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Mean exit time for surface-mediated diffusion: spectral analysis and asymptotic behavior

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Abstract

We consider a model of surface-mediated diffusion with alternating phases of bulk and surface diffusion for two geometries: the disk and rectangles. We develop a spectral approach to derive an exact formula for the mean exit time of a particle through a hole on the boundary. The spectral representation of the mean exit time through the eigenvalues of an appropriate self-adjoint operator is particularly well-suited to investigate the asymptotic behavior in the limit of large desorption rate \(\lambda \). For a point-like target, we show that the mean exit time diverges as \(\sqrt{\lambda }\). For extended targets, we establish the asymptotic approach to a finite limit. In both cases, the mean exit time is shown to asymptotically increase as \(\lambda \) tends to infinity. That implies that the pure bulk diffusion is never an optimal search strategy. We also investigate the influence of rectangle elongation onto the mean exit time, in particular, the dependence of the critical ratio of bulk and surface diffusion coefficients on the rectangle aspect ratio. We show that the intermittent search strategy can significantly outperform pure surface diffusion for elongated rectangles.

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Correspondence to M. Zinsmeister.

Appendices

Appendix A: Asymptotic behavior of the eigenvalues of the operator \(V\tilde{T}V\)

Theorem 5.1

Let \(\lambda _n\) be the decreasing sequence of eigenvalues of the operator \(V\tilde{T}V\), where the operators \(\tilde{T}\) and \(V\) are defined in Eqs. (2.9) and (2.8). We have

$$\begin{aligned} \lambda _n \sim A_{\epsilon } n^{-2} , \end{aligned}$$
(5.1)

where \(A_{\epsilon }\) depends only on \(\epsilon \).

Proof

In order to prove this statement, we first investigate the following problem:

Let \(A\) and \(B\) are two compact, positive, self-adjoint operators. We assume that the eigenvalues of the operator \(A\) are ordered in a decreasing sequence: \(\lambda _1(A)\ge \lambda _2(A)\ge \ldots \ge \lambda _n(A)\ge \ldots \ge 0\). We recall the variational principle as following \(\square \)

Theorem 5.2

$$\begin{aligned} \lambda _n(A)=\max _{F} \min _{O\ne x\in A}\frac{\langle Ax,x\rangle }{\langle x,x\rangle } ;\mathrm{(}\text {where } F\text { is a subspaces of }L^2[a,b];\text { } \dim F=n\mathrm{)}.\nonumber \\ \end{aligned}$$
(5.2)

The \(\max \) is taken over \(F\), the subspace associated with the first \(n\) eigenvectors of \(A\).

We state two following lemmas which will be needed to prove Theorem 5.1.

Lemma 5.3

Let \(\lambda _n(A)\) and \(\lambda _n(A+B)\) be the \(n^{th}\) eigenvalues of the operators \(A\) and \(A+B\). Then, we have

$$\begin{aligned} \lambda _n(A)-\Vert B\Vert \le \lambda _n(A+B)\le \lambda _n(A)+\Vert B\Vert , \end{aligned}$$
(5.3)

where \(\Vert .\Vert \) defines the norm of an operator in \(L^2[a,b]\) space.

Proof

Let \(F(A)\) be the subspace of \(L^2[a,b]\) associated with the first \(n\) eigenvectors of \(A\). For all \(x\in F(A)\), we have

$$\begin{aligned} \langle (A+B)x,x\rangle = \langle Ax,x\rangle + \langle Bx,x\rangle . \end{aligned}$$

According to the variational principle, we have

$$\begin{aligned} \langle Ax,x\rangle \ge \min _{x\in A}\frac{\langle Ax,x\rangle }{\langle x,x\rangle }\Vert x\Vert ^2=\lambda _n(A)\Vert x\Vert ^2. \end{aligned}$$
(5.4)

Besides, we have

$$\begin{aligned} | \langle Bx,x\rangle | \le \Vert B\Vert \Vert x\Vert ^2. \end{aligned}$$
(5.5)

It follows from Eqs. (5.4) and (5.5) that

$$\begin{aligned} \langle (A+B)x,x\rangle \ge \lambda _n(A)\Vert x\Vert ^2-\Vert B\Vert \Vert x\Vert ^2. \end{aligned}$$

This gives

$$\begin{aligned} \min _{0\ne x\in F}\frac{\langle (A+B)x,x\rangle }{\Vert x\Vert ^2}\ge \lambda _n(A)-\Vert B\Vert . \end{aligned}$$

Again, according to the variational principle, we thus get

$$\begin{aligned} \lambda _n(A+B)\ge \lambda _n(A)-\Vert B\Vert . \end{aligned}$$

In the same manner, if we take \(F(A+B)\) be associated to the first \(n\) eigenvectors of \(A+B\), we can get

$$\begin{aligned} \lambda _n(A+B)\le \lambda _n(A)+\Vert B\Vert , \end{aligned}$$

and the Lemma 5.3 follows. \(\square \)

Lemma 5.4

With the notations used in Lemma 5.3, if \(\mathrm{rank}(B) < n < \infty \), then

$$\begin{aligned} \lambda _{n+\mathrm{rank}(B)}(A)\le \lambda _n(A+B)\le \lambda _{n-\mathrm{rank}(B)}(A). \end{aligned}$$
(5.6)

Proof

We call \(F\) the subspace of \(L^2[a,b]\) associated with the first \(n+\mathrm{rank}(B)\) eigenvectors of \(A\).

By the variational principle, we have

$$\begin{aligned} \forall x\in F\cap \ker (B), \Vert x\Vert =1: \langle (A+B)x,x\rangle = \langle Ax,x\rangle \ge \lambda _{n+\mathrm{rank}(B)}(A). \end{aligned}$$

Consequently,

$$\begin{aligned} \min _{x\in F\cap \ker (B); \Vert x\Vert =1}\langle (A+B)x,x\rangle \ge \lambda _{n+\mathrm{rank}(B)}(A). \end{aligned}$$

Since

$$\begin{aligned} \dim (F\cap \ker (B))=\dim F -\dim B(F)\ge n, \end{aligned}$$

we have

$$\begin{aligned} \lambda _n(A+B)\ge \min _{0\ne x\in \ker (B) \cap F; \Vert x\Vert =1}\langle (A+B)x,x\rangle . \end{aligned}$$

So, we conclude that

$$\begin{aligned} \lambda _n(A+B)\ge \lambda _{n+\mathrm{rank}(B)}(A). \end{aligned}$$
(5.7)

The second inequality in (5.6) of this lemma is obtained when we put \(A'=A+B\), \(B'=-B\), \(n'=n-\mathrm{rank}(B)\) and apply the inequality (5.7) for \(A'\), \(B'\) and \(n'\) instead of \(A\), \(B\) and \(n\). \(\square \)

We now call \(\pi _N\) be the orthogonal projection on the first \(N\) eigenvectors of \(B\).

By the property of an orthogonal projection, we can rewrite

$$\begin{aligned} B=\pi _N B\pi _N+(I-\pi _N)B(I-\pi _N), \end{aligned}$$

then

$$\begin{aligned} A+B=A+\pi _N B\pi _N+(I-\pi _N)B(I-\pi _N). \end{aligned}$$

We note that

$$\begin{aligned} \mathrm{rank}(\pi _N B\pi _N)=N, \end{aligned}$$

and

$$\begin{aligned} \Vert (I-\pi _N)B(I-\pi _N)\Vert \le \lambda _N(B). \end{aligned}$$
(5.8)

By applying Lemma 5.3, we get \(\forall ~ n\ge N\),

$$\begin{aligned}&\lambda _n(A+\pi _N B\pi _N)-\Vert (I-\pi _N)B(I-\pi _N)\Vert \le \lambda _n(A+B) , \\&\lambda _n(A+B) \le \lambda _n(A+\pi _N B\pi _N)+\Vert (I-\pi _N)B(I-\pi _N)\Vert . \end{aligned}$$

From (5.8), we obtain

$$\begin{aligned} \lambda _n(A+\pi _N B\pi _N)-\lambda _N(B)\le \lambda _n(A+B)\le \lambda _n(A+\pi _N B\pi _N)+\lambda _N(B). \end{aligned}$$

According to Lemma 5.4,

$$\begin{aligned} \lambda _{n+N}(A)-\lambda _N(B)\le \lambda _n(A+B)\le \lambda _{n-N}(A)+\lambda _N(B). \end{aligned}$$

We can thus conclude that

$$\begin{aligned} \forall N, \text { }\forall n\ge N: \text { } \lambda _{n+N}(A)-\lambda _N(B)\le \lambda _n(A+B)\le \lambda _{n-N}(A)+\lambda _N(B). \end{aligned}$$
(5.9)

Lemma 5.5

Let \(\{\lambda _n(A)\}\) and \(\{\lambda _n(B)\}\) be the eigenvalues of two self-adjoint operators \(A\) and \(B\). If \(\lambda _n(A)\sim c n^{-s}\) and \(\lambda _N(B)=\rho ^N\) where \(A\) and \(\rho \) are some constants, \(0 \le \rho \le 1\), then

$$\begin{aligned} \lambda _n(A+B)\sim c n^{-s}. \end{aligned}$$

Proof

Indeed, if we choose \(N(n)=n^{\delta }\), with \(\delta < 1\), then the inequalities (5.9) imply

$$\begin{aligned} c (n+n^{\delta })^{-s}(1+o(1))-\rho ^{n^{\delta }}\le \lambda _n(A+B)\le c(n-n^{\delta })^{-s}(1+o(1))+\rho ^{n^{\delta }}, \end{aligned}$$

As \(n \rightarrow \infty \), we have

$$\begin{aligned} c n^{-s}(1+o(1))\le \lambda _n(A+B)\le c n^{-s}(1+o(1)). \end{aligned}$$

Hence, we conclude that \(\lambda _n(A+B)\sim c n^{-s}\). \(\square \)

We now turn back to prove Theorem 5.1.

We consider the eigenpairs of the operator \(V\tilde{T}V\), where \(\tilde{T}\) is defined in Eq. (2.9), and \(V\) is defined by

$$\begin{aligned} V(e_n)=\sqrt{1-a_n}~ e_n\simeq \left( 1-\frac{1}{2}a_n\right) e_n \end{aligned}$$

where \(\{e_n\}\) are the eigenvectors of the operator \(V\tilde{T}V\), \(0 \le a_n \le 1\) (the last approximate equality is valid for \(a_n \rightarrow 0\) as \(n\rightarrow \infty \)). Setting

$$\begin{aligned} R(e_n)=\frac{1}{2}a_n e_n, \end{aligned}$$

then,

$$\begin{aligned} V\tilde{T}V=(I-R)\tilde{T}(I-R)=\tilde{T}-R\tilde{T}-\tilde{T}R-R\tilde{T}R. \end{aligned}$$

Let us denote by \(K_N\) the image of the orthonormal projection on the first \(N^\mathrm{th}\) eigenvectors of \(R\) and by \(R_N\) the image of the orthonormal projection on the remaining eigenvectors of \(R\). By definition, \(R=K_N+R_N\). Then,

$$\begin{aligned} V\tilde{T}V&=\tilde{T}-(K_N+R_N)\tilde{T}-\tilde{T}(K_N+R_N)+(K_N+R_N)\tilde{T}(K_N+R_N)\nonumber \\&=\tilde{T}\underbrace{-K_N\tilde{T}-\tilde{T}K_N+K_N\tilde{T}K_N+K_N\tilde{T}R_N+R_N\tilde{T}K_N}_{ \text {these operators have the finite rank which is equal to }N}-R_N\tilde{T}\nonumber \\&\quad -\tilde{T}R_N+R_N\tilde{T}R_N. \end{aligned}$$
(5.10)

We note that \(\mathrm{rank}(K_N)=N\) and in formula (5.10), whenever there is a \(K_N\), we have an operator of rank \(N\). Moreover, \(-R_N\tilde{T}-\tilde{T}R_N+R_N\tilde{T}R_N\) has the norm dominated by the \(N^\mathrm{th}\) eigenvalue of \(R\):

$$\begin{aligned} \Vert -R_N\tilde{T}-\tilde{T}R_N+R_N\tilde{T}R_N\Vert&\le \Vert R_N \tilde{T}\Vert +\Vert \tilde{T} R_N\Vert +\Vert R_N\tilde{T}R_N\Vert \\&\le c \Vert R_N\Vert \le c \lambda _N(R)=\frac{c}{2}a_N. \end{aligned}$$

Since the operator \(\tilde{T}\) is the solution of the Sturm–Liouville problem, \(\lambda _n(\tilde{T})\sim A_{\epsilon } n^{-2}\). Hence, according to Lemma 5.5, we get

$$\begin{aligned} \lambda _n(V\tilde{T}V)\sim A_{\epsilon } n^{-2}. \end{aligned}$$

Appendix B: Numerical computation of spectral characteristics

We briefly present a numerical algorithm to compute the spectral characteristics \(\lambda _n\) and \(\psi _n\). In order to compute the eigenvalues \(\lambda _n\) and the eigenvectors \(e_n\) of the operator \(V\tilde{T}V\), we get an explicit representation of this operator in the basis \(\cos n\theta \). First, we find

$$\begin{aligned} \tilde{T}(\cos n\theta )= {\left\{ \begin{array}{ll} \frac{\cos n\theta -\cos n(\pi -\epsilon )}{n^2}, &{} 0 \le \theta < \pi -\epsilon ,\\ 0, &{} \pi -\epsilon \le \theta \le \pi , \end{array}\right. } \end{aligned}$$
(6.1)

and

$$\begin{aligned} \tilde{T}(1)(\theta )= {\left\{ \begin{array}{ll} \frac{(\pi -\epsilon )^2-\theta ^2}{2}, &{} 0 \le \theta < \pi -\epsilon ,\\ 0, &{} \pi -\epsilon \le \theta \le \pi , \end{array}\right. } \end{aligned}$$
(6.2)

from which the expansion of \(\tilde{T}(\cos n\theta )\) (\(n\ge 0\)) in the basis \(\{\cos n\theta \}\) of \(L_{\text {even}}^{2}{[0,\pi ]}\) is

$$\begin{aligned} \tilde{T}(\cos n\theta ) = \sum _{m\ge 0} \mathbf{T}_{mn}\cos m\theta \quad n\ge 0 , \end{aligned}$$
(6.3)

where the coefficients \(\mathbf{T}_{mn}\) are defined by

(6.4)

In turn, the operator \(V\) has a diagonal representation (we set \(R = 1\)):

$$\begin{aligned} \mathbf{V}_{mn}= {\left\{ \begin{array}{ll} \sqrt{1-(1-a)^m} &{} (m=n;~ m,n\ge 0), \\ 0 &{} (m\ne n;~ m,n\ge 0) . \end{array}\right. } \end{aligned}$$
(6.5)

Combining these results, the operator \(V\tilde{T}V\) is represented by the infinite-dimensional matrix \(\mathbf{V}\mathbf{T}\mathbf{V}\) whose elements are

$$\begin{aligned}{}[\mathbf{V}\mathbf{T}\mathbf{V}]_{m,n}&= \frac{1}{\pi } \sqrt{1-(1-a)^n} ~ \sqrt{1-(1-a)^m}\biggl \{ \delta _{mn}\frac{1}{n^2}\left( \pi -\epsilon + \frac{\sin 2n \epsilon }{2n}\right) \nonumber \\&\quad - (1-\delta _{mn}) \frac{(-1)^{m+n}}{mn} \biggl [\frac{\sin (m-n)\epsilon }{m-n} - \frac{\sin (m+n)\epsilon }{m+n}\biggr ] \biggr \} \quad (m\ge 1;~ n\ge 1), \nonumber \\ \end{aligned}$$
(6.6)

and \([\mathbf{V}\mathbf{T}\mathbf{V}]_{m,n} = 0\) if \(m = 0\) or \(n = 0\). Finding the eigenvalues \(\{\lambda _n\}\) and the eigenvectors \(\{e_n\}\) of the operator \(V\tilde{T}V\) is equivalent to finding the eigenpairs of the associated matrix \(\mathbf{V}\mathbf{T}\mathbf{V}\). Note that this matrix is symmetric.

The matrix \(\mathbf{V}\mathbf{T}\mathbf{V}\) is diagonalized in Matlab that finds the eigenvalues \(\lambda _n\) and the coefficients \(v_{mn}\) determining the orthonormal basis \(\{e_n\}_{n \ge 0}\) as

$$\begin{aligned} e_n(\theta ) = \sqrt{\frac{1}{\pi }}~ v_{0n} + \sqrt{\frac{2}{\pi }} \sum _{m\ge 1} v_{mn} \cos m\theta . \end{aligned}$$
(6.7)

The spectral weights \(\psi _n\) are then given as

$$\begin{aligned} \psi _n = \langle \psi , e_n \rangle = \sqrt{\frac{2}{\pi }} \sum _{m\ge 1} v_{mn} \langle \psi , \cos m\theta \rangle , \end{aligned}$$
(6.8)

where

$$\begin{aligned} \langle \psi ,\cos m\theta \rangle = \sqrt{1-(1-a)^m}~\frac{(-1)^{m+1}}{m^2}\left[ (\pi -\epsilon )\cos m\epsilon + \frac{\sin m\epsilon }{m}\right] , \end{aligned}$$

and \(\langle \psi ,1 \rangle = \langle V\tilde{T}(1), 1 \rangle = \langle \tilde{T}(1), V1 \rangle = 0\).

Appendix C: Pure bulk diffusion phase

In [24], the mean exit time for pure bulk diffusion phase (\(\lambda = \infty \), \(a = 0\)) was found to be

$$\begin{aligned} t_2(r,\theta ) = \frac{R^2}{D_2} \left( \frac{1-(r/R)^2}{4} + \frac{\alpha _0}{2} + \sum \limits _{n=1}^\infty \alpha _n (r/R)^n \cos n\theta \right) , \end{aligned}$$
(7.1)

where

$$\begin{aligned} \alpha _0&= - 2\ln [\sin (\epsilon /2)] , \end{aligned}$$
(7.2)
$$\begin{aligned} \alpha _n&= \frac{(-1)^{n-1}}{2n} \bigl [P_n(\cos \epsilon ) + P_{n-1}(\cos \epsilon )\bigr ], \end{aligned}$$
(7.3)

where \(P_n(z)\) are Legendre polynomials. In the limit \(r\rightarrow R\), one gets

(7.4)

The average of the uniformly distributed starting point \(\theta \) yields

$$\begin{aligned} \langle t_1\rangle _b = \frac{R^2}{\pi D_2} \left( -(\pi -\epsilon )\ln [\sin (\epsilon /2)] + \sum \limits _{n=1}^\infty \frac{\sin n\epsilon }{2n^2} \bigl [P_n(\cos \epsilon ) + P_{n-1}(\cos \epsilon )\bigr ] \right) .\nonumber \\ \end{aligned}$$
(7.5)

For small \(\epsilon \), the first term dominates yielding \(\langle t_1\rangle _b \simeq \frac{R^2}{D_2} \ln (2/\epsilon ) (1 + O(\epsilon ))\).

Figure 11 shows the mean exit times \(\langle t_1\rangle _{\lambda =0}\) and \(\langle t_1\rangle _{\lambda = \infty }\) from Eqs. (2.40) and (7.5) for surface diffusion and pure bulk diffusion, as a function of \(\epsilon \).

Fig. 11
figure 11

Mean exit times \(\langle t_1\rangle _{\lambda =0}\) and \(\langle t_1\rangle _{\lambda = \infty }\) from Eqs. (2.40) and (7.5) for surface diffusion and pure bulk diffusion, as a function of \(\epsilon \). These times are multiplied by the corresponding diffusion coefficients \(D_1\) and \(D_2\)

Appendix D: Transportation case (\(a=R\))

As we earlier discussed, one typically considers small values of the reflection distance \(a\). Nevertheless, the results of this paper are applicable to any value of \(a\) from \(0\) to \(R\). The so-called transportation case \(a = R\) when the particle is reflected to the origin of the disk, was studied by Bénichou et al. [14]. In this case, successive explorations between any two reflections are independent that allows one to get much simpler formulas. For instance, in the limit \(\lambda \rightarrow \infty \), the Laplace-transformed probability density of the exit time, \({\mathcal L}[P(t)](s)\) (with a uniformly chosen initial point on the circle), has a simple expression: \({\mathcal L}[P(t)](s) = q\bigl [1 - (1-q)/I_0\bigl (\sqrt{s R^2/D_2}\bigr )\bigr ]^{-1}\), where \(q = \epsilon /\pi \), and \(1/I_0(\sqrt{s R^2/D_2})\) is the Laplace-transformed probability density for the first passage time to the circle when started from the origin (with \(I_0(z)\) being the modified Bessel function of the first kind). As a consequence, the mean exit is simply

$$\begin{aligned} {\mathcal T}_{a=1} = \frac{R^2(\pi - \epsilon )}{4D_2 \epsilon } . \end{aligned}$$
(8.1)

This limit is clearly seen on Fig. 8.

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Bénichou, O., Grebenkov, D.S., Hillairet, L. et al. Mean exit time for surface-mediated diffusion: spectral analysis and asymptotic behavior. Anal.Math.Phys. 5, 321–362 (2015). https://doi.org/10.1007/s13324-015-0098-0

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