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Write $A=V/D_3$ and $B=L/D_1$. The search time is $A/\lambda +B\lambda$: it decreases from infinity at small positive $\lambda$, reaches one minimum, and increases without bound at large $\lambda$ within this idealized model.

During a facilitated diffusion search, roughly $L/\lambda$ separate DNA landing-and-sliding episodes sample a chromosome. A three-dimensional encounter with any chromosomal landing region has characteristic time $V/(D_{3}L)$, so the total bulk-search contribution is $(L/\lambda )V/(D_{3}L)=V/(D_3\lambda )$. Each one-dimensional diffusion excursion lasts order ${\lambda }^2/D_1$; multiplying by $L/\lambda$ gives $L\lambda /D_1$. Longer sliding therefore reduces the number of bulk relocations but increases time spent repeatedly exploring local DNA. Numerical constants and overlap corrections are already omitted in the given model.

Differentiating and using the positive second derivative givesAt ${\lambda }_{\ast }$ the two time contributions are equal. The optimum is meaningful only if ${\lambda }_{\ast }$ lies between the microscopic capture scale and the available chromosomal contour length.

A decrease in DNA concentration means a decrease in $L/V$, but its effect on the time is not determined by that ratio alone. At fixed chromosome length $L$, dilution by increasing $V$ increases $\tau$ at fixed $\lambda$ and increases ${\tau }_{\min }$ as $\sqrt{V}$; bulk relocation becomes slower. If instead $L$ decreases at fixed $V$, while the target is retained, the stated model predicts a smaller sliding contribution and a smaller optimized time, proportional to $\sqrt{L}$. In both cases ${\lambda }_{\ast }\propto (L/V{)}^{-1/2}$. This distinction prevents attributing a unique time trend to concentration without specifying how it changes.

The left panel normalizes the search law to $\tau /{\tau }_{\min }=(u+u^{-1})/2$, $u=\lambda /{\lambda }_{\ast }$. The other panels illustrate b and c.

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.