Past exam of the mathematics course of the University of Cambridge / 2022 / iii / Paper 344 / 2 / d / Solution

The anchoring conditions are $\theta (0)=0$ modulo $\pi$ and $\theta (L)=\pi /2$ modulo $\pi$, since the nematic director identifies angles differing by $\pi$. Their difference can therefore be $m\pi /2$ for any odd integer $m$. The Euler-Lagrange equation ofis ${\theta }^{\prime \prime }=0$, so every stationary solution has the formIts free energy per unit length in the $y$ direction isThe smallest possible $m^2$ is one, giving exactly the two degenerate global minima $m=1$ and $m=-1$.