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The Mean first-passage time to a target $A$ is $u(x)={\mathbb{E}}_x{T}_A$, which can be infinite. For a diffusion with backward generator $L$, a finite sufficiently regular mean solves the Poisson equationoutside $A$, with $u=0$ on the absorbing target. Reflecting parts of the boundary require the reflecting condition of the generator, rather than an absorbing value. These conditions express different physical rules and cannot be exchanged.

For Brownian diffusion with generator $D{d}^2/d{x}^2$ on $[0,\ell ]$, $D>0$, absorption at both endpoints givesIf zero instead reflects while $\ell$ absorbs, the boundary conditions are $u^{\prime }(0)=0$, $u(\ell )=0$ and the solution is $u(x)=({\ell }^2-x^2)/(2D)$. In particular the mean time from the reflecting endpoint is ${\ell }^2/(2D)$, displaying the squared-distance diffusion scale.

Dynkin's formula explains the equation: before absorption, $u(X_{t\wedge T_A})+(t\wedge T_A)$ has constant expected value. For these bounded-interval nonnegative bounded solutions, this identity first bounds ${\mathbb{E}}_x(t\wedge T_A)$ by $u(x)$, proving integrability of $T_A$; passing to the limit then proves $u(x)={\mathbb{E}}_x{T}_A$. Boundary equations without a finiteness or minimality condition can have extraneous solutions on unbounded domains.

The same geometric series identity givesA direct renewal check gives the identical result: $T_B={\tau }_B+T_I$ and $T_I={\tau }_I+q{T}_B$, so $p{T}_B={\tau }_B+{\tau }_I$. This also checks the units: mean first-passage time has units of time, unlike the dimensionless step count.

The initial slope near one is the short-observation regime $\mathcal{T}(T)\sim T/2$. Successively longer observations include trajectories that enter progressively longer-lived traps before reaching the absorbing minimum. The four marked shoulders represent successive relaxation timescales, not four distinct equilibrium phases. A plateau appears when the contributing faster modes have largely completed but the next slow component has not yet substantially changed the conditional mean. The ultimate plateau is the true mean first-passage time, approximately $e^{32.3}$ in the plotted time units.

To estimate a rate, locate the rounding timescale at which that mode stops growing with the observation window, using ${\nu }_{l}T\sim 1$; the final completion of a crossover requires several lifetimes. Graphical transitions lie approximately around $\ln T\sim 8$, 23, 29 and 35, giving only order-of-magnitude estimatesThese are roughly $3\times 10^{-4}$, $10^{-10}$, $3\times 10^{-13}$ and $6\times 10^{-16}$, with factors of several uncertainty from reading broad crossovers. Plateau heights are weighted sums $\sum A_l/{\nu }_l$ divided by completed probability, not individually $1/{\nu }_l$; in particular the final plateau at $\ln \mathcal{T}\simeq 32.3$ does not require the slowest lifetime to be $e^{32.3}$. The table fixes barriers and basin connectivity, but supplies no temperature or attempt-frequency convention that would justify exact rates directly from the energies.

The mean first-passage time obeys the backward equationFor $\alpha \ne 0$, direct integration givesIn particular,This increases monotonically with drift away from the target, so the constrained optimum isExpanding the exponential at zero drift yields

The adsorption mechanism is uniform along the two vertical sides, and $X$ evolves independently of $Y$. Consequently the mean first-passage time depends only on the initial $x$-coordinate. Its Kolmogorov backward equation for $a_1=1$ is

The forward partially absorbing boundary condition for a diffusion is $\mathbf{J}\cdot \mathbf{n}=\kappa p$. The boundary term in the adjoint relation isbecause the $X$ diffusion coefficient is one. It vanishes for every admissible $p$ precisely whenFor $\kappa =1$, the two Robin boundary conditions are therefore

The general solution of the ordinary differential equation isThe right condition gives $C=2$, and the left gives $D=-2/e$. At the prescribed initial position,

Each addition starts a new statistically identical waiting episode at zero gap. Thus the addition rate is the reciprocal of a mean first-passage time, not the reciprocal of the mean gap:This can also be derived without the steady density. If $T(x)$ is the mean time to reach $\delta$ starting at gap $x$, its backward Smoluchowski equation isThe first boundary is reflecting and the second absorbing. An integrating factor gives $T^{\prime }(x)=[1-e^{\beta Fx}]/(D\beta F)$, whenceAt weak force, $\tau \simeq {\delta }^2/(2D)$ is the time for a freely diffusing wall to move a monomer length, with reflection at the tip. At strong force, opening the gap requires work $F\delta$, with Boltzmann distribution suppression $e^{-\beta F\delta }$. The drift establishes a near-tip thermal length ${\ell }_F=k_{B}T/F$ and a diffusion time ${\ell }_F^2/D$; consequentlyThe exponentially long wait explains why the growth slows. There is no additional chemical waiting time in the imposed absorbing-threshold model; such a time would be needed to describe the separate reaction-limited polymerization Brownian ratchet regime.

At $\alpha =0$ the transverse coordinates are independent Brownian motions with diffusion coefficient $D=1$. The cylinder is infinitely long, so the axial coordinate cannot affect its first contact time. By rotational invariance of Brownian motion, the mean first-passage time $u$ depends only on the transverse radius $\rho =\sqrt{x^2+y^2}$.

The Kolmogorov backward equation for a mean first-passage time is $Lu=-1$. Here $L={\partial }_x^2+{\partial }_y^2$, so the radial boundary value problem isThe absorbing boundary condition for a diffusion is imposed on the circular cross-section. At $\rho =0$, the solution must extend smoothly over the axis. Multiplying by $\rho$ and integrating givesSmoothness at the axis excludes the logarithm term, and $u(R)=0$ fixes $C_0=R^2/4$. ConsequentlyThis also verifies the finiteness assumption behind the Kolmogorov backward equation. Let $T$ be the first contact time and apply the stopped Itô formula to the bounded nonnegative function $u$ just obtained. Its bounded derivatives on the disk make the stopped stochastic integral a martingale, givingBy monotone convergence, $T$ has finite expected value; continuity and dominated convergence then make the final term vanish as $t\to \infty$. Hence $u({\hat{r}}_0)=\mathbb{E}T$. In particular, the mean first-passage time is $R^2/4$ on the axis and tends to zero at the adsorbing wall.

For overdamped Brownian motion at uniform temperature in a potential energy $U$, constant diffusion coefficient $D$ gives the displayed Fokker-Planck equation, with $\beta =(k_{B}T{)}^{-1}$. The probability current is $\mathbf{J}=-D(\nabla p+\beta p\nabla U)$. Zero current recovers the Boltzmann distribution; absorbing boundaries instead permit transport and mean first-passage times. This diffusion equation should be distinguished from the similarly named coagulation equation.