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unimodal map, coupled map, synchronization
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\begin{document}
\baselineskip18pt
\title[diffusively coupled unimodal maps]{Dynamics of two diffusively coupled unimodal maps with zero
topological entropy}
\author[Fernandez and Jiang]{Bastien Fernandez$^1$
and Miaohua Jiang$^{1,2}$}
\maketitle
\begin{center}
$^1$Centre de Physique Th\'eorique, CNRS Luminy Case 907, 13288 Marseille
CEDEX 09, France
$^2$ Department of Mathematics, Wake Forest University, Winston Salem,
NC 27106, USA
Email: bastien@cpt.univ-mrs.fr and jiangm@wfu.edu
\bigskip
{\bf Abstract}
\end{center}
The paper studies the dynamics of a coupled map lattice of two sites
where the local map is a unimodal map for which the period of every
periodic orbit is a power of two. With the help of the existence of
nested subsets of the phase space and their forward invariance after
some iterations, it is shown that every orbit passing such subsets is
eventually periodic and the period is a power of two. A criterion for
an orbit to visit a particular subset is given and a sufficient condition for
the existence of periodic orbits satisfying this criterion is obtained.
AMSCSN: 37C05, 37E99.
\section{Coupled Unimodal Maps}
The aim of this paper is to investigate the effects of a diffusive
coupling on the dynamics of a simple dynamical system with discrete
time, namely the direct product of two unimodal maps.
Physically, models with nonlinear forcing and diffusive coupling can be
viewed as simple models for the time evolution of
reaction-diffusion systems (see e.g.\ the chapter by Kaneko in
\cite{Kaneko}). Understanding the dynamics of these models is then
useful to predict the asymptotic behaviour in these systems, in
particular, to distinguish between chaotic and stable regimes and
to describe the various patterns in the stable case.
Mathematically, we want to describe the topological changes in
the non-wandering set of a class of two-dimensional non-hyperbolic
systems. Just like the H\'enon map, our dynamical systems are derived from a
simple one-dimensional map through coupling and demonstrate rich
dynamics.
The system we consider is a coupled map lattice of two sites. It means
that the dynamics is given by the composition of the direct product of
two identical one-dimensional maps with a convex coupling. The reason for
considering only two sites is simplicity. Moreover, we believe this
analysis will serve as a starting point for the investigation of the dynamics
of lattices with more sites.
Although majority of works on coupled map lattices has considered more than
two sites, analytic and numerical studies of
two diffusively coupled maps can be found in the literature. These
studies determine the parameter regions for the
existence of some periodic orbits and study their bifurcations
\cite{Vilela,Kaneko,Kurten}, or describe the structure of the
attractor \cite{Dobryn}.
Our analysis emphasizes on the global topological aspects of the
dynamics. Instead of considering systems with a specific formula such
as the quadratic map, we study a class of maps with common
topological features: unimodal maps with a family of kneading
sequences. The study focuses on two topics: the asymptotically
periodic behaviour of an orbit passing a sequence of nested squares
and the emergence of non-synchronised orbits when the coupling
gradually relaxes.
We combine specific properties of the kneading sequences with
dynamical properties of the coupling to show the
existence of a sequence of nested squares. Each of these squares is
forward invariant by some iterates of the dynamics. An orbit visiting
such a square is eventually periodic. Moreover, by considering the
relative position of an orbit with respect to the diagonal, we give a
criterion for an orbit to visit a specific square in the sequence.
Finally, we give a condition for existence of periodic orbits satisfying this
criterion.
We consider discrete dynamical systems generated by the
composition of the direct product of two identical maps, the local map,
with a diffusive coupling. The local map is a unimodal map on
$[-1,1]$. It is defined to have the following properties.
\begin{itemize}
\item[(1)] $f(0)=1$;
\item[(2)] $f$ is increasing on $[-1,0]$ and decreasing on $[0,1]$;
\item[(3)] $f$ is $C^1$ and $f'(x)=0$ only if $x=0$;
\item[(4)] $f([f(1),1])\subset [f(1),1]$.
\end{itemize}
A well-known example is the quadratic map $f_a(x)=1-ax^2$ where
$00$ can take any
value smaller than the Feigenbaum's accumulation point of the period-doubling
cascade.
We now recall definitions of itineraries and kneading sequences.
The book by Collet and Eckmann \cite{Collet}
is a good source for more details.
The {\it itinerary} $ I(x)= \{ I_i(x)\}_{i \in \N}$ of a point $x\in [-1,1]$
is either an infinite sequence of $L$'s and $R$'s, or a finite
sequence of $L$'s and $R$'s, followed by $C$. The $i$-th element of $I(x)$ is
defined by
\[ I_i(x)=\left\{\begin{array}{ccl}
L&\hbox{if}&f^i(x)<0\\ C&\hbox{if}&f^i(x)=0\\ R&\hbox{if}&f^i(x)>0
\end{array}\right.\]
The {\it kneading sequence} of a unimodal map is the
itinerary of the point $1$.
The simplest kneading sequences are $I^n:= R^{*n}*R^\infty$, which is
$2^{n+1}$-periodic, and $J^n:=R^{*n}*RC$, which is finite
(see Sections \ref{I-n} and \ref{J-n} for definitions).
As in the case of unimodal maps where the study of the dynamics
of intervals is important in the understanding the dynamics of the map, the
dynamics of certain simple sets (we call them {\it fundamental sets}) under
the coupled map $F_\eps$ plays a major role. We shall see that the
dynamics of these fundamental sets is determined by the kneading sequence.
Among these fundamental sets, we consider squares invariant
under the symmetry map $\theta$.
Given $a, b \in [-1,1]$, let
\[ C_{\{a,b\}}= [ \min\{ a, b\}, \max\{ a, b\} ]^2 .\]
To describe the dynamics of $F_\eps$, we consider the squares
$C_{\{f^{2^k}(0),f^{2^{k+1}}(0)\}}$. Using properties
of itineraries $I^n$ and $J^n$, we first obtain a translation of the interval
dynamics associated to $f$ to the dynamics of the coupled
system.
\begin{Thm}\label{DynFund}
Assume that the kneading sequence of $f$ is $I^n$ or $J^n$ for some $n\in\N$.
Then, the following statements hold.
(i) ${\cal M}=C_{\{ f(0), f^2(0)\}} \supset C_{\{ f^2(0), f^4(0)\}} \supset
\cdots \supset C_{\{ f^{2^k}(0), f^{2^{k+1}}(0)\}}\supset \cdots \supset
C_{\{ f^{2^{n-1}}(0), f^{2^n}(0)\}} \supset C_{\{ 0, f^{2^{n}}(0)\}}$,
if $n\geq 1$.
(ii) If $n\geq 1$ and $0\leq k\leq n-1$, then the set
$C_{\{ f^{2^k}(0), f^{2^{k+1}}(0)\}}$ is forward invariant under the
map $F^{2^k}_\eps$:
\[
F^{2^k}_\eps( C_{\{ f^{2^k}(0), f^{2^{k+1}}(0)\}} )
\subset C_{\{ f^{2^k}(0), f^{2^{k+1}}(0)\}} ,
\]
For any $n\in\N$, the set $C_{\{ 0, f^{2^{n}}(0)\}}$ is forward invariant
under the map $F^{2^n}_\eps$:
\[ F^{2^n}_\eps( C_{\{ 0, f^{2^{n}}(0)\}} )
\subset C_{\{ 0, f^{2^{n}}(0)\}}.
\]
(iii) For any $1\leq k\leq n$ and any $(x,y)\in C_{\{0, f^{2^k}(0)\}}$,
\[
(x-y)(x^{2^k}-y^{2^k})\leq 0.
\]
If $(x,y)\in C_{\{0,f(0)\}}$, then
\[
(x-y)(x^1-y^1)\leq 0\quad\text{if}\quad 0\leq\eps\leq\frac{1}{2},
\]
and
\[
(x-y)(x^1-y^1)\geq 0\quad\text{if}\quad \frac{1}{2}\leq\eps\leq 1.
\]
\end{Thm}
In other words, we have a nested sequence of squares (see Figure 1),
and an orbit starting in one of these sets visits it periodically.
Moreover, the
position of a point in the square $C_{\{0,f^{2^k}(0)\}}$ with respect to the
diagonal alternates after $2^k$ iterations. The case $k=0$ is
different when $\eps\leq \frac{1}{2}$ and
$\eps\geq\frac{1}{2}$ because we consider an odd number of
iterations.
\begin{figure}
\epsfxsize=7truecm
\centerline{\epsfbox{figure1.eps}}
\caption{The squares $C_{\{f^{2^k}(0),f^{2^{k+1}}(0)\}}$ for $k$ up to 2.}
\end{figure}
This alternation of positions gives crucial information about the
asymptotic behaviour of orbits of $F_\eps$. An orbit
$\{(x^t,y^t)\}_{t\in\N}$ is said to be eventually periodic if
there exists $T\geq 1$ such that
$$
\lim_{t\to\infty}(x^t-x^{t+T})=\lim_{t\to\infty}(y^t-y^{t+T})=0.
$$
An orbit is said to be synchronised if $x^t=y^t$ for all $t$ and
eventually synchronised if
\[
\lim_{t\to\infty}(x^t-y^t)=0.
\]
\begin{Thm}\label{Thm:eventualp}
Assume that the kneading sequence of $f$ is $I^n$ or $J^n$ for some
$n\geq 1$.
Then, for any initial condition $(x,y)\in C_{\{f^i(0),f^{i+2^n}(0)\}}$ where
$0\leq i<2^n$, the orbit is eventually periodic of period $2^p$ for
some $1\leq p\leq n+1$. Moreover, if the orbit is not eventually
synchronised, then the period is $2^{n+1}$.
\end{Thm}
The proof of this theorem depends mainly on the alternation of the
sign (Statement {\em (iii)} of Theorem \ref{DynFund}) under the
invariance given by Statement {\em (ii)}. In general, i.e.\ for $k\frac{1}{2}$, a non-synchronised periodic orbit can have
period 1 or 2. For instance, this is the case when $\eps=1$.
For $S$-unimodal maps, i.e.\ unimodal maps with negative Schwarzian
derivative, if the kneading sequence is $I^n$, then $f$ has
a stable periodic orbit of period either $2^n$ or $2^{n+1}$
\cite{Collet}. In both cases, the orbit of the point $1$ is
attracted to the stable periodic orbit. We use this information to
show that if the latter has period $2^n$, then any orbit of the
coupled system initiated from the set $C_{\{0,f^{2^n}(0)\}}$ asymptotically
approaches the diagonal.
\begin{Thm}\label{Synchro}
Assume that the kneading sequence of a $S$-unimodal $f$ is $I^n$ for some
$n\in\N$ and that the $2^n$-periodic orbit in $[f(1),1]$ is stable.
Then, for any initial condition $(x,y)\in
C_{\{f^i(0),f^{i+2^n}(0)\}},\ 0\leq i<2^n$, the orbit is eventually
synchronised and eventually periodic of period $2^n$.
\end{Thm}
When $\eps$ is close to $\frac{1}{2}$, any orbit is asymptotically
synchronised because the distance $|x^t-y^t|$ exponentially
decreases. To understand the mechanism leading the system out of
synchrony when $\eps$ is moving away from $\frac{1}{2}$, we study
the emergence of non-synchronised periodic orbits. A local analysis
showed that the periodic orbits created from the period-doubling bifurcation of
synchronised periodic orbits are symmetric \cite{Vilela}. This
argument shows that such orbits exist provided the system is
sufficiently close to the bifurcation. Here we use the properties of
$I^n$ and $J^n$ to show that the existence of such orbits obtained
from bifurcation arguments is indeed global.
Before proving their existence, we show that such orbits must visit the
squares given in Theorem \ref{DynFund}.
\begin{Thm}\label{Force}
Assume that the kneading sequence of $f$ is $I^n$ or $J^n$ for some
$n\in\N$. Then, the following assertions hold.
(i) Any symmetric $2^{k+1}$-periodic orbit
passes the square $C_{\{f^{2^k}(0),f^{2^{k+1}}(0)\}}$ if $kn$. No symmetric $2$-periodic
orbit exists for $\frac{1}{2}\leq \eps\leq 1$.
\end{Thm}
We denote by $\{x_i\}_{i=1}^{2^k}$ the $2^k$-periodic orbit ($k\in\N$) of
$f$.
\begin{Thm}\label{Exist-2k}
Assume that the kneading sequence of $f$ is $I^n$ or $J^n$ for some
$n\in\N$. If
\begin{equation}\label{Neces}
(1-2\eps)^{2^k}\prod_{i=1}^{2^k}f'(x_i)<-1,
\end{equation}
for some $0\leq k\leq n$,
then $F_\eps$ has a symmetric $2^{k+1}$-periodic
orbit.
\end{Thm}
For any $k\in\N$, we have (see the proof of Lemma \ref{Locuniq})
\[
\prod_{i=1}^{2^k}f'(x_i)<0.
\]
Therefore, if this periodic orbit of $f$ is unstable, the condition
(\ref{Neces}) is satisfied provided that $\eps$ is sufficiently
close to 0, or 1 if $k\geq 1$. However, for $\eps>\frac{1}{2}$
and $k=0$, this condition cannot hold as stated in the previous Theorem.
It is an open question if the inequality (\ref{Neces}) is a necessary
condition for the existence of a symmetric $2^{k+1}$
periodic orbit. In the special case of the quadratic map and
$k=0$, the answer is affirmative \cite{Vilela}. In particular, since
the kneading sequence of this map is $I^0$ or $J^0$ when $a\leq 1$, this
implies the following result.
\begin{Cor}
For the map $f_a(x)=1-ax^2$ with $a\leq 1$, every orbit of
$F_\eps$ is eventually synchronised if and only if
$(1-2\eps)f'(x_1)\geq -1$ where $x_1$ is the fixed point of $f$.
\end{Cor}
Finally, the proof that (\ref{Neces}) with $k=0$ is
a sufficient condition for the existence of a period-2
symmetric orbit holds for any kneading
sequence starting with $R\cdots$, and hence holds for any unimodal
map.
\section{Properties of Unimodal Maps}
Properties of itineraries and their operations were extensively
discussed in \cite{Collet}. We only collect the ones needed
in this article. To make the article as much self-contained as possible, we
provide brief proofs of properties that are not explicitly proved in
\cite{Collet}.
We begin with a lemma on the relative position of iterates of two
points when their itineraries are the same on the first
symbols. Recall that a finite sequence of $L$'s and $R$'s is said to
be {\it odd} if it has an odd number of $R$'s, {\it even}, otherwise.
\begin{Lem}
Let $a\in [-1,1],\ a\neq 0$ and $t\in\N$ be such that
$I_i(a)=I_{i-1}(1)\neq C$, $1\leq i\leq t$ if $t\geq 1$. Then we have
\[
a(f^{t+1}(a)-f^{t+1}(0))>0\quad\text{if}\quad\{I_i(a)\}_{0\leq i\leq
t}\quad\text{is\ even},
\]
and
\[
a(f^{t+1}(a)-f^{t+1}(0))<0\quad\text{if}\quad\{I_i(a)\}_{0\leq i\leq
t}\quad\text{is\ odd}.
\]
\label{ORDERf}
\end{Lem}
{\sl Proof.} We use induction. For $t=0$, the result follows from the piecewise
monotonicity of $f$. Now, given
$t\in\N$, assume that $\{I_i(a)\}_{0\leq i\leq t+1}$ is even. Then
either $\{I_i(a)\}_{0\leq i\leq t}$ is even and $I_{t+1}(a)=I_t(1)=L$, or
$\{I_i(a)\}_{0\leq i\leq t}$ is odd and $I_{t+1}(a)=I_t(1)=R$. In the first
case, we have
$$
(f^{t+1}(a)-f^{t+1}(0))(f^{t+2}(a)-f^{t+2}(0))>0
$$
and using the induction hypothesis, it follows that
$$
a(f^{t+2}(a)-f^{t+2}(0))>0.
$$
Similarly, in the second case, we have
$$
(f^{t+1}(a)-f^{t+1}(0))(f^{t+2}(a)-f^{t+2}(0))<0
$$
and then
$$
a(f^{t+2}(a)-f^{t+2}(0))>0.
$$
Consequently, the statement for $t+1$ holds in the case of an even
sequence. The odd case is proved similarly. \qedd
\subsection{The sequences ${ I}^{n}$}\label{I-n}
We refer to \cite{Collet} for the definition of the $*$-product.
Now, let $A^\infty$ denotes the
infinite concatenation $AAA\cdots$ and let $R^{*n}*A$ denotes the n-fold
$*$-product $R*(R*(\cdots (R*A)))$. According to these notations,
the sequence $I^n=\{I_i^n\}_{i\in\N}=R^{*n}*R^\infty$
is a sequence of $L$'s and $R$'s defined
inductively as follows.
$$
{ I}^{0}=R^\infty\quad\hbox{and}\quad { I}^{n+1}=R*{ I}^{n},\
n\in\N.$$
It follows from the definition of the $*$-product that
\begin{equation}
I_{2i}^{n+1}=R,\ i\in\N,
\label{PAIR}
\end{equation}
and
\begin{equation}
I_{2i+1}^{n+1}=\check{I}_i^{n},\ i\in\N,
\label{IMPAIR}
\end{equation}
where $\check{R}=L$ and $\check{L}=R$. For the convenience of the
reader, we list the first four sequences
\[
\begin{array}{l}
I^0=RR\cdots \\
I^1=RLRL\cdots \\
I^2=RLRRRLRR\cdots \\
I^3=RLRRRLRLRLRRRLRL\cdots
\end{array}
\]
According to expressions (\ref{PAIR}) and (\ref{IMPAIR}),
one can easily verify that
${ I}^n$ is a $2^n$-periodic sequence with the following properties.
\begin{itemize}
\item[(A)] For all $n\in\N$, the sequence $\{I_i^{n}\}_{0\leq i\leq 2^n-1}$ is
odd \cite{Collet}.
\item[(B)] For all $n\geq 1$, $I_{i+2^n}^{n+1}=I_i^{n+1},\ 0\leq
i\leq 2^n -2 $.
\item[(C)] For all $n\in\N$, $I_i^{n+1}=I_i^{n},\ 0\leq i\leq 2^{n+1}-2$.
\item[(D)] For all $n\geq 1$, the sequence $\{I_i^{n+1}\}_{0\leq i\leq 2^n-2}$
is even if $n$ is even and is odd, otherwise.
\item[(E)] For all $n\geq 1$, $I_{2^{n-1}-1}^n=L$ and $I_{2^n-1}^n=R$ if $n$
is even and $I_{2^{n-1}-1}^n=R$ and $I_{2^n-1}^n=L$ if $n$ is odd.
\end{itemize}
{\sl Proof.} All the proofs are done by induction.
(B) We have $I^2=(RLRR)^\infty$. Then
$I^2_{0+2}=I^2_0$. The statement holds for $n=1$.
Now assume the statement holds for some $n\geq 2$. Then,
$\check{I}_{i+2^n}^{n+1}=\check{I}_i^{n+1},\ 0\leq i\leq 2^n-2$. Using
(\ref{IMPAIR}), it follows that $I_{2i+2^{n+1}+1}^{n+2}=I_{2i+1}^{n+2},\ 0\leq
i\leq 2^n-2$. On the other hand, the expression (\ref{PAIR}) implies that
$I_{2i+2^{n+1}}^{n+2}=R=I_{2i}^{n+2},\ 0\leq i\leq 2^n-1$. The
statement then holds for $n+1$.
The proof of (C) is similar.
(D) The sequence $\{I_i^2\}_{0\leq i\leq 0}=R$
is odd and the statement holds for $n=1$.
Now we assume that the statement holds for some $n\in\N$, $n$ odd. We have
$$
\{I^{n+2}_i\}_{0\leq i\leq 2^{n+1}-2}=
\{I^{n+2}_{2i}\}_{0\leq i\leq 2^n-1}\cup
\{I^{n+2}_{2i+1}\}_{0\leq i\leq 2^n-2}.
$$
By (\ref{PAIR}), the sequence $\{I^{n+2}_{2i}\}_{0\leq i\leq 2^n-1}$
is even. By (\ref{IMPAIR}) and by the induction assumption, the sequence
$\{I^{n+2}_{2i+1}\}_{0\leq i\leq 2^n-2}$ is also even. Therefore, the
statement holds for $n+1$ that is even. A similar argument is done in
the case where $n$ is even and this completes the induction.
(E) We have $I_0^1=R$ and $I_1^1=L$ and
then the statement holds for $n=1$. Let $n\geq 1$ be odd and assume
that $I_{2^{n-1}-1}^n=R$ and $I_{2^n-1}^n=L$. By relation
(\ref{IMPAIR}), we have $I_{2^n-1}^{n+1}=\check{I}_{2^{n-1}-1}^n=L$
and $I_{2^{n+1}-1}^{n+1}=\check{I}_{2^n-1}^n=R$ and then the
statement holds for $n+1$ which is even. The case $n$ is even follows
similarly. \qedd
\subsection{The sequences ${ J}^n$}\label{J-n}
The sequences ${ J}^n=\{J_i^n\}_{i=0}^{2^{n+1}-1}=R^{*n}*RC$
satisfy the following induction
$$
{ J}^0=RC\quad\hbox{and}\quad { J}^{n+1}=R*{ J}^n,\
n\in\N. $$
Again the definition of the $*$-product implies that
$$
J_{2i}^{n+1}=R,\ 0\leq i\leq 2^{n+1}-1
$$
and
$$
J_{2i+1}^{n+1}=\check{J}_i^n,\ 0\leq i\leq 2^{n+1}-1,
$$
where $\check{R}=L$, $\check{L}=R$ and $\check{C}=C$. Consequently,
$J^n=\{J^n_i\}_{0\leq i\leq 2^{n+1}-1}$, $J_{2^{n+1}-1}^n=C$ and
\begin{equation}
J^n_i=I^n_i,\quad 0\leq i\leq 2^{n+1}-2.
\label{EGALIJ}
\end{equation}
The proof is similar to that of (B).
\section{Dynamics of Fundamental Sets}
With the properties of the kneading sequences provided, we can now
turn to the analysis of the coupled system.
In this section, we give definitions of fundamental sets and their
properties.
Given $a,b\in [-1,1]$, $a**n$, or $I(1)=J^{m}, m \geq n$.
The next statement shows that the points $f^{2^k}(0)$ and
$f^{2^{k+1}}(0)$, $k0$ when $k$ is even and $f^{2^k}(0)<0$ when $k$ is odd.
\label{NONNUL}
\end{Pro}
{\sl Proof.} When $k=0$, $f(0)=1>0$ holds
whatever the kneading sequence is. When $I(1)=I^k$ for some $k>0$,
the result follows from (E) and the relation
$I_0(f^{2^k}(0))=I_{2^k-1}^k$. If $I(1)=I^n$ with $n>k$,
the result holds because of (C). In the
case $I(1)=J^n$, $n\geq k$, the result can be obtained by using the
expression (\ref{EGALIJ}). \qedd
Statement {\em (i)} of Theorem \ref{DynFund} then follows immediately
from the next Proposition.
\begin{Pro}\label{Inclusion}
If $I(1)=I^n$ or $J^n$ for some $n\geq k+2$ where $k\in\N$, then
\[
C_{\{f^{2^{k+1}}(0),f^{2^{k+2}}(0)\}}\subset
C_{\{f^{2^k}(0),f^{2^{k+1}}(0)\}}.
\]
\end{Pro}
In order to prove this proposition and later results, we use the
following lemma.
\begin{Lem}\label{Lm:31}
If $I(1)=I^n$ or $J^n$ for some $n>k$ where $k\in\N$, then
$$
C_{\{f^{2^k}(0),f^{3\cdot 2^k}(0)\}}\subset C_{\{0,f^{2^k}(0)\}}.
$$
\end{Lem}
{\sl Proof.} We show that $00$ and the result is
proved for $n=k+1$.
The arguments use only the first $3\cdot 2^k-1$ symbols of
$I(1)$. Property (C) implies that the result holds for $I(1)=I^n$
with $n\geq k+2$. Due to the relation (\ref{EGALIJ}), the same is true
in the case $I(1)=J^n$ with $n\geq k+1$. Similar arguments show that
$f^{2^k}(0)0$ and $\{I_{i-1}(1)\}_{1\leq i\leq 2^k-1}$ is even.
Using the previous equality and Lemma \ref{ORDERf}, we conclude that
$f^{2^{k+2}}(0)k+2$ and $J^n$ with
$n\geq k+2$. \qedd
The proof of the remaining assertions of Theorem \ref{DynFund} depends on the
following propositions.
\begin{Pro}
If $I(1)=I^n$ or $J^n$ for some $n\geq k, k\geq 1$, then
$$
F_\eps^{2^k}(\overline{T}_{\{0,f^{2^k}(0)\}})\subset
\underline{T}_{\{f^{2^k}(0),f^{2^{k+1}}(0)\}}.
$$
If $I(1)=I^n$ or $J^n$ for some $n\geq 0$, then
\[
F_\eps(\overline{T}_{\{0,f(0)\}})\subset
\underline{T}_{\{f(0),f^2(0)\}}\quad\text{if}\quad
0\leq\eps\leq\frac{1}{2}\quad\text{and}\quad
F_\eps(\overline{T}_{\{0,f(0)\}})\subset
\overline{T}_{\{f(0),f^2(0)\}}\quad\text{if}\quad
\frac{1}{2}\leq\eps\leq 1.
\]
\label{ITERAT}
\end{Pro}
Obviously, Statement {\em (iii)} follows from this proposition.
{\sl Proof.} Assume that $k\geq 1$.
We proceed similarly as in the proof of Lemma \ref{Lm:31} to check the
conditions of Lemma \ref{ORDERF} with $a=f^{2^k}(0)$ and $t=2^k-1$.
We first consider the case $I(1)=I^k$.
We have
\[
\{I_i(f^{2^k}(0))\}_{1\leq i\leq 2^k-1}=\{I_{i+2^k-1}(1)\}_{1\leq
i\leq 2^k-1}=\{I_{i-1}(1)\}_{1\leq i\leq 2^k-1},
\]
where the second equality is obtained by periodicity. Similarly, we have
\[
\{I_i(f^{2^k}(0))\}_{0\leq i\leq 2^k-1}=
\{I_{i+2^k-1}(1)\}_{0\leq i\leq 2^k-1}.
\]
This sequence is odd by Property (A) and periodicity. One can now
apply Lemma \ref{ORDERF} to obtain the desired result.
The arguments use only the first $2^{k+1}-1$ symbols of
$I(1)$. Property (C) implies that the result holds for $I(1)=I^n$
with $n\geq k+1$. Due to the relation (\ref{EGALIJ}), the same is true
in the case $I(1)=J^n$ with $n\geq k$.
For $k=0$, we have
\[
F_0(\overline{T}_{\{0,f(0)\}})\subset \underline{T}_{\{f(0),f^2(0)\}},
\]
independently of the kneading sequence. The results then follows from
Relation (\ref{PROCOUP}). \qedd
\begin{Pro}
For all $k\in\N$, we have
$\underline{T}_{\{f^{2^k}(0),f^{2^{k+1}}(0)\}}\subset
\underline{T}_{\{0,f^{2^k}(0)\}}$ if $I(1)=I^k$ or $J^k$.
The inclusion is strict except when $I(1)=J^k$.
When $I(1)=I^n$ or $I(1)=J^n, n >k$, we have the strict inclusion
$\underline{T}_{\{0,f^{2^k}(0)\}}\subsetneq
\underline{T}_{\{f^{2^k}(0),f^{2^{k+1}}(0)\}}$.
\label{ITERAT2}
\end{Pro}
{\sl Proof.} We want to apply Lemma \ref{ORDERf} with $a=f^{2^k}(0)$
and $t=2^k-1$. Assume that $I(1)=I^k$ and $k$ is even.
The case when $k$ is odd can be proved similarly.
By periodicity, we have
\[
\{I_i(f^{2^k}(0))\}_{1\leq i\leq 2^k-1}=\{I_{i-1}(1)\}_{1\leq i\leq 2^k-1}.
\]
Similarly, by periodicity and (A),
the sequence $\{I_i(f^{2^k}(0))\}_{0\leq i\leq 2^k-1}$ is odd. Lemma
\ref{ORDERf} then says that the inequality $0k+1$. \qedd
It remains to prove Statement {\em (ii)} of the theorem. Combine
Propositions \ref{ITERAT} and \ref{ITERAT2} to obtain
\[
F_\eps^{2^k}(C_{\{0,f^{2^k}(0)\}})\subset
C_{\{f^{2^k}(0),f^{2^{k+1}}(0)\}}\ \text{if}\ k\leq
n-1\quad\text{and}\quad
F_\eps^{2^n}(C_{\{0,f^{2^n}(0)\}})\subset C_{\{0,f^{2^n}(0)\}}.
\]
Statement {\em (ii)} is then proved for $k=n$. Now if $kk$, the itineraries of $1$, $f^{2^{k+1}}(1)$, and $f^{2^k}(1)$
are the same from $0$ to $2^k -1$. Thus, we have
\[
F_\eps^{2^k-1} (\underline{T}_{f^{2^k+1}(0),1})
\subset F_\eps^{2^k-1} (C_{\{f^{2^k+1}(0),f(0)}\}) \subset
C_{\{f^{2^{k+1}}(0),f^{2^k}(0)\}}
\]
and
\[
F_\eps^{2^k-1}(\underline{T}_{f^{2^{k+1} + 1}(0),1})
\subset F_\eps^{2^k-1}(C_{\{f^{2^{k+1}+1}(0),f(0)\}}) \subset
C_{\{f^{3 \cdot 2^{k}}(0),f^{2^k}(0)\}} \subset C_{\{0,f^{2^k}(0)\}}.
\]
The last inclusion is from Lemma \ref{Lm:31}.
This completes the proof of Theorem \ref{DynFund}.
{\bf Remark.} Assume that $k$ is even and $n>k$. Since $L_\eps$ is
a convex combination of coordinates, we have
\[
F_\eps^{2^k-1}(\underline{T}_{f^{2^k}(1),1})\subset
F_0^{2^k-1}(\underline{T}_{f^{2^k}(1),1})=F_0^{2^k}(
\overline{T}_{0,f^{2^k}(0)}).
\]
Proposition \ref{ITERAT} then shows that
\begin{equation}\label{Nice}
F_\eps^{2^k-1}(\underline{T}_{f^{2^k}(1),1})
\subset
\underline{T}_{f^{2^{k+1}}(0),f^{2^k}(0)}.
\end{equation}
Therefore, the conclusion of Statement {\em (iii)}
also applies
to any point in $\overline{R}_{f^{2^{k+1}}(0),f^{2^k}(0)}$ if the
first iterate of the point belongs to
$\underline{T}_{f^{2^k}(1),1}$. On the other hand, if the first
iterate of such a point belongs to $\overline{T}_{f^{2^{k+1}}(1),1}$,
then, the inclusion (\ref{Nice}) applied to $k+1$ shows that the
$2^{k+1}$-iterate belongs to
$T_{\{f^{2^{k+2}}(0),f^{2^{k+1}}(0)\}}$.
\subsection{Proof of Results on Asymptotic Behaviour}
{\sl Proof of Theorem \ref{Thm:eventualp}.}
By Statement {\em (ii)} of Theorem \ref{DynFund}, we may assume that
$i=0$. Otherwise, iterate the point $(x,y)$ $2^n - i$ times to obtain
\[ F^{2^n - i}_\eps(x,y) \in C_{\{ f^{2^n}(0),f^{2^{n+1}}(0) \}} \subset
C_{\{0,f^{2^n}(0)\}}.\]
Let $\{(x^t,y^t)\}_{t\in\N}$ be an orbit with initial condition
$(x,y)\in C_{\{0,f^{2^n}(0)\}}$. If $(x,y)=(x^{2^{n+1}},y^{2^{n+1}})$,
there is nothing to prove. Therefore, we assume that
$(x,y)\neq (x^{2^{n+1}},y^{2^{n+1}})$. We divide the situation into
four cases:
\begin{itemize}
\item[(a)] $(x\leq x^{2^{n+1}})(y\leq y^{2^{n+1}})$.
\item[(b)] $(x^{2^{n+1}}\leq x)(y^{2^{n+1}}\leq y)$.
\item[(c)] $(x\leq x^{2^{n+1}})(y^{2^{n+1}}\leq y)$.
\item[(d)] $(x^{2^{n+1}}\leq x)(y\leq y^{2^{n+1}})$.
\end{itemize}
The notation $(x_1\leq x_2)(y_1\leq y_2)$ means that both
inequalities hold and at least one inequality is strict.
In the cases (a) and (b), we need the following lemma.
\begin{Lem}\label{Lm:order1}
Let $a\in [-1,1],\ a\neq 0$ and $t\in\N$ be such that
$I_i(a)=I_{i-1}(1)\neq C$, $1\leq i\leq t$ if $t\geq 1$.
Then, for any $(x_1,y_1),(x_2,y_2)\in C_{\{0,a\}}$
such that $(x_1\leq x_2)(y_1\leq y_2)$, we have
\[
(x_1^{t+1}\leq x_2^{t+1})(y_1^{t+1}\leq y_2^{t+1})\quad\text{if}\quad
\{I_i(a)\}_{0\leq i\leq t}\quad\text{is\ even};
\]
\[
(x_2^{t+1}\leq x_1^{t+1})(y_2^{t+1}\leq y_1^{t+1})\quad\text{if}\quad
\{I_i(a)\}_{0\leq i\leq t}\quad\text{is odd}.
\]
\end{Lem}
{\sl Proof of the lemma:} We prove it inductively. Using the piecewise
monotonicity of $f$, we have
\[
(x_1^1\leq x_2^1)(y_1^1\leq y_2^1)\quad\text{if}\quad I_0(a)=L
\quad\text{and}\quad
(x_2^1\leq x_1^1)(y_2^1\leq y_1^1)\quad\text{if}\quad
I_0(a)=R.
\]
Hence, the statement holds for t=0. Now, given $t\in\N$, let us assume that
$\{I_i(a)\}_{0\leq i\leq t+1}$
is even. Then either $\{I_i(a)\}_{0\leq i\leq t}$ is even and
$I_{t+1}(a)=I_t(1)=L$, or $\{I_i(a)\}_{0\leq i\leq t}$ is odd and
$I_{t+1}(a)=I_t(1)=R$.
In the first case, by induction hypothesis, we have
$(x_1^{t+1}\leq x_2^{t+1})(y_1^{t+1}\leq y_2^{t+1})$ and
\[ (x^{t+1}_1,y^{t+1}_1),(x^{t+1}_2,y^{t+1}_2)\in
C_{\{f^{t+1}(0),f^{t+1}(a)\}}\]
by Lemma \ref{ORDERF}. Using the same argument
as the one in the case $t=0$, we obtain
\begin{equation}
(x_1^{t+2}\leq x_2^{t+2})(y_1^{t+2}\leq y_2^{t+2}).
\label{ULTIM}
\end{equation}
In the second case, the inequalities
$(x_2^{t+1}\leq x_1^{t+1})(y_2^{t+1}\leq y_1^{t+1})$ also imply
(\ref{ULTIM}).
Therefore, the statement holds for $t+1$ in the even case. The odd case
follows similarly. \qedd
In the case (a), we apply the previous lemma twice with $a=f^{2^n}(0)$ and
$t=2^n-1$ (see proof of Proposition \ref{ITERAT}), to obtain by induction,
$$
(x^{t\cdot 2^{n+1}}\leq
x^{(t+1)\cdot 2^{n+1}})(y^{t\cdot 2^{n+1}}\leq y^{(t+1)\cdot 2^{n+1}}),
\ t\geq 1,
$$
and the orbit cannot be periodic. Since the sequences
$\{x^{t.2^{n+1}}\}$ and $\{y^{t.2^{n+1}}\}$ are bounded and monotonic, they
must converge. By continuity of $f$, the limit is a
periodic orbit of $F_\eps$ whose period is a factor of $2^{n+1}$.
The case (b) is proved analogously.
In the cases (c) and (d), we use the following lemma. Note that if
$\eps=\frac{1}{2}$, these two cases cannot occur. So we assume
$\eps\neq\frac{1}{2}$ in the following statements.
\begin{Lem}\label{Lm:order2}
Let $a\in [-1,1],\ a\neq 0$ and $t\in\N$ be such that
$I_i(a)=I_{i-1}(1)\neq C$, $1\leq i\leq t$, if $t\geq 1$. Then, for any
$(x_1,y_1),(x_2,y_2)\in C_{\{0,a\}}$ such that $(x_1\leq x_2)(y_2\leq
y_1)$, At least one of the following properties holds.
\begin{itemize}
\item[(i)] There exists $1\leq i\leq t+1$ such that
$$
(x_1^i\leq x_2^i)(y_1^i\leq y_2^i)\quad\text{or}\quad
(x_2^i\leq x_1^i)(y_2^i\leq y_1^i).
$$
\item[(ii)] $(x_1^{t+1}\leq x_2^{t+1})(y_2^{t+1}\leq y_1^{t+1})$
if $\{I_i(a)\}_{0\leq i\leq t}$ is even and
$0\leq\eps <\frac{1}{2}$, or $\{I_i(a)\}_{0\leq i\leq t}$ is odd
and $\frac{1}{2}<\eps\leq 1$.
\item[(iii)] $(x_2^{t+1}\leq x_1^{t+1})(y_1^{t+1}\leq y_2^{t+1})$ if
$\{I_i(a)\}_{0\leq i\leq t}$ is odd and
$0\leq\eps < \frac{1}{2}$, or $\{I_i(a)\}_{0\leq i\leq t}$ is even
and $\frac{1}{2}<\eps\leq 1$.
\end{itemize}
\end{Lem}
{\sl Proof of the lemma:} We again use induction. Assume that $t=0$. If
$I_0(a)=L$, then, we have by assumption $(f(x_1)\leq f(x_2))(f(y_2)\leq
f(y_1))$ and therefore for $0\leq\eps <\frac{1}{2}$,
$$
x_1^1-y_1^1=(1-2\eps)(f(x_1)-f(y_1))<(1-2\eps)(f(x_2)-f(y_2))=
x_2^1-y_2^1.
$$
There are three cases: $(x_1^1\leq x_2^1)(y_1^1\leq y_2^1)$,
$(x_2^1\leq x_1^1)(y_2^1\leq y_1^1)$ or $(x_1^1\leq x_2^1)(y_2^1\leq
y_1^1)$, and the conclusion holds for $t=0$ in the case
$\{I_i(a)\}_{0\leq i\leq 0}$ is even. If $\eps>\frac{1}{2}$, then
the three cases are $(x_1^1\leq x_2^1)(y_1^1\leq y_2^1)$,
$(x_2^1\leq x_1^1)(y_2^1\leq y_1^1)$ and $(x_2^1\leq x_1^1)(y_2^1\leq
y_1^1)$ and the conclusion also holds. The odd case is proved similarly.
Now, given $t\in\N$, we assume that the statement holds. In case {\em
(i)}, there is nothing to prove. In the cases {\em (ii)} or {\em
(iii)}, the statements follow from arguments similar to $t=0$.
\qedd
Now, we go back to the proof of the theorem.
In the case (c), we apply Lemma \ref{Lm:order2} with $a=f^{2^n}(0)$ and
$t=2^n-1$. Then, if we are in the case {\em (i)}, we conclude as in the
cases (a) and (b). Otherwise, we are in the case {\em (iii)} and we
repeat the argument.
By induction, we are either in the case (a) or (b) after a finite number of
steps or we have
$$
(x^{t.2^{n+1}}\leq x^{(t+1).2^{n+1}})(y^{(t+1).2^{n+1}}\leq
y^{t.2^{n+1}}),\quad t\in\N
$$
In this case, the sequences $\{x^{t.2^{n+1}}\}$ and
$\{y^{t.2^{n+1}}\}$ also converge and the orbit is asymptotically
periodic with a period a factor of $2^{n+1}$. The case (d)
is proved similarly.
Moreover, by Statement {\em (iii)} of Theorem \ref{DynFund},
if a periodic orbit in $C_{\{0,f^{2^n}(0)\}}$ is such that
$x^t\neq y^t$, $t\in\N$, then its period must be a multiple of $2^{n+1}$.
To see this, assume otherwise that the period is
$2^t\cdot m$ where $0\leq t\leq n$
and $m$ is odd. Then after $2^n\cdot m$ iterations, the point lies on
the other side of the diagonal according the statement {\em
(iii)} of Theorem \ref{DynFund}.
But it should lie on the same side on the diagonal
because $2^n\cdot m$ is a multiple of the period. We have a
contradiction. \qedd
{\sl Proof of Theorem \ref{Synchro}.}
By assumption we have
$I(f^t(0))=I(f^{t+2^n}(0))\neq C,\ $ for any $t\geq 1$. It follows from
applying Lemma \ref{IMAGE} repeatedly that
$$
F_\eps^{m\cdot 2^n}(C_{\{f^t(0),f^{t+2^n}(0)\}})\subset
C_{\{f^{t+m\cdot 2^n}(0),f^{t+(m+1)\cdot 2^n}(0)\}}, \ m\geq 1.
$$
By Proposition \ref{ITERAT}, this inclusion also holds for
$t=0$.
But, as said before, 0 is attracted to the periodic orbit of $f$.
It follows that the orbit of any initial condition in
$C_{\{f^i(0),f^{i+2^n}(0)\}}$ must be eventually synchronised.
\subsection{Proof of Theorem \ref{Force}}
The main property of a symmetric $2^{k+1}$-periodic orbit
we use in the proof is that the difference $x^t-y^t$ changes its sign
every $2^k$ iterations. Hence, we start by collecting results from the
proof of Theorem \ref{DynFund} on the changes of signs of $x^t-y^t$.
\begin{Lem}
Assume that $I(1)=I^n$ or $J^n$ for some $n\geq 1$. Let
$\{(x^t,y^t)\}_{t\in\N}$ be an orbit of $F_\eps$. Assume there
exists $0n$.
This completes the proof of the theorem. \qedd
{\bf Remark.} The same arguments can be applied to show that a
symmetric $2^{k+1}$-periodic orbit cannot pass the
square at the next level $C_{\{f^{2^{k+1}}(0),f^{2^{k+2}}(0)\}}$ when
$k 0$ and
$f^{2^{k+1}}(0)<0$ if $k0$ and $h_k(0,y)>0$. This completes the proof.
\qedd
{\bf Remark.} The last part of the proof shows no symmetric
$2^{k+1}$-periodic orbit can
pass $C_{ \{0, f^{2^{k+1}}(0)\} }$ since the equation
$h_k(x,y)=0$ has no solution in this square.
By Implicit Function Theorem, for any point $(x_0,y_0)$ not on the
boundary of the square $C_{\{0,f^{2^k}(0)\}}$ or
$C_{\{f^{2^k}(0),f^{2^{k+1}}(0)\}}$ satisfying
$h_k(x_0,y_0)=0$, there is an open neighbourhood of $(x_0,y_0)$
such that its intersection with the set
$\{(x,y): h_k(x,y)=0 \}$ is diffeomorphic to an open interval. If
$(x_0,y_0)$ is on the boundary of these squares, the
intersection is diffeomorphic to a semi-closed interval.
\begin{Lem}\label{Locuniq}
Assume the kneading sequence of $f$ is $I^n$ or $J^n$ for some
$n\in\N$. Then for any $0\leq k\leq n$, $f$ has a unique periodic
point $x^*_k$ of period $2^k$ in the interior of $[0,f^{2^k}(0)]$.
\end{Lem}
{\sl Proof.} Assume for simplicity, that $k$ is even, the odd case can
be proved analogously. We have $f^{2^k}(0)>0$ and
$f^{2^{k+1}}(0)**