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Periodic solutions for a kind of neutral functional differential systems

Abstract

In this paper, we analyze some properties of the linear difference operator A: C T C T , [Ax](t)=x(t)V(t)x(tτ), and then, by using the coincidence degree theory of Mawhin, a kind of neutral differential systems with non-constant matrix is studied. Some new results on the existence of periodicity are obtained. It is worth noting that V(t) is no longer a constant matrix, which is different from the corresponding ones of past work.

1 Introduction

The field of neutral functional equations (in short NFDEs) is making significant breakthroughs in its practice; it is no longer only a specialist’s field. In many practical systems, models of systems are described by NFDEs in which the models depend on the delays of state and state derivatives. Practical examples for neutral systems include population ecology, heat exchanges, mechanics, and economics; see [1]–[4]. In particular, qualitative analysis such as periodicity and stability of solutions of NFDEs has been studied extensively by many authors. We refer to [5]–[12] for some recent work on the subject of periodicity and stability of neutral equations.

In the last few years, the stability of neutral systems of various classes with time delays has received an ever-growing interest from many authors. Many sufficient conditions have been proposed to guarantee the asymptotic stability for neutral time delay systems. We only mention the work of some authors [13]–[15]. It is well known that the existence of periodic solutions of neutral equations and neutral systems is a very basic and important problem, which plays a role similar to stability. Thus, it is reasonable to seek conditions under which the resulting periodic neutral system would have a periodic solution. Much progress has been seen in this direction and many criteria are established based on different approaches. However, there is no paper for investigating the existence of periodic solutions of neutral system with non-constant matrix. In addition, to the best of our knowledge, most of the existing results deal with scalar NFEDs or neutral systems with a constant matrix. For example, in papers [16]–[20], based on Mawhin’s continuation theorem, several types of scalar neutral equations have been studied:

d 2 d t 2 ( u ( t ) k u ( t τ ) ) = f ( u ( t ) ) u ( t ) + α ( t ) g ( u ( t ) ) + j = 1 n β j ( t ) g ( u ( t γ j ( t ) ) ) + p ( t ) , d N d t = N ( t ) [ α ( t ) β ( t ) N ( t ) j = 1 n b j ( t ) N ( t σ j ( t ) ) i = 1 m c i ( t ) N ( t τ i ( t ) ) ] , d N d t = N ( t ) [ r ( t ) j = 1 n α j ( t ) ln N ( t σ j ( t ) ) i = 1 m b i ( t ) d d t ln ( t τ i ( t ) ) ] , x ( t ) + α x ( t τ ) = f ( t , x ( t ) ) , ( u ( t ) + B u ( t τ ) ) = g 1 ( t , u ( t ) ) g 2 ( t , u ( t τ 1 ) ) + p ( t ) .

For a neutral system, we note that Lu and Ge [21] studied the following system:

d 2 d t 2 ( x ( t ) C x ( t τ ˜ ) ) + d d t gradF ( x ( t ) ) +gradG ( x ( t τ ( t ) ) ) =p(t).

But C is a constant symmetric matrix. The purpose of this paper is to investigate the existence of periodic solutions to the nonlinear neutral system with non-constant matrix of the form

d 2 d t 2 ( x ( t ) C ( t ) x ( t τ ) ) + d d t gradF ( x ( t ) ) +gradG ( x ( t γ ( t ) ) ) =p(t),
(1.1)

where x R n , C(t)=diag( c 1 (t), c 2 (t),, c n (t)), C(t+T)=C(t); F(x) C 2 ( R n ,R), G(x) C 1 ( R n ,R); p(R, R n ), p(t+T)=p(t); γC(R,R), γ(t+T)=γ(t); T, and τ are given constants with T>0.

Throughout this paper, we use some notation:

  1. (1)

    I n ={1,2,,n}; a= ( a 1 , a 2 , , a n ) T R n , |a|= ( i = 1 n | a i | 2 ) 1 2 ;

  2. (2)

    C T ={x:xC(R, R n ),x(t+T)=x(t),tR} with the norm

    | φ | 0 = max t [ 0 , T ] | φ ( t ) | ,φ C T ;
  1. (3)

    C T 1 ={x:x C 1 (R, R n ),x(t+T)=x(t),tR} with the norm

    φ= max t [ 0 , T ] { | φ | 0 , | φ | 0 } ,φ C T 1 .

Clearly, C T and C T 1 are Banach spaces.

2 Main lemmas

Lemma 2.1

[22]

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

( 1 ) [ A 1 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 ) 0 T | [ A 1 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 , ( 3 ) A 1 1 f 0 { | f | 0 1 c 0 , c 0 < 1 , f C T , | f | 0 σ 1 , σ > 1 , f C T .

Here

c 0 = max t [ 0 , T ] | c ( t ) | ,σ= min t [ 0 , T ] | c ( t ) | .

Let

A: C T C T ,[A](t)=x(t)V(t)x(tτ),

wheretR, V(t) C T 1 is a real symmetric matrix.

We will give some properties of A.

Lemma 2.2

Suppose that λ 1 (t), λ 2 (t),, λ n (t)are eigenvalues ofV(t). Then the operatorAhas continuous inverse A 1 on C T , satisfying

( 1 ) 0 T | [ A 1 f ] ( t ) | d t { ( i = 1 n 1 ( 1 λ i , L ) 2 ) 1 2 0 T | f ( t ) | d t , λ i , L < 1 , f C T , ( i = 1 n 1 ( 1 λ i , l ) 2 ) 1 2 0 T | f ( t ) | d t , λ i , l > 1 , f C T , ( 2 ) | [ A 1 f ] | 0 { ( i = 1 n 1 ( 1 λ i , L ) 2 ) 1 2 | f | 0 , λ i , L < 1 , f C T , ( i = 1 n 1 ( 1 λ i , l ) 2 ) 1 2 | f | 0 , λ i , l > 1 , f C T ,

where

λ i , L = max t [ 0 , T ] | λ i ( t ) | , λ i , l = min t [ 0 , T ] | λ i ( t ) | ,i I n .

Proof

  1. (1)

    Since V(t) is a real symmetric matrix, there exists an orthogonal matrix U(t) such that

    U(t)V(t) U T (t)= E λ (t)=diag ( λ 1 ( t ) , λ 2 ( t ) , , λ n ( t ) ) .

Consider the system

x(t)V(t)x(tτ)=f(t),

where we have equivalence to

y(t) E λ (t)y(tτ)= f ˜ (t),
(2.1)

where f ˜ (t)=U(t)f(t), y(t)=U(t)x(t). On the other hand, a component of the vector in system (2.1) is

y i (t) λ i (t) y i (tτ)= f ˜ i (t),i I n .

From Lemma 2.1, we have

y i (t)={ f ˜ i ( t ) + j = 1 k = 1 j λ i ( t ( k 1 ) τ ) f ˜ i ( t j τ ) , λ i , L < 1 , f ˜ i ( t + τ ) λ i ( t + τ ) j = 1 k = 1 j + 1 1 λ i ( t + k τ ) f ˜ i ( t + j τ + τ ) , λ i , l > 1 .
(2.2)

Thus, A 1 exists and

A 1 : C T C T , A 1 f(t)=x(t)= U T (t)y(t),t[0,T].
(2.3)

When λ i , L <1, by (2.2) we get

| y i ( t ) | max t [ 0 , T ] | f ˜ i ( t ) | 1 λ i , L ,i I i ,

i.e.,

max t [ 0 , T ] | y i ( t ) | max t [ 0 , T ] | f ˜ i ( t ) | 1 λ i , L ,i I i .

Thus, by (2.3) we have

| A 1 f | 0 = max t [ 0 , T ] | U T ( t ) y ( t ) | = max t [ 0 , T ] | y ( t ) | = max t [ 0 , T ] ( i = 1 n y i 2 ( t ) ) 1 2 ( i = 1 n max t [ 0 , T ] y i 2 ( t ) ) 1 2 ( i = 1 n max t [ 0 , T ] | f ˜ i ( t ) | 2 ( 1 λ i , L ) 2 ) 1 2 ( i = 1 n 1 ( 1 λ i , L ) 2 ) 1 2 | f ˜ | 0 = ( i = 1 n 1 ( 1 λ i , L ) 2 ) 1 2 | U f | 0 = ( i = 1 n 1 ( 1 λ i , L ) 2 ) 1 2 | f | 0 .

Obviously,

0 T | A 1 f ( t ) | dt ( i = 1 n 1 ( 1 λ i , L ) 2 ) 1 2 0 T | f ( t ) | dt.
  1. (2)

    Similar to the above proof, when λ i , l >1, we get

    | A 1 f | 0 ( i = 1 n 1 ( 1 λ i , l ) 2 ) 1 2 | f | 0 , 0 T | A 1 f ( t ) | d t ( i = 1 n 1 ( 1 λ i , l ) 2 ) 1 2 0 T | f ( t ) | d t .

 □

Let X and Y be two Banach spaces and let L:D(L)XY be a linear operator, a Fredholm operator with index zero (meaning that ImL is closed in Y and dimKerL=codimImL<+). If L is a Fredholm operator with index zero, then there exist continuous projectors P:XX, Q:YY such that ImP=KerL, ImL=KerQ=Im(IQ), and L D ( L ) Ker P :(IP)XImL is invertible. Denote by K p the inverse of L P .

Let Ω be an open bounded subset of X, a map N: Ω ¯ Y is said to be L-compact in Ω ¯ if QN( Ω ¯ ) is bounded and the operator K p (IQ)N( Ω ¯ ) is relatively compact. We first give the famous Mawhin continuation theorem.

Lemma 2.3

[23]

Suppose that X and Y are two Banach spaces andL:D(L)XYis a Fredholm operator with index zero. Furthermore, ΩXis an open bounded set andN: Ω ¯ Yis L-compact on Ω ¯ . If all the following conditions hold:

  1. (1)

    LxλNx, xΩD(L), λ(0,1),

  2. (2)

    NxImL, xΩKerL,

  3. (3)

    deg{QN,ΩKerL,0}0,

then the equationLx=Nxhas a solution on Ω ¯ D(L).

3 Main results

Theorem 3.1

Suppose that 0 T p(t)dt=0, φ(t)is a nonzero periodic solution of (3.1) and there exists a constantM>0such that

(H1):i I n , G x i is bounded in the set 1 (or 2 ), where

1 = { x : x = ( x 1 , x 2 , , x n ) R n , x i ( , M ] , x j R , j I n { i } } , 2 = { x : x = ( x 1 , x 2 , , x n ) R n , x i [ M , ) , x j R , j I n { i } } .

(H2): x i G x i >0 (or <0), whenever| x i |>M, i I n .

(H3):Suppose that μ 1 , μ 2 ,, μ n are eigenvalues of 2 F ( v ) x 2 , v R n , and there exists a constant λ F 0such that

max { | μ 1 | , | μ 2 | , , | μ n | } λ F .

Then system (1.1) has at least one T-periodic solution, if λ 0 , i < 1 2 (or σ 0 , i >1), ( λ 2 , i Tn+n λ 1 , i n )T+ λ 0 , i <1, andτ=mT, mZ, where

λ 0 , i = max t [ 0 , T ] { | c i ( t ) | , i I n } , λ 1 , i = max t [ 0 , T ] { | c i ( t ) | , i I n } , λ 2 , i = max t [ 0 , T ] { | c i ( t ) | , i I n } , σ 0 , i = min t [ 0 , T ] { | c i ( t ) | , i I n } .

Proof

Define

A : C T C T , [ A x ] ( t ) = x ( t ) C ( t ) x ( t τ ) , t R , N : C T 1 C T , ( N x ) ( t ) = d d t grad F ( x ( t ) ) grad G ( x ( t γ ( t ) ) ) + p ( t ) , L : D ( L ) C T C T 1 , L x = ( A x ) ,

where D(L)={x:x C T 1 , x C(R, R n )}. Then system (1.1) obeys the operator equation Lx=Nx. We have ( x ( t ) C ( t ) x ( t τ ) ) =0, xKerL. Then

x(t)C(t)x(tτ)= c ˜ 1 t+ c ˜ 2 ,

where c ˜ 1 , c ˜ 2 R n . Since x(t)C(t)x(tτ) C T , we have c ˜ 1 =0. Let φ(t)C(R, R n ) be a nonzero periodic solution of

x(t)C(t)x(tτ)=I,
(3.1)

then | φ ( t ) | 2 >0 and 0 T φ 2 (t)dt0, where I is an unit matrix. We get

KerL= { a 0 φ ( t ) : a 0 R } ,ImL= { y : y C T , 0 T y ( s ) d s = 0 } .

Obviously, ImL is closed in C T and dimKerL=codimImL=n. So L is a Fredholm operator with index zero. Define continuous projectors P, Q:

P: C T KerL,(Px)(t)= 0 T x ( t ) φ ( t ) d t 0 T φ 2 d t φ(t)

and

Q: C T C T /ImL,Qy= 1 T 0 T y(s)ds.

Let

L P =L | D ( L ) Ker P :D(L)KerPImL,

then

L P 1 = K P :ImLD(L)KerP.

Since ImL C T and D(L)KerP C T 1 , K P is an embedding operator. Hence K P is a completely continuous operator in ImL. By the definitions of Q and N, one knows that QN( Ω ¯ ) is bounded on Ω ¯ . Hence the nonlinear operator N is L-compact on Ω ¯ . We complete the proof by three steps.

Step 1. Let Ω 1 ={xD(L) C T 1 :Lx=λNx,λ(0,1)}. We show that Ω 1 is a bounded set. We have Lx=λNxx Ω 1 , i.e.,

d 2 d t 2 ( x ( t ) C ( t ) x ( t τ ) ) +λ d d t gradF ( x ( t ) ) +λgradG ( x ( t γ ( t ) ) ) =λp(t).
(3.2)

Integrating both sides of (3.2) over [0,T], we have

0 T gradG ( x ( t γ ( t ) ) ) dt=0,

i.e., i I n ,

0 T G ( x ( t γ ( t ) ) ) x i dt=0.
(3.3)

Let G x i be bounded in 1 and

x i G x i >0,whenever | x i |>M.
(3.4)

Let

E 1 = { t : t [ 0 , T ] , x ( t γ ( t ) ) M } , E 2 = { t : t [ 0 , T ] , x ( t γ ( t ) ) > M } .

By assumption (H1), if x i M, there exists a constant M 1 >0 such that | G x i | M 1 . From (3.3) and (3.4), we get

E 2 | G ( x ( t γ ( t ) ) ) x i | dt= E 2 G ( x ( t γ ( t ) ) ) x i dt E 1 | G ( x ( t γ ( t ) ) ) x i | dt M 1 T.

Thus

0 T | G ( x ( t γ ( t ) ) ) x i | dt= E 1 | G ( x ( t γ ( t ) ) ) x i | dt+ E 2 | G ( x ( t γ ( t ) ) ) x i | dt2 M 1 T,

i.e.,

0 T | grad G ( x ( t γ ( t ) ) ) | d t = 0 T [ i = 1 n ( G ( x ( t γ ( t ) ) ) x i ) 2 ] 1 2 d t 0 T [ i = 1 n | G ( x ( t γ ( t ) ) ) x i | ] d t 2 n M 1 T .
(3.5)

We claim that there exists a point t 1 R such that

| x i ( t 1 ) | M.
(3.6)

In fact, for t[0,T], we have x i (tγ(t))>M, and by (3.4), we have 0 T G ( x ( t γ ( t ) ) ) x i dt>0, which is a contradiction; see (3.3). So there must be a point ξ[0,T] such that

x i ( ξ λ ( ξ ) ) M.
(3.7)

Similar to the above proof, there must be a point η[0,T] such that

x i ( η γ ( η ) ) M.
(3.8)
  1. (1)

    If x i (ξγ(ξ))M, by (3.7) we have

    | x i ( ξ λ ( ξ ) ) | M.

Let t 1 =ξγ(ξ). This proves (3.6).

  1. (2)

    If x i (ξγ(ξ))<M, from (3.8) and the fact that x i (t) is continuous on , there is a point t 1 between ξγ(ξ) and ηγ(η) such that | x i ( t 1 )|M. This also proves (3.6). Let t 1 =kπ+ t 2 , kZ, t 2 [0,T]. Then | x i ( t 1 )|=| x i ( t 2 )|M. Hence we get

    | x i ( t ) | = max t [ 0 , T ] | x i ( t 2 ) + t 2 t x i ( s ) d s | | x i ( t 2 ) | + 0 T | x i ( s ) | d s M + 0 T | x ( s ) | d s , | x | 0 n ( M + 0 T | x ( s ) | d s ) n ( M + T 1 2 ( 0 T | x ( s ) | 2 d s ) 1 2 ) .

Multiplying the two sides of system (3.2) by x T (t) and integrating them over [0,T], combining with τ=mT, by (3.9) we have

0 T | x ( t ) | 2 d t + λ 2 , i 0 T | x ( t ) | 2 d t + n λ 1 , i | x | 0 0 T | x ( t ) | d t + λ 0 , i 0 T | x ( t ) | 2 d t + λ 0 T x T ( t ) grad G ( x ( t γ ( t ) ) d t λ 0 T x T ( t ) p ( t ) d t 0 ,

i.e.,

( 1 λ 0 , i ) 0 T | x ( t ) | 2 d t λ 2 , i T n ( M + 0 T | x ( t ) | d t ) 2 + n λ 1 , i n ( M + 0 T | x ( t ) | d t ) 0 T | x ( t ) | d t + ( | p | 0 + 2 n M 1 ) T ( M + 0 T | x ( t ) | d t ) = ( λ 2 , i T n + n λ 1 , i n ) ( 0 T | x ( t ) | d t ) 2 + ( 2 λ 2 , i T n M + n λ 1 , i n M + | p | 0 T + 2 n M 1 T ) 0 T | x ( t ) | d t + λ 2 , i T n M 2 + ( | p | 0 + 2 n M 1 ) T M ( 2 λ 2 , i T n + n λ 1 , i n ) T 0 T | x ( t ) | 2 d t + ( 2 λ 2 , i T n M + n λ 1 , i n M + | p | 0 T + 2 n M 1 T ) T 1 2 × ( 0 T | x ( t ) | 2 d t ) 1 2 + λ 2 , i T n M 2 + ( | p | 0 + 2 n M 1 ) T M .
(3.9)

From (3.9) and ( λ 2 , i Tn+n λ 1 , i n )T+ λ 0 , i <1, there is a constant M 2 >0 such that

0 T | x ( t ) | 2 dt M 2 .
(3.10)

In view of (3.9) and (3.10), we get

| x | 0 n ( M + T 1 2 M 2 1 2 ) := M 3 .
(3.11)

From Lemma 2.2, ( A x ( t ) ) =A x (t)2 C (t) x (tτ) C (t)x(tτ) and (3.2), if λ 0 , i < 1 2 , we have

x ( t ) + A 1 [ λ d d t grad F ( x ( t ) ) + λ grad G ( x ( t γ ( t ) ) ) ] = A 1 [ 2 C ( t ) x ( t τ ) + C ( t ) x ( t τ ) + A 1 ( λ p ( t ) ) ] , 0 T | x ( t ) | d t ( i = 1 n 1 ( 1 λ 0 , i ) 2 ) 1 2 0 T | x ( t ) | d t × ( 0 T | 2 F ( x ( t ) x 2 | | x ( t ) | d t + 0 T | grad G ( x ( t γ ( t ) ) ) | d t 0 T | x ( t ) | d t + 2 T λ 1 , i 0 T | x ( t ) | d t + T λ 2 , i | x | 0 + T | p | 0 ) .
(3.12)

From assumption (H3) and (3.10)-(3.12), we get

0 T | x ( t ) | d t ( i = 1 n 1 ( 1 λ 0 , i ) 2 ) 1 2 ( λ F T 1 2 M 2 1 2 + 2 n M 1 T + 2 T λ 1 , i T 1 2 M 2 1 2 + T λ 2 , i M 3 + T | p | 0 ) .

So there exists a constant M 4 >0 such that

0 T | x ( t ) | dt M 4 .
(3.13)

Since x(t) C T 1 , 0 T x (t)dt=0, there is a constant vector α R n such that x (α)=0; then by (3.13) we get

| x ( t ) | 0 T | x ( t ) | dt M 4 .

Thus

| x | 0 M 4 .

Step 2. Let Ω{xKerL:QNx=0}, we shall prove that Ω 2 is a bounded set. We have x(t)= a 0 φ(t), a 0 Rx Ω 2 ; then

0 T gradG ( a 0 φ ( t γ ( t ) ) ) dt= 0 T gradG ( a 0 φ ( t ) ) dt=0.
(3.14)

When λ 0 , i < 1 2 , i I n , we have

φ i ( t ) = A 1 ( 1 ) = 1 + j = 1 k = 1 j c i ( t ( k 1 ) τ ) 1 j = 1 k = 1 j λ 0 , i = 1 λ 0 , i 1 λ 0 , i = 1 2 λ 0 , i 1 λ 0 , i : = δ > 0 .

Then we have

| φ ( t ) | n δ.

Thus

a 0 M n δ .

Otherwise, if, t[0,T], | a 0 φ(t)|>M, then from assumption (H2), we have

G ( a 0 φ ( t ) ) x i >0(or <0),i I n ,

which is a contradiction; see (3.14). When σ 0 , i >1, i I n , we have

φ i ( t ) = A 1 ( 1 ) = 1 c i ( t + τ ) j = 1 k = 1 j 1 c i ( t + k τ ) 1 λ i , l j = 1 k = 1 j + 1 1 λ 0 , i = 1 λ i , l 1 : = γ < 0 .

Then we have

| φ ( t ) | n |γ|.

Thus

a 0 M n | γ | .

Otherwise, if t[0,T], | a 0 φ(t)|>M, then from assumption (H2), we have

G ( a 0 φ ( t ) ) x i >0(or <0),i I n ,

which is a contradiction; see (3.14). Hence Ω 2 is a bounded set.

Step 3. Let Ω={x C T 1 : | x | 0 <n M 2 +1, | x | 0 <n M 4 +1}, then Ω 1 Ω 2 Ω, (x,λ)Ω×(0,1), and from the above proof, LxλNx is satisfied. Obviously, condition (2) of Lemma 2.3 is also satisfied. Now we prove that condition (3) of Lemma 2.3 is satisfied. We have | x 0 |= | a 0 φ | 0 , a 0 R, x 0 ΩKerL. There at least exists a i I n such that | x i 0 |>M. When x i 0 >M, take the homotopy

H(x,μ)=μx+(1μ)QNx,x Ω ¯ KerL,μ[0,1].

Then, by using assumption (H2), we have H(x,μ)0. When x i 0 <M, take the homotopy

H(x,μ)=μx(1μ)QNx,x Ω ¯ KerL,μ[0,1].

We also have H(x,μ)0. Then by degree theory,

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

Applying Lemma 2.3, we reach the conclusion. □

Remark 3.1

When 1 2 λ 0 , i <1 or σ 0 , i <1, there are no existence results for periodic solutions for system (1.1). We hope that there is interest in doing further research on this issue.

As an application, we consider the following system:

d 2 d t 2 ( x ( t ) C ( t ) x ( t 4 π ) ) + d d t gradF ( x ( t ) ) +gradG ( x ( t 5 cos t ) ) =p(t),
(3.15)

where

x ( t ) = ( x 1 ( t ) , x 2 ( t ) ) T , τ = 4 π , γ ( t ) = 5 cos t , C ( t ) = diag ( sin t 1 , 000 , cos t 1 , 000 ) , F ( x ) = 1 2 π ( x 1 2 + 2 x 1 x 2 + x 2 2 + 2 x 1 + 3 x 2 + 1 ) , G ( x ) = 1 2 π ( x 1 + x 2 ) , p ( t ) = ( sin t , cos t ) T .

Clearly, system (3.15) is a particular case of system (1.1). Obviously,

gradG(x)= 1 2 π ( x 1 , x 2 ) T , 2 F ( v ) x 2 = ( 1 π 1 π 1 π 1 π ) .

Here assumptions (H1)-(H2) are all satisfied. In addition,

T = 2 π , λ 0 , i = λ 1 , i = λ 2 , i = 1 1 , 000 , n = 2 , ( λ 2 , i T n + n λ 1 , i n ) T + λ 0 , i 0.0976 < 1 .

By using Theorem 3.1, when λ 0 , i < 1 2 , we know that system (3.15) has at least one 2π-periodic solution.

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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).

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Correspondence to Zhi Min He.

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The authors declare that they have no competing interests.

Authors’ contributions

ZM performed all the steps of proof in this research and also wrote the paper. BD suggested many good ideas that made this paper possible and helped to improve the manuscript. All authors read and approved the final manuscript.

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He, Z.M., Du, B. Periodic solutions for a kind of neutral functional differential systems. Bound Value Probl 2014, 151 (2014). https://doi.org/10.1186/s13661-014-0151-1

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Keywords

  • periodic solutions
  • neutral
  • Mawhin’s continuation theorem