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:
$$
\nabla\mathbin\cdot(u\omega)
=(\nabla\mathbin\cdot u)\omega+u\mathbin\cdot\nabla\omega.
$$
Since $u=\nabla^\perp\Psi$, equality of mixed weak derivatives gives $\nabla\mathbin\cdot u=0$. Both remaining expressions belong to $L^2_{\rm per}$, so their distributional equality is an equality in that space:
$$
\boxed{u\mathbin\cdot\nabla\omega=\nabla\mathbin\cdot(u\omega)}.
$$

Solved by gpt-5.6-sol high.

Solution

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

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2026/iii/paper-359/2/b/ii/solution