# On the cyclicity of Kolmogorov polycycles 00footnotetext: 2010 AMS Subject
Classification: 34C07; 34C20; 34C23. 00footnotetext: Key words and phrases:
limit cycle, polycycle, cyclicity, asymptotic expansion. 00footnotetext: This
work has been partially funded by the Ministry of Science, Innovation and
Universities of Spain through the grants PGC2018-095998-B-I00 and
MTM2017-86795-C3-2-P and by the Agency for Management of University and
Research Grants of Catalonia through the grants 2017SGR1725 and 2017SGR1617.

D. Marín and J. Villadelprat
Departament de Matemàtiques, Edifici Cc, Universitat Autònoma de Barcelona,
08193 Cerdanyola del Vallès (Barcelona), Spain
Centre de Recerca Matemàtica, Edifici Cc, Campus de Bellaterra,
08193 Cerdanyola del Vallès (Barcelona), Spain
Departament d’Enginyeria Informàtica i Matemàtiques, ETSE,
Universitat Rovira i Virgili, 43007 Tarragona, Spain

###### Abstract

In this paper we study planar polynomial Kolmogorov’s differential systems

$X_{\mu}\quad\left\\{\\!\begin{array}[]{l}\dot{x}=xf(x,y;\mu),\\\\[2.0pt]
\dot{y}=yg(x,y;\mu),\end{array}\right.$

with the parameter $\mu$ varying in an open subset
$\Lambda\subset\mathbb{R}^{N}$. Compactifying $X_{\mu}$ to the Poincaré disc,
the boundary of the first quadrant is an invariant triangle $\Gamma$, that we
assume to be a hyperbolic polycycle with exactly three saddle points at its
vertices for all $\mu\in\Lambda.$ We are interested in the cyclicity of
$\Gamma$ inside the family $\\{X_{\mu}\\}_{\mu\in\Lambda},$ i.e., the number
of limit cycles that bifurcate from $\Gamma$ as we perturb $\mu.$ In our main
result we define three functions that play the same role for the cyclicity of
the polycycle as the first three Lyapunov quantities for the cyclicity of a
focus. As an application we study two cubic Kolmogorov families, with $N=3$
and $N=5$, and in both cases we are able to determine the cyclicity of the
polycycle for all $\mu\in\Lambda,$ including those parameters for which the
return map along $\Gamma$ is the identity.

##  1 Introduction and main results

The present paper is motivated by the results obtained by Gasull, Mañosa and
Mañosas [8] with regard to the _stability_ of an unbounded polycycle $\Gamma$
in the Kolmogorov’s polynomial differential systems

$\left\\{\\!\begin{array}[]{l}\dot{x}=xf(x,y),\\\\[2.0pt]
\dot{y}=yg(x,y).\end{array}\right.$

These systems are widely used in ecology to describe the interaction between
two populations, see [19] for instance. That being said, the stability of the
polycycle is not the main issue to which this paper is addressed. Indeed,
assuming that the coefficients of the polynomials $f$ and $g$ depend
analytically on a parameter $\mu$, we are interested in the _cyclicity_ of the
polycycle (see Definition 1 below), which roughly speaking is the number of
limit cycles that can bifurcate from $\Gamma$ as we perturb $\mu$. In our main
result (Theorem A) we define three functions, $d_{1}(\mu)$, $d_{2}(\mu)$ and
$d_{3}(\mu)$, that play the same role for the cyclicity of the polycycle as
the first three _Lyapunov quantities_ for the cyclicity of a focus. Recall
that the displacement map can be analytically extended to a focus and that the
Lyapunov quantities are the coefficients of its Taylor’s series. On the
contrary the displacement map has no smooth extension to a polycycle. At best
one can hope that it has some asymptotic expansion. This is indeed the case
for the polycycle that we study in the present paper and in order to obtain it
we strongly rely in our previous results [14, 15, 16] about the asymptotic
expansion of the Dulac map of an unfolding of hyperbolic saddles. The
principal part of the asymptotic expansion of the displacement map is given in
a monomial scale containing a deformation of the logarithm, the so-called
Ecalle-Roussarie compensator, and the remainder is uniformly flat with respect
to the parameters. The functions $d_{i}(\mu)$ in Theorem A are essentially the
coefficients of the first three monomials in the principal part, which
explains their relation with the cyclicity. For other results regarding the
cyclicity of polycycles and more general limit periodic sets the reader is
referred to [5, 6, 11, 20] and references therein.

Most of the work on planar polynomial differential systems, including this
paper, is related to the questions surrounding Hilbert’s 16th problem (see for
instance [10, 22, 23] and references therein) and its various weakened
versions. In this setting it is worth to mention that, using a compactness
argument, Roussarie [21] showed that to prove the existential part of
Hilbert’s 16 problem in the family $\mathcal{P}_{n}$ of all polynomial vector
fields of degree $\leqslant n$ it is sufficient to show that each _limit
periodic set_ in $\mathcal{P}_{n}$ has finite cyclicity.

  * Definition 1.1.

Let $X$ be a vector field on $\mathbb{R}^{2}$ (or $\mathbb{S}^{2}$). A
_graphic_ $\Gamma$ for $X$ is a compact, non-empty invariant subset which is a
continuous image of $\mathbb{S}^{1}$ and consists of a finite number of
isolated singular points $\\{p_{1},\ldots,p_{m},p_{m+1}=p_{1}\\}$ (not
necessarily distinct) and compatibly oriented separatrices
$\\{s_{1},\ldots,s_{m}\\}$ connecting them (i.e., such that the $\alpha$-limit
set of $s_{j}$ is $p_{j}$ and the $\omega$-limit set of $s_{j}$ is $p_{j+1}$).
A graphic is said to be _hyperbolic_ if all its singular points are hyperbolic
saddles. A _polycyle_ is a graphic with a return map defined on one of its
sides. $\square$

The polycycle that we aim to study is unbounded. In order to investigate the
behaviour of the trajectories of a polynomial vector field $Y$ near infinity
we can consider its Poincaré compactification $p(Y)$, see [2, §5] for details,
which is an analytically equivalent vector field defined on the sphere
$\mathbb{S}^{2}$. The points at infinity of $\mathbb{R}^{2}$ are in bijective
correspondence with the points of the equator of $\mathbb{S}^{2}$, that we
denote by $\ell_{\infty}$. Furthermore the trajectories of $p(Y)$ in
$\mathbb{S}^{2}$ are symmetric with respect to the origin and so it suffices
to draw its flow in the closed northern hemisphere only, the so called
Poincaré disc.

  * Definition 1.2.

Let $\\{X_{\mu}\\}_{\mu\in\Lambda}$ be a family of vector fields on
$\mathbb{S}^{2}$ and suppose that $\Gamma$ is a polycyle for $X_{\mu_{0}}$. We
say that $\Gamma$ has finite cyclicity in the family
$\\{X_{\mu}\\}_{\mu\in\Lambda}$ if there exist $\kappa\in\mathbb{N}$,
$\varepsilon>0$ and $\delta>0$ such that any $X_{\mu}$ with
$\|\mu-\mu_{0}\|<\delta$ has at most $\kappa$ limit cycles $\gamma_{i}$ with
$\text{dist}_{H}(\Gamma,\gamma_{i})<\varepsilon.$ The minimum of such $\kappa$
when $\varepsilon$ and $\delta$ go to zero is called the _cyclicity_ of
$\Gamma$ in $\\{X_{\mu}\\}_{\mu\in\Lambda}$ and denoted by
$\mathrm{Cycl}\big{(}(\Gamma,X_{\mu_{0}}),X_{\mu}\big{)}.$ $\square$

In this paper we consider the family of vector fields
$\\{X_{\mu}\\}_{\mu\in\Lambda}$ given by

$X_{\mu}\\!:=f(x,y;\mu)x\partial_{x}+g(x,y;\mu)y\partial_{y}$ (1)

where $\Lambda$ is an open subset of $\mathbb{R}^{N}$ and $f$ and $g$ are
polynomials in $x$ and $y$ of degree $n\in\mathbb{N}$ with the coefficients
depending analytically on $\mu$. The standing hypothesis on the family
$\\{X_{\mu}\\}_{\mu\in\Lambda}$ are the following:

  1. H1

$f(z,0;\mu)>0$, $g(0,z;\mu)<0$ and $\big{(}f_{n}-g_{n}\big{)}(1,z;\mu)<0$ for
all $z>0$ and $\mu\in\Lambda.$

  2. H2

$\lambda_{1}(\mu)\\!:=\left(\frac{f_{n}}{g_{n}-f_{n}}\right)(1,0;\mu)$,
$\lambda_{2}(\mu)\\!:=\left(\frac{f_{n}-g_{n}}{g_{n}}\right)(0,1;\mu)$ and
$\lambda_{3}(\mu)\\!:=-\left(\frac{g}{f}\right)(0,0;\mu)$, are well defined
and strictly positive for all $\mu\in\Lambda.$

Here, and in what follows, $f_{n}$ and $g_{n}$ denote, respectively, the
homogeneous part of degree $n$ of $f$ and $g.$ Conditions H1 and H2 guarantee
that, after compactifying the polynomial vector field $X_{\mu}$ to the
Poincaré disc, the boundary of the first quadrant is a polycyle with three
hyperbolic saddles, see Figure 1,

[l(0mm),r(0mm),t(0mm),b(5mm)]dib3(0.75) [l]18,13;$\ell_{\infty}$
[l]38,41;$s_{3}$ [l]40,90;$s_{2}$ [l]90,45;$s_{1}$ [l]70,82;$\Gamma$

Figure 1: Placement of the hyperbolic saddles and the polycycle $\Gamma$ in
the Poincaré disc.

$s_{1}\\!:=\\{y=0\\}\cap\ell_{\infty}$, $s_{2}\\!:=\\{x=0\\}\cap\ell_{\infty}$
and $s_{3}\\!:=(0,0)$.

From now on we shall denote this polycyle by $\Gamma$, that we remark is a
compact subset of the Poincaré disc. The hyperbolicity ratios of the saddles
at its vertices are precisely the ones given in ${\bf H2}$. We also define:

$\begin{array}[]{ll}\displaystyle
L_{11}(u)=\exp\left(\int_{0}^{u}\left(\left(\frac{f-g}{f}\right)(1/z,0)+\frac{1}{\lambda_{1}}\right)\frac{dz}{z}\right)&\displaystyle
L_{12}(u)=\exp\left(\int_{0}^{u}\left(\left(\frac{f_{n}}{f_{n}-g_{n}}\right)(1,z)+\lambda_{1}\right)\frac{dz}{z}\right)\\\\[15.0pt]
\displaystyle
L_{21}(u)=\exp\left(\int_{0}^{u}\left(\left(\frac{g_{n}}{g_{n}-f_{n}}\right)(z,1)+\frac{1}{\lambda_{2}}\right)\frac{dz}{z}\right)&\displaystyle
L_{22}(u)=\exp\left(\int_{0}^{u}\left(\left(\frac{g-f}{g}\right)(0,1/z)+\lambda_{2}\right)\frac{dz}{z}\right)\\\\[15.0pt]
\displaystyle
L_{31}(u)=\exp\left(\int_{0}^{u}\left(\left(\frac{g}{f}\right)(z,0)+\lambda_{3}\right)\frac{dz}{z}\right)&\displaystyle
L_{32}(u)=\exp\left(\int_{0}^{u}\left(\left(\frac{f}{g}\right)(0,z)+\frac{1}{\lambda_{3}}\right)\frac{dz}{z}\right)\\\
\end{array}$

together with

$M_{1}(u)=-\frac{L_{11}(u)}{u}\,\partial_{2}\\!\left(\frac{g}{f}\right)(1/u,0)$
and $M_{3}(u)=L_{31}(u)\,\partial_{2}\\!\left(\frac{g}{f}\right)(u,0)$. (2)

We point out that all these functions depend on the parameter $\mu.$ This
dependence is omitted for the sake of shortness when there is no risk of
confusion.

We can now state our main result, which is addressed to the cyclicity of the
polycycle $\Gamma$ inside the polynomial family
$\\{X_{\mu}\\}_{\mu\in\Lambda}.$ More formally, we should refer to the
compactified family $\\{p(X_{\mu})\\}_{\mu\in\Lambda}$ of vector fields on
$\mathbb{S}^{2}$ but for the simplicity in the exposition we commit an abuse
of language by identifying both families. It is clear that the number of limit
cycles of $p(X_{\mu})$ and $X_{\mu}$ is the same. In the statement
$\mathscr{R}(\,\cdot\,;\mu)$ stands for the return map of the vector field
$X_{\mu}$ around the polycycle $\Gamma$ (see Figure 1) and we use the notion
of functional independence that is given in Definition 2.

######  Theorem A.

Let us consider the family of Kolmogorov polynomial vector fields
$\\{X_{\mu}\\}_{\mu\in\Lambda}$ given in $(\ref{kol})$ and verifying the
assumptions H1 and H2. Then, for any $\mu_{0}\in\Lambda,$ the following
assertions hold with regard to the cyclicity of the polycycle $\Gamma$ inside
the family:

  1. $(a)$

If $d_{1}(\mu)\\!:=1-\lambda_{1}\lambda_{2}\lambda_{3}$ does not vanish at
$\mu_{0}$ then $\mathrm{Cycl}\big{(}(\Gamma,X_{\mu_{0}}),X_{\mu}\big{)}=0.$

  2. $(b)$

If $d_{1}$ vanishes and is independent at $\mu_{0}$ and
$\mathscr{R}(\,\cdot\,;\mu_{0})\not\equiv\text{Id}$ then
$\mathrm{Cycl}\big{(}(\Gamma,X_{\mu_{0}}),X_{\mu}\big{)}\geqslant 1.$

  3. $(c)$

If
$d_{2}(\mu)\\!:=\log\left(\left(\frac{L_{12}}{L_{21}}\right)^{\lambda_{2}}\left(\frac{L_{31}}{L_{11}}\right)^{\lambda_{1}\lambda_{2}}\frac{L_{22}}{L_{32}}\right)(1)$
does not vanish at $\mu_{0}$ then
$\mathrm{Cycl}\big{(}(\Gamma,X_{\mu_{0}}),X_{\mu}\big{)}\leqslant 1.$

  4. $(d)$

If $d_{1}$ and $d_{2}$ vanish and are independent at $\mu_{0}$ and
$\mathscr{R}(\,\cdot\,;\mu_{0})\not\equiv\text{Id}$ then
$\mathrm{Cycl}\big{(}(\Gamma,X_{\mu_{0}}),X_{\mu}\big{)}\geqslant 2.$

In case that $\lambda_{1}(\mu_{0})<1$, $\lambda_{2}(\mu_{0})>1$ and
$\lambda_{3}(\mu_{0})>1$ then the following assertions hold as well:

  1. $(e)$

If
$d_{3}(\mu)\\!:=\hat{M}_{3}(\lambda_{3},1)L_{11}(1)-\hat{M}_{1}\big{(}\frac{1}{\lambda_{1}},1\big{)}L_{31}(1)$
does not vanish at $\mu_{0}$ then
$\mathrm{Cycl}\big{(}(\Gamma,X_{\mu_{0}}),X_{\mu}\big{)}\leqslant 2.$

  2. $(f)$

If $d_{1},$ $d_{2}$ and $d_{3}$ vanish and are independent at $\mu_{0}$ and
$\mathscr{R}(\,\cdot\,;\mu_{0})\not\equiv\text{Id}$ then
$\mathrm{Cycl}\big{(}(\Gamma,X_{\mu_{0}}),X_{\mu}\big{)}\geqslant 3.$

Let us make some remarks with regard to the regularity of the functions
$d_{1}$, $d_{2}$ and $d_{3}$ defined in the statement. On account of the
hypothesis H1 and H2 it is evident that $d_{1}$ is analytic on the whole
parameter space $\Lambda$. On the other hand, $d_{2}$ is defined in terms of
the functions $\mu\mapsto L_{ij}(1)$, which in turn are given by some
(apparently) improper integrals. By applying the Weierstrass Division Theorem
one can easily show that each $L_{ij}(1)$ is an analytic strictly positive
function, so that $d_{2}$ is also analytic on $\Lambda.$ Finally, $d_{3}$ is
given by means of a sort of incomplete Mellin transform (which is defined in
Proposition 2.5) of the functions $M_{1}$ and $M_{3}$ in $(\ref{def_M})$. One
can show that the hypothesis $\mathbf{H1}$ and $\mathbf{H2}$ imply that each
$M_{i}(u;\mu)$ is analytic on $(-\varepsilon,+\infty)\times\Lambda$ for some
$\varepsilon>0.$ Taking this into account, by applying $(d)$ in Proposition
2.5 it follows that $d_{3}$ is a meromorphic function on $\Lambda$ having
poles only at those $\mu_{0}$ such that $1/\lambda_{1}(\mu_{0})\in\mathbb{N}$
or $\lambda_{3}(\mu_{0})\in\mathbb{N}$.

Also with regard to the statement of Theorem A, the assertions $(e)$ and $(f)$
hold under the assumptions $\lambda_{1}(\mu_{0})<1$, $\lambda_{2}(\mu_{0})>1$
and $\lambda_{3}(\mu_{0})>1$. However one can always reduce to this case
provided that $\lambda_{i}(\mu_{0})\neq 1$ for $i=1,2,3$ by means of a
rescaling of time and a projective change of coordinates that permute
conveniently the three singular points of the polycycle.

The paper [8] constitutes an important previous contribution to the study of
Kolmogorov polycycles that should be referred. Indeed, following our notations
and definitions, the authors prove (see [8, Theorem 1]) that if
$d_{1}(\mu_{0})=0$ then the return map of $X_{\mu_{0}}$ around the polycycle
$\Gamma$ is of the form

$\mathscr{R}(s;\mu_{0})=\Delta s+\mbox{\rm o}(s),$ (3)

cf. $(b)$ in Theorem 2.6, and they also provide the explicit expression of the
coefficient $\Delta.$ This coefficient is given as the limit of a sum of three
improper integrals, which computed separately diverge. An easy manipulation of
the integrals shows that these divergences cancel each other, yielding to the
expression of $d_{2}$ given in Theorem A. It is important to remark that the
expansion in $(\ref{v1})$ can not be used to obtain an upper bound for
$\mathrm{Cycl}\big{(}(\Gamma,X_{\mu_{0}}),X_{\mu}\big{)}$ because the
remainder is not uniform with respect to the parameters. It is possible
however to use it to obtain lower bounds. In this direction the authors prove
in [8, Corollary 5] that if $d_{1}$ vanishes and is independent at $\mu_{0}$
and $d_{2}(\mu_{0})\neq 0$ then
$\mathrm{Cycl}\big{(}(\Gamma,X_{\mu_{0}}),X_{\mu}\big{)}\geqslant 1$. Since
$d_{2}(\mu_{0})\neq 0$ implies $\mathscr{R}(\,\cdot\,;\mu_{0})\not\equiv Id$
by Theorem 2.6, this lower bound follows by applying $(b)$ in Theorem A.

The paper is organised in the following way. Section 2 is entirely devoted to
prove Theorem A and for that purpose we rely in our previous results about the
asymptotic expansion of the Dulac map of a hyperbolic saddle that we obtain in
[14, 15, 16]. For this reason, before starting the proof of Theorem A we first
state these results and introduce the necessary definitions. The asymptotic
expansion of the displacement map near the polycycle is given in Theorem 2.6
and constitutes the fundamental tool in order to prove Theorem A. As a by-
product of this expansion we obtain a method to study the stability of the
polycycle, see Remark 2. Section 3 is addressed to the applications. The first
one is Theorem 3.1, where we consider a Kolmogorov’s cubic system depending on
three parameters that was previously studied in [8]. The authors in that paper
show that there exist parameters for which the cyclicity of the polycycle is
at least 1. In the present paper we obtain the exact cyclicity of all the
parameters in the family (that can be 0 or 1), including the case in which the
return map along the polycycle is the identity. We also show that there exists
exactly one singularity in the first quadrant, which can be a focus or a
center, and we compute its cyclicity. Finally we prove that it is not possible
a simultaneous bifurcation of limit cycles from the polycycle and that
singularity. We give our second application in Theorem 3.2, where we consider
a Kolmogorov’s cubic system depending on five parameters. In this case we also
provide the exact cyclicity of $\Gamma$ for all the parameters in the family,
which again can be 0 or 1.

##  2 Proof of Theorem A

In order to tackle the proof of Theorem A we will appeal to some previous
results from [14, 15, 16] about the asymptotic expansion of the Dulac map. For
reader’s convenience we gather these results in Proposition 2.4. To this end
it is first necessary to introduce some new notation and definitions.

Setting $\hat{\nu}\\!:=(\lambda,\nu)\in\hat{W}\\!:=(0,+\infty)\times W$ with
$W$ an open set of $\mathbb{R}^{N},$ we consider the family of vector fields
$\\{X_{\hat{\nu}}\\}_{{\hat{\nu}}\in\hat{W}}$ with

$X_{\hat{\nu}}({x_{1}},{x_{2}})={x_{1}}P_{1}({x_{1}},{x_{2}};{\hat{\nu}})\partial_{x_{1}}+{x_{2}}P_{2}({x_{1}},{x_{2}};{\hat{\nu}})\partial_{x_{2}}$
(4)

where

  * •

$P_{1}$ and $P_{2}$ belong to
$\mathscr{C}^{\omega}(\mathscr{U}\\!\times\\!\hat{W})$ for some open set
$\mathscr{U}$ of $\mathbb{R}^{2}$ containing the origin,

  * •

$P_{1}({x_{1}},0;{\hat{\nu}})>0$ and $P_{2}(0,{x_{2}};{\hat{\nu}})<0$ for all
$({x_{1}},0),(0,{x_{2}})\in\mathscr{U}$ and ${\hat{\nu}}\in\hat{W},$

  * •

$\lambda=-\frac{P_{2}(0,0;\nu)}{P_{1}(0,0;\nu)}$.

Thus, for all $\hat{\nu}\in\hat{W}$, the origin is a hyperbolic saddle of
$X_{\hat{\nu}}$ with the separatrices lying in the axis. We point out that
here the hyperbolicity ratio of the saddle is an independent parameter,
although in the proof of Theorem A we will have $\lambda=\lambda(\nu)$. The
reason for this is that the hyperbolicity ratio turns out to be the ruling
parameter in our results and, besides, having it uncoupled from the rest of
parameters simplifies the notation in the statements. Moreover, for $i=1,2,$
we consider a $\mathscr{C}^{\omega}$ transverse section
${\sigma_{i}}\\!:{(-\varepsilon,\varepsilon)\times\hat{W}}\longrightarrow{\Sigma_{i}}$
to $X_{{\hat{\nu}}}$ at $x_{i}=0$ defined by

$\sigma_{i}(s;{\hat{\nu}})=\bigl{(}\sigma_{i1}(s;{\hat{\nu}}),\sigma_{i2}(s;{\hat{\nu}})\bigr{)}$

such that $\sigma_{1}(0,{\hat{\nu}})\in\\{(0,x_{2});x_{2}>0\\}$ and
$\sigma_{2}(0,{\hat{\nu}})\in\\{(x_{1},0);x_{1}>0\\}$ for all
${\hat{\nu}}\in\hat{W}.$ We denote the Dulac map of $X_{\hat{\nu}}$ from
$\Sigma_{1}$ to $\Sigma_{2}$ by $D(\,\cdot\,;{\hat{\nu}})$, see Figure 2.

[l(0mm),r(0mm),t(0mm),b(5mm)]dib1 [l]5,15.5;$0$ [l]12,16;$s$
[l]18,32;$\sigma_{1}$ [l]28,40;$\Sigma_{1}$ [l]31.5,45;$x_{2}$
[l]38,42;$\sigma_{1}(s)$ [l]52,25;$\varphi(\,\cdot\,,\sigma_{1}(s))$
[l]75.5,12;$\sigma_{2}(D(s))$ [l]75,-2;$\Sigma_{2}$ [l]82,1;$x_{1}$
[l]80,26;$\sigma_{2}$ [l]85,33.5;$0$ [l]96,33.5;$D(s)$

Figure 2: Definition of the Dulac map $D(\,\cdot\,;{\hat{\nu}})$, where
$\varphi(t,p;{\hat{\nu}})$ is the solution of $X_{\hat{\nu}}$ passing through
the point $p\in\mathscr{U}$ at time $t=0.$

The asymptotic expansion of $D(s;\hat{\nu})$ at $s=0$ consists of a remainder
and a principal part. The principal part is given in a monomial scale that
contains a deformation of the logarithm, the so-called Ecalle-Roussarie
compensator, whereas the remainder has good flatness properties with respect
to the parameters. We next give precise definitions of these key notions.

  * Definition 2.1.

The function defined for $s>0$ and $\alpha\in\mathbb{R}$ by means of

$\omega(s;\alpha)\>=\left\\{\begin{array}[]{ll}\frac{s^{-\alpha}-1}{\alpha}&\text{if
$\alpha\neq 0,$}\\\\[2.0pt] -\log s&\text{if $\alpha=0,$}\end{array}\right.$

is called the _Ecalle-Roussarie compensator_. $\square$

  * Definition 2.2.

Consider an open subset $U\subset\hat{W}\subset\mathbb{R}^{N+1}.$ We say that
a function $\psi(s;{\hat{\nu}})$ belongs to the class
$\mathscr{C}^{\infty}_{s>0}(U)$ if there exist an open neighbourhood $\Omega$
of

$\\{(s,{\hat{\nu}})\in\mathbb{R}^{N+2};s=0,{\hat{\nu}}\in U\\}=\\{0\\}\times
U$

in $\mathbb{R}^{N+2}$ such that $(s,{\hat{\nu}})\mapsto\psi(s;{\hat{\nu}})$ is
$\mathscr{C}^{\infty}$ on $\Omega\cap\big{(}(0,+\infty)\times U\big{)}$.
$\square$

More formally, the definition of $\mathscr{C}^{\infty}_{s>0}(U)$ must be
thought in terms of germs with respect to relative neighbourhoods of
$\\{0\\}\times U$ in $(0,+\infty)\times U$. In doing so
$\mathscr{C}^{\infty}_{s>0}(U)$ becomes a ring.

We can now introduce the notion of flatness that we shall use in the sequel.

  * Definition 2.3.

Consider an open subset $U\subset\hat{W}\subset\mathbb{R}^{N+1}.$ Given
$L\in\mathbb{R}$ and ${\hat{\nu}}_{0}\in U$, we say that a function
$\psi(s;{\hat{\nu}})\in\mathscr{C}^{\infty}_{s>0}(U)$ is _$L$ -flat with
respect to $s$ at ${\hat{\nu}}_{0}$_, and we write
$\psi\in\mathcal{F}_{L}^{\infty}({\hat{\nu}}_{0})$, if for each
${\hat{\nu}}=({\hat{\nu}}_{0},\ldots,{\hat{\nu}}_{N+1})\in\mathbb{Z}_{\geq
0}^{N+2}$ with
$|{\hat{\nu}}|={\hat{\nu}}_{0}+\cdots+{\hat{\nu}}_{N+1}\leqslant K$ there
exist a neighbourhood $V$ of ${\hat{\nu}}_{0}$ and $C,s_{0}>0$ such that

$\left|\frac{\partial^{|{\hat{\nu}}|}\psi(s;{\hat{\nu}})}{\partial
s^{{\hat{\nu}}_{0}}\partial{\hat{\nu}}_{1}^{{\hat{\nu}}_{1}}\cdots\partial{\hat{\nu}}_{N+1}^{{\hat{\nu}}_{N+1}}}\right|\leqslant
Cs^{L-{\hat{\nu}}_{0}}\text{ for all $s\in(0,s_{0})$ and ${\hat{\nu}}\in V$.}$

If $W$ is a (not necessarily open) subset of $U$ then define
$\mathcal{F}_{L}^{\infty}(W)\\!:=\bigcap_{{\hat{\nu}}_{0}\in
W}\mathcal{F}_{L}^{\infty}({\hat{\nu}}_{0}).$ $\square$

Apart from the remainder and the monomial order, the most important ingredient
for our purposes is the explicit expression of the coefficients in the
asymptotic expansion. In order to give them we introduce next some additional
notation, where for the sake of shortness the dependence on
${\hat{\nu}}=(\lambda,\nu)$ is omitted. We define the functions:

$\begin{array}[]{ll}\displaystyle
L_{1}(u)\\!:=\exp\int_{0}^{u}\left(\frac{P_{1}(0,z)}{P_{2}(0,z)}+\frac{1}{\lambda}\right)\frac{dz}{z}&\displaystyle
L_{2}(u)\\!:=\exp\int_{0}^{u}\left(\frac{P_{2}(z,0)}{P_{1}(z,0)}+{\lambda}\right)\frac{dz}{z}\\\\[15.0pt]
\displaystyle
M_{1}(u)\\!:=L_{1}(u)\partial_{1}\\!\left(\frac{P_{1}}{P_{2}}\right)(0,u)&\displaystyle
M_{2}(u)\\!:=L_{2}(u)\partial_{2}\\!\left(\frac{P_{2}}{P_{1}}\right)(u,0)\\\\[15.0pt]
\end{array}$ (5)

On the other hand, for shortness as well, we use the compact notation
$\sigma_{ijk}$ for the $k$th derivative at $s=0$ of the $j$th component of
$\sigma_{i}(s;{\hat{\nu}})$, i.e.,

$\sigma_{ijk}({\hat{\nu}})\\!:=\partial^{k}_{s}\sigma_{ij}(0;{\hat{\nu}}).$

Taking this notation into account we also introduce the following real values,
where once again we omit the dependence on ${\hat{\nu}}$:

$\begin{array}[]{l}\displaystyle
S_{1}\\!:=\frac{\sigma_{112}}{2\sigma_{111}}-\frac{\sigma_{121}}{\sigma_{120}}\left(\frac{P_{1}}{P_{2}}\right)\\!(0,\sigma_{120})-\frac{\sigma_{111}}{L_{1}(\sigma_{120})}\hat{M}_{1}(1/\lambda,\sigma_{120})\\\\[20.0pt]
\displaystyle
S_{2}\\!:=\frac{\sigma_{222}}{2\sigma_{221}}-\frac{\sigma_{211}}{\sigma_{210}}\left(\frac{P_{2}}{P_{1}}\right)\\!(\sigma_{210},0)-{\frac{\sigma_{221}}{L_{2}(\sigma_{210})}}\hat{M}_{2}(\lambda,\sigma_{210}).\end{array}$
(6)

Here $\hat{M}_{i}$ stands for a sort of incomplete Mellin transform of $M_{i}$
that will be defined by Proposition 2.5 below. We can now state the following
result, which gathers Theorem A and Theorem 4.1 in [16] and that it will
constitute the key tool in order to prove the main result in the present
paper.

######  Proposition 2.4.

Let $D(s;{\hat{\nu}})$ be the Dulac map of the hyperbolic saddle $(\ref{X})$
from $\Sigma_{1}$ and $\Sigma_{2}$ and define

$\Delta_{0}(\hat{\nu})=\frac{\sigma_{111}^{\lambda}\sigma_{120}}{L_{1}^{\lambda}(\sigma_{120})}\frac{L_{2}(\sigma_{210})}{\sigma_{221}\sigma_{210}^{\lambda}},\text{
}\Delta_{1}(\hat{\nu})=\Delta_{0}\lambda S_{1}\text{ and
}\Delta_{2}(\hat{\nu})=-\Delta_{0}^{2}S_{2},$

where $\hat{\nu}=(\lambda,\nu)\in\hat{W}=(0,+\infty)\times W.$ Then
$\Delta_{0}$ is analytic and strictly positive on $\hat{W}$, $\Delta_{1}$ is
meromorphic on $\hat{W}$ with poles only at $\lambda\in\frac{1}{\mathbb{N}}$
and $\Delta_{2}$ is meromorphic on $\hat{W}$ with poles only at
$\lambda\in\mathbb{N}.$ Moreover the following assertions hold:

  1. $(1)$

If $\lambda_{0}<1$ then
$D(s;{\hat{\nu}})=s^{\lambda}\big{(}\Delta_{0}({\hat{\nu}})+\Delta_{2}({\hat{\nu}})s^{\lambda}+\mathcal{F}_{\ell}^{\infty}(\\{\lambda_{0}\\}\times
W)\big{)}$ for any ${\ell}\in\big{[}\lambda_{0},\min(2\lambda_{0},1)\big{)}$.

  2. $(2)$

If $\lambda_{0}=1$ then
$D(s;{\hat{\nu}})=s^{\lambda}\big{(}\Delta_{0}({\hat{\nu}})+\boldsymbol{\Delta}^{\lambda_{0}}(\omega;{\hat{\nu}})s+\mathcal{F}_{\ell}^{\infty}(\\{\lambda_{0}\\}\times
W)\big{)}$ for any ${\ell}\in[1,2),$ where

$\boldsymbol{\Delta}^{\lambda_{0}}(\omega;{\hat{\nu}})=\Delta_{1}({\hat{\nu}})+\Delta_{2}({\hat{\nu}})(1+\alpha\omega),$

$\alpha=1-\lambda$ and $\omega=\omega(s;\alpha)$.

  3. $(3)$

If $\lambda_{0}>1$ then
$D(s;{\hat{\nu}})=s^{\lambda}\big{(}\Delta_{0}({\hat{\nu}})+\Delta_{1}({\hat{\nu}})s+\mathcal{F}_{\ell}^{\infty}(\\{\lambda_{0}\\}\times
W)\big{)}$ for any ${\ell}\in\big{[}1,\min(\lambda_{0},2)\big{)}$.

In particular
$D(s;{\hat{\nu}})=s^{\lambda}\big{(}\Delta_{0}({\hat{\nu}})+\mathcal{F}_{\ell}^{\infty}(\\{\lambda_{0}\\}\times
W)\big{)}$ for any ${\ell}\in\big{(}0,\min(\lambda_{0},1)\big{)}$.

The flatness $\ell$ of the remainder can range in a certain interval depending
on $\lambda_{0}.$ The left endpoint of this interval is only given for
completeness to guarantee that all the monomials in the principal part are
relevant (i.e., they cannot be included in the remainder). The important
information about the flatness is given by the right endpoint. A key tool in
order to give a closed expression of the coefficients $\Delta_{i}$ is the use
of a sort of incomplete Mellin transform, which is accurately defined in the
next result. For a proof of this result the reader is referred to [16,
Appendix B].

######  Proposition 2.5.

Let us consider an open interval $I$ of $\mathbb{R}$ containing $x=0$ and an
open subset $U$ of $\mathbb{R}^{M}$.

  1. $(a)$

Given $f(x;{\upsilon})\in\mathscr{C}^{\infty}(I\times U)$, there exits a
unique
$\hat{f}(\alpha,x;{\upsilon})\in\mathscr{C}^{\infty}((\mathbb{R}\setminus\mathbb{Z}_{\geq
0})\times I\times U)$ such that

$x\partial_{x}\hat{f}({\alpha},x;{\upsilon})-\alpha\hat{f}({\alpha},x;{\upsilon})=f(x;{\upsilon}).$

  2. $(b)$

If $x\in I\setminus\\{0\\}$ then
$\partial_{x}(\hat{f}({\alpha},x;{\upsilon})|x|^{-\alpha})=f(x;{\upsilon})\frac{|x|^{-\alpha}}{x}$
and, taking any $k\in\mathbb{Z}_{\geq 0}$ with $k>\alpha$,

$\hat{f}(\alpha,x;{\upsilon})=\sum_{i=0}^{k-1}\frac{\partial_{x}^{i}f(0;{\upsilon})}{i!(i-\alpha)}x^{i}+|x|^{\alpha}\int_{0}^{x}\\!\left(f(s;{\upsilon})-T_{0}^{k-1}f(s;{\upsilon})\right)|s|^{-\alpha}\frac{ds}{s},$

where
$T_{0}^{k}f(x;{\upsilon})=\sum_{i=0}^{k}\frac{1}{i!}\partial_{x}^{i}f(0;{\upsilon})x^{i}$
is the $k$-th degree Taylor polynomial of $f(x;{\upsilon})$ at $x=0$.

  3. $(c)$

For each $(i_{0},x_{0},{\upsilon}_{0})\in\mathbb{Z}_{\geq 0}\times I\times W$
the function
$(\alpha,x,{\upsilon})\mapsto(i_{0}-\alpha)\hat{f}(\alpha,x;{\upsilon})$
extends $\mathscr{C}^{\infty}$ at $(i_{0},x_{0},{\upsilon}_{0})$ and,
moreover, it tends to
$\frac{1}{i_{0}!}\partial_{x}^{i_{0}}f(0;{\upsilon}_{0})x_{0}^{i_{0}}$ as
$(\alpha,x,{\upsilon})\to(i_{0},x_{0},{\upsilon}_{0}).$

  4. $(d)$

If $f(x;{\upsilon})$ is analytic on $I\times U$ then
$\hat{f}(\alpha,x;{\upsilon})$ is analytic on
$(\mathbb{R}\setminus\mathbb{Z}_{\geq 0})\times I\times U$. Finally, for each
$(\alpha_{0},x_{0},{\upsilon}_{0})\in\mathbb{Z}_{\geq 0}\times I\times U$ the
function
$(\alpha,x,{\upsilon})\mapsto(\alpha_{0}-\alpha)\hat{f}(\alpha,x;{\upsilon})$
extends analytically to $(\alpha_{0},x_{0},{\upsilon}_{0})$.

On account of this result for each $M_{i}(u;\hat{\nu})$ in $(\ref{def_fun})$
we have that $(\alpha,u;\hat{\nu})\mapsto\hat{M}_{i}(\alpha,u;\hat{\nu})$ is a
well defined meromorphic function with poles only at
$\alpha\in\mathbb{Z}_{\geq 0}$. Accordingly, see $(\ref{def_S})$,
$\hat{M}_{1}(1/\lambda,\sigma_{120})$ and $\hat{M}_{2}(\lambda,\sigma_{210})$
are the values (depending on $\hat{\nu}$) that we obtain by taking
$\hat{M}_{1}(\alpha,u;\hat{\nu})$ with $\alpha=1/\lambda$ and
$u=\sigma_{120}(\hat{\nu})$ and by taking $\hat{M}_{2}(\alpha,u;\hat{\nu})$
with $\alpha=\lambda$ and $u=\sigma_{210}(\hat{\nu}),$ respectively.

At this point we get back to the setting treated in the present paper and from
now on we recover the original notation for the parameters in the family under
consideration, see $(\ref{kol})$. In order to study the Dulac maps of the
hyperbolic saddles at the vertices of the polycycle $\Gamma$ we take three
local transverse sections $\Sigma_{1}$, $\Sigma_{2}$, and $\Sigma_{3}$
parametrised, respectively, by $s\mapsto(1,s)$, $s\mapsto(1/s,1/s)$ and
$s\mapsto(s,1)$ with $s>0.$ We define $D_{1}(s;\mu)$ to be the Dulac map of
$X_{\mu}$ from $\Sigma_{1}$ to $\Sigma_{2},$ $D_{2}(s;\mu)$ to be the Dulac
map of $X_{\mu}$ from $\Sigma_{2}$ to $\Sigma_{3}$ and, finally,
$D_{3}(s;\mu)$ to be the Dulac map of $-X_{\mu}$ from $\Sigma_{1}$ to
$\Sigma_{3},$ see Figure 3

[l(0mm),r(0mm),t(0mm),b(5mm)]dib2(0.75) [l]22,12;$\Sigma_{1}$
[l]10,25;$\Sigma_{3}$ [l]55,55;$\Sigma_{2}$ [l]55,77;$\ell_{\infty}$
[l]65,20;$D_{1}$ [l]9,11;$D_{3}$ [l]20,67;$D_{2}$ [l]94,0;$s_{1}$
[l]-5,94;$s_{2}$ [l]-5,0;$s_{3}$

Figure 3: Auxiliary Dulac maps for the definition of $\mathscr{D}=D_{2}\circ
D_{1}-D_{3}$ in Theorem 2.6. The return map in Theorem A, with respect to the
transverse section $\Sigma_{1}$, would be $\mathscr{R}=D_{3}^{-1}\circ
D_{2}\circ D_{1}=D_{3}^{-1}\circ\mathscr{D}+\text{Id}$.

x It is then clear that the limit cycles of $X_{\mu}$ near $\Gamma$ are in one
to one correspondence with the isolated positive zeroes of

$\mathscr{D}(s;\mu)\\!:=\big{(}D_{2}\circ D_{1}-D_{3}\big{)}(s;\mu)$

near $s=0.$ The proof of Theorem A strongly relies in our next result, where
we get the asymptotic expansion of $\mathscr{D}(s;\mu)$ at $s=0$ and we
compute its coefficients. In its statement $d_{i}(\mu)$, for $i=1,2,3,$ are
the functions defined in Theorem A and $\mathbb{R}\\{\mu\\}_{\mu_{0}}$ stands
for the local ring of convergent power series at $\mu_{0}.$

######  Theorem 2.6.

Let us fix any $\mu_{0}\in\Lambda$ and set
$\lambda_{i}^{0}\\!:=\lambda_{i}(\mu_{0})$ for $i=1,2,3.$

  1. $(a)$

If $\lambda_{1}^{0}\lambda_{2}^{0}\lambda_{3}^{0}\neq 1$ then, for any
$\ell_{1}\in\big{(}\min(\lambda_{1}^{0}\lambda_{2}^{0},1/\lambda_{3}^{0}),\min(\lambda_{1}^{0}+\lambda_{1}^{0}\lambda_{2}^{0},1+\lambda_{1}^{0}\lambda_{2}^{0},2\lambda_{1}^{0}\lambda_{2}^{0},1+1/\lambda_{3}^{0},2/\lambda_{3}^{0})\big{)}$,

$\mathscr{D}(s;\mu)=a_{1}(\mu)s^{\lambda_{1}\lambda_{2}}-a_{2}(\mu)s^{1/\lambda_{3}}+\mathcal{F}_{\ell_{1}}^{\infty}(\mu_{0}),$

where $a_{1}$ and $a_{2}$ are analytic and strictly positive functions on
$\Lambda.$

  2. $(b)$

If $\lambda_{1}^{0}\lambda_{2}^{0}\lambda_{3}^{0}=1$ then, for any
$\ell_{2}\in(0,\min(1,\lambda_{1}^{0},\lambda_{1}^{0}\lambda_{2}^{0}))$,

$\mathscr{D}(s;\mu)=\big{(}b_{1}(\mu)\,\omega\big{(}s;\alpha(\mu)\big{)}+b_{2}(\mu)+\mathcal{F}_{\ell_{2}}^{\infty}(\mu_{0})\big{)}s^{1/\lambda_{3}},$

where $\alpha=1/\lambda_{3}-\lambda_{1}\lambda_{2}$, $b_{1}=\alpha a_{1}$ and
$b_{2}=a_{1}-a_{2}.$ Moreover the equalities

$(b_{1})=(d_{1})\text{ and }(b_{1},b_{2})=(d_{1},d_{2})$

between ideals over the ring $\mathbb{R}\\{\mu\\}_{\mu_{0}}$ are verified.
Assuming $\lambda_{1}^{0}>1$, $\lambda_{2}^{0}>1$ and $\lambda_{3}^{0}<1$
additionally then, for any
$\ell_{3}\in\big{(}1,\min(2,\lambda_{1}^{0},1/\lambda_{3}^{0})\big{)}$,

$\mathscr{D}(s;\mu)=\big{(}b_{1}(\mu)\,\omega\big{(}s;\alpha(\mu)\big{)}+b_{2}(\mu)+b_{3}(\mu)s+\mathcal{F}_{\ell_{3}}^{\infty}(\mu_{0})\big{)}s^{1/\lambda_{3}}U(s;\mu),$

where $b_{3}$ is an analytic function at $\mu_{0}$ verifying that

$(b_{1},b_{2},b_{3})=(d_{1},d_{2},d_{3})$

over the ring $\mathbb{R}\\{\mu\\}_{\mu_{0}}$ and $U$ is an analytic function
such that $U(0;\mu_{0})=1$.

  * Proof.

In order to study the Dulac map $D_{1}$ from $\Sigma_{1}$ to $\Sigma_{2}$ we
compactify $X_{\mu}$ by means of the coordinate change
$\\{x_{1}=\frac{y}{x},x_{2}=\frac{1}{x}\\}.$ One can easily verify that the
new vector field is orbitally conjugated to $(\ref{X})$ particularised with
$P_{1}(x_{1},x_{2})=x_{2}^{n}\big{(}f-g)(\frac{1}{x_{2}},\frac{x_{1}}{x_{2}})$
and $P_{2}(x_{1},x_{2})=x_{2}^{n}f(\frac{1}{x_{2}},\frac{x_{1}}{x_{2}}),$
whereas in these coordinates the transverse sections $\Sigma_{1}$ and
$\Sigma_{2}$ are parametrised by $\sigma_{1}(s)=(s,1)$ and
$\sigma_{2}(s)=(1,s)$, respectively. The hyperbolicity ratio of the saddle at
the origin is
$\lambda_{1}=-\big{(}\frac{P_{2}}{P_{1}}\big{)}(0,0)=\big{(}\frac{f_{n}}{g_{n}-f_{n}}\big{)}(1,0).$
Therefore, by applying Proposition 2.4 we can assert that

$D_{1}(s)=\Delta_{10}s^{\lambda_{1}}\big{(}1+\mathcal{F}^{\infty}_{{\ell}_{1}}(\mu_{0})\big{)},\text{
with $\Delta_{10}\\!:=\big{(}L_{12}L_{11}^{-\lambda_{1}}\big{)}(1)$}$ (7)

and any $0<{\ell}_{1}<\min(1,\lambda_{1}^{0})$. Recall here that
$\lambda_{i}^{0}=\lambda_{i}(\mu_{0})$ for $i=1,2,3$ by definition.

Next, to analyse the Dulac map $D_{2}$ from $\Sigma_{2}$ to $\Sigma_{3}$ we
compactify $X_{\mu}$ performing the change of coordinates given by
$\\{x_{1}=\frac{1}{y},x_{2}=\frac{x}{y}\\}.$ One can check that the new vector
field is orbitally conjugated to $(\ref{X})$ with
$P_{1}(x_{1},x_{2})=x_{1}^{n}g(\frac{x_{2}}{x_{1}},\frac{1}{x_{1}})$ and
$P_{2}(x_{1},x_{2})=x_{1}^{n}\big{(}g-f)(\frac{x_{2}}{x_{1}},\frac{1}{x_{1}})$
and that in these coordinates the transverse sections $\Sigma_{2}$ and
$\Sigma_{3}$ are parametrised by $\sigma_{1}(s)=(s,1)$ and
$\sigma_{2}(s)=(1,s)$, respectively. The hyperbolicity ratio of the saddle at
the origin is
$\lambda_{2}=-\big{(}\frac{P_{2}}{P_{1}}\big{)}(0,0)=\big{(}\frac{f_{n}-g_{n}}{g_{n}}\big{)}(0,1).$
Thus, by Proposition 2.4 again,

$D_{2}(s)=\Delta_{20}s^{\lambda_{2}}\big{(}1+\mathcal{F}^{\infty}_{\ell_{2}}(\mu_{0})\big{)},\text{
with $\Delta_{20}\\!:=\big{(}L_{22}L_{21}^{-\lambda_{2}}\big{)}(1)$}$ (8)

and any $0<{\ell}_{2}<\min(1,\lambda_{2}^{0})$.

Finally, to study the Dulac map $D_{3}$ of from $\Sigma_{1}$ to $\Sigma_{3}$
we make the reflection $\\{x_{1}=y,x_{2}=x\\}$, which brings $-X_{\mu}$ to
$(\ref{X})$ with $P_{1}(x_{1},x_{2})=-g(x_{2},x_{1})$ and
$P_{2}(x_{1},x_{2})=-f(x_{2},x_{1})$. In these coordinates the transverse
sections $\Sigma_{1}$ and $\Sigma_{3}$ are parametrised by
$\sigma_{1}(s)=(s,1)$ and $\sigma_{2}(s)=(1,s)$, respectively, and the
hyperbolicity ratio of the saddle is
$\frac{1}{\lambda_{3}}=-\big{(}\frac{P_{2}}{P_{1}}\big{)}(0,0)=-\big{(}\frac{f}{g}\big{)}(0,0).$
Hence by Proposition 2.4 once again,

$D_{3}(s)=\Delta_{30}s^{1/\lambda_{3}}\big{(}1+\mathcal{F}^{\infty}_{\ell_{3}}(\mu_{0})\big{)},\text{
with $\Delta_{30}\\!:=\big{(}L_{32}L_{31}^{-1/\lambda_{3}}\big{)}(1)$}$ (9)

and any $0<{\ell}_{3}<\min(1,1/\lambda_{3}^{0})$. Consequently, from
$(\ref{prop1eq1})$, $(\ref{prop1eq2})$ and $(\ref{prop1eq3})$ we get that

$\displaystyle\mathscr{D}(s)=\big{(}D_{2}\circ D_{1}-D_{3}\big{)}(s)$
$\displaystyle=\Delta_{20}\Delta_{10}^{\lambda_{2}}s^{\lambda_{1}\lambda_{2}}\big{(}1+\mathcal{F}^{\infty}_{\ell_{1}}(\mu_{0})\big{)}^{\lambda_{2}}\big{(}1+\mathcal{F}^{\infty}_{\ell_{4}}(\mu_{0})\big{)}-\Delta_{30}s^{1/\lambda_{3}}\big{(}1+\mathcal{F}^{\infty}_{\ell_{3}}(\mu_{0})\big{)}$
$\displaystyle=\Delta_{20}\Delta_{10}^{\lambda_{2}}s^{\lambda_{1}\lambda_{2}}\big{(}1+\mathcal{F}^{\infty}_{\ell_{1}}(\mu_{0})\big{)}\big{(}1+\mathcal{F}^{\infty}_{\ell_{4}}(\mu_{0})\big{)}-\Delta_{30}s^{1/\lambda_{3}}\big{(}1+\mathcal{F}^{\infty}_{\ell_{3}}(\mu_{0})\big{)}$
$\displaystyle=\Delta_{20}\Delta_{10}^{\lambda_{2}}s^{\lambda_{1}\lambda_{2}}-\Delta_{30}s^{1/\lambda_{3}}+\mathcal{F}^{\infty}_{\ell_{5}}(\mu_{0}),$

where, by [15, Lemma A.2], we can take $\ell_{4}$ and $\ell_{5}$ to be any
numbers such that
$0<\ell_{4}<\min(\lambda_{1}^{0},\lambda_{1}^{0}\lambda_{2}^{0})$ and
$\min(\lambda_{1}^{0}\lambda_{2}^{0},1/\lambda_{3}^{0}))<\ell_{5}<\min(\lambda_{1}^{0}+\lambda_{1}^{0}\lambda_{2}^{0},1+\lambda_{1}^{0}\lambda_{2}^{0},2\lambda_{1}^{0}\lambda_{2}^{0},1+1/\lambda_{3}^{0},2/\lambda_{3}^{0})$,
respectively. Thus, setting $a_{1}\\!:=\Delta_{20}\Delta_{10}^{\lambda_{2}}$
and $a_{2}\\!:=\Delta_{30}$, we obtain

$\mathscr{D}(s)=a_{1}s^{\lambda_{1}\lambda_{2}}-a_{2}s^{1/\lambda_{3}}+\mathcal{F}^{\infty}_{\ell_{5}}(\mu_{0})$
(10)

and this proves $(a)$ because each $\Delta_{i0}$ is a strictly positive
analytic function by Proposition 2.4.

Let us proceed next with the proof of the assertions in $(b)$, thus from now
on we assume that $\lambda_{1}^{0}\lambda_{2}^{0}\lambda_{3}^{0}=1$. Then,
setting $\alpha=1/\lambda_{3}-\lambda_{1}\lambda_{2}$ and taking any
$\ell_{6}\in(0,\min(1,\lambda_{1}^{0},\lambda_{1}^{0}\lambda_{2}^{0}))$, from
the above expression we get

$\displaystyle\mathscr{D}(s)$
$\displaystyle=s^{1/\lambda_{3}}\big{(}a_{1}s^{\lambda_{1}\lambda_{2}-1/\lambda_{3}}-a_{2}+\mathcal{F}^{\infty}_{\ell_{6}}(\mu_{0})\big{)}$
$\displaystyle=s^{1/\lambda_{3}}\big{(}a_{1}(1+\alpha\omega(s;\alpha))-a_{2}+\mathcal{F}^{\infty}_{\ell_{6}}(\mu_{0})\big{)}$
$\displaystyle=s^{1/\lambda_{3}}\big{(}b_{1}\omega(s;\alpha)+b_{2}+\mathcal{F}^{\infty}_{\ell_{6}}(\mu_{0})\big{)},$

where $b_{1}\\!:=\alpha a_{1}$ and $b_{2}\\!:=a_{1}-a_{2}$. Here we also take
Definition 2 into account. Since $a_{1}$ is an analytic non-vanishing function
at $\mu_{0}$ and $d_{1}=\lambda_{3}\alpha$, we obtain the equality
$(b_{1})=(d_{1})$ between ideals over the local ring
$\mathbb{R}\\{\mu\\}_{\mu_{0}}$. In order to show that
$(b_{1},b_{2})=(d_{1},d_{2})$ holds as well we note that, from
$(\ref{prop1eq1})$, $(\ref{prop1eq2})$ and $(\ref{prop1eq3})$ again,

$\displaystyle b_{2}(\mu)$
$\displaystyle=a_{1}-a_{2}=\Delta_{20}\Delta_{10}^{\lambda_{2}}-\Delta_{30}$
$\displaystyle=(L_{22}L_{21}^{-\lambda_{2}}L_{12}^{\lambda_{2}}L_{11}^{-\lambda_{1}\lambda_{2}}-L_{32}L_{31}^{-1/\lambda_{3}})\big{|}_{u=1}$
$\displaystyle=L_{32}L_{31}^{-1/\lambda_{3}}\left(\frac{L_{22}}{L_{32}}\frac{L_{31}^{1/\lambda_{3}}}{L_{11}^{\lambda_{1}\lambda_{2}}}\frac{L_{12}^{\lambda_{2}}}{L_{21}^{\lambda_{2}}}-1\right)\Big{|}_{u=1}$
$\displaystyle=L_{32}L_{31}^{-1/\lambda_{3}}\left(L_{31}^{\alpha}\left(\frac{L_{22}}{L_{32}}\frac{L_{31}^{\lambda_{1}\lambda_{2}}}{L_{11}^{\lambda_{1}\lambda_{2}}}\frac{L_{12}^{\lambda_{2}}}{L_{21}^{\lambda_{2}}}-1\right)+(L_{31}^{\alpha}-1)\right)\Big{|}_{u=1}$
$\displaystyle=\kappa_{1}(\mu)d_{2}(\mu)+\alpha(\mu)\kappa_{2}(\mu),$

where
$\kappa_{1}=\frac{L_{32}}{L_{31}^{1/\lambda_{3}}}\big{|}_{u=1}\frac{e^{d_{2}}-1}{d_{2}}$
and
$\kappa_{2}=\frac{L_{32}}{L_{31}^{1/\lambda_{3}}}\frac{L_{31}^{\alpha}-1}{\alpha}\big{|}_{u=1}$
are analytic functions at $\mu_{0}$ because so it is each $L_{ij}$ by [16,
Lemma 2.3]. Thus, on account of $d_{1}=\lambda_{3}\alpha$, we get that
$b_{2}=\kappa_{1}d_{2}+\kappa_{3}d_{1}$. Hence $(b_{1},b_{2})=(d_{1},d_{2})$
over the ring $\mathbb{R}\\{\mu\\}_{\mu_{0}}$ since $\kappa_{1}(\mu_{0})>0$
and we have already shown that $(b_{1})=(d_{1}).$

So far we have proved the first assertion in $(b)$. To show the second one,
besides $\lambda_{1}^{0}\lambda_{2}^{0}\lambda_{3}^{0}=1$, we assume
$\lambda_{1}^{0}>1$, $\lambda_{2}^{0}>1$ and $\lambda_{3}^{0}<1$. On account
of this we can apply point $(3)$ in Proposition 2.4 to conclude that

$\displaystyle D_{1}(s)$
$\displaystyle=s^{\lambda_{1}}\big{(}\Delta_{10}+\Delta_{11}s+\mathcal{F}_{\ell_{7}}^{\infty}(\mu_{0})\big{)}\text{
for any }{\ell_{7}}\in\big{[}1,\min(\lambda_{1}^{0},2)\big{)},$ (11)
$\displaystyle D_{2}(s)$
$\displaystyle=s^{\lambda_{2}}\big{(}\Delta_{20}+\Delta_{21}s+\mathcal{F}_{\ell_{8}}^{\infty}(\mu_{0})\big{)}\text{
for any }{\ell_{8}}\in\big{[}1,\min(\lambda_{2}^{0},2)\big{)},$ (12) and
$\displaystyle D_{3}(s)$
$\displaystyle=s^{1/\lambda_{3}}\big{(}\Delta_{30}+\Delta_{31}s+\mathcal{F}_{\ell_{9}}^{\infty}(\mu_{0})\big{)}\text{
for any }{\ell_{9}}\in\big{[}1,\min(1/\lambda_{3}^{0},2)\big{)}.$ (13)

Here the first order coefficients $\Delta_{10}$, $\Delta_{20}$ and
$\Delta_{30}$ are the ones already defined in $(\ref{prop1eq1})$,
$(\ref{prop1eq2})$ and $(\ref{prop1eq3})$, respectively. With regard to the
second order coefficients, only the ones of $D_{1}$ and $D_{3}$ are relevant
for our purposes, which are given by

$\Delta_{11}\\!:=-\frac{\lambda_{1}\Delta_{10}\hat{M}_{1}(1/\lambda_{1},1)}{L_{11}(1)}\text{
and
}\Delta_{31}\\!:=-\frac{\Delta_{30}\hat{M}_{3}(\lambda_{3},1)}{\lambda_{3}L_{31}(1)},$
(14)

respectively. In each case, on account of $(\ref{def_M})$, this follows easily
from the formula $\Delta_{1}=\lambda\Delta_{0}S_{1}$ given in Proposition 2.4
and taking $S_{1}$ in $(\ref{def_S})$ particularised to $\sigma_{1}(s)=(s,1)$.

From $(\ref{prop1eq61})$ and $(\ref{prop1eq7})$, by applying [15, Lemma A.2]
we can assert that

$\displaystyle\big{(}D_{2}\circ D_{1}\big{)}(s)$
$\displaystyle=s^{\lambda_{1}\lambda_{2}}\big{(}\Delta_{10}+\Delta_{11}s+\mathcal{F}_{\ell_{7}}^{\infty}\big{)}^{\lambda_{2}}\big{(}\Delta_{20}+\Delta_{21}s^{\lambda_{1}}\big{(}\Delta_{10}+\Delta_{11}s+\mathcal{F}_{\ell_{7}}^{\infty}\big{)}+\mathcal{F}_{\ell_{10}}^{\infty}\big{)}$
$\displaystyle=s^{\lambda_{1}\lambda_{2}}\big{(}\Delta_{10}^{\lambda_{2}}+{\lambda_{2}}\Delta_{10}^{\lambda_{2}-1}\Delta_{11}s+\mathcal{F}_{\ell_{7}}^{\infty}\big{)}\big{(}\Delta_{20}+\Delta_{21}s^{\lambda_{1}}\big{(}\Delta_{10}+\Delta_{11}s\big{)}+\mathcal{F}_{\ell_{11}}^{\infty}+\mathcal{F}_{\ell_{10}}^{\infty}\big{)}$
$\displaystyle=s^{\lambda_{1}\lambda_{2}}\big{(}\Delta_{10}^{\lambda_{2}}+{\lambda_{2}}\Delta_{10}^{\lambda_{2}-1}\Delta_{11}s+\mathcal{F}_{\ell_{7}}^{\infty}\big{)}\big{(}\Delta_{20}+\Delta_{10}\Delta_{21}s^{\lambda_{1}}+\Delta_{11}\Delta_{21}s^{\lambda_{1}+1}+\mathcal{F}_{\ell_{12}}^{\infty}\big{)}$

for any
$\ell_{10}\in[\lambda_{1}^{0},\lambda_{1}^{0}\min(\lambda_{2}^{0},2)\big{)}$
in the first equality, any
$\ell_{11}\in\big{[}\lambda_{1}^{0}+1,\lambda_{1}^{0}+\min(\lambda_{1}^{0},2)\big{)}$
in the second one and any
$\ell_{12}\in\big{[}\lambda_{1}^{0}+1,\min(2\lambda_{1}^{0},\lambda_{1}^{0}+2,\lambda_{1}^{0}\lambda_{2}^{0})\big{)}$
in the third one. Furthermore in the second equality we use that, for any
$\eta=\eta(\mu),$

$(1+as+\mathcal{F}_{\ell})^{\eta}=(1+as)^{\eta}+\mathcal{F}_{\ell}=1+a\eta
s+\mathcal{F}_{2-\varepsilon}+\mathcal{F}_{\ell},$ (15)

for any $\varepsilon>0,$ which in turn follows noting that

$\displaystyle(1+as+\mathcal{F}_{\ell})^{\eta}-(1+as)^{\eta}$
$\displaystyle=\textstyle(1+as)^{\eta}\big{(}(1+\frac{\mathcal{F}_{\ell}}{1+as})^{\eta}-1\big{)}$
$\displaystyle=(1+as)^{\eta}\big{(}(1+\mathcal{F}_{\ell})^{\eta}-1\big{)}=(1+as)^{\eta}\mathcal{F}_{\ell}=\mathcal{F}_{\ell}.$

Consequently

$\big{(}D_{2}\circ
D_{1}\big{)}(s)=s^{\lambda_{1}\lambda_{2}}\big{(}\Delta_{20}\Delta_{10}^{\lambda_{2}}+{\lambda_{2}}\Delta_{11}\Delta_{20}\Delta_{10}^{\lambda_{2}-1}s+\mathcal{F}_{\ell_{7}}^{\infty}\big{)},$

where we use again [15, Lemma A.2] taking $\lambda_{1}^{0}>1$ and
$\lambda_{2}^{0}>1$ into account. Hence, from $(\ref{prop1eq4})$ and plug in
$s^{-\alpha}=1+\alpha\omega(s;\alpha)$ as before, we get

$\displaystyle\mathscr{D}(s)$ $\displaystyle=\big{(}D_{2}\circ
D_{1}-D_{3}\big{)}(s)$
$\displaystyle=s^{1/\lambda_{3}}\Big{(}s^{-\alpha}\big{(}\Delta_{20}\Delta_{10}^{\lambda_{2}}+{\lambda_{2}}\Delta_{11}\Delta_{20}\Delta_{10}^{\lambda_{2}-1}s+\mathcal{F}_{\ell_{7}}^{\infty}\big{)}-\Delta_{30}-\Delta_{31}s-\mathcal{F}_{\ell_{9}}^{\infty}\Big{)}$
$\displaystyle=s^{1/\lambda_{3}}\big{(}(1+\alpha\omega(s;\alpha))\Delta_{20}\Delta_{10}^{\lambda_{2}}U(s)-\Delta_{30}-\Delta_{31}s-\mathcal{F}_{\ell_{9}}^{\infty}\Big{)},$

where we define
$U(s)=1+{\lambda_{2}}\Delta_{11}\Delta_{10}^{-1}s+\mathcal{F}_{\ell_{7}}^{\infty}$.
The application of the formula given in $(\ref{prop1eq8})$ with $\eta=-1$
shows that
$U(s)^{-1}=1-\lambda_{2}\Delta_{11}\Delta_{10}^{-1}s+\mathcal{F}_{\ell_{7}}$.
Thus one can easily verify that the above expression yields to

$\mathscr{D}(s)=s^{1/\lambda_{3}}U(s)\big{(}b_{1}\omega(s;\alpha)+b_{2}+b_{3}s+\mathcal{F}_{\ell_{13}}^{\infty}\big{)}$

with $\ell_{13}\in[1,\min(\lambda_{1}^{0},1/\lambda_{3}^{0},2))$ and
$b_{3}\\!:=\lambda_{2}\Delta_{11}\Delta_{30}\Delta_{10}^{-1}-\Delta_{31}$. Let
us recall here that $b_{1}=\alpha
a_{1}=\alpha\Delta_{20}\Delta_{10}^{\lambda_{2}}$ and
$b_{2}=a_{1}-a_{2}=\Delta_{20}\Delta_{10}^{\lambda_{2}}-\Delta_{30}$, where
$a_{1}$ and $a_{2}$ are the analytic and strictly positive functions in
$(\ref{prop1eq6})$. On account of the assumptions $\lambda_{1}^{0}>1$ and
$\lambda_{3}^{0}<1$ we have $\lambda_{1}(\mu)\notin\frac{1}{\mathbb{N}}$ and
$\lambda_{3}(\mu)\notin\mathbb{N}$ for $\mu\approx\mu_{0},$ which imply
respectively that $\Delta_{11}$ and $\Delta_{31}$ are analytic at $\mu_{0}$ by
Proposition 2.4. Consequently $b_{3}$ is an analytic function at $\mu_{0}$.
That being said we claim that the equality
$(b_{1},b_{2},b_{3})=(d_{1},d_{2},d_{3})$ between ideals over the local ring
$\mathbb{R}\\{\mu\\}_{\mu_{0}}$ is true. In order to prove this, for the sake
of shortness in the next computation we follow the convention that $\kappa$
stands for an analytic function at $\mu_{0}$ and $\hat{\kappa}$ stands for an
analytic strictly positive function at $\mu_{0}.$ Some easy computations
following this convention yield

$\displaystyle b_{3}$
$\displaystyle=\Delta_{10}^{-1}({\lambda_{2}}\Delta_{11}\Delta_{30}-\Delta_{31}\Delta_{10})=\Delta_{10}^{-1}({\lambda_{2}}\Delta_{11}\Delta_{30}-\Delta_{31}\Delta_{10})$
$\displaystyle=\textstyle-\Delta_{30}\Big{(}{\lambda_{2}}\lambda_{1}\frac{\hat{M}_{1}(1/\lambda_{1},1)}{L_{11}(1)}-\frac{1}{\lambda_{3}}\frac{\hat{M}_{3}(\lambda_{3},1)}{L_{31}(1)}\Big{)}=\textstyle-\hat{\kappa}\Big{(}{\lambda_{1}}\lambda_{2}\lambda_{3}\hat{M}_{1}\big{(}\frac{1}{\lambda_{1}},1\big{)}L_{31}(1)-\hat{M}_{3}(\lambda_{3},1)L_{11}(1)\Big{)}$
$\displaystyle=\textstyle-\hat{\kappa}\Big{(}(1-d_{1})\hat{M}_{1}\big{(}\frac{1}{\lambda_{1}},1\big{)}L_{31}(1)-\hat{M}_{3}(\lambda_{3},1)L_{11}(1)\Big{)}=\textstyle\hat{\kappa}\Big{(}\hat{M}_{3}(\lambda_{3},1)L_{11}(1)-\hat{M}_{1}\big{(}\frac{1}{\lambda_{1}},1\big{)}L_{31}(1)\Big{)}+\kappa
d_{1}.$

where in the third and fifth equalities we use $(\ref{prop1eq5})$ and
$d_{1}\\!:=1-{\lambda_{1}}\lambda_{2}\lambda_{3}$, respectively. Hence
$b_{3}=\hat{\kappa}d_{3}+\kappa d_{1}$ since
$d_{3}\\!:=\hat{M}_{3}(\lambda_{3},1)L_{11}(1)-\hat{M}_{1}(\frac{1}{\lambda_{1}},1)L_{31}(1).$
On account of $(b_{1})=(d_{1})$ and $(b_{1},b_{2})=(d_{1},d_{2})$, this shows
the validity of the claim and completes the proof of the result.

  * Remark 2.7.

There are two important observations to be made about Theorem 2.6:

    1. $(a)$

The statement claims that the equalities $(b_{1})=(d_{1})$,
$(b_{1},b_{2})=(d_{1},d_{2})$ and $(b_{1},b_{2},b_{3})=(d_{1},d_{2},d_{3})$
between ideals over the local ring $\mathbb{R}\\{\mu\\}_{\mu_{\star}}$ are
satisfied. As a matter of fact, in the proof we show a stronger property,
namely that the following holds:

$\left(\begin{array}[]{c}d_{1}\\\ d_{2}\\\
d_{3}\end{array}\right)=\left(\begin{array}[]{cccc}\kappa_{1}&0&0\\\
&\kappa_{2}&0\\\
&*&\kappa_{3}\end{array}\right)\left(\begin{array}[]{c}b_{1}\\\ b_{2}\\\
b_{3}\end{array}\right),$

where all the entries in the matrix are analytic functions on $\Lambda$ and
each $\kappa_{i}$ is strictly positive.

    2. $(b)$

From the dynamical point of view it is interesting to point out that the
asymptotic expansion of the displacement map $\mathscr{D}(s;\mu_{0})$ given in
Theorem 2.6 provides a method to study the stability of the polycycle
$\Gamma$. Indeed, taking also the previous observation into account, it shows
that

      1. 1.

if $d_{1}(\mu_{0})<0$ (respectively, $>0$) then $\Gamma$ is asymptotically
stable (respectively, unstable),

      2. 2.

if $d_{1}(\mu_{0})=0$ and $d_{2}(\mu_{0})<0$ (respectively, $>0$) then
$\Gamma$ is asymptotically stable (respectively, unstable), and

      3. 3.

if $d_{1}(\mu_{0})=d_{2}(\mu_{0})=0$ and $d_{3}(\mu_{0})<0$ (respectively,
$>0$) then $\Gamma$ is asymptotically stable (respectively, unstable).

Of course this is relevant because we have an explicit expression of these
functions by Theorem A. In this regard let us note that the first assertion is
well known since $d_{1}(\mu_{0})<0$ is equivalent to require that
$\lambda_{1}^{0}\lambda_{2}^{0}\lambda_{3}^{0}>1$, while the second assertion
was already proved by Gasull _et al._ in [8], see Theorem 1. On the contrary
the third assertion constitutes a new result to the best of our knowledge.

$\square$

We give at this point the precise definition of independence of functions that
we use in the present paper.

  * Definition 2.8.

Consider the functions ${g_{i}}\\!:{\Lambda}\longrightarrow{\mathbb{R}}$ for
$i=1,2,\ldots,k$. The _real variety_ $V(g_{1},g_{2},\ldots,g_{k})$ is defined
to be the set of $\mu\in\Lambda$ such that $g_{i}(\mu)=0$ for
$i=1,2,\ldots,k.$ We say that $g_{1},g_{2},\ldots,g_{k}$ are _independent_ at
$\mu_{\star}\in V(g_{1},g_{2},\ldots,g_{k})$ if the following conditions are
satisfied:

    1. $(1)$

Every neighbourhood of $\mu_{\star}$ contains two points $\mu_{1},\mu_{2}\in
V(g_{1},\ldots,g_{k-1})$ such that $g_{k}(\mu_{1})g_{k}(\mu_{2})<0$ (if $k=1$
then we set $V(g_{1},\ldots,g_{k-1})=V(0)=\Lambda$ for this to hold).

    2. $(2)$

The varieties $V(g_{1},\ldots,g_{i})$, $2\leqslant i\leqslant k-1,$ are such
that if $\mu_{0}\in V(g_{1},\ldots,g_{i})$ then every neighbourhood of
$\mu_{0}$ contains two points $\mu_{1},\mu_{2}\in V(g_{1},\ldots,g_{i-1})$
such that $g_{i}(\mu_{1})g_{i}(\mu_{2})<0$.

    3. $(3)$

If $\mu_{0}\in V(g_{1})$ then every neighbourhood of $\mu_{0}$ contains two
points $\mu_{1},\mu_{2}$ such that $g_{1}(\mu_{1})g_{1}(\mu_{2})<0.$

It is clear that if $g_{i}\in\mathscr{C}^{1}(\Lambda)$ for $i=1,2,\ldots,k$
and the gradients $\nabla g_{1}(\mu_{\star}),\nabla
g_{2}(\mu_{\star})\dots,\nabla g_{k}(\mu_{\star})$ are linearly independent
vectors of $\mathbb{R}^{N+1}$ then there exists a neighbourhood $U_{\star}$ of
$\mu_{\star}$ such that the restrictions of $g_{1},g_{2},\ldots,g_{k}$ to
$U_{\star}$ are independent at $\mu_{\star}$. $\square$

######  Lemma 2.9.

Suppose that the equalities $(c_{1},\ldots,c_{k})=(d_{1},\ldots,d_{k})$
between ideals over the local ring $\mathbb{R}\\{\mu\\}_{\mu_{\star}}$ hold
for $k=1,2,\ldots,n$, where $\mu_{\star}\in
V(c_{1},\ldots,c_{n})=V(d_{1},\ldots,d_{n}).$ Then $c_{1},\ldots,c_{n}$ are
independent at $\mu_{\star}$ if, and only if, $d_{1},\ldots,d_{n}$ are
independent at $\mu_{\star}$.

  * Proof.

Let us assume for instance that $c_{1},\ldots,c_{n}$ are independent at
$\mu_{\star}$ and prove that then $d_{1},\ldots,d_{n}$ are also independent.
To this aim we note that the equalities
$(c_{1},\ldots,c_{k})=(d_{1},\ldots,d_{k})$ for $k=1,2,\ldots,n$ imply the
existence of two triangular matrices $A=(a_{ij})$ and $B=(b_{ij})$ with
coefficients in $\mathbb{R}\\{\mu\\}_{\mu_{\star}}$ such that

$\left(\begin{array}[]{c}d_{1}\\\ \vdots\\\
d_{n}\end{array}\right)=\left(\begin{array}[]{cccc}a_{11}&0&\cdots&0\\\
&a_{22}&\cdots&0\\\ \vdots&\vdots&\ddots&\vdots\\\
&\cdots&*&a_{nn}\end{array}\right)\left(\begin{array}[]{c}c_{1}\\\ \vdots\\\
c_{n}\end{array}\right)\text{ and }\left(\begin{array}[]{c}c_{1}\\\ \vdots\\\
c_{n}\end{array}\right)=\left(\begin{array}[]{cccc}b_{11}&0&\cdots&0\\\
&b_{22}&\cdots&0\\\ \vdots&\vdots&\ddots&\vdots\\\
&\cdots&*&b_{nn}\end{array}\right)\left(\begin{array}[]{c}d_{1}\\\ \vdots\\\
d_{n}\end{array}\right).$

Clearly $R=(r_{ij})\\!:=BA$ is also a triangular matrix with coefficients in
the local ring $\mathbb{R}\\{\mu\\}_{\mu_{\star}}$ and

$\left(\begin{array}[]{c}c_{1}\\\ \vdots\\\
c_{n}\end{array}\right)=\left(\begin{array}[]{cccc}r_{11}&0&\cdots&0\\\
&r_{22}&\cdots&0\\\ \vdots&\vdots&\ddots&\vdots\\\
&\cdots&*&r_{nn}\end{array}\right)\left(\begin{array}[]{c}c_{1}\\\ \vdots\\\
c_{n}\end{array}\right).$

We claim that, since $c_{1},\ldots,c_{n}$ are independent at $\mu_{\star}$,
then $r_{kk}(\mu_{\star})=1$ for all $k=1,2,\ldots,n.$ The fact that this is
true for $k=1$ follows easily by continuity. Let us prove by contradiction
that this is also true for $k\geqslant 2.$ So assume that
$r_{kk}(\mu_{\star})\neq 1$ for some $k\in\\{2,\ldots,n\\}.$ Then the equality
$c_{k}=r_{k1}c_{1}+\ldots+r_{kk}c_{k}$ implies that
$c_{k}=\alpha_{1}c_{1}+\ldots+\alpha_{k-1}c_{k-1}$ where each
$\alpha_{i}\\!:=\frac{r_{ki}}{1-r_{kk}}$ is an analytic function at
$\mu_{\star}.$ This clearly contradicts the assumption that
$c_{1},\ldots,c_{n}$ are independent at $\mu_{\star}$ (see Definition 2).
Hence the claim is true and, consequently,
$\text{det}(R)=\text{det}(A)\text{det}(B)=1$ at $\mu=\mu_{\star}.$ This shows,
in particular, that $A$ is an invertible matrix in the local ring
$\mathbb{R}\\{\mu\\}_{\mu_{\star}}$ and so there exists a neighbourhood $U$ of
$\mu_{\star}$ such that $a_{kk}(\mu)\neq 0$ for all $\mu\in U$ and
$k=1,2,\ldots,n$. On account of this, the fact that $d_{1},\ldots,d_{n}$ are
independent at $\mu_{\star}$ follows easily noting that if we take any two
points $\mu_{1},\mu_{2}\in U\cap V(c_{1},\ldots,c_{i-1})=U\cap
V(d_{1},\ldots,d_{i-1})$ verifying $c_{i}(\mu_{1})c_{i}(\mu_{2})<0$ then we
have that
$d_{i}(\mu_{1})d_{i}(\mu_{2})=a_{ii}(\mu_{1})a_{ii}(\mu_{2})c_{i}(\mu_{1})c_{i}(\mu_{2})<0$.
This completes the proof of the result.

  * Proof of Theorem A.

Let us fix any $\mu_{0}\in\Lambda$ and set
$\lambda_{i}^{0}\\!:=\lambda_{i}(\mu_{0})$ for $i=1,2,3.$ Recall that the
limit cycles of $X_{\mu}$ near $\Gamma$ are in one to one correspondence with
the isolated positive zeros of

$\mathscr{D}(s;\mu)=\big{(}D_{2}\circ D_{1}-D_{3}\big{)}(s;\mu)$

near $s=0.$ If
$d_{1}(\mu_{0})=1-\lambda_{1}^{0}\lambda_{2}^{0}\lambda_{3}^{0}$ is not zero
then by applying $(a)$ in Theorem 2.6 we have that, for any
$\ell_{1}\in\big{(}\min(\lambda_{1}^{0}\lambda_{2}^{0},1/\lambda_{3}^{0}),\min(\lambda_{1}^{0}+\lambda_{1}^{0}\lambda_{2}^{0},1+\lambda_{1}^{0}\lambda_{2}^{0},2\lambda_{1}^{0}\lambda_{2}^{0},1+1/\lambda_{3}^{0},2/\lambda_{3}^{0})\big{)}$,

$\mathscr{D}(s;\mu)=a_{1}(\mu)s^{\lambda_{1}\lambda_{2}}+a_{2}(\mu)s^{1/\lambda_{3}}+\mathcal{F}_{\ell_{1}}^{\infty}(\mu_{0}),$

where $a_{1}$ and $a_{2}$ are analytic and strictly positive functions on
$\Lambda.$ Thus

$\displaystyle\lim_{(s,\mu)\to(0,\mu_{0})}s^{-\lambda_{1}\lambda_{2}}\mathscr{D}(s;\mu)=a_{1}(\mu_{0})\text{
in case that $\lambda_{1}^{0}\lambda_{2}^{0}<\lambda_{3}^{0}$}$ and
$\displaystyle\lim_{(s,\mu)\to(0,\mu_{0})}s^{-1/\lambda_{3}}\mathscr{D}(s;\mu)=a_{2}(\mu_{0})\text{
in case that $\lambda_{1}^{0}\lambda_{2}^{0}>\lambda_{3}^{0}$}.$

Since $a_{i}(\mu_{0})\neq 0$ for $i=1,2$, this implies the existence of an
open neighbourhood $U$ of $\mu_{0}$ and $\varepsilon>0$ small enough such that
$\mathscr{D}(s;\mu)\neq 0$ for all $\mu\in U$ and $s\in(0,\varepsilon)$ when
$\lambda_{1}^{0}\lambda_{2}^{0}\lambda_{3}^{0}\neq 1.$ Hence
$\mathrm{Cycl}\big{(}(\Gamma,X_{\mu_{0}}),X_{\mu}\big{)}=0$ and the assertion
in $(a)$ is true.

In order to prove $(b)$ we note that if
$d_{1}(\mu_{0})=1-\lambda_{1}^{0}\lambda_{2}^{0}\lambda_{3}^{0}$ is equal to
zero then, by $(b)$ in Theorem 2.6, we can assert that, for any
$\ell_{2}\in(0,\min(1,\lambda_{1}^{0},\lambda_{1}^{0}\lambda_{2}^{0}))$,

$s^{-1/\lambda_{3}}\mathscr{D}(s;\mu)=b_{1}(\mu)\,\omega\big{(}s;\alpha(\mu)\big{)}+b_{2}(\mu)+\mathcal{F}_{\ell_{2}}^{\infty}(\mu_{0}),$
(16)

where $\alpha=1/\lambda_{3}-\lambda_{1}\lambda_{2}$ and $b_{1}$ and $b_{2}$
are analytic functions at $\mu_{0}$ such that $(b_{1})=(d_{1})$ and
$(b_{1},b_{2})=(d_{1},d_{2})$ over the ring $\mathbb{R}\\{\mu\\}_{\mu_{0}}$.
The assumptions in this case imply that $\mathscr{D}(s;\mu_{0})\not\equiv 0$,
$b_{1}(\mu_{0})=0$ and, thanks to Lemma 2.9, that $b_{1}$ is independent at
$\mu_{0}.$ Thus, given any $\varepsilon>0$, there exists
$s_{1}\in(0,\varepsilon)$ such that $\mathscr{D}(s_{1};\mu_{0})\neq 0.$ Let us
assume for instance that $\mathscr{D}(s_{1};\mu_{0})>0,$ the other case
follows verbatim. Then, thanks to $(1)$ in Definition 2, there exists
$\mu_{1}\approx\mu_{0}$ with $b_{1}(\mu_{1})<0$ and, by continuity, such that
$\mathscr{D}(s_{1};\mu_{1})>0.$ Moreover by applying Lemmas A.3 and A.4 in
[14],

$Z_{1}(s;\mu)\\!:=\frac{s^{-1/\lambda_{3}}\mathscr{D}(s;\mu)}{\omega\big{(}s;\alpha(\mu)\big{)}}=b_{1}(\mu)+\frac{b_{2}(\mu)}{\omega\big{(}s;\alpha(\mu)\big{)}}+\mathcal{F}_{\ell_{2}-\delta}^{\infty}(\mu_{0})\to\kappa(\mu)\text{
as $s\to 0$,}$ (17)

where

$\kappa(\mu)\\!:=b_{1}(\mu)-b_{2}(\mu)\min(\alpha(\mu),0)).$

Here we use that
$1/{\omega(s;\alpha(\mu))}\in\mathcal{F}_{-\delta}^{\infty}(\mu_{0})$ for any
$\delta>0$ and that $\lim_{s\to 0}1/{\omega(s;\alpha)}=\max(-\alpha,0)$ by
assertions $(a)$ and $(b)$ in [14, Lemma A.4], respectively. Note on the other
hand that, by $(b)$ in Theorem 2.6, $b_{1}=\alpha a_{1}$ and
$b_{2}=a_{1}-a_{2}$, where each $a_{i}$ is an analytic strictly positive
function. Thus $\kappa(\mu)=\alpha(\mu)a_{2}(\mu)$ if $\alpha(\mu)<0$ and
$\kappa(\mu)=\alpha(\mu)a_{1}(\mu)$ if $\alpha(\mu)\geqslant 0.$ Therefore,
since $b_{1}=\alpha a_{1}$, we can write

$\kappa=b_{1}\eta\text{ with $\eta>0$.}$ (18)

Hence, on account of $b_{1}(\mu_{1})<0$, we can assert that
$\kappa(\mu_{1})<0,$ which in turn, from $(\ref{thAeq1})$, guarantees the
existence of some $s_{2}\in(0,s_{1})$ such that $Z_{1}(s_{2};\mu_{1})<0.$ Thus
$\mathscr{D}(s_{1};\mu_{1})\mathscr{D}(s_{2};\mu_{1})<0$ and, by continuity,
$\mathscr{D}(\hat{s};\mu_{1})=0$ for some
$\hat{s}\in(s_{2},s_{1})\subset(0,\varepsilon).$ This shows that
$\mathrm{Cycl}\big{(}(\Gamma,X_{\mu_{0}}),X_{\mu}\big{)}\geqslant 1$ as
desired.

Let us prove $(c)$ next, i.e., that if $d_{2}(\mu_{0})\neq 0$ then
$\mathrm{Cycl}\big{(}(\Gamma,X_{\mu_{0}}),X_{\mu}\big{)}\leqslant 1$. Note
first that to this end we can also assume that $d_{1}(\mu_{0})=0$, otherwise
$\mathrm{Cycl}\big{(}(\Gamma,X_{\mu_{0}}),X_{\mu}\big{)}=0.$ Consequently the
expression in $(\ref{thAeq2})$ is valid. That being said, the proof will
follow by applying the well-known derivation-division algorithm. In doing so,
from $(\ref{thAeq1})$ and the fact that
$\partial_{s}\omega(s;\alpha)=-s^{-\alpha-1},$

$\partial_{s}Z_{1}(s;\mu)=\frac{b_{2}(\mu)}{s^{\alpha(\mu)+1}\omega^{2}\big{(}s;\alpha(\mu)\big{)}}+\mathcal{F}_{\ell_{2}-1-\delta}^{\infty}(\mu_{0}),$

where the flatness of the remainder follows from $(f)$ in Lemma A.3 of [14].
Therefore

$s^{\alpha(\mu)+1}\omega^{2}\big{(}s;\alpha(\mu)\big{)}\partial_{s}Z_{1}(s;\mu)=b_{2}(\mu)+\mathcal{F}_{\ell_{2}-4\delta}^{\infty}(\mu_{0})\to
b_{2}(\mu_{0})\text{ as $(s,\mu)\to(0,\mu_{0}).$}$

Recall at this point that $(b_{1})=(d_{1})$ and $(b_{1},b_{2})=(d_{1},d_{2})$
over the local ring $\mathbb{R}\\{\mu\\}_{\mu_{0}}$ by Theorem 2.6. Thus, the
assumptions $d_{1}(\mu_{0})=0$ and $d_{2}(\mu_{0})\neq 0$ imply that
$b_{2}(\mu_{0})\neq 0$. Accordingly, on account of the above limit and by
Bolzano’s Theorem, we obtain $\varepsilon>0$ such that if
$\|\mu-\mu_{0}\|<\varepsilon$ then $Z_{1}(\,\cdot\,;\mu)$, and so
$\mathscr{D}(\,\cdot\,;\mu)$, has at most one zero for $s\in(0,\varepsilon)$,
multiplicities taking into account. Hence
$\mathrm{Cycl}\big{(}(\Gamma,X_{\mu_{0}}),X_{\mu}\big{)}\leqslant 1$ and $(c)$
follows.

Let us turn next to the proof of $(d)$, in which the assumptions are
$\mathscr{R}(\,\cdot\,;\mu_{0})\not\equiv\text{Id}$ and that $d_{1}$ and
$d_{2}$ vanish and are independent at $\mu_{0}$. Then
$\mathscr{D}(\,\cdot\,;\mu_{0})\not\equiv 0$ and, due to $(b_{1})=(d_{1})$ and
$(b_{1},b_{2})=(d_{1},d_{2})$ once again, $b_{1}$ and $b_{2}$ vanish and are
independent at $\mu_{0}$ by Lemma 2.9. Thus, given any $\varepsilon>0$, there
exists $s_{1}\in(0,\varepsilon)$ such that, for instance,
$\mathscr{D}(s_{1};\mu_{0})<0.$ Then, by continuity and condition $(1)$ in
Definition 2, there exists $\mu_{1}\approx\mu_{0}$ such that
$b_{2}(\mu_{1})>0$, $b_{1}(\mu_{1})=0$ and $\mathscr{D}(s_{1};\mu_{1})<0.$
Hence, from $(\ref{thAeq2})$,

$s^{-1/\lambda_{3}}\mathscr{D}(s;\mu_{1})=b_{2}(\mu_{1})+\mathcal{F}_{\ell_{2}}^{\infty}(\mu_{0})\to
b_{2}(\mu_{1})\text{ as $s\to 0,$}$

which shows the existence of $s_{2}\in(0,s_{1})$ such that
$\mathscr{D}(s_{2};\mu_{1})>0.$ For the same reasons we can choose
$\mu_{2}\approx\mu_{1}$ satisfying $\mathscr{D}(s_{1};\mu_{2})<0$ and
$\mathscr{D}(s_{2};\mu_{2})>0$ together with $b_{1}(\mu_{2})<0.$ Then, from
$(\ref{thAeq1})$ and $(\ref{thAeq3})$, $\lim_{s\to
0}Z_{1}(s;\mu_{2})=b_{1}(\mu_{2})\eta(\mu_{2})<0$ and so there exists
$s_{3}\in(0,s_{2})$ verifying that $\mathscr{D}(s_{3};\mu_{2})<0.$ By
continuity there exist $\hat{s}_{1},\hat{s}_{2}\in(0,\varepsilon)$ with
$D(\hat{s}_{1};\mu_{2})=D(\hat{s}_{2};\mu_{2})=0.$ Accordingly
$\mathrm{Cycl}\big{(}(\Gamma,X_{\mu_{0}}),X_{\mu}\big{)}\geqslant 2$.

From now on, in order to prove $(e)$ and $(f)$, we assume $\lambda_{1}^{0}<1$,
$\lambda_{2}^{0}>1$ and $\lambda_{3}^{0}>1$. Then by applying Theorem 2.6, for
any $\ell_{3}\in\big{(}1,\min(2,\lambda_{1}^{0},1/\lambda_{3}^{0})\big{)}$,

$\mathscr{D}(s;\mu)=\big{(}b_{1}(\mu)\,\omega\big{(}s;\alpha(\mu)\big{)}+b_{2}(\mu)+b_{3}(\mu)s+\mathcal{F}_{\ell_{3}}^{\infty}(\mu_{0})\big{)}s^{1/\lambda_{3}}U(s;\mu),$
(19)

where $b_{3}$ is an analytic function at $\mu_{0}$ verifying that
$(b_{1},b_{2},b_{3})=(d_{1},d_{2},d_{3})$ over the ring
$\mathbb{R}\\{\mu\\}_{\mu_{0}}$ and $U$ is an analytic function such that
$U(0;\mu_{0})=1$. Hence

$Z_{2}(s;\mu)\\!:=\frac{\mathscr{D}(s;\mu)}{s^{1/\lambda_{3}}\omega(s;\alpha(\mu))U(s)}=b_{1}(\mu)+b_{2}(\mu)\frac{1}{\omega(s;\alpha(\mu))}+b_{3}(\mu)\frac{s}{\omega(s;\alpha(\mu))}+\mathcal{F}_{\ell_{3}-\delta}^{\infty}(\mu_{0}),$
(20)

where we use once again that
$1/{\omega(s;\alpha(\mu))}\in\mathcal{F}_{-\delta}^{\infty}(\mu_{0})$ for any
$\delta>0$. Note furthermore that, for $\mu\approx\mu_{0}$, the positive zeros
of $\mathscr{D}(\,\cdot\,;\mu)$ and $Z_{2}(\,\cdot\,;\mu)$ near $s=0$ are in
one to one correspondence because
$\frac{1}{s^{1/\lambda_{3}}\omega(s;\alpha(\mu))}$ tends to $+\infty$ as
$(s,\mu)\to(0,\mu_{0}).$ That being stablished we begin first with the proof
of assertion $(e)$ and to this aim, besides $d_{3}(\mu_{0})\neq 0,$ we can
also suppose $d_{1}(\mu_{0})=d_{2}(\mu_{0})=0$, otherwise
$\mathrm{Cycl}\big{(}(\Gamma,X_{\mu_{0}}),X_{\mu}\big{)}\leqslant 1$ by $(a)$
or $(b)$, which already have been proved. In this case, since
$(b_{1})=(d_{1})$, $(b_{1},b_{2})=(d_{1},d_{2})$ and
$(b_{1},b_{2},b_{3})=(d_{1},d_{2},d_{3})$, it turns out that
$b_{3}(\mu_{0})\neq 0.$ As in the proof of $(c)$, we will apply to steps of
the derivation-division algorithm in $(\ref{thAeq5})$. In doing so we obtain
that

$Z_{3}(s;\mu)\\!:=s^{\alpha(\mu)+1}\omega^{2}(s;\alpha(\mu))\partial_{s}Z_{2}(s;\mu)=b_{2}(\mu)+b_{3}(\mu)\big{(}s+s^{\alpha(\mu)+1}\omega(s;\alpha(\mu))\big{)}+\mathcal{F}_{\ell_{3}-4\delta}^{\infty}(\mu_{0}),$

where the flatness of the remainder follows by applying Lemmas A.3 and A.4 in
[14] as before and we use that $\partial_{s}\omega(s;\alpha)=-s^{-\alpha-1}.$
Note also that the positive zeros of $Z_{3}(\,\cdot\,;\mu)$ and
$\partial_{s}Z_{2}(\,\cdot\,;\mu)$ near $s=0$ are in one to one correspondence
for $\mu\approx\mu_{0}$ because $\omega(s;\alpha(\mu))$ tends to $+\infty$ as
$(s,\mu)\to(0,\mu_{0}).$ Finally

$\frac{\partial_{s}Z_{3}(s;\mu)}{s^{\alpha(\mu)}\omega(s;\alpha(\mu))}=(\alpha(\mu)+1)b_{3}(\mu)+\mathcal{F}_{\ell_{3}-1-6\delta}^{\infty}(\mu_{0})\to
b_{3}(\mu_{0})\neq 0\text{ as $(s,\mu)\to(0,\mu_{0})$.}$

By applying twice Bolzano’s Theorem, we can assert the existence of some
$\varepsilon>0$ such that if $\|\mu-\mu_{0}\|<\varepsilon$ then
$Z_{2}(\,\cdot\,;\mu)$, and so $\mathscr{D}(\,\cdot\,;\mu)$, has at most two
zeros for $s\in(0,\varepsilon)$, multiplicities taking into account. Hence
$\mathrm{Cycl}\big{(}(\Gamma,X_{\mu_{0}}),X_{\mu}\big{)}\leqslant 2$, which
proves $(e)$.

Finally, in order to prove $(f)$ we suppose that
$\mathscr{R}(\,\cdot\,;\mu_{0})\not\equiv\text{Id}$ and that $d_{1}$, $d_{2}$
and $d_{3}$ vanish and are independent at $\mu_{0}$. Consequently
$\mathscr{D}(\,\cdot\,;\mu_{0})\not\equiv 0$ and, due to the equality between
the corresponding ideals over the local ring, $b_{1}$, $b_{2}$ and $b_{3}$
vanish and are independent at $\mu_{0}$ by Lemma 2.9. Thus, given any
$\varepsilon>0$, there exists $s_{1}\in(0,\varepsilon)$ such that, for
instance, $\mathscr{D}(s_{1};\mu_{0})<0.$ Then, by continuity and condition
$(1)$ in Definition 2, there exists $\mu_{1}\approx\mu_{0}$ such that
$b_{3}(\mu_{1})>0$, $b_{1}(\mu_{1})=b_{2}(\mu_{1})=0$ and
$\mathscr{D}(s_{1};\mu_{1})<0.$ Hence, from $(\ref{thAeq4})$,

$\frac{\mathscr{D}(s;\mu_{1})}{s^{1+1/\lambda_{3}}U(s;\mu)}=b_{3}(\mu_{1})+\mathcal{F}_{\ell_{3}-1}^{\infty}(\mu_{0})\to
b_{3}(\mu_{1})>0\text{ as $s\to 0,$}$

which shows the existence of $s_{2}\in(0,s_{1})$ such that
$\mathscr{D}(s_{2};\mu_{1})>0.$ For the same reasons we can choose
$\mu_{2}\approx\mu_{1}$ satisfying $\mathscr{D}(s_{1};\mu_{2})<0$ and
$\mathscr{D}(s_{2};\mu_{2})>0$ together with $b_{1}(\mu_{2})=0$ and
$b_{2}(\mu_{2})<0.$ Accordingly, from $(\ref{thAeq4})$ again,

$\frac{\mathscr{D}(s;\mu_{2})}{s^{1/\lambda_{3}}U(s;\mu)}=b_{2}(\mu_{2})+b_{3}(\mu_{2})s+\mathcal{F}_{\ell_{3}-1}^{\infty}(\mu_{0})\to
b_{2}(\mu_{2})<0\text{ as $s\to 0,$}$

which shows the existence of $s_{3}\in(0,s_{2})$ such that
$\mathscr{D}(s_{3};\mu_{2})<0.$ In the final step we take
$\mu_{3}\approx\mu_{2}$ satisfying $\mathscr{D}(s_{1};\mu_{3})<0$ and
$\mathscr{D}(s_{2};\mu_{3})>0$ and $\mathscr{D}(s_{3};\mu_{3})<0$ together
with $b_{1}(\mu_{3})>0.$ Then, from $(\ref{thAeq1})$ and $(\ref{thAeq3})$,
$\lim_{s\to
0}s^{-1/\lambda_{3}}Z_{1}(s;\mu_{3})=b_{1}(\mu_{3})\eta(\mu_{3})>0$ and so
there exists $s_{4}\in(0,s_{3})$ such that $\mathscr{D}(s_{3};\mu_{3})>0.$ By
continuity there exist $\hat{s}_{1},\hat{s}_{2},\hat{s}_{3}\in(0,\varepsilon)$
with $D(\hat{s}_{i};\mu_{3})=0$ for $i=1,2,3.$ Hence
$\mathrm{Cycl}\big{(}(\Gamma,X_{\mu_{0}}),X_{\mu}\big{)}\geqslant 3$ and this
completes the proof of the result.

##  3 Applications

We begin this section by revisiting in Theorem 3.1 a family of Kolmogorov
differential systems that was first studied in [8], where the authors
(following the notation in our statement) prove that if
$\mu_{0}=(a_{0},p_{0},q_{0})$ verifies $p_{0}+q_{0}=0$ and $a_{0}\neq 0$ then
$\mathrm{Cycl}\big{(}(\Gamma,X_{\mu_{0}}),X_{\mu}\big{)}\geqslant 1$, cf.
assertion $(b)$.

######  Theorem 3.1.

Consider the family of Kolmogorov differential systems

$X_{\mu}\quad\left\\{\\!\begin{array}[]{l}\dot{x}=x(1+x+x^{2}+axy+py^{2}),\\\\[2.0pt]
\dot{y}=y(-1-y+qx^{2}+axy-y^{2}),\end{array}\right.$

where $\mu=(a,p,q)\in\mathbb{R}^{3}$ with $p<-1$ and $q>1$ and let us fix any
$\mu_{0}=(a_{0},p_{0},q_{0}).$ Then, compactifying $X_{\mu}$ to the Poincaré
disc, the boundary of the first quadrant is a polycycle $\Gamma$ such that:

  1. $(a)$

$\mathrm{Cycl}\big{(}(\Gamma,X_{\mu_{0}}),X_{\mu}\big{)}=0$ if
$p_{0}+q_{0}\neq 0.$

  2. $(b)$

$\mathrm{Cycl}\big{(}(\Gamma,X_{\mu_{0}}),X_{\mu}\big{)}=1$ if $p_{0}+q_{0}=0$
and $a_{0}\neq 0$.

  3. $(c)$

The return map of $X_{\mu_{0}}$ along $\Gamma$ is the identity if, and only
if, $a_{0}=p_{0}+q_{0}=0$. In this case $\Gamma$ is the outer boundary of the
period annulus of a center at $(x_{0},y_{0})$ with
$x_{0}=y_{0}=-\frac{1+\sqrt{-3-4p_{0}}}{2(1+p_{0})}$ that foliates the first
quadrant and, moreover,
$\mathrm{Cycl}\big{(}(\Gamma,X_{\mu_{0}}),X_{\mu}\big{)}=1$.

On the other hand the vector field $X_{\mu}$ has a unique singularity
$Q_{\mu}=(\upsilon_{1},\upsilon_{2})$ in the first quadrant, which is either a
focus or a center, and has trace equal to
$\tau(\mu)=\upsilon_{1}+2\upsilon_{1}^{2}+2a\upsilon_{1}\upsilon_{2}-\upsilon_{2}-2\upsilon_{2}^{2}$.
Furthermore the following holds:

  1. $(d)$

If $\tau(\mu_{0})\neq 0$ then
$\mathrm{Cycl}\big{(}(Q_{\mu_{0}},X_{\mu_{0}}),X_{\mu}\big{)}=0$ and a
sufficient condition for $\tau(\mu_{0})\neq 0$ to hold is that $p_{0}+q_{0}=0$
and $a_{0}\neq 0.$

  2. $(e)$

If $\tau(\mu_{0})=0$ and $p_{0}+q_{0}\neq 0$ then $Q_{\mu_{0}}$ is a weak
focus of order 1 and
$\mathrm{Cycl}\big{(}(Q_{\mu_{0}},X_{\mu_{0}}),X_{\mu}\big{)}=1.$

  3. $(f)$

If $\tau(\mu_{0})=0$ and $p_{0}+q_{0}=0$ then $a_{0}=0$. In addition
$Q_{\mu_{0}}$ is a center if, and only if, $p_{0}+q_{0}=a_{0}=0$ and, in this
case, $\mathrm{Cycl}\big{(}(Q_{\mu_{0}},X_{\mu_{0}}),X_{\mu}\big{)}=1.$

Finally it is not possible a simultaneous bifurcation of limit cycles from
$\Gamma$ and $Q_{\mu}$.

  * Proof.

The assertions in $(a)$ and $(b)$ follow directly by applying Theorem A.
Indeed, in this case, following the notation in $(\ref{kol})$,
$f(x,y)=1+x+x^{2}+axy+py^{2}$ and $g(x,y)=-1-y+qx^{2}+axy-y^{2}$, so that

$f_{2}(x,y)=x^{2}+axy+py^{2}\text{ and }g_{2}(x,y)=qx^{2}+axy-y^{2}.$

Taking this into account, together with $p<-1$ and $q>1,$ one can easily check
that the assumptions H1 and H2 are verified. As a matter of fact the first
assumption holds not only for $z>0$ but for all $z\in\mathbb{R}$, and this
implies that the boundary of each quadrant is a monodromic polycycle for the
compactified vector field. Hence, by the Poincaré-Bendixson theorem (see [22]
for instance), there exists at least one singularity of $X_{\mu}$ inside each
one of the four quadrants. Due to $\deg(f)=\deg(g)=2$, by Bézout’s theorem
there exists exactly one in each quadrant. From now on we denote the
singularity of $X_{\mu}$ in the first quadrant by $Q_{\mu}$. That being said,
the hyperbolicity ratios of the saddles at $\Gamma$ are
$\lambda_{1}=\frac{1}{q-1}$, $\lambda_{2}=-(p+1)$ and $\lambda_{3}=1.$
Consequently the first assertion follows from $(a)$ in Theorem A because

$d_{1}(\mu)=1-\lambda_{1}\lambda_{2}\lambda_{3}=\frac{p+q}{q-1}.$ (21)

The second assertion will follow by applying $(b)$ and $(c)$ in Theorem A. To
show this we first recall that

$d_{2}(\mu)=\lambda_{2}\log\left(\frac{L_{12}}{L_{21}}\right)\\!(1)+\log\left(\frac{L_{22}}{L_{32}}\right)\\!(1)+\lambda_{1}\lambda_{2}\log\left(\frac{L_{31}}{L_{11}}\right)\\!(1)$

and this leads us to the computation of the following improper integrals:

$\displaystyle\Delta_{1}(\mu)\\!:=\log\left(\frac{L_{12}}{L_{21}}\right)\\!(1)$
$\displaystyle=\int_{0}^{1}\left(\left(\frac{f_{2}}{f_{2}-g_{2}}\right)(1,z)+\lambda_{1}-\left(\frac{g_{2}}{g_{2}-f_{2}}\right)(z,1)-\frac{1}{\lambda_{2}}\right)\frac{dz}{z}$
$\displaystyle\Delta_{2}(\mu)\\!:=\log\left(\frac{L_{22}}{L_{32}}\right)\\!(1)$
$\displaystyle=\int_{0}^{1}\left(\left(\frac{g-f}{g}\right)(0,1/z)+\lambda_{2}-\left(\frac{f}{g}\right)(0,z)-\frac{1}{\lambda_{3}}\right)\frac{dz}{z}$
$\displaystyle\Delta_{3}(\mu)\\!:=\log\left(\frac{L_{31}}{L_{11}}\right)\\!(1)$
$\displaystyle=\int_{0}^{1}\left(\left(\frac{g}{f}\right)(z,0)+\lambda_{3}-\left(\frac{f-g}{f}\right)(1/z,0)-\frac{1}{\lambda_{1}}\right)\frac{dz}{z}$

These expressions have to be computed assuming that $p+q=0,$ i.e.,
$\lambda_{1}\lambda_{2}=1.$ In doing so we obtain that

$\displaystyle\Delta_{1}(a,p,-p)=\frac{2a}{1-q}\int_{0}^{1}\frac{zdz}{z^{2}+1}=\frac{a\pi}{2(p+1)}$
and
$\displaystyle\Delta_{2}(a,p,-p)=-\Delta_{3}(a,p,-p)=-(p+1)\int_{0}^{1}\frac{zdz}{z^{2}+z+1}=-\frac{(p+1)\pi}{3\sqrt{3}}.$

Therefore $d_{2}(a,p,-p)=-\frac{a\pi}{2}$ is zero if, and only if, $a=0.$
Taking this into account, the combination of $(b)$ and $(c)$ in Theorem A
shows that $\mathrm{Cycl}\big{(}(\Gamma,X_{\mu_{0}}),X_{\mu}\big{)}=1$ for any
$\mu_{0}=(a_{0},p_{0},-p_{0})$ with $a_{0}\neq 0$, as desired. It is important
to remark for the forthcoming analysis that by applying the Weierstrass
Division Theorem (see for instance [9, 12]) we can assert that

$d_{2}(\mu)=-\frac{a\pi}{2}+(p+q)h(\mu)$ (22)

for some analytic function $h.$

Next we proceed with the proof of $(c)$.To this aim we fix any
$\mu_{0}=(a_{0},p_{0},q_{0})$ and apply Theorem 2.6, which gives the
asymptotic expansion of $\mathscr{D}(s;\mu)$ at $s=0$ for $\mu\approx\mu_{0}$.
This result, taking $(\ref{ex1eq1})$ and $(\ref{ex1eq2})$ into account, shows
that if $\mathscr{D}(s;\mu_{0})\equiv 0$ then $a_{0}=p_{0}+q_{0}=0$. In order
to prove the converse observe that if $\mu_{0}=(0,p_{0},-p_{0})$ then the
vector field $X_{\mu_{0}}$ writes as

$\left\\{\\!\begin{array}[]{l}\dot{x}=x(1+x+x^{2}+p_{0}y^{2}),\\\\[2.0pt]
\dot{y}=-y(1+y+p_{0}x^{2}+y^{2}).\end{array}\right.$

One can easily check that $Q_{\mu_{0}}$, the only singularity of $X_{\mu_{0}}$
in the first quadrant, is a weak focus at the point $(x_{0},y_{0})$ with
$x_{0}=y_{0}=-\frac{1+\sqrt{-3-4p_{0}}}{2(1+p_{0})}$. Furthermore, setting
$\sigma(x,y)=(y,x)$, it turns out that
$\sigma^{\star}X_{\mu_{0}}=-X_{\mu_{0}}$ and so the vector field is reversible
with respect to the straight line $y=x.$ Hence $Q_{\mu_{0}}$ is a center and a
straightforward application of the Poincaré-Bendixson theorem shows that its
period annulus fills the first quadrant, which in particular implies that
$\mathscr{D}(s;\mu_{0})\equiv 0$.

So far we have proved that the return map of $X_{\mu_{0}}$ along $\Gamma$ is
the identity if, and only if, $\mu_{0}=(0,p_{0},-p_{0}).$ Our next task is to
show that, in this case,
$\mathrm{Cycl}\big{(}(\Gamma,X_{\mu_{0}}),X_{\mu}\big{)}=1$. With this aim in
view we apply $(b)$ in Theorem 2.6, which shows that if $\mu\approx\mu_{0}$
then

$\mathscr{D}(s;\mu)=\big{(}b_{1}(\mu)\,\omega\big{(}s;\alpha(\mu)\big{)}+b_{2}(\mu)+r(s;\mu)\big{)}s^{1/\lambda_{3}},$
(23)

where $\alpha=1/\lambda_{3}-\lambda_{1}\lambda_{2}$,
$r\in\mathcal{F}_{\ell}^{\infty}(\mu_{0})$ with
$\ell\in\big{(}0,\min(1,\frac{-1}{p+1})\big{)}$ and, in addition,

$(b_{1})=(d_{1})\text{ and }(b_{1},b_{2})=(d_{1},d_{2})$

over the local ring $\mathbb{R}\\{\mu\\}_{\mu_{0}}$. Consequently if
$\mu=(a,p,q)$ satisfies $a=p+q=0$ then $b_{1}(\mu)=b_{2}(\mu)=0$ and
$r(s;\mu)\equiv 0.$ Furthermore, since the vectors $\nabla d_{1}(\mu_{0})$ and
$\nabla d_{2}(\mu_{0})$ are linearly independent, see $(\ref{ex1eq1})$ and
$(\ref{ex1eq2})$, the above equalities between ideals show that this is also
the case of $\nabla b_{1}(\mu_{0})$ and $\nabla b_{2}(\mu_{0}).$ We can thus
take $(\eta_{1},\eta_{2},\eta_{3})\notin\langle\nabla b_{1}(\mu_{0}),\nabla
b_{2}(\mu_{0})\rangle$ and define
$b_{3}(\mu)\\!:=\eta_{1}a+\eta_{2}(p-p_{0})+\eta_{3}(q+p_{0})$ so that
$\nu=\Psi(\mu)\\!:=\big{(}b_{1}(\mu),b_{2}(\mu),b_{3}(\mu)\big{)}$ is a local
analytical change of coordinates in a neighbourhood of $\mu=\mu_{0}$. Note
that $\Psi$ maps $\mu_{0}$ to $0_{3}\\!:=(0,0,0)$ and $\\{a=p+q=0\\}$ to
$\\{\nu_{1}=\nu_{2}=0\\}$ and in addition

$\mathscr{R}_{1}(s;\nu)\\!:=\left.s^{-1/\lambda_{3}}\mathscr{D}(s;\mu)\right|_{\mu=\Psi^{-1}(\nu)}=\nu_{1}\,\omega(s;\hat{\alpha})+\nu_{2}+\hat{r}(s;\nu),$

where $\hat{\alpha}=\hat{\alpha}(\nu)\\!:=\alpha\big{(}\Psi^{-1}(\nu)\big{)}$
and
$\hat{r}(s;\nu)\\!:=r\big{(}s;\Psi^{-1}(\nu)\big{)}\in\mathcal{F}_{\ell}^{\infty}(0_{3}).$
The key point is that $\hat{r}(s;0,0,\nu_{3})\equiv 0$ implies, thanks to [17,
Lemma 4.1], that $\hat{r}(s;\nu)=\nu_{1}h_{1}(s;\nu)+\nu_{2}h_{2}(s;\nu)$ with
$h_{i}\in\mathcal{F}_{\ell}^{\infty}(0_{3}).$ Accordingly

$\mathscr{R}_{1}(s;\nu)=\nu_{1}\big{(}\omega(s;\hat{\alpha})+h_{1}(s;\nu)\big{)}+\nu_{2}\big{(}1+h_{2}(s;\nu)\big{)}.$
(24)

Observe that if $s\to 0^{+}$ and $\nu\to(0,0,0)$ then the factor multiplying
$\nu_{1}$ tends to $+\infty$, whereas the factor multiplying $\nu_{2}$ tends
to $1.$ Here we use Definition 2 and that
$\lim_{(s,\alpha)\to(0,0)}\omega(s;\alpha)=+\infty.$ We claim that there
exists $s_{0}>0$ and an open neighbourhood $U$ of $\nu=(0,0,0)$ such that

$\mathscr{R}_{2}(s;\nu)\\!:=\frac{\mathscr{R}_{1}(s;\nu)}{\omega(s;\hat{\alpha})+h_{1}(s;\nu)}=\nu_{1}+\nu_{2}\frac{1+h_{2}(s;\nu)}{\omega(s;\hat{\alpha})+h_{1}(s;\nu)}$

has at most one zero on $(0,s_{0})$, counted with multiplicities, for all
$\nu=(\nu_{1},\nu_{2},\nu_{3})\in U$ with $\nu_{1}^{2}+\nu_{2}^{2}\neq 0.$
This will imply that
$\mathrm{Cycl}\big{(}(\Gamma,X_{\mu_{0}}),X_{\mu}\big{)}\leqslant 1$ because
$\mathscr{R}_{2}(s;0,0,\nu_{3})\equiv 0$, so that it has not any isolated
zero. The claim is clear in case that $\nu_{2}=0.$ To tackle the case
$\nu_{2}\neq 0$ we compute the derivative with respect to $s$ to obtain that

$\displaystyle\mathscr{R}_{2}^{\prime}(s;\nu)$
$\displaystyle=\nu_{2}\partial_{s}\left(\frac{1+\mathcal{F}_{\ell}^{\infty}}{\omega(s;\hat{\alpha})+\mathcal{F}_{\ell}^{\infty}}\right)=\nu_{2}\partial_{s}\left(\frac{1+\mathcal{F}_{\ell}^{\infty}}{\omega(s;\hat{\alpha})(1+\mathcal{F}_{\ell-\varepsilon}^{\infty})}\right)$
$\displaystyle=\nu_{2}\partial_{s}\left(\frac{1+\mathcal{F}_{\ell-\varepsilon}^{\infty}}{\omega(s;\hat{\alpha})}\right)=\frac{\nu_{2}}{s^{\hat{\alpha}+1}\omega^{2}(s;\hat{\alpha})}(1+\mathcal{F}_{\ell-\varepsilon}^{\infty})+\frac{\nu_{2}}{\omega(s;\hat{\alpha})}\mathcal{F}_{\ell-\varepsilon-1}^{\infty}$
$\displaystyle=\frac{\nu_{2}}{s^{\hat{\alpha}+1}\omega^{2}(s;\hat{\alpha})}\big{(}1+\mathcal{F}_{\ell-\varepsilon}^{\infty}+s^{\hat{\alpha}+1}\omega(s;\hat{\alpha})\mathcal{F}_{\ell-\varepsilon-1}^{\infty}\big{)}=\frac{\nu_{2}}{s^{\hat{\alpha}+1}\omega^{2}(s;\hat{\alpha})}\big{(}1+\mathcal{F}_{\ell-3\varepsilon}^{\infty}\big{)}.$

Here, in the second equality we apply first assertion $(c)$ of Lemma A.4 in
[14] to get that
$1/\omega(s;\hat{\alpha})\in\mathcal{F}^{\infty}_{-\varepsilon}(0_{3})$ for
all $\varepsilon>0$ small enough, due to $\hat{\alpha}(0_{3})=0$, and use next
that
$\mathcal{F}_{-\varepsilon}^{\infty}\mathcal{F}_{\ell}^{\infty}\subset\mathcal{F}_{\ell-\varepsilon}^{\infty}$
from $(g)$ of Lemma A.3 in [14] . In the third equality, on account of
$\frac{1}{1+s}-1\in\mathcal{F}_{1}^{\infty}$ and by $(h)$ of Lemma A.3 in [14]
, we use first the inclusion
$\frac{1}{1+\mathcal{F}^{\infty}_{\ell-\varepsilon}}\subset
1+\mathcal{F}^{\infty}_{\ell-\varepsilon}$. Then, by using $(d)$ and $(g)$ of
Lemma A.3 in [14] , we expand the numerator to get that
$(1+\mathcal{F}_{\ell}^{\infty})(1+\mathcal{F}_{\ell-\varepsilon}^{\infty})\subset
1+\mathcal{F}_{\ell-\varepsilon}^{\infty}$. Next, in the fourth equality we
use that $\partial_{s}\omega(s;\alpha)=s^{-\alpha-1}$ and assertion $(f)$ of
Lemma A.3 in [14] to deduce that
$\partial_{s}\mathcal{F}_{\ell-\varepsilon}^{\infty}\subset\mathcal{F}_{\ell-\varepsilon-1}^{\infty}$.
Finally in the last equality we apply $(c)$ of Lemma A.4 in [14] to get that
$s^{\hat{\alpha}+1}\omega(s;\hat{\alpha})\in\mathcal{F}^{\infty}_{1-2\varepsilon}$
and we use again that
$\mathcal{F}_{1-2\varepsilon}^{\infty}\mathcal{F}_{\ell-\varepsilon-1}^{\infty}\subset\mathcal{F}_{\ell-3\varepsilon}^{\infty}$.
On account of Definition 2 we can assert the existence of some $s_{0}\in(0,1)$
and a neighbourhood $U$ of $\nu=(0,0,0)$ such that
$\mathscr{R}_{2}^{\prime}(s;\nu)\neq 0$ for all $s\in(0,s_{0})$ and $\nu\in U$
with $\nu_{2}\neq 0.$ Hence the application of Rolle’s theorem shows that the
claim is true for $\nu_{2}\neq 0$ as well. So far we have proved that
$\mathrm{Cycl}\big{(}(\Gamma,X_{\mu_{0}}),X_{\mu}\big{)}\leqslant 1$. The fact
that this upper bound is attained follows by applying the assertion in $(b)$
taking $\mu_{0}=(a_{0},p_{0},-p_{0})$ with $a_{0}\approx 0$ but different from
zero. This completes the proof of $(c)$.

Let us turn now to the proof of the assertions regarding the singularity of
$X_{\mu}$ at $Q_{\mu}.$ The approach here is rather standard and the technical
difficulty is that we do not dispose of a feasible expression of the
coordinates of $Q_{\mu}$. To overcome this problem we shall parametrise the
family of vector fields more conveniently. For reader’s convenience we
summarise the chain of reparametrisations that we shall perform:

$\mu=(a,p,q)\to(a,p,\varepsilon)\to(\upsilon_{1},\upsilon_{2},\varepsilon)\to(\upsilon_{1},\upsilon_{2},\tau)\to\hat{\mu}=(\rho,\sigma,\tau).$

For the first one we simply introduce $\varepsilon=p+q.$ In the second one we
take the coordinates of the singular point
$Q_{\mu}=(\upsilon_{1},\upsilon_{2})$ as new parameters, i.e., we isolate $a$
and $p$ from

$\left\\{\\!\begin{array}[]{l}1+\upsilon_{1}+\upsilon_{1}^{2}+a\upsilon_{1}\upsilon_{2}+p\upsilon_{2}^{2}=0,\\\\[2.0pt]
1+\upsilon_{2}+(p-\varepsilon)\upsilon_{1}^{2}-a\upsilon_{1}\upsilon_{2}+\upsilon_{2}^{2}=0,\end{array}\right.$
(25)

to obtain

$a=-\frac{\upsilon_{1}^{2}\upsilon_{2}^{2}\varepsilon+\upsilon_{1}^{4}-\upsilon_{2}^{4}+\upsilon_{1}^{3}-\upsilon_{2}^{3}+\upsilon_{1}^{2}-\upsilon_{2}^{2}}{\upsilon_{1}\upsilon_{2}\left(\upsilon_{1}^{2}+\upsilon_{2}^{2}\right)}\text{
and
}p=\frac{\upsilon_{1}^{2}\varepsilon-\upsilon_{1}^{2}-\upsilon_{2}^{2}-\upsilon_{1}-\upsilon_{2}-2}{\upsilon_{1}^{2}+\upsilon_{2}^{2}}.$

In this respect we point out that $\upsilon_{1}$ and $\upsilon_{2}$ are
strictly positive because $Q_{\mu}$ is inside the first quadrant for all
admissible $\mu$. More important, the map
$\varphi:(a,p,\varepsilon)\mapsto(\upsilon_{1},\upsilon_{2},\varepsilon)$ is
smooth and, taking $(\ref{ex1eq4})$ into account, injective. The smoothness
follows by the Inverse Function Theorem since one can check that the
determinant of the Jacobian of
$(\upsilon_{1},\upsilon_{2},\varepsilon)\mapsto(a,p,\varepsilon)$ is non-zero
at the image by $\varphi$ of any admissible parameter. Then one can check that
the trace of the Jacobian of the vector field at
$(x,y)=(\upsilon_{1},\upsilon_{2})$ is

$\displaystyle
xf_{x}(x,y)+yg_{y}(x,y)\big{|}_{(x,y)=(\upsilon_{1},\upsilon_{2})}$
$\displaystyle=\upsilon_{1}+2\upsilon_{1}^{2}+2a\upsilon_{1}\upsilon_{2}-\upsilon_{2}-2\upsilon_{2}^{2}\Big{|}_{a=a(\upsilon_{1},\upsilon_{2},\varepsilon)}$
$\displaystyle=-{\frac{2\varepsilon\upsilon_{1}^{2}\upsilon_{2}^{2}+(\upsilon_{1}-\upsilon_{2})(\upsilon_{1}+2+\upsilon_{{2}})(\upsilon_{1}+\upsilon_{2})}{\upsilon_{2}^{2}+\upsilon_{1}^{2}}},$

and we introduce $\tau=\tau(\upsilon_{1},\upsilon_{2},\varepsilon)$ isolating
$\varepsilon$ from

$2\varepsilon\upsilon_{1}^{2}\upsilon_{2}^{2}+(\upsilon_{1}-\upsilon_{2})(\upsilon_{1}+2+\upsilon_{{2}})(\upsilon_{1}+\upsilon_{2})=\tau.$
(26)

In other words, $\tau$ is (up to a non-vanishing factor) the trace of the
vector field. Finally, for convenience, we define
$\rho=\frac{\upsilon_{1}-\upsilon_{2}}{2}$ and
$\sigma=\frac{\upsilon_{1}+\upsilon_{2}}{2}.$ Observe then that
$\\{p+q=a=0\\}$ becomes $\\{\rho=\tau=0\\}.$ In what follows, setting
$\hat{\mu}=(\rho,\sigma,\tau)$ for shortness, we denote the vector field by
$X_{\hat{\mu}}$. Let us also remark that the map $\mu\mapsto\hat{\mu}$ is
smooth and injective as a consequence of the previous discussion.

At this point we claim that $Q_{\mu}$ is either a focus or a center. To show
this we will check that the discriminant $D_{\mu}$ of the characteristic
polynomial of the Jacobian matrix of $X_{\mu}$ at $Q_{\mu}$ is strictly
negative for all admissible parameter. Indeed, one can verify that $D_{\mu}$
expressed in terms of $(\upsilon_{1},\upsilon_{2},\varepsilon)$ can be written
as

$D_{\mu}=\frac{4(\upsilon_{1}\upsilon_{2})^{4}\varepsilon^{2}+A_{1}(\upsilon_{1},\upsilon_{2})\varepsilon+A_{0}(\upsilon_{1},\upsilon_{2})}{(\upsilon_{1}^{2}+\upsilon_{2}^{2})^{2}},$

where $A_{i}$ are polynomials of degree 7. Thus $D_{\mu}=0$ gives two roots
$\varepsilon=\hat{\varepsilon}_{i}(\upsilon_{1},\upsilon_{2})$, for $i=1,2,$
that one can check to be well-defined continuous functions on
$V\\!:=\\{(\upsilon_{1},\upsilon_{2})\in\mathbb{R}^{2}:\upsilon_{1}>0,\upsilon_{2}>0\\}.$
To see the claim we first prove that, for $i=1,2,$

$(p+1)(q-1)\big{|}_{\varepsilon=\hat{\varepsilon}_{i}}>0\text{ for all
$(\upsilon_{1},\upsilon_{2})\in V.$}$ (27)

This implies that $D_{\mu}$ can not vanish at an admissible parameter due to
the assumptions $p<-1$ and $q>1.$ In this regard we note that the product
$(p+1)(q-1)$ expressed in terms of $(\upsilon_{1},\upsilon_{2},\varepsilon)$
is given by

$(p+1)(q-1)=\frac{(\upsilon_{1}\upsilon_{2})^{2}\varepsilon^{2}+B_{1}(\upsilon_{1},\upsilon_{2})\varepsilon+B_{0}(\upsilon_{1},\upsilon_{2})}{(\upsilon_{1}^{2}+\upsilon_{2}^{2})^{2}},$

where $\deg(B_{1})=3$ and $\deg(B_{0})=2.$ A computation shows that the
_resultant_ (see [4, Chapter 3] for instance) between the numerators of
$D_{\mu}$ and $(p+1)(q-1)$ with respect to $\varepsilon$ is a polynomial in
$\upsilon_{1}$ and $\upsilon_{2}$ with all the coefficients being natural
numbers. Consequently the resultant does not vanish on $V$ and, accordingly,
$(p+1)(q-1)|_{\varepsilon=\hat{\varepsilon}_{i}}\neq 0$ for all
$(\upsilon_{1},\upsilon_{2})\in V.$ Thus, since $V$ is arc-connected and the
function
$(\upsilon_{1},\upsilon_{2})\mapsto(p+1)(q-1)|_{\varepsilon=\hat{\varepsilon}_{i}}$
is continuous on $V$, it suffices to verify $(\ref{ex1eq5})$ at some
particular choice of parameter. For instance, taking
$\upsilon_{1}=\upsilon_{2}=1$ we obtain that
$(p+1)(q-1)|_{\varepsilon=\hat{\varepsilon}_{i}}=92$ for $i=1,2.$ Therefore
$D_{\mu}\neq 0$ at any admissible parameter. Thus, exactly as before, since
$\mu\mapsto D_{\mu}$ is continuous and the set of admissible parameters is
arc-connected, the claim will follow once we verify its validity at some
particular parameter. For instance the choice $\mu=(0,-2,2)$ yields
$D_{\mu}=-\frac{67+25\sqrt{5}}{2}<0.$

We proceed now with the study of the cyclicity of $Q_{\mu}$. The fact that
$\mathrm{Cycl}\big{(}(Q_{\mu_{0}},X_{\mu_{0}}),X_{\mu}\big{)}=0$ when
$\tau(\mu_{0})\neq 0$ is well-known. On the other hand, if $p+q=0$ then
$\varepsilon=0$ and

$\tau(\upsilon_{1},\upsilon_{2},0)=(\upsilon_{1}-\upsilon_{2})(\upsilon_{1}+2+\upsilon_{{2}})(\upsilon_{1}+\upsilon_{2}),$

which vanishes at an admissible parameter if, and only if,
$\upsilon_{1}=\upsilon_{2}$. For this to happen, see $(\ref{ex1eq4})$ with
$\varepsilon=0$, it is necessary that $a=0.$ This proves the validity of
$(d)$.

We shall next solve the center-focus problem in the family. With this aim in
view, taking a local transversal section at $Q_{\mu}$ we consider the
displacement map $\mathscr{D}(s;\hat{\mu})$, which extends analytically to
$s=0$, so that we can compute its Taylor’s expansion

$\mathscr{D}(s;\hat{\mu})=\eta_{1}(\hat{\mu})s+\eta_{2}(\hat{\mu})s^{2}+\eta_{3}(\hat{\mu})s^{3}+sR(s;\hat{\mu}),$
(28)

where the remainder $R$ is $\mbox{\rm o}(s^{2})$. Recall that the trace of
$X_{\hat{\mu}}$ at $Q_{\hat{\mu}}$ is equal to $\tau u_{1}(\hat{\mu})$, where
$u_{1}$ is a unity. The coefficients $\eta_{i}$ are called the _Lyapunov
quantities_ of the focus. We have in particular (see for instance [18, p. 94])
that $\eta_{1}(\hat{\mu})=e^{\tau u_{1}(\hat{\mu})}-1=\tau u_{2}(\hat{\mu})$,
where $u_{2}$ is again a unity. Since the first nonzero coefficient of the
expansion is the coefficient of an odd power of $s$, see [18, p. 94] again, we
get that $\eta_{2}(\hat{\mu})=\tau\ell_{1}(\hat{\mu})$ for some analytic
function $\ell_{1}$. In order to obtain $\eta_{3}$ we shall appeal to the
well-known relation between the Lyapunov and focus quantities which, following
the notation in [18, Theorem 6.2.3], we denote by $g_{ii}$. The first ones are
the coefficients in the Taylor’s expansion of the displacement map that we
already introduced, while the second ones are the obstructions for the
existence of a first integral. It occurs that $\eta_{2i+1}-\pi
g_{ii}\in(g_{11},\ldots,g_{i-1,i-1})$ and, more important for our purposes,
that $\eta_{3}=\pi g_{11}.$ On account of this we can compute $g_{11}$ instead
of $\eta_{3}$, which is easier to obtain, and in doing so (see [3, p. 29]) we
get that

$\eta_{3}(\hat{\mu})\big{|}_{\tau=0}=\pi{\frac{2\rho\left(\rho-\sigma\right)\left(\sigma+1\right)\left(4+34\sigma+29{\sigma}^{2}+8{\sigma}^{3}-3{\rho}^{2}\right)}{3\left(\rho+\sigma\right)^{3}\left(\rho+2\sigma+2{\sigma}^{2}+2{\rho}^{2}+2\right)^{2}}}.$

In this respect we claim that $\eta_{3}(\hat{\mu})\big{|}_{\tau=0}=\rho
h(\rho,\sigma)$ with $h(\rho,\sigma)\neq 0$ in case that $|\rho|<\sigma$,
which corresponds to the admissible values $\upsilon_{1},\upsilon_{2}>0$ due
to $\rho=\frac{\upsilon_{1}-\upsilon_{2}}{2}$ and
$\sigma=\frac{\upsilon_{1}+\upsilon_{2}}{2}.$ Indeed, it is clear that the
factor $(\rho-\sigma)(\sigma+1)$ does not vanish inside the admissible set,
while the other one does not vanish neither because

$4+34\sigma+29{\sigma}^{2}+8{\sigma}^{3}-3{\rho}^{2}>4+34\sigma+26{\sigma}^{2}+8{\sigma}^{3}>0,$

where the first inequality follows using that $|\rho|<\sigma$ and the second
one the fact that $\sigma>0.$ Hence the claim is true. Therefore, if $Q_{\mu}$
is a center then $\tau=\rho=0$, and the assertion in $(c)$ shows that these
two conditions are also sufficient because $\\{p+q=a=0\\}=\\{\tau=\rho=0\\}.$
Observe moreover that we can write
$\eta_{3}(\hat{\mu})=\tau\ell_{2}(\hat{\mu})+\rho h(\rho,\sigma)$ for some
analytic function $\ell_{2}.$ On the other hand, due to
$R(s;\hat{\mu})|_{\rho=\tau=0}\equiv 0$, we can also write $R=\tau R_{1}+\rho
R_{2}$ with $R_{i}\in\mbox{\rm o}(s^{2})$ and, accordingly,

$\mathscr{D}(s;\mu)=\tau s\big{(}u_{2}+\ell_{1}s+\ell_{2}s^{2}+R_{1})+\rho
s\big{(}hs^{2}+R_{2}\big{)}.$ (29)

Note that if $\tau(\mu)=0$ and $\varepsilon=p+q\neq 0$ then
$\rho=\frac{\upsilon_{1}-\upsilon_{2}}{2}$ must be different from zero because
otherwise, from $(\ref{ex1eq6})$, we would get that $\varepsilon=0.$
Consequently, due to $h(\rho,\sigma)\neq 0$ for all admissible $\rho$ and
$\sigma,$ the equality in $(\ref{ex1eq3})$ implies
$\mathrm{Cycl}\big{(}(Q_{\mu_{0}},X_{\mu_{0}}),X_{\mu}\big{)}\leqslant 1$ in
case that $\tau(\mu_{0})=0$ and $p_{0}+q_{0}\neq 0.$ The fact that this upper
bound is attained follows by means of an easy perturbative argument using that
$\partial_{\varepsilon}\tau(\mu)=-2\upsilon_{1}^{2}\upsilon_{2}^{2}\neq 0.$
This proves the validity of the assertion in $(e).$

In order to prove $(f)$ note that if $\tau(\mu)=0$ and $\varepsilon=p+q=0$
then, from $(\ref{ex1eq6})$, $\rho=\frac{\upsilon_{1}-\upsilon_{2}}{2}=0.$
Hence, from $(\ref{ex1eq4})$, $2a\upsilon_{1}\upsilon_{2}=0$, which implies
$a=0$ and shows the first assertion. That being stablished, we have already
proved that $Q_{\mu_{0}}$ is a center if, and only if $p_{0}+q_{0}=a_{0}=0.$
We show next that, in this case,
$\mathrm{Cycl}\big{(}(Q_{\mu_{0}},X_{\mu_{0}}),X_{\mu}\big{)}\leqslant 1$.
Indeed, since $u_{2}$ is a unity we can consider

$\mathscr{D}_{1}(s;\hat{\mu})\\!:=\frac{\mathscr{D}(s;\hat{\mu})}{s(u_{2}+\ell_{1}s+\ell_{2}s^{2}+R_{1})}=\tau+\rho\frac{hs^{2}+R_{2}}{u_{2}+\ell_{1}s+\ell_{2}s^{2}+R_{1}}.$

The upper bound for the cyclicity of $Q_{\mu_{0}}$ in the center case will
follow once we prove that there exist $s_{0}>0$ and an open neighbourhood $U$
of $(\rho,\sigma,\tau)=(0,\hat{}\sigma,0)$ such that
$\mathscr{D}_{1}(s;\hat{\mu})$ has at most one zero on $(0,s_{0})$, counted
with multiplicities, for all $\hat{\mu}\in U$ with $\rho^{2}+\tau^{2}\neq 0.$
Recall in this regard that $\mathscr{D}_{1}(s;\hat{\mu})|_{\rho=\tau=0}\equiv
0$ and that, on account of $(c)$,
$\hat{}\sigma=-\frac{1+\sqrt{-3-4p_{0}}}{2(1+p_{0})}>0$ is the first component
of $Q_{\mu_{0}}.$ The idea to show this is exactly the same as in the proof of
$(c)$ but with less technicalities because the involved functions are analytic
at $s=0$. The desired property is evident when $\rho=0.$ In case that
$\rho\neq 0$ we compute the derivative of $\mathscr{D}_{1}$ with respect to
$s$ to obtain

$\partial_{s}\mathscr{D}_{1}(s;\hat{\mu})=\rho s\big{(}2h/u_{2}+\mbox{\rm
o}(1)\big{)}.$

Since $h(\rho,\sigma)\neq 0$ in case that $|\rho|<\sigma$ and
$u_{2}=u_{2}(\hat{\mu})$ is a unity, the existence of the desired $s_{0}>0$
and the open neighbourhood $U$ follows by Rolle’s theorem. So far we have
proved that
$\mathrm{Cycl}\big{(}(Q_{\mu_{0}},X_{\mu_{0}}),X_{\mu}\big{)}\leqslant 1$ if
$p_{0}+q_{0}=a_{0}=0$. The fact that this upper bound is attained follows
noting that we can take $\mu=(a,p,q)$ with $\tau(\mu)=0$ and $p+q\neq 0$
arbitrarily close to $\mu_{0}=(0,p_{0},-p_{0})$ and apply then the assertion
in $(e).$ This proves $(f)$.

Let us turn now to the proof of the last assertion in the statement. Observe
in this respect that the combination of $(a)$ and $(b)$ together with $(d)$
and $(e)$ shows that a simultaneous bifurcation of limit cycles from $\Gamma$
and $Q_{\mu}$ can only occur if we perturb some
$\mu_{\star}=(a_{\star},p_{\star},q_{\star})$ with
$a_{\star}=p_{\star}+q_{\star}=0.$ We shall prove by contradiction that this
is neither possible. So assume that for each $n\in\mathbb{N}$ there exist
$\mu_{n}=(a_{n},p_{n},q_{n})$ and two limit cycles $\gamma_{n}$ and
$\gamma_{n}^{\prime}$ of the vector field $X_{\mu_{n}}$ in the first quadrant
such that the Hausdorff distances $d_{H}(\gamma_{n},\Gamma)$ and
$d_{H}(\gamma_{n}^{\prime},Q_{\mu_{n}})$ tend to zero and $\mu_{n}$ tends to
$\mu_{\star}$ as $n\to+\infty.$ Let us consider the asymptotic expansion of
the displacement map of $X_{\mu}$ at the polycycle $\Gamma$ that we compute in
$(\ref{ex1eq7})$ and denote it by $\mathscr{D}_{p}(s;\mu)$. We also consider
its Taylor’s expansion near the focus $Q_{\mu}$ given in $(\ref{ex1eq3})$ and
denote it by $\mathscr{D}_{c}(s^{\prime};\mu).$ Then the assumption implies
the existence of two sequences $s_{n}\to 0^{+}$ and $s_{n}^{\prime}\to 0^{+}$
such that $\mathscr{D}_{p}(s_{n};\mu_{n})=0$ and
$\mathscr{D}_{c}(s^{\prime}_{n};\mu_{n})=0$ for all $n\in\mathbb{N}.$ We claim
that the first equality implies that

$\lim_{n\to+\infty}\frac{p_{n}+q_{n}}{a_{n}}=0.$ (30)

Indeed, from $(\ref{ex1eq10})$ we have that

$\mathscr{R}_{1}(s_{n};\nu)\big{|}_{\nu=\Psi(\mu_{n})}=b_{1}(\mu_{n})\,\big{(}\omega(s_{n};\alpha(\mu_{n}))+h_{1}(s_{n};\nu)\big{)}+b_{2}(\mu_{n})\big{(}1+h_{2}(s;\nu)\big{)}\big{|}_{\nu=\Psi(\mu_{n})}=0\text{
for all $n\in\mathbb{N}.$}$

Thus, due to $\lim_{n\to+\infty}\omega(s_{n};\alpha(\mu_{n}))=+\infty$ and
$h_{i}\in\mathcal{F}_{\ell}^{\infty}(0_{3})$, we obtain that
$\frac{b_{2}(\mu_{n})}{b_{1}(\mu_{n})}\to-\infty$ as $n\to+\infty$. Moreover,
since $(b_{1})=(d_{1})$ and $(b_{1},b_{2})=(d_{1},d_{2})$ with $\nabla d_{1}$
and $\nabla d_{2}$ independent at $\mu=\mu_{\star},$ we can write

$\frac{b_{2}}{b_{1}}=\frac{\kappa_{1}d_{1}+\hat{\kappa}_{2}d_{2}}{\hat{\kappa}_{3}d_{1}}$

with $\hat{\kappa}_{i}(\mu_{\star})\neq 0$ and, consequently,
$\lim_{n\to+\infty}\frac{d_{2}(\mu_{n})}{d_{1}(\mu_{n})}=\infty$. This, on
account of $(\ref{ex1eq1})$ and $(\ref{ex1eq2})$, gives the limit in
$(\ref{ex1eq8})$ and so the claim is true. Recall on the other hand that in
order to study the displacement map near the focus $Q_{\mu}$ we use a more
convenient parametrisation given by
$\hat{\mu}\\!:=(\rho,\sigma,\tau)=\phi(\mu)$. That being said, setting
$(\rho_{n},\sigma_{n},\tau_{n})\\!:=\phi(a_{n},p_{n},q_{n})$, similarly as we
argue before, the fact that $\mathscr{D}_{c}(s^{\prime}_{n};\mu_{n})=0$ for
all $n\in\mathbb{N}$ implies from $(\ref{ex1eq3})$ that

$\lim_{n\to+\infty}\frac{\tau_{n}}{\rho_{n}}=0.$ (31)

Let us remark that here we also take into account that $u_{2}$ is a unity. We
next arrive to contradiction showing that $(\ref{ex1eq8})$ and
$(\ref{ex1eq9})$ cannot hold simultaneously. Indeed, one can verify that,
setting $\sigma_{\star}=-\frac{1+\sqrt{-3-4p_{\star}}}{2(1+p_{\star})},$

$\displaystyle\left.\frac{p_{n}+q_{n}}{a_{n}}\right|_{\mu_{n}=\phi^{-1}(\hat{\mu}_{n})}$
$\displaystyle=4\,\frac{\rho_{n}^{2}+\sigma_{n}^{2}}{\rho_{n}^{2}-\sigma_{n}^{2}}\,\frac{\tau_{n}+\rho_{n}\sigma_{n}(\sigma_{n}+2)}{2\tau_{n}-\rho_{n}(2\sigma_{n}+1)(\rho_{n}^{2}+\sigma_{n}^{2})}$
$\displaystyle=4\,\frac{\rho_{n}^{2}+\sigma_{n}^{2}}{\rho_{n}^{2}-\sigma_{n}^{2}}\,\frac{\tau_{n}/\rho_{n}+\sigma_{n}(\sigma_{n}+2)}{2\tau_{n}/\rho_{n}-(2\sigma_{n}+1)(\rho_{n}^{2}+\sigma_{n}^{2})}\to\frac{4(\sigma_{\star}+2)}{\sigma_{\star}(2\sigma_{\star}+1)}\neq
0\text{ as $n\to+\infty$.}$

Here, in addition to $(\ref{ex1eq9})$, we use that if $p+q$ and $a$ tend to
zero then $\rho\to 0$ and $\sigma\to\sigma_{\star},$ where $\sigma_{\star}$ is
precisely the first component of the center at $Q_{\mu_{\star}}$ (which is in
the diagonal of the first quadrant). This shows that $(\ref{ex1eq8})$ and
$(\ref{ex1eq9})$ cannot occur simultaneously, which yields to the desired
contradiction and finishes the proof of the result.

The following is our second example of application of Theorem A. In this case
the family of Kolmogorov’s systems is five-parametric and to the best of our
knowledge it has not been studied previously.

######  Theorem 3.2.

Consider the family of Kolmogorov differential systems

$X_{\mu}\quad\left\\{\\!\begin{array}[]{l}\dot{x}=x(c+x^{2}+axy-(p+1)y^{2}),\\\\[2.0pt]
\dot{y}=y(-1+(q+1)x^{2}+(a-b)xy-y^{2}),\end{array}\right.$

where $\mu=(a,b,c,p,q)\in\mathbb{R}^{5}$ with $c>0$, $p>0$, $q>0$ and
$b<2\sqrt{pq}$ and let us fix any $\mu_{0}=(a_{0},b_{0},c_{0},p_{0},q_{0}).$
Then there exists a unique singular point $Q_{\mu}$ in the first quadrant,
which is either a center, a focus or a node. Moreover, compactifying $X_{\mu}$
to the Poincaré disc, the boundary of the first quadrant is a polycycle
$\Gamma$ such that:

  1. $(a)$

$\mathrm{Cycl}\big{(}(\Gamma,X_{\mu_{0}}),X_{\mu}\big{)}=0$ if
$p_{0}-c_{0}q_{0}\neq 0.$

  2. $(b)$

$\mathrm{Cycl}\big{(}(\Gamma,X_{\mu_{0}}),X_{\mu}\big{)}=1$ if
$p_{0}-c_{0}q_{0}=0$ and $2c_{0}q_{0}a_{0}-(c_{0}q_{0}-c_{0}+1)b_{0}\neq 0$.

  3. $(c)$

The return map of $X_{\mu_{0}}$ along $\Gamma$ is the identity if, and only
if, $p_{0}-c_{0}q_{0}=2c_{0}q_{0}a_{0}-(c_{0}q_{0}-c_{0}+1)b_{0}=0$. In this
case $Q_{\mu_{0}}$ is a center with first integral

$H(x,y)=\frac{q_{0}(x^{2}+c_{0}(y^{2}+1))-b_{0}xy}{(xy^{c_{0}})^{\frac{2}{c_{0}q_{0}+c_{0}+1}}},$

which foliates the first quadrant. Moreover $\Gamma$ is the outer boundary of
its period annulus and, in addition,
$\mathrm{Cycl}\big{(}(\Gamma,X_{\mu_{0}}),X_{\mu}\big{)}=1$.

  * Remark 3.3.

In contrast to the family of Kolmogorov’s cubic systems studied in Theorem
3.1, for the family in Theorem 3.2 there exist parameters $\mu_{0}$ with
$d_{1}(\mu_{0})=0$ and $d_{2}(\mu_{0})\neq 0$, so that
$\mathrm{Cycl}\big{(}(\Gamma,X_{\mu_{0}}),X_{\mu}\big{)}=1$, and satisfying
additionally that the unique singular point $Q_{\mu_{0}}$ in the first
quadrant is a non-degenerate node. Hence, for appropiate $\mu\approx\mu_{0}$
we will have a limit cycle $\gamma_{\mu}$ with a non-monodromic singular point
$Q_{\mu}$ as unique singularity in its interior. For instance, the choice
$\mu_{0}=(-800.01,-900.99999,1000,1,0.001)$ leads to this phenomenon with
$Q_{\mu_{0}}=(0.1,10)$. A similar occurrence is observed in [2, p. 203] to
take place in the family of cubic Lienard systems studied in [7]. $\square$

  * Proof of Theorem 3.2.

In this case, following the notation in $(\ref{kol})$, we have that

$f(x,y;\mu)\\!:=c+x^{2}+axy-(p+1)y^{2}\text{ and
}g(x,y;\mu)\\!:=-1+(q+1)x^{2}+(a-b)xy-y^{2}.$

Since $f(z,0;\mu)=c+z^{2}$, $g(0,z;\mu)=-1-z^{2}$ and
$(f_{2}-g_{2})(1,z;\mu)=-q+bz-pz^{2}$, one can check that the hypothesis
$\mathbf{H1}$ and $\mathbf{H2}$ are satisfied for the admissible parameters,
i.e., $c>0$, $p>0$, $q>0$ and $b<2\sqrt{pq}$. Moreover the hyperbolicity
ratios are

$\lambda_{1}=1/q$, $\lambda_{2}=p$ and $\lambda_{3}=1/c$. (32)

Then $\Gamma$ is a polycycle and by applying the Poincaré-Bendixson theorem we
deduce the existence of at least one singular point of $X_{\mu}$ in the first
quadrant. We claim that there exists exactly one. In order to show this we
suppose that $(\upsilon_{1},\upsilon_{2})$ is a singular point of $X_{\mu}$ in
the first quadrant and solve $f(\upsilon_{1},\upsilon_{2};\mu)=0$ and
$g(\upsilon_{1},\upsilon_{2};\mu)=0$ for $a$ and $b$ as a function of $c,$
$p,$ $q,$ $\upsilon_{1}$ and $\upsilon_{2}$. In doing so we obtain that

$a={\frac{p\upsilon_{2}^{2}-\upsilon_{1}^{2}+\upsilon_{{2}}^{2}-c}{\upsilon_{{1}}\upsilon_{{2}}}}\text{
and
}b={\frac{p\upsilon_{{2}}^{2}+q\upsilon_{{1}}^{2}-c-1}{\upsilon_{{1}}\upsilon_{{2}}}}.$

The substitution of these values in $f+cg$, which is homogeneous of degree 2
in $x$ and $y$, yields

$\left.\big{(}f+cg\big{)}(x,y;\mu)\right|_{y=rx}=\frac{x^{2}(\upsilon_{2}-r\upsilon_{1})\big{(}r\upsilon_{2}(c+p+1)+\upsilon_{1}(cq+c+1)\big{)}}{\upsilon_{1}\upsilon_{2}}.$

It is clear then that the vanishing of the above numerator provides the
possible values of $r$ such that $X_{\mu}$ has a singular point at the
straight-line $y=rx,$ namely,

$r_{1}=\frac{\upsilon_{2}}{\upsilon_{1}}>0\text{ and
}r_{2}=-\frac{\upsilon_{1}(cq+c+1)}{\upsilon_{2}(c+p+1)}<0.$

Since $x\mapsto f(x,r_{1}x)=c+x^{2}(1+ar_{1}-(p+1)r_{1}^{2})$ vanishes at
$x=\upsilon_{1}>0$, it must have another real zero, which has to be negative.
Therefore $X_{\mu}$ has exactly one singular point in the first quadrant and
exactly one singular point in the third quadrant, showing in particular the
validity of the claim. An easy computation shows that the determinant of the
Jacobian of $X_{\mu}$ at $Q_{\mu}=(\upsilon_{1},\upsilon_{2})$ is equal to
$2\upsilon_{1}^{2}(cq+c+1)+2\upsilon^{2}_{2}(p+c+1)>0$, so that it can be a
center, a focus or a node.

So far we have proved that the first assertion in the statement is true. Let
us turn to the proof of the assertions in $(a)$, $(b)$ and $(c)$. The first
one follows from $(a)$ in Theorem A because

$d_{1}(\mu)=1-\lambda_{1}\lambda_{2}\lambda_{3}=\frac{cq-p}{cq}.$ (33)

The second assertion will follow by applying $(b)$ and $(c)$ in Theorem A. In
this regard let us recall that

$d_{2}(\mu)=\lambda_{2}\log\left(\frac{L_{12}}{L_{21}}\right)\\!(1)+\log\left(\frac{L_{22}}{L_{32}}\right)\\!(1)+\lambda_{1}\lambda_{2}\log\left(\frac{L_{31}}{L_{11}}\right)\\!(1).$
(34)

On account of the definition of each $L_{ij}$, see $(\ref{def_M})$, we easily
obtain that

$\displaystyle\log L_{11}(1)$ $\displaystyle=\frac{1+c(q+1)}{2c}\log(c+1),$
$\displaystyle\log L_{31}(1)$
$\displaystyle=\frac{1+c(q+1)}{2c}\log\left(1/c+1\right),$ (35)
$\displaystyle\log L_{22}(1)$ $\displaystyle=\log L_{32}(1)=\frac{\log
2}{2}(p+c+1).$

Moreover

$\log L_{12}(1)=\frac{1}{q}\int_{0}^{1}\frac{mz+n}{-pz^{2}+bz-q}dz\text{ and
}\log L_{21}(1)=\frac{1}{p}\int_{0}^{1}\frac{mz+n^{\prime}}{-qz^{2}+bz-p}dz,$

where

$m\\!:=-(pq+p+q),\text{ }n\\!:=qa+b\text{ and }n^{\prime}\\!:=p(b-a)+b.$ (36)

The explicit integration of these functions leads to several cases depending
on the parameters. To avoid this we note that

$\frac{mz+n}{-pz^{2}+bz-q}=-\frac{m}{2p}\frac{-2pz+b}{-pz^{2}+bz-q}+\frac{1}{2p}\frac{mb+2np}{-pz^{2}+bz-q},$

so that

$\log
L_{12}(1)=-\frac{m}{2pq}\log\left(\frac{p+q-b}{q}\right)+\frac{mb+2np}{2pq}\int_{0}^{1}\frac{dz}{-pz^{2}+bz-q}.$

It is clear that the same formula holds for $\log L_{21}(1)$ replacing $p,$
$q$ and $n$ by $q$, $p$ and $n^{\prime}$, respectively. On account of this and
the fact that, from $(\ref{ex2eq3})$, $mb+2n^{\prime}q=-mb-2np$, we get

$\log\left(\frac{L_{12}}{L_{21}}\right)\\!(1)=\log L_{12}(1)-\log
L_{21}(1)=-\frac{m}{2pq}\log\left(\frac{p}{q}\right)+(mb+2np)\Phi(\mu),$ (37)

where

$\Phi(\mu)\\!:=\frac{1}{2pq}\int_{0}^{1}\left(\frac{1}{-pz^{2}+bz-q}+\frac{1}{-qz^{2}+bz-p}\right)dz.$

Notice, and this is the key point in the forthcoming arguments, that $\Phi$ is
a non-vanishing function because, thanks to property H1, $\Phi(\mu)<0$ at any
admissible parameter $\mu$. On the other hand, from $(\ref{ex2eq6})$,

$\log\left(\frac{L_{31}}{L_{11}}\right)\\!(1)=-\frac{(1+c+cq)}{2c}\log c\text{
and }\log\left(\frac{L_{22}}{L_{32}}\right)\\!(1)=0.$ (38)

Accordingly the substitution of $(\ref{ex2eq4})$ and $(\ref{ex2eq5})$ in
$(\ref{ex2d2})$ yields

$\displaystyle d_{2}(\mu)$
$\displaystyle=-\frac{m}{2q}\log\left(\frac{p}{q}\right)+p(mb+2np)\Phi(\mu)-\frac{p(1+c+cq)}{2qc}\log
c$
$\displaystyle=\frac{pq+p+q}{2q}\log\left(\frac{p}{q}\right)+p\big{(}2pqa-(pq-p+q)b\big{)}\Phi(\mu)-\frac{p(1+c+cq)}{2qc}\log
c,$

where in the first equality we set the values of the hyperbolicity ratios
given in $(\ref{ex2eq0})$ and in the second one the expressions of $m$ and $n$
defined in $(\ref{ex2eq3})$. Observe at this point, see $(\ref{ex2eq1})$, that
$d_{1}(\mu)=0$ if, and only if, $p=cq.$ Moreover the two logarithmic summands
in the above expression of $d_{2}(\mu)$ cancel each other after the
substitution $p=cq$, so that

$d_{2}(\mu)\big{|}_{p=cq}=cq^{2}\big{(}2cqa-(cq-c+1)b\big{)}\Phi(a,b,c,cq,q).$

Thus, by the Weierstrass Division Theorem (see [9, 12]), there exists an
analytic function $\kappa_{1}$ such that

$d_{2}(\mu)=d_{1}(\mu)\kappa_{1}(\mu)+\big{(}2cqa-(cq-c+1)b\big{)}\kappa_{2}(\mu),$
(39)

where $\kappa_{2}(\mu)\\!:=cq^{2}\Phi(\mu)$ is a unity in the admissible set.
This expression shows that if we take an admissible parameter
$\mu_{0}=(a_{0},b_{0},c_{0},p_{0},q_{0})$ such that $p_{0}-c_{0}q_{0}=0$ and
$2c_{0}q_{0}a_{0}-(c_{0}q_{0}-c_{0}+1)b_{0}\neq 0$ then $d_{2}(\mu_{0})\neq
0$, which by $(c)$ in Theorem A implies that
$\mathrm{Cycl}\big{(}(\Gamma,X_{\mu_{0}}),X_{\mu}\big{)}\leqslant 1.$ The fact
that $\mathrm{Cycl}\big{(}(\Gamma,X_{\mu_{0}}),X_{\mu}\big{)}=1$ follows by
applying $(b)$ in Theorem A because $\nabla d_{1}(\mu_{0})$ is not the zero
vector, see $(\ref{ex2eq1})$. This proves the validity of the assertion in
$(b)$.

To show $(c)$ we take any $\mu_{0}=(a_{0},b_{0},c_{0},p_{0},q_{0})$ satisfying
$p_{0}-c_{0}q_{0}=2c_{0}q_{0}a_{0}-(c_{0}q_{0}-c_{0}+1)b_{0}=0$. Then one can
verify that the function

$H(x,y)=\frac{q_{0}(x^{2}+c_{0}(y^{2}+1))-b_{0}xy}{(xy^{c_{0}})^{\frac{2}{c_{0}q_{0}+c_{0}+1}}}$

is a first integral of $X_{\mu_{0}}$, which is clearly analytic on the whole
first quadrant. (For reader’s convenience let us mention that we found this
first integral looking for an integrating factor of the form $x^{r}y^{s}$ with
$r,s\in\mathbb{R}.$) Thus, since the determinant of the Jacobian of
$X_{\mu_{0}}$ at $Q_{\mu_{0}}$ is strictly positive, we can assert that it is
a center. A straightforward application of the Poincaré-Bendixson theorem
shows that $\Gamma$ is the outer boundary of its period annulus, which fills
the first quadrant. This proves that if
$p_{0}-c_{0}q_{0}=2c_{0}q_{0}a_{0}-(c_{0}q_{0}-c_{0}+1)b_{0}=0$ then the
displacement map $\mathscr{D}(\,\cdot\,;\mu_{0})$ of $X_{\mu_{0}}$ along
$\Gamma$ is identically zero. The converse follows by Theorem 2.6 noting that
$d_{1}(\mu)=d_{2}(\mu)=0$ if, and only if, $p-cq=2cqa-(cq-c+1)b=0$.

It only remains to be proved that
$\mathrm{Cycl}\big{(}(\Gamma,X_{\mu_{0}}),X_{\mu}\big{)}=1$. This follows
verbatim the proof of the same fact in assertion $(c)$ of Theorem 3.1 and so
we shall omit the details for the sake of shortness. Indeed, by $(b)$ in
Theorem 2.6 we have that if $\mu\approx\mu_{0}$ then

$\mathscr{D}(s;\mu)=(b_{1}(\mu)\omega(s;\alpha(\mu))+b_{2}(\mu)+r(s;\mu))s^{c},$

where $\alpha=c-p/q$ and $r\in\mathcal{F}_{\ell}^{\infty}(\mu_{0})$ with
$\ell\in(0,\min(1,p))$. Here we use that the three hyperbolicity ratios are
$\lambda_{1}=1/q$, $\lambda_{2}=p$ and $\lambda_{3}=1/c$. We also have that
$(b_{1})=(d_{1})$ and $(b_{1},b_{2})=(d_{1},d_{2})$ over the local ring
$\mathbb{R}\\{\mu\\}_{\mu_{0}}$. Therefore, if $\mu=(a,b,c,p,q)$ verifies
$p-cq=2cqa-(cq-c+1)b=0$ then $b_{1}(\mu)=b_{2}(\mu)=0$ and $r(s;\mu)\equiv 0$.
Moreover, since $\nabla d_{1}(\mu_{0})$ and $\nabla d_{2}(\mu_{0})$ are
linearly independent, see $(\ref{ex2eq1})$ and $(\ref{d2div})$, this is also
the case of $\nabla b_{1}(\mu_{0})$ and $\nabla b_{2}(\mu_{0})$. We can thus
take three linear functions, say $b_{3}(\mu)$, $b_{4}(\mu)$ and $b_{5}(\mu)$,
such that
$\nu=\Psi(\mu)\\!:=\big{(}b_{1}(\mu),b_{2}(\mu),b_{3}(\mu),b_{4}(\mu),b_{5}(\mu)\big{)}$
is a local analytic change of coordinates in a neighbourhood of $\mu=\mu_{0}$
with $\Psi(\mu_{0})=0_{5}\\!:=(0,0,0,0,0).$ Notice then that $\Psi$ maps
$\\{p-cq=2cqa-(cq-c+1)b=0\\}$ to $\\{\nu_{1}=\nu_{2}=0\\}$ and, moreover,

$\mathscr{R}_{1}(s;\nu):=s^{-c}\mathscr{D}(s;\mu)\Big{|}_{\mu=\Psi^{-1}(\nu)}=\nu_{1}\omega(s;\hat{\alpha})+\nu_{2}+\hat{r}(s;\nu),$

where $\hat{\alpha}=\hat{\alpha}(\nu)\\!:=\alpha(\Psi^{-1}(\nu))$ and
$\hat{r}(s;\nu)\\!:=r(s;\hat{\Psi}^{-1}(\nu))\in\mathcal{F}_{\ell}^{\infty}(0_{5})$.
Due to $\hat{r}(s;0,0,\nu_{3},\nu_{4},\nu_{5})\equiv 0$, by applying [17,
Lemma 4.1] we can write the remainder as
$\hat{r}(s;\nu)=\nu_{1}h_{1}(s;\nu)+\nu_{2}h_{2}(s;\nu)$ with
$h_{i}\in\mathcal{F}_{\ell}^{\infty}(0_{5})$ and, consequently,

$\mathscr{R}_{1}(s;\nu)=\nu_{1}\big{(}\omega(s;\hat{\alpha})+h_{1}(s;\nu)\big{)}+\nu_{2}\big{(}1+h_{2}(s;\nu)\big{)}.$

From this expression we conclude that there exists $s_{0}>0$ and an open
neighbourhood $U$ of $\nu=0_{5}$ such that $\mathscr{R}_{1}(s;\nu)$ that has
at most one zero on $(0,s_{0})$, counted with multiplicities, for all
$\nu=(\nu_{1},\nu_{2},\nu_{3},\nu_{4},\nu_{5})\in U$ with
$\nu_{1}^{2}+\nu_{2}^{2}\neq 0$, which implies that
$\mathrm{Cycl}\big{(}(\Gamma,X_{\mu_{0}}),X_{\mu}\big{)}\leqslant 1$. The
proof of this follows exactly as we argue to show the same fact in Theorem
3.1, cf. $(\ref{ex1eq10})$, and it is omitted for brevity. Finally the fact
that $\mathrm{Cycl}\big{(}(\Gamma,X_{\mu_{0}}),X_{\mu}\big{)}\geqslant 1$
follows taking $\mu_{1}\approx\mu_{0}$ with $p_{1}-c_{1}q_{1}=0$ and
$2c_{1}q_{1}a_{1}-(c_{1}q_{1}-c_{1}+1)b_{1}\neq 0$, and applying the assertion
in $(b)$. This completes the proof of the result.

  * Remark 3.4.

In order to prove Theorem 3.2 it is only necessary to compute the functions
$d_{1}$ and $d_{2}$ in Theorem A, which give the conditions for cyclicity 0
and 1, respectively. Let us explain that, as a matter of fact, we computed the
function $d_{3}$ as well, realizing that it vanishes when $d_{1}=d_{2}=0.$ It
was this fact that lead us to investigate if the return map along the
polycycle is the identity in that case. For completeness let us explain
succinctly the computations that involve the obtention of $d_{3}$ for the
Kolmogorov’s family considered in Theorem 3.2. Recall, see $(e)$ in Theorem A,
that

$d_{3}(\mu)\\!:=\hat{M}_{3}(\lambda_{3},1)L_{11}(1)-\hat{M}_{1}\big{(}1/\lambda_{1},1\big{)}L_{31}(1).$

In this case, cf. $(\ref{ex2eq6})$, we have that

$L_{11}(u)=(1+cu^{2})^{\frac{1+(q+1)c}{2c}}\text{ and
}L_{31}(u)=\left(1+u^{2}/c\right)^{\frac{1+(q+1)c}{2c}}$

and then, from the definition in $(\ref{def_M})$,

$\displaystyle M_{1}(u)$
$\displaystyle=(1+cu^{2})^{\frac{1+(q-3)c}{2c}}\big{(}aq+b+(cb-(c+1)a)u^{2}\big{)}$
and $\displaystyle M_{3}(u)$
$\displaystyle=-u\left(1+u^{2}/c\right)^{\frac{1+(q-3)c}{2c}}\big{(}(aq+b)u^{2}+cb-(c+1)a\big{)}/c^{2}.$

In order to proceed with the computation of
$\hat{M}_{1}\big{(}\frac{1}{\lambda_{1}},1\big{)}$ and
$\hat{M}_{3}\big{(}\lambda_{3},1\big{)}$ we note that if
$J(x;\eta,r)\\!:=(1+\eta x^{2})^{r}$ with $\eta>0$ and $r\in\mathbb{R}$ then

$\hat{J}(\alpha,1;\eta,r)=\int_{0}^{1}(1+\eta
x^{2})^{r}x^{-\alpha-1}dx=-\frac{1}{\alpha}\,{}_{2}F_{1}(-r,-\alpha/2;1-\alpha/2;-\eta)\text{
for all $\alpha<0$},$

where in the first equality we apply $(b)$ in Proposition 2.5 with $k=0$ and
in the second one we use the equality in [1, 15.3.1] to express the definite
integral as a hypergeometric function. In principle the above equality is only
true provided that $\alpha<0$. However its validity can be extended to any
$\alpha\notin\mathbb{N}$ thanks to the meromorphic properties of the functions
${}_{2}F_{1}$ and $\hat{J}$ stablished, respectively, by [17, Lemma B.2] and
$(d)$ in Proposition 2.5. Consequently, thanks to this observation and
applying twice the above formula, we get

$\hat{M}_{1}(1/\lambda_{1},1)=-\frac{aq+b}{q}\,\varphi_{1}(c,q)+\frac{cb-(c+1)a}{2-q}\,\varphi_{2}(c,q),$

where

$\displaystyle\varphi_{1}(c,q)$
$\displaystyle\\!:={}_{2}F_{1}\big{(}(3-q)/2-1/(2c),-q/2;1-q/2;-c\big{)},$
$\displaystyle\varphi_{2}(c,q)$
$\displaystyle\\!:={}_{2}F_{1}\big{(}(3-q)/2-1/(2c),1-q/2;2-q/2;-c\big{)}.$

Here we also use that if $h=f+g$ then
$\hat{h}_{\alpha}=\hat{f}_{\alpha}+\hat{g}_{\alpha}$ and that if
$f(x)=x^{n}g(x)$ then $\hat{f}_{\alpha}(x)=x^{n}\hat{g}_{\alpha-n}(x),$ see
[16, Corollary B3]. Similarly

$\hat{M}_{3}(\lambda_{3},1)=\frac{aq+b}{c(1-3c)}\,\varphi_{3}(c,q)+\frac{cb-(c+1)a}{c(1-c)}\,\varphi_{4}(c,q),$

where

$\displaystyle\varphi_{3}(c,q)$
$\displaystyle\\!:={}_{2}F_{1}\big{(}(3-q)/2-1/(2c),3/2-1/(2c);5/2-1/(2c);-1/c\big{)},$
$\displaystyle\varphi_{4}(c,q)$
$\displaystyle\\!:={}_{2}F_{1}\big{(}(3-q)/2-1/(2c),1/2-1/(2c);3/2-1/(2c);-1/c\big{)}.$

In the proof of Theorem 3.2 we show that $d_{1}(\mu)=d_{2}(\mu)=0$ if, and
only if, $\mu=(a,b,c,p,q)$ verifies $p=cq$ and $a=\frac{b(1-c+cq)}{2cq}$. Long
but easy computations show that, under these two conditions, $d_{3}(\mu)=0$
if, and only if,

$\frac{q}{1-3c}\varphi_{3}(c,q)-\varphi_{4}(c,q)+c^{-\frac{1+c+cq}{2c}}\left(\varphi_{1}(c,q)+\frac{c-1}{q-2}\varphi_{2}(c,q)\right)=0.$

This is an equation for $b,$ $c$ and $q$ that involves four hypergeometric
functions. Surprisingly enough it turns out, by applying the formula in [1,
15.3.7], that the function on the left hand side of the above equation is
identically zero. In other words, $d_{1}(\mu_{0})=d_{2}(\mu_{0})=0$ implies
$d_{3}(\mu_{0})=0.$ $\square$

  * Acknowledgements.

The authors want to thank Joan Torregrosa, who kindly lend us his Maple [13]
procedure to compute the focus quantities that we use to prove $(e)$ in
Theorem 3.1.

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