The escape dynamics of sticky particles from textured surfaces is poorly understood despite importance to various scientific and technological domains. In this work, we address this challenge by investigating the escape time of adsorbates from prevalent surface topographies, including holes/pits, pillars, and grooves. Analytical expressions for the probability density function and the mean of the escape time are derived. A particularly interesting scenario is that of very deep and narrow confining spaces within the surface. In this case, the joint effect of the entrapment and stickiness prolongs the escape time, resulting in an effective desorption rate that is dramatically lower than that of the untextured surface. This rate is shown to abide a universal scaling law, which couples the equilibrium constants of adsorption with the relevant confining length scales. While our results are analytical and exact, we also present an approximation for deep and narrow cavities based on an effective description of one-dimensional diffusion that is punctuated by motionless adsorption events. This simple and physically motivated approximation provides high-accuracy predictions within its range of validity and works relatively well even for cavities of intermediate depth. All theoretical results are corroborated with extensive Monte Carlo simulations.

The fabrication of nanoscale surface topographies has seen rapid development in the past two decades.1–3 In particular, controlled fabrication of nano-arrays can be achieved, where common structured surface topographies include nano-arrays of pillars,4–8 arrays of holes/pits,9,10 and grooves.11–13 Alternatively, the surface can be rugged and possess random roughness.14 

The effect of surface topography has proved to be a key aspect when considering heterogeneous catalysis14,15 and the passivation of catalytic surfaces.16,17 It also plays a cardinal role when considering living cell behavior, as protein adsorption to a textured surface mediates the cell attachment to the surface.1,2,18 Topographical features can affect the adsorption properties of a protein by inducing conformational changes, or by other forms of surface–protein interactions.19 When the length scale of the topographical features is larger than the protein size, additional effects come into play. Often, adsorption is increased as a larger number of active sites for protein adsorption are available. Another crucial effect is the entrapment of the proteins inside confined spaces.20,21

The entrapment effect was vividly illustrated in a series of on-chip devices made by the Patolsky group.22–24 These devices utilized entrapment in sticky confined spaces of textured surfaces for the purpose of selective separation of required protein analytes from raw biosamples. The selective stickiness was achieved by attaching specific antibodies to the surface. The surface was textured by a vertical array of nanopillars, albeit other topographies, such as grooves, are expected to exhibit a similar behavior. The Patolsky group demonstrated that the target proteins are entrapped in the surface for extremely long times (weeks and even months). This came as a surprise, since the same antibody, if used on a flat surface, would bind the biomolecules for a few milliseconds only. Similarly, using the nanopillar vertical array without antibodies leads to fast diffusive escape. The dramatic effect of prolonged escape times is hence due to a combination of topography and adsorption/stickiness. A semi-quantitative explanation of the experimental results was given in Ref. 22.

Qualitatively, when the confining space is deep and narrow, the escaping particle is forced to collide with the confining walls a large number of times before it can escape. Each collision can result in an adsorption event, and these add up and eventually culminate in extremely prolonged escape. Thus, despite the relatively short dissociation time from the antibody, and due to the multitude of adsorption events, the textured surface appears as if it has a very high affinity to the protein. This observation is an important first step, but a more detailed quantitative understanding is currently missing. The challenge is thus to determine how exactly does surface topography and texture affect the escape from a surface. Specifically, how does the mean escape time scale with the depth and width of confining spaces within the surface and how does sticky entrapment affect the statistics of the escape time, as characterized by its probability density function (PDF)?

Recently, we have developed an analytical approach that allows one to provide an exact solution to the aforementioned problems.25 We considered the escape of a diffusing particle from a domain of arbitrary shape, size, and surface reactivity. The escape time from the adsorbing confining spaces of a textured surface can be computed using this formalism. Here, we perform this calculation for three different topographies of adsorbing surfaces: (i) a surface perforated with pits/holes is considered in Sec. II, (ii) a surface textured by an array of pillars is considered in Sec. III, and (iii) a surface textured by grooves is considered in Sec. IV.

For each of the above-mentioned cases, we aim to find the escape time of a particle initially entrapped in the confining spaces of the textured surface. We assume that the surface is effectively infinite and homogeneously textured, i.e., all the confining spaces are of the same size and repeated periodically. Thus, the escape time out of the periodic cell, in fact, is equal to the escape time from the textured surface. Note that in some cases, e.g., a surface perforated with holes, the assumption of periodicity can be easily relaxed—it is the homogeneity which is important. Finally, while in this work we consider homogeneously textured surfaces, the same formalism can be used when dealing with heterogeneous textured surfaces with a known size distribution of the confining spaces: The escape time from the textured surface will then be the appropriately weighted sum of the escape times from cells of different sizes.

As mentioned, we assume that the surface is effectively infinite, namely, large in comparison to the unit cells that comprise it. For example, the area of the on-chip devices that motivated our study is ∼1 cm2, while their unit cells are only ∼0.5 µ m2, i.e., orders of magnitude smaller.22 In this limit, one can safely neglect the finiteness of the device, and identify the escape time from the surface with the escape time from a single unit cell with periodic boundary conditions. Note that this approach provides excellent approximations even for small finite textured patches, given, e.g., reflecting boundary conditions at the lateral edges of the surface. Corrections to the results derived below should be introduced only when dealing with small textured patches in which lateral escape through the surface edges is possible. These are not included in the analysis below.

Alongside exact results, we also present an insightful approximation. In Sec. II C, we introduce a two-state switching diffusion approximation for the diffusive escape from sticky nanocavities. This approximation is appropriate for deep and narrow cavities, where diffusion is effectively one-dimensional. The adsorption to the surface is then effectively accounted for by the introduction of an immobile state for which the diffusion coefficient vanishes. We illustrate that this approximation is very accurate and works well even for cavities of intermediate depth. We utilize this approximation yet again in Sec. IV E, where we calculate the asymptotic decay rate of the escape time PDF. Indeed, the approximation is expected to work in the limit of very deep cavities regardless of the lateral geometry. Thus, a central benefit of the two-state switching diffusion approximation is that it captures the essential physics of the problem at hand and simplifies the analysis without losing much in accuracy.

The three problems solved here abide similar laws and show a similar characteristic behavior. In Sec. V, we discuss a general form of the equation for the mean escape time and for its inverse, which is the effective desorption rate from the textured surface. This suggests that the results presented here are universal in nature and can be applied, even if approximately, when considering more complicated scenarios.

We consider a perforated surface with cylindrical holes, illustrated in Fig. 1. We aim to find the escape time of a particle initially entrapped inside one hole of the surface. Our task is thus calculating the escape time from the periodic cell. For the example under consideration here, the periodic cell is a cylindrical hole. A representative hole is enlarged in Fig. 1. Escaping from the textured surface is thus equivalent to the escape problem in three dimensions when the particle diffuses inside a cylinder of radius L capped by parallel planes at z = 0 and z = H. The top disk is absorbing, whereas the remaining boundary of the domain is adsorbing. We assume that the adsorption kinetics is linear and homogeneous on each surface, but we allow for heterogeneity in the sense that the adsorption and desorption rates on the bottom disk and on the curved cylindrical surface can differ. We are interested in finding the PDF of the first-passage time to the top disk, which can also be thought of as the escape time from the cylindrical compartment. We denote this PDF as Jab(t|r, z), where (r, z) ∈ Ω is the initial location of the particle inside the cylindrical domain Ω. In Ref. 25, we have derived, for the general case, the partial differential equation and boundary conditions governing the Laplace transform of this PDF, J̃ab(s|r,z)=0dtetsJab(t|r,z). For the specific geometry considered here, these equations simplify to
(sDΔ)J̃ab(s|r,z)=0(r,zΩ),
(1a)
J̃ab(s|r,z)=1(z=H),
(1b)
(z+qs)J̃ab(s|r,z)=0(z=0),
(1c)
(r+qs)J̃ab(s|r,z)=0(r=L),
(1d)
where Δ=r2+(1/r)r+z2 is the Laplace operator in cylindrical coordinates (without the angular part) and D is the diffusion coefficient. The surfaces are characterized by the parameters qs and qs,
qs=kaD(1+kd/s),qs=kaD(1+kd/s),
(2)
where ka and kd are the adsorption and desorption constants for the bottom disk and ka and kd are the adsorption and desorption constants for the cylindrical surface.
FIG. 1.

A perforated surface with cylindrical holes. A small part of the surface is shown, in which one of the holes is enlarged: A cylinder of radius L capped by parallel planes at z = 0 and z = H. The top disk at z = H is absorbing (escape region in red), whereas the bottom disk at z = 0 and the cylindrical wall are adsorbing (green), with reversible binding kinetics. Here, ka and kd are the adsorption and desorption constants for the bottom disk, and ka and kd are the adsorption and desorption constants for the cylindrical surface.

FIG. 1.

A perforated surface with cylindrical holes. A small part of the surface is shown, in which one of the holes is enlarged: A cylinder of radius L capped by parallel planes at z = 0 and z = H. The top disk at z = H is absorbing (escape region in red), whereas the bottom disk at z = 0 and the cylindrical wall are adsorbing (green), with reversible binding kinetics. Here, ka and kd are the adsorption and desorption constants for the bottom disk, and ka and kd are the adsorption and desorption constants for the cylindrical surface.

Close modal
We search the solution of Eq. (1a) under the boundary conditions (1b)(1d) as
J̃ab(s|r,z)=2n=0cn(s)J0(αn(s)r̄)gn(s)(z)gn(s)(H),
(3)
where r̄=r/L, Jν(⋅) is the Bessel function of the first kind of order ν, and
gn(s)(z)=α̂n(s)cosh(α̂n(s)z̄)+qsLsinh(α̂n(s)z̄)
(4)
to respect the boundary condition (1c), with z̄=z/L and
α̂n(s)=[αn(s)]2+L2s/D.
(5)
From the boundary condition (1d), we find that αn(s) satisfy the transcendental equation
αn(s)J1(αn(s))J0(αn(s))=qsL.
(6)
For any s ≥ 0, there are infinitely many solutions of this equation that we enumerate by n = 0, 1, 2, … in an increasing order. The unknown coefficients cn(s) are found by multiplying the boundary condition (1b) by rJ0(αk(s)r/L) and integrating over r from 0 to L. This gives
2ck(s)L2J02(αk(s))+J12(αk(s))2=0LdrrJ0(αk(s)r̄)=L2J1(αk(s))/αk(s),
(7)
from which we get
cn(s)=J1(αn(s))/αn(s)J02(αn(s))+J12(αn(s)).
(8)
To obtain these relations, we used the orthogonality of the Bessel functions together with
01drrJ02(αr)=J02(α)+J12(α)2
(9)
and
01drrJ0(αr)=J1(α)α.
(10)

It is worth noting that the numerical inversion of the Laplace transform is challenging here; in fact, one needs to evaluate the solution at complex s, which, in turn, requires an improved algorithm for finding the roots αn(s), given that qs and qs become complex as well.

To facilitate further analysis, we introduce the following dimensionless quantities:
κa=kaLD,κd=kdL2D,κa=kaLD,κd=kdL2D.
(11)
For convenience, in Table I, we collect the definitions of all dimensionless quantities defined so far and also define h = H/L and ρ = l/L that will be used later.
TABLE I.

Summary of dimensionless quantities.

h = H/L 
ρ = l/L 
z̄=z/L 
r̄=r/L 
κa = kaL/D 
κd = kdL2/D 
κa=kaL/D 
κd=kdL2/D 
α̂n(s)=[αn(s)]2+L2s/D 
h = H/L 
ρ = l/L 
z̄=z/L 
r̄=r/L 
κa = kaL/D 
κd = kdL2/D 
κa=kaL/D 
κd=kdL2/D 
α̂n(s)=[αn(s)]2+L2s/D 

Here, we compute the mean escape time by studying the asymptotic behavior of J̃ab(s|r,z) as s → 0. In the spectral expansion (3), we first analyze the term n = 0 and then discuss the other terms with n > 0.

As s → 0, one has qs0 so that α0(s)0. Using the Taylor series expansion of Bessel functions in Eq. (6), one gets in the leading order
α0(s)2qsLs2κa/kd(s0).
(12)
As a consequence, one has
c0(s)=121+[α0(s)]28+=12+κa8kds+O(s2)
(13)
and
J0(α0(s)r/L)=1r2κa2L2kds+O(s2).
(14)
We also get α̂0(s)Ls/D1+2κa/κd in the leading order, from which we obtain
g0(s)(z)g0(s)(H)=1κakdHzL+L22D+κakdH2z2L2s+O(s2).
(15)
Let us now consider the terms with n > 0. Denoting the left-hand side of Eq. (6) as F1(z) = zJ1(z)/J0(z), we apply the Taylor expansion near αn(0)>0,
F1(αn(s))F1(αn(0))+F1(αn(0))(αn(s)αn(0)).
(16)
According to Eq. (6), F1(αn(s))=qsL0 as s → 0 and thus αn(0)=j1,n, where j1,n denote the zeros of the Bessel function J1(z). Plugging in z = j1,n into the relation F1(z)=z(1+J12(z)/J02(z)), we find that F1(j1,n)=j1,n. All that remains is to compare the right-hand side of Eqs. (6) and (16) in the limit s → 0, which gives
αn(s)=j1,n+κakdj1,ns+O(s2).
(17)
Similarly, setting F2(z)=J1(z)/z/J02(z)+J12(z) such that according to Eq. (8) we have cn(s)=F2(αn(s)), and using Eq. (17), we obtain
cn(s)=F2(αn(0))+F2(αn(0))(αn(s)αn(0))+O(s2)=1j1,nJ0(j1,n)κakdj1,ns+O(s2)
(18)
and
α̂n(s)=j1,n+1+2κa/κd2j1,nL2Ds+O(s2).
(19)
Since cn(s)s for n > 0, the other factors in Eq. (3) can be found to the lowest order in s. This gives
gn(s)(z)gn(s)(H)cosh(j1,nz̄)cosh(j1,nh),
(20)
where h = H/L and
J0(αn(s)r̄)J0(j1,nr̄).
(21)
Substituting these expressions into Eq. (3), we get
J̃ab(s|r,z)=1sT(r,z)+O(s2),
(22)
where
T(r,z)=H2z22D+ka(Hz)kdD+kaLkdDH2z2L2+2r2L24L22n=1J0(j1,nr̄)j1,n2J0(j1,n)cosh(j1,nz̄)cosh(j1,nh)
(23)
is the mean escape time.
In the limit ka0 or kd, namely when the cylindrical surface is not sticky but reflecting, we retrieve the mean escape time for a one-dimensional box with a sticky surface,25 
T(r,z)=12D(H2z2)+kakdD(Hz).
(24)
Averaging Eq. (23) over the cross section at a fixed height z yields
T̄(z)=2ππL20LdrrT(r,z)=kakdHzD+1+2kakdLH2z22D,
(25)
where we used 0LdrJ0(j1,nr/L)=0. This remarkably simple expression quantifies the effect of adsorption/desorption mechanisms onto the mean escape time. Further averaging over z, we obtain
Tu=H23D1+kakd32H+kakd2L,
(26)
where the subscript “u” denotes a uniform distribution of the initial position within the cylinder, which is a common experimental condition. Indeed, zooming out and assuming that the particle’s initial position is distributed uniformly inside the holes of the textured surface, it is equally likely to find the particle in any of the holes. Since the surface is homogeneous and the holes are all the same, the mean escape time from the textured surface is thus given by Eq. (26).

In Sec. II D, we will analytically invert Eq. (3) to get the PDF of the escape time. As this inversion is quite involved, it is worthwhile to first consider a simple approximation for the problem at hand: a model of two-state switching diffusion. This model is expected to approximate the escape from a perforated surface in the limit LH, i.e., when the holes are very narrow and deep. In this limit, assuming that the reactivities of all surfaces are comparable, the area of the lower disk becomes negligible compared to the area of the cylindrical surface. The bottom disk is thus expected to have little effect on the escape time, and, so, we can treat it as an inert reflecting surface (ka = 0).

When LH, one can expect that diffusion in the radial direction is not relevant and try to reduce the original model to a two-state switching diffusion model when the particle diffuses in the bulk with a diffusion coefficient D in the state 1, or remains immobile in state 2 (mimicking its adsorbed state). The transition between two states is a random first-order kinetics with rates k12 from the free state to the adsorbed state and k21=kd in the reverse direction (see below). A model of two-state switching free diffusion was introduced by Kärger,26 and more general models were studied in Refs. 27 and 28 (see references therein). In particular, the subordination concept was used in Ref. 28 to show that the PDF of the escape time admits a general spectral expansion,
Jsd(t|x)=n=0tϒ(t;Λn)un(x)Ωdxun(x),
(27)
where Λn and un(x) are the eigenvalues and L2(Ω)-normalized eigenfunctions of the Laplace operator with appropriate boundary conditions and the function ϒ(t; Λ) accounts for switching dynamics. An explicit form of this function for the two-state model26,28 reads in our setting as
ϒ(t;Λ)=eγ+t(DΛγ)eγt(DΛγ+)γ+γ,
(28)
where
γ±=12DΛ+k12+k21±(DΛ+k21+k12)24DΛk21.
(29)
Here, we consider diffusion on the interval (0, H) with an absorbing endpoint z = H and a reflecting endpoint z = 0, for which Λn = π2(n + 1/2)2/H2 and un(z)=2/Hcos(π(n+1/2)z/H) such that
Jsd(t|z)=n=0tϒ(t;Λn)2(1)n+1π(n+1/2)cos(π(n+1/2)z/H).
(30)
From this PDF, we calculate the mean escape time,
Tsd(z)=0dttJsd(t|z)=n=02(1)nπ(n+1/2)×cos(π(n+1/2)z/H)0dtϒ(t;Λn)=(k12+k21)H2Dk12n=02(1)ncos(π(n+1/2)z/H)π3(n+1/2)3=1+k12/k21H2z22D,
(31)
where we note that the sum in the third line is the Fourier series of (1 − (z/H)2)/2. A comparison of Eqs. (25) and (31) suggests a way to assign the transition rates as k12=2ka/L and k21=kd so that the MFPTs are identical in both cases (for ka = 0). Note that for ka > 0, the approximation identifies with the exact result to the leading order in H/L, which was hereby considered large.

In Fig. 2, we assume ka = 0 and demonstrate how the two-state diffusion approximation Jsd(t|z) captures the PDF Jab(t|r, z) in the limit LH. We plot the density and its approximation for three heights H of the cylinder with unit radius L = 1. We see that even when HL, the two-state approximation turns out to be remarkably accurate at long times. Surprisingly, it accurately captures even the short-time behavior.

FIG. 2.

PDF Jab(t|r, z) of the escape time from a cylindrical hole (see Fig. 1) with L = 1, ka = 0, D = 1, ka=1, kd=1, r = 0, z = H/2, and with H = 10 (left), H = 2 (center), and H = 0.5 (right). The solid lines represent the exact solution from Eq. (40) truncated to 30 × 30 = 900 terms. The dashed lines give the two-state switching diffusion approximation Jsd(t|z) from Eq. (30) truncated to 200 terms. The circles give estimates based on 106 particles whose motion was simulated according to the protocol in Appendix D of Ref. 25, with simulation time step Δt = 10−6.

FIG. 2.

PDF Jab(t|r, z) of the escape time from a cylindrical hole (see Fig. 1) with L = 1, ka = 0, D = 1, ka=1, kd=1, r = 0, z = H/2, and with H = 10 (left), H = 2 (center), and H = 0.5 (right). The solid lines represent the exact solution from Eq. (40) truncated to 30 × 30 = 900 terms. The dashed lines give the two-state switching diffusion approximation Jsd(t|z) from Eq. (30) truncated to 200 terms. The circles give estimates based on 106 particles whose motion was simulated according to the protocol in Appendix D of Ref. 25, with simulation time step Δt = 10−6.

Close modal

The desorption kinetics implies the s-dependence of the parameters qs and qs in the Robin boundary condition and thus leads to a convolution-type boundary condition in time domain, rendering the problem much more difficult than that with the ordinary Robin boundary condition for irreversible binding. Nevertheless, as diffusion is restricted in a bounded domain, the PDF of the escape time is still expected to admit a spectral expansion. Moreover, the presence of an absorbing boundary at z = H ensures that the survival probability vanishes exponentially in the long-time limit.

In mathematical terms, the inversion of the Laplace transform J̃ab(s|r,z) can be performed by evaluating its Bromwich integral representation via the residue theorem,
Jab(t|r,z)=12πiγdsestJ̃ab(s|r,z)=jesjtRessj{J̃ab(s|r,z)},
(32)
where {sj} are the poles of J̃ab(s|r,z), Ressj{J̃ab(s|r,z)} is its residue at sj, and γ is a contour in the complex plane of s chosen such that all the poles are to the left of it. As the poles are determined by the zeros of the function gn(s)(H) in Eq. (3), it is more convenient to employ a double index (n, m) instead of a single index j. In fact, the first index n = 0, 1, 2, … refers to the function gn(s)(H), whereas the second index m = 0, 1, 2, … enumerates all positive zeros sn,m of this function,
gn(sn,m)(H)=0.
(33)
The numerical computation of the poles sn,m will be discussed in Sec. II E.
To compute the residues, we need to find
dgn(s)(H)ds=dα̂n(s)dscosh(α̂n(s)h)+hα̂n(s)sinh(α̂n(s)h)+Ldqsdssinh(α̂n(s)h)+qsdα̂n(s)dshcosh(α̂n(s)h),
(34)
where
dqsds=kakdD(kd+s)2
(35)
and
dα̂n(s)ds=12α̂n(s)L2D+2αn(s)dαn(s)ds.
(36)
If there are higher-order poles, one would need to evaluate higher-order derivatives with respect to s. However, we did not observe numerically higher-order poles for all examples considered in this work. The derivative of αn(s) can be obtained by differentiating Eq. (6),
dαn(s)dsαn(s)1+J12(αn(s))J02(αn(s))=kakdLD(kd+s)2.
(37)
Substituting sn,m = −n,m/L2, we get
dgn(s)(H)dss=sn,m=ihβn,msin(βn,mh)cos(βn,mh)1hκaλn,mκdλn,m×L2/D+2αn,mdαn(sn,m)ds2βn,m+isin(βn,mh)κaκdL2/D(κdλn,m)2,
(38)
where αn,m=αn(sn,m), βn,m=iα̂n(sn,m), and
αn,mdαn(s)dss=sn,m=κaκdL2D(κdλn,m)21+J12(αn,m)J02(αn,m),
(39)
which is obtained by the use of Eq. (37). We, therefore, get
Jab(t|r,z)=n,m=0eDtλn,m/L2cn,mJ0(αn,mr/L)×βn,mcos(βn,mz/L)+κa1κd/λn,msin(βn,mz/L),
(40)
where
cn,m=2icn(sn,m)dgn(s)(H)dss=sn,m=2idgn(s)(H)dss=sn,mJ1(αn,m)/αn,mJ02(αn,m)+J12(αn,m).
(41)
The poles of J̃ab(s|r,z) are determined by the zeros of the function gn(s)(H) in Eq. (3). Let us fix n and introduce the shorthand notations αn(s)=α and α̂n(s)=iβ so that
α2+β2=λ=sL2/D>0.
(42)
Here, α is the solution of Eq. (6) that we rewrite explicitly as
αJ1(α)J0(α)=κa1κd/λ.
(43)
Equation (33) then reads
0=iβcos(βh)+κa1κd/λsin(βh),
(44)
which can also be written as
tan(βh)β=1κd/λκa.
(45)
We thus get a system of three nonlinear equations (42), (43), and (45) for the unknown parameters α, β, and λ. For each n, these equations have infinitely many solutions, and the main practical difficulty in their search is to ensure that they are all found. Doing so analytically is not possible. Thus, while the representation of Eq. (40) is applicable for the general case, in what follows, we focus on two limiting cases where the numerical solution is feasible via standard methods.

1. No adsorption on the cylinder wall

We first discuss the simple limit where there is no adsorption on the cylinder wall, ka=0, such that Eq. (43) is reduced to J1(α) = 0, which has infinitely many solutions αn = j1,n, where j1,n are the zeros of J1(z), including α0 = j1,0 = 0. When n = 0, one has λ = β2 and Eq. (45) can be written as
βtan(βh)=κdβ2κa.
(46)
This equation has infinitely many solutions that we denote β0,m. For each n > 0, Eq. (45) also has infinitely many solutions (denoted as βn,m), but they do not contribute because the coefficient cn,m is proportional to J1(αn) according to Eq. (41) and thus vanishes. In other words, when there is no lateral adsorption, the solution J̃ab(s|r,z) and thus Jab(t|r, z) do not depend on the radial coordinate, and one retrieves the one-dimensional problem that was solved in Ref. 25 [see the paragraph below Eq. (9) therein]. In this particular case, αn,m = αn is independent of the second index m and is actually decoupled from βn,m.

2. No adsorption on the bottom disk

Now, we discuss the more interesting limit where there is no adsorption on the bottom disk. If ka = 0, Eq. (45) has infinitely many solutions βnh = π/2 + πn, with n = 0, 1, 2, …. We can rewrite Eq. (43) as
J0(α)αJ1(α)=1κd/(βn2+α2)κa.
(47)
The left-hand side decreases piecewise monotonously on the intervals (j1,l, j1,l+1), while the right-hand side increases monotonously from (1κd/βn2)/κa at α = 0 to 1/κa as α. As a consequence, for each n, there are infinitely many solutions that we denote by αn,l+1 for each l = 0, 1, 2, ….
In conventional diffusion problems without desorption, transcendental equations obtained from the boundary condition usually admit only real solutions. In contrast, the desorption mechanism and the consequent s-dependence in the Robin boundary condition allow for a purely imaginary solution of Eq. (47). In fact, setting α=iᾱ, one gets
I0(ᾱ)ᾱI1(ᾱ)=κd/(β2ᾱ2)1κa,
(48)
where Iν(⋅) is the modified Bessel function of the first kind of order ν. The left-hand side monotonously decreases from + to 0 as ᾱ goes from 0 to +. In turn, the right-hand side monotonously increases from (κd/βn21)/κa at ᾱ=0 to + as ᾱβn and then from − to 1/κa. As a consequence, there exists a single solution of this equation on the interval (0, βn), for each βn, that we denote ᾱn. This solution determines αn,0=iᾱn that contributes to the list of poles. Note also that this solution results in small λn,0=βn2ᾱn2; in particular, λ0,0 determines the pole with the smallest absolute value, which, in turn, determines the asymptotic decay rate of the survival probability,29 
S(t|r,z)=10tdtJab(t|r,z).
(49)
Since βnh = π/2 + πn, some earlier expressions for the residues are simplified,
dgn(s)(H)dss=sn,m=i(1)nhL22D×1+2κaκd(κdλn,m)21+J12(αn,m)J02(αn,m),
(50)
such that
cn,m=4D(1)nJ1(αn,m)hL2αn,mJ02(αn,m)+J12(αn,m)+2κaκdJ02(αn,m)(κdλn,m)2,
(51)
and the PDF of the escape time becomes
Jab(t|r,z)=n,m=0eDtλn,m/L2cn,mJ0(αn,mr/L)×βn,mcos(βn,mz/L).
(52)
The decay time is determined by the smallest eigenvalue and is hence given by
T=L2Dλ0,0.
(53)
The value of λ0,0 can be determined numerically, as described in Sec. II E. In turn, the decay time determines the long-time exponential decay of the survival probability and of the PDF,29 
S(t|r,z)et/T,Jab(t|r,z)et/T(t).
(54)
While one can always find T numerically, in certain cases, we can approximate it analytically. For example, let us assume that LH such that h = H/L ≪ 1. This case corresponds to a very wide and shallow compartment. We recall that for the case of ka = 0, one has β0 = π/(2h). A solution of Eq. (48) can be searched (and then validated with simulations) by setting ᾱ0=π/(2h)ϵ with ϵ ≪ 1. Substituting this expression and expanding to the leading order in ϵ, we get
ϵκdπ/h+2κaI0(π/(2h))I1(π/(2h)),
(55)
where we note that ϵ is, indeed, small when h ≪ 1; thus, the approximation is self-consistent. We then compute λ0,0=β02ᾱ02πϵ/h and thus
TL2D1+2πκahI0(π/(2h))I1(π/(2h))κd=1kd+2πkaHkdDI0(πL/(2H))I1(πL/(2H))1kd1+2πkaHD,
(56)
where we have used I0(πL/(2H))/I1(πL/(2H)) ≈ 1 for h ≪ 1. Note that in this limit, the decay time does not depend on L.

Here, we study a different textured surface, which is covered by a periodic array of nanopillars of radius l and height H, separated by a distance d (Fig. 3). The survival of a diffusing particle in the presence of absorbing nanopillars has recently been studied in Refs. 30 and 31. Here, we take a step forward and consider a more challenging situation when the cylindrical walls and the bottom base are adsorbing. Following the rationales presented in Ref. 30, we approximate a periodic cell of the structure, a rectangular cuboid, by a cylindrical shell of inner radius l and outer radius L, capped by parallel planes at z = 0 and z = H. In this way, the periodic conditions on the cuboid are replaced by a reflecting boundary condition on the outer cylinder, whose radius L is chosen to be L=d/π to get the same cross-sectional area of the true rectangular cuboid cell, i.e., to preserve the volume of the periodic cell.

FIG. 3.

A surface with a periodic array of nanopillars. A small part of the surface is shown, in which the periodic cell, drawn in dashed lines, is a rectangular cuboid that we approximate by a cylindrical cell. Such a cylindrical cell is drawn around one of the pillars and enlarged. The radius of the cell is L=d/π, where d is the distance between adjacent pillars. The pillar is a cylinder of radius l capped by parallel planes at z = 0 and z = H. The top annulus at z = H is absorbing (escape region in red), whereas the bottom annulus at z = 0 and the inner cylindrical wall are adsorbing (green). Here, ka and kd are the adsorption and desorption constants for the bottom annulus, respectively, and ka and kd are the adsorption and desorption constants for the pillar surface, respectively.

FIG. 3.

A surface with a periodic array of nanopillars. A small part of the surface is shown, in which the periodic cell, drawn in dashed lines, is a rectangular cuboid that we approximate by a cylindrical cell. Such a cylindrical cell is drawn around one of the pillars and enlarged. The radius of the cell is L=d/π, where d is the distance between adjacent pillars. The pillar is a cylinder of radius l capped by parallel planes at z = 0 and z = H. The top annulus at z = H is absorbing (escape region in red), whereas the bottom annulus at z = 0 and the inner cylindrical wall are adsorbing (green). Here, ka and kd are the adsorption and desorption constants for the bottom annulus, respectively, and ka and kd are the adsorption and desorption constants for the pillar surface, respectively.

Close modal

In summary, we consider the escape problem from the above cylindrical shell, in which the top annulus is absorbing, the outer cylinder is reflecting, whereas the inner cylinder and the bottom annulus are adsorbing. The adsorption and desorption rates of the bottom annulus and the inner cylinder can differ. We are interested in finding the PDF of the first-passage time to the top annulus, which can also be thought of as the escape time from the textured surface. We denote this PDF as Jab(t|r, z), where (r, z) ∈ Ω is the initial location of the particle inside the cylindrical shell.

Repeating the same considerations as in Eqs. (1a)(1d), we obtain the boundary value problem
(sDΔ)J̃ab(s|r,z)=0(r,zΩ),
(57a)
J̃ab(s|r,z)=1(z=H),
(57b)
(z+qs)J̃ab(s|r,z)=0(z=0),
(57c)
(r+qs)J̃ab(s|r,z)=0(r=l),
(57d)
rJ̃ab(s|r,z)=0(r=L),
(57e)
where Δ=r2+(1/r)r+z2 is the Laplace operator in cylindrical coordinates (without the angular part). The surfaces are characterized by the parameters qs and qs that were defined in Eq. (2), where ka and kd are the adsorption and desorption constants for the bottom annulus and ka and kd are the adsorption and desorption constants for the inner cylindrical surface.
In analogy to Sec. II, we search the solution for Eq. (57a) under the boundary conditions (57b)(57e) as
J̃ab(s|r,z)=n=0cn(s)ω0(αn(s),r̄)gn(s)(z)gn(s)(H),
(58)
where r̄=r/L, with gn(s)(z) and α̂n(s) being defined in Eqs. (4) and (5), respectively. The prefactors in gn(s)(z) were determined to ensure the boundary conditions (57b) and (57c). We further introduce
ων(αn(s),r̄)J1(αn(s))Yν(αn(s)r̄)Y1(αn(s))Jν(αn(s)r̄),
(59)
where Yν(⋅) is the Bessel function of the second kind of order ν such that condition (57e) is satisfied. Indeed, the relation rω0(αn(s),r̄)=αn(s)ω1(αn(s),r̄)/L implies rω0(αn(s),r̄)|r̄=1=0. From the boundary condition (57d), we find that αn(s) satisfy the transcendental equation
αn(s)ω1(αn(s),ρ)ω0(αn(s),ρ)=qsL,
(60)
where ρ = l/L. For any s ≥ 0, there are infinitely many solutions that we enumerate by n = 0, 1, 2, … in an increasing order. The unknown coefficients cn(s) are found by multiplying the boundary condition (57b) by rω0(αn(s),r) and integrating over r from l to L. This gives
ck(s)2ω0(αn(s),1)2ρω0(αn(s),ρ)2ρω1(αn(s),ρ)2=ραk(s)ω1(αn(s),ρ),
(61)
from which we get
cn(s)=2ρω1(αn(s),ρ)/αn(s)ρω1(αn(s),ρ)2+ρω0(αn(s),ρ)2ω0(αn(s),1)2.
(62)
To obtain the left-hand side of Eq. (61), we used the orthogonality of the Bessel functions together with
ck(s)ρ1dr̄r̄ω0(αn(s),r̄)2=ck(s)2r̄ω0(αn(s),r̄)2+rω1(αn(s),r̄)2ρ1,
(63)
and noted that ω1(αn(s),1)=0. To obtain the right-hand side of Eq. (61), we used
ρ1dr̄r̄ω0(αn(s),r̄)=ραk(s)ω1(αn(s),ρ).
(64)
To facilitate further analysis, we use the dimensionless quantities in Table I.

Here, we compute the mean escape time by analyzing the asymptotic behavior of J̃ab(s|r,z) as s → 0. We employ the same procedure that was described in Sec. II B.

In the spectral expansion (58), we first analyze the term n = 0 and then discuss the other terms with n > 0. As s → 0, one has qs0 (with an asymptotic form κaκdLDs), and so, according to Eq. (60), α0(s)0. Using the Taylor expansion of the Bessel functions in Eq. (59), one gets
ω0(αn(s),r̄)=2παn(s)+αn(s)1r̄2+2ln(r̄)2π+O((αn(s))2)
(65)
and
ω1(αn(s),r̄)1r̄2πr̄+O((αn(s))2).
(66)
Therefore, for s → 0, we obtain
α0(s)s2κa/kdρ1ρ2(s0).
(67)
As a consequence, we have
c0(s)π2α0(s)1+ρ1ρ2κakd1ρ4+4ρ2ln(ρ)4(1ρ2)s+O(s5/2).
(68)
We also get α̂0(s)sL2/D1+2(ρ/(1ρ2))κa/κd in the leading order, from which we obtain
g0(s)(z)g0(s)(H)=1κakdHzL+L22D+ρ1ρ2κakdH2z2L2s+O(s2).
(69)
Let us now consider the terms with n > 0. Denoting the left-hand side of Eq. (60) as
F1(x)=xω1(x,ρ)ω0(x,ρ),
(70)
we Taylor expand
F1(αn(s))F1(αn(0))+F1(αn(0))(αn(s)αn(0)).
(71)
According to Eq. (60), F1(αn(s))=qsL0 as s → 0, and thus, αn(0) are the zeros of the function ω1(x, ρ) defined in Eq. (59). Taking the derivative of Eq. (70) and plugging in x=αn(0), we find
F1(αn(0))=αn(0)ρ4π2αn(0)ρω0(αn(0),ρ)2.
(72)
Comparing the right-hand side of Eqs. (71) and (60) in the limit s → 0, we obtain
αn(s)=αn(0)κakd1F1(αn(0))s+O(s2).
(73)
Similarly, setting
F2(x)=2ρω1x,ρ/xρω1x,ρ2+ρω0x,ρ2ω0(x,1)2,
(74)
we have cn(s)=F2(αn(s)) according to Eq. (62). Taking the derivative of Eq. (74) and plugging in x=αn(0), we find
F2(αn(0))=J0(αn(0))Y1(αn(0)ρ)+Y2(αn(0))J1(αn(0)ρ)ρY1(αn(0))J0(αn(0)ρ)+J1(αn(0))Y2(αn(0)ρ)2π2αn(0)ρ+αn(0)ρ2ω0(αn(0),ρ)2,
(75)
and using Eq. (73), we get
cn(s)=F2(αn(0))+F2(αn(0))(αn(s)αn(0))+O(s2)=F2(αn(0))F1(αn(0))κakds+O(s2)
(76)
and
α̂n(s)=αn(0)+12αn(0)κaκdF1(αn(0))L2Ds+O(s2).
(77)
As a consequence, we find that to the leading order in s,
gn(s)(z)gn(s)(H)cosh(αn(0)z̄)cosh(αn(0)h)
(78)
and
ω0(αn(s),r̄)ω0(αn(0),r̄),
(79)
with z̄=z/L. Substituting these expressions into Eq. (58), we get
J̃ab(s|r,z)=1sT(r,z)+O(s2),
(80)
where
T(r,z)=H2z22D+ka(Hz)kdD+kakdLDρ1ρ2h2z̄21r̄2+2ln(r̄)21ρ4+4ρ2ln(ρ)4(1ρ2)+n=1ω0(αn(0),r̄)cosh(αn(0)z̄)cosh(αn(0)h)F2(αn(0))F1(αn(0))
(81)
is the mean escape time, in which h = H/L is the dimensionless cylinder’s height. In Fig. 4, we plot the mean escape time from a surface textured by an array of pillars as given by Eq. (81), with varying pillar height and varying inter-pillar distance.
FIG. 4.

Mean escape time from a surface textured by an array of pillars (Fig. 3). The solid lines are drawn using the exact solution of Eq. (81). Each line represents a different adsorption equilibrium constant for the pillar, K=ka/kd, where we set kd=1 and vary ka accordingly. The marker symbols represent the mean escape time of 104 particles simulated according to the protocol in Appendix D of Ref. 25, with simulation time step Δt = 10−4. We set D = 0.7, l = 0.9, ka = 0.7, kd = 6.1, z = 0.1, and r = 1. The mean escape time is plotted as a function of (a) the height of the pillars H, where we set L = 5, and (b) the radius of the unit cell L (divided by the radius of the pillars l), where we set H = 10.

FIG. 4.

Mean escape time from a surface textured by an array of pillars (Fig. 3). The solid lines are drawn using the exact solution of Eq. (81). Each line represents a different adsorption equilibrium constant for the pillar, K=ka/kd, where we set kd=1 and vary ka accordingly. The marker symbols represent the mean escape time of 104 particles simulated according to the protocol in Appendix D of Ref. 25, with simulation time step Δt = 10−4. We set D = 0.7, l = 0.9, ka = 0.7, kd = 6.1, z = 0.1, and r = 1. The mean escape time is plotted as a function of (a) the height of the pillars H, where we set L = 5, and (b) the radius of the unit cell L (divided by the radius of the pillars l), where we set H = 10.

Close modal
As in Sec. II B, in the limit ka0 or kd, when the pillars are not sticky but inert, we retrieve the mean escape time for a one-dimensional box with a sticky surface,25 
T(r,z)=12D(H2z2)+kakdD(Hz),
(82)
which is identical to Eq. (24).
Averaging Eq. (81) over the cross section at height z yields
T̄(z)=2ππ(L2l2)lLdrrT(r,z)=kakdHzD+1+2kakdLρ1ρ2H2z22D,
(83)
where we used lLdrω0(αn(0),r̄)=0. Further averaging over z, we obtain
Tu=H23D1+kakd32H+kakdρ1ρ22L,
(84)
where the subscript “u” denotes a uniform distribution of the initial position.
The solution in time domain can be found via the residue theorem. The computation is very similar to the case of the capped cylinder in Sec. II D; here, we differentiate Eq. (60) to get
αn(s)dαn(s)ds=κakd(kd+s)2×ω1(αn(s),ρ)αn(s)ω0(αn(s),ρ)+ω0(αn(s),ρ)dω1(αn(s),ρ)dαn(s)ω1(αn(s),ρ)dω0(αn(s),ρ)dαn(s)ω0(αn(s),ρ)21.
(85)
Overall, we obtain
Jab(t|x,z)=n,m=0cn,meDtλn,m/L2ω0(αn,mr/L)×βn,mcos(βn,mz/L)κaλn,mκdλn,msin(βn,mz/L),
(86)
where αn,m=αn(sn,m), λn,m=αn,m2+βn,m2=sn,mL2/D,
cn,m=icn(sn,m)dgn(s)(H)dss=sn,m=idgn(s)(H)dss=sn,m×2ρω1αn,m,ρ/αn,mρω1αn,m,ρ2+ρω0αn,m,ρ2ω0(αn,m,1)2,
(87)
and dgn(H)ds|s=sn,m is given by Eq. (38), in which αn(s)dαn(s)ds is substituted from Eq. (85).
The poles of J̃ab(s|x,z) are determined by the zeros of gn(s)(H), as previously (see Sec. II E). We use the former notations: αn(s)=α, α̂n(s)=iβ, and α2 + β2 = λ = −sL2/D. In this case, Eq. (60) reads
ω0(α,ρ)αω1(α,ρ)=1κd/(α2+β2)κa.
(88)
We focus on the case ka = 0, for which βnh = π/2 + πn. For any fixed βn, the left-hand side of Eq. (88) increases piecewise monotonously from − to + on the intervals (αn(0),αn+1(0)), while the right-hand side is a monotonously decreasing function of α. As a consequence, there is a single solution on each interval (αn(0),αn+1(0)) that we denote as αn,m+1, for m = 0, 1, 2, …. In addition, there is a purely imaginary solution, which can be found by setting α=iᾱ, with ᾱ satisfying
1ᾱI1(ᾱ)K0(ᾱρ)+K1(ᾱ)I0(ᾱρ)K1(ᾱ)I1(ᾱρ)I1(ᾱ)K1(ᾱρ)=1κd/(β2ᾱ2)κa,
(89)
where Iν(⋅) and Kν(⋅) are the modified Bessel functions of the first and second kind of order ν. The left-hand side monotonously increases from − to 0 as ᾱ goes from 0 to +, whereas the right-hand side for any fixed βn decreases monotonously on (0, βn) from (κd/βn21)/κa to − and on (βn, + ) from to 1/κa. As a consequence, there exists only one solution on the interval (0, βn) that we denote ᾱn. This solution determines αn,0=iᾱn that contributes to the list of poles.
The general discussion in Sec. II F is valid here. Let us find the approximation for the decay time T in the limit ka = 0 and h = H/L ≪ 1 such that β0 = π/(2h) ≫ 1. A solution of Eq. (89) can be searched (and then validated with simulations) by setting ᾱ0=π/(2h)ϵ with ϵ ≪ 1. Substituting this expression and expanding to the leading order in ϵ, we get
ϵπhκd+2κaκdI1π2hK0πρ2h+K1π2hI0πρ2hI1π2hK1πρ2hK1π2hI1πρ2h1,
(90)
where we note that ϵ is, indeed, small when h ≪ 1; thus, the approximation is self-consistent. We then compute λ0,0=β02ᾱ02πϵ/h and find
T=L2Dλ0,01kd+2hκaπkdI1π2hK0πρ2h+K1π2hI0πρ2hI1π2hK1πρ2hK1π2hI1πρ2h.
(91)
To proceed, we note that when h ≪ 1, one has Iν(π2h)heπ/(2h)/π, whereas Kν(π2h)heπ/(2h) and is thus negligible. We then get
T1kd+2HkaπDkdK0(πl/(2H))K1(πl/(2H))1kd1+2πkaHD,
(92)
where we noted that πρ/(2h) = πl/(2H) and used K0(πl/(2H))/K1(πl/(2H)) ≈ 1 for h ≪ 1. We thus see that in this limit, the decay time identifies with the decay time in Eq. (56) and that it does not depend on L.

We consider a grooved surface, as illustrated in Fig. 5. This problem is equivalent to diffusion with a diffusion coefficient D in a rectangular domain Ω = (−L, L) × (0, H). The top edge of the domain is absorbing (with a Dirichlet boundary condition), and the three other edges are adsorbing, with reversible binding. As in the previous examples, we allow for different adsorption kinetics on the bottom edge. We search the probability density function Jab(t|x, z) of the escape time through the top edge. Note that the problem of escape from the domain Ω is equivalent to an escape from a twice smaller domain Ω′ = (0, L) × (0, H), where the left edge is reflecting (with a Neumann boundary condition); see Fig. 6. We thus focus on the latter setting.

FIG. 5.

Adsorbing grooved surface. A small part of the surface is shown. The walls separating the grooves are of height H, and the distance between any two walls is 2L. One of the grooves is enlarged, and the problem is effectively the escape from a two-dimensional rectangular compartment. The top side at z = H is absorbing (an escape region in red), whereas the other three sides are adsorbing (green). Here, ka and kd are the adsorption and desorption constants for the bottom edge, respectively, and ka and kd are the adsorption and desorption constants for the left and right edges, respectively.

FIG. 5.

Adsorbing grooved surface. A small part of the surface is shown. The walls separating the grooves are of height H, and the distance between any two walls is 2L. One of the grooves is enlarged, and the problem is effectively the escape from a two-dimensional rectangular compartment. The top side at z = H is absorbing (an escape region in red), whereas the other three sides are adsorbing (green). Here, ka and kd are the adsorption and desorption constants for the bottom edge, respectively, and ka and kd are the adsorption and desorption constants for the left and right edges, respectively.

Close modal
FIG. 6.

(a) A schematic illustration of a rectangular domain Ω=(L,L)×(0,H)R2 with one absorbing edge (an escape region on the top) and three adsorbing edges with reversible binding kinetics characterized by ka and kd (bottom) and ka and kd (left and right). (b) An equivalent twice smaller domain Ω′ = (0, L) × (0, H) with a reflecting edge replacing the adsorbing edge on the left (Neumann boundary condition denoted by N).

FIG. 6.

(a) A schematic illustration of a rectangular domain Ω=(L,L)×(0,H)R2 with one absorbing edge (an escape region on the top) and three adsorbing edges with reversible binding kinetics characterized by ka and kd (bottom) and ka and kd (left and right). (b) An equivalent twice smaller domain Ω′ = (0, L) × (0, H) with a reflecting edge replacing the adsorbing edge on the left (Neumann boundary condition denoted by N).

Close modal
Repeating the same considerations as in Eqs. (1a)(1d), we obtain the boundary value problem
(sDΔ)J̃ab(s|x,z)=0(x,zΩ),
(93a)
J̃ab(s|x,z)=1(z=H),
(93b)
(z+qs)J̃ab(s|x,z)=0(z=0),
(93c)
xJ̃ab(s|x,z)=0(x=0),
(93d)
(x+qs)J̃ab(s|x,z)=0(x=L),
(93e)
where Δ=x2+z2 is the Laplace operator in Cartesian coordinates. The surfaces are characterized by the parameters qs and qs that were defined in Eq. (2), where ka and kd are the adsorption and desorption constants for the bottom edge, respectively, and ka and kd are the adsorption and desorption constants for the right edge, respectively.
The solution of Eq. (93a) under the boundary conditions (93b)(93e) is
J̃ab(s|x,z)=n=0cn(s)cos(αn(s)x/L)gn(s)(z)gn(s)(H),
(94)
where gn(s)(z) and α̂n(s) were defined in Eqs. (4) and (5). The prefactors in gn(s)(z) were determined by the boundary conditions (93b) and (93c).
We have used the reflecting boundary condition (93d) to determine the form of the x-dependent part in Eq. (94). From the boundary condition (93e), we find that αn(s) satisfy the transcendental equation
αn(s)tan(αn(s))=qsL.
(95)
For any s ≥ 0, there are infinitely many solutions that we enumerate by n = 0, 1, 2, … in an increasing order. The unknown coefficients cn(s) are found by multiplying the boundary condition (93b) by cos(αk(s)x/L) and integrating over x from 0 to L. This gives
cn(s)=2sin(αn(s))αn(s)1+sin(2αn(s))2αn(s).
(96)

To facilitate further analysis, we use the dimensionless quantities in Table I. In the solution process, we will employ the same procedure that was described in Sec. II. We will thus skip some of the details of the calculation.

Here, we compute the mean escape time by analyzing the asymptotic behavior of J̃ab(s|r,z) as s → 0. In the limit s → 0, one has
αn(s)sκa/kd(n=0),πn+κas/(kdπn)(n>0),
(97)
where we used αn(0)=πn. We then deduce
cn(s)1+κa6kds(n=0),2s(1)nκa/(kd[πn]2)(n>0).
(98)
Similarly, we deduce
g0(s)(z)g0(s)(H)1κakdHzL+L2D+κakdH2z22L2s
(99)
and
gn(s)(z)gn(s)(H)cosh(πnz/L)cosh(πnH/L).
(100)
Substituting these expressions in Eq. (94), we get J̃ab(s|x,z)=1sT(x,z)+O(s2), where the mean escape time is found to be
T(x,z)=H2z22D+kakdHzD+kaLkdDx22L216+H2z22L22n=1(1)nπ2n2cos(πnx/L)cosh(πnz/L)cosh(πnH/L).
(101)
Averaging over the cross section at height z yields
T̄(z)=1L0LdxT(x,z)=ka(Hz)kdD+1+kakdLH2z22D.
(102)
Further averaging over z, we obtain
Tu=1H0HdzT̄(z)=H23D1+kakd32H+kakd1L,
(103)
where the subscript “u” denotes a uniform distribution of the initial position.

If one uses the two-state switching diffusion approximation instead of the exact solution (see Sec. II C), a comparison with Eq. (31) for ka = 0 suggests that k12=ka/L.

The solution in time domain can be found via the residue theorem. The computation is similar to the case of the capped cylinder in Sec. II D; here, we differentiate Eq. (95) to get
αn(s)dαn(s)ds=κakd(kd+s)2tan(αn(s))αn(s)+1+tan2(αn(s))1.
(104)
Therefore, we obtain
Jab(t|x,z)=n,m=0cn,meDtλn,m/L2cos(αn,mx/L)×βn,mcos(βn,mz/L)κaλn,mκdλn,msin(βn,mz/L),
(105)
where αn,m=αn(sn,m), λn,m=αn,m2+βn,m2=sn,mL2/D,
cn,m=icn(sn,m)dgn(s)(H)dss=sn,m=idgn(s)(H)dss=sn,m2sin(αn,m)/αn,m1+sin(2αn,m)2αn,m,
(106)
and dgn(H)ds|s=sn,m is given by Eq. (38), in which αn(s)dαn(s)ds is substituted from Eq. (104).

Figure 7 illustrates the behavior of the PDF Jab(t|x, z) for H = 10. For small adsorption rates κa (panels a and b), the two-state switching diffusion model yields an excellent approximation. In contrast, when κa=10 (panel c), the two-state model accurately describes the long-time behavior but fails at short times. Similarly, when H = 1, an approximation by the two-state model is less accurate (figure not shown).

FIG. 7.

PDF Jab(t|x, z) of the escape time from a groove with D = 1, L = 1, H = 10, ka = 0, for three different values of κd (see the legend), and κa=0.1 (panel a), κa=1 (panel b), and κa=10 (panel c). The solid lines give the exact solution from Eq. (105), while the dashed lines represent the two-state switching diffusion approximation. The marker symbols give estimates based on 106 particles whose motion was simulated according to the protocol in Appendix D of Ref. 25, with simulation time step Δt = 10−6.

FIG. 7.

PDF Jab(t|x, z) of the escape time from a groove with D = 1, L = 1, H = 10, ka = 0, for three different values of κd (see the legend), and κa=0.1 (panel a), κa=1 (panel b), and κa=10 (panel c). The solid lines give the exact solution from Eq. (105), while the dashed lines represent the two-state switching diffusion approximation. The marker symbols give estimates based on 106 particles whose motion was simulated according to the protocol in Appendix D of Ref. 25, with simulation time step Δt = 10−6.

Close modal
The poles of J̃ab(s|x,z) are determined by the zeros of gn(s)(H), as previously (see Sec. II E). We use the former notations: αn(s)=α, α̂n(s)=iβ, and α2 + β2 = λ = −sL2/D. In this case, Eq. (95) reads
cos(α)αsin(α)=1κd/(α2+β2)κa.
(107)
We focus on the case ka = 0, for which βnh = π/2 + πn with h = H/L. For any fixed βn, the left-hand side of Eq. (107) decreases piecewise monotonously on the intervals (πm, π(m + 1)), while the right-hand side is a monotonously increasing function of α. As a consequence, there is a single solution on each interval (πm, π(m + 1)) that we denote as αn,m+1, for m = 0, 1, 2, …. In addition, there is a purely imaginary solution, which can be found by setting α=iᾱ, with ᾱ satisfying
cosh(ᾱ)ᾱsinh(ᾱ)=κd/(βn2ᾱ2)1κa.
(108)
As ᾱ goes from 0 to +, the left-hand side monotonously decreases from + to 0, whereas the right-hand side for any fixed βn increases monotonously on (0, βn) from (κd/βn21)/κa to + and on (βn, + ) from − to 1/κa. As a consequence, there exists only one solution on the interval (0, βn) that we denote ᾱn. This solution determines αn,0=iᾱn that contributes to the list of poles.
The general discussion in Sec. II F is valid here. Let us find the approximation for the decay time T in the limit ka = 0 and h = H/L ≪ 1 such that β0 = π/(2h) ≫ 1. A solution of Eq. (108) can be searched (and then validated with simulations) by setting ᾱ0=π/(2h)ϵ with ϵ ≪ 1. Substituting this expression and expanding in the leading order to ϵ, we get
ϵκdπh+2κactanh(π/(2h)),
(109)
from which λ0,0=β02ᾱ02πϵ/h and thus
T=L2Dλ0,0L2D1+2πκahctanh(π/(2h))κd=1kd+2πka(H/D)ctanh(πL/(2H))kd1kd1+2πkaHD,
(110)
where we used ctanh(πL/(2H)) ≈ 1 when h ≪ 1. In this limit, the decay time does not depend on L. Note that the decay time in this limit is the same as in the previous two examples; see Eqs. (56) and (92).

For this example, let us also consider the opposite limit h ≫ 1. In fact, we have already derived in Sec. II C the two-state switching diffusion approximation for this case. The decay time of this approximation is determined by γ with the lowest eigenvalue Λ0 = (π/2H)2, according to Eq. (29). At the end of Sec. IV B, we have already seen that for this example, k12=ka/L and k21=kd. Finally, we have Tsd = L2/().

The behavior of the decay time and the validity of the approximation in Eq. (110) and the two-state switching diffusion approximation Tsd are explored in Fig. 8, where we plot the actual decay time as a function of κa and κd and the relative errors that are obtained by using the approximations. While it can be seen that both approximations work very well in their ranges of validity (h ≪ 1 for Eq. (110) and h ≫ 1 for two-state diffusion), we observe that the two-state diffusion approximation also provides fair results for grooves of intermediate depth (h = 1).

FIG. 8.

Left column: The decay time T for the escape from a groove with h = H/L = 0.1 (top), h = 1 (middle), and h = 10 (bottom), for ka = 0, L = 1, and D = 1. Since ka = 0, we have β0 = π/(2h). We attain ᾱ0 by numerically finding the roots of Eq. (108). We then use λ0,0=β02ᾱ02 and T = L2/(0,0). Center column: The relative error (in per cents) of the approximation for T given by Eq. (110). Right column: The relative error (in per cents) of the two-state switching diffusion approximation for T.

FIG. 8.

Left column: The decay time T for the escape from a groove with h = H/L = 0.1 (top), h = 1 (middle), and h = 10 (bottom), for ka = 0, L = 1, and D = 1. Since ka = 0, we have β0 = π/(2h). We attain ᾱ0 by numerically finding the roots of Eq. (108). We then use λ0,0=β02ᾱ02 and T = L2/(0,0). Center column: The relative error (in per cents) of the approximation for T given by Eq. (110). Right column: The relative error (in per cents) of the two-state switching diffusion approximation for T.

Close modal

Since the early studies of Langmuir in the beginning of the twentieth century, and to this day, the vast majority of theoretical studies on adsorption dynamics have dealt with flat surfaces immersed in an infinite bulk of adsorbates32–50 (which is effectively a one-dimensional setting), or with smooth curved surfaces.51–59 In this context, two main models are usually considered: linear kinetics and Langmuir kinetics. While the latter accounts for saturation of the surface under high adsorbate concentration, it is not linear and does not admit an analytical solution.

Given a flat surface located at L and a concentration profile c(x, t), the surface concentration Γ(t) under linear adsorption kinetics follows59,
dΓ(t)dt=kac(L,t)kdΓ(t).
(111)
At equilibrium, i.e., when dΓ(t)dt=0, one gets the Henry isotherm Γ(t) = Kc(L, t), with the equilibrium constant K = ka/kd. Trivially, the desorption rate kd is inversely proportional to the mean time spent being adsorbed to the surface: kd=T1. But what happens when the surface is textured? On the microscopic scale, the desorption rate is still determined by kd. Yet, on a scale comparable to that of the surface roughness, multiple events of adsorption and desorption can give rise to a completely different effective (or macroscopic) desorption rate. In this work, we employed the theoretical framework in Ref. 25 to determine this effective desorption rate.
For this purpose, we adopted a single-particle perspective and calculated the PDF and the mean of the escape time from textured surfaces of three different common topographies: holes, pillars, and grooves. Such solutions are valuable when studying the adsorption–desorption dynamics of surfaces. In particular, we obtained the mean escape time for the three surface topographies in a common experimental setting where the initial position of the particle is uniformly distributed inside the surface cavities. Let us rewrite Eqs. (26), (84), and (103) here as follows:
Tholes=H23D1+K32H+K2L,
(112a)
Tpillars=H23D1+K32H+Kρ1ρ22L,
(112b)
Tgrooves=H23D1+K32H+K1L,
(112c)
where K = ka/kd and K=ka/kd are the equilibrium constants of the bottom and lateral surfaces, respectively, and ρ = l/L.
For all the considered examples, the first term in the parentheses corresponds to the mean escape time in the absence of adsorption (reflecting surfaces): Tref=H2/(3D). The second and third terms correspond to the mean time spent adsorbed to the bottom surface and the lateral surface, respectively. We can easily quantify how the introduction of stickiness affects the mean escape time. For example, dividing Eq. (112a) by Tref, we obtain
TholesTref=1+K32H+K2L.
(113)
It is apparent that there are two contributions to the deviation of the mean escape time from its non-sticky benchmark. The contribution from the bottom surface scales like K/H and the contribution from the lateral surface scales like K′/L, where L and H are the characteristic lengths in the direction of the axes. As first mentioned in Ref. 25, this scaling seems to be universal, and the geometry of the domain determines the effective length ξ. It can be easily verified to hold in Eqs. (112b) and (112c).
In this work, we considered surfaces with canonical topographies. However, the approach we employed is general and we expect that a similar behavior will also be found for other geometries with perpendicular sticky surfaces, up to the geometrical prefactor ξ. For a general (even rugged) sticky surface, the expected form of the mean escape time is
Tu=Tref1+nKnξn,
(114)
where ξn is the effective length scale of the nth surface element and Kn=ka(n)/kd(n) is the equilibrium constant with ka(n) and kd(n) standing for the adsorption and desorption rates, respectively.
Finally, the mean escape time is inversely proportional to the effective desorption rate from the textured surface,
kdeff=Tu1=Tref1+nKnξn1.
(115)
Note that Eq. (115) represents an effective macroscopic description of the system and that it does not imply exponential escape times from the surface. Importantly, kdeff can be orders of magnitude smaller than the characteristic (microscopic) desorption rates in the system, as it vanishes with lateral confining length scales. Equation (115) can guide those who wish to predict and control desorption from textured adsorbing surfaces.

The results established herein can also be used to coarse-grain microscopic desorption kinetics when building macroscopic models of diffusive dynamics near textured surfaces. In fact, former studies on boundary homogenization (see Refs. 30 and 31 and references therein) provided efficient tools for estimating the macroscopic adsorption constant for textured surfaces. This work complements the former results by quantifying the desorption step and thus opening a way to describe the adsorption–desorption kinetics of textured surfaces.

We thank Samyuktha Ganesh for help with the graphical design. This project has received funding from the European Research Council (ERC) under the European Union’s Horizon 2020 research and innovation program (Grant No. 947731). D.G. acknowledges the Alexander von Humboldt Foundation for the support within a Bessel Prize award.

The authors have no conflicts to disclose.

Yuval Scher: Conceptualization (lead); Formal analysis (equal); Writing – original draft (equal); Writing – review & editing (supporting). Shlomi Reuveni: Conceptualization (equal); Formal analysis (supporting); Supervision (lead); Writing – original draft (supporting); Writing – review & editing (lead). Denis S. Grebenkov: Conceptualization (equal); Formal analysis (lead); Supervision (equal); Writing – original draft (equal); Writing – review & editing (equal).

The data that support the findings of this study are available within the article.

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