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

Vary the integral on the right-hand side, imposing $\nabla \cdot \mathbf{u}=0$ with multiplier $2p$. After integration by parts, independent variations of $\mathbf{u}$, $\theta$, and $p$ giveDot the first equation with $\mathbf{u}$ and the second with $\sigma \mathrm{Ra}\theta$, 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.