}
%%

\begin{frame}\frametitle{Domain}

\begin{center}
\resizebox{1.00\textwidth}{!}{\includegraphics{semiinfinitewave}}
\end{center}
%\label{eq:green:wave:semiinfinitewavesoln}
\begin{align*}
u(x, t)
	= {} & \int_{0}^{t} \int_{0}^{\infty} F(\xi, \tau) G(x, t;
        \xi, \tau) \, d\xi \, d\tau	\\
	&+ c^{2} \int_{0}^{t} h(\tau) G_{\xi}(x, t; 0, \tau) \, d\tau	\\
	&+ \int_{0}^{\infty} \left( f(\xi) G_{\tau}(x, t; \xi, 0) - g(\xi)
          G(x, t; \xi, 0) \right) \, d\xi	.
\end{align*}

\end{frame}
%%

\begin{frame}\frametitle{Bounded String}

Consider the forced wave equation on the finite domain $[0, L]$.
%\label{eq:green:wave:finitewaveeqn}
\begin{align*}
u_{tt} - c^{2} u_{xx}
	&= F(x, t)
\mbox{ for $0 < x < L$ and $t > 0$}	\\
\alpha_{1} u(0, t) + \beta_{1} u_{x}(0, t)
	&= 0
\mbox{ for $0 < t$}	\\
\alpha_{2} u(L, t) + \beta_{2} u_{x}(L, t)
	&= 0
\mbox{ for $0 < t$}	\\
u(x, 0)
	&= 0
\mbox{ for $0 < x < L$}	\\
u_{t}(x, 0)
	&= 0
\mbox{ for $0 < x < L$.}
\end{align*}

\end{frame}
%%

\begin{frame}\frametitle{Eigenfunction Expansion}

Suppose the solution can be
written in the form
\[
u(x, t)
	= \sum_{n=1}^{\infty} T_{n}(t) \phi_{n}(x)	,
\]
where $\phi_{n}(x)$ is the $n$th orthonormal eigenfunction of the Sturm-Liouville
boundary value problem,
\begin{align*}
\phi''(x) + \lambda^{2} \phi(x)
	&= 0
\mbox{ for $0 < x < L$}	\\
\alpha_{1} \phi(0) + \beta_{1} \phi'(0)
	&= 0	\\
\alpha_{2} \phi(L) + \beta_{2} \phi'(L)
	&= 0	.
\end{align*}

\end{frame}
%%

\begin{frame}\frametitle{Eigenfunction Expansion}
\fontsize{9pt}{10.8}\selectfont

Substitute this solution into
the differential equation, use the eigenfunctions and
eigenvalues, and integrate over the interval $0 \le x \le L$ to produce
\begin{align*}
\sum_{n=1}^{\infty} T_{n}''(t) \phi_{n}(x) - c^{2} \sum_{n=1}^{\infty}
  T_{n}(t) \phi_{n}''(x)
	&= F(x, t)	\\
\int_{0}^{L} \sum_{n=1}^{\infty} \left( T_{n}''(t) + c^{2} \lambda_{n}^{2} T_{n}(t)
  \right) \phi_{n}(x) \phi_{m}(x) \, dx
	&= \int_{0}^{L} F(x, t) \phi_{m}(x) \, dx	\\
T_{m}''(t) + c^{2} \lambda_{m}^{2} T_{m}(t)
	&= F_{m}(t)	.
\end{align*}
This time dependent function must satisfy the initial conditions
$T_{m}(0) = T_{m}'(0) = 0$.

\vspace*{0.10in}

Using the method of variation of parameters the appropriate solution is
\[
T_{m}(t)
	= \int_{0}^{t} \frac{F_{m}(\tau) \sin (c \lambda_{m} (t -
          \tau))}{c \lambda_{m}} \, d\tau	.
\]
Substituting this into the series solution reveals
\[
u(x, t)
	= \int_{0}^{t} \int_{0}^{L} F(\xi, \tau) \left(
   \sum_{n=1}^{\infty} \frac{\sin (c \lambda_{n} (t - \tau))
   \phi_{n}(\xi) \phi_{n}(x)}{c \lambda_{n}} \right) \, d\xi \, d\tau	.
\]

\end{frame}
%%

\begin{frame}\frametitle{Green's Function}

%\label{eq:green:wave:finitewaveinit}
\begin{align*}
u_{tt} - c^{2} u_{xx}
	&= F(x, t)
\mbox{ for $0 < x < L$ and $t > 0$}	\\
u(0, t)
	&= 0
\mbox{ for $t > 0$}	\\
u(L, t)
	&= 0
\mbox{ for $t > 0$}	\\
u(x, 0)
	&= f(x)
\mbox{ for $0 < x < L$}	\\
u_{t}(x, 0)
	&= g(x)
\mbox{ for $0 < x < L$,}
\end{align*}
The Green's function is
\[
%\label{eq:green:wave:greenfiniteinit}
G(x, t; \xi, \tau)
	= \frac{2}{L} \sum_{n=1}^{\infty} \sin \left( \frac{n \pi x}{L}
        \right) \sin \left( \frac{n \pi \xi}{L} \right) \frac{L}{n \pi
          c} \sin \left( \frac{n \pi c (t - \tau)}{L} \right)	.
\]
The solution can be written as
%\label{eq:green:wave:finiteinitsoln}
\begin{align*}
u(x, t)
	= {} & \int_{0}^{t} \int_{0}^{L} G(x, t; \xi, \tau) F(\xi, \tau) \,
        d\xi \, d\tau	\\
	&+ \int_{0}^{L} G_{\tau}(x, t; \xi, 0) f(\xi) \,
          d\xi + \int_{0}^{L} G(x, t; \xi, 0) g(\xi) \, d\xi	.
\end{align*}

\end{frame}
%%

\begin{frame}\frametitle{Nonhomogeneous Boundary Conditions}

%\label{eq:green:wave:finitewavefullynonhomogeneous}
\begin{align*}
u_{tt} - c^{2} u_{xx}
	&= F(x, t)
\mbox{ for $0 < x < L$ and $t > 0$}	\\
u(0, t)
	&= \alpha(t)
\mbox{ for $t > 0$}	\\
u(L, t)
	&= \beta(t)
\mbox{ for $t > 0$}	\\
u(x, 0)
	&= f(x)
\mbox{ for $0 < x < L$}	\\
u_{t}(x, 0)
	&= g(x)
\mbox{ for $0 < x < L$.}
\end{align*}

\end{frame}
%%

\begin{frame}\frametitle{Reference Solution}

A reference solution is $r(x, t) = (\beta(t) - \alpha(t)) x/L + \alpha(t)$.
If $u(x, t) = r(x, t) + v(x, t)$ then the unknown function $v(x, t)$
satisfies the initial boundary value problem
\begin{align*}
v_{tt} - c^{2} v_{xx}
	&= F(x, t) + (\alpha''(t) - \beta''(t)) \frac{x}{L} - \alpha''(t)
\mbox{ for $0 < x < L$ and $t > 0$}	\\
v(0, t)
	&= 0
\mbox{ for $t > 0$}	\\
v(L, t)
	&= 0
\mbox{ for $t > 0$}	\\
v(x, 0)
	&= f(x) + (\alpha(0) - \beta(0)) \frac{x}{L} - \alpha(0)
\mbox{ for $0 < x < L$}	\\
v_{t}(x, 0)
	&= g(x) + (\alpha'(0) - \beta'(0)) \frac{x}{L} - \alpha'(0)
\mbox{ for $0 < x < L$,}
\end{align*}
which is of the form shown previously and
thus $v(x, t)$ can calculated using the previous formula.

\end{frame}
%%

%\begin{frame}\frametitle{Homework}

%\begin{itemize}
%\item Read Section $x.y$
%\item Exercises: 
%\end{itemize}

%\end{frame}
%%

%%%
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%%%%
