%%%%		-*- 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:      03/19/2025
%%%%
%%%% Purpose:   Solving PDEs Using the Fourier Transform
%%%%

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\title{Solving PDEs Using the Fourier Transform}
\subtitle{\textit{Partial Differential Equations}} % (optional)


\author % (optional, use only with lots of authors)
{J~Robert~Buchanan}
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%   affiliation.

\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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\begin{document}

\begin{frame}
  \titlepage
\end{frame}

%\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
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% 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 use the Fourier transform and inverse Fourier transform to solve
  various partial differential equations.
\end{itemize}

\end{frame}
%%

\begin{frame}\frametitle{Solving the Unbounded Heat Equation}

Consider the heat equation with its spatial domain as the
entire real number line.
\begin{align*}
u_{t}
	&= \kappa u_{xx}
\mbox{ for $-\infty < x < \infty$ and $t > 0$}	\\
u(x, 0)
	&= e^{-x^{2}} \sin x
\mbox{ for $-\infty < x < \infty$}
\end{align*}

\end{frame}
%%

\begin{frame}\frametitle{Apply Fourier Transform}

\begin{align*}
u_{t}
	&= \kappa u_{xx}	\\
\mathcal{F}[ u_{t}(x, t) ](\omega)
	&= \kappa \mathcal{F}[ u_{xx}(x, t) ](\omega)	\\
\hat{u}_{t}(\omega, t)
	&= \kappa (i \omega)^{2} \hat{u}(\omega, t)	\\
\hat{u}_{t}(\omega, t)
	&= -\kappa \omega^{2} \hat{u}(\omega, t)
\end{align*}
This is now a first-order, linear ordinary differential equation with
unknown function $\hat{u}(\omega, t)$.

\end{frame}
%%

\begin{frame}\frametitle{Solve the Ordinary Differential Equation}

\begin{align*}
\hat{u}_{t}(\omega, t)
	&= -\kappa \omega^{2} \hat{u}(\omega, t)	\\
\hat{u}(\omega, t)
	&= \hat{u}(\omega, 0) e^{-\kappa \omega^{2} t}
\end{align*}
where $\hat{u}(\omega, 0)$ is the Fourier transform of the initial
condition $u_{0}(x)$.
\[
\hat{u}(\omega, 0)
	= \frac{i}{2 \sqrt{2}} \left( e^{-(\omega + 1)^{2}/4} -
   e^{-(\omega - 1)^{2}/4} \right)
\]

\end{frame}
%%

\begin{frame}\frametitle{Apply Inverse Fourier Transform}

\begin{align*}
\hat{u}(\omega, t)
	&= \frac{i}{2 \sqrt{2}} \left( e^{-(\omega + 1)^{2}/4} -
   e^{-(\omega - 1)^{2}/4} \right) e^{-\kappa \omega^{2} t}	\\
u(x, t)
	&= \frac{1}{\sqrt{2 \pi}} \int_{-\infty}^{\infty}
   \hat{u}(\omega, t) e^{i \omega x} \, d\omega	\\
	&= \frac{1}{\sqrt{1 + 4 \kappa t}} e^{-\frac{x^{2} + \kappa
   t}{1 + 4 \kappa t}} \sin
   \left( \frac{x}{1 + 4 \kappa t} \right)
\end{align*}

\end{frame}
%%

\begin{frame}\frametitle{Solving the Wave Equation}

Consider the initial value problem for the unbounded string.
\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*}
Solve this initial value problem using the Fourier transform.

\end{frame}
%%

\begin{frame}\frametitle{Apply Fourier Transform}

\begin{align*}
\mathcal{F}[ u_{tt} ](\omega)
	&= c^{2} \mathcal{F}[ u_{xx} ](\omega)	\\
\hat{u}_{tt}(\omega, t)
	&= -c^{2} \omega^{2} \hat{u}(\omega, t)	\\
\hat{u}_{tt}(\omega, t) + c^{2} \omega^{2} \hat{u}(\omega, t)
	&= 0
\end{align*}
This ordinary differential equation models simple harmonic motion
without damping.

\end{frame}
%%

\begin{frame}\frametitle{Solve the Ordinary Differential Equation}

\begin{align*}
\hat{u}_{tt}(\omega, t) + c^{2} \omega^{2} \hat{u}(\omega, t)
	&= 0	\\
\hat{u}(\omega, t)
	&= \hat{A} e^{-i c \omega t} + \hat{B} e^{i c \omega t}	,
\end{align*}
where $\hat{A}$ and $\hat{B}$ are functions of $\omega$ but constant with respect
to variable $t$.

\end{frame}
%%

\begin{frame}\frametitle{Apply Inverse Fourier Transform}

\begin{align*}
\hat{u}(\omega, t)
	&= \hat{A} e^{-i c \omega t} + \hat{B} e^{i c \omega t}	\\
\mathcal{F}^{-1}[ \hat{u}(\omega, t) ](x)
	&= \mathcal{F}^{-1} [ \hat{A}(\omega) e^{-i c \omega t} +
   \hat{B}(\omega) e^{i c \omega t} ](x)	\\
u(x, t)
	&= A(x - c t) + B(x + c t)
\end{align*}
Note that the inverse Fourier transform made use of the shifting
property of the Fourier transform

\end{frame}
%%

\begin{frame}\frametitle{Apply the Initial Conditions}

\begin{align*}
u(x, t)
	&= A(x - c t) + B(x + c t)	\\
u(x, 0)
	&= A(x) + B(x)
	= f(x)	\\
u_{t}(x, 0)
	&= -c A'(x) + c B'(x)
	= g(x)	.
\end{align*}
Integrating the second equation from $0$ to $x$ yields,
\[
-A(x) + B(x)
	= \frac{1}{c} \int_{0}^{x} g(s) \, ds	.
\]
Solving for $A(x)$ and $B(x)$ yields
\begin{align*}
A(x)
	&= \frac{1}{2} f(x) - \frac{1}{2 c} \int_{0}^{x} g(s) \, ds	\\
B(x)
	&= \frac{1}{2} f(x) + \frac{1}{2 c} \int_{0}^{x} g(s) \, ds	.
\end{align*}
Thus we find the d'Alembertian solution.
\[
u(x, t)
	= \frac{1}{2} \left( f(x-c t) + f(x+c t) \right) + \frac{1}{2
          c} \int_{x-c t}^{x+c t} g(s) \, ds
\]

\end{frame}
%%

\begin{frame}\frametitle{Solving Laplace's Equation}

Consider the boundary value problem below.
\begin{align*}
u_{xx} + u_{yy}
	&= 0
\mbox{ for $-\infty < x < \infty$ and $y > 0$}	\\
u(x, 0)
	&= \frac{1}{1 + x^{2}}
\mbox{ for $-\infty < x < \infty$}	\\
\lim_{x \to \infty} u(x, y)
	&= 0
\mbox{ for $y > 0$}	\\
\lim_{x \to -\infty} u(x, y)
	&= 0
\mbox{ for $y > 0$}	\\
\lim_{y \to \infty} u(x, y)
	&= 0
\mbox{ for $-\infty < x < \infty$}
\end{align*}
Note that the boundary conditions at infinity force the solution to be bounded.

\end{frame}
%%

\begin{frame}\frametitle{Apply Fourier Transform}

The Fourier transform can be applied to the partial differential
equation with respect to $x$.
\[
\mathcal{F}[ u_{xx}(x, y) + u_{yy}(x, y) ](\omega)
	= -\omega^{2} \hat{u}(\omega, y) + \hat{u}_{yy}(\omega, y)
	= 0
\]

\end{frame}
%%

\begin{frame}\frametitle{Solve the Ordinary Differential Equation}

\begin{align*}
\hat{u}_{yy}(\omega, y) - \omega^{2} \hat{u}(\omega, y)
	&= 0	\\
\hat{u}(\omega, y)
	= \hat{A}(\omega) e^{\omega y} + \hat{B}(\omega) e^{-\omega y}
\end{align*}
Functions $\hat{A}$ and $\hat{B}$ are constant with respect to $y$.

\end{frame}
%%

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

The boundary condition, $\lim_{y \to \infty} u(x, y) = 0$
implies $\lim_{y \to \infty} \hat{u}(\omega, y) = 0$.
Thus when $\omega > 0$, $\hat{A}(\omega) = 0$ and when $\omega < 0$,
$\hat{B}(\omega) = 0$.
Hence the Fourier transform of the solution may be written as a
piecewise-defined function,
\[
\hat{u}(\omega, y)
	= \left\{ \begin{array}{rl}
\hat{A}(\omega) e^{\omega y}	& \mbox{if $\omega < 0$}	\\
\hat{B}(\omega) e^{-\omega y}	& \mbox{if $\omega > 0$.}
\end{array} \right.
\]
The boundary condition along the $x$-axis implies
\[
\hat{u}(\omega, 0)
	= \mathcal{F}\left[ \frac{1}{1 + x^{2}} \right](\omega)
 	= \frac{\sqrt{\pi} e^{-|\omega|}}{\sqrt{2}}	.
\]
Combining this condition with the piecewise definition of
$\hat{u}(\omega, y)$ when $y = 0$ implies
\begin{align*}
\hat{A}(\omega)
	&= \left\{ \begin{array}{rl}
\frac{\sqrt{\pi} e^{\omega}}{\sqrt{2}}	& \mbox{if $\omega < 0$}	\\
0	& \mbox{if $\omega > 0$}
\end{array} \right.	\\
\hat{B}(\omega)
	&= \left\{ \begin{array}{rl}
0	& \mbox{if $\omega < 0$}	\\
\frac{\sqrt{\pi} e^{-\omega}}{\sqrt{2}}	& \mbox{if $\omega > 0$.}
\end{array} \right.
\end{align*}

\end{frame}
%%

\begin{frame}\frametitle{Apply Inverse Fourier Transform}

\begin{align*}
%\label{eq:FT:laplaceproduct}
\hat{u}(\omega, y)
	&= \frac{\sqrt{\pi} e^{-(1+y) |\omega|}}{\sqrt{2}}	\\
%\label{eq:FT:laplaceconvolution}
u(x, y)
	&= \mathcal{F}^{-1} \left[ \frac{\sqrt{\pi} e^{-(1+y)
   |\omega|}}{\sqrt{2}} \right](x)
	= \frac{1+y}{x^{2} + (1+y)^{2}}
\end{align*}

\end{frame}
%%

%\begin{frame}\frametitle{Homework}

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

%\end{frame}
%%

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