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

Authors: Shunjun Jiang; Yan Ding
Journal: Abstract and Applied Analysis
 (2013)

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## 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.


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## Body

## 1. Introduction and Main Result

In 1991, Levi [1] considered the following equation:
(1)x′′+V′(x,t)=0,
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:
(2)x′′+V′(x)=F(x,t),
wherF(x,t)=p(t)is aπ-periodic function andV(x)=(1/2)x+2+(1/(1-x-2)ν)-1, wherex+=max{x,0},  x-=max{-x,0}, andνis a positive integer. Under the Lazer-Leach assumption that
(3)1+12∫0π‍p(t0+θ)sinθdθ>0,∀t0∈R,
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:
(4)x′′+V′(x)+p(x,t)=0,
where
(5)V=12x+2+11-x-2-1,x>-1.In order to state our main results, we give some notation and assumptions. Letα∈(0,1)be some fixed constant. Let
(6)p¯=p(x,t)|x|α,P(x,t)=∫0x‍p(s,t)ds.(A1)
Assumep(x,t)∈C7,6(S1×R) andlimx→+∞p¯(x,t)=p¯+(t)uniformly int.(A2)
limx→+∞xm(∂m+np¯(x,t)/∂xm∂tn)=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:(7)∫0π‍p¯+(t0+θ)(sinθ)1+αdθ>0,∀t0∈R.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 hassupt∈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(xω(t),xω′(t))of Mather type with rotation numberω. More precisely,Case  1(ω=p/qis rational). The solutions(xω(t+2iπ),xω′(t+2iπ)),1≤i≤q-1, are mutually unlinked periodic solution of periodicqπ; moreover, in this case,
(8)limq→∞mint∈R|xω(t)|+|xω′(t)|=+∞.Case  2(ωis irrational). The solution(xω(t),xω′(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:
(9)x′=∂H∂y,y′=-∂H∂x
with the Hamiltonian function
(10)H(x,y,t)=12y2+V(x)+P(x,t).
In order to introduce action and angle variables, we first consider the auxiliary autonomous equation:
(11)x′=y,y′=-V′(x),
which is an integrable Hamiltonian system with Hamiltonian function
(12)H1(x,y,t)=12y2+V(x).
The closed curvesH1(x,y,t)=h>0are just the integral curves of (11).Denote byT0(h)the time period of the integral curveΓhof (11) defined byH1(x,y,t)=hand byIthe area enclosed by the closed curveΓhfor everyh>0.Let-1<-αh<0<βhbe such thatV(-αh)=V(βh)=h.It is easy to see that
(13)I0(h)=2∫-αhβh‍2(h-V(s))ds,∀h>0,T0(h)=I0′(h)=2∫αhβh‍12(h-V(s))ds,∀h>0.
By direct computation we get
(14)I0(h)=2∫0βh‍2(h-V(s))ds+2∫-αh0‍2(h-V(s))ds=πh+2∫0αh‍2(h-V(-s))ds,
so
(15)T0(h)=π+∫0αh‍12(h-V(-s))ds.
We then have
(16)I0(h)=I-(h)+I+(h),T0(h)=T-(h)+T+(h),
where
(17)I-(h)=2∫0-αh‍2(h-V(s))ds,I+(h)=πh,T-(h)=2∫0-αh‍12(h-V(-s))ds,T+(h)=π.We now give the estimates on the functionsI_andT_.Lemma 3.
One has(18)hn|dnT_(h)dhn|≤Ch-1/2,hn|dnI_(h)dhn|≤Ch1/2,
wheren=0,1,…,6,h→∞. Note that here and below one always uses C,C0, orC0′to indicate some constants.Proof.
Now we estimate the first inequality. We choseV(s)/h=ηas the new variable of integration; then we have
(19)T_(h)=∫-αh012(h-V(s))ds=∫01hV′(s(η,h))12(1-η)dη.
SinceV(s)=(1/(1-s2))-1andV(s)/h=η, we haves=ηh/(1+ηh).By direct computation, we have
(20)V′(s)=2s(1-s2)2=2ηh(1+ηh)21+ηh,
and then we get
(21)T-(n)(h)=(-3/2)!((-3/2)-n)!∫01‍ηn2η(1-η)(1+ηh)(3/2)+ndη,n=0,1,…,6.
When0≤η≤h-1andhis sufficiently large, there exitsC0such that1-η>C0, so we have
(22)∫0h-1‍ηn2η(1-η)(1+ηh)(3/2)+ndη≤C∫0h-1‍ηn2η(1-η)dη≤CC0∫0h-1‍ηn-(1/2)dη≤Ch-(1/2)-n.
Sinceh-2/3≤η≤1, we have
(23)h1/3<1+h1/3≤1+ηh≤1+h,
and then
(24)∫h-2/31‍ηn2η(1-η)(1+ηh)(3/2)+ndη≤C∫h-2/31‍ηnhn2η(1-η)hn(1+ηh)n(1+ηh)3/2dη≤C∫h-2/31‍12η(1-η)hn(1+ηh)3/2dη≤C∫h-2/31‍12η(1-η)hnh1/2dη≤Ch(-1/2)-n∫01‍12η(1-η)dη≤Ch(-1/2)-n.
Observing that there isC0>0such that1-η≥C0whenh-1≤η≤h-2/3andh→+∞, we have
(25)∫h-1h-2/3‍ηn2η(1-η)(1+ηh)(3/2)+ndη≤C1h(-3/2)-n∫h-1h-2/3‍12η(1-η)η3/2dη≤C1C0h(-3/2)-n∫h-1h-2/3‍1η2dη=C1C0h(-3/2)-n1η|h-1h-2/3=C1C0h(-3/2)-n(h-h2/3)≤Ch-(1/2)-n.
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
(26)∂s∂h=ηV′,s=ηh1+ηh,V′(s)=2s(1-s2)2=2ηh(1+ηh)21+ηh.
By direct computation, we have
(27)I-(h)=2∫-αh0‍2(h-V(s))ds=h∫01‍2(1-η)η(1+ηh)3/2dη.
By (27), we can easily get
(28)I-(n)(h)=I-1(n)(h)+I-2(n)(h)=n(-3/2)!((-3/2)-n+1)!×∫01‍2(1-η)ηηn-1(1+ηh)(3/2)+n-1dη+(-3/2)!((-3/2)-n)!h∫01‍2(1-η)ηηn(1+ηh)(3/2)+ndη,
wheren=0,1,…,6.
By a similar way to that used in estimatingT-(n)(h), we get
(29)I-1(n)(h)≤Ch(1/2)-n,I-2(n)(h)≤Ch(1/2)-n,
which means that
(30)I-(n)(h)≤Ch(1/2)-n,n=0,1,…,6.
Thus Lemma3 is proved.Remark 4.
It follows from the definitions ofT+(h),T-(h)and Lemma 3 that
(31)limh→+∞T-(h)=0,limh→+∞T+(h)=π.
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
(32)|hndnI0(h)dhn|≤C0I0(h),forn≥1.Remark 6.
Note thath=h0(I0)is the inverse function ofI0. By Remark 5, we have
(33)|Indnh(I)dIn|≤C0h(I)forn≥1.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 curveΓhconnecting the point on they-axis and point(x,y).We define the well-known map(θ,I)→(x,y)by
(34)y=∂S∂x(x,I),θ=∂S∂I(x,I),
which is symplectic since
(35)dx∧dy=dx∧(Sxxdx+SxIdI)=SxIdx∧dI,dθ∧dI=(SIxdx+SIIdI)∧dI=SIxd∧dI.From the above discussion, we can easily get(36)θ={πT0(h(x,y))×(T-(h(x,y))2+arcsinx2(h(x,y))),ifx>0,y>0,πT0(h(x,y))×(T-(h(x,y))2+π+arcsinx2(h(x,y))),iifx>0,y<0,πT0(h(x,y))×(∫-αhx12(h(x,y)+1-(1-s2)-1)ds),ifx<0,y>0,πT0(h(x,y))×(T0(h(x,y))-∫-αhx12(h(x,y)+1-(1-s2)-1)ds),ifx<0,y<0,I(x,y)=I0(h(x,y))=2∫-αhβh‍2(h(x,y)-V(s))ds.In the new variables(θ,I), the system (9) becomes
(37)θ′=∂H∂I,I′=-∂H∂θ,
where
(38)H(θ,I,t)=πh0(I)+πP(x(I,θ),t).In order to estimateπP(I,θ), we need the estimate on the functionx(I,θ).Lemma 7.
ForIsufficiently large and-αh≤x<0, the following estimates hold:
(39)|In∂nx(I,θ)∂In|≤cI,for0≤n≤6.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
(40)Idθ-Hdt=-(Hdt-Idθ).
This means that if one can solveIfrom (38) as a function ofH(usingθandtas parameters), then
(41)dHdθ=-∂I∂t(t,H,θ),dtdθ=∂I∂H(t,H,θ)
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
(42)∂H∂I⟶1,asI⟶+∞.
So we assume thatIcan be written as
(43)I=I0(Hπ+R(H,t,θ)),
whereRsatisfies|R|<H/π.Recalling thath0is the inverse function ofI0, we have
(44)Hπ+R(H,t,θ)=h0(I),
which implies that
(45)R(H,t,θ)=P(x(I,θ),t).
As a consequence,Ris implicitly defined by
(46)R(H,t,θ)=P(x(I0(Hπ+R(H,t,θ)),θ),t).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)
fx′(x,t)/gx′(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
(47)|fx′(x,t)gx′(x,t)-h(t)|≤ϵ2,∀t∈I.
Letx0=X1+1.Then by Lagrangian differential mean value theorem, it follows that, for allx>X1, we have
(48)|f(x,t)-f(x0,t)g(x,t)-g(x0,t)-h(t)|≤ϵ2,∀t∈I.
Moreover, there exists a constantM>0such that
(49)|g(x0,t)f(x,t)-f(x0,t)g(x,t)-g(x0,t)-f(x0,t)|≤M,∀t∈I.
By condition (A1), there existsX>X1; we have|g(x,t)|>2M/ϵ.
Thus(50)|f(x,t)g(x,t)-h(t)|≤|f(x,t)-f(x0,t)g(x,t)-g(x0,t)-h(t)|+|1g(x,t)(g(x0,t)f(x,t)-f(x0,t)g(x,t)-g(x0,t)-f(x0,t))|<ϵ.Lemma 9.
Under the assumptions (A1) and (A2), the following results hold:(1)
limx→+∞(xm(∂m+np¯(x,t)/∂xm∂tn))=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)
limx→+∞(xm(∂m+n/∂xm∂tn)(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,
(51)limx→+∞xm(∂m+n/∂xm∂tn)p(t,x)|x|σ=σ⋯(σ-m+1)p¯+(n)(t).
Form=1and0≤n≤7, by result (1), we have that
(52)limx→+∞x{-σ·(∂n/∂tn)p(t,x)|x|σ·x+(∂1+n/∂x∂tn)p(t,x)|x|σ}=0,
uniformly int. It follows that
(53)limx→+∞x·(∂1+n/∂x∂tn)p(t,x)|x|σ=σ·p¯+(n)(t).
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(54)∂n∂tnP(t,x)|x|σ·x-11+α·p¯+(n)(t)=∫0x‍[(∂n/∂tn)p(t,x)-p¯+(n)(t)|s|σ]ds|x|σ·x.
Using Lemma 8 and the first result (1) form=0, it follows that
(55)limx→+∞∂n∂tnP(t,x)|x|σ·x=1σ+1p¯+(n)(t),
for0≤n≤7. In a similar way to the proof of (51), we have
(56)limx→+∞xm∂m+n∂xm∂tnP(x,t)|x|σx=1σ+1p¯+,m,n(t),for0≤m≤6,0≤n≤7,
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:
(57)|∂m+lR(H,t,θ)∂Hm∂tl|≤H(α+1)/2,form+l≤6.Moreover, by the implicit function theorem, there exists a functionR1=R1(t,H,θ)such that
(58)R(H,t,θ)=P(x(H,θ),t)+R1(H,t,θ).
Since
(59)R1(H,t,θ)=R(H,t,θ)-P(x(H,θ),t)=P{x[I0(Hπ+R(H,t,θ)),θ],t}-P(x(H,θ),t)=∫01‍p{x[H+s(πR+I-),θ],t}·∂x∂I(H+s(πR+I-),θ)·(πR+I-)ds,by Lemmas3 and 10, we have the estimates onR1(H,t,θ).Lemma 11.
Consider the following:(60)|∂k+lR1(H,t,θ)∂kH∂lt|<Hα/2fork+l≤6.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:(61)|∂k+lI-((H/π)+R)∂kH∂lt|<H1/2fork+l≤6.Now the new Hamiltonian functionI=I(t,H,θ)is written in the form
(62)I=I0(Hπ+R)=I+(Hπ+R)+I-(Hπ+R)=H+πR(H,t,θ)+I-(Hπ+R)=H+πP(x(H,θ),t)+R1(H,t,θ)+I-(Hπ+R).
The system (41) is of the form
(63)dtdθ=∂I∂H=1+π∂x∂H(H,θ)p(x(H,θ),t)+∂R1∂H(H,t,θ)+∂I-∂H(H,t,θ),dHdθ=-∂I∂t=-π∂P∂t(x(θ,H),t)-∂R1∂t(t,H,θ)-∂I-∂t(H,t,θ).
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
(64)dtdθ=∂I∂H=1+π∂x∂H(H,θ)p(x(H,θ),t)+∂R1∂H(H,t,θ)+∂I-∂H(H,t,θ),dρdθ=-∂I∂t=-ϵ2[π∂P∂t(x(θ,H),t)+∂R1∂t(H,t,θ)+∂I-∂t(H,t,θ)],
which is also Hamiltonian system with the new Hamiltonian function
(65)Γ(t,ρ,θ;ϵ)=ρ+πϵ-2P(x(θ,ϵ-2ρ),t)+ϵ-2R1(ϵ-2ρ,θ,t)+ϵ-2I-(ϵ-2ρ,θ,t).
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 notationOk(1)andok(1). We say a functionf(t,ρ,θ,ϵ)∈Ok(1)iffis smooth in(t,ρ)and, fork1+k2≤k,
(66)|∂k1+k2∂tk1∂ρk2f(t,ρ,θ,ϵ)|≤C,
for some constantC>0which is independent of the argumentst,ρ,θ, and ϵ.Similarly, we sayf(t,ρ,θ,ϵ)∈ok(1)iffis smooth in(t,ρ)and, fork1+k2≤k,
(67)limϵ→0|∂k1+k2∂tk1∂ρk2f(t,ρ,θ,ϵ)|=0,
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
(68)f¯:A⟶𝕊×ℝ.
One assumes that the map has the intersection property. Suppose thatf:A→ℝ×ℝ,(τ0,ν0)→(τ1,ν1)is a lift off¯and it has the form
(69)τ1=τ0+2Nπ+δl1(τ0,ν0)+δg~1(τ0,ν0),ν1=ν0+δl2(τ0,ν0)+δg~2(τ0,ν0),
whereNis an integer and δ∈(0,1)is a parameter. The functionsl1,  l2, g~1, andg~2satisfy
(70)l1∈C6(A),l1(τ0,ν0)>0,∂l1∂ν0(τ0,ν0)>0,∀(τ0,ν0)∈A,l2(·,·),g~1(·,·,ϵ),g~2(·,·,ϵ)∈C5(A).
In addition, one assumes that there is a functionI:A→Rsatisfying
(71)I∈C6(A),∂I∂ν0(τ0,ν0)>0,∀(τ0,ν0)∈A,l1(τ0,ν0)·∂I∂τo(τ0,ν0)+l2(τ0,ν0)·∂I∂ν0(τ0,ν0)=0,∀(τ0,ν0)∈A.
Moreover, suppose that there are two numbersa~ andb~such thata<a~<b~<band
(72)IM(a)<Im(a~)≤IM(a~)<Im(b~)≤IM(b~)<Im(b),
where
(73)IM(r)=maxρ∈S1I(ρo,τo),Im(r)=minρ∈S1I(ρo,τo).
Then there existϵ>0andΔ>0such that, ifδ<Δand
(74)∥g~1(·,·,ϵ)∥C5(A)+∥g~2(·,·,ϵ)∥C5(A)<ϵ,
the mappingf¯has an invariant curve inΓA. 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
(75)t=t0+θ+ϵ1-αΣ1(t0,ρ0,θ;ϵ),ρ=ρ0+ϵ1-αΣ2(t0,ρ0,θ;ϵ).
Then, the Poincaré map of (64) is
(76)P:t1=t0+2π+ϵ1-αΣ1(t0,ρ0,2π;ϵ),ρ1=ρ0+ϵ1-αΣ2(t0,ρ0,2π;ϵ).
The functionsΣ1andΣ2satisfy
(77)Σ1=πϵα-1∫0θ‍∂x∂H(ϵ-2ρ,θ)p(x(ϵ-2ρ,θ),t)dθ+ϵα-1∫0θ‍(∂R1∂H(ϵ-2ρ,t,θ)+∂I-∂H(ϵ-2ρ,t,θ))dθ,Σ2=-πϵα+1∫0θ‍∂P∂t(x(ϵ-2ρ,θ),t)dθ-ϵα+1∫0θ‍(∂R1∂t(ϵ-2ρ,t,θ)-∂I-∂t(ϵ-2ρ,t,θ))dθ,
wheret=t0+θ+ϵ1-αΣ1andρ=ρ0+ϵ1-αΣ2.By Lemmas 9, 11, and 12, we know that
(78)|Σ1|+|Σ2|≤Cforθ∈[0,2π].
Hence, forρ0∈[1,2], we may chooseϵsufficiently small such that
(79)ρ0+ϵΣ2≥ρ02≥12.
Moreover we can prove that
(80)Σ1,Σ2∈O6(1).In a similar way to that used for estimatingR1, by direct calculation we have the following lemma.Lemma 15.
The following estimates hold:(81)P(x(ϵ-2ρ,θ),t)-P(x(ϵ-2ρ0,θ),t0)∈ϵ-αO6(1),∂P∂t(x(ϵ-2ρ,θ),t)-∂P∂t(x(ϵ-2ρ0,θ),t0)∈ϵ2-αO6(1).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
(82)Θ+(I)=meas{θ∈[0,π],x(H0,θ)>0},Θ-(I)=T0-Θ+(I),
whereH0=ϵ-2ρ0.Lemma 16.
Consider the following:(83)Θ+(I)=π+ϵO6(1),Θ-(I)=ϵO6(1).Proof.
This Lemma was proved in [5], so we omit the details.
For estimateΣ1andΣ2, we need the estimates ofxandxH.
We recall that, whenx<0, we have
(84)|x(H0,θ)|=O6(1),|xH(H0,θ)|=ϵ2O5(1).
Whenx>0, by the definition ofθ, we have
(85)arcsinx(H0,θ)2h=T0(h)πθ-T-(h)2=θ+ϵ2O5(1),
which yields that
(86)x(H0,θ)=2H0πsinθ+O5(1),xH(H0,θ)=12H0πsinθ+ϵ2O5(1).
Now we can give the estimates ofΣ1andΣ2.Lemma 17.
The following estimates hold true:(87)Σ1(t0,ρ0,2π;ϵ)=(π2ρ0)(α-1)/2×∫0π(sinθ)1+αp¯+(t0+θ)dθ+o6(1),(88)Σ2(t0,ρ0,2π;ϵ)=-π(1-α)/2(2ρ0)(α+1)/2×∫0π(sinθ)1+αp¯+′(t0+θ)dθ+o6(1),
forϵ→0.Proof.
Firstly we considerΣ1.By Lemmas 11, 12, and 15 and (77), we have
(89)Σ1(t0,ρ0,2π;ϵ)=πϵα-1∫0π‍∂x∂H(ϵ-2ρ,θ)p(x(ϵ-2ρ,θ),t)dθ+ϵα-1∫0π‍∂R1∂H(x(ϵ-2ρ,θ),t)+∂I-∂H(x(ϵ-2ρ,θ),t)dθ=πϵα-1∫0π‍∂x∂H(ϵ-2ρ0,θ)p(x(ϵ-2ρ0,θ),t0+θ)dθ+ϵαO6(1)=πϵα-1∫Θ+‍∂x∂H(ϵ-2ρ0,θ)p(x(ϵ-2ρ0,θ),t0+θ)dθ+πϵα-1∫Θ-‍∂x∂H(ϵ-2ρ0,θ)p(x(ϵ-2ρ0,θ),t0+θ)dθ+ϵαO6(1).
By result (2) of Lemma 9, asϵ→0which meansx→∞,we have
(90)πϵα-1∫Θ+‍∂x∂H(ϵ-2ρ0,θ)p(x(ϵ-2ρ0,θ),t0+θ)dθ=πϵα-1∫Θ+‍∂x∂H(θ,ϵ-2ρ)|x|αp¯+(t0+θ)dθ+ϵαO6(1).
By the measure ofΘ-, we have
(91)πϵα-1∫Θ-‍∂x∂H(ϵ-2ρ0,θ)×p(x(ϵ-2ρ0,θ),t0+θ)dθ=ϵαO6(1).
By (90) and (91), we have
(92)Σ1(t0,ρ0,2π;ϵ)=πϵα-1∫Θ+‍∂x∂H(θ,ϵ-2ρ)|x|αp¯+(t0+θ)dθ+ϵαO6(1)=πϵα-1∫0π‍∂x∂H(θ,ϵ-2ρ)|x|αp¯+(t0+θ)dθ+ϵαO6(1)=(π2ρ0)(1-α)/2∫0π‍(sinθ)α+1p¯+(t0+θ)dθ+o6(1).
Now we considerΣ2.By Lemmas 11, 12, and 15 and (77), we have
(93)Σ2(t0,ρ0,2π;ϵ)=-πϵα+1∫0π‍∂P∂t(x(θ,ϵ-2ρ),t)dθ-ϵα+1∫0π‍[∂R1∂t(x(θ,ϵ-2ρ),t)+∂I-∂t(x(θ,ϵ-2ρ),t)]dθ=-πϵα+1∫0π‍∂P∂t(x(θ,ϵ-2ρ0),t0+θ)dθ+ϵαO6(1)=-πϵα+1∫Θ+‍∂P∂t(x(θ,ϵ-2ρ0),t0+θ)dθ-πϵα+1∫Θ-‍∂P∂t(x(θ,ϵ-2ρ0),t0+θ)dθ+ϵαO6(1).
By result (2) of Lemma9, asϵ→0, we have
(94)-πϵα+1∫Θ+‍∂P∂t(x(θ,ϵ-2ρ0),t0+θ)dθ=-πϵα+1α+1∫Θ+‍|x(θ,ϵ-2ρ0)|αx(θ,ϵ-2ρ0)p¯+′(t0+θ)dθ+ϵαO6(1).
By the measure ofΘ-, we have
(95)-πϵα+1∫Θ-‍∂P∂t(x(θ,ϵ-2ρ0),t0+θ)dθ=ϵαO6(1).
By (94) and (95), we have
(96)Σ2=-πϵα+1α+1∫Θ+‍|x(θ,ϵ-2ρ0)|αx(θ,ϵ-2ρ0)p¯+′×(t0+θ)dθ+ϵαO6(1)=-πϵα+1α+1∫0π‍|x(θ,ϵ-2ρ0)|αx(θ,ϵ-2ρ0)p¯+′×(t0+θ)dθ+ϵαO6(1)=-1α+1π(1-α)/2(2ρ0)(α+1)/2∫0π‍(sinθ)1+αp¯+′×(t0+θ)dθ+o6(1).
Thus Lemma17 is proved.


### 2.4. Proof of Theorem1

Let(97)Ψ1(t0,ρ0)=(π2ρ0)(1-α)/2×∫0π‍(sinθ)1+αp¯+(t0+θ)dθ,Ψ2(t0,ρ0)=-1α+1π(1-α)/2(2ρ0)(α+1)/2×∫0π(sinθ)1+αp¯+′(t0+θ)dθ.
Then there are two functions ϕ1 and ϕ2 such that the Poincaré map of (64), given by (76) of the form
(98)P:t1=t0+2π+ϵ1-αΨ1(t0,ρ0)+ϵ1-αϕ1,ρ1=ρ0+ϵ1-αΨ2(t0,ρ0)+ϵ1-αϕ2,
whereϕ1,ϕ2∈O6(1).Since∫0π‍p¯+(t0+θ)sinθdθ>0,∀t0∈R,we have
(99)Ψ1>0,∂Ψ1∂ρ0≠0.
Let
(100)L=ρ0-(1+α)/2∫0π‍(sinθ)1+αp¯+(t0+θ)dθ.
Then
(101)∂L∂t0Ψ1(t0,ρ0)+∂L∂ρ0Ψ2(t0,ρ0)=0.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 Theorem2

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.


## 2.1. Action-Angle Variables and Some Estimates

Observe that (4) is equivalent to the following Hamiltonian system:
(9)x′=∂H∂y,y′=-∂H∂x
with the Hamiltonian function
(10)H(x,y,t)=12y2+V(x)+P(x,t).
In order to introduce action and angle variables, we first consider the auxiliary autonomous equation:
(11)x′=y,y′=-V′(x),
which is an integrable Hamiltonian system with Hamiltonian function
(12)H1(x,y,t)=12y2+V(x).
The closed curvesH1(x,y,t)=h>0are just the integral curves of (11).Denote byT0(h)the time period of the integral curveΓhof (11) defined byH1(x,y,t)=hand byIthe area enclosed by the closed curveΓhfor everyh>0.Let-1<-αh<0<βhbe such thatV(-αh)=V(βh)=h.It is easy to see that
(13)I0(h)=2∫-αhβh‍2(h-V(s))ds,∀h>0,T0(h)=I0′(h)=2∫αhβh‍12(h-V(s))ds,∀h>0.
By direct computation we get
(14)I0(h)=2∫0βh‍2(h-V(s))ds+2∫-αh0‍2(h-V(s))ds=πh+2∫0αh‍2(h-V(-s))ds,
so
(15)T0(h)=π+∫0αh‍12(h-V(-s))ds.
We then have
(16)I0(h)=I-(h)+I+(h),T0(h)=T-(h)+T+(h),
where
(17)I-(h)=2∫0-αh‍2(h-V(s))ds,I+(h)=πh,T-(h)=2∫0-αh‍12(h-V(-s))ds,T+(h)=π.We now give the estimates on the functionsI_andT_.Lemma 3.
One has(18)hn|dnT_(h)dhn|≤Ch-1/2,hn|dnI_(h)dhn|≤Ch1/2,
wheren=0,1,…,6,h→∞. Note that here and below one always uses C,C0, orC0′to indicate some constants.Proof.
Now we estimate the first inequality. We choseV(s)/h=ηas the new variable of integration; then we have
(19)T_(h)=∫-αh012(h-V(s))ds=∫01hV′(s(η,h))12(1-η)dη.
SinceV(s)=(1/(1-s2))-1andV(s)/h=η, we haves=ηh/(1+ηh).By direct computation, we have
(20)V′(s)=2s(1-s2)2=2ηh(1+ηh)21+ηh,
and then we get
(21)T-(n)(h)=(-3/2)!((-3/2)-n)!∫01‍ηn2η(1-η)(1+ηh)(3/2)+ndη,n=0,1,…,6.
When0≤η≤h-1andhis sufficiently large, there exitsC0such that1-η>C0, so we have
(22)∫0h-1‍ηn2η(1-η)(1+ηh)(3/2)+ndη≤C∫0h-1‍ηn2η(1-η)dη≤CC0∫0h-1‍ηn-(1/2)dη≤Ch-(1/2)-n.
Sinceh-2/3≤η≤1, we have
(23)h1/3<1+h1/3≤1+ηh≤1+h,
and then
(24)∫h-2/31‍ηn2η(1-η)(1+ηh)(3/2)+ndη≤C∫h-2/31‍ηnhn2η(1-η)hn(1+ηh)n(1+ηh)3/2dη≤C∫h-2/31‍12η(1-η)hn(1+ηh)3/2dη≤C∫h-2/31‍12η(1-η)hnh1/2dη≤Ch(-1/2)-n∫01‍12η(1-η)dη≤Ch(-1/2)-n.
Observing that there isC0>0such that1-η≥C0whenh-1≤η≤h-2/3andh→+∞, we have
(25)∫h-1h-2/3‍ηn2η(1-η)(1+ηh)(3/2)+ndη≤C1h(-3/2)-n∫h-1h-2/3‍12η(1-η)η3/2dη≤C1C0h(-3/2)-n∫h-1h-2/3‍1η2dη=C1C0h(-3/2)-n1η|h-1h-2/3=C1C0h(-3/2)-n(h-h2/3)≤Ch-(1/2)-n.
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
(26)∂s∂h=ηV′,s=ηh1+ηh,V′(s)=2s(1-s2)2=2ηh(1+ηh)21+ηh.
By direct computation, we have
(27)I-(h)=2∫-αh0‍2(h-V(s))ds=h∫01‍2(1-η)η(1+ηh)3/2dη.
By (27), we can easily get
(28)I-(n)(h)=I-1(n)(h)+I-2(n)(h)=n(-3/2)!((-3/2)-n+1)!×∫01‍2(1-η)ηηn-1(1+ηh)(3/2)+n-1dη+(-3/2)!((-3/2)-n)!h∫01‍2(1-η)ηηn(1+ηh)(3/2)+ndη,
wheren=0,1,…,6.
By a similar way to that used in estimatingT-(n)(h), we get
(29)I-1(n)(h)≤Ch(1/2)-n,I-2(n)(h)≤Ch(1/2)-n,
which means that
(30)I-(n)(h)≤Ch(1/2)-n,n=0,1,…,6.
Thus Lemma3 is proved.Remark 4.
It follows from the definitions ofT+(h),T-(h)and Lemma 3 that
(31)limh→+∞T-(h)=0,limh→+∞T+(h)=π.
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
(32)|hndnI0(h)dhn|≤C0I0(h),forn≥1.Remark 6.
Note thath=h0(I0)is the inverse function ofI0. By Remark 5, we have
(33)|Indnh(I)dIn|≤C0h(I)forn≥1.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 curveΓhconnecting the point on they-axis and point(x,y).We define the well-known map(θ,I)→(x,y)by
(34)y=∂S∂x(x,I),θ=∂S∂I(x,I),
which is symplectic since
(35)dx∧dy=dx∧(Sxxdx+SxIdI)=SxIdx∧dI,dθ∧dI=(SIxdx+SIIdI)∧dI=SIxd∧dI.From the above discussion, we can easily get(36)θ={πT0(h(x,y))×(T-(h(x,y))2+arcsinx2(h(x,y))),ifx>0,y>0,πT0(h(x,y))×(T-(h(x,y))2+π+arcsinx2(h(x,y))),iifx>0,y<0,πT0(h(x,y))×(∫-αhx12(h(x,y)+1-(1-s2)-1)ds),ifx<0,y>0,πT0(h(x,y))×(T0(h(x,y))-∫-αhx12(h(x,y)+1-(1-s2)-1)ds),ifx<0,y<0,I(x,y)=I0(h(x,y))=2∫-αhβh‍2(h(x,y)-V(s))ds.In the new variables(θ,I), the system (9) becomes
(37)θ′=∂H∂I,I′=-∂H∂θ,
where
(38)H(θ,I,t)=πh0(I)+πP(x(I,θ),t).In order to estimateπP(I,θ), we need the estimate on the functionx(I,θ).Lemma 7.
ForIsufficiently large and-αh≤x<0, the following estimates hold:
(39)|In∂nx(I,θ)∂In|≤cI,for0≤n≤6.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
(40)Idθ-Hdt=-(Hdt-Idθ).
This means that if one can solveIfrom (38) as a function ofH(usingθandtas parameters), then
(41)dHdθ=-∂I∂t(t,H,θ),dtdθ=∂I∂H(t,H,θ)
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
(42)∂H∂I⟶1,asI⟶+∞.
So we assume thatIcan be written as
(43)I=I0(Hπ+R(H,t,θ)),
whereRsatisfies|R|<H/π.Recalling thath0is the inverse function ofI0, we have
(44)Hπ+R(H,t,θ)=h0(I),
which implies that
(45)R(H,t,θ)=P(x(I,θ),t).
As a consequence,Ris implicitly defined by
(46)R(H,t,θ)=P(x(I0(Hπ+R(H,t,θ)),θ),t).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)
fx′(x,t)/gx′(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
(47)|fx′(x,t)gx′(x,t)-h(t)|≤ϵ2,∀t∈I.
Letx0=X1+1.Then by Lagrangian differential mean value theorem, it follows that, for allx>X1, we have
(48)|f(x,t)-f(x0,t)g(x,t)-g(x0,t)-h(t)|≤ϵ2,∀t∈I.
Moreover, there exists a constantM>0such that
(49)|g(x0,t)f(x,t)-f(x0,t)g(x,t)-g(x0,t)-f(x0,t)|≤M,∀t∈I.
By condition (A1), there existsX>X1; we have|g(x,t)|>2M/ϵ.
Thus(50)|f(x,t)g(x,t)-h(t)|≤|f(x,t)-f(x0,t)g(x,t)-g(x0,t)-h(t)|+|1g(x,t)(g(x0,t)f(x,t)-f(x0,t)g(x,t)-g(x0,t)-f(x0,t))|<ϵ.Lemma 9.
Under the assumptions (A1) and (A2), the following results hold:(1)
limx→+∞(xm(∂m+np¯(x,t)/∂xm∂tn))=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)
limx→+∞(xm(∂m+n/∂xm∂tn)(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,
(51)limx→+∞xm(∂m+n/∂xm∂tn)p(t,x)|x|σ=σ⋯(σ-m+1)p¯+(n)(t).
Form=1and0≤n≤7, by result (1), we have that
(52)limx→+∞x{-σ·(∂n/∂tn)p(t,x)|x|σ·x+(∂1+n/∂x∂tn)p(t,x)|x|σ}=0,
uniformly int. It follows that
(53)limx→+∞x·(∂1+n/∂x∂tn)p(t,x)|x|σ=σ·p¯+(n)(t).
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(54)∂n∂tnP(t,x)|x|σ·x-11+α·p¯+(n)(t)=∫0x‍[(∂n/∂tn)p(t,x)-p¯+(n)(t)|s|σ]ds|x|σ·x.
Using Lemma 8 and the first result (1) form=0, it follows that
(55)limx→+∞∂n∂tnP(t,x)|x|σ·x=1σ+1p¯+(n)(t),
for0≤n≤7. In a similar way to the proof of (51), we have
(56)limx→+∞xm∂m+n∂xm∂tnP(x,t)|x|σx=1σ+1p¯+,m,n(t),for0≤m≤6,0≤n≤7,
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:
(57)|∂m+lR(H,t,θ)∂Hm∂tl|≤H(α+1)/2,form+l≤6.Moreover, by the implicit function theorem, there exists a functionR1=R1(t,H,θ)such that
(58)R(H,t,θ)=P(x(H,θ),t)+R1(H,t,θ).
Since
(59)R1(H,t,θ)=R(H,t,θ)-P(x(H,θ),t)=P{x[I0(Hπ+R(H,t,θ)),θ],t}-P(x(H,θ),t)=∫01‍p{x[H+s(πR+I-),θ],t}·∂x∂I(H+s(πR+I-),θ)·(πR+I-)ds,by Lemmas3 and 10, we have the estimates onR1(H,t,θ).Lemma 11.
Consider the following:(60)|∂k+lR1(H,t,θ)∂kH∂lt|<Hα/2fork+l≤6.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:(61)|∂k+lI-((H/π)+R)∂kH∂lt|<H1/2fork+l≤6.Now the new Hamiltonian functionI=I(t,H,θ)is written in the form
(62)I=I0(Hπ+R)=I+(Hπ+R)+I-(Hπ+R)=H+πR(H,t,θ)+I-(Hπ+R)=H+πP(x(H,θ),t)+R1(H,t,θ)+I-(Hπ+R).
The system (41) is of the form
(63)dtdθ=∂I∂H=1+π∂x∂H(H,θ)p(x(H,θ),t)+∂R1∂H(H,t,θ)+∂I-∂H(H,t,θ),dHdθ=-∂I∂t=-π∂P∂t(x(θ,H),t)-∂R1∂t(t,H,θ)-∂I-∂t(H,t,θ).
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
(64)dtdθ=∂I∂H=1+π∂x∂H(H,θ)p(x(H,θ),t)+∂R1∂H(H,t,θ)+∂I-∂H(H,t,θ),dρdθ=-∂I∂t=-ϵ2[π∂P∂t(x(θ,H),t)+∂R1∂t(H,t,θ)+∂I-∂t(H,t,θ)],
which is also Hamiltonian system with the new Hamiltonian function
(65)Γ(t,ρ,θ;ϵ)=ρ+πϵ-2P(x(θ,ϵ-2ρ),t)+ϵ-2R1(ϵ-2ρ,θ,t)+ϵ-2I-(ϵ-2ρ,θ,t).
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 notationOk(1)andok(1). We say a functionf(t,ρ,θ,ϵ)∈Ok(1)iffis smooth in(t,ρ)and, fork1+k2≤k,
(66)|∂k1+k2∂tk1∂ρk2f(t,ρ,θ,ϵ)|≤C,
for some constantC>0which is independent of the argumentst,ρ,θ, and ϵ.Similarly, we sayf(t,ρ,θ,ϵ)∈ok(1)iffis smooth in(t,ρ)and, fork1+k2≤k,
(67)limϵ→0|∂k1+k2∂tk1∂ρk2f(t,ρ,θ,ϵ)|=0,
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
(68)f¯:A⟶𝕊×ℝ.
One assumes that the map has the intersection property. Suppose thatf:A→ℝ×ℝ,(τ0,ν0)→(τ1,ν1)is a lift off¯and it has the form
(69)τ1=τ0+2Nπ+δl1(τ0,ν0)+δg~1(τ0,ν0),ν1=ν0+δl2(τ0,ν0)+δg~2(τ0,ν0),
whereNis an integer and δ∈(0,1)is a parameter. The functionsl1,  l2, g~1, andg~2satisfy
(70)l1∈C6(A),l1(τ0,ν0)>0,∂l1∂ν0(τ0,ν0)>0,∀(τ0,ν0)∈A,l2(·,·),g~1(·,·,ϵ),g~2(·,·,ϵ)∈C5(A).
In addition, one assumes that there is a functionI:A→Rsatisfying
(71)I∈C6(A),∂I∂ν0(τ0,ν0)>0,∀(τ0,ν0)∈A,l1(τ0,ν0)·∂I∂τo(τ0,ν0)+l2(τ0,ν0)·∂I∂ν0(τ0,ν0)=0,∀(τ0,ν0)∈A.
Moreover, suppose that there are two numbersa~ andb~such thata<a~<b~<band
(72)IM(a)<Im(a~)≤IM(a~)<Im(b~)≤IM(b~)<Im(b),
where
(73)IM(r)=maxρ∈S1I(ρo,τo),Im(r)=minρ∈S1I(ρo,τo).
Then there existϵ>0andΔ>0such that, ifδ<Δand
(74)∥g~1(·,·,ϵ)∥C5(A)+∥g~2(·,·,ϵ)∥C5(A)<ϵ,
the mappingf¯has an invariant curve inΓA. 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
(75)t=t0+θ+ϵ1-αΣ1(t0,ρ0,θ;ϵ),ρ=ρ0+ϵ1-αΣ2(t0,ρ0,θ;ϵ).
Then, the Poincaré map of (64) is
(76)P:t1=t0+2π+ϵ1-αΣ1(t0,ρ0,2π;ϵ),ρ1=ρ0+ϵ1-αΣ2(t0,ρ0,2π;ϵ).
The functionsΣ1andΣ2satisfy
(77)Σ1=πϵα-1∫0θ‍∂x∂H(ϵ-2ρ,θ)p(x(ϵ-2ρ,θ),t)dθ+ϵα-1∫0θ‍(∂R1∂H(ϵ-2ρ,t,θ)+∂I-∂H(ϵ-2ρ,t,θ))dθ,Σ2=-πϵα+1∫0θ‍∂P∂t(x(ϵ-2ρ,θ),t)dθ-ϵα+1∫0θ‍(∂R1∂t(ϵ-2ρ,t,θ)-∂I-∂t(ϵ-2ρ,t,θ))dθ,
wheret=t0+θ+ϵ1-αΣ1andρ=ρ0+ϵ1-αΣ2.By Lemmas 9, 11, and 12, we know that
(78)|Σ1|+|Σ2|≤Cforθ∈[0,2π].
Hence, forρ0∈[1,2], we may chooseϵsufficiently small such that
(79)ρ0+ϵΣ2≥ρ02≥12.
Moreover we can prove that
(80)Σ1,Σ2∈O6(1).In a similar way to that used for estimatingR1, by direct calculation we have the following lemma.Lemma 15.
The following estimates hold:(81)P(x(ϵ-2ρ,θ),t)-P(x(ϵ-2ρ0,θ),t0)∈ϵ-αO6(1),∂P∂t(x(ϵ-2ρ,θ),t)-∂P∂t(x(ϵ-2ρ0,θ),t0)∈ϵ2-αO6(1).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
(82)Θ+(I)=meas{θ∈[0,π],x(H0,θ)>0},Θ-(I)=T0-Θ+(I),
whereH0=ϵ-2ρ0.Lemma 16.
Consider the following:(83)Θ+(I)=π+ϵO6(1),Θ-(I)=ϵO6(1).Proof.
This Lemma was proved in [5], so we omit the details.
For estimateΣ1andΣ2, we need the estimates ofxandxH.
We recall that, whenx<0, we have
(84)|x(H0,θ)|=O6(1),|xH(H0,θ)|=ϵ2O5(1).
Whenx>0, by the definition ofθ, we have
(85)arcsinx(H0,θ)2h=T0(h)πθ-T-(h)2=θ+ϵ2O5(1),
which yields that
(86)x(H0,θ)=2H0πsinθ+O5(1),xH(H0,θ)=12H0πsinθ+ϵ2O5(1).
Now we can give the estimates ofΣ1andΣ2.Lemma 17.
The following estimates hold true:(87)Σ1(t0,ρ0,2π;ϵ)=(π2ρ0)(α-1)/2×∫0π(sinθ)1+αp¯+(t0+θ)dθ+o6(1),(88)Σ2(t0,ρ0,2π;ϵ)=-π(1-α)/2(2ρ0)(α+1)/2×∫0π(sinθ)1+αp¯+′(t0+θ)dθ+o6(1),
forϵ→0.Proof.
Firstly we considerΣ1.By Lemmas 11, 12, and 15 and (77), we have
(89)Σ1(t0,ρ0,2π;ϵ)=πϵα-1∫0π‍∂x∂H(ϵ-2ρ,θ)p(x(ϵ-2ρ,θ),t)dθ+ϵα-1∫0π‍∂R1∂H(x(ϵ-2ρ,θ),t)+∂I-∂H(x(ϵ-2ρ,θ),t)dθ=πϵα-1∫0π‍∂x∂H(ϵ-2ρ0,θ)p(x(ϵ-2ρ0,θ),t0+θ)dθ+ϵαO6(1)=πϵα-1∫Θ+‍∂x∂H(ϵ-2ρ0,θ)p(x(ϵ-2ρ0,θ),t0+θ)dθ+πϵα-1∫Θ-‍∂x∂H(ϵ-2ρ0,θ)p(x(ϵ-2ρ0,θ),t0+θ)dθ+ϵαO6(1).
By result (2) of Lemma 9, asϵ→0which meansx→∞,we have
(90)πϵα-1∫Θ+‍∂x∂H(ϵ-2ρ0,θ)p(x(ϵ-2ρ0,θ),t0+θ)dθ=πϵα-1∫Θ+‍∂x∂H(θ,ϵ-2ρ)|x|αp¯+(t0+θ)dθ+ϵαO6(1).
By the measure ofΘ-, we have
(91)πϵα-1∫Θ-‍∂x∂H(ϵ-2ρ0,θ)×p(x(ϵ-2ρ0,θ),t0+θ)dθ=ϵαO6(1).
By (90) and (91), we have
(92)Σ1(t0,ρ0,2π;ϵ)=πϵα-1∫Θ+‍∂x∂H(θ,ϵ-2ρ)|x|αp¯+(t0+θ)dθ+ϵαO6(1)=πϵα-1∫0π‍∂x∂H(θ,ϵ-2ρ)|x|αp¯+(t0+θ)dθ+ϵαO6(1)=(π2ρ0)(1-α)/2∫0π‍(sinθ)α+1p¯+(t0+θ)dθ+o6(1).
Now we considerΣ2.By Lemmas 11, 12, and 15 and (77), we have
(93)Σ2(t0,ρ0,2π;ϵ)=-πϵα+1∫0π‍∂P∂t(x(θ,ϵ-2ρ),t)dθ-ϵα+1∫0π‍[∂R1∂t(x(θ,ϵ-2ρ),t)+∂I-∂t(x(θ,ϵ-2ρ),t)]dθ=-πϵα+1∫0π‍∂P∂t(x(θ,ϵ-2ρ0),t0+θ)dθ+ϵαO6(1)=-πϵα+1∫Θ+‍∂P∂t(x(θ,ϵ-2ρ0),t0+θ)dθ-πϵα+1∫Θ-‍∂P∂t(x(θ,ϵ-2ρ0),t0+θ)dθ+ϵαO6(1).
By result (2) of Lemma9, asϵ→0, we have
(94)-πϵα+1∫Θ+‍∂P∂t(x(θ,ϵ-2ρ0),t0+θ)dθ=-πϵα+1α+1∫Θ+‍|x(θ,ϵ-2ρ0)|αx(θ,ϵ-2ρ0)p¯+′(t0+θ)dθ+ϵαO6(1).
By the measure ofΘ-, we have
(95)-πϵα+1∫Θ-‍∂P∂t(x(θ,ϵ-2ρ0),t0+θ)dθ=ϵαO6(1).
By (94) and (95), we have
(96)Σ2=-πϵα+1α+1∫Θ+‍|x(θ,ϵ-2ρ0)|αx(θ,ϵ-2ρ0)p¯+′×(t0+θ)dθ+ϵαO6(1)=-πϵα+1α+1∫0π‍|x(θ,ϵ-2ρ0)|αx(θ,ϵ-2ρ0)p¯+′×(t0+θ)dθ+ϵαO6(1)=-1α+1π(1-α)/2(2ρ0)(α+1)/2∫0π‍(sinθ)1+αp¯+′×(t0+θ)dθ+o6(1).
Thus Lemma17 is proved.


## 2.4. Proof of Theorem1

Let(97)Ψ1(t0,ρ0)=(π2ρ0)(1-α)/2×∫0π‍(sinθ)1+αp¯+(t0+θ)dθ,Ψ2(t0,ρ0)=-1α+1π(1-α)/2(2ρ0)(α+1)/2×∫0π(sinθ)1+αp¯+′(t0+θ)dθ.
Then there are two functions ϕ1 and ϕ2 such that the Poincaré map of (64), given by (76) of the form
(98)P:t1=t0+2π+ϵ1-αΨ1(t0,ρ0)+ϵ1-αϕ1,ρ1=ρ0+ϵ1-αΨ2(t0,ρ0)+ϵ1-αϕ2,
whereϕ1,ϕ2∈O6(1).Since∫0π‍p¯+(t0+θ)sinθdθ>0,∀t0∈R,we have
(99)Ψ1>0,∂Ψ1∂ρ0≠0.
Let
(100)L=ρ0-(1+α)/2∫0π‍(sinθ)1+αp¯+(t0+θ)dθ.
Then
(101)∂L∂t0Ψ1(t0,ρ0)+∂L∂ρ0Ψ2(t0,ρ0)=0.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 Theorem2

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.



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