Exit and Occupation times for Brownian Motion on Graphs with General Drift and Diffusion Constant
aa r X i v : . [ c ond - m a t . s t a t - m ec h ] O c t Exit and Occupation times for Brownian Motion on Graphs withGeneral Drift and Diffusion Constant
Olivier B´enichou and Jean Desbois October 31, 2018 Laboratoire de Physique Th´eorique de la Mati`ere Condens´ee, Universit´e Pierre et Marie Curie, 4, placeJussieu, 75005 Paris, France. Laboratoire de Physique Th´eorique et Mod`eles Statistiques. Universit´e Paris-Sud, Bˆat. 100, F-91405Orsay Cedex, France.
Abstract
We consider a particle diffusing along the links of a general graph possessing some absorbingvertices. The particle, with a spatially-dependent diffusion constant D ( x ) is subjected to adrift U ( x ) that is defined in every point of each link.We establish the boundary conditions to be used at the vertices and we derive generalexpressions for the average time spent on a part of the graph before absorption and, also,for the Laplace transform of the joint law of the occupation times. Exit times distributionsand splitting probabilities are also studied and several examples are discussed. For many years, graphs have interested physicists as well as mathematicians. For instance,equilibrium statistical physics widely uses model systems defined on lattices, the most popularbeing certainly the Ising model [1]. On another hand, in solid-state physics, tight-bindingmodels (see, for instance, [2]) involve discretized versions of Schr¨odinger operators on graphs.For all those models, the thermodynamic properties of the system heavily depend on geometricalcharacteristics of the lattice such as the connectivity and the dimensionality of the embeddingspace. However, in general, they don’t depend explicitly on the lengths of the edges. Randomwalks on graphs, where a particle hops from one vertex to one of its nearest-neighbours, havealso been studied by considering discrete Laplacian operators on graphs [3].Such Laplacian operators can also be useful if they are defined on each link of the graph. Forexample, in the context of organic molecules [4], they can describe free electrons on networksmade of one-dimensional wires. Many other applications can be found in the physical literature.Let us simply cite the study of vibrational properties of fractal structures such as the Sierpin-ski gasket [5] or the investigation of quantum transport in mesoscopic physics [6, 7]. Weaklydisordered systems can also be studied in that context [8]. It appears that weak localizationcorrections in the presence of an eventual magnetic field are related to a spectral determinant onthe graph. This last quantity is actually of central importance and interesting by itself [9, 10].In particular, it allows to recover a trace formula that was first derived by Roth [11]. Moreover,the spectral determinant, when computed with generalized boundary conditions at the vertices,is useful to enumerate constrained random walks on a general graph [12], a problem that hasbeen addressed many times in the mathematical literature [13].1rownian motion on graphs is also worthwhile to be investigated from, both, the physical andmathematical viewpoints. For instance, the probability distribution of the time spent on a link(the so-called occupation time) was first studied by P Levy [14] who considered the time spenton an infinite half-line by a one-dimensional brownian motion stopped at some time t . Thiswork allowed Levy to discover in 1939 one of his famous arc-sine laws [15]. Since that time, thisresult has been generalized to a star-graph [16] and also to a quite general graph [17]. Localtime distributions have also been obtained in [18].It has been pointed out since a long time that first-passage times and, more generally, occupationtimes are of special interest in the context of reaction-diffusion processes [19, 20]. Computationsof such quantities in the presence of a constant external field have already been performed forone-dimensional systems with absorbing points (see, for example, [21]). This was done with thehelp of a linear Fokker-Planck equation [19, 22].The purpose of the present work is to extend those results on a general graph with some absorbingvertices. We will consider a brownian particle diffusing with a spatially-dependent diffusionconstant and subjected to a drift that is defined in every point of each link. The paper isorganized as follows. In section 2, we present the notations that will be used throughout thepaper. We discuss the boundary conditions to be used at each vertex in section 3 and, also, inthe Appendices. More precisely, we analyse in details specific graphs in the Appendices A andB. The obtained results allow to deal with a general graph in Appendix C. Section 4 is devotedto the computation of the average time spent, before absorption, by a brownian particle on apart of the graph. In this section, we also calculate the Laplace transform of the joint law of theoccupation times on each link. In the following section, we present additional results, especiallyconcerning conditional and splitting probabilities. Various examples are discussed all along thedifferent sections. Finally, a brief summary is given in section 6. Let us consider a general graph G made of V vertices linked by B bonds of finite lengths. Oneach bond [ αβ ], of length l αβ , we define the coordinate x αβ that runs from 0 (vertex α ) to l αβ (vertex β ). (We have, of course, x βα = l αβ − x αβ ).Moreover, we suppose that, among all the vertices, N of them are absorbing. (A particle getstrapped if it reaches such a vertex).We will study the motion on G of a brownian particle that starts at t = 0 from some non-absorbing point x . The particle with a spatially-dependent diffusion constant D ( x ) is subjectedto a drift U ( x ) defined on the bonds of G . More precisely, D ( x ) and U ( x ) are differentiablefunctions of x on each link. In particular, on each link [ αβ ], the following limits D ( αβ ) ≡ lim x αβ → D ( x αβ ), D ′ ( αβ ) ≡ lim x αβ → ∂∂x αβ D ( x αβ ), U ( αβ ) ≡ lim x αβ → U ( x αβ ), ..., are well defined.Such notations will be used extensively throughout the paper.The continuity properties of D ( x ) and U ( x ) at each vertex will be discussed in the followingsection.We also specify the motion of the particle when it reaches some vertex α . Let us call β i ( i = 1 , . . . , m α ) the nearest neighbours of α . We assume that the particle will come out towards β i with some arbitrary probability p αβ i ( P i p αβ i = 1 – see [16] for a rigourous mathematicaldefinition). Of course, p λµ = 0 if λ is an absorbing vertex or if [ λµ ] is not a bond of G .Let P ( yt | x
0) be the probability density to find the particle at point y at time t ( P ( y | x
0) = δ ( y − x )). It satisfies on each link [ αβ ] the backward and forward Fokker-Planck equations:2 β ix i x’ i ∆ x Figure 1: The vertex α with its nearest-neighbours β i , i = 1 , ..., m α ; each link is discretized withsteps of lengths ∆ x . ∂P ( yt | x αβ ∂t = D ( x αβ ) ∂ P∂x αβ − ∂U ( x αβ ) ∂x αβ ∂P∂x αβ ≡ L + ( x αβ ) P ( yt | x αβ
0) (1) ∂P ( y αβ t | x ∂t = ∂∂y αβ " ∂∂y αβ ( D ( y αβ ) P ) + ∂U ( y αβ ) ∂y αβ P ≡ L ( y αβ ) P ( y αβ t | x
0) (2) P ( αβ ) will mean lim x αβ → P ( yt | x αβ
0) when we use the backward Fokker-Planck equation andlim y αβ → P ( y αβ t | x
0) when we use the forward equation. The derivatives P ′ ( αβ ) will be definedin a similar way. Let us define the following two situations that can occur at some non absorbing vertex α :(A) : D ( x ) is continuous but U ( x ) is not(B) : U ( x ) is continuous but D ( x ) is notThe main purpose of this section is to establish the boundary conditions for P and its derivativesthat result from those discontinuities. We will not consider the case when both U ( x ) and D ( x )are discontinuous at the same vertex because, in our opinion, it is ill-defined. Let us start by considering a graph G with U ( x ) and D ( x ) constant (standard brownian motion).In Figure 1, we display a given vertex α and its nearest-neighbours β i , i = 1 , ..., m α on G .The transition probabilities from α , p α β i , are not supposed to be all equal. In order to establishthe boundary conditions for the backward equation, we discretize all the links [ αβ i ] (and alsothe time) with steps of length ∆ x (resp. ∆ t ). It is easy to realize that:3 ( y, ( N + 1)∆ t | α,
0) = m α X i =1 p αβ i P ( y, N ∆ t | x i ,
0) (3) P ( y, ( N + 1)∆ t | x i ,
0) = 12 P ( y, N ∆ t | α,
0) + 12 P ( y, N ∆ t | x ′ i ,
0) (4)Taking the limit ∆ x →
0, ∆ t ∝ (∆ x ) , N → ∞ , N ∆ t ≡ t , we obtain, with (4): P ( αβ i ) = 12 P ( y, t | α,
0) + 12 P ( αβ i ) (5)Thus P ( αβ i ) = P ( y, t | α, ∀ i (6)This shows that P is continuous in α .Moreover, expanding (3) at order ∆ x , we get: P ( y, t | α,
0) = m α X i =1 p αβ i P ( αβ i ) + ∆ x m α X i =1 p αβ i P ′ ( αβ i ) ! + O ((∆ x ) ) (7)With (6) and P m α i =1 p αβ i = 1, we show that: m α X i =1 p αβ i P ′ ( αβ i ) = 0 (8)On the other hand, for p αβ = p αβ = ... = 1 /m α , the equation (1) on the link [ αβ ] can bewritten in the form: ∂∂t D ( x αβ ) e − Φ( x αβ ) P ( yt | x αβ ! = ∂∂x αβ e − Φ( x αβ ) ∂∂x αβ P ( yt | x αβ ! (9)with Φ( x αβ ) = Z x αβ x ∂U ( x ′ ) ∂x ′ d x ′ D ( x ′ ) (10) x is some point on the graph.We are aware that Φ( x αβ ) could be multi-valued because, in general, a graph is multiply-connected . However, this is not the case if we restrict ourselves to the vicinity of the givenvertex α . Choosing x located on a link starting from α , the integral in (10) involves in a uniqueway, at most, two integrals along links starting from α . It is well defined if U ( x ) and D ( x )are not discontinuous at the same point. So, with this definition of Φ( x αβ ), the equation (9) iswell-suited to study the boundary conditions at vertex α .Let us consider the case (A).We assume, first, that the p αβ i ’s are all equal. It can also occur that Φ( x ) is not well defined, for instance, when U ( x ) and D ( x ) are both discontinuous atthe same point x . Indeed, 1 D ( x ′ ) ∂U ( x ′ ) ∂x ′ is then proportional to δ ( x ′ − x ) /D ( x ); thus, this quantity is notdefined if D ( x ) is discontinuous at x . α , we get, with D ( x ) continuous at α : m α X i =1 e − U ( αβi ) /D ( α ) P ′ ( αβ i ) = 0 (11)Moreover, P must be continuous at α if we want the quantity e − Φ ∂P∂x to be properly defined.In the Appendix A, those boundary conditions are directly established, for the Laplace Transformof P , on a simple graph.Still for the case (A), let us now assume that the p αβ i ’s are not all equal.In Appendix C, we establish the following boundary condition that must hold for a generalgraph: m α X i =1 p αβ i e − U ( αβi ) /D ( α ) P ′ ( αβ i ) = 0 (12)We also show in this Appendix that P is continuous in α .Turning to the case (B), we can follow the same line as for (A) and use (9) and also theAppendices A and B.Remark that, when the p αβ i ’s ( i = 1 , , ..., m α ) are all equal, the integration of (9) on thevicinity of α will not produce an exponential factor as in (11). This is because, this time, U ( x ) is continuous in α . This fact is confirmed by the direct computations performed in theAppendices. Finally, with the p αβ i ’s not all equal, we get: m α X i =1 p αβ i P ′ ( αβ i ) = 0 (13)Moreover, P is continuous for the cases (A) and (B).In summary, for the backward Fokker-Planck equation, the condition on the derivatives can bewritten: m α X i =1 p ′ αβ i P ′ ( αβ i ) = 0 (14)with p ′ αβ i = p αβ i e − U ( αβi ) D ( α ) case (A) (15)= p αβ i case (B) (16)This notation will show especially useful in the following sections where the backward Fokker-Planck equation is widely used. 5 .2 Forward Fokker-Planck equation Coming back to Figure 1 (with x i , x ′ i and ∆ x replaced by y i , y ′ i and ∆ y ), we consider thediscretized version of the forward equation with D ( x ) and U ( x ) constant and, also, the p αβ i ’snot all equal. We have, on the link [ αβ i ] P ( y i , ( N + 1)∆ t | x,
0) = p αβ i P ( α, N ∆ t | x,
0) + 12 P ( y ′ i , N ∆ t | x,
0) (17) P ( α, ( N + 1)∆ t | x,
0) = 12 m α X i =1 P ( y i , N ∆ t | x,
0) (18)With the limit ∆ y →
0, ∆ t ∝ (∆ y ) , N → ∞ , N ∆ t ≡ t , (17) leads to: P ( αβ i ) = p αβ i P ( α, t | x,
0) + 12 P ( αβ i ) (19)Thus P ( αβ ) p αβ = P ( αβ ) p αβ = ... = P ( αβ mα ) p αβ mα = 2 P ( α, t | x,
0) (20)We see that, in general, P is not continuous in α .Moreover, expanding (18) at order ∆ y , we get: P ( α, t | x,
0) = 12 m α X i =1 (cid:16) P ( αβ i ) + ∆ yP ′ ( αβ i ) (cid:17) + O ((∆ y ) ) (21)With (20), we can write: m α X i =1 P ′ ( αβ i ) = 0 (22)So, the current conservation doesn’t involve the p αβ i ’s.Now, for p αβ = p αβ = ... = 1 /m α , the equation (2) on the link [ αβ ] can be written: ∂∂t P ( y αβ t | x
0) = ∂∂y αβ e − Φ( y αβ ) ∂∂y αβ (cid:16) D ( y αβ ) e Φ( y αβ ) P ( y αβ t | x (cid:17)! (23)with Φ( y αβ ) = Z y αβ y ∂U ( y ′ ) ∂y ′ d y ′ D ( y ′ ) (24) y is some point on the graph; the discussion of section 3.1 for the definition of Φ is, of course,still relevant.Following the same lines as in section 3.1, we get the current conservation at each non absorbingvertex α : m α X i =1 (cid:0) ( DP ) ′ + U ′ P (cid:1) ( αβ i ) = 0 (25)if α is not the starting point. Otherwise, the right-hand side of (25) should be replaced by − δ ( t ).(25) doesn’t depend on the continuity properties of D ( x ) and U ( x ).6ow, let us consider the case (A) and call D ( α ) the diffusion constant at α .When p αβ = ... = p αβ mα , following [22], we can show that e U ( αβ D ( α ) P ( αβ ) = e U ( αβ D ( α ) P ( αβ ) = ... = e U ( αβmα ) D ( α ) P ( αβ mα ) (26)When the p αβ i ’s are not all equal, we use the same approach as the one of Appendix C and get:(A) : e U ( αβ D ( α ) P ( αβ ) p αβ = e U ( αβ D ( α ) P ( αβ ) p αβ = ... = e U ( αβmα ) D ( α ) P ( αβ mα ) p αβ mα (27)Similarly, for the case (B), we obtain:(B) : D ( αβ ) P ( αβ ) p αβ = D ( αβ ) P ( αβ ) p αβ = ... = D ( αβ mα ) P ( αβ mα ) p αβ mα (28)When p αβ = ... = p αβ mα , it is worth to notice that the conditions (27) and (28) are simplyobtained if we want the quantity ∂∂y αβ (cid:16) D ( y αβ ) e Φ( y αβ ) P ( y αβ t | x (cid:17) , that appears in (23), to beproperly defined.In the following section, we will check the consistency of those boundary conditions (for theBackward and the Forward equations) on several examples. Let us first study the average time, h τ ( x ) i , spent on a part D of G by the particle beforeabsorption. We have [21]: h τ ( x ) i = Z ∞ d t Z D d y P ( yt | x
0) (29)( h τ ( x ) i is the sum of the infinitesimal residence times weighted by the probability of presencein D at time t ).With (1), we get for h τ ( x ) i the equation: L + ( x αβ ) h τ ( x αβ ) i = − D ( x αβ ) (30) D ( x ) is the characteristic function of the domain D : D ( x ) = 1 if x ∈ D , = 0 otherwise.The solution writes: h τ ( x αβ ) i = Z x αβ d u αβ φ ( u αβ ) + A αβ Z x αβ d u αβ I ( u αβ ) + B αβ (31)with φ ( x αβ ) = − I ( x αβ ) Z x αβ d z αβ D ( z αβ ) D ( z αβ ) I ( z αβ ) ! (32) I ( x αβ ) = exp Z x αβ d z αβ D ( z αβ ) ∂U∂z αβ ! (33)7he constants A αβ and B αβ are determined by imposing the boundary conditions at each vertex.Continuity implies: lim x αβi → h τ ( x αβ i ) i ≡ h τ α i = B αβ i (34)and lim x αβi → l αβi h τ ( x αβ i ) i ≡ h τ β i i = K ( αβ i ) + A αβ i J ( αβ i ) + B αβ i (35)where K ( αβ ) = Z l αβ d u αβ φ ( u αβ ) (36) J ( αβ ) = Z l αβ d u αβ I ( u αβ ) (37)In general, K ( αβ ) = K ( βα ) , J ( αβ ) = J ( βα ) .Of course, h τ λ i = 0 if λ is absorbing. So, we will determine h τ α i only for α non-absorbing. Insuch a vertex, the current conservation leads to (14-16): X i p ′ αβ i A αβ i = 0 (38)The equations for the h τ α i ’s can be written in a matrix form and finally the average time spenton D , before absorption, by a particle starting from α is given by: h τ α i = det M det M (39) M and M are two ( V − N ) × ( V − N ) matrices with the elements: M ii = X m p ′ im J ( im ) (40) M ij = − p ′ ij J ( ij ) (41)(In (40,41), i and j run only over non-absorbing vertices but m labels all kinds of vertices). M = M except for the α th column:( M ) iα = − X m p ′ im K ( im ) J ( im ) (42)We observe that, for a graph without any absorbing vertex, we have P j M ij = 0 ∀ i . Thus,det M vanishes and h τ α i becomes infinite as expected.Remark also that, when D = G , τ becomes the survival time or, accordingly, the first-passagetime in any absorbing vertex. EXAMPLE 1.As a first example, let us consider the case D = G for the graphs of Figure 2:– in a), the vertex 1 is absorbing and the particle starts from 0. We get h τ i = − K (01) (43)8 (cid:0)(cid:0)(cid:0)(cid:1)(cid:1)(cid:1)(cid:1)(cid:0)(cid:0)(cid:0)(cid:0)(cid:1)(cid:1)(cid:1)(cid:1) (cid:0)(cid:0)(cid:0)(cid:0)(cid:1)(cid:1)(cid:1)(cid:1) nc)a)b) λ Figure 2: Three simple graphs that are used to compute survival times (parts a) and b)) orcovering time (c)). In a) and b), the vertex 1 is absorbing. The brownian particle starts from 0(parts a) and c)) and from λ (part b)). For further explanations, see the text. ∆ U x x λ U Figure 3: The potential is symmetric around 0 except for a step of magnitude ∆ U .9 in b), 1 is still absorbing and the particle starts from λ . Its probability transition in 0 is p = q ( p λ = 1 − q ). Moreover, we assume that l = l λ , U ( x ) = U ( x λ ) + ∆ U (a stepof magnitude ∆ U occurs in 0, see Figure 3) and D ( x ) = D ( x λ ). The mean survival time isgiven by: h τ λ ( q ) i = − (cid:18) K (01) + K (10) q (cid:19) (cid:18) q + (1 − q ) e ∆ UD (0) (cid:19) (44)– c) represents a symmetric star with n legs of length l originating from 0 ( p i = 1 n and U ( x i ) = U ( x j ) (no step in 0 this time) , D ( x i ) = D ( x j ) ∀ i, j = 1 , . . . , n ). Let us computethe average covering time h τ c i (smallest time necessary to reach all the points of the star at leastonce). With the notations of (43) and (44), we get, with ∆ U = 0: h τ c i = h τ i + n − X k =1 h τ λ (1 − k/n ) i = (45)= − K (01) − ( K (01) + K (10) ) n n − X k =1 k (46) ∼ − ( K (01) + K (10) ) n ln n when n → ∞ (47)(Recall that for random walks on a star, h τ c i ∼ n ln n [23])EXAMPLE 2.Let us consider the graph of Figure 4. Among the five vertices, two are absorbing (3 and 4). Thestarting point is 0. U ( x ) is discontinuous in vertices 1 and 2, D ( x ) is discontinuous in vertex0. All the links have the same length l . With the backward equation, we get the mean survivaltime (39): h τ i = l (cid:18) p D + p D (cid:19) (cid:30) (cid:18) p p p + p e ( U − U ) /D + p p p + p e ( U − U ) /D (cid:19) (48)Let us now compute h τ i with the forward equation. Defining P ( y ) ≡ R ∞ d t P ( yt | P ( y ) = a y + a (49)on the link [10] : P ( y ) = b y + b (50)on the link [02] : P ( y ) = c y + c (51)on the link [24] : P ( y ) = d y + d (52)with the boundary conditions P (31) = 0 P (42) = 0 D P (01) p = D P (02) p (53) e U /D P (13) p = e U/D P (10) p e U /D P (24) p = e U/D P (20) p (54) P ′ (13) + P ′ (10) = 0 P ′ (24) + P ′ (20) = 0 D P ′ (02) + D P ′ (01) = − h τ i = R G P ( y )d y = ( b + c ) l , we recover the resultequation (48).EXAMPLE 3. 10 D D U U l l l l
13 0 42
U xxU(x)D(x)
Figure 4: The graph G consists in five vertices (3 and 4: absorbing, 0: starting point), U ( x ) isdiscontinuous in 1 and 2 and D ( x ) is discontinuous in 0.For the graph of Figure 5, the starting point is 0, a pure reflection occurs in 1 ( p = 1) and thevertex 2 is absorbing. The potential is U ( x ) = a x , U ( x ) = − a x . With the notations ofthe Figure and using the backward result (39), we get the mean residence time:on the link [10] h τ i = p p D a (cid:16) e al /D − (cid:17) (cid:16) e al /D − (cid:17) (56)on the link [02] h τ i = D a (cid:18) e al /D − − al D (cid:19) (57)Taking the limit a → h τ i = p p l l D (58)on the link [02] h τ i = l D (59)Turning to the forward equation, still with P ( y ) ≡ R ∞ d t P ( yt | P ( y ) = c e ay /D + c (60)on the link [02] : P ( y ) = c e − ay /D + c (61)with the conditions P (20) = 0 D P (01) p = D P (02) p D P ′ (10) + a P (10) = 0 (62) D P ′ (02) + D P ′ (01) + a (cid:16) P (02) − P (01) (cid:17) = − h τ i = R D P ( y ) d y = c D a (cid:16) e al /D − (cid:17) + c l for the link [10], = c D a (cid:16) − e − al /D (cid:17) + c l for the link [02], we recover the solution (56,57).11 D l l Figure 5: The graph G consists in three vertices (2: absorbing, 0: starting point, pure reflectionin 1); the drift U ( x ) is linear and D ( x ) is discontinuous in 0. Still with the same conditions (drift, spatially-dependent diffusion constant, absorbing ver-tices,. . . ), we now study a Brownian motion stopped at t and call T αβ the time spent up to t on the link [ αβ ]. T αβ is, of course, a random variable depending on t and we will call P t ( { T αβ } ) the joint distri-bution of the T αβ ’s.In the following, we will focus our attention on the quantity S ( t | x
0) =
Z Y [ αβ ] d T αβ P t ( { T αβ } ) e − P [ ij ] ξ ij T ij (64) ≡ (cid:28) e − P [ ij ] ξ ij T ij ( x ) (cid:29) (65)where the ξ ij ’s are positive constants and h . . . ( x ) i stands for averaging over all the browniantrajectories starting from x and developing up to t in the presence of the drift.In order to stick to S ( t | x αβ ], a loss term proportional to ξ αβ ( ξ αβ may be interpreted as a reaction rate per unitlength and time). Thus, we consider the equation: ∂Q∂t = (cid:0) L + ( x αβ ) − ξ αβ (cid:1) Q (66)where Q ( yt | x
0) represents the probability of finding the particle in y at time t , each path from x to y being, this time, weighted by a factor e − P [ αβ ] ξ αβ T αβ . (Notice that Q ( yt | x
0) = δ ( y − x )if x is absorbing). 12t is now easy to realize that we have the relationship: S ( t | x
0) = Z G d y Q ( yt | x
0) (67)In the following, we will be especially interested in the Laplace Transform of S ( t | x b S ( γ | x ≡ b S ( x ) = Z ∞ d t e − γt S ( t | x
0) (68)On the bond [ αβ ], b S ( x αβ ) satisfies the following equation ( γ αβ ≡ γ + ξ αβ ): (cid:0) L + ( x αβ ) − γ αβ (cid:1) b S = − b S ( x αβ ) = 1 γ αβ + Φ( x αβ ) and performing the transformationΦ( x αβ ) = q I ( x αβ ) χ ( x αβ ) (70)( I is defined in (33)), we are left with the following equation for χ : − D ∂ χ∂x αβ + D ∂U ( x αβ ) ∂x αβ ! − D ∂∂x αβ D ∂U ( x αβ ) ∂x αβ ! + γ αβ χ = 0 (71)Let us call χ αβ and χ βα two solutions of (71) such that χ αβ ( α ) = χ βα ( β ) = 1, χ βα ( α ) = χ αβ ( β ) = 0. So, b S writes: b S = 1 γ αβ + A αβ exp Z x αβ d z αβ D ( z αβ ) ∂U ( z αβ ) ∂z αβ ! χ αβ ( x αβ )+ A βα exp Z x αβ l αβ d z αβ D ( z αβ ) ∂U ( z αβ ) ∂z αβ ! χ βα ( x αβ ) (72)The constants A ij are determined by imposing the boundary conditions. Continuity of b S ateach vertex α implies: lim x αβi → b S ( x αβ i ) ≡ b S α = 1 γ αβ i + A αβ i (73)Moreover, if α is absorbing then T λµ = 0 ∀ [ λµ ] and, from (64) and (68), we get: b S α = 1 γ = 1 γ αβ i + A αβ i (74)Following the same lines as for h τ α i , it is easy to show that the b S α ’s ( α non-absorbing) satisfya matrix equation with the solution: b S α = det R det R (75) R and R are again ( V − N ) × ( V − N ) matrices with elements R ii = X m p ′ im C im (76) R ij = p ′ ij W ij (77)13he quantities C im and W ij being given by: C im = lim x im → ∂χ im ∂x im + ∂U∂x im D ( x im ) ! (78) W ij = exp Z l ij ∂U∂x ij D ( x ij ) d x ij lim x ij → ∂χ ji ∂x ij (79) R = R except for the α th column:( R ) iα = X j p ′ ij γ ij ( C ij + W ij ) − γ X k abs. p ′ ik W ik (80)(The last summation is performed only over absorbing vertices).Setting ξ ij ≡
0, we establish that det R = 1 γ det R , showing that the probability distribution P t ( { T αβ } ) is properly normalized (see (64) and (68)).Now, let us call T ′ αβ the time spent on [ αβ ] before absorption. Thus, for a particle starting from α , we can write (cid:28) e − P [ ij ] ξ ij T ′ ij ( α ) (cid:29) = lim t →∞ (cid:28) e − P [ ij ] ξ ij T ij ( α ) (cid:29) (81)and from (65,68) we deduce (cid:28) e − P [ ij ] ξ ij T ′ ij ( α ) (cid:29) = lim γ → (cid:16) γ b S α (cid:17) (82)Finally, we get: (cid:28) e − P [ ij ] ξ ij T ′ ij ( α ) (cid:29) = det R (0)1 det R (0) (83)with R (0) = lim γ → R and R (0)1 = R (0) except for the α th column that is given by (cid:16) R (0)1 (cid:17) iα = − lim γ → X k abs. p ′ ik W ik (84)(84) holds because lim γ → γγ ij p ′ ij ( C ij + W ij ) ! = 0This is obvious when ξ ij = 0. But when ξ ij = 0, this is still true because, in that case,lim γ → ( C ij + W ij ) = 0.Remark that (84) implies that (cid:28) e − P [ ij ] ξ ij T ′ ij ( α ) (cid:29) = 0 if G has no absorbing vertex (in thatcase, T ′ ij = ∞ ).Setting ξ ij ≡ ξ and X [ ij ] T ′ ij ≡ τ (survival time), (83) gives the expression of the Laplace Transformof the probability distribution of τ . 14 .2.1 Example To illustrate this work, let us go back to the example of Figure 2 and study the survival time(parts a) and b) – with ∆ U = 0) and covering time (part c)) distributions.With a), we get D e − ξτ E = − W C (85)( W αβ , C αβ ,. . . are computed with γ αβ = ξ ).b) leads to D e − ξτ λ ( q ) E = − qZ − q (86)with Z = C C W W For the covering time τ c in c) we get (with the notations of (85) and (86)): D e − ξτ c E = − D e − ξτ E n − Y k =1 D e − ξτ λ (1 − k/n ) E (87)= − W C ( n − n n − n − Y k =1 Z − kn (88)Computation of the first moments shows that when n → ∞ : h τ c i ∝ n ln n in agreement with (47) (89) D τ c E − h τ c i ∝ n (90)Then the probability distribution of the scaling variable τ c / h τ c i becomes more and more peakedat its mean value when n → ∞ : P (cid:18) τ c h τ c i = X (cid:19) → n →∞ δ ( X −
1) (91)
This section is essentially devoted to a detailed study of exit (or survival) times. First, wewill compute exit times distributions and, after, in the last parts, we will focus on quantitiesconcerning exit through a given absorbing vertex.
The density of exit time P ( t | x
0) (without specifying the absorbing vertex) is defined throughthe loss of probability: P ( t | x
0) d t = Z G d y P ( yt | x − Z G d y P ( y, t + d t | x
0) (92)Thus: P ( t | x
0) = − ∂∂t Z G d y P ( yt | x ≡ X µ abs. P µ ( t | x
0) (93)15 µ ( t | x
0) is the density of exit time by the absorbing vertex µ .With the forward Fokker-Planck equation, (93) leads to: P ( t | x
0) = − Z G d y L ( y ) P ( yt | x
0) = X µ abs. , i J µβ i (94) J µβ i = lim y µβi → ∂∂y µβ i ( DP ) + ∂U∂y µβ i P ! (95)= lim y µβi → D ∂P∂y µβ i ! (96)because P ( µβ i ) = 0 when µ is absorbing. (The β i ’s are the nearest-neighbours of µ on G .)We deduce that P µ ( t | x
0) is the probability current at vertex µ : P µ ( t | x
0) = X i J µβ i .Let us compute the Laplace transform, b P ( γ | x P ( t | x b P ( γ | x
0) = Z ∞ d te − γt (cid:18) − ∂∂t Z G d y P ( yt | x (cid:19) ≡ X µ abs. b P µ ( γ | x
0) (97)= 1 − γ Z ∞ d te − γt Z G d y P ( yt | x
0) (98)The backward Fokker-Planck equation gives: L + b P ( γ | x
0) = γ b P ( γ | x
0) (99)and L + c P µ ( γ | x
0) = γ b P µ ( γ | x
0) (100)Defining b P µ,α = lim x αi → b P µ ( γ | x αi λ : b P µ,λ = δ µλ (101)This is because P µ ( t | λ
0) = δ ( t ) if λ = µ (102)= 0 otherwise (103)Following section 4.2, we readily find, for a particle starting from α (non absorbing), the Laplacetransform of the exit (by the vertex µ ) time distribution: b P µ,α = det R ( µ,α )2 det R (104)where R ( µ,α )2 = R except for the α th column: (cid:16) R ( µ,α )2 (cid:17) iα = − p ′ iµ W iµ (105) R and W iµ are defined in (76-79) and computed, here, with ξ αβ ≡
0. ( p ′ iµ is defined in (15-16)).16 .2 Splitting probabilities For a particle starting at x , we define the splitting probability Π µ ( x ) as the probability of everreaching the absorbing vertex µ (rather than any other absorbing vertex). Such a quantityhas already been considered for one-dimensional systems [24] (see also [25] for extensions tohigher-dimensional systems).We have, obviously: Π µ ( x ) = Z ∞ d t P µ ( t | x
0) = b P µ ( γ = 0 | x
0) (106)Setting γ = 0 in (100) we see that Π µ ( x ) satisfies, on [ αβ ], the backward equation: L + ( x αβ )Π µ ( x αβ ) = 0 (107)The probability, Π µ,α , for a particle starting from the vertex α to be absorbed by µ , is definedas Π µ,α = lim x αi → Π µ ( x αi ).With (106) and (101), it is easy to realize thatfor an absorbing vertex λ : Π µ,λ = δ µλ (108)Following the same lines as previously (see section 4.1), we find that Π µ,α ( α non absorbing) isagain written as the ratio of two determinants:Π µ,α = det M ( µ,α )2 det M (109)where M ( µ,α )2 = M except for the α th column: (cid:16) M ( µ,α )2 (cid:17) iα = p ′ iµ J ( iµ ) (110)( M is defined in (40,41) and J ( iµ ) in (37)).With simple manipulations on determinants, we check the normalization condition X i abs. Π i,α =1.Let us, for one moment, comment the case when there is no drift ( U ( x ) constant but D ( x )variable, eventually discontinuous at some vertices). In that case p ′ ij = p ij and, also, J ( ij ) = l ij .We conclude that the splitting probabilities don’t depend on the varying diffusion constantwhen there is no drift. This fact can be understood in the following way. Let us consider adiscretization of each link and a continuous time. Modifying the diffusion constant amounts tochange the waiting time at each site of the discretized graph. But, this would not change thetrajectories if there is no drift. Only the time spent is changed. Finally, the splitting probabilitiesremain unaffected by a change of D ( x ). For a graph G without drift, we could expect, with the same argument, that the average time spent on a part D of G would not depend on the diffusion constant on the rest, G\D , of the graph. This is exactly what can bechecked with the formulae (39-42) of section 4.1 and, also, with the formulae (58,59) of the example 3 of section4.1.1. .2.1 Example Let us consider a star-graph without drift. The root 0 has m neighbours, all absorbing. The m links have lengths l i , i = 1 , ..., m . With (109), we obtainΠ i, = (cid:18) p i l i (cid:19) (cid:30) m X m =1 p m l m ! (111) We now turn to the study of the conditional mean first passage time h t µ ( x ) i , which is definedas the mean exit time, given that exit is through the absorbing vertex µ (rather than any otherabsorbing vertex). We set h t µ,α i = lim x αi → h t µ ( x αi ) i .Actually, it is simpler to first compute the quantity θ µ ( x ) ≡ Π µ ( x ) h t µ ( x ) i .Indeed, we have: θ µ ( x ) = Z ∞ d t t P µ ( t | x
0) (112) L + θ µ ( x ) = Z ∞ d t t ∂∂t P µ ( t | x
0) = − Z ∞ d t P µ ( t | x
0) (113)So, we get: L + θ µ ( x ) = − Π µ ( x ) (114)Moreover, for any absorbing vertex λ , we get: θ µ,λ = Π µ,λ h t µ,λ i = δ µ,λ h t µ,µ i = 0 because h t µ,µ i = 0.Thus, comparing this equation with eq.(30), we find for a particle starting from the vertex α Π µ,α h t µ,α i = det M ( µ,α )3 det M (115)where M ( µ,α )3 = M except for the α th column: (cid:16) M ( µ,α )3 (cid:17) iα = X m p ′ im f K ( µ )( im ) J ( im ) (116)with f K ( µ )( ij ) = Z l ij d u ij I ( u ij ) Z u ij d z ij Π µ ( z ij ) D ( z ij ) I ( z ij ) ! (117)In this last equation, Π µ ( z ij ) has to be computed by equationΠ µ ( z ij ) = Π µ,i + Π µ,j − Π µ,i J ( ij ) Z z ij d u ij I ( u ij ) (118)with Π µ,i given by eq.(109). 18 .3.1 Example With the same star-graph as in section 5.2.1 (no drift) and a diffusion constant equal to D i oneach link [0 i ], we get (111,115) h t µ, i = 16 l µ D µ + 13 m X m =1 p m l m D m ! (cid:30) m X m =1 p m l m ! (119) We have used the Fokker-Planck Equation (mainly the backward one) to compute some quan-tities relevant for the study of a Brownian particle moving on a graph. This particle, movingwith a varying diffusion constant, is also subjected to a drift.Being aware that this point is somewhat controversial (see, for instance, [26]), we have discussedin great details the boundary conditions in vertices where either U ( x ) or D ( x ) is discontinuous.Two Appendices support the study of those boundary conditions.Those preliminaries allowed us to deal with various quantities such as the mean residence time,the occupation times probabilities, the splitting probabilities and the conditional mean firstpassage time. In each case, we have established general formulae that can be used for all kindsof graphs.Finally, we have checked, on several examples, the consistency of the boundary conditions wehave put forward. A Backward Equation: boundary conditions when p = p For a direct computation of the boundary conditions, it is simpler to consider the LaplaceTransform: b S ( γ | x
0) = Z ∞ d t e − γt Z G d yP ( yt | x
0) (120)that satisfies the backward equation ( b S ( γ | x ≡ b S ( x )):( L + − γ ) b S ( x ) = − D ( x ) is continuous and U ( x ) is discontinuous at the vertex0. In particular U (01) ≡ U = U (02) ≡ U ; D (0) ≡ D . Moreover, vertices 1 and 2 are absorbingand, in 0, the transition probabilities are equal: p = p = 1 / L (see Figure 6 b)) in such a way thatnothing is changed for the rest (for example, the potential on the link [02] of the originalgraph is the same as the one on the link [32] of the modified graph, ...). On [03], we define U ( x ) = (cid:16) U − U L (cid:17) x + U and D ( x ) ≡ D . When L →
0, we recover Figure 6 a).Working with this modified graph, the solution of (121) writes:19 U U l l L+l L+l U −l −l −l −l Ll l l l b)a) U(x) U(x) D(x)0 000 D Dx xxxD(x) LL
Figure 6: In part a) a simple graph with the absorbing vertices 1 and 2; the potential isdiscontinuous in 0. In part b), we add the link [03] (length L ) and, on this link, a potential thatinterpolates linearly between U and U ; D ( x ) is constant on [03].on the link [10] b S ( x ) = 1 /γ + a φ ( x ) + a φ ( x ) (122)on the link [03] b S ( x ) = 1 /γ + c e r + x + c e r − x (123)on the link [32] b S ( x ) = 1 /γ + b ψ ( x ) + b ψ ( x ) (124)with r ± = A ± p A + 4 γD D (125) A = U − U L (126) φ , and ψ , are solutions of the equation ( L + − γ ) φ = 0 that satisfy ( φ ( αβ ) ≡ lim x αβ → φ ( x αβ )): φ = 0 φ = 1 φ = 1 φ = 0 (127) ψ = 1 ψ = 0 ψ = 0 ψ = 1 (128) U ( x ) and D ( x ) are continuous everywhere on the graph. So, we write standard boundaryconditions at the vertices (continuity of b S and its derivative at vertices 0 and 3; b S = 0 atvertices 1 and 2). We obtain ( φ ′ ( αβ ) ≡ lim x αβ → ∂φ ( x αβ ) ∂x αβ ): a = b = − γ a = c + c b = c e r + L + c e r − L (129)20 D D D l l L+l L+l −l −l −l −l Ll l l l b)a) U(x) U(x)U U00 00xx xxD(x) D(x) LL
Figure 7: a) The same graph as in Figure 6 a) but, this time, D ( x ) is discontinuous in 0; in partb), a parabolic interpolation is used to make D ( x ) continuous; U ( x ) is constant on the link [03]. a φ ′ + a φ ′ + c r + + c r − = 0 (130) b ψ ′ + b ψ ′ − c r + e r + L − c r − e r − L = 0 (131)After some algebra, we get the results: b S (32) b S (01) = γ + b γ + a → L → b S ′ (32) b S ′ (01) = b ψ ′ + b ψ ′ a φ ′ + a φ ′ → L → − e ( U − U ) /D (133)When L →
0, the vertex 3 moves to 0 and equations (132) and (133) give, for the original graph: b S (01) = b S (02) (134) b S ′ (01) e − U (01) /D (0) + b S ′ (02) e − U (02) /D (0) = 0 (135)In particular, we observe that b S is continuous at 0 and that exponential factors appear in thecondition involving the derivatives.Let us now turn to the case (B), still with the same graph and p = p = 1 / D ( x ) is discontinuous at 0 and U ( x ) is continuous. We set: D (01) ≡ D = D (02) ≡ D , U (0) ≡ U . 21s we already did for the case (A), we modify the graph and obtain Figure 7 b). Between vertices0 and 3, we choose: U ( x ) = U and D ( x ) = √ D − √ D L ! x + p D ! ≡ ( ax + b ) .Thus, in the modified graph, D ( x ) and U ( x ) are continuous everywhere on the graph.On the links [10] and [32], the solution of equation (121) is still given by (122) and (124) (withthe conditions (127) and (128)). But, on the link [03], (123) has to be replaced by b S ( x ) = 1 /γ + c ( a x + b ) λ + + c ( a x + b ) λ − (136)with: λ ± = 12 ± r γa (137) a = √ D − √ D L (138)Standard boundary conditions imply: a = b = − γ a = c D λ + / + c D λ − / b = c D λ + / + c D λ − / (139) a φ ′ + a φ ′ + a (cid:16) c λ + D ( λ + − / + c λ − D ( λ − − / (cid:17) = 0 (140) b ψ ′ + b ψ ′ − a (cid:16) c λ + D ( λ + − / + c λ − D ( λ − − / (cid:17) = 0 (141)Finally, we get: b S (32) b S (01) = γ + b γ + a → L → b S ′ (32) b S ′ (01) = b ψ ′ + b ψ ′ a φ ′ + a φ ′ → L → − L →
0, we get, for the original graph, b S continuous in 0 and b S ′ (01) + b S ′ (02) = 0 (144) B Backward Equation: boundary conditions when p = p We want to study the case (A) ( U ( x ) discontinuous in 0) for the graph of Figure 8 a) with, thistime, p = p .Modifying this graph, we obtain a new one, consisting in five vertices, displayed in Figure 8b). Vertices 1 and 2 are still absorbing. We set p = p = p = p = 1 / p = p ( p = p (original graph) and p = p (original graph)). U ( x )(resp. D ( x )) is set equal to some constant U (resp. D ) on the added links [40] and [03]. U ( x ) is discontinuous at vertices 4 and 3. With the modified graph, we can take advantage ofthe computations of the boundary conditions performed at the beginning of section 3 and alsoin Appendix A.The solution of equation (121) writes: 22 U U U l l L’+l L’+l −l −l −L−l −L−l Ll l l b)a) l L’ DD U(x)U(x) U0 000 −L−L L’L’xx xxD(x) D(x)
Figure 8: a) The same graph as in Figure 6 a) but, this time, p = p ; in b), U ( x ) and D ( x )are constant on the added links [40] and [03]; D ( x ) is still continuous on the modified graph.on the link [14] b S ( x ) = 1 /γ + a φ ( x ) + a φ ( x ) (145)on the link [40] b S ( x ) = 1 /γ + b sinh( q γ/Dx ) + b sinh( q γ/D ( L − x )) (146)on the link [03] b S ( x ) = 1 /γ + c sinh( q γ/Dx ) + c sinh( q γ/D ( L ′ − x ))(147)on the link [32] b S ( x ) = 1 /γ + d ψ ( x ) + d ψ ( x ) (148)As before, φ , and ψ , are solutions of the equation ( L + − γ ) φ = 0 that satisfy: φ = 0 φ = 1 φ = 1 φ = 0 (149) ψ = 1 ψ = 0 ψ = 0 ψ = 1 (150)The boundary conditions in 0 are given by the beginning of section 3.1. b S is continuous and p b S ′ (04) + p b S ′ (03) = 0 (151)For vertices 4 and 3, we use the result of Appendix A (case (A)). b S is again continuous and b S ′ (41) e − U /D + b S ′ (40) e − U/D = 0 (152) b S ′ (30) e − U/D + b S ′ (32) e − U /D = 0 (153)We get the relationships: 23 = d = − γ a = b sinh( q γ/DL ) d = c sinh( q γ/DL ′ ) (154) b sinh( q γ/DL ) = c sinh( q γ/DL ′ ) (155) p (cid:18) − b cosh( q γ/DL ) + b (cid:19) + p (cid:18) c − c cosh( q γ/DL ′ ) (cid:19) = 0 (156) (cid:16) a φ ′ + a φ ′ (cid:17) e − U /D + q γ/D (cid:18) b − b cosh( q γ/DL ) (cid:19) e − U/D = 0 (157) (cid:16) d ψ ′ + d ψ ′ (cid:17) e − U /D − q γ/D (cid:18) c cosh( q γ/DL ′ ) − c (cid:19) e − U/D = 0 (158)Solving and taking the limit
L, L ′ →
0, we are left with: b S (32) b S (41) = γ + d γ + a → L,L ′ → b S ′ (32) b S ′ (41) = d ψ ′ + d ψ ′ a φ ′ + a φ ′ → L,L ′ → e U − U D (cid:18) − p p (cid:19) (160)When L, L ′ →
0, the vertices 3 and 4 move to 0 and we get : b S (02) = b S (01) (161) p b S ′ (02) e − U (02) /D (0) + p b S ′ (01) e − U (01) /D (0) = 0 (162)We also observe that: b S (03) b S (30) → L,L ′ → b S ′ (03) b S ′ (30) → L,L ′ → − p = p ) is immediate. Indeed, we see that forthe modified graph (Figure 9 b)), we have, formally, the same solution as equations (145)-(148)and the same boundary conditions but, this time, with U = U = U . It amounts to dropthe exponential factors in (152) and in all the equations that follow. Finally, we get that b S iscontinuous and p b S ′ (02) + p b S ′ (01) = 0 (165)Moreover, (163) and (164) still hold. Recall that p = p (original graph) and p = p (original graph). l D D D D l −l − l −L− l −L− L’+l L’+l Ll l l b)a) L’ D
00 0−L L’L’xx xx
U(x) U(x)D(x) D(x)U U l Figure 9: a) The same graph as in Figure 7 a) but, this time, p = p ; in b), U ( x ) and D ( x )are constant on the added links [40] and [03]; U ( x ) is still continuous on the modified graph. C Backward Equation: general boundary conditions
For the case (A), let us assume that the p αβ i ’s are not all equal. In Figure 10 a), where a givenvertex α and its m α nearest-neighbours are shown, we suppose that U ( x ) is discontinuous in α ( D ( x ) is continuous, D ( x ) ≡ D ( α ) in α ) .In part b), we slightly modify the graph along the same lines as in Appendix B. We add vertices µ i in such a way that each new link [ µ i β i ] is identical to the original link [ αβ i ] (same length, samepotential and diffusion constant). Moreover, in the added subgraph (Figure 10 b), heavy linesof lengths L ) the potential and the diffusion constant are assumed to be constant (respectivelyequal to some value U and to D ( α )). So, in the vicinity of α , the discontinuities of U ( x ) willoccur at the µ i ’s ( D ( x ) will be continuous in the same domain). Of course, for the transitionprobabilities from α , we choose p αµ i = p αβ i . By taking the limit L →
0, we will recover theoriginal graph.Now, for the small subgraph where U ( x ) and D ( x ) are constant, we can take advantage of theresult, equation (8), to write: m α X i =1 p αµ i P ′ ( αµ i ) = 0 (166)Moreover, for the vertex µ i , where p µ i α = p µ i β i = 1 /
2, we can use (11) and also (135) (directlyestablished in Appendix A) to get: e − U/D ( α ) P ′ ( µ i α ) + e − U ( µiβi ) /D ( α ) P ′ ( µ i β i ) = 0 (167)25 µ βα β i ii a) b) Figure 10: a) the vertex α with its nearest-neighbours β i , i = 1 , ..., m α ; b) we have added theheavy lines where the potential is set equal to some constant U and the diffusion constant isequal to D ( α ); for the rest of the graph, nothing has been changed. For further explanations,see the text.Weighting with p αβ i ( ≡ p αµ i ) and summing over i , we deduce: m α X i =1 p αβ i e − U ( µiβi ) /D ( α ) P ′ ( µ i β i ) = − e − U/D ( α ) m α X i =1 p αµ i P ′ ( µ i α ) ! (168)Now, taking the limit L →
0, we have from (164) (Appendix B): P ′ ( µ i α ) P ′ ( αµ i ) → L → − µ i moves to α and we recover the original graph. Finally, with(166,168,169), we obtain, for the case (A), the boundary condition : m α X i =1 p αβ i e − U ( αβi ) /D ( α ) P ′ ( αβ i ) = 0 (170)Moreover, for the modified graph, P is continuous in α and in µ i . From Appendix B, we knowthat (163): P ( µ i α ) P ( αµ i ) → L → P is continuous in α . References [1] Baxter R J 1982
Exactly Solved Models in Statistical Mechanics (London: Academic Press)[2] Economou E N 1983
Green’s Functions in Quantum Physics (Berlin: Springer) Remark that (170) is unchanged when we add a constant to U ( x ). Now, if we want to consider the case when U ( x ) and D ( x ) are both discontinuous at some vertex α , we must add, on each link, vertices µ i and µ ′ i whereeither D ( x ) or U ( x ) are discontinuous. The resulting boundary condition will depend on the repartition of thoseadditional vertices. Moreover, inconsistencies will appear when we add a constant to U ( x ). This is why we saythat, in our opinion, this problem is ill-defined. Spectra of Graphs, Theory and Application (Academic, New York)[4] Rudenberg K and Scherr C 1953 J. Chem. Phys. L63[11] Roth J P 1983 C. R. Acad. Sc. Paris Processus Stochastiques et Mouvement Brownien (Paris: Editions JacquesGabay)[15] Levy P 1939 Compositio Math. S´em. Probabilit´es XXIII (Lecture Notes in Maths vol1372) (Berlin: Springer) p 294[17] Desbois J 2002 J. Phys. A: Math. Gen. L673[18] Comtet A, Desbois J and Majumdar S N 2002 J. Phys. A: Math. Gen. L687[19] van Kampen N G 1981
Stochastic processes in Physics and Chemistry (Amsterdam: Else-vier)[20] Redner S 2001
A Guide to First-Passage Processes (Cambridge: Cambridge UniversityPress)B´enichou O, Coppey M, Klafter J, Moreau M and Oshanin G 2005 J. Phys. A: Math. Gen. The Fokker-Planck Equation (Berlin: Springer)[23] Aldous D and Fill J 1999
Reversible Markov chains and random walks on graphs
Handbook of Stochastic Methods for Physics, Chemistry and NaturalSciences (Berlin: Springer).[25] Condamin S, Tejedor V, Voituriez R, B´enichou O and Klafter J 2008 Proc. Natl. Acad.Sci. USA85