\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{equation}
\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}
\end{equation}
where $N>2$, $\Delta _{p}u$, $(1<p<\infty )$ is the $p$-Laplacian operator
and the function $a(x)$ satisfies the following hypotheses:

\begin{itemize}
\item[(A1)] $a(x)\in C_{\mathrm{loc}}^{0,\alpha }(\mathbb{R}^{N})$ for some 
$\alpha \in (0,1)$;

\item[(A2)] $a(x)>0$ in $\mathbb{R}^{N}$;

\item[(A3)] For $\Phi (r)=\max_{|x|=r}a(x)$ and $p<N$, 
\begin{gather*}
\int_{0}^{\infty }r^{1/(p-1)}\Phi ^{1/(p-1)}(r)dr<\infty \quad \mbox{if }%
1<p\leq 2 \\
\int_{0}^{\infty }r^{\frac{(p-2)N+1}{p-1}}\Phi (r)dr<\infty \quad \mbox{if }%
2\leq p<\infty \,.
\end{gather*}
\end{itemize}

This problem has been studied extensively in the case $p=2$ and 
$f(u)=u^{-\gamma }$, with $\gamma >0$. 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 $0<q<1$ as 
$x\to\infty $ if

\begin{enumerate}
\item $a(x)\in C_{\mathrm{loc}}^{\alpha }(\mathbb{R}^{N})$, $a(x)>0$ for 
$x\in \mathbb{R}^{N}\backslash \{0\}$;

\item There exists $0<c<1$ such that $c\Phi (|x|)\leq a(x)\leq \Phi (|x|)$
where $\Phi (r):=\max_{|x|=r}a(x)$ , $r\in \lbrack 0,\infty )$;

\item $\int_{1}^{\infty }r^{N-1+\gamma (N-2)}(\max_{|x|=t}a(x))dt<\infty $.
\end{enumerate}

Lair and Shaker continued in \cite{LS} the study of \eqref{e1} for $p=2$ and 
$f(u)=u^{-\gamma }$, $\gamma >0$. 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 $1<p<\infty $. More exactly, let $f:(0,\infty )\to (0,\infty )$ be a 
$C^{1}$ function that satisfies the following assumptions:

\begin{itemize}
\item[(F1)] There exists $\beta >0$ 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 }1<p\leq 2.
\end{equation*}
Then  \eqref{e1} has no positive radial solution.
\end{theorem}

\section{Uniqueness}

Suppose $u$ and $v$ are arbitrary solutions of problem \eqref{e1}. Let us
show that $u\leq v$ or, equivalently, $\ln (u(x)+\beta )\leq \ln (v(x)+\beta
)$, for any $x\in \mathbb{R}^{N}$. Assume the contrary. Since, we have 
\begin{equation*}
\lim_{|x|\to \infty }(\ln (u(x)+\beta )-\ln (v(x)+\beta))=0,
\end{equation*}
we deduce that 
\begin{equation*}
\max_{\mathbb{R}^{N}}(\ln (u(x)+\beta )-\ln (v(x)+\beta ))
\end{equation*}
exists and is positive. At that point, say $x_{0}$, we have 
\begin{equation*}
\nabla (\ln (u(x_{0})+\beta )-\ln (v(x_{0})+\beta ))=0,
\end{equation*}
so 
\begin{equation*}
\frac{1}{u(x_{0})+\beta }\cdot \nabla u(x_{0}) 
=\frac{1}{v(x_{0})+\beta }\cdot \nabla v(x_{0}),
\end{equation*}
and 
\begin{equation}
\frac{1}{(u(x_{0})+\beta )^{p-2}}\cdot |\nabla u(x_{0})| ^{p-2}
=\frac{1}{(v(x_{0})+\beta ) ^{p-2}}\cdot |\nabla v(x_{0})|^{p-2}.  \label{e2}
\end{equation}
By (f1) we obtain 
\begin{equation}
\frac{f(u(x_{0}))}{(u(x_{0})+\beta )^{p-1}}<\frac{f(v(x_{0}))}{
(v(x_{0})+\beta )^{p-1}}.  \label{e3}
\end{equation}
Since $0\geq \Delta (\ln (u(x_{0})+\beta )-\ln (v(x_{0})+\beta ))$, it
follows that 
\begin{equation*}
\frac{\Delta u(x_{0})}{u(x_{0})+\beta }\leq \frac{\Delta v(x_{0})}{
v(x_{0})+\beta },
\end{equation*}
so 
\begin{equation}
\frac{1}{(u(x_{0})+\beta )^{p-1}}\cdot |\nabla u(x_{0})|^{p-2}\Delta
u(x_{0}) \leq \frac{1}{(v(x_{0})+\beta )^{p-1}}\cdot |\nabla v(x_{0})|
^{p-2} \Delta v(x_{0})\,.  \label{e4}
\end{equation}
Since 
\begin{equation*}
|\nabla \ln (u(x_{0})+\beta )|^{p-2}=\frac{1}{(u(x_{0})+\beta )^{p-2}} \cdot
|\nabla u(x_{0})|^{p-2},
\end{equation*}
it follows that 
\begin{align*}
&\nabla (|\nabla \ln (u(x_{0})+\beta )| ^{p-2}) \\
&= -(p-2)\frac{|\nabla u(x_{0})|^{p-2}(u(x_{0})+\beta )^{p-3}}{%
(u(x_{0})+\beta )^{2(p-2)}}\cdot \nabla u(x_{0}) +\frac{\nabla (|\nabla
u(x_{0})|^{p-2})}{(u(x_{0}) +\beta)^{p-2}}.
\end{align*}
Then 
\begin{equation}  \label{e5}
\begin{aligned} &\nabla (|\nabla \ln (u(x_{0})+\beta )|^{p-2}) \cdot \nabla
(\ln (u(x_{0})+\beta ))\\ &=-(p-2)\frac{|\nabla u(x_{0})|^{p-2}|\nabla
u(x_{0})| ^{2}}{(u(x_{0}) +\beta)^{p}}+\frac{\nabla (| \nabla
u(x_{0})|^{p-2})\cdot \nabla u(x_{0})}{(u(x_{0})+\beta )^{p-1}} 
\end{aligned}
\end{equation}
and 
\begin{equation*}
|\nabla \ln (u(x_{0})+\beta )|^{p-2}\Delta (\ln (u(x_{0})+\beta )) 
= \frac{|\nabla u(x_{0})|^{p-2}\Delta u(x_{0})}{( u(x_{0})+\beta )^{p-1}}
-\frac{|\nabla u(x_{0})|^{p}}{ (u(x_{0})+\beta )^{p}}.
\end{equation*}
So, by \eqref{e2}, \eqref{e3}, \eqref{e4} and \eqref{e5} we have 
\begin{align*}
&0\geq \Delta _{p}(\ln (u(x_{0})+\beta ))-\Delta _{p}(\ln (v(x_{0})+\beta ))
\\
&=\frac{\Delta _{p}u(x_{0})}{(u(x_{0})+\beta )^{p-1}} -(p-1)\frac{|\nabla
u(x_{0})|^{p}}{(u(x_{0})+\beta )^{p}} 
-\frac{\Delta_{p}v(x_{0})}{(v(x_{0})+\beta )^{p-1}} \\
&\quad +(p-1)\frac{|\nabla v(x_{0})|^{p}}{(v(x_{0})+\beta )^{p}} \\
&=\frac{\Delta _{p}u(x_{0})}{(u(x_{0})+\beta )^{p-1}} -\frac{\Delta
_{p}v(x_{0})}{(v(x_{0})+\beta )^{p-1}} \\
&=-a(x_{0})(\frac{f(u(x_{0}))}{(u(x_{0})+\beta )^{p-1}} 
-\frac{f(v(x_{0}))}{(v(x_{0})+\beta )^{p-1}})>0
\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{equation}
\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}
\end{equation}
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{equation}
\begin{gathered} -\Delta _{p}u_{k}=a(x)f(u_{k}),\quad \text{if }| x|<k, \\
u_{k}(x)=0, \quad\text{if }|x|=k. \end{gathered}  \label{e10}
\end{equation}
The following two distinct situations may occur:

\noindent \emph{Case 1:} $f$ is bounded on $(0,+\infty )$. In this case, as
we have initially observed, $\lim_{u\searrow 0}f(u)$ exists and is finite,
so $f$ can be extended by continuity at the origin. To obtain a solution to 
\eqref{e10}, it is sufficinet to verify that the hypotheses of the 
Diaz-Sa\`{a} theorem are fulfilled.

* Since $f\in C^{1}((0,\infty ),(0,\infty ))$ it follows that the mapping 
$u\to a(x)f(u)$ is continuous in $[0,\infty )$.

* From $a(x)\frac{f(u)}{u^{p-1}}=a(x)\frac{f(u)}{(u+\beta )^{p-1}} \cdot 
\frac{(u+\beta )^{p-1}}{u^{p-1}}$, using positivity of $a$ and (F1) we
deduce that the function $u\to a(x)\frac{f(u)}{u^{p-1}}$ is decreasing on 
$(0,\infty )$.

* For all $u\geq 0$, since $a(x)\in C_{\mathrm{loc}}^{0,\alpha }(\mathbb{R}
^{N})$, we obtain $x\to a(x)f(u)$ belongs to $L^{\infty}(\Omega )$.

* By $\lim_{u\to \infty }\frac{f(u)}{u^{p-1}+1} 
=\lim_{u\to \infty }\frac{f(u)}{u^{p-1}}\cdot \frac{u^{p-1}}{u^{p-1}+1}=0$ 
and $f\in C^{1}((0,\infty),(0,\infty ))$, there exists $C>0$ 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 $|x|<k$.

\subsection*{Case 2}
$\lim_{u\searrow 0}f(u)=+\infty $. We will apply the method of sub- and
supersolutions in order to find a solution to the problem \eqref{e10}. We
first observe that $0$ is a subsolution for this problem.

We construct in what follows a positive supersolution. By the boundedness of 
$f$ in a neighbourhood of $+\infty $, there exists $A>0$ 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  } 0<u\leq 1, \\ 
A, & \mbox{if }u>1.
\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, \\
U=0,\quad \mbox{if }|x|=k.
\end{gather*}
has a positive solution. Now, since $h\geq f$ on $(0,+\infty )$, it follows
that $U$ is supersolution of \eqref{e10}.

In both cases studied above we define $u_{k}=0$ for $|x|>k$. 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 $a,b\in I$, $a<b$  and $1\leq h<+\infty $ (resp $0<h\leq 1$)
\end{theorem}

\noindent \textit{Case 1:} Let $1<p\leq 2$, so $0<p-1\leq 1$, follows that 
$1\leq \frac{1}{p-1}<+\infty $. By Theorem \ref{thmE} for any $r>0,$ 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 }1<p<2,\mbox{ or} \\
K=\frac{p-1}{N-p}\cdot \int_{0}^{\infty }\xi ^{\frac{(p-2)N+1}{p-1}}\Phi
(\xi )d\xi \quad \text{if }2\leq p<+\infty .
\end{gather*}%
Clearly, for all $r>0$, 
\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 }1<p\leq 2,\mbox{ or} \\
w(r)<\frac{p-1}{N-p}\cdot \int_{0}^{\infty }\xi ^{\frac{(p-2)N+1}{p-1}}\Phi
(\xi )d\xi \quad \text{if }2\leq p<+\infty ,
\end{gather*}

An upper-solution to \eqref{e1} will be constructed. Consider the function 
$\overline{f}(u)=(f(u)+1)^{1/(p-1)}$, for $u>0$.

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{equation}
\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}
\end{equation}
From \eqref{e11} we deduce that 
\begin{equation}
\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}
\end{equation}
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 
\begin{equation}
\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}
\end{equation}
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\; 1<p\leq 2 \\
\lim_{r\to 0} \frac{\int_{0}^{r}\sigma ^{N-1}p(\sigma )f(u(\sigma ))d\sigma 
}{r^{N}}=0 \\
\lim_{r\to 0} \frac{\int_{0}^{r}\sigma ^{N-1}p(\sigma )f(u(\sigma ))d\sigma 
}{r^{N-1}}=0
\end{gather*}
it is easy to prove that $u(r)\in C^{2}(\mathbb{R}^{N})$ if $1<p\leq 2$
because $\lim_{r\to \infty } u^{\prime\prime}(r)$ is finite and $u(r)\in
C^{1}(\mathbb{R}^{N})$ if $2<p<\infty $ because $\lim_{r\to \infty
}u'(r)$ is finite.

\section{Proof of Theorem \protect\ref{thm2}}

Suppose \eqref{e1} has a solution $u(r)$, then 
\begin{equation*}
(r^{N-1}|u'(r)|^{p-2}u'(r))'=-r^{N-1}f(u(r))a(r),
\end{equation*}
integrating from $0$ to $r$, we have 
\begin{equation*}
|u'(r)|^{p-2}u'(r)=-r^{1-N}\int_{0}^{r}\sigma
^{N-1}f(u(\sigma ))a(\sigma )d\sigma ,
\end{equation*}
hence $u'(r)<0$. We put $\ln (u(r)+1):=\overline{u}(r)>0$ 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 
\begin{equation}
\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}
\end{equation}
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 
\begin{equation}
(-\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}
\end{equation}
Multiplying equation \eqref{e15} by $\xi ^{1-N}$, we deduce 
\begin{equation}
(-\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}
\end{equation}
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 
\begin{equation}
-\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}
\end{equation}
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 
$u(r)<u(0)$, for all $r>0$.

If $\beta \geq 1$, then the function $u\mapsto \frac{f(u)}{(u+\beta )^{p-1}}$
is decreasing on $(0,+\infty )$. This implies 
\begin{equation}
\frac{f(u(\sigma ))}{(u(\sigma )+1)^{p-1}}>\frac{f(u(0))}{(u(0)+1)^{p-1}}.
\label{e18}
\end{equation}
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}}

Assume $u$ 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 
$u(r)<u(0)$, for all $r>0$. If $\beta \geq 1$ then the function $u\mapsto 
\frac{f(u)}{(u+\beta )^{p-1}}$ is decreasing on $(0,+\infty )$. This implies 
\begin{equation}
\frac{f(u(\sigma ))}{(u(\sigma )+1)^{p-1}}> \frac{f(u(0))}{(u(0)+1)^{p-1}},
\label{e19}
\end{equation}
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 $1<p<2$, so $0<p-1<1$, it follows that $1<\frac{1}{p-1}<+\infty $. Set 
\begin{gather*}
\int_{0}^{\xi }r^{N-1}a(r)dr<1\quad \text{for }\xi >0, \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$ 1<p\leq 2$. Then $1\leq \frac{1}{p-1}<+\infty$. From the above
proofs we observe that
\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\\
\leq \frac{p-1}{N-2}\int_{0}^{\infty
}t^{1/(p-1)}a^{1/(p-1)}(t)dt.
\end{equation*}
If
$\int_{0}^{\xi }\sigma ^{N-1}a(\sigma )d\sigma \geq 1$,
then
\begin{align*}
&\frac{N-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\\
&\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 \,.
\end{align*}
Then if $ \int_{0}^{\infty }t^{\frac{(p-2)N+1}{p-1}}a(t)dt=\infty
$ we have $\int_{0}^{\infty }t^{1/(p-1)}a^{1/(p-1)}(t)dt=\infty $.
\end{remark}

\subsection*{Acknowledgments}

The author thanks Professor V. Radulescu for proposing this problem, as well
as for his valuable suggestions on this subject.

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\smallskip


 Dragos-Patru Covei \\
Constantin Br\^{a}ncu\c{s}i' University of T\^{a}rgu-Jiu,
Bulevardul Republicii, nr. 1, 210152 T\^{a}rgu-Jiu, Romania\\
Email: dragoscovei77@yahoo.com

\section*{Corrigendum posted on October 8, 2007}

The author wants to correct some misprints and to clarify the existence
and the non-existence of solutions to the problem considered in this article.

Page 2, line 8. In the formula
\begin{equation*}
\int_{1}^{\infty }r^{N-1+\gamma (N-2)}(\max_{|x|=t}a(x))dt<\infty ,
\end{equation*}
replace the factor $r^{N-1+\gamma (N-2)}$ with $t^{N-1+\gamma (N-2)}$.

Page 2, line 17. In the formula
\begin{equation*}
\int_{1}^{\infty }r^{N-1+\gamma (N-2)}(\max_{|x|=t}a(x))dt<\infty ,
\end{equation*}
replace the factor $r^{N-1+\gamma (N-2)}$ with $t^{N-1+\gamma (N-2)}$.

Page 3, line 1. Replace ``\textit{Problem \eqref{e1} has no positive radial
solution if }$p\geq N$.'' with 
``\textit{Suppose $a(r)$ is a positive radial function which is continuous on 
$\mathbb{R}^{N}$.
Then problem \eqref{e1} has no positive radial solution if $p\geq N$.}''

Page 4, line 16. Replace ``bounded domain'' with ``smooth open
bounded domain''.

Page 4, line 21. Replace ``bounded domain with smooth boundary" with
``smooth open bounded domain".

Page 5, line 25-26. Replace ``\cite[Theorem 1.3]{GS}'' with 
``\cite[Theorem 1.3 and lower-upper solutions method]{GS}''

Page 5, line 27. In the equation
\begin{equation*}
-\Delta _{p}U=p(x)h(U),\quad \mbox{if }|x|<k,
\end{equation*}
 replace $p(x)$ with $a(x)$.

Page 7, line 17. In the formula
\begin{equation*}
K=\frac{p-1}{N-2}\cdot \int_{0}^{\infty }\xi ^{\frac{1}{p-1}}\Phi
^{1/(p-1)}(\xi )d\xi 
\end{equation*}
replace the symbol `` $=$ '' with `` $\leq $ ''.

Page 7, line 18. Replace
\begin{equation*}
K=\frac{p-1}{N-p}\cdot \int_{0}^{\infty }\xi ^{\frac{(p-2)N+1}{p-1}}\Phi
(\xi )d\xi \quad \text{if }2\leq p<+\infty .
\end{equation*}
with ``
\begin{equation*}
K\leq \frac{p-1}{N-p}\cdot \int_{0}^{\infty }\xi ^{\frac{(p-2)N+1}{p-1}}\Phi
(\xi )d\xi \quad  \text{or} \quad
 K\leq {\rm Const.}+\int_{1}^{\infty }\xi ^{\frac{1-N}{p-1}}d\xi ,
\end{equation*}
if $2\leq p<+\infty $ and for the above considered cases.''

Page 7, line 21. Replace
\begin{equation*}
w(r)<\frac{p-1}{N-p}\cdot \int_{0}^{\infty }\xi ^{\frac{(p-2)N+1}{p-1}}\Phi
(\xi )d\xi \quad \text{if }2\leq p<+\infty .
\end{equation*}
with `` 
\begin{equation*}
w(r)\leq \frac{p-1}{N-p}\cdot \int_{0}^{\infty }\xi ^{\frac{(p-2)N+1}{p-1}
}\Phi (\xi )d\xi \quad   \text{or} \quad
 w(r)\leq {\rm Const.}+\int_{1}^{\infty }\xi ^{\frac{1-N}{p-1}}d\xi .
\end{equation*}
if $2\leq p<+\infty $ and for the above considered cases.''

Page 8, line 1. ``(F2')'' must be replaced by ``(F2') 
$ \lim_{u\searrow 0}\overline{f}(u)/u=\infty $, 
$\lim_{u\nearrow \infty }\overline{f}(u)/u=0$ and 
$u\mapsto \overline{f}(u)/u^{p-1}$ is decreasing on 
$(0,\infty )$.''

Page 9. line 10. Replace ``Using'' with 
``If $u(x)$ is radially symmetric
solution (see \cite{GS} for conditions to $a(x)$ and $f(u(x))$) then, using''

Page 9. line 15. Replace
\begin{equation*}
\lim_{r\to 0} \frac{\int_{0}^{r}\sigma
^{N-1}p(\sigma )f(u(\sigma ))d\sigma }{r^{N}}=0
\]
with 
\[ 
\lim_{r\to 0} \frac{\int_{0}^{r}\sigma ^{N-1}a(\sigma )f(u(\sigma ))
d\sigma }{r^{N}}=\frac{a(0)f(u(0))}{N}
\end{equation*}

Page 9. line 16. Replace $p(\sigma )$ with $a(\sigma )$.

Page 9. After line 18, insert ``If $u(x)$ is a weak solution, then
applying the regularity theory for quasilinear elliptic equations (see for
example \cite{DB} or  \cite[Theorem 1.3]{CD}) we find that 
$u\in C^{1,\alpha }(\mathbb{R}^{N})$.''

Page 12. line 5. Replace $r^{-\frac{1-N}{p-1}}$ with $r^{\frac{1-N}{p-1}}$.

Page 12. line 7. Replace $r^{-\frac{1-N}{p-1}}$ with $r^{\frac{1-N}{p-1}}$.

Page 13. line 5. Replace $p(t)$ with $a(t)$.

Page 14. After line 18 insert ``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*}
we observe that $\int_{0}^{\infty }t^{\frac{(p-2)N+1}{p-1}}a(t)dt\neq \infty 
$ as in Case 2, Theorem 1.1.''

Also the author wants to present an alternative proof for the existence of
solutions (see section 3. Existence of a solution).

\begin{proof} The existence of solutions will be established by solving
the approximate problems
\begin{equation} \label{er}
\begin{gathered}
-\Delta _{p}u = a(x)f(u+\varepsilon ), \quad \text{if } |x| <k,    \\
u(x) =0, \quad\text{if } |x| =k,  
\end{gathered}
\end{equation}
for $\varepsilon >0$ 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}


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