Open Access

Existence of periodic solutions for p-Laplacian neutral Rayleigh equation

Advances in Difference Equations20142014:67

https://doi.org/10.1186/1687-1847-2014-67

Received: 3 May 2013

Accepted: 30 January 2014

Published: 14 February 2014

Abstract

In this paper, we use topological degree theory to establish new results on the existence of periodic solutions for a p-Laplacian neutral Rayleigh equation.

Keywords

periodic solutionsneutral Rayleigh equationp-Laplacian equationtopological degree

1 Introduction

In this paper, we consider the following second-order p-Laplacian neutral functional differential equation:
( ϕ p ( x ( t ) c x ( t σ ) ) ) + f ( x ( t ) ) + β ( t ) g ( x ( t τ ( t ) ) ) = e ( t ) ,
(1.1)

where ϕ p ( x ) = | x | p 2 x for x 0 and p > 1 ; σ and c are given constants with | c | 1 ; ϕ p ( 0 ) = 0 , f ( 0 ) = 0 . The conjugate exponent of p is denoted by q, i.e. 1 p + 1 q = 1 . f, g, β, e, and τ are real continuous functions on ; τ, β, and e are periodic with periodic T, T > 0 is a constant; 0 T e ( t ) d t = 0 , 0 T β ( t ) 0 .

As we know, the p-Laplace Rayleigh equation with a deviating argument τ ( t ) is applied in many fields such as physics, mechanics, engineering technique fields, and so on. The existence of a periodic solution for the second-order p-Laplacian Rayleigh equations with a deviating argument as follows:
( ϕ p ( x ( t ) ) ) + f ( x ( t ) ) x ( t ) + g ( x ( t τ ( t ) ) ) = e ( t )
(1.2)
and
( ϕ p ( x ( t ) ) ) + f ( x ( t ) ) + g ( x ( t τ ( t ) ) ) = e ( t )
(1.3)

has been extensively studied in [14]. In recent years, Lu et al. [57] used Mawhin’s continuation theory to do research to the existence of a periodic solution for p-Laplacian neutral Rayleigh equation. They obtained some existence results of periodic solutions to p-Laplacian neutral Rayleigh equations.

In the research mentioned above, the corresponding nonlinear terms of the second-order p-Laplacian Rayleigh equation did not include a variable coefficient. Only little literature discussed this kind of p-Laplacian Rayleigh equation. For more details refer to [811]. Here, we focus on [11] by Liang Feng et al. They discussed the existence of the solution to the following equation:
( ϕ p ( x ( t ) c x ( t r ) ) ) = f ( x ( t ) ) x ( t ) + β ( t ) g ( x ( t τ ( t ) ) ) + e ( t ) .
(1.4)

They established sufficient conditions for the existence of a T-periodic solution of (1.4). But their conclusions are founded on the prerequisite 0 T ( g ( x ( t τ ( t ) ) ) + e ( t ) ) d t = 0 , which does not satisfy (1.1). Another significance of the paper is that the result is related to the deviating argument τ ( t ) , while the conclusions in those existing papers mentioned above have no relation with τ ( t ) .

2 Preliminary results

For convenience, throughout this paper, we will adopt the following assumptions:

(H1) x p = ( 0 T | x ( t ) | p d t ) 1 p , x = max t [ 0 , T ] | x ( t ) | , x = max { x , x ( t ) } ;

(H2) m 0 = min t [ 0 , T ] | β ( t ) | , m 1 = max t [ 0 , T ] | β ( t ) | ;

(H3) C T = { x | x C ( R , R ) , x ( t + T ) = x ( t ) , t R } ;

(H4) C T 1 = { x | x C 1 ( R , R ) , x ( t + T ) = x ( t ) , x ( t + T ) = x ( t ) , t R } .

It is obvious that C T with norm x and C T 1 with norm x are two Banach spaces.

Now we define a linear operator A : C T C T , ( A x ) ( t ) = x ( t ) c x ( t σ ) .

According to [12, 13], we know that the operator A has the following properties.

Lemma 2.1 [12, 13]

If | c | 1 , then A has continuous bounded inverse on C T and
  1. (1)

    A 1 x = x | 1 | C | | , x C T ,

     
  2. (2)

    ( A 1 x ) ( t ) = { j 0 c j x ( t j σ ) , | c | < 1 , j 0 c j x ( t + j σ ) , | c | > 1 ,

     
  3. (3)

    0 T | ( A 1 x ) ( t ) | d t 1 | 1 | c | | 0 T | x ( t ) | d t , x C T .

     
Lemma 2.2 If | c | 1 and p > 1 , then
A 1 x ( t ) p 1 1 | c | x ( t ) p , x C T .
(2.1)
Proof We know that x ( t ) is a periodic function. So 0 T | x ( t j σ ) | d t = 0 T | x ( t ) | d t for j 0 . When | c | < 1 , from Lemma 2.1, we have
0 T | A 1 x ( t ) | p d t = 0 T | A 1 x ( t ) | p 1 | A 1 x ( t ) | d t = 0 T | A 1 x ( t ) | p 1 | j 0 c j x ( t j σ ) | d t j 0 | c j | 0 T | A 1 x ( t ) | p 1 | x ( t j σ ) | d t j 0 | c j | ( 0 T | A 1 x ( t ) | ( p 1 ) q d t ) 1 q ( 0 T | x ( t j σ ) | p d t ) 1 p = j 0 | c j | ( 0 T | A 1 x ( t ) | p d t ) 1 q ( 0 T | x ( t ) | p d t ) 1 p = 1 1 | c | ( 0 T | A 1 x ( t ) | p d t ) 1 q ( 0 T | x ( t ) | p d t ) 1 p

which implies ( 0 T | A 1 x ( t ) | p d t ) 1 p 1 1 | c | ( 0 T | x ( t j σ ) | p d t ) 1 p . That is to say (2.1) holds. If | c | > 1 , we can also prove that (2.1) is true in the same way. Thus Lemma 2.2 is proved.

Now we consider the following equation in C T 1 :
( ϕ p ( u ( t ) ) ) = F ( u ) .
(2.2)

F : C T 1 C T is continuous and takes a bounded set into bounded set.

Let us define P : C T 1 C T , u | u ( 0 ) , Q : C T C T , h | 1 T 0 T h ( s ) d s and
H ( h ( t ) ) = 0 t h ( s ) d s , h C T .
It is clear that if u C T 1 is the solution to (2.2), then u satisfies the abstract equation
u = P u + Q F ( u ) + K ( F ( u ) ) ,
where the operator K : C T C T 1 is given by
K ( h ( t ) ) = H { ϕ q [ α ( ( I Q ) h ) + H ( ( I Q ) ) ] } ( t ) , t R ,

α is a continuous function which sends bounded sets of C T into bounded sets of , and it is a completely continuous mapping. For more details as regards the meaning of α, please refer to [14]. □

Lemma 2.3 [14]

Let Ω be an open bounded set in C T 1 . Suppose that the following conditions hold:
  1. (i)

    For each λ ( 0 , 1 ) , the equation ( ϕ p ( u ) ) = λ F ( u ) has no solution on Ω.

     
  2. (ii)

    The equation Ϝ ( u ) = 1 T 0 T F ( u ( t ) ) d t = 0 has no solution on Ω R .

     
  3. (iii)

    The Brouwer degree of Ϝ, deg { Ϝ , Ω R , 0 } 0 .

     

Then (2.2) has at least one T-periodic solution in Ω ¯ .

Lemma 2.4 [15]

Ω R n is open bounded and symmetric with respect to 0 Ω . If f C ( Ω ¯ , R n ) and f ( x ) μ f ( x ) , x Ω , μ [ 0 , 1 ] , then deg { f , Ω , 0 } is an odd number.

3 Main results

Theorem 3.1 Suppose that the following conditions hold:

(A1) τ ( t ) C 1 ( R , R ) , τ ( t ) > 1 or τ ( t ) < 1 , m 2 = min t [ 0 , T ] 1 1 τ ( t ) .

(A2) The sign of β ( t ) 1 τ ( t ) is unchanged in the interval [ 0 , T ] .

(A3) There exist constants r 1 0 , r 2 > 0 and k > 0 such that
  1. (1)

    | f ( x ) | k + r 1 | x | p 1 , x R ,

     
  2. (2)

    lim | x | | g ( x ) | | x | p 1 r 2 .

     

(A4) There exists a constant d > 0 such that x g ( x ) > 0 , | x | > d .

Then (1.1) has at least one solution with periodic T if there exists a constant ε > 0 such that the following condition holds:
| 1 + | c | | ( a + T 1 q ) [ m 1 T ( r 2 + ε ) + r 1 T 1 p ] < | 1 | c | | p ,
(3.1)

where a is defined in (3.8).

Proof Consider the homotopic equation of (1.1) as follows:
( ϕ p ( x ( t ) c x ( t σ ) ) ) + λ f ( x ( t ) ) + λ β ( t ) g ( x ( t τ ( t ) ) ) = λ e ( t ) , λ ( 0 , 1 ) .
(3.2)
Let x ( t ) be a possible T-periodic solution to (3.2). By integrating both sides of (3.2) over [0, T], we have
0 T [ f ( x ( t ) ) + β ( t ) g ( x ( t τ ( t ) ) ) ] d t = 0 .
(3.3)
Let u ( t ) = t τ ( t ) , by the condition (A1), we know that u ( t ) has a unique inverse denoted by t = γ ( u ) ; noting that τ ( 0 ) = τ ( T ) , we get
0 T β ( t ) g ( x ( t τ ( t ) ) ) d t = τ ( 0 ) T τ ( T ) β ( γ ( u ) ) g ( u ) 1 τ ( γ ( u ) ) d u = 0 T β ( γ ( u ) ) g ( u ) 1 τ ( γ ( u ) ) d u .
(3.4)
Based on the condition of (A2) and the integral mean value theorem, there exists ξ [ 0 , T ] such that
g ( x ( ξ ) ) 0 T β ( γ ( u ) ) 1 τ ( γ ( u ) ) d u = 0 T f ( x ( t ) ) d t .
(3.5)
By the condition (A3)(2) for a given ε > 0 , ρ > d > 0 when | x ( t ) | > ρ such that
| g ( x ) | | x | p 1 r 2 ε > 0 , | g ( x ) | | x | p 1 r 2 + ε .
(3.6)
Now we can claim that there are two constants a and b such that
| x ( ξ ) | a x p + b ,
(3.7)
where
a = { [ r 1 ( r 2 ε ) m 0 m 2 T 1 q ] 1 p 1 2 2 p p 1 , p 1 < 1 , [ r 1 ( r 2 ε ) m 0 m 2 T 1 q ] 1 p 1 , p 1 > 1 ,
(3.8)
b = { [ k ( r 2 ε ) m 0 m 2 ] 1 p 1 2 2 p p 1 , p 1 < 1 , [ k ( r 2 ε ) m 0 m 2 ] 1 p 1 , p 1 > 1 .
(3.9)

In the following, we prove the above claim in two cases.

Case 1. If | x ( ξ ) | ρ , ξ [ 0 , T ] , then (3.7) holds.

Case 2. If | x ( ξ ) | > ρ , ξ [ 0 , T ] , by (3.5), (3.6), and the condition (A3)(1), we have
| r 2 ε | | x ( ξ ) | p 1 m 0 m 2 T | g ( x ( ξ ) ) | 0 T β ( γ ( u ) ) 1 τ ( γ ( u ) ) d u | r 2 ε | | x ( ξ ) | p 1 m 0 m 2 T 0 T | f ( x ( t ) ) | d t k T + r 1 0 T | x ( t ) | p 1 d t | r 2 ε | | x ( ξ ) | p 1 m 0 m 2 T k T + r 1 T 1 p ( 0 T | x ( t ) | p d t ) p 1 p = k T + r 1 T 1 p x p p 1 , | x ( ξ ) | p 1 k ( r 2 ε ) m 0 m 2 + r 1 ( r 2 ε ) m 0 m 2 T 1 q x p p 1 .
When 0 < p 1 1 , according to the Minkowski inequality, we have
| x ( ξ ) | [ k ( r 2 ε ) m 0 m 2 ] 1 p 1 2 2 p p 1 + [ r 1 ( r 2 ε ) m 0 m 2 T 1 q ] 1 p 1 2 2 p p 1 x p .
When p 1 > 1 , from ( a + b ) 1 m ( a ) 1 m + ( b ) 1 m , a , b [ 0 , + ) , m > 1 , we have
| x ( ξ ) | [ k ( r 2 ε ) m 0 m 2 ] 1 p 1 + [ r 1 ( r 2 ε ) m 0 m 2 T 1 q ] 1 p 1 x p .

Therefore, (3.7) is also satisfied for case 2.

For any t R , there exists t 0 [ 0 , T ] , such that t = k T + t 0 , where k is an integer. Then
| x ( t ) | = | x ( t 0 ) | | x ( ξ ) | + 0 T | x ( s ) | d s , ξ [ 0 , T ] .
By (3.7), we have
x a x p + b + T 1 q x p = ( a + T 1 q ) x p + b .
(3.10)
At first, we prove that there is a constant R 1 such that
x R 1 .
(3.11)

By (3.10), we only need to prove that x p is bounded in order to prove (3.11).

If x p = 0 , then x p is obviously bounded.

If b a + T 1 q x p h , then x p b a h h T 1 q , that is, x p is bounded as well.

If b a + T 1 q x p < h , we prove that x p is bounded in the following.

By multiplying both sides of (3.2) by A ( x ( t ) ) = x ( t ) c x ( t σ ) and integrating them over [ 0 , T ] , we have
| 0 T [ φ p ( A x ) ] A ( x ( t ) ) d t | = | φ p ( A x ) A ( x ( t ) ) | 0 T 0 T φ p ( A x ) A x d t | = 0 T | A x | p d t = A x p p = | λ 0 T A ( x ( t ) ) [ f ( x ( t ) ) + β ( t ) g ( x ( t τ ( t ) ) ) e ( t ) ] d t | ( 1 + | c | ) x 0 T [ | f ( x ( t ) | + | β ( t ) g ( x ( t τ ( t ) ) ) | + | e ( t ) | ] d t .
(3.12)
Let E 1 = { t [ 0 , T ] : | x ( t τ ( t ) ) | ρ } , E 2 = { t [ 0 , T ] : | x ( t τ ( t ) ) | > ρ } , then
0 T | β ( t ) g ( x ( t τ ( t ) ) ) | d t = E 1 | β ( t ) g ( x ( t τ ( t ) ) ) | d t + E 2 | β ( t ) g ( x ( t τ ( t ) ) ) | d t m 1 m 3 T + m 1 T ( r 2 + ε ) x p 1 ,
(3.13)
where
m 3 = max | x | ρ | g ( x ) | .
(3.14)
By (3.12) and (3.13), we get
A x p p ( 1 + | c | ) x × [ m 1 m 3 T + m 1 T ( r 2 + ε ) x p 1 + k T + r 1 T 1 p x p p 1 + 0 T | e ( t ) | d t ] = ( 1 + | c | ) ( m 1 m 3 T + k T + 0 T | e ( t ) | d t ) x + ( 1 + | c | ) m 1 T ( r 2 + ε ) x p + ( 1 + | c | ) r 1 T 1 p x x p p 1 a 1 [ b + ( a + T 1 q ) ] x p + a 2 [ b + ( a + T 1 q ) x p ] p + a 3 [ b + ( a + T 1 q ) x p ] x p p 1 ,
(3.15)

where a 1 = ( 1 + | c | ) ( m 1 m 3 T + k T + 0 T | e ( t ) | d t ) , a 2 = ( 1 + | c | ) m 1 T ( r 2 + ε ) , a 3 = ( 1 + | c | ) r 1 T 1 p .

By elementary analysis, we know that there is a constant h > 0 which satisfies b a h > 0 such that
( 1 + u ) p 1 + ( 1 + p ) u , u ( 0 , h ] .
(3.16)
By (3.16), one has
[ b + ( a + T 1 q ) x p ] p = ( a + T 1 q ) p x p p ( 1 + b ( a + T 1 q ) x p ) p ( a + T 1 q ) p x p p + b ( 1 + p ) ( a + T 1 q ) p 1 x p p 1 .
(3.17)
By Lemma 2.2, together with (3.15) and (3.17), we can derive
| 1 | c | | p x p p = | 1 | c | | p A 1 A x p p A x p p a 1 [ b + ( a + T 1 q ) ] x p + a 2 [ ( a + T 1 q ) p x p p + b ( 1 + p ) ( a + T 1 q ) p 1 x p p 1 ] + a 3 [ b + ( a + T 1 q ) x p ] x p p 1 = ( a 2 + a 3 ) ( a + T 1 q ) x p p + [ a 2 b ( 1 + p ) ( a + T 1 q ) p 1 + a 3 b ] x p p 1 + a 1 ( a + T 1 q ) x p + a 1 b .
(3.18)

From (3.1) and (3.18), we know that x p also is bounded in this case. Based on the above, we can derive the result that x p has a bound; therefore, (3.11) holds.

Secondly, we prove that there is a constant R 2 such that
x R 2 .
(3.19)
Based on (1.1), together with (3.13) and the condition (A2)(1), we get
0 T | ( ϕ p ( ( A x ( t ) ) ) ) | 0 T [ | f ( x ( t ) ) | + | β ( t ) g ( x ( t τ ( t ) ) ) | + | e ( t ) | ) ] d t k T + r 1 T 1 p x p p 1 + m 1 m 3 T + m 1 T ( r 2 + ε ) x p 1 + 0 T | e ( t ) | d t : = R 3 .
Because A ( x ( 0 ) ) = A ( x ( T ) ) , there exists t 0 [ 0 , T ] such that ( A x ( t 0 ) ) = A ( x ( t 0 ) ) = 0 . Noting that ϕ p ( 0 ) = 0 , we have
| ϕ p ( A x ( t ) ) | = | t 0 t ( ϕ p ( A ( x ( t ) ) ) ) d t | 0 T | ϕ p ( A ( x ( t ) ) ) | d t R 3 ,
then
A x ϕ q ( R 3 ) .
From Lemma 2.1, we derive
x = A 1 A x A x | 1 | c | | ϕ q ( R 3 ) | 1 | c | | : = R 2 ,

therefore, (3.19) is satisfied.

Let y ( t ) = ( A x ( t ) ) , then (3.2) is equivalent to the following equation:
( ϕ p ( y ( t ) ) ) + λ f ( A 1 y ( t ) ) + λ g ( A 1 y ( t τ ( t ) ) ) = λ e ( t ) .
(3.20)
Then x = A 1 y is a T-periodic solution of (3.2) if y is a T-periodic solution of (3.20). Let
F ( y ( t ) ) = e ( t ) f ( ( A 1 y ( t ) ) ) g ( A 1 ( y ( t τ ( t ) ) ) ) ,
(3.21)

since f, g are continuous and A has a continuous inverse, the mapping F : C T 1 C T in (3.21) is continuous and takes bounded sets into bounded sets.

In addition, (3.20) can be represented as
( ϕ p ( y ( t ) ) ) = λ F ( y ( t ) ) .
(3.22)

Let R = M max { R 1 , R 2 , ρ } ; M > 1 + | c | is a constant, Ω = { y ( t ) C T 1 , y < R , y < R } , then (3.22) has no solution on Ω for λ ( 0 , 1 ) . In fact, if y = A x is a solution to (3.22) on Ω, then y = R or y = R . If y = R , then y = A x = x ( t ) c x ( t σ ) ( 1 + | c | ) x . That is to say, x y 1 + | c | > R 1 . This is a contradiction with (3.11). Similarly, y R . Then (3.22) satisfies the condition (i) of Lemma 2.3.

If y Ω R , y is a constant and | y | = y = R , then f ( ( A 1 y ) ) = 0 and | y | ( 1 + | c | ) | x | , | x | | y | 1 + | c | > ρ > d . By (A4), we obtain g ( A 1 ( y ( t τ ( t ) ) ) ) = g ( A 1 y ) = g ( x ) 0 . Therefore
Ϝ ( y ) = 1 T 0 T F ( y ) d t = g ( A 1 ( y ( t τ ( t ) ) ) ) = g ( x ) 0
(3.23)

on Ω R . This indicates that (3.22) satisfies the condition (ii) of Lemma 2.3.

We know ( Ω R ) = { R , R } , then for y ( Ω R ) , we have y = R > d or y = R < d . By (3.23) and the condition (A4), we conclude that Ϝ ( y ) μ Ϝ ( y ) , μ [ 0 , 1 ] , y ( Ω R ) . Based on Lemma 2.4, we get deg { Ϝ , Ω R , 0 } 0 .

Based on the above, (3.22) satisfies all the conditions of Lemma 2.3. So does (3.20). By Lemma 2.3, (3.20) has at least one T-periodic solution, then (1.1) has also at least a periodic solution. □

4 Example

Consider the following equation:
( ϕ 4 ( x ( t ) 1 10 x ( t 1 2 ) ) ) + f ( x ( t ) ) + β ( t ) g ( x ( t 1 2 sin t ) ) = 1 400 cos t ,
(4.1)

where p = 4 , c = 1 10 , σ = 1 2 , τ ( t ) = 1 2 sin t , e ( t ) = 1 400 cos t , T = 2 π . Obviously we get τ ( t ) < 1 , m 2 = 2 3 .

If we take f ( x ) = { 1 100 x , | x | 1 , 1 100 x 3 , | x | > 1 , g ( x ) = x 100 + x 3 10 , β ( t ) = 1 10 ( 1 + sin 2 t ) , then the condition (A4) of Theorem 3.1 is satisfied and
| f ( x ) | 1 100 + 1 100 | x 3 | , lim | x | | g ( x ) | | x | 3 1 10 , m 0 = 1 20 , m 1 = 1 10 .
(4.2)

By (4.2), we obtain k = 1 100 , r 1 = 1 100 , r 2 = 1 10 .

If we choose ε = 0.01 , ρ > 1 , then when | x | > ρ , we have
| g ( x ) | | x | 3 r 2 ε , | g ( x ) | | x | 3 r 2 + ε .
We calculate easily that
a = 0.843 , | 1 + | c | | ( a + T 1 q ) [ m 1 T ( r 2 + ε ) + r 1 T 1 p ] = 0.4496 < | 1 | c | | 4 = 0.6561 .

Based on the above, we know that (4.1) satisfies all conditions included in Theorem 3.1; therefore, (4.1) has at least one T-periodic solution.

Declarations

Acknowledgements

The authors wish to thank the anonymous referee for his/her valuable suggestions to this paper. This work is supported by the NSFC of China (11171085).

Authors’ Affiliations

(1)
Department of Mathematics, Tianmu College, Zhejiang A and F University
(2)
Hangzhou Normal University

References

  1. Peng SG, Zhu SM: Brca1 protein products: functional motifs. Nat. Genet. 1996, 13: 266–267. 10.1038/ng0796-266View ArticleGoogle Scholar
  2. Xiao B, Liu BW: Periodic solutions for type Rayleigh p -Laplacian equations with a deviating argument. Nonlinear Anal. 2009, 10: 16–22. 10.1016/j.nonrwa.2007.08.010MathSciNetView ArticleGoogle Scholar
  3. Cheung WS, Ren JL: Periodic solutions for p -Laplacian equations. Nonlinear Anal. 2006, 65: 2003–2012. 10.1016/j.na.2005.11.002MathSciNetView ArticleGoogle Scholar
  4. Liu BW: Existence and uniqueness of periodic solutions for a kind of Liénard type p -Laplacian equations. Nonlinear Anal. 2009, 69: 724–729.View ArticleGoogle Scholar
  5. Zhu YL, Lu SP: Periodic solutions for p -Laplacian neutral functional differential equation with a deviating arguments. J. Math. Anal. Appl. 2007, 325: 377–385. 10.1016/j.jmaa.2005.10.084MathSciNetView ArticleGoogle Scholar
  6. Lu SP: On the existence of periodic solutions to a p -Laplacian neutral differential equations in the critical case. Nonlinear Anal. 2009, 10: 2884–2893. 10.1016/j.nonrwa.2008.09.005View ArticleGoogle Scholar
  7. Lu SP: Periodic solutions to a second p -Laplacian neutral differential system. Nonlinear Anal. 2008, 69: 4215–4229. 10.1016/j.na.2007.10.049MathSciNetView ArticleGoogle Scholar
  8. Cheung WS, Ren JL: On the existence periodic solutions for p -Laplacian generalized Liénard equation. Nonlinear Anal. 2005, 60: 65–75.MathSciNetGoogle Scholar
  9. Feng L, Xiang GL, Lu SP: New results of periodic solutions for a Rayleigh type p -Laplacian equation with a variable. Nonlinear Anal. 2009, 70: 2072–2077. 10.1016/j.na.2008.02.107MathSciNetView ArticleGoogle Scholar
  10. Gao FB, Lu SP, Zhang W: Periodic solutions for a Rayleigh type p -Laplacian equation with sign variable coefficient ahead of the nonlinear term. Appl. Math. Comput. 2010, 216: 2010–2015. 10.1016/j.amc.2010.03.031MathSciNetView ArticleGoogle Scholar
  11. Feng L, Guo LX, Lu SP: Existence of periodic solutions for a p -Laplacian neutral functional equation. Nonlinear Anal. 2009, 71: 427–436. 10.1016/j.na.2008.10.122MathSciNetView ArticleGoogle Scholar
  12. Zhang MR: Periodic solutions to linear and quasilinear neutral functional differential equation. J. Math. Anal. Appl. 1995, 189: 378–392. 10.1006/jmaa.1995.1025MathSciNetView ArticleGoogle Scholar
  13. Lu SP, Re J, Ge W: Problems of periodic solutions for a kind of second order neutral functional differential equations. Appl. Anal. 2003, 82: 393–410.View ArticleGoogle Scholar
  14. Manásevich R, Mawhin J: Periodic solutions for nonlinear systems p -Laplacian-like. J. Differ. Equ. 1998, 145: 367–393. 10.1006/jdeq.1998.3425View ArticleGoogle Scholar
  15. Zhong C, Fan X, Chen W: Introduction to Nonlinear Functional Analysis. Lanzhou University Press, Lan Zhou; 2004. (in Chinese)Google Scholar

Copyright

© Min He and Shen; licensee Springer. 2014

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.