Past exam of the mathematics course of the University of Cambridge / 2026 / iii / Paper 359 / 2 / b / ii

The regularity from part (i) makes $u\omega$ and its first weak derivatives square-integrable. The product rule therefore holds in the distributional derivative sense:Since $u={\nabla }^{\perp }\Psi$, equality of mixed weak derivatives gives $\nabla \cdot u=0$. Both remaining expressions belong to $L_{\mathrm{per}}^2$, so their distributional equality is an equality in that space:

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