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©Hindawi Publishing Corp.

ABOUT THE EXISTENCE OF THE THERMODYNAMIC LIMIT

FOR SOME DETERMINISTIC SEQUENCES

OF THE UNIT CIRCLE

STEFANO SIBONI

(Received 20 December 1999)

Abstract.We show that in the set Ω=R+×(1,+∞)⊂R2+, endowed with the usual Lebesgue measure, for almost all(h,λ)∈Ωthe limit limn→+∞(1/n)ln|h(λn−λ−n)mod

12,12|exists and is equal to zero. The result is related to a characterization of relaxation to equilibrium in mixing automorphisms of the two-torus. It is nothing but a curiosity, but maybe you will find it nice.

Keywords and phrases. Thermodynamic limit, Borel-Cantelli argument.

2000 Mathematics Subject Classification. Primary 11B83; Secondary 37A25.

1. Introduction. In the analysis of relaxation to equilibrium of mixing automor-phisms of the two-torus [1, 2, 3] one encounters the following problem. Suppose that the one-torus is parameterized by the unit interval12,12and for appropriate con-stantsh >0 andλ >1 consider the real sequence

xn=hλnλ−nmod1 2,

1 2

∀n∈N. (1.1)

A significant definition of an exponential “relaxation rate” can be given if the so-called “thermodynamic” limit [3],

lim n→+∞−

1

nln|xn| (1.2)

exists and is equal to zero. Existence of (1.2) is clearly not obvious, since the xn’s typically wander through the whole interval12,12but every so often they visit a small neighborhood of zero, where the logarithm is singular. Actually, not even if one replaces the ordinary limit in (1.2) with a supremum limit the finiteness of the result is assured.

This note is devoted to a measure theoretical discussion of the previous problem. One can show that existence to zero of limit (1.2) occurs almost surely, for almost any choice of the parametershandλ, with respect to a measure suitably defined.

2. Results. Our goal is to prove the statement below.

Theorem2.1. Consider the set= R+×(1,+∞)⊂R2+ endowed with the usual Lebesgue measureµ. Then, forµalmost all(h,λ)∈there holds

lim n→+∞

1 nln

hλnλ−nmod1 2,

1 2

(2)

This result can be easily deduced by means of standard arguments of measure the-ory once the following main theorem is proved.

Theorem2.2. Leth >0andQ∈N,Q >1, some fixed constants. Consider the set Gof allλ∈[1,Q]for which a (possiblyλ-dependent) real sequence(an)n∈N and an

integernNexist such that (a) an>0∀n > n;

(b) an≤ |h(λnλ−n)mod1 2,12

| ∀n > n; (c) limn→+∞(1/n)lnan=0.

Then, ifµdenotes the Lebesgue measure onR: (1) the setG⊆[1,Q]is actually nonempty;

(2) Gisµ-measurable and its measure holdsµ(G)=Q−1.

As a consequence, the setB=[1,Q]\G, where conditions (a), (b), and (c) are not simul-taneously satisfied, is alsoµ-measurable and of vanishing measure.

We firstly prove the result by considering values ofλin the interval[1+η,Q], with ηsmall positive number arbitrarily fixed (η <1/2). We therefore look for the subset ofλ∈[1+η,Q], where hypotheses (a), (b), and (c) are satisfied by a suitable choice of the sequence(an)n∈Nand of the integern∈N. The basic idea of the proof is that

theµ-measure ofturns out to beQ−1−ηeven if we confine ourselves to choose the sequence(an)n∈Nin the form

an= 1

n2 ∀n∈N, (2.2)

which certainly fulfills requirements (a) and (c), and enable us to deal with theonly condition (b) onλ.

Let us then takean=1/n2for allnNand an arbitrarily given value ofnN. Before tackling the real proof, we need some definitions.

Definition2.3. We introduce the setBn⊆[1+η,Q]

Bn=λ∈[1+η,Q]:an>hλnλ−nmod1 2,

1 2

, (2.3)

that is, the set ofλ∈[1+η,Q], where the conditionan≤ |h(λnλ−n)mod1 2,12

| is not satisfied for the assignedn∈N.

Notice thatBnis a finite union of intervals because the functionΦn(λ)=λnλ−nis strictly increasing in[+1,+∞)at fixedn. In fact

Φ n(λ)=n

λn−1+λ−(n+1)>0 λ[1,+∞). (2.4) Consequently,Bnisµ-measurable as a finite union of bounded intervals.

Definition2.4. We further introduce the set ˆBn⊆[1+η,Q],n∈N, given by

ˆ Bn=

λ∈[1+η,Q]:an>hλnλ−nmod1 2,

1 2

, n > n=

n>nBn (2.5)

(3)

Definition2.5. We finally introduce the “bad” setBη⊆[1+η,Q]

Bη=

n=1 ˆ

Bn, (2.6)

where condition (b) is not satisfied—with this particular choice of the sequence (an)n∈N. is also aµ-measurable set, as a countable intersection ofµ-measurable

sets.

An immediate consequence of the previous definitions is that[1+η,Q]\Gη=Bη. Our goal is to prove thatµ(Bη)=0. To this end, since for all nN, Bη Bnˆ by definition, it is enough to show that

lim n→+∞µ

ˆ

Bn=0. (2.7)

Therefore, we can confine ourselves to consider values ofnNlarge enough, and owing to Definition 2.4, we can also assume values ofn∈Ngreater that n. More precisely, we impose the following technical requirements on the size ofnandn. We taken > nNsuch that:

(i) an=1/n2< η⇒an< η∀n > n.

(ii) an−h/(1+η)n=1/n2h/(1+η)n>0n > n.

(iii) h[(1+η)n(1+η)−n] >3/2 andh[QnQ−n] >5/2n > n. Under the previous conditions we can state the following lemmas. Lemma2.6. Theµ-measure ofBn,nas above, admits the upper bound

µBn≤2εn 1

h 1/n

hQnQ−n1/n, (2.8) whereεn=an+h/(1+η)n>0.

Proof. We firstly notice that 1+an<1by (i); on the other hand, sinceη <1/2 by hypothesis, (i) implies an <1/2, so that all the intervals(p−an,p+an), p= 2,...,h(QnQ−n)+1 are disjoint.

By using (iii) and denoted withIn the integer set {2,3,...,h(QnQ−n) +1}, we deduce

Bn⊆    

λ∈[1+η,Q]:h

λnλ−nh(Q

nQ−n)+1 p=2

p−an,p+an      =λ∈[1+η,Q]:p−an< hλnλ−n< p+an, pIn

=λ∈[1+η,Q]:p−an−hλ−n< hλn< p+an+−n, pIn.

(2.9)

Now it is clear that for allλ∈[1+η,Q],

an−(1+hη)n ≤an−hλ−n< an+−nan+ h

(1+η)n (2.10) and by (ii),

an− h

(4)

from which we obtain

0< an−(1+hη)n ≤an−hλ−n< an+−nan+ h

(1+η)n ∀λ∈[1+η,Q]. (2.12)

By enlarging each covering interval in (2.9), we are then led to the inclusion

Bn⊆

λ∈[1+η,Q]:p−

an+(1+hη)n

< hλn< p+an+ h

(1+η)n, p∈In

(2.13)

and recalling the definition ofεn,

Bn⊆

λ∈[1+η,Q]:p−εn h

1/n

< λ <p+εn h

1/n , p∈In

h(QnQ−n)+1 p=2

pεn h

1/n ,

p+εn h

1/n ,

(2.14)

the final set beingµ-measurable as a finite union of intervals. Whence

µBn≤

h(QnQ−n)+1

p=2 1

h 1/n

p+εn1/n−p−εn1/n. (2.15)

Moreover, for allp=2,3,...,h(QnQ−n)+1 Lagrange mean value theorem implies the equalities below

p+εn1/n−p−εn1/n= 1 n

p+ξp(1/n)12εn (2.16)

for someξp∈(−εn,εn), and since

p+ξp(1/n)1= 1

(p+ξp)1−(1/n)≤ 1

(p−1)1−(1/n) (2.17) we conclude that

µBn≤

h(QnQ−n)+1

p=2

1 h

1/n1 n

1

(p−1)1−(1/n)2εn

=2nεn 1

h

1/nh(QnQ−n)

p=1

1 p1−(1/n).

(2.18)

As(p1−(1/n))1is a decreasing function ofp, the following upper estimate holds h(QnQ−n)

p=1

1 p1−(1/n)≤

h(QnQ−n)

0 p

(1/n)−1dp=np1/nh(Qn−Q−n) 0

=nhQnQ−n1/n

(2.19)

(5)

Lemma2.7. Ifn>0(satisfying (i), (ii), and (iii)) is sufficiently large, theµ-measure ofBnˆ is bounded by

µBnˆ 2(Q+ε)

n=n+1

εn (2.20)

for some smallε >0.

Proof. Because of the identity ˆBn= ∪n>nBnand using Lemma 2.6, we have the following estimate

µBnˆ

n=n+1

µBn≤2

n=n+1 εn1

h 1/n

hQnQ−n1/n. (2.21)

Notice that for allh >0, andQ∈N,Q >1

lim n→+∞

1 h

1/n

hQnQ−n1/n=Q (2.22)

so that for someε >0,εQ, andnsufficiently large there holds

Q−ε < 1

h 1/n

hQnQ−n1/n< Q+ε nN, n > n. (2.23)

Whence fornas above

µBnˆ 2

n=n+1εn(Q+ε)=2(Q+ε)

n=n+1εn (2.24) which is finite, owing to

n=1 εn=

n=1 1

n2+ h (1+η)n

62+hη. (2.25)

Lemma2.8. The measure ofBηis zero

µBη=0. (2.26)

Proof. Since for allnNwe have thatBnˆ , in particular this will be true for allnNlarge enough to satisfy the requirements of the previous lemmas. Thus

µBη≤µBnˆ 2(Q+ε)

n=n+1

εn (2.27)

and therefore

µBη≤ lim

n→+∞2(Q+ε)

(6)

Proof of Theorem2.2. As a consequence of Lemma 2.8, the “good” set Gη= [1+η,Q]\Bηofλ-values in[1+η,Q]satisfying condition (b) for the particular choice of(an)n∈N,an=1/n2, is of courseµ-measurable and with Lebesgue measure

µGη=Q−1−η−µBη=Q−1−η. (2.29)

If we now consider anarbitrary choice of the sequence(an)n∈N, compatible again

with conditions (a) and (c), the previous set will maybe “grow” by a subset ˜Gη⊆ [1+η,Q]\Gη:

→Go

η=Gη∪Gη.˜ (2.30)

But asµ([1+η,Q]\Gη)=0 it follows that ˜is alsoµ-measurable and of vanishing µ-measure. Hence we finally conclude that thefullsetGo

η, corresponding to arbitrary (a)- and (c)-conditioned sequences(an)n∈N, isµ-measurable with measure

µGo η

=Q−1−η (2.31)

and that the correspondingfullsetBo

η=[1+η,Q]\Goηofλvalues where condition (b) is not fulfilled forany(a)- and (c)-conditioned sequence(an)n∈Nis in turnµ-measurable

with vanishingµ-measure:

µBo

η=Q−1−η−µGoη=0. (2.32)

So far we have proved that is a set of vanishing measure in any closed interval [1+η,Q]withη >0. Consider nowB,G in[1,Q], that is, according to the previous notation

B=B0, G=G0. (2.33)

We firstly notice thatBandGare bothµ-measurable because

G=

n=1

n>nSn, (2.34)

where Sn is the finite union of subintervals in [1,Q] (dependent on n), and B = [1,Q]\G. Then take

B=B∩[1,1+η)∪B∩[1+η,Q] (2.35)

union of disjoint sets, for some fixedη∈0,12. Theµ-measurable setB∩[1+η,Q]is the “bad” set in[1+η,Q], so that by Lemma 2.8,

µB∩[1+η,Q]=0. (2.36)

As for theµ-measurable setB∩[1,1+η), we have the identity

B∩[1,1+η)=B∩

{1}∪

k=1

1+kη+1,1+ηk

=B∩{1}∪

k=1

B∩1+ η k+1,1+

η k

(7)

union of disjoint sets. But theµ-measurable set

B∩

1+kη+1,1k

, k∈N\{0}, (2.38)

satisfies

B∩1+ η k+1,1+

η k

⊆B∩1+ η k+1,Q

(2.39)

and since 1/2> η/(k+1) >0 for any givenk∈N\{0}, we obtain

µ

B∩

1+k+η1,1+ηk

=0. (2.40)

On the other hand, there trivially holdsµ(B∩{1})=0, so that

µB∩[1,1+η)=µB∩{1}+

k=1

µB∩1+ η k+1,1+

η k

=0. (2.41)

Whence, finally,

µ(B)=µB∩[1,1+η)+µB∩[1+η,Q)=0, (2.42)

that is,µ(B)=0 andµ(G)=Q−1. The proof is complete.

References

[1] A. Bazzani, S. Siboni, G. Turchetti, and S. Vaienti,From dynamical systems to local diffusion processes, Chaotic Dynamics (Patras, 1991), pp. 139–144, NATO Adv. Sci. Inst. Ser. B Phys., vol. 298, Plenum, New York, 1992. CMP 1 209 873.

[2] S. Siboni,Rilassamento all’equilibrio in un sistema mixing e analisi di un modello di dif-fusione modulata, Tesi di Dottorato di Ricerca dell’Università degli Studi, Bologna (Italy), 1991.

[3] S. Siboni, G. Turchetti, and S. Vaienti,Thermodynamic limit and relaxation to equilibrium in toral area-preserving transformations, Jour. Stat. Phys.75(1994), no. 1–2, 167.

Stefano Siboni: Dipartimento di Ingegneria dei Materiali, Facoltà di Ingegneria, Università di Trento, Mesiano38050, Trento, Italy

References

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