The Controllability for Second-Order Semilinear Impulsive Systems

Abstract

This paper studies the controllability of the initial value problems of linear and semilinear second-order impulsive systems. Necessary and sufficient conditions of controllability for linear problems are obtained, and a new rank criterion is presented. We also show semilinear problems are controllable via Krasnoselskii’s fixed point theorem. Finally, two examples are provided to verify the theoretically results.

1 Introduction

Many evolution processes in science and technology, such as mechanics, population dynamics, pharmacokinetics, industrial robotics, biotechnology, economics and so on, may change their state rapidly, or the duration of the change is negligible. We describe these processes with impulsive effects by impulsive differential equations and the theory of impulsive differential equations is an important branch of differential equation theory; see [1] and the references therein.

Control theory is an important branch in applied mathematics and engineering and modern control theory was developed by Kalman. Roughly speaking, the object of control theory is to find a control function that can steer the state function to the desired result at the end (terminal). Numerous papers are devoted to the controllability of differential equations in Banach space [2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22], such as exact controllability, approximate controllability and null controllability, and the main techniques are based on fixed point theorems [3, 4, 14, 18, 23], variational methods [5, 24], semigroup theory [2, 8], and so on.

Second-order systems capture the dynamic behavior of many natural phenomena and have applications in many fields such as mathematical physics, electrical power systems, quantum mechanics, biology, long transmission lines and finance [25, 26]. Numerous papers focus on the controllability of second-order impulsive systems (see [2, 3, 6, 7, 11, 27] for cosine family theory and [2, 8, 10, 28] where the corresponding operators of the cosine family are compact). However, as noted by Travis and Webb [29], some of these results work only to finite-dimensional spaces. We refer the reader also to [3, 30] for other results on the controllability of second-order impulsive systems.

For the controllability of initial value problems for second-order differential equations

$$\begin{aligned} \left\{ \begin{aligned}&x''(t)=Ax(t)+Bu(t),\quad t\in {[0,b]},\\&x(0)=x_{0},\quad x'(0)=y_{0}, \end{aligned} \right. \end{aligned}$$

(1.1)

many authors consider the controllability of the solution x(t) i.e., one finds a control u which makes the state function x(t) arrive at the value that we wish at the terminal. As mentioned in [7], it is unreasonable to regard the damped term $x'(t)$ in the controllability. Recently, the authors in [7, 10, 11, 27] consider the controllability of x(t) and $x'(t)$.

In [7], Li et al. consider the approximate controllability of system (1.1). Let $J=[0,b]$, the state $x(\cdot )$ takes values in a Banach space X, $u(\cdot )\in L^{2}(J, U)$ is the control function where U is a Banach space, the definition of controllability defined as follows: Systems  (1.1)  are  said  to be  approximately controllable  on  J  if  ${\overline{D}}=X\times X$,  where  $D=\{{(x(b,x_{0},y_{0},u),y(b,x_{0},y_{0},u))}:u\in L^{2}(J,U)\}$,  $y(\cdot ,x_{0},y_{0},u)=x'(\cdot ,x_{0},y_{0},u)$  and $x(\cdot ,x_{0},y_{0},u)$  is a  mild  solution  of  (1.1).

Their aim is to pick a control function u which controls both x(t) and $x'(t)$. In [10, 11, 27], the following two assumptions are used,

(A1) The linear operator $G_{1}:L^{2}(J,U)\rightarrow X$, defined by

$$\begin{aligned} G_{1}u:=\int _{0}^{b}S(b-s)Bu(s)ds, \end{aligned}$$

has an invertible operator $G_{1}^{-1}$ which takes the values in $L^{2}(J,U)/\ker G_{1}$ and there exists positive constant $M_{1}$ such that $\Vert G_{1}^{-1}\Vert \le M_{1}$.

(A2) The linear operator $G_{2}:L^{2}(J,U)\rightarrow X$, defined by

$$\begin{aligned} G_{2}u:=\int _{0}^{b}C(b-s)Bu(s)ds, \end{aligned}$$

has an invertible operator $G_{2}^{-1}$ which takes the values in $L^{2}(J,U)/\ker G_{2}$ and there exists positive constant $M_{2}$ such that $\Vert G_{2}^{-1}\Vert \le M_{2}$.

As pointed by Balachandran and Kim [31] the control function defined in [11, 27] can not steer the value of the state function to what we want at the terminal unless the condition

(H) $G_{1}G_{2}^{-1}=G_{2}G_{1}^{-1}=0$ is satisfied.

For the second-order systems in finite dimensional space, (A1) or (A2) will lead to a contradiction with the definition of controllability. Since if we assume system (1.1) is controllable. Then for any $(x_{1},y_{1})\in X\times X$, there exists a control $u_{1}$ such that $x(b)=x_{1}$, and $x'(b)=y_{1}$ under the control $u_{1}$. For another point $(x_{1},y_{2})$, since $y_{1}\ne y_{2}$, there exists a control $u_{2}$ such that $x(b)=x_{1}$, and $x'(b)=y_{2}$ under the control $u_{2}$ as well. Then if $u_{1}=u_{2}$, we have $y_{1}=y_{2}$, a contradiction; if $u_{1}\ne u_{2}$, since A is the infinitesimal generator of a strongly continuous cosine family C(t) on X, hence, the Cauchy problem (1.1) is well posed. Then from the expression of the solution for (1.1),

$$\begin{aligned} x(t)=C(t)x_{0}+S(t)y_{0}+\int _{0}^{t}S(t-s)Bu(s)ds, \end{aligned}$$

and we get

$$\begin{aligned} x_{1}=C(b)x_{0}+S(b)y_{0}+\int _{0}^{b}S(b-s)Bu_{1}(s)ds, \end{aligned}$$

and

$$\begin{aligned} x_{1}=C(b)x_{0}+S(b)y_{0}+\int _{0}^{b}S(b-s)Bu_{2}(s)ds. \end{aligned}$$

Combining these two equalities with conditions (A1), we find

$$\begin{aligned} u_{1}=G_{1}^{-1}(x_{1}-C(b)x_{0}-S(b)y_{0})=u_{2}, \end{aligned}$$

a contradiction to the assumption $u_{1}\ne u_{2}$. Hence, if assumptions (A1) or (A2) hold, we cannot obtain the controllability result of system (1.1) under the definition of controllability defined in [7]. In view of this, we introduce a weaker definition of controllability in Sect. 2.

To the best of our knowledge, there are only a few articles on the controllability of second-order linear systems, and we note that, for finite-dimensional linear systems, all the concepts of controllability are equivalent (exact controllability, approximate controllability and null controllability). In this paper, we consider the controllability of the following initial value problems for second-order impulsive differential equations

$$\begin{aligned} \left\{ \begin{aligned}&x''(t)=Ax(t)+Bu(t),\quad t\in J=[0,b],\quad t\ne t_{i},\\&\Delta x(t_{i})=B_{1}x(t_{i}^{-}),\quad i=1,2,\ldots ,m,\\&\Delta x'(t_{i})=B_{2}x'(t_{i}^{-}),\quad i=1,2,\ldots ,m,\\&x(0)=x_{0},\quad x'(0)=y_{0}, \end{aligned} \right. \end{aligned}$$

(1.2)

and semilinear second-order impulsive differential equations

$$\begin{aligned} \left\{ \begin{aligned}&x''(t)=Ax(t)+Bu(t)+f(t,x(t)),\quad t\in J'=J{\setminus }\{t_{i}\},\quad i=1,2,\ldots ,m,\\&x(t^{+}_{i})=x(t_{i}^{-})+B_{1}x(t_{i}^{-}),\quad i=1,2,\ldots ,m,\\&x'(t_{i}^{+})=x'(t_{i}^{-})+B_{2}x'(t_{i}^{-}),\quad i=1,2,\ldots ,m,\\&x(0)=x_{0},\quad x'(0)=y_{0}, \end{aligned} \right. \end{aligned}$$

(1.3)

where A, $B_{1}$ and $B_{2}$ are constant $n\times n$ matrices satisfying $AB_{1}=B_{1}A$, $AB_{2}=B_{2}A$, $B_{1}B_{2}=B_{2}B_{1}$, $0=t_{0}<t_{1}<\cdots<t_{k}<t_{k+1}=b$ are impulsive points, $u\in L^{2}(J,{\mathbb {R}}^{n})$ is a control function, and $f\in C(J\times {\mathbb {R}}^{n};{\mathbb {R}}^{n})$.

The contributions of this paper are as follows:

• (1) We introduce a weaker definition of controllability with respect to the state function x(t) and the damped term $x'(t)$.

• (2) We present a new algebraic method to obtain a rank criterion, and a rank criterion of controllability for second-order impulsive linear systems is given.

• (3) Based on the controllability of the linear systems, we give a sufficient condition to guarantee the controllability of the semilinear second-order impulsive systems.

The paper is structured in the following way. In Sect. 2, we give a weaker definition of controllability and some associated notations and essential lemmas. In Sect. 3, instead of converting a second-order system into a first order system, we obtain a new rank criterion of controllability of system (1.2) by direct analysis of the second-order system itself. In Sect. 4, we give a sufficient condition of the controllability of the system (1.3). Finally, in Sect. 5, some examples are provided to illustrate the suitability of our results.

2 Preliminaries

In this section, we modify the definition of controllability and list some notations and properties needed to establish our main results.

Let$PC(J,{\mathbb {R}}^{n})$ denote the Banach space of piecewise continuous functions on the interval J, that is $PC(J,{\mathbb {R}}^{n})=\{v:J\rightarrow {\mathbb {R}}^{n}|u\in C((t_{k-1},t_{k}],{\mathbb {R}}^{n})$ for $k\in \{1, \ldots ,m+1\}$ and there exists $v(t_{k}^{-})$ and $v(t_{k}^{+}), k\in \{1, \ldots ,m\}$ with $v(t_{k})=v(t_{k}^{-}) \}$ equipped with the Chebyshev PC-norm $||v||_{PC}:=\sup \{||v(t)||:t\in J\}$. Let $PC^{1}(J,{\mathbb {R}}^{n}):=\{x\in PC(J,{\mathbb {R}}^{n}):x'\in PC(J,{\mathbb {R}})\}$ equipped with the norm $\Vert x\Vert _{PC^{1}}=\max \{\Vert x\Vert _{PC},\Vert x'\Vert _{PC}\}$. Obviously, $PC(I,{\mathbb {R}}^{n})$ endowed with the norm $\Vert \cdot \Vert _{PC^{1}}$ is also a Banach space. We use the notation

$$\begin{aligned} A_{1}=I+\frac{B_{1}+B_{2}}{2},\quad A_{2}=\frac{B_{1}-B_{2}}{2}. \end{aligned}$$

Let $m=i(t,0)$ denote the number of impulsive points on (0, t), and assume $AB=BA$.

Definition 2.1

The system (1.2) is said to be exact controllability in ${\mathbb {R}}^{n}$, if for each pair $(x_{0},y_{0})\in {\mathbb {R}}^{n}\times {\mathbb {R}}^{n}$, there exists a pair of control functions $(u_{1}(\cdot ),u_{2}(\cdot ))\in L^{2}([0,b],{\mathbb {R}}^{n})\times L^{2}([0,b],{\mathbb {R}}^{n})$ such that for any $(x_{1},y_{1})\in {\mathbb {R}}^{n}\times {\mathbb {R}}^{n}$,

$$\begin{aligned} x(b)=x_{1},\quad y'(b)=y_{1}, \end{aligned}$$

here $x(\cdot )$ is the solution of (1.2) under the control $u_{1}$, $y(\cdot )$ is the solution of (1.2) under the control $u_{2}$, and $y'(t)=dy(t)/dt$.

Remark 2.2

In [4, 5, 9, 14, 18], the definition of controllability imply that one find a control function which steer the state function $x(\cdot )$ to the target value, and in [7, 10, 11, 27], which imply that one find a control function which steer both the state function $x(\cdot )$ and damped term $x'(\cdot )$ to the value we wanted. However, Definition 2.1 indicates that one pick a pair of control functions $(u_{0},u_{1})$ such that $u_{0}$ control the state function x(t) and $u_{1}$ control the damped term $y'(t)$. Notice that at this moment except for a constant difference, the antiderivative of damped term $y'(t)$ may be different with the state function x(t).

The following Lemmas is crucial to our proof of main results.

Lemma 2.3

(see [32]) Let $|\cdot |$ be a norm on ${\mathbb {R}}^{n}$ and B be an $n\times n$ matrix. Then for any $\varepsilon >0$ there exist $T_{B,\varepsilon }\ge 1$ such that $||B^{k}||\le T_{B,\varepsilon }(\rho (B)+\varepsilon )^{k}$, where $\rho (B)$ is the spectral radius of B.

Lemma 2.4

(Krasnoselskii’s fixed point theorem) Let B be a bounded closed and convex subset of a Banach space X and let $F_{1},F_{2}$ be maps of B into X such that $F_{1}x+F_{2}y\in B$ for every $x,y\in B$. If $F_{1}$ is a contraction and $F_{2}$ is compact and continuous, then the equation $F_{1}x+F_{2}x=x$ has a solution on B.

Lemma 2.5

(PC-type Ascoli–Arzela theorem, see [33]) Let $Q\subset PC(\Omega ,X)$ where X is a Banach space. Then Q is a relatively compact subset of $PC(\Omega ,X)$ if, (a) Q is uniformly bounded subset of $PC(\Omega ,X)$; (b) Q is equicontinuous in $(t_{i},t_{i+1})$, $i=0,1,\ldots ,k$; and (c) $Q(t)=\{v(t)|v\in Q,t\in \Omega \backslash \{t_{i}\},i=0,1,\ldots ,k\}$, $Q(t_{i}^{+})=\{v(t_{i}^{+})|v\in Q\}$ and $Q(t_{i}^{-})=\{v(t_{i}^{-})|v\in Q\}$ are relatively compact subsets of X.

Lemma 2.6

(see [34]) For $t\in (t_{m},t_{m+1}]$, $m=0,1,\ldots ,k$, the solution of (1.2) is given by

$$\begin{aligned} \begin{aligned} x(t)=&\,W(A,t,x_{0},y_{0})+\sum _{i=0}^{m-1}\int _{t_{i}}^{t_{i+1}}W_{i}(A,t,s)Bu(s)ds\\&+A^{-\frac{1}{2}}\int _{t_{m}}^{t}\sinh A^{\frac{1}{2}}(t-s)Bu(s)ds, \end{aligned} \end{aligned}$$

where $W(A,t,x_{0},y_{0})$ is the solution of the homogeneous initial value problem of (1.2), and

$$\begin{aligned} \begin{aligned}&W_{i}(A,t,s)\\&\quad =A_{1}^{m-i}A^{-\frac{1}{2}}\sinh A^{\frac{1}{2}}(t-s)-A_{1}^{m-i-1}A_{2}A^{-\frac{1}{2}}\sum \limits _{i+1\le i_{11}\le m}\sinh A^{\frac{1}{2}}(t-2t_{i_{11}}+s)\\&\qquad +A_{1}^{m-i-2}A_{2}^{2}A^{-\frac{1}{2}}\sum \limits _{i+1\le i_{21}<i_{22}\le m}\sinh A^{\frac{1}{2}}(t-2t_{i_{22}}+2t_{i_{21}}-s)\\&\qquad +\cdots +(-1)^{m-i-1}A_{1}A_{2}^{m-i-1}A^{-\frac{1}{2}}{\cdot }\\&\qquad \sum \limits _{i+1\le i_{m-i-1,1}< i_{m-i-1,2}<\cdots <i_{m-i-1,m-i-1}\le m}\sinh A^{\frac{1}{2}}(t-2t_{i_{m-i-1,m-i-1}}\\&\qquad 2t_{i_{m-i-1,m-i-2}}-\cdots \pm 2t_{i_{m-i-1,1}}\mp s)+(-1)^{m-i}A_{2}^{m-i}A^{-\frac{1}{2}}\sinh A^{\frac{1}{2}}\\&\qquad (t-2t_{m}+2t_{m-1}-\cdots \pm 2t_{i+1}\mp s),\quad i=0,1,\ldots ,m-1. \end{aligned} \end{aligned}$$

Consider the notation

$$\begin{aligned} Q_{m}(t,s)=\left\{ \begin{aligned}&W_{0}(A,t,s),\quad t_{0}\le s\le t_{1},\\&W_{1}(A,t,s),\quad t_{1}< s\le t_{2},\\&\cdots \\&W_{m-1}(A,t,s),\quad t_{m-1}< s\le t_{m},\\&A^{-\frac{1}{2}}\sinh A^{\frac{1}{2}}(t-s),\quad t_{m}<s\le t, \end{aligned} \right. \end{aligned}$$

then the solution of (1.2) can be expressed by

$$\begin{aligned} x(t)=W(A,t,x_{0},y_{0})+\int _{0}^{t}Q_{m}(t,s)Bu(s)ds. \end{aligned}$$

(2.1)

Lemma 2.7

For any $t_{m}<\tau _{1}\le \tau _{2}\le b$, and $t_{m}<t\le t_{m+1}\le b$, we have

$$\begin{aligned} \Big \Vert \int _{0}^{t}\big (Q_{m}(\tau _{2},s)-Q_{m}(\tau _{1},s)\big )ds\Big \Vert \le \theta _{1}|\tau _{2}-\tau _{1}|, \end{aligned}$$

and

$$\begin{aligned} \Big \Vert \int _{0}^{t}\big (Q_{m}'(\tau _{2},s)-Q_{m}'(\tau _{1},s)\big )ds\Big \Vert \le \theta _{2}|\tau _{2}-\tau _{1}|, \end{aligned}$$

where $\theta _{1}$ and $\theta _{2}$ are positive constants, and $Q_{m}'(t,s)$ denotes the function that takes derivative with respect to t.

Proof

Since

$$\begin{aligned}{} & {} A^{\frac{1}{2}}\sinh A^{\frac{1}{2}}t=A\sum _{n=0}^{\infty }\frac{A^{n}t^{2n+1}}{(2n+1)!}, \\{} & {} \cosh A^{\frac{1}{2}}t=\sum _{n=0}^{\infty }\frac{A^{n}t^{2n}}{(2n)!}, \end{aligned}$$

combining this with the Lemma 2.3, we have

$$\begin{aligned} \Vert A^{\frac{1}{2}}\sinh A^{\frac{1}{2}}t\Vert \le T_{A,\varepsilon }\sqrt{\rho (A)+\varepsilon }e^{\sqrt{\rho (A)+\varepsilon }t}, \end{aligned}$$

(2.2)

and

$$\begin{aligned} \Vert \cosh A^{\frac{1}{2}}t\Vert \le T_{A,\varepsilon }e^{\sqrt{\rho (A)+\varepsilon }t}. \end{aligned}$$

(2.3)

According to the definition of $Q_{m}$, inequality (2.3), and the mean value theorem, we find

$$\begin{aligned}&\Big \Vert \int _{0}^{t}{\big (}Q_{m}(\tau _{2},s) -Q_{m}(\tau _{1},s){\big )}ds\Big \Vert \\&\quad \le \sum _{i=0}^{m}\int _{t_{i}}^{t_{i+1}}\Vert W_{i}(A,\tau _{2},s)-W_{i}(A,\tau _{1},s)\Vert ds\\&\quad \le \sum _{i=0}^{m}\int _{t_{i}}^{t_{i+1}}\Vert A_{1}^{m-i}\cosh A^{\frac{1}{2}}\varsigma _{0}\Vert +\Vert A_{1}^{m-i-1}A_{2}\sum _{j=1}^{C_{m-i}^{1}}\cosh A^{\frac{1}{2}}\varsigma _{1,j}\Vert \\&\qquad +\cdots +\Vert A_{1}A_{2}^{m-i-1}\sum _{j=1}^{C_{m-i}^{m-i-1}}\cosh A^{\frac{1}{2}}\varsigma _{m-i-1,j}\Vert \\&\qquad +\Vert A_{2}^{m-i}\cosh A^{\frac{1}{2}}\varsigma _{m-i}\Vert ds\cdot |\tau _{2}-\tau _{1}|\\&\quad \le T_{A,\varepsilon }e^{\sqrt{\rho (A)+\varepsilon }b}\sum _{i =0}^{m}\int _{t_{i}}^{t_{i+1}}T_{\varepsilon }^{2}(\rho (A_{1})+\rho (A_{2}) +2\varepsilon )^{m-i}ds\\&\qquad \cdot |\tau _{2}-\tau _{1}|\\&\quad \le \sum _{i=0}^{m}\big (\rho (A_{1})+\rho (A_{2})+2\varepsilon \big ) ^{m-i}bT_{\varepsilon }^{3}e^{\sqrt{\rho (A)+\varepsilon }b}|\tau _{2}-\tau _{1}|\\&\quad =:\theta _{1}|\tau _{2}-\tau _{1}|, \end{aligned}$$

where $\varsigma _{0},\varsigma _{1,j},\ldots ,\varsigma _{m-i-1,j},\varsigma _{m-i}$ are selected by the mean value theorem located in $[-b,b]$, $T_{\varepsilon }=\max \{T_{A,\varepsilon },T_{A_{1},\varepsilon },T_{A_{2},\varepsilon }\}$. Similarly, by virtue of the definition of $Q_{m}$, inequality (2.2), and the mean value theorem, we have

$$\begin{aligned} \begin{aligned}&\Big \Vert \int _{0}^{t}{\big (}Q_{m}'(\tau _{2},s)-Q_{m}'(\tau _{1},s){\big )}ds\Big \Vert \\&\quad \le \sum _{i=0}^{m}\int _{t_{i}}^{t_{i+1}}\Vert W_{i}'(A,\tau _{2},s)-W_{i}'(A,\tau _{1},s)\Vert ds\\&\quad \le \sum _{i=0}^{m}\int _{t_{i}}^{t_{i+1}}\Vert A_{1}^{m-i}A^{\frac{1}{2}}\sinh A^{\frac{1}{2}}\xi _{0}\Vert +\Vert A_{1}^{m-i-1}A_{2}A^{\frac{1}{2}}\sum _{j=1}^{C_{m-i}^{1}}\sinh A^{\frac{1}{2}}\xi _{1,j}\Vert \\&\qquad +\cdots +\Vert A_{1}A_{2}^{m-i-1}A^{\frac{1}{2}}\sum _{j=1}^{C_{m-i}^{m-i-1}}\sinh A^{\frac{1}{2}}\xi _{m-i-1,j}\Vert \\&\qquad +\Vert A_{2}^{m-i}A^{\frac{1}{2}}\sinh A^{\frac{1}{2}}\xi _{m-i}\Vert ds\cdot |\tau _{2}-\tau _{1}|\\ \end{aligned} \\ \begin{aligned}&\quad \le T_{A,\varepsilon }\sqrt{\rho (A)+\varepsilon }e^{\sqrt{\rho (A)+\varepsilon }b}\sum _{i=0}^{m} \int _{t_{i}}^{t_{i+1}}T_{\varepsilon }^{2}(\rho (A_{1})+\rho (A_{2})+2\varepsilon )^{m-i}ds\\&\qquad \cdot |\tau _{2}-\tau _{1}|\\&\quad \le \sum _{i=0}^{m}\big (\rho (A_{1})+\rho (A_{2})+2\varepsilon \big )^{m-i}bT_{\varepsilon }^{3}\sqrt{\rho (A) +\varepsilon }e^{\sqrt{\rho (A)+\varepsilon }b}|\tau _{2}-\tau _{1}|\\&\quad =:\theta _{2}|\tau _{2}-\tau _{1}|, \end{aligned} \end{aligned}$$

where $\xi _{0},\xi _{1,j},\ldots ,\xi _{m-i-1,j},\xi _{m-i}$ are selected by the mean value theorem located in $[-b,b]$. $\square$

3 The controllability of linear systems

In this section, we present some controllability criteria for systems (1.2) by using an algebraic method.

Theorem 3.1

The following statements are equivalent: $1^{\circ }$The system (1.2) is exact controllability; $2^{\circ }$ The matrix $\Gamma _{0}^{b}=\int _{0}^{b}Q_{k}(b,s)BB^{*}Q_{k}^{*}(b,s)ds$ and $\Lambda _{0}^{b}=\int _{0}^{b}Q'_{k}(b,s)BB^{*}Q_{k}'^{*}(b,s)ds$ are nonsingular; $3^{\circ }$ There at least exists a pair of integers $0\le i\le k$, $0\le j\le k$ such that both $\int _{t_{i}}^{t_{i+1}}W_{i}(b,s)BB^{*}W_{i}^{*}(b,s)ds$ and $\int _{t_{j}}^{t_{j+1}}W_{j}'(b,s)BB^{*}W_{j}'^{*}(b,s)ds$ are nonsingular.

Proof

First, we show the equivalence of $1^{\circ }$ and $2^{\circ }$. Assume the systems are exact controllability. We show that the matrix $\Gamma _{0}^{b}$ and $\Lambda _{0}^{b}$ both are nonsingular. If the result is not true, then at least one of matrices $\Gamma _{0}^{b}$ and $\Lambda _{0}^{b}$ is singular. Suppose $\Gamma _{0}^{b}$ is singular. Then there exists a nonzero vector ${\overline{x}}_{0}\in {\mathbb {R}}^{n}$ such that

$$\begin{aligned} \int _{0}^{b}{\overline{x}}_{0}^{T}Q_{k}(b,s)BB^{*}Q_{k}^{*}(b,s){\overline{x}}_{0}ds=0. \end{aligned}$$

Hence we have

$$\begin{aligned} \int _{0}^{b}\Vert B^{*}Q_{k}^{*}(b,s){\overline{x}}_{0}\Vert ^{2}ds=0, \end{aligned}$$

that is

$$\begin{aligned} B^{*}Q_{k}^{*}(b,s){\overline{x}}_{0}={\textbf{0}},\quad \forall s\in (0,b]. \end{aligned}$$

(3.1)

On the other hand, since the systems are exact controllability, then because of the definition of exactly controllability, there exists a pair of control functions $(u_{1},u_{2})$ such that for ${\overline{x}}_{0}+W(A,b,x_{0},y_{0})\in {\mathbb {R}}^{n}$, the solution $x(\cdot )$ of systems (1.2) under the control $u_{1}(\cdot )$ arrives at ${\overline{x}}_{0}+W(A,b,x_{0},y_{0})\in {\mathbb {R}}^{n}$ at the terminal b, i.e.

$$\begin{aligned} {\overline{x}}_{0}+W(A,b,x_{0},y_{0})=W(A,b,x_{0},y_{0})+\int _{0}^{b}Q_{k}(b,s)Bu_{1}ds. \end{aligned}$$

(3.2)

Now (3.1) with (3.2) allows us to affirm that

$$\begin{aligned} \Vert {\overline{x}}_{0}\Vert ^{2}={\overline{x}}_{0}^{T}{\overline{x}}_{0} =\int _{0}^{b}u_{1}^{T}B^{*}Q_{k}^{*}(b,s){\overline{x}}_{0}ds=0, \end{aligned}$$

which implies ${\overline{x}}_{0}={\textbf{0}}$ and this contradicts the hypothesis. Hence $\Gamma _{0}^{b}$ is nonsingular. In a similar way, we obtain that $\Lambda _{0}^{b}$ is nonsingular,

If both the matrices $\Gamma _{0}^{b}$ and $\Lambda _{0}^{b}$ are nonsingular, we prove that systems (1.2) are exactly controllability, that is for any fixed $(x_{1},y_{1})\in {\mathbb {R}}^{n}\times {\mathbb {R}}^{n}$, we show that there exists a pair of control functions $(u_{1}(\cdot ),u_{2}(\cdot ))\in L^{2}([0,b],{\mathbb {R}}^{n})\times L^{2}([0,b],{\mathbb {R}}^{n})$ such that the solution x(t) of systems (1.2) satisfies $x(b)=x_{1}$ under the control $u_{1}(\cdot )$ and $y'(b)=y_{1}$ under the control $u_{2}(\cdot )$. We choose the control functions by

$$\begin{aligned} u_{1}(t)=B^{*}Q_{k}^{*}(b,t)(\Gamma _{0}^{b})^{-1}(x_{1}-W(A,b,x_{0},y_{0})), \end{aligned}$$

(3.3)

and

$$\begin{aligned} u_{2}(t)=B^{*}Q_{k}'^{*}(b,t)(\Lambda _{0}^{b})^{-1}(y_{1}-W'(A,b,x_{0},y_{0})). \end{aligned}$$

(3.4)

Then we have

$$\begin{aligned} x(t)=W(A,t,x_{0},y_{0})+\int _{0}^{t}Q_{k}(t,s)BB^{*}Q_{k}^{*} (b,s)(\Gamma _{0}^{b})^{-1}(x_{1}-W(A,b,x_{0},y_{0}))ds. \end{aligned}$$

Obviously, $x(b)=x_{1}$. Similarly, under the control function $u_{2}(t)$, $y'(t)$ satisfies

$$\begin{aligned} y'(t)=W'(A,t,x_{0},y_{0})+\int _{0}^{t}Q'_{k}(t,s)BB^{*}Q_{k}'^{*} (b,s)(\Lambda _{0}^{b})^{-1}(y_{1}-W'(A,b,x_{0},y_{0}))ds, \end{aligned}$$

and we have $y'(b)=y_{1}$. Hence the systems (1.2) are exact controllability.

Next, we show the equivalence of $2^{\circ }$ and $3^{\circ }$. Assume the matrix $\Gamma _{0}^{b}$ is singular. Then there exists a nonzero vector ${\overline{x}}_{0}\in {\mathbb {R}}^{n}$ such that

$$\begin{aligned} \int _{0}^{b}{\overline{x}}_{0}^{T}Q_{k}(b,s)BB^{*}Q_{k}^{*}(b,s){\overline{x}}_{0}ds=0, \end{aligned}$$

that is

$$\begin{aligned} \begin{aligned}&\int _{0}^{t_{1}}{\overline{x}}_{0}^{T}W_{0}(b,s)BB^{*}W_{0}^{*}(b,s){\overline{x}}_{0}ds +\int _{t_{1}}^{t_{2}}{\overline{x}}_{0}^{T}W_{1}(b,s)BB^{*}W_{1}^{*}(b,s){\overline{x}}_{0}ds\\&\quad +\cdots +\int _{t_{k-1}}^{t_{k}} {\overline{x}}_{0}^{T}W_{k-1}(b,s)BB^{*}W_{k-1}^{*}(b,s){\overline{x}}_{0}ds\\&\quad +\int _{t_{k}}^{b}{\overline{x}}_{0}^{T}A^{-\frac{1}{2}}\sinh A^{\frac{1}{2}}(b-s)BB^{*} (\sinh A^{\frac{1}{2}}(b-s))^{*}(A^{-\frac{1}{2}})^{*}{\overline{x}}_{0}ds=0, \end{aligned} \end{aligned}$$

which is equivalent to

$$\begin{aligned} \int _{t_{i}}^{t_{i+1}}{\overline{x}}_{0}^{T}W_{i}(b,s)BB^{*}W_{i}^{*}(b,s) {\overline{x}}_{0}ds=0,\quad \forall 0\le i\le k, \end{aligned}$$

that is $\int _{t_{i}}^{t_{i+1}}W_{i}(b,s)BB^{*}W_{i}^{*}(b,s)ds$ is singular for all $0\le i\le k$. Hence $\Gamma _{0}^{b}$ is nonsingular iff there at least exists a constant $0\le i\le k$ such that $\int _{t_{i}}^{t_{i+1}}W_{i}(b,s)BB^{*}W_{i}^{*}(b,s)ds$ is nonsingular.

By the same argument, we also can show that $\Lambda _{0}^{b}$ is nonsingular iff there at least exists a constant $0\le j\le k$ such that the matrix $\int _{t_{j}}^{t_{j+1}}W_{j}'(b,s)BB^{*}W_{j}'^{*}(b,s)ds$ is nonsingular. $\square$

Remark 3.2

Theorem 3.1 shows that initial value problems of second-order linear impulsive systems (1.2) are controllable iff there exist constants $\lambda >0$ and $\gamma >0$ such that for all $x\in {\mathbb {R}}^{n}$,

$$\begin{aligned} (\Gamma _{0}^{b}x,x)\ge \gamma \Vert x\Vert ^{2}, \end{aligned}$$

and

$$\begin{aligned} (\Lambda _{0}^{b}x,x)\ge \lambda \Vert x\Vert ^{2}. \end{aligned}$$

Then

$$\begin{aligned} \Vert (\Gamma _{0}^{b})^{-1}\Vert \le \frac{1}{\gamma },\quad \Vert (\Lambda _{0}^{b})^{-1}\Vert \le \frac{1}{\lambda }. \end{aligned}$$

(3.5)

Since the conditions which guarantee the controllability in Theorem 3.1 are formal and are hard to verify. In what follows, we give a new rank criterion of controllability of systems (1.2). For convenience in writing, in what follows, we use the notation

$$\begin{aligned} \Psi (t)=A^{-\frac{1}{2}}\sinh A^{\frac{1}{2}}t. \end{aligned}$$

Theorem 3.3

Systems (1.2) are exact controllability iff there exists a pair of integers $l_{1},l_{2}\in \{0,1,2,\ldots ,k\}$ such that

$$\begin{aligned} \text {Rank} \begin{pmatrix} A_{1}^{l_{1}}B&\cdots&A_{1}^{l_{1}-i}A_{2}^{i}B&\cdots&A_{2}^{l_{1}}B \end{pmatrix} =n, \end{aligned}$$

and

$$\begin{aligned} \text {Rank} \begin{pmatrix} A_{1}^{l_{2}}B&\cdots&A_{1}^{l_{2}-i}A_{2}^{i}B&\cdots&A_{2}^{l_{2}}B \end{pmatrix} =n. \end{aligned}$$

Proof

Theorem 3.1 shows that systems (1.2) are exact controllability iff there is a pair of integers $0\le i\le k$, $0\le j\le k$ such that both $\int _{t_{i}}^{t_{i+1}}W_{i}(b,s)BB^{*}W_{i}^{*}(b,s)ds$ and $\int _{t_{j}}^{t_{j+1}}W_{j}'(b,s)BB^{*}W_{j}'^{*}(b,s)ds$ are nonsingular. We subdivide the proof into several cases.

Case 1 If $i=j=k$, that is both $\int _{t_{k}}^{b}W_{k}(b,s)BB^{*}W_{k}^{*}(b,s)ds$ and $\int _{t_{k}}^{b}W_{k}'(b,s)BB^{*}W_{k}'^{*}(b,s)ds$ are nonsingular. Now $\int _{t_{k}}^{b}W_{k}(b,s)BB^{*}W_{k}^{*}(b,s)ds$ is singular iff there exists a nonzero vector $x_{0}\in {\mathbb {R}}^{n}$ such that $x_{0}^{T}W_{k}(b,s)B={\textbf{0}}$ for all $t_{k}\le s<b$. Since A is a nonsingular matrix, we have Rank $W_{k}(b,s)=n$, hence Rank $B<n$. Likewise, we can show that $\int _{t_{k}}^{b}W_{k}'(b,s)BB^{*}W_{k}'^{*}(b,s)ds$ is singular iff Rank $B<n$. Hence both $\int _{t_{k}}^{b}W_{k}(b,s)BB^{*}W_{k}^{*}(b,s)ds$ and $\int _{t_{k}}^{b}W_{k}'(b,s)BB^{*}W_{k}'^{*}(b,s)ds$ are nonsingular iff Rank $B=n$.

Case 2 If $i=j=k-1$, we show that both $\int _{t_{k-1}}^{t_{k}}W_{k-1}(b,s)BB^{*}W_{k-1}^{*}(b,s)ds$ and

$\int _{t_{k-1}}^{t_{k}}W_{k-1}'(b,s)BB^{*}W_{k-1}'^{*}(b,s)ds$ are nonsingular iff

$$\begin{aligned} \text {Rank} \begin{pmatrix} A_{1}B&A_{2}B \end{pmatrix} =n. \end{aligned}$$

Assume $\int _{t_{k-1}}^{t_{k}}W_{k-1}(b,s)BB^{*}W_{k-1}^{*}(b,s)ds$ is nonsingular. Then we show that

$$\begin{aligned} \text {Rank} \begin{pmatrix} A_{1}B&A_{2}B \end{pmatrix} =n. \end{aligned}$$

If this is not true, that is Rank $\begin{pmatrix} A_{1}B&A_{2}B \end{pmatrix} <n$, then there exists a nonzero vector $x_{0}\in {\mathbb {R}}^{n}$ such that

$$\begin{aligned} x_{0}^{T} \begin{pmatrix} A_{1}B&A_{2}B \end{pmatrix} ={\textbf{0}}_{1\times 2n}, \end{aligned}$$

i.e.,

$$\begin{aligned} x_{0}^{T}A_{1}B={\textbf{0}},\quad x_{0}^{T}A_{2}B={\textbf{0}}. \end{aligned}$$

Hence, for all $t_{k-1}< s\le t_{k}$,

$$\begin{aligned} \begin{aligned} x_{0}^{T}W_{k-1}(b,s)B&=x_{0}^{T}[A_{1}A^{-\frac{1}{2}}\sinh A^{\frac{1}{2}}(b-s)B-A_{2}A^{-\frac{1}{2}}\sinh A^{\frac{1}{2}}(b-2t_{k}+s)B]\\&=x_{0}^{T}[A_{1}BA^{-\frac{1}{2}}\sinh A^{\frac{1}{2}}(b-s)-A_{2}BA^{-\frac{1}{2}}\sinh A^{\frac{1}{2}}(b-2t_{k}+s)]\\ {}&={\textbf{0}}, \end{aligned} \end{aligned}$$

which implies $\int _{t_{k-1}}^{t_{k}}W_{k-1}(b,s)BB^{*}W_{k-1}^{*}(b,s)ds$ is singular. This contradicts the hypothesis. Therefore, Rank $\begin{pmatrix} A_{1}B&A_{2}B \end{pmatrix} =n$.

Assume $\text {Rank} \begin{pmatrix} A_{1}B&A_{2}B \end{pmatrix} =n$. We will show that $\int _{t_{k-1}}^{t_{k}}W_{k-1}(b,s)BB^{*}W_{k-1}^{*}(b,s)ds$ is nonsingular. Assume $\int _{t_{k-1}}^{t_{k}}W_{k-1}(b,s)BB^{*}W_{k-1}^{*}(b,s)ds$ is singular. First, we prove that there exists a number sequence $(\lambda _{1},\lambda _{2})\in (t_{k-1},t_{k}]^{2}$, where $\lambda _{1}\ne \lambda _{2}$, such that the matrix

$$\begin{aligned} \begin{pmatrix} \Psi (b-\lambda _{1})&{}\quad \Psi (b-\lambda _{2})\\ \Psi (b-2t_{k}+\lambda _{1})&{}\quad \Psi (b-2t_{k}+\lambda _{2}) \end{pmatrix} \end{aligned}$$

(3.6)

is nonsingular. Suppose for every $(\lambda _{1},\lambda _{2})\in (t_{k-1},t_{k}]^{2}$, we have

$$\begin{aligned} \left| \begin{matrix} \Psi (b-\lambda _{1})&{}\quad \Psi (b-\lambda _{2})\\ \Psi (b-2t_{k}+\lambda _{1})&{}\quad \Psi (b-2t_{k}+\lambda _{2}) \end{matrix} \right| =0. \end{aligned}$$

(3.7)

Take $\lambda _{2}=t_{k}$ in (3.7), since $|\Psi (b-\lambda _{2})|\ne 0$, we find

$$\begin{aligned} |\Psi (b-2t_{k}+\lambda _{1})-\Psi (b-\lambda _{1})|=0, \end{aligned}$$

(3.8)

however, by the Jordan decomposition, we find zero is not an eigenvalue of $\Psi (b-2t_{k}+\lambda _{1})-\Psi (b-\lambda _{1})$, hence, (3.8) is not valid, that is there exists a number sequence $(\lambda _{1},\lambda _{2})\in (t_{k-1},t_{k}]^{2}$ such that (3.6) is nonsingular.

Suppose $\int _{t_{k-1}}^{t_{k}}W_{k-1}(b,s)BB^{*}W_{k-1}^{*}(b,s)ds$ is singular. Then there exists a nonzero vector $x_{0}\in {\mathbb {R}}^{n}$ such that

$$\begin{aligned} \int _{t_{k-1}}^{t_{k}}x_{0}^{T}W_{k-1}(b,s)BB^{*}W_{k-1}^{*}(b,s)x_{0}ds=0, \end{aligned}$$

that is

$$\begin{aligned} \int _{t_{k-1}}^{t_{k}}\Vert x_{0}^{T}W_{k-1}(b,s)B\Vert ^{2}ds=0, \end{aligned}$$

which implies

$$\begin{aligned}{} & {} x_{0}^{T}\Big [A_{1}A^{-\frac{1}{2}}\sinh A^{\frac{1}{2}}(b-s)B-A_{2}A^{-\frac{1}{2}}\sinh A^{\frac{1}{2}}(b-2t_{k}+s)B\Big ]\nonumber \\ {}{} & {} \quad ={\textbf{0}},\quad \forall t_{k-1}<s\le t_{k}. \end{aligned}$$

(3.9)

Take $(\lambda _{1},\lambda _{2})\in (t_{k-1},t_{k}]^{2}$ such that (3.6) is nonsingular, and then by (3.9), we find

$$\begin{aligned} x_{0}^{T} \begin{pmatrix} A_{1}B&-A_{2}B \end{pmatrix} \begin{pmatrix} \Psi (b-\lambda _{1})&{}\quad \Psi (b-\lambda _{2})\\ \Psi (b-2t_{k}+\lambda _{1})&{}\quad \Psi (b-2t_{k}+\lambda _{2}) \end{pmatrix} ={\textbf{0}}_{1\times 2n}, \end{aligned}$$

therefore,

$$\begin{aligned} x_{0}^{T} \begin{pmatrix} A_{1}B&-A_{2}B \end{pmatrix} ={\textbf{0}}_{1\times 2n}, \end{aligned}$$

that is

$$\begin{aligned} \text {Rank} \begin{pmatrix} A_{1}B&-A_{2}B \end{pmatrix} <n, \end{aligned}$$

which contradict the hypothesis.

Thus $\int _{t_{k-1}}^{t_{k}}W_{k-1}(b,s)BB^{*}W_{k-1}^{*}(b,s)ds$ is nonsingular iff

$$\begin{aligned} \text {Rank} \begin{pmatrix} A_{1}B&A_{2}B \end{pmatrix} =n. \end{aligned}$$

Using the same argument we can establish that $\int _{t_{k-1}}^{t_{k}}W_{k-1}'(b,s)BB^{*}W_{k-1}'^{*}(b,s)ds$ is nonsingular iff

$$\begin{aligned} \text {Rank} \begin{pmatrix} A_{1}B&A_{2}B \end{pmatrix} =n. \end{aligned}$$

Case 3 If $i=j=k-2$, we show that both $\int _{t_{k-2}}^{t_{k-1}}W_{k-2}(b,s)BB^{*}W_{k-2}^{*}(b,s)ds$ and $\int _{t_{k-2}}^{t_{k-1}}W_{k-2}'(b,s)BB^{*}W_{k-2}'^{*}(b,s)ds$ are nonsingular iff

$$\begin{aligned} \text {Rank} \begin{pmatrix} A_{1}^{2}B&A_{1}A_{2}B&A_{2}^{2}B \end{pmatrix} =n. \end{aligned}$$

To do this, we first show an auxiliary result. For $s_{1},s_{2},s_{3}\in (t_{k-2},t_{k-1}]$, let

$$\begin{aligned} \Sigma _{3}=\begin{pmatrix} \Psi (b-s_{1}) &{} \quad \Psi (b-s_{2})&{}\quad \Psi (b-s_{3})\\ \sum \limits _{k-1\le i_{11}\le k}\Psi (b-2t_{i_{11}}+s_{1})&{}\quad \sum \limits _{k-1\le i_{11}\le k}\Psi (b-2t_{i_{11}}+s_{2})&{}\quad \sum \limits _{k-1\le i_{11}\le k}\Psi (b-2t_{i_{11}}+s_{3})\\ \Psi (b-2t_{k}+2t_{k-1}-s_{1})&{}\quad \Psi (b-2t_{k}+2t_{k-1}-s_{2})&{}\quad \Psi (b-2t_{k}+2t_{k-1}-s_{3}) \end{pmatrix}, \end{aligned}$$

and

$$\begin{aligned} \Sigma _{3}'=\begin{pmatrix} \Psi '(b-s_{1}) &{} \quad \Psi '(b-s_{2})&{}\quad \Psi '(b-s_{3})\\ \sum \limits _{k-1\le i_{11}\le k}\Psi ' (b-2t_{i_{11}}+s_{1})&{}\quad \sum \limits _{k-1\le i_{11}\le k}\Psi ' (b-2t_{i_{11}}+s_{2})&{}\quad \sum \limits _{k-1\le i_{11}\le k}\Psi ' (b-2t_{i_{11}}+s_{3})\\ \Psi '(b-2t_{k}+2t_{k-1}-s_{1})&{}\quad \Psi '(b-2t_{k}+2t_{k-1}-s_{2})&{}\quad \Psi '(b-2t_{k}+2t_{k-1}-s_{3}) \end{pmatrix}. \end{aligned}$$

We claim that

$$\begin{aligned} \Sigma _{3}x_{0}={\textbf{0}}_{3n\times n},\quad \forall (s_{1},s_{2},s_{3})\in (t_{k-2},t_{k-1})^{3}, \end{aligned}$$

(3.10)

or

$$\begin{aligned} \Sigma _{3}'x_{0}={\textbf{0}}_{3n\times n},\quad \forall (s_{1},s_{2},s_{3})\in (t_{k-2},t_{k-1})^{3}, \end{aligned}$$

(3.11)

where $x_{0}=(\lambda _{1}I,\lambda _{2}I,\lambda _{3}I)^{T}\in {\mathbb {R}}^{3n\times n}$, implies $x_{0}={\textbf{0}}_{3n\times n}$. We only show that (3.10) implies that $x_{0}={\textbf{0}}_{3n\times n}$, and the other case can be treated similarly. For any $s\in (t_{k-2},t_{k-1})$ and $\varepsilon _{1}, \varepsilon _{2}>0$ small enough, let

$$\begin{aligned} s_{1}=s,\quad s_{2}=(1+\varepsilon _{1})s,\quad s_{3}=(1+\varepsilon _{2})s. \end{aligned}$$

By the first row of equality (3.10), we have

$$\begin{aligned} \lambda _{1}\Psi (b-s)+\lambda _{2}\Psi (b-(1+\varepsilon _{1})s) +\lambda _{3}\Psi (b-(1+\varepsilon _{2})s)={\textbf{0}}_{n\times n}, \end{aligned}$$

(3.12)

take the second and fourth derivatives with respect to s in (3.12) respectively and we have

$$\begin{aligned}{} & {} \lambda _{1}A\Psi (b-s)+\lambda _{2}A(1+\varepsilon _{1})^{2}\Psi (b-(1+\varepsilon _{1})s) +\lambda _{3}A(1+\varepsilon _{2})^{2}\Psi (b-(1+\varepsilon _{2})s)\nonumber \\{} & {} \quad ={\textbf{0}}_{n\times n}, \end{aligned}$$

(3.13)

$$\begin{aligned}{} & {} \lambda _{1}A^{2}\Psi (b-s)+\lambda _{2}A^{2}(1+\varepsilon _{1})^{4}\Psi (b-(1+\varepsilon _{1})s) +\lambda _{3}A^{2}(1+\varepsilon _{2})^{4}\Psi (b-(1+\varepsilon _{2})s)\nonumber \\{} & {} \quad ={\textbf{0}}_{n\times n}.\nonumber \\ \end{aligned}$$

(3.14)

Combine (3.12), (3.13) and (3.14) to obtain

$$\begin{aligned} \begin{pmatrix} \lambda _{1}&{}\quad \lambda _{2}&{}\quad \lambda _{3}\\ \lambda _{1}A&{}\quad \lambda _{2}A(1+\varepsilon _{1}) ^{2}&{}\quad \lambda _{3}A(1+\varepsilon _{2})^{2}\\ \lambda _{1}A^{2}&{}\quad \lambda _{2}A^{2}(1+\varepsilon _{1})^{4} &{}\quad \lambda _{3}A^{2}(1+\varepsilon _{2})^{4}\\ \end{pmatrix} \begin{pmatrix} \Psi (b-s)\\ \Psi (b-(1+\varepsilon _{1})s)\\ \Psi (b-(1+\varepsilon _{2})s)\\ \end{pmatrix} ={\textbf{0}}_{3n\times n}, \end{aligned}$$

and since $(\Psi (b-s),\Psi (b-(1+\varepsilon _{1})s),\Psi (b-(1+\varepsilon _{2})s))^{T}$ is a nonzero vector therefore

$$\begin{aligned} \left| \begin{matrix} \lambda _{1}&{}\quad \lambda _{2}&{}\quad \lambda _{3}\\ \lambda _{1}A&{}\quad \lambda _{2}A(1+\varepsilon _{1})^{2}&{}\quad \lambda _{3}A(1+\varepsilon _{2})^{2}\\ \lambda _{1}A^{2}&{}\quad \lambda _{2}A^{2}(1+\varepsilon _{1})^{4}&{}\quad \lambda _{3}A^{2}(1+\varepsilon _{2})^{4}\\ \end{matrix} \right| =0, \end{aligned}$$

which implies that at least one of $\lambda _{1}$, $\lambda _{2}$, and $\lambda _{3}$ is zero.

Let $\lambda _{1}=0$, then by the second row of (3.10), we find

$$\begin{aligned}{} & {} \lambda _{2}\sum \limits _{k-1\le i_{11}\le k}\Psi (b-2t_{i_{11}}+(1+\varepsilon _{1})s)+\lambda _{3}\sum \limits _{k-1\le i_{11}\le k}\Psi (b-2t_{i_{11}}+(1+\varepsilon _{2})s)\nonumber \\ {}{} & {} \quad ={\textbf{0}}_{n\times n}. \end{aligned}$$

(3.15)

Take the second derivative with respect to s in (3.15) to get

$$\begin{aligned} \begin{aligned} \lambda _{2}A(1+\varepsilon _{1})^{2}&\sum \limits _{k-1\le i_{11}\le k}\Psi (b-2t_{i_{11}}+(1+\varepsilon _{1})s)\\&+\lambda _{3}A(1+\varepsilon _{2})^{2}\sum \limits _{k-1\le i_{11}\le k}\Psi (b-2t_{i_{11}}+(1+\varepsilon _{2})s)={\textbf{0}}_{n\times n}. \end{aligned} \end{aligned}$$

(3.16)

Combine (3.15) with (3.16) and we have

$$\begin{aligned} \begin{pmatrix} \lambda _{2}&{}\quad \lambda _{3}\\ \lambda _{2}A(1+\varepsilon _{1})^{2}&{}\quad \lambda _{3}A(1+\varepsilon _{2})^{2}\\ \end{pmatrix} \begin{pmatrix} \sum \limits _{k-1\le i_{11}\le k}\Psi (b-2t_{i_{11}}+(1+\varepsilon _{1})s)\\ \sum \limits _{k-1\le i_{11}\le k}\Psi (b-2t_{i_{11}}+(1+\varepsilon _{2})s)\\ \end{pmatrix} ={\textbf{0}}_{2n\times n}, \end{aligned}$$

since

$$\begin{aligned} \begin{pmatrix} \sum \limits _{k-1\le i_{11}\le k}\Psi (b-2t_{i_{11}}+(1+\varepsilon _{1})s)\\ \sum \limits _{k-1\le i_{11}\le k}\Psi (b-2t_{i_{11}}+(1+\varepsilon _{2})s)\\ \end{pmatrix}\ne {\textbf{0}}_{n\times n}, \end{aligned}$$

hence,

$$\begin{aligned} \left| \begin{matrix} \lambda _{2}&{}\quad \lambda _{3}\\ \lambda _{2}A(1+\varepsilon _{1})^{2} &{}\quad \lambda _{3}A(1+\varepsilon _{2})^{2}\\ \end{matrix} \right| =0, \end{aligned}$$

which implies that at least one of $\lambda _{2}$ and $\lambda _{3}$ is zero.

Let $\lambda _{2}=0$, then by the third row of (3.10), we have

$$\begin{aligned} \lambda _{3}\Psi (b-2t_{k}+2t_{k-1}-(1+\varepsilon _{2})s)={\textbf{0}}_{n\times n}, \end{aligned}$$

(3.17)

obviously, (3.17) implies $\lambda _{3}=0$. Thus, $x_{0}={\textbf{0}}_{3n\times n}$.

Suppose $\int _{t_{k-2}}^{t_{k-1}}W_{k-2}(b,s)BB^{*}W^{*}_{k-2}(b,s)ds$ is nonsingular and assume Rank $\begin{pmatrix} A_{1}^{2}B&A_{1}A_{2}B&A_{2}^{2}B \end{pmatrix} <n$. Then there exists a nonzero vector $x_{0}\in {\mathbb {R}}^{n}$ such that

$$\begin{aligned} x_{0}^{T} \begin{pmatrix} A_{1}^{2}B&A_{1}A_{2}B&A_{2}^{2}B \end{pmatrix} ={\textbf{0}}, \end{aligned}$$

that is, for all $s\in (t_{k-2},t_{k-1}]$,

$$\begin{aligned} \begin{aligned} x_{0}^{T}W_{k-2}(b,s)B=&\, x_{0}^{T}\Big (A_{1}^{2}B\Psi (b-s) -A_{1}A_{2}B\big (\Psi (b-2t_{k}+s)\\&+A^{-\frac{1}{2}}\Psi (b-2t_{k-1}+s)\big )+A_{2}^{2}BA^{ -\frac{1}{2}}\Psi (b-2t_{k}+2t_{k-1}-s)\Big )\\ =&\, {\textbf{0}},\\ \end{aligned} \end{aligned}$$

which contradicts the fact that $\int _{t_{k-2}}^{t_{k-1}}W_{k-2}(b,s)BB^{*}W^{*}_{k-2}(b,s)ds$ is nonsingular.

Suppose Rank $\begin{pmatrix} A_{1}^{2}B&A_{1}A_{2}B&A_{2}^{2}B \end{pmatrix} =n$ and assume

$$\begin{aligned} \int _{t_{k-2}}^{t_{k-1}}W_{k-2}(b,s)BB^{*}W^{*}_{k-2}(b,s)ds \end{aligned}$$

is singular. There exists a nonzero vector $x_{0}\in {\mathbb {R}}^{n}$ such that

$$\begin{aligned} \int _{t_{k-2}}^{t_{k-1}}x_{0}^{T}W_{k-2}(b,s)BB^{*}W^{*}_{k-2}(b,s)x_{0}ds=0. \end{aligned}$$

which implies

$$\begin{aligned} x_{0}^{T}W_{k-2}(b,s)B={\textbf{0}},\quad \forall s\in (t_{k-2},t_{k-1}]. \end{aligned}$$

(3.18)

For $s_{1},s_{2},s_{3}$ selected in the auxiliary result, by equation (3.18), we have

$$\begin{aligned} \begin{aligned} x_{0}^{T}\Big (A_{1}^{2}&BA^{-\frac{1}{2}}\Psi (b-s_{1}) -A_{1}A_{2}B\big (A^{-\frac{1}{2}}\Psi (b-2t_{k}+s_{1})\\&+A^{-\frac{1}{2}}\Psi (b-2t_{k-1}+s_{1})\big ) +A_{2}^{2}BA^{-\frac{1}{2}}\Psi (b-2t_{k}+2t_{k-1}-s_{1})\Big )={\textbf{0}},\\ \end{aligned} \\ \begin{aligned} x_{0}^{T}\Big (A_{1}^{2}&BA^{-\frac{1}{2}}\Psi (b-s_{2}) -A_{1}A_{2}B\big (A^{-\frac{1}{2}}\Psi (b-2t_{k}+s_{2})\\&+A^{-\frac{1}{2}}\Psi (b-2t_{k-1}+s_{2})\big )+A_{2}^{2}BA^{ -\frac{1}{2}}\Psi (b-2t_{k}+2t_{k-1}-s_{2})\Big )={\textbf{0}},\\ \end{aligned} \\ \begin{aligned} x_{0}^{T}\Big (A_{1}^{2}&BA^{-\frac{1}{2}}\Psi (b-s_{3}) -A_{1}A_{2}B\big (A^{-\frac{1}{2}}\Psi (b-2t_{k}+s_{3})\\&+A^{-\frac{1}{2}}\Psi (b-2t_{k-1}+s_{3})\big )+A_{2}^{2}BA^{ -\frac{1}{2}}\Psi (b-2t_{k}+2t_{k-1}-s_{3})\Big )={\textbf{0}},\\ \end{aligned} \end{aligned}$$

that is

$$\begin{aligned} x_{0}^{T} \begin{pmatrix} A_{1}^{2}B&-A_{1}A_{2}B&A_{2}^{2}B \end{pmatrix} \Sigma _{3}={\textbf{0}}, \end{aligned}$$

then by the auxiliary result, we find

$$\begin{aligned} x_{0}^{T} \begin{pmatrix} A_{1}^{2}B&-A_{1}A_{2}B&A_{2}^{2}B \end{pmatrix}={\textbf{0}}, \end{aligned}$$

which implies

$$\begin{aligned} \text {Rank} \begin{pmatrix} A_{1}^{2}B&-A_{1}A_{2}B&A_{2}^{2}B \end{pmatrix}<n, \end{aligned}$$

which contradict the hypothesis.

Thus $\int _{t_{k-2}}^{t_{k-1}}W_{k-2}(b,s)BB^{*}W_{k-2}^{*}(b,s)ds$ is nonsingular iff

$$\begin{aligned} \text {Rank} \begin{pmatrix} A_{1}^{2}B&A_{1}A_{2}B&A_{2}^{2}B \end{pmatrix} =n. \end{aligned}$$

In similar way, we obtain $\int _{t_{k-2}}^{t_{k-1}}W_{k-2}'(b,s)BB^{*}W_{k-2}'^{*}(b,s)ds$ is nonsingular iff

$$\begin{aligned} \text {Rank} \begin{pmatrix} A_{1}^{2}B&A_{1}A_{2}B&A_{2}^{2}B \end{pmatrix} =n. \end{aligned}$$

Thus, we have case 3.

Similarly, taking the proof method of auxiliary result in case 3, for any $0\le m\le k-2$, we can show an auxiliary result for $i=j=m$, i.e.

$$\begin{aligned} \Sigma _{k-m+1}x_{0}={\textbf{0}}_{(k-m+1)n\times n}, \end{aligned}$$

or

$$\begin{aligned} \Sigma _{k-m+1}'x_{0}={\textbf{0}}_{(k-m+1)n\times n} \end{aligned}$$

implies $x_{0}={\textbf{0}}_{(k-m+1)n\times n}$, where $\Sigma _{k-m+1}$ is constructed the same way as $\Sigma _{3}$. Making use of this auxiliary result and proceeding as the technique in case 3, we can show that for $i=j=0$, both $\int _{t_{0}}^{t_{1}}W_{0}(b,s)BB^{*}W_{0}^{*}(b,s)ds$ and $\int _{t_{0}}^{t_{1}}W_{0}'(b,s)BB^{*}W_{0}'^{*}(b,s)ds$ are nonsingular iff

$$\begin{aligned} \text {Rank} \begin{pmatrix} A_{1}^{k}B&\cdots&A_{1}^{k-i}A_{2}^{i}B&\cdots&A_{2}^{k}B \end{pmatrix} =n. \end{aligned}$$

By Theorem 3.1, obviously, if there exists an integer $l\in \{0,1,2,\ldots ,k\}$ such that

$$\begin{aligned} \text {Rank} \begin{pmatrix} A_{1}^{l}B&\cdots&A_{1}^{l-i}A_{2}^{i}B&\cdots&A_{2}^{l}B \end{pmatrix} =n, \end{aligned}$$

then system (1.2) is controllable.

On the other hand, if system (1.2) is controllable. Then, by Theorem 3.1, there at least exists a pair of integers $0\le {\mathfrak {i}}\le k$, $0\le {\mathfrak {j}}\le k$ such that both $\int _{t_{{\mathfrak {i}}}}^{t_{{\mathfrak {i}}+1}}W_{{\mathfrak {i}}}(b,s)BB^{*}W_{{\mathfrak {i}}}^{*}(b,s)ds$ and $\int _{t_{{\mathfrak {j}}}}^{t_{{\mathfrak {j}}+1}}W_{{\mathfrak {j}}}'(b,s)BB^{*}W_{{\mathfrak {j}}}'^{*}(b,s)ds$ are nonsingular, that is

$$\begin{aligned} \text {Rank} \begin{pmatrix} A_{1}^{k-{\mathfrak {i}}}B&\cdots&A_{1}^{k-{\mathfrak {i}}-i}A_{2}^{i}B&\cdots&A_{2}^{k-{\mathfrak {i}}}B \end{pmatrix} =n, \end{aligned}$$

and

$$\begin{aligned} \text {Rank} \begin{pmatrix} A_{1}^{k-{\mathfrak {j}}}B&\cdots&A_{1}^{k-{\mathfrak {j}}-i}A_{2}^{i}B&\cdots&A_{2}^{k-{\mathfrak {j}}}B \end{pmatrix} =n. \end{aligned}$$

$\square$

4 The controllability of semilinear systems

In this section, we consider the controllability of the initial value problems of second-order semilinear systems (1.3).

For convenience in writing, let us introduce the notation $\Vert B\Vert =K$, and the following assumptions.

$(H_{1})$ $f:J\times {\mathbb {R}}^{n}\rightarrow {\mathbb {R}}^{n}$ is a continuous function and there exist a positive constant L such that

$$\begin{aligned} \Vert f(t,x)-f(t,y)\Vert \le L\Vert x-y\Vert , \end{aligned}$$

for every $x,y\in {\mathbb {R}}^{n}$, and $N=\max _{t\in [0,b]} \Vert f(t,0)\Vert$

$(H_{2})$ The linear systems (1.2) are exactly controllable.

$(H_{3})$ Let

$$\begin{aligned} \rho (A_{1})+\rho (A_{2})<1. \end{aligned}$$

Theorem 4.1

Let $x_{0},y_{0}\in {\mathbb {R}}^{n}$ and assume the condition $(H_{1})$-$(H_{3})$ are satisfied. Then the initial value problems of semilinear second-order impulsive systems (1.3) are exactly controllable provided that

$$\begin{aligned} L \left( K^{2}\frac{T_{\varepsilon }^{9}}{2\gamma \rho (A)^{5/2}}e^{3\sqrt{\rho (A)}b} +\frac{T_{\varepsilon }^{3}}{\sqrt{\rho (A)}}e^{\sqrt{\rho (A)}b} \right) <1 \end{aligned}$$

(4.1)

and

$$\begin{aligned} L \left( K^{2}T_{\varepsilon }^{9}\frac{1}{2\lambda }\mu e^{3\sqrt{\rho (A)}b} +{T_{\varepsilon }^{3}e^{\sqrt{\rho (A)}b}}\right) <\min \{\rho (A)^{1/2},1\}, \end{aligned}$$

(4.2)

where ${\mu =\max \{\frac{1}{\rho (A)^{1/2}},\frac{1}{\rho (A)^{3/2}}\}}$.

Proof

From Lemma 2.6, for $t\in (t_{k},b]$, (1.3) are equivalent to the integral equation

$$\begin{aligned} \begin{aligned} x(t)&=\,W(A,t,x_{0},y_{0})+\sum _{i=0}^{k-1}\int _{t_{i}}^{t_{i +1}}W_{i}(A,t,s)(Bu(s)+f(s,x(s)))ds\\&\quad +A^{-\frac{1}{2}}\int _{t_{k}}^{t}\sinh A^{\frac{1}{2}}(t-s)(Bu(s)+f(s,x(s)))ds\\&:=\,W(A,t,x_{0},y_{0})+\int _{0}^{t}Q_{k}(t,s)(Bu(s)+f(s,x(s)))ds. \end{aligned} \end{aligned}$$

In light of $(H_{3})$, we choose $\varepsilon >0$ small enough such that

$$\begin{aligned} \rho (A_{1})+\rho (A_{2})+2\varepsilon <1. \end{aligned}$$

Combine with Lemma 2.3 and it follows that for $t\in (t_{k},b]$,

$$\begin{aligned} \begin{aligned} \Vert W_{i}(A,t,s)\Vert&\le \frac{T_{\varepsilon }^{3}}{\sqrt{\rho (A) +\varepsilon }}(\rho (A_{1})+\rho (A_{2})+2\varepsilon )^{k-i}e^{\sqrt{\rho (A) +\varepsilon }(t-s)}\\&\le \frac{T_{\varepsilon }^{3}}{\sqrt{\rho (A) +\varepsilon }}e^{\sqrt{\rho (A)+\varepsilon }(t-s)}, \end{aligned} \end{aligned}$$

(4.3)

$$\begin{aligned} \begin{aligned} \Vert W_{i}'(A,t,s)\Vert&\le T_{A,\varepsilon }^{3}(\rho (A_{1}) +\rho (A_{2})+2\varepsilon )^{k-i}e^{\sqrt{\rho (A)+\varepsilon }(t-s)}\\&\le T_{A,\varepsilon }^{3}e^{\sqrt{\rho (A)+\varepsilon }(t-s)}, \end{aligned} \end{aligned}$$

(4.4)

and

$$\begin{aligned} \begin{aligned} \Vert W(A,t,x_{0},y_{0})\Vert&\le \Delta _{1}e^{\sqrt{\rho (A) +\varepsilon }b}(\rho (A_{1})+\rho (A_{2})+2\varepsilon )^{k}\\&\le \Delta _{1}e^{\sqrt{\rho (A)+\varepsilon }b}, \end{aligned} \end{aligned}$$

(4.5)

$$\begin{aligned} \begin{aligned} \Vert W'(A,t,x_{0},y_{0})\Vert&\le \Delta _{1}e^{\sqrt{\rho (A) +\varepsilon }b}(\rho (A_{1})+\rho (A_{2})+2\varepsilon )^{k}\\&\le \Delta _{1}e^{\sqrt{\rho (A)+\varepsilon }b}, \end{aligned} \end{aligned}$$

(4.6)

where

$$\begin{aligned} \Delta _{1}=2\max \{\frac{1}{\sqrt{\rho (A) +\varepsilon }},\sqrt{\rho (A)+\varepsilon }\}T_{\varepsilon }^{3}, \end{aligned}$$

$T_{\varepsilon }=\max \{T_{A,\varepsilon }, T_{A_{1},\varepsilon },T_{A_{2},\varepsilon }\}$ and $\varepsilon >0$ small enough.

We show the controllability of the solutions of (1.3). Define the feedback control function

$$\begin{aligned} u_{1x}(t)=B^{*}Q_{k}^{*}(b,t)(\Gamma _{0}^{b})^{ -1}\left( x_{1}-W(A,b,x_{0},x_{1})-\int _{0}^{b}Q_{k}(b,s)f(s,x(s))ds\right) .\nonumber \\ \end{aligned}$$

(4.7)

and the operator ${\mathcal {F}}:PC(J,{\mathbb {R}}^{n})\rightarrow PC(J,{\mathbb {R}}^{n})$ as follows,

$$\begin{aligned} ({\mathcal {F}}x)(t)=W(A,t,x_{0},y_{0})+\int _{0}^{t}Q_{k}(t,s)\big (Bu_{1x}(s)+f(s,x(s))\big )ds,\quad t\in (t_{k},b]. \end{aligned}$$

Let

$$\begin{aligned} B_{r}=\{v\in PC(J,{\mathbb {R}}^{n})\big |\Vert v\Vert _{PC}\le r\}, \end{aligned}$$

and we use the notation

$$\begin{aligned}{} & {} (F_{1}x)(t)=W(A,t,x_{0},y_{0})+\int _{0}^{t}Q_{k}(t,s)Bu_{1x}(s)ds,\quad t\in (t_{k},b], \\{} & {} (F_{2}x)(t)=\int _{0}^{t}Q_{k}(t,s)f(s,x(s))ds, \quad t\in (t_{k},b]. \end{aligned}$$

Now ${\mathcal {F}}=F_{1}+F_{2}$.

We subdivide the proof into several steps.

Step 1 We show that for every $x,y\in B_{r}$, $F_{1}x+F_{2}y\in B_{r}$. In fact, for every $x,y\in B_{r}$, from (4.3), (4.5), (4.7) and $(H_{1})$, we have

$$\begin{aligned} \begin{aligned}&\Vert F_{1}x+F_{2}y\Vert _{PC}\\&\quad \le \Vert W(A,t,x_{0},y_{0})\Vert +\Big \Vert \int _{0}^{t}Q_{k}(t,s)Bu_{1x}(s)ds\Big \Vert +\Big \Vert \int _{0}^{t}Q_{k}(t,s)f(s,y(s))ds\Big \Vert \\&\quad \le \Vert W(A,t,x_{0},y_{0})\Vert +K^{2}\frac{1}{\gamma }\int _{0}^{b} \frac{T_{\varepsilon }^{6}}{\rho (A)+\varepsilon }e^{2\sqrt{\rho (A)+\varepsilon }(b-s)}ds\\&\qquad \cdot \left( \Vert x_{1}\Vert +\Vert W(A,b,x_{0},y_{0})\Vert +(N+Lr)\int _{0}^{b}\frac{T_{\varepsilon }^{3}}{\sqrt{\rho (A) +\varepsilon }}e^{\sqrt{\rho (A)+\varepsilon }(b-s)}ds\right) \\&\qquad +(N+Lr)\int _{0}^{b}\frac{T_{\varepsilon }^{3}}{\sqrt{\rho (A)+\varepsilon }}e^{\sqrt{\rho (A)+\varepsilon }(b-s)}ds\\&\quad \le \Vert W(A,t,x_{0},y_{0})\Vert + K^{2}\frac{1}{\gamma }\frac{T_{\varepsilon }^{6}}{2(\rho (A) +\varepsilon )^{3/2}}e^{2\sqrt{\rho (A)+\varepsilon }b}\\&\qquad \cdot \left( \Vert x_{1}\Vert +\Vert W(A,b,x_{0},y_{0})\Vert +(N+Lr)\frac{T_{\varepsilon }^{3}}{\rho (A) +\varepsilon }e^{\sqrt{\rho (A)+\varepsilon }b}\right) \\&\qquad +(N+Lr)\frac{T_{\varepsilon }^{3}}{\rho (A)+\varepsilon }e^{\sqrt{\rho (A)+\varepsilon }b}\\&\quad \le \Delta _{1}e^{\sqrt{\rho (A)+\varepsilon }b}(\Vert x_{0}\Vert +\Vert y_{0}\Vert ) +K^{2}\frac{1}{\gamma }\frac{T_{\varepsilon }^{6}}{2(\rho (A)+\varepsilon )^{3/2}}e^{2\sqrt{\rho (A)+\varepsilon }b}\\&\qquad \cdot \left( \Vert x_{1}\Vert +\Delta _{1}e^{\sqrt{\rho (A)+\varepsilon }b}(\Vert x_{0}\Vert +\Vert y_{0}\Vert )+(N+Lr)\frac{T_{\varepsilon }^{3}}{\rho (A)+\varepsilon }e^{\sqrt{\rho (A)+\varepsilon }b}\right) \\&\qquad +(N+Lr)\frac{T_{\varepsilon }^{3}}{\rho (A)+\varepsilon }e^{\sqrt{\rho (A)+\varepsilon }b}\\&\quad \le \Delta +L(K^{2}\frac{1}{\gamma }\frac{T_{\varepsilon }^{9}}{2\rho (A)^{5/2}}e^{3\sqrt{\rho (A) +\varepsilon }b}+\frac{T_{\varepsilon }^{3}}{\rho (A)}e^{\sqrt{\rho (A)+\varepsilon }b})r,\\ \end{aligned} \end{aligned}$$

where

$$\begin{aligned} \begin{aligned} \Delta =&\,\Delta _{1}e^{\sqrt{\rho (A)+\varepsilon }b}(\Vert x_{0}\Vert +\Vert y_{0}\Vert ) +N\left( \frac{T_{\varepsilon }^{3}}{\rho (A)}e^{\sqrt{\rho (A)+\varepsilon }b} +K^{2}\frac{1}{\gamma }\frac{T_{\varepsilon }^{9}}{2\rho (A)^{5/2}}e^{3\sqrt{\rho (A)+\varepsilon }b}\right) \\&+K^{2}\frac{1}{\gamma }\frac{T_{\varepsilon }^{6}}{2\rho (A)^{3/2}}e^{2\sqrt{\rho (A) +\varepsilon }b}\big (\Vert x_{1}\Vert +\Delta _{1}e^{\sqrt{\rho (A)+\varepsilon }b}(\Vert x_{0}\Vert +\Vert y_{0}\Vert )\big ). \end{aligned} \end{aligned}$$

By inequality (4.1), we can pick

$$\begin{aligned} r\ge \frac{\Delta }{1- L(K^{2}\frac{1}{\gamma }\frac{T_{\varepsilon }^{9}}{2\rho (A)^{5/2}}e^{3\sqrt{\rho (A) +\varepsilon }b}+\frac{T_{\varepsilon }^{3}}{\rho (A)}e^{\sqrt{\rho (A)+\varepsilon }b})}, \end{aligned}$$

(4.8)

Then, we have

$$\begin{aligned} \Vert F_{1}x+F_{2}y\Vert _{PC}\le r, \end{aligned}$$

that is

$$\begin{aligned} F_{1}x+F_{2}y\in B_{r}. \end{aligned}$$

Step 2 We claim that $F_{1}:B_{r}\rightarrow PC(J,{\mathbb {R}}^{n})$ is a contraction mapping. For every $x,y\in B_{r}$, by (3.5), (4.1), (4.3), and $(H_{1})$, we have

$$\begin{aligned} \begin{aligned} \Vert F_{1}x-F_{1}y\Vert _{PC}&=\Big \Vert \int _{0}^{t}Q_{k}(t,s)B(u_{1x}(s)-u_{1y}(s))ds\Big \Vert \\&\quad \le \frac{T_{\varepsilon }^{6}}{\rho (A)+\varepsilon }K^{2}\frac{1}{\gamma }\int _{0}^{t}e^{2\sqrt{\rho (A) +\varepsilon }(b-s)}ds\\ {}&\qquad \cdot \Big \Vert \int _{0}^{b}Q_{k}(b,s)\big (f(s,x(s))-f(s,y(s))\big )ds\Big \Vert \\&\quad \le \frac{T_{\varepsilon }^{9}K^{2}L}{2(\rho (A)+\varepsilon )^{5/2}}\frac{1}{\gamma }e^{3\sqrt{\rho (A) +\varepsilon }b}\Vert x-y\Vert _{PC}, \end{aligned} \end{aligned}$$

so $F_{1}$ is a contraction mapping.

Step 3 We show that $F_{2}$ is compact and continuous. For any $x,y\in B_{r}$, by the inequality (4.3), we have

$$\begin{aligned} \begin{aligned} \Vert F_{2}x-F_{2}y\Vert _{PC}\le&\int _{0}^{t}\Vert Q_{k}(t,s)\Vert \cdot \Vert f(s,x(s))-f(s,y(s))\Vert ds\\ \le&\frac{bLT_{\varepsilon }^{3}}{\rho (A)+\varepsilon }e^{\sqrt{\rho (A)+\varepsilon }b}\Vert x-y\Vert _{PC}, \end{aligned} \end{aligned}$$

therefore, $F_{2}:B_{r}\rightarrow PC(J,{\mathbb {R}})$ is continuous. To check the compactness of $F_{2}$, we prove that $F_{2}$ is uniformly bounded and equicontinuous. In fact, for any $x\in B_{r}$, by the inequality (4.3), we have

$$\begin{aligned} \begin{aligned} \Vert F_{2}x\Vert _{PC}=&\Big \Vert \int _{0}^{t}Q_{k}(t,s)f(s,x(s))ds\Big \Vert \\ \le&\frac{T_{\varepsilon }^{3}}{\rho (A)+\varepsilon }e^{\sqrt{\rho (A)+\varepsilon }b}(N+Lr), \end{aligned} \end{aligned}$$

that is $F_{2}B_{r}=\{F_{2}x\big |x\in B_{r}\}$ is uniformly bounded. Next, we show that $F_{2}$ is equicontinuous. For any $t_{k}<\tau _{1}\le \tau _{2}\le b$, by Lemma 2.7, and (4.3), we have

$$\begin{aligned} \begin{aligned} \Vert (F_{2}x)(\tau _{2})-(F_{2}x)(\tau _{1})\Vert \le&\,\Big \Vert \int _{0}^{\tau _{2}}Q_{k}(\tau _{2},s) f(s,x(s))ds-\int _{0}^{\tau _{1}}Q_{k}(\tau _{1},s)f(s,x(s))ds\Big \Vert \\ \le&\,\Big \Vert \int _{0}^{\tau _{1}}\big (Q_{k}(\tau _{2},s)-Q_{k}(\tau _{1},s)\big )f(s,x(s))ds\Big \Vert \\&+\Big \Vert \int _{\tau _{1}}^{\tau _{2}}Q_{k}(\tau _{2},s)f(s,x(s))ds\Big \Vert \\ \le&\,\theta _{1}(N+Lr)|\tau _{2}-\tau _{1}|+\frac{T_{\varepsilon }^{3}}{\sqrt{\rho (A)+\varepsilon }}e^{\sqrt{\rho (A)+\varepsilon }b}(N+Lr)\\&\cdot |\tau _{2}-\tau _{1}|, \end{aligned} \end{aligned}$$

therefore, $F_{2}B_{r}$ is the equicontinuous family of functions in $PC(J,{\mathbb {R}}^{n})$. From Lemma 2.5, $F_{2}B_{r}$ is relatively compact in $PC(J,{\mathbb {R}}^{n})$.

From Krasnoselskii’s fixed point theorem, we obtain that F has a fixed point x in $B_{r}$, which is the solution of (1.3) and satisfies $x(b)=x_{1}$.

In what follows, we show the controllability of the derivative of solutions for systems (1.3). Define the feedback control function

$$\begin{aligned} u_{2x}(t)=B^{*}Q_{k}'^{*}(b,t)(\Lambda _{0}^{b})^{-1}\big (y_{1} -W'(A,b,x_{0},y_{0})-\int _{0}^{b}Q_{k}'(b,s)f(s,x(s))ds\big ).\nonumber \\ \end{aligned}$$

and the operator ${\mathcal {H}}:PC^{1}(J,{\mathbb {R}}^{n})\rightarrow PC^{1}(J,{\mathbb {R}}^{n})$ as follows,

$$\begin{aligned} ({\mathcal {H}}x)(t)=W(A,t,x_{0},y_{0})+\int _{0}^{t}Q_{k}(t,s)\ \big (Bu_{2x}(s)+f(s,x(s))\big )ds,\quad t\in (t_{k},b]. \end{aligned}$$

Let

$$\begin{aligned}{} & {} (H_{1}x)(t)=W(A,t,x_{0},y_{0})+\int _{0}^{t}Q_{k}(t,s)Bu_{2x}(s)ds, \\{} & {} (H_{2}x)(t)=\int _{0}^{t}Q_{k}(t,s)f(s,x(s))ds, \end{aligned}$$

and ${\mathcal {H}}=H_{1}+H_{2}$. Let

$$\begin{aligned} D_{\ell }=\{x\in PC^{1}(J,{\mathbb {R}}^{n})\big |\Vert x\Vert _{PC^{1}}\le \ell \}. \end{aligned}$$

We show that ${\mathcal {H}}:D_{\ell }\rightarrow PC^{1}(J,{\mathbb {R}}^{n})$ has a fixed point. Proceeding as before, we subdivide the proof into several steps.

Step 1 We show that $H_{1}x+H_{2}y\in D_{\ell }$, for any $x,y\in D_{\ell }$.

In fact, for any $x,y\in D_{\ell }$, proceeding as in the proof for the operator ${\mathcal {F}}$, and by inequalities (3.5), (4.3)–(4.6) and condition $(H_{1})$, we have

$$\begin{aligned} \begin{aligned}&\Vert H_{1}x+H_{2}y\Vert _{PC}\\&\quad \le \Vert W(A,t,x_{0},y_{0})\Vert +\Big \Vert \int _{0}^{t}Q_{k}(t,s)Bu_{2x}(s)ds\Big \Vert +\Big \Vert \int _{0}^{t}Q_{k}(t,s)f(s,y(s))ds\Big \Vert \\&\quad \le \Vert W(A,t,x_{0},y_{0})\Vert +K^{2}\frac{1}{\lambda }\int _{0}^{b}\frac{T_{\varepsilon }^{6}}{\sqrt{\rho (A) +\varepsilon }}e^{2\sqrt{\rho (A)+\varepsilon }(b-s)}ds\\&\qquad \cdot \left( \Vert y_{1}\Vert +\Vert W'(A,b,x_{0},y_{0})\Vert +(N+Lr)\int _{0}^{b}T_{\varepsilon }^{3}e^{\sqrt{\rho (A)+\varepsilon }(b-s)}ds\right) \\&\qquad +(N+Lr)\int _{0}^{b}\frac{T_{\varepsilon }^{3}}{\sqrt{\rho (A)+\varepsilon }}e^{\sqrt{\rho (A) +\varepsilon }(b-s)}ds\\&\quad \le \Vert W(A,t,x_{0},y_{0})\Vert + K^{2}\frac{1}{\lambda }\frac{T_{\varepsilon }^{6}}{2(\rho (A) +\varepsilon )}e^{2\sqrt{\rho (A)+\varepsilon }b}\\&\qquad \cdot \left( \Vert y_{1}\Vert +\Vert W'(A,b,x_{0},y_{0})\Vert +(N+Lr)\frac{T_{\varepsilon }^{3}}{\sqrt{\rho (A) +\varepsilon }}e^{\sqrt{\rho (A)+\varepsilon }b}\right) \\&\qquad +(N+Lr)\frac{T_{\varepsilon }^{3}}{\rho (A)+\varepsilon }e^{\sqrt{\rho (A)+\varepsilon }b}\\&\quad \le \Delta _{1}e^{\sqrt{\rho (A)+\varepsilon }b}(\Vert x_{0}\Vert +\Vert y_{0}\Vert )+K^{2}\frac{1}{\lambda }\frac{T_{\varepsilon }^{6}}{2(\rho (A)+\varepsilon )}e^{2\sqrt{\rho (A)+\varepsilon }b}\\&\qquad \cdot \left( \Vert y_{1}\Vert +\Delta _{1}e^{\sqrt{\rho (A)+\varepsilon }b}(\Vert x_{0}\Vert +\Vert y_{0}\Vert ) +(N+Lr)\frac{T_{\varepsilon }^{3}}{\sqrt{\rho (A)+\varepsilon }}e^{\sqrt{\rho (A)+\varepsilon }b}\right) \\&\qquad +(N+Lr)\frac{T_{\varepsilon }^{3}}{\rho (A)+\varepsilon }e^{\sqrt{\rho (A)+\varepsilon }b}\\&\quad \le {\overline{\Delta }}+L\left( K^{2}\frac{1}{\lambda }\frac{T_{\varepsilon }^{9}}{2\rho (A)^{3/2}}e^{3\sqrt{\rho (A) +\varepsilon }b}+\frac{T_{\varepsilon }^{3}}{\rho (A)}e^{\sqrt{\rho (A)+\varepsilon }b} \right) r, \end{aligned} \end{aligned}$$

where

$$\begin{aligned} \begin{aligned} {\overline{\Delta }}=&\,\Delta _{1}e^{\sqrt{\rho (A) +\varepsilon }b}(\Vert x_{0}\Vert +\Vert y_{0}\Vert )+N\left( \frac{T_{\varepsilon }^{3}}{\rho (A)}e^{\sqrt{\rho (A) +\varepsilon }b}+K^{2}\frac{1}{\lambda }\frac{T_{\varepsilon }^{9}}{2\rho (A)^{3/2}}e^{3\sqrt{\rho (A)+\varepsilon }b}\right) \\&+K^{2}\frac{1}{\lambda }\frac{T_{\varepsilon }^{6}}{2\rho (A)^{3/2}}e^{2\sqrt{\rho (A) +\varepsilon }b}\big (\Vert y_{1}\Vert +\Delta _{1}e^{\sqrt{\rho (A)+\varepsilon }b}(\Vert x_{0}\Vert +\Vert y_{0}\Vert )\big ). \end{aligned} \end{aligned}$$

By inequality (4.2), we can pick

$$\begin{aligned} r\ge \frac{{\overline{\Delta }}}{1- L \left( K^{2}\frac{1}{\lambda } \frac{T_{\varepsilon }^{9}}{2\rho (A)^{3/2}}e^{3\sqrt{\rho (A) +\varepsilon }b}+\frac{T_{\varepsilon }^{3}}{\rho (A)}e^{\sqrt{\rho (A)+\varepsilon }b} \right) }, \end{aligned}$$

(4.9)

then, we have

$$\begin{aligned} \Vert H_{1}x+H_{2}y\Vert _{PC}\le r. \end{aligned}$$

It follows that for any $x,y\in D_{r}$, $H_{1}x+H_{2}y\in D_{r}$.

Similarly, making use of (3.5), (4.4), (4.6), (4.7) and $(H_{1})$, we get

$$\begin{aligned} \begin{aligned}&\Vert (H_{1}x)'+(H_{2}y)'\Vert _{PC}\\&\quad \le \Vert W'(A,t,x_{0},y_{0})\Vert +\Big \Vert \int _{0}^{t}Q_{k}'(t,s)Bu_{2x}(s)ds\Big \Vert +\Big \Vert \int _{0}^{t}Q_{k}'(t,s)f(s,y(s))ds\Big \Vert \\&\quad \le \Vert W'(A,t,x_{0},y_{0})\Vert +K^{2}T_{\varepsilon }^{6}\frac{1}{\lambda }\int _{0}^{b}e^{2\sqrt{\rho (A)+\varepsilon }(b-s)}ds\\&\qquad \cdot \left( \Vert y_{1}\Vert +\Vert W'(A,b,x_{0},y_{0})\Vert +(N+Lr)\int _{0}^{b}T_{\varepsilon }^{3}e^{\sqrt{\rho (A)+\varepsilon }(b-s)}ds\right) \\&\qquad +(N+Lr)\int _{0}^{b}T_{\varepsilon }^{3}e^{\sqrt{\rho (A)+\varepsilon }(b-s)}ds\\&\quad \le \Delta _{1}e^{\sqrt{\rho (A)+\varepsilon }b}(\Vert x_{0}\Vert +\Vert y_{0}\Vert )+\frac{K^{2}T_{\varepsilon }^{6}}{2\sqrt{\rho (A) +\varepsilon }}\frac{1}{\lambda }e^{2\sqrt{\rho (A)+\varepsilon }b}\\&\qquad \cdot \left( \Vert y_{1}\Vert +\Delta _{1}e^{\sqrt{\rho (A) +\varepsilon }b}(\Vert x_{0}\Vert +\Vert y_{0}\Vert )+\frac{T_{\varepsilon }^{3}}{\sqrt{\rho (A) +\varepsilon }}(N+Lr)e^{\sqrt{\rho (A)+\varepsilon }b}\right) \\&\qquad +(N+Lr)\frac{T_{\varepsilon }^{3}}{\sqrt{\rho (A) +\varepsilon }}e^{\sqrt{\rho (A)+\varepsilon }b}\\&\quad \le \Gamma +L \left( K^{2}\frac{1}{\lambda }\frac{T_{\varepsilon }^{9}}{2\rho (A)}e^{3\sqrt{\rho (A) +\varepsilon }b}+\frac{T_{\varepsilon }^{3}}{\sqrt{\rho (A)}}e^{\sqrt{\rho (A)+\varepsilon }b} \right) r,\\ \end{aligned} \end{aligned}$$

where

$$\begin{aligned} \begin{aligned} \Gamma =&\,K^{2}\frac{1}{\lambda }\frac{T_{\varepsilon }^{6}}{2\sqrt{\rho (A)}}e^{2\sqrt{\rho (A) +\varepsilon }b}\big (\Vert y_{1}\Vert +\Delta _{1}e^{\sqrt{\rho (A)+\varepsilon }b}(\Vert x_{0}\Vert +\Vert y_{0}\Vert )\big )\\&+\Delta _{1}e^{\sqrt{\rho (A)+\varepsilon }b}(\Vert x_{0}\Vert +\Vert y_{0}\Vert ) +N\left( \frac{T_{\varepsilon }^{3}}{\sqrt{\rho (A)}}e^{\sqrt{\rho (A) +\varepsilon }b}+K^{2}\frac{T_{\varepsilon }^{9}}{2\rho (A)}\frac{1}{\lambda }e^{3\sqrt{\rho (A)+\varepsilon }b}\right) . \end{aligned} \end{aligned}$$

By inequality (4.2), we can pick

$$\begin{aligned} r\ge \frac{\Gamma }{1- L(K^{2}\frac{1}{\lambda } \frac{T_{\varepsilon }^{9}}{2\rho (A)^{2}}e^{3\sqrt{\rho (A)+\varepsilon }b} +\frac{T_{\varepsilon }^{3}}{\sqrt{\rho (A)}}e^{\sqrt{\rho (A)+\varepsilon }b})}, \end{aligned}$$

(4.10)

Then, we have

$$\begin{aligned} \Vert (H_{1}x)'+(H_{2}y)'\Vert _{PC}\le r. \end{aligned}$$

Take r be the maximum of the right hand of (4.8) and (4.9), then we obtain

$$\begin{aligned} \Vert H_{1}x+H_{2}y\Vert _{PC^{1}}\le r, \end{aligned}$$

that is

$$\begin{aligned} H_{1}x+H_{2}y\in B_{r}. \end{aligned}$$

Step 2 We state that $H_{1}$ is a contraction.

For any $x,y\in D_{r}$, by inequalities (3.5), (4.2), (4.4), and $(H_{1})$, we have

$$\begin{aligned} \begin{aligned} \Vert H_{1}x-H_{1}y\Vert _{PC}=&\,\Big \Vert \int _{0}^{t}Q_{k}(t,s)B(u_{2x}(s)-u_{2y}(s))ds\Big \Vert \\ \le&\,LK^{2}\frac{T_{\varepsilon }^{6}}{\sqrt{\rho (A)+\varepsilon }}\frac{1}{\lambda } \int _{0}^{t}e^{2\sqrt{\rho (A)+\varepsilon }(b-s)}ds\cdot \int _{0}^{b}\Vert Q_{k}'(b,s)\Vert \\&\cdot \Vert x-y\Vert _{PC}ds\\ \le&\,LK^{2}T_{\varepsilon }^{9}\frac{1}{\lambda }\frac{1}{2\rho (A)^{3/2}}e^{3\sqrt{\rho (A) +\varepsilon }b}\Vert x-y\Vert _{PC}, \end{aligned} \end{aligned}$$

and

$$\begin{aligned} \begin{aligned} \Vert (H_{1}x)'-(H_{1}y)'\Vert _{PC}&=\,\Big \Vert \int _{0}^{t}Q_{k}'(t,s)B(u_{2x}(s)-u_{2y}(s))ds\Big \Vert \\&\le \, LT_{\varepsilon }^{6}K^{2}\frac{1}{\lambda }\int _{0}^{t}e^{2\sqrt{\rho (A) +\varepsilon }(b-s)}ds\\ {}&\quad \cdot \int _{0}^{b}\Vert Q_{k}'(b,s)\Vert \cdot \Vert x-y\Vert _{PC}ds\\&\le \,\frac{T_{\varepsilon }^{9}K^{2}L}{2(\rho (A)+\varepsilon )}\frac{1}{\lambda }e^{3\sqrt{\rho (A) +\varepsilon }b}\Vert x-y\Vert _{PC}, \end{aligned} \end{aligned}$$

hence according to (4.2), there exists $\varepsilon >0$ small enough such that

$$\begin{aligned} LK^{2}T_{\varepsilon }^{9}\frac{1}{\lambda }\frac{1}{2\rho (A)^{3/2}}e^{3\sqrt{\rho (A)+\varepsilon }b}<1, \end{aligned}$$

and

$$\begin{aligned} \frac{T_{\varepsilon }^{9}K^{2}L}{2(\rho (A)+\varepsilon )} \frac{1}{\lambda }e^{3\sqrt{\rho (A)+\varepsilon }b}<1. \end{aligned}$$

Therefore, $H_{1}$ is a contraction mapping.

Step 3 We show that $H_{2}$ is compact and continuous. Since, for every $x,y\in D_{\ell }$, by (4.3) and (4.4), we have

$$\begin{aligned} \begin{aligned} \Vert H_{2}x-H_{2}y\Vert _{PC}\le&\int _{0}^{t}\Vert Q_{k}(t,s)\Vert \cdot \Vert f(s,x(s))-f(s,y(s))\Vert ds\\ \le&\frac{bLT_{\varepsilon }^{3}}{\rho (A)+\varepsilon }e^{\sqrt{\rho (A)+\varepsilon }b}\Vert x-y\Vert _{PC^{1}}, \end{aligned} \end{aligned}$$

and

$$\begin{aligned} \begin{aligned} \Vert (H_{2}x)'-(H_{2}y)'\Vert _{PC}\le&\int _{0}^{t}\Vert Q_{k}'(t,s)\Vert \cdot \Vert f(s,x(s))-f(s,y(s))\Vert ds\\ \le&\frac{bLT_{\varepsilon }^{3}}{\sqrt{\rho (A) +\varepsilon }}e^{\sqrt{\rho (A)+\varepsilon }b}\Vert x-y\Vert _{PC^{1}}, \end{aligned} \end{aligned}$$

therefore, $H_{2}:D_{\ell }\rightarrow PC^{1}(J,{\mathbb {R}})$ is continuous. To check the compactness of $H_{2}$, we consider the mapping

$$\begin{aligned} (H_{2}x)'(t)=\int _{0}^{t}Q_{k}'(t,s)f(s,x(s))ds. \end{aligned}$$

For every $x\in D_{\ell }$, by inequality (4.4) and $(H_{1})$, we obtain

$$\begin{aligned} \begin{aligned} \Vert (H_{2}x)'\Vert _{PC}\le&\int _{0}^{t}\Vert Q_{k}'(t,s)\Vert \cdot \Vert f(s,x(s))\Vert ds\\ \le&\int _{0}^{t}T_{\varepsilon }^{3}e^{\sqrt{\rho (A)+\varepsilon }(t-s)}(N+Lr)ds\\ \le&\, bT_{\varepsilon }^{3}(N+Lr)e^{\sqrt{\rho (A)+\varepsilon }b}, \end{aligned} \end{aligned}$$

which implies $(H_{2}D_{\ell })'=\{(H_{2}x)'\big |x\in D_{\ell }\}$ is uniformly bounded in $PC(J,{\mathbb {R}})$. We prove that for any $x\in D_{\ell }$, $(H_{2}x)'$ is equicontinuous. In fact, for any $t_{k}<\tau _{1}<\tau _{2}\le b$, in term of inequality (4.4), $(H_{1})$ and Lemma 2.7, we have

$$\begin{aligned} \begin{aligned}&\Vert (H_{2}x)'(\tau _{2})-(H_{2}x)'(\tau _{1})\Vert \\ {}&\quad =\Big \Vert \int _{0}^{\tau _{2}}Q_{k}'(\tau _{2},s)f(s,x(s))ds -\int _{0}^{\tau _{1}}Q_{k}'(\tau _{1},s)f(s,x(s))ds\Big \Vert \\&\quad \le \int _{0}^{\tau _{1}}\Vert Q_{k}'(\tau _{2},s)-Q_{k}'(\tau _{1},s)\Vert \cdot \Vert f(s,x(s))\Vert ds\\&\qquad +\int _{\tau _{1}}^{\tau _{2}}\Vert Q_{k}'(\tau _{2},s)f(s,x(s))\Vert ds\\&\quad \le \,\theta (N+Lr)|\tau _{2}-\tau _{1}|+T_{\varepsilon }^{3}e^{\sqrt{\rho (A) +\varepsilon }b}(N+Lr)\cdot |\tau _{2}-\tau _{1}|, \end{aligned} \end{aligned}$$

therefore, $(H_{2}D_{\ell })'$ is the equicontinuous family of functions in $PC(J,{\mathbb {R}}^{n})$. From Lemma 2.5, $(H_{2}D_{\ell })'$ is relatively compact in $PC(J,{\mathbb {R}}^{n})$. Hence, for any sequence $\{x_{n}\}\subset D_{\ell }$, there exists a subsequence of $\{x_{n}\}$, again denoted by $\{x_{n}\}$, such that

$$\begin{aligned} (H_{2}x_{n})'\rightarrow \phi \quad \text {in} \quad PC(J,{\mathbb {R}}^{n}) \quad \text {as} \quad n\rightarrow \infty . \end{aligned}$$

(4.11)

Obviously,

$$\begin{aligned} \Vert x\Vert _{PC}\le b\Vert x'\Vert _{PC}, \end{aligned}$$

for any $x\in PC^{1}(J,{\mathbb {R}}^{n})$. Let ${\overline{\phi }}$ be the antiderivative of $\phi$, combining this inequality with (4.11), we have

$$\begin{aligned} \begin{aligned} \Vert H_{2}x_{n}-{\overline{\phi }}\Vert _{PC^{1}}=&\max \{\Vert H_{2}x_{n} -{\overline{\phi }}\Vert _{PC},\Vert (H_{2}x_{n})'-\phi \Vert _{PC}\}\\ \le&\max \{b,1\}\Vert (H_{2}x_{n})'-\phi \Vert _{PC}\\ <&\varepsilon , \end{aligned} \end{aligned}$$

as n is large enough, which implies that for any $\{H_{2}x_{n}\}\subset H_{2}D_{\ell }$, there exists a subsequence $\{H_{2}x_{n_{k}}\}$ which is convergence in $PC^{1}(J,{\mathbb {R}})$ . Thus, $H_{2}:D_{\ell }\rightarrow PC^{1}(J,{\mathbb {R}}^{n})$ is a compact and continuous operator.

Hence, by the Krasnoselskii’s fixed point theorem, we obtain that ${\mathcal {H}}$ has a fixed point x in $D_{\ell }$ which is the solution of (1.3) and satisfies $x'(b)=y_{1}$.

In conclusion, second-order impulsive systems (1.3) are exactly controllable.

$\square$

5 Examples

In this section, we give some examples to illustrate the effectiveness of our results.

Example 5.1

For the simplicity of calculation, we consider the controllability of systems (1.2) with

$$\begin{aligned} A= \begin{pmatrix} 1 &{}\quad 0\\ 0 &{}\quad 2 \end{pmatrix}, \quad B_{1}=B_{2}= \begin{pmatrix} 1 &{}\quad 2\\ 0 &{}\quad 3 \end{pmatrix}, \quad B= \begin{pmatrix} 2 &{}\quad 0\\ 0 &{}\quad 3 \end{pmatrix}, \quad x_{0}= \begin{pmatrix} 0\\ 1 \end{pmatrix}, \quad y_{0}= \begin{pmatrix} 1\\ 0 \end{pmatrix}, \end{aligned}$$

and $0=t_{0}<1=t_{1}< 2=t_{2}=b$. Obviously, B is nonsingular, then Theorem 3.1 holds for $l=0$, that is system (1.2) is controllable. For the sake of convenience in calculating, we consider $x_{1}=(30~40)^{T}$, then we show that we can choose a control function $u_{1}(t)$ such that, under $u_{1}(t)$, $x(2)=(30~40)^\textrm{T}$. By Theorem 3.1 in [34], we obtain that, for $t\in (1,2]$, the solution $W(A,t,x_{0},y_{0})$ of the homogeneous initial value problems of (1.2) is expressed as follows,

$$\begin{aligned} \begin{aligned} W(A,t,x_{0},y_{0})&= \begin{pmatrix} 2\cosh t &{}\quad 2\cosh \sqrt{2}t\\ 0 &{}\quad 4\cosh \sqrt{2}t \end{pmatrix}x_{0} + \begin{pmatrix} 2\sinh t &{}\quad \sqrt{2}\sinh \sqrt{2}t\\ 0 &{}\quad 2\sqrt{2}\sinh \sqrt{2}t \end{pmatrix}y_{0}\\&=\begin{pmatrix} 2\sinh t+2\cosh \sqrt{2}t\\ 4\cosh \sqrt{2}t \end{pmatrix}. \end{aligned} \end{aligned}$$

(5.1)

By the calculation, we find

$$\begin{aligned} Q_{1}(t,s)=\left\{ \begin{aligned}&W_{0}(A,t,s),\quad 0\le s\le 1,\\&W_{1}(A,t,s),\quad 1< s\le t, \end{aligned} \right. \end{aligned}$$

here

$$\begin{aligned} W_{0}(A,t,s)= \begin{pmatrix} 2\sinh (t-s) &{}\quad \sqrt{2}\sinh \sqrt{2}(t-s)\\ 0 &{}\quad 2\sqrt{2}\sinh \sqrt{2}(t-s) \end{pmatrix}, \\ W_{1}(A,t,s)= \begin{pmatrix} \sinh (t-s) &{} 0\\ 0 &{} \frac{1}{\sqrt{2}}\sinh \sqrt{2}(t-s) \end{pmatrix}. \end{aligned}$$

and

$$\begin{aligned} \Gamma _{0}^{2}= \begin{pmatrix} \begin{aligned} -3\sinh 2+4\sinh 4-\frac{9}{2\sqrt{2}}\sinh 2\sqrt{2}&{}+\frac{9}{2\sqrt{2}}\sinh 4\sqrt{2}-19 &{}\\ &{}-\frac{9}{\sqrt{2}}\sinh 2\sqrt{2}+\frac{9}{\sqrt{2}}\sinh 4\sqrt{2}-18\\ -\frac{9}{\sqrt{2}}\sinh 2\sqrt{2}+\frac{9}{\sqrt{2}}\sinh 4\sqrt{2}-18&{} \\ &{}-\frac{18}{\sqrt{2}}\sinh 2\sqrt{2}+\frac{18}{\sqrt{2}}\sinh 4\sqrt{2}-36 \end{aligned} \end{pmatrix}_{2\times 2}. \end{aligned}$$

(5.2)

Fig. 1

A The control function $u_{1}(t)$, where blue line denote the first component of $u_{1}(t)$, red line denote the second component of $u_{1}(t)$. B The state function X(t) of (1.2) without any control, similarly, where blue line denote the first component of X(t), red line denote the second component of X(t). C Figure C1 denote the first component of state function $x_{1}(t)$ under the control function $u_{1}(t)$, and Figure C2 denote the second component of state function $x_{1}(t)$ under the control function $u_{1}(t)$. It should be noted that, in figure C1 and C2, on the left side of impulsive point $t=1$, we refer to the left scale, and on the right side of impulsive point $t=1$, we refer to the right scale (color figure online)

Fig. 2

E1 The state function X(t) of (1.2) without any control function. The blue line denote the first component of X(t), and the red line denote the second component of X(t). E2 The derivative of state function X(t) without any control function. Similarly, the blue line denote the first component, and the red line denote the second component. F The control function $u_{2}$ we picked to control the derivative function, the blue line denote the first component of $u_{2}$, and the red line denote the second component $u_{2}$. G1 The state function $x_{2}(t)$ under the control $u_{2}$. The blue line denote the first component of $x_{2}(t)$, and the red line denote the second component $x_{2}(t)$. G2 The derivative of state function $x_{2}(t)$. The blue line denote the first component, and the red line denote the second component (color figure online)

Hence, by (3.3), we can define the control function $u_{1}(t)$ by a piecewise function,piecewise function,

$$\begin{aligned} u_{1}(t)=\left\{ \begin{aligned}&\begin{pmatrix} 4\sinh (2-t) &{}\quad 0\\ 3\sqrt{2}\sinh \sqrt{2}(2-t) &{}\quad 6\sqrt{2}\sinh \sqrt{2}(2-t) \end{pmatrix} (\Gamma _{0}^{2})^{-1}\big (x_{1}-W(A,2,x_{0},y_{0})\big ),\quad t\in (0,1],\\&\begin{pmatrix} 2\sinh (2-t) &{}\quad 0\\ 0 &{}\quad \frac{3}{\sqrt{2}}\sinh \sqrt{2}(2-t) \end{pmatrix} (\Gamma _{0}^{2})^{-1}\big (x_{1}-W(A,2,x_{0},y_{0})\big ),\quad t\in (1,2], \end{aligned} \right. \end{aligned}$$

here $W(A,t,x_{0},y_{0})$ is expressed by (5.1), $\Gamma _{0}^{2}$ is expressed by (5.2), $x_{1}$ is the state we want to arrive. Therefore, under the control $u_{1}$, we have $x(2)=x_{1}$, see Fig. 1. Similarly, take $y_{1}=(0~0)^\textrm{T}$, then we can take the control $u_{2}$ as follows,

$$\begin{aligned} u_{2}(t)=\left\{ \begin{aligned}&\begin{pmatrix} 4\cosh (2-t) &{}\quad 0\\ 6\cosh \sqrt{2}(2-t) &{}\quad 12\cosh \sqrt{2}(2-t) \end{pmatrix} (\Lambda _{0}^{2})^{-1}\big (y_{1}-W'(A,2,x_{0},y_{0})\big ),\quad t\in (0,1],\\&\begin{pmatrix} 2\cosh (2-t) &{}\quad 0\\ 0 &{}\quad 3\cosh \sqrt{2}(2-t) \end{pmatrix} (\Lambda _{0}^{2})^{-1}\big (y_{1}-W'(A,2,x_{0},y_{0})\big ),\quad t\in (1,2], \end{aligned} \right. \end{aligned}$$

and under this control, we can steer the derivative of the solution of systems (1.2) to $(0~0)^{T}$ at terminal, see Fig. 2.

Example 5.2

Consider the systems (1.3) with

$$\begin{aligned} A= \begin{pmatrix} 2 &{}\quad 0\\ 0 &{}\quad 2 \end{pmatrix}, \quad B_{1}=B_{2}= \begin{pmatrix} -\frac{1}{2} &{}\quad \frac{1}{2}\\ 0 &{}\quad -\frac{1}{3} \end{pmatrix}, \quad B= \begin{pmatrix} \frac{1}{2} &{}\quad 0\\ 0 &{}\quad \frac{1}{2} \end{pmatrix}, \end{aligned}$$

$0=t_{0}<\frac{1}{2}=t_{1}< 1=t_{2}=b$, and $f(t,x(t))=\frac{1}{37}\sin x(t)$. Since B is nonsingular, by Theorem 3.3, we find condition $(H_{2})$ is satisfied. Obviously, $(H_{1})$ is satisfied with $L=\frac{1}{37}$. Then we show that (4.1) and (4.2) hold. Define a new matrix norm $\Vert \cdot \Vert '$ by

$$\begin{aligned} \Vert x\Vert '=\Vert Qx\Vert , \quad \forall x\in {\mathbb {R}}^{n}, \end{aligned}$$

where

$$\begin{aligned} Q=\begin{pmatrix} 1&{}\quad 0\\ 0&{}\quad 2\varepsilon \end{pmatrix}, \end{aligned}$$

and $\Vert A\Vert =\max \limits _{1\le j\le n}\sum \nolimits _{i=1}^{n}|a_{ij}|$. According the Theorem 2.2.8 of [35], we find for $A_{1}=I+\frac{B_{1}+B_{2}}{2}$,

$$\begin{aligned} \Vert A_{1}\Vert '=\Vert Q^{-1}A_{1}Q\Vert \le \rho (A_{1})+\varepsilon . \end{aligned}$$

(5.3)

Let $\varepsilon =\frac{49}{100}$, then we have, for all $x\in {\mathbb {R}}^{n}$,

$$\begin{aligned} \frac{98}{100}\Vert x\Vert \le \Vert x\Vert '\le \Vert x\Vert , \end{aligned}$$

hence,

$$\begin{aligned} \Vert A_{1}\Vert =\sup _{x\ne 0}\left\{ \frac{\Vert A_{1}x\Vert }{\Vert x\Vert } \right\} \le \sup _{x\ne 0} \left\{ \frac{100}{98}\frac{\Vert A_{1}x\Vert '}{\Vert x\Vert '}\right\} =\frac{100}{98}\Vert A_{1}\Vert ', \end{aligned}$$

combining this with (5.3), we have

$$\begin{aligned} \Vert A_{1}\Vert \le \frac{100}{98}(\rho (A)+\varepsilon ), \end{aligned}$$

that is $T_{A_{1},\frac{49}{100}}=\frac{100}{98}$. Obviously, $T_{A,\varepsilon }=1$, therefore, $T_{\frac{49}{100}}=\frac{100}{98}$. With a simple calculation, we find

$$\begin{aligned} \Gamma _{0}^{1}= \begin{pmatrix} a_{11}&{}\quad a_{12}\\ a_{21}&{}\quad a_{22} \end{pmatrix}, \end{aligned}$$

here

$$\begin{aligned}{} & {} a_{11}=\frac{\sinh \sqrt{2}\cosh \sqrt{2}+\frac{1}{2} -\sinh \frac{\sqrt{2}}{2}\cosh \frac{\sqrt{2}}{2}}{32\sqrt{2}}+\frac{\frac{1}{2} +\sinh \frac{\sqrt{2}}{2}\cosh \frac{\sqrt{2}}{2}}{16\sqrt{2}}, \\{} & {} a_{12}=\frac{\sinh \sqrt{2}\cosh \sqrt{2}+\frac{1}{2} -\sinh \frac{\sqrt{2}}{2}\cosh \frac{\sqrt{2}}{2}}{48\sqrt{2}}, \\{} & {} a_{21}= \frac{\sinh \sqrt{2}\cosh \sqrt{2}+\frac{1}{2} -\sinh \frac{\sqrt{2}}{2}\cosh \frac{\sqrt{2}}{2}}{48\sqrt{2}}, \\{} & {} a_{22}=\frac{\sinh \sqrt{2}\cosh \sqrt{2}+\frac{1}{2} -\sinh \frac{\sqrt{2}}{2}\cosh \frac{\sqrt{2}}{2}}{36\sqrt{2}} +\frac{\frac{1}{2}+\sinh \frac{\sqrt{2}}{2}\cosh \frac{\sqrt{2}}{2}}{16\sqrt{2}}, \end{aligned}$$

hence, for all $x\in {\mathbb {R}}^{n}$,

$$\begin{aligned} (\Gamma _{0}^{1}x,x)>\frac{1}{10}\Vert x\Vert ^{2}, \end{aligned}$$

that is we can pick $\gamma =\frac{1}{10}$. Hence we obtain

$$\begin{aligned}{} & {} L \left( K^{2}\frac{T_{\varepsilon }^{9}}{2\gamma \rho (A)^{5/2}}e^{3\sqrt{\rho (A)}b} +\frac{T_{\varepsilon }^{3}}{\sqrt{\rho (A)}}e^{\sqrt{\rho (A)}b}\right) \nonumber \\ {}{} & {} \quad =\frac{1}{37}\left( \frac{(\frac{100}{98})^{9}e^{3\sqrt{2}}}{3.2\sqrt{2}} +\frac{(\frac{100}{98})^{2}e^{\sqrt{2}}}{\sqrt{2}}\right) <1. \end{aligned}$$

(5.4)

Similarly, we can get

$$\begin{aligned} \Lambda _{0}^{1}= \begin{pmatrix} b_{11}&{}\quad b_{12}\\ b_{21}&{}\quad b_{22} \end{pmatrix}, \end{aligned}$$

here

$$\begin{aligned}{} & {} b_{11}= \frac{\sinh \sqrt{2}\cosh \sqrt{2}-\frac{1}{2} -\sinh \frac{\sqrt{2}}{2}\cosh \frac{\sqrt{2}}{2}}{16\sqrt{2}} +\frac{\sinh \frac{\sqrt{2}}{2}\cosh \frac{\sqrt{2}}{2}-\frac{1}{2}}{8\sqrt{2}}, \\{} & {} b_{12}=\frac{\sinh \sqrt{2}\cosh \sqrt{2}-\frac{1}{2} -\sinh \frac{\sqrt{2}}{2}\cosh \frac{\sqrt{2}}{2}}{24\sqrt{2}}, \\{} & {} b_{21}= \frac{\sinh \sqrt{2}\cosh \sqrt{2}-\frac{1}{2} -\sinh \frac{\sqrt{2}}{2}\cosh \frac{\sqrt{2}}{2}}{24\sqrt{2}}, \\{} & {} b_{22}= \frac{\sinh \sqrt{2}\cosh \sqrt{2}-\frac{1}{2} -\sinh \frac{\sqrt{2}}{2}\cosh \frac{\sqrt{2}}{2}}{18\sqrt{2}} +\frac{\sinh \frac{\sqrt{2}}{2}\cosh \frac{\sqrt{2}}{2}-\frac{1}{2}}{8\sqrt{2}}, \end{aligned}$$

hence, for all $x\in {\mathbb {R}}^{n}$,

$$\begin{aligned} (\Lambda _{0}^{1}x,x)>\frac{1}{10}\Vert x\Vert ^{2}, \end{aligned}$$

i.e., we can pick $\lambda =\frac{1}{10}$. Put these constants into (4.2), we obtain

$$\begin{aligned} L\left( K^{2}T_{\varepsilon }^{9}\frac{1}{2\lambda }\mu e^{3\sqrt{\rho (A)}b} +\frac{T_{\varepsilon }^{3}}{\sqrt{\rho (A)}}e^{\sqrt{\rho (A)}b}\right) <1. \end{aligned}$$

(5.5)

Therefore, all the assumptions of Theorem 4.1 are satisfied, that is the systems (1.3) are exactly controllable.

6 Conclusion

In this paper, we introduce a new definition of controllability, and obtain some sufficient and necessary conditions of second-order linear impulsive systems, the rank criterion of impulsive systems is obtained as well. Then, we present some sufficient conditions of nonlinear impulsive systems provided the linear systems are controllable. Finally, some examples are presented to illustrate our results.