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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.