%%%%		-*- 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:      02/12/2025
%%%%
%%%% Purpose:   Heat Equation on Unbounded Intervals
%%%%

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\title{Heat Equation on Unbounded Intervals}
\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]
% (optional, but mostly needed)
{
  Department of Mathematics
}
% - Use the \inst command only if there are several affiliations.
% - Keep it simple, no one is interested in your street address.

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

\subject{Talks}
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% out. 



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% the beginning of each subsection:
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{
  \begin{frame}<beamer>
    \frametitle{Outline}
    \tableofcontents[currentsection,currentsubsection]
  \end{frame}
}


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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 learn about:
\begin{itemize}
\item the fundamental solution to the heat equation,
\item solutions to the heat equation for $0 \le x < \infty$, and
\item solutions to the heat equation for $-\infty < x < \infty$.
\end{itemize}

\end{frame}
%%

\begin{frame}\frametitle{Fundamental Solution}

For $t > 0$ define the function
\[
U(x, t)
	= \frac{1}{\sqrt{4 \pi \kappa t}} e^{-x^{2}/(4 \kappa t)}	.
\]

\vspace*{0.10in}

\textbf{Remarks}:
\begin{itemize}
\item $U(x, t)$ is related to the probability density function for a
  normally distributed random variable.
\item While defined only for $t > 0$, the limit as $t \to 0^{+}$
  exists.
\item $U(x, t)$ solves the heat equation.
\end{itemize}

\end{frame}
%%

\begin{frame}\frametitle{Connection to Normal Distribution}

A normally distributed, continuous random variable $X$
with mean $\mu$ and
standard deviation $\sigma$ has a probability distribution of
\[
f_{X}(x)
	= \frac{1}{\sigma \sqrt{2 \pi}} e^{-(x - \mu)^{2}/(2
  \sigma^{2})}
\mbox{ for $-\infty < x < \infty$.}
\]

\uncover<2->{%
Consider the fundamental solution to the heat equation,
\[
U(x, t)
	= \frac{1}{\sqrt{4 \pi \kappa \, t}} e^{-x^{2}/(4 \kappa t)}
	= \frac{1}{\sqrt{2 \kappa \, t} \, \sqrt{2 \pi}} e^{-x^{2}/(2 (\sqrt{2 \kappa \, t})^{2})}
	.
\]
For every $t > 0$ the heat energy is distributed normally with mean
$\mu = 0$ and standard deviation $\sigma = \sqrt{2 \kappa \, t}$.
}

\end{frame}
%%

\begin{frame}\frametitle{Graph}

\begin{center}
\resizebox{0.80\textwidth}{!}{\includegraphics{fundsoln}}
\end{center}

\end{frame}
%%

\begin{frame}\frametitle{$\displaystyle \lim_{t \to 0^{+}} U(x, t)$}

If $x \ne 0$ then
\[
\lim_{t \to 0^{+}} U(x, t)
	= \lim_{t \to 0^{+}} \frac{1}{\sqrt{4 \pi \kappa t}} e^{-x^{2}/(4 \kappa t)}
	= 0	.
\]

\vspace*{0.10in}

\uncover<2->{%
If $x = 0$ then
\[
\lim_{t \to 0^{+}} U(0,t)
	= \lim_{t \to 0^{+}} \frac{1}{\sqrt{4 \pi \kappa t}}
	= \infty	.
\]
}

\end{frame}
%%

\begin{frame}\frametitle{Justification}

Suppose $x \ne 0$ then
\begin{align*}
\lim_{t \to 0^{+}} U(x, t)
	&= \lim_{t \to 0^{+}} \frac{1}{\sqrt{4 \pi \kappa t}} e^{-x^{2}/(4 \kappa t)}	\\
	&= \lim_{t \to 0^{+}} \frac{1/\sqrt{t}}{\sqrt{4 \pi \kappa} e^{x^{2}/(4 \kappa t)}}
\mbox{ (indeterminate $\infty/\infty$)}	\\
	&= \lim_{t \to 0^{+}} \frac{-1/(2 t^{3/2})}{-\frac{x^{2}}{4 \kappa t^{2}}
              \sqrt{4 \pi \kappa} e^{x^{2}/(4 \kappa t)}}	\\
	&= \lim_{t \to 0^{+}} \frac{1}{\frac{x^{2}}{4 \kappa t^{1/2}}
              \sqrt{\pi \kappa} e^{x^{2}/(4 \kappa t)}}	\\
	&= 0	.
\end{align*}

\end{frame}
%%

\begin{frame}\frametitle{Area Under the Curve}

Assume that for fixed $t > 0$ the improper integral
\[
\int_{-\infty}^{\infty} U(x, t) \, dx
\]
converges.
Find the value of the integral.

\end{frame}
%%

\begin{frame}\frametitle{Solution}

\begin{align*}
\mbox{If } S
	&= \int_{-\infty}^{\infty} U(x, t) \, dx	\\
S^{2} &= \left( \int_{-\infty}^{\infty} U(x, t) \, dx \right)
\left( \int_{-\infty}^{\infty} U(y,t) \, dy	\right)	\\
	&= \frac{1}{4 \pi \kappa t} \int_{-\infty}^{\infty} e^{-x^{2}/(4 \kappa t)} \, dx
\int_{-\infty}^{\infty} e^{-y^{2}/(4 \kappa t)} \, dy	\\
	&= \frac{1}{4 \pi \kappa t} \int_{-\infty}^{\infty} \int_{-\infty}^{\infty}
e^{-(x^{2}+y^{2})/(4 \kappa t)} \, dx \, dy	\\
	&= \frac{1}{4 \pi \kappa t} \int_{0}^{2 \pi} \int_{0}^{\infty}
r e^{-r^{2}/(4 \kappa t)} \, dr \, d\theta	\\
	&= \frac{1}{2 \kappa t} \int_{0}^{\infty}
r e^{-r^{2}/(4 \kappa t)} \, dr	\\
S^{2}
	&= 1
\implies
S = 1	.
\end{align*}

\end{frame}
%%

\begin{frame}\frametitle{Dirac Delta Function}

Since
\begin{itemize}
\item for $x \ne 0$, $\displaystyle \lim_{t \to 0^{+}} U(x, t) = 0$ and
\item $\displaystyle \int_{-\infty}^{\infty} U(x, t) \, dx = 1$ for all
  $t > 0$
\end{itemize}
then
\[
\lim_{t \to 0^{+}} \frac{1}{\sqrt{4 \pi \kappa t}} e^{-x^{2}/(4 \kappa t)}
	= \delta(x)
\]
the \textbf{Dirac delta function}.

\end{frame}
%%

\begin{frame}\frametitle{Logarithmic Differentiation (1 of 2)}

\begin{align*}
U(x, t)
	&= \frac{1}{\sqrt{4 \pi \kappa t}} e^{-x^{2}/(4\kappa t)}	\\
\ln U
	&= -\frac{1}{2} \ln (4 \pi \kappa t) - \frac{x^{2}}{4 \kappa t}	\\
\frac{\partial}{\partial t} \left[ \ln U \right]
	&= \frac{\partial}{\partial t} \left[
-\frac{1}{2} \ln (4 \pi \kappa t) - \frac{x^{2}}{4 \kappa t}
\right]	\\
\uncover<2->{%
\frac{U_{t}}{U}
	&= -\frac{1}{2 t} + \frac{x^{2}}{4 \kappa t^{2}}	\\
U_{t}
	&= \frac{1}{\sqrt{4 \pi \kappa t}} e^{-x^{2}/(4\kappa t)} \left(
-\frac{1}{2 t} + \frac{x^{2}}{4 \kappa t^{2}} \right)
}
\end{align*}

\end{frame}
%%

\begin{frame}\frametitle{Logarithmic Differentiation (2 of 2)}

\begin{align*}
U(x, t)
	&= \frac{1}{\sqrt{4 \pi \kappa t}} e^{-x^{2}/(4\kappa t)}	\\
%\ln U &= -\frac{1}{2} \ln (4 \pi \kappa t) - \frac{x^{2}}{4 \kappa t}	\\
\frac{\partial}{\partial x} \left[ \ln U \right]
	&= \frac{\partial}{\partial x} \left[
-\frac{1}{2} \ln (4 \pi \kappa t) - \frac{x^{2}}{4 \kappa t}
\right]	\\
\frac{U_{x}}{U}
	&= -\frac{x}{2 \kappa t}	\\
U_{x}
	&= -U \left( \frac{x}{2 \kappa t} \right)	\\
U_{xx}
	&= -U_{x} \left( \frac{x}{2 \kappa t} \right) - U \left(
             \frac{1}{2 \kappa t} \right)	\\
	&= U \left( \frac{x^{2}}{4 \kappa^{2} t^{2}} \right) - U \left(
              \frac{1}{2 \kappa t} \right)	\\
U_{xx}
	&= \frac{1}{\sqrt{4 \pi \kappa t}} e^{-x^{2}/(4\kappa t)} \left( \frac{x^{2}}{4
              \kappa^{2} t^{2}} - \frac{1}{2 \kappa t} \right)
\end{align*}

\end{frame}
%%

\begin{frame}\frametitle{Solution to the Heat Equation}

\begin{align*}
U_{t}(x, t)
	&= \frac{1}{4 \sqrt{\kappa \pi t}} e^{-x^{2}/(4 \kappa t)} \left(
\frac{x^{2}}{2 \kappa t^{2}} - \frac{1}{t} \right)	\\
U_{xx}(x, t)
	&= \frac{1}{4 \kappa \sqrt{\kappa \pi t}} e^{-x^{2}/(4 \kappa t)} \left(
\frac{x^{2}}{2 \kappa t^{2}} - \frac{1}{t} \right)
\end{align*}
and thus $U_{t} = \kappa U_{xx}$.

\vspace*{0.10in}

\textbf{Remark}:
since the fundamental solution is defined for $-\infty < x < \infty$,
no boundary conditions need be considered.

\vspace*{0.10in}

\uncover<2->{%
If $u(x, 0) = f(x)$ is an initial condition defined on $-\infty < x <
\infty$, how do we form a solution to the IVP?
}

\end{frame}
%%

\begin{frame}\frametitle{Solving the IVP}

\begin{theorem}
Consider the initial value problem
\begin{align*}
u_{t}
	&= \kappa \, u_{xx}
\mbox{ for $-\infty < x < \infty$ and $t > 0$}	\\
u(x, 0)
	&= f(x)	,
\mbox{ for $-\infty < x < \infty$.}
\end{align*}
If $f(x)$ is continuous and if
$\displaystyle \int_{-\infty}^{\infty} |f(x)| \, dx$ converges,
then the piecewise defined function
\[
u(x, t)
	= \left\{ \begin{array}{ll}
\displaystyle
\int_{-\infty}^{\infty} U(x-y, t) f(y) \, dy	& \mbox{if $t > 0$,}	\\
\displaystyle f(x)	& \mbox{if $t = 0$}
\end{array}
\right.
\]
solves the heat equation and satisfies the initial condition in the sense that
\[
\lim_{(x, t) \to (x_{0}, 0^{+})} u(x, t)
	= f(x_{0})	.
\]
\end{theorem}

\end{frame}
%%

\begin{frame}\frametitle{Uniqueness}

\begin{theorem}%[Uniqueness]
Consider the initial value problem with conditions imposed as
$x \to \pm \infty$,
\begin{align*}
u_{t}
	&= \kappa u_{xx}	,
\mbox{ for $-\infty < x < \infty$ and $t > 0$}	\\
u(x, 0)
	&= f(x)	,
\mbox{ for $-\infty < x < \infty$}	\\
\lim_{x \to \pm \infty} \left( \max_{0 \le t \le T} | u(x, t) | \right)
	&= 0
\end{align*}
for $-\infty < x < \infty$, $t > 0$, and $T > 0$.
If $f(x)$ is continuous, if $\displaystyle \lim_{x \to \pm \infty} f(x) = 0$, and if
$\displaystyle \int_{-\infty}^{\infty} |f(x)| \, dx$ converges,
then
\[
u(x, t)
	= \left\{ \begin{array}{ll}
\displaystyle
\int_{-\infty}^{\infty} U(x-y, t) f(y) \, dy	& \mbox{if $t > 0$,}	\\
\displaystyle f(x)	& \mbox{if $t = 0$}
\end{array}
\right.
\]
is the unique, continuous solution to the initial value problem above on
$(-\infty, \infty) \times [0,\infty)$.
\end{theorem}

\end{frame}
%%

\begin{frame}\frametitle{Example}

Find the unique solution to the following initial boundary value
problem.
\begin{align*}
u_{t}
	&= u_{xx}	,
\mbox{ for $-\infty < x < \infty$ and $t > 0$}	\\
u(x, 0)
	&= e^{-x^{2}} \cos x	,
\mbox{ for $-\infty < x < \infty$}	\\
\lim_{x \to \pm \infty} \left( \max_{0 \le t \le T} | u(x, t) |\right)
	&= 0
\end{align*}

\end{frame}
%%

\begin{frame}\frametitle{Solution (1 of 4)}

Since $\kappa = 1$ then
\begin{align*}
u(x, t)
	&= \int_{-\infty}^{\infty} U(x-y,t) e^{-y^{2}} \cos y \, dy	\\
	&= \int_{-\infty}^{\infty} \frac{1}{\sqrt{4 \pi t}} e^{-(x-y)^{2}/(4 
              t)} e^{-y^{2}} \Re{e^{i y}} \, dy	\\
\uncover<2->{%
	&= \frac{1}{\sqrt{4 \pi t}} e^{-x^{2}/(4 t)} \int_{-\infty}^{\infty}
              e^{(2 x y - y^{2})/(4 t)} e^{-y^{2}} \Re{e^{i y}} \, dy	\\
\uncover<3->{%
	&= U(x, t) \int_{-\infty}^{\infty}
              e^{(2 x y - [1 + 4 t] y^{2})/(4 t)} \Re{e^{i y}} \,
              dy	\\
\uncover<4->{%
	&= U(x, t) \int_{-\infty}^{\infty}
              \Re{e^{(2 x y - [1 + 4 t] y^{2})/(4 t)} e^{i y}} \,
              dy	\\
\uncover<5->{%
u(x, t)
	&= U(x, t) \Re{\int_{-\infty}^{\infty}
              e^{([2 x + i 4 t] y - [1 + 4 t] y^{2})/(4  t)} \, dy}
}
}
}
}
\end{align*}

\end{frame}
%%

\begin{frame}\frametitle{Solution (2 of 4)}

\[
u(x, t)
	= U(x, t) \Re{\int_{-\infty}^{\infty}
              e^{([2 x + i 4 t] y - [1 + 4 t] y^{2})/(4 t)} \, dy}
\]
Complete the square in the exponent.
\begin{align*}
u(x, t)
	&= U(x, t) \Re{\int_{-\infty}^{\infty}
              e^{-\frac{1+4 t}{4 t} \left( y^{2} - \frac{2 x + i 4
              t}{1+4 t} y \right)} \, dy}	\\
	&= U(x, t) \Re{\int_{-\infty}^{\infty}
              e^{-\frac{1+4 t}{4 t} \left( y^{2} - \frac{2 x + i 4
              t}{1+4 t} y + \left[ \frac{x + i 2
              t}{1+4 t} \right]^{2} - \left[ \frac{x + i 2
              t}{1+4 t} \right]^{2} \right)} \, dy}	\\
	&= U(x, t) \Re{e^{\frac{1+4 t}{4 t} \, \left[ \frac{x + i 2
              t}{1+4 t} \right]^{2}} \int_{-\infty}^{\infty}
              e^{-\frac{1+4 t}{4 t} \left( y - \frac{x + i 2 t}{1+4 t} \right)^{2}} \, dy}
\end{align*}

\end{frame}
%%

\begin{frame}\frametitle{Solution (3 of 4)}

\[
u(x, t)
	= U(x, t) \Re{e^{\frac{(x + i 2 t)^{2}}{4 t (1+4 t)}} \int_{-\infty}^{\infty}
              e^{-\frac{1+4 t}{2 t} \left( y - \frac{x + i 2 t}{1+4 t} \right)^{2}/2} \, dy}
\]
Substitute $\displaystyle z = \sqrt{\frac{1 + 4 t}{2 t}} \left( y - \frac{x + i 2
  t}{1+4t} \right)$.
\begin{align*}
u(x, t)
	&= U(x, t) \Re{e^{\frac{(x + i 2 t)^{2}}{4 t (1+4 t)}} \sqrt{\frac{2
              t}{1+4 t}} \int_{-\infty}^{\infty}
              e^{-z^{2}/2} \, dz}	\\
	&= U(x, t) \Re{e^{\frac{(x + i 2 t)^{2}}{4 t (1+4 t)}} \sqrt{\frac{4 \pi t}{1+4 t}}}	\\
	&= \sqrt{\frac{4 \pi t}{1+4 t}} U(x, t) \Re{e^{\frac{x^{2} + i 4 x t - 4 t^{2}}{4 t (1+4 t)}}}	\\
\end{align*}

\end{frame}
%%

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

\begin{align*}
u(x, t)
	&= \sqrt{\frac{4 \pi t}{1+4 t}} U(x, t) \Re{e^{\frac{x^{2} - 4
              t^{2}}{4 t (1+4 t)}} e^{\frac{i 4 x t}{4 t (1+4 t)}}}	\\
	&= \sqrt{\frac{4 \pi t}{1+4 t}} U(x, t) e^{\frac{x^{2} - 4
              t^{2}}{4 t (1+4 t)}} \cos \left( \frac{x}{1 + 4 t} \right)	\\
	&= \sqrt{\frac{4 \pi t}{1+4 t}} \frac{1}{\sqrt{4 \pi t}}
              e^{-\frac{x^{2}}{4 t}} e^{\frac{x^{2} - 4 t^{2}}{4 t
              (1+4 t)}} \cos \left( \frac{x}{1 + 4 t} \right)	\\
	&= \frac{1}{\sqrt{1+4 t}} e^{-\frac{x^{2}+t}{1 + 4 t}} \cos
   \left( \frac{x}{1 + 4 t} \right)
\end{align*}

\end{frame}
%%

\begin{frame}\frametitle{Illustration}

\[
u(x, t)
	= \frac{1}{\sqrt{1+4 t}} e^{-\frac{x^{2}+t}{1 + 4 t}} \cos
        \left( \frac{x}{1 + 4 t} \right)
\]
\begin{center}
\resizebox{0.75\textwidth}{!}{\includegraphics{example01}}
\end{center}

\end{frame}
%%

\begin{frame}\frametitle{The Error Function}

\begin{definition}
The \textbf{error function} of $z$ denoted $\erf(z)$ is
\[
\erf(z)
	= \int_{0}^{z} \frac{2}{\sqrt{\pi}} e^{-t^{2}} \, dt	.
\]
\end{definition}

\textbf{Properties}:
\begin{itemize}
\item $\erf(-z) = -\erf(z)$
\item $\displaystyle \lim_{z \to \infty} \erf(z) = 1$
\item $\displaystyle \frac{d}{dz} \left[ \erf(z) \right] =
  \frac{2}{\sqrt{\pi}} e^{-z^{2}}$
\end{itemize}

\uncover<2->{%
\begin{definition}
The \textbf{complementary error function} denoted $\erfc(z)$ is
\[
\erfc(z)
	= 1 - \erf(z)	.
\]
\end{definition}
}

\end{frame}
%%

\begin{frame}\frametitle{Semi-Infinite Intervals}

\begin{theorem}
Suppose $f(x)$ is continuous on $[0,\infty)$, $f(0) = 0$,
$\displaystyle \lim_{x \to \infty} f(x) = 0$, and
$\displaystyle \int_{0}^{\infty} |f(x)| \, dx$ converges.
The initial boundary value problem
\begin{align*}
u_{t}
	&= \kappa \, u_{xx}
\mbox{ for $0 < x < \infty$ and $t > 0$}	\\
u(0+,t)
	&= 0	\\
u(x,0+)
	&= f(x)	\\
\lim_{x \to \infty} \left( \max_{0 \le t \le T} | u(x, t) |\right)
	&= 0
\end{align*}
has a unique, continuous solution defined for $t > 0$,
\[
u(x, t)
	= \int_{0}^{\infty} \left( U(x-y, t) - U(x+y, t) \right) f(y) \, dy	.
\]
\end{theorem}

\end{frame}
%%

\begin{frame}\frametitle{Example}

Solve the initial, boundary value problem:
\begin{align*}
u_{t}
	&= u_{xx}
\mbox{ for $0 < x < \infty$ and $t > 0$}	\\
u(0+,t)
	&= 0	\\
u(x,0+)
	&= x \, e^{-x}	\\
\lim_{x \to \infty} \left( \max_{0 \le t \le T} | u(x, t) |\right)
	&= 0	.
\end{align*}

\end{frame}
%%

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

For simplicity $\kappa = 1$ and thus
\begin{align*}
u(x, t)
	= {} & \int_{0}^{\infty} \frac{1}{\sqrt{4 \pi t}} \left[ e^{-(x-y)^{2}/(4 t)}
              - e^{-(x+y)^{2}/(4 t)} \right] y e^{-y} \, dy	\\
	= {} & \frac{1}{\sqrt{4 \pi t}} \int_{0}^{\infty} y \left[
e^{\frac{-(x^{2} - 2 (x-2t) y + y^{2})}{4t}} -
e^{\frac{-(x^{2} + 2 (x+2t) y + y^{2})}{4t}}
\right] \, dy	\\
	= {} & \frac{1}{\sqrt{4 \pi t}} e^{-\frac{x^{2}}{4t}} \int_{0}^{\infty} y
              \left[
e^{\frac{-(y^{2} - 2 (x-2t) y)}{4t}} -
e^{\frac{-(y^{2} + 2 (x+2t) y)}{4t}} \right] \, dy	\\
	= {} & \frac{1}{\sqrt{4 \pi t}} e^{-\frac{x^{2}}{4t}}
              e^{\frac{(x-2t)^{2}}{4t}} \int_{0}^{\infty}
y \, e^{\frac{-(y - x + 2t))^{2}}{4t}} \, dy	\\
	&- \frac{1}{\sqrt{4 \pi t}} e^{-\frac{x^{2}}{4t}}
            e^{\frac{(x+2t)^{2}}{4t}} \int_{0}^{\infty}
y \, e^{\frac{-(y + x+2t)^{2}}{4t}} \, dy	\\
\end{align*}

\end{frame}
%%

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

\begin{align*}
u(x, t)
	= {} & \frac{1}{\sqrt{4 \pi t}} e^{-x+t} \int_{0}^{\infty} (y - (x-2t) + (x-2t))
e^{\frac{-(y - x + 2t)^{2}}{4t}} \, dy	\\
	& - \frac{1}{\sqrt{4 \pi t}} e^{x+t} \int_{0}^{\infty} (y + (x+2t) - (x+2t))
e^{\frac{-(y + x + 2t)^{2}}{4t}} \, dy	\\
	= {} & \frac{e^{-x+t}}{\sqrt{\pi}} \int_{0}^{\infty}
\frac{y - x + 2t}{\sqrt{4 t}} e^{\frac{-(y - x + 2t)^{2}}{4t}} \, dy	\\
	&+ \frac{(x-2t) e^{-x+t}}{\sqrt{4 \pi t}} \int_{0}^{\infty} 
e^{-\frac{(y - x + 2t)^{2}}{4 t}} \, dy	\\
	&- \frac{e^{x+t}}{\sqrt{\pi}} \int_{0}^{\infty}
\frac{y + x+2t}{\sqrt{4 t}} e^{-\frac{(y + x + 2t)^{2}}{4t}} \, dy	\\
	&+ \frac{(x+2t) e^{x+t}}{\sqrt{4 \pi t}} \int_{0}^{\infty} 
e^{-\frac{(y + x + 2t)^{2}}{4 t}} \, dy
\end{align*}

\end{frame}
%%

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

Integrate by substitution.
\begin{align*}
u(x, t)
	= {} & \sqrt{\frac{t}{\pi}} e^{-\frac{x^{2}}{4t}} +
\frac{(x-2t) e^{-x+t}}{\sqrt{\pi}} \int_{-\infty}^{\frac{x-2t}{2
               \sqrt{t}}} e^{-z^{2}} \, dz	\\
	&- \sqrt{\frac{t}{\pi}} e^{-\frac{x^{2}}{4t}}  +
\frac{(x+2t) e^{-x+t}}{\sqrt{\pi}} \int_{-\infty}^{\frac{x+2t}{2
               \sqrt{t}}} e^{-z^{2}} \, dz	\\
	= {} & \frac{(x-2t) e^{-x+t}}{\sqrt{\pi}} \int_{-\infty}^{0} e^{-z^{2}} \, dz
+ \frac{(x-2t) e^{-x+t}}{\sqrt{\pi}} \int_{0}^{\frac{x-2t}{2
               \sqrt{t}}} e^{-z^{2}} \, dz	\\
	&+ \frac{(x+2t) e^{-x+t}}{\sqrt{\pi}} \int_{-\infty}^{0} e^{-z^{2}} \, dz
+ \frac{(x+2t) e^{-x+t}}{\sqrt{\pi}} \int_{0}^{\frac{x+2t}{2
               \sqrt{t}}} e^{-z^{2}} \, dz 	\\
	= {} & \frac{(x-2t) e^{-x+t}}{2} \left( 1 +
              \erf \frac{x-2t}{2 \sqrt{t}} \right)	\\
	&+ \frac{(x+2t) e^{x+t}}{2} \left( 1 -
              \erf \frac{x+2t}{2 \sqrt{t}} \right)
\end{align*}

\end{frame}
%%

\begin{frame}\frametitle{Illustration}

\begin{align*}
u(x, t)
	= {} & \frac{(x-2t) e^{-x+t}}{2} \left( 1 +
              \erf \frac{x-2t}{2 \sqrt{t}} \right)	\\
	&+ \frac{(x+2t) e^{x+t}}{2} \left( 1 -
              \erf \frac{x+2t}{2 \sqrt{t}} \right)
\end{align*}
\begin{center}
\resizebox{0.60\textwidth}{!}{\includegraphics{example02}}
\end{center}

\end{frame}
%%

\begin{frame}\frametitle{Semi-Infinite Interval, Neumann BCs}

\begin{theorem}
Suppose $f(x)$ is continuous on $[0,\infty)$, $f(0) = 0$,
$\displaystyle \lim_{x \to \infty} f(x) = 0$, and
$\displaystyle \int_{0}^{\infty} |f(x)| \, dx$ converges.
The initial boundary value problem
\begin{align*}
u_{t}
	&= \kappa \, u_{xx}
\mbox{ for $0 < x < \infty$ and $t > 0$}	\\
u_{x}(0+,t)
	&= 0	\\
u(x,0+)
	&= f(x)	\\
\lim_{x \to \infty} \left( \max_{0 \le t \le T} | u(x, t) |\right)
	&= 0
\end{align*}
has a unique, continuous solution defined for $t > 0$,
\[
u(x, t)
	= \int_{0}^{\infty} \left( U(x-y, t) + U(x+y, t) \right) f(y) \, dy	.
\]
\end{theorem}

\end{frame}
%%

\begin{frame}\frametitle{Example}

Solve the initial, boundary value problem:
\begin{align*}
u_{t}
	&= u_{xx}
\mbox{ for $0 < x < \infty$ and $t > 0$}	\\
u_{x}(0+,t)
	&= 0	\\
u(x,0+)
	&= e^{-x^{2}} \cos x	\\
\lim_{x \to \infty} \left( \max_{0 \le t \le T} | u(x, t) |\right)
	&= 0	.
\end{align*}

\end{frame}
%%

\begin{frame}\frametitle{Solution}

\[
u(x, t)
	= \frac{1}{\sqrt{1+4t}} e^{-\frac{x^{2}+t}{1+4t}} \cos \left(
  \frac{x}{1+4t} \right)
\]
\begin{center}
\resizebox{0.70\textwidth}{!}{\includegraphics{example03}}
\end{center}

\end{frame}
%%

%\begin{frame}\frametitle{Homework}

%\begin{itemize}
%\item Read Section $4.4$
%\item Exercises: 17, 18, 19
%\end{itemize}

%\end{frame}
%%

%%%
\end{document}

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