# Lagrangian Stability of a Class of Second-Order Periodic Systems

Authors: Shunjun Jiang; Fang Fang
Journal: Abstract and Applied Analysis
 (2011)

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## Abstract

We study the following second-order periodic system:x′′+V′(x)+p(t)=0 where V(x) has a singularity and p(t)=p(t+1). Under some assumptions on the V(x) and p(t), by Moser's twist theorem we obtain the existence of quasiperiodic solutions and boundedness of all the solutions.


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## Body

## 1. Introduction and Main Result

In the early 1960s, Littlewood [1] asked whether or not the solutions of the Duffing-type equations
(1.1)x′′+g(x)=e(t),wheree(t+1)=e(t)
are bounded for all time, that is, whether there are resonances that might cause the amplitude of the oscillations to increase without bound. Littlewood suggested studying the following two cases: (i)
superlinear case:g(x)/x→+∞ as x→±∞,(ii)
sublinear case:sign(x)·x→+∞ and g(x)/x→0 as x→±∞.The first positive result of boundedness of solutions in the superlinear case (i) was due to Morris [2]. By means of KAM theorem, Morris proved that every solution of the second-order system (1.1) is bounded if g(x)=2x3 and e(t) is piecewise continuous and periodic. This result relies on the fact that the nonlinearity 2x3 can guarantee the twist condition of KAM theorem. Later, several authors (see [3–5]) improved the Morris’s result and obtained similar results for a large class of superlinear function g(x).In 1999, the first result in the sublinear case was proved by Küpper and You [6] in the study of
(1.2)x′′+|x|α-1x=e(t),
where 0<α<1 and e(t)∈C∞(T). The authors transform (1.2) into a perturbation of an integrable Hamiltonian system and then prove that the Poincaré map of the transformed system is close to a so-called twist map. So, the Moser’s twist theorem guarantees the boundedness of all solutions of (1.2). The general sublinear case was considered by Liu [7] under certain reasonable conditions.The Littlewood problem for singular potentials is known to be challenging, and there are only very a few results. Recently, Capietto et al. [8] studied
(1.3)x′′+V′(x)=p(t),
with p(t) is a π-periodic function and V=(1/2)x+2+(1/(1-x-2)ν)-1, where x+=max{x,0},x-=max{-x,0} and ν>2 is a positive integer. Under the Lazer-Leach assumption that
(1.4)1+12∫0πp(t0+θ)sin⁡θdθ>0,∀t0∈R,
they prove the boundedness of solutions and the existence of quasiperiodic solution by Moser’ twist theorem. It is the first time that the equation of the boundedness of all solution is treated in case of a singular potential.In this paper, We consider the following sublinearly growing potential:(1.5)x′′+V′(x)=p(t),
where V(x)=x+α+1+(1/1-x-2)-1,0<α<1.Our main result is the following theorem.Theorem 1.1.
Ifp(t)∈C6 is 1-periodic continuous, then all the solutions of (1.5) are bounded.The idea for proving the boundedness of solutions of (1.5) is as follows. By means of transformation theory, (1.5) is, outside of a large disc 𝒟={(x,ẋ)∈ℝ2:x2+ẋ2≤r2} in the (x,ẋ)-plane, transformed into a perturbation of an integrable Hamiltonian system. Then, Poincaré map of the transformed system is close to a so-called twist map in ℝ∖𝒟. The Moser’s twist theorem [9] guarantees the existence of arbitrarily large invariant curves diffeomorphic to circles and surrounding the origin in the (x,ẋ)-plane. Every such curves is the base of a time-periodic and flow-invariant cylinder in the extended phase space (x,ẋ,t)∈ℝ2×ℝ, which confines the solutions in the interior and which leads to a bound of these solutions.The paper is organized as follows. In Sections2.1 and 2.2, we give action-angle variables and some estimates which is useful for our proof. In Section 2.3, we will give an asymptotic expression of the Poincaré map and prove the main result by Moser’s twist theorem [9].


## 2. Proof of Theorem

### 2.1. Action-Angle Variables and Some Estimates

Without loss of generality and for brevity of arguments, we assume that the average value ofp(t) vanishes; that is, ∫01p(t)dt=0. Hence the function P(t)=∫0sp(s)ds is also 1-periodic in t and is in C6.System (1.5) is equivalent to the planar Hamiltonian system
(2.1)x′=y+P(t),y′=-V′(x),
where Hamiltonian is ℋ(x,y,t)=(1/2)y2+V(x)+yP(t).In order to introduce action and angle variables, we first consider the auxiliary autonomous system(2.2)x′=y,y′=-V′(x),
which is integrable with the Hamiltonian
(2.3)H0(x,y)=12y2+V(x).
The closed curves H0(x,y)=h>0 are just the integral curves of (2.2). Denote by T0(h) the time period of the integral curve Γh: H0(x,y)=h and by I0 the area enclosed by the closed curve Γh. Let
(2.4)αh=hh+1,βh=h1/(1+α).
Then, V(-αh)=V(βh)=h.It is easy to see that(2.5)I0(h)=2∫-αhβh2(h-V(s))ds,∀h>0,T0(h)=I0′(h)=2∫-αhβh12(h-V(s))ds,∀h>0.
Denote
(2.6)I+(h)=2∫0βh2(h-V(s))ds,I-(h)=2∫0αh2(h-V(-s))ds,T+(h)=2∫0βh12(h-V(s))ds,T-(h)=2∫0αh12(h-V(-s))ds.
Then,
(2.7)I0(h)=I-(h)+I+(h),T0(h)=T-(h)+T+(h).The following estimates on the functionsI+, I-, and I0 and T+, T-, and T0 are crucial for this paper. We first estimate I+ and T+. Since I+ is the area enclosed by the closed curve Γh and y-axis when x≥0, we can easily prove that
(2.8)I+(h)=2∫0βh2(h-xα+1)dx=2∫0h1/(1+α)2(h-xα+1)dx.Letx=th1/(1+α), then we get
(2.9)I+(h)=2∫012h(1-t1+α)h1/(1+α)dt=22h(1/2)+1/(1+α)∫011-t1+αdt.
Since T+(h)=I+′(h), we have
(2.10)T+(h)=22(12+1α+1)h-(1/2)+1/(1+α)∫011-t1+αdt.We now give the estimates on the functionI- and T-.Lemma 2.1.
One has(2.11)hn|dnT-(h)dhn|≤Ch-1/2,hn|dnI-(h)dhn|≤Ch1/2,
where n=0,1,…,6, h→+∞. Note that here and below, one always uses C, C0, or C0′ to indicate some constants.Proof.
Now, we estimate the first inequality. We chooseV(s)/h=η as the new variable of integration, then we have
(2.12)T-(h)=∫-αh012(h-V(s))ds=∫01hV′(s(η,h))12(1-η)dη.
Since V(s)=(1/(1-s2))-1 and V(s)/h=η, we have s=ηh/(1+ηh). By direct computation, we have
(2.13)V′(s)=2s(1-s2)2=2ηh(1+ηh)21+ηh,
then we get
(2.14)T-(n)(h)=(-3/2)!(-(3/2)-n)!∫01ηn2η(1-η)(1+ηh)(3/2)+ndη,n=0,1,…,6.
When 0≤η≤h-1 and h sufficient large, there exits C0 such that 1-η>C0, so we have
(2.15)∫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.
Since h-2/3≤η≤1, we have
(2.16)h1/3<1+h1/3≤1+ηh≤1+h,
then
(2.17)∫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 is C0>0 such that 1-η≥C0 when h-1≤η≤h-2/3 and h→+∞, we have
(2.18)∫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 (2.15)–(2.18), we have T-(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 the brief proof.
We chooseV(s)/h=η as the new variable of integration, so we have
(2.19)∂s∂h=ηV′,s=ηh1+ηh,V′(s)=2s(1-s2)2=2ηh(1+ηh)21+ηh.
By direct computation, we have
(2.20)I-(h)=2∫-αh02(h-V(s))ds=h∫012(1-η)η(1+ηh)3/2dη.
By (2.20), we can easily get
(2.21)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,
where n=0,1,…,6.
By the similar way in estimatingT-(n)(h), we get
(2.22)I-1(n)(h)≤Ch(1/2)-n,I-2(n)(h)≤Ch(1/2)-n,
which means that
(2.23)I-(n)(h)≤Ch(1/2)-n,n=0,1,…,6.
Thus, we complete the proof of Lemma2.1.Remark 2.2.
It follows from (2.10) and Lemma 2.1 that
(2.24)lim⁡h→+∞T-(h)=0,lim⁡h→+∞T+(h)=+∞.
Thus, the time period T0(h) is dominated by T+(h) when h is sufficiently large. By T0(h)=I0′(h), we know I0(h) is dominated by I+(h) when h is sufficiently large.Remark 2.3.
It also follows from the definition ofI0(h), I-(h), I+(h) and Remark 2.2 that
(2.25)|hndnI0(h)dhn|≤C0I0(h),forn≥1.
In particular,
(2.26)ch(1/2)+1/(1+α)≤I0(h)≤Ch(1/2)+1/(1+α).Remark 2.4.
Note thath=h0(I0) is the inverse function of I0. By Remark 2.3, we have
(2.27)|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)=∫Γ2(h-V(s))ds, where Γ is the part of the the closed curve Γh connecting the point on the y-axis and point (x,y).We define the well-known map(θ,I)→(x,y) by
(2.28)y=∂S∂x(x,I),θ=∂S∂I(x,I).
It is well-known that the map is symplectic, since
(2.29)dx∧dy=dx∧(Sxxdx+SxIdI)=SxIdx∧dI,dθ∧dI=(SIxdx+SIIdI)∧dI=SIxd∧dI.
From the above discussion, we can easily get
(2.30)θ={1T0(h(x,y))(∫-αhx12(h(x,y)-V(s))ds),ify>0,1-1T0(h(x,y))(∫-αhx12(h(x,y)-V(s))ds),ify<0.(2.31)I(x,y)=I0(h(x,y))=2∫-αhβh2(h(x,y)-V(s))ds.
In the new variables (θ,I), the system (2.1) becomes
(2.32)θ′=∂H∂I,I′=-∂H∂θ,
where
(2.33)H(θ,I,t)=h0(I)+H1(I,θ,t),
where h0(I) is the inverse function of I0(H) and H1(I,θ,t)=y(I,θ)P(t).In order to estimateH1(I,θ,t), we need the estimate on the functions x(I,θ) and y(I,θ). For this purpose, we first give some definitions which are very similar to those in [4].Define a functionL in terms of h and V by
(2.34)L(x,I)=-hIIhI+hIh(W′-12),
and a linear differential operator acting on functions of x,I, according to
(2.35)L(f)=hIh{(fVVx)x-12f}+fI,
where f(x,I) is a smooth function, and we denote ℒn=ℒ∘⋯∘ℒ︸n.The following equality (its proof can be found in [4]) is crucial for the proof of the following lemmas:
(2.36)ddI∫-αhxf(s,I)1h-V(s)ds=∫-αhxL(f)1h-V(s)ds+f(x,I)∫-αhxL(s,I)1h-V(s)ds.Before giving the estimates onx(I,θ) and y(I,θ), we now prove some lemmas which will be used frequently in the following proof.Lemma 2.5.
Suppose that there is a constantC0 such that |g(x,I)|≤C0I-k, then one can find a constant C0′ and C1′ such that, for -αh≤x≤0,
(2.37)|V′(x)∫-αhxg(s,I)h-V(s)ds|≤C0′I-kh-V(x),(2.38)|h-V(x)∫-αhxg(s,I)h-V(x)|≤C1′I-k(αh+x).Proof.
We now prove (2.37). Let
(2.39)G(x,I)=I-kh-V(x)V′(x),F(x,I)=∫-αhxg(s,I)1h-V(s)ds,
then
(2.40)F(-αh,I)=lim⁡x→-αhG(x,I)=0.
By direct calculation,
(2.41)|∂F(x,I)∂x|=|g(x,I)1h-V(x)|≤C0I-k1h-V(x),∂G(x,I)∂x=I-k((V′(x))2+(h-V(x))V′′(x)(V′(x))2)1h-V(x).
Since
(2.42)V′′(x)=(1-x2)2+4x2(1-x2)(1-x2)4>0,
we have
(2.43)(V′(x))2+(h-V(x))V′′(x)(V′(x))2>0.
By |g(x,I)|≤C0I-k, there is C0′=2C0 such that
(2.44)-C′0I-k((V′(x))2+(h-V(x))V′′(x)(V′(x))2)1(h-V(x))≤∂F(x,I)/∂x2h-V(x)≤C0′I-k((V′(x))2+(h-V(x))V′′(x)(V′(x))2)1h-V(x),
that is
(2.45)-C0′∂G(x,I)∂x≤∂F(x,I)∂x≤C0′∂G(x,I)∂x,
which means that
(2.46)-C0′G(x,I)≤F(x,I)≤C0′G(x,I).
That is,
(2.47)|V′(x)∫-αhxg(s,I)1h-V(x)|≤CI-kh-V(x).
Thus, we complete the proof (2.37).
Now, we prove (2.38). Let
(2.48)G(x,I)=I-k(αh+x)h-V(x),F(x,I)=∫-αhxg(s,I)1h-V(s)ds.
Then, we have
(2.49)F(-αh,I)=lim⁡x→-αh⁡G(x,I)=0.
By direct computation, we have
(2.50)|∂F(x,I)∂x|=|g(x,I)1(h-V(x))|≤C0I-k1h-V(x),∂G(x,I)∂x=I-k(1+V′(x)(αh+x)2(h-V(x)))1h-V(x).
Since V′′(x)>0 and h=V(-αh+x), it follows that for
(2.51)|V′(x)(αh)2(h-V(x))|≤12,
so for C1′>2C0+1, we have
(2.52)-C1′∂G(x,I)∂x≤∂F(x,I)∂x≤C1′∂G(x,I)∂x,
which means that
(2.53)-C1′G(x,I)≤F(x,I)≤C1′G(x,I).
By the definition of G(x,I), we have
(2.54)|h-V(x)F(x,I)|≤C1′I-k(αh+x).
Thus, we complete the proof of (2.38) and Lemma 2.5.By Lemma2.5, we have the following Lemma which is important to our estimation.Lemma 2.6.
One can find a constantC such that, for -αh≤x≤0,
(2.55)|∂Ik(∫-αhxL(x,I)h-V(s)ds)|≤CV′(x)I-(k+1)h-V(x),(2.56)|∂Ik(∫-αhxL(x,I)h-V(s)ds)|≤Ch-V(x)I-(k+1)(αh+x),
where 0≤k≤6.Proof.
Whenk=1, we have
(2.57)∂I(∫-αhxL1h-V(s)ds)=∫-αhxL(L)1h-V(s)ds+L∫-αhxL1h-V(s)ds.
By the definitions of L and ℒ, we have |L|≤CI-1 and |ℒ(L)|≤CI-2. By (2.37), we obtain
(2.58)∫-αhxL1h-V(s)ds≤CV′I-2h-V.
Suppose that k=l, we have
(2.59)∂Il(∫-αhxL1h-V(s)ds)≤CV′I-(l+1)h-V.
We now proven that when k=l+1,
(2.60)∂Il+1(∫-αhxL1h-V(s)ds)≤CV′I-(l+2)h-V.
By direct computation, we have
(2.61)∂Il+1(∫-αhxLh-V(s)ds)={∂In(Lm(f))∂Il-m-n(∫-αhxLh-V(s)ds),∫-αhxLl+1(L)h-V(s)ds},
where {f1,f2,…,fn} denotes linear combination of functions f1,…,fn with integer coefficients and 0≤m+n≤l.
Since|ℒ(L)|≤CI-2, we have ℒl+1(L)≤CI-(l+2). By (2.37), we obtain
(2.62)|∫-αhxLl+1(L)h-V(s)ds|≤CV′I-(l+2)h-V.
By direct computation, we get
(2.63)∂n(Lm(L))≤CI-(m+n+1).
By assumption (2.59), we have
(2.64)|∂Il-m-n(∫-αhxLh-V(s)ds)|≤CV′I-(l-m-n+1)h-V.
So, we have
(2.65)|∂In(Lm(f))∂Il-m-n(∫-αhxLh-V(s)ds)|≤CV′I-(l+2)h-V.
By (2.62) and (2.65), we have
(2.66)|∂Il+1(∫-αhxLh-V(s)ds)|≤CV′I-(l+2)h-V.
Thus, we have proved (2.55).
The inequality (2.56) can be proved by (2.38), and the process of proof is similar to that of (2.55), so we omit it.
Thus, we have proved Lemma2.6.Now, we give the estimates ofx(I,θ) and y(I,θ).Lemma 2.7.
ForI sufficient large and -αh≤x<0,  the following estimates hold:
(2.67)|In∂nx(I,θ)∂In|≤C|x(I,θ)+1|,|In∂ny(I,θ)∂In|≤C|y(I,θ)|,for0≤n≤6.Proof.
We now prove the first inequality. It is sufficient to prove that(2.68)In∂nx(I,θ)∂In|≤C(1+x).
Case k=1. Differentiating (2.30) by I and noting 1/T0=hI, we have
(2.69)0=hII∫-αhx12(h(x,y)-V(s))ds+hI∂I(∫-αhx12(h(x,y)-V(s))ds).
Now, we choose V(s)/h=η as the new variable of integration, so
(2.70)∂I(∫-αhx12(h(x,y)-V(s))ds)=∂I(∫0V(x)/hhV′(s(η,t,h))12(1-η)dη)=V′xIh-VhIh2h1/2V′12(1-η)+∫0V(x)/h(1/2)h-1/2V′hI-h1/2V′′(∂s/∂I)(V′)212(1-η)dη=xI2(h-V)-hIh2(h-V)VV′+hIh+∫0V(x)/h(1/2)h-1/2V′hI-h1/2V′′(∂s/∂I)(V′)2h2(h-V)V′hds.
Observing that
(2.71)∂s∂I=hIηV′,W′=1-VV′′(V′)2,
and simplifying, we have
(2.72)(2.70)=xI2(h-V)-hIh2(h-V)VV′+hIh∫0x(W′-12)ds2(h-V).
By (2.69)–(2.72), we have
(2.73)xI(θ,I)=2(h-V)∫-αhxL(s,I)12(h-V(s))ds+hIhW(x).
Since |L|≤CI-1, by Lemma 2.6, we have
(2.74)|(2(h-V))∫-αhxL(s,I)12(h-V(s))ds|≤CI-1(αh+x).
We observe that
(2.75)W(x)=12x(1-x)(1+x),
where -1<x<0, so |W(x)|≤C(1+x), which means that
(2.76)|hIhW(x)|≤CI-1(1+x).
By (2.74) and (2.76), we have |xI|≤I-1(1+x).
We suppose that(2.77)|∂Ik-1x|≤I-(k-1)(1+x),
where 1≤k≤6. We will prove |∂Ikx|≤I-k(1+x).
For this purpose, we firstly estimate2(h-V(x)). We differentiate 2(h-V(x)) in (2.73) and using (2.73), then we obtain
(2.78)ddI2(h-V(x))=hIh2(h-V(x))-V′(x)2∫-αhxL(s,I)12(h-V(s))ds.
Since |hI/h|≤CI-1, we have
(2.79)|hIh2(h-V(x))|≤CI-12(h-V(x)).
By (2.37), we have
(2.80)|V′(x)2∫-αhxL(s,I)12(h-V(s))ds|≤CI-12(h-V(x)).
By (2.78)–(2.80), we have
(2.81)|∂I(2(h-V(x)))|≤CI-12(h-V(x)).
We suppose that whenn≤k-1,
(2.82)|∂In(2(h-V(x)))|≤CI-n2(h-V(x)).
We will prove that whenn=k,
(2.83)|∂In(2(h-V(x)))|≤CI-n2(h-V(x)).
By direct computation, we have
(2.84)∂Ik(2(h-V(x)))={V(k1)∂Il1x⋯∂Ilk1-1x∂Ik3(∫-αhxL(s,I)12(h-V(s))ds),∂Im(hIh)∂Ik-m-1(2(h-V(x)))(∫-αhxL(s,I)12(h-V(s)ds)},
where l1+⋯+lk1-1=k2,k2+k3=k-1 and k2<k1<k,k3<k.
By the assume (2.77), we have
(2.85)|∂Il1x⋯∂Ilk1-1x|≤(1+x)k1-1I-k2.
By (2.55), we get
(2.86)|∂Ik3(∫-αhxL(s,I)12(h-V(s))ds)|≤CV′I-k32(h-V(x))
By (2.85), (2.86) and noting the fact that
(2.87)(1+x)k-1∂Ik(V)≤C(|V′|+(1+x)k1+1),-1<x<0,
we obtain
(2.88)|∑V(k1)∂Il1x⋯∂Ilk1-1x∂Ik3(∫-αhxL(s,I)12(h-V(s))ds)|≤∑CV′V(k1)(1+x)k1-1I-(k2+k3+1)2(h-V(x))≤∑[|V′|+(1+x)k1+1]V′I-(k)2(h-V(x))≤CI-(k)2(h-V(x)).
By assumption (2.82), we have
(2.89)|∂Ik-m-1(2(h-V(x)))|≤CIk-m-1(2(h-V(x))).
By (2.89) and the fact that |∂Im(hI/h)|≤CI-(m+1), we have
(2.90)|∑∂Im(hIh)∂Ik-m-1(2(h-V(x)))|≤CI-k2(h-V(x)).
So, by (2.88) and (2.90), we have proved (2.83).
By (2.56), we have
(2.91)|∂Im(∫-αhxL(x,I)2(h-V(s))ds)|≤C2(h-V(x))I-(m+1)(αh+x).
By (2.83) and (2.91), we have
(2.92)|∂n(2(h-V(x)))∂m(∫-αhxL(s,I)12(h-V(s))ds)|≤CI-(m+n+1)(αh+x)≤CI-(k+1)(αh+x).
By the assumption (2.77), the fact that ∂Im3(hI/h)≤CI-(m3+1), and noting that
(2.93)∂I(hIhW)=∑kW(m1)∂Il1x⋯∂Ilm1x∂Im3(hIh),
where m2<m1,l1+⋯+lm1=m2,m2+m3=k, we have
(2.94)∂I(hIhW)≤∑kW(m1)I-(l1+⋯+lm1)(1+x)m1I-(m3+1)≤C∑kW(m1)(1+x)m1I-(m2+m3+1)≤C(1+x)I-(k+1).
By (2.92) and (2.94), we have
(2.95)∂IkxI=∑k∂n(2(h-V(x)))∂m(∫-αhxL(s,I)12(h-V(s))ds)+∂k(hIhW(x))≤CI-(k+1)(1+x)≤CI-(k+1)(1+x),
which means
(2.96)∂Ikx≤CI-k(1+x).
We now prove(2.97)|In∂ny(I,θ)∂In|≤cI.
Since
(2.98)h(I)=12y2+V(x),
we have
(2.99)y=±2(h(I)-V(x)).
we have proved (2.83), so we have
(2.100)|In∂ny(I,θ)∂In|≤C⋅I-ny,
which means that
(2.101)|In∂ny(I,θ)∂In|≤C⋅h0(I).
The proof of Lemma 2.7 is complete.Remark 2.8.
Lemma2.7 also holds when x≥0. Since the idea and the process of the proof is more easily than that of Lemma 2.7, we omit the details.Now, we give the estimate ofH1(I,θ,t).Lemma 2.9.
(2.102)Ik|∂Ik∂tl∂θiH1(I,θ,t)∂Ik∂tl∂θi|≤C⋅h0(I),k+i≤7,i=0,1.Proof.
This lemma can be proved easily form the definition ofH1 and |y(θ,I)|≤2h0(I).


### 2.2. New Action and Angle Variables

Now, we are concerned with the Hamiltonian system (2.32) with Hamiltonian function H(θ,I,t) given by (2.33). Note that
(2.103)Idθ-Hdt=-(Hdt-Idθ).
This means that if one can solve I form (2.32) as a function of H (θ and t as parameters), then
(2.104)dHdθ=-∂I∂t(t,H,θ),dtdθ=-∂I∂H(t,H,θ)
is also a Hamiltonian system with Hamiltonian function I, and now, the action, angle, and time variables are H, t, and θ.Form Remarks2.3 and 2.4, we have
(2.105)∂H∂I⟶1,asI⟶+∞.
Hence, by the implicit function theorem, there is a function I(t,H,θ) such that
(2.106)H(θ,I(t,H,θ),t)=H.
By Lemma 2.9, we have
(2.107)H1(θ,I,t)H0(I)⟶0,asI→+∞.
So, there is a function R(t,H,θ) with |R|≤(1/2)H such that
(2.108)I(t,H,θ)=I0(H-R(t,H,θ)),forH⟶+∞.
Let
(2.109)I1(t,H,θ)=I0(H-R(t,H,θ))-I0(H)=∫0πI0′(H-sR(t,H,θ))R(t,H,θ)dθ.
Then,
(2.110)I(t,H,θ)=I0(H)+I1(t,H,θ).
From Remark 2.3, we have known the estimate of I0(H), so we need to give the estimate of I1(t,H,θ). For this propose, we need firstly the following Lemma on the estimate of R(t,H,θ).Lemma 2.10.
The functionR(t,H,θ) possesses the following estimates:
(2.111)Hk|∂k+lR(H,t,θ)∂kH∂lt|≤H1/2,
for k+l≤6.Proof.
From (2.108) and (2.110), it follows that
(2.112)R(t,H,θ)=H1(θ,I0(H-R),t).
When k+l=0, by Lemma 2.9, we have
(2.113)|R(t,H,θ)|=|H1(θ,I0(H-R),t)|≤C⋅h0(I0(H-R))≤C⋅H-R≤C⋅H.
When k+l=1, we first denote
(2.114)Δ=∂H1∂I(θ,I0(H-R),t)I0′(H-R).
By Remark 2.3, we observe that I0 is increasing and
(2.115)I0(12H)≤I0(H)≤I0(32H)≤C⋅I0(12H).
By Lemma 2.9 and (2.115), we have
(2.116)|Δ|=|∂H1∂I(θ,I0(H-R),t)⋅I0′(H-R)|≤C⋅1I0(H-R)⋅h0(I0(H-R))⋅I0′(H-R)≤C⋅1I0(H/2)⋅32H⋅1(3/2)H⋅I0(32H)≤C⋅H-1/2≤12.
So,
(2.117)1+Δ≥12.
By direct computation, we have
(2.118)|∂R∂H|=|(∂H1/∂I)(θ,I0(H-R),t)I0′(H-R)1+Δ|≤C⋅|∂H1∂I(θ,I0(H-R),t)I0′(H-R)|≤C⋅H-1/2,|∂R∂t|=|(∂H1/∂t)(I0(H-R))1+Δ|≤C⋅|∂H1∂t(I0(H-R))|≤C⋅H1/2.
By (2.118), we can easily get
(2.119)Hk|∂k+lR(H,t,θ)∂kH∂lt|≤H1/2.
When k+l≥2, one may get
(2.120)∂k+lR(H,t,θ)∂kH∂lt=∑cn,j1⋯jn∂nH1∂nI0j1∂j1I0(H-R)∂Hj1⋯∂jnI0(H-R)∂Hjn,
where 1≤n≤k, j1+⋯+jn<k. It is easy to verify
(2.121)|∂k+lR(H,t,θ)∂kH∂lt|≤C⋅H(1/2)-k,
for k+l≥2. This complete the proof.Now, we give the estimates ofI1(t,H,θ).Lemma 2.11.
The functionI1(t,H,θ) possesses the following estimates:
(2.122)Hk|∂k+lI1(H,t,θ)∂kH∂lt|≤H1/(1+α),
for k+l≤6.Proof.
Whenk=l=0. By Remark 2.3 and |R|≤(1/2)H, we have
(2.123)I0′(H-sR(t,H,θ))≤C⋅(H-sR)1/(1+α)(H-sR)1/2≤C⋅((3/2)H)1/(1+α)((1/2)H)1/2≤C⋅H1/(1+α)-(1/2).
By Lemma 2.10, we know
(2.124)|R|≤C⋅H1/2.
Since
(2.125)I1(t,H,θ)=∫0πI0′(H-sR(t,H,θ))R(t,H,θ)dθ,
it is easy to get that |I1|≤C·H1/(1+α).
Whenk+l≥1. By direct computation, we have
(2.126)∂k+lI1(t,H,θ)∂Hk∂tl=∑∫0π∂k1+l1I0′(t,H,θ)∂Hk1∂tl1∂k2+l2R(t,H,θ)∂Hk2∂tl2ds,
where k1+k2=k, l1+l2=l. Now, we need to estimate the first term of the integrand. The following equality is important:
(2.127)∂k1+l1I0′(t,H,θ)∂Hk1∂tl1=∑I0(p+q+1)(H-sR)∂m1u∂Hm1⋯∂mpu∂Hmp⋅∂j1+n1u∂Hj1∂tn1⋯∂jq+nqu∂Hjq∂tnq,
where u=H-sR, p≤m, q≤n, n1,…,nq>0, m1,…,mp>0, n1+⋯+nq=n, m1+⋯+mp+j1+⋯+jq=m. Assume that there are β(≤p) members: m1,…,mβ in {m1,…,mp} which equal to 1. Noting that
(2.128)|∂u∂H|=|1+s∂R∂H|≤C,|∂ku∂Hk|≤|∂kR∂Hk|≤C⋅H(1/2)-k,k>1,|∂k+lu∂Hk∂tl|≤|∂k+lR∂Hk∂tl|≤C⋅H(1/2)-k,l>0.
By the above discussions, we have
(2.129)|∂k1+l1I0′(t,H,θ)∂Hk1∂tl1|≤C⋅(H-sR)1/(1+α)+(1/2)(H-sR)p+q+1⋅H(1/2)(p-β)-(mβ+1+⋯+mp)⋅H(1/2)q-(j1+⋯+jq)≤CH(1/2)+1/(1+α)Hp+q+1⋅H(1/2)(p+q-β)H-(m-β)≤CH-(1/2)+1/(1+α)-m⋅H(p+q-β)((1/2)-1)≤CH-(1/2)+1/(1+α)-m.
By Lemma 2.10, we have known
(2.130)Hk|∂k+lR(H,t,θ)∂kH∂lt|≤H1/2,
then we have
(2.131)Hk|∂k+lI1(H,t,θ)∂kH∂lt|≤CH1/(1+α).


### 2.3. Proof of the Main Result

Up to now, we have given an equivalent form of (1.5), that is, the system (2.32), which is expressed in the action and angle variables (H,t). In this section, we first introduce some transformations such that in the transformed system, the perturbation terms of (2.32) depending on the new angle variable are very small if the new action variable is sufficiently large and then prove, by Moser’s twist theorem, the statement of Theorem 1.1.Lemma 2.12.
There is a canonical transformationΨ:(λ,τ)→(H,t) of the form
(2.132)Ψ:H=λ+U(τ,λ,θ),t=τ+V(τ,λ,θ),
where the functions U and V are 1-periodic in θ and satisty
(2.133)U(τ,λ,θ)λ,V(τ,λ,θ)⟶0asλ⟶∞,
uniformly for (τ,θ)∈T2 such that under this mapping, the system and the Hamiltonian function I in (2.110) is changed into the form
(2.134)dλdθ=-∂K∂τ(τ,λ,θ),dτdθ=∂K∂H(τ,λ,θ),
where
(2.135)K(τ,λ,θ)=I0(λ)+[I1](λ,θ)+M(τ,λ,θ),
with
(2.136)[I1](λ,θ)=∫01I1(t,λ,θ)dt.
Moreover, the new perturbation ℳ possesses the estimate
(2.137)|∂k+l∂λk∂τlM(τ,λ,θ)|≤C⋅λ-k+(1/2).Proof.
We will look for the required transformationΨ by a generating function ℱ(t,λ,θ) in the following way:
(2.138)H=λ+∂F∂t(t,λ,θ),τ=t+∂F∂λ(t,λ,θ),
where the function ℱ will be given later. Under Ψ, the transformed system of (2.104) is of the form
(2.139)dλdθ=-∂K∂τ(τ,λ,θ),dτdθ=-∂K∂λ(τ,λ,θ),
where
(2.140)K(τ,λ,θ)=I0(λ+∂F∂t)+I1(t,λ+∂F∂t,θ)+∂F∂θ.
By Taylor’s formula, one can write
(2.141)K(τ,λ,θ)=I0(λ)+I0′(λ)∂F∂t+I1(t,λ,θ)+M(τ,λ,θ),
where
(2.142)M(τ,λ,θ)=∂F∂θ+∫01(1-s)I0′′(λ+s∂F∂t)⋅(∂F∂θ)2ds+∫01∂I1∂H(t,λ+s∂F∂t,θ)⋅∂F∂tds.
We chooseℱ(2.143)F(t,λ,θ)=-∫0t1I0′̅(λ)⋅(I1(t,λ,θ)-[I1](λ,θ)).
Then, 𝒦 is of the form (2.135).
We now show thatℳ satisfies (2.137). From Remark 2.3 and Lemma 2.11, it follows that
(2.144)|∂k+k+i∂λk∂tl∂θiF(t,λ,θ)|≤C⋅λ-k+(1/2),
for k+i+l≤6 and i=0,1. In particular
(2.145)|∂2∂λ∂tF(t,λ,θ)|≤Cλ-1/2≤12,
if λ≫1. So we can solve the second equation of (2.138) for t,
(2.146)t=τ+V(τ,λ,θ),
where the function V satisfies
(2.147)V(τ,λ,θ)=-∂F∂λ(τ+V,λ,θ).
Set
(2.148)U(τ,λ,θ)=∂F∂t(τ+V,λ,θ).
Then, the canonical transformation Ψ is of the form (2.132). Moreover, similar to the proof of [5, Lemma  2], we can verify that
(2.149)|∂k+l∂λk∂τlU(τ,λ,θ)|≤C⋅λ-k+(1/2),|∂k+l∂λk∂τlV(τ,λ,θ)|≤C⋅λ-k-(1/2),
for k+l≤5 and U/λ,V→0 as λ→+∞. Let
(2.150)ϕ1(τ,λ,θ)=∂F∂θ(τ+V,λ,θ),ϕ2(τ,λ,θ)=∫01(1-s)I0′′(λ+sU)⋅U2ds,ϕ3(τ,λ,θ)=∫01∂I1∂H(τ+V,λ+sU,θ)⋅Uds.
It is not difficult to prove that
(2.151)|∂k+l∂λk∂τlϕ1(τ,λ,θ)|≤C⋅λ-k+(1/2),|∂k+l∂λk∂τlϕ1(τ,λ,θ)|≤C⋅λ-k+(1/(1+α))-(1/2),|∂k+l∂λk∂τlϕ1(τ,λ,θ)|≤C⋅λ-k+(1/(1+α))-(1/2),
for k+l≤5. Note that 0<α<1, we have 1/(1+α)-1/2<1/2.
Hence, we have(2.152)|∂k+l∂λk∂τlM(τ,λ,θ)|≤λ-k+(1/2).
The proof of Lemma 2.12 is complete.
Forλ0>0, we denote by Aλ0 the domain
(2.153)Aλ0={(λ,τ,θ)∣λ≥λ0,(τ,θ)∈T2}.Lemma 2.13.
The Poincaré mappingP of (2.134) has the intersection property on AH0; that is, if Γ is an embedded circle in AH0 homotopic to a circle λ= const. in AH0, then P(Γ)⋂Γ≠∅.Proof.
The proof can be found in [5].
Definite a diffeomorphismΨ1:AH0×𝕊1→AH0×𝕊1(2.154)ν=I0′(H),τ=τ,θ=θ.
Then, the system (2.134) under the transformation Ψ1 becomes
(2.155)dνdθ=f1(λ,t,θ),dtdθ=ν+f2(λ,t,θ),
where
(2.156)f1(τ,ν,θ)=-I0′′(λ)∂M∂τ(τ,λ,θ),f2(τ,ν,θ)=∂[I1]∂λ(λ,θ)+∂M∂λ(τ,λ,θ),
with λ=λ(ν) defined through the transformation Ψ1.
Now, we estimatef1 and f2. Since
(2.157)c⋅λ1/(1+α)-(1/2)≤I0′(λ)≤C⋅λ1/(1+α)-(1/2),|∂[I1]∂λ|≤C⋅λ1/(1+α)-1,
we have
(2.158)c⋅λ2(1+α)/(1-α)≤λ(ν)≤C⋅λ2(1+α)/(1-α),
then
(2.159)λ≫1⟺ν≫1.
Moreover, we have
(2.160)|∂k∂νkλ(ν)|≤C⋅ν-kλ(ν).
When 0<α<1, we have
(2.161)|∂k+l∂λk∂τlM(τ,λ,θ)|≤C⋅λ-k+(1/2),12<11+α<1,
so
(2.162)|∂k+l∂νk∂τlf1(τ,ν,θ)|≤C⋅ν-k⋅λ-2⋅ν-kI0(λ)⋅λ1/2≤C⋅ν-kλ-α/(1+α)≤C⋅ν-k+(-4α/(1-α))≤C⋅ν-k-σ,|∂k+l∂νk∂τlf2(τ,ν,θ)|≤C⋅ν-k⋅(λ-1/2+λ1/(1+α)-1)≤C⋅ν-k⋅λ1/(1+α)-1≤C⋅ν-k-(2α/(1-α))≤C⋅ν-k-σ,
for all k+l≤4, where σ=min{4α/(1-α),2α/(1-α)}=2α/(1-α)>0.
Sincef1 and f2 are sufficiently small as ν≫1, all solution of (2.155) exist for 0≤θ≤1 when the initial values ν(0)=λ are sufficiently large. Hence, the Poincaré map Φ associated to (2.155) is well defined on Aλ0 as λ0≫1. In fact, by integrating (2.155) from θ=0 to θ=π, we see that Φ has the form
(2.163)Φ:τ1=τ0+μ0+Ξ1(t0,λ0),ν1=ν0+Ξ2(t0,λ0),
where Ξ1 and Ξ2 satisfy the same estimates as those of f1 and f2; that is,
(2.164)∂νk∂τlΞi≤ν-σ,
where i=1,2,0≤k+l≤4.
SinceΦ satisfies all the assumptions of Moser’s twist theorem [9], from which we conclude that for any ω≫1 satisfying
(2.165)|ω-pq|≥c0|q|-5/2,pq∈Q.
There is an invariant curve Γω of Φ which is conjugated to pure rotation of the circle with rotation number ω. Tracing back to the system (2.32), Γω gives rise to an invariant closed curve of the Poincaré map Φ of (2.32) with rotation number 1/ω which surrounds and is arbitrarily far away form the origin. Hence, all solutions of (1.5) are bounded. This completes the proof of the Theorem.


## 2.1. Action-Angle Variables and Some Estimates

Without loss of generality and for brevity of arguments, we assume that the average value ofp(t) vanishes; that is, ∫01p(t)dt=0. Hence the function P(t)=∫0sp(s)ds is also 1-periodic in t and is in C6.System (1.5) is equivalent to the planar Hamiltonian system
(2.1)x′=y+P(t),y′=-V′(x),
where Hamiltonian is ℋ(x,y,t)=(1/2)y2+V(x)+yP(t).In order to introduce action and angle variables, we first consider the auxiliary autonomous system(2.2)x′=y,y′=-V′(x),
which is integrable with the Hamiltonian
(2.3)H0(x,y)=12y2+V(x).
The closed curves H0(x,y)=h>0 are just the integral curves of (2.2). Denote by T0(h) the time period of the integral curve Γh: H0(x,y)=h and by I0 the area enclosed by the closed curve Γh. Let
(2.4)αh=hh+1,βh=h1/(1+α).
Then, V(-αh)=V(βh)=h.It is easy to see that(2.5)I0(h)=2∫-αhβh2(h-V(s))ds,∀h>0,T0(h)=I0′(h)=2∫-αhβh12(h-V(s))ds,∀h>0.
Denote
(2.6)I+(h)=2∫0βh2(h-V(s))ds,I-(h)=2∫0αh2(h-V(-s))ds,T+(h)=2∫0βh12(h-V(s))ds,T-(h)=2∫0αh12(h-V(-s))ds.
Then,
(2.7)I0(h)=I-(h)+I+(h),T0(h)=T-(h)+T+(h).The following estimates on the functionsI+, I-, and I0 and T+, T-, and T0 are crucial for this paper. We first estimate I+ and T+. Since I+ is the area enclosed by the closed curve Γh and y-axis when x≥0, we can easily prove that
(2.8)I+(h)=2∫0βh2(h-xα+1)dx=2∫0h1/(1+α)2(h-xα+1)dx.Letx=th1/(1+α), then we get
(2.9)I+(h)=2∫012h(1-t1+α)h1/(1+α)dt=22h(1/2)+1/(1+α)∫011-t1+αdt.
Since T+(h)=I+′(h), we have
(2.10)T+(h)=22(12+1α+1)h-(1/2)+1/(1+α)∫011-t1+αdt.We now give the estimates on the functionI- and T-.Lemma 2.1.
One has(2.11)hn|dnT-(h)dhn|≤Ch-1/2,hn|dnI-(h)dhn|≤Ch1/2,
where n=0,1,…,6, h→+∞. Note that here and below, one always uses C, C0, or C0′ to indicate some constants.Proof.
Now, we estimate the first inequality. We chooseV(s)/h=η as the new variable of integration, then we have
(2.12)T-(h)=∫-αh012(h-V(s))ds=∫01hV′(s(η,h))12(1-η)dη.
Since V(s)=(1/(1-s2))-1 and V(s)/h=η, we have s=ηh/(1+ηh). By direct computation, we have
(2.13)V′(s)=2s(1-s2)2=2ηh(1+ηh)21+ηh,
then we get
(2.14)T-(n)(h)=(-3/2)!(-(3/2)-n)!∫01ηn2η(1-η)(1+ηh)(3/2)+ndη,n=0,1,…,6.
When 0≤η≤h-1 and h sufficient large, there exits C0 such that 1-η>C0, so we have
(2.15)∫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.
Since h-2/3≤η≤1, we have
(2.16)h1/3<1+h1/3≤1+ηh≤1+h,
then
(2.17)∫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 is C0>0 such that 1-η≥C0 when h-1≤η≤h-2/3 and h→+∞, we have
(2.18)∫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 (2.15)–(2.18), we have T-(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 the brief proof.
We chooseV(s)/h=η as the new variable of integration, so we have
(2.19)∂s∂h=ηV′,s=ηh1+ηh,V′(s)=2s(1-s2)2=2ηh(1+ηh)21+ηh.
By direct computation, we have
(2.20)I-(h)=2∫-αh02(h-V(s))ds=h∫012(1-η)η(1+ηh)3/2dη.
By (2.20), we can easily get
(2.21)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,
where n=0,1,…,6.
By the similar way in estimatingT-(n)(h), we get
(2.22)I-1(n)(h)≤Ch(1/2)-n,I-2(n)(h)≤Ch(1/2)-n,
which means that
(2.23)I-(n)(h)≤Ch(1/2)-n,n=0,1,…,6.
Thus, we complete the proof of Lemma2.1.Remark 2.2.
It follows from (2.10) and Lemma 2.1 that
(2.24)lim⁡h→+∞T-(h)=0,lim⁡h→+∞T+(h)=+∞.
Thus, the time period T0(h) is dominated by T+(h) when h is sufficiently large. By T0(h)=I0′(h), we know I0(h) is dominated by I+(h) when h is sufficiently large.Remark 2.3.
It also follows from the definition ofI0(h), I-(h), I+(h) and Remark 2.2 that
(2.25)|hndnI0(h)dhn|≤C0I0(h),forn≥1.
In particular,
(2.26)ch(1/2)+1/(1+α)≤I0(h)≤Ch(1/2)+1/(1+α).Remark 2.4.
Note thath=h0(I0) is the inverse function of I0. By Remark 2.3, we have
(2.27)|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)=∫Γ2(h-V(s))ds, where Γ is the part of the the closed curve Γh connecting the point on the y-axis and point (x,y).We define the well-known map(θ,I)→(x,y) by
(2.28)y=∂S∂x(x,I),θ=∂S∂I(x,I).
It is well-known that the map is symplectic, since
(2.29)dx∧dy=dx∧(Sxxdx+SxIdI)=SxIdx∧dI,dθ∧dI=(SIxdx+SIIdI)∧dI=SIxd∧dI.
From the above discussion, we can easily get
(2.30)θ={1T0(h(x,y))(∫-αhx12(h(x,y)-V(s))ds),ify>0,1-1T0(h(x,y))(∫-αhx12(h(x,y)-V(s))ds),ify<0.(2.31)I(x,y)=I0(h(x,y))=2∫-αhβh2(h(x,y)-V(s))ds.
In the new variables (θ,I), the system (2.1) becomes
(2.32)θ′=∂H∂I,I′=-∂H∂θ,
where
(2.33)H(θ,I,t)=h0(I)+H1(I,θ,t),
where h0(I) is the inverse function of I0(H) and H1(I,θ,t)=y(I,θ)P(t).In order to estimateH1(I,θ,t), we need the estimate on the functions x(I,θ) and y(I,θ). For this purpose, we first give some definitions which are very similar to those in [4].Define a functionL in terms of h and V by
(2.34)L(x,I)=-hIIhI+hIh(W′-12),
and a linear differential operator acting on functions of x,I, according to
(2.35)L(f)=hIh{(fVVx)x-12f}+fI,
where f(x,I) is a smooth function, and we denote ℒn=ℒ∘⋯∘ℒ︸n.The following equality (its proof can be found in [4]) is crucial for the proof of the following lemmas:
(2.36)ddI∫-αhxf(s,I)1h-V(s)ds=∫-αhxL(f)1h-V(s)ds+f(x,I)∫-αhxL(s,I)1h-V(s)ds.Before giving the estimates onx(I,θ) and y(I,θ), we now prove some lemmas which will be used frequently in the following proof.Lemma 2.5.
Suppose that there is a constantC0 such that |g(x,I)|≤C0I-k, then one can find a constant C0′ and C1′ such that, for -αh≤x≤0,
(2.37)|V′(x)∫-αhxg(s,I)h-V(s)ds|≤C0′I-kh-V(x),(2.38)|h-V(x)∫-αhxg(s,I)h-V(x)|≤C1′I-k(αh+x).Proof.
We now prove (2.37). Let
(2.39)G(x,I)=I-kh-V(x)V′(x),F(x,I)=∫-αhxg(s,I)1h-V(s)ds,
then
(2.40)F(-αh,I)=lim⁡x→-αhG(x,I)=0.
By direct calculation,
(2.41)|∂F(x,I)∂x|=|g(x,I)1h-V(x)|≤C0I-k1h-V(x),∂G(x,I)∂x=I-k((V′(x))2+(h-V(x))V′′(x)(V′(x))2)1h-V(x).
Since
(2.42)V′′(x)=(1-x2)2+4x2(1-x2)(1-x2)4>0,
we have
(2.43)(V′(x))2+(h-V(x))V′′(x)(V′(x))2>0.
By |g(x,I)|≤C0I-k, there is C0′=2C0 such that
(2.44)-C′0I-k((V′(x))2+(h-V(x))V′′(x)(V′(x))2)1(h-V(x))≤∂F(x,I)/∂x2h-V(x)≤C0′I-k((V′(x))2+(h-V(x))V′′(x)(V′(x))2)1h-V(x),
that is
(2.45)-C0′∂G(x,I)∂x≤∂F(x,I)∂x≤C0′∂G(x,I)∂x,
which means that
(2.46)-C0′G(x,I)≤F(x,I)≤C0′G(x,I).
That is,
(2.47)|V′(x)∫-αhxg(s,I)1h-V(x)|≤CI-kh-V(x).
Thus, we complete the proof (2.37).
Now, we prove (2.38). Let
(2.48)G(x,I)=I-k(αh+x)h-V(x),F(x,I)=∫-αhxg(s,I)1h-V(s)ds.
Then, we have
(2.49)F(-αh,I)=lim⁡x→-αh⁡G(x,I)=0.
By direct computation, we have
(2.50)|∂F(x,I)∂x|=|g(x,I)1(h-V(x))|≤C0I-k1h-V(x),∂G(x,I)∂x=I-k(1+V′(x)(αh+x)2(h-V(x)))1h-V(x).
Since V′′(x)>0 and h=V(-αh+x), it follows that for
(2.51)|V′(x)(αh)2(h-V(x))|≤12,
so for C1′>2C0+1, we have
(2.52)-C1′∂G(x,I)∂x≤∂F(x,I)∂x≤C1′∂G(x,I)∂x,
which means that
(2.53)-C1′G(x,I)≤F(x,I)≤C1′G(x,I).
By the definition of G(x,I), we have
(2.54)|h-V(x)F(x,I)|≤C1′I-k(αh+x).
Thus, we complete the proof of (2.38) and Lemma 2.5.By Lemma2.5, we have the following Lemma which is important to our estimation.Lemma 2.6.
One can find a constantC such that, for -αh≤x≤0,
(2.55)|∂Ik(∫-αhxL(x,I)h-V(s)ds)|≤CV′(x)I-(k+1)h-V(x),(2.56)|∂Ik(∫-αhxL(x,I)h-V(s)ds)|≤Ch-V(x)I-(k+1)(αh+x),
where 0≤k≤6.Proof.
Whenk=1, we have
(2.57)∂I(∫-αhxL1h-V(s)ds)=∫-αhxL(L)1h-V(s)ds+L∫-αhxL1h-V(s)ds.
By the definitions of L and ℒ, we have |L|≤CI-1 and |ℒ(L)|≤CI-2. By (2.37), we obtain
(2.58)∫-αhxL1h-V(s)ds≤CV′I-2h-V.
Suppose that k=l, we have
(2.59)∂Il(∫-αhxL1h-V(s)ds)≤CV′I-(l+1)h-V.
We now proven that when k=l+1,
(2.60)∂Il+1(∫-αhxL1h-V(s)ds)≤CV′I-(l+2)h-V.
By direct computation, we have
(2.61)∂Il+1(∫-αhxLh-V(s)ds)={∂In(Lm(f))∂Il-m-n(∫-αhxLh-V(s)ds),∫-αhxLl+1(L)h-V(s)ds},
where {f1,f2,…,fn} denotes linear combination of functions f1,…,fn with integer coefficients and 0≤m+n≤l.
Since|ℒ(L)|≤CI-2, we have ℒl+1(L)≤CI-(l+2). By (2.37), we obtain
(2.62)|∫-αhxLl+1(L)h-V(s)ds|≤CV′I-(l+2)h-V.
By direct computation, we get
(2.63)∂n(Lm(L))≤CI-(m+n+1).
By assumption (2.59), we have
(2.64)|∂Il-m-n(∫-αhxLh-V(s)ds)|≤CV′I-(l-m-n+1)h-V.
So, we have
(2.65)|∂In(Lm(f))∂Il-m-n(∫-αhxLh-V(s)ds)|≤CV′I-(l+2)h-V.
By (2.62) and (2.65), we have
(2.66)|∂Il+1(∫-αhxLh-V(s)ds)|≤CV′I-(l+2)h-V.
Thus, we have proved (2.55).
The inequality (2.56) can be proved by (2.38), and the process of proof is similar to that of (2.55), so we omit it.
Thus, we have proved Lemma2.6.Now, we give the estimates ofx(I,θ) and y(I,θ).Lemma 2.7.
ForI sufficient large and -αh≤x<0,  the following estimates hold:
(2.67)|In∂nx(I,θ)∂In|≤C|x(I,θ)+1|,|In∂ny(I,θ)∂In|≤C|y(I,θ)|,for0≤n≤6.Proof.
We now prove the first inequality. It is sufficient to prove that(2.68)In∂nx(I,θ)∂In|≤C(1+x).
Case k=1. Differentiating (2.30) by I and noting 1/T0=hI, we have
(2.69)0=hII∫-αhx12(h(x,y)-V(s))ds+hI∂I(∫-αhx12(h(x,y)-V(s))ds).
Now, we choose V(s)/h=η as the new variable of integration, so
(2.70)∂I(∫-αhx12(h(x,y)-V(s))ds)=∂I(∫0V(x)/hhV′(s(η,t,h))12(1-η)dη)=V′xIh-VhIh2h1/2V′12(1-η)+∫0V(x)/h(1/2)h-1/2V′hI-h1/2V′′(∂s/∂I)(V′)212(1-η)dη=xI2(h-V)-hIh2(h-V)VV′+hIh+∫0V(x)/h(1/2)h-1/2V′hI-h1/2V′′(∂s/∂I)(V′)2h2(h-V)V′hds.
Observing that
(2.71)∂s∂I=hIηV′,W′=1-VV′′(V′)2,
and simplifying, we have
(2.72)(2.70)=xI2(h-V)-hIh2(h-V)VV′+hIh∫0x(W′-12)ds2(h-V).
By (2.69)–(2.72), we have
(2.73)xI(θ,I)=2(h-V)∫-αhxL(s,I)12(h-V(s))ds+hIhW(x).
Since |L|≤CI-1, by Lemma 2.6, we have
(2.74)|(2(h-V))∫-αhxL(s,I)12(h-V(s))ds|≤CI-1(αh+x).
We observe that
(2.75)W(x)=12x(1-x)(1+x),
where -1<x<0, so |W(x)|≤C(1+x), which means that
(2.76)|hIhW(x)|≤CI-1(1+x).
By (2.74) and (2.76), we have |xI|≤I-1(1+x).
We suppose that(2.77)|∂Ik-1x|≤I-(k-1)(1+x),
where 1≤k≤6. We will prove |∂Ikx|≤I-k(1+x).
For this purpose, we firstly estimate2(h-V(x)). We differentiate 2(h-V(x)) in (2.73) and using (2.73), then we obtain
(2.78)ddI2(h-V(x))=hIh2(h-V(x))-V′(x)2∫-αhxL(s,I)12(h-V(s))ds.
Since |hI/h|≤CI-1, we have
(2.79)|hIh2(h-V(x))|≤CI-12(h-V(x)).
By (2.37), we have
(2.80)|V′(x)2∫-αhxL(s,I)12(h-V(s))ds|≤CI-12(h-V(x)).
By (2.78)–(2.80), we have
(2.81)|∂I(2(h-V(x)))|≤CI-12(h-V(x)).
We suppose that whenn≤k-1,
(2.82)|∂In(2(h-V(x)))|≤CI-n2(h-V(x)).
We will prove that whenn=k,
(2.83)|∂In(2(h-V(x)))|≤CI-n2(h-V(x)).
By direct computation, we have
(2.84)∂Ik(2(h-V(x)))={V(k1)∂Il1x⋯∂Ilk1-1x∂Ik3(∫-αhxL(s,I)12(h-V(s))ds),∂Im(hIh)∂Ik-m-1(2(h-V(x)))(∫-αhxL(s,I)12(h-V(s)ds)},
where l1+⋯+lk1-1=k2,k2+k3=k-1 and k2<k1<k,k3<k.
By the assume (2.77), we have
(2.85)|∂Il1x⋯∂Ilk1-1x|≤(1+x)k1-1I-k2.
By (2.55), we get
(2.86)|∂Ik3(∫-αhxL(s,I)12(h-V(s))ds)|≤CV′I-k32(h-V(x))
By (2.85), (2.86) and noting the fact that
(2.87)(1+x)k-1∂Ik(V)≤C(|V′|+(1+x)k1+1),-1<x<0,
we obtain
(2.88)|∑V(k1)∂Il1x⋯∂Ilk1-1x∂Ik3(∫-αhxL(s,I)12(h-V(s))ds)|≤∑CV′V(k1)(1+x)k1-1I-(k2+k3+1)2(h-V(x))≤∑[|V′|+(1+x)k1+1]V′I-(k)2(h-V(x))≤CI-(k)2(h-V(x)).
By assumption (2.82), we have
(2.89)|∂Ik-m-1(2(h-V(x)))|≤CIk-m-1(2(h-V(x))).
By (2.89) and the fact that |∂Im(hI/h)|≤CI-(m+1), we have
(2.90)|∑∂Im(hIh)∂Ik-m-1(2(h-V(x)))|≤CI-k2(h-V(x)).
So, by (2.88) and (2.90), we have proved (2.83).
By (2.56), we have
(2.91)|∂Im(∫-αhxL(x,I)2(h-V(s))ds)|≤C2(h-V(x))I-(m+1)(αh+x).
By (2.83) and (2.91), we have
(2.92)|∂n(2(h-V(x)))∂m(∫-αhxL(s,I)12(h-V(s))ds)|≤CI-(m+n+1)(αh+x)≤CI-(k+1)(αh+x).
By the assumption (2.77), the fact that ∂Im3(hI/h)≤CI-(m3+1), and noting that
(2.93)∂I(hIhW)=∑kW(m1)∂Il1x⋯∂Ilm1x∂Im3(hIh),
where m2<m1,l1+⋯+lm1=m2,m2+m3=k, we have
(2.94)∂I(hIhW)≤∑kW(m1)I-(l1+⋯+lm1)(1+x)m1I-(m3+1)≤C∑kW(m1)(1+x)m1I-(m2+m3+1)≤C(1+x)I-(k+1).
By (2.92) and (2.94), we have
(2.95)∂IkxI=∑k∂n(2(h-V(x)))∂m(∫-αhxL(s,I)12(h-V(s))ds)+∂k(hIhW(x))≤CI-(k+1)(1+x)≤CI-(k+1)(1+x),
which means
(2.96)∂Ikx≤CI-k(1+x).
We now prove(2.97)|In∂ny(I,θ)∂In|≤cI.
Since
(2.98)h(I)=12y2+V(x),
we have
(2.99)y=±2(h(I)-V(x)).
we have proved (2.83), so we have
(2.100)|In∂ny(I,θ)∂In|≤C⋅I-ny,
which means that
(2.101)|In∂ny(I,θ)∂In|≤C⋅h0(I).
The proof of Lemma 2.7 is complete.Remark 2.8.
Lemma2.7 also holds when x≥0. Since the idea and the process of the proof is more easily than that of Lemma 2.7, we omit the details.Now, we give the estimate ofH1(I,θ,t).Lemma 2.9.
(2.102)Ik|∂Ik∂tl∂θiH1(I,θ,t)∂Ik∂tl∂θi|≤C⋅h0(I),k+i≤7,i=0,1.Proof.
This lemma can be proved easily form the definition ofH1 and |y(θ,I)|≤2h0(I).


## 2.2. New Action and Angle Variables

Now, we are concerned with the Hamiltonian system (2.32) with Hamiltonian function H(θ,I,t) given by (2.33). Note that
(2.103)Idθ-Hdt=-(Hdt-Idθ).
This means that if one can solve I form (2.32) as a function of H (θ and t as parameters), then
(2.104)dHdθ=-∂I∂t(t,H,θ),dtdθ=-∂I∂H(t,H,θ)
is also a Hamiltonian system with Hamiltonian function I, and now, the action, angle, and time variables are H, t, and θ.Form Remarks2.3 and 2.4, we have
(2.105)∂H∂I⟶1,asI⟶+∞.
Hence, by the implicit function theorem, there is a function I(t,H,θ) such that
(2.106)H(θ,I(t,H,θ),t)=H.
By Lemma 2.9, we have
(2.107)H1(θ,I,t)H0(I)⟶0,asI→+∞.
So, there is a function R(t,H,θ) with |R|≤(1/2)H such that
(2.108)I(t,H,θ)=I0(H-R(t,H,θ)),forH⟶+∞.
Let
(2.109)I1(t,H,θ)=I0(H-R(t,H,θ))-I0(H)=∫0πI0′(H-sR(t,H,θ))R(t,H,θ)dθ.
Then,
(2.110)I(t,H,θ)=I0(H)+I1(t,H,θ).
From Remark 2.3, we have known the estimate of I0(H), so we need to give the estimate of I1(t,H,θ). For this propose, we need firstly the following Lemma on the estimate of R(t,H,θ).Lemma 2.10.
The functionR(t,H,θ) possesses the following estimates:
(2.111)Hk|∂k+lR(H,t,θ)∂kH∂lt|≤H1/2,
for k+l≤6.Proof.
From (2.108) and (2.110), it follows that
(2.112)R(t,H,θ)=H1(θ,I0(H-R),t).
When k+l=0, by Lemma 2.9, we have
(2.113)|R(t,H,θ)|=|H1(θ,I0(H-R),t)|≤C⋅h0(I0(H-R))≤C⋅H-R≤C⋅H.
When k+l=1, we first denote
(2.114)Δ=∂H1∂I(θ,I0(H-R),t)I0′(H-R).
By Remark 2.3, we observe that I0 is increasing and
(2.115)I0(12H)≤I0(H)≤I0(32H)≤C⋅I0(12H).
By Lemma 2.9 and (2.115), we have
(2.116)|Δ|=|∂H1∂I(θ,I0(H-R),t)⋅I0′(H-R)|≤C⋅1I0(H-R)⋅h0(I0(H-R))⋅I0′(H-R)≤C⋅1I0(H/2)⋅32H⋅1(3/2)H⋅I0(32H)≤C⋅H-1/2≤12.
So,
(2.117)1+Δ≥12.
By direct computation, we have
(2.118)|∂R∂H|=|(∂H1/∂I)(θ,I0(H-R),t)I0′(H-R)1+Δ|≤C⋅|∂H1∂I(θ,I0(H-R),t)I0′(H-R)|≤C⋅H-1/2,|∂R∂t|=|(∂H1/∂t)(I0(H-R))1+Δ|≤C⋅|∂H1∂t(I0(H-R))|≤C⋅H1/2.
By (2.118), we can easily get
(2.119)Hk|∂k+lR(H,t,θ)∂kH∂lt|≤H1/2.
When k+l≥2, one may get
(2.120)∂k+lR(H,t,θ)∂kH∂lt=∑cn,j1⋯jn∂nH1∂nI0j1∂j1I0(H-R)∂Hj1⋯∂jnI0(H-R)∂Hjn,
where 1≤n≤k, j1+⋯+jn<k. It is easy to verify
(2.121)|∂k+lR(H,t,θ)∂kH∂lt|≤C⋅H(1/2)-k,
for k+l≥2. This complete the proof.Now, we give the estimates ofI1(t,H,θ).Lemma 2.11.
The functionI1(t,H,θ) possesses the following estimates:
(2.122)Hk|∂k+lI1(H,t,θ)∂kH∂lt|≤H1/(1+α),
for k+l≤6.Proof.
Whenk=l=0. By Remark 2.3 and |R|≤(1/2)H, we have
(2.123)I0′(H-sR(t,H,θ))≤C⋅(H-sR)1/(1+α)(H-sR)1/2≤C⋅((3/2)H)1/(1+α)((1/2)H)1/2≤C⋅H1/(1+α)-(1/2).
By Lemma 2.10, we know
(2.124)|R|≤C⋅H1/2.
Since
(2.125)I1(t,H,θ)=∫0πI0′(H-sR(t,H,θ))R(t,H,θ)dθ,
it is easy to get that |I1|≤C·H1/(1+α).
Whenk+l≥1. By direct computation, we have
(2.126)∂k+lI1(t,H,θ)∂Hk∂tl=∑∫0π∂k1+l1I0′(t,H,θ)∂Hk1∂tl1∂k2+l2R(t,H,θ)∂Hk2∂tl2ds,
where k1+k2=k, l1+l2=l. Now, we need to estimate the first term of the integrand. The following equality is important:
(2.127)∂k1+l1I0′(t,H,θ)∂Hk1∂tl1=∑I0(p+q+1)(H-sR)∂m1u∂Hm1⋯∂mpu∂Hmp⋅∂j1+n1u∂Hj1∂tn1⋯∂jq+nqu∂Hjq∂tnq,
where u=H-sR, p≤m, q≤n, n1,…,nq>0, m1,…,mp>0, n1+⋯+nq=n, m1+⋯+mp+j1+⋯+jq=m. Assume that there are β(≤p) members: m1,…,mβ in {m1,…,mp} which equal to 1. Noting that
(2.128)|∂u∂H|=|1+s∂R∂H|≤C,|∂ku∂Hk|≤|∂kR∂Hk|≤C⋅H(1/2)-k,k>1,|∂k+lu∂Hk∂tl|≤|∂k+lR∂Hk∂tl|≤C⋅H(1/2)-k,l>0.
By the above discussions, we have
(2.129)|∂k1+l1I0′(t,H,θ)∂Hk1∂tl1|≤C⋅(H-sR)1/(1+α)+(1/2)(H-sR)p+q+1⋅H(1/2)(p-β)-(mβ+1+⋯+mp)⋅H(1/2)q-(j1+⋯+jq)≤CH(1/2)+1/(1+α)Hp+q+1⋅H(1/2)(p+q-β)H-(m-β)≤CH-(1/2)+1/(1+α)-m⋅H(p+q-β)((1/2)-1)≤CH-(1/2)+1/(1+α)-m.
By Lemma 2.10, we have known
(2.130)Hk|∂k+lR(H,t,θ)∂kH∂lt|≤H1/2,
then we have
(2.131)Hk|∂k+lI1(H,t,θ)∂kH∂lt|≤CH1/(1+α).


## 2.3. Proof of the Main Result

Up to now, we have given an equivalent form of (1.5), that is, the system (2.32), which is expressed in the action and angle variables (H,t). In this section, we first introduce some transformations such that in the transformed system, the perturbation terms of (2.32) depending on the new angle variable are very small if the new action variable is sufficiently large and then prove, by Moser’s twist theorem, the statement of Theorem 1.1.Lemma 2.12.
There is a canonical transformationΨ:(λ,τ)→(H,t) of the form
(2.132)Ψ:H=λ+U(τ,λ,θ),t=τ+V(τ,λ,θ),
where the functions U and V are 1-periodic in θ and satisty
(2.133)U(τ,λ,θ)λ,V(τ,λ,θ)⟶0asλ⟶∞,
uniformly for (τ,θ)∈T2 such that under this mapping, the system and the Hamiltonian function I in (2.110) is changed into the form
(2.134)dλdθ=-∂K∂τ(τ,λ,θ),dτdθ=∂K∂H(τ,λ,θ),
where
(2.135)K(τ,λ,θ)=I0(λ)+[I1](λ,θ)+M(τ,λ,θ),
with
(2.136)[I1](λ,θ)=∫01I1(t,λ,θ)dt.
Moreover, the new perturbation ℳ possesses the estimate
(2.137)|∂k+l∂λk∂τlM(τ,λ,θ)|≤C⋅λ-k+(1/2).Proof.
We will look for the required transformationΨ by a generating function ℱ(t,λ,θ) in the following way:
(2.138)H=λ+∂F∂t(t,λ,θ),τ=t+∂F∂λ(t,λ,θ),
where the function ℱ will be given later. Under Ψ, the transformed system of (2.104) is of the form
(2.139)dλdθ=-∂K∂τ(τ,λ,θ),dτdθ=-∂K∂λ(τ,λ,θ),
where
(2.140)K(τ,λ,θ)=I0(λ+∂F∂t)+I1(t,λ+∂F∂t,θ)+∂F∂θ.
By Taylor’s formula, one can write
(2.141)K(τ,λ,θ)=I0(λ)+I0′(λ)∂F∂t+I1(t,λ,θ)+M(τ,λ,θ),
where
(2.142)M(τ,λ,θ)=∂F∂θ+∫01(1-s)I0′′(λ+s∂F∂t)⋅(∂F∂θ)2ds+∫01∂I1∂H(t,λ+s∂F∂t,θ)⋅∂F∂tds.
We chooseℱ(2.143)F(t,λ,θ)=-∫0t1I0′̅(λ)⋅(I1(t,λ,θ)-[I1](λ,θ)).
Then, 𝒦 is of the form (2.135).
We now show thatℳ satisfies (2.137). From Remark 2.3 and Lemma 2.11, it follows that
(2.144)|∂k+k+i∂λk∂tl∂θiF(t,λ,θ)|≤C⋅λ-k+(1/2),
for k+i+l≤6 and i=0,1. In particular
(2.145)|∂2∂λ∂tF(t,λ,θ)|≤Cλ-1/2≤12,
if λ≫1. So we can solve the second equation of (2.138) for t,
(2.146)t=τ+V(τ,λ,θ),
where the function V satisfies
(2.147)V(τ,λ,θ)=-∂F∂λ(τ+V,λ,θ).
Set
(2.148)U(τ,λ,θ)=∂F∂t(τ+V,λ,θ).
Then, the canonical transformation Ψ is of the form (2.132). Moreover, similar to the proof of [5, Lemma  2], we can verify that
(2.149)|∂k+l∂λk∂τlU(τ,λ,θ)|≤C⋅λ-k+(1/2),|∂k+l∂λk∂τlV(τ,λ,θ)|≤C⋅λ-k-(1/2),
for k+l≤5 and U/λ,V→0 as λ→+∞. Let
(2.150)ϕ1(τ,λ,θ)=∂F∂θ(τ+V,λ,θ),ϕ2(τ,λ,θ)=∫01(1-s)I0′′(λ+sU)⋅U2ds,ϕ3(τ,λ,θ)=∫01∂I1∂H(τ+V,λ+sU,θ)⋅Uds.
It is not difficult to prove that
(2.151)|∂k+l∂λk∂τlϕ1(τ,λ,θ)|≤C⋅λ-k+(1/2),|∂k+l∂λk∂τlϕ1(τ,λ,θ)|≤C⋅λ-k+(1/(1+α))-(1/2),|∂k+l∂λk∂τlϕ1(τ,λ,θ)|≤C⋅λ-k+(1/(1+α))-(1/2),
for k+l≤5. Note that 0<α<1, we have 1/(1+α)-1/2<1/2.
Hence, we have(2.152)|∂k+l∂λk∂τlM(τ,λ,θ)|≤λ-k+(1/2).
The proof of Lemma 2.12 is complete.
Forλ0>0, we denote by Aλ0 the domain
(2.153)Aλ0={(λ,τ,θ)∣λ≥λ0,(τ,θ)∈T2}.Lemma 2.13.
The Poincaré mappingP of (2.134) has the intersection property on AH0; that is, if Γ is an embedded circle in AH0 homotopic to a circle λ= const. in AH0, then P(Γ)⋂Γ≠∅.Proof.
The proof can be found in [5].
Definite a diffeomorphismΨ1:AH0×𝕊1→AH0×𝕊1(2.154)ν=I0′(H),τ=τ,θ=θ.
Then, the system (2.134) under the transformation Ψ1 becomes
(2.155)dνdθ=f1(λ,t,θ),dtdθ=ν+f2(λ,t,θ),
where
(2.156)f1(τ,ν,θ)=-I0′′(λ)∂M∂τ(τ,λ,θ),f2(τ,ν,θ)=∂[I1]∂λ(λ,θ)+∂M∂λ(τ,λ,θ),
with λ=λ(ν) defined through the transformation Ψ1.
Now, we estimatef1 and f2. Since
(2.157)c⋅λ1/(1+α)-(1/2)≤I0′(λ)≤C⋅λ1/(1+α)-(1/2),|∂[I1]∂λ|≤C⋅λ1/(1+α)-1,
we have
(2.158)c⋅λ2(1+α)/(1-α)≤λ(ν)≤C⋅λ2(1+α)/(1-α),
then
(2.159)λ≫1⟺ν≫1.
Moreover, we have
(2.160)|∂k∂νkλ(ν)|≤C⋅ν-kλ(ν).
When 0<α<1, we have
(2.161)|∂k+l∂λk∂τlM(τ,λ,θ)|≤C⋅λ-k+(1/2),12<11+α<1,
so
(2.162)|∂k+l∂νk∂τlf1(τ,ν,θ)|≤C⋅ν-k⋅λ-2⋅ν-kI0(λ)⋅λ1/2≤C⋅ν-kλ-α/(1+α)≤C⋅ν-k+(-4α/(1-α))≤C⋅ν-k-σ,|∂k+l∂νk∂τlf2(τ,ν,θ)|≤C⋅ν-k⋅(λ-1/2+λ1/(1+α)-1)≤C⋅ν-k⋅λ1/(1+α)-1≤C⋅ν-k-(2α/(1-α))≤C⋅ν-k-σ,
for all k+l≤4, where σ=min{4α/(1-α),2α/(1-α)}=2α/(1-α)>0.
Sincef1 and f2 are sufficiently small as ν≫1, all solution of (2.155) exist for 0≤θ≤1 when the initial values ν(0)=λ are sufficiently large. Hence, the Poincaré map Φ associated to (2.155) is well defined on Aλ0 as λ0≫1. In fact, by integrating (2.155) from θ=0 to θ=π, we see that Φ has the form
(2.163)Φ:τ1=τ0+μ0+Ξ1(t0,λ0),ν1=ν0+Ξ2(t0,λ0),
where Ξ1 and Ξ2 satisfy the same estimates as those of f1 and f2; that is,
(2.164)∂νk∂τlΞi≤ν-σ,
where i=1,2,0≤k+l≤4.
SinceΦ satisfies all the assumptions of Moser’s twist theorem [9], from which we conclude that for any ω≫1 satisfying
(2.165)|ω-pq|≥c0|q|-5/2,pq∈Q.
There is an invariant curve Γω of Φ which is conjugated to pure rotation of the circle with rotation number ω. Tracing back to the system (2.32), Γω gives rise to an invariant closed curve of the Poincaré map Φ of (2.32) with rotation number 1/ω which surrounds and is arbitrarily far away form the origin. Hence, all solutions of (1.5) are bounded. This completes the proof of the Theorem.



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*Source: 106214-2011-09-18.xml*