 Solution

For a uniform distribution of phases along one period, average the ensemble force $B{F}_{\mathrm{bound}}$:Using the negligible-carry-over form in b givesConsequentlywhere $\boxed{B_0=k_1/(k_1+k_2),\alpha =(k_1+k_2)/v,\beta =k_2/v}$. This is the printed expression to the accuracy justified by the velocity condition.

It is not an exact identity for the periodic problem at arbitrary $v$. Keeping $b_{\mathrm{in}}$ instead givesFor a concrete counterexample outside the small-velocity approximation, choose $k_1=k_2=P=G_0=v=1$ and $x_1=1/2$. Then the exact periodic value is about $+0.01976$, whereas the printed reset expression is about $-0.06478$. The discrepancy is due to nonzero incoming occupancy, not an integration sign change.

As $v\to 0^{+}$, both $\alpha$ and $\beta$ diverge, and both formulas giveThe bound population occupies the downhill binding region, while occupancy on the uphill region is confined to a vanishing boundary layer. A nonzero limiting stall force is compatible with zero mechanical power $v⟨F⟩$. The spatially restricted binding and uniform unbinding do not obey local detailed balance for this sawtooth free energy: a passive equilibrium model would require $k_1(x)/k_2(x)$ to reflect the local energy difference and not vanish on an interval of finite energy. This is a driven or prescribed-kinetics motor model; the equilibrium relation in a cannot be imposed on it without an energy supply and additional states.