\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 $$-\Delta _{q}u-\Delta _{p}u=f(x,u)\quad\text{in }\Omega \label{eq1}$$ with Dirichlet boundary condition $$u=0\quad\text{on }\partial \Omega , \label{Dbc}$$ 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 $$u_{t}=\operatorname{div}(A(u)\nabla u)+f(x,u), \label{schr}$$ 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 $$f(x,u)=g(x,u)-b(x)|u|^{m-2}u-c(x)|u|^{p-2}u \label{part}$$ 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)] $10\} \] 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 $$r^{p-q}\|\nabla \upsilon \|_{p}^{p}+r^{2-q}\|\nabla \upsilon \|_2^2+r^{s-q}B(\upsilon )=A(\upsilon ), \label{BF1}$$ 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} $$r(\upsilon _n)^{p-q}\|\nabla \upsilon _n\|_{p}^{p}+r(\upsilon _n)^{2-q}\|\nabla \upsilon _n\|_2^2+r(\upsilon _n)^{s-q}B(\upsilon _n)=A(\upsilon _n). \label{n}$$ By taking the limit as$n\to +\infty $, we obtain $$0<\tilde{r}^{p-q}\|\nabla \tilde{\upsilon}\|_{p}^{p}+\tilde{r}^{2-q}\|\nabla \tilde{\upsilon}\|_2^2+\tilde{r}^{s-q}B(\tilde{\upsilon})\leq A(\tilde{ \upsilon}), \label{LTE}$$ which implies that$\tilde{\upsilon}\in G_1$. In view of \eqref{1}, $$r(\tilde{\upsilon})^{p-q}\|\nabla \tilde{\upsilon}\|_{p}^{p}+r(\tilde{ \upsilon})^{2-q}\|\nabla \tilde{\upsilon}\|_2^2+r(\tilde{\upsilon} )^{s-q}B(\tilde{\upsilon})=A(\tilde{\upsilon}), \label{E}$$ and so \eqref{LTE} shows that$\tilde{r}\leq r(\tilde{\upsilon})$. If we assume that$\tilde{r}p$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 $$\|\nabla \upsilon \|_{p}^{p}<\frac{p}{q}\frac{2-q}{2-p}r_{\ast }(\upsilon )^{q-p}A(\upsilon ), \label{f1}$$ and so $\|\nabla \upsilon \|_{p}^{p}<\frac{p}{q}\frac{2-q}{2-p}r(\upsilon )^{q-p}A(\upsilon ).$ Thus $$\frac{2-p}{p}r(\upsilon )^{p}\|\nabla \upsilon \|_{p}^{p}+\frac{q-2}{q} r(\upsilon )^{q}A(\upsilon )<0. \label{f2}$$ By \eqref{f2} and \eqref{e4} we conclude that \hat{\Phi}(\upsilon )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{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} implying that $$\|\nabla \tilde{\upsilon}\|_{p}^{p} \leq Q(r_{\ast }(\tilde{\upsilon}),\tilde{\upsilon}). \label{ineq}$$ If we assume the equality $$\|\nabla \tilde{\upsilon}\|_{p}^{p}=Q(r_{\ast }(\tilde{\upsilon}),\tilde{ \upsilon}), \label{u2}$$ then by using \eqref{BE} for \upsilon =\upsilon _n and passing to the limit as n\to +\infty , we obtain \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} 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. Sincer_{\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 $$\left( q-p\right) A(\tilde{\upsilon})=\left( s-p\right) r_{\ast }(\tilde{ \upsilon})^{s-q}B(\tilde{\upsilon})+\left( 2-p\right) r_{\ast }(\tilde{ \upsilon})^{2-q}\|\nabla \tilde{\upsilon}\|_2^2. \label{k1}$$ Thus, $$A(\tilde{\upsilon})=\frac{s-p}{q-p}\tilde{r}^{s-q}B(\tilde{\upsilon})+\frac{ 2-p}{q-p}\tilde{r}^{2-q}\|\nabla \tilde{\upsilon}\|_2^2. \label{a}$$ 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$$$r_{\ast }(t\tilde{\upsilon})t\tilde{\upsilon} =r_{\ast }(\tilde{\upsilon}) \tilde{\upsilon}, \label{mr}$$ by \eqref{BE2}, \eqref{s1} and \eqref{mr}, we have $\|\nabla \tilde{\upsilon}\|_{p}^{p}\Phi (r(\tilde{\upsilon})\tilde{ \upsilon})=\Phi (r(t\tilde{\upsilon})t\tilde{\upsilon})=\hat{\Phi}(t\tilde{ \upsilon}),$ 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,$p0$. Since$r_{\ast }(\upsilon )^{2-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{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} On the other hand, \eqref{BF1} implies that $$r(\upsilon )\leq \Big( \frac{A(\upsilon )}{B(\upsilon )}\Big) ^{1/(s-q)}, \label{rg}$$ 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 $$\frac{p}{q}\frac{2-q}{2-p} \Big( \frac{q-p}{s-p}\frac{A(\upsilon )}{ B(\upsilon )}-\frac{2-p}{s-p}A(\upsilon )^{(2-q)/(s-q)}\frac{\|\nabla \upsilon \|_2^2}{B(\upsilon )^{\frac{2-q}{s-q}+1}}\Big) ^{(q-p)/(s-q)}A(\upsilon )>\|\nabla \upsilon \|_{p}^{p}, \label{rd}$$ 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 $$r_{\ast }(\upsilon )=\Big( \frac{q-p}{2-p}\frac{A(\upsilon )}{\|\nabla \upsilon \|_2^2}\Big) ^{1/(2-q)}, \label{z2}$$ 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, $$\frac{p}{q}\frac{2-q}{2-p}\Big( \frac{q-p}{2-p}\Big) ^{\frac{q-p}{2-q} }A(\upsilon )^{\frac{2-p}{2-q}}>\|\nabla \upsilon \|_2 ^{2(2-p)/(2-q)}, \label{rem3}$$ implying that $G_{_3}\neq \emptyset$. \textbf{Case 3:} $p0$. For $r\geq 0$ let $$F(r,\upsilon ):=r^{p-s}Q(r,\upsilon )=r^{q-s}A(\upsilon )-B(\upsilon )-\|\nabla \upsilon \|_2^2r^{2-s}. \label{bi}$$ 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 $$\bar{r}(\upsilon )=\Big( \frac{q-s}{2-s}\frac{A(\upsilon )}{\|\nabla \upsilon \|_2^2}\Big) ^{1/(2-q)} \label{by}$$ with $$F(\bar{r}(\upsilon ),\upsilon )=\frac{2-q}{2-s}\bar{r}^{q-s}A(\upsilon )-B(\upsilon ). \label{bz}$$ Consequently, $Q(r,\upsilon )>0$ for some $r>0$ if and only if $F(\bar{r}(\upsilon ),\upsilon )>0$, and this holds if $$\bar{r}(\upsilon )>\hat{r}(\upsilon ) :=(\frac{2-s}{2-q}\frac{B(\upsilon )}{ A(\upsilon )})^{1/(q-s)}. \label{ba}$$ 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 $r_{1\ast }(\upsilon )0$ 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 $$r_{\ast }(\upsilon ):=\Big( \frac{q-p}{2-p}\frac{A(\upsilon )}{\|\nabla \upsilon \|_2^2}\Big) ^{1/(2-q)} \label{bn}$$ with $$Q(r_{\ast }(\upsilon ),\upsilon )=\frac{2-q}{2-p}r_{\ast }(\upsilon )^{q-p}A(\upsilon ). \label{bc}$$ Since $r_{\ast }(\upsilon )$ satisfies the equation $Q_{r}(\cdot,\upsilon )=0$, that is $$(q-p)A(\upsilon )r_{\ast }(\upsilon )^{q-s}=(s-p)B(\upsilon )+(2-p)\|\nabla \upsilon \|_2^2r_{\ast }(\upsilon )^{2-s}, \label{hu}$$ we have that $$r_{\ast }(\upsilon )\leq \Big( \frac{q-p}{2-p}\frac{A(\upsilon )}{ \|\nabla \upsilon \|_2^2}\Big) ^{1/(2-q)}. \label{ml}$$ 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 $r_1(\upsilon )r_{\ast }(\upsilon )$, by the implicit function theorem, $r\in C^1(G_2)$. We will assume that the set $$\label{xz} G_4:=\{\upsilon \in G_1:\|\nabla \upsilon \|_{p}^{p} \leq \frac{p}{s}\frac{2-s}{2-p}\big( \frac{s}{q}\frac{2-q}{2-s}\bar{r }(\upsilon )^{q-s}A(\upsilon )-B(\upsilon )\big) \bar{r}(\upsilon )^{s-p}\}$$ 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 )\\ &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 $$\tilde{r}\geq \tilde{r}_{\ast }\geq \hat{r}(\tilde{\upsilon}):=(\frac{2-s}{ 2-q}\frac{B(\tilde{\upsilon})}{A(\tilde{\upsilon})})^{1/(q-s)}. \label{df}$$ 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 $$(q-p)A(\tilde{\upsilon})\tilde{r}^{q-s}-(s-p)B(\tilde{\upsilon} )=(2-p)\|\nabla \tilde{\upsilon}\|_2^2\tilde{r}^{2-s}. \label{vr}$$ 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,$p0. From \eqref{by} we obtain \begin{align*} &\frac{p}{q}\frac{2-q}{2-p}\bar{r}(\upsilon )^{q-p}A(\upsilon ) -\frac{p}{s}\frac{2-s}{2-p}B(\upsilon )\bar{r}(\upsilon )^{s-p} \\ &=\frac{p}{q}\frac{2-q}{2-p}\Big( \frac{q-s}{2-s}\frac{A(\upsilon )}{ \|\nabla \upsilon \|_2^2}\Big) ^{(q-p)/(2-q)}A(\upsilon )-\frac{p}{ s}\frac{2-s}{2-p}B(\upsilon )\Big( \frac{q-s}{2-s}\frac{A(\upsilon )}{ \|\nabla \upsilon \|_2^2}\Big) ^{(s-p)/(2-q)} \\ &=\frac{p}{q}\frac{2-q}{2-p}\Big( \frac{q-s}{2-s}\frac{1}{\|\nabla \upsilon \|_2^2}\Big) ^{(q-p)/(2-q)}A(\upsilon )^{\frac{q-p}{2-q}+1} \\ &\quad -\frac{p}{s}\frac{2-s}{2-p}B(\upsilon )\Big( \frac{q-s}{2-s}\frac{1}{ \|\nabla \upsilon \|_2^2}\Big) ^{(s-p)/(2-q)}A(\upsilon )^{\frac{ s-p}{2-q}}. \end{align*} If we assume that \label{sw} \begin{aligned} &\frac{p}{q}\frac{2-q}{2-p}\Big( \frac{q-s}{2-s}\frac{1}{\|\nabla \upsilon \|_2^2}\Big) ^{(q-p)/(2-q)}A(\upsilon )^{\frac{q-p}{2-q}+1} \\ &-\frac{p}{s}\frac{2-s}{2-p}B(\upsilon )\Big( \frac{q-s}{2-s}\frac{1}{ \|\nabla \upsilon \|_2^2}\Big) ^{(s-p)/(2-q)}A(\upsilon ) ^{(s-p)/(2-q)}>\|\nabla \upsilon \|_{p}^{p}, \end{aligned} then\upsilon\in G_4$. 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 $$\frac{p}{q}\frac{2-q}{2-p}\Big( \frac{q-s}{2-s}\frac{1}{\|\nabla \upsilon \|_2^2}\Big) ^{(q-p)/(2-q)}A(\upsilon )^{\frac{q-p}{2-q} +1}>\|\nabla \upsilon \|_{p}^{p}, \label{rem26}$$ then\upsilon \in G_4$. Note that if$a^{+}(\cdot)$is large enough then \eqref{rem26} holds. \textbf{Case 4:}$p<20$a.e. in$\Omega $. \end{itemize} Let $$Q(r,\upsilon ):=r^{q-2}A(\upsilon )-r^{s-2}B(\upsilon )-r^{p-2}\|\nabla \upsilon \|_{p}^{p}. \label{BE3}$$ Then \eqref{BE} is equivalent to $$Q(r,\upsilon )=\|\nabla \upsilon \|_2^2. \label{BE4}$$ 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 $$(q-2)r_{\ast }^{q-p}A(\upsilon )+(2-p)\|\nabla \upsilon \|_{p}^{p}=(s-2)r_{\ast }^{s-p}B(\upsilon ). \label{eqr*}$$ Thus, $$r_{\ast }(\upsilon )\geq (\frac{q-2}{s-2}\frac{A(\upsilon )}{B(\upsilon )})^{ \frac{1}{s-q}}. \label{rab*}$$ 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 \|_2^21 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}, $$\|\nabla \upsilon \|_2^2<\frac{2}{q}\frac{s-q}{s-2}A(\upsilon )r(\upsilon )^{q-2}-\frac{2}{p}\frac{s-p}{s-2}\|\nabla \upsilon \|_{p}^{p}r(\upsilon )^{p-2}. \label{ad1}$$ 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 $$r(\upsilon )\leq (\frac{A(\upsilon )}{B(\upsilon )})^{1/(s-q)}, \label{rab}$$ while hypothesis (H3) implies $$A(\upsilon )\leq cB(\upsilon )^{q/s} \label{aabb}$$ for every$\upsilon \in E$and some$c>0$. At the same time if$\upsilon \in \tilde{G}_2$, then for some$\theta >0$, $$\theta <\|\nabla \upsilon \|_2^2<\frac{q-p}{2-p}r_{\ast }(\upsilon )^{q-2}A(\upsilon )<\frac{q-p}{2-p}r(\upsilon )^{q-2}A(\upsilon ). \label{rad}$$ From \eqref{rab} and \eqref{rad} we deduce $$\theta <\frac{q-p}{2-p}\Big( \frac{A(\upsilon )}{B(\upsilon )} \Big) ^{(q-2)/(s-q)}A(\upsilon ). \label{abth}$$ Next, by using \eqref{aabb} and \eqref{abth}, we have $$\theta <\frac{q-p}{2-p}c^{\frac{s-2}{s-q}}B(\upsilon )^{\frac{2}{s}}, \label{posB}$$ 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{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} we see that $$\|\nabla \tilde{\upsilon}\|_2^2\leq Q(r_{\ast }(\tilde{\upsilon}),\tilde{ \upsilon}). \label{h2}$$ We shall show that strict inequality holds. Indeed, let us suppose $$\|\nabla \tilde{\upsilon}\|_2^2=Q(r_{\ast }(\tilde{\upsilon}),\tilde{ \upsilon}). \label{h3}$$ Since\tilde{r}>0$, by applying \eqref{BE4} for$\upsilon =\upsilon _nand passing to the limit, we also obtain \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} 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*}, $$\frac{q-2}{2s}\tilde{r}^{q}A(\tilde{\upsilon})+\frac{2-p}{2s}\tilde{r} ^{p}\|\nabla \tilde{\upsilon}\|_{p}^{p}=\frac{s-2}{2s}\tilde{r}^{s}B(\tilde{ \upsilon}). \label{rlim8}$$ 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}\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, $p<2\frac{p-2}{s-q}$, $G_5\neq \emptyset$ for $a^{+}(\cdot)$ large enough. \end{remark} \textbf{Case 5:} $sr_{\ast }(\upsilon )$, by \eqref{g6} we get $$\|\nabla \upsilon \|_{p}^{p}<\frac{p}{q}\frac{2-q}{2-p}r(\upsilon )^{q-p}A(\upsilon )-\frac{p}{s}\frac{2-s}{2-p}r(\upsilon )^{s-p}B(\upsilon ). \label{ap2}$$ 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} $$\frac{q-p}{2p}\tilde{r}^{s}A(\tilde{\upsilon}) +\frac{p-s}{2p}\tilde{r}^{s}B( \tilde{\upsilon}) =\frac{2-p}{2p}\tilde{r}^2\|\nabla \tilde{\upsilon} \|_2^2. \label{rlim2}$$ 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, $s0$. From (\eqref{eqr2} $$\left( \frac{p-s}{2-p}\frac{B(\upsilon )}{\|\nabla \upsilon \|_2^2} \right) ^{1/(2-s)}\leq r_{\ast }(\upsilon ), \label{coci}$$ 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{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} 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 $$\Big( \frac{q-p}{2-p}\frac{A(\upsilon )}{\|\nabla \upsilon \|_2^2} \Big) ^{1/(2-q)}=r_{\ast }(\upsilon ), \label{ram1}$$ 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 $$A(\upsilon )^{\frac{2-p}{2-q}}>\frac{q}{p}\frac{2-p}{2-q} \Big( \frac{q-p}{2-p}\Big) ^{(p-q)/(2-q)}\|\nabla \upsilon \|_2^2, \label{ram}$$ and so if $a^{+}(\cdot)$ large enough the condition \eqref{ram} is valid implying that $G_6\neq \emptyset .$ \textbf{Case 6:} $s0$, 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 $$(p-s)B(\upsilon )=(p-q)r_{\ast }^{q-s}A(\upsilon )+(2-p)r_{\ast }^{2-s}\|\nabla \upsilon \|_2^2. \label{gm}$$ Clearly, if $\upsilon \in G_2$, then \eqref{BE} has exactly two positive solutions $r_1(\upsilon )$ and $r_2(\upsilon )$ with $r_1(\upsilon )0$. 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}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, $$\|\nabla \upsilon _n\|_{p}^{p}\geq Q(r_{\ast }(\upsilon _n),\upsilon _n)\text{ for every }n\in\mathbb{N}. \label{sa}$$ 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 $sp$, 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 $$\frac{2-p}{p}r(\upsilon )^{p}\|\nabla \upsilon \|_{p}^{p}+\frac{q-2}{q} r(\upsilon )^{q}A(\upsilon )<0. \label{rf5}$$ Combining \eqref{rf5} with \eqref{e3}, we conclude that $\hat{\Phi}(\upsilon )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 $$A(\tilde{\upsilon})=\frac{s-p}{q-p}\tilde{r}^{s-q}B(\tilde{\upsilon}) +\frac{2-p}{q-p}\tilde{r}^{2-q}\|\nabla \tilde{\upsilon}\|_2^2. \label{rfm}$$ 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, $p0$. Since r_{\ast }(\upsilon )^{2-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{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} 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*} Ifa^{+}(\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 Ewe have \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} If\upsilon =u_nin \eqref{ps1}, then $$\int a(x)u_n{}^{q}dx\leq \varepsilon \|u_n\|_{1,k}+\int |\nabla u_n|^{p}dx+\int |\nabla u_n|^2dx+\int b(x)u_n{}^{s}dx. \label{ps2}$$ By hypothesis $$\frac{1}{p}\|\nabla u_n\|_{p}^{p}+\frac{1}{2}\|\nabla u_n\|_2^2- \frac{1}{q}\int a(x)|u_n|^{q}dx+\frac{1}{s}\int b(x)|u_n|^{s}dx\leq C. \label{ps3}$$ 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 $(\frac{1}{p}-\frac{1}{q})\|\nabla u_n\|_{p}^{p}+(\frac{1}{2}-\frac{1}{q} )\|\nabla u_n\|_2^2+(\frac{1}{s}-\frac{1}{q})\int b(x)|u_n|^{s}dx\leq C+\frac{1}{q}\varepsilon \|u_n\|_{E}.$ Sinceq>\max \{p,2,s\}$, we deduce that $$(\frac{1}{p}-\frac{1}{q})\|\nabla u_n\|_{p}^{p}+(\frac{1}{2}-\frac{1}{q} )\|\nabla u_n\|_2^2\leq C+\frac{1}{q}\varepsilon \|u_n\|_{E} \label{pl}$$ 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, $$\lim_{n\to \infty } \langle \Phi '(u_n)-\Phi'(u),u_n-u\rangle =0. \label{poo}$$ By taking$\upsilon =u_n-uin \eqref{ps1} we have \begin{aligned} &\int \left( |\nabla u_n|^{p-2}\nabla u_n-|\nabla u|^{p-2}\nabla u\right) (\nabla u_n-\nabla u)dx +\int \left( \nabla u_n-\nabla u\right) (\nabla u_n-\nabla u)dx\\ &=\langle \Phi '(u_n)-\Phi '(u),u_n-u\rangle - \int |\nabla u_n|^{p-2}\nabla u_n\nabla (u_n-u)dx\\ &\quad -\int \nabla u_n\nabla (u_n-u)dx +\int |\nabla u|^{p-2}\nabla u\nabla (u_n-u)dx +\int \nabla u\nabla (u_n-u)dx\\ &\quad -\int a(x)|u|^{q-2}u(u_n-u)dx+\int b(x)|u_n|^{s-2}u_n(u_n-u)dx\\ &\quad +\int a(x)|u_n{}|^{q-2}u_n(u_n-u)dx+\int b(x)|u|^{s-2}u(u_n-u)dx. \end{aligned}\label{pooo} Since, at least for a subsequence,u_n\to u$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. 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