\documentclass[reqno]{amsart} \usepackage{amsfonts,hyperref} \AtBeginDocument{{\noindent\small {\em Electronic Journal of Differential Equations}, Vol. 2005(2005), No. 139, pp. 1--15.\newline ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu \newline ftp ejde.math.txstate.edu (login: ftp)} \thanks{\copyright 2005 Texas State University - San Marcos.} \vspace{9mm}} \begin{document} \title[\hfilneg EJDE-2005/139\hfil Quasilinear elliptic problems] {Existence and uniqueness of positive solutions to a quasilinear elliptic problem in $\mathbb{R}^{N}$} \author[D. P. Covei\hfil EJDE-2005/139\hfilneg] {Dragos-Patru Covei} \date{} \thanks{Submitted June 8, 2005. Published December 5, 2005.} \thanks{Supported by grant ET 65/2005 from ANCS-MEDC } \subjclass[2000]{35J60, 35J70} \keywords{Quasilinear elliptic problem; uniqueness; existence; nonexistence; \hfill\break\indent lower-upper solutions} \begin{abstract} We prove the existence of a unique positive solution to the problem \begin{equation*} -\Delta _{p}u=a(x)f(u) \end{equation*} in $\mathbb{R}^{N}$, $N>2$. Our result extended previous works by Cirstea-Radulescu and Dinu, while the proofs are based on two theorems on bounded domains, due to Diaz-Sa\{a} and Goncalves-Santos. \end{abstract} \maketitle \numberwithin{equation}{section} \newtheorem{theorem}{Theorem}[section] \newtheorem{lemma}[theorem]{Lemma} \newtheorem{remark}[theorem]{Remark} \section{Introduction} Our purpose in this paper is to study the problem $$\begin{gathered} -\Delta _{p}u=a(x)f(u)\quad \mbox{in }\mathbb{R}^{N}\\ u>0\quad \mbox{in }\mathbb{R}^{N} \\ u(x)\to 0\quad \mbox{as }|x|\to \infty \,, \end{gathered} \label{e1}$$ where $N>2$, $\Delta _{p}u$, $(10$ in $\mathbb{R}^{N}$; \item[(A3)] For $\Phi (r)=\max_{|x|=r}a(x)$ and $p0$. Lazer and McKenna \cite{LM} studied the special case when $\Omega \subset \mathbb{R}^{N}$ $(N\geq 1)$ is a bounded domain with smooth boundary. They proved the existence and the uniqueness of a positive solution $u\in C^{2+\alpha }(\Omega )\cap C( \overline{\Omega })$ with homogeneous Dirichlet boundary condition, provided that $a(x)\in C^{\alpha }(\overline{\Omega })$ and $a(x)>0$ for all $x\in \overline{\Omega }$. The existence of entire positive solutions on $\mathbb{R}^{N}$ for $\gamma \in (0,1)$ and under certain additional hypotheses has been established by Edelson \cite{E} and Kusano-Swanson \cite{KS}. Kusano-Swanson proved that the problem \eqref{e1} has an entire positive solution in $\mathbb{R}^{2}$ with logarithmic growth at $\infty$ if $a(x)>0$, $x>0$, $a(x)\in C(0,\infty )$ and \begin{equation*} \int_{e}^{\infty }t(Logt)^{-\gamma }\big(\max_{|x|=t}a(x)\big)dt<\infty . \end{equation*} Edelson proved the existence of a solution provided that \begin{equation*} \int_{1}^{\infty }r^{N-1+\gamma (N-2)}(\max_{|x|=t} a(x))dt<\infty, \end{equation*} for some $\gamma \in (0,1)$. This result is generalized for any $\gamma >0$ via the sub- and super solutions method in Shaker \cite{S} and by other methods by Dalmasso \cite{D}. Shaker proved that problem \eqref{e1} with $p=2$ and $f(u)=u^{-\gamma }$, $\gamma >0$ has an entire positive solution $u(x)$ such that $c_{1}\leq u(x)|x|^{q|N-2|}\leq c_{2}$ for some $c_{1}$, $c_{2}$ and $00$ for $x\in \mathbb{R}^{N}\backslash \{0\}$; \item There exists $00$. Under the above conditions the authors proved the existence of a unique positive solution $u\in C_{\mathrm{loc}}^{2,\alpha }(\mathbb{R}^{N})$ vanishing at infinity to this special problem. Zhang \cite{Z}, imposed the following condition to guarantee the existence of positive solutions to problem \eqref{e1}: \begin{itemize} \item[(A4)] $f\in C^{1}((0,\infty ),(0,\infty ))$, $\lim_{s\searrow 0+}{\lim }f(s)=\infty$, and $f'(s)<0$, for all $s\in (0,\infty )$, namely, $f$ is strictly decreasing in $(0,\infty )$. \end{itemize} Under the above condition Zhang's proved that problem \eqref{e1} has a unique positive solution, $u\in C_{\mathrm{loc}}^{2+\alpha}(\mathbb{R}^{N})$, vanishing at infinity. Cirstea-Radulescu \cite{CR} and Dinu \cite{DI} extended the results of Lair, Shaker and Zhang for the case of a nonlinearity that is not necessarily decreasing on $(0,\infty )$. Our aim is to extend the results of Cirstea-Radulescu and Dinu in the sense that $10$ such that the mapping $u\mapsto f(u)/(u+\beta )^{p-1}$ is decreasing on $(0,\infty )$ \item[(F2)] $\lim_{u\searrow 0}f(u)/u^{p-1}=+\infty$ and $f$ is bounded in a neighbourhood of $+\infty$. \end{itemize} Our main results are the following: \begin{theorem} \label{thm1} Under hypotheses (F1), (F2), (A1), (A2), (A3), problem \eqref{e1} has a unique positive global solution vanishing at infinity. \end{theorem} \begin{theorem} \label{thm2} Suppose $a(r)$ is a positive radial function which is continuous on $\mathbb{R}^{N}$ and fulfills \begin{equation*} \int_0^\infty {r}^{1/(p-1)} a^{1/(p-1)} (r)dr=\infty \quad\text{if } 2\leq p<\infty \end{equation*} Then \eqref{e1} has no positive radial solution. \end{theorem} \begin{theorem} \label{thm3} Problem \eqref{e1} has no positive radial solution if $p\geq N$. \end{theorem} \begin{theorem} \label{thm4} Suppose $a(r)$ is a positive radial function which is continuous on $\mathbb{R}^{N}$ and \begin{equation*} \int_0^\infty {r}^{\frac{{(p-2)N+1}}{{p-1}}} a(r)dr=\infty \quad\mbox{if }10 \end{align*} which is a contradiction. Hence $u\leq v$. By symmetry we also have $v\leq u$, and the proof is complete. \section{Existence of a solution} We first show that our hypothesis (F1) implies $\lim_{u\searrow 0}f(u)$ exists, finite or $+\infty$. Indeed, since $\frac{f(u)}{(u+\beta )^{p-1}}$ is decreasing, there exists $L:=\lim_{u\searrow 0}\frac{f(u)}{(u+\beta )^{p-1}}\in (0,+\infty ]$. It follows that $\lim_{u\searrow 0}f(u)=L\beta ^{p-1}$. To prove the existence of a solution to Problem \eqref{e1}, we need to employ a corresponding result by Diaz-Sa\{a} \cite{DS} for bounded domains. They considered the problem $$\begin{gathered} -\Delta _{p}u=g(x,u)\quad \text{in } \Omega \\ u\geq 0\quad \text{in }\Omega \\ u(x)=0\quad \text{on }\partial \Omega , \end{gathered} \label{e6}$$ where $\Omega \subset \mathbb{R}^{N}$ is a bounded domain with smooth boundary and $g(x,u):\Omega \times [ 0,\infty )\to \mathbb{R}$. Assume that \begin{align} & \mbox{-for a.e. $x\in \Omega$ the function $u\to g(x,u)$ is continuous on $[0,\infty )$} \notag \\ & \mbox{and the function $u\to g(x,u)/u^{p-1}$ is decreasing on $(0,\infty )$;} \label{e7} \\ & \mbox{-for each $u\geq 0$ the function $x\to g(x,u)$ belongs to $L^{\infty }(\Omega )$;} \label{e8} \\ & \mbox{-there exists $C>0$ such that $g(x,u)\leq C(u^{p-1}+1)$ a.e. $x\in \Omega$, for all $u\geq 0$.} \label{e9} \end{align} Under these hypotheses on $g$, Diaz-Sa\{a} \cite{DS} proved that there is at most one solution of \eqref{e1}. Let us consider the problem \begin{gathered} -\Delta _{p}u_{k}=a(x)f(u_{k}),\quad \text{if }| x|0$such that$f(u)\leq C(u^{p-1}+1)$for all$u\geq 0$. Therefore,$a(x)f(u)\leq C(u^{p-1}+1)$for all$u\geq 0$. * Observe that$a_{0}(x)=\lim_{u\searrow 0}\frac{p(x)f(u)}{u^{p-1}} =+\infty $and$a_{\infty }(x)=\lim_{u\to +\infty }\frac{p(x)f(u)}{u^{p-1}}=0$. Thus by Diaz-Saa, problem \eqref{e10} has a unique solution$u_{k}$which, by the maximum principle, is positive in$|x|0$such that$f(u)\leq A,$for any$u\in (1,+\infty )$. Let$f_{0}:(0,1]\to (0,+\infty )$be a continuous nonincreasing function such that$f_{0}\geq f $on$(0,1]$. We can assume without loss of generality that$f_{0}(1)=A$. Set \begin{equation*} g(u)= \begin{cases} f_{0}(u), & \mbox{if } 01. \end{cases} \end{equation*} Then$g$is a continuous nonincreasing function on$(0,+\infty )$. Let$h:(0,+\infty )\to (0,+\infty )$be a$C^{1}$nonincreasing function such that$h\geq g$. Thus by in \cite[Theorem 1.3]{GS} the problem \begin{gather*} -\Delta _{p}U=p(x)h(U),\quad \mbox{if }|x| k$. Using a comparision principle argument as already done above for proving the uniqueness, we can show that $u_{k}\leq u_{k+1}$ on $\mathbb{R}^{N}$. We now justify the existence of a continuous function $v:\mathbb{R}^{N}\to \mathbb{R}$ such that $u_{k}\leq v$ in $\mathbb{R}^{N}$. We first construct a positive radially symmetric function $w$ such that $-\Delta _{p}w=\Phi (r),$ $(r=|x|)$ on $\mathbb{R}^{N}$ and $\lim_{r\to \infty }w(r)=0$. A straightforward computation shows that \begin{equation*} w(r):= K-\int_{0}^{r}\Big[ \xi ^{1-N}\int_{0}^{\xi }\sigma ^{N-1}\Phi (\sigma )d\sigma \Big] ^{1/(p-1)}d\xi , \end{equation*} where \begin{equation*} K=\int_{0}^{\infty }\Big[ \xi ^{1-N}\int_{0}^{\xi }\sigma ^{N-1}\Phi (\sigma )d\sigma \Big] ^{1/(p-1)}d\xi . \end{equation*} We first show that (A3) implies that \begin{equation*} \int_{0}^{+\infty }\Big[ \xi ^{1-N}\int_{0}^{\xi }\sigma ^{N-1}\Phi (\sigma )d\sigma \Big] ^{1/(p-1)}d\xi , \end{equation*} is finite. \begin{theorem} \label{thmE} If $j:I\subseteq R\to R$ is a locally integrable nonnegative function, then \begin{equation*} \Big(\frac{1}{b-a}\int_{a}^{b}j(x)dx\Big)^{h}\leq \mbox{ (resp. $\geq$) } \frac{1}{b-a}\int_{a}^{b}j^{h}(x)dx \end{equation*} for all $a,b\in I$, $a0,$ we have \begin{align*} & \int_{0}^{r}\xi ^{\frac{1-N}{p-1}}\Big[\frac{\xi }{\xi }\int_{0}^{\xi }\sigma ^{N-1}\Phi (\sigma )d\sigma \Big]^{1/(p-1)}d\xi \\ & =\int_{0}^{r}\xi ^{\frac{1-N}{p-1}}\xi ^{1/(p-1)}\Big[\frac{1}{\xi } \int_{0}^{\xi }\sigma ^{N-1}\Phi (\sigma )d\sigma \Big]^{1/(p-1)}d\xi \\ & \leq \int_{0}^{r}\xi ^{\frac{2-N}{p-1}}\frac{1}{\xi }\int_{0}^{\xi }\sigma ^{\frac{N-1}{p-1}}\Phi ^{1/(p-1)}(\sigma )d\sigma d\xi \\ & =\int_{0}^{r}\xi ^{\frac{2-N}{p-1}-1}\int_{0}^{\xi }\sigma ^{\frac{N-1}{p-1 }}\Phi ^{1/(p-1)}(\sigma )d\sigma d\xi \\ & =-\frac{p-1}{N-2}\int_{0}^{r}\frac{d}{d\xi }\xi ^{\frac{2-N}{p-1}% }\int_{0}^{\xi }\sigma ^{\frac{N-1}{p-1}}\Phi ^{1/(p-1)}(\sigma )d\sigma d\xi \\ & =\frac{p-1}{N-2}\Big[-r^{\frac{2-N}{p-1}}\int_{0}^{r}\sigma ^{\frac{N-1}{ p-1}}\Phi ^{1/(p-1)}(\sigma )d\sigma +\int_{0}^{r}\xi ^{1/(p-1)}\Phi ^{1/(p-1)}(\xi )d\xi \Big]. \end{align*} Now, by L'H\^{o}pital's rule, we have \begin{align*} & \lim_{r\to \infty }\Big[-r^{\frac{2-N}{p-1}}\int_{0}^{r}\sigma ^{ \frac{N-1}{p-1}}\Phi ^{1/(p-1)}(\sigma )d\sigma +\int_{0}^{r}\xi ^{1/(p-1)}\Phi ^{1/(p-1)}(\xi )d\xi \Big] \\ & =\lim_{r\to \infty }\frac{-\int_{0}^{r}\sigma ^{\frac{N-1}{p-1} }\Phi ^{1/(p-1)}(\sigma )d\sigma +r^{\frac{N-2}{p-1}}\int_{0}^{r}\xi ^{1/(p-1)}\Phi ^{1/(p-1)}(\xi )d\xi }{r^{\frac{N-2}{p-1}}} \\ & =\lim_{r\to \infty }\int_{0}^{r}\xi ^{\frac{1}{p-1}}\Phi ^{1/(p-1)}(\xi )d\xi \\ & =\int_{0}^{\infty }\xi ^{1/(p-1)}\Phi ^{1/(p-1)}(\xi )d\xi <\infty , \end{align*} \emph{Case 2:} Let $2\leq p<+\infty$, so $1\leq p-1$, it follows that $1\geq \frac{1}{p-1}>0$. Set \begin{gather*} \int_{0}^{\xi }\sigma ^{N-1}\Phi (\sigma )d\sigma \leq 1\quad \text{for }\xi >0,\quad \text{or} \\ \int_{0}^{\xi }\sigma ^{N-1}\Phi (\sigma )d\sigma >1\quad \text{for }\xi >0, \end{gather*}% In the first case \begin{equation*} \Big[\int_{0}^{\xi }\sigma ^{N-1}\Phi (\sigma )d\sigma \Big]^{1/(p-1)}\leq 1, \end{equation*}% so \begin{equation*} \int_{0}^{r}\xi ^{\frac{1-N}{p-1}}\Big[\int_{0}^{\xi }\sigma ^{N-1}\Phi (\sigma )d\sigma \Big]^{1/(p-1)}d\xi \leq \int_{0}^{r}\xi ^{\frac{1-N}{p-1}}d\xi \end{equation*} is finite as $r\to \infty$ and $N>p$. In the second case, \begin{equation*} \Big[\int_{0}^{\xi }\sigma ^{N-1}\Phi (\sigma )d\sigma \Big]^{1/(p-1)}\leq \int_{0}^{\xi }\sigma ^{N-1}\Phi (\sigma )d\sigma \end{equation*} for $\xi \geq 0$, so \begin{equation*} \int_{0}^{r}\xi ^{\frac{1-N}{p-1}}\Big[\int_{0}^{\xi }\sigma ^{N-1}\Phi (\sigma )d\sigma \Big]^{1/(p-1)}d\xi \leq \int_{0}^{r}\xi ^{\frac{1-N}{p-1} }\int_{0}^{\xi }\sigma ^{N-1}\Phi (\sigma )d\sigma d\xi \,. \end{equation*} Integration by parts gives \begin{align*} & \int_{0}^{r}\xi ^{\frac{1-N}{p-1}}\int_{0}^{\xi }\sigma ^{N-1}\Phi (\sigma )d\sigma d\xi \\ & =-\frac{p-1}{N-p}\int_{0}^{r}\frac{d}{d\xi }\xi ^{\frac{p-N}{p-1} }\int_{0}^{\xi }\sigma ^{N-1}\Phi (\sigma )d\sigma d\xi \\ & =\frac{p-1}{N-p}(-r^{\frac{p-N}{p-1}}\int_{0}^{r}\sigma ^{N-1}\Phi (\sigma )d\sigma +\int_{0}^{r}\xi ^{\frac{(p-2)N+1}{p-1}}\Phi (\xi )d\xi )\,. \end{align*} Now, by L' H\^{o}pital's rule, we have \begin{align*} & \lim_{r\to \infty }\Big[-r^{\frac{p-N}{p-1}}\int_{0}^{r}\sigma ^{N-1}\Phi (\sigma )d\sigma +\int_{0}^{r}\xi ^{\frac{(p-2)N+1}{p-1}}\Phi (\xi )d\xi \Big] \\ & =\lim_{r\to \infty }\frac{-\int_{0}^{r}\sigma ^{N-1}\Phi (\sigma )d\sigma +r^{\frac{N-p}{p-1}}\int_{0}^{r}\xi ^{\frac{(p-2)N+1}{p-1}}\Phi (\xi )d\xi }{r^{\frac{N-p}{p-1}}} \\ & =\lim_{r\to \infty }\int_{0}^{r}\xi ^{\frac{(p-2)N+1}{p-1}}\Phi (\xi )d\xi \\ & =\int_{0}^{\infty }\xi ^{\frac{(p-2)N+1}{p-1}}\Phi (\xi )d\xi <\infty , \end{align*} From cases 1 and 2 above, it follows that \begin{gather*} K=\frac{p-1}{N-2}\cdot \int_{0}^{\infty }\xi ^{\frac{1}{p-1}}\Phi ^{1/(p-1)}(\xi )d\xi \text{ if }10$, \begin{gather*} w(r)<\frac{p-1}{N-2}\cdot \int_{0}^{\infty }\xi ^{1/(p-1)}\Phi ^{1/(p-1)}(\xi )d\xi \quad \text{if }10$. Note that the hypothesis $u\to f(u)/(u+\beta )^{p-1}$ is a decreasing function on $(0,\infty )$ implies that $u\to f(u)/u^{p-1}$ is a decreasing function on $(0,\infty )$, because $\frac{v+\beta }{v}\leq \frac{u+\beta }{u} \Leftrightarrow$ $vu+\beta u\leq vu+v\beta \Leftrightarrow$ $\beta (u-v)\leq 0$, is true $\forall u\leq v$ and $\beta >0$. We have \begin{itemize} \item[(F1')] $\overline{f}(u)\geq f(u)^{1/(p-1)}$ \item[(F2')] $\lim_{u\searrow 0}\overline{f}(u)/u=\infty$ and $u\mapsto \overline{f}(u))/u^{p-1}$ is decreasing on $(0,\infty )$. \end{itemize} Let $v$ be a positive function such that $w(r)=\frac{1}{C} \int_{0}^{v(r)}t^{p-1}/\overline{f}(t)\, dt$, where $C$ is a positive constant such that $KC\leq \int_{0}^{C^{1/(p-1)}} t^{p-1}/\overline{f} (t)\,dt$. We prove that we can find $C>0$ with this property. From our hypothesis (F2') we obtain that $\lim_{x\to +\infty }\int_{0}^{x}t^{p-1}/\overline{f}(t)\,dt=+\infty$. Now using L'H\^{o}pital's rule we have \begin{equation*} \lim_{x\to \infty }\frac{1}{x^{p-1}} \int_{0}^{x}\frac{t^{p-1}}{\overline{f} (t)}dt =\lim_{x\to \infty }\frac{x}{(p-1)\overline{f}(x)}=+\infty . \end{equation*} This means that there exists $x_{1}>0$ such that $\int_{0}^{x} t^{p-1}/ \overline{f}(t)\,dt\geq Kx^{p-1}$, for all $x\geq x_{1}$. It follows that for any $C\geq x_{1}$, \begin{equation*} KC\leq \int_{0}^{C^{1/(p-1)}}\frac{t^{p-1}}{\overline{f}(t)}dt. \end{equation*} But $w$ is a decreasing function, and this implies that $v$ is a decreasing function too. Then \begin{equation*} \int_{0}^{v(r)}\frac{t^{p-1}}{\overline{f}(t)}dt \leq \int_{0}^{v(0)}\frac{t^{p-1}}{\overline{f}(t)}dt =C\cdot w(0) =C\cdot K\leq \int_{0}^{C^{1/(p-1)}} \frac{t^{p-1}}{\overline{f}(t)}dt. \end{equation*} It follows that $v(r)\leq C^{1/(p-1)}$ for all $r>0$. From $w(r)\to 0$ as $r\to +\infty$ we deduce $v(r)\to 0$ as $r\to +\infty$. By the choice of $v$ we have \begin{equation*} \nabla w=\frac{1}{C}\cdot \frac{v^{p-1}}{\overline{f}(v)}\nabla v \quad \mbox{and}\quad \Delta w=\frac{1}{C}\cdot \frac{v^{p-1}}{\overline{f}(v)} \Delta v +\frac{1}{C}\big(\frac{v^{p-1}}{\overline{f}(v)}\big)% '|\nabla v|^{2}. \end{equation*} so \begin{equation*} |\nabla w|^{p-2}=\frac{1}{C^{p-2}} \big(\frac{v^{p-1}}{\overline{f}(v)}\big) ^{p-2}|\nabla v|^{p-2}\,. \end{equation*} It follows that \begin{align*} |\nabla w|^{p-2}\Delta w &=\frac{1}{C^{p-2}}(\frac{v^{p-1} }{\overline{f}(v)} )^{p-2}|\nabla v|^{p-2} \Big(\frac{1}{C}\frac{v^{p-1}}{\overline{f}(v)} \Delta v+\frac{1}{C}(\frac{v^{p-1}}{\overline{f}(v)})'|\nabla v|^{2} \Big) \\ & =\frac{1}{C^{p-1}}(\frac{v^{p-1}}{\overline{f}(v)}) ^{p-1}| \nabla v|^{p-2}\Delta v+\frac{1}{C^{p-1}} (\frac{v^{p-1}}{\overline{f}(v)}) ^{p-2}( \frac{v^{p-1}}{\overline{f}(v)}) '|\nabla v|^{p}, \end{align*} so \begin{align*} &\nabla (|\nabla w|^{p-2})\cdot \nabla w \\ &=\Big\{\frac{1}{C^{p-2}}(\frac{v^{p-1}}{\overline{f}(v)}) ^{p-2}\nabla (|\nabla v|^{p-2})+\frac{1}{C^{p-2}} \Big[ (\frac{v^{p-1}}{\overline{f}(v)} )^{p-2}\Big] '|\nabla v|^{p-2} \nabla v\Big\} \cdot [ \frac{1}{C} \cdot \frac{v^{p-1}}{\overline{f}(v)}\nabla v] \\ & =\frac{1}{C^{p-1}}(\frac{v^{p-1}}{\overline{f}(v)}) ^{p-2}\nabla (|\nabla v|^{p-2})\cdot \nabla v+\frac{1}{C^{p-1}} \Big[ (\frac{v^{p-1}}{\overline{f} (v)})^{p-2}\Big]'\frac{v^{p-1}}{\overline{f}(v)}|\nabla v|^{p}, \end{align*} so that \begin{aligned} \Delta _{p}w &=\frac{1}{C^{p-1}}\big(\frac{v^{p-1}}{\overline{f}(v)}\big) ^{p-1}\Big[ |\nabla v|^{p-2}\Delta v+\nabla ( | \nabla v|^{p-2})\cdot \nabla v\Big] \\ &\quad + \frac{1}{C^{p-1}}\big(\frac{v^{p-1}}{\overline{f}(v)}\big) ^{p-2}| \nabla v|^{p}( \frac{v^{p-1}}{\overline{f}(v)}) '+\frac{1}{C^{p-1}} \Big[ \big(\frac{v^{p-1}}{\overline{f}(v)}\big)^{p-2}\Big] ' \frac{v^{p-1}}{\overline{f}(v)}|\nabla v|^{p}\\ &=\frac{1}{C^{p-1}}(\frac{v^{p-1}}{\overline{f}(v)}) ^{p-1}\Delta _{p}v+(p-1)\frac{1}{C^{p-1}}|\nabla v|^{p}( \frac{v^{p-1}}{\overline{f}(v)})^{p-2}(\frac{v^{p-1}}{\overline{f}(v)})'. \end{aligned} \label{e11} From \eqref{e11} we deduce that $$\Delta _{p}w=\frac{1}{C^{p-1}}(\frac{v^{p-1}}{\overline{f}(v)}) ^{p-1}\Delta _{p}v+(p-1)\frac{1}{C^{p-1}}|\nabla v| ^{p}(\frac{v^{p-1}}{\overline{f}(v)} )^{p-2}(\frac{v^{p-1}}{\overline{f}(v)})'. \label{e12}$$ From \eqref{e12} and the fact that $u\to \frac{\overline{f}(u)}{u^{p-1}}$ is a decreasing function on $(0,+\infty)$, we deduce that $$\Delta _{p}v\leq C^{p-1}\big(\frac{\overline{f}(v)}{v^{p-1}}\big) % ^{p-1}\Delta _{p}w =-C^{p-1}\big(\frac{\overline{f}(v)}{v^{p-1}}\big)^{p-1} \Phi (r)\leq -f(v)\Phi (r). \label{e13}$$ By \eqref{e12} and \eqref{e13} and using in an essential manner the hypothesis (F1), as already done for proving the uniqueness, we obtain that $u_{k}\leq v$ for $|x|\leq k$ and, hence, for all $\mathbb{R}^{N}$. Now we have a bounded increasing sequence $u_{1}\leq u_{2}\leq \dots \leq u_{k}\leq dots \leq v$ with $v$ vanishing at infinity. Thus there exists a function, say $u\leq v$ such that $u_{k}\to u$ pointwise in $\mathbb{R}^{N}$. Using \begin{gather*} u'(r)=\Big[ r^{1-N}\int_{0}^{r}\sigma ^{N-1}p(\sigma )f(u(\sigma ))d\sigma \Big] ^{1/(p-1)}, \\ u^{\prime\prime}(r)=-\frac{p(r)f(u(r))+(1-N)r^{-N}\int_{0}^{r}\sigma ^{N-1}p(\sigma )f(u(\sigma ))d\sigma }{p-1} \\ \times \Big[ r^{1-N}\int_{0}^{r}\sigma ^{N-1}p(\sigma )f(u(\sigma ))d\sigma \Big] ^{\frac{2-p}{p-1}}, \\ \frac{2-p}{p-1}\geq 0\;\Longleftrightarrow\; 10$for all$r>0$. Then we have \begin{equation*} \Delta _{p}\overline{u}(r)=\frac{\Delta _{p}u(r)}{(u(r)+1)^{p-1}}-(p-1) \frac{|\nabla u(r)|^{p}}{(u(r)+1)^{p}}. \end{equation*} Then$\overline{u}(r)$satisfies $$\frac{1}{r^{N-1}}\Big(r^{N-1}(-\overline{u}'(r))^{p-2}\overline{u} '(r)\Big)'+(p-1)\frac{|\nabla u(r)|^{p}}{(u(r)+1)^{p}}=- \frac{f(u(r))a(r)}{(u(r)+1)^{p-1}}\,. \label{e14}$$ Multiplying \eqref{e14} by$r^{N-1}$and integrating on$(0,\xi )yield \begin{align*} & \int_{0}^{\xi }\Big((-\overline{u}'(\sigma ))^{p-1}\sigma ^{N-1} \Big)'d\sigma -(p-1)\int_{0}^{\xi }\frac{\sigma ^{N-1}|\nabla u(\sigma )|^{p}}{(u(\sigma )+1)^{p}}d\sigma \\ & =\int_{0}^{\xi }\frac{f(u(\sigma ))a(\sigma )\sigma ^{N-1}}{(u(\sigma )+1)^{p-1}}d\sigma , \end{align*} equivalently $$(-\overline{u}'(\xi ))^{p-1}\xi ^{N-1}-\int_{0}^{\xi }(p-1)\frac{ \sigma ^{N-1}|\nabla u(\sigma )|^{p}}{(u(\sigma )+1)^{p}}d\sigma =\int_{0}^{\xi }\frac{f(u(\sigma ))a(\sigma )\sigma ^{N-1}}{(u(\sigma )+1)^{p-1}}d\sigma \,. \label{e15}$$ Multiplying equation \eqref{e15} by\xi ^{1-N}$, we deduce $$(-\overline{u}'(\xi ))^{p-1}-\xi ^{1-N}(p-1)\int_{0}^{\xi }\frac{ \sigma ^{N-1}|\nabla u(\sigma )|^{p}}{(u(\sigma )+1)^{p}}d\sigma =\xi ^{1-N}\int_{0}^{\xi }\frac{f(u(\sigma ))a(\sigma )\sigma ^{N-1}}{(u(\sigma )+1)^{p-1}}d\sigma . \label{e16}$$ From \eqref{e16}, we have \begin{equation*} \big(-\overline{u}'(\xi )\big)^{p-1}\geq \xi ^{1-N}\int_{0}^{\xi } \frac{f(u(\sigma ))a(\sigma )\sigma ^{N-1}}{(u(\sigma )+1)^{p-1}}d\sigma , \end{equation*} so $$-\overline{u}'(\xi )\geq \xi ^{\frac{^{1-N}}{p-1}}\Big[ \int_{0}^{\xi }\frac{f(u(\sigma ))a(\sigma )\sigma ^{N-1}}{(u(\sigma )+1)^{p-1}}d\sigma \Big]^{1/(p-1)}, \label{e17}$$ integrating \eqref{e17} on$(0,r)$, we have \begin{equation*} \overline{u}(0)-\overline{u}(r)\geq \int_{0}^{r}\xi ^{\frac{^{1-N}}{p-1}} \Big[\int_{0}^{\xi }\frac{f(u(\sigma ))a(\sigma )\sigma ^{N-1}}{(u(\sigma )+1)^{p-1}}d\sigma \Big]^{1/(p-1)}d\xi . \end{equation*} We observe that$\overline{u}(r)<\overline{u}(0),$for all$r>0$implies$u(r)0$. If$\beta \geq 1$, then the function$u\mapsto \frac{f(u)}{(u+\beta )^{p-1}}$is decreasing on$(0,+\infty )$. This implies $$\frac{f(u(\sigma ))}{(u(\sigma )+1)^{p-1}}>\frac{f(u(0))}{(u(0)+1)^{p-1}}. \label{e18}$$ Since$\overline{u}$is positive, we have \begin{equation*} \int_{0}^{r}\xi ^{\frac{^{1-N}}{p-1}}\Big[\int_{0}^{\xi }\frac{f(u(\sigma ))a(\sigma )\sigma ^{N-1}}{(u(\sigma )+1)^{p-1}}d\sigma \Big]^{1/(p-1)}d\xi \leq \overline{u}(0),\quad \forall r>0, \end{equation*} substituting \eqref{e18} into this expression, we obtain \begin{equation*} \int_{0}^{r}\xi ^{\frac{^{1-N}}{p-1}}\Big[\int_{0}^{\xi }a(\sigma )\sigma ^{N-1}d\sigma \Big]^{1/(p-1)}d\xi \leq \frac{u(0)+1}{f(u(0))^{\frac{1}{p-1}}} \overline{u}(0)<\infty . \end{equation*} Let$2\leq p<+\infty $, so$1\leq p-1$, follows that$1\geq \frac{1}{p-1}>0. We have \begin{align*} & \int_{0}^{r}\xi ^{\frac{1-N}{p-1}}\Big[\frac{\xi }{\xi }\int_{0}^{\xi }\sigma ^{N-1}a(\sigma )d\sigma \Big]^{1/(p-1)}d\xi \\ & =\int_{0}^{r}\xi ^{\frac{1-N}{p-1}}\xi ^{1/(p-1)}\Big[\frac{1}{\xi } \int_{0}^{\xi }\sigma ^{N-1}a(\sigma )d\sigma \Big]^{1/(p-1)}d\xi \\ & \geq \int_{0}^{r}\xi ^{\frac{2-N}{p-1}}\frac{1}{\xi }\int_{0}^{\xi }\sigma ^{\frac{N-1}{p-1}}a^{1/(p-1)}(\sigma )d\sigma d\xi \\ & =\int_{0}^{r}\xi ^{\frac{2-N}{p-1}-1}\int_{0}^{\xi }\sigma ^{\frac{N-1}{p-1 }}a^{1/(p-1)}(\sigma )d\sigma d\xi \\ & =-\frac{p-1}{N-2}\int_{0}^{r}\frac{d}{d\xi }\xi ^{\frac{2-N}{p-1} }\int_{0}^{\xi }\sigma ^{\frac{N-1}{p-1}}a^{1/(p-1)}(\sigma )d\sigma d\xi \\ & =\frac{p-1}{N-2}(-r^{\frac{2-N}{p-1}}\int_{0}^{r}\sigma ^{\frac{N-1}{p-1} }a^{1/(p-1)}(\sigma )d\sigma +\int_{0}^{r}\xi ^{1/(p-1)}a(\xi )^{1/(p-1)}d\xi ) \\ & \geq \frac{p-1}{N-2}\frac{1}{r^{\frac{N-2}{p-1}}}\int_{0}^{r}\Big[r^{\frac{ N-2}{p-1}}-(t)^{\frac{N-2}{p-1}}\Big]t^{1/(p-1)}a^{1/(p-1)}(t)dt \\ & \geq \frac{p-1}{N-2}\frac{1}{r^{\frac{N-2}{p-1}}}\Big(r^{\frac{N-2}{p-1}}-( \frac{r}{2})^{\frac{N-2}{p-1}}\Big)\int_{0}^{r/2}t^{1/(p-1)}a^{1/(p-1)}(t)dt \\ & =\frac{p-1}{N-2}(1-(\frac{1}{2})^{\frac{N-2}{p-1}})% \int_{0}^{r/2}t^{1/(p-1)}a^{1/(p-1)}(t)dt\to \infty \quad \text{as } r\to \infty . \end{align*} So \begin{equation*} \infty >\frac{u(0)+1}{f(u(0))^{1/(p-1)}}\overline{u}(0)\geq \infty , \end{equation*} which is a contradiction. If\beta <1$then the function$u\mapsto \frac{(u+\beta )^{p-1}}{(u+1)^{p-1}}$is increasing on$(0,+\infty ). In this case \begin{align*} \overline{u}(0)&\geq \int_{0}^{r}\xi ^{\frac{^{1-N}}{p-1}} \Big[ \int_{0}^{\xi }\frac{f(u(\sigma ))a(\sigma )\sigma ^{N-1}}{(u(\sigma )+1)^{p-1}}d\sigma \Big] ^{1/(p-1)}d\xi \\ &= \int_{0}^{r}\xi ^{\frac{^{1-N}}{p-1}}\Big[ \int_{0}^{\xi }\frac{ f(u(\sigma ))a(\sigma )(u(\sigma )+\beta )^{p-1}\sigma ^{N-1}}{(u(\sigma )+\beta )^{p-1}(u(\sigma )+1)^{p-1}}d\sigma \Big] ^{1/(p-1)}d\xi \\ &\geq \frac{f(u(0))^{1/(p-1)}}{u(0)+\beta }\beta \int_{0}^{r} \xi ^{\frac{1-N }{p-1}} \Big[ \int_{0}^{\xi }\sigma ^{N-1}a(\sigma )d\sigma \Big] ^{1/(p-1)} d\xi , \end{align*} which implies \begin{equation*} \infty >\frac{u(0)+\beta }{f(u(0))^{1/(p-1)}\beta }\overline{u}(0)\geq \int_{0}^{r}\xi ^{\frac{1-N}{p-1}}\Big[ \int_{0}^{\xi }\sigma ^{N-1}a(\sigma )d\sigma \Big] ^{1/(p-1)}d\xi \geq \infty , \end{equation*} which is a contradiction. \section{Proof of Theorem \protect\ref{thm3}} Assumeu$is positive for$r>0$and satisfies \begin{equation*} (r^{N-1}|u'(r)|^{p-2}u'(r))'=-r^{N-1}f(u(r))a(r). \end{equation*} Since$f(u(r))a(r)$is positive for$r>0$, follows that \begin{equation*} (r^{N-1}|u'(r)|^{p-2}u'(r))'<0,\quad \text{for }r>0, \end{equation*} and that$r^{N-1}|u'(r)|^{p-2}u'(r)$is a decreasing function. Because this function is decreasing and$u'<0$, \begin{equation*} r^{N-1}|u'(r)|^{p-2}u'(r)\leq -C,\quad \text{for } r\geq R, \end{equation*} where$C$is positive constant. As a consequence \begin{equation*} -u'(r)\geq C_{1}r^{-\frac{1-N}{p-1}},\quad \text{with }C_{1}>0. \end{equation*} Integrating this inequality from$R$to$r$we have \begin{equation*} u(R)-u(r)\geq C_{1}\int_{R}^{r}r^{-\frac{1-N}{p-1}}dr, \quad \text{for } r\geq R. \end{equation*} Letting$r\to \infty $, we arrive at a contradiction. \section{Proof of Theorem \protect\ref{thm4}} As in proof of Theorem \ref{thm2}, we have \begin{equation*} \overline{u}(0)-\overline{u}(r)\geq \int_{0}^{r} \xi ^{\frac{^{1-N}}{p-1}} \Big[ \int_{0}^{\xi } \frac{f(u(\sigma ))a(\sigma )\sigma ^{N-1}}{(u(\sigma )+1)^{p-1}}d\sigma \Big] ^{1/(p-1)}d\xi , \end{equation*} We observe that$\overline{u}(r)<\overline{u}(0)$, for all$r>0$implies$u(r)0$. If$\beta \geq 1$then the function$u\mapsto \frac{f(u)}{(u+\beta )^{p-1}}$is decreasing on$(0,+\infty )$. This implies $$\frac{f(u(\sigma ))}{(u(\sigma )+1)^{p-1}}> \frac{f(u(0))}{(u(0)+1)^{p-1}}, \label{e19}$$ Since$\overline{u}$is positive, we have \begin{equation*} \int_{0}^{r}\xi ^{\frac{^{1-N}}{p-1}}\Big[ \int_{0}^{\xi }\frac{f(u(\sigma ))a(\sigma )\sigma ^{N-1}}{(u(\sigma )+1)^{p-1}}d\sigma \Big] ^{1/(p-1)}d\xi \leq \overline{u}(0),\quad \forall r>0 \end{equation*} substituting \ref{e19} into this expression we obtain \begin{equation*} \int_{0}^{r}\xi ^{\frac{^{1-N}}{p-1}}\Big[ \int_{0}^{\xi }a(\sigma )\sigma ^{N-1}d\sigma \Big] ^{1/(p-1)}d\xi \leq \frac{u(0)+1}{f(u(0))^{1/(p-1)}} \overline{u}(0)<\infty . \end{equation*} Let$10, \quad\text{or} \\ \int_{0}^{\xi }r^{N-1}a(r)dr\geq 1\quad \text{for }\xi >0, \end{gather*} In the second case, we have \begin{equation*} \Big[ \int_{0}^{\xi }\sigma ^{N-1}a(\sigma )d\sigma \Big]^{1/(p-1)} \geq \int_{0}^{\xi }\sigma ^{N-1}a(\sigma )d\sigma , \end{equation*} so \begin{equation*} \int_{0}^{r}\xi ^{\frac{1-N}{p-1}}\Big[ \int_{0}^{\xi }\sigma ^{N-1}a(\sigma )d\sigma \Big] ^{1/(p-1)}d\xi \geq \int_{0}^{r}\xi ^{\frac{1-N}{p-1} }\int_{0}^{\xi }\sigma ^{N-1} a(\sigma )d\sigma d\xi \,. \end{equation*} Integration by parts gives \begin{align*} &\int_{0}^{r}\xi ^{\frac{1-N}{p-1}}\int_{0}^{\xi }\sigma ^{N-1}a(\sigma )d\sigma d\xi \\ &=-\frac{p-1}{N-p}\int_{0}^{r}\frac{d}{d\xi }\xi ^{\frac{p-N}{p-1} }\int_{0}^{\xi }\sigma ^{N-1}a(\sigma )d\sigma d\xi \\ &= \frac{p-1}{N-p}(-r^{\frac{p-N}{p-1}}\int_{0}^{r}\sigma ^{N-1}a(\sigma )d\sigma +\int_{0}^{r}\xi ^{\frac{(p-2)N+1}{p-1}}a(\xi )d\xi ) \\ &\geq \frac{p-1}{N-p}\frac{1}{r^{\frac{N-p}{p-1}}}\int_{0}^{r} \Big[ r^{\frac{N-p}{p-1}}-(t)^{\frac{N-p}{p-1}}\Big] t^{\frac{(p-2)N+1}{p-1}}p(t)dt \\ &\geq \frac{p-1}{N-p}\frac{1}{r^{\frac{N-p}{p-1}}} (r^{\frac{N-p}{p-1}} -(\frac{r}{2})^{\frac{N-p}{p-1}}) \int_{0}^{r/2}t^{\frac{(p-2)N+1}{p-1}}a(t)dt \\ &= \frac{p-1}{N-p}(1-(\frac{1}{2})^{\frac{N-p}{p-1}}) \int_{0}^{r/2}t^{\frac{(p-2)N+1}{p-1}}a(t)dt \\ &=\infty \quad \text{as }r\to \infty . \end{align*} Then \begin{equation*} \infty >\frac{u(0)+1}{f(u(0))^{1/(p-1)}}\overline{u}(0)\geq \infty , \end{equation*} which is a contradiction. If $\beta <1$ we have $\frac{u+\beta }{u+1}>\beta \Longleftrightarrow u+\beta >\beta u+\beta \Longleftrightarrow (1-\beta )u>0$ is true. In this case we have \begin{align*} \overline{u}(0) &\geq \int_{0}^{r}\xi ^{\frac{^{1-N}}{p-1}} \Big[ % \int_{0}^{\xi }\frac{f(u(\sigma ))a(\sigma )\sigma ^{N-1}}{(u(\sigma )+1)^{p-1}}d\sigma \Big] ^{1/(p-1)}d\xi \\ &=\int_{0}^{r}\xi ^{\frac{^{1-N}}{p-1}}\Big[ \int_{0}^{\xi }\frac{f(u(\sigma ))a(\sigma )(u(\sigma )+\beta )^{p-1}\sigma ^{N-1}}{(u(\sigma )+\beta )^{p-1}(u(\sigma )+1)^{p-1}}d\sigma \Big] ^{1/(p-1)}d\xi \\ &\geq \frac{f(u(0))^{1/(p-1)}}{u(0)+\beta }\beta \int_{0}^{r} \xi ^{\frac{1-N}{p-1}} \Big[ \int_{0}^{\xi }\sigma ^{N-1}a(\sigma )d\sigma \Big] ^{1/(p-1)}d\xi , \end{align*} which implies \begin{equation*} \infty >\frac{u(0)+\beta }{f(u(0))^{1/(p-1)}\beta }\overline{u}(0)\geq \int_{0}^{r}\xi ^{\frac{1-N}{p-1}}\Big[ \int_{0}^{\xi }\sigma ^{N-1}a(\sigma )d\sigma \Big] ^{1/(p-1)}d\xi \geq \infty , \end{equation*} which is a contradiction. In the first case we observe that we can not have $\int_{0}^{\xi }r^{\frac{(p-2)N+1}{p-1}}a(r)dr=\infty$ because \begin{align*} \int_{0}^{r}\xi ^{\frac{1-N}{p-1}}d\xi & >\int_{0}^{r}\xi ^{\frac{1-N}{p-1}} \int_{0}^{\xi }\sigma ^{N-1}a(\sigma )d\sigma d\xi \\ &\geq \frac{p-1}{N-p}(1-(\frac{1}{2})^{\frac{N-p}{p-1}}) \int_{0}^{r/2}t^{ \frac{(p-2)N+1}{p-1}}a(t)dt \to \infty \quad \text{as }r\to \infty \end{align*} which is a contradiction. \begin{remark} \label{rmk1} \rm Let $2\leq p<+\infty$. Then $1\geq \frac{1}{p-1}>0$. From the above proofs we observe if that \begin{equation*} \Big(\int_{0}^{\xi }\sigma ^{N-1}a(\sigma )d\sigma \Big)^{1/(p-1)}\leq 1\, \end{equation*} then \begin{align*} &\frac{p-1}{N-2}(1-(\frac{1}{2})^{\frac{N-2}{p-1}}) \int_{0}^{r/2}t^{1/(p-1)}a^{1/(p-1)}(t)dt\\ &\leq \int_{0}^{r}\xi ^{\frac{1-N}{p-1}}\Big[ \int_{0}^{\xi }\sigma^{N-1}a(\sigma )d\sigma \Big] ^{1/(p-1)}d\xi \\ &\leq \int_{0}^{r}\xi ^{\frac{1-N}{p-1}}d\xi . \end{align*} As $r\to \infty$, we have $\int_{0}^{\infty }t^{\frac{1}{p-1}}a^{1/(p-1)}(t)dt\neq \infty$. \end{remark} On the other hand, if \begin{equation*} \Big(\int_{0}^{\xi }\sigma ^{N-1}a(\sigma )d\sigma \Big)^{1/(p-1)}>1, \end{equation*} then \begin{align*} &\frac{p-1}{N-2}(1-(\frac{1}{2})^{\frac{N-2}{p-1}}) \int_{0}^{r/2}t^{1/(p-1)}a^{1/(p-1)}(t)dt \\ &\leq \int_{0}^{r}\xi ^{\frac{1-N}{p-1}}\Big[ \int_{0}^{\xi }\sigma ^{N-1}a(\sigma )d\sigma \Big] ^{1/(p-1)}d\xi \\ &\leq \frac{p-1}{N-p}\int_{0}^{r/2}t^{\frac{(p-2)N+1}{p-1}}a(t)dt. \end{align*} Then if $\int_{0}^{\infty}t^{1/(p-1)}a^{1/(p-1)}(t)dt=\infty$ we have $\int_{0}^{\infty }t^{\frac{(p-2)N+1}{p-1}}a(t)dt=\infty$. \begin{remark} \label{rmk2} \rm Let$10$ and then showing the convergence of $u_{\varepsilon }$ as $\varepsilon \to +0$ to a solution $u$. It is clear that the problems \eqref{er} has a unique solution which is due to Diaz-Sa\{a}. In the next steps we established some properties for such solution. For this, let $\varepsilon :=\varepsilon _{n}$ be a decreasing sequence converging to $0$ and set $u_{n}:=u_{\varepsilon _{n}}$ with $n>k\geq 1$ in \eqref{er}. By \cite{CD} we see that $u_{n}\geq c_{0,B_{k}}\varphi _{1,B_{k}}$ and there exists some function $u_{k}\in C(\overline{B}_{k})$ such that \begin{itemize} \item[((i)] $u_{n}\to u_{k}$ a.e. in $B_{k}$ as $n\to \infty$, \item[(ii)] $u_{k}\geq c_{0,k}\varphi _{1}$ a.e. in $B_{k}$, \end{itemize} where $\varphi _{1}:=\varphi _{1,B_{k}}$ is the first eigenfunction for the eigenvalue $\lambda _{1}$ of $(-\Delta _{p})$ in $W_{0}^{1,p}(B_{k})$ and $B_{k}:=\{x\in \mathbb{R}^{N}:| x| \}$. Moreover using Diaz-Sa\`{a}'s comparison lemma we have a sequence $\{u_{k}\}$ (which is $0$ for $|x|>k$), as in the present paper, such that \begin{equation*} u_{1}\leq u_{2}\leq \dots \leq u_{k}\leq \dots \leq v\quad \text{in }\mathbb{R}^{N}, \end{equation*} where $v$ is the same function as above and so the existence of solution $u$ to the problem \eqref{e1} is proved. \end{proof} This alternative proof is treated more generally in \cite{CD}. With this alternative proof we observe that it is sufficient to apply the rest the reference \cite{DS}. This technique is inspired by \cite{CRT} and by another results due to Goncalves and Santos, which are treated more generally in \cite{CD}. \begin{thebibliography}{9} \bibitem[15]{CD} D. P. Covei, Existence and asymptotic behavior of positive solution to a quasilinear elliptic problem in $R^{N}$, Nonlinear Analysis: TMA, DOI 10.1016/j.na.2007.08.039. \bibitem[16]{DB} E. DiBenedetto, $C^{1,\alpha }$- local regularity of weak solutions of degenerate elliptic equations, Nonlinear Anal. 7 (1983), 827-850. \end{thebibliography} \end{document} \end{document}