 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 $\omega P_A^{\prime }=-k_0{P}_A$. In the second half-turn the rates are interchanged, giving $\omega P_A^{\prime }=k_0(1-P_A)$. HenceContinuity at $\pi$ gives $1-C=\alpha e^{-\beta \pi }$, while periodicity gives $\alpha =1-C{e}^{-\beta \pi }$. Solving the two linear conditions gives $C=\alpha$ andIn 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 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.