\documentclass[reqno]{amsart}
\usepackage{hyperref}

\AtBeginDocument{{\noindent\small
\emph{Electronic Journal of Differential Equations},
Vol. 2012 (2012), No. 240, pp. 1--10.\newline
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu
\newline ftp ejde.math.txstate.edu}
\thanks{\copyright 2012 Texas State University - San Marcos.}
\vspace{9mm}}

\begin{document}
\title[\hfilneg EJDE-2012/240\hfil Asymptotic behavior of positive solutions]
{Asymptotic behavior of positive solutions for the radial p-Laplacian
equation}

\author[S. Ben Othman, H. M\^aagli\hfil EJDE-2012/240\hfilneg]
{Sonia Ben Othman, Habib M\^aagli}  % in alphabetical order

\address{Sonia Ben Othman \newline
D\'epartement de Math\'ematiques, 
Facult\'e des Sciences de Tunis, 
Campus universitaire, 
2092 Tunis, Tunisia}
\email{Sonia.benothman@fsb.rnu.tn}

\address{Habib M\^aagli \newline
King Abdulaziz University,
College of Sciences and Arts, Rabigh Campus,
Department of Mathematics,
P.O. Box 344, Rabigh 21911, Saudi Arabia}
\email{habib.maagli@fst.rnu.tn}

\thanks{Submitted September 23, 2012. Published December 28, 2012.}
\subjclass[2000]{34B15, 35J65}
\keywords{p-Laplacian; asymptotic behavior; positive solutions;
\hfill\break\indent  Schauder's fixed point theorem}

\begin{abstract}
 We study the existence, uniqueness and asymptotic
 behavior of positive  solutions to the  nonlinear problem
 \begin{gather*}
 \frac{1}{A}(A\Phi _p(u'))'+q(x)u^{\alpha}=0,\quad \text{in }(0,1),\\ 
 \lim_{x\to 0}A\Phi _p(u')(x)=0,\quad u(1)=0,
 \end{gather*}
 where $\alpha <p-1$, $\Phi _p(t)=t|t| ^{p-2}$, $A$
 is a positive differentiable function and $q$ is a positive measurable
 function in $(0,1)$ such that for some $c>0$,
 \[
 \frac{1}{c}\leq q(x)(1-x)^{\beta }\exp \Big(
 -\int_{1-x}^{\eta }\frac{z(s)}{s}ds\Big)\leq c.
 \]
 Our arguments combine monotonicity methods with Karamata regular variation
 theory.
\end{abstract}

\maketitle
\numberwithin{equation}{section}
\newtheorem{theorem}{Theorem}[section]
\newtheorem{lemma}[theorem]{Lemma}
\newtheorem{proposition}[theorem]{Proposition}
\allowdisplaybreaks

\section{Introduction}

Let $p>1$ and $\alpha <p-1$. We consider the boundary-value problem
\begin{equation} \label{eP}
\begin{gathered}
-\frac{1}{A}(A\Phi _p(u'))'+q(x)u^{\alpha }=0,\quad  \text{in }(0,1) \\
A\Phi _p(u')(0):=\lim_{x\to 0}A\Phi _p(u')(x)=0, \quad
u(1)=0.
\end{gathered}
\end{equation}
Here, $A$ is a continuous function in $[0,1)$, differentiable and positive
on $(0,1)$ and for all $t\in\mathbb{R}$, $\Phi _p(t)=t| t| ^{p-2}$.
Our goal in this paper is
to study problem \eqref{eP} under appropriate conditions on $q$. We
obtain the existence of a unique positive continuous solution to \eqref{eP}
and  establish estimates on such solution.

Several articles have been devoted to the study of the differential equation
\[
-\frac{1}{A}(A\Phi _p(u'))'+q(x)u^{\alpha }=0,\quad \text{in }(0,1)
\]
with various boundary conditions, especially for the one-dimensional
$p$-Laplacian equation (see \cite{a1,a2,a3,b1,b2,g3,h1,h2,k1}).
For  $\alpha <0$,
problem \eqref{eP} has been studied in \cite{b1}, where  the existence and
uniqueness of  positive solutions and some estimates for the solutions
have been obtained.
Thus, it is interesting to
know the exact asymptotic behavior of such solution as $x\to 1$ and
to extend the study of \eqref{eP} to $0\leq \alpha <p-1$.


Asymptotic behavior of solutions of the semilinear elliptic
equation
\begin{equation} \label{e1.1}
-\Delta u=q(x)u^{\alpha },\quad \alpha <1,\; x\in \Omega ,
\end{equation}
for $\Omega $  bounded or an unbounded  in $\mathbb{R}^{n}$ 
$(n\geq 2)$, with homogeneous Dirichlet boundary conditions,
has been investigated by several authors;
see for example  \cite{b3,c1,c2,g1,g2,g4,m1,m2,z1} and the references therein.
Applying Karamata regular variation theory, M\^{a}agli \cite{m1}  studied 
\eqref{e1.1}, when $\Omega $ is a bounded $C^{1,1}$-domain. 
He showed that \eqref{e1.1} has a unique positive classical solution that
satisfies homogeneous Dirichlet boundary conditions and gave sharp estimates
on such solution. This studied extended the estimates stated in 
\cite{g4,m2,z1}. 
In this work, we  extend the result established in \cite{m1}
to the radial case associated to problem \eqref{eP}.

To simplify our statements, we need to fix some notation and make some
assumptions. Throughout this paper, we shall use $\mathcal{K}$ to denote the
set of Karamata functions $L$ defined on $(0,\eta ]$ by
\[
L(t):=c\exp \Big(\int_{t}^{\eta }\frac{z(s)}{s}ds\Big),
\]
for some positive constants $\eta,c$, and a function $z\in C([0,\eta ])$
such that $z(0)=0$.
Recall that $L\in \mathcal{K}$ if and only if $L$ is a positive
function in $C^{1}((0,\eta ])$, for some $\eta >0$, such that
\begin{equation}
\lim_{t\to 0}\frac{tL'(t)}{L(t)}=0.
\end{equation}
For two nonnegative functions $f$ and $g$ defined on a set $S$, we write
$f(x)\approx g(x)$,  if there exists a constant $c>0$ such that
$\frac{1}{c}g(x)\leq f(x)\leq cg(x)$, for each $x\in S$. Furthermore, we
refer to $G_pf$, as the function defined on $(0,1)$ by
\[
G_pf(x):=\int_{x}^{1}\Big(\frac{1}{A(t)}
\int_0^{t}A(s)f(s)ds\Big)^{\frac{1}{p-1}}dt,
\]
where $f$ is a nonnegative measurable function in $(0,1)$.
 We point out that
if $f$ is a nonnegative continuous function such that the mapping
$x\mapsto A(x)f(x)$ is integrable in a neighborhood of $0$,
 then $G_pf$ is the solution of the problem
\begin{equation} \label{e1.3}
\begin{gathered}
-\frac{1}{A}(A\Phi _p(u'))'=f,\quad \text{in } (0,1), \\
A\Phi _p(u')(0)=0,\quad u(1)=0.
\end{gathered}
\end{equation}

As it is mentioned above, our main purpose in this paper is to establish
existence and global behavior of a positive solution for problem \eqref{eP}. Let
us introduce our hypotheses.

The function $A$ is continuous in $[0,1)$, differentiable
and positive in $(0,1)$ such that
\[
A(x)\approx x^{\lambda }(1-x)^{\mu }
\]
with $\lambda \geq 0$ and $\mu <p-1. $

The function $q$ is required to satisfy
\begin{itemize}
\item[(H1)] $q$ is a positive measurable function on $(0,1)$ such that
\[
q(x)\approx (1-x)^{-\beta }L(1-x),
\]
with $\beta \leq p$ and $L\in \mathcal{K}$ defined on $(0,\eta ]$ $(\eta >1)$
such that 
\[
\int_0^{\eta }t^{\frac{1-\beta }{p-1}}(
L(t))^{\frac{1}{p-1}}dt<+\infty .
\]
\end{itemize}

We need to verify the condition
\[
\int_0^{\eta }t^{\frac{1-\beta }{p-1}}(L(t))^{
\frac{1}{p-1}}dt<+\infty
\]
in hypothesis (H1), only if $\beta =p$ (See Lemma \ref{lem2} below).

As a typical example of function $q$ satisfying (H1), we have 
\[
q(x):=(1-x)^{-\beta }(\log \frac{2}{1-x})^{-\nu }, \quad x\in
[ 0,1).
\]
Then for $\beta <p$ and $\nu \in \mathbb{R} $ or $\beta =p$ and $\nu >p-1$,
 the function $q$ satisfies (H1).


Our main result is as follows.

\begin{theorem} \label{thm1}
Assume {\rm (H1)}. Then problem \eqref{eP} has a unique positive and continuous 
solution $u $ satisfying, for $x\in (0,1)$,
\begin{equation} \label{e1.4}
u(x)\approx \theta _{\beta }(x),
\end{equation}
where $\theta _{\beta }$ is the function defined on $[0,1)$
by
\begin{equation} \label{e1.5}
\theta _{\beta }(x):=
\begin{cases}
\Big(\int_0^{1-x}\frac{(L(s))^{\frac{1}{p-1}}
}{s}ds\Big)^{\frac{p-1}{p-1-\alpha }}, & \text{if }  \beta =p
 \\
(1-x)^{\frac{p-\beta }{p-1-\alpha }}(L(1-x))^{\frac{1}{
p-1-\alpha }}, & \text{if }  \frac{(\mu +1)(p-1-\alpha
)+\alpha p}{p-1}<\beta <p,  \\
(1-x)^{\frac{p-1-\mu }{p-1}}, & \text{if }  \beta <\frac{(\mu
+1)(p-1-\alpha )+\alpha p}{p-1}  \\
(1-x)^{\frac{p-1-\mu }{p-1}}(\int_{1-x}^{\eta }\frac{L(s)
}{s}ds)^{\frac{1}{p-1-\alpha }}, & \text{if }  \beta =\frac{(
\mu +1)(p-1-\alpha )+\alpha p}{p-1}.
\end{cases}
\end{equation}
\end{theorem}

The article is organized as follows. 
In Section 2, we prove some basic estimates and  recall some results 
on functions belonging to  $\mathcal{K}$.
In Section 3, we prove Theorem \ref{thm1}. 
In the last section, we present some applications.

\section{Estimates}

In what follows, we  give estimates on the functions $G_pq$
and $G_p(q\theta _{\beta }^{\alpha })$, where $q$ is a function satisfying
(H1) and $\theta _{\beta }$ is the function given by \eqref{e1.5}. To this end,
we recall some fundamental properties of functions belonging to the class
$\mathcal{K}$, taken from \cite{c1,m3,s1}.

\begin{lemma}[\cite{m3,s1}] \label{lem1}
 Let $L_1,L_2\in \mathcal{K}$, $m\in\mathbb{R}$ and $\epsilon >0$. 
Then  $L_1L_2\in \mathcal{K}$, $L_1^{m}\in
\mathcal{K}$, and $\lim_{t\to 0^{+}}t^{\epsilon }L_1(t)=0$.
\end{lemma}

\begin{lemma}[\cite{m3,s1}] \label{lem2} 
Let $L\in \mathcal{K}$ and $\delta \in\mathbb{R}$. Then we have the following:
\begin{itemize}
\item[(i)] If $\delta <2$, then $\int_0^{\eta}t^{1-\delta }L(t)dt$ converges and 
\[
\int_0^{s}t^{1-\delta}L(t)dt\sim \frac{s^{2-\delta }L(s)}{2-\delta }\quad
\text{as } s\to 0^{+}.
\]

\item[(ii)] If $\delta >2$, then $\int_0^{\eta }t^{1-\delta }L(t)dt$ diverges 
and 
\[
\int_{s}^{\eta }t^{1-\delta }L(t)dt \sim \frac{s^{2-\delta
}L(s)}{\delta -2}\quad \text{as } s\to 0^{+}.
\]
\end{itemize}
\end{lemma}

\begin{lemma}[\cite{c1}] \label{lem3} 
Let $L\in \mathcal{K}$ be defined on $(0,\eta ]$, then we have
\[
t\mapsto \int_{t}^{\eta }\frac{L(s)}{s}ds\in \mathcal{K}.
\]
If further $\int_0^{\eta }\frac{L(s)}{s}ds$ converges,
then 
\[
t\mapsto \int_0^{t}\frac{L(s)}{s}ds\in \mathcal{K}.
\]
\end{lemma}

\begin{proposition} \label{prop1}
Assume $q$ satisfies (H1). Then for $x\in (0,1)$, we have
\[
G_pq(x)\approx \Psi (1-x),
\]
where $\psi $ is the function defined  on $(0,1]$ by
\begin{equation} \label{e2.1}
\Psi (t)=\begin{cases}
\int_0^{t}\frac{(L(s))^{\frac{1}{p-1}}}{s}ds, &\text{if }\beta =p,  \\
t^{\frac{p-\beta }{p-1}}(L(t))^{\frac{1}{p-1}},&\text{if }\mu +1<\beta <p,  \\
t^{\frac{p-1-\mu }{p-1}},&\text{if }\beta <\mu +1 \\
t^{\frac{p-1-\mu }{p-1}}(\int_{t}^{\eta }\frac{L(s)}{s}
ds)^{\frac{1}{p-1}},&\text{if }\beta =\mu +1.
\end{cases}
\end{equation}
\end{proposition}

\begin{proof}
For $x\in (0,1)$, we have
\[
G_pq(x)\approx \int_{x}^{1}\frac{1}{t^{\frac{\lambda }{p-1}
}(1-t)^{\frac{\mu }{p-1}}}\Big(\int_0^{t}s^{\lambda
}(1-s)^{\mu -\beta }L(1-s)ds\Big)^{\frac{1}{p-1}}dt.
\]
Put
\[
h(x):=\int_{x}^{1}\frac{1}{t^{\frac{\lambda }{p-1}}(1-t)^{
\frac{\mu }{p-1}}}\Big(\int_0^{t}s^{\lambda }(1-s)^{\mu
-\beta }L(1-s)ds\Big)^{\frac{1}{p-1}}dt,\text{ }x\in (0,1).
\]
We shall estimate $h(x)$. Since $h$ is continuous and positive on
$[0,1/2]$, it follows that $h(x)\approx 1$, for $x\in [ 0,1/2]$.
 Now, assume that $x\in [1/2,1)$. Then
\[
h(x)\approx \int_{x}^{1}\frac{1}{(1-t)^{\frac{\mu }{p-1}}}
\Big(\int_0^{t}s^{\lambda }(1-s)^{\mu -\beta
}L(1-s)ds\Big)^{\frac{1}{p-1}}dt.
\]
Moreover, for $t\in [ x,1)$, we have
\begin{align*}
&\int_0^{t}s^{\lambda }(1-s)^{\mu -\beta }L(1-s)ds\\
&=\int_0^{1/2}s^{\lambda }(1-s)^{\mu -\beta
}L(1-s)ds+\int_{\frac{1}{2}}^{t}s^{\lambda }(1-s)^{\mu
-\beta }L(1-s)ds \\
&\approx 1+\int_{1-t}^{1/2}s^{\mu -\beta }L(s)ds.
\end{align*}
Then we distinguish the following cases:

\noindent $\bullet $ If $\beta <\mu +1$, then by Lemma \ref{lem2}, 
$\int_0^{1/2}s^{\mu -\beta }L(s)ds<\infty $. So,
since $\mu <p-1$, we obtain
\[
h(x)\approx (1-x)^{\frac{p-1-\mu }{p-1}}.
\]

\noindent $\bullet $ If $p>\beta >\mu +1$, then by Lemma \ref{lem2},
\[
\int_{1-t}^{1/2}s^{\mu -\beta }L(s)ds\approx
(1-t)^{\mu +1-\beta }L(1-t).
\]
So,
\[
(1+\int_{1-t}^{1/2}s^{\mu -\beta
}L(s)ds)^{\frac{1}{p-1}}\approx (1-t)^{\frac{\mu +1-\beta }{p-1}}L^{
\frac{1}{p-1}}(1-t).
\]
 Thus, using the fact that $\beta <p$ and again Lemma \ref{lem2}, we obtain that
\[
h(x)\approx (1-x)^{\frac{p-\beta }{p-1}}L^{\frac{1}{p-1}}(1-x).
\]

\noindent $\bullet $ If $\beta =\mu +1$, then
\[
h(x)\approx \int_0^{1-x}\frac{1}{t^{\frac{\mu }{p-1}}}\Big(
\int_{1-t}^{1}\frac{L(s)}{s}ds\Big)^{\frac{1}{p-1}}dt.
\]
So, using Lemma \ref{lem3} and the fact that $\mu <p-1$, by Lemma \ref{lem2}
 it follows that
\[
h(x)\approx (1-x)^{\frac{p-1-\mu }{p-1}}(\int_{1-x}^{1}
\frac{L(s)}{s}ds)^{\frac{1}{p-1}}.
\]

\noindent $\bullet $ If $\beta =p$, we deduce by Lemma \ref{lem2} that
\[
\int_{1-t}^{1/2}s^{\mu -\beta }L(s)ds\approx
(1-t)^{\mu +1-p}L(1-t),
\]
hence
\[
h(x)\approx \int_0^{1-x}\frac{(L(s))^{\frac{1}{p-1}}}{s}ds.
\]
This completes the proof.
\end{proof}

The following proposition plays a crucial role in this article.

\begin{proposition} \label{prop2}
Let $q$  satisfy (H1) and let $\theta _{\beta }$ be the
function given in \eqref{e1.5}. Then for $x\in (0,1)$, we have
\[
G_p(q\theta _{\beta }^{\alpha })(x)\approx \theta _{\beta }(x).
\]
\end{proposition}

\begin{proof}
Let $\beta \leq p$ and $\mu <p-1$, a straightforward computation shows that
for $x\in (0,1)$,
\[
q(x)\theta _{\beta }^{\alpha }(x)\approx \widetilde{q}(x),
\]
where
\[
\widetilde{q}(x):=\begin{cases}
\frac{L(1-x)}{(1-x)^{p}}\Big(\int_0^{1-x}\frac{(
L(s))^{\frac{1}{p-1}}}{s}ds\Big)^{\frac{\alpha (p-1)}{p-1-\alpha }}
&\text{if }  \beta =p
\\
\frac{(L(1-x))^{\frac{p-1}{p-1-\alpha }}}{(1-x)^{(\beta -
\frac{\alpha (p-\beta )}{p-1-\alpha })}}
& \text{if } \frac{(\mu +1)(p-1-\alpha )+\alpha p}{p-1}<\beta <p,
 \\
\frac{L(1-x)}{(1-x)^{(\beta -\frac{\alpha (p-1-\mu )}{p-1})}}
& \text{if }  \beta <\frac{(\mu +1)(p-1-\alpha
)+\alpha p}{p-1}
\\
\frac{L(1-x)}{(1-x)^{(\mu +1)}}\Big(
\int_{1-x}^{\eta }\frac{L(s)}{s}ds\Big)^{\frac{\alpha }{
p-1-\alpha }} & \text{if }  \beta =\frac{(\mu +1)(p-1-\alpha
)+\alpha p}{p-1}.
\end{cases}
\]
So, we deduce that
\[
\widetilde{q}(x)=(1-x)^{-\delta }\widetilde{L}(1-x),
\]
where $\delta \leq p$. Then, using Lemmas \ref{lem1} and \ref{lem3},
 we verify that
$ \widetilde{L}\in \mathcal{K}$ and
$\int_0^{\eta }t^{\frac{ 1-\delta }{p-1}}(\widetilde{L}(t))^{\frac{1}{p-1}}dt
<+\infty $.
Hence, by Proposition \ref{prop1},
\[
G_p(q\theta _{\beta }^{\alpha })(x)\approx G_p\widetilde{q}(x)\approx
\widetilde{\psi }(1-x),\quad x\in (0,1),
\]
where $\widetilde{\psi }$ is the function defined in \eqref{e2.1}
 by replacing $L $ by $\widetilde{L}$ and $\beta $ by $\delta $.
This completes the proof.
\end{proof}

\section{Proof of Theorem \ref{thm1}}

 \subsection{Existence and asymptotic behavior}
Let $q$ satisfy (H1) and let $\theta _{\beta }$ be the function given
 by \eqref{e1.5}.
By Proposition \ref{prop2}, there exists a constant $m\geq 1$ such that for each 
$x\in (0,1)$,
\begin{equation} \label{e3.1}
\frac{1}{m}\theta _{\beta }(x)\leq G_p(q\theta _{\beta }^{\alpha })(x)\leq
m\theta _{\beta }(x).
\end{equation}
Now we look  at the existence of positive solution of problem \eqref{eP}
satisfying \eqref{e1.4}. For the case $\alpha <0$, we refer to 
\cite{b1}.
So prove the existence result only for the case $0\leq \alpha <p-1$,
 and then give the precise asymptotic behavior of
such solution for $\alpha <p-1$. We will split the proof into two
cases.

\noindent\textbf{Case 1: $\alpha <0$.}
Let $u$ be a positive continuous solution of \eqref{eP}. To
obtain estimates \eqref{e1.4} on the function $u$, we need the following
comparison result.

\begin{lemma} \label{lem4}
Let $\alpha <0$ and $u_1,u_2 \in C^{1}((0,1))\cap C([0,1])$ 
be two positive functions such that
\begin{equation} \label{e3.2}
\begin{gathered}
-\frac{1}{A}(A\Phi _p(u_1'))'\leq q(x)u_1^{\alpha },
\quad \text{in }(0,1), \\
A\Phi _p(u_1')(0)=0,\quad u_1(1)=0
\end{gathered}
\end{equation}
and
\begin{equation} \label{e3.3}
\begin{gathered}
-\frac{1}{A}(A\Phi _p(u_2'))'\geq q(x)u_2^{\alpha },
\quad \text{in }(0,1), \\
A\Phi _p(u_2')(0)=0,\quad u_2(1)=0.
\end{gathered}
\end{equation}
Then $u_1\leq u_2$.
\end{lemma}

\begin{proof}
Suppose that $u_1(x_0)>u_2(x_0)$ for some $x_0\in (0,1)$. Then
there exists $x_1,x_2\in [ 0,1]$, such that 
$0\leq x_1<x_0<x_2\leq 1$ and for $x_1<x<x_2$, $u_1(x)>u_2(x)$ with
$u_1(x_2)=u_2(x_2)$, $u_1(x_1)=u_2(x_1)$ or $x_1=0$.

We deduce that
\begin{equation} \label{e3.4}
A\Phi _p(u_2')(x_1)\leq A\Phi _p(u_1')(x_1).
\end{equation}
On the other hand, since $\alpha <0$, we have
 $u_1^{\alpha }(x)<u_2^{\alpha }(x)$, for each $x\in (x_1,x_2)$. This yields 
\[
\frac{1}{A}(A\Phi _p(u_1'))'-\frac{1}{A}(A\Phi
_p(u_2'))'\geq q(u_2^{\alpha }-u_1^{\alpha
})\geq 0\quad \text{on }(x_1,x_2).
\]
Using further \eqref{e3.4}, we deduce that the function
$\omega (x):=(A\Phi _p(u_1')-A\Phi _p(u_2'))(x)$ is
nondecreasing on $(x_1,x_2)$ with $\omega (x_1)\geq 0$. Hence, from
the monotonicity of $\Phi _p$, we obtain that the function $x\mapsto
(u_1-u_2)(x)$ is nondecreasing on $(x_1,x_2)$ with $
(u_1-u_2)(x_1)\geq 0$ and $(u_1-u_2)(x_2)=0$. This yields to a
contradiction. The proof is complete.
\end{proof}

Now, we are ready to prove \eqref{e1.4}. Put
 $c=m^{-\frac{\alpha }{p-1-\alpha }}$
and $v:=G_p(q\theta _{\beta }^{\alpha })$. It follows from \eqref{e1.3} that
the function $v$ satisfies
\[
-\frac{1}{A}(A\Phi _p(v'))'=q\theta _{\beta }^{\alpha },
\quad \text{in }(0,1).
\]
According to \eqref{e3.1}, we obtain by simple calculation that $\frac{1}{c}v$
and $cv$ satisfy respectively \eqref{e3.2} and \eqref{e3.3}.
Thus, we deduce by Lemma \ref{lem4} that
\[
\frac{1}{c}v(x)\leq u(x)\leq cv(x),\text{ }x\in (0,1).
\]
This proves the result.

\noindent\textbf{Case 2: $0\leq \alpha <p-1$.}
Put $c_0=m^{\frac{p-1}{p-1-\alpha }}$ and let
\[
\Lambda :=\big\{ u\in C([0,1]);\text{ }\frac{1}{c_0}\theta _{\beta }\leq
u\leq c_0\theta _{\beta }\big\}.
\]
Obviously, the function $\theta _{\beta }$ belongs to $C([0,1])$ and so
$\Lambda $ is not empty. We consider the integral operator $T$ on $\Lambda $
defined by
\[
Tu(x):=G_p(qu^{\alpha })(x),\quad x\in [ 0,1].
\]
We prove that $T$ has a fixed point in $\Lambda $, in order to construct a
solution of problem \eqref{eP}. For this aim, first we observe that
$T\Lambda \subset \Lambda $. Let $u\in \Lambda $, then for each $x\in [ 0,1)$
\[
\frac{1}{c_0^{\alpha }}(q\theta _{\beta }^{\alpha })(x)\leq q(x)u^{\alpha
}(x)\leq c_0^{\alpha }(q\theta _{\beta }^{\alpha })(x).
\]
This together with \eqref{e3.1} implies that
\[
\frac{1}{mc_0^{\frac{\alpha }{p-1}}}\theta _{\beta }\leq Tu\leq mc_0^{
\frac{\alpha }{p-1}}\theta _{\beta }.
\]
Since $mc_0^{\frac{\alpha }{p-1}}=c_0$ and $T\Lambda \subset C([0,1])$,
then $T$ leaves invariant the convex $\Lambda $. Moreover, since
$\alpha \geq 0$, then the operator $T$ is nondecreasing on $\Lambda $.
Now, let $\{u_k\}_k$ be a sequence of functions in $C([0,1])$ defined by
\[
u_0=\frac{1}{c_0}\theta _{\beta }, \quad u_{k+1}=Tu_k,\quad
\text{for } k\in\mathbb{N}.
\]
Since $T\Lambda \subset \Lambda $, we deduce from the monotonicity of $T$
that for $k\in \mathbb{N}$, we have
\[
u_0\leq u_1\leq \dots \leq u_k\leq u_{k+1}\leq c_0\theta _{\beta }.
\]
Applying the monotone convergence theorem, we deduce that the sequence
$\{u_k\}_k$ converges to a function $u\in \Lambda $ which satisfies
\[
u(x)=G_p(qu^{\alpha })(x),\text{ }x\in [ 0,1].
\]
We conclude that $u$ is a positive continuous solution of problem \eqref{eP}
which satisfies \eqref{e1.4}.

\subsection{Uniqueness}
Assume that $q$ satisfies (H1). For $\alpha <0$, the uniqueness of solution
 to problem \eqref{eP} follows from Lemma \ref{lem4}. Thus, we look at the case 
$0\leq \alpha <p-1$. Let
\[
\Gamma =\{u\in C([0,1]): u(x)\approx \theta _{\beta }(x)\}.
\]
Let $u$ and $v$ be two positives solutions of problem \eqref{eP} in $\Gamma $.
Then there exists a constant $k\geq 1$ such that
\[
\frac{1}{k}\leq \frac{v}{u}\leq k.
\]
This implies that the set
\[
J=\{t\in (1,+\infty ): \frac{1}{t}u\leq v\leq tu\}
\]
is not empty. Now, put $c:=\inf J$, then we aim to show that $c=1$.
Suppose that $c>1$, then
\begin{gather*}
-\frac{1}{A}(A\Phi _p(v'))'+\frac{1}{A}(A\Phi _p(c^{
\frac{-\alpha }{p-1}}u'))'=q(x)(v^{\alpha
}-c^{-\alpha }u^{\alpha }),\quad \text{in }(0,1),
\\
\lim_{x\to 0^{+}}(A\Phi _p(v')-A\Phi _p(c^{
\frac{-\alpha }{p-1}}u'))(x)=0,
\\
(v-c^{\frac{-\alpha }{p-1}}u)(1)=0.
\end{gather*}
So, we have
\[
-\frac{1}{A}(A\Phi _p(v'))'+\frac{1}{A}
(A\Phi _p(c^{\frac{-\alpha }{p-1}}u'))'\geq 0\quad\text{in }(0,1),
\]
which implies that the function $\theta (x):=(A\Phi _p(c^{
\frac{-\alpha }{p-1}}u')-A\Phi _p(v'))(x)$ is
nondecreasing on $(0,1)$ with $\lim_{x\to 0^{+}}\theta (x)=0$.
Hence from the monotonicity of $\Phi _p$, we obtain that the function $
x\mapsto (c^{\frac{-\alpha }{p-1}}u-v)(x)$ is nondecreasing on $[0,1)$ with
$(c^{-\frac{\alpha }{p-1}}u-v)(1)=0$.
This implies that $c^{\frac{-\alpha }{p-1}}u\leq v$.
On the other hand, we deduce by symmetry that
$v\leq c^{\frac{\alpha }{p-1}}u$. Hence $c^{\frac{\alpha }{p-1}}\in J$. Now,
since $\alpha <p-1$ and $c>1$, we have $c^{\frac{\alpha }{p-1}}<c$. This
yields to a contradiction with the fact that $c:=\inf J$. Hence, $c=1$ and
then $u=v$.

\section{Applications}

\subsection*{First application} 
Let $q$ be a positive measurable function in
$[0,1)$ satisfying for $x\in [ 0,1)$
\[
q(x)\approx (1-x)^{-\beta }\Big(\log \frac{3}{1-x}\Big)^{-\sigma },
\]
where the real numbers $\beta $ and $\sigma $ satisfy one of the following
two conditions:
\begin{itemize}
\item  $\beta <p$ and $\sigma \in\mathbb{R}$,

\item  $\beta =p$ and $\sigma >p-1$.
\end{itemize}
Using Theorem \ref{thm1}, we deduce that problem \eqref{eP} has a positive
continuous solution\ $u$ in $[0,1]$ satisfying
\begin{itemize}
\item[(i)] If $\beta <\frac{(\mu +1)(p-1-\alpha )+\alpha p}{p-1}$, then
 for $x\in (0,1)$, 
\[
u(x)\approx (1-x)^{\frac{p-1-\mu }{p-1}}.
\]

\item[(ii)] If $\beta =\frac{(\mu +1)(p-1-\alpha )+\alpha p}{
p-1}$ and $\sigma =1$, then for $x\in (0,1)$,
\[
u(x)\approx (1-x)^{\frac{p-1-\mu }{p-1}}\Big(\log \log \frac{3}{1-x}
\Big)^{\frac{1}{p-1-\alpha }}.
\]

\item[(iii)] If $\beta =\frac{(\mu +1)(p-1-\alpha )+\alpha p}{
p-1}$ and $\sigma <1$, then for $x\in (0,1)$,
\[
u(x)\approx (1-x)^{\frac{p-1-\mu }{p-1}}\Big(\log \frac{3}{1-x}\Big)^{
\frac{1-\sigma }{p-1-\alpha }}.
\]

\item[(iv)] If $\beta =\frac{(\mu +1)(p-1-\alpha )+\alpha p}{
p-1}$ and $\sigma >1$, then for $x\in (0,1)$,
\[
u(x)\approx (1-x)^{\frac{p-1-\mu }{p-1}}.
\]

\item[(v)] If $\frac{(\mu +1)(p-1-\alpha )+\alpha p}{p-1}
<\beta <p$, then for $x\in (0,1)$,
\[
u(x)\approx (1-x)^{\frac{p-\beta }{p-1-\alpha }}\Big(\log \frac{3}{1-x}
\Big)^{\frac{-\sigma }{p-1-\alpha }}.
\]

\item[(vi)] If $\beta =p$ and $\sigma >p-1$, then for $x\in (0,1)$,
\[
u(x)\approx \Big(\log \frac{3}{1-x}\Big)^{\frac{p-1-\sigma }{p-1-\alpha}}
\]
\end{itemize}

\subsection*{Second application}
Let $q$ be a function satisfying (H1) and let $\alpha ,\gamma <p-1$.
 We are interested in the  nonlinear problem
\begin{equation} \label{e4.1}
\begin{gathered}
-\frac{1}{A}(A\Phi _p(u'))'+\frac{\gamma }{u}\Phi
_p(u')u'=q(x)u^{\alpha },\quad \text{in }(0,1), \\
A\Phi _p(u')(0)=0,\quad u(1)=0.
\end{gathered}
\end{equation}
Put $v=u^{1-\frac{\gamma }{p-1}}$; then $v$ satisfies
\begin{equation} \label{e4.2}
\begin{gathered}
-\frac{1}{A}(A\Phi _p(v'))'=(\frac{p-1-\gamma }{p-1
})^{p-1}q(x)v^{\frac{(\alpha -\gamma )(p-1)}{p-1-\gamma }},\quad \text{in }(0,1),
 \\
A\Phi _p(v')(0)=0,\quad v(1)=0.
\end{gathered}
\end{equation}
Using Theorem \ref{thm1}, we deduce that  \eqref{e4.2} has a unique solution $v$
such that $v(x)\approx \widetilde{\theta }_{\beta }(x)$, where
\[
\widetilde{\theta }_{\beta }(x)=
\begin{cases}
\Big(\int_0^{1-x}\frac{(L(s))^{\frac{1}{p-1}}}{s}
ds\Big)^{\frac{p-1-\gamma }{p-1-\alpha }}
&  \text{if }  \beta =p ,
\\
(1-x)^{\frac{(p-\beta )(p-1-\gamma )}{(
p-1)(p-1-\alpha )}}(L(1-x))^{\frac{p-1-\gamma }{(
p-1)(p-1-\alpha )}},
& \text{if }  \frac{(\mu +1)
(p-1-\alpha )+(\alpha -\gamma )p}{p-1-\gamma }\\
&\quad <\beta <p,
\\
(1-x)^{\frac{p-1-\mu }{p-1}}
&  \text{if }  \beta <\frac{(\mu
+1)(p-1-\alpha )+(\alpha -\gamma )p}{p-1-\gamma },
 \\
(1-x)^{\frac{p-1-\mu }{p-1}}(\int_{1-x}^{\eta }\frac{L(s)}{s}
ds)^{\frac{p-1-\gamma }{(p-1)(p-1-\alpha )}}
& \text{if }  \beta =\frac{(\mu +1)(p-1-\alpha )+(\alpha -\gamma )p}{
p-1-\gamma }.
\end{cases}
\]
Consequently,  \eqref{e4.1} has a unique solution $u$
satisfying
\[
u(x)\approx \begin{cases}
\Big(\int_0^{1-x}\frac{(L(s))^{\frac{1}{p-1}
}}{s}ds\Big)^{\frac{p-1}{p-1-\alpha }}, & \text{if }  \beta =p
 \\
(1-x)^{\frac{p-\beta }{p-1-\alpha }}(L(1-x))^{\frac{1}{
p-1-\alpha }}, &\text{if } \frac{(\mu +1)(p-1-\alpha
)+(\alpha -\gamma )p}{p-1-\gamma }<\beta <p,
 \\
(1-x)^{\frac{p-1-\mu }{p-1-\gamma }}, &\text{if } \beta <\frac{(
\mu +1)(p-1-\alpha )+(\alpha -\gamma )p}{p-1-\gamma }
\\
(1-x)^{\frac{p-1-\mu }{p-1-\gamma }}(\int_{1-x}^{\eta }
\frac{L(s)}{s}ds)^{\frac{1}{p-1-\alpha }},
&\text{if } \beta = \frac{(\mu +1)(p-1-\alpha )
 +(\alpha -\gamma )p}{p-1-\gamma }.
\end{cases}
\]


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\end{document}