= Solution
Write the deterministic drift of the <Adler phase equation> as $f(\theta)=\omega-\epsilon\sin\theta$. Its minimum is $\omega-\epsilon$ and its maximum is $\omega+\epsilon$. For $0<\omega<\epsilon$, the zero condition has exactly two solutions in the specified interval:
$$
\boxed{\theta_s=\alpha=\arcsin(\omega/\epsilon),\qquad \theta_u=\pi-\alpha.}
$$
The <linearization of a dynamical system> at an <equilibrium point> gives $\delta\dot\theta=f'(\theta_*)\delta\theta$, with $f'=-\epsilon\cos\theta$. Defining $\kappa=\sqrt{\epsilon^2-\omega^2}$ gives \b[$f'(\theta_s)=-\kappa<0$, so $\theta_s$ is stable; $f'(\theta_u)=+\kappa>0$, so $\theta_u$ is unstable]. For $\omega>\epsilon$, $f$ 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 $-V'$. Therefore the <tilted washboard potential> is
$$
\boxed{V(\theta)=-\omega\theta-\epsilon\cos\theta+C.}
$$
For $\omega<\epsilon$, 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 $\omega>\epsilon$, $V'=\epsilon\sin\theta-\omega<0$ 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.
\Image[/past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2015/iii/paper-75-phase-dynamics.png]
{title=Locked and running Adler phase dynamics, with drift zeros and the corresponding tilted potentials}
{height=600}
At the transition $\omega=\epsilon$, the two <equilibrium points> merge at $\theta=\pi/2$ in a <saddle-node bifurcation>. There $f\simeq\epsilon(\theta-\pi/2)^2/2$, so the point is attracting from the left and repelling from the right; a zero linear derivative alone does not establish stable trapping.
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