= Solution
Take $v>0$ and the displayed <potential energy> to be the position-dependent mechanical <free energy> $G(x)$ of a bound motor. On the descending ramp $G'= -G_0/x_1$, and on the ascending ramp $G'=G_0/(P-x_1)$. Thus the <force> conditional on being bound is
$$
\boxed{F_{\rm bound}(x)=-G'(x)=\begin{cases}
G_0/x_1,&0<x<x_1,\\
-G_0/(P-x_1),&x_1<x<P.
\end{cases}}
$$
It is periodic, piecewise constant and undefined at idealized sharp corners; the corners have zero measure in averages. An unbound motor has zero <force>.
A cohort transported through position $x$ obeys $d/dt=v\,d/dx$. Its bound <probability> therefore satisfies
$$
vB'=k_1(1-B)-k_2B\quad(0<x<x_1),\qquad
vB'=-k_2B\quad(x_1<x<P).
$$
Define $B_0=k_1/(k_1+k_2)$, $\alpha=(k_1+k_2)/v$, $\beta=k_2/v$, $a=e^{-\alpha x_1}$ and $q=e^{-\beta(P-x_1)}$. The exact periodic <steady state> includes a small carry-over $b_{\rm in}=B(0)=B(P)$:
$$
b_{\rm in}=\frac{qB_0(1-a)}{1-aq},\qquad
B(x)=\begin{cases}
B_0+(b_{\rm in}-B_0)e^{-\alpha x},&0<x<x_1,\\
\displaystyle\frac{B_0(1-a)}{1-aq}e^{-\beta(x-x_1)},&x_1<x<P.
\end{cases}
$$
These expressions follow by solving each first-order <ordinary differential equation>, requiring continuity at $x_1$, and imposing periodicity.
The stipulated $v\ll k_2(P-x_1)$ implies $q\ll1$, so the incoming bound fraction is exponentially small. Dropping it gives the simpler intended forms
$$
\boxed{B(x)\simeq\begin{cases}
B_0(1-e^{-\alpha x}),&0<x<x_1,\\
B_0(1-e^{-\alpha x_1})e^{-\beta(x-x_1)},&x_1<x<P.
\end{cases}}
$$
The occupancy rises towards $B_0$ on the binding interval, then decays over length $v/k_2$ outside it. The mean <force> per member of the full ensemble at a given position is $B(x)F_{\rm bound}(x)$, distinct from the <force> per bound motor. The right panels of <1/c> show both the occupancy and the conditional <force>.
Back to article page
