%%%%		-*- 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:      01/26/2016
%%%%
%%%% Purpose:   First-Order PDEs
%%%%

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\title{First-Order 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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% Since this a solution template for a generic talk, very little can
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%\section{Introduction}

\begin{frame}\frametitle{Objectives}

In this lesson we will learn:
\begin{itemize}
\item solutions to linear and quasilinear first-order partial
  differential equations,
\item applications of first-order PDEs as mathematical models of
  traffic flow, structured populations, \textit{etc}.
\end{itemize}

\end{frame}
%%

\begin{frame}\frametitle{Transport Equations}

The general form of a first-order scalar PDE is
\[
u_{t} + \nabla \cdot f(x,t,u,\nabla u) = g(x,t,u)
\]
where
\begin{itemize}
\item $\nabla \cdot f$ denotes the divergence of $f$, and
\item $\nabla u$ denotes the gradient of $u$.
\end{itemize}

\end{frame}
%%

\begin{frame}\frametitle{Simple Case}

Let $c$ be a constant and consider
\begin{eqnarray*}
u_{t} + c \, u_{x} & = & 0	\\
\uncover<2->{%
\langle u_{x}, u_{t} \rangle \cdot \langle c, 1 \rangle & = & 0	\\
D_{\langle c, 1 \rangle} u(x,t) & = & 0	.
}
\end{eqnarray*}
\uncover<2->{%
\textbf{Remarks}:
\begin{itemize}
\item This can be interpreted as stating the directional derivative of
$u(x,t)$ in the direction of vector $\langle c, 1 \rangle$ is $0$.
\item<3-> Function $u(x,t)$ is constant along lines of the form $x - c
  \, t = k$.
\item<4-> Function $u(x,t) \equiv f(k)$ where $f$ is an arbitrary
  differentiable function.
\end{itemize}
}

\end{frame}
%%

\begin{frame}\frametitle{Confirmation}

Let $u(x,t) = f(x - c \, t)$ then
\begin{eqnarray*}
u_{t} & = & -c \, f'(x - c \, t)	\\
u_{x} & = & f'(x - c \, t)
\end{eqnarray*}
\uncover<2->{%
and
\[
u_{t} + c \, u_{x} = -c \, f'(x - c \, t) + c \, f'(x - c \, t) = 0	.
\]
\uncover<3->{%
Functions of the form $u(x,t) = f(x - c \, t)$ are referred to as the
general solution to this simple PDE.
}
}

\end{frame}
%%

\begin{frame}\frametitle{Initial Conditions}

If the value of $u(x,t)$ along the line where $t=0$ is specified, then
an initial condition of the form $u(x,0) = \phi(x)$ further specifies
the solution.
\begin{eqnarray*}
u(x,t) & = & f(x - c \, t)
\quad\mbox{(general solution)}	\\
u(x,0) & = & f(x) = \phi(x)
\quad\mbox{(initial condition)}	\\
\uncover<2->{%
u(x,t) & = & \phi(x - c \, t)
\quad\mbox{(solution to IVP)}
}
\end{eqnarray*}

\end{frame}
%%

\begin{frame}\frametitle{General First-Order Linear PDE}

The solution technique used previously can be extended to the
\textbf{general first-order linear PDE} of the form
\[
a(x,y) u_{x} + b(x,y) u_{y} + c(x,y) u = d(x,y)	,
\]
for $(x,y) \in D \subset \mathbb{R}^{2}$.

\vspace*{0.10in}

\textbf{Note}: symbol $y$ is used in place of $t$ as in the previous
example.
This change of variable is made so that $t$ can be used as a parameter
in the solution procedure to be outlined next.

\end{frame}
%%

\begin{frame}\frametitle{Vector Field (1 of 2)}

Suppose $(x,y) \equiv (x(t),y(t))$ where $t$ is a parameter, then
\[
\frac{d}{dt} u(x,y) = u_{x} x'(t) + u_{y} y'(t)
\]
and we can think of
\begin{eqnarray*}
a(x,y) u_{x} + b(x,y) u_{y} + c(x,y) u & = & d(x,y)
\quad\mbox{as}	\\
u_{x} x'(t) + u_{y} y'(t) & = & -c(x,y) u + d(x,y)
\quad\mbox{where}	\\
\frac{dx}{dt} & = & a(x,y)	\\
\frac{dy}{dt} & = & b(x,y)	.
\end{eqnarray*}
\uncover<2->{%
\textbf{Remark}: the parametric curve $(x(t), y(t))$ is tangent to the
vector field $\langle a(x,y), b(x,y) \rangle$ for all $(x,y) \in D$.
}

\end{frame}
%%

\begin{frame}\frametitle{Vector Field (2 of 2)}

If $\displaystyle \frac{dx}{dt} = a(x,y)$ and $\displaystyle
\frac{dy}{dt} = b(x,y)$ then
\[
\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{b(x,y)}{a(x,y)}	.
\]
Suppose the implicit form of the solution of this ODE is $\phi(x,y) =
k$ where $k$ is an arbitrary constant.

\vspace*{0.10in}

Along the curves defined by $\phi(x,y) = k$ the general first-order
linear PDE
\[
a(x,y) u_{x} + b(x,y) u_{y} = -c(x,y) u + d(x,y)
\]
becomes
\begin{eqnarray*}
u_{x} + u_{y} \, \frac{b(x,y)}{a(x,y)} & = & \frac{-c(x,y) u + d(x,y)}{a(x,y)}	\\
u_{x} + u_{y} \, \frac{dy}{dx} & = & \frac{-c(x,y) u + d(x,y)}{a(x,y)}
  \\
\frac{du}{dx} & = & \frac{-c(x,y) u + d(x,y)}{a(x,y)}	.
\end{eqnarray*}

\end{frame}
%%

\begin{frame}\frametitle{Ordinary Differential Equation}

\[
\frac{du}{dx} = \frac{-c(x,y) u + d(x,y)}{a(x,y)}
\]
is an ordinary differential equation for $u$ as a function of variable
$x$ (since $y \equiv y(x)$) and arbitrary constant $k$.

\vspace*{0.10in}

Solving this ODE yields $u \equiv u(x,k) \equiv u(x,y)$ since $k = \phi(x,y)$.

\end{frame}
%%

\begin{frame}\frametitle{Characteristics}

\begin{itemize}
\item The ordinary differential equations
\begin{eqnarray*}
\frac{dy}{dx} & = & \frac{b(x,y)}{a(x,y)}	\\
\frac{du}{dt} & = & -c(x,y) u + d(x,y)
\end{eqnarray*}
are called the \textbf{characteristic equations}.
\item<2-> The solution curves $(x(t), y(t), u(x(t), y(t)))$ to the
  characteristic equations are called the \textbf{characteristic
    curves}.
\item<3-> The curves in the $xy$-plane of the form $(x(t), y(t))$ are
  called \textbf{characteristics}.
\end{itemize}

\end{frame}
%%

\begin{frame}\frametitle{Example}

Consider the initial value problem:
\begin{eqnarray*}
2 u_{x} + 3 u_{y} - 4 u & = & 0	\\
u(x,0) & = & \sin x	.
\end{eqnarray*}
\begin{enumerate}
\item Find the general solution to the PDE using the method of characteristics.
\item Find the solution to the IVP.
\end{enumerate}

\end{frame}
%%

\begin{frame}\frametitle{Solution (1 of 2)}

\[
2 u_{x} + 3 u_{y} - 4 u = 0
\]
In this PDE we can let $a(x,y) = 2$ and $b(x,y) = 3$, thus
\[
\frac{dy}{dx} = \frac{3}{2}
\implies
y = \frac{3}{2} x + \hat{k}
\iff
2 y - 3 x = k	.
\]
\textbf{Comment}: 
the characteristics for this PDE are straight lines of the form $2 y -
3 x = k$.

\vspace*{0.10in}

\uncover<2->{%
Think of $u(x,y) = u(x,y(x))$ so that
\[
\frac{du}{dx} = u_{x} + u_{y} \frac{dy}{dx} =
u_{x} + \frac{3}{2} u_{y} = 2 u
\]
which implies
\[
u(x,y) = f(k) e^{2 x} = f(2 y - 3 x) e^{2 x}
\]
where $f$ is an arbitrary differentiable function.
}

\end{frame}
%%

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

In order to satisfy the initial condition
\begin{eqnarray*}
u(x,0) & = & \sin x	\\
f(-3x) e^{2 x} & = & \sin x	\\
f(-3x) & = & e^{-2 x} \sin x	\\
f(z) & = & -e^{2 z/3} \sin (z/3)	.
\end{eqnarray*}
Thus the solution to the IVP is
\[
u(x,y) = -e^{2(2y-3x)/3} \sin((2y-3x)/2) e^{2 x} =
-e^{4 y/3} \sin \left( y - \frac{3}{2} x \right)	.
\]

\end{frame}
%%

\begin{frame}\frametitle{Illustration}

\[
u(x,y) = -e^{4 y/3} \sin \left( y - \frac{3}{2} x \right)
\]
\begin{center}
\resizebox{0.80\textwidth}{!}{\includegraphics{example01}}
\end{center}

\end{frame}
%%

\begin{frame}\frametitle{Example}

Find the general solution to the first-order linear PDE:
\[
-y \, u_{x} + x \, u_{y} = 0	.
\]

\end{frame}
%%

\begin{frame}<handout:0>\frametitle{Solution (1 of 2)}

If we let $a(x,y) = -y$ and $b(x,y) = x$ then the characteristics
satisfy the ODE
\begin{eqnarray*}
\frac{dy}{dx} & = & -\frac{x}{y}	\\
y \, dy & = & - x \, dx	\\
\int y \, dy & = & - \int x \, dx	\\
\frac{1}{2} y^{2} & = & -\frac{1}{2} x^{2} + \hat{k}	\\
x^{2} + y^{2} & = & k	.
\end{eqnarray*}
In this example the characteristics are circles of radius $\sqrt{k}$
for $k \ge 0$.

\end{frame}
%%

\begin{frame}<handout:0>\frametitle{Solution (2 of 2)}

Assuming $u(x,y) = u(x,y(x))$ then
\begin{eqnarray*}
\frac{du}{dx} & = & u_{x} + u_{y} \frac{dy}{dx}	\\
	& = & u_{x} - \frac{x}{y} u_{y}	\\
\frac{du}{dx} & = & 0	\\
u	& = & f(k)	\\
u(x,y) & = & f(x^{2} + y^{2})
\end{eqnarray*}
where $f$ is an arbitrary differentiable function.

\end{frame}
%%

\begin{frame}\frametitle{Example}

Solve the initial value problem:
\begin{eqnarray*}
2 x \, y \, u_{x} + u_{y} - u & = & 0	\\
u(x,0) & = & x	,
\end{eqnarray*}
for $x > 0$ and $y > 0$.

\end{frame}
%%

\begin{frame}<handout:0>\frametitle{Solution (1 of 2)}

Let $a(x,y) = 2 x \, y$ and $b(x,y) = 1$ then
\begin{eqnarray*}
\frac{dy}{dx} & = & \frac{1}{2 x \, y} 	\\
y \, dy & = & \frac{1}{2 x} \, dx	\\
\int y \, dy & = & \int \frac{1}{2 x} \, dx	\\
\frac{1}{2} y^{2} & = & \frac{1}{2} \ln x + \hat{k}	\\
y^{2} - \ln x & = & k	.
\end{eqnarray*}
These are the characteristics of the solution.

\end{frame}
%%

\begin{frame}<handout:0>\frametitle{Solution (2 of 2)}
\fontsize{10pt}{12}\selectfont

If $u(x,y) = u(x,y(x))$ then
\begin{eqnarray*}
\frac{du}{dx} & = & u_{x} + u_{y} \, \frac{dy}{dx}	\\
	& = & u_{x} + \frac{1}{2 x \, y} u_{y}	\\
\frac{du}{dx} & = & \frac{u}{2 x \, y} = \frac{u}{2 x \sqrt{k + \ln x}}	\\
\frac{1}{u} \, du & = & \frac{1}{2 x \sqrt{k + \ln x}} \, dx	\\
\int \frac{1}{u} \, du & = & \int \frac{1}{2 x \sqrt{k + \ln x}} \, dx
  \\
\ln u & = & \sqrt{k + \ln x} + \ln f(k)	\\
u(x,y)	& = & f(k) e^{\sqrt{k + \ln x}} = f(y^{2} - \ln x) e^{y}
\end{eqnarray*}
where $f$ is an arbitrary differentiable function.

\vspace*{0.05in}

\uncover<2->{%
Since $u(x,0) = x$ then
\[
x = f(-\ln x)
\iff f(x) = e^{-x}
\implies
u(x,y) = e^{-y^{2} + \ln x} e^{y} = x \, e^{y - y^{2}}	.
\]
}

\end{frame}
%%

\begin{frame}<handout:0>\frametitle{Illustration}

\[
u(x,y) = x \, e^{y - y^{2}}
\]
\begin{center}
\resizebox{0.80\textwidth}{!}{\includegraphics{example02}}
\end{center}

\end{frame}
%%

\begin{frame}\frametitle{Homework}

\begin{itemize}
\item Read Section $2.1$
\item Exercises: 1--5
\end{itemize}

\end{frame}
%%

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

A first-order PDE is called \textbf{quasilinear} if it has the form
\[
a(x,y,u) u_{x} + b(x,y,u) u_{y} - c(x,y,u) = 0
\]
(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
\[
a(x,y) u_{x} + b(x,y) u_{y} - c(x,y,u) = 0	.
\]

\vspace*{0.10in}

We will generalize the method of characteristics in order to solve
quasilinear PDEs.

\end{frame}
%%

\begin{frame}\frametitle{Method of Lagrange}

Solutions of the quasilinear PDE
\[
a(x,y,u) u_{x} + b(x,y,u) u_{y} - c(x,y,u) = 0
\]
can be expressed implicitly as $\varphi(x,y,z) = 0$ when
$\varphi(x,y,z)$ is a solution to the PDE
\[
a(x,y,z) \varphi_{x} + b(x,y,z) \varphi_{y} + c(x,y,z) \varphi_{z} = 0	.
\]

\end{frame}
%%

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

\begin{itemize}
\item Suppose $u \equiv u(x,y)$ is a solution to
\[
a(x,y,u) u_{x} + b(x,y,u) u_{y} - c(x,y,u) = 0	.
\]
\item Define $\varphi(x,y,z) = u(x,y) - z$, then
\[
\varphi_{x} = u_{x}	,
\quad
\varphi_{y} = u_{y}	,
\quad
\varphi_{z} = -1	.
\]
\item Substitute into the expression
\begin{eqnarray*}
\lefteqn{a(x,y,z) \varphi_{x} + b(x,y,z) \varphi_{y} + c(x,y,z) \varphi_{z}}	\\
\uncover<2->{%
	& = &
a(x,y,u(x,y)) u_{x} + b(x,y,u(x,y)) u_{y} + c(x,y,u(x,y)) (-1)	\\
	& = & 0	,
}
\end{eqnarray*}
\uncover<2->{%
when $z = u(x,y)$ or equivalently when $\varphi(x,y,z) = 0$.
}
\end{itemize}

\end{frame}
%%

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

\begin{itemize}
\item Suppose $\varphi(x,y,z)$ is a solution of
\[
a(x,y,z) \varphi_{x} + b(x,y,z) \varphi_{y} + c(x,y,z) \varphi_{z} = 0
\]
and let $(x_{0},y_{0},z_{0})$ be a point on the surface
$\varphi(x,y,z) = 0$ for which $\varphi_{z}(x_{0},y_{0},z_{0}) \ne 0$.
\item Near point $(x_{0},y_{0},z_{0})$ the surface $\varphi(x,y,z) =
  0$ is the graph of some function, call it $u(x,y)$.
\[
\varphi(x,y,u(x,y)) = 0
\]
\item Differentiate the equation above.
\begin{eqnarray*}
\varphi_{x} + \varphi_{z} u_{x} & = & 0
\iff \varphi_{x} = -\varphi_{z} u_{x}	\\
\varphi_{y} + \varphi_{z} u_{y} & = & 0
\iff \varphi_{y} = -\varphi_{z} u_{y}
\end{eqnarray*}
\end{itemize}

\end{frame}
%%

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

\begin{itemize}
\item So far we have $\varphi(x,y,z) = 0$ when $z = u(x,y)$ where
  $\varphi(x,y,z)$ is a solution of
\[
a(x,y,z) \varphi_{x} + b(x,y,z) \varphi_{y} + c(x,y,z) \varphi_{z} = 0
\]
\item We also know $\varphi_{x} = -\varphi_{z} u_{x}$ and $\varphi_{y}
  = -\varphi_{z} u_{y}$ where $\varphi_{z} \ne 0$.
\item Substitute into the PDE above.
\begin{eqnarray*}
-a(x,y,u) \varphi_{z} u_{x} - b(x,y,u) \varphi_{z} u_{y} + c(x,y,z)
  \varphi_{z} & = & 0	\\
a(x,y,u) u_{x} + b(x,y,u) u_{y} - c(x,y,u) & = & 0
\end{eqnarray*}
So $u(x,y)$ solves the quasilinear PDE.
\end{itemize}

\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{eqnarray*}
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{eqnarray*}
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{eqnarray*}
\frac{dx}{dt} & = & a(x,y,u)	\\
\frac{dy}{dt} & = & b(x,y,u)	\\
\frac{du}{dt} & = & c(x,y,u)	
\end{eqnarray*}
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{0.80\textwidth}{!}{\includegraphics{family}}
\end{center}

\end{frame}
%%

\begin{frame}\frametitle{Solution Surface}

The union of all the characteristic curves through points on $\Gamma$
is an solution (integral) surface to the quasilinear PDE.
\begin{center}
\resizebox{0.80\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{eqnarray*}
x & = & A \cos t + B \sin t	\\
y & = & B \cos t - A \sin t	\\
u & = & -\ln (C - t)	.
\end{eqnarray*}

\vspace*{0.10in}

\uncover<2->{%
The non-characteristic curve $\Gamma$ is parameterized as
\begin{eqnarray*}
x & = & s	\\
y & = & \sin s	\\
u & = & 0	.
\end{eqnarray*}
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{eqnarray*}
x(0) & = & s = A	\\
y(0) & = & \sin s = B	\\
u(0) & = & 0 = -\ln C
\iff C = 1
\end{eqnarray*}

\vspace*{0.10in}

The characteristic curve intersecting $\Gamma$ at $t = 0$ can be
parameterized as
\begin{eqnarray*}
x(t) & = & s \cos t + \sin s \sin t	\\
y(t) & = & \sin s \cos t - s \sin t	\\
u(t) & = & -\ln (1 - t)	.
\end{eqnarray*}
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{eqnarray*}
x & = & s \cos t + \sin s \sin t	\\
y & = & \sin s \cos t - s \sin t	\\
u & = & -\ln (1 - t)	.
\end{eqnarray*}
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{eqnarray*}
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{eqnarray*}
Multiply the first equation by $\sin t$ and the second equation by
$\cos t$ and add them together.
\begin{eqnarray*}
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{eqnarray*}
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{eqnarray*}
\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{eqnarray*}

\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}<handout:0>\frametitle{Solution (1 of 3)}

The characteristic system is:
\begin{eqnarray*}
\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{eqnarray*}
\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{eqnarray*}
x(0) & = & A = s^{2}	\\
y(0) & = & B = \sin s	\\
u(0) & = & 0 = 0 + C
\implies C = 0
\end{eqnarray*}
}

\end{frame}
%%

\begin{frame}<handout:0>\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}<handout:0>\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}<handout:0>\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, quasi linear PDE with side
condition.
\begin{eqnarray*}
u_{x} + u \, u_{y} & = & 6 x	\\
u(0,y) & = & 3 y
\end{eqnarray*}

\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{eqnarray*}
\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{eqnarray*}

\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{eqnarray*}
x(0) & = & 0 = A	\\
y(0) & = & s = C	\\
u(0) & = & 3 s = B
\end{eqnarray*}
Consequently the characteristic curves are in parametric form:
\begin{eqnarray*}
x(t) & = & t	\\
y(t) & = & t^{3} + s (3 t + 1)	\\
u(t) & = & 3 t^{2} + 3 s	.
\end{eqnarray*}

\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{eqnarray*}
s	& = & \frac{u}{3} - x^{2}	\\
t	& = & x	.
\end{eqnarray*}
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--8
\end{itemize}

\end{frame}
%%

%%%
\end{document}

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