Configuration of Ten Limit Cycles in a Class of Cubic Switching Systems

Abstract

The bifurcation of limit cycles is an important part in the study of switching systems. The investigation of limit cycles includes the number and configuration, which are related to Hilbert’s 16th problem. Most researchers studied the number of limit cycles, and only few works focused on the configuration of limit cycles. In this paper, we develop a general method to determine the configuration of limit cycles based on the Lyapunov constants. To show our method by an example, we study a class of cubic switching systems, which has three equilibria: $(0,0)$ and $(\pm 1,0)$, and compute the Lyapunov constants based on Poincaré return map, then find at least 10 small-amplitude limit cycles that bifurcate around $(1,0)$ or $(-1,0)$. Using our method, we determine the location distribution of these ten limit cycles.

1. Introduction

The bifurcation of limit cycles in planar polynomial dynamical systems is a part of Hilbert’s 16th problem. In recent decades, many researchers have carried out studies on this problem and made great achievements, see, e.g., [1,2,3,4,5,6,7,8,9]. However, how to obtain the numerical simulation and determine the location distribution of limit cycles has always been a difficult problem to overcome. At present, it is difficult to realize the numerical simulation even for three limit cycles. In this paper, we present the numerical method with simulations and realizations of 10 limit cycles, and then determine the location distribution of these 10 limit cycles.

Switching dynamical systems describes many problems in mechanical, electrical engineering and automatic control. In the past few decades, more and more researchers focused on the qualitative analysis of switching systems [10,11,12,13]. Filippov [14] established basic qualitative theory for switching systems. Gasull and Torregrosa [15] constructed five limit cycles in a quadratic switching system, while only four limit cycles have been constructed for planar quadratic continuous differential systems [2,16]. Tian and Yu [17] provided a complete classification on the conditions of a singular point as the center in the switching Bautin system. Guo et al. [18] studied a class of $Z_2$-equivariant cubic switching systems, and showed the existence of 18 limit cycles. Very recently, Gouveia and Torregrosa [19] found 24 limit cycles in a cubic switching polynomial system with Degenerated Hopf and pseudo-Hopf bifurcations, by perturbing a single Darboux center. Yu et al. [20] constructed a cubic planar switching polynomial system with $Z_2$-symmetry, and proved that such a system could exhibit at least 9 small-amplitude limit cycles around each of two symmetric foci, giving a total 18 limit cycles. In addition, many researchers have tried to extend the classical methods of studying smooth systems to the study of switching (non-smooth) systems [21,22].

However, few works studied the configuration or location distribution of the limit cycles in switching systems. It is necessary because it is not only a part of Hilbert’s 16th problem, but also the basis for the image simulation of limit cycles, which plays an important role when practical systems are studied. In this paper, we develop a method for determining the configuration of limit cycles.

Specially, we consider a class of cubic switching systems, which is also studied in [23]:

$$\begin{array}{cc}\left\{\begin{array}{cc}{\displaystyle \frac{dx}{dt}}=\hfill & b_{0}y-(b_0+2)x^{2}y-2b_{2}x{y}^2+2b_3{y}^3\hfill \\ \hfill & -ϵ[\delta (x^3-x)+p_{8}y+p_7(x^3-x)+p_1{x}^{2}y+p_{2}x{y}^2+p_3{y}^3],\hfill \\ {\displaystyle \frac{dy}{dt}}=\hfill & -x+x^3+2b_{5}x{y}^2-2b_6{y}^3\hfill \\ \hfill & -ϵ[2\delta y+p_{10}y+p_9(x^3-x)+p_4{x}^{2}y+p_{5}x{y}^2+p_6{y}^3],\hfill \end{array}\right.\hfill & (y>0);\hfill \\ \\ \left\{\begin{array}{cc}{\displaystyle \frac{dx}{dt}}=\hfill & b_{0}y-(b_0+2)x^{2}y+2b_{2}x{y}^2+2b_3{y}^3\hfill \\ \hfill & -ϵ[\delta (x^3-x)+q_{8}y+q_7(x^3-x)+q_1{x}^{2}y+q_{2}x{y}^2+q_3{y}^3],\hfill \\ {\displaystyle \frac{dy}{dt}}=\hfill & -x+x^3+2b_{5}x{y}^2+2b_6{y}^3\hfill \\ \hfill & -ϵ[2\delta y+q_{10}y+q_9(x^3-x)+q_4{x}^{2}y+q_{5}x{y}^2+q_6{y}^3],\hfill \end{array}\right.\hfill & (y<0),\hfill \end{array}$$

(1)

where $\delta$, $p_i$’s and $q_i$’s are real parameters, satisfying $\left|\delta \right|\ll 1$ and $0<ϵ\ll 1$.

The authors of [23] focused on the computation of Lyapunov constants, however, in this paper, we are going to find the configuration of the limit cycles in System (1).

Obviously, this system has three equilibria: $(0,0)$ and $(\pm 1,0)$. Note that the separate upper or lower system has $Z_2$-equivariant symmetry, but the whole System (1) is not $Z_2$-equivariant. We noticed that the substitutions $(x,y)\to (-x,-y)$ and,

$$(b_0,b_2,b_3,b_5,b_6,p_i,q_i)\to (b_0,-b_2,b_3,b_5,-b_6,q_i,p_i)$$

to System (1) yield the same system. We say the properties of limit cycles around $(1,0)$ are the same with that around $(-1,0)$ in System (1). Thus, though the limit cycles around $(1,0)$ and that around $(-1,0)$ cannot exist simultaneously, we only need to consider the limit cycles around $(1,0)$.

2. Preliminary Results

In this section, we present some basic methods and preliminary results.

The general switching differential system has the form:

$$\left(\begin{array}{c}\dot{x}\hfill \\ \dot{y}\hfill \end{array}\right)=\left\{\begin{array}{c}\left(\begin{array}{c}\delta x-y+{\displaystyle \sum _{k=2}^n}{X}_k^{+}(x,y)\hfill \\ x+\delta y+{\displaystyle \sum _{k=2}^n}{Y}_k^{+}(x,y)\hfill \end{array}\right),\mathrm{if}y>0,\hfill \\ \left(\begin{array}{c}\delta x-y+{\displaystyle \sum _{k=2}^n}{X}_k^{-}(x,y)\hfill \\ x+\delta y+{\displaystyle \sum _{k=2}^n}{Y}_k^{-}(x,y)\hfill \end{array}\right),\mathrm{if}y<0,\hfill \end{array}\right.$$

(2)

where $X_k^{\pm }(x,y)$ and $Y_k^{\pm }(x,y)$ are homogeneous polynomials in x and y. By the polar coordinates transformation, x = rcos$\theta$ and y = rsin$\theta$, System (2) can be rewritten as:

$$\begin{array}{c}{\displaystyle \frac{dr}{d\theta }}=\left\{\begin{array}{c}{\displaystyle \frac{\delta r+{\sum }_{k=2}^n{\mathsf{{\rm Y}}}_k^{+}\left(\theta \right)r^k}{1+{\sum }_{k=2}^n{\Theta }_k^{+}\left(\theta \right)r^{k-1}}},\mathrm{for}\theta \in (0,\pi ),\hfill \\ {\displaystyle \frac{\delta r+{\sum }_{k=2}^n{\mathsf{{\rm Y}}}_k^{-}\left(\theta \right)r^k}{1+{\sum }_{k=2}^n{\Theta }_k^{-}\left(\theta \right)r^{k-1}}},\mathrm{for}\theta \in (-\pi ,0).\hfill \end{array}\right.\hfill \end{array}$$

(3)

Note that ${\mathsf{{\rm Y}}}_k^{\pm }\left(\theta \right)$ and ${\Theta }_k^{\pm }\left(\theta \right)$ are polynomials expressed by sin$\theta$ and cos$\theta$ with degrees $k+1$. As described in [24], Chapter 3, we define the dynamic behavior of the switching system on the switching boundary as the passing between the upper and lower subsystems. Specifically, with the passage of time, each solution trajectory of the lower system is connected with a solution trajectory of the upper system on the switching boundary. Based on the theory of small parameters of Poincaré, we give the forms of the solutions of the upper and lower systems in (3):

$$r^{+}(h,\theta )=\sum _{k\ge 1}{u}_k\left(\theta \right)h^k,r^{-}(h,\theta )=\sum _{k\ge 1}{v}_k\left(\theta \right)h^k,$$

(4)

where $u_1\left(0\right)=v_1\left(0\right)=1$, $u_k\left(0\right)=v_k\left(0\right)=0$, $\forall k\ge 2$. Substituting the above solutions into (3), we can solve $u_k\left(\theta \right)$ or $v_k\left(\theta \right)$ by integral operations successively. Suppose the successive functions for the upper and lower systems of (3) are:

$${▵}^{+}\left(h\right)=r^{+}(h,\pi )-h,{▵}^{-}\left(h\right)=h-r^{-}(h,-\pi ),$$

respectively. Then, we define the successive function for the switching System (2):

$$\begin{array}{c}▵\left(h\right)={▵}^{+}\left(h\right)+{▵}^{-}\left(h\right)=r^{+}(h,\pi )-r^{-}(h,-\pi ).\hfill \end{array}$$

(5)

In [15], we know the displacement function $▵\left(h\right)$ can be expanded as:

$$▵\left(h\right)=\sum _{k=1}^n(u_k\left(\pi \right)-v_k(-\pi ))h^k=\sum _{k=0}^{n-1}{V}_k{h}^{k+1}.$$

(6)

Here, we call $V_k$ the kth-order Lyapunov constant of the switching System (2). By simple computation, we get $u_1\left(\theta \right)=v_1\left(\theta \right)=e^{\delta \theta }$, yielding $V_0=\frac{1}{e^{\pi \delta }}(e^{2\pi \delta }-1)$.

The sufficient conditions for determining the existence of small-amplitude limit cycles in the switching system (2) are given in the following theorem. (The proof can be found in [17].)

Theorem 1

([17]). Suppose that there exists a sequence of Lyapunov constants of System (2), $V_{i_0},V_{i_1},\dots ,V_{i_k}$, with $1=i_0<i_1<\cdots <i_k$, such that $V_j=O\left(\right|V_{i_0},\dots ,V_{i_l}\left|\right)$ for any $i_l<j<i_{l+1}$. Further, if at the critical point $\mathit{C}$, $V_{i_0}=V_{i_1}=\cdots =V_{i_k-1}=0,V_{i_k}\ne 0$, and:

$$det{\left[{\displaystyle \frac{\partial (V_{i_0},V_{i_1},\cdots ,V_{i_{k-1}})}{\partial (c_1,c_2,\cdots ,c_k)}}\right]}_{\mathit{C}}\ne 0,$$

(7)

then System (2) has exactly k limit cycles in a δ-ball whose center is at the origin.

3. Lyapunov Constants

In this section, we compute the Lyapunov constants of System (1), and then determine the number of limit cycles around the point (1, 0) or (−1, 0). Note that the computation process is also presented in [23]. In order to clearly describe in the next section, we present the core steps and formulas for calculating the Lypunov constant here.

We set $p_{10}=-p_4-2p_7,q_{10}=-q_4-2q_7$ to make System (1) have Hopf singular points at $(1,0)$. For a further simplification in computation, we set $p_7=p_9=q_7=q_9=0$, $p_8=-p_1$, $q_8=-q_1$ and $b_0=1$. Then, we further compute the Lyapunov constants at $(1,0)$. We give the following transformation,

$$x=1-X,y=Y,t\to -{\displaystyle \frac{1}{2}}t,$$

into System (1). Then the singular point $(1,0)$ of (1) becomes the origin of the following system,

$$\begin{array}{cc}\left\{\begin{array}{cc}{\displaystyle \frac{dX}{dt}}=\hfill & ϵ\delta X-Y-\frac{3}{2}ϵ\delta X^2+(ϵp_1+3)XY-(b_2+\frac{1}{2}ϵp_2)Y^2+\frac{1}{2}ϵ\delta X^3\hfill \\ \hfill & -(\frac{3}{2}+\frac{1}{2}ϵp_1)X^{2}Y+(b_2+\frac{1}{2}ϵp_2)X{Y}^2+(b_3-\frac{1}{2}ϵp_3)Y^3,\hfill \\ {\displaystyle \frac{dY}{dt}}=\hfill & X+ϵ\delta Y-\frac{3}{2}{X}^2-ϵp_{4}XY-(b_5-\frac{1}{2}ϵp_5)Y^2+\frac{1}{2}{X}^3+\frac{1}{2}ϵp_4{X}^{2}Y\hfill \\ \hfill & +(b_5-\frac{1}{2}ϵp_5)X{Y}^2+(b_6+\frac{1}{2}ϵp_6)Y^3,\hfill \end{array}\right.\hfill & (Y>0);\hfill \\ \left\{\begin{array}{cc}{\displaystyle \frac{dX}{dt}}=\hfill & ϵ\delta X-Y-\frac{3}{2}ϵ\delta X^2+(ϵq_1+3)XY+(b_2-\frac{1}{2}ϵq_2)Y^2+\frac{1}{2}ϵ\delta X^3\hfill \\ \hfill & -(\frac{3}{2}+\frac{1}{2}ϵq_1)X^{2}Y-(b_2-\frac{1}{2}ϵq_2)X{Y}^2+(b_3-\frac{1}{2}ϵq_3)Y^3,\hfill \\ {\displaystyle \frac{dY}{dt}}=\hfill & X+ϵ\delta Y-\frac{3}{2}{X}^2-ϵq_{4}XY-(b_5-\frac{1}{2}ϵq_5)Y^2+\frac{1}{2}{X}^3+\frac{1}{2}ϵq_4{X}^{2}Y\hfill \\ \hfill & +(b_5-\frac{1}{2}ϵq_5)X{Y}^2-(b_6-\frac{1}{2}ϵq_6)Y^3.\hfill \end{array}\right.\hfill & (Y<0).\hfill \end{array}$$

(8)

Thus, we can compute the $ϵ$-order Lyapunov constants around $(1,0)$ for System (1) based on Section 2. First, we have $V_0=2\pi \delta$, thus setting $\delta =0$ yields $V_0=0$. By a direct computation in higher Lyapunov constants, we may set the non-used parameters $q_1=q_2=q_3=q_4=q_5=q_6=0$, and choose $p_4$ as a free parameter. Then, without loss of generality, we let $p_4=1$. We have $V_1=\frac{2}{3}(p_1+p_5)$. Letting $V_1=0$ results in $p_5=-p_1$, and then $V_2$ is simplified to:

$$V_2={\displaystyle \frac{\pi }{16}}\left[2b_5(p_2-1)-2p_2+3p_6-2\right].$$

Setting,

$$p_6=-\frac{2}{3}\left[b_5(p_2-1)-p_2-1\right],$$

we have $V_2=0$. In order to obtain the maximal number of small-amplitude limit cycles bifurcating from the origin of system (8), we assume that:

$$\begin{array}{cc}{F}_0=\hfill & (2b_5-3)(b_2-b_6)\left[5b_5(b_2-15b_6)-8b_2-69b_6\right][(10b_2^2+30b_2{b}_6+60b_3+21)b_5\hfill \\ \hfill & -12b_5^3-(20b_3-20)b_5^2-10b_2^2-15b_2{b}_6+45b_6^2-45b_3-36\left]\right\{4(b_2+b_6)b_5^3\hfill \\ & +4\left[(b_3-3)b_2+3b_3{b}_6\right]b_5^2-\left[2b_2^2(b_2+6b_6)+3(6b_6^2+4b_3-3)b_2+9(4b_3+3)b_6\right]b_5\hfill \\ & +2b_2^3+9b_2^2{b}_6+9b_2{b}_3-27b_6(b_6^2-b_3-1)\}\ne 0.\hfill \end{array}$$

Then, we have:

$$\begin{array}{cc}{V}_3=\hfill & -\frac{2}{45}\left[3p_1(4b_5^2-9)+2b_2(4b_5{p}_2+2b_5+1)+4p_2(3b_6-2b_2)+3p_3(3-2b_5)\right],\hfill \\ V_4=\hfill & \frac{\pi }{96(3-2b_5)}\{{(3-2b_5)}^2[3p_1(b_2-2b_6)-p_2(3b_5+5b_3+4)-2b_3-3]\hfill \\ \hfill & +(3-2b_5)[3p_1(b_2+3b_6)+4b_3{b}_5+15]+5[2b_2(b_5-1)+3b_6][p_2(b_2+3b_6)+2b_2]\},\hfill \\ V_5=\hfill & \frac{32}{1575[(10b_2^2+30b_2{b}_6+60b_3+21)b_5-12b_5^3-(20b_3-20)b_5^2-10b_2^2-15b_2{b}_6+45b_6^2-45b_3-36]}{F}_5,\hfill \\ V_6=\hfill & \frac{\pi }{1152(b_2-b_6)}\{[105b_2(b_2-b_6)-b_3(5b_5-8)-b5(14-5b_5)-15](b_2-b_6)\hfill \\ \hfill & +b_6(10b_5+11)(5b_5+7b_3+6)\},\hfill \end{array}$$

(9)

where $F_5$ is a polynomial in $b_2,b_3,b_5,b_6$ and $p_1$.

Next, the polynomial equations in (9) are linearly solved one by one. We use $p_3$ to solve for $V_3=0$, $p_2$ for $V_4=0$, $p_1$ for $V_5=0$ and $b_3$ for $V_6=0$. Then, the higher Lyapunov constants are obtained:

$$\begin{array}{cc}\hfill V_7=& -{\displaystyle \frac{64F_7}{33075[5b_5(b_2-15b_6)-8b_2-69b_6]}},\hfill \\ \hfill V_8=& {\displaystyle \frac{F_8}{20321280{[5b_5(b_2-15b_6)-8b_2-69b_6]}^2(b_2-b_6)}},\hfill \\ \hfill V_9=& -{\displaystyle \frac{F_9}{6706022400{[5b_5(b_2-15b_6)-8b_2-69b_6]}^2(b_2-b_6)}},\hfill \\ \hfill V_{10}=& {\displaystyle \frac{F_{10}}{804722688000{[5b_5(b_2-15b_6)-8b_2-69b_6]}^3(b_2-b_6)}},\hfill \end{array}$$

where $F_7$, $F_8$, $F_9$ and $F_{10}$ are polynomials in $b_2$, $b_5$ and $b_6$, which are omitted here. Thus, if we want to obtain the maximal number of limit cycles, we need to find the solutions of $b_2,b_5$ and $b_6$ such that $F_7=F_8=F_9=0$, but $F_0{F}_{10}\ne 0$. This choice yields at most 10 small-amplitude limit cycles bifurcating from the origin of System (8).

In order to find the solutions of $F_7=F_8=F_9=0$, we use the Maple built-in command “rem” to simplify their expressions, yielding:

$$F_{87}=\mathrm{rem}(F_8,F_7,b_5)={\displaystyle \frac{27\pi }{2{(23b_2^2-60b_2{b}_6+45b_6^2)}^3}}{F}_{87a},$$

where $F_{87a}$ is a polynomial in $b_2,b_5$ and $b_6$, and linear about $b_5$, which are omitted here. According to the Remainder Theorem, $F_7=F_8=F_9=0$ is equivalent to $F_7=F_{87a}=F_9=0$. Solving $b_5$ from $F_{87a}=0$, we obtain $b_5=-{\displaystyle \frac{b_{5N}}{b_{5D}}}$, where $b_{5N}$, $b_{5D}$ are lengthy polynomials in $b_2$ and $b_6$.

We substitute the solution $b_5$ into $F_7$ and $F_9$ to obtain:

$$\begin{array}{cc}{F}_7=\hfill & {\displaystyle \frac{392{(b_2-b_6)}^3{(23b_2^2-60b_2{b}_6+45b_6^2)}^3}{b_{5D}^2}}(175b_2^4-5425b_2^3{b}_6+44625b_2^2{b}_6^2-39375b_2{b}_6^3\hfill \\ \hfill & -41b_2^2+1986b_2{b}_6-153b_6^2{)}^2{F}_{7a},\hfill \\ F_9=\hfill & -{\displaystyle \frac{4{(b_2-b_6)}^3}{b_{5D}^5}}(175b_2^4-5425b_2^3{b}_6+44625b_2^2{b}_6^2-39375b_2{b}_6^3-41b_2^2+1986b_2{b}_6\hfill \\ & -153b_6^2{)}^2{F}_{9a},\hfill \end{array}$$

where $F_{7a}$ and $F_{9a}$ are lengthy polynomials in $b_2$ and $b_6$, which are omitted here. Now, we need to find solutions of $b_2$ and $b_6$ such that $F_{7a}=F_{9a}=0$, but $F_0{b}_{5D}{F}_{10}\ne 0$. Again, we use the Maple command “rem” and “resultant” to obtain $\mathrm{rem}({\mathrm{F}}_{9\mathrm{a}},{\mathrm{F}}_{7\mathrm{a}},{\mathrm{b}}_6)={\mathrm{F}}_{97}$, and then:

$$\begin{array}{cc}\hfill F_{7a97}& =\mathrm{resultant}(F_{7a},F_{97},b_6)\hfill \\ \hfill & =C{b}_2^{296}(53630051863608150000000000b_2^{10}+5264245855401885313684240000b_2^8\hfill \\ \hfill & +544451168318211521548337100b_2^6-387765757092266273103490296b_2^4\hfill \\ \hfill & -255257807648772965284666341b_2^2-32345418979891744705150434)\hfill \\ \hfill & \times (741986003638863750976562500000000b_2^{16}+1047364945197617238766601562500000b_2^{14}\hfill \\ \hfill & +541377423879749549575886962890625b_2^{12}+124762729270469072104017978515625b_2^{10}\hfill \\ \hfill & +13054805934302191749817586484375b_2^8+731541816166608967815824681250b_2^6\hfill \\ \hfill & +16914737223581362822481430375b_2^4+9078812025969674245082805b_2^2\hfill \\ \hfill & {+2479608114583719233097)}^8(615601902373365446219763960910156250000b_2^{10}\hfill \\ \hfill & -731907810934317520907061383859175170625b_2^8\hfill \\ \hfill & -468737289830045377055886038650808852550b_2^6\hfill \\ \hfill & -16937175919452904610229612666535720785b_2^4\hfill \\ \hfill & -353690927917112505903442270661328588b_2^2\hfill \\ \hfill & {-47839883903858245269575882615966208)}^{10},\hfill \end{array}$$

where C is a constant. We solve $F_{7a97}=0$ to find the solutions of $b_2$. This polynomial has five real roots, which yield five corresponding solutions for $b_6$. By checking that $F_{7a}=F_{9a}=0$ and $F_0{b}_{5D}{F}_{10}\ne 0$, we found that only two of them satisfy the original equations. We take one of the solutions:

$$b_2=-0.6499316542\cdots ,b_6=-1.4007939402\cdots .$$

The corresponding values of other perturbation parameters are listed as follows,

$$\begin{array}{ccc}{p}_5=-0.6593001227\cdots ,\hfill & p_6=-0.3425204361\cdots ,\hfill & p_3=-0.7686180478\cdots ,\hfill \\ p_2=0.1486175491\cdots ,\hfill & p_1=0.6593001227\cdots ,\hfill & b_3=0.6172899175\cdots ,\hfill \\ b_5=-1.9525868799\cdots .\hfill & \hfill \end{array}$$

We define a critical point by the forementioned critical values, called $p_c$, by which the $ϵ$-order Lyapunov constants are obtained:

$$V_i=0,i=1,2,\cdots ,9,V_{10}=0.0157763313\cdots \ne 0.$$

By a direct calculation, we show that:

$$det\left[{\displaystyle \frac{\partial (V_7,V_8,V_9)}{\partial (b_2,b_5,b_6)}}\right]=0.0003020524\cdots \ne 0.$$

By Theorem 1, we know that System (8) has 10 small-amplitude limit cycles around the origin. Thus, System (1) can have 10 limit cycles around $(1,0)$ or $(-1,0)$.

4. The Configuration of 10 Limit Cycles

In this section, we develop a method to realize 10 small-amplitude limit cycles bifurcating from singular point of $(1,0)$ or $(-1,0)$ of switching Systems (1), thus we can determine the configuration of limit cycles.

In general, it is simple and direct to determine the location distribution of a single limit cycle, and determining the location distribution of two limit cycles is also relatively simple. However, it is relatively challenging to determine the location distribution of three limit cycles, so determining the location distribution of ten limit cycles is much more difficult. The main difficulty of the problem is how to choose the perturbation of parameters properly at the critical point, such that the truncated Poincaré successive function has 10 positive real roots. The process is still simple if the perturbation can be carried out step by step and we can choose one parameter at every step. However, it is very difficult to find perturbations if that polynomial equation which is obtained from Poincaré successive function is coupled. Here, we present a method motivated by Liu [25,26] to determine the location distribution of 10 limit cycles.

We assume that the successive function is defined as:

$$▵\left(h\right)=h(\sum _{i=0}^{10}{v}_i{h}^i+\cdots ),$$

(10)

where $v_i$ are called the ith Lyapunov constants. As described in Section 3, $v_i$=$ϵ$$V_i$+$o\left(ϵ\right)$, $i=0,1,2,\cdots$. Now, we rewrite Function (10) as:

$$▵\left(h\right)=ϵh(\sum _{i=0}^{10}{V}_i{h}^i+\cdots )+o\left(ϵ\right)=ϵh\overline{▵}\left(h\right)+o\left(ϵ\right).$$

(11)

Because we are only interested in small-amplitude limit cycles, we submit the scaling transformation $h\to \epsilon h(0<\epsilon \ll 1)$ to Function (11), and we obtain:

$$\overline{▵}\left(\epsilon h\right)=\sum _{i=0}^{10}{V}_i{\epsilon }^i{h}^i+\cdots$$

(12)

when the $ϵ$ is small enough, the positive real roots h obtained from $▵\left(\epsilon h\right)=0$ are equivalent to the positive real roots h obtained from $\overline{▵}\left(\epsilon h\right)=0$.

Next, we assume perturbed Lyapunov constants as following,

$$V_i=K_i{\epsilon }^{10-i}+o\left({\epsilon }^{10-i}\right),i=0,1,2,\cdots ,9,V_j=K_j+o\left(1\right),j\ge 10,$$

(13)

based on the above expression, function $\overline{▵}\left(\epsilon h\right)=0$ becomes:

$${\epsilon }^{10}[\sum _{i=0}^{10}{K}_i{h}^i+\epsilon G(\epsilon ,h)]=0$$

(14)

where $G(\epsilon ,h)$ is an analytic function in $(0,0)$. By implicit function theorem, when $\epsilon$ is small enough, we can obtain it from (14), if function:

$$\sum _{i=0}^{10}{K}_i{x}^i=0$$

(15)

has 10 real positive roots $h_i,i=1,2,\cdots ,10$, then function $\overline{▵}\left(\epsilon h\right)=0$ has 10 real positive roots. The 10 real positive roots are close enough to $h_i,i=1,2,\cdots ,10$, then System (1) can exist with 10 limit cycles, which are close to the periodic ring, respectively: ${(x-1)}^2+y^2={\epsilon }^2{h}_i^2$ or ${(x+1)}^2+y^2={\epsilon }^2{h}_i^2,i=1,2,\cdots ,10$. More precisely, we have the following theorem,

Theorem 2.

When there are the following perturbation parameters:

$$\begin{array}{cc}\delta =\hfill & (9111.485413\cdots ){\epsilon }^{10},\hfill \\ p_5=\hfill & -(2.515214199\cdots )\times {10}^5{\epsilon }^9+(1.784762660\cdots )\times {10}^8{\epsilon }^5+(1.078714264\cdots )\times {10}^6{\epsilon }^4\hfill \\ \hfill & +(49591.37283\cdots ){\epsilon }^3-(209.9605836\cdots ){\epsilon }^2-(1.008688029\cdots )\epsilon -(0.659300122\cdots ),\hfill \\ p_6=\hfill & (3.415756075\cdots )\times {10}^5{\epsilon }^8-(2.118886286\cdots )\times {10}^9{\epsilon }^7-(1.897497184\cdots )\times {10}^{10}{\epsilon }^6\hfill \\ \hfill & -(3.670336529\cdots )\times {10}^8{\epsilon }^5+(1.659560201\cdots )\times {10}^6{\epsilon }^4+(24680.05637\cdots ){\epsilon }^3\hfill \\ \hfill & +(2125.310729\cdots ){\epsilon }^2+(12.63057273\cdots )\epsilon -(0.342520436\cdots ),\hfill \\ p_3=\hfill & (9.068284943\cdots )\times {10}^{10}{\epsilon }^7+(3.611179371\cdots )\times {10}^8{\epsilon }^6+(2.306098530\cdots )\times {10}^8{\epsilon }^5\hfill \\ \hfill & +(1.439119873\cdots )\times {10}^6{\epsilon }^4+(74926.34227\cdots ){\epsilon }^3+(261.2478972\cdots ){\epsilon }^2\hfill \\ \hfill & +(1.648778517\cdots )\epsilon -(0.768618047\cdots ),\hfill \\ p_2=\hfill & -(9.618959849\cdots )\times {10}^9{\epsilon }^6-(1.865579656\cdots )\times {10}^8{\epsilon }^5+(8.417152117\cdots )\times {10}^5{\epsilon }^4\hfill \\ \hfill & +(12417.43671\cdots ){\epsilon }^3+(1078.990903\cdots ){\epsilon }^2+(6.511976302\cdots )\epsilon +(0.1486175491\cdots ),\hfill \\ p_1=\hfill & -(1.784762660\cdots )\times {10}^8{\epsilon }^5-(1.078714264\cdots )\times {10}^6{\epsilon }^4-(49591.37283\cdots ){\epsilon }^3\hfill \\ \hfill & +(209.9605836\cdots ){\epsilon }^2+(1.008688029\cdots )\epsilon +(0.6593001227\cdots ),\hfill \\ b_3=\hfill & (9.978626412\cdots )\times {10}^5{\epsilon }^4+(22599.66143\cdots ){\epsilon }^3-(1624.678559\cdots ){\epsilon }^2\hfill \\ \hfill & -(10.54120577\cdots )\epsilon +(0.6172899175\cdots ),\hfill \\ b_2=\hfill & (67493.38048\cdots ){\epsilon }^3+(1411.157799\cdots ){\epsilon }^2+(9.315816922\cdots )\epsilon -(0.6499316542\cdots ),\hfill \\ b_5=\hfill & -(0.330423401\cdots )\epsilon -(1.952586879\cdots ),\hfill \\ b_6=\hfill & (2988.179279\cdots ){\epsilon }^2+(18.75876352\cdots )\epsilon -(1.400793940\cdots ),\hfill \end{array}$$

(16)

then function $\overline{▵}\left(ϵh\right)=0$ has 10 real roots, which are close enough to $1,2,3,\cdots ,10$, System (1) has 10 limit cycles, which are close to periodic ring ${(x-1)}^2+y^2=j^2{ϵ}^2,j=1,2,\cdots ,10$ correspondingly.

Proof. 

Without loss of generality, we suppose the 10 real roots of Function (15) are $x=1,2,3,\cdots ,10$. Note that, in practice, any value can be set according to the actual needs to determine the configuration of the limit cycle. We can find $K_i$ by Function (15) as:

$$\begin{array}{ccc}{K}_0=57249.15127\cdots ,\hfill & K_1=-(1.676809466\cdots )\times {10}^5,\hfill & \\ K_2=(2.012046411\cdots )\times {10}^5,\hfill & K_3=-(1.326710586\cdots )\times {10}^5,\hfill & \\ K_4=53906.61995\cdots ,\hfill & K_5=-14231.11859\cdots ,\hfill & \\ K_6=2489.079129\cdots ,\hfill & K_7=-286.3404143\cdots ,\hfill & \\ K_8=20.82475740\cdots ,\hfill & K_9=-0.867698225\cdots ,\hfill & K_{10}=0.015776331\cdots .\hfill \end{array}$$

(17)

Here, $\delta$, $p_5$, $p_3$, $p_2$, $p_1$ and $b_3$ can be obtained by linearly solving the first seven Lyapunov constants function $V_i=0,i=0,1,2,\cdots ,6$, respectively. Thus, we first consider function $V_7=V_8=V_9=0$ to determine the perturbed parameters $b_2$, $b_6$ and $b_5$. Then, we can use these values to find the correct perturbation of rest parameters: use $V_6$ to find $b_3$, use $V_5$ to find $p_1$, use $V_4$ to find $p_2$, use $V_3$ to find $p_3$, use $V_2$ to find $p_6$, use $V_1$ to find $p_5$, use $V_0$ to find $\delta$. Without loss of generality, we suppose:

$$\begin{array}{c}{b}_2=b_{2c}+k_{11}\epsilon +k_{12}{\epsilon }^2+k_{13}{\epsilon }^3,\hfill \\ b_6=b_{5c}+k_{21}\epsilon +k_{22}{\epsilon }^2+k_{23}{\epsilon }^3,\hfill \\ b_5=b_{6c}+k_{31}\epsilon +k_{32}{\epsilon }^2+k_{33}{\epsilon }^3,\hfill \end{array}$$

(18)

where $b_{2c}$, $b_{5c}$ and $b_{6c}$ are critical value which are satisfied $V_7=V_8=V_9=0$, submitting (18) to $V_7$, $V_8$ and $V_9$, then expand them in Taylor series, we obtain:

$$\begin{array}{c}{V}_7=E_{70}+E_{71}\epsilon +E_{72}{\epsilon }^2+E_{73}{\epsilon }^3+o\left({\epsilon }^3\right),\hfill \\ V_8=E_{80}+E_{81}\epsilon +E_{82}{\epsilon }^2+o\left({\epsilon }^2\right),\hfill \\ V_9=E_{90}+E_{91}\epsilon +o\left(\epsilon \right),\hfill \end{array}$$

(19)

where $E_{ij}$ is a function in $b_2$, $b_5$ and $b_6$, therefore, it is a function in $k_{ij},i,j=1,2,3$.

Combining (13) and (17), balancing the coefficient of the same degree about $\epsilon$, we obtain:

$$\begin{array}{cc}{E}_{70}=E_{71}=E_{72}=0,\hfill & E_{80}=E_{81}=E_{90}=0,E_{73}=K_7=-286.3404143\cdots ,\hfill \\ E_{82}=K_8=20.82475740\cdots ,\hfill & E_{91}=K_9=-0.867698225\cdots .\hfill \end{array}$$

(20)

Solving equation set (20), we obtain:

$$\begin{array}{ccc}{k}_{31}=-0.330423401\cdots ,\hfill & k_{22}=2988.179279\cdots ,\hfill & k_{13}=67493.38048\cdots ,\hfill \\ k_{11}=9.315816922\cdots ,\hfill & k_{21}=18.75876352\cdots ,\hfill & \\ k_{12}=1411.157799\cdots ,\hfill & k_{23}=k_{33}=k_{32}=0.\hfill \end{array}$$

(21)

Therefore, we can get the perturbation of parameters $b_2$, $b_5$ and $b_6$ from (18) as the following,

$$\begin{array}{c}{b}_2=(67493.38048\cdots ){\epsilon }^3+(1411.157799\cdots ){\epsilon }^2+(9.315816922\cdots )\epsilon -(0.6499316542\cdots ),\hfill \\ b_6=(2988.179279\cdots ){\epsilon }^2+(18.75876352\cdots )\epsilon -(1.400793940\cdots ),\hfill \\ b_5=-(0.330423401\cdots )\epsilon -(1.952586879\cdots ).\hfill \end{array}$$

(22)

Next, we suppose:

$$b_3=k_{40}+k_{41}\epsilon +k_{42}{\epsilon }^2+k_{43}{\epsilon }^3+k_{44}{\epsilon }^4.$$

(23)

Submitting (22) and (23) to $V_6$ and expanding them in Taylor series of $\epsilon$-forth order,

$$V_{16}=E_{60}+E_{61}ϵ+E_{62}{ϵ}^2+E_{63}{ϵ}^3+E_{64}{ϵ}^4+o\left({ϵ}^4\right),$$

(24)

where $E_{ij}$ is a function in $b_3$, therefor, it is a function in $k_{4i},i=0,1,2,3,4$. Then, combining (14) and (15), balancing coefficients of the same degree about $\epsilon$, we obtain:

$$E_{60}=E_{61}=E_{62}=E_{63}=0,E_{64}=K_6=2489.079129\cdots .$$

(25)

Solving equation set (25), we obtain:

$$\begin{array}{ccc}{k}_{40}=0.617289917\cdots ,\hfill & k_{41}=-10.54120577\cdots ,\hfill & k_{42}=-1624.678559\cdots ,\hfill \\ k_{43}=22599.66143\cdots ,\hfill & k_{44}=(9.978626412\cdots )\times {10}^5,\hfill \end{array}$$

(26)

Then, $b_3$ can be expressed as:

$$\begin{array}{cc}{b}_3=\hfill & (9.978626412\cdots )\times {10}^5{\epsilon }^4+(22599.66143\cdots ){\epsilon }^3-(1624.678559\cdots ){\epsilon }^2\hfill \\ \hfill & -(10.54120577\cdots )\epsilon +(0.6172899175\cdots ).\hfill \end{array}$$

(27)

Repeating the above steps, we can also obtain:

$$\begin{array}{cc}{p}_1=\hfill & -(1.784762660\cdots )\times {10}^8{\epsilon }^5-(1.078714264\cdots )\times {10}^6{\epsilon }^4-(49591.37283\cdots ){\epsilon }^3\hfill \\ \hfill & +(209.9605836\cdots ){\epsilon }^2+(1.008688029\cdots )\epsilon +(0.6593001227\cdots ),\hfill \\ p_2=\hfill & -(9.618959849\cdots )\times {10}^9{\epsilon }^6-(1.865579656\cdots )\times {10}^8{\epsilon }^5+(8.417152117\cdots )\times {10}^5{\epsilon }^4\hfill \\ \hfill & +(12417.43671\cdots ){\epsilon }^3+(1078.990903\cdots ){\epsilon }^2+(6.511976302\cdots )\epsilon +(0.1486175491\cdots ),\hfill \\ p_3=\hfill & (9.068284943\cdots )\times {10}^{10}{\epsilon }^7+(3.611179371\cdots )\times {10}^8{\epsilon }^6+(2.306098530\cdots )\times {10}^8{\epsilon }^5\hfill \\ \hfill & +(1.439119873\cdots )\times {10}^6{\epsilon }^4+(74926.34227\cdots ){\epsilon }^3+(261.2478972\cdots ){\epsilon }^2\hfill \\ \hfill & +(1.648778517\cdots )\epsilon -(0.768618047\cdots ),\hfill \\ p_6=\hfill & (3.415756075\cdots )\times {10}^5{\epsilon }^8-(2.118886286\cdots )\times {10}^9{\epsilon }^7-(1.897497184\cdots )\times {10}^{10}{\epsilon }^6\hfill \\ \hfill & -(3.670336529\cdots )\times {10}^8{\epsilon }^5+(1.659560201\cdots )\times {10}^6{\epsilon }^4+(24680.05637\cdots ){\epsilon }^3\hfill \\ \hfill & +(2125.310729\cdots ){\epsilon }^2+(12.63057273\cdots )\epsilon -(0.342520436\cdots ),\hfill \\ p_5=\hfill & -(2.515214199\cdots )\times {10}^5{\epsilon }^9+(1.784762660\cdots )\times {10}^8{\epsilon }^5+(1.078714264\cdots )\times {10}^6{\epsilon }^4\hfill \\ \hfill & +(49591.37283\cdots ){\epsilon }^3-(209.9605836\cdots ){\epsilon }^2-(1.008688029\cdots )\epsilon -(0.659300122\cdots ).\hfill \end{array}$$

(28)

Finally, by solving function $V_0=K_0{ϵ}^{10}=(57249.15127\cdots ){ϵ}^{10}$, we obtain the value of the perturbed parameter $\delta$: $\delta =(9111.485413\cdots ){\epsilon }^{10}$.

Now, we have obtained all the values of the perturbed parameters, which are given in (16), we compute the Lyapunov constants after perturbing as the following,

$$\begin{array}{cc}{V}_{10}=(57249.15127\cdots ){ϵ}^{10}+o\left({ϵ}^{10}\right),\hfill & V_{11}=-(1.676809466\cdots )\times {10}^5{ϵ}^9+o\left({ϵ}^9\right),\hfill \\ V_{12}=(2.012046411\cdots )\times {10}^5{ϵ}^8+o\left({ϵ}^8\right),\hfill & V_{13}=-(1.326710586\cdots )\times {10}^5{ϵ}^7+o\left({ϵ}^7\right),\hfill \\ V_{14}=(53906.61995\cdots ){ϵ}^6+o\left({ϵ}^6\right),\hfill & V_{15}=-(14231.11859\cdots ){ϵ}^5+o\left({ϵ}^5\right),\hfill \\ V_{16}=(2489.079129\cdots ){ϵ}^4+o\left({ϵ}^4\right),\hfill & V_{17}=-(286.3404143\cdots ){ϵ}^3+o\left({ϵ}^3\right),\hfill \\ V_{18}=(20.82475740\cdots ){ϵ}^2+o\left({ϵ}^2\right),\hfill & V_{19}=-(0.867698225\cdots )ϵ+o\left(ϵ\right),\hfill \\ V_{110}=(0.015776331\cdots )+o\left(1\right).\hfill \end{array}$$

(29)

Based on the above Lyapunov constants after perturbing, Function $\overline{▵}\left(ϵh\right)=0$ in (12) has 10 real roots, which are close enough to 1, 2, 3, 4, 5, 6, 7, 8, 9, 10. Therefore, System (1) has 10 limit cycles that are close to periodic rings: ${(x-1)}^2+y^2=j^2{ϵ}^2$, $j=1,2,...,10$. In order to give the exact approximate value of the limit cycle, we use $\epsilon ={10}^{-8}$,

$$\begin{array}{cc}{v}_0=(5.724915127\cdots )\times {10}^{-76},\hfill & v_1=-(1.676809466\cdots )\times {10}^{-67},\hfill \\ v_2=(2.012046411\cdots )\times {10}^{-59},\hfill & v_3=-(1.326710586\cdots )\times {10}^{-51},\hfill \\ v_4=(5.390661995\cdots )\times {10}^{-44},\hfill & v_5=-(1.423111859\cdots )\times {10}^{-36},\hfill \\ v_6=(2.489079129\cdots )\times {10}^{-29},\hfill & v_7=-(2.863404143\cdots )\times {10}^{-22},\hfill \\ v_8=(2.082475740\cdots )\times {10}^{-15},\hfill & v_9=-(0.867698225\cdots )\times {10}^{-8},\hfill \\ v_{10}=0.015776331\cdots ,\hfill & \hfill \end{array}$$

then, $▵\left(h\right)=0$ has 10 real roots:

$$\begin{array}{cc}{h}_1=(1.000000014\cdots )\times {10}^{-8},\hfill & h_2=(1.999999085\cdots )\times {10}^{-8},\hfill \\ h_3=(3.000015248\cdots )\times {10}^{-8},\hfill & h_4=(3.999895408\cdots )\times {10}^{-8},\hfill \\ h_5=(5.000336853\cdots )\times {10}^{-8},\hfill & h_6=(5.999583523\cdots )\times {10}^{-8},\hfill \\ h_7=(6.999654257\cdots )\times {10}^{-8},\hfill & h_8=(8.001583007\cdots )\times {10}^{-8},\hfill \\ h_9=(8.998357200\cdots )\times {10}^{-8},\hfill & h_{10}=(1.000057666\cdots )\times {10}^{-8},\hfill \end{array}$$

This is the approximate amplitude of the 10 limit cycles.

Therefore, without loss of generality, we obtain a configuration distribution about 10 limit cycles around $(1,0)$: ${(x-1)}^2+y^2=h_j^2$, $j=1,2,\dots ,10$. In the same way, we can obtain the configuration distribution about 10 limit cycles around $(-1,0)$. □

5. Conclusions

In this paper, we present a method for the achievement of simulation and realization with limit cycles, hence, determining the configuration of limit cycles. This method is not only suitable for finding 10 limit cycles in our paper, but also suitable for finding any number of limit cycles in other general (switching or smoothing) systems. By our method, the general perturbation form of parameters is obtained, and this ensures that we get more than one location relationship of limit cycles, so that we can find the best set of parameter values for the image simulation.

The numerical simulation of the limit cycle is always a very difficult problem in the dynamic system. When the number of limit cycles is large, how to carry out the parameter and numerical realization of the limit cycle requires a more systematic and general method. Our method is to solve this problem. Based on the knowledge of symbolic computation, this method can be efficiently implemented on the computer.