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Take the leading real field to be $A{e}^{i(k_{c}x+{\omega }_{0}t)}+B{e}^{i(-k_{c}x+{\omega }_{0}t)}+\text{complex conjugate}$. A spatial translation gives $(A,B)\mapsto (e^{i{k}_c\delta }A,e^{-i{k}_c\delta }B)$; a time translation gives $(A,B)\mapsto (e^{i\psi }A,e^{i\psi }B)$; spatial reflection exchanges $A$ and $B$.

For a monomial $A^j{\overline{A}}^k{B}^l{\overline{B}}^m$ in the first amplitude equation, covariance requires $j-k+l-m=1$ and $j-k-l+m=1$. Hence $j-k=1$, $l-m=0$. At linear order only $A$ is allowed, no quadratic monomial satisfies these conditions, and the cubic possibilities are $A^2\overline{A}$ and $AB\overline{B}$. Reflection fixes the coefficients in the second equation. Removing the small common linear frequency correction by a rotating frame makes $\mu$ real. This derives the two cubic amplitude equations, rather than merely postulating them.

Let $r=|A{|}^2$, $s=|B{|}^2$. Taking real parts givesA travelling wave has $(r,s)=(\mu /{\beta }_r,0)$, or its reflection, whenever $\mu /{\beta }_r>0$. Its phase evolves at $-{\beta }_i\mu /{\beta }_r$, shifting the physical oscillation frequency. The nonzero-amplitude radial eigenvalue is $-2\mu$, while the absent mode has real growth rate $\mu (1-{\gamma }_r/{\beta }_r)$. Therefore, on the supercritical side,The phase direction is neutral because of the symmetry, so stability means orbital stability. A branch with $\mu <0$ is radially unstable, whatever the transverse coefficient.

A standing wave has $r=s=h=\mu /({\beta }_r+{\gamma }_r)>0$. Its two phases evolve together at $-({\beta }_i+{\gamma }_i)h$; the relative phase fixes the spatial position of the standing pattern. In the two intensity directions the eigenvalues are $-2\mu$ and $-2h({\beta }_r-{\gamma }_r)$. ThusAgain the symmetry phases are neutral and the conclusion is orbital stability. Equality ${\beta }_r={\gamma }_r$ leaves a continuum of mixed intensities at cubic order; equality ${\beta }_r+{\gamma }_r=0$ removes cubic standing-wave saturation. These degeneracies require higher-order terms and cannot be classified by the cubic truncation. The zero state is linearly stable for $\mu <0$ and unstable for $\mu >0$.

Forcing at $2{\omega }_0$ preserves translation by one forcing period, which acts on the critical amplitudes as $(A,B)\mapsto (-A,-B)$. It destroys the continuous common phase symmetry. Under the remaining spatial translation, $\overline{B}$ transforms as $e^{i{k}_c\delta }\overline{B}$, just like $A$, and $\overline{A}$ transforms just like $B$. Both change sign under the remaining temporal sign reversal. They are therefore allowed linear terms in the respective amplitude equations. Spatial reflection requires their coefficients to coincide.

The resonant physical-frequency calculation gives the same result: multiplying the forcing factor $e^{2i{\omega }_{0}t}$ by the conjugate left-going mode $\overline{B}{e}^{i(k_{c}x-{\omega }_{0}t)}$ produces $\overline{B}{e}^{i(k_{c}x+{\omega }_{0}t)}$, resonant with the $A$ mode. The other term follows by reflection. Thus the leading parametrically forced counterpropagating waves obeywith the reflected equation for $B$. The forcing phase can be absorbed into a common amplitude phase so that $\nu$ is real; its magnitude measures the resonant forcing strength. Higher-order forcing corrections are beyond this leading weak-forcing truncation.

Write $\psi =\theta +\phi$, $U={\beta }_r+{\gamma }_r$ and $V={\beta }_i+{\gamma }_i$. The polar amplitude equations arewith their reflected counterparts. On the equal-amplitude standing wave branch $R=S>0$, equilibrium point requiresSquaring and adding eliminates the phase:For $U^2+V^2>0$ the candidate squared amplitudes areFor each positive root, set $\sin \psi =-Vz/\nu$ and $\cos \psi =(Uz-\mu )/\nu$. The quadratic identity ensures these define a phase. Conversely these phase equations imply the quadratic, so a nonzero phase-locked branch exists exactly when the discriminant is nonnegative and at least one root is positive.

Since $|\cos \chi |=|V|/\sqrt{U^2+V^2}$, the discriminant condition is the printed $|\mu \cos \chi /\nu |\le 1$ for $\nu \ne 0$. It is sufficient under the usual above-onset supercritical assumptions $\mu >0,U>0$, because then the plus root is positive. Without these assumptions the condition is not sufficient: $\mu =-1,U=1,V=0,\nu =1/2$ satisfies it but gives $z=-1\pm 1/2<0$. This is the positivity condition for a phase-locked standing wave, not a removable algebraic detail. The zero solution always exists and is distinct from the nonzero branch. For $\nu =0$, the undivided equations must be used; a nonzero equilibrium point in this fixed rotating frame requires $V=0$ and $z=\mu /U>0$. If $U=V=0$, nonzero equilibrium points require ${\mu }^2={\nu }^2$ and a compatible phase, with no cubic amplitude selection.

Strong enough forcing can overcome the nonlinear frequency shift, lock the sum phase to the external signal and produce a standing wave at half the forcing frequency. The difference phase remains free through spatial translation. The amplitude inequality is an existence condition, not a stability theorem: attraction requires the linearized forced amplitude equations, and nonlinear coefficients have not been restricted enough here to promise stability of every root.

Write $y=\dot{x}$ and $E=(x^2+y^2)/2$. Along a trajectory,Use the unperturbed orbit $x=R\sin t,y=R\cos t$ to compute the leading energy change over one period:Zero averaged energy change gives the nonzero amplitudeThe supplied sign condition makes this real and nonzero. The averaged amplitude equation is $\dot{R}=ϵ[\gamma R/2-(\alpha +3\beta )R^3/8]+O({ϵ}^2)$, whose derivative at that root is $-ϵ\gamma +O({ϵ}^2)$. Its nonzero derivative also makes the root persist as an isolated periodic orbit for sufficiently small positive $ϵ$.

Directly, the planar divergence is $-ϵ(\alpha x^2+3\beta y^2-\gamma )$. On the leading orbit its integral over $2\pi$ is $-ϵ[\pi R^2(\alpha +3\beta )-2\pi \gamma ]=-2\pi ϵ\gamma$. ThusThe cycle is attracting for $\gamma >0$ and repelling for $\gamma <0$. Both sign possibilities are allowed by $(\alpha +3\beta )\gamma >0$.

Magnetoconvection is the competition between thermally driven convection and the force and transport of a conducting fluid's magnetic field. An imposed magnetic field resists motions that bend its field lines through magnetic tension, while magnetic diffusion permits slippage. The competition changes the onset threshold, preferred spatial scale, temporal behaviour and finite-amplitude organization. The geometry matters: a disturbance nearly invariant along a horizontal imposed magnetic field can avoid much of its tension, whereas an impermeable layer with a vertical imposed magnetic field cannot generally avoid bending it.

A useful quantitative example is a nonrotating Boussinesq layer of depth $d$, heated from below with temperature drop $\Delta T>0$ and threaded by a vertical magnetic field $B_0\hat{\mathbf{z}}$. Let $\nu$, $\kappa$, $\eta$ denote kinematic viscosity, thermal diffusivity and magnetic diffusivity. In thermal-diffusion units define the Rayleigh number, Prandtl number, magnetic-to-thermal diffusivity ratio and Chandrasekhar number byUse stress-free impermeable boundaries, fixed temperature perturbation zero, and magnetic boundary conditions keeping the field vertical. More explicitly, the normal modes below satisfy $w={\partial }_z^{2}w=\theta =b_x=b_y={\partial }_z{b}_z=0$ at $z=0,1$. For a horizontal wavenumber $a>0$, take $w,\theta \propto \sin \pi z$, $b_z\propto \cos \pi z$, and put $s=a^2+{\pi }^2$. The linear temperature and resistive induction equation give $(\lambda +s)\theta =w$ and $(\lambda +\zeta s)b_z={\partial }_{z}w$. Applying the solenoidal projection to the magnetohydrodynamic momentum equation removes pressure; the magnetic coupling is $Q\zeta {\partial }_z\mathbf{b}$, and the vertical projected buoyancy has coefficient $R{a}^2/s$. Substituting the temperature and magnetic responses gives the vertical-field magnetoconvection dispersion relationThese stated boundary conditions make the single vertical mode exact in this model; other boundaries change numerical thresholds.

For steady marginal convection, set $\lambda =0$. The threshold curve isThe imposed magnetic field adds a positive penalty. Differentiating with respect to $x=a^2$ shows that the minimizing wavenumber obeysAt $Q=0$ this gives $a_c=\pi /\sqrt{2}$ and $R_c=27{\pi }^4/4$. With increasing $Q$, the root moves to larger $x$ because the left-hand side is strictly increasing for $x\ge {\pi }^2/2$. The strong-field wavenumber selection in magnetoconvection limit isThus a strong vertical imposed field raises the steady threshold and selects narrower convection cells. At small horizontal wavenumber the incompressibility constraint makes a given vertical motion require large horizontal displacement and magnetic bending. Increasing horizontal wavenumber reduces this magnetic penalty, but increases viscous and thermal diffusion. Their balance selects the narrow cells. This statement is specific to the vertical-field geometry: wavenumber selection in oblique-field magnetoconvection can instead favour nearly field-invariant motions.

Oscillatory marginality in vertical-field magnetoconvection occurs because magnetic and temperature perturbations need not relax at the same rate. To exhibit the phase balance, substitute $\lambda =i\omega$ with $\omega \ne 0$ and write the neutral condition asThe imaginary part must vanish because $R$ is real. Dividing that imaginary part by $\omega s$ givessoA nonzero oscillation consequently requires $0<\zeta <1$ and a sufficiently strong imposed magnetic field. In this regime thermal diffusion can erase the temperature anomaly faster than magnetic diffusion relaxes bent field lines. The magnetic restoring force can outlast the buoyant anomaly, reverse the motion, and maintain the phase lag needed for oscillatory instability. The real part gives the oscillatory neutral curveOnly its portions with ${\omega }^2>0$ are physical. The first onset is the lower of the minimized steady and admissible oscillatory thresholds; the existence of an oscillatory neutral solution at a given wavenumber alone does not establish that it is first. For example, at fixed $P>0$ and $0<\zeta <1$, both minimizing wavenumbers grow like $Q^{1/6}$ for large $Q$, while $R_s\sim {\pi }^{2}Q$ and $R_o\sim D{\pi }^{2}Q$ with $D=\zeta (P+\zeta )/(1+P)<1$. At those oscillatory minima, the positive order-$Q$ contribution to ${\omega }^2$ dominates its order-$Q^{2/3}$ negative contribution, so the oscillatory branch is genuinely admissible and becomes the lower threshold. An overstability therefore precedes steady onset in this limit. More generally nonlinear convection can oscillate even when linear onset is steady; the linear calculation is not a classification of every time-dependent state.

The nonlinear regime introduces a different feedback. The magnetic Reynolds number $\mathrm{Rm}=Ud/\eta$ measures induction relative to diffusion for a finite convective velocity $U$. At large but finite $\mathrm{Rm}$, persistent convection rolls can redistribute magnetic flux. In a two-dimensional incompressible roll, a flux function $\mathcal{A}$ obeysCirculation repeatedly stretches the flux pattern, building gradients on which even small magnetic diffusivity acts efficiently. Over the flux-rearrangement time this permits flux expulsion: the interior of a vigorous roll becomes weakly magnetized, while magnetic flux is concentrated near its edges. The weak-field interior then suffers less magnetic inhibition and can support broader, stronger convection than the linear narrow-cell prediction for the original uniform magnetic field. Exact zero resistivity would preserve material flux and cannot by itself justify the relaxed flux-expelled state.

The weak interior field also follows from a steady leading-order balance for a coherent two-dimensional roll. Let $\psi$ be its streamfunction and suppose its interior streamlines are nested closed curves with $\nabla \psi \ne 0$. In the small-diffusivity interior, $\mathbf{u}\cdot \nabla {\mathcal{A}}_0=0$ implies ${\mathcal{A}}_0=F(\psi )$. Integrate the full steady flux equation over the region enclosed by one streamline. Because $\mathbf{u}\cdot \mathbf{n}=0$ there, the advection integral vanishes; the divergence theorem then gives $\oint {\partial }_n\mathcal{A}ds=0$. In a regular interior expansion this requires $F^{\prime }(\psi )\oint {\partial }_n\psi ds=0$. The normal derivative of $\psi$ has one nonzero sign around a regular nested streamline, so $F^{\prime }(\psi )=0$. Hence the leading interior flux function is constant, and its spatial gradient, which gives the two-dimensional magnetic field, vanishes. The mismatch with external flux is accommodated by magnetic boundary layers. This argument assumes the coherent steady roll and a regular interior; it does not prove that every turbulent flow expels flux.

The mean imposed vertical magnetic flux is conserved for horizontally periodic boundaries. In a schematic two-region picture, let a fraction $f$ have weak field $B_c$ in vigorous convection and the remainder have field $B_q$. ThenThe intensified field outside the active cells inhibits surrounding convection. This gives magnetic flux separation: vigorous weak-field patches coexist with strong-field regions whose motion is small or occurs on a finer scale. The redistribution is a consequence of nonlinear induction and back reaction, rather than disappearance of the net imposed flux.

This feedback can permit subcritical magnetoconvection. Infinitesimal perturbations decay below the conductive state's linear threshold because they encounter the full imposed field. A finite-amplitude roll can nevertheless persist by first expelling enough flux to reduce its own magnetic restraint. A sufficient finite disturbance can then switch between the stable conductive state and a convective state, producing hysteresis. A schematic amplitude equation makes the distinction between local linear stability and finite-amplitude persistence explicit:Here $\mu$ measures distance from linear onset; this is an illustrative normal form, not a claim that these coefficients have been derived for every magnetic layer. Writing $r=|A{|}^2$, the nonzero equilibria satisfyThey appear at a saddle-node bifurcation at $\mu =-c^2/(4d)$. For $-c^2/(4d)<\mu <0$, the zero state is stable, the lower radial-amplitude branch is unstable, and the upper branch is stable because its radial derivative has sign $c-2d{r}_{+}<0$. Thus finite-amplitude convection can coexist with a linearly stable conductive state below the onset threshold. Large $\mathrm{Rm}$ facilitates the flux-expulsion mechanism but is not a universal sufficient condition: geometry, the imposed magnetic field, the capacity to form coherent recirculating cells, and magnetic back reaction all matter. The broad nonlinear cells in weak-field regions can coexist with much narrower cells in magnetically dominated regions, linking the nonlinear flux distribution back to scale selection.

Write $Q=x^2{y}^2+y^2{z}^2+z^2{x}^2$ and $V=xyz$. Direct differentiation of the cubic amplitude equations givesWhen $c=e$, these reduce to $\dot{X}=2X(\mu -aX)$ and $\dot{V}=3V(\mu -aX)$. Therefore, wherever $X>0$,This conserved directional quantity is nonconstant in every neighborhood of a nonzero equilibrium. If such an equilibrium were asymptotically stable, all sufficiently nearby initial points would approach it, forcing their conserved values to equal its value. That is impossible. The origin is unstable since its linearization is $\mu I$ with $\mu >0$. Hence there are no asymptotically stable equilibria when $c=e$ in the displayed cubic system. For $c=e=0$ the sphere is entirely equilibria, so its individual points still cannot be asymptotically stable.

The radial equation attracts every nonzero radius toward $X=R^2=\mu /a$. On this invariant sphere, with $c=e$, the remaining motion isFor $c\ne 0$, the six coordinate-axis points are saddles on the sphere. The eight equal-magnitude points $|x|=|y|=|z|=R/\sqrt{3}$ are centers of the restricted motion. In each open octant the regular level curves of $V$ are closed loops around that octant's center. The coordinate planes, where $V=0$, form heteroclinic orbits joining the axis saddles. In the positive octant, for $c>0$ the boundary orientation is $x\text{-axis}\to y\text{-axis}\to z\text{-axis}\to x\text{-axis}$; for $c<0$ all arrows reverse. Sign reflections give the other octants.

The following sketch uses normalized squared coordinates $U=x^2/R^2$, $W=y^2/R^2$, $Z=z^2/R^2$, with $U+W+Z=1$. This is a one-to-one coordinate representation of each closed spherical octant, so its triangle depicts the sphere's trajectories without hiding a hemisphere. The center is $U=W=Z=1/3$, and the interior loops are contours of $UWZ$. This is the conserved octant dynamics of a cyclic amplitude system.

The equivariant Hopf theorem states that, under the usual nondegeneracy and transverse-crossing assumptions at an equivariant Hopf point, each complex axial isotropy subgroup of the spatial symmetry group combined with temporal phase produces a branch of periodic solutions with that spatiotemporal symmetry. Its common fixed-point space must have real dimension two, or complex dimension one. Restricting the Hopf normal form of a dynamical system to that plane gives a scalar complex amplitude equation; a nonzero real cubic coefficient provides a small periodic branch on the corresponding parameter side. The theorem supplies existence and symmetry, not stability.

Put $\zeta =e^{2\pi i/3}$ and write $T_{\theta }z=e^{i\theta }z$, so $T_{\theta }={\tau }_{\theta /{\omega }_0}$. Three complex axial fixed lines areFor the first generator, its eigenvalues are $1,{\zeta }^2$, so the fixed line is exactly $z_2=0$. Reflection supplies the conjugate line $z_1=0$, representing the opposite rotation direction. For the other two generators, swapping coordinates, with or without the common minus sign, gives the displayed one-complex-dimensional fixed spaces.

These subgroups are inequivalent: the rotating-wave spatial projection has order three, while the standing-wave projections have order two; the two reflection types differ by the central phase $T_{\pi }$, which conjugation cannot remove. Rotating the two standing lines by $\rho$ gives three orientations of each type, not new symmetry types. Therefore there are three distinct generic periodic branch types: rotating waves and two inequivalent standing-wave families. Their members have a period near $2\pi /{\omega }_0$, with the combined spatial–temporal operations determining their relative phase relations.

The unforced onset is a stationary pattern-forming instability. Under a horizontal translation, the critical Fourier mode transforms as $A\mapsto e^{i\varphi }A$. A cubic amplitude equation without forcing must have the same phase weight: its leading terms are $\tilde{\mu }A-c|A{|}^{2}A$. Reflection of the unforced spatial pattern conjugates $A$, permitting real coefficients in this stationary problem. They are determined by a weakly nonlinear expansion and projection onto the adjoint eigenfunction; symmetry alone does not calculate their values or guarantee a nonzero coupling.

Represent the third-harmonic forcing by a complex coefficient $F$ multiplying $e^{3i{k}_{c}x}$, whose phase weight is three. The product $F{\overline{A}}^2$ has weight $3-2=1$ and therefore resonates with the critical positive harmonic. Neither a direct third-harmonic term nor $F\overline{A}$ has the required wavenumber balance. A travelling boundary pattern makesA response phase and the sign of its coefficient can be incorporated into $\delta$. The resulting three-to-one spatially forced amplitude equation isThis retains the leading resonant forcing term, linear detuning and cubic saturation, while dropping higher powers and nonresonant harmonics. The forcing is weak, the unforced critical eigenvalue is near zero, and the amplitude varies on a slow time; very high forcing frequency outside that slow scaling would require a different averaging argument. Reduction to the specific saturating canonical form in part (i) additionally assumes $c>0$ and nonzero forcing.

Apply the method of multiple scales with independent fast time $t$ and slow time $T={ϵ}^{2}t$. A preliminary slow time $ϵt$ produces no resonant forcing at first order, so its leading amplitudes are constant; the first nontrivial modulation is at $T$. Write $x=x_0+ϵx_1+\cdots$, withAt order $ϵ$, the inhomogeneous linear differential equation is $({\partial }_t^2+1)x_1=-\cos t{x}_0$. Absorbing its homogeneous part into the amplitudes givesAt order ${ϵ}^2$, the forcing is $-2{\partial }_t{\partial }_T{x}_0-k{x}_0-\cos t{x}_1$. Its resonant Fourier series coefficients areThe solvability condition in the method of multiple scales sets both coefficients to zero, giving the amplitude equations. DefineFor bounded initial conditions $x(0)=x_{\ast }$, $\dot{x}(0)=v_{\ast }$, the requested leading approximation isThis holds on bounded intervals of $T$, hence $t=O({ϵ}^{-2})$, with the constant in the error allowed to depend on the bounded $T$ interval. If desired, retain the displayed $ϵx_1$ and adjust $A(0),B(0)$ by $O(ϵ)$ to match initial conditions at that order. No particular initial conditions were supplied.

Since $M^2=abI$, the matrix exponential is $I\cosh (\sigma T)+(M/\sigma )\sinh (\sigma T)$ when ${\sigma }^2=ab>0$. When $ab<0$, put $\omega =\sqrt{-ab}$ and replace this with $I\cos (\omega T)+(M/\omega )\sin (\omega T)$. At $ab=0$ it is $I+TM$. Thus the leading second instability tongue of a weak Mathieu oscillator has exponentially growing solutions for $-1/12<k<5/12$, and the leading modulation is bounded for every initial condition precisely whenAt either displayed endpoint, $M$ is a nonzero nilpotent operator; some initial conditions give a secular term proportional to $T$. This leading calculation alone must not be mistaken for an exact finite-$ϵ$ stability decision at the endpoints.

For an exact small-$ϵ$ interpretation, put $t=2s$. The standard Mathieu equation parameters are $a_{\mathrm{M}}=4+4k{ϵ}^2$ and $q=-2ϵ$. Its two relevant Mathieu characteristic values are obtained by projecting $-d^2/d{s}^2+2q\cos 2s$ onto even and odd Fourier series modes. Near the unperturbed value $4$, the even modes $1,\cos 2s,\cos 4s,\cos 6s$ have diagonal entries $0,4,16,36$ and adjacent couplings $\sqrt{2}q,q,q$ in an orthonormal basis; the odd modes $\sin 2s,\sin 4s,\sin 6s$ have diagonal entries $4,16,36$ and couplings $q,q$. Substituting $4+c_2{q}^2+c_4{q}^4$ into their characteristic polynomials givesHigher modes of the Fourier series first contribute at higher order. Therefore the fourth-order edges of the second Mathieu instability tongue areFloquet theory gives boundedness of every solution outside the closed gap, $k<k_{-}(ϵ)$ or $k>k_{+}(ϵ)$, in this fixed-$k=O(1)$ neighborhood. At the actual gap edges the second independent solution is generically unbounded. In particular, if $k$ is fixed exactly at $-1/12$ or $5/12$, it lies just outside the true gap and is stable for each sufficiently small positive $ϵ$. The strict boxed criterion is the robust leading-order, uniformly separated criterion; the endpoints require this higher-order refinement, and their amplification bounds need not remain uniform as $ϵ\to 0$.

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 $\pi$.

For a real steady amplitude, integrate the conserved-field equation twice. Periodicity forces the integration slope to vanish, and the zero mean fixes the remaining constant. HenceSubstituting this into the steady amplitude equation gives the nonlocal ordinary differential equationFor $A=R\mathrm{sech}(cX)$, use the hyperbolic secant identity $A^{\prime \prime }=c^{2}A-2c^2{A}^3/R^2$. Vanishing of the two independent coefficients givesChoose $c>0$ and $R>0$; an overall negative sign is another solution. Thus $\kappa >1$. The exact average of the proposed profile over the finite interval isFor $cL\gg 1$, the hyperbolic tangent differs from one by $O(e^{-cL})$, so $cL⟨A^2⟩\simeq 2R^2$. Eliminating $R^2=2c^2/(\kappa -1)$ gives the requested leading relationEquivalently $c+c^{-1}=K$, with $K=4\kappa /[L(\kappa -1)]$. Positive real roots exist precisely when $K\ge 2$; they are $(K\pm \sqrt{K^2-4})/2$ and are reciprocal. Each candidate must additionally satisfy $cL\gg 1$. For $L>2$, the leading necessary range is $1<\kappa \le L/(L-2)$, with equality at the coalescence $c=1$. Away from this leading-order fold, the inequality is strict.

The strict finite-window condition can also be seen by retaining the average exactly. Its consistency relation is $c+c^{-1}=K\tanh (cL/2)$. Since $\tanh (cL/2)<1$, it implies $K>2$, and henceThe strict endpoint distinction is exponentially small in the large-$cL$ approximation; the leading polynomial alone permits its double root. When $0<L\le 2$, this algebra imposes no finite upper bound on $\kappa >1$. The printed interval therefore needs $L>2$: for example $L=1$, $\kappa =1.1$ gives a large positive root $c\simeq 43.98$ of the leading equation and $cL\gg 1$, despite the printed upper endpoint being negative.

Finally, the single localized pulse of a conserved-field convection model is an approximate periodic solution. Its values agree at the interval ends, but its derivatives have opposite signs there; their magnitude is exponentially small, $O(cR{e}^{-cL/2})$. Periodic-tail corrections are needed for an exact smooth periodic boundary condition. The range derived from the truncated profile is a consistency condition, not a proof that every permitted parameter gives an exact stable periodic pulse.

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.

The linearized rotating Rayleigh-Bénard convection equations about conduction areThe Coriolis force does no work on the fluid, but it couples vertical motion to vertical vorticity and thereby changes the balance of buoyancy and viscous damping. On the plates require $w=\theta =0$ and ${\partial }_z{u}_x={\partial }_z{u}_y=0$. Write $\zeta =(\nabla \times \mathbf{u}{)}_z$. Taking a curl and a double curl removes pressure and givesA normal mode with horizontal wavenumber $k>0$, vertical index $n\ge 1$ and complex growth rate $s$ has $w=W\sin (n\pi z)e^{st+i\mathbf{k}\cdot \mathbf{x}}$, $\theta =\Theta \sin (n\pi z)e^{st+i\mathbf{k}\cdot \mathbf{x}}$ and $\zeta =Z\cos (n\pi z)e^{st+i\mathbf{k}\cdot \mathbf{x}}$. Put $a=k^2+n^2{\pi }^2$. The three equations become $a(s/P+a)W+\sqrt{\mathrm{Ta}}n\pi Z=R{k}^2\Theta$, $(s/P+a)Z=\sqrt{\mathrm{Ta}}n\pi W$, and $(s+a)\Theta =W$. Their determinant gives the rotating-convection growth-rate polynomialThis cubic includes viscous and thermal modes as well as the convective instability. It avoids excluding a root by division by $s/P+a$. Linear stability means every root has negative real part; stationary onset has $s=0$, while overstability has $s=\pm i\omega$ with nonzero angular frequency.

Setting $s=0$ gives the stationary neutral curve of rotating convectionFor oscillatory convection, set $s=i\omega$ and equate real and imaginary parts. Eliminating $R$ yieldsThe oscillatory neutral curve of rotating convection is physically admissible only when ${\omega }^2>0$. Thus $P<1$ and sufficiently rapid rotation are necessary; a formal minimum with ${\omega }^2\le 0$ is not a Hopf bifurcation. Convection begins at the smaller of the stationary minimum and the admissible oscillatory minimum. The angular frequency approaches zero where a fixed-wave-number stationary and oscillatory threshold meet; that is a double-zero limit of the cubic, requiring a different slow-time reduction.

For wavenumber selection in rotating convection, the lowest vertical index is $n=1$. To compare indices, set $k^2=n^2{\pi }^{2}y$ at fixed $y>0$. Both neutral thresholds have a positive term proportional to $n^4$ and a rotation term independent of $n$, while the admissible angular frequency squared decreases with $n$. Their minima therefore cannot improve on $n=1$. Set $t=k^2$, $m={\pi }^2$. Minimizing each neutral curve givesFor the latter equation also check ${\omega }^2>0$, or minimize over the admissible set instead. Without rotation, $k_c=\pi /\sqrt{2}$ and $R_c=27{\pi }^4/4$. At rapid rotation, $k_s\sim (\mathrm{Ta}{\pi }^2/2{)}^{1/6}$ and $R_{s,c}\sim 3(\mathrm{Ta}{\pi }^2/2{)}^{2/3}$: rotation narrows the horizontal convection rolls and raises their threshold. In a finite box only discrete wavenumbers are permitted, so the minimum must be taken over those allowed modes.

The rapid-rotation comparison has $R_{o,c}/R_{s,c}\sim 2P^{4/3}/(1+P{)}^{1/3}$. Its equality gives $8P^4-P-1=0$, with positive root about $0.67660$. Below this value, sufficiently rapid rotation can make oscillatory convection the primary instability; low Prandtl number allows inertial motions to interact with the thermal field before viscosity damps them. This estimate describes the selected neutral-curve comparison in the stated free-slip asymptotic model, rather than claiming every $P<1$ has oscillatory primary onset at every rotation rate.

Linear selection does not determine the nonlinear planform or its stability. A weakly nonlinear expansion with a solvability condition gives amplitude equations. A stationary convection roll amplitude has the Landau amplitude equation $A_T=rA-g|A{|}^{2}A$; a supercritical branch requires $g>0$ and has $|A{|}^2=r/g$, whereas $g<0$ requires higher-order saturation and permits subcritical behaviour. The coefficient depends on the physical parameters and boundary conditions, and should not be assumed positive merely because the linear threshold is known.

A nonzero-frequency Hopf bifurcation supplies oppositely travelling convection roll amplitudes $Z_{+},Z_{-}$. At a fixed orientation, a cubic normal form iswith complex $g,h$ and the understood symmetry exchanging propagation directions. For $r>0$ and $g_r=\mathrm{Re}g>0$, a travelling-roll branch has one nonzero amplitude, of squared modulus $r/g_r$; it is stable to the opposite travelling amplitude when $h_r=\mathrm{Re}h>g_r$. A standing-roll branch has equal squared moduli $r/(g_r+h_r)$, and is stable within this pair when $g_r>h_r$ and $g_r+h_r>0$. The imaginary parts shift angular frequencies. These conditions concern amplitude perturbations in that reduced subspace; orientation and modulation modes still have to be tested. This is the competition of travelling and standing convection rolls.

Near simultaneous stationary and oscillatory thresholds, a codimension-two bifurcation requires retaining both types of mode. After selecting an orientation and fixing a steady spatial phase, a schematic nonresonant steady–Hopf mode interaction isHere $a$ is real, $z$ is the chosen complex oscillatory amplitude, $g_s,h_s$ are real, and $g_o,h_o$ may be complex. Assume $r_s>0$, $g_s>0$ for a radially stable pure steady branch, and $r_o>0$, $g_{o,r}=\mathrm{Re}(g_o)>0$ for a radially stable pure oscillatory branch. Steady convection rolls suppress the oscillatory mode if $r_o-h_{o,r}{r}_s/g_s<0$, where $h_{o,r}=\mathrm{Re}(h_o)$; oscillatory convection rolls suppress the steady mode if $r_s-h_s{r}_o/g_{o,r}<0$. These are transverse tests for branches whose existence and radial stability have already been checked. Mixed states have positive intensities $u=a^2$, $v=|z{|}^2$ solving $g_{s}u+h_{s}v=r_s$, $h_{o,r}u+g_{o,r}v=r_o$. If $D=g_s{g}_{o,r}-h_s{h}_{o,r}\ne 0$, then $u=(r_s{g}_{o,r}-h_s{r}_o)/D$, $v=(g_s{r}_o-h_{o,r}{r}_s)/D$. For positive $u,v$ and the positive self-saturation coefficients above, the intensity Jacobian matrix has negative trace and determinant $4uvD$, so the mixed state is attracting in intensities when $D>0$ and is a saddle equilibrium when $D<0$; phase and other-mode perturbations remain separate tests. Depending on the cross-couplings, one obtains coexistence or competition and bistability; degeneracies or resonances require extra terms. A full travelling/standing-wave competition must retain both $Z_{\pm }$, not just this illustrative one-mode $z$. If the Hopf angular frequency tends to zero, averaging over fast oscillations fails and a double-zero normal form must keep the corresponding two slow variables. A crossing of global minima at distinct wavenumbers is instead an interaction of distinct modes, not automatically the same double-zero problem.