Boundedness of Solutions for a Class of Second-Order Periodic Systems

Abstract

In this paper we study the following second-order periodic system: 𝑥^′′+𝑉^′(𝑥)+𝑝(𝑥,𝑡)=0, where 𝑉(𝑥) has a singularity. Under some assumptions on the 𝑉(𝑥) and 𝑝(𝑥,𝑡) by Ortega’ small twist theorem, we obtain the existence of quasi-periodic solutions and boundedness of all the solutions.

1. Introduction and Main Result

In 1991, Levi [1] considered the following equation: 𝑥^′′+𝑉^′(𝑥,𝑡)=0, (1) where𝑉(𝑥,𝑡)satisfies some growth conditions and𝑉(𝑥,𝑡)=𝑉(𝑥,𝑡+1). The author reduced the system to a normal form and then applied Moser twist theorem to prove the existence of quasi-periodic solution and the boundedness of all solutions. This result relies on the fact that the nonlinearity𝑉(𝑥,𝑡)can guarantee the twist condition of KAM theorem. Later, several authors improved Levi's result; we refer to [2–4] and the references therein.

Recently, Capietto et al. [5] studied the following equation: 𝑥^′′+𝑉^′(𝑥)=𝐹(𝑥,𝑡), (2) wher𝐹(𝑥,𝑡)=𝑝(𝑡)is a𝜋-periodic function and𝑉(𝑥)=(1/2)𝑥²₊+(1/(1−𝑥²₋)^𝜈)−1, where𝑥₊=max{𝑥,0},??𝑥₋=max{−𝑥,0}, and𝜈is a positive integer. Under the Lazer-Leach assumption that 11+2^𝜋₀𝑝𝑡₀+𝜃sin𝜃𝑑𝜃>0,∀𝑡₀∈𝑅, (3) they prove the boundedness of solutions and the existence of quasi-periodic solution by Moser twist theorem. It is the first time that the equation of the boundedness of all solutions is treated in case of a singular potential.

We observe that𝐹(𝑥,𝑡)=𝑝(𝑡)in (2) is smooth and bounded, so a natural question is to find sufficient conditions on𝐹(𝑥,𝑡)such that all solutions of (2) are bounded when𝐹(𝑥,𝑡)is unbounded. The purpose of this paper is to deal with this problem.

Motivated by the papers [1, 5, 6], we consider the following equation: 𝑥^′′+𝑉^′(𝑥)+𝑝(𝑥,𝑡)=0, (4) where 1𝑉=2𝑥²₊+11−𝑥²₋−1,𝑥>−1. (5)

In order to state our main results, we give some notation and assumptions. Let𝛼∈(0,1)be some fixed constant. Let 𝑝()𝑝=𝑥,𝑡|𝑥|^𝛼,𝑃(𝑥,𝑡)=^𝑥₀𝑝(𝑠,𝑡)𝑑𝑠. (6)(A1)Assume𝑝(𝑥,𝑡)∈𝐶^7,6(𝑆¹×𝑅) andlim_𝑥→+∞𝑝(𝑥,𝑡)=𝑝₊(𝑡)uniformly in𝑡.(A2)lim_𝑥→+∞𝑥^𝑚(𝜕^𝑚+𝑛𝑝(𝑥,𝑡)/𝜕𝑥^𝑚𝜕𝑡^𝑛)=𝑝_+,𝑚,𝑛(𝑡)uniformly in𝑡for(𝑚,𝑛)=(6,0),(6,7),(0,7),where𝑝_+,𝑚,0(𝑡)≡0and𝑝_+,6,7(𝑡)≡0.(A3)We suppose Lazer-Leach assumption holds: ^𝜋₀𝑝₊𝑡₀+𝜃(sin𝜃)^1+𝛼𝑑𝜃>0,∀𝑡₀∈𝑅. (7)

Our main result is the following theorem.

Theorem 1. Under the assumptions (A1)–(A3), all the solutions of (4) are defined for all𝑡∈(−∞,+∞), and for each solution𝑥(𝑡), one hassup_𝑡∈𝑅(|𝑥(𝑡)|+|𝑥^′(𝑡)|)<+∞.

The main idea of our proof is acquired from [6]. The proof of Theorem 1 is based on a small twist theorem due to Ortega [7]. The hypotheses (A1)–(A3) of our theorem are used to prove that the Poincaré mapping of (4) satisfies the assumptions of Ortega's theorem.

Moreover, we have the following theorem on solutions of Mather type.

Theorem 2. Assume that𝑝(𝑡)∈𝐶satisfies (7); then, there is an𝜖₀>0such that, for any𝜔∈(1/𝜋,1/(𝜋+𝜖₀)), (4) has a solution(𝑥_𝜔(𝑡),𝑥^′_𝜔(𝑡))of Mather type with rotation number𝜔. More precisely,

Case??1( 𝜔=𝑝/𝑞is rational). The solutions(𝑥_𝜔(𝑡+2𝑖𝜋),𝑥^′_𝜔(𝑡+2𝑖𝜋)),1≤𝑖≤𝑞−1, are mutually unlinked periodic solution of periodic𝑞𝜋; moreover, in this case, lim_𝑞→∞min_𝑡∈𝑅𝑥_𝜔(𝑡)+𝑥^′_𝜔(𝑡)=+∞. (8)

Case??2 ( 𝜔 is irrational). The solution(𝑥_𝜔(𝑡),𝑥^′_𝜔(𝑡))is either a usual quasi-periodic solution or a generalized one.

2. Proof of Theorem

2.1. Action-Angle Variables and Some Estimates

Observe that (4) is equivalent to the following Hamiltonian system: 𝑥^′=𝜕𝐻𝜕𝑦,𝑦^′=−𝜕𝐻𝜕𝑥 (9) with the Hamiltonian function 𝐻=1𝑥,𝑦,𝑡2𝑦²+𝑉(𝑥)+𝑃(𝑥,𝑡). (10) In order to introduce action and angle variables, we first consider the auxiliary autonomous equation: 𝑥^′=𝑦,𝑦^′=−𝑉^′(𝑥), (11) which is an integrable Hamiltonian system with Hamiltonian function 𝐻₁=1𝑥,𝑦,𝑡2𝑦²+𝑉(𝑥). (12) The closed curves𝐻₁(𝑥,𝑦,𝑡)=ℎ>0are just the integral curves of (11).

Denote by𝑇₀(ℎ)the time period of the integral curveΓ_ℎof (11) defined by𝐻₁(𝑥,𝑦,𝑡)=ℎand by𝐼the area enclosed by the closed curveΓ_ℎfor everyℎ>0.Let−1<−𝛼_ℎ<0<𝛽_ℎbe such that𝑉(−𝛼_ℎ)=𝑉(𝛽_ℎ)=ℎ.It is easy to see that 𝐼₀(ℎ)=2^𝛽ℎ_−𝛼ℎ2((𝑠𝑇ℎ−𝑉))𝑑𝑠,∀ℎ>0,₀(ℎ)=𝐼^′₀(ℎ)=2^𝛽ℎ_𝛼ℎ12(ℎ−𝑉(𝑠))𝑑𝑠,∀ℎ>0. (13) By direct computation we get 𝐼₀(ℎ)=2^𝛽ℎ₀2(ℎ−𝑉(𝑠))𝑑𝑠+2⁰_−𝛼ℎ2(ℎ−𝑉(𝑠))𝑑𝑠=𝜋ℎ+2^𝛼ℎ₀2(ℎ−𝑉(−𝑠))𝑑𝑠, (14) so 𝑇₀(ℎ)=𝜋+^𝛼ℎ₀12(ℎ−𝑉(−𝑠))𝑑𝑠. (15) We then have 𝐼₀(ℎ)=𝐼₋(ℎ)+𝐼₊(ℎ),𝑇₀(ℎ)=𝑇₋(ℎ)+𝑇₊(ℎ), (16) where 𝐼₋(ℎ)=2^−𝛼ℎ₀2(ℎ−𝑉(𝑠))𝑑𝑠,𝐼₊(ℎ)𝑇=𝜋ℎ,₋(ℎ)=2^−𝛼ℎ₀12(ℎ−𝑉(−𝑠))𝑑𝑠,𝑇₊(ℎ)=𝜋. (17)

We now give the estimates on the functions𝐼_and𝑇_.

Lemma 3. One has ℎ^𝑛𝑑^𝑛𝑇_(ℎ)𝑑ℎ^𝑛≤𝐶ℎ^−1/2,ℎ^𝑛𝑑^𝑛𝐼_(ℎ)𝑑ℎ^𝑛≤𝐶ℎ^1/2, (18) where𝑛=0,1,…,6,ℎ→∞. Note that here and below one always uses 𝐶,𝐶₀, or𝐶^′₀to indicate some constants.

Proof. Now we estimate the first inequality. We chose𝑉(𝑠)/ℎ=𝜂as the new variable of integration; then we have 𝑇_(ℎ)=⁰_−𝛼ℎ12((𝑠=ℎ−𝑉))𝑑𝑠₁₀√ℎ𝑉^′𝑠1𝜂,ℎ21−𝜂𝑑𝜂. (19) Since𝑉(𝑠)=(1/(1−𝑠²))−1and𝑉(𝑠)/ℎ=𝜂, we have𝑠=𝜂ℎ/(1+𝜂ℎ).By direct computation, we have 𝑉^′(𝑠)=2𝑠1−𝑠²²=2𝜂ℎ1+𝜂ℎ²1+𝜂ℎ, (20) and then we get 𝑇₋^(𝑛)(ℎ)=(−3/2)!((−3/2)−𝑛)!¹₀𝜂^𝑛2𝜂1−𝜂1+𝜂ℎ^(3/2)+𝑛𝑑𝜂,𝑛=0,1,…,6. (21) When0≤𝜂≤ℎ⁻¹andℎis sufficiently large, there exits𝐶₀such that1−𝜂>𝐶₀, so we have ^ℎ−1₀𝜂^𝑛2𝜂1−𝜂1+𝜂ℎ^(3/2)+𝑛𝑑𝜂≤𝐶^ℎ−1₀𝜂^𝑛≤𝐶2𝜂1−𝜂𝑑𝜂𝐶₀^ℎ−1₀𝜂^𝑛−(1/2)𝑑𝜂≤𝐶ℎ^{−(1/2)−𝑛}. (22) Sinceℎ^−2/3≤𝜂≤1, we have ℎ^1/3<1+ℎ^1/3≤1+𝜂ℎ≤1+ℎ, (23) and then ¹_ℎ−2/3𝜂^𝑛2𝜂1−𝜂1+𝜂ℎ^(3/2)+𝑛𝑑𝜂≤𝐶¹_ℎ−2/3𝜂^𝑛ℎ^𝑛2𝜂1−𝜂ℎ^𝑛1+𝜂ℎ^𝑛1+𝜂ℎ^3/2𝑑𝜂≤𝐶¹_ℎ−2/312𝜂1−𝜂ℎ^𝑛1+𝜂ℎ^3/2𝑑𝜂≤𝐶¹_ℎ−2/312𝜂1−𝜂ℎ^𝑛ℎ^1/2𝑑𝜂≤𝐶ℎ^{(−1/2)−𝑛1}₀12𝜂1−𝜂𝑑𝜂≤𝐶ℎ^{(−1/2)−𝑛}. (24) Observing that there is𝐶₀>0such that1−𝜂≥𝐶₀whenℎ⁻¹≤𝜂≤ℎ^−2/3andℎ→+∞, we have ^ℎ−2/3_ℎ−1𝜂^𝑛2𝜂1−𝜂1+𝜂ℎ^(3/2)+𝑛𝑑𝜂≤𝐶₁ℎ^{(−3/2)−𝑛ℎ−2/3}_ℎ−112𝜂1−𝜂𝜂^3/2≤𝐶𝑑𝜂₁𝐶₀ℎ^{(−3/2)−𝑛ℎ−2/3}_ℎ−11𝜂²𝐶𝑑𝜂=¹𝐶₀ℎ^{(−3/2)−𝑛}1𝜂^ℎ−2/3_ℎ−1=𝐶₁𝐶₀ℎ^{(−3/2)−𝑛}ℎ−ℎ^2/3≤𝐶ℎ^{−(1/2)−𝑛}. (25) By (22)–(25) we have𝑇₋^(𝑛)(ℎ)≤𝐶ℎ^{(−1/2)−𝑛},𝑛=0,1,…,6.
The proof of the second inequality is similar to the first one, so we only give a brief proof.
We choose𝑉(𝑠)/ℎ=𝜂as the new variable of integration, so we have 𝜕𝑠=𝜂𝜕ℎ𝑉^′,𝑠=𝜂ℎ,𝑉1+𝜂ℎ_′(𝑠)=2𝑠1−𝑠²²=2𝜂ℎ1+𝜂ℎ².1+𝜂ℎ (26) By direct computation, we have 𝐼₋(ℎ)=2⁰_−𝛼ℎ2(ℎ−𝑉(𝑠))𝑑𝑠=ℎ¹₀21−𝜂√𝜂1+𝜂ℎ^3/2𝑑𝜂. (27) By (27), we can easily get 𝐼₋^(𝑛)(ℎ)=𝐼^(𝑛)₋₁(ℎ)+𝐼^(𝑛)₋₂(ℎ)()!=𝑛−3/2))!×((−3/2−𝑛+1₁₀21−𝜂√𝜂𝜂^𝑛−11+𝜂ℎ^{(3/2)+𝑛−1}+()!𝑑𝜂−3/2))!ℎ((−3/2−𝑛¹₀21−𝜂√𝜂𝜂^𝑛1+𝜂ℎ^(3/2)+𝑛𝑑𝜂, (28) where𝑛=0,1,…,6.
By a similar way to that used in estimating𝑇₋^(𝑛)(ℎ), we get 𝐼^(𝑛)₋₁(ℎ)≤𝐶ℎ^(1/2)−𝑛,𝐼^(𝑛)₋₂(ℎ)≤𝐶ℎ^(1/2)−𝑛, (29) which means that 𝐼₋^(𝑛)(ℎ)≤𝐶ℎ^(1/2)−𝑛,𝑛=0,1,…,6. (30)
Thus Lemma 3 is proved.

Remark 4. It follows from the definitions of??𝑇₊(ℎ),𝑇₋(ℎ)and Lemma 3 that lim_ℎ→+∞𝑇₋(ℎ)=0,lim_ℎ→+∞𝑇₊(ℎ)=𝜋. (31) Thus the time period𝑇₀(ℎ)is dominated by𝑇₊(ℎ)whenℎis sufficiently large. From the relation between𝑇₋(ℎ)and𝐼₋(ℎ), we know𝐼₀(ℎ)is dominated by𝐼₊(ℎ)whenℎis sufficiently large.

Remark 5. It also follows from the definition of𝐼(ℎ),𝐼₋(ℎ),𝐼₊(ℎ)and Remark 4 that ℎ^𝑛𝑑^𝑛𝐼₀(ℎ)𝑑ℎ^𝑛≤𝐶₀𝐼₀(ℎ),for𝑛≥1. (32)

Remark 6. Note thatℎ=ℎ₀(𝐼₀)is the inverse function of𝐼₀. By Remark 5, we have 𝐼^𝑛𝑑^𝑛ℎ(𝐼)𝑑𝐼^𝑛≤𝐶₀ℎ(𝐼)for𝑛≥1. (33)

We now carry out the standard reduction to the action-angle variables. For this purpose, we define the generating function𝑆(𝑥,𝐼)=∫_𝐶2(ℎ−𝑉(𝑠))𝑑𝑠, where𝐶is the part of the closed curveΓ_ℎconnecting the point on the𝑦-axis and point(𝑥,𝑦).

We define the well-known map(𝜃,𝐼)→(𝑥,𝑦)by 𝑦=𝜕𝑆()𝜕𝑥𝑥,𝐼,𝜃=𝜕𝑆(),𝜕𝐼𝑥,𝐼 (34) which is symplectic since 𝑑𝑥∧𝑑𝑦=𝑑𝑥∧𝑆_𝑥𝑥𝑑𝑥+𝑆_𝑥𝐼𝑑𝐼=𝑆_𝑥𝐼𝑆𝑑𝑥∧𝑑𝐼,𝑑𝜃∧𝑑𝐼=_𝐼𝑥𝑑𝑥+𝑆_𝐼𝐼𝑑𝐼∧𝑑𝐼=𝑆_𝐼𝑥𝑑∧𝑑𝐼. (35)

From the above discussion, we can easily get𝜋𝜃=𝑇₀ℎ×𝑇𝑥,𝑦₋ℎ𝑥,𝑦2𝑥+arcsin2ℎ,𝜋𝑥,𝑦if𝑥>0,𝑦>0,𝑇₀ℎ×𝑇𝑥,𝑦₋ℎ𝑥,𝑦2𝑥+𝜋+arcsin2ℎ,𝜋𝑥,𝑦if𝑥>0,𝑦<0,𝑇₀ℎ×∫𝑥,𝑦_{𝑥−𝛼ℎ}12ℎ𝑥,𝑦+1−1−𝑠²⁻¹,𝜋𝑑𝑠if𝑥<0,𝑦>0,𝑇₀ℎ×𝑇𝑥,𝑦₀ℎ𝑥,𝑦−∫^𝑥_−𝛼ℎ12ℎ𝑥,𝑦+1−1−𝑠²⁻¹,𝐼𝑑𝑠if𝑥<0,𝑦<0,𝑥,𝑦=𝐼₀ℎ𝑥,𝑦=2^𝛽ℎ_−𝛼ℎ2ℎ𝑥,𝑦−𝑉(𝑠)𝑑𝑠. (36)

In the new variables(𝜃,𝐼), the system (9) becomes 𝜃^′=𝜕𝐻𝜕𝐼,𝐼^′=−𝜕𝐻,𝜕𝜃 (37) where 𝐻(𝜃,𝐼,𝑡)=𝜋ℎ₀(𝐼)+𝜋𝑃(𝑥(𝐼,𝜃),𝑡). (38)

In order to estimate𝜋𝑃(𝐼,𝜃), we need the estimate on the function𝑥(𝐼,𝜃).

Lemma 7. For𝐼sufficiently large and−𝛼_ℎ≤𝑥<0, the following estimates hold: 𝐼^𝑛𝜕^𝑛𝑥(𝐼,𝜃)𝜕𝐼^𝑛≤𝑐√𝐼,𝑓𝑜𝑟0≤𝑛≤6. (39)

The Lemma was first proved in [1], and later [5] gives a different proof; [8] using the method of induction hypothesis also gives another proof. So, for concision, we omit the proof.

2.2. New Action and Angle Variables

Now we are concerned with the Hamiltonian system (37) with Hamiltonian function𝐻(𝜃,𝐼,𝑡)given by (38). Note that 𝐼𝑑𝜃−𝐻𝑑𝑡=−(𝐻𝑑𝑡−𝐼𝑑𝜃). (40) This means that if one can solve𝐼from (38) as a function of𝐻(using𝜃and𝑡as parameters), then 𝑑𝐻𝑑𝜃=−𝜕𝐼(),𝜕𝑡𝑡,𝐻,𝜃𝑑𝑡=𝑑𝜃𝜕𝐼()𝜕𝐻𝑡,𝐻,𝜃 (41) is also a Hamiltonian system with Hamiltonian function𝐼and now the action, angle, and time variables are𝐻,𝑡, and𝜃.

From (38) and Lemma 3, we have 𝜕𝐻𝜕𝐼⟶1,as𝐼⟶+∞. (42) So we assume that𝐼can be written as 𝐼=𝐼₀𝐻𝜋+𝑅(𝐻,𝑡,𝜃), (43) where𝑅satisfies|𝑅|<𝐻/𝜋.Recalling thatℎ₀is the inverse function of𝐼₀, we have 𝐻𝜋+𝑅(𝐻,𝑡,𝜃)=ℎ₀(𝐼), (44) which implies that 𝑅(𝐻,𝑡,𝜃)=𝑃(𝑥(𝐼,𝜃),𝑡). (45) As a consequence,𝑅is implicitly defined by 𝑅(𝐻,𝑡,𝜃)=𝑃𝑥𝐼₀𝐻𝜋+𝑅(𝐻,𝑡,𝜃),𝜃,𝑡. (46)

For the estimates of𝑅, we need the following lemmas.

Lemma 8. Let𝑓(𝑥,𝑡) and 𝑔(𝑥,𝑡)be continuously differentiable for(𝑥,𝑡)∈[0,+∞)×𝐼, where𝐼is an interval of𝑅. If(1)𝑔(𝑥,𝑡)→∞as𝑥→+∞, uniformly with respect to𝑡∈𝐼,(2)𝑓^′_𝑥(𝑥,𝑡)/𝑔^′_𝑥(𝑥,𝑡)→ℎ(𝑡)as𝑥→+∞, uniformly with respect to𝑡∈𝐼,then one has 𝑓(𝑥,𝑡)/𝑔(𝑥,𝑡)→ℎ(𝑡)as𝑥→+∞, uniformly with respect to𝑡∈𝐼.

Proof. For any0<𝜖<1, there exits𝑋₁, such that if𝑥>𝑋₁, we have 𝑓^′_𝑥(𝑥,𝑡)𝑔^′_𝑥()(𝑡≤𝜖𝑥,𝑡−ℎ)2,∀𝑡∈𝐼. (47) Let𝑥₀=𝑋₁+1.Then by Lagrangian differential mean value theorem, it follows that, for all𝑥>𝑋₁, we have 𝑓(𝑥,𝑡)−𝑓𝑥₀,𝑡𝑔(𝑥,𝑡)−𝑔𝑥₀(𝑡≤𝜖,𝑡−ℎ)2,∀𝑡∈𝐼. (48)
Moreover, there exists a constant𝑀>0such that 𝑔𝑥₀𝑓()𝑥,𝑡𝑥,𝑡−𝑓₀,𝑡𝑔(𝑥,𝑡)−𝑔𝑥₀𝑥,𝑡−𝑓₀,𝑡≤𝑀,∀𝑡∈𝐼. (49) By condition (A1), there exists𝑋>𝑋₁; we have|𝑔(𝑥,𝑡)|>2𝑀/𝜖.
Thus 𝑓(𝑥,𝑡)𝑔()(𝑡≤𝑓()𝑥𝑥,𝑡−ℎ)𝑥,𝑡−𝑓₀,𝑡𝑔(𝑥,𝑡)−𝑔𝑥₀(𝑡+1,𝑡−ℎ)𝑔()𝑔𝑥𝑥,𝑡⁰𝑓()𝑥,𝑡𝑥,𝑡−𝑓₀,𝑡𝑔(𝑥,𝑡)−𝑔𝑥₀𝑥,𝑡−𝑓₀,𝑡<𝜖. (50)

Lemma 9. Under the assumptions (A1) and (A2), the following results hold:(1)lim_𝑥→+∞(𝑥^𝑚(𝜕^𝑚+𝑛𝑝(𝑥,𝑡)/𝜕𝑥^𝑚𝜕𝑡^𝑛))=𝑝_+,𝑚,𝑛(𝑡)uniformly in𝑡, for0≤𝑚≤6and0≤𝑛≤7, where𝑝_+,𝑚,𝑛(𝑡)=0for1≤𝑚≤6, 0≤𝑛≤7and𝑝_+,0,𝑛(𝑡)=𝑝^(𝑛)(𝑡)for0≤𝑛≤7.(2)lim_𝑥→+∞(𝑥^𝑚(𝜕^𝑚+𝑛/𝜕𝑥^𝑚𝜕𝑡^𝑛)(𝑃(𝑥,𝑡)/|𝑥|^𝜎𝑥))=(1/(𝜎+1))𝑝_+,𝑚,𝑛(𝑡), for0≤𝑚≤6and0≤𝑛≤7, where𝑝_+,7,𝑛(𝑡)=0for0≤𝑛≤7.

Proof. Result (1) is similar to Lemma2.1in [9], so we omit the proof.
For result (2), we first prove that, for1≤𝑚≤6 and 0≤𝑛≤7, lim_𝑥→+∞𝑥^𝑚𝜕^𝑚+𝑛/𝜕𝑥^𝑚𝜕𝑡^𝑛𝑝(𝑡,𝑥)|𝑥|^𝜎=𝜎⋯(𝜎−𝑚+1)𝑝₊^(𝑛)(𝑡). (51) For𝑚=1and0≤𝑛≤7, by result (1), we have that lim_𝑥→+∞𝑥𝜕−𝜎⋅^𝑛/𝜕𝑡^𝑛𝑝(𝑡,𝑥)|𝑥|^𝜎+𝜕⋅𝑥^1+𝑛/𝜕𝑥𝜕𝑡^𝑛𝑝(𝑡,𝑥)|𝑥|^𝜎=0, (52) uniformly in𝑡. It follows that lim_𝑥→+∞𝜕𝑥⋅^1+𝑛/𝜕𝑥𝜕𝑡^𝑛𝑝(𝑡,𝑥)|𝑥|^𝜎=𝜎⋅𝑝₊^(𝑛)(𝑡). (53)
This means that (51) holds for𝑚=1.For𝑚=2, combining this with result (1) and the result of (51) for𝑚=1, we have that (51) still holds. Inductively we can prove that (51) holds for all1≤𝑚≤6.
Obviously, it follows that 𝜕^𝑛𝜕𝑡^𝑛𝑃(𝑡,𝑥)|𝑥|^𝜎−1⋅𝑥⋅1+𝛼𝑝₊^(𝑛)(𝑡)=∫^𝑥₀𝜕^𝑛/𝜕𝑡^𝑛𝑝(𝑡,𝑥)−𝑝₊^(𝑛)(𝑡)|𝑠|^𝜎𝑑𝑠|𝑥|^𝜎.⋅𝑥 (54) Using Lemma 8 and the first result (1) for𝑚=0, it follows that lim_𝑥→+∞𝜕^𝑛𝜕𝑡^𝑛𝑃(𝑡,𝑥)|𝑥|^𝜎=1⋅𝑥𝜎+1𝑝₊^(𝑛)(𝑡), (55) for0≤𝑛≤7. In a similar way to the proof of (51), we have lim_𝑥→+∞𝑥^𝑚𝜕^𝑚+𝑛𝜕𝑥^𝑚𝜕𝑡^𝑛𝑃(𝑥,𝑡)|𝑥|^𝜎𝑥=1𝜎+1𝑝_+,𝑚,𝑛(𝑡),for0≤𝑚≤6,0≤𝑛≤7, (56) where𝑝_+,7,𝑛(𝑡)=0.Thus we proved Lemma 9.

Now we give the estimates of𝑅.By Lemma 9, in a similar way to that for Lemma 2.3 in [5], we have the following lemma.

Lemma 10. The function𝑅(𝐻,𝑡,𝜃)satisfies the following estimates: 𝜕^𝑚+𝑙𝑅(𝐻,𝑡,𝜃)𝜕𝐻^𝑚𝜕𝑡^𝑙≤𝐻^(𝛼+1)/2,for𝑚+𝑙≤6. (57)

Moreover, by the implicit function theorem, there exists a function𝑅₁=𝑅₁(𝑡,𝐻,𝜃)such that 𝑅(𝐻,𝑡,𝜃)=𝑃(𝑥(𝐻,𝜃),𝑡)+𝑅₁(𝐻,𝑡,𝜃). (58) Since 𝑅₁()()(𝑥())𝑥𝐼𝐻,𝑡,𝜃=𝑅𝐻,𝑡,𝜃−𝑃𝐻,𝜃,𝑡=𝑃₀𝐻𝜋()(𝑥())=+𝑅𝐻,𝑡,𝜃,𝜃,𝑡−𝑃𝐻,𝜃,𝑡₁₀𝑝𝑥𝐻+𝑠𝜋𝑅+𝐼₋⋅,𝜃,𝑡𝜕𝑥𝜕𝐼𝐻+𝑠𝜋𝑅+𝐼₋,𝜃⋅𝜋𝑅+𝐼₋𝑑𝑠, (59)

by Lemmas 3 and 10, we have the estimates on𝑅₁(𝐻,𝑡,𝜃).

Lemma 11. Consider the following: 𝜕^𝑘+𝑙𝑅₁(𝐻,𝑡,𝜃)𝜕^𝑘𝐻𝜕^𝑙𝑡<𝐻^𝛼/2𝑓𝑜𝑟𝑘+𝑙≤6. (60)

For the estimate of𝐼((𝐻/𝜋)+𝑅), we need the estimate on𝐼₋((𝐻/𝜋)+𝑅).By Lemma 3 and noticing that|𝑅|<𝐻/𝜋, we have the following lemma.

Lemma 12. Consider the following: 𝜕^𝑘+𝑙𝐼₋((𝐻/𝜋)+𝑅)𝜕^𝑘𝐻𝜕^𝑙𝑡<𝐻^1/2𝑓𝑜𝑟𝑘+𝑙≤6. (61)

Now the new Hamiltonian function𝐼=𝐼(𝑡,𝐻,𝜃)is written in the form 𝐼=𝐼₀𝐻𝜋+𝑅=𝐼₊𝐻𝜋+𝑅+𝐼₋𝐻𝜋()+𝑅=𝐻+𝜋𝑅𝐻,𝑡,𝜃+𝐼₋𝐻𝜋(𝑥())+𝑅=𝐻+𝜋𝑃𝐻,𝜃,𝑡+𝑅₁(𝐻,𝑡,𝜃)+𝐼₋𝐻𝜋+𝑅. (62) The system (41) is of the form 𝑑𝑡=𝑑𝜃𝜕𝐼𝜕𝐻=1+𝜋𝜕𝑥()𝑝(𝑥())+𝜕𝐻𝐻,𝜃𝐻,𝜃,𝑡𝜕𝑅₁()+𝜕𝐻𝐻,𝑡,𝜃𝜕𝐼₋(),𝜕𝐻𝐻,𝑡,𝜃𝑑𝐻𝑑𝜃=−𝜕𝐼𝜕𝑡=−𝜋𝜕𝑃(𝑥())−𝜕𝑡𝜃,𝐻,𝑡𝜕𝑅₁()−𝜕𝑡𝑡,𝐻,𝜃𝜕𝐼₋().𝜕𝑡𝐻,𝑡,𝜃 (63) Introduce a new action variable𝜌∈[1,2]and a parameter𝜖>0by𝐻=𝜖⁻²𝜌. Then,𝐻≫1⇔0<𝜖≪1.Under this transformation, the system (63) is changed into the form 𝑑𝑡=𝑑𝜃𝜕𝐼𝜕𝐻=1+𝜋𝜕𝑥()𝑝(𝑥())+𝜕𝐻𝐻,𝜃𝐻,𝜃,𝑡𝜕𝑅₁()+𝜕𝐻𝐻,𝑡,𝜃𝜕𝐼₋(),𝜕𝐻𝐻,𝑡,𝜃𝑑𝜌𝑑𝜃=−𝜕𝐼𝜕𝑡=−𝜖²𝜋𝜕𝑃(𝑥())+𝜕𝑡𝜃,𝐻,𝑡𝜕𝑅₁()+𝜕𝑡𝐻,𝑡,𝜃𝜕𝐼₋(,𝜕𝑡𝐻,𝑡,𝜃) (64) which is also Hamiltonian system with the new Hamiltonian function Γ𝑡,𝜌,𝜃;𝜖=𝜌+𝜋𝜖⁻²𝑃𝑥𝜃,𝜖⁻²𝜌,𝑡+𝜖⁻²𝑅₁𝜖⁻²𝜌,𝜃,𝑡+𝜖⁻²𝐼₋𝜖⁻²𝜌,𝜃,𝑡. (65) Obviously, if𝜖≪1, the solution(𝑡(𝜃,𝑡₀,𝜌₀),𝜌(𝜃,𝑡₀,𝜌₀))of (64) with the initial date(𝑡₀,𝜌₀)∈𝑅×[1,2]is defined in the interval𝜃∈[0,2𝜋]and𝜌(𝜃,𝑡₀,𝜌₀)∈[(1/2),3]. So the Poincaré map of (64) is well defined in the domain𝑅×[1,2].

Lemma 13 (see [6] Lemma5.1). The Poincaré map of (64) has intersection property.

The proof is similar to the corresponding one in [6].

For convenience we introduce the notation𝑂_𝑘(1)and𝑜_𝑘(1). We say a function𝑓(𝑡,𝜌,𝜃,𝜖)∈𝑂_𝑘(1)if𝑓is smooth in(𝑡,𝜌)and, for𝑘₁+𝑘₂≤𝑘, 𝜕^𝑘1+𝑘2𝜕𝑡^𝑘1𝜕𝜌^𝑘2𝑓𝑡,𝜌,𝜃,𝜖≤𝐶, (66) for some constant𝐶>0which is independent of the arguments𝑡,𝜌,𝜃, and 𝜖.

Similarly, we say𝑓(𝑡,𝜌,𝜃,𝜖)∈𝑜_𝑘(1)if𝑓is smooth in(𝑡,𝜌)and, for𝑘₁+𝑘₂≤𝑘, lim_𝜖→0𝜕^𝑘1+𝑘2𝜕𝑡^𝑘1𝜕𝜌^𝑘2𝑓𝑡,𝜌,𝜃,𝜖=0, (67) uniformly in(𝑡,𝜌,𝜃).

2.3. Poincaré Map and Twist Theorems

We will use Ortega' small twist theorem to prove that the Pioncaré map𝑃has an invariant closed curve, if𝜖is sufficiently small. Let us first recall the theorem in [7].

Lemma 14 (Ortega’s Theorem). Let𝐴=𝕊¹×[𝑎,𝑏]be a finite cylinder with universal cover𝔸=ℝ×[𝑎,𝑏].The coordinate in𝔸is denoted by(𝜏,𝜈).Consider a map 𝑓∶𝐴⟶𝕊×ℝ. (68) One assumes that the map has the intersection property. Suppose that𝑓∶𝐴→ℝ×ℝ,(𝜏₀,𝜈₀)→(𝜏₁,𝜈₁)is a lift of𝑓and it has the form 𝜏₁=𝜏₀+2𝑁𝜋+𝛿𝑙₁𝜏₀,𝜈₀+𝛿𝑔₁𝜏₀,𝜈₀,𝜈₁=𝜈₀+𝛿𝑙₂𝜏₀,𝜈₀+𝛿𝑔₂𝜏₀,𝜈₀, (69) where𝑁is an integer and 𝛿∈(0,1)is a parameter. The functions𝑙₁,??𝑙₂, 𝑔₁, and𝑔₂satisfy 𝑙₁∈𝐶⁶(𝐴),𝑙₁𝜏₀,𝜈₀>0,𝜕𝑙₁𝜕𝜈₀𝜏₀,𝜈₀∀𝜏>0,₀,𝜈₀𝑙∈𝐴,₂(⋅,⋅),𝑔₁(⋅,⋅,𝜖),𝑔₂(⋅,⋅,𝜖)∈𝐶⁵(𝐴). (70) In addition, one assumes that there is a function𝐼∶𝐴→𝑅satisfying 𝐼∈𝐶⁶(𝐴),𝜕𝐼𝜕𝜈₀𝜏₀,𝜈₀>0,∀𝜏₀,𝜈₀𝑙∈𝐴,₁𝜏₀,𝜈₀⋅𝜕𝐼𝜕𝜏_𝑜𝜏₀,𝜈₀+𝑙₂𝜏₀,𝜈₀⋅𝜕𝐼𝜕𝜈₀𝜏₀,𝜈₀∀𝜏=0,₀,𝜈₀∈𝐴. (71) Moreover, suppose that there are two numbers𝑎 and𝑏̃such that𝑎<𝑎<𝑏̃<𝑏and 𝐼_𝑀(𝑎)<𝐼_𝑚(𝑎)≤𝐼_𝑀(𝑎)<𝐼_𝑚𝑏̃≤𝐼_𝑀𝑏̃<𝐼_𝑚(𝑏), (72) where 𝐼_M(𝑟)=max_𝜌∈𝑆1𝐼𝜌_𝑜,𝜏_𝑜,𝐼_𝑚(𝑟)=min_𝜌∈𝑆1𝐼𝜌_𝑜,𝜏_𝑜. (73) Then there exist𝜖>0andΔ>0such that, if𝛿<Δand 𝑔₁(⋅,⋅,𝜖)_𝐶5(𝐴)+𝑔₂(⋅,⋅,𝜖)_𝐶5(𝐴)<𝜖, (74) the mapping𝑓has an invariant curve inΓ_𝐴. The constant𝜖is independent of𝛿.

We make the ansatz that the solution of (64) with the initial condition(𝑡(0),𝜌(0))=(𝑡₀,𝜌₀)is of the form 𝑡=𝑡₀+𝜃+𝜖^1−𝛼Σ₁𝑡₀,𝜌₀,𝜃;𝜖,𝜌=𝜌₀+𝜖^1−𝛼Σ₂𝑡₀,𝜌₀,𝜃;𝜖. (75) Then, the Poincaré map of (64) is 𝑃∶𝑡₁=𝑡₀+2𝜋+𝜖^1−𝛼Σ₁𝑡₀,𝜌₀,𝜌,2𝜋;𝜖₁=𝜌₀+𝜖^1−𝛼Σ₂𝑡₀,𝜌₀,2𝜋;𝜖. (76) The functionsΣ₁andΣ₂satisfy Σ₁=𝜋𝜖^𝛼−1𝜃₀𝜕𝑥𝜖𝜕𝐻⁻²𝜌,𝜃𝑝𝑥𝜖⁻²𝜌,𝜃,𝑡𝑑𝜃+𝜖^𝛼−1𝜃₀𝜕𝑅₁𝜖𝜕𝐻⁻²𝜌,𝑡,𝜃+𝜕𝐼₋𝜖𝜕𝐻⁻²Σ𝜌,𝑡,𝜃𝑑𝜃,₂=−𝜋𝜖^𝛼+1𝜃₀𝜕𝑃𝑥𝜖𝜕𝑡⁻²𝜌,𝜃,𝑡𝑑𝜃−𝜖^𝛼+1𝜃₀𝜕𝑅₁𝜖𝜕𝑡⁻²𝜌,𝑡,𝜃−𝜕𝐼₋𝜖𝜕𝑡⁻²𝜌,𝑡,𝜃𝑑𝜃, (77) where𝑡=𝑡₀+𝜃+𝜖^1−𝛼Σ₁and𝜌=𝜌₀+𝜖^1−𝛼Σ₂.By Lemmas 9, 11, and 12, we know that Σ₁+Σ₂≤𝐶for𝜃∈[0,2𝜋]. (78) Hence, for𝜌₀∈[1,2], we may choose𝜖sufficiently small such that 𝜌₀+𝜖Σ₂≥𝜌₀2≥12. (79) Moreover we can prove that Σ₁,Σ₂∈𝑂₆(1). (80)

In a similar way to that used for estimating𝑅₁, by direct calculation we have the following lemma.

Lemma 15. The following estimates hold: 𝑃𝑥𝜖⁻²𝜌,𝜃,𝑡−𝑃𝑥𝜖⁻²𝜌₀,𝜃,𝑡₀∈𝜖^−𝛼𝑂₆(1),𝜕𝑃𝑥𝜖𝜕𝑡⁻²𝜌,𝜃,𝑡−𝜕𝑃𝑥𝜖𝜕𝑡⁻²𝜌₀,𝜃,𝑡₀∈𝜖^2−𝛼𝑂₆(1). (81)

Now we give an asymptotic expression of Poincaré map of (63); that is, we study the behavior of the functionsΣ₁andΣ₂at𝜃=𝜋as𝜖→0.In order to estimateΣ₁andΣ₂, we need to introduce the following definition and lemma. Let Θ₊(𝐼)=meas𝜃∈[0,𝜋],𝑥𝐻₀,Θ,𝜃>0₋(𝐼)=𝑇₀−Θ₊(𝐼), (82) where𝐻₀=𝜖⁻²𝜌₀.

Lemma 16. Consider the following: Θ₊(𝐼)=𝜋+𝜖𝑂₆(1),Θ₋(𝐼)=𝜖𝑂₆(1). (83)

Proof. This Lemma was proved in [5], so we omit the details.
For estimateΣ₁andΣ₂, we need the estimates of𝑥and𝑥_𝐻.
We recall that, when𝑥<0, we have 𝑥𝐻₀,𝜃=𝑂₆(1),𝑥_𝐻𝐻₀,𝜃=𝜖²𝑂₅(1). (84) When𝑥>0, by the definition of𝜃, we have 𝑥𝐻arcsin⁰,𝜃√=𝑇2ℎ⁰(ℎ)𝜋𝑇𝜃−⁻(ℎ)2=𝜃+𝜖²𝑂₅(1), (85) which yields that 𝑥𝐻₀,𝜃=2𝐻₀𝜋sin𝜃+𝑂₅(1),𝑥_𝐻𝐻₀,𝜃=12𝐻₀𝜋sin𝜃+𝜖²𝑂₅(1). (86)
Now we can give the estimates ofΣ₁andΣ₂.

Lemma 17. The following estimates hold true: Σ₁𝑡₀,𝜌₀=𝜋,2𝜋;𝜖2𝜌₀^(𝛼−1)/2×^𝜋₀(sin𝜃)^1+𝛼𝑝₊𝑡₀+𝜃𝑑𝜃+𝑜₆(1),Σ₂𝑡₀,𝜌₀,2𝜋;𝜖=−𝜋^(1−𝛼)/22𝜌₀^(𝛼+1)/2×^𝜋₀(sin𝜃)^1+𝛼𝑝^′₊𝑡₀+𝜃𝑑𝜃+𝑜₆(1), (87)(88) for𝜖→0.

Proof. Firstly we considerΣ₁.By Lemmas 11, 12, and 15 and (77), we have Σ₁𝑡₀,𝜌₀,2𝜋;𝜖=𝜋𝜖^𝛼−1𝜋₀𝜕𝑥𝜖𝜕𝐻⁻²𝜌,𝜃𝑝𝑥𝜖⁻²𝜌,𝜃,𝑡𝑑𝜃+𝜖^𝛼−1𝜋₀𝜕𝑅₁𝑥𝜖𝜕𝐻⁻²+𝜌,𝜃,𝑡𝜕𝐼₋𝑥𝜖𝜕𝐻⁻²𝜌,𝜃,𝑡𝑑𝜃=𝜋𝜖^𝛼−1𝜋₀𝜕𝑥𝜖𝜕𝐻⁻²𝜌₀,𝜃𝑝𝑥𝜖⁻²𝜌₀,𝜃,𝑡₀+𝜃𝑑𝜃+𝜖^𝛼𝑂₆(1)=𝜋𝜖^𝛼−1_Θ+𝜕𝑥𝜖𝜕𝐻⁻²𝜌₀,𝜃𝑝𝑥𝜖⁻²𝜌₀,𝜃,𝑡₀+𝜃𝑑𝜃+𝜋𝜖^𝛼−1_Θ−𝜕𝑥𝜖𝜕𝐻⁻²𝜌₀,𝜃𝑝𝑥𝜖⁻²𝜌₀,𝜃,𝑡₀+𝜃𝑑𝜃+𝜖^𝛼𝑂₆(1). (89) By result (2) of Lemma 9, as𝜖→0which means𝑥→∞,we have 𝜋𝜖^𝛼−1_Θ+𝜕𝑥𝜖𝜕𝐻⁻²𝜌₀,𝜃𝑝𝑥𝜖⁻²𝜌₀,𝜃,𝑡₀+𝜃𝑑𝜃=𝜋𝜖^𝛼−1_Θ+𝜕𝑥𝜕𝐻𝜃,𝜖⁻²𝜌|𝑥|^𝛼𝑝₊𝑡₀+𝜃𝑑𝜃+𝜖^𝛼𝑂₆(1). (90) By the measure ofΘ₋, we have 𝜋𝜖^𝛼−1_Θ−𝜕𝑥𝜖𝜕𝐻⁻²𝜌₀𝑥𝜖,𝜃×𝑝⁻²𝜌₀,𝜃,𝑡₀+𝜃𝑑𝜃=𝜖^𝛼𝑂₆(1). (91) By (90) and (91), we have Σ₁𝑡₀,𝜌₀,2𝜋;𝜖=𝜋𝜖^𝛼−1_Θ+𝜕𝑥𝜕𝐻𝜃,𝜖⁻²𝜌|𝑥|^𝛼𝑝₊𝑡₀+𝜃𝑑𝜃+𝜖^𝛼𝑂₆(1)=𝜋𝜖^𝛼−1𝜋₀𝜕𝑥𝜕𝐻𝜃,𝜖⁻²𝜌|𝑥|^𝛼𝑝₊𝑡₀+𝜃𝑑𝜃+𝜖^𝛼𝑂₆(1)=𝜋2𝜌₀^{(1−𝛼)/2𝜋}₀(sin𝜃)^𝛼+1𝑝₊𝑡₀+𝜃𝑑𝜃+𝑜₆(1). (92)
Now we considerΣ₂.By Lemmas 11, 12, and 15 and (77), we have Σ₂𝑡₀,𝜌₀,2𝜋;𝜖=−𝜋𝜖^𝛼+1𝜋₀𝜕𝑃𝑥𝜕𝑡𝜃,𝜖⁻²𝜌,𝑡𝑑𝜃−𝜖^𝛼+1𝜋₀𝜕𝑅₁𝑥𝜕𝑡𝜃,𝜖⁻²𝜌+,𝑡𝜕𝐼₋𝑥𝜕𝑡𝜃,𝜖⁻²𝜌,𝑡𝑑𝜃=−𝜋𝜖^𝛼+1𝜋₀𝜕𝑃𝑥𝜕𝑡𝜃,𝜖⁻²𝜌₀,𝑡₀+𝜃𝑑𝜃+𝜖^𝛼𝑂₆(1)=−𝜋𝜖^𝛼+1_Θ+𝜕𝑃𝑥𝜕𝑡𝜃,𝜖⁻²𝜌₀,𝑡₀+𝜃𝑑𝜃−𝜋𝜖^𝛼+1_Θ−𝜕𝑃𝑥𝜕𝑡𝜃,𝜖⁻²𝜌₀,𝑡₀+𝜃𝑑𝜃+𝜖^𝛼𝑂₆(1). (93)
By result (2) of Lemma 9, as𝜖→0, we have −𝜋𝜖^𝛼+1_Θ+𝜕𝑃𝑥𝜕𝑡𝜃,𝜖⁻²𝜌₀,𝑡₀+𝜃𝑑𝜃=−𝜋𝜖^𝛼+1𝛼+1_Θ+𝑥𝜃,𝜖⁻²𝜌₀^𝛼𝑥𝜃,𝜖⁻²𝜌₀𝑝^′₊𝑡₀+𝜃𝑑𝜃+𝜖^𝛼𝑂₆(1). (94)
By the measure ofΘ₋, we have −𝜋𝜖^𝛼+1_Θ−𝜕𝑃𝑥𝜕𝑡𝜃,𝜖⁻²𝜌₀,𝑡₀+𝜃𝑑𝜃=𝜖^𝛼𝑂₆(1). (95)
By (94) and (95), we have Σ₂=−𝜋𝜖^𝛼+1𝛼+1_Θ+𝑥𝜃,𝜖⁻²𝜌₀^𝛼𝑥𝜃,𝜖⁻²𝜌₀𝑝^′₊×𝑡₀+𝜃𝑑𝜃+𝜖^𝛼𝑂₆(1)=−𝜋𝜖^𝛼+1𝛼+1^𝜋₀𝑥𝜃,𝜖⁻²𝜌₀^𝛼𝑥𝜃,𝜖⁻²𝜌₀𝑝^′₊×𝑡₀+𝜃𝑑𝜃+𝜖^𝛼𝑂₆(1)1=−𝜋𝛼+1^(1−𝛼)/22𝜌₀^{(𝛼+1)/2𝜋}₀(sin𝜃)^1+𝛼𝑝^′₊×𝑡₀+𝜃𝑑𝜃+𝑜₆(1). (96)
Thus Lemma 17 is proved.

2.4. Proof of Theorem 1

Let Ψ₁𝑡₀,𝜌₀=𝜋2𝜌₀^(1−𝛼)/2×^𝜋₀(sin𝜃)^1+𝛼𝑝₊𝑡₀Ψ+𝜃𝑑𝜃,₂𝑡₀,𝜌₀1=−𝜋𝛼+1^(1−𝛼)/22𝜌₀^(𝛼+1)/2×^𝜋₀(sin𝜃)^1+𝛼𝑝^′₊𝑡₀+𝜃𝑑𝜃. (97) Then there are two functions 𝜙₁ and 𝜙₂ such that the Poincaré map of (64), given by (76) of the form 𝑃∶𝑡₁=𝑡₀+2𝜋+𝜖^1−𝛼Ψ₁𝑡₀,𝜌₀+𝜖^1−𝛼𝜙₁,𝜌₁=𝜌₀+𝜖^1−𝛼Ψ₂𝑡₀,𝜌₀+𝜖^1−𝛼𝜙₂, (98) where𝜙₁,𝜙₂∈𝑂₆(1).

Since∫^𝜋₀𝑝₊(𝑡₀+𝜃)sin𝜃𝑑𝜃>0,∀𝑡₀∈𝑅,we have Ψ₁>0,𝜕Ψ₁𝜕𝜌₀≠0. (99) Let 𝜌𝐿=^{0−(1+𝛼)/2}∫^𝜋₀(sin𝜃)^1+𝛼𝑝₊𝑡₀.+𝜃𝑑𝜃 (100) Then 𝜕𝐿𝜕𝑡₀Ψ₁𝑡₀,𝜌₀+𝜕𝐿𝜕𝜌₀Ψ₂𝑡₀,𝜌₀=0. (101)

The other assumptions of Ortega’s theorem are easily verified. Hence, there is an invariant curve of𝑃in the annulus(𝑡₀,𝜌₀)∈𝑆¹×[1.2]which implies the boundedness of our original equation (4). Then Theorem 1 is proved.

2.5. Proof of Theorem 2

We apply Aubry-Mather theory. By Theorem B in [10] and the monotone twist property of the Pioncarémap𝑃guaranteed by𝜕Ψ₁/𝜕𝜌₀<0, it is straightforward to check that Theorem 2 is correct.