%%%%		-*- 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:   Fourier Sine and Cosine Transform
%%%%

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\title{Fourier Sine and Cosine 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
%Spring
%Summer
\the\year}

\subject{Talks}
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% the beginning of each subsection:
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  \begin{frame}<beamer>
    \frametitle{Outline}
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\begin{document}

\begin{frame}
  \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
% 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 define the Fourier cosine and sine transforms,
\item define the inverse Fourier cosine and sine transforms, and
\item use the Fourier cosine and sine transforms to solve partial
  differential equations on the half line and half plane.
\end{itemize}

\end{frame}
%%

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

\begin{definition}
If $f(x)$ is defined on $(0, \infty)$, then the \textbf{Fourier cosine
  transform} and the \textbf{inverse Fourier cosine transform} are defined
respectively as follows.
\begin{align*}
%\label{eq:FT:costrans}
\mathcal{F}_{c}[ f ](\omega)
	&= \hat{f}^{c}(\omega)
	= \sqrt{\frac{2}{\pi}} \int_{0}^{\infty} f(x) \cos (\omega x)
   \, dx	\\
%\label{eq:FT:invcostrans}
\mathcal{F}_{c}^{-1}[ F ](x)
	&= \sqrt{\frac{2}{\pi}} \int_{0}^{\infty} F(\omega) \cos (\omega x)
   \, d\omega
\end{align*}
\end{definition}
Both transforms require that the improper integrals converge.

\end{frame}
%%

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

\begin{definition}
If $f(x)$ is defined on $(0, \infty)$, then
the \textbf{Fourier sine transform} and the \textbf{inverse Fourier
  sine transform} are defined as
\begin{align*}
%\label{eq:FT:sintrans}
\mathcal{F}_{s}[ f ](\omega)
	&= \hat{f}^{s}(\omega)
	= \sqrt{\frac{2}{\pi}} \int_{0}^{\infty} f(x) \sin (\omega x)
   \, dx	\\
%\label{eq:FT:invsintrans}
\mathcal{F}_{s}^{-1}[ F ](x)
	&= \sqrt{\frac{2}{\pi}} \int_{0}^{\infty} F(\omega) \sin (\omega x)
   \, d\omega
\end{align*}
\end{definition}
\textbf{Remark}: the transformed function is even-extended for
the Fourier cosine transform and odd-extended for the Fourier sine
transform.

\end{frame}
%%

\begin{frame}\frametitle{Example}

Calculate the Fourier sine transform of $f(x) = e^{-x}$.

\end{frame}
%%

\begin{frame}\frametitle{Solution}

\begin{align*}
\mathcal{F}_{s}[ e^{-x} ](\omega)
	&= \sqrt{\frac{2}{\pi}} \int_{0}^{\infty} e^{-x} \sin (\omega
   x) \, dx	\\
	&= \sqrt{\frac{2}{\pi}} \left[ \frac{-e^{-x} (\omega \cos
   (\omega x) - \sin (\omega x))}{1 + \omega^{2}} \right]_{x=0}^{x \to \infty}	\\
	&= \sqrt{\frac{2}{\pi}} \frac{\omega}{1+\omega^{2}}	.
\end{align*}

\end{frame}
%%

\begin{frame}\frametitle{Properties}

\begin{align*}
\mathcal{F}_{c}[ f' ](\omega)
	&= \omega \mathcal{F}_{s}[ f ](\omega) - \sqrt{\frac{2}{\pi}} f(0)	\\
\mathcal{F}_{s}[ f' ](\omega)
	&= -\omega \mathcal{F}_{c}[ f ](\omega)	\\
%\label{eq:FT:costransfxx}
\mathcal{F}_{c}[ f'' ](\omega)
	&= -\omega^{2} \mathcal{F}_{c}[ f ](\omega) -
   \sqrt{\frac{2}{\pi}} f'(0)	\\
%\label{eq:FT:sintransfxx}
\mathcal{F}_{s}[ f'' ](\omega)
	&= -\omega^{2} \mathcal{F}_{s}[ f ](\omega) +
   \sqrt{\frac{2}{\pi}} \omega f(0)
\end{align*}

\end{frame}
%%

\begin{frame}\frametitle{Example}

Consider the heat equation on the half-line.
\begin{align*}
u_{t}
	&= \kappa u_{xx}
\mbox{ for $0 < x < \infty$ and $t > 0$}	\\
u(x, 0)
	&= x^{2} e^{-x}
\mbox{ for $0 < x < \infty$}	\\
u(0, t)
	&= 0
\mbox{ for $t > 0$}	\\
\end{align*}
Use a transform method to find the solution to the initial boundary
value problem.

\end{frame}
%%

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

Since the boundary value $u(0, t)$ is known, the Fourier sine
transform can be applied to this IBVP.
\begin{align*}
\mathcal{F}_{s}[ u_{t}(x, t) ](\omega)
	&= \mathcal{F}_{s}[ \kappa u_{xx}(x, t) ](\omega)	\\
\hat{u}_{t}^{s}(\omega, t)
	&= \kappa \left( -\omega^{2} \hat{u}^{s}(\omega, t) +
   \sqrt{\frac{2}{\pi}} \omega u(0, t) \right)	\\
	&= -\kappa \omega^{2} \hat{u}^{s}(\omega, t)	.
\end{align*}

\end{frame}
%%

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

\[
\hat{u}_{t}^{s}(\omega, t)
	= -\kappa \omega^{2} \hat{u}^{s}(\omega, t)
\]
The general solution to this first-order, linear ordinary differential
equation in variable $t$ with unknown $\hat{u}^{s}$ is
$\hat{u}^{s}(\omega, t) = A(\omega) e^{-\kappa \omega^{2} t}$.
The expression
\[
A(\omega)
	= \mathcal{F}_{s}[ u(x, 0) ](\omega)
	= \mathcal{F}_{s}[ x^{2} e^{-x} ](\omega)
	= 2 \sqrt{\frac{2}{\pi}} \left( \frac{3-\omega^{2}}{(1+\omega^{2})^{3}} \right)	.
\]

\end{frame}
%%

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

\begin{align*}
\hat{u}_{s}(\omega, t)
	&= 2 \sqrt{\frac{2}{\pi}} \left(
   \frac{3-\omega^{2}}{(1+\omega^{2})^{3}} \right) e^{-\kappa \omega^{2} t}	\\
u(x, t)
	&= \mathcal{F}_{s}^{-1}\left[ 2 \sqrt{\frac{2}{\pi}} \left(
   \frac{3-\omega^{2}}{(1+\omega^{2})^{3}} \right) e^{-\kappa \omega^{2} t} \right](x)	\\
	&= \frac{4}{\pi} \int_{0}^{\infty}
   \frac{(3-\omega^{2})}{(1+\omega^{2})^{3}} e^{-\kappa \omega^{2} t} \sin
   (\omega x) \, d\omega
\end{align*}

\end{frame}
%%

\begin{frame}\frametitle{Example}

Consider the following boundary value problem for Laplace's equation
on the positive quadrant of the Cartesian plane.
\begin{align*}
u_{xx} + u_{yy}
	&= 0
\mbox{ for $0 < x < \infty$ and $0 < y < \infty$}	\\
u(0, y)
	&= e^{-y} \sin y
\mbox{ for $0 < y < \infty$}	\\
u_{y}(x, 0)
	&= e^{-x^{2}} \sin x
\mbox{ for $0 < x < \infty$}	\\
\lim_{x \to \infty} u(x, y)
	&= 0
\mbox{ for $0 < y < \infty$}	\\
\lim_{y \to \infty} u(x, y)
	&= 0
\mbox{ for $0 < x < \infty$}
\end{align*}

\end{frame}
%%

\begin{frame}\frametitle{Solution (1 of 6)}

Let $u(x, y) = u_{1}(x, y) + u_{2}(x, y)$ where $u_{1}(x, y)$
satisfies the following boundary value problem.
\begin{align*}
(u_{1})_{xx} + (u_{1})_{yy}
	&= 0
\mbox{ for $0 < x < \infty$ and $0 < y < \infty$}	\\
u_{1}(0, y)
	&= e^{-y} \sin y
\mbox{ for $0 < y < \infty$}	\\
(u_{1})_{y}(x, 0)
	&= 0
\mbox{ for $0 < x < \infty$}	\\
\lim_{x \to \infty} u_{1}(x, y)
	&= 0
\mbox{ for $0 < y < \infty$}	\\
\lim_{y \to \infty} u_{1}(x, y)
	&= 0
\mbox{ for $0 < x < \infty$}
\end{align*}

\end{frame}
%%

\begin{frame}\frametitle{Solution (2 of 6)}

Calculate the Fourier cosine transform of the partial differential
equation with respect to the variable $y$.
\begin{align*}
\mathcal{F}_{c}[ (u_{1})_{xx} + (u_{1})_{yy} ](\omega)
	&= \mathcal{F}_{c}[ 0 ](\omega)	\\
\frac{d^{2}}{dx^{2}} [ \hat{u}_{1}^{c}(x, \omega) ] - \omega^{2}
  \hat{u}_{1}^{c}(x, \omega)
	&= 0
\end{align*}
The general solution to this ordinary differential equation is
\[
\hat{u}_{1}^{c}(x, \omega)
	= A(\omega) e^{-\omega x} + B(\omega) e^{\omega x}	,
\]
where $A$ and $B$ are currently functions of $\omega$ to be determined by
the boundary conditions.

\end{frame}
%%

\begin{frame}\frametitle{Solution (3 of 6)}

$\hat{u}_{1}^{c}(x, \omega) \to 0$ as $x \to \infty$ since
$\lim_{x \to \infty} u_{1}(x, y) = 0$.
This implies $B(\omega) = 0$ for all $\omega > 0$.
Hence $\hat{u}_{1}^{c}(x, \omega) = A(\omega) e^{-\omega x}$ for all
$x \ge 0$.

\vspace*{0.10in}

In particular when $x = 0$,
\[
\hat{u}_{1}(0, \omega)
	= A(\omega)
	= \mathcal{F}_{c}[ e^{-y} \sin y ](\omega)
	= \frac{4 - 2 \omega^{2}}{(4 + \omega^{4}) \sqrt{2 \pi}}
\]
which implies
\[
\hat{u}_{1}^{c}(x, \omega)
	= \frac{4 - 2 \omega^{2}}{(4 + \omega^{4}) \sqrt{2 \pi}}
        e^{-\omega x}	.
\]
Taking the inverse cosine transform yields
\[
u_{1}(x, y)
	= \mathcal{F}_{c}^{-1} [ \hat{u}_{1}^{c}(x, \omega) ](y)
	= \frac{1}{\pi} \int_{0}^{\infty} \frac{4 - 2
          \omega^{2}}{4 + \omega^{4}} e^{-\omega x}
        \cos (\omega y) \, d\omega	.
\]

\end{frame}
%%

\begin{frame}\frametitle{Solution (4 of 6)}

Function $u_{2}(x, y)$ must satisfy the following boundary value problem.
\begin{align*}
(u_{2})_{xx} + (u_{2})_{yy}
	&= 0
\mbox{ for $0 < x < \infty$ and $0 < y < \infty$}	\\
u_{2}(0, y)
	&= 0
\mbox{ for $0 < y < \infty$}	\\
(u_{2})_{y}(x, 0)
	&= e^{-x^{2}} \sin x
\mbox{ for $0 < x < \infty$}	\\
\lim_{x \to \infty} u_{2}(x, y)
	&= 0
\mbox{ for $0 < y < \infty$}	\\
\lim_{y \to \infty} u_{2}(x, y)
	&= 0
\mbox{ for $0 < x < \infty$.}
\end{align*}

\end{frame}
%%

\begin{frame}\frametitle{Solution (5 of 6)}

Taking the Fourier sine transform with respect to $x$ results in a
transformed general solution of the form,
\[
\hat{u}_{2}^{s}(\omega, y)
	= A(\omega) e^{-\omega y} + B(\omega) e^{\omega y}	.
\]
If $\omega > 0$ and
the solution is to remain bounded and asymptotically
approach zero as $y \to \infty$, then $B(\omega) = 0$ for all $\omega$.
Taking the partial derivative of $\hat{u}_{2}^{s}(\omega, y)$ with
respect to $y$, evaluating this partial derivative at $y = 0$,
and setting it equal to the Fourier sine transform of the
nonhomogeneous Neumann boundary condition results in
\begin{align*}
-\omega A(\omega)
	&= \mathcal{F}_{s}[ e^{-x^{2}} \sin x ](\omega)	\\
A(\omega)
	&= \frac{-1}{\omega \sqrt{2}} e^{-\frac{\omega^{2} + 1}{4}} \sinh \frac{\omega}{2}
\end{align*}
The solution $u_{2}(x, y)$ is the inverse Fourier sine transform of
the product of this expression and $e^{-\omega y}$.
\[
u_{2}(x, y)
	= \mathcal{F}_{s}^{-1}[ A(\omega) e^{-\omega y} ](x)
	= \frac{-1}{\sqrt{\pi}} \int_{0}^{\infty} \frac{1}{\omega}
        e^{-\frac{\omega^{2} + 4 y \omega + 1}{4}} \sinh \frac{\omega}{2} \sin
        (\omega x) \, d\omega
\]

\end{frame}
%%

\begin{frame}\frametitle{Solution (6 of 6)}

The solution to the original boundary value problem is assembled as
\begin{align*}
u(x, y)
	= {} & u_{1}(x, y) + u_{2}(x, y)	\\
	= {} & \frac{1}{\pi} \int_{0}^{\infty} \frac{4 - 2
          \omega^{2}}{4 + \omega^{4}} e^{-\omega x}
        \cos (\omega y) \, d\omega	\\
	&- \frac{1}{\sqrt{\pi}} \int_{0}^{\infty} \frac{1}{\omega}
        e^{-\frac{\omega^{2} + 4 y \omega + 1}{4}} \sinh \frac{\omega}{2} \sin
        (\omega x) \, d\omega	.
\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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