The existence of affine-periodic solutions for nonlinear impulsive differential equations

Shuai Wang

doi:10.1186/s13661-018-1033-8

Research
Open Access

The existence of affine-periodic solutions for nonlinear impulsive differential equations

Shuai Wang¹Email author View ORCID ID profile

Boundary Value Problems20182018:113

https://doi.org/10.1186/s13661-018-1033-8

Received: 12 March 2018
Accepted: 6 July 2018
Published: 18 July 2018

Abstract

In this paper, we study the existence of affine-periodic solutions of nonlinear impulsive differential equations. The affine-periodic solutions have the form $x(t+T)=Qx(t)$ with some nonsingular matrix Q. We give a theorem on the existence of the affine-periodic solutions, respectively, depending on wether $\operatorname{det}(I-Q)$ (I= identity matrix) is equal to 0 or not.

Keywords

Nonlinear impulsive differential equations
Affine-periodic solutions
Boundary value problem
Topological degree

1 Introduction

The periodicity is a very important property in the study of the impulsive differential equations [1, 2]. However, not all natural phenomena can be described alone by periodicity. Some differential equations often exhibit certain symmetries rather than periodicity. For example, consider the system

$$ \dot{x}=f(t,x), $$

(1)

where $f:R^{1}\times R^{n}\rightarrow R^{n}$ is continuous, and for some $Q\in GL_{n}(R)$ (general linear group), satisfies the following affine symmetry:

$$ f(t+T,x)=Qf \bigl(t,Q^{-1}x \bigr). $$

We call it a $(Q,T)$-affine-periodic system. For this $(Q,T)$-affine-periodic system, we are concerned with the existence of $(Q,T)$-affine-periodic solutions $x(t)$ with

$$ x(t+T)=Qx(t), \quad \forall t. $$

It should be pointed out that when $Q=I$ (identity matrix) or $Q=-I$, the solutions are just the pure periodic solutions or antiperiodic ones; when $Q\in SO_{n}$ (special orthogonal group), the solutions correspond to the solutions with Q-rotating symmetry, particularly to some special quasi-periodic solutions. So the interest to particular kinds of periodic solutions that we are going to study is not purely theoretical. The antiperiodicity property or some quasi-periodicity property, which is obviously a particular case of affine-periodic solutions, has drawn wide attention from physicists and astronomers [3, 4].

Recently, these conceptions and existence results of the solutions have been introduced and proved by Li and his coauthors; see [5] for Levinson’s problem, [6] for Lyapunov function type theorems, [7] for averaging methods of affine-periodic solutions, and [8] for some dissipative dynamical systems. The aim of this paper is to touch such a topic for affine-periodic solutions of nonlinear impulsive differential equations.

The paper is organized as follows. We first change the affine-periodic solutions problem to the boundary value problem in Sect. 2. In Sect. 3, when $\operatorname{det}(I-Q)\neq0$, we give an unique affine-periodic solution by using the Banach contraction mapping principle. Furthermore, via the topological degree theory, we prove the existence of affine-periodic solutions for nonlinear impulsive system when $\operatorname{det}(I-Q)=0$ in Sect. 4. We give two examples by numerical simulation in Sect. 5.

2 Nonlinear impulsive differential system

In this paper, we investigate the following system:

$$ \begin{aligned} &\dot{x}=f(t,x), \quad t\neq t_{k} , t\in R, \\ &\Delta x=I_{k}(x), \quad t=t _{k},k \in Z. \end{aligned}$$

(2)

The system satisfies the following hypotheses H:

(1)

$f(\cdot)\in C(R\times R^{n},R^{n})$ and $f(t+T,x)= Qf(t,Q^{-1}x)$ for some $G\in SO_{n}(R)$.
(2)

$I_{k}(\cdot)\in C(R^{n},R^{n})$, $t_{k}< t_{k+1}$ ($k \in Z$).
(3)

There exists $q\in N$ such that $I_{k+q}(x)=Q I_{k}(Q^{-1}x)$ and $t_{k+q}=t_{k}+T$ ($k \in Z$).

In system (2), the continuous part corresponds to a nonlinear $(Q,T)$-affine-periodic system. The discrete component models the affine-periodic impulsive change of $x(t)$.

Lemma 2.1

The existence of Q-affine-periodic solutions of equation (2) is equivalent to the existence of the boundary value problem (2) with $x(T)=Qx(0)$.

Proof

Let $x(t)$ be a solution of equation (2) defined on $t\in[0,T]$. Then

$$ u(t)= \textstyle\begin{cases} x(t),& t\in(0,T], \\ Q^{j}x(t-jT),& t\in(jT,jT+T], \end{cases} $$

(3)

is a Q-affine-periodic solution of (2). Indeed, if $t\in(jT,jT+T]$ and $t\neq t_{k}$, then $t-jT \in(0,T]$, and

$$ \begin{aligned}[b] \frac{d u(t)}{dt} &=Q^{j} \frac{dx(t-jT)}{dt} \\ &=Q^{j}f \bigl(t-jT,x(t-jT) \bigr) \\ &=Q^{j}\cdot Q^{-j}f \bigl(t,Q^{j}x(t-jT) \bigr) \\ &=f \bigl(t,u(t) \bigr), \end{aligned} $$

(4)

and if $t_{k}\in(jT,jT+T]$, then $t_{k-jq}=t_{k}-jT\in(0,T]$ and

$$ \begin{aligned}[b] \Delta u(t_{k})&=Q^{j}\Delta x(t_{k}-jT) \\ &=Q^{j}I_{k-jq} \bigl(x(t_{k}-jT) \bigr) \\ &=Q^{j}\cdot Q^{-j}I_{k} \bigl(Q^{j}x(t_{k}-jT) \bigr) \\ & =I_{k} \bigl(u(t_{k}) \bigr). \end{aligned} $$

(5)

Let $x(t)$ be any solution of (2) with $x(T)=Qx(0)$. Then $x(t)$ has the form

$$ x(t)=x(0)+ \int_{0}^{t}f \bigl(s,x(s) \bigr)\,ds+\sum _{0\leq t_{k}< t}I_{k} \bigl(x(t_{k}) \bigr). $$

Denote $x(0)$ by $x_{0}$. Then we have

$$ (I-Q)x_{0}=- \biggl[ \int_{0}^{T}f \bigl(s,x(s) \bigr)\,ds+\sum _{0\leq t_{k}< T}I_{k} \bigl(x(t _{k}) \bigr) \biggr]. $$

(6)

□

3 Noncritial case

$$ \operatorname{det}(I-Q)\neq0. $$

In this case, $(I-Q)^{-1}$ exists. Then

$$ x_{0}=-(I-Q)^{-1} \biggl[ \int_{0}^{T}f \bigl(s,x(s) \bigr)\,ds+\sum _{0\leq t_{k}< T}I_{k} \bigl(x(t _{k}) \bigr) \biggr]. $$

(7)

So, the existence of Q-affine-periodic solutions of equation (2) is equivalent to the existence of solutions of the following impulsive integral equation:

$$\begin{aligned} x(t) =&-(I-Q)^{-1} \biggl[ \int_{0}^{T}f \bigl(s,x(s) \bigr)\,ds+\sum _{0\leq t_{k}< T}I_{k} \bigl(x(t _{k}) \bigr) \biggr] \\ &{}+ \int_{0}^{t}f \bigl(s,x(s) \bigr)\,ds+\sum _{0\leq t_{k}< t}I_{k} \bigl(x(t_{k}) \bigr). \end{aligned}$$

(8)

Let

$$ \textrm{X}= \bigl\{ x:[0,T]\rightarrow R^{n}: x(t) \text{ is continuous on } [0,T] \bigr\} , $$

and define the norm $\|x\|=\sup_{t\in[0,T]}|x(t)|$. It is easy to see that X is a Banach space with norm $\|x\|$. We also define the norm of the matrix $\|X(t)\|=\|(x_{1}(t),x_{2}(t),\ldots,x_{n}(t))\|= \max_{i=1,2,\ldots,n}\|x_{i}\|$. Then we have the following theorem.

Theorem 3.1

Let a function $p\in L([0,T],R^{+})$ and nonnegative constants $\alpha_{k}$ ($k=1,2,\ldots,q$) be such that

$$\begin{aligned}& \bigl\vert f(t,y)-f(t,x) \bigr\vert \leq p(t) \vert y-x \vert , \quad \forall t\in[0,T], x,y\in R^{n}, \\& \bigl\vert I_{k}(y)-I_{k}(x) \bigr\vert \leq a_{k} \vert y-x \vert ,\quad \alpha_{k}\in R(k=1,2, \ldots,q), x,y \in R^{n}, \end{aligned}$$

and

$$ \Biggl( \int_{0}^{T}p(s)\,ds+\sum _{k=1}^{q}a_{k} \Biggr)< \frac{1}{ \Vert (I-Q)^{-1} \Vert +1}. $$

Then system (2) has an unique Q-affine-periodic solution.

Proof

Define

$$\begin{aligned} A \bigl(x(t) \bigr) =&-(I-Q)^{-1} \biggl[ \int_{0}^{T}f \bigl(s,x(s) \bigr)\,ds+\sum _{0\leq t_{k}< T}I_{k} \bigl(x(t _{k}) \bigr) \biggr] \\ &{}+ \int_{0}^{t}f \bigl(s,x(s) \bigr)\,ds+\sum _{0\leq t_{k}< t}I_{k} \bigl(x(t_{k}) \bigr). \end{aligned}$$

Then

$$\begin{aligned}& \bigl\vert A \bigl(y(t) \bigr)-A \bigl(x(t) \bigr) \bigr\vert \\& \quad = \biggl\vert -(I-Q)^{-1} \biggl[ \int_{0}^{T} \bigl(f \bigl(s,y(s) \bigr)-f \bigl(s,x(s) \bigr) \bigr)\,ds+ \sum_{0\leq t_{k}< T}(I_{k} \bigl(y(t_{k})-I_{k} \bigl(x(t_{k}) \bigr) \bigr) \biggr] \\& \qquad {}+ \int_{0} ^{t} \bigl(f \bigl(s,y(s) \bigr)-f \bigl(s,x(s) \bigr) \bigr)\,ds+\sum_{0\leq t_{k}< t} \bigl(I_{k} \bigl(y(t_{k}) \bigr)-I _{k} \bigl(x(t_{k}) \bigr) \bigr) \biggr\vert \\& \quad \leq \bigl\Vert (I-Q)^{-1} \bigr\Vert \Biggl( \int_{0}^{T}p(s)\,ds+\sum _{k=1} ^{q}a_{k} \Biggr) \vert y-x \vert \\& \qquad {}+ \Biggl( \int_{0}^{t}p(s)\,ds+\sum _{k=1}^{q^{,}}a_{k} \Biggr) \vert y-x \vert \\& \quad \leq \bigl( \bigl\Vert (I-Q)^{-1} \bigr\Vert +1 \bigr) \Biggl( \int_{0}^{T}p(s)\,ds+\sum _{k=1}^{q}a_{k} \Biggr) \vert y-x \vert . \end{aligned}$$

(9)

So, if $(\int_{0}^{T}p(s)\,ds+\sum_{k=1}^{q}a_{k})<\frac{1}{\|(I-Q)^{-1} \|+1}$, then by the Banach contraction mapping principle system (2) has a unique Q-affine-periodic solution. □

4 Critial case

$$ \operatorname{det}(I-Q)=0. $$

To investigate the existence of solutions of system (2), the following auxiliary equation is often considered:

$$\begin{aligned}& \begin{aligned} &\dot{x}=\lambda f(t,x), \quad t\neq t_{k} , t\in R, \\ &\Delta x=\lambda I _{k}(x), \quad t=t_{k},k \in Z. \end{aligned} \end{aligned}$$

(10)

Then we give the following existence theorem for (Q,T)-affine-periodic solutions by using the topological degree theory [6, 7, 9–11].

Theorem 4.1

Let $D\subset R^{n}$ be a bounded open set. Assume that the following hypotheses hold for system (10):

(H1)

For each $\lambda\in(0,1]$, every Q-affine-periodic solution $x(t)$ of system (10) satisfies
$$ x(t)\notin\partial D \quad \textit{for all }t; $$
(H2)

the Brouwer degree,
$$ \operatorname{deg}\bigl(g,D\cap \operatorname{Ker}(I-Q),0 \bigr)\neq0 \quad \textit{if } \operatorname{Ker}(I-Q) \neq{0}, $$

where

$$ g(a)=\frac{1}{T} \biggl[ \int_{0}^{T}Pf(s,a)\,ds+\sum _{0\leq t_{k}< T}PI_{k} \bigl(x(t _{k}) \bigr) \biggr], $$

with an orthogonal projection $P:R^{n}\rightarrow \operatorname{Ker}(I-Q)$.

Then system (2) has at least one Q-affine-periodic solution $x_{*}(t)\in D$ for all t.

Proof

Consider the auxiliary equation (10) with the boundary value condition $x(T)=Qx(t)$, where $\lambda\in(0,1]$. Let $x(t)$ be any solution of (10) with $x(T)=Qx(0)$. Then

$$\begin{aligned}& (I-Q)x_{0} \\& \quad =-\lambda \biggl[ \int_{0}^{T}f \bigl(s,x(s) \bigr)\,ds+\sum _{0\leq t_{k}< T}I _{k} \bigl(x(t_{k}) \bigr) \biggr]. \end{aligned}$$

(11)

In this case, $(I-Q)^{-1}$ does not exist. By coordinate transformation, without loss of generality, we can just let

$$ Q= \left ( \begin{matrix} I & 0 \\ 0 & Q_{1} \end{matrix} \right ) , $$

(12)

where $(I-Q_{1})^{-1}$ exists. Here $Q=Q_{1}\oplus I$.

Let $P:R^{n}\rightarrow{ \operatorname{Ker}(I-Q)}$ be the orthogonal projection. Then

$$ \begin{aligned}[b] (I-Q)x_{0}={} &(I-Q) \bigl(x^{0}_{\ker }+x^{0}_{\bot} \bigr) \\ ={}&{-}\lambda \biggl[ \int _{0}^{T}f \bigl(s,x(s) \bigr)\,ds+\sum _{0\leq t_{k}< T}I_{k} \bigl(x(t_{k}) \bigr) \biggr] \\ ={}&{-}\lambda \biggl[ \int_{0}^{T}Pf \bigl(s,x(s) \bigr)\,ds+\sum _{0\leq t_{k}< T}PI_{k} \bigl(x(t_{k}) \bigr) \biggr] \\ &{}- \lambda \biggl[ \int_{0}^{T}(I-P)f \bigl(s,x(s) \bigr)\,ds+\sum _{0\leq t_{k}< T}(I-P)I_{k} \bigl(x(t _{k}) \bigr) \biggr], \end{aligned} $$

(13)

where $x^{0}_{\ker }\in \operatorname{Ker}(I-Q)$, $x^{0}_{\bot}\in \operatorname{Im}(I-Q)$ and $x_{0}=x^{0}_{\ker }+x^{0}_{\bot}$.

Let $L_{p}=(I-Q)|_{\operatorname{Im}(I-Q)}$. It is easy to see that $L^{-1}_{p}$ exists. Thus equation (13) is equivalent to

$$\begin{aligned}& (I-Q)x^{0}_{\ker }=-\lambda \biggl[ \int_{0}^{T}Pf \bigl(s,x(s) \bigr)\,ds+ \sum _{0\leq t_{k}< T}PI_{k} \bigl(x(t_{k}) \bigr) \biggr]=0, \\& (I-Q)x^{0}_{\bot}=-\lambda \biggl[ \int_{0}^{T}(I-P)f \bigl(s,x(s) \bigr)\,ds+ \sum _{0\leq t_{k}< T}(I-P)I_{k} \bigl(x(t_{k}) \bigr) \biggr]. \end{aligned}$$

Thus we have

$$ x^{0}_{\bot}=\lambda L^{-1}_{p} \biggl[ \int_{0}^{T}(I-P)f \bigl(s,x(s) \bigr)\,ds+ \sum _{0\leq t_{k}< T}(I-P)I_{k} \bigl(x(t_{k}) \bigr) \biggr]. $$

For $x\in\textrm{X}$ such that $x(t)\in\overline{D}$ for all $t\in[0,T]$, we define the operator $\mathtt{T}(x^{0}_{\ker },x, \lambda)$ by

$$ \mathtt{T} \bigl(x^{0}_{\ker },x,\lambda \bigr) = \left ( \begin{matrix} {x^{0}_{\ker }+ \frac{1}{T} [ \int_{0}^{T}Pf (s,x(s) )\,ds+\sum_{0\leq t _{k}< T}PI_{k} (x(t_{k}) ) ]} \\ {x^{0}_{\ker }-\lambda L^{-1}_{p} [ \int_{0}^{T}(I-P)f (s,x(s) )\,ds+\sum_{0\leq t_{k}< T}(I-P)I_{k} (x(t_{k}) ) ]} \\ {{}+\lambda [ \int_{0}^{t}f (s,x(s) )\,ds+\sum_{0\leq t_{k}< t}I_{k} (x(t _{k}) ) ]} \end{matrix} \right ), $$

(14)

where $\lambda\in[0,1]$. We claim that each fixed point x of T in X is a solution of (10) with $x(T)=Qx(0)$.

In fact, if x is a fixed point of T, we have

$$ \left ( \begin{matrix} {x^{0}_{\ker}} \\ {x(t)} \end{matrix} \right ) = \left ( \begin{matrix} {x^{0}_{\ker}+\frac{1}{T} [ \int_{0}^{T}Pf (s,x(s) )\,ds+\sum_{0\leq t_{k}< T}PI _{k} (x(t_{k}) ) ]} \\ {x^{0}_{\ker}-\lambda L^{-1}_{p} [ \int_{0}^{T}(I-P)f (s,x(s) )\,ds+ \sum_{0\leq t_{k}< T}(I-P)I_{k} (x(t_{k}) ) ]} \\ {{}+\lambda [ \int_{0}^{t}f (s,x(s) )\,ds+ \sum_{0\leq t_{k}< t}I_{k} (x(t_{k}) ) ]} \end{matrix} \right ) . $$

Thus

$$\begin{aligned}& \frac{1}{T} \biggl[ \int_{0}^{T}Pf \bigl(s,x(s) \bigr)\,ds+\sum _{0\leq t_{k}< T}PI_{k} \bigl(x(t _{k}) \bigr) \biggr]=0, \end{aligned}$$

(15)

$$\begin{aligned}& x(t)= x^{0}_{\ker } -\lambda L^{-1}_{p} \biggl[ \int_{0}^{T}(I-P)f \bigl(s,x(s) \bigr)\,ds+ \sum _{0\leq t_{k}< T}(I-P)I_{k} \bigl(x(t_{k}) \bigr) \biggr] \\& \hphantom{x(t)=}{}+\lambda \biggl[ \int_{0}^{t}f \bigl(s,x(s) \bigr)\,ds+ \sum _{0\leq t_{k}< t}I_{k} \bigl(x(t_{k}) \bigr) \biggr]. \end{aligned}$$

(16)

By equation (16) we know that

$$ x_{0}=x^{0}_{\ker }-\lambda L^{-1}_{p} \biggl[ \int_{0}^{T}(I-P)f \bigl(s,x(s) \bigr)\,ds+ \sum _{0\leq t_{k}< T}(I-P)I_{k} \bigl(x(t_{k}) \bigr) \biggr]. $$

According to $(I-Q)x^{0}_{\ker }=0$, we have

$$ \begin{aligned} Qx_{0} &=Qx^{0}_{\ker }- \lambda QL^{-1}_{p} \biggl[ \int_{0}^{T}(I-P)f \bigl(s,x(s) \bigr)\,ds+ \sum _{0\leq t_{k}< T}(I-P)I_{k} \bigl(x(t_{k}) \bigr) \biggr] \\ &=x^{0}_{\ker }-\lambda QL^{-1}_{p} \biggl[ \int_{0}^{T}(I-P)f \bigl(s,x(s) \bigr)\,ds+\sum _{0\leq t_{k}< T}(I-P)I _{k} \bigl(x(t_{k}) \bigr) \biggr]. \end{aligned} $$

Since equation (15) holds, we have

$$\begin{aligned}& (I-Q)L^{-1}_{p} \biggl[ \int_{0}^{T}(I-P)f \bigl(s,x(s) \bigr)\,ds+\sum _{0\leq t_{k}< T}(I-P)I _{k} \bigl(x(t_{k}) \bigr) \biggr] \\& \quad = \biggl[ \int_{0}^{T}(I-P)f \bigl(s,x(s) \bigr)\,ds+\sum _{0\leq t_{k}< T}(I-P)I _{k} \bigl(x(t_{k}) \bigr) \biggr] \\& \quad = \biggl[ \int_{0}^{T}(I-P)f \bigl(s,x(s) \bigr)\,ds+\sum _{0\leq t_{k}< T}(I-P)I _{k} \bigl(x(t_{k}) \bigr) \biggr] \\& \qquad {} + \biggl[ \int_{0}^{T}Pf \bigl(s,x(s) \bigr)\,ds+\sum _{0\leq t_{k}< T}PI_{k} \bigl(x(t_{k}) \bigr) \biggr] \\& \quad = \int_{0}^{T}f \bigl(s,x(s) \bigr)\,ds+\sum _{0\leq t_{k}< T}I_{k} \bigl(x(t_{k}) \bigr). \end{aligned}$$

Thus

$$\begin{aligned}& \lambda QL^{-1}_{p} \biggl[ \int_{0}^{T}(I-P)f \bigl(s,x(s) \bigr)\,ds+\sum _{0\leq t_{k}< T}(I-P)I _{k} \bigl(x(t_{k}) \bigr) \biggr] \\& \quad =\lambda L^{-1}_{p} \biggl[ \int_{0}^{T}(I-P)f \bigl(s,x(s) \bigr)\,ds+ \sum _{0\leq t_{k}< T}(I-P)I_{k} \bigl(x(t_{k}) \bigr) \biggr] \\& \qquad {}-\lambda \biggl[ \int_{0}^{T}f \bigl(s,x(s) \bigr)\,ds+ \sum _{0\leq t_{k}< T}I_{k} \bigl(x(t_{k}) \bigr) \biggr]. \end{aligned}$$

Then

$$\begin{aligned} Qx_{0} = &x^{0}_{\ker }- \lambda QL^{-1}_{p} \biggl[ \int_{0}^{T}(I-P)f \bigl(s,x(s) \bigr)\,ds+ \sum _{0\leq t_{k}< T}(I-P)I_{k} \bigl(x(t_{k}) \bigr) \biggr] \\ =&x^{0}_{\ker }-\lambda L^{-1}_{p} \biggl[ \int_{0}^{T}(I-P)f \bigl(s,x(s) \bigr)\,ds+\sum _{0\leq t_{k}< T}(I-P)I _{k} \bigl(x(t_{k}) \bigr) \biggr] \\ &{}+\lambda \biggl[ \int_{0}^{T}f \bigl(s,x(s) \bigr)\,ds+\sum _{0\leq t _{k}< T}I_{k} \bigl(x(t_{k}) \bigr) \biggr] =x(T). \end{aligned}$$

(17)

By equations (16) and (17), equation (11) holds. Thus,

$$ x^{0}_{\bot}=-\lambda L^{-1}_{p} \biggl[ \int_{0}^{T}(I-P)f \bigl(s,x(s) \bigr)\,ds+ \sum _{0\leq t_{k}< T}(I-P)I_{k} \bigl(x(t_{k}) \bigr) \biggr]. $$

Then,

$$\begin{aligned} x(t) = &x^{0}_{\ker }-\lambda L^{-1}_{p} \biggl[ \int_{0}^{T}(I-P)f \bigl(s,x(s) \bigr)\,ds+ \sum _{0\leq t_{k}< T}(I-P)I_{k} \bigl(x(t_{k}) \bigr) \biggr] \\ &{}+\lambda \biggl[ \int_{0}^{t}f \bigl(s,x(s) \bigr)\,ds+ \sum _{0\leq t_{k}< t}I \bigl(x(t_{k}) \bigr) \biggr] \\ =&x^{0}_{\ker }+x^{0}_{\bot}+ \lambda \biggl[ \int_{0}^{t}f \bigl(s,x(s) \bigr)\,ds+\sum _{0\leq t_{k}< t}I_{k} \bigl(x(t_{k}) \bigr) \biggr] \\ =&x_{0}+\lambda \biggl[ \int_{0}^{t}f \bigl(s,x(s) \bigr)\,ds+\sum _{0\leq t_{k}< t}I_{k} \bigl(x(t _{k}) \bigr) \biggr]. \end{aligned}$$

This means that the fixed point x is a solution of (10) with $x(T)=Qx(0)$.

Now, we need to prove the existence of the fixed point of T. Take a constant M such that $M> \sup_{t\in[0,T],x\in\overline{D}}|f(t,x)|$, and let

$$ \textrm{X}_{\lambda}= \biggl\{ x\in\textrm{X}: \biggl\vert \frac{x(t)-x(r)}{t-r} \biggr\vert \leq \lambda M \text{ for all } t,r \in(t_{k},t_{k+1}],t\neq r \biggr\} . $$

Then, it is easy to make a retraction $\alpha_{\lambda}:\textrm{X} \rightarrow\textrm{X}_{\lambda}$.

Define an operator $\widehat{\mathtt{T}}(x^{0}_{\ker },x,\lambda)$ by

$$ \begin{aligned} &\widehat{\mathtt{T}} \bigl(x^{0}_{\ker },x, \lambda \bigr) \\ &\quad = \left ( \begin{matrix} {x^{0}_{\ker} +\frac{1}{T} [ \int_{0}^{T}Pf (s,\alpha_{\lambda}\circ x(s) )\,ds+ \sum_{0\leq t_{k}< T}PI_{k} ( \alpha_{\lambda}\circ x(t_{k}) ) ]} \\ {\alpha_{\lambda}\circ x^{0}_{\ker }-\lambda L^{-1}_{p} [ \int_{0} ^{T}(I-P)f (s,\alpha_{\lambda} \circ x(s) )\,ds+\sum_{0\leq t_{k}< T}(I-P)I _{k} (\alpha_{\lambda}\circ x(t_{k}) ) ]} \\ {{}+\lambda [ \int_{0}^{t}f (s, \alpha_{\lambda}\circ x(s) )\,ds+\sum_{0\leq t_{k}< t}I_{k} ( \alpha_{ \lambda}\circ x(t_{k}) ) ]} \end{matrix} \right ). \end{aligned} $$

(18)

Since $P:R^{n}\rightarrow \operatorname{Ker}(I-Q)$, it is easy to see that

$$ \frac{1}{T} \biggl[ \int_{0}^{T}Pf \bigl(s, x(s) \bigr)\,ds+\sum _{0\leq t_{k}< T}PI_{k} \bigl(x(t _{k}) \bigr) \biggr] \in \operatorname{Ker}(I-Q). $$

Also,

$$ \frac{1}{T} \biggl[ \int_{0}^{T}Pf \bigl(s,\alpha_{\lambda}\circ x(s) \bigr)\,ds+ \sum_{0\leq t_{k}< T}PI_{k} \bigl( \alpha_{\lambda}\circ x(t_{k}) \bigr) \biggr]\in \operatorname{Ker}(I-Q). $$

Let us consider the homotopy

$$\begin{aligned}& H \bigl(x^{0}_{\ker },x,\lambda \bigr)= \widehat{ \mathtt{T}} \bigl(x^{0}_{\ker },x, \lambda \bigr), \end{aligned}$$

(19)

$$\begin{aligned}& \bigl(x^{0}_{\ker },x,\lambda \bigr)\in \bigl(D \cap \operatorname{Ker}(I-Q)\times\widetilde{D} \times[0,1] \bigr), \end{aligned}$$

(20)

where $\widetilde{D}=\{x\in X:x(t)\in D \text{ for all } t \in[0,T]\}$.

We claim that

$$ 0\notin(id-H) (\partial \bigl( \bigl(D\cap \operatorname{Ker}(I-Q)\times \widetilde{D} \bigr)\times[0,1] \bigr). $$

(21)

Suppose, on the contrary, that there exists $(\widehat{x}^{0}_{\ker }, \widehat{x},\widehat{\lambda})\in\partial((D\cap \operatorname{Ker}(I-Q)\times \widetilde{D})\times[0,1]$ such that $(id-H)(\widehat{x}^{0}_{\ker }, \widehat{x},\widehat{\lambda})=0$. Since $\widehat{x}^{0}_{\ker } \in\partial D$ is contradictory to ($H_{1}$) and since $\partial(D \cap \operatorname{Ker}(I-Q))\subset\partial D$, we have that $\widehat{x}^{0}_{ \ker }\notin\partial(D\cap \operatorname{Ker}(I-Q))$. In other words, $\widehat{x}\in\partial D$. Then (21) can be proved as follows.

(i) When $\widehat{\lambda}=0$, by the definition of the set $\textrm{X}_{\lambda}$ we have

$$ \textrm{X}_{0}= \biggl\{ x\in X: \biggl\vert \frac{x(t)-x(r)}{t-r} \biggr\vert \leq0 \text{ for all } t,r\in(t_{k},t_{k+1}],t \neq r \biggr\} . $$

Hence $\alpha_{0}\circ x(t)\equiv\alpha_{0}\circ x(t_{k+1})$ for all $t\in(t_{k},t_{k+1}]$. Since $(id-H)(\widehat{x}^{0}_{\ker}, \widehat{x},0)=0$, we have

$$ \left ( \begin{matrix} {\widehat{x}^{0}_{\ker }} \\ {\widehat{x}(t)} \end{matrix} \right ) = \left ( \begin{matrix} { \widehat{x}^{0}_{\ker } +\frac{1}{T} [ \int_{0}^{T}Pf (s,\alpha_{ \lambda}\circ x(s) )\,ds+\sum_{0\leq t_{k}< T}PI_{k} ( \alpha_{\lambda} \circ x(t_{k}) ) ]} \\ {\alpha_{0} \circ\widehat{x}^{0}_{\ker }} \end{matrix} \right ) . $$

(22)

This means that $\widehat{x}(t)\equiv\widehat{x}(0)$ for all $t\in[0,T]$. Taking $\widehat{x}(t)=p$, we have $\alpha_{0}\circ \widehat{x}^{0}_{\ker }=\widehat{x}(t)=p$. Consequently,

$$ \frac{1}{T} \biggl[ \int_{0}^{T}Pf \bigl(s,\alpha_{\lambda}\circ x(s) \bigr)\,ds+ \sum_{0\leq t_{k}< T}PI_{k} \bigl( \alpha_{\lambda}\circ x(t_{k}) \bigr) \biggr]=0, $$

and this is equivalent to $g(p)=0$ by the definition of $g(a)$. Notice that $\widehat{x}\in\partial\widetilde{D}$ and $\widetilde{D}=\{x \in D \text{ for all } t\in[0,T]\}$. Then there exists $t_{0}\in[0,T]$ such that $\widehat{x(t)}_{0}\in\partial D$. Since $\widehat{x}(t)\equiv p$ for all $t\in[0,T]$, we obtain that $p\in\partial D$. Thus, we have $p\in\partial D$ and $g(p)=0$. It is contradictory to ($H_{2}$) because the Brouwer degree $\operatorname{deg}(g,D,0) \neq0$.

(ii) When $\widehat{\lambda}\in(0,1]$, as $0=(id-H)(\widehat{x}^{0} _{\ker },\widehat{x},\widehat{\lambda})$, we have

$$\begin{aligned}& \left ( \begin{matrix} {\widehat{x}^{0}_{\ker }} \\ {\widehat{x}(t)} \end{matrix} \right ) \\& \quad = \left ( \begin{matrix} {\widehat{x}^{0}_{\ker } +\frac{1}{T} [ \int_{0}^{T}Pf (s, \alpha_{\widehat{\lambda}}\circ x(s) )\,ds+\sum_{0\leq t_{k}< T}PI_{k} ( \alpha_{\widehat{\lambda}}\circ x(t_{k}) ) ]} \\ {\alpha_{\widehat{\lambda}}\circ x^{0}_{\ker }-\widehat{\lambda} L ^{-1}_{p} [ \int_{0}^{T}(I-P)f (s,\alpha_{\widehat{\lambda}} \circ x(s) )\,ds+ \sum_{0\leq t_{k}< T}(I-P)I_{k} ( \alpha_{\widehat{\lambda}}\circ x(t _{k}) ) ]} \\ {{}+\widehat{\lambda} [ \int_{0}^{t}f (s,\alpha_{ \widehat{\lambda}}\circ x(s) )\,ds+\sum_{0\leq t_{k}< t}I_{k} ( \alpha_{\widehat{\lambda}}\circ x(t_{k}) ) ]} \end{matrix} \right ) . \end{aligned}$$

Thus

$$ \frac{1}{T} \biggl[ \int_{0}^{T}Pf \bigl(s,\alpha_{\widehat{\lambda}}\circ x(s) \bigr)\,ds+ \sum_{0\leq t_{k}< T}PI_{k} \bigl( \alpha_{\widehat{\lambda}}\circ x(t_{k}) \bigr) \biggr]=0 $$

and

$$ \begin{aligned}[b] \widehat{x}(t)={} & \alpha_{\widehat{\lambda}} \circ x^{0}_{\ker }- \widehat{\lambda} L^{-1}_{p} \biggl[ \int_{0}^{T}(I-P)f \bigl(s, \alpha_{\widehat{\lambda}} \circ x(s) \bigr)\,ds+\sum_{0\leq t_{k}< T}(I-P)I _{k} \bigl(\alpha_{\widehat{\lambda}}\circ x(t_{k}) \bigr) \biggr] \\ &{}+ \widehat{\lambda} \biggl[ \int_{0}^{t}f \bigl(s,\alpha_{\widehat{\lambda}}\circ x(s) \bigr)\,ds+ \sum_{0\leq t_{k}< t}I_{k} \bigl( \alpha_{\widehat{\lambda}}\circ x(t_{k}) \bigr) \biggr]. \end{aligned} $$

(23)

Note that

$$\begin{aligned}& \biggl\vert \frac{x(t)-x(r)}{t-r} \biggr\vert \\& \quad = \frac{1}{ \vert t-r \vert } \biggl\vert \widehat{\lambda} \int _{0}^{t}f \bigl(s,\alpha_{\widehat{\lambda}}\circ \widehat{x}(s) \bigr)\,ds- \widehat{\lambda} \int_{0}^{r}f \bigl(s,\alpha_{\widehat{\lambda}}\circ \widehat{x}(s) \bigr)\,ds \biggr\vert \\& \quad =\frac{1}{ \vert t-r \vert } \biggl\vert \widehat{\lambda} \int _{r}^{t}f \bigl(s,\alpha_{\widehat{\lambda}}\circ \widehat{x}(s) \bigr)\,ds \biggr\vert \\& \quad \leq \lambda M. \end{aligned}$$

By the definition of $\textrm{X}_{\lambda}$ we obtain $\widehat{x} \in\textrm{X}_{\widehat{\lambda}}$, which means that $\alpha_{\widehat{\lambda}}\circ\widehat{x}=\widehat{x}$. Now we can rewrite equation (23) as

$$\begin{aligned} \widehat{x}(t) =& x^{0}_{\ker } - \widehat{ \lambda} L^{-1}_{p} \biggl[ \int _{0}^{T}(I-P)f \bigl(s, x(s) \bigr)\,ds+\sum _{0\leq t_{k}< T}(I-P)I_{k} \bigl(x(t_{k}) \bigr) \biggr] \\ &{}+\widehat{\lambda} \biggl[ \int_{0}^{t}f \bigl(s, x(s) \bigr)\,ds+\sum _{0\leq t_{k}< t}I _{k} \bigl(x(t_{k}) \bigr) \biggr]. \end{aligned}$$

By a similar discussion of equation (16) we can prove that $\widehat{x}(t)$ is a solution of equation (10). By hypothesis ($H_{1}$) we know that $\widehat{x}(t)\notin\partial\widetilde{D}$ for any $t\in[0,T]$. This is a contradiction to $\widehat{x}\in\partial \widetilde{D}$.

By (i) and (ii) we obtain that

$$ 0\notin(id-H) \bigl(\partial \bigl( \bigl(D\cap \operatorname{Ker}(I-Q) \bigr)\times \widetilde{D} \bigr)\times [0,1] \bigr). $$

Therefore, by the homotopy invariance and the theory of Brouwer degree we have

$$ \begin{aligned} &\operatorname{deg}\bigl(id-H \bigl(x^{0}_{\ker}, \cdot,1 \bigr), \bigl(D\cap \operatorname{Ker}(I-Q) \bigr)\times\widetilde{D},0 \bigr) \\ &\quad =\operatorname{deg}\bigl(id-H \bigl(x^{0}_{\ker},\cdot,0 \bigr), \bigl(D \cap \operatorname{Ker}(I-Q) \bigr)\times \widetilde{D},0 \bigr) \\ &\quad =\operatorname{deg}\bigl(g,D\cap \operatorname{Ker}(I-Q),0 \bigr)\neq0. \end{aligned} $$

This means that there exists $\widehat{x}_{*}\in\widetilde{D}$ such that

$$ \left ( \begin{matrix} {\widehat{x}^{0}_{*\ker }} \\ {\widehat{x}_{*}(t)} \end{matrix} \right ) =\widehat{ T} \bigl(\widehat{x}^{0}_{*\ker }, \widehat{x} _{*}(t),1 \bigr). $$

(24)

Similarly to the proof in (ii), we get $\widehat{x}_{*}\in\textrm{X} _{\lambda}$. Then

$$ { } \widehat{T} \bigl(\widehat{x}^{0}_{*\ker }, \widehat{x}_{*}(t),1 \bigr)= T \bigl(\widehat{x}^{0}_{*\ker }, \widehat{x}_{*}(t),1 \bigr). $$

(25)

By equations (24) and (25) we obtain that $\widehat{x}_{*}$ is a fixed point of T in X. Thus, $\widehat{x}_{*}$ is a solution of system (2) with boundary value condition $x(T)=Qx(0)$. □

5 Numerical simulation

Example 1

Consider the system

$$ \begin{aligned} &\dot{x}=- \vert x \vert ^{2}x+(\sin\pi t,\cos\pi t)^{T}, \quad t\neq N, \\ &\Delta x= \biggl(\frac{1}{e}-1 \biggr)x,\quad t=N. \end{aligned} $$

(26)

Set

$$ Q= \left ( \begin{matrix} -1& 0 \\ 0& -1 \end{matrix} \right ) . $$

In this example, $Q=-I$. System (26) has an antiperiodic solution (see Fig. 1).

Figure 1
The antiperiodic solution of system (26). The green line is the trajectory of $x(t)$ for $t\in(0,1]$, the red line corresponds to the trajectory of $Qx(t)$ for $t\in(0,1]$

Example 2

Consider the system

$$ \begin{aligned} &\dot{x}=- \vert x \vert ^{2}x+(\sin t,\cos t,1)^{T},\quad t\neq N, \\ &\Delta x= \biggl(\frac{1}{e}-1 \biggr)x, \quad t=N. \end{aligned} $$

(27)

Set

$$ Q= \left ( \begin{matrix} {\cos(2\pi-1)}& {-\sin(2\pi-1)} &0 \\ {\sin(2\pi-1)}& {\cos(2\pi-1)} &0 \\ 0 &0 &1 \end{matrix} \right ) . $$

Similarly to Example 1, system (26) has a $(Q,1)$-affine-periodic solution., which is a quasi-periodic solution (see Fig. 2).

Figure 2
The $(Q,T)$-affine-periodic solution of system (27). The black line is the trajectory of $x(t)$ for $t\in(0,1]$, the red line corresponds to the trajectory of $Qx(t)$ for $t\in(0,1]$, and the green line corresponds to the trajectory of $Q^{2}x(t)$ for $t\in(0,1]$. It is easy to see that $x(t)$ is a quasi-periodic solution of system (27)

Declarations

Acknowledgements

The author expresses sincere thanks to the anonymous referees for their comments.

Availability of data and materials

Data sharing not applicable to this paper as no datasets were generated or analyzed during the current study.

Funding

The author is supported by the fund of the “Thirteen Five” Scientific and Technological Research Planning Project of the Department of Education of Jilin Province (JJKH20170301KJ).

Author’s contributions

The author declares that the work described was an original research, which has not been published previously and is not under consideration for publication elsewhere. Author read and approved the final manuscript.

Competing interests

The author has no competing interests. There is no conflict of interest exists in the submission of this manuscript.

Open Access This article is distributed under the terms of the Creative Commons Attribution 4.0 International License (http://creativecommons.org/licenses/by/4.0/), which permits unrestricted use, distribution, and reproduction in any medium, provided you give appropriate credit to the original author(s) and the source, provide a link to the Creative Commons license, and indicate if changes were made.

Authors’ Affiliations

(1)

School of Science, Changchun University of Science and Technology, Changchun, China

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