Past exam of the mathematics course of the University of Cambridge 2022 iii Paper 344 2 e Solution 2026-09-28
Traverse the large rectangle counterclockwise, taking its lower edge at and upper edge at . Along the lower edge the director angle changes by ; along the upper edge, traversed from to , it changes by another . 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 , soA nonsingular director field on the enclosed disk would have zero winding. At least one nematic disclination must therefore lie inside.
Past exam of the mathematics course of the University of Cambridge 2022 iii Paper 344 2 f Solution 2026-09-28
The least costly configuration contains one charge- nematic disclination, placed near by the reflection symmetries of the boundary data. Away from its small core, the director interpolates as smoothly as possible between the wall anchoring and the two far-field textures. Locally around the defect one may sketchThe elastic energy of an isolated defect scales as . Splitting the required total charge into additional allowed half-charge defects is impossible without also adding compensating defects, which raises both the logarithmic elastic energy and the positive core energy. The single centered defect is therefore the lowest-energy topology, up to smooth distortions and symmetry-related core placement.