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

\AtBeginDocument{{\noindent\small
\emph{Electronic Journal of Differential Equations},
Vol. 2014 (2014), No. 22, pp. 1--9.\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-2014/22\hfil Existence, uniqueness and other properties]
{Existence, uniqueness and other properties of the limit cycle 
 of a generalized \\ Van der Pol equation}

\author[X. Ioakim \hfil EJDE-2014/22\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 June 22, 2013. Published January 10, 2014.}
\subjclass[2000]{34C07, 34C23, 34C25}
\keywords{Generalized Van der Pol equation; limit cycles; existence; uniqueness}

\begin{abstract}
 In this article, we study the bifurcation of limit cycles from
 the linear oscillator $\dot{x}=y$,  $\dot{y}=-x$ in  the class
 $$
 \dot{x}=y,\quad \dot{y}=-x+\varepsilon y^{p+1}\big(1-x^{2q}\big),
 $$
 where $\varepsilon$ is a small positive parameter tending to 0,
 $p \in \mathbb{N}_0$ is even and $q \in \mathbb {N}$.
 We prove that the above differential system, in the global plane
 where $p \in \mathbb{N}_0$ is even and $q \in \mathbb{N}$,
 has a unique limit cycle. More specifically, the existence
 of a limit cycle, which is the main result in this work,
 is obtained by using the Poincar\'{e}'s method, and the uniqueness
 can be derived from the work of Sabatini and Villari \cite{SV}.
 We also investigate and some other properties of this unique
 limit cycle for some special cases of this differential system.
 Such special cases have been studied by Minorsky \cite{M} and
 Moremedi et al.~\cite{MMG}.
\end{abstract}

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


\section{Introduction}

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 y^{p+1}\big(1-x^{2q}\big),
\end{gathered} \label{gVdP1}
\end{equation}
where $p \in \mathbb{N}_0$ is even, $q \in \mathbb{N}$ and
$0 < \varepsilon \ll  1$. System \eqref{gVdP1} reduces to
the Van der Pol equation for $p=0$ and $q=1$.
 Our purpose here is to find an upper bound for the number of limit 
cycles for system \eqref{gVdP1}, depending only on the degree of
its polynomials.

System \eqref{gVdP1} is the generalized Van der Pol equation of the form
\begin{equation}
\ddot{x}-\varepsilon(\dot{x})^{p+1}\big(1-x^{2q}\big)+x=0,
\label{gVdPeq 1}
\end{equation}
where $p \in \mathbb{N}_0$ is even, $q \in \mathbb{N}$ and
$0 < \varepsilon \ll  1$. We search to find an upper bound for the
number of limit cycles for equation \eqref{gVdPeq 1},
depending only on $p$ and $q$. We prove that the generalized 
Van der Pol equation \eqref{gVdPeq 1} has a unique limit cycle,
and it is simple and stable. We also examine the manner in which 
the position and size of the limit cycle depend on $p$ and $q$.

Several other generalizations of the Van der Pol equation have been 
considered in the literature. Minorsky \cite{M} has considered a 
generalized Van der Pol equation of the form
\begin{equation}
\ddot{x}-\varepsilon\dot{x}\big(1-x^{2q}\big)+x=0,
\label{gVdPeq M}
\end{equation}
where $q \in \mathbb{N}$ and $0 < \varepsilon \ll 1$. For $q=1$, 
equation \eqref{gVdPeq M} reduces to the Van der Pol equation.
For $p=0$ equations \eqref{gVdPeq 1} and \eqref{gVdPeq M} are identical.
By applying a perturbation method, he showed for \eqref{gVdPeq M}
that the stationary amplitude $A_0$, to first order in $\varepsilon$, is
\begin{equation}
A_0=\Big(\frac{\int_0^{2\pi}\sin^{2}(t)\,dt}{\int_0^{2\pi}
\sin ^{2}(t)\cos^{2q}(t)\,dt}\Big)^{1/(2q)}.
\label{ampl}
\end{equation}
For $q=1, \, 2$ and 3, Minorsky found from \eqref{ampl} that 
$A_0=2, \, 1.68$ and 1.53, respectively.

The solution of the generalized Rayleigh equation
\begin{equation}
\ddot{y}-\varepsilon\dot{y}\Big(1-\frac{1}{2q+1}(\dot{y}) ^{2q}\Big) +y=0,
\label{g Rayleigh eq}
\end{equation}
where $q \in \mathbb{N}$, is closely related to the solution 
of \eqref{gVdPeq M}. For, if we differentiate \eqref{g Rayleigh eq}
with respect to $t$ and let $\dot{y}=x$, then $x$ satisfies \eqref{gVdPeq M}.
Hence, results for \eqref{g Rayleigh eq} can be derived from the 
corresponding results for \eqref{gVdPeq M}.

Holmes and Rand \cite{HR} have examined the qualitative
behaviour of the non-linear oscillations governed by a differential 
equation of the form
\[
\ddot{x}+\dot{x}\big(\alpha+\gamma x^{2}\big)+\beta x+\delta x^{3}=0,
\]
where $\alpha,\,\beta,\,\gamma$ and $\delta$ are constants;
$\alpha=-1$, $\beta=1$, $\gamma=1$ and $\delta=0$ corresponds to the
Van der Pol equation. They investigated the presence of local and
 global bifurcations and considered their physical significance.

A more general class of equations, containing \eqref{gVdPeq 1}
as a special case, has the form
\begin{equation}
\ddot{x}+\dot{x}\phi(x,\dot{x})+x=0,
\label{g cl eq}
\end{equation}
and was studied in \cite{S} and \cite{S1}. They  obtained 
conditions about the existence and uniqueness of limit cycles 
of \eqref{g cl eq}. In general, we observe that the existence 
and uniqueness theorem for limit cycles of \eqref{g cl eq} proved 
there does not apply for equation \eqref{gVdPeq 1}.

The plan of this paper is as follows. 
In Section 2 we will make some elementary remarks about small 
perturbation of a Hamiltonian system. Section 3 will be devoted 
to study system \eqref{gVdP1}.

\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{align*}
\dot{x}=y, \\
\dot{y}=-x,
\end{align*}
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} \label{lem2.1}
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{thm2.1}
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{The non-linear equation 
$\ddot{x}-\varepsilon(\dot{x})^{p+1}\big(1-x^{2q}\big)+x=0$}

In this section, we prove that system \eqref{gVdP1} has a unique limit
cycle, and it is simple and stable. We present this main result in 
Theorem \ref{thm3.1}. In Proposition \ref{prop3.1} we study the system \eqref{gVdP1},
with $p \in \mathbb{N}_0$ is even, $q \in \mathbb{N}$ 
satisfying $p + 2 = 2q$. The system \eqref{gVdP1},
in the case where $p = 0$ and $q \to \infty$ will be studied
in Proposition \ref{prop3.2} and in the case where $q = 1$ and 
$p \to \infty$ will be studied in Proposition \ref{prop3.3}.

Our main result in this section is given in the following theorem.

\begin{theorem} \label{thm3.1}
System \eqref{gVdP1}, where $p \in \mathbb{N}_0$ is even,
 $q \in \mathbb{N}$ and $0 < \varepsilon \ll 1$ has the unique limit cycle
\[
x^{2}+y^{2}=\Big[\frac{(p+2q+2)(p+2q)\ldots(2q+2)}{(p+2)p 
\ldots4\cdot2}\frac{2q(2q-2)\ldots4\cdot2} {(2q-1)(2q-3)\ldots3\cdot1} 
\Big]^{1/q}+O(\varepsilon),
\]
and it is simple and stable.
\end{theorem}

\begin{proof}
 From \eqref{energy} we have 
\begin{equation}
\dot{E}(x,y)=y^{p+2}\big(1-x^{2q}\big),
\label{energy 1}
\end{equation}
where $p \in \mathbb{N}_0$ is even and $q \in \mathbb{N}$.
Substituting \eqref{energy 1} into \eqref{FA}, we obtain that
\begin{equation}
F(A)=\int_0^{2\pi}(y^{0}(t))^{p+2}\big(1-(x^{0}(t))^{2q}\big)dt,
\label{FA 1}
\end{equation}
where $p \in \mathbb{N}_0$ is even and $q \in \mathbb{N}$.
Substituting $x^{0}(t) = A\cos(t-t_0)$ and $y^{0}(t) = -A\sin(t-t_0)$ 
into \eqref{FA 1}, and using the assumption that $p \in \mathbb{N}_0$ 
is even we get
\begin{equation}
F(A)=A^{p+2}\Big[\int_0^{2\pi}\sin^{p+2}(t-t_0)dt-A^{2q} 
\int_0^ {2\pi}\sin^{p+2}(t-t_0)\cos^{2q}(t-t_0)dt\Big].
\label{FA 2}
\end{equation}
Let
\begin{gather*}
c_1:=\int_0^{2\pi}\sin^{p+2}(t-t_0)dt,\\
c_2:=\int_0^{2\pi}\sin^{p+2}(t-t_0)\cos^{2q}(t-t_0)dt,
\end{gather*}
where $p \in \mathbb{N}_0$ is even and $q \in \mathbb{N}$.
Using the fact that
\[
c_1=4\int_0^{\pi/2}\sin^{p+2}(t-t_0)dt,
\]
from Proposition \ref{prop4.2}, we obtain
\[
c_1=2\frac{(p+1)(p-1)\ldots3\cdot1}{(p+2)p\ldots4\cdot2}\pi.
\]
Using the fact that
\[
c_2=4\int_0^{\pi/2}\sin^{p+2}(t-t_0)\cos^{2q} (t-t_0)dt,
\]
from Proposition \ref{prop4.1}, we obtain
\[
c_2=2\frac{(p+1)(p-1)\ldots5\cdot3\cdot1}{(p+2q+2)(p+2q)
\ldots(2q+2)}\frac{(2q-1)(2q-3)\ldots3\cdot1}{2q(2q-2) \ldots4\cdot2}\pi.
\]
Substituting $c_1$ and $c_2$ given as above into \eqref{FA 2} it follows that
\begin{align*}
F(A)&=2\pi A^{p+2}\Big[\frac{(p+1)(p-1)\ldots3\cdot1}{(p+2)p\ldots 4\cdot2}\\ 
&\quad -\frac{(p+1)(p-1)\ldots5\cdot3\cdot1}{(p+2q+2)(p+2q)\ldots (2q+2)}
 \frac{(2q-1)\ldots3\cdot1}{2q\ldots4\cdot2}A^{2q}\Big].
\end{align*}
Now, for $A > 0$ the polynomial $F$ has the root
\[
A=\Big[\frac{(p+2q+2)(p+2q)\ldots(2q+2)}{(p+2)p\ldots4\cdot2} 
\frac{2q(2q-2)\ldots4\cdot2}{(2q-1)(2q-3)\ldots3\cdot1} \Big]^{1/(2q)}.
\]
Let
\begin{equation}
A_0=A_0(p,q):=\Big[\frac{(p+2q+2)(p+2q)\ldots(2q+2)}{(p+2)p 
\ldots4\cdot2}\frac{2q(2q-2)\ldots4\cdot2}{(2q-1)(2q-3)\ldots3 \cdot1}\Big]
^{1/(2q)},
\label{ampl 1}
\end{equation}
where $p \in \mathbb{N}_0$ is even and $q \in \mathbb{N}$.

For the derivative of $F$ we have that
\[
\begin{aligned}
F'(A)&=2\pi A^{p+1}\Big[\frac{(p+1)(p-1)\ldots3\cdot1}{p(p-2)\ldots 4\cdot2}\\
 &\quad -\frac{(p+1)(p-1)\ldots3\cdot1}{(p+2q)(p+2q-2)\ldots(2q+2)} 
 \frac{(2q-1)\ldots3\cdot1}{2q\ldots4\cdot2}A^{2q}\Big].
\end{aligned}
\]
We compute the derivative of $F$ at $A_0$ and we get
\[
F'(A_0)=-4\pi A_0^{p+1}\frac{(p+1)(p-1)\ldots3\cdot1} {(p+2)p\ldots4\cdot2}\cdot 
q \neq 0,
\]
using the assumptions that $p \in \mathbb{N}_0$ is even, 
$q \in \mathbb{N}$ and $A_0 > 0$. So, from Theorem \ref{thm2.1}, it follows that 
\eqref{gVdP1} has a limit cycle close to the circle $x^{2} + y^{2} = A_0^{2}$.
Moreover, since $F'(A_0) < 0$, this limit cycle is simple and stable.

Let now prove that the number of limit cycles for system \eqref{gVdP1},
with $\varepsilon$ small is exactly one. The proof of this can be derived 
from the work of Sabatini and Villari \cite{SV} using Corollary 1 proved there. 
We first note that the system \eqref{gVdP1} can be written and in the form
\begin{gather*}
\dot{x}=y-\varepsilon x^{p+1}\big(y^{2q}-1\big), \\
\dot{y}=-x,
\end{gather*}
where $p \in \mathbb{N}_0$ is even, $q \in \mathbb{N}$ and 
$0 < \varepsilon \ll  1$. As we already saw, Poincar\'{e}'s method 
(see Theorem \ref{thm2.1}) ensures the existence of a limit cycle for \eqref{gVdP1}.
Since $a=-1, b=1, G(x)=\frac{x^2}{2}$, 
one has $G(a)=G(b)$, so the hypotheses of Corollary 1 hold (see \cite{SV}), 
and the system \eqref{gVdP1} has exactly one limit cycle. This completes 
the proof that \eqref{gVdP1} has exactly one limit cycle.

So, we prove that \eqref{gVdP1} has a unique limit cycle, and it
is simple and stable.
\end{proof}

\begin{remark} \rm
The expression \eqref{ampl} obtained by Minorsky, is a special case 
of the expression \eqref{ampl 1} which we found. Indeed, 
for $p=0$ it can be verified that \eqref{ampl 1} equals \eqref{ampl}.
 This may be done by evaluating the integral in the denominator of 
\eqref{ampl}, using the Proposition \ref{prop4.1} from the appendix.
\end{remark}

\begin{proposition} \label{prop3.1}
System \eqref{gVdP1}, with $p \in \mathbb{N}_0$ is even,
$q \in \mathbb{N}$ satisfying $p + 2 = 2q$, and $0 < \varepsilon \ll  1$ 
has the unique limit cycle $x^{2} + y^{2} = 4 + O(\varepsilon)$, 
and it is simple and stable.
\end{proposition}

\begin{proof}
 From Theorem \ref{thm3.1} it follows that system \eqref{gVdP1}, with
$p \in \mathbb{N}_0$ is even, $q \in \mathbb{N}$ and $0 < \varepsilon \ll  1$ 
has a unique limit cycle, and it is simple and stable. It remains to prove that
\begin{equation}
\Big[\frac{(p+2q+2)(p+2q)\ldots(2q+2)}{(p+2)p\ldots4\cdot2} 
\frac{2q(2q-2)\ldots4\cdot2}{(2q-1)(2q-3)\ldots3\cdot1}\Big]^{1/q}=4,
\label{remains}
\end{equation}
when $p + 2 = 2q$.

By the assumption that $p + 2 = 2q$ the left-hand side of \eqref{remains} gives
\[
\Big[\frac{2^{q}(2q)(2q-1)(2q-2)\ldots(q+2)(q+1)} 
{(2q-1)(2q-3)(2q-5)\ldots5\cdot3\cdot1}\Big]^{1/q}
=2\Big[\frac{2q(2q-1)\ldots(q+2)(q+1)} {(2q-1)(2q-3)\ldots5\cdot3\cdot1}\Big]^{1/q}.
\]
Hence it suffices to show that
\[
\Big[\frac{2q(2q-1)(2q-2)\ldots(q+2)(q+1)} {(2q-1)(2q-3)(2q-5)
\ldots5\cdot3\cdot1}\Big]^{1/q}=2.
\]

\noindent\textbf{Claim.} It is valid that
\[
\frac{2q(2q-1)(2q-2)\ldots(q+2)(q+1)}{(2q-1)(2q-3)(2q-5)
\ldots 5\cdot3\cdot1}=2^{q}, \quad \forall  q \in \mathbb{N}.
\]

\begin{proof}
 It will be proved by induction on $q$. 
For $q = 1$, we have $\frac{2}{1} = 2^{1}$, therefore the claim is valid 
for $q = 1$. Supposing that the claim is valid for $q$, we will prove 
that it is true and for $q+1$, namely
\begin{equation}
\frac{\big[2(q+1)\big](2q+1)(2q)(2q-1)\ldots(q+3)(q+2)} 
{(2q+1)(2q-1)(2q-3)(2q-5)\ldots5\cdot3\cdot1}=2^{q+1}.
\label{claim}
\end{equation}
The left-hand side of \eqref{claim} is equal to
\[
2(q+1)\frac{2q(2q-1)(2q-2)\ldots(q+2)}{(2q-1)(2q-3)\ldots5\cdot 3\cdot1}
=2\cdot2^{q}=2^{q+1},
\]
which is the right-hand side of \eqref{claim}. Therefore, the claim 
is valid for every $q \in \mathbb{N}$. 
\end{proof}
This completes the proof of  the proposition.
\end{proof}

\begin{remark} \rm
It is well known that the Van der Pol equation with $0 < \varepsilon \ll  1$ 
has the unique limit cycle $x^{2} + y^{2} = 4 + O(\varepsilon)$, and 
it is simple and stable. This arises and from Proposition \ref{prop3.1}
 with $p = 0$  and $q = 1$.
\end{remark}

In the next proposition, we give a different proof, much more elementary 
than the proof has been given by Moremedi et al.~\cite{MMG}, concerning the 
decreases of the amplitude of the limit cycle of system \eqref{gVdP1}
with $p = 0$ and $0 < \varepsilon \ll  1$, as $q$ increases.

\begin{proposition} \label{prop3.2}
System \eqref{gVdP1}, with $p = 0, \, q \in \mathbb{N}$ and
 $0 < \varepsilon \ll  1$ has a unique limit cycle which is simple, 
stable and its amplitude decreases monotonically from $2$ to $1$ as $q$ 
increases from $q=1$. Therefore, the unique limit cycle of the 
system \eqref{gVdP1}, with $p = 0$ has the equation
$x^{2} + y^{2} = 1 + O(\varepsilon)$ as $q \to \infty$.
\end{proposition}

\begin{proof}
 From Theorem \ref{thm3.1} it follows that system \eqref{gVdP1}, with
$p = 0, \, q \in \mathbb{N}$ and $0 < \varepsilon \ll  1$ has a unique 
limit cycle, and it is simple and stable. From \eqref{ampl 1} when $p = 0$ it follows that
\[
A_0=\Big[\frac{2q+2}{2}\frac{2q(2q-2)\ldots4\cdot2} {(2q-1)(2q-3)
\ldots3\cdot1}\Big]^{1/(2q)}.
\]
Let
\begin{equation}
A_0(q):=\Big[\frac{2q+2}{2}\frac{2q(2q-2)\ldots4\cdot2} 
{(2q-1)(2q-3)\ldots3\cdot1}\Big]^{1/(2q)}, \quad q\in \mathbb{N}.
\label{ampl p0}
\end{equation}
Clearly, $A_0(1) = 2$. In order to prove that the sequence
$A_0(q), \, q\in \mathbb{N}$ given by \eqref{ampl p0} is strictly 
decreasing we must show that $A_0(q+1) < A_0(q)$ for all  $q \in \mathbb{N}$.

We have that
\begin{align*}
A_0(q+1)
&=\big[\frac{2q+4}{2}\frac{(2q+2)(2q)\ldots4\cdot 2}{(2q+1)(2q-1)\ldots3\cdot1}
\big]^{\frac{1}{2(q+1)}} \\
&=\big[\frac{2q+4}{2q+1}\big]^{\frac{1}{2(q+1)}} 
\big[\frac{2q+2}{2}\frac{2q(2q-2)\ldots4\cdot2}{(2q-1)(2q-3)\ldots 3\cdot1}
\big]^{\frac{1}{2q}-\frac{1}{2q(q+1)}} \\
&=\big[s(q)\big]^{\frac{1}{2(q+1)}}A_0(q),
\end{align*}
where
\[
s(q)=\frac{2q+4}{2q+1}\big[\frac{1}{q+1}\frac{(2q-1)(2q-3)\ldots3 
\cdot1}{2q(2q-2)\ldots4\cdot 2}\big]^{1/q}, \quad q\in \mathbb{N}.
\]
Now, in order to show that $A_0(q+1) < A_0(q)$, it suffices to show that
$s(q) < 1$ for all $ q \in \mathbb{N}$.
We have that
\[
s(q)<\frac{2q+4}{2q+1}\frac{1}{(q+1)^{1/q}}.
\]

\noindent\textbf{Claim I.}
It is valid that
\begin{equation}
\frac{2q+4}{2q+1}\le(q+1)^{1/q}, \quad \forall \, q \in \mathbb{N}.
\label{claim I}
\end{equation}

\begin{proof}
 The inequality \eqref{claim I} is valid for $q=1,\ldots,5$, 
as it can easily be checked. In order to prove \eqref{claim I} 
for $q \in \mathbb{N}, \, q \geq 6$ we will show that
\begin{equation}
1+\frac{2}{q} < q^{1/q} \Longleftrightarrow \big(1+\frac{2}{q}\big)^{q}<q, 
\quad \forall  q\in \mathbb{N}, \, q \geq 6.
\label{q 6}
\end{equation}
One can easily check that the inequality \eqref{q 6} is valid for $q=6$ 
and 7. Since $\lim_{q\to\infty}\big(1+\frac{2}{q}\big)^ {q} = {\mathrm{e}}^{2}$, 
in order to prove \eqref{q 6} for $q \in \mathbb{N}$,  $ q \geq 8$, 
it suffices to show that the sequence 
$\big(1+\frac{2}{q}\big)^{q}$, $q\in \mathbb{N}$, is strictly increasing.
Notice that
\begin{align*}
\big(1+\frac{2}{q}\big)^{q}<\big(1+\frac{2}{q+1}\big)^{q+1} 
&\Longleftrightarrow \frac{q+1}{q+3}<[\frac{q(q+3)}{(q+1)(q+2)}]^{q} \\ 
&\Longleftrightarrow 1-\frac{2}{q+3}<[1-\frac{2}{(q+1)(q+2)}]^{q}.
\end{align*}
Now, using Bernoulli's inequality, we have for $q \in \mathbb{N}$ that
\[
[1-\frac{2}{(q+1)(q+2)}]^{q} \geq 1-\frac{2q}{(q+1)(q+2)}.
\]
Since is valid that
\[
\frac{1}{q+3}>\frac{q}{(q+1)(q+2)},
\]
the proof that the sequence $\big(1+\frac{2}{q}\big)^{q}$, $q \in \mathbb{N}$ 
is strictly increasing is complete.

So, we have proved the inequality \eqref{claim I} for every $q \in \mathbb{N}$. 
Therefore,
\[
s(q) < 1, \quad \forall  q \in \mathbb{N},
\]
which proves that the sequence $A_0(q)$, $q \in \mathbb{N}$ is strictly 
decreasing.

Now, note that \eqref{ampl p0} gives
\begin{equation}
A_0(q)=[(q+1)^{1/q}]^{1/2} [(2q+1)^{1/(2q)}]^{1/2} 
\Big[\frac{1}{2q+1}\Big(\frac{2q(2q-2)\ldots4\cdot2} {(2q-1)(2q-3)\ldots3\cdot1}
\Big)^{2}\Big]^{1/(4q)}.
\label{ampl p0 1}
\end{equation}
\end{proof}

\noindent\textbf{Claim II.} It is valid that
\begin{equation}
\lim_{q\to\infty}\Big[\frac{1}{2q+1} 
\Big(\frac{2q(2q-2)\ldots4\cdot2}{(2q-1)(2q-3)\ldots3\cdot1} \Big)^{2}
\Big]^{1/(4q)}=1.
\label{claim II}
\end{equation}

\begin{proof} From the inequality $0 < \sin t < 1$, $t \in (0,\pi/2)$ 
(with induction) we have that $\sin^{2q+1}t < \sin^{2q}t < \sin^{2q-1}t$, 
for every $t \in (0,\pi/2)$ and $q \in \mathbb{N}$. 
So, we have that
\begin{equation}
\int_0^{\pi/2}\sin^{2q+1}t\,dt<\int_0^{\pi/2}\sin^{2q}t\,dt
<\int_0^{\pi/2}\sin^{2q-1}t\,dt.
\label{int}
\end{equation}
Using Proposition \ref{prop4.2} from the appendix, \eqref{int} leads to
\begin{equation}
\frac{1\cdot3\ldots(2q-1)}{2\cdot4\ldots(2q-2)}
<\frac{2\cdot4\ldots (2q-2)2q}{1\cdot3\ldots(2q-3)(2q-1)}\frac{2}{\pi}
<\frac{1\cdot3 \ldots(2q+1)}{2\cdot4\ldots2q}.
\label{leads to}
\end{equation}
Multiplying \eqref{leads to} by
\[
\frac{2\cdot4\ldots(2q-2)2q}{1\cdot3\ldots(2q-1)(2q+1)}\frac{\pi }{2},
\]
we get
\begin{equation}
\frac{2q}{2q+1}\frac{\pi}{2}<\frac{1}{2q+1}
\big[\frac{2\cdot 4\ldots(2q-2)2q}{1\cdot3\ldots(2q-3)(2q-1)}\big]^{2}
<\frac{\pi }{2},
\label{we get}
\end{equation}
and then the inequality
\[
\Big(\frac{2q}{2q+1}\Big)^{1/(4q)}\Big(\frac{\pi}{2} \Big)^{1/(4q)}
<\Big[\frac{1}{2q+1}\Big(\frac{2\cdot 4\ldots(2q-2)2q}{1\cdot3\ldots(2q-3)(2q-1)}
\Big)^{2}\Big]^{1/(4q)}<\Big(\frac{\pi}{2}\Big)^{1/(4q)},
\]
implies \eqref{claim II}.
\end{proof}

Using \eqref{claim II}, from \eqref{ampl p0 1}, we easily obtain 
 $\lim_{q\to\infty}A_0(q) = 1$.
The proof of Proposition \ref{prop3.2} is complete.
\end{proof}

\begin{remark} \rm
The uniqueness of the limit cycle for the system \eqref{gVdP1},
with $p = 0, \, q \in \mathbb{N}$ studied in Proposition \ref{prop3.2} follows 
and from the fact that the function $\phi(x,y)=-\varepsilon (1-x^{2q})$
 is strictly star-shaped (see \cite{S},\cite{S1}).
\end{remark}

\begin{remark} \rm
 From \eqref{we get} it follows that
\[
\lim_{q\to\infty}\frac{1}{2q+1} 
\Big[\frac{2\cdot4\ldots(2q-2)2q}{1\cdot3\ldots(2q-3)(2q-1)} \Big]^{2} 
= \frac{\pi}{2},
\]
which is the Wallis's product. It is exciting and unexpected how this 
limit of Wallis appears in the proof of Proposition \ref{prop3.2}.
\end{remark}

\begin{proposition} \label{prop3.3}
System \eqref{gVdP1}, with $p \in \mathbb{N}_0$ is even, $q = 1$
and $0 < \varepsilon \ll  1$ has a unique limit cycle which is simple, 
stable and its amplitude increases monotonically from $2$ to infinity 
as $p$ increases from $p=0$.
\end{proposition}

\begin{proof}
 From Theorem \ref{thm3.1} it follows that system \eqref{gVdP1},
with $p \in \mathbb{N}_0$ is even, $q = 1$ and $0 < \varepsilon \ll 1$ 
has a unique limit cycle, and it is simple and stable. From \eqref{ampl 1} 
when $q = 1$ it follows that
\[
A_0=\Big[\frac{(p+4)(p+2)p\ldots6\cdot4}{(p+2)p\ldots4\cdot2}
\cdot \frac{2}{1}\Big]^{1/2}=(p+4)^{1/2}.
\]
Let $A_0(p):=(p+4)^{1/2}$, $p\in \mathbb{N}_0$ is even. 
Clearly, $A_0(0) = 2$. Obviously $A_0(p) < A_0(p+1)$, 
for all $p\in \mathbb{N}_0$ is even and $A_0(p) \to \infty$ as 
$p \to \infty$ and so the proof is complete.
\end{proof}

\begin{remark} \rm
We make now an observation on the type of the bifurcation phenomenon 
of limit cycles encountered in Proposition \ref{prop3.3}. 
Not the ``large amplitude limit cycle" is encountered in 
Proposition \ref{prop3.3}
 but the ``medium amplitude limit cycle". For given $p$ the limit cycle 
of \eqref{gVdP1}, with $q = 1$, has a finite limiting radius
and therefore is called ``medium amplitude limit cycle". 
When increasing $p$ also the radius of the limiting circle increases; 
in particular when $p \to \infty$ then the limiting radius also
 tends to $\infty$. The ``large amplitude limit cycle" 
would disappear at $\infty$ when the bifurcation parameter $\varepsilon$ 
tends to $0$.
\end{remark}

\section{Appendix}
Here we list some important formulas used in Section 3 (see \cite{NGG}).

\begin{proposition} \label{prop4.1}
For each $m, \, n \in \mathbb{N}$ and even,
\[
\int_0^{\pi/2}\sin^{m}(t)\cos^{n}(t)\,dt 
=\frac{(m-1)(m-3)\ldots5\cdot3\cdot1}{(m+n)(m+n-2)\ldots(n+2)} 
\frac{(n-1)(n-3)\ldots3\cdot1}{n(n-2)\ldots4\cdot2}\frac{\pi}{2}.
\]
\end{proposition}

\begin{proposition} \label{prop4.2}
For each $n \in \mathbb{N}$
\begin{gather*}
\int_0^{\pi/2}\sin^{2n-1}(t)\,dt
=\frac{2\cdot4\ldots (2n-2)}{1\cdot3\ldots(2n-1)},\\
\int_0^{\pi/2}\sin^{2n}(t)\,dt
=\frac{1\cdot3\ldots (2n-1)}{2\cdot4\ldots2n}\frac{\pi}{2}.
\end{gather*}
\end{proposition}

\subsection*{Acknowledgements}
The author wishes to thank Nikolaos Alikakos for considering the
 generalized Van der Pol equation \eqref{gVdPeq 1}.
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 suggestions.


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\end{document}