Chiral hexagonal amplitude equations (${\dot{A}}_i=\mu A_i+\alpha A_{i+1}^{\ast }{A}_{i+2}^{\ast }-A_i(a|A_i{|}^2+b|A_{i+1}{|}^2+c|A_{i+2}{|}^2)$)

Three Fourier wavevectors sum to zero and form a hexagonal star. Translations give each amplitude the phase of its own wavevector; cyclic rotations permute them, and a half-turn conjugates them. These symmetries permit the resonant quadratic term and force its coefficients to be real. A spatial reflection would interchange the two cross-couplings and require $b=c$, but chiral patterns need not have that extra symmetry. The equations describe rolls, hexagons, phase locking and oscillatory changes in dominant orientation.

Assume $\alpha >0$. For equal positive amplitudes $r$, let $s=a+b+c$ and use $\mu +\alpha r-s{r}^2=0$. The radial Jacobian eigenvalue is $\alpha r-2s{r}^2$. The conjugate pair has real part $(b+c-2a)r^2-2\alpha r$ and imaginary parts $\pm \sqrt{3}(b-c)r^2$. When $b+c>2a>0$ and $b\ne c$, the upper hexagon branch loses amplitude stability through this pair at the displayed radius. The associated parameter is ${\mu }_H=2{\alpha }^2(4a+b+c)/(b+c-2a{)}^2$. Nonlinear terms decide the direction and stability of the resulting periodic branch.

Assume $\alpha >0$, as in the stable phase-locked branch. At a positive equilibrium set $x_i=r_i^2$ and $p=r_1{r}_2{r}_3$. The equations imply $a{x}_i+b{x}_{i+1}+c{x}_{i+2}-\alpha p/x_i=\mu$. Suppose $b>a>c$ and relabel cyclically so $x_1$ is maximal. If $x_2\ge x_3$, subtracting equations one and two makes the cubic side positive unless all are equal, whereas $\alpha p(1/x_1-1/x_2)$ is nonpositive. If $x_2\le x_3$, subtracting equations two and three gives the opposite sign contradiction. Interchanging the orientation handles $c>a>b$. Thus all positive equilibria have equal amplitudes. With $\alpha \ne 0$, a state with exactly two nonzero amplitudes cannot be an equilibrium because it immediately forces the third.

For $A_i=r_i{e}^{i{\varphi }_i}$ with all $r_i>0$, the translation-invariant total phase is $\Phi ={\varphi }_1+{\varphi }_2+{\varphi }_3$. With $\alpha >0$, the displayed equation makes $\Phi =0$ stable and $\Phi =\pi$ unstable. A spatial translation can remove two phases, so the stable locked states have all amplitudes positive real in a suitable origin. This is a statement modulo translation, not that every translated representative is real; the zero solution and rolls belong to boundary strata.