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\begin{document}
\title{Attraction properties of unbounded solutions for a nonlinear
  Abel integral equation} 

\author{M.R. Arias\thanks{This paper has been
  partially supported by Project MTM2004-06226 of CICYT}
  \qquad R. Benítez$^*$ \qquad V. J. Bolós$^*$\\ Dpto. de Matemáticas,
  Universidad de Extremadura\\
  Avda. de Elvas s/n. 06071 Badajoz (Spain) \\ e-mail:
  \texttt{arias@unex.es} \qquad \texttt{rbenitez@unex.es}\qquad
  \texttt{vjbolos@unex.es}}
\date{}
\maketitle


%\author[Badajoz]{M. R. Arias\corauthref{cor}}\ead{arias@unex.es},
%\author[Plasencia]{R. Benítez}
%\ead{rbenitez@unex.es},
%\author[Badajoz]{V. J. Bolós}
%\ead{vjbolos@unex.es}


%\address[Badajoz]{Dep. Matemáticas, Facultad de Ciencias,
%  Universidad de Extremadura. Avda. de Elvas s/n. 06071--Badajoz,
%  Spain.}  \address[Plasencia]{Dep. Matemáticas, C. U. de Plasencia,
%  Universidad de Extremadura, Avda. Virgen del Puerto, 2.
%  10600--Plasencia (Cáceres), Spain.}  \corauth[cor]{Corresponding
%  author. Fax: 34-924-272911} \thanks[proy]{This paper has been
%  partially supported by Project MTM2004-06226 of CICYT}
\begin{abstract}
  In this paper a nonlinear Abel integral equation with a power
  nonlinearity is considered. This equation has a solution which is
  unbounded in zero, i.e., is unbounded on $[0,\delta)$, for every
  positive $\delta$. Attraction properties for this solution
  are studied. We show that it is always possible to find functions
  attracted by the unbounded solution, as well as functions not
  attracted by such solution.
\end{abstract}

%\begin{keyword}
\noindent 
\textbf{Keywords:} Nonlinear Abel integral equations, attracting behaviour, attraction basins. \\\noindent
%\MSC 45D05 \sep 45Gxx \sep 45M20
%\end{keyword}
\section{Introduction}

In this paper we consider the Abel integral equation
\begin{equation}
u(x)=\int_{0}^{x}(x-s)^{\alpha }u(s)^{\beta } ds, \label{eq:abel1}
\end{equation}%
where $x\geq 0$ and $(\alpha ,\beta )\in (-1,0)\times (-1/\alpha
,+\infty )$. This equation is a particular case of a nonlinear
Volterra integral equation with convolution kernel

\begin{equation}\label{eq:vol1}
u(x)=\int_0^xk(x-s)g(u(s)) ds;
\end{equation}
where the kernel $k$ and the nonlinearity $g$ satisfy the
following properties:
\begin{itemize}
\item $k$ is a positive function of $L^1_{loc}(\mathbb{R}^+)$, such
that $K(x)=\int_0^xk(s) ds$ is a strictly increasing function.
\item $g$ is a continuous strictly increasing function, such that
$g(0)=0$, $g'>0$ almost everywhere, and transforms null sets into
null sets.
\end{itemize}
Function zero is a solution of (\ref{eq:vol1}), known as {\it trivial
  solution}. Let $u$ be a solution of equation (\ref{eq:vol1}). For
every $c>0$, function
\[
 u_c(x)=
\begin{cases}
  0&\text{ if }x\in [0,c)\\
u(x-c)&\text{ if }x\geq c,\\
\end{cases}
\]
is also a solution of (\ref{eq:vol1}); this kind of solutions is known
as {\it horizontally translated solution}. In this paper we are only
interested in {\it nontrivial solutions}, i.e., neither in the trivial
solution nor in horizontally translated solutions. Therefore when we
refer to the uniqueness of solutions, we must not forget that we are
just considering nontrivial solutions.

It is known that equation (\ref{eq:abel1}) has a positive solution,
$u$, such that $%
\lim_{x\rightarrow 0+}u(x)=+\infty $, see \cite{AriBen03a}. Indeed, $u$ has
the form
\begin{equation}
u(x)=Kx^{\gamma },  \label{eq:sol1}
\end{equation}%
being $K=B(\alpha +1,\gamma -\alpha )^{1/(1-\beta )}$, $\gamma
=(\alpha +1)/(1-\beta )$ and $B$ is the Euler beta function. Since
$(\alpha ,\beta )\in (-1,0)\times (-1/\alpha ,+\infty )$, we have that
$K>0$ and $\gamma <0$.

Solutions of equation (\ref{eq:abel1}) are the fixed points of the
nonlinear integral operator
\[
Tf(x)=\int_0^x(x-s)^\alpha f(s)^\beta\,ds.
\]
Therefore, the integral equation (\ref{eq:abel1}) can be considered as
the fixed point equation $u=Tu$.  Since $\beta$ is positive, it is
immediate that $T$ is a monotone increasing operator in the following
sense: if $f_1\leq f_2$ on $\mathbb{R}^+$, then $Tf_1\leq Tf_2$ on
$\mathbb{R }^+$.

In the study of fixed point equations, it is interesting to know
whether solutions are attractors or not.  Recall that it is said that
a fixed point $u$, of the operator $T$, is a {\it global attractor} of
a family of functions $\mathcal{F}$ if, for every $f\in\mathcal{F}$,
$lim_{n\to\infty}T^nf(x)=u(x)$, on $\mathbb{R}^+$; here $T^n$ denotes,
as usual, the composition of $T$ with itself, $n$ times. Analogously,
it is said that $u$ is a {\it local attractor} of $\mathcal{F}$ if we
can only assure the convergence of
$\left(T^nf(x)\right)_{n\in\mathbb{N}}$ to $u(x)$, \textit{near zero}.
In this paper we say that a property $P$ is held \textit{near zero} if
there exists a positive $\delta$ such that $P$ holds on $\left(
  0,\epsilon\right)$, for every $0 <\epsilon<\delta $.


In \cite{AriBen03b} it was compared the attractive behaviour of the
solution defined in (\ref {eq:sol1}), and the attractive behaviour of
continuous solutions for Volterra integral equation (\ref{eq:vol1})
when the kernel is bounded near zero (see \cite{Ari00,AriBen01}). In
this case, the solutions of (\ref{eq:vol1}) are global attractors of
all positive and measurable functions different from zero near zero.
When kernels are just positive measurable functions, continuous
solutions of Volterra integral equation (\ref{eq:vol1}) are global
attractors of all positive and locally bounded functions not vanishing
near zero (see \cite{AriBen03a}).
In \cite{AriBen03b} it is shown that
there is a family of functions not attracted by $u$. In light of that
result, it was pointed out that solution $u$ could actually be a {\it
  repellor}, that is, a solution such that for any positive and
measurable function $f\not= u$ almost everywhere, the sequence
$\left(T^nf\right)_{n\in\mathbb{N}}$ converges to the trivial
solution, or diverges to $+\infty$.


In this paper we are going to study the attractive character of
the unbounded solution $u$, and we shall show that it is not a
repellor, by determining a part of its attraction basin. To that aim
we introduce the following definition.

\begin{definition}
  A positive function $f$, defined on $\mathbb{R}^{+}$ is
  $u$-separable, if there exists a positive constant $a\not=1$ such
  that

\begin{itemize}
\item $f>au$ near zero, if $a>1$,

\item $f<au$ near zero, if $0<a<1$.
\end{itemize}
\end{definition}

Next result was already proved in \cite[theorem 2]{AriBen03b}. The
proof presented here is much shorter and easier.

\begin{lemma}
\label{lem:1} Let $f=au$, where $a>0$. Then, for every $x>0$,

\begin{itemize}
\item $\lim_{n\to\infty}T^nf(x)=+\infty$, for $a>1$.

\item $\lim_{n\to\infty}T^nf(x)=0$, for $0<a<1$.
\end{itemize}
\end{lemma}

\begin{proof}
  The proof is straightforward. Note that
  \[
    T^nf(x)= T^{n-1}[a^\beta u](x)=
    a^{\beta^n} u(x)=a^{\beta^n -1}f(x).
  \]
Hence, since $\beta>1$, the proposition is proved.
\end{proof}

From Lemma \ref{lem:1} and the monotone increasing character
of $T$, we can assure that functions $u$-separable, are
repelled by $u$. This fact suggests that the unbounded solution for
equation (\ref{eq:abel1}) is a repellor of all positive and
measurable functions. Nevertheless, we are going to show in this
paper that this statement is false; that is, there exist functions,
different from the solution, that are attracted by it.

As far as we have seen in the literature
\cite{AriCas99,Buc97,KilSai99,Myd99,RobMan51,Pad58,Szw92}, solutions for
nonlinear homogeneous Volterra integral equations, either attract all
measurable and positive functions (in case they are attractors), or
are the only element in its attraction basin. Therefore, the results
we present in this work form an example of an integral equation with a
solution not satisfying this rule.
\section{Functions non $u$-separable}

Through this paper $u$ will denote the unbounded solution of
(\ref{eq:vol1}), given in (\ref{eq:sol1}). As it was indicated in the
introduction, from Lemma \ref{lem:1}, it is clear that functions
attracted by $u$ are not $u$-separable. Let us consider the family of
functions
\[
  \mathcal{V}=\left\{v:\lim_{x\to 0+}\frac{v(x)}{u(x)}=1\right\}.
\]
Functions from $\mathcal{V}$ are not $u$-separable. However, note that
$\mathcal{V}$ is not the total set of non $u$-separable functions.
Indeed, function $v(x)=u(x)(\sin(1/x)+1)/2$ is not $u$-separable and
does not belong to $\mathcal{V}$.

It is not difficult to show that $T(\mathcal{V})\subset\mathcal{V}$.
Now, we are going to study the iterations by $T$ of the set
\begin{equation*}
  \mathcal{W}= \{v: v(x)=u(x)+cx^p;\quad c\in\mathbb{R}
,\,p\in (\gamma,+\infty)\}.
\end{equation*}
Note that $\mathcal{W}\subset\mathcal{V}$. Our aim in this section
is to prove that some functions of $\mathcal{W}$ are in fact
attracted by $u$. To do so, for any $v\in\mathcal{W}$, we need to
find a suitable expression of $Tv$ that allows us to compare it
with $v$.

First, in order to simplify the notation, we shall introduce the
following function
\[ F(\alpha,\gamma,p):=\left(1-\frac{\alpha+1}{\gamma}\right)
    \frac{B(\alpha+1,p-\alpha)}{B(\alpha+1,\gamma-\alpha)},
\]
with $\left( \alpha ,\gamma ,p\right) \in \left(
-1,0\right) \times \left( \alpha ,0\right) \times \left( \gamma
,+\infty \right)$.

\begin{proposition}
\label{prop1} For every  $c\in \mathbb{R}$ and $\left( \alpha
,\gamma ,p\right) \in \left( -1,0\right) \times \left( \alpha
,0\right) \times \left( \gamma ,+\infty \right)$, we have
\[
Tv\left( x\right) =u\left( x\right)
+F(\alpha,\gamma,p)cx^{p}+\sum_{m=2}^{+\infty }k_{m1}x^{m\left(
p-\gamma \right) +\gamma },
\]
near zero, being $v(x)=u(x)+cx^p$ and $k_{m1}$ real constants.
\end{proposition}
\begin{proof}
Taking into account the definition of $v$, we have
\[
  Tv(x)=\int_0^x(x-s)^\alpha (u(s)+cs^p)^\beta\,ds.
\]
Since $\lim_{x\to 0+}\frac{cx^p}{u(x)}=0$, there exists a
neighborhood $(0,\delta)$ such that
$\left|\frac{cx^p}{u(x)}\right|<1$, and therefore the series
\[
  1+\sum_{m=1}^{+\infty}\frac{1}{m!}\left[\prod_{i=0}^{m-1}(\beta-i)\right]
\left(\frac{cx^p}{u(x)}\right)^m
\]
converges uniformly to $\left(1+\frac{cx^p}{u(x)}\right)^\beta$ on
$(0,\delta)$ (see \cite[theorem 7.46]{Strom}). Thus, on $(0,\delta)$,
it is verified that
%% \begin{equation}
%%   \label{eq:chunga1}
%%   \begin{split}
%%     Tv(x)
%%     &=\int_0^x(x-s)^\alpha u(s)^\beta\,ds+c\beta\int_0^x(x-s)^\alpha
%% u(s)^{\beta-1}s^p\,ds\\
%% &+\sum_{m=2}^{+\infty}c^m\frac{\beta(\beta-1)\cdots(\beta-m+1)}{m!}
%% \int_0^x(x-s)^\alpha u(s)^{\beta-m}s^{pm}\,ds
%%   \end{split}
%% \end{equation}

\begin{eqnarray}
  Tv\left( x\right)&=&\int_{0}^{x}\left( x-s\right)^{\alpha
  }\left( u\left( s\right) +cs^{p}\right) ^{\beta }ds  \label{suma}\nonumber\\
  &=&\int_{0}^{x}\left( x-s\right) ^{\alpha }u\left( s\right) ^{\beta
  }ds \label{suma_1} \\
  &+&c\beta \int_{0}^{x}\left( x-s\right) ^{\alpha }u\left( s\right)
  ^{\beta -1}s^{p}ds  \label{suma_2} \\
  &+&\sum_{m=2}^{+\infty }c^{m}\frac{1}{m!}\left[\prod_{i=0}^{m-1}(\beta-i)\right]
  \int_{0}^{x}\left( x-s\right)
  ^{\alpha }u\left( s\right) ^{\beta -m}s^{pm}ds \label{suma_3}
\end{eqnarray}%


The term (\ref{suma_1}) is $u(x)$, because $u$ is a solution of
(\ref{eq:abel1}).

Moreover, taking into account that
$u(x)=B(\alpha+1,\gamma-\alpha)^{1/(1-\beta)}x^\gamma$, 
(\ref{suma_2}) can be written as
\[
\begin{split}
  c\beta\int_0^x(x-s)^\alpha u(s)^{\beta-1}s^p\,ds&
  =\frac{c\beta}{B(\alpha+1,\gamma-\alpha)}
  \int_0^x(x-s)^\alpha s^{p-\alpha-1}\,ds\\
  =c\beta\frac{B(\alpha+1,p-\alpha)}
    {B(\alpha+1,\gamma-\alpha)}x^p
&=\left(1-\frac{\alpha+1}{\gamma}\right)
    \frac{B(\alpha+1,p-\alpha)}{B(\alpha+1,\gamma-\alpha)}cx^p\\
    =F(\alpha,\gamma,p)cx^p.&\\
\end{split}
\]

To complete the proof it suffices to show that, near zero,
(\ref{suma_3}) can be written as $\sum_{m=2}^{+\infty
}k_{m1}x^{m\left( p-\gamma \right) +\gamma }$, for some constants
$k_{m1}$.

We have
\[
  \begin{split}
    &\sum_{m=2}^{+\infty}\frac{c^m}{m!}\left[\prod_{i=0}^{m-1}(\beta-i)\right]
    \int_0^x(x-s)^\alpha u(s)^{\beta-m}s^{pm}\,ds\\
    =&\sum_{m=2}^{+\infty}\frac{c^m}{m!B(\alpha+1,\gamma-\alpha)
      ^{\frac{\beta-m}{\beta-1}}}
    \left[\prod_{i=0}^{m-1}(\beta-i)\right]\int_0^x(x-s)^\alpha
    s^{\gamma(\beta-m)+pm}\,ds\\
    =&\sum_{m=2}^{+\infty}\frac{c^m}{m!}\left[\prod_{i=0}^{m-1}(\beta-i)\right]
    \frac{B(\alpha+1,m(p-\gamma)+\gamma-\alpha)}{B(\alpha+1,\gamma-\alpha)^
      {\frac{\beta-m}{\beta-1}}}x^{m(p-\gamma)+\gamma}\\
    =&\sum_{m=2}^{+\infty }k_{m1}x^{m\left( p-\gamma \right) +\gamma
    }.
  \end{split}
\]
\end{proof}

\begin{lemma}
\label{lem:2} Under the hypotheses and notation of last
proposition, expression $\sum_{m=2}^{+\infty }k_{m1}x^{m\left(
p-\gamma \right) +\gamma }$ verifies that:
\begin{enumerate}
\item[(a)] it is of order  $x^{p+\frac{1}{2}\left( p-\gamma
\right) }$, \item[(b)] it is positive,
\end{enumerate}
near zero.
\end{lemma}

\begin{proof}
\begin{enumerate}
\item[(a)] Considering the convergence orders, we can find a
constant $M>1$ such that, for every $m\in\mathbb{N}$,
\[
\left|k_{m1}\right|=\left|\frac{c^m}{m!}\left[\prod_{i=0}^{m-1}(\beta-i)\right]
\frac{B(\alpha+1,m(p-\gamma)+\gamma-\alpha)}{B(\alpha+1,\gamma-\alpha)^
{\frac{\beta-m}{\beta-1}}}\right|<M^m.
\]
Thus, for every $x\in \left( 0,\left( 2M\right)
^{\frac{-4}{p-\gamma }}\right)$, we obtain
\begin{equation}\label{desp}
\begin{split}
 \sum_{m=2}^{+\infty }\left|k_{m1}\right|x^{m\left(
p-\gamma \right) +\gamma }<& \sum_{m=2}^{+\infty }M^{m}x^{m\left(
p-\gamma \right) +\gamma }\\
  =& \sum_{m=2}^{+\infty }M^{m}x^{\frac{1}{4}m\left( p-\gamma \right)}
x^{\frac{3}{4}m\left( p-\gamma \right) +\gamma}\\
<&\sum_{m=2}^{+\infty }
2^{-m}x^{%
  \frac{3}{4}m\left( p-\gamma \right) +\gamma }
<x^{p+\frac{1}{2}\left( p-\gamma \right) }\sum_{m=2}^{+\infty
}2^{-m}\\
=&\mathcal{O}\left( x^{p+\frac{1}{2}\left( p-\gamma
\right)
  }\right).\\
\end{split}
\end{equation}
\item[(b)] In a similar way it can be proved that
\[
\sum_{m=3}^{+\infty }\left\vert k_{m1}x^{m\left( p-\gamma \right)
+\gamma}\right\vert=\mathcal{O}\left(x^{p+\frac{3}{2}(p-\gamma)}\right),
\]
and hence the sign of $\sum_{m=2}^{+\infty }k_{m1}x^{m\left(
p-\gamma \right) +\gamma }$ equals the sign of its first term near
zero, which is always positive.
\end{enumerate}
\end{proof}

From Lemma \ref{lem:2}, $\sum_{m=2}^{+\infty }k_{m1}x^{m\left(
p-\gamma \right) +\gamma }$ is positive and negligible in
comparison with $u(x)+F(\alpha,\gamma,p)cx^p$. Hence, taking into
account Proposition \ref{prop1}, the following properties hold
near zero:
\begin{itemize}
\item If $F\left( \alpha ,\gamma ,p\right) >1$ then, if $c>0$,
  $Tv>v>u$, while if $c<0$, then $Tv<v<u$.
%$\left\vert Tv\left(
%      x\right) -u\left( x\right) \right\vert >\left\vert v\left(
%      x\right) -u\left( x\right) \right\vert$.
\item If $F\left( \alpha ,\gamma ,p\right) =1$ and $c>0$ then $u<v<Tv$.
%  $\left\vert Tv\left( x\right) -u\left( x\right) \right\vert
%  >\left\vert v\left( x\right) -u\left( x\right) \right\vert$.
\item If $F\left( \alpha ,\gamma ,p\right) =1$ and $c<0$ then $v<Tv<u$.
%  $\left\vert Tv\left( x\right) -u\left( x\right) \right\vert
%  <\left\vert v\left( x\right) -u\left( x\right) \right\vert$.
\item If $F\left( \alpha ,\gamma ,p\right) <1$ then, if $c>0$,
  $u<Tv<v$, while if $c<0$, then $v<Tv<u$.
%$\left\vert
%$    Tv\left( x\right) -u\left( x\right) \right\vert <\left\vert
%    v\left( x\right) -u\left( x\right) \right\vert$.
\end{itemize}
So, taking into account the monotony of $T$, in the first two cases,
the sequence $\left(|T^nv(x)-u(x)|\right)_{n\in\mathbb{N}}$ is
increasing and therefore $v$ is not attracted by $u$ near zero; while
in the last two cases, $\left(|T^nv(x)-u(x)|\right)_{n\in\mathbb{N}}$
is decreasing and hence $v$ is ``getting closer" to $u$ near zero.

Now the importance of the function $F$ in the study of the attracting
behaviour of $u$ becomes clear.

Let us see some properties of $F$. Note that for any fixed $\left(
  \alpha _{0},\gamma _{0}\right) \in \left( -1,0\right) \times \left(
  \alpha _{0},0\right)$, the map $p\mapsto F(\alpha_0,\gamma_0,p)$ is
strictly decreasing on $(\gamma_0,+\infty)$. Moreover,

\begin{itemize}
\item $\lim_{p\rightarrow \gamma _{0}^{+}}F\left( \alpha _{0},\gamma _{0},p\right)
= 1-(\alpha _{0}+1)/\gamma_{0} >1,$ and
\item $\lim_{p\rightarrow +\infty }F\left( \alpha _{0},\gamma _{0},p\right) =0$.
\end{itemize}

Hence, the existence of a unique $\tilde{p}\in \left( \gamma
  _{0},+\infty \right)$ such that $F\left( \alpha _{0},\gamma
  _{0},\tilde{p}\right) =1$ follows from the continuity of $F$ on
$\left( -1,0\right) \times \left( \alpha ,0\right) \times \left(
  \gamma ,+\infty \right) $.  Therefore, on a small neighborhood of
zero, we have:
\begin{itemize}
\item If $\gamma_0<p<\tilde{p}$ then $F(\alpha_0,\gamma_0,p)>1$, and
  so $v$ is not attracted by $u$.
\item If $p=\tilde{p}$ then, for $c>0$, $v$ is not attracted by
$u$; and for $c<0$, $Tv$ is getting closer to $u$. \item If
$p>\tilde{p}$, then $Tv$ is getting closer to $u$.
\end{itemize}

Consequently, there can always be found functions different from $u$,
whose iterants are getting closer to $u$.  In next section it will be
shown that some of these functions are actually attracted by $u$.

On the other hand, given any fixed $\left( \alpha
_{0},p_{0}\right)\in\left(
  -1,0\right) \times \left( \alpha _{0},+\infty \right)$, the map
$\gamma\mapsto F(\alpha_0,\gamma,p_0)$ is strictly increasing on
$\left( \alpha _{0},\min \left( 0,p_{0}\right) \right)$. Moreover,

\begin{itemize}
\item $\lim _{\gamma \rightarrow \alpha _{0}^{+}}F\left( \alpha
    _{0},\gamma ,p_{0}\right)=0$ and
\item $\lim _{\gamma \rightarrow \min \left( 0,p_{0}\right) ^{-}}F\left(
\alpha _{0},\gamma ,p_{0}\right)=+\infty$.
\end{itemize}
Hence, it follows that there exists a unique
$\tilde{\gamma}\in\left(\alpha_{0},\min\left(0,p_{0}\right)\right)$
such that $F\left(\alpha_{0},\tilde{\gamma},p_{0}\right)=1$. Then,
equation $F\left(\alpha,\gamma,p\right)=1$ with $-1<\alpha<0$,
$\alpha<p$ and $\alpha<\gamma<\min\left(0,p\right)$, represents a
uniparametric family of curves (with parameter $p$), in
$\mathbb{R}^2$ (see Figure \ref{fig1}).


Given $\alpha$, $\beta$ (and therefore $\gamma$), and considering
$p>\gamma$, we shall see in next section that if the point
$(\alpha,\gamma)$ lies below the level curve $F(\cdot,\cdot,p)=1$, the
solution $u$ attracts all functions $u(x)+cx^p$. On the other hand, if
$(\alpha,\gamma)$ lies above of the such level curve, it will be shown
that functions $u(x)+cx^p$ are not attracted by $u$. Finally, if
$(\alpha,\gamma)$ is on the level curve, it can be only assured that
$u(x)+cx^p$ will get closer to $u$ if $c<0$, or will not be attracted
by $u$ if $c>0$.

\begin{figure}[tbp]
\begin{center}
  \includegraphics[width=0.6\textwidth]{grafica_F}
\end{center}
\caption{Level curves $F(\alpha,\gamma,p)=1$ with
$p=-0.8,-0.5,0,2,100$,
  defined on the region $\left(\alpha,\gamma\right)
  \in\left(-1,0\right)\times \left(\alpha,0\right)$.}
\label{fig1}
\end{figure}

\section{Attraction}
We have already seen that there exist functions getting closer to
the solution $u$. In this section, we are going to see that some
of these functions are actually attracted by $u$.

First we will need the following lemma, which generalizes
Proposition \ref{prop1} in the case $F(\alpha,\gamma,p)<1$.
\begin{lemma}\label{lem:3}
Given $c\in \R$, and $(\alpha ,\gamma, p)\in \left( -1,0\right)
\times \left( \alpha ,0\right) \times \left( \gamma ,+\infty
\right)$ such that $F(\alpha,\gamma,p)<1$, we have
\[
  T^{n}v\left( x\right) =u\left( x\right)
+F(\alpha,\gamma,p)^{n}cx^{p}+\sum_{m=2}^{+\infty
}k_{mn}x^{m\left( p-\gamma \right) +\gamma },
\]
being $v(x)=u(x)+cx^p$ and $k_{mn}$ real constants for every
natural $n,m$.
\end{lemma}
\begin{proof} 
Let us assume that $c>0$ (the case $c<0$ is analogous). We
have
\begin{equation}\label{eq:chunga6}
\begin{split}
T^2v(x)=&T(Tv(x))=\int_0^x(x-s)^\alpha\left(Tv(s)\right)^\beta ds\\
=&\int_0^x(x-s)^\alpha
u(s)^\beta\left(1+\frac{Tv(s)-u(s)}{u(s)}\right)^\beta ds.
\end{split}
\end{equation}


From the definition of $v$, we can assure that
$\frac{v\left(x\right)-u\left(x\right)}{u\left(x\right)}<1$. Since
$F(\alpha,\gamma,p)<1$ then
$\frac{Tv\left(x\right)-u\left(x\right)}{u\left(x\right)
}<\frac{v\left(x\right)-u\left(x\right)}{u\left(x\right)}<1$, and
hence the series
\[
1+\sum_{m=1}^{+\infty
}\frac{1}{m!}\left[\prod_{i=0}^{m-1}(\beta-i)\right]\left(
\frac{Tv\left(x\right) -u\left( x\right) }{u\left( x\right)
}\right) ^{m}
\]
converges uniformly to
$\left(1+\frac{Tv\left(x\right)-u\left(x\right)}{u\left( x\right)
  }\right) ^{\beta }$. Then (\ref{eq:chunga6}) takes the following
  form
  \begin{eqnarray}
T^2v(x)&=&\int_0^x(x-s)^\alpha u(s)^\beta ds \label{eq:chunga71}\\
&+&\int_0^x(x-s)u(s)^\beta\beta\frac{Tv(s)-u(s)}{u(s)} ds\label{eq:chunga72}\\
&+&\int_0^x(x-s)^\alpha u(s)^\beta\sum_{m=2}^{+\infty
}\frac{1}{m!}\left[\prod_{i=0}^{m-1}(\beta-i)\right]\left(
\frac{Tv\left(s\right) -u\left( s\right) }{u\left( s\right)
}\right)^{m} ds \nonumber \\ \label{eq:chunga73}
  \end{eqnarray}

  Since $u$ is a solution of (\ref{eq:abel1}), (\ref{eq:chunga71}) is
  $u(x)$.

Taking into account the expression (\ref{eq:sol1}) and Proposition
\ref{prop1}, it can be shown that (\ref{eq:chunga72}) can
be written as $F(\alpha,\gamma,p)^2cx^p+
\sum_{j=2}^{+\infty}\hat{k}_{j2}x^{j(p-\gamma)+\gamma}$, for some
constants $\hat{k}_{j2}$.

Now, from (\ref{eq:sol1}) and Proposition \ref{prop1}, and since
the convolution of $x^\alpha$ and
$x^{\gamma-\alpha-1+m(p-\gamma)}$ is a power of exponent
$m(p-\gamma)+\gamma$, it can be shown that the term
(\ref{eq:chunga73}) can be written as
$\sum_{m=2}^{+\infty}\tilde{k}_{m2}x^{m(p-\gamma)+\gamma}$, for
some constants $\tilde{k}_{m2}$.

Therefore,
\[
T^2v(x)=u(x)+F(\alpha,\gamma,p)^2cx^p+
\sum_{m=2}^{+\infty}k_{m2}x^{m(p-\gamma)+\gamma},
\]
for some constants $k_{m2}$.

Repeating the above reasoning we can assure that for every natural
$n$, the statement of the lemma holds.
\end{proof}


\begin{theorem} \label{teo}
Given $c\in \R$, and $(\alpha ,\gamma, p)\in \left( -1,0\right)
\times \left( \alpha ,0\right) \times \left( \gamma ,+\infty
\right)$ such that $F(\alpha,\gamma,p)<1$, the function
$v(x)=u(x)+cx^p$ is attracted by the solution $u$.
\end{theorem}

%\begin{proof}
\begin{proof} Let us assume that $c>0$ (the case $c<0$ is analogous). Then,
$\left(T^nv\right)_{n\in\mathbb{N}}$ is a strictly decreasing
sequence, bounded from below by $u$. Hence it converges
pointwisely to a function $\omega$ that, by means of the Monotone
Convergence Theorem, is a fixed point of the operator $T$. We will
see next that $\omega=u$.

From Lemma \ref{lem:3} we have
\[
  T^{n}v\left( x\right) =u\left( x\right)
+F(\alpha,\gamma,p)^{n}cx^{p}+\sum_{m=2}^{+\infty }k_{mn}
x^{m\left( p-\gamma \right) +\gamma
},
\]
being $k_{mn}$ some constants for every natural $n,m$. Thus,
\[
\begin{split}
  \omega(x)=&\lim_{n\to+\infty }T^{n}v\left( x\right)\\
  =&u\left( x\right) +\lim_{n\to +\infty }\left(
    F(\alpha,\gamma,p)^{n}cx^{p}+\sum_{m=2}^{+\infty
    }k_{mn}x^{m\left( p-\gamma \right) +\gamma }\right)\\
=&u(x)+\lim_{n\to +\infty}\sum_{m=2}^{+\infty
    }k_{mn}x^{m\left( p-\gamma \right) +\gamma }\\
\end{split}
\]
Since the convergence of last series is uniform, defining
$k'_m:=\lim_{n\to\infty}k_{mn}$, it follows that
\begin{eqnarray}\label{eq:chunga4}
\omega(x) &=&u\left( x\right) +\sum_{m=2}^{+\infty }k_{m}^{\prime
}x^{m\left( p-\gamma \right)
+\gamma } \nonumber\\
&=&u\left( x\right) +k_{2}^{\prime }x^{2\left( p-\gamma \right)
+\gamma }+\sum_{m=3}^{+\infty }k_{m}^{\prime }x^{m\left( p-\gamma
\right) +\gamma }.
\end{eqnarray}%

Since $2\left( p-\gamma \right) +\gamma >p$, from the strictly
decreasing character of the map $p\mapsto F(\alpha,\gamma,p)$, we
can infer that $0<F\left( \alpha ,\gamma ,2\left( p-\gamma \right)
+\gamma \right) <F\left( \alpha ,\gamma ,p\right) <1$, therefore,
using similar arguments as those of Lemma \ref{lem:2}, it can be
shown that
\begin{equation}\label{eq:chunga5}
T\omega(x) =u\left( x\right) +F\left( \alpha ,\gamma ,2\left(
p-\gamma \right) +\gamma \right) k_{2}^{\prime }x^{2\left(
p-\gamma \right) +\gamma }+\sum_{m=3}^{+\infty }k_{m}^{\prime
\prime }x^{m\left( p-\gamma \right) +\gamma }; \end{equation}
where $k_{m}^{\prime \prime }$ are some constants. Comparing
(\ref{eq:chunga4}) and (\ref{eq:chunga5}) term by term, we can
infer that  $ k_{m}^{\prime }=0$ for every  natural $m\geq 2$, so
$\omega=u$. 
\end{proof}



\section{Final remarks}

It has been proved that solution $u$, given by (\ref{eq:sol1}), attracts
a family of positive and measurable functions on the form
$v(x)=u(x)+cx^p$, for certain $c$ and $p$.

When $u$ is the unique nontrivial solution of equation
(\ref{eq:abel1}), the following statements are held near zero:
\begin{enumerate}
\item[(i)] $\lim_{n\to \infty }T^nv=u$ if
\begin{itemize}
\item $F(\alpha ,\gamma, p)<1$, or \item $F(\alpha ,\gamma, p)=1$
and $c<0$.
\end{itemize}
\item[(ii)] $\lim_{n\to \infty }T^nv=+\infty$ if
 $F(\alpha ,\gamma, p)\geq 1$ and $c>0$.
\item[(iii)] $\lim_{n\to \infty }T^nv=0$ if $F(\alpha ,\gamma,
p)>1$ and $c<0$.
\end{enumerate}
This problem can be regarded under the scope of the Discrete Dynamical
Systems Theory. Indeed, for any $\epsilon>0$, we have a nonlinear
monotone operator $T:L_+^1(0,\epsilon)\mapsto L_+^1(0,\epsilon)$,
where $L^1_+(0,\epsilon)$ denotes the space of positive functions of
$L^1(0,\epsilon)$. Such operator defines the infinite-dimensional
discrete dynamical system
\[
\begin{cases}
  u_0\in L^1_+(0,\epsilon)&\\
  u_n=T^n u_0,&n\in\mathbb{N}.\\
\end{cases}
\]
This system has a fixed point $u$, which is unstable; that is, for
every ball, centered on $u$, $\mathcal{B}(u)$ (with the
$L^1_+(0,\epsilon)$ topology), we can find functions $u_0\in
\mathcal{B}(u)$, such that $\left(T^nu_0\right)_{n\in\mathbb{N}}$
converges to $u$, converges to the trivial solution, or diverges to
$+\infty$.


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\end{document}
