= Solution
For the <driven two-state rotary motor>, use positive <angular velocity> $\omega>0$ and extend all angular functions with period $2\pi$. In the first half-turn the source figure has $k_{AB}=k_0$, $k_{BA}=0$, so the stationary equation is $\omega P_A'=-k_0P_A$. In the second half-turn the rates are interchanged, giving $\omega P_A'=k_0(1-P_A)$. Hence
$$
P_A(\theta)=\begin{cases}
\alpha e^{-\beta\theta},&0\le\theta\le\pi,\\
1-Ce^{-\beta(\theta-\pi)},&\pi\le\theta\le2\pi,
\end{cases}\qquad\beta=\frac{k_0}\omega.
$$
<Continuity> at $\pi$ gives $1-C=\alpha e^{-\beta\pi}$, while <periodicity> gives $\alpha=1-Ce^{-\beta\pi}$. Solving the two linear conditions gives $C=\alpha$ and
$$
\boxed{\alpha=\frac1{1+e^{-\pi k_0/\omega}},\qquad
\beta=\frac{k_0}{\omega},\qquad
P_A(\theta+\pi)=1-P_A(\theta),\qquad P_B=1-P_A.}
$$
In particular $P_A(0)=P_A(2\pi)=\alpha$ and $P_A(\pi)=1-\alpha$, so the two pieces fill in the apparent endpoint gap in the printed piecewise formula. The <probability> decreases exponentially during the detachment half-cycle and rises exponentially during the attachment half-cycle; $P_B$ is its complement. Both remain between zero and one. They are state <probabilities> \i[conditional on rotor angle], not separate angular densities normalized individually to one.
For $\omega\ll k_0$, attachment and detachment almost finish within their allotted half-turns: $P_A$ is nearly zero through most of $0<\theta<\pi$ and nearly one through most of $\pi<\theta<2\pi$, with angular <boundary layers> of width $\omega/k_0$. For $\omega\gg k_0$, it stays near one half with a small nearly triangular modulation. The upper panels of the original figure in part (b) sketch both limits. At exactly zero speed the advective stationary problem is singular; these statements concern the positive-speed limit.
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