 Solution

Take the leading real field to be $A{e}^{i(k_{c}x+{\omega }_{0}t)}+B{e}^{i(-k_{c}x+{\omega }_{0}t)}+\text{complex conjugate}$. A spatial translation gives $(A,B)\mapsto (e^{i{k}_c\delta }A,e^{-i{k}_c\delta }B)$; a time translation gives $(A,B)\mapsto (e^{i\psi }A,e^{i\psi }B)$; spatial reflection exchanges $A$ and $B$.

For a monomial $A^j{\overline{A}}^k{B}^l{\overline{B}}^m$ in the first amplitude equation, covariance requires $j-k+l-m=1$ and $j-k-l+m=1$. Hence $j-k=1$, $l-m=0$. At linear order only $A$ is allowed, no quadratic monomial satisfies these conditions, and the cubic possibilities are $A^2\overline{A}$ and $AB\overline{B}$. Reflection fixes the coefficients in the second equation. Removing the small common linear frequency correction by a rotating frame makes $\mu$ real. This derives the two cubic amplitude equations, rather than merely postulating them.

Let $r=|A{|}^2$, $s=|B{|}^2$. Taking real parts givesA travelling wave has $(r,s)=(\mu /{\beta }_r,0)$, or its reflection, whenever $\mu /{\beta }_r>0$. Its phase evolves at $-{\beta }_i\mu /{\beta }_r$, shifting the physical oscillation frequency. The nonzero-amplitude radial eigenvalue is $-2\mu$, while the absent mode has real growth rate $\mu (1-{\gamma }_r/{\beta }_r)$. Therefore, on the supercritical side,The phase direction is neutral because of the symmetry, so stability means orbital stability. A branch with $\mu <0$ is radially unstable, whatever the transverse coefficient.

A standing wave has $r=s=h=\mu /({\beta }_r+{\gamma }_r)>0$. Its two phases evolve together at $-({\beta }_i+{\gamma }_i)h$; the relative phase fixes the spatial position of the standing pattern. In the two intensity directions the eigenvalues are $-2\mu$ and $-2h({\beta }_r-{\gamma }_r)$. ThusAgain the symmetry phases are neutral and the conclusion is orbital stability. Equality ${\beta }_r={\gamma }_r$ leaves a continuum of mixed intensities at cubic order; equality ${\beta }_r+{\gamma }_r=0$ removes cubic standing-wave saturation. These degeneracies require higher-order terms and cannot be classified by the cubic truncation. The zero state is linearly stable for $\mu <0$ and unstable for $\mu >0$.