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, ${(u(t)-Cu(t-\tau ))}^{″}+\beta (t)x^{\prime }(t)+g(u(t-\gamma (t)))=p(t)$, 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, $C={[c_{ij}]}_{n\times n}$ is a constant symmetrical matrix and $\beta (t)$ is allowed to change sign.

1 Introduction

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

$${(u(t)-Cu(t-\tau ))}^{″}+\beta (t)x^{\prime }(t)+g\left(u(t-\gamma (t))\right)=p(t),$$

(1.1)

where $\beta \in C^1(R,R)$ with $\beta (t+T)\equiv \beta (t)$, $g\in C(R^n,R^n)$, $p\in C(R,R^n)$, and $\gamma (t)$ is a continuous T-periodic function with $\gamma (t)\ge 0$; $T>0$ and τ are given constants; $C={[c_{ij}]}_{n\times n}$ is a constant symmetrical matrix and $\beta (t)$ is allowed to change sign.

As is well known, a solution $u(t)$ of Eq. (1.1) is called homoclinic (to O) if $u(t)\to 0$ and $u^{\prime }(t)\to 0$ as $|t|\to +\mathrm{\infty }$. In addition, if $u\ne 0$, 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

$$u^{″}(t)+\beta (t)x^{\prime }(t)+g(t,u(t-\gamma (t)))=p(t),$$

(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:

$${(u(t)-Cu(t-\tau ))}^{″}+\frac{d}{dt}\mathrm{\nabla }F(u(t))+\mathrm{\nabla }G(u(t))=e(t),$$

(1.3)

where $F\in C^2(R^n,R)$; $G\in C^1(R^n,R)$; $e\in C(R,R^n)$; $C=diag(c_1,c_2,\dots ,c_n)$, $c_i$ ($i=1,2,\dots ,n$) 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:

$${(u(t)-Cu(t-\tau ))}^{″}+\beta (t)u^{\prime }(t)+g\left(u(t-\gamma (t))\right)=p_k(t),$$

(1.4)

where $k\in N$, $p_k:R\to R^n$ is a $2kT$-periodic function such that

$$p_k(t)=\{\begin{array}{ll}p(t),& t\in [-kT,kT-{\epsilon }_0),\\ p(kT-{\epsilon }_0)+\frac{p(-kT)-p(kT-{\epsilon }_0)}{{\epsilon }_0}(t-kT+{\epsilon }_0),& t\in [kT-{\epsilon }_0,kT],\end{array}$$

(1.5)

${\epsilon }_0\in (0,T)$ is a constant independent of k. However, the approaches to show $u^{\prime }(t)\to 0$ as $|t|\to +\mathrm{\infty }$ 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 [3–5]. Furthermore, $C={[c_{ij}]}_{n\times n}$ is a constant symmetrical matrix and $\beta (t)$ is allowed to change sign, different from the corresponding ones of [2].

2 Preliminary

Throughout this paper, $〈\cdot ,\cdot 〉:R^n\times R^n\to R$ denotes the standard inner product, and $|\cdot |$ denotes the absolute value and the Euclidean norm on $R^n$. For each $k\in N$, let $C_{2kT}=\{x|x\in C(R,R^n),x(t+2kT)\equiv x(t)\}$, $C_{2kT}^1=\{x|x\in C^1(R,R^n),x(t+2kT)\equiv x(t)\}$ and ${|x|}_0={max}_{t\in [0,2kT]}|x(t)|$. If the norms of $C_{2kT}$ and $C_{2kT}^1$ are defined by ${\parallel \cdot \parallel }_{C_{2kT}}={|\cdot |}_0$ and ${\parallel \cdot \parallel }_{C_{2kT}^1}=max\{{|x|}_0,{|x^{\prime }|}_0\}$, respectively, then $C_{2kT}$ and $C_{2kT}^1$ are all Banach spaces. Furthermore, for $\phi \in C_{2kT}$, ${\parallel \phi \parallel }_r={({\int }_{-kT}^{kT}{|\phi (t)|}^{r}dt)}^{\frac{1}{r}}$, $r>1$.

Define the linear operator

$$A:C_T\to C_T,[Ax](t)=x(t)-Cx(t-\tau ).$$

Lemma 2.1 [6]

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

[A₁] For each $\lambda \in (0,1)$, the equation

$${(u(t)-Cu(t-\tau ))}^{″}+\lambda \beta (t)u^{\prime }(t)+\lambda g\left(u(t-\gamma (t))\right)=\lambda p_k(t)$$

has no solution on ∂ Ω.

[A₂] The equation

$$\mathrm{\Delta }(a):=\frac{1}{2kT}{\int }_{-kT}^{kT}[g(a)-p_k(t)]dt=0$$

has no solution on $\partial \mathrm{\Omega }\cap R^n$.

[A₃] The Brouwer degree

$$d_B\{\mathrm{\Delta },\mathrm{\Omega }\cap R^n,0\}\ne 0.$$

Equation (1.4) has a $2kT$-periodic solution in $\overline{\mathrm{\Omega }}$.

Lemma 2.2 [7]

If set $P_T=\{x|x\in C(R,R),x(t+T)\equiv x(t)\}$ and $A_0:P_T\to P_T$, $[A_{0}x](t)=x(t)-cx(t)$, where $c\in R$ is a constant with $|c|\ne 1$, then operator $A_0$ has continuous inverse $A_0^{-1}$ on $P_T$, satisfying

$$\left[A_0^{-1}f\right](t)=\{\begin{array}{ll}{\sum }_{j\ge 0}{c}^{j}f(t-j\tau ),& |c|<1,\mathrm{\forall }f\in P_T,\\ -{\sum }_{j\ge 1}{c}^{-j}f(t+j\tau ),& |c|>1,\mathrm{\forall }f\in P_T.\end{array}$$

Lemma 2.3 [5]

If $u:R\to R^n$ is continuously differentiable on R, $a>0$, $\mu >1$, and $p>1$ are constants, then for every $t\in R$, the following inequality holds:

$$|u(t)|\le {(2a)}^{-\frac{1}{\mu }}{\left({\int }_{t-a}^{t+a}{|u(s)|}^{\mu }ds\right)}^{\frac{1}{\mu }}+a{(2a)}^{-\frac{1}{p}}{\left({\int }_{t-a}^{t+a}{|u^{\prime }(s)|}^{p}ds\right)}^{\frac{1}{p}}.$$

This lemma is a special case of Lemma 2.2 in [5].

Lemma 2.4 [6]

Suppose that $c_1,c_2,\dots ,c_n$ are eigenvalues of matrix C. If $|c_i|\ne 1$ ($i=1,2,\dots ,n$), then A has a continuous bounded inverse with the following relationships:

• (1) $\parallel A^{-1}f\parallel \le ({\sum }_{i=1}^n\frac{1}{|1-|c_i||})\parallel f\parallel$, $\mathrm{\forall }f\in C_T$,

• (2) ${\int }_0^T{|(A^{-1}f)(t)|}^{p}dt\le \alpha {\int }_0^T{|f(t)|}^{p}dt$, $\mathrm{\forall }f\in C_T$, $p\ge 1$, where

  $$\alpha =\{\begin{array}{ll}max(\frac{1}{{(1-|c_i|)}^2}),& p=2,\\ {({\sum }_{i=1}^n\frac{1}{(1-|c_i|)\frac{2p}{2-p}})}^{\frac{2-p}{2}},& p\in [1,2),\\ {({\sum }_{i=1}^n\frac{1}{1-{|c_i|}^q})}^{\frac{p}{q}},& p\in [2,+\mathrm{\infty }),\end{array}$$

  q is a constant with $\frac{1}{p}+\frac{1}{q}=1$.

• (3) ${(Ax)}^{\prime }=A{x}^{\prime }$, $\mathrm{\forall }x\in C_T^1$.

Lemma 2.5 [7]

Let $s\in C(R,R)$ with $s(t+\omega )\equiv s(t)$ and $s(t)\in [0,\omega ]$, $\mathrm{\forall }t\in R$. Suppose $p\in (1,+\mathrm{\infty })$, ${|s|}_0={max}_{t\in [0,\omega ]}s(t)$ and $u\in C^1(R,R)$ with $u(t+\omega )\equiv u(t)$. Then

$${\int }_0^{\omega }{|u(t)-u(t-s(t))|}^{p}dt\le {|s|}_0^p{\int }_0^{\omega }{|u^{\prime }(t)|}^{p}dt.$$

Throughout this paper, we suppose in addition that $c_m=max\{|c_i|\}$, $i=1,2,\dots ,n$, where $c_1,c_2,\dots ,c_n$ are eigenvalues of matrix C with $|c_i|\ne 1$ and let ${\beta }_L^{\prime }=min|{\beta }^{\prime }(t)|$, ${\beta }_M=max|\beta (t)|$, $\mathrm{\forall }t\in [0,T]$.

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

$$|g(x_1)-g(x_2)|\le L|x_1-x_2|,\text{for all }{x}_1,x_2\in R^n,$$

and

$$〈(E-C)x,g(x)〉\le -m{|x|}^2,\text{for all }x\in R^n,$$

[H₂] $p\in C(R,R^n)$ is a bounded function with $p(t)\ne O={(0,0,\dots ,0)}^{\mathrm{\top }}$ and

$$B:={\left({\int }_R{|p(t)|}^{2}dt\right)}^{\frac{1}{2}}+\underset{t\in R}{sup}|p(t)|<+\mathrm{\infty }.$$

Remark 2.1 [8]

From (1.5), we see that $|p_k(t)|\le {sup}_{t\in R}|p(t)|$. So if assumption [H₂] holds, for each $k\in \mathbf{N}$, ${({\int }_{-kT}^{kT}{|p_k(t)|}^{2}dt)}^{\frac{1}{2}}<B$.

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:

$${(x(t)-Cx(t-\tau ))}^{″}+\lambda \beta (t)x^{\prime }(t)+\lambda g\left(x(t-\gamma (t))\right)=\lambda p_k(t),\lambda \in (0,1].$$

(3.1)

For each $k\in \mathbf{N}$, let $\mathrm{\Sigma }\subset C_{2kT}^1$ represent the set of all the $2kT$-periodic solutions to system (3.1).

Theorem 3.1 Suppose assumptions [H₁]-[H₂] hold, ${\beta }_L^{\prime }>-2m$, and

$$\frac{\alpha {[c_m^{\frac{1}{2}}L({|\gamma |}_0+|\tau |)+L{|\gamma |}_0+c_m^{\frac{1}{2}}{\beta }_M]}^2}{(\frac{1}{2}{\beta }_L^{\prime }+m)}<1,$$

then for each $k\in \mathbf{N}$, if $u\in \mathrm{\Sigma }$, then there are positive constants $A_0$, $A_1$, ${\rho }_0$, and ${\rho }_1$ which are independent of k and λ, such that

$${\parallel u\parallel }_2\le A_0,{\parallel u^{\prime }\parallel }_2\le A_1,{|u|}_0\le {\rho }_0,{\left|u^{\prime }\right|}_0\le {\rho }_1.$$

Proof For each $k\in \mathbf{N}$, if $u\in \mathrm{\Sigma }$, then u must satisfy

$${(u(t)-Cu(t-\tau ))}^{″}+\lambda \beta (t)u^{\prime }(t)+\lambda g\left(u(t-\gamma (t))\right)=\lambda p_k(t),\lambda \in (0,1].$$

(3.2)

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

$$\begin{array}{r}-{\parallel A{u}^{\prime }\parallel }_2^2+\lambda {\int }_{-kT}^{kT}〈[Au](t),\beta (t)u^{\prime }(t)〉dt+\lambda {\int }_{-kT}^{kT}〈[Au](t),g\left(u(t-\gamma (t))\right)〉dt\\ =\lambda {\int }_{-kT}^{kT}〈[Au](t),p_k(t)〉dt.\end{array}$$

(3.3)

Clearly, ${\int }_{-kT}^{kT}〈u(t),\beta (t)u^{\prime }(t)〉dt=-\frac{1}{2}{\int }_{-kT}^{kT}{\beta }^{\prime }(t)u^2(t)dt$, then we have

$$\begin{array}{r}\lambda {\int }_{-kT}^{kT}〈[Au](t),p_k(t)〉dt\\ =-{\parallel A{u}^{\prime }\parallel }_2^2-\lambda \frac{1}{2}{\int }_{-kT}^{kT}{\beta }^{\prime }(t)u^2(t)dt+\lambda {\int }_{-kT}^{kT}〈C{u}^{\prime }(t-\tau ),\beta (t)u^{\prime }(t)〉dt\\ +\lambda {\int }_{-kT}^{kT}〈u(t),g\left(u(t-\gamma (t))\right)-g(u(t))〉dt+\lambda {\int }_{-kT}^{kT}〈u(t),g(u(t))〉dt\\ -\lambda {\int }_{-kT}^{kT}〈Cu(t-\tau ),g\left(u(t-\gamma (t))\right)-g(u(t-\tau ))〉dt\\ -\lambda {\int }_{-kT}^{kT}〈Cu(t-\tau ),g(u(t-\tau ))〉dt\end{array}$$

(3.4)

and from (3.4) and [H₁] that

$$\begin{array}{r}{\parallel A{u}^{\prime }\parallel }_2^2+\lambda (\frac{1}{2}{\beta }_L^{\prime }+m){\parallel u\parallel }_2^2\\ \le \lambda {\int }_{-kT}^{kT}\left|〈Cu(t-\tau ),\beta (t)u^{\prime }(t)〉\right|dt\\ +\lambda {\int }_{-kT}^{kT}\left|〈u(t),g\left(u(t-\gamma (t))\right)-g(u(t))〉\right|dt\\ +\lambda {\int }_{-kT}^{kT}\left|〈Cu(t-\tau ),g\left(u(t-\gamma (t))\right)-g(u(t-\tau ))〉\right|dt\\ +\lambda {\int }_{-kT}^{kT}\left|〈Au(t),p_k(t)〉\right|dt.\end{array}$$

(3.5)

By using [H₁] and Lemma 2.5, we get

$$\begin{array}{r}{\int }_{-kT}^{kT}\left|〈u(t),g\left(u(t-\gamma (t))\right)-g(u(t))〉\right|dt\\ \le {\left({\int }_{-kT}^{kT}{|u(t)|}^{2}dt\right)}^{\frac{1}{2}}{\left({\int }_{-kT}^{kT}{|g\left(u(t-\gamma (t))\right)-g(u(t))|}^{2}dt\right)}^{\frac{1}{2}}\\ \le L{|\gamma |}_0{\parallel u\parallel }_2{\parallel u^{\prime }\parallel }_2.\end{array}$$

(3.6)

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

$${\int }_{-kT}^{kT}\left|〈Cu(t-\tau ),g\left(u(t-\gamma (t))\right)-g(u(t-\tau ))〉\right|dt\le c_m^{\frac{1}{2}}L({|\gamma |}_0+|\tau |){\parallel u\parallel }_2{\parallel u^{\prime }\parallel }_2.$$

(3.7)

By using [H₂], we get

$$\begin{array}{rl}{\int }_{-kT}^{kT}\left|〈[Au](t),p_k(t)〉\right|dt& \le {\parallel e_k\parallel }_2{\parallel u\parallel }_2+c_m^{\frac{1}{2}}{\parallel p_k\parallel }_2{\parallel u\parallel }_2\\ \le B(1+c_m^{\frac{1}{2}}){\parallel u\parallel }_2\end{array}$$

(3.8)

and

$${\int }_{-kT}^{kT}\left|〈Cu(t-\tau ),\beta (t)u^{\prime }(t)〉\right|dt\le c_m^{\frac{1}{2}}{\beta }_M{\parallel u\parallel }_2{\parallel u^{\prime }\parallel }_2.$$

(3.9)

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

$$\begin{array}{rl}{\parallel A{u}^{\prime }\parallel }_2^2+\lambda (\frac{1}{2}{\beta }_L^{\prime }+m){\parallel u\parallel }_2^2\le & \lambda [c_m^{\frac{1}{2}}L({|\gamma |}_0+|\tau |)+L{|\gamma |}_0+c_m^{\frac{1}{2}}{\beta }_M]{\parallel u\parallel }_2{\parallel u^{\prime }\parallel }_2\\ +\lambda B(1+c_m^{\frac{1}{2}}){\parallel u\parallel }_2.\end{array}$$

(3.10)

Thus, from (3.10)

$$\begin{array}{rl}(\frac{1}{2}{\beta }_L^{\prime }+m){\parallel u\parallel }_2^2\le & [c_m^{\frac{1}{2}}L({|\gamma |}_0+|\tau |)+L{|\gamma |}_0+c_m^{\frac{1}{2}}{\beta }_M]{\parallel u\parallel }_2{\parallel u^{\prime }\parallel }_2\\ +B(1+c_m^{\frac{1}{2}}){\parallel u\parallel }_2.\end{array}$$

(3.11)

By using Lemma 2.4, we have ${\parallel u^{\prime }\parallel }_2={\parallel A^{-1}A{u}^{\prime }\parallel }_2\le {\alpha }^{\frac{1}{2}}{\parallel A{u}^{\prime }\parallel }_2$, and from (3.10)-(3.11)

$$\begin{array}{rl}{\parallel A{u}^{\prime }\parallel }_2^2\le & \frac{\alpha {[c_m^{\frac{1}{2}}L({|\gamma |}_0+|\tau |)+L{|\gamma |}_0+c_m^{\frac{1}{2}}{\beta }_M]}^2}{(\frac{1}{2}{\beta }_L^{\prime }+m)}{\parallel A{u}^{\prime }\parallel }_2^2\\ +\frac{2{\alpha }^{1/2}B(1+c_m^{\frac{1}{2}}[c_m^{\frac{1}{2}}L({|\gamma |}_0+|\tau |)+L{|\gamma |}_0+c_m^{\frac{1}{2}}{\beta }_M]}{(\frac{1}{2}{\beta }_L^{\prime }+m)}{\parallel A{u}^{\prime }\parallel }_2\\ +\frac{B^2{(1+c_m^{\frac{1}{2}})}^2}{(\frac{1}{2}{\beta }_L^{\prime }+m)}.\end{array}$$

(3.12)

Since

$$\frac{\alpha {[c_m^{\frac{1}{2}}L({|\gamma |}_0+|\tau |)+L{|\gamma |}_0+c_m^{\frac{1}{2}}{\beta }_M]}^2}{(\frac{1}{2}{\beta }_L^{\prime }+m)}<1,$$

there is a constant $M>0$ such that

$${\parallel A{u}^{\prime }\parallel }_2\le M,$$

(3.13)

$${\parallel u^{\prime }\parallel }_2\le {\alpha }^{\frac{1}{2}}{\parallel A{u}^{\prime }\parallel }_2\le {\alpha }^{\frac{1}{2}}M:=A_1,$$

(3.14)

and by (3.11)

$${\parallel u\parallel }_2\le \frac{[c_m^{\frac{1}{2}}L({|\gamma |}_0+|\tau |)+L{|\gamma |}_0+c_m^{\frac{1}{2}}{\beta }_M]A_1+B(1+c_m^{\frac{1}{2}})}{(\frac{1}{2}{\beta }_L^{\prime }+m)}:=A_0.$$

(3.15)

Obviously, $A_0$ and $A_1$ are constants independent of k and λ. Thus by using Lemma 2.2, for all $t\in [-kT,kT]$, we get

$$\begin{array}{rl}|u(t)|& \le {(2T)}^{-\frac{1}{2}}{\left({\int }_{t-T}^{t+T}{|u(s)|}^{2}ds\right)}^{\frac{1}{2}}+T{(2T)}^{-\frac{1}{2}}{\left({\int }_{t-T}^{t+T}{|u^{\prime }(s)|}^{2}ds\right)}^{\frac{1}{2}}\\ \le {(2T)}^{-\frac{1}{2}}{\left({\int }_{t-kT}^{t+kT}{|u(s)|}^{2}ds\right)}^{\frac{1}{2}}+T{(2T)}^{-\frac{1}{2}}{\left({\int }_{t-kT}^{t+kT}{|u^{\prime }(s)|}^{2}ds\right)}^{\frac{1}{2}}\\ ={(2T)}^{-\frac{1}{2}}{\left({\int }_{-kT}^{kT}{|u(s)|}^{2}ds\right)}^{\frac{1}{2}}+T{(2T)}^{-\frac{1}{2}}{\left({\int }_{-kT}^{kT}{|u^{\prime }(s)|}^{2}ds\right)}^{\frac{1}{2}}.\end{array}$$

From (3.14) and (3.15), we obtain

$${|u|}_0\le {(2T)}^{-\frac{1}{2}}{\parallel u\parallel }_2+T{(2T)}^{-\frac{1}{2}}{\parallel u^{\prime }\parallel }_2\le {(2T)}^{-\frac{1}{2}}{A}_0+T{(2T)}^{-\frac{1}{2}}{A}_1:={\rho }_0,$$

(3.16)

where ${\rho }_0$ is a constant independent of k and λ.

For $i=-k,-k+1,\dots ,k-1$, from the continuity of $[A{u}^{\prime }](t)$, one can find that there is a $t_i\in [iT,(i+1)T]$ such that

$$|\left[A{u}^{\prime }\right](t_i)|=|\frac{1}{T}{\int }_{iT}^{(i+1)T}\left[A{u}^{\prime }\right](s)ds|=\left|\frac{[Au]((i+1)T)-[Au](iT)}{T}\right|\le \frac{2}{T}(1+c_m^{\frac{1}{2}}){\rho }_0,$$

and it follows from (3.14) that for $t\in [iT,(i+1)T]$, $i=-k,-k+1,\dots ,k-1$,

$$\begin{array}{rl}|\left[A{u}^{\prime }\right](t)|=& |{\int }_{t_i}^t{[Au]}^{″}(s)ds+\left[A{u}^{\prime }\right](t_i)|\\ \le & {\int }_{t_i}^t|{[Au]}^{″}(s)|ds+\frac{2}{T}(1+c_m^{\frac{1}{2}}){\rho }_0\\ \le & {\int }_{iT}^{(i+1)T}|{[Au]}^{″}(s)|ds+\frac{2}{T}(1+c_m^{\frac{1}{2}}){\rho }_0\\ \le & {\int }_{iT}^{(i+1)T}|\beta (s)u^{\prime }(s)|ds+{\int }_{iT}^{(i+1)T}\left|g\left(u(s-\gamma (s))\right)\right|ds\\ +{\int }_{iT}^{(i+1)T}|p_k(s)|ds+\frac{2}{T}(1+c_m^{\frac{1}{2}}){\rho }_0\\ \le & {\beta }_M{T}^{\frac{1}{2}}{\left({\int }_{-kT}^{kT}{|u^{\prime }(s)|}^{2}ds\right)}^{\frac{1}{2}}+T{g}_M+TB+\frac{2}{T}(1+c_m^{\frac{1}{2}}){\rho }_0\\ \le & {\beta }_M{T}^{\frac{1}{2}}{A}_1+T{g}_M+TB+\frac{2}{T}(1+c_m^{\frac{1}{2}}){\rho }_0:=\rho ,\end{array}$$

i.e.,

$${\left|A{u}^{\prime }\right|}_0\le \rho ,$$

(3.17)

where $g_M={max}_{{|u|}_0\le {\rho }_0}|g(u(t-\tau (t)))|$.

By Lemma 2.4 and (3.17), we get

$${\left|u^{\prime }\right|}_0={\left|A^{-1}A{u}^{\prime }\right|}_0\le \left(\sum _{i=1}^n\frac{1}{|1-|c_i||}\right){\left|A{u}^{\prime }\right|}_0\le \left(\sum _{i=1}^n\frac{1}{|1-|c_i||}\right)\rho :={\rho }_1.$$

Clearly, ${\rho }_1$ 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 $k\in N$, Eq. (3.2) has at least one $2kT$-periodic solution $u_k(t)$ such that

$${\parallel u_k\parallel }_2\le A_0,{\parallel u_k^{\prime }\parallel }_2\le A_1,{|u_k|}_0\le {\rho }_0,{\left|u_k^{\prime }\right|}_0\le {\rho }_1,$$

where $A_0$, $A_1$, ${\rho }_0$, and ${\rho }_1$ are constants defined by Theorem 3.1.

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

$${(u(t)-Cu(t-\tau ))}^{″}+\lambda \beta (t)u^{\prime }(t)+\lambda g\left(u(t-\gamma (t))\right)=\lambda p_k(t),\lambda \in (0,1).$$

(3.18)

Let ${\mathrm{\Omega }}_1\subset C_{2kT}^1$ represent the set of all the $2kT$-periodic of system (3.18), since $(0,1)\subset (0,1]$, then ${\mathrm{\Omega }}_1\subset \mathrm{\Sigma }$, where Σ is defined by Theorem 3.1. If $u\in {\mathrm{\Omega }}_1$, by using Theorem 3.1, we have

$${|u|}_0\le {\rho }_0,{\left|u^{\prime }\right|}_0\le {\rho }_1.$$

Let ${\mathrm{\Omega }}_2=\{x:x\in KerL,QNx=0\}$, where

$L:D(L)\subset C_{2kT}\to C_{2kT}$, $Lu={(Au)}^{″}$,

$N:C_{2kT}\to C_{2kT}^1$, $Nu=-\beta (t)u^{\prime }(t)-g(u(t-\gamma (t)))+p_k(t)$,

$Q:C_{2kT}\to C_{2kT}/ImL$, $Qy=\frac{1}{2kT}{\int }_{-kT}^{kT}y(s)ds$.

If $x\in {\mathrm{\Omega }}_2$, then $x=a\in R^n$ (constant vector) and by [H₁], we see that

$$2kTm{|a|}^2\le {\int }_{-kT}^{kT}\left|〈(E-C)a,p_k(t)〉\right|dt\le B|a|(1+c_m){(2kT)}^{\frac{1}{2}},$$

i.e.,

$$|a|\le m^{-1}B{T}^{\frac{-1}{2}}(1+c_m):=B_0.$$

Now, if we set $\mathrm{\Omega }=\{x:x\in C_{2kT}^1,{|x|}_0<{\rho }_0+B_0,{|x^{\prime }|}_0<{\rho }_1+1\}$, then $\mathrm{\Omega }\supset {\mathrm{\Omega }}_1\cup {\mathrm{\Omega }}_2$. 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

$$H(x,\mu ):(\mathrm{\Omega }\cap R^n)\times [0,1]⟶R^n:H(x,\mu )=-\mu x+(1-\mu )\mathrm{\Delta }(x),$$

where $\mathrm{\Delta }(x)=\frac{1}{2kT}{\int }_{-kT}^{kT}[g(x)-p_k(t)]dt$ is determined by Lemma 2.1. From assumption [H₁], we have

$$H(x,\mu )\ne 0,\mathrm{\forall }(x,\mu )\in \left[\partial (\mathrm{\Omega }\cap R^n)\right]\times [0,1].$$

Hence

$$\begin{array}{rl}deg\{JQN,\mathrm{\Omega }\cap KerL,0\}& =deg\{H(x,0),\mathrm{\Omega }\cap KerL,0\}\\ =deg\{H(x,1),\mathrm{\Omega }\cap KerL,0\}\\ \ne 0.\end{array}$$

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 $u_k\in \overline{\mathrm{\Omega }.}$ Evidently, $u_k(t)$ is a $2kT$-periodic solution to Eq. (3.1) for the case of $\lambda =1$, so $u_k\in \mathrm{\Sigma }$. Thus, by using Theorem 3.1, we get

$${\parallel u_k\parallel }_2\le A_0,{\parallel u_k^{\prime }\parallel }_2\le A_1,{|u_k|}_0\le {\rho }_0,{\left|u_k^{\prime }\right|}_0\le {\rho }_1.$$

(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 $k\in N$, there exists a $2kT$-periodic solution $u_k(t)$ to Eq. (1.2). So for every $k\in N$, $u_k(t)$ satisfies

$${(u_k(t)-C{u}_k(t-\tau ))}^{″}+\beta (t)u_k^{\prime }(t)+g\left(u_k(t-\gamma (t))\right)=p_k(t).$$

(3.20)

Let $y_k=(A{u}_k^{\prime })$ for $k>k_0$. By (3.17),

$${|y_k|}_0\le \rho$$

and, by (3.20),

$${\left|y_k^{\prime }\right|}_0\le {\beta }_M{\left|u_k^{\prime }\right|}_0+g_M+\underset{t\in R}{sup}|p(t)|:={\rho }_2.$$

Obviously, ${\rho }_2$ is a constant independent of k. Similar to the proof of Lemma 2.4 in [5], we see that there exists a $u_0\in C^1(R,R^n)$ such that for each interval $[c,d]\subset R$, there is a subsequence $\{u_{k_j}\}$ of $\{u_k\}$ with R, $u_{k_j}(t)\to u_0(t)$ and $u_{k_j}^{\prime }(t)\to u_0^{\prime }(t)$ uniformly on $[c,d]$.

For all $a,b\in R$ with $a<b$, there must be a positive integer $j_0$ such that for $j>j_0$, $[-k_{j}T,k_{j}T-{\epsilon }_0]\supset [a-{|\gamma |}_0,b+{|\gamma |}_0]$. So for $t\in [a-{|\gamma |}_0,b+{|\gamma |}_0]$, from (1.5) and (3.20) we see that

$${(u_{k_j}(t)-C{u}_{k_j}(t-\tau ))}^{″}=-\beta (t)u_{k_j}^{\prime }(t)-g\left(u_{k_j}(t-\gamma (t))\right)+p(t).$$

(3.21)

By (3.21),

$$\begin{array}{rl}{y}_k^{\prime }& ={\left(A{u}_{k_j}^{\prime }\right)}^{\prime }\\ =-\beta (t)u_{k_j}^{\prime }(t)-g\left(u_{k_j}(t-\gamma (t))\right)+p(t)\\ \to -\beta (t)u_0^{\prime }(t)-g\left(u_0(t-\gamma (t))\right)+p(t)\\ :=\chi (t),\end{array}$$

uniformly on $[a,b]$.

By the fact that $y_{k_j}^{\prime }(t)$ is a continuous differential on $(a,b)$, for $j>j_0$, $y_{k_j}^{\prime }(t)\to \chi (t)$ uniformly $[a,b]$. We have $\chi (t)={(u_0(t)-C{u}_0(t-\tau ))}^{″}$, $t\in R$, in view of $a,b\in R$ being arbitrary, that is, $u_0(t)$ is a solution to system (1.1).

Now, we will prove $u_0(t)\to 0$ and $u_0^{\prime }(t)\to 0$ for $|t|\to +\mathrm{\infty }$. We have

$$\begin{array}{rl}{\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}({|u_0(t)|}^2+{|u_0^{\prime }(t)|}^2)dt& =\underset{i\to +\mathrm{\infty }}{lim}{\int }_{-iT}^{iT}({|u_0(t)|}^2+{|u_0^{\prime }(t)|}^2)dt\\ =\underset{i\to +\mathrm{\infty }}{lim}\underset{j\to +\mathrm{\infty }}{lim}{\int }_{-iT}^{iT}({|u_{k_j}(t)|}^2+{|u_{k_j}^{\prime }(t)|}^2)dt.\end{array}$$

Clearly, for every $i\in N$ if $k_j>i$, by (3.14) and (3.15), we get

$${\int }_{-iT}^{iT}({|u_{k_j}(t)|}^2+{|u_{k_j}^{\prime }(t)|}^2)dt\le {\int }_{-k_{j}T}^{k_{j}T}({|u_{k_j}(t)|}^2+{|u_{k_j}^{\prime }(t)|}^2)dt\le A_0^2+A_1^2.$$

Let $i\to +\mathrm{\infty }$ and $j\to +\mathrm{\infty }$; we have

$${\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}({|u_0(t)|}^2+{|u_0^{\prime }(t)|}^2)dt\le A_0^2+A_1^2,$$

(3.22)

and then

$${\int }_{|t|\ge r}({|u_0(t)|}^2+{|u_0^{\prime }(t)|}^2)dt\to 0,$$

(3.23)

as $r\to +\mathrm{\infty }$.

From (3.13), in a similar way we get

$${\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}{|u_0^{\prime }(t)-C{u}_0^{\prime }(t-\tau )|}^{2}dt\le M^2.$$

(3.24)

So, by using Lemma 2.3,

$$\begin{array}{rl}|u_0(t)|& \le {(2T)}^{-\frac{1}{2}}{\left({\int }_{t-T}^{t+T}{|u_0(s)|}^{2}ds\right)}^{\frac{1}{2}}+T{(2T)}^{-\frac{1}{2}}{\left({\int }_{t-T}^{t+T}{|u_0^{\prime }(s)|}^{2}ds\right)}^{\frac{1}{2}}\\ \le max\{{(2T)}^{-\frac{1}{2}},T{(2T)}^{-\frac{1}{2}}\}{\int }_{t-T}^{t+T}({|u_0(t)|}^2+{|u_0^{\prime }(t)|}^2)dt\to 0,|t|\to +\mathrm{\infty }.\end{array}$$

Finally, in order to obtain

$$|u_0^{\prime }(t)|\to 0,|t|\to +\mathrm{\infty },$$

we show that

$$|{\left[\tilde{A}{u}^{\prime }\right]}_0(t)|:=|u_0^{\prime }(t)-C{u}_0^{\prime }(t-\tau )|\to 0,|t|\to +\mathrm{\infty }.$$

(3.25)

From (3.16), we have ${|u|}_0\le {\rho }_0$ and by (1.1), we get

$$\begin{array}{rl}\left|{(\left[\tilde{A}{u}_0^{\prime }\right](t))}^{\prime }\right|& \le |\beta (t)u_0(t)|+\left|g\left(u_0(t-\gamma (t))\right)\right|+\underset{t\in R}{sup}|p(t)|\\ \le {\beta }_M{\rho }_0+\underset{|u|\le {\rho }_0}{sup}|g(u)|+\underset{t\in R}{sup}|p(t)|:=\tilde{M},\text{for }t\in R.\end{array}$$

If (3.25) does not hold, then there exist ${\epsilon }_0\in (0,\frac{1}{2})$ and a sequence $\{t_k\}$ such that

$$|t_1|<|t_2|<|t_3|<\cdots <|t_k|+1<|t_{k+1}|,k=1,2,\dots ,$$

and

$$|\left[\tilde{A}{u}_0^{\prime }\right](t_k)|\ge 2{\epsilon }_0,k=1,2,\dots .$$

From this, we have, for $t\in [t_k,t_k+{\epsilon }_0/(1+\tilde{M})]$,

$$|\left[\tilde{A}{u}_0^{\prime }\right](t)|=|\left[\tilde{A}{u}_0^{\prime }\right](t_k)+{\int }_{t_k}^t{(\left[\tilde{A}{u}_0^{\prime }\right](s))}^{\prime }ds|\ge |\left[\tilde{A}{u}_0^{\prime }\right](t_k)|-{\int }_{t_k}^t\left|{(\left[\tilde{A}{u}_0^{\prime }\right](s))}^{\prime }\right|ds\ge {\epsilon }_0.$$

It follows that

$${\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}{|\left[\tilde{A}{u}_0^{\prime }\right](t_k)|}^{2}dt\ge \sum _{k=1}^{\mathrm{\infty }}{\int }_{t_k}^{t_k+{\epsilon }_0/(1+\tilde{M})}{|\left[\tilde{A}{u}_0^{\prime }\right](t_k)|}^{2}dt=\mathrm{\infty },$$

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 $TC{T}^{\mathrm{\top }}=E_c=diag(c_1,c_2,\dots ,c_n)$.

Let $y_{k_j}^{\prime }(t)=T{u}_{k_j}^{\prime }(t)=(y_{k_j}^{\prime (1)}(t),y_{k_j}^{\prime (2)}(t),\dots ,y_{k_j}^{\prime (n)}(t))=T{(u_{k_j}^{\prime (1)}(t),u_{k_j}^{\prime (2)}(t),\dots ,u_{k_j}^{\prime (n)}(t))}^{\mathrm{\top }}$, then we get $y_0^{\prime }(t)=(y_0^{\prime (1)}(t),y_0^{\prime (2)}(t),\dots ,y_0^{\prime (n)}(t))=T{u}_0^{\prime }(t)=T{(u_0^{\prime (1)}(t),u_0^{\prime (2)}(t),\dots ,u_0^{\prime (n)}(t))}^{\mathrm{\top }}$ as $j\to \mathrm{\infty }$. By (3.25), we have

$$|y_0^{\prime }(t)-E_c{y}_0^{\prime }(t-\tau )|\to 0,|t|\to +\mathrm{\infty }.$$

(3.26)

By using (3.19), we see that $|A{u}_k^{\prime }|<(1+c_m^{\frac{1}{2}}){\rho }_1:=\tilde{B}$, which implies

$$\left|TA{u}_k^{\prime }\right|={\left|〈TA{u}_k^{\prime },TA{u}_k^{\prime }〉\right|}^{\frac{1}{2}}<\tilde{B},$$

i.e.,

$$|y_k^{\prime }(t)-E_c{y}_k^{\prime }(t-\tau )|<\tilde{B},\mathrm{\forall }t\in R.$$

(3.27)

For all $\epsilon >0$, there exists $N=[{log}_{|c_i|}^{\frac{\epsilon (1-|c_i|)}{2\tilde{B}}}]>0$ such that ${\sum }_{h=N+1}^{\mathrm{\infty }}{|c_i|}^h<\frac{\epsilon }{2\tilde{B}}$ ($|c_i|<1$), for $t>N$. Similarly, by (3.26), we see that there is a constant $G>0$ such that $|y_{0_i}^{\prime }(t)-c_i{y}_{0_i}^{\prime }(t-\tau )|<\frac{\epsilon }{2(N+1)}$, for $t>G$.

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

$$\begin{array}{rl}|y_0^{\prime (i)}(t)|& =\underset{j\to +\mathrm{\infty }}{lim}|\left[A_0^{-1}{A}_0{y}_{k_j}^{\prime (i)}\right](t)|\\ \le |\underset{j\to \mathrm{\infty }}{lim}\sum _{h\ge 0}^N{c}_i^h\left[A_0{y}_{k_j}^{\prime (i)}\right](t-h\tau )+\sum _{h=N+1}^{\mathrm{\infty }}{c}_i^h\left[A_0{y}_{k_j}^{\prime (i)}\right](t-h\tau )|\\ \le |\underset{j\to \mathrm{\infty }}{lim}\sum _{h\ge 0}^N{c}_i^h\left[A_0{y}_{k_j}^{\prime (i)}\right](t-h\tau )|+|\underset{j\to \mathrm{\infty }}{lim}\sum _{h=N+1}^{\mathrm{\infty }}{c}_i^h\left[A_0{y}_{k_j}^{\prime (i)}\right](t-h\tau )|\\ \le \underset{j\to \mathrm{\infty }}{lim}\sum _{h\ge 0}^N{|c_i|}^h|\left[A_0{y}_{k_j}^{\prime (i)}\right](t-h\tau )|+\tilde{B}\sum _{h=N+1}^{\mathrm{\infty }}{|c_i|}^h\\ =\sum _{h\ge 0}^N{|c_i|}^h\left|(y_0^{\prime (i)}(t-h\tau )-c_i{y}_0^{\prime (i)}(t-(h+1)\tau ))\right|+\tilde{B}\sum _{h=N+1}^{\mathrm{\infty }}{|c_i|}^h.\end{array}$$

(3.28)

Now, by (3.27) and (3.28), we conclude that $\mathrm{\forall }\epsilon >0$, there exists $\overline{N}=G+N$ such that for $t>\overline{N}$,

$$\begin{array}{rl}|y_{0_i}^{\prime }(t)|& \le \sum _{h\ge 0}^N{|c_i|}^h\left|(y_0^{\prime (i)}(t-h\tau )-c_i{y}_0^{\prime (i)}(t-(h+1)\tau ))\right|+\left|\tilde{B}\sum _{h=N+1}^{\mathrm{\infty }}{c}_i^h\right|\\ <(N+1)\frac{\epsilon }{2(N+1)}+\tilde{B}\frac{\epsilon }{2\tilde{B}}\\ =\epsilon .\end{array}$$

Thus, we get $|y_0^{\prime (i)}(t)|\to 0$, as $|t|\to +\mathrm{\infty }$.

In the similar way, when $|c_i|>1$, we can proof $|y_0^{\prime (i)}(t)|\to 0$, as $|t|\to +\mathrm{\infty }$.

Therefore, $|y_0^{\prime }(t)|\to 0$, as $|t|\to +\mathrm{\infty }$; i.e.,

$$T{(\underset{|t|\to +\mathrm{\infty }}{lim}{u}_0^{\prime (1)}(t),\underset{|t|\to +\mathrm{\infty }}{lim}{u}_0^{\prime (2)}(t),\dots ,\underset{|t|\to +\mathrm{\infty }}{lim}{u}_0^{\prime (n)}(t))}^{\mathrm{\top }}=O,$$

we know T is an orthogonal matrix, then $u_0^{\prime (i)}(t)\to 0$ as $|t|\to +\mathrm{\infty }$.

Thus, we have

$$|u_0^{\prime }(t)|\to 0,|t|\to +\mathrm{\infty }.$$

Clearly, $u_0(t)\ne 0$; otherwise, $p(t)=0$, which contradicts the assumption [H₂].

As an application, we consider the following equation:

$${(u(t)-Cu(t-0.01))}^{″}+sin(t)x^{\prime }(t)+g\left(u(t-{cos}^{2}t)\right)=p(t),$$

(3.29)

where $C=\left(\begin{array}{cc}26& 3\\ 3& 17\end{array}\right)$, $u(t)={(u_1(t),u_1(t))}^{\mathrm{\top }}$, $g(x)=x={(x_1,x_2)}^{\mathrm{\top }}$ and $p(t)={(p_1(t),p_2(t))}^{\mathrm{\top }}={(\frac{1}{\sqrt{1+t^2}},\frac{2}{\sqrt{1+t^2}})}^{\mathrm{\top }}$. Clearly, ${\lambda }_{1,2}=\frac{43\pm \sqrt{117}}{2}\ne \pm 1$. Also, $〈(E-C)x,g(x)〉=-25x_1^2-6x_1{x}_2-16x_2^2<-10(x_1^2+x_2^2)$ and $g(x)=x$, which implies that assumption [H₁] is satisfied with $L=2$, $m=10$. $p(t)={(\frac{1}{\sqrt{1+t^2}},\frac{2}{\sqrt{1+t^2}})}^{\mathrm{\top }}$ is a bounded function and ${({\int }_R{|p(t)|}^{2}dt)}^{\frac{1}{2}}+{sup}_{t\in R}|p(t)|=\sqrt{5}(1+\frac{\sqrt{2}}{2}\pi )$, which implies that assumption [H₂] holds. Furthermore, we can choose $\alpha =\frac{4}{{(\sqrt{117}-41)}^2}$, $c_m=\frac{43+\sqrt{117}}{2}$, ${|\gamma |}_0=1$, ${\beta }_M=1$ and ${\beta }_L^{\prime }>-20$, then

$$\frac{\frac{1}{{(\sqrt{117}-41)}^2}{[{(\frac{43+\sqrt{117}}{2})}^{\frac{1}{2}}2(1+0.01)+2+{(\frac{43+\sqrt{117}}{2})}^{\frac{1}{2}}]}^2}{-\frac{1}{2}+10}<1.$$

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