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

For a normal mode, ${\partial }_t\mapsto -i\omega$, ${\partial }_x\mapsto ik$, and ${\partial }_x^2\mapsto -k^2$. Substitution into the linear complex Ginzburg-Landau equation givesTherefore the dispersion relation isFor real $k$, its temporal growth rate is ${\omega }_i=\mu -k^2$, with unique maximum $\mu$ at $k=0$. Its real angular frequency is $Uk+c_d{k}^2$, displaying the separate advection and wave dispersion terms.

For a zonal basic geostrophic flow $U(y)$, take ${\psi }_0^{\prime }(y)=-U(y)$ and write $\psi ={\psi }_0+\varphi$. The flat-bottom shallow-water quasi-geostrophic potential vorticity gradient of the basic state isLinearizing its material conservation gives the Rossby-wave equation for a sheared zonal current,For a general nonuniform jet, a global two-dimensional plane wave is not an exact normal mode: the coefficients depend on $y$. The exact zonal normal mode problem, $\varphi =\hat{\varphi }(y)e^{i(kx-\omega t)}$, iswith appropriate transverse boundary or radiation conditions. If the jet varies slowly compared with a wavelength, a local plane wave freezes these coefficients at $y_0$. Its local dispersion relation isIn the nondivergent Barotropic Rossby wave model under a rigid-lid approximation, this becomes $\omega =kU-k(\beta -U^{\prime \prime })/(k^2+l^2)$. It is exact for uniform $U$ and otherwise a local relation. Retaining the free-surface term also requires retaining the basic surface slope in $Q_0^{\prime }$; simply adding $R_D^{-2}$ to the denominator while dropping $R_D^{-2}U$ from the numerator would describe a different prescribed-background model.

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.

Let the free surface be at $r=R_1$ and take the Newtonian gravitational potential to vanish at infinity. Mass conservation fixes $R_1=(3M_1/(4\pi {\rho }_1){)}^{1/3}$. The enclosed mass is $m(r)=4\pi {\rho }_1{r}^3/3$, and hydrostatic equilibrium requires $d{p}_0/dr=-{\rho }_{1}Gm(r)/r^2$. With vacuum outside, $p_0(R_1)=0$, givingThe pressure is zero outside. A prescribed constant external pressure would simply add that constant to $p_0$.

The bulk Eulerian and Lagrangian fluid perturbations obey $\delta \rho =0$ because the mass density is uniform and the Lagrangian displacement has zero divergence. The linearized Euler momentum equation in the interior isTaking the curl gives ${\omega }^2\nabla \times \mathbf{\xi }=0$. Thus every nonzero-frequency normal mode has an irrotational flow displacement; the simply connected space formed by the interior admits $\mathbf{\xi }=\nabla U$ and incompressibility gives the Laplace equation ${\nabla }^{2}U=0$. For a regular center, one spherical harmonic component is $U=A_l{r}^l{Y}_l^m$, with $|m|\le l$.

It is essential to retain the surface gravity perturbation of a uniform-density sphere. Although the bulk mass density perturbation is zero, the displaced density discontinuity givesThe Poisson equation implies that $\delta \Phi$ is continuous and its outward radial derivative has the jumpUsing regularity at the center and decay at infinity, writeThe derivative jump is $-(2l+1)C_l{R}_1^{l-1}{Y}_l^m$, so $C_l=-4\pi G{\rho }_{1}l{A}_l/(2l+1)$.

Integrating the bulk Euler momentum equation gives $\delta p/{\rho }_1={\omega }^{2}U-\delta \Phi$ for $l\ge 1$. At the free surface the Lagrangian pressure perturbation vanishes:Combining these relations yields the Kelvin stellar mode frequencyThe $2l+1$ choices of $m$ are degenerate because of spherical symmetry. For $l=1$, $r{Y}_1^m$ is a linear combination of coordinates in a Cartesian coordinate system, so $\nabla U$ is a constant displacement: the zero frequency is rigid translation of the whole isolated body. Such a translation has no restoring force. The constant $l=0$ potential generates no displacement; a radial breathing motion is excluded by incompressibility and a regular center.

The printed irrotational assertion needs the nonzero-frequency qualification, or restriction to the potential incompressible stellar surface modes. It is false for all neutral displacements: $\mathbf{\xi }=\mathbf{\Omega }\times \mathbf{r}$ has zero divergence, zero normal displacement at the surface, and $\delta p=\delta \Phi =0$, so it is a zero-frequency displacement, but $\nabla \times \mathbf{\xi }=2\mathbf{\Omega }\ne 0$. This does not change the Kelvin stellar mode spectrum just derived.

The separation of the two bodies is $a$, as is fixed by the supplied Kepler third law ${\omega }_o^2=G(M_1+M_2)/a^3$. The phrase “orbit of radius $a$” cannot denote the first body's distance from the center of mass in that formula.

Expand the companion's Newtonian potential of a point mass about the first body's center. The constant term produces no force; the linear term produces the uniform acceleration of the center and disappears in the translating frame. The leading differential acceleration therefore comes from the quadrupolar tidal forcingwhere $\mathbf{n}(t)$ points toward the companion. It is a degree-two solid spherical harmonic. For a circular nonrotating-frame orbit, its time-dependent components have $m=\pm 2$ and frequency $2{\omega }_o$, because the squared direction cosines repeat twice per orbit. There is also a static $m=0$ component. The real part of one complex harmonic represents the pair of time-dependent components; its normalization and orientation are absorbed into the order-one coefficient $\psi$. This gives $\Psi =\psi (G{M}_2/a^3)r^2{Y}_2^m$ and $\omega =2{\omega }_o$. The expansion is the leading term for $R_1/a$ small; spin is being neglected.

For a forced potential flow with the same temporal frequency, the Euler momentum equation becomesUse $U=A{r}^2{Y}_2^m$. The surface gravity perturbation of a uniform-density sphere is unchanged, as is the vanishing Lagrangian pressure perturbation boundary condition. Integrating the equation and applying that condition givesSince ${\xi }_r(R_1)=2A{R}_1{Y}_2^m$, the real surface displacement isIndeed $ϵ{\omega }_2^2=2G{M}_2/a^3$, so the normalization is consistent. This is a particular forced response away from the resonance; arbitrary free normal modes can be added.

Put $\mu =M_2/M_1$ and $q=(R_1/a{)}^3$. The tidal resonance of a stellar oscillation condition is $4{\omega }_o^2={\omega }_2^2$, henceThe resonance gives $0\le ϵ<1/2$ for finite nonnegative $\mu$, with actual forcing requiring $M_2>0$. At exact resonance the undamped linearization has no bounded response at the forcing frequency: the time-domain particular solution grows proportionally to $t$ times a sinusoid. Thus the frequency-response formula has a pole, rather than describing a finite resonant displacement. Away from that pole, linearization requires the displayed surface displacement to be small compared with $R_1$; the leading quadrupolar tidal forcing approximation separately requires $R_1/a$ small.

For a normal mode proportional to $e^{ikx-i\omega t}$, the linear equations giveFor $\omega \ne 0$, eliminate $U_y=-i\Omega U_x/(2\omega )$ and $\hat{\sigma }=k{\Sigma }_0{U}_x/\omega$. The density-wave dispersion relation in a non-self-gravitating disk isThe ${\Omega }^2$ term is the radial epicyclic frequency squared for Keplerian rotation: Coriolis force and background shear restore an orbital displacement. The $c_s^2{k}^2$ term is acoustic restoration by a pressure gradient. Long wavelengths are mainly epicyclic motion; short wavelengths are mainly acoustic waves. The given model contains no disk self-gravity term.

For completeness, the full three-variable determinant is proportional to $\omega ({\omega }^2-{\Omega }^2-c_s^2{k}^2)$. Its additional $\omega =0$ branch has $U_x=0$ and $2\Omega U_y=ik{c}_s^2\hat{\sigma }/{\Sigma }_0$. This is a stationary pressure/Coriolis balance, the zero-frequency balanced mode of an axisymmetric disk, rather than a propagating density wave. At $k=0$ it reduces to a uniform surface-density change. Dividing by $\omega$ is legitimate only for the wave branches just derived.

Use ${\varphi }_j=U_{j}x-\frac{1}{2}{U}_j^{2}t+{\varphi }_j^{\prime }$ and retain terms linear in the displacement and perturbation potentials. Evaluation at $z=\eta$ may then be replaced by evaluation at $z=0$: the correction to a perturbation is second order. The kinematic boundary conditions becomeThe linearized pressure continuity isThe decaying normal modes give $-k{A}_1=ik(U_1-c)B$ and $k{A}_2=ik(U_2-c)B$. Thus $A_1=-i(U_1-c)B$, $A_2=i(U_2-c)B$. Substitution into the dynamic condition yields the Kelvin-Helmholtz dispersion relation with gravity,Put $\rho ={\rho }_1+{\rho }_2$ and $\overline{U}=({\rho }_1{U}_1+{\rho }_2{U}_2)/\rho$. Completing the square givesFor $k>0$, the temporal growth rate is $k{c}_i$. A growing complex conjugate branch therefore exists precisely when the radicand of the square root is negative:At equality the two phase velocities coincide and there is no exponential growth. Below it the roots describe real internal gravity waves. The denser lower fluid gives stable density stratification, but the tangential shear can overcome it. The assumption $U_1\ne U_2$ is needed for the displayed threshold; equal streams have no such shear instability.

Separate the uniform stream and an arbitrary time-dependent gauge from each velocity potential: ${\varphi }_j=U_{j}x+f_j(t)+{\varphi }_j^{\prime }$. Only gradients of the velocity potential determine the velocity field. The remote fluid must remain in its unperturbed uniform stream, soAfter fixing the harmless gauge, the nonzero-wavenumber perturbation velocity potentials themselves vanish at the corresponding infinities. Incompressible flow and irrotational flow give the Laplace equation for each perturbation. For a normal mode with $k>0$, the vertical amplitude solves ${\Phi }_j^{\prime \prime }-k^2{\Phi }_j=0$, henceThe opposite exponentials are excluded because they make the perturbation grow at infinity. It is the perturbation, rather than the total potential $U_{j}x+f_j(t)$, that decays.

Let $D=\overline{U}-c$ and $F=\hat{w}/D^{1/2}$. For $c_i>0$, $D$ never vanishes, so an analytic branch of a square root exists along the real flow domain. Take real smooth velocity and buoyancy frequency coefficients, nonzero $k$ (chosen positive for the growing-wave convention), a nontrivial regular normal mode, and boundary decay strong enough to remove the integration-by-parts term.

Substitution into the Taylor–Goldstein equation gives the half-power case of the power-transformed Taylor–Goldstein energy identity:For completeness, differentiating $\hat{w}=D^{1/2}F$ yields ${\hat{w}}^{\prime \prime }=D^{1/2}{F}^{\prime \prime }+{\overline{U}}^{\prime }{D}^{-1/2}{F}^{\prime }+[\frac{1}{2}{\overline{U}}^{\prime \prime }{D}^{-1/2}-\frac{1}{4}({\overline{U}}^{\prime }{)}^2{D}^{-3/2}]F$, which explains the coefficient $1/4$.

Multiply by $F^{\ast }$ and integrate over $z$. The boundary term $[F^{\ast }D{F}^{\prime }]$ is zero for the given homogeneous endpoint conditions, or for sufficiently decaying finite-energy modes at infinity. ThereforeSince $\mathrm{Im}D=-c_i$, $\mathrm{Im}(1/D)=c_i/|D{|}^2$, and $S,{\overline{U}}^{\prime \prime }$ are real, taking the imaginary part and dividing by $-c_i$ givesThe first two terms have strictly positive integral for a nonzero mode. Thus $S$ cannot be nonnegative everywhere:This proves the Miles–Howard theorem in contrapositive form. Where ${\overline{U}}^{\prime }\ne 0$, the corresponding gradient Richardson number must fall below $1/4$ somewhere. Failure of this sufficient-stability criterion does not prove instability. The derivation uses $c_i>0$; it cannot be applied unchanged to a neutral singular critical layer.

For the equal-width Hazel model with stable density stratification $J\ge 0$,The Miles–Howard theorem therefore excludes exponentially growing regular normal modes for $J\ge 1/4$. The candidate neutral modes of the equal-width Hazel model instead lie onThe parabola touches the sufficient-stability threshold at its maximum. There is no contradiction: neutrality has $c_i=0$, whereas the proof in part (a) assumes $c_i>0$. Moreover, the $k=1/2$ profile has a singular critical-layer derivative and logarithmically divergent horizontal kinetic energy; it is not a regular growing mode satisfying the proof's hypotheses.

For $J<1/4$, the local gradient Richardson number condition permits instability but does not establish it. Substitution of a neutral ansatz alone also does not determine on which side of the curve unstable eigenvalues lie. It identifies the formal neutral curve; concluding a full stability boundary requires additional continuation analysis of the eigenvalues. The nondecaying $k=0$ endpoint and the smooth $k=1,J=0$ endpoint have the distinct qualifications described above.

When $R\ll Q$, the ratio of cubic damping to linear amplification is $R^2/Q^2\ll 1$. The cubic amplitude saturation equation therefore giveswhile the small-amplitude regime lasts. With $c_d=0$, the matching real-wavenumber normal mode of the linear complex Ginzburg-Landau equation has $\omega =Uk+i(\mu -k^2)$ and hence amplitude proportional to $e^{(\mu -k^2)t}$. ThusThis is the temporal growth rate of the same Fourier component. It is not the fixed-position absolute frequency rate $\mu -U^2/4$ of a localized impulse. Nonlinear damping becomes important when $R$ is comparable with $Q$ and then arrests exponential amplification.

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.