%%%%		-*- 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:      04/01/2025
%%%%
%%%% Purpose:   Burgers' Equation
%%%%

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\title{Burgers' Equation}
\subtitle{\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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%\section{Introduction}

\begin{frame}\frametitle{Objectives}

In this lesson we will:
\begin{itemize}
\item introduce Burgers' equation,
\item examine conservation laws' role in determining solutions,
\item develop traveling wave solutions to Burgers' equation,
\item introduce the Cole-Hopf transformation,
\item introduce the inviscid form of Burgers' equation, and
\item recognize conditions forming shocks and rarefactions in
  solutions to PDEs.
\end{itemize}

\end{frame}
%%

\begin{frame}\frametitle{Standard Form of Burgers' Equation}

\[
%\label{eq:nlpde:burgers}
\frac{\partial u}{\partial t} + u \frac{\partial u}{\partial x} -
\nu \frac{\partial^{2} u}{\partial x^{2}}
	= 0	,
\]
where the diffusion coefficient $\nu > 0$ is a constant.
\begin{itemize}
\item Burgers' equation was first proposed to model the motion of
  turbulent fluid flow.
\item The nonlinear term $u \, u_{x}$ models a convection process.
\item The term $\nu \, u_{xx}$ models the effect of viscosity or the diffusion
process.
\item If $\nu = 0$, the equation is known as the
inviscid Burgers' equation.
\end{itemize}

\end{frame}
%%

\begin{frame}\frametitle{Conservation Laws}

\begin{itemize}
\item The solution to Burgers' equation is sometimes not unique or is
multi-valued.
\item If a quantity must be conserved in Burgers' equation, that
  quantity can be used to determine which of the
multiple values (if any) is the ``correct'' solution to
Burgers' equation.
\end{itemize}

Integrate the viscous Burgers'
equation with respect to $x$ over an
arbitrary interval $[a, b]$.
\begin{align*}
\int_{a}^{b} u_{t} \, dx + \int_{a}^{b} u \, u_{x} \, dx - \nu
  \int_{a}^{b} u_{xx} \, dx
	&= 0	\\
%\label{eq:nlpde:Burgersconserve}
\int_{a}^{b} u_{t} \, dx + \frac{1}{2} (u(b, t))^{2} - \frac{1}{2}
  (u(a, t))^{2} - \nu u_{x}(b, t) + \nu u_{x}(a, t)
	&= 0
\end{align*}
By Leibniz's rule,
\[
%\label{eq:nlpde:BurgersconserveLeibniz}
\frac{d}{dt} \int_{a}^{b} u(x, t) \, dx
	= \int_{a}^{b} u_{t} \, dx + u(b, t) \frac{d b}{dt} - u(a, t) \frac{d a}{d t}	.
\]

\end{frame}
%%

\begin{frame}\frametitle{Conserved Quantity}

%\label{eq:nlpde:conserve}
\begin{align*}
\frac{d}{d t} \int_{a}^{b} u(x, t) \, dx
	= {} & \left( u(b, t) \frac{d b}{d t} - \frac{1}{2} (u(b, t))^{2} + \nu
  u_{x}(b, t) \right)	\\
	& - \left( u(a, t) \frac{d a}{d t} - \frac{1}{2} (u(a, t))^{2} + \nu
  u_{x}(a, t) \right)	.
\end{align*}
The rate of change of $\int_{a}^{b} u(x, t) \, dx$ is the quantity,
\[
%\label{eq:nlpde:boundary}
\left[ u(x, t) \frac{d x}{d t} - \frac{1}{2} (u(x, t))^{2} + \nu \,
  u_{x}(x, t) \right]_{x=a}^{x=b}	.
\]
The rate of change of the integral of $u$ over $[a, b]$ is
dependent only on the values at the endpoints of the integral (namely
$a$ and $b$).
Thus the integral of $u$ is said to be \textit{conserved}.

\end{frame}
%%

\begin{frame}\frametitle{Traveling Wave Solutions}

Set $\xi = x - c \, t$ and substitute $u(x, t) = U(\xi)$ into
Burgers' equation.
\[
-c \, \frac{dU}{d\xi} + u \, \frac{dU}{d\xi} - \nu \, \frac{d^{2} U}{d \xi^{2}}
	= 0
\]
Integrate both sides of the equation with respect to $\xi$.
\[
%\label{eq:nlpde:burgerstrav2}
-c \, U + \frac{1}{2} U^{2} - \nu \, \frac{d U}{d \xi}
	= A	,
\]
where $A$ is a constant.
Therefore,
\[
%\label{eq:nlpde:burgerstrav3}
\frac{d U}{d \xi}
	= \frac{1}{2 \nu} (U^{2} - 2 c \, U - 2 A)	.
\]

\end{frame}
%%

\begin{frame}\frametitle{Remarks}

\begin{itemize}
\item There is no bounded solution if the quadratic equation
$U^{2} - 2 c \, U - 2 A = 0$ has only one
real root (other than the constant solution) or two complex roots.
\item Assume $U^{2} - 2 c \, U - 2 A = 0$ has two distinct
real roots, denoted $U_{1} = c - \sqrt{c^{2} + 2 A}$
and $U_{2} = c + \sqrt{c^{2} + 2 A}$.
\item The constant functions $U(\xi) = U_{1}$ and $U(\xi) = U_{2}$ are
equilibrium solutions.
\item If $U(\xi)$ is a solution to the ODE with
initial value between $U_{1}$ and $U_{2}$, then by the uniqueness of
solutions to the ordinary differential equation, $U(\xi)$ will be
bounded and stay between $U_{1}$ and $U_{2}$.
\end{itemize}

\end{frame}
%%

\begin{frame}\frametitle{Separation of Variables (1 of 2)}

Separate the variables and rewrite the ODE as
\[
\frac{1}{(U - U_{1})(U - U_{2})} \frac{dU}{d\xi}
	= \frac{1}{2 \nu}
\]
Integrate with respect to $\xi$.
\[
\ln \left| \frac{U - U_{1}}{U - U_{2}} \right|
	= \frac {(U_{1} - U_{2})}{2 \nu} \xi + B	.
\]
where $B$ is an arbitrary constant of integration.
Use the fact that $U_{1} < U(\xi) < U_{2}$,
\[
\frac{U - U_{1}}{U_{2} - U}
	= e^{\alpha \xi + B}.
\]
where $\alpha = (U_{1} - U_{2})/(2 \nu) < 0$.
Solve this equation for $U$.
\[
%\label{eq:nlpde:Ufunction}
U(\xi)
	= \frac{U_{1} + U_{2} e^{\alpha \xi+B}}{1+ e^{\alpha \xi+B}}	.
\]

\end{frame}

%%
\begin{frame}\frametitle{Separation of Variables (2 of 2)}

The solution $U$ satisfies
\[
\lim_{\xi \to -\infty} U(\xi)
	= U_{2}
\mbox{ and }
\lim_{\xi \to \infty} U(\xi)
	= U_{1}	.
\]
The traveling wave solution for Burgers' equation:
\[
%\label{eq:nlpde:burgerstrav}
u(x, t)
	= U(x - c \, t)
	= \frac{U_{1} + U_{2} e^{\alpha (x - c \, t) + B}}{1+
          e^{\alpha (x - c \, t) + B}}	,
\]

\end{frame}
%%

\begin{frame}\frametitle{Special Case}

If $B = 0$, the traveling wave solution becomes
\[
u(x, t)
	= U(x - c \, t)
	= \frac{U_{1} + U_{2} e^{\alpha (x - c \, t)}}{1+ e^{\alpha (x - c \, t)}}
\]
and the initial wave profile is obtained by setting $t = 0$,
\[
u(x, 0)
	= U(x)
	= \frac{U_{1} + U_{2} e^{\alpha x}}{1+ e^{\alpha x}}	.
\]
Only when the initial condition is given by the function above or one
of its horizontal translations does this lead to a traveling wave solution.

\end{frame}
%%

\begin{frame}\frametitle{Properties of Traveling Wave Solutions}

\begin{enumerate}
\item For any fixed $t$, a traveling wave solution decreases from
  $U_{2}$ to $U_{1}$ as $x$ increases from $-\infty$ to $\infty$.
%
\item The traveling waves travel from the left to right with a speed
  $c = (U_{1} + U_{2})/2$ since $U_{1}$ and $U_{2}$ are solutions of the
  quadratic equation $U^{2} - 2 c \, U - 2 A = (U - U_{1})(U - U_{2})
  = 0$.
%
\item As the viscosity $\nu$ approaches $0$, for fixed $t$ and $x$, if
  $x < c \, t$, the solution $u(x, t) \to U_{2}$ and if $x > c \, t$,
  $u(x, t) \to U_{1}$.
In the limit, a shock wave develops.
\end{enumerate}

\end{frame}
%%

\begin{frame}\frametitle{Arbitrary Initial Conditions}

\textbf{Question}:  does
Burgers' equation have a solution for any given initial condition,
\[
%\label{eq:nlpde:burgersIV}
u(x, 0)
	= f(x)
\mbox{ for $-\infty < x < \infty$.}
\]
If there exists a solution to the initial value problem, is the
solution unique and is there a way to find the solution explicitly?

\end{frame}
%%

\begin{frame}\frametitle{Cole-Hopf Transformation (1 of 2)}

Rewrite Burgers' equation as
\[
%\label{eq:nlpde:burgers1}
\frac{\partial u}{\partial t} +
\frac{\partial }{\partial x} \left(
\frac{1}{2} u^{2} - \nu \frac{\partial u}{\partial x}
\right)
	= 0
\mbox{ for $-\infty < x < \infty$ and $t > 0$.}
\]
The Cole-Hopf transformation is defined by
\[
%\label{eq:nlpde:cole-hopf}
u
	= -2 \nu \frac{v_{x}}{v}
\iff
u
	= -2 \nu \frac{\partial}{\partial x} \left[ \ln v \right]	,
\]
where it is implicitly assumed that the unknown function
$v(x, t) > 0$.
The chain rule implies 
\begin{align*}
\frac{\partial u}{\partial t}
	&= -2 \nu \frac{v \, v_{xt} - v_{x} v_{t}}{v^{2}}
	= -2 \nu \frac{v \, v_{tx} - v_{t} v_{x}}{v^{2}}
	= -2 \nu \frac{\partial}{\partial x} \left[ \frac{v_{t}}{v} \right] 	\\
\frac{\partial u}{\partial x}
	&= -2 \nu \frac{v \, v_{xx} - (v_{x})^{2}}{v^{2}}	.
\end{align*}

\end{frame}
%%

\begin{frame}\frametitle{Cole-Hopf Transformation (2 of 2)}

This allows Burgers' equation to be written as
\[
-2 \nu \frac{\partial}{\partial x} \left[ \frac{v_{t}}{v} \right] +
\frac{\partial}{\partial x} \left[ 2 \nu^{2} \frac{v_{xx}}{v} \right]
	= 0	,
\]
which simplifies to the following:
\[
%\label{eq:nlpde:veqn}
%\frac{\partial}{\partial x} \left[ \frac{v_{t}}{v} \right] -
%\nu \frac{\partial}{\partial x} \left[ \frac{v_{xx}}{v} \right]
%	=
\frac{\partial}{\partial x} \left[
\frac{v_{t} - \nu \, v_{xx}}{v}
\right]
	= 0	.
\]
\textbf{Remark}: if a function  $v(x, t) > 0$ is a solution of the linear,
homogeneous heat equation
\[
%\label{eq:nlpde:heat}
v_{t} = \nu \, v_{xx}	,
\]
the numerator of the term inside the partial derivative
will be zero and thus $u(x, t)$ is a solution to Burgers' equation.

\vspace*{0.10in}

Conversely if $u(x, t)$ is any  solution to Burgers' equation, the
function
\[
%\label{eq:nlpde:inversecole-hopf}
v(x, t) = e^{-\frac{1}{2 \nu} \int_{0}^{x} u(y, t) \, dy}
\]
solves the transformed equation.

\end{frame}
%%

\begin{frame}\frametitle{Example}

Function $v(x, t) = A + B e^{-\nu t} \sin x$ is a solution of the heat equation
for any constants $A$ and $B$,.

\vspace*{0.10in}

If $A$ is large enough to guarantee that
$v(x, t)$ is positive everywhere,
\[
u(x, t)
	= \frac{-2 \nu B e^{-\nu t}\cos x}{A + B e^{-\nu t} \sin x}
\]
is a solution to Burgers's equation.

\end{frame}
%%

\begin{frame}\frametitle{Burgers' Equation IVP}

Suppose the initial condition for Burgers' equation is $u(x, 0) =
f(x)$, then
\[
%\label{eq:nlpde:heatIV}
v(x, 0)
	= g(x)
	= e^{-\frac{1}{2 \nu} \int_{0}^{x} f(y) \, dy}	.
\]
and the solution to the heat equation is
\[
%\label{eq:nlpde:vfunction}
v(x, t)
	= \frac{1}{\sqrt{4 \pi \nu t}}
\int_{-\infty}^{\infty} g(y) e^{-\frac{(x-y)^{2}}{4 \nu t}} \, dy
\mbox{ for $t > 0$.}
\]
Therefore
\[
u(x, t)
%\label{eq:nlpde:IVsolution}
	= \frac{\int_{-\infty}^{\infty} (x-y) e^{-\frac{1}{2 \nu}
\left( \int_{0}^{y} f(s) \, ds + \frac{(x-y)^{2}}{2 t} \right)} \, dy}%
{t \int_{-\infty}^{\infty} e^{-\frac{1}{2 \nu}
\left( \int_{0}^{y} f(s) \, ds + \frac{(x-y)^{2}}{2 t} \right)} \, dy}	.
\]

\end{frame}
%%

\begin{frame}\frametitle{Inviscid Burgers' Equation}

If $\nu = 0$, then
Burgers' equation becomes
\[
%\label{eq:nlpde:inviscid}
\frac{\partial u}{\partial t} + u \, \frac{\partial u}{\partial x}
	= 0	,
\]
known as the inviscid Burgers' equation.

\vspace*{0.10in}

Use the method of characteristics to solve the inviscid Burgers' equation.
\[
\frac{d x}{d s}
	= u	,
\quad
\frac{d t}{d s}
	= 1	,
\mbox{ and }
\frac{d u}{d s}
	= 0	.
\]
The 3rd characteristic equation indicates that
$u$ is constant along characteristics.
The 2nd characteristic equation allows the
parameter $s$ to be identified with the independent variable $t$.
Hence the 1st characteristic equation may be rewritten as $dx/dt =
u$ and thus
\begin{align*}
x
	&= u \, t + \gamma_{1}	\\
u
	&= \gamma_{2}
\end{align*}
where $\gamma_{1}$ and $\gamma_{2}$ are arbitrary constants which may be
determined from initial conditions imposed on the inviscid Burgers'
equation.

\end{frame}
%%

\begin{frame}\frametitle{Inviscid Burgers' Equation IVP}

If the initial condition is $u(x, 0) = f(x)$, then along the
characteristic through $(x_{0}, 0)$ it is the case that $\gamma_{1} =
x_{0}$ and thus $x_{0} = x - u \, t$ on the characteristic.
Therefore the solution can be written in implicit form as
\[
%\label{eq:nlpde:Burgersimplicitsoln}
u(x, t)
	= f(x_{0})
	= f(x - u(x, t) t)	.
\]
For example if $u(x, 0) = e^{-x^{2}}$,
the implicit form of the solution is
\[
u(x, t)
	= e^{-(x - u(x, t) t)^{2}}	.
\]

\end{frame}
%%

\begin{frame}\frametitle{Intersection of Characteristics}

The characteristics are straight lines with slopes that depend
on $u$, so unless $u$ is a constant, these characteristics may intersect
for some $t > 0$.
\begin{center}
\resizebox{0.50\textwidth}{!}{\includegraphics{charintersect}}
\end{center}
The thick curve is the locus of points where characteristics intersect
and is called a shock.

\end{frame}
%%

\begin{frame}\frametitle{Shocks and Rarefactions}

\begin{itemize}
\item If the $t$-axis is oriented vertically and the $x$-axis is oriented
horizontally, the slope of the
characteristics is $1/u(x, 0)$, thus if the initial condition $u(x, 0)
= f(x)$ is a decreasing function
of $x$ in some interval, the slopes of the characteristics will
increase, leading to the intersection of the characteristics at some
$t > 0$.
At the location and time of intersection of two characteristics with
different values of $u$, a phenomenon known as a
\textbf{shock} occurs.
\item If the initial condition contains a jump discontinuity at some $x_{0}$
where $f(x_{0}^{-}) < f(x_{0}^{+})$ there is a region between the
characteristics $x(t) = f(x_{0}^{-}) t + x_{0}$ and $x(t) =
f(x_{0}^{+}) t + x_{0}$ that appears to have no solution.
This phenomenon is known as a \textbf{rarefaction}.
\end{itemize}

\end{frame}
%%

\begin{frame}\frametitle{Quasilinear Cauchy IVP}

%\label{eq:nlpde:burgers:jumpdata}
\begin{align*}
u_{t} + c(u) u_{x}
	&= 0
\mbox{ for $-\infty < x < \infty$ and $t > 0$}	\\
u(x, 0)
	&= \left\{ \begin{array}{rl}
u_{L}	& \mbox{if $x < x_{0}$}	\\
u_{R}	& \mbox{if $x > x_{0}$}
\end{array} \right.
\end{align*}
Suppose $u_{L} > u_{R}$.
Imagine the fluid to the left of $x_{0}$ is traveling faster than the
fluid to the right of $x_{0}$ and thus is overtaking it resulting in a shock.

\vspace*{0.10in}

In order to illustrate the handling of a shock, rewrite the
quasilinear partial differential in conservation form as
\[
u_{t} + (C(u))_{x}
	= 0
\]
where $C'(u) = c(u)$.

\end{frame}
%%

\begin{frame}\frametitle{Rankine-Hugonoit Condition}

Assume there is a smooth solution with the property that $u(x, t) =
h(x - c_{s} t)$ where $c_{s}$ denotes the speed of the shock and
$\lim_{x \to -\infty} u(x, t) = u_{L}$ and $\lim_{x \to \infty} u(x, t) = u_{R}$.
\begin{align*}
-c_{s} h'(x - c_{s} t) + C(h(x - c_{s} t))'
	&= 0	\\
\int_{-\infty}^{\infty} \left[ -c_{s} h'(x - c_{s} t) + C(h(x - c_{s}
  t))' \right] \, dx
	&= 0	\\
-c_{s} (u_{R} - u_{L}) + C(u_{R}) - C(u_{L})
	&= 0	\\
c_{s}
	&= \frac{C(u_{R}) - C(u_{L})}{u_{R} - u_{L}}
\end{align*}
The shock speed is a function of $t$.
The last equation is often concisely written as
\[
%\label{eq:nlpde:burgers:RankineHugonoit}
\frac{dx_{s}}{dt}
	= \frac{[ C(u) ]}{[ u ]}
\]
and is known as the \textbf{Rankine-Hugonoit}
condition.

\end{frame}
%%

\begin{frame}\frametitle{Determining the Location of a Shock}

Consider the IVP:
\begin{align*}
\frac{dx_{s}}{dt}
	&= \frac{[ C(u) ]}{[ u ]}	\\
x_{s}(0)
	&= x_{0}	.
\end{align*}
To the left of the shock, \textit{i.e.}, where $x < x_{s}(t)$, the
solution is defined as $u(x, t) = u_{L}$ while to the right of
the shock $u(x, t) = u_{R}$.
The shock itself is at
\[
x_{s}(t)
	= \left( \frac{C(u_{R}) - C(u_{L})}{u_{R} - u_{L}} \right) t + x_{0}	.
\]

\end{frame}
%%

\begin{frame}\frametitle{Illustration}

\begin{center}
\resizebox{1.00\textwidth}{!}{\includegraphics{quasilinear-shock}}
\end{center}
The typical solution to a quasilinear partial differential
  equation admitting a shock due to a jump discontinuity in the
  initial data.

\end{frame}
%%

\begin{frame}\frametitle{Rarefaction}

When $u_{L} < u_{R}$, the characteristics to
the left of $x_{0}$ angle away from the characteristics to the right.
As before the characteristics are described by the equations $x = c(f(x_{i}))
t + x_{i}$ where $f$ is a function denoting the initial data.
This leaves a region between the lines $x = c(u_{L}) t + x_{0}$ and $x =
c(u_{R}) t + x_{0}$ which appears to contain no characteristics.
A solution does exist in this equation if the initial data is
modified to include a set-valued condition.
Suppose the initial condition is written as
\[
f(x)
	= \left\{ \begin{array}{rl}
u_{L}	& \mbox{if $x < x_{0}$}	\\{}
[u_{L}, u_{R}]	& \mbox{if $x = x_{0}$}	\\
u_{R}	& \mbox{if $x > x_{0}$.}
\end{array} \right.
\]
Intuitively the initial data takes on all values in the interval
$[u_{L}, u_{R}]$ when $x = x_{0}$.

\end{frame}
%%

\begin{frame}\frametitle{Illustration}

In the rarefaction region the characteristics all begin at $x_{0}$ and
have the form
\[
x = c(u) t + x_{0}
\mbox{ for $u_{L} < u < u_{R}$.}
\]
\begin{center}
\resizebox{0.90\textwidth}{!}{\includegraphics{quasilinear-rarefaction}}
\end{center}

\end{frame}
%%

\begin{frame}\frametitle{Inviscid Burgers' Equation and Shocks}

If the inviscid Burgers' equation is thought of as a model of fluid
flow, then the downstream fluid is moving slower than the upstream
fluid, or equivalently, the upstream fluid is overtaking the
downstream fluid.
Since the dependent variable $u$ is constant along each
characteristic, the intersection of  characteristics implies $u$ is
multivalued, which means the solution is not well-defined.
\begin{center}
\resizebox{0.75\textwidth}{!}{\includegraphics{breakingwave}}
\end{center}

\end{frame}
%%

\begin{frame}\frametitle{Example}

The shock will form at the earliest time the surface $u(x, t)$ has a
vertical tangent.
Using implicit differentiation on the general implicit solution
to the inviscid
Burgers' equation, the slope of the solution in the $x$-direction is
\[
u_{x}
	= \frac{f'(z)}{1 + t \, f'(z)}	.
\]
A vertical tangent occurs when $u_{x}$
is undefined, or equivalently when $t = -1/f'(z)$.
Thus the earliest time this occurs is the time for which $f'(z)$ takes on
its negative minimum.

\end{frame}
%%

\begin{frame}\frametitle{Case: $f(x) = e^{-x^{2}}$}

\begin{itemize}
\item The shock forms at $t_{0} = \sqrt{e/2}$ when $f(x) = e^{-x^{2}}$.
\item The $x$-coordinate at which the shock forms is found by solving the
equation $1 + t_{0} f'(x_{0}) = 0$ which makes $u_{x}$ undefined and
gives the coordinate on the
$t = 0$ line where the characteristic passing through the beginning of
the shock passes.
\item Solving the equation,
\[
0
	= 1 - 2 \sqrt{\frac{e}{2}} x_{0} e^{-x_{0}^{2}}
\implies
x_{0}
	= \frac{1}{\sqrt{2}}	.
\]
\item The shock begins at the point with coordinates:
\[
(x, t)
	= \left( x_{0} + t_{0} f(x_{0}), t_{0} \right)
	= \left( \sqrt{2}, \sqrt{\frac{e}{2}} \right)
\]
\end{itemize}

\end{frame}
%%

\begin{frame}\frametitle{Illustration}

\begin{center}
\resizebox{0.75\textwidth}{!}{\includegraphics{inviscidBurgers}}
\end{center}

\end{frame}
%%

\begin{frame}\frametitle{Locating the Shock}

\begin{itemize}
\item Suppose the shock is located at position $x_{s}$ at time $t$.
\item The integral of $u$ is conserved even when passing through the shock
and since the shock has infinitesimal width, the definite integral must
be zero.
\item The correct value of $u$ can be found by placing the shock at
such as position that the shaded area to the right of $x_{s}$ and left
of the curve $u(x, t) = f(x - u(x, t) t)$ equals the shaded area to the
left of $x_{s}$ and right of the curve.
Since one region is on either side of the vertical line and the
regions are of equal area, their sum is $0$.
\item The value of $u$ where $x_{s}$ intersects the curve
$u(x, t) = f(x - u(x, t) t)$ and which separates the left and right
regions is then the ``correct'' value of
$u(x, t)$.
\end{itemize}

\end{frame}
%%

\begin{frame}\frametitle{Speed of the Shock}

Using the conservation of the integral of $u$,
by choosing $a < x_{s}$ and $b > x_{s}$ and setting $\nu = 0$, in the
limit as $a \to x_{s}^{-}$ and $b \to x_{s}^{+}$,
\begin{align*}
u(x_{s}+,t) \frac{d x_{s}}{d t} - \frac{1}{2} (u(x_{s}+,t))^{2}
	&= u(x_{s}-,t) \frac{d x_{s}}{d t} - \frac{1}{2} (u(x_{s}-,t))^{2}	\\
\frac{d x_{s}}{d t}
	&= \frac{1}{2} (u(x_{s}+,t) + u(x_{s}-,t))
\end{align*}
where $u(x_{s}+,t)$ and $u(x_{s}-,t)$ denote the values of the solution.
Therefore the shock moves at a velocity which is the average of the
values of $u$ immediately on either side of the shock.

\end{frame}
%%

%\begin{frame}\frametitle{Homework}

%\begin{itemize}
%\item Read Section $x.y$
%\item Exercises: 
%\end{itemize}

%\end{frame}
%%

%%%
\end{document}

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