 Solution

To find the threefold phase-locked equilibria, write $C=R{e}^{i\theta }$ with $R>0$; separating real and imaginary parts givesA steady state satisfiessoWriting $y=R_0^2$ and $D=1+4\mu -4{\omega }^2$, the amplitude branches areThere are two distinct positive roots for $\mu >{\omega }^2-1/4$, except at $(\mu ,\omega )=(0,0)$, where $y_{-}=0$ and only the upper root is nonzero. Indeed their sum is positive in this range and their product is ${\mu }^2+{\omega }^2$. Below the fold condition there are none. At equality, there is one positive repeated root $y={\omega }^2+1/4$, the saddle-node bifurcation limit.

For each positive amplitude, the sine and cosine determine $3{\theta }_0$ modulo $2\pi$, giving three phases separated by $2\pi /3$. Thus the source's “two states” means two amplitude branches modulo the threefold spatial symmetry. Generically there are six nonzero complex equilibria, three on each branch, not literally two.

The polar Jacobian matrix at an equilibrium isOn the lower branch, $\det J=-3y_{-}\sqrt{D}<0$, so it is a saddle equilibrium and unstable. On the upper branch, $\det J=3y_{+}\sqrt{D}>0$ andbecause existence implies $\mu >-1/4$. Therefore every upper-branch equilibrium is asymptotically stable, and every lower-branch equilibrium is a saddle equilibrium, away from the degenerate endpoints. The eigenvalues are unchanged by the smooth polar coordinate transformation at $R_0>0$. The origin, not covered by those coordinates, has eigenvalues $\mu \pm i\omega$ and is stable for $\mu <0$ and unstable for $\mu >0$.