%%%%		-*- 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:      11/15/2017
%%%%
%%%% Purpose:   First-Order Quasilinear PDEs
%%%%

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\title{First-Order Quasilinear PDEs}
\subtitle{MATH 467 \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{frame}
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% Since this a solution template for a generic talk, very little can
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% - Exactly two or three sections (other than the summary).
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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 learn:
\begin{itemize}
\item to solve semilinear and quasilinear first-order partial
  differential equations.
\end{itemize}

\end{frame}
%%

\begin{frame}\frametitle{Quasilinear and Semilinear PDEs}

A first-order PDE is called \textbf{quasilinear} if it has the form
\[
a(x, y, u) u_{x} + b(x, y, u) u_{y} = c(x, y, u)
\]
(assuming $u \equiv u(x, y)$ for $(x, y) \in D \subset \mathbb{R}^{2}$).

\vspace*{0.10in}

A first-order PDE is called \textbf{semilinear} if it has the form
\[
a(x, y) u_{x} + b(x, y) u_{y} = c(x, y, u)	.
\]

\vspace*{0.10in}

We will generalize the method of characteristics in order to solve
quasilinear PDEs (of which semilinear PDEs are a special case).

\end{frame}
%%

\begin{frame}\frametitle{Integral Surface}

Suppose $u(x, y)$ solves the quasilinear PDE:
\[
a(x, y, u) u_{x} + b(x, y, u) u_{y} - c(x, y, u)  = 0
\]
for $(x, y) \in D$, then the surface
\[
S = \{ (x, y, u(x, y)) \, : \, (x, y) \in D \}
\]
is called an \textbf{integral surface}.
The vector $\langle u_{x}, u_{y}, -1 \rangle$ is normal to the
integral surface for all $(x, y) \in D$.

\vspace*{0.10in}

\uncover<2->{%
Since
\begin{align*}
a(x, y, u) u_{x} + b(x, y, u) u_{y} - c(x, y, u)  &= 0	\\
\langle u_{x}, u_{y}, -1 \rangle \cdot
\langle a(x, y, u), b(x, y, u), c(x, y, u) \rangle &= 0
\end{align*}
then vector $\langle a(x, y, u), b(x, y, u), c(x, y, u) \rangle$ is
perpendicular to the normal to the integral surface, \textit{i.e.},
tangent to the integral surface.
}

\end{frame}
%%

\begin{frame}\frametitle{Characteristic System}

The vector $\langle a(x, y, u), b(x, y, u), c(x, y, u) \rangle$ defines a
vector field in $xyu$-space.
The integral curves $(x(t), y(t), u(t))$ defined by the
\textbf{characteristic system} of ODEs
\begin{align*}
\frac{dx}{dt} &= a(x, y, u)	\\
\frac{dy}{dt} &= b(x, y, u)	\\
\frac{du}{dt} &= c(x, y, u)	
\end{align*}
are called the \textbf{characteristic curves}.
The projections of the characteristic curves in the $xy$-plane will be
called \textbf{characteristics}.

\end{frame}
%%

\begin{frame}\frametitle{Constructing a Solution}

Suppose $\Gamma$ is a \textit{non-characteristic} curve in $xyu$-space
and we construct a family of characteristic curves through points of
$\Gamma$.
\begin{center}
\resizebox{0.60\textwidth}{!}{\includegraphics{family}}
\end{center}

\end{frame}
%%

\begin{frame}\frametitle{Solution Surface}

The union of all the characteristic curves through points on $\Gamma$
is a solution (integral) surface to the quasilinear PDE.
\begin{center}
\resizebox{0.60\textwidth}{!}{\includegraphics{integralsurface}}
\end{center}

\end{frame}
%%

\begin{frame}\frametitle{Example}

Find the solution of
\[
y \, u_{x} - x \, u_{y} - e^{u} = 0
\]
that passes through the curve $\Gamma = \{ (x, y, u) = (s, \sin s, 0) \, : \, s
\in \mathbb{R} \}$.

\vspace*{0.10in}

\textbf{Remarks}:
\begin{itemize}
\item $\Gamma$ is parameterized by $s$ for clarity, this parameter will
  not be used for any other purpose.
\item The solution $u$ must satisfy the condition
\[
u(x, \sin x) = 0
\]
for all $x$.
\end{itemize}

\end{frame}
%%

\begin{frame}\frametitle{Solution (1 of 5)}

The characteristic system for this example is
\[
\frac{dx}{dt} = y	,	\quad
\frac{dy}{dt} = -x	,	\quad
\frac{du}{dt} = e^{u}	.
\]

\vspace*{0.10in}

\uncover<2->{%
From the last equation, $-e^{-u} = t - C$ where $C$ is a constant.

\vspace*{0.10in}

\uncover<3->{%
From the second equation,
\[
\frac{dy}{dt} = -x = \frac{d}{dt} \left[ \frac{dx}{dt} \right] =
\frac{d^{2} x}{d t^{2}}
\iff
x''(t) + x = 0	.
\]
Thus $x(t) = A \cos t + B \sin t$ and consequently $y(t) = x'(t) = -A
\sin t + B \cos t$ where $A$ and $B$ are constants.
}
}

\end{frame}
%%

\begin{frame}\frametitle{Solution (2 of 5)}

The characteristic curves of the solution can be parameterized as
\begin{align*}
x &= A \cos t + B \sin t	\\
y &= B \cos t - A \sin t	\\
u &= -\ln (C - t)	.
\end{align*}

\vspace*{0.10in}

\uncover<2->{%
The non-characteristic curve $\Gamma$ is parameterized as
\begin{align*}
x &= s	\\
y &= \sin s	\\
u &= 0	.
\end{align*}
For each point on $\Gamma$ we want to find a characteristic curve
which passes through the point when $t = 0$ (arbitrary choice).
}

\end{frame}
%%

\begin{frame}\frametitle{Solution (3 of 5)}

When $t = 0$,
\begin{align*}
x(0) &= s = A	\\
y(0) &= \sin s = B	\\
u(0) &= 0 = -\ln C
\iff C = 1
\end{align*}

\vspace*{0.10in}

The characteristic curve intersecting $\Gamma$ at $t = 0$ can be
parameterized as
\begin{align*}
x(t) &= s \cos t + \sin s \sin t	\\
y(t) &= \sin s \cos t - s \sin t	\\
u(t) &= -\ln (1 - t)	.
\end{align*}
These equations also produce the integral surface for the solution of
the quasilinear PDE.

\end{frame}
%%

\begin{frame}\frametitle{Solution (4 of 5)}
\fontsize{10pt}{12}\selectfont

\begin{align*}
x &= s \cos t + \sin s \sin t	\\
y &= \sin s \cos t - s \sin t	\\
u &= -\ln (1 - t)	.
\end{align*}
We can eliminate $s$ and $t$ from the equations.
Multiply the first equation by $\cos t$ and the second equation by $-\sin
t$ and add them together.
\begin{align*}
x \cos t - y \sin t &= s \cos^{2} t + \sin s \cos t \sin t -
\sin s \cos t \sin t + s \sin^{2} t	\\
	&= s
\end{align*}
Multiply the first equation by $\sin t$ and the second equation by
$\cos t$ and add them together.
\begin{align*}
x \sin t + y \cos t &=
s \cos t \sin t + \sin s \sin^{2} t +
\sin s \cos^{2} - s \cos t \sin t	\\
	&= \sin s
\end{align*}
Combining the results produces:
\[
\sin (x \cos t - y \sin t) = x \sin t + y \cos t	.
\]

\end{frame}
%%

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

We managed to eliminate parameter $s$.
Using $u = -\ln (1-t) \iff t = 1 - e^{-u}$ we can eliminate parameter
$t$ as well.

\vspace*{0.10in}

The solution to the quasilinear PDE can be expressed in implicit form
as
\begin{align*}
\lefteqn{\sin \left( x \cos \left[ 1 - e^{-u} \right] - y \sin \left[
  1 - e^{-u} \right] \right)}	\\
	&=
x \sin \left[ 1 - e^{-u} \right] + y \cos \left[ 1 - e^{-u} \right]	.
\end{align*}

\end{frame}
%%

\begin{frame}\frametitle{Illustration}

\begin{center}
\resizebox{0.70\textwidth}{!}{\includegraphics{implicit}}
\end{center}

\end{frame}
%%

\begin{frame}\frametitle{Example}

Consider the quasilinear first-order PDE
\[
x \, u_{x} + y \, u_{y} - \sec u = 0	.
\]
Find the solution passing through the parametric curve $\Gamma$ given
by $\Gamma = \{ (x, y, u) = (s^{2}, \sin s , 0) \, : \, s \in \mathbb{R} \}$.

\end{frame}
%%

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

The characteristic system is:
\begin{align*}
\frac{dx}{dt} &= x
\uncover<2->{\implies x(t) = A \, e^{t}}	\\
\frac{dy}{dt} &= y
\uncover<2->{\implies y(t) = B \, e^{t}}	\\
\frac{du}{dt} &= \sec u
\uncover<3->{\implies \sin u = t + C}
\end{align*}
\uncover<3->{where $A$, $B$, and $C$ are constants.}

\vspace*{0.10in}

\uncover<4->{%
Suppose the characteristics pass through $\Gamma$ at $t = 0$
(arbitrary choice).
\begin{align*}
x(0) &= A = s^{2}	\\
y(0) &= B = \sin s	\\
u(0) &= 0 = 0 + C
\implies C = 0
\end{align*}
}

\end{frame}
%%

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

The characteristic curves take the form:
\[
x = s^{2} e^{t}	,	\quad
y = e^{t} \sin s ,	\quad
u = \arcsin t	.
\]
\begin{center}
\resizebox{0.60\textwidth}{!}{\includegraphics{example03chars}}
\end{center}

\end{frame}
%%

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

We can eliminate parameters $s$ and $t$ from the characteristics
curves.
Since $t = \sin u$ then
\[
y = e^{\sin u} \sin s
\iff
\sin s = y \, e^{-\sin u}
\implies
s = \arcsin \left( y \, e^{-\sin u} \right)	.
\]
Consequently the implicit form of the integral surface can be
expressed as
\[
x = \left[ \arcsin \left( y \, e^{-\sin u} \right) \right]^{2} e^{\sin u}	.
\]

\end{frame}
%%

\begin{frame}\frametitle{Illustration}

\[
x = \left[ \arcsin \left( y \, e^{-\sin u} \right) \right]^{2} e^{\sin u}
\]
\begin{center}
\resizebox{0.60\textwidth}{!}{\includegraphics{example03surf}}
\end{center}

\end{frame}
%%

\begin{frame}\frametitle{Example}

Find the solution to the first-order, quasilinear PDE with side
condition.
\begin{align*}
u_{x} + u \, u_{y} &= 6 x	\\
u(0,y) &= 3 y
\end{align*}

\vspace*{0.10in}

\uncover<2->{%
\textbf{Remark}: since no non-characteristic curve $\Gamma$ is
mentioned we are free to create one.
A natural one to use would be $\Gamma = \{ (x, y, u) = (0,s,3s) \, : \,
s \in \mathbb{R} \}$.
}

\end{frame}
%%

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

Re-write the PDE as
\[
u_{x} + u \, u_{y} - 6 x = 0
\]
which has the characteristic system:
\begin{align*}
\frac{dx}{dt} &= 1
\uncover<2->{\implies x = t + A}	\\
\frac{dy}{dt} &= u
\uncover<4->{= 3 t^{2} + 6 A \, t + B \implies y = t^{3} + 3 A \,
                    t^{2} + B \, t + C}	\\
\frac{du}{dt} &= 6 x
\uncover<3->{= 6 t + 6 A \implies u = 3 t^{2} + 6 A \, t + B}	.
\end{align*}

\end{frame}
%%

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

For each point on $\Gamma$ find the characteristic curve which passes
through the point when $t=0$ (arbitrary choice).
\begin{align*}
x(0) &= 0 = A	\\
y(0) &= s = C	\\
u(0) &= 3 s = B
\end{align*}
Consequently the characteristic curves are in parametric form:
\begin{align*}
x(t) &= t	\\
y(t) &= t^{3} + s (3 t + 1)	\\
u(t) &= 3 t^{2} + 3 s	.
\end{align*}

\end{frame}
%%

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

Now eliminate $s$ and $t$ so as to write the solution in implicit
form.
Taking the parametric equations for $x$ and $u$ we can solve for $s$
and $t$ to obtain
\begin{align*}
s	&= \frac{u}{3} - x^{2}	\\
t	&= x	.
\end{align*}
Substituting into the equation for $y$ we get
\[
y = \frac{u}{3} + x \, u - x^{2} - 3 x^{3}
\]
which can be solved explicitly for
\[
u(x, y) = \frac{3 (2 x^{3} + x^{2} + y)}{3x+1}	.
\]

\end{frame}
%%

\begin{frame}\frametitle{Illustration}

\begin{center}
\resizebox{0.70\textwidth}{!}{\includegraphics{example04}}
\end{center}

\end{frame}
%%

\begin{frame}\frametitle{Homework}

\begin{itemize}
\item Read Section $2.2$
\item Exercises: 6--11
\end{itemize}

\end{frame}
%%

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