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

\AtBeginDocument{{\noindent\small
\emph{Electronic Journal of Differential Equations},
Vol. 2010(2010), No. 156, pp. 1--8.\newline
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu
\newline ftp ejde.math.txstate.edu}
\thanks{\copyright 2010 Texas State University - San Marcos.}
\vspace{9mm}}

\begin{document}
\title[\hfilneg EJDE-2010/156\hfil Properties of the first eigenvalue]
{Properties of the first eigenvalue of a model for
non Newtonian fluids}

\author[O. Chakrone, O. Diyer, D. Sbibih \hfil EJDE-2010/156\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 6, 2010. Published October 28, 2010.}
\subjclass[2000]{74S05, 76T10}
\keywords{k-Laplacian; eigenvalue; minimization}

\begin{abstract}
 We consider a nonlinear Stokes problem on a  bounded domain.
 We prove the existence of the first eigenvalue which
 is given by a minimization formula. Some properties such as
 strict monotony and the Fredholm alternative are established.
\end{abstract}

\maketitle
\numberwithin{equation}{section}
\newtheorem{theorem}{Theorem}[section]
\newtheorem{proposition}[theorem]{Proposition}
\newtheorem{remark}[theorem]{Remark}

\section{Introduction}

In studies of semi-linear elliptic equations such as
\begin{gather*}
 -\Delta u = f(x,u)+h(x) \quad \text{in } \Omega, \\
 u=0 \quad \text{on } \partial \Omega
\end{gather*}
where $\Omega$ is a bounded domain of $\mathbb{R}^{n}$.
It is usual to impose conditions on
the asymptotic behavior of the nonlinearity $f (x, u)$ in relation
to the spectrum of the linear part of $-\Delta$. In the simplest
situations, we consider $f (x, u)$ as a perturbation of $\lambda
u$. According to that $\lambda$ being or not an eigenvalue of
$-\Delta$, the results of such resonance or non-resonance are
then obtained. Among the classical references on this subject,
we can cite \cite{H} $(\lambda<\lambda_1)$, \cite{D} ($\lambda$
between two consecutive eigenvalues), \cite{L}
$(\lambda=\lambda_1)$. We also cite the Dirichlet problem
\begin{gather*}
-\operatorname{div}(|\nabla u|^{k-2}\nabla u)
= \lambda m(x)|u|^{k-2}u \quad \text{in } \Omega, \\
u=0 \quad \text{on } \partial \Omega.
\end{gather*}
 The first eigenvalue $\lambda_1$ of the Dirichlet
problem is simple and isolated. It was proved that it is the unique
positive eigenvalue having a non negative eigenfunction, see
\cite{a2}.

 Now we consider the eigenvalue problem of a non-linear operator
k-Laplacian. Let $\Omega \subset \mathbb{R}^{2}$ be a bounded
domain with boundary
$\Gamma=\bigcup_{i=1}^{4}\overline{\Gamma_{i}}$, 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
\ref{fig1}. In the interior of this domain, a non-Newtonian liquid
is subjected to pressures of known differences between the two sides
$\Gamma_1$ and $\Gamma_2$.


\begin{figure}[ht] 
\begin{center} 
\setlength{\unitlength}{1mm}
\begin{picture}(70,40)(-7,0)
\put(0,0){\line(0,1){40}}
\put(60,0){\line(0,1){40}}
\qbezier(0,40)(30,25)(60,40)
\put(0,20){\line(1,0){60}}
\qbezier(0,0)(30,15)(60,0)
\multiput(0,18.5)(12,0){6}{\line(0,1){3}}
\put(10.2,16){\scriptsize 0.2}
\put(22.2,16){\scriptsize0.4}
\put(34.2,16){\scriptsize0.6}
\put(46.2,16){\scriptsize0.8}
\put(-5,16){$\Gamma_1$}
\put(62,16){$\Gamma_2$}
\put(28,36){$\Gamma_3$}
\put(28,0.2){$\Gamma_4$}
\put(29,12){$x$}
\put(-4.5,-0.2){\scriptsize $-1$}
\put(-7,9.8){\scriptsize $-0.5$}
\put(-4.5,29.8){\scriptsize 0.5}
\put(-2.5,39.8){\scriptsize 1}
\end{picture} 
\end{center}
\caption{Geometry of channel $\Omega$}
\label{fig1}
\end{figure}


 We denote by $V$ the closure of $\mathcal{V}$ in the space
$W^{1,k}(\Omega)$, where
\begin{align*}
\mathcal{V}=\big\{&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 $[-1,1]$}\\
&\text{for $i=1,2$ and $u=0$ on } \Gamma_{3}\cup \Gamma_{4}\big\}.
\end{align*}
For given $\alpha\in \mathbb{R}$,
 we consider the eigenvalue problem:
 Find $(\lambda,u,p)\in \mathbb{R}\times V\setminus\{0\}\times
 L^{2}(\Omega)$ such that
\begin{equation} \label{VPa}
\begin{gathered}
-\Delta_{k}u_1+\frac{\partial p}{\partial x}
 = \lambda m(x,y) |u_1|^{k-2}u_1 \quad\text{in }\Omega,\\
-\Delta_{k}u_2+\frac{\partial p}{\partial y}
 =\lambda m(x,y) |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)=-\alpha \quad \text{on } [-1,1]
\end{gathered}
\end{equation}
 where the weight function $m(x,y)\in L^{\infty}(\Omega)$
can change the sign and it is positive in a subset of $\Omega$,
$$
-\Delta_{k}u_{i}=-\operatorname{div}(|\nabla u_{i}|^{k-2}\nabla u_{i})
$$
is a k-Laplacian, $i=1,2$ and $1<k<\infty$. In the particular case
$k=2$; i.e., $\Delta_{k}=\Delta$, and $\lambda =0$, the above problem
has been studied by many authors, we cite for example  Amrouche et al.
\cite{a1}. Here, we give an extension to previous work in the
nonlinear case by applying new methods to characterize the first
eigenvalue for this kind of problem such as minimization and as
application is to solving the problem of Fredholm alternative.
 This note is organized as follows. In Section 2, we
give the existence and the characterization of the first
eigenvalue. In Section 3, we prove the Fredholm alternative and we
justify all the given properties. In Section 4, we give a
conclusion.

\section{Existence and characterization of the first eigenvalue}

\begin{theorem}\label{thm1}
There exists one principal eigenvalue $\lambda_1$ for Problem
\eqref{VPa}. It is characterized by
\begin{equation}\label{e1}
 \lambda_1=k\beta+(k-1)\alpha\int_{-1}^1(\varphi^1)_1(0,y)dy,
\end{equation}
 where
$\varphi^1$ is the principal corresponding eigenfunction and
\begin{align*}
\beta=\min\big\{&\frac{1}{k}\int_{\Omega}|\nabla u_1|^k
 +|\nabla u_2|^k-\alpha
\int_{-1}^1u_1(0,y)dy;\\
&\int_{\Omega}m(x,y)(|u_1|^{k}+|u_2|^{k})=1, u\in V\big\},
\end{align*}
\begin{gather*}
\beta=\frac{1}{k}\int_{\Omega}|\nabla (\varphi^1)_1|^k+|\nabla (\varphi^1)_2|^k-\alpha
\int_{-1}^1(\varphi^1)_1(0,y)dy,\\
\int_{\Omega}m(x,y)[|(\varphi^1)_1|^k+|(\varphi^1)_2|^k]=1.
\end{gather*}
Furthermore, for all $\alpha$, $\alpha'\in \mathbb{R}$ such that
$\alpha\alpha'>0$, $\lambda_1(\alpha)$ is an eigenvalue of Problem
\eqref{VPa}.
\end{theorem}

For the sake of simplicity,  in what follows, we denote
$\lambda_1=\lambda_1(m)=\lambda_1(\alpha,m)=\lambda_1(\alpha)$.

\begin{theorem}\label{thm2}

(i) $\lambda_1$ defined by
\begin{equation}\label{2}
 \frac{1}{\lambda_1}=\max\{\int_{\Omega}m(x,y)(|u_1|^k+|u_2|^k);
 \ \int_{\Omega}|\nabla u_1|^k+|\nabla u_2|^k=1, u\in V\}
\end{equation}
is the first eigenvalue of Problem \eqref{VPa} with $\alpha=0$
 in the sense $\Sigma\subset[\lambda_1,+\infty[$, where
$\Sigma$ is the set of the positive eigenvalues.
Moreover, $u$ is the eigenfunction associated
with $\lambda_1$ if and only if $\int_{\Omega}|\nabla
u_1|^k+|\nabla
u_2|^k-\lambda_1\int_{\Omega}m(x,y)(|u_1|^k+|u_2|^k)=
 \inf\{\int_{\Omega}|\nabla v_1|^k+|\nabla v_2|^k-\lambda_1\int_{\Omega}m(x,y)(|v_1|^k+|v_2|^k)\ \ v\in V \}
=0$.

(ii) $\lambda_1(.)$ is strictly monotone in
$L^{\infty}(\Omega)$; i.e., if $m_1, m_2$ are in the set
$$
\big\{m\in
L^{\infty}(\Omega);\operatorname{measure}\{(x,y)\in \Omega;\;
 m(x,y)>0\}\neq 0\big\}
$$
such that $m_1(x,y)<m_2(x,y)$
a.e., then $\lambda_1(m_1)>\lambda_1(m_2)$.

(iii) $\lambda_1(.)$ is continuous in $L^{\infty}(\Omega)$.
\end{theorem}

\begin{theorem}[Fredholm alternative] \label{thm3}
 Suppose that $\lambda<\lambda_1$,
then for $f\in (C(\overline{\Omega}))^2$ the problem:
 Find $(u,p)\in V\times L^{2}(\Omega)$ such that
\begin{equation} \label{Q}
\begin{gathered}
-\Delta_{k}u_1+\frac{\partial p}{\partial x}
 =\lambda m(x,y) |u_1|^{k-2}u_1+f_1 \quad\text{in }\Omega,\\
-\Delta_{k}u_2+\frac{\partial p}{\partial y}
= \lambda m(x,y) |u_2|^{k-2}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 \quad \text{on } [-1,1]
\end{gathered}
\end{equation}
has a solution.
\end{theorem}

\section{Proof of the main theorems}

For proving Theorem \ref{thm1}, we need the following results.

\begin{proposition}\label{prop3.1}
$u=(u_1,u_2)^{t}$ is a solution of problem:
Find $(u,p)\in V\setminus\{0\}\times L^{2}(\Omega)$  such that
\begin{equation} \label{P1}
\begin{gathered}
-\Delta_{k}u_1+\frac{\partial p}{\partial x}=f_1 \quad
\text{in }\Omega,\\
-\Delta_{k}u_2+\frac{\partial p}{\partial y}=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 \quad\text{on } [-1,1]
\end{gathered}%
\end{equation}
where $f=(f_1,f_2)^t\in (C(\overline{\Omega}))^{2}$,
if and only if $u$ is a solution of problem:
Find $u\in V$  such that
\begin{equation} \label{P1a}
\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}(f_1v_1+f_2v_2)
\end{equation}
for all $v\in V$.
\end{proposition}

\begin{remark} \label{rmk3.2} \rm
If we take $f_{i}=\lambda m(x,y)|u_{i}|^{k-2}u_{i}$, $i=1,2$. Then
$(\lambda,u,p)$ is a solution of \eqref{VPa} if and only if
$$
\sum_{i=1}^{2}\int_{\Omega}|\nabla u_{i}|^{k-2}\nabla
u_{i}\nabla v_{i}-
\alpha\int_{-1}^1v_1(0,y)dy=\lambda\sum_{i=1}^{2}
\int_{\Omega}m(x,y)|u_{i}|^{k-2}u_{i}v_{i}
$$
 for all $v\in V$. For a proof of this remark see \cite{c}.
\end{remark}


\begin{proof}[Proof of Theorem \ref{thm1}]
 Since for all $v\in V$, $v=0$
on $\Gamma_{3}\cup \Gamma_{4} $, $u\in V\to(\int_{\Omega}|\nabla
u_1|^k+|\nabla u_2|^k)^{1/k}$ define a norm in $V$
according to the Poincar\'e inequality in the space $V$:
There exists $c>0$  such that
\begin{equation}\label{E1}
c\int_{\Omega} |u_1|^k+|u_2|^k\leq
\int_{\Omega}|\nabla u_1|^k+|\nabla u_2|^k.
\end{equation}
Suppose by contradiction that for all $n\in \mathbb{N}^{*}$
there exists $u_{n}=(u_1^{n},u_2^{n})^{t}\in V$ such
that
$\frac{1}{n}\int_{\Omega}|u_1^{n}|^{k}+|u_2^{n}|^{k}>\int_{\Omega}|\nabla
u_1^{n}|^{k}+|\nabla u_2^{n}|^{k}$, then we put
$$
v_{n}=(v_1^{n},v_2^{n})^{t}
$$
where
$v_{i}^{n}=\frac{u_{i}^{n}}{(\int_{\Omega}|u_1^{n}|^{k}
+|u_2^{n}|^{k})^{1/k}}$,
$i=1,2$. Thus $\int_{\Omega}|v_1^{n}|^{k}+|v_2^{n}|^{k}=1$, so
\begin{equation}\label{E4}
 \frac{1}{n}>\int_{\Omega}|\nabla v_1^{n}|^{k}+|\nabla
v_2^{n}|^{k}.
\end{equation}
 As $(v_{n})_{n}$ is bounded in $V$, we have for a
subsequence also denoted $(v_{n})_{n}$, $v_{n}\rightharpoonup v$ in
$V$ and $v_{n}\rightarrow v$ in $L^{k}(\Omega)$. Therefore
$\|v\|_{L^{k}(\Omega)}=1$, so $v\neq 0$. By passing to the limit in
\eqref{E4}, we have
$$
0\geq \liminf_{n}\int_{\Omega}|\nabla
v_1^{n}|^{k}+|\nabla v_2^{n}|^{k}\geq \int_{\Omega}|\nabla
v_1|^{k}+|\nabla v_2|^{k}.
$$
So $\int_{\Omega}|\nabla v_1|^{k}
=\int_{\Omega}|\nabla v_2|^{k}=0$, hence $v=cst$,
therefore $v=0$ because $v=0$ on $\Gamma_{3}\cup \Gamma_{4}$, is a
contradiction.
 By using \eqref{E1} and the
Holder's inequality, we easily prove that $\beta$ is well defined.
Let $(u_{n})=((u_{n1},u_{n2}))$ be a suitable minimization of
$\beta$, then we have
$$
\beta=\lim_{n\to\infty}\frac{1}{k}\int_{\Omega}|\nabla u_{n1}|^k+|\nabla u_{n2}|^k-\alpha\int_{-1}^1u_{n1}(0,y)dy
$$
and
$$
\int_{\Omega}m(x,y)(| u_{n1}|^k+| u_{n2}|^k)=1.
$$
The sequence $(X_{n}):=(\frac{1}{k}\int_{\Omega}|\nabla
u_{n1}|^k+|\nabla u_{n2}|^k)$ is bounded, if we have not for a
subsequence, also denoted $(X_{n})$, $X_{n}\rightarrow +\infty$.
Using the Holder's inequality and the fact that $V \hookrightarrow
L^{k}(\Gamma_1)$ we get
$$
\alpha \int_{-1}^1u_{n1}(0,y)dy\leq
|\alpha|c(\frac{1}{k}\int_{\Omega}|\nabla u_{n1}|^k+|\nabla
u_{n2}|^k)^{1/k}=|\alpha|cX_{n}^{1/k}
$$
where $c\in\mathbb{R}$. Thus
$\frac{1}{k}\int_{\Omega}\sum_{i=1}^{2}|\nabla
u_{ni}|^{k}-\alpha\int_{-1}^1u_{n1}(0,y)dy\geq
X_{n}-|\alpha|cX_{n}^{1/k}$, this prove that
$\beta=+\infty$, which is impossible.
 According to the reflexivity of the space $V$ and the
 compact injections
$V\hookrightarrow L^k(\Omega)$  and
$V\hookrightarrow L^k(\Gamma_1)$, there exists a subsequence of
$(u_{n})=((u_{n1},u_{n2}))$, which is also denoted by
$(u_{n})=((u_{n1},u_{n2}))$, such that
\begin{gather*}
u_{n}=(u_{n1},u_{n2}) \rightharpoonup \varphi^1
 =((\varphi^1)_1,(\varphi^1)_2) \quad\text{in } V,\\
u_{n}=(u_{n1},u_{n2}) \rightarrow \varphi^1
 =((\varphi^1)_1,(\varphi^1)_2)\quad \text{in }L^k(\Omega),\\
u_{n1}|_{\Gamma_1} \rightarrow (\varphi^1)_1|_{\Gamma_1}
\quad \text{in }L^k(\Gamma_1).
\end{gather*}
Hence
$\int_{\Omega}m(x,y)(|(\varphi^1)_1|^k+|(\varphi^1)_2|^k)=1$,
consequently $\varphi^1\neq 0$ and
\begin{align*}
\beta
&\leq \frac{1}{k}\int_{\Omega}|\nabla
 (\varphi^1)_1|^k+|\nabla (\varphi^1)_2|^k
-\alpha\int_0^1(\varphi^1)_1(0,y)dy\\
&\leq \frac{1}{k}\int_{\Omega}|\nabla u_{n1}|^k
 +|\nabla u_{n2}|^k-\alpha\int_0^1u_{n1}(0,y)dy,
\end{align*}
so
$$
\beta= \frac{1}{k}\int_{\Omega}|\nabla
(\varphi^1)_1|^k+|\nabla
(\varphi^1)_2|^k-\alpha\int_0^1(\varphi^1)_1(0,y)dy.
$$
 On the other hand, for all $t>0$, $v=(v_1,v_2)\in V$, we
 put $w_{t}=(w_{thm1},w_{t2})$ where
\begin{gather*}
w_{thm1}=\frac{(\varphi^1)_1+tv_1}{(\int_\Omega m(x,y)(|(\varphi^1)_1+tv_1|^k+
|(\varphi^1)_2+tv_2|^k))^{1/k}},\\
w_{t2}=\frac{(\varphi^1)_2+tv_2}{(\int_\Omega
 m(x,y)(|(\varphi^1)_1+tv_1|^k+
|(\varphi^1)_2+tv_2|^k))^{1/k}},
\end{gather*}
 so that $\int_\Omega m(x,y)(|w_{thm1}|^k+|w_{t2}|^k)=1$ and
\begin{align*}
\beta&=\frac{1}{k}\int_{\Omega}|\nabla (\varphi^1)_1|^k+|\nabla
(\varphi^1)_2|^k-\alpha\int_0^1(\varphi^1)_1(0,y)dy\\
&\leq \frac{1}{k}\int_{\Omega}|\nabla w_{thm1}|^k+|\nabla
w_{t2}|^k-\alpha\int_0^1w_{thm1}(0,y)dy.
\end{align*}
By developing to order 1 for $t\to 0$ and by applying the same
reasoning  to $(-v)$, we obtain
\begin{align*}
&\sum_{i=1}^{2}\int_{\Omega}|\nabla
(\varphi^1)_{i}|^{k-2}\nabla (\varphi^1)_{i}\nabla
v_{i}-\alpha\int_0^1v_1(0,y)dy\\
&=(k\beta+(k-1)\alpha\int_0^1(\varphi^1)_1(0,y)dy) \times
(\sum_{i=1}^{2}\int_{\Omega}m(x,y)|(\varphi^1)_{i}|^{k-2}
(\varphi^1)_{i}v_{i}).
\end{align*}
Now we suppose that $\alpha\alpha'> 0$. We put
$\overline{\varphi^1}=(\overline{\varphi^1}_1,
\overline{\varphi^1}_2)$,
where $\overline{\varphi^1}_{i}=\eta \varphi^1_{i}$ with
$\eta^{k-1}=\frac{\alpha'}{\alpha}$. Then, by replacing in the
equation $(P_1(\alpha))$, we obtain
$$
\sum_{i=1}^{2}\int_{\Omega}|\nabla
\overline{\varphi^1}_{i}|^{k-2}\nabla
\overline{\varphi^1}_{i}\nabla
v_{i}-\alpha'\int_0^1v_1(0,y)dy=\lambda_1(\alpha)\sum_{i=1}^{2}\int_{\Omega}m(x,y)
|\overline{\varphi^1}_{i}|^{k-2}\overline{\varphi^1}_{i}v_{i},
$$
which completes the proof of Theorem \ref{thm1}.
\end{proof}

\begin{proof}[Proof of Theorem \ref{thm2}]
 (i) It is easy to prove that
 for $\alpha=0$, $\lambda_1$ is an eigenvalue of
Problem \eqref{VPa} with $\alpha=0$
 and $u\neq 0$ is a eigenfunction
if and only if $ \sum_{i=1}^{2}\int_{\Omega}|\nabla
u_{i}|^k-\lambda_1(m)\int_{\Omega}m(x,y)(|u_1|^k+|u_2|^k)=0=\inf\{\sum_{i=1}^{2}\int_{\Omega}|\nabla
v_{i}|^k-\lambda_1(m)\int_{\Omega}m(x,y)(|v_1|^k+|v_2|^k);\ \
v\in V \}$.
 The proofs of (ii) and (iii) follow from (i).
\end{proof}

\begin{proof}[Proof of Theorem \ref{thm3}]
It is clear that Problem \eqref{Q} is equivalent to the weak
formulation: Find $u\in V$ such that
\begin{equation} \label{Q'}
\begin{aligned}
&\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\\
&=\lambda\sum_{i=1}^{2}\int_{\Omega}m(x,y)|u_{i}|^{k-2}u_{i}v_{i}
+\sum_{i=1}^{2}\int_{\Omega}f_{i}v_{i} \quad
\forall  v\in V.
\end{aligned}
\end{equation}
We consider the energy functional defined on $V$,
\begin{equation}\label{E3}
 \Phi(u)=\frac{1}{k}\sum_{i=1}^{2}\int_{\Omega}|\nabla u_{i}|^k-\alpha
\int_{-1}^1u_1(0,y)dy-\frac{\lambda}{k}\sum_{i=1}^{2}\int_{\Omega}m(x,y)|u_{i}|^k-
\sum_{i=1}^{2}\int_{\Omega}f_{i}u_{i}.
\end{equation}
We verify that $u$ is a solution of Problem \eqref{Q'} if and
only if $u$ is a critical point of the function $\Phi$. For the
existence, it suffices to prove that there exists $u\in V$ such
that
$$
\Phi(u)=\inf_{v\in V}\, \Phi(v).
$$
The functional $\Phi$ is continuous and convex, it suffices to show
that $\Phi$ is coercive, indeed for all $u\in V$, using Theorem
\ref{thm2}, we obtain
\begin{equation}\label{t6}
\lambda_1\int_{\Omega}m(x,y)(|u_1|^k+|u_2|^k )
\leq \int_{\Omega}|\nabla u_1|^k+|\nabla u_2|^k.
\end{equation}
Since the function $u\mapsto (\int_{\Omega}|\nabla
u_1|^{k})^{1/k}+(\int_{\Omega}|\nabla
u_2|^{k})^{1/k}:=\|u\|_{V}$ defines a norm in $V$, then we have
successively
\begin{align*}
 \sum_{i=1}^{2}\int_{\Omega}f_{i}u_{i}
&\leq \sum_{i=1}^{2}\|f_{i}\|_{L^{k'}}
 \|u_{i}\|_{L^{k}} \\
&\leq  c\sum_{i=1}^{2}\|\nabla u_{i}\|_{L^{k}}
= c\|u\|_{V},
\end{align*}
where $c>0$.
 \begin{align*}
 \alpha\int_{-1}^1u_1(0,y)dy
&\leq |\alpha|\int_{\partial\Omega}|u_1|d\sigma \\
&\leq  |\alpha| c'(\int_{\partial\Omega}|u_1|^{k}d\sigma)^{1/k} \quad
\text{(Holder's inequality)} \\
&\leq  |\alpha| c'(\int_{\Omega}|\nabla u_1|^{k})^{1/k} \quad
( V \hookrightarrow L^{k}(\partial\Omega)
 \text{ a continuous injection})\\
&= c''\|u\|_{V},
\end{align*}
where $c'$ and $c''$ are positive.
\begin{align*}
 \frac{\lambda}{k}\sum_{i=1}^{2}\int_{\Omega}m(x,y)|u_{i}|^k
&\leq \frac{\tilde{\lambda}}{k}\sum_{i=1}^{2}
 \int_{\Omega}m(x,y)|u_{i}|^k \\
&\leq  \frac{\widetilde{\lambda}}{\lambda_1k}\int_{\Omega}
|\nabla u_1|^{k}+ |\nabla u_2|^{k},
 \end{align*}
where $\widetilde{\lambda}:= \begin{cases}
 0 & \text{if } \lambda < 0 \\
 \lambda & \text{if } \lambda \geq 0.
 \end{cases}$
According to \eqref{E3}, we obtain
$$
\Phi(u)\geq
\frac{1}{k}(1-\frac{\tilde{\lambda}}{\lambda_1})
\int_{\Omega}|\nabla u_1|^k+|\nabla u_2|^k
-c'''\|u\|_{V},
$$
 where $c'''>0$.
Thus
$$
\Phi(u)\geq \frac{1}{k}(1-\frac{\tilde{\lambda}}{\lambda_1})
\|u\|_{V}^{k}-c'''\|u\|_{V},
$$
where $c'''>0$.
Since $\lambda<\lambda_1$, we deduce that $\Phi(u)\to +\infty$
when $\|u\|_V\to +\infty$, so we have proved the existence.
\end{proof}

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\end{document}