Skip to main content

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

Abstract

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

x + V (x)+p(t) | x | α =0,

where V(x) has a singularity. Under some assumptions on the V(x) and p(t) by Ortega’s small twist theorem, we obtain the existence of quasi-periodic solutions and boundedness of all the solutions.

1 Introduction and main result

In 1991, Levi [1] considered the following equation:

x + V (x,t)=0,
(1.1)

where V(x,t) satisfies some growth conditions and V(x,t)=V(x,t+1). The author reduced the system to a normal form and then applied Moser twist theorem to prove the existence of quasi-periodic solution and the boundedness of all solutions. This result relies on the fact that the nonlinearity V(x,t) 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),
(1.2)

with F(x,t)=p(t) is a π-periodic function and V(x)= 1 2 x + 2 + 1 ( 1 x 2 ) ν 1, where x + =max{x,0}, x =max{x,0} and ν is a positive integer. Under the Lazer-Leach assumption that

1+ 1 2 0 π p( t 0 +θ)sinθdθ>0, t 0 R,
(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) in (1.2) is smooth and bounded, so a natural question is to find sufficient conditions on F(x,t) such that all solutions of (1.2) are bounded when F(x,t) is unbounded. The purpose of this paper is to deal with this problem.

Motivated by the papers [1, 5, 6], we consider the following equation:

x + V (x)+p(t) | x | α =0,
(1.4)

where p(t) is a π-periodic function,

V(x)= 1 2 x + 2 + 1 1 x 2 1,0<α<1,x>1.
(1.5)

We suppose Lazer-Leach assumption hold:

0 π p( t 0 +θ) ( sin θ ) 1 + α dθ>0, t 0 R.
(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) satisfies (1.6); then, there is an ϵ 0 >0 such that, for any ω( 1 π , 1 π + ϵ 0 ), t equation (1.4) has a solution ( x ω (t), x ω (t)) of the Mather type with rotation number ω. More precisely:

Case 1: ω= p q is rational. The solutions ( x ω (t+2iπ), x ω (t+2iπ)), 1iq1 are independent periodic solutions of periodic ; moreover, in this case,

lim q min t R ( | x ω ( t ) | + | x ω ( t ) | ) =+.

Case 2: ω is irrational. The solution ( x ω (t), x ω (t)) 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
(2.1)

with the Hamiltonian function

H(x,y,t)= 1 2 y 2 +V(x)+ p ( t ) ( α + 1 ) | x | α x.

In order to introduce action and angle variables, we first consider the auxiliary autonomous equation:

x =y, y = V (x),
(2.2)

which is an integrable Hamiltonian system with Hamiltonian function

H 1 (x,y,t)= 1 2 y 2 +V(x).

The closed curves H 1 (x,y,t)=h>0 are just the integral curves of (2.2).

Denote by T 0 (h) the time period of the integral curve Γ h of (2.2) defined by H 1 (x,y,t)=h and by I the area enclosed by the closed curve Γ h for every h>0. Let 1< α h <0< β h be such that V( α h )=V( β h )=h. It is easy to see that

I 0 (h)=2 α h β h 2 ( h V ( s ) ) ds,h>0

and

T 0 (h)= I 0 (h)=2 α h β h 1 2 ( h V ( s ) ) ds,h>0.

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 ,

so

T 0 (h)=π+ 0 α h 1 2 ( h V ( s ) ) ds.

We then have

I 0 (h)= I (h)+ I + (h), T 0 (h)= T (h)+ T + (h),

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 .

Similar in estimating in [5], we have the estimation of functions I and T .

Lemma 1 We have

h n | d n T ( h ) d h n | C h 1 2

and

h n | d n I ( h ) d h n | C h 1 2 ,

where n=0,1,,6, h+. Note that here and below we always use C, C 0 or C 0 to indicate some constants.

Remark 1 It follows from the definitions of T + (h), T (h) and Lemma 1 that

lim h + T (h)=0, lim h + T + (h)=π.

Thus the time period T 0 (h) is dominated by T + (h) when h is sufficiently large. From the relation between T (h) and I (h), we know I 0 (h) is dominated by I + (h) when h is sufficiently large.

Remark 2 It also follow from the definition of I(h), I (h), I + (h) and Remark 1 that

| h n d n I 0 ( h ) d h n | C 0 I 0 (h)for n1.

Remark 3 Note that h= h 0 ( I 0 ) is the inverse function of I 0 . By Remark 2, we have

| I n d n h ( I ) d I n | C 0 h(I)for n1.

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 ) ) ds, where C is the part of the closed curve Γ h connecting the point on the y-axis and point (x,y).

We define the well-know map (θ,I)(x,y) by

y= S x (x,I),θ= S I (x,I),

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 .

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
(2.3)

and

I(x,y)= I 0 ( h ( x , y ) ) =2 α h β h 2 ( h ( x , y ) V ( s ) ) ds.
(2.4)

In the new variables (θ,I), the system (2.1) is

θ = H I , I = H θ ,
(2.5)

where

H(θ,I,t)=π h 0 (I)+π p ( t ) ( α + 1 ) | x ( I , θ ) | α x(I,θ).
(2.6)

In order to estimate π p ( t ) ( α + 1 ) | x ( I , θ ) | α x(I,θ), we need the following lemma.

Lemma 2 [[5], Lemma 2.2]

For I sufficient large and α h x<0, 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 0n6.

2.2 New action and angle variables

Now we are concerned with the Hamiltonian system (2.5) with Hamiltonian function H(θ,I,t) given by (2.6). Note that

IdθHdt=(HdtIdθ).

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),
(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 1as I+.

So, we assume that I can be written as

I= I 0 ( H π + R ( H , t , θ ) ) ,

where R satisfies |R|< H π . Recalling that h 0 is the inverse function of I 0 , we have

H π +R(H,t,θ)= h 0 (I),

which implies that

R(H,t,θ)= p ( t ) ( α + 1 ) | x ( I , θ ) | α x(I,θ).

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 ) .
(2.8)

Lemma 3 The function R(H,t,θ) satisfies the following estimates:

| m R ( H , t , θ ) H m | H α + 1 2 for m+l6.

Proof Case m=0. By (2.8), Lemma 2 and noticing that H I 1 as I+, 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 .

Case m1. 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 ) .

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 .

Since

| 1 x I I 0 ( H π + R ) p ( t ) | 1as H+,

we have

H | R H | C H 1 + α 2 .

We suppose that

| m R ( H , t , θ ) H m | H α + 1 2
(2.9)

holds where m=k1. We will prove (2.9) also holds where m=k, k6.

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 ) ,
(2.10)

where 1nk, j 1 ++ j n =k, 1 j 1 ,, j n <k.

Since

j n I 0 ( H π + R ) H j n = I 0 ( n ) [ ( H π + R ) ] n ++ I 0 ( H π + R ) ( n ) ,

by Lemma 1 and (2.9), when j n 2, 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 .
(2.11)

When j n =1, we have

| j n I 0 ( H π + R ) H j n | C.
(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 ,
(2.13)

where 1 j n <k.

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 .
(2.14)

By (2.10), (2.14) and Lemma 2, we have (2.9) holds where m=k. 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) satisfies the following estimates:

| m + l R ( H , t , θ ) H m t l | H α + 1 2 for m+l6.

Moreover, by the implicit function theorem, there exists a function R 1 = R 1 (t,H,θ) such that

R(H,t,θ)= p ( t ) ( α + 1 ) | x ( H , θ ) | α x(H,θ)+ R 1 (H,t,θ).

Since

R 1 (H,t,θ)=R(H,t,θ) p ( t ) ( α + 1 ) | x ( H , θ ) | α x(H,θ)

for x0, 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 | .

For x<0, 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 | .

By Lemmas 1 and 4, we have the estimates on R 1 (H,t,θ).

For concision, in the estimates and the calculation below, we only consider the case x0, since the case x<0 have the similar result.

Lemma 5 | k + l R 1 ( H , t , θ ) k H l t |< H α 2 for k+l6.

For the estimates of I( H π +R), we need the estimates on I ( H π +R). By Lemmas 1 and 5, noticing that |R|< H π , we have the following lemma.

Lemma 6 | k + l I ( H π + R ) k H l t |< H 1 2 for k+l6.

Now the new Hamiltonian function I=I(t,H,θ) 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 ) .

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 ) .
(2.15)

Introduce a new action variable ρ[1,2] and a parameter ϵ>0 by H= ϵ 2 ρ. Then H10<ϵ1. 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 ) ] ,
(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 ) .

Obviously, if ϵ1, the solution (t(θ, t 0 , ρ 0 ),ρ(θ, t 0 , ρ 0 )) of (2.16) with the initial date ( t 0 , ρ 0 )R×[1,2] is defined in the interval θ[0,2π] and ρ(θ, t 0 , ρ 0 )[ 1 2 ,3]. So the Poincaré map of (2.16) is well defined in the domain R×[1,2].

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) and o k (1). We say a function f(t,ρ,θ,ϵ) O k (1) if f is smooth in (t,ρ) and for k 1 + k 2 k,

| k 1 + k 2 t k 1 ρ k 2 f ( t , ρ , θ , ϵ ) | C

for some constant C>0 which is independent of the arguments t, ρ, θ, ϵ.

Similarly, we say f(t,ρ,θ,ϵ) o k (1) if f is smooth in (t,ρ) and for k 1 + k 2 k,

lim ϵ 0 | k 1 + k 2 t k 1 ρ k 2 f ( t , ρ , θ , ϵ ) | =0,

uniformly in (t,ρ,θ).

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] be a finite cylinder with universal cover A=R×[a,b]. The coordinate in A is denoted by (τ,ν). Consider a map

f ¯ :AS×R.

We assume that the map has the intersection property. Suppose that f:AR×R, ( τ 0 , ν 0 )( τ 1 , ν 1 ) is a lift of f ¯ 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 ) ,
(2.17)

where N is an integer, δ(0,1) is a parameter. The functions l 1 , l 2 , g ˜ 1 and g ˜ 2 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 ) .
(2.18)

In addition, we assume that there is a function I:AR satisfying

I C 6 (A), I ν 0 ( τ 0 , ν 0 )>0,( τ 0 , ν 0 )A
(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.
(2.20)

Moreover, suppose that there are two numbers a ˜ , and b ˜ such that a< a ˜ < b ˜ <b and

I M (a)< I m ( a ˜ ) I M ( a ˜ )< I m ( b ˜ ) I M ( b ˜ )< I m (b),
(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 ).

Then there exist ϵ>0 and Δ>0 such that, if δ<Δ and

g ˜ 1 ( , , ϵ ) C 5 ( A ) + g ˜ 2 ( , , ϵ ) C 5 ( A ) <ϵ,

the mapping f ¯ has an invariant curve in Γ A . 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 ) 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).

Then the Poincaré map of (2.16) is

P t 1 = t 0 +2π+ ϵ 1 α Σ 1 ( t 0 , ρ 0 ,2π;ϵ), ρ 1 = ρ 0 + ϵ 1 α Σ 2 ( t 0 , ρ 0 ,2π;ϵ).
(2.22)

The functions Σ 1 and Σ 2 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 ,
(2.23)

where t= t 0 +θ+ ϵ 1 α Σ 1 , ρ= ρ 0 + ϵ 1 α Σ 2 . By Lemmas 4, 6 and 7, we know that

| Σ 1 |+| Σ 2 |Cfor θ[0,2π].
(2.24)

Hence, for ρ 0 [1,2], we may choose ϵ sufficiently small such that

ρ 0 +ϵ Σ 2 ρ 0 2 1 2 .
(2.25)

Moreover, we can prove that

Σ 1 , Σ 2 O 6 (1).
(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 ) .

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 .

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 .

Take the derivative with respect to ρ 0 in the both sides of Δ( t 0 , ρ 0 ,θ), 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 .

Using Lemma 2 and noticing |Δ|C ϵ α , we have

| Δ ρ 0 | C ϵ α .

Analogously, one may obtain, by a direct but cumbersome commutation that

| k + l Δ ρ 0 k t 0 l | C ϵ α ,

which means that

x α + 1 ( θ , ϵ 2 ρ ) x α + 1 ( θ , ϵ 2 ρ 0 ) ϵ α O 6 (1).

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 ) 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 and Σ 2 at θ=π as ϵ0. In order to estimate Σ 1 and Σ 2 , 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),

where H 0 = ϵ 2 ρ 0 .

Lemma 10

Θ + (I)=π+ϵ O 6 (1), Θ (I)=ϵ O 6 (1).

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

For estimate Σ 1 and Σ 2 , we need the estimates of x and x H .

We recall that when x<0, we have

| x ( H 0 , θ ) | = O 6 (1), | x H ( H 0 , θ ) | = ϵ 2 O 5 (1).

When x>0, 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),

which yields that

x( H 0 ,θ)= 2 H 0 π sinθ+ O 5 (1), x H ( H 0 ,θ)= 1 2 H 0 π sinθ+ ϵ 2 O 5 (1).

Now we can give the estimates of Σ 1 and Σ 2 .

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 )

for ϵ0.

Proof Firstly, we consider Σ 1 . 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 ) .

Now we consider Σ 2 .

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 ) .

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 .

Then there are two functions ϕ 1 and ϕ 2 such that the Poincaré map of (2.16), given by (2.22), is of the form

P t 1 = t 0 +2π+ 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 ,

where ϕ 1 , ϕ 2 o 6 (1).

Since 0 π p( t 0 +θ)sinθdθ>0, t 0 R, we have

Ψ 1 >0, Ψ 1 ρ 0 0.

Let

L= ρ 0 α + 1 2 0 π ( sin θ ) 1 + α p ( t 0 + θ ) d θ .

Then

L t 0 Ψ 1 ( t 0 , ρ 0 )+ L ρ 0 Ψ 2 ( t 0 , ρ 0 )=0.

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] 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. 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),

where g(x) 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) satisfies the sublinear condition.

References

  1. Levi M: Quasiperiodic motions in superquadratic time-periodic potential. Commun. Math. Phys. 1991, 144: 43-82.

    Article  Google Scholar 

  2. Kupper T, You J: Existence of quasiperiodic solutions and Littlewood’s boundedness problem of Duffing equations with subquadratic potentials. Nonlinear Anal. 1999, 35: 549-559. 10.1016/S0362-546X(97)00709-8

    MathSciNet  Article  Google Scholar 

  3. Li X: Boundedness of solutions for Duffing’s equations with semilinear potentials. J. Differ. Equ. 2001, 176: 248-268. 10.1006/jdeq.2001.3953

    Article  Google Scholar 

  4. Wang Y: Boundedness for sublinear Duffing equations with time-dependent potential. J. Differ. Equ. 2009, 247: 104-118. 10.1016/j.jde.2008.12.024

    Article  Google Scholar 

  5. Capietto A, Dambrosio W, Liu B: On the boundedness of solutions to a nonlinear singular oscillator. Z. Angew. Math. Phys. 2009, 60(6):1007-1034. 10.1007/s00033-008-8094-y

    MathSciNet  Article  Google Scholar 

  6. Liu B: Quasi-periodic solutions of forced isochronous oscillators at resonance. J. Differ. Equ. 2009, 246: 3471-3495. 10.1016/j.jde.2009.02.015

    Article  Google Scholar 

  7. Ortega R: Boundedness in a piecewise linear oscillator and a variant of the small twist theorem. Proc. Lond. Math. Soc. 1999, 79: 381-413. 10.1112/S0024611599012034

    Article  Google Scholar 

  8. Pei ML: Aubry-Mather sets for finite-twist maps of a cylinder and semilinear Duffing equations. J. Differ. Equ. 1994, 113: 106-127. 10.1006/jdeq.1994.1116

    Article  Google Scholar 

  9. Wang Z, Ma T: Existence and multiplicity of periodic solutions of semilinear resonant Duffing equations with singularities. Nonlinearity 2012, 25: 279-307. 10.1088/0951-7715/25/2/279

    MathSciNet  Article  Google Scholar 

Download references

Acknowledgements

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

Author information

Authors and Affiliations

Authors

Corresponding author

Correspondence to Shunjun Jiang.

Additional information

Competing interests

The author declares that they have no competing interests.

Rights and permissions

Open Access This article is distributed under the terms of the Creative Commons Attribution 2.0 International License (https://creativecommons.org/licenses/by/2.0), which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

Reprints and Permissions

About this article

Cite this article

Jiang, S. Boundedness of solutions for a class of second-order differential equation with singularity. Bound Value Probl 2013, 84 (2013). https://doi.org/10.1186/1687-2770-2013-84

Download citation

  • Received:

  • Accepted:

  • Published:

  • DOI: https://doi.org/10.1186/1687-2770-2013-84

Keywords

  • boundedness of solutions
  • singularity
  • small twist theorem