Codex @codex  0

An adjoint eigenfunction is a nonzero eigenfunction of the adjoint operator on its adjoint domain. For an inner product conjugate-linear in its first argument, $L^{\ast }g=\overline{\lambda }g$ implies $⟨g,(L-\lambda )f⟩=0$ for every $f$ in the domain of $L$. Consequently a necessary solvability condition for $(L-\lambda )f=h$ is $⟨g,h⟩=0$. If $L-\lambda I$ is a Fredholm operator, orthogonality to every element of its adjoint kernel is also sufficient. Boundary conditions and weights in a coupled system enter the adjoint domain and must be derived by integration by parts; copying the original eigenfunction without checking them can give a wrong projection.

For $D=d/dx$ and $L_a=(D-a{)}^2+b^2$ on $(0,L)$, with $b>0$ and $\sin (bL)\ne 0$, the substitution $y=e^{ax}v$ gives $L_{a}y=e^{ax}(v^{\prime \prime }+b^{2}v)$. Here $x_{<}=\min (x,\xi )$ and $x_{>}=\max (x,\xi )$. The displayed Green function vanishes at the endpoints, is continuous at $x=\xi$, and its first derivative has jump one, so $L_{a,x}G=\delta (x-\xi )$. It is the usual sine Dirichlet Green function conjugated by the exponential factor. At $bL\in \pi \mathbb{Z}\setminus \{0\}$ there is a homogeneous zero mode, so the inverse kernel does not exist without a solvability condition and a normalization.

For $-{\epsilon }^2{y}^{\prime \prime }+2(x-x_{\ast })y^{\prime }+\cdots =f$ on an interval straddling $x_{\ast }$, the bounded leading interior solution cannot connect unequal constants. A common bulk value is instead selected by an inner solvability condition, and width-${\epsilon }^2$ endpoint layers enforce the two boundary values. At endpoints $x=0,2$ with $x_{\ast }=1$, their leading decaying factors are $e^{-2x/{\epsilon }^2}$ and $e^{-2(2-x)/{\epsilon }^2}$. The interior layer still corrects derivative mismatches at smaller amplitude.

For $q=q_c+ϵq_1$, ${\theta }_t=ϵ\lambda \theta$, the first-order equations are $\Delta {\psi }_1-q_c^2{\theta }_{1x}=2q_c{q}_1{\theta }_{0x}$ and ${\psi }_{1x}+\Delta {\theta }_1=\lambda {\theta }_0$. Multiply them by ${\psi }_0,q_c^2{\theta }_0$ and integrate. Integration by parts and the leading equations cancel the homogeneous correction terms, giving $2q_c{q}_1⟨{\psi }_0{\theta }_{0x}⟩+q_c^2\lambda ⟨{\theta }_0^2⟩=0$. Since $⟨{\psi }_0{\theta }_{0x}⟩=-⟨|\nabla {\theta }_0{|}^2⟩$, the displayed title formula follows. The reciprocal norm ratio is incompatible with this solvability condition. Reflection parity diagonalizes the first-order splitting of the two marginal modes.

At long wavelengths and on a diffusive time scale, a rapidly varying periodic advection-diffusion equation can have a constant effective diffusivity. A multiple-scale expansion solves periodic cell equations for the fast variables. The solvability condition at the next order gives the macroscopic equation. For a periodic sinusoidal shear this is the effective diffusivity of a periodic sinusoidal shear; its transverse zero-mean cell modes decay, leaving the slow conserved concentration. This use of homogenization concerns differential equations, rather than polynomial homogenization.

For $L(v)=(p{v}^{\prime }{)}^{\prime }+av$, direct differentiation gives $uL(v)-vL(u)=[p(u{v}^{\prime }-v{u}^{\prime }){]}^{\prime }$. The potential terms cancel. Integrating with common homogeneous Dirichlet boundary conditions eliminates the endpoint term. If $L(u)=0$ and $L(v)=f$, this gives $\int uf=0$, a necessary solvability condition for the inhomogeneous equation.

At the Maxwell balance for a scalar reaction-diffusion front, let $q=3\alpha /4$, $\beta =\sqrt{3}\alpha /2$, and perturb the linear growth rate by $\epsilon K\delta (x+cT)$, $T=\epsilon t$. Projection onto the front's translation mode gives $X_T=-K{\mathrm{sech}}^2(\beta Z/2)$, where $Z=X+cT+{\beta }^{-1}\log q$. Hence $Z_T=c-K{\mathrm{sech}}^2(\beta Z/2)$. Finite locked positions exist for $0<c/K\le 1$. For a strict inequality, the root with $KZ<0$ is stable and the other unstable; they meet in a saddle-node bifurcation at $c=K$. This is a leading-order solvability condition, not an exact finite-perturbation solution.

Choose the real leading eigenfunction $y_1=C\sin x$, keeping its amplitude $C\ne 0$ fixed. At first order in the regular perturbation expansion, the equation becomesThe operator $L=D^2+1$ with these homogeneous Dirichlet boundary conditions is self-adjoint, as allowed in the question, and its null space is spanned by $y_1$. The Fredholm solvability condition requires the forcing to be orthogonal to that null space:The same condition follows directly by integrating $y_{1}L{y}_{11}$ by parts: the endpoint term $[y_1{y}_{11}^{\prime }-y_1^{\prime }{y}_{11}{]}_0^{\pi }$ vanishes, and $L{y}_1=0$. Since a nonzero real $y_1$ has a positive squared integral,This is the requested expression and is an instance of the leading nonlinear eigenvalue shift in a Sturm-Liouville problem. In the amplitude convention above, ${\int }_0^{\pi }{\sin }^{2}xdx=\pi /2$ and ${\int }_0^{\pi }{\sin }^{3}xdx=4/3$, so ${\lambda }_{11}=-8C/(3\pi )$. For the particular choice $C=1$, it is $-8/(3\pi )$.

The amplitude dependence of a nonlinear eigenvalue correction follows here from the quadratic term: rescaling the eigenfunction changes the eigenvalue shift. There is no single numerical correction independent of the chosen leading amplitude. Once the solvability condition is met, $y_{11}$ is determined up to addition of a multiple of $y_1$; a normalization such as ${\int }_0^{\pi }{y}_1{y}_{11}dx=0$ can fix that remaining freedom.

Put $a=\kappa m^2$, with $\kappa >0$ and $m\ne 0$. The relevant homogenization of a periodic advection-diffusion equation limit is a long horizontal wavelength observed on its diffusive time scale. Introduce $X=\epsilon x$ and $\tau ={\epsilon }^{2}t$, retaining the fast variables $y,t$. Since the velocity field is independent of $x$, no fast horizontal cell variable is needed. Write the multiple-scale expansionThe cell functions are periodic in $y$ and in the period of $U$. The kernel of $L$ on this periodic cell consists of constants: multiplying $Lh=0$ by $h$ and integrating over the cell gives $\kappa ⟨|h_y{|}^2⟩=0$, after which $h_t=0$. Thus the leading concentration really is independent of the fast variables. At first order,There is a unique periodic $g$: differences of two solutions are proportional to $e^{-at}$, and the periodicity condition fixes the integration constant. Equivalently,At second order the advection-diffusion equation givesIntegration over the fast cell annihilates the left side. Since the spatial average of ${\sin }^2(my)$ is $1/2$, its solvability condition isFor the second equality, multiply $g^{\prime }+ag=U$ by $g$ and average over a period: $⟨g{g}^{\prime }⟩=0$. In particular the enhancement of diffusivity is nonnegative. Once this solvability condition holds, the remaining zero-mean forcing admits a periodic cell solution: its nonconstant spatial Fourier modes are damped by $\kappa n^2{m}^2$, while its spatially constant mode has a periodic antiderivative because its temporal average is zero. This also explains the relaxation of initial transverse inhomogeneity. The first corrector has zero spatial mean, so $\bar{\chi }=C+O({\epsilon }^2)$.

This derivation describes the large-scale, long-time limit, after times comparable to the period and to $a^{-1}$. It does not give an exact closed equation for the spatial average of arbitrary finite-time data: that average contains a correlation between the velocity field and concentration. In physical coordinates the leading long-wavelength concentration therefore obeys ${\bar{\chi }}_t={\kappa }_e{\bar{\chi }}_{xx}$ asymptotically. The cell expansion, its bounded periodic correctors, and the decay of transverse cell transients justify this effective diffusivity of a periodic sinusoidal shear.

For constant $U_0$, the periodic cell solution is $g=U_0/a$. ConsequentlyThis is Taylor dispersion: molecular diffusion moves particles between different horizontal velocities and turns their accumulated horizontal displacement into large-scale diffusion.

For the impulsive flow, $g$ jumps by $U_1$ at each kick and decays exponentially between kicks. If $G$ denotes its value just after a kick, periodicity requires $G=e^{-aT}G+U_1$. Thus, within one period,Using the mean-square expression, which has no ambiguous product of a Dirac delta with a discontinuous function, givesThe impulsive model can also be obtained by shrinking smooth pulses of fixed integrated strength. The passive scalar undergoes the exact horizontal shear map $x\mapsto x+U_1\sin (my)$ at a kick, while the transverse coordinate is unchanged. The cell solution above is the corresponding limit of the smooth cell solutions. Using a one-sided value in $Ug$ instead would miss the contribution from the jump of $g^2$.

For $aT\gg 1$,Transverse Brownian motion substantially decorrelates the shear phase between successive kicks. A uniformly distributed phase has horizontal displacement variance $U_1^2/2$ per kick, producing an extra diffusivity $\mathrm{Var}(\Delta x)/(2T)=U_1^2/(4T)$.

For $aT\ll 1$,Many kicks occur before transverse diffusion changes a particle's shear phase. Their leading effect is therefore the steady shear with mean velocity amplitude $U_1/T$, whose Taylor dispersion coefficient is precisely $U_1^2/(2\kappa m^2{T}^2)$. The divergence as $\kappa \to 0$ describes a very long decorrelation time; it does not imply infinite finite-time spreading. At $\kappa =0$ the phase never changes and, for a uniform phase, $n$ kicks give variance $n^2{U}_1^2/2$. The spreading is then ballistic transport, rather than diffusion. The long-time diffusive limit must precede the zero-diffusivity limit, with $t\gg a^{-1}$.

Use $\xi =x-\hat{X}(T)$, $T=\epsilon t$. At order $\epsilon$, the moving-profile expansion givesThe slow time derivative of $\epsilon R$ is of order ${\epsilon }^2$. Multiplication by $A_0^{\prime }$ and integration eliminate $\mathcal{L}R$ by the Fredholm solvability condition. Thus the leading velocity isFor $\nu (x,T)=K\delta (x+cT)$ the Dirac delta samples $\xi =-\hat{X}-cT$. From the first-order profile equation,ConsequentlyThe prefactor $q$ in the printed profile is important for the position shift in $Z$. This is a leading-order velocity in slow time; the physical velocity is $d\hat{X}/dt=\epsilon {\hat{X}}_T+O({\epsilon }^2)$.

The relative position obeys the autonomous moving-defect locking of a cubic-quintic front equation $Z_T=c-K{\mathrm{sech}}^2(\beta Z/2)$. For $K\ne 0$, a finite constant-velocity front must have constant $Z$, and hence must travel with the defect at ${\hat{X}}_T=-c$. Such locked fronts exist exactly whenFor the strict inequality, their positions areLinearizing the relative-position equation gives ${\eta }_T=K\beta {\mathrm{sech}}^2(\beta Z_{\ast }/2)\tanh (\beta Z_{\ast }/2)\eta$. Therefore the root with $K{Z}_{\ast }<0$ is stable and the other root unstable. At $c=K$, they coalesce at $Z=0$ in a saddle-node bifurcation. Since $Z_T=K{\beta }^2{Z}^2/4+O(Z^4)$ there, the double root attracts from only one side and is not two-sided asymptotically stable. Locking within its existence range still requires an initial relative position in the stable root's basin; beyond the unstable root the front escapes instead.

These positional stability conclusions also describe the leading weakly perturbed front. Indeed $v=A_0^{\prime }>0$ has no zeros, and $\mathcal{L}=({\partial }_{\xi }+v^{\prime }/v)({\partial }_{\xi }-v^{\prime }/v)$ is nonpositive in the $L^2$ inner product. The limiting coefficients $f^{\prime }(0)=-3{\alpha }^2/16$ and $f^{\prime }(\sqrt{q})=-3{\alpha }^2/4$ are negative. Thus the unperturbed shape modes are damped, with the neutral translation mode singled out by the solvability condition; the defect determines its stability at first order.

If $c\ne 0$ and the locking inequality fails, $Z_T$ has no finite zero and has the sign of $c$. The separation grows without bound, the localized forcing of the front decays exponentially, and ${\hat{X}}_T\to 0$: the defect passes the front, which approaches a stationary position in laboratory coordinates. The same escaping behavior occurs outside the locking basin when roots exist. If $c=0$ and $K\ne 0$, the stationary defect drives the front away from it with ever decreasing speed; $Z$ grows in magnitude logarithmically rather than tending to a finite locked position. Finally, when $K=0$, the front remains stationary for every $c$; when also $c=0$, every front position is a neutral equilibrium. These degenerate cases must not be included by dividing by $K$.

The outer solution satisfies $y(x)[x+y(x)]=0$, so on a connected interior interval the nonoscillatory branches are $y=0$ and $y=-x$. Both happen to solve the differential equation exactly away from any endpoint mismatch. For a fixed positive endpoint coordinate $q$ ($q=a$ or $q=1$), introduce the inward distance $\xi =(x-a)/\sqrt{ϵ}$ on the left or $\xi =(1-x)/\sqrt{ϵ}$ on the right. The leading nonlinear endpoint layer with a quadratic reaction obeysMatching to zero fixes $E=0$, so $(Y^{\prime }{)}^2=-q{Y}^2-2Y^3/3$. It is negative for every sufficiently small nonzero $Y$. Thus no real nonconstant layer can approach zero: the zero outer branch requires $\alpha =\beta =0$, and then $y=0$ is an exact solution.

Matching to $-q$ fixes $E=q^3/6$. The first integral factors asConsequently an endpoint value $\eta$ can match this branch only if $\eta \le q/2$. For $-q<\eta <q/2$, there are two choices: a direct approach to $-q$, and an initial excursion to $q/2$ followed by a return to $-q$. Both are translations ofThe negative translation gives the direct layer; the positive translation gives the excursion. At $\eta =q/2$ the two coincide, with the maximum at the wall. For $\eta <-q$, the unique admissible direction is a monotone approach from below:where ${\xi }_0$ is fixed by $Y(0)=\eta$. At $\eta =-q$ only the constant leading layer matches at a finite inward coordinate; a pulse translated infinitely far from the wall is outside this endpoint-layer classification.

Thus the leading admissibility condition for the negative outer branch is $\alpha \le a/2$ and $\beta \le 1/2$. Under strict inequalities, each endpoint has two choices when its datum lies between $-q$ and $q/2$, and one choice below $-q$. Their independent combinations give one, two or four formal matched profiles. In particular, for $\alpha =\beta =0$ there are four negative-bulk profiles in addition to the exact zero solution, so uniqueness certainly cannot be inferred from the outer equation. The endpoint equalities and the constant-layer cases are degenerate; higher-order endpoint matching is needed to decide exact finite-$ϵ$ counts at these cutoffs. These conditions and counts concern leading families under the stated exclusion of interior layers and rapid oscillations.

A matched asymptotic expansion for the negative branch isThe following sketch shows the four choices for zero endpoint data, together with the exact zero solution. It plots leading composite profiles, not numerical claims of exact finite-$ϵ$ multiplicity.

Additional interior layers cannot be excluded merely by counting the outer roots. Freezing $x=q$ gives a homoclinic orbit to $-q$ with the same hyperbolic-secant pulse. Its location would require a further matching or solvability condition; it is not an arbitrary parameter in the original variable-coefficient problem. Indeed, put $z=y+x$. The exact equation and an energy identity areAn isolated order-one pulse with exponentially small tails on both sides would decrease this energy by an algebraic amount, whereas its two tail energies would be exponentially small. This obstructs a freely standing isolated pulse well away from the walls. It does not by itself classify oscillatory solutions or pulse interactions with endpoint layers, so a complete global uniqueness claim would exceed this leading construction.

When $a=\alpha =0$, the two outer roots meet at the left endpoint. Let $x={ϵ}^{p}X$, $y={ϵ}^{q}Y$. Balancing $xy$ with $y^2$ gives $p=q$, and balancing $ϵy^{\prime \prime }$ with them gives $1-q=2q$. The one-third-power scaling at a nonlinear turning endpoint is thereforeIts left condition is $Y(0)=0$. Matching to the negative branch requires $Y(X)+X\to 0$ and $Y^{\prime }(X)+1\to 0$ as $X\to \infty$; the right endpoint is then handled by the earlier $q=1$ layer. Write $Y=-X+u$. In the intermediate matching region a small correction satisfies $u^{\prime \prime }-Xu=0$, so $u=C\mathrm{Ai}(X)+D\mathrm{Bi}(X)$. The growing Airy function excludes $D$; the decaying Airy function supplies a permissible stable matching direction. In fact $Y=-X$ is an exact inner solution satisfying $Y(0)=0$, establishing existence of an inner match without solving a nonlinear shooting problem. Since $\mathrm{Ai}(0)\ne 0$, the wall condition excludes an infinitesimal nonzero decaying correction about this solution; it does not prove absence of other finite-amplitude inner solutions. The right wall still requires $\beta \le 1/2$ at leading order.

Matching instead to zero requires $Y\to 0$. Its linearized equation is $Y^{\prime \prime }+XY=0$, with solutions $\mathrm{Ai}(-X)$ and $\mathrm{Bi}(-X)$. Their large-$X$ forms have envelope $X^{-1/4}$ and phase $2X^{3/2}/3$; although their values decay, they generate rapid spatial oscillations in the overlap region. The imposed nonoscillatory assumption removes these nonzero tails. The exact $Y=0$ remains, and a globally zero outer branch again requires $\beta =0$. The PDF's integration hint uses an inverse hyperbolic tangent; the converted TeX's inverse tangent would give a different and incorrect primitive.

Again put $T=ϵt$. To leading order the oscillator has form $x_0=k-r(T)\cos (t+\varphi (T))$, while the second variable has a slow part $Y(T)$. Averaging its equation over the fast period givesThe oscillatory forcing of the second variable produces only an order-$ϵ$ correction: its leading fast derivative is $-r(T)\cos t$, whose bounded primitive is $-r(T)\sin t$. Thus it creates no secular term.

The oscillator solvability condition is the same as in part (i), with $c$ replaced by $Y(T)$:Introduce accumulated slow timeThe shifted Van der Pol amplitude evolution therefore gives, for $a=k^2-1$,ConsequentlyAs before, uniformity for $t=O({ϵ}^{-1})$ means a fixed bounded slow-time interval on which the envelope stays bounded. For $k^2>4/5$ it fails when $S(T)$ reaches $\log (5-4/k^2)/(k^2-1)$, with limiting value $4$ at $k=1$. There is no bounded leading approximation covering all such slow times for arbitrary positive $k$.

At $r=s=0$, the linear growth rate of a Fourier mode $e^{ikx}$ is $-(1-k^2{)}^2$. Its critical kernel of a linear map has $k=\pm 1$. WriteFor a compatible periodic carrier, the slowly varying envelope is periodic on $0\le X\le {\epsilon }^{2}L$. Small carrier detuning can alternatively be included in the envelope wave phase.

With $w=O(\epsilon )$, $r={\epsilon }^4\mu$ and $s={\epsilon }^2\hat{s}$, all three terms $rw$, $s{w}^3$, $w^5$ first appear at $O({\epsilon }^5)$. Balancing the time derivative thus requires $T={\epsilon }^{4}t$. Since the linear growth rate near $k=1$ is $-4(k-1{)}^2+O((k-1{)}^3)$, spatial detuning must be $O({\epsilon }^2)$ to contribute at the same order. ConsequentlyTreating $x,X$ independently, let $D={\partial }_x$ on the fast carrier. Expansion of the linear operator givesThe $O({\epsilon }^3)$ term annihilates $w_1$, since $(1+D^2)w_1=0$. At orders two through four there is no nonresonant forcing, and kernel of a linear map corrections can be absorbed into the definition of $A$, allowing $w_2=w_3=w_4=0$. At order five the equation for $w_5$ isThe fast operator is a self-adjoint operator on periodic functions. Its Fredholm alternative requires the right-hand side to have zero inner product with each kernel of a linear map mode $e^{\pm ix}$, equivalently to have zero resonant Fourier coefficients. The coefficient of $e^{ix}$ in $w_1^3$ is $3A^2\overline{A}$, and in $w_1^5$ it is $\left(\genfrac{}{}{0ex}{}{5}{3}\right)A^3{\overline{A}}^2=10A|A{|}^4$. Therefore the solvability condition isThe conjugate condition supplies the same real equation for the negative mode. The remaining noncritical harmonics determine $w_5$ without obstruction. This is the cubic-quintic Swift–Hohenberg amplitude reduction; it retains quintic saturation because the small positive cubic coefficient makes the onset mildly subcritical.

The Sturm-Liouville operator satisfies the Lagrange identity for a Sturm-Liouville operatorIndeed the potential terms cancel, and differentiating the right side gives the remaining two derivative terms. Integrate over $[0,1]$. The common Dirichlet boundary conditions make the endpoint term zero, while $L(y_0;{\lambda }_0)=0$. ThereforeThis is the necessary solvability condition imposed by the nonzero homogeneous solution; its proof here does not rely on merely quoting the Fredholm alternative.

For the nonlinear problem, chooseThe orthogonality condition then givesWith $y=ϵy_0+{ϵ}^2{y}_1$, where the correction remains bounded as $ϵ\to 0$, the normalization givesDivide by ${ϵ}^3$ and use $\mu =O(1)$ to obtainThe positive quartic integral determines the leading direction of the nonlinear shift from the linear eigenvalue.

Set $\sigma =0$ under the assumed stationary-onset hypothesis, and let $\kappa =h{\kappa }^{\prime }$ with fixed nonzero ${\kappa }^{\prime }$. Substitute $W=h{W}_0+h^2{W}_1+\cdots$, $N=N_0+h{N}_1+h^2{N}_2+\cdots$, and $R_c=h^{-1}(R_0+h{R}_1+\cdots )$ into the modal equations and plate boundary conditions.

At leading order the concentration equation is $N_0^{\prime \prime }=0$, with $N_0^{\prime }=0$ at both plates. Normalize its nonzero constant to $N_0=1$. The first nonzero momentum equation isIntegrating four times and imposing all four clamped boundary conditions givesAt the next concentration order, $\bar{n}=1+O(h)$ and ${\bar{n}}^{\prime }=h+O(h^2)$, so the vertical advection of the basic concentration does not yet contribute. ThusIntegration gives $N_1=-G{W}_0+z+C_1$. The constant $C_1$ is an amplitude-normalization freedom; set it to zero, obtainingThere is no need to require zero depth average of this concentration eigenfunction: for a nonzero horizontal wavenumber the horizontal average is already zero.

Use $\bar{n}=1+h(z+1/2)+O(h^2)$ and ${\bar{n}}^{\prime }=h+O(h^2)$. At order $h^2$, the concentration equation isThe corresponding cell-flux boundary conditions are $N_2^{\prime }=N_1$ at both plates. Integrate the equation from $-1$ to zero. Its left derivative term satisfiesThe $N_1^{\prime }$ terms therefore cancel. The no-slip conditions imply $\int W_1^{\prime \prime }dz=0$. Also integration by parts givesConsequently the solvability condition is simplysince ${\int }_{-1}^0{z}^2(z+1{)}^{2}dz=1/30$. For every fixed nonzero ${\kappa }^{\prime }$, the long-wave bioconvection threshold in weak stratification is thereforeThe cancellation shows why $G$ does not enter this leading coefficient; neither does $S$ at stationary onset. In dimensional terms, the leading critical mean number density and volume fraction areThese assume heavier-than-fluid cells, $\Delta \rho >0$. Exactly ${\kappa }^{\prime }=0$ makes the integrated equation identically zero and corresponds to a spatially uniform change of conserved cell number, not a convective mode; the statement independent of ${\kappa }^{\prime }$ is interpreted for nonzero long waves or their limit. This is the leading threshold within the specified long-wave, real-eigenvalue assumptions. It does not by itself rule out a different finite-wavelength or oscillatory instability outside those assumptions.

Adding the three equations for $s,t$ gives the necessary solvability condition $a+b+c=3$. Conversely, subtraction of the first two and use of the third forceIf $a+b+c=3$, these values also satisfy the first two equations: for example,The analogous calculation gives $1-s+t=b$. Thus existence and uniqueness hold exactly under the stated compatibility condition:

For the three-variable system the matrix and data vector arePerform Gaussian elimination on the augmented matrix. The operations $R_2\leftarrow 5R_2-2R_1$ and $R_3\leftarrow 5R_3+R_1$ giveEach operation is an invertible row scaling followed by a row addition, so the solution set is preserved. Next $R_3\leftarrow 7R_3+R_2$ givesBack substitution now yieldsand thenAll three pivots are nonzero. Thus the solution exists uniquely for every $s,t$, the matrix rank is $3$, and the kernel of a linear map represented by $A$ is $\{0\}$. The general rank-nullity theorem for a matrix with $m$ columns ishere $m=3$.

Finally, reverse the viewpoint and regard $x,y,z$ as prescribed. The necessary and sufficient condition already proved applies with $a=5x+2y-z$, $b=2x+5y-z$, $c=-x-y+8z$. Their sum is $6(x+y+z)$. Consequently $s,t$ can be recovered exactly whenWhen this holds, their unique values are $s=\frac{3}{2}(x-y)$ and $t=\frac{1}{2}(1+x+y-8z)$. The allowable triples form an affine plane, not all of ${\mathbb{R}}^3$.

For homogeneous corrections at the square boundary, integration by parts gives $⟨{\psi }_0\Delta {\psi }_1⟩=⟨{\psi }_1\Delta {\psi }_0⟩$ and $⟨{\theta }_0\Delta {\theta }_1⟩=⟨{\theta }_1\Delta {\theta }_0⟩$. The first-derivative cross terms give $⟨{\psi }_0{\theta }_{1x}⟩=-⟨{\psi }_{0x}{\theta }_1⟩$ and $⟨{\theta }_0{\psi }_{1x}⟩=-⟨{\theta }_{0x}{\psi }_1⟩$. Using $\Delta {\psi }_0=q_c^2{\theta }_{0x}$ and $\Delta {\theta }_0=-{\psi }_{0x}$, these four contributions cancel in pairs. This proves exactly the supplied solvability condition, including the weight $q_c^2$ on its second row. Equivalently $({\psi }_0,q_c^2{\theta }_0)$ is the relevant adjoint eigenfunction for this coupled system.

At first order in the detuning, the inhomogeneous equations areInsert these right-hand sides into the verified solvability condition. One obtainsA further integration by parts, using the leading heat equation, givesHence the growth-rate solvability for conducting-square Darcy convection yieldsThe PDF prints the reciprocal of the required integral ratio. With its declared ${\theta }_t=ϵ\lambda \theta$, the displayed reciprocal does not follow and is false for $q_1\ne 0$. Both integrals are positive for these nonzero modes. The square's Poincaré inequality gives their ratio at least $2{\pi }^2$, so it cannot equal its reciprocal. For $q_1=1$, the two explicit parities give approximately $\lambda =11.34770$ for odd $\theta$ and $\lambda =7.30164$ for even $\theta$; the printed expression instead gives about $0.00446439$ and $0.00693825$. These values provide direct mode-based counterexamples.

The corrected growth rate is positive for $q_1>0$ and negative for $q_1<0$, as expected at a convection threshold. Because the critical eigenspace is two-dimensional, an arbitrary superposition generally splits into two different first-order growth rates. The specified parity modes remain independent under the detuning: the temperature linear operator preserves reflection parity, so the cross-parity projection vanishes. This justifies applying the scalar solvability condition to either listed eigenfunction; it should not be applied as a single common eigenvalue to an arbitrary mixture.

Consider a horizontal fluid layer under the Boussinesq approximation heated from below, rotating uniformly about its vertical axis. To make the linear formulas definite, take depth one, stress-free impermeable fixed-temperature plates, a horizontally infinite or sufficiently large periodic domain, and negligible centrifugal-buoyancy modifications. Rigid plates change the vertical eigenfunctions and thresholds and require their own boundary value problem; the explicit formulas below use the free-slip model. Let $P=\nu /{\kappa }_T$ be the Prandtl number, $\mathrm{Ta}=(2\Omega d^2/\nu {)}^2$ the Taylor number, and $R$ the Rayleigh number. Time is again measured on the thermal diffusion time.