Dynamics Analysis of Llibre-Menezes Piecewise Linear Systems

Abstract

This paper presents a qualitative analysis of discontinuous Llibre-Menezes piecewise linear systems. We obtain the explicit parameter conditions for the existence of limit cycles and the stable sliding segments. In addition, we prove that if a Llibre-Menezes piecewise system is continuous, then this system has a global asymptotically stable equilibrium point. Some examples are given to illustrate the main results.

Appendix: A Complete Proof of Theorem 7 of [17]

We note that the proof of Theorem 7 of [17] is incomplete, especially in the analysis of the zeros of $g_2(t)$. Before giving a complete proof, we recall Theorem 7 of [17] and some notations.

Theorem 3

([17]) Assume that the eigenvalues of vector fields $\mathbf{X}$ and $\mathbf{Y}$ have negative real parts, then the discontinuous piecewise linear differential system (6.1) has at most one limit cycle where system (6.1) is given by

$$\begin{aligned} \dot{\mathbf{z }} = \left\{ \begin{aligned}&\mathbf{X} (\mathbf{z} )= \begin{pmatrix} a &{} 1 \\ 1 &{} a \end{pmatrix} \begin{pmatrix} x-p_1 \\ y-p_2 \end{pmatrix},\ \text {if}\ x >0, \\&\mathbf{Y} (\mathbf{z} )= \begin{pmatrix} b_{11} &{} b_{12} \\ b_{21} &{} b_{22} \end{pmatrix} \begin{pmatrix} x \\ y- q_2 \end{pmatrix},\ \text {if}\ x < 0, \end{aligned}\right. \end{aligned}$$

(6.1)

where $\mathbf{z} =(x,y)^T, a<-1, b_{12}>0, b_{11}+b_{22}<0, p_1<0, 4b_{12}b_{21}+(b_{11}-b_{22})^2<0$.

Let $\varphi (t,\bar{x},\bar{y})$ denote the solution of system (6.1) where

$$\begin{aligned} \varphi (t,\bar{x},\bar{y}) = \left\{ \begin{aligned}&\varphi ^+(t,\bar{x},\bar{y}),\ \text {if}\ \bar{x} >0, \\&\varphi ^-(t,\bar{x},\bar{y}),\ \text {if}\ \bar{x} <0, \end{aligned}\right. \end{aligned}$$

satisfying $\varphi (0,\bar{x},\bar{y})=(\bar{x},\bar{y})$, where $\varphi ^+(t,\bar{x},\bar{y})$ and $\varphi ^-(t,\bar{x},\bar{y})$ are the solutions of vector fields $\mathbf{X}$ and $\mathbf{Y}$. Set $t^+(\bar{y})>0$ the smallest positive time such that $\varphi ^+(t^+(\bar{y}),0,\bar{y})\in \Sigma$.

The first part of the proof of this theorem is in the proof of Proposition 14 of [18] and in the proof of Theorem 7 of [17]. We add the part after the function $g_2(t)$ and thus complete the proof.

Proof

Doing a translation to system (6.1) that preserves the half-plane $x>0$ and the discontinuity line $\Sigma$, we can assume that $p_2 = 0$. Clearly the point $(0,a p_1 ) \in \Sigma$ is an invisible fold point, and $(p_1,0)$ is the singularity of $\mathbf{X}$. Furthermore the invariant straight lines of the node intersect the line $\Sigma$ at the points $(0, y^s)$ and $(0, y^{ss})$, respectively, where $y^s=-p_1 < a p_1$ and $y^{ss}=p_1< -p_1$. It follows that the function $t^+(y)>0$ is defined for every $y >a p_1$.

By definition of $t^+(\bar{y})>0$, we get that

$$\begin{aligned} p_1+e^{at^+(\bar{y})}(-p_1\cosh t^+(\bar{y})+\bar{y} \sinh t^+(\bar{y}))=0 \end{aligned}$$

for every $\bar{y} > a p_1$. Thus defining $y^+(t)=-p_1G(t)$ for $t>0$, with G(t) given by

$$\begin{aligned} G(t)=\frac{e^{-at}-\cosh t}{\sinh t}, \end{aligned}$$

we have that $y^+(t^+(\bar{y}))=\bar{y}$ for every $\bar{y} >a p_1$. Computing implicitly the derivative in the variable $\bar{y}$ of the identity $y^+(t^+(\bar{y}))=\bar{y}$ we obtain

$$\begin{aligned} \frac{d t^+(\bar{y})}{d \bar{y}}(\bar{y})=-\frac{\sinh ^2 t^+(\bar{y})}{p_1[1-e^{-a t^+(\bar{y})}(a\sinh t^+(\bar{y}) +\cosh t^+(\bar{y}))]}>0, \end{aligned}$$

for $t^+(\bar{y})>0$, because $a <-1$ and $p_1<0$.

We claim that $t^+(y^+(t))=t$ for every $t >0$. Consider $y_0=y^+(t_0)$ for some $t_0 > 0$. From Lemma 3 of [18] we have that the function $y^+(t)$ is injective on $\mathbb {R}^+$ and $y^+(t)>a p_1$. Therefore $y_0>a p_1$ and from the previous results $y_0=y^+(t^+(y_0))$. Thus $y^+(t_0)=y^+(t^+(y_0))$ from which it follows that $t_0 =t^+(y_0)=t^+(y^+(t_0))$. Since $t_0 > 0$ was arbitrarily chosen we conclude that $t^+(y^+(t))=t$ for every $t > 0$. Therefore the function $t^+:(ap_1,+\infty )\rightarrow \mathbb {R}^+$ is invertible with inverse equal to $y^+:\mathbb {R}^+ \rightarrow (ap_1,+\infty )$.

Note that the periodic orbits are in correspondence with the zeros of the following function

$$\begin{aligned} g_2(t)=-q_2(1+\delta )+\frac{p_1(e^{at}-\delta e^{-at})-p_1(1-\delta )\cosh t}{\sinh t} \end{aligned}$$

(6.2)

for $t\in t^+((Y_M),+\infty )\subset \mathbb {R}^+$ and $\delta =e^{-(b_{11}+b_{22})\pi /\Gamma }$.

Computing the zeros of the function (6.2) is equivalent to compute the zeros of the function

$$\begin{aligned} h(t)=A \sinh t +(1-\delta )\cosh t-e^{a t}+\delta e^{-a t} \end{aligned}$$

(6.3)

where $A=q_2(1+\delta )/p_1$. It is easy to obtain that

$$\begin{aligned} h(t)\ge \frac{\delta +1+A}{2}(e^t-e^{-t}). \end{aligned}$$

(6.4)

Thus if $A>-(1+\delta )$, then

$$\begin{aligned} h(t)>0,\ \text {for all}\ t>0. \end{aligned}$$

Since h(t) is analytic, we get

$$\begin{aligned} h(t)=\sum _{n=0}^{+\infty }h^{(n)}(0)\frac{t^n}{n!}, \end{aligned}$$

where

$$\begin{aligned} h^{(2n)}(0)=&\,(a^{2n}-1)(\delta -1)>0,\\ h^{(2n-1)}(0)=&\,-a^{2n-1}(\delta +1)+A, \end{aligned}$$

for all $n\in \mathbb {N}$. If

$$\begin{aligned} a(1+\delta )\le A \le -(1+\delta ), \end{aligned}$$

then

$$\begin{aligned} h^{(2n+1)}(0)>h^{(2n-1)}(0)\ge h'(0)\ge 0. \end{aligned}$$

Combining this inequality and $h^{(2n)}(0)>0$, we have

$$\begin{aligned} h(t)>0,\ \text {for all}\ t>0. \end{aligned}$$

If $A< a(1+\delta )$, then from

$$\begin{aligned} h(0)=0, \ h'(0)<0, \end{aligned}$$

we get that there exists a $t_1>0$ such that $h(t)<0$ for $0<t<t_1$. It is easy to obtain that

$$\begin{aligned} \lim _{t\rightarrow +\infty }\frac{h_3(t)}{h_4(t)}=+\infty . \end{aligned}$$

According to the intermediate value theorem, there exists a $t_2>t_1$ such that $h(t_2)=0$ and $h(t)<0, \ 0<t<t_2$. It is apparent that $h'(t_2)>0$.

Owing to $h''(t)>h(t)$, we have $h''(t_2)>0$. Hence from continuity of the function $h''(t)$ there exists a $\epsilon >0$ such that $h''(t)>0$ for $t_2-\epsilon<t<t_2+\epsilon$. It follows that $h'(t)>0$ and $h(t)>0$ for $t\in (t_2-\epsilon ,t_2+\epsilon )$. Therefore, we get $h(t)>0$ for $t>t_2$ by contradiction. If there exists $t_3>t_2$ such that $h(t_3)=0$ and $h(t)>0, \ t_2<t<t_3$. It is apparent to see that $h'(t_3)<0$. This leads to a contradiction since $h''(t)>0,\ t_2<t<t_3$.

In conclusion, function (6.3) has only one zero if $A< a(1+\delta )$ and no zeros if $A\ge a(1+\delta )$. This means system (6.1) has at most one limit cycle and thus Theorem 7 of [17] is proved. $\square$

We remark that the condition $A< a(1+\delta )$ is equivalent to $q_2> a p_1$. From the above proof, we get that if $A< a(1+\delta )$ then system (6.1) has a limit cycle which is consistent with statement (a) of Proposition 3.