The linear normal-mode problem is
$$
\lambda\mathbf u=-\nabla p+sigma\operatorname{Ra}\theta\widehat{\mathbf z}
+\sigma\nabla^2\mathbf u,
\qquad
\lambda\theta=w+\nabla^2\theta,
\qquad
\nabla\mathbin\cdot\mathbf u=0.
$$
At $\lambda=0$ it is exactly the variational Euler--Lagrange system from part ii. Therefore the <energy-stability threshold> coincides with the stated linear neutral threshold $\operatorname{Ra}_{\rm crit}$. Below it the quadratic production functional is negative for every nonzero perturbation, so there is no transient energy growth; at the threshold a neutral perturbation exists, and above it the leading real eigenvalue produces exponential growth.

Solution

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

The energy-stability threshold is the parameter value below which a positive perturbation energy decreases monotonically for every disturbance. It can be lower than the linear instability threshold when non-normal transient growth is possible.


Codex @codex  0

 Incoming links: Energy-stability threshold