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Put $a=a_1>0$, so $a_2=4a/5$ and $a_3=8a/5$. The second equilibrium equation factors as $B({\lambda }_2+a_{3}A)=0$. The resulting equilibria areThe mixed pair exists strictly when ${\lambda }_2({\lambda }_1-5{\lambda }_2/8)>0$; equality describes its merger with an axial equilibrium. At $O$ the eigenvalues of the Jacobian matrix are $({\lambda }_1,{\lambda }_2)$, and at $P$ they are $(-{\lambda }_1,{\lambda }_2-8{\lambda }_1/5)$. At a nonzero mixed equilibrium,Thus both mixed equilibria are saddle equilibria. The axial equilibria are nodes when their two real eigenvalues have the same sign and saddles when the signs differ. In particular,There are exactly three local bifurcation lines away from their common codimension-two intersection. On ${\lambda }_1=0$, $O$ and $P$ meet in a transcritical bifurcation, visible in $\dot{A}={\lambda }_{1}A+a{A}^2$ on the invariant axis $B=0$. On ${\lambda }_2=0$, the mixed pair meets $O$ in a pitchfork bifurcation. Eliminating $A$ near $O$, for fixed nonzero ${\lambda }_1$, gives $\dot{B}={\lambda }_{2}B-(a_2{a}_3/{\lambda }_1)B^3+\cdots$: this pitchfork is supercritical for ${\lambda }_1>0$ and subcritical for ${\lambda }_1<0$. On ${\lambda }_2=8{\lambda }_1/5$, the mixed pair meets $P$. Writing $A=-{\lambda }_1/a+u$ and $\eta ={\lambda }_2-8{\lambda }_1/5$, elimination gives $\dot{B}=\eta B+(a_2{a}_3/{\lambda }_1)B^3+\cdots$. This pitchfork is subcritical for ${\lambda }_1>0$ and supercritical for ${\lambda }_1<0$. A supercritical pitchfork here need not produce a stable mixed state: its other eigenvalue can already be positive. The negative mixed Jacobian determinant settles the full stability question. There is no Hopf bifurcation in this quadratic truncation.

Along ${\lambda }_2={\lambda }_1-\Delta$, the axial curves are $A=0$ and $A=-{\lambda }_1/a$. The mixed pair has the same plotted amplitude for its two signs of $B$:Consequently the mixed curves exist outside the interval whose endpoints are $\Delta$ and $-5\Delta /3$. The stable portions of the axial curves areFor $\Delta >0$, stability passes directly from $O$ to $P$ at the transcritical bifurcation ${\lambda }_1=0$; the two pitchfork bifurcations lie at $-5\Delta /3$ and $\Delta$. For $\Delta <0$, $O$ loses stability at ${\lambda }_1=\Delta$ and $P$ becomes stable only at ${\lambda }_1=-5\Delta /3$. Between these points none of the equilibria is stable. This local statement does not prescribe a bounded attractor of the quadratic truncation. All mixed portions are unstable.

At the double threshold, $L{\psi }_0=L{\psi }_1=0$. For $j\ge 2$, $q_j^2\ge 9{\pi }^2/4$, so $|K^{\ast }-q_j^2|\ge 13{\pi }^2/8>3{\pi }^2/8$ and every remaining eigenvalue is strictly negative. There is a spectral gap separating this two-dimensional kernel from the stable modes. The centre manifold theorem therefore gives a two-dimensional state centre manifold, withIf the two detuning parameters are adjoined as variables with zero time derivatives, the extended centre manifold has total dimension four; its state fibers still have dimension two.

The only nontrivial spatial symmetry of the bounded interval is the reflection $x\mapsto -x$. Since ${\psi }_0$ is even and ${\psi }_1$ odd, its group action on the amplitudes is $(A,B)\mapsto (A,-B)$. Thus equivariance requires $\dot{A}$ to be even and $\dot{B}$ odd in $B$. There is no sign symmetry $\psi \mapsto -\psi$, because the quadratic nonlinearity breaks it. The general smooth form is $\dot{A}=F(A,B^2)$, $\dot{B}=BG(A,B^2)$, with the zero solution preserved. The quadratic even-odd mode interaction follows on truncation:Writing $m=\mu -{\mu }^{\ast }$ and $\kappa =K-K^{\ast }$, the exact linear growth rates are $m-3{\pi }^2\kappa /4-{\kappa }^2$ and $m+3{\pi }^2\kappa /4-{\kappa }^2$. To first order in detuning,The orthogonal projection of the quadratic term onto the neutral eigenfunctions determines the leading coefficients. Both eigenfunctions have squared $L^2$ norm one, and parity kills the unwanted products. The quadratic correction $h$ contributes no term through $Lh$, since $L$ is self-adjoint and the projected modes lie in its kernel. ConsequentlyThese projections are also a concrete way of computing the normal form. In this normalization they give $a_1=8/(3\pi )$, $a_2=32/(15\pi )$ and $a_3=64/(15\pi )$, consistent with the stated coefficient ratios.

Use a real physical field, so the three positive wavevectors are accompanied by their complex-conjugate modes. Modulo lattice translations, the effective symmetry group is $G={\mathbb{T}}^2⋊C_6$. Its translation group action isA $120^{\circ }$ rotation cyclically permutes the amplitudes; a half-turn $\kappa$ acts by simultaneous complex conjugation. Together these generate the sixfold rotation group. These formulas specify the group action independently of the choice of orientation of the three wavevectors.

For a nonzero real hexagon $(h,h,h)$, its stabilizer subgroup is the rotation group $C_6$. A translation fixes it only when all three phase factors equal one. Cyclic permutation forces a fixed vector to have equal components, and the half-turn forces those components to be real. Therefore $\mathrm{Fix}(C_6)=\{(a,a,a):a\in \mathbb{R}\}$. For a roll $(r,0,0)$ with nonzero real $r$, the stabilizer subgroup consists of the translations $T_{0,\theta }$ and the half-turn, with $\kappa T_{0,\theta }\kappa =T_{0,-\theta }$. It is $S^1⋊C_2$, isomorphic to $O(2)$ as an abstract group even though spatial reflections are absent. The circle eliminates $B,C$ from its fixed-point subspace of a group action, and the half-turn makes $A$ real. ThusIf unreduced physical translations are used, the full lattice kernel is included in both stabilizer subgroups. The real representation on the six critical components is absolutely irreducible: translations distinguish the three Fourier pairs, rotations permute them, and the half-turn rules out a complex scalar in the commutant. The equivariant branching lemma consequently supplies generic steady branches of both axial types at a transverse crossing of the common linear eigenvalue. It guarantees existence, not stability or exhaustiveness of the possible branches.

Translation equivariance severely restricts the monomials in $\dot{A}$. Through cubic order they are $A$, $\overline{B}\overline{C}$, $|A{|}^{2}A$, $|B{|}^{2}A$ and $|C{|}^{2}A$: each has the same translation weight as $A$, using ${\mathbf{k}}_1+{\mathbf{k}}_2+{\mathbf{k}}_3=0$. The half-turn forces their coefficients to be real; the $120^{\circ }$ rotation gives the two cyclic counterpart equations. A spatial reflection would interchange $B,C$ and require equal cross-couplings, but it is not a symmetry here. After a real amplitude rescaling makes the generic nonzero quadratic coefficient one, the rotational hexagon amplitude equations arewith the other two equations obtained by cyclic permutation. All four coefficients are real; $\delta$ records the absence of spatial reflection. The normalization assumes the resonant quadratic coefficient is nonzero; its vanishing would be an additional degeneracy.

For the roll, $r^2=\mu /{\nu }_1$. The radial eigenvalue is $-2\mu$ and the phase of the active mode is a neutral translation. The transverse variables $(b,\overline{c})$ have matrixIts two real eigenvalues, each occurring twice in the full real linearization, are $-({\nu }_2-{\nu }_1)r^2\pm \sqrt{{\delta }^2{r}^4+r^2}$. Define $d={\nu }_2-{\nu }_1$ and $D_r=d^2-{\delta }^2$. Strict attraction modulo translation is therefore equivalent toIndeed the larger transverse eigenvalue is negative precisely when $d{r}^2>\sqrt{{\delta }^2{r}^4+r^2}$, and squaring is legitimate only after requiring $d>0$. If ${\delta }^2>d^2$, that inequality is impossible. Equality ${\delta }^2=d^2$ is also insufficient because of the additional positive term $r^2$. Any nonzero roll that exists with ${\nu }_1<0$ already has a positive radial eigenvalue; the degenerate case ${\nu }_1=0$ has no ordinary isolated nonzero roll branch.

For hexagons put $N={\nu }_1+2{\nu }_2$. A nonzero real amplitude $h$ satisfies $\mu =N{h}^2-h$, or $h=(1\pm \sqrt{1+4N\mu })/(2N)$ when $N\ne 0$. Separate real and imaginary disturbances. The real Jacobian matrix is cyclic, with diagonal $-h-2{\nu }_1{h}^2$ and off-diagonal entries $h-2({\nu }_2\pm \delta )h^2$. Its eigenvalues areThe imaginary Jacobian matrix has every entry equal to $-h$, giving eigenvalues $-3h,0,0$. The two zero modes translate the hexagonal pattern. Thus the general strict stability conditions for any nonzero root $h$ areThese conditions state orbital stability transverse to translations, not asymptotic convergence to one prescribed spatial phase. Equalities require a nonlinear bifurcation calculation.

When ${\nu }_2>{\nu }_1>0$, both $d,N$ are positive. The upper positive hexagon branch is radially stable after its saddle-node bifurcation at $\mu =-1/(4N)$, $h=1/(2N)$; it remains stable until $h=1/d$. At this second threshold the pair ${\lambda }_{\pm }$ crosses the imaginary axis with nonzero frequency if $\delta \ne 0$. The crossing is transverse, since $dh/d\mu =1/(2Nh-1)$ and the derivative of $-2h+2d{h}^2$ with respect to $h$ is two there. Hence, subject to the usual nonzero cubic coefficient, this is a Hopf bifurcation:For $\delta =0$ the same threshold has two zero real eigenvalues, so it is not a Hopf bifurcation.

A possible phase portrait when neither steady branch is stable has an attracting periodic oscillation surrounding the unstable hexagon, with the three amplitudes waxing and waning cyclically. This is realized, for example, by ${\nu }_1=1$, ${\nu }_2=2$, $\delta =2$ just above ${\mu }_H=4$. Rolls are unstable for every positive $\mu$ because ${\delta }^2=4>d^2=1$. At the hexagon Hopf point $h=1$, the real spectrum is $-9,\pm 4i\sqrt{3}$. Evaluation of the cubic Hopf normal form with a unit-norm critical eigenvector gives $\mathrm{Re}{G}_{21}=-64/27<0$, so this example has a supercritical Hopf bifurcation and an attracting small limit cycle in the real invariant subspace. The figure shows such a cycle at $\mu =4.2$ and an approaching trajectory; the positive octant is forward invariant because an amplitude's derivative at zero is the product of the other two. This is a possible portrait, rather than a claim that every parameter set with unstable rolls and hexagons has the same attractor. Stability asserted for the periodic orbit here is within the stipulated real subspace.

Near a spatially extended Hopf bifurcation, a critical oscillatory normal mode has a slowly varying complex amplitude. For a selected travelling-wave branch, a method of multiple scales expansion takes the formHere the distance from onset is of order ${\epsilon }^2$, $v_0$ is the critical eigenvector, and $v_g$ is the group velocity. The comoving coordinate removes the first spatial derivative of the envelope. Projection at the first resonant order gives $A_T=\rho A+D_0{A}_{XX}-G_0|A{|}^{2}A$. Spatial translation and the temporal phase of the Hopf oscillation give a constant-phase symmetry $A\mapsto e^{i\theta }A$, which permits the cubic term $|A{|}^{2}A$ but forbids a generic $A^2$ term. The coefficients are complex because the envelope has both growth and frequency detuning. For a supercritical branch with positive real diffusion, real rescalings and a uniform phase rotation give the complex Ginzburg–Landau equationA subcritical Hopf branch requires higher saturation terms instead. If spatial reflection makes opposite travelling waves critical together, two coupled amplitude equations are required initially; the scalar equation applies to a selected single-wave branch when its competing amplitude is stable.

Distant boundaries can still determine which travelling pattern is observed. They can select its phase and wavenumber, reflect a travelling disturbance, inject a competing wave, or let a growing wave packet leave the domain. In particular, amplification in a comoving frame can be convective wave-packet instability rather than absolute wave-packet instability at a fixed laboratory point. A finite-domain outcome then need not reflect intrinsic bulk modulation dynamics. Periodic boundary conditions are the appropriate idealization when studying bulk instabilities without end selection or reflected-wave forcing: they eliminate physical end layers and make the allowed envelope Fourier modes discrete. They do not assert that real distant boundaries are always negligible. Choose a periodic length compatible with the carrier, fix the phase winding, and compare perturbations within that same periodic problem. On an open physical domain the group velocity and actual boundary conditions must be restored.

The optimum carrier wavenumber corresponds to the spatially uniform envelope $A_f=e^{-icT}$, often called a flat state. Its phase is arbitrary. Write $A=e^{-icT}(1+r)e^{i\varphi }$ with real small amplitude and phase disturbances. To first order,A Fourier mode $e^{\lambda T+ikX}$ therefore has matrix and dispersion relationThe trace is negative. The amplitude mode at $k=0$ has eigenvalue $-2$, while the phase mode is neutral because constant phase shifts preserve the solution. Put $D=1+bc$. For every nonzero real $k$, the determinant is positive if $D\ge 0$, whereas $D<0$ makes it negative for $0<k^2<-2D/(1+b^2)$. ThusThis is the Benjamin-Feir stability condition. At $D=0$, nonzero modes still decay, but the leading long-wave decay is fourth order rather than diffusive. On a periodic envelope interval of length $L$, the smallest nonzero wavenumber is $2\pi /L$, so strict stability of all allowed nonzero modes requiresA sufficiently short periodic domain can therefore exclude the unstable band even when the infinite-domain Benjamin-Feir stability condition fails.

The amplitude is damped while the phase is slow. For a slowly varying phase, the leading slaved amplitude is $r=-\frac{1}{2}\{b{\varphi }_{XX}+({\varphi }_X{)}^2\}+\cdots$. Substitution in the phase equation gives the diffusion term $D{\varphi }_{XX}$ and the nonlinear frequency shift $(c-b)({\varphi }_X{)}^2$. The latter also follows directly from a constant phase gradient: an envelope with gradient $q$ has amplitude squared $1-q^2$ and frequency $c+(b-c)q^2$. Constant phase symmetry rules out undifferentiated $\varphi$; translation invariance gives constant coefficients. The leading comoving Ginzburg-Landau equation also has spatial parity, so odd linear derivatives are absent at this order. Its neutral phase eigenvalue expands asThis coefficient follows by inserting a power series for $\lambda$ in the quadratic dispersion relation. Close to $D=0$, it becomes $E=(1+b^2)/2>0$ at leading order. The long-wave phase reduction of the complex Ginzburg-Landau equation is consequentlyThe fourth derivative regularizes negative phase diffusion. Terms omitted here have higher order in the long-wave expansion. First-order advection has been removed by the group-velocity frame; higher odd derivatives in a more general travelling-wave problem are beyond this leading amplitude approximation.

Take $D=-d$ with $d>0$ small. At the instability boundary $c=-1/b$, so $g=-(1+b^2)/b$ is nonzero. Set $\xi =\sqrt{d/E}X$, $\tau =d^{2}T/E$, and $\varphi =(d/g)\Phi$. At leading order the phase dynamics become the potential Kuramoto-Sivashinsky equation,With $u=-2{\Phi }_{\xi }$ this isThe variable $u$ is a phase-gradient disturbance, not the original physical field. Periodicity of the phase fixes ${\int }_0^{P}ud\xi =0$ for the zero-winding branch; the Kuramoto-Sivashinsky equation conserves this mean. The scaled period is $P=L\sqrt{d/E}$.

At fixed period $P$, the flat phase has growth rates ${\lambda }_n=k_n^2-k_n^4$, $k_n=2\pi n/P$. Its first loss of stability occurs at $P=2\pi$. To find what replaces it, put $k=2\pi /P$ and consider $P$ just above $2\pi$. After choosing an origin by translation, expand a small steady modulation as $u=a\sin (k\xi )+b_2\sin (2k\xi )+\cdots$. Projection of $-u{u}_{\xi }$ onto the first two harmonics yieldsSince ${\lambda }_2=-12$ at onset, this harmonic is slaved to $b_2=k{a}^2/(2{\lambda }_2)+\cdots$. The fundamental amplitude obeysThe primary periodic bifurcation of the Kuramoto-Sivashinsky equation is therefore supercritical. For $0<P-2\pi \ll 1$,The radial eigenvalue of this nonzero branch is $-2{\lambda }_1+O({\lambda }_1^2)<0$; all higher harmonics remain damped and the modulation's spatial phase is neutral. Thus the small finite-amplitude modulation is stable in the sense of orbital stability, at fixed mean and fixed period, despite the flat state's instability. The two signs of $a$ represent translations of the same modulation. In the potential formulation the spatial mean of $\Phi$ advances at $⟨({\Phi }_{\xi }{)}^2⟩$: the original wave acquires a nonlinear frequency correction rather than requiring a time-independent absolute phase.

For larger $P$, further linear modes become active and the small-amplitude argument no longer determines stability. A periodic modulation $\bar{u}$ must then be tested with the operator $v_{\tau }=-v_{\xi \xi }-v_{\xi \xi \xi \xi }-{\partial }_{\xi }(\bar{u}v)$ on mean-zero periodic disturbances. Its translation mode is always neutral, and secondary oscillatory or more complicated modulation can occur. Moreover, stability against perturbations of the same period does not prove stability against disturbances of a larger period; those require a separate sideband or Floquet theory calculation. The controlled conclusion near the first threshold is the existence of a stable small modulation family, not universal stability of all periodic Kuramoto-Sivashinsky solutions.

Put $h=1/(m+1)$ and impose homogeneous Dirichlet boundary conditions, as required by the stated continuum eigenvalues. Let $T=\mathrm{tridiag}(1,-2,1)$ be the unscaled one-dimensional Dirichlet discrete Laplacian. For $1\le k\le m$, the vectors $v_j^{(k)}=\sin (jk\pi h)$ satisfyThe identity follows by adding the two neighboring sines; the values at $j=0,m+1$ vanish. These $m$ mutually orthogonal vectors form the discrete sine transform basis. Consequently their tensor products give all $m^2$ two-dimensional modes, not merely a few candidate eigenvectors.

In a compatible grid ordering the axial operator is $A_5=h^{-2}(T\otimes I+I\otimes T)$. The nine-point operator isIndeed, the added Kronecker product supplies the diagonal neighbors with weight $1/6$, subtracts $1/3$ from each axial neighbor and adds $2/3$ to the central coefficient. Thus it has exactly the coefficients of the second discretization. On $v^{(k)}\otimes v^{(l)}$ the eigenvalues areandEquivalently, $h^2{\lambda }_{9,kl}=-10/3+4[\cos (k\pi h)+\cos (l\pi h)]/3+2\cos (k\pi h)\cos (l\pi h)/3$.

For fixed mode numbers $k,l$, expansion of the sines giveswhereas the mixed Kronecker product correction givesThus both eigenvalue approximations are second order for fixed modes. The leading nine-point error includes an additional positive term ${\pi }^4{h}^2{k}^2{l}^2/6$.

There is also an exact comparison, not just an asymptotic one. Since $\sin x<x$ for $0<x<\pi /2$, and both sine factors in the correction are positive,Both estimates lie on the less-negative side of the continuum Laplacian eigenvalue, so the five-point formula is the better eigenvalue estimate for every corresponding interior mode. The five-point versus nine-point Laplacian eigenvalue accuracy comparison does not contradict the useful isotropy or special Poisson accuracy properties of a nine-point stencil. Those properties concern a different error criterion. The fixed-mode expansion is not uniform for modes with $k$ or $l$ comparable to $h^{-1}$; wavelengths close to the mesh scale need not have small relative spectral error.

The original PDF has a sign defect in the lower-left stage coefficient. Put $d=c_2-c_1$. With the printed negative sign, the row sums and the relevant Butcher order conditions areThus the claimed universal second order is false. For example $c_1=0,c_2=1$ gives, on $y^{\prime }=y$, two stages equal to $y$ and the update $Y=(1+h)y$. This is the first-order Forward Euler method. For the literal tableau, order at least two holds exactly when $c_1=1/2$ or $c_2=0$, subject to distinct allowed nodes.

For completeness the literal tableau's A-stability can also be classified algebraically. DefineThe stability function obtained from $R(z)=1+z{b}^T(I-zA{)}^{-1}e$ isIf $D\ge 0$, the denominator has no zero in the closed left half-plane: its reciprocal roots are the stage-matrix eigenvalues, whose positive sum and nonnegative product give positive real parts for nonzero eigenvalues. On the imaginary axis,Consequently for the literal tableau with uniquely solvable stages throughout the left half-plane, the complete parameter criterion isThese explicit inequalities in $c_1,c_2$ are necessary by the small and large imaginary parameters, and sufficient by the maximum modulus principle, including the bound at infinity. For $D=0$ the last inequality forces $r=0$, leaving the stable linear-fractional case. For $D<0$ there is a negative real stage pole. If one defines stability solely through an analytically continued scalar stability function, the only additional possibilities areIndeed these are precisely the cases where the left pole cancels and $R(z)=[1+(1-q)z]/[1-qz]$ is an A-stable theta method. The original stage equations are nevertheless singular at $z=-1/(q-a)$, so they do not give uniquely determined stages there. This distinction prevents a canceled rational factor from concealing an ill-defined stage solve.

The natural intended repair is to replace just the lower-left coefficient by $+c_2^2/(2d)$. This makes the method the collocation Runge-Kutta method on the two nodes. To derive the repaired entries, takeIntegration gives the other three printed entries and the weights unchanged, with the positive lower-left entry. Since $L_1+L_2=1$ and $c_1{L}_1+c_2{L}_2=t$, it follows that $Ae=c$, $b^{T}e=1$ and $b^{T}c=1/2$. Hence the corrected family is always of order at least two.

For the corrected family let $a=(c_1+c_2)/2$, $D=c_1{c}_2/2$ and $r=(1-c_1)(1-c_2)/2$. Its stability function has the same displayed rational form, now with $q=1/2$. The denominator again has no left-half-plane zero, and on the imaginary axis the squared-modulus difference is $(D^2-r^2)y^4$. Because $D,r\ge 0$, the necessary and sufficient inequality is $D\ge r$, orFor sufficiency, $|R|\le 1$ on the imaginary axis and at infinity; apply the maximum modulus principle on growing left half-discs to get the same bound inside. If $D=0$, the condition leaves the endpoint trapezoidal rule and its linear denominator. If $c_1+c_2<1$, every nonzero imaginary argument gives $|R|>1$. This is the two-node collocation A-stability criterion, explicitly distinguished from the literal defective tableau.