= Solution
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.
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