 Solution

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.