Past exam of the mathematics course of the University of Cambridge / 2009 / iii / Paper 73 / 4 / vi / Solution

Divide the equilibrium equation by $q>0$ to get $b{q}^2-q+a=0$. Hence both positive branches, when $0<4ab<1$, areAt equilibrium, ${\mathrm{Ri}}_f=a/q=1-bq$. Therefore the larger-speed branch $q_{+}$ gives the printed flux Richardson numberwhile the smaller-speed branch gives the same formula with a plus sign before the square root. Solving the quadratic alone does not discard that second branch.

The larger-speed branch is selected by local kinetic-energy stability. If the same closures are used in the local energy evolution, $\frac{1}{2}d(q^2)/dt=\mathcal{P}-B-ϵ$. For $q>0$, divide by $q$ to obtainIts linearization at $q_{\pm }$ has derivative $C_{u}S-2q_{\pm }/L_u=\mp C_{u}S\sqrt{1-4ab}$. The upper branch is therefore stable and the lower unstable in this local closure model.

The discriminant gives the critical gradient Richardson numberAt criticality the branches meet and ${\mathrm{Ri}}_f=1/2$. At zero stratification the stable positive solution is $q=C_{u}S{L}_u$ with ${\mathrm{Ri}}_f=0$; the lower algebraic root is $q=0$. This closure-dependent critical value is not automatically the inviscid $1/4$ criterion of the Miles–Howard theorem.