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)i∈Z
∈RZ. The setRZwill be endowed with the product topology and the partial order de- fined byx < yif and only ifxi< yifor alli∈Z.
We have an order-preserving actionT:Z2×RZ→RZdefined by T(k,x)=Tkx=y⇐⇒k=
k1,k2
∈Z2,
x,y∈RZ, yi=xi−k1+k2 ∀i∈Z. (1.1) Letx,y∈RZ. We say that yis a translation ofx if there isn=(n1,n2)∈Z2 such that y=Tnx.
Leth:R2→R1be a continuous function. We extend hto arbitrary finite segments (xj,. . .,xk),j < k, of configurationsx∈RZby
hxj,. . .,xk
:=
k−1 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
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 hξ1,η1
+hξ2,η2
< hξ1,η2
+hξ2,η1
; (1.3)
(H4) if (x−1,x0,x1)=(y−1,y0,y1) are (h)-minimal segments andx0=y0, then x−1−y−1
x1−y1
<0. (1.4)
A configurationx∈RZis (h)-minimal if, for each pair of integers jandksatisfying j < kand each finite segment{yi}ki=j⊂R1satisfying 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 configurationsx∈RZandx∗∈RZcross (a) ati∈Zifxi=x∗i and (xi−1−x∗i−1)(xi+1−xi+1∗ )<0, (b) betweeniandi+ 1 if (xi−x∗i )(xi+1−x∗i+1)<0.
Definition 1.2. The configurationx∈RZis periodic with period (q,p)∈(Z\ {0})×Zif T(q,p)x=x.
Remark 1.3. Assume thath=h(ξ1,ξ2)∈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 h∈C2(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].
Proposition1.4. There exists a continuous functionα(h):ᏹ(h)→R1with the following properties:
(i)for allx∈ᏹ(h),i∈Z,
xi−x0−iα(h)(x)<1; (1.6) (ii)ifx∈ᏹ(h)is periodic with period(q,p)∈Z2, thenα(h)(x)=p/q;
(iii)for allα∈R1, the set{x∈ᏹ(h) :α(h)(x)=α} = ∅. Remark 1.5. We callα(h)(x) the rotation number ofx∈ᏹ(h).
For eachα∈R1, define
ᏹ(h,α)= x∈ᏹ(h) :α(h)(x)=α. (1.7) We studyᏹ(h,α) with rationalα∈R1.
Let a rationalα=p/qbe an irreducible fraction, whereq≥1 andpare integers. De- note byᏹper(h,α) the set of all periodic (h)-minimal configurationsx∈ᏹ(h,α) which satisfyT(q,p)x=x, equivalently,xi−q+p=xi, for alli∈Z.
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 x∈ ᏹper(h,α), thenxis a minimizer ofhqp:Pqp→R1, where
hqp(x)=hx0,. . .,xq
, Pqp= x∈RZ: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 −x−i =0, lim
i→∞yi(1)−x+i =0,
i→−∞lim y(2)i −x+i =0, lim
i→∞yi(2)−x−i =0.
(1.9)
Suppose thatx−< x+are (h)-neighboring elements ofᏹper(h,α). Define ᏹ+h,α,x−,x+=
x∈ᏹ(h,α) : lim
i→−∞xi−x−i =0, lim
i→∞xi−x+i =0, ᏹ−h,α,x−,x+=
x∈ᏹ(h,α) : lim
i→−∞xi−xi+=0, lim
i→∞xi−xi−=0. (1.10) We denote byᏹ+(h,α) (resp.,ᏹ−(h,α)) the union of the setsᏹ+(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∈ᏹ−(h,α,x−,x+)∪ᏹ+(h,α,x−,x+), wherex−,x+∈ᏹper(h,α) are(h)-neighboring andx−< x+, thenx−< x < x+.
(2)ᏹ(h,α)=ᏹper(h,α)∪ᏹ+(h,α)∪ᏹ−(h,α).
(3)The setsᏹper(h,α)∪ᏹ+(h,α)andᏹper(h,α)∪ᏹ−(h,α)are totally ordered.
(4)ᏹ+(h,α)= {x∈ᏹ(h,α) : x > T(q,p)x}andᏹ−(h,α)= {x∈ᏹ(h,α) :x < T(q,p)x}. Letk≥2 be an integer. In this paper, we consider a complete metric space of functions h:R2→R1which belong toCk(R2). This space is defined inSection 2and is denoted by Mk. We establish the existence of a setᏲ⊂Mkwhich is a countable intersection of open everywhere dense subsets ofMkand such that for eachh∈Ᏺand each rationalα=p/q withpandqrelatively prime, the following properties hold:
(i) there exist (h)-minimal configurationsx(+),x(−), andx(0)with rotation number αsuch thatx(+)i−q+p > x(+)i ,x(i−−q)+p < xi(−), andx(0)i−q+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
Letk≥2 be an integer. For f = f(x1,x2)∈Ck(R2) andq=(q1,q2)∈ {0,. . .,k}2, satisfy- ingq1+q2≤k, we set
|q| =q1+q2, Dqf = ∂|q|f
∂x1q1∂xq22
. (2.1)
Denote byMkthe set of allh∈Ck(R2) which have the property (H1), satisfying ∂2h
∂x1∂x2
ξ1,ξ2
≤0 ∀ ξ1,ξ2
∈R2, (2.2)
and have the following property:
(H5) there existδh∈(0, 1) andch>0 such that hx1,x2
≥δhx1−x2
2
−ch ∀ x1,x2
∈R2. (2.3)
Clearly (H5) implies (H2).
Denote byMk0the set of allh∈Mksuch that ∂2h
∂x1∂x2
ξ1,ξ2
<0 ∀ ξ1,ξ2
∈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,x2≤N
∩ 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.
Lemma2.1. Leta,b∈R1,∈(0, 1), and|a−b|<+max{|a|,|b|}. Then
|a−b|<+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 setᏲ0⊂Mk0which is a countable intersection of open (in the weak topology) everywhere dense (in the strong topology) subsets ofMk.
Proof. Forh∈Mkandγ∈(0, 1), definehγ:R2→R1by hγ
x1,x2
=hx1,x2
+γx1−x22
, x1,x2
∈R2. (2.7)
It is easy to see that forh∈Mkandγ∈(0, 1),hγ∈Mk0and ∂2hγ
∂x1∂x2
ξ1,ξ2
≤ −2γ, ξ1,ξ2
∈R2, (2.8)
andhγ→hasγ→0+in the strong topology.
Let f ∈Mk, letγ∈(0, 1), and leti≥1 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 eachg∈ᐁ(f,γ,i) and each (ξ1,ξ2)∈R2satisfying|ξ1|,|ξ2| ≤i, the inequality
∂2g/∂x1∂x2(ξ1,ξ2)≤ −γholds.
DefineᏲ0= ∩∞n=1∪ {ᐁ(f,γ,i) : f ∈Mk, γ∈(0, 1), i≥n}.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 Ᏺ0⊂Mk0. Leth∈Ᏺ0, (ξ1,ξ2)∈R2. Choose a natural numbernsuch that|ξ1|+|ξ2|< n. There exist f ∈Mk,γ∈(0, 1), and an inte- geri≥nsuch thath∈ᐁ(f,γ,i). It follows from property (P1) and the choice ofnthat (∂2h/∂x1∂x2)(ξ1,ξ2)≤ −γ. Therefore,h∈Mk0. This completes the proof ofTheorem 2.2.
3. The main results
We will prove the following result.
Theorem 3.1. Let k≥2be an integer and αa rational number. Then there exists a set Ᏺα⊂Mk0which is a countable intersection of open (in the weak topology) everywhere dense (in the strong topology) subsets of Mk such that for each f ∈Ᏺα, the following assertions hold:
(1)ifx,y∈ᏹper(f,α), then there exist integersm,nsuch thatyi=xi−m+nfor alli∈Z;
(2)ifx,y∈ᏹ+(f,α), then there exist integersm,nsuch thatyi=xi−m+nfor alli∈Z; (3)ifx,y∈ᏹ−(f,α), then there exist integersm,nsuch thatyi=xi−m+nfor alli∈Z. It is not difficult to see thatTheorem 3.1implies the following result.
Theorem3.2. Letk≥2be an integer. Then there exists a setᏲ⊂Mk0which 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 k≥2be an integer andα a rational number. Then there exists a setᏲα+⊂Mk0which is a countable intersection of open (in the weak topology) everywhere dense (in the strong topology) subsets ofMk such that for each f ∈Ᏺα+, assertions (1) and (2) of Theorem 3.1hold.
Proposition3.4. Let k≥2be an integer andα a rational number. Then there exists a setᏲα−⊂Mk0which is a countable intersection of open (in the weak topology) everywhere dense (in the strong topology) subsets of Mksuch that for each f ∈Ᏺα−, 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
Letm≥1 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. LetΩbe a closed subset of (R1/Z)2. Then there exists a nonnegative functionφ∈C∞((R1/Z)2)such thatΩ= {x∈(R1/Z)2: φ(x)=0}.
Corollary4.2. LetΩbe a closed subset ofR1/Z. Then there exists a nonnegative function φ∈C∞(R1/Z)such thatΩ= {x∈R1/Z:φ(x)=0}.
In this section, we assume thatk≥2 is an integer andα=p/qis an irreducible frac- tion, whereq≥1 andpare integers.
For each f ∈Mk0, define Eα(f)=
q−1 i=0
fxi,xi+1, x∈ᏹper(f,α), (4.1) (seeProposition 1.6).
Proposition4.3. Let f ∈Mk, letQ be a natural number, and letD,>0. Then there exists a neighborhoodᐁof f inMkwith the weak topology such that for eachg∈ᐁ, each pair of integersn1,n2∈[n1+ 1,n1+Q], and each sequence{xi}ni=2n1⊂R1which satisfies
min n2−1
i=n1
fxi,xi+1
,
n2−1 i=n1
gxi,xi+1
≤D, (4.2)
the inequality
n2−1 i=n1
fxi,xi+1−
n2−1 i=n1
gxi,xi+1≤ (4.3)
holds.
Proof. By (H5), there existδ0∈(0, 1) andc0>0 such that fx1,x2
≥δ0
x1−x22
−c0 ∀ x1,x2
∈R2. (4.4)
Choose a positive number1for which 1
Q+c0Q+D<4−1min{1,} (4.5) and a positive number0<1 which satisfies
0+20
1−0
−1
+0
1−0
−1
<4−11. (4.6) Define
ᐁ= g∈Mk: (f,g)∈Ek 1,0
(4.7)
(see (2.5)).
Assume that g∈ᐁ, n1,n2∈Z,n2∈[n1+ 1,n1+Q], {xi}ni=2n1⊂R1, 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
1−0
−1
+0
1−0
−1
min fz1,z2,gz1,z2
<4−11+ 4−11min fz1,z2,gz1,z2.
(4.9)
Formulas (4.4) and (4.9) imply that for every (z1,z2)∈R2, gz1,z2
≥ fz1,z2
−4−11−4−11fz1,z2≥ −4−11−2c0. (4.10) Set
λi=min fxi,xi+1,gxi,xi+1, i=n1,. . .,n2−1. (4.11) It follows from (4.4), (4.9), (4.10), and (4.11) that fori=n1,. . .,n2−1,
fxi,xi+1−gxi,xi+1
<4−11+ 4−11min fxi,xi+1
+ 2c0,gxi,xi+1
+ 4c0+ 2
≤4−11+ 4−11λi+c01+1 2.
(4.12)
By these inequalities, (4.2), (4.5), and (4.11),
n2−1 i=n1
fxi,xi+1−gxi,xi+1
≤ n2−n1
4−11+ 2−11+1c0
+ 4−11 n2−1 i=n1
λi
≤ n2−n1
1+1c0
+ 4−11D
≤Q1+1c0
+ 4−11D <.
(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 eachg∈ᐁ∩Mk0,Eα(g)≤Eα(f) +.
Proposition4.5. Assume that f ∈Mk0,fn∈Mk0,n=1, 2,. . .,limn→∞fn= f in the weak topology,
x(n)∈ᏹfn
, n=1, 2,. . .,x∈RZ,
nlim→∞x(n)i =xi ∀i∈Z. (4.14) Thenx∈ᏹ(f).
Proof. We assume the converse. Then, there exist integersi1< i2and a sequence{yi}ii2=i1⊂ R1such that
yi1=xi1, yi2=xi2,
i2−1 i=i1
fyi,yi+1
<
i2−1 i=i1
fxi,xi+1
. (4.15)
Set
∆=
i2−1 i=i1
fxi,xi+1
−fyi,yi+1
. (4.16)
For each integern≥1, define a finite sequence{yi(n)}ii2=i1⊂R1as 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→∞
i2−1
i=i1
fx(n)i ,xi+1(n)−
i2−1 i=i1
fyi(n),yi+1(n)
=
i2−1 i=i1
fxi,xi+1−
i2−1 i=i1
fyi,yi+1=∆>0.
(4.18)
Formulas (4.14) and (4.18) imply that the sequences i2−1
i=i1
fx(n)i ,xi+1(n)
∞ n=1
,
i2−1
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→∞
i2−1
i=i1
fx(n)i ,xi+1(n)−
i2−1 i=i1
fn
x(n)i ,x(n)i+1=0,
nlim→∞
i2−1
i=i1
fy(n)i ,y(n)i+1−
i2−1 i=i1
fn
yi(n),yi+1(n)=0.
(4.20)
Formulas (4.18) and (4.20) imply that
nlim→∞
i2−1
i=i1
fn
x(n)i ,xi+1(n)−
i2−1 i=i1
fn
yi(n),yi+1(n)=∆>0. (4.21)
There is an integern0≥1 such that for each integern≥n0,
i2−1 i=i1
fn
x(n)i ,xi+1(n)−
i2−1 i=i1
fn
yi(n),yi+1(n)>∆
2. (4.22)
This fact contradicts the (fn)-minimality ofx(n)for alln≥n0. The contradiction we have
reached provesProposition 4.5.
Proposition4.6. Let f ∈Mk0, fn∈Mk0,n=1, 2,. . .,limn→∞fn=f in the weak topol- ogy,x(n)∈ᏹper(fn,α),n=1, 2,. . .,and let the sequence {x0(n)}∞n=1 be bounded. Then the following assertions hold:
(1)there existx∈RZand a strictly increasing sequence of natural numbers{nj}∞j=1such that
xi+q=xi+p, i∈Z, (4.23)
x(ni j)−→xi asj−→ ∞,∀i∈Z; (4.24) (2)assume thatx∈RZand{nj}∞j=1is a strictly increasing sequence of natural numbers
such that (4.23) and (4.24) hold. Thenx∈ᏹper(f,α)and Eα(f)=
q−1 i=0
fxi,xi+1
=lim
j→∞
q−1 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 alli∈Z. This fact im- plies that there exist a strictly increasing sequence of natural numbers{nj}∞j=1andx∈RZ such that (4.23) and (4.24) are valid. Therefore, assertion (1) is true.
We will prove assertion (2). Assume thatx∈RZ 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), x∈ᏹper(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{q−1
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→∞
q−1 i=0
fn
x(n)i ,xi+1(n)−
q−1 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)≤
q−1 i=0
fxi,xi+1=lim
j→∞
q−1 i=0
fx(ni j),xi+1(nj)
=lim
j→∞
q−1 i=0
fnjx(ni j),xi+1(nj)=lim
j→∞Eαfnj≤Eα(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)∈ᏹper(f,α), then there existsn=(n1,n2)∈Z2such thatx(2)=Tnx(1). Then there existsn¯=( ¯n1, ¯n2)∈Z2such that for eachx∈ᏹper(f,α),
Tn¯x > x, y∈ᏹper(f,α) :x < y < Tn¯x= ∅. (4.28) Proof. Let ¯x∈ᏹper(f,α). Then
ᏹper(f,α)= Tnx¯:n= n1,n2
∈Z2
= Tnx¯:n= n1,n2
∈Z2, 0≤n1≤q−1. (4.29) Formula (4.29) implies that the set
y∈ᏹper(f,α) : ¯x < y < T(0,1)x¯ (4.30) is either finite or empty. Therefore, there exists ¯x+∈ᏹper(f,α) such that
¯
x <x¯+, y∈ᏹper(f,α) : ¯x < y <x¯+= ∅. (4.31) There exists ¯n=( ¯n1, ¯n2)∈Z2such that
Tn¯x¯=x¯+. (4.32)