\documentclass[reqno]{amsart} \AtBeginDocument{{\noindent\small {\em Electronic Journal of Differential Equations}, Vol. 2002(2002), No. 34, pp. 1--7.\newline ISSN: 1072-6691. URL: http://ejde.math.swt.edu or http://ejde.math.unt.edu \newline ftp ejde.math.swt.edu (login: ftp)} \thanks{\copyright 2002 Southwest Texas State University.} \vspace{1cm}} \begin{document} \title[\hfilneg EJDE--2001/34\hfil Centering conditions] {Centering conditions for planar septic systems} \author[Evgenii P. Volokitin\hfil EJDE--2001/34\hfilneg] {Evgenii P. Volokitin} \address{Sobolev Institute of Mathematics, Novosibirsk, 630090, Russia} \email{volok@math.nsc.ru} \date{} \thanks{Submitted August 22, 2002. Published April 3, 2003.} \thanks{Partially supported by Grant 02-01-00194 from the Russian Foundation for Basic Research} \subjclass[2000]{34C05, 34C25} \keywords{centering conditions, isochronicity, commutativity} \begin{abstract} We find centering conditions for the following $O$-symmetric system of degree 7: \begin{gather*} \dot x=y+ x (H_2 (x,y)+H_6 (x,y)),\\ \dot y=-x+ y (H_2 (x,y)+H_6 (x,y)), \end{gather*} where $H_2 (x,y)$ and $H_6 (x,y)$ are homogeneous polynomials of degrees 2 and 6, respectively. In some cases, we can find commuting systems and first integrals for the original system. We also study the geometry of the central region. \end{abstract} \maketitle \newtheorem{theorem}{Theorem}[section] \newtheorem{lemma}[theorem]{Lemma} \numberwithin{equation}{section} \section{introduction} Consider the planar autonomous system of ordinary differential equations $$\label{1} \begin{gathered} \dot x = y + x R_{n-1} (x,y),\\ \dot y = - x +y R_{n-1} (x,y),\\ \end{gathered}$$ where $R_{n-1}(x,y)$ is a polynomial in $x$ and $y$, of degree $n-1$, and $R_{n-1}(0,0)=0$. This system has only one singular point at $O(0,0)$ which is the center of the linear part of the system. The orbits of this system move around the origin with constant angular speed, and the origin is so a uniformly isochronous singular point. Such systems have been studied in many papers; see \cite{1}--\cite{5} and references therein. The following problem was stated as Problem 19.1. in \cite{2}: \begin{quote} Identify systems \eqref{1} of odd degree which are $O$-symmetric (not necessarily quasi homogeneous) having $O$ as a (uniformly isochronous) center. \end{quote} In this article we solve this problem for some systems of degree 7. In particular, we find necessary and sufficient conditions for system \eqref{1} with \label{2} \begin{aligned} R_{n-1} (x,y)=& a_0 x^2 + a_1 x y + a_2 y^2 + c_0 x^6 + c_1 x^5 y + c_2 x^4 y^2 \\ &+c_3 x^3 y^3 + c_4 x^2 y^4 + c_5 x y^5 + c_6 y^6, \end{aligned} where $a_0, a_1, a_2, c_0, c_1, c_2, c_3, c_4, c_5, c_6$ are real numbers. The plan for this paper is as follows: In Section 2, we present centering conditions. In Section 3, we investigate some properties of systems in the presence of a center. In particular, we discuss the question of existence of a polynomial commuting system. When this system exists, we give a first integral of the original system. We also study the geometry of the central region. \section{Results} \begin{theorem} \label{thm1} The origin is a center of (\ref{2}) if and only if one of the following two conditions is satisfied: \begin{itemize} \item[(i)] $a_0=a_1=a_2=0$, $5c_{0} + c_{2} + c_{4} + 5c_{6}=0$; \item[(ii)] \begin{gather*} a_0+a_2=0,\\ 5c_0 + c_2 + c_4 + 5c_6 = 0,\\ a_1(15c_0 + c_2 - c_4 - 15c_6) + 2a_0(5c_1 + 3c_3 + 5c_5) = 0,\\ (a_1^2 - 4a_0^2)(3c_0 - c_2 - c_4 + 3c_6) + 8a_0 a_1(c_1 - c_5)= 0,\\ a_1(a_1^2-12a_0^2)(c_0 - c_2 + c_4 - c_6) + 2a_0(3a_1^2 -4a_0^2)(c_1-c_3+c_5) =0. \end{gather*} \end{itemize} \end{theorem} Before proving this theorem, we consider the instance of (\ref{2}) in which $a_0=a_2=0$, that is $$\label{4} \begin{gathered} \dot x=y+x(a_1 x y + c_0 x^6 + c_1 x^5 y + c_2 x^4 y^2 + c_3 x^3 y^3 + c_4 x^2 y^4 + c_5 x y^5 + c_6 y^6),\\ \dot y=-x+y(a_1 x y + c_0 x^6 + c_1 x^5 y + c_2 x^4 y^2 + c_3 x^3 y^3 + c_4 x^2 y^4 + c_5 x y^5 + c_6 y^6). \end{gathered}$$ \begin{lemma} \label{lm1} The origin is a center of (\ref{4}) if and only if one of the following two conditions is satisfied: \begin{gather}\label{5i} a_1=0, \quad 5c_{0} + c_{2} + c_{4} + 5c_{6}=0;\\ c_0=c_2=c_4=c_6=0. \label{5ii} \end{gather} \end{lemma} \begin{proof} We used the software package {\it Mathematica} to find the first six Poincar\'{e}-Lyapunov constants of (\ref{4}) (see more details about our method in \cite{6}). Up to a positive scalar factor they are \begin{align*} l_1 &=0, l_2 = 0,\\ l_3 &= 5c_0 + c_2 + c_4 + 5c_6,\\ l_4 &= -a_1(5c_0 + 3c_2 + 5c_4 + 35c_6),\\ l_5 &= a_1^2(-101c_0 - 17c_2 - 9c_4 + 19c_6),\\ l_6 &= 15a_1^3(621 c_0 + 367c_2 + 565c_4 + 3367c_6)\\ &\quad - 56(31c_1 + 20c_3 + 49c_5)(5c_0 + c_2 + c_4 + 5c_6). \end{align*} Necessity of conditions \eqref{5i}--\eqref{5ii} result from solving the simultaneous equations $l_3=l_4=l_5=l_6=0$. In the case \eqref{5i}, the sufficiency part of the lemma follows from the fact that (\ref{4}) is a quasi homogeneous system of degree 7 whose coefficients satisfy the equation $5c_0+c_2+c_4+5c_6=0$ representing a necessary and sufficient centering condition \cite{1}. In the case \eqref{5ii}, system (\ref{4}) is reversible and its trajectories are symmetric with respect to both coordinate axes. It is well known that if the linear part of a reversible system has a center at the origin, then the origin is also a center of the system itself. Thus, under conditions \eqref{5i}-\eqref{5ii}, the origin is a center of system (\ref{4}). This completes the proof of the lemma. \end{proof} \begin{proof}[Proof of Theorem \ref{thm1}] In the previous lemma we considered a particular case. Now, we consider the general system (\ref{2}). The first Poincar\'{e}-Lyapunov constant $l_1$ of (\ref{2}) is $$l_1=2(a_0 + a_2).$$ If $a_0+a_2=0$ then the change of variables $x\mapsto x \cos \vartheta + y \sin \vartheta$, $y\mapsto -x \sin \vartheta + y \cos \vartheta$, with $\vartheta$ defined from the condition $$\label{6} a_0 \cos^2 \vartheta + a_1 \sin \vartheta \cos \vartheta -a_0 \sin^2 \vartheta=0,$$ reduces (\ref{2}) to a system of the form (\ref{4}): \begin{gather*} \dot x=y+x(a_{1}' x y + c_{0}' x^6 + c_{1}' x^5 y + c_{2}' x^4 y^2 + c_{3}' x^3 y^3 + c_{4}' x^2 y^4 + c_{5}' x y^5 + c_{6}' y^6),\\ \dot y=-x+y(a_{1}' x y + c_{0}' x^6 + c_{1}' x^5 y + c_{2}' x^4 y^2 + c_{3}' x^3 y^3 + c_{4}' x^2 y^4 + c_{5}' x y^5 + c_{6}' y^6) \end{gather*} whose coefficients are expressible in terms of the coefficients of (\ref{2}). In particular, we have \label{7} \begin{aligned} &a_{1}'=a_1 \cos^2\vartheta - 4 a_0 \cos\vartheta \sin\vartheta - a_1 \sin^2\vartheta,\\ &c_{0}' = (2d_0 - d_2 + 2d_4 - d_6)/32,\\ &c_{2}' = (6d_0 - d_2 - 10 d_4 + 15d_6)/32,\\ &c_{4}' = (6d_0+d_2-10d_4-15d_6)/32,\\ &c_{6}' = (2d_0+d_2+2d_4+d_6)/32, \end{aligned} where \begin{align*} &d_0 = 5c_0 + c_2 + c_4 + 5c_6,\\ &d_2 = (15c_0 + c_2 - c_4 - 15c_6)\cos2\vartheta + (5c_1 + 3c_3 + 5c_5)\sin2\vartheta,\\ &d_4 = (3c_0 - c_2 - c_4 + 3c_6)\cos4\vartheta + 2(c_1 - c_5)\sin4\vartheta,\\ &d_6 = (c_0 - c_2 + c_4 - c_6)\cos6\vartheta + (c_1 - c_3 + c_5)\sin6\vartheta, \end{align*} and $\vartheta$ is defined in (\ref{6}). Using Lemma \ref{lm1}, we see that the origin is a center of system (\ref{2}) if and only if one of the following two conditions is satisfied: \begin{gather} \label{8i} a_0+a_2=0, a_{1}'=0, \quad 5c_{0}' + c_{2}' + c_{4}' + 5c_{6}'=0;\\ a_0+a_2=0, \quad c_{0}'=c_{2}'=c_{4}'=c_{6}'=0. \label{8ii} \end{gather} In the case \eqref{8i}, $a_1'=0$ and (\ref{6}) amount to $a_0=a_1=0$. Next, (\ref{7}) yields $5c_{0}' + c_{2}' + c_{4}' + 5c_{6}'=d_0=5c_{0} + c_{2} + c_{4} + 5c_{6}$. This proves the theorem in the case (\mbox{i}). Next, $c_{0}'=c_{2}'=c_{4}'=c_{6}'=0$ amounts to $d_0=d_2=d_4=d_6=0$. Using (\ref{6}), we eliminate $\vartheta$ from the last equations, thus arriving at the conditions of the case \eqref{8ii} expressed in terms of the coefficients of the original system (\ref{2}). These conditions coincide with those in the case (ii) of Theorem \ref{thm1} and the proof is complete. \end{proof} \section{Properties of systems with center} Consider system (\ref{4}) with coefficients that satisfy the centering conditions \eqref{5i}--\eqref{5ii}. For a planar polynomial system, the presence of an isochronous center is well known to be equivalent to the existence of a transverse analytic system commuting with it \cite{7}. It is proved in \cite{5} that in the case \eqref{5i} system (\ref{4}) commutes with a polynomial system of the form $$\label{9} \begin{gathered} \dot x=x+x Q(x,y),\\ \dot y=y+y Q(x,y), \end{gathered}$$ where $Q(x,y)=q_0 x^6 + q_1 x^5 y + q_2 x^4 y^2 + q_3 x^3 y^3 + q_4 x^2 y^4 + q_5 x y^5 + q_6 y^6$ which is a homogeneous polynomial of degree 6 satisfying $$\label{10} y Q_x(x,y) - x Q_y(x,y) = 6 P_6(x,y)$$ with $P_6(x,y)=c_0 x^6 + c_1 x^5 y + c_2 x^4 y^2 + c_3 x^3 y^3 + c_4 x^2 y^4 + c_5 x y^5 + c_6 y^6\,.$ To satisfy this equation, we can take $$\label{11} \begin{gathered} q_0=0, \quad q_1=-6c_0, \quad q_2=-3c_1, q_3=-2(5c_0+c_2), \\ q_4=-3(c_1+c_3/2),\quad q_5=6c_6, \quad q_6=-(c_1+c_3/2+c_5). \end{gathered}$$ Note that if we add $c(x^2+y^2)^3$ to the polynomial $Q(x,y)$ with the coefficients (\ref{11}) then the resultant system of the form (\ref{9}) commutes with (\ref{4}). Following \cite{8}, we say that a function $C: R^2 \to R$ and the curve $C=0$ are invariants for a system $\dot x=p(x,y)$, $\dot y=q(x,y)$ if there is a polynomial $L$ such that $\dot C=C L$, where $\dot C=C_x p+C_y q$. The polynomial $L$ is called the cofactor of $C$. Note that the functions $$C_1=x^2+y^2, \quad C_2=1+Q(x,y)$$ are invariants for (\ref{4}). This enables us to find the first Darboux integral $$\label{12} H(x,y)=\frac{(x^2+y^2)^3}{1+Q(x,y)}$$ for this system. The Darboux method is presented, for instance, in \cite{8,9}. By \cite{2}, the center of (\ref{4}) is of type $B^k$, and the boundary of the center domain is the union of $k$ open unbounded trajectories ($1\leq k\leq 6$). In the case under study, we can describe this boundary explicitly and indicate the possible values of $k$ more precisely. Passing to the polar coordinates $x=\varrho \cos \varphi$, $y=\varrho \sin \varphi$ in (\ref{12}), we can show that in the case \eqref{5i}, the boundary of the central region is defined by the equation $$\varrho=\frac{1}{(c_0-Q(\cos \varphi, \sin \varphi))^{1/6}}$$ with $c_0=\max_{[0,2\pi]} Q(\cos \varphi, \sin \varphi)$. The central region is a curvilinear $k$-polygon whose vertices are points at infinity in the intersection of the equator of the Poincar\'{e} sphere with the rays $x=r \cos\varphi_i, y=r \sin\varphi_i, r>0$, where the values of $\varphi_i$ are determined from the conditions $$Q(\cos \varphi_i, \sin \varphi_i)=c_0, 0\leq \varphi_i < 2\pi.$$ Note that the values $\varphi_i$ are solutions of the equation $P_6(\cos \varphi, \sin \varphi)=0$. Indeed, from (\ref{10}) we deduce: \begin{align*} 0&=\frac{d}{d \varphi} Q(\cos \varphi, \sin \varphi)|_{\varphi=\varphi_i}\\ &=-Q_x(\cos \varphi_i, \sin \varphi_i) \sin \varphi_i +Q_y(\cos \varphi_i, \sin \varphi_i)\cos \varphi_i\\ &=6P_6(\cos \varphi_i, \sin \varphi_i). \end{align*} The trigonometric polynomial $Q(\cos \varphi, \sin \varphi)$ of degree 6 satisfies the condition $Q(\cos(\varphi+\pi), \sin(\varphi+\pi))=Q(\cos \varphi, \sin \varphi)$ and takes its every value, $c_0$ inclusively, on the interval $[0,2\pi)$ an even number of times. Thus, in the case \eqref{5i} the central region is symmetric about the origin and its boundary is the union of an even number of unbounded trajectories. Therefore, the center is of type $B^k$, where $k=2, 4, 6$. Moreover, a generic" system has a center of type $B^2$. For the center to be of type $B^4$ or $B^6$, the trigonometric polynomial $Q(\cos \varphi, \sin \varphi)$ must take its greatest value $c_0$ on the interval $[0, 2\pi)$ more than twice. This requires extra restrictions on the coefficients of the system. In the case \eqref{5ii} system (\ref{4}) takes the form \begin{gather*} \dot x = y+x(a_1 x y + c_1 x^5 y + c_3 x^3 y^3 + c_5 x y^5),\\ \dot y = -x+y(a_1 x y + c_1 x^5 y + c_3 x^3 y^3 + c_5 x y^5). \end{gather*} If $a_1=0$ then we arrive at the case \eqref{5i}. If $a_1\neq 0$ then we may assume that $a_1=1$. The general case is reduced to this by the change of variables $x \mapsto x/\sqrt{a}, y \mapsto y/\sqrt{a}$ for $a>0$ or $x \mapsto y/\sqrt{-a}, y \mapsto x/\sqrt{-a}, t \mapsto -t$ for $a<0.$ According to \cite{5}, the system $$\label{13} \begin{gathered} \dot x = y+x(x y + c_1 x^5 y + c_3 x^3 y^3 + c_5 x y^5)\equiv y+ x P(x,y),\\ \dot y = -x+y(x y + c_1 x^5 y + c_3 x^3 y^3 + c_5 x y^5)\equiv -x+ y P(x,y) \end{gathered}$$ commutes with an analytic system of the form $$\label{14} \dot x = x Q(x,y),\quad \dot y= y Q(x,y),$$ where the function $Q(x,y)$ meets the equation \label{15} \begin{aligned} &x(Q_y(x,y)+P_x(x,y)Q(x,y)-P(x,y)Q_x(x,y))\\ &+y(-Q_x(x,y)+P_y(x,y)Q(x,y)-P(x,y)Q_y(x,y))=0. \end{aligned} Suppose that the function $Q(x,y)$ is a polynomial in the variables $x$ and $y$ which has degree $N$ in $y$: $Q(x,y)=Q_0(x) + Q_1(x)y+\ldots+Q_N(x)y^N$. After insertion of $Q(x,y)$ in (\ref{15}), the left-hand side becomes a polynomial of degree at most $N+5$ and the coefficient of $y^{N+5}$ is $$c_5 x((6-N)Q_N(x)-x Q_N'(x)).$$ Therefore, we must have $x Q_N'(x)=(6-N) Q_N(x)$ or $c_5 =0$. If $c_5\neq0$, this yields $N\leq6$. The same bound of $N$ can be shown to be true also in the case when $c_5=0$. Likewise, we can show that the degree of $Q(x,y)$ in $x$ is at most 6. Substituting the polynomial $Q(x,y)=\sum_{i,j=0}^6 q_{ij}x^i y^j$ in (\ref{15}) yields a system of linear equations in the coefficients $q_{ij}$. We derived this system and investigated its properties by using the software package {\it Mathematica}. In this way we found out that the necessary solvability condition is $$\label{16} c_3^2-4c_1c_5=0.$$ This condition is also sufficient. As an example, assume that $c_1=\alpha^2$, $c_5=\beta^2$, and $c_3=2\alpha\beta$. Then (\ref{4}) takes the form $$\label{17} \begin{gathered} \dot x=y+x^2 y (1 + (\alpha x^2 +\beta y^2)^2),\\ \dot y=-x+x y^2 (1 + (\alpha x^2 +\beta y^2)^2). \end{gathered}$$ Straightforward calculations show that (\ref{17}) commutes with the polynomial system \begin{gather*} \dot x=x(\alpha-\beta + (\alpha x^2 +\beta y^2) +(\alpha x^2 +\beta y^2)^3),\\ \dot y=y(\alpha-\beta + (\alpha x^2 +\beta y^2) +(\alpha x^2 +\beta y^2)^3). \end{gather*} Likewise, we can study the case when $c_1=\alpha^2$, $c_5=\beta^2$, $c_3=-2\alpha\beta$ and the case when $c_1=-\alpha^2$, $c_5=-\beta^2$, and $c_3=\pm 2\alpha\beta$. We have thus proved that system (\ref{13}) commutes with a polynomial system of the form (\ref{14}) if and only if (\ref{16}) holds. Recall that every uniformly isochronous $O$-symmetric quintic system satisfying the center conditions commutes with some polynomial system of the same degree \cite{6}. At the same time, an arbitrary uniformly isochronous (not necessarily $O$-symmetric) quintic system with a center may fail to commute with any polynomial system \cite{4}. The functions $$C_1=x^2+y^2, C_2=\alpha-\beta + (\alpha x^2 +\beta y^2) +(\alpha x^2 +\beta y^2)^3$$ are invariants for (\ref{17}) with the respective cofactors $$L_1=2xy(1+(\alpha x^2+\beta y^2)^2), \quad L_2=2xy(1+3(\alpha x^2+\beta y^2)^2).$$ Moreover, if $\alpha\neq\beta$ then the function $$C_3=\exp\Big(\int_0^{\alpha x^2+\beta y^2} \frac{dt}{\alpha-\beta+t+t^3}\Big)$$ is an invariant with the cofactor $L_3=2xy$. We have $3L_1-L_2-2L_3=0.$ In this case the function \begin{align*} H(x,y)&=\frac{C_1^3}{C_2 C_3^2}\\ &=\frac{(x^2+y^2)^3}{\alpha-\beta + (\alpha x^2 +\beta y^2) +(\alpha x^2 +\beta y^2)^3}\\ &\quad \times\frac1{ \exp(2\int_0^{\alpha x^2+\beta y^2} dt/(\alpha-\beta+t+t^3)} \end{align*} is a first Darboux integral of (\ref{17}). When $\alpha=\beta$, the change of variables $x=\sqrt{r} \cos \varphi$, $y=\sqrt{r} \sin \varphi$ reduces (\ref{17}) to the system $$\dot r=r^2(1+\alpha^2 r^2) \sin 2 \varphi, \dot \varphi=-1$$ for which the function $$G(x,y)=\frac{1}{r}+\sin^2 \varphi + \arctan \alpha r$$ is a first integral. Therefore, the function $$H(x,y)=\frac{x^2+y^2}{1+x^2+\alpha(x^2+y^2)\arctan \alpha (x^2+y^2)}$$ is a first integral of (\ref{17}) for $\alpha=\beta$. As in the case \eqref{5i}, system (\ref{17}) has a center of type $B^k$ $(1\leq k\leq 6)$. Since the system under study is reversible, the central region is symmetric with respect to both coordinate axes. Therefore, the center is of type $B^2$, $B^4$, or $B^6$. The singular points on the equator of the Poincar\'{e} sphere, vertices of the symmetric boundary of the central region, are saddle points. They have common separatrices only in exceptional cases. So in the case \eqref{8i} the center is generically" of type $B^2$. 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