\documentclass{article}
\usepackage{amsmath}
\usepackage{amssymb}
\usepackage{amsfonts}

\begin{document}


\begin{center}
Conservation Laws
\end{center}

Let $u$ be a density and $f$ be a flux. We define a conservation law (1D) as

\begin{equation*}
\frac{\partial u}{\partial t}+\frac{\partial f(u)}{\partial x}=0
\end{equation*}

Integrating with respect to x we get%
\begin{equation*}
\frac{d}{dt}\int_{a}^{b}u(x,t)dx=-\int_{a}^{b}\frac{\partial f(u)}{\partial x%
}dx=f(a)-f(b)
\end{equation*}

Hence, the change in the total "mass" changes only through fluxes that enter
and leave the boundary. In several dimesnions this generalizes to

\begin{equation*}
\frac{\partial u}{\partial t}+div(f)=0
\end{equation*}%
and%
\begin{equation*}
\frac{d}{dt}\int \int_{D}u(\overset{\rightarrow }{x},t)d\overset{\rightarrow 
}{x}=-\int \int div(f)d\overset{\rightarrow }{x}=\doint fds
\end{equation*}

\newpage

\begin{center}
Generalized solution
\end{center}

For a linear problem discontinuities occur only across characteristics. For
nonlinear problems we can have discontinuities anywhere. We need to define a
solution when it is discontinuous

\begin{equation*}
\int_{x_{L}}^{x_{R}}\left( \frac{\partial u}{\partial t}+\frac{\partial f}{%
\partial x}\right) \ dx=\frac{d}{dt}\int_{x_{L}}^{x_{R}}\frac{\partial u}{%
\partial t}\ dx+f_{R}-f_{L}
\end{equation*}%
Hence, space derivatives no longer appear. More generally let $\phi \in
C^{\infty }$ \newline
\noindent (compact support and infinite number of derivatives). Then%
\begin{equation*}
(\ast )\qquad \int_{-\infty }^{\infty }\left( \frac{\partial u}{\partial t}+%
\frac{\partial f}{\partial x}\right) \phi \ dx=\int_{-\infty }^{\infty }%
\frac{\partial u}{\partial t}\phi \ dx-\int_{-\infty }^{\infty }\frac{%
\partial \phi }{\partial x}f\ dx
\end{equation*}

\bigskip

\noindent \textit{Def}: A solution is a generalized solution if (*) is true
for every $\phi $ in some class of functions

\medskip

\noindent \textit{Note}: A classical solution is a generalized solution but
not necessarily conversely. \newline
\noindent Uniqueness may be lost when including generalized solutions !!

\begin{center}
\textbf{Rankine-Hugoniot}
\end{center}

\noindent Define:%
\begin{equation*}
I(t)=\int_{a}^{b}u(x,t)dx=\int_{a}^{y(t)}u(x,t)dx+\int_{y(t)}^{b}u(x,t)dx
\end{equation*}%
Denote:\qquad $u_{l}=u(y^{-})\quad u_{r}=u(y^{+})\quad s=\frac{dy}{dt}$

\begin{equation*}
\frac{dI}{dt}=\int_{a}^{y}\frac{\partial u}{\partial t}(x,t)dx+u_{l}s+%
\int_{y}^{b}\frac{\partial u}{\partial t}(x,t)dx-u_{r}s
\end{equation*}

but $\frac{du}{dt}=\frac{\partial f}{\partial x}$ . So

\begin{align*}
\frac{dI}{dt} &=-\int_{a}^{y}\frac{\partial f}{\partial x}%
(x,t)dx-\int_{y}^{b}\frac{\partial f}{\partial x}(x,t)dx+\left(
u_{l}-u_{r}\right) s \\
&= f_{a}-f_{l}-f_{b}+f_{r}+\left( u_{l}-u_{r}\right) s
\end{align*}

but $\frac{dI}{dt}=\int_{a}^{b}\frac{\partial f}{\partial x}%
(x,t)dx=f_{a}-f_{b}$

So%
\begin{align*}
f_{r}-f_{l} &= (u_{r}-u_{l})s \\
s\left[ u\right] &= \left[ f\right]
\end{align*}

\newpage

\begin{center}
example

\begin{equation*}
f(u)=\frac{u^{2}}{2}
\end{equation*}

\begin{equation*}
u_{t}+\left( \frac{u^{2}}{2}\right) _{x}=u_{t}+u\ u_{x}=0
\end{equation*}
\end{center}

So the slope of the characteristic curves are: $\frac{dt}{dx}=\frac{1}{u}$ .

\bigskip

Choose as initial data 
\begin{equation*}
u_0(x)=%
\begin{cases}
1 & x\leq 0 \\ 
1-x & 0\leq x\leq 1 \\ 
0 & 1\leq x%
\end{cases}%
\end{equation*}

\bigskip

Until t=1 the solution is single valued.

To choose the solution after t=1 we use the Rankine-Hugoniot condition 
\begin{equation*}
s=\frac{\left[ f\right] }{\left[ u\right] }=\frac{\frac{u_{r}^{2}}{2}-\frac{%
u_{l}^{2}}{2}}{u_{r}-u_{l}}=\frac{\frac{0-1}{2}}{0-1}=\frac{1}{2}
\end{equation*}%
So for $t>1$ we choose the solution%
\begin{equation*}
u(x,t)=%
\begin{cases}
1 & x\leq \frac{1+t}{2} \\ 
0 & \frac{1+t}{2}\leq x%
\end{cases}%
\end{equation*}

\newpage

\begin{center}
Entropy Condition
\end{center}

Characteristics on either side of the discontinuity in the direction of
increasing t intersect the line of discontinuity. Hence, we can connect
every point on the discontiuity back to the initial line. Equivalently%
\begin{equation*}
a(u_{l})>s>a(u_{r})\qquad a=\frac{\partial f}{\partial u}
\end{equation*}

\bigskip 

\noindent An equivalent formulation is%
\begin{equation*}
u_t^{\varepsilon }+f(u^{\varepsilon})_x =
(\varepsilon u_x)_x \text{\quad and \qquad} \varepsilon \rightarrow 0
\end{equation*}

\begin{center}
example
\end{center}

\begin{equation*}
f(u)=\frac{u^2}{2}
\end{equation*}

\bigskip

\noindent Choose as initial data 
\begin{equation*}
u_0(x)=%
\begin{cases}
0 & x\leq 0 \\ 
1 & x>1%
\end{cases}%
\end{equation*}

\noindent Instead of characteristics colliding we have a gap. Choose %
\begin{equation*}
u(x,t)=%
\begin{cases}
0 & x \leq \frac{t}{2} \\ 
1 & x >\frac{t}{2}%
\end{cases}%
\end{equation*}

\noindent This satisfies the Rankine-Hugoniot condition. 
\newline \noindent Consider%
\begin{equation*}
u(x,t)=\frac{x}{t}
\end{equation*}

\noindent This also satisfies the R-H condition. However, only the 
second one satisfies the entropy condition.

\bigskip 

Definition: A solution is a shock if it satisfies bothe Rankine-Hugoniot
condition and the entropy condition


\end{document}
