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\title{Existence and non-existence of a positive solution \\for $(p,q)$-Laplacian with singular weights }

\author{{\sc Abdellah ZEROUALI\thanks{abdellahzerouali@yahoo.fr}} \\
        Centre Régional des Métiers de l'\'{E}ducation et de la Formation, Fès, Maroc\\
        {\sc Belhadj KARIM\thanks{karembelf@gmail.com}}\\
                 Faculté des Sciences et Techniques,
                  Erraachidia, Maroc}

\maketitle

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\begin{abstract}
We use the Hardy-Sobolev inequality to study existence and
non-existence results for a positive solution of the quasilinear
elliptic problem $$ -\triangle_{p}u-\mu\triangle_{q}u = \lambda[
m_{p}(x)|u|^{p-2}u+\mu m_{q}(x)|u|^{q-2}u]\hbox{ in }\Omega$$
driven by nonhomogeneous operator
 $(p,q)$-Laplacian with singular weights under the Dirichlet boundary
 condition. We also prove that in the case where  $\mu>0$ and with
 $1<q<p<\infty$ the results are completely different from those
 for the usual eigenvalue problem for the $p$-Laplacian with
 singular weight under the Dirichlet boundary
 condition, which is retrieved when $\mu=0$. Precisely, we show
 that when $\mu>0$ there exists an interval of eigenvalues for our eigenvalue problem.


\end{abstract}

\medskip \noindent\textbf{Keywords:} Nonlinear eigenvalue problem;
$(p,q)$-Laplacian; singular weight; indefinite weight the
Hardy-Sobolev inequality; Harnack inequality .


\medskip

\noindent\textbf{2000 Mathematics Subject Classification:} 35J20,
35J62, 35J70, 35P05, 35P30.


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\section {Introduction}


\hspace{1cm}Consider the $(p,q)$-Laplacian eigenvalue problem
 \begin{equation*}
(P_{\lambda, \mu})\left\{
\begin{array}{rclclll}
\hbox{To find } (u, \lambda)&\in& (W_{0}^{1,
p}(\Omega)\setminus\{0\})\times\mathbb{R} &\hbox{ such that}\\
-\triangle_{p}u-\mu\triangle_{q}u &=& \lambda[
m_{p}(x)|u|^{p-2}u+\mu m_{q}(x)|u|^{q-2}u]      &\hbox{ in }\Omega,\\
u &=& 0    &\hbox{ on } \partial \Omega
 \end{array}
\right.
\end{equation*}
where $\Omega$ is a bounded domain in $\mathbb{R}^{N}$ with
piecewise $C^{1}$ boundary $\partial \Omega$,
$\lambda\in\mathbb{R}$, $\mu\geq 0$ and $1<q<p<\infty$. For
$r=p,q$, $\triangle_{r}u= div(|\nabla u|^{r-2}\nabla u)$ indicate
the $r$-Laplacian and the weight $m_{r}$ may be unbounded and
change sign. As in \cite{M}, we assume for $r=p,q$ that
$m_{r}\delta^{\tau}\in L^{a}(\Omega)$ with
$\delta(x)=dist(x,\partial\Omega)$ and $m_{r}^{+}\not \equiv 0$,
where $a$, $r$ and $\tau$ satisfy one of the following conditions:

\begin{itemize}
\item [(H1):]    $\partial\Omega$ is piecewise $C^{1}$, $0<\tau
<1$, $\frac{r}{1-\tau }\leq a$ and $a\leq \frac{Nr}{N-\tau r}$ if
$N>\tau r$;
\item[(H2)] $\partial\Omega$ is piecewise $C^{1}$, $0<\tau
<1$, $r< \frac{N}{1-\tau}\leq a$;
\item[(H3):]    $\partial\Omega$ is piecewise $C^{1}$, $\tau =1$  and
$a=\infty$;
\item[(H4):]    $\Omega$ is any bounded domain, $\tau =0$  and
$a=\infty$.
\end{itemize}

The problem $(P_{\lambda, \mu})$ comes, for example, from a
general reaction diffusion system

\begin{equation} \label{P2}
u_{t}= div(D(u)\nabla u) + c(x,u),
\end{equation}
where $D(u)= (|\nabla u|^{p-2}+\mu|\nabla u|^{q-2})$. This system
has a wide range of applications in physics and related sciences
like chemical reaction design \cite{Aris}, biophysics \cite{Fife}
and plasma physics \cite{Struwe}. In such applications, the
function u describes a concentration, the first term on the
right-hand side of \eqref{P2} corresponds to the diffusion with a
diffusion coefficient $D(u)$; whereas the second one is the
reaction and relates to source and loss processes. Typically, in
chemical and biological applications, the reaction term $c(x; u)$
has a polynomial form with respect to the concentration.

Our problem was addressed in \cite{MT} for domains with boundary
$C^{2}$ and bounded weights, when only the condition (H4) holds
true. These work proved that in the case where $\mu>0$, there
exists an interval of eigenvalues. The authors proved the
existence of positive solutions in resonant cases. A non-existence
result is also given. Here we will assume that the boundary
$\partial\Omega$ is a piecewise $C^{1}$ and singular weights
$m_{r}$ $(r=p,q)$ which satisfy one of the conditions (H1), (H2),
(H3) or (H4). Our work represent developments of the study
performed in \cite{MT} because we prove all results of this paper
by considering others conditions that represent the singularity of
the domain and the weights. Our main tool is the Hardy-Sobolev
inequality, see Lemma \ref{L1} in preliminary section.\\
 Many authors have studied the
nonhomogeneous operator $(p,q)$-Laplacian (see \cite{MP, Si, Y,
Z}). However, there are few results one the eigenvalue problems
for the $(p,q)$-Laplacian. In \cite{BB, BBe}, the authors
established the existence of the principal eigenvalue and of a
continuous family of  eigenvalues for problem
$$-\triangle_{p}u -\triangle_{q}u= \lambda
g(x)|u|^{p-2}u \hbox{ in } \mathbb{R}^{N}.$$
 where $g$ is a bounded positive weight. Eigenvalue
problem for a $(p,2)$-Laplacian was studied in \cite{B}. The
existence of non trivial solution for the following Dirichlet
equation is proved in \cite{CD}
$$-\triangle_{p}u -\mu\triangle u= \lambda
|u|^{p-2}u+g(u) \hbox{ in } \Omega,~~u=0 \mbox{ on
}\partial\Omega,$$ in the case where $p>2$, $g\in C^{1}$ and
$\lambda\not\in \sigma(-\Delta_{p})$, where $\sigma(-\Delta_{p})$
is the spectrum of $(-\Delta_{p})$. Under the Neumann boundary
condition, \cite{Mih} determined the set of eigenvalues for the
equation
$$-\triangle_{p}u -\triangle u= \lambda
u \hbox{ in } \Omega,$$ where $p>2$. In \cite{Ta}, M. Tanaka has
completely described the generalized eigenvalue $\lambda$  for
which the following equation
$$-\triangle_{r}u -\mu\triangle u_{r*}= \lambda m_{r}
|u|^{r-2}u \hbox{ in } \Omega, ~~u=0 \mbox{ on }\partial\Omega.$$
has a positive solution, where $1<r\neq r*<\infty$ and $\mu >0$.

We recall that a value $\lambda\in \mathbb{R}$ is an eigenvalue of
problem $(P_{\lambda, \mu})$ if and only if there exists $u \in
W_{0}^{1, p}(\Omega)\backslash \{0\}$  such that
\begin{equation}\label{P5}
\int_{\Omega}(|\nabla u|^{p-2}+ \mu|\nabla u|^{q-2})\nabla u
\nabla \varphi dx = \lambda\left[\int_{\Omega}(m_{p}(x)|u|^{p-2} +
\mu m_{q}(x)|u|^{q-2}) u\varphi dx\right]
\end{equation}
for all $\varphi\in W_{0}^{1, p}(\Omega)$. $u$ is then called an
eigenfunction of $\lambda.$

Letting $\mu \rightarrow 0^{+}$, our problem $(P_{\lambda, \mu})$
turns into the $(p-1)$-homogeneous problem known as the usual
weighted eigenvalue problem for the $p$-Laplacian with singular
weight $m_{p}$:

\begin{equation*}
(P_{\lambda, m_{p}})\left\{
\begin{array}{rclclll}
-\triangle_{p}u &=& \lambda m_{p}(x)|u|^{p-2}u      &\hbox{ in
}\Omega,\\ u &=& 0    &\hbox{ on } \partial \Omega
 \end{array}
\right.
\end{equation*}
Moreover, after multiplying our equation $(P_{\lambda, \mu})$ by
$1/\mu$ and then letting $\mu\rightarrow +\infty$, we obtain the
$(q-1)$-homogeneous equation:
\begin{equation*}
(P_{\lambda, m_{q}})\left\{
\begin{array}{rclclll}
-\triangle_{q}u &=& \lambda m_{q}(x)|u|^{q-2}u      &\hbox{ in
}\Omega,\\ u &=& 0    &\hbox{ on } \partial \Omega
 \end{array}
\right.
\end{equation*}

Nonlinear eigenvalue problem $(P_{\lambda, m_{r}})$, where $r=p,q$
and with bounded weight have been studied by several authors, for
example (see \cite{A, C, G, O, S, Struwe}). These works proved
that there exists a first eigenvalue $\lambda_{1}(r,m_{r})>0,$
where
\begin{equation} \label{equ0}
\lambda_{1}(r, m_{r}):=\inf\left\{\frac{1}{r}\int_{\Omega}|\nabla
u|^{r}dx; u\in W^{1,r}_{0}(\Omega) \mbox{ and
}\frac{1}{r}\int_{\Omega}m_{r}(x)|u|^{r}dx=1\right\},
\end{equation}

which is simple in the sense that two eigenfunctions corresponding
to it are proportional. Moreover, the corresponding first
eigenfunction $\phi_{1}(r, m_{r})$ can be assumed to be positive.
It was also shown (see \cite{A}) that $\lambda_{1}(r, m_{r})$ is
simple and isolated. Recently, the problem $(P_{\lambda, m_{r}})$
with singular weight $m_{r}$ satisfying the conditions (H1), (H2),
(H3) or (H4), was studied in \cite{M}. The authors use the
Hardy-Sobolev inequality to characterize the first eigenvalue. In
some cases it is shown that $\lambda_{1}(r, m_{r})>0$ is positive
simple, isolated and has a nonnegative corresponding eigenfunction
$\phi_{1}(r, m_{r})\in L^{\infty}(\Omega)$. Higher eigenvalues, in
particular the second one, are also determined.

The plan of this paper is the following. In Section 2, which has a
preliminary character, we collect some results concerning the
first eigenvalue $\lambda_{1}(r, m_{r})$ of problem  $(P_{\lambda,
m_{r}})$, where $r=p,q$. In Section 3, we study Rayleigh quotient
for our problem $(P_{\lambda, \mu})$. In contrast to homogeneous
case, we prove that if $\lambda_{1}(p, m_{p})\neq \lambda_{1}(q,
m_{q})$ or $\phi_{1}(p, m_{p})\neq k\phi_{1}(q, m_{q})$ for every
$k>0$, then the infimum in Rayleigh quotient is not attained. We
also show nonexistence results for positive solutions of the
eigenvalue problem $(P_{\lambda, \mu})$ formulated as Theorem
\ref{thm2}. Our existence results for positive solutions of the
eigenvalue problem $(P_{\lambda, \mu})$ are presented in Section
4. After studying the non-resonant cases (Theorem \ref{thm21})
which prove that when $\mu>0$ there exists an interval of positive
eigenvalues for the problem $(P_{\lambda, \mu})$, we present the
resonant cases in Theorem \ref{thm3}.

\section{Preliminaries}

Throughout this paper $\Omega$ will be a bounded domain of
$\mathbb{R}^{N}$ with piecewise $C^{1}$ boundary,\\ $1 <q <p
<\infty$ and $r=p \mbox{ or }q$. We will always assume for $r=p,q$
that $m_{r}\delta^{\tau}\in L^{a}(\Omega)$ with
$\delta(x)=dist(x,\partial\Omega)$ and $m_{r}^{+}\not \equiv 0$,
where $a$, $r$ and $\tau$ satisfy one of the conditions
 (H1), (H2), (H3) or (H4).
\begin{rk}\rm
Condition (H4) implies $m_{r}\delta^{\tau}=m_{_{r}}\in
L^{\infty}(\Omega)$, including results of the previously cited
paper \cite{MT}. Here $\partial\Omega$ is piecewise $C^{1}$ except
for (H4).
\end{rk}
We will write
$\|u\|_{r}:=\left(\int_{\Omega}|u|^{r}dx\right)^{1/r}$ for the
$L^{r}(\Omega)-$norm and $W_{0}^{1,r}(\Omega)$ will denote the
usual Sobolev
space with usual norm $\|\nabla u\|_{r}$.\\

In the sequel, we collects some results relative to the first
eigenvalue $\lambda_{1}(r, m_{r})$ defined by \eqref{equ0} and its
corresponding normalized eigenfunction $\phi_{1}(r, m_{r})$.
 The following lemma concerns the
Hardy-Sobolev inequality proved in \cite{K}, which characterize
the first eigenvalue $\lambda_{1}(r, m_{r})$ of problem
$(P_{\lambda, m_{r}})$. This inequality is our main tool in this
paper.

\begin{lem} \label{L1} \cite{K}
Let $0\leq\tau \leq1$ and $s$  such that
$\frac{1}{s}=\frac{1}{r}-\frac{1-\tau}{N}$ for $r<N $ and
$\frac{1}{s}=\frac{\tau}{r}$ for $r\geq N$ . If $\partial\Omega$
is piecewise $C^{1}$, then
$\left\|\frac{u}{\delta^{\tau}}\right\|_{L^{s}(\Omega)}\leq C
\|\nabla u\|_{L^{r}(\Omega)}$ for all $u\in W_{0}^{1, r}(\Omega),$
where $\delta(x)=dist(x,\partial\Omega)$ and $C=C(N, r, \tau)>0 $
is a constant. In the case $s=r=p,q$, no regularity on
$\partial\Omega$ is needed.
\end{lem}
 We give now an example of the weight $m_{r}$ such that $m_{r}\delta^{\tau}\in L^{a}(\Omega)$ with $m_{r}^{+}\not\equiv 0$,
 where $a$, $\tau$ and $r$ satisfying the condition (H2).
 \begin{example}\label{Ex1} \rm
The weight $m_{r}(x)=\delta(x)^{-\beta}= (1 - |x|)^{-\beta}$ is
admissible in the open unit ball
 of $\mathbb{R}^{N}$ (i.e. $\Omega=B_{1}(0)$). For
 $1/2<\beta<25/42,$ $p=3/2,$ $N=3,$ $\tau=1/2$ and $a=21/2,$ we
 have $m_{r}\not\in L^{N/r}(\Omega)=L^{2}(\Omega),$ but
 $m_{r}\delta^{\tau}\in L^{a}(\Omega)= L^{21/2}(\Omega)$.
 \end{example}


To use Harnack inequality as in \cite{M} and \cite{Tr}, we make
now the following definitions involving locally integrable
weights. Let $\epsilon(\rho)$ be a smooth function defined for
$\rho
> 0$ such that
\begin{equation}\label{equ2}
\lim_{\rho \rightarrow 0^{+}}\epsilon(\rho)=0 \mbox{ and }
\int_{0}^{\rho*}\frac{\epsilon(\rho)}{\rho}d\rho<\infty,
\end{equation}
for some $\rho*>0.$ We denote by $K_{x_{0}}(\rho)$ an
$N$-dimensional cube contained in $\Omega$ whose edges are of
length $\rho$ and are parallel to the coordinate axes. We define
\begin{equation}\label{equ3}
L^{t}_{\epsilon(\rho)}(\Omega)= \{u\in L^{t}(\Omega):
\|u\|_{t,\epsilon(\rho),\Omega}<\infty\}, \mbox{ where }
\|u\|_{t,\epsilon(\rho),\Omega}= \sup_{x_{0}\in \Omega,
\rho>0}\frac{\|u\|_{L^{t}(K_{x_{0}}(\rho)\cap
\Omega)}}{\epsilon(\rho)}.
\end{equation}

\begin{rk} \rm
The weight $m_{r}$ in Example \ref{Ex1} is such that $m_{r}\in
L^{N/r}_{\epsilon(\rho)}(\Omega)$, but $m_{r} \not\in
L^{s}(\Omega)$ for $s>N/r$ if $1<r\leq N.$
\end{rk}

The following theorem guarantees the simplicity and isolation of
$\lambda_{1}(r, m_{r})$, where $r=p,q$. This result is proved by
M. Montenegro and S. Lorca in \cite{M}. To ensure positiveness of
$\phi_{1}(r, m_{r})$, the authors apply the Harnack inequality of
\cite{Tr}.

\begin{thm}\label{thm1} \cite{M}
If one supposes $\partial\Omega$ is piecewise $C^{1}$ and
$m_{r}\delta^{\tau}\in L^{a}(\Omega)$ with $m_{r}^{+}\not\equiv
0,$ where $a$, $\tau$ and $r$ satisfy (H1), (H2), (H3) or (H4),
then the number $\lambda_{1}(r, m_{r})$ is attained by some
$\phi_{1}(r, m_{r})\in W_{0}^{1,p}(\Omega)$, where we may assume
that $\phi_{1}(r, m_{r})\geq 0$ a.e. in $\Omega$, $\phi_{1}(r,
m_{r})^{+}\not\equiv
0.$ Moreover $\lambda_{1}(r, m_{r})$ is positive and isolated.\\
If in addition one assumes $ m_{r}\in L^{1}(\Omega)$ for $r>N$ or
$m_{r}\in L^{N/p}_{\epsilon(\rho)}(\Omega)$ for $1<r\leq N$, then
the first eigenvalue $\lambda_{1}(r, m_{r})$ is simple and  any
positive eigenvalue other than $\lambda_{1}(r, m_{r})$ has no
positive eigenfunctions.
\end{thm}


\section{Rayleigh quotient and non-existence results}
\subsection{Rayleigh quotient for the problem $(P_{\lambda,
\mu})$} This subsection concerns the Rayleigh quotient for our
problem $(P_{\lambda, \mu})$.

\begin{rk}\rm \label{rk1}
We start by pointing out to find a solution for the problem
$(P_{\lambda, \mu})$ is equivalent to seek a solution in the case
$\mu=1$, that is to solve the problem $(P_{\lambda, 1})$.\\Indeed,
if $u$ is a solution of $(P_{\lambda, 1})$, then multiplying
equation $(P_{\lambda, 1})$ by $s^{p-1}$ for $s>0$ we deduce that
$v=su$ is a solution for problem $(P_{\lambda, \mu =s^{p-q}})$.\\
Conversely, let $u$ be a solution of problem $(P_{\lambda, \mu})$.
Then it follows that $v=\mu^{1/q-p}u$ is a solution of
$(P_{\lambda, 1})$.
\end{rk}
We introduce now the functionals $A$ and $B$ on
$W^{1,p}_{0}(\Omega)$ by
\begin{equation}\label{equ11}
A(u):=\frac{1}{p}\int_{\Omega}|\nabla
u|^{p}dx+\frac{1}{q}\int_{\Omega}|\nabla u|^{q}dx
\end{equation}
\begin{equation}\label{equ12}
B(u):=\frac{1}{p}\int_{\Omega}m_{p}(x)|u|^{p}dx+\frac{1}{q}\int_{\Omega}m_{q}(x)|u|^{q}dx
\end{equation}
for all $u\in W^{1,p}_{0}(\Omega)$.
\begin{prop}\label{prop1}
(i) The functional $A$ is well defined and sequently weakly lower
semi-continuous.\\
(ii) If  $m_{r}\delta^{\tau}\in L^{a}(\Omega)$ and $m_{r}^{+}\not
\equiv 0$ ($r=p,q$), where $a$, $r$ and $\tau$ satisfy one of the
conditions
 (H1), (H2), (H3) or (H4), then the
functional $B$ are also well defined and weakly continuous.
\end{prop}

\begin{proof}
(i) The functional $A$ is well defined. indeed, since $\Omega$
bounded and $q<p$, we have $W^{1,p}_{0}(\Omega)\subset
W^{1,q}_{0}(\Omega)$. Then for all $u\in W^{1,p}_{0}(\Omega)$,
$\frac{1}{p}\int_{\Omega}|\nabla u|^{p}dx<\infty$ and
$\frac{1}{q}\int_{\Omega}|\nabla u|^{q}dx<\infty$. It follows that
$A(u)<\infty$. It is clear that $A$ is sequently weakly lower
semi-continuous.\\
(ii) The functional $B$ is also well defined. Indeed, for $u\in
W^{1,p}_{0}(\Omega)$, by H\"{o}lder's inequality with
$\frac{1}{a}+\frac{1}{b}+\frac{r-1}{r}=1$, where $r=p,q$ and
$b=b(r)=\frac{ar}{a-r}$ if $a<\infty$ and $b=b(r)=r$ if
$a=\infty$, we obtain
\begin{align*}
\frac{1}{r}\int_{\Omega}m_{r}(x)|u|^{r}dx &\leq
\frac{1}{r}\int_{\Omega}m_{r}\delta^{\tau}\frac{|u|}{\delta^{\tau}}|u|^{r-1}dx\\&\leq
\frac{1}{r}\|m_{r}\delta^{\tau}\|_{a}\left\|\frac{u}{\delta^{\tau}}\right\|_{b}\||u|^{r-1}\|_{\frac{r}{r-1}}.
\end{align*}
Under assumption (H1) and Lemma \ref{L1}, we have
$$\left\|\frac{u}{\delta^{\tau}}\right\|_{b}\|\leq C\|\nabla u\|_{\tau
b}<\infty, \mbox{ because } \tau b\leq r.$$ Condition (H2) and
Lemma \ref{L1} imply
$$\left\|\frac{u}{\delta^{\tau}}\right\|_{b}\|\leq C\|\nabla u\|_{\frac{bN}{N+b(1-\tau)}}<\infty, \mbox{ because } bN<r(N+b(1-\tau).$$
By virtue  Lemma \ref{L1} and (H3) or (H4),
$$\left\|\frac{u}{\delta^{\tau}}\right\|_{b}\|\leq C\|\nabla u\|_{r}<\infty.$$
Finally, in each case $B(u)<\infty$ and $C=C(r,N,a,\tau)>0$ is a
constant that may differ in each case except if $a=\infty$,
$C=C(N,r)>0$.\\ Let us now show that $B$ is weakly continuous. If
$u_{n}\rightarrow u$ weakly in $W^{1, p}_{0}(\Omega)$, up to a
subsequence, $u_{n}\rightarrow u$ strongly in $L^{r}(\Omega)$ and
$|u_{n}|^{r-1}\rightarrow |u|^{r-1}$ strongly in
$L^{r/r-1}(\Omega)$ with $r=p,q$, because $\|\nabla u_{n}\|_{p}$
is bounded and the embedding $W^{1, p}_{0}(\Omega)\subset
L^{r}(\Omega)$ is compact. Hence by H\"{o}lder's inequality, we
have
\begin{align*}
|B(u_{n})-B(u)|&\leq\frac{1}{p}\left|\int_{\Omega}m_{p}(x)(|u_{n}|^{p}-|u|^{p})dx\right|+\frac{1}{q}\left|\int_{\Omega}m_{q}(x)(|u_{n}|^{q}-|u|^{q})dx\right|\\
&\leq
C_{p}\|m_{p}\delta^{\tau}\|_{a}\left\|\frac{|u_{n}|+|u|}{\delta^{\tau}}\right\|_{b(p)}\left\||u_{n}|^{p-1}-|u|^{p-1}\right\|_{p/p-1}\\&+
C_{q}\|m_{q}\delta^{\tau}\|_{a}\left\|\frac{|u_{n}|+|u|}{\delta^{\tau}}\right\|_{b(q)}\left\||u_{n}|^{q-1}-|u|^{q-1}\right\|_{q/q-1}\\&
\rightarrow 0.
\end{align*}
because $\left\||u_{n}|^{r-1}-|u|^{r-1}\right\|_{r/r-1}\rightarrow
0$ and under (H1), (H2), (H3) or (H4) the norms
$\|m_{r}\delta^{\tau}\|_{a}$ and
$\left\|\frac{|u_{n}|+|u|}{\delta^{\tau}}\right\|_{b(r)}$ are
bounded. The constant $C(r)=C/r>0$, where $C$ comes from the
inequality $|\alpha^{r}-\beta^{r}|\leq C(\alpha
+\beta)|\alpha^{r-1}-\beta^{r-1}|$ for positive numbers $\alpha$
and $\beta$. To be precise, $C=1$ if $r\geq 2$ and $C>r/(r-1)$ if
$1<r<2$. Thus $B$ is weakly continuous.
\end{proof}

Define now the Rayleigh quotient
\begin{equation}\label{equ}
\lambda^{*}=\inf\left\{\frac{A(u)}{B(u)}; u\in
W^{1,p}_{0}(\Omega), B(u)>0\right\}.
\end{equation}

\begin{prop}\label{prop2}
One assumes the same conditions as for Theorem \ref{thm1}. \\If
$\lambda_{1}(p, m_{p})\neq \lambda_{1}(q, m_{q})$ or $\phi_{1}(p,
m_{p})\neq k\phi_{1}(q, m_{q})$, for every $k>0$. Then  the
infimum in \ref{equ} is not attained.

\end{prop}
For the proof of Proposition \ref{prop2}, we will need to use the
following lemma.

\begin{lem} \label{L2} The
infimum in \ref{equ} verifies
$$\lambda^{*}=\min\{\lambda_{1}(p, m_{p}), \lambda_{1}(q,
m_{q})\}$$

\end{lem}
\begin{proof} For sufficiently large $k>0$, using \eqref{equ11} et \eqref{equ12}, we have
$$B(k\phi_{1}(p,m_{p}))=
k^{q}\left(k^{p-q}+\frac{1}{q}\int_{\Omega}m_{q}(x)\phi_{1}^{q}(p,m_{p})dx\right)>0.$$
and $$A(k\phi_{1}(p,m_{p}))=k^{p}\left(\lambda_{1}(p,
m_{p})+\frac{1}{q}k^{q-p}\int_{\Omega}|\nabla
\phi_{1}(p,m_{p})|^{q}dx\right).$$ By \eqref{equ}, we find
\begin{align*}
\lambda^{*}&\leq\frac{A(k\phi_{1}(p,m_{p}))}{B(k\phi_{1}(p,m_{p}))}\\&=\frac{\lambda_{1}(p,
m_{p})+\frac{1}{q}k^{q-p}\int_{\Omega}|\nabla
\phi_{1}(p,m_{p})|^{q}dx}{1+\frac{1}{q}k^{q-p}\int_{\Omega}m_{q}(x)\phi_{1}^{q}(p,m_{p})dx}\\
&\rightarrow\lambda_{1}(p, m_{p}) \mbox{ as }k\rightarrow +\infty,
\mbox{ because } q<p.
\end{align*}
It follows that $\lambda^{*}\leq \lambda_{1}(p, m_{p})$. On the
other hand, we also have
\begin{align*}
\lambda^{*}&\leq\frac{A(k\phi_{1}(q,m_{q}))}{B(k\phi_{1}(q,m_{q}))}\\&=\frac{\lambda_{1}(q,
m_{q})+\frac{1}{p}k^{p-q}\int_{\Omega}|\nabla
\phi_{1}(q,m_{q})|^{p}dx}{1+\frac{1}{p}k^{p-q}\int_{\Omega}m_{p}(x)\phi_{1}^{p}(q,m_{q})dx}\\
&\rightarrow\lambda_{1}(q, m_{q}) \mbox{ as }k\rightarrow 0^{+},
\mbox{ because } q<p.
\end{align*}
Thus, we obtain $\lambda^{*}\leq \lambda_{1}(q, m_{q})$, which
implies that $$\lambda^{*}\leq\min\{\lambda_{1}(p, m_{p}),
\lambda_{1}(q, m_{q})\}$$ Conversely, suppose by contradiction
that $\lambda^{*}<\min\{\lambda_{1}(p, m_{p}), \lambda_{1}(q,
m_{q})\}$. Then, by \eqref{equ}, there exists $ u\in
W^{1,p}_{0}(\Omega)$ such that $B(u)>0$ and
$$\frac{A(u)}{B(u)}<\min\{\lambda_{1}(p, m_{p}), \lambda_{1}(q,
m_{q})\}.$$ We distinguish three cases.\\ \textbf{Case (i)}:
Suppose that $\int_{\Omega}m_{p}|u|^{p}dx>0$ and
$\int_{\Omega}m_{q}|u|^{q}dx\leq 0.$ There hold $pB(u)\leq
\int_{\Omega}m_{p}|u|^{p}dx$ and $pA(u)\geq\|\nabla u\|_{p}^{p}$.
Using the definition of $\lambda_{1}(p, m_{p})$, we arrive at the
contradiction.
\begin{equation}\label{equ13}
\min\{\lambda_{1}(p, m_{p}), \lambda_{1}(q,
m_{q})\}>\frac{A(u)}{B(u)}\geq\frac{\|\nabla
u\|_{p}^{p}}{\int_{\Omega}m_{p}|u|^{p}dx}\geq \lambda_{1}(p,
m_{p}).
\end{equation}
\textbf{Case (ii)}: Suppose that
$\int_{\Omega}m_{p}|u|^{p}dx\leq0$ and
$\int_{\Omega}m_{q}|u|^{q}dx
>0.$ Using the definition of $\lambda_{1}(q,
m_{q})$, we also arrive at contradiction
\begin{equation}\label{equ14}
\min\{\lambda_{1}(p, m_{p}), \lambda_{1}(q,
m_{q})\}>\frac{A(u)}{B(u)}\geq\frac{\|\nabla
u\|_{q}^{q}}{\int_{\Omega}m_{q}|u|^{q}dx}\geq \lambda_{1}(q,
m_{q}).
\end{equation}
\textbf{Case (iii)}: Suppose now that
$\int_{\Omega}m_{p}|u|^{p}dx>0$ and $\int_{\Omega}m_{q}|u|^{q}dx >
0.$ It follows from the definition of $\lambda_{1}(r, m_{r}),$
where $r=p,q$ that $$\|\nabla u\|_{r}^{r}\geq \lambda_{1}(r,
m_{r})\int_{\Omega}m_{r}|u|^{r}dx.$$ Hence we get
\begin{equation}\label{equ15}
\begin{aligned}
A(u)&\geq \frac{\lambda_{1}(p,
m_{p})}{p}\int_{\Omega}m_{p}|u|^{p}dx + \frac{\lambda_{1}(q,
m_{q})}{q}\int_{\Omega}m_{q}|u|^{q}dx\\&\geq\min\{\lambda_{1}(p,
m_{p}), \lambda_{1}(q, m_{q})\}B(u).
\end{aligned}
\end{equation}
Against the assumption in our reasoning by contradiction.

\end{proof}

\begin{proof}[Proof of Proposition \ref{prop2}.] By contradiction,
we suppose that there exists $ u\in W^{1,p}_{0}(\Omega)$ such that
$B(u)>0$ and $\frac{A(u)}{B(u)}=\lambda^{*}$. Using Lemma
\ref{L2}, we give
\begin{equation}\label{equ16}
\frac{A(u)}{B(u)}=\lambda^{*}=\min\{\lambda_{1}(p, m_{p}),
\lambda_{1}(q, m_{q})\}.
\end{equation}
We argue by considering the three cases in the proof of Lemma
\ref{L2}.\\
\textbf{Case (i)}: By \eqref{equ13}, \eqref{equ16} and
$\int_{\Omega}m_{q}|u|^{q}dx \leq 0,$ we have
$$\lambda^{*}=\frac{A(u)}{B(u)}\geq \frac{\|\nabla
u\|_{p}^{p}+\frac{p}{q}\|\nabla
u\|_{q}^{q}}{\int_{\Omega}m_{p}|u|^{p}dx}\geq\frac{\|\nabla
u\|_{p}^{p}}{\int_{\Omega}m_{p}|u|^{p}dx}\geq \lambda_{1}(p,
m_{p})\geq\lambda^{*}.$$ We deduce that
$$\|\nabla
u\|_{p}^{p}=\lambda_{1}(p, m_{p})\int_{\Omega}m_{p}|u|^{p}dx
\mbox{ and } \|\nabla u\|_{q}=0.$$ This contradicts the fact that
$u\neq 0$.\\
\textbf{Case (ii):} similarly, By \eqref{equ14}, \eqref{equ16} and
$\int_{\Omega}m_{p}|u|^{p}dx \leq 0,$ we get
$$\|\nabla
u\|_{q}^{q}=\lambda_{1}(q, m_{q})\int_{\Omega}m_{q}|u|^{q}dx
\mbox{ and } \|\nabla u\|_{p}=0.$$ Which contradicts $u\neq 0$.\\
\textbf{Case (iii):} In this case, using \eqref{equ15} and
\eqref{equ16}, we find
$$A(u)= \lambda^{*}B(u)= \frac{\lambda_{1}(p,
m_{p})}{p}\int_{\Omega}m_{p}|u|^{p}dx + \frac{\lambda_{1}(q,
m_{q})}{q}\int_{\Omega}m_{q}|u|^{q}dx. $$ It follows
$$[\lambda_{1}(p,
m_{p})-\lambda^{*}]\int_{\Omega}m_{p}|u|^{p}dx + [\lambda_{1}(q,
m_{q})-\lambda^{*}]\int_{\Omega}m_{q}|u|^{q}dx=0.
$$
Since $\int_{\Omega}m_{p}|u|^{p}dx> 0$,
$\int_{\Omega}m_{q}|u|^{q}dx>0$ and
$\lambda^{*}=\min\{\lambda_{1}(p, m_{p}),\lambda_{1}(q, m_{q})\}$,
we have $$\lambda^{*}=\lambda_{1}(p, m_{p})=\lambda_{1}(q,
m_{q}).$$ We deduce that $$\frac{\|\nabla u\|_{p}^{p}
}{\int_{\Omega}m_{p}|u|^{p}dx}=\lambda_{1}(p,
m_{p})=\lambda_{1}(q, m_{q})=\frac{\|\nabla u\|_{q}^{q}
}{\int_{\Omega}m_{q}|u|^{q}dx}.$$ Hence, the simplicity of
eigenvalue $\lambda_{1}(r, m_{r})$ (for $r=p,q$), given by Theorem
\ref{thm1}, guarantees that $u=t\phi_{1}(p, m_{p})=s\phi_{1}(q,
m_{q})$ for some $t\neq 0 $ and $s\neq 0$. The hypothesis of
proposition is thus contradicted.
\end{proof}

\subsection{Non-existence results} This subsection studies a non-existence results for the problem
$(P_{\lambda, 1})$ , so for the problem $(P_{\lambda, \mu})$. This
work is inspired from \cite{MT}. The following theorem is the main
result of this section.
\begin{thm} \label{thm2}
One assumes the same conditions as for Theorem \ref{thm1}.
\begin{enumerate}
\item If it holds $0<\lambda<\lambda^{*}$, then the problem $(P_{\lambda, 1})$ has no non-trivial
solutions.
\item Moreover, if one of the following conditions holds
\begin{itemize}
\item [(i)] $\lambda_{1}(p, m_{p})\neq \lambda_{1}(q, m_{q})$;
\item [(ii)] $\phi_{1}(p, m_{p})\neq k\phi_{1}(q, m_{q})$, for every $k>0,$
\end{itemize}
then the problem $(P_{\lambda, 1})$, with $\lambda=\lambda^{*}$
has no non-trivial solutions.
\end{enumerate}
\end{thm}

\begin{rk}\rm It is easy to see that if $\lambda_{1}(p, m_{p})=\lambda_{1}(q, m_{q})$ and $\phi_{1}(p,
m_{p})= k\phi_{1}(q, m_{q})$, for some $k>0,$ then $\phi_{1}(p,
m_{p})$ and $\phi_{1}(q, m_{q})$ are positive solutions of problem
$(P_{\lambda, 1})$, with $\lambda=\lambda_{1}(p, m_{p})=
\lambda_{1}(q, m_{q})$.
\end{rk}
\begin{proof}[Proof of Theorem \ref{thm2}.] Assume by
contradiction that there exists a non-trivial solution $u$ of
problem $(P_{\lambda, 1}).$ Then, for every $s>0$, we have that
$v=su$ is a non-trivial solution of problem $(P_{\lambda,
s^{p-q}})$ (see Remark \ref{rk1}). Choose $s^{p-q}=p/q$ and then
act with $su$ as test function on the problem $(P_{\lambda,
s^{p-q}})$. We arrive at
\begin{equation}\label{equ17}
0<pA(su)=p\lambda B(su).
\end{equation}
From the estimate \eqref{equ17} and according to Lemma \ref{L2},
we obtain
$$\lambda=\frac{A(su)}{B(su)}\geq \lambda^{*}=\min\{\lambda_{1}(p, m_{p}), \lambda_{1}(q,
m_{q})\}.$$ This contradiction yields the first assertion of the
theorem.\\ The second part of the Theorem \ref{thm2} follows by
Proposition \ref{prop2}.
\end{proof}
\section{Existence results}
\subsection{ Non-resonant cases} The following theorem is
our main existence result for problem $(P_{\lambda, 1})$ (or
$(P_{\lambda, \mu})$) in the non-resonant cases. This result prove
that there exists an interval of positive eigenvalues for the
problem $(P_{\lambda, 1})$ (or $(P_{\lambda, \mu})$, with
$\mu>0$).
\begin{thm} \label{thm21}
In addition to the hypotheses of Theorem \ref{thm1} one supposes
that $\lambda_{1}(p, m_{p})\neq \lambda_{1}(q, m_{q})$. If
$$\min\{\lambda_{1}(p, m_{p}), \lambda_{1}(q, m_{q})\}<\lambda
<\max\{\lambda_{1}(p, m_{p}), \lambda_{1}(q, m_{q})\},$$ then the
problem $(P_{\lambda, 1})$ has at least one positive solution.
\end{thm}
\begin{rk} \label{rk20}\rm
The proof of Theorem \ref{thm21} reduces to provide a non-trivial
critical point of the functional $I_{\lambda, m_{p}, m_{q}}$
defined for all $ u\in W^{1,p}_{0}(\Omega)$ by
$$I_{\lambda, m_{p}, m_{q}}(u):=A(u)-\lambda B(u^{+}),$$ where  $u^{+}=\max\{u,0\}$ and $A,$
$B$ are the functionals defined by \eqref{equ11} and
\eqref{equ12}. This non-trivial critical point $u$ of $I_{\lambda,
m_{p}, m_{q}}$ is a non-negative solution of the problem
$(P_{\lambda, 1})$. We can check that $u\in L^{\infty}(\Omega)$
(see Remark 1.7 in \cite{M}). Then the Harnack inequality of
\cite{Tr} can be applied to ensure positiveness of $u$.

The argument will be
separately developed in two cases: \\
(a) $\lambda_{1}(q, m_{q})<\lambda <\lambda_{1}(p, m_{p})$.\\
(b)$\lambda_{1}(p, m_{p})<\lambda <\lambda_{1}(q, m_{q})$.\\
In case (a), we apply the minimum principle and in case (b), we
use the mountain pass theorem.
\end{rk}

\textbf{Proof of case (a).} By Proposition \ref{prop1},  $A$ is
sequently weakly lower semi-continuous and $B$ is weakly
continuous. It follows that $I_{\lambda, m_{p}, m_{q}}$ is
sequently weakly lower semi-continuous. It is remains to show that
$I_{\lambda, m_{p},
m_{q}}$ is coercive and bounded from below.\\
We distinguish two cases:\\
(i) For $u\in W_{0}^{1,p}(\Omega)$ such that
$\int_{\Omega}m_{p}(u^{+})^{p}dx\leq 0$. A calculation similar to
that in the proof of Proposition \ref{prop1} for $r=q$ gives
\begin{equation} \label{equ175}
\begin{aligned}
I_{\lambda, m_{p}, m_{q}}(u)&\geq \frac{1}{p}\|\nabla
u\|^{p}_{p}-\frac{\lambda}{q}\|m_{q}\delta^{\tau}\|_{a}\left\|\frac{u^{+}}{\delta^{\tau}}\right\|_{b(q)}\left\|(u^{+})^{q-1}\right\|_{q/q-1}\\&\geq
\frac{1}{p}\|\nabla
u\|^{p}_{p}-\frac{C\lambda}{q}\|m_{q}\delta^{\tau}\|_{a}\|\nabla
u\|_{q}\|u\|_{q}^{q-1}\\&\geq \frac{1}{p}\|\nabla
u\|^{p}_{p}-\frac{CC'\lambda}{q}\|m_{q}\delta^{\tau}\|_{a}\|\nabla
u\|_{q}^{q}\\&\geq \frac{1}{p}\|\nabla
u\|^{p}_{p}-\frac{CC'C''\lambda}{q
}\|m_{q}\delta^{\tau}\|_{a}\|\nabla u\|^{q}_{p},
\end{aligned}
\end{equation}
where $C, C', C''>0$ are the constants given respectively by the
Hardy-Sobolev inequality (see Lemma \ref{L1}), the
compact embedding $W^{1, q}_{0}(\Omega)\subset L^{q}(\Omega)$ and the continuous embedding $W^{1, p}_{0}(\Omega)\subset W^{1, q}_{0}(\Omega)$.\\
(ii) For $u\in W_{0}^{1,p}(\Omega)$ such that
$\int_{\Omega}m_{p}(u^{+})^{p}dx> 0$. Fix $\epsilon >0$ such that
\begin{equation} \label{equ18}
(1-\epsilon)\lambda_{1}(p,m_{p})>\lambda,
\end{equation}
which is possible due to the assumption in case (a). By the
definition of $\lambda_{1}(p,m_{p})$ we have
$$\|\nabla
u^{+}\|^{p}_{p}\geq
\lambda_{1}(p,m_{p})\int_{\Omega}m_{p}(u^{+})^{p}dx.$$ Then taking
into account \eqref{equ18}, we derive
\begin{equation}\label{equ19}
\begin{aligned}
I_{\lambda, m_{p}, m_{q}}(u)&\geq \frac{\epsilon}{p}\|\nabla
u\|^{p}_{p}+
\frac{(1-\epsilon)\lambda_{1}(p,m_{p})-\lambda}{p}\int_{\Omega}m_{p}(u^{+})^{p}dx\\&
-\frac{C\lambda}{q}\|m_{q}\delta^{\tau}\|_{a}\|\nabla
u\|_{q}\|u\|_{q}^{q-1}\\&\geq \frac{1}{p}\|\nabla
u\|^{p}_{p}-\frac{CC'C''\lambda}{q
}\|m_{q}\delta^{\tau}\|_{a}\|\nabla u\|^{q}_{p}.
\end{aligned}
\end{equation}
Since $q<p$, it follows from \eqref{equ175} and \eqref{equ19} that
the functional $I_{\lambda, m_{p}, m_{q}}$ is coercive and bounded
from below. Consequently, by minimum principle, there exists a
global minimizer $u_{0}$ of $I_{\lambda, m_{p}, m_{q}}$. Finally,
$u_{0}\neq 0$, indeed it suffices to prove that $I_{\lambda,
m_{p}, m_{q}}(u_{0})=\min_{W^{1,p}_{0}(\Omega)}I_{\lambda, m_{p},
m_{q}}<0$. For sufficiently small $k>0$, we have

$$I_{\lambda, m_{p}, m_{q}}(k\phi_{1}(q, m_{q}))= k^{q}\left
(\frac{k^{p-q}}{p}\|\nabla \phi_{1}(q, m_{q})
\|^{p}_{p}-\frac{\lambda
k^{p-q}}{p}\int_{\Omega}m_{p}\phi_{1}^{p}(q,
m_{p})dx+\frac{\lambda_{1}(q, m_{q})-\lambda}{q}\right).$$ Then
$I_{\lambda, m_{p}, m_{q}}(k\phi_{1}(q, m_{q}))<0$, because
$\lambda_{1}(q, m_{q})<\lambda,$ which completes the proof of case
(a).\\

\textbf{Proof of case (b).} We organize the proof of this case in
several lemmas. In the sequel, we design by $o(1)$ a quantity
tending to 0 as $n\rightarrow \infty$.

\begin{lem}\label{lem1}
 Suppose that $m_{r}\delta^{\tau}\in L^{a}(\Omega)$ and $m_{r}^{+}\not \equiv 0$
($r=p,q$), where $a$, $r$ and $\tau$ satisfy one of the conditions
 (H1), (H2), (H3) or (H4). In addition, we assume that $ m_{r}\in L^{1}(\Omega)$ for $r>N$ or
$m_{r}\in L^{N/p}_{\epsilon(\rho)}(\Omega)$ for $1<r\leq N$. If
$\lambda\neq\lambda_{1}(p, m_{p})$, then the functional
$I_{\lambda, m_{p}, m_{q}}$ satisfies the Palais-Smale condition
on $W^{1,p}_{0}(\Omega)$.
\end{lem}
\begin{proof}
Let $(u_{n})\subset W^{1,p}_{0}(\Omega)$ be a sequence such that
$$I_{\lambda, m_{p}, m_{q}}(u_{n})\rightarrow c \mbox{ for }
c\in\mathbb{R} \mbox{ and }I'_{\lambda, m_{p},
m_{q}}(u_{n})\rightarrow 0 \mbox{ in }(W^{1,p}_{0}(\Omega))^{*}
\mbox{ as } n\rightarrow \infty.$$ Let us first show that the
sequence $u_{n}$ is bounded.  It is sufficient only to prove the
boundedness of $\|u_{n}\|_{p}$ because
\begin{equation}\label{EE}
\|\nabla u_{n}\|_{p}^{p}\leq pc+ o(1)+ C_{p}\|\nabla u_{n}\|_{p}
\|u_{n}\|_{p}^{p-1}+ \frac{p\alpha\beta C_{q}}{q}\|\nabla
u_{n}\|_{p}\|u_{n}\|_{p}^{q-1},
\end{equation}
where  $\alpha$, $\beta$ and $C$ are respectively the constants in
inequalities $\|u\|_{q}\|\leq \alpha\|u\|_{p}$, $\|\nabla
u\|_{q}\|\leq \beta\|\nabla u\|_{p}$(since $\Omega$ bounded and
$q<p$) and $\left\|\frac{u}{\delta^{\tau}}\right\|_{b(r)}\|\leq
C\|\nabla u\|_{r}$ (in each case (H1), (H2), (H3) or (H4), see the
proof of Proposition \ref{prop1}) and $C_{r}=\lambda
C\|m_{r}\delta^{\tau}\|_{a}$ ($r=p,q$). Suppose by contradiction
that $\|u_{n}\|_{p}\rightarrow\infty$ and let
$v_{n}:=\frac{u_{n}}{\|u_{n}\|_{p}}$. The sequence $v_{n}$ is
bounded in $W^{1,p}_{0}(\Omega)$. Indeed, dividing \eqref{EE} by
$\|u_{n}\|_{p}^{p}$, we have
\begin{equation}\label{EEE}
\begin{aligned}
\|\nabla v_{n}\|_{p}^{p}&\leq  o(1)+ C_{p}\|\nabla v_{n}\|_{p}+
\frac{p\alpha\beta}{q\|u_{n}\|_{p}^{p-q}}C_{q}\|\nabla
v_{n}\|_{p}\\&= o(1)+ (C_{p}+o(1))\|\nabla v_{n}\|_{p}.
\end{aligned}
\end{equation}
Since $p>1$, the inequality \eqref{EEE} implies the boundedness of
$v_{n}$ in $W^{1,p}_{0}(\Omega)$. For a subsequence,
$v_{n}\rightarrow v$ weakly in $W^{1,p}_{0}(\Omega)$. By the
compact embedding $W^{1,p}_{0}(\Omega)\subset L^{r}(\Omega)$
($r=p,q$), we have $v_{n}\rightarrow v$ strongly in
$L^{r}(\Omega)$ ($r=p,q$). First we, observe that $v^{-}\equiv 0$
in $\Omega$. In fact, acting with $-u_{n}^{-}$ as test function,
we have
\begin{equation}\label{E}
o(1)\|\nabla (u_{n}^{-})\|_{p}=\langle I'_{\lambda, m_{p},
m_{q}}(u_{n}), -u_{n}^{-}\rangle=\|\nabla
(u_{n}^{-})\|_{p}^{p}+\|\nabla (u_{n}^{-})\|_{q}^{q}\geq \|\nabla
(u_{n}^{-})\|_{p}^{p}.
\end{equation}
Because $p>1$, the inequality \eqref{E} guarantees the boundedness
of $\|\nabla (u_{n}^{-})\|_{p}$ and so $\|\nabla
v_{n}^{-}\|_{p}=\|\nabla
(u_{n}^{-})\|_{p}/\|u_{n}\|_{p}\rightarrow 0$. Thus $v^{-}\equiv 0$ holds, hence $v\geq 0$ in $\Omega$.\\
Now, by taking $(v_{n}-v)/\|u_{n}\|_{p}^{p-1}$ as test function,
we have
\begin{equation}\label{E1}
\begin{aligned}
o(1)&= \left\langle I'_{\lambda, m_{p}, m_{q}}(u_{n}),\frac{
v_{n}-v}{\|u_{n}\|_{p}^{p-1}}\right
\rangle\\&=\int_{\Omega}|\nabla v_{n}|^{p-2}\nabla v_{n}\nabla
(v_{n}- v)dx + \frac{1}{\|u_{n}\|_{p}^{p-q}}\int_{\Omega}|\nabla
v_{n}|^{q-2}\nabla v_{n}\nabla (v_{n}-v)dx\\
&-\lambda\int_{\Omega}m_{p}|v_{n}|^{p-2}v_{n}(v_{n}-v)dx-
\frac{\lambda}{\|u_{n}\|_{p}^{p-q}}\int_{\Omega}m_{q}|v_{n}|^{q-2}v_{n}(v_{n}-v)dx\\&=\int_{\Omega}|\nabla
v_{n}|^{p-2}\nabla v_{n}\nabla (v_{n}- v)dx+ o(1),
\end{aligned}
\end{equation}
because $q<p$, $\|u_{n}\|_{p} \rightarrow +\infty$, $v_{n}$ is
bounded in $W^{1,p}_{0}(\Omega)$ and converges to $v$ strongly in
$L^{r}(\Omega)$ (r=p,q). Thus by \eqref{E1} and $(S_{+})$ property
of $(-\Delta_{p})$ on $W^{1,p}_{0}(\Omega)$, we deduce that
$v_{n}\rightarrow v$ strongly in $W^{1,p}_{0}(\Omega)$. For any
$\varphi\in W^{1,p}_{0}(\Omega)$, by taking
$\varphi/\|u_{n}\|_{p}^{p-1}$ as test function, we obtain
\begin{equation}\label{eq}
\begin{aligned}
o(1)&= \left\langle I'_{\lambda, m_{p}, m_{q}}(u_{n}),\frac{
\varphi}{\|u_{n}\|_{p}^{p-1}}\right
\rangle\\&=\int_{\Omega}|\nabla v_{n}|^{p-2}\nabla v_{n}\nabla
\varphi dx + \frac{1}{\|u_{n}\|_{p}^{p-q}}\int_{\Omega}|\nabla
v_{n}|^{q-2}\nabla v_{n}\nabla \varphi dx\\
&-\lambda\int_{\Omega}m_{p}|v_{n}|^{p-2}v_{n}\varphi dx-
\frac{\lambda}{\|u_{n}\|_{p}^{p-q}}\int_{\Omega}m_{q}|v_{n}|^{q-2}v_{n}\varphi
dx.
\end{aligned}
\end{equation}
Passing to the limit in \eqref{eq}, we see that $v$ is a
non-negative and non-trivial solution of problem
$(P_{\lambda,m_{p}})$ (note $v\geq 0$ and $\|\nabla v\|_{p}=1$).
According to the Harnack inequality (see Remark 1.7 in \cite{M}),
we have $v>0$ in $\Omega$. This implies that
$\lambda=\lambda_{1}(p, m_{p})$ because any positive eigenvalue
other than $\lambda_{1}(p, m_{p})$ has no positive eigenfunctions
(see Theorem \ref{thm1}). Therefore, we obtain a contradiction
since we assumed $\lambda\neq\lambda_{1}(p, m_{p}).$ Hence $u_{n}$
is bounded in $W^{1,p}_{0}(\Omega)$. For a subsequence,
$u_{n}\rightarrow u$ weakly in $W^{1,p}_{0}(\Omega)$ and
$u_{n}\rightarrow u$ strongly in $L^{r}(\Omega)$ ($r=p,q$).


We claim now that $u_{n}\rightarrow u$ strongly in
$W^{1,p}_{0}(\Omega)$. It suffices to prove that $\|\nabla
u_{n}\|_{p}\rightarrow\|\nabla u\|_{p}$, because
$W^{1,p}_{0}(\Omega)$ is uniformly convex. It is clear that
\begin{equation}\label{EPS5}
\begin{aligned}
o(1)=&\langle I'_{\lambda, m_{p}, m_{q}}(u_{n})-I'_{\lambda,
m_{p}, m_{q}}(u),u_{n}-u\rangle\\&= \int_{\Omega}(|\nabla
u_{n}|^{p-2}\nabla u_{n}-|\nabla u|^{p-2}\nabla u)\nabla
(u_{n}-u)dx\\& + \int_{\Omega}(|\nabla u_{n}|^{q-2}\nabla
u_{n}-|\nabla u|^{q-2}\nabla u)\nabla (u_{n}-u)dx+o(1).
\end{aligned}
\end{equation}
Using H\"{o}lder inequality and for $r=p,q$, we have
\begin{equation}\label{EPS6}
\begin{aligned}
&\int_{\Omega}(|\nabla u_{n}|^{r-2}\nabla u_{n}-|\nabla
u|^{r-2}\nabla u)\nabla (u_{n}-u)dx\\& = \int_{\Omega}|\nabla
u_{n}|^{r}dx+\int_{\Omega}|\nabla u|^{r}dx-\int_{\Omega}|\nabla
u_{n}|^{r-2}\nabla u_{n}\nabla udx-\int_{\Omega}|\nabla
u|^{r-2}\nabla u\nabla u_{n}dx\\&\geq \int_{\Omega}|\nabla
u_{n}|^{r}dx+\int_{\Omega}|\nabla
u|^{r}dx-\left(\int_{\Omega}|\nabla
u_{n}|^{r}dx\right)^{(r-1)/r}\left(\int_{\Omega}|\nabla
u|^{r}dx\right)^{1/r}\\&-\left(\int_{\Omega}|\nabla
u_{n}|^{r}dx\right)^{1/r}\left(\int_{\Omega}|\nabla
u|^{r}dx\right)^{(r-1)/r}\\&=(\|\nabla u_{n}\|_{r}^{r-1} -\|\nabla
u\|_{r}^{r-1})(\|\nabla u_{n}\|_{r} -\|\nabla u\|_{r})\\&\geq 0.
\end{aligned}
\end{equation}
Moreover, \eqref{EPS5} and \eqref{EPS6} imply that $\|\nabla
v_{n}\|_{r}\rightarrow\|\nabla v\|_{r}$ (for $r=p,q$). Thus
 $u_{n}\rightarrow u$ strongly in
$W^{1,p}_{0}(\Omega)$.


\end{proof}
\begin{lem}\label{lem2}
Suppose $m_{r}\delta^{\tau}\in L^{a}(\Omega)$ and $m_{r}^{+}\not
\equiv 0$ ($r=p,q$), where $a$, $r$ and $\tau$ satisfy one of the
conditions
 (H1), (H2), (H3) or (H4) If  $\lambda
<\lambda_{1}(q, m_{q})$, then there exist $\alpha
>0$ and $\beta>0$ such that
\begin{equation}\label{equ20}
I_{\lambda, m_{p}, m_{q}}(u)\geq\alpha \mbox{ whenever }
\|u\|_{q}=\beta.
\end{equation}
\end{lem}
 To prove the Lemma \ref{lem2}, we need the following lemma.
\begin{lem} \label{lem3}
Suppose $\tau$, $a$ and $p$ as in Lemma \ref{lem2} and let $b$ be
such that $b=\frac{ap}{a-p}$ if $a<\infty$ and $b=p$ if
$a=\infty$. Set $X(d):=\{u\in W^{1,p}_{0}(\Omega); \|\nabla
u\|_{p}^{p}\leq
d\|\frac{u}{\delta^{\tau}}\|_{b}\|u\|_{p}^{p-1}\}$, for $d>0$.
Then there exists $\alpha(d)>0$ such that $\|\nabla u\|_{p}\leq
\alpha(d)\|u\|_{q}$, for all $u\in X(d)$.
\end{lem}
\begin{rk}\rm
Conditions (H4) implies that the Lemma \ref{lem3} includes the
Lemma 11 of the previously cited paper \cite{MT}.
\end{rk}
\begin{proof} Suppose, by contradiction, that
\begin{equation}\label{equ22}
(\forall n\in \mathbb{N}^{*})~(\exists u_{n}\in X(d)):~
\frac{1}{n}\|\nabla u_{n}\|_{p}> \|u_{n}\|_{q}.
\end{equation}
Because of $\|\nabla u_{n}\|_{p}\neq 0$, we set
$v_{n}:=\frac{u_{n}}{\|\nabla u_{n}\|_{p}}$. Thus, by
\eqref{equ22}, $v_{n}\rightarrow 0$ strongly in $L^{q}(\Omega)$.
Since $\|\nabla v_{n}\|_{p}=1$, the sequence $v_{n}$ is bounded in
$W^{1,p}_{0}(\Omega)$. For a subsequence, $v_{n}\rightarrow v$
weakly in $W^{1,p}_{0}(\Omega)$. By the compact embedding
$W^{1,p}_{0}(\Omega)\subset L^{r}(\Omega)$ ($r=p,q$), we have
$v_{n}\rightarrow v$ strongly in $L^{r}(\Omega)$ ($r=p,q$). Hence,
 $\|v_{n}\|_{q}\rightarrow
\|v\|_{q}$ and $\|v_{n}\|_{p}\rightarrow \|v\|_{q}$. By uniqueness
of limit, we deduce that $\|v\|_{r}=0$. It follows that $v=0$. As
$u_{n}\in X(d)$, we obtain
\begin{equation}\label{equ23}
\frac{1}{d}\leq\frac{\|\frac{u_{n}}{\delta^{\tau}}\|_{b}}{\|\nabla
u_{n}\|_{p}}\|v_{n}\|_{p}^{p-1}.
\end{equation}
By Lemma \ref{L1} and under one of the properties (H1), (H2), (H3)
or (H4), the sequence
$\frac{\|\frac{u_{n}}{\delta^{\tau}}\|_{b}}{\|\nabla u_{n}\|_{p}}$
is bounded (see the proof of Proposition \ref{prop1}). Passing to
the limit in \eqref{equ23}, we get $\frac{1}{d}\leq 0$. This
contradiction completes the proof of Lemma \ref{lem3}.
\end{proof}
\begin{proof}[Proof of Lemma \ref{lem2}.]
According to Lemma \ref{lem3}, choose
\begin{equation}\label{dequ}
d > \max\{1, \lambda\|m_{p}\delta^{\tau}\|_{a},
C\lambda\|m_{p}\delta^{\tau}\|_{a}\},
\end{equation}
 where $C>0$ is a constant
that may differ in each case (H1), (H2), (H3) or (H4) (see the
proof of Proposition \ref{prop1}).\\ For any $u\in X(d)$
satisfying $\int_{\Omega}m_{q}(u^{+})^{q}dx \leq 0$, there exists
$\alpha(d)>0$ such that $\|\nabla u\|_{p}\leq \alpha(d)\|u\|_{q}$
and we have
\begin{equation}\label{equ27}
\begin{aligned}
I_{\lambda, m_{p}, m_{q}}(u)&\geq\frac{1-d}{p}\|\nabla
u\|_{p}^{p}+\frac{\lambda_{1}(q,1)}{q}\|u\|_{q}^{q}+\frac{d}{p}\|\nabla
u\|_{p}^{p}-\frac{C\lambda\|m_{p}\delta^{\tau}\|_{a}}{p}\|\nabla
u\|_{p}^{p}\\& \geq\frac{1-d}{p}\|\nabla
u\|_{p}^{p}+\frac{\lambda_{1}(q,1)}{q}\|u\|_{q}^{q}+\frac{1}{p}\|\nabla
u\|_{p}^{p}\left(d-C\lambda\|m_{p}\delta^{\tau}\|_{a}\right)\\
&\geq\frac{(1-d)[\alpha(d)]^{p}}{p}\|u\|_{q}^{p}+\frac{\lambda_{1}(q,1)}{q}\|u\|_{q}^{q}.
\end{aligned}
\end{equation}
For any $u\not\in X(d)$ satisfying
$\int_{\Omega}m_{q}(u^{+})^{q}dx \leq 0$, we have $\|\nabla
u\|_{p}^{p}> d\|\frac{u}{\delta^{\tau}}\|_{b}\|u\|_{p}^{p-1}.$ It
follows that
\begin{equation}\label{equ29}
\begin{aligned}
I_{\lambda, m_{p},
m_{q}}(u)&\geq\left(\frac{d-\lambda\|m_{p}\delta^{\tau}\|_{a}}{p}\right)\left\|\frac{u}{\delta^{\tau}}\right\|_{b}\|u\|_{p}^{p-1}
+\frac{\lambda_{1}(q,1)}{q}\|u\|_{q}^{q}\\&\geq\frac{\lambda_{1}(q,1)}{q}\|u\|_{q}^{q}.
\end{aligned}
\end{equation}
If $u\in W^{1,p}_{0}(\Omega)$ satisfying
$\int_{\Omega}m_{q}(u^{+})^{q}dx >0$, by the definition of
$\lambda_{1}(q, m_{q})$, we get
\begin{equation}\label{equ24}
\|\nabla u\|_{q}^{q}\geq\|\nabla u^{+}\|_{q}^{q}\geq\lambda_{1}(q,
m_{q})\int_{\Omega}m_{q}(u^{+})^{q}dx.
\end{equation}

Our assumption in $\lambda$ enables us to fix $0<\epsilon <1$ with
\begin{equation}\label{equ25}
(1-\epsilon)\lambda_{1}(q, m_{q})> \lambda.
\end{equation}

If in addition $u\not\in X(d)$, then due to \eqref{dequ},
\eqref{equ24} and \eqref{equ25} we have
\begin{equation}\label{equ26}
\begin{aligned}
I_{\lambda, m_{p},
m_{q}}(u)&\geq\left(\frac{d-\lambda\|m_{p}\delta^{\tau}\|_{a}}{p}\right)\left\|\frac{u}{\delta^{\tau}}\right\|_{b}\|u\|_{p}^{p-1}
+\frac{\epsilon}{q}\|\nabla
u\|_{q}^{q}+\frac{1}{q}[(1-\epsilon)\lambda_{1}(q,
m_{q})-\lambda]\int_{\Omega}m_{q}(u^{+})^{q}dx\\&\geq\frac{\lambda_{1}(q,1)}{q}\|u\|_{q}^{q}.
\end{aligned}
\end{equation}
Finally, if $u\in X(d)$ and $\int_{\Omega}m_{q}(u^{+})^{q}dx >0,$
then \eqref{dequ}, \eqref{equ24} and \eqref{equ25} imply
\begin{equation}\label{equ28}
\begin{aligned}
I_{\lambda, m_{p}, m_{q}}(u)&\geq\frac{1-d}{p}\|\nabla u\|_{p}^{p}
+\frac{\epsilon}{q}\|\nabla
u\|_{q}^{q}+\frac{1}{q}[(1-\epsilon)\lambda_{1}(q,
m_{q})-\lambda]\int_{\Omega}m_{q}(u^{+})^{q}dx\\&+\frac{d}{p}\|\nabla
u\|_{p}^{p}-\frac{\lambda\|m_{p}\delta^{\tau}\|_{a}}{p}\left\|\frac{u}{\delta^{\tau}}\right\|_{b}\|u\|_{p}^{p-1}\\&
\geq\frac{(1-d)[\alpha(p)]^{p}}{p}\|u\|_{q}^{p}+\frac{\epsilon\lambda_{1}(q,1)}{q}\|u\|_{q}^{q}+\frac{d-C\lambda\|m_{p}\delta^{\tau}\|_{a}}{p}\|\nabla
u\|_{p}^{p}\\&\geq\frac{(1-d)[\alpha(p)]^{p}}{p}\|u\|_{q}^{p}+\frac{\epsilon\lambda_{1}(q,1)}{q}\|u\|_{q}^{q}.
\end{aligned}
\end{equation}
Using that $q<p$, the claim \eqref{equ20} in Lemma \ref{lem2}
follows from \eqref{equ27}, \eqref{equ29}, \eqref{equ26} and
\eqref{equ28}.

\end{proof}

\begin{lem}\label{lem4}
 Suppose $m_{r}\delta^{\tau}\in L^{a}(\Omega)$ and $m_{r}^{+}\not
\equiv 0$ ($r=p,q$), where $a$, $r$ and $\tau$ satisfy one of the
conditions
 (H1), (H2), (H3) or (H4). If $\lambda_{1}(p,m_{p})<\lambda$, then there is $R>0$ such that
\begin{equation}\label{equ21}
\|R\phi_{1}(p, m_{p})\|_{q}>\beta \mbox{ and } I_{\lambda, m_{p},
m_{q}}(R\phi_{1}(p, m_{p}))<0,
\end{equation}
where $\beta > 0$ is the constant in \eqref{equ20}.
\end{lem}

\begin{proof}
 For sufficiently large $R>0$, taking into account that $q<p$
and $\lambda_{1}(p,m_{p})<\lambda$, we have
$$  \frac{I_{\lambda, m_{p}, m_{q}}(R\phi_{1}(p, m_{p}))}{R^{\beta}}= \frac{\lambda_{1}(p,m_{p})-\lambda}{p}+\frac{1}{qR^{p-q}}
\left( \|\phi_{1}(p,
m_{p})\|_{q}^{q}-\int_{\Omega}m_{q}\phi_{1}^{q}(p,
m_{p})dx\right)<0.$$
\end{proof}

Recalling that the functional $I_{\lambda, m_{p}, m_{q}}$
satisfies the Palais-Smale condition by virtue of Lemma
\ref{lem1}, the properties \eqref{equ20} of Lemma \ref{lem2} and
\eqref{equ21} of Lemma \ref{lem4} allow us to apply the mountain
pass theorem, which guarantees the existence of a critical value
$c\geq \alpha$ of $I_{\lambda, m_{p}, m_{q}}$, with $\alpha > 0$
in \eqref{equ20}, namely $c:=\inf\limits_{\gamma \in
\Sigma}\max\limits_{t\in [0,1]}I_{\lambda, m_{p},
m_{q}}(\gamma(t)),$ where $\Sigma :=\{\gamma\in C([0,1],
W^{1,p}_{0}(\Omega)); \gamma(0)=0, \gamma(1)=R\phi_{1}(p,
m_{p})\}$. This completes the proof of Theorem \ref{thm21}.


\subsection{Resonant cases} In this
section, we study the existence result for problem $(P_{\lambda,
1})$ (or $(P_{\lambda, \mu})$) in the resonant cases. The
following theorem is our main result in this section.
\begin{thm}\label{thm3}
One assumes the same conditions as for Theorem \ref{thm1}. If one
of the following assertions holds
\begin{itemize}
\item [(i)] $\lambda=\lambda_{1}(p,m_{p})>\lambda_{1}(q,m_{q})$ and
$\int_{\Omega}|\nabla
\phi_{1}(p,m_{p})|^{q}dx-\lambda\int_{\Omega}m_{q}|
\phi_{1}(p,m_{p})|^{q}dx>0$;
\item [(ii)] $\lambda=\lambda_{1}(q,m_{q})>\lambda_{1}(p,m_{p})$ and
$\int_{\Omega}|\nabla
\phi_{1}(q,m_{q})|^{p}dx-\lambda\int_{\Omega}m_{p}|
\phi_{1}(q,m_{q})|^{p}dx>0,$
\end{itemize}
then the problem $(P_{\lambda, 1})$ (or $(P_{\lambda, \mu})$) has
at least one positive solution.
\end{thm}

\begin{proof}
Proof of case (i): Since $m_{p}^{+}\not\equiv 0$, the Lebesgue
mesure of $\{x\in \Omega; m_{p}(x)>0\}$ is positive. Thus there
exists an $n_{0}\in\mathbb{N}$ such that
$(m_{p}-1/n_{0})\not\equiv 0$. For $n\geq n_{0}$, we define, as in
\cite{MT}, the functional $I_{n}$ on $W^{1,p}_{0}(\Omega)$ by
$$I_{n}(u)=I_{\lambda, m_{p},
m_{q}}(u)+\frac{\lambda}{pn}\|u^{+}\|_{p}^{p}=I_{\lambda,
m_{p}-1/n, m_{q}}(u).$$ Using the strict monotonicity of the first
eigenvalue of problem $(P_{\lambda, m_{p}})$ (see \cite{M}), we
obtain $\lambda_{1}(p,m_{p}-1/n)>\lambda_{1}(p,m_{p})=\lambda.$
Thus we are able to apply Theorem \ref{thm2} obtaining a positive
solution $u_{n}$ of the problem
 \begin{equation*} \left\{
\begin{array}{rclclll}
-\triangle_{p}u-\triangle_{q}u &=& \lambda[
(m_{p}(x)-1/n)|u|^{p-2}u+ m_{q}(x)|u|^{q-2}u]      &\hbox{ in }\Omega,\\
u &=& 0    &\hbox{ on } \partial \Omega.
 \end{array}
\right.
\end{equation*}
We may assume that $u_{n}$ is a global minimizer of $I_{n}$ and
$I_{n}(u_{n})<0$ (see the case (a) in the proof of Theorem
\ref{thm2}). In addition, observing that $I_{n}\leq I_{n_{0}}$
provided $n\geq n_{0}$, we infer that for all $n\geq n_{0}$,
$$I_{n}(u_{n})=\min_{W^{1,p}_{0}(\Omega)}I_{n}\leq I_{n}(u_{n_{0}})\leq
I_{n_{0}}(u_{n_{0}})<0.$$ We claim that  if $u_{n}$ is bounded in
$W^{1,p}_{0}(\Omega)$, then $u_{n}$ is a bounded Palais-Smale
sequence of $I_{\lambda, m_{p}, m_{q}}$. Indeed,  there exists
$c\in\mathbb{R}$ such that  $I_{\lambda, m_{p},
m_{q}}(u_{n})\rightarrow c$ because $I_{n}(u_{n})$ is a convergent
sequence and $\|\nabla u_{n}\|_{p}$ is bounded. On the other hand,
since $I'_{n}(u_{n})=0$, we have
$$\|I'_{\lambda, m_{p}, m_{q}}(u_{n})\|_{(W^{1,p}_{0}(\Omega))^{*}}=\|I'_{\lambda, m_{p}, m_{q}}(u_{n})-I'_{n}(u_{n})\|_{(W^{1,p}_{0}(\Omega))^{*}}
\leq \frac{\lambda[\lambda_{1}(p,1)]^{-1/p}}{n}\|\nabla
(u_{n})^{+}\|^{p-1}_{p}.$$ As $\|\nabla u_{n}\|_{p}$ is bounded,
we obtain
$\|I'_{\lambda}(u_{n}\|_{(W^{1,p}_{0}(\Omega))^{*}}\rightarrow 0$.
This completes the proof of our claim.\\
We prove now the boundedness of $u_{n}$ in $W^{1,p}_{0}(\Omega)$
by way of contradiction. Suppose that $\|\nabla
u_{n}\|_{p}\rightarrow \infty$ and let $v_{n}:=u_{n}/\|\nabla
u_{n}\|_{p}$. For a subsequence, $v_{n}\rightarrow v$ weakly in
$W^{1,p}_{0}(\Omega)$ and $v_{n}\rightarrow v$ strongly in
$L^{p}(\Omega)$. Following the same steps of the proof of Lemma
\ref{lem1}, we show that $v$ is a positive solution of problem
$(P_{\lambda,m_{p}})$. This entails $v$ is a positive
eigenfunction corresponding to $\lambda_{1}(p, m_{p})$. Thus the
simplicity of $\lambda_{1}(p, m_{p})$ implies that $v=\phi_{1}(p,
m_{p})$. The facts that $I_{n}(u_{n})<0$ for all $n\geq n_{0}$ and
$u_{n}$ is a critical point of $I_{n}$ result in
$$ 0>\frac{I_{n}(u_{n})}{\|\nabla
u_{n}\|_{p}^{q}}=\left(\frac{1}{q}-\frac{1}{p}\right)\left
(\|\nabla
v_{n}\|_{q}^{q}-\lambda\int_{\Omega}m_{q}v_{n}^{q}dx\right).
$$
Passing to the limit, we obtain $\int_{\Omega}|\nabla
\phi_{1}(p,m_{p})|^{q}dx-\lambda\int_{\Omega}m_{q}|
\phi_{1}(p,m_{p})|^{q}dx\leq 0$ which contradicts second point of
assertion (i). Thus $u_{n}$ is a bounded Palais-Smale sequence of
$I_{\lambda, m_{p}, m_{q}}$. Since $\lambda\neq\lambda_{1}(p,
m_{p})$, $I_{\lambda, m_{p}, m_{q}}$ satisfies Pais-Smale
condition (see Lemma \ref{lem1}). It follows that $u_{n}$ has a
subsequence converging to some critical point $u_{0}$ of
$I_{\lambda, m_{p}, m_{q}}$.\\ We note that $u_{0}\neq 0$ because
$I_{\lambda, m_{p}, m_{q}}(u_{0})=\lim\limits_{n}I_{n}(u_{n})\leq
I_{n_{0}}(u_{0})<0.$ Therefore $u_{0}$ is a positive solution of
problem $(P_{\lambda, 1})$ (see Remark \ref{rk20}).

Proof of case (ii): As in the proof of case (i), we can choose
$n_{0}\in \mathbb{N}$ such that $(m_{q}-1/n_{0})\neq 0$. For
$n\geq n_{0}$, we define the functional $J_{n}$ on
$W^{1,p}_{0}(\Omega)$ by
$$J_{n}(u)=I_{\lambda, m_{p},
m_{q}}(u)+\frac{\lambda}{pn}\|u^{+}\|_{q}^{q}=I_{\lambda, m_{p},
m_{q}-1/n}(u).$$ Using the strict monotonicity of the first
eigenvalue of problem $(P_{\lambda, m_{q}})$, we obtain
$\lambda_{1}(q,m_{q}-1/n)>\lambda_{1}(q,m_{q})=\lambda,$ for any
$n\geq n_{0}$. Thus we may apply Theorem \ref{thm2} obtaining a
positive solution $u_{n}$ of the problem
 \begin{equation*} \left\{
\begin{array}{rclclll}
-\triangle_{p}u-\triangle_{q}u &=& \lambda[
m_{p}(x)|u|^{p-2}u+ (m_{q}(x)-1/n)|u|^{q-2}u]      &\hbox{ in }\Omega,\\
u &=& 0    &\hbox{ on } \partial \Omega.
 \end{array}
\right.
\end{equation*}
Following the pattern of the proof of case (i), this time
proceeding as in case (b) in the proof of Theorem \ref{thm2}, we
deduce that $J_{n}(u_{n})>0$ for all $n\geq n_{0}$. For the
boundedness of $u_{n}$ in $W^{1,p}_{0}(\Omega)$, proceeding as in
the proof of case (i) and the contradiction follows from the
condition $\lambda=\lambda_{1}(q,m_{q})>\lambda_{1}(p,m_{p})$. The
bounded sequence $u_{n}$ is a Palais-Smale sequence for the
functional $I_{\lambda, m_{p}, m_{q}}$ as can be seen from the
estimate $$\|I'_{\lambda, m_{p},
m_{q}}(u_{n})\|_{(W^{1,p}_{0}(\Omega))^{*}}=\|I'_{\lambda, m_{p},
m_{q}}(u_{n})-J'_{n}(u_{n})\|_{(W^{1,p}_{0}(\Omega))^{*}} \leq
\frac{c}{n}\|u_{n}\|^{q-1}_{q},$$ where $c$ is a positive constant
independent of $n$. It follows that $u_{n}$ has a subsequence
converging to some critical point $u_{0}$ of $I_{\lambda, m_{p},
m_{q}}$. In order to complete the proof, due to the Harnack
inequality, it suffices to justify that $u_{0}\neq 0$. We assume,
by contradiction, that $u_{n}\rightarrow 0$ strongly in
$W^{1,p}_{0}(\Omega)$. Set $v_{n}:=\frac{u_{n}}{\|\nabla
u_{n}\|_{p}}$. Then, for a subsequence, $v_{n}\rightarrow v$
weakly in $W^{1,p}_{0}(\Omega),$ weakly in $W^{1,q}_{0}(\Omega)$
and strongly in $L^{p}(\Omega)$. It is easy to see that $v\geq 0$
in $\Omega$. Using $\frac{v_{n}-v}{\|\nabla u_{n}\|_{p}^{q-1}}$
as test function, we obtain
$$0=\left\langle J'_{n}(u_{n}), \frac{v_{n}-v}{\|u_{n}\|^{q-1}}\right\rangle=\int_{\Omega}|\nabla
v_{n}|^{q-2}\nabla v_{n}\nabla(v_{n}-v)dx + o(1).$$ By $(S_{+})$
property for $-\Delta_{q}$ on $W^{1,q}_{0}(\Omega)$, we have
$v_{n}\rightarrow v$ strongly in $W^{1,q}_{0}(\Omega).$ We claim
that $v\neq 0$ in $\Omega$. Using $u_{n}$ as test function, we may
write
\begin{equation}\label{EEE1}
\begin{aligned}
0&=\langle J'_{n}(u_{n}), u_{n}\rangle\\&=\|\nabla
u_{n}\|_{p}^{p}-\lambda\int_{\Omega}m_{p}u_{n}^{p}dx+\|\nabla
u_{n}\|_{q}^{q}-\lambda\int_{\Omega}(m_{q}-\frac{1}{n})u_{n}^{q}dx\\&\geq
\|\nabla u_{n}\|_{p}^{p}-\lambda\int_{\Omega}m_{p}u_{n}^{p}dx,
\end{aligned}
\end{equation}
because $\lambda=\lambda_{1}(q, m_{q})$ and by the definition of
$\lambda_{1}(q, m_{q})$, we have $\|\nabla
u_{n}\|_{q}^{q}-\lambda\int_{\Omega}m_{q}u_{n}^{q}dx\geq 0$. Thus
according (H1), (H2), (H3) or (H4), we have $\|\nabla
u\|_{p}^{p}\leq
\lambda\|m_{p}\delta^{\tau}\|_{a}\|\frac{u}{\delta^{\tau}}\|_{b}\|u\|_{p}^{p-1}$,
where $\tau$, $a$ and $b$ as in Lemma \ref{lem3}. Whence $u_{n}\in
X(d)$ with $d=\lambda\|m_{p}\delta^{\tau}\|_{a}$. Therefore, Lemma
\ref{lem3} guarantees the existence of a constant $\alpha(d)>0$
such that $$\|\nabla
u\|_{p}\leq\alpha(d)\|u_{n}\|_{q}\leq\alpha(d)[\lambda_{1}(q,
1)]^{-1/q}\|\nabla u_{n}\|_{q}, \mbox{ for all  } n\geq n_{0}.$$
Consequently, recalling that $v_{n}\rightarrow v$ strongly in
$W^{1,q}_{0}(\Omega),$ we have
$$\|\nabla v\|_{q}=\lim \|\nabla v_{n}\|_{q}=\lim\frac{\|\nabla u_{n}\|_{q}}{\|\nabla u_{n}\|_{p}}\geq\frac{1}{\alpha(d)[\lambda_{1}(q,
1)]^{-1/q}}>0, $$ thus proving our claim.\\ For any $\varphi\in
W^{1,p}(\Omega)$, by using $\frac{\varphi}{\|\nabla
u_{n}\|_{p}^{q-1}}$ as test function we show, as in proof of case
(i), that $v$ is a nonnegative nontrivial solution of
$(P_{\lambda, m_{q}})$. The simplicity of $\lambda_{1}(q, m_{q})$
guarantees that $v=\phi_{1}(q, m_{q})$ is a positive eigenfunction
corresponding to $\lambda_{1}(q, m_{q})$.\\ Using that
$J_{n}(u_{n})>0$ for all $n\geq n_{0}$ in conjunction with
\eqref{EEE1}, we have
$$0\leq\frac{J_{n}(u_{n})}{\|\nabla
u_{n}\|_{p}^{p}}=\left(\frac{1}{p}-\frac{1}{q}\right)\left(\|\nabla
v_{n}\|_{p}^{p}-\lambda\int_{\Omega}m_{p}v_{n}^{p}dx\right).$$
Since $q<p$, it follows that
$$ \|\nabla v_{n}\|_{p}^{p}-\lambda\int_{\Omega}m_{p}v_{n}^{p}dx\leq 0.$$
By passing to the limit inferior we obtain
$$ \|\nabla \phi_{1}(q, m_{q})\|_{p}^{p}-\lambda\int_{\Omega}m_{p}[\phi_{1}(q, m_{q})]^{p}dx\leq 0.$$
 By this contradiction the proof of Theorem \ref{thm3} is
achieved.

\end{proof}

\hspace{-0.6cm}\textbf{Acknowledgments}\\

\hspace{-0.6cm}The authors would like to express their sincere
thanks to Professor D. Motreanu, Professeur M. Tanaka, Professor
M. Montenegro and Professor S. Lorca for their extraordinary ideas
which we are inspired.

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