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 therefore
$$
\Delta\theta=-\pi.
$$
The <topological charge of a two-dimensional nematic disclination> enclosed by a circuit is $q=\Delta\theta/(2\pi)$, so
$$
\boxed{q_{\rm total}=-\frac12.}
$$
A nonsingular director field on the enclosed disk would have zero winding. At least one <nematic disclination> must therefore lie inside.

Solution

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

{title2=$q=\Delta\theta/(2\pi)$}

For a continuous director angle around a closed counterclockwise circuit, the enclosed topological charge is $q=\Delta\theta/(2\pi)$. Since $\theta$ is defined modulo $\pi$, elementary nematic disclinations can have $q=\pm1/2$.


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 Incoming links: Topological charge of a two-dimensional nematic disclination