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

\AtBeginDocument{{\noindent\small
\emph{Electronic Journal of Differential Equations},
Vol. 2010(2010), No. 162, pp. 1--23.\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-2010/162\hfil Existence of solutions]
{Existence of solutions to indefinite quasilinear
elliptic problems of p-q-Laplacian type}

\author[N. E. Sidiropoulos\hfil EJDE-2010/162\hfilneg]
{Nikolaos E. Sidiropoulos}

\address{Nikolaos E. Sidiropoulos \newline
Department of Sciences, Technical University of Crete,
73100 Chania, Greece}
\email{niksidirop@gmail.com}

\thanks{Submitted July 7, 2010. Published November 12, 2010.}
\subjclass[2000]{35J60, 35J62, 35J92}
\keywords{Indefinite quasilinear elliptic problems;
 subcritical nonlinearities; \hfill\break\indent
 p-Laplacian; p-q-Laplacian; fibering
 method; mountain pass theorem}

\begin{abstract}
 We study the indefinite quasilinear elliptic problem
 \begin{gather*}
 -\Delta u-\Delta _{p}u=a(x)|u|^{q-2}u-b(x)|u|^{s-2}u
 \quad\text{in }\Omega , \\
 u=0\quad\text{on }\partial \Omega ,
 \end{gather*}
 where $\Omega $ is a bounded domain in
 $\mathbb{R}^{N}$, $N\geq 2$, with a sufficiently smooth boundary,
 $q,s$ are subcritical exponents, $a(\cdot)$ changes
 sign and $b(x)\geq 0$ a.e. in $\Omega $. Our proofs are variational
 in character and are based either on the fibering method or the
 mountain pass theorem.
\end{abstract}

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

\section{Introduction}

 Let $\Omega $ be a bounded domain in $\mathbb{R}^{N}$, $N\geq 2$,
with a sufficiently smooth boundary $\partial \Omega $. We
consider the  stationary nonlinear equation
\begin{equation}
-\Delta _{q}u-\Delta _{p}u=f(x,u)\quad\text{in }\Omega  \label{eq1}
\end{equation}
with Dirichlet boundary condition
\begin{equation}
u=0\quad\text{on }\partial \Omega ,  \label{Dbc}
\end{equation}
where $p,q\in (1,N)$, $\Delta _{p}u
=\operatorname{div}(|\nabla u|^{p-2}\nabla u)$ is
the $p$-Laplacian operator and
$f:\Omega \times\mathbb{R}\to\mathbb{R}$
is a Caratheodory function.

 Solutions to  \eqref{eq1} are the steady state solutions
of the reaction diffusion system
\begin{equation}
u_{t}=\operatorname{div}(A(u)\nabla u)+f(x,u),  \label{schr}
\end{equation}
where $A(u)=(|\nabla u|^{q-2}+|\nabla u|^{p-2})$.  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}. The function $u$ describes the
concentration of a substance, $\operatorname{div} (A(u)\nabla u)$
corresponds to the diffusion with diffusion coefficient $A(u) $
and $f(\cdot,\cdot)$ represents the reaction.

 Equation \eqref{eq1} also arises in the study of soliton-like
solutions of the nonlinear Schr\"{o}dinger equation
\[
i\psi _{t}=-\Delta \psi -\Delta _{p}\psi +f(x,\psi )
\]
which was considered by Derrick \cite{Derr} as a model for elementary
particles.

 When $p=q=2$, \eqref{eq1} is a normal Schrodinger equation which
has been extensively studied, we refer to
\cite{Bahri-Li,Ber-Lion1,Ber-Lion2}. Recently, the problem when
$m=2\neq q$ and

\[
f(x,u)=V'(u)
\]
was studied in \cite{B-A-F-P} where it is proved that
\eqref{eq1}-\eqref{Dbc} admits a weak solution with a prescribed
value of topological charge. The eigenvalue problem
\[
-\Delta u+V(x)u+\varepsilon ^{r}(-\Delta _{p}u+W'(u))=\mu u
\]
was considered in \cite{B-M-V} and the behavior of the eigenvalues as
$\varepsilon \to 0$ was examined. In \cite{Cher-Il} the case where
$m\neq p$ and
\[
f(x,u)=\lambda a(x)|u|^{\gamma -2}u-b(x)|u|^{m-2}u-c(x)|u|^{p-2}u
\]
is studied and a bifurcation result is also presented. A solution is also
provided in \cite{He-Li} under the assumption that
\begin{equation}
f(x,u)=g(x,u)-b(x)|u|^{m-2}u-c(x)|u|^{p-2}u  \label{part}
\end{equation}
where the function $g(\cdot,\cdot)$ does not satisfy the
Ambrosetti-Rabinowitz condition. The $C^{1,\delta }$-regularity of
the solutions of problem \eqref{eq1} was shown in \cite{He-Li2}.
Constraint minimization is employed in \cite{Wu-Y} with constraint
functional
\[
\int_{ \mathbb{R} ^{N}}[b(x)|u|^{q}-c(x)|u|^{p}u]dx=\lambda
\]
when $f(\cdot,\cdot)$ satisfies \eqref{part}, in order to show
that \eqref{eq1} admits a solution for $\lambda \in (0,\lambda
_0)$, $\lambda _0>0$. Sufficient conditions for the existence of
two solutions to problem (\eqref {eq1} are provided in
\cite{Li-Liang}.

In this article we study the problem
\begin{gather}
-\Delta u-\Delta _{p}u=a(x)|u|^{q-2}u-b(x)|u|^{s-2}u\quad\text{in
}\Omega , \label{1}\\
 u=0\quad\text{on }\partial \Omega ,
\label{2}
\end{gather}
where the exponents $q,s$ are subcritical and $a(\cdot),b(\cdot)$
are essentially bounded functions, $a(\cdot)$ changes sign while
$b(\cdot)\geq 0$ a.e. in $\Omega $. Our proofs are variational in
character and rely either on the fibering method of Pohozaev
\cite{Poho} or on the mountain pass theorem of
Ambrosetti-Rabinowitz \cite{Amb-Rab}.

By symmetry, we will only consider the cases where $p<2$.

\section{Preliminaries and main results}

 We make the following hypotheses concerning the data of problem
\eqref{1}-\eqref{2}:
\begin{itemize}
\item[(H0)]  $1<s,q<2^{\ast }$.

\item[(H1)] $a(\cdot)\in L^{\infty }(\Omega )$ and
$a_{+}:=\max \{a,0\}\neq 0$.

\item[(H2)]  $b(\cdot)\in L^{\infty }(\Omega )$ and $b(x)\geq 0$
a.e. in $\Omega $.
\end{itemize}

 We will seek weak solutions in the space
\[
E:=H_0^1(\Omega ),
\]
supplied with the norm
$\|\upsilon \|_{E}=\|\nabla \upsilon \|_2$.
The energy functional $\Phi :E\to \mathbb{R}$ associated with
\eqref{1}-\eqref{2} is
\begin{equation}
\ \Phi (\upsilon ):=\frac{1}{p}\|\nabla \upsilon \|_{p}^{p}+\frac{1}{2}
\|\nabla \upsilon \|_2^2-\frac{1}{q}A(\upsilon )+\frac{1}{s}B(\upsilon ),
\label{EF}
\end{equation}
where
\[
A(\upsilon ):=\int_{\Omega }a(x)|\upsilon |^{q}dx\text{ and }B(\upsilon
):=\int_{\Omega }b(x)|\upsilon |^{s}dx.
\]
to find nonnegative critical points for $\Phi (\cdot)$ we
use the fibering method. So we decompose the function $u\in
E$ as $u=r\upsilon $, where $r\in\mathbb{R}$,
$v\in E$, and define the extended functional $F(\cdot,\cdot)$
associated with $\Phi (\cdot)$ as
\begin{equation}
F(r,\upsilon ):=\Phi (r\upsilon )=\frac{|r|^{p}}{p}\|\nabla \upsilon
\|_{p}^{p}+\frac{|r|^2}{2}\|\nabla \upsilon \|_2^2-\frac{|r|^{q}}{q}
A(\upsilon )+\frac{|r|^{s}}{s}B(\upsilon ).  \label{Fib}
\end{equation}
If $u=r\upsilon $ is a critical point of $\Phi (\cdot)$, then we
must have
\begin{equation}
F_{r}(r,\upsilon )=0.  \label{BE0}
\end{equation}
Clearly, \eqref{BE0} is equivalent to
\begin{equation}
r^2\|\nabla \upsilon \|_2^2+r^{p}\|\nabla \upsilon
\|_{p}^{p}=r^{q}A(\upsilon )-r^{s}B(\upsilon ).  \label{BE}
\end{equation}
Let $r:=r(\upsilon )$ be a positive solution of \eqref{BE}.
We define the reduced functional
$\hat{\Phi}(\upsilon ):=\Phi (r(\upsilon )\upsilon )$,
$\upsilon \in E$, which, in view of \eqref{BE}, has the
following equivalent expressions
\begin{align}
\hat{\Phi}(\upsilon )
&:=r^2(\frac{1}{2}-\frac{1}{p})\|\nabla \upsilon
\|_2^2+r^{q}(\frac{1}{p}-\frac{1}{q})A(\upsilon )+r^{s}(\frac{1}{s}-
\frac{1}{p})B(\upsilon )  \label{e1}
\\
&=r^{q}(\frac{1}{p}-\frac{1}{q})\|\nabla \upsilon \|_{p}^{p}+r^2(\frac{1}{2}
-\frac{1}{q})\|\nabla \upsilon \|_2^2+r^{s}(\frac{1}{s}-\frac{1}{q}
)B(\upsilon )  \label{e2}
\\
&=r^{p}(\frac{1}{p}-\frac{1}{s})\|\nabla \upsilon \|_{p}^{p}+r^2(\frac{1}{2}
-\frac{1}{s})\|\nabla \upsilon \|_2^2+r^{q}(\frac{1}{s}-\frac{1}{q}
)A(\upsilon )  \label{e3}
\\
&=r^{p}(\frac{1}{p}-\frac{1}{2})\|\nabla \upsilon \|_{p}^{p}+r^{q}(\frac{1}{2}
-\frac{1}{q})A(\upsilon )+r^{s}(\frac{1}{s}-\frac{1}{2})B(\upsilon ).
\label{e4}
\end{align}
The fibering method is based on the following fact.

\begin{lemma} \label{FibLemma}
Let $H:E\to\mathbb{R}$ be a functional which is continuously
Fr\'{e}chet-differentiable in $E\backslash \{0\}$ and satisfies
the conditions:
\[
\langle H'(\upsilon ),\upsilon \rangle \neq 0\quad
\text{if }H(\upsilon )=1,
\]
and $H(0)=0$. If $\upsilon \neq 0$ is a conditional critical point
of $\hat{\Phi}(\cdot)$ under the constraint $H(\upsilon )=1$,
then $u:=r(\upsilon )\upsilon $ is a nonzero critical point of
$\Phi (\cdot)$.
\end{lemma}

 For more details we refer to \cite{Dra-Poh}.
The constraint functional we are going to use is
\[
H(\upsilon ):=\|\nabla \upsilon \|_{p}^{p}+\|\nabla \upsilon \|_2^2
\]
which clearly satisfies the two conditions in Lemma \ref{FibLemma}. Let
\begin{equation}
S^1:=\{\upsilon \in E:H(\upsilon )=1\}.  \label{ff}
\end{equation}
Note that, because of assumption $(H_1)$, the set
\[
G_1:=\{\upsilon \in E:A(\upsilon )>0\}
\]
is nonempty.

We distinguish the following cases:

 \textbf{Case 1:} $q<\min \{p,s,2\}$.
We will work as in \cite{dkan,dkan-alyb}. From \eqref{BE}
 we see that
\begin{equation}
r^{p-q}\|\nabla \upsilon \|_{p}^{p}+r^{2-q}\|\nabla \upsilon
\|_2^2+r^{s-q}B(\upsilon )=A(\upsilon ),  \label{BF1}
\end{equation}
which admits a unique solution $r(\upsilon )>0$ for every
$\upsilon \in G_1$. It is easy to check that
$r(\upsilon )\upsilon =r(k\upsilon )k\upsilon $ for every $k>0$.
The implicit
function theorem, see \cite{Zeid} , shows that $r(\cdot)$ $\in $
$C^1(G_1)$. If $\upsilon $ $\in $ $S^1$ then the H\"{o}lder
inequality implies that $\|\nabla \upsilon \|_2^2\geq \theta$
for some $\theta >0$ and so, by \eqref{BF1}, $r(\cdot)$ is
bounded on $G_1\cap S^1$ because $A(\cdot)$ is bounded on
$S^1$ by the Rellich theorem. Consequently, $\hat{\Phi}(\cdot)$
is bounded on $G_1\cap S^1$. Let
\[
M=\inf_{u\in G_1\cap S^1} \hat{\Phi}(u).
\]
By \eqref{e2}, $M<0$. Suppose that $\{\upsilon _n\}
$ is a minimizing sequence for $\hat{\Phi}(\cdot)$ in
$G_1\cap S^1$. Then, at least for a subsequence, we have that
$\upsilon_n\to \tilde{\upsilon}$ weakly in $E$, and so we may assume
that $A(\upsilon _n)\to A( \tilde{\upsilon})$ and
$B(\upsilon _n)\to B(\tilde{\upsilon})$. Exploiting the weak lower
semicontinuity of the norms we get that
\[
0\leq \|\nabla \tilde{\upsilon}\|_2^2\leq \lim \inf \|\nabla \upsilon
_n\|_2^2,\quad
0\leq \|\nabla \tilde{\upsilon}\|_{p}^{p}\leq
\lim \inf \|\nabla \upsilon _n\|_{p}^{p}.
\]
Since $\{r(\upsilon _n)\}_{n\in\mathbb{N}}$ is bounded we may
also assume that $r(\upsilon _n)\to \tilde{r}$.  Therefore,
\[
\Phi (\tilde{r}\tilde{\upsilon})\leq \liminf \Phi (r_n\upsilon _n)=M<0,
\]
implying that $\tilde{r}>0$ and $\tilde{\upsilon}\neq 0$. On
the other hand, by \eqref{BF1}
\begin{equation}
r(\upsilon _n)^{p-q}\|\nabla \upsilon _n\|_{p}^{p}+r(\upsilon
_n)^{2-q}\|\nabla \upsilon _n\|_2^2+r(\upsilon _n)^{s-q}B(\upsilon
_n)=A(\upsilon _n).  \label{n}
\end{equation}
By taking the limit as $n\to +\infty $, we obtain
\begin{equation}
0<\tilde{r}^{p-q}\|\nabla \tilde{\upsilon}\|_{p}^{p}+\tilde{r}^{2-q}\|\nabla
\tilde{\upsilon}\|_2^2+\tilde{r}^{s-q}B(\tilde{\upsilon})\leq A(\tilde{
\upsilon}),  \label{LTE}
\end{equation}
which implies that $\tilde{\upsilon}$ $\in $ $G_1$.
In view of \eqref{1},
\begin{equation}
r(\tilde{\upsilon})^{p-q}\|\nabla \tilde{\upsilon}\|_{p}^{p}+r(\tilde{
\upsilon})^{2-q}\|\nabla \tilde{\upsilon}\|_2^2+r(\tilde{\upsilon}
)^{s-q}B(\tilde{\upsilon})=A(\tilde{\upsilon}),  \label{E}
\end{equation}
and so \eqref{LTE} shows that $\tilde{r}\leq r(\tilde{\upsilon}$ $)$.
If we assume that $\tilde{r}$ $<r(\tilde{\upsilon})$, then, since
the function
$t\to \Phi (t\tilde{\upsilon})$, $t\in (0,r(\tilde{\upsilon}))$,
is strictly decreasing, we have
\begin{equation}
\hat{\Phi}(\tilde{\upsilon})=\Phi (r(\tilde{\upsilon})\tilde{\upsilon})<\Phi
(\tilde{r}\tilde{\upsilon})\leq M.  \label{Less}
\end{equation}
Then
\[
\hat{\Phi}(\frac{\tilde{\upsilon}}{\|\tilde{\upsilon}\|_{E}})=\hat{\Phi}(
\tilde{\upsilon})=M,
\]
a contradiction. Therefore, $\tilde{r}=r(\tilde{\upsilon})$. Then, by
\eqref{n} and \eqref{E},
\begin{equation}
\lim_{n\to \infty } \{\|\nabla \upsilon
_n\|_{p}^{p}+r(\upsilon _n)^{2-p}\|\nabla \upsilon
_n\|_2^2\}=\|\nabla \tilde{\upsilon}\|_{p}^{p}+r(\tilde{\upsilon
})^{2-p}\|\nabla \tilde{\upsilon}\|_2^2,  \label{fe}
\end{equation}
which implies that
$\|\nabla \upsilon _n\|_{p}^{p}\to
\|\nabla \tilde{\upsilon}\|_{p}^{p}$ and
$\|\nabla \upsilon _n\|_2^2\to \|\nabla \tilde{\upsilon}\|_2^2$.
Consequently, $\tilde{\upsilon}\in $ $S^1$ and
$\hat{\Phi}(\tilde{\upsilon})=M$. Since $|\tilde{\upsilon}|$ is
also a minimizer of $\hat{\Phi}(\cdot)$, we may assume that
$\tilde{\upsilon} \geq 0$. Lemma \ref{FibLemma} implies that
$u:=r(\tilde{\upsilon})\tilde{ \upsilon}$ is a solution to
\eqref{1}-\eqref{2}. By \cite[Theorem 1]{He-Li2}, $u\in
C^{1,\delta }(\Omega )$ for some $\delta \in (0,1)$. Therefore we
have the following result.

\begin{theorem} \label{thm2}
Assume that  {\rm (H0)-(H2)} are satisfied and
$q<\min \{p,s,2\}$. Then \eqref{1}-\eqref{2} admits a non-negative
solution $u\in C^{1,\delta }(\Omega )$ for some $\delta \in (0,1)$.
\end{theorem}

 \textbf{Case 2:} $p<q<2<s$.
Let
\begin{equation}
Q(r,\upsilon ):=r^{q-p}A(\upsilon )-r^{s-p}B(\upsilon )-r^{2-p}\|\nabla
\upsilon \|_2^2.  \label{BE1}
\end{equation}
Then \eqref{BE} is equivalent to
\begin{equation}
Q(r,\upsilon )=\|\nabla \upsilon \|_{p}^{p}.  \label{BE2}
\end{equation}
We see that for $\upsilon $ $\in $ $G_1$ the function
$Q(\cdot,\upsilon )$
has a unique critical point $r_{\ast }:=r_{\ast }(\upsilon )$
satisfying
\begin{equation}
(q-p)A(\upsilon )=(s-p)r_{\ast }(\upsilon )^{s-q}B(\upsilon )
+(2-p)r_{\ast}(\upsilon )^{2-q}\|\nabla \upsilon \|_2^2.  \label{cp1}
\end{equation}
In view of \eqref{BE1}, we get the following equivalent expressions
for \eqref {cp1}, that will be needed in the sequel,
\begin{gather}
Q(r_{\ast }(\upsilon ),\upsilon )=\frac{2-q}{2-p}r_{\ast }(\upsilon
)^{q-p}A(\upsilon )+\frac{s-2}{2-p}r_{\ast }(\upsilon )^{s-p}B(\upsilon ),
\label{mm2}
\\
Q(r_{\ast }(\upsilon ),\upsilon )=\frac{s-q}{s-p}r_{\ast }(\upsilon
)^{s-p}A(\upsilon )+\frac{2-s}{s-p}r_{\ast }(\upsilon )^{2-p}\|\nabla
\upsilon \|_2^2 , \label{mm1}
\\
Q(r_{\ast }(\upsilon ),\upsilon )=\frac{s-q}{q-p}r_{\ast }(\upsilon
)^{s-p}B(\upsilon )+\frac{2-q}{q-p}r_{\ast }(\upsilon )^{s-p}\|\nabla
\upsilon \|_2^2.  \label{mm3}
\end{gather}
Let
\begin{equation}
G_2:=\{\upsilon \in G_1:\|\nabla \upsilon \|_{p}^{p}<Q(r_{\ast
}(\upsilon ),\upsilon )\}.  \label{s}
\end{equation}
Equation \eqref{BE2} has two positive solutions $r_1(\upsilon )$,
$r_2(\upsilon )$ with
$r_1(\upsilon )<r_{\ast }(\upsilon )<r_2(\upsilon)$ for every
$\upsilon \in G_2$. Let $r:=r_2(\upsilon )$. Then
\[
r^{p-q+1}Q_{r}(r,\upsilon )=(q-p)A(\upsilon )-(s-p)r^{s-q}B(\upsilon
)-(2-p)r^{2-q}\|\nabla \upsilon \|_2^2,
\]
which, combined with \eqref{cp1}, gives
\[
r^{p-q+1}Q_{r}(r,\upsilon )=(2-p)\|\nabla \upsilon \|_2^2(r_{\ast
}^{2-q}-r^{2-q})+(s-p)B(\upsilon )(r_{\ast }^{s-q}-r^{s-q})<0.
\]
By the implicit function theorem $r(\cdot)$ is continuously
differentiable. Let
\begin{equation}
G_3:=\big\{\upsilon \in G_1:\|\nabla \upsilon \|_{p}^{p}<\frac{p}{q}
\frac{2-q}{2-p}r_{\ast }(\upsilon )^{q-p}A(\upsilon )\big\}\label{s1}
\end{equation}
and assume that $G_3\neq \emptyset $. Since $q>p$ and
$r(\upsilon)>r_{\ast }(\upsilon )$, we see that $G_3\subseteq G_2$
and so $G_2\neq \emptyset $. If $\upsilon \in $ $G_3$, then
\begin{equation}
\|\nabla \upsilon \|_{p}^{p}<\frac{p}{q}\frac{2-q}{2-p}r_{\ast }(\upsilon
)^{q-p}A(\upsilon ),  \label{f1}
\end{equation}
and so
\[
\|\nabla \upsilon \|_{p}^{p}<\frac{p}{q}\frac{2-q}{2-p}r(\upsilon
)^{q-p}A(\upsilon ).
\]
Thus
\begin{equation}
\frac{2-p}{p}r(\upsilon )^{p}\|\nabla \upsilon \|_{p}^{p}+\frac{q-2}{q}
r(\upsilon )^{q}A(\upsilon )<0.  \label{f2}
\end{equation}
By \eqref{f2} and \eqref{e4} we conclude that
\[
\hat{\Phi}(\upsilon )<r^{p}(\frac{1}{p}-\frac{1}{2})\|\nabla \upsilon
\|_{p}^{p}+r^{q}(\frac{1}{2}-\frac{1}{q})A(\upsilon )<0.
\]
On the other hand, if $\upsilon \in G_2\cap S^1$, by \eqref{BF1}
\begin{equation}
r(\upsilon )\leq \Big( \frac{A(\upsilon )}{\|\nabla \upsilon \|_2^2}
\Big) ^{1/(2-q)},  \label{br}
\end{equation}
and so $r(\cdot)$ is bounded on $G_2\cap S^1$. Consequently,
$\hat{\Phi} (\upsilon )$ is also bounded on $G_2\cap S^1$. Let
\[
M:=\inf_{\upsilon \in G_2\cap S^1} \hat{\Phi}(\upsilon )<0.
\]
Suppose that $\{\upsilon _n\}_{n\in\mathbb{N}}$ is a minimizing
sequence in $G_2\cap S^1$. Then there exists
$\tilde{\upsilon}$ $\in $ $E$ such that, at least for a subsequence,
$A(\upsilon_n)\to A(\tilde{\upsilon})$, $B(\upsilon _n)\to B(\tilde{
\upsilon})$,
\begin{gather*}
0\leq \|\nabla \tilde{\upsilon}\|_2\leq \lim \inf \|\nabla \upsilon
_n\|_2\leq 1,\\
0\leq \|\nabla \tilde{\upsilon}\|_{p}\leq \lim \inf \|\nabla \upsilon
_n\|_{p}\leq 1.
\end{gather*}
We must have $\tilde{\upsilon}\neq 0$ because, otherwise, $0=\Phi (0)\leq
\lim \inf_{n\to \infty }\Phi (r(\upsilon _n)\upsilon _n)=M$, a
contradiction. Since $\{r(\upsilon _n)\}_{n\in\mathbb{N}}$ is bounded
and $r_{\ast }(\upsilon _n)<r(\upsilon _n)$, $n\in\mathbb{N}$, we may
assume that $r_{\ast }(\upsilon _n)\to \tilde{r}_{\ast }$ and
$r(\upsilon _n)\to \tilde{r}>0$.
If $A(\tilde{\upsilon})=0$, then, by \eqref{br}, we obtain that
$\tilde{r}=0$ which is a
contradiction. Thus, $A(\tilde{\upsilon})>0$ and so
$\tilde{\upsilon}\in G_1$. Also, $\tilde{r}_{\ast }>0$
by \eqref{cp1}. We claim that $\tilde{\upsilon}\in G_3$. Indeed,
by \eqref{BE2},
\begin{equation}
\begin{aligned}
\|\nabla \tilde{\upsilon}\|_{p}^{p}
&\leq \limsup_{n\to \infty } \|\nabla \upsilon _n\|_{p}^{p}
\leq \limsup_{n\to \infty } Q(r_{\ast }(\upsilon _n),\upsilon _n) \\
&\leq \limsup_{n\to \infty } \{r_{\ast }(\upsilon _n)^{q-p}
 A(\upsilon _n)-r_{\ast }(\upsilon _n)^{s-p}B(\upsilon
_n)\}-\liminf_{n\to \infty } r_{\ast }(\upsilon_n)^{2-p}\|\nabla
\upsilon _n\|_2^2\\
&\leq \tilde{r}_{\ast }^{q-p}A(\tilde{\upsilon})-\tilde{r}_{\ast }^{s-p}B(
\tilde{\upsilon})-\tilde{r}_{\ast }^{2-p}\|\nabla \tilde{\upsilon}
\|_2^2=Q(\tilde{r}_{\ast },\tilde{\upsilon}),
\end{aligned} \label{u1}
\end{equation}
implying that
\begin{equation}
\|\nabla \tilde{\upsilon}\|_{p}^{p}
\leq Q(r_{\ast }(\tilde{\upsilon}),\tilde{\upsilon}).  \label{ineq}
\end{equation}
If we assume the equality
\begin{equation}
\|\nabla \tilde{\upsilon}\|_{p}^{p}=Q(r_{\ast }(\tilde{\upsilon}),\tilde{
\upsilon}),  \label{u2}
\end{equation}
then by using \eqref{BE} for $\upsilon =\upsilon _n$ and passing
to the limit as $n\to +\infty $, we obtain
\begin{equation}
\begin{aligned}
&\|\nabla \tilde{\upsilon}\|_{p}^{p}\\
&\leq \limsup_{n\to \infty } \|\nabla \tilde{\upsilon}_n\|_{p}^{p}
\leq \limsup_{n\to \infty } Q(r(\upsilon _n),\upsilon _n)\\
&\leq \limsup_{n\to \infty } \{r(\upsilon _n)^{q-p}A(\upsilon _n)
-r(\upsilon _n)^{s-p}B(\upsilon _n)\}
- \liminf_{n\to \infty } r(\upsilon _n)^{2-p}\|\nabla
\upsilon _n\|_2^2 \\
&\leq \tilde{r}^{q-p}A(\tilde{\upsilon})-\tilde{r}^{s-p}B(\tilde{\upsilon})-
\tilde{r}^{2-p}\|\nabla \tilde{\upsilon}\|_2^2=Q(\tilde{r},\tilde{
\upsilon}).
\end{aligned} \label{u3}
\end{equation}
In view of \eqref{u1}, \eqref{u2} and \eqref{u3}, we conclude
that $\tilde{r} =\tilde{r}_{\ast }=\tilde{r}_{\ast }(\tilde{\upsilon})$. On the other hand,
by replacing $\upsilon $ by $\upsilon _n$ in \eqref{cp1} and passing to
the limit we obtain
\[
(q-p)A(\tilde{\upsilon})\geq (s-p)r_{\ast }(\tilde{\upsilon})^{s-q}B(\tilde{
\upsilon})+(2-p)r_{\ast }(\tilde{\upsilon})^{2-q}\|\nabla \tilde{\upsilon}
\|_2^2.
\]
Since $r_{\ast }(\tilde{\upsilon})$ satisfies
\[
(q-p)A(\tilde{\upsilon})=(s-p)r_{\ast }(\tilde{\upsilon})^{s-q}B(\tilde{
\upsilon})+(2-p)r_{\ast }(\tilde{\upsilon})^{2-q}\|\nabla \tilde{\upsilon}
\|_2^2,
\]
we deduce that $\|\nabla \upsilon _n\|_2^2\to \|\nabla \tilde{
\upsilon}\|_2^2$ and
\begin{equation}
\left( q-p\right) A(\tilde{\upsilon})=\left( s-p\right) r_{\ast }(\tilde{
\upsilon})^{s-q}B(\tilde{\upsilon})+\left( 2-p\right) r_{\ast }(\tilde{
\upsilon})^{2-q}\|\nabla \tilde{\upsilon}\|_2^2.  \label{k1}
\end{equation}
Thus,
\begin{equation}
A(\tilde{\upsilon})=\frac{s-p}{q-p}\tilde{r}^{s-q}B(\tilde{\upsilon})+\frac{
2-p}{q-p}\tilde{r}^{2-q}\|\nabla \tilde{\upsilon}\|_2^2.  \label{a}
\end{equation}
On the other hand, \eqref{e1} and \eqref{a} imply that
\[
M=\lim_{n\to \infty }\hat{\Phi}(\upsilon _n)=\frac{(s-q)(s-p)}{pqs}
\tilde{r}^{s}B(\tilde{\upsilon})+\frac{(2-p)(2-q)}{2pq}\tilde{r}^2\|\nabla
\tilde{\upsilon}\|_2^2>0,
\]
a contradiction. Therefore, $\tilde{\upsilon}\in G_3$ proving the claim. We
shall show next that $\tilde{r}=r(\tilde{\upsilon})$. Let $t>0$ be such
that $t\tilde{\upsilon}\in S^1$. Since for $t>0$
\begin{equation}
r_{\ast }(t\tilde{\upsilon})t\tilde{\upsilon}
=r_{\ast }(\tilde{\upsilon})
\tilde{\upsilon},  \label{mr}
\end{equation}
by \eqref{BE2}, \eqref{s1} and \eqref{mr}, we have
\[
\|\nabla \tilde{\upsilon}\|_{p}^{p}<Q(r_{\ast }(\tilde{\upsilon}),\tilde{
\upsilon})=Q(tr_{\ast }(t\tilde{\upsilon}),\tilde{\upsilon})=t^{-p}Q(r_{\ast
}(t\tilde{\upsilon}),t\tilde{\upsilon}).
\]
Thus
\[
\|t\nabla \tilde{\upsilon}\|_{p}^{p}\leq Q(r_{\ast }(t\tilde{\upsilon}),t
\tilde{\upsilon}),
\]
which implies $t\tilde{\upsilon}$ $\in $ $G_2\cap S^1$.
Furthermore, by \eqref{BE2}, $r(t\tilde{\upsilon})$ satisfies
\begin{equation}
Q(tr(t\tilde{\upsilon}),\tilde{\upsilon})=\|\nabla \tilde{\upsilon}
\|_{p}^{p}=Q(r(\tilde{\upsilon}),\tilde{\upsilon}),  \label{th}
\end{equation}
which gives
\begin{equation}
tr(t\tilde{\upsilon})=r(\tilde{\upsilon}).  \label{mr1}
\end{equation}
In view of \eqref{u3},
\[
Q(r(\tilde{\upsilon}),\tilde{\upsilon})=\|\nabla \tilde{\upsilon}
\|_{p}^{p}\leq Q(\tilde{r},\tilde{\upsilon}),
\]
implying that $\tilde{r}\leq r(\tilde{\upsilon})$.
If we assume that $\tilde{r}<r(\tilde{\upsilon})$, then, since
the function $z\to \Phi (z\tilde{\upsilon})$ is strictly decreasing
in $(\tilde{r},r(\tilde{\upsilon}))$, by \eqref{mr1} we obtain
\[
M=\liminf_{n\to \infty } \Phi (r(\upsilon _n)\upsilon
_n)\geq \Phi (\tilde{r}\tilde{\upsilon})>\Phi (r(\tilde{\upsilon})\tilde{
\upsilon})=\Phi (r(t\tilde{\upsilon})t\tilde{\upsilon})=\hat{\Phi}(t\tilde{
\upsilon}),
\]
which is a contradiction. Thus $\tilde{r}=r(\tilde{\upsilon})$.
Then \eqref{fe} holds, and so $\tilde{\upsilon}\in S^1$ and
 $\hat{\Phi}(\tilde{\upsilon})=M$. As in the previous case we may
assume that $\tilde{\upsilon}\geq 0$. Lemma \ref{FibLemma}
implies that $u:=r(\tilde{\upsilon})
\tilde{\upsilon}$ is a solution to \eqref{1}-\eqref{2}.

Therefore, we have proved the following result.

\begin{theorem} \label{thm3}
Assume that  conditions {\rm (H0)-(H2)} are satisfied,
$p<q<2<s$ and the set $G_3$ defined in \eqref{s1} is not empty.
Then the problem \eqref{1}-\eqref{2} admits a non-negative
solution $u\in C^{1,\delta }(\Omega )$ for some $\delta \in
(0,1)$.
\end{theorem}

\begin{remark} \label{rmk1} \rm
We will now give some conditions which guarantee that
$G_3\neq \emptyset $. Suppose that
$\operatorname{supp}a^{+})\subseteq \operatorname{supp}b)$.
Then there exists $\upsilon $ $\in S^1$ such that $B(\upsilon )>0$.
Since $r_{\ast }(\upsilon )^{2-q}<r(\upsilon )^{2-q}$,
 \eqref{cp1} yields
\begin{equation}
(q-p)A(\upsilon )<(s-p)r_{\ast }(\upsilon )^{s-q}B(\upsilon
)+(2-p)r(\upsilon )^{2-q}\|\nabla \upsilon \|_2^2,  \label{arb1}
\end{equation}
and so
\[
r_{\ast }(\upsilon )^{s-q}>\frac{q-p}{s-p}\frac{A(\upsilon )}{B(\upsilon )}-
\frac{2-p}{s-p}r(\upsilon )^{2-q}\frac{\|\nabla \upsilon \|_2^2}{
B(\upsilon )}.
\]
Consequently,
\begin{equation}
\begin{aligned}
&\frac{p}{q}\frac{2-q}{2-p}r_{\ast }(\upsilon )^{q-p}A(\upsilon )\\
&>\frac{p}{q}\frac{2-q}{2-p}\Big( \frac{q-p}{s-p}\frac{A(\upsilon )}{
B(\upsilon )}-\frac{2-p}{s-p}r(\upsilon )^{2-q}\frac{\|\nabla \upsilon
\|_2^2}{B(\upsilon )}\Big) ^{(q-p)/(s-q)} A(\upsilon ).
\end{aligned}\label{con1}
\end{equation}
On the other hand, \eqref{BF1} implies that
\begin{equation}
r(\upsilon )\leq \Big( \frac{A(\upsilon )}{B(\upsilon )}\Big)
 ^{1/(s-q)},  \label{rg}
\end{equation}
which combined with \eqref{con1} gives
\begin{align*}
&\frac{p}{q}\frac{2-q}{2-p}\Big( \frac{q-p}{s-p}\frac{A(\upsilon )}{
B(\upsilon )}-\frac{2-p}{s-p}r(\upsilon )^{2-q}\frac{\|\nabla \upsilon
\|_2^2}{B(\upsilon )}\Big) ^{(q-p)/(s-q)}A(\upsilon )
\\
&>\frac{p}{q}\frac{2-q}{2-p}\Big( \frac{q-p}{s-p}\frac{A(\upsilon )}{
B(\upsilon )}-\frac{2-p}{s-p}\Big( \frac{A(\upsilon )}{B(\upsilon )}
\Big)^{(2-q)/(s-q)}\frac{\|\nabla \upsilon \|_2^2}{B(\upsilon )}\Big) ^{
\frac{q-p}{s-q}}A(\upsilon ).
\end{align*}
If $a^{+}(\cdot)$ is large enough then
\begin{equation}
\frac{p}{q}\frac{2-q}{2-p}
\Big( \frac{q-p}{s-p}\frac{A(\upsilon )}{
B(\upsilon )}-\frac{2-p}{s-p}A(\upsilon )^{(2-q)/(s-q)}\frac{\|\nabla
\upsilon \|_2^2}{B(\upsilon )^{\frac{2-q}{s-q}+1}}\Big)
^{(q-p)/(s-q)}A(\upsilon )>\|\nabla \upsilon \|_{p}^{p},  \label{rd}
\end{equation}
implying that $\upsilon \in G_3$.
\end{remark}

 Suppose now that
$\big( \operatorname{supp}a^{+})\backslash \operatorname{supp}b)\big)
^o\neq \emptyset $. Then there exists $\upsilon $ $\in S^1$ with
$B(\upsilon )=0$. From \eqref{cp1} we see that
\begin{equation}
r_{\ast }(\upsilon )=\Big( \frac{q-p}{2-p}\frac{A(\upsilon )}{\|\nabla
\upsilon \|_2^2}\Big) ^{1/(2-q)},  \label{z2}
\end{equation}
and so
\[
\frac{p}{q}\frac{2-q}{2-p}r_{\ast }(\upsilon )^{q-p}A(\upsilon )
=\frac{p}{q}
\frac{2-q}{2-p}\Big( \frac{q-p}{2-p}\frac{A(\upsilon )}{\|\nabla \upsilon
\|_2^2}\Big) ^{(q-p)/(2-q)}A(\upsilon ).
\]
Consequently, if $a^{+}(\cdot)$ is large enough,
\begin{equation}
\frac{p}{q}\frac{2-q}{2-p}\Big( \frac{q-p}{2-p}\Big) ^{\frac{q-p}{2-q}
}A(\upsilon )^{\frac{2-p}{2-q}}>\|\nabla \upsilon \|_2
^{2(2-p)/(2-q)},
\label{rem3}
\end{equation}
implying that $G_{_3}\neq \emptyset$.

 \textbf{Case 3:} $p<s<q<2$.
In this case we define
\[
Q(r,\upsilon ):=r^{q-p}A(\upsilon )-r^{s-p}B(\upsilon )-r^{2-p}\|\nabla
\upsilon \|_2^2.
\]
Let $\upsilon \in G_1$ and assume that $B(\upsilon )>0$.
For $r\geq 0$ let
\begin{equation}
F(r,\upsilon ):=r^{p-s}Q(r,\upsilon )=r^{q-s}A(\upsilon )-B(\upsilon
)-\|\nabla \upsilon \|_2^2r^{2-s}.  \label{bi}
\end{equation}
Then, $F(0,\upsilon )=-B(\upsilon )<0$ and
$\lim_{r\to +\infty } F(r,\upsilon )=-\infty $. It is easy to see
that $F(\cdot,\upsilon )$ attains its maximum at
\begin{equation}
\bar{r}(\upsilon )=\Big( \frac{q-s}{2-s}\frac{A(\upsilon )}{\|\nabla
\upsilon \|_2^2}\Big) ^{1/(2-q)}  \label{by}
\end{equation}
with
\begin{equation}
F(\bar{r}(\upsilon ),\upsilon )=\frac{2-q}{2-s}\bar{r}^{q-s}A(\upsilon
)-B(\upsilon ).  \label{bz}
\end{equation}
Consequently, $Q(r,\upsilon )>0$ for some $r>0$ if and only if
$F(\bar{r}(\upsilon ),\upsilon )>0$, and this holds if
\begin{equation}
\bar{r}(\upsilon )>\hat{r}(\upsilon )
:=(\frac{2-s}{2-q}\frac{B(\upsilon )}{
A(\upsilon )})^{1/(q-s)}.  \label{ba}
\end{equation}
Suppose that \eqref{ba} holds. Then it is easy to see that the
function
\[
r\mapsto r^{p-s+1}Q_{r}(r,\upsilon )=(q-p)r^{q-s}A(\upsilon
)-(2-p)\|\nabla \upsilon \|_2^2r(\upsilon )^{2-s}-(s-p)B(\upsilon ),
\]
has two positive roots $r_{1\ast }(\upsilon )$ and
$r_{2\ast }(\upsilon )$ with
$r_{1\ast }(\upsilon )<r_{2\ast }(\upsilon )$. Clearly,
$r_{1\ast}(\upsilon )$ is a point of local minimum of
$Q(.,\upsilon )$ while
$r_{2\ast }(\upsilon )$ is a point of global maximum of $Q(.,\upsilon )$.
Define $r_{\ast }(\upsilon ):=r_{2\ast }(\upsilon )$. We claim that
\begin{equation}
\bar{r}(\upsilon )<r_{\ast }(\upsilon ).  \label{bv}
\end{equation}
Indeed,
\[
r^{s-p}F_{r}(r,\upsilon )=Q_{r}(r,\upsilon )+(p-s)\frac{Q(r,\upsilon )}{r},
\]
and since $F_{r}(\bar{r}(\upsilon ),\upsilon )=0$ and
$Q(\bar{r}(\upsilon),\upsilon )
=\bar{r}(\upsilon )^{s-p}F(\bar{r}(\upsilon ),\upsilon )>0$ we
get
\[
Q_{r}(\bar{r}(\upsilon ),\upsilon )=(s-p)\frac{Q(\bar{r}(\upsilon ),\upsilon
)}{\bar{r}(\upsilon )}>0,
\]
proving the claim.

 Next, let $\upsilon \in G_1$ and assume that $B(\upsilon )=0$.
Clearly $Q(\cdot,\upsilon )$ attains its maximum at
\begin{equation}
r_{\ast }(\upsilon ):=\Big( \frac{q-p}{2-p}\frac{A(\upsilon )}{\|\nabla
\upsilon \|_2^2}\Big) ^{1/(2-q)}  \label{bn}
\end{equation}
with
\begin{equation}
Q(r_{\ast }(\upsilon ),\upsilon )=\frac{2-q}{2-p}r_{\ast }(\upsilon
)^{q-p}A(\upsilon ).  \label{bc}
\end{equation}
Since $r_{\ast }(\upsilon )$ satisfies the equation
$Q_{r}(\cdot,\upsilon )=0$,
that is
\begin{equation}
(q-p)A(\upsilon )r_{\ast }(\upsilon )^{q-s}=(s-p)B(\upsilon )+(2-p)\|\nabla
\upsilon \|_2^2r_{\ast }(\upsilon )^{2-s},  \label{hu}
\end{equation}
we have that
\begin{equation}
r_{\ast }(\upsilon )\leq \Big( \frac{q-p}{2-p}\frac{A(\upsilon )}{
\|\nabla \upsilon \|_2^2}\Big) ^{1/(2-q)}.  \label{ml}
\end{equation}
If $\upsilon \in G_2$ and the condition \eqref{ba} is satisfied,
then \eqref{BE} has two positive solutions $r_1(\upsilon )$,
$r_2(\upsilon )$ with
$r_1(\upsilon )<r_{\ast }(\upsilon )<r_2(\upsilon )$.
Define
$r(\upsilon ):=r_2(\upsilon )$. Since $Q_{r}(r,\upsilon )<0$ for all
$r>r_{\ast }(\upsilon )$, by the implicit function theorem,
$r\in C^1(G_2)$. We will assume that the set
\begin{equation}  \label{xz}
G_4:=\{\upsilon \in G_1:\|\nabla \upsilon \|_{p}^{p}
\leq \frac{p}{s}\frac{2-s}{2-p}\big( \frac{s}{q}\frac{2-q}{2-s}\bar{r
}(\upsilon )^{q-s}A(\upsilon )-B(\upsilon )\big) \bar{r}(\upsilon
)^{s-p}\}
\end{equation}
is not empty. Thus,
\[
\bar{r}(\upsilon )>\Big( \frac{q}{s}\frac{2-s}{2-q}\frac{B(\upsilon )}{
A(\upsilon )}\Big) ^{1/(q-s)}.
\]
We will show that $G_4\subseteq G_2$. Indeed,
let $\upsilon \in G_5$
and assume first that $B(\upsilon )>0$. Then, since
$\frac{p}{s},\frac{2-s}{2-p}$ and $\frac{s}{q}$ are less
than $1$, \eqref{bi}, \eqref{bz}, \eqref{bv}
 and \eqref{xz} imply that
\begin{align*}
\|\nabla \upsilon \|_{p}^{p}
&<\Big( \frac{s}{q}\frac{2-q}{2-s}\bar{r}
(\upsilon )^{q-s}A(\upsilon )-B(\upsilon )\Big)
\bar{r}(\upsilon )^{s-p} \\
&<\Big( \frac{2-q}{2-s}\bar{r}(\upsilon )^{q-s}A(\upsilon )-B(\upsilon
)\Big) \bar{r}(\upsilon )^{s-p}\\
&=F(\bar{r}(\upsilon ),\upsilon )\bar{r}(\upsilon )^{s-p}
=Q(\bar{r}(\upsilon),\upsilon )\\
&<Q(r_{\ast }(\upsilon ),\upsilon ),
\end{align*}
and so $\upsilon \in G_2$. Next, let $\upsilon \in $ $G_4$
and assume $B(\upsilon )=0$. Then, from \eqref{bv},
\[
\|\nabla \upsilon \|_{p}^{p}<\frac{p}{q}\frac{2-q}{2-p}\bar{r}(\upsilon
)^{q-p}A(\upsilon )<\frac{2-q}{2-p}r_{\ast }(\upsilon )^{q-p}A(\upsilon
)=Q(r_{\ast }(\upsilon ),\upsilon ),
\]
which shows that $\upsilon \in G_2$. Notice also that
$G_4\cap S^1\neq \emptyset $. Since
$\bar{r}(\upsilon )<r_{\ast }(\upsilon )<r(\upsilon )$
for any $\upsilon \in G_4$, we get
\[
\|\nabla \upsilon \|_{p}^{p}\leq \frac{p}{s}\frac{2-s}{2-p}
\Big( \frac{s}{q} \frac{2-q}{2-s}r(\upsilon )^{q-s}A(\upsilon )
-B(\upsilon )\Big) r(\upsilon)^{s-p},
\]
which, in view of \eqref{e4}, implies that $\hat{\Phi}(\upsilon )<0$
for $\upsilon \in G_{4.}$ On the other hand, if
$\upsilon \in G_2\cap S^1$,
then \eqref{BF1} implies that
\begin{equation}
r(\upsilon )\leq \Big( \frac{A(\upsilon )}{\|\nabla \upsilon \|_2^2}
\Big) ^{1/(2-q)},  \label{ds}
\end{equation}
and so $r(\cdot)$ is bounded on $G_2\cap S^1$. Therefore
$\hat{\Phi} (\upsilon )$ is bounded on $G_2\cap S^1$. Let
\[
M:=\inf_{\upsilon \in G_2\cap S^1} \hat{\Phi}(\upsilon )<0.
\]
Suppose that $\{\upsilon _n\}_{n\in\mathbb{N}}$ is a minimizing
sequence for $\hat{\Phi}(\cdot)$ in
$\widetilde{G}_2\cap S^1$. Then, there exist
$\tilde{\upsilon}\in E$ such that, at least for a
subsequence, $A(\upsilon _n)\to A(\tilde{\upsilon})$,
$B(\upsilon _n)\to B(\tilde{\upsilon})$. We must have
$\tilde{\upsilon}\neq 0$ because, otherwise,
$0=\Phi (0)\leq \liminf_{n\to \infty }\Phi (r(\upsilon _n)
\upsilon _n)=M$, a
contradiction. Since $\{r(\upsilon _n)\}_{n\in\mathbb{N}}$ is bounded
we get $r(\upsilon _n)\to $ $\tilde{r}$, and
$r_{\ast }(\upsilon _n)\to \tilde{r}_{\ast }$. On the other
hand, $\tilde{r}>0$ because
$M=\liminf_{n\to \infty }\hat{\Phi}(\upsilon _n)<0$.
If we assume that $A(\tilde{\upsilon})=0$,
then, by \eqref{ds}, we should have $\tilde{r}=0$, a contradiction.
Thus, $\tilde{\upsilon}\in G_1$. Also, by \eqref{ba} and \eqref{bv},
we have
\begin{equation}
\tilde{r}\geq \tilde{r}_{\ast }\geq \hat{r}(\tilde{\upsilon}):=(\frac{2-s}{
2-q}\frac{B(\tilde{\upsilon})}{A(\tilde{\upsilon})})^{1/(q-s)}.
\label{df}
\end{equation}
We will show that $\tilde{\upsilon}\in G_2$. Indeed, if not, then,
as in proof of the previous Theorem,
$\tilde{r}=\tilde{r}_{\ast }=r_{\ast }(\tilde{ \upsilon})$ where
$r_{\ast }(\tilde{\upsilon})$ is the point of global
maximum of $Q(\cdot,\tilde{\upsilon})$ which satisfies
\[
(q-p)A(\tilde{\upsilon})r_{\ast }(\tilde{\upsilon})^{q-s}
=(s-p)B(\tilde{ \upsilon})+(2-p)\|\nabla \tilde{\upsilon}\|_2^2
r_{\ast }(\tilde{\upsilon})^{2-s}.
\]
Consequently, by passing to the limit in \eqref{hu}, where we have
replaced $\upsilon $ by $\upsilon _n$, $n\in\mathbb{N}$,
we get $\|\nabla \upsilon _n\|_2^2\to \|\nabla \tilde{
\upsilon}\|_2^2$, where
\begin{equation}
(q-p)A(\tilde{\upsilon})\tilde{r}^{q-s}-(s-p)B(\tilde{\upsilon}
)=(2-p)\|\nabla \tilde{\upsilon}\|_2^2\tilde{r}^{2-s}.  \label{vr}
\end{equation}
This, however, leads to a contradiction since, \eqref{e1}, \eqref{vr}
and \eqref{df},
\[
M=\lim_{n\to \infty }\hat{\Phi}(\upsilon _n)
=\frac{(q-p)(2-q)}{2pq}
\Big( \tilde{r}^{q-s}A(\tilde{\upsilon})-\frac{q}{s}\frac{s-p}{q-p}
\frac{2-s}{2-q}\frac{B(\tilde{\upsilon})}{A(\tilde{\upsilon})}\Big)
\tilde{r}^{s}A(\tilde{\upsilon})>0.
\]
Therefore, $\tilde{\upsilon}\in G_2$ as claimed. A similar reasoning
as in Case 2 shows that $\tilde{r}$ $=r(\tilde{\upsilon})$.
Finally, by passing to the limit in \eqref{BE2} we conclude
that $\tilde{\upsilon}\in $ $S^1$ and
$\hat{\Phi}(\tilde{\upsilon})=M$. Lemma \ref{FibLemma} implies
that $u:=r(\tilde{\upsilon})\tilde{\upsilon}\geq 0$ is a solution
to \eqref{1}-\eqref{2}. Therefore, we have the following result.


\begin{theorem} \label{thm4}
Assume that {\rm (H0)-(H2)} are satisfied,
$p<s<q<2$ and the set $G_4$ defined in \eqref{xz} is not empty.
Then  \eqref{1} -\eqref{2} admits a non-negative
solution $u\in C^{1,\delta }(\Omega )$ for some $\delta \in
(0,1)$.
\end{theorem}

\begin{remark} \label{rmk2}\rm
 We will now give some conditions which guarantee that
$G_4\neq \emptyset $. Suppose that
$\operatorname{supp}a^{+}\subseteq \operatorname{supp}b$.
Then there exists $\upsilon \in S^1$ such that $B(\upsilon )>0$.
From \eqref{by} we obtain
\begin{align*}
&\frac{p}{q}\frac{2-q}{2-p}\bar{r}(\upsilon )^{q-p}A(\upsilon )
 -\frac{p}{s}\frac{2-s}{2-p}B(\upsilon )\bar{r}(\upsilon )^{s-p}
\\
&=\frac{p}{q}\frac{2-q}{2-p}\Big( \frac{q-s}{2-s}\frac{A(\upsilon )}{
\|\nabla \upsilon \|_2^2}\Big) ^{(q-p)/(2-q)}A(\upsilon )-\frac{p}{
s}\frac{2-s}{2-p}B(\upsilon )\Big( \frac{q-s}{2-s}\frac{A(\upsilon )}{
\|\nabla \upsilon \|_2^2}\Big) ^{(s-p)/(2-q)}
\\
&=\frac{p}{q}\frac{2-q}{2-p}\Big( \frac{q-s}{2-s}\frac{1}{\|\nabla \upsilon
\|_2^2}\Big) ^{(q-p)/(2-q)}A(\upsilon )^{\frac{q-p}{2-q}+1} \\
&\quad -\frac{p}{s}\frac{2-s}{2-p}B(\upsilon )\Big( \frac{q-s}{2-s}\frac{1}{
\|\nabla \upsilon \|_2^2}\Big) ^{(s-p)/(2-q)}A(\upsilon )^{\frac{
s-p}{2-q}}.
\end{align*}
If we assume that
\begin{equation} \label{sw}
\begin{aligned}
&\frac{p}{q}\frac{2-q}{2-p}\Big( \frac{q-s}{2-s}\frac{1}{\|\nabla \upsilon
\|_2^2}\Big) ^{(q-p)/(2-q)}A(\upsilon )^{\frac{q-p}{2-q}+1}
 \\
&-\frac{p}{s}\frac{2-s}{2-p}B(\upsilon )\Big( \frac{q-s}{2-s}\frac{1}{
\|\nabla \upsilon \|_2^2}\Big) ^{(s-p)/(2-q)}A(\upsilon )
^{(s-p)/(2-q)}>\|\nabla \upsilon \|_{p}^{p},
\end{aligned}
\end{equation}
then $\upsilon\in G_4$. It is easy to see that if
$a^{+}(\cdot)$ is large enough then \eqref{sw} is true.
\end{remark}

 On the other hand, suppose that
$\left( \operatorname{supp}a^{+})\backslash
\operatorname{supp}b)\right) ^{o}\neq \emptyset $. Then there exists
$\upsilon \in G_1$
with $B(\upsilon )=0$. From \eqref{by} we obtain
\begin{align*}
\frac{p}{q}\frac{2-q}{2-p}\bar{r}(\upsilon )^{q-p}A(\upsilon ) 
&=\frac{p}{q}\frac{2-q}{2-p}\Big( \frac{q-s}{2-s}\frac{A(\upsilon )}{\|\nabla
\upsilon \|_2^2}\Big) ^{(q-p)/(2-q)}A(\upsilon )\\
&=\frac{p}{q}\frac{
2-q}{2-p}\Big( \frac{q-s}{2-s}\frac{1}{\|\nabla \upsilon \|_2^2}\Big)
^{(q-p)/(2-q)}A(\upsilon )^{\frac{q-p}{2-q}+1}.
\end{align*}
If we assume that
\begin{equation}
\frac{p}{q}\frac{2-q}{2-p}\Big( \frac{q-s}{2-s}\frac{1}{\|\nabla \upsilon
\|_2^2}\Big) ^{(q-p)/(2-q)}A(\upsilon )^{\frac{q-p}{2-q}
+1}>\|\nabla \upsilon \|_{p}^{p},  \label{rem26}
\end{equation}
then $\upsilon \in G_4$. Note that if $a^{+}(\cdot)$ is large
enough then \eqref{rem26} holds.

\textbf{Case 4:} $p<2<q<s$.
In this case we make the additional assumption:
\begin{itemize}
\item[(H3)] $b(x)\geq bo>0$ a.e. in $\Omega $.
\end{itemize}
 Let
\begin{equation}
Q(r,\upsilon ):=r^{q-2}A(\upsilon )-r^{s-2}B(\upsilon )-r^{p-2}\|\nabla
\upsilon \|_{p}^{p}.  \label{BE3}
\end{equation}
Then \eqref{BE} is equivalent to
\begin{equation}
Q(r,\upsilon )=\|\nabla \upsilon \|_2^2.  \label{BE4}
\end{equation}

 For every $\upsilon \in G_1$ the function $Q(\cdot,\upsilon )$
has a unique critical point $r_{\ast }:=r_{\ast }(\upsilon )$ which
corresponds to a global maximum with
\begin{equation}
(q-2)r_{\ast }^{q-p}A(\upsilon )+(2-p)\|\nabla \upsilon
\|_{p}^{p}=(s-2)r_{\ast }^{s-p}B(\upsilon ).  \label{eqr*}
\end{equation}
Thus,
\begin{equation}
r_{\ast }(\upsilon )\geq (\frac{q-2}{s-2}\frac{A(\upsilon )}{B(\upsilon )})^{
\frac{1}{s-q}}.  \label{rab*}
\end{equation}
On combining \eqref{BE3} with \eqref{eqr*} we get
\begin{align}
Q(r_{\ast }(\upsilon ),\upsilon )
&=\frac{q-p}{2-p}r_{\ast }(\upsilon
)^{q-2}A(\upsilon )-\frac{s-p}{2-p}r_{\ast }(\upsilon )^{s-2}B(\upsilon )
  \label{m44} \\
&=\frac{s-q}{s-2}r_{\ast }(\upsilon )^{q-2}A(\upsilon )-\frac{s-p}{s-2}
r_{\ast }(\upsilon )^{p-2}\|\nabla \upsilon \|_{p}^{p}.  \label{m4}
\end{align}
Let
\[
\tilde{G}_2:=\{\upsilon \in G_1:\|\nabla \upsilon \|_2^2<Q(r_{\ast
}(\upsilon ),\upsilon )\}.
\]
Clearly, if $\upsilon \in \tilde{G}_2$, then \eqref{BE} has exactly two
positive solutions $r_1(\upsilon )$ and $r_2(\upsilon )$ with
$r_1(\upsilon )<r_{\ast }(\upsilon )<r_2(\upsilon )$.
As before, let $r:=r_2(\upsilon )$. Since
\[
r^{2-q+1}Q_{r}(r,\upsilon )=(q-2)A(\upsilon )-(s-2)r^{s-q}B(\upsilon
)-(p-2)r^{p-q}\|\nabla \upsilon \|_{p}^{p},
\]
in view of \eqref{eqr*}, we obtain
\[
r^{2-q+1}Q_{r}(r,\upsilon )=(s-2)B(\upsilon )(r_{\ast
}^{s-q}-r^{s-q})+(2-p)\|\nabla \upsilon \|_{p}^{p}(r^{p-q}-r_{\ast
}^{p-q})<0,
\]
which implies that $r(\cdot)$ is continuously differentiable. We
now define
\begin{equation} \label{g5}
G_5:=\{\upsilon \in G_1:\|\nabla \upsilon \|_2^2<\frac{2}{q}
\frac{s-q}{s-2}A(\upsilon )r_{\ast }(\upsilon )^{q-2}
 -\frac{2}{p}\frac{s-p}{s-2}\|\nabla \upsilon \|_{p}^{p}r_{\ast
}(\upsilon )^{p-2}\},
\end{equation}
and assume that $G_5\neq \emptyset $. Since $\frac{2}{q}<1$ and
$\frac{2}{p}>1$ we see that $G_5\subseteq \tilde{G}_2$, and so
 $\tilde{G}_2\neq \emptyset $ as well. Furthermore,
$G_5\cap S^1\neq \emptyset $ because $r\ $satisfies \eqref{mr1}.
If $\upsilon \in G_5$, then by \eqref{g5},
\begin{equation}
\|\nabla \upsilon \|_2^2<\frac{2}{q}\frac{s-q}{s-2}A(\upsilon
)r(\upsilon )^{q-2}-\frac{2}{p}\frac{s-p}{s-2}\|\nabla \upsilon
\|_{p}^{p}r(\upsilon )^{p-2}.  \label{ad1}
\end{equation}
On the other hand, \eqref{e3} and \eqref{ad1} show that
\[
r^{p}(\frac{1}{p}-\frac{1}{s})\|\nabla \upsilon \|_{p}^{p}+r^2(\frac{1}{2}-
\frac{1}{s})\|\nabla \upsilon \|_2^2+r^{q}(\frac{1}{s}-\frac{1}{q}
)A(\upsilon )<0,
\]
and so $\hat{\Phi}(\upsilon )<0$. We claim that $r(\cdot)$ is
bounded above on $\tilde{G}_2\cap S^1$. Indeed, from
\eqref{BF1} we have
\begin{equation}
r(\upsilon )\leq (\frac{A(\upsilon )}{B(\upsilon )})^{1/(s-q)},
\label{rab}
\end{equation}
while hypothesis (H3) implies
\begin{equation}
A(\upsilon )\leq cB(\upsilon )^{q/s}  \label{aabb}
\end{equation}
for every $\upsilon \in E$ and some $c>0$. At the same time if
$\upsilon \in \tilde{G}_2$, then for some $\theta >0$,
\begin{equation}
\theta <\|\nabla \upsilon \|_2^2<\frac{q-p}{2-p}r_{\ast }(\upsilon
)^{q-2}A(\upsilon )<\frac{q-p}{2-p}r(\upsilon )^{q-2}A(\upsilon ).
\label{rad}
\end{equation}
From \eqref{rab} and \eqref{rad} we deduce
\begin{equation}
\theta <\frac{q-p}{2-p}\Big( \frac{A(\upsilon )}{B(\upsilon )}
\Big) ^{(q-2)/(s-q)}A(\upsilon ).  \label{abth}
\end{equation}
Next, by using \eqref{aabb} and \eqref{abth}, we have
\begin{equation}
\theta <\frac{q-p}{2-p}c^{\frac{s-2}{s-q}}B(\upsilon )^{\frac{2}{s}},
\label{posB}
\end{equation}
and so $B(\cdot)$ is bounded away from $0$. The claim is proved by
reverting to ( \ref{rab}). Accordingly, $\hat{\Phi}(\upsilon )$ is
also bounded on $\tilde{G }_2\cap S^1$. Consider the
variational problem
\[
M=\inf_{\tilde{G}_2\cap S^1} \hat{\Phi}(\upsilon )<0
\]
and assume that $\{\upsilon _n\}_{n\in\mathbb{N}}$ is a minimizing
sequence in $\tilde{G}_2\cap S^1$. Since
$\{\upsilon _n\}_{n\in\mathbb{N}}$ is bounded, there exists
$\tilde{\upsilon}$ $\in $ $E$ such that, at
least for a subsequence, $A(\upsilon _n)\to A(\tilde{\upsilon}
)\geq 0\ $and $B(\upsilon _n)\to B(\tilde{\upsilon})$. By
\eqref{posB}, $\tilde{\upsilon}\neq 0$. We may also assume
that $r_{\ast}(\upsilon _n)\to $ $\tilde{r}_{\ast }$ and
$r(\upsilon_n)\to $ $\tilde{r}$. Clearly, $\tilde{r}>0$ since
$M=\liminf_{n\to \infty } \hat{\Phi}(\upsilon _n)<0$. On the
other hand, $A(\tilde{\upsilon})>0$ because, otherwise, this
would imply $\tilde{r}=0$. Furthermore $\tilde{r}_{\ast }>0$
by \eqref{rab*}. We claim
that $\tilde{\upsilon}\in G_5$. Since
\begin{equation}
\begin{aligned}
\|\nabla \tilde{\upsilon}\|_2^2
&\leq \limsup_{n\to \infty } \|\nabla \tilde{\upsilon}_n\|_2^2
\leq \limsup_{n\to \infty } Q(r_{\ast }(\upsilon _n),\upsilon _n) \\
&\leq \limsup_{n\to \infty } \{r_{\ast }(\upsilon
_n)^{q-2}A(\upsilon _n)-r_{\ast }(\upsilon _n)^{s-2}B(\upsilon_n)\}\\
&\quad -\lim\inf_{n\to \infty } r_{\ast }(\upsilon_n)^{p-2}
\|\nabla \upsilon _n\|_{p}^{p} \\
&\leq \tilde{r}_{\ast }^{q-2}A(\tilde{\upsilon})-\tilde{r}_{\ast }^{s-2}B(
\tilde{\upsilon})-\tilde{r}_{\ast }^{p-2}\|\nabla \tilde{\upsilon}
\|_2^2
=Q(\tilde{r}_{\ast },\tilde{\upsilon}),
\end{aligned} \label{h1}
\end{equation}
we see that
\begin{equation}
\|\nabla \tilde{\upsilon}\|_2^2\leq Q(r_{\ast }(\tilde{\upsilon}),\tilde{
\upsilon}).  \label{h2}
\end{equation}
We shall show that strict inequality holds. Indeed, let us suppose
\begin{equation}
\|\nabla \tilde{\upsilon}\|_2^2=Q(r_{\ast }(\tilde{\upsilon}),\tilde{
\upsilon}).  \label{h3}
\end{equation}
Since $\tilde{r}>0$, by applying \eqref{BE4} for
$\upsilon =\upsilon _n$
and passing to the limit, we also obtain
\begin{equation}
\begin{aligned}
\|\nabla \tilde{\upsilon}\|_2^2
&\leq \limsup_{n\to \infty } \|\nabla \tilde{\upsilon}_n\|_2^2
\leq \limsup_{n\to \infty } Q(r(\upsilon _n),\upsilon _n) \\
&\leq \limsup_{n\to \infty } \{r(\upsilon_n)^{q-2}A(\upsilon _n)
-r(\upsilon _n)^{s-2}B(\upsilon _n)\}\\
&\quad -\liminf_{n\to \infty } r(\upsilon _n)^{p-2}\|\nabla
\upsilon _n\|_{p}^{p}\\
&\leq \tilde{r}^{q-2}A(\tilde{\upsilon})-\tilde{r}^{s-2}B(\tilde{\upsilon})-
\tilde{r}^{p-2}\|\nabla \tilde{\upsilon}\|_{p}^{p}
=Q(\tilde{r},\tilde{\upsilon}).
\end{aligned}  \label{h4}
\end{equation}
Consequently, by \eqref{h1}, \eqref{h3} and \eqref{h4},
we should have
$\tilde{r}=\tilde{r}_{\ast }=\tilde{r}_{\ast }(\tilde{\upsilon})$.
On the other hand, by replacing $\upsilon $ by $\upsilon _n$ in
 \eqref{eqr*} and passing to the limit we obtain
\[
(q-2)r_{\ast }(\tilde{\upsilon})^{q-p}A(\tilde{\upsilon})+(2-p)\|\nabla
\tilde{\upsilon}\|_{p}^{p}\leq (s-2)r_{\ast }(\tilde{\upsilon})^{s-p}B(
\tilde{\upsilon}).
\]
Since $r_{\ast }(\tilde{\upsilon})$ satisfies
\[
(q-2)r_{\ast }(\tilde{\upsilon})^{q-p}A(\tilde{\upsilon})+(2-p)\|\nabla
\tilde{\upsilon}\|_{p}^{p}=(s-2)r_{\ast }(\tilde{\upsilon})^{s-p}B(\tilde{
\upsilon}),
\]
we deduce that $\|\nabla \upsilon _n\|_{p}^{p}\to \|\nabla \tilde{
\upsilon}\|_{p}^{p}$ where, by \eqref{eqr*},
\begin{equation}
\frac{q-2}{2s}\tilde{r}^{q}A(\tilde{\upsilon})+\frac{2-p}{2s}\tilde{r}
^{p}\|\nabla \tilde{\upsilon}\|_{p}^{p}=\frac{s-2}{2s}\tilde{r}^{s}B(\tilde{
\upsilon}).  \label{rlim8}
\end{equation}
Then, \eqref{e4} and \eqref{rlim8} yield
\[
M=\lim_{n\to \infty }\hat{\Phi}(\upsilon _n)=\frac{(2-p)(s-p)}{2ps}
\tilde{r}^{p}\|\nabla \tilde{\upsilon}\|_{p}^{p}+\frac{(q-2)(s-q)}{2ps}
\tilde{r}^{q}A(\tilde{\upsilon})>0,
\]
which is a contradiction. Therefore, $\tilde{\upsilon}\in \tilde{G}_2$
as claimed. A similar reasoning as in Case 2 shows that
$\tilde{r}\leq r(\tilde{\upsilon})$. If we assume that
$\tilde{r}<r(\tilde{\upsilon})$, then,
since the function
\begin{equation}
\psi (z):=\frac{\partial }{\partial z}\Phi (z\tilde{\upsilon})=z\{
\|\nabla \tilde{\upsilon}\|_2^2-Q(z,\tilde{\upsilon})\},
\label{miti}
\end{equation}
is strictly negative for $z\in (\tilde{r},r(\tilde{\upsilon}))$,
by \eqref{mr1} we obtain
\[
M=\liminf_{n\to \infty } \Phi (r(\upsilon _n)\upsilon
_n)\geq \Phi (\tilde{r}\tilde{\upsilon})>\Phi (r(\tilde{\upsilon})\tilde{
\upsilon})=\Phi (r(t\tilde{\upsilon})t\tilde{\upsilon})
=\hat{\Phi}(t\tilde{\upsilon}),
\]
contradicting the definition of $M$. Consequently,
$\tilde{\upsilon}\in S^1$ and $\hat{\Phi}(\tilde{\upsilon})=M$.
Therefore $u:=r(\tilde{\upsilon})\tilde{\upsilon}$ is a solution
of \eqref{1}-\eqref{2}.

Thus, we have proved the following result.

\begin{theorem} \label{thm5}
Assume that conditions {\rm (H0)--(H3)} are satisfied,
$p<2<q<s$ and the set $G_5$ defined in \eqref{g5} is not empty.
Then  \eqref{1})-\eqref{2} admits a non-negative
solution $u\in C^{1,\delta }(\Omega )$ for some $\delta \in
(0,1).$
\end{theorem}

 \begin{remark} \label{rmk3} \rm
We will present a condition which guarantees
that $G_5\neq \emptyset $. From \eqref{eqr*},
\[
\Big( \frac{q-2}{s-2}\frac{A(\upsilon )}{B(\upsilon )}\big)
^{1/(s-q)}\leq r_{\ast }(\upsilon ),
\]
and so
\begin{align*}
&\frac{2}{q}\frac{s-q}{s-2}A(\upsilon )r_{\ast }(\upsilon )^{q-2}
-\frac{2}{p} \frac{s-p}{s-2}\|\nabla \upsilon \|_{p}^{p}r_{\ast }
(\upsilon )^{p-2}\\
&\geq \frac{2}{q}\frac{s-q}{s-2}A(\upsilon )
 \Big( \frac{q-2}{s-2}\frac{A(\upsilon )}{B(\upsilon )}\Big)
 ^{(q-2)/(s-q)}-\frac{2}{p}\frac{s-p}{s-2}\|\nabla
 \upsilon \|_{p}^{p}\Big( \frac{q-2}{s-2}\frac{A(\upsilon )}{
 B(\upsilon )}\Big) ^{\frac{p-2}{s-q}}\\
&=\frac{2}{q}\frac{s-q}{s-2}\Big(\frac{q-2}{s-2}\Big)^{(q-2)/(s-q)}
A(\upsilon )^{(s-2)/(s-q)}B(\upsilon) ^{(2-q)/(s-q)} \\
&\quad -\frac{2}{p}\frac{s-p}{s-2}\Big(\frac{q-2}{s-2}\Big) ^{\frac{p-2}{s-q}}\|\nabla \upsilon \|_{p}^{p}B(\upsilon) ^{\frac{2-p}{s-q}
}A(\upsilon )^{\frac{p-2}{s-q}}.
\end{align*}
Since $\frac{s-2}{s-q}>\frac{p-2}{s-q}$, $G_5\neq \emptyset $
for $a^{+}(\cdot)$ large enough.
\end{remark}

 \textbf{Case 5:} $s<p<q<2$.
In this case we define
\[
Q(r,\upsilon ):=r^{q-p}A(\upsilon )-r^{s-p}B(\upsilon )-r^{2-p}\|\nabla
\upsilon \|_2^2.
\]
For $\upsilon \in G_1$, $Q(\cdot,\upsilon )$ has a unique
critical point $r_{\ast }:=r_{\ast }(\upsilon )$ which corresponds
to a global maximum and satisfies
\begin{equation}
(q-p)r_{\ast }^{q-s}A(\upsilon )+(p-s)B(\upsilon )=(2-p)r_{\ast
}^{2-s}\|\nabla \upsilon \|_2^2  \label{eqr2}
\end{equation}
and
\[
Q(r_{\ast }(\upsilon ),\upsilon )=\frac{2-q}{2-p}r_{\ast }(\upsilon
)^{q-p}A(\upsilon )-\frac{2-s}{2-p}r_{\ast }(\upsilon )^{s-p}B(\upsilon ).
\]
 From \eqref{eqr2} we get
\begin{equation}
r_{\ast }(\upsilon )\geq \Big( \frac{q-p}{2-p}\frac{A(\upsilon )}{\|\nabla
\upsilon \|_2^2}\Big) ^{1/(2-q)}.  \label{rab2}
\end{equation}
Clearly, if $\upsilon \in G_2$ then \eqref{BE1} has exactly two
positive solutions $r_1(\upsilon )$, $r_2(\upsilon )$ with
$r_1(\upsilon )<r_{\ast }(\upsilon )<r_2(\upsilon )$. We set
$r:=r(\upsilon )$ to be the greater solution. We have
\[
r^{p-1}Q_{r}(r,\upsilon )=(q-p)A(\upsilon )r^{q-2}-(s-p)r^{s-2}B(\upsilon
)-(2-p)\|\nabla \upsilon \|_2^2,
\]
which, on account of \eqref{eqr2}, yields
\[
r^{p-1}Q_{r}(r,\upsilon )=(q-p)A(\upsilon )(r^{q-2}-r_{\ast
}^{q-2})+(p-s)B(\upsilon )(r^{s-2}-r_{\ast }^{s-2})<0.
\]
Therefore, $r(\cdot)$ is continuously differentiable. Let
\begin{equation}
G_6:=\{\upsilon \in G_1:\|\nabla \upsilon \|_{p}^{p}<\frac{p}{q}
\frac{2-q}{2-p}r_{\ast }(\upsilon )^{q-p}A(\upsilon )-\frac{p}{s}\frac{2-s}{
2-p}r_{\ast }(\upsilon )^{s-p}B(\upsilon )\}\label{g6}
\end{equation}
and assume that $G_6\neq \emptyset $. We immediately see that
$G_6\subseteq G_2$, since $\frac{p}{q}<1$ and so $G_2\neq \emptyset $
as well. Moreover, $G_6\cap S^1\neq \emptyset $ and $\hat{\Phi}
(\upsilon )<0$ for any $\upsilon \in G_6$. Indeed, since
$r(\upsilon)>r_{\ast }(\upsilon )$, by \eqref{g6} we get
\begin{equation}
\|\nabla \upsilon \|_{p}^{p}<\frac{p}{q}\frac{2-q}{2-p}r(\upsilon
)^{q-p}A(\upsilon )-\frac{p}{s}\frac{2-s}{2-p}r(\upsilon )^{s-p}B(\upsilon ).
\label{ap2}
\end{equation}
At the same time, \eqref{e4} and \eqref{ap2} yield
\[
r^{q}\frac{2-p}{2p}\|\nabla \upsilon \|_{p}^{p}+r^{q}\frac{q-2}{2q}
A(\upsilon )+r^{s}\frac{2-s}{2s}B(\upsilon )<0,
\]
which proves the assertion. Next, because $2>q$, \eqref{br} shows
that $r(\cdot)$ is bounded above on $G_2\cap S^1$.
Consequently, $\hat{\Phi}(\upsilon )$ is also bounded on
$G_2\cap S^1$. Consider the variational problem
\[
M=\inf_{\upsilon \in G_2\cap S^1} \hat{\Phi}(\upsilon )<0.
\]
If $\{\upsilon _n\}_{n\in \mathbb{N}}$ is a minimizing sequence in
$G_2\cap S^1$ then, there exist $\tilde{\upsilon}\in E$ such that,
at least for a subsequence,
$A(\upsilon _n)\to A(\tilde{\upsilon})\geq 0$ and
$B(\upsilon _n)\to B(\tilde{\upsilon})\geq 0$, while by \eqref{ff}
we get
\[
0<\|\nabla \tilde{\upsilon}\|_2^2\leq \liminf \|\nabla \upsilon
_n\|_2^2\leq 1.
\]
Since $r(\cdot)$ is bounded on $G_2\cap S^1$ we may assume
that $r_{\ast }(\upsilon _n)\to \tilde{r}_{\ast }$ and
$r(\upsilon _n)\to \tilde{r}$. Again $\tilde{r}>0$ because,
otherwise, $M= \liminf_{n\to \infty } \hat{\Phi}(\upsilon _n)=0$,
a contradiction. We also have that
$A(\tilde{\upsilon})>0$, because, if we assume the contrary,
\eqref{BE2} yields
\[
r(\upsilon _n)^{2-q}\|\nabla \upsilon _n\|_2^2\leq A(\upsilon _n),
\]
and by passing to the limit,
\[
\tilde{r}^{2-q}\|\nabla \tilde{\upsilon}\|_2^2
\leq \liminf_{n\to \infty } (r(\upsilon _n)^{2-q}\|\nabla
\upsilon _n\|_2^2)\leq \lim_{n\to \infty }
A(\upsilon _n)=A(\tilde{\upsilon}).
\]
Thus, $\tilde{r}=0$, a contradiction. Furthermore
$\tilde{r}_{\ast }>0$ due
to \eqref{rab2}. We claim that $\tilde{\upsilon}\in G_6$. Indeed, if not,
then, by applying the same arguments as in the proof of Case 2, we would
have $\tilde{r}=\tilde{r}_{\ast }=r_{\ast }(\tilde{\upsilon})$, while,
along a subsequence, $\|\nabla \upsilon _n\|_2^2\to \|\nabla
\tilde{\upsilon}\|_2^2$ where, by \eqref{eqr2}
\begin{equation}
\frac{q-p}{2p}\tilde{r}^{s}A(\tilde{\upsilon})
+\frac{p-s}{2p}\tilde{r}^{s}B( \tilde{\upsilon})
=\frac{2-p}{2p}\tilde{r}^2\|\nabla \tilde{\upsilon}
\|_2^2.  \label{rlim2}
\end{equation}
Then \eqref{e1} and \eqref{rlim2} yield
\[
M=\lim_{n\to \infty }\hat{\Phi}(\upsilon _n)=\frac{(q-p)(2-q)}{2pq}
\tilde{r}^{q}A(\tilde{\upsilon})+\frac{(p-s)(2-s)}{2ps}\tilde{r}^{s}
B(\tilde{\upsilon})>0.
\]
Therefore, $\tilde{\upsilon}\in G_2$ as claimed. A similar reasoning
as in Case 2 shows that $\tilde{r}$ $=r(\tilde{\upsilon})$. Finally,
by passing to the limit in \eqref{BE2} we rederive \eqref{fe} which
implies that $\tilde{\upsilon}\in $ $S^1$ and
$\hat{\Phi}(\tilde{\upsilon})=M$. Thus
$u:=r(\tilde{\upsilon})\tilde{\upsilon}$ is a solution to
\eqref{1}-\eqref{2}.

Therefore we have proved the following result.

\begin{theorem} \label{thm6}
Assume that conditions {\rm (H0)--(H2)} are satisfied,
$s<p<q<2$ and the set $G_6$ defined in \eqref{g6} is not empty.
Then  \eqref{1}-\eqref{2} admits a non-negative
solution $u\in C^{1,\delta }(\Omega )$ for some $\delta \in
(0,1)$.
\end{theorem}

\begin{remark} \label{rmk4}\rm
 We will give some conditions which guarantee that
$G_6\neq \emptyset $. Suppose that
$\operatorname{supp}a^{+})\subseteq \operatorname{supp}b)$.
Then there exists $\upsilon \in S^1$ such that $B(\upsilon )>0$.
 From (\eqref{eqr2}
\begin{equation}
\left( \frac{p-s}{2-p}\frac{B(\upsilon )}{\|\nabla \upsilon \|_2^2}
\right) ^{1/(2-s)}\leq r_{\ast }(\upsilon ),  \label{coci}
\end{equation}
and so, in view of \eqref{coci},
\begin{align*}
&\frac{p}{q}\frac{2-q}{2-p}r_{\ast }(\upsilon )^{q-p}A(\upsilon )
-\frac{p}{s} \frac{2-s}{2-p}r_{\ast }(\upsilon )^{s-p}B(\upsilon )
\\
&\geq \frac{p}{q}\frac{2-q}{2-p}
\Big( \frac{q-p}{2-p}\frac{A(\upsilon )}{
\|\nabla \upsilon \|_2^2}\Big) ^{(q-p)/(2-s)}A(\upsilon )
-\frac{p}{s}\frac{2-s}{2-p}\Big( \frac{p-s}{2-p}
 \frac{B(\upsilon )}{\|\nabla \upsilon
\|_2^2}\Big) ^{(s-p)/(2-s)}B(\upsilon )
\\
&\geq \frac{p}{q}\frac{2-q}{2-p}\Big( \frac{q-p}{2-p}\frac{1}{\|\nabla
\upsilon \|_2^2}\Big) ^{(q-p)/(2-s)}A(\upsilon )^{\frac{2-p}{2-s}+1}\\
&\quad -\frac{p}{s}\frac{2-s}{2-p}\Big( \frac{p-s}{2-p}\frac{1}{\|\nabla
\upsilon \|_2^2}\Big) ^{(s-p)/(2-s)}B(\upsilon )^{(2-p)/(2-s)}.
\end{align*}
Note that if
\begin{equation}
\begin{aligned}
&\frac{p}{q}\frac{2-q}{2-p}\Big( \frac{q-p}{2-p}\frac{1}{\|\nabla \upsilon
\|_2^2}\Big) ^{(q-p)/(2-s)}A(\upsilon )^{\frac{2-p}{2-s}+1}
\label{coco3} \\
&-\frac{p}{s}\frac{2-s}{2-p}\Big( \frac{p-s}{2-p}\frac{1}{\|\nabla \upsilon
\|_2^2}\Big) ^{(s-p)/(2-s)}B(\upsilon )^{\frac{2-p}{2-s}}
>\|\nabla \upsilon \|_2^2,
\end{aligned}
\end{equation}
then $G_6\neq \emptyset $. It is clear that if $a^{+}(\cdot)$
is large compared to $b(\cdot)$ then \eqref{coco3} is satisfied.
\end{remark}

Suppose now that
$( \operatorname{supp}a^{+})\backslash \operatorname{supp}b)) ^{o}\neq
\emptyset $. Then there exists $\upsilon $ $\in S^1$ with $B(\upsilon
)=0 $. From \eqref{eqr2} we have
\begin{equation}
\Big( \frac{q-p}{2-p}\frac{A(\upsilon )}{\|\nabla \upsilon \|_2^2}
\Big) ^{1/(2-q)}=r_{\ast }(\upsilon ),  \label{ram1}
\end{equation}
and so, in view of \eqref{ram1},
\begin{align*}
\frac{p}{q}\frac{2-q}{2-p}r_{\ast }(\upsilon )^{q-p}A(\upsilon )
&=\frac{p}{q}\frac{2-q}{2-p}\Big( \frac{q-p}{2-p}\frac{A(\upsilon )}{
\|\nabla \upsilon \|_2^2}\Big) ^{(q-p)/(2-q)}A(\upsilon )
\\
&=\frac{p}{q}\frac{2-q}{2-p}\Big( \frac{q-p}{2-p}\frac{1}{\|\nabla \upsilon
\|_2^2}\Big) ^{(q-p)/(2-q)}A(\upsilon )^{\frac{2-p}{2-q}}.
\end{align*}
If we assume that
\[
\frac{p}{q}\frac{2-q}{2-p}\Big( \frac{q-p}{2-p}\frac{1}{\|\nabla \upsilon
\|_2^2}\Big) ^{(q-p)/(2-q)}A(\upsilon )^{\frac{2-p}{2-q}}>\|\nabla
\upsilon \|_2^2,
\]
we have
\begin{equation}
A(\upsilon )^{\frac{2-p}{2-q}}>\frac{q}{p}\frac{2-p}{2-q}
\Big( \frac{q-p}{2-p}\Big) ^{(p-q)/(2-q)}\|\nabla \upsilon \|_2^2,
 \label{ram}
\end{equation}
and so if $a^{+}(\cdot)$ large enough the condition \eqref{ram} is
valid implying that $G_6\neq \emptyset .$

 \textbf{Case 6:} $s<q<p<2$.
In this case we assume that the following condition holds:
\begin{itemize}
\item[(H4)] $V:=( \operatorname{supp} a^{+}\backslash
\operatorname{supp} b) ^{o}\neq \emptyset $.
\end{itemize}
 We define
\begin{equation}
Q(r,\upsilon ):=r^{q-p}A(\upsilon )-r^{s-p}B(\upsilon )-r^{2-p}\|\nabla
\upsilon \|_2^2.  \label{qq}
\end{equation}
 Let $\upsilon \in G_1$. If $B(\upsilon )=0$, the equation
\eqref{BF1} has a unique solution $r(\upsilon )>0$, while
if $B(\upsilon )>0$, the function $Q(\cdot,\upsilon )$ has a unique
critical point $r_{\ast}:=r_{\ast }(\upsilon )$ which corresponds
to a global maximum and satisfies
\begin{equation}
(p-s)B(\upsilon )=(p-q)r_{\ast }^{q-s}A(\upsilon )+(2-p)r_{\ast
}^{2-s}\|\nabla \upsilon \|_2^2.  \label{gm}
\end{equation}
Clearly, if $\upsilon \in G_2$, then \eqref{BE} has exactly two positive
solutions $r_1(\upsilon )$ and $r_2(\upsilon )$ with $r_1(\upsilon
)<r_{\ast }(\upsilon )<r_2(\upsilon )$. Let $r:=r(\upsilon )$ be the
unique solution of \eqref{BE} in case $B(\upsilon )=0$ or the greater
solution $r_2$ in case $B(\upsilon )>0$. Note that, if $B(\upsilon )>0$
then
\[
r^{p-s+1}Q_{r}(r,\upsilon )=(q-p)A(\upsilon )r^{q-s}-(s-p)B(\upsilon
)-(2-p)r^{2-s}\|\nabla \upsilon \|_2^2
\]
and so, in view of \eqref{gm}, we obtain
\[
r^{p-s+1}Q_{r}(r,\upsilon )=(p-q)A(\upsilon )(r_{\ast
}^{q-s}-r^{q-s})-(p-2)\|\nabla \upsilon \|_2^2(r_{\ast
}^{2-s}-r^{2-s})<0,
\]
while if $B(\upsilon )=0$, then
\[
r^{p+1}Q_{r}(r,\upsilon )=(q-p)A(\upsilon )r^{q}-(2-p)\|\nabla \upsilon
\|_2^2r^2<0.
\]
Thus $r(\cdot)$ is continuously differentiable by the implicit
function theorem. We now define
\[
G_7
=\{\upsilon \in G_1:B(\upsilon )=0\}
\cup \{\upsilon \in G_1:B(\upsilon )>0\text{ and }\|\nabla \upsilon
\|_{p}^{p}<Q(r_{\ast }(\upsilon ),\upsilon )\}.
\]
In view of (H1) and (H4), we see that $G_7\neq \emptyset$
since for any $\upsilon \in E$ with
$\operatorname{supp}\upsilon \subseteq V$ there holds
$A(\upsilon )>0$ and $B(\upsilon )=0$. We claim that $G_7$ is open.
Indeed, let $\hat{\upsilon}$ $\in $ $G_7$ and assume that there
exists a sequence
$\{\upsilon _n\}_{n\in\mathbb{N}}\subseteq E\backslash G_7$ with
$\upsilon _n\to \hat{\upsilon}$ strongly in $E$. Suppose, without
loss of generality, that $B(\hat{\upsilon})=0$ while
$B(\hat{\upsilon})>0$ for every $n\in\mathbb{N}$. Therefore,
\begin{equation}
\|\nabla \upsilon _n\|_{p}^{p}\geq Q(r_{\ast }(\upsilon _n),\upsilon
_n)\text{ for every }n\in\mathbb{N}.  \label{sa}
\end{equation}
Since $A(\hat{\upsilon})>0$, on account of \eqref{gm},
$r_{\ast }(\upsilon_n)\to 0$. Combining \eqref{gm} and \eqref{qq}
we obtain
\[
Q(r_{\ast }(\upsilon ),\upsilon )=\frac{q-s}{p-s}r_{\ast }(\upsilon
)^{q-p}A(\upsilon )-\frac{2-s}{p-s}r_{\ast }(\upsilon )^{2-p}\|\nabla
\upsilon \|_2^2,
\]
and so $\lim_{n\to \infty }Q(r_{\ast }(\upsilon _n),\upsilon
_n)=+\infty $, contradicting \eqref{sa}.
 It follows from \eqref{BE} that $r(\cdot)$ is bounded and so
$\hat{\Phi}(\cdot)$ is also bounded on $G_7\cap S^1$. On account of
\eqref{e1} and (H4), $M<0$.

 Consider the variational problem
\[
M=\inf_{\upsilon \in G_2\cap S^1} \hat{\Phi}(\upsilon )<0
\]
and assume that $\{\upsilon _n\}_{n\in \mathbb{N}}$ is a minimizing
sequence in $G_7\cap S^1$. Then there exists
$\tilde{\upsilon}\in E$ so that $A(\upsilon _n)\to A(\tilde{\upsilon}
)\geq 0$, $B(\upsilon _n)\to B(\tilde{\upsilon})\geq 0$ and
\[
0\leq \|\nabla \tilde{\upsilon}\|_{p}^{p}
\leq \liminf \|\nabla \upsilon _n\|_{p}^{p}\leq 1.
\]
Furthermore, $r(\upsilon _n)\to \tilde{r}$ for a new
subsequence. In particular, $\tilde{r}>0$ because if $\tilde{r}=0$
then, by \eqref{e1}), $M=\lim_{n\to \infty }\hat{\Phi}(\upsilon _n)=0$;
a contradiction. We claim that $A(\tilde{\upsilon})>0$.
Indeed, from \eqref{BF1} we have
\[
\|\nabla \upsilon _n\|_{p}^{p}r(\upsilon _n)^{p-q}\leq A(\upsilon _n),
\]
and by passing to the limit,
\[
\|\nabla \tilde{\upsilon}\|_{p}^{p}r(\tilde{\upsilon})^{p-q}\leq
\liminf_{n\to \infty } \|\nabla \upsilon _n\|_{p}^{p}r(\upsilon
_n)^{p-q}
\leq \lim_{n\to \infty } A(\upsilon _n)=A(\tilde{\upsilon}).
\]
Thus, if $A(\tilde{\upsilon})=0$ then $\tilde{\upsilon}=0$. However,
this leads to a contradiction because by \eqref{Fib}, we should have
$0=\Phi(0)\leq \liminf_{n\to \infty } \Phi (r(\upsilon_n)\upsilon _n)=M$.

 We shall show next that $\tilde{\upsilon}\in G_7$. Let us
assume that $B(\tilde{\upsilon})>0$. Since
\[
(p-s)B(\upsilon _n)=(p-q)r_{\ast }^{q-s}A(\upsilon _n)+(2-p)r_{\ast
}^{2-s}\|\nabla \upsilon _n\|_2^2,
\]
we see that the sequence $\{r_{\ast }(\upsilon _n)\}_{n\in\mathbb{N}}$
 is bounded. Thus, up to a further subsequence,
$r_{\ast }(\upsilon_n)\to \tilde{r}_{\ast }>0$. As before,
$\tilde{r}=\tilde{r}_{\ast }=r_{\ast }(\tilde{\upsilon})$.
On the other hand, by passing to the
limit in \eqref{gm} we see that
$\|\nabla \upsilon _n\|_2^2\to \|\nabla \tilde{\upsilon}\|_2^2$ and
\[
B(\tilde{\upsilon})
=\frac{p-q}{p-s}r_{\ast }^{q-s}(\tilde{\upsilon})A(\tilde{\upsilon})
+\frac{2-p}{p-s}r_{\ast }^{2-s}(\tilde{\upsilon})\|\nabla \tilde{
\upsilon}\|_2^2.
\]
Thus,
\[
M=\lim_{n\to \infty } \hat{\Phi}(\upsilon _n)
=\frac{(2-s)(2-p)}{2ps}\tilde{r}^2\|\nabla
\tilde{\upsilon}\|_2^2+\tilde{r}^{q}A(\tilde{\upsilon})
\frac{(q-s)(p-q)}{psq}>0,
\]
which is a contradiction. Therefore, $\tilde{\upsilon}\in G_7$ as
claimed. On the other hand, if $B(\tilde{\upsilon})=0$ then it
is obvious that $\tilde{\upsilon}$ $\in G_7$. Working as in Case 2
we are lead to the following result.

\begin{theorem} \label{thm7}
Assume that conditions {\rm (H0)-(H2),  (H4)} are
satisfied and $s<q<p<2$. Then \eqref{1}-\eqref{2}
admits a non-negative solution $u\in C^{1,\delta }(\Omega )$ for
some $\delta \in (0,1)$.
\end{theorem}

\textbf{Case 7:} $p<q<s<2$.
In this case we define
\[
Q(r,\upsilon ):=r^{q-p}A(\upsilon )-r^{s-p}B(\upsilon )
-r^{2-p}\|\nabla \upsilon \|_2^2.
\]
We see that for $\upsilon $ $\in $ $G_1$ the function
$Q(\cdot,\upsilon )$ has a unique critical point
$r_{\ast }:=r_{\ast }(\upsilon )$ satisfying
\begin{equation}
(q-p)A(\upsilon )=(s-p)r_{\ast }(\upsilon )^{s-q}B(\upsilon )
+(2-p)r_{\ast}(\upsilon )^{2-q}\|\nabla \upsilon \|_2^2.  \label{rf1}
\end{equation}
It is clear that \eqref{BE} has two positive solutions
$r_1(\upsilon )$, $r_2(\upsilon )$ with
$r_1(\upsilon )<r_{\ast }(\upsilon )<r_2(\upsilon)$ for every
$\upsilon \in G_2$. Let $r:=$ $r_2(\upsilon )$. Then
\[
r^{p-q+1}Q_{r}(r,\upsilon )=(q-p)A(\upsilon )-(s-p)r^{s-q}B(\upsilon
)-(2-p)r^{2-q}\|\nabla \upsilon \|_2^2,
\]
which combined with \eqref{rf1}, gives
\[
r^{p-q+1}Q_{r}(r,\upsilon )=(2-p)\|\nabla \upsilon \|_2^2(r_{\ast
}^{2-q}-r^{2-q})+(s-p)B(\upsilon )(r_{\ast }^{s-q}-r^{s-q})<0.
\]
Therefore, the implicit function theorem implies that $r(\cdot)$
is continuously differentiable. Assume that the set
\[
G_8:=\{\upsilon \in G_1:\|\nabla \upsilon \|_{p}^{p}<\frac{p}{q}
\frac{s-q}{s-p}r_{\ast }(\upsilon )^{q-p}A(\upsilon )\}
\]
is not empty. Since $q>p$, and
$r(\upsilon )^{q-p}>r_{\ast }(\upsilon)^{q-p}$, we see that
$G_8\subseteq G_2$ and so $G_2\neq \emptyset $.
If $\upsilon \in G_8$, then
\[
\|\nabla \upsilon \|_{p}^{p}
<\frac{p}{q}\frac{s-q}{s-p}r_{\ast }(\upsilon)^{q-p}A(\upsilon )
<\frac{p}{q}\frac{s-q}{s-p}r(\upsilon )^{q-p}A(\upsilon )
\]
and so
\begin{equation}
\frac{2-p}{p}r(\upsilon )^{p}\|\nabla \upsilon \|_{p}^{p}+\frac{q-2}{q}
r(\upsilon )^{q}A(\upsilon )<0.  \label{rf5}
\end{equation}
Combining \eqref{rf5} with \eqref{e3}, we conclude that
\[
\hat{\Phi}(\upsilon )<r^{p}(\frac{1}{p}-\frac{1}{s})\|\nabla \upsilon
\|_{p}^{p}+r^{q}(\frac{1}{s}-\frac{1}{q})A(\upsilon )<0.
\]
On the other hand, if $\upsilon \in G_2\cap S^1$, then \eqref{BF1}
implies
\[
r(\upsilon )\leq \Big( \frac{A(\upsilon )}{\|\nabla \upsilon \|_2^2}
\Big) ^{1/(2-q)}
\]
and so $r(\cdot)$ is bounded on $G_2\cap S^1$. Consequently,
$\hat{\Phi} (\upsilon )$ is also bounded on $G_2\cap S^1$. Let
\[
M:=\inf_{\upsilon \in G_2\cap S^1} \hat{\Phi}(\upsilon )<0
\]
and assume that $\{\upsilon _n\}_{n\in\mathbb{N}}$ is a minimizing
sequence in $G_2\cap S^1$. Then, there exist $\tilde{\upsilon}\in E$
such that, at least for a subsequence,
$A(\upsilon_n)\to A(\tilde{\upsilon})\geq 0$,
$B(\upsilon _n)\to B(\tilde{\upsilon})\geq 0$,
\begin{gather*}
0\leq \|\nabla \tilde{\upsilon}\|_2\leq \lim \inf \|\nabla \upsilon
_n\|_2\leq 1,\\
0\leq \|\nabla \tilde{\upsilon}\|_{p}\leq \lim \inf \|\nabla \upsilon
_n\|_{p}\leq 1.
\end{gather*}
We must have $\tilde{\upsilon}\neq 0$ because, otherwise,
$0=\Phi (0)\leq \lim \inf_{n\to \infty }\Phi (r(\upsilon _n)
\upsilon _n)=M$, a
contradiction. Since $\{r(\upsilon _n)\}_{n\in\mathbb{N}}$ is bounded
and $r_{\ast }(\upsilon _n)<r(\upsilon _n)$, $n\in\mathbb{N}$,
we may assume that $r_{\ast }(\upsilon _n)\to \tilde{r}_{\ast }$
and $r(\upsilon _n)\to \tilde{r}$.
Since $M=\liminf_{n\to\infty } \hat{\Phi}(\upsilon _n)<0$
we obtain $\tilde{r}>0$. We also have that $A(\tilde{\upsilon})>0$,
because, if we assume the opposite,
then by
\[
\tilde{r}^{2-q}\|\nabla \tilde{\upsilon}\|_2^2
\leq \liminf_{n\to \infty } (r(\upsilon _n)^{2-q}\|\nabla \upsilon
_n\|_2^2)
\leq \lim_{n\to \infty } A(\upsilon _n)=A(\tilde{\upsilon})
\]
we would get $\tilde{r}=0$, a contradiction. Therefore,
$\tilde{\upsilon}\in G_1$. Also, $\tilde{r}_{\ast }>0$ by \eqref{rf1}.
We will show that $\tilde{\upsilon}\in G_2.$Working as in Case 2
we conclude that
$\tilde{r}= \tilde{r}_{\ast }=\tilde{r}_{\ast }(\tilde{\upsilon})$.
On the other hand, replacing $\upsilon $ by $\upsilon _n$
in \eqref{rf1} and passing to the limit leads to
\[
(q-p)A(\tilde{\upsilon})\geq (s-p)r_{\ast }(\tilde{\upsilon})^{s-q}B(\tilde{
\upsilon})+(2-p)r_{\ast }(\tilde{\upsilon})^{2-q}\|\nabla \tilde{\upsilon}
\|_2^2.
\]
However,  $r_{\ast }(\tilde{\upsilon})$ satisfies
\[
(q-p)A(\tilde{\upsilon})=(s-p)r_{\ast }(\tilde{\upsilon})^{s-q}B(\tilde{
\upsilon})+(2-p)r_{\ast }(\tilde{\upsilon})^{2-q}\|\nabla \tilde{\upsilon}
\|_2^2,
\]
so we deduce that $\|\nabla \upsilon _n\|_2^2\to \|\nabla
\tilde{\upsilon}\|_2^2$. From \eqref{k1} we get
\begin{equation}
A(\tilde{\upsilon})=\frac{s-p}{q-p}\tilde{r}^{s-q}B(\tilde{\upsilon})
+\frac{2-p}{q-p}\tilde{r}^{2-q}\|\nabla \tilde{\upsilon}\|_2^2.  \label{rfm}
\end{equation}
Thus, \eqref{e3} and \eqref{rfm} yield
\[
M=\lim_{n\to \infty }\hat{\Phi}(\upsilon _n)=\frac{(s-q)(s-p)}{pqs}
\tilde{r}^{s}B(\tilde{\upsilon})+\frac{(2-p)(2-q)}{2pq}\tilde{r}^2\|\nabla
\tilde{\upsilon}\|_2^2>0,
\]
a contradiction, proving the claim. Working as in Case 2 we
have  $\tilde{r}=r(\tilde{\upsilon})$. Finally, by passing to
the limit in \eqref{BE2} we have \eqref{fe}, which implies
$\tilde{\upsilon}\in $ $S^1$ and $\hat{\Phi}(\tilde{\upsilon})=M$.
Therefore, we have the following theorem.

\begin{theorem} \label{thm8}
Assume that  conditions {\rm (H0)-(H2)} are satisfied,
$p<q<s<2$ and the set $G_3$ defined in \eqref{s1} is not empty.
Then  \eqref{1}-\eqref{2} admits a non-negative
solution $u\in C^{1,\delta }(\Omega )$ for some $\delta \in
(0,1)$.
\end{theorem}

\begin{remark} \label{rmk5} \rm
We will now give some conditions which guarantee that
$G_3\neq \emptyset $. Suppose that
$\operatorname{supp}a^{+})\subseteq  \operatorname{supp}b)$.
Then there exists $\upsilon $ $\in G_1$ such that $B(\upsilon )>0$.
Since $r_{\ast }(\upsilon )^{2-q}<r(\upsilon )^{2-q}$, \eqref{rf1}
yields
\begin{equation}
(q-p)A(\upsilon )<(s-p)r_{\ast }(\upsilon )^{s-q}B(\upsilon
)+(2-p)r(\upsilon )^{2-q}\|\nabla \upsilon \|_2^2,  \label{v1}
\end{equation}
and so
\[
r_{\ast }(\upsilon )^{s-q}
>\frac{q-p}{s-p}\frac{A(\upsilon )}{B(\upsilon )}-
\frac{2-p}{s-p}r(\upsilon )^{2-q}\frac{\|\nabla \upsilon \|_2^2}{
B(\upsilon )}.
\]
Consequently,
\begin{equation}
\begin{aligned}
&\frac{p}{q}\frac{s-q}{s-p}r_{\ast }(\upsilon )^{q-p}A(\upsilon )\\
&>\frac{p}{q}\frac{s-q}{s-p}\bigg( \frac{q-p}{s-p}\frac{A(\upsilon )}{
B(\upsilon )}-\frac{2-p}{s-p}r(\upsilon )^{2-q}\frac{\|\nabla \upsilon
\|_2^2}{B(\upsilon )}\bigg) ^{(q-p)/(s-q)}A(\upsilon ).
\end{aligned} \label{v2}
\end{equation}
On the other hand, \eqref{BF1} implies
\[
r(\upsilon )\leq \Big( \frac{A(\upsilon )}{B(\upsilon )}\Big)^{1/(s-q)},
\]
which combined with \eqref{v2} gives
\begin{align*}
&\frac{p}{q}\frac{s-q}{s-p}\Big( \frac{q-p}{s-p}\frac{A(\upsilon )}{
B(\upsilon )}-\frac{2-p}{s-p}r(\upsilon )^{2-q}\frac{\|\nabla \upsilon
\|_2^2}{B(\upsilon )}\Big) ^{(q-p)/(s-q)}A(\upsilon )
\\
&>\frac{p}{q}\frac{s-q}{s-p}\Big( \frac{q-p}{s-p}\frac{A(\upsilon )}{
B(\upsilon )}-\frac{2-p}{s-p}\Big( \frac{A(\upsilon )}{B(\upsilon )}
\Big)^{(2-q)/(s-q)}\frac{\|\nabla \upsilon \|_2^2}{B(\upsilon )}\Big) ^{
\frac{q-p}{s-q}}A(\upsilon ).
\end{align*}
If $a^{+}(\cdot)$ is large enough, then
\[
\frac{p}{q}\frac{s-q}{s-p}\Big( \frac{q-p}{s-p}\frac{A(\upsilon )}{
B(\upsilon )}-\frac{2-p}{s-p}A(\upsilon )^{(2-q)/(s-q)}\frac{\|\nabla
\upsilon \|_2^2}{B(\upsilon )^{\frac{2-q}{s-q}+1}}\Big)
^{(q-p)/(s-q)}A(\upsilon )>\|\nabla \upsilon \|_{p}^{p},
\]
implying that $\upsilon \in G_8$. Thus $G_8\neq \emptyset $.
\end{remark}

 Suppose next that
$( \operatorname{supp}a^{+})\backslash \operatorname{supp}b))
^{o}\neq \emptyset $. Then there exists $\upsilon \in S^1$ with
$B(\upsilon )=0$. By \eqref{rf1}
\[
r_{\ast }(\upsilon )=\Big( \frac{q-p}{2-p}\frac{A(\upsilon )}{\|\nabla
\upsilon \|_2^2}\Big) ^{1/(2-q)},
\]
and so
\[
\frac{p}{q}\frac{s-q}{s-p}r_{\ast }(\upsilon )^{q-p}A(\upsilon )=\frac{p}{q}
\frac{s-q}{s-p}\Big( \frac{q-p}{2-p}\frac{A(\upsilon )}{\|\nabla \upsilon
\|_2^2}\Big) ^{(q-p)/(2-q)}A(\upsilon ).
\]
Therefore, if $a^{+}(\cdot)$ is large enough, then
\[
\frac{p}{q}\frac{s-q}{s-p}\Big( \frac{q-p}{2-p}\frac{A(\upsilon )}{\|\nabla
\upsilon \|_2^2}\Big) ^{(q-p)/(2-q)}A(\upsilon )>\|\nabla \upsilon
\|_{p}^{p},
\]
implying that $G_{_8}\neq \emptyset$.

\textbf{Case 8:} $q>\max \{p,s,2\}$.
In this case we shall use the mountain pass theorem.

\begin{lemma}
$\Phi (\cdot)$ satisfies the Palais-Smale condition.
\end{lemma}

\begin{proof}
Let $\{u_n\}_{n=1}^{\infty }$ be a sequence in $E$ such that
$|\Phi(u_n)|\leq C$ for some $C>0$ and every $n\in\mathbb{N}$ and
 $\Phi '(u_n)\to 0$ in $H^{-1}(\Omega )$. For $\varepsilon >0$ and
$\upsilon \in E$ we have
\begin{equation} \label{ps1}
\begin{aligned}
\vert \langle \Phi '(u_n),\upsilon \rangle \vert
& =\Big\vert \int |\nabla u_n|^{p-2}\nabla u_n\nabla \upsilon dx+\int
\nabla u_n\nabla \upsilon dx   \\
&\quad  -\int a(x)u_n{}^{q-1}\upsilon dx+\int b(x)u_n{}^{s-1}\upsilon
dx\Big\vert \\
&\leq \varepsilon \|\upsilon \|_{E}.
\end{aligned}
\end{equation}
If $\upsilon =u_n$ in \eqref{ps1}, then
\begin{equation}
\int a(x)u_n{}^{q}dx\leq \varepsilon \|u_n\|_{1,k}+\int |\nabla
u_n|^{p}dx+\int |\nabla u_n|^2dx+\int b(x)u_n{}^{s}dx.  \label{ps2}
\end{equation}
By hypothesis
\begin{equation}
\frac{1}{p}\|\nabla u_n\|_{p}^{p}+\frac{1}{2}\|\nabla u_n\|_2^2-
\frac{1}{q}\int a(x)|u_n|^{q}dx+\frac{1}{s}\int b(x)|u_n|^{s}dx\leq C.
\label{ps3}
\end{equation}
On combining \eqref{ps2} and \eqref{ps3} we obtain
\begin{align*}
&\frac{1}{p}\|\nabla u_n\|_{p}^{p}+\frac{1}{2}\|\nabla u_n\|_2^2+
\frac{1}{s}\int b(x)|u_n|^{s}dx-\frac{1}{q}\varepsilon \|u_n\|_{E}
 \\
&-\frac{1}{q}\int |\nabla u_n|^{p}dx-\frac{1}{q}\int |\nabla u_n|^2dx-
\frac{1}{q}\int b(x)u_n{}^{s}dx\leq C,
\end{align*}
and so
\[
(\frac{1}{p}-\frac{1}{q})\|\nabla u_n\|_{p}^{p}+(\frac{1}{2}-\frac{1}{q}
)\|\nabla u_n\|_2^2+(\frac{1}{s}-\frac{1}{q})\int
b(x)|u_n|^{s}dx\leq C+\frac{1}{q}\varepsilon \|u_n\|_{E}.
\]
Since $q>\max \{p,2,s\}$, we deduce that
\begin{equation}
(\frac{1}{p}-\frac{1}{q})\|\nabla u_n\|_{p}^{p}+(\frac{1}{2}-\frac{1}{q}
)\|\nabla u_n\|_2^2\leq C+\frac{1}{q}\varepsilon \|u_n\|_{E}
\label{pl}
\end{equation}
which implies that the sequence $\{u_n\}_{n=1}^{\infty }$ is bounded
in $E$. By passing to a subsequence if necessary, we may assume that
$u_n\to u$ weakly in $E$. Consequently,
\begin{equation}
\lim_{n\to \infty } \langle \Phi '(u_n)-\Phi'(u),u_n-u\rangle =0.  \label{poo}
\end{equation}
By taking $\upsilon =u_n-u$ in \eqref{ps1} we have
\begin{equation}
\begin{aligned}
&\int \left( |\nabla u_n|^{p-2}\nabla u_n-|\nabla u|^{p-2}\nabla u\right)
(\nabla u_n-\nabla u)dx
+\int \left( \nabla u_n-\nabla u\right) (\nabla u_n-\nabla u)dx\\
&=\langle \Phi '(u_n)-\Phi '(u),u_n-u\rangle
- \int |\nabla u_n|^{p-2}\nabla u_n\nabla (u_n-u)dx\\
&\quad -\int \nabla u_n\nabla (u_n-u)dx
 +\int |\nabla u|^{p-2}\nabla u\nabla (u_n-u)dx
+\int \nabla u\nabla (u_n-u)dx\\
&\quad -\int a(x)|u|^{q-2}u(u_n-u)dx+\int b(x)|u_n|^{s-2}u_n(u_n-u)dx\\
&\quad +\int a(x)|u_n{}|^{q-2}u_n(u_n-u)dx+\int b(x)|u|^{s-2}u(u_n-u)dx.
\end{aligned}\label{pooo}
\end{equation}
Since, at least for a subsequence, $u_n\to u$ in $L^{p}(\Omega )$
and $L^2(\Omega )$, \eqref{pooo} yields
\begin{align*}
&\lim_{n\to \infty } \Big\{\int \left( |\nabla
u_n|^{p-2}\nabla u_n-|\nabla u|^{p-2}\nabla u\right) (\nabla
u_n-\nabla u)dx \\
&\quad +\int \left( \nabla u_n-\nabla u\right) (\nabla u_n-\nabla
u)dx\Big\}=0.
\end{align*}
We now use the inequality
\begin{align*}
0&\leq \Big\{\Big( \int |\varphi |^{k}dx\Big) ^{1/k'}
-\Big(\int |\psi |^{k}dx\Big) ^{1/k'}\Big\}
\Big\{\Big( \int |\varphi |^{k}dx\Big) ^{1/k}
-\Big( \int |\psi|^{k}dx\Big) ^{1/k}\Big\} \\
&\leq \int \left( |\varphi |^{k-2}\varphi -|\psi |^{k-2}\psi \right)
(\varphi-\psi )dx,
\end{align*}
which holds for $\varphi ,\psi \in L^{k}(\Omega )$ and
$k'=k/(k-1)$, see \cite{Dra-Her}, to conclude that $u_n\to u$ in
$E$.
\end{proof}

\begin{lemma} \label{lem10}
(i) There exist $\rho ,\alpha >0$ such that $\Phi (u)\geq \alpha $ if
$\|u\|_{E}=\rho $.

(ii) There exists $u\in E$ with $\|u\|>\rho $ and $\Phi (u)<0$.
\end{lemma}

\begin{proof}
(i) Fix $u\in E\backslash \{0\}$. Then
\[
\Phi (u)\geq \frac{1}{2}\|\nabla u\|_2^2-\frac{1}{q}\int a(x)|u|^{q}dx.
\]
By the Sobolev embedding and the fact that $q>2$ we have
\[
\Phi (u)\geq \frac{1}{p}\|u\|_{E}^2-\frac{c}{q}\|u\|_{E}^{q}
\geq \alpha >0,
\]
whenever $\|u\|_{E}=\rho $ and $\rho >0$ is small enough.
Now fix $v\in G_1$. Then for $t>0$
\[
\Phi (tv)=\frac{t^{p}}{p}\|\nabla v\|_{p}^{p}+\frac{t^2}{2}\|\nabla
v\|_2^2-\frac{t^{q}}{q}\int a(x)|v|^{q}dx+\frac{t^{s}}{s}\int
b(x)|v|^{s}dx,
\]
and so $\lim_{t\to \infty }\Phi (tv)=-\infty $. Thus $\Phi (tv)<0$
for large enough $t$.
\end{proof}

By an application of the mountain pass theorem we obtain the following
result.

\begin{theorem} \label{thm11}
Assume that conditions {\rm (H0)--(H4)} hold with
$q>\max \{p,s,2\}$. Then  \eqref{1}-\eqref{2}
admits a solution.
\end{theorem}



\subsection*{Acknowledgements}
The author wishes to thank Professor A. N.
Lyberopoulos, at the University of the Aegean, for suggesting
the problem, and Professor D. A. Kandilakis, at Technical University
of Crete, for his support.

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