Put $x=R^2>0$. Substituting the plane wave into the quintic real Ginzburg-Landau equation givesso the two nonzero branches areThe upper branch has $x_{+}\ge \alpha /2$ and exists for $Q^2\le \mu +{\alpha }^2/4$. The lower branch is positive when $\mu <Q^2\le \mu +{\alpha }^2/4$; at $Q^2=\mu$ it reaches the zero solution. The two branches meet at the fold $x=\alpha /2$, $\mu =Q^2-{\alpha }^2/4$. Thus for a given $\mu$, nonzero waves exist only if $\mu \ge -{\alpha }^2/4$, withThe zero solution exists for every $\mu$ independently of $Q$, because its spatial phase has no meaning.

For stability, write $A=e^{iQX}(R+u+iv)$ with real small $u,v$. Linearizing and using the steady-wave condition givesDefine $\Lambda =2R^2(2R^2-\alpha )$. A sideband of wavenumber $p$ has matrixand exact growth ratesOn the lower branch $\Lambda <0$, the uniform amplitude perturbation already grows at rate $-\Lambda >0$, so that branch is unstable. On the upper branch $\Lambda >0$, the fast amplitude mode decays and the slow phase mode hasThe coefficient of phase diffusion gives the Eckhaus boundary for a subcritical quintic amplitude equation:for robust stability to long wavelengths. At equality, ${\sigma }_{+}=-p^4/\Lambda +O(p^6)$; the leading diffusion coefficient is marginal, while nonzero sufficiently long-wave linear perturbations still decay at the next spatial order. Indeed when $4Q^2\le \Lambda$, the matrix trace is negative and its determinant is $p^2(p^2+\Lambda -4Q^2)\ge 0$, proving linear non-growth at every $p$, not just asymptotically small $p$. The neutral $p=0$ phase mode reflects the constant phase symmetry.

At the fold, $\Lambda =0$. If $Q\ne 0$, ${\sigma }_{+}=-p^2+2|Qp|$ is positive for small $p$, so the fold wave is unstable. At $Q=0$, the fold is a marginal linear point, but uniform amplitude perturbations have the one-sided nonlinear instability of a saddle-node; it is not a robust stable state. For the trivial state the linear growth rate is $\mu -p^2$, so it is stable for $\mu <0$ and only linearly marginal at $\mu =0$.

The requested regions can be expressed without eliminating $R$. In the $(\mu ,R)$ plane, nonzero waves exist wherever $\mu \ge R^4-\alpha R^2$. The robust stable part isThe left boundary is the zero-detuning wave $Q=0$; the right boundary is the marginal Eckhaus curve. Both meet at $(\mu ,R)=(-{\alpha }^2/4,\sqrt{\alpha /2})$. At fixed $\mu >-{\alpha }^2/4$, an equivalent description iswith the lower endpoint added if only marginal linear stability is required.

In the $(\mu ,Q)$ plane, eliminate $R$ from the marginal condition. It gives $R^2=[\alpha +\sqrt{{\alpha }^2+16Q^2}]/4$, henceThe equality curve is the marginal boundary; the existence boundary is $\mu =Q^2-{\alpha }^2/4$. They touch at $Q=0,\mu =-{\alpha }^2/4$ but separate for nonzero detuning, leaving an unstable wave band. Equivalently, for fixed $\mu >-{\alpha }^2/4$ the robust stable interval isThe stable amplitude is the upper root throughout this interval. The sketch uses scaled coordinates $\mu /{\alpha }^2,R/\sqrt{\alpha },Q/\alpha$, so it applies to every fixed positive $\alpha$. Green denotes stable nonzero waves; the marginal curves form its boundaries, while the surrounding gray region contains existing but unstable nonzero waves.

The six critical Fourier modes are at $\pm {\mathbf{k}}_1,\pm {\mathbf{k}}_2,\pm {\mathbf{k}}_3$, with ${\mathbf{k}}_1+{\mathbf{k}}_2+{\mathbf{k}}_3=0$. Reality of the pattern pairs each positive-wavevector coefficient with the conjugate coefficient at the negative wavevector. The leading critical part of the extended centre manifold parametrization is consequentlyThe displayed six-mode expression is its leading-order projection; nonlinear harmonics on the full centre manifold need not vanish identically.

A translation $\mathbf{d}$ acts by $A_i\mapsto e^{i{\mathbf{k}}_i\cdot \mathbf{d}}{A}_i$. Each monomial in the equation for $A_i$ must therefore carry wavevector ${\mathbf{k}}_i$. Since $-{\mathbf{k}}_{i+1}-{\mathbf{k}}_{i+2}={\mathbf{k}}_i$, the allowed quadratic term is $A_{i+1}^{\ast }{A}_{i+2}^{\ast }$. At cubic order the allowed terms are $A_i|A_i{|}^2,A_i|A_{i+1}{|}^2,A_i|A_{i+2}{|}^2$. Rotation through $60$ degrees acts, with one choice of orientation, asIts square cyclically permutes the amplitudes, equating the coefficients across the three equations. Its cube conjugates every amplitude; equivariance under this half-turn makes all coefficients real. It does not interchange the two cross-coupling coefficients. A reflection would do that and force $b=c$, but that symmetry is expressly absent. Thus the chiral hexagonal amplitude equations areat the retained order, with cyclic indices and real coefficients. The missing opening modulus bar in the last cubic term of the printed formula must be supplied; phase equivariance requires $|A_{i+2}{|}^2$.

Write $A_i=r_i{e}^{i{\varphi }_i}$, with all three amplitudes nonzero, and let $\Phi ={\varphi }_1+{\varphi }_2+{\varphi }_3$. Separating real and imaginary parts givesHence the phase locking of a resonant hexagonal triad obeysFor $\alpha >0$, $\Phi =0$ is phase stable and $\Phi =\pi$ is phase unstable. Two independent translation coordinates remove two phases; since the wavevectors sum to zero, the remaining phase is exactly $\Phi$. Thus a stable fully active phase-locked pattern can be represented by three positive real amplitudes after choosing the spatial origin. The statement is modulo translations, not a restriction on every representative. Boundary states must also be retained: the zero solution is stable for $\mu <0$, and rolls have two zero amplitudes. These are the necessary qualifications to a literal claim that all stable solutions are strictly positive.

In the real nonnegative subspace, a roll isor a cyclic permutation. For hexagons let $s=a+b+c$, which is positive under the assumptions. The equal-amplitude branches solve $\mu +\alpha r-s{r}^2=0$, givingwith only positive roots retained. The upper branch exists for $\mu \ge -{\alpha }^2/(4s)$; the lower branch is positive for $-{\alpha }^2/(4s)\le \mu <0$. They meet at a fold.

For a purported rectangle $(p,p,q)$ with $p,q>0$, subtract its first two equilibrium equations after dividing each by its own amplitude. The quadratic terms agree, and the difference is $(c-b)(p^2-q^2)=0$. Since $(b-a)(c-a)<0$ implies $b\ne c$, it forces $p=q$. A state with exactly two nonzero amplitudes is also impossible because the quadratic term drives the zero third amplitude. Thus there are no nontrivial rectangles.

In fact no other unequal positive equilibria are possible under the straddling cross-couplings. Here is the uniqueness of positive hexagon equilibria with straddling cross-couplings argument. Put $x_i=r_i^2$, $p=r_1{r}_2{r}_3>0$. The equilibrium equations implyConsider $b>a>c$, write $B=b-a>0$, $C=c-a<0$, and relabel cyclically so $x_1$ is maximal. If $x_2\ge x_3$, subtracting equations one and two givesThe left side is positive unless all three are equal, while the right side is nonpositive. If $x_2\le x_3$, subtract equations two and three instead:Now the left side is negative unless all are equal and the right side is nonnegative. Both cases force equality. For $c>a>b$, reverse the cyclic orientation and exchange $b,c$. Thus the only nonzero equilibria in this subspace are the stated rolls and hexagons.

At a roll the active radial eigenvalue is $-2\mu$. The two inactive amplitudes have real-part linearizationIts determinant is ${\mu }^2(a-c)(a-b)/a^2-{\alpha }^2\mu /a<0$. It therefore has a positive eigenvalue: the rolls are unstable for every $\mu >0$, which is stronger than the large-$\mu$ conclusion. In the complex system there is also a neutral translation phase, and the imaginary transverse block has the same determinant.

At a positive hexagon the real-amplitude Jacobian is circulant, with diagonal $d=-2a{r}^2-\alpha r$ and off-diagonals $p=\alpha r-2b{r}^2$, $q=\alpha r-2c{r}^2$. Its uniform eigenvector $(1,1,1)$ has eigenvalueThe other two eigenvectors lie in the zero-sum subspace; using the two nontrivial cube roots of unity givesFor sufficiently large $\mu$, $r\sim \sqrt{\mu /s}$, and the positive term $(b+c-2a)r^2$ dominates the quadratic-coupling term $2\alpha r$. Thus the hexagons are unstable at large $\mu$, while the uniform radial direction on the upper branch is stable. The two translation phases are neutral and the total phase has eigenvalue $-3\alpha r$, so the amplitude instability is genuine.

More precisely, the upper branch is amplitude stable between its fold and the chiral hexagon Hopf thresholdAt this threshold the conjugate pair crosses with nonzero frequency because $b\ne c$, permitting a Hopf bifurcation. The linear calculation locates the crossing; the nonlinear normal form decides the stability and direction of its periodic branch.

The dynamics are bounded in the retained model. For $S=r_1^2+r_2^2+r_3^2$ in the positive phase-locked subspace,Since $r_1{r}_2{r}_3\le (S/3{)}^{3/2}$, the negative quadratic-in-$S$ term dominates at large amplitude. For large positive $\mu$, the origin, rolls and hexagons are all unstable, so generic bounded dynamics cannot end at an asymptotically stable steady pattern. The chiral difference $b-c$ favors successive changes in dominant roll orientation, in the spirit of the Küppers–Lortz instability. Oscillatory competition or periodic switching is the natural behavior near the Hopf transition; linear stability alone does not establish a unique global limit cycle. With $\alpha =0$, coordinate planes are invariant and heteroclinic roll switching is possible. For $\alpha >0$ a missing third mode is regenerated when two others are present, so that exact boundary-plane heteroclinic picture is broken. Finally, sufficiently large $\mu$ is a conclusion of the cubic truncation; far from onset, omitted higher-order terms can change the physical pattern dynamics.

Use the modulus of continuity on the circle, $\omega (f,\delta )={\sup }_{|h|\le \delta }\Vert f(\cdot +h)-f{\Vert }_{\infty }$. Splitting a displacement into steps of length at most $\delta$ and applying the triangle inequality givesThe Fejér kernel is nonnegative and has integral one, soWe therefore need its first absolute moment. The inequalities $|\sin (nt/2)|\le n|\sin (t/2)|$ and $\sin (|t|/2)\ge |t|/\pi$ on $[-\pi ,\pi ]$ implySplitting at $1/n$ givesTaking $\delta ={\delta }_n=(\ln n)/n$ in the preceding error bound proves the Fejér first-moment approximation bound:The constants do not depend on $f$ or $n$. By the Heine-Cantor theorem, a continuous function on the compact circle is uniformly continuous, hence $\omega (f,{\delta }_n)\to 0$ as ${\delta }_n\to 0$. Thus the Fejér sums converge uniformly to $f$. The index restriction is necessary: at $n=1$ the printed scale is zero, while ${\sigma }_{1}f$ is the mean of $f$ and generally differs from it.

In a thin disk dominated by a central stellar mass, the vertical gravitational potential is harmonic to leading order, so its vertical acceleration is $-{\Omega }_0^{2}z$. For a locally isothermal background, $p=c_s^2\rho$ with $c_s$ independent of height. Vertical hydrostatic equilibrium therefore givesThis is the isothermal vertical structure in stellar gravity. On the upper half of the disk the downward gravity magnitude is $g={\Omega }_0^{2}z>0$; the density profile extends evenly through the midplane.

For an adiabatically displaced parcel kept in pressure equilibrium, the parcel's logarithmic density gradient is ${\gamma }^{-1}d\log p/dz$, whereas the environmental density gradient is $d\log \rho /dz$. A small upward displacement $\eta$ consequently produces a density contrast $\delta \rho /\rho =[{\gamma }^{-1}d\log p/dz-d\log \rho /dz]\eta$. Its buoyancy acceleration is $-g\delta \rho /\rho =-N^2\eta$. Thus the Brunt–Väisälä frequency isBoth background gradients equal $-z/H^2$ in the isothermal atmosphere, soAn isothermal background is therefore stably stratified for adiabatic perturbations with $\gamma >1$, but not buoyant for isothermal perturbations. In applying the adiabatic equation of state to a stratified atmosphere, its proportionality relates Lagrangian parcel changes: $\Delta p=(\gamma p/\rho )\Delta \rho$. Eulerian perturbations generally obey $p^{\prime }+\eta p_z=(\gamma p/\rho )({\rho }^{\prime }+\eta {\rho }_z)$ instead. Thus the printed simple proportionality must be interpreted as a parcel relation for the buoyancy calculation. For $\gamma =1$ in this isothermal background, the gradient terms cancel and the Eulerian proportionality holds too.

Now take isothermal perturbations, so $\gamma =1$ and $N^2=0$. The two supplied wave equations reduce toFor the nonsingular oscillatory case $\omega \ne 0$ and ${\omega }^2\ne {\Omega }_0^2$, substitute $u_z={\xi }_z/{\omega }^2$. With $x=z/H$, the resulting Hermite differential equation isThe exceptional frequencies must be treated through the original velocity-pressure equations rather than division by these factors.

The physical vertical condition is finite perturbation energy when integrated with the Gaussian density. In particular the pressure contribution is proportional to $\int e^{-x^2/2}|\xi {|}^{2}dx$ for nonzero frequency. Generic nonterminating solutions have an exponentially growing component at at least one vertical end and fail this condition; acceptable modes are the terminating Hermite polynomials. This is a finite-energy condition, not pointwise boundedness of the velocity in the vanishing-density atmosphere.

To prove the allowed values of $\alpha$, let the nonzero polynomial have degree $n$ and leading coefficient $a_n$. The highest power in ${\xi }_{xx}-x{\xi }_x+\alpha \xi$ has coefficient $(\alpha -n)a_n$. Hence $\alpha =n\in \{0,1,2,\dots \}$. Conversely, writing $\xi ={\sum }_j{a}_j{x}^j$ gives the recurrenceFor $\alpha =n$, choose the parity of $n$ and iterate until the series terminates at degree $n$. Thus all allowed Hermite vertical modes of an isothermal disk havewhere ${\mathrm{He}}_n$ is the Probabilists' Hermite polynomial. In particular, ${\mathrm{He}}_0=1$ and ${\mathrm{He}}_1=x$.

Let $K=kH$ and $s={\omega }^2/{\Omega }_0^2$. Eliminating the denominator for regular modes givesFor $n=0$, the nonzero-frequency solution isIts pressure variable is independent of height and $u_z=0$. These are inertial-acoustic waves, restored by epicyclic motion and pressure; at short radial wavelength they become ordinary acoustic waves. The multiplied dispersion polynomial also has $s=0$. This is not an extra oscillatory solution represented by nonzero $\xi$, since $\xi =i\omega p^{\prime }/\rho$ vanishes at zero frequency for finite pressure perturbation. The full horizontal equations have a zero-frequency balanced mode of an axisymmetric disk, with no radial velocity and azimuthal Coriolis acceleration balancing a radial pressure gradient. It must be recovered without dividing by $\omega$, rather than counted as a propagating wave. At $k=0$, the propagating branch limits to an epicyclic oscillation at $|\omega |={\Omega }_0$, another value at which the eliminated form must be interpreted by a limit.

For $n=1$, solve $(s-1{)}^2=K^{2}s$. The two branches areBoth signs of frequency occur on each squared-frequency branch. The upper branch is acoustic in the large-$|K|$ limit, $|{\omega }_{+}|\sim |k|c_s$; the lower branch is an inertial wave with $|{\omega }_{-}|\sim {\Omega }_0/|K|$. At $|K|\ll 1$, both frequencies approach ${\Omega }_0$, reflecting the equality of the radial and vertical epicyclic frequencies in a Keplerian disk. Here $\xi \propto z$, so $u_z={\xi }_z/{\omega }^2$ is independent of height: these modes have a coherent vertical displacement of the disk column, rather than a vertical breathing motion.

For a local non-axisymmetric disturbance proportional to $e^{i(kR+m\varphi -\omega t)}$, the material frequency is $\hat{\omega }=\omega -m{\Omega }_0$. In the same local radial-wave approximation, replace $\omega$ by $\hat{\omega }$. For $n=1,m=1$, this givesThe four frequencies can equivalently be written $\omega /{\Omega }_0=1\pm (\sqrt{K^2+4}\pm |K|)/2$, with independent choices of the two signs. This is a local dispersion relation, not a claim that Doppler substitution supplies an exact global non-axisymmetric solution in a differentially rotating disk. The radial-wave ordering neglects azimuthal gradients compared with radial gradients; $|k|R\gg 1$ can coexist with $|k|H\ll 1$ in a thin disk.

Write $w=\omega /{\Omega }_0$. For the two low inertial-frame frequency branches, $|w|\ll 1$, while the material frequency remains near $-{\Omega }_0$. Expanding the dispersion relation givesThereforeFor signed $k$, the two leading branches are conventionally labelled by $\pm kH/2$. These are bending waves of a Keplerian disk: an $m=1$ vertical displacement varies once around an annulus and corresponds to a small tilt of its orbital plane. A radial variation of the tilt is a warp. The waves transport the warp radially at group speeds $c_s/2$ in opposite directions. At zero radial wavenumber their low-frequency limit is a stationary rigid tilt; the apparent slow pattern does not mean that the orbiting gas has slow vertical motion in its own frame.