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Let the constant mass density be $\rho >0$ and put ${\mathbf{v}}_a=\mathbf{B}/\sqrt{{\mu }_0\rho }$, the Alfvén velocity. Since both the velocity and the magnetic field have zero divergence, the magnetic tension and magnetic pressure decomposition of the Lorentz force givesThe Newtonian gravitational potential $\Phi$ remains in $\psi$; uniform mass density does not justify dropping a prescribed gravitational acceleration. Adding and subtracting the two equations proves the Elsässer variable equationsHere ${\mathbf{z}}^{\pm }=\mathbf{u}\pm {\mathbf{v}}_a$; each Elsässer variable is transported by the other.

Define $I_{\pm }={\int }_V|{\mathbf{z}}^{\pm }{|}^{2}dV$. Taking the dot product with $2{\mathbf{z}}^{\pm }$ gives the Elsässer energy invariant balanceThe divergence theorem proves $d{I}_{\pm }/dt=0$ whenever the net boundary flux vanishes. In particular, $\mathbf{u}\cdot \mathbf{n}=\mathbf{B}\cdot \mathbf{n}=0$ makes ${\mathbf{z}}^{\pm }\cdot \mathbf{n}=0$, so both fluxes vanish pointwise. Periodic boundary conditions or decay at infinity are also sufficient. These boundary conditions matter: fixed volume alone gives no conservation.

The kinetic energy plus magnetic energy isExpanding $|\mathbf{u}\pm {\mathbf{v}}_a{|}^2$ and using the cross-helicity definition yieldsBoth energy and cross-helicity therefore follow from the two Elsässer energy invariants. The quantity $E$ here is exactly the requested kinetic energy plus magnetic energy, without an additional gravitational term.

Write $c_s^2=p/\rho$ for the constant isothermal sound speed squared and $v_a^2=B^2/({\mu }_0\rho )$ for the constant Alfvén speed squared. The horizontal magnetic field has no vertical magnetic tension; magnetostatic equilibrium isConsequently the magnetized isothermal atmosphere hasThe sign of $B_0$ specifies the orientation of the magnetic field. The adiabatic sound speed is $v_s^2=\gamma p/\rho =\gamma c_s^2$, also constant.

For the proposed normal mode, let $A=\exp (i\mathbf{k}\cdot \mathbf{x}+z/(2H)-i\omega t)$, $a=1/(2H)$, and temporarily write $\mathbf{\xi }$ for its constant complex amplitude. The divergence amplitude is $\Delta =i{k}_x{\xi }_x+i{k}_y{\xi }_y+(i{k}_z+a){\xi }_z$. Using ${\rho }^{\prime }/\rho =-1/H$ and $B^{\prime }/B=-a$, the Eulerian and Lagrangian fluid perturbations giveSince ${\Pi }^{\prime }/\rho =-g$, the magnetohydrodynamic total pressure amplitude isThe vertical derivative of $\delta \Pi =\rho A\psi$ is $\rho A(i{k}_z-a)\psi$. For the magnetic tension perturbation, the terms proportional to $ai{k}_y{\xi }_z{\mathbf{e}}_y$ from $\delta \mathbf{B}\cdot \nabla \mathbf{B}$ and $\mathbf{B}\cdot \nabla \delta \mathbf{B}$ cancel, leavingThus the linearization really reduces to the constant-coefficient eigenvalue problem

For a static ideal magnetohydrodynamics equilibrium with conservative gravitational acceleration, the magnetohydrodynamic energy principle gives a Hermitian operator for the restoring force in the mass density weighted inner product, provided the surface terms vanish. The factor $e^{z/(2H)}$ removes the exponential mass density weight. This can be checked directly, without assuming stability: set $s=v_s^2$, $b=v_a^2$, and $q=s+b$. Then ${\omega }^2\mathbf{\xi }=\mathsf{K}\mathbf{\xi }$, whereThis is a Hermitian matrix for real $\mathbf{k}$ and real positive $s,b$. Therefore every squared frequency ${\omega }^2$ is real; negative values give exponential growth, rather than an oscillation with a complex squared frequency. In an infinite atmosphere these are generalized plane waves, not finite-total-energy global modes.

At ${\omega }^2=0$ and $k_y\ne 0$, the $y$ equation, after cancelling its magnetic terms, gives $s\Delta =g{\xi }_z$. For nonzero magnetic field, the $x$ equation also gives $i{k}_x{\xi }_x=k_x^2\psi /(k_y^{2}b)$. Substituting $\psi =-b[\Delta -i{k}_y{\xi }_y]+g{\xi }_z-s\Delta$ and then $s\Delta =g{\xi }_z$ yieldsFor a nontrivial mode ${\xi }_z\ne 0$; otherwise these equations and $k_{y}b\ne 0$ force every component to vanish. Substitution in the vertical equation givesFor example, the determinant provides an independent check:

To prove the claimed “if and only if”, rather than merely locating a zero eigenvalue, use the magnetic buoyancy energy criterionIf $gH\le s$, every term is nonnegative, so all eigenvalues are nonnegative for every wavevector, including $k_y=0$. If $gH>s$, choose $k_z=0$, any fixed $k_x\ne 0$, and ${\xi }_z=1$, ${\xi }_x=-a/(i{k}_x)$, ${\xi }_y=g/(si{k}_y)$. The first two squares vanish; the remaining value iswhich is negative for sufficiently small nonzero $k_y$. The Rayleigh quotient of this displacement is therefore negative, proving a negative eigenvalue and an unstable normal mode. This uses precisely the assumption that boundary conditions do not exclude the chosen wavelengths. Equality is a neutral threshold, not exponential growth.

Finally the plasma beta is $\beta =p/[B^2/(2{\mu }_0)]=2c_s^2/v_a^2$. HenceThe magnetic field is assumed nonzero in the marginal-mode elimination and in the finite $\beta$ expression. The positive-sound-speed, nonmagnetic limit remains covered by the energy identity; for the usual $\gamma >1$ it is stable. Parker instability occurs because magnetic pressure supports part of the atmosphere's weight while the gas can drain along gently bent magnetic field lines. The resulting buoyant displacement can release Newtonian gravitational potential energy; choosing long wavelengths along the magnetic field makes the stabilizing magnetic tension small. The completed-square calculation quantifies this competition, including the adiabatic process restoring force.

Linearity of Stokes flow and rotational symmetry, including reflection in mathematics invariance of the spherical geometry, determine the possible rigid motions before any calculation. Translation is a polar vector; its linear dependence on a single vector $\mathbf{A}$ must be $c\mathbf{A}$. A traceless second-rank tensor $\mathbf{B}$ cannot produce a polar vector by an isotropic linear map: tensor contraction with the identity matrix vanishes and tensor contraction with the totally antisymmetric tensor vanishes because $\mathbf{B}$ is a symmetric second-rank tensor. Angular velocity is an axial vector. Neither $\mathbf{A}$ nor a symmetric $\mathbf{B}$ can produce one by an isotropic linear map. Products such as $\mathbf{A}\times \mathbf{BA}$ are excluded by Linearity of Stokes flow. Thus $\mathbf{\Omega }=0$.

Use the Unscaled Papkovich–Neuber representation, with harmonic functions $\mathbf{\Phi }$ and $\chi$:Indeed $\nabla \cdot \mathbf{u}=0$ and $\mu {\nabla }^2\mathbf{u}=\nabla p$ because each potential is a harmonic function. Put $r=|\mathbf{x}|$ and $S=\mathbf{x}\cdot \mathbf{Bx}$. The decaying harmonic functions with the necessary angular dependence are $\mathbf{A}\cdot \mathbf{x}/r^3$, $\mathbf{Bx}/r^3$ and $S/r^5$. The last is a harmonic function precisely because $\mathrm{tr}\mathbf{B}=0$. A $1/r$ Stokeslet is excluded by the force-free condition, and a rotlet by the torque-free condition. TryThe resulting Stokes flow isAt $r=a$, the coefficient of $\mathbf{n}(\mathbf{A}\cdot \mathbf{n})$ fixes $c=a^3/3$; the remaining constant vector gives $\mathbf{U}=-2\mathbf{A}/3$. In the $\mathbf{B}$ mode, matching $\mathbf{Bn}-(\mathbf{n}\cdot \mathbf{Bn})\mathbf{n}$ gives $e=a^4/2$ and $d=-a^2/2$. Therefore the complete two-mode tensorial squirmer flow isThe boundary condition is satisfied in the laboratory frame, so the Stokes flow tends to zero at infinity. The potential dipole in the $\mathbf{A}$ mode gives an irrotational flow and decays as $r^{-3}$; the leading $\mathbf{B}$ mode is a stresslet, decaying as $r^{-2}$. Neither carries a net force or torque. Direct surface integral averaging with the surface slip velocity formula gives the same translation, providing an independent check. Uniqueness of Stokes flow then identifies this decaying solution.

Write $D=R_e-R_i$, $S=R_e+R_i$, $A=R_e^2-R_i^2=DS$, and $\Delta p=p_i-p_e$. Incompressible flow with axial strain rate $E$ givesIts radial Stokes equation gives $p_r=0$, because $u_{r,rr}+u_{r,r}/r-u_r/r^2=0$. The radial Newtonian fluid stress tensor component is ${\sigma }_{rr}=-p-\mu E-2\mu C/r^2$. The outer and inner Young–Laplace equation conditions have opposite curvature signs:Subtracting and eliminating $C$ givesIn particular, the boxed pressure is equivalent to the printed identity for $(-p-\mu E)A$. For equal gas pressures, $C=-\gamma R_i{R}_e/(2\mu D)$, so surface tension draws both interfaces inward in addition to the imposed extension.

The kinematic boundary condition is ${\dot{R}}_j=-E{R}_j/2+C/R_j$ at either interface. Taking the difference, and then the difference of squared radii, givesThe cancellation of $C$ in the area equation is mass conservation: axial stretching reduces the liquid area while radial redistribution changes the hole size. For $E=0$, equal pressures, and $R_i(0)=b$, $R_e(0)=2b$, one has $A=3b^2$ and $D=b+\gamma t/(2\mu )$. Since $S=A/D$,Closure occurs at $D=\sqrt{A}=\sqrt{3}b$, soThis capillary collapse of an annular viscous cylinder has finite closure time for $\gamma >0$ and $b>0$. For $\gamma =0$ it does not close in this case. The annular formula is used only up to closure; the singular cylindrical inner curvature is not a boundary condition on the subsequent solid cylinder.

The three distinguished limits are the outer expansion at $x=O(1)$, an intermediate asymptotic region at $x=O(ϵ)$, and an inner expansion at $x=O({ϵ}^2)$. The middle scale resolves the coefficient change from $x$ to $x+ϵ$; the narrower scale resolves the unmet left boundary condition. The original equation is understood on $x>0$ with a continuous extension to the boundary. As explained below, imposing a classical second derivative at $x=0$ would be too strong for this degenerate ordinary differential equation.

In the outer expansion, set $y=y_0+ϵy_1+\cdots$. The leading equation is $x(x+1)y_0^{\prime }-x{y}_0=0$. Enforcing the right boundary condition gives $y_0=x+1$. Because $y_0^{\prime \prime }=0$, the next equation is particularly simple:An integrating factor, or the substitution $y_1=(x+1)u$, gives $u^{\prime }=-1/[x(x+1)]$. ThusThe error statement is for $x$ bounded away from zero. The left limiting form is $1+x-ϵ\log (2x)$ through the terms needed for matching. A logarithm in the correction makes a pure integer-power inner ansatz inadequate.

For the intermediate asymptotic region, put $x=ϵs$ and $y=V(s)$. The transformed equation isThe leading $V_0$ is a constant fixed to $1$ by matching. There is no order-$\sqrt{ϵ}$ forcing, so that homogeneous constant is zero after matching. At order $ϵ$, $(s+1)V_1^{\prime }=s$, giving $V_1=s-\log (1+s)+C(ϵ)$. Matching for $1\ll s\ll 1/ϵ$ with the outer expansion fixes $C=-\log (2ϵ)$. HenceThe switchback term $-ϵ\log ϵ$ is larger than an ordinary $O(ϵ)$ constant correction and must be retained. The highest derivative term first changes this middle solution at order ${ϵ}^{3/2}$, so it does not alter the displayed result.

To locate the inner expansion, compare the derivative terms for $x\ll ϵ$. Their ratio at a layer width $\delta$ is $ϵ/\sqrt{\delta }$, so the square-root-degenerate endpoint layer has $\delta ={ϵ}^2$. Put $x={ϵ}^{2}X$ and $y=U(X)$. After dividing by the common leading factor, the equation through order $ϵ$ isBoth orders therefore have the same homogeneous operator $\sqrt{X}{d}^2/d{X}^2+d/dX$. Its derivative mode is proportional to $e^{-2\sqrt{X}}$. Using $U(0)=0$ and the supplied elementary integral, defineThe intermediate asymptotic region gives the overlap value $1-ϵ\log (2ϵ)$ as $s\to 0$. ThereforeIt satisfies the left boundary condition exactly to the retained orders, and its large-$X$ limit matches the middle result in $1\ll X\ll 1/ϵ$. The outer and middle limits match with $ϵ\ll x\ll 1$. These three formulae include all terms through $O(ϵ)$, including the logarithmic switchback; the omitted terms in the middle and inner regions need not be $O({ϵ}^2)$.

For a check on that last point, continue the middle equation by one order. Its ${ϵ}^{3/2}$ term obeysThe integration constant is selected by the outer limit. Its value at $s=0$ is $\pi /2$, so the inner matching amplitude has a next correction $\pi {ϵ}^{3/2}/2$. This independently explains why an $o(ϵ)$ remainder is appropriate.

If a single additive composite expansion is useful, combine the three formulae and subtract their two common overlaps. Writing $A=1-ϵ\log (2ϵ)$ and interpreting $x\log x=0$ at zero givesIt is zero at $x=0$, gives $2+O({ϵ}^2)$ at $x=1$, and reproduces all three retained expansions. It does not assert an $O({ϵ}^2)$ error uniformly through the nested regions.

Finally, near zero $H(X)=2X-\frac{8}{3}{X}^{3/2}+O(X^2)$. Thus the solution has a finite nonzero first derivative but generally a second derivative diverging like $x^{-1/2}$. This is compatible with the equation for $x>0$: the vanishing coefficient of $y^{\prime \prime }$ multiplies that singular derivative to give a finite limit. If one instead required $y\in C^2[0,1]$ and the differential equation pointwise at $0$, the left condition would force $y^{\prime }(0)=0$. The normalized equation's other coefficients have only integrable $x^{-1/2}$ singularities, so uniqueness for the initial value problem with $y(0)=y^{\prime }(0)=0$ would give the zero solution, contradicting the right condition. Explicitly, write the normalized equation as a first-order system for $(y,y^{\prime })$ with an integrable coefficient matrix; its integral equation and the Gronwall inequality force a solution with zero initial vector to vanish. The continuous-endpoint interpretation is therefore essential.

Take $h=2k_c$ and $D=d/dz$. Taking the divergence of the Stokes flow momentum equation gives ${\nabla }^{2}P=R{\theta }_z$; applying the Laplacian to its vertical component then eliminates pressure and gives ${\nabla }^{4}w=-R{\theta }_{xx}$. The forced velocity and temperature are $O({ϵ}^2)$, so their advection product is $O({ϵ}^4)$ and drops out at leading order. The steady heat equation reduces to $w_0+{\nabla }^2{\theta }_0=0$.

Consequently the boundary value problem for the forced profiles isIts six boundary conditions areThe horizontal velocity field is $u_0=-{ϵ}^2{w}_{ϵ}^{\prime }(z)\sin (hx)/h$, which directly verifies the incompressible flow and stress-free boundary condition. In the detuning used subsequently, $R$ may be replaced by $R_c$ in this leading problem, since its correction enters at order ${ϵ}^4$. The vertical profiles depend linearly on $\gamma$.

For reference, the unforced free-slip convection neutral curve is $R_0(k)=(k^2+{\pi }^2{)}^3/k^2$, with $k_c=\pi /\sqrt{2}$ and $R_c=27{\pi }^4/4$. At $h=2k_c$ the neutral curve has $R_0(h)=2R_c$, so the forced second horizontal harmonic is noncritical at $R_c$, and the stated linear problem has no critical-mode resonance.

Take the layer depth as the length unit, its thermal diffusion time as the time unit, and its imposed temperature difference as the temperature unit. Then $\mathbf{u}=(u,0,w)$ is the velocity field, $P$ the pressure after the conductive hydrostatic pressure has been subtracted, $\theta$ the departure from the conductive temperature $T_0=1-z$, ${\mathbf{e}}_z$ the upward unit vector, and $R=g\alpha \Delta T{d}^3/(\nu {\kappa }_T)$ the Rayleigh number. Here $d$ is the dimensional depth, $\nu$ the kinematic viscosity, ${\kappa }_T$ the thermal diffusivity and $\alpha$ the thermal expansion coefficient. The gradient, divergence and Laplacian act on the dimensionless coordinates. The term $w$ in the heat equation is advection of the background gradient, since $-\mathbf{u}\cdot \nabla T_0=w$. Mass density variations enter only the buoyancy term under the Boussinesq approximation; the incompressible flow constraint is $\nabla \cdot \mathbf{u}=0$.

For fixed, impermeable, stress-free plates, the no-penetration boundary condition, stress-free boundary condition and perfectly conducting thermal boundary condition areFor a nonzero horizontal Fourier mode, incompressibility gives $u=i{w}_z/k$, so the velocity conditions are equivalently $w=w_{zz}=0$ on both plates. A uniform horizontal velocity is not fixed by these boundary conditions; take its mean to be zero when discussing the stationary forced pattern.

The printed small-Prandtl-number label does not generally justify these equations. In thermal-time units the full momentum equation contains ${\mathrm{Pr}}^{-1}({\partial }_t\mathbf{u}+\mathbf{u}\cdot \nabla \mathbf{u})$, where the Prandtl number is $\mathrm{Pr}=\nu /{\kappa }_T$. Dropping that term gives infinite-Prandtl-number convection or another explicitly justified Stokes flow approximation. It is not the general $\mathrm{Pr}\to 0$ limit. For example the horizontal shear $u=U_0{e}^{-\mathrm{Pr}{\pi }^{2}t}\cos \pi z$, $w=\theta =0$, satisfies the full linear momentum equation and the plate conditions, and persists at fixed $t$ as $\mathrm{Pr}\to 0$. The printed inertia-free equation would instead require this shear to vanish. The following parts solve the equations actually displayed; applying them at small Prandtl number needs an additional slaving or creeping-flow ordering.

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.

Weakly nonlinear convection rolls can also lose stability spatially. The real Ginzburg–Landau equation $A_T=rA+\xi A_{XX}-g|A{|}^{2}A$, with $r,\xi ,g>0$, has detuned convection rolls with $|A{|}^2=(r-\xi Q^2)/g$. Linear phase modulation gives diffusion coefficient $\xi (r-3\xi Q^2)/(r-\xi Q^2)$, so the Eckhaus instability excludes $Q^2>r/(3\xi )$ even though convection rolls exist up to $Q^2<r/\xi$. Transverse bending and mean-flow couplings impose additional restrictions in the actual rotating layer; the scalar equation is a local longitudinal example, not its complete stability theory.

Most distinctively, rotation breaks mirror symmetry between competing convection roll orientations. For two sets of convection rolls at relative angle $\vartheta$, write ${\dot{A}}_2=r{A}_2-g_0|A_2{|}^2{A}_2-g(\vartheta )|A_1{|}^2{A}_2$ and interchange indices with $\vartheta \mapsto -\vartheta$ for the other equation. The established convection roll $|A_1{|}^2=r/g_0$ is unstable to the new orientation when its linear growth rate $r[1-g(\vartheta )/g_0]$ is positive. The inequality need not be symmetric under $\vartheta \mapsto -\vartheta$ because the imposed rotation supplies handedness. This is the Küppers–Lortz instability: sufficiently rapid rotation can destabilize steady convection rolls immediately above their stationary onset to oblique convection rolls of another orientation. Successive replacements can produce time-dependent convection roll switching or heteroclinic cycles, instead of a stable single convection roll pattern. It requires three-dimensional perturbations even when the original convection roll solution is independent of its axial coordinate.

At finite Prandtl number and stress-free plates, nearly parallel convection rolls can couple resonantly to a slowly damped large-scale Eulerian mean flow. This small-angle instability of rotating convection rolls makes a regular two-roll expansion nonuniform as the angle tends to zero; one must retain the mean-flow mode. It should not be identified automatically with the finite-angle Küppers–Lortz instability or assigned a universal threshold from a one-amplitude equation. Consequently the onset type and selected scale follow from the admissible neutral curves, while persistent convection roll patterns require a separate nonlinear and sideband stability calculation.

With periodic or sufficiently decaying boundary conditions, integration by parts givesAssume $\kappa \ge 0$ for a stable quadratic free energy; the diffusive smoothing discussed below requires $\kappa >0$. In a periodic box of volume $V$, use the unitary convention ${\varphi }_{\mathbf{q}}=V^{-1/2}{\int }_V\varphi (\mathbf{r})e^{-i\mathbf{q}\cdot \mathbf{r}}d\mathbf{r}$. A real field has ${\varphi }_{-\mathbf{q}}={\varphi }_{\mathbf{q}}^{\ast }$. Since the Fourier transform of a derivative converts ${\nabla }^2$ to $-q^2$, each Fourier mode obeysIt is an Ornstein-Uhlenbeck process. Multiplication by the integrating factor $e^{r(q)t}$ givesThe noise integral is understood as an Itô integral with a deterministic kernel; writing $\eta ds$ is shorthand for the corresponding Wiener increment. Direct differentiation in the mild stochastic sense checks both the equation and the initial value. The stated Kronecker-delta noise covariance corresponds to this finite-volume normalization. Unlike a conserved order parameter, the zero wavenumber mode decays at the nonzero rate $\Gamma a$.

The field $\varphi$ is a compositional order parameter in a binary fluid mixture. Its spatial integral is conserved in a closed sample, so advection and diffusion enter a continuity equation; the material derivative follows the incompressible fluid. The diffusive current $\mathbf{J}=-M\nabla \mu$, with positive order-parameter mobility $M$, moves material down the chemical potential gradient. The chemical potential of a composition field is the functional derivative of the symmetric quartic Landau free energyso $\mu =a\varphi +b{\varphi }^3-\kappa {\nabla }^2\varphi$. Positive $b,\kappa$ stabilize large amplitudes and penalize sharp interfaces; $a<0$ permits phase separation.

The momentum equation is the Navier-Stokes equation with dynamic viscosity $\eta$, constant mass density $\rho$, and the Korteweg force $-\varphi \nabla \mu$ arising from composition stresses. The pressure enforces incompressible flow, $\nabla \cdot \mathbf{v}=0$, and can absorb pure gradient contributions to that thermodynamic force. For example, $-\varphi \nabla \mu =\mu \nabla \varphi -\nabla (\varphi \mu )$ gives an equivalent force after redefining the pressure. These equations constitute deterministic Model H dynamics.