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

Dot the momentum equation with $\mathbf{u}$, multiply the temperature equation by $\sigma \mathrm{Ra}\theta$, and integrate over $V$. The pressure term vanishes by incompressibility, while skew-symmetry of incompressible transport removes both nonlinear advection terms. Integration by parts and the boundary conditions givewhereFor $\mathrm{Ra}=0$, this reduces toso viscosity monotonically dissipates kinetic energy.

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.

The linear normal-mode problem isAt $\lambda =0$ it is exactly the variational Euler--Lagrange system from part ii. Therefore the energy-stability threshold coincides with the stated linear neutral threshold ${\mathrm{Ra}}_{\mathrm{crit}}$. 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.

After the pressure enforces incompressibility, let $L$ denote the displayed linear operator. Integration by parts givesThe pressure terms vanish by incompressibility and the boundary conditions. The expression is symmetric under $i\leftrightarrow j$, soThus $L$ is a self-adjoint operator in the energy inner product and hence a normal operator. Its orthogonal eigenmodes cannot generate non-normal transient growth.