= Solution
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, giving
$$
\boxed{\theta_1=\frac1{2\pi}\sin(\pi x)\sin(\pi z).}
$$
At second order,
$$
\nabla^2\psi_2+R_c\theta_{2x}=0,\qquad
\nabla^2\theta_2-\psi_{2x}=J(\psi_1,\theta_1)=\frac\pi4\sin(2\pi z).
$$
The forcing is horizontally uniform. Taking no additional critical-mode component, a solution satisfying the homogeneous boundaries is
$$
\boxed{\psi_2=0,\qquad\theta_2=-\frac1{16\pi}\sin(2\pi z).}
$$
This <mean-temperature correction in weakly nonlinear Darcy convection> modifies the vertical temperature gradient and provides the feedback that saturates the convection.
\Image[/past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2019/iii/paper-331-convection-rolls.png]
{title=Critical Darcy convection rolls and their temperature perturbation}
{description=Arrows show the <velocity> generated by the <streamfunction> $\psi_1=\cos(\pi x)\sin(\pi z)$. Colour shows $\theta_1$ on a horizontal period of length two: warm perturbations rise and cool perturbations sink.}
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