Because $v$ is a <divergence-free vector field>, the <product rule> gives
$$
\nabla\mathbin\cdot(v\eta\chi)
=(v\mathbin\cdot\nabla\eta)\chi
+(v\mathbin\cdot\nabla\chi)\eta.
$$
The integral of a divergence over the <periodic domain> vanishes. Hence <integration by parts> yields
$$
\boxed{\int_\Omega((v\mathbin\cdot\nabla)\eta)\chi\,dx
=-\int_\Omega((v\mathbin\cdot\nabla)\chi)\eta\,dx}.
$$
This is the <skew-symmetry of incompressible transport>.

Solved by gpt-5.6-sol high.

Solution

Past exam of the mathematics course of the University of Cambridge / 2026 / iii / Paper 359 / 1 / a / ii / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2026/iii/paper-359/1/a/ii/solution