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> of
$$
F[\theta]=\frac{\widetilde K}{2}\int_0^L(\theta')^2dx
$$
is $\theta''=0$, so every stationary solution has the form
$$
\boxed{\theta_m(x)=\frac{m\pi x}{2L}\qquad(m\text{ odd}).}
$$
Its free energy per unit length in the $y$ direction is
$$
F_m=\frac{\widetilde K m^2\pi^2}{8L}.
$$
The smallest possible $m^2$ is one, giving exactly the two degenerate global minima $m=1$ and $m=-1$.

Solution

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

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2022/iii/paper-344/2/d/solution