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journal of statistical physics, vol. 123, no. 1, april 2006 ( c?2006) doi: 10.1007/s10955-005-9012-8 around multicolour disordered lattice gas azzouz dermoune1and servet martinez2 received june 23, 2005; accepted november 22, 2005 published online: april 5, 2006 weconsiderasystemofmulticolourdisorderedlatticegas,followingcloselythe(mono- colour) introduced by faggionato and martinelli(3,4). we study the projection on the monocoloursystemandwederiveanestimateoftheclosenessbetweengrandcanonical and canonical gibbs measures. key words: multicolour, gibbs measure, lattice gas. ams classifi cation: primary: 60k35, 82c20, 82c22 1. introduction faggionato and martinelli(3,4)have studied the hydrodynamic limit for a dynamics on lattice gas which depends on a collection of real bounded i.i.d. random variables = (x: x zd) corresponding to some external quenched disorder fi eld. the hydrodynamic limit was obtained for a monocolour particles, an important byproduct of their study being the equivalence of ensembles shown in lemma a.4.(4) the aim of our work is to give an equivalence of ensembles result with sharp bounds for multicolour particles. our proof uses the projection on the monocolour system, and then the equivalence of ensembles result of the monocolour lattice gas.(3)after, we study the variation of the canonical gibbs measures of the mul- ticolour problem with respect to the monocolour one, showing that it does not add new terms. this is done by elementary computations. we point out that the 1universit e de lille i, umr-cnrs 8524, laboratoire paul painlev e, universit e des sciences et tech- nologies de lille, 59655 villeneuve dascq, cedex, france; e-mail: azzouz.dermounemath.univ- lille1.fr 2universidad de chile, departamento ingenier a matem atica-centro modelamiento matem atico, umr2071 cnrs-uchile, casilla 170 - 3, correo 3, santiago, chile; e-mail: smartinedim. uchile.cl 181 0022-4715/06/0400-0181/0c?2006 springer science+business media, inc. 182dermoune and martinez equivalence of ensembles of multicolour is an important step towards obtaining the hydrodynamic limit for the multicolour case. on the other hand, as it is shown in ref. 5, the hydrodynamic limit for the multicolour case is necessary to obtain propagation of chaos. the model introduced in refs. 3 and 4 has been previously extended to the two colours in refs. 1 and 2 and an equivalence of ensembles result was proven by using similar techniques as those in ref. 3 which are different from ours. let us introduce the problem. ? zd denotes a fi nite set of sites, each of the sites could be empty or has a colour (or spin). the fi nite set of colours is denoted by i and it is useful to consider the set i0= i 0, where we have added the value 0 that expresses that the site has no colour (or that it is empty). we denote |i| = 1(i i)fori i0(whereweput1(x a)=1if x a,vanishingotherwise). the set of confi gurations is i? 0 , and its elements are denoted by = (x: x ?). for each x ? x= ? iif there is a particle with colour i i at x 0if there is no particle at x we will use the notation | := (|x| : x ?) 0,1?. 1.1. gibbs measures for monocolour lattice gas let r and the disorder be fi xed. the gibbs measure | := ,or the monocolour problem is a probability measure on 0,1?characterized by: all the projections x,x ? are independent |() := |( : =) = ? x? |( : x=x)for 0,1?;(1.1) and the marginals verify |( : x= 1) = e+x 1 + e+x .(1.2) the measure | is symmetric and invariant for the dynamics studied in break refs. 3 and 4. briefl y this dynamics is given by the following rate of change c x,x+e() = fe(x,x,x+e,x+e) for 0,1 ? and x,x + e ?. here e is anelementofasymmetricneighbourhood? oftheoriginandthefunctions( fe: e ?) verify the conditions specifi ed in ref. 4 (symmetry, exclusion, lower uniform boundedness and detailed balance). for 0,1?put = x ? : x= 1 and we denote by n() = | the number of elements of (that is, |a| denotes the cardinality of a). observe that for each 0 0 : i i) be fi xed with ? ii ni= n. recall ni() = |x : x= i|. let (ni= ni: i i) = i? 0 : ni() = ni,i i. the canonical gibbs measure for the multicolour lattice gas is the following conditional measure on i? 0 : () = ( | ni= ni: i i). before giving our main result we need the following background. for i? 0 and ? ? put ?= (x: x ?). let g : i? 0 r. the function g depends on ? if for any couple ,? i? 0 such that ?= ?, it holds g() = g(?). this is equivalent tothe existence ofa function g?: i? 0 rsuch that g() = g?(?).if gdepends on?1andon?2,thenitiseasytoshowthatgdepends on?1 ?2.let d(g) = ? ? : g depends on ?, and put ?(g) = ?d(g)?. then g depends on ?(g) and this set is the smallest one verifying this property and we call it the support of g. so, ?(g)= ?(g)implies g() = g(?) and by abuse of notation we put g() = g(?(g). lemma 1.1.let be a gibbs measure. then for every g : i? 0 r it holds ?(s(g) ?(g) and |s(g)| |g|. proof:from s(g)() = ? :|=g() () the inequality between the norms |is direct. let us show the inclusion of the supports. observe that = p for all 0,1? , for a fi xed probability vector p = (pi: i i) p(i). put around multicolour disordered lattice gas185 ? = ?(g) and partition = (?,?), = (?,?). since pis a product measure and g() = g(?) we have s(g)() = ? ?:|?|=? g(?)? p (?) ? ?:|?|=? ? p (?) = ? ?:|?|=? g(?)? p (?) . therefore s(g)() only depends on ?, and the result is shown.? now we can announce our main result. theorem 1.2.let p = (pi: i i) with pi= ni/n,i i, be a gibbs measure such that = p, and |(n) = n. suppose that minni: i i := (n) 1 for every n. then, for any l 1 fi xed, there exists constant c = c(|i|, l) such that for all g : i? 0 r verifying |?(g)| (n)/l it holds |(s(g) (g) | c|g| |?(g)| (n) .(1.6) more precisely |(s(g) (g) | c|g| (n) ? 0,1?:n()=n ?|?(g) |?|( |n = n) (1.7) in particular, we get that, in the framework of the theorem, there exists a constant d = d(|i|, l) such that |(s(g) (g)| d |g|?(g)| (n) (1.8) from (1.3), (g) = |(s(g). we deduce (g) (g) = |(s(g) |(s(g) + |(s(g) (g).(1.9) now, lemma 1.1. gives the inclusion ?(s(g) ?(g). then, from lemma a.4 in ref. 4, we get that: for any (0,1) and any g with support ?(g) ? and such that |?(g)| |?|1, for large ? it holds |(s(g) |(s(g) c |g| |?(g)| |?| .(1.10) therefore, by combining theorem 1.2 with (1.8), (1.9) and (1.10), we get the following result. 186dermoune and martinez corollary 1.3.let p = (pi: i i) with pi= ni/n, i i, and a gibbs measure such that = p. suppose that there exists 0 such that minni: i i n for every n. then, for any (0,1) there exists constant c?= c?(|i|,) such that for all g : i? 0 r verifying |?(g)| n1it holds |(g) (g)| c?|g|?(g)| n .(1.12) the rest of this paper is consecrated to the proof of theorem 1.2. we point out that the main technical point is the control of |(sg) (g). from (2.1) and (2.3) below, we are led to study a quotient that is dominated by the usual entropy estimates, see the proof of lemma 2.6. 2. proof of the main result let be a gibbs measure for the multicolour lattice gas. recall that the monocolour canonical gibbs measure is defi ned on 0,1?by |() = |( | n = n) , so |() = 1(n() = n) |() |(n=n) for 0,1?. therefore, |( f ) |( f ) = ? 0,1? f ()|() ? 1 1(n() = n) |(n = n) ? for f : 0,1? r . the canonical gibbs measure for the multicolour lattice gas is defi ned on i? 0 by () = ( | ni= ni: i i). hence, (g) = ? i? 0 g()()1(ni() = ni : i i) (ni= ni: i i) forg : i? 0 r , so, (g) (g) = ? i? 0 g()() ? 1 1(ni() = ni: i i) (ni= ni: i i) ? . let us fi rst establish some general relations which do not depend on the fact that p verifi es the requirements of the theorem. the hypothesis on p will be only used at the end of the proof, in lemma 2.3. we have |(s(g) = ? 0,1? ? :|= g()() 1(n() = n) |(n = n) |() . around multicolour disordered lattice gas187 since i? 0 : ni= ni,i i i? 0 : n(|) = n and ( : n(|) = n) = |( : n() = n), we deduce (g) = ? 0,1? ? :|= g()() 1(ni() = ni: i i) (ni= ni,i i | n = n) 1(n() = n) |(n = n) |(). therefore |(s(g) (g) = ? 0,1?:n()=n ?g 1() ? g 2() ? |( | n = n) ,(2.1) where ?g 1() = ? :|= g()()and?g 2() = ? :|=,ni()=ni,ii g() () (ni= ni,i i | n = n) . from now on, we fi x = pwith p = (pi: i i) a probability vector in i. since for every i? 0 such that ni() = nifor i i, it holds () = ? ii pni i , we get ?g 2() = ? ii pni i (ni= ni,i i | n = n) ? :|=,ni()=ni,ii g() . let ? n ni,ii ? = n! ? iini!. we have (ni= ni,i i | n = n) = ? n ni,i i ? ii pni i , so, ? ii pni i (ni= ni,i i | n = n) = ? n ni, i i ?1 . then, ?g 2() = ? :|=,ni()=ni,ii g()()with() = ? n ni, i i ?1 . in the sequel, for every 0,1?with n() = n we put ?= ?(g) and = ?(g). then = ? is a partition. every i? 0 with | = can 188dermoune and martinez be represented by = (?,). in fact in the complement we have x= 0 for x ? . with this notation and since g() = g(?(g), we obtain g() = g(?) for every i? 0 with | = . then, by using pis a product measure we get ?g 1() = ? :|= g()p() = ? ?i? ? i g(?) ?(g) p (?) ?(g) p () = ? ?i? g(?) ?(g) p (?) . in a similar way, we fi nd ?g 2() = ? ?i? g(?) ? i:ni()=nini(?), ii ? n ni, i i ?1 = ? ?i? g(?) ? n |?| ni ni(?),i i ? n ni, i i ?1 . then, ?g 1()? g 2() = ? ?i? g(?) ? ?(g) p (?) ? n |?| ni ni(?),i i ? ? n ni, i i ?1? (2.2) and we obtain, |?g 1() ? g 2()| |g|m with(2.3) m= ? ?i? ? ? ? ? ? ?(g) p (?) ? n |?| ni ni(?),i i ? n ni, i i ?1? ? ? ? . we observe that m= |q r|is the total variation of two probability mea- sures q and r defi ned on i?by q(?) = ?(g) p (?) and r(?) = ? n |?| ni ni(?),i i ? n ni, i i ?1 . around multicolour disordered lattice gas189 therefore, m= 2 ? ?i?: r(?)q(?) (r(?) q(?) . letusdenote?= |?|andj= ?l = (li: i i) ni: li 0 for i i, ? iili = ?. to every ?we associate?l(?) = (li: i i) jwith li= ni(?) = |x ?: x= i|. we observe that q(?) = q(?l(?) and r(?) = r(?l(? ). then, by defi ning q(?l) = ? ii pli i and r(?l) = ? n ? ni li,i i ? n ni, i i ?1 (2.4) we fi nd, m= 2 ? ?lj: r(?l)q(?l) q(?l) ? r(?l) q(?l) 1 ? .(2.5) the proof of the theorem 1.2 will be deduced from the following result. lemma 2.3.let p = (pi: i i) with pi= ni/n,i i. assume that minni: i i := (n) 1, and |?(g)| l1(n) for some l 1. then m e? () ? 1 ? (n) ?|i|/2 1(2.7) where ?(?) = ? 11n(n ?) (2.8) proof: from relation (2.5) it suffi ces to show that max ? r(?l) q(?l) :?l j ? e? () ? 1 ? (n) ?|i|/2 .(2.9) from (2.4) we have r(?l) q(?l) = n?(n ?)! ? ii ni! n! ? ii ?(n i li)!n li i ? . (2.10) 190dermoune and martinez by using the stirling formula m! = em(logm1)(2m)1/2e(m)with 1 12m + 1 (m) 1 12m , we fi nd, r(?l) q(?l) = z(?l)eh(? l)e?(?l) , where z(?l) = ?(n ? ) ? ii ni n ? ii(ni li) ?1/2 = ? 1 ? n ?1/2 ? ii ? 1 li ni ?1/2 , ?(?l) = (n ?) (n) + ? ii (ni) (ni li) , h(?l) = ?log n + (n ?)(log(n ?) 1) n(log n 1) + ? ii (ni(log ni 1) (ni li)(log(ni li) 1) lilog ni). then, h(?l) = (n ?)log(1 ? n) ? ii(ni li)log(1 li ni). let us consider the maximization problem max ? h(?l) :?l ?p(i) ? where ?p(i) = ?l ri +: li 0,i i, ? ii li= ?. it is easy to show that the maximum is attained at li/ni= a constant, then by the constraints ? iili = ? we fi nd that li/ni= ?/n for all i i, and at this point h(?l) = 0. since j ?p(i), we deduce max ? h(?l) :?l j ? 0. on the one hand from min(ni: i i) (n) and li ?, we have ? ii ? 1 li ni ?1/2 (1 ? (n) |i|/2 then z(?l) (1 ? (n) |i|/2. a simple computation shows that (n ?) (n) 1 12(n ?) 1 12n + 1 = 1 + 12? 12(n ?)(12n + 1) 13? 144(n ?)n ? 11(n ?)n on the other hand for all i,(ni) (ni li) = 0 for li= 0, and for li 1, (ni) (ni li) 12li+ 1 12ni(12(ni li) + 1) 0. around multicolour disordered lattice gas191 from the latter estimates we get ?(?l) ? 11n(n ?) = ?(?) . this shows (2.9), and the lemma follows.? let us fi nish the proof of theorem 1.2. from relations (2.1), (2.3) and (2.7), and by using the hypothesis of the theorem, we obtain |(s(g) (g)| ? 0,1?:n()=n |g|m|( | n = n) |g| ? 0,1?:n()=n ? e? () ? 1 ? (n) ?|i|/2 1 ? |( | n = n) . from |?(g)| l1(n) and (n) n we have ? l1n. below we use (1 x)1 1 + hx when 0 x (h 1)/h, to deduce that ?(?) (2l 1)? 11(l 1)n2 . from the inequality ? l1(n),