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

\AtBeginDocument{{\noindent\small
\emph{Electronic Journal of Differential Equations},
Vol. 2012 (2012), No. 37, pp. 1--10.\newline
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu
\newline ftp ejde.math.txstate.edu}
\thanks{\copyright 2012 Texas State University - San Marcos.}
\vspace{9mm}}

\begin{document}
\title[\hfilneg EJDE-2012/37\hfil  A non-resonance problem]
{A non-resonance problem for non-Newtonian fluids}

\author[O. Chakrone, O. Diyer, D. Sbibih \hfil EJDE-2012/37\hfilneg]
{Omar Chakrone, Okacha Diyer, Driss Sbibih}  % in alphabetical order

\address{Omar Chakrone \newline
Universit\'e Mohammed I, Facult\'e des sciences\\
Laboratoire LANOL, Oujda, Maroc}
\email{chakrone@yahoo.fr}

\address{Okacha Diyer \newline
 Universit\'e Mohammed I, Ecole Sup\'erieure de Technologie\\
 Laboratoire MATSI, Oujda, Maroc}
\email{odiyer@yahoo.fr}

\address{Driss Sbibih \newline
 Universit\'e Mohammed I, Ecole Sup\'erieure de Technologie\\
 Laboratoire MATSI, Oujda, Maroc}
\email{sbibih@yahoo.fr}

\thanks{Submitted May 17, 2011. Published March 7, 2012.}
\subjclass[2000]{74S05, 76T10}
\keywords{Non-Newtonian fluid; k-Laplacian operator; eigenvalues;
\hfill\break\indent variational method}

\begin{abstract}
 In this article we study a highly nonlinear problem which describes
 a non-Newtonian fluid in a specific domain (symmetric channel).
 This fluid is subjected to pressure of known differences between two
 parallel plates. We establish the existence and uniqueness of a
 weak solution. Our solution method is based on a minimization technique
 when the nonlinearity is asymptotically on the left of the first eigenvalue
 of the operator k-Laplacian.
\end{abstract}

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


\section{Introduction}

Let $\Omega \subset \mathbb{R}^2$ be a bounded domain with
boundary $\partial\Omega=\Gamma=\cup_{i=1}^{4}\overline{\Gamma_i}$,
where $\Gamma_1=\{0\}\times ]-1,1[$, $\Gamma_2=\{1\}\times
]-1,1[$ and $\Gamma_3$, $\Gamma_4$ are symmetrical to the
$x$-axis, see Figure \eqref{fig1}. In the interior of  this domain, a
non-Newtonian fluid is subjected to pressures of known differences
between the two sides $\Gamma_1$ and $\Gamma_2$.


 We note by $u=(u_1,u_2)^{T}\in (C^2(\Omega)\cap C^1(\bar{\Omega}))^2$ and
$-\vec{\Delta}_{k}u=(-\Delta_{k}u_1,-\Delta_{k}u_2)^{T}$,
where $-\Delta_{k}u_i=-div(|\nabla u_i|^{k-2}\nabla u_i)$
is the operator k-Laplacian $i=1,2$ and $1< k < \infty$, which is a
nonlinear operator, (if $k=2$, there is the usual Laplacian).
$\Delta_{k}$ has been used on Sobolev spaces by several authors
we cite for example \cite{a6,a8}, we extend some results of
existence and uniqueness relative to the first eigenvalue of a
Stokes problem. Let $ p\in L^2(\Omega)$, we note
$\vec{g}(x,y,s_1,s_2)=(g_1(x,y,s_1),g_2(x,y,s_2))^{T}$,
where $(x,y)^{T}\in \Omega$, $(s_1,s_2)^{T}\in \mathbb{R}^2$,
$\vec{g}\in C(\bar{\Omega}\times \mathbb{R}^2,\mathbb{R}^2)$ and
$f=(f_1,f_2)^{T}\in (C(\bar{\Omega}))^2$.


\begin{figure}[t] 
\begin{center}
\setlength{\unitlength}{1mm} \scriptsize
\begin{picture}(60,60)(0,0)
\put(60,0){\line(0,1){60}}
\put(0,0){\line(0,1){60}}
\put(0,60){\line(1,0){60}}
\put(0,0){\line(1,0){60}}
\put(3,29){$\Gamma_1$}
\put(53,29){$\Gamma_2$}
\put(28,49){$\Gamma_3$}
\put(28,11){$\Gamma_4$}
\put(10,10){\line(0,1){40}}
\put(5,9){$-1$}
\put(3,19){$-0.5$}
\put(7,29){$0$}
\put(5,39){$0.5$}
\put(7,49){$1$}
\put(50,10){\line(0,1){40}}
\put(10,30){\line(1,0){40}}
\multiput(18,29)(8,0){4}{\line(0,1){2}}
\put(10,26){$0$}
\put(16,26){$0.2$}
\put(24,26){$0.4$}
\put(32,26){$0.6$}
\put(40,26){$0.8$}
\put(48.5,26){$1$}
\qbezier(10,50)(30,40)(50,50)
\qbezier(10,10)(30,20)(50,10)
\put(28,22){$x$}
\end{picture}
\end{center} 
\caption{Geometry of channel} \label{fig1}
\end{figure}


 For $\alpha \in \mathbb{R}$, we consider the nonlinear Stokes problem
\begin{equation} \label{eP}
\begin{gathered}
-\Delta_{k}u_1+\frac{\partial p}{\partial x}=g_1(x,y,u_1)+f_1  \quad \text{in  }\Omega,\\
-\Delta_{k}u_2+\frac{\partial p}{\partial y}=g_2(x,y,u_2)+f_2   \quad\text{in  }\Omega,\\
\operatorname{div}u=\frac{\partial u_1}{\partial x}+\frac{\partial u_2}{\partial y}=0
\quad \text{in  }\Omega,\\
u_1(0,y)= u_1(1,y)   \quad\text{on  } [-1,1],\\
u_2(0,y)= u_2(1,y)  \quad \text{on  } [-1,1],\\
\frac{\partial u_1}{\partial x}(0,y)=\frac{\partial u_1}{\partial x}(1,y)   \quad
 \text{on  } [-1,1],\\
|\nabla u_2(0,y)|^{k-2}\frac{\partial u_2}{\partial x}(0,y)
=|\nabla u_2(1,y)|^{k-2}\frac{\partial u_2}{\partial x}(1,y)   \quad\text{on  } [-1,1],\\
p(1,y)-p(0,y)=-\alpha   \text{on  } [-1,1].
\end{gathered}
\end{equation}
 We assume also the growth condition:
\begin{equation}\label{g}
   |g_i(x,y,s)|\leq c|s|^{k-1}+d(x,y)\quad \forall (x,y)^{T} \in \Omega,\; \forall s\in
\mathbb{R},
\end{equation}
where $c \in \mathbb{R}$ and $d\in L^{k'}(\Omega)$,
 with $\frac{1}{k}+\frac{1}{k'}=1$.

 Note that the second member of \eqref{eP} depends on $u$ and since the pressure difference
is  constant between two parallel plates of the specific domain, we prove that we can
 associate to \eqref{eP} an energy functional $\psi$.
   So a critical point of $\psi$ is a solution of \eqref{eP}.
We denote by $V$ the closure of $\mathcal{V}$ in the space
$(W^{1,k}(\Omega))^2$, where $\mathcal{V}=\{u=(u_1,u_2)^{T}\in
(C^1(\bar{\Omega}))^2| \operatorname{div} u=0, u_i(0,y)=u_i(1,y)
 \text{on }[-1,1]$ for  $i=1,2$ and $u=0$ on $\Gamma_3\cup \Gamma_4\}$.
 We want to extend the work done by  Amrouche, Batchi
and Batina in the linear case with the Laplacian operator see
\cite{a1}, which showed equivalence between the classical and
variational problem, existence and uniqueness of the solution in a
linear case where $f = g = 0$. In this paper we introduce the
k-Laplacian operator to describe the movement of non-Newtonian fluid
with a nonlinear second member, the technique used for the
resolution is a minimization and is completely different to that
given in \cite{a1,a2,a3,a7}. In the case $f=0$,
$g_1(x,y,s_1)=\lambda |s_1|^{k-2}s_1$ and
$g_2(x,y,s_1)=\lambda |s_2|^{k-2}s_2$, we have established
in \cite{c1} that the first eigenfunction $\lambda_1$ of \eqref{eP} is well defined, strictly positive and characterized
by $\lambda_1^{-1}=\sup\{ \int_{\Omega}|u_1|^{k}+|u_2|^{k};
    \int_{\Omega}|\nabla u_1|^{k}+|\nabla u_2|^{k}=1, u\in V\}$.

This article is organized as follows. In Section 2, we prove that
$ u $ is a weak solution of \eqref{eP} if and only if $ u $
satisfies a weak formulation independent of pressure $p$.
In Section 3, we introduce the first eigenvalue of the operator
$-\vec{\Delta}_{k}u+\nabla p$ and as an application, we prove the
existence of solution where the primitive of the nonlinear function
$\vec{g}$ is asymptotically in the left of the first eigenvalue. In
Section 4, we add a condition of monotony for the function
$ \vec{g}$ and we prove the uniqueness of the solution, then we give an
example of such a function $\vec{ g} $ which satisfies the
conditions. Finally we give in Section 5 a conclusion.

\section{Weak formulation of \eqref{eP}}

We establish the equivalence between the classical problem and weak
formulation of problem which is independent of pressure p, This
allows us to find the existence of the weak solution of \eqref{eP} by a new method of minimization.
\begin{definition} \label{def2.1} \rm
A classical solution of \eqref{eP} is a function
$(u,p)^{T}\in (C^2(\Omega)\cap C^1(\bar{\Omega}))^2\times
L^2(\Omega)$ and $\nabla p\in (C(\overline{\Omega}))^2$ which
verify \eqref{eP}.
\end{definition}

\begin{theorem} \label{thm2.2}
If $(u,p)^{T}$ is a classical solution of \eqref{eP}, then
\begin{equation}\label{EE1}
    \sum_{i=1}^2\int_{\Omega}|\nabla u_i|^{k-2}\nabla
u_i.\nabla
v_i-\alpha\int_{-1}^1v_1(0,y)dy=\int_{\Omega}\vec{g}(x,y,u).v
+\int_{\Omega}f.v\quad \forall v\in \mathcal{V}
\end{equation}
\end{theorem}

\begin{proof}
If $(u,p)^{T}$ is a classical solution of \eqref{eP} where
$u=(u_1,u_2)^{T}$, then for $v=(v_1,v_2)^{T}\in \mathcal{V}
$, we multiply the first equation by $v_1$, the second equation by
$v_2$ of \eqref{eP} and we integrate on $\Omega$, we
obtain
$$
 \int_{\Omega}-(\Delta_{k}u_1)v_1+\int_{\Omega}-(\Delta_{k}u_2)v_2+\int_{\Omega}(\nabla p).v=\int_{\Omega}
\vec{g}(x,y,u_1,u_2).v+\int_{\Omega}f.v .
$$
 According to Green's formula, we have for all $1\leq i \leq 2$,
 $$
\int_{\Omega}-(\Delta_{k}u_i)v_i=\int_{\Omega}|\nabla u_i|^{k-2}\nabla u_i.\nabla
 v_i-\int_{\partial\Omega}|\nabla u_i|^{k-2}\nabla u_i.\vec{\eta}v_id\sigma,
 $$
where $\vec{\eta}$ is the unit outward normal to $\partial\Omega$.
On the one hand, we have
\begin{align*}
\int_{\partial\Omega}|\nabla u_i|^{k-2}\nabla
u_i.\vec{\eta}v_id\sigma
&=\int_{\Gamma_1}|\nabla u_i|^{k-2}\nabla
u_i.\vec{\eta}v_id\sigma+\int_{\Gamma_2}|\nabla
u_i|^{k-2}\nabla u_i.\vec{\eta}v_id\sigma
\\
&\quad +\int_{\Gamma_3\cup \Gamma_4}|\nabla u_i|^{k-2}\nabla
u_i.\vec{\eta}v_id\sigma.
\end{align*}
As $v\in \mathcal{V} $,
$$
\int_{\Gamma_3\cup \Gamma_4}|\nabla u_i|^{k-2}\nabla
u_i.\vec{\eta}v_id\sigma=0,
$$
we have on $\Gamma_1, \vec{\eta}=-(1,0)^{T}$ and on $\Gamma_2, \vec{\eta}=(1,0)^{T}$,
thus
$$
\int_{\Gamma_1}|\nabla u_i|^{k-2}\nabla
u_i.\vec{\eta}v_id\sigma=-\int_{-1}^1|\nabla
u_i(0,y)|^{k-2}\frac{\partial u_i}{\partial
x}(0,y)v_i(0,y)dy,
$$
 and
 $$
\int_{\Gamma_2}|\nabla
u_i|^{k-2}\nabla u_i.\vec{\eta}v_id\sigma=\int_{-1}^1|\nabla
u_i(1,y)|^{k-2}\frac{\partial u_i}{\partial
x}(1,y)v_i(1,y)dy.
$$
As $v\in \mathcal{V}$, we have
$v_i(0,y)=v_i(1,y)$, for all $-1\leq y \leq 1$, $i=1,2$.
According to \eqref{eP}, we have
\begin{equation}\label{EE2}
    \int_{\partial\Omega}|\nabla
u_2|^{k-2}\nabla u_2.\vec{\eta}v_2=0.
\end{equation}
On the other hand, as $\frac{\partial
u_1}{\partial y}(0,y)=\frac{\partial u_1}{\partial y}(1,y)$,
thus $\nabla u_1(0,y)=\nabla u_1(1,y)$, we deduce that
$$
\int_{\partial\Omega}|\nabla
u_1|^{k-2}\nabla u_1.\vec{\eta}v_1=0.
$$
 Then, by Green's formula and $v\in V$, we have
$$
\int_{\Omega}\nabla p.v=\int_{\partial \Omega}pv.\vec{\eta}-\int_{\Omega}p \operatorname{div}v,$$
and
\begin{align*}
\int_{\partial\Omega} p v.\vec{\eta}
&= \int_{\Gamma_1}p v.\vec{\eta}+\int_{\Gamma_2}p v.\vec{\eta}
 + \int_{\Gamma_3\cup \Gamma_4}p v.\vec{\eta}\\
&= -\int_{-1}^1p(0,y)v_1(0,y)dy+\int_{-1}^1p(1,y)v_1(1,y)dy\\
&=  \int_{-1}^1(p(1,y)-p(0,y))v_1(0,y)dy\\
&= -\alpha \int_{-1}^1v_1(0,y)dy.
\end{align*}
This proves \eqref{EE1}.
\end{proof}

Now, we study the reciprocal problem; i.e., if $u$ is a weak solution
of \eqref{eP} with some regularity, then $u$ is a
classical solution of \eqref{eP}.

\begin{definition} \label{def2.3} \rm
A weak solution of \eqref{eP} is a function $u\in V$ satisfying
\eqref{EE1}.
\end{definition}

\begin{theorem} \label{thm2.4}
If $u$ is a weak solution of  \eqref{eP} with $u\in
(C^2(\Omega)\cap C^1(\bar{\Omega}))^2$, then there exists
$p\in L^2(\Omega)$ such that $(u, p)^{T}$ is a classical solution
of  \eqref{eP}. Furthermore we have $\nabla p\in
(C(\bar{\Omega}))^2$ and
$-\alpha=p(1,y)-p(0,y)=\int_{0}^1\frac{\partial p}{\partial
x}(t,y)dt$.
\end{theorem}

\begin{proof}
Let $u\in (C^2(\Omega)\cap C^1(\bar{\Omega}))^2$ which
satisfies \eqref{eP}, by Green's formula, we have
\begin{equation}\label{EE3}
    \int_{\Omega}(-\vec{\Delta}_{k}u-\vec{g}(x,y,u_1,u_2)-f).v
+\sum_{i=1}^2\int_{\partial\Omega}|\nabla
u_i|^{k-2}\nabla u_i.\vec{\eta}v_id\sigma
-\alpha\int_{-1}^1v_1(0,y)dy=0
\end{equation}
for all $v \in \mathcal{V}$.
We put $F=\{v\in (\mathfrak{D}(\Omega))^2|\quad \operatorname{div}v=0\}$ where
$\mathfrak{D}(\Omega)$ is the set of all infinitely differentiable
functions with compact support in $\Omega$.
\eqref{EE3} becomes
$$
\int_{\Omega}(-\vec{\Delta}_{k}u-\vec{g}(x,y,u_1,u_2)-f).v=0 \quad \forall v \in F.
$$
By \eqref{g}, as $u\in (C^2(\Omega)\cap
C^1(\bar{\Omega}))^2$ and $f\in (C(\bar{\Omega}))^2$, we have
$-\vec{\Delta}_{k}u-\vec{g}(x,y,u_1,u_2)-f\in
(C(\bar{\Omega}))^2\subset (L^2(\Omega))^2$, according to
Rham's theorem see \cite{a4,a5}, there exists $p\in L^2(\Omega)$
such that $-\vec{\Delta}_{k}u+\nabla
p=\vec{g}(x,y,u_1,u_2)+f \text{ in } (C(\bar{\Omega}))^2$.
Thus
\begin{equation}\label{EE4}
    \sum_{i=1}^2\int_{\partial \Omega}|\nabla
u_i|^{k-2}\nabla u_i.\vec{\eta}v_id\sigma- \alpha
\int_{-1}^1v_1 (0,y)dy= \int_{\Omega}(\nabla p).v\quad \forall v\in
\mathcal{V},
\end{equation}
 where $\nabla p\in (C(\bar{\Omega}))^2$ and
$y\mapsto p(1,y)-p(0,y)=\int_{-1}^1\frac{\partial p}{\partial
x}(t,y)dt \in C^1([-1,1]).$\\
As  $u_i(0,y)=u_i(1,y)$ for all $y\in [-1,1]$, we have
 $ \frac{\partial u_i}{\partial y}(0,y)=\frac{\partial u_i}{\partial y}(1,y)$.
Moreover we know that $\operatorname{div}u=0  \text{ in } \Omega$ and $u\in
(C^1(\bar{\Omega}))^2$, we conclude that $\frac{\partial
u_1}{\partial x}(0,y)=-\frac{\partial u_2}{\partial
y}(0,y)=-\frac{\partial u_2}{\partial y}(1,y)$ for all $y\in
[-1,1]$. Thus $\frac{\partial u_1}{\partial x}(0,y)=\frac{\partial
u_1}{\partial x}(1,y)$ and $\nabla u_1(0,y)=\nabla u_1(1,y)$.
Hence $\int_{\partial \Omega}|\nabla u_1|^{k-2}\nabla
u_1.\vec{\eta}v_1d\sigma=0$.

On the other hand, according to \eqref{EE4}, we have
\begin{align*}
\int_{\partial \Omega}|\nabla
u_2|^{k-2}\nabla u_2.\vec{\eta}v_2d\sigma- \alpha
\int_{-1}^1v_1 (0,y)dy
&=  \int_{\Omega}(\nabla p).v\quad \forall v\in\mathcal{V}\\
&=  \int_{\partial\Omega}p v.\vec{\eta}\\
&=  \int_{-1}^1(p(1,y)-p(0,y))v_1(0,y)dy.
\end{align*}
Therefore,
\begin{equation} \label{EE5}
\begin{split}
&\int_{-1}^1-|\nabla u_2(0,y)|^{k-2}\frac{\partial  u_2}{\partial x}(0,y)v_2(0,y)dy\\
&+\int_{-1}^1|\nabla u_2(1,y)|^{k-2}\frac{\partial  u_2}{\partial x}(1,y)v_2(1,y)dy
   - \alpha \int_{-1}^1v_1(0,y)dy
\\
& = \int_{-1}^1(p(1,y)-p(0,y))v_1(0,y)dy.
\end{split}
\end{equation}
Let $H^{1/2}_{00}(\Gamma_1)$ \cite{a1} be the space defined by
$$
H^{1/2}_{00}(\Gamma_1)=\{ \varphi \in L^2(\Gamma _1);
\exists v\in H^1(\Omega),\text{ with } v|_{\Gamma_3\cup
\Gamma_4}=0,v|_{\Gamma_{ 1} \cup \Gamma _2} =\varphi \}.
$$
Let $\mu\in H^{1/2}_{00}(\Gamma_1)$,
we put $\nu=(0,\mu_2)^{T} $ where
$\mu_2= \begin{cases}
              \mu & \text{on }\Gamma _1\cup \Gamma _2\\
              0 & \text{on } \Gamma_3\cup \Gamma _4.
            \end{cases}
$.
It is clear that $\nu \in (H^{1/2}(\Gamma))^2$ and
$\int_{\partial\Omega}\nu.\vec{\eta}d\sigma=0$,
so  there exists $v\in (H^1(\Omega))^2$ such that
 $\operatorname{div}v=0$ in $\Omega$ and $v=\nu$ on $\Gamma$ (see \cite{a1});
 therefore $v \in V$.
 According to \eqref{EE5}, we have for all $\mu \in H_{00}^{1/2}(\Gamma_1)$,
$$
\int_{-1}^1|\nabla u_2(0,y)|^{k-2}\frac{\partial u_2}{\partial x}(0,y)\mu dy=
\int_{-1}^1|\nabla u_2(1,y)|^{k-2}\frac{\partial u_2}{\partial
x}(1,y)\mu dy,
$$
 thus
 $$ 
|\nabla u_2(0,y)|^{k-2}\frac{\partial u_2}{\partial x}(0,y)=
|\nabla u_2(1,y)|^{k-2}\frac{\partial u_2}{\partial x}(1,y).
$$
 According to \eqref{EE5}, we have
\begin{equation}\label{EE6}
    -\alpha\int_{-1}^1v_1(0,y)dy=\int_{-1}^1(p(1,y)-p(0,y))v_1(0,y)dy.
\end{equation}

On the other hand, let $\gamma \in H_{00}^{1/2}(\Gamma_1)$. Now we consider
$\beta=(\gamma_1,0)^{T}$ where
$\gamma_1= \begin{cases}
              \gamma & \text{on }\Gamma _1\cup \Gamma _2\\
              0 & \text{on } \Gamma_3\cup \Gamma _4.
            \end{cases}$.
We have $\beta \in (H^{1/2}(\Gamma))^2$ and
$\int_{\partial\Omega}\beta.\vec{\eta}d\sigma=0$, so there exists
$v\in (H^1(\Omega))^2$ such that
 $\operatorname{div}v=0$  in $\Omega$ and $v=\beta$ on
 $\Gamma$  \cite{a1}; therefore, $v \in V$. By \eqref{EE6}
 $-\alpha\int_{-1}^1\gamma dy=\int_{-1}^1(p(1,y)-p(0,y))\gamma dy$.
Finally we  prove $p(1,y)-p(0,y)=-\alpha$.
\end{proof}


\section{Existence of a solution}

Let us introduce the energy functional associated with
\eqref{EE1}, $\psi:V\to \mathbb{R}$:
\begin{equation} \label{E6}
\begin{split}
    \psi(u)&=\frac{1}{k}\int_{\Omega}|\nabla u_1|^{k}
   + \frac{1}{k}\int_{\Omega}|\nabla u_2|^{k}
-\alpha \int_{-1}^1u_1(1,y)dy \\
&-\int_{\Omega}F_1(x,y,u_1)-\int_{\Omega}F_2(x,y,u_2)
-\int_{\Omega}f_1u_1-\int_{\Omega}f_2u_2,
\end{split}
\end{equation}
where $F: \Omega\times \mathbb{R}^2\to \mathbb{R}$;
$ F(x,y,u)=F_1(x,y,u_1)+F_2(x,y,u_2)$ and
$F_i(x,y,s)=\int_{0}^{s}g_i(x,y,t)dt$, $i=1, 2$. It is clear that
$\psi$ is well defined,  $C^1\mbox { on } V$ and for all $v\in
\mathcal{V}$
\begin{equation}\label{E7}
   \langle \psi'(u),v\rangle=\sum_{i=1}^2\int_{\Omega}|\nabla u_i|^{k-2}\nabla u_i.\nabla v_i
   - \alpha\int_{-1}^1v_1(0,y)dy
-\int_{\Omega}\vec{g}(x,y,u).v-\int_{\Omega}f.v.
\end{equation}
We know that a critical point of the function $\psi$ is a weak
solution of  \eqref{eP} and reciprocally.
 We assume that the nonlinearity is asymptotically in the left of the first eigenvalue of
 k-Laplacian; i.e.,
\begin{equation}\label{E8}
      F(x,y,s_1,s_2)\leq \frac{\lambda}{k}(|s_1|^{k}+|s_2|^{k})+\rho(x,y),
\end{equation}
where $\rho \in L^1(\Omega) \text{ and } \lambda <\lambda_1$,
$\lambda_1$ is the first eigenvalue of the problem
\begin{equation} \label{eVP}
\begin{gathered}
-\Delta_{k}u_1+\frac{\partial p}{\partial x}= \lambda |u_1|^{k-2}u_1    \quad
 \text{in  }\Omega,\\
-\Delta_{k}u_2+\frac{\partial p}{\partial y}= \lambda |u_2|^{k-2}u_2     \quad
 \text{in  }\Omega,\\
\operatorname{div}u=\frac{\partial u_1}{\partial x}+\frac{\partial u_2}{\partial y}= 0  \quad
 \text{in  }\Omega,\\
u_1(0,y)= u_1(1,y)   \quad\text{on  } [-1,1],\\
u_2(0,y)= u_2(1,y)    \quad \text{on  } [-1,1],\\
\frac{\partial u_1}{\partial x}(0,y)= \frac{\partial u_1}{\partial x}(1,y)
  \quad\text{on  } [-1,1],\\
|\nabla u_2(0,y)|^{k-2}\frac{\partial u_2}{\partial x}(0,y)=
|\nabla u_2(1,y)|^{k-2}\frac{\partial u_2}{\partial x}(1,y)   \quad\text{on  } [-1,1],\\
p(1,y)-p(0,y)= 0 \quad \text{on  } [-1,1].
\end{gathered}
\end{equation}
In \cite{c1}, we have proved that the first eigenvalue $\lambda_1$ of  \eqref{eVP}
is well defined, strictly positive and characterized by
\begin{equation}\label{E9}
    \lambda_1^{-1}=\sup\big\{ \int_{\Omega}|u_1|^{k}+|u_2|^{k};
    \int_{\Omega}|\nabla u_1|^{k}+|\nabla u_2|^{k}=1, u\in V\big\}.
\end{equation}
 So
\begin{equation}\label{E22}
    \lambda_1\int_{\Omega}|u_1|^{k}+|u_2|^{k}\leq
\int_{\Omega}|\nabla u_1|^{k}+|\nabla u_2|^{k}
\quad  \forall u\in V.
\end{equation}

\begin{theorem} \label{thm3.1}
Assume that \eqref{g}$ and \eqref{E8}$ are satisfied, then there
exists $u\in V$ such that $\psi(u)=\inf_{v\in V}\psi(v)$.
Consequently, $u$ is the weak solution of \eqref{eP}.
\end{theorem}

\begin{proof}
As $\psi$ is convex and class $C^1$, it suffices to show that
$\psi$ is coercive; i.e., $\psi(u)\to +\infty$ when
$\|u\|_{W^{1,k}}\to +\infty$. According to \eqref{E22}, the
function $u\mapsto (\int_{\Omega}|\nabla
u_1|^{k})^{1/k}+(\int_{\Omega}|\nabla
u_2|^{k})^{1/k}:=\|u\|_V$ define a norm in $V$.
 We have successively
\begin{equation}\label{EE7}
    \psi(u)=\frac{1}{k}\int_{\Omega}|\nabla u_1|^{k}
   + \frac{1}{k}\int_{\Omega}|\nabla u_2|^{k}
-\alpha \int_{-1}^1u_1(1,y)dy -\int_{\Omega}F(x,y,u) -\langle f,u\rangle,
\end{equation}
\begin{align*}
  \langle f,u\rangle
=\int_{\Omega}f_1u_1+\int_{\Omega}f_2u_2
&\leq  \sum_{i=1}^2\|f_i\|_{L^{k'}}   \|u_i\|_{L^{k}} \\
&\leq  c\sum_{i=1}^2\|\nabla u_i\|_{(L^{k})^2},~\text{ where } c>0\\
&=  c\|u\|_V.
\end{align*}
\begin{align*}
  \lambda\int_{-1}^1u_1(1,y)dy
&\leq |\lambda|\int_{\partial\Omega}|u_1(1,y)|dy \\
   &\leq  |\lambda| c'(\int_{\partial\Omega}|u_1(1,y)|^{k})^{1/k}dy\quad
   \text{(Holder's inequality), where } c'>0 \\
   &\leq  |\lambda| c'(\int_{\Omega}|\nabla u_1|^{k})^{1/k}\quad
   (V \to (L^{k}(\partial\Omega))^2
  \quad \text{ trace theorem })\\
   &=  c''\|u\|_V,\quad  \text{where } c''>0,
\end{align*}
the trace theorem is because $V\subset W_{div}^{1,k}(\Omega)$ and
$W_{div}^{1,k}(\Omega)\to (L^{k}(\partial\Omega))^2$ with
continuous injection.
\begin{align*}\label{E21}
  \int_{\Omega}F(x,y,u)
&\leq \frac{\alpha}{k}\int_{\Omega}|u_1|^{k}+|u_2|^{k}+\int_{\Omega}\rho(x,y) \\
&\leq  \frac{\widetilde{\alpha}}{\lambda_1k}\int_{\Omega}|\nabla u_1|^{k}
+ |\nabla u_2|^{k}+   \int_{\Omega}\rho(x,y),
\end{align*}
 where $\widetilde{\alpha}:=\begin{cases}
  0      & \text{if  } \alpha < 0 \\
  \alpha & \text{if  } \alpha \geq 0.
\end{cases}$
 It follows that
\begin{equation*}
    \psi(u)\geq \frac{1}{k}(1-\frac{\widetilde{\alpha}}{\lambda_1})\int_{\Omega}|\nabla u_1|^{k}+|\nabla u_2|^{k}
-c\|u\|_V-c'\|u\|_V-\int_{\Omega}\rho(x,y).
\end{equation*}
Hence
\begin{equation}\label{E15}
    \psi(u)\geq \frac{1}{k}(1-\frac{\widetilde{\alpha}}{\lambda_1})\|u\|_V^{k}
-c''\|u\|_V-\int_{\Omega}\rho(x,y), \text{ where } c''>0.
\end{equation}
Finally, as $(1-\frac{\widetilde{\alpha}}{\lambda_1})>0$, the
property is proved.
\end{proof}

\section{Uniqueness of the solution}

We assume again that the function $\vec{g}$ is decreasing in the
following sense:
\begin{equation}\label{E19}
(\vec{g}(x,y,\xi)-\vec{g}(x,y,\xi'),\xi-\xi')\leq 0 \quad\text{for all }
\xi,\xi' \in \mathbb{R}^2.
\end{equation}

\begin{theorem} \label{T}
Problem \eqref{EE1} has  a unique solution.
\end{theorem}

\begin{proof}
Let $u$ and $\widetilde{u}$ be two solutions of problem \eqref{EE1}.
 For all $v\in V$, we have
\begin{equation}\label{E16}
    \sum_{i=1}^2\{\int_{\Omega}[(|\nabla u_i|^{k-2}\nabla u_i-|\nabla
\widetilde{u}_i|^{k-2}\nabla \widetilde{u}_i).\nabla v_i
-(g_i(x,y,u_i)-g_i(x,y,\widetilde{u}_i))v_i]\Big\}=0.
\end{equation}
In particular for $v=u-\widetilde{u}$, we have
\begin{equation}\label{E17}
\begin{split}
&\sum_{i=1}^2\Big\{\int_{\Omega}[(|\nabla u_i|^{k-2}\nabla u_i-|\nabla
\widetilde{u}_i|^{k-2}\nabla \widetilde{u}_i).\nabla(
u_i-\tilde{u}_i) \\
&-(g_i(x,y,u_i)-g_i(x,y,\widetilde{u_i}))(u_i-\tilde{u}_i)]\}=0.
\end{split}
\end{equation}
As $ (|\xi|^{k-2}\xi-|\xi'|^{k-2}\xi').(\xi-\xi')> 0$  for all
$\xi\neq \xi'\in \mathbb{R}^2$ and \eqref{E19},
 we deduce that
\begin{equation}\label{E24}
 \sum_{i=1}^2\int_{\Omega}(|\nabla u_i|^{k-2}\nabla u_i-|\nabla
\widetilde{u}_i|^{k-2}\nabla \widetilde{u}_i).\nabla
(u_i-\tilde{u}_i)=0.
\end{equation}
Thus $\nabla u_i=\nabla \tilde{u}_i$, $i=1,2$, therefore
$u_i=\widetilde{u}_i+\epsilon$, where $\epsilon \in \mathbb{R}$.
As $u_i,\tilde{u}_i \in V$, we have $\epsilon =0$, this
completes the proof.
\end{proof}

\subsection*{Example of function $\vec{g}$}
 We consider $\vec{g}(x,y,s)=(g_1(s_1),g_2(s_2))$ for all
$s=(s_1,s_2)\in \mathbb{R}^2$ and $(x,y)^{T}\in \mathbb{R}^2$, where
$$
g_i(s_i)= \begin{cases}
    \frac{\alpha}{2}(\frac{s_i^{k-1}}{1+(s_i^{k})}) 
& \text{if }s_i\geq (k-1)^{1/k} \\
  \frac{\alpha}{2k}(k-1)^{\frac{k-1}{k}} 
& \text{if }  -(k-1)^{1/k}\leq s_i\leq (k-1)^{1/k}  \\
   \frac{\alpha}{2}(\frac{(-s_i)^{k-2}s_i}{1+(-s_i)^{k}})
  +\frac{\alpha}{k}(k-1)^{(k-1)/k}
& \text{if }  s_i\leq -(k-1)^{1/k}.
\end{cases}
$$
We have $g_i(x,y,.)$ is a continuous function, so it has a
primitive $F_i$, for $i=1,2$.


\begin{figure}[ht] 
\begin{center}
\setlength{\unitlength}{1mm} \scriptsize
\begin{picture}(60,54)(0,-4)
\put(0,0){\line(1,0){60}}
\multiput(0,-1)(10,0){7}{\line(0,1){2}}
\put(-2.5,-4){-15}
\put(7.5,-4){-10}
\put(18.5,-4){-5}
\put(29.5,-4){0}
\put(39.5,-4){5}
\put(49,-4){10}
\put(59,-4){15}
\put(30,0){\line(0,1){45}}
\multiput(29,10)(0,10){4}{\line(1,0){2}}
\put(25,9.5){0.1}
\put(25,19.5){0.2}
\put(25,29.5){0.3}
\put(25,39.5){0.4}
\qbezier(0,48.5)(23,46.5)(25,25)
\put(25,25){\line(1,0){10}}
\qbezier(35,25)(37,3.5)(60,1.5)
\end{picture}
\end{center} 
\caption{Graph of $g_i$} \label{fig2}
\end{figure}

For $k=2$ and $\alpha=1$, Figure \eqref{fig2},  we have
$$
g_i(s)=\begin{cases}
    \frac{1}{2}(\frac{s}{1+s^2}) & \text{if }s\geq 1 \\
    \frac{1}{4}   & \text{if }  -1\leq s\leq 1  \\
    \frac{1}{2}(\frac{s}{1+s^2})+\frac{1}{2}
& \text{if }  s\leq 1.
  \end{cases}
$$
\begin{itemize}
  \item [(i)] $g$ satisfies  \eqref{g}, indeed:
If $s_i\geq (k-1)^{1/k}$, then
\[
|g_i(s)|=\frac{\alpha}{2}(\frac{|s_i|^{k-1}}{1+(|s_i|^{k})})
\leq \frac{\alpha}{2}|s_i|^{k-1}.
\]
If $s_i\leq  -(k-1)^{1/k}$, then
\begin{align*}
|g_i(s)|=\frac{\alpha}{2}(\frac{|s_i|^{k-1}}{1+(|s_i|^{k})})+c\leq
\frac{\alpha}{2}|s_i|^{k-1}+c
       \leq \frac{\alpha}{2}|s|^{k-1}+c' \text{ for all } s\in
\mathbb{R}^2,
\end{align*}
 where  $c\in \mathbb{R}$ and 
$c'=c+\frac{\alpha}{2k}(k-1)^{\frac{k-1}{k}}$.

\item [(ii)] We have $F(x,y,s_1,s_2)=F_1(x,y,s_1)+F_2(x,y,s_2)$,
where $F_i(x,y,s)=\int_{0}^{s}g_i(x,y,t)dt$, i=1,2.
So
 \begin{align*}
 F_i(x,y,s)=\int_{0}^{s}g_i(t)dt&\leq
\int_{0}^{s}\frac{\alpha}{2}(\frac{|t|^{k-2}t}{1+|t|^{k}})dt+c, \text{ where } c\in \mathbb{R}.\\\\
&\leq \frac{\alpha}{2k}\ln(1+|s|^{k})+c',\text{ where } c'\in \mathbb{R}.\\
\end{align*}
Thus
\begin{align*}
F(x,y,s_1,s_2)=F_1(x,y,s_1)+F_2(x,y,s_2)
&\leq \frac{\alpha}{2k}\ln(1+|s_1|^{k})+\frac{\alpha}{2k}\ln(1+|s_2|^{k})+c'\\
&\leq  \frac{\alpha}{2k}(|s_1|^{k}+|s_2|^{k})+c'  \\
&\leq \frac{\alpha}{k}(s_1^2+s_1^2)^{\frac{k}{2}}+c',
\end{align*}
consequently $F$ satisfies condition \eqref{E8}.

 \item [(iii)] Finally $\vec{g}$ is decreasing.
For $s_i\geq (k-1)^{1/k}$,
\begin{align*}
g'_i(s_i)&= \frac{\alpha}{2}(\frac{(k-1)s_i^{k-2}(1+(s_i)^{k})-s_i^{k-1}(ks_i^{k-1})}
{(1+(s_i)^{k})^2})\\
&= \frac{\alpha}{2}(\frac{(ks_i^{k-2}+ks_i^{2k-2}-s_i^{k-2}-s_i^{2k-2}-ks_i^{2k-2}}{(1+(s_i)^{k})^2})\\
&= \frac{\alpha}{2}(\frac{s_i^{k-2}(k-1-s_i^{k})}{(1+(s_i)^{k})^2})\leq
0 .
\end{align*}
For $s_i\leq -(k-1)^{1/k}$,
\begin{align*}
g'_i(s)&= \frac{\alpha}{2}
\Big[(-(k-2)(-s_i)^{k-3}s_i+(-s_i)^{k-2})(1+(-s_i)^{k})\\
&\quad +(-s_i)^{k-2}s_i(k(-s_i)^{k-1})\Big] /(1+(-s_i)^{k})^2\\
        &= \frac{\alpha}{2}
\Big[-(k-2)(-s_i)^{k-3}s_i-(k-2)(-s_i)^{2k-3}s_i+(-s_i)^{k-2}\\
&\quad +(-s_i)^{2k-2}+k(-s_i)^{2k-3}s_i\Big] /(1+(-s_i)^{k})^2
\\
        &= \frac{\alpha}{2}
\Big[-(k-2)(-s_i)^{k-3}s_i+2(-s_i)^{2k-3}s_i+(-s_i)^{k-2}\\
&\quad +(-s_i)^{2k-2}\Big] /(1+(-s_i)^{k})^2\\
        &= \frac{\alpha}{2}(-s_i)^{k-3}
\big[-(k-2)s_i+2(-s_i)^{k}s_i+(-s_i)+(-s_i)^{k+1}\big]
/(1+(-s_i)^{k})^2\\
        &= \frac{\alpha}{2}(-s_i)^{k-3}
\big[-ks_i+s_i+(-s_i)^{k}(2s_i-s_i)\big] /(1+(-s_i)^{k})^2 \\
        &= \frac{\alpha}{2}(-s_i)^{k-3}
\big[-(k-1)s_i+(-s_i)^{k}s_i\big] /(1+(-s_i)^{k})^2\\
        &= \frac{\alpha}{2}(-s_i)^{k-3}s_i
\big[-(k-1)+(-s_i)^{k}\big] /(1+(-s_i)^{k})^2 \leq 0.
\end{align*}
\end{itemize}

\subsection*{Conclusion}
We have shown the existence and uniqueness of a solution by a
minimization method. We can also define the other eigenvalues and 
placed them between two consecutive eigenvalues, in this case we must
consider using saddle points.

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 \end{document}