Conducting-square Darcy convection ($R_c=8{\pi }^2$)

In a unit square with prescribed conductive boundary temperatures and impermeable boundaries, the stream function convention $\mathbf{u}=(-{\psi }_z,0,{\psi }_x)$ gives linear equations $\Delta \psi =R{\theta }_x$, ${\theta }_t={\psi }_x+\Delta \theta$, with $\psi =\theta =0$ on all sides. The velocity equation follows by taking the vertical-plane curl of Darcy law, and the temperature equation by linearizing the heat equation about $1-z$. Unlike a laterally unbounded Darcy layer, the conducting side boundaries produce a two-dimensional real marginal eigenspace at $R_c=8{\pi }^2$.

For $q=q_c+ϵq_1$, ${\theta }_t=ϵ\lambda \theta$, the first-order equations are $\Delta {\psi }_1-q_c^2{\theta }_{1x}=2q_c{q}_1{\theta }_{0x}$ and ${\psi }_{1x}+\Delta {\theta }_1=\lambda {\theta }_0$. Multiply them by ${\psi }_0,q_c^2{\theta }_0$ and integrate. Integration by parts and the leading equations cancel the homogeneous correction terms, giving $2q_c{q}_1⟨{\psi }_0{\theta }_{0x}⟩+q_c^2\lambda ⟨{\theta }_0^2⟩=0$. Since $⟨{\psi }_0{\theta }_{0x}⟩=-⟨|\nabla {\theta }_0{|}^2⟩$, the displayed title formula follows. The reciprocal norm ratio is incompatible with this solvability condition. Reflection parity diagonalizes the first-order splitting of the two marginal modes.

At zero growth rate put $P=\psi +iq\theta$, $q=\sqrt{R}$. The coupled Darcy equations become $\Delta P+iq{P}_x=0$. Substituting $P=e^{-iq(x-1/2)/2}F$ gives $\Delta F+(q^2/4)F=0$ with homogeneous Dirichlet boundary conditions. The square Dirichlet Laplacian eigenvalues give $q^2=4{\pi }^2(m^2+n^2)$, $m,n\ge 1$, whose minimum is $8{\pi }^2$. For $S=\sin \pi x\sin \pi z$, $a=q_c/2$, the complex multiples $F=S,iS$ give real pairs $(S\cos (a\xi ),-S\sin (a\xi )/q_c)$ and $(S\sin (a\xi ),S\cos (a\xi )/q_c)$, $\xi =x-1/2$. Their reflection parities are opposite, yielding two independent physical modes.