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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(tγ(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), gC( R n , R n ), pC(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 u0, 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. [1] 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 [2] 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); eC(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 [3], Izydorek and Janczewska in [4] and Tan and Xiao in [5], 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 kN, 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 [35]. 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].

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 kN, let C 2 k T ={x|xC(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(tτ).

Lemma 2.1 [6]

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 [7]

If set P T ={x|xC(R,R),x(t+T)x(t)} and A 0 : P T P T , [ A 0 x](t)=x(t)cx(t), where cR 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 [5]

If u:R R n is continuously differentiable on R, a>0, μ>1, and p>1 are constants, then for every tR, 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 [5].

Lemma 2.4 [6]

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 dtα 0 T | f ( t ) | p dt, f C T , p1, 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 [7]

Let sC(R,R) with s(t+ω)s(t) and s(t)[0,ω], tR. 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

| g ( x 1 ) g ( x 2 ) | L| x 1 x 2 |,for all  x 1 , x 2 R n ,

and

( E C ) x , g ( x ) m | x | 2 ,for all x R n ,

[H2] pC(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 [8]

From (1.5), we see that | p k (t)| sup t R |p(t)|. So if assumption [H2] holds, for each kN, ( k T k T | p k ( t ) | 2 d t ) 1 2 <B.

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 kN, 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 kN, if uΣ, 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 kN, if uΣ, 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,,k1, 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,,k1,

| [ 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(tτ(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 kN, 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 kN, 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:xKerL,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(tγ(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 ) | dtB|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+(1μ)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 kN, there exists a 2kT-periodic solution u k (t) to Eq. (1.2). So for every kN, 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 [5], 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,bR with a<b, 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 ) ) , tR, in view of a,bR 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 iN 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 ) dt0,
(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

| ( [ A ˜ u 0 ] ( t ) ) | | β ( t ) u 0 ( t ) | + | g ( u 0 ( t γ ( t ) ) ) | + sup t R | p ( t ) | β M ρ 0 + sup | u | ρ 0 | g ( u ) | + sup t R | p ( t ) | : = M ˜ , for  t R .

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 ˜ ,tR.
(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 (tτ)|< 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, (EC)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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Chen, W. Homoclinic solutions for a class of neutral Duffing differential systems. Adv Differ Equ 2014, 121 (2014). https://doi.org/10.1186/1687-1847-2014-121

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Keywords

  • homoclinic solution
  • continuation theorem
  • periodic solution