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A completely regular Hausdorff space is a Hausdorff space in which each point $x$ outside a closed set $F$ is separated from $F$ by a continuous function $u:X\to [0,1]$ with $u(x)=1$ and $u{|}_F=0$. Put $E=C_b(X)$ and define the evaluation character by ${\delta }_x(g)=g(x)$. The weak-star topology on $E^{\prime }$ is the topology of pointwise convergence on $E$: its coordinate maps $p\mapsto p(g)$ are continuous, and generate the topology. It is Hausdorff, and the dual unit ball is compact in it by the Banach-Alaoglu theorem.

For nonempty $X$, $\Vert {\delta }_x\Vert =1$ by the supremum norm bound and the constant function $1$. All the coordinates of $x\mapsto {\delta }_x$ are continuous, so $\delta$ is continuous. Complete regularity separates distinct points, proving injectivity. To prove continuity of the inverse onto its image, let $x\in O$ with $O$ open in $X$. Separating $x$ from $X\setminus O$ gives a bounded continuous $u$ withSuch neighborhoods prove that $\delta$ is a homeomorphism onto $\delta (X)$. This qualification is essential: it is not onto the whole dual, since the zero functional is not an evaluation on a nonempty space.

Define the Stone-Čech compactification to beThe Banach-Alaoglu theorem makes this a compact Hausdorff space, with $X$ identified with its dense evaluation copy. For $g\in C_b(X)$, the coordinate function $\tilde{g}(p)=p(g)$ is a continuous extension. Density gives $\Vert \tilde{g}{\Vert }_{\infty }=\Vert g{\Vert }_{\infty }$. Conversely any $h\in C(\beta X)$ restricts to an element $g$ of $C_b(X)$, and $h=\tilde{g}$ because they agree on the dense subset $X$. Restriction and coordinate extension are inverse linear isometries, henceThey also preserve products: the corresponding continuous functions agree on $X$ and hence everywhere.

For the universal extension property, let $f:X\to K$ be continuous and $K$ a compact Hausdorff space. Pullback defines a bounded linear operator $T:C(K)\to C_b(X)$ by $Tg=g\circ f$. Its dual map $T^{\prime }:E^{\prime }\to C(K{)}^{\prime }$ is weak-star continuous, since $(T^{\prime }p)(g)=p(Tg)$. Evaluation ${\delta }_K:K\to C(K{)}^{\prime }$ is a homeomorphism onto a compact, thus closed, subset: a compact Hausdorff space is completely regular, by the Urysohn lemma. Now $T^{\prime }{\delta }_x={\delta }_K(f(x))$, soTherefore $\overline{f}={\delta }_K^{-1}\circ T^{\prime }{|}_{\beta X}$ is the required continuous extension. Two continuous maps into the Hausdorff space $K$ agreeing on the dense subset $X$ agree everywhere, proving uniqueness.

If $X$ is open in $\beta X$, then it is a locally compact space: around any $x$, regularity of the compact Hausdorff space $\beta X$ gives an open neighborhood whose closure is compact and lies in $X$. Conversely, suppose $X$ is locally compact. Choose an open neighborhood $U$ of $x$ in $X$ with compact closure $C\subseteq X$. There is an open $O\subseteq \beta X$ with $O\cap X=U$. Density of $X$ implies $O\subseteq {\overline{U}}^{\beta X}$, while compactness of $C$ makes it closed in $\beta X$, so ${\overline{U}}^{\beta X}\subseteq C$. Thus $x\in O\subseteq X$. This provesThe empty-space cases are immediate.

The final printed assertion, for an arbitrary subset of $\beta X$, is false. Take the infinite discrete space $X=\mathbb{N}$. It is not compact, so its compactification has a point $p\in \beta \mathbb{N}\setminus \mathbb{N}$. The singleton $\{p\}$ is closed, hence equals its own closure. It is not open, since every nonempty open subset of $\beta \mathbb{N}$ meets the dense copy of $\mathbb{N}$. This is a counterexample.

The valid result holds for subsets $D\subseteq X$, and more generally for open subsets of $\beta X$. For $D\subseteq X$, its indicator is bounded continuous because $X$ is discrete. Its extension $\widetilde{1_D}$ is still idempotent by density, so it takes only the values zero and one. The set $H=\{\widetilde{1_D}=1\}$ is clopen and has $H\cap X=D$. Density then gives $H={\overline{D}}^{\beta X}$. If $O$ is open in $\beta X$, the set $O\cap X$ is dense in $O$, so $\overline{O}=\overline{O\cap X}$ is clopen by the result just proved. Consequently $\beta X$ is extremally disconnected, which means the closure of every open set is open. This is the discrete Stone-Čech compactifications are extremally disconnected theorem, with the essential open-set qualification.

A sufficient set of assumptions is that $L$ is a densely defined closed operator on the Hilbert space $H$, generating an analytic semigroup $S(t)$, and that $f\in C^{0,\theta }([0,T],H)$ for some $0<\theta \le 1$. For an operator criterion, one may assume that $A=\omega I-L$ is a sectorial operator of angle less than $\pi /2$, with dense domain. The analytic semigroup then satisfies, on a fixed finite time interval,The Hölder continuity assumption means $\Vert f(t)-f(s){\Vert }_H\le K|t-s{|}^{\theta }$. It supplies time regularity of the forcing, not spatial membership of $f(t)$ in $D(L)$.

With initial datum $u_0\in H$, a classical solution of an abstract Cauchy problem meanswith the evolution identity holding in $H$ at positive times and $u(t)\to u_0$ in $H$. Equivalently, the positive-time domain continuity is in the graph norm. If the definition requires differentiability also at zero, assume additionally $u_0\in D(L)$; the proof below then gives $u\in C^1([0,T],H)\cap C([0,T],D(L))$.

The variation-of-constants formula gives the candidateThe integral is a Bochner integral. Boundedness of $S$ and continuity of $f$ make the expression continuous into $H$, and its integral term tends to zero with $t$. Thus it is a mild solution of an abstract Cauchy problem with the correct initial value. The issue is showing that its integral term lies in the domain of the unbounded linear operator $L$.

Write $v(t)={\int }_0^{t}S(r)f(t-r)dr$. For $\epsilon >0$, the truncated integral $v_{\epsilon }(t)={\int }_{\epsilon }^{t}S(r)f(t-r)dr$ lies in $D(L)$. Subtract the current forcing before applying $L$:Here the last term follows by integrating $LS(r)f(t)=\frac{d}{dr}S(r)f(t)$ away from zero; it does not assume that $f(t)\in D(L)$. The first integrand satisfiesSince $v_{\epsilon }(t)\to v(t)$ in $H$ and the displayed images also converge, the closed operator property givesThis is the reason for the Hölder continuity assumption: the unsplit estimate would have the nonintegrable factor $r^{-1}$. Merely knowing that $f$ is continuous does not provide the integrable bound needed here. More generally, a modulus of continuity $\eta$ satisfying ${\int }_0^1\eta (r)dr/r<\infty$ suffices for this argument; a positive Hölder exponent is a convenient sufficient hypothesis.

The formula also proves continuity of $Lv$ at positive times. On compact positive-time intervals, split the integral at a small $r=\epsilon$: its short part is uniformly bounded by $C{\epsilon }^{\theta }$, while the remaining part is continuous by the strongly continuous semigroup property and ordinary dominated convergence. Together with analytic smoothing of $S(t)u_0$, this makes $Lu$ continuous at positive times. The semigroup property givesBecause $u(t)\in D(L)$, division of this identity by $h$ after subtracting $u(t)$ gives the continuous right derivative $Lu(t)+f(t)$. The left derivative agrees by the same identity and continuity of $Lu+f$. Thus $u^{\prime }=Lu+f$, proving existence of the classical solution of an abstract Cauchy problem on the entire given finite interval, and in particular local existence.

For uniqueness, the difference $h$ of two such classical solutions of an abstract Cauchy problem satisfies $h^{\prime }=Lh$ and $h(0)=0$. For $0<\epsilon <t$, differentiation of $S(t-s)h(s)$ on $[\epsilon ,t]$ gives zero, so $h(t)=S(t-\epsilon )h(\epsilon )$. Letting $\epsilon \downarrow 0$ yields $h(t)=0$. Finally, if $u_0\in D(L)$, then $LS(t)u_0=S(t)L{u}_0\to L{u}_0$, and the expression for $Lv(t)$ tends to zero. Hence $u^{\prime }(t)\to L{u}_0+f(0)$ and the stronger time-zero regularity follows. The solution is unique; the forcing regularity controls the singularity in the differentiated integral.

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.

Positivity means $u>0$ almost everywhere, since the real logarithm must belong to $L^1$. A positivity-preserving operator gives $s=Tu\ge 0$. For fixed $x$, the scalar shifted Poisson data fidelity is $f_x(s)=s-g(x)\log (1+s)$, withComposition with the linear operator $T$ proves convexity in $u$, and strict convexity holds along pairs whose forward images differ on a set of positive measure. The admissible class itself is convex: the concavity of the logarithm gives $\log (\theta u+(1-\theta )v)\ge \theta \log u+(1-\theta )\log v$, controlling its negative part, while ${\log }^{+}w\le w$ controls its positive part.

Because $|\Omega |=1$ and $\log (1+s)\ge 0$, the Jensen inequality gives the requested bound:The final equality uses nonnegativity and the unit area. Since $t-G\log (1+t)\to \infty$, this controls the forward-image norm on energy sublevels.

The printed strictly positive problem has no minimizer. This is nonattainment under strict positivity for shifted Poisson fidelity, rather than a failure of the coercivity calculation. Indeed, with $0<g<1$ and $s>0$, $\log (1+s)<s$ implies $f_x(s)>0$. Also $Tu$ cannot vanish identically for a strictly positive $u$. To see this, let $E_n=\{u\ge 1/n\}$. If $Tu=0$, positivity and $0\le {\chi }_{E_n}\le nu$ give $T{\chi }_{E_n}=0$. But ${\chi }_{E_n}\to {\chi }_{\Omega }$ in $L^1$ and continuity of $T$ would imply $T{\chi }_{\Omega }=0$, contradicting the hypothesis. Thus every admissible $u$ has positive fidelity and hence positive total energy.

Conversely, the constants $u_{\epsilon }=\epsilon {\chi }_{\Omega }$ are admissible, have zero total variation, and satisfyTheir logarithms are integrable for each $\epsilon >0$, but the limit is excluded. ThereforeA bounded minimizing sequence and bounded-variation compactness do not repair a nonclosed positivity/logarithm constraint. In particular, a literal existence or uniqueness proof for that domain is impossible.

The natural correction is to minimize over $BV(\Omega )$ with $u\ge 0$, omitting the unnecessary $\log u\in L^1$ condition: the fidelity only contains $\log (1+Tu)$, which is already integrable. Here is the full existence for nonnegative shifted Poisson regularization argument, also valid for any bounded nonnegative data $g$. Let $G=\Vert g{\Vert }_{\infty }$, $m_G={\inf }_{t\ge 0}\{t-G\log (1+t)\}>-\infty$, and take a minimizing sequence of energy at most $C$. The displayed bound gives $\Vert T{u}_n{\Vert }_1\le C_1$ and $\alpha \mathrm{TV}(u_n)\le C-m_G$. Write $c_n={\int }_{\Omega }{u}_n\ge 0$. The Poincaré inequality for total variation and mean control for positive imaging operators yieldThe denominator is nonzero, so the full $BV$ norm is bounded. Bounded-variation compactness gives $u_n\to u$ in $L^1$ along a subsequence, with $u\ge 0$. Continuity gives $T{u}_n\to Tu$ in $L^1$. On $s\ge 0$, $|f_x^{\prime }(s)|\le 1+G$, so the fidelity converges in the integral; lower semicontinuity of variation completes the direct method in the calculus of variations.

For the actual printed data $0<g<1$, the corrected problem has the unique minimizer $u=0$, even if $T$ is not injective. Zero attains energy zero. Any other zero-energy candidate would have both $Tu=0$ and $\mathrm{TV}(u)=0$; on the connected square, zero variation makes $u$ a nonnegative constant, and $T{\chi }_{\Omega }\ne 0$ forces that constant to vanish. For general positive bounded data, an injective $T$ is a sufficient uniqueness condition, because its fidelity is strictly convex; injectivity is not a necessary condition in every instance.

In the finite-dimensional interpretation, let ${\lambda }_{ij}=1+(Tu{)}_{ij}$. Independent Poisson observations with these intensities have negative log-likelihood ${\sum }_{ij}[{\lambda }_{ij}-g_{ij}\log {\lambda }_{ij}]+C(g)$. Removing the constant $\sum 1$ gives precisely the stated fidelity. Thus the model is Poisson counting noise with a unit background intensity, or an approximate version of it for rescaled/continuous grey values. Literal Poisson counts are integers; the constraint $0<g<1$ is a grey-value normalization, not a literal unscaled count sample.

If $x=0$, the generated subspace is zero and invariant. Otherwise, let $r$ be the first index for which $x,Tx,\dots ,T^{r}x$ are linearly dependent. Such an index exists with $1\le r\le n$, since there are $n+1$ vectors. The preceding vectors are independent, so the coefficient of $T^{r}x$ in this first relation cannot vanish. HenceThe linear operator $T$ sends each of the first $r-1$ generators to the next and sends the last into their span. This proves that their span is an invariant subspace. Every later iterate lies in the same subspace by induction, so it equals the span of the first $n$ iterates. Therefore the indicated subspace is $T$-invariant. This proof of a cyclic subspace uses only finite dimension and linear dependence.

Apply the direct method in the calculus of variations to $J(u)=\Vert f-Ku{\Vert }_1+\alpha |Du|(\Omega )$. Since $J(0)=\Vert f{\Vert }_1$, take a minimizing sequence with $J(u_k)\le C:=\Vert f{\Vert }_1+1$. ThenThus the sequence is bounded in the bounded-variation space. The bounded-variation compactness gives a subsequence converging in the strong convergence sense in $L^1$ to $u\in BV(\Omega )$. Boundedness of the linear operator $K$ gives $K{u}_k\to Ku$ strongly in $L^1$, so the residual norm converges. The variation term is sequentially lower semicontinuous, henceTherefore $u$ attains the infimum. The useful inverse hypothesis is the lower bound $\Vert u{\Vert }_1\le C_K\Vert Ku{\Vert }_1$; a bounded inverse on the range already suffices for this proof. No uniqueness follows from the nonsquared $L^1$ fidelity.

For a bounded linear operator $A:X\to Y$ between Hilbert spaces, let $N=\ker A$ and $R=\mathrm{ran}A$. The Moore–Penrose inverse of an operator is defined onFor $y=Ax+z$, with $z\in R^{\perp }$, defineThis is independent of the chosen preimage $x$, since two preimages differ by a vector in $N$. Equivalently, it is the inverse of the restriction of $A$ to $N^{\perp }$, applied to the component of the data in $R$, and is zero on $R^{\perp }$.

The generalized solution $x^{†}=A^{†}y$ is the unique minimum-norm least-squares solution. Its residual is orthogonal to the range, giving the operator normal equationThe operator normal equation alone leaves an arbitrary null-space component; the second condition fixes the minimum-norm representative. The identities $A^{†}A=P_{N^{\perp }}$ and $A{A}^{†}=P_{\overline{R}}$ hold on their appropriate domains.

For an infinite-rank compact operator, the range need not be closed, and $A^{†}$ is generally unbounded. Data outside $R\oplus R^{\perp }$ need not have any least-squares solution at all, even though they can be approximated by range elements. This domain qualification is crucial in part (d); the Moore–Penrose notation does not turn an ill-posed inverse into an everywhere-defined bounded operator.

On each finite interval use the Banach space $X_T=C([0,T];E)$ with the supremum norm. The free term $g(t)=U_t{f}_0$ belongs to $X_T$, and the time-integral operator $\tau$ maps $X_T$ to itself with norm at most $2T$. The strongly continuous semigroup property and boundedness of $B$ justify continuity of the Bochner integral.

DefineThe factorial bound for a Volterra iterate gives $\Vert {\tau }^{j}g{\Vert }_{X_T}\le (2T{)}^j\Vert f_0{\Vert }_1/j!$. The series therefore converges absolutely in $X_T$. Since $\tau$ is a bounded linear operator on $X_T$, it can be passed through the convergent sum, givingThus $f=g+\tau f$, the required integral formulation, and $f(0)=f_0$. Each term on a larger interval restricts to the identical term on a smaller interval, so these constructions define a single global solution without having to restart at successive times. This is an integrable Volterra solution for normalized velocity relaxation. It is a mild solution of an abstract Cauchy problem in $L^1$ and hence an $L^1$ weak solution in the paper's integral-formulation sense. No smallness condition such as $2T<1$ is needed.

We prove the finite-dimensional spectral theorem for normal operators by induction on dimension. A complex linear operator has an eigenvector $v$ with eigenvalue $\lambda$, by the fundamental theorem of algebra. For a normal operator $T$, the operator $T-\lambda I$ is also normal, andfor every $w$. Applying this to $v$ shows $T^{\ast }v=\overline{\lambda }v$.

Normalize $v$. Its orthogonal complement $W=v^{\perp }$ is invariant under both $T$ and $T^{\ast }$: the adjoint operator identities pair $Tw$ with $v$ through $T^{\ast }v$, and $T^{\ast }w$ with $v$ through $Tv$. The restricted operator on $W$ remains normal, and its adjoint is the restriction of $T^{\ast }$. Induction supplies an orthonormal eigenbasis of $W$, which together with $v$ gives one for $V$.

Conversely, in an orthonormal basis of eigenvectors, $T$ is diagonal and $T^{\ast }$ is the diagonal matrix of the conjugate eigenvalues. The two diagonal operators commute. Thus normality is equivalent to an orthonormal eigenbasis. The zero-dimensional case is immediate.

The orbital angular momentum linear operator is the cross product of the position operator and the momentum operator:In position space, $\hat{\mathbf{p}}=-i\hslash \nabla$, so $\hat{\mathbf{L}}=-i\hslash \mathbf{r}\times \nabla$. The Levi-Civita symbol gives ${\hat{L}}_1={\hat{x}}_2{\hat{p}}_3-{\hat{x}}_3{\hat{p}}_2$ and its cyclic analogues.

The useful object is the Choi state of the entanglement-breaking channel. Put $d_A=\dim {\mathcal{H}}_A$, introduce a reference $R$ with the same computational basis as $A$, and use the normalized maximally entangled stateThe entanglement-breaking channel makes $C_{RB}$ a separable quantum state. Its partial trace is ${\mathrm{Tr}}_B{C}_{RB}=I_R/d_A$, because $\mathcal{E}$ is a quantum channel. Write a separable positive operator decompositionDiscard any term with $\mathrm{Tr}{\beta }_y=0$: a positive operator with zero trace is zero. DefineEach ${\tau }_y$ is a density operator. Each $E_y$ is a positive operator, and taking the matrix transpose of the partial trace identity givesThus the $E_y$ form a POVM. In this reference-first convention, the Choi reconstruction formula isThe factor $d_A$ is essential because we used a normalized Choi state. A measurement channel storing the outcome $y$, followed by the prescribed quantum state preparation of ${\tau }_y$, has precisely this action on every linear operator $L$. The desired factorization is therefore

Write $|\zeta ⟩={\sum }_{i,j}{M}_{ij}|i{⟩}_A|j{⟩}_B$. Normalization says $\Vert M{\Vert }_{\mathrm{HS}}=1$, and the Schmidt rank equals the matrix rank of $M$. Its nonzero singular values are therefore $s_1,\dots ,s_r$ with ${\sum }_j{s}_j^2=1$.

The overlap with the maximally entangled state is $⟨{\varphi }^{+}|\zeta ⟩=\mathrm{Tr}(M)/\sqrt{d}$. The trace norm estimate of Q3(i), with test linear operator $I$, followed by the Cauchy-Schwarz inequality, givesFor two pure states, the unsquared quantum fidelity is the absolute value of their overlap. Hence the maximally entangled overlap bound from Schmidt rank isThe bound is attained by $|\zeta ⟩=r^{-1/2}{\sum }_{j=0}^{r-1}|jj⟩$. Thus uniform Schmidt coefficients on $r$ aligned basis pairs give the greatest possible overlap for that Schmidt rank.

The structure theorem for finitely generated modules over a principal ideal domain applies in particular to a Euclidean domain $R$: every finitely generated module has an isomorphismwith each $d_i$ nonzero and a nonunit. The rank of a free module $r$ and the invariant factors of a finitely generated module $d_i$ are unique up to multiplication by units; empty sums are allowed. This is the classification theorem being used, without a proof of that theorem.

The rational canonical form theorem states that for a linear operator $T$ on a finite-dimensional vector space over any field $F$, there are unique monic nonconstant polynomials $p_1\mid \cdots \mid p_k$ such that some basis givesHere, for $p(t)=t^d+a_{d-1}{t}^{d-1}+\cdots +a_0$, the companion matrix has ones on its subdiagonal and last column $(-a_0,\dots ,-a_{d-1}{)}^T$. The empty matrix covers a zero-dimensional space. Two linear operators have matrix similarity exactly when they have the same invariant factors of a linear operator.

For the proof, make $V$ an $F[t]$-module by $t\cdot v=T(v)$. It is a finitely generated module, since a vector-space basis also generates it as a module. It is a torsion module: the powers of $T$ have linear dependence in the finite-dimensional space ${\mathrm{End}}_F(V)$, so some nonzero polynomial annihilates every vector. Because $F[t]$ is a Euclidean domain, the classification theorem givesThere is no free summand, since a nonzero free module over $F[t]$ is not a torsion module. Choose monic generators of the ideals. On $F[t]/(p_i)$, the classes of $1,t,\dots ,t^{d_i-1}$ form an $F$-basis by polynomial division. Multiplication by $t$ is exactly $C(p_i)$ in that basis, proving existence. Uniqueness in the classification theorem proves uniqueness of the rational canonical form. Finally, an $F[t]$-module isomorphism is precisely an invertible $F$-linear map intertwining the operators, establishing the assertion about matrix similarity. In particular the characteristic polynomial is ${\prod }_i{p}_i$ and the minimal polynomial is $p_k$ for nonzero $V$.

Eliminate the instantaneous Stokes flow velocity in favour of temperature. On a horizontal Fourier mode $k$, the Stokes temperature-slaving operator maps $\theta$ to $w={\mathcal{W}}_R\theta$, where $(D^2-k^2{)}^{2}w=R{k}^2\theta$ and $w=w^{\prime \prime }=0$. The temperature evolution has linear operator ${\mathcal{L}}_R=D^2-k^2+{\mathcal{W}}_R$ and bilinear map $\mathcal{N}(\varphi ,\chi )=-{\mathbf{u}}_R[\varphi ]\cdot \nabla \chi$. Under the homogeneous thermal Dirichlet boundary conditions, ${\mathcal{L}}_{R_c}$ is self-adjoint. Normalize its critical eigenfunction as $g(z)=\sin \pi z$ and set $s=k_c^2+{\pi }^2$; the critical vertical velocity is $f=sg$.

At order $ϵ$, the critical eigenfunction equation gives ${\mathcal{L}}_{R_c}(g{e}^{i{k}_{c}x})=0$. At order ${ϵ}^2$, the weakly nonlinear expansion contains the imposed second harmonic and the quadratic products of the critical mode: a horizontally uniform temperature correction proportional to $|A{|}^2$ and, in a general vertical-mode calculation, a second harmonic proportional to $A^2$. These corrections are found by solving the noncritical boundary value problems, with homogeneous thermal data except for the imposed forcing.

At order ${ϵ}^3$, the method of multiple scales produces the slow derivative $A_{\tau }$, the detuning term $R_2{\partial }_R{\mathcal{L}}_R$, and the two cross-advection terms involving first- and second-order fields. Project the $e^{i{k}_{c}x}$ component onto the adjoint eigenfunction using the vertical inner product. This is the solvability condition in the method of multiple scales: divide each resonant projection by $⟨g,g{⟩}_z$. The detuning supplies $\mu A$ with $\mu \propto R_2$; interactions of horizontal wavenumbers $2k_c$ and $-k_c$ permit $\beta A^{\ast }$ with $\beta \propto \gamma$; self-interaction through the slaved mean and second harmonic supplies $-\lambda |A{|}^{2}A$. Other products have the wrong horizontal wavenumber. Reflection $x\mapsto -x$ permits real coefficients with this cosine forcing. Thus the symmetry-allowed spatially forced convection amplitude equation isThere is a useful specialization that should not be silently missed. For the literal one-vertical-mode Stokes flow problem, the vanishing two-to-one forcing coefficient for Stokes convection makes $\beta =0$ at this order. To see this, write a positive second-harmonic forcing component as $(w,\theta )=(W(z),\Theta (z))e^{2i{k}_{c}x}$, incorporating the cosine's factor $1/2$. Its coupling to the negative critical harmonic has projected integrand, apart from sign and its factor $A^{\ast }$,The integral vanishes because $g=0$ at both plates, even though $\Theta (0)$ is nonzero. This proves the cancellation without solving the forced profiles. The permitted coefficient is therefore zero times $\gamma$; symmetry alone does not establish nonzero phase pinning for the equations actually supplied.

The same normalization makes the remaining coefficients explicit. Since ${\mathcal{W}}_{R}g=(R{k}_c^2/s^2)g$, $\mu =R_2{k}_c^2/s^2$. The quadratic second harmonic cancels for $f=sg$, while the uniform correction is ${\theta }_{20}=-s|A{|}^2\sin (2\pi z)/(2\pi )$. Projecting $-w_1{\partial }_z{\theta }_{20}$ gives $-s^2|A{|}^{2}A/2$. Thus for the literal model and this temperature normalization,A generic nonzero $\beta$ would require a nonvanishing projection in an amended physical model or a different forcing structure. It is still meaningful to classify the real-coefficient amplitude equation requested independently.

Write $A=x+iy$. Then $x_{\tau }=(\mu +\beta -\lambda (x^2+y^2))x$ and $y_{\tau }=(\mu -\beta -\lambda (x^2+y^2))y$. These are a gradient flow for $V=-\frac{1}{2}(\mu +\beta )x^2-\frac{1}{2}(\mu -\beta )y^2+\frac{\lambda }{4}(x^2+y^2{)}^2$, so local minima give stable equilibrium points. At the origin the two eigenvalues are $\mu +\beta$ and $\mu -\beta$. The origin has exponential asymptotic stability if $\mu <-|\beta |$, retains asymptotic stability with algebraic decay at $\mu =-|\beta |$, and is unstable if $\mu >-|\beta |$. At equality, ${\rho }^2=x^2+y^2$ obeys $({\rho }^2{)}_{\tau }\le -2\lambda {\rho }^4$, since both linear coefficients are nonpositive. Integrating this inequality proves attraction even in the zero-eigenvalue direction.

For $\beta >0$ the stable nonzero equilibrium points are real; for $\beta <0$ they are imaginary:The real branch has Jacobian matrix eigenvalues $-2(\mu +\beta ),-2\beta$; the imaginary branch has $-2(\mu -\beta ),2\beta$. The oppositely aligned branch, when it exists, is a saddle equilibrium. No mixed real-imaginary nonzero equilibrium is possible when $\beta \ne 0$.

For $\beta =0$, the origin is stable for $\mu \le 0$, with algebraic decay at zero. If $\mu >0$, the circle $|A{|}^2=\mu /\lambda$ is radially attracting. Each point has Lyapunov stability but has a neutral phase direction, so it does not have individual asymptotic stability; the circle has orbital stability. This is the literal model's unpinned family. The general nonzero-$\beta$ branches instead exhibit phase locking to one of two phases separated by $\pi$.

Assume $R^2=1-q^2>0$. The symbol $R$ in this part is the convection roll amplitude, not the Rayleigh number of the previous part. Set $r=R^2$ for the algebra. The periodic boundary condition requires $qL\in 2\pi \mathbb{Z}$ and modulation wavenumbers $lL\in 2\pi \mathbb{Z}$. The conserved-field equation preserves the imposed zero mean of $B$, since the integral of an $X$ derivative over a period vanishes.

Equating the $e^{ilX+\lambda T}$ coefficients after linearization of the convection amplitude coupled to a conserved field givesThe conjugate sideband is needed because the variation of $|A{|}^{2}A$ couples a perturbation to its complex conjugate. Put $S=a_1+a_2$, $D=a_1-a_2$. The linear operator on $(S,D,b)$ isIts characteristic polynomial, expressed conveniently without expanding every coefficient, isAt $l=0$ it has roots $-2r,0,0$. The phase root is neutral; the uniform $B$ root is excluded by the prescribed zero-flux-perturbation mean. Nonzero arbitrarily long-wave modes remain allowed on a sufficiently large domain. For fixed $r>0$, the two roots tending to zero have $\lambda =l^2\nu +O(l^4)$ at simple limiting roots. Dividing the polynomial by $2l^4$ yieldsRepeated limiting roots may require a further expansion, but do not affect the strict instability criterion. If $(\mu -\sigma )r+2\sigma q^2>0$, the constant term is negative and the two real roots have opposite signs. The positive root gives a growing long-wave sideband instability. Using $r=1-q^2$ givesThis is sufficient for instability when a sufficiently small nonzero allowed $l$ is available. It is not an unconditional instability theorem for every fixed finite period. For example $q=0$, $\mu =2$, $\sigma =1$, $L=1$ satisfy the inequality, but the nonzero allowed $l$ obey $l^2\ge 4{\pi }^2$. Their roots are $-l^2$ and $-l^2-1\pm \sqrt{1+4l^2}$, all negative. The prescribed zero mean removes the uniform flux root and the remaining uniform phase is merely neutral. The corrected conclusion is the long-wave instability with a conserved mean field, with domain size and mode availability stated.

For Darcy-Bénard convection, the dimensionless Darcy law is $\mathbf{u}=-\nabla p+R\theta {\mathbf{e}}_z$, with incompressibility and a heat equation about $T_0=1-z$. Here $R$ is a Darcy thermal-time Rayleigh number, with permeability replacing the squared-depth dependence in the fluid-layer Rayleigh number. The stated stream function convention gives $\mathbf{u}=(-{\psi }_z,0,{\psi }_x)$, so $w={\psi }_x$. The vertical-plane curl of Darcy law gives ${\nabla }^2\psi =R{\theta }_x$; omitting the quadratic perturbation advection from the heat equation gives ${\theta }_t={\psi }_x+{\nabla }^2\theta$.