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\begin{center}
Traffic flow
\end{center}

\begin{equation*}
\frac{\partial \rho }{\partial t}+\frac{\partial f}{\partial x}=0
\end{equation*}

$\rho $ = density of cars = cars/km

$u$ = speed of cars

$f=\rho u$ = flux = cars/hours past a given point

\bigskip

Assumptions:

\begin{itemize}
\item flux is low at low speeds (bumper to bumper)

\ \ \ \ \ \ \ \ \ \ \ \ \ \ and high speed (distance between cars)

\item length of car = 5m

\item highest density is $u_{b}=200$ \ \ cars/km

\item $f_{\max }=$ 1500 cars/hour

\item $f_{\max }$ is achieved at 35 km / hour

\item optimal density is about 43 cars /km
\end{itemize}

\noindent So f(u) is a concave function. \newline
\noindent Assume we have a red light at $x=x_{j}$. We have cars stretched
from $x_s$ to $x_j$. So cars approach the traffic jam from $x<x_{s}$ \ and
they slow down. 
\begin{equation*}
s=\frac{\left[ f\right] }{\left[ u\right] }=\frac{\text{negative}}{\func{%
positive}}<0
\end{equation*}%
i.e. the shock wave travels backward. As drivers enter the shock wave they
have to decelerate rapidly. When the cars pass the traffic light $x>x_s$ we
have an expansion wave.

\bigskip

If there is a sequence of red lights separated by a distance $L$ then the
time for the shock to reach the previous light is 
\begin{equation*}
T=\frac{L(u_{b}-u_{\text{max}})}{f_{\text{max}}}\simeq \frac{L\text{ km }(200%
\text{ cars/km \ - }43\text{ cars/km)}}{1500\text{ cars/hr}}\simeq \frac{L}{%
10}\text{ hr}
\end{equation*}%
If the light is red longer than $T$ hours the traffic will come to a halt!

\newpage

\begin{center}
Another example
\end{center}

\noindent Consider $u=1-\rho $ \ \ ($\rho $ is relative density). So as the
density increases the velocity decreases. Then $f=\rho (1-\rho )=\rho -\rho
^{2}$. \quad $f^{^{\prime }}=1-2\rho $ .

\noindent So the characteristics are straight lines with slope $1-2\rho $ .

\noindent From the Rankine Hugoniot relationship we have

\begin{equation*}
s=\frac{\left[ \rho (1-\rho )\right] }{\rho }=\frac{\rho _{r}(1-\rho
_{r})-\rho _{l}(1-\rho _{l})}{\rho _{r}-\rho _{l}}
\end{equation*}

\bigskip

\noindent Assume we have a stop light at $x=1$ at $t=0$. Then the initial
condition is 
\begin{equation*}
\rho (x,0)=%
\begin{cases}
0 & x\leq 0\quad \text{and}\quad x\geq 1 \\ 
x & 0<x<1%
\end{cases}%
\end{equation*}%
The characteristics are given by

\begin{center}
$\left\{ 
\begin{array}{ccc}
x\leq t & x-s=t & \rho =0 \\ 
t<x<1-t & x-s=(1-2s)t & \rho =s \\ 
1-t<x<1+t & x-1=(1-2s)t & \rho =\frac{t-x+1}{2t} \\ 
x>t+1 & x-s=t & \rho =0%
\end{array}%
\right. $

\bigskip 

They first "meet" at $x=\frac{1}{2}$ \ $t=\frac{1}{2}$. The speed of the
shock is then given by%
\begin{equation*}
S=\frac{s+t-1}{2t}
\end{equation*}
\end{center}

\end{document}
