= Solution
For <Darcy-Bénard convection>, the dimensionless <Darcy law> is $\mathbf u=-\nabla p+R\theta\mathbf e_z$, with <incompressibility> and a <heat equation> about $T_0=1-z$. Here $R$ is a <Darcy thermal-time Rayleigh number>, with permeability replacing the squared-depth dependence in the fluid-layer <Rayleigh number>. The stated <stream function> convention gives $\mathbf u=(-\psi_z,0,\psi_x)$, so $w=\psi_x$. The vertical-plane <curl> of <Darcy law> gives $\nabla^2\psi=R\theta_x$; omitting the quadratic perturbation <advection> from the <heat equation> gives $\theta_t=\psi_x+\nabla^2\theta$.
On the impermeable connected square boundary, zero normal <velocity> makes $\psi$ constant along the boundary; choose that constant as zero. The prescribed conductive boundary <temperature> fixes $\theta=0$ on all four sides. Thus the perfect-conductor wording is interpreted through the supplied homogeneous perturbation data: the side <temperatures> retain the background vertical profile, rather than imposing a different isothermal side state.
At marginal stability put $q=\sqrt R$ and $P=\psi+iq\theta$. The two coupled equations imply
$$
\nabla^2P+iqP_x=0,\qquad P=0\text{ on the boundary}.
$$
The <gauge transformation for conducting-square Darcy onset> $P=e^{-iq(x-1/2)/2}F$ removes the first derivative:
$$
\nabla^2F+\frac{q^2}{4}F=0,\qquad F=0\text{ on the boundary}.
$$
The square's <Dirichlet Laplacian eigenfunctions> are $\sin(m\pi x)\sin(n\pi z)$, with <eigenvalues> of $-\nabla^2$ equal to $\pi^2(m^2+n^2)$, $m,n\geq1$. The first marginal value is therefore
$$
\boxed{q_c=2\sqrt2\pi,\qquad R_c=8\pi^2.}
$$
It is genuinely the first instability threshold, not just an isolated neutral value. Eliminating $\psi$ gives the <temperature> <linear operator> $L_R=\Delta_D+R\partial_x\Delta_D^{-1}\partial_x$, where $\Delta_D$ is the <Dirichlet realization of an elliptic operator> for the <Laplacian>. <Integration by parts> shows the second term is <self-adjoint> and nonnegative, since its quadratic form is $-\langle\theta_x,\Delta_D^{-1}\theta_x\rangle\geq0$. Thus the spectrum is real, starts negative at $R=0$, and its leading <eigenvalue> cannot cross into growth without a zero <eigenvalue>. The smallest zero value is the one found above.
Let $S=\sin\pi x\sin\pi z$, $\xi=x-1/2$ and $a=q_c/2=\sqrt2\pi$. Multiplying the lowest real $F=S$ by independent real and imaginary constants gives two real physical <eigenfunctions>:
$$
\boxed{\begin{aligned}(\psi_e,\theta_o)&=(S\cos(a\xi),-S\sin(a\xi)/q_c),\\(\psi_o,\theta_e)&=(S\sin(a\xi),S\cos(a\xi)/q_c).\end{aligned}}
$$
The subscripts indicate even and odd parity about $x=1/2$. Because $S$ is even in $\xi$, the first has even $\psi$ and odd $\theta$, while the second reverses those parities. They are linearly independent and satisfy all four <Dirichlet boundary conditions>. The lowest complex $F$ <eigenfunction> is simple over the complex numbers, but its arbitrary complex multiplier supplies two independent real pairs; this explains the physical degeneracy.
\Image[/past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2017/iii/paper-337-porous-modes.png]
{title=Neutral and oscillatory modes in porous convection}
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