%%%%		-*- 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:      03/20/2025
%%%%
%%%% Purpose:   Fourier Transform in Higher Dimensions
%%%%

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\title{Fourier Transform in Higher Dimensions}
\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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\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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% - 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 extend the definitions of the Fourier transform and its inverse
  to multidimensional settings,
\item explore the properties of the extended Fourier transform, and
\item use the Fourier transform in several examples.
\end{itemize}

\end{frame}
%%

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

\begin{definition}
If $f(x_{1}, x_{2}, \ldots, x_{n})$ is absolutely integrable over
$\mathbb{R}^{n}$ then the multidimensional Fourier transform of $f$ is
defined as
%\label{eq:FT:higherdim}
\begin{align*}
\hat{f}(\bm{\omega})
	&= \mathcal{F}[ f ](\bm{\omega})	\\
	&= \left( \frac{1}{\sqrt{2 \pi}} \right)^{n}
        \int_{\mathbb{R}^{n}}
        f(x_{1}, \ldots, x_{n}) e^{-i \omega_{1} x_{1} - \cdots - i \omega_{n} x_{n}} \, dx_{1}
        \cdots dx_{n}	.
\end{align*}
The inverse transform is
%\label{eq:FT:higherdiminv}
\begin{align*}
f(\bm{x})
	&= \mathcal{F}^{-1}[ \hat{f} ](\bm{x})	\\
	&= \left( \frac{1}{\sqrt{2 \pi}} \right)^{n}
        \int_{\mathbb{R}^{n}}
        \hat{f}(\omega_{1}, \ldots, \omega_{n}) e^{i \omega_{1} x_{1}
          + \cdots + i \omega_{n} x_{n}} \, d\omega_{1} \cdots d\omega_{n}	.
\end{align*}
\end{definition}

\end{frame}
%%

\begin{frame}\frametitle{Vector Notation}

Adopting the vector notation enables the transform and its inverse to
be written concisely as follows
\begin{align*}
%\label{eq:FT:higherdimvector}
\hat{f}(\bm{\omega})
	&= \left( \frac{1}{\sqrt{2 \pi}} \right)^{n}
        \int_{\mathbb{R}^{n}} f(\bm{x}) e^{-i \bm{\omega}
   \cdot \bm{x}} \, d\bm{x}	\\
%\label{eq:FT:higherdiminvvector}
f(\bm{x})
	&= \left( \frac{1}{\sqrt{2 \pi}} \right)^{n}
        \int_{\mathbb{R}^{n}} \hat{f}(\bm{\omega}) e^{i \bm{\omega}
   \cdot \bm{x}} \, d\bm{\omega}	.
\end{align*}
The expression $\bm{\omega} \cdot \bm{x}$ is the dot product or
inner product of the vectors $\bm{\omega}$ and $\bm{x}$.

\end{frame}
%%

\begin{frame}\frametitle{Example}

Find the Fourier transform of the function
\[
f(x_{1}, x_{2})
	= \left\{ \begin{array}{rl}
1	& \mbox{for $(x_{1}, x_{2}) \in (-1, 1) \times (-1, 1)$}	\\
0	& \mbox{otherwise.}
\end{array} \right.
\]

\end{frame}
%%

\begin{frame}\frametitle{Solution}

\begin{align*}
\hat{f}(\omega_{1}, \omega_{2})
	&= \frac{1}{2 \pi} \int_{-\infty}^{\infty}
   \int_{-\infty}^{\infty} f(x_{1}, x_{2}) e^{-i \omega_{1} x_{1} - i
   \omega_{2} x_{2}} \, dx_{1} \, dx_{2}	\\
	&= \frac{1}{2 \pi} \int_{-1}^{1}
   \int_{-1}^{1} e^{-i \omega_{1} x_{1} - i
   \omega_{2} x_{2}} \, dx_{1} \, dx_{2}	\\
	&= \frac{1}{2 \pi} \int_{-1}^{1} e^{-i \omega_{1} x_{1}} \, dx_{1}
\int_{-1}^{1} e^{-i \omega_{2} x_{2}} \, dx_{2}	\\
	&= \frac{2}{\pi} \frac{\sin \omega_{1} \sin
   \omega_{2}}{\omega_{1} \omega_{2}}	.
\end{align*}

\end{frame}
%%

\begin{frame}\frametitle{Illustration}

\begin{center}
\resizebox{0.48\textwidth}{!}{\includegraphics{unitstep}}%
\hfill%
\resizebox{0.48\textwidth}{!}{\includegraphics{unitsteptransform}}
\end{center}
The surface on the left is the graph of the two-dimensional unitstep
  function and
the surface on the right is the graph of its Fourier transform in the
$(\omega_{1}, \omega_{2})$-plane.

\end{frame}
%%

\begin{frame}\frametitle{Shifting Property}

\begin{theorem}
%\label{thm:FT:higherdim:shifting}
Suppose function $f(\bm{x})$ has Fourier transform
$\hat{f}(\bm{\omega})$ and that $\bm{a}$ is a
constant vector in $\mathbb{R}^{n}$, then
\begin{align*}
%\label{eq:FT:higherdim:shiftf}
\mathcal{F}[ f(\bm{x}-\bm{a}) ](\bm{\omega})
	&= e^{-i \bm{\omega} \cdot \bm{a}} \hat{f}(\bm{\omega})	\\
%\label{eq:FT:higherdim:shiftfhat}
\mathcal{F}[ f(\bm{x}) e^{-i \bm{a} \cdot \bm{x}} ](\bm{\omega})
	&= \hat{f}(\bm{\omega}+\bm{a})	.
\end{align*}
\end{theorem}

\end{frame}
%%

\begin{frame}\frametitle{Derivative Property}

\begin{theorem}
%\label{thm:FT:higherdim:derivative}
Suppose that $f \, : \, \mathbb{R}^{n} \to \mathbb{R}$ is absolutely
integrable over $\mathbb{R}^{n}$, that all necessary partial
derivatives of $f$ are as well, and that $| f(\bm{x}) |$ and its
partial derivatives vanish as $\| \bm{x} \| \to \infty$.
\[
%\label{eq:FT:higherdim:derivative}
\mathcal{F}\left[
\frac{\partial^{k_{1} + k_{2} + \cdots + k_{n}}}{\partial x_{1}^{k_{1}}
  \partial x_{2}^{k_{2}} \cdots \partial x_{n}^{k_{n}}} [
f(\bm{x}) ] \right](\bm{\omega})
	= i^{k_{1} + k_{2} + \cdots + k_{n}} \omega_{1}^{k_{1}}
        \omega_{2}^{k_{2}} \cdots \omega_{n}^{k_{n}} \hat{f}(\bm{\omega})
\]
\end{theorem}

\end{frame}
%%

\begin{frame}\frametitle{Fourier Convolution}

This Fourier convolution is defined as
\[
%\label{eq:FT:higherdim:convolution}
(f \ast g)(\bm{x})
	= \left( \frac{1}{\sqrt{2 \pi}} \right)^{n}
        \int_{\mathbb{R}^{n}} f(\bm{x}-\bm{z}) g(\bm{z})
        \, d\bm{z}	.
\]
%%
\begin{theorem}[Fourier Convolution Theorem]
%\label{thm:FT:higherdim:convolution}
Suppose functions $f$ and $g$ have Fourier transforms $\hat{f}$ and
$\hat{g}$ respectively, then
\[
%\label{eq:FT:higherdim:transconvolution}
\mathcal{F}[ f \ast g ](\bm{\omega})
	= \hat{f}(\bm{\omega}) \hat{g}(\bm{\omega})	,
\]
in other words the Fourier transform of the Fourier
convolution of $f$ and $g$ is the product of the Fourier transforms of
$f$ and $g$.
\end{theorem}

\end{frame}
%%

\begin{frame}\frametitle{Application: Newton's Theory of Gravitation}

Let $\bm{x} \in \mathbb{R}^{3}$ and suppose $\phi(\bm{x})$ is
the gravitational potential at position $\bm{x}$.
The partial differential equation modeling this situation is Poisson's
equation taking the form:
\[
\Delta \phi(\bm{x})
	= -4 \pi G \rho(\bm{x})	.
\]
The notation $\Delta \phi(\bm{x})$ is a shorthand for $\phi_{xx}
+ \phi_{yy} + \phi_{zz}$.
The constant $G$ is the universal gravitational
constant.
The function $\rho(\bm{x})$ is the mass density at position
$\bm{x}$.

\end{frame}
%%

\begin{frame}\frametitle{Fourier Transform}

\begin{align*}
\phi_{xx} + \phi_{yy} + \phi_{zz}
	&= -4 \pi G \rho(\bm{x})	\\
\mathcal{F}[\phi_{xx} + \phi_{yy} + \phi_{zz}](\omega)
	&= -4 \pi G \mathcal{F}[\rho(\bm{x})](\omega)	\\
-(\omega_{1}^{2} + \omega_{2}^{2} + \omega_{3}^{2}) \hat{\phi}(\bm{\omega})
	&= -4 \pi G \hat{\rho}(\bm{\omega})	\\
\hat{\phi}(\bm{\omega})
	&= \frac{4 \pi G}{\bm{\omega} \cdot \bm{\omega}} \hat{\rho}(\bm{\omega})
\end{align*}

\end{frame}
%%

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

\begin{align*}
\hat{\phi}(\bm{\omega})
	&= \frac{4 \pi G}{\bm{\omega} \cdot \bm{\omega}} \hat{\rho}(\bm{\omega})	\\
\phi(\bm{x})
	&= \left( \frac{1}{\sqrt{2 \pi}} \right)^{3}
        \int_{\mathbb{R}^{3}} \frac{4 \pi G}{\bm{\omega} \cdot
          \bm{\omega}} \hat{\rho}(\bm{\omega}) e^{i \bm{\omega} \cdot
          \bm{x}} \, d\bm{\omega}
\end{align*}

\end{frame}
%%

\begin{frame}\frametitle{Example: Heat Equation}

Consider the initial value problem for heat diffusion in
three-dimensional space.
\begin{align*}
u_{t}
	&= \kappa \Delta u
\mbox{ for $\bm{x} \in \mathbb{R}^{3}$ and $t > 0$}	\\
u(\bm{x}, 0)
	&= \delta(\bm{x})
\mbox{ for $\bm{x} \in \mathbb{R}^{3}$}
\end{align*}

\end{frame}
%%

\begin{frame}\frametitle{Solution (1 of 3)}

The Fourier transform of the partial differential equation results in
the ordinary differential equation,
\[
\hat{u}_{t}(\bm{\omega}, t)
	= -\kappa (\omega_{1}^{2} + \omega_{2}^{2} + \omega_{3}^{2})
        \hat{u}(\bm{\omega}, t)	.
\]
This is a first-order, linear ordinary differential equation whose
general solution is
\[
\hat{u}(\bm{\omega}, t)
	= \hat{u}(\bm{\omega}, 0) e^{-\kappa (\bm{\omega}  \cdot
          \bm{\omega}) t}	.
\]
When $t=0$ the left-hand side of this equation will equal the Fourier
transform of the Dirac delta function.
\[
\hat{u}(\bm{\omega}, 0)
	= \frac{1}{(2 \pi)^{3/2}}
\]

\end{frame}
%%

\begin{frame}\frametitle{Solution (2 of 3)}

\[
\hat{u}(\bm{\omega}, t)
	= \frac{1}{(2 \pi)^{3/2}}
e^{-\kappa (\bm{\omega}  \cdot \bm{\omega}) t}	.
\]
Inverse transform this expression.
\begin{align*}
u(\bm{x}, t)
	&= \left( \frac{1}{\sqrt{2 \pi}} \right)^{3}
   \int_{\mathbb{R}^{3}} \frac{1}{(2 \pi)^{3/2}}
e^{-\kappa (\bm{\omega}  \cdot \bm{\omega}) t} e^{i \bm{\omega} \cdot
   \bm{x}} \, d\bm{\omega}	\\
	&= \left( \frac{1}{2 \pi} \right)^{3} e^{-\frac{\bm{x}
   \cdot \bm{x}}{4 \kappa t}}
   \int_{\mathbb{R}^{3}} e^{-\kappa t \left( \omega_{1} - \frac{i x_{1}}{2 \kappa t} \right)^{2}}
e^{-\kappa t \left( \omega_{2} - \frac{i x_{2}}{2 \kappa t} \right)^{2}}
e^{-\kappa t \left( \omega_{3} - \frac{i x_{3}}{2 \kappa t} \right)^{2}} \, d\bm{\omega}
\end{align*}

\end{frame}
%%

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

\begin{align*}
u(\bm{x}, t)
	&= \left( \frac{1}{2 \pi} \right)^{3} e^{-\frac{\bm{x}
   \cdot \bm{x}}{4 \kappa t}}
   \int_{\mathbb{R}^{3}} e^{-\kappa t \left( \omega_{1} - \frac{i x_{1}}{2 \kappa t} \right)^{2}}
e^{-\kappa t \left( \omega_{2} - \frac{i x_{2}}{2 \kappa t} \right)^{2}}
e^{-\kappa t \left( \omega_{3} - \frac{i x_{3}}{2 \kappa t} \right)^{2}} \, d\bm{\omega}	\\
U(\bm{x}, t)
	&= \frac{1}{(4 \pi \kappa t)^{3/2}} e^{-\frac{\bm{x} \cdot \bm{x}}{4 \kappa t}}
\end{align*}
\textbf{Remark}: this is the fundamental solution to the heat equation
in three dimensions.

\end{frame}
%%

%\begin{frame}\frametitle{Homework}

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

%\end{frame}
%%

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