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RETRACTED ARTICLE: Landesman-Lazer type condition for second-order differential equations at resonance with impulsive effects

This article was retracted on 04 July 2015

Abstract

In this paper, we study the existence of periodic solutions of second-order impulsive differential equations at resonance. We prove the existence of periodic solutions under a generalized Landesman-Lazer type condition by using the variational method. The impulses can generate a periodic solution.

1 Introduction

We are concerned with periodic boundary value problem of second-order impulsive differential equations at resonance

{ x ( t ) + m 2 x ( t ) + f ( t , x ( t ) ) = e ( t ) , a.e.  t [ 0 , 2 π ] , x ( 0 ) x ( 2 π ) = x ( 0 ) x ( 2 π ) = 0 , x ( t j + ) = x ( t j ) , Δ x ( t j ) : = x ( t j + ) x ( t j ) = I j ( t j , x ( t j ) ) , j = 1 , 2 , , p ,
(1.1)

where mN, f:[0,2π]×RR is a Carathéodory function, e L 1 (0,2π), 0< t 1 < t 2 << t p <2π, and I j :[0,2π]×RR is continuous for every j.

When Δ x ( t j )0, problem (1.1) becomes to the well-known periodic boundary value problem at resonance

{ x ( t ) + m 2 x ( t ) + f ( t , x ( t ) ) = e ( t ) , a.e.  t [ 0 , 2 π ] , x ( 0 ) x ( 2 π ) = x ( 0 ) x ( 2 π ) = 0 .
(1.2)

There are many existence results for problem (1.2) in the literature. Let us mention some pioneering works by Lazer [1], Lazer and Leach [2], and Landesman and Lazer [3]. In [3], a key sufficient condition for the existence of solutions of problem (1.2) is the so-called Landesman-Lazer condition,

0 2 π e ( t ) sin ( m t + θ ) d t < 0 2 π [ ( lim inf x + f ( t , x ) ) sin + ( m t + θ ) ( lim sup x f ( t , x ) ) sin ( m t + θ ) ] d t , θ R ,
(1.3)

where sin ± (mt+θ)=max{±sin(mt+θ),0}.

It is well known that the theory of impulsive differential equations has been recognized to not only be richer than that of differential equations without impulses, but also to provide a more adequate mathematical model for numerous processes and phenomena studied in physics, biology, engineering, etc. We refer the reader to the book [4]. Recently, the Dirichlet and periodic boundary conditions problems for second-order differential equations with impulses in the derivative and without impulses are studied by some authors via variational method [511]. In this paper, we will investigate problem (1.1) under a more general Landesman-Lazer type condition. Define

F(t,x)= 0 x f(t,s)ds, F + (t)= lim inf x + F ( t , x ) x , F (t)= lim sup x F ( t , x ) x

and for j=1,2,,p,

J j (t,x)= 0 x I j (t,s)ds, J j + (t)= lim sup x + J j ( t , x ) x , J j (t)= lim inf x J j ( t , x ) x .

Throughout this paper, we give the following fundamental assumptions.

( H 1 ) There exists p L 1 ([0,2π],[0,+)) such that |f(t,x)|p(t), for a.e. t[0,2π] and for all xR.

( H 2 ) There exist positive constants c 1 , c 2 ,, c p such that for all t,xR,

| I j ( t , x ) | c j ,j=1,2,,p.

( H 3 ) For all θR,

j = 1 p J j + ( t j ) sin + ( m t j + θ ) j = 1 p J j ( t j ) sin ( m t j + θ ) + 0 2 π e ( t ) sin ( m t + θ ) d t < 0 2 π ( F + ( t ) sin + ( m t + θ ) F ( t ) sin ( m t + θ ) ) d t .

We now can state the main theorem of this paper.

Theorem 1.1 Assume that the conditions ( H 1 ), ( H 2 ), and ( H 3 ) hold. Then problem (1.1) has at least one 2π-periodic solution.

To demonstrate the impulsive effects clearly, we can take

I j (t,x) d j ,j=1,2,,p,
(1.4)

where d 1 , d 2 ,, d p are constants. Hence, J j ± (t)= d j .

From Theorem 1.1, we obtain the following result.

Corollary 1.2 Assume that we have the conditions ( H 1 ), (1.4), and the following.

( H 3 ) For all θR,

j = 1 p d j sin ( m t j + θ ) + 0 2 π e ( t ) sin ( m t + θ ) d t < 0 2 π ( F + ( t ) sin + ( m t + θ ) F ( t ) sin ( m t + θ ) ) d t

hold. Then problem (1.1) has at least one 2π-periodic solution.

Moreover, we have the following corollary.

Corollary 1.3 Assume that we have the conditions ( H 1 ) and the following.

( H 3 ) For all θR,

0 2 π e(t)sin(mt+θ)dt< 0 2 π ( F + ( t ) sin + ( m t + θ ) F ( t ) sin ( m t + θ ) ) dt
(1.5)

holds. Then problem (1.2) has at least one 2π-periodic solution.

Remark 1.4 By a simple calculation, one can easily derive

F + (t)= lim inf x + F ( t , x ) x lim inf x + f(t,x), F (t)= lim sup x F ( t , x ) x lim sup x f(t,x).

A simple example f(t,x)=sint+cosx illustrates it. Thus condition ( H 3 ) generalizes condition (1.3). Hence, our results improve the related results in the literature mentioned above. Moreover, since we consider the problem with impulses, Theorem 1.1 is also a complement of the pioneering works.

Remark 1.5 It is remarkable that Landesman-Lazer condition ( H 3 ) is an ‘almost’ necessary and sufficient condition when F + and F are replaced by f + and f , where f + = lim x + f(t,x), f = lim x f(t,x), and f (t)f(t,x) f + (t) (see [[12], p.70]). If the condition (1.5) is not satisfied, i.e., θR,

0 2 π e(t)sin(mt+θ)dt 0 2 π ( F + ( t ) sin + ( m t + θ ) F ( t ) sin ( m t + θ ) ) dt,

problem (1.2) cannot be guaranteed to have periodic solution. For example, we consider resonant differential equation

x + m 2 x+(1+sinmt)arctanx=8sinmt.
(1.6)

Obviously, f(t,x)=(1+sinmt)arctanx, e(t)=8sinmt, and F + (t)= π 2 (1+sinmt), F (t)= π 2 (1+sinmt). Taking θ=0, we have

0 2 π e ( t ) sin m t d t 0 2 π ( F + ( t ) sin + m t F ( t ) sin m t ) d t = 8 π π 2 0 2 π ( 1 + sin m t ) | sin m t | d t 8 π 2 π 2 > 0 .

Then ( H 3 ) is not satisfied. From now on, we prove that (1.6) has not 2π-periodic solution by contradiction. Assume that (1.6) has 2π-periodic solution. Multiplying both sides of (1.6) by sinmt and integrating over [0,2π], we get

8 π = 0 2 π ( 1 + sin m t ) arctan x sin m t d t 0 2 π | ( 1 + sin m t ) arctan x cos m t | d t π 0 2 π d t = 2 π 2 ,

which is impossible. Hence, problem (1.2) may have no solution if the condition ( H 3 ) is not satisfied. However, as long as ( H 3 ) holds, problem (1.1) will have at least one periodic solution. Therefore, the impulses can generate a periodic solution.

The rest of the paper is organized as follows. In Section 2, we shall state some notations, some necessary definitions, and a saddle theorem due to Rabinowitz. In Section 3, we shall prove Theorem 1.1.

2 Preliminaries

In the following, we introduce some notations and some necessary definitions.

Define

H= { x H 1 ( 0 , 2 π ) : x ( 0 ) = x ( 2 π ) } ,

with the norm

x= ( 0 2 π ( x 2 ( t ) + x 2 ( t ) ) d t ) 1 2 .

Consider the functional φ(x) defined on H by

φ ( x ) = 1 2 0 2 π x 2 ( t ) d t m 2 2 0 2 π x 2 ( t ) d t 0 2 π F ( t , x ( t ) ) d t + 0 2 π e ( t ) x ( t ) d t + j = 1 p J j ( t j , x ( t j ) ) .
(2.1)

Similarly as in [7], φ(x) is continuously differentiable on H, and

φ ( x ) v ( t ) = 0 2 π x ( t ) v ( t ) d t m 2 0 2 π x ( t ) v ( t ) d t 0 2 π f ( t , x ( t ) ) v ( t ) d t + 0 2 π e ( t ) v ( t ) d t + j = 1 p I j ( t j , x ( t j ) ) v ( t j ) , for  v ( t ) H .
(2.2)

Now, we have the following lemma.

Lemma 2.1 If xH is a critical point of φ, then x is a 2π-periodic solution of (1.1).

The proof of Lemma 2.1 is similar to Lemma 2.1 in [6], so we omit it.

We say that φ satisfies (PS) if every sequence ( x n ) for which φ( x n ) is bounded in and φ ( x n )0 (as n) possesses a convergent subsequence.

To prove the main result, we will use the following saddle point theorem due to Rabinowitz [13] (or see [12]).

Theorem 2.2 Let φ C 1 (H,R) and H= H H + , dim( H )<, dim( H + )=. We suppose that:

  1. (a)

    There exists a bounded neighborhood D of 0 in H and a constant α such that φ | D α;

  2. (b)

    there exists a constant β>α such that φ | H + β;

  3. (c)

    φ satisfies (PS).

Then the functional φ has a critical point in H.

3 The proof of Theorem 1.1

In this section, we first show that the functional φ satisfies the Palais-Smale condition.

Lemma 3.1 Assume that the conditions ( H 1 ), ( H 2 ), and ( H 3 ) hold. Then φ defined by (2.1) satisfies (PS).

Proof Let M>0 be a constant and { x n }H be a sequence satisfying

| φ ( x n ) | = | 1 2 0 2 π x n 2 d t m 2 2 0 2 π x n 2 d t 0 2 π F ( t , x n ) d t + 0 2 π e ( t ) x n ( t ) d t + j = 1 p J j ( t j , x n ( t j ) ) | M
(3.1)

and

lim n φ ( x n ) =0.
(3.2)

We first prove that { x n } is bounded in H by contradiction. Assume that { x n } is unbounded. Let { z k } be an arbitrary sequence bounded in H. It follows from (3.2) that, for any kN,

lim n | φ ( x n ) z k | lim n φ ( x n ) z k =0.

Thus

lim n φ ( x n ) z k =0uniformly for kN.

Hence,

lim n ( 0 2 π ( x n z k m 2 x n z k ) d t 0 2 π ( f ( t , x n ) z k e ( t ) z k ) d t + j = 1 p I j ( t j , x n ( t j ) ) z k ( t j ) ) = 0 .
(3.3)

By ( H 1 ) and ( H 2 ), we have

lim n ( 0 2 π f ( t , x n ) z k e ( t ) z k x n d t j = 1 p I j ( t j , x n ( t j ) ) z k ( t j ) x n ) =0.
(3.4)

From (3.3) and (3.4), we obtain

lim n 0 2 π ( x n x n z k m 2 x n x n z k ) dt=0.
(3.5)

Set

y n = x n x n .

Then we have

lim n 0 2 π ( y n z k m 2 y n z k ) dt=0,

and furthermore,

lim n i 0 2 π [ ( y n y i ) z k m 2 ( y n y i ) z k ] dt=0.
(3.6)

Replacing z k in (3.6) by ( y n y i ), we get

lim n i ( y n y i 2 ( m 2 + 1 ) y n y i 2 2 ) =0.

Due to the compact embedding H L 2 (0,2π), going to a subsequence,

y n y 0 weakly in H, y n y 0 in  L 2 (0,2π).

Therefore,

lim n i y n y i 2 2 =0.

Furthermore, we have

lim n i y n y i 2 =0,

which implies ( y n ) is Cauchy sequence in H. Thus, y n y 0 in H. It follows from (3.5) and the usual regularity argument for ordinary differential equations (see [14]) that

y 0 = k 1 sinmt+ k 2 cosmt,
(3.7)

where k 1 2 + k 2 2 = 1 ( m 2 + 1 ) π ( y 0 =1). (Different subsequences of { y n } correspond to different k 1 and k 2 .)

Write (3.7) as

y 0 = 1 ( m 2 + 1 ) π sin(mt+θ),

where θ satisfies sinθ= k 2 k 1 2 + k 2 2 and cosθ= k 1 k 1 2 + k 2 2 .

Taking z k = 1 ( m 2 + 1 ) π sin(mt+θ), we get, for any nN,

0 2 π ( x n z k m 2 x n z k ) dt=0.
(3.8)

Thus, it follows from (3.3) and (3.8) that

lim n [ 0 2 π ( f ( t , x n ) e ( t ) ) 1 ( m 2 + 1 ) π sin ( m t + θ ) d t j = 1 p I j ( t j , x n ( t j ) ) 1 ( m 2 + 1 ) π sin ( m t j + θ ) ] = 0 .
(3.9)

By ( H 1 ) and ( H 2 ), we obtain

lim n [ 0 2 π ( f ( t , x n ) e ( t ) ) ( 1 ( m 2 + 1 ) π sin ( m t + θ ) y n ) d t j = 1 p I j ( t j , x n ( t j ) ) ( 1 ( m 2 + 1 ) π sin ( m t j + θ ) y n ( t j ) ) ] = 0 .
(3.10)

It follows from (3.9) and (3.10) that

lim n [ 0 2 π ( f ( t , x n ) e ( t ) ) y n d t j = 1 p I j ( t j , x n ( t j ) ) y n ( t j ) ] =0.

Hence, replacing z k in (3.3) by y n , we have

lim n 0 2 π ( x n x n x n m 2 x n x n x n ) dt=0.
(3.11)

Now, dividing (3.1) by x n , we get

M x n 1 2 0 2 π ( x n 2 x n m 2 x n 2 x n ) d t 0 2 π F ( t , x n ) e ( t ) x n x n + j = 1 p J j ( t j , x n ( t j ) ) x n M x n ,

which yields

0 2 π F ( t , x n ) e ( t ) x n x n M x n + 1 2 0 2 π ( x n 2 x n m 2 x n 2 x n ) d t + j = 1 p J j ( t j , x n ( t j ) ) x n .
(3.12)

Note that x n x n 1 ( m 2 + 1 ) π sin(mt+θ) in H. Due to the compact embedding HC(0,2π) and | x n (t)|+, we have x n x n 1 ( m 2 + 1 ) π sin(mt+θ) in C(0,2π). Furthermore,

lim n x n (t)={ + , t I + : = { t [ 0 , 2 π ] | sin ( m t + θ ) > 0 } , , t I : = { t [ 0 , 2 π ] | sin ( m t + θ ) < 0 } .

Hence, from (3.11) and (3.12), we have

lim inf n 0 2 π F ( t , x n ) e ( t ) x n x n d t lim inf n j = 1 p J j ( t j , x n ( t j ) ) x n ( t j ) x n + ( t j ) x n ( t j ) x n lim sup n j = 1 p J j ( t j , x n ( t j ) ) x n ( t j ) x n + ( t j ) x n lim inf n j = 1 p J j ( t j , x n ( t j ) ) x n ( t j ) x n ( t j ) x n = 1 ( m 2 + 1 ) π j = 1 p J j + ( t j ) sin + ( m t j + θ ) 1 ( m 2 + 1 ) π j = 1 p J j ( t j ) sin ( m t j + θ ) .
(3.13)

Using Fatou’s lemma, we get

lim inf n 0 2 π F ( t , x n ) x n d t = lim inf n [ I + F ( t , x n ) x n x n x n d t I F ( t , x n ) x n x n x n d t ] I + lim inf n F ( t , x n ) x n x n x n d t I lim sup n F ( t , x n ) x n x n x n d t .

Thus, by a simple computation, we have

lim inf n 0 2 π F ( t , x n ) x n d t 1 ( m 2 + 1 ) π 0 2 π [ F + ( t ) sin + ( m t + θ ) F ( t ) sin ( m t + θ ) ] d t .
(3.14)

Hence, it follows from (3.13) and (3.14) that

j = 1 p J j + ( t j ) sin + ( m t j + θ ) j = 1 p J j ( t j ) sin ( m t j + θ ) + 0 2 π e ( t ) sin ( m t + θ ) d t 0 2 π [ F + ( t ) sin + ( m t + θ ) F ( t ) sin ( m t + θ ) ] d t .

This contradicts ( H 3 ). It implies that the sequence ( x n ) is bounded. Thus, there exists x 0 H such that x n x 0 weakly in H. Due to the compact embedding H L 2 (0,2π) and HC(0,2π), going to a subsequence,

x n x 0 in  L 2 (0,2π), x n x 0 in C(0,2π).

From (3.3), we obtain

lim n i ( 0 2 π ( ( x n x i ) z k m 2 ( x n x i ) z k ) d t 0 2 π ( f ( t , x n ) f ( t , x i ) ) z k d t + j = 1 p ( I j ( t j , x n ( t j ) ) I j ( t j , x i ( t j ) ) ) z k ( t j ) ) = 0 .

Replacing z k by x n x i in the above equality, we get

lim n i ( 0 2 π ( ( x n x i ) 2 m 2 ( x n x i ) 2 ) d t 0 2 π ( f ( t , x n ) f ( t , x i ) ) ( x n x i ) d t + j = 1 p ( I j ( t j , x n ( t j ) ) I j ( t j , x i ( t j ) ) ) ( x n ( t j ) x i ( t j ) ) ) = 0 .
(3.15)

By ( H 1 ) and ( H 2 ), we have

lim n i 0 2 π ( f ( t , x n ) f ( t , x i ) ) ( x n x i )dt=0
(3.16)

and

lim n i j = 1 p ( I j ( t j , x n ( t j ) ) I j ( t j , x i ( t j ) ) ) ( x n ( t j ) x i ( t j ) ) =0.
(3.17)

Thus, it follows from (3.15), (3.16), and (3.17) that

lim n i 0 2 π [ ( x n x i ) 2 m 2 ( x n x i ) 2 ] dt=0.

Therefore,

lim n i x n x i 2 =0,

which implies x n x 0 in H. It shows that φ satisfies (PS). □

Now, we can give the proof of Theorem 1.1.

Proof of Theorem 1.1 Denote

H =Rspan{sint,cost,sin2t,cos2t,,sinmt,cosmt}

and

H + =span { sin ( m + 1 ) t , cos ( m + 1 ) t , } .

We first prove that

lim inf x φ(x)=,for x H ,
(3.18)

by contradiction. Assume that there exists a sequence ( x n ) H such that x n (as n) and there exists a constant c satisfying

lim inf n φ( x n ) c .
(3.19)

By ( H 1 ), we have

lim n 0 2 π F ( t , x n ) e ( t ) x n x n 2 dt=0.
(3.20)

By ( H 2 ), we get

lim n j = 1 p J j ( t j , x n ( t j ) ) x n 2 =0.
(3.21)

From (3.19) and the definition of φ, we obtain

lim inf n [ 1 2 0 2 π x n 2 m 2 x n 2 x n 2 d t 0 2 π F ( t , x n ) e ( t ) x n x n 2 d t + j = 1 p J j ( t j , x n ( t j ) ) x n 2 ] 0 .
(3.22)

For x H , we have

0 2 π ( x 2 m 2 x 2 ) dt= x 2 ( m 2 + 1 ) x 2 2 0.
(3.23)

The equality in (3.23) holds only for

x= 1 ( m 2 + 1 ) π sin(mt+θ),θR.

Set y n = x n x n . Since dim H <, going to a subsequence, there exists y 0 H such that y n y 0 in H and y n y 0 in L 2 (0,2π). Then (3.20), (3.21), (3.22), and (3.23) imply that

y 0 = 1 ( m 2 + 1 ) π sin(mt+θ),θR.

By (3.19), we have, for n large enough,

1 2 0 2 π x n 2 m 2 x n 2 x n dt 0 2 π F ( t , x n ) e ( t ) x n x n dt+ j = 1 p J j ( t j , x n ( t j ) ) x n c x n .
(3.24)

It follows from x n H that

0 2 π x n 2 m 2 x n 2 x n 0.
(3.25)

From (3.24) and (3.25), we get, for n large enough,

c x n 0 2 π F ( t , x n ) e ( t ) x n x n dt+ j = 1 p J j ( t j , x n ( t j ) ) x n .

Thus,

lim inf n 0 2 π ( F ( t , x n ) x n e ( t ) ) x n x n dt lim inf n j = 1 p J j ( t j , x n ( t j ) ) x n .

Using an argument similar to the proof of Lemma 3.1, we get

j = 1 p J j + ( t j ) sin + ( m t j + θ ) j = 1 p J j ( t j ) sin ( m t j + θ ) + 0 2 π e ( t ) sin ( m t + θ ) d t 0 2 π ( F + ( t ) sin + ( m t + θ ) F ( t ) sin ( m t + θ ) ) d t ,

which is a contradiction to ( H 3 ).

Then (3.18) holds.

Next, we prove that

lim x φ(x)=,for all x H + ,

and φ is bounded on bounded sets.

Because of the compact embedding of HC(0,2π) and H L 2 (0,2π), there exists constants m 1 , m 2 such that

x m 1 x, x 2 m 2 x.

Then by ( H 1 ) and ( H 2 ), one has

| φ ( x ) | = | 1 2 0 2 π x 2 d t m 2 2 0 2 π x 2 d t 0 2 π [ F ( t , x ) e ( t ) x ] d t + j = 1 p J j ( t j , x ( t j ) ) | 1 2 x 2 + m 2 2 m 2 2 x 2 + 0 2 π ( | p ( t ) | | x | + | e ( t ) | | x | ) d t + j = 1 p c j | x ( t j ) | 1 + m 2 m 2 2 2 x 2 + m 1 ( p 1 + e 1 ) x + j = 1 p c j m 1 x .
(3.26)

Hence, φ is bounded on bounded sets of H.

Since x H + , we have

x 2 ( ( m + 1 ) 2 + 1 ) x 2 2 .
(3.27)

Thus, from (3.26) and (3.27), we obtain

φ ( x ) = 1 2 0 2 π x 2 d t m 2 2 0 2 π x 2 d t 0 2 π [ F ( t , x ) e ( t ) x ] d t + j = 1 p J j ( t j , x ( t j ) ) 2 m + 1 2 ( ( m + 1 ) 2 + 1 ) x 2 m 1 ( p 1 + e 1 + j = 1 p c j ) x ,

which implies

lim x φ(x)=,for all x H + .

Up to now, the conditions (a) and (b) of Theorem 2.2 are satisfied. According to Lemma 3.1, (c) is also satisfied. Hence, by Theorem 2.2, (1.1) has at least one solution. This completes the proof. □

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Acknowledgements

The authors would like to express their thanks to the editor of the journal and the referees for their carefully reading of the first draft of the manuscript and making many helpful comments and suggestions which improved the presentation of the paper.

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The first author has contributed in obtaining new results and written the whole article. The second author has written the references with BibTeX and formatted the manuscript such that it conforms to the journal style. All authors have also read and approved the final manuscript.

This article has been retracted by Professor Ravi P Agarwal, Editor-in-Chief of Advances in Difference Equations. Following publication of this article, it was brought to the attention of the editorial and publishing staff that this article has substantial overlap with an article by Jin Li, Jianlin Luo and Zaihong Wang, published in November 2014 in Mathematical Modelling and Analysis. This is a violation of publication ethics which, in accordance with the Springer Policy on Publishing Integrity, warrants a retraction of the article and a notice to this effect to be published in the journal.

An erratum to this article can be found online at http://dx.doi.org/10.1186/s13662-015-0446-2.

A retraction note to this article can be found online at http://dx.doi.org/10.1186/s13662-015-0446-2.

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Li, J., Zheng, M. RETRACTED ARTICLE: Landesman-Lazer type condition for second-order differential equations at resonance with impulsive effects. Adv Differ Equ 2014, 235 (2014). https://doi.org/10.1186/1687-1847-2014-235

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  • DOI: https://doi.org/10.1186/1687-1847-2014-235

Keywords

  • impulsive differential equations
  • Landesman-Lazer type condition
  • variational method