Past exam of the mathematics course of the University of Cambridge / 2019 / iii / Paper 331 / 4

The zero streamfunction gives zero velocity. The temperature ${\theta }_0=-z$ obeys both prescribed temperature values, has no horizontal gradient, and has zero Laplacian. Thus it satisfies both equations with ${\psi }_0=0$:This is the conductive state of Darcy-Bénard convection.

Linearizing about the conductive state givesThe perturbations vanish at $z=0,1$. For a normal mode with vertical dependence $\sin (n\pi z)$, put $q=k^2+n^2{\pi }^2$. The two modal equations are $-q\hat{\psi }=-ikR\hat{\theta }$ and $s\hat{\theta }+ik\hat{\psi }=-q\hat{\theta }$. Eliminating $\hat{\psi }$ givesNeutrality requires $R=(k^2+n^2{\pi }^2{)}^2/k^2$. The arithmetic-geometric mean inequality bounds this below by $4n^2{\pi }^2$, with equality at $k=n\pi$. The first unstable vertical mode is therefore $n=1$, givingThe wavelength statement applies at the first onset. At a general larger $R$, maximizing the growth rate within a fixed vertical mode gives $k^2=n\pi \sqrt{R}-n^2{\pi }^2$, so its preferred wavelength need not remain exactly two. This is the onset of convection in a horizontal Darcy layer.

The full perturbation equations arewhere the streamfunction advection bracket is $J(f,g)=f_z{g}_x-f_x{g}_z$. Put $R=R_c+{ϵ}^2$ and introduce the slow time $T={ϵ}^{2}t$, so ${\partial }_t={ϵ}^2{\partial }_T$ on the amplitude-dependent fields. ThenThe weakly nonlinear expansion balances the small linear growth against the cubic saturation on this slow time scale.

At first order, ${\nabla }^2{\psi }_1+R_c{\theta }_{1x}=0$ and ${\nabla }^2{\theta }_1-{\psi }_{1x}=0$. Since ${\nabla }^2{\psi }_1=-2{\pi }^2{\psi }_1$ and $R_c=4{\pi }^2$, the first equation fixes ${\theta }_{1x}={\psi }_1/2$. The homogeneous temperature boundary conditions and the second equation remove any horizontally uniform term, givingAt second order,The forcing is horizontally uniform. Taking no additional critical-mode component, a solution satisfying the homogeneous boundaries isThis mean-temperature correction in weakly nonlinear Darcy convection modifies the vertical temperature gradient and provides the feedback that saturates the convection.

Let ${\Psi }_3=A^3{\psi }_3$ and ${\Theta }_3=A^3{\theta }_3$ denote the full third-order coefficients. This notation keeps the detuning term, proportional to $A$, and the slow-time term, proportional to $A_T$, visible. DefineThe equations at order ${ϵ}^3$ areEquivalently, where $A\ne 0$, the question's coefficients satisfy ${\nabla }^2{\psi }_3+R_c{\theta }_{3x}=-A^{-2}{\theta }_{1x}$ and ${\nabla }^2{\theta }_3-{\psi }_{3x}=(A_T/A^3){\theta }_1+B$. Eliminate ${\Theta }_3$ from the full coefficients:Use the Fredholm solvability condition for a self-adjoint operator. With $⟨f,g⟩={\int }_0^1{\int }_0^{2}fgdxdz$, projection onto the critical eigenfunction ${\psi }_1$ makes the left side vanish. Therefore the Landau amplitude equation isThese integrals already answer the requested coefficient calculation. Evaluating them gives $⟨{\psi }_1,{\theta }_{1x}⟩=1/4$, $⟨{\psi }_1,{\partial }_x{\nabla }^2{\theta }_1⟩=-{\pi }^2/2$, and $⟨{\psi }_1,{\partial }_{x}B⟩={\pi }^2/32$. Hence, for the prescribed normalization,The cubic term opposes growth, so the nonzero steady amplitudes $A=\pm 2/\pi$ are stable equilibria within the chosen roll phase. The conductive state $A=0$ is unstable for this positive detuning. This is supercritical saturation of Darcy convection rolls.