1 Introduction

At present, the stability of the nonlinear system control methods are: intermittent control [1–7], adaptive fuzzy control [8, 9], alternate control [10, 11], impulsive control [12–20], nonimpulsive control, continuous control, and so on [21–28] and so on. Our goal is a good system control. We will design some better control systems by studying the system control methods that are currently commonly used.

To make a nonlinear system stable, in this paper, first of all, each system period is divided into two equal parts. In the first part of the period, there are continuous inputs \(Cx(t)\), and in the other part, there is no input. We call it intermittent control system. Figure 1 provides the working principles of the intermittent control system. Next, we can add the impulsive control in the intermittent control system to control its better stability. We add an input of the impulsive J1 to the system at the middle of each period. Similarly, we add an impulse J2 at the end of each period of the system. In this way, there are a continuous control \(Cx(t)\) and impulsive controls J1 and J2 in each period of the system. We call it an alternate-continuous-control system with double-impulse. Figure 2 provides the working principles of alternate-continuous-control systems with double-impulse. The mathematical model proposed in this paper can be applied in many fields. In the medical field, it can be used in large-scale surgery to control patient’s life characteristics. In the field of electronics, it can be used to control a variety of chaotic circuits, such as circuits that incorporate memristors. In the field of intelligent robots, it can be used to control the robot walking gait in real time. It can also play a great advantage in the field of environmental pollution control [29–34].

Figure 1
figure 1

Working principle of the intermittent control system: in the first part of the period, there are continuous inputs \(\pmb{Cx(t)}\) , and in the other part, there is no input (we can write 0).

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Figure 2
figure 2

Alternate-continuous-control systems with double-impulse: in the first part of the period, there are continuous inputs \(\pmb{Cx(t)}\) , and in the other part, there is no input (we can write 0). We add an impulse J1 to the system at the middle of each period and an impulse J2 at the end of each period.

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In this paper, we construct alternate-continuous-control systems with double-impulse. We investigate the stability of the systems and get an exponential stability criterion in terms of a set of linear matrix inequalities. Some mathematical proofs make our conclusions reliable. At last, the Chua oscillator is controlled by using the results obtained.

The rest of this paper is organized as follows. In Section 3, we introduce some basic mathematical knowledge and mathematical symbols about system control, and two commonly used mathematical lemmas will be involved. In Section 4, we give the results of this paper and an exponential stability criterion, the main theory of this paper. In Section 5, we use this conclusion to control the Chua oscillator. Lastly, we give a summary of this paper.

2 Problem formulation and preliminaries

A classic nonlinear system can be described as

$$ \textstyle\begin{cases} \dot{x}(t) = Hx(t)+f(x(t))+u(t), \\ x(t_{0}) = x_{0}, \end{cases} $$
(1)

where \(x\in R^{n}\) is a state vector, \(H\in R^{n\times n}\) is a constant matrix, \(f:R^{n}\rightarrow R^{n}\) is a continuous nonlinear function satisfying \(f(0)=0\); we assume that there exists a diagonal matrix \(L=\operatorname{diag}(a_{1},a_{2},\ldots,a_{n})\geq0\) such that \(\Vert f(x) \Vert ^{2}\leq x^{T}Lx\) for all \(x\in R^{n}\). \(Hx(t)\) is the linear part of the system, \(f(x(t))\) is the nonlinear interference, and \(u(t)\) is the external input to system (1).

To stabilize the origin of system (1) by means of alternate-continuous-control systems with double-impulse, we impose two kinds of control, that is, assuming that the period of the system is T and m is a nonnegative integer, from mT to \(mT+\frac{T}{2}\), we set \(u(t)=Cx(t)\), where \(C\in R^{n\times n}\) is a constant matrix, at time \(mT+\frac{T}{2}\), an impulse J1 given, from \(mT+\frac{T}{2}\) to \((m+1)T\), no input to the system is given, and an impulse J2 is given to the system at time \((m+1)T\).

So system (1) can be redefined as

$$ \textstyle\begin{cases} \dot{x}(t) = Hx(t)+f(x(t))+Cx(t), & mT< t< mT+\frac{T}{2},\\ {x}(t) = x(t^{-})+J1x(t^{-}), & t=mT+\frac{T}{2},\\ \dot{x}(t) = Hx(t)+f(x(t)), & mT+\frac{T}{2}< t< (m+1)T,\\ {x}(t) = x(t^{-})+J2x(t^{-}), & t=(m+1)T,\\ {x}(t_{0}) = x_{0}, & t_{0}=0, \end{cases} $$
(2)

where \(C,J1,J2\in R^{n\times n}\) are constant matrices, and \(T>0\) denotes the period of control.

We will use the following two mathematical lemmas.

Lemma 1

Sanchez and Perez [35]

For any three real matrices \(\Phi_{1},\Phi_{2},\Phi_{3}\) of appropriate dimensions and a scalar \(\epsilon\geq0\) such that \(0<\Phi_{3}=\Phi_{3}^{T}\), we have the following inequality:

$$ \Phi_{1}^{T}\Phi_{2}+ \Phi_{2}^{T}\Phi_{1}\leq \epsilon \Phi_{1}^{T}\Phi_{3}\Phi_{1}+ \epsilon^{-1}\Phi_{2}^{T}\Phi_{3}^{-1} \Phi_{2}. $$
(3)

Lemma 2

Boyd et al. [36]

The LMI

$$ \begin{bmatrix} W(x) & Z(x) \\ Z^{T}(x) & R(x) \end{bmatrix} >0, $$

where \(W(x)=W^{T}(x)\), \(R(x)=R^{T}(x)\), and \(Z(x)\) depend affinely on x, is equivalent to

$$R(x)>0,\quad W(x)-Z(x)R^{-1}(x)Z^{T}(x)>0. $$

We denote by \(\lambda_{m}(D)\), \(\lambda_{M}(D)\), and \(D^{T}\) the minimum eigenvalue, the maximum eigenvalue, and the transpose of a square matrix D, respectively. The Euclidean norm of the vector x is denoted \(\Vert x \Vert \). The matrix norm \(\Vert \cdot \Vert \) is also referred to the Euclidean norm. We will use \(D>0\) to display a symmetric positive definite matrix D, \(D<0\) to display a symmetric negative definite matrix D, \(D\leq0\) to display a symmetric seminegative definite matrix D, and \(D\geq0\) to display a symmetric semi-positive definite matrix D. We denote \(f(x(a^{-}))=\lim_{t\rightarrow a}f(x(t))\).

3 Main results

Theorem 1

Suppose that a symmetric and positive definite matrix \(D\in R^{n\times n}\) and positive scalar constants \(h_{1}>0\), \(h_{2}>0\), \(\epsilon_{1}>0\), and \(\epsilon_{2}>0\) satisfy the following conditions:

  1. (1)

    \(DH+H^{T}D+DC+C^{T}D+\epsilon_{1}D^{2}+\epsilon_{1}^{-1}L+h_{1}D\leq 0\),

  2. (2)

    \(DH+H^{T}D+\epsilon_{2}D^{2}+\epsilon_{2}^{-1}L-h_{2}D\leq 0\),

  3. (3)

    \(h_{1}\frac{T}{2}-h_{2}\frac{T}{2}-\ln\lambda_{1}-\ln\lambda_{2}>0\),

where \(\lambda_{1}=\lambda_{M}(D^{-1}(I+J1)^{T}D(I+J1))\), \(\lambda_{2}=\lambda _{M}(D^{-1}(I+J2)^{T}D(I+J2))\). Then the origin of system (2) is exponentially stable.

Proof

First, we construct the Lyapunov function

$$ V \bigl(x(t) \bigr)=x^{T}(t)Dx(t), $$
(4)

so that

$$ \lambda_{m}(D) \bigl\Vert x(t) \bigr\Vert ^{2}\leq V \bigl(x(t) \bigr)\leq\lambda _{M}(D) \bigl\Vert x(t) \bigr\Vert ^{2}. $$
(5)

If \(mT< t<mT+\frac{T}{2}\), then by (2), (3), and (4) we get

$$\begin{aligned} \dot{V}(x) =&2x^{T}D\dot{x} \\ =&2x^{T}D \bigl[Hx+f(x)+Cx \bigr] \\ =&2x^{T}DHx+2x^{T}Df(x)+2x^{T}DCx \\ =& x^{T}[2DH+2DC]x+2x^{T}Df(x) \\ =& x^{T} \bigl[DH+H^{T}D+DC+C^{T}D \bigr]x+2x^{T}Df(x) \\ \leq&x^{T} \bigl[DH+H^{T}D+DC+C^{T}D \bigr]x \\ &{}+\epsilon_{1} x^{T}D^{2}x+ \epsilon_{1}^{-1}x^{T}Lx \\ =&x^{T} \bigl[DH+H^{T}D+DC+C^{T}D+ \epsilon_{1}D^{2}+\epsilon_{1}^{-1}L \bigr]x \\ =&-h_{1}V(x)+x^{T} \bigl[DH+H^{T}D+DC+C^{T}D \\ &{}+\epsilon_{1}D^{2}+\epsilon_{1}^{-1}L+h_{1}D \bigr]x \\ \leq&-h_{1}V(x), \end{aligned}$$

where \(DH+H^{T}D+DC+C^{T}D+\epsilon_{1}D^{2}+\epsilon_{1}^{-1}L+h_{1}D\leq 0\). We get

$$ V \bigl(x(t) \bigr)\leq V \bigl(x \bigl((mT)^{+} \bigr) \bigr)\exp \bigl(-h_{1}(t-mT) \bigr), $$
(6)

where \(mT< t<mT+\frac{T}{2}\).

If \(t=mT+\frac{T}{2}\), then we get

$$\begin{aligned} V(x)|_{t=mT+\frac{T}{2}} =& \bigl(x \bigl(t^{-} \bigr)+J1x \bigl(t^{-} \bigr) \bigr)^{T}D \bigl(x \bigl(t^{-} \bigr)+J1x \bigl(t^{-} \bigr) \bigr) \\ =&x \bigl(t^{-} \bigr)^{T}(I+J1)^{T}D (I+J1)x \bigl(t^{-} \bigr) \\ \leq& \lambda_{1} V \bigl(x \bigl(t^{-} \bigr) \bigr). \end{aligned}$$
(7)

If \(mT+\frac{T}{2}< t<(m+1)T\), then we get

$$\begin{aligned} D^{+}V(x) =&2x^{T}D\dot{x} \\ =&2x^{T}D \bigl[Hx+f(x) \bigr] \\ =& 2x^{T}DHx+2x^{T}Df(x) \\ \leq& x^{T} \bigl[DH+H^{T}D \bigr]x+\epsilon_{2} x^{T}D^{2}x+\epsilon_{2}^{-1}x^{T}Lx \\ =&x^{T} \bigl[DH+H^{T}D+\epsilon_{2}D^{2}+ \epsilon_{2}^{-1}L \bigr]x \\ =&h_{2}V(x) \\ &{}+x^{T} \bigl[DH+H^{T}D+\epsilon_{2}D^{2}+ \epsilon_{2}^{-1}L-h_{2}D \bigr]x \\ \leq& h_{2}V(x), \end{aligned}$$

where \(DH+H^{T}D+\epsilon_{2}D^{2}+\epsilon_{2}^{-1}L-h_{2}D\leq 0\). We get

$$ V \bigl(x(t) \bigr)\leq\lambda_{1}V \biggl(x \biggl( \biggl(mT+\frac{T}{2} \biggr)^{-} \biggr) \biggr)\exp \biggl(h_{2} \biggl(t-mT-\frac {T}{2} \biggr) \biggr), $$
(8)

where \(mT+\frac{T}{2}< t<(m+1)T\).

If \(t=(m+1)T\), then we get

$$\begin{aligned} V(x)|_{t=(m+1)T} =& \bigl(x \bigl(t^{-} \bigr)+J2x \bigl(t^{-} \bigr) \bigr)^{T}D \bigl(x \bigl(t^{-} \bigr)+J2x \bigl(t^{-} \bigr) \bigr) \\ =&x \bigl(t^{-} \bigr)^{T}(I+J2)^{T}D (I+J2)x \bigl(t^{-} \bigr) \\ \leq& \lambda_{2} V \bigl(x \bigl(t^{-} \bigr) \bigr). \end{aligned}$$
(9)

We can do the following mathematical induction through (6), (7), (8), and (9).

Case 1: \(m=0\).

Subcase 1. If \(0< t<\frac{T}{2}\), then we get

$$V \bigl(x(t) \bigr)\leq V(x_{0})\exp(-h_{1}t). $$

So

$$V \biggl(x \biggl(\frac{T}{2}^{-} \biggr) \biggr)\leq V(x_{0}) \exp \biggl(-h_{1}\frac{T}{2} \biggr). $$

Subcase 2. If \(\frac{T}{2}\leq t< T\), then we get

$$\begin{aligned} V \bigl(x(t) \bigr) \leq&\lambda_{1}V \biggl(x \biggl( \frac{T}{2}^{-} \biggr) \biggr)\exp \biggl(h_{2} \biggl(t- \frac{T}{2} \biggr) \biggr) \\ \leq& \lambda_{1}V(x_{0})\exp \biggl(-h_{1} \frac{T}{2}+h_{2} \biggl(t-\frac{T}{2} \biggr) \biggr) \end{aligned}$$

and

$$V \bigl(x \bigl(T^{-} \bigr) \bigr)\leq\lambda_{1}V(x_{0}) \exp \biggl(-h_{1}\frac{T}{2}+h_{2}\frac{T}{2} \biggr). $$

Subcase 3. If \(t=T\), then we get

$$\begin{aligned} V \bigl(x(T) \bigr) \leq&\lambda_{2} V \bigl(x \bigl(T^{-} \bigr) \bigr) \\ \leq& \lambda_{1} \lambda_{2} V(x_{0})\exp \biggl(-h_{1}\frac{T}{2}+h_{2}\frac{T}{2} \biggr). \end{aligned}$$

Case 2: \(m=1\).

Subcase 1. If \(T< t< T+\frac{T}{2}\), then we get

$$\begin{aligned} V \bigl(x(t) \bigr) \leq&V \bigl(x \bigl(T^{+} \bigr) \bigr)\exp \bigl(-h_{1}(t-T) \bigr) \\ \leq&V \bigl(x(T) \bigr)\exp \bigl(-h_{1}(t-T) \bigr) \\ \leq&\lambda_{1} \lambda_{2} V(x_{0})\exp \biggl(-h_{1} \biggl(t-\frac{T}{2} \biggr)+h_{2} \frac{T}{2} \biggr) \end{aligned}$$

and

$$V \biggl(x \biggl( \biggl(T+\frac{T}{2} \biggr)^{-} \biggr) \biggr)\leq \lambda_{1} \lambda_{2} V(x_{0})\exp \biggl(-h_{1}T+h_{2}\frac{T}{2} \biggr). $$

Subcase 2. If \(T+\frac{T}{2}\leq t<2T\), then we get

$$V\bigl(x(t)\bigr)\leq\lambda_{1}^{2} \lambda_{2}V(x_{0}) \exp\bigl(-h_{1}T+h_{2}(t-T)\bigr). $$

So

$$V \bigl(x \bigl((2T)^{-} \bigr) \bigr)\leq\lambda_{1}^{2} \lambda_{2}V(x_{0})\exp(-h_{1}T+h_{2}T). $$

Subcase 3. If \(t=2T\), then we get

$$\begin{aligned} V\bigl(x(2T)\bigr) \leq&\lambda_{2} V\bigl(x\bigl((2T)^{-}\bigr)\bigr) \\ \leq& \lambda_{1}^{2} \lambda_{2}^{2} V(x_{0})\exp(-h_{1}T+h_{2}T). \end{aligned}$$

Case 3: \(m=2\).

Subcase 1. If \(2T< t<2T+\frac{T}{2}\), then we get

$$\begin{aligned} V\bigl(x(t)\bigr) \leq&V\bigl(x\bigl((2T)\bigr)\bigr)\exp\bigl(-h_{1}(t-2T) \bigr) \\ \leq& \lambda_{1}^{2} \lambda_{2}^{2} V(x_{0})\exp\bigl(-h_{1}(t-T)+h_{2}T\bigr). \end{aligned}$$

So

$$V\biggl(x\biggl(\biggl(2T+\frac{T}{2}\biggr)^{-}\biggr)\biggr)\leq \lambda_{1}^{2} \lambda_{2}^{2} V(x_{0})\exp \biggl(-h_{1}\biggl(T+\frac{T}{2} \biggr)+h_{2}T\biggr). $$

Subcase 2. If \(2T+\frac{T}{2}\leq t<3T\), then we get

$$V\bigl(x(t)\bigr) \leq \lambda_{1}^{3} \lambda_{2}^{2}V(x_{0})\exp\biggl(-h_{1} \biggl(T+\frac {T}{2}\biggr)+h_{2}\biggl(t-\frac{3T}{2} \biggr)\biggr). $$

So

$$V\bigl(x\bigl((3T)^{-}\bigr)\bigr) \leq \lambda_{1}^{3} \lambda_{2}^{2}V(x_{0})\exp\biggl(-h_{1} \biggl(T+\frac {T}{2}\biggr)+h_{2}\frac{3T}{2}\biggr). $$

Subcase 3. If \(t=3T\), then we get

$$V\bigl(x(3T)\bigr) \leq \lambda_{1}^{3} \lambda_{2}^{3}V(x_{0})\exp\biggl(-h_{1} \biggl(T+\frac {T}{2}\biggr)+h_{2}\frac{3T}{2}\biggr). $$

Case 4: \(m=3\).

Subcase 1. If \(3T< t<3T+\frac{T}{2}\), then we get

$$\begin{aligned} V\bigl(x(t)\bigr) \leq&V\bigl(x\bigl((3T)\bigr)\bigr)\exp\bigl(-h_{1}(t-3T) \bigr) \\ \leq& \lambda_{1}^{3} \lambda_{2}^{3} V(x_{0})\exp\biggl(-h_{1}\biggl(t-\frac{3T}{2} \biggr)+h_{2}\frac {3T}{2}\biggr). \end{aligned}$$

So

$$V\biggl(x\biggl(\biggl(3T+\frac{T}{2}\biggr)^{-}\biggr)\biggr) \leq \lambda_{1}^{3} \lambda_{2}^{3} V(x_{0})\exp \biggl(-h_{1}2T+h_{2} \frac{3T}{2}\biggr). $$

Subcase 2. If \(3T+\frac{T}{2}\leq t<4T\), then we get

$$V\bigl(x(t)\bigr) \leq \lambda_{1}^{4} \lambda_{2}^{3}V(x_{0})\exp\bigl(-h_{1}2T+h_{2}(t-2T) \bigr). $$

So

$$V\bigl(x\bigl((4T)^{-}\bigr)\bigr) \leq \lambda_{1}^{4} \lambda_{2}^{3}V(x_{0})\exp(-h_{1}2T+h_{2}2T). $$

Subcase 3. If \(t=4T\), then we get

$$V\bigl(x(4T)\bigr) \leq \lambda_{1}^{4} \lambda_{2}^{4}V(x_{0})\exp(-h_{1}2T+h_{2}2T). $$

Through the above induction, we get the following.

Case \(k+1\): \(m=k\).

Subcase 1. If \(kT< t<kT+\frac{T}{2}\), then we get

$$ V\bigl(x(t)\bigr) \leq \lambda_{1}^{k} \lambda_{2}^{k} V(x_{0})\exp\biggl(-h_{1} \biggl(t-\frac{kT}{2}\biggr)+h_{2}\frac {kT}{2}\biggr). $$
(10)

Subcase 2. If \(kT+\frac{T}{2}\leq t<(k+1)T\), then we get

$$ V\bigl(x(t)\bigr) \leq \lambda_{1}^{{k + 1}} \lambda_{2}^{k} V(x_{0})\exp\biggl(-h_{1} \frac {(k+1)T}{2}+h_{2}\biggl(t-\frac{(k+1)T}{2}\biggr)\biggr). $$
(11)

Subcase 3. If \(t=(k+1)T\), then we get

$$ V\bigl(x(t)\bigr)|_{t=(k+1)T} \leq \lambda_{1}^{{k + 1}} \lambda_{2}^{{k + 1}}V(x_{0}) \exp\biggl(-h_{1}\frac{(k+1)T}{2}+h_{2}\frac{(k+1)T}{2} \biggr). $$
(12)

From (10) we get that if \(kT< t<kT+\frac{T}{2}\), then we let \(t=kT\), so that

$$\begin{aligned} V\bigl(x(t)\bigr) \leq& \lambda_{1}^{k} \lambda_{2}^{k}V(x_{0})\exp\biggl( -h_{1}\biggl(t-\frac {kT}{2}\biggr)+h_{2} \frac{kT}{2}\biggr) \\ \leq&\lambda_{1}^{k} \lambda_{2}^{k}V(x_{0})\exp\biggl(-h_{1} \frac{kT}{2}+h_{2}\frac {kT}{2}\biggr) \\ \leq&\exp(k\ln\lambda_{1}+k\ln\lambda_{2})V(x_{0}) \exp\biggl(-h_{1}\frac {kT}{2}+h_{2}\frac{kT}{2} \biggr) \\ \leq&V(x_{0})\exp\biggl(-\biggl(h_{1} \frac{T}{2}-h_{2}\frac{T}{2}-\ln\lambda_{1}-\ln \lambda _{2}\biggr)k\biggr) \end{aligned}$$
(13)

for \(kT< t<kT+\frac{T}{2}\).

From (11) we get that if \(kT+\frac{T}{2}\leq t<(k+1)T\), then we let \(t=(k+1)T\), so that

$$\begin{aligned} V\bigl(x(t)\bigr) \leq&\lambda_{1}^{{k + 1}} \lambda_{2}^{k}V(x_{0})\exp\biggl(-h_{1} \frac{(k+1)T}{2}+h_{2}\biggl(t-\frac{(k+1)T}{2}\biggr)\biggr) \\ \leq&\lambda_{1}^{{k + 1}} \lambda_{2}^{k}V(x_{0}) \exp\biggl(-h_{1}\frac {(k+1)T}{2}+h_{2}\frac{(k+1)T}{2} \biggr) \\ \leq&V(x_{0})\exp\biggl(-h_{1}\frac{(k+1)T}{2}+h_{2} \frac{(k+1)T}{2}+(k+1)\ln \lambda_{1}+k\ln\lambda_{2} \biggr) \\ \leq&V(x_{0})\exp\biggl(-h_{1}\frac{T}{2}-h_{1} \frac{kT}{2}+h_{2}\frac {kT}{2}+h_{2} \frac{T}{2}+\ln\lambda_{1}+k\ln\lambda_{1}+k\ln \lambda_{2}\biggr) \\ \leq&V(x_{0})\exp\biggl(-h_{1}\frac{T}{2}+h_{2} \frac{T}{2}+\ln\lambda_{1}-\biggl(h_{1} \frac {T}{2}-h_{2}\frac{T}{2}-\ln\lambda_{1}-\ln \lambda_{2}\biggr)k\biggr) \end{aligned}$$
(14)

for \(kT+\frac{T}{2}\leq t<(k+1)T\).

From (12) we get that, for \(t=(k+1)T\),

$$\begin{aligned} V\bigl(x(t)\bigr)|_{t=(k+1)T} \leq& \lambda_{1}^{{k + 1}} \lambda_{2}^{{k + 1}}V(x_{0}) \exp\biggl(-h_{1}\frac{(k+1)T}{2}+h_{2}\frac{(k+1)T}{2} \biggr) \\ \leq&\exp\biggl((k+1)\ln\lambda_{1}+(k+1)\ln\lambda_{2}-h_{1} \frac{(k+1)T}{2}+h_{2}\frac{(k+1)T}{2}\biggr)V(x_{0}) \\ \leq&V(x_{0})\exp\biggl(-\biggl(h_{1} \frac{T}{2}-h_{2}\frac{T}{2}-\ln\lambda_{1}-\ln \lambda _{2}\biggr) (k+1)\biggr). \end{aligned}$$
(15)

From (13), (14), (15), and the conditions of Theorem 1 we conclude that \(k\rightarrow\infty\) as \(t\rightarrow \infty\). So

$$\lim_{t\rightarrow\infty}V \bigl(x(t) \bigr)=0, $$

which ends the proof. □

Corollary 1

As a consequence of Lemma  2, the first two conditions of Theorem  1 are equivalent to the following two LIMs:

$$\begin{aligned}& \begin{bmatrix} DH+H^{T}D+DC+C^{T}D +\epsilon_{1}^{-1}L+h_{1}D & -D \\ -D & -\epsilon_{1}^{-1}I \end{bmatrix} \leq0, \end{aligned}$$
(16)
$$\begin{aligned}& \begin{bmatrix} DH+H^{T}D +\epsilon_{2}^{-1}L-h_{2}D & -D \\ -D & -\epsilon_{2}^{-1}I \end{bmatrix} \leq0. \end{aligned}$$
(17)

4 Numerical example

Studying system examples, we can define the original and dimensionless form of Chua’s oscillator [37] as follows:

$$ \textstyle\begin{cases} \dot{x}_{1} = \omega(x_{2}-x_{1}-g(x_{1})), \\ \dot{x}_{2} = x_{1}-x_{2}+x_{3}, \\ \dot{x}_{3} = -\gamma x_{2}, \end{cases} $$
(18)

where ω and γ are two parameters, \(g(x)\) is a piecewise linear characteristic of Chua’s diode. It can be defined by

$$ g(x)=bx+0.5(a-b) \bigl( \vert x+1 \vert - \vert x-1 \vert \bigr), $$
(19)

where a and b are two constants such that \(0>b>a\).

Next, we choose the parameters \(\omega=9.1196\), \(\gamma=15.8946\), \(a=-1.44905\), and \(b=-0.79735\), which make Chua’s circuit (18) chaotic [37]. Figure 3 shows that Chua’s oscillator with the initial condition \(x(0)=(22,-2,-15)'\) is a chaotic phenomenon.

Figure 3
figure 3

The Chua’s oscillator with the initial condition \(\pmb{x(0)=(22,-2,-15)'}\) is a chaotic phenomenon.

Full size image

We can redefine system (18) as follows:

$$ \dot{\bf x}=H{\bf x}+f({\bf x}), $$
(20)

where

$$H= \begin{bmatrix} -\omega-\omega b & \omega& 0 \\ 1 & -1 & 1 \\ 0 & -\gamma& 0 \end{bmatrix} $$

and

$$f(x)= \begin{bmatrix} -0.5\omega(a-b)( \vert x_{1}+1 \vert - \vert x_{1}-1 \vert ) \\ 0 \\ 0 \end{bmatrix}. $$

So we get

$$\begin{aligned} \bigl\Vert f(x) \bigr\Vert ^{2} =&0.25\omega^{2}(a-b)^{2} \bigl[(x_{1}+1)^{2} +(x_{1}-1)^{2}-2 \bigl\vert (x_{1}+1) (x_{1}-1) \bigr\vert \bigr] \\ =& 0.5\omega^{2}(a-b)^{2}\bigl(x_{1}^{2}+1- \bigl\vert x_{1}^{2}-1 \bigr\vert \bigr) \\ =& \textstyle\begin{cases} \omega^{2}(a-b)^{2},& x_{1}^{2}>1,\\ \omega^{2}(a-b)^{2}x_{1}^{2},&x_{1}^{2}\leq1 \end{cases}\displaystyle \\ \leq& \omega^{2}(a-b)^{2}x_{1}^{2}. \end{aligned}$$

Thus we set \(L=\operatorname{diag}(\omega^{2}(a-b)^{2},0,0)\) and choose

$$\begin{aligned}& C=\operatorname{diag}(-49, -42, -32), \\& J1= \begin{bmatrix} -1 & 0 & -0.5\\ 0 & 0.2 & 0.4\\ 1.5 & 0 & 0.5 \end{bmatrix}, \\& J2= \begin{bmatrix} -1 & 0 & -0.8\\ 0 & 0.3 & 0.9\\ 1.8 & 0 & 1.5 \end{bmatrix}. \end{aligned}$$

With \(T=0.10\), solving LMIs (16) and (17) and the inequality \(h_{1}\frac{T}{2}-h_{2}\frac{T}{2}-\ln\lambda_{1}-\ln\lambda_{2}>0\), we obtain a feasible solution

$$\epsilon_{1}=0.5,\qquad \epsilon_{2}=0.5,\qquad h_{1}=21,\qquad h_{2}=20, $$

and

$$D= \begin{bmatrix} 3.7422 & 1.3317 & 0.4360\\ 1.3317 & 3.8921 & -0.4290\\ 0.4360 & -0.4290 & 2.1051 \end{bmatrix}. $$

Thus, by Theorem 1 we know that the origin of system (18) becomes exponentially stable. The time response curves of Chua’s oscillator with the proposed method is shown in Figure 4.

Figure 4
figure 4

Time response curves of Chua’s oscillator with alternate-continuous-control system with double-impulse .

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5 Conclusions

We proposed a new model of a control system named an alternate-continuous-control system with double-impulse. The introduction of an impulse input has played a positive role in the stability control of the system. Theorem 1 gives stability criteria of the current new system control. Moreover, the chaotic Chua circuit can be controlled by the new method.

Through the control method of this paper, we can control most of the nonlinear systems. The method can be used in physics, electronics, robotics, and other fields. Later, we can design more control methods to stabilize nonlinear systems by combining impulse control and intermittent control.