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Use the homogeneous equilibrium point condition $k({\rho }_0){\rho }_0=a_{0}f({\rho }_0)$. Evaluate $D_1,D_2$ at $(a_0,{\rho }_0)$ and abbreviateThe quantity $c$ is the chemical relaxation rate with the cell density held fixed. Derivatives of $D_1,D_2$ do not enter this linear stability analysis: they multiply spatial derivatives of the homogeneous background or products of perturbations. For a Fourier mode with time dependence $e^{st}$, put $z=|\mathbf{q}{|}^2$ and writeConsequentlyAt $z=0$, the eigenvalues are $0$ and $-c$. The neutral cell-density mode reflects mass conservation, not decay of every homogeneous perturbation.

Write $a=D_2>0$, $b=D_{\rho }>0$, $C=D_1{f}_0$, and interpret production physically as $f_0\ge 0$. If $c\ge 0$, the trace-determinant stability criterion shows that an unstable nonzero wavenumber exists precisely when $C>ac$. The unstable band isIf $c<0$, the chemical field is already unstable at zero wavenumber; arbitrarily small positive wavenumbers are unstable too. Since $C\ge 0$, this also implies $C>ac$. Thus, on the infinite plane, the undivided Keller--Segel aggregation threshold isFor $\kappa >0$, dividing by $D_2\kappa$ gives the requested formAt equality there is no strictly growing mode when $c\ge 0$; nonzero spatial modes decay. On a finite domain, a permitted nonzero wavenumber must actually lie in the unstable band. If one allows a signed production function, the displayed matrix and trace-determinant stability criterion remain valid, but the simplification using $C\ge 0$ must be revisited.

The printed hypotheses do not ensure $\kappa >0$. Positivity of the degradation rate $k$ alone is insufficient. For a concrete counterexample, take $k(\rho )=e^{-2\rho }$, $f(\rho )=e^{-2}$, $a_0={\rho }_0=1$, and $D_1=D_2=D_{\rho }=1$. The homogeneous equilibrium point condition holds, all rates and transport coefficients are positive, but $\kappa =c=-e^{-2}$. The printed left-hand side equals $-1$, although $\det M=z(z-2e^{-2})<0$ for $0<z<2e^{-2}$. These spatial modes grow. If $\kappa =0$, the printed expression is undefined. The undivided criterion and the matrix above resolve both cases.

To find the fastest-growing Keller--Segel mode, use the larger eigenvalueFor $C>0$ the square root is real. Put $p=a+b$ and $d=b-a$. Differentiation givesA positive maximizing wavenumber exists in either of two growing cases: $c\ge 0$, $C>ac$, or $c<0$, $C>-bc$. Equivalently, $C>\max (ac,-bc)$. In this range $C+dc>0$, andThus the stationary point is the unique maximum. The derivative condition yieldsSolving it, with the root that satisfies the unsquared derivative equation, givesFor $c=0$, the unequal-diffusivity expression simplifies to $z_{\ast }=C/[\sqrt{ab}(\sqrt{a}+\sqrt{b}{)}^2]$, also agreeing with the equal-diffusivity limit. The formula maximizes the full two-field growth rate; it makes no instantaneous-chemical approximation. Every direction of $\mathbf{q}$ with this length is equivalent by rotational symmetry.

If $c<0$ but $0\le C\le -bc$, the maximum instead occurs at $z=0$: the fastest mode is homogeneous, with $s_{\ast }=-c$ and infinite wavelength. For $C>0$, its initial derivative is $-b+C/(-c)\le 0$. If $C+dc\ge 0$ the derivative thereafter decreases, while if $C+dc<0$ it increases towards the still-negative large-$z$ limit; either way no positive-$z$ maximum is missed. At $C=0$, the two eigenvalues are simply $-az$ and $-c-bz$, giving the same conclusion. In a finite box, maximize $s_{+}$ over the allowed Fourier modes; excluding the homogeneous mode can change the selected length. In a stable parameter range there is no fastest-growing mode.

The first ratio compares the positive feedback loop “more cells produce more attractant, which draws in more cells” with spreading by cell diffusion and removal of chemical perturbations. Its numerator $D_1{f}_0$ measures chemotaxis together with attractant production; its denominator $D_2\kappa$ measures dispersal together with incremental degradation. This competition between directed chemotaxis and cell diffusion produces spatial aggregation. The second ratio compares the concentration dependence of chemical production, $a_0{f}_0^{\prime }$, with incremental degradation $\kappa$. Positive $f_0^{\prime }$ amplifies chemical fluctuations, whereas negative $f_0^{\prime }$ suppresses them. It changes the chemical relaxation available to the chemotaxis feedback loop and can itself destabilize the homogeneous chemical field. Chemical diffusion $D_{\rho }$ suppresses short scales and sets the selected wavelength, but does not change the infinite-plane long-wave threshold. These interpretations as ratios of stabilizing and destabilizing processes assume $\kappa >0$.

For the motionless uniform layer, the heat equation reduces to $T_{zz}=0$. The lower temperature and upper heat-transfer condition determine the quasistatic temperature of a conductively cooled film:When $\alpha h/\kappa \ll 1$, $T_s=T_0-\Delta T\alpha h/\kappa +O(\Delta T(\alpha h/\kappa {)}^2)$. For a varying film let its vertical velocity scale as $Uh/L$, as required by incompressible flow. Relative to vertical thermal diffusion, horizontal thermal diffusion is smaller by $(h/L{)}^2$, and either horizontal or vertical thermal advection is smaller by $U{h}^2/(\kappa L)$. Time variation on the film's transport timescale $L/U$ has the same small factor. Thus the leading heat equation is again $T_{zz}=0$ at each $x,t$ when both specified dimensionless parameters are small. This argument retains the small $\alpha h/\kappa$ assumption. If thickness were externally varied on a faster timescale $t_h$, one would additionally need $h^2/(\kappa t_h)\ll 1$; the spatial conditions alone do not control arbitrary rapid time dependence.

Define $b={\gamma }^{\prime }\Delta T\alpha /\kappa >0$. The surface tension at the surface is ${\gamma }_s={\gamma }_0+bh$, so the Marangoni stress is ${\gamma }_{s,x}=b{h}_x$. With capillary pressure $p=-{\gamma }_0{h}_{xx}$ to leading order, horizontal Stokes flow in the film solves $\mu u_{zz}=p_x$, $u(0)=0$, $\mu u_z(h)={\gamma }_{s,x}$. HenceThe continuity equation gives the thermocapillary thin-film equationTo justify using constant surface tension in the capillary pressure, compare the two volume fluxes on horizontal scale $L$: their magnitudes are ${\gamma }_0{h}^4/(3\mu L^3)$ and $b{h}^3/(2\mu L)$. When both mechanisms are retained at leading order, their balance gives $bh/{\gamma }_0=O((h/L{)}^2)\ll 1$. Thus replacing ${\gamma }_s$ by ${\gamma }_0$ in the curvature term incurs only a relative small correction. Equivalently, small $\alpha h/\kappa$ ensures this if ${\gamma }^{\prime }\Delta T/{\gamma }_0$ is bounded; if it were parametrically large, the extra small surface-tension-variation condition would have to be imposed separately.

The capillary pressure is higher beneath a crest than beneath a trough, so pressure-driven volume flux drains the crest and smooths thickness. The Marangoni effect acts oppositely: a thicker region has a colder surface and therefore larger surface tension. Surface velocity is pulled towards that region, supplying more liquid and amplifying the thickness variation. The arrows below represent these two mechanisms separately.

Put $a_c={\gamma }_0/(3\mu )$ and $a_m=b/(2\mu )$. Substitution of $h=\hat{h}H$, $x=\hat{h}X/ϵ$, $t=\hat{t}\tau$ gives coefficients $a_m{ϵ}^2\hat{t}$ and $a_c{ϵ}^4\hat{t}/\hat{h}$. Making both equal to one yieldsIn these scales the fractional surface tension change is $b\hat{h}/{\gamma }_0=2{ϵ}^2/3$, making the previous approximation explicit.

The linear stability analysis about $H=1$ gives ${\eta }_{\tau }+{\eta }_{XX}+{\eta }_{XXXX}=0$. For a normal mode $e^{s\tau +ikX}$,Long waves with $0<|k|<1$ grow, $|k|>1$ decay, and $k=0,\pm 1$ are neutral. The neutral $k=0$ mode changes the mean film thickness. The dispersion relation is even in $k$, with its two maxima at $\pm 1/\sqrt{2}$.

For a steady positive film with zero volume flux, divide $H^2{H}_X+H^3{H}_{XXX}=0$ by $H^3$ and integrate to obtain $H_{XX}+\ln H=C$. Choosing $\hat{h}$ as specified fixes $C=0$. Multiplication by $H_X$ and another integration then give the zero-flux thermocapillary film profiles first integralFor $H>0$, $V^{\prime }=\ln H$ and $V^{\prime \prime }=1/H>0$: the unique minimum is $V(1)=-1$, while $V(0^{+})=0$ and $V(e)=0$. This potential energy interpretation classifies the profiles without assuming a sinusoidal shape at finite amplitude.

For $E=-1$, the only profile is the uniform film $H=1$. For $-1<E<0$, there are two positive turning heights $H_{-}<1<H_{+}<e$ and a smooth periodic film oscillates between them. Its period isFor $E=0$, the maximum is $e$ and the lower endpoint is zero. A drop reaches zero thickness in finite distance, with zero limiting slope but unbounded curvature since $H_{XX}=-\ln H$. Its half-width isThus the formal limiting profile consists of drops of peak height $e$ and footprint width $2\sqrt{\pi e}$, with zero limiting contact angle. Identical drops can touch, or a zero-thickness region can separate them in the degenerate zero-flux model. These are limiting wet-region solutions, not everywhere positive twice differentiable films; the singular contact region is not resolved by lubrication theory. A useful parametric drop profile isFinally let $E=-1+\delta E$ with small positive $\delta E$. Writing $H=1+\eta$ gives $V=-1+{\eta }^2/2+O({\eta }^3)$ and $H_{XX}=-\eta +O({\eta }^2)$. The small-amplitude steady profile is $H=1+\sqrt{2\delta E}\cos (X-X_c)+O(\delta E)$, with period tending to $2\pi$. Its wavenumber tends to the neutral boundary $|k|=1$ of the unstable band, not to the fastest-growing wavenumber.

Use the well-mixed natural ventilation model for a distributed floor source. Let the building height be $H_b$, its volume be $\mathcal{V}$, its temperature excess be $\theta$, and the exterior mass density be $\rho$. Let ${\alpha }_T$ be the coefficient of thermal expansion, so reduced gravity is $g^{\prime }=g{\alpha }_T\theta$. Convert the total heat input $Q$ to the temperature-volume flux $\mathcal{H}=Q/(\rho c_p)$, where $c_p$ is specific heat capacity. If the quoted heat flux is per floor area, first multiply it by that area.

For effective opening area for pressure-driven ventilation, define the equal-opening hydraulic coefficient $K_A$ such that the signed throughflow $q$ obeys $q|q|=K_A\Pi$, with $\Pi$ the total driving pressure divided by mass density. In the effective-area convention of question 6, each opening has discharge $A\sqrt{\delta p/\rho }$, so $K_A=A^2/2$. The usual physical-area orifice law instead gives $K_A=C_d^2{A}^2$; that prefactor does not affect the stability conclusions.

Set $W=\Delta P/\rho >0$ and $G_T=g{\alpha }_T{H}_b$. Positive $q$ denotes entry at the floor and exit at the roof. hydrostatic pressure and wind combine to giveBecause the inflowing air is at exterior temperature in either direction, the heat budget isThe well-mixed ventilation temperature balance at steady state is thereforeAll roots must satisfy $\theta >0$ and their appropriate flow-direction inequality; squaring a pressure relation without imposing those conditions could add spurious branches.

The assisting case has a unique positive solution and it is stable. For opposing wind, define ${\theta }_w=W/G_T$. There is always one solution with $\theta >{\theta }_w$, a buoyancy-dominated upward flow. Below ${\theta }_w$, the heat-removal curve is $R(\theta )=\sqrt{K_A}\theta \sqrt{W-G_T\theta }$. It is zero at both endpoints, and its maximum occurs at $\theta =2{\theta }_w/3$:For $0<\mathcal{H}<R_{\max }$ there are two downward-flow roots as well as the upward-flow root. Equivalently, setting $y=G_T\theta /W$ and $\epsilon =G_T\mathcal{H}/(\sqrt{K_A}{W}^{3/2})$, the equilibrium curve isFor linear stability analysis of the opposing-wind ventilation bistability, linearize the heat budget: the eigenvalue is $-R^{\prime }(\theta )/\mathcal{V}$. The cooler downward root has $0<\theta <2{\theta }_w/3$ and $R^{\prime }>0$, so it is stable. The warmer downward root has $2{\theta }_w/3<\theta <{\theta }_w$ and $R^{\prime }<0$, so it is unstable and separates the two basins. The upward root has $R^{\prime }>0$ and is stable. At $\mathcal{H}=R_{\max }$ the two downward roots merge at a saddle-node bifurcation; above that value only the upward equilibrium remains.

For ventilation switching under wind and heating changes, increasing wind pressure or decreasing heat input both reduce $\epsilon$, so both can create the pair of wind-driven equilibria. They nevertheless have different physical and transient effects. On the hot upward branch, implicit differentiation of $\mathcal{H}=\sqrt{K_A}\theta \sqrt{G_T\theta -W}$ shows that increasing $W$ increases $\theta$, while decreasing $\mathcal{H}$ decreases $\theta$. Both reduce the upward ventilation rate, but only a wind change shifts ${\theta }_w$.

Under slow parameter variation, a state follows its current stable branch while that branch persists; the upward branch does not disappear when wind increases or heating decreases. A sufficiently large abrupt wind increase can raise ${\theta }_w$ beyond the current temperature and place that state below the new unstable threshold, causing flow reversal and cooling towards the wind-dominated equilibrium. A jump insufficient to cross its basin boundary returns to the upward equilibrium.

At fixed $W$, reducing a still-positive heat input cannot force an initially upward-flow state through ${\theta }_w$: at that boundary the temperature balance has $\dot{\theta }=\mathcal{H}/\mathcal{V}>0$. Thus heat reduction alone does not cause that reversal in this model. Conversely, decreasing wind or increasing heat from the stable downward branch eventually destroys it at the fold and forces a switch to the upward branch, giving history dependence. Wind strengthening and heat reduction have the same effect on the steady dimensionless control parameter, but not the same temperature change or switching dynamics. These stability statements use quasi-steady opening flow and a single well-mixed temperature; other stratification or airflow-inertia models require their own stability analysis.

Validity and the complete angular equation. The Lagrange planetary equations here describe a planar, circular-planet, small-orbital eccentricity, weakly perturbed exterior mean-motion resonance. Require $M_{\mathrm{pl}}/M_{\star }\ll 1$, $e\ll 1$, one isolated slowly varying resonant argument, perturbation times long compared with an orbital period, and no close planetary encounters. Other resonant harmonics and short-period terms must be negligible; coefficients may be frozen only across a narrow range of semi-major axis. Take $p>0$, $q=1$ or $2$ in a reduced integer ratio. Exactly $e=0$ makes the longitude of periapsis undefined, so the displayed angular variables must then be replaced by nonsingular eccentricity vector components.

Neglecting the perturbation to the accumulated mean longitude gives $\dot{\lambda }\simeq n(a)$; this is an accumulated-phase approximation, not differentiation of a fictitious expression $n(t)t$ while discarding $t\dot{n}$. With $n^2{a}^3=G{M}_{\star }$,Differentiating, using $\dot{n}=-3n\dot{a}/(2a)$ and the supplied orbital element equations, gives the full result within this constant-coefficient model:An equivalent formula replaces the last two terms in brackets by $q^2{C}_r{e}^{q-2}[(p+q)n-p{n}_{\mathrm{pl}}-2q{C}_s]-2q^3{C}_r^2{e}^{2q-4}\cos \varphi$. No derivatives of $C_r,C_s$ are included because they were specified as constants.

A full fixed point requires $\sin \varphi =0$ and $\dot{\varphi }=0$. Hence ${\varphi }_{\ast }=0$ or $\pi$, with its corresponding detuned semi-major axis fixed byAt either point $\dot{a}=\dot{e}=0$. The longitude of periapsis still precesses, so the fixed point refers to the reduced resonant dynamics.

Linear stability analysis and encounter geometry. Linearizing the full equation about a fixed point givesFor $q=2$, $\pi$ is a center equilibrium and zero is a saddle equilibrium. For $q=1$, $\pi$ is always a center equilibrium, while zero is a saddle equilibrium ifAt sufficiently small orbital eccentricity the full truncated equations instead admit a center at zero as well; equality is a degenerate case requiring higher-order analysis. Thus an unconditional instability claim at zero does not follow from the full equation. In the usual fixed-eccentricity weak-resonance regime the inequality holds and the stable libration center is $\pi$ for either order.

The physical explanation is resonance protection. At equal mean longitudes, the conjunction direction relative to periapsis obeys $q(\Lambda -\varpi )=\varphi$ modulo $2\pi$. For $q=1$, $\varphi =0$ places conjunction near periapsis, where the exterior particle comes closest to the planet; $\varphi =\pi$ places it near apoapsis. For $q=2$, zero includes both apsidal conjunctions, one of them near periapsis, whereas $\pi$ places the two conjunction branches near quadrature. The protected arrangements give restoring kicks in the positive-$C_r$ leading-harmonic model. This geometric explanation is conditional on its non-crossing, small-orbital eccentricity approximation; it does not override the low-$e$ term retained in the full stability calculation.

Small resonant-argument librations. Define the nominal resonant semi-major axis and the displaced center byThis follows by expanding $n=n_r[1-3(a-a_r)/(2a_r)]$ and taking $\cos (\pi +\delta )=-1+O({\delta }^2)$. It gives the requested initial semi-major axis to first order in the precession-induced detuning; $e_0=e(0)$ and $\delta =\Delta \varphi$.

For the pendulum approximation of a mean-motion resonance, also require fractional changes in $e$ small across a libration, and precession terms small compared with $n$. In particular $C_r/(n_r{e}_0^3)\ll 1$ for $q=1$ makes the extra full-equation curvature negligible; for $q=2$, small fractional eccentricity changes require $C_r/(n_r{e}_0^2)\ll 1$. Freeze $e=e_0$ in the leading restoring coefficient and setThen $u=\varphi -\pi$ obeys simple harmonic motion. Using $u(0)=\delta$, $\dot{u}(0)=0$ and integrating the remaining Lagrange planetary equations givesThe omitted apsidal modulation comes from higher perturbative orders and $u^2$. All expressions are leading resonant approximations, valid while the omitted fractional changes remain small. The small-libration phase portrait iswith period $2\pi /{\omega }_0$. The semi-major axis half-width is $2a_r{\omega }_0|\delta |/(3p{n}_{\mathrm{pl}})$ and the orbital eccentricity half-width is $q{C}_r{e}_0^{q-1}|\delta |/{\omega }_0$.

Dividing the original $\dot{e}$ and $\dot{a}$ equations gives the exact invariant of their retained terms,Thus the $e$–$a$ plot is a segment of this increasing curve traversed back and forth, not a closed ellipse. Locally its slope is $q/[2(p+q)e_0{a}_c]$.

Finite amplitude and the separatrix. Keep the same weak-resonance approximation but allow $\delta$ to be finite. The leading angular equation is $\ddot{\varphi }={\omega }_0^2\sin \varphi$, or $\ddot{u}=-{\omega }_0^2\sin u$. Multiplication by $\dot{u}$ givesThis resonant pendulum energy is constant in the pendulum approximation of a mean-motion resonance. It is not an exact integral of the earlier full equation when $e$ and its restoring coefficient vary: the $\dot{\varphi }\sin \varphi$ term alone prevents that conclusion in general.

For $|\delta |<\pi$ the phase portrait has closed libration curves around $\pi$, withwhere $K$ is the complete elliptic integral of the first kind. The largest speed is $2{\omega }_0\sin (|\delta |/2)$ and the leading semi-major axis half-width is $4a_r{\omega }_0\sin (|\delta |/2)/(3p{n}_{\mathrm{pl}})$. As $|\delta |\to \pi$, the energy approaches $2{\omega }_0^2$, the separatrix through the saddles at zero and $2\pi$; the particle spends increasingly long intervals near those saddles. With $ϵ=\pi -|\delta |$,At exactly $\delta =\pi$, $\dot{\varphi }=0$, the particle stays at the unstable fixed point in the ideal model. Nonstationary separatrix trajectories approach the saddle only in infinite time; above separatrix energy, the resonant argument circulates and resonance protection is lost. These distinctions matter when describing the limiting motion.

Rotating Rayleigh-Bénard convection combines buoyancy-driven instability with the Coriolis force. Consider a plane layer of depth $d$ rotating uniformly about the vertical axis, heated from below. Adopt the Boussinesq approximation, fixed boundary temperatures and, for explicit formulas, impermeable stress-free boundary conditions. The conductive state of Rayleigh-Bénard convection is motionless with a linear temperature profile. The dimensionless controls are the Rayleigh number, Prandtl number and Taylor number:Here $\nu$ is kinematic viscosity and $\kappa$ thermal diffusivity. In thermal-diffusion time units, linear perturbations satisfyRotation does no direct mechanical work, since $\mathbf{u}\cdot ({\mathbf{e}}_z\times \mathbf{u})=0$, but couples vertical motion to vertical vorticity and changes the damping and oscillation balance.

For horizontal wavenumber $k$ and vertical mode $n$, put $m=n^2{\pi }^2$ and $a=k^2+m$. Use $w=W\sin (n\pi z)$, $\theta =\Theta \sin (n\pi z)$ and vertical vorticity $\zeta =Z\cos (n\pi z)$ with growth rate $\sigma$. Curling the momentum equation and eliminating pressure givesTheir determinant, without dividing by a possibly zero factor, is the rotating-convection growth-rate polynomialThis makes the linear stability analysis question precise: onset occurs when a root reaches zero real part and all other modes still decay.

A stationary neutral root has $\sigma =0$, givingRotation raises this stationary threshold. An oscillatory neutral root has $\sigma =i\varpi$ with $\varpi \ne 0$. Real-imaginary separation gives the oscillatory neutral curve of rotating convectionThis branch is admissible only when ${\varpi }^2>0$, requiring $P<1$ and sufficiently strong rotation. The restoring Coriolis force coupling permits an inertial/thermal oscillation whose phase-lagged buoyancy can overcome dissipation. At large $P$, temperature and momentum diffusion do not permit that overstability mechanism at primary onset, so the exchange of stabilities is stationary. The actual threshold is the minimum of the stationary and admissible oscillatory curves over all allowed modes, not an arbitrary formal value of $R_o$. Rigid plates require a different vertical eigenproblem and Ekman layers, so the explicit free-slip numbers are not universal.

At large Taylor number, the first vertical mode is selected in the ideal plane layer. Let $t=k^2,m={\pi }^2$. Minimizing the stationary curve givesThus the stationary neutral curve of rotating convection hasThe physical horizontal wavelength is $2\pi d/k_c$, hence decreases as ${\mathrm{Ta}}^{-1/6}$; its prefactor depends on the boundary convention. Thin nearly vertical cells reconcile the strong Coriolis force constraint with viscous and thermal diffusion. The oscillatory minimization replaces the right side of the wavenumber equation by $P^2\mathrm{Ta}m/(1+P{)}^2$, giving the same Taylor number exponent at fixed positive $P$. Where its frequency remains admissible,Equality is $8P^4-P-1=0$, whose positive root is approximately $P_{\ast }=0.6766$. Accordingly, for sufficiently rapid rotation in this free-slip problem, $P<P_{\ast }$ selects oscillatory onset and $P>P_{\ast }$ stationary onset. The weaker condition $P<1$ is only necessary for an oscillatory neutral mode; it does not by itself identify the first instability. Finite Taylor number, finite lateral geometry, allowed discrete wave numbers and plate conditions change the selection.

For the counterpropagating Hopf amplitudes in rotating convection near a simple oscillatory onset, the Hopf bifurcation produces slow complex amplitudes for counterpropagating roll waves. After separating the fast carrier oscillation, symmetry permits the cubic equationswith generally complex coefficients; an $i{\omega }_0{Z}_{\pm }$ term restores the fast frequency if desired. The real parts govern amplitude saturation and the imaginary parts give nonlinear frequency shifts. Write $a_s=\mathrm{Re}{g}_s$, $a_c=\mathrm{Re}{g}_c$. For a travelling wave from a supercritical bifurcation with only one amplitude nonzero, $|Z_{\pm }{|}^2=r/a_s$ with $r,a_s>0$, and the competing wave's growth rate is $r(1-a_c/a_s)$. It is amplitude-stable against that competitor when $a_c>a_s$. A standing wave has equal intensities $r/(a_s+a_c)$; provided this is positive, its intensity-difference mode is stable when $a_s>a_c$. Temporal and spatial phase symmetries leave neutral phase directions, so these are orbital/amplitude stability statements, not decay of every phase displacement.

These coefficients follow from nonlinear interactions and the Fredholm solvability condition obtained by projection onto an adjoint eigenfunction; symmetry alone cannot decide their signs. A negative saturating coefficient gives subcritical bifurcation behavior requiring higher-order terms. Spatial modulation leads to coupled complex Ginzburg–Landau equations with group velocities and diffusion; phase instabilities, mean-flow coupling and differently oriented rolls can destabilize a wave stable in the restricted two-amplitude system. A weakly nonlinear expansion of oscillations therefore predicts travelling waves or standing waves, frequency shifts, modulation and possible secondary mode competition, not a unique universal periodic state.

The Küppers–Lortz instability is a different route to time dependence: it destabilizes steady saturated rolls against oblique roll perturbations. For stationary-roll amplitudes of orientations ${\varphi }_j$, a leading competition system hasA pure roll has $|A_i{|}^2=r/g_0$ with $g_0>0$. An infinitesimal new roll at relative angle $\vartheta$ grows atFor sufficiently strong rotation in appropriate boundary and Prandtl number regimes, some finite oblique angle has $g(\vartheta )<g_0$, so a steady roll is unstable arbitrarily close above its stationary onset. Rotation is handed and allows $g(\vartheta )\ne g(-\vartheta )$, so replacement of one roll by another can favor a definite cyclic sense. Three or more competing orientations can form a heteroclinic cycle; whether it attracts depends on contraction/expansion rates and other modes. Noise, spatially varying domains and modulation can turn this competition into repeated orientation switching and irregular patterns.

The invading rolls are three-dimensional disturbances even when the original straight roll is described by a two-dimensional section. The finite-angle Küppers–Lortz instability mechanism should also be distinguished from the small-angle instability of rotating convection rolls mediated by large-scale mean flow at finite Prandtl number. Numerical thresholds and favored angles depend on mechanical boundaries and material parameters; the essential criterion is the cross-coupling relative to self-saturation. Rotation both changes primary onset and wavelength, and can prevent the resulting steady roll pattern from remaining a stable nonlinear state.

Expand the square root and quotient in the logarithmic-height plume equations:The linear stability analysis givesIn particular, the second row contains $-2m{M}^{\prime }$, since the reciprocal of $\hat{M}$ contributes another $-M^{\prime }$.

For the growing mode of power-law plume similarity, the characteristic polynomial isWithin the physical interval $q,m>0$ and $f<0$, $P(0)<0$ while $P(\lambda )\to +\infty$ as $\lambda \to +\infty$. There is a positive real eigenvalue, so the similarity equilibrium is unstable with increasing height. Perturbations behave as $e^{\lambda \xi }\propto z^{\lambda }$; the result is not stability in physical time. Thus this exact positive-buoyancy similarity trajectory is not a generic attracting far field.

At the upper endpoint,The limiting polynomial is $\lambda (\lambda +1)(\lambda +10/3)$, reproducing the three stated eigenvalues. The positive eigenvalue tends to zero, with $\lambda \sim -2f/3$. The two negative modes decay as $z^{-1}$ and $z^{-10/3}$; the remaining mode is neutral at the limiting linear system. Its eigenvector is $(1,2,3)$ in relative-flux coordinates.

The neutral buoyancy-flux mode of a pure plume has a simple interpretation: a constant surviving buoyancy flux fixes the amplitudes of a pure plume, with $Q\propto F^{1/3}{z}^{5/3}$ and $M\propto F^{2/3}{z}^{4/3}$. A small fractional change in $F$ produces changes in $Q,M$ in the ratio $1:2:3$. It is retained rather than damped out.

There is a qualification at the exact endpoint. For fixed nonzero $K$ and $\beta =-8/3$, taking $Q\propto z^{5/3}$ gives $F_z\propto -z^{-1}$, not zero. The flux coefficients above also diverge as $f\to 0$. Thus the constant-flux powers are the formal limiting scaling, not an exact constant-buoyancy solution in a nonzero endpoint stratification; cumulative ambient effects can require logarithmic corrections.

Under the convention $e^{i(kx-\omega t)}$, a mode's amplitude grows as $e^{{\omega }_{i}t}$. The linear stability analysis of real-wavenumber disturbances is therefore summarized byAt ${\omega }_{i,\max }=0$ there is marginal or neutral hydrodynamic stability: no normal mode grows exponentially, but some may fail to decay. If “linearly stable” includes neutral modes, its condition is ${\omega }_{i,\max }\le 0$. These modal statements do not rule out transient growth from non-normal modes or algebraic growth.