\documentclass[11pt,leqno]{article} 
\title{Well-posedness of a degenerate parabolic equation\\
issuing from two-dimensional perfect fluid dynamics} 
\author{Carole Rosier*, Lionel Rosier**}
\date{}
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\def\N{{\mathbb N}}
\def\Z{{\mathbb Z}}
\def\Q{{\mathbb Q}}
\def\R{{\mathbb R}}
\def\C{{\mathbb C}}
\def\DS{\displaystyle}
\def\va{\hfill\break}
\def\sou#1{$\underline{\smash{\hbox{#1}}}$}
\def\proof{\noindent {\sc Proof:}\qquad}
\def\to{\rightarrow}
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\def\be{\begin{equation}}
\def\ee{\end{equation}}
\def\bee{\begin{equation*}}
\def\eee{\end{equation*}}
\def\ba{\begin{array}}
\def\ea{\end{array}}
\def\ep{\epsilon}
\def\o{\omega}
\def\p{\psi}
\def\U{{\cal U}}
\def\B{{\cal B}}
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\def\H{\tilde{H}}
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\begin{document} 
\vspace{-4.cm}
\maketitle
\vspace*{-1cm}
\begin{center}
* UMR CNRS 5585, Analyse Num\'erique, \\
Universit\'e Lyon 1, b\^at. 101,\\
69622 Villeurbanne Cedex, France. \\
{\footnotesize
e-mail: rosier@lan.univ-lyon1.fr}\\
\vspace{0.5cm}
** Laboratoire d'Analyse Num\'erique et EDP,\\
Universit\'e Paris 11, b\^at. 425,\\ 
91405 Orsay Cedex, France.\\ 
{\footnotesize
e-mail: Lionel.Rosier@math.u-psud.fr}\\
\end{center}

\begin{abstract}
A degenerate parabolic equation of convection-diffusion
type has been proposed by Robert and Sommeria in \cite{RS2} to describe the
relaxation towards statistical equilibrium states in 
2D incompressible perfect fluid dynamics. 
The paper is concerned with the Cauchy problem for this
equation. The local existence of a variational solution is obtained in using
the decrease of the (negative) mixing entropy and Schauder theorem.
A smoothing effect is used when proving the
uniqueness of the variational solution by Youdovitch's method.
Finally, global existence of solutions and their asymptotic convergence 
towards Gibbs states are shown for a large class of initial data.
\va\va
{\bf Key Words:} Degenerate parabolic equation,
statistical equilibrium, Youdovitch's method
\va
{\bf AMS subject classification:} 35K65, 35Q35, 76C05, 76F99
\va
\end{abstract}
\section{Introduction}

The motion of an incompressible perfect fluid in a periodic  
domain ${\cal Y}=\left( \R / \Z \right) ^2$ is described by Euler 
equations, which in the stream function-vorticity formulation 
read as follows:
\begin{equation}
\label{eq:200}
\left\{
\begin{array}{l}
\DS\frac{\partial \omega }{\partial t}+
\hbox{div}(\omega \nabla\times \psi )=0 \hbox{ in } {\cal Y}, \\
\\
-\triangle \psi =\omega \hbox{ in }{\cal Y},\ 
\int_{\cal Y}\psi\, dy=\int_{\cal Y}\omega\, dy=0,\ \psi \hbox{ is }
1-\hbox{periodic in }y_1,y_2.
\end{array}
\right.
\end{equation}
Here $\omega$ is the {\em vorticity}, $\psi$ is the {\em stream function}
and $\nabla\times \psi=(\frac{\partial \psi}{\partial y_2},  
-\frac{\partial \psi}{\partial y_1})$ 
stands for the {\em velocity}.
A famous result by Youdovitch \cite{Y} asserts the existence and unicity of 
a weak solution $\omega\in L^{\infty}((0,+\infty )\times{\cal Y})$ of
(\ref{eq:200}) together with $\omega _{\vert _{t=0}}=\omega _0$, for any initial
state $\omega _0\in L^{\infty}({\cal Y})$ such that 
$\int_{\cal Y}\omega _0\, dy=0$. 
>From now on we shall assume (for the sake of simplicity) 
that $\omega _0$ is a {\em vortex patch}, meaning 
that $\omega _0(y)$ takes only
two values, say $-1$ and $1$. Due to the conservation  of all generalized
enstrophies  $\int_{\cal Y}f(\omega (t,y ))\, dy$ (where $f$ denotes
any continuous function defined on $\R$), 
$\omega (t,\cdot )$ also takes values $-1$ and $1$ only, for almost every
$t\ge 0$. Moreover it is well known that the vorticity 
$\omega (t,\cdot )$ oscillates (between $-1$ and $1$) on increasingly small 
scales over time, whereas large scale organized structures (the so-called
{\em coherent structures}) seem to emerge in the long-time behavior, as it
has been shown by both physical and computational 
experiments (see \cite{IC},
\cite{Ma}, \cite{SMS}). 
A statistical equilibrium theory has been developped
to explain the emergence of these coherent structures
(see \cite{M}, \cite{MWC}, \cite{R1}, \cite{RS1}). 
Roughly speaking, a statistical equilibrium state 
is obtained by solving a constrained maximization problem, 
the objective functional being the {\em mixing entropy}
(namely $-\frac{1}{2}\int_{\cal Y}\{ (1+\omega ) \hbox{ln} (1+\omega )+
(1-\omega ) \hbox{ln} (1-\omega )\}\, dy$) and 
the constraint functionals being the {\em energy} 
(namely $\frac{1}{2}\int_{\cal Y}\omega\psi\, dy$) and the generalized
{\em enstrophies}. An extensive study of the links between the nature of the
equilibrium state and the value of the energy is carried out in \cite{TW1}. 
In this paper the numerical determination of the equilibrium
state is performed by directly solving previous 
constrained maximum entropy problem. Another approach (\cite{RS2}, 
\cite{RR}) to determine the statistical equilibrium
rests on numerical integration over a sufficiently long-time interval of
a {\em macroscopic} evolution equation. To be precise, a relaxation
equation of convection-diffusion type, which is designed to conserve above  
constraint functionals but also to maximize the {\em production} of
entropy, is proposed by Robert and Sommeria 
in \cite{RS2} to serve as a model for the dynamics of the 
{\em macroscopic} vorticity 
$\overline{\omega}$, which is derived from the {\em microscopic} vorticity
$\omega$ by taking local mean value. (Notice that $\overline{\omega}$ is 
expected to take values in $[-1,1]$.) An advantage of this approach is to 
give, in addition to the statistical equilibrium state
(which corresponds to a
(strong) limit in $L^2({\cal Y})$ for $\overline{\omega}(t)$) a 
(coarse-grained) description of the transient relaxation process 
towards  the equilibrium state. A stream function-vorticity 
formulation (see \cite{RR}) of above
relaxation equation with periodic boundary conditions is as follows:
\def\oo{{\overline{\omega}}}
\def\pp{{\overline{\psi}}}
\begin{equation}
\label{eq:202}
\left\{
\begin{array}{l}
\DS\frac{\partial \oo }{\partial t}+
\hbox{div}(\oo \nabla\times \pp ) 
-\hbox{div}\big( A(\oo )\nabla \oo +
\beta (\oo ) A(\oo )(1-\oo ^2)\nabla\pp \big)
=0 \hbox{ in } {\cal Y}, \\
\\
-\triangle \pp =\oo \hbox{ in }{\cal Y},\ 
\int_{\cal Y}\pp \, dy=\int_{\cal Y}\oo \, dy=0,\ \pp \hbox{ is } 
1-\hbox{periodic in } y_1,y_2, 
\end{array}
\right.
\end{equation}
where
\begin{equation}
\label{eq:203}
A(\oo ):=\nu (1-\oo ^2),\ \nu >0 \hbox{ being constant}
\end{equation}
and
\begin{equation}
\label{eq:204}
\beta (\oo ):=-\frac{\int_{\cal Y} \nabla\pp\cdot\nabla\oo A(\oo )\, dy}
{\int_{\cal Y} \vert\nabla \pp\vert ^2 (1-\oo ^2)A(\oo )\, dy }\cdot
\end{equation}
We emphasize that in contrast to Navier-Stokes equation (in the 
stream function-vorticity formulation) the viscosity coefficient $A(\oo )$
fails to be constant (it vanishes when there is no mixing of vorticity 
at small scales) and that a supplementary convection term 
$-\beta (\oo )\hbox{div}(A(\oo )(1-\oo ^2)\nabla \pp )$ has been incorporated
in (\ref{eq:202}) in order to ensure the conservation of energy. 

The aim of this paper is to study the well-posedness of (\ref{eq:202}). There
are two reasons for wishing to accomplish such a task:\va
$\bullet$ Numerical implementation of (\ref{eq:202}) works very well (notice
that a blow-up of the inverse temperature $\beta (\oo )$ never occurs) and
numerical experiments are in good agreement with those carried out for 
Navier-Stokes equation with a high Reynolds number (see \cite{RR},
\cite{RSRS}). On the other hand
it is well-known that the equilibrium state we get by integrating Navier-Stokes
equation with high Reynolds number (instead of Euler equation) suffers from
energy dissipation. This drawback of Navier-Stokes equation does not
affect the relaxation equation (\ref{eq:202}), which has been designed to
conserve the energy. \va
$\bullet$ From a technical viewpoint  this equation presents two interesting
features: 1) it is a degenerate parabolic equation with a (non-local)
convectif term; 2) the (non-local) singular coefficient $\beta (\oo )$,
which appears in a crucial manner in most of the estimates, is shown to
be bounded provided that $\omega$ is bounded in $H^1({\cal Y})$, whereas
it is at most infered from physical considerations (e.g. the conservation
of energy, the decrease of the negative entropy, etc.) that
$\Vert \omega\Vert _{L^{\infty}({\cal Y})} \le 1$.

The particular form of the variable coefficient $A$ in (\ref{eq:203})
has been derived from dynamical considerations 
on the diffusion currents in \cite{RR}. When we replace the coefficient
$A$ by some constant $\nu >0$ in (\ref{eq:202}) and (\ref{eq:204}), we get a 
``simplified'' relaxation equation which has been studied in great details in  
\cite{MR}. Whereas numerical simulations using
the relaxation equation (\ref{eq:202})-(\ref{eq:203})
or high Reynolds number Navier-Stokes equation yield almost identical results,
noticeably different results are obtained with the
simplified relaxation equation: the final state proves to be deeply
altered by the diffusion \cite{RR}.
The goal of this paper is to show that the main results
in \cite{MR} also hold true when $A$ has the good expression given by
(\ref{eq:203}).
It turns out that local existence and unicity 
in the degenerate case require specific
(and somewhat technical) treatments, in contrast to \cite{MR} where
(local) well-posedness is obtained as a direct consequence of sectorial
operators theory.

The paper is organized as follows. In section 2 we
introduce notations and tools 
which will be subsequently used. In section 3 local existence of a solution
of a variational problem derived from (\ref{eq:202}) is proved 
by means
of Schauder fixed-point theorem together with an estimate related to the
decrease of the negative entropy.  Such a solution is shown to take values
in $[-1,1]$. Next the main physical features of the solution, namely  
the conservation of the energy, the decrease of the negative entropy and 
also two useful identities (\ref{eq:28}), (\ref{eq:29}) (see below) are
collected in prop. 1. In section 4 a (uniform in time) estimate for
$\Vert A(\oo )\nabla \oo \Vert _{L^2({\cal Y}) ^2}$,
which is used later in the proof of uniqueness,
is established by applying a regularization procedure 
from \cite{BG}. In section 5 uniqueness of variational solution is  
shown by adapting to our degenerate situation  
the clever proof in \cite{Y} (see also \cite{Mi})
of Youdovitch's theorem.
When the initial state $\oo _0$ is assumed to belong to a (non-trivial) set
${\cal U}(p,\epsilon )$ (see below), then global existence of the solution
together with the boundedness of $\beta(\oo )$ are obtained in  
prop. 3. Furthermore it is proved in section 6 that for such initial state 
the $\omega$-limit set is made up of Gibbs states, which represent the
coherent structures mentioned above.

Being from now on {\em only} concerned with 
{\em macroscopic} functions $\oo$,$\pp$ 
we shall omit the bars over $\o$,$\p$
for the sake of simplicity. Thus 
$\omega (t,y)$ will stand for the local average of the (microscopic)
vorticity, instead of the microscopic vorticity itself, and $\psi (t,y)$ 
will denote the corresponding stream function. Notice that for a smooth 
enough and periodic vorticity $\omega$, we have
\begin{equation}
\label{eq:2005}
\int_{\cal Y}\nabla\psi\cdot\nabla\omega \, (1-\omega ^2)\, dy 
=\int_{\cal Y}\omega ^2(1-\frac{\omega ^2}{3})\, dy,
\end{equation} 
hence
\begin{equation}
\label{eq:2006}
\beta (\omega )=-
\frac{\int_{\cal Y}\omega ^2(1-\frac{\omega ^2}{3})dy}{
\int_{\cal Y}\vert\nabla \psi\vert ^2(1-\omega ^2)^2dy}\cdot
\end{equation}
Finally (\ref{eq:202}) may be written as follows
\begin{equation}
\label{eq:207}
\left\{
\begin{array}{l}
\DS\frac{\partial \omega }{\partial t}
-\nu\, \hbox{div}\big( (1-\omega ^2)\nabla\omega \big) +
\hbox{div}(\omega\nabla\times \psi )-\nu \beta (\omega )
\hbox{div}\big( (1-\omega ^2)^2 \nabla\psi\big) =0 
\hbox{ in } {\cal Y}, \\
\\
-\triangle \psi =\omega \hbox{ in }{\cal Y},\ 
\int_{\cal Y}\psi\,  dy=\int_{\cal Y}\omega\,  dy=0,\ \psi \hbox{ is }
1-\hbox{periodic in }y_1,y_2. 
\end{array}
\right.
\end{equation}
\section{Some auxiliary results}

In this section we introduce some notations and we recall some useful facts.
Setting ${\cal Y}:=(-\frac{1}{2},\frac{1}{2})^2$, let $H_{per}^n ({\cal Y})$
denote the usual Sobolev space of {\em periodic} functions, that is 
$H_{per}^n ({\cal Y})$ is the completion of the space 
\begin{equation*}
C_{per}^{\infty}(\overline{\cal Y}):=\{ \phi\in C^{\infty}(\R ^2);\
\phi (y_1+1,y_2)=\phi (y_1,y_2)= \phi (y_1,y_2+1)\hbox{ for every }
y\in \R ^2\}
\end{equation*}
for the norm
\begin{equation*}
\Vert \phi \Vert _{H_{per}^n({\cal Y}) }:=
\left( \sum_{
\alpha\in \N ^2,
\vert \alpha \vert \le n}
\Vert \partial ^{\alpha} \phi \Vert _{L^2({\cal Y})}^2 \right) ^{\frac{1}{2}}.
\end{equation*}
\def\y{{\cal Y}}
\def\b{\beta}
Let
\begin{equation*}
\tilde{L}^2(\y ):=\{ u\in L^2(\y );\ \int_{\y} u(y)\, dy=0\} ,\quad
\tilde{H}_{per}^n(\y ):=  H_{per}^n(\ Y)\cap \tilde{L}^2(\y )
\end{equation*}
and ${\cal A}(u):=-\triangle u\in \tilde{L}^2(\y )$ for 
$u\in \tilde{H}_{per}^2(\y )$. Let
a {\em vorticity} state $\o\in \tilde{L}^2(\y )$ be given. The 
{\em stream function}  $\p$ associated with $\o$ is defined by
\begin{equation}
\label{eq:1}
\left\{
\begin{array}{rl}
-\triangle \p&=\o \hbox{ in }\y ;\\
\p \hbox{ is } &1-\hbox{periodic in } y_1 \hbox{ and } y_2  ;\\
\int_{\y}\p (y)\, dy&=0.
\end{array}
\right.
\end{equation}
It is easy to see that $\p\in \tilde{H}_{per}^2(\y )$ and that 
$\Vert \p \Vert _{H_{per}^2(\y )}\le C \Vert \o \Vert _{L^2(\y )}$
for some constant $C>0$. In other words ${\cal A}^{-1}$ is a continuous 
operator $\tilde{L}^2(\y )\to \tilde{H}_{per}^2(\y )$ and
$\p ={\cal A}^{-1}(\o )$. The {\em velocity} $u$ associated with $\o$ may
be defined by means of $\p$ as follows:
\begin{equation*}
u=(u_1,u_2)
:=\left(
\frac{\partial \p}{\partial y_2},
-\frac{\partial \p}{\partial y_1}
\right) \cdot
\end{equation*}
It follows that 
$u=\left(
\frac{\partial}{\partial y_2}, 
-\frac{\partial}{\partial y_1}
\right)  \circ {\cal A}^{-1}(\o )=:G(\o )$, where 
$G:\tilde{L}^2(\y )\to \tilde{H}_{per}^1(\y )^2$ is a continuous operator. 
The incompressibility equation
\begin{equation}
\label{eq:10000}
\hbox{div}\, G(\o )=0
\end{equation}
follows at once. 
In the sequel the space 
$\tilde{H}_{per}^1(\y )$ will be endowed with the 
Hilbertian norm
\begin{equation*}
\Vert \phi\Vert _{ \tilde{H}_{per}^1(\y )} :=
\left( \int_{\y}\vert \nabla \phi (y)\vert ^2\, dy\right) ^{\frac{1}{2}}
\end{equation*}
which is equivalent to the norm $\Vert \cdot\Vert _{H_{per}^1(\y )}$, thanks
to Poincar\'e's second inequality. Let $\tilde{H}_{per}^{-1}(\y )$ denote
the dual space of $\tilde{H}_{per}^1(\y )$, endowed with the (dual) norm
\begin{equation*}
\Vert f\Vert _{ \tilde{H}_{per}^{-1}(\y )}:=
\sup_{\phi\in  \tilde{H}_{per}^1(\y ) } 
\frac{<f,\phi >}
{\left( \int _{\y}\vert \nabla \phi\vert ^2\, dy\right) ^{\frac{1}{2}} } ,
\end{equation*}
where $<\cdot ,\cdot >$ denotes the duality pairing. Notice 
that if $\o$ is merely 
assumed to be in  $\tilde{H}_{per}^{-1}(\y )$, the solution $\p$ of 
(\ref{eq:1}) satisfies 
$\int_{\y} \nabla\p \cdot \nabla \phi\, dy =<\o ,\phi >$
for any $\phi\in \tilde{H}_{per}^{1}(\y )$, hence (by Cauchy-Schwarz 
inequality)
\begin{equation*}
\Vert \o \Vert _{\tilde{H}_{per}^{-1}(\y )}=
\left( \int _{\y}\vert \nabla \psi\vert ^2\, dy\right) ^{\frac{1}{2}}=
\Vert\psi \Vert _{  \tilde{H}_{per}^{1}(\y )}.
\end{equation*}
Also if $\o\in \tilde{L}^2(\y )$,
$\Vert \o \Vert _{\tilde{H}_{per}^{-1}(\y )}^2= 2{\cal E}(\o)$, where the 
{\em energy functional } ${\cal E}(\o )$ is given by
\begin{equation*}
{\cal E}(\o ):=\frac{1}{2}\int_{\y}\p (y)\o (y)\, dy.
\end{equation*}
For any open set $\Omega\subset\R ^2$, any $n\in \N ^*$ and  any 
$p\in [2,+\infty ]$, let $W^{n,p}(\Omega )$ denote the usual Sobolev space
endowed with the norm
\begin{equation*}
\Vert \phi \Vert _{W^{n,p}(\Omega )}:=
\sum_{
\alpha \in \N ^2,
\vert \alpha \vert\le n}
\Vert \partial ^{\alpha }\phi \Vert _{L^p(\Omega )} .
\end{equation*}
The following lemma is a periodic version of a famous 
result by Youdovitch (see \cite{Y}). Its proof is omitted.
\begin{lem}
There exists a constant $C>0$ such that for every $p\in(2,+\infty )$
and every $\omega\in L^p(\y )\cap \tilde{L}^2(\y )$ the solution $\p$ of
(\ref{eq:1}) satisfies
\begin{equation*}
\Vert \p \Vert _{W^{2,p}(\y )} \le 
C\cdot p\cdot \Vert \o \Vert _{L^p(\y )} .
\end{equation*}
\end{lem}
\va
For the sake of brevity we shall write
\begin{equation*}
V=H^1_{per}(\y ),\ V'={H^1_{per}(\y )}',\ H=L^2(\y )\quad (resp.\  
\tilde{V}=\tilde{H}^1_{per}(\y ),\
\tilde{V}'=\tilde{H}^{-1}_{per}(\y ),\  \tilde{H}=\tilde{L}^2(\y )).
\end{equation*}
\noindent The symbol $<\cdot ,\cdot >$ will denote the duality pairing
between $V$ and $V'$, or between $\V$ and $\V '$. It
is clear that each embedding in
$V\subset H=H'\subset V'$ (resp. $\V\subset \H =\H '\subset\V '$) 
is dense and compact. For any $T>0$ let $W(0,T)$ denote the
space 
\begin{equation*}
W(0,T):=\{ \o\in L^2(0,T,\V ),\ \frac{d \o }{dt}\in L^2(0,T,\V ')\}
\end{equation*}
endowed with the Hilbertian norm
\begin{equation*}
\Vert \o \Vert _{W(0,T)}=
\left( \Vert \o \Vert ^2_{L^2(0,T,\V )}+
\Vert \frac{d\o }{dt}\Vert ^2_{ L^2(0,T,\V ')} \right) ^{\frac{1}{2}}.
\end{equation*}
Notice that for every $\o\in W(0,T)$,
$\frac{d \o}{dt}\in L^2(0,T,V')$, as well. Indeed, we easily infer
from the relation
$\int_{\y }\o (t)\, dy=0 \hbox{ for a.e. } t\in (0,T)$
that the distributional derivative of $\o$ in
${\cal D}'(0,T,V')$ agrees with the function
$f\in L^2(0,T,V')$ defined by
$<f(t),v>_{V',V}
:=<\frac{d\o}{dt},v-\int_{\y}v\, dy>_{\V ',\V }$
for a.e. $t\in (0,T)$ and every $v\in V$.
On the other hand it follows from \cite[prop. 2.1 and thm
3.1, chapter 1]{LM} that
$W(0,T)\subset C([0,T],[\V ,\V ']_{\frac{1}{2}})=C([0,T],\H )$
with continuous embedding. Finally we infer from Aubin's Lemma
(see e.g. \cite{S}) that 
\begin{equation}
\label{eq:10}
\hbox{the embedding }W(0,T)\subset L^2(0,T,\H  ) \hbox{ is } compact.
\end{equation}
The following result by F. Mignot (see \cite{GM}) will be used many times
in the sequel.
\begin{lem}
Let $f:\R \to \R$ be a continuous and nondecreasing function such that
\begin{equation*}
\limsup_{\vert\lambda \vert \to +\infty} 
\left\vert \frac{f(\lambda )}{\lambda }\right\vert <+\infty .
\end{equation*}
Let $\o\in L^2(0,T,H)$ be such that
$\frac{d\o}{dt}\in L^2(0,T,V')$ and
$f(\o )\in L^2(0,T,V)$.
Then  
\begin{equation*}
<\frac{d\o }{dt},f(\o )>_{V',V}
=\frac{d}{dt}\int_{\y}\left( \int_0^{\o (\cdot ,y)}f(r)\, dr\right)\, dy
\quad \hbox{ in } {\cal D}'(0,T).
\end{equation*}
Hence for all $0\le t_1<t_2\le T$
\begin{equation*}
\int_{t_1}^{t_2}
<\frac{d\o }{dt},f(\o )>_{V',V}\, dt
=\int_{\y}\left( \int_{\o (t_1,y)}^{\o (t_2,y)}f(r)\, dr\right)\, dy.
\end{equation*}
\end{lem}
\va
The following functions will play a great role in what follows.
\begin{equation*}
\varphi (x):=\int_{-1}^x (1-s^2)\, ds,\quad 
S(x):=\frac{1+x}{2}\, \hbox{ln}(1+x)+\frac{1-x}{2}\, \hbox{ln}(1-x),\quad
x\in [-1,1].
\end{equation*}
Notice that the function $\varphi ^{-1}$  is
H\"older continuous with exponent $\frac{1}{2}$ 
on $[0,\frac{4}{3}]$.
\section{Local existence}

The aim of this section is to prove the following
\begin{thm}
Let $\o _0\in \H$ be such that
\begin{equation} 
\label{eq:2000}
-1\le \o _0(y)\le 1\hbox{ for a.e. } y\in\y \hbox{ and }
\int_{\y}\vert \nabla \p _0\vert ^2(1-{\o _0}^2)^2\, dy>0,
\end{equation}
where $\p _0$ solves (\ref{eq:1}) for $\o =\o _0$. Then there exist
a time $T>0$ and a solution $\o\in W(0,T)$ of
the following variational equation
\begin{equation}
\label{eq:2}
\left\{
\begin{array}{l}
\forall v\in L^2(0,T,V) \\
\\
\int_0^T<\frac{d\o}{dt},v>_{V',V}\, dt\ + \
\nu\int_0^T\int_{\y}
\{ \nabla \varphi (\o )-\frac{1}{\nu}\o G(\o )+\b (\o )(1-\o ^2)^2
\nabla \p \} \cdot \nabla v\, dydt\ = \ 0,
\end{array}
\right.
\end{equation}
such that
\begin{equation}
\label{eq:3}
\o (0,\cdot )=\o _0;
\end{equation}
\begin{equation}
\label{eq:4}
-1\le \o (t,y)\le 1 \hbox{ for a.e. }(t,y)\in [0,T]\times \y ;
\end{equation}
\begin{equation}
\label{eq:5}
\b (\o (t))\hbox{ is a (bounded) continuous
function on }[0,T].
\end{equation}
\end{thm}

Before proceeding to the proof of Theorem 1, let us do some comments.\va
i) $\o\in W(0,T)\subset C([0,T],\tilde{L}^2(\y ))$, hence (\ref{eq:3}) is
meaningfull. Moreover it follows from \cite[lem. 8.1, chapter 3]{LM} that 
\begin{equation} 
\label{eq:500}
\o\in C([0,T],L^p(\y )) 
\ \hbox{  for every } p\in [2,+\infty )
\end{equation}
and
\begin{equation}
\label{eq:501}
\Vert\o (t)\Vert _{L^{\infty}(\y )}\le 1 
\ \hbox{ for {\em every}  }  t\in [0,T].
\end{equation}
ii) Setting
$ <\hbox{div } u,b>:=-\int_{\y}u\cdot \nabla b\, dy\quad \hbox{for }
u\in H\times H,\ b\in V$,
(\ref{eq:2}) may be written
\begin{equation*}
\frac{d\o }{dt}-\nu\, \hbox{div}(\nabla \varphi (\o )) +
\hbox{div}\, \left( \o  G(\o )-\nu\b (\o )(1-\o ^2)^2\nabla \p \right)=0
\hbox{ in } L^2(0,T,V ').
\end{equation*}
\noindent{\sc Proof of Theorem 1:}\qquad
Set
$x^+:=\max (x,0)$ and $T_l(x):=\max (-l,\min (x,l))$
for $x\in\R$ and $l>0$.
Pick a constant
$M> \vert \beta (\omega _0 )\vert$,
%\frac{\int_{\y}{\o _0}^2\left( 1 -\frac{\o _0^2}{3}\right)\, dy}
%{\int_{\y}\vert \nabla \p _0\vert ^2(1-\o _0^2)^2\, dy},
set
$N:=\sup_{\Vert\o \Vert _{H}\le 2}
\Vert \nabla \p \Vert _{H\times H}$
%=\sup_{\Vert \o\Vert _{L^2(\y )}\le 2}
%\left( \int_{\y}\{ \left( \frac{\partial \p}{\partial y_1}\right) ^2
%+\left( \frac{\partial \p}{\partial y_2}\right) ^2 \} dy\right) ^{\frac{1}{2}}
and
$h_N(x):=\min (1,\frac{N}{x})$ for $x\ge 0$.
Let $(\ep _k)_{k\ge 0}$ be a decreasing sequence which tends to $0$ 
and let 
\begin{equation*}
\varphi _k(x):=\ep _kx+\int_{-1}^x (1-s^2)^+\, ds, \quad x\in [-1,1].
\end{equation*}
We replace (\ref{eq:2}) by the following variational equation
\begin{equation}
\label{eq:6}
\left\{
\begin{array}{l}
\forall v\in L^2(0,T,\V )  \\
\int_0^T <\frac{d\o }{dt},v>_{\V ',\V }
\, dt + \nu\int_0^T\int_{\y}
\bigg\{\nabla \varphi _k(\o )-h_N(\Vert \nabla \p \Vert _{H\times H})
\bigg( \frac{1}{\nu}T_1(\o )G(\o )+ \\
T_M\left( \frac{\int_{\y}\o ^2\left( 1-\frac{\o ^2}{3}\right) ^+\, dy}
{\ep _k +\int_{\y}\vert\nabla \p \vert ^2(1-\o ^2)(1-\o ^2)^+\, dy}\right)
(1-\o ^2)(1-\o ^2)^+ \nabla \p \bigg) \bigg\} \cdot \nabla v\, dydt\ = \ 0.
\end{array}
\right.
\end{equation}
The proof is outlined as follows: In the first step, using 
Schauder theorem, we prove that for every $T>0$ 
and every $k\in\N$ (\ref{eq:6})
has a solution $\o _k\in W(0,T)$ such that $\o _k(0,\cdot )=\o _0$.
Next we show that for every $k$, $-1\le \o _k (t,x)\le 1$
for a.e. $(t,x)\in [0,T]\times \y$ (hence the subscript $+$ and the
functions $T_1$, $h_N$ may be dropped in (\ref{eq:6})) and we observe that
the sequence $(\o _k)$ is bounded in $L^2(0,T,\V )$. Finally we prove that any
(weak) limit point $\o$ in $W(0,T)$ for the sequence $(\o _k)$ 
satisfies (\ref{eq:2})-(\ref{eq:5}), provided that $T>0$ is small enough.
\va\va
\sou{Step 1} \va
Let $k\in\N$ be fixed. 
For any $\o \in \H$ set $\p ={\cal A}^{-1}(\o )$ and 
\begin{equation*}
\begin{array}{l}
d_k(\o ):=
-h_N(\Vert \nabla \p \Vert _{H\times H})
\bigg( \frac{1}{\nu}T_1(\o )G(\o )+ \\
T_M\left( \frac{\int_{\y}\o ^2\left( 1-\frac{\o ^2}{3}\right) ^+\, dy}
{\ep _k +\int_{\y}\vert\nabla \p \vert ^2(1-\o ^2)(1-\o ^2)^+\, dy}\right)
(1-\o ^2)(1-\o ^2)^+ \nabla \p \bigg) .
\end{array}
\end{equation*}
Then for any $\o =\o (t)\in L^2(0,T,\H )$ 
\begin{equation}
\label{eq:8}
\Vert d_k(\o )\Vert _{L^{\infty}(0,T,H\times H)} 
\le \left( \frac{1}{\nu}+M \right) N.
\end{equation}
On the other hand for any $g\in \H$
\begin{equation*}
%\begin{array}{rl}
\forall \o\in \V\quad 
\nu\int_{\y}\varphi '_k(g)\nabla\o \cdot \nabla\o \, dy 
= \nu \int_{\y}\{ \ep _k+(1-g(y)^2)^+ \} \vert\nabla \o (y)\vert ^2 \, dy
\ge \nu \ep _k\Vert \o \Vert _{\V} ^2.
%\end{array}
\end{equation*}
It follows from \cite[thm 4.1, chapter 3]{LM} that for every 
$g\in L^2(0,T,\H )$ 
there exists a unique solution $\o =:{\cal F}_k (g)\in W(0,T)$ of the
following linear parabolic problem
\begin{equation*}
\left\{
\begin{array}{l}
\frac{d\o}{dt}-\hbox{div}(\nu \, \varphi _k'(g)
\nabla \o )=\nu \, \hbox{div}(d_k(g))\ 
(\in L^{\infty}(0,T,\V ')),\\
\o _{\vert _{t=0}}=\o _0.
\end{array}
\right.
\end{equation*}
In other words we have $\o (0,\cdot )=\o _0$ and 
\begin{equation}
\label{eq:7}
\forall v\in L^2(0,T,\V )\quad 
\int_0^T \{ <\frac{d\o }{dt},v>_{\V ',\V }+
\nu\int_{\y}\big( \varphi _k'(g)\nabla \o 
+d_k(g)\big) \cdot \nabla v \, dy\ \} dt=0.
\end{equation}
Obviously, every fixed-point of ${\cal F}_k$ is a solution of (\ref{eq:6}).  
In order to apply Schauder fixed-point theorem we have to show that
${\cal F}_k$ maps a bounded
closed convex subset $W_k$ of $W(0,T)$ into itself and that ${\cal F}_k$ 
is weakly sequentially continuous on $W_k$. 

Letting $v=\o $ in (\ref{eq:7}), using Lemma 2 and (\ref{eq:8}) we get 
\begin{equation*}
\Vert\o (T,\cdot )\Vert _H^2 +2\nu \ep _k\Vert \o \Vert ^2_{L^2(0,T,\V )}
\le \Vert \o _0\Vert _H^2+2(1+\nu M)N\sqrt{T}\Vert \o \Vert _{L^2(0,T,\V )} ,
\end{equation*}
hence there exists a constant $C_k>0$ (which depends on $\ep _k$, but does
not depend on $g$) such that
\begin{equation*}
\forall g\in L^2(0,T,\H )\quad 
\Vert{\cal F}_k(g)\Vert _{L^2(0,T,\V )}\le C_k.
\end{equation*}
On the other hand
\begin{equation*}
\begin{array}{rl}
\Vert \frac{d\o }{dt} \Vert _{L^2(0,T,\V ')} &=
\sup _{{\Vert v\Vert}_{L^2(0,T,\V )}\le 1}
\, \int_0^T <\frac{d\o }{dt},v>_{\V ',\V }\, dt \\
&\le \nu (\ep _k+1)C_k + (1+\nu M)N\sqrt{T}=:C_k'.
\end{array}
\end{equation*}
Let 
\begin{equation}
\label{eq:9}
W_k:=\{ g\in W(0,T);\ g(0)=\o _0,\ \Vert g\Vert _{L^2(0,T,\V )}\le C_k
\hbox{ and } \Vert \frac{dg}{dt} \Vert _{L^2(0,T,\V ')}\le C'_k \} .
\end{equation}
Then $W_k$ is clearly a nonempty (strongly) closed convex bounded set
in $W(0,T)$, hence it is  compact for the 
{\em weak} topology. ${\cal F}_k$ maps $W_k$ into itself. The classical proof 
of the weak sequential continuity of ${{\cal F}_k}_{\vert _{W_k}}$  
is omitted. (We refer to \cite{RSpre} for the details.)
It follows from Schauder theorem \cite[cor. 9.7]{Z}
that there exists
$\o _k\in W_k$ such that ${\cal F}_k(\o _k)=\o _k$. Notice that
(\ref{eq:6}) also holds true for every $v\in L^2(0,T,V)$, since
$<\frac{d\o _k}{dt},v> _{V',V} 
= <\frac{d\o _k}{dt},v-\int_{\y}v\, dy >_{\V ',\V }$ 
and $\nabla v=\nabla (v-\int_{\y}v\, dy  )$,
$v-\int_{\y}v\, dy  \in L^2(0,T,\V )$.
\va\va
\sou{Step 2} 
\va
We claim that for every $k\ge 0$,
\begin{equation}
\label{eq:26}
\hbox{for a.e. }(t,y)\in (0,T)\times  \y \quad
-1\le \o _k(t,y)\le 1.
\end{equation}
Fix $k\ge 0$ and write (for brevity) $\o$ for $\o _k$, $\p$ for
${\cal A}^{-1}(\o _k )$, etc. 
Let $T'$ be the lowest $t\in (0,T]$ such that 
$\Vert \o (t)\Vert _{H} =2$ (if any), and $T'=T$ otherwise. Let 
$\eta >0$ and $\o _\eta (t,y):=(\o (t,y)-\eta -1)^+$.
\def\oe{{\o _\eta}}
Clearly $\oe \in L^2(0,T,V )$ and 
$\nabla \oe =\chi _{\{ \o >\eta +1\} } \nabla \o$. 
(The letter $\chi$ will always denote a characteristic function.) 
Let $\tau \in (0,T']$.
Letting $v(t,y) =\oe (t,y)\chi _{(0,\tau )}(t)$ in (\ref{eq:6})
 results in
\begin{equation}
\begin{array}{rl}
\label{eq:15}
\int_0^{\tau } &<\frac{d\o }{dt},\oe >_{V ',V}\, dt+
\nu\int_0^{\tau }\int_{\y}\big( \ep _k+(1-\o ^2)^+\big) 
\chi _{\{ \o >\eta +1\} } \vert \nabla \o\vert ^2\, dydt \\
&+\nu\int_0^{\tau } \int_{\y}d_k(\o ) \cdot \nabla \oe\, dydt =0.
\end{array}
\end{equation}
It follows from Lemma 2 (applied to $f(\lambda ):=(\lambda -\eta -1)^+$)
that
\begin{equation}
\label{eq:16}
\int_0^{\tau} <\frac{d\o }{dt},\oe >_{V ',V}\, dt =
\frac{1}{2}\int_{\y}\oe ^2(\tau )\, dy -0.
\end{equation}
(Recall that $\o _0(y)\le 1$ for a.e. $y\in\y$.) We infer from 
(\ref{eq:15}) and (\ref{eq:16}) that 
\begin{equation*}
\begin{array}{rl}
\frac{1}{2} \int_{\y}\oe ^2(\tau )dy 
&\le -\nu \int_0^{\tau}\int_{\y}d_k(\o )\cdot \nabla \oe \, dydt \\
&\le \int_0^{\tau}\int_{\y}T_1(\o )G(\o )\cdot\nabla \oe \, dydt \\ 
&+\int_0^{\tau}\int_{\y}
T_M\left( \frac{\int_{\y}\o ^2\left( 1-\frac{\o ^2}{3}\right) ^+\, dy}
{\ep _k +\int_{\y}\vert\nabla \p \vert ^2(1-\o ^2)(1-\o ^2)^+\, dy}\right)
(1-\o ^2)(1-\o ^2)^+ \nabla\p \cdot \nabla \oe \, dydt\\
&=:I_1+I_2.
\end{array}
\end{equation*}
(Recall that for any $t\in [0,T']$, $\Vert \o (t)\Vert _{H}\le 2$,
hence $h_N \left( \Vert\nabla \psi (t)\Vert _{H\times H}\right) =1$.)
Using (\ref{eq:10000}), 
noticing that
$T_1(\o )\chi _{\{ \o > \eta +1\} }=
\chi _{\{ \o > \eta +1\} }$
and
$(1-\o ^2)^+\chi _{\{ \o > \eta +1\} }=0$,
we get at once $I_1=I_2=0$.
Hence for every $\tau \in [0,T']$
$\ \int_{\y}\oe ^2(\tau )\, dy\le 0$, that is
$\o (\tau ,y)\le \eta +1 \hbox{ for a.e. } y\in \y $.
Letting $\eta\to 0^+$ yields
\begin{equation*}
\hbox{for every } \tau\in [0,T'],
\hbox{ for a.e. } y\in \y \quad \o (\tau ,y)\le 1.
\end{equation*}
Notice that equation (\ref{eq:6}) holds true with $-\o$ (resp. $-\p$)
instead of $\o$ (resp. $\p$), provided that the sign $-$ 
in front of the term
\begin{equation*}
\int_0^T\int_{\y}h_N\left( \Vert \nabla \p \Vert  _{H\times H}\right)
T_1(\o )G(\o )\cdot\nabla v\, dydt
\end{equation*}
is changed into $+$. Since $-\o _0(y)\le 1$ for a.e. $y\in\y$ and $I_1=0$
in the proof above, we infer from above that $-\o (\tau ,y)\le 1$ for
every $\tau \in [0,T']$ and a.e. $y\in\y$. Hence 
$\Vert \o (\tau ,\cdot )\Vert _{H}\le 
\Vert \o (\tau ,\cdot )\Vert _{L^{\infty}(\y )} \le 1$ for
$\tau\in [0,T']$ and
then $T'=T$. The claim is proved. As a consequence, equation 
(\ref{eq:6}) (applied to $\o _k$) may be simplified as follows
(with $\p _k:={\cal A}^{-1}(\o _k)$):
\begin{equation}
\label{eq:17}
\left\{
\begin{array}{l}
\forall v\in L^2(0,T,V)  \\
\int_0^T <\frac{d\o _k}{dt},v>\, dt + \nu\int_0^T\int_{\y}
\big( \ep _k + (1-\o _k^2)\big) \nabla \o _k\cdot \nabla v\, dydt
-\int_0^T\int_{\y} \o _k G(\o _k)\cdot\nabla v \, dydt\\
-\nu\int_0^T\int_{\y}
T_M\left( \frac{\int_{\y}\o _k^2\left( 1-\frac{\o _k^2}{3}\right)\, dy}
{\ep _k +\int_{\y}\vert\nabla \p _k\vert ^2(1-\o _k^2)^2\, dy}\right)
(1-\o _k^2)^2\nabla \p _k\cdot \nabla v\, dydt\ = \ 0.
\end{array}
\right.
\end{equation}
\va
\sou{Step 3}\va
We now claim that there exists a constant $C>0$ such that 
\begin{equation}
\label{eq:22}
\forall k\ge 0 \quad \Vert \o _k\Vert _{L^2(0,T,\V )}\le C.
\end{equation}
We fix $k\ge 0$ and we write $\o $ (resp. $\p$) instead of
$\o _k$ (resp. $\p _k$).
Formally, the variation of the negative entropy functional 
$\int_{\y}S(\o )\, dy$ may be evaluated by letting $v=S'(\o )$ in  
(\ref{eq:17}). However, $S'(x)$ is unbounded as $\vert x\vert \to 1^-$,
so we are led (as in \cite{MR}) to substitute a function $S_\delta$ for
$S$, defined for any $\delta >0$ by
\begin{equation}
\label{eq:34}
S_{\delta}(x):= 
\frac{1+\delta +x}{2} \, \hbox{ln} (1+\delta +x) + 
\frac{1+\delta -x}{2} \, \hbox{ln} (1+\delta -x),\ x\in [-1,1].
\end{equation}
%Notice that 
%\begin{equation*}
%S_\delta '(x)=\frac{1}{2}\hbox{ln}
%\left( \frac{1+\delta +x}{1+\delta -x}\right) ,\quad
%S_\delta ''(x)=
%\frac{1+\delta}{(1+\delta )^2 -x^2}\hbox{ on } [-1,1].
%\end{equation*}
Letting $v=S_\delta '(\o )\in L^2(0,T,V )$ in (\ref{eq:17}) yields 
\begin{equation}
\label{eq:18}
\begin{array}{l}
\int_0^T <\frac{d\o}{dt},S_\delta '(\o )>\, dt + 
\nu (1+\delta )\int_0^T\int_{\y}
\frac{\epsilon _k+1-\o ^2}{(1+\delta )^2 -\o ^2}
\vert \nabla \o \vert ^2\, dydt 
- \int_0^T\int_{\y}\o S_\delta ''(\o )\nabla\o \cdot G(\o )\ dydt \\
-\nu\int_0^T\int_{\y}
T_M\left( \frac{\int_{\y}\o ^2\left( 1-\frac{\o ^2}{3}\right)\, dy}
{\ep _k +\int_{\y}\vert\nabla \p \vert ^2(1-\o ^2)^2\, dy}\right)
(1-\o ^2)^2 S_{\delta} ''(\o )\nabla \o \cdot \nabla \p\, dydt\ = \ 0.
\end{array}
\end{equation}
We infer from Lemma 2 that
\begin{equation}
\label{eq:19}
\int_0^T <\frac{d\o}{dt},S_\delta '(\o )>\, dt 
=\int_{\y}S_\delta (\o (T))\, dy -\int_{\y}S_\delta (\o _0)dy.
\end{equation}
Using (\ref{eq:10000}) we obtain 
\begin{equation}
\label{eq:20}
\int_0^T\int_{\y} \o S_\delta ''(\o )\nabla \o \cdot G(\o )\, dydt
=0.
\end{equation}
Finally
\begin{equation}
\label{eq:21}
\begin{array}{rl}
\int_{\y} (1-\o ^2)^2 S_\delta ''(\o )&\nabla \o \cdot \nabla\p\, dydt\\
&=(1+\delta )\int_{\y}\left(   
-\frac{\o ^4}{3} +\big( 2-(1+\delta )^2 \big) \o ^2 +
\frac{\delta ^2(\delta +2 )^2}{2(1+\delta )} 
\o \hbox{ln}\big( \frac{1+\delta +\o}{1+\delta -\o} \big) \right) \, dy.
\end{array}
\end{equation}
Letting $\delta \to 0^+$ in (\ref{eq:18}), 
applying Lebesgue theorem and  using 
(\ref{eq:19})-(\ref{eq:21}) and the classical property that 
$\chi _{ \{ \o =\pm 1\} } \nabla\o  =0$ a.e. in $\y$, 
we infer that 
\begin{equation}
\label{eq:35}
\begin{array}{rl}
\int_{\y}S(\o (T))\, dy 
&+\nu \int_0^T\int_{\y}\vert\nabla \o \vert ^2\, dydt
\le \int_{\y}S(\o (0))\, dy\\
&+\nu \int_0^T
T_M\left( \frac{\int_{\y}\o ^2 \left( 1-\frac{\o ^2}{3}\right) \, dy}
{\ep _k +\int_{\y}\vert\nabla \p \vert ^2(1-\o ^2)^2\, dy}\right)
\left(\int_{\y}\o ^2\left( 1-\frac{\o ^2}{3}\right)\ dy\right) \, dt
\end{array}
\end{equation}
(see \cite{RSpre} for the details), hence 
\begin{equation*}
\int_0^T\int_{\y} \vert\nabla \o \vert ^2\, dydt
\le \frac{\hbox{ln}(2)}{\nu} +\frac{2}{3}TM=:C^2.
\end{equation*}
We now proceed to the last step in the proof of Theorem 1, namely we let
$k\to +\infty$.\va\va
\sou{Step 4}         
\va
We infer from (\ref{eq:17}) and (\ref{eq:22}) that
$\o _k$ is bounded in $W(0,T)$. Extracting a
subsequence of $(\ep _k)$ if needed, we may assume that
(for some $\o \in W(0,T)$)
\begin{equation}
\label{eq:23}
\o _k\cf \o \ \hbox{ in }\ W(0,T)\  \hbox{ as }\  k\to\infty .
\end{equation}
(\ref{eq:4}) readily follows from (\ref{eq:10}), (\ref{eq:26}) 
and (\ref{eq:23}). 
Since $\int_{\y}\vert\nabla \p _0\vert ^2 (1-\o _0^2)^2\, dy >0$, we see
that if $T>0$ is small enough the function 
$t\in [0,T]\mapsto \b (\o (t))$ is well-defined, continuous and it satisfies
\begin{equation}
\label{eq:25}
\forall t\in [0,T] \quad \vert \b (\o (t))\vert \le M.
\end{equation}
Letting $k\to \infty$ in (\ref{eq:17}) and using (\ref{eq:26}),
(\ref{eq:23}) and (\ref{eq:25}), we get at once (\ref{eq:2}).
Finally (\ref{eq:3}) holds true
since the map $\o \in W(0,T)\mapsto \o _{\vert _{t=0}}\in \H$ is
continuous. The proof of Theorem 1 is complete. \qed
\va

Several useful properties of the solution $\o$ constructed above are collected
in the following proposition.
\begin{prop} Let $\o _0$, $T$, and $\o$ as given in Theorem 1. Then
for all $t\in [0,T]$
\begin{equation*}
%\label{eq:255}
{\cal E}(\o (t))\ (:=\frac{1}{2} \Vert \o (t) \Vert _{\V '}^2 
=\frac{1}{2} \Vert \p (t)\Vert _{\V}^2 )
={\cal E}(\o _0);
\end{equation*}
\begin{equation*}
%\label{eq:27}
{\cal S}(\o (t))\ (:= \int_{\y}S(\o (t))\, dy )\le {\cal S}(\o _0);
\end{equation*}
and for all $0\le t_1\le t_2\le T$
\begin{equation}
\label{eq:28}
\begin{array}{rl}
&\frac{1}{2}\int_{\y}\o ^2(t_2)\, dy 
+\nu\int_{t_1}^{t_2}\int_{\y} (1-\o ^2)\vert\nabla\o \vert ^2\, dydt \\
=&\frac{1}{2}\int_{\y}\o ^2(t_1)\, dy
+\nu\int_{t_1}^{t_2}\vert\b (\o )\vert 
\{ \int_{\y} \o ^2\left( \frac{\o ^4}{5}-\frac{2}{3}\o ^2 +1\right) \,dy
\} \, dt;
\end{array}
\end{equation}
\begin{equation}
\label{eq:29}
\int_{\y}S(\o (t_2))\, dy 
+\nu\int_{t_1}^{t_2}\int_{\y} \vert\nabla\o \vert ^2\, dydt =
\int_{\y}S(\o (t_1))\, dy 
+\nu\int_{t_1}^{t_2}\vert\b (\o )\vert 
\{ \int_{\y} \o ^2\left( 1-\frac{\o ^2}{3} \right) \,dy \} \, dt.
\end{equation}
\end{prop}
\proof Let us first show the conservation of the functional energy
${\cal E}(\o (t))$. Let $t_2\in (0,T]$ and let 
$v=\psi\chi _{(0,t_2)} \in L^2(0,T,\V )$ in (\ref{eq:2}).
We get
\begin{equation}
\label{eq:30}
\begin{array}{rl}
\int_0^{t_2}<\frac{d\o}{dt},\p > _{\V ',\V }
&+\nu \int_0^{t_2}\int_{\y}
\{ (1-\o ^2)\nabla\o \cdot \nabla\p
+\b (\o )(1-\o ^2)^2\vert\nabla\p \vert ^2 \} \, dydt\\
&-\int_0^{t_2}\int_{\y}\o G(\o )\cdot \nabla\p \, dydt =0.
\end{array}
\end{equation}
Due to the definitions of $\beta (\o )$ and $G(\o )$
\begin{equation*}
\int_0^{t_2} \int_{\y}
\{ (1-\o ^2)\nabla\o \cdot\nabla\p +\b (\o )(1-\o ^2)^2
\vert\nabla \p \vert ^2 \} \, dydt =0=
\int_0^{t_2} \int_{\y} \o \, G(\o )\cdot \nabla\p \, dydt.
\end{equation*}
Next it is easy to see that
\begin{equation}
\label{eq:33}
\int_0^{t_2}<\frac{d\o }{dt},\p >_{\V ',\V  } \, dt=
\frac{1}{2}\left( \int_{\y}\vert\nabla\p (t_2)\vert ^2\, dy
-\int_{\y}\vert\nabla\p _0\vert ^2\, dy \right) .
\end{equation}
It follows that 
$\int_{\y}\vert\nabla\p (t_2)\vert ^2\, dy =
\int_{\y}\vert\nabla\p _0\vert ^2\, dy$
for every $t_2\in [0,T]$, that is
\begin{equation*}
{\cal E}(\o (t_2))={\cal E}(\o _0)\quad \forall t_2\in [0,T].
\end{equation*}
We now proceed to the decrease of the negative entropy functional. Let
$v=S_\delta '(\o )\chi _{(t_1,t_2)}$ in (\ref{eq:2}), where 
$0\le t_1<t_2\le T$ and $S_\delta$ is defined in (\ref{eq:34}).
(\ref{eq:29}) is proved by letting $\delta \to 0$, in the same 
way as for (\ref{eq:35}). Using Cauchy-Schwarz inequality,
(\ref{eq:2005}) and (\ref{eq:2006}) we readily get 
\begin{equation}
\label{eq:36}
\vert \b (\o )\vert \le \frac{\int_{\y} \vert\nabla \o \vert ^2\, dy}
{\int_{\y}\o ^2\left( 1-\frac{\o ^2}{3}\right) \, dy} ,\ \  
\forall t\in [0,T].
\end{equation}
Since (by (\ref{eq:29})) for every $t\in [0,T]$
\begin{equation*}
{\cal S}(\o (t))-{\cal S}(\o _0)=\nu \int_0^t
\left( \vert \b (\o )\vert
\int_{\y}\o ^2 \left( 1-\frac{\o ^2}{3}\right) \, dy 
-\int_{\y}\vert\nabla \o \vert ^2\, dy \right) d\tau ,
\end{equation*}
we get ${\cal S}(\o (t))-{\cal S}(\o _0)\le 0$, by (\ref{eq:36}).
Finally to prove ({\ref{eq:28}) it is sufficient to take 
$v=\o \chi _{(t_1,t_2)}$ in
(\ref{eq:2}) and to observe that
$\int_{\y} \o G(\o )\cdot \nabla\o \, dy=0$ and that  
\begin{equation}
\label{eq:3000}
\int_{\y}(1-\o ^2)^2\nabla\o \cdot \nabla \p \, dy=
\int_{\y}\o ^2\left(\frac{\o ^4}{5} - \frac{2}{3}\o ^2 +1\right) \, dy.
\end{equation}
The proof of Proposition 1 is complete.\qed
\section{An $L^{\infty}(\ep ,T,H\times H)$ estimate for
$(1-\o ^2)\nabla\o$}

Let $\o _0\in \H$, $T>0$ and $\o \in W(0,T)$ be as in Theorem 1.
Let 
\begin{equation*}
%\label{eq:39}
u:=\varphi (\o )=\int_{-1}^{\o}(1-s^2)\, ds.
\end{equation*}
Then $u\in L^2(0,T,V)$ and $\nabla u=(1-\o ^2)\nabla \o $. Since for
some constant $C>0$ $\Vert \o \Vert _{\V '}\le C\Vert\o \Vert _H$, hence
\begin{equation*}
%\label{eq:40}
\int_{\y}\o ^2\left( 1-\frac{\o ^2}{3}\right) \, dy 
\ge \frac{2}{3}\int_{\y}\o ^2\, dy
\ge \frac{2}{3C^2}\Vert \o \Vert ^2_{\V '}
= \frac{2}{3C^2}\Vert \o _0\Vert ^2_{\V '} 
\end{equation*}
we infer from (\ref{eq:36}) that
\begin{equation*}
%\label{eq:41}
\vert \b (\o (t))\vert \le \frac{3C^2}{2\Vert \o _0\Vert ^2_{\V '}}
\Vert \nabla\o (t)\Vert ^2_{H\times H} \cdot
\end{equation*}
Actually, an estimate of this kind holds true when $\nabla u(t)$ is
substituted for $\nabla \o (t)$. Indeed it follows from Cauchy-Schwarz
inequality that
\begin{equation*}
\left( \int_{\y}\nabla\p \cdot\nabla\o (1-\o ^2)^2\, dy\right) ^2
\le \int_{\y}\vert\nabla\p \vert ^2 (1-\o ^2 )^2\, dy \cdot
\int_{\y}\vert\nabla\o \vert ^2 (1-\o ^2 )^2\, dy .
\end{equation*}
hence, using (\ref{eq:2006}) and (\ref{eq:3000}),  
we get
\begin{equation*}
\begin{array}{rl}
\vert\b (\o (t))\vert 
&\le \frac{\int_{\y}\o ^2\left( 1-\frac{\o ^2}{3}\right)\, dy}
{\left( \int_{\y} \o ^2\left( 
\frac{\o ^4}{5}-\frac{2}{3}\o ^2+1\right) \, dy \right) ^2}
\cdot
\int_{\y}\vert\nabla u\vert ^2\, dy  \\
\\
&\le \left( \frac{15}{8}\right) ^2
\frac{\int_{\y}\vert\nabla u\vert ^2\, dy}{\int_{\y}\o ^2\, dy}
\le \left( \frac{15}{8}\right) ^2 \frac{C^2}{\Vert \o _0\Vert ^2_{\V '}}
\Vert \nabla u\Vert ^2_{H\times H} \cdot
\end{array}
\end{equation*}
Roughly speaking, as long as $\Vert\nabla u (t)\Vert _{H\times H}$
is bounded, $\vert \b (\o (t))\vert$ is also bounded and (the) solution
exists. Conversely, we claim that whatever be $\o _0\in \H$,  if $\b$ is
bounded on some interval $[0,T]$, then
\begin{equation*}
\forall \epsilon\in (0,T)\quad \nabla u\in 
L^{\infty}(\epsilon ,T,H\times H ).
\end{equation*}
A more accurate statement of this {\em smoothing} effect is given in
next proposition.
\va
\begin{prop} Let $\o _0$, $T$ and $\o\in W(0,T)$ be as in Theorem 1.
Let $u=\varphi (\o )$. Then the map 
$t\in (0,T)\mapsto \nabla u(t)\in H\times H $ is weakly continuous 
and there exists a constant $C>0$ such that
\begin{equation}
\label{eq:42}
\forall t\in (0,T)\quad \Vert\nabla u(t)\Vert _{H\times H} \le
\frac{C}{\sqrt{t}} \cdot
\end{equation}
\end{prop}
Notice that the value $\frac{1}{2}$ of the power of $t$ in 
(\ref{eq:42}) plays a great role in the proof of Theorem 2 (which deals with
unicity of variational solution): Indeed, given any estimate 
$\Vert\nabla u(t)\Vert _{H\times H} \le \frac{C}{t^{\alpha}}$
instead of (\ref{eq:42}), it is required in the proof of Theorem 2
that the function $t^{-\alpha}$ be integrable on $(0,1)$.   
\va

The rest of this section is devoted to the 
\va
\noindent{\sc Proof of Proposition 2:}\qquad Set $Q:=(0,T)\times\y$ and
$Q_h:=(0,T-h) \times \y$ for $0<h<T$. Let
\begin{equation*}
%\label{eq:43}
F(\o ):=-\hbox{div}(\o G(\o ))+\nu\b (\o )
\, \hbox{div}( (1-\o ^2)^2\nabla\p ) .
\end{equation*}
Since $\o\in L^2(0,T,\V )$ and $\o$, $G(\o )$, $\b$, $\nabla\p$ are bounded
on $Q$, we see that 
\begin{equation}
\label{eq:44}
F(\o )\in L^2(Q ).
\end{equation}
On the other hand we readily infer from (\ref{eq:2}) that $\o$ and 
$u$ satisfy (in $L^2(0,T,V')$)
\begin{equation}
\label{eq:45}
\frac{d\o}{dt}-\nu\triangle u=F(\o ).
\end{equation}
(For any $u\in V$, $\triangle u\in V'$ is defined by:
$<\triangle u ,v>_{V',V}:=-\int_{\cal Y}\nabla u\cdot\nabla v\, dy$ for
all $v\in V$.)
We now proceed as in \cite[proof of Thm 3.1]{BG}. For any function 
$v=v(t,y)$ in $L^1(Q)$ and any $h\in (0,T)$, we set
\begin{equation*}
%\label{eq:46}
v^h(t,y):=\frac{1}{h}\int_t^{t+h}v(\tau ,y)d\tau
\hbox{ for } (t,y)\in Q_h.
\end{equation*}
Straightforward computations show that for every $v\in L^2(Q)$
\begin{equation}
\label{eq:47}
\Vert v^h\Vert _{L^2(Q_h)}\le \Vert v\Vert _{L^2(Q)} \cdot
\end{equation}
Since $u\in L^2(0,T,V)$, we see that 
$u^h\in H^1(0,T-h,V)$ and that
\begin{equation}
\label{eq:48}
\frac{du^h}{dt}=\frac{1}{h}\big( u(t+h)-u(t)\big)
\in L^{\infty}(Q_h)\cap L^2(0,T-h,V).
\end{equation}
In the same way, $\o ^h\in H^1(0,T-h,\V )$, 
$\frac{d\o ^h}{dt}=\frac{1}{h}\big( \o (t+h)-\o (t)\big)\in L^{\infty}(Q_h)$.
We readily infer from (\ref{eq:45}) that for every $h\in (0,T)$
\begin{equation}
\label{eq:49}
\frac{d\o ^h}{dt}-\nu \triangle (u^h) =F(\o )^h \hbox{ in } L^2(0,T-h,V').
\end{equation}
Since $\frac{d\o ^h}{dt}\in L^2(0,T-h,\V )$ and
$F(\o )^h\in L^2(Q_h)$ (by (\ref{eq:44}) and (\ref{eq:47})) we see that
$\triangle (u^h)\in L^2(Q_h)$. Let $0\le t_1<t_2\le T-h$. Multiplying
each term in (\ref{eq:49}) by $\frac{du^h}{dt}$ ($\in L^{\infty}(Q_h)$)
and integrating on $(t_1,t_2)\times \y$, we get
\begin{equation}
\label{eq:50}
\int_{t_1}^{t_2}\int_{\y}\frac{d\o ^h}{dt}\frac{du^h}{dt}\, dydt +
\nu \int_{t_1}^{t_2}\int_{\y}(-\triangle u^h)\frac{du^h}{dt}\, dydt =
\int_{t_1}^{t_2}\int_{\y} F(\o )^h\frac{du^h}{dt}\, dydt.
\end{equation}
We now claim that
\begin{equation}
\label{eq:51}
\int_{t_1}^{t_2}\int_{\y} \left( \frac{du^h}{dt}\right) ^2\, dydt
\le 
\int_{t_1}^{t_2}\int_{\y} \frac{d\o ^h}{dt}\frac{du^h}{dt} dydt.
\end{equation}
Indeed, for almost every $t\in (t_1,t_2)$
\begin{equation*}
\frac{du^h}{dt}(t)=\frac{u(t+h)-u(t)}{h}
=\frac{\varphi\circ\o (t+h) -\varphi\circ\o (t)}{h}\cdot
\end{equation*}
Since $\Vert\varphi '\Vert _{L^{\infty}(-1,1)}=1$, we infer from the 
mean value theorem that for a.e. $t\in (t_1,t_2)$ 
\begin{equation*}
\left\vert \frac{du^h}{dt}(t)\right\vert 
\le \left\vert \frac{\o (t+h)-\o (t)}{h}\right\vert 
= \left\vert \frac{d\o ^h}{dt}(t)\right\vert .
\end{equation*}
Since $\varphi$ is increasing, $\frac{du^h}{dt}$ and $\frac{d\o ^h}{dt}$
have the same sign, hence for a.e. $t\in (t_1,t_2)$
\begin{equation}
\label{eq:52}
\left( \frac{du^h}{dt}\right) ^2 \le
\left\vert \frac{du^h}{dt}\right\vert \cdot
\left\vert \frac{d\o ^h}{dt}\right\vert =
\frac{du^h}{dt} \cdot \frac{d\o ^h}{dt}
\end{equation}
and (\ref{eq:51}) results from (\ref{eq:52}). 

Next we prove that
\begin{equation}
\label{eq:53}
\int_{t_1}^{t_2}\int_{\y} (-\triangle u^h)\frac{du^h}{dt}\, dydt 
=\frac{1}{2} \left( \Vert \nabla u^h(t_2)\Vert ^2_{H\times H} -
\Vert \nabla u^h(t_1)\Vert ^2_{H\times H} \right) \cdot
\end{equation}
It is easy to see that $\int_{\y}F(\o (t,y))\, dy =0$ for a.e. $t\in (0,T)$,
hence 
\begin{equation*}
\int_{\y}\triangle u^h\, dy=\frac{1}{\nu}\int_{\y}
\{ \frac{\o (t+h)-\o (t)}{h} -F(\o )^h \} \, dy=0 \hbox{ for a.e. }
t\in (0,T-h).
\end{equation*}
On the other hand it is a straightforward exercise to show that any $v\in V$
such that $\triangle v\in \H$ satisfies $v\in H^2_{per}(\y )$
and $\Vert v\Vert _{H^2_{per}(\y )} \le C\left( 
\Vert\triangle v\Vert _{H} +\Vert v \Vert _{H}\right) $
for some constant $C>0$. Applying this inequality to 
$v=u^h(t)$ (for a.e. $t\in (0,T-h)$) we get 
\begin{equation*}
\int_0^{T-h}\Vert u^h(t)\Vert ^2_{H^2_{per}(\y )}\, dt 
\le 2C^2 \int_0^{T-h}\left( \Vert \triangle u^h(t)\Vert ^2_{H} +
\Vert u^h(t)\Vert ^2_{H}\right) dt <+\infty ,
\end{equation*}
hence $u^h\in L^2(0,T-h,H^2_{per}(\y ))$. On the other hand 
(by (\ref{eq:48})) $\frac{du^h}{dt}\in L^2(0,T-h,V)$, so we get 
(by Green's identity)
\begin{equation}
\label{eq:54}
\int_{t_1}^{t_2}\int_{\y} (-\triangle u^h)\frac{du^h}{dt}\, dydt
=\int_{t_1}^{t_2}\int_{\y} \nabla u^h\cdot\nabla\frac{du^h}{dt}\, dydt.
\end{equation}
Since $\nabla u^h\in H^1(0,T-h,H\times H)$ and 
$\nabla \frac{du^h}{dt}=\frac{d}{dt}\nabla u^h$, we may write
\begin{equation}
\label{eq:55}
\int_{t_1}^{t_2}\int_{\y} \nabla u^h\cdot\nabla\frac{du^h}{dt}\, dydt
=\frac{1}{2} \left( \Vert \nabla u^h(t_2)\Vert ^2_{H\times H} -
\Vert \nabla u^h(t_1)\Vert ^2_{H\times H} \right) \cdot
\end{equation}
(\ref{eq:53}) follows from (\ref{eq:54}) and (\ref{eq:55}). We infer from
(\ref{eq:47}) that
\begin{equation}
\begin{array}{rl}
\int_{t_1}^{t_2}\int_{\y} F(\o )^h\, \frac{du^h}{dt}\, dydt
&\le \Vert F(\o )\Vert _{L^2(Q)}\cdot 
\Vert \frac{du^h}{dt}\Vert _{L^2((t_1,t_2)\times \y )}\\
\label{eq:56}
&\le \frac{1}{2}\left( \Vert F(\o )\Vert ^2_{L^2(Q)} +
\Vert \frac{du^h}{dt}\Vert ^2_{L^2((t_1,t_2)\times \y )} \right) .
\end{array}
\end{equation}
Now it follows from (\ref{eq:50}), (\ref{eq:51}), (\ref{eq:53}) and
(\ref{eq:56}) that for all $0\le t_1<t_2\le T-h$
\begin{equation}
\label{eq:57}
\Vert \frac{du^h}{dt} \Vert ^2_{L^2((t_1,t_2)\times\y )}
+\nu \Vert \nabla u^h(t_2)\Vert ^2_{H\times H} 
\le \nu\Vert \nabla u^h(t_1)\Vert ^2_{H\times H}
+\Vert F(\o )\Vert ^2_{L^2(Q)} \cdot
\end{equation}
Next it is clear that for all $0\le t_1<t_2<T$,
 $\nabla (u^h)=(\nabla u)^h\to\nabla u$ in 
$L^2((t_1,t_2)\times \y )^2$ as $h\to 0^+$. Fix some $b\in (0,T)$. 
Since 
\begin{equation*}
\int_0^b \left( \int_{\y}
\vert \nabla (u^h)(t,y)-\nabla u(t,y)\vert ^2\, dy\right) dt\to 0
\hbox{ as } h\to 0^+,
\end{equation*}
there exist a decreasing sequence $h_n\to 0$ and a zero-measure set
$N\subset (0,b)$ such that 
\begin{equation}
\label{eq:58}
\forall t\in (0,b)\setminus N,\ 
\int_{\y}\vert \nabla (u^{h_n})(t,y)-\nabla u(t,y)\vert ^2\, dy \to 0
\hbox{ as } n\to\infty .
\end{equation}
Set
%\label{eq:59}
$C':=\int_0^T\Vert \nabla u(t,\cdot )\Vert ^2_{H\times H} dt <+\infty$.
Let $t_2\in (0,b)\setminus N$ be s.t.
$\int_{\y}\vert \nabla u(t_2,y)\vert ^2\, dy <+\infty$.
We claim that there exists 
$t_1\in (0,t_2)\setminus N$ such that
\begin{equation*}
%\label{eq:60}
\Vert \nabla u(t_1,\cdot )\Vert ^2_{H\times H} \le \frac{C'}{t_2} \cdot
\end{equation*}
Otherwise, 
$\Vert \nabla u(t,\cdot )\Vert ^2_{H\times H} > \frac{C'}{t_2}$ for
a.e. $t\in (0,t_2)$, hence
\begin{equation*}
\int_0^{t_2}\Vert \nabla u(t,\cdot )\Vert ^2_{H\times H} dt >
\int_0^{t_2}\frac{C'}{t_2}\, dt =C'
=\int_0^T\Vert \nabla u(t,\cdot )\Vert ^2_{H\times H}dt,
\end{equation*}
a contradiction. Letting $h=h_n\to 0$ in (\ref{eq:57}) and
applying (\ref{eq:58}) with $t=t_1$ (resp. $t=t_2$), we get
\begin{equation}
\label{eq:61}
\Vert \nabla u(t_2)\Vert ^2_{H\times H} 
\le \Vert \nabla u(t_1)\Vert ^2_{H\times H} +
\frac{1}{\nu}\Vert F(\o )\Vert ^2_{L^2(Q)} 
\le \frac{C'}{t_2}+\frac{1}{\nu}\Vert F(\o )\Vert ^2_{L^2(Q )} 
\le \frac{C^2}{t_2},
\end{equation}
where
$C^2:=C'+\frac{T}{\nu}\Vert F(\o )\Vert ^2_{L^2(Q)}$.
It follows that $\nabla u\in L^{\infty}(a,b,H\times H)$ for all
$0<a<b<T$. On the other hand it may be seen (thanks to Lebesgue theorem)
that $u\in C([0,T],H)$, hence 
$\nabla u\in C([0,T],{V'}^2)$.
Then we infer from \cite[lem. 8.1, chapter 3]{LM} 
that for all $0<a<b<T$, $\nabla u$ is weakly continuous
as a map $[a,b]\to H\times H$ and that 
\begin{equation}
\label{eq:62} \forall t\in [a,b]\quad 
\Vert \nabla u(t)\Vert _{H\times H}
\le\Vert\nabla u\Vert _{L^{\infty}(a,b,H\times H)}\cdot 
\end{equation}
It follows from (\ref{eq:61}) and (\ref{eq:62}) that 
\begin{equation*}
%\label{eq:63}
\forall t\in (0,T)\quad \Vert \nabla u(t) \Vert _{H\times H } \le
\frac{C}{\sqrt{t}},
\end{equation*}
as required. The proof of Proposition 2 is complete.\qed
\section{Uniqueness}

This section is devoted to the proof of the following uniqueness result:
\begin{thm} Let $\o _0\in H$ be such that (\ref{eq:2000}) is
fulfilled. Let $T>0$ and let $\o _i\in W(0,T)$, $i=1,2$ be functions
such that, setting $\p _i={\cal A}^{-1}(\o _i)$, (\ref{eq:2})-(\ref{eq:5})
hold true for $(\o , \p )=(\o _i,\p _i)$, $i=1,2$. Then $\o _1=\o _2$.
\end{thm}
\proof Let
\begin{equation*}
%\label{eq:64}
C_1:=\sup_{\Vert \o \Vert _{L^{\infty}(\y )}\le 1,\ \int_{\y}\o\, dy=0}
\left( \Vert{\cal A}^{-1}\o \Vert _{L^{\infty}(\y )}
+\Vert\, \vert\nabla ({\cal A}^{-1}\o )
\vert \,\Vert _{L^{\infty}(\y )} \right) \cdot
\end{equation*}
Clearly $C_1<\infty$ and for every 
$i\in\{ 1,2\}$ and almost every $t\in (0,T)$
\begin{equation}
\label{eq:65}
\Vert \p _i(t)\Vert _{L^{\infty}(\y )} +
\Vert \,\vert \nabla \p _i(t)\vert \, \Vert _{L^{\infty}(\y )} \le C_1.
\end{equation}
Set $\o =\o _1-\o _2$, $\p =\p _1-\p _2$ and
\begin{equation*}
%\label{eq:66}
g(t):=\frac{1}{2} \int_{\y}\vert\nabla \p (t)\vert ^2\, dy
=\frac{1}{2} \Vert\o (t)\Vert ^2_{\V '} \cdot
\end{equation*}
We aim to show that $g=0$, which is equivalent to $\o _1=\o _2$. It
follows from (\ref{eq:33}) that $g\in W^{1,1}(0,T)$ and that for
a.e. $t\in (0,T)$ 
\begin{equation}
\label{eq:67}
g'(t)=<\frac{d\o }{dt}(t),\p (t)>_{\V ',\V } \cdot
\end{equation}
Letting $v=\chi _{(0,t)}\p$ in (\ref{eq:2}) with $\o =\o _1$ (resp.
with $\o =\o _2$) and taking the difference of the two equations, we get  
(for every $t\in (0,T)$)
\begin{equation}
\label{eq:68}
\begin{array}{rl}
&\int_0^t <\frac{d\o }{dt},\p >_{\V ',\V } d\tau 
+\nu\int_0^t\int_{\y}\{ \nabla\left( \o _1-\frac{\o _1^3}{3}\right) -
\nabla\left( \o _2-\frac{\o _2^3}{3}\right) \} \cdot \nabla \p\, dyd\tau \\
=&\int_0^t\int_{\y} \{
\o _1G(\o _1)-\o _2G(\o _2)\}\cdot \nabla\p\, dyd\tau\\
&-\nu \int_0^t\int_{\y}\{ \b (\o _1)(1-\o _1^2)^2\nabla\p _1 -
\b (\o _2)(1-\o _2^2)^2\nabla\p _2\} \cdot \nabla\p \, dydt .
\end{array}
\end{equation}
Differentiating (\ref{eq:68}) yields
\begin{equation}
\label{eq:69}
\begin{array}{l}
\hbox{for a.e. } t\in (0,T)\quad
<\frac{d\o }{dt},\p >_{\V ',\V} 
+\nu\int_{\y}\{ \nabla\left( \o _1-\frac{\o _1^3}{3}\right) -
\nabla\left( \o _2-\frac{\o _2^3}{3}\right) \} \cdot \nabla \p\, dy =\\
\\
\int_{\y} \{ \o _1G(\o _1)-\o _2G(\o _2)\}\cdot \nabla\p\, dy
-\nu \int_{\y}\{ \b (\o _1)(1-\o _1^2)^2\nabla\p _1 -
\b (\o _2)(1-\o _2^2)^2\nabla\p _2\} \cdot \nabla\p \, dy.
\end{array}
\end{equation}
The following computations hold true for {\em almost} every $t\in (0,T)$.
(We recall that 
$\o _i\in L^2(0,T,\V )\cap L^{\infty}(0,T,L^{\infty}(\y ))$,
$\p _i\in L^2(0,T,\tilde{H}^2_{per}(\y ))$.) By Green's identity 
\begin{equation}
\label{eq:70}
\begin{array}{rl}
\int_{\y}\{ \nabla\left( \o _1-\frac{\o _1^3}{3}\right) -
\nabla\left( \o _2-\frac{\o _2^3}{3}\right) \} \cdot \nabla \p\, dy
&= \int_{\y}\{ \o _1-\o _2
-\frac{1}{3}(\o _1^3-\o _2^3) \} (-\triangle \p )\, dy \\
&=\int_{\y} \o ^2\left( 1-\frac{1}{3}
(\o _1^2+\o _1\o _2+\o _2^2)\right) \, dy
\end{array}
\end{equation}
hence this term is nonnegative for a.e. $t\in (0,T)$. Let $(u_1,u_2)$ denote
the components of $G(\o _1)$. Then
\begin{equation*}
\begin{array}{rl}
\nabla\p \cdot \nabla \big( G(\o _1)\cdot \nabla\p \big)
%&= \sum_{i,j=1}^2
%\frac{\partial\p}{\partial y_i}\cdot \frac{\partial}{\partial y_i}
%\left( u_j\frac{\partial\p}{\partial y_j} \right) \\
%&= \sum_{i,j=1}^2
%\{  \frac{\partial u_j}{\partial y_i} \frac{\partial\p}{\partial y_i}
%\frac{\partial\p}{\partial y_j} + u_j \frac{\partial}{\partial y_j}
%\left( \frac{1}{2} \big( \frac{\partial\p}{\partial y_i}\big) ^2 \right)
%\} \\
= \sum_{i,j=1}^2
\frac{\partial u_j}{\partial y_i} \frac{\partial\p}{\partial y_i}
\frac{\partial\p}{\partial y_j} + G(\o _1)\cdot\nabla
\left( \frac{1}{2} \vert\nabla\p \vert ^2 \right) ,
\end{array}
\end{equation*}
and since
$\int_{\y}G(\o _1)\cdot\nabla
\left( \frac{1}{2} \vert\nabla\p \vert ^2 \right)\, dy=0$
(by (\ref{eq:10000})) and
$G(\o )\cdot\nabla \p =0$
we infer that 
%$\frac{1}{2} \vert\nabla\p \vert ^2\in V$,
%$G(\o _1)\in V^2$ and $\hbox{div}\, G(\o _1)=0$).
\begin{equation*}
\int_{\y}\{ \o _1G(\o _1)-\o _2G(\o _2) \} \cdot\nabla\p\, dy
=\sum_{i,j=1}^2\int_{\y}
\frac{\partial u_j}{\partial y_i} \frac{\partial\p}{\partial y_i}
\frac{\partial\p}{\partial y_j} \, dy.
\end{equation*}
%Classical computations (using (\ref{eq:10000})) yields
%\begin{equation*}
%\int_{\y}\{ \o _1G(\o _1)-\o _2G(\o _2) \} \cdot\nabla\p\, dy
%=\sum_{i,j=1}^2\int_{\y}
%\frac{\partial u_j}{\partial y_i} \frac{\partial\p}{\partial y_i}
%\frac{\partial\p}{\partial y_j} \, dy.
%\end{equation*}
Denote by $C_2>0$ the constant $C$ given in Lemma 1. Notice that for every
$i,j\in \{ 1,2 \}$, every $p\in (2,+\infty )$ and a.e. $t\in (0,T)$ 
\begin{equation*}
\Vert\frac{\partial u_j}{\partial y_i}\Vert _{L^p(\y )}
\le \Vert \p _1\Vert _{W^{2,p}(\y )} 
\le C_2\, p\, \Vert\o _1\Vert _{L^p(\y )} \le C_2\, p,
\end{equation*}
(we recall that $u_1=\frac{\partial \p _1}{\partial y_2}$, 
$u_2=-\frac{\partial \p _1}{\partial y_1}$) hence for a.e. $t\in (0,T)$
and every $p\in (2,+\infty )$, by H\"older inequality and (\ref{eq:65}), 
\begin{equation}
\label{eq:71}
\begin{array}{rl}
\left\vert \int_{\y}\{ \o _1G(\o _1)-\o _2G(\o _2) \}
\cdot \nabla\p\, dy\right\vert
&\le \sum_{i,j=1}^2
\left\Vert \frac{\partial u_j}{\partial y_i} \right\Vert _{L^p(\y )}\cdot
\left\Vert \frac{\partial \p}{\partial y_i}
\frac{\partial \p}{\partial y_j} \right\Vert _{L^{p'}(\y )} \\
&\le 4C_2p\left( \int_{\y}\vert\nabla\p\vert ^{2p'}\, dy
\right) ^{\frac{1}{p'} }\\
&\le 4C_2p(2C_1)^{\frac{2p'-2}{p'} }\cdot \big( 2g(t) \big) ^{\frac{1}{p'}}
=2^{4-\frac{1}{p'}}C_1^{\frac{2}{p}}C_2pg(t)^{\frac{1}{p'}},
\end{array}
\end{equation}
where $\frac{1}{p}+\frac{1}{p'}=1$.
We finally have a look at the last integral term in (\ref{eq:69}):
\begin{equation}
\label{eq:710}
\begin{array}{l}
\int_{\y} \{ \b (\o _1)(1-\o _1^2)^2\nabla \p _1 
-\b (\o _2)(1-\o _2^2)^2\nabla \p _2 \} \cdot\nabla \p \, dydt=\\
\\
\big( \b (\o _1)-\b (\o _2) \big)
\int_{\y}(1-\o _1^2)^2\nabla\p _1\cdot\nabla \p \, dy 
\ +\ \b (\o _2)\int_{\y}\{ (1-\o _1^2)^2-(1-\o _2^2)^2 \} 
\nabla\p _1\cdot\nabla\p \, dy\\ 
+\ \b (\o _2)\int_{\y}(1-\o _2^2)^2\vert\nabla\p \vert ^2\, dy 
\quad =\quad :I_1+I_2+I_3.
\end{array}
\end{equation}
We successively estimate  $\vert I_1\vert$, $\vert I_2\vert$ and 
$\vert I_3\vert$. Let $C_3>0$ be such that for every $i=1,2$  and
every $t\in [0,T]$
\begin{equation}
\label{eq:72}
\left( \int_{\y} \vert\nabla\p _i\vert ^2 (1-\o _i^2)^2\, dy \right) ^{-1}
<C_3 .
\end{equation}
(Recall that (\ref{eq:5}) holds true for $\o =\o _i$, $i=1,2$.)
Setting 
\begin{equation*}
\begin{array}{rl}
J_1&:=\int_{\y} \vert\nabla\p _1\vert ^2(1-\o _1^2)^2\, dy -
\int_{\y} \vert\nabla\p _2\vert ^2(1-\o _2^2)^2\, dy, \\
J_2&:=\int_{\y}\nabla\p _1\cdot
\nabla \left( \o _1-\frac{\o _1^3}{3}\right) \, dy
-\int_{\y}\nabla\p _2\cdot
\nabla \left( \o _2-\frac{\o _2^3}{3}\right) \, dy,
\end{array}
\end{equation*}
we readily get
\begin{equation}
\label{eq:75}
\vert\b (\o _1) -\b (\o _2 )\vert \le \frac{2}{3}C_3^2\vert J_1\vert  +
C_3\vert J_2\vert \cdot
\end{equation}
We may write
\begin{equation*}
\begin{array}{rl}
J_1&=\int_{\y}\nabla (\p _1+\p _2)\cdot\nabla (\p _1-\p _2)(1-\o _1^2)^2\, dy
+\int_{\y}\vert \nabla \p _2\vert ^2
\{ (1-\o _1^2)^2 -(1-\o _2^2)^2\} \, dy \\
&=\int_{\y}\nabla\p\cdot\nabla(\p _1+\p _2)(1-\o _1^2)^2\, dy
-\int_{\y}\vert \nabla \p _2\vert ^2
\o (\o _1+\o _2)(2-\o _1^2-\o _2^2) \, dy. 
\end{array}
\end{equation*}
Let 
\begin{equation*}
%\label{eq:76}
C_4^2:=\sup_{(\o _1,\o _2)\in (-1,1)^2}\
\frac{(2-\o _1^2-\o _2^2)^2}{1-\frac{1}{3}(\o _1^2+\o _1\o _2+\o _2^2)} \,
<+\infty \cdot
\end{equation*}
Using (\ref{eq:4}), (\ref{eq:65}) and 
Cauchy-Schwarz inequality, we get
\begin{equation}
\label{eq:77}
\begin{array}{rl}
\vert J_1\vert &\le 2C_1 \left( \int_{\y}\vert\nabla\p \vert ^2\, dy 
\right) ^{\frac{1}{2}} + \\
&\left( \int_{\y} \vert\nabla\p _2\vert ^4(\o _1+\o _2)^2
\frac{(2-\o _1^2-\o _2^2)^2}{1-\frac{1}{3}(\o _1^2+\o _1\o _2+\o _2^2)}
\, dy\right) ^{\frac{1}{2}} 
\left( \int_{\y}
\o ^2\left( 1-\frac{1}{3}(\o _1^2+\o _1\o _2+\o _2^2)\right) \, 
dy\right) ^{\frac{1}{2}} \cdot  \\
&\le 2C_1 \left( \int_{\y}\vert\nabla\p \vert ^2\, dy 
\right) ^{\frac{1}{2}} +
2C_1^2C_4\left( \int_{\y}
\o ^2\left( 1-\frac{1}{3}(\o _1^2+\o _1\o _2+\o _2^2)\right) \, 
dy\right) ^{\frac{1}{2}} \cdot  
\end{array}
\end{equation}
For estimating $\vert J_2\vert$, we apply Proposition 2 with
$\o =\o _1$ and we infer that there exists a constant $C_5>0$ such that
\begin{equation*}
\forall t\in (0,T)\quad 
\Vert \nabla \left(\o _1 -\frac{\o _1^3}{3}\right) (t) \Vert _{H\times H}
=\Vert \nabla \varphi (\o _1)(t)\Vert _{H\times H} 
\le \frac{C_5}{\sqrt{t}}\cdot
\end{equation*}
Next we write 
\begin{equation*}
\begin{array}{rl}
J_2&=\int_{\y}\nabla\p \cdot
\nabla \left( \o _1-\frac{\o _1^3}{3}\right) \, dy
+\int_{\y}\nabla\p _2\cdot
\nabla \left( (\o _1-\frac{\o _1^3}{3}) 
-(\o _2-\frac{\o _2^3}{3}) \right) \, dy \\
&=\int_{\y}\nabla\p \cdot
\nabla \left( \o _1-\frac{\o _1^3}{3}\right) \, dy
+\int_{\y} \o _2\, \o
\left( 1-\frac{1}{3}(\o _1^2+\o _1\o _2+\o _2^2)\right) \, dy.
\end{array}
\end{equation*}
Hence, by Cauchy-Schwarz inequality,
\begin{equation}
\label{eq:78}
\vert J_2\vert \le 
\frac{C_5}{\sqrt{t}}\left(\int_{\y}\vert\nabla\p \vert ^2\, dy 
\right) ^{\frac{1}{2}} +
\left( \int_{\y}
\o ^2\left( 1-\frac{1}{3}(\o _1^2+\o _1\o _2+\o _2^2)\right) \, 
dy\right) ^{\frac{1}{2}} \cdot  
\end{equation}
Finally we infer from (\ref{eq:65}) and (\ref{eq:75})-(\ref{eq:78}) that
\begin{equation}
\begin{array}{rl}
\vert I_1\vert 
&\le\left\{ \left( \frac{4}{3}C_1C_3^2+\frac{C_3C_5}{\sqrt{t}}\right)
\left(\int_{\y}\vert\nabla\p \vert ^2\, dy \right) ^{\frac{1}{2}} \right. \\
&+\left. \left( \frac{4}{3}C_1^2 C_3^2 C_4+C_3 \right) \left( \int_{\y}
\o ^2\left( 1-\frac{1}{3}(\o _1^2+\o _1\o _2+\o _2^2)\right) \, 
dy\right) ^{\frac{1}{2}} \right\} C_1
\left( \int_{\y}\vert\nabla\p \vert ^2\, dy \right) ^{\frac{1}{2}} \\
\label{eq:79}
&\le \frac{C_6}{\sqrt{t}} \int_{\y}\vert\nabla\p \vert ^2\, dy 
+\frac{1}{2}\int_{\y}
\o ^2\left( 1-\frac{1}{3}(\o _1^2+\o _1\o _2+\o _2^2)\right) \, dy,
\end{array}
\end{equation}
for some constant $C_6>0$, which may easily be expressed in terms  of
$C_1$, $C_3$, $C_4$, $C_5$ and $T$. 

Now we have a look at $I_2$. By (\ref{eq:72}) (which implies 
$\vert\b (\o _2)\vert\le\frac{2}{3} C_3$) and Cauchy-Schwarz inequality,
\begin{equation}
\begin{array}{l}
\vert I_2\vert 
=\left\vert \b (\o _2)\int_{\y}\o (\o _1+\o _2)(2-\o _1^2-\o _2^2)
\nabla\p _1\cdot\nabla\p \, dy\right\vert \\
\le \frac{2}{3} C_3 \bigg( \int_{\y}\o ^2
\left( 1-\frac{1}{3}(\o _1^2+\o _1\o _2+\o _2^2)\right) \, dy 
\bigg) ^{\frac{1}{2}}
\left( \int_{\y} (\o _1+\o _2)^2
\frac{(2-\o _1^2-\o _2^2)^2}{1-\frac{1}{3}(\o _1^2+\o _1\o _2+\o _2^2)}
\vert\nabla\p _1\cdot\nabla\p \vert ^2
\, dy\right) ^{\frac{1}{2}} \\
\label{eq:80}
\le \frac{1}{2} \int_{\y}\o ^2
\left( 1-\frac{1}{3}(\o _1^2+\o _1\o _2+\o _2^2)\right) \, dy 
+C_7\int_{\y}\vert\nabla\p\vert ^2\, dy
\end{array}
\end{equation}
for some constant $C_7>0$. 
Finally 
\begin{equation}
\label{eq:81}
\vert I_3\vert \le\frac{2}{3}C_3\int_{\y}\vert\nabla\p \vert ^2\, dy.
\end{equation}
It follows from (\ref{eq:67}), (\ref{eq:69})-(\ref{eq:710}),
(\ref{eq:79})-(\ref{eq:81}) that for every $p\in (2,+\infty )$ and for
a.e. $t\in (0,T)$
\begin{equation}
\label{eq:82}
\begin{array}{rl}
g'(t)&\le 16 C_1 ^{\frac{2}{p}}C_2\, p\, g(t)^{\frac{1}{p'}} +
\nu \left( \frac{C_6}{\sqrt{t}}+C_7+\frac{2}{3}C_3\right) 
\int_{\y}\vert\nabla \p \vert ^2\, dy \\
&\le \frac{C_8}{2\sqrt{t}}\, p\, g(t)^{\frac{1}{p'}}
\end{array}
\end{equation}
for some constant $C_8>0$. (Indeed 
$\frac{1}{2} \int_{\y}\vert\nabla \p (t)\vert ^2\, dy =g(t)
\le \left( 1+\frac{C_1^2}{2}\right) g(t)^{\frac{1}{p'}}$.)
Integrating in (\ref{eq:82})
and using the fact that $g(0)=0$ we get
\begin{equation}
\label{eq:820}
\forall t\in [0,T]\quad g(t) \le 
\left( C_8\sqrt{t}\right) ^p.
\end{equation}
Letting $p\to \infty$ in (\ref{eq:820})
yields $g(t)=0$ for $0< t<C_8^{-2}$. Successive applications of this result
show that $g(t)=0$ for all $t\in [0,T]$,
that is $\o _1(t)=\o _2(t)$ for all $t\in [0,T]$. The proof of Theorem 2
is complete. \qed
\va

Let $\o _0\in\H$ be such that (\ref{eq:2000}) is fulfilled.
A map $\o :[0,T')\times \y\to \R$, where $0<T'\le +\infty$, will be
 called a {\em solution} of (\ref{eq:207})
 with $\o _0$ as an initial value if for every $T\in (0,T')$,
$\o _{\vert _{[0,T]}}\in W(0,T)$ and (\ref{eq:2})-(\ref{eq:5}) 
hold true. 
The following result is an obvious consequence of Theorem 1 and Theorem 2.
\begin{cor} Let $\o _0\in \H$ be such that (\ref{eq:2000})
holds true. Then there exists a unique solution $\o$ of (\ref{eq:207})
(with $\o _0$ as an initial value)
defined on a right-maximal interval of existence $[0,T_{max})$, where
$0<T_{max}\le +\infty$. If $\b (\o ) \in L ^{\infty}(0,T_{max})$, then 
$T_{max}=+\infty$. 
\end{cor}
\noindent
(For the last assertion, it is sufficient to remark that the last step 
in the proof of Theorem 1 works for every $T>0$ - hence there exists a 
solution $\o\in W(0,T)$ for every $T>0$ - provided that 
$M>\Vert \b \Vert _{L^{\infty}(0,T_{max})}$. It follows that 
$T_{max}=+\infty$.) 

The issue whether $T_{max}=+\infty$ for {\em every} $\o _0\in \H$
such that (\ref{eq:2000}) holds (as it seems to be shown by numerical
evidences) remains open at the present time. Nevertheless, following 
\cite{MR}, we may define whole a family 
$\left( {\cal U}_{p,\ep}\right) _{p>2,0<\ep <1}$ of 
sets in $\H$ such that 
$T_{max}=+\infty$ whenever $\o _0\in {\cal U}_{p,\ep}$. To be precise,
we set for $p\in (2,+\infty )$ and $0<\epsilon <1$
\begin{equation}
\label{eq:83}
{\cal U}_{p,\ep} :=\bigg\{ \o _0\in \H  ;\ -1\le \o _0\le 1
\hbox{ a.e. in }\y \hbox{ and }  
\Vert \o _0\Vert ^2_{\V '} \ge 
\frac{2C(p)}{1-\ep}\left( \int_{\y}S(\o _0)\, dy\right) ^{1+\frac{2}{p}}
\bigg\} ,
\end{equation}
where 
\begin{equation*}
%\label{eq:84}
C(p):=\frac{2}{\sqrt{\pi}}\left( \frac{4p(p-1)\pi ^{\frac{3}{2}}}{p-2} 
\right) ^{\frac{2}{p}} \cdot 
\end{equation*}
It follows from Proposition 1 that ${\cal U}_{p,\ep}$ is 
conserved by the flow of equation (\ref{eq:207}). Moreover
\begin{equation*}
{\cal U}_{p,\ep}\supset {\cal O}_{p,\ep }:=
\left\{ \o _0\in \H ;\ \Vert \o _0\Vert _{L^{\infty}(\y )} < 1
\hbox{ and }
\Vert \o _0\Vert ^2_{\V '} > 
\frac{2C(p)}{1-\ep}\left( \int_{\y}S(\o _0)\, dy\right) ^{1+\frac{2}{p}}
\right\} ,
\end{equation*}
and ${\cal O}_{p,\ep }$ is a {\em nonempty} (see \cite[prop. 3.11]{MR}) 
{\em open} set in the space 
$\tilde{L}^{\infty}(\y ):=L^{\infty}(\y )\cap  \H$ endowed with the
$L^\infty$ norm. The introduction of ${\cal U}_{p,\ep }$ is motivated
by the following result:
\begin{prop}
Let $p>2$, $0<\ep <1$ and $\o _0\in {\cal U}_{p,\ep}$. Then 
\begin{equation*}
%\label{eq:85}
\int_{\y} (1-\o _0^2)^2\vert\nabla {\cal A}^{-1}\o _0\vert ^2\, dy \ge
\ep \Vert \o _0\Vert ^2 _{\V '} \cdot
\end{equation*}
As a consequence, if $\o$ and $T_{max}$ are as given in Corollary 1,
then $T_{max}=+\infty$ and \va
$\vert\b (\o (t))\vert\le\frac{2}{3\ep \Vert \o _0\Vert ^2_{\V '} }$
for all $t\ge 0$. 
\end{prop}
The proof of Proposition 3 is (word by word) the same as that of  
\cite[prop. 3.12]{MR}. However, notice that the definition of 
${\cal U}_{p,\ep}$ in (\ref{eq:83}) is slightly different from the
one in \cite{MR} ($2C(p)$ is substituted for $C(p)$), due to the 
following estimation from below for the term 
$\int_{\y} (1-\o _0^2)^2\vert\nabla {\cal A}^{-1}\o _0\vert ^2\, dy$
(instead of 
$\int_{\y} (1-\o _0^2)\vert\nabla {\cal A}^{-1}\o _0\vert ^2\, dy$
in \cite{MR})
\begin{equation*}
\begin{array}{rl}
\int_{\y} (1-\o _0^2)^2\vert\nabla {\cal A}^{-1}\o _0\vert ^2\, dy
&= \int_{\y}\vert\nabla {\cal A}^{-1}\o _0\vert ^2\, dy -
\int_{\y} (2-\o _0^2)\o _0^2\vert\nabla {\cal A}^{-1}\o _0\vert ^2\, dy \\
&\ge \Vert \o _0\Vert ^2_{\V}
-2\int_{\y} \o _0^2 \vert\nabla {\cal A}^{-1}\o _0\vert ^2\, dy \cdot
\end{array}
\end{equation*}
Next, 
$\int_{\y} \o _0^2 \vert\nabla {\cal A}^{-1}\o _0\vert ^2\, dy$ is
estimated from above as in \cite{MR}. 
%\begin{rmq}
%As in \cite{MR} we may associate with each pair $(p,\ep )$ a
%{\em dynamical system} in the following way: Let $X$ denote the
%(complete) metric space ${\cal U}_{p,\ep }$ for the
%$L^2({\y})$ metric and $X^*=X\setminus \{ 0\} $.
%Set $T(t)\o _0 :=\o (t)$ for $\o _0\in X^*$, $t\ge 0$ (where $\o (\cdot )$
%is as in Proposition 3), and $T(t)0=0$ for all $t\ge 0$. Obviously,
%$T(0)=Id_{X}$, $T(s+t)=T(s)\circ T(t)$ for all $s,t \ge 0$ and
%$T(\cdot ) \o _0\in C([0,+\infty ),X)$ for any $\o _0\in X$. Hence the family
%$\big( T(t)\big) _{t\ge 0}$ is a continuous nonlinear semigroup 
%(a dynamical system) on  $X$ 
%(see \cite{H}) provided that for every $t>0$ the map 
%$\o _0\mapsto T(t)\o _0$ be continuous. The continuity at $\o _0=0$ follows
%from the decrease of the negative entropy functional together 
%with the obvious estimates 
%$\frac{1}{2}\Vert \o \Vert ^2_H 
%\le {\cal S}(\o )\le C \Vert \o \Vert ^2_H$ for some constant $C>0$, 
%if  $\Vert \o \Vert _{L^{\infty}(\y )}\le 1$. Let now $\bar{t}>0$ and
%$\o _0\in X^*$ be given. We aim to show that if $(\o ^n_0)$ is any sequence
%in $X^*$ such that $\o ^n_0\to \o _0$ (in $X$), then 
%$T(\bar{t})\o ^n_0\to T(\bar{t})\o _0$ also. Pick any $T>\bar{t}$. 
%Since $\inf_{n}\Vert \o ^n_0\Vert _{\V '}>0$, we easily infer from 
%Proposition 3 that (\ref{eq:72}) holds true for some $C_3>0$ with 
%$T(\cdot )\o ^n_0$ instead of $\o _i$. Next  we infer from  (\ref{eq:820}) 
%that (for some constant $C_8>0$) we have
%\begin{equation}
%\label{eq:999}
%g_n(t)\le \left( C_8\sqrt{t} +g_n(0)^{\frac{1}{p}} \right) ^p,
%\end{equation}
%where $g_n(t):=\frac{1}{2}\Vert T(t)\o ^n_0-T(t)\o _0\Vert ^2_{\V '}$.  
%We may assume that $\o ^n_0\ne \o _0$ for all $n$. Letting 
%\begin{equation*}
%p=p_n:=\frac{\hbox{ln}  (g_n(0)^{-1})}{\hbox{ln} (4)}
%\end{equation*}
%in (\ref{eq:999}) for $n$ large enough yields
%\begin{equation} 
%g_n(t)\le \left( \frac{3}{4}\right) ^{p_n}\quad\hbox{for }
%0\le t\le (2C_8)^{-2}.
%\end{equation}
%Hence $g_n(t)\to 0$ as $n\to \infty$ uniformly w.r.t. t on $[0,(2C_8)^{-2}]$.
%Applying this argument a finite number of times we get that $g_n(t)\to 0$
%uniformly w.r.t. t on $[0,T]$. 
%Thus $T(\bar{t})\o ^n_0\to T(\bar{t})\o _0$ in $\V '$. On the other hand
%it follows from Propositions 2 and 3 that the sequence 
%$\big( \varphi \circ (T(\bar{t})\o ^n_0)\big)$ is bounded in $V$. 
%Applying \cite[Lemma 2.2]{GM}, we infer that $T(\bar{t})\o ^n_0$
%is bounded in the Sobolev space $W^{\frac{1}{4},4}(\y )$, which is 
%{\em compactly} embedded in $L^4(\y )$. We conclude that 
%$T(\bar{t})\o ^n_0\to T(\bar{t}) \o _0$ in  $L^4(\y )$ (and in $X$). 
%The proof that the family  $\big( T(t)\big) _{t\ge 0}$ 
%be a dynamical system on $X$ is complete.
%\end{rmq}
\section{Asymptotic behavior}

As in \cite{MR} we show that if $\b (\o )$ remains bounded (hence 
$T_{max}=+\infty$), then the $\o -$limit set for the $L^2$ norm is 
made up of Gibbs states. In contrast to \cite{MR} we cannot assert
that $\o \in L^{\infty}(\epsilon ,+\infty ,\V )$
for all $\epsilon >0$ - but it may be
proved that $\varphi (\o )\in L^{\infty}(\epsilon ,+\infty ,V )$ - and the 
proof of the following result differs somewhat from that of
\cite[thm 4.6]{MR}.
\def\ob{\overline{\omega}}
\def\pb{\overline{\psi}}
\def\yb{\overline{\cal Y}}
\begin{thm} Let $\o _0\in \H$ be such that
(\ref{eq:2000}) holds true and
$\b (\o )\in L^{\infty}(0,T_{max})$ (hence $T_{max}=+\infty$). 
Let $(t_n)_{n\ge 0}$ be a sequence of nonnegative 
numbers such that $t_n\nearrow \infty$. Then there exist a subsequence 
$(t_{n_k})_{k\ge 0}$, a constant $C\in\R$ and a function  
$\ob\in C^{\infty}_{per}(\yb )$ such that ($\pb$ denoting 
${\cal A}^{-1}(\ob )$)
\begin{equation}
\label{eq:86}
\ob = \tanh \big( -\b (\ob ) \pb +C\big)
\end{equation}
and 
\begin{equation}
\label{eq:87}
\begin{array}{rl}
\o (t_{n_k})&\to \ob \hbox{ in } H, \\
\b (\o (t_{n_k} )) &\to \b (\ob ) \hbox{ as } k\to \infty .
\end{array}
\end{equation}
\end{thm}
\proof Let $M>0$ be such that 
\begin{equation}
\label{eq:89}
\forall t\ge 0\quad \vert\b (\o (t)) \vert \le M.
\end{equation}
It follows from (\ref{eq:501}) that
\begin{equation}
\label{eq:90}
\forall t\ge 0\quad \Vert\o (t)\Vert _{L^{\infty}(\y )} \le 1.
\end{equation}
Set
\begin{equation*}
\o _n(t,y):=\o (t_n+t,y), \quad
\p _n(t,\cdot ):={\cal A}^{-1} \o _n(t,\cdot ).
\end{equation*}
We infer from (\ref{eq:29}), (\ref{eq:89}) and (\ref{eq:90}) that
\begin{equation*}
\label{eq:91}
\sup_{n\ge 0} \Vert \o _n\Vert _{L^2(0,1,\V  )} <+\infty
\end{equation*}
which, together with (\ref{eq:2}), implies
\begin{equation*}
%\label{eq:92}
\sup_{n\ge 0} \Vert \frac{d\o _n}{dt}
\Vert _{L^2(0,1,\V ')} <+\infty \cdot
\end{equation*}
It follows that the sequence $(\o _n)_{n\ge 0}$ is bounded in the 
Hilbert space $W(0,1)$, 
hence there exists a map $\ob \in W(0,1)\subset 
C([0,1],\H )$ 
and a subsequence $(\o _{n_k})_{k\ge 0}$ such that, as $k\to\infty$, 
\begin{eqnarray}
\label{eq:93}
\o _{n_k} & \cf &\ob \hbox{ in } L^2(0,1,\V ),\\
\label{eq:94}
\frac{d\o _{n_k}}{dt} & \cf & \frac{d\ob }{dt} \hbox{ in } L^2(0,1,\V ').
\end{eqnarray}
Thanks to Aubin's Lemma \cite{S}, 
extracting subsequences we may (without loss
of generality) assume that  
\begin{equation}
\label{eq:95}
\o _{n_k}\to \ob \hbox{ in } L^2(0,1,\H )\cap C([0,1],\V ')
\end{equation}
and also
\begin{equation}
\label{eq:96}
\o _{n_k} (t,y)\to \ob (t,y) \hbox{ for all }
t\in [0,1]\setminus N \hbox{ and a.e. }y\in\y ,
\end{equation}
where $N\subset [0,1]$ denotes some set of (Lebesgue) measure 0. 
We now infer from (\ref{eq:90}) and (\ref{eq:96}) that for
every $t\in [0,1]\setminus N$, as $k\to\infty$,  
\begin{equation*} 
\o _{n_k}(t)\to\ob (t)\hbox{ in } L^3(\y ),
\end{equation*}
hence 
\begin{equation*} 
\p _{n_k}(t)\to \pb (t):={\cal A}^{-1}\big(  \ob (t)\big)  
\hbox{ in } W^{2,3}(\y )\subset C^1(\yb ).
\end{equation*}
Thus
\begin{equation}
\label{eq:97}
\left\{ 
\begin{array}{rll}
\p _{n_k}(t,y)&\to \pb (t,y)&\hbox{ for every } y\in\yb ;\\
\nabla \p _{n_k}(t,y)&\to \nabla \pb (t,y)&\hbox{ for every } y\in\yb ;\\
G(\o _{n_k})(t,y)&\to G(\ob )(t,y)&\hbox{ for every } y\in\yb .
\end{array}
\right.
\end{equation}
On the other hand, for every $t\in [0,1]\setminus N$
\begin{equation*}
\int_{\y}S(\o _{n_k}(t,y))\, dy \to \int_{\y}S(\ob (t,y))\, dy
\end{equation*}
and 
\begin{equation}
\label{eq:98}
\b (\o _{n_k}(t)) \to \b (\ob (t)) \quad (\hbox{in } [-M,0]).
\end{equation}
Since $\o $ satisfies (\ref{eq:2}) for all $T>0$, we may write for 
every $k\ge 0$ and every $v\in L^2(0,1,\V )$
\begin{equation}
\label{eq:99}
\begin{array}{l}
\int_0^1 <\frac{d\o _{n_k}}{dt},v>\, dt + \nu\int_0^1\int_{\y}
\big( 1-\o _{n_k}^2\big) \nabla v\cdot \nabla \o _{n_k}\, dydt \\
+\int_0^1 \int_{\y} \{ -\o _{n_k} G(\o _{n_k}) +\nu \b (\o _{n_k})
\big( 1-\o _{n_k}^2\big) ^2\nabla\p _{n_k}\} \cdot \nabla v\, dydt
=0.
\end{array}
\end{equation}
Letting $k\to \infty$ we readily infer from 
(\ref{eq:89})-(\ref{eq:99}) and Lebesgue theorem that
\begin{equation}
\label{eq:100}
\int_0^1 \{ <\frac{d\ob}{dt},v>+\nu \int_{\y}\left( \nabla \varphi (\ob )
-\frac{1}{\nu} \ob G(\ob )+\b (\ob )(1-\ob ^2)^2\nabla\pb \right)
\cdot\nabla v\, dy \} \, dt =0.
\end{equation}
As in the proof of Theorem 1, we easily see that (\ref{eq:100}) also holds 
true when $v\in L^2(0,1,V)$. Thus, $\ob$ is on $[0,1]$ the solution 
of (\ref{eq:207}) associated with the initial value
$\ob (0)$. On the other hand, the function ${\cal S}(\o (t))$ is 
nonincreasing and nonnegative. Let
\begin{equation*}
L:=\lim_{t\to\infty}{\cal S}(\o (t)).
\end{equation*}
Then for every $t\in (0,1)\setminus N$,
${\cal S}(\ob (t))=\lim_{k\to\infty }{\cal S}(\o (t_{n_k}+t)) =L$,
and we infer from the continuity of the map ${\cal S}(\ob (\cdot ))$ that
the equality ${\cal S}(\ob (t))=L$ 
holds true for {\em each} $t\in [0,1]$,
hence, by (\ref{eq:29}),  
\begin{equation*}
\forall t\in [0,1]\quad
\int_0^t\int_{\y} \vert\nabla\ob \vert ^2\, dyd\tau = -\int_0^t \b (\ob )
\{ \int_{\y} \ob ^2\left( 1-\frac{\ob ^2}{3}\right)\, dy\} \, d\tau . 
\end{equation*}
Differentiating w.r.t. $t$ we get
\begin{equation*}
\hbox{ for a.e. } t\in [0,1] \quad 
\int_{\y} \vert\nabla\ob (t)\vert ^2\, dy =-\b (\ob (t))
\int_{\y} \ob ^2(t) \left( 1-\frac{\ob ^2(t)}{3}\right) \, dy,
\end{equation*}
that is, by (\ref{eq:2005})-(\ref{eq:2006})
\begin{equation}
\label{eq:101}
\left( \int_{\y} \nabla\pb (t)\cdot \nabla\ob (t) 
(1-\ob ^2(t))\, dy\right) ^2
=\int_{\y} \vert\nabla \pb (t)\vert ^2 (1-\ob ^2(t))^2 \, dy \cdot 
\int_{\y} \vert\nabla\ob (t)\vert ^2\, dy.
\end{equation}
Since
$\int_{\cal Y}\nabla\pb (t) \cdot\nabla\ob (t)(1-\ob ^2(t))dy 
=\int_{\cal Y}\ob ^2(t)(1-\frac{\ob ^2(t)}{3})\, dy>0$ 
(hence $(1-\ob ^2(t))\nabla\pb (t)\ne 0$), we infer from Cauchy-Schwarz
inequality and (\ref{eq:101}) that for a.e. $t\in [0,1]$ there exists
$\lambda (t)>0$ such that
\begin{equation}
\label{eq:102}
\nabla\ob (t) =\lambda (t) (1-\ob ^2(t))\nabla \pb (t),
\end{equation}
which immediately gives
\begin{equation}
\label{eq:103}
\b (\ob (t))=-\lambda (t)
\end{equation}
and
\begin{equation}
\label{eq:104}
\hbox{div}\big( \ob (t)G(\ob (t))\big) =\nabla\ob (t)\cdot G(\ob (t))=0.
\end{equation}
Now, thanks to (\ref{eq:102})-(\ref{eq:104}), (\ref{eq:100}) 
simplifies to $\int_0^1 <\frac{d\ob}{dt},v>\, dt=0$ for all 
$v\in L^2(0,T,\V )$, hence $\frac{d\ob}{dt}=0$ and $\ob$ takes on the 
interval $[0,1]$ a constant value, also denoted by $\ob$. Arguing as in
\cite{MR}, we easily infer from (\ref{eq:102}) that 
$\ob\in C^{\infty}_{per}(\yb )$, $\Vert \ob \Vert _{L^{\infty}(\y )} <1$
and that (\ref{eq:86}) holds for some constant $C\in\R$. 
It remains to prove (\ref{eq:87}). By (\ref{eq:90}) and (\ref{eq:95}) 
we get at once 
\begin{equation*}
%\label{eq:105}
\o (t_{n_k})\cf \ob \hbox{ in } H,
\end{equation*}
and (\ref{eq:87}) holds true provided that
\begin{equation}
\label{eq:106}
\limsup_{k\to\infty}\int_{\y}\o (t_{n_k},y)^2\, dy 
\le \int_{\y} \ob (y)^2\, dy.
\end{equation}
We now prove (\ref{eq:106}). Let $\ep >0$ be given. An application of
(\ref{eq:29}) with $\o =\o _{n_k}$, $t_1=0$, $t_2=t\in (0,1)$ yields
\begin{equation*}
\begin{array}{rl}
\nu\int_0^t\int_{\y} \vert\nabla \o _{n_k}\vert ^2\, dyd\tau  =  
&\int_{\y} \{ S(\o _{n_k}(0))-S(\o _{n_k}(t)) \} dy \\ 
&+\nu \int_0^t\vert\b (\o _{n_k})\vert\{ 
\int_{\y}\o _{n_k}^2\big( 1-\frac{\o _{n_k}^2}{3}\big) \, dy\} d\tau ,
\end{array}
\end{equation*}
hence 
\begin{equation}
\begin{array}{rl}
\limsup_{k\to\infty} \nu\int_0^t\int_{\y} (1-\o _{n_k}^2)
\vert \nabla \o _{n_k}\vert ^2\, dyd\tau &\le \limsup_{k\to\infty} 
\nu\int_0^t\int_{\y}\vert\nabla \o _{n_k}\vert ^2\, dyd\tau \\
\label{eq:107}
&\le 0+\frac{2}{3}\nu Mt \le \ep
\end{array}
\end{equation}
provided that $0<t<\delta := \frac{3\ep}{2\nu M}$.
Now let $t\in (0,\min \{ \delta , 1\} )\setminus N$, hence 
$\o (t_{n_k}+t)\to \ob\hbox{ in }H$ as $k\to\infty$. An application of
(\ref{eq:28}) with $\o =\o _{n_k}$, $t_1=0$ and $t_2=t$ gives
\begin{equation}
\label{eq:108}
\begin{array}{rl}
\frac{1}{2} \int_{\y} \o (t_{n_k},y)^2\, dy
&=\nu\int_0^t\int_{\y} (1-\o _{n_k}^2)\vert\nabla \o _{n_k}\vert ^2\, dy d\tau
+\frac{1}{2}\int_{\y}\o (t_{n_k}+t,y)^2\, dy \\
&-\nu \int_0^t\vert\b (\o _{n_k})\vert 
\{ \int_{\y}\o _{n_k}^2\big( \frac{\o _{n_k}^4}{5}-\frac{2}{3}\o _{n_k}^2 
+1\big) \, dy \} \, d\tau .
\end{array}
\end{equation}
Letting $k\to\infty$ in (\ref{eq:108}) and using (\ref{eq:107}) 
we get
\begin{equation*}
\limsup_{k\to\infty}\int_{\y}\o (t_{n_k},y)^2\, dy 
\le 2\ep + \int_{\y} \ob (y)^2\, dy,
\end{equation*}
which in turn implies (\ref{eq:106}). The proof of Theorem 3 is
complete. \qed
\va\va
%\begin{rmq} It follows from Theorem 3 that for every 
%$\o _0\in \H$ satisfying (\ref{eq:2000})  and such that $\b$ 
%remains bounded, the
%orbit $\gamma (\o _0):=\{ \o (t);\ t\ge 0\}$ is {\em precompact} in $H$. As
%a consequence of \cite[thm 4.3.3]{H}, the $\o -$limit set 
%\begin{equation*}
%\Omega (\o _0):=\{ \ob \in \H;\ \exists t_n\to\infty,\ \o (t_n)\to\ob
%\hbox{ in }H\}
%\end{equation*}
%is a nonempty compact connected subset of $H$ such that
%\begin{equation*}
%\lim_{t\to\infty}\hbox{dist}_H \big( \o (t),\Omega (\o _0)\big) =0.
%\end{equation*}
%Theorem 3 tells us that $\Omega (\o _0)$ is made up of Gibbs states. 
%\end{rmq}
{\bf Acknowledgments.} The authors wish to acknowledge the 
contribution of Andro Mikelic in offering valuable suggestions and 
stimulating discussion during the course of this work 
and are deeply indebted to Raoul Robert
for bringing Youdovitch's method to their attention.
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\end{document}

