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Let $L=\Delta +c$ with real $c$, and letThe Fredholm alternative for an elliptic Dirichlet problem says that $N$ is finite-dimensional. If $N=0$, all the prescribed data admit a unique solution. In general the exact boundary compatibility in the self-adjoint Fredholm alternative isThese conditions are necessary and sufficient; when they hold, all solutions form $u_0+N$. Thus in the second branch some data are incompatible, while compatible data have nonunique solutions. The outward normal derivative fixes the displayed sign.

To prove the alternative, put $X=\{w\in C^{2,\alpha }(\overline{\Omega }):w{|}_{\partial \Omega }=0\}$ and $Y=C^{0,\alpha }(\overline{\Omega })$. The standard zero-boundary Poisson equation theorem makes $\Delta :X\to Y$ an isomorphism. Multiplication by $c$, followed by ${\Delta }^{-1}$, is a compact operator $K:X\to X$: the embedding $C^{2,\alpha }\hookrightarrow C^{0,\alpha }$ is compact, and multiplication is bounded. Writing $u=\psi +w$ givesThe Fredholm alternative for a compact operator implies finite kernel and cokernel of equal dimension, and invertibility exactly when the kernel vanishes.

For the explicit compatibility conditions, use the self-adjoint Dirichlet realization of $L$ on $L^2$. It has compact resolvent: a sufficiently large positive shift of $-L$ is coercive, its inverse gains two derivatives, and the Rellich-Kondrachov compactness theorem makes that inverse compact. The Fredholm solvability condition for a self-adjoint operator is $F\perp N$. Standard boundary elliptic regularity upgrades the resulting weak solution to $C^{2,\alpha }$ for these data and boundary. Thus it gives the same kernel and solvability as the preceding Hölder-space formulation.

Finally Green second identity, with $z=0$ on the boundary and $Lz=0$, givesConsequently $\int zF=\int zf+\int \psi {\partial }_{\nu }z$, proving both necessity and sufficiency with the stated sign. If using complex data and real $c$, replace $z$ by $\overline{z}$ in the pairings.

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.

For $R=4{\pi }^2+{ϵ}^2$, fixed roll phase, $T={ϵ}^{2}t$ and normalization $\psi =ϵA(T)\cos (\pi x)\sin (\pi z)+\cdots$, the Landau amplitude equation isThe mean-temperature correction in weakly nonlinear Darcy convection supplies the cubic negative feedback. Projection of the third-order equations onto the critical eigenfunction via the Fredholm solvability condition for a self-adjoint operator fixes both coefficients. The stable nonzero amplitudes are $\pm 2/\pi$ within the chosen phase, giving a supercritical pitchfork bifurcation in this real-amplitude reduction.

A weakly nonlinear expansion writes a small disturbance as powers of an amplitude parameter while resolving its evolution on a slow time. Near a critical eigenvalue, linear detuning and nonlinear interactions enter the same perturbation order. The Fredholm solvability condition for a self-adjoint operator, or its adjoint form for a general operator, yields an amplitude equation without requiring the full higher-order correction.