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\heading{Traffic, the ASEP model}
\auname{dekel shapira}



\begin{document}
\part*{Traffic}
\subsection*{The ASEP model}
The ASEP (Asymmetric Exclusion Process) model consists of a one dimensional lattice of \textit{N} cells. Each cell may contain a particle ($ n_i = 1 $) or be empty ($ n_i = 0 $). Time is discrete.\\
The dynamics is defined as follows: at time \textit{t}, each one of the particles checks its right and left neighbors. If the left cell is empty, it will hop to it with probability \textit{q} at time \textit{t + 1}, whereas if the right cell is empty it will hop with probability $1 - q$ at time $t+1$.\\
At the boundaries ($ n_1$ or $n_N $) the dynamics is different: if the cell at time \textit{t} is empty at the left (or respectively right) boundary, a particle will be injected with probability $ \alpha $ (or respectively $ \delta $) at time \textit{t+1}. If, on the other hand, the left (or respectively right) boundary is occupied at time \textit{t}, the particle will be removed from the left (respectively right) cell with probability $ \gamma $ (or respectively $ \beta $), see figure \ref{fig_asep1}.\\
\begin{figure}[h!b]
\center
\includegraphics[width=8cm]{asep.eps}
\caption{particles in the ASEP model (picture taken from \cite{derrida_08})}
\label{fig_asep1}
\end{figure}
This model can be seen as a heat transfer model, where each particle carries an energy unit $ \epsilon$, and the left and right side remain with constant temperatures $ \tau_a,\ \tau_b $ where $ exp[-\epsilon/\tau_a] = \alpha/\gamma,\ exp[-\epsilon/\tau_b] = \delta/\beta $. In addition, each boundary has a density of
\[ \rho_a = \dfrac{\alpha}{\alpha + \gamma},\ \rho_b = \dfrac{\delta}{\delta + \beta}
\] and
\textit{q} is a parameter which determines the external field magnitude (gravitation, electric field).
For our purposes we will set $ q=\gamma=\delta \rightarrow 0,\ \epsilon = 1 $, which can be seen as $\tau_a \rightarrow 0^+,\ \tau_b \rightarrow 0^-,\ \rho_a = 1,\ \rho_b = 0 $ and a strong external field.
\subsubsection*{Traffic}
Applying ASEP to traffic we can replace each particle with a car, driving from left to right on a one-lane road. Following this concept, drivers avoid collision by stopping if the next cell is occupied.\\
The $ \alpha $ parameter models the occupancy of the road segment which proceeds the one  referred to in our model, whereas the $1 - \beta$ parameter refers to the one following it.\\
Our aim is to find the current $ I( \alpha, \beta) $ in a steady state.\\
\begin{figure}[h!tb]
\center
\includegraphics[width=12cm]{car1.eps}
\caption{cars in the restricted ASEP model}
\label{fig_asep}
\end{figure}
One can solve the above model exactly, however, it is easier to make the approximation of a \emph{random} update process: instead of simultaneously updating the cells, at each time point, an integer $ 0 \leq i \leq N$ will be chosen with equal probability. The above rules would then apply to the chosen cell (where $0$ stands for the case of transition to \textit{$ i=1 $} with probability $\alpha$).
\subsection*{Steady state and mean-field approximation}
For a steady state we will require the average occupancy $ \langle n_i(t) \rangle $ to remain unchanged (in time), while the mean field approximation consists of the  $ n_i's  $ to be  uncorrelated : $ \langle n_i n_j \rangle = \langle n_i \rangle \langle n_j \rangle \equiv p_i p_j\ (i \neq j)$.\\
Considering one of the cells $ 2 \leq i \leq N-2 $ at time $ t $ with occupancy $ n_i(t) $, we can write an expression for the occupancy at time $ t+1 $:
%BUG: find better solution for split
\begin{align}
\begin{split}
n_i(t+1) =\ & n_i(t) \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \mbox{with probability $1 - \dfrac{2}{N+1}$}\\
& n_i(t) + [1 - n_{i}(t)]n_{i-1}(t)\ \qquad \mbox{with probability $\dfrac{1}{N+1}$}\\
& n_i(t)n_{i+1}(t) \quad \qquad \qquad \quad \quad \ \ \mbox{with probability $\dfrac{1}{N+1}$}
\end{split}
\label{eq_prob_site}
\end{align}
The first expression relates to the case where neither $ i $ nor $ i-1 $ are chosen. the second, to the case where $ i-1 $ is chosen, and the third, to the case where $ i $ is chosen.\\
Averaging \eqref{eq_prob_site}, we can rewrite it as a rate equation:
\begin{align}
\begin{split}
\dot{p_i}(t) & \equiv \dfrac{\langle n_i(t+1) \rangle - \langle n_i(t) \rangle}{1} = - [ I_i(t) - I_{i-1}(t)] \\
I_i(t) & = \dfrac{1}{N+1}\Big ( \langle n_i \rangle - \langle n_i n_{i+1}\rangle \Big ) \\
\end{split}
\label{eq_change_of_p}
\end{align}
where similar results are found for $ p_1$ and $ p_N $.
Using the mean field approximation, the current can be written as
\begin{align}
I = -p \nabla p
\end{align}
Which is different from Fick's law:
\begin{align}
I = -D \nabla p, \qquad D = const
\end{align}
We now invoke the steady state conditions $\dot{p_i}(t) = 0 $. Finding $ I_{i} = I_{i-1} = I $, we can write \eqref{eq_change_of_p} as a recursion with the appropriate boundary conditions:
\begin{align}
p_{i+1} & = 1 - \dfrac{C}{p_i} \qquad \qquad 0 \leq i \leq N \label{eq_recursion}\\
p_0 & = \alpha \label{eq_boudary_0}\\
p_{N+1} & = 1 - \beta \label{eq_boudary_N}
\end{align}
where the 'normalized' current $ C = (N+1)I $ is our main interest.
%
\subsection*{Analyzing the flow}
%
Looking at \eqref{eq_recursion} we next try to find the \textit{flow} of the system. That is, the sign of $ p_{i+1} - p_i $ near $ p_i $. Defining the \textit{fixed points} to be the solutions of $ p_{i+1}(C,p_i) - p_i = 0$ we can characterize the flow according to the fixed points positions.\\
For $ C < 1/4$ there are two fixed points (Figure $\ref{fig_t}a$):
\begin{align}
p_{\pm} = \dfrac{1}{2}[1 \pm \sqrt{1 - 4C}] \label{eq_p_plus_minus}
\end{align}
Because of the flow directions $ p_- $ is called unstable while $ p_+ $ is called stable.\\
For $ C = 1/4 $ there is only one fixed point (Figure $\ref{fig_t}b$) which is \textit{marginal}, while for $ C > 1/4 $ there are no real fixed points (Figure $\ref{fig_t}c$).\\
\begin{figure}[h!tb]
\center
\includegraphics[height=7cm,width=8cm]{t_i_vs_t_i_1.eps}
\caption{Drawing fixed points and flow direction for various C (picture taken from \cite{derrida_92})}
\label{fig_t}
\end{figure}\\
Using the above analysis, it can be seen that there are \textit{a priori} four types of solutions:
\begin{itemize}
\item \textit{Type 1}: $C<1/4,\ p_- < p_0 < p_{N+1} < p_+ $
\item \textit{Type 2}: $C<1/4,\ p_+ < p_{N+1} < p_0$
\item \textit{Type 3}: $C<1/4,\ p_{N+1} < p_0 < p_-$
\item \textit{Type 4}: $C > 1/4,\ p_{N+1} < p_0 $
\end{itemize}

\subsection*{Solutions for the $ N \rightarrow \infty $ limit}
We now find $ C(\alpha, \beta) $ for the above solutions as $ N \rightarrow \infty $:
\begin{itemize}
\item \textit{Type 1}: $C<1/4,\ p_- < p_0 < p_{N+1} < p_+ $\\
Since the interval $ [p_-,p_+] $ is finite, the $ p_i's $ must accumulate as $ N \rightarrow \infty $. This can be done either at $ p_- $ or at $ p_+ $. In accordance, \textit{Type 1} solution is split into two possible solutions:\\
\textit{Type 1a}: $C<1/4,\quad p_- < p_0 < p_{N+1},\quad p_{N+1} = p_+ + 0^-$

Using \eqref{eq_boudary_0}, \eqref{eq_boudary_N} and the above conditions we
get the restrictions for $ \alpha,\ \beta$:
\[
\alpha > \beta,\ \alpha + \beta < 1
\]
And the current
\[
C = p_N (1 - p_{N+1}) = \beta (1 - \beta)
\]\\
\textit{Type 1b}: $C<1/4,\quad p_0 < p_{N+1} < p_+,\quad p_0 = p_- + 0^+$\\
Again using \eqref{eq_boudary_0}, \eqref{eq_boudary_N} we get
\[
\alpha < \beta,\ \alpha + \beta < 1
\]
And the current
\[
C = p_0 (1 - p_{1}) = \alpha(1 - \alpha)
\]
\item \textit{Type 2}: $C<1/4,\quad p_+ < p_{N+1} < p_0$\\
For this solution to exist as $ N \rightarrow \infty $, we must have $ p_{N+1} = p_+ + 0^+ $, otherwise we can find $ i $ such that $ p_i > 1 $ (in fact it will be true for any $p_j,\ j < i$).\\
The conditions above and \eqref{eq_boudary_0}, \eqref{eq_boudary_N} results in the restrictions for $ \alpha, \beta$:
\[
\alpha + \beta > 1,\beta < 1/2
\]
The current is
\[
C = p_N (1 - p_{N+1}) = \beta (1 - \beta)
\]
\item \textit{Type 3}: $C<1/4,\quad p_{N+1} < p_0 < p_-$\\
For a similar reasoning to the above, we require $ p_0 = p_- + 0^- $. Again using \eqref{eq_boudary_0}, \eqref{eq_boudary_N} we get:\\
\[
\alpha < 1/2,\ \alpha + \beta > 1
\]
The current is
\[
C = p_0(1 - p_1) = \alpha(1 - \alpha)
\]
\item \textit{Type 4}: $C > 1/4,\quad p_{N+1} < p_0 $\\
For this case we must have $ C = 1/4 + 0^+ $ and $ p_0 \geq 1/2, p_{N+1} \leq 1/2 $. otherwise we will have either $ p_i > 1 $ or $ p_i < 0 $ for some $ i $. Using this conditions we find
\[
\alpha \geq 1/2,\ \beta \geq 1/2
\]
and
\[
C = 1/4
\]
\end{itemize}
%
%
Collecting all of the results, it is possible to draw the phase diagram (Figure \ref{fig_final}).\\
\begin{figure}[h!tb]
\center
\includegraphics[height=7cm,width=8cm]{final.eps}
\caption{Phase diagram}
\label{fig_final}
\end{figure}
\begin{thebibliography}{9}
\bibitem{derrida_92}
B. Derrida, E. Domany and D. Mukamel, J. Stat. Phys., 69, 667 (1992)
\bibitem{derrida_08}
B. Derrida, \textit{Non equilibrium steady states: fluctuations and large deviations of the density and of
the current}, arXiv:0703762
\end{thebibliography}
\end{document}
\end{document}
