%%%%		-*- 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:      06/30/2022
%%%%
%%%% Purpose:   Preliminary Definitions and Examples
%%%%

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\title{Preliminary Definitions and Examples}
\subtitle{MATH 467 \textit{Partial Differential Equations}} % (optional)


\author % (optional, use only with lots of authors)
{J~Robert~Buchanan}
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{
  Department of Mathematics
}
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Fall
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\subject{Talks}
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%\section{Introduction}

\begin{frame}\frametitle{What is a PDE?}

\begin{definition}
A \textbf{partial differential equation} (PDE) is an equation
involving an unknown function $u$, of two or more independent
variables $t$, $x$, $y$, \textit{etc} and the partial derivatives of $u$.
\end{definition}

\end{frame}
%%

\begin{frame}\frametitle{Examples}

\begin{align*}
u_{t} + u \, u_{y} &= 0	\\
u_{xx} + u_{yy} &= f(x,y)	\\
u_{xx} + x \, y^{3} u_{yy} - e^{x} u_{z} + x \, y \, u &= 0	\\
u_{t} - u_{xx} + (u_{x})^{2} &= 0	\\
u_{t} + c \, u \, u_{x} + u_{xxx} &= 0
\end{align*}

\end{frame}
%%

\begin{frame}\frametitle{Partial Derivative Notation}

\begin{align*}
u_{t} &= \frac{\partial u}{\partial t}	\\
u_{xx} &= \frac{\partial^{2} u}{\partial x^{2}}	\\
u_{xy} &= \frac{\partial^{2} u}{\partial y \, \partial x}	\\
{}	&\vdotswithin{=} {}
\end{align*}


\end{frame}
%%

\begin{frame}\frametitle{Description of PDEs}

PDEs can be described as being
\begin{itemize}
\item linear or nonlinear (or in some cases semilinear or
  quasilinear),
\item homogeneous or nonhomogeneous,
\item first order, second order, third order, \textit{etc}.
\end{itemize}

\end{frame}
%%

\begin{frame}\frametitle{Order}

\begin{definition}
The \textbf{order} of a PDE is the highest order partial derivative
present in the PDE.
\end{definition}

\vspace*{0.05in}

\uncover<2->{%
Find the order of each of the following PDEs.
\begin{align*}
u_{t} + u \, u_{y} &= 0	\\
u_{xx} + u_{yy} &= f(x,y)	\\
u_{xx} + x \, y^{3} u_{yy} - e^{x} u_{z} + x \, y \, u &= 0	\\
u_{t} - u_{xx} + (u_{x})^{2} &= 0	\\
u_{t} + c \, u \, u_{x} + u_{xxx} &= 0
\end{align*}
}

\end{frame}
%%

\begin{frame}\frametitle{Linearity}

\begin{definition}
A PDE is \textbf{linear} if all the terms of the PDE are linear in the
unknown and its partial derivatives.
Otherwise the PDE is \textbf{nonlinear}.
\end{definition}

\vspace*{0.05in}

\uncover<2->{%
Determine which of the following PDEs are linear or nonlinear.
\begin{align*}
u_{t} + u \, u_{y} &= 0	\\
u_{xx} + u_{yy} &= f(x,y)	\\
u_{xx} + x \, y^{3} u_{yy} - e^{x} u_{z} + x \, y \, u &= 0	\\
u_{t} - u_{xx} + (u_{x})^{2} &= 0	\\
u_{t} + c \, u \, u_{x} + u_{xxx} &= 0
\end{align*}
}

\end{frame}
%%

\begin{frame}\frametitle{Solution to a PDE}

\begin{definition}
A \textbf{solution} to a PDE is a function defined on an appropriate
domain which makes the PDE a true equation when substituted for the
unknown of the PDE.
\end{definition}

\vspace*{0.05in}

\uncover<2->{%
Consider the second-order, linear PDE
\[
u_{xy} = 0
\quad\mbox{for $(x,y) \in \mathbb{R}^{2}$.}
\]
Verify that $u(x,y) = F(x) + G(y)$ is a solution to the PDE
where $F$ and $G$ are arbitrary
differentiable functions defined for all real numbers.
}

\end{frame}
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\begin{frame}\frametitle{Common Generic PDEs}

We will primarily focus on properties and solutions of the following
two types of PDE.
\begin{description}
\item[First-order Linear:]
\[
a(x,t) u_{x} + b(x,t) u_{t} + c(x,t) u = g(x,t)
\]
\item[Second-order Linear:]
\[
A u_{tt} + B u_{tx} + C u_{xx} + D u_{t} + E u_{x} + F u = G
\]
where $A$, $B$, $C$, $D$, $E$, $F$, and $G$ are functions of $(x,t)$.
\end{description}

\vspace*{0.05in}

\textbf{Note}: a linear PDE is called \textbf{homogeneous} if $g(x,t)
\equiv 0$ or $G(x,t) \equiv 0$, otherwise it is \textbf{nonhomogeneous}.

\end{frame}
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\begin{frame}\frametitle{Principle of Superposition}

\begin{theorem}%[Principle of Superposition]
\label{thm:intro:superposition}
If $u_{i}$ is a solution of the second-order, linear partial
differential equation
\[
A u_{tt} + B u_{tx} + C u_{xx} + D u_{t} + E u_{x} + F u = G_{i}
\]
for $i = 1, 2, \ldots, n$ on a domain $\Omega \subset \mathbb{R}^{2}$,
then for any constants $c_{1}, c_{2}, \ldots, c_{n}$ the function $u =
c_{1} u_{1} + c_{2} u_{2} + \cdots + c_{n} u_{n}$ is a solution of
\[
A u_{tt} + B u_{tx} + C u_{xx} + D u_{t} + E u_{x} + F u =
\sum_{i=1}^{n} c_{i} G_{i}
\]
on $\Omega$.
\end{theorem}

\end{frame}
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\begin{frame}\frametitle{Principle of Subtraction}

\begin{theorem}
If $u_{1}(x,t)$ and $u_{2}(x,t)$ are solutions to
\[
A u_{tt} + B u_{tx} + C u_{xx} + D u_{t} + E u_{x} + F u = G
\]
then $u(x,t) \equiv u_{1}(x,t) - u_{2}(x,t)$ is a solution to
\[
A u_{tt} + B u_{tx} + C u_{xx} + D u_{t} + E u_{x} + F u = 0	.
\]
\end{theorem}

\vspace*{0.05in}

\textbf{Remark}: any two solutions to a nonhomogeneous PDE differ by a
solution to the associated homogeneous PDE.

\end{frame}
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\begin{frame}\frametitle{Initial and Boundary Conditions}

\begin{itemize}
\item If $u(x,t)$ is the solution to a PDE which satisfies the
  requirement that $u(x,0) = f(x)$, then this requirement is called an
  \textbf{initial condition}.
\item If the domain of the solution $u(x,t)$ is $D = \{ (x,t) \, : \,
  t \ge 0, \, 0 \le x \le L \}$ and
\begin{align*}
u(0,t) &= g(t)	\\
u(L,t) &= h(t)	,
\end{align*}
then these requirements are called \textbf{boundary conditions}.
\end{itemize}

\end{frame}
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\begin{frame}\frametitle{Illustration}

\begin{center}
\resizebox{0.65\textwidth}{!}{\includegraphics{domain}}
\end{center}

\end{frame}
%%

\begin{frame}\frametitle{Initial Boundary Value Problems}

\begin{definition}
A PDE along with specified initial and boundary conditions is referred
to as an \textbf{initial boundary value problem} (IBVP).
If only boundary conditions are specified (perhaps because $u$ is
independent of $t$) the PDE is called a \textbf{boundary value
  problem} (BVP).
\end{definition}

\vspace*{0.05in}

\begin{definition}
An IBVP is \textbf{well-posed} if it has a unique solution and that
solution depends continuously on the initial and boundary conditions.
Otherwise the PDE is \textbf{ill-posed}.
\end{definition}

\end{frame}
%%

\begin{frame}\frametitle{Example}

Consider the PDE:
\[
u_{xx} + u_{yy} = 0	.
\]
Verify that the following functions solve the PDE.
\begin{align*}
u_{1}(x,y) &= x+y	\\
u_{2}(x,y) &= x^{2} - y^{2}	\\
u_{3}(x,y) &= e^{x} \cos y	\\
u_{4}(x,y) &= \ln (x^{2}+y^{2})	\\
u_{5}(x,y) &= \frac{x}{x^{2}+y^{2}}
\end{align*}

\end{frame}
%%

\begin{frame}\frametitle{Example: BVP}

Consider the BVP:
\begin{align*}
u_{xx} + u_{yy} &= 0	\\
u(x,y) &= 0
\quad\mbox{for $x^{2} + y^{2} = 1$.}
\end{align*}
Find its solution.

\end{frame}
%%

\begin{frame}\frametitle{Example: IBVP}

Consider the IBVP:
\begin{align*}
u_{t} - u_{xx} &= 0
\quad\mbox{for $t \ge 0$, $0 \le x \le \pi$}	\\
u(x,0) &= 100 + \sin x
\quad\mbox{(IC)}	\\
u(0,t) &= 100
\quad\mbox{(BC for $x = 0$)}	\\
u(\pi,t) &= 100
\quad\mbox{(BC for $x= \pi$)}
\end{align*}
\begin{enumerate}
\item Show that $u_{1}(x,t) = t + x^{2}/2$ solves the PDE.
\item<2-> Show that $u_{2}(x,t) = A + B e^{-t} \sin x$ solves the PDE.
\item<3-> Choose $A$ and $B$ appropriately so that $u_{2}(x,t)$ solves
  the IBVP. 
\end{enumerate}

\end{frame}
%%

\begin{frame}\frametitle{Classification of PDEs}

Recall the general second-order PDE:
\[
A u_{tt} + B u_{tx} + C u_{xx} + D u_{t} + E u_{x} + F u = G
\]
where $A$, $B$, $C$, $D$, $E$, $F$, and $G$ are functions of $(x,t)
\in D$.
The PDE is said to be
\begin{description}
\item[\textbf{elliptic}] on $D$ if $4 A C - B^{2} > 0$ for all $(x,t)
  \in D$,
\item[\textbf{parabolic}] on $D$ if $4 A C - B^{2} = 0$ for all $(x,t)
  \in D$,
\item[\textbf{hyperbolic}] on $D$ if $4 A C - B^{2} < 0$ for all $(x,t)
  \in D$.
\end{description}

\end{frame}
%%

\begin{frame}\frametitle{Examples}

Classify each of the following PDEs as either \textbf{elliptic},
\textbf{parabolic}, or \textbf{hyperbolic}.
\begin{align*}
u_{tt} + u_{xx} + u_{t} - 2 u_{x} &= f(x,t)	\\
u_{tt} - u_{tx} + 2 u_{xx} + e^{x} u_{t} - \sin (t \, x) u_{x} &=
                                                                     f(x,t)	\\
u_{tt} - 2 u_{tx} + u_{xx} + 2 u_{t} - u_{x} &= 0	\\
u_{tt} + t \, u_{xx} &= 0	\\
u_{xx} - \beta \, u_{t} + \alpha \, u_{x} + \rho \, u + f(x,t) &= u_{t}
\end{align*}

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

\begin{itemize}
\item Read Sections $1.1$ and $1.2$
\item Exercises: 1--6
\end{itemize}

\end{frame}
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