\documentclass[reqno]{amsart} \usepackage{hyperref} \AtBeginDocument{{\noindent\small \emph{Electronic Journal of Differential Equations}, Vol. 2016 (2016), No. 191, pp. 1--16.\newline ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu} \thanks{\copyright 2016 Texas State University.} \vspace{8mm}} \begin{document} \title[\hfilneg EJDE-2016/191\hfil Existence of nonnegative solutions] {Existence of nonnegative solutions for singular elliptic problems} \author[T. Godoy, A. J. Guerin \hfil EJDE-2016/191\hfilneg] {Tomas Godoy, Alfredo J. Guerin} \address{Tomas Godoy \newline FaMAF, Universidad Nacional de C\'ordoba, Ciudad Universitaria, 5000 C\'ordoba, Argentina} \email{tomasgodo@gmail.com} \address{Alfredo J. Guerin \newline FaMAF, Universidad Nacional de C\'ordoba, Ciudad Universitaria, 5000 C\'ordoba, Argentina} \email{guerin.alfredojose@gmail.com} \thanks{Submitted March 8, 2016. Published July 13, 2016.} \subjclass[2010]{35J75, 35D30, 35J20} \keywords{Singular elliptic problem; variational problems; nonnegative solution; \hfill\break\indent positive solution; sub-supersolution} \begin{abstract} We prove the existence of nonnegative nontrivial weak solutions to the problem \begin{gather*} -\Delta u=au^{-\alpha}\chi_{\{ u>0\} }-bu^p\quad\text{in }\Omega, \\ u=0\quad\text{on }\partial\Omega, \end{gather*} where $\Omega$ is a bounded domain in $\mathbb{R}^{n}$. A sufficient condition for the existence of a continuous and strictly positive weak solution is also given, and the uniqueness of such a solution is proved. We also prove a maximality property for solutions that are positive a.e. in $\Omega$. \end{abstract} \maketitle \numberwithin{equation}{section} \newtheorem{theorem}{Theorem}[section] \newtheorem{lemma}[theorem]{Lemma} \newtheorem{remark}[theorem]{Remark} \newtheorem{corollary}[theorem]{Corollary} \newtheorem{example}[theorem]{Example} \allowdisplaybreaks \section{Introduction and statement of the problem} Let $\Omega$ be a bounded domain in $\mathbb{R}^{n}$ with $C^{1,1} $ boundary, let $a$ and $b$ be nonnegative functions on $\Omega$, and let $\alpha$ and $p$ be positive real numbers. Consider the following singular elliptic problem \begin{equation} \begin{gathered} -\Delta u=au^{-\alpha}-bu^p\quad \text{in }\Omega,\\ u=0\quad \text{on }\partial\Omega, \\ u>0\quad \text{in }\Omega \end{gathered}\label{Problem 0} \end{equation} Problems like \eqref{Problem 0} appear in chemical catalysts process, non-Newtonian fluids, and in models for the temperature of electrical devices (see e.g., \cite{cohen,callegari,Diaz-Morel-Oswald,Fulks MayBe}). Several works can be found concerning the existence of positive solutions to \eqref{Problem 0} for the case $b=0$, i.e., for the problem $-\Delta u=au^{-\alpha}$ in $\Omega$, $u=0$ on $\partial\Omega$, $u>0$ in $\Omega$; let us mention a few: Classical solutions $u\in C^2(\Omega) \cap C(\overline{\Omega}) $ satisfying $u(x) >0$ for all $x\in\Omega$ were obtained by Crandall, Rabinowitz and Tartar \cite{Crandall} under the following hypothesis: $a\in C^1(\overline{\Omega}) $ and $\min_{\overline{\Omega}}a>0$. Lazer and McKenna \cite{Lazer-Makenna} proved the existence of positive weak solutions $u\in H_0^1(\Omega) $ to \eqref{Problem 0} assuming that $a\in C^{\gamma}(\overline{\Omega}) $, $\gamma\in(0,1) $, and, again, $a$ strictly positive on $\overline{\Omega}$. The case $0\leq a\in L^{\infty}(\Omega)$, $a\not \equiv 0$ (that is: $| \{ x\in\Omega:a(x) >0\} |>0$) was studied by Del Pino \cite{del-Pino}. Situations where $a$ is singular on the boundary $\partial\Omega$ were considered by Bougherara, Giacomoni and Hern\'{a}ndez \cite{Jesus2}. The existence of classical solutions to problem \eqref{Problem 0} was proved by Coclite and Palmieri \cite{Coclite-Palmieri} for $a$ and $b$ in $C^1(\overline{\Omega}) $, $00$ in $\Omega$, where $\Omega$ is a smooth bounded domain in $\mathbb{R}^{n}$, $\lambda,\mu\geq0$, $00$ in $\Omega$, where $\Omega$ is a smooth bounded domain in $\mathbb{R}^{n}$, $d_{\Omega}(x)=\operatorname{dist}(x,\partial \Omega) $, $\lambda>0$, $\mu\in\mathbb{R}$, $0<\beta\leq2$, $p( d_{\Omega}(x) ) $ is a positive weight possibly singular at $\partial\Omega$, $\ g\in C^1(0,\infty) $ is a positive decreasing function such that $\lim_{s\to0^{+}}g(s) =\infty,\ f:\overline{\Omega}\times[ 0,\infty) \to[ 0,\infty) $ is a H\"{o}lder continuous function which is positive on $\Omega\times(0,\infty) $ and satisfies that $s\to f(x,s) $ is nondecreasing and also that $f(x,s) $ is either linear or sublinear with respect to $s$. R\u{a}dulescu \cite{Radulescu3} states existence, nonexistence and uniqueness results for blow-up boundary solutions of logistic equations and for Lane-Emden-Fowler equations with singular nonlinearities and subquadratic convection term. Existence and nonexistence results for solutions to the inequality $Lu\geq K(x) u^p$ in $\Omega$, $u>0$ in $\Omega$ were obtained by Ghergu, Liskevich and Sobol \cite{Ghergu} for the case where $\Omega$ is a punctured ball $B_{R}(0) \backslash\{ 0\} $, $p\in\mathbb{R}$, $K\in L_{\rm loc}^{\infty}(B_{R}(0) \backslash\{ 0\} ) $, $ess\inf K>0$, and $Lu:=\sum_{1\leq i,j\leq n}a_{ij}(x) \frac{\partial^2u}{\partial x_i\partial x_j}+\sum_{1\leq j\leq n}b_j(x) \frac{\partial u}{\partial x_j}$, where the matrix $\mathbf{a}=\{ a_{ij}(x) \} _{1\leq i,j\leq n}$ is symmetric, uniformly elliptic on $\Omega$, with each $a_{ij}\in L^{\infty}(B_{R}(0) ) $, and each $b_j$ is a measurable function and satisfies $ess\sup_{x\in B_{R}(0) \backslash\{ 0\} }| x| b_j(x) <\infty$. Existence and uniqueness results were obtained by Bougherara and Giacomoni \cite{Boughe-Giaco} for mild solutions to singular initial value parabolic problems involving the p-Laplacian operator of the form $u_{t}-\Delta_{p}u=u^{-\alpha}+f(x,u) $ in $Q_{T}:=( 0,T) \times\Omega$, $u=0$ on $(0,T) \times\partial\Omega $, $u>0$ in $Q_{T}$, $u(0,x) =u_0(x) $ in $\Omega$ where $\Omega$ is a regular bounded domain in $\mathbb{R}^{n}$, $f:\Omega\times\mathbb{R}\to\mathbb{R}$ is a bounded below Carath\'{e}odory function and nonincreasing with respect to the second variable, $\Delta_{p}u:=\operatorname{div}(| \nabla u| ^{p-2}\nabla u) $, $10$, $T>0$, and $u_0$ in a suitable functional space. Singularly perturbed elliptic problems on an annulus whose solutions concentrate in a circle were studied by Manna and Srikanth \cite{Manna}. Let us mention also that Loc and Schmitt \cite{LocSchmitt}, \cite{LocSchmitt1}, extended the method of sub and supersolutions to deal with singular elliptic problems. A comprehensive treatment of the subject can be found in Ghergu and R\u{a}dulescu's book \cite{Radulescu} (see also \cite{Radulescu3}), and in the survey article \cite{Diaz}, by D\'iaz and Hern\'{a}ndez. Let us state the problem that we will consider from now on: Let $\Omega$ be a bounded domain in $\mathbb{R}^{n}$ with $C^{1,1}$ boundary, $\alpha\in(0,1) $, and $p\in(0,2^{\ast}-1) $, where $2^{\ast}$ is defined by $\frac{1}{2^{\ast}}=\frac{1}{2}-\frac{1}{n}$ if $n>2$ and $2^{\ast}=\infty$ if $n\leq2$. Let $a$ and $b$ be nonnegative functions such that $a$ belongs to $L^{\infty}(\Omega) $, $a\not \equiv 0$, and $b$ is in $L^{r}(\Omega) $, with $r=\frac{2}{1-p}$ if $p<1$, and $r=\infty$ otherwise. We are concerned with weak solutions to the problem \begin{equation} \begin{gathered} -\Delta u=au^{-\alpha}\chi_{\{ u>0\} }-bu^p\quad \text{in }\Omega,\\ u=0\quad \text{on }\partial\Omega,\text{ }\\ u\geq0\quad \text{in }\Omega \end{gathered} \label{Problem} \end{equation} where $au^{-\alpha}\chi_{\{ u>0\} }$ stands for the function defined by $au^{-\alpha}\chi_{\{ u>0\} }(x)=a(x) u(x) ^{-\alpha}$ if $u(x) \neq0$, and $au^{-\alpha}\chi_{\{ u>0\} }(x) =0$ if $u(x) =0$. By a weak solution to \eqref{Problem} we mean a nonnegative function $u\in H_0^1(\Omega) $ such that, for all $\varphi$ in $H_0^1(\Omega) \cap L^{\infty}(\Omega) $, $(au^{-\alpha}\chi_{\{ u>0\} }-bu^p) \varphi\in L^1(\Omega) $, and the following holds \begin{equation} \int_{\Omega}\langle \nabla u,\nabla\varphi\rangle =\int_{\Omega }(au^{-\alpha}\chi_{\{ u>0\} }-bu^p) \varphi. \label{weak form} \end{equation} The main aim of this work is to prove the existence of at least one nonnegative weak solution $u\not \equiv 0$ to the stated problem (see Theorem \ref{thm3.1}). Additionally, we give a condition on $a,b$ that guarantees the existence of a strictly positive weak solution to \eqref{Problem} (see Theorem \ref{thm3.5}). In Theorem \ref{thm3.8} we prove that there is at most one solution that is positive a.e. in $\Omega$, and give a maximality property for such a solution. Examples of non-existence of strictly positive solutions, and of non-uniqueness of the nonnegative solutions, are also provided. To prove Theorem \ref{thm3.1}, we show that the energy functional $J$ associated with \eqref{Problem} attains its minimum at some nonnegative nontrivial $u\in H_0^1(\Omega) \cap L^{\infty}(\Omega) $. Note that $J$ may fail to be Gateaux differentiable at $u$; despite this fact, we manage to prove that the said minimizer is indeed a weak solution of problem \eqref{Problem}. Theorem \ref{thm3.5} is proved using the sub and supersolutions method for singular elliptic problems developed in \cite{LocSchmitt}. \section{Preliminary lemmas} Let $J:H_0^1(\Omega) \to\mathbb{R}$ be the energy functional associated with \eqref{Problem}, \begin{equation} J(u) :=\frac{1}{2}\int_{\Omega}| \nabla u| ^2-\frac{1}{1-\alpha}\int_{\Omega}a| u| ^{1-\alpha} +\frac{1}{1+p}\int_{\Omega}b| u| ^{1+p}. \label{energy} \end{equation} Let us start with the following lemma. \begin{lemma} \label{lem2.1} The following statements hold: \begin{itemize} \item[(i)] $J$ is coercive on $H_0^1(\Omega)$. \item[(ii)] $\inf_{u\in H_0^1(\Omega) }J(u) >-\infty$. \item[((iii)] $\inf_{u\in H_0^1(\Omega) }J(u) $ is achieved at some $u\in H_0^1(\Omega) $. \end{itemize} \end{lemma} \begin{proof} Let $u\in H_0^1(\Omega) $. Since $0<1-\alpha<1$, the H\"{o}lder's and Poincare's inequalities give \[ \frac{1}{1-\alpha}\int_{\Omega}a| u| ^{1-\alpha}\leq c\| \nabla u\| _2^{1-\alpha}\] for some positive constant $c$ independent of $u$, and so $J(u) \geq\frac{1}{2}\|\nabla u\| _2^2-c\| \nabla u\| _2^{1-\alpha}$, which clearly implies (i) and (ii). To prove (iii), let $\beta=\inf_{u\in H_0^1(\Omega) }J(u) $, and consider a sequence $\{u_j\} _{j\in\mathbb{N}}\subset H_0^1(\Omega) $ such that $\lim_{j\to\infty}J(u_j) =\beta$. Then, by i), $\{ u_j\} _{j\in\mathbb{N}}$ is bounded in $H_0^1(\Omega) $. Let $q$ be in $(p+1,2^{\ast}) $. Since the inclusion $H_0^1(\Omega) \hookrightarrow L^{q}(\Omega) $ is a compact map, we can assume (taking a subsequence if necessary) that $\{ u_j\} _{j\in\mathbb{N}}$ converges strongly to some $u\in L^{q}(\Omega) $. Since $\{ u_j\}_{k\in\mathbb{N}}$ is bounded in $H_0^1(\Omega) $, there exists $v\in H_0^1(\Omega) $, and a subsequence $\{u_{j_k}\} _{k\in\mathbb{N}}$, such that the subsequence converges strongly to $v$ in $L^2(\Omega) $, and $\{ \nabla u_{j_k}\} _{k\in\mathbb{N}}$ converges weakly to $\nabla v$ in $L^2(\Omega,\mathbb{R}^{n}) $. Thus $v=u$, $\{ u_{j_k}\} _{k\in\mathbb{N}}$ converges to $u$ in $L^{q}(\Omega)$, and \begin{equation} \| \nabla u\| _2\leq\lim\inf_{k\to\infty}\|\nabla u_{j_k}\| _2. \label{limite debil1} \end{equation} On the other hand, the Nemytskii operators $f(u):=| u| ^{1-\alpha}$ and $g(u) :=|u| ^{1+p}$ are continuous from $L^2(\Omega)$ into $L^{\frac {2}{1-\alpha}}(\Omega)$, and from $L^{q}(\Omega)$ into $L^{\frac{q}{1+p} }(\Omega)$, respectively \cite[Theorem 1.2.1]{Ambrosetti-Arcoya} and so, since $a\in L^{\infty}(\Omega)$ and $b\in L^{r}(\Omega)$, \begin{equation} \label{limite fuerte} \begin{aligned} & \lim_{j\to\infty}\int_{\Omega}\Big(\frac{1}{1-\alpha}a| u_{j_k}| ^{1-\alpha}-\frac{1}{1+p}b| u_{j_k}| ^{1+p}\Big) \\ & =\int_{\Omega}(\frac{1}{1-\alpha}a| u| ^{1-\alpha}-\frac{1}{1+p}b| u| ^{1+p}) \end{aligned} \end{equation} which, combined with \eqref{limite debil1}, gives $J(u) \leq \lim\inf_{k\to\infty}J(u_{j_k}) =\beta$, therefore (iii) holds (since $\beta\leq J(u)$). \end{proof} \begin{corollary} \label{coro2.2} $\inf_{u\in H_0^1(\Omega)}J(u) $ is achieved at some nonnegative $u\in H_0^1(\Omega) $. \end{corollary} \begin{proof} Lemma \ref{lem2.1} states that $J$ attains its minimum at some $u\in H_0^1(\Omega) $. Since $J(u) =J(| u| ) $, a nonnegative minimizer exists. \end{proof} For the rest of this article, we fix a nonnegative minimizer for $J$ on $H_0^1(\Omega) $, and denote it by $\mathbf{u}$. \begin{lemma} \label{lem2.3} The equality \begin{equation} \int_{\Omega}\langle \nabla\mathbf{u},\nabla(\mathbf{u} \varphi) \rangle =\int_{\Omega}(a\mathbf{u}^{1-\alpha }-b\mathbf{u}^{1+p}) \varphi \label{weak sing} \end{equation} holds for any $\varphi\in H^1(\Omega) \cap L^{\infty }(\Omega) $ such that $\varphi\mathbf{u}\in H_0^1(\Omega)$. \end{lemma} \begin{proof} Let $\varphi\in H^1(\Omega) \cap L^{\infty}(\Omega) $ be such that $\varphi\mathbf{u}\in H_0^1(\Omega) $; satisfying, in addition, $\|\varphi\| _{\infty}\leq\frac{1}{2}$. Let $\tau\in R$ such that $| \tau| <1$. Then $\mathbf{u}+\tau\mathbf{u} \varphi\geq0$, and $J(\mathbf{u}) \leq J(\mathbf{u}+\tau\mathbf{u}\varphi) $. A computation shows that this inequality can be written as \begin{equation} \label{weak sing 1} \begin{aligned} & \tau\int_{\Omega}\langle \nabla\mathbf{u},\nabla( \mathbf{u}\varphi) \rangle \\ & \geq\frac{1}{1-\alpha}\int_{\Omega}a\mathbf{u}^{1-\alpha}\big( (1+\tau\varphi) ^{1-\alpha}-1\big) -\frac{1}{1+p}\int_{\Omega}b\mathbf{u}^{1+p}\big((1+\tau\varphi) ^{1+p}-1\big) \\ &\quad -\frac{\tau^2}{2}\int_{\Omega}\mathbf{u}^2| \nabla \varphi| ^2-\frac{\tau^2}{2}\int_{\Omega}\varphi^2| \nabla\mathbf{u}| ^2-\tau^2\int_{\Omega}\mathbf{u} \varphi\langle \nabla\mathbf{u},\nabla\varphi\rangle . \end{aligned} \end{equation} Note that, for $\gamma>0$, the second-order Taylor expansion of the function $h(t) =(1+t) ^{\gamma}-1$ gives \begin{equation} \label{weak sing 2} (1+\tau\varphi) ^{\gamma}-1=\gamma\tau\varphi-\frac{\tau^2} {2}\gamma(\gamma-1) (1+\zeta_{\tau,\gamma})^{\gamma-2}\varphi^2 \end{equation} for some measurable function $\zeta_{\tau,\gamma}:\Omega\to\mathbb{R}$ satisfying $| \zeta_{\tau,\gamma}| \leq| \tau\varphi| \leq\frac{1}{2}$. Inserting \eqref{weak sing 2} (used with $\gamma=1-\alpha$ and $\gamma=1+p$) in \eqref{weak sing 1}, we obtain \begin{equation} \label{weak sing 3} \begin{aligned} & \tau\int_{\Omega}\langle \nabla\mathbf{u},\nabla( \mathbf{u}\varphi) \rangle \\ & \geq\tau\int_{\Omega}a\mathbf{u}^{1-\alpha}\varphi-\frac{\tau^2} {2}\alpha\int_{\Omega}a\mathbf{u}^{1-\alpha}(1+\zeta_{\tau ,1-\alpha}) ^{-\alpha-1}\varphi^2\\ &\quad -\big(\tau\int_{\Omega}b\mathbf{u}^{1+p}\varphi+\frac{\tau^2} {2}p\int_{\Omega}b\mathbf{u}^{1+p}(1+\zeta_{\tau,1+p}) ^{p-1}\varphi^2\Big) \\ &\quad -\frac{\tau^2}{2}\int_{\Omega}\mathbf{u}^2| \nabla \varphi| ^2-\frac{\tau^2}{2}\int_{\Omega}\varphi^2| \nabla\mathbf{u}| ^2-\tau^2\int_{\Omega}\mathbf{u} \varphi\langle \nabla\mathbf{u},\nabla\varphi\rangle . \end{aligned} \end{equation} Also, $1+\zeta_{\tau,1-\alpha}\geq\frac{1}{2}$ and $1+\zeta_{\tau,1+p}\geq\frac{1}{2}$, and thus \begin{gather*} | \int_{\Omega}a\mathbf{u}^{1-\alpha}(1+\zeta _{\tau,1-\alpha}) ^{-\alpha-1}\varphi^2| \leq c,\\ | \int_{\Omega}b\mathbf{u}^{1+p}(1+\zeta_{\tau ,1+p}) ^{p-1}\varphi^2| \leq c \end{gather*} for some positive constant $c$ independent of $\tau$. Now we take $\tau$ positive in \eqref{weak sing 3}. Dividing by $\tau$, and then letting $\tau\to0^{+}$, from \eqref{weak sing 3} we obtain \[ \int_{\Omega}\langle \nabla\mathbf{u},\nabla(\mathbf{u} \varphi) \rangle \geq\int_{\Omega}a\mathbf{u}^{1-\alpha }\varphi-\int_{\Omega}b\mathbf{u}^{1+p}\varphi. \] We note that this inequality holds if we put $-\varphi$ instead of $\varphi$; therefore we obtain also the reverse inequality, and we conclude that \eqref{weak sing} is valid for $\| \varphi\| _{\infty} \leq\frac{1}{2}$. Finally, since both sides in \eqref{weak sing} are linear on $\varphi$, the assumption $\| \varphi\| _{\infty}\leq\frac {1}{2}$ can be removed. \end{proof} \begin{lemma} \label{lem2.4} There exists $v\in H_0^1(\Omega) $ such that $J(v) <0$. \end{lemma} \begin{proof} It is sufficient to show that there exists a function $\Phi\in H_0^1(\Omega) $ such that $\int_{\Omega}a| \Phi| ^{1-\alpha}>0$. Indeed, if such a $\Phi$ exists, then, for $t>0$, we have \begin{align*} J(t\Phi) & =\frac{t^2}{2}\| \nabla\Phi\| _2^2-\frac{t^{1-\alpha}}{1-\alpha}\int_{\Omega}a| \Phi| ^{1-\alpha}+\frac{t^{1+p}}{1+p}\int_{\Omega}b| \Phi| ^{1+p}\\ & =t^{1-\alpha}(\frac{t^{1+\alpha}}{2}\| \nabla\Phi\| _2^2-\frac{1}{1-\alpha}\int_{\Omega}a| \Phi| ^{1-\alpha}+\frac{t^{p+\alpha}}{1+p}\int_{\Omega}b| \Phi| ^{1+p}) \end{align*} which gives that $J(t\Phi) $ is negative for $t$ positive and small enough. Such a $\Phi$ can be constructed as follows: Let $h\in C_{c}^{\infty}(\mathbb{R}^{n}) $ be a nonnegative radial function with support in the unit ball $B=\{ x\in\mathbb{R} ^{n}:| x| <1\} $, and such that $\int_{B}h=1$. For $\varepsilon>0$ let $h_{\varepsilon}(x) :=\frac{1} {\varepsilon^{n}}h(\frac{x}{\varepsilon}) $. For $\delta>0$ let $\Omega_{\delta}:=\{ x\in\Omega:\operatorname{dist}(x,\partial\Omega) >\delta\} $. Since $| \{ x\in\Omega:a(x) >0\} | >0$, we have $| \{ x\in\Omega:a( x) >0\} \cap\Omega_{\delta}| >0$ for $\delta$ positive and small enough. We fix such a $\delta$, and set $E=\{ x\in \Omega:a(x) >0\} \cap\Omega_{\delta}$. For $\varepsilon >0$ we define $\Phi_{\varepsilon}:=h_{\varepsilon}\ast\chi_{E}$. Then $\Phi_{\varepsilon}\in C^{\infty}(\mathbb{R}^{n}) $ and $\operatorname{supp}(\Phi_{\varepsilon}) \subset\Omega$ for $\varepsilon <\delta$. Thus $\Phi_{\varepsilon}\in C_{c}^{\infty}(\Omega) $ for $\varepsilon<\delta$. Also, $\lim_{\varepsilon\to0^{+}} \Phi_{\varepsilon}=\chi_{E}$ with convergence in $L^2(\Omega) $ (see \cite[Theorem 4.22]{Brezis}). Then $\lim_{\varepsilon \to0^{+}}a\Phi_{\varepsilon}^{1-\alpha}=a\chi_{E}$ with convergence in $L^1(\Omega) $ (see \cite[Theorem 1.2.1]{Ambrosetti-Arcoya}), therefore \[ \lim_{\varepsilon\to0^{+}}\int_{\Omega}a\Phi_{\varepsilon}^{1-\alpha }=\int_{\Omega}a(\chi_{E}) ^{1-\alpha}=\int_{\Omega}a\chi _{E}>0. \] Then $\int_{\Omega}a| \Phi_{\varepsilon}| ^{1-\alpha}>0$ for $\varepsilon$ small enough. \end{proof} \begin{corollary} \label{coro2.5} $\mathbf{u}\not \equiv 0$. \end{corollary} \begin{remark} \label{rmk2.6} \rm Let us observe that $\nabla(v^2) =2v\nabla(v) $ for any (possibly unbounded) $v\in H^1(\Omega) $. Indeed, for $k\in\mathbb{N}$, let $v_k$ be the truncation of $v$, defined by $v_k(x) =v(x) $ if $| v(x) | \leq k$, and by $v_k(x) =k\ sign(v(x) ) $ otherwise. Then $\{ v_k\} _{k\in\mathbb{N}}$ converges to $v$ in $H^1(\Omega) $ as $k$ tends to $\infty$, and, since each $v_k$ is bounded, it follows from the chain rule (as stated e.g. in \cite[Lemma 7.5]{Gilbarg-Trudinger}) that $\frac{\partial}{\partial x_i}(v_k^2)=2v_k\frac{\partial v_k}{\partial x_i}$, $i=1,2,\dots ,n$. Since $\{v_k\} _{k\in\mathbb{N}}$ converges to $v$ in $L^2(\Omega) $, we have that $\{ v_k^2\} _{k\in\mathbb{N}}$ converges to $v^2$ in $L^1(\Omega) $, and so also in $D'(\Omega) $. Then $\{ \frac{\partial}{\partial x_i}(v_k^2) \} _{k\in\mathbb{N}}$ converges to $\frac{\partial}{\partial x_i}(v^2) $ in $D'(\Omega) $. Since $\{ 2v_k\frac{\partial v_k}{\partial x_i}\} _{k\in\mathbb{N}}$ converges to $2v\frac{\partial v}{\partial x_i}$ in $L^1(\Omega) $, and therefore in $D'(\Omega) $, we obtain that, for each $i$, $\frac{\partial}{\partial x_i}(v^2) =2v\frac{\partial v}{\partial x_i}$. \end{remark} \begin{lemma} \label{lem2.7} $\mathbf{u}\in L^{\infty}(\Omega) $. \end{lemma} \begin{proof} Let $\Omega'$ be a bounded $C^{0,1}$ domain such that $\overline{\Omega}\subset\Omega'$, and let $\widetilde{\mathbf{u}}$, $\widetilde{a}:\mathbb{R}^{n}\to\mathbb{R}$ be the extensions by zero of $\mathbf{u}$ and $a$ respectively. We consider first the case $n>2$. Let $r=\frac{1-\alpha}{2}$, $\eta=\frac{2^{\ast}}{1-\alpha}$. Then $01$, and $a\mathbf{u}^{2r}\in L^{\eta}(\Omega) $. Let $z\in W^{2,\eta}(\Omega')\cap W_0^{1,\eta}(\Omega') $ be the solution of \begin{equation} \begin{gathered} -\Delta z=2\widetilde{a}\widetilde{\mathbf{u}}^{2r}\quad \text{in } \Omega',\\ z=0\quad \text{on }\partial\Omega'. \end{gathered} \label{bound0} \end{equation} Let $\widetilde{z}:\mathbb{R}^{n}\to\mathbb{R}$ be the extension by zero of $z$ and let $\varphi$ be a nonnegative function in $C_{c}^{\infty}(\Omega') $ By Remark \ref{rmk2.6} and Lemma \ref{lem2.3} we have \begin{equation} \label{bound1} \begin{aligned} \int_{\Omega'}\langle \nabla(\widetilde{\mathbf{u}}^2) ,\nabla\varphi\rangle & =\int_{\Omega'}\langle 2\widetilde{\mathbf{u}}\nabla\widetilde{\mathbf{u} },\nabla\varphi\rangle \\ & =\int_{\Omega}2\mathbf{u}\langle \nabla\mathbf{u} ,\nabla\varphi\rangle \leq\int_{\Omega}2\langle \nabla( \mathbf{u}\varphi) ,\nabla\mathbf{u}\rangle \\ & =2\int_{\Omega}(a\mathbf{u}^{1-\alpha}-b\mathbf{u} ^{p+1}) \varphi\\ & \leq2\int_{\Omega'}\widetilde{a}\widetilde{\mathbf{u}} ^{2r}\varphi=\int_{\Omega'}\langle \nabla z,\nabla\varphi \rangle \end{aligned} \end{equation} For $\varepsilon>0$ let $h_{\varepsilon}$ be the mollifiers defined as in the proof of Lemma \ref{lem2.3}. For $\varepsilon$ small enough we have $0\leq\varphi\ast h_{\varepsilon}\in C_{c}^{\infty}(\Omega') $, and so, by \eqref{bound1}, \begin{align*} \int_{\Omega'}\langle \nabla(h_{\varepsilon}\ast \widetilde{\mathbf{u}}^2) ,\nabla\varphi\rangle & =\int_{\Omega'}\langle \nabla(\widetilde{\mathbf{u} }^2) ,h_{\varepsilon}\ast\nabla\varphi\rangle \\ & =\int_{\Omega'}\langle \nabla(\widetilde{\mathbf{u} }^2) ,\nabla(h_{\varepsilon}\ast\varphi) \rangle \\ &\leq\int_{\Omega'}\langle \nabla z,\nabla(h_{\varepsilon }\ast\varphi) \rangle \end{align*} where we have used that, since $h_{\varepsilon}$ is an even function, the convolution operator with kernel $h_{\varepsilon}$ is self-adjoint in $L^2(\mathbb{R}^{n}) $. Recall that $\widetilde{z}\in W^{1,\eta}(\mathbb{R}^{n}) $ and $\operatorname{supp}(\widetilde {z}) \subset\overline{\Omega'}$. Also, $\nabla\widetilde {z}=\nabla z$ a.e. in $\Omega'$, and $\nabla\widetilde{z}=0$ a.e. in $\mathbb{R}^{n}-\Omega'$. Thus \begin{align*} \int_{\Omega'}\langle \nabla z,\nabla(h_{\varepsilon} \ast\varphi) \rangle & =\int_{\mathbb{R}^{n}}\langle \nabla\widetilde{z},\nabla(h_{\varepsilon}\ast\varphi) \rangle \\ & =\int_{\mathbb{R}^{n}}\langle \nabla(h_{\varepsilon} \ast\widetilde{z}) ,\nabla\varphi\rangle \\ &=\int_{\Omega'}\langle \nabla(h_{\varepsilon}\ast\widetilde{z}) ,\nabla\varphi\rangle . \end{align*} Then \[ \int_{\Omega'}\langle \nabla(h_{\varepsilon}\ast \widetilde{\mathbf{u}}^2) ,\nabla\varphi\rangle \leq \int_{\Omega'}\langle \nabla(h_{\varepsilon}\ast \widetilde{z}) ,\nabla\varphi\rangle \] and so the divergence theorem gives \[ -\int_{\Omega'}\varphi\Delta(h_{\varepsilon}\ast\widetilde {\mathbf{u}}^2) \leq-\int_{\Omega'}\varphi\Delta( h_{\varepsilon}\ast\widetilde{z}) . \] Since this inequality holds for all nonnegative $\varphi\in C_{c}^{\infty}(\Omega') $ we obtain \[ -\Delta(h_{\varepsilon}\ast\widetilde{\mathbf{u}}^2) \leq-\Delta(h_{\varepsilon}\ast\widetilde{z}) \text{ in } \Omega'. \] We have also $h_{\varepsilon}\ast\widetilde{\mathbf{u}}^2=0\leq h_{\varepsilon}\ast\widetilde{z}$ on $\partial\Omega'$. Thus, the classical maximum principle gives $h_{\varepsilon}\ast\widetilde {\mathbf{u}}^2\leq h_{\varepsilon}\ast\widetilde{z}$ in $\Omega'$. Now, $\widetilde{\mathbf{u}}^2$ and $\widetilde{z}$ belong to $L^{\frac{2^{\ast}}{2}}(\mathbb{R}^{n}) $, and so $\lim _{\varepsilon\to0^{+}}(h_{\varepsilon}\ast\widetilde {\mathbf{u}}^2) =\widetilde{\mathbf{u}}^2$, and $\lim_{\varepsilon\to0^{+}}(h_{\varepsilon}\ast\widetilde {z}) =\widetilde{z}$, in both cases with convergence in $L^{\frac {2^{\ast}}{2}}(\mathbb{R}^{n}) $. Then, $\lim_{\varepsilon \to0^{+}}(h_{\varepsilon}\ast\widetilde{\mathbf{u}} ^2) |_{\Omega}=\widetilde{\mathbf{u}}_{|_{\Omega}}^2$; and $\lim_{\varepsilon\to0^{+}}(h_{\varepsilon}\ast \widetilde{z}) |_{\Omega}=\widetilde{z}|_{\Omega}$, in each case with convergence in $L^{\frac{2^{\ast}}{2}}(\Omega) $. Then $\mathbf{u}^2\leq z$ in $\Omega$. Now the lemma follows from the following standard bootstrap argument: Let $\{ \eta_j\} _{j\in\mathbb{N}}$ be recursively defined by $\eta_1=\eta^{\ast}$ and by $\eta_{j+1}=\eta_j^{\ast}$. We can see inductively that $\mathbf{u}\in L^{2\eta_j}(\Omega) $ for all $j$. Indeed, $z\in W^{2,\eta}(\Omega') $, and so $z\in L^{\eta^{\ast}}(\Omega') $. Then $\mathbf{u} ^2\in L^{\eta^{\ast}}(\Omega) $, and thus $\mathbf{u}\in L^{2\eta^{\ast}}(\Omega) =L^{2\eta_1}(\Omega) $. Suppose now that $\mathbf{u}\in L^{2\eta_j}(\Omega) $, then $2\widetilde{a}\widetilde{\mathbf{u}}^{2r}\in L^{\frac{\eta_j}{p} }(\Omega') \subset L^{\eta_j}(\Omega^{\prime }) $, and so $z\in W^{2,\eta_j}(\Omega') \subset L^{\eta_j^{\ast}}(\Omega') =L^{\eta_j^{\ast }}(\Omega') $, which gives $\mathbf{u}\in L^{2\eta_j^{\ast}}(\Omega) =L^{2\eta_{j+1}}( \Omega) $. Thus $\mathbf{u}\in L^{2\eta_j}(\Omega) $ for all $j$, and so, taking $j$ large enough, we obtain $\mathbf{u}\in L^{s}(\Omega) $ for some $s>2n$, then $2\widetilde{a} \widetilde{\mathbf{u}}^{2r} \in L^{\frac{s}{2r}}(\Omega') \subset L^{\frac{s}{2}}(\Omega') $. Thus $z\in W^{2,\frac{s}{2}}(\Omega') \subset L^{\infty}( \Omega') $. Since $\mathbf{u}^2\leq z$ in $\Omega$, we obtain $\mathbf{u}\in L^{\infty}(\Omega) $. Finally, if $n\leq2$, we have $\mathbf{u}\in L^{s}( \Omega) $ for all $s\in[ 1,\infty) $. We take $\eta>n$ and, for $r$, $z$, $\overline{z}$ and $\widetilde{\mathbf{u}}$ defined as above, we have $a\mathbf{u}^{2r}\in L^{\eta}(\Omega) $. Thus $\widetilde{z}\in W^{2,\eta}(\Omega') \subset C( \overline{\Omega'}) $ and, as before, $\mathbf{u}^2\leq z$ in $\Omega$. Then $\mathbf{u}\in L^{\infty}(\Omega) $ also in this case. \end{proof} \begin{lemma} \label{lem2.8} \begin{equation} \int_{\Omega}\langle \nabla\mathbf{u},\nabla(\mathbf{u} \varphi) \rangle =\int_{\Omega}(a\mathbf{u}^{1-\alpha }-b\mathbf{u}^{1+p}) \varphi \label{lema 2.9} \end{equation} \emph{for all }$\varphi\in H^1(\Omega) \cap L^{\infty }(\Omega) $. \end{lemma} \begin{proof} Let $\varphi\in H^1(\Omega) \cap L^{\infty}(\Omega) $. By Lemma \ref{lem2.7} we have $\mathbf{u}\in L^{\infty}(\Omega) $ and so $\mathbf{u}\varphi\in H_0 ^1(\Omega) $. Thus Lemma \ref{lem2.3} gives \eqref{lema 2.9}. \end{proof} \section{Main results} \begin{theorem} \label{thm3.1} There exists a nonnegative weak solution $0\not \equiv u\in H_0^1(\Omega) \cap L^{\infty}(\Omega) $ of problem \eqref{Problem}. \end{theorem} \begin{proof} Let $\mathbf{u}$ be the nonnegative minimizer of $J$ considered in the previous section. Let $\psi$ be a nonnegative function in $H_0^1(\Omega) \cap L^{\infty}(\Omega) $, and let $\varepsilon>0$. Note that $\frac{\psi}{\mathbf{u}+\varepsilon}\in H^1(\Omega) \cap L^{\infty}(\Omega) $, and that $\nabla(\mathbf{u}\frac{\psi}{\mathbf{u}+\varepsilon}) =\varepsilon\frac{\nabla\mathbf{u}}{(\mathbf{u}+\varepsilon ) ^2}\psi+\frac{\mathbf{u}}{\mathbf{u}+\varepsilon}\nabla \psi$, and so Lemma \ref{lem2.8} gives \begin{equation} \label{Lemma2.9 1} \varepsilon\int_{\Omega}\psi\frac{| \nabla\mathbf{u}| ^2}{(\mathbf{u}+\varepsilon) ^2}+\int_{\Omega} \frac{\mathbf{u}}{\mathbf{u}+\varepsilon}\langle \nabla \mathbf{u},\nabla\psi\rangle =\int_{\Omega}(a\mathbf{u} ^{1-\alpha}-b\mathbf{u}^{1+p}) \frac{1}{\mathbf{u}+\varepsilon}\psi. \end{equation} Since $\nabla\mathbf{u}=0$ a.e. on the set $\{ x\in\Omega:\mathbf{u}(x) =0\}$, and since $a\mathbf{u}^{1-\alpha}=b\mathbf{u}^{1+p}=0$ on the same set, \eqref{Lemma2.9 1} can be written as \begin{equation} \label{Lemma2.9.3} \begin{aligned} & \varepsilon\int_{\{ \mathbf{u}>0\} }\psi\frac{| \nabla\mathbf{u}| ^2}{(\mathbf{u}+\varepsilon ) ^2}+\int_{\{ \mathbf{u}>0\} }\frac{\mathbf{u} }{\mathbf{u}+\varepsilon}\langle \nabla\mathbf{u},\nabla \psi\rangle +\int_{\{ \mathbf{u}>0\} }b\mathbf{u}^p \frac{\mathbf{u}}{\mathbf{u}+\varepsilon}\psi\\ & =\int_{\{ \mathbf{u}>0\} }a\mathbf{u}^{-\alpha} \frac{\mathbf{u}}{\mathbf{u}+\varepsilon}\psi. \end{aligned} \end{equation} Also \[ \lim_{\varepsilon\to0^{+}}(\frac{\mathbf{u} }{\mathbf{u}+\varepsilon}\langle \nabla\mathbf{u},\nabla \psi\rangle ) =\chi_{\{ \mathbf{u}>0\} }\langle \nabla\mathbf{u},\nabla\psi\rangle =\langle \nabla\mathbf{u},\nabla\psi\rangle \] a.e. in $\Omega$, and $| \frac{\mathbf{u}}{\mathbf{u}+\varepsilon}\langle \nabla\mathbf{u},\nabla\psi\rangle | \leq| \langle \nabla\mathbf{u},\nabla\psi\rangle | \in L^1(\Omega) $, and so Lebesgue's dominated convergence theorem gives \begin{equation} \lim_{\varepsilon\to0^{+}}\int_{\{ \mathbf{u}>0\} }\frac{\mathbf{u}}{\mathbf{u}+\varepsilon}\langle \nabla \mathbf{u},\nabla\psi\rangle =\int_{\Omega}\langle \nabla\mathbf{u},\nabla\psi\rangle . \label{Lemma2.9.4} \end{equation} On the other hand, $\lim_{\varepsilon\to0^{+}}a\mathbf{u} ^{-\alpha}\frac{\mathbf{u}}{\mathbf{u}+\varepsilon}\psi =a\mathbf{u}^{-\alpha}\psi$ on the set $\{ x\in\Omega:\mathbf{u} (x) >0\} $ and, since $a\mathbf{u}^{-\alpha} \frac{\mathbf{u}}{\mathbf{u}+\varepsilon}\psi$ is non-increasing in $\varepsilon$, the monotone convergence theorem gives \begin{equation} \lim_{\varepsilon\to0^{+}}\int_{\{ \mathbf{u}>0\} }a\mathbf{u}^{-\alpha}\frac{\mathbf{u}}{\mathbf{u}+\varepsilon }\psi=\int_{\{ \mathbf{u}>0\} }a\mathbf{u}^{-\alpha} \psi=\int_{\Omega}a\mathbf{u}^{-\alpha}\chi_{\{ \mathbf{u} >0\} }\psi \label{Lemma2.9.5} \end{equation} Also \begin{equation} \lim_{\varepsilon\to0^{+}}\int_{\{ \mathbf{u}>0\} }b\mathbf{u}^p\frac{\mathbf{u}}{\mathbf{u}+\varepsilon}\psi =\int_{\Omega}b\mathbf{u}^p\psi \label{Lemma2.9.6} \end{equation} Then, from \eqref{Lemma2.9.3}, \eqref{Lemma2.9.4}, \eqref{Lemma2.9.5} and \eqref{Lemma2.9.6}, we obtain \begin{align*} & \int_{\Omega}\langle \nabla\mathbf{u},\nabla\psi\rangle +\int_{\Omega}b\mathbf{u}^p\psi\\ & =\lim_{\varepsilon\to0^{+}}\Big(\int_{\{ \mathbf{u} >0\} }\frac{\mathbf{u}}{\mathbf{u}+\varepsilon}\langle \nabla\mathbf{u},\nabla\psi\rangle +\int_{\{ \mathbf{u} >0\} }b\mathbf{u}^p\frac{\mathbf{u}}{\mathbf{u} +\varepsilon}\psi\Big) \\ & \leq\limsup _{\varepsilon\to0^{+}}\Big(\int_{\{ \mathbf{u}>0\} }\frac{\varepsilon\psi| \nabla \mathbf{u}| ^2}{(\mathbf{u}+\varepsilon) ^2}+\int_{\{ \mathbf{u}>0\} }\frac{\mathbf{u} }{\mathbf{u}+\varepsilon}\langle \nabla\mathbf{u},\nabla \psi\rangle +\int_{\{ \mathbf{u}>0\} }b\mathbf{u} ^p\frac{\mathbf{u}}{\mathbf{u}+\varepsilon}\psi\Big) \\ & =\limsup _{\varepsilon\to0^{+}}\int_{\{ \mathbf{u} >0\} }a\mathbf{u}^{-\alpha}\frac{\mathbf{u}}{\mathbf{u} +\varepsilon}\psi\\ & =\int_{\Omega}a\mathbf{u}^{-\alpha}\chi_{\{ \mathbf{u} >0\} }\psi. \end{align*} Thus \begin{equation} \int_{\Omega}\langle \nabla\mathbf{u},\nabla\psi\rangle +\int_{\Omega}b\mathbf{u}^p\psi\leq\int_{\Omega}a\mathbf{u} ^{-\alpha}\chi_{\{ \mathbf{u}>0\} }\psi. \label{theorem1} \end{equation} Let us see that the reverse inequality in \eqref{theorem1} holds: A computation gives, for $t>0$, \begin{align*} 0 & \leq\frac{1}{t}(J(\mathbf{u}+t\psi) -J( \mathbf{u}) ) \\ & =\int_{\Omega}\langle \nabla\mathbf{u},\nabla\psi\rangle +\frac{t}{2}\int_{\Omega}| \nabla\psi| ^2-\frac {1}{(1-\alpha) t}\int_{\Omega}a((\mathbf{u} +t\psi) ^{1-\alpha}-\mathbf{u}^{1-\alpha}) \\ & \quad +\frac{1}{(1+p) t}\int_{\Omega}b(( \mathbf{u}+t\psi) ^{1+p}-\mathbf{u}^{1+p}) , \end{align*} and so \begin{equation} \label{theorem 3} \begin{aligned} & \frac{1}{(1-\alpha) t}\int_{\Omega}a(( \mathbf{u}+t\psi) ^{1-\alpha}-\mathbf{u}^{1-\alpha})\\ & \leq\int_{\Omega}\langle \nabla\mathbf{u},\nabla\psi\rangle +\frac{1}{(1+p) t}\int_{\Omega}b((\mathbf{u} +t\psi) ^{1+p}-\mathbf{u}^{1+p}) +\frac{t}{2}\int_{\Omega}| \nabla\psi| ^2. \end{aligned} \end{equation} The mean value theorem gives $(\mathbf{u}+t\psi) ^{1-\alpha}-\mathbf{u}^{1-\alpha}=(1-\alpha) (\mathbf{u} +\sigma_{t}) ^{-\alpha}t\psi$ for some measurable function $\sigma_{t}$ such that $0<\sigma_{t}0\} \cap\{ \psi>0\} }a((\mathbf{u}+t\psi) ^{1-\alpha}-\mathbf{u}^{1-\alpha}) \\ & =\int_{\{ a>0\} \cap\{ \psi>0\} }a( \mathbf{u}+\sigma_{t}) ^{-\alpha}\psi \end{align*} Now, $\lim_{t\to0^{+}}a(\mathbf{u}+\sigma_{t}) ^{-\alpha}\psi=a\mathbf{u}^{-\alpha}\psi$ $a.e$ on the set $\{a>0\} \cap\{ \psi>0\} $. Then, by Fatou's Lemma, \begin{equation} \label{theorem 4} \begin{aligned} & \liminf _{t\to0^{+}}\frac{1}{(1-\alpha) t}\int_{\Omega}a((\mathbf{u}+t\psi) ^{1-\alpha }-\mathbf{u}^{1-\alpha}) \\ & =\liminf _{t\to0^{+}}\int_{\{ a>0\} \cap\{ \psi>0\} }a(\mathbf{u}+\sigma_{t}) ^{-\alpha}\psi\\ & \geq\int_{\{ a>0\} \cap\{ \psi>0\} } a\mathbf{u}^{-\alpha}\psi\geq\int_{\Omega}a\mathbf{u}^{-\alpha} \chi_{\{ \mathbf{u}>0\} }\psi. \end{aligned} \end{equation} Again by the mean value theorem, we have \[ \frac{1}{(1+p) t}\int_{\Omega}b((\mathbf{u} +t\psi) ^{1+p}-\mathbf{u}^{1+p}) =\int_{\Omega}b( \mathbf{u}+\sigma_{t}) ^p\psi. \] Note that, for $00\} }\psi \label{theorem6} \end{equation} Since $b\mathbf{u}^p\psi\in L^1(\Omega) $, \eqref{theorem6} implies that $a\mathbf{u}^{-\alpha}\chi_{\{ \mathbf{u}>0\} }\psi\in L^1(\Omega) $. We apply \eqref{theorem6}, combined with \eqref{theorem1}, to complete the proof. \end{proof} \begin{remark} \label{rmk3.2} \rm It is well known (see e.g., \cite{Djairo}) that, for $m\in L^{\infty}(\Omega) $ such that $| \{ x\in\Omega:m(x) >0\} | >0$, there exists a unique $\lambda=\lambda_1(-\Delta,\Omega,m) $ such that the problem \begin{gather*} -\Delta\varphi_1=\lambda m\varphi_1\quad \text{in }\Omega,\\ \varphi_1=0\quad \text{on }\partial\Omega,\\ \varphi_1>0\quad \text{in }\Omega \end{gather*} has a solution $\varphi_1\in H_0^1(\Omega) $. This solution is unique up to a multiplicative constant, belongs to $C^{1.\gamma }(\overline{\Omega}) $ for some $0<\gamma<1$, satisfies that $| \nabla\varphi| (x) >0$ for all $x\in\partial\Omega$, and there are positive constants $c_1$, $c_2$ such that $c_1d_{\Omega}\leq\varphi\leq c_2d_{\Omega}$ in $\Omega$, where $d_{\Omega}:\Omega\to\mathbb{R}$ is the function defined by \[ d_{\Omega}(x) =\operatorname{dist}(x,\partial\Omega) . \] $\lambda_1$ and $\varphi_1$ are called, respectively, the principal eigenvalue and a positive principal eigenfunction for $-\Delta$ in $\Omega$, with Dirichlet boundary condition and weight $m$. \end{remark} \begin{remark} \label{rmk3.3} \rm It is well known that, under our assumptions on $\Omega,\alpha$, and $a$, the problem \begin{gather*} -\Delta\theta=a\theta^{-\alpha}\text{ in }\Omega,\\ \theta=0\text{ on }\partial\Omega,\\ \theta>0\text{ in }\Omega \end{gather*} has a unique weak solution $\theta\in H_0^1(\Omega) $. Moreover, $\theta$ is in $C(\overline{\Omega}) $, and $\theta\geq c'd_{\Omega}$ for some positive constant $c'$ (see \cite{del-Pino, aranda}). A computation shows that (in weak sense) $-\Delta(\theta^{\alpha+1}) =-(\alpha+1) \theta^{\alpha}\Delta\theta-(\alpha+1) \alpha\theta^{\alpha -2}| \nabla\theta| ^2\leq(\alpha+1) \| a\| _{\infty}$ in $\Omega$, and so we have $\theta\leq c''d_{\Omega}^{\frac{1}{\alpha+1}}$ in $\Omega$, for some constant $c''>0$. \end{remark} \begin{remark} \label{rmk3.4} \rm Following \cite{LocSchmitt}, we say that $w\in W_{\rm loc}^{1,2}(\Omega) $ is a subsolution (supersolution) to the problem \begin{equation} -\Delta z=az^{-\alpha}-bz^p\text{ in }\Omega \label{Remark3.4.1} \end{equation} in the sense of distributions, if, and only if: $w>0$ a.e. in $\Omega$, $aw^{-\alpha}-bw^p\in L_{\rm loc}^1(\Omega) $, and for all nonnegative $\varphi\in C_{c}^{\infty}(\Omega) $, it holds that \[ \int_{\Omega}\langle \nabla w,\nabla\varphi\rangle \leq(\geq) \int_{\Omega}(aw^{-\alpha}-bw^p) \varphi. \] We also say that $z\in W_{\rm loc}^{1,2}(\Omega) $ is a solution, in the sense of distributions, of \eqref{Remark3.4.1} if, and only if, $z>0$ a.e. in $\Omega$, and, for all $\varphi\in C_{c}^{\infty}(\Omega) $ it holds that \[ \int_{\Omega}\langle \nabla z,\nabla\varphi\rangle =\int_{\Omega}(az^{-\alpha}-bz^p) \varphi. \] For subsolutions, supersolutions and solutions defined in the above sense, \cite[Theorem 2.4]{LocSchmitt} says that, if \eqref{Remark3.4.1} has a subsolution $\underline{z}$ and a supersolution $\overline{z}$ (in the sense of distributions), both in $L_{\rm loc}^{\infty}(\Omega) $, and such such that $0<\underline{z}(x) \leq\overline{z}(x)$ a.e. $x\in\Omega $, and if there exists $k\in L_{\rm loc}^{\infty}(\Omega) $ such that $|as^{-\alpha}-bs^p|\leq k(x) $ a.e. $x\in\Omega$ for all $s\in[ \underline{z}(x) ,\overline{z}(x)] $; then \eqref{Remark3.4.1} has a solution $z$ in the sense of distributions, and $z$ satisfies $\underline{z}\leq z\leq\overline{z}$ a.e. in $\Omega$. \end{remark} \begin{theorem} \label{thm3.5} Suppose that $a\geq\varepsilon b$ for some $\varepsilon>0$. Then there exists a weak solution $v\in H_0^1(\Omega) \cap L^{\infty}(\Omega) $ of \eqref{Problem} such that $v\geq cd_{\Omega}$ in $\Omega$ for some $c>0$, and $v\in C_{\rm loc}^1(\Omega) \cap C(\overline{\Omega})$. \end{theorem} \begin{proof} Suppose that $a\geq\varepsilon b$ for some $\varepsilon>0$. Let $\varphi_1\in H_0^1(\Omega) $ be the positive principal eigenfunction associated to the weight function $a$, normalized by $\| \varphi_1\| _{\infty}=1$ (see Remark \ref{rmk3.2}). Note that (in weak sense), for $t$ positive and small enough, \begin{equation} -\Delta(t\varphi_1) \leq a(t\varphi_1) ^{-\alpha}-b(t\varphi_1) ^p\text{ in }\Omega. \label{subsolution} \end{equation} Indeed, $-\Delta(t\varphi_1) =\lambda_1at\varphi_1$, and so \eqref{subsolution} is equivalent to $(1-\lambda_1(t\varphi_1) ^{1+\alpha}) a\geq(t\varphi_1) ^{p+\alpha}b$ in $\Omega$. But, for $t$ small enough, we have $b( t\varphi_1) ^{p+\alpha}\leq bt^{p+\alpha}\leq\frac{1}{2}\varepsilon b\leq\frac{1}{2}a\leq(1-\lambda_1(t\varphi_1)^{1+\alpha}) a$ in $\Omega$. Since $t\varphi_1>0$ in $\Omega$, it follows that, for such a $t$, $t\varphi_1$ is a subsolution of \eqref{Problem}, in the sense of Remark \ref{rmk3.4}. On the other hand, let $\theta\in H_0^1(\Omega) \cap C(\overline{\Omega}) $ be the solution of the problem $-\Delta\theta=a\theta^{-\alpha} $ in $\Omega$, $\theta=0$ on $\partial\Omega$. Since $\theta\geq c'd_{\Omega}$ in $\Omega$ for some $c'>0$, we have that $\theta$ is strictly positive in $\Omega$, and, by diminishing $t$ if necessary, we can assume, that $t\varphi_1\leq\theta$. Clearly (in weak sense) $-\Delta\theta\geq a\theta^{-\alpha}-b\theta^p$ in $\Omega$, and so $\theta$ is a supersolution of \eqref{Problem}, again in the sense of Remark \ref{rmk3.4}). Since $t\varphi_1\geq c_1 td_{\Omega}$ in $\Omega$ for some $c_1>0$, and since $\theta\leq c''d_{\Omega}^{\frac{1}{\alpha+1}}$ in $\Omega$ for some $c''>0$, we have $[ t\varphi_1(x) ,\theta(x) ] \subset[ c_1td_{\Omega}( x) ,c''d_{\Omega}^{\frac{1}{\alpha+1}}(x)] $ for $x\in\Omega$. Therefore a.e. $x\in \Omega$, for all $s\in[ t\varphi_1(x) ,\theta(x)] $, the following holds \[ |as^{-\alpha}-bs^p|\leq\| a\| _{\infty}( c_1t) ^{-\alpha}d_{\Omega}(x) ^{-\alpha}+\| b\| _{\infty}(c'') ^pd_{\Omega} ^{\frac{p}{\alpha+1}}(x) :=k(x) . \] Since $k\in L_{\rm loc}^{\infty}(\Omega) $, \cite[Theorem 2.4]{LocSchmitt} (see Remark \ref{rmk3.4}), says that there exists $v\in W_{\rm loc}^{1,2}( \Omega) $ such that $t\varphi_1\leq v\leq\theta$ in $\Omega$, and such that, for any $\varphi\in C_{c}^{\infty}(\Omega) $, \begin{equation} \int_{\Omega}\langle \nabla v,\nabla\varphi\rangle =\int_{\Omega }(av^{-\alpha}-bv^p) \varphi. \label{Loc Schmitt} \end{equation} Note that $v\in H_0^1(\Omega) $: Indeed, let $\Omega'$ be a subdomain of $\Omega$ such that $\overline{\Omega'}\subset\Omega$. Since $v\geq c''d_{\Omega}$ in $\Omega$ for some $c''>0$, we have $av^{-\alpha}-bv^p\in L^{\infty}( \Omega') $. Therefore, from \eqref{Loc Schmitt}, a density argument, and Lebesgue's dominated convergence theorem give that, for any $\varphi\in H_0^1(\Omega') $, it holds \begin{equation} \int_{\Omega'}\langle \nabla v,\nabla\varphi\rangle =\int_{\Omega'}(av^{-\alpha}-bv^p) \varphi. \label{Loc Schmitt 2} \end{equation} Let $\varepsilon>0$. Since $v\leq\theta\leq c''d_{\Omega} ^{\frac{1}{1+\alpha}}$ for some $c''>0$, we have that $\operatorname{supp}(v-\varepsilon) ^{+}\subset\Omega'$ for some subdomain $\Omega'$ such that $\overline{\Omega'}\subset \Omega$. Also $(v-\varepsilon) ^{+}\in H^1( \Omega) $ and so $(v-\varepsilon) ^{+}\in H_0 ^1(\Omega) $. Thus, from \eqref{Loc Schmitt 2}, we obtain \begin{equation} \label{Loc Schmitt 3} \begin{aligned} \int_{\Omega}\chi_{\{ v>\varepsilon\} }\nabla v.\nabla v &=\int_{\Omega'}\nabla v.\nabla(v-\varepsilon)^{+}\\ & =\int_{\Omega'}(av^{-\alpha}-bv^p) (v-\varepsilon) ^{+}\\ & =\int_{\Omega}(av^{-\alpha}-bv^p) (v-\varepsilon) \chi_{\{ v>\varepsilon\} }. \end{aligned} \end{equation} The monotone convergence theorem gives \[ \lim_{\varepsilon\to0^{+}}\int_{\Omega}\chi_{\{ v>\varepsilon \} }\nabla v.\nabla v=\int_{\Omega}\nabla v.\nabla v \] and, since $av^{-\alpha}-bv^p\in L^1(\Omega) $, and $v\in L^{\infty}(\Omega) $, Lebesgue's dominated convergence theorem gives \[ \lim_{\varepsilon\to0^{+}}\int_{\Omega}(av^{-\alpha} -bv^p) (v-\varepsilon) \chi_{\{ v>\varepsilon\} } =\int_{\Omega}(av^{1-\alpha}-bv^{1+p}) . \] Taking limits in \eqref{Loc Schmitt 3}, we obtain \[ \int_{\Omega}\nabla v.\nabla v=\int_{\Omega}(av^{1-\alpha}-bv^{1+p}) <\infty. \] Thus $v\in H^1(\Omega) $ and, since $t\varphi_1\leq v\leq\theta$, we have $v\in H_0^1(\Omega) $. Note also that $av^{-\alpha}-bv^p\in L^1(\Omega) $ and so, again by a density argument, and applying Lebesgue's dominated convergence theorem, we conclude that \eqref{Loc Schmitt} holds for all $\varphi$ in $H_0^1( \Omega) \cap L^{\infty}(\Omega) $. Let $\Omega'$ be an arbitrary subdomain of $\Omega$ such that $\overline{\Omega'}\subset\Omega$, and let $\Omega''$ be such that $\overline{\Omega'}\subset\Omega'' \subset\overline{\Omega''}\subset\Omega$. Since $v\in L^{\infty}(\Omega'') $ and $(av^{-\alpha} -bv^p) |_{\Omega''}\in L^{\infty}(\Omega'') $, we have $v|_{\Omega'}\in W^{2,s}(\Omega') $ for all $s\in[ 1,\infty)$ (see e.g., Proposition 4.1.2 in \cite{cazenave}) and so $v|_{\Omega'}\in C^1(\overline{\Omega'}) $. Thus $v\in C_{\rm loc}^1(\Omega) $ and, since $t\varphi_1\leq v\leq\theta$, $v$ is continuous on $\partial\Omega$. \end{proof} \begin{example} \label{examp3.6}\rm Let $\Omega=(0,2\pi) $, $\alpha=1/3$, and $p\in(0,1/5)$. Let $a$ and $b $ be the functions defined on $\Omega$ by $a=2(1-\cos(2x)) \sqrt[3]{\sin^2(x) }$, $b(x)=2| \sin^2(x) | ^{-p}$. Then $a\geq0$, $b\geq0$, $0\not \equiv a\in L^{\infty}(\Omega) $ and $b\in L^{\frac{2}{1-p}}(\Omega) $. Consider now the following three functions in $C^1(\overline{\Omega}) $: $u(x) =\sin^2(x) \chi_{(0,\pi) }$, $v(x)=\sin^2(x) \chi_{(0,2\pi) }$, and $w(x) =\sin^2(x) \chi_{(\pi,2\pi) }$. A computation shows that $u$, $v$, and $w$ are all weak solutions of \eqref{Problem} ($v$ is in fact a classical solution). Therefore (without additional assumptions on $a$ and $b$) uniqueness is not to be expected for nonnegative nontrivial weak solutions of \eqref{Problem}. Notice that $w\equiv0$ on $(0,\pi) $. Note also that $v(x)>0$ for $x\in\Omega-\{ \pi\} $ and $v(\pi) =0$, therefore, by Theorem \ref{thm3.8} below, there is no continuous and strictly positive solution to \eqref{Problem}. \end{example} \begin{example} \label{examp3.7} Let $\Omega=(0,2) $, let $\alpha\in(0,1) $, $p\in(0,1) $, let $b:=\chi_{(0,1) }$ and let $a:=\chi_{(1,1+\delta)}$, with \[ 0<\delta\leq(\frac{1-\alpha}{2}) ^{\frac{1}{1-\alpha}} \Big((\frac{2}{p+1}) ^{\frac{1}{1-p}}(\frac{1-p}{2}) ^{\frac{1+p}{1-p}}\Big) ^{\frac{1+\alpha}{1-\alpha}}. \] Let us show that the problem \begin{equation} \begin{gathered} -u''=au^{-\alpha}-bu^p\quad \text{in }\Omega,\\ u=0\quad \text{on }\partial\Omega \end{gathered} \label{the equationn} \end{equation} has no weak solution $u\in H_0^1(\Omega) $ such that $u>0$ a.e. in $\Omega$. Let us suppose, for the sake of contradiction, that $u$ is a weak solution such that $u>0$ a.e. in $\Omega$. Since $H_0^1(\Omega) \subset C^{\gamma}(\overline{\Omega}) $ for some $\gamma\in(0,1) $, we have $u\in C^{\gamma}(\overline{\Omega}) $ for such a $\gamma$. Throughout this example, unless there is risk of confusion, the restrictions of $u$ to $(0,1) $, $(1,1+\delta) $, and $(1+\delta,2)$, will be still denoted by $u$. Since $u$ belongs to $C^{\gamma}([ 0,1] ) $, and $| u^p(x) -u^p(y) | \leq| u(x) -u(y) | ^p$ for any $x,y\in[ 0,1] $, we have $u^p\in C^{.\gamma p}([ 0,1] ) $. Let $A=u(1) $. Since \begin{equation} \begin{gathered} -u''=-u^p\quad \text{in }(0,1),\\ u(0) =0,\\ u(1) =A \end{gathered} \label{equation on (0,1)} \end{equation} we have that $u$ is a classical solution of \eqref{equation on (0,1)} that belongs to $C^2([ 0,1] ) \cap C([0,1] ) $ and so $-u''=-u^p$ in $[0,1]$. (see, e.g., \cite[Theorem 6.14]{Gilbarg-Trudinger}). Note also that \begin{equation} u(x) \geq\big(\frac{1-p}{2}\big) ^{\frac{2}{1-p}}\big( \frac{2}{p+1}\big) ^{\frac{1}{1-p}}x^{\frac{2}{1-p}}\quad \text{for all } x\in[ 0,1] .\label{inequation in (0,1)} \end{equation} Indeed, multiplying \eqref{equation on (0,1)} by $u'$ we obtain $\frac{1}{2}((u') ^2) '=\frac {1}{p+1}(u^{p+1}) '$ on $[ 0,1] $, and so $\frac{1}{2}(u'(x) ) ^2-\frac{1}{p+1}u(x) ^{p+1}=\frac{1}{2}(u'(0) ) ^2\geq0$ for all $x\in[ 0,1] $. Thus \begin{equation} (u') ^2\geq\frac{2}{p+1}u^{p+1}\quad \text{in }[0,1] . \label{inequation altra1} \end{equation} As $u\geq0$ on $[ 0,1] $ and $u(0) =0$, we have $u'(0) \geq0$. Observe also that \eqref{equation on (0,1)} implies $u''\geq0$ on $[0,1] $, and so $u$ is a convex function on $[ 0,1] $. Thus $u'$ is nondecreasing on $[ 0,1] $ and, since $u'(0) \geq0$, we have $u'\geq0$ in $[ 0,1] $, and so, from \eqref{inequation altra1}, we conclude \begin{equation} u'\geq(\frac{2}{p+1}) ^{1/2}u^{\frac{p+1}{2}}\quad \text{in }[0,1] . \label{inequation altra2} \end{equation} If $u(\overline{x}) =0$ for some $\overline{x}\in( 0,1) $ we would have $u(x) =0$ for all $x\in(0,\overline{x}) $, which contradicts the assumption that $u>0$ a.e. in $\Omega$. Thus $u(x) >0$ for all $x\in[ 0,1] $, therefore \eqref{inequation altra2} can be rewritten as $u^{-\frac{p+1}{2} }u'\geq(\frac{2}{p+1}) ^{1/2}$ on $[0,1] $. By integrating this inequality over $(0,x)$ we obtain $\frac{2}{1-p}(u(x) ) ^{\frac{1-p}{2}} \geq(\frac{2}{p+1}) ^{1/2}x$ for all $x\in[ 0,1] $, and so \eqref{inequation in (0,1)} holds. In particular we have \begin{equation} u(1) \geq\big(\frac{1-p}{2}\big) ^{\frac{2}{1-p}}\big( \frac{2}{p+1}\big) ^{\frac{1}{1-p}}x^{\frac{2}{1-p}} \label{inequation altra3} \end{equation} and then, by \eqref{inequation altra2}, \begin{equation} u'(1) \geq\big(\frac{2}{p+1}\big) ^{\frac{1}{1-p} }\big(\frac{1-p}{2}\big) ^{\frac{1+p}{1-p}}. \label{inequation altra4} \end{equation} Consider now the restriction of $u$ to $(1,1+\delta) $; $u\in H^1(1,1+\delta) \subset C([ 1,1+\delta]) $, and solves \begin{gather*} -u''=u^{-\alpha}\text{ in }(1,1+\delta) \\ u(1) \geq0,\text{ }u(1+\delta) \geq0. \end{gather*} Let $\zeta\in H_0^1(1,1+\delta) \subset C([1,1+\delta] ) $ be the solution to the problem \begin{gather*} -\zeta''=\zeta^{-\alpha}\quad \text{in }(1,1+\delta) \\ \zeta>0\quad \text{in }(1,1+\delta) \\ \zeta(1) =0,\quad \zeta(1+\delta) =0. \end{gather*} Observe that $u\geq\zeta$ on $(1,1+\delta) $. To prove this, suppose, for the sake of contradiction, that $\{ x\in( 1,1+\delta) :u(x) <\zeta(x) \} \neq\emptyset$, and let $U$ be one of its connected components. Note that $U$ is an open interval, since $u$ and $\zeta$ are continuous on $(1,1+\delta) $. Since $-\zeta''=\zeta^{-\alpha}\leq u^{-\alpha}=-u''$ on $U$, and $\zeta=u$ on $\partial U$, the maximum principle gives $\zeta\leq u$ on $U$, which is a contradiction. Thus $u\geq\zeta$ on $(1,1+\delta) $ as claimed. Recall that there exists $c>0$ such that $\zeta\geq cd$ on $(1,1+\delta) $, where $d(x) =\operatorname{dist}(x,\partial(1,1+\delta) ) $ for all $x\in(1,1+\delta) $ (see Remark \ref{rmk3.3}); therefore $u\geq cd$ on $(1,1+\delta) $. Note also that $u(1+\delta) >0$. If not, since $u(2) =0$ and $u''=0$ in $(1,2) $, we would have $u=0$ in $(1,2) $; which would contradict $u>0$ a.e. in $\Omega$. Since $u(1) >0$, $u(1+\delta) >0$, and $u\geq cd$ on $(1,1+\delta) $, it follows that $u(x) >0$ for any $x\in[ 1,1+\delta] $ and, since $u$ is continuous on $[1,1+\delta] $, we have $u\geq const>0$ on $[ 1,1+\delta] $. Now \begin{align*} | u^{-\alpha}(x) -u^{-\alpha}(y)| & =(u(x) u(y) )^{-\alpha}| u(x) ^{\alpha}-u(y) ^{\alpha}| \\ & \leq(u(x) u(y) ) ^{-\alpha}| u(x) -u(y) | ^{\alpha} \end{align*} and so, since $u\in C^{\gamma}(\overline{\Omega}) $, we have $u^{-\alpha}\in C^{\alpha\gamma}([ 1,1+\delta] ) $. Let $A=u(1) $, $B=u(1+\delta) $. Since $u$ solves \begin{equation} \begin{gathered} -u''=u^{-\alpha}\quad \text{in }(1,1+\delta) \\ u(1) =A,\quad u(1+\delta) =B, \end{gathered} \label{equation on (1,1+delta)} \end{equation} it follows that $u$ is a classical solution of \eqref{equation on (1,1+delta)} that belongs to $C^2([ 1,1+\delta] ) \cap C([ 1,1+\delta] ) $ (see \cite[Theorem 6.14]{Gilbarg-Trudinger}). On the other hand, since $u''=0$ on $(1+\delta,2) $ and $u(2) =0$, we have \begin{equation} u(x) =\frac{u(1+\delta) }{1-\delta}(2-x) \quad \text{for all }x\in(1+\delta,2) \label{u en (1+delta, 2)} \end{equation} Since $u^{-\alpha}\in C^{\alpha\gamma}([ 1,1+\delta]) $ and $u\in H_0^1(\Omega) \subset C(\overline{\Omega}) $, we have $au^{-\alpha}-bu^p\in L^2(\Omega) $, and thus, from \eqref{the equationn}, it follows that $u\in W^{2,2}(\Omega) \subset C^1(\overline{\Omega})$. Multiplying \eqref{equation on (1,1+delta)} by $u'$ we obtain \begin{equation} \Big(\frac{1}{2}(u') ^2\Big) '=-\frac {1}{1-\alpha}(u^{1-\alpha}) ' \quad \text{on }(1,1+\delta) \label{derivatives} \end{equation} and so $\frac{1}{2}(u') ^2+\frac{1}{1-\alpha }u^{1-\alpha}={\rm const}=\frac{1}{2}(u'(1) ) ^2+\frac{1}{1-\alpha}u(1) ^{1-\alpha}$. Therefore, for $x\in(1,1+\delta) $: $u'(x) =0$ if, and only if, $\frac{1}{1-\alpha}u^{1-\alpha}(x) =\frac{1}{2}( u'(1) ) ^2+\frac{1}{1-\alpha}u(1) ^{1-\alpha}$. If there were no $x$ in $(1,1+\delta) $ such that $\frac{1}{1-\alpha}u^{1-\alpha}(x) =\frac{1}{2}( u'(1) ) ^2+\frac{1}{1-\alpha}u(1) ^{1-\alpha}$, we would have $u'(x) \neq0$ for all $x\in(1,1+\delta) $; which would imply that $u'(x) >0$ for all $x\in(1,1+\delta) $ (since $u'$ is continuous on $[ 1,1+\delta] $, and since $u'(1) >0$). Thus $u'(1+\delta) \geq0$, but, by \eqref{u en (1+delta, 2)}, $u'(1+\delta) =-\frac {u(1+\delta) }{1-\delta}<0$, which is a contradiction. Therefore $\{ x\in(1,1+\delta) :\frac{1}{1-\alpha}u^{1-\alpha }(x) =\frac{1}{2}(u'(1) ) ^2+\frac{1}{1-\alpha}u(1) ^{1-\alpha}\} \neq \emptyset$; let $x_1$ be its infimum. Since $u$ is continuous, $x_1$ is a minimum, therefore we have $u(x_1) =(\frac{1-\alpha}{2}(u'(1) ) ^2+u(1) ^{1-\alpha}) ^{\frac{1}{1-\alpha}}$. Note that $u'(x) >0$ for all $x\in[ 1,x_1) $. Moreover, \eqref{equation on (1,1+delta)} gives that $u$ is concave on $[1,1+\delta] $, and so $\frac{u(x_1) -u(1)}{x_1-1}\leq u'(1) $. Then, recalling \eqref{inequation altra4}, \begin{align*} x_1-1 & \geq\frac{u(x_1) -u(1) }{u'(1) } =\frac{\big(\frac{1-\alpha}{2}(u'(1) ) ^2+u(1) ^{1-\alpha}\big )^{\frac{1}{1-\alpha}}-u(1) }{u'(1) }\\ & \geq\frac{\big(\frac{1-\alpha}{2}(u'(1) ) ^2\big) ^{\frac{1}{1-\alpha}}+(u(1) ^{1-\alpha}) ^{\frac{1}{1-\alpha}}-u(1) }{u'(1) }\\ & =\frac{(\frac{1-\alpha}{2}(u'(1) ) ^2) ^{\frac{1}{1-\alpha}}}{u'(1)} =\big(\frac{1-\alpha}{2}\big) ^{\frac{1}{1-\alpha}} \big(u'(1) \big) ^{\frac{1+\alpha}{1-\alpha}}\\ & \geq\big(\frac{1-\alpha}{2}\big) ^{\frac{1}{1-\alpha}}\Big(\big( \frac{2}{p+1}\big) ^{\frac{1}{1-p}}\big(\frac{1-p}{2}\big) ^{\frac{1+p}{1-p}}\Big) ^{\frac{1+\alpha}{1-\alpha}}\geq\delta, \end{align*} which contradicts $x_1<1+\delta$. \end{example} \begin{theorem} \label{thm3.8} There is at most one weak solution $v\in H_0^1(\Omega) \cap L^{\infty}(\Omega)$ of \eqref{Problem} such that $v(x) >0$ a.e. in $\Omega$; and, if it exists, it satisfies $v\geq u$ for any other nonnegative weak solution $u\in H_0^1(\Omega) \cap L^{\infty}(\Omega) $ of \eqref{Problem}. \end{theorem} \begin{proof} Since $s\to f(s) :=as^{-\alpha}-bs^p$ is nondecreasing, the uniqueness assertion of the theorem follows from a standard argument: If $w$ is another solution which is positive a.e. in $\Omega$, take $\varphi:=v-w$ as a test function in the weak form of the equation \begin{gather*} -\Delta(v-w) =f(v) -f(w) \quad \text{in }\Omega,\\ v-w=0\quad \text{on }\partial\Omega \end{gather*} to obtain $\int_{\Omega}| \nabla(v-w) |^2=\int_{\Omega}(f(v) -f(w) ) (v-w) \leq0$, which implies $v=w$. Let $u\in H_0^1(\Omega) \cap L^{\infty}( \Omega) $ be a nonnegative solution of \ref{Problem}. Therefore, for any $\varphi\in H_0^1(\Omega) \cap L^{\infty}(\Omega) $, we have \begin{equation} \label{final} \begin{aligned} & \int_{\Omega}\langle \nabla(u-v) ,\nabla\varphi \rangle \\ & =\int_{\Omega}(au^{-\alpha}\chi_{\{ u>0\} } -bu^p-(av^{-\alpha}-bv^p) ) \varphi\\ & =\int_{\{ u>0\} }(f(u) -f(v) ) \varphi +\int_{\{ u=0\} }(-av^{-\alpha}+bv^p) \varphi. \end{aligned} \end{equation} Now, we take $\varphi=(u-v) ^{+}$. Since $v>0$ a.e. in $\Omega$, we have \[ \int_{\{ u=0\} }(-av^{-\alpha}+bv^p) (u-v) ^{+}=0. \] Thus, from \eqref{final}, we obtain $\int_{\Omega}| \nabla( u-v) ^{+}| ^2\leq0$, and so $u\leq v$ in $\Omega$. \end{proof} \begin{thebibliography}{99} \bibitem{Ambrosetti-Arcoya} A. Ambrosetti, D. Arcoya; \emph{An Introduction to Nonlinear Functional Analysis and Elliptic Problems}, Birkh\"{a}user, 2011. \bibitem{Aranda2} C. Aranda, T. 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