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GENERIC UNIQUENESS OF MINIMAL CONFIGURATIONS WITH RATIONAL ROTATION NUMBERS

IN AUBRY-MATHER THEORY

ALEXANDER J. ZASLAVSKI Received 18 November 2002

We study (h)-minimal configurations in Aubry-Mather theory, wherehbelongs to a com- plete metric space of functions. Such minimal configurations have definite rotation num- ber. We establish the existence of a set of functions, which is a countable intersection of open everywhere dense subsets of the space and such that for each elementhof this set and each rational numberα, the following properties hold: (i) there exist three different (h)-minimal configurations with rotation numberα; (ii) any (h)-minimal configuration with rotation numberαis a translation of one of these configurations.

1. Introduction

Let Z be a set of all integers. A configuration is a bi-infinite sequence x=(xi)iZ

RZ. The setRZwill be endowed with the product topology and the partial order de- fined byx < yif and only ifxi< yifor alliZ.

We have an order-preserving actionT:Z2×RZRZdefined by T(k,x)=Tkx=y⇐⇒k=

k1,k2

Z2,

x,yRZ, yi=xik1+k2 iZ. (1.1) Letx,yRZ. We say that yis a translation ofx if there isn=(n1,n2)Z2 such that y=Tnx.

Leth:R2R1be a continuous function. We extend hto arbitrary finite segments (xj,. . .,xk),j < k, of configurationsxRZby

hxj,. . .,xk

:=

k1 i=j

hxi,xi+1

. (1.2)

A segment (xj,. . .,xk) is called (h)-minimal ifh(xj,. . .,xk)h(yj,. . .,yk) whenever xj=yjandxk=yk.

Copyright©2004 Hindawi Publishing Corporation Abstract and Applied Analysis 2004:8 (2004) 691–721 2000 Mathematics Subject Classification: 37J45, 37E45, 70K75 URL:http://dx.doi.org/10.1155/S1085337504310067

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We assume thathhas the following properties [3,4]:

(H1) for all (ξ,η)R2,h(ξ+ 1,η+ 1)=h(ξ,η);

(H2) lim|η|→∞h(ξ,ξ+η)= ∞uniformly inξ;

(H3) ifξ1< ξ2andη1< η2, then hξ11

+hξ22

< hξ1,η2

+hξ2,η1

; (1.3)

(H4) if (x1,x0,x1)=(y1,y0,y1) are (h)-minimal segments andx0=y0, then x1y1

x1y1

<0. (1.4)

A configurationxRZis (h)-minimal if, for each pair of integers jandksatisfying j < kand each finite segment{yi}ki=jR1satisfying yj=xjandyk=xk, the inequality h(xj,. . .,xk)h(yj,. . .,yk) holds. Denote byᏹ(h) the set of all (h)-minimal configura- tions. It is known that the setᏹ(h) is closed [2,3].

The notion of global minimizers ((h)-minimal configurations in the present paper) is crucial to the Aubry-Mather theory. The works by Aubry and Mather were begun inde- pendently and with different motivations but led to similar results by different methods.

While Mather [12] studied area-preserving annulus mappings as they occur as section mappings for Hamiltonian systems of two degrees of freedom, Aubry [1] investigated cer- tain models of solid state physics related to dislocations in one-dimensional crystals. For more details on Aubry-Mather theory, see [1,2,3,4,12,13,14,15,18,19]. For the usage of the notion of global minimizers in the related topics of calculus of variations, partial differential equations, and geometry, see also [3,4,5,6,7,8,10,11,16,17,20,21].

We briefly review the definitions, notions, and some basic results from Aubry-Mather theory [2,3].

Definition 1.1. The configurationsxRZandxRZcross (a) atiZifxi=xi and (xi1xi1)(xi+1xi+1 )<0, (b) betweeniandi+ 1 if (xixi )(xi+1xi+1)<0.

Definition 1.2. The configurationxRZis periodic with period (q,p)(Z\ {0})×Zif T(q,p)x=x.

Remark 1.3. Assume thath=h(ξ12)C2(R2) and (∂2h/∂ξ1∂ξ2)(u,v)<0 for all (u,v) R2. It is not difficult to show that (H3) and (H4) hold. Moreover, we can show that if hC2(R2), then (H3) holds if and only if

(u,v)R2: 2h

∂ξ1∂ξ2

(u,v)<0

(1.5) is an everywhere dense subset ofR2.

We have the following result [3, Corollary 3.16, Theorem 3.17].

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Proposition1.4. There exists a continuous functionα(h):ᏹ(h)R1with the following properties:

(i)for allxᏹ(h),iZ,

xix0(h)(x)<1; (1.6) (ii)ifxᏹ(h)is periodic with period(q,p)Z2, thenα(h)(x)=p/q;

(iii)for allαR1, the set{xᏹ(h) :α(h)(x)=α} = ∅. Remark 1.5. We callα(h)(x) the rotation number ofxᏹ(h).

For eachαR1, define

ᏹ(h,α)= xᏹ(h) :α(h)(x)=α. (1.7) We studyᏹ(h,α) with rationalαR1.

Let a rationalα=p/qbe an irreducible fraction, whereq1 andpare integers. De- note byᏹper(h,α) the set of all periodic (h)-minimal configurationsxᏹ(h,α) which satisfyT(q,p)x=x, equivalently,xiq+p=xi, for alliZ.

For the proof of the following result, see [2,3].

Proposition 1.6. ᏹper(h,α)is a nonempty closed totally ordered set. Moreover, if xper(h,α), thenxis a minimizer ofhqp:PqpR1, where

hqp(x)=hx0,. . .,xq

, Pqp= xRZ:T(q,p)x=x. (1.8)

Two elements ofᏹper(h,α) are called (h)-neighboring if there does not exist an ele- ment ofᏹper(h,α) between them. The following two propositions describe the structure of the setᏹ(h,α). For their proofs, see [3].

Proposition1.7. Suppose thatx< x+are(h)-neighboring elements of the setᏹper(h,α).

Then there existy(1),y(2)ᏹ(h,α)such that

x< y(1)< x+, x< y(2)< x+,

i→−∞lim y(1)i xi =0, lim

i→∞yi(1)x+i =0,

i→−∞lim y(2)i x+i =0, lim

i→∞yi(2)xi =0.

(1.9)

Suppose thatx< x+are (h)-neighboring elements ofᏹper(h,α). Define ᏹ+h,α,x,x+=

xᏹ(h,α) : lim

i→−∞xixi =0, lim

i→∞xix+i =0, ᏹh,α,x,x+=

xᏹ(h,α) : lim

i→−∞xixi+=0, lim

i→∞xixi=0. (1.10) We denote byᏹ+(h,α) (resp.,ᏹ(h,α)) the union of the sets+(h,α,x,x+) (resp., ᏹ(h,α,x,x+)) extended over all pairs of (h)-neighboring elementsx< x+ofᏹper(h,α).

Proposition1.8. (1)Ifx(h,α,x,x+)+(h,α,x,x+), wherex,x+per(h,α) are(h)-neighboring andx< x+, thenx< x < x+.

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(2)ᏹ(h,α)=per(h,α)+(h,α)(h,α).

(3)The setsper(h,α)+(h,α)andper(h,α)(h,α)are totally ordered.

(4)ᏹ+(h,α)= {xᏹ(h,α) : x > T(q,p)x}and(h,α)= {xᏹ(h,α) :x < T(q,p)x}. Letk2 be an integer. In this paper, we consider a complete metric space of functions h:R2R1which belong toCk(R2). This space is defined inSection 2and is denoted by Mk. We establish the existence of a setᏲMkwhich is a countable intersection of open everywhere dense subsets ofMkand such that for eachhᏲand each rationalα=p/q withpandqrelatively prime, the following properties hold:

(i) there exist (h)-minimal configurationsx(+),x(), andx(0)with rotation number αsuch thatx(+)iq+p > x(+)i ,x(iq)+p < xi(), andx(0)iq+p=xi(0)for all integersi;

(ii) any (h)-minimal configuration with rotation numberαis a translation of one of the configurationsx(+),x(), andx(0).

2. Spaces of functions

Letk2 be an integer. For f = f(x1,x2)Ck(R2) andq=(q1,q2)∈ {0,. . .,k}2, satisfy- ingq1+q2k, we set

|q| =q1+q2, Dqf = |q|f

∂x1q1∂xq22

. (2.1)

Denote byMkthe set of allhCk(R2) which have the property (H1), satisfying 2h

∂x1∂x2

ξ12

0 ξ1,ξ2

R2, (2.2)

and have the following property:

(H5) there existδh(0, 1) andch>0 such that hx1,x2

δhx1x2

2

ch x1,x2

R2. (2.3)

Clearly (H5) implies (H2).

Denote byMk0the set of allhMksuch that 2h

∂x1∂x2

ξ12

<0 ξ12

R2. (2.4)

For eachN,>0, we set Ek(N,)= h1,h2

Mk×Mk:Dqh1

x1,x2

Dqh2

x1,x2 for eachq∈ {0,. . .,k}2satisfying|q| ≤k

and eachx1,x2

R2satisfyingx1,x2N

h1,h2

Mk×Mk:h1

x1,x2

h2

x1,x2<

+max h1

x1,x2,h2

x1,x2 x1,x2

R2.

(2.5)

Using the following simple lemma, we can easily show that for the setMkthere exists the uniformity which is determined by the baseEk(N,),N,>0.

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Lemma2.1. Leta,bR1,(0, 1), and|ab|<+max{|a|,|b|}. Then

|ab|<+2(1)1+(1)1min |a|,|b|

. (2.6)

It is not difficult to see that the uniformity determined by the baseEk(N,),N,>0, is metrizable (by a metricdk) and complete [9]. For the setMk, we consider the topology induced by the metricd2, which is called the weak topology, and the topology induced by the metricdk, which is called the strong topology.

The following result shows that a generic function in Mk belongs to Mk0 and by Remark 1.3has the properties (H1), (H2), (H3), and (H4).

Theorem2.2. There exists a set0Mk0which is a countable intersection of open (in the weak topology) everywhere dense (in the strong topology) subsets ofMk.

Proof. ForhMkandγ(0, 1), definehγ:R2R1by hγ

x1,x2

=hx1,x2

+γx1x22

, x1,x2

R2. (2.7)

It is easy to see that forhMkandγ(0, 1),hγMk0and 2hγ

∂x1∂x2

ξ12

≤ −2γ, ξ12

R2, (2.8)

andhγhasγ0+in the strong topology.

Let f Mk, letγ(0, 1), and leti1 be an integer. By (2.5) and (2.8), there exists an open neighborhoodᐁ(f,γ,i) of fγinMkwith the weak topology such that the following property holds:

(P1) for eachgᐁ(f,γ,i) and each (ξ12)R2satisfying|ξ1|,|ξ2| ≤i, the inequality

2g/∂x1∂x212)≤ −γholds.

DefineᏲ0= ∩n=1∪ {ᐁ(f,γ,i) : f Mk, γ(0, 1), in}.Clearly, Ᏺ0 is a count- able intersection of open (in the weak topology) everywhere dense (in the strong topol- ogy) subsets ofMk. We will show that Ᏺ0Mk0. Leth0, (ξ1,ξ2)R2. Choose a natural numbernsuch that|ξ1|+|ξ2|< n. There exist f Mk,γ(0, 1), and an inte- gerinsuch thathᐁ(f,γ,i). It follows from property (P1) and the choice ofnthat (∂2h/∂x1∂x2)(ξ12)≤ −γ. Therefore,hMk0. This completes the proof ofTheorem 2.2.

3. The main results

We will prove the following result.

Theorem 3.1. Let k2be an integer and αa rational number. Then there exists a setαMk0which is a countable intersection of open (in the weak topology) everywhere dense (in the strong topology) subsets of Mk such that for each f α, the following assertions hold:

(1)ifx,yper(f,α), then there exist integersm,nsuch thatyi=xim+nfor alliZ;

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(2)ifx,y+(f,α), then there exist integersm,nsuch thatyi=xim+nfor alliZ; (3)ifx,y(f,α), then there exist integersm,nsuch thatyi=xim+nfor alliZ. It is not difficult to see thatTheorem 3.1implies the following result.

Theorem3.2. Letk2be an integer. Then there exists a setMk0which is a countable intersection of open (in the weak topology) everywhere dense (in the strong topology) subsets ofMksuch that for each rational numberαand each f Ᏺ, assertions (1), (2), and (3) of Theorem 3.1hold.

Theorem 3.1follows from the next two propositions.

Proposition3.3. Let k2be an integer andα a rational number. Then there exists a setα+Mk0which is a countable intersection of open (in the weak topology) everywhere dense (in the strong topology) subsets ofMk such that for each f α+, assertions (1) and (2) of Theorem 3.1hold.

Proposition3.4. Let k2be an integer andα a rational number. Then there exists a setαMk0which is a countable intersection of open (in the weak topology) everywhere dense (in the strong topology) subsets of Mksuch that for each f α, assertions (1) and (3) of Theorem 3.1hold.

Our goal is to proveProposition 3.3.Proposition 3.4is proved analogously.

4. Preliminary results for assertion (1) ofTheorem 3.1

Letm1 be an integer. Consider the manifold (R1/Z)mand the canonical mappingPm: Rm(R1/Z)m. We have the following result [21, Proposition 6.2].

Proposition 4.1. Letbe a closed subset of (R1/Z)2. Then there exists a nonnegative functionφC((R1/Z)2)such that= {x(R1/Z)2: φ(x)=0}.

Corollary4.2. Letbe a closed subset ofR1/Z. Then there exists a nonnegative function φC(R1/Z)such that= {xR1/Z:φ(x)=0}.

In this section, we assume thatk2 is an integer andα=p/qis an irreducible frac- tion, whereq1 andpare integers.

For each f Mk0, define Eα(f)=

q1 i=0

fxi,xi+1, xper(f,α), (4.1) (seeProposition 1.6).

Proposition4.3. Let f Mk, letQ be a natural number, and letD,>0. Then there exists a neighborhoodof f inMkwith the weak topology such that for eachgᐁ, each pair of integersn1,n2[n1+ 1,n1+Q], and each sequence{xi}ni=2n1R1which satisfies

min n21

i=n1

fxi,xi+1

,

n21 i=n1

gxi,xi+1

D, (4.2)

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the inequality

n21 i=n1

fxi,xi+1

n21 i=n1

gxi,xi+1 (4.3)

holds.

Proof. By (H5), there existδ0(0, 1) andc0>0 such that fx1,x2

δ0

x1x22

c0 x1,x2

R2. (4.4)

Choose a positive number1for which 1

Q+c0Q+D<41min{1,} (4.5) and a positive number0<1 which satisfies

0+20

10

1

+0

10

1

<411. (4.6) Define

= gMk: (f,g)Ek 1,0

(4.7)

(see (2.5)).

Assume that gᐁ, n1,n2Z,n2[n1+ 1,n1+Q], {xi}ni=2n1R1, and that (4.2) holds. By (2.5) and (4.7) for every (z1,z2)R2,

fz1,z2

gz1,z2<0+0max fz1,z2,gz1,z2. (4.8) It follows from (4.6), (4.8), andLemma 2.1that for every (z1,z2)R2,

fz1,z2

gz1,z2

<0+02

10

1

+0

10

1

min fz1,z2,gz1,z2

<411+ 411min fz1,z2,gz1,z2.

(4.9)

Formulas (4.4) and (4.9) imply that for every (z1,z2)R2, gz1,z2

fz1,z2

411411fz1,z2≥ −4112c0. (4.10) Set

λi=min fxi,xi+1,gxi,xi+1, i=n1,. . .,n21. (4.11) It follows from (4.4), (4.9), (4.10), and (4.11) that fori=n1,. . .,n21,

fxi,xi+1gxi,xi+1

<411+ 411min fxi,xi+1

+ 2c0,gxi,xi+1

+ 4c0+ 2

411+ 411λi+c01+1 2.

(4.12)

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By these inequalities, (4.2), (4.5), and (4.11),

n21 i=n1

fxi,xi+1gxi,xi+1

n2n1

411+ 211+1c0

+ 411 n21 i=n1

λi

n2n1

1+1c0

+ 411D

Q1+1c0

+ 411D <.

(4.13)

This completes the proof ofProposition 4.3.

Corollary4.4. Let f Mk0and>0. Then there exists a neighborhoodᐁof f inMk

with the weak topology such that for eachgMk0,Eα(g)Eα(f) +.

Proposition4.5. Assume that f Mk0,fnMk0,n=1, 2,. . .,limn→∞fn= f in the weak topology,

x(n)fn

, n=1, 2,. . .,xRZ,

nlim→∞x(n)i =xi iZ. (4.14) Thenxᏹ(f).

Proof. We assume the converse. Then, there exist integersi1< i2and a sequence{yi}ii2=i1 R1such that

yi1=xi1, yi2=xi2,

i21 i=i1

fyi,yi+1

<

i21 i=i1

fxi,xi+1

. (4.15)

Set

=

i21 i=i1

fxi,xi+1

fyi,yi+1

. (4.16)

For each integern1, define a finite sequence{yi(n)}ii2=i1R1as follows:

yi(n)1 =xi(n)1 , y(n)i2 =x(n)i2 , yi(n)=yi, i i1,. . .,i2

\ i1,i2

. (4.17)

It follows from (4.14), (4.15), (4.16), (4.17), and the continuity of f that

nlim→∞

i21

i=i1

fx(n)i ,xi+1(n)

i21 i=i1

fyi(n),yi+1(n)

=

i21 i=i1

fxi,xi+1

i21 i=i1

fyi,yi+1=>0.

(4.18)

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Formulas (4.14) and (4.18) imply that the sequences i21

i=i1

fx(n)i ,xi+1(n)

n=1

,

i21

i=i1

fyi(n),yi+1(n)

n=1

(4.19) are bounded. It follows from this fact,Proposition 4.3, and the equality f =limn→∞fnin the weak topology that

nlim→∞

i21

i=i1

fx(n)i ,xi+1(n)

i21 i=i1

fn

x(n)i ,x(n)i+1=0,

nlim→∞

i21

i=i1

fy(n)i ,y(n)i+1

i21 i=i1

fn

yi(n),yi+1(n)=0.

(4.20)

Formulas (4.18) and (4.20) imply that

nlim→∞

i21

i=i1

fn

x(n)i ,xi+1(n)

i21 i=i1

fn

yi(n),yi+1(n)=>0. (4.21)

There is an integern01 such that for each integernn0,

i21 i=i1

fn

x(n)i ,xi+1(n)

i21 i=i1

fn

yi(n),yi+1(n)>

2. (4.22)

This fact contradicts the (fn)-minimality ofx(n)for allnn0. The contradiction we have

reached provesProposition 4.5.

Proposition4.6. Let f Mk0, fnMk0,n=1, 2,. . .,limn→∞fn=f in the weak topol- ogy,x(n)per(fn,α),n=1, 2,. . .,and let the sequence {x0(n)}n=1 be bounded. Then the following assertions hold:

(1)there existxRZand a strictly increasing sequence of natural numbers{nj}j=1such that

xi+q=xi+p, iZ, (4.23)

x(ni j)−→xi asj−→ ∞,iZ; (4.24) (2)assume thatxRZand{nj}j=1is a strictly increasing sequence of natural numbers

such that (4.23) and (4.24) hold. Thenxper(f,α)and Eα(f)=

q1 i=0

fxi,xi+1

=lim

j→∞

q1 i=0

fnj

x(ni j),x(ni+1j)=lim

j→∞Eα fnj

. (4.25)

Proof. ByProposition 1.4, the sequence{x(n)i }n=1is bounded for alliZ. This fact im- plies that there exist a strictly increasing sequence of natural numbers{nj}j=1andxRZ such that (4.23) and (4.24) are valid. Therefore, assertion (1) is true.

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We will prove assertion (2). Assume thatxRZ and {nj}j=1 is a strictly increas- ing sequence of natural numbers such that (4.23) and (4.24) hold. ByProposition 4.5 and (4.23), xper(f,α). Since limn→∞fn= f in the weak topology, it follows from Corollary 4.4that the sequence{Eα(fn)}n=1 is bounded from above. Therefore, the se- quence{q1

i=0 fn(x(n)i ,xi+1(n))}n=1is also bounded from above. It follows from this fact, the equality limn→∞fn= f in the weak topology, andProposition 4.3that

nlim→∞

q1 i=0

fn

x(n)i ,xi+1(n)

q1 i=0

fxi(n),x(n)i+1=0. (4.26) By (4.1), (4.23), (4.24), (4.26), andCorollary 4.4,

Eα(f)

q1 i=0

fxi,xi+1=lim

j→∞

q1 i=0

fx(ni j),xi+1(nj)

=lim

j→∞

q1 i=0

fnjx(ni j),xi+1(nj)=lim

j→∞EαfnjEα(f).

(4.27)

These relations imply (4.25).Proposition 4.6is proved.

Proposition 4.6andCorollary 4.4imply the following result.

Proposition4.7. The function f Eα(f)is continuous onMk0with the relative weak topology.

Proposition4.8. Assume that f Mk0and that the following property holds:

Ifx(1),x(2)per(f,α), then there existsn=(n1,n2)Z2such thatx(2)=Tnx(1). Then there existsn¯=( ¯n1, ¯n2)Z2such that for eachxper(f,α),

Tn¯x > x, yper(f,α) :x < y < Tn¯x= ∅. (4.28) Proof. Let ¯xper(f,α). Then

per(f,α)= Tnx¯:n= n1,n2

Z2

= Tnx¯:n= n1,n2

Z2, 0n1q1. (4.29) Formula (4.29) implies that the set

yper(f,α) : ¯x < y < T(0,1)x¯ (4.30) is either finite or empty. Therefore, there exists ¯x+per(f,α) such that

¯

x <x¯+, yper(f,α) : ¯x < y <x¯+= ∅. (4.31) There exists ¯n=( ¯n1, ¯n2)Z2such that

Tn¯x¯=x¯+. (4.32)

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