Théo Girard

One-dimensional Hughes model

In the one-dimensional setting, the Hughes' model simulate the evacuation of a corridor ( here we choose the corridor to be the domain \( [-1,1] \) ) with exits on the right and the left. It can be rewritten as: \[\left\{ \begin{matrix} \partial_t \rho(t,x) + \partial_x \left[ \textrm{sign} (x - \xi(t)) f(\rho(t,x)) \right] = 0 \\ \int_{-1}^{\xi(t)} c(\rho(t,x)) \textrm{d} x = \int_{\xi(t)}^1 c(\rho(t,x)) \textrm{d} x. \end{matrix} \right.\] Here \( f(\rho) = \rho v(\rho) \) is the flux function corresponding to Lighthill-Whitham-Richards (LWR) model and \(v\) corresponds to the speed of pedestrians. Also, \( c(\rho) \) is the running cost paid by agents to travel in a location with density \( \rho \). The unknown \( \xi \) named the "turning curve" corresponds to the point in space where pedestrian pay the same total cost to leave the corridor through the left or the right exit.
Below, two simulations of this model using a finite volume scheme. The turning curve \(\xi\) is in red and the density is plotted on \([-1.5,1.5]\) even if the dynamics of the turning curve only depends on what's inside \([-1,1]\).

2D Hughes' model

In the two-dimensional setting, the Hughes' model takes the following form: \[\left\{ \begin{matrix} \partial_t \rho(t,x) + \textrm{div} \left( \vec{V}(x) f(\rho(t,x)) \right) = 0 \\ \left|\nabla u(x) \right| = c(\rho(t,x)) \\ \vec{V}(x) = - \frac{\nabla u(x)}{|\nabla u (x)|}. \end{matrix}\right.\] Here \( u \) is a solution of the Eikonal equation with a Dirichlet condition \(u=0\) at the exits.
Below is a simulation of the 2D Hughes model where the exits are represented in green and the density is represented by a gradient from red (high dentity) to black (low density).