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The linearized complex equation couples a perturbation and its conjugate, so a single complex Fourier mode by itself is not a closed perturbation ansatz. Equivalently use real amplitude and wave phase perturbations, with Fourier modes paired to make each real field. Writing $A=A_0+\alpha +i\beta$ and $B=A_0^2$ givesA real amplitude mode of slow wavenumber $\ell >0$ therefore becomes neutral whenFor a nonzero uniform state the equilibrium point condition eliminates $\mu$, leaving amplitude growth $6\hat{s}B-40B^2-4{\ell }^2$. The wave phase mode growth is simply $-4{\ell }^2$, so it cannot give a nonzero-wavenumber bifurcation from these undetuned states.

The concave quadratic has maximumThus a nonzero uniform state needs $\ell \le 3\hat{s}/(4\sqrt{10})$ to have a modulation bifurcation. The slow domain length is ${\epsilon }^{2}L$, so ${\ell }_n=2\pi n/({\epsilon }^{2}L)$. If its smallest positive wavenumber already exceeds this bound, none can become neutral. Since $s={\epsilon }^2\hat{s}$, the conclusion isThis is the modulation cutoff of a cubic-quintic uniform pattern. If the bound is reversed, candidate squared amplitudes are $B=(3\hat{s}\pm \sqrt{9{\hat{s}}^2-160{\ell }^2})/40$. They lie on the lower branch; the upper branch is already damped in amplitude and cannot undergo this stationary modulation bifurcation.

The word “nonzero” is necessary in the final claim. At $A_0=0$, the equilibrium point exists independently of $\mu$, and the mode growth is $\mu -4{\ell }^2$. It becomes neutral at $\mu =4{\ell }_n^2$ for every finite domain. For example, take $\hat{s}=1$, ${\epsilon }^{2}L=20$: then $L=20/{\epsilon }^2<8\pi \sqrt{10}/(3s)$, but the zero state has a first-sideband threshold at $\mu ={\pi }^2/25$. Therefore the printed assertion, if read to include the trivial state, is false. The derived domain-size bound is the correct statement for nonzero pattern branches.