Isochronicity and linearizability of a planar cubic system

TL;DR

This paper studies when certain cubic planar differential systems can be simplified to linear systems and explores conditions for isochronous centers, providing a classification based on system parameters.

Contribution

It offers a classification of linearizable cubic systems and conditions for isochronous centers, advancing understanding of their dynamics.

Findings

Derived parameter conditions for linearizability

Identified coexistence criteria for isochronous centers

Classified families of cubic systems based on linearizability

Abstract

In this paper we investigate the problem of linearizability for a family of cubic complex planar systems of ordinary differential equations. We give a classification of linearizable systems in the family obtaining conditions for linearizability in terms of parameters. We also discuss coexistence of isochronous centers in the systems.

Equations

$$\dot{x}=-y+P(x,y),\dot{y}=x+Q(x,y),$$

$$\dot{x}$$

$$\dot{y}$$

$$p_2=$$

$$q_2=$$

$$r_2=$$

$$\dot{x}=-y+\sum _{p+q\ge 2}^n{a}_{p,q}{x}^p{y}^q=P(x,y),\dot{y}=x+\sum _{p+q\ge 2}^n{b}_{p,q}{x}^p{y}^q=Q(x,y),$$

$$x_1=x+\sum _{m+n\ge 2}{c}_{m,n}{x}^m{y}^n:=H_1(x,y),y_1=y+\sum _{m+n\ge 2}{d}_{m,n}{x}^m{y}^n:=H_2(x,y),$$

$$\dot{z}=iz+Z(z,\bar{z}),$$

$${\dot{x}}_1=$$

$${\dot{y}}_1=$$

$$\sum _{m+n\ge 2}{d}_{m,n}{x}^m{y}^n+\sum _{p+q\ge 2}^n{a}_{p,q}{x}^p{y}^q+\left(\sum _{m+n\ge 2}m{c}_{m,n}{x}^{m-1}{y}^n\right)\left(-y+\sum _{p+q\ge 2}^n{a}_{p,q}{x}^p{y}^q\right)$$

$$+\left(\sum _{m+n\ge 2}n{c}_{m,n}{x}^m{y}^{n-1}\right)\left(x+\sum _{p+q\ge 2}^n{b}_{p,q}{x}^p{y}^q\right)\equiv 0,$$

$$-\sum _{m+n\ge 2}{c}_{m,n}{x}^m{y}^n+\sum _{p+q\ge 2}^n{b}_{p,q}{x}^p{y}^q+\left(\sum _{m+n\ge 2}m{d}_{m,n}{x}^{m-1}{y}^n\right)\left(-y+\sum _{p+q\ge 2}^n{a}_{p,q}{x}^p{y}^q\right)$$

$$+\left(\sum _{m+n\ge 2}n{d}_{m,n}{x}^m{y}^{n-1}\right)\left(x+\sum _{p+q\ge 2}^n{b}_{p,q}{x}^p{y}^q\right)\equiv 0.$$

$$z=x+iy,w=x-iy$$

$$\dot{z}=i(z+X(z,w)),\dot{w}=-i(w+Y(z,w)),$$

$$\dot{z}=z+X(z,w),\dot{w}=-w-Y(z,w).$$

$$\frac{\partial f}{\partial z}\dot{z}+\frac{\partial f}{\partial w}\dot{w}=Kf,$$

$$Z_1(z,w)=\prod _{j=0}^m{f}_j^{{\alpha }_j}(z,w)=z+{\tilde{Z}}_1(z,w),\text{leavevmode}\text{leavevmode}{W}_1(z,w)=\prod _{j=0}^n{g}_j^{{\beta }_j}(z,w)=w+{\tilde{W}}_1(z,w),$$

$$K_0+{\alpha }_1{K}_1+\cdot \cdot \cdot +{\alpha }_s{K}_s=1\text{mbox}and{L}_0+{\beta }_1{L}_1+\cdot \cdot \cdot +{\beta }_t{L}_t=-1.$$

$$z_1=H_1(z,w)=f_0{f}_1^{{\alpha }_1}\cdot \cdot \cdot f_s^{{\alpha }_s},y_1=H_2(z,w)=g_0{g}_1^{{\beta }_1}\cdot \cdot \cdot g_t^{{\beta }_t}.$$

$$i_1=$$

$$j_1=$$

$${\tilde{i}}_2=$$

$$+6a_{11}^2{b}_{11}^2+490a_{02}{a}_{20}{b}_{11}^2+464a_{20}^2{b}_{11}^2-96a_{02}{b}_{11}^3-110a_{20}{b}_{11}^3+6b_{11}^4-170a_{11}^3{b}_{20}$$

$$-720a_{11}{a}_{20}^2{b}_{20}+720a_{11}{a}_{20}{b}_{11}{b}_{20}-146a_{11}{b}_{11}^2{b}_{20}-550a_{11}^2{b}_{20}^2+2600a_{02}{a}_{20}{b}_{20}^2$$

$$+3080a_{20}^2{b}_{20}^2-1580a_{02}{b}_{11}{b}_{20}^2-2060a_{20}{b}_{11}{b}_{20}^2+154b_{11}^2{b}_{20}^2-160a_{11}{b}_{20}^3+520b_{20}^4$$

$$-100a_{11}{a}_{20}{r}_{02}+560a_{11}{b}_{11}{r}_{02}-6800a_{20}{b}_{20}{r}_{02}+2180b_{11}{b}_{20}{r}_{02}+5250r_{02}^2$$

$$-55a_{11}^2{r}_{11}+50a_{02}{a}_{20}{r}_{11}-170a_{20}^2{r}_{11}+5a_{02}{b}_{11}{r}_{11}+225a_{20}{b}_{11}{r}_{11}-55b_{11}^2{r}_{11}$$

$$-220a_{11}{b}_{20}{r}_{11}-220b_{20}^2{r}_{11}-100a_{11}{a}_{20}{r}_{20}+560a_{11}{b}_{11}{r}_{20}+800a_{02}{b}_{20}{r}_{20}$$

$$-6000a_{20}{b}_{20}{r}_{20}+2180b_{11}{b}_{20}{r}_{20}+8500r_{02}{r}_{20}+3250r_{20}^2),$$

$${\tilde{j}}_2=$$

$$+48a_{02}{a}_{20}^2{b}_{20}-6a_{11}^2{b}_{11}{b}_{20}+16a_{02}{a}_{20}{b}_{11}{b}_{20}+56a_{20}^2{b}_{11}{b}_{20}-4a_{02}{b}_{11}^2{b}_{20}$$

$$-8a_{20}{b}_{11}^2{b}_{20}-2b_{11}^3{b}_{20}-40a_{11}{a}_{20}{b}_{20}^2-12a_{11}{b}_{11}{b}_{20}^2-48a_{20}{b}_{20}^3-8b_{11}{b}_{20}^3$$

$$-24a_{11}^2{r}_{02}-16a_{20}^2{r}_{02}+64a_{20}{b}_{11}{r}_{02}+4b_{11}^2{r}_{02}-64a_{11}{b}_{20}{r}_{02}-32b_{20}^2{r}_{02}$$

$$+128a_{02}{b}_{20}{r}_{11}+128a_{20}{b}_{20}{r}_{11}-128r_{02}{r}_{11}-8a_{11}^2{r}_{20}-32a_{02}{a}_{20}{r}_{20}+16a_{20}^2{r}_{20}$$

$$-80a_{02}{b}_{11}{r}_{20}-80a_{20}{b}_{11}{r}_{20}+20b_{11}^2{r}_{20}+32b_{20}^2{r}_{20}-128r_{11}{r}_{20}).$$

$$a_{02}+a_{20}=a_{11}+2b_{20}=b_{11}-2a_{20}=r_{02}+r_{20}=0.$$

$$J_1=⟨a_{02}+a_{20},a_{11}+2b_{20},b_{11}-2a_{20},r_{02}+r_{20}⟩,$$

$$\overline{\mathbf{V}(I)\setminus \mathbf{V}(H)}\subset \mathbf{V}(I:H),$$

Full text

1. Introduction

For planar real analytic differential systems of the form

[TABLE]

where $P$ and $Q$ are polynomials without constant and linear terms, it is well known that the origin can be either a center or a focus. In the first case all solutions in a neighbourhood of the origin are periodic and their trajectories are closed curves. If the origin is a center, there arises the problem to determine whether all periodic solutions in a neighbourhood of the origin have the same period. This problem is known as the isochronicity problem.

The studies of isochronicity of polynomial differential systems go back to Lloud [20], who found the necessary and sufficient conditions for isochronicity of system (1.1) when $P$ and $Q$ are homogeneous polynomials of degree two. Latter on, Pleshkan [22] solved the isochroniciy problem in the case when $P$ and $Q$ are homogeneous polynomials of degree three (see also [17]). In the case when $P$ and $Q$ are homogeneous polynomials of degree five the problem was solved in [23], however, the case of homogeneous polynomials of degree four is still unsolved. A number of works is devoted to the investigation of some other particular families (see, e.g [2, 5, 6, 7, 8, 9, 15, 18, 19, 21, 25, 27] and references given there).

The following family of planar cubic systems

[TABLE]

where

[TABLE]

has been studied in [3, 4, 16, 19] for the case when all parameters are real.

In [3] and [4] the authors have shown that real system (1.2) has a center and an isochronous center, respectively, if and only if in polar coordinates after some transformations it can be written in one of four and five forms, respectively. However from their results it is difficult to determine the conditions on parameters of polynomials $p_{2},q_{2},r_{2}$ for existence of centers and isochronous centers. Conditions on parameters of $p_{2},q_{2},r_{2}$ for the existence of a center were obtained in [19] and latter on using another approach in [16].

In the work [19] published in 1997 the authors obtained the necessary and sufficient conditions for existence of isochronous center of system (1.2) represented by four series of condition on coefficients of the system, however in the more recent paper [4] published in 1999 the authors gave five conditions for existence of isochronous center of system (1.2). One of aims of this paper is to clarify the conditions for isochronicity of system (1.2). For this purpose we use the approach different from the ones of [19] and [4], namely we consider system (1.2) as system with complex coefficients and find conditions for linearization of the system. We obtain five series of conditions for linearizability of (1.2) and show that all linearizable systems are Darboux linearizable.

The paper is organized as follows. In Section 2 we recall some definitions and describe briefly a procedure to study the isochronicity and linearizability of polynomial systems. Applying this procedure, in Section 3 we present our main result, Theorem 3.1, which gives conditions for linearizability of system (1.2). In Section 4 we present the relation between the results obtained in Theorem 3.1 (and in [19]) and the results of [4]. Finally, in the last section we discuss the coexistence of isochronous centers in system (1.2).

2. Linearizability quantities and Darboux linearization

In this section we remind some statements related to isochronicity and linearizability of polynomial differential systems and describe an approach to compute the linearizability quantities for the system

[TABLE]

where $a_{p,q},b_{p,q}$ are real or complex parameters.

If the equilibrium point at the origin of real system (2.1) is known to be a center it is said that this center is isochronous if all periodic solutions of (2.1) in a neighbourhood of the origin have the same period. System (2.1) is said to be linearizable if there is an analytic change of coordinates

[TABLE]

that reduces (2.1) to the linear system $\dot{x}_{1}=-y_{1}$, $\dot{y}_{1}=x_{1}$.

The following theorem, which goes back to Poincaré and Lyapunov, shows that the linearizability and isochronicity problems are equivalent. A proof can be found e.g. in [25].

Theorem 2.1**.**

The origin of real system (2.1) is an isochronous center if and only if the system is linearizable.

Two most common ways to compute isochronicity quantities (obstacles for isochronicity) are passing to polar coordinates (the approach used in [4]) or writing real system (2.1) in the complex form

[TABLE]

using the change $z=x+iy$ and then looking for a linearization of equation (2.3) (the approach used in [19]). Since we would like to perform the investigation differently from [4] and [19] we use another computational approach. Namely, we look for conditions for linearizability of system (2.1) arising from applying transformation (2.2).

Taking the derivatives with respect to $t$ on both sides of each equation of (2.2) we obtain

[TABLE]

Hence, the change of coordinates (2.2) linearizes system (2.1) if it holds that

[TABLE]

Obstacles for the fulfilment of equations in (2.4) give us necessary conditions for existence of a linearizing change of coordinates (2.2) of system (2.1). Thus, a computational procedure to find necessary conditions for linearizability can be described as follows.

(1) Write the left hand sides of two equations in (2.4) in the form $\sum_{k,l\geq 2}h_{1}^{(k,l)}x^{k}y^{l}$, and $\sum_{k,l\geq 2}h_{2}^{(k,l)}x^{k}y^{l}$, respectively, where $h_{1}^{(k,l)}$ and $h_{2}^{(k,l)}$ are polynomials in the parameters $a_{p,q},b_{p,q}$ ($p+q\geq 2$) of system (2.1) and $c_{m,n}$, $d_{m,n}$ ($m+n\geq 2$) of (2.2).

(2) Solve the polynomial system $h_{i}^{(k,l)}=0$ ($i=1,2$, $k+l=2$) for the coefficients $c_{m,n}$, $d_{m,n}$ ($m+n=2$) of (2.2).

(3) Solve the polynomial system $h_{i}^{(k,l)}=0$ ($i=1,2$, $k+l=3$) for the coefficients $c_{m,n}$, $d_{m,n}$ ($m+n=3$) of (2.2). In general case the system cannot be solved. However dropping from it two suitable equations we obtain a system that has a solution. We denote the two dropped polynomials on the left hand sides of these two equations by $i_{1}$ and $j_{1}$.

(4) Proceed step-by-step solving the polynomial systems $h_{i}^{(k,l)}=0$ ($i=1,2$, $k+l=r$, $r>3$). Generally speaking, at all steps when $r=k+l$ is an odd number the polynomial system $h_{i}^{(k,l)}=0$ ($i=1,2$, $k+l=r$) cannot be solved. Dropping on each such step two suitable equations (and denoting by $i_{(r-1)/2}$ and $j_{(r-1)/2}$ the corresponding polynomials), we obtain a system that has a solution.

This procedure yields the polynomials $i_{k}$ and $j_{k}$ which are polynomials in the parameters $a_{p,q}$ and $b_{p,q}$ of system (2.1) called the linearizability quantities. It is clear that system (2.1) admits a linearizing change of coordinates (2.2) if and only if $i_{k}=j_{k}=0$ for all $k>1$. Thus, the simultaneous vanishing of all linearizability quantities provide conditions which characterize when the system (2.1) is linearizable (equivalently it has an isochronous center at the origin). The ideal $\mathcal{L}=\langle i_{1},j_{1},i_{2},j_{2},...\rangle\subset\mathbb{C}[a,b]$ defined by the linearizability quantities is called the linearizability ideal and its affine variety, $V_{\mathcal{L}}={\bf V}(\mathcal{L})$, is called the linearizability variety. Therefore, the linearizability problem will be solved finding the variety $V_{\mathcal{L}}$.

By the Hilbert Basis Theorem there exists a positive integer $k$ such that $\mathcal{L}=\mathcal{L}_{k}=\langle i_{1},j_{1},...,i_{k},j_{k}\rangle$. Note that the inclusion $V_{\mathcal{L}}\subset{\bf V}(\mathcal{L}_{k})$ holds for any $k\geq 1$. The opposite inclusion is verified finding the irreducible decomposition of the variety ${\bf V}(\mathcal{L}_{k})$ and then checking that any point of each component of the decomposition corresponds to a linearizable system. The irreducible decomposition can be found using the routine minAssGTZ [12] (which is based on the algorithm of [13]) of the computer algebra system Singular [11], however it involves extremely laborious calculations.

One of the most efficient method to find a linearizing change of coordinates is the Darboux linearization method. To construct a Darboux linearization it is convenient to perform the substitution

[TABLE]

obtaining from (2.1) a system of the form

[TABLE]

and, after the rescaling of time by $i$, the system

[TABLE]

Since the change of coordinates (2.5) is analytic, system (2.1) is linearizable if and only if system (2.7) is linearizable.

We remind that a Darboux factor of system (2.7) is a polynomial $f(z,w)$ satisfying

[TABLE]

where $K(z,w)$ is a polynomial called the cofactor of $f$. A Darboux linearization of system (2.7) is an analytic change of coordinates $z_{1}=Z_{1}(z,w)$, $w_{1}=W_{1}(z,w)$, such that

[TABLE]

which linearizes (2.7), where $f_{j},g_{j}\in\mathbb{C}[z,w]$, $\alpha_{j},\beta_{j}\in\mathbb{C}$, and $\tilde{Z}_{1}$ and $\tilde{W}_{1}$ have neither constant nor linear terms. System (2.7) is said to be Darboux linearizable if it admits a Darboux linearization. The next theorem provides a way to construct a Darboux linearization using Darboux factors (see e.g. [25] for a proof).

Theorem 2.2**.**

System (2.7) is Darboux linearizable if and only if there exit $s+1\geq 1$ Darboux factors $f_{0},...,f_{s}$ with corresponding cofactors $K_{0},...,K_{s}$ and $t+1\geq 1$ Darboux factors $g_{0},...,g_{t}$ with corresponding cofactors $L_{0},...,L_{t}$ with the following properties:

1. a.

$f_{0}(z,w)=z+\cdot\cdot\cdot\mbox{ but }f_{j}(0,0)=1\mbox{ for }j\geq 1$; 2. b.

$g_{0}(z,w)=w+\cdot\cdot\cdot\mbox{ but }g_{j}(0,0)=1\mbox{ for }j\geq 1$; and 3. c.

there are $s+t$ constants $\alpha_{1},...,\alpha_{s},\beta_{1},...,\beta_{t}\in\mathbb{C}$ such that

[TABLE]

The Darboux linearization is then given by

[TABLE]

Sometimes we cannot find enough Darboux factors to construct Darboux linearizations of both equations of the system. Let say that we can find only transformation $z_{1}$, which linearizes the first equation of (2.7). If we can find a first integral of system (2.7) of the form $\Psi=xy+h.o.t.$ then the second equation of (2.7) can be linearized by the transformation $w_{1}=\dfrac{\Psi}{z_{1}}$. We note also that if system (2.7) has $p$ irreducible Darboux factors $f_{1},...,f_{p}$ with associated cofactors $K_{1},...,K_{p}$, satisfying $s_{1}K_{1}+...+s_{p}K_{p}=0$, then the function $H=f_{1}^{s_{1}}...f_{p}^{s_{p}}$ is a first integral of (2.7).

3. Linearizability of system (1.2)

In this section we obtain the necessary and sufficient conditions for linearizability of system (1.2) with complex parameters.

Without loss of generality, we suppose that $b_{02}=-b_{20}$ in system (1.2). Otherwise, if $a_{02}+a_{20}=0$, we can apply the transformation $\tilde{x}=x+(b_{02}+b_{20})y/(a_{02}+a_{20})$ and $\tilde{y}=y-(b_{02}+b_{20})x/(a_{02}+a_{20})$. When $a_{02}+a_{20}=0$, we only need to make the change $(x,y)\to(y,x)$ together with the time scaling $dt=-d\tau$ and obtain the same effect.

The following theorem gives the conditions for linearizability of system (1.2).

Theorem 3.1**.**

Complex system (1.2) with $b_{02}=-b_{20}$ is linearizable at the origin if one of the following conditions holds:

1. (1)

$4a_{20}^{2}+a_{11}^{2}+4a_{11}b_{20}+4b_{20}^{2}-4a_{20}b_{11}+b_{11}^{2}=r_{20}+r_{02}=a_{02}+a_{20}=0$*, * 2. (2)

$a_{02}=r_{02}=a_{11}+2b_{20}=b_{11}-4a_{20}=r_{11}+b_{20}^{2}=r_{20}-a_{20}b_{20}=0$, 3. (3)

$4a_{02}+a_{20}=a_{11}+2b_{20}=2b_{11}-a_{20}=4r_{02}+a_{20}b_{20}=r_{11}+b_{20}^{2}=r_{20}-a_{20}b_{20}=0$, 4. (4)

$a_{02}=r_{02}=a_{11}+2b_{20}=b_{11}-a_{20}=r_{20}-a_{20}b_{20}=0$, 5. (5)

$9a_{11}^{2}-12a_{11}b_{20}+4b_{20}^{2}+4b_{11}^{2}=-6a_{11}b_{20}+4b_{20}^{2}+2a_{20}b_{11}-b_{11}^{2}=6a_{20}a_{11}-4a_{20}b_{20}-3a_{11}b_{11}+10b_{20}b_{11}=4a_{20}^{2}-12a_{11}b_{20}+24b_{20}^{2}-b_{11}^{2}=-\frac{4}{3}b_{20}^{2}-\frac{b_{11}^{2}}{3}+r_{11}=\frac{4}{9}a_{20}b_{20}+\frac{a_{11}b_{11}}{6}-\frac{b_{20}b_{11}}{9}+r_{02}=\frac{a_{20}a_{11}}{6}-\frac{a_{20}b_{20}}{3}+\frac{a_{11}b_{11}}{12}-\frac{b_{20}b_{11}}{6}+r_{20}+r_{02}=a_{02}+\frac{a_{20}}{3}-\frac{b_{11}}{3}=0$.

Proof.

Using the computer algebra system Mathematica following the computational procedure described in the previous section we computed the first eight pairs of the linearizability quantities for system (1.2). The first pair is

[TABLE]

and the second pair reduced by the Groebner basis of $\langle i_{1},j_{1}\rangle$ is

[TABLE]

The other polynomials have very long expressions, so we do not present them here, however, the reader can easily compute them using any available computer algebra system.

To find conditions for linearizability we have to “solve” the system $i_{1}=\dots=i_{8}=j_{1}=\dots=j_{8}=0$, or, more precisely, to find the irreducible decomposition of the variety ${\bf V}(\mathcal{L}_{8})$ of the ideal $\mathcal{L}_{8}=\langle i_{1},j_{1},...,i_{8},j_{8}\rangle$. Although nowadays there are few algorithms for computing such decompositions the calculations seldom can be completed over the field of rational numbers for non-trivial ideals due to high complexity of Groebner bases computations. We tried to perform the decomposition of the variety of ${\bf V}(\mathcal{L}_{8})$ using the routine minAssGTZ [12] of Singular [11], however we have not succeeded to complete computations neither over $\mathbb{Q}$ nor over the field $\mathbb{Z}_{32003}$.

To find the decomposition we proceed as follows. First, we use the conditions for isochronicity of real system (1.2) obtained in [19], which are conditions (2)–(4) of the statement of the theorem and the condition

[TABLE]

It is clear that under condition (3.1) and conditions (2)–(4) of the theorem complex system (1.2) is linearizable as well.

Denote by $J_{1}$ the ideal generated by polynomials defining condition (3.1), that is,

[TABLE]

and by $J_{2},J_{3},J_{4}$ ideals generated by polynomials of conditions (2)–(4) of the theorem.

As we have mentioned above we are not able to compute the decomposition of the variety ${\bf V}(\mathcal{L}_{8})$ of $\mathcal{L}_{8}$ (that is, to find the minimal associate primes of $\mathcal{L}_{8}$) even over fields of finite characteristic. However using the ideals $J_{1}-J_{4}$ we can find the decomposition of the variety ${\bf V}(\mathcal{L}_{8})$. The idea is to subtract from ${\bf V}(\mathcal{L}_{8})$ the components defined by the ideals $J_{1}-J_{4}$ and then find the decomposition of the remaining variety. For this aim we use the theorem (see, e.g. [10, Chapter 4]) for the proof), which says that given two ideals $I$ and $H$ in $k[x_{1},\dots,x_{n}]$,

[TABLE]

where the overline indicates the Zariski closure. Moreover, if $k=\mathbb{C}$ and $I$ is a radical ideal, then

[TABLE]

Thus, to remove the components ${\bf V}(J_{1}),...,{\bf V}(J_{4})$ from ${\bf V}(\mathcal{L}_{8})$, we compute over the field $\mathbb{Z}_{32003}$ with the intersect of Singular the intersection $J=J_{1}\cap J_{2}\cap J_{3}\cap{J}_{4}$ (clearly, ${\bf V}(J)={\bf V}(\ J_{1})\cup{\bf V}(\ J_{2})\cup{\bf V}(J_{3})\cup{\bf V}({J}_{4})$), then with the radical we compute $R=\sqrt{\mathcal{L}_{8}}$, then with quotient we compute the ideal $G=R:J$ and, finally, with minAssGTZ we compute the minimal associate primes of $G$, obtaining that $G=G_{1}\cap G_{2}$, where

$G_{1}=\langle r_{20}+r_{02},a_{02}+a_{20}a_{20}^{2}+8001a_{11}^{2}+a_{11}b_{20}+b_{20}^{2}-a_{20}b_{11}+8001b_{11}^{2}\rangle$

and

$G_{2}=\langle a_{02}+10668a_{20}-10668b_{11},a_{11}r_{02}-10667b_{20}r_{02}+14224a_{20}r_{11}-14224b_{11}r_{11},a_{20}r_{02}+16000b_{11}r_{02}+16001a_{11}r_{11}-b_{20}r_{11},r_{20}^{2}+6r_{20}r_{02}+9r_{02}^{2}+r_{11}^{2},b_{11}r_{20}+3b_{11}r_{02}-16000a_{11}r_{11}-b_{20}r_{11},b_{20}r_{20}+3b_{20}r_{02}-16001a_{20}r_{11}-8001b_{11}r_{11},a_{11}r_{20}-2b_{20}r_{02}-a_{20}r_{11}-16001b_{11}r_{11},a_{20}r_{20}-15997b_{11}r_{02}+8003a_{11}r_{11}-16001b_{20}r_{11},a_{11}b_{11}-2b_{20}b_{11}+4r_{20}+6r_{02},b_{20}^{2}+8001b_{11}^{2}+8000r_{11},\\ a_{11}b_{20}-10668a_{20}b_{11}+10668b_{11}^{2}+16001r_{11},a_{20}b_{20}-16001b_{20}b_{11}+16000r_{20},a_{11}^{2}-14224a_{20}b_{11}-7111b_{11}^{2}-10668r_{11},a_{20}a_{11}+b_{20}b_{11}+r_{20}+3r_{02},a_{20}^{2}-a_{20}b_{11}+8000b_{11}^{2}+3r_{11},a_{20}b_{11}r_{11}-16001b_{11}^{2}r_{11}-9r_{20}r_{02}-27r_{02}^{2}-6r_{11}^{2},b_{20}b_{11}r_{02}-8001b_{11}^{2}r_{11}+8003r_{02}^{2}+r_{11}^{2},a_{20}b_{11}^{2}-16001b_{11}^{3}-18b_{20}r_{02}-6a_{20}r_{11},b_{11}^{2}r_{02}r_{11}+b_{20}b_{11}r_{11}^{2}+15997r_{20}r_{02}^{2}+15988r_{02}^{3}-r_{20}r_{11}^{2}+15997r_{02}r_{11}^{2},b_{11}^{3}r_{02}+b_{20}b_{11}^{2}r_{11}-9b_{20}r_{02}^{2}-3b_{11}r_{02}r_{11}-4b_{20}r_{11}^{2}\rangle.$

Since $8001\equiv\frac{1}{4}{\ \rm mod\ }32003$, lifting the ideal $G_{1}$ from the ring of polynomials over the field $\mathbb{Z}_{32003}$ to the ring of polynomials over the field $\mathbb{Q}$ we obtain polynomials given in condition (1) of the theorem.

Similarly, lifting the ideal $G_{2}$ we obtain the ideal which we denote by $J_{5}$ (the lifting can be performed algorithmically using the algorithm of [26] and the Mathematica code of [14]). Simple computations show that ${\bf V}(J_{5})$ is the same set as the set given by conditions (5) of the theorem.

To check the correctness of the obtained conditions we use the procedure described in [24]. First, we computed the ideal $\tilde{J}=J_{1}\cap\ J_{2}\cap J_{3}\cap{J}_{4}\cap{J}_{5}$, which defines the union of all five components of the theorem and checked that Groebner bases of all ideals $\langle\tilde{J},1-wi_{k}\rangle$, $\langle\tilde{J},1-wj_{k}\rangle$ (where $k=1,\dots,8$ and $w$ is a new variable) computed over $\mathbb{Q}$ are $\{1\}$. By the Radical Membership Test (see e.g. [10, 25]) it means that

[TABLE]

To check the opposite inclusion it is sufficient to check that

[TABLE]

for all polynomials $f$ from a basis of $\tilde{J}$. Unfortunately, we were not able to perform the check over $\mathbb{Q}$, however we have checked that (3.5) holds over few fields of finite characteristic. It yields that (3.5) holds with high probability [1] 111For this reason we say in the statement of Theorem 3.1 that conditions (1)-(5) are only necessary, but not necessary and sufficient conditions for linearizability of system (1.2)..

We now prove that under each of conditions (1)-(5) of the theorem the system is linearizable.

$Case\hskip 2.84544pt(1):$ In this case $a_{11}=-2b_{20}\pm(2a_{20}-b_{11})i$. We consider only the case $a_{11}=-2b_{20}+(2a_{20}-b_{11})i$, since when $a_{11}=-2b_{20}-(2a_{20}-b_{11})i$ the consideration is analogous. In this case system (1.2) becomes

[TABLE]

After the substitution (2.5) we obtain from (3.6) the system

[TABLE]

System (3.7) has Darboux factors

[TABLE]

where $\xi=\sqrt{b_{11}^{2}-4ib_{11}b_{20}-4b_{20}^{2}-4r_{11}+8ir_{20}}$ and

[TABLE]

It is easy to verify that the first of conditions (2.8) is satisfied with $f_{0}=l_{1}$, $f_{1}=l_{4}$, $f_{2}=l_{5}$, $f_{3}=l_{6}$, and

[TABLE]

Moreover, the system has the Darboux first integral

[TABLE]

where $s_{1}=1,\leavevmode\nobreak\ s_{2}=-1,\leavevmode\nobreak\ s_{3}=-\frac{\eta_{-}}{\eta_{+}},\leavevmode\nobreak\ s_{4}=\frac{\eta_{-}}{\eta_{+}},$ $f_{1}=l_{3}$, $f_{2}=l_{4}$, $f_{3}=l_{5}$, and $f_{4}=l_{6}$.

Therefore, the system is linearizable by the substitution

[TABLE]

$Case\hskip 2.84544pt(2):$ In this case system (1.2) becomes

[TABLE]

and after substitution (2.5) we have the system

[TABLE]

System (3.9) has the Darboux factors

[TABLE]

which when $a_{20}\neq 0$ allow to construct the Darboux linearization

[TABLE]

where

[TABLE]

Since the set of linearizable system is an affine variety and therefore it is a closed set in the Zariski topology, the system is linearizable also when $a_{20}=0$.

$Case\hskip 2.84544pt(3):$ In this case system (1.2) becomes

[TABLE]

and the corresponding system of the form (2.7) is

[TABLE]

System (3.12) has the following Darboux factors

[TABLE]

which when $a_{20}\neq 0$ allow to construct the Darboux linearization

[TABLE]

where

[TABLE]

If $a_{20}=0$, case (3) is equivalent to case (2). Thus system (3.12) is linearizable.

$Case\hskip 2.84544pt(4):$ In this case system (1.2) becomes

[TABLE]

and after the substitution (2.5) we obtain the system

[TABLE]

which admits the Darboux factors

[TABLE]

where $C=\sqrt{a_{20}^{2}-4b_{20}^{2}-4r_{11}}$. When $C\neq 0$ we obtain the Darboux linearization

[TABLE]

where

[TABLE]

Using the same argument as in case (2) we conclude that the system is linearizable also when $C=0$.

$Case\hskip 2.84544pt(5):$ If $b_{20}\neq 0$, we can rewrite the condition as

[TABLE]

We only consider the case $a_{20}=3a_{02}+4b_{20}i$, since when $a_{20}=3a_{02}-4b_{20}i$, the consideration is analogous. Under this condition after the substitution (2.5) system (1.2) becomes

[TABLE]

System (3.9) has the Darboux factors

[TABLE]

which allow to construct the Darboux linearization

[TABLE]

Similarly as above, using the Zariski closure argument we conclude that the system is linearizable also when $b_{20}=0$. ∎

4. Relation between isochronicity conditions of [4] and Theorem 3.1

In [4] the authors presented conditions for isochronicity of system (1.2) when all parameters of the system are real. We investigate the relation between their conditions and the conditions presented in Theorem 3.1 and in [19]. The following result is obtained in [4].

Theorem 4.1** (Theorem 1 of [4]).**

The origin of system (1.2) is an isochronous center if and only if (1.2) can be transformed in one of the following forms in polar coordinates:

1. (a)

$\begin{cases}\dot{r}=r^{2}(\cos 3\theta-\frac{7}{3}\cos\theta-k_{1}\sin\theta)+r^{3}(-\frac{2k_{1}}{3}-\frac{2k_{1}}{3}\cos 2\theta-\frac{k_{1}^{2}}{2}\sin 2\theta),\\ \dot{\theta}=1+r(-\sin 3\theta+k_{1}\cos\theta-\sin\theta),\end{cases}$** 2. (b)

$\begin{cases}\dot{r}=r^{2}(\cos 3\theta+\frac{13}{3}\cos\theta-k_{1}\sin\theta)+r^{3}(2k_{1}+\frac{10k_{1}}{3}\cos 2\theta-\frac{k_{1}^{2}}{2}\sin 2\theta),\\ \dot{\theta}=1+r(-\sin 3\theta+k_{1}\cos\theta+\frac{1}{3}\sin\theta),\end{cases}$** 3. (c)

$\begin{cases}\dot{r}=r^{2}k_{1}\cos\theta+r^{3}(k_{2}\cos 2\theta+k_{3}\sin 2\theta),\\ \dot{\theta}=1+rk_{1}\cos\theta,\end{cases}$** 4. (d)

$\begin{cases}\dot{r}=r^{2}(k_{1}\cos\theta+k_{2}\sin\theta)+r^{3}(\frac{k_{1}k_{2}}{2}-\frac{k_{1}k_{2}}{2}\cos 2\theta+k_{3}\sin 2\theta),\\ \dot{\theta}=1+rk_{1}\sin\theta\end{cases}$* and* 5. (e)

$\begin{cases}\dot{r}=r^{2}(k_{1}\cos\theta+k_{2}\sin\theta)+r^{3}(k_{3}+k_{4}\cos 2\theta+k_{5}\sin 2\theta),\\ \dot{\theta}=1,\end{cases}$**

where $k_{j}$’s in each system are independent and are functions of original parameters in system (1.2).

As it is mentioned in the previous section by the result of [19] real system (1.2) is linearizable (equivalently, it has isochronous center) if and only if condition (3.1) or one of conditions (2)-(4) of Theorem 3.1 hold. The following theorem gives the relation of the results of [19] (and Theorem 3.1) and [4].

Theorem 4.2**.**

System (1.2) under conditions (3.1), $(2)$, $(3)$, and $(4)$ of Theorem 3.1 can be changed into system (c), (a), (b) and (d) of Theorem 4.1, respectively.

Proof.

System (1.2) under condition (3.1) becomes

[TABLE]

Applying the linear transformation

[TABLE]

and a time scaling $dt=-d\tilde{t}$, we change system (4.1) to

[TABLE]

where $k_{2}=\frac{a_{20}b_{20}r_{11}+a_{20}^{2}r_{20}+b_{20}^{2}r_{20}}{\left(a_{20}^{2}+b_{20}^{2}\right)^{2}}$, $k_{3}=\frac{a_{20}^{2}r_{11}-b_{20}^{2}r_{11}+4a_{20}b_{20}r_{20}}{2\left(a_{20}^{2}+b_{20}^{2}\right)^{2}}$, and below we write $x$ and $y$ instead of $\tilde{x}$ and $\tilde{y}$. System (4.2) in polar coordinates $x=r\cos\theta$, $y=r\sin\theta$ becomes system (c).

System (1.2) under condition (2) becomes system (3.8). The transformation $x=\frac{4}{3a_{20}}\tilde{x},\leavevmode\nobreak\ \leavevmode\nobreak\ y=-\frac{4}{3a_{20}}\tilde{y}$ and the time scaling $dt=-d\tilde{t}$ change system (3.8) to

[TABLE]

where we write $x$ and $y$ instead of $\tilde{x}$ and $\tilde{y}$, and $k_{1}=\frac{4b_{20}}{3a_{20}}$. System (4.3) in polar coordinates $x=r\cos\theta$, $y=r\sin\theta$ becomes system (a).

System (1.2) under condition (3) becomes system (3.11). Applying the transformation $x=\frac{16}{3a_{20}}\tilde{x},\leavevmode\nobreak\ \leavevmode\nobreak\ y=\frac{16}{3a_{20}}\tilde{y},$ we transform (3.11) to the system

[TABLE]

where we write $x$ and $y$ instead of $\tilde{x}$ and $\tilde{y}$, and $k_{1}=\frac{16b_{20}}{3a_{20}}$. System (4.4) in polar coordinates $x=r\cos\theta$, $y=r\sin\theta$ becomes system (b).

System (1.2) under condition (4) becomes system (3.14). The transformation $x=\tilde{y}$, $y=\tilde{x}$ and a time scaling $dt=-d\tilde{t}$ change system (3.14) to

[TABLE]

where $k_{1}=b_{20}$, $k_{2}=-a_{20}$, $k_{3}=-\frac{r_{11}}{2}$, and we write $x$ and $y$ instead of $\tilde{x}$ and $\tilde{y}$. System (4.5) in polar coordinates $x=r\cos\theta$, $y=r\sin\theta$ becomes system (d). ∎

However system (e) from Theorem 4.1 does not have an isochronous center at the origin, since, generally speaking, the origin of the system is not a center, but a focus. Indeed, system (e) can be written in the Cartesian coordinates $x=r\cos(\theta),y=r\sin(\theta)$ as

[TABLE]

We computed the first two Lyapunov quantities for system (4.6) and obtained $\eta_{1}=k_{3}$ and $\eta_{2}=2k_{1}k_{2}k_{5}+k_{4}(k_{1}^{2}-k_{2}^{2})$. Thus, the origin of system (e) is a focus, which is stable if $k_{3}<0$ or $k_{3}=0,\eta_{2}<0$, and unstable if $k_{3}>0$ or $k_{3}=0,\eta_{2}>0$. So, the necessary condition for existence of a center and a isochronous center at the origin of system (e) is $k_{3}=\eta_{2}=0$.

When $k_{3}=\eta_{2}=0$, by the linear transformation $x_{1}=x+\frac{k_{2}}{k_{1}}y$, $y_{1}=y-\frac{k_{2}}{k_{1}}x$, system (4.6) is changed into

[TABLE]

System (4.7) is a special case of system (3.14) when $b_{20}=0$, which is system (1.2) under condition (4) after adding the condition $b_{20}=0$. Therefore, when $k_{3}=\eta_{2}=0$, the origin of system (4.6), and thus of system (e), is an isochronous center.

It appears the authors of [4] made the following mistake in their reasoning. They obtained system (e) from the condition of vanishing of two period constants (period constants are an analogue of linearizability quantities when computing in polar coordinates). Then observing that the second equation of the system is $\dot{\theta}=1$, they concluded that the system has an isochronous center at the origin. However, as we have shown, unless $k_{3}=\eta_{2}=0$, the origin of the system is an isochronous focus.

5. Coexistence of isochronous centers

In this section we present our study on existence of few isochronous centers in real system (1.2).

Theorem 5.1**.**

System (1.2) has at most two isochronous centers including the origin when all parameters are real. More precisely, under condition (3.1) and conditions (2), (3) and (4) of Theorem 3.1, system (1.2) has at most two, one, two and two isochronous centers, respectively.

Proof.

We first consider the simplest case, case (2) of Theorem 3.1. In this situation, system (1.2) has the form (3.8). From the first equation of (3.8), we know that the coordinates of equilibria must satisfy $b_{20}x+1=0$ or $a_{20}x^{2}-b_{20}xy-y=0$. Substituting $y=a_{20}x^{2}/(1+b_{20}x)$ into the right hand of the second equation of (3.8) we obtain $4a_{20}^{2}x^{2}+(b_{20}x+1)^{2}=0$. Then, we get $x=0$ or $x=-1/b_{20}$. On the other hand, substituting $x=-1/b_{20}$ into the right hand of the second equation of (3.8), we have $-3a_{20}y/b_{20}=0$. Thus, other than the origin $O:(0,0)$ we get the equilibrium $A:(-1/b_{20},0)$ when $a_{20}b_{20}\neq 0$, no equilibria exist when $b_{20}=0$ and $a_{20}\neq 0$, or the line $x=-1/b_{20}$ is filled by equilibria when $a_{20}=0$ and $b_{20}\neq 0$.

Computing the determinant of linear matrix for system (3.8) at the equilibrium $A:(-1/b_{20},0)$, we find that it is equal to $-3a_{20}^{2}/b_{20}^{2}<0$, indicating that the equilibrium $A$ is a saddle if it exists. Clearly, any equilibrium on the line $x=-1/b_{20}$ cannot be an isochronous center when $a_{20}=0$. Therefore, in the case (2) of Theorem 3.1, system (1.2) has only one isochronous center at the origin.

Consider now case (3) of Theorem 3.1. In this case system (1.2) can be written as

[TABLE]

From the first equation of (5.1) we see that the coordinates of equilibria must satisfy $b_{20}x+1=0$ or $g_{3}(x,y):=4b_{11}x^{2}-b_{11}y^{2}-2b_{20}xy-2y=0$. Substituting $x=-1/b_{20}$ into the right hand side of the second equation of (5.1), we have $-yb_{11}(b_{20}^{2}y^{2}-2)/b_{20}=0$. Thus, we find three equilibria $A:(-1/b_{20},0)$ and $A_{\pm}:(-1/b_{20},\pm\sqrt{2}/b_{20})$ if $b_{20}\neq 0$.

If we solve $g_{3}(x,y)=0$ and substitute the solution into the right hand side of the second equation of (5.1) a very complicated expression arises. However, we only need to find the coordinates of centers of system (5.1) and for a singular point of the center type at the trace of linear matrix is zero. We calculate

[TABLE]

Computing a Groebner basis of the ideal $\langle g_{3},Q_{3},T_{3}\rangle$ we got the basis

[TABLE]

When $b_{11}=0$ we obtain the equilibrium $O:(0,0)$ or the line $b_{20}x+1$ is filled with equilibria. When $b_{11}\neq 0$, we obtain the equilibrium $B:(0,-2/b_{11})$.

Notice that all equilibria on the line $b_{20}x+1$ are degenerate when $b_{11}=0$, because the determinant of linear matrix at each equilibrium is zero. Thus, an equilibrium on the line $b_{20}x+1$ cannot be isochronous centers when $b_{11}=0$. By calculations, among all equilibria $A:(-1/b_{20},0)$, $A_{\pm}:(-1/b_{20},\pm\sqrt{2}/b_{20})$ and $B:(0,-2/b_{11})$, only at $B$ the trace of linear part is zero and the determinant of linear part is positive at the same time. So we only need check the isochronicity of equilibrium $B:(0,-2/b_{11})$. Moving the equilibrium $B$ to the origin and making the change

[TABLE]

together with the time scaling $dt=-d\tau$, we obtain from (5.1) the system

[TABLE]

where we still write $x,y$ instead of $u,v$. It is easy to show that system (5.3) is Darboux linearizable. Therefore, the system has isochronous centers at the origin and at the point $B:(0,-2/b_{11})$ if $b_{11}\neq 0$.

Now consider case (4) of Theorem 3.1. In this case system (1.2) has the form

[TABLE]

It is difficult to find the coordinates of equilibria of system (5.4) explicitly. However, we can calculate

[TABLE]

to find only coordinates of centers. Computing a Groebner basis of $\langle P_{4},Q_{4},T_{4}\rangle$ we obtained

[TABLE]

Letting the first and the second polynomials in $\mathcal{G}_{4}$ be zeros, we get $y=-4(b_{20}x+1)/a_{20}$ when $a_{20}\neq 0$ or $x=y=0$. Substituting $y=-4(b_{20}x+1)/a_{20}$ into $\mathcal{G}_{4}$, we have

[TABLE]

Using the first polynomial in (5.7), we obtain $b_{20}x+1=0$, $y=0$ or $y=-4(b_{20}x+1)/a_{20},3a_{20}^{2}-16b_{20}^{2}-16r_{11}=0$. Substituting them in (5.7), we obtain

[TABLE]

and

[TABLE]

respectively. From the first and the second polynomials in above two sets, we see that on the line $y=-4(b_{20}x+1)/a_{20}$ no center type equilibria exist when $a_{20}\neq 0$.

When $a_{20}=0$, the basis $\mathcal{G}_{4}$ becomes $\{b_{20}x^{2}+x,b_{20}xy+y,b_{20}^{2}y+r_{11}y\}$, and we obtain that $b_{20}x+1=0,(b_{20}^{2}+r_{11})y=0$ or $x=0,y=0$. When $a_{20}=0$, $b_{20}x+1=0$ and $b_{20}^{2}+r_{11}=0$, the line $b_{20}x+1=0$ is full of equilibria, none of which can be an isochronous center of system (5.4). Hence, we only get the unique possible center $A:(-1/b_{20},0)$ if $a_{20}=0$, at which the trace of the linear matrix for system (5.4) is zero and the determinant is $r_{11}/b_{20}^{2}+1$. If $a_{20}=0$, after moving the origin to the point $(-1/(2b_{20}),0)$, system (5.4) is changed into

[TABLE]

which is symmetric with respect to the $x$-axis. Moreover, equilibria $(\pm 1/(2b_{20}),0)$ of system (5.8) correspond to equilibria $A$ and $O$ of system (5.4) respectively. Thus, except of the origin $O:(0,0)$ we get another isochronous center at the equilibrium $A:(-1/b_{20},0)$ when $a_{20}=0$, $r_{11}/b_{20}^{2}+1>0$ and $b_{20}\neq 0$. Therefore, in case (4) system (1.2) has at most two isochronous centers.

At last, we study the case when condition (3.1) is fulfilled. In this situation let the vector field of system (1.2) be $(P_{1}(x,y),Q_{1}(x,y))$, as shown in (4.1). Similarly to case (4), we only consider equilibria of center type avoiding complicated calculations of coordinates of all equilibria. We calculate

[TABLE]

to find coordinates of centers. The Groebner basis of $\langle P_{1},Q_{1},T_{1}\rangle$ is

[TABLE]

If $a_{20}=b_{20}=0$, system (4.1) cannot have other centers except of the origin. Without loss of generality we suppose $b_{20}\neq 0$. If $a_{20}\neq 0$ the discussion is similar and we only need to make the change $(x,y)\to(y,x)$ with the time rescaling $dt=-d\tau$. From the first polynomial in $\mathcal{G}_{1}$, we get $x=-(a_{20}y+1)/b_{20}$. Substituting it into $\mathcal{G}_{1}$, we have

[TABLE]

where $a_{0}=a_{20}b_{20}-r_{20}$, $a_{1}=a_{20}^{2}b_{20}+b_{20}^{3}-2a_{20}r_{20}+b_{20}r_{11}$ and $a_{2}=-a_{20}^{2}r_{20}+a_{20}b_{20}r_{11}+b_{20}^{2}r_{20}$. Thus, from (5.11) we find two roots $y_{\pm}=(-a_{1}\pm\sqrt{a_{1}^{2}-4a_{2}a_{0}})/(2a_{2})$ and then get two equilibria $C_{\pm}:(-(a_{20}y_{\pm}+1)/b_{20},y_{\pm})$ when $d_{0}:=a_{1}^{2}-4a_{2}a_{0}>0$ and $a_{2}\neq 0$. At $C_{\pm}$ the trace of linear matrix for system (4.1) is zero and the determinant of that is

[TABLE]

Moreover,

[TABLE]

implying that at most one of $C_{+}$ and $C_{-}$ is a center. Actually, when $r_{20}=a_{20}b_{20}$, we find that the equilibrium $B:(0,-2/b_{11})$ is an isochronous center, since it is easy to show that system (4.1) is Darboux linearizable at this point. Therefore, if (3.1) holds, then system (1.2) has at most two isochronous centers. ∎

To conclude, we have found conditions for isochronicity and linearizability of system (1.2) and clarified conditions of isochronicity obtained in Chavarriga et al [4]. An important feature of our approach is the treatment of coefficients of system (1.2) as complex parameters, since this has allowed us to use formula (3.3) for finding the decomposition of the integrability variety and to use the Radical Membership Test in order to check the correctness computations involved modular arithmetics.

Acknowledgements

The first author is partially supported by a CAPES grant. The second author is supported by the Slovenian Researcher Agency. The fourth author is supported by Marie Sklodowska-Curie Actions grant 655212-UBPDS-H2020-MSCA-IF-2014, and partially supported by the National Natural Science Foundations of China (No. 11431008). The first and second authors are also supported by Marie Curie International Research Staff Exchange Scheme Fellowship within the 7th European Community Framework Programme, FP7-PEOPLE-2012-IRSES-316338.