 Solution

Write the deterministic drift of the Adler phase equation as $f(\theta )=\omega -ϵ\sin \theta$. Its minimum is $\omega -ϵ$ and its maximum is $\omega +ϵ$. For $0<\omega <ϵ$, the zero condition has exactly two solutions in the specified interval:The linearization of a dynamical system at an equilibrium point gives $\delta \dot{\theta }=f^{\prime }({\theta }_{\ast })\delta \theta$, with $f^{\prime }=-ϵ\cos \theta$. Defining $\kappa =\sqrt{{ϵ}^2-{\omega }^2}$ gives $f^{\prime }({\theta }_s)=-\kappa <0$, so ${\theta }_s$ is stable; $f^{\prime }({\theta }_u)=+\kappa >0$, so ${\theta }_u$ is unstable. For $\omega >ϵ$, $f$ is positive everywhere and there are no equilibrium points: the phase runs continuously. This is the distinction between phase locking and running phase dynamics.

With mobility scaled to one, the effective force is $-V^{\prime }$. Therefore the tilted washboard potential isFor $\omega <ϵ$, its alternating local minima and maxima trap noise-free trajectories in wells. Minima coincide with ${\theta }_s+2\pi j$, and maxima with ${\theta }_u+2\pi j$. For $\omega >ϵ$, $V^{\prime }=ϵ\sin \theta -\omega <0$ everywhere: there are no wells and the particle slides down the tilt. The potential is defined on the unwrapped phase and obeys $V(\theta +2\pi )=V(\theta )-2\pi \omega$; it is not a single-valued periodic equilibrium potential on the circle.

At the transition $\omega =ϵ$, the two equilibrium points merge at $\theta =\pi /2$ in a saddle-node bifurcation. There $f\simeq ϵ(\theta -\pi /2{)}^2/2$, so the point is attracting from the left and repelling from the right; a zero linear derivative alone does not establish stable trapping.