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If $e_1,\dots ,e_n$ are orthonormal, any relation ${\sum }_i{c}_i{e}_i=0$ gives $c_j=0$ on taking the dot product with $e_j$. Thus the vectors are linearly independent. A linearly independent set of $n$ vectors in the $n$-dimensional vector space ${\mathbb{R}}^n$ is a basis, proving the claim.

Expand both $x$ and $f$ in this orthonormal basis, writing $f_i=e_i\cdot f$. Since $A{e}_i={\lambda }_i{e}_i$, the linear equation becomesComparing the unique basis coefficients gives, when $\mu$ is not an eigenvalue,If $\mu ={\lambda }_1$, the equation is solvable exactly when $e_i\cdot f=0$ for every $i$ with ${\lambda }_i=\mu$. For those indices $a_i$ is arbitrary; for all other indices the displayed formula still applies. Thus repeated eigenvalues require orthogonality to the entire corresponding eigenspace, not merely to one chosen eigenvector. This is the finite-dimensional Fredholm solvability condition for a self-adjoint operator.

For the specific real symmetric matrix, chooseTheir dot products with $f$ are $-1/\sqrt{2}$, $3/\sqrt{2}$ and $3$. At $\mu =2$ this givesOne can check $(A-2I)x=f$ directly. At $\mu =1$, the first coefficient equation would require $0\cdot a_1=-1/\sqrt{2}$, which is impossible. Hence there is no solution when $\mu =1$; the nonzero projection onto its eigenspace is the obstruction.

Write $P=P_0+d{P}_1+O(d^2)$ and $r=r_0+d{r}_1+O(d^2)$, with each coefficient satisfying the zero Dirichlet boundary condition. The steady complex equation, including the fluctuating coupling, is $0={\mathcal{L}}_{r}P-dB\sin (2\pi x/\ell )$. Since $B_0=(P_0-P_0^{\ast })/(2i)$, its order-$d$ terms giveThis is a regular perturbation equation for the threshold shift. The right projection uses the complex inner product, not a bilinear product without conjugation.

The linear operator ${\mathcal{L}}_{r_0}$ is self-adjoint on the domain $H^2(0,\ell )\cap H_0^1(0,\ell )$ in complex L2 space. For functions $v,w$ in that domain, two integration by parts steps giveThe boundary term vanishes. In particular $i{r}_0{\partial }_x$ is formally self-adjoint: both complex conjugation of $i$ and integration by parts contribute a minus sign. Taking $v=P_0$, $w=P_1$ and using ${\mathcal{L}}_{r_0}{P}_0=0$ provesThis is the Fredholm solvability condition for a self-adjoint operator. Define $s(x)=\sin (2\pi x/\ell )$ and $N={\int }_0^{\ell }|P_0{|}^{2}dx$. The gauged sine mode hassince the real envelope contributes an integral of its own derivative, whose endpoint term is zero. In particular the denominator in the solvability condition is nonzero. ThereforeFor either parity choice this expression is real. Indeed $|P_0{|}^2$, $A_0^2$ and $B_0^2$ are even about the midpoint, while $s$ is odd. Hence $\int |P_0{|}^{2}s=\int B_0^{2}s=0$, and the last numerator reduces to $\int A_0{B}_{0}s$.

No integral evaluation is needed to compare the two shifts. The phase change $P_Q=i{P}_D$ gives $A_Q=-B_D$, $B_Q=A_D$, and leaves $N$ and $I$ unchanged. Thus $A_Q{B}_Q=-A_D{B}_D$, andEquivalently, the term involving $|P_0{|}^2$ vanishes by parity, while $(P_Q^{\ast }{)}^2=-(P_D^{\ast }{)}^2$ reverses the remaining perturbation numerator. This is parity splitting of a mean-field dynamo threshold: the perturbation breaks the phase degeneracy and lowers one onset threshold by the amount it raises the other.

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.