\documentclass[reqno]{amsart}
\usepackage{hyperref}

\AtBeginDocument{{\noindent\small
\emph{Electronic Journal of Differential Equations},
Vol. 2014 (2014), No. 120, pp. 1--22.\newline
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu
\newline ftp ejde.math.txstate.edu}
\thanks{\copyright 2014 Texas State University - San Marcos.}
\vspace{9mm}}

\begin{document}
\title[\hfilneg EJDE-2014/120\hfil Generalized Van der Pol equation]
{Generalized Van der Pol equation and \\ Hilbert's 16th problem}

\author[X. Ioakim \hfil EJDE-2014/120\hfilneg]
{Xenakis Ioakim}  % in alphabetical order

\address{Xenakis Ioakim \newline
Department of Mathematics and Statistics,
University of Cyprus, P.O. Box 20537, 1678 Nicosia, Cyprus}
\email{xioaki01@ucy.ac.cy}

\thanks{Submitted September 23, 2013. Published May 2, 2014.}
\subjclass[2000]{34C07, 34C23, 34C25}
\keywords{Generalized Van der Pol equation; Hilbert's 16th problem;
limit cycle; 
\hfill\break\indent existence; sinusoidal-type number; sinusoidal-type set;
dependent radius; $\Lambda$-point}

\begin{abstract}
 In this article, we study the bifurcation of limit cycles from the
 harmonic oscillator $\dot{x}=y$,  $\dot{y}=-x$ in the system
 $$
 \dot{x}=y,\quad \dot{y}=-x+\varepsilon f(y)\big(1-x^2\big),
 $$
 where $\varepsilon$ is a small positive parameter tending
 to 0 and $f$ is an odd polynomial of degree $2n + 1$, with $n$ an arbitrary
 but fixed natural number. We prove that, the above differential system,
 in the global plane, for particularly chosen odd polynomials $f$ of degree
 $2n + 1$ has exactly $n + 1$ limit cycles and that this number is an
 upper bound for the number of limit cycles for every case of an arbitrary
 odd polynomial $f$ of degree $2n + 1$. More specifically, the existence of
 the limit cycles, which is the first of the main results in this work,
 is obtained by using the Poincar\'e's method, and the upper bound for the
 number of limit cycles can be derived from the work of Iliev \cite{I}.
 We also investigate the possible relative positions of the limit cycles
 for this differential system, which is the second main problem studying
 in this work. In particular, we construct differential systems with $n$ given
 limit cycles and one limit cycle whose position depends on the position of
 the previous $n$ limit cycles. Finally, we give some examples in order to
 illustrate the general theory presented in this work.
\end{abstract}

\maketitle
\numberwithin{equation}{section}
\newtheorem{theorem}{Theorem}[section]
\newtheorem{lemma}[theorem]{Lemma}
\newtheorem{proposition}[theorem]{Proposition}
\newtheorem{definition}[theorem]{Definition}
\newtheorem{example}[theorem]{Example}
\newtheorem{remark}[theorem]{Remark}
\allowdisplaybreaks

\section{Introduction}

\subsection{Generalized Van der Pol equation and statement of the main results}

In this article, we study the second part of Hilbert's 16th problem 
for a generalized Van der Pol equation. 
More specifically, we consider the system
\begin{equation}
\begin{gathered}
\dot{x}=y, \\
\dot{y}=-x+\varepsilon f(y)\big(1-x^2\big),
\end{gathered}
\label{g VdP}
\end{equation}
where $f$ is an odd polynomial of degree $2n + 1$, with $n$ an arbitrary but
fixed natural number and $0 < \varepsilon \ll 1$.
 System \eqref{g VdP} reduces to the Van der Pol equation for $f(y) = y$.
 Our purpose here is to find an upper bound for the number of limit cycles 
for system \eqref{g VdP}, depending only on the degree of its polynomials 
and investigate their relative positions.

System \eqref{g VdP} is the generalized Van der Pol equation of the form
\begin{equation}
\ddot{x}-\varepsilon f(\dot{x})\big(1-x^2\big)+x=0,
\label{g VdP eq}
\end{equation}
where $f$ is an odd polynomial of degree $2n + 1$, with $n$ an 
arbitrary but fixed natural number and $0 < \varepsilon \ll 1$. 
The problem is to find an upper bound for the number of limit cycles 
for equation \eqref{g VdP eq}, depending only on the degree $2n + 1$ of 
the odd polynomial $f$ and investigate their relative positions. 
We prove that the generalized Van der Pol equation \eqref{g VdP eq} has 
exactly $n + 1$ limit cycles for particularly chosen odd polynomials 
$f$ of degree $2n + 1$ and that this number is an upper bound for the 
number of limit cycles for every case of an arbitrary odd polynomial $f$ 
of degree $2n+1$. Furthermore, we show how to construct these polynomials 
of equation \eqref{g VdP eq} which attain that upper bound. 
On the possible relative positions of the $n + 1$ limit cycles we show 
that there exists a limit cycle whose position depends on the position 
of the rest $n$ limit cycles (actually, this limit cycle is close to the 
circle with the dependent radius (see Definition \ref{dfn:dependent radius})).

Now, we state the main results of this article, which are the following theorems. 
The proofs of these theorems will be given in Section 3. For the definitions 
appear in these theorems, like the sets $V^n$, $V_{n+1}^n$,
 $n \in {\mathbb N}$, $n\geq2$,  $V^{1}$ and the dependent radius, 
see the next subsection. The first and second of our results, consider 
the system \eqref{g VdP}, with $f$ an odd polynomial of degree $2n + 1$, 
where $n \in {\mathbb N}$, $n\geq2$.

\begin{theorem}
\label{thm:criterion}
Let $(\lambda_1,\lambda_2,\dots,\lambda_{n}) \in V^n$ be such that
$(\lambda_1,\lambda_2,\dots,\lambda_{n},\lambda_{n+1})  \in  V_{n+1}^n$,
where $\lambda_{n+1}$ is the dependent radius given by \eqref{dependent radius n}, 
if $n \in {\mathbb N}$, $n\geq2$.  Then the system \eqref{g VdP}, with 
$0<\varepsilon\ll 1$ and
\begin{equation}
\begin{aligned}
f(y)&=\tau y^{2n+1}+\dots +\tau(2n-2k+3)\dots(2n+1)\\
&\quad \times \Big[1-\frac{1}{2(n+2)}\sum_{i_1=1}^{n+1}\lambda _{i_1}
+\frac{1}{4(n+1)(n+2)}\sum_{\substack{i_1,i_2=1\\i_1 <i_2}}^{n+1}
 \lambda_{i_1}\lambda_{i_2} +\dots\\
&\quad +\frac{1}{2^{k}(n-k+3)\dots(n+2)}(-1)^{k}\sum_{\substack{i_1 ,\dots,i_{k}=1
\\i_1<\dots<i_{k}}}^{n+1}\lambda_{i_1}\dots \lambda_{i_{k}}\Big]
y^{2(n-k)+1} \\
&\quad +\dots+\tau\Big[\frac{1\cdot3\dots(2n+1)}{2^{n+1}(n+2)!} 
(-1)^n\prod_{i_1=1}^{n+1}\lambda_{i_1}\Big]y,
\end{aligned}
\label{definition of f}
\end{equation}
where $\tau \in {\mathbb R} \setminus \{0\}$ and $1 \leq k \leq n-1$,
 has exactly the following $n+1$ limit cycles:
\[
x^2+y^2=\lambda_1+O(\varepsilon), \quad
x^2+y^2=\lambda_2+ O(\varepsilon),\dots, \quad
x^2+y^2=\lambda_{n+1}+O(\varepsilon).
\]
Furthermore, (assuming from now on an ordering such that
$\lambda_1<\lambda_2<\dots< \lambda_{n}<\lambda_{n+1}$, where now
$\lambda_{n+1}$ is not necessary the dependent radius) we have for the
 stability of the limit cycles that, if $\tau>0$ (respectively $\tau< 0$),
\[
x^2+y^2=\lambda_1+O(\varepsilon), \quad
x^2+y^2=\lambda_3+O(\varepsilon), \dots, \quad
x^2+y^2=\lambda_{n+1}+O(\varepsilon)
\]
are stable (respectively unstable) and
\[
x^2+y^2=\lambda_2+O(\varepsilon), \quad
x^2+y^2=\lambda_{4}+ O(\varepsilon),\dots, \quad
x^2+y^2=\lambda_{n}+O(\varepsilon)
\]
 are unstable (respectively stable) for $n$ even; and
\[
x^2+y^2=\lambda_1+O(\varepsilon), \quad
x^2 +y^2=\lambda_3+O(\varepsilon),\dots, \quad
x^2+y^2=\lambda_{n}+ O(\varepsilon)
\]
 are unstable (respectively stable) and
\[
x^2+y^2=\lambda_2+O(\varepsilon), \quad
x^2+y^2=\lambda_{4}+ O(\varepsilon),\dots, \quad
x^2+y^2=\lambda_{n+1}+O(\varepsilon)
\]
 are stable (respectively unstable) for $n$ odd.
\end{theorem}

\begin{theorem}
\label{thm:criterion extra}
For system \eqref{g VdP}, where $\varepsilon$ is small and $f$ is an arbitrary 
odd polynomial of degree $2n+1$ we have that the number of $n+1$ limit cycles 
is an upper bound for the number of limit cycles. Moreover, from the set 
of all the odd polynomials, the polynomials $f$ given by \eqref{definition of f}, 
are the only that attain that upper bound.
\end{theorem}

Our third and fourth results, concern the system \eqref{g VdP}, with $f$ 
an odd polynomial of degree 3.

\begin{theorem}
\label{thm:criterion 1}
Let $\lambda_1  \in V^{1}$. Then the system \eqref{g VdP}, with $0<\varepsilon\ll 1$
 and
\begin{equation}
f(y)=\tau y^3-\tau \frac{1}{8}\lambda_1\lambda_2y,
\label{definition of f 1}
\end{equation}
where $\tau \in {\mathbb R} \setminus \{0\}$ and $\lambda_2$ 
is the dependent radius given by \eqref{dependent radius 1}, 
has exactly the following $2$ limit cycles:
\[
x^2+y^2=\lambda_1+O(\varepsilon), \quad x^2+y^2=\lambda_2+ O(\varepsilon).
\]
Furthermore, (assuming from now on an ordering such that $\lambda_1<\lambda_2$, 
where now $\lambda_2$ is not necessary the dependent radius)
 we have for the stability of the limit cycles that, if $\tau>0$ 
(respectively $\tau< 0$)
$x^2+y^2=\lambda_1+O(\varepsilon)$ is unstable (respectively stable) and
$x^2+y^2=\lambda_2+O(\varepsilon)$ is stable (respectively unstable).
\end{theorem}

\begin{theorem}
\label{thm:criterion 1 extra}
For system \eqref{g VdP}, where $\varepsilon$ is small and $f$ is an arbitrary
 odd polynomial of degree $3$ we have that the number of $2$ limit cycles 
is an upper bound for the number of limit cycles. Moreover, from the set 
of all the odd polynomials, the polynomials $f$ given by \eqref{definition of f 1}, 
are the only that attain that upper bound.
\end{theorem}

\begin{remark} \rm 
It is important to note that the above theorems don't inform us which limit 
cycles we have for a differential equation of the form \eqref{g VdP}. 
That we succeed through these theorems is to construct differential equations 
of the form \eqref{g VdP} with $n$ given limit cycles and one limit cycle 
which is close to the circle with the dependent radius, for particularly 
chosen odd polynomials $f$ of degree $2n+1$. So, we show how to construct 
differential equations of the form \eqref{g VdP} that attain the upper 
bound of $n+1$ limit cycles, when the odd polynomial $f$ is of degree $2n+1$. 
Evenly important it is still and one negative result which can be obtained by 
these theorems, that we know a priori which limit cycles we can't have for 
system \eqref{g VdP} with odd polynomials $f$ of degree $2n+1$. Substantially,
 we construct the set of all the possible limiting radii of limit cycles for 
the system \eqref{g VdP} with odd polynomials $f$ of degree $2n+1$. 
This is the set $V_{n+1}^n$ which contains the $\varLambda$-points
(see Definition \ref{dfn:lambda points}).
\end{remark}

\begin{remark} \rm
It is surprising the connection between the dependent radius for a circle 
(see Definition \ref{dfn:dependent radius}) and the existence of one branch 
which can not separate from the rest branches for an algebraic curve. 
More specifically, relatively to the existence of such branch we refer the 
following of Hilbert's speech about the first part of Hilbert's 16th 
problem ``As of the curves of degree $6$, I have -admittedly in a rather 
elaborate way- convinced myself that the $11$ branches, that they can have
 according to Harnack, never all can be separate, rather there must exist 
one branch, which have another branch running in its interior and nine branches 
running in its exterior, or opposite". Here, we have for the relative positions 
of limit cycles that the limit cycle which is close to the circle with the 
dependent radius can not lie wherever, contrary the position of this limit 
cycle depends on the position of the rest limit cycles. In this sense, 
we can say that the first and second part of Hilbert's 16th problem come closer.
\end{remark}

\begin{remark} \rm
I would mention for system \eqref{g VdP} that by forcing the coefficients of 
an arbitrary odd polynomial to be those given in the Theorem \ref{thm:criterion} 
when $n \in {\mathbb N}$, $n \geq 2$ (respectively in the Theorem 
\ref{thm:criterion 1} when $n=1$), do not allow us to put $n + 1$ (respectively 2) 
limit cycles in arbitrary placements. The reason for this is the 
Theorem \ref{thm:criterion extra} (respectively the 
Theorem \ref{thm:criterion 1 extra}); in the statement of these theorems we see 
that the proposing polynomials $f$ (given in Theorems \ref{thm:criterion} 
and \ref{thm:criterion 1}) are the only that attain the upper bound of the 
$n+1$ limit cycles. Now it is easy to see that in the coefficients of these 
polynomials (unless in the first monomial in each case) appears the dependent 
radius, and this observation in turn implies that one limit cycle do not lie 
in arbitrary placements.

In order to see this more clearly consider for the system \eqref{g VdP} the 
case where $n=1$. Once we chose $\lambda_1$ from $V^{1}$, the dependent 
radius $\lambda_2$ follows from \eqref{dependent radius 1} will be positive 
(see Proposition \ref{prp:criterion radius}) and different from the associated 
$\lambda_1$ (see Remark \ref{rmrk:n 1}), and then for the system \eqref{g VdP} 
with $n=1$, the polynomial $f$ given by \eqref{definition of f 1} is the only 
that realizing the maximal number of 2 limit cycles, and are asymptotic to 
the circles $x^2+y^2=\lambda_1$ and $x^2+y^2=\lambda_2$ (note that for this 
circle the placement is not arbitrary, it depends on $\lambda_1$) as 
$\varepsilon \to 0$ (see Theorems \ref{thm:criterion 1} and 
\ref{thm:criterion 1 extra}). (See and Example \ref{example l=7}.)
\end{remark}

\subsection{Definitions}

In this subsection, we introduce some new definitions. These definitions
 are obtained by using technical integral expressions 
(see Remark \ref{rmrk:integral}) and properties of symmetric functions of 
the roots of polynomials (Vieta's formulas). The first of these definitions
 has an important role in the construction of the sinusoidal-type sets and 
also the advantage played by this definition along with the 
Definition \ref{dfn:sets} is going to be understandable in the proof of 
Proposition \ref{prp:criterion radius}.

\begin{remark} \rm
\label{rmrk:integral}
I adopt the sinusoidal terminology for the next two definitions, 
due to the formula
\begin{equation}
\int_{0}^{2\pi}\sin^{2n}t\,dt=\frac{1\cdot 3\dots(2n-3)(2n-1)}{2^{n-1}n!}\,\pi, 
\quad \text{for } \ n\in {\mathbb N},
\label{sinusoidal}
\end{equation}
(see \cite{NGG}) which gives the coefficients of the sums and products.
\end{remark}

\begin{definition}[sinusoidal-type numbers] \rm
\label{dfn:numbers}
Let $\lambda_1,\lambda_2,\dots,\lambda_{n-1},\lambda_{n}$ be distinct
 positive real numbers, where $n  \in  {\mathbb N}$, $n  \geq  2$.
 We define for $n  \in  {\mathbb N}$, $n  \geq  3$, the sinusoidal-type numbers 
of order $n$, associated to the 
$\lambda_1,\lambda_2,\dots,\lambda_{n-1},\lambda_{n}$
\begin{align*}
\bar{s}^n&:= 2(n+2)+\dots +\frac{(-1)^{k}}{2^{k-1}(n-k+3)\dots(n+1)}
\sum_{\substack{i_1 ,\dots,i_{k}=1\\i_1<\dots<i_{k}}}^n\lambda_{i_1}\dots 
\lambda_{i_{k}}+\dots \\
&\quad +\frac{(-1)^n}{2^{n-2}(n+1)!}\prod_{i_1 =1}^n\lambda_{i_1}, \\
\hat{s}^n&:= 2(n+1)+\dots +\frac{(-1)^{k}}{2^{k-1}(n-k+2)(n-k+3)\dots n}\sum _{\substack{i_1,\dots,i_{k}=1\\i_1<\dots<i_{k}}}^n\lambda _{i_1}\dots\lambda_{i_{k}}+\dots \\
&\quad +\frac{(-1)^n}{2^{n-1}n!} \prod_{i_1=1}^n\lambda_{i_1}, \\
\tilde{s}^n&:= \frac{1}{4n(n+1)}\sum_{\substack{i_1,i_2=1 \\i_1<i_2}}^n
 \lambda_{i_1}\lambda_{i_2}+\dots \\
&\quad +\frac{(-1)^{k}(k-1)}{2^{k}(n-k+2)(n-k+3)\dots(n+1)}
\sum _{\substack{i_1,\dots,i_{k}=1\\i_1<\dots<i_{k}}}^n\lambda _{i_1}
 \dots\lambda_{i_{k}}+\dots \\
&\quad +\frac{(-1)^n(n-1)}{2^n (n+1)!}\prod_{i_1=1}^n\lambda_{i_1},
\end{align*}
where $2 \leq k \leq n-1$ for the $\bar{s}^n, \hat{s}^n$ and 
$3 \leq k \leq n-1$ for $\tilde{s}^n$. In the special case where $n = 2$, 
we define the sinusoidal-type numbers of second order, associated to the 
$\lambda_1,\lambda_2$
\begin{gather*}
\bar{s}^2:= 8+\frac{1}{6}\lambda_1\lambda_2, \\
\hat{s}^2:= 6+\frac{1}{4}\lambda_1\lambda_2, \\
\tilde{s}^2:= \frac{1}{24}\lambda_1\lambda_2.
\end{gather*}
For $n=1$, we define the sinusoidal-type numbers of first order
\[
\bar{s}^{1}:= 6, \quad
\hat{s}^{1}:= 4.
\]
\end{definition}

For the sinusoidal-type numbers we have the following result. 
The proof is given in the Appendix.

\begin{lemma}
\label{lemma:criterion numbers}
For $n\in {\mathbb N}$, $n\geq2$, we have that
\begin{gather*}
\bar{s}^n=\hat{s}^n\Longleftrightarrow\tilde{s}^n=1, \\ 
\bar{s}^n>\hat{s}^n\Longleftrightarrow\tilde{s}^n<1, \\ 
\bar{s}^n<\hat{s}^n\Longleftrightarrow\tilde{s}^n>1,
\end{gather*}
where $\bar{s}^n$,  $\hat{s}^n$ and $\tilde{s}^n$ are the  sinusoidal-type 
numbers of order $n$, with $n\in {\mathbb N}$, $n\geq2$, of the Definition
 \ref{dfn:numbers}.
\end{lemma}

We continue with another definition. The role played by this definition 
is that the square roots of the coordinates $\lambda_1,\lambda_2,\dots,\lambda_{n}$ 
of the points $(\lambda_1,\lambda_2,\dots,\lambda_{n})$, where
 $n\in {\mathbb N}$, $n\geq2$, are going to be the limiting radii of the limit 
cycles which are asymptotic to circles of radii $\sqrt{\lambda_{i}}$ for 
$i=1,2,\dots,n$ centered at the origin when the small positive parameter 
of our system tending to 0. This will be done by forcing 
$\sqrt{\lambda_1},\sqrt{\lambda_2}, \dots,\sqrt{\lambda_{n}}$ to be simple 
roots of the polynomial $F$ defined in \eqref{FA} (see Theorem \ref{thm:method}). 
For this reason, we assume that the given $\lambda_1,\lambda_2,\dots, \lambda_{n}$ 
are all positive and with $\lambda_{i}\ne\lambda_{j}$ for all $i\ne j$ where 
$i,j=1,2, \dots,n$. In this way, we are going to construct $n$ limit cycles 
for system \eqref{g VdP}. But Iliev in \cite{I} proved that the maximal number 
of limit cycles due to polynomial perturbations of degree $n$ of the harmonic
 oscillator is equal to $[\frac{n-1}{2}]$ (the largest integer less than or 
equal to $\frac{n-1}{2}$). Since in our case the polynomial perturbations are 
of degree $2n+3$ we can achieve $n+1$ limit cycles. Now, we see that we can 
have an additional limit cycle. About the position of this limit cycle we 
later give the definition of the dependent radius.

The same observation of all the above is valid and in the case where $n=1$.

\begin{definition}[sinusoidal-type sets] \rm
\label{dfn:sets}
For $n \in  {\mathbb N}$, $n\geq2$, we define the sinusoidal-type sets of order $n$
as
\begin{align*}
S_1^n:=\Big\{&(\lambda_1,\lambda_2,\dots,\lambda_{n}): 
\lambda_{i}\ne\lambda_{j}\; \forall i\ne j \text{ where } i,j=1,2, \dots,n
\text{ with } \lambda_{i}>0 \\
& \forall i=1,2,\dots,n  \text{ and } \sum_{i=1}^n\lambda_{i}<\bar{s}^n 
\text{ when } \tilde{s}^n>1\Big\},
 \\
S_2^n:=\Big\{&(\lambda_1,\lambda_2,\dots,\lambda_{n}):
\lambda_{i}\ne\lambda_{j}\;\forall  i\ne j \text{ where } i,j=1,2, \dots,n
\text{ with } \lambda_{i}>0 \\
& \forall   i=1,2,\dots,n \text{ and }\sum_{i=1}^n\lambda_{i}<\hat{s}^n \text{ when } \tilde{s}^n<1\Big\}, \\
S_3^n:=\Big\{&(\lambda_1,\lambda_2,\dots,\lambda_{n}): \lambda_{i}\ne\lambda_{j}\;\forall  i\ne j \text{ where } i,j=1,2, \dots,n\text{ with } \lambda_{i}>0 \\&
\forall   i=1,2,\dots,n \text{ and }\sum_{i=1}^n\lambda_{i}<\bar{s}^n
=\hat{s}^n \text{ when } \tilde{s}^n=1\Big\}, \\
S_{4}^n:=\Big\{&(\lambda_1,\lambda_2,\dots,\lambda_{n}): \lambda_{i}\ne\lambda_{j}
\;\forall  i\ne j \text{ where } i,j=1,2, \dots,n\text{ with } \lambda_{i}>0 \\
&\forall   i=1,2,\dots,n \text{ and }\sum_{i=1}^n\lambda_{i}>\bar{s}^n 
 \text{ when } \tilde{s}^n<1\Big\}, \\
S_{5}^n:=\Big\{&(\lambda_1,\lambda_2,\dots,\lambda_{n}): 
 \lambda_{i}\ne\lambda_{j}\;\forall  i\ne j \text{ where } i,j=1,2, \dots,n
 \text{ with } \lambda_{i}>0 \\
&\forall   i=1,2,\dots,n \text{ and }\sum_{i=1}^n\lambda_{i}>\hat{s}^n 
\text{ when } \tilde{s}^n>1\Big\}, \\
S_{6}^n:=\Big\{&(\lambda_1,\lambda_2,\dots,\lambda_{n})
: \lambda_{i}\ne\lambda_{j}\;\forall  i\ne j \text{ where } i,j=1,2, 
\dots,n\text{ with } \lambda_{i}>0 \\
&\forall   i=1,2,\dots,n \text{ and }\sum_{i=1}^n\lambda_{i}>\bar{s}^n
=\hat{s}^n \text{ when } \tilde{s}^n=1\Big\},
\end{align*}
where $\bar{s}^n$, $\hat{s}^n$ and $\tilde{s}^n$ are the  sinusoidal-type 
numbers of order $n$, with $n\in {\mathbb N}$, $n\geq2$, of the Definition 
\ref{dfn:numbers}. For $n=1$, we define the sinusoidal-type sets of first order
\begin{gather*}
S_1^{1}:=\big\{\lambda_1:\lambda_1 \in (0,4)\big\}, \\
S_2^{1}:=\big\{\lambda_1:\lambda_1 \in (6,+\infty)\big\}.
\end{gather*}
\end{definition}

\begin{definition} \rm
We define for $n\in {\mathbb N}$, $n\geq2$, the set $V^n$ as the set
\[
V^n:=\cup_{i=1}^{6}S_{i}^n.
\]
For $n=1$, we define the set $V^{1}$ as the set
\[
V^{1}:=S_1^{1}\cup S_2^{1}.
\]
\end{definition}

Now, we continue with the last statement of the observation that we made 
before the Definition \ref{dfn:sets}. The positions of the $n$ limit cycles 
have to satisfy an algebraic relation in order that there is an odd 
polynomial $f$, realizing the maximal number of limit cycles, and the 
position of the $(n+1)$-th limit cycle is estimated in terms of the positions 
of these $n$ limit cycles. On this we have the following definition.

\begin{definition}[dependent radius] \rm
\label{dfn:dependent radius}
Let $(\lambda_1,\lambda_2,\dots,\lambda_{n})\in {\mathbb R}^n$, 
$n\in {\mathbb N}$,  $n\geq2$. We call dependent radius representing with 
$\lambda_{n+1}$ the quantity (when is defined) given by the formula
\begin{equation}
\lambda_{n+1}=\lambda_{n+1}(\lambda_1,\lambda_2,\dots, \lambda_{n-1},\lambda_{n})
:=\frac{\Xi}{\Psi},
\label{dependent radius n}
\end{equation}
where
\begin{align*}
\Xi&=2^{n+1}(n+2)!+\dots+(-1)^{k}2^{n-k+1}(n-k+2)!
\sum_ {\substack{i_1,\dots,i_{k}=1\\i_1<\dots<i_{k}}}^n\lambda _{i_1}\dots
\lambda_{i_{k}} \\ 
&\quad +\dots+4(-1)^n\prod_{i_1=1} ^n\lambda_{i_1}
\end{align*}
and
\begin{align*}
\Psi&=2^n(n+1)!+\dots+(-1)^{k}2^{n-k}(n-k+1)!
\sum_{\substack{i_1,\dots,i_{k}=1\\i_1<\dots<i_{k}}}^n \lambda_{i_1}
\dots\lambda_{i_{k}} \\ 
&\quad +\dots+(-1)^n\prod_{i_1=1} ^n\lambda_{i_1},
\end{align*}
where $1\leq k\leq n-1$. So, the dependent radius is the $(n+1)$-th 
radius associated to the radii $\lambda_1,\lambda_2,\dots,
\lambda_{n-1},\lambda_{n}$.

For $n=1$, let $\lambda_1\in {\mathbb R}$, then we call dependent radius 
representing with $\lambda_2$ the quantity (when is defined) given by the formula
\begin{equation}
\lambda_2=\lambda_2(\lambda_1):=\frac{24-4\lambda_1} {4-\lambda_1}.
\label{dependent radius 1}
\end{equation}
So, in this case the dependent radius is the second radius associated to 
the radius $\lambda_1$.
\end{definition}

For the dependent radius we have the following result. 
The proof is given in the Appendix.

\begin{proposition}
\label{prp:criterion radius}
If $(\lambda_1,\lambda_2,\dots,\lambda_{n-1},\lambda_{n})\in V^n$, 
$n\in {\mathbb N}$, $n\geq2$, then the dependent radius 
$\lambda_{n+1}=\lambda_{n+1}(\lambda_1,\lambda_2,\dots, \lambda_{n-1},\lambda_{n})$, 
$n\in {\mathbb N}$, $n\geq2$, is positive. 
If $n=1$ and suppose that $\lambda_1\in V^{1}$, then the dependent radius 
$\lambda_2=\lambda_2(\lambda_1)$ is positive.

On the other hand if $\lambda_1,\lambda_2,\dots,\lambda_{n-1},\lambda_{n}$, 
$n\in {\mathbb N}$, $n\geq2$, are distinct positive real numbers so that the 
dependent radius $\lambda_{n+1}$, $n\in {\mathbb N}$, $n\geq2$, associated with
 the radii $\lambda_1,\lambda_2,\dots,\lambda_{n-1},\lambda_{n}$ is positive, 
then $(\lambda_1,\lambda_2,\dots,\lambda_{n-1},\lambda_{n})\in V^n$, 
$n\in {\mathbb N}$, $n\geq2$. If $\lambda_1$ is a positive real number so that 
the dependent radius $\lambda_2$ associated to the radius $\lambda_1$ is positive, 
then $\lambda_1\in V^{1}$.
\end{proposition}

\begin{remark} \rm
\label{rmrk:radius}
According to Proposition \ref{prp:criterion radius}, the set $V^n$ is 
the biggest set from which we can choose the points 
$(\lambda_1,\lambda_2,\dots, \lambda_{n})$ so that the corresponding dependent 
radius $\lambda_{n+1}$ given by \eqref{dependent radius n}, is positive if 
$n\in {\mathbb N}, \, n\geq2$ and the set $V^{1}$ is the biggest set from which 
we can choose the numbers $\lambda_1$ so that the corresponding dependent radius 
$\lambda_2$ given by \eqref{dependent radius 1}, is positive.
\end{remark}

Now, is following the definition which has the central role. The advantage 
played by this definition is that the square roots of the coordinates 
$\lambda_1,\lambda_2,\dots,\lambda_{n},\lambda_{n+1}$ (where $\lambda_{n+1}$ 
is the dependent radius associated to the radii 
$\lambda_1,\lambda_2,\dots, \lambda_{n-1},\lambda_{n}$) of the points 
$(\lambda_1,\lambda_2, \dots,\lambda_{n},\lambda_{n+1})$, where 
$n \in  {\mathbb N}$, $n \geq 2$, are going to be the limiting radii of the 
limit cycles which are asymptotic to circles of radii $\sqrt{\lambda_{i}}$ 
for $i=1,2,\dots,n,n+1$ centered at the origin when the small positive 
parameter of our system tending to 0. This will be done by forcing 
$\sqrt{\lambda_1},\sqrt{\lambda_2},\dots, \sqrt{\lambda_{n}},
\sqrt{\lambda_{n+1}}$ to be all the simple roots of the polynomial $F$ 
defined in \eqref{FA} (see Theorem \ref{thm:method}). For this reason, 
we assume that the given points $(\lambda_1,\lambda_2,\dots,\lambda_{n})$
 belong to $V^n$ and we want for the corresponding dependent radius 
$\lambda_{n+1}=\lambda_{n+1}(\lambda_1,\lambda_2,\dots, \lambda_{n-1},\lambda_{n})$ 
(which from Proposition \ref{prp:criterion radius} is positive) to satisfy that 
$\lambda_{n+1}\ne\lambda_{j}$ for all $j=1,2,\dots,n$. In this way, we construct 
$n+1$ limit cycles for system \eqref{g VdP}, and so we achieve the maximal
 number of limit cycles due to polynomial perturbations of degree $2n+3$ of 
the harmonic oscillator (see \cite{I}).

The same observation of all the above is valid and in the case where $n=1$.

\begin{definition} \rm
We define now for $n\in {\mathbb N}$, $n\geq2$, the sets
\begin{align*}
S_{1,n+1}^n:=\big\{&(\lambda_1,\lambda_2,\dots,\lambda_{n}, 
 \lambda_{n+1}):(\lambda_1,\lambda_2,\dots,\lambda_{n}) \in S_1^n,\, 
 \lambda_{n+1}=\lambda_{n+1}(\lambda_1,\dots, \lambda_{n}), \\
& \lambda_{n+1}\ne\lambda_{j}\;\forall  j=1,2,\dots,n \big\}, \\
S_{2,n+1}^n:=\big\{&(\lambda_1,\lambda_2,\dots,\lambda_{n}, 
 \lambda_{n+1}):(\lambda_1,\lambda_2,\dots,\lambda_{n}) \in S_2^n,\, 
 \lambda_{n+1}=\lambda_{n+1}(\lambda_1,\dots, \lambda_{n}), \\
& \lambda_{n+1}\ne\lambda_{j}\;\forall  j=1,2,\dots,n \big\}, \\
S_{3,n+1}^n:=\big\{&(\lambda_1,\lambda_2,\dots,\lambda_{n}, 
 \lambda_{n+1}):(\lambda_1,\lambda_2,\dots,\lambda_{n}) \in S_3^n,\,
 \lambda_{n+1}=\lambda_{n+1}(\lambda_1,\dots, \lambda_{n}), \\
& \lambda_{n+1}\ne\lambda_{j}\;\forall  j=1,2,\dots,n \big\}, \\
S_{4,n+1}^n:=\big\{&(\lambda_1,\lambda_2,\dots,\lambda_{n}, 
 \lambda_{n+1}):(\lambda_1,\lambda_2,\dots,\lambda_{n}) \in S_4^n,\, 
 \lambda_{n+1}=\lambda_{n+1}(\lambda_1,\dots, \lambda_{n}), \\
& \lambda_{n+1}\ne\lambda_{j}\;\forall  j=1,2,\dots,n \big\}, \\
S_{5,n+1}^n:=\big\{&(\lambda_1,\lambda_2,\dots,\lambda_{n}, 
 \lambda_{n+1}):(\lambda_1,\lambda_2,\dots,\lambda_{n}) \in S_5^n,\,
 \lambda_{n+1}=\lambda_{n+1}(\lambda_1,\dots, \lambda_{n}), \\
& \lambda_{n+1}\ne\lambda_{j}\;\forall  j=1,2,\dots,n \big\}, \\
S_{6,n+1}^n:=\big\{&(\lambda_1,\lambda_2,\dots,\lambda_{n}, 
 \lambda_{n+1}):(\lambda_1,\lambda_2,\dots,\lambda_{n}) \in S_6^n,\, 
 \lambda_{n+1}=\lambda_{n+1}(\lambda_1,\dots, \lambda_{n}), \\
& \lambda_{n+1}\ne\lambda_{j}\;\forall  j=1,2,\dots,n \big\},
\end{align*}
where $S_1^n, S_2^n, S_3^n, S_4^n, S_5^n$ and $S_6^n$ are the  sinusoidal-type 
sets of order $n$, with $n\in {\mathbb N}$, $n\geq2$, of the 
Definition \ref{dfn:sets} and $\lambda_{n+1}$ is the dependent radius 
given by \eqref{dependent radius n}. For $n=1$, we define the sets
\begin{gather*}
S_{1,2}^{1}:=\big\{(\lambda_1,\lambda_2):\lambda_1 \in S_1^1,\,
  \lambda_2=\lambda_2(\lambda_1),\, \lambda_2\ne \lambda_1 \big\}, \\
S_{2,2}^{1}:=\big\{(\lambda_1,\lambda_2):\lambda_1 \in S_2^1,\, 
 \lambda_2=\lambda_2(\lambda_1),\, \lambda_2\ne \lambda_1 \big\},
\end{gather*}
where $S_1^1$ and $S_2^1$ are the sinusoidal-type sets of first order of the 
Definition \ref{dfn:sets} and $\lambda_2$ is the dependent radius given 
by \eqref{dependent radius 1}.
\end{definition}

\begin{remark} \rm
\label{rmrk:n 1}
Notice that, if $\lambda_1 \in (0,4)$, then the dependent radius $\lambda_2$ 
given by \eqref{dependent radius 1}, belongs to $(6,+\infty)$ and so we have 
that $(\lambda_1,\lambda_2) \in S_{1,2}^{1}$. If $\lambda_1 \in (6,+\infty)$, 
then the dependent radius $\lambda_2$ given by \eqref{dependent radius 1}, 
belongs to $(0,4)$ and so we have that $(\lambda_1,\lambda_2) \in S_{2,2}^{1}$.
\end{remark}

\begin{remark} \rm
According to Remark \ref{rmrk:n 1}, the sets $S_{1,2}^{1}$ and $S_{2,2}^{1}$, 
which defined as above, take the more simple form
\begin{gather*}
S_{1,2}^{1}=\big\{(\lambda_1,\lambda_2): \lambda_1 \in S_1^1,\,
  \lambda_2=\lambda_2(\lambda_1) \big\}, \\
S_{2,2}^{1}=\big\{(\lambda_1,\lambda_2): \lambda_1 \in S_2^1,\, 
 \lambda_2=\lambda_2(\lambda_1) \big\},
\end{gather*}
where $S_1^1$ and $S_2^1$ are the sinusoidal-type sets of first order of the
 Definition \ref{dfn:sets} and $\lambda_2$ is the dependent radius given 
by \eqref{dependent radius 1}.
\end{remark}

\begin{definition} \rm
We define for $n\in {\mathbb N}$, $n\geq2$, the set $V_{n+1}^n$ as the set
\[
V_{n+1}^n:=\cup_{i=1}^{6}S_{i,n+1}^n.
\]
For $n=1$, we define the set $V_2^{1}$ as the set
\[
V_2^{1}:=S_{1,2}^{1}\cup S_{2,2}^{1}.
\]
\end{definition}

\begin{remark} \rm
In the set $V_{n+1}^n, \, n\in {\mathbb N}$, the positive dependent radius 
$\lambda_{n+1}, \, n\in {\mathbb N}$, is obviously different from 
$\lambda_1,\lambda_2,\dots,\lambda_{n-1},\lambda_{n}, \, n\in {\mathbb N}$.
\end{remark}

\begin{remark} \rm
It is possible, for $n  \in  {\mathbb N}$, $n\geq2$, the point 
$(\lambda_1,\lambda_2,\dots, \lambda_{n})  \in  V^n$ but the point 
$(\lambda_1,\lambda_2,\dots,\lambda_{n},\lambda_{n+1}) \notin V_{n+1}^n$,
 where $\lambda_{n+1}$ is the dependent radius given by \eqref{dependent radius n}, 
associated to the radii $\lambda_1,\lambda_2,\dots,\lambda_{n}$. 
(For example, it is easy to see that the point $(4,6)  \in V^2$; in particular
 belongs to $S_3^2$. We calculate the dependent radius $\lambda_3$,
associated to the $4,6$, which from Proposition \ref{prp:criterion radius} 
is positive and we have that $\lambda_3=6$. Now, the point
$(4,6,6) \notin V_3^2$.)

For $n=1$, according to Remark \ref{rmrk:n 1}, if $\lambda_1 \in V^{1}$,
 then $(\lambda_1,\lambda_2) \in V_2^{1}$, where $\lambda_2$ is the dependent 
radius given by \eqref{dependent radius 1}, associated to the radius $\lambda_1$.
\end{remark}

\begin{definition}[$\varLambda$-points (lambda points)] \rm
\label{dfn:lambda points}
We will call the points
\[
(\lambda_1,\lambda_2,\dots,\lambda _{n},\lambda_{n+1}) 
\in V_{n+1}^n, \quad n\in {\mathbb N},
\]
the $\varLambda$-points.
\end{definition}

\subsection{Known results on Li\'enard equations}

We now continue with known results on Li\'enard equations. 
Such equations are very challenging and many questions about these are still open. 
Note that Smale \cite{S} proposed to study Hilbert's 16th problem restricted 
to these special classes. This is Smale's 13th problem. We mention several 
interesting works here. The Li\'enard equation
\begin{equation}
\ddot{x}+g(x)\dot{x}+x=0,
\label{Lienard}
\end{equation}
where $g$ is a polynomial, is another generalization of the Van der Pol equation. 
Equation \eqref{Lienard} can be studied in a phase plane as a system
\begin{equation}
\begin{gathered}
\dot{x}=y, \\
\dot{y}=-x-g(x)y,
\end{gathered}
\label{Lienard s}
\end{equation}
or in the so-called Li\'enard plane as
\begin{equation}
\begin{gathered}
\dot{x}=y-G(x), \\
\dot{y}=-x,
\end{gathered}
\label{Lienard p}
\end{equation}
where $G(x)=\int_{0}^{x}g(s)\,ds$. The systems \eqref{Lienard s} and 
\eqref{Lienard p} are analytically conjugate. We observe that system 
\eqref{g VdP} is not of the form of Li\'enard's equation \eqref{Lienard s}, 
except when $f(y) = y$. Obviously, for $f(y) \ne y$, \eqref{Lienard s} can 
not reduce to \eqref{g VdP}. So, in general, \eqref{g VdP} is not a special 
case of \eqref{Lienard s} and \eqref{Lienard s} is not a special case of 
\eqref{g VdP}.

Li\'enard \cite{L} proved that, if $G$ is a continuous odd function, which 
has a unique positive root at $x = a$ and is monotone increasing for $x \geq a$, 
then \eqref{Lienard p} has a unique limit cycle. Rychkov \cite{R} proved that,
 if $G$ is an odd polynomial of degree 5, then \eqref{Lienard p} has at most two 
limit cycles.

Lins, de Melo and Pugh \cite{LdeMP} have studied the Li\'enard equation
 \eqref{Lienard p}, where $G$ is a polynomial of degree $d$. They proved that, 
if $G(x) = a_3x^3 + a_2x^2 + a_1x$, then \eqref{Lienard p} has at most
one limit cycle. In fact, they gave a complete classification of the phase 
space of the cubic Li\'enard's equation, in terms of some explicit algebraic 
conditions on the coefficients of $G$. Also, using a method due to Poincar\'e 
they proved that, if $d=2n+1$ or $2n+2$, then for any $k \in {\mathbb N}_{0}$ 
with $0 \le k \le n$ there exists a polynomial $G(x) = a_{d}x^{d} + \dots + a_1x$ 
such that the system \eqref{Lienard p} has exactly $k$ closed orbits. 
Motivated by this, they conjectured that the maximum number of limit cycles 
for system \eqref{Lienard p}, where $G$ is a polynomial of degree $n$ would 
be equal to $[\frac{n-1}{2}]$.

However, in \cite{DPR} it has been proven by Dumortier, Panazzolo and Roussarie 
the existence of classical Li\'enard equations \eqref{Lienard p} of degree 7
 with at least 4 limit cycles. This easily implied the existence of classical 
Li\'enard equations of degree $n, \, n \geq 7$, with $[\frac{n-1}{2}]+1$ 
limit cycles. The counterexamples were proven to occur in systems
\[
\begin{gathered}
\dot{x}=y-\Big(x^{7}+\sum_{i=2}^{6}c_{i}x^{i}\Big), \\
\dot{y}=\varepsilon(b-x),
\end{gathered}
\]
for small $\varepsilon > 0$. Recently, in \cite{MD} it has been proven 
by De Maesschalck and Dumortier the existence of classical Li\'enard 
equations \eqref{Lienard p} of degree 6 having 4 limit cycles. 
It implies the existence of classical Li\'enard equations of degree 
$n$, $n \geq 6$, having at least $[\frac{n-1}{2}]+2$ limit cycles.

Ilyashenko and Panov \cite{IP} proved that, if
\[
G(x)=x^n+\sum_{i=1}^{n-1}a_{i}x^{i}, \quad |a_{i}| \leq C, \quad C \geq 4, 
\quad n \geq 5,
\]
and suppose that $n$ is odd, then the number $L(n,C)$ of limit cycles 
of \eqref{Lienard p} admits the  upper estimate
\[
L(n,C) \leq \exp(\exp\,C^{14n}).
\]
Caubergh and Dumortier \cite{CD} proved that the maximal number of 
limit cycles for \eqref{Lienard p} of even degree is finite when 
restricting the coefficients to a compact, thus proving the existential 
part of Hilbert's 16th problem for Li\'enard equations when restricting 
the coefficients to a compact set.

\section{Elementary remarks about small perturbation of a Hamiltonian system}

We consider the system
\begin{equation}
\begin{gathered}
\dot{x}=y+\varepsilon f_1(x,y), \\
\dot{y}=-x+\varepsilon f_2(x,y),
\end{gathered}
\label{f1,f2}
\end{equation}
where $0 < \varepsilon \ll 1$ and $f_1, \, f_2$ are $C^{1}$ functions 
of $x$ and $y$, which is a perturbation of the linear harmonic oscillator
\[
\begin{gathered}
\dot{x}=y, \\
\dot{y}=-x,
\end{gathered}
\]
which has all the solutions periodic with:
\[
x^{0}(t) = A\cos(t-t_{0}) \quad \text{and} \quad y^{0}(t) = -A\sin(t-t_{0}).
\]

In general, the phase curves of \eqref{f1,f2} are not closed and it is possible 
to have the form of a spiral with a small distance of order $\varepsilon$ 
between neighboring turns. In order to decide if the phase curve approaches 
the origin or recedes from it, we consider the function (mechanic energy)
\[
E(x,y)=\frac{1}{2}\big(x^2+y^2\big).
\]
It is easy to compute the derivative of the energy and it is proportional 
to $\varepsilon$:
\begin{equation}
\frac{d}{dt}E(x,y)=x\dot{x}+y\dot{y}=\varepsilon\big(xf_1(x,y)
+ yf_2(x,y)\big)=:\varepsilon\dot{E}(x,y).
\label{energy}
\end{equation}
We want information for the sign of the quantity
\begin{equation}
\int_{0}^{T(\varepsilon)}\varepsilon\dot{E}\big(x^{\varepsilon}(t), 
y^{\varepsilon}(t)\big)dt=:\Delta E,
\label{quantity}
\end{equation}
which corresponds to the change of energy of 
$(x^{\varepsilon}(t),y^{\varepsilon}(t))$ in one complete turn:
$y^{\varepsilon}(0) = y^{\varepsilon}(T(\varepsilon))=0.$
Using the theorem of continuous dependence on parameters in ODEs, one can 
prove the following lemma (see \cite{A}):

\begin{lemma}
For \eqref{quantity} we have
\begin{equation}
\Delta E=\varepsilon\int_{0}^{2\pi}\dot{E}\big(A\cos(t-t_{0}),
 -A\sin(t-t_{0})\big)dt+o(\varepsilon).
\label{quantity 1}
\end{equation}
\end{lemma}

Let
\begin{equation}
F(A):=\int_{0}^{2\pi}\dot{E}\big(x^{0}(t),y^{0}(t)\big)dt,
\label{FA}
\end{equation}
and we write \eqref{quantity 1} as
\[
\Delta E = \varepsilon\Big[F(A)+\frac{o(\varepsilon)}{\varepsilon}\Big].
\]

Using the implicit function theorem, one can prove the following theorem, 
which is the Poincar\'e's method (see \cite{A}):

\begin{theorem}
\label{thm:method}
If the function $F$ given by \eqref{FA}, has a positive simple root $A_{0}$, namely
\[
F(A_{0}) = 0 \quad \text{and} \quad F'(A_{0}) \neq 0,
\]
then \eqref{f1,f2} has a periodic solution with amplitude 
$A_{0} + O(\varepsilon)$ for $0 < \varepsilon \ll 1$.
\end{theorem}

\section{Proofs of Theorems \ref{thm:criterion}, \ref{thm:criterion extra}, 
\ref{thm:criterion 1} and \ref{thm:criterion 1 extra}}


\begin{proof}[Proof of Theorem \ref{thm:criterion}]
 From \eqref{energy} we have
\begin{equation}
\dot{E}(x,y)=yf(y)\big(1-x^2\big),
\label{energy 1}
\end{equation}
where $f$ is the polynomial introduced in \eqref{definition of f}. 
Substituting \eqref{energy 1} into \eqref{FA}, we obtain that
\begin{equation}
F(A)=\int_{0}^{2\pi}y^{0}(t)f(y^{0}(t))\big(1-(x^{0}(t))^2\big)dt,
\label{FA 1}
\end{equation}
where $f$ is the polynomial introduced in \eqref{definition of f}. 
We insert the definition of $f$ given by \eqref{definition of f} 
in \eqref{FA 1} to obtain
\begin{equation}
\begin{aligned}
F(A)&=\tau  \int_{0}^{2\pi}\Big[(y^{0}(t))^{2(n+1)}+(2n+1)
\Big(1- \frac{1}{2(n+2)}\sum_{i_1=1}^{n+1}\lambda_{i_1}\Big) (y^{0}(t))^{2n} \\ 
&\quad +\dots+\Big(\frac{1\cdot3\dots(2n+1)}{2^{n+1}(n+2)!}(-1)^n
\prod _{i_1=1}^{n+1}\lambda_{i_1}\Big)(y^{0}(t))^2\Big]\big(1- (x^{0}(t))^2\big)dt.
\label{FA 2}
\end{aligned}
\end{equation}
Substituting $x^{0}(t)=A\cos(t-t_{0})$ and $y^{0}(t)=-A\sin(t-t_{0})$ 
into \eqref{FA 2} we get
\begin{align*}
F(A)&=\tau A^2\int_{0}^{2\pi}\Big[A^{2n}\sin^{2(n+1)}(t-t_{0})
\\ 
&\quad +(2n+1)\Big(1-\frac{1}{2(n+2)}\sum_{i_1=1}^{n+1}\lambda_{i_1 }
 \Big)A^{2(n-1)}\sin^{2n}(t-t_{0}) \\ 
&\quad +\dots+\Big(\frac{1\cdot3\dots(2n+1)}{2^{n+1}(n+2)!}(-1)^n
 \prod _{i_1=1}^{n+1}\lambda_{i_1}\Big)\sin^2(t-t_{0})\Big] \\ 
&\quad \times \big[1-A^2+A^2\sin^2(t-t_{0})\big]dt,
\end{align*}
whence, after multiplying the terms in the two brackets we get
\begin{align*}
F(A)&=\tau A^2\int_{0}^{2\pi}\Big[A^{2n}\sin^{2(n+1)}(t-t_{0}) 
 -A^{2(n+1)}\sin^{2(n+1)}(t-t_{0}) \\ 
&\quad  +A^{2(n+1)}\sin^{2(n+2)}(t-t_{0}) \\ 
& \quad +(2n+1)\Big(1-\frac{1}{2(n+2)}\sum_{i_1=1}^{n+1}\lambda_{i_1 }
 \Big)A^{2(n-1)}\sin^{2n}(t-t_{0}) \\ 
&\quad -(2n+1)\Big(1-\frac{1}{2(n+2)}\sum_{i_1=1}^{n+1}\lambda_{i_1 }\Big)
 A^{2n}\sin^{2n}(t-t_{0}) \\ 
&\quad +(2n+1)\Big(1-\frac{1}{2(n+2)}\sum_{i_1=1}^{n+1}\lambda_{i_1 }\Big)
 A^{2n}\sin^{2(n+1)}(t-t_{0}) \\ 
&\quad +\dots+\Big(\frac{1\cdot3\dots(2n+1)}{2^{n+1}(n+2)!}(-1)^n
 \prod _{i_1=1}^{n+1}\lambda_{i_1}\Big)\sin^2(t-t_{0}) \\
&\quad -\Big(\frac{1\cdot3\dots(2n+1)}{2^{n+1}(n+2)!}(-1)^n\prod_{i_1 =1}^{n+1}
 \lambda_{i_1}\Big)A^2\sin^2(t-t_{0}) \\ 
&\quad +\Big(\frac{1\cdot3\dots(2n+1)}{2^{n+1}(n+2)!}(-1)^n
 \prod_{i_1 =1}^{n+1}\lambda_{i_1}\Big)A^2\sin^{4}(t-t_{0})\Big]dt.
\end{align*}
Using now \eqref{sinusoidal}, we finally obtain
\begin{align*}
F(A)&= \pi\tau A^2\frac{1\cdot3\dots(2n+1)}{2^{n+1}(n+2)!}
  \Big[-A^{2(n+1)}+\sum_{i_1=1}^{n+1}\lambda_{i_1}A^{2n}
 -\sum _{\substack{i_1,i_2=1\\i_1<i_2}}^{n+1}\lambda_{i_1}  \lambda_{i_2}A^{2(n-1)} \\
 & \quad +\dots-(-1)^{k}\sum_{\substack{i_1,\dots,i_{k}=1\\i_1<\dots <i_{k}}}^{n+1}
 \lambda_{i_1}\dots\lambda_{i_{k}}A^{2(n-k+1)} -\dots \\
 &\quad -(-1)^n\sum_{\substack{i_1,\dots,i_{n}=1\\i_1<\dots <i_{n}}}^{n+1}
 \lambda_{i_1}\dots\lambda_{i_{n}}A^2+(-1)^n \prod_{i_1=1}^{n+1}\lambda_{i_1}\Big].
\end{align*}

We show now that $\sqrt{\lambda_1},\sqrt{\lambda_2},\dots ,
\sqrt{\lambda_{n}},\sqrt{\lambda_{n+1}}$ are roots of the polynomial $F$. Let
\begin{align*}
W(A)&:= A^{2(n+1)}-\sum_{i_1=1}^{n+1}\lambda_{i_1}A^{2n}  +\sum_{\substack{i_1,i_2=1\\i_1<i_2}}^{n+1}\lambda_{i_1}  \lambda_{i_2}A^{2(n-1)}-\dots
\\ 
&\quad +(-1)^n\sum_{\substack{i_1, \dots,i_{n}=1\\i_1<\dots<i_{n}}}^{n+1}
\lambda_{i_1}\dots \lambda_{i_{n}}A^2-(-1)^n\prod_{i_1=1}^{n+1}\lambda_{i_1},
\end{align*}
namely, we write
\[
F(A)=-\pi\tau A^2\frac{1\cdot3\dots(2n+1)}{2^{n+1}(n+2)!}\cdot W(A).
\]
Now, it suffices to show that $\sqrt{\lambda_1},\sqrt{\lambda_2 },\dots,
\sqrt{\lambda_{n}},\sqrt{\lambda_{n+1}}$ 
are roots of the polynomial $W$. Without loss of generality we consider 
the quantity $\sqrt{\lambda _1}$. For $W\big(\sqrt{\lambda_1}\,\big)$ we have
\begin{align*}
W\big(\sqrt{\lambda_1}\,\big) &=  \lambda_1^{n+1}-\lambda_1^n \sum_{i_1=1}^{n+1}
\lambda_{i_1}+\lambda_1^{n-1}\sum _{\substack{i_1,i_2=1\\i_1<i_2}}^{n+1}
\lambda_{i_1} \lambda_{i_2}-\dots-(-1)^n\prod_{i_1=1}^{n+1}\lambda_{i_1} \\
&= \lambda_1^{n+1}-\lambda_1^{n+1}-\lambda_1^n\sum_{i_1 =2}^{n+1}\lambda_{i_1}
 +\lambda_1^n\sum_{i_1=2}^{n+1}\lambda _{i_1}+\lambda_1^{n-1}
 \sum_{\substack{i_1,i_2=2\\i_1 <i_2}}^{n+1}\lambda_{i_1}\lambda_{i_2} \\ 
&\quad -\lambda_1^{n-1} \sum_{\substack{i_1,i_2=2\\i_1<i_2}}^{n+1}\lambda_{i_1} 
\lambda_{i_2} -\lambda_1^{n-2}\sum_{\substack{i_1,i_2,i_3=2\\i_1<i_2 <i_3}}^{n+1}
 \lambda_{i_1}\lambda_{i_2}\lambda_{i_3} \\
&\quad +\lambda_1^{n-2}\sum_{\substack{i_1,i_2,i_3=2\\i_1<i_2 <i_3}}^{n+1}
 \lambda_{i_1}\lambda_{i_2}\lambda_{i_3}+\dots +(-1)^n\prod_{i_1=1}^{n+1}
 \lambda_{i_1}-(-1)^n\prod_{i_1=1}^{n+1}\lambda_{i_1}\\
&=  0.
\end{align*}
So, $\sqrt{\lambda_1},\sqrt{\lambda_2},\dots,\sqrt{\lambda_{n}} ,
\sqrt{\lambda_{n+1}}$ are roots of the polynomial $W$ and therefore and for 
the polynomial $F$.

Now, using Theorem \ref{thm:method}, it suffices to show that 
$\sqrt{\lambda_1},\sqrt{\lambda_2},\dots,\sqrt{\lambda_{n}}, \sqrt{\lambda_ {n+1}}$ 
are not roots of the polynomial $W'$; therefore they are not roots and for
 polynomial $F'$. For the derivative of $W$ we have that
\begin{align*}
W'(A)&=2A\Big[(n+1)A^{2n}-n\sum_{i_1=1}^{n+1}\lambda_{i_1}  
 A^{2(n-1)}+(n-1)\sum_{\substack{i_1,i_2=1\\i_1<i_2}}^ {n+1}\lambda_{i_1}
 \lambda_{i_2}A^{2(n-2)} \\
&\quad -\dots+(-1)^n\sum_{\substack{i_1,\dots,i_{n}=1\\i_1<\dots 
<i_{n}}}^{n+1}\lambda_{i_1}\dots\lambda_{i_{n}}\Big].
\end{align*}
Now, we have that one root of $W'$ is $A=0$ and we also have another $2n$ roots. 
From those $2n$ roots, $n$ are positive and the other $n$ are negative 
(these roots are opposite numbers).
Let
\begin{align*}
G(A)&:= (n+1)A^{2n}-n\sum_{i_1=1}^{n+1}\lambda_{i_1}A^{2(n-1)} 
 +(n-1)\sum_{\substack{i_1,i_2=1\\i_1<i_2}}^{n+1}\lambda _{i_1}
 \lambda_{i_2}A^{2(n-2)} \\
&\quad -\dots+(-1)^n\sum_{\substack{i_1,\dots,i_{n}=1\\i_1<\dots 
<i_{n}}}^{n+1}\lambda_{i_1}\dots\lambda_{i_{n}},
\end{align*}
namely, we write
\[
W'(A)=2A\cdot G(A).
\]
Now, we check if the roots 
$\sqrt{\lambda_1},\sqrt{\lambda_2}, \dots,\sqrt{\lambda_{n}},\sqrt{\lambda_{n+1}}$ 
of the polynomial $W$ are possible to be roots and for the polynomial $W'$, 
therefore and for the polynomial $G$. Without loss of generality we consider 
the root $\sqrt{\lambda_1}$ of $W$. For $G\big(\sqrt{\lambda_1}\,\big)$ we have
\begin{align*}
G\big(\sqrt{\lambda_1}\,\big)
&= (n+1)\lambda_1^n-n\lambda_1 ^{n-1}\sum_{i_1=1}^{n+1}\lambda_{i_1}+(n-1)
 \lambda_1^{n-2} \sum_{\substack{i_1,i_2=1\\i_1<i_2}}^{n+1}\lambda_{i_1} 
  \lambda_{i_2} \\
&\quad -\dots+(-1)^n\sum_{\substack{i_1,\dots,i_{n}=1\\i_1<\dots <i_{n}}}^{n+1}
 \lambda_{i_1}\dots\lambda_{i_{n}} \\
&=  \lambda_1^n-\lambda_1^{n-1}\sum_{i_1=2}^{n+1}\lambda _{i_1}
 +\lambda_1^{n-2}\sum_{\substack{i_1,i_2=2\\i_1< i_2}}^{n+1}
 \lambda_{i_1}\lambda _{i_2}-\lambda_1^{n-3}\sum _{\substack{i_1,i_2,i_3
 =2\\i_1<i_2<i_3}}^{n+1}\lambda _{i_1}\lambda_{i_2}\lambda_{i_3} \\
&\quad +\dots-(-1)^n\lambda_1\sum_{\substack{i_1,\dots,i_{n}=2 
 \\i_1<\dots<i_{n}}}^{n+1}\lambda_{i_1}\dots\lambda_{i_{n}} 
 +(-1)^n\lambda_2\lambda_3\dots\lambda_{n}\lambda_{n+1} \\
&=  (\lambda_1-\lambda_2)(\lambda_1-\lambda_3)\dots
(\lambda_1-\lambda_{n})(\lambda_1-\lambda_{n+1}).
\end{align*}
Obviously, $W'\big(\sqrt{\lambda_1}\big)$ is not zero since in the set 
$V_{n+1}^n$ we have that $\lambda_1\ne\lambda_{j}$ for  $j=2,3,\dots,n,n+1$. 
Similarly, none of the 
$\sqrt{\lambda_2},\sqrt{\lambda_3},\dots,\sqrt{\lambda_{n}}, \sqrt{\lambda_{n+1}}$
is a root of $W'$.

Therefore, we have that the roots 
$\sqrt{\lambda_1},\sqrt {\lambda_2},\dots,\sqrt{\lambda_{n}},\sqrt{\lambda_{n+1}}$ 
of  $W$ are not roots of $W'$. Finally, none of the roots 
$\sqrt{\lambda_1},\sqrt{\lambda_2},\dots,\sqrt {\lambda_{n}},\sqrt{\lambda_{n+1}}$ 
of $F$ is a root of $F'$. That is essential so that the $n+1$ simple roots of $F$
 create $n+1$ limit cycles. Hence, from Poincar\'e's method 
(see Theorem \ref{thm:method}) it follows that \eqref{g VdP}, with $f$ be 
the polynomial introduced in \eqref{definition of f}, has at least $n+1$ 
limit cycles, and are asymptotic to circles of radius $\sqrt{\lambda_{i}}$ 
for $i=1,2,\dots,n+1$ centered at the origin as $\varepsilon \to 0$.

Let now prove that the number of limit cycles for system \eqref{g VdP}, with 
$\varepsilon$ small and $f$ be the polynomial introduced in \eqref{definition of f}, 
is exactly $n+1$. The proof of this can be derived from the work of Iliev \cite{I} 
since it constitutes a special case of the Theorem 1 proved there. Actually, 
applying this theorem from \cite{I} for the special case $k=1$, since the degree 
of \eqref{g VdP} is $2n+3$ we can obtain at most $n+1$ limit cycles. 
Finally, combining this result with the result that \eqref{g VdP}, with $f$ 
be the polynomial introduced in \eqref{definition of f}, has at least $n+1$ 
limit cycles we get the desired result, namely that the number of limit cycles 
for system \eqref{g VdP}, with $\varepsilon$ small and $f$ be the polynomial 
introduced in \eqref{definition of f} is exactly $n+1$.

Now, concerning the stability of the limit cycles we have the following.

From now on we will suppose for the coordinates
 $\lambda_1,\lambda_2,\dots, \lambda_{n},\lambda_{n+1}$ of the points 
$(\lambda_1,\lambda_2,\dots, \lambda_{n},\lambda_{n+1})  \in  V_{n+1}^n$, 
$ n  \in  {\mathbb N}$, $n \geq 2$, an ordering such that 
$\lambda_1 <\lambda_2<\dots< \lambda_{n}<\lambda_{n+1}$. We can always achieve 
this since the positive real numbers 
$\lambda_1,\lambda_2,\dots, \lambda_{n},\lambda_{n+1}$ are distinct.
 Note that in this order with $\lambda_{n+1}$ we do not necessary mean 
the dependent radius. Since
\[
G\big(\sqrt{\lambda_1}\,\big)=(\lambda_1-\lambda_2)
(\lambda_1 -\lambda_3)\dots(\lambda_1-\lambda_{n})(\lambda_1-\lambda_ {n+1}),
\]
we have that $W'\big(\sqrt{\lambda_1}\big)<0$ if $n$ is odd and that
 $W'\big(\sqrt{\lambda_1}\big)>0$ if $n$ is even.

So using the fact that, if $F'\big(\sqrt{\lambda_{i}}\big)<0$ the limit cycle
 $x^2+y^2=\lambda_{i}+O(\varepsilon)$ is stable and if 
$F'\big(\sqrt{\lambda_{i}}\big)>0$ the limit cycle is unstable we have for the 
stability of the $n+1$ limit cycles that, if $\tau>0$ (respectively $\tau<0$)
\[
x^2+y^2=\lambda_1+O(\varepsilon), \quad x^2+y^2=\lambda_3+O(\varepsilon),
\dots, \quad x^2+y^2=\lambda_{n+1}+O(\varepsilon)
\]
 are stable (respectively unstable) and
\[
x^2+y^2=\lambda_2+O(\varepsilon), \quad
x^2+y^2=\lambda_{4}+ O(\varepsilon),\dots, \quad
x^2+y^2=\lambda_{n}+O(\varepsilon)
\]
are unstable (respectively stable) for $n$ even; and
\[
x^2+y^2=\lambda_1+O(\varepsilon), \quad
x^2 +y^2=\lambda_3+O(\varepsilon),\dots, \quad
x^2+y^2=\lambda_{n}+ O(\varepsilon)
\]
 are unstable (respectively stable) and
\[
x^2+y^2=\lambda_2+O(\varepsilon), \quad
x^2+y^2=\lambda_{4}+ O(\varepsilon),\dots, \quad
x^2+y^2=\lambda_{n+1}+O(\varepsilon)
\]
 are stable (respectively unstable) for $n$ odd.
The proof is complete.
\end{proof}



\begin{proof}[Proof of Theorem \ref{thm:criterion extra}]
 As we already saw, according to Theorem 1 from \cite{I} the number
 of $n+1$ limit cycles is an upper bound for the number of limit cycles 
for system \eqref{g VdP}, where $\varepsilon$ is small and $f$ is an 
arbitrary odd polynomial of degree $2n+1$.

Now, it is easy to see that for system \eqref{g VdP}, where $f$ is an arbitrary 
odd polynomial of degree $2n+1$,  the associated $F$ given by \eqref{FA} 
is an even polynomial of degree $2n+4$, with 0 as a double root. 
Therefore, in general the polynomial $F$ has at most $n+1$ simple positive roots. 
Furthermore, since $V^n, \, n\in {\mathbb N}, \, n\geq2$ is the biggest set 
from which we can choose the points $(\lambda_1,\lambda_2,\dots, \lambda_{n})$ 
so that the dependent radius $\lambda_{n+1}$ given by \eqref{dependent radius n}, 
is positive if $n\in {\mathbb N}, \, n\geq2$ (see Remark \ref{rmrk:radius}) 
and $F$ as we showed has at most $n+1$ simple positive roots, we must choose 
the points $(\lambda_1,\lambda_2,\dots, \lambda_{n},\lambda_{n+1}) \in V_{n+1}^n$, 
in order the polynomial $F$ has exactly $n+1$ simple positive roots, and thus, 
from the set of all the odd polynomials, the polynomials $f$ given by 
\eqref{definition of f} are the only such that the system \eqref{g VdP} 
attains the upper bound of the $n+1$ limit cycles.
The proof is complete.
\end{proof}

\begin{proof}[Proof of Theorem \ref{thm:criterion 1}]
The proof is identical as in Theorem \ref{thm:criterion};  the only 
modification is that the polynomial $f$ given by \eqref{definition of f} 
will be replaced by the polynomial $f$ introduced in \eqref{definition of f 1}.
\end{proof}


\begin{proof}[Proof of Theorem \ref{thm:criterion 1 extra}]
 The proof is identical as in Theorem \ref{thm:criterion extra};
 the only modification is that the case where $n\in {\mathbb N}$, $n\geq2$ will
 be replaced by $n=1$.
\end{proof}

\section{Examples}

In this section we illustrate the general theory of this work by some examples.

\begin{example} \rm
We consider $\lambda_1=4$, $\lambda_2=5$. These $\lambda_1,\lambda_2$ 
are distinct and positive. We have according to Definition \ref{dfn:numbers} 
that the sinusoidal-type numbers of second order, associated to  the
values $4,5$ are
\begin{gather*}
\bar{s}^2:=8+\frac{1}{6}\lambda_1\lambda_2= 8+\frac{1}{6}\cdot4\cdot5 =\frac{34}{3}, \\
\hat{s}^2:=6+\frac{1}{4}\lambda_1\lambda_2= 6+\frac{1}{4}\cdot4\cdot5=11, \\
\tilde{s}^2:=\frac{1}{24}\lambda_1\lambda_2= \frac{1}{24}\cdot4\cdot5=\frac{5}{6}.
\end{gather*}
Since $\lambda_1+\lambda_2=4+5=9$, we have that $(4,5)\in S_2^2$ and 
therefore $(4,5)\in V^2$. We calculate the dependent radius $\lambda_3$,
associated to the values $4,5$, which from Proposition \ref{prp:criterion radius} 
is positive and we have that
\[
\lambda_3:=\frac{192-24(\lambda_1+\lambda_2)+4\lambda_1 \lambda_2}{24-4(\lambda_1+\lambda_2)+\lambda_1\lambda_2} =\frac{192-216+80}{24-36+20}=7.
\]
So, we have the $\varLambda$-point $(4,5,7)$ which belongs to the set $V_3^2$.

Now, using Theorem \ref{thm:criterion}, for $\tau=16$, we have that the system
\begin{equation}
\begin{gathered}
\dot{x}=y, \\
\dot{y}=-x+\varepsilon\big(16y^{5}-80y^3+175y\big)\big(1-x^2\big),
\end{gathered}
\label{4,5}
\end{equation}
with $0<\varepsilon\ll1$ has exactly, the limit cycles 
$x^2+y^2=4+O(\varepsilon)$, $x^2+y^2=5+O(\varepsilon)$ and 
$x^2+y^2=7+O(\varepsilon)$.

Since $4<5<7$, the limit cycles $x^2+y^2=4+O(\varepsilon)$,
 $x^2+y^2=7+O(\varepsilon)$ are stable and the limit cycle 
$x^2+y^2=5+O(\varepsilon)$ is unstable.

From Theorem \ref{thm:criterion} we have for the system \eqref{4,5} that, 
if we change $\tau$ from $16$ to $-16$ the unstable limit cycle 
$x^2+y^2=5+O(\varepsilon)$ becomes stable and the stable limit cycles 
$x^2+y^2=4+O(\varepsilon), \, x^2+y^2=7+ O(\varepsilon)$ become unstable.
\end{example}

\begin{example} \rm
We consider $\lambda_1=4,\lambda_2=16$. These $\lambda_1, \lambda_2$ 
are distinct and positive. We have according to Definition \ref{dfn:numbers} 
that the sinusoidal-type numbers of second order, associated to the $4,16$ are
\begin{gather*}
\bar{s}^2:=8+\frac{1}{6}\lambda_1\lambda_2= 8+\frac{1}{6}\cdot4\cdot16 =\frac{56}{3}, \\
\hat{s}^2:=6+\frac{1}{4}\lambda_1\lambda_2= 6+\frac{1}{4}\cdot4\cdot16=22, \\
\tilde{s}^2:=\frac{1}{24}\lambda_1\lambda_2= \frac{1}{24}\cdot4\cdot16=\frac{8}{3}.
\end{gather*}
Since $\lambda_1+\lambda_2=4+16=20$, we have that $(4,16)\notin S_1^2$, 
$(4,16)\notin S_{5}^2$ and therefore $(4,16)\notin V^2$. Therefore from 
Proposition \ref{prp:criterion radius} the dependent radius $\lambda_3$,
associated to the $4,16$ is not positive.

So, according to the Theorem \ref{thm:criterion} it does not exist a system 
of the form
\[
\begin{gathered}
\dot{x}=y, \\
\dot{y}=-x+\varepsilon\big(a_{0}y^{5}+a_1y^3+a_2y\big) \big(1-x^2\big),
\end{gathered}
\]
where $0 < \varepsilon \ll 1$ and $a_{0},a_1,a_2 \in {\mathbb R}$,
 which has exactly three limit cycles whereof the two of them have the 
equations $x^2+y^2=4+O(\varepsilon)$,  $x^2+y^2=16+ O(\varepsilon)$.
\end{example}

\begin{example} \rm
We consider $\lambda_1=1$, $\lambda_2=2$, $\lambda_3=3$. These
$\lambda_1,\lambda_2,\lambda_3$ satisfy our assertions, since
$\lambda_{i}\ne\lambda_{j}$ for all  $i\ne j$ where $i,j=1,2,3$ and are positive. 
We have according to Definition \ref{dfn:numbers} that the sinusoidal-type 
numbers of third order, associated to the $1,2,3$ are
\begin{gather*}
\bar{s}^3:=10+\frac{1}{8}(\lambda_1\lambda_2+\lambda_1 \lambda_3
 +\lambda_2\lambda_3)-\frac{1}{48}\lambda_1 \lambda_2\lambda_3
=10+\frac{11}{8}-\frac{1}{8}=\frac{45}{4}, \\
\hat{s}^3:=8+\frac{1}{6}(\lambda_1\lambda_2+\lambda_1 \lambda_3
+\lambda_2\lambda_3)-\frac{1}{24}\lambda_1
\lambda_2\lambda_3=8+\frac{11}{6}-\frac{1}{4}=\frac{115}{12}, \\
\tilde{s}^3:=\frac{1}{48}(\lambda_1\lambda_2+ \lambda_1\lambda_3
+\lambda_2\lambda_3)-\frac{1}{96} \lambda _1\lambda_2\lambda_3
=\frac{11}{48}-\frac{1}{16}=\frac{1}{6}.
\end{gather*}
Since $\lambda_1+\lambda_2+\lambda_3=1+2+3=6$, we have that $(1,2,3)\in S_2^3$
and therefore $(1,2,3)\in V^3$. We calculate the dependent radius $\lambda_{4}$,
 associated to the $1,2,3$, which from Proposition \ref{prp:criterion radius} 
is positive and we have that
\[
\lambda_{4}:=\frac{1920-192(\lambda_1+\lambda_2+\lambda_3)+
24(\lambda_1\lambda_2+\lambda_1 \lambda_3+\lambda_2 \lambda_3)
 -4\lambda _1\lambda_2\lambda_3} {192-24(\lambda_1+\lambda_2
 +\lambda_3)+4(\lambda_1
\lambda_2+\lambda_1 \lambda_3+\lambda_2 \lambda_3)
-\lambda _1\lambda_2\lambda_3}=\frac{504}{43}.
\]
So, we have the $\varLambda$-point $(1,2,3,504/43)$ which belongs to the set 
$V_{4}^3$.

Now, using Theorem \ref{thm:criterion}, for $\tau=43/8$, we have that the system
\begin{equation}
\begin{gathered}
\dot{x}=y, \\
\dot{y}=-x+\varepsilon\Big(\frac{43}{8}y^{7}
-\frac{581}{20}y^{5} +\frac{5887}{128}y^3-\frac{1323}{64}y\Big)\big(1-x^2\big),
\end{gathered}
\label{1,2,3}
\end{equation}
with $0<\varepsilon\ll 1$ has exactly, the limit cycles 
$x^2+y^2 =1+O(\varepsilon)$, $x^2 +y^2 =2+ O(\varepsilon)$, 
$x^2 +y^2 =3+O(\varepsilon)$ and $\,x^2+y^2=(504/43)+ O(\varepsilon)$.

Since $1<2<3<504/43$, the limit cycles $x^2+y^2=1+ O(\varepsilon)$, 
$x^2 +y^2 =3+ O(\varepsilon)$ are unstable and the limit cycles 
$x^2 +y^2 =2+ O(\varepsilon), \, x^2 +y^2 =(504/43) + O(\varepsilon)$ are stable.

From Theorem \ref{thm:criterion} we have for the system \eqref{1,2,3} that,
 if we change $\tau$ from $43/8$ to $-43/8$ the unstable limit cycles 
$x^2+y^2 =1+O(\varepsilon), \, x^2+ y^2=3+O(\varepsilon)$ become stable 
and the stable limit cycles $x^2+y^2=2+O(\varepsilon)$, 
$x^2+y^2=(504/43)+ O(\varepsilon)$ become unstable.
\end{example}

\begin{example} \rm
We consider $\lambda_{i}=i$ for $i=1,2,\dots,6$. These 
$\lambda_1,\lambda_2, \lambda_3,\lambda_{4},\lambda_{5},\lambda_{6}$
satisfy our assertions, since $\lambda_{i}\ne\lambda_{j}$ for all  $i\ne j$
where $i,j=1,2,3,4,5,6$ and are positive. It is easy to show, after some 
calculations,  that $(1,2,3,4,5,6)\in V^{6}$. 
We calculate the dependent radius $\lambda_{7}$, 
associated to the $1,2,3,4,5,6$, which from Proposition \ref{prp:criterion radius} 
is positive and we have that $\lambda_{7}=13337/690$. So, we have the 
$\varLambda$-point $(1,2,3,4,5,6,13337/690)$ which belongs to the set $V_{7}^{6}$.

Therefore, from Theorem \ref{thm:criterion} exists a system of the form 
\eqref{g VdP}, where $0 < \varepsilon \ll 1$ and $f$ is an odd polynomial 
of degree $13$, which has exactly the limit cycles: 
$x^2+y^2=1+O(\varepsilon)$, $x^2+y^2=2+O(\varepsilon )$, 
$x^2+y^2=3+O(\varepsilon)$, $x^2+y^2=4+O(\varepsilon)$, $x^2+y^2=5+O(\varepsilon)$,
$x^2+y^2=6+O(\varepsilon)$, $x^2+y^2=(13337/690)+O(\varepsilon)$.
\end{example}

\begin{example} \rm
We consider $\lambda_1=7,\lambda_2=701/100$. It is easy to show, after 
some calculations, that $(7,701/100)\in V^2$. We calculate the dependent 
radius $\lambda_3$, associated to the $7,701/100$, which from Proposition
\ref{prp:criterion radius} is positive and we have that $\lambda_3=5204/1703$.
So, we have the $\varLambda$-point $(7,701/100,5204/1703)$ which belongs 
to the set $V_3^2$.

Now, using Theorem \ref{thm:criterion}, for $\tau=-2179840$, we have that the system
\begin{equation}
\begin{gathered}
\dot{x}=y, \\
\dot{y}=-x+\varepsilon\big( -2179840y^{5}+12351224y^3 -25536028y\big)\big(1-x^2\big),
\end{gathered}
\label{7,701:100}
\end{equation}
with $0<\varepsilon\ll1$ has exactly, the limit cycles 
$x^2+y^2=7+O(\varepsilon)$, $x^2+y^2=(701/100)+O(\varepsilon)$ and 
$x^2+y^2=(5204/1703)+O(\varepsilon)$.

Since $5204/1703<7<701/100$, the limit cycles 
$x^2+y^2=(5204/1703)+O(\varepsilon)$, $x^2+y^2=(701/100)+O(\varepsilon)$ are 
unstable and the limit cycle $x^2+y^2=7+O(\varepsilon)$ is stable.

Since $\sqrt{\lambda_1}$ and $\sqrt{\lambda_2}$ have very small difference
 ($|\sqrt{\lambda_1} - \sqrt{\lambda_2}| 
= | \frac{ 10\sqrt{7}-\sqrt{701}}{10}|\simeq0.0019$), 
the qualitative and quantitative image that one gets using a program, 
may give the misimpression that system \eqref{7,701:100} has a semistable 
limit cycle. This happens because the stable limit cycle 
$x^2 +y^2 =7+ O(\varepsilon)$ lies close enough to the unstable limit 
cycle $x^2 +y^2 =(701/100) + O(\varepsilon)$. This of course is prospective 
since a priori we have chosen the $\lambda _1$ and $\lambda _2$ so as to be 
close enough the one to the other. So, the two limit cycles 
$x^2 +y^2 =7+ O(\varepsilon)$ and $x^2 +y^2 =(701/100) + O(\varepsilon)$, 
create ``one system with one pseudosemistable limit cycle" as we can say, 
since the two limit cycles together behave like a semistable limit cycle.
\end{example}

\begin{remark} \rm
It is easy to see, according to Remark \ref{rmrk:n 1}, that system \eqref{g VdP}
 with $n=1$ can't have ``a system with a pseudosemistable limit cycle" 
as we mean above ``the system with a pseudosemistable limit cycle".
\end{remark}

\begin{example} \rm
\label{example l=7}
We consider $\lambda_1=7$. This $\lambda_1$ belongs to $S_2^{1}$. 
We know from Remark \ref{rmrk:n 1} that the point $(7,\lambda_2) \in S_{2,2}^{1}$, 
where $\lambda_2$ is the dependent radius associated to the $7$.

We calculate the dependent radius $\lambda_2$, associated to the $7$, 
(which from Proposition \ref{prp:criterion radius} is positive) and we have that
\[
\lambda_2:=\frac{24-4\lambda_1}{4-\lambda_1} =\frac{24-28}{4-7}=\frac{4}{3}.
\]
So, we have the $\varLambda$-point $(7,4/3)$ which belongs to the set $V_2^{1}$.

Now, using Theorem \ref{thm:criterion 1}, for $\tau=6$, we have that the system
\begin{equation}
\begin{gathered}
\dot{x}=y, \\
\dot{y}=-x+\varepsilon\big(6y^3-7y\big)\big(1-x^2\big),
\end{gathered}
\label{7}
\end{equation}
with $0<\varepsilon\ll1$ has exactly, the limit cycles $x^2+y^2=7+O(\varepsilon)$
 and $x^2+y^2=(4/3)+O(\varepsilon)$.

Since $4/3<7$, the limit cycle $x^2+y^2=(4/3)+O(\varepsilon)$ is unstable and 
the limit cycle $x^2+y^2=7+O(\varepsilon)$ is stable.

From Theorem \ref{thm:criterion 1} we have for the system \eqref{7} that, 
if we change $\tau$ from $6$ to $-6$ the unstable limit cycle 
$x^2+y^2=(4/3)+O(\varepsilon)$ becomes stable and the stable limit cycle 
$x^2+y^2=7+O(\varepsilon)$ becomes unstable.
\end{example}

\section*{Appendix}

\begin{proof}[Proof of Lemma \ref{lemma:criterion numbers}]
 Clearly, for $n\in {\mathbb N}$, $n \geq 3$,
\begin{align*}
&2(n+2)+\frac{1}{2(n+1)}\sum_{\substack{i_1,i_2 =1\\i_1<i_2}}^n
 \lambda_{i_1}\lambda_{i_2}-\frac{1} {4n(n+1)}
 \sum_{\substack{i_1,i_2,i_3=1\\i_1<i_2<i_3}} ^n
 \lambda_{i_1}\lambda_{i_2}\lambda_{i_3}+\dots \\
&+\frac{(-1)^{k}}{2^{k-1}(n-k+3)\dots(n+1)}
 \sum_{\substack{i_1 ,\dots,i_{k}=1\\i_1<\dots<i_{k}}}^n
 \lambda_{i_1}\dots \lambda_{i_{k}}+\dots+\frac{(-1)^n}{2^{n-2}(n+1)!}
 \prod_{i_1 =1}^n\lambda_{i_1} \\
&= 2(n+1)+\frac{1}{2n}\sum_{\substack{i_1,i_2=1 \\i_1<i_2}}^n
 \lambda_{i_1}\lambda_{i_2}-\frac{1}{4(n-1)n} \sum_{\substack{i_1,i_2,i_3
 =1\\i_1<i_2<i_3}}^n\lambda _{i_1}\lambda_{i_2}\lambda_{i_3}+\dots \\
&\quad +\frac{(-1)^{k}}{2^{k-1}(n-k+2)(n-k+3)\dots n}
 \sum _{\substack{i_1,\dots,i_{k}=1\\i_1<\dots<i_{k}}}^n
 \lambda _{i_1}\dots\lambda_{i_{k}}+\dots+\frac{(-1)^n}{2^{n-1}n!} 
 \prod_{i_1=1}^n\lambda_{i_1}
\end{align*}
if and only if
\begin{align*} 
&\frac{1}{4n(n+1)}
 \sum_{\substack{i_1,i_2=1 \\i_1<i_2}}^n\lambda_{i_1}\lambda_{i_2}
 -\frac{2} {8(n-1)n(n+1)}\sum_{\substack{i_1,i_2,i_3=1\\i_1<i_2 <i_3}}^n
 \lambda_{i_1}\lambda_{i_2}\lambda_{i_3}+\dots \\
&\quad +\frac{(-1)^{k}(k-1)}{2^{k}(n-k+2)(n-k+3)\dots(n+1)}
 \sum _{\substack{i_1,\dots,i_{k}=1\\i_1<\dots<i_{k}}}^n
 \lambda _{i_1}\dots\lambda_{i_{k}}+\dots \\
&\quad +\frac{(-1)^n(n-1)}{2^n (n+1)!}\prod_{i_1=1}^n\lambda_{i_1}=1,
\end{align*}
and the first equivalence has been proved. Similarly, one can prove and
 the rest two equivalences. For $n=2$, (i.e. for the sinusoidal-type numbers 
of second order), it is easy to see that the above equivalences hold.
The proof of the lemma is complete.
\end{proof}

\begin{proof}[Proof of Proposition \ref{prp:criterion radius}]
To prove that the dependent radius $\lambda_{n+1}$, $n\in {\mathbb N}$, $n\geq2$, 
associated to the radii $\lambda_1,\lambda_2,\dots,\lambda_{n}$ is positive 
when $(\lambda_1,\lambda_2,\dots,\lambda_{n})\in V^n$, $n\in {\mathbb N}$, $n\geq2$, 
it suffices to show that both numerator and denominator of 
\eqref{dependent radius n} are of the same sign.

We will check the case where 
$(\lambda_1,\lambda_2,\dots, \lambda_{n-1},\lambda_{n})\in S_1^n$, 
$n\in {\mathbb N}$, $n\geq2$. Similarly, one can prove and the other cases.

In the set $S_1^n$, $n\in {\mathbb N}$, $n\geq2$, we have 
$\sum_{i_1=1}^n\lambda _{i_1}<\bar{s}^n$. Now, for 
$n\in {\mathbb N}$, $n\geq3$, using the definition of $\bar{s}^n$ and 
multiplying the last inequality by $-2^n (n+1)!$ we have
\[
2^{n+1}(n+2)!-2^n(n+1)!\sum_{i_1=1}^n\lambda_{i_1}+2^{n-1} n!
\sum_{\substack{i_1,i_2=1\\i_1<i_2}}^n\lambda_{i_1}  
\lambda_{i_2}-\dots+4(-1)^n\prod_{i_1=1}^n\lambda_{i_1} >0,
\]
which shows that the numerator of \eqref{dependent radius n} is positive.

Since $\sum_{i_1=1}^n\lambda _{i_1}<\bar{s}^n$ in the set $S_1^n$, we have 
that $-\sum_{i_1=1}^n\lambda_{i_1} >-\bar{s}^n$. Using this observation 
we obtain the first inequality for the denominator of \eqref{dependent radius n}
\begin{align*}
&2^n(n+1)!-2^{n-1}n!\sum_{i_1=1}^n\lambda_{i_1}+2^{n-2} (n-1)!
\sum_{\substack{i_1,i_2=1\\i_1<i_2}}^n\lambda _{i_1}\lambda_{i_2}
-\dots+(-1)^n\prod_{i_1=1}^n\lambda _{i_1} \\
&> 2^n(n+1)!-2^{n-1}n!\bar{s}^n+2^{n-2}(n-1)!\sum _{\substack{i_1,i_2=1\\i_1<i_2}}^n
 \lambda_{i_1}\lambda _{i_2}-\dots+(-1)^n\prod_{i_1=1}^n\lambda_{i_1} \\
&= 2^n(n+1)!-2^nn!(n+2)-2^{n-2}\frac{n!}{n+1}\sum _{\substack{i_1,i_2=1\\i_1<i_2}}^n
 \lambda_{i_1}\lambda _{i_2}+\dots-\frac{2(-1)^n}{n+1}\prod_{i_1=1}^n\lambda _{i_1} \\
&\quad +2^{n-2}(n-1)!\sum_{\substack{i_1,i_2=1\\i_1 <i_2}}^n\lambda_{i_1}
 \lambda_{i_2}-\dots+(-1)^n\prod _{i_1=1}^n\lambda_{i_1} \\
&=-2^nn!+2^nn!\tilde{s}^n>-2^nn!+2^nn!=0.
\end{align*}
Here, we have used in the first equality the definition of $\bar{s}^n$ for 
$n\in {\mathbb N}, \, n\geq3$ and in the last inequality that $\tilde{s}^n>1$ 
in the set $S_1^n$.

So, we proved that both numerator and denominator of \eqref{dependent radius n}
 when $n\in {\mathbb N}$, $n\geq3$, are positive, which show that the dependent 
radius $\lambda_{n+1}$ is positive if
 $(\lambda_1,\dots,\lambda_{n})\in S_1^n$, $n\in {\mathbb N}$, $n\geq3$.

In the case where $n=2$, it is easy to show that the dependent radius 
$\lambda_3$ associated to the radii $\lambda_1, \lambda_2$ is positive if
$(\lambda_1,\lambda_2) \in S_1^2$, since in that case both numerator and 
denominator of \eqref{dependent radius n} with $n=2$, are positive.

If $n=1$, it is easy to show that the dependent radius $\lambda_2$ associated 
to the radius $\lambda_1$ is positive if $\lambda_1 \in V^{1}$, since in that 
case both numerator and denominator of \eqref{dependent radius 1} are of 
the same sign.

Let us now show the inverse. 
Let $\lambda_1,\lambda_2, \dots,\lambda_{n-1},\lambda_{n}$, 
$n\in {\mathbb N}$, $n\geq2$, be distinct positive real numbers so that 
the dependent radius $\lambda_{n+1}$, $n\in {\mathbb N}$, $n\geq2$, 
associated to the radii $\lambda_1,\lambda_2,\dots,\lambda_{n}$ is positive.

First, we examine the case where both numerator and denominator of $\lambda_{n+1}$ 
are positive.

Since we suppose that the numerator of $\lambda_{n+1}$ is positive, we have that
\[
2^{n+1}(n+2)!-2^n(n+1)!\sum_{i_1=1}^n\lambda_{i_1}+2^{n-1} n!
\sum_{\substack{i_1,i_2=1\\i_1<i_2}}^n\lambda_{i_1}  
\lambda_{i_2}-\dots+4(-1)^n\prod_{i_1=1}^n\lambda_{i_1} >0,
\]
and dividing this inequality by $-2^n(n+1)!$ we have that 
$\sum_{i_1=1}^n\lambda_{i_1}<\bar{s}^n$.

Since we suppose that the denominator of $\lambda_{n+1}$ is positive, 
we have that
\[ 
2^n(n+1)!-2^{n-1}n!\sum_{i_1=1}^n\lambda_{i_1}+2^{n-2} (n-1)!
\sum_{\substack{i_1,i_2=1\\i_1<i_2}}^n\lambda _{i_1}
\lambda_{i_2}-\dots+(-1)^n\prod_{i_1=1}^n\lambda _{i_1}>0,
\]
and dividing this inequality by $-2^{n-1}n!$ we have that 
$\sum_{i_1=1}^n\lambda_{i_1}<\hat{s}^n$.

Now, we have the following possibilities: 
$\bar{s}^n=\hat{s}^n$ or $\bar{s}^n>\hat{s}^n$ or $\bar{s}^n<\hat{s}^n$, 
where $n \in {\mathbb N}$, $ n\geq2$.

In the case where $\bar{s}^n=\hat{s}^n$ we know from Lemma 
\ref{lemma:criterion numbers} that  $\tilde{s}^n=1$ and hence 
$(\lambda_1,\lambda_2, \dots,\lambda_{n})\in S_3^n$, $n\in {\mathbb N}$,
$ n\geq2$.

In the case where $\bar{s}^n>\hat{s}^n$ we know from Lemma 
\ref{lemma:criterion numbers} that  $\tilde{s}^n<1$ and hence
 $(\lambda_1,\lambda_2, \dots,\lambda_{n})\in S_2^n$, $n\in {\mathbb N}$, $n\geq2$.

In the case where $\bar{s}^n<\hat{s}^n$ we know from Lemma 
\ref{lemma:criterion numbers} that  $\tilde{s}^n>1$ and hence 
$(\lambda_1,\lambda_2, \dots,\lambda_{n})\in S_1^n$, $n\in {\mathbb N}$, $n\geq2$.

Let us now examine the case where both numerator and denominator of 
$\lambda_{n+1}$ are negative.

Since we suppose that the numerator of $\lambda_{n+1}$ is negative, we have that
\[
2^{n+1}(n+2)!-2^n(n+1)!\sum_{i_1=1}^n\lambda_{i_1}
+2^{n-1} n!\sum_{\substack{i_1,i_2=1\\i_1<i_2}}^n\lambda_{i_1}  
\lambda_{i_2}-\dots+4(-1)^n\prod_{i_1=1}^n\lambda_{i_1} <0,
\]
and dividing this inequality by $-2^n(n+1)!$ we have that 
$\sum_{i_1=1}^n\lambda_{i_1}>\bar{s}^n$.

Since we suppose that the denominator of $\lambda_{n+1}$ is negative, we have that
\[ 
2^n(n+1)!-2^{n-1}n!\sum_{i_1=1}^n\lambda_{i_1}+2^{n-2} (n-1)!
\sum_{\substack{i_1,i_2=1\\i_1<i_2}}^n\lambda _{i_1}\lambda_{i_2}-\dots
+(-1)^n\prod_{i_1=1}^n\lambda _{i_1}<0,
\]
and dividing this inequality by $-2^{n-1}n!$ we have that 
$\sum_{i_1=1}^n\lambda_{i_1}>\hat{s}^n$.

Now, we have the following possibilities: 
$\bar{s}^n=\hat{s}^n$ or $\bar{s}^n>\hat{s}^n$ or $\bar{s}^n<\hat{s}^n$, 
where $n \in {\mathbb N}, \, n\geq2$.

In the case where $\bar{s}^n=\hat{s}^n$ we know from Lemma 
\ref{lemma:criterion numbers} that  $\tilde{s}^n=1$ and hence 
$(\lambda_1,\lambda_2, \dots,\lambda_{n})\in S_{6}^n$, $n\in {\mathbb N}$, $n\geq2$.

In the case where $\bar{s}^n>\hat{s}^n$ we know from Lemma 
\ref{lemma:criterion numbers} that  $\tilde{s}^n<1$ and hence 
$(\lambda_1,\lambda_2, \dots,\lambda_{n})\in S_{4}^n$, $n\in {\mathbb N}$, $n\geq2$.

In the case where $\bar{s}^n<\hat{s}^n$ we know from Lemma 
\ref{lemma:criterion numbers} that  $\tilde{s}^n>1$ and hence 
$(\lambda_1,\lambda_2, \dots,\lambda_{n})\in S_{5}^n$, $n\in {\mathbb N}$, $ n\geq2$.

We have thus proved the inverse of Proposition \ref{prp:criterion radius} 
for $n\in {\mathbb N}$, $n\geq2$. In fact we proved a stronger result. 
Let $\lambda_1,\lambda_2,\dots,\lambda_{n}$, $n\in {\mathbb N}$, $n\geq2$, 
be distinct positive real numbers. Supposing that both numerator and denominator 
of the positive dependent radius $\lambda_{n+1}$, $n\in {\mathbb N}$, $n\geq2$, 
associated to the radii $\lambda_1,\lambda_2,\dots,\lambda_{n}$ are positive, 
then $(\lambda_1,\lambda_2,\dots,\lambda_{n}) \in \cup_{i=1}^3S_{i}^n$,
$n \in {\mathbb N}$, $n \geq 2$. If we suppose that both numerator and denominator 
of the positive dependent radius $\lambda_{n+1}$, $n\in {\mathbb N}$, $n\geq2$, 
associated to the radii $\lambda_1,\lambda_2,\dots,\lambda_{n}$ are negative, 
then $(\lambda_1,\lambda_2,\dots,\lambda_{n}) \in \cup_{i=4}^{6}S_{i}^n$, 
$n \in {\mathbb N}$, $n \geq 2$.

If $n=1$, let $\lambda_1$ be a positive real number so that the dependent
 radius $\lambda_2$ associated to the radius $\lambda_1$ is positive.

First, we examine the case where both numerator and denominator of
 $\lambda_2$ are positive. In that case we have for the numerator 
that $24-4\lambda_1>0$ which implies that $\lambda_1<6$ and for 
the denominator that $4-\lambda_1>0$ which implies that $\lambda_1<4$. 
Combining the last two results about $\lambda_1$, we have that $0<\lambda_1<4$ 
and hence $\lambda_1 \in S_1^{1}$.

Let now examine the case where both numerator and denominator of 
$\lambda_2$ are negative. In that case we have for the numerator that 
$24-4\lambda_1<0$ which implies that $\lambda_1>6$ and for the denominator 
that $4-\lambda_1<0$ which implies that $\lambda_1>4$. Combining the last 
two results about $\lambda_1$, we have that $6<\lambda_1<+\infty$ and hence 
$\lambda_1 \in S_2^{1}$.

So, we proved that, if $\lambda_1$ is a positive real number so that the 
dependent radius $\lambda_2$ associated to the radius $\lambda_1$ is positive, 
then $\lambda_1 \in V^{1}$.
The proof of the proposition is complete.
\end{proof}

\subsection*{Acknowledgements}
The author wishes to thank Nikolaos Alikakos for considering the generalized 
Van der Pol equation \eqref{g VdP eq} in the particular case where $f$ 
is an odd polynomial of degree 5. I am grateful to Charalambos Evripidou 
for several helpful and stimulating discussions and comments. 
The author would also like to thank Yiorgos-Sokratis Smyrlis for his assistance 
in technical matters in the preparation of this paper and Dimitris Ioakim 
for improving the use of the English language. I would also like to express 
my sincere gratitude to the anonymous referees for their useful comments 
and valuable suggestions, which led to the improvement of this article.

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\end{document}