= Solution
For the intended nontrivial interface, take the limiting amplitude at the patterned end to be $R_->0$; the identically zero solution is otherwise a trivial exception. Splitting the stationary <real-coefficient cubic-quintic amplitude equation> into real and imaginary parts gives
$$
R''-R(\phi')^2+\mu R+\alpha R^3-R^5=0,\qquad 2R'\phi'+R\phi''=0.
$$
The imaginary equation implies $R^2\phi'=J$, a constant. The radial first integral is
$$
\frac12(R')^2+\frac{J^2}{2R^2}+V(R)=E,\qquad V(R)=\frac\mu2R^2+\frac\alpha4R^4-\frac16R^6.
$$
A regular solution has finite $E$. As $R\to0$, the nonnegative $J^2/(2R^2)$ term would diverge unless $J=0$. Hence \b[$\phi'=0$ wherever the nontrivial front has positive amplitude]. This conclusion does not assume in advance that the patterned end has no phase gradient.
For a <heteroclinic orbit> connecting two constant states, $R'$ tends to zero at both ends. Thus $E=V(0)=0$, while the nonzero end must obey both $\mu+\alpha q-q^2=0$ and $V(\sqrt q)=0$. Eliminating $\mu$ gives
$$
0=\frac{q^3}{3}-\frac{\alpha q^2}{4},\qquad \boxed{q=\frac{3\alpha}{4},\quad\mu_M=-\frac{3\alpha^2}{16}<0.}
$$
This is the <stationary front of a cubic-quintic amplitude equation>: a unique equal-potential parameter in the bistable interval.
Existence, not just a necessary condition, follows from the first integral. At $\mu_M$, $V(R)=-R^2(R^2-q)^2/6$. The descending solution obeys $R'=-R(q-R^2)/\sqrt3$, which integrates to
$$
\boxed{A(x)=e^{i\phi_0}\sqrt{\frac{3\alpha/4}{1+\exp[\sqrt3\alpha(x-x_0)/2]}}.}
$$
It connects the stable upper pattern at $x\to-\infty$ to conduction at $x\to+\infty$, with arbitrary translation $x_0$ and constant phase $\phi_0$.
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