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Let ${\Psi }_3=A^3{\psi }_3$ and ${\Theta }_3=A^3{\theta }_3$ denote the full third-order coefficients. This notation keeps the detuning term, proportional to $A$, and the slow-time term, proportional to $A_T$, visible. DefineThe equations at order ${ϵ}^3$ areEquivalently, where $A\ne 0$, the question's coefficients satisfy ${\nabla }^2{\psi }_3+R_c{\theta }_{3x}=-A^{-2}{\theta }_{1x}$ and ${\nabla }^2{\theta }_3-{\psi }_{3x}=(A_T/A^3){\theta }_1+B$. Eliminate ${\Theta }_3$ from the full coefficients:Use the Fredholm solvability condition for a self-adjoint operator. With $⟨f,g⟩={\int }_0^1{\int }_0^{2}fgdxdz$, projection onto the critical eigenfunction ${\psi }_1$ makes the left side vanish. Therefore the Landau amplitude equation isThese integrals already answer the requested coefficient calculation. Evaluating them gives $⟨{\psi }_1,{\theta }_{1x}⟩=1/4$, $⟨{\psi }_1,{\partial }_x{\nabla }^2{\theta }_1⟩=-{\pi }^2/2$, and $⟨{\psi }_1,{\partial }_{x}B⟩={\pi }^2/32$. Hence, for the prescribed normalization,The cubic term opposes growth, so the nonzero steady amplitudes $A=\pm 2/\pi$ are stable equilibria within the chosen roll phase. The conductive state $A=0$ is unstable for this positive detuning. This is supercritical saturation of Darcy convection rolls.