Vary the integral on the right-hand side, imposing $\nabla\mathbin\cdot\mathbf u=0$ with multiplier $2p$. After integration by parts, independent variations of $\mathbf u$, $\theta$, and $p$ give
$$
\boxed{
0=-\nabla p+\sigma\operatorname{Ra}\theta\widehat{\mathbf z}
+\sigma\nabla^2\mathbf u,
\qquad
0=w+\nabla^2\theta,
\qquad
\nabla\mathbin\cdot\mathbf u=0}.
$$
Dot the first equation with $\mathbf u$ and the second with $\sigma\operatorname{Ra}\theta$, then integrate. Their sum says exactly that the energy-production functional is zero. Hence any nonzero stationary point is a perturbation whose energy initially neither grows nor decays.


Solution

Past exam of the mathematics course of the University of Cambridge / 2023 / iii / Paper 331 / 2 / b / ii / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2023/iii/paper-331/2/b/ii/solution