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Let the basic temperature be $\overline{T}=T_{\ast }-\beta z$ with $\beta >0$, the density be ${\rho }_0$, gravitational acceleration be $-g\hat{\mathbf{z}}$, and thermal expansion coefficient be $a_T$. Denote kinematic viscosity, thermal diffusivity and magnetic diffusivity by $\nu ,\kappa ,\eta$. Write the dimensional magnetic perturbation as ${\mathbf{b}}_d$, the temperature perturbation as ${\theta }_d$, and use constant permeability ${\mu }_0$. Linearizing the Boussinesq equations and the resistive induction equation about the static state givesThe Lorentz force identity is $(\nabla \times {\mathbf{b}}_d)\times B_0\mathbf{c}=B_0(\mathbf{c}\cdot \nabla ){\mathbf{b}}_d-\nabla (B_0\mathbf{c}\cdot {\mathbf{b}}_d)$. Its gradient term is absorbed into $P_d$, the perturbation magnetohydrodynamic total pressure. The temperature source follows from ${\mathbf{u}}_d\cdot \nabla \overline{T}=-\beta u_{d,z}$.

Choose any reference length $d$, since no vertical boundaries are prescribed. Scale length by $d$, time by $d^2/\kappa$, velocity by $\kappa /d$, temperature perturbation by $\beta d$, magnetic perturbation by $B_0$, and total pressure by ${\rho }_0\nu \kappa /d^2$. The dimensionless parameters areThey are the Prandtl number, magnetic-to-thermal diffusivity ratio, Rayleigh number, and Chandrasekhar number, respectively. Division by the chosen scales givesIn particular $Q\zeta =B_0^2{d}^2/({\mu }_0{\rho }_0\nu \kappa )$, so the diffusivity convention agrees with the momentum coefficient.

For a plane wave put $\mathbf{l}=(k,0,m)$, $q=|\mathbf{l}{|}^2=k^2+m^2$, and $d_{\parallel }=\mathbf{c}\cdot \mathbf{l}=k\sin \alpha +m\cos \alpha$. The induction and temperature equations imply $\hat{\mathbf{b}}=i{d}_{\parallel }\hat{\mathbf{u}}/(s+\zeta q)$ and $\hat{\theta }=\hat{w}/(s+q)$. The projection $P=I-\mathbf{l}{\mathbf{l}}^{\mathsf{T}}/q$ removes pressure. Because $(P\hat{\mathbf{z}}{)}_z=k^2/q$, its vertical momentum equation isThus the oblique-field plane-wave magnetoconvection dispersion relation isIf out-of-plane velocity is also allowed, its decoupled magnetic branch satisfies $(s/\sigma +q)(s+\zeta q)+Q\zeta d_{\parallel }^2=0$; it has no buoyant driving. At stationary marginality, for $k\ne 0$,These are stationary thresholds; an oscillatory marginal mode, if present for particular diffusivity ratios, is a different onset calculation.

For $m=1$, write $R_{mathrmstat}=F(k)+QG(k)$, where $F=(1+k^2{)}^3/k^2$ and $G=(1+k^2)(\cos \alpha +k\sin \alpha {)}^2/k^2$. Direct differentiation gives $F^{\prime }(k)=-2k^{-3}+6k+4k^3$, so its two minima are $\pm 1/\sqrt{2}$, with $F=27/4$ and $F^{\prime \prime }=36$. For $Q>0$ the negative sign has the smaller magnetic penalty: at every positive $k$, $G(-k)<G(k)$ because both sine and cosine of the inclination are positive. The wavenumber selection in oblique-field magnetoconvection therefore starts from $a=-1/\sqrt{2}$. Expanding $F^{\prime }(k_c)+Q{G}^{\prime }(k_c)=0$ gives $k_c=a+C_{1}Q+O(Q^2)$, where $C_1=-G^{\prime }(a)/F^{\prime \prime }(a)$. Taylor expansion of the value then givesThis proves the small-field forms, including the order of the correction.

For large $Q$, $G$ is nonnegative and vanishes only at $b=-\cot \alpha$. The candidate $k=b$ has finite threshold $F(b)$, so a global minimizer must satisfy $QG(k_c)\le F(b)$ and hence tend to $b$. Here $G^{\prime \prime }(b)=2(1+b^2){\sin }^2\alpha /b^2>0$. Expanding the stationarity condition at $b$ yieldsThe value expansion isThe leading coefficient follows from $1+{\cot }^2\alpha ={\csc }^2\alpha$. The expansion holds at fixed $0<\alpha <\pi /2$.

Finally choose $\cot \alpha =1/\sqrt{2}$. Then the nonmagnetic minimizing wavevector also has $d_{\parallel }=0$, and its magnetic penalty vanishes. Since $F(k)\ge 27/4$ and $QG(k)\ge 0$, this is the global minimum for every nonnegative field strength:This field-independent stationary threshold in oblique magnetoconvection has a simple physical explanation: the disturbance is constant along the imposed field, so it does not bend field lines.

Let the layer depth be $d$, its imposed temperature drop $\Delta T$, and its dimensional vertical field $B_0\hat{\mathbf{z}}$. Use length $d$, thermal time $d^2/\kappa$, velocity $\kappa /d$, temperature $\Delta T$ and field $B_0$ as units. Write the total dimensionless field as $\hat{\mathbf{z}}+\mathbf{b}$, and temperature as the conductive profile plus $\theta$. The Boussinesq equations coupled to the resistive induction equation areThe dimensionless parameters in this normalization areHere ${\alpha }_T$ is the thermal expansion coefficient, $\nu$ is kinematic viscosity, $\kappa$ is thermal diffusivity, and $\eta$ is magnetic diffusivity. The first parameter is the Rayleigh number and $\sigma$ is the Prandtl number.

The displayed dispersion relation fixes the magnetic normalization: its $Q$ must be the thermal-diffusion magnetic-field parameter above. The conventional Chandrasekhar number is instead $Q_{\mathrm{Ch}}=B_0^2{d}^2/({\mu }_0\rho \nu \eta )$, so $Q=\zeta Q_{\mathrm{Ch}}$. If $Q$ is reserved for that conventional definition, the magnetic term in the printed polynomial requires an extra factor $\zeta$. We keep $Q$ for the printed polynomial throughout the following calculation.

Linearize about the conductive state. Absorbing linear magnetic pressure into $p$, the magnetic force becomes $\sigma Q{\partial }_z\mathbf{b}$. The pressure-eliminated vertical momentum equation isThe normal mode implicit in the given relation has $w,\theta \propto \sin \pi z$ and $b_z\propto \cos \pi z$, as for stress-free, fixed-temperature boundaries with the corresponding vertical-field magnetic conditions. Put $x=k^2$, $p_0={\pi }^2$ and $m=x+p_0={\beta }^2$. Taking growth proportional to $e^{st+i\mathbf{k}\cdot {\mathbf{x}}_h}$ gives the amplitude equationsTheir determinant isThis is the stated vertical-field magnetoconvection dispersion relation after accounting for its field-strength convention. The determinant derivation remains valid even at values where an intermediate division by $s+m$ or $s+\zeta m$ would not be allowed.

For steady marginality set $s=0$. One obtainsFor oscillatory marginality write $s=i\omega$ with $\omega \ne 0$. Dividing the polynomial by $m$ gives a real cubic $s^3+A{s}^2+Bs+C_0$, whereIts real and imaginary parts imply ${\omega }^2=B$ and $C_0=AB$. Solving for the oscillatory convection threshold givesandA physical oscillatory branch therefore requires $0<\zeta <1$ and positive frequency squared. In particular an algebraic continuation of $R_o$ into ${\omega }^2<0$ is not an oscillatory onset.

The steady-Hopf merger in vertical-field magnetoconvection occurs at zero frequency:Substitution verifies $R_s=R_o$. Equivalently the merged horizontal wavenumber satisfiesA finite-wavenumber intersection exists only when this expression is positive. At the intersection the growth polynomial has a double zero root: the nonzero-frequency neutral branch ends there.

For the requested merger at a magnetoconvection neutral-curve minimum, first note that a curve of the form $R=[A_{\ast }{m}^3+C_{\ast }m]/x$ has derivative zero exactly whenIts unique positive minimum has $x>p_0/2$, because the left side increases there from zero to infinity. At this minimum $R_c=2A_{\ast }{m}^3/p_0$.

Merger at the steady minimum. For $R_s$, the stationarity condition is $m_s^2(2x_s-p_0)=Q{p}_0^2/\zeta$. Combining it with the merger equation givesThe corresponding critical Rayleigh number is $R_{s,c}=2(x_s+p_0{)}^3/p_0$. For fixed positive $\sigma$ and large $Q$, set $\zeta =1-{\delta }_s$ and retain the leading powers. ThenMore generally, before taking $\zeta \to 1$, the steady minimum obeys $k_s^2\sim [Q{\pi }^4/(2\zeta ){]}^{1/3}$.

Merger at the oscillatory minimum. Define $A_o=(\sigma +\zeta )(1+\zeta )/\sigma$. The stationarity condition for $R_o$ isCombining it with the zero-frequency relation instead givesHere $R_{o,c}=2A_o(x_o+p_0{)}^3/p_0$. For fixed $\sigma >0$ and $Q\gg 1$, setting $\zeta =1-{\delta }_o$ givesThese two merger conditions are distinct parameter surfaces, not simultaneous assertions about the same minimum. Both have the strong-field wavenumber selection in magnetoconvection scaling $k=O(Q^{1/6})$.

If the first subleading Rayleigh corrections are desired, let $X_s=[Q{p}_0^2/(2\zeta ){]}^{1/3}$ and $X_o=[\sigma Q{p}_0^2/(2(1+\sigma )(1+\zeta )){]}^{1/3}$. With parameters bounded in the above regime,These retain the $O(Q^{2/3})$ corrections, including the effect of the $O(Q^{-1/3})$ diffusivity deficit at a merger.

Use the flat-surface Boussinesq approximation, with small thermal expansion, constant properties, negligible surface deformation, and constant surface tension. The top is impermeable and shear-free, while the solid walls are no-slip. Temperature-dependent surface tension would introduce a separate Marangoni stress, which is not included in this buoyancy-driven model. The stated time scale requires $\kappa$ to mean thermal diffusivity; if the material property given is a dimensional thermal conductivity $k$, use $\kappa =k/({\rho }_0{c}_p)$.

Take $z$ positive downwards and define the nondimensional streamfunction by $u={\psi }_z$, $w=-{\psi }_x$. Velocity is scaled by $\kappa /L$, streamfunction by $\kappa$, and temperature is measured from the cold-end surface value. Define the Prandtl number and horizontal-length Rayleigh number bywhere ${\alpha }_T$ is the thermal expansion coefficient. Put $J(\psi ,f)={\psi }_z{f}_x-{\psi }_x{f}_z$, and let ${\nabla }^2={\partial }_x^2+{\partial }_z^2$. Taking the curl of the Boussinesq equations eliminates pressure and gives the full nonlinear vorticity-temperature systemThe buoyancy sign follows from the downward coordinate: warmer fluid has body acceleration $-g{\alpha }_T\Delta TT{\mathbf{e}}_z$.

Choose the same streamfunction constant, zero, on the connected boundary. The mechanical and thermal boundary conditions areThe upper stress-free boundary condition reduces to ${\psi }_{zz}=0$, since $\psi =0$ there already gives ${\psi }_{xx}=0$. An arbitrary additive constant in $T$ changes only the reference hydrostatic pressure.

In steady infinite-Prandtl-number convection, momentum inertia disappears but thermal advection remains:For a shallow lake, with fixed $\mathrm{Ra}$ as $ϵ\to 0$, the leading interior temperature is $T_0=x$. This is an outer solution: it does not satisfy the insulated sidewall derivative right at $x=0,1$. The leading global conductive solution, which resolves those end regions, isIt is harmonic, has $T_c(x,0)=x$, and satisfies the insulated sides and bottom. Away from sidewall layers of horizontal width $O(ϵ)$ it reduces to $x$ with exponentially small end corrections. These layers also turn the flow and enforce the sidewall no-slip boundary condition.

In the shallow interior the dominant momentum derivatives are vertical, so ${\psi }_{0,zzzz}=-\mathrm{Ra}$. Integrating with the surface and bottom conditions gives the shallow horizontal convection with an insulated bottom solution, with $y=z/ϵ$:Surface flow is toward the colder end, with a deeper return flow. The depth-integrated volume flux is zero because ${\psi }_0(ϵ)={\psi }_0(0)=0$. The bulk expressions apply away from the turning regions, not unchanged on the vertical walls. A useful small-advection condition in the end layers is $\mathrm{Ra}{ϵ}^4\ll 1$, automatic for fixed Rayleigh number in the stated thin limit.

The heat flux in these units is $\mathbf{q}=(uT-T_x,wT-T_z)$. For the leading bulk temperature, the horizontal transport through a vertical section per unit out-of-plane width isThus leading advective horizontal transport vanishes: both layers have the same temperature at a given $x$, and their opposing volume fluxes cancel. The leading conductive transport is from the hot end toward the cold end. In dimensional form it is $-{\rho }_0{c}_p\kappa \Delta Tϵ=-k\Delta Tϵ$ per unit width. Heat enters through the warmer portion of the maintained upper surface and leaves through its colder portion, with the conductive end adjustments particularly important near the two insulated sidewalls. It cannot enter through the insulated sides or bottom.

For the advection correction in shallow horizontal convection, set $T=x+\Theta (z)$ in the bulk. Since $w_0=0$ and $T_x=1$, the steady heat equation gives ${\Theta }_{zz}=u_0={\psi }_{0,z}$. The conditions $\Theta (0)=0$ and ${\Theta }_z(ϵ)=0$ imply ${\Theta }_z={\psi }_0$ and henceFor example the bottom bulk temperature is $T(x,ϵ)=x-\mathrm{Ra}{ϵ}^5/320$. The correction makes the returning deeper fluid slightly colder. Its horizontal advective heat transport isThus the transport including this bulk correction is $\boxed{{\mathcal{H}}_x=-ϵ-19{\mathrm{Ra}}^2{ϵ}^9/1451520}$, both terms directed toward the colder end.

Vertical transport must distinguish local flux from net transport. In the leading interior $w_0=T_{0,z}=0$, whereas the temperature correction gives a local downward conductive flux $q_z=-{\Theta }_z=-{\psi }_0>0$. The end circulation gives an upward advective transport: with impermeable sidewalls and $T_0\simeq x$, integration by parts gives ${\int }_0^{1}w{T}_{0}dx=-{\int }_0^{1}x{\psi }_{x}dx={\int }_0^1\psi dx\simeq {\psi }_0(z)<0$. It compensates that downward conduction. More generally, steady heat conservation and insulated sides give $d[{\int }_0^1{q}_{z}dx]/dz=0$, and the insulated impermeable bottom sets the constant to zero. ThereforeThere is no net vertical heat transport through the lake, despite its local recirculating heat flux and its nonzero horizontal heat transport.

First distinguish concentration conventions. If $C$ is cell number density, the excess mixture density is $v\Delta \rho C$. If $\phi =vC$ is cell volume fraction, it is $\Delta \rho \phi$. The printed definition of $R$ is dimensionless for number density, whereas the prose calls ${\hat{n}}_0$ a volume fraction. Use $C_0$ for mean number density and ${\phi }_0=v{C}_0$ for mean volume fraction. Then the consistent Rayleigh number isThe dimensionless concentration $n=C/C_0=\phi /{\phi }_0$ is the same in both conventions. The extra factor $v$ must be removed if the source symbol is literally a volume fraction.

Remove the hydrostatic basic pressure and use a dilute Boussinesq approximation. With ${\hat{C}}_e=\hat{C}-C_0\bar{n}$ and perturbation fluid velocity $\hat{\mathbf{u}}$, the linear incompressible Newtonian fluid equations areThe cell conservation equation linearizes toAt this order the vertical orientation perturbation is zero, because $|\mathbf{p}|=1$. The instantaneous-orientation approximation from the introductory calculation givesIncompressibility implies ${\hat{\nabla }}_{\perp }\cdot (\hat{\mathbf{\omega }}\times {\mathbf{e}}_3{)}_{\perp }=-\hat{\Delta }\hat{w}$. This identity fixes the sign of the gyrotactic term.

Apply the length, viscous-time, velocity, concentration and pressure scales in the paper. Writing $n_e=n-\bar{n}$, the dimensionless equations arewhere $S=\nu /D$ and $G=V_{s}B/(2H)$. The factor $S$ appears in the cell equation because its diffusive time differs from the viscous time used for $t$.

For a normal mode $[n_e,w]=[N,W]e^{\sigma t+i\kappa x}$, set $\mathcal{L}=d^2/d{z}^2-{\kappa }^2$. Taking the divergence of the momentum equation gives $\mathcal{L}{P}_e=-R{N}^{\prime }$. Applying $\mathcal{L}$ to its vertical component eliminates pressure and yieldsThe concentration equation becomesThese reproduce the requested modal equations, with the physically consistent sign of the orientation balance and concentration convention made explicit.

At both plates, the no-slip boundary condition gives $w=0$ and horizontal perturbation velocity zero. For a nonzero horizontal wavenumber, incompressibility then gives $W^{\prime }=0$. The normal cell flux is $\bar{n}w+hN-N^{\prime }$; the orientation correction has no vertical component at this order. Impermeability therefore givesIf orientation lag is retained rather than the given instantaneous balance, ${\mathbf{p}}_{\perp }$ acquires the relaxation response $(1+B{\partial }_{\hat{t}}{)}^{-1}$ and the modal gyrotactic term changes accordingly. The displayed equations are the quasistatic-orientation model requested here.

Use the layer depth $d$, the thermal diffusion time $d^2/\kappa$, velocity $\kappa /d$, and the imposed temperature difference $\Delta T$ as scales. Let $\theta$ be the temperature departure from the conductive profile, $\mathbf{u}$ the velocity, $w=\mathbf{u}\cdot \hat{\mathbf{z}}$, and let pressure absorb hydrostatic terms. The dimensionless Boussinesq equations for Rayleigh-Bénard convection areThe Prandtl number and Rayleigh number are $\sigma =\nu /\kappa$ and $R=g{\alpha }_T\Delta T{d}^3/(\nu \kappa )$, with kinematic viscosity $\nu$, thermal diffusivity $\kappa$ and coefficient of thermal expansion ${\alpha }_T$. At $z=0,1$, the stress-free boundary condition and perfectly conducting thermal boundary condition give $w=0$, ${\partial }_z{u}_x={\partial }_z{u}_y=0$, and $\theta =0$.

For the fundamental vertical normal mode, take $w=W\sin (\pi z)e^{i\mathbf{k}\cdot \mathbf{x}+\lambda t}$ and $\theta =\vartheta \sin (\pi z)e^{i\mathbf{k}\cdot \mathbf{x}+\lambda t}$. Here $k=|\mathbf{k}|>0$ is the horizontal wavenumber. With $s=k^2+{\pi }^2$, projection onto incompressible flow eliminates pressure and givesThus the stress-free convection growth-rate polynomial is $(\lambda +\sigma s)(\lambda +s)=\sigma R{k}^2/s$.

For the infinite-Prandtl number limit, the finite thermal growth rate isThe other, viscously damped root tends to $-\sigma s$. For $\sigma =1$, both roots are explicit:The plus root determines instability. A higher vertical harmonic replaces ${\pi }^2$ by $n^2{\pi }^2$; the fundamental is the most favorable one.

For the large-$R$ maximum in the infinite-Prandtl number limit, put $x=k^2$ and $p={\pi }^2$. Differentiating gives $d{\lambda }_{\infty }/dx=R(p-x)/(p+x{)}^3-1$. The unique maximum for sufficiently large $R$ therefore satisfies $R(p-x)=(p+x{)}^3$, whenceThis is fastest growth in infinite-Prandtl convection. The large-$R$ expansion is taken after the infinite-Prandtl number limit; at finite $\sigma$, neglecting inertia additionally requires $|\lambda |\ll \sigma s$. It selects the fastest growing mode far above onset, rather than the neutral-curve minimum $k^2={\pi }^2/2$.

The Lorenz model of thermosolutal convection is a Galerkin method truncation retaining a roll velocity and the leading temperature and concentration modes. The variable $a$ measures the roll velocity; $b$ is the temperature mode correlated with the vertical motion, while $c$ is the horizontally averaged temperature-profile distortion generated by that motion. The corresponding concentration modes are $d$ and $e$. The symbol $d$ in this model is a solute amplitude, not the dimensional layer depth. The terms $ab$ and $ad$ change the mean scalar profiles; $a(1-c)$ and $a(1-e)$ express advection of the remaining background gradients.

The positive parameters are the Prandtl number $\sigma =\nu /\kappa$, the ratio $\tau ={\kappa }_s/\kappa$ of solutal diffusivity to thermal diffusivity, and the geometric vertical-mode decay ratio $\varpi$. In the usual fundamental-roll truncation with time normalized to the thermal roll-mode decay rate, $\varpi =4{\pi }^2/(k^2+{\pi }^2)$; treating it as an independently specified positive truncation parameter leaves the arguments below unchanged.

The parameter $r$ is the destabilizing thermal Rayleigh number divided by the free-slip critical-mode value, and $r_s>0$ measures stabilizing solutal buoyancy in the same thermal-diffusive normalization. For example, using a positive bottom-to-top stabilizing solute difference $\Delta S$ and density coefficient ${\beta }_S$,If a solutal Rayleigh number is instead defined with ${\kappa }_s$ in its denominator, the coefficient $r_s$ here is $\tau$ times that normalized number. Thermal buoyancy drives the roll; solutal buoyancy opposes it and diffuses at a different rate. This is the essential mechanism of thermosolutal convection.

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.