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\markboth{\hfil Minimizing pseudo-harmonic maps in manifolds \hfil
EJDE--2001/37}
{EJDE--2001/37\hfil Yuxin Ge \hfil}
\begin{document}
\title{\vspace{-1in}\parbox{\linewidth}{\footnotesize\noindent
{\sc  Electronic Journal of Differential Equations},
Vol. {\bf 2001}(2001), No. 37, pp. 1--15. \newline
ISSN: 1072-6691. URL: http://ejde.math.swt.edu or http://ejde.math.unt.edu
\newline ftp  ejde.math.swt.edu  (login: ftp)}
 \vspace{\bigskipamount} \\
 %
  Minimizing pseudo-harmonic maps in manifolds
 %
\thanks{ {\em Mathematics Subject Classifications:} 53C43, 58E20.
\hfil\break\indent
{\em Key words:} harmonic maps, diffeomorphism, uniqueness, regularity.
\hfil\break\indent
\copyright 2001 Southwest Texas State University. \hfil\break\indent
Submitted March 30, 2001. Published May 21, 2001.} }
\date{}
%
\author{Yuxin Ge}
\maketitle

\begin{abstract}
In this work, we show some regularity and uniqueness results for
generalized harmonic maps on  target manifolds which
are graphs of real-valued functions defined on ellipsoids. As an
application, we prove a diffeomorphism property for such harmonic
maps in two dimensions.
\end{abstract}

\newcommand{\mint}{-\!\!\!\!\!\!\int}
\newtheorem{theorem}{Theorem}
\newtheorem{lemma}{Lemma}

\section{Introduction}

Harmonic maps with values in a ``convex" ball of a Riemannian manifold enjoy
nice properties such as the regularity of weak solutions and their uniqueness.
Such results were obtained in \cite{HKW} concerning the regularity of a weakly
harmonic map and in \cite{Ja} for the uniqueness, under the hypothesis that the
map takes its values in a geodesically convex ball whose radius is strictly
less
than $\frac{\pi}{2\sqrt{K}}$, where $K$ is the upper bound for the sectional
curvature.  It turns out that these ``convexity" conditions on the target are
not necessary as shown by the result in \cite{H} where similar conditions are
obtained for maps into the upper hemisphere of a ``flattened" ellipsoid of
revolution.

Here we want to show that these regularity and uniqueness results are true also
for more general targets than spheres or ellipsoids of revolution, namely for
target manifolds which are graphs of real-valued functions defined on
ellipsoids
satisfying some property.

We also prove a diffeomorphism property for harmonic maps between a
two-dimensional ball and such a two-dimensional manifold under suitable
boundary
conditions.

Our approach relies on a method of E.  Sandier and J.  Shafrir \cite{SS}, which
allows simplest proofs for the uniqueness and regularity theorems.

The class of target manifolds that we consider is defined as follows. Let
$(,)$ be some scalar product in $\mathbb{R}^n$, different from the standard
Euclidean scalar product $\langle,\rangle$ in general. For simplicity,
we assume that $|y|^2=(y,y) =\sum_{i=1}^n a_i y_i^2$ with $a_i>0$.
Denote $A = \{y\in\mathbb{R}^n/ (y,y)<1\}$. Clearly, $A$
is a convex set.
Let ${\cal N}$ be a hypersurface in $\mathbb{R}^{n+1}$, defined by
${\cal N} = \{(y, y_{n+1})
\in \mathbb{R}^{n+1}/ y\in A, y_{n+1} = f(|y|^2)\}$, where $f$ is a
function in $C^\infty([0,1))\cap C^0([0,1])$ such that $f(1) = 0$.
Let $\Omega\subset\mathbb{R}^N$
be a bounded domain and let $g:\partial\Omega\to  {\cal N}$
be a prescribed $C^{2,\gamma}$ mapping with $\gamma >0$. We
consider the space $H^1_g(\Omega, {\cal N}) = \{ v\in H^1(\Omega,
\mathbb{R}^{n+1})/ v(x)\in
{\cal N}\mbox{ a.e. } x\in \Omega \mbox { and } v|_{\partial \Omega} =
g\}$. Let
$\langle,\rangle_{\mathbb{R}^{n+1}}$ be the standard Euclidean inner product on
$\mathbb{R}^{n+1}$. We
define on $H^1_g(\Omega, {\cal N})$ the energy functional
$$ E(u) = \frac{1}{2} \int_\Omega e(u) dx= \frac{1}{2} \int_\Omega
\sum^N_{i, j = 1} a_{ij}(x){\Big\langle \frac{\partial u}{\partial x_i},
\frac{\partial u}{\partial x_j} \Big\rangle_{\mathbb{R}^{n+1}}} dx,
\eqno{(1)}$$
where $a_{ij}(x)$ satisfy the following conditions:
$$\displaylines{
\hfill \exists \;\alpha > 0, \mbox{ such that } a_{ij}(x) \xi^i\xi^j
\geq \alpha {\mid \xi \mid}^2, \quad \forall \; x \in \Omega, \forall \; \xi
\in \mathbb{R}^N; \hfill\llap{(2)} \cr
\hfill a_{ij}(x) \in C^{1,\gamma}(\bar{\Omega}, \mathbb{R}),
 \quad\mbox{for some }\gamma>0
\mbox { and }1 \leq i, j \leq N ; \hfill\llap(3)\cr
\hfill a_{ij} = a_{ji}, \quad 1 \leq i, j \leq N.  \hfill
\llap(4)
}$$
The critical points of $E$ satisfy in the sense of distributions the
Euler equation
$$\displaylines{
\hfill L u+\sum^N_{i, j = 1} a_{ij}(x)
C\left( \frac{\partial u}{\partial x_i}(x),
\frac{\partial u}{\partial x_j}(x) \right) =0 , \quad \mbox{in }\Omega
\hfill\llap(5)\cr
u=g, \quad \mbox{on } \partial{\Omega},
}$$
where $L = \sum^N_{i, j = 1} {\frac {\partial} {\partial
x_i}}{\left(a_{ij}(x){\frac{\partial }{\partial x_j}}\right)}$ and
$C(,)$ is the second fundamental form of ${\cal N}$.
We remark that if $N\neq 2$,
this problem is equivalent to studying harmonic maps from $(\Omega, g_{ij})$
on ${\cal N}$, where $(g_{ij}) = (\det(a_{ij}))^{\frac{1}{N-2}}
(a_{ij})^{-1}$. However, if $N=2$, this problem concerns a more general
problem which agrees to harmonic maps if and only if $\det(a_{ij})=1$.

Our first result is a uniqueness principle. Let ${\cal M}$ be a
hypersurface such that $\overline{\cal N} \subset {\cal M}$. We will show
the following result.

\begin{theorem} \label{thm1}
 Assume that $g(\partial \Omega)\subset \overline{\cal N}$
and the following conditions are verified,
$$\displaylines{
\hfill f^{(1)}<0 \mbox{ on } [0,1);\hfill\llap(6)\cr
\hfill f^{(2)}\leq 2(f^{(1)})^3 \mbox{ and } 2(f^{(1)}f^{(3)}-
3(f^{(2)})^2) + f^{(2)}f^{(1)}- 2(f^{(1)})^4\geq 0\mbox { on } [0,1);
\hfill\llap(7)\cr
\hfill  a_i <1 \quad\forall i=1,\dots,n. \hfill\llap(8)
}$$
Then, the minimizer of $E$ in $H^1_g(\Omega, \overline{\cal N})$ is unique.
Moreover, if
$u$ is a critical point of $E$, whose image lies in a compact subset of
${\cal N}$, then
$u$ is a minimizer.
\end{theorem}

This result is a generalized variant of \cite{H}. In fact, if we set
$f=\sqrt{1-t}$, we find that ${\cal N}$ is the upper hemisphere of the
ellipsoid. In this paper, we will adopt the same strategy as
in \cite{SS}. Our approach relies on a convex inequality for the energy
functional $E$.

The second result is a regularity theorem. We will show the following.

\begin{theorem} \label{thm2}
 Under the same hypotheses as in Theorem 1, assume that
$u\in H^1_g(\Omega,{\cal N})$, whose image belongs to a compact subset
of ${\cal N}$, and that $u$ is a critical point of $E$,
i. e., $u$ is a weak solution of (5). Then $u$ is $C^2$ on
$\Omega$.\end{theorem}

This theorem is a variant of a result of S. Hildebrandt, H. Kaul and K-O.
Widman, who have proved in \cite{HKW} the same result for a harmonic map in
a geodesic ball
of radius $r<\frac{\pi}{2\sqrt{K}}$ where $K$ is an upper bound of the
sectional curvature
of the manifold. By the uniqueness principle, we need only prove it for the
minimizing maps.
So it is relatively easy to obtain the regularity property with the help of
$\epsilon$-regularity due to R. Schoen and Uhlenbeck \cite{SU1}. Note that
if ${\cal N}$ is
a hemisphere of a flattened ellipsoid, we find again the result
of \cite{H}.

In the third part, we will use these two first results to study the problem
of diffeomorphism. Let $n=N=2$ and $\Omega = B$ be the unit disc in
$\mathbb{R}^2$. Assume that $g:\partial \Omega\to  {\cal N}$ is a
convex Jordan curve ${\cal N}\cap \{x_3 = \alpha>0\}$. We show the
following theorem.

\begin{theorem} \label{thm3}
Under the above assumptions, assume that $u\in H^1_g(\Omega, {\cal N})$
is a critical point of $E$. Then, $u$ is a
diffeomorphism.\end{theorem}

Here, we argue by a continuity method, due to J. Jost \cite{J}.
We connect the critical point with a harmonic map. Using a result of
Hartman and Wintner \cite{HW}, we conclude our claim.
In the last section, we will treat the limit case. We set
$f(t) = \sqrt{1-t}$ and $a_1<\dots<a_r<a_{r+1}=\dots = a_n = 1$.
In general, the uniqueness principle fails, as E. Sandier
and I. Shafrir have proved for the case where $a_1 = \dots =a_n = 1$ and
$a_{ij} = \delta_{ij}$. With the same procedure as in \cite{SS}, we will
establish a criterion on the boundary condition for which the uniqueness
principle holds.

\section {Proof of Theorem 1}
First, we give a basic inequality, which is a variant of the
inequality in \cite{SS}.

\begin{lemma} \label{lm1}
Assume that (6) and (7) in the theorem 1 hold and that
$$
 a_i\leq 1. \eqno(8')
$$
Then function
$$
G(v,w) = (w, w) + (2f'(|v|^2)(v,w))^2
$$
is convex over the set
$A=\{(v,w)\in \mathbb{R}^n \times \mathbb{R}^n/ (v,v)<1\}$.
Moreover, we have
that  for any $(v^0, w^0)$, $(v^1, w^1)\in A$, with $(v^0, w^0)\neq (v^1,
w^1)$, $$
G\left( \frac{ v^0 + v^1}{2},  \frac{ w^0 + w^1}{2}\right) = \frac{1}{2}(
G(v^0, w^0) +
G(v^1, w^1))
$$
holds only if for any $0\leq t \leq 1$ we have
$$\displaylines{
\rlap{i)}\hfill \delta w + 4(f'(|v^t|^2))^2 (v^t, w^t) \delta v =0 ,
\hfill\llap(9)\cr
\rlap{ii)}\hfill ( \delta v, w^t)f'(|v^t|^2) +
4[f^{(2)}(|v^t|^2)-f'(|v^t|^2)^3](v^t,w^t) (\delta v, v^t)  = 0\hfill
}$$
and then $G(v^t,w^t) = G(v^0,w^0)$ for $0\leq t \leq 1$. Here we wrote
$(v^t,w^t) =
(1-t) (v^0,w^0) + t (v^1,w^1)$ and $(\delta v, \delta w) = (v^0 - v^1, w^0
- w^1)$.
Furthermore, if $f^{(2)}= 2(f^{(1)})^3 $ and $f^{(1)}f^{(3)} =
3(f^{(2)})^2$, i) and ii)
are also the sufficient conditions.   \end{lemma}

\paragraph{Proof.} Define $F(s) = G(v^0 + s\delta v, w^0 + s\delta w)$. A
calculation leads to
\begin{eqnarray*}
F'(t) &=& 2(w^t, \delta w) + 8[((\delta v, w^t) + (v^t,
\delta w))f'(|v^t|^2)\\
&&+ 2f^{(2)}(|v^t|^2)(v^t, \delta v)(v^t, w^t)]f'( |v^t|^2)(v^t, w^t)
\end{eqnarray*}
and
\begin{eqnarray*}
2F^{(2)}(t) &= &|\delta w + 4f'(|v^t|^2)(v^t, w^t)\delta v|^2\\
&&+4[4f^{(2)}(|v^t|^2)(v^t, \delta v)(v^t, w^t)+((\delta v, w^t) + (v^t,
\delta w))f'(|v^t|^2)]^2\\
&&+8[f'(|v^t|^2)f^{(2)}(|v^t|^2) - 2(f'(|v^t|^2))^4](v^t, w^t)^2(\delta
v,\delta v)\\
&&+16[f'(|v^t|^2)f^{(3)}(|v^t|^2) - 3(f^{(2)}(|v^t|^2))^2](v^t,
w^t)^2(v^t,\delta v)^2.
\end{eqnarray*}
Hence, $F^{(2)}(t)\geq 0$ on $[0,1]$ since $(v^t,\delta v)^2 \leq (\delta
v, \delta v)(v^t, v^t)\leq (\delta v, \delta v)$. The convexity of $G$
follows. Finally, note that if $F(s) = (1-s) F(0) + sF(1)$ for some $s\in
(0,1)$, then $F^{(2)}(t)\equiv 0$ for all $t\in [0,1]$. Hence, we achieve
the proof.\hfill$\diamondsuit$\smallskip

We will adapt the notation in \cite{SS}. So that every
$u\in\overline {\cal N}$ can be written in the form
$u = u^h + u^v e_{n+1}$, where $u^h\in A$
and $u^v = f(|u^h|^2)$. Then for $u_0$, $u_1\in H^1(\Omega, \overline{\cal
N})$ and $0\leq t\leq 1$, we define the map $ u^t(x) = (1-t) u^0
\oplus t u^1$ by
$$[(1-t) u^0 \oplus t u^1]^h(x) = (1-t) (u^0)^h(x) + t (u^1)^h(x),
$$
and  $[(1-t) u^0 \oplus t u^1]^v(x) = f( { |(u^t)^h|^2})$.

\begin{lemma} \label{lm2}
For any $u^0, u^1 \in H^1(B, \overline{\cal N})$, let $u^t
= (1-t)u^0 \oplus t u^1$ for $0 \leq t \leq 1$. Then $u_t \in H^1(B,
\overline{\cal N})$ and
$$
e(u^t)\leq (1-t)e(u^0) +t e(u^1)\quad\mbox{a. e. on }\Omega. \eqno{(10)}
$$\end{lemma}

\paragraph{Remark.}
Here, $H^1(\Omega,\overline {\cal N}) = \{ u\in
H^1(\Omega,{\cal M})$  and $u(x)\in \overline {\cal N}$ a. e.
on $\Omega\}$, where ${\cal M}$ is a hypersurface in $\mathbb{R}^{n+1}$
such that $\overline {\cal N}\subset {\cal M}$.

\paragraph{Proof.} We will decompose $e$ into two parts:
$e(v)=  F_1(v) + F_2(v)$,
where
$$
F_1(v)(x) =  \sum^N_{i, j = 1} a_{ij}(x)\left[\left\langle \frac{\partial
v^h}{\partial x_i}, \frac{\partial v^h}{\partial x_j}
\right\rangle_{\mathbb{R}^{n}}- \left(\frac{\partial v^h}{\partial x_i},
\frac{\partial v^h}{\partial x_j} \right)\right]
$$
and
$$
F_2(v)(x) =  \sum^N_{i, j = 1} a_{ij}(x)\left[ \left(\frac{\partial
v^h}{\partial x_i}, \frac{\partial v^h}{\partial x_j} \right) +
\frac{\partial v^v}{\partial x_i} \frac{\partial v^v}{\partial x_j} \right].
$$
Clearly,
$F_1(u^t)(x)\leq (1-t) F_1(u^0)(x) + t F_1(u^1)(x)$, for all $x\in\Omega$,
since $\langle,\rangle_{\mathbb{R}^n} - ( , )$ is a positive bilinear form on
$\mathbb{R}^n$. Now we fix $x\in \Omega$, then there exists $C(x) =
(c_{ij}(x))\in GL(N,\mathbb{R})$ such that
$$
\sum^N_{i, j = 1} a_{ij}(x)\xi^i\xi^j =
\sum^N_{i=1}\Big(\sum^N_{j=1}c_{ij}(x)\xi^j\Big)^2
\mbox{ for all } \xi \in \mathbb{R}^N.
$$
Let us first suppose that for some $\epsilon > 0$,
$$
|(u^0)^h| , |(u^1)^h| \leq 1 - \epsilon.
$$
It is clear that  $|(u^t)^h| \leq 1 - \epsilon$ for $0 \leq t \leq
1$, or that
$$
 \frac{\partial (u^t)^v}{\partial x_j} = 2f'( |(u^t)^h|^2)((u^t)^h ,
\partial (u^t)^h/{\partial x_j}), \mbox{ for } j = 1,\dots,N.
$$
Hence,
$$
F_2(u^t)(x) =  \sum_{i=1}^N G\Big((u^t)^h , \sum_{j=1}^Nc_{ij}(x)\partial
(u^t)^h/\partial x_j\Big).
$$
Therefore, (10) follows by lemma 1. In the general case, for getting the
result, we use an
approximation argument. Define a map $P_{\lambda} : \overline {\cal
N}\to \overline {\cal N}$ depending on $\lambda > 0$ by
$$
P_{\lambda}(u^h, u^v) = ((1-\lambda)u^h, f((1-\lambda)^2|u^h|^2)).
$$
Obviously, $P_{\lambda}(\overline {\cal N}) \subset  \overline {\cal N}\cap
\{u_{n+1} \geq \epsilon(\lambda)\}$, where $\epsilon(\lambda)$ is strictly
positive and $P_{\lambda}$ converges in $C^1(\overline {\cal N},\overline
{\cal N})$ norm  to the identity mapping as $\lambda$ goes to 0. For any
$\lambda > 0$, let $u^0_{\lambda} = P_{\lambda} \circ u^0$ and
$u^1_{\lambda} = P_{\lambda} \circ u^1$. Setting $u^t_{\lambda} = (1-t)
u^0_{\lambda} \oplus  t u^1_{\lambda}$, we have that for any $0 \leq t \leq
1$ and a.e. $x$ in $\Omega$,
$$
F_2(u^t_{\lambda})(x) \leq (1-t)F_2(u^0_{\lambda})(x) + t
F_2(u^1_{\lambda})(x). \eqno(11)
$$
Now we pass to the limit as $\lambda$ goes to 0. The right hand side
converges in $L^1$ to $(1-t)F_2(u^0) + t F_2(u^1)$. By coerciveness and
(11), $u^t_{\lambda}$ remains bounded in $H^1$. After choosing a
subsequence, we may assume that $u^t_{\lambda} \to  v^t$ weakly
in $H^1$ for some $v^t \in H^1$. But $u^t_{\lambda} \to  u^t$
a.e. in $B$ and hence $u^t = v^t$. Let $K \subset \Omega$ a subset of
$\Omega$. By weak lower semi-continuity, we deduce from (11) that
\begin{eqnarray*}
\int_K F_2(u^t) & \leq & \liminf_{\lambda \rightarrow 0} \int_K
F_2(u^t_\lambda) \\
& \leq & \lim_{\lambda \rightarrow 0}  \int_K
(1-t)F_2(u^0_\lambda)+tF_2(u^1_\lambda)\\
& = &  \int_K (1-t)F_2(u^0)+tF_2(u^1).
\end{eqnarray*}
But it is well known that
$$
\mint_{B(x,r)}v \to  v(x) \mbox{ a.e. in } B \mbox{ as } r
\to  0 \mbox{ for any } v \in L^1,
$$
where $B(x,r) = \{ y \in \mathbb{R}^N | |y-x| <r \}$.
Consequently, (10) holds.\hfill$\diamondsuit$



\paragraph{Proof of Theorem 1.}
We assume that $u^0$ and $u^1$ are two distinct minimizers with
the same boundary data $g$. Thanks to  lemma 2, we obtain
$$
E(u^t)\leq (1-t) E(u^0) + t E(u^1) = E(u^0),
$$
which implies for  $0\leq t\leq 1$,
$$
F_1(u^t)(x)= (1-t) F_1(u^0)(x) + t F_1(u^1)(x) \quad\mbox{a. e. } x\in\Omega,
$$
that is,
$$
d(u^0)^h(x) = d(u^1)^h(x)\quad\mbox{a. e. } x\in\Omega,
$$
since $a_i <1$ for $1\leq i \leq n$. Hence we conclude $u^0 = u^1$.
This contradiction terminates the first part of our claim. Now let $u^0$ be
the minimizer and $u^1$ be a critical point of $E$ which agrees with $u^0$
on $\partial\Omega$. Assume that $u^0$ and $u^1$ lie in a  compact subset
of ${\cal N}$. Denote $I(t) = E(u^t)$ for any $0\leq t \leq 1$. Obviously,
$I(t)\in C^1([0,1])$ and is convex by lemma 2. But $I'(0)\geq 0 = I'(1)$.
Thus, it follows that $I(t)\equiv I(0)$, that is, $u^1$ is the
minimizer.\hfill$\diamondsuit$

\paragraph{Remarks}  1.)
$u\in H^1_g(\Omega, \overline{\cal N})$ is a minimizer in
$H^1_g(\Omega,\overline{\cal N})$. Thus $u$ verifies the Euler-Lagrange
equation (5). Indeed, let $C$ be a contraction from ${\cal M}$ onto
$\overline{\cal N}$ such that $C|_{\overline{\cal N}} = Id$. So we have
$E(C(v))\leq E(v)$, that is,
$$\inf_{v \in H^1_g(\Omega, \overline{\cal N})} E(v)
= \inf_{v \in H^1_g(\Omega, {\cal M})} E(v). $$
2.) The existence of a minimizer is obtained by the minimizing
method.

\section {Proof of Theorem 2}

Now we consider a weak pseudo-harmonic map $u$, that is a solution of (5),
whose image lies into a compact subset of ${\cal N}$. So there exists
$\alpha_0>0$ with
$$u_{n+1}\geq \alpha_0\quad\mbox{a.e.\quad on }\Omega.$$
In view of theorem 1, it is sufficient to prove it for the minimizing maps.
Thanks to a result due to R. Schoen and K. Ulenbeck \cite{SU1}, there
exists $\epsilon_0>0$ such that $\int_{B_r(x)} e(u) \leq \epsilon_0$
for any $x\in \Omega$ and for all $r>0$ such that $B_r(x)\subset \Omega$,
then $u$ is regular on $B_{r/2}(x)$. Here, we will use the arguments in
\cite{HKW} ( see  also in \cite{H2}). For any bounded domain $\Omega$ in
which the relations of coefficients $(a_{ij})$ are defined as above, then
we have an associate Green function $G$, which satisfies
$$
\forall \xi \in C_c^\infty (\Omega,\mathbb{R}), \forall y\in \Omega,
\quad \xi(y) =- \int_\Omega L \xi(x)G(x,y)dx.
\eqno{(12)}
$$
Fixed $y\in \Omega$, there exists $\tau_0>0$ such that
$B_{\tau_0}(y)\subset \Omega$. Then we can consider
$$
G^{\tau}(x,y) = \mint_{B_{\tau}(y)} G(x,z)dz,\quad \mbox{for all }\tau\in
(0,\tau_0].
\eqno{(13)}
$$
It follows from (12) that
$$
\mint_{B_{\tau}(y)} \xi(z)dz = -\int_\Omega
\sum_{ij=1}^N\partial_i(a_{ij}(x)\partial_j \xi(x))G^\tau (x,y)dx,
$$
where $G^\tau$ is an approximation to the Green function in the sense that
$$
\lim_{\tau\rightarrow 0}G^\tau = G(x,y) \quad\mbox{for } x\neq y, x, y\in
\Omega.
\eqno{(14)}
$$
(see \cite{LSW}, \cite{HKW} and \cite{H2}). Moreover, there exists strictly
positive constants $K_1$, $K_2$ and $K_3$ such that
$$\displaylines{
\hfill  0\leq G(x,y)\leq K_1|x-y|^{2-N};\hfill\llap(15) \cr
\hfill G(x,y)\geq K_2|x-y|^{2-N}\quad\mbox{if } |x-y|\leq \frac{3}{4} d
(y,\partial\Omega);\hfill\llap(16) \cr
\hfill |\nabla_x G(x,y)|\leq K_3|x-y|^{1-N};\hfill\llap(17) \cr
\hfill G^\tau (x,y)\leq 2^{N-2}K_1|x-y|^{2-N}\quad\mbox{if }
\tau<\frac{1}{2}|x-y|;\hfill\llap(18) \cr
\hfill \mbox{if }d (y,\partial\Omega)>\tau, x\to
G^\tau(x,y)\in H^1_0(\Omega,\mathbb{R})\cap L^\infty (\Omega,\mathbb{R}),
\hfill\llap(19)
}$$
where $x$ and $y$ are in $\Omega$. Without loss of generality, assume that
$B(0,r)\subset\Omega$ for some $r>0$. Taking the map $G^\tau(x,0)$ as a
test function in (5), we obtain
$$
\int_\Omega  G^\tau(x,0)Lu^v(x) + \lambda(x)G^\tau(x,0) dx= 0,
\eqno{(20)}
$$
where
\begin{eqnarray*}
\lambda
&=&\Big( -2f'(|u^h|^2)\sum_{ij=1}^N a_{ij}
\langle\partial_i u^h, \partial_j u^h\rangle_{\mathbb{R}^n}\\
&& -4f^{(2)}(|u^h|^2)\sum_{ij=1}^N a_{ij} \langle  u^h, \partial_i
u^h\rangle_{\mathbb{R}^n}\langle  u^h, \partial_j u^h\rangle_{\mathbb{R}^n}
\Big) \\
&&\div \Big(1 +4(f')^2(|u^h|^2)\sum_{i=1}^n a_i^2 u_i^2\Big).
\end{eqnarray*}
Let $\omega\in H^1(\Omega,
\mathbb{R})$ be the solution of the equation
$$\displaylines{
L \omega = 0, \quad \mbox{in }\Omega\cr
\omega = u^v, \quad \mbox{on }\partial\Omega.
}$$
Clearly, $u^v-\omega\in H^1_0(\Omega)$ and from (12) we deduce that
$$
-\int_\Omega G^\tau(x,0)Lu^v(x)dx = -\int_\Omega G^\tau(x,0)L(u^v-
\omega)dx = \mint_{B_\tau(0)} (u^v- \omega)dx.
$$
Consequently,
$$
\int_\Omega G^\tau(x,0)\lambda(x) = \mint_{B_\tau(0)} (u^v- \omega)dx \leq
f(0).
$$
Hence, we obtain
$$
\int_\Omega G^\tau(x,0)|\nabla u^h|^2 \leq C f(0),
\eqno{(21)}
$$
since $\lambda(x)\geq \alpha |\nabla u^h|^2(x)$. Using Fatou's lemma and
passing to the limit in (21) as $\tau\rightarrow 0$, we deduce
$$
\int_\Omega G(x,0)|\nabla u^h|^2 \leq C f(0).
$$
It follows from (16) that
$$
\lim_{\epsilon\rightarrow 0} \int_{B_\epsilon(0)}|\nabla u^h(x)|^2
|x|^{2-N} dx = 0.
$$
On the other hand, remark that $u$ belongs to a compact subset of ${\cal
N}$, which implies
$$
\lim_{\epsilon\rightarrow 0}\frac{1}{\epsilon^{N-2}}\int_{B_\epsilon(0)}
|\nabla u(x)|^2  dx = 0.
$$
This completes the present proof.


\paragraph{Remarks} 1.) If $f\in C^2([0,1])$ and $f'(1) < 0$, we have the same
conclusion in $H_g^1(\Omega,\overline{\cal N})$. Moreover, replacing (8) by
(8') and (7) by $f^{(2)}\leq 0$, our result is also right for minimizing
maps.
\\
2.) For $N=2$, we will give the proof in the following section  (see also
\cite{G}).


\section{Proof of Theorem 3}

In this part, we will use a similar strategy as in \cite{J} (see also
\cite{G}). In order to prove our result, we will consider the following
energy functional
\setcounter{equation}{21}
\begin{eqnarray}
E_t(v) &=& \frac{1}{2} \int_B \sum^2_{i, j = 1}\Big[(1-t)\delta_{ij} + t
a_{ij}(x)\Big]{\left\langle \frac{\partial v}{\partial x_i}, \frac{\partial
v}{\partial x_j} \right\rangle} dx \\
&=& \frac{1}{2} \int_B \sum^2_{i, j = 1}
a_{ij}(t,x){\left\langle \frac{\partial v}{\partial x_i}, \frac{\partial
v}{\partial x_j} \right\rangle} dx. \nonumber
\end{eqnarray}
We consider $G:[0,1]\times \partial B\rightarrow {\cal N}$ to be a
$C^{2,\gamma}$ map such that
\begin{eqnarray}
 &  G(t, . )\mbox{ is a diffeomorphism };& \\
 &  G(1, . ) = g;& \\
 &  G(t, . ) = \{u_3 = \alpha(t)\}\cap {\cal N}, \mbox{ where }
\alpha(0) \mbox{ is a number near to } f(0).&
\end{eqnarray}
Let $I_t = \inf_{v\in H^1(B,{\cal N})} E_t(v) $. Denote $u^t\in H^1(B,{\cal
N})$ the unique minimum of $E_t$ in $H^1_{G(t,.)}(B,{\cal N})$ given by
Theorem 1, then $u^t$ satisfies:
$$\displaylines{
\hfill {\sum^2_{i, j = 1} {\frac {\partial} {\partial
x_i}}{\left(a_{ij}(t,x){\frac{\partial u^t}{\partial
x_j}}\right)}}+\lambda_t u^t = 0 , \quad \mbox{in }B \hfill\llap (26)\cr
u^t = G_t, \quad \mbox{on}\; \partial B,
}$$
where $\lambda_t  ={\sum^2_{i, j = 1} a_{ij}(t,x){C\left\langle
\frac{\partial u^t}{\partial x_i}(x), \frac{\partial u^t}{\partial x_j}(x)
\right\rangle} }$. First we show the following lemma.

\begin{lemma} \label{lm3}
We have $u^t \in C^{2,\gamma}(\bar B, \mathbb{R}^3)$ and
$$
\| u^t\|_{C^{2,\gamma}}\leq C_1.
$$
\end{lemma}

\paragraph{Proof.} We write $E_t$ as follows
$$E_t(v)
 =\frac{1}{2} \int_B \sum^2_{i, j = 1}a_{ij}(t,x)\Big[\left\langle
\frac{\partial v^h}{\partial x_i}, \frac{\partial v^h}{\partial x_j}
\right\rangle +(f'(|v^h|^2))^2\big{(}v^h, \frac{\partial v^h}{\partial
x_i}\big{)}\big(v^h, \frac{\partial v^h}{\partial x_j}\big)\Big] dx.
$$
Obviously, there exists some $\beta>0$ such that $\left((u^t)^h,
(u^t)^h\right)\leq 1-\beta$ for any $t\in [0,1]$ and $x\in \bar B$.
Thanks to a result of Jost and Meier (see \cite{JM}), we deduce that
$$
\| u^t\|_{W^{1,q}}\leq C,\mbox{ for some } q>2.
$$
Recalling that $u^t$ satisfies the equations (26), it follows that
$$
\| u^t\|_{W^{2,q}}\leq C.
$$
Consequently,
$$
\| u^t\|_{W^{1,\frac{2q}{4-q}}}\leq C \| u^t\|_{W^{2,q}}\leq C,
\quad\mbox{if } q<4.
$$
Now iterating the above procedure and using Sobolev's embedding theorem, we
obtain
$$
\| u^t\|_{C^{1,\gamma}}\leq C.
$$
Hence, using Schauder's estimates, we complete the proof.\hfill$\diamondsuit$

\begin{lemma} \label{lm4}
With the above notation, we have $\mathop{\rm rank}(\nabla u^t(x))=2$,
for any $t \in [0, 1]$  and $x \in \partial{B}$.
\end{lemma}

\paragraph{Proof.} Denote
$L_t =  \sum^2_{i, j =1}\partial_i(a_{ij}(t,x)\partial_j)$.
Using  (26), we state
$L_t((u^t)_3) \leq 0$, and the strong maximum principle implies
$$
(u^t)_3(x) > \alpha(t) \quad \forall x \in B, \mbox{  or  } (u^t)_3 \equiv
\alpha(t).
$$
The latter is incompatible with $L_t((u^t)_3) < 0$ on $\partial B$. Hence,
the claim follows from Hopf's maximum principle. \hfill$\diamondsuit$


\begin{lemma}[\cite{S}] \label{lm5}
Assume that $U$ and $V$ are smooth, bounded
domains in $\mathbb{R}^2$, diffeomorphic to $B$. Let $ \varphi : \bar{U}
\to  \bar{V}$ be a $C^1$ map such that
$$
\det (\nabla\varphi(x) ) >0, \quad \forall x \in{U}.
$$
Moreover, suppose that $\varphi|_{\partial U}$ is a diffeomorphism from
$\partial U$ to
$\partial V$. Then $\varphi$ is a diffeomorphism.
\end{lemma}

\begin{lemma}[{\cite[theorem 5.1.1]{J}}]  \label{lm6}
 We have $\mathop{\rm rank} (\nabla u^0(x)) =2$,
for all $x\in \bar{B}$, and $u_0$ is a diffeomorphism.
\end{lemma}

\begin{lemma}[\cite{G}] \label{lm7} the mapping
$F_*:  [0, 1] \to  C^{1, \beta}(\bar{B}, {\cal N})\cap H^1(B,
{\cal N})$ such that $t \mapsto u^t$ is continuous.
\end{lemma}
The proof of this lemma is a consequence of Theorem 1 and lemma 3.\smallskip

Now, we define the set
$$
T_1 = \{ t \in [0, 1], \; u^t \mbox{ is a diffeomorphism } \}.  \eqno{(27)}
$$
It suffices to prove that $T_1$ is open, closed and not empty.

\noindent\textbf{ Step 0.} $0 \in T_1$.
This is just the statement of lemma 6.

\noindent\textbf{ Step 1.} $T_1$ is open. It follows from lemma 5 and 7.

\noindent\textbf{ Step 2.} $T_1$ is also closed.
Let $\{ t_n \}_{n \in \mathbb{N}}$ be a sequence converging to $t$. Assume
that $u^{t_n}$ are diffeomorphic, for all $n \in \mathbb{N}$. We assume that
$$
\exists \;x_0 \in \bar{B}, \quad \mbox{such that}\quad
\mathop{\rm Rank}(\nabla ( u^t) (x_0))\leq 1. $$
We shall use geodesic parallel coordinates based on a geodesic arc $c$
through $q= u^t(x_0)$ as in \cite{J}. In these coordinates $(v_1, v_2)$,
for $v^2\equiv 0$, $v_1$ is the arclength parameter of $c$, where as the
curves $v_1\equiv const$ are geodesics normal to $c$ parametrized by
arclength $v_2$, consequently the curves $v_2\equiv const$ are parallel
curves of $c$. Moreover, we can choose the coordinates such that
$\partial_z (v_2\circ u^t)(q) = 0$. In these coordinates, we have for the
metric tensor
$$
g_{11}(v_1, 0 ) = 1,\quad
g_{12}(v_1, v_2 ) = 0,\quad
g_{22}(v_1, v_2 ) = 1,
$$
therefore, the only non-vanishing Christoffel symbols are
$$\displaylines{
\Gamma^1_{11} = \frac{1}{2} g^{11}\partial_{x_1}g_{11},\cr
\Gamma^1_{12} =\Gamma^1_{21}= \frac{1}{2} g^{11}\partial_{x_2}g_{11},\cr
\Gamma^2_{11} = -\frac{1}{2} g^{22}\partial_{x_2}g_{11},
}$$
Hence, equations (26) for $u^t$ take the form
$$\displaylines{
L_t v_1 = - \Gamma^1_{11}\left(\sum^2_{ij=1}a_{ij}\partial_{x_i}
v_1\partial_{x_j} v_1\right) -
\Gamma^1_{12}\left(\sum^2_{ij=1}a_{ij}\partial_{x_i} v_1\partial_{x_j}
v_2\right),\cr L_t v_2 = - \Gamma^2_{11}\sum^2_{ij=1}a_{ij}\partial_{x_i}
v_1\partial_{x_j} v_1.
}$$
Applying a result due to Hartman and Wintner \cite{HW}, we obtain that
$$
\partial_z (v_1\circ u^{t})(z) = a_1(z-z_1)^m + o(|z-z_1|^m), \mbox{ for
some } m\geq 1\mbox{ and } a\in \mathbb{C}^*,
$$
where $z_1$ are coordinates of $q$. Consequently,
$$
\deg(\partial_z (v_1\circ u^t), B(z_1, r_1), 0) = m \geq 1,\mbox{ for
some }  r_1>0,
$$
which implies
$\deg(\partial_z (v_1\circ u^{t_n}), B(z_1, r_1), 0) = m \geq 1$,
for some sufficiently large $n$.
Hence, there exists $z_2\in B(z_1, r_1)$ such that $\partial_z (v_1\circ
u^{t_n})(z_2) = 0$ by the property of degree. This contradicts that
$u^{t_n}$ is diffeomorphic. Therefore, the assertion follows from Lemma 5.

\paragraph{Remark.} We have a more general result, that is, our result
also holds for a convex curve in $\overline{\cal N}$.

\section{The limit case}

In this section, we consider the limit case; that is, $f(t) =
\sqrt{1-t}$. Hence,
$$\overline{\cal N} = \Big\{x\in \mathbb{R}^{n+1};
x_{n+1}\geq 0\mbox { and }x_{n+1}^2 + \sum_{i=1}^n a_i^2 x_i^2 = 1\Big\}
$$
is an upper hemisphere of a $n$-dimensional ellipsoid. Assume that
for some $r$ with $1\leq r\leq n$,
$$
a_1\leq a_2\leq \dots\leq a_r <1\quad  a_{r+1} = \dots = a_n =1\,. \eqno(28)
$$
In general, Theorem 1 fails. We will show that uniqueness depends on the
boundary data as in \cite{SS}. Let $\{ e_1,\dots,e_{n+1}\}$ denote a basis of
$\mathbb{R}^{n+1}$ and $P$ (resp. $P_1$) the projection from
$\mathbb{R}^{n+1}$ onto
$\mathbb{R}^{n-r+1}$ (resp. $\mathbb{R}^r$) defined as follows
$$\displaylines{
P(x_1,\dots,x_{n+1}) = (x_{r+1},\dots, x_{n},x_{n+1}), \cr
P_1(x_1,\dots,x_{n+1}) = (x_1,\dots, x_r).
}$$
For any map $g$ with values in $\overline{\cal N}$, we define $Rank(P\circ
g)$ to be the smallest integer $k$ with $0\leq k \leq n-r+1$ such that the
image of $g$ lies in a $k$-dimensional vector subspace of $\mathbb{R}^{n-r+1}$.
With the same procedure as in \cite{SS}, we will prove the following
result.

\begin{theorem} \label{thm4}
 Let $g:\partial \Omega\to  \overline{\cal N}$
be a $C^{2,\gamma}$ map for some $\gamma >0$. Then uniqueness of the
minimizer for the boundary data $g$ fails if and only if\\
(I) $ k = Rank (P\circ g) \leq n-r -1\mbox{ and } Image ( g)\subset
\overline{\cal N}\cap \{x_{n+1} =0\}$,\\
(II)  the $\overline{\cal N}^k$ valued minimization problem
$$
(P^k_g)\quad \inf\{E(v), v\in H^1_g(\Omega, \overline{\cal N}^k)\}
$$
has a solution $u$ such that $\mathop{\rm Rank }(P\circ u) = k+1$, where
$$\overline{\cal N}^k= \big\{x\in \mathbb{R}^{r+k+1}; x_{r+k+1}\geq 0
\mbox { and }\sum_{i=1}^r a_i^2 x_i^2+\sum_{i=r+1}^{r+k+1} x_i^2 = 1\big\}\,.
$$
Moreover, when uniqueness fails, let $u^0$ be any one of the minimizers,
the set of the minimizers is obtained by composition of $u^0$ with any
rotation of $\mathbb{R}^{n+1}$ that leaves the $k$-dimensional vector subspace
containing $P\circ g(\partial\Omega)$ and $P_1(\mathbb{R}^{n+1})$ invariant.
\end{theorem}


We divide our proof in several steps. \\
\textbf{ Step 1} consists of the
following  lemma.

\begin{lemma} \label{lm8}
We assume that $u^0$ and $u^1$ are two distinct
$\overline{\cal N}$-valued minimizers with same boundary data $g \in
C^{2,\gamma}$. Then for every $0<t<1$, $u^t(x) = [(1-t) u^0 \oplus t
u^1](x)$ is also a minimizer which is $C^{2,\gamma}$ in $\Omega$ and
$(u^t)^v>0$ in $\Omega$.
\end{lemma}

\paragraph{Proof.}
Clearly, it follows  from lemma 2 that $u^t$ is a minimizer for any
$0<t<1$. Fix some $0<t<1$ and so by results of R. Schoen and K. Uhlenbeck
\cite{SU1} and \cite{SU2}, it is $C^{2,\gamma}$ near the boundary and in
$\Omega$ outside of a closed set $M$ of Hausdorff dimension at most $N-3$.
We know  that $-\triangle (u^t)_{n+1} \geq 0$ and $(u^t)_{n+1} \geq 0$ in
$\Omega\setminus M$. Applying the strong maximum principle in
$\Omega\setminus M$, we have either  $(u^t)_{n+1} > 0$ in $\Omega\setminus
M$ or $(u^t)_{n+1} \equiv 0$ in $\Omega\setminus M$. The latter would imply
$u^0\equiv u^1$ in $\Omega\setminus M$ since $\Omega\setminus M$  is
connect, and then in $\Omega$, which contradicts our assumptions. Thus,
$(u^t)^v>0$ in $\Omega\setminus M$. However, $\Omega\setminus M$ contains a
neighborhood of $\partial\Omega$. Therefore, using theorem 1, we conclude
the claim.\hfill$\diamondsuit$

\textbf{ Step 2.} consists of the following lemma.

\begin{lemma} \label{lm9} Under the above assumptions,
$$
P_1\circ u^0 = P_1\circ u^1,
\eqno{(29)}
$$
and for $0<t<1$ and $i=1,\dots, N$,
$$
\frac{\partial^2 u}{\partial t \partial x_i} =  - \frac{{(u^h ,
{\partial u^h}/{\partial x_i} )}}{1-|u^h|^2} \frac{\partial
u}{\partial t}.\eqno(30)
$$
\end{lemma}

\paragraph{Proof.}
We fix $0<t_0<t<t_1<1$ and denote ${\tilde u}^0 = (1-t_0) u^0
\oplus t_0 u^1$, ${\tilde u}^1 = (1-t_1) u^0 \oplus t_1 u^1$. Then we can
write ${u^t = \frac{t_1 - t}{t_1 - t_0} {\tilde u}^0 \oplus \frac{t -
t_0}{t_1 - t_0} {\tilde u}^1}$ and $({\tilde u}^0)^v, ({\tilde u}^1)^v >0$
in $\Omega$. Obviously,
$$
F_1(\tilde u^t) = (1-t)F_1(\tilde u^0) + t F_1(\tilde u^1),
\eqno{(31)}
$$
and
$$
F_2(\tilde u^t) = (1-t)F_2(\tilde u^0) + t F_2(\tilde u^1),
\eqno{(32)}
$$
Thus, from (31), we obtain (29). Using lemma 1, we deduce that
$$
 \sum_{j=1}^N c_{ij}(x) \frac{\partial^2 u^h}{\partial t \partial x_j}
=  - \frac{\left(u^h , \sum_{j=1}^N c_{ij}(x) \frac{\partial
u^h}{\partial x_j} \right)}{1-|u^h|^2} {\frac{\partial
u^h}{\partial t}},\mbox{ for } 1\leq i \leq N,
$$
that is,
$$
{ \frac{\partial^2 u^h}{\partial t \partial x_j}} =  -
\frac{{\left(u^h , {\partial u^h}/{\partial x_j}
\right)}}{{1-|u^h|^2}} {\frac{\partial u^h}{\partial t}}.
$$
Therefore, as the same arguments as in \cite{SS} (see also \cite{G}), we get
$$
{ \frac{\partial^2 u^v}{\partial t \partial x_j}} =  - \frac{{(u^h ,
{\partial u^h}/{\partial x_j} )}}{{1-|u^h|^2}} {\frac{\partial
u^v}{\partial t}}
$$


\textbf{ Step 3.} In fact, (30) can be written as
$$
{\frac{\partial}{\partial x_j}} \left( \frac{{{\partial u}/{\partial
t}}}{{u^v}} \right) = 0,\mbox{ for } j=1,..,N,
$$
which implies
$$
{\frac{\partial u}{\partial t}(t,x)} = u^v(t,x) \alpha(t),\eqno(33)
$$
for some map $\alpha: (0,1) \to  \mathbb{R}^{n+1}$. On the other hand,
$$
\left(\frac{\partial u^h}{\partial t}, u^h\right) + u^v\frac{\partial
u^v}{\partial t} = 0,
$$
and so we get
$$
\left(\alpha^h, u^h(t,x)\right) + \alpha^v u^v(t,x) = 0.
\eqno{(34)}
$$
Thus,
$$
\langle P\circ u(t,x), P\circ \alpha(t)\rangle  = 0\mbox{ and }   P_1\circ
\alpha(t)  = 0,
\eqno{(35)}
$$
since $\frac{\partial P_1\circ u}{\partial t} = 0$. Or, from (34), we
claim that $e_{n+1}$ and $\alpha(t)$ are not proportional. Then, by
considering $x\in\partial \Omega$, it follows from (33) and (35) that
$$
u^v(x,t) = 0 \mbox{ and }  \langle P\circ u(t,x), P\circ
\alpha(\frac{1}{2})\rangle  = 0\quad\mbox{ on }\partial\Omega.
$$

\begin{lemma} \label{lm10}
 Suppose that $u^0$ and $u^1$ are two distinct minimizers
for the same boundary data $g$. Then
$$
Rank(P\circ u^0) = Rank(P\circ u^1) \leq n-r.
$$
\end{lemma}

\paragraph{Proof.}
Suppose that $\mathop{\rm Rank}(P\circ u^0) = m_0+1$ and
$\mathop{\rm Rank}(P\circ u^1)= m_1 +1\leq m_0$, and that
$P\circ u^i\subset S_i$ where $S_i$ is a $m_i$-dimensional sub-sphere.
Without loss of generality, we can assume that
$S_1\subset S_0$ after a rotation. Then $u^0$ and $u^1$ are also two
$\overline {\cal N}^{m_0}$-valued minimizers (here we will replace
$u^0$ by ${(u^0)^h + |(u^0)^v|e_{m_0+1}}$ if it is necessary). From
(35), we deduce that $\langle P\circ u^0, P\circ \alpha(0)\rangle  = 0$,
that is, $\mathop{\rm Rank}(P\circ u^0)\leq m_0$, a contradiction.
\hfill$\diamondsuit$

\textbf{  Step 4.} In the following, ${\overline {\cal N}}^k$ is a submanifold
of ${\overline{\cal N}}$ in a natural way. In fact, let $u$ be a minimizer
for the problem $P^k_g$. Denote $g_\lambda = P_\lambda\circ g$.
Let $u_\lambda $ be the unique pseudo-harmonic map with the boundary
data $g_\lambda $ and in particular
$u_\lambda(\Omega)\subset {\overline {\cal N}}^k$. Assume that
$u_\lambda\rightharpoonup \tilde u$ in $H^1$ and let $w$ be a $\overline
{\cal N}$-minimizer for the boundary $g$. Therefore, we obtain
$$
E(\tilde u)\leq \liminf_{\lambda\rightarrow 0} E(u_\lambda)\leq
\liminf_{\lambda\rightarrow 0} E(P_\lambda\circ w) = E(w).
$$
This means that ${\overline {\cal N}}^k$-valued map $\tilde u$ is
minimizing among $\overline {\cal N}$-valued maps. Suppose that the
minimizer $u$ of the problem $P^k_g$ has rank $k$. Then by lemma 10 and the
above result, all $\overline {\cal N}$-valued minimizers have rank $k$. If
the $\overline {\cal N}$-valued minimization problem  has two solutions,
then so does the ${\overline {\cal N}}^k$-valued problem. But this means
$Rank(P\circ g)\leq k-1$ which is false. Therefore $P\circ u$ has rank
$k+1$. So we complete the part of necessity.

\textbf{ Step 5.}
Let $u$ be the unique minimizer of the problem $P^k_g$. Let $R$ be a
rotation which leaves the vector space containing $Image(P\circ g)$ and
$P_1(\mathbb{R}^{n+1})$invariant. So $R\circ u$ is another minimizer.
Hence, this
terminate the part of sufficiency. The rest of the theorem is evident.

\paragraph{Acknowledgments.} The author thanks Professor F. H\'elein for his
constant support.

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\noindent\textsc{Yuxin Ge} \\
D\'epartement de Math\'ematiques\\
Facult\'e de Sciences et  Technologie\\
Universit\'e Paris XII-Val de Marne\\
61, avenue du G\'en\'eral de Gaulle\\
94010 Cr\'eteil Cedex, France\\
e-mail: ge@univ-paris12.fr





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