\documentclass[reqno]{amsart}

\AtBeginDocument{{\noindent\small
{\em Electronic Journal of Differential Equations},
Vol. 2003(2003), No. 96, pp. 1--24.\newline
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu
\newline ftp ejde.math.txstate.edu  (login: ftp)}
\thanks{\copyright 2003 Texas State University-San Marcos.}
\vspace{9mm}}

\begin{document}

\title[\hfilneg EJDE--2003/96\hfil Large energy simple modes]
{Large energy simple modes for a class of Kirchhoff equations}

\author[Marina Ghisi\hfil EJDE--2003/96\hfilneg]
{Marina Ghisi}

\address{Marina Ghisi \newline
Dipartimento di Matematica\\
Universit\`a degli Studi di Pisa\\
via M. Buonarroti 2, 56127 Pisa, Italy}
\email{ghisi@dm.unipi.it}

\date{}
\thanks{Submitted April 28, 2003. Published September 17, 2003.}
\subjclass[2000]{35L70, 37J40, 70H08}
\keywords{Kirchhoff equations, orbital stability, Hamiltonian systems,
\hfill\break\indent
Poincar\'{e} map, KAM theory}


\begin{abstract}
 It is well known that the Kirchhoff equation admits infinitely many
 simple modes, i.e., time periodic solutions with only one Fourier
 component in the space variable(s). We prove that for some form of
 the nonlinear term these simple modes are stable provided that
 their energy is large enough. Here stable means orbitally stable
 as solutions of the two-modes system obtained considering initial
 data with two Fourier components.
\end{abstract}

\maketitle
\numberwithin{equation}{section}
\newtheorem{thm}{Theorem}[section]
\newtheorem{rmk}[thm]{Remark}
\newtheorem{prop}[thm]{Proposition}
\newtheorem{defn}[thm]{Definition}
\newtheorem{lemma}[thm]{Lemma}

\section{Introduction}

Let	$H$	be a real Hilbert space, with norm $|\cdot|$ and scalar
product	$\langle \cdot,\cdot\rangle$. Let $A$ be a self-adjoint	linear
positive operator on $H$ with dense	domain $D(A)$ (i.e., $\langle
Au,u\rangle> 0$	for	all	$u\in D(A)$). We consider the evolution	problem
\begin{equation}
	u''(t)+	m(|A^{1/2}u(t)|^2)Au(t)=0,	   \label{dissipeq}
\end{equation}
where $m:[0,+\infty)\to	(0,+\infty)$ is	a $C^1$	function.
Equation (\ref{dissipeq}) is an	abstract setting of	the	hyperbolic PDE
with a non-local non-linearity of Kirchhoff	type
\begin{equation}
	u_{tt}-m\Big(\int_{\Omega}|\nabla u|^2\,dx\Big)\Delta u=0,
	\quad\mbox{in }\Omega\times\mathbb{R},	   \label{concrete}
\end{equation}
where $\Omega\subseteq\mathbb{R}^n$	is an open set,	$\nabla
u$ is the gradient of $u$ with respect to space	variables, and
$\Delta$ is	the	Laplace	operator.
When $\Omega$ is an	interval of	the	real line, this	equation is	a model
for	the	small transversal vibrations of	an elastic string.

When $H$ admits	a complete orthogonal system made of
eigenvectors of	$A$	(this is the case e.g. in (\ref{concrete}) if $\Omega$
is bounded), (\ref{dissipeq})
may	be thought as a	system of ODEs with	infinitely many	unknowns,
namely the components of $u$.

 Many papers have been written about equations (\ref{dissipeq})	and
(\ref{concrete}) after Kirchhoff's monograph \cite{k}. The interested
reader can find	appropriate	references in the surveys \cite{arosio}
and	\cite{spagnolo}.  We just state	that, at the present, the
existence of global	solutions for all initial data in $C^\infty$ or	in
Sobolev	spaces is still	an open	problem.

In this	paper, we consider a particular	class of global	solutions of
(\ref{dissipeq}). Let us assume	that $\lambda$ is an eigenvalue	of
$A$, and $e_{\lambda}$ is a	corresponding eigenvector, which we	assume
normalized so that $|e_{\lambda}|=1$. If the initial data are
multiples of $e_{\lambda}$,	say
$u(0)=w_{0}e_{\lambda}$, $u'(0)=w_{1}e_{\lambda}$,
then the solution of (\ref{dissipeq}) remains a	multiple of
$e_{\lambda}$ for every	$t\in\mathbb{R}$; i.e.,	we have	that
$u(t)=w(t)e_{\lambda}$,	where $w(t)$ is	the	solution of	the	ODE
$$
w''(t)+\lambda m(\lambda w^{2}(t))w(t)=0, \quad
w(0)=w_{0},	\quad w'(0)=w_{1}.
$$
Such solutions are called \emph{simple modes} of equation
(\ref{dissipeq}), and are known	to be time periodic	under very general
assumptions	on $m$.

In this	paper, we are interested to	stability of high energy simple	modes for
particular choices of $m$. This	program	is too optimist	since
stability problems are often hard already for systems with 3 unknowns, and we
have seen that (\ref{dissipeq})	has	infinitely
many degrees of	freedom. For this reason we	 limit ourselves to	consider
the	two-mode system
\begin{equation}
	\begin{gathered}
		w''(t)+\lambda\,m(\lambda w^{2}(t)+\mu z^{2}(t))w(t) = 0,  \\
		z''(t)+\mu\,m(\lambda w^{2}(t)+\mu z^{2}(t))z(t) = 0,
	\end{gathered}	  \label{system:primitivo}
\end{equation}
where $\mu\neq \lambda$	 is	another	eigenvalue of $A$, corresponding to	an
eigenvector	$e_{\mu}$ such that	$|e_{\mu}|=1$, and
$u(t)=w(t)e_{\lambda}+z(t)e_{\mu}$.
It is clear	that simple	modes are particular solutions of this system,
corresponding to initial data with $z(0)=z'(0)=0$. What	we actually
study is the stability of simple modes as solutions	of
(\ref{system:primitivo}).

To simplify	the	notation, let us set
$$
\nu:=\frac{\mu}{\lambda},\quad
u(t):=\sqrt{\lambda}\,w\big(\frac{t}{\sqrt{\lambda}}\big), \quad
v(t):=\sqrt{\mu}\,z\big(\frac{t}{\sqrt{\lambda}}\big),
$$
so that	(\ref{system:primitivo}) is	equivalent to
\begin{equation}
	\begin{gathered}
		u''(t)+m(u^{2}(t)+v^{2}(t))u(t)	= 0,  \\
		v''(t)+\nu\,m(u^{2}(t)+v^{2}(t))v(t) = 0.
	\end{gathered} \label{system:main}
\end{equation}
This system	(as	well as	(\ref{system:primitivo}) and
(\ref{dissipeq})) are Hamiltonian, with	conserved energy
\begin{equation}
	H(u,u',v,v'):=\frac{1}{2}\Big\{
[u']^{2}+\frac{[v']^{2}}{\nu}+M(u^{2}+v^{2})\Big\},
	\label{energia}
\end{equation}
where $M(r)	= \int_{0}^r m(s)\,	ds$.
As far as we know, stability of	simple modes was studied in	at least
four papers	(see section \ref{sec:stability} for
	precise	definitions).
\begin{itemize}
	\item Dickey \cite{dickey} proved that simple modes	are
	\emph{linearly stable} provided	that their energy is \emph{small}
	enough.	 Roughly speaking, linearly	stable means that $v(t)\equiv
	0$ is a	stable solution	for	the	linearization of the second
	equation in	(\ref{system:main}).

	\item In \cite{ghisi-gobbino} it was proved	that simple	modes as solutions of
(\ref{system:primitivo}) are \emph{orbitally stable}  provided that	their energy is
	\emph{small} enough. Roughly speaking, orbitally stable	means that every solution
$(u(t),v(t))$ of system	(\ref{system:main})	with initial data near
$(u_{0},u_{1},0,0)$	remains	close to the periodic orbit	of the simple
mode for every $t\in\mathbb{R}$.

	\item  Cazenave	and	Weissler \cite{cazenave}
	assumed	that there exists $\alpha>0$ such that
	$$
	\lim_{\sigma\to	+\infty}\frac{m(\sigma r)}{m(\sigma)}=r^{\alpha},
	$$
	uniformly on bounded intervals (\emph{e. g.} $m(r)=1+r^{\alpha}$).
	They showed	that if
	$$
	\nu\in\bigcup_{m\in\mathbb{N}}^{}
	\left((m+1)\left((\alpha+1)m+1\right),
	(m+1)\left((\alpha+1)m+1+2\alpha\right)\right),
	$$
	then every simple mode of (\ref{system:main}) with \emph{large}
	enough energy is \emph{unstable}.  If $\alpha=1$,
	and	$\Omega$ is	an interval	of the real	line, this result implies
	the	instability	of every simple	mode of	(\ref{concrete}) with
	large enough energy.

	\item In \cite{ggexp} it was proved	that if
 $m$ is	nondecreasing and	for	every $r\in[0,1)$ one has
		$$\lim_{\sigma\to +\infty}\frac{m(\sigma r)}{m(\sigma)}=0,$$
		then every simple mode of (\ref{dissipeq}) with	large enough energy
  is unstable.
\end{itemize}

\begin{rmk}	\label{rmk1.1} \rm
	Let	$m > 0$	be a continuous	function such that for all $r\in(0,1)$ there exists
	$$
	\lim_{\sigma\to	+\infty}\frac{m(\sigma r)}{m(\sigma)}.
	$$
	Since this limit is	a multiplicative
		function, then there are only three
		possibilities:
		\begin{itemize}
			\item  The limit is	$r^{\alpha}$ for some $\alpha>0$.

			\item  The limit is	0 for every	$r\in(0,1)$.

			\item  The limit is	1 for every	$r\in(0,1)$.
		\end{itemize}
\end{rmk}

In \cite{cazenave}	and	in \cite{ggexp}, the first two cases
were treated  (and proved instability).	Here we	treat the third
case and prove orbital stability. Our main result is the
following.

\begin{thm}\label{thm:main}
	Let	$\nu\neq 1$	be a positive real number. Let
	$m:[0,+\infty)\to (0,+\infty)$ be a	smooth function	with $m'(x)	> 0$
	 for all $x	> 0$ such that
\begin{enumerate}
	\renewcommand{\labelenumi}{\emph{(H\arabic{enumi})}}
 %	   \item \label{H1}	we have	$m'(x) > 0$,
	\item \label{H3} There exists a	constant $c$ such that,	for	every
	real $k	> 0$,
	$\sup_{y \in (0,1)}	\frac{ym'(ky)}{m'(k)} \leq c$
	\item \label{H4} For every $y \in (0,1)$
	$ \lim_{k\to +\infty} \frac{m'(ky)}{m'(k)} = \frac{1}{y}$.
\end{enumerate}
	Then there exists $k_{0}>0$	such that, if
	$H(u_{0},u_{1},0,0)>k_{0}$,	then the simple	mode of
	(\ref{system:main})	with $u(0)=u_{0}$, $u'(0)=u_{1}$ is	orbitally
	stable.
\end{thm}

Let	us remark that any function	with $m'>0$	such that $xm'(x) \to l	> 0$
as $x \to +\infty$ (\emph{e. g.} $\log(2 + x^{2})$)	satisfies
(H\ref{H3})	- (H\ref{H4}).

We conclude	with a few comments	on Theorem \ref{thm:main}.
		\begin{itemize}
			\item Since	there exists $\lim_{r\to +\infty} m(r)$,
			by (H\ref{H4}) we get $\lim_{k\to +\infty}
			m(ky)/m(k) = 1$	for	every $y > 0$.

			\item Assumption $m'(x)	> 0$ for all $x	> 0$  is not essential
but	this would only	complicate proofs without introducing new ideas.

			\item To prove orbital stability we	use	KAM	theory to
			Poincar\'{e} map (see section \ref{Ks}).
			Smoothness of $m$ is used only to give the
			smoothness required	by KAM theory.	To this	end, $m\in
			C^{5}$ is enough.

			\item Since	(\ref{system:main})	is reversible (if
			$(u(t),v(t))$ is any solution, then	$(u(-t),v(-t))$	is
			another	solution), then	a consequence of Theorem
			\ref{thm:main} is the following: ``if the energy of	a
			two-mode solution of (\ref{system:main}) is	large enough,
			then it	is \emph{not} possible that	asymptotically all
			this energy	is absorbed	by one of the two components''.
		\end{itemize}

This paper is organized	as follows:	in section 2 we	rescal the problem
and	give preliminaries on stability	and	Poincar\'{e}
map; in	section	3 we state our results;	in section 4 we	give the
proofs.	 Section 4 is divided in four parts: in	the	first two parts	we
prove all we need about	function $m$ and simple	mode;
in the third part we get the proof of Theorem
\ref{thm:1}	and	in the last	one	we prove
Theorem	\ref{thm:2}.

\section{Preliminaries}
\subsection{Rescaling}
In this	section	we rescale the solutions of	(\ref{system:main})	and
find an	equivalent system, in which	is simpler to work.
Given $k>0$, let us	consider the simple	mode $\overline{u}_{k}$	of system
(\ref{system:main})	which solves
\begin{equation}
	\overline{u}_{k}''(t)+m(\overline{u}_{k}^{2}(t))\overline{u}_{k}(t)=0,\quad
	\overline{u}_{k}(0)=k,\quad\overline{u}_{k}'(0)=0.
	\label{defn:uk}
\end{equation}
We recall that $\overline{u}_{k}$ is a periodic	function, and so we	can	assume
$\overline{u}_{k}(0)>0$	and	$\overline{u}_{k}'(0)=0$ without loss of generality.  Moreover
assuming that $k$ is large is equivalent to	assuming that the energy
of $\overline{u}_{k}$ is large.

Let	$\tau_{k}$ be the period of	$\overline{u}_{k}$.	Applying
conservation low (\ref{energia}) it	is easy	to see that
\begin{equation}
	\tau_{k} = 4\int_{0}^{k} \frac{1}{\sqrt{M(k^{2}) - M(y^{2})}}\,	dy =
	4k\int_{0}^{1} \frac{1}{\sqrt{M(k^{2}) - M(k^{2}y^{2})}}\, dy.
	\label{deft}
\end{equation}
Now	let
 $(u_{k},v_{k})$ be	the	solution of	 (\ref{system:main}) with
 initial data $u_{k}(0)	= a_{1}k, \, u_{k}'(0) = b_{1}k$, $v_{k}(0)	=x_{1},\,
  v_{k}'(0)	= y_{1}$.
  Setting
  $$
 w_{k}(t)=\frac{u_{k}(\tau_k t)}{k},
\quad
 z_{k}(t)=v_{k}(\tau_k t),
$$
it turns out that $(w_{k},z_{k})$ is the solution of
 \begin{equation}
  \begin{gathered}
   w_{k}''(t)+\tau_k^{2}m(k^{2}w_{k}^{2}(t)	+ z_{k}^{2}(t))w_{k}(t)=0,\quad
	   w_{k}(0)=a,\quad	w_{k}'(0)=b	   \\
	z_{k}''(t)+\nu \tau_k^{2}m(k^{2}w_{k}^{2}(t) + z_{k}^{2}(t)) z_{k}(t)=0,\quad
	   z_{k}(0)=x,\quad	z_{k}'(0)=y
  \end{gathered} \label{rescaled}
  \end{equation}
  where	$a = a_{1}$, $b	= \tau_{k}b_{1}$, $x = x_{1}$ and $y =
  \tau_{k}y_{1}$.
  In the sequel	we study the stability of simple modes of (\ref{rescaled}).	Indeed
following result holds,	whose simple proof is omitted
(see also Definition \ref{defn:os} below).

  \begin{thm} \label{thm:res}
	Let	$k > 0$	be fixed and let $\overline{u_{k}}$	be defined in
	(\ref{defn:uk}).  If $U_{k}(t) = \overline{u_{k}} (\tau_{k}t)/k$
	is orbitally stable	as solution	of (\ref{rescaled})	then
	$\overline{u_{k}}$ is orbitally	stable as solution of (\ref{system:main}).
\end{thm}

We remark that for (\ref{rescaled}),
  we can write the conserved energy	as
  $$
H_{k}(w_{k},w_{k}',z_{k},z_{k}') = \frac{1}{2}\left\{
[w_{k}']^{2}+\frac{[z_{k}']^{2}}{k^{2}\nu}+\frac{\tau_{k}^{2}}{k^{2}} M(k^{2}w_{k}^{2} +
z_{k}^{2})\right\}.
$$

\subsection{Kam	Theory and stability} \label{Ks}
We recall the notion of	stability, and then	we
describe the Poincar\'{e} map $P_{k}$ associated with a	simple mode
$U_{k}$	of system (\ref{rescaled}).

We refer to	\cite{hale}	for	general	facts about	dynamical and
Hamiltonian	systems, and to	\cite{cazenave,ghisi-gobbino} for specific
results	related	to the particular system (\ref{rescaled}).
Before we enter	into the details, we fix some notation.

We assume that $m:[0,+\infty)\to(0,+\infty)$ is	a nondecreasing	function of	class
$C^{5}$.
We denote by $M_{2\times 2}$ the set of	$2\times 2$	matrices.  For
each $A\in M_{2\times 2}$, $a_{ij}$	is the element in the $i$-th row
and	$j-th$ column, unless otherwise	stated,	and
$\mathrm{Tr}A=a_{11}+a_{22}$ is	the	trace of $A$.  For every
$\omega\in\mathbb{R}$, $R_{\omega}$	denotes	the	rotation matrix
$$R_{\omega}=
\begin{pmatrix}
	\cos \omega	& \sin \omega  \\
	-\sin \omega & \cos\omega
\end{pmatrix}.$$

\subsubsection{Stability}
\label{sec:stability}
In this	section	we recall some definitions of stability	from the
classical theory of	Hamiltonian	systems. For the sake of simplicity,
we adapt definitions to	the	case of	simple modes for system
(\ref{rescaled}).
In the phase space $\mathbb{R}^{4}$	we consider	the	energy level
$$\mathcal{H}_{k}:=\left\{(x_{1},x_{2},x_{3},x_{4})\in\mathbb{R}^{4}:\
H_{k}(x_{1},x_{2},x_{3},x_{4})=H_{k}(1,0,0,0)\right\},$$
and	the	orbit
$$\Gamma_{k}:=\left\{
(U_{k}(t),U_{k}'(t),0,0):\ t\in\mathbb{R}\right\}.$$

\begin{defn}\label{defn:os}	\rm
		The	simple mode	$U_{k}$	is called \emph{orbitally
		stable}	if,	for	every $\epsilon>0$ there exists	$\delta>0$ such
		that for every solution	$(w(t),z(t))$ of system
		(\ref{rescaled}), the following	property holds:	if the
		initial	datum $(w(0),w'(0),z(0),z'(0))$	belongs	to a $\delta$
		neighborhood of	$(1,0,0,0)$, then for every	$t\in\mathbb{R}$ the point
		$(w(t),w'(t),z(t),z'(t))$ lies in an $\epsilon$	neighborhood of
		$\Gamma_{k}$.
\end{defn}

\begin{defn} \label{defn:ios} \rm
		The	simple mode	$U_{k}$	is called
		\emph{isoenergetically orbitally stable} if	the	condition of
		Definition \ref{defn:os} is	satisfied with the restriction
	 $(w(0),w'(0),z(0),z'(0))\in\mathcal{H}_{k}$.
\end{defn}

\begin{defn}   \rm
		The	simple mode	$U_{k}$	is said	to be \emph{linearly
		stable}	if $z(t)\equiv 0$ is a stable solution of the linear equation
		$z''(t)+\nu\ \tau_{k}^{2}m(k^{2}U_{k}^{2}(t))z(t)=0$,
		(that is the linearization of the second equation
		in (\ref{rescaled})),
		i.e., for every	$\epsilon>0$ there exists
		$\delta>0$ such	that
		$$\|(z(0),z'(0))\|<\delta\ \Longrightarrow\
		\|(z(t),z'(t))\|<\epsilon,\quad\forall t\in\mathbb{R}.$$
\end{defn}

It is obvious that orbital stability implies isoenergetical
	orbital	stability.	In non-degenerate situations, isoenergetical
	orbital	stability implies linear stability.	 Here ``non-degenerate
	situation''	means that $(0,0)$ is not a	parabolic point	for	the
	associated Poincar\'{e}	map	(see section \ref{sec:poincare}	and
	\ref{sec:KAM} below).  It is not essential to explain now such
	condition; we just remark that it is satisfied by our large	energy
	simple modes.

\subsubsection{The	Poincar\'{e} map} \label{sec:poincare}
Let	us consider	the	open set
$\mathcal{U}_{k}\subseteq\mathbb{R}^{2}$ defined by
$$
	\mathcal{U}_{k}:=\left\{(x,y)\in\mathbb{R}^{2}:\ H_{k}(0,0,x,2\pi\sqrt{\nu}y)<
	H_{k}(1,0,0,0)\right\}.
	$$
For	every $(x,y)\in\mathcal{U}_{k}$, let $\alpha(x,y)>0$ be	the	unique
positive number	such that
$$H_{k}(\alpha(x,y),0,x,2\pi\sqrt{\nu}y)=H_{k}(1,0,0,0).
$$
Let	$(w(t),z(t))$ be the solution of system
(\ref{rescaled}) with initial data
$$ w(0)=\alpha(x,y),\quad w'(0)=0,
\quad z(0)=x,\quad z'(0)=2\pi\sqrt{\nu}y.
$$
Finally, let $T:=T(x,y)$ be	the	smallest $t>0$ such	that $w'(t)=0$ and
$w(t)>0$.  The interested reader can verify	that such a	$T$	exists for
every $(x,y) \in \mathcal{U}_{k}$.	On the other hand the existence	of
$T$	is classical up	to restricting $\mathcal{U}_{k}$.

The	Poincar\'{e} map $P_{k}: \mathcal{U}_{k}\to\mathbb{R}^{2}$,
relative to	the	simple mode	$U_{k}$	of (\ref{rescaled}), is	defined	by
$$
P_{k}(x,y):=\left(z(T),(4\pi^{2}\nu)^{-1/2}z'(T)\right).
$$
We point out that both $z$ and $T$ depend on $(x,y)$ and $k$.
When $(x,y)=(0,0)$,	then $w(t)=U_{k}(t)$ and $z(t)=0$ for
every $t\in\mathbb{R}$.	It follows that	$P_{k}(0,0)=(0,0)$,	i.e.,
$(0,0)$	is a fixed point of	the	Poincar\'{e} map.

The	interested reader is referred to the quoted	literature,	and	in
particular to \cite{ghisi-gobbino},	for	a heuristic
description	of the Poincar\'{e}	map.

Now	we recall the classical	definition of stability	of fixed points
for	planar maps.

\begin{defn}\label{defn:P-stable} \rm
		Let	$\mathcal{U}\subseteq\mathbb{R}^{2}$ be	an open	set	containing
		$(0,0)$, and let $P:\mathcal{U}\to\mathbb{R}^{2}$ be a map such
		that $P(0,0)=(0,0)$.  The fixed	point $(0,0)$ is said to be
		\emph{stable} if for every $\epsilon>0$	there exists $\delta>0$
		such that
		$$(x,y)\in\mathcal{U},\	\|(x,y)\|<\delta\
		\Longrightarrow\
		\|P^n(x,y)\|<\epsilon
		\quad\forall n\in\mathbb{N},$$
		where $P^n$	denotes	the	$n$-th iteration of	$P$.
\end{defn}

The	stability of $U_{k}$ as	a
periodic solution is clearly related to	the	stability of $(0,0)$ as	a fixed
point of $P_{k}$.  This	relation is	stated in Theorem
\ref{thm:OSiffPS} below.

\subsubsection{KAM	theory for planar maps}
\label{sec:KAM}
Stability of planar	maps has long been studied.
In this	subsection we sum the basic	results	we need	in the sequel.
Let	$\mathcal{U}\subseteq\mathbb{R}^{2}$ be	an open	set	containing $(0,0)$,
and	let	$P:\mathcal{U}\to\mathcal{U}$. The theory of planar	maps has
been developed for very	general	maps $P$; however we state the results
under suitable assumptions which allow to simplify some	notations, and
are	trivially satisfied	in our case. Therefore let us assume that:
\begin{enumerate}
	\renewcommand{\labelenumi}{(P\arabic{enumi})}
	\item  $P\in C^{5}(\mathcal{U},\mathcal{U})$ and $P(0,0)=(0,0)$;

	\item  $P$ is area-preserving;

	\item  if $P(x,y)=(a,b)$, then $P(a,-b)=(x,-y)$;

	\item  $P(-x,-y)=-P(x,y)$.
\end{enumerate}

The	first object to	look at	in order to	study the stability	of the
fixed point	$(0,0)$	is the differential	of $P$ at $(0,0)$, which we
denote by $L$. It is well known	that the canonical form	of $L$ is one
of the following three.
\begin{itemize}
	\item  $\begin{pmatrix}
		\lambda	& 0	 \\
		0 &	\lambda^{-1}
	\end{pmatrix}$
	for	some $\lambda\in\mathbb{R}$, $|\lambda|>1$.	In this	case $(0,0)$ is
	said to	be \emph{hyperbolic} and it	is \emph{unstable}.

	\item  $\begin{pmatrix}
		\pm	1 &	a  \\
		0 &	\pm	1
	\end{pmatrix}$ for some	$a\neq 0$. In this case	$(0,0)$	is
	said to	be \emph{parabolic}. The map $L$ is	unstable, but nothing can
	be said	about $P$. However,	we will	not	find this degenerate case in
	this paper.

	\item  $R_{\omega}$	for	some $\omega\in\mathbb{R}$.	In this	case $(0,0)$ is
	said to	be \emph{elliptic}.	The	map	$L$	is stable, but this	is in
	general	not	enough to guarantee	the	stability of $P$.
\end{itemize}

Therefore, $L$ gives only necessary	conditions for stability (i.e.,	non
hyperbolicity).	 KAM theory	provides sufficient	conditions in the case
of elliptic	fixed points.  To describe such	conditions,	it is better
to write $P$ in	polar coordinates up to	terms of order three.  If we
choose coordinates where $L$ is	written	in the canonical form of a
rotation, then,	in the corresponding polar coordinates,	$P$	becomes
$$P\begin{pmatrix}
	\rho  \\
	\theta \end{pmatrix}=
\begin{pmatrix}
	\rho+a(\theta)\rho^{3}	\\
	\theta-\omega+b(\theta)\rho^{2}
\end{pmatrix} +o(\rho^{3}),
$$
where $\omega$ is the same as in the linear	term $L$, and $a(\theta)$
and	$b(\theta)$	are	trigonometric polynomials of degree	4.	The
absence	of even	powers of $\rho$ in	the	first component, and of	odd
powers of $\rho$ in	the	second component, is due to	(P4).  Finally we
set
\begin{equation}
	\gamma(P):=\frac{1}{2\pi}\int_{0}^{2\pi}b(\theta)\,d\theta.
	\label{defn:gamma}
\end{equation}
Then we	have the following KAM result.

\begin{thm}\label{thm:KAM}
	Let	$P:\mathcal{U}\to\mathcal{U}$ be a planar map satisfying
	(P1)--(P4).
	Let	$(0,0)$	be an elliptic fixed point,	and	let	$\omega$ and
	$\gamma$ be	defined	as above.
	Let	us assume that
	\begin{description}
		\renewcommand{\labelenumi}{(KAM	\arabic{enumi})}
		\item[\textnormal{(KAM 1)\ }] $e^{hi\omega}\neq	1$ for every
		$h\in\{1,2,3,4\}$;

		\item[\textnormal{(KAM 2)\ }]  $\gamma(P)\neq 0$.
	\end{description}
	Then $(0,0)$ is	stable for $P$ according to	Definition \ref{defn:P-stable}.
\end{thm}

The	following result relates stability,	Poincar\'{e} maps, and KAM theory.
It is the fundamental tool in our analysis.

\begin{thm}\label{thm:OSiffPS}
	Let	$U_{k}$	be a simple	mode of	system
	(\ref{rescaled}), and let $P_{k}$ be the associated
	Poincar\'{e} map.  Then
	\begin{itemize}
		\item $U_{k}$ is \emph{linearly} stable	if and only	if
		$(0,0)$	is an elliptic fixed point of $P_{k}$;

		\item $U_{k}$ is \emph{isoenergetically	orbitally}
		stable if and only if $(0,0)$ is a stable fixed	point of the
		Poincar\'{e} map $P_{k}$;

		\item  if $(0,0)$ is an	elliptic fixed point of	$P_{k}$,
		and	$P_{k}$	satisfies (KAM 1) and (KAM 2), then
		$U_{k}$	is \emph{orbitally}	stable.
	\end{itemize}
\end{thm}

Thanks to Theorem \ref{thm:KAM}	and	Theorem	\ref{thm:OSiffPS}, the
orbital	stability of a periodic	solution in	the	four dimensional space
can	be proved by verifying that	a planar map satisfies two algebraic
conditions.

\section{Statement of results}
Let	us denote by $P_k:\mathcal{U}_{k}\to\mathcal{U}_{k}$ the
Poincar\'{e} map associated	with $U_{k}$ as	in section \ref{sec:poincare},
and	by $L_k$ its differential in the fixed point $(0,0)$.
In the next	result we sum up the main properties of	$P_k$ and $L_k$.

\begin{thm}\label{thm:0}
	For	every $k>0$, let $P_k$ and $L_k$ be	as above. Then
	\begin{enumerate}
	\renewcommand{\labelenumi}{(\arabic{enumi})}
		\item  $P_k$ satisfies (P1)--(P4);

		\item  $\det L_k=1$;

		\item  if $L_k^{ij}$ are the entries of	$L_k$, then
		$L_k^{11}=L_k^{22}$.
	\end{enumerate}
\end{thm}
We do not prove	 such properties, since	they are well known
in the literature  (see	\cite{ghisi-gobbino}).

Thanks to Theorem \ref{thm:res}	the	main result	of this	paper
 (Theorem \ref{thm:main}) is reduced to	prove that $U_{k}$ is
orbitally stable if	$k$	is large. Thanks to	Theorem	\ref{thm:OSiffPS}
and	Theorem	\ref{thm:KAM}, the main	result will	be proved if we	show
that $P_k$ satisfies assumptions (KAM 1) and (KAM 2) of	Theorem	\ref{thm:KAM}.
Assumption (KAM	1) follows from	statements (1)--(3)	of the following
result,	where the behaviour	of $L_k$ for large $k$ is considered.

\begin{thm}
	\label{thm:1}
	Let	$\nu\neq 1$	be a positive real number.
	Then there exist $k_{1}>0$,	$\omega:(k_{1},+\infty)\to\mathbb{R}$, and
	$\delta:(k_{1},+\infty)\to (0,+\infty)$	such that
	\begin{enumerate}
	\renewcommand{\labelenumi}{(\arabic{enumi})}
		\item  for every $k\geq	k_{1} $	the	eigenvalues	of $L_k$ are
		$\left\{e^{\pm i \omega(k)}\right\}$;

		\item  $\omega(k)\to 2\pi\sqrt{\nu}$ as	$k\to +\infty$;

		\item  $\omega(k)\neq 2\pi\sqrt{\nu}$ for $k$ large	enough if
		$2\pi\sqrt{\nu}	=
		h\pi$ for some $h\in\mathbb{Z}$;

		\item  setting $D(k)=\begin{pmatrix}
			1 &	0  \\
			0 &	\delta(k)
		\end{pmatrix}$ we have that
		$\left[D(k)\right]^{-1}L_k D(k)=R_{\omega(k)}$;

		\item  $\delta(k)\to \delta	> 0$ as	$k\to +\infty$.
	\end{enumerate}
\end{thm}

Statements (2) and (3) prevent $e^{i\omega(k)}$	from being a $h$-th
root of	1 for $h\in\{1,2,3,4\}$	and	$k$	large.	Indeed,	if
$e^{2\pi\sqrt{\nu}hi}\neq 1$ for $h\in\{1,2,3,4\}$,	then by	(2)	the
same holds true	for	$e^{\omega(k)hi}$, provided	that $k$ is	large
enough;	if on the contrary $e^{2\pi\sqrt{\nu}i}$ is	a $h$-th root of 1
for	some $h\in\{1,2,3,4\}$,	then for $k$ large $e^{\omega(k)i}$	is
not	because	of (3).	 This shows	in particular that $(0,0)$ is an
elliptic fixed point of	$P_{k}$	for	$k$	large.
Statement (4) says that	$L_k$ can be written in	the
canonical form by a	diagonal matrix	$D(k)$.

The	following result implies that $P_k$	satisfies assumption (KAM 2)
for	$k$	large.

\begin{thm}
	\label{thm:2}
	Let	 $\gamma_{k}:=\gamma(P_k)$ be as in	formula
	(\ref{defn:gamma}).
	Then
	$\gamma_{k}\neq	0$ for $k$	large enough.
\end{thm}

We have	therefore reduced the proof	of Theorem \ref{thm:main} to the
proof of Theorem \ref{thm:1} and Theorem \ref{thm:2}.

\section{Proofs}

Throughout this	section	we denote by
c various constants	depending only on function $m$.

\subsection{Properties of the function $m$}
In the following lemmata we	state all properties of	function $m$ we	need in	the
proofs.
\begin{lemma} \label{lemma0}
	As $k\to +\infty$,
	\begin{equation}
 \lambda_{k} :=\frac{k^{2}m'(k^{2})}{m(k^{2})} \to 0.
	\label{tempo}
\end{equation}
\end{lemma}

\begin{lemma} \label{lemma1}
	For	all	$y \in (0,1)$,
	$$\frac{m(k^{2}y)}{m(k^{2})} = 1 + \lambda_{k}\log y + \phi_{k}(y)$$
	where:
	\begin{enumerate}	\renewcommand{\labelenumi}{(m\arabic{enumi})}
		\item  \label{p1}$|\phi_{k}(y)|	\leq c \lambda_{k}|\log	y|$,

		\item  \label{p2} for all $0 < a < 1$, $\lim_{k\to +\infty}
	\sup_{a	\leq y \leq	1} \lambda_{k}^{-1}|\phi_{k}(y)| =0$.
	\end{enumerate}
\end{lemma}

\begin{lemma} \label{lemma2}
For	all	$x > 0$:
	\begin{enumerate}\renewcommand{\labelenumi}{(M\arabic{enumi})}
		\item  \label{m1} $m(x)\leq	c(1	+ \sqrt[8]{x})$;

		\item	\label{m2} $M(x) = x m(x) -	\int_{0}^x sm'(s) \,ds$;

		\item	\label{m3} $k^{2}m(k^{2})(M(k^{2}))^{-1} = 1 + \lambda_{k} +
		o(\lambda_{k})$;

		\item	\label{m4} $\lim_{k\to + \infty} M(kx)(M(k))^{-1} =	x$;

		\item  \label{m5}  $\sup\{(1 - y^{2})M(k^{2}) (M(k^{2})	- M(k^{2}y^{2}))^{-1}: \; y
		\in	(0,1)\}	\leq c$.
	\end{enumerate}
\end{lemma}

In the next	lemma we state some	properties of  $\tau_{k}$,
as defined in (\ref{deft}).

\begin{lemma} \label{lemma3}
The following equalities hold
	\begin{equation}
		\frac{\tau_{k} \sqrt{M(k^{2})}}{4k}	= \frac{\pi}{2}	+ \frac{\lambda_{k}}{2}
	 \int_{0}^{1} \frac{y^{2}\log y^{2}}{(1	- y^{2})^{3/2}}
	dy + o(\lambda_{k})	=: \frac{\pi}{2} +
	\lambda_{k}	h_{0}  + o(\lambda_{k}).
		\label{defh0}
	 \end{equation}
Moreover
	$$\tau_{k}^{2} m(k^{2})	= 4\pi^{2} + (4\pi^{2} +
	16h_{0}\pi)\lambda_{k} + o(\lambda_{k}).$$
\end{lemma}

 \begin{proof}[Proof of	Lemma \ref{lemma0}]
 Since
 $$\lambda_{k}^{-1}	=  \frac{\int_{0}^{k^{2}}m'(s)\, ds	+
 m(0)}{k^{2}m'(k^{2})} \geq	\int_{0}^{1} \frac{m'(k^{2}y)}{m'(k^{2})}\,
 dy,$$
 for all $\varepsilon >	0$,	using hypotheses (H\ref{H3})--(H\ref{H4}) we get
 $$\liminf_{k\to +\infty}\lambda_{k}^{-1}  \geq
  \liminf_{k\to	+\infty}\int_{\varepsilon}^{1}
  \frac{m'(k^{2}y)}{m'(k^{2})}\,
 dy	= \int_{\varepsilon}^{1}\frac{1}{y}	\, dy.
 $$
  Since	$\varepsilon$ is arbitrary we have obtained	that
  $\lim_{k\to +\infty} \lambda_{k}^{-1}	= +\infty$.	\end{proof}

 \begin{proof}[Proof of	Lemma \ref{lemma1}]
 Firstly let us	observe	that
 $$\frac{m(k^{2}y) - m(k^{2})}{m(k^{2})}  -	\lambda_{k}\log	y =	-
 \int_{y}^{1} \frac{k^{2}m'(k^{2}s)}{m(k^{2})} ds +	\int_{y}^{1}
 \frac{\lambda_{k}}{s} ds =	\phi_{k}(y).$$
 To	prove (m\ref{p1}) it suffices to remark	that by	hypothesis (H\ref{H3}),	we obtain
 $$
 |\phi_{k}(y)| \leq
 \lambda_{k}\int_{y}^{1} \frac{1}{s}\left|\frac{m'(k^{2}s)s}{m'(k^{2})}	-
 1\right| \, ds	\leq -\lambda_{k}c\log y.
 $$
  Moreover	for	$y \geq	a >0$ holds	true
 $$|\phi_{k}(y)| \leq \lambda_{k} \int_{a}^{1}
 \frac{1}{s}\left|\frac{m'(k^{2}s)s}{m'(k^{2})}	- 1\right| \,
 ds.
 $$
 Passing now to	the	limit using	Lebesgue's Theorem for the dominate
 convergence and  (H\ref{H4}) we have (m\ref{p2}).
 \end{proof}

 \begin{proof}[Proof of	Lemma \ref{lemma2}]
 To	prove (M\ref{m1}) it suffices to remark	that by	 (\ref{tempo})
  we have $m'(x)/m(x) \leq 1/(8x)$ for large $x$.
 To	show property (M\ref{m2}) it is	enough
 integrate by parts	the	function $m$.
 Using property	(M\ref{m2})	 and hypothesis	(H\ref{H3})	we then	get:
 \begin{align*}
  \frac{k^{2}m(k^{2})}{M(k^{2})	} -	1 -	\lambda_{k}	&=	\int_{0}^{1}
 \frac{k^{4}y m'(k^{2}y)}{M(k^{2})}	dy - \lambda_{k} \bigskip\\
	 & =  \lambda_{k} \Big(\frac{\int_{0}^{1}
 y m'(k^{2}y)(m'(k^{2}))^{-1}dy	}{M(k^{2})(k^{2}m(k^{2}))^{-1}}	- 1\Big) \\
	 & =  \lambda_{k} \Big(\frac{\int_{0}^{1}
 y m'(k^{2}y)(m'(k^{2}))^{-1}dy	}{1	-\lambda_{k} \int_{0}^{1}
 y m'(k^{2}y)(m'(k^{2}))^{-1}dy	 } - 1\Big)	\\
 &	=	\lambda_{k}\Big(\frac{\int_{0}^{1}
 y m'(k^{2}y)(m'(k^{2}))^{-1}dy	}{1	 + o(1)} - 1\Big) =:   \lambda_{k} R_{k}.
 \end{align*}
 Since by Lebesgue's Theorem $\lim_{k\to +\infty} R_{k}	=
 0$, we	have obtained (M\ref{m3}).

 Property (M\ref{m4}) follows from L' Hopital's	Theorem.
 Finally, to show  property	 (M\ref{m5}) it	is enough to remark	that,
 using	Cauchy's Theorem and the monotonicity of $m$ we	find
 $$\frac{M(k^{2}y^{2})}{M(k^{2}) }=	\frac{m(\xi	y^{2})}{m(\xi) }y^{2}\leq
 y^{2}.
 $$
 \end{proof}

\begin{proof}[Proof	of Lemma \ref{lemma3}]
Let
$$\psi_{k}(y) =	\frac{1}{\sqrt{1 - y^{2}}\sqrt{1 -
M(k^{2}y^{2})/M(k^{2})}	(\sqrt{1 - y^{2}} +\sqrt{1 -
M(k^{2}y^{2})/M(k^{2})}) }.$$
Applying Lemma \ref{lemma2}	and	Lemma \ref{lemma1}	it turns out that
\begin{align*}
& \frac{\tau_{k} \sqrt{M(k^{2})}}{4k}- \frac{\pi}{2}\\
&= \int_{0}^{1}\frac{1}{\sqrt{1	- M(k^{2}y^{2})/M(k^{2})}} -
	 \frac{1}{\sqrt{1 -	y^{2}}}\, dy \\
& =	\int_{0}^{1} \psi_{k}(y) \Big(-y^{2} +\frac{
	 M(k^{2}y^{2})}{M(k^{2})}\Big) \, dy \\
& =	\int_{0}^{1} \Big(y^{2}k^{2}(m(k^{2}y^{2}) - m(k^{2})) +
	 y^{2}\int_{0}^{k^{2}} sm'(s)\,	ds -
	  \int_{0}^{y^{2}k^{2}}	sm'(s)\, ds\Big)
	  \frac{\psi_{k}(y)}{M(k^{2})} \, dy  \\
& =	-\int_{0}^{1} y^{2}\Big[1 -
	 \frac{m(k^{2}y^{2})}{m(k^{2})}\Big] (1	+ o(1))\psi_{k}(y) \, dy +	\\
&\quad + \int_{0}^{1}(1	+o(1))\frac{\lambda_{k}	}{k^{4}}
	 \Big[ y^{2} \int_{0}^{k^{2}} \frac{sm'(s)}{m'(k^{2})}\,ds -
	 \int_{0}^{k^{2}y^{2}}
	 \frac{sm'(s)}{m'(k^{2})}\,ds\Big]\psi_{k}(y) \, dy	 \\
& =	\int_{0}^{1} -y^{2}\left[-\lambda_{k} \log
	 y^{2} - \phi_{k}(y^{2})\right]	(1 +o(1))\psi_{k}(y) \,	dy + \\
&\quad + \lambda_{k}\int_{0}^{1}(1 +o(1))\Big[(y^{2} - 1)
	 \int_{0}^{1} \frac{sm'(k^{2}s)}{m'(k^{2})}\,ds	+
	 \int_{y^{2}}^{1}
	 \frac{sm'(k^{2}s)}{m'(k^{2})}\,ds\Big]\psi_{k}(y) \, dy.
\end{align*}
Therefore,
\begin{align*}
& \Big(\frac{\tau_{k} \sqrt{M(k^{2})}}{4k}-	\frac{\pi}{2} -
	 h_{0}\lambda_{k}\Big)\lambda_{k}^{-1} \\
&=	\int_{0}^{1}y^{2} \log y^{2}\Big(
	 \psi_{k}(y) - \frac{1}{2(1	- y^{2})^{3/2}}\Big) \,	dy +  \\
&\quad + \int_{0}^{1} o(1)y^{2}	\log y^{2}
	 \psi_{k}(y)  +	(1 + o(1))y^{2}\psi_{k}(y)
	 \frac{\phi_{k}(y^{2})}{\lambda_{k}} \,	dy +  \\
&\quad + \int_{0}^{1} (1 + o(1))\psi_{k}(y)	(1 -
	 y^{2})\Big[-\int_{0}^{1}\frac{sm'(k^{2}s)}{m'(k^{2})}\,ds	+
	 \frac{1}{1	- y^{2}} \int_{y^{2}}^{1}\frac{sm'(k^{2}s)}{m'(k^{2})}\,ds
	 \Big].
\end{align*}
Let	us remark that by  Lemma \ref{lemma2}, properties  (M\ref{m4})--(M\ref{m5}),
$ (1 - y^{2})^{3/2}\psi_{k}(y) $ is	a bounded function and converges
for	$y \in (0,1)$ to the function identically  $= 1/2$,	hence, using
once more hypotheses (H\ref{H3})--(H\ref{H4}) and Lemma	\ref{lemma1},
by Lebesgue's Theorem we obtain
$$	\lim_{k\to + \infty} \Big(\frac{\tau_{k} \sqrt{M(k)}}{4k}- \frac{\pi}{2} -
	 h_{0}\lambda_{k}\Big)\lambda_{k}^{-1}	= 0.
$$
	 In	order to prove the second part of the lemma	it suffices	to observe that
	$$
		\tau_{k}^{2} m(k^{2})  =  \Big(\frac{\tau_{k}
		\sqrt{M(k^{2})}}{4k} \Big)^{2} 16 k^{2}	\frac{m(k^{2})}{M(k^{2})}
		 =	 16( \frac{\pi}{2} + h_{0} \lambda_{k} + o(\lambda_{k}))^{2}
		  (	1 +	\lambda_{k}	+ o(\lambda_{k})).
	$$
\end{proof}

\subsection{Properties of the simple mode $U_{k}$}
	 Let us	recall that	$U_{k}$	is the solution	of the problem:
	 \begin{equation}
		U_{k}''	+ \tau_{k}^{2} m(k^{2}U_{k}^{2})U_{k} =	0 \quad
		U_{k}(0) = 1, \:\: U_{k}'(0) = 0.
		\label{sm}
	 \end{equation}
 In	the	sequel we  need	the	following simple properties	of $U_{k}$:
 \begin{enumerate}
	   \renewcommand{\labelenumi}{(U\arabic{enumi})}
	   \item  $U_{k}$ is a 1-periodic function,	and	for	every $t\in[0,1/4]$,
	   $$U_{k}(t)=U_{k}(1-t)=-U_{k}(1/2-t)=-U_{k}(1/2+t);$$

	   \item  $U_{k}$ is decreasing	in $[0,1/2]$ and increasing	in
	   $[1/2,1]$;

	   \item  for every	$t\in[0,1]$	we have	that
	   $$|U_{k}'|^{2} +	\frac{\tau_{k}^{2}}{k^{2}}\int_{0}^{k^{2}U_{k}^{2}}
		m(s)\,
	   ds =	\frac{\tau_{k}^{2}}{k^{2}}\int_{0}^{k^{2}} m(s)\,
	   ds ;$$

	   \item  $|U_{k}(t)|\leq 1$ for every $t\in [0,1]$;

	   \item  ${ |U_{k}'(t)|\leq \frac{\tau_k}{k}\sqrt{M(k^{2})}}$
	   for every $t\in [0,1]$.
  \end{enumerate}
Properties (U1)	and	(U2) follow	from the symmetries	of $U_{k}$;	(U3)
  follows from the conservation	of the Hamiltonian for $U_{k}$,	and
  (U4) and (U5)	are	consequences of	(U3).

The	simple mode	verifies also the following	properties.

	  \begin{lemma}
		\label{lemma4}
		One has:
		\begin{enumerate} \renewcommand{\labelenumi}{(B\arabic{enumi})}
			\item \label{U1} $U_{k}(t) = \cos(2\pi t) +	o(1)$ where	$o(1)$ is
			uniform	in $t$ in bounded intervals;

			\item  \label{U2} there	exist $\lambda > 0$, $0	< t_{0}	< 1/4$
			and	$k_{0} \in \mathbb{R}$
			such that for all
			$k \geq	k_{0}$,
			$|U_{k}'(t)| \geq \lambda$ for all
		$t \in\big[t_{0},\frac{1}{4}\big]$;

			\item \label{U3} there exists $c(t)\in L^{1}([0,1])$ such
			that for $k	$ large: $|\log(U_{k}^{2}(t))| \leq	c(t)$ for
			all	$t \in [0,1]$.
		 \end{enumerate}
		 \end{lemma}

Throughout the paper we	need also  some	properties of integrals	of $U_{k}$.

		\begin{lemma}
			\label{lemma7}
		The following inequalities hold true
			\begin{equation}
			 \int_{0}^{1}	\left|1	-
	 \frac{\tau_{k}^{2}}{4\pi^{2}}m(k^{2}U_{k}(t))\right|\,	dt
	\leq c\lambda_{k};
				\label{U4}
			 \end{equation}
\begin{equation}
	\int_{0}^{1}\frac{|U_{k}(t)|}{1	+ k^{2}U_{k}^{2}(t)} dt	\leq
			c\,	\frac{\log k}{k^{2}};
	\label{int1}
\end{equation}
		\begin{equation}
			\int_{0}^{1} m'(k^{2}U_{k}(s)^{2})\, ds	\leq \frac{c}{k}.
			\label{int2}
		\end{equation}
	\end{lemma}

\begin{proof}[Proof	of Lemma \ref{lemma4}]
 By	Lemma \ref{lemma1} and Lemma \ref{lemma3} we obtain
		$$	U_{k}''	+ (4\pi^{2}	+ o(1))	( 1	+ \lambda_{k}\log
	U_{k}^{2} +	\phi_{k}(U_{k}^{2})	) U_{k}	= 0, $$
 that we can rewrite as
	\[
	U_{k}''	+ 4\pi^{2} U_{k}	=  (4\pi^{2} + o(1))( \lambda_{k}\log
	U_{k}^{2} +	\phi_{k}(U_{k}^{2}))  U_{k}	+  o(1)	U_{k}.
	\]
	Since, thanks to (m\ref{p1}) and (U4)
	$$\big[\log
	U_{k}^{2} +	\frac{\phi_{k}(U_{k}^{2})}{\lambda_{k}})\big] U_{k}$$ is
	a bounded function,	we get
	 $	U_{k}''	+ 4\pi^{2} U_{k} = o(1)	$.
	Moreover, setting  $v_{k}(t) = U_{k}(t ) - \cos(2\pi t)$,
	  we find
	 $v_{k}'' +	4\pi^{2} v_{k}=	o(1)$,
	 hence (B\ref{U1}) follows from
	 $$\sqrt{|v_{k}'(t)|^{2} + 4\pi^{2}|v_{k}(t)|^{2}} \leq	o(1)t.
$$
	 By	 (B\ref{U1}) there exists $t_{0} \in (0,1/4)$ such that
$|U_{k}(t)|	\leq 1/2$ for all $t \in [t_{0},1/4]$ and $k$ large.
To prove (B\ref{U2}) it	is enough to remark	that, by Lemma \ref{lemma1},
 Lemma \ref{lemma3}	we have
$$
\inf_{t_{0}	\leq t	  \leq 1/4}|U_{k}'(t)| = \inf_{t_{0} \leq t	\leq
1/4}\int_{U_{k}^{2}(t)}^{1}	\tau_{k}^{2}
m(k^{2}s)\,ds \, \geq \, \int_{1/4}^{1}	\tau_{k}^{2}
m(k^{2}s)\,	ds	\to	3\pi^{2}.
$$
Let	now	$t_{0}$	be as in  (B\ref{U2}). By  (B\ref{U1}) and (U4)	for
large $k$ we get $1	\geq |U_{k}(t)|	\geq c > 0$	for	all	$t \in [0,t_{0}]$.
Using (B\ref{U2}) and Cauchy's Theorem,	for	large $k$ and  $t \in [t_{0},1/4]$ we get
$$	\frac{U_{k}(t)}{1/4	- t} = - U_{k}'(\xi_{t}) \geq \lambda, $$
hence
$$\big|\frac{\sqrt{U_{k}(t)}\log(U_{k}^{2}(t))}{\sqrt{U_{k}(t)}}\big|
\leq \frac{c}{\sqrt{1/4	- t}}.
$$
Then (B\ref{U3}) follows from the symmetries of	the	function $U_{k}$.
 \end{proof}

 \begin{proof}[Proof of	Lemma \ref{lemma7}]
 To	show (\ref{U4})	it suffices	to observe that, by
  Lemma	\ref{lemma1}, Lemma	\ref{lemma3} and Lemma \ref{lemma4}, we	have that
	 \begin{align*}
		 \int_{0}^{1}	\big|1 -
	 \frac{\tau_{k}^{2}}{4\pi^{2}}m(k^{2}U_{k}(s))\big|\, ds
 & =  \int_{0}^{1} \big|1 -	\big(1 + (1	+ 4\frac{h_{0}}{\pi}) \lambda_{k} +	o(
	\lambda_{k})\big)\frac{m(k^{2}U_{k}(s))}{m(k^{2})}\big|\, ds  \\
 & \leq	  c\lambda_{k} \int_{0}^{1}	1 +	\big|
	 \log (U_{k}(s)^{2}) + \frac{\phi_{k}(U_{k}(s)^{2})
	 }{\lambda_{k}}\big|\, ds \leq c \lambda_{k}.
 \end{align*}
	  Thanks to	(U1), in order to prove	 (\ref{int1}) it is	enough to show that
	  $$\int_{0}^{1/4}\frac{|U_{k}(t)|}{1 +	k^{2}U_{k}^{2}(t)} dt  \leq
	  c\frac{\log k}{k^{2}}.
$$
	 Let us	take $t_{0}$ as	in Lemma \ref{lemma4}, (B\ref{U2}) and let us
divide the integral
 as	follows
 $$\int_{0}^{1/4} \frac{|U_{k}(t)|}{1 +	k^{2}U_{k}^{2}(t)} dt =
 \int_{0}^{t_{0}} \frac{|U_{k}(t)|}{1 +	k^{2}U_{k}^{2}(t)} dt +
 \int_{t_{0}}^{1/4}	\frac{|U_{k}(t)|}{1	+ k^{2}U_{k}^{2}(t)} dt.
 $$
 Now we	can	estimate the two terms separately.
 By	(U4) and Lemma \ref{lemma4}, (B\ref{U1}),  for large $k$ we	have that:
 $1	\geq |U_{k}(t) | \geq c$  for all $t \in [0,t_{0}]$,
 hence
 $$	\int_{0}^{t_{0}} \frac{|U_{k}(t)|}{1 + k^{2}U_{k}^{2}(t)} dt \leq
 \frac{c}{k^{2}}.$$
 Let $\lambda$ be as in	Lemma \ref{lemma4},	(B\ref{U2}). Since
 $U_{k}(1/4) = 0$, we get
\[
  \int_{t_{0}}^{1/4} \frac{|U_{k}(t)|}{1 + k^{2}U_{k}^{2}(t)}
	 dt	= -\int_{t_{0}}^{1/4} \frac{U_{k}(t) U_{k}'(t)}{(1 +
 k^{2}U_{k}^{2}(t))|U_{k}'(t)|}	dt	 \leq	 \frac{1}{\lambda k^{2}}\log(1 +
 k^{2}U_{k}^{2}(t_{0})).
\]
To prove (\ref{int2}) we can proceed as	to prove (\ref{int1}).
Only we	remark that, since by (\ref{tempo})	and	(M\ref{m1})	one	has
$$ m'(z^{2}) \leq \frac{c\,	m(z^{2})}{(1 + z^{2})} \leq	\frac{c}{1 +
z^{7/4}},$$	then in	$[0,t_{0}]$	we get $m'(k^{2}U_{k}^{2}(t)) \leq
c/k^{7/4}$ and in $[t_{0},1/4]$:
\begin{align*}
\int_{t_{0}}^{1/4} m'(k^{2}U_{k}^{2}(s))\, ds  &
=	-\int_{t_{0}}^{1/4}	\frac{m'(k^{2}U_{k}^{2}(s))U_{k}'(s)}{|U_{k}'(s)|}\,
  ds  \leq - c \int_{t_{0}}^{1/4}m'(k^{2}U_{k}^{2}(s))U_{k}'(s)	 \\
&=	-\frac{c}{k}\int_{kU_{k}(t_{0})}^{0} m'(z^{2})\, dz
	 \leq \frac{c}{k}\int_{0}^{+\infty}	m'(z^{2})\,	dz <
	 +\infty.
\end{align*}
\end{proof}

\subsection{Linear stability}

\subsubsection{Preliminary	results	for	$k$	fixed}
  Let $P_k$	be the Poincar\'{e}	map	associated with
$U_{k}$, and let $L_k$ be its differential in $(0,0)$.	Then the
linear operator	$L_k:\mathbb{R}^{2}\to\mathbb{R}^{2}$ can be characterized in the
following way.

Given $(x,y)\in\mathbb{R}^{2}$,	let	$z_k(t)$ be	the	solution of	the
linear problem
\begin{equation}
	z_k''(t)+\nu\,\tau_{k}^{2} m(k^{2}U_{k}^{2}(t))z_k(t)=0,\quad
	z_k(0)=x,\quad z_k'(0)=	2\pi\sqrt{\nu}\,y.
	\label{defn:vep}
\end{equation}
This problem is	the	linearization of the second	equation of	system
(\ref{rescaled}).  Then	we have	that
$$
L_k(x,y):=\left(z_k(1),(4\pi^{2}\nu)^{-1/2}z_k'(1)\right).
$$
We do not give the proof of	this characterization, since it	is completely
analogous to the proof of \cite[Proposition	2.1]{cazenave}.
However, $L_k$ is the first	term in	the	Taylor expansion of	$P_k$
and	in section \ref{orbstab} we	find the first three terms of the
expansion of $P_k$.

The	fundamental	tool in	the	analysis of	the	behaviour of $L_k$ for	$k$
large is the following linear algebra result, whose
proof is omitted since it is analogous of the corresponding	one	in
\cite{ghisi-gobbino} (Proposition 4.1).

\begin{prop}
	\label{prop:linear-algebra}
	Let	$k_{0}>0$, and let $A:(k_{0}, +\infty)\to M_{2\times 2}$. Let us
	assume that
	\begin{enumerate}
	\renewcommand{\labelenumi}{(\roman{enumi})}
		\item  $\det A(k)=1$ for every $k\geq k_{0}$;

		\item  $a_{11}(k)=a_{22}(k)$ for every $k\geq k_{0}$;

		\item  there exist $\omega_{0}\in\mathbb{R}$  and $B\in	M_{2\times 2}$
		such that, for $k\to +\infty$,
		$A(k)=R_{\omega_{0}}+\lambda_{k}B+o(\lambda_{k})$;

		\item  $\omega_{0}$	and	$B$	satisfy	one	of the following conditions:
		\begin{enumerate}
		\renewcommand{\labelenumii}{(\roman{enumi}-\arabic{enumii})}
			\item  $\omega_{0}\neq h\pi$ for every $h\in\mathbb{Z}$;

			\item  $\omega_{0}=	h\pi$ for some $h\in\mathbb{Z}$, $\mathrm{Tr} B	= 0$, and
			$b_{12}\cdot b_{21}	< 0$.
		\end{enumerate}
	\end{enumerate}
 Then there	exist $k_{1}\geq k_{0}$,
	$\omega:(k_{1},	+\infty)\to\mathbb{R}$,	and	$\delta:(k_{1},	+\infty)\to
	(0,+\infty)$ such that
	\begin{enumerate}
	\renewcommand{\labelenumi}{(\arabic{enumi})}
		\item  for every $k\geq	k_{1}$ the eigenvalues of $A(k)$ are
		$\left\{e^{\pm i \omega(k)}\right\}$;

		\item  $\omega(k)\to \omega_{0}$ as	$k\to +\infty$;

		\item  $\omega(k)\neq \omega_{0}$ for $k$ large	enough if
		$\omega_{0}	= h\pi$	for	some $h	\in	\mathbb{Z}$;

		\item  setting $D(k)= \begin{pmatrix}
			1 &	0  \\
			0 &	\delta(k)
		\end{pmatrix} $	we have	that
		$\left[D(k)\right]^{-1}A(k)	D(k)=R_{\omega(k)}$;

		\item  $\delta(k)\to \delta	$ as $k\to +\infty$	when $\delta = 1$
		if $\omega_{0} \neq	h\pi$ for all $h \in \mathbb{Z}$ and $\delta =
		\sqrt{-b_{21}/b_{12}}$ otherwise.
	\end{enumerate}
\end{prop}

We use also	the	following lemma.

\begin{lemma}
	\label{lemmaint} The following equalities hold true.
	\begin{enumerate}\renewcommand{\labelenumi}{(S\arabic{enumi})}
		\item	For	all	$h \geq	1$ we have that
	$$\alpha_{h} :=	\int_{0}^{\pi/2} \sin(2	h x) \frac{\sin	x}{\cos
	x} \, dx = (-1)^{h + 1}\frac{\pi}{2}.$$

		\item If $h_{0}$ is	as in (\ref{defh0})	 then
	$$ h_{0} = -\frac{\pi}{2} -	\frac{1}{2}	\int_{0}^{\pi/2}
		\log(\cos^{2}x)	\, dx.$$

		\item  For all $h \geq 1$, $h \in \mathbb{N}$ one has
		$$\int_{0}^{\pi/2} 2 \sin^{2}(hx) \log(\cos^{2}x) =
		 (-1)^{h}\frac{\pi}{2h}	+ \int_{0}^{\pi/2}
		\log(\cos^{2}x)	\, dx.$$
	\end{enumerate}
\end{lemma}

\begin{proof}[Proof	of Lemma \ref{lemmaint}]
We prove (S1).
Obviously we have $\alpha_{1} =	\pi/2$;	moreover for $h	> 1$ thesis
follows	from
\begin{align*}
&\int_{0}^{\pi/2} \sin(2 (h-1) x + 2x) \frac{\sin x}{\cos x} \,	dx \\
&=\int_{0}^{\pi/2} 2\cos(2(h-1)x)\sin^{2}x \, dx + \int_{0}^{\pi/2} \sin(2(h-1)x)	(2\cos^{2}x	- 1)\frac{\sin
	x}{\cos	x} \, dx \\
& =	\int_{0}^{\pi/2} \cos(2(h-1)x) (1 -	\cos 2x)\, dx +	\int_{0}^{\pi/2}
  \sin(2(h-1)x)	\sin 2x	\, dx -	\alpha_{h-1}	 \\
& =	- \alpha_{h-1} - \int_{0}^{\pi/2} \cos((2(h-1) + 2)x) \, dx
	+ \frac{\sin(2(h-1)x)}{2(h-1)}\Big|_{0}^{\pi/2}	= -
	\alpha_{h-1}
\end{align*}
To prove (S2) it suffices a	change of variables	and	an integration by
parts.
To prove (S3) we need only to remark that
\begin{align*}
&\int_{0}^{\pi/2} 2	\sin^{2}(hx) \log(\cos^{2}x) \,	dx \\
&=\int_{0}^{\pi/2} \log(\cos^{2}x) \, dx - \int_{0}^{\pi/2}	\cos(2h	x)
\log(\cos^{2}x)	\, dx \\
& =	\int_{0}^{\pi/2} \log(\cos^{2}x) \,	dx - \left.
	\frac{\sin(2hx)}{2h}\log(\cos^{2}x)\right|_{0}^{\pi/2} -
	\frac{\alpha_{h}}{h}.
\end{align*}
 \end{proof}

\subsubsection{Polar coordinates for $z_k(t)$.} \label{sec:polar1}
We write (\ref{defn:vep}) as a first order system. To this end we
set
$x_{k}(t)=z_k(t)$, $y_{k}(t)=(\nu 4\pi^{2})^{-1/2}z_k'(t)$,
so that	(\ref{defn:vep}) becomes
$$x_{k}'(t)=\sqrt{\nu 4\pi^{2}}\:y_{k}(t),\quad
y_{k}'(t)=-(\nu	4\pi^{2})^{-1/2}\,\nu\,\tau_{k}^{2}	m(k^{2}U_{k}^{2}(t))\,x_{k}(t),$$
with initial data
$x_{k}(0)=x$, $y_{k}(0)=y$.

If $(x,y)\neq (0,0)$, then $(x_{k}(t),y_{k}(t))\neq	(0,0)$ for
every $t\in\mathbb{R}$.	We can therefore study this	system introducing polar
coordinates	$\rho_k(t)$, $\theta_k(t)$ such	that
\begin{equation}
	x_{k}(t)=\rho_k(t)\cos\theta_k(t),\quad	y_{k}(t)=\rho_k(t)\sin\theta_k(t).
	\label{cambio}
\end{equation}
In a standard way it turns out that	$\rho_k$ and $\theta_k$	solve the
system
\begin{gather}
	\rho_k'	=  2\pi\sqrt{\nu}\,\rho_k\sin\theta_k\cos\theta_k\big\{1-
	\tau_{k}^{2}\frac{m(k^{2}U_{k}^{2})}{4\pi^{2}}\big\},
	\label{eqn:rep}	\\
	\theta_k' =
	-2\pi\sqrt{\nu}\big\{\sin^{2}\theta_k+\tau_{k}^{2}\frac{m(k^{2}U_{k}^{2})}{4\pi^{2}}
	\cos^{2}\theta_k\big\},
	\label{eqn:thep}
\end{gather}
with initial data
$\rho_k(0)=\rho$, $\theta_k(0)=\theta,$
such that $x=\rho\cos\theta$, $y=\rho\sin\theta$.

\subsubsection{Behaviour of $\rho_k$ and $\theta_k$ for large $k$}\label{abrt}
We look	for	functions $\rho_{0,k}$,	$\rho_{2,k}$, $\theta_{0,k}$,
$\theta_{2,k}$ such	that, as $k\to +\infty$:
\begin{gather}
	\rho_k(t)  =  \rho_{0,k}(t)+\rho_{2,k}(t)\lambda_{k}+o(\lambda_{k}),
	\label{exp:rep}	 \\
	\theta_k(t)	 =	\theta_{0,k}(t)+\theta_{2,k}(t)\lambda_{k}+o(\lambda_{k}),
	\label{exp:thep}
\end{gather}
 where $o(\lambda_{k})$	is uniform in $t\in	[0,1]$.
We prove that
 $\rho_{0,k}(t)	\equiv \rho$  and $\rho_{2,k}(t)$ solves
 $$
  \rho_{2,k}'(t) = -\rho\pi\sqrt{\nu}\Big(1	+\frac{4h_{0}}{\pi}	+
 \log(U_{k}^{2}(t))\Big)\sin(2\theta - 4\pi\sqrt{\nu}t)	\quad
 \rho_{2,k}(0) = 0,
 $$
 while $\theta_{0,k}(t)	=\theta	- 2\pi\sqrt{\nu}t$,	and	$\theta_{2,k}$ solves
 $$
 \theta_{2,k}'(t) =	-2\pi\sqrt{\nu}\Big(1 +\frac{4h_{0}}{\pi} +
 \log(U_{k}^{2}(t))\Big)\cos^{2}(\theta	- 2\pi\sqrt{\nu}t) \quad
 \theta_{2,k}(0) = 0.
 $$
 Thanks	to (\ref{eqn:rep}) we find:
$$
 \rho_k(t)	\leq  c\int_{0}^t \rho_k(s)\Big| 1 -
 \tau_{k}^{2}\frac{m(k^{2}U_{k}^{2}(s))}{4\pi^{2}}\Big|	\, ds +	\rho,
 $$
 hence using  (\ref{U4}) we
 obtain	that,  for all $t \in [0,1]$, $|\rho_k(t) -	\rho
 |\leq c\lambda_{k}$. In the same way we also get
 $ \left|\theta_k(t) + 2\pi\sqrt{\nu}t - \theta	\right|\leq	c \lambda_{k}$.
Moreover using	Lemma \ref{lemma1} and Lemma \ref{lemma3} for $t\in[0,1]$,
	\begin{align*}
& \frac{|\rho_k(t) - \rho -	 \lambda_{k}\rho_{2,k}(t)  |}{\lambda_{k}}\\
&\leq  \frac{c}{\lambda_{k}}\int_{0}^t\Big|1 -
 \tau_{k}^{2}\frac{m(k^{2}U_{k}^{2}(s))}{4\pi^{2}}
 + \lambda_{k}(1 +\frac{4h_{0}}{\pi} +
 \log(U_{k}^{2}(s)))\Big|\,	ds \\
&\quad +  \frac{c}{\lambda_{k}}\int_{0}^t\Big|1	-
 \tau_{k}^{2}\frac{m(k^{2}U_{k}^{2}(s))}{4\pi^{2}}\Big|	(|\rho_k(s)	- \rho
  |+  |\sin(2\theta_k(s)) -	\sin(2\theta - 4\pi\sqrt{\nu}s)|) \, ds	 \\
& \leq	\frac{c}{\lambda_{k}}\int_{0}^t|\lambda_{k}o(1)\log
  (U_{k}^{2}(s))|  + o(\lambda_{k})
  +	|\phi_{k}(U_{k}^{2}(s))|\, ds \\
&\quad + \frac{c}{\lambda_{k}}\int_{0}^t(\lambda_{k}
 + |\theta_k(s)	- \theta + 2\pi\sqrt{\nu}s|)\Big|1 -
 \tau_{k}^{2}\frac{m(k^{2}U_{k}^{2}(s))}{4\pi^{2}}\Big|	\, ds \\
& \leq c\int_{0}^{1}\Big|1 -
 \tau_{k}^{2}\frac{m(k^{2}U_{k}^{2}(s))}{4\pi^{2}}\Big|	+
 o(1)(|\log(U_{k}^{2}(s)) |	+ 1) \,	ds+
	c\int_{0}^{1}\frac{\phi_{k}(U_{k}^{2}(s))}{\lambda_{k}}	ds.	 \\
 \end{align*}
By Lemma \ref{lemma1}, (m\ref{p1})--(m\ref{p2})	and	Lemma \ref{lemma4},
(B\ref{U3}), in	a standard way we get
$$\lim_{k\to +\infty}
\int_{0}^{1}\frac{\phi_{k}(U_{k}^{2}(s))}{\lambda_{k}} ds =	0,$$
hence using	once more (B\ref{U3}) and (\ref{U4}) it	turns out that:
$$\lim_{k\to + \infty} \sup_{t\in[0,1]}
\frac{|\rho_k(t) - \rho	-  \lambda_{k}\rho_{2,k}(t)| }{\lambda_{k}}	= 0,$$
that is	(\ref{exp:rep}). In	a similar way one can prove	also
(\ref{exp:thep}).
By Lemma \ref{lemma4} we can now pass to limit using Lebesgue's	Theorem,
then
\begin{gather*}
\lim_{k\to +\infty}	\rho_{2,k}(1) =	-\pi\sqrt{\nu}\rho
\int_{0}^{1}\sin(2\theta - 4\pi\sqrt{\nu}t)(1 +	\frac{4h_{0}}{\pi} +
\log(\cos^{2}(2\pi t) )\, dt:= \rho_{1,\rho,\theta},
\\
\lim_{k\to +\infty}	\theta_{2,k}(1)	= -2\pi\sqrt{\nu}
\int_{0}^{1}\cos^{2}(\theta	- 2\pi\sqrt{\nu}t)(1 + \frac{4h_{0}}{\pi} +
\log(\cos^{2}(2\pi t)) \, dt:= \theta_{1,\rho,\theta},
\end{gather*}
hence
\begin{equation}
	\rho_k(1) =	\rho + \lambda_{k} \rho_{1,\rho,\theta}	+ o(\lambda_{k}), \quad
	\mbox{and} \quad	 \theta_k(1) = \theta -	2\pi \sqrt{\nu}
	  +	\lambda_{k}	\theta_{1,\rho,\theta} + o(\lambda_{k}).
	\label{asintli}
\end{equation}

\subsubsection{Behaviour of $L_{k}$ for large $k$.} \label{abl}
Now	let	us denote by
$L_{k}^{ij}$ the entries of	the	matrix $L_{k}$.	 Then it holds true
that $(L^{11}_{k},L^{21}_{k})=\left(z_k(1),	(4\pi^{2} \nu
)^{-1/2}z_k'(1)\right)$, where $z_k$ has initial data $x=1$, $y=0$,
corresponding to $\rho=1$, $\theta=0$.	By (\ref{cambio}) and
(\ref{asintli})	we obtain that
\begin{gather*}
L^{11}_{k}=\cos(2\pi\sqrt{\nu})+
\lambda_{k}	(\rho_{1,1,0}\cos(2\pi\sqrt{\nu}) +
\theta_{1,1,0}\sin(2\pi\sqrt{\nu}))	+o(\lambda_{k}),
\\
L^{21}_{k}=-\sin(2\pi\sqrt{\nu})+
\lambda_{k}
(-\rho_{1,1,0}\sin(2\pi\sqrt{\nu}) + \theta_{1,1,0}\cos(2\pi\sqrt{\nu}))
+o(\lambda_{k}).
\end{gather*}
Making the same	computations with initial data $x=0$,
$y=1$, corresponding to	$\rho=1$, $\theta=\pi/2$, we find that
$L^{22}_{k}=L^{11}_{k}$, and
$$
L^{12}_{k} =\sin(2\pi\sqrt{\nu})+ \lambda_{k}
(\rho_{1,1,\pi/2}\sin(2\pi\sqrt{\nu}) -	\theta_{1,1,\pi/2}\cos(2\pi\sqrt{\nu}))
+o(\lambda_{k}).
$$
We have	thus proved	that
$$
L_k=R_{\omega_{0}}+\lambda_{k}B+o(\lambda_{k}),
$$
where $\omega_{0}=2\pi\sqrt{\nu}$, and $B$ is a	matrix whose entries
are
\begin{gather*}
b_{11}=b_{22}=(\rho_{1,1,0}\cos(2\pi\sqrt{\nu})	+
\theta_{1,1,0}\sin(2\pi\sqrt{\nu})), \\
b_{12}=(\rho_{1,1,\pi/2}\sin(2\pi\sqrt{\nu}) -
\theta_{1,1,\pi/2}\cos(2\pi\sqrt{\nu})),\\
b_{21}=(-\rho_{1,1,0}\sin(2\pi\sqrt{\nu}) +
\theta_{1,1,0}\cos(2\pi\sqrt{\nu})).
\end{gather*}

\subsubsection{Properties of $B$.}\label{abb} \label{sec:propb}
 If	$\omega_{0}= h\pi$ for
some $h\in\mathbb{Z}$, then	the	matrix $B$ becomes
$$B=\pm	\begin{pmatrix}
	\rho_{1,1,0} & -\theta_{1,1,\pi/2}	\\
	\theta_{1,1,0} & \rho_{1,1,0}
\end{pmatrix}.
$$
In this	case we	have $\rho_{1,1,0} = 0$, indeed	it is the
integral of	a periodic (of period 1) odd function over the
interval $[0,1]$; therefore	$\mathrm{Tr} B=0$. Moreover	it holds
\begin{gather*}
-\theta_{1,1,\pi/2}	= \pi h	\int_{0}^{1} \sin^{2}(\pi h	t)
 \Big(1	+ \frac{4h_{0}}{\pi} +
\log(\cos^{2}(2\pi t)\Big) \, dt
\\
\theta_{1,1,0} = -\pi h	\int_{0}^{1} \cos^{2}(\pi h	t)
\Big(1 + \frac{4h_{0}}{\pi}	+
\log(\cos^{2}(2\pi t)\Big) \, dt.
\end{gather*}
If $h$ is odd, by Lemma	\ref{lemmaint},	(S2), computing, one has
\begin{align*}
-\frac{\theta_{1,1,\pi/2}}{\pi h}  &=
	\int_{0}^{1}\frac{1	- \cos(2h\pi t)}{2}\Big(1 +
	\frac{4h_{0}}{\pi} +\log(\cos^{2}2\pi t)\Big)\,	dx	\\
	& =	 \big(1	+
	\frac{4h_{0}}{\pi}\big)\frac{1}{2} +
	\frac{1}{\pi}\int_{0}^{\pi/2}\log(\cos^{2}x) \,	dx -	\frac{1}{4\pi}
\int_{0}^{2\pi}	\cos (ht) \log(\cos^{2}t)\,	dt \\
&= -1/2;
\end{align*}
indeed in this case
$$
\int_{0}^{\pi} \cos	(ht) \log(\cos^{2}t)\, dt =	- \int_{\pi}^{2\pi}	\cos (ht)
 \log(\cos^{2}t)\, dt. $$
In an analogous	way	we get also
$$\frac{\theta_{1,1,0}}{\pi h} =
	\frac{1}{2}	= -	\frac{\theta_{1,1,\pi/2}}{\pi h}.$$
If $h =	2j$	is even, then using	once more Lemma	\ref{lemmaint} and computing,
one	has
\begin{align*}
-\frac{\theta_{1,1,\pi/2}}{2\pi	j}
& =	 \int_{0}^{1}(1	+	 \frac{4h_{0}}{\pi})\sin^{2}(2\pi j	t) dt\,	+ \frac{1}{2\pi}
	\int_{0}^{2\pi}	\sin^{2}(jx) \log(\cos^{2}x)\, dx  \\
& =	 (1	+	 \frac{4h_{0}}{\pi})\frac{1}{2}	+ \frac{1}{\pi}
  \Big(\frac{(-1)^j	\pi}{2j} + \int_{0}^{\pi/2}\log(\cos^{2}x) \, dx \Big)\\
&= -\frac{1}{2}	+ \frac{(-1)^j }{2j} < 0.
\end{align*}
Since $\nu \neq	1$,	hence we get  $j \neq 1$; then work	as before we
find
\begin{eqnarray*}
	\frac{\theta_{1,1,0}}{2\pi j} &	= &	\frac{1}{2}	+ \frac{(-1)^j}{2j}	>
	0.
\end{eqnarray*}
We have	hence proved that in all cases $B$ satisfies hypothesis	(iv) of	Proposition
\ref{prop:linear-algebra}.

 \subsubsection{Proof of Theorem \ref{thm:1}}
 By	Subsections	\ref{abl} and \ref{abb}	 and Theorem \ref{thm:0} we	get	that
 $L_k$ satisfies all the assumptions of
 Proposition \ref{prop:linear-algebra}.	 Therefore statements (1)--(5)
 of	Theorem	\ref{thm:1}	follow from	the	corresponding statements of
 Proposition \ref{prop:linear-algebra}.

 \subsection{Orbital stability}	\label{orbstab}

 In	proofs,	we need	expansions of solutions	of Cauchy
 problems depending	on some	small parameter. We	will always	work
 formally as follows. Assume that the Cauchy problem is
 \begin{equation}
	 Z'=F(Z,\mu),\quad Z(0)=\Phi(\mu),	\label{defn:cauchy}
 \end{equation}
 where $\mu$ is	the	small parameter, and $Z(t)\in\mathbb{R}^k$ is the
 unknown. Then we look for an expansion	like
 \begin{equation}
	 Z(t)=Z_{0}(t)+Z_{1}(t)\mu+Z_{2}(t)\mu^{2}+\ldots+
	 Z_{h}(t)\mu^{h}+o(\mu^h).
	 \label{defn:exp}
 \end{equation}
 We	replace	$Z$	in (\ref{defn:cauchy}) with	this expression, and using
 the Taylor	formula, we	write also $F(Z,\mu)$ and $\Phi(\mu)$ as
 polynomials of	degree $h$ in $\mu$	(in	the	first case the
 coefficients depend on	$Z_{0},Z_{1},\ldots	Z_{h}$)	plus $o(\mu^h)$.
 Finally, considering the coefficients of $\mu^{0},
 \mu^{1},\ldots,\mu^h$,	we find	the	Cauchy problems	solved by
 $Z_{0},Z_{1},\ldots Z_{h}$.

 It	is well	known that,	if $F$ and $\Phi$ are smooth enough, then this
 procedure can be rigorously justified,	and	that (\ref{defn:exp})
 turns out to be uniform on	bounded	time intervals.
 To	avoid useless terms	in writing expansion
 (\ref{defn:exp}), we always omit from the beginning the terms which \emph{a
 posteriori} would turn	out	to be zero.

\subsubsection{Taylor expansions in $\rho$	for	$k$	fixed.}	\label{sec:thm2part1}

 In	this section $k$ will be fixed.	We compute the
first three	terms in the Taylor	expansion in a neighborhood	of $(0,0)$ of the
Poincar\'{e} map $P_k$ associated with the simple mode $U_{k}$ given in
(\ref{sm}).	In order to	fix	notations, we write	once more the
definition of $P_k$, following section \ref{sec:poincare}.

Given $(x,y)\in\mathcal{U}_{k}$	we consider
the	solution of	the	system
\begin{gather}
	W''+ \tau_{k}^{2}m(k^{2}W^{2}+Z^{2})W=0,  \quad
	W(0)=\alpha,\quad W'(0)=0,
	\label{eqn:U}  \\
	Z''+\nu\, \tau_{k}^{2}m(k^{2}W^{2}+Z^{2})Z=0,  \quad
	Z(0)=x,\quad Z'(0)=2\pi\sqrt{\nu}\,y,
	\label{eqn:V}
\end{gather}
where $\alpha$ is the positive solution	of
$$
	\frac{4\pi^{2}\,y^{2}}{k^{2}}+\frac{\tau_{k}^{2}}{k^{2}}M(k^{2}\alpha^{2}+x^{2})=
	\frac{\tau_{k}^{2}}{k^{2}}M(k^{2}).
	$$
Let	$T$	be the smallest	$t>0$ such that	$W'(t)=0$ and $W(t)>0$.	Then
$$P_k(x,y):=\left(Z(T),(4\pi^{2}\nu	)^{-1/2}Z'(T)\right).
$$
Since we plan to use polar coordinates we assume
that $x=\rho\cos\theta$, $y=\rho\sin\theta$.

Formally this definition is	very similar to	the	definition of $L_k$.
However	the	situation is here much more	complicated, because $\alpha$
depends	on $k$,	$\rho$,	$\theta$, hence	also $W$, $Z$ and $T$
depend on $k$, $\rho$, $\theta$.
We use capital letters to avoid	confusion with the corresponding
functions used in the study	of the linear term.	We also	write
$W(k,\rho,\theta,t)$, $\alpha(k,\rho,\theta)$, and so on, to
recall the dependence on all these variables. The symbol $'$
will always	denote differentiation with	respect	to the time	variable
$t$.

In this	first part of the proof	we consider	the	asymptotic behaviour
of these functions as $\rho\to 0^{+}$ ($k$ fixed). All the terms
$o(\rho^j)$	we introduce are uniform on	$\theta\in[0,2\pi]$, and on
$t$	belonging to any bounded time interval.

\noindent\textbf{Asymptotic	behaviour of $\alpha$.}
We prove that as $\rho\to 0^{+}$, we have that
\begin{equation}
	\alpha(k,\rho,\theta)=1	- \frac{1}{2k^{2}} \Big[\frac{4\pi^{2}
	\sin^{2} \theta}{\tau_{k}^{2}m(k^{2})} - \cos^{2} \theta
	\Big]\rho^{2} +o(\rho^{3}).
	\label{claim:alpha}
\end{equation}
Since	$\alpha(k,0,\theta)	= 1$ we	look for an	expansion of $\alpha$ as
$$	\alpha(k,\rho,\theta) =	1 +	\alpha_{2}(k,\theta)\rho^{2} +
o(\rho^{3}).
$$
Since
$$ \frac{4\pi^{2}\rho^{2}
	\sin^{2}
	\theta}{k^{2}}	+ \int_{0}^{\alpha^{2} + (\rho^{2}\cos^{2}
	\theta)/k^{2}} \tau_{k}^{2}	m(k^{2}s)\,	ds = \int_{0}^{1}
	\tau_{k}^{2} m(k^{2}s)\, ds,$$
	then taking	into account the Taylor	expansions,
$$
		\frac{4\pi^{2} \rho^{2}
	\sin^{2}
	\theta}{k^{2}} +\tau_{k}^{2} m(k^{2}) \left(1 -	\alpha^{2} - \rho^{2}\frac{\cos^{2}
	\theta}{k^{2}} \right) + o(\rho^{3}) = 0.$$
	 Hence thesis
	 follows immediately by
	 $$	\frac{4\pi^{2} \rho^{2}	\sin^{2} \theta}{k^{2}}
	 +\tau_{k}^{2} m(k^{2})	\left(-	2 \alpha_{2}(k,\theta)-	\frac{\cos^{2}
	 \theta}{k^{2}}	\right)\rho^{2}	+ o(\rho^{3}) =	0 .
	 $$
\noindent\textbf{Polar coordinates for $Z$.}
We argue as	in section \ref{sec:polar1}. Setting
$$X(k,\rho,\theta,t)=Z(k,\rho,\theta,t),\quad
Y(k,\rho,\theta,t)=(\nu	4\pi^{2})^{-1/2}Z'(k,\rho,\theta,t),$$
and	using polar	coordinates	$R(k,\rho,\theta,t)$,
$\Theta(k,\rho,\theta,t)$	such that $X=R\cos\Theta$,
$Y=R\sin\Theta$, it	turns out that $R$ and $\Theta$	solve the  system
\begin{gather}
	R'	=  2\pi\sqrt{\nu}\,	R\,\sin\Theta\cos\Theta\big\{1-
	\frac{\tau_{k}^{2}m(k^{2}W^{2}+R^{2}\cos^{2}\Theta)}{4\pi^{2}}\big\},
	\quad R(k,\rho,\theta,0) = \rho,
	\label{eqn:R} \\
	\Theta'	 =	 -2\pi\sqrt{\nu}\big\{\sin^{2}\Theta+
		\frac{\tau_{k}^{2}m(k^{2}W^{2}+R^{2}\cos^{2}\Theta)}{4\pi^{2}}\cos^{2}
	\Theta\big\}, \quad	\Theta(k,\rho,\theta,0)	= \theta.
	\label{eqn:Theta}
 \end{gather}

 \noindent\textbf{Asymptotic	behaviour of $W$, $R$, $\Theta$.}
 \label{sec:exp-URTheta}
 We	look for functions $W_{0}$,	$W_{2}$, $R_{1}$, $R_{3}$, $\Theta_{0}$,
 $\Theta_{2}$ such that, as	$\rho\to 0^{+}$,
 \begin{gather}
	 W(k,\rho,\theta,t)	 =
	 W_{0}(k,\theta,t)+W_{2}(k,\theta,t)\rho^{2}+o(\rho^{3}),
	 \label{exp:U} \\
	 R(k,\rho,\theta,t)	 =
	 R_{1}(k,\theta,t)\rho+R_{3}(k,\theta,t)\rho^{3}+o(\rho^{3}),
	 \label{exp:R} \\
	 \Theta(k,\rho,\theta,t)	=
	 \Theta_{0}(k,\theta,t)+
	 \Theta_{2}(k,\theta,t)\rho^{2}+o(\rho^{3}).
	 \label{exp:Theta}
 \end{gather}
 Using these expansions, we	have
 \begin{equation}
 \begin{aligned}
	 m(k^{2}W^{2}+Z^{2}) & =
	 m\left(k^{2}W_{0}^{2}+(2k^{2}W_{0}W_{2}+R_{1}^{2}\cos^{2}\Theta_{0})
	 \rho^{2}+o(\rho^{3})\right)\\
  &	=  m(k^{2}W_{0}^{2})+m'(k^{2}W_{0}^{2})
	  (2W_{0}W_{2}k^{2}+R_{1}^{2}\cos^{2}\Theta_{0})
	  \rho^{2}+o(\rho^{3}).
 \end{aligned} \label{claim:mrho}
 \end{equation}
 Setting (\ref{exp:U}),	(\ref{claim:mrho}),	and	(\ref{claim:alpha})	in
 equation (\ref{eqn:U}), and looking at	the	terms without $\rho$, we
 find that $W_{0}$ solves
 \begin{equation}
	 W_{0}''+\tau_{k}^{2}m(k^{2}W_{0}^{2})W_{0}=0,
	 \quad
	 W_{0}(k,\theta,0)=1,
	 \quad
	 W_{0}'(k,\theta,0)=0,
	 \label{eqn:U0}
 \end{equation}
 while,	looking	at the terms in	$\rho^{2}$,	we find	that $W_{2}$ solves
 \begin{equation}
	 W_{2}''+\tau_{k}^{2}m(k^{2}W_{0}^{2})W_{2}+\tau_{k}^{2}m'(k^{2}W_{0}^{2})
	 \left(
	 2W_{0}W_{2}k^{2}+R_{1}^{2}\cos^{2}\Theta_{0}\right)W_{0}=0,
	 \label{eqn:U2}
 \end{equation}
 with initial data
 \begin{equation}
	 W_{2}(k,\theta,0)=-\frac{1}{2k^{2}}\Big[\frac{4\pi^{2}
	 \sin^{2}
	 \theta}{\tau_{k}^{2}m(k^{2})} - \cos^{2} \theta \Big],\quad
	 W_{2}'(k,\theta,0)=0.
	 \label{eqn:U2-data}
 \end{equation}
 From (\ref{eqn:U0}) we can see that $W_{0}$ is just the simple mode
 $U_{k}$.	 In	particular,	it is independent on $\theta$, and so from now
 on	we write $U_{k}(t)$, instead of	$W_{0}(k,\theta,t)$.
 Setting (\ref{exp:R}),	(\ref{exp:Theta}), and (\ref{claim:mrho}) in
 equation (\ref{eqn:R}), and looking at	the	terms in $\rho$, we	find
 that $R_{1}$ solves
 \begin{equation}
	 R_{1}'=2\pi\sqrt{\nu}\Big\{1-\frac{\tau_{k}^{2}m(k^{2}U_{k}^{2})}{4\pi^{2}}\Big\}
	 R_{1}\cos\Theta_{0}\sin\Theta_{0},
	 \quad R_{1}(k,\theta,0)=1.
	 \label{eqn:R-1}
 \end{equation}
 In	an analogous way, looking at the terms without $\rho$ in
 (\ref{eqn:Theta}),	we find	that $\Theta_{0}$ solves
 \begin{equation}
	 \Theta_{0}'=-2\pi\sqrt{\nu}\Big\{\sin^{2}\Theta_{0}+
	 \frac{\tau_{k}^{2}m(k^{2}U_{k}^{2})}{4\pi^{2}}
	 \cos^{2}\Theta_{0}\Big\},
	 \quad\Theta_{0}(k,\theta,0)=\theta.
	 \label{eqn:Th-0}
 \end{equation}
 Finally, using	in equation	(\ref{eqn:Theta}) expansions (\ref{exp:R}),
 (\ref{exp:Theta}),	and	(\ref{claim:mrho}),	and	recalling that by
 Taylor	formula
 \begin{gather*}
	 \sin^{2}\Theta	 =	\sin^{2}\Theta_{0}+
	 2\rho^{2}\Theta_{2}\cos\Theta_{0}\sin\Theta_{0}+o(\rho^{2}),  \\
	 \cos^{2}\Theta	 =	\cos^{2}\Theta_{0}-
	 2\rho^{2}\Theta_{2}\cos\Theta_{0}\sin\Theta_{0}+o(\rho^{2}),
 \end{gather*}
 looking at	the	terms in $\rho^{2}$, we	find that $\Theta_{2}$ solves
 \begin{equation}
 \begin{aligned}
 \Theta_{2}'  &=
	 -2\pi\sqrt{\nu}\Big\{2\Theta_{2}\cos\Theta_{0}\sin\Theta_{0}\big[
	 1-\frac{\tau_{k}^{2}m(k^{2}U_{k}^{2})}{4\pi^{2}}
	 \big]	\\
 & \quad +\frac{\tau_{k}^{2}m'(k^{2}U_{k}^{2})}{4\pi^{2}}
	  \cos^{2}\Theta_{0}\left[
	 2k^{2}U_{k}W_{2}+R_{1}^{2}\cos^{2}\Theta_{0}\right] \Big\},
	 \quad\Theta_{2}(k,\theta,0)=0.
 \end{aligned} \label{eqn:Th-2}
 \end{equation}
 We	do not write the equation for $R_{3}$ because we don't need	it in
 the sequel.

 \noindent\textbf{Asymptotic behaviour of $T$.}\label{sec:exp-T}
 We	prove that,	as $\rho\to	0^{+}$,
 \begin{equation}
	 T(k,\rho,\theta)=1+T_{2}(k,\theta)\rho^{2}+o(\rho^{3}),
	 \label{claim:T}
 \end{equation}
 where
 \begin{equation}
	 T_{2}(k,\theta)=\frac{W_{2}'(k,\theta,1)}{
	 \tau_{k}^{2}\,	m(k^{2})}.
	 \label{defn:T2}
 \end{equation}
 It	is natural to look for an expansion	like (\ref{claim:T}) since
 for $\rho=0$, $W$ is exactly the simple mode $U_{k}$. Replacing the
 expansions	of $W$ and $T$ in the condition	$W'(T)=0$ we obtain	that
 \begin{equation}
 \begin{aligned}
	 0 & =	W'(k,\rho,\theta,T(k,\rho,\theta))	\\
	  &	=  U_{k}'(T(k,\rho,\theta))+
	  \rho^{2}W_{2}'(k,\theta,T(k,\rho,\theta))+o(\rho^{3})	 \\
	  &	=  U_{k}'(1)+\rho^{2}\left\{
	  U_{k}''(1)T_{2}(k,\theta)+
	  W_{2}'(k,\theta,1)\right\}+o(\rho^{3}).
 \end{aligned}	   \label{passaggio:TR}
 \end{equation}
  The first	summand	is zero.
 Moreover by equation (\ref{sm})
 $$
 U_{k}''(1)=U_{k}''(0)=-\tau_{k}^{2}m\left(
 k^{2}U_{k}^{2}(0)\right)U_{k}(0)=-\tau_{k}^{2}m(k^{2}).
 $$
 Setting equal to zero the coefficient of $\rho^{2}$ in
 (\ref{passaggio:TR}), and using the last equality,	we get
 (\ref{defn:T2}). It is	easy to	see	that with this choice also
 condition $U(T)>0$	is satisfied for $\rho$	small.

\noindent\textbf{Asymptotic	behaviour of the Poincar\'{e} map.}
Using the expansions in	(\ref{exp:R})-(\ref{exp:Theta})-(\ref{claim:T}),
we obtain that
\begin{align*}
\Theta(k,\rho,\theta,T(k,\rho,\theta)) &=
	\Theta_{0}(k,\theta,T(k,\rho,\theta))+
	\Theta_{2}(k,\theta,T(k,\rho,\theta))\rho^{2}+o(\rho^{3})  \\
	 & =  \Theta_{0}(k,\theta,1)+\left\{
	 \Theta_{0}'(k,\theta,1)T_{2}(k,\theta)+
	 \Theta_{2}(k,\theta,1)\right\}\rho^{2}+o(\rho^{3}),  
\end{align*}
and	similarly
\[
 R(k,\rho,\theta,T(k,\rho,\theta)) =  R_{1}(k,\theta,1)\rho
+\left\{R_{1}'(k,\theta,1)T_{2}(k,\theta)+
	R_{3}(k,\theta,1)\right\}\rho^{3}+o(\rho^{3}).
\]
Therefore, in polar	coordinates	the	Poincar\'{e} map is
$$P_k	 \begin{pmatrix}
		\rho  \\
		\theta
	\end{pmatrix}= \begin{pmatrix}
		R(k,\rho,\theta,T(k,\rho,\theta))  \\
		\Theta(k,\rho,\theta,T(k,\rho,\theta))
	\end{pmatrix} =\begin{pmatrix}
		\alpha_{1}(k,\theta)\rho  \\
		\beta_{0}(k,\theta)
	\end{pmatrix}+\begin{pmatrix}
		\alpha_{3}(k,\theta)\rho^{3}  \\
		\beta_{2}(k,\theta)\rho^{2}
	\end{pmatrix}+o(\rho^{3}),$$
where
\begin{equation}
\begin{gathered}
	\alpha_{1}(k,\theta)  =	 R_{1}(k,\theta,1),	 \\
	\alpha_{3}(k,\theta)  =	 R_{1}'(k,\theta,1)T_{2}(k,\theta)+	R_{3}(k,\theta,1),	\\
	\beta_{0}(k,\theta)	 =	\Theta_{0}(k,\theta,1),	 \\
	\beta_{2}(k,\theta)	 =	\Theta_{0}'(k,\theta,1)T_{2}(k,\theta)+
	\Theta_{2}(k,\theta,1).
\end{gathered}	   \label{eqn:beta2}
\end{equation}
Let	us now recall that if we
choose coordinates where $L_{k}$ is	written	in the canonical form of a
rotation, then,	in the corresponding polar coordinates,	$P_{k}$	becomes
$$
P_{k}\begin{pmatrix}
	I  \\
	\sigma
\end{pmatrix}=\begin{pmatrix}
	I+a(k,\sigma)I^{3}	\\
	\sigma-\omega_{k}+b(k,\sigma)I^{2}
\end{pmatrix}+o(I^{3}).
$$
 The coordinate	change from	the	cartesian coordinates
$(X,Y)$	to the original
coordinates	$(x,y)$	is given by	the	diagonal matrix	$D(k)$
introduced in Theorem \ref{thm:1}. The expression of
the	corresponding change $D_{*}(k)$	from $(I,\sigma)$ to
$(\rho,\theta)$	is not so simple: it is	given by
\begin{equation}
	\rho=\alpha_{*}(k,\sigma)I,\quad
	\theta=\delta_{*}(k,\sigma),
	\label{defn:coordchange}
\end{equation}
where $\alpha_{*}(k,\sigma)=
\left\{\cos^{2}\sigma+
\delta^{2}(k)\sin^{2}\sigma\right\}^{1/2}$,	and
$\delta_{*}(k,\sigma)=\arctan(\delta(k)\tan\sigma)$
for	$\sigma\in (-\pi/2,\pi/2)$,	and	similarly for all other
values of $\sigma$.	  If we	denote the
inverse	change by $D^{*}(k):=\left[D_{*}(k)\right]^{-1}$, then his
components $\alpha^{*}(k,\theta)\rho$ and $\delta^{*}(k,\theta)$
are	defined	in analogy with	$\alpha_{*}$, $\delta_{*}$,	but	with
$\delta^{-1}(k)$ instead of	$\delta(k)$.

Considering	the	second
component of $P_k^{*}= D^{*}(k)P_k D_{*}(k)$
we have	that, up to	$o(I^{3})$,
$$\sigma-\omega(k)+
b(k,\sigma)I^{2}= \delta^{*}\left[k,\beta_{0}(k,\delta_{*}(k,\sigma))
+\beta_{2}(k,\delta_{*}(k,\sigma))\cdot
\alpha_{*}^{2}(k,\sigma)\,I^{2}\right],$$
so that, making	the	Taylor expansion of	the	right hand side	and
looking	at the coefficients	of $I^{2}$,	it turns out that
$$b(k,\sigma)=
\frac{\partial\delta^{*}}{\partial\theta}
\left[k,\beta_{0}(k,\delta_{*}(k,\sigma))\right]\cdot
\beta_{2}(k,\delta_{*}(k,\sigma))\cdot
\alpha_{*}^{2}(k,\sigma).$$


\subsubsection{Behaviour for large	$k$	of coefficients	in Taylor
expansions.}

\textbf{Behaviour of the coordinate	change}
  By statement (5) of Theorem \ref{thm:1} we obtain:
\begin{gather}
\alpha_{*}(k,\sigma)
	\to	(\cos^{2}\sigma	+\delta^{2}	\sin^{2}\sigma)^{1/2} =
	\mu_{0}(\sigma)	\geq c > 0,		\label{eqn:coordchange1} \\
 \delta_{*}(k,\sigma)
	\to	\arctan(\delta \tan(\sigma)) =
	\mu_{1}(\sigma),\quad \sigma
	\in\left(-\frac{\pi}{2},\frac{\pi}{2}\right), \label{eqn:coordchange2}\\
\frac{\partial\delta_{*}}
	{\partial\sigma}(k,\sigma)
	\to	\delta\frac{1 +	\tan^{2}\sigma}{1 +	\delta^{2}\tan^{2}\sigma} =
	\mu_{2}(\sigma)	\geq c > 0,		\label{eqn:coordchange3}
\end{gather}
as $k\to +\infty$, uniformly in	$\sigma$.  The same	holds true for the
inverse	change $D^{*}(k)$.

\noindent\textbf{Behaviour of $W_{2}$}
We prove that
 \begin{equation}
	|W_{2}(k,\theta,t)|	+ |W_{2}(k,\theta,t)'| \leq	c\,\frac{\log k}{k^{2}}	\quad
	\forall	\, t \in [0,1].
	\label{stiw}
 \end{equation}
	Indeed let us set
$$
E(k,t) = |W_{2}(k,\theta,t)|^{2} + |W_{2}(k,\theta,t)'|^{2}.
$$
 Thanks	to (\ref{eqn:U2}), since $m' \geq 0$, in a standard	way	we get
 \begin{align*}
	E'(k,t)	\leq &	E(k,t) [1 +\tau_{k}^{2}	m(k^{2}U_{k}(t)^{2}) + 2\tau_{k}^{2}
 m'(k^{2}U_{k}(t)^{2}) k^{2} U_{k}(t)^{2}]	\\
 &	 +	\tau_{k}^{2}
 m'(k^{2}U_{k}(t)^{2}) R_{1}^{2}(k,\theta,t) |U_{k}(t)|\sqrt{ E(k,t)}.
 \end{align*}
 Now, using	(U4), $m' \geq 0$, Lemma \ref{lemma3} and  (\ref{tempo})
 we	obtain:
 \begin{gather}
	\tau_{k}^{2} m(k^{2}U_{k}(t)^{2}) \leq \tau_{k}^{2}
  m(k^{2}) \leq	c;	   \label{l1} \\
	\tau_{k}^{2} m'(k^{2}U_{k}(t)^{2}) k^{2} U_{k}(t)^{2}
	\leq \tau_{k}^{2}m(k^{2})
	\frac{m'(k^{2}U_{k}(t)^{2})	k^{2}
	U_{k}(t)^{2}}{m(k^{2}U_{k}(t)^{2})}	\leq c;	\label{l2} \\
	 \tau_{k}^{2}
 m'(k^{2}U_{k}(t)^{2})	|U_{k}(t)| \leq	c
 \frac{m'(k^{2}U_{k}(t)^{2})}{m(k^{2}U_{k}(t)^{2})}	|U_{k}(t)|	\leq c\,\frac{|U_{k}(t)|}{1	+
 k^{2}U_{k}(t)^{2}}.	 \label{l3}
  \end{gather}
 Moreover,	by (\ref{eqn:R-1}) we obtain that the function $R_{1}(k,\theta,t) $	is bounded
 in	$[0,1]$.
 Therefore,	integrating	on $[0,t]$ the inequality
 $$E'(k,s) \leq	c\,	E(k,s) + c\,\sqrt{E(k,s)}\frac{|U_{k}(s)|}{1 +
 k^{2}U_{k}(s)^{2}}$$
 and using Lemma \ref{lemma7} and (\ref{eqn:U2-data}) we get the
 desired estimate.

\noindent\textbf{Behaviour of	$\Theta_{2}$}
We prove that
\begin{equation}
	\liminf_{k\to +\infty} -km(k^{2})\int_{0}^{2\pi}
	\frac{\partial\delta^{*}}{\partial\theta}
	\left[k,\beta_{0}(k,\delta_{*}(k,\sigma))\right]
	\Theta_{2}(k,\delta_{*}(k,\sigma),1)\alpha_{*}^{2}(k,\sigma)
	\, d\sigma > 0.
	\label{pinco}
\end{equation}
To this	end	we need	the	following lemmata.

\begin{lemma}\label{lemmatheta}
 The function $\Theta_{2}$ satisfies
	$km(k^{2})|\Theta_{2}(k, \theta, t)| \leq c$
for	all	$t \in [0,1]$, $k \geq \overline{k}$.
\end{lemma}

\begin{proof}
Integrating	(\ref{eqn:Th-2}) we	find
\begin{align*}
\Theta_{2}(k, \theta, t) &=	-2\pi\sqrt{\nu}	\int_{0}^t
	 \sin(2\Theta_{0}(k, \theta, s))\Big(1 -
	 \frac{\tau_{k}^{2}}{4\pi^{2}}m(k^{2}U_{k}^{2}(s))\Big)\Theta_{2}(k, \theta,
	 s)	\, ds \\
& -2\pi	\sqrt{\nu}\int_{0}^t
	  2k^{2}\frac{\tau_{k}^{2}}{4\pi^{2}}m'(k^{2}U_{k}^{2}(s))
	 \cos^{2}(\Theta_{0}(k,	\theta,s))
	U_{k}(s)W_{2}(k,\theta,s) \, ds	 \\
&  -2\pi \sqrt{\nu}\int_{0}^t
	  \frac{\tau_{k}^{2}}{4\pi^{2}}m'(k^{2}U_{k}^{2}(s))
	 \cos^{4}(\Theta_{0}(k,	\theta,s))
	 R_{1}(k,\theta,s)\, ds.
\end{align*}
Hence by  $m'\geq 0$, Lemma	\ref{lemma3}, (\ref{stiw}),	 (\ref{l3})	and	Lemma
\ref{lemma7} we	obtain
\begin{align*}
	|\Theta_{2}(k, \theta, t)|
& \leq	 c\int_{0}^{t}
	\Big|1 + \frac{\tau_{k}^{2}}{4\pi^{2}}m(k^{2})\Big|
	|\Theta_{2}(k, \theta, s)|+
	\frac{m'(k^{2}U_{k}^{2}(s))}{m(k^{2})} (|U_{k}(s)|\log k + 1)  \, ds \\
& \leq	  c\int_{0}^{t}
	|\Theta_{2}(k, \theta, s)|\, ds	 + \frac{c}{km(k^{2})}	+ c\log
	k\int_{0}^{t}
	 \frac{|U_{k}(s)|}{1 + k^{2}U_{k}^{2}(s)}\,ds \\
& \leq	 c\int_{0}^{t} |\Theta_{2}(k, \theta, s)|\,	ds	+ \frac{c}{km(k^{2})}  +
	  \frac{c\log^{2} k}{k^{2}}.
\end{align*}
Hence thesis follows by	Lemma \ref{lemma2},	(M\ref{m1}), and Gronwall's	Lemma.
\end{proof}

\begin{lemma}	  \label{lemmatheta2}
	The next equality holds:
	$$\liminf_{k\to	+\infty}- km(k^{2})	\Theta_{2}(k, \theta, 1) =
	2\pi\sqrt{\nu} \liminf_{k\to
	+\infty}\int_{0}^{1}km'(k^{2}U_{k}(s)^{2})\cos^{4} (\theta -
	2\pi\sqrt{\nu} s) \, ds.$$
\end{lemma}

\begin{proof}
Integrating equation (\ref{eqn:Th-2}) we find:
\begin{align*}
& -km(k^{2}) \Theta_{2}(k, \theta, 1)\\
&=	2\pi\sqrt{\nu} \int_{0}^{1}
	 \sin(2\Theta_{0}(k,\theta,t))km(k^{2})	\Big(1 -
	 \frac{\tau_{k}^{2}}{4\pi^{2}}m(k^{2}U_{k}^{2}(t))\Big)
	 \Theta_{2}(k, \theta, t) \, dt	 \\
&\quad +  2\pi\sqrt{\nu} \int_{0}^{1} 2\frac{\tau_{k}^{2}}{4\pi^{2}}k^{3}
	 m(k^{2}) \cos^{2} (\Theta_{0}(k,\theta,t))
	 m'(k^{2}U_{k}(t)^{2})U_{k}(t) W_{2}(k,\theta,t)\, dt  \\
&\quad +   2\pi\sqrt{\nu} \int_{0}^{1}
  \frac{\tau_{k}^{2}}{4\pi^{2}}m(k^{2})kR_{1}^{2}(k,\theta,t)m'(k^{2}U_{k}(t)^{2})
  \cos^{4}(\Theta_{0}(k,\theta,t))\, dt.
\end{align*}
Now	we estimate	all	terms in previous equality.
 Using Lemma \ref{lemmatheta} and Lemma	\ref{lemma7} we	obtain
 \begin{equation}
 \begin{aligned}
 \int_{0}^{1}	\left|\sin(2\Theta_{0}(k,\theta,t))km(k^{2}) \Big(1	-
	 \frac{\tau_{k}^{2}}{4\pi^{2}}m(k^{2}U_{k}^{2}(t))\Big)
	 \Theta_{2}(k,\theta,t)\right|\, dt	\\
\leq \int_{0}^{1} \Big|1 -	 \frac{\tau_{k}^{2}}{4\pi^{2}}m(k^{2}U_{k}^{2}(t))\Big|
 \,	dt \leq	c\lambda_{k}.
\end{aligned}	
\label{stip1}
 \end{equation}
Thanks to (\ref{stiw}),	(\ref{l3}),	Lemma \ref{lemma7},	Lemma
\ref{lemma2} (property (M\ref{m1}))	we also	obtain
\begin{equation}
\begin{aligned}
  \int_{0}^{1}	\left|\frac{\tau_{k}^{2}}{4\pi^{2}}k^{3}
	 m(k^{2}) \cos^{2} (\Theta_{0}(k,\theta,t))
	 m'(k^{2}U_{k}(t)^{2})U_{k}(t) W_{2}(k,\theta,t)\right|\, dt\\
  \leq	c \int_{0}^{1}k\log	k |U_{k}(t)|m'(k^{2}U_{k}(t)^{2})\,	dt \\
  \leq	c\int_{0}^{1} k\log	k \, m(k^{2}U_{k}^{2}(t))\frac{|U_{k}(t)|}{1 +
	 k^{2}U_{k}^{2}(t)}\, dt\\
\leq c\frac{\log^{2} k}{k}m(k^{2})	 \leq c\frac{1}{\sqrt{k}}.
\end{aligned} \label{stip2}
\end{equation}
Let us remark that, thanks to Lemma \ref{lemma7} and
(\ref{eqn:R-1})--(\ref{eqn:Th-0}),
$R_{1}(k,\theta,t) = 1 + o(1)$, $\Theta_{0}(k,\theta,t) = \theta -
2\pi\sqrt{\nu} t + o(1)$ (where $o(1)$ do not depends on $t \in
[0,1]$), and, by  Lemma \ref{lemma3},
$\tau_{k}^{2}m(k^{2}) = 4\pi^{2} + o(1)$, hence
\begin{align*}
     &\int_{0}^{1}
  \frac{\tau_{k}^{2}}{4\pi^{2}}m(k^{2})kR_{1}^{2}(k,\theta,t)m'(k^{2}U_{k}(t)^{2})
  \cos^{4}(\Theta_{0}(k,\theta,t))\, dt  \\
& =\int_{0}^{1} (1 + o(1))^{3}   km'(k^{2}U_{k}(t)^{2})
  (\cos^{4}(\theta -2\pi\sqrt{\nu} t ) + o(1))\, dt.
\end{align*}
 Now we can use once more Lemma \ref{lemma7} and obtain
\begin{equation}
    \lim_{k\to +\infty} \int_{0}^{1} o(1)  km'(k^{2}U_{k}(t)^{2})
  [1 +\cos^{4}(\theta -
2\pi\sqrt{\nu} t ) ] \, dt = 0.
    \label{stip3}
\end{equation}
\end{proof}
We are now able to prove (\ref{pinco}).
Lemma \ref{lemmatheta2} says that, for large $k$,
$-\Theta_{2}(k,\theta,1) $ is a non negative function,
hence, by (\ref{eqn:coordchange1}) - (\ref{eqn:coordchange3}), and a
change of variable
\begin{equation}
-\int_{0}^{2\pi}
    \frac{\partial\delta^{*}}{\partial\theta}
    \left[k,\beta_{0}(k,\delta_{*}(k,\sigma))\right]
    \Theta_{2}(k,\delta_{*}(k,\sigma),1)\alpha_{*}^{2}(k,\sigma)
    \, d\sigma \geq c \int_{0}^{2\pi}
        \Theta_{2}(k,\theta,1)
    \, d\theta.
    \label{eq:ts}
\end{equation}
As in the proof of Lemma \ref{lemmatheta2} (using
(\ref{stip1}), (\ref{stip2}), (\ref{stip3}) )
we get
    \begin{align*}
         &  \liminf_{k\to +\infty} -km(k^{2})\int_{0}^{2\pi}
    \Theta_{2}(k,\theta,1) \, d\theta   \\
         &   =     \liminf_{k\to +\infty}  2\pi\sqrt{\nu}
        \int_{0}^{1}km'(k^{2}U_{k}(s)^{2})\int_{0}^{2\pi}\cos^{4} (\theta -
    2\pi\sqrt{\nu} s) \, d\theta \, ds  \\
         &   = \frac{3}{2}\pi^{2}\sqrt{\nu}\liminf_{k\to
    +\infty}\int_{0}^{1}km'(k^{2}U_{k}(s)^{2})\, ds.
    \end{align*}
    Let $t_{0}$ be as in Lemma \ref{lemma4}, (B\ref{U2}). Since thanks
    also to (U5) and Lemma \ref{lemma3} we find
    $0 < c_{1} \leq |U'_{k}(t)| < c_{2}$ for $t\in [t_{0},1/4]$, it
    turns out that:
$$\int_{0}^{1}
    km'(k^{2}U_{k}(s)^{2})\, ds \geq \int_{t_{0}}^{1/4}
    km'(k^{2}U_{k}(s)^{2})\frac{-U_{k}'(s)}{|U_{k}'(s)|}\, ds \geq
    c\int_{0}^{kU_{k}(t_{0})}m'(y^{2})\, dy,
$$
    and passing to the limit
    $$ \int_{0}^{kU_{k}(t_{0})}m'(y^{2})\, dy \to
    \int_{0}^{+\infty}m'(y^{2})\, dy > 0 \quad\mbox{as}
    \quad k \to +\infty.$$
    Hence,  we have
    $$\liminf_{k\to +\infty} -km(k^{2})\int_{0}^{2\pi}
    \Theta_{2}(k,\theta,1) \, d\theta \geq c
    \liminf_{k\to +\infty}\int_{t_{0}}^{1/4}
     km'(k^{2}U_{k}(s)^{2}) \, d\theta  > 0.$$
    Thanks to (\ref{eq:ts}) we then obtain (\ref{pinco}).\\                                                                                                    %%
\noindent\textbf{Behaviour of $b(k,\sigma)$.}
We prove that, for $k$ large,
\begin{equation}
\int_{0}^{2\pi}b(k,\sigma) \, d\sigma =
\int_{0}^{2\pi}\frac{\partial\delta^{*}}{\partial\theta}
\left[k,\beta_{0}(k,\delta_{*}(k,\sigma))\right]\cdot
\beta_{2}(k,\delta_{*}(k,\sigma))\cdot
\alpha_{*}^{2}(k,\sigma)\, d\sigma < 0.
    \label{eq:betaas}
\end{equation}
Let us recall that in (\ref{eqn:beta2}),
$\beta_{2}(k,\theta)  =  \Theta_{0}'(k,\theta,1)T_{2}(k,\theta)+
     \Theta_{2}(k,\theta,1)$.
Since $\alpha_{*}$,
$\frac{\partial\delta^{*}}{\partial\theta}$ and $\Theta_{0}'(k,\theta,t)$,
 are  bounded functions, by
(\ref{defn:T2}) and (\ref{stiw}), we obtain
\begin{align*}
& \int_{0}^{2\pi}\Big|\frac{\partial\delta^{*}}{\partial\theta}
 \left[k,\beta_{0}(k,\delta_{*}(k,\sigma))\right]
 \Theta_{0}'(k,\delta_{*}(k,\sigma),1)T_{2}
 (k,\delta_{*}(k,\sigma))\alpha_{*}^{2}(k,\sigma)\Big|\, d\sigma \\
& \leq c \int_{0}^{2\pi}|W_{2}'(k,\delta_{*}(k,\sigma),1)|\, d\sigma
\leq c\frac{\log k}{k^{2}}\,.
\end{align*}
 Since  $\log k/k = o(1/m(k^{2})) $ to get the desired inequality it is now
enough to remark that (\ref{pinco}) says that for large $k$:
 $$
 -\int_{0}^{2\pi}
    \frac{\partial\delta^{*}}{\partial\theta}
    \left[k,\beta_{0}(k,\delta_{*}(k,\sigma))\right]
    \Theta_{2}(k,\delta_{*}(k,\sigma),1)\alpha_{*}^{2}(k,\sigma)
    \, d\sigma  \geq
 \frac{c}{km(k^{2})}.
 $$
\subsubsection*{Proof of Theorem \ref{thm:2}} To proof Theorem \ref{thm:2}, we have  to show that for large $k$,
$\gamma_{k} < 0$, but this is exactly (\ref{eq:betaas}).


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\end{document}