= Solution
For the <real-coefficient cubic-quintic amplitude equation>, the trivial pattern $A=0$ exists for every $\mu$. A nonzero spatially uniform pattern has arbitrary constant phase and intensity $q=|A|^2>0$ satisfying $\mu+\alpha q-q^2=0$. Thus
$$
\boxed{q_\pm=\frac{\alpha\pm\sqrt{\alpha^2+4\mu}}2.}
$$
Nonzero patterns exist for $\mu\ge-\alpha^2/4$. The upper branch $q_+$ exists throughout that range; a distinct positive lower branch $q_-$ exists only for $-\alpha^2/4<\mu<0$ and reaches zero at $\mu=0$.
For a homogeneous perturbation, write $A=Re^{i\phi}$. Then $\dot R=R(\mu+\alpha R^2-R^4)$ and $\dot\phi=0$. The radial eigenvalue at a nonzero pattern is
$$
\lambda_R=\mu+3\alpha q-5q^2=2q(\alpha-2q),
$$
while the phase eigenvalue is zero. Therefore \b[the upper branch is stable modulo its arbitrary phase for $\mu>-\alpha^2/4$; the lower branch is unstable.] The phase-neutral direction prevents asymptotic attraction to one particular complex amplitude, but permits <orbital stability> of the circle of equivalent patterns.
Conduction is asymptotically stable for $\mu<0$ and unstable for $\mu>0$. At $\mu=0$, the positive cubic term destabilizes arbitrarily small nonzero amplitudes, even though its linear eigenvalue vanishes. At $\mu=-\alpha^2/4$, the merged nonzero branch is semistable from above only: $\dot R=-R(R^2-\alpha/2)^2$, so amplitudes just below it eventually decrease toward zero. Thus the bistable interval is \b[$-\alpha^2/4<\mu<0$], with the endpoints treated separately rather than declared stable merely because an eigenvalue is zero.
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