= Solution
Take $h=2k_c$ and $D=d/dz$. Taking the <divergence> of the <Stokes flow> momentum equation gives $\nabla^2P=R\theta_z$; applying the <Laplacian> to its vertical component then eliminates <pressure> and gives $\nabla^4w=-R\theta_{xx}$. The forced <velocity> and <temperature> are $O(\epsilon^2)$, so their <advection> product is $O(\epsilon^4)$ and drops out at leading order. The steady <heat equation> reduces to $w_0+\nabla^2\theta_0=0$.
Consequently the <boundary value problem> for the forced profiles is
$$
\boxed{(D^2-h^2)^2w_\epsilon=R h^2\theta_\epsilon,\qquad (D^2-h^2)\theta_\epsilon=-w_\epsilon.}
$$
Its six <boundary conditions> are
$$
\boxed{w_\epsilon(0)=w_\epsilon(1)=w_\epsilon''(0)=w_\epsilon''(1)=0,\qquad \theta_\epsilon(0)=\gamma,\quad\theta_\epsilon(1)=0.}
$$
The horizontal <velocity field> is $u_0=-\epsilon^2w_\epsilon'(z)\sin(hx)/h$, which directly verifies the <incompressible flow> and <stress-free boundary condition>. In the detuning used subsequently, $R$ may be replaced by $R_c$ in this leading problem, since its correction enters at order $\epsilon^4$. The vertical profiles depend linearly on $\gamma$.
For reference, the unforced <free-slip convection neutral curve> is $R_0(k)=(k^2+\pi^2)^3/k^2$, with $k_c=\pi/\sqrt2$ and $R_c=27\pi^4/4$. At $h=2k_c$ the neutral curve has $R_0(h)=2R_c$, so the forced second horizontal harmonic is noncritical at $R_c$, and the stated linear problem has no critical-mode resonance.
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