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

For a periodic or boundary-tangent <divergence-free vector field> $u$, <integration by parts> gives
$$
\int (u\mathbin\cdot\nabla f)g
=-\int (u\mathbin\cdot\nabla g)f.
$$
Consequently $\int(u\mathbin\cdot\nabla f)f=0$, the cancellation underlying many fluid <energy estimates>.

Skew-symmetry of incompressible transport

{wiki=Solenoidal_vector_field}

A divergence-free vector field satisfies $\nabla\mathbin\cdot u=0$. It preserves volume under its flow, and its transport operator is skew-symmetric when boundary fluxes vanish.


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