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

Traverse the large rectangle counterclockwise, taking its lower edge at $y\to -\infty$ and upper edge at $y\to +\infty$. Along the lower edge the director angle changes by $-\pi /2$; along the upper edge, traversed from $x=L$ to $x=0$, it changes by another $-\pi /2$. The anchored director is constant along the two vertical edges. The net continuous angle change is thereforeThe topological charge of a two-dimensional nematic disclination enclosed by a circuit is $q=\Delta \theta /(2\pi )$, soA nonsingular director field on the enclosed disk would have zero winding. At least one nematic disclination must therefore lie inside.