= Solution
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.
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