 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 givesThe imaginary equation implies $R^2{\varphi }^{\prime }=J$, a constant. The radial first integral isA regular solution has finite $E$. As $R\to 0$, the nonnegative $J^2/(2R^2)$ term would diverge unless $J=0$. Hence ${\varphi }^{\prime }=0$ 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, $R^{\prime }$ 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 $\mu$ givesThis 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 $R^{\prime }=-R(q-R^2)/\sqrt{3}$, which integrates toIt connects the stable upper pattern at $x\to -\infty$ to conduction at $x\to +\infty$, with arbitrary translation $x_0$ and constant phase ${\varphi }_0$.