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

\AtBeginDocument{{\noindent\small
{\em Electronic Journal of Differential Equations},
Vol. 2005(2005), No. 80, pp. 1--22.\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-2005/80\hfil Self similar solutions]
{Self similar solutions of generalized Burgers equation}
\author[A. Gmira, A. Hamydy, S. Ouailal\hfil EJDE-2005/80\hfilneg]
{Abdelilah Gmira, Ahmed Hamydy, Salek Ouailal}  % in alphabetical order

\address{Universite Abdelmalek Essaadi,
Faculte des Sciences Departement \\
de Mathematiques et informatique B. P. 2121 Tetouan - Maroc}
\email[A. Gmira]{gmira@fst.ac.ma}
\email[A. Hamydy]{hamydy@caramail.com}
\email[S. Ouailal]{salekouailal@yahoo.com}

\date{}
\thanks{Submitted December 6, 2004. Published July 15, 2005.}
\subjclass[2000]{35k55, 35k65}
\keywords{Burgers equation; self similar; classification; asymptotic behavior}

\begin{abstract}
 In this paper, we study the initial-value problem
 \begin{gather*}
 (|u'|^{p-2}u')'+\beta r u'+\alpha u-\gamma |u|^{q-1}u|u'|^{p-2}u'=0,
 \quad  r>0, \\
 u(0)=A,\quad u'(0)=0,
 \end{gather*}
 where $A>0$, $p>2$, $q>1$, $\alpha>0$, $\beta>0 $ and $ \gamma \in{\mathbb{R}}$.
 Existence and complete classification of solutions are established.
 Asymptotic behavior for nonnegative solutions is also presented.
\end{abstract}

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

\section{Introduction}

This paper concerns the  nonlinear parabolic equation
\begin{equation}
U_t-( |U_x |^{p-2}U_x)_x=-\frac kt U+|U|^{q-1}U|U_x|^{p-2}U_x \quad
\text{in }{\mathbb{R}\times\mathbb{R}}^{+},\label{E11}
\end{equation}
where $p>2$, $q>1$ and $k>0$.
As is often the case in nonlinear PDE's  of parabolic type the
characteristic properties of an equation, are displayed by means of the
existence of so-called self similar solutions; this is our main interest.
It is worth mentioning that if $p=2$, $q=1$ and $k=0$, we get the classical
one dimensional Burgers equation
\begin{equation}
U_t=U_{xx}+U_xU,
\label{E12}
\end{equation}
which is originally proposed as a simplified model of Navier-Stokes
Turbulence (see \cite{B1}  and  \cite {B2}) .

By design, Burgers equation is the simplest model of hydrodynamic flow that
captures the interaction of nonlinear wave propagation and viscosity.
Burgers turbulence is often viewed as a pared-down model of acoustic
turbulence (see \cite{GM} and \cite{SZ}).

The importance and popularity of equation (\ref{E12}) lie in its simplicity
and in the fact that the well known Hopf-cole substitution $w=\frac{U_{x}}{U}$
reduces it to the linear heat equation. This nonlinear change of variables
permits an explicit description of solutions of  (\ref{E12}) and explains their
essentially nonlinear first order asymptotic as $t$ goes to infinity.

If $p=2$ and $k\neq 0$, equation (\ref{E11}) becomes
\begin{equation}
U_t=U_{xx}+|U|^{q-1}UU_x-\frac ktU
\label{E13}
\end{equation}
which is studied by \cite{SR}. Note that, if $q=1$,
equation (\ref{E13}) describes the propagation of weakly nonlinear longitudinal
waves in gases or liquids from a non planar source
(see  \cite {LS} and \cite {L}).

If $p>2$, equations (\ref{E11}) appears in the description of ice sheet
dynamics (see \cite{F} ) where the reaction term
$-\frac{k}{t}U$ can be considered as a turbulent term. In this case the
selfsimilar solutions of problem (\ref{E11}) take the form
\[
U(x,t)=t^\sigma f(y),\quad \text{where}\quad y= xt^\eta,
\]
with
\[
\sigma=\frac{-1}{pq+p-2} \quad \text{and}\quad  \eta=\frac{- q}{pq+p-2}.
\]
Then the profile $f$ is determined as a solution in of the ODE
\[
( |f'|^{p-2}f') '+\frac{q}{pq+p-2}yf'
-(k-\frac{1}{pq+p-2})f+|f|^{q-1}f|f'|^{p-2}f'=0,\quad
y\in {\mathbb{R}}.
\]
Where the prime denotes the differentiation with respect to $y$. If we set
\[
\ g(y)= \left\{
\begin{array}{l}
f(y)\quad \forall $ y $\in \mathbb {R}^{+},\\
f(-y)\quad \forall $ y $\in \mathbb {R}^{-},
\end{array}
\right .
\]
then, $g$  satisfies
\[
( |g'|^{p-2}g') '+\alpha g +\beta yg'+ \gamma|g|^{q-1}g|g'|^{p-2}g'=0,
\]
 in ${\mathbb{R}}$;  with $\alpha=-k+\frac{1}{pq+p-2}$,
 $\beta=\frac{q}{pq+p-2}$
\[
\alpha=-k+\frac{1}{pq+p-2},   \beta=\frac{q}{pq+p-2}
\]
and $\gamma=-1$, if $y>0$; $\gamma=1$, if $y<0$.
Consequently, we have just to focus  on the study of the
initial-value problem
\begin{equation}
\begin{gathered} \label{Q}
( |u'|^{p-2}u') '+\beta ru'+\alpha u-\gamma |u|^{q-1}u|u'|^{p-2}u'=0,
\quad r>0  \\
u(0)=A,\quad u'(0)=0,
\end{gathered}
\end{equation}
when $\alpha >0$, $\beta >0$ and $\gamma \in \mathbb{R}$. We will mainly discuss:
(i) The existence and uniqueness of solutions for \eqref{Q};
(ii) the asymptotic behavior of positive solutions, and
(iii) a classification of solutions.


The main results of this paper  are the following.

\begin{theorem} \label{thm1}
Assume $p>2$, $q>1$, $\alpha >0$, $\beta >0$, and $\gamma \in \mathbb{R}$.
Then for each $A>0$, there exists a real $R_{\rm max}>0$ such that \eqref{Q}
has a unique solution $u\equiv u(.,A)$ defined in the right open interval
 $[0,R_{\rm max}[$, meaning that $u$ and $|u'|^{p-2}u'$ are a $C^1$ functions
in  $[0,R_{\rm max}[$, satisfying  \eqref{Q}.
\end{theorem}

The following result gives the monotonicity of solutions of problem \eqref{Q}
with respect to initial data.

\begin{theorem}  \label{thm2}
Assume $\alpha>0$, $\beta>0$ and $\gamma<0$. Let $u(.,A)$ and $u(.,B)$ be
two solutions of problem \eqref{Q} with $u(0,A)=A$, $u(0,B)=B$ and $A\neq B$.
Then $u(.,A)$ and $u(.,B)$ can not intersect each other before their first zero.
\end{theorem}

Concerning the asymptotic behavior, we have the following results.

 \begin{theorem} \label{thm3}
Let $u$ be a strictly positive solution of \eqref{Q}. Then
\[
\lim_{r\to +\infty }u(r)=\lim_{r\to+\infty }u'(r)=0.
\]
Furthermore, if $\alpha>0$,  $\beta>0$ and $\gamma\leq0$, then
\[
\lim_{r\to+\infty }r^{\frac \alpha \beta }u(r)=L
\]
exists and lies in $[ 0,+\infty [$. Moreover this limit $L$ is strictly
positive for $0<\alpha\leq\beta$ and $\gamma<0$.
\end{theorem}

Finally, the structure of solutions of problem \eqref{Q} consists of
three families: The set of strictly positive solutions, the set of changing
sign solutions  and finally solutions with compact support.
This classification depends strongly on the sign of $\gamma(\alpha-\beta)$.

\begin{theorem} \label{thm4}  Assume $p>2, q>1$, and $\gamma<0$. Then we have
\begin{itemize}
\item[(i)] For $\alpha>\beta$ there exist two constants $A_1$ and $A_2$ such
that for any $A>A_1$, the solution $u(.,A)$ is strictly positive and for
$A<A_2$,  $u(.,A)$ changes sign. Moreover, there exists at least one solution
with  compact support.
\item[(ii)] For $0<\alpha\leq\beta$, any solution of \eqref{Q} is strictly
positive.
\end{itemize}
\end{theorem}

The situation for  $\gamma>0$ is quite the opposite.

\begin{theorem} \label{thm5} Assume $p>2$, $q>1$ and  $\gamma>0$. Then
\begin{itemize}
\item[(i)] If $\alpha\geq\beta$ any solution of \eqref{Q} change sign.

\item[(ii)] If $\alpha<\beta$, there exist two constants $A_1$ and $A_2$
such that for any $A<A_2$,  the solution $u(.,A)$  is strictly positive and
for $A>A_1, u(.,A)$ changes sign.
\end{itemize}
\end{theorem}
The organization of this paper is as follows.
Theorems \ref{thm1}, \ref{thm2} and \ref{thm3} are proved in section 2. In section 3 a classification
of solutions is investigated and then Theorems \ref{thm4}  and \ref{thm5}
 are established.

\section{Existence and asymptotic behavior of solutions}

In this section, we investigate existence, uniqueness and asymptotic behavior
of solutions of the problem \eqref{Q}.
We start with a local existence and uniqueness result.

\begin{proposition} \label{prop2.1}
Assume $p>2$,  $q>1$, $\alpha >0$, $\beta >0$ and $\gamma \in {\mathbb{R}}$.
Then for each $A>0$,  there exists a right open interval $I=[0,R_{\rm max}[$ and
a unique function $u$ such that, $u$ and $|u'|^{p-2}u'$ lie in $C^1(I)$ and
satisfy \eqref{Q}.
\end{proposition}

First of all, we note that, for a fixed $\alpha,  \beta$ and $\gamma$,
it easy to see that $u(.,\gamma,A)=-u(.,\gamma,-A)$. Therefore,
in the sequel we restrict ourselves to the case of $A>0$.

\begin{remark} \label{rmk2.1} \rm
The first equation in \eqref{Q} can be reduced to  the first order system
\begin{equation}
\begin{gathered}
X'=|Y|^{-\frac{p-2}{p-1}}Y \\
Y'=-\alpha X-\beta|Y|^{-\frac{p-2}{p-1}}Y+\gamma|X|^{q-1}XY.
\end{gathered} \label{E21}
\end{equation}
Since the mapping
\[
(X,Y)\mapsto  \Big( |Y|^{-\frac{p-2}{p-1}}Y ,
-\alpha X-\beta|Y|^{-\frac{p-2}{p-1}}Y+\gamma|X|^{q-1}XY \Big)
\]
is a locally Lipschitz continuous function in the set
$\{ (X,Y)\in \mathbb{R}\times \mathbb{R}^{*}\}$, we deduce that, for any
 $ r_0>0, A\geq0$ and $B\neq0$, there exists a unique solution of \eqref{Q}
in a neighborhood of $r_0$ such that $u(r_0)=A$ and $u'(r_0)=B$.
\end{remark}


Because of the presence of the term $|Y|^{-\frac{p-2}{p-1}}Y$, the above
function is not locally Lipschitz continuous near $r_0$ whenever $u'(r_0)=0$.
Consequently, for our problem \eqref{Q} the above argument does not work.
To avoid this difficulty, we make use an idea from \cite{GV}.
Then the proof becomes similar to that of \cite[proposition 1.1]{GB1}
 and \cite[proposition 2.1]{GB2}. We present  it here for the convenience of the
reader.

\begin{proof}[Proof proposition \ref{prop2.1}]
The idea of the proof is to convert our initial value problem \eqref{Q} to a
fixed point problem of some operator. This will be done in two steps.

\noindent\textbf{Step 1.} Local existence and uniqueness.
It is clear that to solve problem \eqref{Q} is equivalent to find a function
$u \in C^1(I)$ defined in some interval $I=[0,R[$ with $R>0$ such that
$|u'|^{p-2}u'\in C^1(I)$ and satisfies the integral equation
\begin{equation}
u(r)=A-\int_0^rG(F_u)(s)ds, \label{E22}
\end{equation}
where
$G(s)=|s|^{(2-p)/(p-1} s$, for all $s\in \mathbb{R}$, and
\begin{equation}
F_u(s)=\beta su(s) +( \alpha -\beta ) \int_0^s u(\tau)d\tau-\gamma
\int_0^s|u|^{q-1}u(\tau ) |u'|^{p-2}u'(\tau)d\tau.
\label{E23}
\end{equation}
  Now, let us define  on $[0,A]$ the following two functions
\begin{equation}
f_1(X)=\begin{cases}
\alpha (A-X)-|\gamma |X^{p-1}(A+X)^q & \text{if }\alpha \geq \beta , \\
\alpha (A+X)-2\beta X-|\gamma |X^{p-1}(A+X)^q & \text{if }\alpha <\beta,
\end{cases}
\label{E24}
\end{equation}
and
\begin{equation}
f_2(X)=\begin{cases}
(A+X)\left\{ \alpha +|\gamma |X^{p-1}(A+X)^{q-1}\right\}
& \text{if }\alpha \geq \beta , \\
\alpha (A-X)+|\gamma |X^{p-1}(A+X)^q &\text{if }\alpha <\beta .
\end{cases}
\label{E25}
\end{equation}
Since $f_1$ is continuous  and $f_1(0)=\alpha A>0$, then there exists
some interval $[0,A_0]\subset[0,A]$ such that
\[
f_1(X)>0 \quad \forall X\in[0,A_0].
\]
Let us introduce some useful notation for the proof:
\begin{equation}
f_1(A_0)=K_1, \quad  f_2(A_0)=K_2,\quad
R_0=\inf \{ 1, \frac{A_0^{p-1}}{2\Gamma A}, \frac{K_1^{p-2}}{%
(2\Gamma )^{p-1}}\}, \label{E26}
\end{equation}
where
\begin{equation}
\Gamma =\beta +|\alpha -\beta |+(2p-2+q)2^{q-1}|\gamma
|A^{q+p-2}. \label{E27}
\end{equation}
It is easy to see that the function $f_2$ satisfies the estimate
\begin{equation}
f_2(X)\leq2A\Gamma, \quad \forall X\in [0,A].
\label{E28}
\end{equation}
Now, we consider the complete metric space
\begin{equation}
X=\{ \varphi \in C^1([ 0,R_0] ):\| \varphi-A\| _X\leq A_0\}
\label{E29}
\end{equation}
where
\begin{equation}
\|\varphi\|_X=\max(\|\varphi\|_0,\|\varphi'\|_0).
\label{E210}
\end{equation}
and $\|.\|_0$ denotes the sup norm. Next we define the mapping
$\mathcal{T}$ on $X$, by
\begin{equation}
T(\varphi)=A-\int_0^rG(F_\varphi)(s)ds,\quad \forall r\in[0,R_0]
\label{E211}
\end{equation}

\noindent  \textbf{Claim 1.} $\mathcal{T}$ maps $X$ into itself.
In fact, take $\varphi\in X$.
First, it is easy to see that
$\mathcal{T}(\varphi )\in C^1( [ 0,R_0] )$. Also  by a simple calculation
we get
\begin{equation}
0<K_1s< F_\varphi (s)< K_2s\quad \forall s\in[ 0,R_0] .
\label{E212}
\end{equation}
And thereby, $ \mathcal{T}(\varphi )$ satisfies the following estimates
\begin{gather}
|\mathcal{T}(\varphi )(r)-A|\leq
\int_0^rF_\varphi^{1/(p-1)}(s) ds\leq\frac{p-1}pK_2^{1/(p-1)}r^{p/(p-1)},
\label{E213} \\
|\mathcal{T'}(\varphi)(r)|\leq F_\varphi^{1/(p-1)}(r)\leq K_2^{1/(p-1)}r^{1/(p-1)},
\label{E214}
\end{gather}
for any $r\in$$[0,R_0]$. These last two equations, combined with the
 expression of $R_0$ given by (\ref{E26}) imply that
  $\mathcal{T}(\varphi)\in$ $X$.

\noindent\textbf{Claim 2.} $\mathcal{T}$ is a contraction. To prove this,
take $\varphi $ and $ \psi \in X$.
Then
\begin{equation}
|\mathcal{T}(\varphi )(r)-T(\psi )(r)|\leq
\int_0^r|G(F_\varphi (s))-G(F_\psi (s))|ds,
\label{E215}
\end{equation}
  for any $r$$\in [0,R_0]$, where $F_\varphi $ is given by (\ref{E23}).
  In view of estimate (\ref{E212}) (which is also valid for $F_\psi$), we get
\begin{equation}
\begin{aligned}
|G(F_\varphi )(s)-G(F_\psi )(s)|
\leq & |F_\varphi^{1/(p-1)}(s)-F_\psi^{1/(p-1)}(s)|\nonumber\\
\leq & \frac1{p-1}K_1^{(2-p)/(p-1)}|F_\varphi (s)-F_\psi (s)|s^{(2-p)/(p-1)}.
\end{aligned}
\label{E216}
\end{equation}
  Recalling the expression of  $F_\varphi $ and  $F_\psi$, we deduce
\begin{equation}
|F_\varphi (s)-F_\psi (s)|\leq [\beta+|\alpha-\beta|]\| \varphi-\psi \| _0s+|\gamma|I
\label{E217}
\end{equation}
where
\begin{equation}
I=\int_0^s|{\varphi^q(\tau ) |\varphi'|^{p-2}\varphi'(\tau)-\psi^q(\tau ) |\psi'|^{p-2}\psi'(\tau )}|d\tau.
\label{E218}
\end{equation}
But
\begin{equation}
I\leq\int_0^s|\varphi{\prime}(\tau )|^{p-1}|{\varphi^q(\tau)-\psi^q(\tau)}|
d\tau +\int_0^s\psi^q(\tau)\big||\varphi'|^{p-2}\varphi'
-|\psi'|^{p-2}\psi'\big|d\tau.
\label{E219}
\end{equation}
Using the fact that $\varphi$ and $\psi$ are elements of $X$, we get
\begin{equation}
I\leq (q+2p-2)2^{q-1}A^{q+p-2}\|\varphi-\psi\|_Xs.
\label{E220}
\end{equation}
Combining this last equation with (\ref{E217}), we get
\begin{equation}
|F_\varphi (s)-F_\psi (s)|\leq \Gamma\|\varphi-\psi\|_Xs,\quad \forall s\in[0,R_0]
\label{E221}
\end{equation}
where $\Gamma$ is given by (\ref{E27}). Therefore
\begin{equation}
|\mathcal{T}(\varphi )(r)-T(\psi )(r)|\leq \frac {\Gamma}pK_1^{(2-p)/(p-1)}\|\varphi-\psi\|_Xr^{p/(p-1)}.
\label{E222}
\end{equation}
Similarly, one can easily obtain
\begin{equation}
|\mathcal{T'}(\varphi )(r)-\mathcal{T'}(\psi (r)|\leq \frac {\Gamma}{p-1}K_1^{(2-p)/(p-1)}\|\varphi-\psi\|_Xr^{1/(p-1)}.
\label{E223}
\end{equation}
>From the choice of $R_0$, these last two equations imply that
$\mathcal{T}$ is a contraction.
The use of the Banach's Contraction theorem leads to the existence of a
unique function $u$  solving problem \eqref{Q} in $(0,R_0)$.

\noindent\textbf{Step 2.} $|u'|^{p-2}u'\in C^1([0,R_0))$.
We have just to prove the regularity at $r = 0$. For this purpose, note
that the first equation in \eqref{Q} gives
\begin{equation}
\lim_{r\to 0}(|u'|^{p-2}u')'(r)=-\alpha A.
\label{E224}
\end{equation}
Integrating equation \eqref{Q} from $0$ to $r$, and letting r go to 0, we obtain
\begin{equation}
\lim_{r\to 0}\frac{|u'|^{p-2}u'(r)}r=-\alpha A.
\label{E225}
\end{equation}
Hence $|u'|^{p-2}u'\in C^1(([0,R_0[)$.
This completes the proof of Proposition \ref{prop2.1}.
\end{proof}

\begin{remark} \label{rmk2.2} \rm
It is not difficult to see that the solution $u$ of \eqref{Q} is a $C^\infty $
function at any $r>0$ whenever $u'(r)\neq 0$.
\end{remark}

The remaining  of this section is  devoted to the proof of the
Theorem \ref{thm2}. For this purpose we start with the following lemma.

\begin{lemma} \label{lem2.1}
Let $A>0$ and $u$ be the corresponding solution of
\eqref{Q}. Then as long as $u$ is strictly positive we have
$0<u(r)<A$ and $u'(r)<0$.
\end{lemma}

\begin{proof}
Let
\begin{equation}
E(r)=\frac{p-1}p |u'|^p( r) +\frac \alpha
2u^2( r) ,\quad \forall r\in [0,R_{\rm max}[, \label{E226}
\end{equation}
be the induced energy function. We have
\begin{equation}
E'(r)=-\beta r |u'|^2+\gamma |u|
^{q-1}u|u'|^p,\quad \forall  {r}\in [0,R_{\max }[.
\label{E227}
\end{equation}
Then if $\gamma \leq 0$ the energy function  is decreasing as long as $u$
is positive.
Particularly  $u( r) \leq A$. On the other hand since
\begin{equation}
( |u'|^{p-2}u') '(0) =-\alpha A<0,  \label{E228}
\end{equation}
the lemma follows easily when  $\gamma >0$.
\end{proof}

The next result gives the monotonicity of solutions of the problem \eqref{Q}
with respect to initial data. More exactly, we have

\begin{proposition} \label{prop2.2}
Assume $\alpha>0$,  $\beta>0$ and $\gamma<0$. Let $0<A_1<A_2$. Then
$u(.,A_1)$ and $u(.,A_2)$ can not intersect each other before their first zero.
\end{proposition}

\begin{proof}
For ease of notation, we write $u(.,A_1)=u(.)$ and  $u(.,A_2)=v(.)$ and
we denote by $R_1$ (respectively $R_2$) the first zero of $u$ (respectively $v$).
The proof will be done by contradiction: it is based on the idea
of \cite[Lemma 2.4 ]{PT}.  We assume that there exists some point
$R_0\in[0,\min \{R_1,R_2\}[$ such that
\begin{equation}
u(r)<v(r)\quad \text{for}\quad r\in [0,R_0[\quad \text{and }\quad
 u(R_0)=v(R_0).
\label{E229}
\end{equation}
Now, for any $k>0$, we set
\begin{equation}
u_k(r)=k^{-p/(p-2)}u(kr),\quad r\in [0,\frac{R_1}k].
\label{E230}
\end{equation}
Then $u_k$ satisfy the equation
\begin{equation}
 (|u_k'|^{p-2}u_k'(r))'+\beta
ru_k'(r)+\alpha u_k-\gamma k^\mu u_k^q(s)|u_k'|^{p-2}u_k'(r)=0,
\label{E231}
\end{equation}
with
$\mu =1+\frac{pq}{p-2}$. %\label{E234}
Since $u $ is strictly positive and decreasing on $[0,R_1[$, the function
$k\mapsto u_k$ is strictly increasing. Moreover for any $r\in [0,R_0]$
 $\lim_{k\to 0 }u_k(r)=+\infty$.
Then there exists a  small $k_0>$0 such that
\[
u_k(r)>v(r)\quad \text{for } r\in [ 0,R_0]\quad\text{and}  k\in [ 0,k_0]
\]
Therefore, the set
\begin{equation}
\Omega\equiv \left\{ k\in ]0,k_0[; u_k(r)>v(r) \text{for \ }r\in [0,R_0]\right\}
\label{E232}
\end{equation}
is not empty and open. In particular if we denote by $K $ the supremum of
$\Omega $, the real $K\notin\Omega$ and thereby, necessarily there exists
$r_0\in [ 0,R_0]$ such that $u_K(r_0) = v(r_0)$.
As $k_0 $ is small without loss of generality we assume
\begin{equation}
k_0=(\frac{A_1}{2A_2})^{(p-2)/p}. \label{E233}
\end{equation}
If $r_0=R_0$, then
\begin{equation}
u_K( R_0) =K^{- p/(p-2)}u(KR_0)=v( R_0) .
\label{E234}
\end{equation}
But $u( R_0) =v( R_0)$, then using again the strictly increasing of the
function $k\mapsto u_k$ is  strictly increasing we deduce necessarily  $K=1$.
This is a contradiction with the choice of the real $k_0$.
If $r_0=0$ we get
\[
u_K(0)=K^{-p/p-2)}u(0)=K^{-p/(p-2)}A_1=A_2
\]
which contradicts (\ref{E233}). Consequently we deduce that there exists
some point $r_0\in [0,R_0[$ such that
\begin{equation}
u_K > v\quad \mbox{on }] 0,R_0[ \mbox{ and }u_K( r_0) = v(r_0).
\label{E235}
\end{equation}
So $u_K-v$ has a local minimum at the point $r_0$, where the graphs of $u_K$
and $v$are tangent. Moreover, as $v'$ and $u'$ are strictly negative,
the equation satisfied by  $v$ (respectively by $u_K$) can be written in the
form
\begin{equation}
( p-1) |v'|^{p-2}v''+\beta
rv'+\alpha v-\gamma v^q|v'|^{p-2}v'=0,
\label{E236}
\end{equation}
and respectively
\begin{equation}
( p-1) |u_K'|^{p-2}u_K'' +\beta ru_k'+\alpha u_K-\gamma K^\mu u_K^q|u_K'|
^{p-2}u_K'=0.
\label{E237}
\end{equation}
Subtract (\ref{E236}) from (\ref{E237}), we obtain at point $r_0$,
\begin{equation}
( p-1)|v'|^{p-2}( u_K-v)''(x)=\gamma ( K^\mu -1) v^q|v'|^{p-2}v'(r_0).
\label{E238}
\end{equation}
Since $\gamma <0$, $v' <0,K<1$ and $\mu >0$, we get
\begin{equation}
( p-1) |v'|^{p-2}( u_K-v)''(r_0)=\gamma ( k^\mu -1) v^q|
v'|^{p-2}v'( r_0) <0.
\label{E239}
\end{equation}
This is impossible because $( u_K-v)$ has a local minimum at
$x$ and then the proposition is proved.
\end{proof}

In the next result, we investigate the asymptotic behavior of positive solutions.

\begin{proposition} \label{prop2.3}
 Let $u$ be a positive solution of \eqref{Q} defined on $[ 0,+\infty[$.
Then
\[
\lim_{r\to +\infty }u(r)=\lim_{r\to+\infty }u'(r)=0.
\]
\end{proposition}

The proof of this result depends strongly on the sign of $\gamma$.
In fact, if $\gamma \leq 0$, the result follows from the energy function.
However, for $\gamma >0$ we need some information about the monotonicity
of $u'$ this is given in the following lemma.

\begin{lemma} \label{lem2.2}
Assume $\gamma >0$. Let a real $A>0$ and $u(.,A)$ be a strictly positive
solution of \eqref{Q} defined in $[0,+\infty [$. Then there exists a
unique real number $R(A)>0$ such that
\[
u''<0\quad \mbox{on }[ 0,R(A)[\quad\mbox{and}\quad u''\geq 0\quad
\mbox{on} [R(A),+\infty [ .
\]
\end{lemma}


\begin{proof}
First, note that from Lemma \ref{lem2.1}, the solution $u=u(.,A)$
is decreasing and converges to some nonnegative constant. On the other hand
(\ref{E228}),
implies that $( |u'|^{p-2}u') '$ must change sign.
Let $R(A)>0$ its first zero. For simplicity we set  $ R=R(A)$.
Then $( |u'|^{p-2}u') '( r) <0$ for any $r$ in $[0,R[$. Furthermore,
\begin{equation}
\alpha u( R) =-[ \beta R-\gamma u^q|u'|
^{p-2}( R) ] u'(R).  \label{E240}
\end{equation}
As $u$ is strictly positive we deduce that $u'(R) \neq 0$, and then
$u$ is a $C^\infty$ function at the point $R$. So,  the first equation
in \eqref{Q} can be written in some
neighborhood of $R$, say $] R-\varepsilon ,R+\varepsilon [$
($\varepsilon >0$), in the form
\begin{equation}\label{E241}
( p-1) |u'|^{p-2}u''+\beta
ru'+\alpha u-\gamma u^q|u'|^{p-2}u'=0.
\end{equation}
Differentiating this last equality and taking $r=R$, we obtain
\begin{equation}
( p-1) |u'|^{p-2}u^{( 3) }(R) =-( \alpha +\beta ) u'( R) +\gamma
qu^{q-1}|u'|^p( R).
\label{E242}
\end{equation}
But, since $\gamma >0$, $\alpha +\beta \geq 0$ and $u'( R) <0$, then the
left hand side of the last equation is
strictly positive. By continuity of $u^{( 3) }$ we get
\[
u^{( 3) }( r) >0\quad \text{ for any }r\in [R,R+\varepsilon [.
\]
Hence $u''$ is non-negative on\ $[R,R+\varepsilon [ $. Finally,
using \eqref{Q} we deduce
\[
 u''( r) \geq 0\quad \text{ for any $r$ in }[ R,+\infty [,
\]
 which completes the proof.
\end{proof}

\begin{remark} \label{rmk2.3} \rm
Note that the right hand side of (\ref{E242}) satisfies
\[
-( \alpha +\beta ) u'( R) + \gamma
qu^{q-1}|u'|^p( R) =\frac{u'(R)}{
u(R)}\{ \gamma qu^q|u'|^{p-2}u'(R)-(\alpha +\beta ) u(R)\}.
\]
Using \eqref{E240}, the relation (\ref{E242}) becomes
\[
( p-1) |u'|^{p-2}u^{( 3) }(
R) =\beta (1+\frac \beta \alpha )R\frac{|u'(R)|^2
}{u(R)} - \gamma (1+\frac \beta \alpha -q)u^{q-1}|u'|
^p(R).
\]
Hence, if $\gamma <0$  and $q\leq 1+\frac \beta \alpha $, we get
$u^{( 3) }( R) >0$ and thereby the Lemma \ref{lem2.2} also holds in this case.
\end{remark}


\begin{proof}[Proof of Proposition \ref{prop2.3}]
By Lemma \ref{lem2.1}
$ \lim_{r\to+\infty }u(r)=l$ exists and lies in $[ 0,A[$.
We start by establishing the proposition when $\gamma \leq 0$,  in this case
the energy function given by (\ref{E226}) is positive and decreasing.
It then converges, and
$\lim_{r\to +\infty }u'(r) = 0$.
Moreover integrating equation \eqref{Q} between $0$ and $r$, we get
\begin{equation}
|u'|^{p-2}u'( r) +\beta ru(
r) +\int_0^r\left\{ ( \alpha -\beta ) u( s)
-\gamma u^q( s) |u'|^{p-2}u'(
s) \right\} ds=0.  \label{E243}
\end{equation}
Therefore,
\begin{equation}
\lim_{r\to +\infty }\frac 1r\int_0^r\{ (
\alpha -\beta ) u( s) -\gamma u^q( s) |
u'|^{p-2}u'( s)\} ds=- \beta l.
\label{E244}
\end{equation}
On the other hand, if $l\neq 0$ the L'H\^opital rule implies that
\[
\lim_{r\to +\infty }\frac 1r\int_0^r\{ (
\alpha -\beta ) u( s) -\gamma u^q( s) |
u'|^{p-2}u'( s) \} ds=(\alpha -\beta )l.
\]
This contradicts (\ref{E244}) and therefore  $l=0$.

To handle the case $\gamma >0$, we use the above Lemma \ref{lem2.2}. Assume
 that $l\neq 0$ and\ integrate equation \eqref{Q} on $( r,2r)$
 for some $r>0$. We obtain
\begin{equation}
\begin{aligned}
|u'|^{p-2}u'( 2r) =& |u'|^{p-2}u'( r) +\beta ru( r)
-2\beta ru( 2r)  \\
& +( \beta -\alpha ) \int_r^{2r}u( s) ds+\gamma
\int_r^{2r}u^q( s) |u'|^{p-2}u'( s) ds.
\end{aligned}\label{E245}
\end{equation}
Since $\gamma >0$ and $u'\leq 0$, for i$\beta\geq\alpha$, we obtain
\begin{equation}
\frac{|u'|^{p-2}u'( 2r) }{2r}\leq
( \beta -\frac \alpha 2) ( u( r) -u(2r))
-\frac \alpha 2u(2r)\,.  \label{E246}
\end{equation}
On the other hand if $\beta<\alpha$,
\begin{equation}
\frac{|u'|^{p-2}u'( 2r) }{r}\leq
\beta ( u( r) -u(2r))-\alpha u(2r).  \label{E247}
\end{equation}
 Now, observe that
\[
\lim_{r\to +\infty }[ u(r)-u(2r)] =0 \quad \text{and}\quad
\lim_{r\to +\infty }u(2r)=l\neq 0;
\]
therefore, for any $\alpha, \beta \geq 0$, we get
\begin{equation}
\frac{|u'|^{p-2}u'( 2r) }{2r}\leq -\frac \alpha 2u( 2r) \quad
\text{for large }r. \label{E248}
\end{equation}
This gives
\begin{equation}
( u^{(p-2)/(p-1)}) '( r) \leq
-\frac{p-2}{p-1}(\frac {\alpha}{2})^{1/(p-1)}r^{1/(p-1)},
\label{E249}
\end{equation}
which contradicts  that $u$is strictly positive. Consequently $l=0$
and the proof is complete.
\end{proof}

Now, we pass to the asymptotic behavior of positive solutions.

\begin {proposition} \label{prop2.4}
Assume $\alpha >0$, $\beta >0$ and $\gamma \leq 0$.
Let $u$ be a strictly positive solution of \eqref{Q}. Then
$ \lim_{r\to+\infty }r^{\alpha/ \beta }u(r)=L$ exists and lies in
$[ 0,+\infty [$.
\end{proposition}

Some preliminary results are needed for the proof of this proposition.


\begin {lemma} \label{lem2.3}
 Assume $\alpha >0$, $\beta >0$ and $\gamma \leq 0$.
Let $u$ be a strictly positive solution of \eqref{Q} such that
\begin{equation}
u(r)\leq K (1+r)^{-\sigma }\quad \text{for } r\geq 0.
\label{E250}
\end{equation}
Then, there exists a constant $M$ depending on $K$ and $\sigma$ such that
\begin{equation}
|u'(r)|\leq M (1+r)^{-\sigma -1}\quad \text{for }
r\geq 0.  \label{E251}
\end{equation}
\end{lemma}

\begin {proof}
 Without loss of generality we have just to prove (\ref{E251}) for
$r>2$. In fact, as $u'$ is a continuous function, it is bounded
in $[ 0,2]$. So there exists some constant $C>0$
such that
\begin{equation}
|u'(r)|\leq C,\quad \text{for }r \text{ in }[ 0,2].
\label{E252}
\end{equation}
  Hence,  if we take $M \geq C3^{\sigma +1}$, then  (\ref{E251})  holds for
$r$ in $[ 0,2]$.
 For any $r>2$, we set
\begin{equation}
F(r)=\exp \big[ \frac{-\gamma }{p-1}\int_0^ru^q(s)ds\big]
\label{E253}
\end{equation}
  and consider the  function
\begin{equation}
G(r)=\exp \big[ \frac \beta {p-1}\int_0^rs|u'(s)|
^{2-p}ds\big] ,\quad  r > 2.
\label{E254}
\end{equation}
  In view of  (\ref{E228}) we have
\begin{equation}
u'(r)\sim -(\alpha A)^{1/(p-1)}r^{1/(p-1)} \quad  \text{as } r  \to 0.
\label{E255}
\end{equation}
Recalling that $u'$ is strictly negative, we deduce that
the function $G$ is well posed.

Now, we write equation \eqref{Q} in the form
\begin{equation}
(FGu')'(r)+\frac \alpha {\beta }\frac{F(r)}%
ru(r)G'(r)=0.  \label{E256}
\end{equation}
Integrating the above equation, we obtain
\begin{equation}
|u'(r)|= \frac \alpha {\beta F(r)G(r)}\int_0^r%
\frac{F(s)}su(s)G'(s)ds.  \label{E257}
\end{equation}
Since  $\gamma \leq 0 $  the function $F$ is increasing and
\begin{equation}
|u'(r)|\leq \frac \alpha {\beta G(r)}\int_0^r%
\frac{u(s)}sG'(s)ds.  \label{E258}
\end{equation}
Next, we find a bound the right-hand side of the above inequality.
We set
\begin{equation}
I_1=\int_0^1\frac{u(s)}sG'(s)ds,\quad
I_2=\int_1^{r/2}\frac{u(s)}sG'(s)ds,\quad
I_3=\int_{r/2}^r\frac{u(s)}sG'(s)ds,  \label{E259}
\end{equation}
so that
\begin{equation}
\int_0^r\frac{u(s)}sG'(s)ds=I_1+I_2+I_3 \,.\label{E260}
\end{equation}
First, note that (\ref{E255}) implies easily that $I_{1}$ is bounded.
On the other hand, in view of Proposition \ref{prop2.3}, there exists a constant
$K>0$ such that
\begin{equation}
|u'(r)|^{2-p}\geq K\quad \text{for  }r\geq 0. \label{E261}
\end{equation}
Then
\begin{equation}
G(r)\geq \exp (Kr^2)\quad \text{for }r > 2. \label{E262}
\end{equation}
To estimate $I_2$, we use (\ref{E253}) to obtain
\begin{equation}
I_2\leq C\int_1^{r/2}\frac{(s+1)^{-\sigma }}sG'(s)ds\leq
C\int_1^{r/2}G'(s)ds\leq CG(\frac r2).  \label{E263}
\end{equation}
Or
\begin{equation}
\frac 1{G(r)}I_2\leq C\frac{G(r/2)}{G(r)}\leq C\exp [ \frac{%
-\beta }{p-1}\int_{\frac r2}^rs|u'(s)|
^{2-p}ds] .  \label{E264}
\end{equation}
 Now recalling (\ref{E261}), we get
\begin{equation}
\frac 1{G(r)}I_2\leq C\exp (-K_1r^2),\label{E265}
\end{equation}
with $K_1=\frac{3\beta }{8(p-1)}K$.
But as the solution $u$ is decreasing, then
\[
\frac 1{G(r)}I_3=\frac 1{G(r)}\int_{r/2}^r\frac{u(s)}sG'(s)ds\leq
\frac 2ru(r/2).
\]
Using again the estimate (\ref{E250}), we obtain
\begin{equation}
\frac 1{G(r)}I_3\leq C(r+1)^{-\sigma -1}\quad \text{for }r>2.  \label{E266}
\end{equation}
Finally, putting together (\ref{E258}), (\ref{E265}) and (\ref{E266})
the desired estimate (\ref{E254}) follows. This completes the proof of the
lemma.
\end{proof}

\begin {lemma} \label{lem2.4}
Assume $\alpha >0$, $\beta >0$ and $\gamma \leq 0$.
Let $u$ be a strictly positive solution of \eqref{Q}. Then
\begin{equation}
u(r)\leq C r^{-\frac \alpha \beta },\quad \text{ for large }r .
\label{E267}
\end{equation}
\end{lemma}

\begin{proof}
Using  the first equation in \eqref{Q}, the function $u(r)$ satisfies
\begin{equation}
\begin{aligned}
\frac \alpha 2\frac{u^2(r)^{ }}r=&\frac{|u'|^p}{2r}
-\frac \beta 2uu'-\frac 1{2r^2}u|u'|^{p-2}u'\\
& +\frac \gamma {2r}\ u^{q+1}|u'|
^{p-2}u'(r)-\frac 12[ \frac{u|u'|^{p-2}u'}r] '.
\end{aligned}\label{E268}
\end{equation}
Recalling the expression of the energy function given by (\ref{E226}) we
deduce
\begin{equation}
\begin{aligned}
\frac{E(r)}r=&\frac{3p-2}{2p}\frac{|u'|^p}r-\frac{\beta}4(u^2)'
-\frac 1{2r^2}u|u'|^{p-2}u'\\
&-\frac{\ 1}2[ \frac{u|u'|^{p-2}u'}r] '+\frac \gamma {2r}\ u^{q+1}|u'|
^{p-2}u'(r).
\end{aligned}\label{E269}
\end{equation}
Integrating the above inequality on the interval $(r,R)$
 we obtain
\begin{align*}
\int_r^R\frac{E(s)}sds
=&  \frac{3p-2}{2p}\int_r^R\frac{|u'(s)|^p}sds+\frac{u(r)|u'|^{p-2}u'(r)}{2r}\\
& -\frac{u(R)|u'|^{p-2}u'(R)}{2R}+\frac{\beta \ }4u^2(r)-\frac{\beta}4u^2(R) \\
& -\frac 12\int_r^R\frac{u(s)|u'(s)|^{p-2}u'(s)}{s^2}ds
 +\frac \gamma 2\int_r^R\frac{u^{q+1}(s)|u'(s)|^{p-2}u'(s)}sds.
\end{align*}% \label{E273}
Since $u'$ is negative, $\beta \geq 0$ and $\gamma \leq 0$ we get
\begin{equation}
\begin{aligned}
\int_r^R\frac{E(s)}s\,ds
\leq& \frac{3p-2}{2p}\int_r^R\frac{|u'(s)|^p}sds+\frac{u(R)|u'(R)|^{p-1}}{2R}
+\frac{\beta \ }4u^2(r) \\
&+\frac 12\int_r^R\frac{u(s)|u'(s)|^{p-1}}{s^2}ds
 + \frac{|\gamma |}2\int_r^R\frac{u^{q+1}(s)|u'(s)|^{p-1}}sds.
\end{aligned}\label{E270}
\end{equation}
Since $E$ is strictly decreasing and converges to zero when $r$
approaches to infinity,  we deduce that $E' \in L^1(] 0,\infty [)$.
 In particular $r|u'|^2$ and $u^q|u'|^p$ lie in $L^1(] 0,\infty [)$.
Letting $R\to$ $\infty $,
\begin{equation}
\begin{aligned}
\int_r^\infty \frac{E(s)}sds
\leq & \frac \beta 4u^2(r)+\frac{
3p-2}{2p}\int_r^\infty \frac{|u'(s)|^p}s\,ds \\
& +\frac 12\int_r^\infty \frac{u(s)|u'(s)|^{p-1}}{s^2}
ds+\frac{|\gamma |}2\int_r^\infty \frac{u^{q+1}(s)|u'(s)|^{p-1}}s\,ds.
\end{aligned}\label{E271}
\end{equation}
Now, we set
\begin{equation}
H(r)=\int_r^\infty \frac{E(s)}sds.  \label{E272}
\end{equation}
First, using the fact that $u^2(r)\leq \frac 2\alpha E(r)$, we
obtain
\begin{equation}
H(r)\geq \int_r^{2r}\frac{E(s)}sds\geq \frac{E(2r)}2\geq \frac \alpha
4u^2(2r).  \label{E273}
\end{equation}
On the other hand, inequality (\ref{E271}) gives
\begin{equation}
\begin{aligned}
H(r)+\frac \beta {2\alpha }rH'(r)
\leq& \frac{3p-2}{2p}\int_r^\infty \frac{|u'(s)|^p}sds
+\frac{|\gamma|}2\int_r^\infty\frac{u^{q+1}(s)|u'(s)|^{p-1}}s\,ds\\
&+\frac 12\int_r^\infty\frac{u(s)}{s^2}|u'(s)|^{p-1}ds.
\end{aligned}\label{E274}
\end{equation}
 Assume that the function $u$ satisfies
\begin{equation}
u(r)\leq Cr^{-\sigma }\quad \text{for }r\geq 1, \label{E275)}
\end{equation}
for some fixed $\sigma \geq 0$ and some constant $C$ (this is
possible because $u(r)\leq A$ for all $r\geq 0$).
If $\sigma\geq (2\alpha)\beta $ we have obviously (\ref{E267}). Assume now
 that $\sigma< (2\alpha)\beta$.
Lemma \ref{lem2.3} implies $|u'(r)|\leq Cr^{-\sigma -1}$ for any $r\geq 1$.
Consequently
\begin{equation}
[ r^{2\alpha /\beta }H(r)] '\leq Cr^{2\alpha /\beta -1-p(\sigma +1)}
[ 1+r^{1-q\sigma }] \quad \text{for } r\geq 1.  \label{E276)}
\end{equation}
By a simple integration, we obtain
\begin{equation}
H(r)\leq Cr^{-2\alpha /\beta }+Cr^{-p(1+\sigma )}+Cr^{1-p-(p+q)\sigma }
\label{E277}
\end{equation}
when $[p(1+\sigma )-2\alpha/\beta][2\alpha /\beta -p+1-(p+q)\sigma]\neq 0$.
Otherwise if
$[p(1+\sigma )-2\alpha/\beta][2\alpha /\beta -p+1-(p+q)\sigma]= 0$, we have
\begin{equation}
H(r)\leq Cr^{-2\alpha /\beta }+Cr^{-2\alpha /\beta }\ln r
+Cr^{-(1-q\sigma +2\alpha /\beta )}.  \label{E278}
\end{equation}
Combining (\ref{E273}), (\ref{E277}), (\ref{E278}) and using the fact that
$\sigma< (2\alpha)\beta $, we deduce that there exists $m>\sigma $ such that
\begin{equation}
u(r)\leq Cr^{-m}{}^{\quad }\text{for all }r\geq 1.
\label{E279}
\end{equation}
If $m=\alpha /\beta $ we have exactly the estimate (\ref{E267}). Otherwise if
$m\neq \alpha /\beta $, the desired estimate (\ref{E267}) follows by
induction starting with $\sigma =m$.
This completes the proof.
\end{proof}

\begin{proof}[Proof of Proposition \ref{prop2.4}]
Set
\begin{equation}
I(r)=r^{\alpha /\beta }\big[ u+\frac 1{\beta r}|u'|^{p-2}u'\big]\,.
\label{E280}
\end{equation}
Then we have
\begin{equation}
I'(r)=\frac 1\beta r^{ \alpha /\beta -1}\big[ (\frac \alpha
\beta -1)\frac{|u'|^{p-2}u'(r)}r+\gamma u^q|u'|^{p-2}u'\big].
\label{E281}
\end{equation}
In view of Lemma \ref{lem2.3} and Lemma \ref{lem2.4}, the functions
$r\mapsto r^{\alpha /\beta -1}u^q|u'(r)|^{p-1}$ and
$r\mapsto r^{\alpha /\beta -2}|u'(r)|^{p-1}$
are in $L^1(] 0,\infty[)$.
Consequently$, I'(r)\in{L^1(] 0,\infty[ )}$. Moreover (\ref{E255}) implies
I(0)=0,  and therefore
\[
\lim_{r\to +\infty }I(r)=\int_0^\infty I'(s)ds\quad
\]
 exists. Since
$ \lim_{r\to +\infty }r^{\alpha /\beta -1}|u'|^{p-2}u'=0$,
we deduce that
 \[\lim_{r\to +\infty }r^{\alpha /\beta }u(r)=L  \in  [ 0,\infty [.
 \]
 This completes the proof.
\end{proof}

\begin{proposition} \label{prop2.5}
 Assume $\alpha >0$, $\beta >0$, $\gamma <0$ and $L=0$
in Proposition \ref{prop2.4}. Then $r^mu(r)\to 0$ and $
r^mu'(r)\to 0$\ as $r\to +\infty$ for
all positive integers $m$.
\end{proposition}

\begin{proof}
>From the proof of the previous proposition,
$\lim_{r\to +\infty }I(r)=0$.
 Thus,  $I(r)=-\int_r^\infty I'(s)ds$. Therefore ,(\ref{E280}) gives
\begin{equation}
u(r)=\frac{-1}{\beta r}|u'|^{p-2}u'(r)-r^{-\alpha /\beta }\int_r^\infty I'(s)ds.
\label{E282}
\end{equation}
Since $\gamma <0$ and $u'<0$, we deduce from (\ref{E285}) that
\begin{equation}
u(r)\leq \frac 1{\beta r}|u'|^{p-1}+\frac 1\beta
|\frac \alpha \beta -1|r^{- \alpha /\beta }\int_r^\infty
s^{ \alpha /\beta -2}|u'|^{p-1}ds.
\label{E283}
\end{equation}
Then in view of Lemma \ref{lem2.3}, we get
\begin{equation}
u(r)\leq Cr^{-(p+(p-1)\alpha )/\beta} .
\label{E284}
\end{equation}
Define the sequence $(m_k)_{k\in {\mathbb N}} $ by
\begin{equation}
\begin{cases}
m_0=\alpha /\beta, \\
m_k=p+(p-1)m_{k-1}.
\end{cases}\label{E285}
\end{equation}
Then
$\lim_{k\to +\infty }m_k=+ \infty$.
Consequently, the proposition follows by induction
starting with $m_0= \alpha /\beta$. This completes the proof.
\end{proof}

\begin{proposition} \label{prop2.6}
Assume $\alpha \leq \beta $ and $\gamma <0$. Let $u$
be a strictly positive solution of \eqref{Q}. Then
$\lim_{r\to +\infty}r^{ \alpha /\beta }u(r)>0$.
\end{proposition}

\begin{proof}
 By Proposition \ref{prop2.4},
\[
\lim_{r\to +\infty }r^{ \alpha /\beta }u(r)\in [ 0,\infty[.
\]
Suppose that
$\lim_{r\to +\infty }r^{\alpha /\beta }u(r)=0$.
Then Proposition \ref{prop2.5} implies
\[
 \lim_{r\to +\infty }
r^{ \alpha /\beta -1}|u'|^{p-2}u'=0,
\]
and therefore, $\lim_{r\to +\infty }I(r)=0$.
On the other hand, (\ref{E281}) implies that the function $I$ given
by (\ref{E280}) is strictly increasing; this is  a contradiction which
completes the proof.
\end{proof}

\section{ Classification of solutions}

In this section we give a classification of solutions of \eqref{Q}.
For this purpose we Set
\begin{gather*}
P=\{ A>0: u(r,A)>0, \forall  r>0\},\\
N=\{ A>0: \exists\,  r_0>0; u(r,A)> 0 \text{for }r\in [ 0,r_0[, u(r_0,A)=0
\text{ and }u'(r_0,A)<0\},\\
C=\{ A>0;\exists\, r_0>0 ;u(r_0,A)=u'(r_0,A)=0\}.
\end{gather*}
This classification depends strongly on the sign of$ \gamma $ and
$ \alpha -\beta$. First, we start with the following result.

\begin{proposition} \label{prop3.1}
Assume $\gamma >0$ and $ \alpha \geq\beta$. Then any solution of
\eqref{Q} changes sign.
\end{proposition}

\begin{proof}
Let $u$ be a solution of \eqref{Q}. Then
\begin{equation}
|u'|^{p-2}u'( r) =-\beta ru(
r) -( \alpha -\beta ) \int_0^ru(s)ds+\gamma
\int_0^ru^q( s) |u'|^{p-2}u'(s) ds. \label{E31}
\end{equation}
for any $r\in [ 0,R_{\max }$. If the set $C$ is not empty, there
exists a finite $r_0>0 $ such that $u(r_0)= u'(r_0)=0 $ and
$u(r)>0$ on $] 0,r_0[ $. Taking  $r=r_{0}$ in (\ref{E31}) and
using again Lemma \ref{lem2.1}, we get
\begin{equation}
( \alpha -\beta ) \int_0^{r_0}u(s)ds+\gamma \int_0^{r_0}u^q|
u'|^{p-1}ds=0. \label{E32}
\end{equation}
This contradicts $\gamma >0$ and $\alpha -\beta \geq 0$. Hence
$C=\emptyset$.

If the set $P $is not empty, without loss of generality we can assume that
equation \eqref{E31} holds with $u$ strictly positive and $u'$ negative.
Then since $\alpha -\beta \geq 0$ and $\gamma >0$ we deduce
\begin{equation}
|u'|^{p-2}u'( r) \leq - \beta ru( r).\label{E33}
\end{equation}
By integrating this last inequality we get a contradiction.
\end{proof}


\begin{proposition} \label{prop3.2}
 Assume $\gamma >0$ and $\alpha< \beta $.
Then \begin{itemize}
\item[(i)] $ u(.,A)$ is strictly positive for any $A\in ] 0,A_0[$
with
\[
A_0=[ \frac {q-1}\beta( \frac {\beta-\alpha }\gamma) ^{ p/(p-1)}] ^{(p-1)/(p(q+1)
-2)}
\]
\item[(ii)] $u(.,A)$ changes sign for large $A$.
\end{itemize}
\end{proposition}

For the proof we need some preliminary results.
Let $u$ be a solution of problem \eqref{Q} defined in $[ 0,R_{\rm max}[ $.
Set
\begin{equation}
h(r)=( \beta -\alpha ) u(r)+\gamma |u|^{q-1}u|
u'|^{p-2}u'(r)\quad \forall r\in [0,R_{\rm max }[ .\label{E34}
\end{equation}
Then the following result holds.

\begin{lemma} \label{lem3.1}
 Assume $\alpha <\beta $. Let $u$ be a
solution of \eqref{Q}. Then $u$  cannot vanish before the first zero
of $h$.
\end{lemma}

\begin{proof}
On the contrary, suppose that $u$ vanishes before$ h $and let $r_0$
be the first zero of $u$. As $h(0)=( \beta -\alpha ) u(0)>0$, then
$h(r_0)\geq 0$ and $h(r)>0$ for $r\in [0,r_0[$.  %\label{E35}
Integrating \eqref{Q}, we obtain
\begin{equation}
|u'|^{p-2}u'(r_0)=\int_0^{r_0}h(s)ds>0.
\label{E35}
\end{equation}
This contradicts$ u'(r_0)\leq 0$ and then, the Lemma is
proved.
\end{proof}

Now, assume that there exists some initial data $A>0$ such that $u(.,A) (=u) $
is a strictly positive solution of \eqref{Q} and set
\begin{equation}
g(r)=\beta r-\gamma u^q|u'|^{p-2}(r),\quad r>0.
\label{E36}
\end{equation}

\begin{lemma} \label{lem3.2}
  There exists $\rho  (=\rho (A))>0$  such that
\begin{equation}
g(\rho )=0\quad\mbox{and}\quad g(r)<0\mbox{ for $r$ in $] 0,\rho[$}. \label{E37}
\end{equation}
Furthermore,
\begin{equation}
\lim_{A\to +\infty }u(\rho ,A)=0\quad \mbox{and} \quad
\lim_{A\to +\infty }u'(\rho ,A)=-\infty.
\label{E38}
\end{equation}
\end{lemma}

\begin{proof}
 First, we observe that the function $g$ satisfies
\begin{equation}
( |u'|^{p-2}u'(r)) '=-\alpha u(r)-u'(r)g(r)\quad \text{for all }r>0.
\label{E39}
\end{equation}
The proof is divided in 3 steps.

\noindent{\bf Step 1.} $g(\rho )=0$ and $g(r)<0$ for $r\in [ 0,\rho [ $.
Recalling (\ref{E225}) we get
\begin{equation}
|u'|^{p-2}u'(r)\sim - \alpha Ar, \quad \text{as }  r\to 0.\label{E310)}
\end{equation}
Hence,
\begin{equation}
g(r)\sim - \gamma A^q( \alpha A r)^{(p-2)/(p-1)}, \quad\text{as }
r\to 0. \label{E311}
\end{equation}
and therefore $g $ starts with a negative value. If $g$ has
a constant sign for all $r>0$, equation (\ref{E39}) gives
\[
( |u'|^{p-2}u') '(r)<- \alpha u(r)<0,\quad \text{for }r>0.
\]
and then the solution $u(.,A)$ must change sign; this is
a contradiction and then (\ref{E37}) follows.

\noindent{\bf Step 2.} We claim that $\lim_{A\to +\infty }u'(\rho,A)=- \infty $.
In fact, equation \eqref{E39} implies that the solution $u $ satisfies
\begin{equation}
\big[ \frac{p-1}p|u'|^p+\frac \alpha 2u^2\
\big] '(r)=-( u') ^2g(r),\quad  \text{for }r>0. \label{E312}
\end{equation}
Integrating this last equality on $[0,R]\subset[0,\rho [$ and using the fact
that $g$ is negative on $[0,\rho [$, we get
\begin{equation}
|u'|^p( r) \geq \frac{p\alpha }{2(
p-1) }[ A^2-u^2(r)] ,\quad \forall r\in [0,\rho [.
\label{E313}
\end{equation}
Hence, if $u(\rho ,A)$ is bounded, by letting $A$ approach $ \infty $, we deduce
\[
\lim_{A\to +\infty }u'(\rho(A) ,A)=- \infty.
\]
Otherwise if $u(\rho ,A)$ is not bounded, then there exists a subsequence,
 denoted also $\rho (A)$ such that
\[
\lim_{A\to +\infty }u( \rho (A),A) =+ \infty .
\]
Now, recalling (\ref{E37}) and (\ref{E39}) we get
\begin{equation}
( |u'|^{p-2}u'(r)) '\leq
- \alpha u(r)<0,\quad \text{for any }r\in [ 0,\rho [ .
\label{E314}
\end{equation}
In particularly, we deduce that $u$ is concave in $[ 0,\rho [ $ and therefore
\begin{equation}
u(r)\geq A+\frac{u(\rho )-A}\rho r,\quad \text{for any }r\in [ 0,\rho[ .\label{E315}
\end{equation}
Thus integrating (\ref{E314}) on $(0,\rho )$, we obtain
\begin{equation}
|u'|^{p-2}u'(\rho )\leq -\alpha \int_0^\rho
\big( A+\frac{u(\rho )-A}\rho r\big) dr.  \label{E316}
\end{equation}
Hence,
\begin{equation}
|u'|^{p-2}u'( \rho ) \leq -\frac
\alpha 2\rho [ A+u(\rho )] .  \label{E317}
\end{equation}
But $g(\rho )=0$, then
$u^q|u'|^{p-2}(\rho )=\frac \beta \gamma \rho$. %\label{319}
Inserting this last equality in (\ref{E317}) the following estimate
holds
\begin{equation}
|u'( \rho ) |\geq \frac{\gamma \alpha }{
2\beta }u^q( \rho ) [ A+u( \rho ) ] .
\label{E318}
\end{equation}
Consequently,
$\lim_{A\to +\infty }u'(\rho,A)=- \infty $.

\noindent{\bf Step3.} We assert that
$\lim_{A\to +\infty }u(\rho ,A)=0$.
In fact, integrating \eqref{E314} on an interval $] 0,r[ \subset ] 0,\rho[ $
and using (\ref{E315}) we obtain
\begin{equation}
|u'|^{p-2}u'( r) \leq -\alpha
r[ A+\frac{u(\rho )-A}{2\rho }r]  \text{ for any }0<r<\rho.
\label{E319}
\end{equation}
On the other hand (\ref{E314}) implies that $u'$ is decreasing in
 $] 0,\rho [ $,  so (\ref{E319}) gives
\begin{equation}
|u'(\rho )|^{p-2}u'( r) \leq
-\alpha r[ A+\frac{u(\rho )-A}{2\rho }r] \quad \text{for any }
0<r<\rho .  \label{320}
\end{equation}
Integrating this last inequality on $] 0,\rho[ $ we get
\begin{equation}
A[ |u'(\rho )|^{p-2}-\frac \alpha 3\rho ^2]
\geq u(\rho )[ \frac \alpha 6\rho ^2+|u'(\rho )|
^{p-2}].  \label{321}
\end{equation}
Therefore,
\begin{equation}
|u'(\rho )|^{p-2}-\frac \alpha 3\rho ^2\geq 0.
\label{E322}
\end{equation}
Recalling that $g(\rho )=0$, this  means
\begin{equation}
\rho =\frac \gamma \beta u^q(\rho )|u'(\rho )|^{p-2}.
\label{E323}
\end{equation}
Combining (\ref{E322}) and (\ref{E323}) we obtain
\begin{equation}
u(\rho )\leq [ \frac 3\alpha (\frac \beta \gamma )^2] ^{1/2q}
|u'(\rho )|^{(2-p)/2q},  \label{E324}
\end{equation}
Since $\lim_{A\to +\infty }u'(\rho ,A)=- \infty $,
 we deduce that
$\lim_{A\to +\infty }u(\rho ,A)=0$.
The proof is complete.
\end{proof}

\begin{lemma} \label{lem3.3}
Let $u(.,A)$ be a strictly positive solution and let $R(A)$
given by Lemma \ref{lem2.2}. Then,
\begin{equation}
\lim_{A\to +\infty }u(R(A),A)=0\quad  and \quad \lim_{A\to +\infty }u'(R(A),A)=- \infty.
\label{E325}
\end{equation}
\end{lemma}

\begin{proof}
First, note that Lemma \ref{lem2.2} implies that $u''<0$ on $[0,R(A)[ $ and then
from step1 of the proof of Lemma \ref{lem3.2}  we get
\begin{equation}
\rho (A)\leq R(A)\quad  \text{and}\quad u'(R(A),A)\leq u'(\rho (A),A).  \label{E326}
\end{equation}
On the other hand, since the function $r\to u(r,A)$ is decreasing, we deduce
\begin{equation}
u(R(A),A)\leq u(\rho (A),A).  \label{327}
\end{equation}
Letting $A\to\infty $ in there  two inequalities,  (\ref{E325})
holds.
\end{proof}

\begin{proof}[Proof of Proposition \ref{prop3.2}]
 The proof is divided in two steps.

\noindent{\bf Step 1.} the proof of part (i).
Set
\begin{equation}
R_0=\sup\{   r>0; h(s)>0\text{\ on\ }[ 0,r[  \}.
\label{328}
\end{equation}
Since $h(0)=( \beta -\alpha ) u(0)>0$, the set
$ \{ r>0: h(r)>0\text{\ on\ }[ 0,r[  \} $ is not empty.

We claim that $R_0$ is infinite. To the contrary,  assume that $R_0$ is
a real number. Then $h(R_0)=0$ and $h'(R_0)\leq 0$, so from Lemma \ref{lem3.1},
$u(R_0)>0 $. Moreover, by continuity, $u(r)\neq 0$ for r $\in ]
R_0-\varepsilon ,R_0+\varepsilon [ $ (with some $\varepsilon >0)$. Thus, we
can write $h(r) $ in the form
\begin{equation}
h(r)=u^q(r)\widetilde{h}(r),  \label{329}
\end{equation}
for any $r\in ] R_0-\varepsilon ,R_0+\varepsilon [$, with
\begin{equation}
\widetilde{h}(r)=( \beta -\alpha ) u^{1-q}(r)+\gamma |
u'|^{p-2}u'(r).  \label{330}
\end{equation}
We clearly have
\begin{eqnarray}
\begin{aligned}
\widetilde{h}'(R_0)=&u(R_0) [-\gamma \beta
+\gamma \beta R_0( \frac{\beta -\alpha }\gamma )
^{1/(p-1)}u^{-(q-1)(p-1)-1}(R_0) \\
&+ (q-1)(\beta -\alpha )( \frac{\beta -\alpha }\gamma ) ^{
1/(p-1)} u^{(2-p(q+1))/(p-1)}(R_0)].
\end{aligned}
\label{E331}
\end{eqnarray}
Since $u$ is decreasing, $\gamma >0$ and $\beta -\alpha \geq 0$,
\begin{equation}
\widetilde{h}'(R_0)>u(R_0)[ -\gamma \beta +(q-1)(\beta -\alpha
)( \frac{\beta -\alpha }\gamma ) ^{ 1/(p-1)}A^{(2-p(q+1))/(p-1)}].  \label{E332}
\end{equation}
Consequently, for any $A$ such that
\begin{equation}
A^{(p(q+1)-2)/(p-1)}<\frac {q-1}\beta ( \frac{\beta -\alpha }
\gamma ) ^{ p/(p-1)},  \label{333}
\end{equation}
we obtain $\widetilde{h}'(R_0)>0$. This contradicts
$ \widetilde{h}(r)>0 $ for r$\in[ 0,R_0[ $ and $ \widetilde{h}(R_0)=0$. Hence
$ R_0$ is infinite; meaning that the function $h$ is strictly positive and
therefore by Lemma \ref{lem3.1}, $u$ is also strictly positive.

\noindent{\bf Step 2.} The proof of part (ii).
Assume for contradiction that $u$ is positive for all $A$. Since
 $u'(R(A),A)<0$ and $u''(R(A),A)=0;$ then by putting $r=R(A)$ in \eqref{Q}
we get
\begin{equation}
\alpha \frac{u(R(A))}{|u'(R(A)|^{p-2}}=u'(R(A))
\big\{ -\beta \frac{R(A)}{|u'(R(A)|^{p-2}}
+\gamma u^q(R(A))\big\}.  \label{E334}
\end{equation}
Invoking Lemma \ref{lem3.3}, we deduce
\begin{equation}
\lim_{A\to +\infty }\frac{R(A)}{|u'(R(A)|^{p-2}}=0.  \label{E335}
\end{equation}
Integrating equation \eqref{Q} on $] R(A),r[ $, we obtain
\begin{equation}
\begin{aligned}
&|u'(r)|^{p-2}u'(r)-|u'(R(A)|^{p-2}u'(R(A))+ \beta ru(r)\\
&-\beta R(A)u(R(A))+(\alpha -\beta )\int_{R(A)}^ru(s)ds
-\gamma \int_{R(A)}^ru^q|u'|^{p-2}u'(s)ds=0.\label{E336}
\end{aligned}
\end{equation}
Since $u'$ is negative and strictly increasing in $[ R(A),\infty[  $ we get
\begin{equation}
|u'(r)|^{p-2}u'(r)>|u'(R(A)|^{p-2}u'(r),  \label{E337}
\end{equation}
for any $r>R(A)$. Hence, equation \eqref{E336} gives
\begin{equation}
\begin{aligned}
|u'(R(A))|^{p-2}u'(r)
<&|u'(R(A))|^{p-2}u'(R(A))-\beta ru(r)+\beta R(A)u(R(A))\\
&+(\beta-\alpha )\int_{R(A)}^ru(s)ds.\label{E338}
\end{aligned}
\end{equation}
Now using the fact that $u $ is decreasing and that $\beta >\alpha >0$,
we obtain
\begin{equation}
u'(r)<u'(R(A))+\beta \frac{R(A)u(R(A))}{|u'(R(A))|^{p-2}}+(\beta -\alpha )\frac{u(R(A))}{|u'(R(A))|^{p-2}}(r-R(A)).  \label{E339}
\end{equation}
Integrating this last inequality on $] R(A),R(A)+1[ $ we get
\begin{equation}
\begin{aligned}
&u(R(A)+1)\\
&<u(R(A))+u'(R(A))+\beta \frac{R(A)}{|u'(R(A))|^{p-2}}u(R(A))
 +  \frac{\beta -\alpha }2\frac{u(R(A))}{|u'(R(A))|^{p-2}}.
\end{aligned} \label{E340}
\end{equation}
Putting together formula \eqref{E335} and Lemma \ref{lem3.3}  we arrive at
$\lim_{A\to +\infty}u(R(A)+1)=- \infty$ which is not possible.
\end{proof}

The remaining of the paper is devoted to the case $\gamma <0$.

\begin{proposition} \label{prop3.3}
Assume $\gamma <0$ and  $\alpha \leq
\beta$. Then all solutions of \eqref{Q} are strictly positive.
\end{proposition}

\begin{proof}
Assume that there exists a real $r_0>0 $such that $u(r_0)=0$
and $u(r)>0$ in $[ 0,r_0[ $. Then from Lemma \ref{lem2.1}, $u'(r)<0$ in
$] 0,r_0[$ and $u'(r_0)\leq 0$.
 Now integrating equation \eqref{Q} on $] 0,r_0[ $ we get
\begin{equation}
|u'|^{p-2}u'( r_0) =( \beta
-\alpha ) \int_0^{r_0}u(s)ds+\gamma \int_0^{r_0}u^q( s)
|u'|^{p-2}u'( s) ds.  \label{E341}
\end{equation}
This is a contradiction with $\gamma <0$ {\it and\ }$\alpha \leq \beta $.
\end{proof}


In the case of $\gamma <0 $ and $\alpha >\beta >0 $ we will prove that
the three sets $P$, $N$ and $C $ are not empty. More precisely we have
the following statement.

\begin{proposition} \label{prop3.4}
Assume $\gamma <0$  and $\beta <\alpha$. Then there exist two constants
$A_{-}$ and $A_{+}$  such that $u(.,A) $ is strictly positive for any
$A\geq A_{+}$ and $u(.,A)$ changes sign for any $0<A\leq A_{-}$; this
implies $]A_{+},+\infty [ \subset P$ and $] 0,A_{-}[\subset N$.
\end{proposition}

\begin{proof}  The proof is divided in two steps.

\noindent{\bf Step 1.} Let  $A, $a large real positive. We scale the
variables and set
\begin{equation}
u(r)=Av(x),\quad x=A^qr,  \label{342}
\end{equation}
for any  $r\in [ 0,R_{\rm max }[ $. Then in terms of the
new variables, problem \eqref{Q} becomes
\begin{equation}
\begin{gathered}
(|v'|^{p-2}v')'+\beta xA^{2-p(q+1)}v'+\alpha A^{2-p(q+1)}v-\gamma |v|
^{q-1}v|v'|^{p-2}v'=0 \\
v(0)=1,\quad v'(0)=0.
\end{gathered} \label{343}
\end{equation}
Since $\gamma <0$, the energy function $E$ given by (\ref{E226}) is
decreasing. In particular for any $r\in [ 0,R_{\rm max }[ $,
\begin{equation}
0<u(r)<A\quad \text{and}\quad |u'(r)|^p\leq \frac {p\alpha}{2(p-1)}A^2. \label{E344}
\end{equation}
Therefore, $v $ and $v'$ are bounded for all $A\geq 1$. In
fact  for any $x\in [ 0,A^qR_{\rm max }[ $,
\[
0<v(x)<1\quad \text{and}\quad |v'(x)|\leq [ \frac{p\alpha
}{2(p-1)}] ^{ 1/p}A^{2/p-(1+q)}.
\]
Let $A\to +\infty $, it follows from standard O.D.E.
arguments that $v(x)$ converges to the solution of the problem
\begin{equation}
\begin{gathered}
(|V'|^{p-2}V')'-\gamma |V|^{q-1}V|V'|^{p-2}V'=0, \\
V(0)=1,\quad V'(0)=0.
\end{gathered}  \label{E345}
\end{equation}
The first equation of this problem can be written as
\begin{equation}
(|V'|^{p-2}V'exp(-\gamma \int_0^r|V|^{q-1}V(s)ds))'=0.  \label{E346}
\end{equation}
Hence $V\equiv 1$. Consequently, $v(.,A)$ converges to 1 when $A$ approaches
$+\infty $; in particular $u(.,A)$ is strictly positive.

\noindent{\bf Step 2}. As for (i), we introduce new variables.  We set
\begin{equation}
u(r)=Aw(x)\quad x=rA^{-(p-2)/p}.  \label{E347}
\end{equation}
Then $w$ satisfies
\begin{equation}
\begin{gathered}
(|w'|^{p-2}w')'+\beta rw'+\alpha w-\gamma A^{q-1+(2(p-1))/p}|w|^{q-1}w|
w'|^{p-2}w'=0, \\
w(0)=1,\quad w'(0)=0.
\end{gathered} \label{E348}
\end{equation}
By letting $A$ approach $0$, the function  $w(x)$ converges to the solution
of the problem
\begin{equation}
\begin{gathered}
(|W'|^{p-2}W')'+\beta xW'+\alpha W=0, \\
W(0)=1, \quad   W'(0)=0.
\end{gathered}  \label{E349}
\end{equation}
We claim that $W$ changes sign. In fact we have
\begin{equation}
|W'|^{p-2}W'(x)+\beta xW(x)=(\beta -\alpha)\int_0^xW(s)ds.  \label{E350}
\end{equation}
If $W $  is strictly positive for all $x$, by using $\beta -\alpha< 0$,
we get
\begin{equation}
( W^{(p-2)/(p-1)}) '(x) \leq -\frac{p-2}{p-1}\beta
^{\frac 1{p-1}}x^{\frac 1{p-1}},\quad \forall x>0;  \label{E351}
\end{equation}
This is  a contradiction. As if there exists some $x_0>0$ such that\\
{$W(x_0)=W'(x_0)=0$} and $W(x)>0$ in $] 0,x_0[ $ we obtain
\[
(\beta -\alpha )\int_0^{x_0}W(s)ds=0,
\]
this is also a contradiction because $\alpha \neq
\beta . $Thereby $W$ is non positive and consequently $u(.,A)$ changes sign for small $
A$. This completes the proof.
\end{proof}

We have also the following result.

\begin{proposition} \label{prop3.5}
 Assume $\gamma <0$  and $ 0<\beta <\alpha $. Then $N$ and $P$ are
non-empty open sets.
\end{proposition}

Before to start the proof we introduce the function
\begin{equation}
\Gamma(r)=u(r)+|u'|^{p-2}u'(r).  \label{352}
\end{equation}

\begin{lemma} \label{lem3.4}
 Assume $\gamma <0$, $\alpha >0 $, and $\beta >0$. Let $u$ be a strictly
 positive solution of \eqref{Q}.
Then the function $\Gamma(r)$ is strictly positive for large $r$.
\end{lemma}

\begin{proof}
 Since $u(r)>0$, Proposition \ref{prop2.4} implies
\[
 \lim_{r\to +\infty }r^{\alpha /\beta }u(r)=L\in [ 0,+\infty[ .
 \]
 If $L>0$, $u(r)\approx Lr^{-\alpha /\beta }$ for large r and then
Lemma \ref{lem2.3} implies
 \[
 \lim_{r\to +\infty }r^{\alpha /\beta }|
u'|^{p-2}u'=0.
\]
Thus, the function  $\Gamma(r)$   behaves like $Lr^{-\alpha /\beta}$,
 as $r\to\infty$ and therefore, $\Gamma$ is strictly positive.

 For the case $L=0$, the proof will
be done into two steps.

\noindent{\bf Step 1.} $\Gamma(r)$ is monotone for large $r$.
For this purpose we set
\begin{equation}
J(r)=\beta ru'(r)+\alpha u(r).  \label{353}
\end{equation}
We assert that $J(r) $has a constant sign for large $r$. In fact, assume that
there exists a large $r_0$ such that $J(r_0)=0$. According to equation
\eqref{Q}, we obtain
\begin{equation}
(p-1)|u'(r_0)|^{p-2}J'(r_0)=-\beta {(\frac{\alpha}{\beta}) }^{p-1}
\frac{u^{p-1}(r_0)}{r_0^{p-1}}\left\{ (p-1)(\alpha/\beta +1)
+\gamma r_0u^q(r_0)\right\}.
\label{354}
\end{equation}
Since
$\lim_{r\to +\infty }r^{\alpha /\beta }u(r)=0$,
we deduce that $J'(r_0)<0$. Consequently $J(r)$
has the same sign for large $r$.
Now, note that for any $r>0$,
\begin{equation}
\Gamma'(r)=u'(r)-\beta ru'-\alpha u(r)+\gamma u^q|
u'|^{p-2}u'(r).  \label{355}
\end{equation}
Hence $\Gamma'(r)$ and $J(r)$ have the opposite signs, in particular the
function $J$ is monotone for large $r$.

\noindent{\bf Step 2.} We claim that $\Gamma$ is not negative for large $r$.
In fact if not,using the step1 we deduce that there exists a large $R_1$ such
that   $\Gamma (r)=u(r) + |u'|^{p-2}u'\leq0$   for any  $r\geq R_{1}$.
Integrating this last inequality on $(R_{1},r)$ we get
\begin{equation}
u^{(p-2)/(p-1)}(r)\leq u^{(p-2)/(p-1)}(R_1)-(p-2)/(p-1)r+(p-2)/(p-1)R_1.  \label{356}
\end{equation}
By letting $r\to+\infty $, we obtain a contradiction.

Combining step 1 and step 2 we deduce $\Gamma(r)>0$ for large $r$.
This completes the proof.
\end{proof}

Now we use step by step the idea introduced by Brezis et al \cite{BPT},
for studying a very singular solution of the heat equation with absorption.
Ever since,  this idea was used in many papers, see for example \cite{PW}.
In order to do this, we write  \eqref{Q} as the system
\begin{equation} \label{S}
\begin{gathered}
u'=|v|^{-(p-2)/(p-1)}v, \\
v'=-\beta r|v|^{-(p-2)/(p-1)}v-\alpha u+\gamma
|u|^{q-1}uv.
\end{gathered} %\label{359}
\end{equation}
For each $\lambda >0$, we define the set
\begin{equation}
{ L}_\lambda =\{ (f_{1, }f_2): 0<f_1<1, -\lambda f_1<f_2<0\}
\label{E358}
\end{equation}

\begin{lemma} \label{lem3.5}
 For any $\lambda >0$ there exists $
r_\lambda =[\lambda +\alpha \lambda ^{-1/(p-1)}]/\beta;$ such that $
{ L}_\lambda $ is positively invariant for $r\geq r_\lambda $.
That is, if
$(u_0,v_0)\in { L}_\lambda $ and $(u(r),v(r))$ is the solution
of \eqref{S} which satisfies $(u(r_0),v(r_0))=(u_0,v_0) $ for some
$r_0>r_\lambda$, then for any $r\geq r_\lambda $ the orbit
$(u(r),v(r)) $ lies in ${L}_\lambda $ for all $ r\geq r_0$.
\end{lemma}

\begin{proof}
We shall show that, given $\lambda >0$, there
exists $r_\lambda >0$ such that if $r>r_\lambda $, then the vector field
determined by \eqref{S} points into ${ L}_\lambda $, except at the
critical point $(0,0)$.
On the top $(f_2=0)$,
\[
v'=-\alpha u<0\quad \text{for all } r>0.
\]
While on the right side $(u=1)$,
\[
v'=-\beta r|v|^{-(p-2)(p-1)}v<0\quad \text{for all }r>0.
\]
On the line $f_2=- \lambda f_1$ we must prove that
$\frac{v'}{u'}<- \lambda $ for large $r$. This is true because
\[
\frac{v'}{u'}=\frac{v'|v|^{(p-2)/(p-1)}}v=-\beta r+\alpha \lambda ^{-1/(p-1)}u^{(p-2)/(p-1)}+\gamma \lambda ^{(p-2)/(p-1)}u^{q+(p-2)/(p-1)}.
\]
Since $\gamma <0$,  if $r\geq (\lambda +\alpha \lambda ^{-
1/(p-1)})/\beta=r_\lambda $ we obtain $\frac{v'}{u'}<- \lambda$.
\end{proof}

\begin{proof}[Proof of Proposition \ref{prop3.5}]. First, note that from
Proposition \ref{prop3.4}, the sets $P$ and $N$ are not empty.
On the other hand, the continuous dependence of solutions on the
initial value implies that $N$ is an open set.
To prove that $P$ is open, take $A_0\in P$ and let a large $r_0>0$ be fixed.
Then by continuous dependence of solutions on the initial data, there is a
neighborhood $O$ of $A_0$ such that $u(r,A)>0$ for any
$(r,A)\in[0,r_0] \times O$. In particular $u(r_0,A)>0$ for any $A\in O$
and then from Lemma \ref{lem3.4}
\begin{equation}
\Gamma(r_0)=u(r_0,A)+|u'|^{p-2}u'(r_0,A)>0.
\label{E359}
\end{equation}
Since $r_0$ is large then $u(r_0,A)<1$ and consequently
$(u(r_0,A),|u'|^{p-2}u'(r_0,A))$ is in $L_1$.
Recalling Lemma \ref{lem3.5}, we deduce that the trajectory remains in $ L_1$
for any $r\geq r_0$, which implies in particular that
$u(r,A)>0 $ for any $r\geq r_0$ and $A\in O$. Therefore,
 $u(r,A)>0$ for any $r\geq0$ and there by $P$ is open.
The proof is complete.
\end{proof}

The rest of the paper is devoted to the study of solution with compact support

\begin{proposition} \label{prop3.6}
Assume that $\gamma<0$ and $\beta<\alpha$,  then there exists at least one
solution with compact support.
\end{proposition}

\begin{proof} As $P$ and $N$ are open disjoint sets, then there exists
$A\in \mathbb {R}^{+}-(P\cup N)$; that is, $u(.,A)$ has a compact support.
\end{proof}

We conclude this paper with a study of the behavior of solution with a
compact support.

\begin{lemma}
Assume $ \gamma <0. $Let $u $be a solution with compact support $[0,R]$. Then
\begin{equation}
( u^{(p-2)/(p-1)}) '(R)=-\frac{p-2}{p-1}\beta ^{^{1/(p-1)}}R^{1/(p-1)}.
\label{E360}
\end{equation}
\end{lemma}

\begin{proof}
Take $r$ close to $R$.  and integrate \eqref{Q} between $r $and $R$;
using the fact that $u $is decreasing, we get
\[
 |u'|^{p-2}u'(r)=\beta ru(r)-( \alpha
-\beta ) \int_r^Ru(s)ds+\gamma \int_r^Ru^q(s)|u'|
^{p-2}u'(s)ds.
\]
Dividing by $u(r) $, we have
\[
\frac{|u'|^{p-1}( r) }{u( r) }
=\beta r-\frac{\alpha -\beta }{u( r) }\int_r^Ru(s)ds+\frac
\gamma {u( r) }\int_r^Ru^q(s)|u'|
^{p-1}u'(s)ds.
\]
First, note that
\[
0\leq \int_r^Ru(s)ds\leq u(r)( R-r).
\]
Hence
\[
\lim_{r\to R }\frac{\alpha -\beta }{u( r) }
\int_r^Ru(s)ds=0.
\]
On the other hand, since the function $u'(s) $is negative and also $|u'|$
decreasing  near $R$, then
\begin{align*}
\frac \gamma {u( r) }\int_r^Ru^q(s)|u'|^{p-2}u'(s)ds
= & \frac{|\gamma |}{u( r) }\int_r^Ru^q(s)|u'|^{p-1}(s)ds\\
\leq & \frac {-|\gamma |}{u( r) }|u'|^{p-2}(r)\int_r^Ru^q(s)u'(s)ds\\
\leq & \frac{|\gamma |}{q+1}u^q(r)|u'(r)|^{p-2}.
\end{align*}
The last term of this inequality approaches zero as ${r\to R}$ and then
 we get
\[
\lim_{r\to R }\frac{|u'|
^{p-1}( r) }{u( r) }=\beta R.
\]
This is equivalent to (\ref{E360}); thus the proof of the lemma is complete.
\end{proof}

\subsection*{Acknowledgments}.
The authors would like to express their deep gratitude to professor
 M. Kirane for his assistance, and  also thank the anonymous referee for
his/her helpful comments. This work was supported financially by
Centre National de Coordination et de Planification de La Recherche
 Scientifique et Technique PARS MI 29.

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