Open Access

Periodic solutions to a generalized Liénard neutral functional differential system with p-Laplacian

Journal of Inequalities and Applications20122012:270

https://doi.org/10.1186/1029-242X-2012-270

Received: 24 April 2012

Accepted: 8 November 2012

Published: 27 November 2012

Abstract

By means of the generalized Mawhin’s continuation theorem, we present some sufficient conditions which guarantee the existence of at least one T-periodic solution for a generalized Liénard neutral functional differential system with p-Laplacian.

MSC:34B15, 34L30.

Keywords

periodic solutionsneutral equationsgeneralized Mawhin’s continuation theorem

1 Introduction

This paper is devoted to investigating the following p-Laplacian Liénard neutral differential system:
( φ p ( ( x ( t ) B x ( t τ ) ) ) ) + f ( x ( t ) ) x ( t ) + g ( x ( t γ ( t ) ) ) = e ( t ) ,
(1.1)
where x ( t ) = ( x 1 ( t ) , x 2 ( t ) , , x n ( t ) ) ;

e C ( R , R n ) with e ( t + T ) = e ( t ) ; γ C ( R , R ) with γ ( t + T ) = γ ( t ) ; τ is a given constant; B = [ b i j ] n × n is a real matrix with | B | = ( i = 1 n j = 1 n | b i j | 2 ) 1 / 2 .

When the matrix B is a constant, Zhang [1] studied the properties of a difference operator A and obtained the following results: define the operator A on C T
A : C T C T , [ A x ] ( t ) = x ( t ) c x ( t τ ) , t R ,
(1.2)
where C T = { x : x C ( R , R ) , x ( t + T ) x ( t ) } , c is a constant. If | c | 1 , then A has a unique continuous bounded inverse A 1 satisfying
[ A 1 f ] ( t ) = { j 0 c j f ( t j τ ) , if  | c | < 1 , f C T , j 1 c j f ( t + j τ ) , if  | c | > 1 , f C T .
On the basis of Zhang’s work, Lu [2] further studied the properties of the difference operator A and gave the following inequality properties for A:
  1. (1)

    A 1 1 | 1 | k | | ;

     
  2. (2)

    0 T | [ A 1 f ] ( t ) | d t 1 | 1 | k | | 0 T | f ( t ) | d t , f C T ;

     
  3. (3)

    0 T | [ A 1 f ] ( t ) | 2 d t 1 | 1 | k | | 0 T | f ( t ) | 2 d t , f C T .

     
After that, by using the above results, many researchers studied the existence of periodic solutions for some kinds of differential equations; see [37]. In a recent paper [8], when the constant c of (1.2) is a variable c ( t ) , we generalized the results of [1] and obtained the following results. If | c ( t ) | 1 , then the operator A has continuous inverse A 1 on C T , satisfying
  1. (1)
    [ A 1 f ] ( t ) = { f ( t ) + j = 1 i = 1 j c ( t ( i 1 ) τ ) f ( t j τ ) , c 0 < 1 , f C T , f ( t + τ ) c ( t + τ ) j = 1 i = 1 j + 1 1 c ( t + i τ ) f ( t + j τ + τ ) , σ > 1 , f C T .
     
  2. (2)
    0 T | [ A 1 f ] ( t ) | d t { 1 1 c 0 0 T | f ( t ) | d t , c 0 < 1 , f C T , 1 σ 1 0 T | f ( t ) | d t , σ > 1 , f C T .
     

Using the above results, we have obtained some existence results of periodic solutions for first-order, second-order and p-Laplacian neutral equations with a variable parameter; see [911].

However, when B of (1.1) is a matrix, there are few existence results of periodic solutions for neutral differential systems. In [12], when B is a symmetric matrix, the authors studied a second-order p-Laplacian neutral functional differential system and obtained the existence of periodic solutions. In [13], when B is a general matrix, the authors studied a second-order neutral differential system. But for p-Laplacian functional differential system, to the best our knowledge, there are no results on the existence of periodic solutions. Hence, in this paper, we will study system (1.1) and obtain the existence of periodic solutions by using the generalization of Mawhin’s continuation theorem.

2 Main lemmas

In this section, we give some notations and lemmas which will be used in this paper. Let
X = C T 1 with the norm x = max { | x | 0 , | x | 0 } , Z = C T with the norm
| x | 0 = max 0 t T | x ( t ) | , | x ( t ) | = ( i = 1 n x i 2 ) 1 / 2 ,
U is a complex such that
U B U 1 = E λ = diag ( J 1 , J 2 , , J n )
(2.1)
is a Jordan’s normal matrix, where
J i = ( λ i 1 0 0 0 0 λ i 1 0 0 0 0 0 λ i 1 0 0 0 0 λ i ) n i × n i
with i = 1 l n i = n , { λ i : i = 1 , 2 , , l } is the set of eigenvalues of matrix B. Let
A 1 : C T C T , [ A 1 x ] ( t ) = x ( t ) B x ( t τ ) .
(2.2)

Furthermore, we suppose that γ ( t ) C 1 ( R , R ) with γ ( t ) < 1 , t R . It is obvious that the function t γ ( t ) has a unique inverse denoted by μ ( t ) .

Lemma 2.1 ([13])

Suppose that the matrix U and the operator A 1 are defined by (2.1) and (2.2), respectively, and for all i = 1 , 2 , , l , | λ i | 1 . Then A 1 has its inverse A 1 1 : C T C T with the following properties:
  1. (1)

    A 1 1 | U 1 | | U | σ 0 , σ 0 = i = 1 l j = 1 n i k = 1 j 1 | 1 λ i | k .

     
  2. (2)
    For all f C T , 0 T | [ A 1 1 f ] ( s ) | p d s | U 1 | p | U | p σ 1 0 T | f ( s ) | p d s , p [ 1 , + ) , where
    σ 1 = { i = 1 l j = 1 n i ( k = 1 j 1 | 1 λ i | k ) 2 , p = 2 , n 2 p 2 [ i = 1 l j = 1 n i ( k = 1 j 1 | 1 λ i | k ) q ] p q , p [ 1 , 2 ) , [ i = 1 l j = 1 n i ( k = 1 j 1 | 1 λ i | k ) q ] p q , p [ 2 , + )
     
and q > 0 is a constant with 1 / p + 1 / q = 1 .
  1. (3)

    A 1 1 f C T 1 , [ A 1 1 f ] ( t ) = [ A 1 1 f ] ( t ) , for all f C T 1 , t R .

     

Definition 2.1 ([14])

Let X and Z be two Banach spaces with norms X , Z , respectively. A continuous operator
M : X dom M Z
is said to be quasi-linear if
  1. (i)

    Im M : = M ( X dom M ) is a closed subset of Z;

     
  2. (ii)

    Ker M : = { x X dom M : M x = 0 } is linearly homeomorphic to R n , n < .

     

Definition 2.2 ([14])

Let Ω X be an open and bounded set with the origin θ Ω , N λ : Ω ¯ Z , λ [ 0 , 1 ] is said to be M-compact in Ω ¯ if there exists a subset Z 1 of Z satisfying dim Z 1 = dim Ker M and an operator R : Ω ¯ × [ 0 , 1 ] X 2 being continuous and compact such that for λ [ 0 , 1 ] ,
  1. (a)

    ( I Q ) N λ ( Ω ¯ ) Im M ( I Q ) Z ,

     
  2. (b)

    Q N λ x = 0 , λ ( 0 , 1 ) Q N x = 0 , x Ω ,

     
  3. (c)

    R ( , 0 ) 0 and R ( , λ ) | λ = ( I P ) | λ ,

     
  4. (d)

    M [ P + R ( , λ ) ] = ( I Q ) N λ , λ [ 0 , 1 ] ,

     

where X 2 is the complement space of KerM in X, i.e., X = Ker M X 2 ; P, Q are two projectors satisfying Im P = Ker M , Im Q = Z 1 , N = N 1 , λ = { x Ω ¯ : M x = N λ x } .

Lemma 2.2 ([14])

Let X and Z be two Banach spaces with norms X , Z , respectively and Ω X be an open and bounded nonempty set. Suppose
M : X dom M Z

is quasi-linear and N λ : Ω ¯ Z , λ [ 0 , 1 ] is M-compact in Ω ¯ . In addition, if the following conditions hold:

(A1) M x N λ x , ( x , λ ) Ω × ( 0 , 1 ) ;

(A2) Q N x 0 , x Ker M Ω ;

(A3) deg { J Q N , Ω Ker M , 0 } 0 , J : Im Q Ker M is a homeomorphism.

Then the abstract equation M x = N x has at least one solution in dom M Ω ¯ .

Lemma 2.3 ([15])

Let s , σ C ( R , R ) with s ( t + T ) s ( t ) and σ ( t + T ) σ ( t ) . Suppose that the function t σ ( t ) has a unique inverse μ ( t ) , t R . Then s ( μ ( t + T ) ) s ( μ ( t ) ) .

For fixed l Z and a R n , define
G l ( a ) = 1 T 0 T φ p 1 ( a + l ( t ) ) d t .

Lemma 2.4 ([16])

The function G l has the following properties:
  1. (1)
    For any fixed l Z , there must be a unique a ˜ = a ˜ ( l ) such that the equation
    G l ( a ) = 0 .
     
  2. (2)

    The function a ˜ : Z R n defined as above is continuous and sends bounded sets into bounded sets.

     

Lemma 2.5 ([17])

Let p ( 1 , + ) be a constant, s C ( R , R ) such that s ( t ) s ( t + T ) , u X . Then
0 T | u ( t ) u ( t s ( t ) ) | p d t 2 ( max t [ 0 , T ] | s ( t ) | ) p 0 T | u ( t ) | p d t .

3 Main results

For convenience of applying Lemma 2.2, the operators A, M, N λ are defined by
(3.1)
(3.2)
(3.3)
where dom M = { x X : φ p [ ( A x ) ] C T 1 } . For convenience of the proof, let
F ( t , x ) = f ( x ( t ) ) x ( t ) g ( x ( t γ ( t ) ) ) + e ( t ) ,
(3.4)
then ( N λ x ) ( t ) = λ F . By (3.1)-(3.3), Eq. (1.1) is equivalent to the operator equation N x = M x , where N 1 = N . Then we have

Since Ker M R n , ImM is a closed set in Z, then we have the following.

Lemma 3.1 Let M be as defined by (3.2), then M is a quasi-linear operator.

Let

Lemma 3.2 If f, g, e, γ satisfy the above conditions, then N λ is M-compact.

Proof Let Z 1 = Im Q . For any bounded set Ω ¯ X , define R : Ω ¯ × [ 0 , 1 ] Ker P ,
R ( x , λ ) ( t ) = A 1 { 0 t φ q [ a x + 0 s λ ( F ( r , x ( r ) ) ( Q F ) ( r ) ) d r ] d s } , t [ 0 , T ] ,

where F is defined by (3.4) and a x is a constant vector in R n which depends on x. By Lemma 2.4, we know that a x exists uniquely. Hence, R ( x , λ ) ( t ) is well defined.

We first show that R ( , λ ) is completely continuous on Ω ¯ × [ 0 , 1 ] . Let
G λ ( t ) = 0 t φ q [ a x + 0 s λ ( F ( r , x ( r ) ) ( Q F ) ( r ) ) d r ] d s , t [ 0 , T ] ,
we have
R ( x , λ ) ( t ) = [ A 1 G λ ] ( t ) .
From the properties of f, g, e, γ, obviously, x Ω ¯ , G λ ( t ) C T . Then by Lemma 2.1 R ( x , λ ) is uniformly bounded on Ω ¯ × [ 0 , 1 ] . Now, we show R ( x , λ ) is equicontinuous. t 1 , t 2 [ 0 , T ] , ε > 0 is sufficiently small, then there exists δ > 0 , for | t 1 t 2 | < δ , by G λ , A 1 G λ C T we have
| [ A 1 G λ ] ( t 1 ) [ A 1 G λ ] ( t 2 ) | < ε .

Hence, R ( x , λ ) is equicontinuous on Ω ¯ × [ 0 , 1 ] . By using the Arzelà-Ascoli theorem, we have R ( x , λ ) is completely continuous on Ω ¯ × [ 0 , 1 ] .

Secondly, we show that N λ is M-compact in four steps, i.e., the conditions of Definition 2.2 are all satisfied.

Step 1. By Q 2 = Q , we have Q ( I Q ) N λ ( Ω ¯ ) = θ , so ( I Q ) N λ ( Ω ¯ ) Ker Q = Im M , here θ is an n-dimension zero vector. On the other hand, z Im M . Clearly, Q z = θ , so z = z Q z = ( I Q ) z , then z ( I Q ) Z . So, we have
( I Q ) N λ ( Ω ¯ ) Im M ( I Q ) Z .

Step 2. We show that Q N λ x = θ , λ ( 0 , 1 ) Q N x = θ , x Ω . Because Q N λ x = 1 T λ F d r = θ , we get 1 T F d r = θ , i.e., Q N x = θ . The inverse is true.

Step 3. When λ = 0 , from the above proof, we have a x = θ . So, we get R ( , 0 ) = θ . x λ = { x Ω ¯ : M x = N λ x } , we have ( φ p [ ( A x ) ] ) = λ F and Q F = θ . In this case, when a x = φ p [ ( A x ) ( 0 ) ] , we have
G λ ( T ) = 0 T φ q [ a x + 0 s λ ( F ( r , x ( r ) ) ( Q F ) ( r ) ) d r ] d s = 0 T φ q [ φ p [ ( A x ) ( 0 ) ] + 0 s λ F ( r , x ( r ) ) d r ] d s = 0 T φ q [ φ p [ ( A x ) ( 0 ) ] + 0 s ( φ p [ ( A x ) ( r ) ] ) d r ] d s = 0 T ( A x ) ( s ) d s = ( A x ) ( T ) ( A x ) ( 0 ) = θ .
Hence,
R ( x , λ ) ( t ) = A 1 { 0 t φ q [ φ p [ ( A x ) ( 0 ) ] + 0 s λ ( F ( r , x ( r ) ) ( Q F ) ( r ) ) d r ] d s } = A 1 { 0 t φ q [ φ p [ ( A x ) ( 0 ) ] + 0 s λ F ( r , x ( r ) ) d r ] d s } = A 1 { 0 t φ q [ φ p [ ( A x ) ( 0 ) ] + 0 s ( φ p [ ( A x ) ( r ) ] ) d r ] d s } = A 1 { 0 t ( A x ) ( s ) d s } = A 1 [ ( A x ) ( t ) ( A x ) ( 0 ) ] = [ ( I P ) x ] ( t ) .
Step 4. x Ω ¯ , we have

Hence, N λ is M-compact in Ω ¯ . □

Theorem 3.3 Suppose that 0 T e ( s ) d s = θ , λ 1 , λ 2 , , λ l are eigenvalues of the matrix B with | λ i | 1 , i = 1 , 2 , l , and there exist positive constants D > 0 , l > 0 and σ > 0 such that

(H1) x i g i ( x i ) > 0 , x i R , | x i | > D , for each i = 1 , 2 , , n ,

(H2) | g i ( u 1 ) g i ( u 2 ) | l | u 1 u 2 | , u 1 , u 2 R , for each i = 1 , 2 , , n ,

(H3) | f i ( x i ) | σ , x i R , for each i = 1 , 2 , , n .

Then Eq. (1.1) has at least one T-periodic solution if one of the following two conditions is satisfied:

where f R 2 = max | x | R 2 | f ( x ) | , R 2 is defined by (3.14).

Proof We complete the proof in three steps.

Step 1. Let Ω 1 = { x dom M : M x = N λ x , λ ( 0 , 1 ) } . We show that Ω 1 is a bounded set. If x Ω 1 , then M x = N λ x , i.e.,
( φ p [ ( A x ) ] ) = λ f ( x ( t ) ) x ( t ) λ g ( x ( t γ ( t ) ) ) + λ e ( t ) .
(3.5)
Integrating both sides of (3.5) over [ 0 , T ] , we have
0 T g ( x ( t γ ( t ) ) ) d t = θ ,
which together with assumption (H1) leads to the fact that there exists a point ξ i R such that
| x i ( ξ i γ ( ξ i ) ) | D , for each  i = 1 , 2 , , n .
Let ξ i γ ( ξ i ) = k T + η i , k Z , η i [ 0 , T ] . Then
| x i ( η i ) | D , for each  i = 1 , 2 , , n .
Thus,
| x i | D + 0 T | x i ( s ) | d s , for each  i = 1 , 2 , , n .
(3.6)
By (3.6), we have
| x | = ( x 1 2 + x 2 2 + + x n 2 ) 1 2 n ( D + 0 T | x ( s ) | d s )
(3.7)
and
| x | 0 n ( D + 0 T | x ( s ) | d s ) .
(3.8)
On the other hand, multiplying the two sides of Eq. (3.5) by [ x ( t ) ] from the left side and integrating them over [ 0 , T ] , we have
0 T [ x ( t ) ] ( φ p [ ( A x ) ] ) d t = λ 0 T [ x ( t ) ] f ( x ( t ) ) x ( t ) d t λ 0 T [ x ( t ) ] g ( x ( t γ ( t ) ) ) d t + λ 0 T [ x ( t ) ] e ( t ) d t .
(3.9)
Let ω ( t ) = φ p [ ( A x ) ( t ) ] , then
0 T [ x ( t ) ] ( φ p [ ( A x ) ] ) d t = 0 T { A 1 ( φ q ( ω ( t ) ) ) } d ω ( t ) = 0 .
By (3.9), we have
(3.10)
By assumption (H3), we have
σ i = 1 n 0 T | x i ( t ) | 2 d t i = 1 n 0 T | f i ( x i ( t ) ) | [ x i ( t ) ] 2 d t = | i = 1 n 0 T f i ( x i ( t ) ) [ x i ( t ) ] 2 d t | .
(3.11)
From (3.10) and (3.11), we have
σ i = 1 n 0 T | x i ( t ) | 2 d t 0 T | [ x ( t ) ] g ( x ( t γ ( t ) ) ) | d t + 0 T | [ x ( t ) ] e ( t ) | d t .
(3.12)
From 0 T [ x ( t ) ] g ( x ( t ) ) d t = 0 , assumption (H2), Lemma 2.5 and (3.12), we have
(3.13)
Since σ > 2 l max t [ 0 , T ] | γ ( t ) | , by (3.13), there exists a positive constant R 1 such that
0 T | x ( s ) | d s R 1 .
Then by (3.8),
| x | 0 n ( D + R 1 ) : = R 2 .
(3.14)
By (3.5), we have
| ( φ p [ ( A x ) ] ) | f R 2 | x ( t ) | + g R 2 + | e | 0 ,
where f R 2 = max | x | R 2 | f ( x ) | , g R 2 = max | x | R 2 | g ( x ) | . Take φ p [ ( A x ) ( t ) ] = y ( t ) , then
| y | 0 f R 2 | x ( t ) | + g R 2 + | e | 0
(3.15)
and ( A x ) ( t ) = φ q ( y ( t ) ) . Because there exists a t i [ 0 , T ] such that y ( t i ) = 0 , i = 1 , 2 , , n , so by (3.15), we get
| y ( t ) | n T | y | 0 n T f R 2 | x ( t ) | + n T g R 2 + n T | e | 0
and
| ( A x ) ( t ) | ( n T f R 2 | x ( t ) | + n T g R 2 + n T | e | 0 ) q 1 .
(3.16)
By (3.16) and Lemma 2.1, we have
| x ( t ) | = | [ A 1 A x ] ( t ) | | U 1 | | U | σ 0 | ( A x ) ( t ) | | U 1 | | U | σ 0 ( n T f R 2 | x ( t ) | + n T g R 2 + n T | e | 0 ) q 1 .
(3.17)
Now, we consider ( n T f R 2 | x ( t ) | + n T g R 2 + n T | e | 0 ) q 1 . In the formal case, we get
(3.18)
By classical elementary inequalities, we see that there is a constant h ( p ) > 0 , which is dependent on p only, such that
( 1 + x ) p < 1 + ( 1 + p ) x , x ( 0 , h ( p ) ] .
(3.19)
Case 2.1. If n T g R 2 + n T | e | 0 n T f R 2 | x ( t ) | > h , then
| x ( t ) | < n T g R 2 + n T | e | 0 n T f R 2 h : = M 1 .
(3.20)
Case 2.2. If n T g R 2 + n T | e | 0 n T f R 2 | x ( t ) | h , by (3.18) and (3.19), we have
(3.21)
From (3.17) and (3.21), we have
| x ( t ) | | U 1 | | U | σ 0 ( n T f R 2 ) q 1 | x ( t ) | q 1 + | U 1 | | U | σ 0 q ( n T g R 2 + T | e | 0 ) ( n T f R 2 ) q 2 | x ( t ) | q 2 .
(3.22)
When q = 2 , from | U 1 | | U | σ 0 n T f R 2 < 1 , we know that there exists a constant M 2 > 0 such that
| x ( t ) | M 2 .
(3.23)
When 1 < q < 2 , there must be a constant M 3 > 0 such that
| x ( t ) | M 3 .
(3.24)
Hence, from (3.14), (3.20), (3.23) and (3.24), we have
x < max { R 2 , M 1 , M 2 , M 3 } + 1 : = L .

Step 2. Let Ω 2 = { x Ker M : Q N x = θ } , we shall prove that Ω 2 is a bounded set. x Ω 2 , then x = a , a R n , we have g i ( a i ) = 0 for each i = 1 , 2 , , n . By assumption (H1) we have | a i | D and | a | n D . So, Ω 2 is a bounded set.

Step 3. Let Ω = { x X : x < L } , then Ω 1 Ω 2 Ω , ( x , λ ) Ω × ( 0 , 1 ) . From the above proof, M x N λ x is satisfied. Obviously, condition (A2) of Lemma 2.2 is also satisfied. Now, we prove that condition (A3) of Lemma 2.2 is satisfied. Take the homotopy
H ( x , μ ) = μ x ( 1 μ ) J Q N x , x Ω ¯ Ker M , μ [ 0 , 1 ] ,
where J : Im Q Ker M is a homeomorphism with J a = a , a R n . x Ω Ker M , we have x = a 1 R n , | a 1 | = L > D , then
H ( x , μ ) = a 1 μ ( 1 μ ) 1 T 0 T ( g ( a 1 ) + e ( t ) ) d t = a 1 μ + ( 1 μ ) g ( a 1 ) ,
then we have
a 1 H ( x , μ ) = a 1 a 1 μ + ( 1 μ ) a 1 g ( a 1 ) .
By using assumption (H1), we have H ( x , μ ) 0 . And then, by the degree theory,
deg { J Q N , Ω Ker M , 0 } = deg { H ( , 0 ) , Ω Ker M , 0 } = deg { H ( , 1 ) , Ω Ker M , 0 } = deg { I , Ω Ker M , 0 } 0 .

Applying Lemma 2.2, we complete the proof. □

Remark Assumption (H1) guarantees that condition (A2) of Lemma 2.2 is satisfied. Furthermore, using assumptions (H1)-(H3), we can easily estimate prior boud of the solution to Eq. (1.1).

As an application, we consider the following example:
( φ p [ ( x ( t ) B x ( t π ) ) ] ) + f ( x ( t ) ) x ( t ) + g ( x ( t π ) ) = e ( t ) ,
(3.25)
where
x ( t ) = ( x 1 ( t ) x 2 ( t ) ) R 3 , g ( x ) = ( 1 100 x 1 1 100 x 2 ) , B = ( 1 3 4 0 ) ,

e ( t ) = ( sin t , cos t ) , τ = γ = π , p = 1.5 , T = 2 π , f ( x ) = ( 5 + sin x 1 , 10 + cos x 2 ) .

Obviously, λ 1 = 3 ± 1 , λ 2 = 4 ± 1 ,
0 2 π e ( t ) d t = ( 0 2 π cos t d t 0 2 π sin t d t ) = ( 0 0 ) .
Since
x 1 g 1 ( x 1 ) = 1 100 x 1 2 > 0 for | x 1 | > D > 0 , x 2 g 2 ( x 2 ) = 1 100 x 2 2 > 0 for  | x 2 | > D > 0 ,
so assumption (H1) is satisfied. Take l = 1 100 , then
| g i ( u 1 ) g i ( u 2 ) | 1 100 | u 1 u 2 | , u 1 , u 2 R ,  for each  i = 1 , 2 ,
and assumption (H2) is satisfied. Take σ = 4 , then
| f 1 ( x 1 ) | = | 5 + sin x 1 | 4 , | f 2 ( x 2 ) | = | 10 + cos x 2 | 4 ,
and assumption (H3) is satisfied. Hence, assumptions (H1)-(H3) are all satisfied. Take
U = ( 1 1 4 3 ) , U 1 = ( 3 7 1 7 4 7 1 7 )
such that
U B U 1 = ( 3 0 0 4 ) .
Take γ ( t ) = π , then
σ > 2 l max t [ 0 , T ] | γ ( t ) | .

By using Theorem 3.3, we know that Eq. (3.25) has at least one 2π-periodic solution.

Declarations

Acknowledgements

This work was supported by NSF of Jiangsu education office (11KJB110002), Postdoctoral Fundation of Jiangsu (1102096C), Postdoctoral Fundation of China (2012M511296) and Jiangsu province fund (BK2011407).

Authors’ Affiliations

(1)
Department of Mathematics, Huaiyin Normal University

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© Yang and Du; licensee Springer 2012

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