A STUDY ON CONDENSATION IN ZERO RANGE PROCESSES

Transcription

1 J. KSIAM Vol.22, No.3, 37 54, A STUDY ON CONDENSATION IN ZERO RANGE PROCESSES CHEOL-UNG PARK AND INTAE JEON 2 XINAPSE, SOUTH KOREA address: cheol.ung.park0@gmail.com 2 DEPARTMENT OF MATHEMATICS, THE CATHOLIC UNIVERSITY OF KOREA, SOUTH KOREA address: injeon@catholic.ac.kr ABSTRACT. We investigate the condensation transition of a zero range process with jump rate g given by M k, if k an α gk = 0. k, if k > an, α where α > 0 and a0 < a < /2 is a rational number. We show that for any ϵ > 0, there exists M > 0 such that, for any 0 < M M, the maximum cluster size is between a ϵn and a + ϵn for large n.. INTRODUCTION The zero range process introduced by Spitzer in 970, is not only an important example of interacting particle systems in probability theory, but also a model for a wide range of real phenomena such as metastability, granular clustering, wealth condensation, hub formation in complex networks, or even jamming in traffic flow [, 3, 6, 3, 5, 6]. The zero range process describes the dynamics in which m non-distinguishable particles are distributed over n sites and particles move around the site according to rules based on the transition matrix and positive function g defined on natural numbers. Each particle at one site jumps to other site with a probability given by transition matrix, after waiting exponentially distributed amount of time with parameter gk where k is the size of the cluster which contains the particle. For given m and n, this process form a finite state irreducible Markov Chain and the system converges to a unique invariant measure. Let Z =. Z, Z 2,, Z n be the random vector Received by the editors August 5 208; Accepted September 7 208; Published online September Mathematics Subject Classification. 60K35, 82C22. Key words and phrases. zero-range process, condensation, invariant measure. Corresponding author. 37

2 38 C. PARK AND I. JEON corresponding to the invariant measure, and let Z n = max i n Z i.. be the size of the largest cluster. One of the most interesting phenomena about the zero range process is that it accounts for condensation. Condensation is a phenomenon in which very small particles gather to form large clusters as the water vapor becomes a water droplet. A Bose-Einstein condensate which is a state of dilute gaseous bosons cooled to a temperature very close to 0 K, in particular, explains the important phenomenon in physics that a finite fraction of the bosons take on the lowest quantum state to form a large cluster. In the zero range process, condensation, then, can be determined by how large the biggest cluster size of the invariant measure is, compared to the total number of particles m of which we generally assume m = n so that the total density m/n =. Using the above notation, Zn we may say that, if lim n n a where 0 < a <, condensation occurs. It has a great attention after the discovery of existence of condensation transition by Jeon et al. [3] and Evans [5] independently in zero range processes. Indeed, they found that Z n n converges to a, with 0 < a < for the jump rates given by gk = + β, β > 0..2 k Since then, there have been many studies involving condensation [2, 3, 4, 7, 8, 9, 0, 2]. As the rigorous mathematical proof involves a lot of elaboration, Jeon considered a different type of rate function to find a case where the model is more intuitive and easier to apply []. In his model g is given by { k α if a < k/n gk = Mk α.3 if k/n a with α > 0, M > 0 and /2 < a <, and he was able to show with much simpler way that the largest cluster size is close to an, i.e., Z n n converges to a. More precisely, for any ϵ there exists M > 0 such that, for any 0 < M M the maximum cluster size is between a ϵn and a + ϵn for large n. Though providing a case where condensation occurs, he was not able to find out the size of the largest cluster for the case that 0 < a /2, and it was remained as an open problem. In this paper, we prove that the largest cluster size for the case that 0 < a < /2 and a is a rational number is about an. See Theorem 3.9 and Theorem 3.0. This result suggests, at least theoretically, a very simple and intuitive way to create a cluster of size an and/or to maintain the rates in order to limit the cluster size to an or less. This method can be used to control large masses produced naturally in the systems, such as traffic congestion and distribution of wealth, etc. by not exceeding a certain scale. This study is organized as follows: Section 2 briefly introduces the zero range processes and invariant measures; and Section 3 presents proofs of the main theorems.

3 CONDENSATION IN ZERO RANGE PROCESSES ZERO-RANGE PROCESS In this section, we introduce the zero range process rigorously. consider the subset of natural numbers N n = {, 2,, n}, the configuration space Ω n = {0,, 2,... } N n and a symmetric irreducible stochastic matrix {P ij } i,j n. Note that P ij = P ji and n j= P ij = for all i. Let g be a positive rate function which is defined on N = {, 2, }. Now we define a stochastic process on Ω n as follows. Suppose the process is in state η at certain time, which means that at site i there is an ηi- cluster. Then at any site, say i, the ηi-cluster waits for exponentially distributed amount of time with parameter gηi, and picks site j with probability P ij and gives one particle to the cluster at site j. As a result, ηi decreases to ηi, while ηj increases to ηj +. Let. η t = ηt, η t 2,, η t n, 0 t <, be the Markov process which represents such a dynamics. Since η t preserves the total number of particles, i.e., n i= η ti = n i= η 0i for all t, and since P ij is irreducible, if we let n Ω m n = {η Ω n : ηi = m}, m <, 2. i= then there is a unique invariant measure on Ω m n, say ν m n, which gives the steady state of the process. Let Z. = Z, Z 2,, Z n be the random vector corresponding to the invariant measure. The following proposition gives the explicit invariant measure on Ω m n. Lemma 2.. For any rate function gl, and for any η Ω m n, let n µ m n η = {g!ηi}, 2.2 i= where g!l = glgl gl 2 gg0, with convention g!0 = g0 =. Let νn m η = Γ µm n η, where Γ = µ m n Ω m n = µ m n η. 2.3 Then ν m n is the equilibrium measure corresponding to gl [6]. η Ω m n Let Ω m n be the number of elements in Ω m n. Since Ω m n is the set of nonnegative integers of the equation, we have x + x x n = m, and elementary combinatorics gives Ω m n = n+m n 3. THE MAIN THEOREMS AND PROOFS Let a = p/q 0, /2 Q be fixed, where p and q are relatively prime. Then we can express a by p a =, 3. p m + λ p

4 40 C. PARK AND I. JEON where m 2, m N and λ p {, 2,, p }. Choose large an such that n/ an is sufficiently smaller than an, where is a floor function. Since an < an an, we get a + aan > n an a. 3.2 Therefore, for sufficiently large n we have n = an Since an an < an +, it is enough to consider the case that. 3.3 a an = an. 3.4 From 3. and 3.4, we have b = n a 3.5 a p m + λp p = n p p m + λ p p = n m p m + λ p = λ p n. q Consider the rate function g denoted by M k α, if k an gk = 3.6 k α, if k > an, where 0 < a < /2. Let A k be the set of configurations of which the maximum cluster size is less than or equal to k denoted by A k := {η Ω n : max ηj k}. j n Let B k := A k A k be the set of configurations of which the maximum cluster size is exactly k. Define γ := {j : ηj }, γ indicates the total number of occupied sites. Let C k,l be the set of configurations in B k with exactly l occupied sites denoted by C k,l = {η B k : γη = l}. Note that C k,l is bounded by P n n l n l, where P is some polynomial. Lemma 3.. Let m = lk + r, 0 r < k, and let η = k, k,, k, r, 0,, 0 A k, where the k s are repeated l times. Let g be a rate function with corresponding invariant measure ν n on Ω n. Then for any η A k :

5 a ν n η ν n η, if g is decreasing. b ν n η ν n η, if g is increasing. Proof. See Lemma.3 in [3]. CONDENSATION IN ZERO RANGE PROCESSES 4 Proposition 3.2. Suppose rate function g is given by 3.6. Let a = p/q < /2, where p and q are relatively prime. Then ν n A n/2q 0 as n. 3.7 Proof. Choose configurations denoted by η 0 = an,, an, b, 0,, 0 Ω n, η = n/2q,, n/2q, d, 0,, 0 Ω n, where n = a an + b and b = λ p n q, for some λ p {0,, 2,, p } and n = t n/2q + d with 0 d < n/2q. From Lemma 3., µ n η 0 µ n η, for any η A an and µ n η µ n η for any η A n 2q From Stiring s formula. Moreover, we note q n = p ν n A n = µ na n 2q µ n Ω n pn q + λ pn = n q 2q 2q µ nη 2n n µ n η 0 P n 4 Qn 2 α q 2p + 2λ p p n/2q! q/p 2p+2λp pn/q! q/p pn/q! n nα p p/q q/p. α 2n n λ λ p α/q p where α > 2 and degrees of polynomials P and Q are independent of n. Therefore the right hand side goes to zero as n. Lemma 3.3. n, d dx log Γx + = γ + p, 3.8 x + p p= where Γ is Gamma function and γ Euler gamma constant. Lemma 3.4. If n N, then d dx log Γx + x=n = γ + n p= p. 3.9

6 42 C. PARK AND I. JEON Proposition 3.5. Suppose rate function g is given by 3.6. Let a = p/q < /2, where p and q are relatively prime. For any small ε > 0, ν n Aa εn A n/2q 0 as n. 3.0 Proof. Without loss of generality, we may assume n/2q = n/2q. First, n = 2q n/2q 3. n = 2q n/2q + + n/2q 2q n = 2q n/2q n/2q 22q where m is the largest integer with Second, n = 2q n/2q + m + n/2q m 2q, n/2q m 2q > 0. n = 2q 2 n/2q + m + + n/2q 2 m + 2q n = 2q 2 n/2q + m n/2q 2 m + 22q 2 n = 2q 2 n/2q + m + m 2 + n/2q 2 m + m 2 2q 2, where m 2 is the largest integer with Inductively, n/2q 2 m + m 2 2q 2 > 0. n = 2q k n/2q + m + + m k + n/2q k m + + m k 2q k, 3.3 where m k is the largest integer with And we choose l = lε N such that n/2q k m + + m k 2q k > 0. a εn = n/2q + m + + m l + ˆm l, 3.4 where ˆm l m l. Set E = {x N : n/2q + x a εn}.

7 CONDENSATION IN ZERO RANGE PROCESSES 43 Consider the disjoint collection {E i } l i= of E denoted by E = {x N : n/2q + x n/2q + m } 3.5 E 2 = {x N : n/2q + m + x n/2q + m + m 2 } E l = {x N : n/2q + m + + m l 2 + x n/2q + m + + m l 2 + m l } E l = {x N : n/2q + m + + m l + x n/2q + m + + m l + ˆm l }. For j =, 2, l, set G j be the real-valued extension super set of E j denoted by G = {x R : n/2q + x n/2q + m } 3.6 G l = {x R : n/2q + m + + m l 2 + x n/2q + m + + m l 2 + m l } G l = {x R : n/2q + m + + m l + x n/2q + m + + m l + ˆm l }. For any k {, 2,, l}, let x E k, by division algorithm, n = 2q k x + d k x, 3.7 where 0 d k x < x. From 3.7, if x is increasing on E k, d k x is decreasing on E k, but 2q k is constant on E k. In this paper, we use the symbol! as Gamma function notation which means that x R, x! = Γx +. Lemma 3.6. For k {, 2,, l}, x E k, let ηx k = x,, x, d k x, 0,, 0 with n = 2q k times 2q k x + d k x, 0 d k x < x. Suppose rate function g is given by 3.6. Then the unnormalized measure α µ n ηx k := x! 2q k d k x! is increasing on E k, where α > 0. Proof. For x G k, set φ k be denoted by φ k x = x! 2q k d k x!. 3.8 From 3.7, d k x = n 2q k x, if we take derivation by x except for end points of G k, then d dx d kx = 2q k. 3.9 We take a logarithm for φ k, then log φ k x = 2q k log x! + log d k x!. 3.20

8 44 C. PARK AND I. JEON By Lemma 3.3 and from 3.9, we have d dx log φ kx = d d 2q k log x! + dx dx log d kx! 3.2 = 2q k p x + p p. d k x + p p= If x E k, then by Lemma 3.4 and from 3.2, we get d d x k x dx log φ kx = 2q k p > 0, 3.22 p p= p= p= since 0 d k x < x. Thus µ n η k x is increasing on E k. Set x k L be the last element of E k and x k+ F = x k L + the first element of E k+. Then we note that n = 2q k x k L + d k x k L, 3.23 n = 2q k x k+ F + d k+ x k+ F, = 2q k x k L + + d k x k L + x k L 2q k. Lemma 3.7. x k L! 2q k dk x k L! 2q k x k+ F! dk+ x k+ F! Proof. Set x k L = y. From 3.23, we have = = y! 2q k d k y! y +! 2q k d k y + y 2q k! y! y + 2q k d k y + y 2q k d k y + 2q k times y 2q k times {}}{ y y 2q k 2 y 2q k y + y + 2q k times d k y + y 2q k d k y + y 2q k times.

9 CONDENSATION IN ZERO RANGE PROCESSES 45 For n/2q x a εn, consider configurations η x, η Ω n, denoted by η x = x,, x, d x, 0,, 0, 3.25 s times η = x, an x, an,, an, b, 0,, 0, /a times where n = s x + d x, 0 d x < x and n = /a an + b, 0 b < an. Then we have the following Lemma. Lemma 3.8. For x an, choose configurations η x, η denoted by Suppose rate function g is given by 3.6. Then µ n η x µ n η. Proof. By definition of the un-normalized measure, it is enough to show that x! s d x! x! an x! an! /a b! To prove this lemma, we use induction. For x = an, 3.26 is satisfied. i.e. From division algorithm, we assume that an! /a b! an! 0! an! /a b! n = l k k + d k, 0 d k < k, 3.28 n = l k+ k + + d k+, 0 d k+ < k +, 3.29 where k an. From Lemma 3.6 and Lemma 3.7, l k+ is classified into two cases: case: l k+ = l k, set l k = l. case2: l k+ = l k, set l k = l. n = l k+ k + + d k+, 3.30 = l k + + d k l. n = l k+ k + + d k+, 3.3 = l k + + d k + k l +. Set Lx = x! s d x! and Rx = x! an x! an! /a b!. Consider the ratio between Lk + and Lk. From 3.30 and 3.3, Lk + /Lk is classified into two cases: case: Lk + Lk = k +!l d k l! k! l d k! = k + l d k d k d k l + >. 3.32

10 46 C. PARK AND I. JEON case2: Lk + Lk = k +!l d k + k l +! k! l d k! = k + l d k + k l + d k + k l d k + 2 d k + k k k l + 2 k l + k l 2 l times k l times >. Consider the ratio between Rk + and Rk. Rk + Rk If k < an/2, then we have For k < an/2, we get = k +! an k! an! /a b! k! an k! an! /a b! = k + an k. Rk + Rk = k + an k. Lk < Lk + Rk + Rk. Hence 3.26 is true for the case that k < an/2. If an/2 k an, then we get Rk + Rk = k + an k >. Since Lk + > Lk and Rk + > Rk, to prove 3.26, we must show that Lk + Lk Rk +, 3.35 Rk for an/2 k an. From 3.30 and 3.3, 3.35 is classified as follows: case : case2 : k + l d k d k d k l + k + an k k + l d k + k l + d k + k l d k + 2 d k + k k k l + 2 k l + k l 2 l times k l times k + an k. 3.37

11 Case 3.36 becomes CONDENSATION IN ZERO RANGE PROCESSES 47 k + l an k d k d k d k l Set ˆLk = k + lk an k and ˆRk = d k d k d k l k + with n = l k k +d k. To prove 3.38, we use an induction. For k = an/2, 3.38 is satisfied. i.e. an 2/a an 2 + d d d 2/a +, with n = 2/a an/2+d and 0 d < an/2. From 3.30, say case, and l k = l k+ = l > 2, we easily get ˆLk + = k + 2l an k ˆLk k + l > an k And ˆRk + ˆRk From 3.39, 3.40 and 3.4, we note = d k+ d k+ d k+ l + d k d k d k l + = d k l d k l d k 2l + d k d k d k l + <. 3.4 ˆLk + > ˆLk ˆRk > ˆRk Thus 3.38 is satisfied. Let s consider another case now. Then case becomes k + l 2 an k d k + k l + d k + 2d k + k! Set Lk = k + lk 2 an k d k + k l k + d k + 2 d k + and Rk = k! with n = l k k + d k. Similarly, we use an induction. For k = an/2, 3.43 is satisfied. i.e. an 2/a 2 an d 2 + k + an 2 2 2/a d k + an an an 2 2/a d k + 2 d k , with n = 2/a an/2 + d and 0 d < an/2. From 3.3, say case2, and l k+ = l k = l, we easily get Lk + Lk k l+ times {}}{ = k + 2l 3 an k d k + 2k 2l + 4 d k + k l + 2 k + l an k d k + k l + d k + Since the size of k is sufficiently larger that that of l, we get Lk + Lk >. 3.46

12 48 C. PARK AND I. JEON Clearly, Consider the ratio denoted by Then 3.48 becomes Rk + Rk = Lk + Lk k +! k! = k + > / Rk Rk k l+ times {}}{ k + 2 l 3 an k d k + 2k 2l + 4 d k + k l + 2 k + l an k d k + k l + d k + Clearly, this ratio is greater than. Thus Lk + Lk > Rk Rk Therefore from 3.44 and 3.50, 3.43 is satisfied. In all cases, this completes the proof. Consider configurations η 0 = an,, an, b, 0,, 0 Ω n, /a times η x = x,, x, d x, 0,, 0 Ω n, s times where n = s x + d x, 0 d x < x and n/2q < x a εn. For x b, 0 b < an, choose η = x, an x, an,, an, b, 0,, 0 Ω n. From /a times Lemma 3., µ n η x µ n η for any η C x,l. From Lemma 3.8, we get µ n η µ n η x. From Stirling s formula, α µ n η µ n η 0 = x! an x!an! /a b! an! /a 3.5 b! x! an x! α = an! x x an x an x α P n an an αn x/n x/n a x/n a x/n = P n a a P n γ n, 3.52

13 CONDENSATION IN ZERO RANGE PROCESSES 49 for some 0 < γ <. The degree of the polynomial P n which may vary in each expression is independent of n. 2 For b < x b + ε 0 n, where 0 < ε 0 < ε. Choose η 0 = an ε 0 n, b + ε 0 n, an,, an, 0,, 0, /a times η = x, an x ε 0 n, b + ε 0 n, an,, an, 0,, 0, /a times From Lemma 3.6 and Lemma 3.7, we note that a ε 0 n! /a ε 0 d! an! /a b + ε 0 n!, 3.53 where n a ε 0 n = /a ε 0 a ε 0 n + d, 0 d < a ε 0 n. And we should choose small ε 0 > 0 with 0 b + ε 0 n < an. From 3.53, we get a ε 0 n! /a ε 0 d! a ε 0 n! b + ε 0 n! an! /a, 3.54 where n = /a ε 0 a ε 0 n + d, 0 d < a ε 0 n. This fact corresponds to that of 3.27 in Lemma 3.8. Hence, by the similar inductive method of Lemma 3.8, we have x! s d x! x!a ε 0 n x! b + ε 0 n! an! /a b! 3.55 Thus µ n η µ n η x From Lemma 3., µ n η 0 µ n η 0 and µ n η x µ n η for any η C x,l. In all, we have Hence µ n η µ n η for any η C x,l. µ n η µ n η 0 µ nη x!an x µ n η 0 = ε0 n! α 3.57 an ε 0 n! αn x/n x/n a ε 0 x/n a ε 0 x/n P n a ε 0 a ε 0 P n γ n 2, for some 0 < γ 2 <. The degree of polynomial P n which may vary in each expression is independent of n. 3 For x > b + ε 0 n, choose η = x, an x, b, an,, an, 0,, 0 Ω n, /a times η 2 = x, an x + b, an,, an, 0,, 0 Ω n. /a times

14 50 C. PARK AND I. JEON Since an x + b! an b! b!, we get From Lemma 3. and Lemma 3.8, we have In all, µ n η 2 µ n η µ n η µ n η x µ n η for any η C x,l. µ n η 2 µ n η for any η C x,l Hence from Stiring s formula, µ n η 2 x!an x + b! α µ n η 0 = 3.60 an!b! x x an + b x an+b x α P n an an b b x x an+b x x α an+b an + b an + b P n an an an+b an an an+b an + b an + b P n γ3 n, for some 0 < γ 3 <. The degree of polynomial P n which may vary in each expression is independent of n. Since x > b + ε 0 n, we note x an + b an an + b. 3.6 an Moreover, since an + b > 2, x < an and from 3.6, the above last inequality of 3.60 is verified. In all, we choose 0 < γ < with γ > max{γ, γ 2, γ 3 }. Therefore, ν n a εn x=n/2q l C x,l l x l µ nc x,l µ n η 0 n P n l for some polynomial P n. Let ε > 0 be sufficiently small. n l 3.62 γ n, 3.63 a For l < ε n, we have n n. l ε n

15 CONDENSATION IN ZERO RANGE PROCESSES 5 Then ν n a εn x=n/2q l<ε n C x,l n n n 2 P n γ n 3.64 ε n ε n n γ Qn ε ε ε ε 2, 3.65 where ε can be chosen ε ε ε ε 2 > γ. Degrees of P n, Qn are independent of n. The right hand side goes to 0 as n. Then b For l ε n, assume n = s x + d, 0 d < s. i If d < l s, then we choose η = x,, x, n xs l + s,,,, 0,, 0 Ω n s times l s times ii If d l s, then we choose η = x,, x, d l + s,,,, 0,, 0 Ω n s times l s times Clearly, µ n η µ n η for any η C x,l. i d < l s : n xs + s + = d + x + s + εn. µ n η x! s α µ n η 0 = n xs l + s! an! /a 3.68 b! α x! s n xs! an! /a b! n xs n xs l + s + α n! an! /a b! n xs n xs ε n + s + C n ε ε n αε, where C > 0 is some constant. ii d l s : note that d + s l ε n and s is sufficiently smaller than l. Thus we have d! ε n s! τn τn e τn, for some τ > 0.

16 52 C. PARK AND I. JEON µ n η x! s µ n η 0 = d l + s! α an! /a 3.69 b! x! s d! α an! /a b! dd d l + s + x! s d! n! α n! an! /a b! dd d ε n + s + x! s d! n! α n! an! /a b! ε n sε n s n C2 τn τα, where C 2, τ are some positive constants. Therefore, from 3.68 and 3.69, we note that µ n η µ n η 0 C3 where C 3, ξ are some positive constants. Thus a εn ν n C x, l µ n x µ n η 0 x=n/2q l ε n n ξ x n P n l C n n δ, n, 3.70 l C x, l l µ n C x, l µ n η 0 n l C3 n ξ n 3.7 where C, δ are some positive constants and the degree of P n is independent of n. In all cases, ν n Aa εn A n/2q 0 as n. Theorem 3.9. Suppose rate function g is given by 3.6. Let a = p/q < /2, where p and q are relatively prime. For any small ε > 0, there exists M such that Z n a εn, for large n. Theorem 3.0. Suppose rate function g is given by 3.6. Let a = p/q < /2, where p and q are relatively prime. For any small ε > 0, there exists M, 0 < M < such that Z n a + εn, for large n.

17 CONDENSATION IN ZERO RANGE PROCESSES 53 Proof. For any l, l n. Let η 0 = an,, an, b, 0,, 0 Ω n, where n = a an + b, 0 b < an. Consider η, η 2 Ω n where η = l, l, d, 0,, 0, η 2 = l, n l, 0,, 0 Ω n, 3.72 t with n = t l + d, 0 d < l. By Lemma 3., µ n η µ n η for any η B l and µ n η 2 l! n l! α n l! α µ n η = l! t = d! l! t, 3.73 d! since n l = t l + d. Define a function ϕ : Ω n {0,, 2, } which counts the number of Ms contained in µ n η as ϕη = m. Clearly ϕη 0 = n, ϕη 2 < n, since l a + εn. Thus there exists ω > 0 such that For l a + εn, P {Z n B l } = ν n B l ϕη 0 ϕη 2 = ωn. µ nb l µ n η 0 n 2n n µn η 2 µ n η 0 pn 2 n M ϕη 0 ϕη 2 l! n l! pn 2 n M ωn n! pn M ωn λ n, l! n l! α an! /a b! n! an! /a b! where λ > is some constant and the degree of polynomial p is independent of n. Hence P {Z n a + εn} n pn M ωn λ n If M < /λ /ω, the right hand side goes to zero as n. This completes the proof. ACKNOWLEDGMENTS I. Jeon was supported by the Research Fund, 207 of the Catholic University of Korea. REFERENCES [] I. Armendáriz and M. Loulakis, Thermodynamic Limit for the Invariant Measures in Supercritical Zero Range Processes, Prob. Th. Rel. Fields, , no. -2, [2] I. Armendáriz, S. Grosskinsky and M. Loulakis, Metastability in a condensing zero-range process in the thermodynamic limit, Prob. Th. Rel. Fields, , no. -2, [3] J. Beltran and C. Landim, Metastabiltiy of reversible condensed zero range processes on a finite set, Prob. Th. Rel. Fields, , no. 3, α

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