# Obstacle Problem

We consider the setting where there is an obstacle lying above the graph in the coupled system (coupled system) . To do this we reformulate our original motion by mean curvature problem into a variational inequality

$\int_\Omega \frac{(u^{n+1}-u^n) (\varphi-u^{n+1})}{\tau \sqrt{ 1+|\nabla u^n|^2 }} \geq \int_\Omega \frac{\nabla u^{n+1} \cdot \nabla (\varphi-u^{n+1})}{\sqrt{ 1+|\nabla u^n|^2 }} + \int_\Omega \frac{f(c) (\varphi-u^{n+1})}{\sqrt{ 1+|\nabla u^n|^2 }} \qquad \forall \varphi \in K$

which can also be considered as the minimisation of

$J(u^{n+1}) = \min_{v \in K} \int_\Omega \sqrt{1+|\nabla u^n|^2} \left[ \frac{ v^2}{2 \tau} - \frac{u^n v}{\tau} + \frac{|\nabla v|^2}{2} - f(c) v \right ].$

where K is a convex set constraining the graph to lie below the obstacle.To solve this we interpolate the initial curve with cubic splines and then use a relaxtion algorithm to minimise at each time step.

Considering only a constant forcing term in the upward normal direction, that is $f(c) = 2$ , Running this with 400 time steps and nodal width $0.05$, $u^n = 5|sin(x)|$. We see that the curve reaches an equilibrium where the downward forcing of the obstance and mean curvature is enough to overcome to upward forcing on the curve.