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Subharmonic solutions for a class of second-order impulsive Lagrangian systems with damped term

Xingyong Zhang

Author Affiliations

Department of Mathematics, Faculty of Science, Kunming University of Science and Technology, Kunming, Yunnan, 650500, P.R. China

Boundary Value Problems 2013, 2013:218  doi:10.1186/1687-2770-2013-218


The electronic version of this article is the complete one and can be found online at: http://www.boundaryvalueproblems.com/content/2013/1/218


Received:16 July 2013
Accepted:27 August 2013
Published:7 November 2013

© 2013 Zhang; licensee Springer.

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.

Abstract

In this paper, by using the mountain pass theorem, we investigate the existence of subharmonic weak solutions for a class of second-order impulsive Lagrangian systems with damped term under asymptotically quadratic conditions. Some new existence criteria are established. Finally, an example is presented to verify our results.

MSC: 37J45, 34C25, 70H05.

Keywords:
impulsive Lagrangian systems; damped term; subharmonic weak solutions; mountain pass theorem

1 Introduction and main results

In this paper, we investigate the existence of subharmonic weak solutions for the following second-order impulsive Lagrangian system with damped term:

{ d ( P ( t ) u ˙ ( t ) ) d t + ( q ( t ) P ( t ) + B ) u ˙ ( t ) + ( 1 2 q ( t ) B A ( t ) ) u ( t ) + F ( t , u ( t ) ) = 0 , a.e.  t R , Δ ( P ( t j ) u ˙ ( t j ) ) = P ( t j ) u ˙ ( t j + ) P ( t j ) u ˙ ( t j ) = I j ( u ( t j ) ) , j = 1 , , p , (1.1)

where T > 0 , p N , t 0 = 0 < t 1 < t 2 < < t p < t p + 1 = T , u ( t ) = ( u 1 ( t ) , , u N ( t ) ) , I j : R N R , q C ( R , R ) satisfying q ( t + T ) = q ( t ) and 0 T q ( t ) d t = 0 , B is a skew-symmetric N × N constant matrix, P ( t ) and A ( t ) are symmetric and continuous N × N matrix-value functions on ℝ satisfying P ( t + T ) = P ( t ) and A ( t + T ) = A ( t ) , and F : R × R N R satisfies F ( t , x ) = K ( t , x ) + W ( t , x ) , where K, W are T-periodic in their first variable, and the following assumption:

(A) F ( t , x ) is measurable intfor every x R N and continuously differentiable inxfor a.e. t [ 0 , T ] , and there exist a C ( R + , R + ) and b L 1 ( 0 , T ; R + ) with b ( t + T ) = b ( t ) such that

| F ( t , x ) | a ( | x | ) b ( t ) , | F ( t , x ) | a ( | x | ) b ( t ) , | I j ( x ) | a ( | x | ) , | I j ( x ) | a ( | x | )

for all x R N and a.e. t [ 0 , T ] .

Lagrangian systems are applied extensively to study the fluid mechanics, nuclear physics and relativistic mechanics. Especially, as a special case of Lagrangian systems, the following second-order Hamiltonian systems are considered by many authors:

u ¨ ( t ) = F ( t , u ( t ) ) , a.e.  t R , (1.2)

where F : R × R N R satisfies F ( t + T , x ) = F ( t , x ) , and the existence and multiplicity of periodic solutions, subharmonic solutions and homoclinic solutions are obtained via variational methods. We refer readers to [1-14]. Especially, in 2010, under the asymptotically quadratic conditions, Tang and Jiang [10] obtained the following interesting result.

Theorem A (see [10], Theorem 1.1)

Assume thatFsatisfies

(F) F ( t , x ) = K ( t , x ) + W ( t , x ) and K , W C 1 ( R × R N , R ) areT-periodic in their first variable with T > 0 , and thatKandWsatisfy the following assumptions:

(H1) There exist constants b > 0 and γ ( 1 , 2 ] such that

K ( t , 0 ) = 0 , K ( t , x ) b | x | γ for ( t , x ) [ 0 , T ] × R N ;

(H2) ( K ( t , x ) , x ) 2 K ( t , x ) for ( t , x ) [ 0 , T ] × R N ;

(H3) lim sup | x | 0 W ( t , x ) | x | 2 < b uniformly for t [ 0 , T ] ;

(H4) There exists a function g L 1 ( [ 0 , T ] , R ) such that

( W ( t , x ) , x ) 2 W ( t , x ) g ( t ) for ( t , x ) [ 0 , T ] × R N

and

lim | x | [ ( W ( t , x ) , x ) 2 W ( t , x ) ] = + for a.e. t [ 0 , T ] ;

(H5) There exist constants a > 0 and d > 0 such that

W ( t , x ) a | x | 2 + d for ( t , x ) [ 0 , T ] × R N ;

(H6) There exists x 0 R N such that

0 T [ K ( t , x 0 ) W ( t , x 0 ) g ( t ) 2 ] d t < 0 .

Then system (1.2) has a nontrivialT-periodic solution.

In recent years, variational methods have been applied to study the existence and multiplicity of solutions for impulsive differential equations and lots of interesting results have been obtained, see [15-20].

In [15], Nieto and O’Regan considered a one-dimensional Dirichlet boundary value problem with impulses. They obtained that the solutions of the impulsive problem minimize some (energy) functional and the critical points of the functional are indeed solutions of the impulsive problem.

In [16], Nieto introduced a variational formulation for the following one-dimensional damped nonlinear Dirichlet problem with impulses:

{ u ( t ) + g ( t ) u ( t ) + λ u ( t ) = f ( t , u ( t ) ) , t [ 0 , T ] , Δ u ( t j ) = I j ( u ( t j ) ) , j = 1 , 2 , , p , u ( 0 ) = u ( T ) = 0 , (1.3)

and gave the concept of a weak solution for such a problem. They obtained that the weak solutions of problem (1.3) are indeed the critical points of the functional:

φ ( v ) = 1 2 0 T e G ( t ) ( v ( t ) ) 2 d t + λ 2 0 T e G ( t ) v 2 ( t ) d t + j = 1 p e G ( t j ) 0 v ( t j ) I j ( t ) d t 0 T e G ( t ) F ( t , v ( t ) ) d t , (1.4)

where G ( t ) = 0 t g ( t ) d t and F ( t , v ) = 0 v f ( t , s ) d s . In [17] and [18], the authors also dealt with some one-dimensional impulsive problems with damped term by variational methods.

For higher dimensional dynamical systems, some interesting results have also been obtained (see [21-23]). In [21], Zhou and Li investigated the second-order Hamiltonian system with impulsive effects:

{ u ¨ ( t ) = F ( t , u ( t ) ) , a.e.  t [ 0 , T ] , u ( 0 ) u ( T ) = u ˙ ( 0 ) u ˙ ( T ) = 0 , Δ u ˙ i ( t j ) = u ˙ i ( t j + ) u ˙ i ( t j ) = I i j ( u i ( t j ) ) , i = 1 , 2 , , N , j = 1 , 2 , , p . (1.5)

By using the least action principle and the saddle point theorem, they obtained some existence results of solutions under sublinear condition and some reasonable conditions. In [22], system (1.5) with F ( t , u ) = 1 2 A ( t ) u u λ W ( t , u ) μ G ( t , u ) , where λ , μ R , was also investigated. By using variational methods, the authors obtained that system (1.5) has at least three weak solutions. In [23], the authors investigated system (1.5) with F ( t , u ) = 1 2 A ( t ) u u W ( t , u ) . They obtained that system (1.5) has infinitely many solutions under the assumptions that nonlinear term is superquadratic, asymptotically quadratic and subquadratic, respectively.

In recent years, via variational methods, some authors have been interested in studying the existence and multiplicity of periodic solutions and homoclinic solutions for the following Lagrangian systems with damped term:

d ( P ( t ) u ˙ ( t ) ) d t + B u ˙ ( t ) + F ( t , u ( t ) ) = 0 , (1.6)

where P ( t ) is a symmetric and continuous N × N matrix-valued function, B is a skew-symmetric N × N constant matrix and F : R × R N R . They obtained some interesting results. We refer readers to [24-27].

In 2010, Li et al.[28] investigated the following system, more general than system (1.6), with P ( t ) I N :

{ u ¨ ( t ) + ( q ( t ) I N × N + B ) u ˙ ( t ) + ( 1 2 q ( t ) B A ( t ) ) u ( t ) + F ( t , u ( t ) ) = 0 , a.e.  t [ 0 , T ] , u ( 0 ) u ( T ) = u ˙ ( 0 ) u ˙ ( T ) = 0 . (1.7)

Motivated by [28], in [29], we investigated the following system, more general than system (1.7):

{ d ( P ( t ) u ˙ ( t ) ) d t + ( q ( t ) P ( t ) + B ) u ˙ ( t ) + ( 1 2 q ( t ) B A ( t ) ) u ( t ) + F ( t , u ( t ) ) = 0 , a.e.  t [ 0 , T ] , u ( 0 ) u ( T ) = P ( 0 ) u ˙ ( 0 ) P ( T ) u ˙ ( T ) = 0 . (1.8)

By variational methods, under superquadratic or subquadratic conditions, we obtained that system (1.8) has infinitely many solutions. One can see more details of our results and more research background of system (1.8) in [29].

In [32], Luo et al. investigated the existence of subharmonic solutions with prescribed minimal period for the following one-dimensional second-order impulsive differential equation:

{ u ( t ) + f ( t , u ( t ) ) = 0 , a.e.  t J , Δ u ( t m ) = I m ( u ( t m ) ) , m Z 0 , (1.9)

where f C ( R × R , R ) , Z 0 = Z + Z , J = R { t m m Z 0 } , I m C ( R , R + { 0 } ) , 0 < t 1 < t 2 < < t p < T , I m + p = I m and t m = t m + p T if m Z + , while t m = t m + p + 1 T if m Z .

In this paper, motivated by [10,15,16,21,28,29] and [32], we focus on the existence of subharmonic weak solutions for system (1.1), which is of impulsive conditions, and we study the problem under asymptotically quadratic conditions. To the best of our knowledge, there are few papers that consider such a problem for system (1.1). We call a solution u subharmonic if u is kT-periodic for some k N .

Let

A = sup t [ 0 , T ] max | x | = 1 , x R N | A ( t ) x | = sup t [ 0 , T ] max { λ ( t ) : λ ( t )  is the eigenvalue of  A τ ( t ) A ( t ) }

and

B = max | x | = 1 , x R N | B x | = max { λ : λ  is the eigenvalue of  B τ B } .

In this paper, we make the following assumptions:

(P) There exists a constant m > 1 2 such that the matrix P ( t ) satisfies

P ( 0 ) = P ( T ) , ( P ( t ) x , x ) > m ( x , x ) for all  ( t , x ) R × { R N / { 0 } } ;

(K1) There exist constants a > 0 and γ ( 0 , 2 ] such that

K ( t , 0 ) = 0 , K ( t , x ) ( B 2 + 2 A 4 + a ) | x | γ for all  x R N  and a.e.  t [ 0 , T ] ;

(K2) ( K ( t , x ) , x ) 2 K ( t , x ) for all x R N and a.e. t [ 0 , T ] ;

(K3) There exists D 1 > 0 such that

K ( t , x ) D 1 | x | 2 for all  x R N  and a.e.  t [ 0 , T ] ;

(W1) lim | x | 0 W ( t , x ) | x | 2 < a uniformly for a.e. t [ 0 , T ] ;

(W2) There exist constants b > 0 and d > 0 such that

W ( t , x ) b | x | 2 + d for all  x R N  and a.e.  t [ 0 , T ] ;

(W3) There exists a function h L 1 ( 0 , T ; R ) such that

e Q ( t ) [ ( W ( t , x ) , x ) 2 W ( t , x ) ] h ( t ) for all  x R N  and a.e.  t [ 0 , T ]

and

lim | x | e Q ( t ) [ ( W ( t , x ) , x ) 2 W ( t , x ) ] = + for a.e.  t [ 0 , T ] ,

where Q ( t ) = 0 t q ( s ) d s ;

(W4) There exists x 0 R N such that

0 T e Q ( t ) [ 1 2 ( A ( t ) x 0 , x 0 ) + K ( t , x 0 ) W ( t , x 0 ) h ( t ) 2 ] d t < 0 ;

(W5) There exists a constant D 2 > A 2 + D 1 such that

W ( t , x ) D 2 | x | 2 for all  x R N  and a.e.  t [ 0 , T ] ;

(I1) There exist constants l j > 0 ( j = 1 , , p ) such that

I j ( x ) l j ( j = 1 , , p )  for all  x R N ;

(I2) j = 1 p e Q ( t j ) I j ( 0 ) = 0 and j = 1 p e Q ( t j ) I j ( x ) 0 for all x R N ;

(I3) There exists a constant C such that

j = 1 p e Q ( t j ) [ 2 I j ( x ) ( I j ( x ) , x ) ] C for all  x R N .

This paper is organized as follows. In Section 2, we present the definition of a subharmonic classical solution, a subharmonic weak solution and the variational structure for system (1.1) and make some preliminaries. In Section 3, we present our main theorems and their proofs. In Section 4, an example is given to verify our main theorems.

2 Preliminaries

In this section, we present the variational structure of system (1.1), which is motivated by [15-17,28] and [29].

Let

H k T 1 = { u : R R N u  is absolutely continuous,  u ( t + k T ) = u ( t )  and  u ˙ L 2 ( 0 , k T ) } .

Define

u , v = 0 k T e Q ( t ) ( u ( t ) , v ( t ) ) d t + 0 k T e Q ( t ) ( P ( t ) u ˙ ( t ) , v ˙ ( t ) ) d t

and

u = [ 0 k T e Q ( t ) | u ( t ) | 2 d t + 0 k T e Q ( t ) ( P ( t ) u ˙ ( t ) , u ˙ ( t ) ) d t ] 1 / 2

for each u , v H k T 1 . Then ( H k T 1 , , ) is a Hilbert space. It is well known that

u H k T 1 = [ 0 k T | u ( t ) | 2 d t + 0 k T | u ˙ ( t ) | 2 d t ] 1 / 2

is also a norm on H k T 1 . Obviously, if the condition (P) holds, u H k T 1 and u are equivalent. Moreover, there exists C 0 k > 0 such that

u = max t [ 0 , k T ] | u ( t ) | C 0 k u H k T 1

(see Proposition 1.1 in [1]). Hence, there exist positive constants C 1 k , C 2 k such that

u C 1 k u , u H k T 1 C 2 k u . (2.1)

For any a , b R , define

H 2 ( a , b ) = { u : R R N both  u  and  u ˙  are absolutely continuous on  ( a , b ) , H 2 ( a , b ) = and  u ¨ L 2 ( a , b ) } , H 2 [ a , b ) = { u : R R N both  u  and  u ˙  are absolutely continuous on  [ a , b ) , H 2 [ a , b ) = and  u ¨ L 2 ( a , b ) } , H 2 ( a , b ] = { u : R R N both  u  and  u ˙  are absolutely continuous on  ( a , b ] , H 2 [ a , b ) = and  u ¨ L 2 ( a , b ) } .

If u H k T 1 , then u ˙ ( t + ) u ˙ ( t ) = 0 may not hold, which leads to impulsive effects.

Definition 2.1 Assume that u H k T 1 H 2 [ 0 , t 1 ) ( j = 1 p 1 H 2 ( t j , t j + 1 ) ) H 2 ( t p , T ] H 2 [ T , k T ] and the limits u ˙ ( t j + ) and u ˙ ( t j ) ( j = 1 , 2 , , p ) exist. If u satisfies system (1.1), then we say that u is a subharmonic classical solution of system (1.1).

Remark 2.1 In [32], impulsive effects may occur periodically in t j , j { 1 , , p } . In order to obtain a sequence of distinct subharmonic weak solutions (see Theorem 3.2 below), different from [32], in Definition 2.1, we assume that the impulsive effects only occur in t j , j = 1 , , p , which belong to ( 0 , T ) . In other words, u is absolutely continuous on ℝ and u ˙ is absolutely continuous on [ 0 , t 1 ) ( j = 1 p 1 ( t j , t j + 1 ) ) ( t p , T ] [ T , k T ] . Moreover, note that u ( t ) = u ( t + k T ) . Then it is easy to see that u ˙ ( 0 ) = u ˙ ( k T ) .

Note that Q ( t ) = 0 t q ( s ) d s . Then, by T-periodicity of q, we have Q ( k T ) = k 0 T q ( t ) d t = 0 . Moreover, obviously, Q ( t ) is continuous on ℝ. We transform system (1.1) into the following system:

{ d ( e Q ( t ) P ( t ) u ˙ ( t ) ) d t + e Q ( t ) B u ˙ ( t ) + e Q ( t ) ( 1 2 q ( t ) B A ( t ) ) u ( t ) + e Q ( t ) F ( t , u ( t ) ) = 0 , a.e.  t R , Δ ( P ( t j ) u ˙ ( t j ) ) = P ( t j ) u ˙ ( t j + ) P ( t j ) u ˙ ( t j ) = I j ( u ( t j ) ) , j = 1 , , p . (2.2)

Then system (2.2) is equivalent to system (1.1) and its solutions are the solutions of system (1.1).

By the idea in [15], we take v H k T 1 and multiply the two sides of the equality

d ( e Q ( t ) P ( t ) u ˙ ( t ) ) d t + e Q ( t ) B u ˙ ( t ) + e Q ( t ) ( 1 2 q ( t ) B A ( t ) ) u ( t ) + e Q ( t ) F ( t , u ( t ) ) = 0

by v and integrate it from 0 to kT. Then we obtain

0 k T ( d ( e Q ( t ) P ( t ) u ˙ ( t ) ) d t + e Q ( t ) B u ˙ ( t ) + e Q ( t ) ( 1 2 q ( t ) B A ( t ) ) u ( t ) + e Q ( t ) F ( t , u ( t ) ) , v ( t ) ) d t = 0 . (2.3)

Note that P ( t + T ) = P ( t ) , P ( t ) is continuous on ℝ and u ˙ ( 0 ) = u ˙ ( k T ) . By integration by parts and the continuity of v, we obtain

0 k T ( d ( e Q ( t ) P ( t ) u ˙ ( t ) ) d t , v ( t ) ) d t = j = 0 p t j t j + 1 ( d ( e Q ( t ) P ( t ) u ˙ ( t ) ) d t , v ( t ) ) d t + T k T ( d ( e Q ( t ) P ( t ) u ˙ ( t ) ) d t , v ( t ) ) d t = j = 0 p [ ( e Q ( t j + 1 ) P ( t j + 1 ) u ˙ ( t j + 1 ) , v ( t j + 1 ) ) ( e Q ( t j ) P ( t j ) u ˙ ( t j + ) , v ( t j ) ) t j t j + 1 ( e Q ( t ) P ( t ) u ˙ ( t ) , v ˙ ( t ) ) d t ] + ( e Q ( k T ) P ( k T ) u ˙ ( k T ) , v ( k T ) ) ( e Q ( T ) P ( T ) u ˙ ( T ) , v ( T ) ) T k T ( e Q ( t ) P ( t ) u ˙ ( t ) , v ˙ ( t ) ) d t = ( e Q ( T ) P ( T ) u ˙ ( T ) , v ( T ) ) ( e Q ( 0 ) P ( 0 ) u ˙ ( 0 ) , v ( 0 ) ) j = 1 p [ ( e Q ( t j ) P ( t j ) u ˙ ( t j + ) , v ( t j ) ) ( e Q ( t j ) P ( t j ) u ˙ ( t j ) , v ( t j ) ) ] 0 T ( e Q ( t ) P ( t ) u ˙ ( t ) , v ˙ ( t ) ) d t + ( e Q ( k T ) P ( k T ) u ˙ ( k T ) , v ( k T ) ) ( e Q ( T ) P ( T ) u ˙ ( T ) , v ( T ) ) T k T ( e Q ( t ) P ( t ) u ˙ ( t ) , v ˙ ( t ) ) d t = ( e Q ( k T ) P ( k T ) u ˙ ( k T ) , v ( k T ) ) ( e Q ( 0 ) P ( 0 ) u ˙ ( 0 ) , v ( 0 ) ) j = 1 p [ ( e Q ( t j ) P ( t j ) u ˙ ( t j + ) , v ( t j ) ) ( e Q ( t j ) P ( t j ) u ˙ ( t j ) , v ( t j ) ) ] 0 k T ( e Q ( t ) P ( t ) u ˙ ( t ) , v ˙ ( t ) ) d t = j = 1 p e Q ( t j ) ( Δ ( P ( t j ) u ˙ ( t j ) ) , v ( t j ) ) 0 k T ( e Q ( t ) P ( t ) u ˙ ( t ) , v ˙ ( t ) ) d t . (2.4)

Definition 2.2 u H k T 1 is called a subharmonic weak solution of system (1.1) if

0 k T ( e Q ( t ) P ( t ) u ˙ ( t ) , v ˙ ( t ) ) d t + j = 1 p e Q ( t j ) ( I j ( u ( t j ) ) , v ( t j ) ) = 0 k T e Q ( t ) [ 1 2 q ( t ) ( B u ( t ) , v ( t ) ) + ( B u ˙ ( t ) , v ( t ) ) ( A ( t ) u ( t ) , v ( t ) ) + ( F ( t , u ( t ) ) , v ( t ) ) ] d t

holds for any v H k T 1 .

Lemma 2.1If u H k T 1 is a subharmonic weak solution of system (1.1), thenuis a subharmonic classical solution of system (1.1).

Proof Motivated by [15], for j { 0 , 1 , 2 , , p } , choose v H k T 1 with v ( t ) = 0 for every t [ 0 , t j ] [ t j + 1 , k T ] . Then, by Definition 2.2, we obtain

t j t j + 1 ( e Q ( t ) P ( t ) u ˙ ( t ) , v ˙ ( t ) ) d t = t j t j + 1 e Q ( t ) [ 1 2 q ( t ) ( B u ( t ) , v ( t ) ) + ( B u ˙ ( t ) , v ( t ) ) ( A ( t ) u ( t ) , v ( t ) ) + ( F ( t , u ( t ) ) , v ( t ) ) ] d t . (2.5)

Choose v H k T 1 with v ( t ) = 0 for every t [ 0 , T ] . Then we obtain

T k T ( e Q ( t ) P ( t ) u ˙ ( t ) , v ˙ ( t ) ) d t = T k T e Q ( t ) [ 1 2 q ( t ) ( B u ( t ) , v ( t ) ) + ( B u ˙ ( t ) , v ( t ) ) ( A ( t ) u ( t ) , v ( t ) ) + ( F ( t , u ( t ) ) , v ( t ) ) ] d t . (2.6)

Equations (2.5) and (2.6) imply that u H k T 1 H 2 [ 0 , t 1 ) ( j = 1 p 1 H 2 ( t j , t j + 1 ) ) H 2 ( t p , T ] H 2 [ T , k T ] and u satisfies

d ( e Q ( t ) P ( t ) u ˙ ( t ) ) d t + e Q ( t ) B u ˙ ( t ) + e Q ( t ) ( 1 2 q ( t ) B A ( t ) ) u ( t ) + e Q ( t ) F ( t , u ( t ) ) = 0 , a.e.  t [ 0 , k T ] .

Multiplying the above equality by v and integrating between 0 and kT, combining the argument of (2.4) and Definition 2.2, we obtain that

j = 1 p e Q ( t j ) ( Δ ( P ( t j ) u ˙ ( t j ) ) , v ( t j ) ) = j = 1 p e Q ( t j ) ( I j ( u ( t j ) ) , v ( t j ) ) .

Hence, Δ ( P ( t j ) u ˙ ( t j ) ) = I j ( u ( t j ) ) for every j = 1 , 2 , , p . This completes the proof. □

For every k N , define φ k : H k T 1 R by

φ k ( u ) = 0 k T e Q ( t ) [ 1 2 ( P ( t ) u ˙ ( t ) , u ˙ ( t ) ) + 1 2 ( B u ( t ) , u ˙ ( t ) ) + 1 2 ( A ( t ) u ( t ) , u ( t ) ) F ( t , u ( t ) ) ] d t + j = 1 p e Q ( t j ) I j ( u ( t j ) ) .

It follows from assumption (A) and Theorem 1.4 in [1] that the functional φ k is continuously differentiable and

φ k ( u ) , v = 0 k T e Q ( t ) [ ( P ( t ) u ˙ ( t ) , v ˙ ( t ) ) 1 2 q ( t ) ( B u ( t ) , v ( t ) ) ( B u ˙ ( t ) , v ( t ) ) + ( A ( t ) u ( t ) , v ( t ) ) ( F ( t , u ( t ) ) , v ( t ) ) ] d t + j = 1 p e Q ( t j ) ( I j ( u ( t j ) ) , v ( t j ) ) (2.7)

for u , v H k T 1 . Obviously, if u 0 H k T 1 is a critical point of φ k , i.e., φ k ( u 0 ) = 0 , then u 0 is a subharmonic weak solution of system (1.1).

We will use the following mountain pass theorem to prove our results.

Lemma 2.2 (see [30])

LetEbe a real Banach space, and let ϕ C 1 ( E , R ) satisfy the (PS) condition. Ifϕsatisfies the following conditions:

(i) ϕ ( 0 ) = 0 ;

(ii) There exist constants ρ , α > 0 such that ϕ | B ρ ( 0 ) α ;

(iii) There exists e E / B ¯ ρ ( 0 ) such that ϕ ( e ) 0 , thenϕpossesses a critical value c α given by

c = inf g Γ max s [ 0 , 1 ] ϕ ( g ( s ) ) ,

where B ρ ( 0 ) is an open ball inEof radiusρcentered at 0, and

Γ = { g C ( [ 0 , 1 ] , E ) : g ( 0 ) = 0 , g ( 1 ) = e } .

Remark 2.2 As shown in [31], a deformation lemma can be proved by replacing the usual (PS)-condition with the condition (C), and it turns out that Lemma 2.2 holds true under the condition (C). We say that ϕ satisfies the condition (C), i.e., for every sequence { u n } E , { u n } has a convergent subsequence if ϕ ( u n ) is bounded and ( 1 + u n ) ϕ ( u n ) 0 as n .

3 Main results

Theorem 3.1Assume that (P), (K1), (K2), (W1)-(W4) and (I1)-(I3) hold. Then, for every k N , system (1.1) has at least onekT-periodic weak solution in H k T 1 .

Proof We use Lemma 2.2 to prove the theorem. Let E = H k T 1 .

Step 1. We prove that φ k satisfies assumption (ii) of Lemma 2.2. It follows from (W1) and (W2) that there exist 0 < ε < m 2 1 4 , θ > 2 and C 1 > 0 such that

W ( t , x ) ( a ε ) | x | 2 + C 1 | x | θ . (3.1)

Choose 0 < δ < 1 such that ε ( δ C 1 k ) 2 C 1 C 1 k θ ( δ C 1 k ) θ 0 k T e Q ( t ) d t > 0 . Then it follows from (K1), (I2), (3.1) and (2.1) that for all u H k T 1 with u = δ / C 1 k : = ρ k ,

φ k ( u ) = 0 k T e Q ( t ) [ 1 2 ( P ( t ) u ˙ ( t ) , u ˙ ( t ) ) + 1 2 ( B u ( t ) , u ˙ ( t ) ) + 1 2 ( A ( t ) u ( t ) , u ( t ) ) ] d t + 0 k T e Q ( t ) [ K ( t , u ( t ) ) W ( t , u ( t ) ) ] d t + j = 1 p e Q ( t j ) I j ( u ( t j ) ) 0 k T e Q ( t ) [ m 2 | u ˙ ( t ) | 2 1 4 ( B 2 | u ( t ) | 2 + | u ˙ ( t ) | 2 ) A | u ( t ) | 2 2 ] d t + ( B 2 + 2 A 4 + a ) 0 k T e Q ( t ) | u ( t ) | γ d t ( a ε ) 0 k T e Q ( t ) | u ( t ) | 2 d t C 1 0 k T e Q ( t ) | u ( t ) | θ d t ( m 2 1 4 ) 0 k T e Q ( t ) | u ˙ ( t ) | 2 d t + a 0 k T e Q ( t ) | u ( t ) | γ d t ( a ε ) 0 k T e Q ( t ) | u ( t ) | 2 d t C 1 0 k T e Q ( t ) | u ( t ) | θ d t ( m 2 1 4 ) 0 k T e Q ( t ) | u ˙ ( t ) | 2 d t + ε 0 k T e Q ( t ) | u ( t ) | 2 d t C 1 u θ 0 k T e Q ( t ) d t ε u 2 C 1 C 1 k θ u θ 0 k T e Q ( t ) d t = ε ( δ C 1 k ) 2 C 1 C 1 k θ ( δ C 1 k ) θ 0 k T e Q ( t ) d t = : α k > 0 .

Step 2. We prove that φ k satisfies assumption (iii) of Lemma 2.2. Set φ ( s ) = s 2 W ( t , s x 0 ) for s > 0 . By the argument in [10], we know that (W3) implies that

W ( t , s x 0 ) s 2 W ( t , x 0 ) + h ( t ) 2 ( s 2 1 ) for a.e.  t [ 0 , T ] , s > 1 , (3.2)

and (K2) implies that

K ( t , s x 0 ) s 2 K ( t , x 0 ) for a.e.  t [ 0 , T ] , s > 1 . (3.3)

It follows from (3.2), (3.3), (W3) and (I1) that for sufficiently large s,

φ k ( s x 0 ) = s 2 2 0 k T e Q ( t ) ( A ( t ) x 0 , x 0 ) d t + 0 k T e Q ( t ) [ K ( t , s x 0 ) W ( t , s x 0 ) ] d t + j = 1 p e Q ( t j ) I j ( s x 0 ) = s 2 2 k 0 T e Q ( t ) ( A ( t ) x 0 , x 0 ) d t + k 0 T e Q ( t ) [ K ( t , s x 0 ) W ( t , s x 0 ) ] d t + j = 1 p e Q ( t j ) I j ( s x 0 ) s 2 k 0 T e Q ( t ) [ 1 2 ( A ( t ) x 0 , x 0 ) + K ( t , x 0 ) W ( t , x 0 ) h ( t ) 2 ] d t + 1 2 0 k T e Q ( t ) h ( t ) d t + j = 1 p e Q ( t j ) l j .

By (W4), we can choose sufficiently large s 0 k such that s 0 k x 0 > ρ k and φ k ( s 0 k x 0 ) 0 . Let e k = s 0 k x 0 . Then φ k satisfies assumption (iii) of Lemma 2.2.

Step 3. We prove that φ satisfies the (C)-condition on H k T 1 . The proof is motivated by [10]. For every { u n } H k T 1 , assume that there exists a constant C 3 k > 0 such that

| φ k ( u n ) | C 3 k , ( 1 + u n ) φ k ( u n ) C 3 k for all  n N . (3.4)

Then it follows from antisymmetry of B, (K2) and (I3) that

3 C 3 k 2 φ k ( u n ) ( φ k ( u n ) , u n ) = 1 2 0 k T e Q ( t ) q ( t ) ( B u n ( t ) , u n ( t ) ) d t + 0 k T e Q ( t ) [ ( F ( t , u n ( t ) ) , u n ( t ) ) 2 F ( t , u n ( t ) ) ] d t + j = 1 p e Q ( t j ) [ 2 I j ( u n ( t j ) ) ( I j ( u n ( t j ) ) , u n ( t j ) ) ] = 0 k T e Q ( t ) [ 2 K ( t , u n ( t ) ) ( K ( t , u n ( t ) ) , u n ( t ) ) ] d t + 0 k T e Q ( t ) [ ( W ( t , u n ( t ) ) , u n ( t ) ) 2 W ( t , u n ( t ) ) ] d t + j = 1 p e Q ( t j ) [ 2 I j ( u n ( t j ) ) ( I j ( u n ( t j ) ) , u n ( t j ) ) ] 0 k T e Q ( t ) [ ( W ( t , u n ( t ) ) , u n ( t ) ) 2 W ( t , u n ( t ) ) ] d t + C . (3.5)

Next we prove that { u n } is bounded. Assume that u n as n . Let z n = u n u n . Then z n = 1 , and so there exists a subsequence, still denoted by { z n } , such that z n z on H k T 1 . Then, by Proposition 1.2 in [1], we get z n z 0 . Hence, we have 0 k T | z n ( t ) z ( t ) | 2 d t 0 and z n ( t ) z ( t ) for a.e. t [ 0 , k T ] . Thus, by conditions (P), (W2) and (I2), we have

φ k ( u n ) = 0 k T e Q ( t ) [ 1 2 ( P ( t ) u ˙ n ( t ) , u ˙ n ( t ) ) + 1 2 ( B u n ( t ) , u ˙ n ( t ) ) + 1 2 ( A ( t ) u n ( t ) , u n ( t ) ) + K ( t , u n ( t ) ) W ( t , u n ( t ) ) ] d t + j = 1 p e Q ( t j ) I j ( u n ( t j ) ) 0 k T e Q ( t ) [ 1 2 ( P ( t ) u ˙ n ( t ) , u ˙ n ( t ) ) B 2 | u n ( t ) | 2 + | u ˙ n ( t ) | 2 4 A | u n ( t ) | 2 2 ] d t b 0 k T e Q ( t ) | u n ( t ) | 2 d t d 0 k T e Q ( t ) d t ( 1 2 1 4 m ) 0 k T e Q ( t ) [ ( P ( t ) u ˙ n ( t ) , u ˙ n ( t ) ) ] d t B 2 + 2 A + 4 b 4 0 k T e Q ( t ) | u n ( t ) | 2 d t d 0 k T e Q ( t ) d t = ( 1 2 1 4 m ) u n 2 ( 1 2 1 4 m ) 0 k T e Q ( t ) | u n ( t ) | 2 d t B 2 + 2 A + 4 b 4 0 k T e Q ( t ) | u n ( t ) | 2 d t d 0 k T e Q ( t ) d t = ( 1 2 1 4 m ) u n 2 [ B 2 + 2 A + 4 b 4 + ( 1 2 1 4 m ) ] 0 k T e Q ( t ) | u n ( t ) | 2 d t d 0 k T e Q ( t ) d t .

Hence, we have

φ k ( u n ) u n 2 ( 1 2 1 4 m ) [ B 2 + 2 A + 4 b 4 + ( 1 2 1 4 m ) ] 0 k T e Q ( t ) | u n ( t ) | 2 u n 2 d t d 0 k T e Q ( t ) d t u n 2 .

Let n . Then, by (3.4), we get

1 2 1 4 m [ B 2 + 2 A + 4 b 4 + ( 1 2 1 4 m ) ] 0 k T e Q ( t ) | z n ( t ) | 2 d t . (3.6)

Then it follows from m > 1 2 and (3.6) that 0 k T e Q ( t ) | z ( t ) | 2 d t > 0 and so z 0 . Let S = { t [ 0 , k T ] : lim | x | e Q ( t ) [ 2 W ( t , x ) ( W ( t , x ) , x ) ] = + } and S 1 = { t S : z ( t ) 0 } . Then meas S 1 > 0 and

(3.7)

Let f n ( t ) = e Q ( t ) [ ( W ( t , u n ( t ) ) , u n ( t ) ) 2 W ( t , u n ( t ) ) ] . Then (3.7) and T-periodicity of W ( t , x ) in t imply that

(3.8)

It follows from (3.8) and Lemma 1 in [6] that there exists a subset S 2 of S 1 with meas S 2 > 0 such that

lim n f n ( t ) = + uniformly for  t S 2 . (3.9)

By (W3), we have

Let n . Then by Fatou’s lemma and (3.9), we have

0 k T e Q ( t ) [ ( W ( t , u n ( t ) ) , u n ( t ) ) 2 W ( t , u n ( t ) ) ] d t + ,

which contradicts (3.5). Hence { u n } is bounded. Going if necessary to a subsequence, assume that u n u in H k T 1 . Then, by Proposition 1.2 in [1], we have u n u 0 and so 0 k T e Q ( t ) | u n u | 2 d t 0 as n . Similar to the argument of Theorem 3.1 in [28], it is easy to obtain that 0 k T e Q ( t ) ( P ( t ) ( u ˙ n u ˙ ) , u ˙ n u ˙ ) d t 0 . Hence, u n u 0 as n . Hence, φ k satisfies the (C)-condition.

Finally, (K1), (W1) and (I2) imply that φ k ( 0 ) = 0 . Hence, combining Step 1-Step 3 with Lemma 2.2 and Remark 2.2, we obtain that φ k has at least a critical point u k in H k T 1 and φ k ( u k ) = c k α k > 0 . Then system (1.1) has at least one kT-periodic solution u k in H k T 1 . This completes the proof. □

Remark 3.1 It is easy to see that Theorem 3.1 generalizes Theorem A. To be precise, when k = 1 , P ( t ) I N , q ( t ) 0 , B = 0 , A ( t ) 0 and I j ( x ) 0 , Theorem 3.1 reduces to Theorem A.

Theorem 3.2Assume (P), (K1)-(K3), (W1)-(W5) and (I1)-(I3) hold. Then system (1.1) has a sequence of distinct subharmonic weak solutions with period k j T satisfying k j N and k j as j .

Proof By Theorem 3.1, we know that for every k N , system (1.1) has at least one kT-periodic solution u k in H k T 1 and φ k ( u k ) = c k α k > 0 . By Lemma 2.2, we have

φ k ( u k ) = c k = inf g Γ max s [ 0 , 1 ] φ k ( g ( s ) ) ,

where

Γ = { g C ( [ 0 , 1 ] , E ) : g ( 0 ) = 0 , g ( 1 ) = e k } .

Let g ( s ) = s e k = s s 0 k x 0 , s [ 0 , 1 ] . Obviously, g Γ . Hence, by (K3), (W5) and (I1), we have

φ k ( u k ) max s [ 0 , 1 ] φ k ( s s 0 x 0 ) = max s [ 0 , 1 ] { 0 k T e Q ( t ) [ 1 2 ( A ( t ) s s 0 x 0 , s s 0 x 0 ) + K ( t , s s 0 x 0 ) W ( t , s s 0 x 0 ) ] d t + j = 1 p e Q ( t j ) I j ( s s 0 x 0 ) } = max s [ 0 , 1 ] { k 0 T e Q ( t ) [ 1 2 ( A ( t ) s s 0 x 0 , s s 0 x 0 ) + K ( t , s s 0 x 0 ) W ( t , s s 0 x 0 ) ] d t + j = 1 p e Q ( t j ) I j ( s s 0 x 0 ) } max s [ 0 , 1 ] { k 0 T e Q ( t ) [ A 2 | s s 0 x 0 | 2 + D 1 | s s 0 x 0 | 2 D 2 | s s 0 x 0 | 2 ] d t + j = 1 p e Q ( t j ) l j } = max s [ 0 , 1 ] { [ A 2 + D 1 D 2 ] s 2 s 0 2 | x 0 | 2 k 0 T e Q ( t ) d t + j = 1 p e Q ( t j ) l j } j = 1 p e Q ( t j ) l j : = M . (3.10)

Hence, φ k ( u k ) is uniformly bounded for all k N .

Obviously, we can find k 1 N / { 1 } such that k 1 > M α 1 , then we claim that u k is distinct to u 1 for all k k 1 . In fact, if u k = u 1 for some k k 1 , it is easy to check that

φ k ( u k ) = k φ 1 ( u 1 ) k α 1 .

Then, by (3.10), we have k 1 k M α 1 , a contradiction. We can also find k 2 > max { k 1 , k 1 M α k 1 } such that u k 1 k u k 1 for all k k 2 k 1 . Otherwise, if u k 1 k = u k 1 for some k k 1 , we have φ k 1 k ( u k 1 k ) = k φ k 1 ( u k 1 ) k α k 1 . Then by (3.10), we have k 2 k 1 k M α k 1 , a contradiction. In the same way, we can obtain that system (1.1) has a sequence of distinct periodic solutions with period k j T satisfying k j N and k j as j . This completes the proof. □

4 Example

The following example is inspired partially by Example 3.1 in [10]. Let T = 2 π , N = 3 . Consider the following impulsive Lagrangian system with damped term:

{ d ( P ( t ) u ˙ ( t ) ) d t + ( q ( t ) P ( t ) + B ) u ˙ ( t ) + ( 1 2 q ( t ) B A ( t ) ) u ( t ) + F ( t , u ( t ) ) = 0 , a.e.  t R , Δ ( P ( π 2 ) u ˙ ( π 2 ) ) = P ( π 2 ) u ˙ ( π 2 + ) P ( π 2 ) u ˙ ( π 2 ) = I 1 ( u ( π 2 ) ) , Δ ( P ( π ) u ˙ ( π ) ) = P ( π ) u ˙ ( π + ) P ( π ) u ˙ ( π ) = I 2 ( u ( π ) ) , (4.1)

where

P ( t ) = ( sin t + 2 0 0 0 sin t + 2 0 0 0 cos t + 2 ) , A ( t ) = ( cos t 0 0 0 cos t 0 0 0 cos t ) , B = ( 0 1 0 1 0 1 0 1 0 ) , q ( t ) = sin t , K ( t , x ) = 2 ( | sin t 2 | + 6.5 ) | x | 2 , W ( t , x ) = 5 ( | cos t 2 | + 1 ) | x | 2 ( 2 1 ln ( e + | x | 2 ) ) , I 1 ( x ) = | x | 2 1 + | x | 2 , I 2 ( x ) = ln ( 1 + | x | 2 ) | x | 2 .

Obviously, the condition (P) holds and A = 1 , B = 2 and (K1), (K2), (W1) and (W2) hold with a = 11 and γ = 2 .

e Q ( t ) [ ( W ( t , x ) , x ) 2 W ( t , x ) ] = 10 ( | cos t 2 | + 1 ) e 1 cos t | x | 4 ( e + | x | 2 ) [ ln ( e + | x | 2 ) ] 2 .

Then (W3) holds with h ( t ) 0 . Moreover,

0 2 π e Q ( t ) [ 1 2 ( A ( t ) x , x ) + K ( t , x ) W ( t , x ) h ( t ) 2 ] d t = 0 2 π e 1 cos t [ 1 2 cos t | x | 2 + 2 ( | sin t 2 | + 6.5 ) | x | 2 5 ( | cos t 2 | + 1 ) | x | 2 ( 2 1 ln ( e + | x | 2 ) ) ] d t e 2 0 2 π [ 1 2 cos t | x | 2 + 2 ( | sin t 2 | + 6.5 ) | x | 2 5 ( | cos t 2 | + 1 ) | x | 2 ( 2 1 ln ( e + | x | 2 ) ) ] d t = e 2 [ ( 8 + 26 π ) | x | 2 ( 20 + 10 π ) | x | 2 ( 2 1 ln ( e + | x | 2 ) ) ] .

Hence, it is easy to see that there exists x 0 R N such that (W4) holds by the above inequality. Obviously, (I1) and (I2) hold. Note that

j = 1 2 e Q ( t j ) [ 2 I j ( x ) ( I j ( x ) , x ) ] = e Q ( π 2 ) [ 2 | x | 2 1 + | x | 2 2 | x | 2 ( 1 + | x | 2 ) 2 ] + e Q ( π ) [ 2 ln ( 1 + | x | 2 ) | x | 2 2 | x | 2 1 + | x | 2 2 ln ( 1 + | x | 2 ) | x | 2 ] = 2 | x | 2 1 + | x | 2 e Q ( π 2 ) [ 1 1 1 + | x | 2 ] + e Q ( π ) [ 4 ln ( 1 + | x | 2 ) | x | 2 2 1 + | x | 2 ] e Q ( π ) 2 1 + | x | 2 2 e Q ( π ) .

So (I3) holds. Hence, by Theorem 3.1, we obtain that system (4.1) has at least one kT-periodic solution for every k N .

Moreover, it is easy to see that (K3) holds with D 1 = 4 . Since

W ( t , x ) = 5 ( | cos t 2 | + 1 ) | x | 2 ( 2 1 ln ( e + | x | 2 ) ) 5 | x | 2 ( 2 1 ln ( e + | x | 2 ) ) 5 | x | 2 .

Choose D 2 = 5 . Then (W5) holds. Hence, by Theorem 3.2, we obtain that system (4.1) has a sequence of distinct subharmonic weak solutions with period k j T satisfying k j N and k j as j .

Competing interests

The author declares that he has no competing interests.

Author’s contributions

The author read and approved the final manuscript.

Acknowledgements

This work is supported by Tianyuan Fund for Mathematics of the National Natural Science Foundation of China (No. 11226135) and the Fund for Fostering Talents in Kunming University of Science and Technology (No. KKSY201207032).

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