Boundedness of solutions for a class of second-order differential equation with singularity

Boundary Value Problems20132013:84

DOI: 10.1186/1687-2770-2013-84

Received: 14 September 2012

Accepted: 22 March 2013

Published: 10 April 2013

Abstract

In this paper, we study the following second-order periodic system:

x + V ( x ) + p ( t ) | x | α = 0 , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-84/MediaObjects/13661_2012_Article_338_Equa_HTML.gif

where V ( x ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-84/MediaObjects/13661_2012_Article_338_IEq1_HTML.gif has a singularity. Under some assumptions on the V ( x ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-84/MediaObjects/13661_2012_Article_338_IEq1_HTML.gif and p ( t ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-84/MediaObjects/13661_2012_Article_338_IEq2_HTML.gif by Ortega’s small twist theorem, we obtain the existence of quasi-periodic solutions and boundedness of all the solutions.

Keywords

boundedness of solutions singularity small twist theorem

1 Introduction and main result

In 1991, Levi [1] considered the following equation:
x + V ( x , t ) = 0 , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-84/MediaObjects/13661_2012_Article_338_Equ1_HTML.gif
(1.1)

where V ( x , t ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-84/MediaObjects/13661_2012_Article_338_IEq3_HTML.gif satisfies some growth conditions and V ( x , t ) = V ( x , t + 1 ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-84/MediaObjects/13661_2012_Article_338_IEq4_HTML.gif. 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 nonlinearity V ( x , t ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-84/MediaObjects/13661_2012_Article_338_IEq3_HTML.gif can guarantee the twist condition of KAM theorem. Later, several authors improved the Levi’s result; we refer to [24] and the references therein.

Recently, Capietto, Dambrosio and Liu [5] studied the following equation:
x + V ( x ) = F ( x , t ) , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-84/MediaObjects/13661_2012_Article_338_Equ2_HTML.gif
(1.2)
with F ( x , t ) = p ( t ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-84/MediaObjects/13661_2012_Article_338_IEq5_HTML.gif is a π-periodic function and V ( x ) = 1 2 x + 2 + 1 ( 1 x 2 ) ν 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-84/MediaObjects/13661_2012_Article_338_IEq6_HTML.gif, where x + = max { x , 0 } http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-84/MediaObjects/13661_2012_Article_338_IEq7_HTML.gif, x = max { x , 0 } http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-84/MediaObjects/13661_2012_Article_338_IEq8_HTML.gif and ν is a positive integer. Under the Lazer-Leach assumption that
1 + 1 2 0 π p ( t 0 + θ ) sin θ d θ > 0 , t 0 R , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-84/MediaObjects/13661_2012_Article_338_Equ3_HTML.gif
(1.3)

they prove the boundedness of solutions and the existence of quasi-periodic solution by KAM theorem. It is the first time that the equation of the boundedness of all solution is treated in case of a singular potential.

We observe that F ( x , t ) = p ( t ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-84/MediaObjects/13661_2012_Article_338_IEq5_HTML.gif in (1.2) is smooth and bounded, so a natural question is to find sufficient conditions on F ( x , t ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-84/MediaObjects/13661_2012_Article_338_IEq9_HTML.gif such that all solutions of (1.2) are bounded when F ( x , t ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-84/MediaObjects/13661_2012_Article_338_IEq9_HTML.gif 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 ( t ) | x | α = 0 , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-84/MediaObjects/13661_2012_Article_338_Equ4_HTML.gif
(1.4)
where p ( t ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-84/MediaObjects/13661_2012_Article_338_IEq2_HTML.gif is a π-periodic function,
V ( x ) = 1 2 x + 2 + 1 1 x 2 1 , 0 < α < 1 , x > 1 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-84/MediaObjects/13661_2012_Article_338_Equ5_HTML.gif
(1.5)
We suppose Lazer-Leach assumption hold:
0 π p ( t 0 + θ ) ( sin θ ) 1 + α d θ > 0 , t 0 R . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-84/MediaObjects/13661_2012_Article_338_Equ6_HTML.gif
(1.6)

Our main result is the following theorem.

Theorem 1 Under the assumptions (1.5) and (1.6), all the solutions of (1.4) are bounded.

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]. It mainly consists of two steps. The first one is to transform (1.4) into a perturbation of integrable Hamilton system. The second one is to show that Poincaré map of the equivalent system satisfies Ortega’s twist theorem, then some desired result can be obtained.

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

Theorem 2 Assume that p ( t ) C ( R ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-84/MediaObjects/13661_2012_Article_338_IEq10_HTML.gif satisfies (1.6); then, there is an ϵ 0 > 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-84/MediaObjects/13661_2012_Article_338_IEq11_HTML.gif such that, for any ω ( 1 π , 1 π + ϵ 0 ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-84/MediaObjects/13661_2012_Article_338_IEq12_HTML.gif, t equation (1.4) has a solution ( x ω ( t ) , x ω ( t ) ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-84/MediaObjects/13661_2012_Article_338_IEq13_HTML.gif of the Mather type with rotation number ω. More precisely:

Case 1: ω = p q http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-84/MediaObjects/13661_2012_Article_338_IEq14_HTML.gif is rational. The solutions ( x ω ( t + 2 i π ) , x ω ( t + 2 i π ) ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-84/MediaObjects/13661_2012_Article_338_IEq15_HTML.gif, 1 i q 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-84/MediaObjects/13661_2012_Article_338_IEq16_HTML.gif are independent periodic solutions of periodic ; moreover, in this case,
lim q min t R ( | x ω ( t ) | + | x ω ( t ) | ) = + . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-84/MediaObjects/13661_2012_Article_338_Equb_HTML.gif

Case 2: ω is irrational. The solution ( x ω ( t ) , x ω ( t ) ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-84/MediaObjects/13661_2012_Article_338_IEq17_HTML.gif is either a usual quasi-periodic solution or a generalized one.

We will apply Aubry-Mather theory, more precisely, the theorem in [8], to prove this theorem.

2 Proof of theorem

2.1 Action-angle variables and some estimates

Observe that (1.4) is equivalent to the following Hamiltonian system:
x = H y , y = H x http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-84/MediaObjects/13661_2012_Article_338_Equ7_HTML.gif
(2.1)
with the Hamiltonian function
H ( x , y , t ) = 1 2 y 2 + V ( x ) + p ( t ) ( α + 1 ) | x | α x . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-84/MediaObjects/13661_2012_Article_338_Equc_HTML.gif
In order to introduce action and angle variables, we first consider the auxiliary autonomous equation:
x = y , y = V ( x ) , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-84/MediaObjects/13661_2012_Article_338_Equ8_HTML.gif
(2.2)
which is an integrable Hamiltonian system with Hamiltonian function
H 1 ( x , y , t ) = 1 2 y 2 + V ( x ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-84/MediaObjects/13661_2012_Article_338_Equd_HTML.gif

The closed curves H 1 ( x , y , t ) = h > 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-84/MediaObjects/13661_2012_Article_338_IEq18_HTML.gif are just the integral curves of (2.2).

Denote by T 0 ( h ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-84/MediaObjects/13661_2012_Article_338_IEq19_HTML.gif the time period of the integral curve Γ h http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-84/MediaObjects/13661_2012_Article_338_IEq20_HTML.gif of (2.2) defined by H 1 ( x , y , t ) = h http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-84/MediaObjects/13661_2012_Article_338_IEq21_HTML.gif and by I the area enclosed by the closed curve Γ h http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-84/MediaObjects/13661_2012_Article_338_IEq20_HTML.gif for every h > 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-84/MediaObjects/13661_2012_Article_338_IEq22_HTML.gif. Let 1 < α h < 0 < β h http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-84/MediaObjects/13661_2012_Article_338_IEq23_HTML.gif be such that V ( α h ) = V ( β h ) = h http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-84/MediaObjects/13661_2012_Article_338_IEq24_HTML.gif. It is easy to see that
I 0 ( h ) = 2 α h β h 2 ( h V ( s ) ) d s , h > 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-84/MediaObjects/13661_2012_Article_338_Eque_HTML.gif
and
T 0 ( h ) = I 0 ( h ) = 2 α h β h 1 2 ( h V ( s ) ) d s , h > 0 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-84/MediaObjects/13661_2012_Article_338_Equf_HTML.gif
By direct computation, we get
I 0 ( h ) = 2 0 β h 2 ( h V ( s ) ) d s + 2 α h 0 2 ( h V ( s ) ) d s = π h + 2 0 α h 2 ( h V ( s ) ) d s , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-84/MediaObjects/13661_2012_Article_338_Equg_HTML.gif
so
T 0 ( h ) = π + 0 α h 1 2 ( h V ( s ) ) d s . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-84/MediaObjects/13661_2012_Article_338_Equh_HTML.gif
We then have
I 0 ( h ) = I ( h ) + I + ( h ) , T 0 ( h ) = T ( h ) + T + ( h ) , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-84/MediaObjects/13661_2012_Article_338_Equi_HTML.gif
where
I ( h ) = 2 0 α h 2 ( h V ( s ) ) d s , I + ( h ) = π h , T ( h ) = 2 0 α h 1 2 ( h V ( s ) ) d s , T + ( h ) = π . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-84/MediaObjects/13661_2012_Article_338_Equj_HTML.gif

Similar in estimating in [5], we have the estimation of functions I http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-84/MediaObjects/13661_2012_Article_338_IEq25_HTML.gif and T http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-84/MediaObjects/13661_2012_Article_338_IEq26_HTML.gif.

Lemma 1 We have
h n | d n T ( h ) d h n | C h 1 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-84/MediaObjects/13661_2012_Article_338_Equk_HTML.gif
and
h n | d n I ( h ) d h n | C h 1 2 , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-84/MediaObjects/13661_2012_Article_338_Equl_HTML.gif

where n = 0 , 1 , , 6 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-84/MediaObjects/13661_2012_Article_338_IEq27_HTML.gif, h + http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-84/MediaObjects/13661_2012_Article_338_IEq28_HTML.gif. Note that here and below we always use C, C 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-84/MediaObjects/13661_2012_Article_338_IEq29_HTML.gif or C 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-84/MediaObjects/13661_2012_Article_338_IEq30_HTML.gif to indicate some constants.

Remark 1 It follows from the definitions of T + ( h ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-84/MediaObjects/13661_2012_Article_338_IEq31_HTML.gif, T ( h ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-84/MediaObjects/13661_2012_Article_338_IEq32_HTML.gif and Lemma 1 that
lim h + T ( h ) = 0 , lim h + T + ( h ) = π . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-84/MediaObjects/13661_2012_Article_338_Equm_HTML.gif

Thus the time period T 0 ( h ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-84/MediaObjects/13661_2012_Article_338_IEq19_HTML.gif is dominated by T + ( h ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-84/MediaObjects/13661_2012_Article_338_IEq31_HTML.gif when h is sufficiently large. From the relation between T ( h ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-84/MediaObjects/13661_2012_Article_338_IEq32_HTML.gif and I ( h ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-84/MediaObjects/13661_2012_Article_338_IEq33_HTML.gif, we know I 0 ( h ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-84/MediaObjects/13661_2012_Article_338_IEq34_HTML.gif is dominated by I + ( h ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-84/MediaObjects/13661_2012_Article_338_IEq35_HTML.gif when h is sufficiently large.

Remark 2 It also follow from the definition of I ( h ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-84/MediaObjects/13661_2012_Article_338_IEq36_HTML.gif, I ( h ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-84/MediaObjects/13661_2012_Article_338_IEq33_HTML.gif, I + ( h ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-84/MediaObjects/13661_2012_Article_338_IEq35_HTML.gif and Remark 1 that
| h n d n I 0 ( h ) d h n | C 0 I 0 ( h ) for  n 1 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-84/MediaObjects/13661_2012_Article_338_Equn_HTML.gif
Remark 3 Note that h = h 0 ( I 0 ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-84/MediaObjects/13661_2012_Article_338_IEq37_HTML.gif is the inverse function of I 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-84/MediaObjects/13661_2012_Article_338_IEq38_HTML.gif. By Remark 2, we have
| I n d n h ( I ) d I n | C 0 h ( I ) for  n 1 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-84/MediaObjects/13661_2012_Article_338_Equo_HTML.gif

We now carry out the standard reduction to the action-angle variables. For this purpose, we define the generating function S ( x , I ) = C 2 ( h V ( s ) ) d s http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-84/MediaObjects/13661_2012_Article_338_IEq39_HTML.gif, where C is the part of the closed curve Γ h http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-84/MediaObjects/13661_2012_Article_338_IEq20_HTML.gif connecting the point on the y-axis and point ( x , y ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-84/MediaObjects/13661_2012_Article_338_IEq40_HTML.gif.

We define the well-know map ( θ , I ) ( x , y ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-84/MediaObjects/13661_2012_Article_338_IEq41_HTML.gif by
y = S x ( x , I ) , θ = S I ( x , I ) , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-84/MediaObjects/13661_2012_Article_338_Equp_HTML.gif
which is symplectic since
d x d y = d x ( S x x d x + S x I d I ) = S x I d x d I , d θ d I = ( S I x d x + S I I d I ) d I = S I x d d I . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-84/MediaObjects/13661_2012_Article_338_Equq_HTML.gif
From the above discussion, we can easily get
θ = { π T 0 ( h ( x , y ) ) ( T ( h ( x , y ) ) 2 + arcsin x 2 ( h ( x , y ) ) ) , if  x > 0 , y > 0 , π T 0 ( h ( x , y ) ) ( T ( h ( x , y ) ) 2 + π + arcsin x 2 ( h ( x , y ) ) ) , if  x > 0 , y < 0 , π T 0 ( h ( x , y ) ) ( α h x 1 2 ( h ( x , y ) + 1 ( 1 s 2 ) 1 ) d s ) , if  x < 0 , y > 0 , π T 0 ( h ( x , y ) ) ( T 0 ( h ( x , y ) ) α h x 1 2 ( h ( x , y ) + 1 ( 1 s 2 ) 1 ) d s ) , if  x < 0 , y < 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-84/MediaObjects/13661_2012_Article_338_Equ9_HTML.gif
(2.3)
and
I ( x , y ) = I 0 ( h ( x , y ) ) = 2 α h β h 2 ( h ( x , y ) V ( s ) ) d s . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-84/MediaObjects/13661_2012_Article_338_Equ10_HTML.gif
(2.4)
In the new variables ( θ , I ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-84/MediaObjects/13661_2012_Article_338_IEq42_HTML.gif, the system (2.1) is
θ = H I , I = H θ , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-84/MediaObjects/13661_2012_Article_338_Equ11_HTML.gif
(2.5)
where
H ( θ , I , t ) = π h 0 ( I ) + π p ( t ) ( α + 1 ) | x ( I , θ ) | α x ( I , θ ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-84/MediaObjects/13661_2012_Article_338_Equ12_HTML.gif
(2.6)

In order to estimate π p ( t ) ( α + 1 ) | x ( I , θ ) | α x ( I , θ ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-84/MediaObjects/13661_2012_Article_338_IEq43_HTML.gif, we need the following lemma.

Lemma 2 [[5], Lemma 2.2]

For I sufficient large and α h x < 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-84/MediaObjects/13661_2012_Article_338_IEq44_HTML.gif, the following estimates hold:
| I n n x ( I , θ ) I n | c I , | I n n y ( I , θ ) I n | c I for 0 n 6 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-84/MediaObjects/13661_2012_Article_338_Equr_HTML.gif

2.2 New action and angle variables

Now we are concerned with the Hamiltonian system (2.5) with Hamiltonian function H ( θ , I , t ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-84/MediaObjects/13661_2012_Article_338_IEq45_HTML.gif given by (2.6). Note that
I d θ H d t = ( H d t I d θ ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-84/MediaObjects/13661_2012_Article_338_Equs_HTML.gif
This means that if one can solve I from (2.6) as a function of H (θ and t as parameters), then
d H d θ = I t ( t , H , θ ) , d t d θ = I H ( t , H , θ ) , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-84/MediaObjects/13661_2012_Article_338_Equ13_HTML.gif
(2.7)

is also a Hamiltonian system with Hamiltonian function I and now the action, angle and time variables are H, t and θ.

From (2.6) and Lemma 1, we have
H I 1 as  I + . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-84/MediaObjects/13661_2012_Article_338_Equt_HTML.gif
So, we assume that I can be written as
I = I 0 ( H π + R ( H , t , θ ) ) , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-84/MediaObjects/13661_2012_Article_338_Equu_HTML.gif
where R satisfies | R | < H π http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-84/MediaObjects/13661_2012_Article_338_IEq46_HTML.gif. Recalling that h 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-84/MediaObjects/13661_2012_Article_338_IEq47_HTML.gif is the inverse function of I 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-84/MediaObjects/13661_2012_Article_338_IEq38_HTML.gif, we have
H π + R ( H , t , θ ) = h 0 ( I ) , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-84/MediaObjects/13661_2012_Article_338_Equv_HTML.gif
which implies that
R ( H , t , θ ) = p ( t ) ( α + 1 ) | x ( I , θ ) | α x ( I , θ ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-84/MediaObjects/13661_2012_Article_338_Equw_HTML.gif
As a consequence, R is implicitly defined by
R ( H , t , θ ) = p ( t ) ( α + 1 ) | x | α x ( I 0 ( H π + R ( H , t , θ ) ) , θ ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-84/MediaObjects/13661_2012_Article_338_Equ14_HTML.gif
(2.8)
Lemma 3 The function R ( H , t , θ ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-84/MediaObjects/13661_2012_Article_338_IEq48_HTML.gif satisfies the following estimates:
| m R ( H , t , θ ) H m | H α + 1 2 for m + l 6 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-84/MediaObjects/13661_2012_Article_338_Equx_HTML.gif
Proof Case m = 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-84/MediaObjects/13661_2012_Article_338_IEq49_HTML.gif. By (2.8), Lemma 2 and noticing that H I 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-84/MediaObjects/13661_2012_Article_338_IEq50_HTML.gif as I + http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-84/MediaObjects/13661_2012_Article_338_IEq51_HTML.gif, we have
| R ( H , t , θ ) | = | π p ( t ) ( α + 1 ) x α + 1 ( I 0 ( H π + R ( H , t , θ ) ) , θ ) | | I 0 ( H π + R ( H , t , θ ) ) | 1 + α 2 p ( t ) = | I 1 + α 2 | p ( t ) C H 1 + α 2 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-84/MediaObjects/13661_2012_Article_338_Equy_HTML.gif
Case m 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-84/MediaObjects/13661_2012_Article_338_IEq52_HTML.gif. Derivative both sides of (2.8) with respect to H, we have
R H = 1 π 1 1 + 1 x I I 0 ( H π + R ) p ( t ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-84/MediaObjects/13661_2012_Article_338_Equz_HTML.gif
By Remark 2, Lemma 2 and the estimate of R, we have
| x I I 0 ( H π + R ) p ( t ) | C H 1 + α 2 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-84/MediaObjects/13661_2012_Article_338_Equaa_HTML.gif
Since
| 1 x I I 0 ( H π + R ) p ( t ) | 1 as  H + , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-84/MediaObjects/13661_2012_Article_338_Equab_HTML.gif
we have
H | R H | C H 1 + α 2 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-84/MediaObjects/13661_2012_Article_338_Equac_HTML.gif
We suppose that
| m R ( H , t , θ ) H m | H α + 1 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-84/MediaObjects/13661_2012_Article_338_Equ15_HTML.gif
(2.9)

holds where m = k 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-84/MediaObjects/13661_2012_Article_338_IEq53_HTML.gif. We will prove (2.9) also holds where m = k http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-84/MediaObjects/13661_2012_Article_338_IEq54_HTML.gif, k 6 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-84/MediaObjects/13661_2012_Article_338_IEq55_HTML.gif.

By direct calculation, we have
k R H k = c n j 1 j n n x I n j 1 H j 1 ( H π + R ) j n H j n ( H π + R ) 1 x I I 0 ( H π + R ) p ( t ) , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-84/MediaObjects/13661_2012_Article_338_Equ16_HTML.gif
(2.10)

where 1 n k http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-84/MediaObjects/13661_2012_Article_338_IEq56_HTML.gif, j 1 + + j n = k http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-84/MediaObjects/13661_2012_Article_338_IEq57_HTML.gif, 1 j 1 , , j n < k http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-84/MediaObjects/13661_2012_Article_338_IEq58_HTML.gif.

Since
j n I 0 ( H π + R ) H j n = I 0 ( n ) [ ( H π + R ) ] n + + I 0 ( H π + R ) ( n ) , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-84/MediaObjects/13661_2012_Article_338_Equad_HTML.gif
by Lemma 1 and (2.9), when j n 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-84/MediaObjects/13661_2012_Article_338_IEq59_HTML.gif, we have
| j n I 0 ( H π + R ) H j n | C | I 0 ( H π + R ) ( n ) | C H α + 1 2 H j n C H H j n . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-84/MediaObjects/13661_2012_Article_338_Equ17_HTML.gif
(2.11)
When j n = 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-84/MediaObjects/13661_2012_Article_338_IEq60_HTML.gif, we have
| j n I 0 ( H π + R ) H j n | C . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-84/MediaObjects/13661_2012_Article_338_Equ18_HTML.gif
(2.12)
By (2.11) and (2.12), we have
| j n I 0 ( H π + R ) H j n | C H 1 j n , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-84/MediaObjects/13661_2012_Article_338_Equ19_HTML.gif
(2.13)

where 1 j n < k http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-84/MediaObjects/13661_2012_Article_338_IEq61_HTML.gif.

By (2.13), we have
| j n I 0 ( H π + R ) H j n | | j n I 0 ( H π + R ) H j n | C H n k . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-84/MediaObjects/13661_2012_Article_338_Equ20_HTML.gif
(2.14)

By (2.10), (2.14) and Lemma 2, we have (2.9) holds where m = k http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-84/MediaObjects/13661_2012_Article_338_IEq54_HTML.gif. Thus, we prove Lemma 3. □

Analogously, one may obtain, by a direct but cumbersome commutation, the following estimates.

Lemma 4 The function R ( H , t , θ ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-84/MediaObjects/13661_2012_Article_338_IEq48_HTML.gif satisfies the following estimates:
| m + l R ( H , t , θ ) H m t l | H α + 1 2 for m + l 6 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-84/MediaObjects/13661_2012_Article_338_Equae_HTML.gif
Moreover, by the implicit function theorem, there exists a function R 1 = R 1 ( t , H , θ ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-84/MediaObjects/13661_2012_Article_338_IEq62_HTML.gif such that
R ( H , t , θ ) = p ( t ) ( α + 1 ) | x ( H , θ ) | α x ( H , θ ) + R 1 ( H , t , θ ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-84/MediaObjects/13661_2012_Article_338_Equaf_HTML.gif
Since
R 1 ( H , t , θ ) = R ( H , t , θ ) p ( t ) ( α + 1 ) | x ( H , θ ) | α x ( H , θ ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-84/MediaObjects/13661_2012_Article_338_Equag_HTML.gif
for x 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-84/MediaObjects/13661_2012_Article_338_IEq63_HTML.gif, we have
| R 1 ( H , t , θ ) | = | p ( t ) ( α + 1 ) x α + 1 ( I 0 ( H π + R ( H , t , θ ) ) , θ ) p ( t ) ( α + 1 ) x α + 1 ( H , θ ) | = | 0 1 x α ( H + s ( π R + I ) ) x H ( H + s ( π R + I ) ) ( π R + I ) p ( t ) d s | . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-84/MediaObjects/13661_2012_Article_338_Equah_HTML.gif
For x < 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-84/MediaObjects/13661_2012_Article_338_IEq64_HTML.gif, we have
| R 1 ( H , t , θ ) | = | p ( t ) ( α + 1 ) ( x ) α + 1 ( I 0 ( H π + R ( H , t , θ ) ) , θ ) p ( t ) ( α + 1 ) ( x ) α + 1 ( H , θ ) | = | 0 1 ( x ) α ( H + s ( π R + I ) ) x H ( H + s ( π R + I ) ) ( π R + I ) p ( t ) d s | . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-84/MediaObjects/13661_2012_Article_338_Equai_HTML.gif

By Lemmas 1 and 4, we have the estimates on R 1 ( H , t , θ ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-84/MediaObjects/13661_2012_Article_338_IEq65_HTML.gif.

For concision, in the estimates and the calculation below, we only consider the case x 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-84/MediaObjects/13661_2012_Article_338_IEq63_HTML.gif, since the case x < 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-84/MediaObjects/13661_2012_Article_338_IEq64_HTML.gif have the similar result.

Lemma 5 | k + l R 1 ( H , t , θ ) k H l t | < H α 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-84/MediaObjects/13661_2012_Article_338_IEq66_HTML.gif for k + l 6 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-84/MediaObjects/13661_2012_Article_338_IEq67_HTML.gif.

For the estimates of I ( H π + R ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-84/MediaObjects/13661_2012_Article_338_IEq68_HTML.gif, we need the estimates on I ( H π + R ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-84/MediaObjects/13661_2012_Article_338_IEq69_HTML.gif. By Lemmas 1 and 5, noticing that | R | < H π http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-84/MediaObjects/13661_2012_Article_338_IEq70_HTML.gif, we have the following lemma.

Lemma 6 | k + l I ( H π + R ) k H l t | < H 1 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-84/MediaObjects/13661_2012_Article_338_IEq71_HTML.gif for k + l 6 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-84/MediaObjects/13661_2012_Article_338_IEq67_HTML.gif.

Now the new Hamiltonian function I = I ( t , H , θ ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-84/MediaObjects/13661_2012_Article_338_IEq72_HTML.gif is written in the form
I = I 0 ( H π + R ) = I + ( H π + R ) + I ( H π + R ) = H + π R ( H , t , θ ) + I ( H π + R ) = H + π p ( t ) ( α + 1 ) | x ( H , θ ) | α x ( H , θ ) + R 1 ( H , t , θ ) + I ( H π + R ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-84/MediaObjects/13661_2012_Article_338_Equaj_HTML.gif
The system (2.7) is of the form
{ d t d θ = I H = 1 + π x H ( H , θ ) | x ( H , θ ) | α p ( t ) + R 1 H ( H , t , θ ) + I H ( H , t , θ ) , d H d θ = I t = π p ( t ) ( α + 1 ) | x ( θ , H ) | α x ( θ , H ) R 1 t ( t , H , θ ) I t ( H , t , θ ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-84/MediaObjects/13661_2012_Article_338_Equ21_HTML.gif
(2.15)
Introduce a new action variable ρ [ 1 , 2 ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-84/MediaObjects/13661_2012_Article_338_IEq73_HTML.gif and a parameter ϵ > 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-84/MediaObjects/13661_2012_Article_338_IEq74_HTML.gif by H = ϵ 2 ρ http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-84/MediaObjects/13661_2012_Article_338_IEq75_HTML.gif. Then H 1 0 < ϵ 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-84/MediaObjects/13661_2012_Article_338_IEq76_HTML.gif. Under this transformation, the system (2.15) is changed into the form
{ d t d θ = I H = 1 + π x H ( H , θ ) | x | α ( H , θ ) p ( t ) + R 1 H ( H , t , θ ) + I H ( H , t , θ ) , d ρ d θ = I t = ϵ 2 [ π p ( t ) ( α + 1 ) | x ( θ , H ) | α x ( θ , H ) + R 1 t ( t , H , θ ) + I t ( H , t , θ ) ] , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-84/MediaObjects/13661_2012_Article_338_Equ22_HTML.gif
(2.16)
which is also Hamiltonian system with the new Hamiltonian function
Γ ( t , ρ , θ ; ϵ ) = ρ + π ϵ 2 p ( t ) α + 1 | x ( θ , ϵ 2 ρ ) | α x ( θ , ϵ 2 ρ ) + ϵ 2 R 1 ( t , ϵ 2 ρ , θ ) + ϵ 2 I ( t , ϵ 2 ρ , θ ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-84/MediaObjects/13661_2012_Article_338_Equak_HTML.gif

Obviously, if ϵ 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-84/MediaObjects/13661_2012_Article_338_IEq77_HTML.gif, the solution ( t ( θ , t 0 , ρ 0 ) , ρ ( θ , t 0 , ρ 0 ) ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-84/MediaObjects/13661_2012_Article_338_IEq78_HTML.gif of (2.16) with the initial date ( t 0 , ρ 0 ) R × [ 1 , 2 ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-84/MediaObjects/13661_2012_Article_338_IEq79_HTML.gif is defined in the interval θ [ 0 , 2 π ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-84/MediaObjects/13661_2012_Article_338_IEq80_HTML.gif and ρ ( θ , t 0 , ρ 0 ) [ 1 2 , 3 ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-84/MediaObjects/13661_2012_Article_338_IEq81_HTML.gif. So the Poincaré map of (2.16) is well defined in the domain R × [ 1 , 2 ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-84/MediaObjects/13661_2012_Article_338_IEq82_HTML.gif.

Lemma 7 [[6], Lemma 5.1]

The Poincaré map of (2.16) has intersection property.

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

For convenience, we introduce the notation O k ( 1 ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-84/MediaObjects/13661_2012_Article_338_IEq83_HTML.gif and o k ( 1 ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-84/MediaObjects/13661_2012_Article_338_IEq84_HTML.gif. We say a function f ( t , ρ , θ , ϵ ) O k ( 1 ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-84/MediaObjects/13661_2012_Article_338_IEq85_HTML.gif if f is smooth in ( t , ρ ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-84/MediaObjects/13661_2012_Article_338_IEq86_HTML.gif and for k 1 + k 2 k http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-84/MediaObjects/13661_2012_Article_338_IEq87_HTML.gif,
| k 1 + k 2 t k 1 ρ k 2 f ( t , ρ , θ , ϵ ) | C http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-84/MediaObjects/13661_2012_Article_338_Equal_HTML.gif

for some constant C > 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-84/MediaObjects/13661_2012_Article_338_IEq88_HTML.gif which is independent of the arguments t, ρ, θ, ϵ.

Similarly, we say f ( t , ρ , θ , ϵ ) o k ( 1 ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-84/MediaObjects/13661_2012_Article_338_IEq89_HTML.gif if f is smooth in ( t , ρ ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-84/MediaObjects/13661_2012_Article_338_IEq86_HTML.gif and for k 1 + k 2 k http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-84/MediaObjects/13661_2012_Article_338_IEq87_HTML.gif,
lim ϵ 0 | k 1 + k 2 t k 1 ρ k 2 f ( t , ρ , θ , ϵ ) | = 0 , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-84/MediaObjects/13661_2012_Article_338_Equam_HTML.gif

uniformly in ( t , ρ , θ ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-84/MediaObjects/13661_2012_Article_338_IEq90_HTML.gif.

2.3 Poincaré map and twist theorems

We will use Ortega’s small twist theorem to prove that the Poincaré map P has an invariant closed curve, if ϵ is sufficiently small. Let us first recall the theorem in [7].

Lemma 8 (Ortega’s theorem)

Let A = S 1 × [ a , b ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-84/MediaObjects/13661_2012_Article_338_IEq91_HTML.gif be a finite cylinder with universal cover A = R × [ a , b ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-84/MediaObjects/13661_2012_Article_338_IEq92_HTML.gif. The coordinate in A http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-84/MediaObjects/13661_2012_Article_338_IEq93_HTML.gif is denoted by ( τ , ν ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-84/MediaObjects/13661_2012_Article_338_IEq94_HTML.gif. Consider a map
f ¯ : A S × R . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-84/MediaObjects/13661_2012_Article_338_Equan_HTML.gif
We assume that the map has the intersection property. Suppose that f : A R × R http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-84/MediaObjects/13661_2012_Article_338_IEq95_HTML.gif, ( τ 0 , ν 0 ) ( τ 1 , ν 1 ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-84/MediaObjects/13661_2012_Article_338_IEq96_HTML.gif is a lift of f ¯ http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-84/MediaObjects/13661_2012_Article_338_IEq97_HTML.gif and it has the form
{ τ 1 = τ 0 + 2 N π + δ l 1 ( τ 0 , ν 0 ) + δ g ˜ 1 ( τ 0 , ν 0 ) , ν 1 = ν 0 + δ l 2 ( τ 0 , ν 0 ) + δ g ˜ 2 ( τ 0 , ν 0 ) , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-84/MediaObjects/13661_2012_Article_338_Equ23_HTML.gif
(2.17)
where N is an integer, δ ( 0 , 1 ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-84/MediaObjects/13661_2012_Article_338_IEq98_HTML.gif is a parameter. The functions l 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-84/MediaObjects/13661_2012_Article_338_IEq99_HTML.gif, l 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-84/MediaObjects/13661_2012_Article_338_IEq100_HTML.gif, g ˜ 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-84/MediaObjects/13661_2012_Article_338_IEq101_HTML.gif and g ˜ 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-84/MediaObjects/13661_2012_Article_338_IEq102_HTML.gif satisfy
l 1 C 6 ( A ) , l 1 ( τ 0 , ν 0 ) > 0 , l 1 ν 0 ( τ 0 , ν 0 ) > 0 , ( τ 0 , ν 0 ) A , l 2 ( , ) , g ˜ 1 ( , , ϵ ) , g ˜ 2 ( , , ϵ ) C 5 ( A ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-84/MediaObjects/13661_2012_Article_338_Equ24_HTML.gif
(2.18)
In addition, we assume that there is a function I : A R http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-84/MediaObjects/13661_2012_Article_338_IEq103_HTML.gif satisfying
I C 6 ( A ) , I ν 0 ( τ 0 , ν 0 ) > 0 , ( τ 0 , ν 0 ) A http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-84/MediaObjects/13661_2012_Article_338_Equ25_HTML.gif
(2.19)
and
l 1 ( τ 0 , ν 0 ) I τ o ( τ 0 , ν 0 ) + l 2 ( τ 0 , ν 0 ) I ν 0 ( τ 0 , ν 0 ) = 0 , ( τ 0 , ν 0 ) A . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-84/MediaObjects/13661_2012_Article_338_Equ26_HTML.gif
(2.20)
Moreover, suppose that there are two numbers a ˜ http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-84/MediaObjects/13661_2012_Article_338_IEq104_HTML.gif, and b ˜ http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-84/MediaObjects/13661_2012_Article_338_IEq105_HTML.gif such that a < a ˜ < b ˜ < b http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-84/MediaObjects/13661_2012_Article_338_IEq106_HTML.gif and
I M ( a ) < I m ( a ˜ ) I M ( a ˜ ) < I m ( b ˜ ) I M ( b ˜ ) < I m ( b ) , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-84/MediaObjects/13661_2012_Article_338_Equ27_HTML.gif
(2.21)
where
I M ( r ) = max ρ S 1 I ( ρ o , τ o ) , I m ( r ) = min ρ S 1 I ( ρ o , τ o ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-84/MediaObjects/13661_2012_Article_338_Equao_HTML.gif
Then there exist ϵ > 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-84/MediaObjects/13661_2012_Article_338_IEq74_HTML.gif and Δ > 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-84/MediaObjects/13661_2012_Article_338_IEq107_HTML.gif such that, if δ < Δ http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-84/MediaObjects/13661_2012_Article_338_IEq108_HTML.gif and
g ˜ 1 ( , , ϵ ) C 5 ( A ) + g ˜ 2 ( , , ϵ ) C 5 ( A ) < ϵ , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-84/MediaObjects/13661_2012_Article_338_Equap_HTML.gif

the mapping f ¯ http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-84/MediaObjects/13661_2012_Article_338_IEq97_HTML.gif has an invariant curve in Γ A http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-84/MediaObjects/13661_2012_Article_338_IEq109_HTML.gif. The constant ϵ is independent of δ.

We make the ansatz that the solution of (2.16) with the initial condition ( t ( 0 ) , ρ ( 0 ) ) = ( t 0 , ρ 0 ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-84/MediaObjects/13661_2012_Article_338_IEq110_HTML.gif is of the form
t = t 0 + θ + ϵ 1 α Σ 1 ( t 0 , ρ 0 , θ ; ϵ ) , ρ = ρ 0 + ϵ 1 α Σ 2 ( t 0 , ρ 0 , θ ; ϵ ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-84/MediaObjects/13661_2012_Article_338_Equaq_HTML.gif
Then the Poincaré map of (2.16) is
P t 1 = t 0 + 2 π + ϵ 1 α Σ 1 ( t 0 , ρ 0 , 2 π ; ϵ ) , ρ 1 = ρ 0 + ϵ 1 α Σ 2 ( t 0 , ρ 0 , 2 π ; ϵ ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-84/MediaObjects/13661_2012_Article_338_Equ28_HTML.gif
(2.22)
The functions Σ 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-84/MediaObjects/13661_2012_Article_338_IEq111_HTML.gif and Σ 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-84/MediaObjects/13661_2012_Article_338_IEq112_HTML.gif satisfy
{ Σ 1 = π ϵ α 1 0 θ x H ( θ , ϵ 2 ρ ) | x | α p ( t ) d θ Σ 1 = + ϵ α 1 0 θ ( R 1 H ( H , t , θ ) + I H ( H , t , θ ) ) d θ , Σ 2 = π ϵ α + 1 α + 1 0 θ | x ( θ , ϵ 2 ρ ) | α x ( θ , ϵ 2 ρ ) p ( t ) d θ Σ 2 = ϵ α + 1 α + 1 0 θ ( R 1 t ( H , t , θ ) I t ( H , t , θ ) ) d θ , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-84/MediaObjects/13661_2012_Article_338_Equ29_HTML.gif
(2.23)
where t = t 0 + θ + ϵ 1 α Σ 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-84/MediaObjects/13661_2012_Article_338_IEq113_HTML.gif, ρ = ρ 0 + ϵ 1 α Σ 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-84/MediaObjects/13661_2012_Article_338_IEq114_HTML.gif. By Lemmas 4, 6 and 7, we know that
| Σ 1 | + | Σ 2 | C for  θ [ 0 , 2 π ] . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-84/MediaObjects/13661_2012_Article_338_Equ30_HTML.gif
(2.24)
Hence, for ρ 0 [ 1 , 2 ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-84/MediaObjects/13661_2012_Article_338_IEq115_HTML.gif, we may choose ϵ sufficiently small such that
ρ 0 + ϵ Σ 2 ρ 0 2 1 2 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-84/MediaObjects/13661_2012_Article_338_Equ31_HTML.gif
(2.25)
Moreover, we can prove that
Σ 1 , Σ 2 O 6 ( 1 ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-84/MediaObjects/13661_2012_Article_338_Equ32_HTML.gif
(2.26)
Lemma 9 The following estimates hold:
x α + 1 ( θ , ϵ 2 ρ ) x α + 1 ( θ , ϵ 2 ρ 0 ) ϵ α O 6 ( 1 ) , x H ( θ , ϵ 2 ρ ) x α ( θ , ϵ 2 ρ ) x H ( θ , ϵ 2 ρ 0 ) x α ( θ , ϵ 2 ρ 0 ) ϵ 2 α O 6 ( 1 ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-84/MediaObjects/13661_2012_Article_338_Equar_HTML.gif

Proof

Let
Δ ( t 0 , ρ 0 , θ ) = x α + 1 ( θ , ϵ 2 ρ ) x α + 1 ( θ , ϵ 2 ρ 0 ) = 0 1 ( α + 1 ) x α ( θ , ϵ 2 ρ 0 + s ϵ 1 Σ 2 ) x H ( θ , ϵ 2 ρ 0 + s ϵ 1 Σ 2 ) ϵ 1 Σ 2 d s . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-84/MediaObjects/13661_2012_Article_338_Equas_HTML.gif
By Lemma 2 and (2.25), we have
| Δ ( t 0 , ρ 0 , θ ) | C ( ϵ 2 ρ 0 + s ϵ 1 Σ 2 ) α 2 ( ϵ 2 ρ 0 + s ϵ 1 Σ 2 ) 1 2 ϵ 1 Σ 2 C ( ϵ 2 ρ 0 + s ϵ 1 Σ 2 ) α 1 2 ϵ α 1 ϵ α C ϵ α . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-84/MediaObjects/13661_2012_Article_338_Equat_HTML.gif
Take the derivative with respect to ρ 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-84/MediaObjects/13661_2012_Article_338_IEq116_HTML.gif in the both sides of Δ ( t 0 , ρ 0 , θ ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-84/MediaObjects/13661_2012_Article_338_IEq117_HTML.gif, we have
Δ ρ 0 = 0 1 [ ( α + 1 ) α x α 1 x H 1 + s ϵ Σ 2 ρ 0 ϵ 2 x H Σ 2 ϵ + ( α + 1 ) x α 2 x H 2 1 + s ϵ Σ 2 ρ 0 ϵ 2 Σ 2 ϵ + ( α + 1 ) x α x H Σ 2 ρ 0 1 ϵ ] d s . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-84/MediaObjects/13661_2012_Article_338_Equau_HTML.gif
Using Lemma 2 and noticing | Δ | C ϵ α http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-84/MediaObjects/13661_2012_Article_338_IEq118_HTML.gif, we have
| Δ ρ 0 | C ϵ α . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-84/MediaObjects/13661_2012_Article_338_Equav_HTML.gif
Analogously, one may obtain, by a direct but cumbersome commutation that
| k + l Δ ρ 0 k t 0 l | C ϵ α , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-84/MediaObjects/13661_2012_Article_338_Equaw_HTML.gif
which means that
x α + 1 ( θ , ϵ 2 ρ ) x α + 1 ( θ , ϵ 2 ρ 0 ) ϵ α O 6 ( 1 ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-84/MediaObjects/13661_2012_Article_338_Equax_HTML.gif

The estimates for x H ( θ , ϵ 2 ρ ) x α ( θ , ϵ 2 ρ ) x H ( θ , ϵ 2 ρ 0 ) x α ( θ , ϵ 2 ρ 0 ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-84/MediaObjects/13661_2012_Article_338_IEq119_HTML.gif follow from a similar argument, we omit it here. Thus, Lemma 9 is proved. □

Now we turn to give an asymptotic expression of Poincaré map of (2.15), that is, we study the behavior of the functions Σ 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-84/MediaObjects/13661_2012_Article_338_IEq111_HTML.gif and Σ 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-84/MediaObjects/13661_2012_Article_338_IEq112_HTML.gif at θ = π http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-84/MediaObjects/13661_2012_Article_338_IEq120_HTML.gif as ϵ 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-84/MediaObjects/13661_2012_Article_338_IEq121_HTML.gif. In order to estimate Σ 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-84/MediaObjects/13661_2012_Article_338_IEq111_HTML.gif and Σ 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-84/MediaObjects/13661_2012_Article_338_IEq112_HTML.gif, we need introduce the following definition and lemma. Let
Θ + ( I ) = meas { θ [ 0 , π ] , x ( H 0 , θ ) > 0 } , Θ ( I ) = T 0 Θ + ( I ) , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-84/MediaObjects/13661_2012_Article_338_Equay_HTML.gif

where H 0 = ϵ 2 ρ 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-84/MediaObjects/13661_2012_Article_338_IEq122_HTML.gif.

Lemma 10
Θ + ( I ) = π + ϵ O 6 ( 1 ) , Θ ( I ) = ϵ O 6 ( 1 ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-84/MediaObjects/13661_2012_Article_338_Equaz_HTML.gif

Proof This lemma was proved in [5], so we omit the details. □

For estimate Σ 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-84/MediaObjects/13661_2012_Article_338_IEq111_HTML.gif and Σ 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-84/MediaObjects/13661_2012_Article_338_IEq112_HTML.gif, we need the estimates of x and x H http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-84/MediaObjects/13661_2012_Article_338_IEq123_HTML.gif.

We recall that when x < 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-84/MediaObjects/13661_2012_Article_338_IEq64_HTML.gif, we have
| x ( H 0 , θ ) | = O 6 ( 1 ) , | x H ( H 0 , θ ) | = ϵ 2 O 5 ( 1 ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-84/MediaObjects/13661_2012_Article_338_Equba_HTML.gif
When x > 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-84/MediaObjects/13661_2012_Article_338_IEq124_HTML.gif, by the definition of θ, we have
arcsin x ( H 0 , θ ) 2 h = T 0 ( h ) π θ T ( h ) 2 = θ + ϵ 2 O 5 ( 1 ) , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-84/MediaObjects/13661_2012_Article_338_Equbb_HTML.gif
which yields that
x ( H 0 , θ ) = 2 H 0 π sin θ + O 5 ( 1 ) , x H ( H 0 , θ ) = 1 2 H 0 π sin θ + ϵ 2 O 5 ( 1 ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-84/MediaObjects/13661_2012_Article_338_Equbc_HTML.gif

Now we can give the estimates of Σ 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-84/MediaObjects/13661_2012_Article_338_IEq111_HTML.gif and Σ 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-84/MediaObjects/13661_2012_Article_338_IEq112_HTML.gif.

Lemma 11 The following estimates hold true:
Σ 1 ( t 0 , ρ 0 , 2 π ; ϵ ) = ( π 2 ρ 0 ) α 1 2 0 π ( sin θ ) 1 + α p ( t 0 + θ ) d θ + o 6 ( 1 ) , Σ 2 ( t 0 , ρ 0 , 2 π ; ϵ ) = 1 α + 1 π 1 α 2 ( 2 ρ 0 ) α + 1 2 0 π ( sin θ ) 1 + α p ( t 0 + θ ) d θ + o 6 ( 1 ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-84/MediaObjects/13661_2012_Article_338_Equbd_HTML.gif

for ϵ 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-84/MediaObjects/13661_2012_Article_338_IEq121_HTML.gif.

Proof Firstly, we consider Σ 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-84/MediaObjects/13661_2012_Article_338_IEq111_HTML.gif. By Lemmas 2, 6 and (2.23), we have
Σ 1 ( t 0 , ρ 0 , 2 π ; ϵ ) = π ϵ α 1 0 π x H ( θ , ϵ 2 ρ ) | x ( θ , ϵ 2 ρ ) | α p ( t ) d θ + ϵ α 1 0 π ( R 1 H ( ϵ 2 ρ , t , θ ) + I H ( ϵ 2 ρ , t , θ ) ) d θ = π ϵ α 1 0 π x H ( θ , ϵ 2 ρ 0 ) | x ( θ , ϵ 2 ρ 0 ) | α p ( t 0 + θ ) d θ + ϵ α O 6 ( 1 ) = π ϵ α 1 ( Θ + x H ( θ , ϵ 2 ρ ) | x | α p ( t 0 + θ ) d θ + Θ x H ( θ , ϵ 2 ρ ) | x | α p ( t 0 + θ ) d θ ) + ϵ α O 6 ( 1 ) = π ϵ α 1 Θ + x H ( θ , ϵ 2 ρ ) | x | α p ( t 0 + θ ) d θ + ϵ α O 6 ( 1 ) = π ϵ α 1 0 π x H ( θ , ϵ 2 ρ ) | x | α p ( t 0 + θ ) d θ + ϵ α O 6 ( 1 ) = ( π 2 ρ 0 ) 1 α 2 0 π ( sin θ ) α + 1 p ( t 0 + θ ) d θ + o 6 ( 1 ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-84/MediaObjects/13661_2012_Article_338_Eqube_HTML.gif
Now we consider Σ 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-84/MediaObjects/13661_2012_Article_338_IEq112_HTML.gif.
Σ 2 ( t 0 , ρ 0 , 2 π ; ϵ ) = π ϵ α + 1 α + 1 0 π | x ( θ , ϵ 2 ρ ) | α x ( θ , ϵ 2 ρ ) p ( t ) d θ ϵ α + 1 α + 1 0 π ( R 1 t ( ϵ 2 ρ , t , θ ) + I t ( ϵ 2 ρ , t , θ ) ) d θ = π ϵ α + 1 α + 1 0 π | x ( θ , ϵ 2 ρ 0 ) | α x ( θ , ϵ 2 ρ 0 ) p ( t 0 + θ ) d θ + ϵ α O 6 ( 1 ) = π ϵ α + 1 α + 1 ( Θ + | x ( θ , ϵ 2 ρ ) | α x ( θ , ϵ 2 ρ 0 ) p ( t 0 + θ ) d θ + Θ | x ( θ , ϵ 2 ρ 0 ) | α x ( θ , ϵ 2 ρ ) p ( t 0 + θ ) d θ ) + ϵ α O 6 ( 1 ) = π ϵ α + 1 α + 1 Θ + | x ( θ , ϵ 2 ρ ) | α x ( θ , ϵ 2 ρ 0 ) p ( t 0 + θ ) d θ + ϵ α O 6 ( 1 ) = π ϵ α + 1 α + 1 0 π | x ( θ , ϵ 2 ρ 0 ) | α x ( θ , ϵ 2 ρ ) p ( t 0 + θ ) d θ + ϵ α O 6 ( 1 ) = 1 α + 1 π 1 α 2 ( 2 ρ 0 ) α + 1 2 0 π ( sin θ ) 1 + α p ( t 0 + θ ) d θ + o 6 ( 1 ) . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-84/MediaObjects/13661_2012_Article_338_Equbf_HTML.gif

Thus, Lemma 11 is proved. □

2.4 Proof of Theorem 1

Let
Ψ 1 ( t 0 , ρ 0 ) = ( π 2 ρ 0 ) 1 α 2 0 π ( sin θ ) 1 + α p ( t 0 + θ ) d θ , Ψ 2 ( t 0 , ρ 0 ) = 1 α + 1 π 1 α 2 ( 2 ρ 0 ) α + 1 2 0 π ( sin θ ) 1 + α p ( t 0 + θ ) d θ . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-84/MediaObjects/13661_2012_Article_338_Equbg_HTML.gif
Then there are two functions ϕ 1 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-84/MediaObjects/13661_2012_Article_338_IEq125_HTML.gif and ϕ 2 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-84/MediaObjects/13661_2012_Article_338_IEq126_HTML.gif such that the Poincaré map of (2.16), given by (2.22), is of the form
P t 1 = t 0 + 2 π + ϵ 1 α Ψ 1 ( t 0 , ρ 0 ) + ϵ 1 α ϕ 1 , ρ 1 = ρ 0 + ϵ 1 α Ψ 2 ( t 0 , ρ 0 ) + ϵ 1 α ϕ 2 , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-84/MediaObjects/13661_2012_Article_338_Equbh_HTML.gif

where ϕ 1 , ϕ 2 o 6 ( 1 ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-84/MediaObjects/13661_2012_Article_338_IEq127_HTML.gif.

Since 0 π p ( t 0 + θ ) sin θ d θ > 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-84/MediaObjects/13661_2012_Article_338_IEq128_HTML.gif, t 0 R http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-84/MediaObjects/13661_2012_Article_338_IEq129_HTML.gif, we have
Ψ 1 > 0 , Ψ 1 ρ 0 0 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-84/MediaObjects/13661_2012_Article_338_Equbi_HTML.gif
Let
L = ρ 0 α + 1 2 0 π ( sin θ ) 1 + α p ( t 0 + θ ) d θ . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-84/MediaObjects/13661_2012_Article_338_Equbj_HTML.gif
Then
L t 0 Ψ 1 ( t 0 , ρ 0 ) + L ρ 0 Ψ 2 ( t 0 , ρ 0 ) = 0 . http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-84/MediaObjects/13661_2012_Article_338_Equbk_HTML.gif

The other assumptions of Ortega’s theorem are easily verified. Hence, there is an invariant curve of P in the annulus ( t 0 , ρ 0 ) S 1 × [ 1 , 2 ] http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-84/MediaObjects/13661_2012_Article_338_IEq130_HTML.gif which imply that the boundedness of our original equation (1.4). Then Theorem 1 is proved.

2.5 Proof of Theorem 2

We apply Aubry-Mather theory. By Theorem B in [8] and the monotone twist property of the Poincaré map P guaranteed by Ψ 1 ρ 0 < 0 http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-84/MediaObjects/13661_2012_Article_338_IEq131_HTML.gif. It is straightforward to check that Theorem 2 is correct.

Remark 4 In [9], the authors study the multiplicity of positive periodic solutions of singular Duffing equations
x + g ( x ) = p ( t ) , http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-84/MediaObjects/13661_2012_Article_338_Equbl_HTML.gif

where g ( x ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-84/MediaObjects/13661_2012_Article_338_IEq132_HTML.gif satisfies the semilinear condition at infinity and the time map satisfies an oscillation condition, and prove that the given equation possesses infinitely many positive 2π-periodic solutions by using the Poincaré-Birkhoff theorem. By the methods and techniques in [9], we can also prove the existence of 2π-periodic solutions of (1.4) where V ( x ) http://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-84/MediaObjects/13661_2012_Article_338_IEq1_HTML.gif satisfies the sublinear condition.

Declarations

Acknowledgements

Thanks are given to referees whose comments and suggestions were very helpful for revising our paper.

Authors’ Affiliations

(1)
College of Sciences, Nanjing University of Technology

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Copyright

© Jiang; licensee Springer. 2013

This article is published under license to BioMed Central Ltd. This is an Open Access article distributed under the terms of the Creative Commons Attribution License (http://​creativecommons.​org/​licenses/​by/​2.​0), which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.