Past exam of the mathematics course of the University of Cambridge / 2021 / iii / Paper 331 / 1

With ${\mathbf{u}}_0=0$ and ${\theta }_0=-z$, the incompressibility condition and temperature equation hold because ${\nabla }^{2}z=0$. The momentum equation is satisfied by the hydrostatic pressurebecause $\nabla p_0=\sigma Ra{\theta }_0\hat{\mathbf{z}}$. Thus this is the conductive basic state.

Write $\mathbf{u}={\mathbf{u}}^{\prime }$ and $\theta =-z+{\theta }^{\prime }$. Dropping quadratic perturbation terms gives the Linearized Boussinesq equationsandThe fixed temperatures give ${\theta }^{\prime }=0$ at $z=0,1$. Impermeable stress-free boundary conditions give

Apply $\hat{\mathbf{z}}\cdot \nabla \times$ to the linear momentum equation. The pressure and buoyancy terms vanish, while incompressibility gives $\hat{\mathbf{z}}\cdot \nabla \times (\hat{\mathbf{z}}\times {\mathbf{u}}^{\prime })=-DW$. HenceApplying $\hat{\mathbf{z}}\cdot \nabla \times \nabla \times$ and using the identity supplied in the question gives

For a normal mode proportional to $e^{\mu t+ikx}$, put ${\hat{D}}^2=D^2-k^2$. The three scalar equations becomeandMultiplying through by the two scalar operators and eliminating $\hat{\omega },\hat{\theta }$ yields

At stationary onset, $\mu =0$. The lowest stress-free vertical mode is $\hat{W}=\sin \pi z$, for which $D^2=-{\pi }^2$ and ${\hat{D}}^2=-({\pi }^2+k^2)$. Substitution givesThe rotation term is positive, so rotation raises the critical Rayleigh number and is stabilizing.

Put $x=k^2$ and $T=\lambda /\sigma$. Differentiating $Ra$ gives the exact stationarity equationWhen $T\gg 1$, the optimum has $x\gg {\pi }^2$, so $2x^3\sim {\pi }^2{T}^2$. Therefore

Let $r=Ra-R{a}_c$. The amplitude equation is $\dot{A}=A(r-A^2)$. For $r<0$, $A=0$ is the sole equilibrium and every solution tends monotonically to it. At $r=0$, the origin remains attracting but only algebraically. For $r>0$, the origin is unstable and the two equilibriaare stable: positive initial data tend to $A_{+}$, negative initial data tend to $A_{-}$, and $A(0)=0$ remains zero. A plot of $A(t)$ therefore shows a pitchfork bifurcation normal form at $Ra=R{a}_c$.