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With an absorbing boundary condition for a diffusion on each vertical side, the splitting probability $u(x)$ of hitting the left side before the right is harmonic for the Kolmogorov backward equation:For $a_1\ne 0$,The particle starts at $x=0$, soThe continuous limit at zero is $g(0)=1/2$, as required by reflection symmetry. A large positive drift drives the particle toward the right and gives $g(a_1)\to 0$; a large negative drift drives it toward the left and gives $g(a_1)\to 1$.

In this polymerization Brownian ratchet model, a gap of width $\delta$ immediately accommodates one monomer. That event advances the tip by $\delta$ and resets the gap to zero. The threshold is therefore an absorbing boundary condition for a diffusion,All waiting-time probability lies within $0\le x<\delta$, so the steady occupation probability density satisfiesThere is a reinjection flux $j_0$ at $x=0$ equal to the absorption flux at $x=\delta$. The nonpenetration constraint at zero is a reflecting boundary condition for a diffusion during each individual waiting episode. For the stationary reset process it must not be confused with setting the net current $J(0^{+})$ to zero: the source at reset supplies $J(0^{+})=j_0$. This is how a normalized steady density can coexist with a nonzero exit flux.

Put $k=\beta F$ and $q=k\delta$. The absorbing boundary condition for a diffusion gives $A=j_0{e}^q/(Dk)$, henceThis is nonnegative throughout the allowed interval. Normalizing the probability density givesThereforeEquivalently the normalized density isThe denominator is positive for $q>0$, since $e^q>1+q$. The dimensions are consistent: $j_0$ has units of inverse time, $A$ and $p$ of inverse length, and $q$ is dimensionless.