1 Introduction

This paper is devoted to the application of Mawhin’s continuation theorem to investigate the existence of periodic solutions for the following equation:

( x ( t ) p ( t ) x ( t σ ) ) ( n ) +f ( x ( t ) ) x (t)+g ( r 0 x ( t + s ) d α ( s ) ) =q(t),
(1.1)

where p, q are continuous periodic functions with period T>0, p C n (R,R) with |p(t)|1, f,gC(R,R), r>0, n is a positive integer, σR, α:[r,0] R + is a bounded variation function, r 0 (α)=1 and α(0)α(r), where r 0 (α) is the total variation of α(s) over [r,0].

In recent years, the existence of periodic solution for functional differential equations has been studied extensively (see [17]). For example, in [1], the authors studied the following equation with a deviating argument:

x (t)+f ( x ( t ) ) x (t)+g ( x ( t τ ( t , x ( t ) ) ) ) =e(t).
(1.2)

Du et al. [2] studied the second-order neutral equation with variable parameter:

( x ( t ) c ( t ) x ( t τ ) ) +f ( x ( t ) ) x (t)+g ( x ( t γ ( t ) ) ) =e(t).
(1.3)

In [2], the authors proved for the first time the lemma (Lemma 2.1) for the existence of A 1 with (Ax)(t)=x(t)c(t)x(tτ) and some properties of A 1 when c(t) is not a constant. Then they established sufficient conditions for the existence of periodic solutions of (1.3) by using Mawhin’s theorem.

In [3], Wang and Lu discussed a kind of high-order neutral functional differential equation with distributed delay:

( x ( t ) c x ( t σ ) ) ( n ) +f ( x ( t ) ) x (t)+g ( r 0 x ( t + s ) d α ( s ) ) =p(t)
(1.4)

with |c|1 a constant. However, there are several errors in the proof of Theorems 3.1 and 3.2 of [3]. The main purpose of this paper is to improve the results of [3] and modify the errors. Meanwhile, the problem considered in paper [2] is generalized to the higher-order case in our work. Moreover, two examples are given to demonstrate our results.

2 Related lemmas

For convenience we denote p i = max t [ 0 , T ] | p ( i ) (t)|, i=0,1,,n, p ( 0 ) (t)=p(t), τ= min t [ 0 , T ] |p(t)|, and define the spaces

C T = { x C ( R , R ) : x ( t + T ) x ( t ) , t R }

with the norm | x | 0 = max t [ 0 , T ] |x(t)| and

C T 1 = { x C 1 ( R , R ) : x ( t + T ) x ( t ) , t R }

with the norm x=max{ | x | 0 , | x | 0 }. Clearly, C T and C T 1 are all Banach spaces.

Define a linear operator A: C T C T by

(Ax)(t)=x(t)p(t)x(tσ),t[0,T].

Lemma 2.1 [2]

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

( 1 ) ( A 1 x ) ( t ) = { x ( t ) + j = 1 i = 1 j p ( t ( i 1 ) σ ) x ( t j σ ) , p 0 < 1 , x C T , x ( t + σ ) p ( t + σ ) j = 1 i = 1 j + 1 1 p ( t + i σ ) x ( t + ( j + 1 ) σ ) , τ > 1 , x C T ; ( 2 ) | ( A 1 x ) ( t ) | { | x | 0 1 p 0 , p 0 < 1 , x C T , | x | 0 τ 1 , τ > 1 , x C T ; ( 3 ) 0 T | ( A 1 x ) ( t ) | d t { 1 1 p 0 0 T | x ( t ) | d t , p 0 < 1 , x C T , 1 τ 1 0 T | x ( t ) | d t , τ > 1 , x C T .

Let X and Y be real Banach spaces and L:domLXY be a Fredholm operator with index zero, i.e., ImL is closed and dimkerL=codimImL<. If L is a Fredholm operator with index zero, then there exist continuous projections P:XX and Q:YY such that ImP=kerL, ImL=kerQ and L P =L | dom L ker P :(IP)XImL is invertible. Denote by K P the inverse of L P . Define N: Ω ¯ X, where ΩX is an open and bounded set; N is L-compact on Ω ¯ , if QN is continuous and bounded and K P (IQ)N is compact on Ω ¯ .

Lemma 2.2 [8]

Let X and Y be two Banach spaces with norms X and Y , respectively, and ΩX an open and bounded set. Suppose L:XdomLY is a Fredholm operator of index zero and N: Ω ¯ Y is L-compact. In addition, if

  1. (i)

    LxλNx for λ(0,1), x(domLkerL)Ω;

  2. (ii)

    NxImL for xkerLΩ;

  3. (iii)

    deg{JQN | Ω ¯ ker L ,ΩkerL,0}0, where J:ImQkerL is a homeomorphism.

Then the abstract equation Lx=Nx has at least one solution in Ω ¯ .

Define the linear operator L:domL C T 1 C T by Lx= ( A x ) ( n ) with

domL= { x C n ( R , R ) : x ( t + T ) x ( t ) , t [ 0 , T ] } =: C T n

and the nonlinear operator

N: C T 1 C T ,(Nx)(t)=f ( x ( t ) ) x (t)g ( r 0 x ( t + s ) d α ( s ) ) +q(t).

From xkerL, ( x ( t ) p ( t ) x ( t σ ) ) ( n ) =0, we can obtain

x(t)p(t)x(tσ)= c n 1 t n 1 ++ c 1 t+ c 0 ,

where c i R, i=0,1,2,,n1. From x(t)p(t)x(tσ) C T n , we have c j =0, j=1,2,,n1. Let ω C T be a solution of x(t)p(t)x(tσ)=1, from Lemma 2.1, ω(t)0, then

kerL= { x dom L : x ( t ) = c ω ( t ) , t [ 0 , T ] } .

Clearly, ImL={y C T : 0 T y(t)dt=0}. Define continuous projections

P : C T 1 C T 1 , ( P x ) ( t ) = 0 T x ( s ) d s 0 T ω ( s ) d s ω ( t ) , t [ 0 , T ] , Q : C T C T , ( Q x ) ( t ) = 1 T 0 T y ( s ) d s , t [ 0 , T ] .

It is easy to see that ImP=kerL and kerQ=ImL. So dimkerL=1=dimImQ=codimImL. Notice that ImL is closed, then L is a Fredholm operator of index zero.

Set K P = L P 1 :ImLdomLkerP by

( K P y)(t)= A 1 { i = 0 n 1 1 i ! ( A x ) ( i ) ( 0 ) t i + 1 ( n 1 ) ! 0 t ( t s ) n 1 y ( s ) d s } ,t[0,T],

where ( A x ) ( i ) (0), i=1,2,,n1, are defined by the equation CX=B; here

X T = ( ( A x ) ( n 1 ) ( 0 ) , ( A x ) ( n 2 ) ( 0 ) , , ( A x ) ( 0 ) ) , B T = ( b 1 , b 2 , , b n 1 ) , b i = 1 i ! T 0 T ( T s ) i y ( s ) d s , C = ( 1 0 0 0 0 c 1 1 0 0 0 c 2 c 1 1 0 0 c n 3 c n 4 c n 5 1 0 c n 2 c n 3 c n 4 c 1 1 ) ( n 1 ) × ( n 1 )

with

c j = T j ( j + 1 ) ! ,j=1,2,,n2.

Whereafter, since kerP={x C T 1 : 0 T x(s)ds=0}, (Ax)(0) can be determined by

0 T A 1 { i = 0 n 1 1 i ! ( A x ) ( i ) ( 0 ) t i + 1 ( n 1 ) ! 0 t ( t s ) n 1 y ( s ) d s } dt=0.

It follows from Lemma 2.1, the definition of N, Q, K P , and the continuity of f, g, q that N is L-compact.

3 Main results

Theorem 3.1 Suppose n is an even integer and i = 0 k C k i p i T i <1 with k= n 2 . In addition, if there exist constants a0 and M>0 such that

(H1) x(g(x) q ¯ )>0 (or <0), whenever |x|>M, where q ¯ = 1 T 0 T q(t)dt,

(H2) lim sup | x | max t [ 0 , T ] | g ( x ) q ( t ) | | x | a,

then (1.1) has at least one periodic solution provided

a T n 1 i = 0 k C k i p i T i <1.
(3.1)

Proof Without loss of generality, we may assume that x(g(x) q ¯ )>0, when |x|>M. Now, we will complete the proof by three steps.

Step 1. Let Ω 1 ={xdomLkerL:Lx=λNx,λ(0,1)}, x Ω 1 , Lx=λNx, i.e.,

( x ( t ) p ( t ) x ( t σ ) ) ( n ) +λf ( x ( t ) ) x (t)+λg ( r 0 x ( t + s ) d α ( s ) ) =λq(t).
(3.2)

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

0 T [ g ( r 0 x ( t + s ) d α ( s ) ) q ¯ ] dt=0.
(3.3)

By the integral mean value theorem, there exists a constant ξ(0,T) such that g( r 0 x(ξ+s)dα(s)) q ¯ =0. So (H1) implies that | r 0 x(ξ+s)dα(s)|M. Since r 0 (α)=1 and α(r)α(0) on t[r,0], by the properties of the Riemann-Stieltjes integral, there exists a constant η(r,0) such that |x(ξ+η)|M. Because x(t) is a T-periodic function, there exists an integer k 0 such that ξ+η= k 0 T+ t , t [0,T). Then |x( t )|M. Hence,

| x | 0 = max t [ 0 , T ] |x ( t ) + t t x (t)dt|M+ 0 T | x (t)|dt.
(3.4)

Multiplying both sides of (3.2) by x(t) and integrating them over [0,T], one gets

( 1 ) k 0 T | x ( k ) ( t ) | 2 d t = ( 1 ) k 0 T ( p ( t ) x ( t σ ) ) ( k ) x ( k ) ( t ) d t λ 0 T [ g ( r 0 x ( t + s ) d α ( s ) ) q ( t ) ] x ( t ) d t = ( 1 ) k i = 0 k C k i 0 T p ( i ) ( t ) x ( k i ) ( t σ ) x ( k ) ( t ) d t λ 0 T [ g ( r 0 x ( t + s ) d α ( s ) ) q ( t ) ] x ( t ) d t .
(3.5)

Since x(0)=x(T), x (0)= x (T),, x ( n 1 ) (0)= x ( n 1 ) (T), there exist ξ i (0,T), i=1,2,,n, such that x ( ξ 1 )= x ( ξ 2 )== x ( n ) ( ξ n )=0. Thus,

0 T | x (t)|dtT 0 T | x (t)|dt T n 1 0 T | x ( n ) (t)|dt
(3.6)

and

0 T | x (t) | 2 dt T 2 0 T | x (t) | 2 dt T 2 ( n 1 ) 0 T | x ( n ) (t) | 2 dt.
(3.7)

By the Hölder inequality and (3.7), we have

0 T | x ( k ) ( t ) | 2 d t i = 0 k C k i p i ( 0 T | x ( k i ) ( t σ ) | 2 d t ) 1 2 ( 0 T | x ( k ) ( t ) | 2 d t ) 1 2 + | x | 0 0 T | g ( r 0 x ( t + s ) d α ( s ) ) q ( t ) | d t i = 0 k C k i p i T i 0 T | x ( k ) ( t ) | 2 d t + | x | 0 0 T | g ( r 0 x ( t + s ) d α ( s ) ) q ( t ) | d t .
(3.8)

In view of (H2) and the properties of the bounded variation function, ε>0, there exists a constant ρ>M such that

|g ( r 0 x ( t + s ) d α ( s ) ) q(t)|(a+ε)| r 0 x(t+s)dα(s)|(a+ε) | x | 0 ,
(3.9)

when | r 0 x(t+s)dα(s)|>ρ. Let

E 1 = { t [ 0 , T ] : | r 0 x ( t + s ) d α ( s ) | > ρ } , E 2 = { t [ 0 , T ] : | r 0 x ( t + s ) d α ( s ) | ρ } .

Hence,

0 T | g ( r 0 x ( t + s ) d α ( s ) ) q ( t ) | d t = ( E 1 + E 2 ) | g ( r 0 x ( t + s ) d α ( s ) ) q ( t ) | d t T ( a + ε ) | x | 0 + T g ˜ ρ ,

where g ˜ ρ = max t E 2 |g( r 0 x(t+s)dα(s))q(t)|. By (3.4) and (3.6), we get

| x | 0 M + 0 T | x ( t ) | d t M + T k 1 0 T | x ( k ) ( t ) | d t , | x | 0 2 M 2 + 2 M T k 1 0 T | x ( k ) ( t ) | d t + T 2 ( k 1 ) ( 0 T | x ( k ) ( t ) | d t ) 2 | x | 0 2 M 2 + 2 M T k 1 0 T | x ( k ) ( t ) | d t + T 2 k 1 0 T | x ( k ) ( t ) | 2 d t .

Thus,

0 T | x ( k ) ( t ) | 2 d t 1 1 i = 0 k C k i p i T i [ T ( a + ε ) | x | 0 2 + T g ˜ ρ | x | 0 ] 1 1 i = 0 k C k i p i T i [ ( a + ε ) T 2 k 0 T | x ( k ) ( t ) | 2 d t + ( 2 ( a + ε ) M T k + T k g ˜ ρ ) 0 T | x ( k ) ( t ) | d t + ( a + ε ) T M 2 + M T g ˜ ρ ] 1 1 i = 0 k C k i p i T i [ ( a + ε ) T n 0 T | x ( k ) ( t ) | 2 d t + T k + 1 2 ( 2 ( a + ε ) M + g ˜ ρ ) ( 0 T | x ( k ) ( t ) | 2 d t ) 1 2 + ( a + ε ) T M 2 + M T g ˜ ρ ] .

In view of a T n 1 i = 0 k C k i p i T i <1, there exists a constant M 1 >0 such that 0 T | x ( k ) ( t ) | 2 dt M 1 . Then

| x | 0 M + T k 1 0 T | x ( k ) ( t ) | d t M + T k 1 2 M 1 1 2 = : M 2 , | x | 0 0 T | x ( t ) | d t T k 2 0 T | x ( k ) ( t ) | d t T k 3 2 M 1 1 2 = : M 3 .

Step 2. Let Ω 2 ={xkerL:NxImL}. Then for x Ω 2 , x(t)=cω(t), cR, and NxImL=kerQ, that is,

QNx= 1 T 0 T [ g ( r 0 c ω ( t + s ) d α ( s ) ) q ¯ ] dt=0.

Repeating the process of Step 1, we see that Ω 2 is bounded, that is, there exists M 4 >0 such that x M 4 for x Ω 2 .

Step 3. Let Ω={xdomL:x< M ˆ }, where M ˆ =max{ M 2 , M 3 , M 4 }+1, then Ω ¯ 1 Ω ¯ 2 Ω. In view of Step 1 and Step 2, conditions (i) and (ii) in Lemma 2.2 are all satisfied. Next, we will show that (iii) of Lemma 2.2 holds.

Set the isomorphism J:ImQkerL by J(x)=x. For x Ω ¯ kerL, μ[0,1], define the homotopy

H(x,μ)=μx+(1μ)JQNx,

i.e.,

H(x,μ)=μx+ 1 μ T 0 T [ g ( r 0 x ( t + s ) d α ( s ) ) q ¯ ] dt.

Since x>M for xΩkerL, x(g(x) q ¯ )>0. So H(x,μ)0 for (x,μ)(ΩkerL)×[0,1]. Hence, by the homotopy invariance of the Brouwer degree, we obtain

deg { J Q N | Ω ¯ ker L , 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 2.2, we reach the conclusion.

For the case x(g(x) q ¯ )<0, when |x|>M, a similar argument can complete the proof. Here we omit it. □

Remark 3.1 In [3], the calculation of formula (3.5) is wrong. So the bound of 0 T |g( r 0 x(t+s)dα(s))q(t)|dt is not evaluated correctly. The same error appeared again in the next theorem. We modified those errors in this paper.

Theorem 3.2 Suppose that n is an odd integer and i = 0 k C k i p i T i <1, where k= n + 1 2 and k is odd. Moreover, if f(y)0, yR and conditions (H1), (H2) and (3.1) are satisfied. Then (1.1) has at least one periodic solution.

Proof Without loss of generality, we assume that x(g(x) q ¯ )>0, when |x|>M. As in the proof of Theorem 3.1, define the set Ω 1 , for x Ω 1 , λ(0,1), (3.2) can be obtained. Multiplying both sides of (3.2) by x (t) and integrating them over [0,T], we obtain

( 1 ) k 1 0 T | x ( k ) ( t ) | 2 d t = ( 1 ) k 1 0 T ( p ( t ) x ( t σ ) ) ( k ) x ( k ) ( t ) d t λ 0 T f ( x ( t ) ) | x ( t ) | 2 d t λ 0 T [ g ( r 0 x ( t + s ) d α ( s ) ) q ( t ) ] x ( t ) d t ( 1 ) k 1 i = 0 k C k i 0 T p ( i ) ( t ) x ( k i ) ( t σ ) x ( k ) ( t ) d t λ 0 T [ g ( r 0 x ( t + s ) d α ( s ) ) q ( t ) ] x ( t ) d t .

In view of

| x | 0 M+ 0 T | x (t)|dtM+T| x | 0 ,

by the Hölder inequality and (3.6), (3.7) in the proof of Theorem 3.1, one gets

0 T | x ( k ) ( t ) | 2 d t i = 0 k C k i p i ( 0 T | x ( k i ) ( t σ ) | 2 d t ) 1 2 ( 0 T | x ( k ) ( t ) | 2 d t ) 1 2 + | x | 0 0 T | g ( r 0 x ( t + s ) d α ( s ) ) q ( t ) | d t i = 0 k C k i p i T i 0 T | x ( k ) ( t ) | 2 d t + | x | 0 0 T | g ( r 0 x ( t + s ) d α ( s ) ) q ( t ) | d t 1 1 i = 0 k C k i p i T i | x | 0 0 T | g ( r 0 x ( t + s ) d α ( s ) ) q ( t ) | d t 1 1 i = 0 k C k i p i T i [ ( a + ε ) T | x | 0 | x | 0 + T g ˜ ρ | x | 0 ] 1 1 i = 0 k C k i p i T i [ ( a + ε ) T ( M + T | x | 0 ) | x | 0 + T g ˜ ρ | x | 0 ] 1 1 i = 0 k C k i p i T i [ ( a + ε ) T 2 k 1 0 T | x ( k ) ( t ) | 2 d t + ( ( a + ε ) M + g ˜ ρ ) T k 1 0 T | x ( k ) ( t ) | d t ] 1 1 i = 0 k C k i p i T i [ ( a + ε ) T n 0 T | x ( k ) ( t ) | 2 d t + ( ( a + ε ) M + g ˜ ρ ) T k 1 2 ( 0 T | x ( k ) ( t ) | 2 d t ) 1 2 ] .

Since the condition (3.1) holds, 0 T | x ( k ) ( t ) | 2 dt is bounded. The remainder can be proved in the same way as in Theorem 3.1. □

Remark 3.2 In [3], the first inequality in (3.17) cannot be obtained without assuming that k= n + 1 2 is odd. In our paper, we correct this error by adding such a condition.

Remark 3.3 In our paper, we consider the n-order period equation (1.1); in this sense, we generalize the model in [2] to higher order under the equivalent conditions. Moreover, in view of the variable parameter p(t) in (1.1), we develop the results in [3] with the constant coefficient c.

4 Examples

As an application, we list the following examples.

Example 4.1 Consider

( x ( t ) 1 960 sin t x ( t σ ) ) ( 6 ) + 2 e 4 ( x 2 ( t ) 1 ) x ( t ) + π 0 x ( t + s ) d s e ( π 0 x ( t + s ) d s ) 2 = cos t + 1 30 sin 2 t .
(4.1)

Corresponding to (1.1), we have p(t)= 1 960 sint, g(x)=x e x 2 , f(x)=2 e 4 ( x 2 1 ) , q(t)=cost+ 1 30 sin2t, σ=2π, r=π, n=6, k=3, T=2π. Obviously, p i = 1 960 , i=0,1,2,,6; we have

i = 0 3 C 3 i 1 960 ( 2 π ) i 8 π 3 + 12 π 2 + 6 π + 1 960 < 1 , a = lim | x | max t [ 0 , 2 π ] | g ( x ) q ( t ) | | x | = 0

and

a T 2 k 1 i = 0 3 C 3 i 1 960 ( 2 π ) i =0<1.

Therefore, Theorem 3.1 implies that (4.1) has at least one 2π-periodic solution. In fact, x(t)=cost is such a solution.

Example 4.2 Consider

( x ( t ) 1 480 sin t x ( t σ ) ) ( 5 ) + 2 e 4 ( x 2 ( t ) 1 ) x ( t ) + π 0 x ( t + s ) d s e ( π 0 x ( t + s ) d s ) 2 = cos t 1 30 sin 2 t .
(4.2)

Corresponding to (1.1), we have p(t)= 1 480 sint, g(x)=x e x 2 , f(x)=2 e 4 ( x 2 1 ) , q(t)=cost 1 30 sin2t, σ=2π, r=π, n=5, k=3, T=2π. Obviously, p i = 1 480 , i=0,1,2,,5; we have

i = 0 3 C 3 i 1 480 ( 2 π ) i 8 π 3 + 12 π 2 + 6 π + 1 480 < 1 , a = lim | x | max t [ 0 , 2 π ] | g ( x ) q ( t ) | | x | = 0

and

a T 2 k 1 1 i = 0 3 C 3 i 1 480 ( 2 π ) i =0<1.

Therefore, Theorem 3.2 implies that (4.2) has at least one 2π-periodic solution. It is easy to see that x(t)=sint is such a solution.