\documentclass[reqno]{amsart} \usepackage{hyperref} \AtBeginDocument{{\noindent\small \emph{Electronic Journal of Differential Equations}, Vol. 2011 (2011), No. 70, pp. 1--14.\newline ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu \newline ftp ejde.math.txstate.edu} \thanks{\copyright 2011 Texas State University - San Marcos.} \vspace{9mm}} \begin{document} \title[\hfilneg EJDE-2011/70\hfil Structure of ground states solutions] {Structure of ground state solutions for singular elliptic equations with a quadratic gradient term} \author[A. L. Melo, C. A. Santos \hfil EJDE-2011/70\hfilneg] {Antonio Luiz Melo, Carlos Alberto Santos} % in alphabetical order \address{Ant\^onio Luiz Melo \newline Department of Mathematics, University of Bras\'ilia, Planaltina, 70910-990, Brazil} \email{almelo@unb.br} \address{Carlos Alberto Santos \newline Department of Mathematics, University of Bras\'ilia, Bras\'ilia, 70910-900, Brazil} \email{csantos@unb.br, capdsantos@gmail.com} \thanks{Submitted April 26, 2010. Published May 31, 2011.} \thanks{C. A. Santos was supported by CNPq/Brasil, PROCAD-MS, and FAPDF under grant \hfill\break\indent PRONEX 193.000.580/2009. A. L. Melo research was supported by CNPq/Brasil, \hfill\break\indent PROCAD-MS} \subjclass[2000]{35J25, 35J20, 35J67} \keywords{Singular elliptic equations; gradient term; ground state solution} \begin{abstract} We establish results on existence, non-existence, and asymptotic behavior of ground state solutions for the singular nonlinear elliptic problem \begin{gather*} -\Delta u = g(u)| \nabla u |^2 + \lambda\psi(x) f(u) \quad\text{in } \mathbb{R}^N,\\ u > 0 \quad\text{in } \mathbb{R}^N,\quad \lim_{|x| \to \infty} u(x)=0, \end{gather*} where $\lambda \in \mathbb{R}$ is a parameter, $\psi \geq 0 $, not identically zero, is a locally H\"older continuous function; $g:(0,\infty) \to \mathbb{R}$ and $f:(0,\infty) \to (0,\infty)$ are continuous functions, (possibly) singular in $0$; that is, $f(s)\to \infty$ and either $g(s)\to \infty$ or $g(s)\to -\infty$ as $s \to 0$. The main purpose of this article is to complement the main theorem in Porru and Vitolo \cite{pv}, for the case $\Omega=\mathbb{R}^N$. No monotonicity condition is imposed on $f$ or $g$. \end{abstract} \maketitle \numberwithin{equation}{section} \newtheorem{theorem}{Theorem}[section] \newtheorem{lemma}[theorem]{Lemma} \newtheorem{remark}[theorem]{Remark} %\newtheorem{definition}[theorem]{Definition} \allowdisplaybreaks \section{Introduction} In this article, we establish results concerning non-existence, existence and asymptotic behavior of positive ground state solutions; that is, entire positive classical solutions (in $C^2(\mathbb{R}^N)$) vanishing at infinity, for the singular nonlinear elliptic problem \begin{equation}\label{1.1} \begin{gathered} -\Delta u = g(u)| \nabla u |^2 + \lambda\psi(x) f(u) \quad\text{in } \mathbb{R}^N,\\ u > 0 \quad\text{in } \mathbb{R}^N,\quad \lim_{|x| \to \infty} u(x) =0, \end{gathered} \end{equation} where $g: (0, \infty) \to \mathbb{R}$ and $f: (0, \infty) \to (0, \infty)$ are continuous functions, possibly, singular in $0$ in the sense, for example, that either $g(s) \to \infty$ or $g(s) \to-\infty$ and $f(s) \to \infty$ as $s \to 0$; $\psi: \mathbb{R}^N \to [0, \infty)$, $\psi \neq 0$ is a locally H\"older continuous function and $\lambda \in \mathbb{R}$ is a real parameter. The search for classical solutions to \eqref{1.1} with $\lambda=1$ and $g = 0$; that is, for the problem \begin{equation} \label{1.4} \begin{gathered} -\Delta u = \psi(x) f (u) \quad\text{in } \mathbb{R}^N,\\ u > 0 \quad\text{in } \mathbb{R}^N,\quad \lim_{|x| \to \infty} u(x) =0, \end{gathered} \end{equation} where $\psi$ and $f$ are as above with $f$ singular at $0$, has received much attention in recent years; see \cite{5,6,7,GS,LS,3,2,4,8} and references therein. For more general nonlinearities, we refer the reader to Mohammed \cite{MO}, and for nonlinearities including singular terms in the origin and super-linear terms at infinity to Santos \cite{santos}. For further studies on \eqref{1.1}, the reader is referred to \cite{9} and the references therein. However, \cite{9} does not include the nonlinearity in the coefficient of the gradient term. For the version of \eqref{1.1} on bounded domains, \begin{equation} \label{1.2} \begin{gathered} -\Delta u = \lambda g(u)| \nabla u |^2 + \psi(x) f (u) \quad\text{in } \Omega,\\ u > 0 \quad\text{in } \Omega,\quad u(x)= 0 \quad\text{on } \partial\Omega, \end{gathered} \end{equation} where $\Omega \subset \mathbb{R}^N$ is a regular bounded domain, $\lambda $ is a real parameter, $\psi:\Omega \to [0,\infty)$ and $f,g$ are appropriate functions, see for example \cite{BAI, BST, SC, PS, pv} and their references. Problems such as \eqref{1.2} were studied in \cite{BAI, gsm,PS} with $f(s)=1,~s>0$. In \cite{BST} and \cite{PS}, \eqref{1.2} was considered with general terms $f$ but in all cases $g$ is non-singular in $0$, that is, $ g $ is continuously extendable to $ 0 $. In \cite{pv}, \eqref{1.2} was studied with $\psi(x)=1$, in $\Omega$. Under some conditions on $f$ and $g$ the authors showed existence and, in particular cases, asymptotic behavior of solutions to \eqref{1.2}. In most cases, monotonicity conditions are imposed upon $f$ or $g$. To establish our main results regarding problem \eqref{1.1}, we shall denote by $$ G(s)=\int_1^{s} g(t) dt,\quad s>0, $$ a primitive of $g$. We define \begin{gather*} f_{go}=\liminf_{s\to 0}\frac{e^{G(s)}f(s)}{\int_0^s e^{G(t)} dt},\quad f_{g{\infty}}=\limsup_{s\to \infty}\frac{e^{G(s)}f(s)}{\int_0^s e^{G(t)} dt},\\ \underline{f}_{go}=\liminf_{s\to 0}\frac{e^{G(s)}f(s)} {[\int_s^1 e^{G(t)} dt]^q}, \quad \overline{f}_{g{o}}=\limsup_{s\to 0}\frac{e^{G(s)}f(s)} {[\int_s^1 e^{G(t)} dt]^p} \end{gather*} with $1 0 \quad\text{in } \mathbb{R}^{N},\quad \lim_{| x | \to \infty} u(x) =0 \end{gathered} \end{equation} has a unique solution $w_{\psi} \in C^{2,\alpha}_{\rm loc}(\mathbb{R}^N)$, for some $\alpha\in (0,1)$. Also we will say that $\psi$ satisfies the condition $(\psi_{\infty})'$ if \begin{equation} 0< \liminf_{| x | \to \infty}\frac{\psi(x)}{| x |^{\gamma}} \leq \limsup_{| x | \to \infty}\frac{\psi(x)}{| x |^{\gamma}}< \infty, \end{equation} where $\psi>0$, and $\gamma$ is a negative constant such that $\gamma<-2p$ with $p$ given in $\overline{f}_{g{o}}$. \begin{remark} \label{rmk1.1}\rm Concerning the hypothesis $(\psi_{\infty})$, we have: (1) If \begin{equation} \label{1.1.2} \int_0^{\infty}\Big[s^{1-N}\int_0^s t^{N-1} \hat{\psi}(t) dt\Big]ds <\infty, \end{equation} where $\hat{\psi}(r)=\max_{| x |=r} \psi(x), ~r>0$, then $(\psi_{\infty})$ holds. In this case, \begin{equation} \label{1.1.2.1} w_{\psi}(x)\leq \int_{| x |}^{\infty}\Big[s^{1-N}\int_0^s t^{N-1} \hat{\psi}(t) dt\Big]ds :=\hat{w}_{\psi}(| x |),~x\in \mathbb{R}^N, \end{equation} because $\hat{w}_{\psi}(| \cdot |)$ is an upper solution of \eqref{1.1.1.a}. (see details in Santos \cite{santos1}). (2) If we assume $N\geq 3$ and $$ \begin{array}{c} \int_1^{\infty} r \hat{\psi}(r) dr < \infty, \end{array} $$ then \eqref{1.1.2} will be true (see details in Goncalves and Santos \cite{7}). \end{remark} To state our next theorem, we consider the problem \begin{equation} \label{1.3} \begin{gathered} -\Delta u = \lambda {\psi}(x) u \quad\text{in } \Omega,\\ u = 0, \quad\text{on }\partial\Omega, \end{gathered} \end{equation} where $\Omega \subset \mathbb{R}^N$ is a bounded and smooth domain and $\psi$ is a non-negative and suitable function. We know that the first eigenvalue $\lambda_1({\psi},\Omega)$ of \eqref{1.3} is positive and non-increasing in the sense that $\lambda_1({\psi},\Omega_2) \leq \lambda_1({\psi},\Omega_1)$ if $\Omega_1\subseteq\Omega_2 $. So there exists \begin{equation} \label{1.31} \lambda_1({\psi}) = \lim_{k \to \infty}\lambda_1({\psi}, B_k(0))\in [0,\infty), \end{equation} where $B_k(0)$ is the ball centered in the origin of $\mathbb{R}^N$ with radius $k$. For more details concerning the principal eigenvalue $\lambda_1({\psi})$, we refer to Santos \cite{santos1}. Our main results read as follows: \begin{theorem} \label{thm1.1} Assume that $\int_0^{1} e^{G(t)} dt<\infty$, $(\psi_{\infty})$ and $f_{go} \in (0,\infty]$ hold. Then \eqref{1.1} admits a solution $u=u_{\lambda} \in C^2(\mathbb{R}^N)$ if $\lambda_1(\psi)/f_{g{0}} < \lambda < \lambda^{\star}$ for some $\lambda^{\star}>0$. \end{theorem} \begin{remark} \label{rmk1.2}\rm The $\lambda^{\star}>0$ and the solution $u$, given by Theorem \ref{thm1.1} depend on the behavior of $g$ and $f$ at infinity. More specifically, denoting by \begin{equation}\label{1.1.1.1} F(s) = \int_0^s e^{G(t)}dt, s\geq 0 , \quad F_{\infty} = \lim_{s\to \infty}F(s)= \int_0^{\infty} e^{G(t)}dt, \end{equation} we have \begin{itemize} \item [(i)] If $F_{\infty}=\infty$ and \begin{itemize} \item [(1)] $0 \leq f_{g{\infty}} < \infty$, then $\lambda^{\star}\geq \frac{1}{\| w_{\psi} \|_{\infty}f_{g{\infty}}}$, \item [(2)] $f_{g{\infty}} = \infty$, then $\lambda^{\star}$ is a positive constant. \end{itemize} \item [(ii)] If $F_{\infty}<\infty$, then \begin{itemize} \item [(1)] $\lambda^{\star}=\frac{1}{\| w_{\psi} \|_{\infty}}\frac{1}{F_{\infty}}\int_0^{F_{\infty}} \Big( s^{-1}\int_0^s\Big[\sup_{r>F^{-1}(t)} \frac{e^{G(r)}f(r)}{F(r)} \Big]^{-1}dt \Big)ds\in (0,\infty]$, \item [(2)] $ \| u \|_{\infty} \leq F_{\infty}$. \end{itemize} \end{itemize} As an example that satisfies all the assumptions of Theorem \ref{thm1.1}, we have \begin{equation} \label{1.1.1} \begin{gathered} -\Delta u = - \frac{\mu}{u}| \nabla u |^2 + \lambda\psi(x) f(u) \quad\text{in } \mathbb{R}^N,\\ u > 0 \quad\text{in } \mathbb{R}^N,\quad \lim_{|x| \to \infty} u(x) =0, \end{gathered} \end{equation} if $-\infty < \mu<1$, $\lim_{s\to 0}f(s)/s>0$ and $\psi$ satisfies $(\psi_{\infty})$. Furthermore, if we have $\lim_{s\to \infty}f(s)/s=0$, then $\lambda^{\star}=\infty$. \end{remark} In the next result and Theorem \ref{thm1.5}, we assume that $f$ is a $C^1$-function and $N \geq 3$. \begin{theorem} \label{thm1.2} Assume that $\int_0^{1} e^{G(t)}dt=\infty$, $(\psi_{\infty})'$, $\underline{f}_{go}\in (0,\infty]$ and $\overline{f}_{go}\in [0,\infty)$ hold. Then there exists $\lambda^{\star}>0$ such that for all $\lambda\in(0, \lambda^{\star})$ the problem \eqref{1.1} has a solution. \end{theorem} \begin{remark} \label{rmk1.3}\rm Again here $\lambda^{\star}>0$ depends on the behavior of $f$ and $g$ at infinity. That is, if \begin{equation} \label{1.1.3.a} \limsup_{s\to \infty}\frac{e^{G(s)}f(s)}{[\int_s^{s+1}e^{G(t)}dt]^p} <\infty, \end{equation} where $p>1$ is defined in $\overline{f}_{g{o}}$, then, for some positive constant $c$, $$ \lambda^{\star}\geq c\inf_{s>0}\Big[ \int_s^{s+1}e^{G(t)}dt\Big]^{1-p}. $$ Consider the example \begin{equation}\label{1.1.1b} \begin{gathered} -\Delta u = - \frac{\mu}{u}| \nabla u |^2 + \lambda\psi(x) f(u) \quad\text{in } \mathbb{R}^N,\\ u > 0 \quad\text{in } \mathbb{R}^N,\quad \lim_ {|x| \to \infty} u(x)=0. \end{gathered} \end{equation} All hypotheses of Theorem \ref{thm1.2} are satisfied if $\psi$ satisfies $(\psi_{\infty})'$, $\mu\geq 1$ and $f$ satisfies $$ \lim_{s\to 0}\frac{f(s)}{s(\ln 1/s)^p}>0,\quad \text{if }\mu=1, \quad\text{and}\quad \lim_{s\to 0}\frac{f(s)}{s^{\mu(1-p)+p}}>0,\quad \text{if }\mu>1, $$ where $p=q$ is given in $(\psi_{\infty})'$. Besides this, $\lambda^{\star}=\infty$, if $$ \lim_{s\to \infty}\frac{f(s)}{s(\ln 1/s)^p}<\infty,\quad\text{if } \mu=1,\quad \text{and}\quad \lim_{s\to \infty}\frac{f(s)}{s^{\mu(1-p)}}<\infty,\quad \text{if }\mu>1. $$ \end{remark} For the non-existence, we have the following result. \begin{theorem} \label{thm1.3} Assume that $g:(0,\infty) \to \mathbb{R}$, $f:(0,\infty) \to [0,\infty)$, $\psi:\mathbb{R}^N\to [0,\infty)$ are continuous functions and $\lambda \leq 0$. Then \eqref{1.1} has no solution. \end{theorem} Concerning the asymptotic behavior, we have the following result. \begin{theorem} \label{thm1.4} Assume that \eqref{1.1.2} holds and $N\geq 3$, then the solution given by Theorem \ref{thm1.1} (which we shall denote as $u=u_{\lambda}$) satisfies $$ F^{-1}(c| x |^{2-N}) \leq u(x) \leq F^{-1}(d| x |^{2-N}),\quad | x | \geq 1, $$ for some positive constants $c$ and $d$ with $F$ defined in \eqref{1.1.1.1}. In particular, if $g=0$, then $$ c| x |^{2-N} \leq u(x) \leq d| x |^{2-N},\quad | x | \geq 1. $$ \end{theorem} For example the solution of \eqref{1.1.1}, given by Theorem \ref{thm1.1}, satisfies $$ c| x |^{4-2N} \leq u(x) \leq d| x |^{4-2N},\quad | x | \geq 1, $$ if in addition we assume $\lim_{s\to 0}f(s)/s<\infty$. \begin{theorem} \label{thm1.5} The solution given by Theorem \ref{thm1.2} (which we shall denote as $u=u_{\lambda}$) satisfies $$ F_{0}^{-1}(c| x |^\frac{\gamma+2}{1-q}) \leq u(x) \leq F_{0}^{-1}(d| x |^\frac{\gamma+2}{1-p}),\quad | x | \geq R, $$ for some positive constants $c$, $d$ and $R$ with $$ F_{0}(s) = \int_s^{ 1} e^{G(t)}dt,\quad 0 < s < 1. $$ \end{theorem} For example the solution of \eqref{1.1.1b} with $\mu>1$ satisfies $$ \frac{1}{c+ | x |^{\frac{\gamma+2}{(1-p)(\mu-1)}}} \leq u(x) \leq \frac{1}{d+ | x |^{\frac{\gamma+2}{(1-p)(\mu-1)}}},\quad | x | \geq R, $$ for some constants $c,d,R>0$. \begin{remark} \label{rmk1.4}\rm Examples of $\psi:\mathbb{R}^N \to (0,\infty)$ satisfying ($\psi_{\infty}$) with $\nu>2$ are as follows: $$ \psi(x) = \frac{1}{1+| x |^{\nu}},\quad \psi(x) = \frac{1}{2+\sin(| x |^2) +| x|^{\nu}} $$ while $$ \psi(x)=\frac{1}{1+| x |^{2p +1}},\quad x\in \mathbb{R}^N $$ satisfies $(\psi_{\infty})'$, where $p>2$. \end{remark} The proof of Theorem \ref{thm1.1} is based on the suitable diffeomorphisms and in Santos's arguments which showed existence of at least one entire positive solution for the problem \eqref{1.4} in the presence of singular and super linear terms at infinity without imposing any monotonicity condition in $f(s)$ or $f(s)/s$ (for more details see \cite{santos}). \subsection*{Proof of Theorem \ref{thm1.1}} Consider the function defined in \eqref{1.1.1.1}; that is, $F:[0,\infty) \to [0,\infty)$ with $$ F(s) = \int_0^s e^{ G(t)}dt,\quad s\geq 0\quad \text{and}\quad F_{\infty} = \lim_{s \to \infty}\int_0^{s} e^{G(t)} dt. $$ Thus, $F$ is increasing, $F(0)=0$. Now, we will consider two separate cases. \textbf{Case 1:} $F_{\infty}=\infty$. In this case, $F(s) \to \infty$ as $s\to \infty$. Now, let the continuous function $h(s)=F'(F^{-1}(s))f(F^{-1}(s))$, $s>0$ and for each $\tau,\lambda>0$ given, consider the continuous function $\widetilde{H}_{\lambda}:(0,\infty)\times (0,\infty) \to (0,\infty)$ defined by $$ \widetilde{H}_{\lambda}(\tau,s) = \begin{cases} \lambda s\sup_{s \leq t \leq \tau}\frac{h(t)}{t}, &s \leq \tau,\\ \lambda s\frac{h(\tau)}{\tau}, &s \geq \tau. \end{cases} $$ So, it is easy to check that \begin{itemize} \item[(i)] $\widetilde{H}_{\lambda}(\tau,s) \geq \lambda h(s)$, $00$, \item[(iii)] $\lim_{s\to 0^+} \widetilde{H}_{\lambda} (\tau,s)/s=\lambda\sup_{00. \] Besides this, $ \hat{H_{\lambda}}(\tau, \cdot)\in C^1(0,\infty)$, for each $\tau >0$. Using (i)-(iv), it follows that for each $\lambda \geq 0$, $\hat{H_{\lambda}}$ satisfies the following. \begin{lemma} \label{lem2.1a} If $\int_0^{1} e^{G(t)} dt<\infty$, then, for each $\tau>0$, \begin{itemize} \item[(i)] $\hat{H}_{\lambda}(\tau,s)/s$ is non-increasing for $s>0$, \item[(ii)] $\lim_{s\to 0}\hat{H}_{\lambda}(\tau,s)/s =\lambda \sup_{00, $$ where $w_{\psi}$ is given by the hypothesis $(\psi_{\infty})$. Hence, \begin{equation}\label{2.4} H_{\lambda}(\tau) = \frac{1}{\lambda}H_{1}(\tau),\quad \tau,\lambda>0. \end{equation} Let $$ \lambda^{\star} = \sup_{\tau \geq 1}H_{1}(\tau)>0. $$ Since $$ \liminf_{\tau\to \infty}H_{1}(\tau) =\frac{1}{\| w_{\psi} \|_{\infty}f_{g{\infty}}} $$ it follows that $$ \frac{1}{\| w_{\psi} \|_{\infty}f_{g{\infty}}} \leq \lambda^{\star} \leq \infty. $$ This proves Remark \ref{rmk1.2} part (i)(1). So, from \eqref{2.4}, for each $0<\lambda < \lambda^{\star}$, we can take a $\tau_{\infty}=\tau_{\lambda}\geq 1$ such that $ H_{\lambda}(\tau_{\infty})>1$. That is, \begin{equation} \label{2.5} \frac{1}{\tau_{\infty}}\int_0^{\tau_{\infty}} \frac{t}{\hat{H_{\lambda}}(\tau_{\infty},t)}\,dt > \| w_{\psi} \|_{\infty}. \end{equation} Now, defining the $C^2$- increasing function $$ \hat{h}_{\lambda}(s) = \frac{1}{\tau_{\infty}} \int_0^{s}\frac{t}{\hat{H_{\lambda}}(\tau_{\infty},t)}\,dt,\quad s\geq 0 $$ and defining $v(x)=\hat{h}_{\lambda}^{-1}(w_{\psi}(x))$, $x\in \mathbb{R}^N$, we obtain, using \eqref{2.5}, $$ v(x)=\hat{h}_{\lambda}^{-1}(w_{\psi}(x)) \leq \hat{h}_{\lambda}^{-1}(\| w_{\psi} \|_{\infty}) <\hat{h}_{\lambda}^{-1}(\hat{h}_{\lambda}(\tau_{\infty})) = \tau_{\infty},\quad x \in \mathbb{R}^N $$ and after some calculations, we obtain that $v\in C^2(\mathbb{R}^N)$, $v(x)\to 0$ as $| x | \to \infty$ and that it satisfies \begin{gather*} -\Delta v \geq \psi(x)\hat{H_{\lambda}}(\tau_{\infty},v) \geq \lambda \psi(x) h(v) \quad\text{in } \mathbb{R}^{N},\\ v > 0 \quad\text{in } \mathbb{R}^{N},\quad \lim_{| x | \to \infty} v(x)= 0. \end{gather*} On the other hand, given $\lambda_1(\psi)/f_{g{0}} < \lambda < \lambda^{\star}$ (we point out that $\lambda_1(\psi)/f_{g{0}}=0$ if either $f_{g{0}}=\infty$ or $\lambda_1(\psi)=0$) we can take from \eqref{1.31} a $k_{\lambda}>1$ such that $$ \frac{\lambda_1(\psi)}{f_{g{0}}} \leq \frac{\lambda_1(\psi,k)} {f_{g{0}}} < \lambda <\lambda^{\star},\quad \text{for all } k\geq k_{\lambda}. $$ As a consequence of this, there exists a $s_0=s_{0,\lambda,k} \in(0,1)$ such that $$ \lambda h(s) \geq \lambda_1(\psi,k) s,\quad \text{for all } 00$ satisfies $$ \varepsilon_{\lambda,k}\max\{\psi_k(x):x\in \overline{B_k(0)}\} \leq s_0, $$ it follows that $v_k$ satisfies \begin{gather*}%\label{2.6} -\Delta v_k \leq \lambda \psi(x) h(v_k) \quad\text{in } B_k(0),\\ v > 0 \quad\text{in } B_k(0),\quad v(x) = 0 \quad\text{on } \partial B_k(0). \end{gather*} Following the arguments of either Mohammed \cite{MO} or Santos \cite{santos}, we have a $v\in C^2(\mathbb{R}^N)$ satisfying \begin{gather*}%\label{2.5.1} -\Delta v = \lambda\psi(x) h(v) \quad\text{in } \mathbb{R}^N,\\ v > 0 \quad\text{in } \mathbb{R}^N,\quad \lim_{|x| \to \infty} v(x)=0. \end{gather*} Let $$ u(x)=u_{\lambda}(x)=F^{-1}(v(x)),~x\in \mathbb{R}^N. $$ Such that $$ 00$ given, we consider the continuous functions $\widetilde{H}_{\lambda},\hat{H_{\lambda}}$ defined by $$ \widetilde{H}_{\lambda}(s) = \lambda s\sup_{s \leq t \leq F_{\infty}}\frac{h(t)}{t},\quad 00, $$ where $w_{\psi}$ is given by hypothesis $(\psi_{\infty})$. Hence, \begin{equation} \label{2.4.1} H_{\lambda}(\tau) = \frac{1}{\lambda}H_{1}(\tau),\quad \tau,\;\lambda>0. \end{equation} Define \begin{align*} \lambda^{\star} & = \lim_{\tau \to F_{\infty}}H_1(\tau)\\ & = \lim_{\tau \to F_{\infty}}\frac{1}{\| w_{\psi} \|_{\infty}\tau}\int_0^{\tau}\frac{t}{\hat{H_{1}}(t)}dt\\ &=\frac{1}{\| w_{\psi} \|_{\infty}F_{\infty}} \int_0^{F_{\infty}}\frac{t}{\hat{H_{1}}(t)}dt \\ & = H_1(F_{\infty})>0. \end{align*} Such that, from \eqref{2.4.1}, for each $0<\lambda < \lambda^{\star}$, we have $$ H_{\lambda}(F_{\infty})=\frac{1}{\lambda} H_{1}(F_{\infty}) =\frac{\lambda^{\star}}{\lambda}>1. $$ That is, \begin{equation} \label{2.5.1} \frac{1}{F_{\infty}}\int_0^{F_{\infty}} \frac{t}{\hat{H_{\lambda}}(t)}\,dt > \| w_{\psi} \|_{\infty}. \end{equation} Now, defining the $C^2$ increasing function $$ \hat{h}_{\lambda}(s) = \frac{1}{F_{\infty}}\int_0^{s} \frac{t}{\hat{H_{\lambda}}(t)}\,dt,\quad 00$ for $t>0$, $\psi$ as before and the problem \begin{equation}\label{2.6} \begin{gathered} \Delta u = \psi(x) k (u) \quad\text{in } \mathbb{R}^N,\\ u > 0 \quad\text{in } \mathbb{R}^N,\quad \lim_{|x|\to \infty} u(x)= \infty. \end{gathered} \end{equation} \begin{lemma} \label{lem3.1} Let $\psi\in C^{\nu}_{\rm loc}(\mathbb{R}^N)$ for some $\nu \in (0,1)$ and $\psi(x)>0,~\forall ~x\in \mathbb{R}^N$, $N \geq 3$. Assume that there exist $1 0}\frac{k(s)}{s^p}\in (0,\infty) \end{gather} and condition $(\psi_{\infty})'$ holds with $\gamma < -2 p$. Then \eqref{2.6} admits at least one solution $u \in C^2(\mathbb{R}^N)$ such that \begin{equation} \label{2.6.1} u(x)\geq a_{\psi}{S_{\infty}^{\frac{1}{1-p}}}>0 \quad \text{for all } x \in \mathbb{R}^N, \end{equation} for some positive constant $a_{\psi}$. \end{lemma} \begin{remark} \label{rmk3.2}\rm The main novelty in Lemma \ref{lem3.1} is the lower limit of solution $u$ of \eqref{2.6} by a positive constant throughout $\mathbb{R}^N$. A similar result was proved in \cite{GR} without Claim \ref{2.6.1}. \end{remark} \begin{proof}[Proof of Lemma \ref{lem3.1}] We follows similar arguments as those in \cite[Theorem 1.1]{GR}. In fact, from conditions \eqref{kinf} and $(\psi_{\infty})'$ there exists a $R_{\psi}>0$ such that $$ b_1=\frac{1}{2}\liminf_{|x| \rightarrow \infty}\frac{\psi(x)}{|x|^\gamma} \leq \frac{\psi(x)}{|x|^\gamma} \leq 2 \liminf_{|x| \rightarrow \infty}\frac{\psi(x)}{|x|^\gamma} = b_2,\quad \forall | x | \geq R_{\psi} $$ and \begin{equation}\label{3.1} k(s) \geq \frac{\ell_\infty}{2}s^q , ~ \forall ~ s \geq R_{\psi}. \end{equation} Now, defining \begin{gather*} \alpha =\frac{\gamma+2}{1-p}>2,\quad \beta=\frac{\gamma+2}{1-q}>2, \\ A_{\psi}=\max_{[0,R_{\psi}]}\frac{\big[\frac{t^\alpha}{(N+\alpha-2) \alpha}+1+\frac{t^2}{2N}\big]^p}{1+t^{\alpha-2}},\quad B_{\psi}=\max_{[R_{\psi},\infty)} \frac{\big[\frac{1}{(N+\alpha-2)\alpha}+\frac{1}{t^\alpha}+ \frac{t^{2-\alpha}}{2N}\big]^p}{1+t^{-\gamma-\alpha p}}, \\ C_{\psi}=\min_{[0,R_{\psi}]}\frac{\big[\frac{t^\beta}{(N+\beta-2)\beta}+1+\delta+ \frac{t^2}{2N}\big]^q}{1+t^{\beta-2}},\quad D_{\psi}=\min_{[R_{\psi},\infty)}\frac{\big[\frac{1}{(N+\beta-2)\beta}+\frac{1+\delta}{t^\beta}+ \frac{t^{2-\beta}}{2N}\big]^q}{1+t^{-\gamma-\beta q}}, \\ \delta=\begin{cases} 0, & \text{if }\alpha=\beta,\\ \frac{[\alpha \beta (N + \beta - \alpha )]^{\frac{\alpha}{\beta-\alpha}}}{(\beta-\alpha)^{\frac{\alpha}{\beta-\alpha}}[\alpha (\alpha + 1) (N + \alpha -2)]^{\frac{\beta}{\beta-\alpha}}}, &\text{if } \alpha<\beta, \end{cases} \end{gather*} we have \begin{align*} 0&<\tilde{\lambda} = \min\big\{\big(M_{\psi} S_\infty A_{\psi}\big)^{\frac{1}{1-p}}, \big(b_2 S_{\infty} B_{\psi}\big)^{\frac{1}{1-p}} \big\}\\ &\leq \tilde{\Lambda} = \max\big\{\tilde{\lambda}, \frac{R_{\psi}}{1+\delta}, \big(\frac{m_{\psi} \ell_{\infty} C_{\psi}}{2}\big)^{\frac{1}{1-q}},~\big(\frac{b_1 \ell_\infty D_{\psi}}{2}\big)^{\frac{1}{1-q}}\big\}<\infty, \end{align*} where $$ M_{\psi}=\max_{|x|\leq R_{\psi}}\psi(x)\quad \text{and}\quad m_{\psi}=\min_{|x|\leq R_{\psi}}\psi(x). $$ In the sequel, we use the notation \begin{gather*} \underline{u}(x)=\tilde{\lambda}\Big[ \frac{| x |^{\alpha}} {(N + \alpha-2)\alpha} + \frac{| x |^2}{2N}+1\Big],\quad x\in \mathbb{R}^N, \\ \overline{u}(x)=\tilde{\Lambda}\Big[ \frac{| x |^{\beta}}{(N + \beta-2)\beta} + \frac{| x |^2}{2N}+\delta +1\Big],\quad x\in \mathbb{R}^N \end{gather*} and separately considering the cases $| x | \leq R_{\psi}$ and $| x | \geq R_{\psi}$. We obtain by direct computations, using \eqref{3.1}, that \begin{gather*} \underline{u}(x) \leq \overline{u}(x),\quad x\in \mathbb{R}^N, \\ \underline{u}(x),\quad \overline{u}(x) \to \infty \quad \text{as }{|x| \to \infty} \\ \Delta \underline{u}(x) \leq \psi(x) k(\underline{u}(x)), \quad \Delta \overline{u}(x) \geq \psi(x) k(\overline{u}(x)). \end{gather*} Now, by applying \cite[Theorem 2.1]{GR}, we have a solution $u \in C^2(R^N)$ of \eqref{2.6} with $$ 00. $$ This completes the proof. \end{proof} \begin{proof}[Proof of Theorem \ref{thm1.2}] For each $\tau>0$ given, define $F_{\tau}:(0,\tau]\to (0,\infty)$ by $$ F_{\tau}(s)=\int_s^{\tau+1}e^{G(t)}dt. $$ So, $F_{\tau}$ is a decreasing continuous function. From $\int_0^{1} e^{G(t)} dt=\infty$, we have $$ \lim_{s\to 0}F_{\tau}(s)=\infty\quad\text{and}\quad \lim_{s\to \tau}F_{\tau}(s)=F_{\tau}(\tau). $$ Now, we consider the $C^1$-function $k_{\tau}:[0,\infty)\to[0,\infty)$ defined by $$ k_{\tau}(s) = \begin{cases} s^p, &0\leq s <\frac{F_{\tau}(\tau)}{2}, \\ \xi_{\tau}(s), & \frac{F_{\tau}(\tau)}{2}\leq s \leq F_{\tau}(\tau), \\ e^{G(F_{\tau}^{-1}(s))}f(F_{\tau}^{-1}(s)), & s\geq F_{\tau}(\tau), \end{cases} $$ for appropriate function $\xi_{\tau}$, and the $\tau$-problems family \begin{equation} \label{2.9} \begin{gathered} \Delta v = \lambda\psi(x) k_{\tau} (v) \quad\text{in } \mathbb{R}^N,\\ v > 0 \quad\text{in } \mathbb{R}^N,\quad \lim_{|x|\to \infty} v(x) \infty. \end{gathered} \end{equation} We claim that $$ \ell_{\infty, \tau}=\liminf_{s\to \infty}\frac{k_{\tau}(s)}{s^q} \in (0,\infty]. $$ In fact, making $t=F_{\tau}^{-1}(s)$, $0 < s \leq \tau$, we have \begin{align*} \ell_{\infty, \tau} &=\liminf_{s\to \infty}\frac{k_{\tau} (s)}{s^q} = \liminf_{t\to 0}\frac{e^{G_{\tau} (t)}f(t)}{F_{\tau}(t)^q}\\ &= \liminf_{t\to 0}\frac{e^{G (t)}f(t)}{\big[\int_t^1 e^{G (s)} ds + b_2(\tau) \big]^q} = \liminf_{t\to 0}\frac{e^{G (t)}f(t)} {\big[\int_t^1 e^{G (s)} ds\big]^q [1 + \frac{b_2(\tau)}{\int_t^1 e^{G (s)} ds} ]^q} \\ & = \liminf_{t\to 0}\frac{e^{G (t)}f(t)} {\big[\int_t^1 e^{G (s)} ds \big]^q }= \underline{f}_{go}, \end{align*} where $b_2(\tau)$ denotes a real positive constant. Since $\underline{f}_{go}\in (0,\infty]$, we obtain \eqref{kinf} of Lemma \ref{lem3.1}. Also, since $$ \limsup_{s \to 0}\frac{k_{\tau}(s)}{s^p} = 1 $$ and making $t=F_{\tau}^{-1}(s), ~0 < s \leq \tau$, we have \begin{align*} \limsup_{s \to \infty}\frac{k_{\tau}(s)}{s^p} & = \limsup_{t \to 0}\frac{e^{G(t)}f(t)} {\big[\int_t^{\tau+1} e^{G (r)} dr\big]^p}\\ &= \limsup_{t \to 0}\frac{e^{G (t)}f(t)}{\big[\int_t^1 e^{G (r)} dr + \int_1^{\tau+1} e^{G(r)} dr \big]^p}\\ & \leq \limsup_{t \to 0}\frac{e^{G (t)}f(t)} {\big[\int_t^1 e^{G (r)} dr \big]^p} = \overline{f}_{go}. \end{align*} By hypothesis the $\overline{f}_{go}\in [0,\infty)$, we have that \begin{equation} \label{2.8.1} S_{\infty,\tau} = \sup_{s> 0}\frac{k_{\tau}(s)}{s^p}\in (0,\infty), \end{equation} for each $\tau>0$. Hence, \eqref{kzero} of Lemma \ref{lem3.1} is satisfied. Let \begin{equation}\label{2.8.2} \lambda^{\star}: = \sup_{\tau>0} \frac{a_{\psi}^{p-1}}{F_{\tau}(\tau)^{p-1}S_{\infty,\tau}} >0, \end{equation} where $a_{\psi}>0$ is the constant of Lemma \ref{lem3.1}. Given $0<\lambda<\lambda^{\star}$, pick a $\tau=\tau(\lambda)>1$ such that \begin{equation} \label{2.8} F_{\tau}(\tau) < \frac{1}{\lambda^{\frac{1}{p-1}}} \big[\frac{1}{S_{\infty,\tau}}\big]^{\frac{1}{p-1}}a_{\psi}. \end{equation} and apply Lemma \ref{lem3.1} to the problem \eqref{2.9}. That is, there exists a $v=v_{\tau}=v_{\tau(\lambda)}$ solution of \eqref{2.9} satisfying, by \eqref{2.6.1} and \eqref{2.8}, $$ v_{\tau}(x)\geq a_{\psi} {[\lambda S_{\infty,\tau}]^{\frac{-1}{p-1}}} > F_{\tau}(\tau) ,\quad \text{for all }x \in \mathbb{R}^N. $$ Define $$ u_{\tau}(x) = F_{\tau}^{-1}(v_{\tau}(x)),\quad x \in \mathbb{R}^N. $$ Thus, of $F_{\tau}^{-1}$ decreasing, we have $$ u_{\tau}(x) = F_{\tau}^{-1}(v_{\tau}(x)) \leq F_{\tau}^{-1}(F_{\tau}(\tau))=\tau,~x \in \mathbb{R}^N $$ and from the regularity of $F_{\tau}^{-1}$, it follows that $$ 00}\frac{e^{G(s)}f(s)}{\big[\int_s^{s+1}e^{G(t)}dt\big]^p}, $$ where $M$ is finite by \eqref{1.1.3.a}, and, if necessary redefine, $\xi_{\tau}$ in $k_{\tau}$ such that $0<\xi_{\tau}(s)\leq (M+1)s^p$, $\frac{1}{2}F_{\tau}(\tau) \leq s \leq F_{\tau}(\tau)$. This is possible because $(M+1)F_{\tau}(\tau)^p> e^{G(\tau)}f(\tau)$ for each $\tau>0$ given. So, it is easy to verify that $S_{\infty,\tau}$, defined in \eqref{2.8.1}, satisfies $$ S_{\infty,\tau} \leq M+1,\quad \text{for all }\tau>0. $$ Hence, from \eqref{2.8.2}, it follows the claim with $c=a_{\psi}^{p-1}/(M+1)>0$. \end{proof} \section{Proofs of main results} \begin{proof}[Proof of Theorem \ref{thm1.3}] Assume, by contradiction, that \eqref{1.1} admits one solution, say $u \in C^2(\mathbb{R}^N)$. Since $u(x)>0$ for all $x\in \mathbb{R}^N$ and $u(x) \to 0$ as $| x | \to \infty$ it follows that $u$ achieves its maximum $M>0$ in $x_0$. That is, $01$ such that $$ \eta(r) \leq \phi(r), \quad0 \leq r \leq r_0\quad \text{and}\quad \eta(r_0) = \phi(r_0). $$ Using D\'iaz and Saa's \cite{DS} inequality on $B_{r_0}(0)$ - ball centered in $0$ and radius $r_0$ -, it follows that \begin{align*} 0 & \leq \int_{B_{r_0}(0)} \big(\frac{-\Delta \phi}{\phi} + \frac{\Delta \eta}{\eta}\big) (\phi(| x |)^2 - \eta(| x |)^2) d x \\ &\leq - \int_{B_{r_0}(0)} \hat{\psi}(| x |) h( \eta(| x |)) (\phi(| x |)^2 - \eta(| x |)^2) \,dx\, . \end{align*} This is impossible, because the last term is negative. This proves the claim. On the other hand, using classical estimates (see for example Serrin and Zou \cite{SZ}), we obtain a $c>0$ constant such that $$ v(x) \geq c| x |^{2-N},~~ | x | \geq 1. $$ As a consequence of the last inequality, the prior claim and of $F^{-1}$ being increasing, we have $$ F^{-1}(c| x |^{2-N})\leq u(x) = F^{-1}(v(x)) \leq F^{-1}(d| x |^{2-N}),~~| x | \geq 1. $$ In a similar manner, we reach this conclusion, if $F_{\infty}<\infty$ holds. This completes the proof. \end{proof} \begin{proof}[Proof of Theorem \ref{thm1.5}] Consider $u = u_{\lambda} \in C^2(\mathbb{R}^N)$ the solution of \eqref{1.1} given by Theorem \ref{thm1.2}. So, from the demonstration of Theorem \ref{thm1.2}, there exists a $\tau=\tau(\lambda)>0$ such that $u$ satisfies \begin{equation} \label{4.2} \underline{u}(x) \leq \int_{u(x)}^{\tau +1} e^{G(t)} dt \leq \overline{u}(x), \quad x \in \mathbb{R}^N, \end{equation} where $\underline{u}$ and $\overline{u}$ were defined in the proof of Lemma \ref{lem3.1}. As a consequence of the definition of $\underline{u}$ and $\overline{u}$ there are $c, d$ and $R$ positive constants such that \begin{equation} \label{4.3} d| x |^{\alpha}\leq {\underline{u}(x)- \int_{1}^{\tau +1} e^{G(t)} dt}, \quad | x | \geq R, \end{equation} and \begin{equation} \label{4.4} {\overline{u}(x)- \int_{1}^{\tau +1} e^{G(t)} dt} \leq | x |^{\beta}{c}, \quad | x | \geq R. \end{equation} Hence from \eqref{4.2}, \eqref{4.3}, \eqref{4.4} and some calculations, we obtain $$ d | x |^{\alpha} \leq \int_{u(x)}^{1} e^{G(t)} dt \leq c | x |^{\beta}, ~~| x | \geq R. $$ This completes the proof of Theorem \ref{thm1.5}, remembering that $F_{0}$ is decreasing. \end{proof} \begin{thebibliography}{00} \bibitem{BAI} B. Abdellaoui, A. Dall'Aglio, I. Peral; {Some remarks on elliptic problems with critical growth in the gradient}, {\it J. Differential Equations} { 222} (2006), 21-62. \bibitem{BST} L. Boccardo, S. Segura de Le\'on, C. 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