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Some results for high-order generalized neutral differential equation

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Abstract

In this paper, we consider the following high-order p-Laplacian generalized neutral differential equation

( φ p ( x ( t ) c x ( t δ ( t ) ) ) ) ( n 1 ) +g ( t , x ( t ) , x ( t τ ( t ) ) , x ( t ) ) =e(t),

where p2, φ p (x)= | x | p 2 x for x0 and φ p (0)=0; g: R 4 R is a continuous periodic function with g(t+T,,,)g(t,,,), and g(t,a,a,0)e(t)0 for all aR. e:RR is a continuous periodic function with e(t+T)e(t) and 0 T e(t)dt=0, c is a constant and |c|1, δ C 1 (R,R) and δ is a T-periodic function, T is a positive constant; n is a positive integer. By applications of coincidence degree theory and some analysis skills, sufficient conditions for the existence of periodic solutions are established.

MSC:34K13, 34K40, 34C25.

1 Introduction

In recent years, there has been a good amount of work on periodic solutions for neutral differential equations (see [19] and the references cited therein). For example, in [1], Cao and He investigated a class of high-order neutral differential equations

( x ( p ) ( t ) + b p x ( p ) ( t h p ) ) + i = 0 p 1 ( a i x ( i ) + b i x ( i ) ( t h i ) ) =f(t).
(1.1)

By using the Fourier series method and inequality technique, they obtained the existence of a periodic solution for (1.1). In [8], applying Mawhin’s continuation theorem, Wang and Lu studied the existence of a periodic solution for a high-order neutral functional differential equation with distributed delay as follows:

( x ( t ) c x ( t σ ) ) ( n ) +f ( x ( t ) ) x (t)+g ( r 0 x ( t + s ) d α ( s ) ) =p(t),
(1.2)

here |c|1. Recently, in [5] and [6], Ren et al. observed the high-order p-Laplacian neutral differential equation

( φ p ( x ( t ) c x ( t σ ) ) ( l ) ) ( n l ) =F ( t , x ( t ) , x ( t ) , , x ( l 1 ) ( t ) )
(1.3)

and presented sufficient conditions for the existence of periodic solutions for (1.3) in the critical case (i.e., |c|=1) and in the general case (i.e., |c|1), respectively.

In this paper, we consider the following high-order p-Laplacian generalized neutral differential equation

( φ p ( x ( t ) c x ( t δ ( t ) ) ) ) ( n 1 ) +g ( t , x ( t ) , x ( t τ ( t ) ) , x ( t ) ) =e(t),
(1.4)

where p2, φ p (x)= | x | p 2 x for x0 and φ p (0)=0; g: R 4 R is a continuous periodic function with g(t+T,,,)g(t,,,), and g(t,a,a,0)e(t)0 for all aR. e:RR is a continuous periodic function with e(t+T)e(t) and 0 T e(t)dt=0, c is a constant and |c|1, δ C 1 (R,R) and δ is a T-periodic function, T is a positive constant; n is a positive integer.

In (1.4), the neutral operator A=x(t)cx(tδ(t)) is a natural generalization of the operator A 1 =x(t)cx(tδ), which typically possesses a more complicated nonlinearity than A 1 . For example, A 1 is homogeneous in the following sense ( A 1 x ) (t)=( A 1 x )(t), whereas A in general is inhomogeneous. As a consequence, many of the new results for differential equations with the neutral operator A will not be a direct extension of known theorems for neutral differential equations.

The paper is organized as follows. In Section 2, we first give qualitative properties of the neutral operator A which will be helpful for further studies of differential equations with this neutral operator; in Section 3, by applying Mawhin’s continuation theory and some new inequalities, we obtain sufficient conditions for the existence of periodic solutions for (1.4), an example is also given to illustrate our results.

2 Lemmas

Let C T ={ϕC(R,R):ϕ(t+T)ϕ(t)} with the norm | ϕ | = max t [ 0 , T ] |ϕ(t)|. Define difference operators A and B as follows:

A : C T C T , ( A x ) ( t ) = x ( t ) c x ( t δ ( t ) ) ; B : C T C T , ( B x ) ( t ) = c ( t δ ( t ) ) .

Lemma 2.1 (see [10])

If |c|1, then the operator A has a continuous inverse A 1 on C T , satisfying

(1) ( A 1 f ) ( t ) = { f ( t ) + j = 1 c j f ( s i = 1 j 1 δ ( D i ) ) for | c | < 1 , f C T , f ( t + δ ( t ) ) c j = 1 1 c j + 1 f ( s + δ ( t ) + i = 1 j 1 δ ( D i ) ) for | c | > 1 , f C T . (2) | ( A 1 f ) ( t ) | f | 1 | c | | , f C T . (3) 0 T | ( A 1 f ) ( t ) | d t 1 | 1 | c | | 0 T | f ( t ) | d t , f C T .

Let X and Y be real Banach spaces and let L:D(L)XY be a Fredholm operator with index zero, here D(L) denotes the domain of L. This means that ImL is closed in Y and dimKerL=dim(Y/ImL)<+. Consider supplementary subspaces X 1 , Y 1 of X, Y respectively such that X=KerL X 1 , Y=ImL Y 1 . Let P:XKerL and Q:Y Y 1 denote the natural projections. Clearly, KerL(D(L) X 1 )={0} and so the restriction L P :=L | D ( L ) X 1 is invertible. Let K denote the inverse of L P .

Let Ω be an open bounded subset of X with D(L)Ω. A map N: Ω ¯ Y is said to be L-compact in Ω ¯ if QN( Ω ¯ ) is bounded and the operator K(IQ)N: Ω ¯ X is compact.

Lemma 2.2 (Gaines and Mawhin [11])

Suppose that X and Y are two Banach spaces, and L:D(L)XY is a Fredholm operator with index zero. Let ΩX be an open bounded set and N: Ω ¯ Y be L-compact on Ω ¯ . Assume that the following conditions hold:

  1. (1)

    LxλNx, xΩD(L), λ(0,1);

  2. (2)

    NxImL, xΩKerL;

  3. (3)

    deg{JQN,ΩKerL,0}0, where J:ImQKerL is an isomorphism.

Then the equation Lx=Nx has a solution in Ω ¯ D(L).

Lemma 2.3 (see [12])

If x C n (R,R) and x(t+T)x(t), then

0 T | x ( t ) | p dt ( T π p ) p ( n 1 ) 0 T | x ( n ) ( t ) | p dt,
(2.1)

where π p =2 0 ( p 1 ) / p d s ( 1 s p p 1 ) 1 / p = 2 π ( p 1 ) 1 / p p sin ( π / p ) , and p is a fixed real number with p>1.

Remark 2.1 When p=2, π 2 =2 0 ( 2 1 ) / 2 d s ( 1 s 2 2 1 ) 1 / 2 = 2 π ( 2 1 ) 1 / 2 2 sin ( π / 2 ) =π, then (2.1) is transformed into 0 T | x ( t ) | 2 dt ( T π ) 2 ( n 1 ) 0 T | x ( n ) ( t ) | 2 dt.

In order to apply Mawhin’s continuation degree theorem, we rewrite (1.4) in the form

{ ( A x 1 ) ( t ) = φ q ( x 2 ( t ) ) x 2 ( n 1 ) ( t ) = g ( t , x 1 ( t ) , x 1 ( t τ ( t ) ) , x 1 ( t ) ) + e ( t ) ,
(2.2)

where 1 p + 1 q =1. Clearly, if x(t)= ( x 1 ( t ) , x 2 ( t ) ) is a T-periodic solution to (2.2), then x 1 (t) must be a T-periodic solution to (1.4). Thus, the problem of finding a T-periodic solution for (1.4) reduces to finding one for (2.2).

Now, set X={x=( x 1 (t), x 2 (t))C(R, R 2 ):x(t+T)x(t)} with the norm | x | =max{ | x 1 | , | x 2 | }; Y={x=( x 1 (t), x 2 (t)) C 1 (R, R 2 ):x(t+T)x(t)} with the norm x=max{ | x | , | x | }. Clearly, X and Y are both Banach spaces. Meanwhile, define

L:D(L)= { x C n ( R , R 2 ) : x ( t + T ) = x ( t ) , t R } XY

by

(Lx)(t)= ( ( A x 1 ) ( t ) x 2 ( n 1 ) ( t ) )

and N:XY by

(Nx)(t)= ( φ q ( x 2 ( t ) ) g ( t , x 1 ( t ) , x 1 ( t τ ( t ) ) , x 1 ( t ) ) + e ( t ) ) .
(2.3)

Then (2.2) can be converted into the abstract equation Lx=Nx. From the definition of L, one can easily see that

KerL R 2 ,ImL= { y Y : 0 T ( y 1 ( s ) y 2 ( s ) ) d s = ( 0 0 ) } .

So, L is a Fredholm operator with index zero. Let P:XKerL and Q:YImQ R 2 be defined by

Px= ( ( A x 1 ) ( 0 ) x 2 ( 0 ) ) ;Qy= 1 T 0 T ( y 1 ( s ) y 2 ( s ) ) ds,

then ImP=KerL, KerQ=ImL. Setting L P =L | D ( L ) Ker P and L P 1 :ImLD(L) denotes the inverse of L P , then

[ L P 1 y ] ( t ) = ( ( A 1 G y 1 ) ( t ) ( G y 2 ) ( t ) ) , [ G y 1 ] ( t ) = 0 t y 1 ( s ) d s , [ G y 2 ] ( t ) = j = 1 n 2 1 j ! x 2 ( j ) ( 0 ) t j + 1 ( n 2 ) ! 0 t ( t s ) n 2 y 2 ( s ) d s ,
(2.4)

where x 2 ( j ) (0) (j=1,2,,n2) are defined by the following

B = ( 1 0 0 0 0 b 1 1 0 0 0 b 2 b 1 1 0 0 b n 3 b n 4 b n 5 1 0 b n 2 b n 3 b n 4 b 1 0 ) ( n 2 ) × ( n 2 ) , X = ( x ( n 2 ) ( 0 ) , , x ( 0 ) , x ( 0 ) ) , C = ( C 1 , C 2 , , C n 2 ) , C j = 1 j ! T 0 T ( T s ) j y 2 ( s ) d s , b k = T k ( k + 1 ) ! , k = 1 , 2 , , n 3 .

From (2.3) and (2.4), it is clear that QN and K(IQ)N are continuous, QN( Ω ¯ ) is bounded and then K(IQ)N( Ω ¯ ) is compact for any open bounded ΩX which means N is L-compact on Ω ¯ .

3 Existence of periodic solutions for (1.4)

For the sake of convenience, we list the following assumptions which will be used repeatedly in the sequel:

(H1) There exists a constant D>0 such that

v 1 g(t, v 1 , v 2 , v 3 )>0(t, v 1 , v 2 , v 3 )[0,T]× R 3  with | v 1 |>D;

(H2) There exists a constant D 1 >0 such that

v 1 g(t, v 1 , v 2 , v 3 )<0(t, v 1 , v 2 , v 3 )[0,T]× R 3  with | v 1 |> D 1 ;

(H3) There exist non-negative constants α 1 , α 2 , α 3 , m such that

| g ( t , v 1 , v 2 , v 3 ) | α 1 | v 1 | p 1 + α 2 | v 2 | p 1 + α 3 | v 3 | p 1 +m(t, v 1 , v 2 , v 3 )[0,T]× R 3 ;

(H4) There exist non-negative constants γ 1 , γ 2 , γ 3 such that

| g ( t , u 1 , u 2 , u 3 ) g ( t , v 1 , v 2 , v 3 ) | γ 1 | u 1 v 1 |+ γ 2 | u 2 v 2 |+ γ 3 | u 3 v 3 |

for all (t, u 1 , u 2 , u 3 ),(t, v 1 , v 2 , v 3 )[0,T]× R 3 .

Theorem 3.1 Assume that (H1) and (H3) hold, then (1.4) has at least one non-constant T-periodic solution if |1|c|||c| δ 1 >0 and [ ( α 1 + α 2 ) T p + 1 + 2 p 1 α 3 T 2 ] 2 p + 1 ( | 1 | c | | | c | δ 1 ) p 1 ( T π ) 2 ( n 3 ) <1, here δ 1 = max t [ 0 , T ] | δ (t)|.

Proof Consider the equation

Lx=λNx,λ(0,1).

Set Ω 1 ={x:Lx=λNx,λ(0,1)}. If x(t)= ( x 1 ( t ) , x 2 ( t ) ) Ω 1 , then

{ ( A x 1 ) ( t ) = λ φ q ( x 2 ( t ) ) x 2 ( n 1 ) ( t ) = λ g ( t , x 1 ( t ) , x 1 ( t τ ( t ) ) , x 1 ( t ) ) + λ e ( t ) .
(3.1)

Substituting x 2 (t)= λ 1 p φ p [ ( A x 1 ) (t)] into the second equation of (3.1), we get

( φ p ( A x 1 ) ( t ) ) ( n 1 ) + λ p g ( t , x 1 ( t ) , x 1 ( t τ ( t ) ) , x 1 ( t ) ) = λ p e(t).
(3.2)

Integrating both sides of (3.2) from 0 to T, we have

0 T g ( t , x 1 ( t ) , x 1 ( t τ ( t ) ) , x 1 ( t ) ) dt=0.
(3.3)

From (3.3), there exists a point ξ[0,T] such that

g ( ξ , x 1 ( ξ ) , x 1 ( ξ τ ( ξ ) ) , x 1 ( ξ ) ) =0.

In view of (H1), we obtain

| x 1 ( ξ ) | D.

Then we have

| x 1 ( t ) | =| x 1 (ξ)+ ξ t x 1 (s)ds|D+ ξ t | x 1 ( s ) | ds,t[ξ,ξ+T],

and

| x 1 ( t ) | = | x 1 ( t T ) | = | x 1 ( ξ ) t T ξ x 1 ( s ) d s | D + t T ξ | x 1 ( s ) | d s , t [ ξ , ξ + T ] .

Combining the above two inequalities, we obtain

| x 1 | = max t [ 0 , T ] | x 1 ( t ) | = max t [ ξ , ξ + T ] | x 1 ( t ) | max t [ ξ , ξ + T ] { D + 1 2 ( ξ t | x 1 ( s ) | d s + t T ξ | x 1 ( s ) | d s ) } D + 1 2 0 T | x 1 ( s ) | d s .
(3.4)

Since (A x 1 )(t)= x 1 (t)c x 1 (tδ(t)), we have

( A x 1 ) ( t ) = ( x 1 ( t ) c x 1 ( t δ ( t ) ) ) = x 1 ( t ) c x 1 ( t δ ( t ) ) + c x 1 ( t δ ( t ) ) δ ( t ) = ( A x 1 ) ( t ) + c x 1 ( t δ ( t ) ) δ ( t ) ,

and

( A x 1 ) (t)= ( A x 1 ) (t)c x 1 ( t δ ( t ) ) δ (t).

By applying Lemma 2.1, we have

| x 1 | = max t [ 0 , T ] | A 1 A x 1 ( t ) | max t [ 0 , T ] | ( A x 1 ) ( t ) c x 1 ( t δ ( t ) ) δ ( t ) | | 1 | c | | φ q ( | x 2 | ) + | c | δ 1 | x 1 | | 1 | c | | ,

where δ 1 = max t [ 0 , T ] | δ (t)|. Since |1|c|||c| δ 1 >0, then

| x 1 | φ q ( | x 2 | ) | 1 | c | | | c | δ 1 .
(3.5)

On the other hand, from x 2 ( n 3 ) (0)= x 2 ( n 3 ) (T), there exists a point t 1 [0,T] such that x 2 ( n 2 ) ( t 1 )=0, which together with the integration of the second equation of (3.1) on interval [0,T] yields

2 | x 2 ( n 2 ) ( t ) | 2 ( x 2 ( n 2 ) ( t 1 ) + 1 2 0 T | x 2 ( n 1 ) ( t ) | d t ) = λ 0 T | g ( t , x 1 ( t ) , x 1 ( t τ ( t ) ) , x 1 ( t ) ) + e ( t ) | d t α 1 0 T | x 1 ( t ) | p 1 d t + α 2 0 T | x 1 ( t τ ( t ) ) | p 1 d t + α 3 0 T | x 1 ( t ) | p 1 d t + ( m + | e | ) T ( α 1 + α 2 ) T ( D + 1 2 0 T | x 1 ( t ) | d t ) p 1 + α 3 0 T | x 1 ( t ) | p 1 d t + ( m + | e | ) T = ( α 1 + α 2 ) T 2 p 1 ( 2 D 0 T | x 1 ( t ) | d t + 1 ) p 1 ( 0 T | x 1 ( t ) | d t ) p 1 + α 3 0 T | x 1 ( t ) | p 1 d t + ( m + | e | ) T .
(3.6)

For a given constant δ>0, which is only dependent on k>0, we have

( 1 + x ) k 1+(1+k)xfor x[0,δ].

From (3.5) and (3.6), we have

2 | x 2 ( n 2 ) ( t ) | ( α 1 + α 2 ) T 2 p 1 ( 2 D 0 T | x 1 ( t ) | d t + 1 ) p 1 ( 0 T | x 1 ( t ) | d t ) p 1 + α 3 0 T | x 1 ( t ) | p 1 d t + ( m + | e | ) T ( α 1 + α 2 ) T 2 p 1 ( 1 + 2 D p 0 T | x 1 ( t ) | d t ) ( 0 T | x 1 ( t ) | d t ) p 1 + α 3 0 T | x 1 ( t ) | p 1 d t + ( m + | e | ) T ( α 1 + α 2 ) T 2 p 1 T p 1 | x 1 | p 1 + ( α 1 + α 2 ) T D p 2 p 2 T p 2 | x 1 | p 2 + α 3 | x 1 | p 1 T + ( m + | e | ) T ( ( α 1 + α 2 ) T p 2 p 1 + α 3 T ) ( φ q | x 2 | ) p 1 ( | 1 | c | | | c | δ 1 ) p 1 + ( α 1 + α 2 ) T p 1 D p 2 p 2 ( φ q | x 2 | ) p 2 ( | 1 | c | | | c | δ 1 ) p 2 + ( m + | e | ) T = ( ( α 1 + α 2 ) T p 2 p 1 + α 3 T ) | x 2 | ( | 1 | c | | | c | δ 1 ) p 1 + ( α 1 + α 2 ) T p 1 D p 2 p 2 | x 2 | 2 q ( | 1 | c | | | c | δ 1 ) p 2 + ( m + | e | ) T .
(3.7)

Since 0 T φ q ( x 2 (t))dt= 0 T ( A x 1 ) (t)dt=0, there exists a point t 2 [0,T] such that x 2 ( t 2 )=0. From (3.4) and Remark 2.1, we can easily get

| x 2 | 1 2 0 T | x 2 ( t ) | d t T 2 ( 0 T | x 2 ( t ) | 2 d t ) 1 2 T 2 ( T π ) 2 ( n 3 ) ( 0 T | x 2 ( n 2 ) ( t ) | 2 d t ) 1 2 T 2 ( T π ) 2 ( n 3 ) | x 2 ( n 2 ) | .
(3.8)

Combination of (3.7) and (3.8) implies

| x 2 | T 2 ( T π ) 2 ( n 3 ) | x 2 ( n 2 ) | T 2 2 ( T π ) 2 ( n 3 ) [ ( ( α 1 + α 2 ) T p 2 p 1 + α 3 T ) | x 2 | ( | 1 | c | | | c | δ 1 ) p 1 + ( α 1 + α 2 ) T p 1 D p 2 p 2 | x 2 | 2 q ( | 1 | c | | | c | δ 1 ) p 2 ] + T 2 2 ( T π ) 2 ( n 3 ) ( m + | e | ) T .

Since p2 and [ ( α 1 + α 2 ) T p + 1 + 2 p 1 α 3 T 2 ] 2 p + 1 ( | 1 | c | | | c | δ 1 ) p 1 ( T π ) 2 ( n 3 ) <1, there exists a positive constant M 1 (independent of λ) such that

| x 2 | M 1 .
(3.9)

From (3.5) and (3.9), we obtain that

| x 1 | φ q ( | x 2 | ) | 1 | c | | | c | δ 1 M 1 q 1 | 1 | c | | | c | δ 1 := M 2 .

Hence

| x 1 | D+ 1 2 0 T | x 1 ( t ) | dtD+ T M 2 2 := M 3 .

From (3.6), we know

| x 2 ( n 2 ) | 1 2 max | 0 T x 2 ( n 1 ) ( t ) d t | 1 2 0 T | g ( t , x 1 ( t ) , x 1 ( t τ ( t ) ) , x 1 ( t ) ) + e ( t ) | d t 1 2 [ ( α 1 + α 2 ) T | x 1 | p 1 + α 3 T | x 1 | p 1 + ( m + | e | ) T ] 1 2 [ ( α 1 + α 2 ) T M 3 p 1 + α 3 T M 2 p 1 + ( m + | e | ) T ] : = M n 2 .

From (3.8), we can get

| x 2 | T 2 ( T π ) 2 ( n 4 ) | x 2 ( n 2 ) | T 2 ( T π ) 2 ( n 4 ) M n 2 := M 4 .

Let M=max{ M 1 , M 2 , M 3 , M 4 }+1, Ω={x= ( x 1 , x 2 ) :x<M} and Ω 2 ={x:xΩKerL}, then xΩKerL

QNx= 1 T 0 T ( φ q ( x 2 ( t ) ) g ( t , x 1 ( t ) , x 1 ( t τ ( t ) ) , x 1 ( t ) ) + e ( t ) ) dt.

If QNx=0, then x 2 (t)=0, x 1 =M or −M. But if x 1 (t)=M, we know

0= 0 T g(t,M,M,0)dt.

From assumption (H1), we have MD, which yields a contradiction. Similarly, in the case x 1 =M, we also have QNx0, that is, xΩKerL, xImL. So, conditions (1) and (2) of Lemma 2.2 are both satisfied. Define the isomorphism J:ImQKerL as follows:

J ( x 1 , x 2 ) = ( x 2 , x 1 ) .

Let H(μ,x)=μx+(1μ)JQNx, (μ,x)[0,1]×Ω, then (μ,x)(0,1)×(ΩKerL),

H(μ,x)= ( μ x 1 ( t ) 1 μ T 0 T [ g ( t , x 1 ( t ) , x 1 ( t τ ( t ) ) , x 1 ( t ) ) e ( t ) ] d t μ x 2 ( t ) ( 1 μ ) φ q ( x 2 ( t ) ) ) .

We have 0 T e(t)dt=0 and then

H ( μ , x ) = ( μ x 1 ( t ) 1 μ T 0 T [ g ( t , x 1 ( t ) , x 1 ( t τ ( t ) ) , x 1 ( t ) ) ] d t μ x 2 ( t ) ( 1 μ ) φ q ( x 2 ( t ) ) ) , ( μ , x ) ( 0 , 1 ) × ( Ω Ker L ) .

From (H1), it is obvious that x H(μ,x)<0, (μ,x)(0,1)×(ΩKerL). Hence,

deg { J Q N , Ω Ker L , 0 } = deg { H ( 0 , x ) , Ω Ker L , 0 } = deg { H ( 1 , x ) , Ω Ker L , 0 } = deg { I , Ω Ker L , 0 } 0 .

So, condition (3) of Lemma 2.2 is satisfied. By applying Lemma 2.2, we conclude that equation Lx=Nx has a solution x= ( x 1 , x 2 ) on Ω ¯ D(L), i.e., (2.2) has a T-periodic solution x 1 (t).

Finally, observe that y 1 (t) is not a constant. For if y 1 a (constant), then from (1.4) we have g(t,a,a,0)e(t)0, which contradicts the assumption that g(t,a,a,0)e(t)0. The proof is complete. □

Similarly, we can get the following result.

Theorem 3.2 Assume that (H2) and (H3) hold, then (1.4) has at least one non-constant T-periodic solution if |1|c|||c| δ 1 >0 and [ ( α 1 + α 2 ) T p + 1 + 2 p 1 α 3 T 2 ] 2 p + 1 ( | 1 | c | | | c | δ 1 ) p 1 ( T π ) 2 ( n 3 ) <1.

We illustrate our results with an example.

Example 3.1 Consider the following neutral functional differential equation

( φ p ( x ( t ) 15 x ( t 1 60 sin 4 t ) ) ) ( 5 ) + 1 3 π x 5 ( t ) + 1 6 π sin x ( t cos 4 t ) + 1 8 π sin 4 t cos x ( t ) = sin 4 t .
(3.10)

Here p=6. It is clear that T= π 2 , c=15, δ(t)= 1 60 sin4t, τ(t)=cos4t, e(t)=sin4t, δ 1 = max t [ 0 , T ] | 1 15 cos4t|= 1 15 , then we can get |1|c|||c| δ 1 =13>0, g(t, v 1 , v 2 , v 3 )= 1 3 π v 1 5 + 1 6 π sin v 2 + 1 8 π cos v 3 sin4t, and g(t,a,a,0)e(t)= 1 3 π a 5 + 1 6 π sina 8 π 1 8 π sin4t0. Choose D=3π such that (H1) holds. Now we consider the assumption (H3), it is easy to see

| g ( t , z 1 , z 2 , z 3 ) | 1 3 π | z 1 | 5 +1,

which means that (H3) holds with α 1 = 1 3 π , α 2 =0, α 3 =0, m=1. Obviously,

[ ( α 1 + α 2 ) T p + 1 + 2 p 1 α 3 T 2 ] 2 p + 1 ( | 1 | c | | | c | δ 1 ) p 1 ( T π ) 2 ( n 3 ) = 1 3 π ( π 2 ) 7 + 0 + 0 2 6 + 1 ( | 1 | c | | c | δ 1 ) 6 1 ( 1 2 ) 2 ( 6 3 ) = π 6 3 × 2 20 × 13 5 < 1 .

By Theorem 3.1, (3.10) has at least one nonconstant π 2 -periodic solution.

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Acknowledgements

CZB and RJL would like to thank the referee for invaluable comments and insightful suggestions. This work was supported by NSFC project (No. 11271339) and NCET Program (No. 10-0141).

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Correspondence to Jingli Ren.

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CZB and RJL worked together in the derivation of the mathematical results. Both authors read and approved the final manuscript.

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Keywords

  • periodic solution
  • high-order
  • generalized neutral differential equation