# Homoclinic solutions for a class of neutral Duffing differential systems

## Abstract

By using an extension of Mawhin’s continuation theorem and some analysis methods, the existence of a set with $2kT$-periodic for a n-dimensional neutral Duffing differential systems, $( u ( t ) − C u ( t − τ ) ) ″ +β(t) x ′ (t)+g(u(t−γ(t)))=p(t)$, is studied. Some new results on the existence of homoclinic solutions is obtained as a limit of a certain subsequence of the above set. Meanwhile, $C= [ c i j ] n × n$ is a constant symmetrical matrix and $β(t)$ is allowed to change sign.

## 1 Introduction

The aim of this paper is to consider a kind of neutral Duffing differential systems as follows:

$( u ( t ) − C u ( t − τ ) ) ″ +β(t) x ′ (t)+g ( u ( t − γ ( t ) ) ) =p(t),$
(1.1)

where $β∈ C 1 (R,R)$ with $β(t+T)≡β(t)$, $g∈C( R n , R n )$, $p∈C(R, R n )$, and $γ(t)$ is a continuous T-periodic function with $γ(t)≥0$; $T>0$ and τ are given constants; $C= [ c i j ] n × n$ is a constant symmetrical matrix and $β(t)$ is allowed to change sign.

As is well known, a solution $u(t)$ of Eq. (1.1) is called homoclinic (to O) if $u(t)→0$ and $u ′ (t)→0$ as $|t|→+∞$. In addition, if $u≠0$, then u is called a nontrivial homoclinic solution.

Under the condition of $C=O$, system (1.1) transforms into a classic second-order Duffing equation

$u ″ (t)+β(t) x ′ (t)+g ( t , u ( t − γ ( t ) ) ) =p(t),$
(1.2)

which has been studied by Li et al.  and some new results on the existence and uniqueness of periodic solutions for (1.2) are obtained. Very recently, by using Mawhin’s continuation theorem, Du  studied the following neutral differential equations:

$( u ( t ) − C u ( t − τ ) ) ″ + d d t ∇F ( u ( t ) ) +∇G ( u ( t ) ) =e(t),$
(1.3)

where $F∈ C 2 ( R n ,R)$; $G∈ C 1 ( R n ,R)$; $e∈C(R, R n )$; $C=diag( c 1 , c 2 ,…, c n )$, $c i$ ($i=1,2,…,n$) and τ are given constants, obtaining the existence of homoclinic solutions for (1.3).

In this paper, like in the work of Rabinowitz in , Izydorek and Janczewska in  and Tan and Xiao in , the existence of a homoclinic solution for (1.1) is obtained as a limit of a certain sequence of $2kT$-periodic solutions for the following equation:

$( u ( t ) − C u ( t − τ ) ) ″ +β(t) u ′ (t)+g ( u ( t − γ ( t ) ) ) = p k (t),$
(1.4)

where $k∈N$, $p k :R→ R n$ is a $2kT$-periodic function such that

$p k (t)= { p ( t ) , t ∈ [ − k T , k T − ε 0 ) , p ( k T − ε 0 ) + p ( − k T ) − p ( k T − ε 0 ) ε 0 ( t − k T + ε 0 ) , t ∈ [ k T − ε 0 , k T ] ,$
(1.5)

$ε 0 ∈(0,T)$ is a constant independent of k. However, the approaches to show $u ′ (t)→0$ as $|t|→+∞$ are different from the corresponding ones used in the past and the existence of $2kT$-periodic solutions to Eq. (1.4) is obtained by using an extension of Mawhin’s continuation theorem, which is quite different from the approach of . Furthermore, $C= [ c i j ] n × n$ is a constant symmetrical matrix and $β(t)$ is allowed to change sign, different from the corresponding ones of .

## 2 Preliminary

Throughout this paper, $〈⋅,⋅〉: R n × R n →R$ denotes the standard inner product, and $|⋅|$ denotes the absolute value and the Euclidean norm on $R n$. For each $k∈N$, let $C 2 k T ={x|x∈C(R, R n ),x(t+2kT)≡x(t)}$, $C 2 k T 1 ={x|x∈ C 1 (R, R n ),x(t+2kT)≡x(t)}$ and $| x | 0 = max t ∈ [ 0 , 2 k T ] |x(t)|$. If the norms of $C 2 k T$ and $C 2 k T 1$ are defined by $∥ ⋅ ∥ C 2 k T = | ⋅ | 0$ and $∥ ⋅ ∥ C 2 k T 1 =max{ | x | 0 , | x ′ | 0 }$, respectively, then $C 2 k T$ and $C 2 k T 1$ are all Banach spaces. Furthermore, for $φ∈ C 2 k T$, $∥ φ ∥ r = ( ∫ − k T k T | φ ( t ) | r d t ) 1 r$, $r>1$.

Define the linear operator

$A: C T → C T ,[Ax](t)=x(t)−Cx(t−τ).$

Lemma 2.1 

Suppose that Ω is an open bounded set in X such that the following conditions are satisfied:

[A1] For each $λ∈(0,1)$, the equation

$( u ( t ) − C u ( t − τ ) ) ″ +λβ(t) u ′ (t)+λg ( u ( t − γ ( t ) ) ) =λ p k (t)$

has no solution on Ω.

[A2] The equation

$Δ(a):= 1 2 k T ∫ − k T k T [ g ( a ) − p k ( t ) ] dt=0$

has no solution on $∂Ω∩ R n$.

[A3] The Brouwer degree

$d B { Δ , Ω ∩ R n , 0 } ≠0.$

Equation (1.4) has a $2kT$-periodic solution in $Ω ¯$.

Lemma 2.2 

If set $P T ={x|x∈C(R,R),x(t+T)≡x(t)}$ and $A 0 : P T → P T$, $[ A 0 x](t)=x(t)−cx(t)$, where $c∈R$ is a constant with $|c|≠1$, then operator $A 0$ has continuous inverse $A 0 − 1$ on $P T$, satisfying

$[ A 0 − 1 f ] (t)= { ∑ j ≥ 0 c j f ( t − j τ ) , | c | < 1 , ∀ f ∈ P T , − ∑ j ≥ 1 c − j f ( t + j τ ) , | c | > 1 , ∀ f ∈ P T .$

Lemma 2.3 

If $u:R→ R n$ is continuously differentiable on R, $a>0$, $μ>1$, and $p>1$ are constants, then for every $t∈R$, the following inequality holds:

$| u ( t ) | ≤ ( 2 a ) − 1 μ ( ∫ t − a t + a | u ( s ) | μ d s ) 1 μ +a ( 2 a ) − 1 p ( ∫ t − a t + a | u ′ ( s ) | p d s ) 1 p .$

This lemma is a special case of Lemma 2.2 in .

Lemma 2.4 

Suppose that $c 1 , c 2 ,…, c n$ are eigenvalues of matrix C. If $| c i |≠1$ ($i=1,2,…,n$), then A has a continuous bounded inverse with the following relationships:

1. (1)

$∥ A − 1 f∥≤( ∑ i = 1 n 1 | 1 − | c i | | )∥f∥$, $∀f∈ C T$,

2. (2)

$∫ 0 T | ( A − 1 f ) ( t ) | p dt≤α ∫ 0 T | f ( t ) | p dt$, $∀f∈ C T$, $p≥1$, where

$α= { max ( 1 ( 1 − | c i | ) 2 ) , p = 2 , ( ∑ i = 1 n 1 ( 1 − | c i | ) 2 p 2 − p ) 2 − p 2 , p ∈ [ 1 , 2 ) , ( ∑ i = 1 n 1 1 − | c i | q ) p q , p ∈ [ 2 , + ∞ ) ,$

q is a constant with $1 p + 1 q =1$.

3. (3)

$( A x ) ′ =A x ′$, $∀x∈ C T 1$.

Lemma 2.5 

Let $s∈C(R,R)$ with $s(t+ω)≡s(t)$ and $s(t)∈[0,ω]$, $∀t∈R$. Suppose $p∈(1,+∞)$, $| s | 0 = max t ∈ [ 0 , ω ] s(t)$ and $u∈ C 1 (R,R)$ with $u(t+ω)≡u(t)$. Then

$∫ 0 ω | u ( t ) − u ( t − s ( t ) ) | p dt≤ | s | 0 p ∫ 0 ω | u ′ ( t ) | p dt.$

Throughout this paper, we suppose in addition that $c m =max{| c i |}$, $i=1,2,…,n$, where $c 1 , c 2 ,…, c n$ are eigenvalues of matrix C with $| c i |≠1$ and let $β L ′ =min| β ′ (t)|$, $β M =max|β(t)|$, $∀t∈[0,T]$.

For convenience, we list the following assumptions which will be used to study the existence of homoclinic solutions to Eq. (1.1) in Section 3.

[H1] There are constants $L>0$ and $m>0$ such that

and

[H2] $p∈C(R, R n )$ is a bounded function with $p(t)≠O= ( 0 , 0 , … , 0 ) ⊤$ and

$B:= ( ∫ R | p ( t ) | 2 d t ) 1 2 + sup t ∈ R | p ( t ) | <+∞.$

Remark 2.1 

From (1.5), we see that $| p k (t)|≤ sup t ∈ R |p(t)|$. So if assumption [H2] holds, for each $k∈N$, $( ∫ − k T k T | p k ( t ) | 2 d t ) 1 2 .

## 3 Main results

In order to investigate the existence of $2kT$-periodic solutions to system (1.4), we need to study some properties of all possible $2kT$-periodic solutions to the following system:

$( x ( t ) − C x ( t − τ ) ) ″ +λβ(t) x ′ (t)+λg ( x ( t − γ ( t ) ) ) =λ p k (t),λ∈(0,1].$
(3.1)

For each $k∈N$, let $Σ⊂ C 2 k T 1$ represent the set of all the $2kT$-periodic solutions to system (3.1).

Theorem 3.1 Suppose assumptions [H1]-[H2] hold, $β L ′ >−2m$, and

$α [ c m 1 2 L ( | γ | 0 + | τ | ) + L | γ | 0 + c m 1 2 β M ] 2 ( 1 2 β L ′ + m ) <1,$

then for each $k∈N$, if $u∈Σ$, then there are positive constants $A 0$, $A 1$, $ρ 0$, and $ρ 1$ which are independent of k and λ, such that

$∥ u ∥ 2 ≤ A 0 , ∥ u ′ ∥ 2 ≤ A 1 , | u | 0 ≤ ρ 0 , | u ′ | 0 ≤ ρ 1 .$

Proof For each $k∈N$, if $u∈Σ$, then u must satisfy

$( u ( t ) − C u ( t − τ ) ) ″ +λβ(t) u ′ (t)+λg ( u ( t − γ ( t ) ) ) =λ p k (t),λ∈(0,1].$
(3.2)

Multiplying both sides of Eq. (3.2) by $[Au](t)$ and integrating on the interval $[−kT,kT]$, we have

$− ∥ A u ′ ∥ 2 2 + λ ∫ − k T k T 〈 [ A u ] ( t ) , β ( t ) u ′ ( t ) 〉 d t + λ ∫ − k T k T 〈 [ A u ] ( t ) , g ( u ( t − γ ( t ) ) ) 〉 d t = λ ∫ − k T k T 〈 [ A u ] ( t ) , p k ( t ) 〉 d t .$
(3.3)

Clearly, $∫ − k T k T 〈u(t),β(t) u ′ (t)〉dt=− 1 2 ∫ − k T k T β ′ (t) u 2 (t)dt$, then we have

$λ ∫ − k T k T 〈 [ A u ] ( t ) , p k ( t ) 〉 d t = − ∥ A u ′ ∥ 2 2 − λ 1 2 ∫ − k T k T β ′ ( t ) u 2 ( t ) d t + λ ∫ − k T k T 〈 C u ′ ( t − τ ) , β ( t ) u ′ ( t ) 〉 d t + λ ∫ − k T k T 〈 u ( t ) , g ( u ( t − γ ( t ) ) ) − g ( u ( t ) ) 〉 d t + λ ∫ − k T k T 〈 u ( t ) , g ( u ( t ) ) 〉 d t − λ ∫ − k T k T 〈 C u ( t − τ ) , g ( u ( t − γ ( t ) ) ) − g ( u ( t − τ ) ) 〉 d t − λ ∫ − k T k T 〈 C u ( t − τ ) , g ( u ( t − τ ) ) 〉 d t$
(3.4)

and from (3.4) and [H1] that

$∥ A u ′ ∥ 2 2 + λ ( 1 2 β L ′ + m ) ∥ u ∥ 2 2 ≤ λ ∫ − k T k T | 〈 C u ( t − τ ) , β ( t ) u ′ ( t ) 〉 | d t + λ ∫ − k T k T | 〈 u ( t ) , g ( u ( t − γ ( t ) ) ) − g ( u ( t ) ) 〉 | d t + λ ∫ − k T k T | 〈 C u ( t − τ ) , g ( u ( t − γ ( t ) ) ) − g ( u ( t − τ ) ) 〉 | d t + λ ∫ − k T k T | 〈 A u ( t ) , p k ( t ) 〉 | d t .$
(3.5)

By using [H1] and Lemma 2.5, we get

$∫ − k T k T | 〈 u ( t ) , g ( u ( t − γ ( t ) ) ) − g ( u ( t ) ) 〉 | d t ≤ ( ∫ − k T k T | u ( t ) | 2 d t ) 1 2 ( ∫ − k T k T | g ( u ( t − γ ( t ) ) ) − g ( u ( t ) ) | 2 d t ) 1 2 ≤ L | γ | 0 ∥ u ∥ 2 ∥ u ′ ∥ 2 .$
(3.6)

In a similar way as in the proof of (3.6), we have

$∫ − k T k T | 〈 C u ( t − τ ) , g ( u ( t − γ ( t ) ) ) − g ( u ( t − τ ) ) 〉 | dt≤ c m 1 2 L ( | γ | 0 + | τ | ) ∥ u ∥ 2 ∥ u ′ ∥ 2 .$
(3.7)

By using [H2], we get

$∫ − k T k T | 〈 [ A u ] ( t ) , p k ( t ) 〉 | d t ≤ ∥ e k ∥ 2 ∥ u ∥ 2 + c m 1 2 ∥ p k ∥ 2 ∥ u ∥ 2 ≤ B ( 1 + c m 1 2 ) ∥ u ∥ 2$
(3.8)

and

$∫ − k T k T | 〈 C u ( t − τ ) , β ( t ) u ′ ( t ) 〉 | dt≤ c m 1 2 β M ∥ u ∥ 2 ∥ u ′ ∥ 2 .$
(3.9)

By applying (3.6)-(3.9), we see that

$∥ A u ′ ∥ 2 2 + λ ( 1 2 β L ′ + m ) ∥ u ∥ 2 2 ≤ λ [ c m 1 2 L ( | γ | 0 + | τ | ) + L | γ | 0 + c m 1 2 β M ] ∥ u ∥ 2 ∥ u ′ ∥ 2 + λ B ( 1 + c m 1 2 ) ∥ u ∥ 2 .$
(3.10)

Thus, from (3.10)

$( 1 2 β L ′ + m ) ∥ u ∥ 2 2 ≤ [ c m 1 2 L ( | γ | 0 + | τ | ) + L | γ | 0 + c m 1 2 β M ] ∥ u ∥ 2 ∥ u ′ ∥ 2 + B ( 1 + c m 1 2 ) ∥ u ∥ 2 .$
(3.11)

By using Lemma 2.4, we have $∥ u ′ ∥ 2 = ∥ A − 1 A u ′ ∥ 2 ≤ α 1 2 ∥ A u ′ ∥ 2$, and from (3.10)-(3.11)

$∥ A u ′ ∥ 2 2 ≤ α [ c m 1 2 L ( | γ | 0 + | τ | ) + L | γ | 0 + c m 1 2 β M ] 2 ( 1 2 β L ′ + m ) ∥ A u ′ ∥ 2 2 + 2 α 1 / 2 B ( 1 + c m 1 2 [ c m 1 2 L ( | γ | 0 + | τ | ) + L | γ | 0 + c m 1 2 β M ] ( 1 2 β L ′ + m ) ∥ A u ′ ∥ 2 + B 2 ( 1 + c m 1 2 ) 2 ( 1 2 β L ′ + m ) .$
(3.12)

Since

$α [ c m 1 2 L ( | γ | 0 + | τ | ) + L | γ | 0 + c m 1 2 β M ] 2 ( 1 2 β L ′ + m ) <1,$

there is a constant $M>0$ such that

$∥ A u ′ ∥ 2 ≤M,$
(3.13)
$∥ u ′ ∥ 2 ≤ α 1 2 ∥ A u ′ ∥ 2 ≤ α 1 2 M:= A 1 ,$
(3.14)

and by (3.11)

$∥ u ∥ 2 ≤ [ c m 1 2 L ( | γ | 0 + | τ | ) + L | γ | 0 + c m 1 2 β M ] A 1 + B ( 1 + c m 1 2 ) ( 1 2 β L ′ + m ) := A 0 .$
(3.15)

Obviously, $A 0$ and $A 1$ are constants independent of k and λ. Thus by using Lemma 2.2, for all $t∈[−kT,kT]$, we get

$| u ( t ) | ≤ ( 2 T ) − 1 2 ( ∫ t − T t + T | u ( s ) | 2 d s ) 1 2 + T ( 2 T ) − 1 2 ( ∫ t − T t + T | u ′ ( s ) | 2 d s ) 1 2 ≤ ( 2 T ) − 1 2 ( ∫ t − k T t + k T | u ( s ) | 2 d s ) 1 2 + T ( 2 T ) − 1 2 ( ∫ t − k T t + k T | u ′ ( s ) | 2 d s ) 1 2 = ( 2 T ) − 1 2 ( ∫ − k T k T | u ( s ) | 2 d s ) 1 2 + T ( 2 T ) − 1 2 ( ∫ − k T k T | u ′ ( s ) | 2 d s ) 1 2 .$

From (3.14) and (3.15), we obtain

$| u | 0 ≤ ( 2 T ) − 1 2 ∥ u ∥ 2 +T ( 2 T ) − 1 2 ∥ u ′ ∥ 2 ≤ ( 2 T ) − 1 2 A 0 +T ( 2 T ) − 1 2 A 1 := ρ 0 ,$
(3.16)

where $ρ 0$ is a constant independent of k and λ.

For $i=−k,−k+1,…,k−1$, from the continuity of $[A u ′ ](t)$, one can find that there is a $t i ∈[iT,(i+1)T]$ such that

$| [ A u ′ ] ( t i ) | = | 1 T ∫ i T ( i + 1 ) T [ A u ′ ] ( s ) d s | = | [ A u ] ( ( i + 1 ) T ) − [ A u ] ( i T ) T | ≤ 2 T ( 1 + c m 1 2 ) ρ 0 ,$

and it follows from (3.14) that for $t∈[iT,(i+1)T]$, $i=−k,−k+1,…,k−1$,

$| [ A u ′ ] ( t ) | = | ∫ t i t [ A u ] ″ ( s ) d s + [ A u ′ ] ( t i ) | ≤ ∫ t i t | [ A u ] ″ ( s ) | d s + 2 T ( 1 + c m 1 2 ) ρ 0 ≤ ∫ i T ( i + 1 ) T | [ A u ] ″ ( s ) | d s + 2 T ( 1 + c m 1 2 ) ρ 0 ≤ ∫ i T ( i + 1 ) T | β ( s ) u ′ ( s ) | d s + ∫ i T ( i + 1 ) T | g ( u ( s − γ ( s ) ) ) | d s + ∫ i T ( i + 1 ) T | p k ( s ) | d s + 2 T ( 1 + c m 1 2 ) ρ 0 ≤ β M T 1 2 ( ∫ − k T k T | u ′ ( s ) | 2 d s ) 1 2 + T g M + T B + 2 T ( 1 + c m 1 2 ) ρ 0 ≤ β M T 1 2 A 1 + T g M + T B + 2 T ( 1 + c m 1 2 ) ρ 0 : = ρ ,$

i.e.,

$| A u ′ | 0 ≤ρ,$
(3.17)

where $g M = max | u | 0 ≤ ρ 0 |g(u(t−τ(t)))|$.

By Lemma 2.4 and (3.17), we get

$| u ′ | 0 = | A − 1 A u ′ | 0 ≤ ( ∑ i = 1 n 1 | 1 − | c i | | ) | A u ′ | 0 ≤ ( ∑ i = 1 n 1 | 1 − | c i | | ) ρ:= ρ 1 .$

Clearly, $ρ 1$ is a constant independent of k and λ. Hence the conclusion of Theorem 3.1 holds. □

Theorem 3.2 Assume that the conditions of Theorem 3.1 are satisfied. Then for each $k∈N$, Eq. (3.2) has at least one $2kT$-periodic solution $u k (t)$ such that

$∥ u k ∥ 2 ≤ A 0 , ∥ u k ′ ∥ 2 ≤ A 1 , | u k | 0 ≤ ρ 0 , | u k ′ | 0 ≤ ρ 1 ,$

where $A 0$, $A 1$, $ρ 0$, and $ρ 1$ are constants defined by Theorem 3.1.

Proof In order to use Lemma 2.1, for each $k∈N$, we consider the following equation:

$( u ( t ) − C u ( t − τ ) ) ″ +λβ(t) u ′ (t)+λg ( u ( t − γ ( t ) ) ) =λ p k (t),λ∈(0,1).$
(3.18)

Let $Ω 1 ⊂ C 2 k T 1$ represent the set of all the $2kT$-periodic of system (3.18), since $(0,1)⊂(0,1]$, then $Ω 1 ⊂Σ$, where Σ is defined by Theorem 3.1. If $u∈ Ω 1$, by using Theorem 3.1, we have

$| u | 0 ≤ ρ 0 , | u ′ | 0 ≤ ρ 1 .$

Let $Ω 2 ={x:x∈KerL,QNx=0}$, where

$L:D(L)⊂ C 2 k T → C 2 k T$, $Lu= ( A u ) ″$,

$N: C 2 k T → C 2 k T 1$, $Nu=−β(t) u ′ (t)−g(u(t−γ(t)))+ p k (t)$,

$Q: C 2 k T → C 2 k T /ImL$, $Qy= 1 2 k T ∫ − k T k T y(s)ds$.

If $x∈ Ω 2$, then $x=a∈ R n$ (constant vector) and by [H1], we see that

$2kTm | a | 2 ≤ ∫ − k T k T | 〈 ( E − C ) a , p k ( t ) 〉 | dt≤B|a|(1+ c m ) ( 2 k T ) 1 2 ,$

i.e.,

$|a|≤ m − 1 B T − 1 2 (1+ c m ):= B 0 .$

Now, if we set $Ω={x:x∈ C 2 k T 1 , | x | 0 < ρ 0 + B 0 , | x ′ | 0 < ρ 1 +1}$, then $Ω⊃ Ω 1 ∪ Ω 2$. So condition [A1] and condition [A2] of Lemma 2.1 are satisfied. What remains is verifying condition [A3] of Lemma 2.1. In order to do this, let

$H(x,μ): ( Ω ∩ R n ) ×[0,1]⟶ R n :H(x,μ)=−μx+(1−μ)Δ(x),$

where $Δ(x)= 1 2 k T ∫ − k T k T [g(x)− p k (t)]dt$ is determined by Lemma 2.1. From assumption [H1], we have

$H(x,μ)≠0,∀(x,μ)∈ [ ∂ ( Ω ∩ R n ) ] ×[0,1].$

Hence

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

So condition [A3] of Lemma 2.1 is satisfied. Therefore, by using Lemma 2.1, we see that Eq. (1.2) has a $2kT$-periodic solution $u k ∈ Ω . ¯$ Evidently, $u k (t)$ is a $2kT$-periodic solution to Eq. (3.1) for the case of $λ=1$, so $u k ∈Σ$. Thus, by using Theorem 3.1, we get

$∥ u k ∥ 2 ≤ A 0 , ∥ u k ′ ∥ 2 ≤ A 1 , | u k | 0 ≤ ρ 0 , | u k ′ | 0 ≤ ρ 1 .$
(3.19)

□

Theorem 3.3 Suppose that the conditions in Theorem 3.1 hold, then Eq. (1.1) has a nontrivial homoclinic solution.

Proof From Theorem 3.2, we see that for each $k∈N$, there exists a $2kT$-periodic solution $u k (t)$ to Eq. (1.2). So for every $k∈N$, $u k (t)$ satisfies

$( u k ( t ) − C u k ( t − τ ) ) ″ +β(t) u k ′ (t)+g ( u k ( t − γ ( t ) ) ) = p k (t).$
(3.20)

Let $y k =(A u k ′ )$ for $k> k 0$. By (3.17),

$| y k | 0 ≤ρ$

and, by (3.20),

$| y k ′ | 0 ≤ β M | u k ′ | 0 + g M + sup t ∈ R | p ( t ) | := ρ 2 .$

Obviously, $ρ 2$ is a constant independent of k. Similar to the proof of Lemma 2.4 in , we see that there exists a $u 0 ∈ C 1 (R, R n )$ such that for each interval $[c,d]⊂R$, there is a subsequence ${ u k j }$ of ${ u k }$ with R, $u k j (t)→ u 0 (t)$ and $u k j ′ (t)→ u 0 ′ (t)$ uniformly on $[c,d]$.

For all $a,b∈R$ with $a, there must be a positive integer $j 0$ such that for $j> j 0$, $[− k j T, k j T− ε 0 ]⊃[a− | γ | 0 ,b+ | γ | 0 ]$. So for $t∈[a− | γ | 0 ,b+ | γ | 0 ]$, from (1.5) and (3.20) we see that

$( u k j ( t ) − C u k j ( t − τ ) ) ″ =−β(t) u k j ′ (t)−g ( u k j ( t − γ ( t ) ) ) +p(t).$
(3.21)

By (3.21),

$y k ′ = ( A u k j ′ ) ′ = − β ( t ) u k j ′ ( t ) − g ( u k j ( t − γ ( t ) ) ) + p ( t ) → − β ( t ) u 0 ′ ( t ) − g ( u 0 ( t − γ ( t ) ) ) + p ( t ) : = χ ( t ) ,$

uniformly on $[a,b]$.

By the fact that $y k j ′ (t)$ is a continuous differential on $(a,b)$, for $j> j 0$, $y k j ′ (t)→χ(t)$ uniformly $[a,b]$. We have $χ(t)= ( u 0 ( t ) − C u 0 ( t − τ ) ) ″$, $t∈R$, in view of $a,b∈R$ being arbitrary, that is, $u 0 (t)$ is a solution to system (1.1).

Now, we will prove $u 0 (t)→0$ and $u 0 ′ (t)→0$ for $|t|→+∞$. We have

$∫ − ∞ + ∞ ( | u 0 ( t ) | 2 + | u 0 ′ ( t ) | 2 ) d t = lim i → + ∞ ∫ − i T i T ( | u 0 ( t ) | 2 + | u 0 ′ ( t ) | 2 ) d t = lim i → + ∞ lim j → + ∞ ∫ − i T i T ( | u k j ( t ) | 2 + | u k j ′ ( t ) | 2 ) d t .$

Clearly, for every $i∈N$ if $k j >i$, by (3.14) and (3.15), we get

$∫ − i T i T ( | u k j ( t ) | 2 + | u k j ′ ( t ) | 2 ) dt≤ ∫ − k j T k j T ( | u k j ( t ) | 2 + | u k j ′ ( t ) | 2 ) dt≤ A 0 2 + A 1 2 .$

Let $i→+∞$ and $j→+∞$; we have

$∫ − ∞ + ∞ ( | u 0 ( t ) | 2 + | u 0 ′ ( t ) | 2 ) dt≤ A 0 2 + A 1 2 ,$
(3.22)

and then

$∫ | t | ≥ r ( | u 0 ( t ) | 2 + | u 0 ′ ( t ) | 2 ) dt→0,$
(3.23)

as $r→+∞$.

From (3.13), in a similar way we get

$∫ − ∞ + ∞ | u 0 ′ ( t ) − C u 0 ′ ( t − τ ) | 2 dt≤ M 2 .$
(3.24)

So, by using Lemma 2.3,

$| u 0 ( t ) | ≤ ( 2 T ) − 1 2 ( ∫ t − T t + T | u 0 ( s ) | 2 d s ) 1 2 + T ( 2 T ) − 1 2 ( ∫ t − T t + T | u 0 ′ ( s ) | 2 d s ) 1 2 ≤ max { ( 2 T ) − 1 2 , T ( 2 T ) − 1 2 } ∫ t − T t + T ( | u 0 ( t ) | 2 + | u 0 ′ ( t ) | 2 ) d t → 0 , | t | → + ∞ .$

Finally, in order to obtain

$| u 0 ′ ( t ) | →0,|t|→+∞,$

we show that

$| [ A ˜ u ′ ] 0 ( t ) | := | u 0 ′ ( t ) − C u 0 ′ ( t − τ ) | →0,|t|→+∞.$
(3.25)

From (3.16), we have $| u | 0 ≤ ρ 0$ and by (1.1), we get

If (3.25) does not hold, then there exist $ε 0 ∈(0, 1 2 )$ and a sequence ${ t k }$ such that

$| t 1 |<| t 2 |<| t 3 |<⋯<| t k |+1<| t k + 1 |,k=1,2,…,$

and

$| [ A ˜ u 0 ′ ] ( t k ) | ≥2 ε 0 ,k=1,2,….$

From this, we have, for $t∈[ t k , t k + ε 0 /(1+ M ˜ )]$,

$| [ A ˜ u 0 ′ ] ( t ) | = | [ A ˜ u 0 ′ ] ( t k ) + ∫ t k t ( [ A ˜ u 0 ′ ] ( s ) ) ′ d s | ≥ | [ A ˜ u 0 ′ ] ( t k ) | − ∫ t k t | ( [ A ˜ u 0 ′ ] ( s ) ) ′ | ds≥ ε 0 .$

It follows that

$∫ − ∞ + ∞ | [ A ˜ u 0 ′ ] ( t k ) | 2 dt≥ ∑ k = 1 ∞ ∫ t k t k + ε 0 / ( 1 + M ˜ ) | [ A ˜ u 0 ′ ] ( t k ) | 2 dt=∞,$

which contradicts (3.24), so (3.25) holds.

Since C is symmetrical, it is easy to see that there is an orthogonal matrix T such that $TC T ⊤ = E c =diag( c 1 , c 2 ,…, c n )$.

Let $y k j ′ (t)=T u k j ′ (t)=( y k j ′ ( 1 ) (t), y k j ′ ( 2 ) (t),…, y k j ′ ( n ) (t))=T ( u k j ′ ( 1 ) ( t ) , u k j ′ ( 2 ) ( t ) , … , u k j ′ ( n ) ( t ) ) ⊤$, then we get $y 0 ′ (t)=( y 0 ′ ( 1 ) (t), y 0 ′ ( 2 ) (t),…, y 0 ′ ( n ) (t))=T u 0 ′ (t)=T ( u 0 ′ ( 1 ) ( t ) , u 0 ′ ( 2 ) ( t ) , … , u 0 ′ ( n ) ( t ) ) ⊤$ as $j→∞$. By (3.25), we have

$| y 0 ′ ( t ) − E c y 0 ′ ( t − τ ) | →0,|t|→+∞.$
(3.26)

By using (3.19), we see that $|A u k ′ |<(1+ c m 1 2 ) ρ 1 := B ˜$, which implies

$| T A u k ′ | = | 〈 T A u k ′ , T A u k ′ 〉 | 1 2 < B ˜ ,$

i.e.,

$| y k ′ ( t ) − E c y k ′ ( t − τ ) | < B ˜ ,∀t∈R.$
(3.27)

For all $ε>0$, there exists $N=[ log | c i | ε ( 1 − | c i | ) 2 B ˜ ]>0$ such that $∑ h = N + 1 ∞ | c i | h < ε 2 B ˜$ ($| c i |<1$), for $t>N$. Similarly, by (3.26), we see that there is a constant $G>0$ such that $| y 0 i ′ (t)− c i y 0 i ′ (t−τ)|< ε 2 ( N + 1 )$, for $t>G$.

Then, by using Lemma 2.2 and (3.27), when $| c i |<1$, we get

$| y 0 ′ ( i ) ( t ) | = lim j → + ∞ | [ A 0 − 1 A 0 y k j ′ ( i ) ] ( t ) | ≤ | lim j → ∞ ∑ h ≥ 0 N c i h [ A 0 y k j ′ ( i ) ] ( t − h τ ) + ∑ h = N + 1 ∞ c i h [ A 0 y k j ′ ( i ) ] ( t − h τ ) | ≤ | lim j → ∞ ∑ h ≥ 0 N c i h [ A 0 y k j ′ ( i ) ] ( t − h τ ) | + | lim j → ∞ ∑ h = N + 1 ∞ c i h [ A 0 y k j ′ ( i ) ] ( t − h τ ) | ≤ lim j → ∞ ∑ h ≥ 0 N | c i | h | [ A 0 y k j ′ ( i ) ] ( t − h τ ) | + B ˜ ∑ h = N + 1 ∞ | c i | h = ∑ h ≥ 0 N | c i | h | ( y 0 ′ ( i ) ( t − h τ ) − c i y 0 ′ ( i ) ( t − ( h + 1 ) τ ) ) | + B ˜ ∑ h = N + 1 ∞ | c i | h .$
(3.28)

Now, by (3.27) and (3.28), we conclude that $∀ε>0$, there exists $N ¯ =G+N$ such that for $t> N ¯$,

$| y 0 i ′ ( t ) | ≤ ∑ h ≥ 0 N | c i | h | ( y 0 ′ ( i ) ( t − h τ ) − c i y 0 ′ ( i ) ( t − ( h + 1 ) τ ) ) | + | B ˜ ∑ h = N + 1 ∞ c i h | < ( N + 1 ) ε 2 ( N + 1 ) + B ˜ ε 2 B ˜ = ε .$

Thus, we get $| y 0 ′ ( i ) (t)|→0$, as $|t|→+∞$.

In the similar way, when $| c i |>1$, we can proof $| y 0 ′ ( i ) (t)|→0$, as $|t|→+∞$.

Therefore, $| y 0 ′ (t)|→0$, as $|t|→+∞$; i.e.,

$T ( lim | t | → + ∞ u 0 ′ ( 1 ) ( t ) , lim | t | → + ∞ u 0 ′ ( 2 ) ( t ) , … , lim | t | → + ∞ u 0 ′ ( n ) ( t ) ) ⊤ =O,$

we know T is an orthogonal matrix, then $u 0 ′ ( i ) (t)→0$ as $|t|→+∞$.

Thus, we have

$| u 0 ′ ( t ) | →0,|t|→+∞.$

Clearly, $u 0 (t)≠0$; otherwise, $p(t)=0$, which contradicts the assumption [H2].

As an application, we consider the following equation:

$( u ( t ) − C u ( t − 0.01 ) ) ″ +sin(t) x ′ (t)+g ( u ( t − cos 2 t ) ) =p(t),$
(3.29)

where $C= ( 26 3 3 17 )$, $u(t)= ( u 1 ( t ) , u 1 ( t ) ) ⊤$, $g(x)=x= ( x 1 , x 2 ) ⊤$ and $p(t)= ( p 1 ( t ) , p 2 ( t ) ) ⊤ = ( 1 1 + t 2 , 2 1 + t 2 ) ⊤$. Clearly, $λ 1 , 2 = 43 ± 117 2 ≠±1$. Also, $〈(E−C)x,g(x)〉=−25 x 1 2 −6 x 1 x 2 −16 x 2 2 <−10( x 1 2 + x 2 2 )$ and $g(x)=x$, which implies that assumption [H1] is satisfied with $L=2$, $m=10$. $p(t)= ( 1 1 + t 2 , 2 1 + t 2 ) ⊤$ is a bounded function and $( ∫ R | p ( t ) | 2 d t ) 1 2 + sup t ∈ R |p(t)|= 5 (1+ 2 2 π)$, which implies that assumption [H2] holds. Furthermore, we can choose $α= 4 ( 117 − 41 ) 2$, $c m = 43 + 117 2$, $| γ | 0 =1$, $β M =1$ and $β L ′ >−20$, then

$1 ( 117 − 41 ) 2 [ ( 43 + 117 2 ) 1 2 2 ( 1 + 0.01 ) + 2 + ( 43 + 117 2 ) 1 2 ] 2 − 1 2 + 10 <1.$

By applying Theorem 3.3, we see that Eq. (3.29) has a nontrivial homoclinic solution. □

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The author would like to express the sincere gratitude to Editor for handling the process of reviewing the paper, as well as to the reviewers who carefully reviewed the manuscript.

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