%%%%		-*- 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:      03/21/2025
%%%%
%%%% Purpose:   Green's Functions for Initial Value Problems
%%%%

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\title{Green's Functions for Initial Value Problems}
\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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% - Keep it simple, no one is interested in your street address.

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

\subject{Talks}
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  \titlepage
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%\begin{frame}
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%\section{Introduction}

\begin{frame}\frametitle{Objectives}

In this lesson we will:
\begin{itemize}
\item develop the Green's function for the $n$th-order initial value problem,
\item use Green's functions to solve $n$th-order initial value problems.
\end{itemize}

\end{frame}
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\begin{frame}\frametitle{$n$th-Order Initial Value Problem}

For $n \ge 2$ the general $n$th order initial value problem is assumed
here to have the form,
%\label{eq:green:IVP:generalIVP}
\begin{align*}
u^{(n)}(t) + a_{n-1}(t) u^{(n-1)}(t) +
\cdots + a_{1}(t) u'(t) + a_{0}(t) u(t)
	&= f(t)	\\
u(0)
	= u'(0)
	= \cdots
	= u^{(n-1)}(0)
	&= 0	.
\end{align*}
The left-hand side of the ordinary
differential equation will be denoted $L[u](t)$, another linear
differential operator.
If the coefficient functions $a_{0}(t)$, $a_{1}(t)$, \ldots,
$a_{n-1}(t)$, and $f(t)$ are continuous on an interval, say $[0, T]$,
then a unique solution to the IVP exists on that interval.

\vspace*{0.10in}

\uncover<2->{%
\textbf{Remarks}:
\begin{itemize}
\item This is not the most general $n$th order IVP, but more general forms
can usually be re-written in this form.
\item The initial conditions are chosen to be all zero so that the solution
to this IVP will be what is often called a ``particular'' solution.
\end{itemize}
}

\end{frame}
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\begin{frame}\frametitle{Green's Function}
\fontsize{9pt}{10.8}\selectfont

A Green's function for the initial value problem will be
\[
G(t; \tau)
	= \left\{ \begin{array}{rl}
G_{1}(t; \tau)
	& \mbox{if $0 \le t < \tau < \infty$,}	\\
G_{2}(t; \tau)
	& \mbox{if $0 \le \tau < t < \infty$.}
\end{array} \right.
\]
\begin{itemize}
\item $G(t; \tau)$ and its first $n-2$ partial derivatives with
respect to $t$ must be continuous on $(0, \infty) \times (0, \infty)$.
\[
G_{2}(\tau; \tau)
	= \partial_{t}G_{2}(\tau; \tau)
	= \cdots
	= \partial_{t}^{n-2}G_{2}(\tau; \tau)
	= 0	.
\]
The $(n-1)$st partial derivative with respect to $t$ is continuous on
the region where $0 \le t < \tau < \infty$, is continuous on the
region where $0 \le \tau < t < \infty$, and has a jump
discontinuity of magnitude $1$ across the line $t = \tau$, since
\[
1
	= \lim_{t \to \tau^{+}} \partial_{t}^{n-1}G(t; \tau) -
\lim_{t \to \tau^{-}} \partial_{t}^{n-1}G(t; \tau)
	= \partial_{t}^{n-1}G(\tau; \tau) - 0
	= \partial_{t}^{n-1}G(\tau; \tau)	.
\]
Thus $L^{\ast}[G] = \delta(\tau - t)$.
\end{itemize}

\end{frame}
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\begin{frame}\frametitle{Solution Using a Green's Function}

The solution to the initial value problem is determined by an integral.
\begin{align*}
u(t)
	&= \int_{0}^{\infty} \delta(\tau - t) u(\tau) \, d\tau	\\
	&= \int_{0}^{\infty} L^{\ast}[G] u(\tau) \, d\tau	\\
	&= \int_{0}^{\infty} G(t; \tau) L[u](\tau) \, d\tau	\\
	&= \int_{0}^{\infty} G(t; \tau) f(\tau) \, d\tau
\end{align*}

\end{frame}
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\begin{frame}\frametitle{Finding a Green's Function}

There exists a fundamental set of solutions $\{ u_{1}, u_{2}, \ldots,
u_{n} \}$ such that 
\[
G(t; \tau)
	= C_{1} u_{1}(t) + C_{2} u_{2}(t) + \cdots + C_{n} u_{n}(t)
\]
where the coefficients $C_{1}$, $C_{2}$, \ldots, $C_{n}$ are functions
of $\tau$ chosen so that $G(t; \tau)$ satisfies the continuity and
jump conditions.

\vspace*{0.10in}

Once the Green's function is determined, the solution to the original
initial value problem is found via an integral.
\[
u(t)
	= \int_{0}^{\infty} G(t; \tau) f(\tau) \, d\tau
	= \int_{0}^{t} G(t; \tau) f(\tau) \, d\tau
\]
The second equality holds since $G(t; \tau) = 0$ for $t < \tau$ by
construction.

\end{frame}
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\begin{frame}\frametitle{Causality Condition}

The solution $u(t)$ depends only on the values of the
nonhomogeneous forcing function $f$ at times prior to $t$.
The solution is said to obey a \textbf{causality
  condition} which agrees with physical
intuition.
\begin{center}
\resizebox{1.00\textwidth}{!}{\includegraphics{causality}}
\end{center}

\end{frame}
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\begin{frame}\frametitle{Example}

Use the Green's function approach to solve the following initial value problem.
\begin{align*}
u'' - 4 u
	&= \cos t
\mbox{ for $t > 0$}	\\
u(0)
	= u'(0)
	&= 0
\end{align*}

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

\begin{itemize}
\item On the interval $0 \le \tau < t$ the Green's function
is $G(t; \tau) = C_{1} e^{2 t} + C_{2} e^{-2 t}$ where the coefficients $C_{1}$
and $C_{2}$ are functions of $\tau$.
\item If $G(\tau; \tau) = 0$ then $C_{1} e^{2 \tau} = -C_{2} e^{-2 \tau}$.
\item To satisfy the jump discontinuity requirement,
\begin{align*}
2 C_{1} e^{2 \tau} - 2 C_{2} e^{-2 \tau}
	&= 1	\\
2 C_{1} e^{2 \tau} + 2 C_{1} e^{2 \tau}
	&= 1	\\
C_{1}
	&= \frac{1}{4} e^{-2 \tau}
\end{align*}
which implies $C_{2} = \frac{-1}{4} e^{2 \tau}$.
\item The Green's function can be expressed as
\[
G(t; \tau)
	= \left\{ \begin{array}{rl}
0
	& \mbox{if $0 \le t < \tau < \infty$,}	\\
-\frac{1}{2} \sinh(2 (\tau - t))
	& \mbox{if $0 \le \tau < t < \infty$.}
\end{array} \right.
\]
\end{itemize}

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

The solution to the original initial value problem can be found as
\begin{align*}
u(t)
	&= \int_{0}^{\infty} G(t; \tau) \cos \tau \, d\tau	\\
	&= -\frac{1}{2} \int_{0}^{t} \sinh (2(\tau-t)) \cos \tau \, d\tau	\\
	&= \left[ \frac{-1}{5} \cosh (2(\tau-t)) \cos \tau -
   \frac{1}{10} \sinh (2(\tau-t)) \sin \tau \right]_{\tau=0}^{\tau=t}	\\
	&= \frac{1}{5} \cosh (2 t) - \frac{1}{5} \cos t	.
\end{align*}

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

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

%\end{frame}
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