In this paper, we show the existence of heteroclinic orbits between two different static classes of the Aubry set of a Tonelli Lagrangian when the Aubry set has only finite static classes (which is a generic condition in the sense of Mañé) and the Mañé set satisfies certain isolated condition.
Citation: Guowei Yu. Heteroclinic orbits between static classes of time periodic Tonelli Lagrangian systems[J]. Electronic Research Archive, 2022, 30(6): 2283-2302. doi: 10.3934/era.2022116
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In this paper, we show the existence of heteroclinic orbits between two different static classes of the Aubry set of a Tonelli Lagrangian when the Aubry set has only finite static classes (which is a generic condition in the sense of Mañé) and the Mañé set satisfies certain isolated condition.
Let M be a closed connected smooth Riemannian manifold, and L∈C2(TM×R,R) be a Lagrangian satisfying the following conditions introduced by Mather [1]:
1. Periodicity: L is 1-periodic in time, i.e., L(x,v,t)=L(x,v,t+1) for all (x,v,t)∈TM×R;
2. Positive definiteness: ∂2L/∂v2(x,v,t) is positive definite, as a quadratic form, for all (x,v,t)∈TM×R;
3. Superlinear growth: lim‖v‖x→+∞L(x,v,t)‖v‖x=+∞, in each fiber, i.e., for each A∈R, there exists B(A)∈R such that L(x,v,t)≥A‖v‖x−B(A), for all (x,v,t)∈TM×R.
4. Completeness: Every orbit of the Euler-Lagrange flow introduced by L is defined for all time.
Remark 1.1. ‖⋅‖x denotes the norm associated to the Riemannian metric on M. Since M is compact, condition (3) in the above definition is independent of which Riemannian metric is chosen on M. For any x,y∈M, we use d(x,y) denotes the distance defined by the Riemannian metric.
For any compact interval [a,b]⊂R, we denote C([a,b],M) by the set of all absolutely continuous curves defined on [a,b]. The action AL of L on any γ∈C([a,b],M) is defined by
AL(γ):=∫baL(dγ(t),t)dt,wheredγ(t)=(γ(t),˙γ(t)). |
Given x,y∈M, we set C[a,b](x,y):={γ∈C([a,b],M):γ(a)=x,γ(b)=y} and CT(x,y):=C[0,T](x,y) for any T>0. The extremals of AL in C[a,b](x,y) are solutions of the Euler-Lagrange equation which in local coordinates is given by
ddt∂L∂v(γ(t),˙γ(t),t)=∂L∂x(γ(t),˙γ(t),t).(EL) |
Furthermore we assume L is a critical Lagrangian. A Lagrangian is called critical if the infimum of the actions of all closed curves is 0. Every Tonelli Lagrangian can be made critical by adding a proper constant, see [2].
For any (x,a),(y,b)∈M×R with a<b, we define
h((x,a),(y,b))=inf{∫baL(dγ(t),t)dt:γ∈C[a,b](x,y)}. |
The infimum above is actually a minimum, see Theorem 2.1.
For any (x,α),(y,β)∈M×T, we define
h∞((x,α),(y,β))=lim infT∈Z+,T→+∞h((x,α),(y,β+T)), |
Φ((x,α),(y,β))=infT∈Z+h((x,α),(y,β+T)). |
The following propositions of the functions we just defined will be needed in this paper and we refer the interested readers to [3] for their proofs.
Proposition 1.1. 1. The function h:M×R×M×R→R;((x,a),(y,b))→h((x,a),(y,b)) is Lipschitz continuous and bounded on {b−a≥1}.
2. For any α,β∈T, h∞((∗,α),(∗,β)):M×M→R is Lipschitz continuous and satisfies triangle inequality
h∞((x,α),(y,β))+h∞((y,β),(z,ρ))≥h∞((x,α),(z,ρ)), |
for any (x,α),(y,β),(z,ρ)∈M×T.
3. We set ˜d:M×T×M×T:→R as
˜d((x,α),(y,β))=Φ((x,α),(y,β))+Φ((y,β),(x,α)), |
then ˜d is non-negative.
Let I⊂R be an interval of time, a curve γ∈C(I,M) is called semi-static if
AL(γ|[a,b])=Φ((γ(a),amod1),(γ(b),bmod1)),∀[a,b]⊂I. |
When I=R, we say γ is a global semi-static curve. When I=[a,+∞) (or I=(−∞,a]) for some a∈R, we say γ is a forward (or backward) semi-static curve. Obviously a semi-static curve is a solution of (EL).
If γ satisfies
AL(γ|[a,b])=−Φ((γ(b),bmod1),(γ(a),amod1)),∀[a,b]⊂I, |
we say it is static. When I=R, we say γ is a global static curve. By Proposition 1.1 it is not hard to see a static curve must be semi-static.
We define the Mañé set ˜N⊂TM×T and Aubry set ˜A⊂TM×T as
˜N:={(dγ(t),tmod1):γ∈C(R,M)isglobalsemi−static.}; |
˜A:={(dγ(t),tmod1):γ∈C(R,M)isglobalstatic.}. |
Similarly we define
˜N+:={(dγ(t),tmod1):γisforwardsemi−static.}; |
˜N−:={(dγ(t),tmod1):γisbackwardsemi−static.}. |
It is easy to see
˜A⊂˜N⊂˜N±. |
Let π:TM×T→M×T be the usual projection. We set N=π(˜N) and A=π(˜A). The famous Mather's graph theorem (see [1]) tells us π|˜A:˜A→A is bijective and its inverse (π|˜A)−1:A→˜A is Lipschitz.
Following Mather's graph theorem, for any (x,α)∈A⊂M×T, there is a unique global static curve, denoted by γ(x,α), satisfying γ(x,α)(α)=x. Furthermore using ˜d defined in Proposition 1.1, we can define an equivalence relation on ˜A by saying (x,v,a),(y,w,b)∈˜A are equivalent iff ˜d(π(x,v,a),π(y,w,b))=˜d((x,a),(y,b))=0.
By this equivalence relation, we break ˜A into classes will be called static classes. Let ˜A be the set of static classes, through the entire paper ˜Λ,˜Ω,˜Γ and ˜Δ with or without super-index will be used to represent static classes.
Under the assumption that
˜Acontainsonlyfiniteelements,(∗1) |
we will prove the following two theorems.
Remark 1.2. (∗1) is a generic condition in the sense of Mañé, see [4].
Theorem 1.1. For any two different static classes ˜Λ1,˜Λ2∈˜A, one of the following must be true:
1. There is a global semi-static curve γ with the α-limit set α(dγ)⊂˜Λ1 and the ω-limit set ω(dγ)⊂˜Λ2;
2. There is a finite set of static classes {˜Ω1,…,˜Ωn}⊂˜A∖{˜Λ1,˜Λ2} and global semi-static curves γi:i=0,…,n with α(dγi)⊂˜Ωi and ω(dγi)⊂˜Ωi+1 for i=0,…,n, where ˜Ω0=˜Λ1 and ˜Ωn+1=˜Λ2.
By α(dγ) (or ω(dγ)), we mean the α-limit set (or ω-limit set) of the orbit {(γ(t),˙γ(t),tmod1):t∈R} of the Euler-Lagrange flow. Similarly we let α(γ) (or ω(γ)) be the α-limit set (or ω-limit set) of {γ(t):t∈Z} in M.
When L is time-independent the above theorem has been proved by Contreras and Paternain in [5].
Remark 1.3. Through the entire paper, for any subset U⊂M and δ>0, by U(δ) we mean U(δ):={x∈M:d(x,U)≤δ}.
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Let N0:=N∩M×{0} and A0:=A∩M×{0}. Under an further assumption
N0∖A0(δ)istotallyisolated,forsomeδ>0smallenough.(∗2) |
We will prove along a chain of heteroclinic orbits obtained in Theorem 1.1, there is a real heteroclinic orbit connecting the two given static classes.
Theorem 1.2. If (∗1),(∗2) are true, for any two different static classes ˜Λ1,˜Λ2∈˜A, there is a curve γ∈C2(R,M), such that (dγ(t),tmod1) is an orbit of the Euler-Lagrange system introduced by L with α(dγ)⊂˜Λ1 and ω(dγ)⊂˜Λ2.
Similar variational method has been used in [6] and [7], where the authors established the existence of homoclinic orbits to the Aubry set under various conditions different from ours.
By Mather's graph theorem, we can break A into a set of equivalent classes A:={Λ=π(˜Λ):˜Λ∈˜A}. By abusing of notation, they will also be called static classes and capital Greek letters Λ,Ω with or without super-index will be reserved to represent such static classes throughout this paper.
For all the sets ˜A,˜N,˜Λ (A,N,Λ), by putting a sub-index t∈T to them, we mean it is their intersection with TM×{t}(M×{t}), for example ˜At=˜A∩TM×{t}, At=A∩M×{t}.
For any n∈N, we define hn:M×M→R by
hn(x,y)=h((x,0),(y,n)), |
and h∞:M×M→R, Φ:M×M→R by
h∞(x,y)=lim infn∈N,n→+∞hn(x,y); |
Φ(x,y)=Φ((x,0),(y,0)). |
In some cases, we need to shift the time parameterization of a given curve for which we introduce the following operator.
Definition 2.1. Given a<b∈R for any c∈R, we define an operator τc:C([a,b],M)→C([a+c,b+c],M) by
τc(γ)(t)=γ(t−c),foranyγ∈C([a,b],M)andt∈[a+c,b+c]. |
The variational study of Tonelli Lagrangian L depends on some standard results proved by Mather in [1].
Lemma 2.1. Given a real number K and a compact interval [a,b],
{γ∈C([a,b],M):AL(γ)≤K} |
is compact for the topology of uniform convergence.
Theorem 2.1. (Tonelli Theorem) Given two points x,y∈M and a compact interval [a,b], there is a γ∈C[a,b](x,y) with AL(γ)=h((x,a),(y,b)) and γ is a C2 solution of (EL).
If a curve γ∈C[a,b](x,y) satisfies AL(γ)=h((x,a),(y,b)) we will call it a minimizer. Obviously a semi-static curve is a minimizer, but a minimizer is not necessarily a semi-static curve.
The next lemma is well-known to experts, however we can not locate a complete proof in the literature, therefore we give one at here.
Lemma 2.2. Given any p,q∈M and two sequences of positive integers {T1k}↗+∞,{T2k}↗+∞, if {γk∈C[−T1k,T2k](p,q)} is a sequence of minimizers, i.e., AL(γk)=hT1k+T2k(γk(−T1k),γk(T2k)), for any k∈N, satisfying
limk→+∞AL(γk)=lim infn→+∞hn(p,q)=h∞(p,q), |
then there is a global semi-static curve γ, such that γk converges to γ uniformly on any compact interval along a subsequence.
Proof. By Proposition 1.1 and Lemma 2.1, it is not hard to see for any T>0, along a subsequence γk converges uniformly to a γT∈C([−T,T],M) on [−T,T]. Apply this to a sequence of positive integers {Tn}↗+∞, then by a diagonal extraction, we can find a subsequence of γk, which we rename as γk, and a γ∈C(R,M), such that γk converges uniformly to γ on any compact interval.
We claim for any T∈Z+, AL(γ|[−T,T])=Φ(γ(−T),γ(T)), so γ is a global semi-static curve.
Assume this is not true, then there is a T∈Z+, such that
ε=AL(γ|[−T,T])−Φ((γ(−T),γ(T))>0. | (2.1) |
Since γk|[−T,T] converges uniformly to γ|[−T,T], the lower semi-continuity of AL implies
AL(γk|[−T,T])≥AL(γ|[−T,T])−ε4forklargeenough. | (2.2) |
By the definition of Φ there is a S∈Z+, such that
hS(γ(−T),γ(T))≤Φ(γ(−T),γ(T))+ε4. | (2.3) |
Because hS is Lipschitz,
|hS(γk(−T),γk(T))−hS(γ(−T),γ(T))|≤C[d(γk(−T),γ(−T))+d(γk(T),γ(T))]. |
As limk→+∞d(γk(±T),γ(±T))=0, for k large enough we have
|hS(γk(−T),γk(T))−hS(γ(−T),γ(T))|≤ε4. | (2.4) |
By Tonelli Theorem, for any k∈N, there are ξk∈C[0,S](γk(−T),γk(T)) with
AL(ξk)=hS(γk(−T),γk(T)), |
so by (2.4),
AL(ξk)≤hS(γ(−T),γ(T))+ε4. | (2.5) |
Combine this with (2.1), (2.2) and (2.3), we get
AL(ξk)≤Φ(γ(−T),γ(T))+ε2=AL(γ|[−T,T])−ε2≤AL(γk|[−T,T])−ε4, |
for k large enough.
We define a new sequence of curves {˜γk∈CT1k+T2k−2T+S(p,q)} by
˜γk(t)={γk(t−T1k)ift∈[0,T1k−T]ξk(t−T1k+T)ift∈[T1k−T,T1k−T+S]γk(t−T1k+2T−S)ift∈[T1k−T+S,T1k+T2k−2T+S]. |
For all k large enough, AL(˜γk)≤AL(γk)−ε4. Since {T2k+T1k−2T+S} goes to infinity as k→+∞, we have
h(p,q)≤lim infk→+∞hT2k+T1k−2T+S(p,q)≤lim infk→+∞AL(˜γk)≤limk→+∞AL(γk)−ε4=h(p,q)−ε4, |
which is absurd and we proved our claim.
Remark 2.1. In the previous lemma, if we assume {T1k} (resp. {T2k}) is bounded, by the same arguments it is not hard to see there is a forward semi-static curve (resp. backward semi-static curve) γ, such that, γk converges uniformly to γ on any compact interval in the domain of γ along a subsequence.
In this paper a stronger version of Lemma 2.2 will be needed.
Lemma 2.3. Given any p,q∈M and a sequence of positive integers {Tk}↗+∞, let {γk∈CTk(p,q)} be a sequence of minimizers, i.e., AL(γk)=h((p,0),(q,Tk)), satisfies limk→+∞AL(γk)=h∞(p,q).
If {ak}↗+∞,{bk}↗+∞ are two sequence of positive integers satisfying
1. 0≤ak≤bk≤Tk, for all k∈N;
2. bk−ak→+∞ as k→+∞;
3. there are x,y∈M, such that along a subsequences d(γk(ak),x)→0 and d(γk(bk),y)→0 as k→+∞,
then along a subsequence, limk→+∞AL(γk|[ak,bk])=h∞(x,y).
Proof. Noticing that {AL(γk|[ak,bk])} has a finite upper bound, passing to a subsequence if necessary, we can say
limk→+∞hbk−ak(γk(ak),γk(bk))=limk→+∞AL(γk|[ak,bk])=B<+∞. |
We claim B=h∞(x,y) and the lemma follows immediately from this claim.
By Lipschitz continuity of hn,
|hbk−ak(γk(ak),γk(bk))−hbk−ak(x,y)|≤C(d(γk(ak),x)+d(γk(bk),y)), |
by passing to a proper subsequence, we can say d(γk(ak),x)+d(γk(bk),y))→0 as k→+∞, then
limk→+∞hbk−ak(x,y)=limk→+∞hbk−ak(γk(ak),γk(bk)). |
Hence,
B=limk→+∞hbk−ak(x,y)≥lim infn→+∞hn(x,y)=h∞(x,y). |
Assume ε=B−h∞(x,y)>0, then there are non-negative integers a′k<b′k with b′k−a′k→+∞ as k→+∞ satisfying
limk→+∞hb′k−a′k(x,y)=h∞(x,y)=B−ε. |
Then for k large enough, we have
hb′k−a′k(x,y)≤B−34ε. | (2.6) |
On the other hand, B=limk→+∞hbk−ak(γk(ak),γk(bk)), so for k large enough
hbk−ak(γk(ak),γk(bk))≥B−ε4. | (2.7) |
Combine (2.6) and (2.7), we get
hbk−ak(γk(ak),γk(bk))≥hb′k−a′k(x,y)+ε2. | (2.8) |
Again by the Lipschitz continuity of hn and d(γk(ak),x)+d(γk(bk),y))→0 as k→+∞, for k large enough we have
hb′k−a′k(γk(ak),γk(bk))≤hb′k−a′k(x,y)+ε4. | (2.9) |
(2.8) and (2.9) imply
hbk−ak(γk(ak),γk(bk))≥hb′k−a′k(γk(ak),γk(bk))+ε4. |
Now following the same argument as in the proof of Lemma 2.2, we can construct a new sequence {γ∗k∈CTk−(bk−ak)+(b′k−a′k)(p,q)} with
h∞(p,q)≤lim infk→+∞AL(γ∗k)≤lim infk→+∞AL(γk)−ε4=h∞(p,q)−ε4, |
which is a contradiction.
In next lemma we will strengthen the triangle inequality of h∞ given in Proposition 1.1.
Lemma 2.4. Given a Λ∈A and a compact set U⊂M with Λ0∩U=∅, then we can find an ε>0 small enough and a δ=δ(ε)>0, such that
h∞(x,y)+h∞(y,z)≥h∞(x,z)+ε,∀x,z∈Λ0(δ),∀y∈U. |
Proof. Because Λ0∩U=∅ and U is compact, there is an ε′>0, such that
˜d(p,y)=h∞(p,y)+h∞(y,p)≥ε′>0,∀p∈Λ0,∀y∈U. |
Then for any q∈Λ0, as h∞ satisfies triangle inequality,
h∞(p,y)+h∞(y,q)+h∞(q,p)≥h∞(p,y)+h∞(y,p)≥ε′, |
so
h∞(p,y)+h∞(y,q)≥−h∞(q,p)+ε′=h∞(p,q)+ε′, | (2.10) |
the last equality is because of h∞(p,q)+h∞(q,p)=˜d(p,q)=0, as p,q∈Λ0 are in the same static class.
Let ε=ε′4 and δ=ε8C, then for any x,z∈Λ0(δ), there are p′,q′∈Λ0 with d(x,p′)≤δandd(z,q′)≤δ. By the Lipschitz continuity of h∞, we have
h∞(x,y)≥h∞(p′,y)−Cδ, |
h∞(y,z)≥h∞(y,q′)−Cδ, |
h∞(x,z)≤h(p′,q′)+2Cδ, |
combine these with (2.10), we get
h∞(x,y)+h∞(y,z)≥h∞(p′,y)+h∞(y,q′)−2Cδ≥ h(p′,q′)+ε′−2Cδ≥h∞(x,z)−2Cδ+2ε−2Cδ=h∞(x,z)+2ε−4Cδ=h∞(x,z)+32ε |
Given an arbitrary (x,α)∈M×T, by Fathi's weak KAM theory [8], there are at least one forward semi-static curve γ+∈C2([a,+∞),M) with γ+(a)=x, amod1=α and one backward semi-static curve γ−∈C2((−∞,a],M) with γ−(a)=x, amod1=α.
One of the most important feature of these forward (or backward) semi-static curves is that the must asymptotic to a unique static class of ˜A.
Lemma 2.5. If γ is a forward (or backward) semi-static curve, then there is a unique static class ˜Λ∈˜A, such that ω(dγ) (or α(dγ)) ⊂˜Λ.
Generally speaking there may be more than one forward (or backward) semi-static curves starting from (or ending at) a given point (x,α)∈M×T, however if (x,α) belongs to A, then they are unique.
Lemma 2.6. If (x,α)∈A, then there is a unique v∈TxM, such that (x,v,α)∈˜A⊂˜N±.
As a consequence there is a unique forward semi-static curve γ+∈C([α,+∞),M) with γ+(α)=x and a unique backward semi-static curve γ−∈C((−∞,α],M) with γ−(α)=x, and γ+,γ− must be static, with
γ+(t)=γ(x,α)(t),ift∈[α,+∞); |
γ−(t)=γ(x,α)(t),ift∈(−∞,α], |
where γ(x,α) is the unique global static curve passing (x,α).
The proofs of the above two lemmas can be found in [2,5,8], which we will not repeat here.
We will give the proof of Theorem 1.1 in this section. By assumption (∗1), we can fix a δ∗>0 small enough, such that for any two different static classes Λ1,Λ2∈A, Λ10(δ∗)∩Λ20(δ∗)=∅. In order to distinguish the two different cases in Theorem 1.1, we introduce the following definition
Definition 3.1. Given a δ>0 and a curve γ∈C([a,b],M), for any two static classes Λ1,Λ2∈A, we defined the following set
K(γ,δ,Λ1,Λ2):={Λ∈A∖{Λ1,Λ2}:mint∈[a,b]∩Zd(γ(t),Λ0)≤δ}. |
A similar idea was used by Maxwell [9] and Rabinowitz [10] on a special class of Hamiltonian systems including periodic forced multiple pendulum under time-reversibility assumption.
For the remainder of this section, we fix two arbitrary static classes Λ1≠Λ2∈A, and two points p∈Λ10,q∈Λ20. By Tonelli Theorem, there is a sequence of minimizers {γk∈C[−T1k,T2k](p,q)} satisfying
1. {T1k∈Z+}↗+∞,{T2k∈Z+}↗+∞ as k→+∞;
2. limk→+∞AL(γk)=limk→+∞hT2k+T1k(p,q)=h∞(p,q).
Let's consider the collection of sets K(γk,δ,Λ1,Λ2), for all k∈N and δ>0, there are two possibilities:
Case 1: There is a δ0>0 small enough, such that for each m∈N, there is a k>m with K(γk,δ0,Λ1,Λ2)=∅;
Case 2: For each δ>0, there is a m∈N, such that K(γk,δ,Λ1,Λ2)≠∅,∀k>m.
First we shall assume case 1 hold, then by passing {γk} to a subsequence, we may assume K(γk,δ0,Λ1,Λ2)=∅ for all k∈N.
Proposition 3.1. If Case 1 hold, there is a global semi-static curve γ satisfying
α(dγ)⊂˜Λ1andω(dγ)⊂˜Λ2. |
Proof. For each k∈N, we set Sk:=min{t∈[−T1k,T2k]∩Z:d(γk(t),Λ10)>δ∗}. It is not hard to see, for each k∈N, Sk is a well-defined integer and d(γk(t),Λ10)≤δ∗,forallt∈[−T1k,Sk)∩Z.
Lemma 3.2. Both {Sk+T1k} and {T2k−Sk} go to infinity as k goes to infinity.
Proof. (Lemma 3.2) We will only give the detailed proof for Sk+T1k→+∞, while T2k−Sk→+∞ can be proven similarly.
Suppose {Sk+T1k} is bounded, then passing {γk} to a subsequence, we can say Sk+T1k≡T∈Z+ for all k∈N. For each k∈N, we define γ∗k∈C([0,T1k+T2k],M) by γ∗k=τT1k(γk), where τT1k is defined as in Definition 2.1, i.e., we shift the time parameterization on γk forward by T1k.
Replace γ∗k by a proper subsequence, by Remark 2.1, there is a forward semi-static curve γ∗∈C([0,+∞),M), such that γ∗k converges to γ∗ on any compact interval of [0,+∞). Since
d(γ∗(T),Λ10)=limk→+∞d(γ∗k(T),Λ10)=limk→+∞d(γk(Sk),Λ10)≥δ∗, |
γ∗(T)∉Λ10. At the same time γ∗(0)=limγ∗k(0)=p∈Λ10, then γ∗ can not be a static curve.
Because p∈Λ10⊂A0, there must be a global static curve ξ with ξ(0)=p. As a result we have two different forward semi-static curves starting from p and this is a contradiction to Lemma 2.6.
By Lemma 3.2, without loss of generality, we may assume Sk≡0 for all k∈N. Passing γk to a subsequence, by Lemma 2.2, there is a global semi-static curve γ, such that γk converges uniformly to γ on any compact interval.
Since d(γk(t),Λ10)≤δ∗, for all t∈[−T1k,0)∩Z and k∈N, we have d(γ(t),Λ10)≤δ∗, for all t∈(−∞,0)∩Z.
By Lemma 2.5, there is a unique ˜Λ∈˜A, such that α(dγ)⊂˜Λ, hence we must have ˜Λ=˜Λ1 and α(dγ)⊂˜Λ1.
Because Case 1 is true, either ω(dγ)⊂˜Λ2 or ω(dγ)⊂˜Λ1 must be true. Assume ω(dγ)⊂˜Λ1, then there is a sequence of positive integers {Tj}↗+∞ with limj→+∞d(γ(Tj),x)=0 for some x∈Λ10.
On the other hand we can find a subsequence {γkj∈C[−T1kj,T2kj](p,q)} of {γk} with Tj≤T2kj for all j∈N and d(γkj(Tj),γ(Tj)) approaches to 0 as j→+∞. Therefore d(γkj(Tj),x) approaches to 0 as j→+∞.
Replacing {γkj} by a proper subsequence, by Lemma 2.3
h∞(p,x)=limj→+∞AL(γkj|[−T1kj,Tj]). | (3.1) |
Since {T1kj} and {Tj} goes to infinity, as j→+∞, both sequences {γkj|[−T1kj,0]} and {γkj|[0,Tj]} satisfy conditions of Lemma 2.3, passing {γkj} to a proper subsequence, we have
h∞(p,γ(0))=limj→+∞AL(γkj|[−T1kj,0]);h∞(γ(0),x)=limj→+∞AL(γkj|[0,Tj]). | (3.2) |
By (3.1) and (3.2), h∞(p,γ(0))+h∞(γ(0),x)=h∞(p,x).
However it is easy to see d(γ(0),Λ10)≥δ∗, at the same time p,x∈Λ10, by Lemma 2.4, h∞(p,γ(0))+h∞(γ(0),x)>h∞(p,x), which is a contradiction and we finished our proof.
Now we assume Case 2 is true, because of (∗1), by passing {γk} to a subsequence, we can say that for a sequence of positive real numbers {δk∈(0,δ∗)}↘0, there is a finite set of static classes {Ω1,…,Ωn}⊂A∖{Λ1,Λ2} satisfying K(γk,δk,Λ1,Λ2}≡{Ω1,…,Ωn} for all k∈N.
We set Ω0=Λ1 and Ωn+1=Λ2.
Proposition 3.3. If Case 2 is true, there is a chain of global semi-static curves γi:i=0,…,n satisfying α(dγi)⊂˜Ωi and ω(dγi)⊂˜Ωi+1 for i=0,…,n.
Proof. By Definition 3.1, for every i=1,…,n and k∈N, we can find an Sik∈(−T1k,T2k)∩Z with d(γk(Sik),Ωi0)≤δk and we set S0k=−T1k and Sn+1k=T2k for all k∈N.
Without loss of generality, we can assume Sik≤Si+1k for all i=0,…,n and k∈N. Furthermore by passing {γk} to a proper subsequence, we can say there are xi∈Ωi0 for i=0,…,n+1 with
limk→+∞d(γk(Sik),xi)=0, |
where x0=p and xn+1=q.
Lemma 3.4. For every i=0,…,n, {Si+1k−Sik}→+∞ as k→+∞.
Proof. (Lemma 3.4) It can be proven similarly as we did in the proof of Lemma 3.2 and we will not repeat it here.
Lemma 3.5. There is a ˜δ∈(0,δ∗) small enough, such that for every k large enough
K(γk|[Sik,Si+1k],˜δ,Ωi,Ωi+1)=∅,∀i=0,…,n. |
Proof. (Lemma 3.5) If not, without loss of generality we can assume there are 0≤i0≠i1≤n and a sequence of positive numbers {˜δk∈(0,δ∗)}↘0 satisfying
Ωi1∈K(γk|[Si0k,Si0+1k],˜δk,Ωi0,Ωi0+1) |
for all k large enough.
By the definition of K, we must have i1≠i0+1. Let's assume i1>i0+1 (the case of i1<i0 can be proven similarly), then there is a sequence of integers {Sk∈(Si0k,Si0+1k)}, such that d(γk(Sk),Ωi10)≤˜δk.
First we will show that {Si0+1k−Sk} approaches positive infinity as k goes to +∞. If this is not true, replacing {γk} by a proper subsequence, we may assume Si0+1k−Sk≡S∈R for all k∈N.
We define a new sequence of minimizers {γ∗k=τ−Sk(γk)}, then
d(γ∗k(0),Ωi10)=d(τ−Sk(γk)(0),Ωi10)=d(γk(Sk),Ωi10)≤˜δk. |
Since ˜δk→0 as k→+∞, passing {γ∗} to a proper subsequence, we can say limk→+∞γ∗k(0)=x for some x∈Ωi10.
Again passing {γ∗k} to a proper subsequence, by Lemma 2, there is a global semi-static curve γ∗ such that γ∗k converges to γ∗ on any compact interval. Hence
γ∗(0)=limk→+∞γ∗k(0)=x∈Ωi10. |
On the other hand
γ∗(S)=limk→+∞γ∗k(S)=limk→+∞γk(S+Sk)=limk→+∞γk(Si0+1k)=xi0+1∈Ωi0+10. |
As Ωi0+10∩Ωi10=∅, γ∗ can not be a static curve. Since (x,0)∈Ωi1⊂A, there is a global static curve γ(x,0) with γ(x,0)(0)=x. Obviously γ(x,0)≠γ∗ and we have two different forward semi-static curves starting from x, which is a contradiction to Lemma 2.6. Therefore {Si0+1k−Sk}→+∞ as k→+∞.
At the same time by Lemma 3.4, {Si1k−Si0+1k}→+∞ as k→+∞. Noticing that {γk|[Sk,Si1k]} is a sequence of minimizers with
limk→+∞γk(Sk)=x,limk→+∞γ(Si0+1k)=xi0+1andlimk→+∞γk(Si1k)=xi1, |
by Lemma 2.3, passing {γk} to a proper subsequence, we have
h∞(x,xi1)=limk→+∞AL(γk|[Sk,Si1k])=limk→+∞AL(γk|[Sk,Si0+1k])+limk→+∞AL(γk|[Si0+1k,Sk1k])=h∞(x,xi0+1)+h∞(xi0+1,xi1). |
However x,xi1∈Ωi10 and xi0+1∈Ωi0+10, by Lemma 2.4,
h∞(x,xi1)<h∞(x,xi0+1)+h∞(xi0+1,xi1), |
which is absurd, so we are done.
Now we resume our proof of Proposition 3.3.
For each i=0,…,n, by Lemma 3.4 and Lemma 3.5, we showed that {γ|[Sik,Si+1k]} is a sequence of minimizers satisfying the following:
1. Si+1k−Sik→+∞ as k→+∞;
2. limk→+∞d(γk(Sik),xi)=0 and limk→+∞d(γk(Si+1k),xi+1)=0;
3. K(γk|[Sik,Si+1k],˜δ,Ωi,Ωi+1)=∅, for all k large enough.
Let {γik=τck(γ|[Sik,Si+1k])}, where ck=−Sik+Si+1k2, then following the same argument as in the proof of Proposition 3.1, we can show that along a subsequence {γik} converges uniformly on any compact interval to a global semi-static curve γi satisfying α(dγi)⊂˜Ωi and ω(dγi)⊂˜Ωi+1.
Hence {γi:i=0,…,n} form a chain of heteroclinic orbits as we wanted and {xi:i=0,…,n} satisfies xi∈ω(γi−1)∩α(γi)∩Ωi0 for i=1,…,n, and x0∈α(γ0)∩Ω00, xn+1∈ω(γn)∩Ωn+10.
Furthermore we have
h∞(xi,xi+1)=limk→+∞AL(γk|[Sik,Si+1k]),foralli=0,…,n, | (3.3) |
Hence,
h∞(p,q)=limk→+∞AL(γk|[S0k,Sn+1k])=n∑i=0limk→+∞AL(γk|[Sik,Si+1k])=n∑i=0h∞(xi,xi+1). | (3.4) |
Obviously Theorem 1.1 follows directly from Proposition 3.1 and Proposition 3.3.
This section will be devoted to the proof of Theorem 1.2. In the previous section we proved that for any two different static classes, there is a chain of heteroclinic orbits connecting them. Under the assumption (∗2), we will show that there is a real heteroclinic orbit connecting those two static classes along such a chain of heteroclinic orbits.
Such a heteroclinic orbits will be found as a constraint minimizer of the action of Lagrangian L. Using the minimizing properties of static curves and semi-static curves, we will show that the constraint minimizers we find will not bump up to the boundary conditions we posted and it is a real orbit.
First we will introduce some technical lemmas.
Lemma 4.1. There is a ˆδ∈(0,δ∗) small enough, such that if p,q∈Λ0(ˆδ), for some Λ∈A, then
h∞(p,x)+h∞(x,q)=h∞(p,q),∀x∈Λ0, | (4.1) |
and
h∞(p,x)+h∞(x,q)>h∞(p,q),ifx∈A0∖Λ0. | (4.2) |
Proof. Choose an arbitrary p,q∈Λ0(δ∗), first let's show h∞(p,x)+h∞(x,q)=h∞(p,x′)+h∞(x′,q) if x,x′∈Λ0. By the triangle inequality of h∞,
h∞(p,x)+h∞(x,q)≤h∞(p,x′)+h∞(x′,x)+h∞(x,x′)+h∞(x′,q)=h∞(p,x′)+h∞(x′,q), |
the last equality is due to the fact that x,x′ are contained in the same static class, which means
h∞(x,x′)+h∞(x′,x)=˜d(x,x′)=0. |
Therefore h∞(p,x)+h∞(x,q)≤h∞(p,x′)+h∞(x′,q), the other direction of the inequality can be proven similarly.
First we will prove (4.1) under the assumption that both p,q are contained in Λ0.
For any x∈Λ0, by triangle inequality
h∞(p,x)+h∞(x,q)≥h∞(p,q). |
On the other hand, since q,x are contained in the same static class Λ0, ˜d(q,x)=h∞(q,x)+h∞(x,q)=0, then
h∞(p,q)−h∞(x,q)=h∞(p,q)+h∞(q,x)≥h∞(p,x), |
so
h∞(p,q)≥h∞(p,x)+h∞(x,q). |
As a result, we proved equality (4.1) under the assumption p,q∈Λ0.
Now we will drop our previous assumption. Without loss of generality, let's say q∉Λ0. Then there is a sequence of minimizers {γk∈C[0,Tk](p,q)} with {Tk∈Z+}↗+∞, and a forward semi-static curve γ∈C([0,+∞),M), such that γk converges uniformly to γ on any compact sub-interval of [0,+∞).
Then ω(γ)⊂Ω for some Ω∈A. We can find a sequence of positive integers {Sk}↗+∞ and a x∈Ω0, such that limk→+∞d(γ(Sk),x)=0, then
limk→+∞d(γk(Sk),x)=0 |
. 0 We claim Tk−Sk→+∞ as k→+∞, if this is not true, passing to a subsequence, we may assume Tk−Sk≡T∈Z+. Let ξk∈C[−Tk,0](p,q);ξk=τ−Tk(γk) be a new sequence of minimizers.
Then passing to a subsequence ξk converges uniformly to a backward semi-static curve ξ on any compact sub-interval of (−∞,0].
Noticing that ξk(−T)=ξk(Sk−Tk)=γk(Sk) approaches to x as k goes to +∞, so ξ(−T)=limk→+∞ξk(−T)=x∈Ω0. Then by Lemma 2.6, ξ must be part of a global static curve and (ξ(t),0)∈A0 for any t∈(−∞,0]∩Z. However ξ(0)=limξk(0)=limγk(Tk)=q∉Λ0, which is a contradiction.
As we just showed, when k goes to +∞, {Tk−Sk} approaches to +∞, {Sk} approaches to +∞ and γk(Sk) approaches to x, by Lemma 2.3, passing {γk} to a proper subsequence, we have
h∞(p,q)=limk→+∞AL(γk|[0,Tk])=limk→+∞AL(γk|[0,Sk])+limk→+∞AL(γk|[Sk,Tk])=h∞(p,x)+h∞(x,q). |
Now we will show that for a small enough ˆδ>0, we must have Ω=Λ, then x∈Λ0 and (4.1) follows immediately from that.
Assume Ω≠Λ, then x∈A0∖Λ0, which is a compact set without any intersection with Λ0, then by Lemma 2.4, we can find an ˆδ∈(0,δ∗), such that for any p,q∈Λ0(ˆδ) and y∈A0∖Λ0, we have
h∞(p,y)+h∞(y,q)>h∞(p,q). |
This is a contradiction to to what we just proved. Therefore we must have Ω=Λ.
Finally (4.2) follows directly from Lemma 2.4.
In section 2, we have mentioned that for any (x,α)∈M×T, there is at least one forward semi-static curve γ+ (or backward semi-static curve γ−) starting from (or ending at) (x,α) with its ω-limit set (or α-limit set) contained in a unique static class. Generally we can not determine which static class it will approach to, however by the above lemma we can determine where it asymptotic to in some special case.
Lemma 4.2. Let ˆδ be defined as in Lemma 4.1, if p∈Λ0(ˆδ), q∈Λ0 for some Λ∈A and {γk∈CTk(p,q)} is a sequence of minimizers with limk→+∞AL(γk)=h∞(p,q) and {Tk}↗+∞, then there is a forward semi-static curve γ+∈C2([0,+∞),M) satisfying the following conditions
1. γ+(0)=p;
2. {γk} converges uniformly to γ+ along a subsequence on any compact interval;
3. ω(dγ+)⊂˜Λ.
Proof. By Remark 2.1, there is a forward semi-static curve γ satisfying conditions (1),(2), we claim it also satisfies condition (3). If not, by Lemma 2.5, there is another static class Ω≠Λ, such that ω(dγ)⊂˜Ω.
Then there is a x∈Ω0 and a sequence of integers {Sk∈(0,Tk)}↗+∞, such that
limk→+∞d(γ(Sk),x)=0. |
Hence passing {γk} to a subsequence, we have
limk→+∞d(γk(Sk),x)=0. |
Similar to the proof of Lemma 4.1, we have {Tk−Sk}→+∞ as k→+∞.
Since {Tk−Sk}→+∞, {Sk}→+∞ and limk→+∞d(γk(Sk),x)=0 as k→+∞, Lemma 2.3 tells us that along a subsequence of {γk},
h∞(p,x)=limk→+∞AL(γk|[0,Sk]); |
h∞(x,q)=limk→+∞AL(γk|[Sk,Tk]). |
Therefore
h∞(p,q)=limk→+∞AL(γk|[0,Tk])=h∞(p,x)+h∞(x,q). | (4.3) |
If x∈Ω0≠Λ0, Lemma 4.1 tells us
h∞(p,q)<h∞(p,x)+h∞(x,q), |
which is a contradiction to (4.3). As a result we proved our claim and γ is the forward semi-static curve γ+ we are looking for.
Remark 4.1. It is not hard to see that similar argument can to be used to show the existence of a backward semi-static curve γ− with γ−(0)=p and α(dγ−)⊂˜Λ.
Lemma 4.3. Given a global semi-static curve γ, if p∈α(γ)∩A0 and q∈ω(γ)∩A0, then
h∞(p,q)=h∞(p,γ(S))+AL(γ|[S,T])+h∞(γ(T),q),foranyS≤T∈Z. |
Proof. For arbitrary p∈α(γ)∩A0 and q∈ω(γ)∩A0, there are two sequences of integers {Sk}↘−∞ and {Tk}↗+∞, such that limk→+∞γ(Sk)=p and limk→+∞γ(Tk)=q.
Following Lemma 2.3, without loss of generality, we can say
h∞(p,q)=limk→+∞AL(γ|[Sk,Tk]), |
h∞(p,γ(S))=limk→−∞AL(γ|[Sk,S]), |
h∞(γ(T),q)=limk→+∞AL(γ|[T,Tk]). |
Therefore
h∞(p,q)=limk→+∞[AL(γ|[Sk,S])+AL(γ|[S,T])+AL(γ|[T,Tk])]=limk→+∞AL(γ|[Sk,S])+AL(γ|[S,T])+limk→+∞AL(γ|[T,Tk])=h∞(p,γ(S))+AL(γ|[S,T])+h∞(γ(T),q). |
Lemma 4.4. Given a global semi-static curve γ, if there are two compact sets U,V⊂M satisfying
1. γ(S)∈U,γ(T)∈V for some S≤T∈Z;
2. ∂U∩N0=∅,∂V∩N0=∅, where ∂U,∂V are the boundaries of U,V.
Then for any p∈α(γ)∩A0, q∈ω(γ)∩A0 and (x,y)∈U×V., we have
Δ(p,x,y,q;γ):=h∞(p,x)+hT−S(x,y)+h∞(y,q)−h∞(p,q)≥0. |
Furthermore, if x∈U∖N0 or y∈V∖N0, then
Δ(p,x,y,q;γ)>0 |
and there is a ε>0 such that
min{Δ(p,x,y,q;γ):(x,y)∈U×V∖Int(U)×Int(V)}≥ε, |
where Int(U),Int(V) are the interiors of U,V.
Proof. Assume there are x∈U, y∈V with Δ(p,x,y,q;γ)<0, then
h∞(p,x)+hT−S(x,y)+h∞(y,q)<h∞(p,q). | (4.4) |
By Remark 2.1, we can find two sequences of minimizers
{ξk∈C[0,Sk](p,x)}with{Sk∈Z}↗+∞; |
{ζk∈C[0,Tk](y,q)}with{Tk∈Z}↗+∞, |
satisfying
h∞(p,x)=limk→+∞AL(ξk);h∞(y,q)=limk→+∞AL(ζk). |
Let η∈C[S,T](x,y) be a minimizer, i.e., AL(η)=hT−S(x,y). We define a new sequence of curves {γk∈C[−Sk,T−S+Tk](p,q)} by
γk(t)={ξk(t+Sk),ift∈[−Sk,0];η(t+S),ift∈[0,T−S];ζk(t−T+S),ift∈[T−S,T−S+Tk]. |
Obviously Sk+Tk+T−S→+∞ as k→+∞, and
limk→+∞AL(γk)=h∞(p,x)+hT−S(x,y)+h∞(y,q). |
However,
h∞(p,q)=lim infn→+∞hn(p,q)≤lim infk→+∞hSk+Tk+T−S(p,q)≤limk→+∞AL(γk)=h∞(p,x)+hT−S(x,y)+h∞(y,q)<h∞(p,q), |
where the last inequality follows from (4.4). However this is absurd and we proved the first part of the lemma.
For the second part, without loss of generality we can assume there are x∈U∖N0 and y∈V with Δ(p,x,y,q;γ)=0. Let {γk} be defined as above then
limk→+∞AL(γk)=h∞(p,x)+hT−S(x,y)+h∞(y,q)=h∞(p,q). | (4.5) |
Although γk is not necessary a minimizer, by (4.5), it is not hard to see, for any T′∈Z+, {AL(γk|[−T′,T′])} has a finite upper bound, and the argument in the proof of Lemma 2.2 will still hold. Therefore along a subsequence {γk} converges uniformly to a global semi-static curve γ∗ on any compact interval. Which means
γ∗(0)=limk→+∞γk(0)=η(S)=x∈N0, |
and this is a contradiction to our assumption.
Hence Δ(p,x,y,q;γ)>0, if x∈U∖N0 or y∈V∖N0. By the Lipschitz continuity of hn and h∞, there is a ε>0 such that
min{Δ(p,x,y,q;γ):(x,y)∈U×V∖Int(U)×Int(V)}≥ε>0. |
Now we are ready to prove Theorem 1.2.
Proof. (Theorem 1.2) We will follow the notations from the previous section, let p∈Λ10,q∈Λ20 and {γk∈C[−T1k,T2k](p,q)} is a sequence of minimizers satisfying
h∞(p,q)=limk→+∞AL(γk). |
Again there are two different cases as we discussed in the previous section.
If Case 1 is true, nothing needs to be done here.
If Case 2 is true, by the proof of Proposition 3.3, we have a set of finite static classes {Ω1,…,Ωn}⊂A∖{Λ1,Λ2}, a chain of global semi-static curves {γi:i=1,…,n} and {xi∈Ωi0:i=0,1,…,n+1}, where x0=p,xn+1=q, satisfying
1. α(dγi)⊂˜Ωi and ω(dγi)⊂˜Ωi+1, for i=0,…,n, where Ω0=Λ1 and Ωn+1=Λ2;
2. xi∈ω(γi−1)∩α(γi)∩Ωi0, for i=1,…,n, and x0=p∈α(γ0)∩Ω00, xn+1∈ω(γn)∩Ωn+10.
3. h∞(x0,xn+1)=∑ni=0h∞(xi,xi+1).
Let ˆδ>0 be defined as in Lemma 4.1, for each γi, we can choose Si<Ti∈Z satisfying
γi(Si)∈Int(Ωi0(ˆδ))andγi(Ti)∈Int(Ωi+10(ˆδ)). |
Then by assumption (∗2), for each i=0,…,n there are compact sets Ui,Vi⊂M satisfying
γi(Si)∈Int(Ui),Ui⊂Ωi0(ˆδ),∂Ui∩N0=∅; |
γi(Ti)∈Int(Vi),Vi⊂Ωi+10(ˆδ),∂Vi∩N0=∅. |
For each (Y,Z):={(y0,z0),…,(yn,zn)}∈∏ni=0Ui×Vi, we define a function J by
J(Y,Z):=h∞(x0,y0)+n∑i=0hTi−Si(yi,zi)+n−1∑i=0h∞(zi,yi+1)+h∞(zn,xn+1). | (4.6) |
For each i=0,…,n−1, zi∈Vi⊂Ωi+10(ˆδ) and yi+1∈Ui+1⊂Ωi+10(ˆδ), by Lemma 4.1,
h∞(zi,yi+1)=h∞(zi,xi+1)+h∞(xi+1,yi+1). |
Hence the function J can be rewritten as
J(Y,Z)=n∑i=0[h∞(xi,yi)+hTi−Si(yi,zi)+h∞(zi,xi+1)]. | (4.7) |
For any τ=(τi)n−1i=0 with each τi∈Z+ define a Jτ:∏ni=0Ui×Vi→R, by
Jτ(Y,Z)=h∞(x0,y0)+n∑i=0hTi−Si(yi,zi)+n−1∑i=0hτi(zi,yi+1)+h∞(zn,xn+1). | (4.8) |
Let
cτ:=inf{Jτ(Y,Z):(Y,Z)={(y0,z0),…,(yn,zn)}∈n∏i=0Ui×Vi}, |
by the compactness of Ui,Vi and Lipschitz continuity of hn,h∞, it is easy to see the above infimum is in fact a minimum.
Lemma 4.5. There is a τ, such that if (Y′,Z′)={(y′0,z′0),…,(y′n,z′n)}∈∏ni=0Ui×Vi satisfies JT(Y′,Z′)=cT, then
(Y′,Z′)={(y′0,z′0),…,(y′n,z′n)}∈n∏i=0Int(Ui)×Int(Vi). |
We postpone the proof of the above lemma for a moment. Now there is a
(Y′,Z′)={(y′0,z′0),…,(y′n,z′n)}∈n∏i=0Int(Ui)×Int(Vi) |
with Jτ(Y′,Z′)=cτ. Meanwhile for each i=0,…,n, there is a minimizer ζi∈C[0,Ti−Si](y′i,z′i) with AL(ζi)=hTi−Si(y′i,z′i), and for each j=0,…,n−1, there is a minimizer ηj∈C[0,τj](z′j,y′j+1) such that AL(ηj)=hτj(z′j+1,y′j+1).
By Lemma 4.2, there is a forward semi-static γ+∈C([0,+∞),M) with γ+(0)=z′n and ω(dγ+)⊂˜Ωn+1, and a backward semi-static curve γ−∈C((−∞,0],M) with γ−(0)=y′0 and α(dγ−)⊂˜Ω0.
Gluing these curves together by the following order
γ−∗ζ0∗η0∗⋯∗ηn−1∗ζn∗γ+, |
we get a new curve γ∈C(R,M) with α(dγ)⊂˜Λ1 and ω(dγ)⊂˜Λ2.
By the standard variational argument, it is not hard to see γ is a classical solution of (EL) and we are done.
Proof of Lemma 4.5. Set {γi(Si),γi(Ti)}:={(γ0(S0),γ0(T0)),…,(γn(Sn),γn(Tn))}. Since {γi(Si),γi(Ti)}∈∏ni=0Int(Ui)×Int(Vi), it is enough to show, there is a τ such that
Jτ(Y,Z)>Jτ({γi(Si),γi(Ti)}), |
for all (Y,Z)={(y0,z0),…,(yn,zn)}∈∏ni=0Ui×Vi∖∏ni=0Int(Ui)×Int(Vi). Using the expression of J given in (4.7), we get
J(Y,Z)−J({γi(Si),γi(Ti)})=n∑i=0{h∞(xi,yi)+hTi−Si(yi,zi)+h∞(zi,xi+1)−[h∞(xi,γi(Si))+hTi−Si(γi(Si),γi(Ti))+h∞(γi(Ti),xi+1)]}. |
Meanwhile by Lemma 4.3, for each i=0,…,n,
h∞(xi,γi(Si))+hTi−Si(γi(Si),γi(Ti))+h∞(γi(Ti),xi+1)=h∞(xi,xi+1). |
Therefore
J(Y,Z)−J({γi(Si),γi(Ti)})=n∑0Δ(xi,yi,zi,xi+1;γi). |
By Lemma 4.4, there is an ε>0 independent of the choice of (Y,Z), such that
J(Y,Z)−J({γi(Si),γi(Ti)})≥ε. | (4.9) |
By the compactness of Ui,Vi and Lipschitz continuity of hn,h∞, there is a τ′∈Z+, such that for all i=0,…,n−1, when τi≥τ′,
hτi(zi,yi+1)−h∞(zi,yi+1)≥−ε4(n−1). |
Then (4.6) and (4.8) imply
Jτ(Y,Z)≥J(Y,Z)−ε4. | (4.10) |
Meanwhile for each i=0,…,n−1, we can always find a τi≥τ′, such that
hτi(γi(Ti),γi+1(Si+1))−h∞(γi(Ti),γi+1(si+1)≤ε4(n−1). |
Again by (4.6) and (4.8), we get
Jτ({γi(Si),γi(Ti)})≤J({γi(Si),γi(Ti)})+ε4. | (4.11) |
Combining the above inequality with (4.9) and (4.10), we have
Jτ(Y,Z)≥J(Y,Z)−ε4≥J({γi(Si),γi(Ti)})+3ε4≥Jτ({γi(Si),γi(Ti)})+ε2. |
This finishes our proof of the lemma.
This work is supported by the National Key R & D Program of China (2020YFA0713303), NSFC (No. 12171253), the Fundamental Research Funds for the Central Universities and Nankai Zhide Fundation.
The authors declare there is no conflicts of interest.
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