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

\AtBeginDocument{{\noindent\small
\emph{Electronic Journal of Differential Equations},
Vol. 2009(2009), No. 136, pp. 1--9.\newline
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu
\newline ftp ejde.math.txstate.edu}
\thanks{\copyright 2009 Texas State University - San Marcos.}
\vspace{9mm}}

\begin{document}
\title[\hfilneg EJDE-2009/136\hfil Existence of solutions]
{Existence of local and global solutions to some impulsive
fractional differential equations}

\author[R. Atmania, S. Mazouzi\hfil EJDE-2009/136\hfilneg]
{Rahima Atmania, Said Mazouzi} % in alphabetical order

\address{Rahima Atmania \newline
Laboratory of Applied Mathematics (LMA) \\
Department of Mathematics, University of Badji Mokhtar Annaba \\
P.O. Box 12, Annaba 23000, Algeria}
\email{atmanira@yahoo.fr}

\address{Said Mazouzi \newline
Laboratory of Applied Mathematics (LMA) \\
Department of Mathematics, University of Badji Mokhtar Annaba \\
P.O. Box 12, Annaba 23000, Algeria}
\email{mazouzi\_sa@yahoo.fr}

\thanks{Submitted July 3, 2009. Published October 21, 2009.}
\thanks{Supported by the LMA, University of Badji Mokhtar
Annaba, Algeria}
\subjclass[2000]{26A33, 34A12, 34A37}
\keywords{Fractional derivative; impulsive conditions; fixed point;
\hfill\break\indent
local solution; global solution}

\begin{abstract}
 First, by using Schauder's fixed-point theorem we establish the
 existence uniqueness of locals for some fractional differential
 equation with a finite number of impulses. On the other hand,
 by using Brouwer's fixed-point theorem, we establish existence of
 the global solutions under suitable assumptions.
\end{abstract}

\maketitle
\numberwithin{equation}{section}
\newtheorem{theorem}{Theorem}[section]
\newtheorem{lemma}[theorem]{Lemma}
\newtheorem{example}[theorem]{Example}


\section{Introduction}

The concept of fractional calculus can be considered as a
generalization of ordinary differentiation and integration to
arbitrary (non-integer) order. However, great efforts must be done
before the ordinary derivatives could be truly interpreted as a
special case of the fractional derivatives. For more details, we
refer to the books by  Oldham and  Spanier \cite{o1} and by
Miller and  Ross \cite{m1}.

Actually, fractional derivatives have been extensively applied in
many fields, for example in Probability, Viscoelasticity,
Electronics, Economics, Mechanics as well as Biology.

Some results on quantitative and qualitative theory of some
fractional differential equations are obtained, we may cite the
references \cite{d1,l1,m1,o1,y1}. On the other hand, the theory of
impulsive differential  equations is also an important area of
research which has been investigated in the last few years by
great number of mathematicians. We recall that the impulsive
differential equations may better model phenomena and dynamical
processes subject to a great changes in short times issued, for
instance, in Physics, Biotechnology, Automatics and Robotics. To
learn more about the most recent used techniques for this kind of
problems we refer to the book of Benchohra  et al \cite{b1}.

So, we propose to study fractional differential equation subject
to a finite number of impulses. As we know there just few authors
have  investigated this subject \cite{m2}. We have obtained some results
regarding local existence and uniqueness for some fractional
integrodifferential problem with a finite number of impulses. For
the  existence and uniqueness of local solutions we use the
Schauder's fixed-point theorem, while we use Brouwer's fixed-point
theorem for the global solutions.

\section{Preliminaries}

Among the definitions of fractional derivatives we recall the
Riemann-Liouville definiton as follows.
\[
D^{\alpha }u(t) =\frac{1}{\Gamma (n-\alpha ) }\frac{
d^{n}}{dt^{n}}\int_{t_0}^t(t-s) ^{-\alpha +n-1}u(s) \,ds
\]
where $\Gamma (\cdot) $ is the well known gamma function and
$\alpha \in (n-1,n) $, with $n$  being an integer. One may observe
that the derivative of a constant is not at all equal to zero
which can cause serious problems in both views, theoretical and
practical. For this reason we prefer to use Caputo's definition
which gives better results than those of Riemann-Liouville. So we
define Caputo's derivative of order $\alpha \in (n-1,n) $ of a
function $u(t)$ by
\[
D^{\alpha }u(t) =\frac{1}{\Gamma (n-\alpha ) } \int_{t_0}^t(t-s)
^{-\alpha +n-1}\frac{ d^{n}}{ds^{n}}u(s) \,ds.
\]
Also, we use the fractional integral operator of order $\alpha >0$
given by
\[
D^{-\alpha }u(t) =\frac{1}{\Gamma (\alpha ) }
\int_{t_0}^ t (t-s) ^{\alpha -1}u( s) \,ds.
\]

We shall consider the fractional differential equation
\begin{equation}
D^{\alpha }u(t) =f(t,u(t) ) ;\quad t\in [ t_{0},t_{0}+\tau ],\;
t\neq t_{k},\; k=1,\dots ,m;  \label{1}
\end{equation}
with the initial condition
\begin{equation}
D^{\alpha -1}u(t_{0}) =u_{0};\quad
( t-t_{0}) ^{1-\alpha }u(t) \big|_{t=t_{0}}
=\frac{ u_{0}}{\Gamma (\alpha ) };
\label{2}
\end{equation}
subject to the impulsive conditions
\begin{equation} \label{3}
\begin{gathered}
D^{\alpha -1}(u(t_{k}^{+}) -u(t_{k}^{-}) ) = I_{k}(t) ;\quad
t=t_{k},\; k=1,\dots ,m;  \\
 (t-t_{k}) ^{1-\alpha }u(t) \big|_{t=t_{k}}
=\frac{I_{k}(t_{k}) }{\Gamma (\alpha ) } ,\quad k=1,\dots ,m.
\end{gathered}
\end{equation}
We set the following assumptions
\begin{itemize}
\item[(A1)]  $t>t_{0}\geq 0$,  $\alpha $ is a real number
such that $0<\alpha \leq 1$, $u_{0}$ is a real constant vector of
$\mathbb{R}^{n}$  (the usual real $n$-dimensional Euclidean
space equipped with its Euclidean norm $\|.\|$);

\item[(A2)] $f(t,u) :I\times \mathbb{R}^{n}\to
\mathbb{R}^{n}$; $I_{k}(t) :I\to \mathbb{R}^{n}$,
$k=1,\dots ,m$, with $I=[t_{0},t_{0}+\tau ]$;

\item[(A3)]  $t_{k}\in I$, $k=1,\dots ,m$  and
$t_{0}<t_{1}<\dots <t_{k}<\dots <t_{m}$.
\end{itemize}
We introduce the following spaces:
\\
$\mathcal{PC}(I,\mathbb{R}^{n}) =\{ u:I\to \mathbb{R}^{n}:u(t)$
 is continuous at $t\neq t_{0}$,  $t\neq t_{k}$, and left continuous
at $t=t_{k}$, and $(t_{0}^{+}) $ and $u(t_{k}^{+}) $
 exist for $k=1,\dots ,m \}$;
\\
$\mathcal{PC}_{\alpha }(I,\mathbb{R}^{n})
=\{ u\in \mathcal{PC}(I,\mathbb{R}^{n}):
\lim_{t\to t_{0}^{+}}(t-t_{0}) ^{\alpha }u(t)$  and
$\lim_{t\to t_{k}^{+}} (t-t_{k})^{\alpha }u(t)$ exist and are finite
for $k=1,\dots ,m$, $\alpha >0\}$.
This is a Banach space with respect to the norm
\[
\|u\| _{\alpha }=\sup_{t\in I'}( t-t_{0})
^{\alpha +1} \prod_{i=1}^m (t-t_{i}) ^{\alpha +1}\|u(t) \| ,
\]
where $I'=(t_{0},t_{0}+\tau ]\backslash \{ t_{k}\}_{k=1,2,\dots }$.

We begin with the following Lemma.

\begin{lemma}\label{lem1}
If $f$ and $I_{k}$, $k=1,\dots m$ are continuous functions, then
$u(t) $  is a solution to problem  \eqref{1}-\eqref{3}
 in $\mathcal{PC}_{1-\alpha }([t_{0},t_{0}+\tau
] ,\mathbb{R}^{n}) $ if and only if $u(t) $ satisfies the
 integrodifferential equation
\begin{equation}  \label{4}
\begin{aligned}
u(t) &=\frac{u_{0}}{\Gamma (\alpha ) }( t-t_{0}) ^{\alpha
-1}+\frac{1}{\Gamma (\alpha ) }
\int_{t_0}^t (t-s) ^{\alpha -1}f( s,u(s) )\,ds  \\
&\quad +\frac{1}{\Gamma (\alpha ) }\sum_{t_{0}<t_{k}\leq t}
(t-t_{k}) ^{\alpha -1}I_{k}(t_{k}).
\end{aligned}
\end{equation}
\end{lemma}

\begin{proof}
Let $u(t) $ be a solution of problem \eqref{1}-\eqref{3}. Using
the fractional integral of order $\alpha >0$ and the properties of
derivative of order $\alpha >0$, and then applying $D^{-1}$ to
\eqref{1}  we obtain
\begin{align*}
D^{-1}(D^{\alpha }u(t) )
&=\int_{t_0}^t  f(s,u(s) ) \,ds
=\int_{t_0}^t \frac{d}{dt}D^{-(1-\alpha )}u(s)\,ds \\
&= D^{\alpha -1}u(t) -u_{0}-\sum_{t_{0}<t_{k}\leq t}
I_{k}(t_{k}) .
\end{align*}
So
\begin{equation}  \label{5}
D^{\alpha -1}u(t) =u_{0}+\int_{t_0}^t f(s,u(s) )
\,ds+\sum_{t_{0}<t_{k}\leq t}  I_{k}(t_{k}).
\end{equation}
Next, applying the operator $D^{1-\alpha }$ to $D^{\alpha -1}u(
t) $ we obtain
\begin{align*}
u(t) &= D\Big(\frac{1}{\Gamma (\alpha ) }
\int_{t_0}^t (t-s) ^{\alpha -1}u_{0}\,ds+\frac{1}{\Gamma (\alpha ) }
\int_{t_0}^t(t-s) ^{\alpha -1}f(s,u(s) ) \,ds \\
&\quad +\frac{1}{\Gamma (\alpha ) }
\int_{t_0}^t (t-s) ^{\alpha -1} \sum_{t_{0}<t_{k}\leq s}
I_{k}(t_{k}) \,ds\Big) ,
\end{align*}
 which gives the integral equation \eqref{4}. On the other hand,
from (\ref{2}) and \eqref{3}, it follows that
\begin{equation} \label{6a}
\begin{aligned}
\lim_{t\to t_0^+} (t-t_{0}) ^{1-\alpha }u(t)
=\frac{u_{0}}{\Gamma (\alpha ) }, \\
\lim_{t\to t_k^+} (t-t_{k}) ^{\alpha }u(t)
=\frac{I_{k}(t_{k}) }{\Gamma (\alpha )},\quad k=1,2,\dots ,
\end{aligned}
\end{equation}
 which proves that $u(t) \in \mathcal{PC}_{1-\alpha}([t_{0},t_{0}+\tau ])$.

Let $u(t) $ be a solution to the integral equation
\eqref{4} in $\mathcal{PC}_{\alpha }([t_{0},t_{0}+\tau ] ) $.
Performing $D^{\alpha }$ to the integral equation \eqref{4} we get
for $t\neq t_{0}$ and $t\neq t_{k}$,  $k=1,\dots ,m$;
\[
 D^{\alpha }u(t) =DD^{\alpha }D^{-\alpha }u_{0}+DD^{\alpha
}D^{-\alpha }\sum_{t_0<t_k\leq t} I_{k}(t_{k})
+D^{\alpha }D^{-\alpha }f(t,u(t) )
=f(t,u( t) ),
\]
and for  $t=t_{0}$, and  $t=t_{k}$, $k=1,\dots ,m$,  we
apply $D^{\alpha -1}$ to the integral equation \eqref{4} to obtain
\eqref{5} which  in turn gives
\begin{gather*}
D^{\alpha -1}u(t_{0}) = u_{0} \\
D^{\alpha -1}(u(t_{k}^{+}) -u(t_{k}^{-}) ) = I_{k}(t_{k}),
\quad k=1,\dots ,m.
\end{gather*}
Now, since $u(t) \in \mathcal{PC}_{1-\alpha }([ t_{0},t_{0}+\tau
] ) $,  it satisfies the limits (\ref{6a}) from which we
get conditions (\ref{2}) and \eqref{3}.
\end{proof}

\section{Existence and uniqueness of a local solution}

We need the following Schauder's fixed-point theorem.

\begin{theorem}\label{thm1}
 If $U$ is a closed , bounded, convex subset
of a Banach space $X$ and the mapping $A:U\to U$ is
completely continuous, then $A$ has a fixed point in $U$.
\end{theorem}

 Let us denote the right hand side of \eqref{4} by
$Au(t) $ which we write as  $Au(t) =u_{1}( t) +Bu(t) $, where
\begin{gather*}
u_{1}(t) =\frac{u_{0}}{\Gamma (\alpha ) }( t-t_{0}) ^{\alpha
-1}+\frac{1}{\Gamma (\alpha ) }
\sum_{ t_{0}<t_{k}\leq t}(t-t_{k}) ^{\alpha -1}I_{k}(t_{k}), \\
Bu(t) =\frac{1}{\Gamma (\alpha ) }\int_{t_0}^t
(t-s) ^{\alpha -1}f(s,u( s) ) \,ds.
\end{gather*}

\begin{theorem} \label{thm2}
If $f\in C([t_{0},t_{0}+\tau ] \times
\mathbb{R}^{n},\mathbb{R}^{n}) $ and there exist positive
constants $N$, $b_{k}$, $k=1,\dots ,m$; such that
$\| f(t,u)\| \leq N$,
$\|I_{k}(t_{k})\| \leq b_{k}$, $k=1,\dots ,m$, then
there exists at least one solution $u(t) $ of the problem
\eqref{1}-\eqref{3} in
$\mathcal{PC}_{1-\alpha }([t_{0},t_{0}+\mu ] ) $
 for some positive constant
\[
\mu =\min \Big(\tau ,\, \big(\frac{\alpha \beta }{N}\Gamma (\alpha ) \big)
^{1/(m(2-\alpha ) +2) }\Big) .
\]
\end{theorem}

\begin{proof}
It is easy to see that the set
\[
U_{1-\alpha }^{\beta }=\{ u\in \mathcal{PC}_{1-\alpha
}([t_{0},t_{0}+\mu ] ,\mathbb{R}^{n}) : \| u(t)
-u_{1}(t) \| _{1-\alpha }\leq \beta \}
\]
is not empty because $u_{1}(t) \in U_{1-\alpha }^{\beta }$; on the
other hand, it is a closed, bounded, convex subset of
the Banach space
$\mathcal{PC}_{1-\alpha }([t_{0},t_{0}+\mu] , \mathbb{R}^{n}) $.

Next, we define the operator $A$ on $U_{1-\alpha }^{\beta }$ by
\begin{equation} \label{6b}
\begin{aligned}
Au(t) &= \frac{u_{0}}{\Gamma (\alpha ) }( t-t_{0}) ^{\alpha
-1}+\frac{1}{\Gamma (\alpha ) }
\int{t_0}^t (t-s) ^{\alpha -1}f( s,u(s) ) \,ds   \\
&\quad +\frac{1}{\Gamma (\alpha ) }\sum_{t_0<t_k\leq t}
(t-t_{k}) ^{\alpha -1}I_{k}(t_{k}) .
\end{aligned}
\end{equation}
To prove that $A$ maps $U_{1-\alpha }^{\beta }$ into itself we see that,
for every $u\in U_{1-\alpha }^{\beta }$, we have
\begin{align*}
(t-t_{0}) ^{2-\alpha }\prod_{i=1}^m (t-t_{i})
^{2-\alpha }\|Au(t) -u_{1}( t) \|
&\leq  \frac{\mu ^{(m+1) (2-\alpha ) }}{\Gamma (\alpha ) }N
\int_{t_0}^t  (t-s) ^{\alpha -1}\,ds \\
&\leq \frac{\mu ^{m(2-\alpha ) +2}}{\alpha \Gamma ( \alpha )}N\leq \beta .
\end{align*}
Hence
\begin{equation}
\|Au(t) -u_{1}(t) \| _{1-\alpha }\leq \beta , \label{7}
\end{equation}
which implies  $AU_{1-\alpha }^{\beta }\subset U_{1-\alpha
}^{\beta }$.

To see that $AU_{1-\alpha }^{\beta }$ is uniformly bounded in
$\mathcal{PC} _{1-\alpha }([t_{0},t_{0}+\mu ] ) $ we note
that
\begin{align*}
&(t-t_{0}) ^{2-\alpha }\prod_{i=1}^m
(t-t_{i}) ^{2-\alpha }\|u_{1}(t) \|
\\
&\leq \frac{\|u_{0}\| }{\Gamma (\alpha ) } (t-t_{0})
\prod_{i=1}^m (
t-t_{i}) ^{2-\alpha } \\
&\quad +\frac{(t-t_{0}) ^{2-\alpha }}{\Gamma (\alpha ) }
\sum_{t_0<t_k\leq t} \prod_{i=1}^m ( t-t_{i}) ^{2-\alpha
}(t-t_{k}) ^{\alpha -1}\|I_{k}(t_{k}) \| ;
\\
&\leq \frac{\|u_{0}\| }{\Gamma (\alpha ) }\mu ^{m(2-\alpha )
+1}+\frac{\mu ^{2-\alpha }\mu ^{\text{(} m-1)(2-\alpha
)+1}}{\Gamma (\alpha ) }\sum_{k=1}^m b_{k}.
\end{align*}
Hence,
\begin{equation}
\|u_{1}(t) \| _{1-\alpha }\leq \frac{\mu ^{m(2-\alpha )
+1}}{\Gamma (\alpha ) }( b+\|u_{0}\| ) , \label{8}
\end{equation}
where $b=\sum_{t_0<t_k\leq t} b_{k}$.
 From (\ref {7}) and (\ref{8}), we deduce that
\[
\|u(t) \| _{1-\alpha }\leq \frac{\mu ^{m(2-\alpha ) +1}}{\Gamma
(\alpha ) }( \|u_{0}\| +b) +\frac{\mu ^{m(2-\alpha ) +2}}{\alpha
\Gamma (\alpha ) }N,
\]
for every $u\in U_{1-\alpha }^{\beta }$.

We shall prove in the next step that  $AU_{1-\alpha
}^{\beta }$ is equicontinuous in
$\mathcal{PC}_{1-\alpha }([t_{0},t_{0}+\mu ] )$. We observe on
the one hand that the
derivative of $u_{1}(t) $ is uniformly bounded in
$\mathcal{PC}_{1-\alpha }([t_{0},t_{0}+\mu ] ) $ because
\begin{align*}
&(t-t_{0}) ^{2-\alpha }\prod_{i=1}^m
(t-t_{i}) ^{2-\alpha }\|u_{1}'( t)\| \\
&\leq \frac{1-\alpha }{\Gamma (\alpha ) }\| u_{0}\|
\prod_{i=1}^m ( t-t_{i})^{2-\alpha } \\
&\quad +\frac{1-\alpha }{\Gamma (\alpha ) }(t-t_{0}) ^{2-\alpha
}\sum_{t_0<t_k\leq t}
\prod_{i=1}^m (t-t_{i}) ^{2-\alpha }(t-t_{k}) ^{\alpha -2}\|
I_{k}(t_{k}) \|
\end{align*}
giving
\[
\|u_{1}'(t) \| _{1-\alpha }\leq \frac{ 1-\alpha }{\Gamma
(\alpha ) }\mu ^{m(2-\alpha ) }(\| u_{0}\| +b\mu ^{-(2-\alpha )}).
\]
 On the other hand, we have for $t_{0}<s_{1}<s_{2}<t_{0}+\mu
$,
\begin{align*}
&(t-t_{0}) ^{2-\alpha }\prod_{i=1}^m
(t-t_{i}) ^{2-\alpha }\|Bu(s_{2}) -Bu(s_{1}) \| \\
&\leq \frac{\mu ^{(m+1)(2-\alpha )}}{\Gamma (\alpha ) }
N|\overset{s_{2}}{\underset{t_{0}}{\int }}(s_{2}-s) ^{\alpha
-1}-\overset{s_{1}}{\underset{t_{0}}{\int }}(s_{1}-s) ^{\alpha
-1}\,ds|
\\
&\leq \frac{\mu ^{(m+1)(2-\alpha )}}{\Gamma (\alpha ) } N
\Big| \overset{s_{1}}{\underset{t_{0}}{\int }}(( s_{2}-s) ^{\alpha
-1}-(s_{1}-s) ^{\alpha -1}) \,ds+
\overset{s_{2}}{\underset{s_{1}}{\int }}(s_{2}-s) ^{\alpha
-1}\,ds\Big|,
\end{align*}
so that
\begin{equation}  \label{9}
\|Bu(s_{2}) -Bu(s_{1}) \|_{1-\alpha }
\leq \frac{\mu ^{(m+1)(2-\alpha )}}{\Gamma (\alpha ) }N[
2(s_{2}-s_{1}) ^{\alpha }+|( s_{2}-t_{0}) ^{\alpha }-(s_{1}-t_{0})
^{\alpha }|];
\end{equation}
that is, $Bu(t) $ is equicontinuous, and so $Au( t) $ is
equicontinuous in $\mathcal{PC}_{1-\alpha }([t_{0},t_{0}+\mu
] ,\mathbb{R}^{n}) $. Hence,
$\overline{AU_{1-\alpha}^{\beta }}$ is compact in
$\mathcal{PC}_{1-\alpha }([t_{0},t_{0}+\mu ] ) $ showing that $A$
 is completely continuous. Therefore, we conclude by Schauder's
theorem that $A$ has at least one fixed-point in
$U_{1-\alpha }^{\beta }$ which is exactly a solution to
 (\ref{1} )-\eqref{3} in view of lemma \ref{lem1}.
The proof is now complete.
\end{proof}

\begin{theorem} \label{thm3}
Besides the hypotheses of theorem \ref{thm1}, we suppose that
there exists a constant $L$ such that
\begin{equation}
0<L<\frac{\alpha \Gamma (\alpha ) }{\mu ^{\alpha }}, \label{10}
\end{equation}
where $\mu $ is defined as in theorem \ref{thm2}, and
\[
\|f(t,u) -f(t,w) \| \leq L\|u-w\| ,\quad \text{for every }
u,w\in \mathbb{R}^n.
\]
Then, the solution $u(t) $ of  \eqref{1}-\eqref{3} is
unique in
$\mathcal{PC}_{1-\alpha }([t_{0},t_{0}+\mu ] ,\mathbb{R}
^{n}) $.
\end{theorem}

\begin{proof}
In virtue of theorem \ref{thm1} there exists at least one solution
$u( t) $ of \eqref{1}-\eqref{3} in
$\mathcal{PC} _{1-\alpha }([t_{0},t_{0}+\mu ],\mathbb{R}^{n}) $.

First, suppose to the contrary that there exist two different
solutions $u$ and $w$ in $\mathcal{PC}_{1-\alpha }([
t_{0},t_{0}+\mu ] , \mathbb{R}^{n}) $
which satisfy the integral equation \eqref{4}. It is easy to see that
\begin{align*}
\|u(t) -w(t) \| &\leq  \frac{1}{ \Gamma (\alpha ) }\int_{t_0}^t(
t-s) ^{\alpha -1}\| f(s,u(s) )
-f(s,w(s) ) \| \,ds \\
&\leq  \frac{\mu ^{\alpha }}{\alpha \Gamma (\alpha ) } L\| u(t)
-w(t) \| .
\end{align*}
It follows  that
\[
\|u(t) -w(t) \| _{1-\alpha }\leq \frac{\mu ^{\alpha }}{\alpha
\Gamma (\alpha ) }L\|u(t) -w(t) \| _{1-\alpha },
\]
and taking into account  condition (\ref{10}) we obtain
\begin{equation*}
\|u(t) -w(t) \| _{1-\alpha }=0.
\end{equation*}
So, the two solutions are identical in $\mathcal{PC}_{1-\alpha
}([t_{0},t_{0}+\mu ] ,\mathbb{R}^{n}) $ which completes the
proof.
\end{proof}

To illustrate the foregoing results we propose the following example:

\begin{example} \label{exa3.4} \rm
 On the interval $[0,1]$, Consider  the impulsive
fractional differential initial-value problem
\begin{equation} \label{11}
\begin{gathered}
D^{1/2}u(t) =\frac{e^{-t}}{t+2}\sin u(t) ;\quad
t\neq \frac{k}{k+1},\; k=1,2; \\
D^{-1/2}u(0) =0;\quad  t^{1/2}u(t)\big|_{t=0}=0; \\
D^{-1/2}(u(t_{k}^{+}) -u(t_{k}^{-}) )
=t_{k}\cos t_{k};\quad t_{k}=\frac{k}{k+1},\; k=1,2 \\
(t-\frac{k}{k+1})^{1/2}u(t) \big|_{t= \frac{1}{k+1}}
=\frac{k}{\sqrt{\pi }(k+1) }\cos (\frac{k}{ k+1}),
\quad k=1,2.
\end{gathered}
\end{equation}
\end{example}

We see that $f(t,u) =\frac{e^{-t}}{t+2}\sin u\in C( [ 0,1]
\times \mathbb{R};\mathbb{R})$; $I_{k}( t) \in C([ 0,1]
;\mathbb{R})$, and since $\Gamma (1/2) =\sqrt{\pi }$, then the
solution of (\ref{11}) satisfies the  integral equation
\begin{equation} \label{12}
\begin{aligned}
u(t) &= \frac{1}{\sqrt{\pi }}\int_{t_0}^t (t-s)
^{-1/2}\frac{e^{-s}}{s+2}\sin u(s) \,ds\\
&\quad +\frac{1}{\sqrt{\pi }}\sum_{0<t_{k}\leq t}
\big(t-\frac{k}{k+1} \big) ^{-1/2}\frac{k}{k+1}\cos \big(\frac{k}{k+1}\big) .
\end{aligned}
\end{equation}
Since $|f(t,u)|=|\frac{e^{-t}}{( t+2) }\sin u|\leq
\frac{1}{2}$, for every $t\in [ 0,1 ] $ and $u\in \mathbb{R}$,
and
\[
|f(t,u)-f(t,w)|\leq \frac{1}{2}| u-w|\text{, for every }u,w\in
\mathbb{R},
\]
 condition (\ref{10}) is easily satisfied and so
 in view of theorems \ref{thm2} and \ref{thm3},
(\ref{11}) admits a unique solution $u(t) $  in
$\mathcal{PC}_{1/2}([0,1] ,\mathbb{R})$.

\section{Existence of a global solution}

In this part we shall prove the  existence of a global solution
to  \eqref{1}-\eqref{3} under suitable assumptions, by
using the following Brouwer's fixed-point theorem.

\begin{theorem}\label{thm4}
 Set $\Omega $ be a closed, bounded, convex
non empty subset of $X$ a Banach space and let
$A:\Omega \to X$ be a continuous mapping.
If $A(\Omega ) \subset \Omega$, then $A$ has a fixed-point in
$\Omega $.
\end{theorem}

Consider the scalar fractional differential equation
\begin{equation}
D^{\alpha }v(t) =g(t,v(t) ) ;\quad t\in [t_{0},+\infty[;
\; t\neq t_{k},\; k=1,\dots ,m; \; 0<\alpha \leq
1; \label{13}
\end{equation}
subject to the initial conditions
\begin{equation}
D^{\alpha -1}v(t) =v_{0};\quad
( t-t_{0}) ^{1-\alpha}v(t) \big|_{t=t_{0}}
=\frac{1}{\Gamma (\alpha ) }v_{0};
\label{14}
\end{equation}
and the impulsive conditions
\begin{equation} \label{15}
\begin{gathered}
D^{\alpha -1}v(t_{k}^{+}) -D^{\alpha -1}v( t_{k}^{-})
= J_{k}(t_{k}) ;\quad k=1,\dots ,m  \\
 (t-t_{k}) ^{1-\alpha }v(t) \big|_{t=t_{k}}
= \frac{1}{\Gamma (\alpha ) }J_{k}(v( t_{k}) ) ,\quad k=1,\dots ,m.
\end{gathered}
\end{equation}
We assume that $v_{0}$ is a positive constant;
$f(t,u) \in C([t_{0},+\infty [\times \mathbb{R}^{n},\mathbb{R}^{n}) $;
$g(t,v) \in C([t_{0},+\infty [ \times \mathbb{R}_{+},\mathbb{R}_{+})$,
$I_{k}(t) \in C([t_{0},+\infty [,\mathbb{R} ^{n})$ and
$J_{k}(t) \in C([t_{0},+\infty [,\mathbb{R}_{+})$; $k=1,\dots ,m$.

In view of theorem \ref{thm1} the solution of \eqref{13}-\eqref{15}
 satisfies the integral equation
\begin{equation} \label{16}
\begin{aligned}
v(t) &= \frac{v_{0}}{\Gamma (\alpha ) }( t-t_{0}) ^{\alpha
-1}+\frac{1}{\Gamma (\alpha ) }\overset{t }{\underset{t_{0}}{\int
}}(t-s) ^{\alpha -1}g( s,v(
s) ) \,ds   \\
&\quad +\frac{1}{\Gamma (\alpha ) }\sum_{t_0<t_k\leq t}
(t-t_{k}) ^{\alpha -1}J_{k}(t_{k}) .
\end{aligned}
\end{equation}
in the Banach space $\mathcal{PC}_{1-\alpha }([t_{0},+\infty[,\mathbb{R})$
  endowed with the norm
\[
|v|_{1-\alpha }=\sup_{t\in [ t_{0},+\infty [\backslash \{
t_{k}\}_{k=0,1,\dots m}} ( t-t_{0}) ^{2-\alpha
}\prod_{i=1}^m (t-t_{i}) ^{2-\alpha }|v(t) |.
\]

\begin{theorem} \label{thm5}
Assume that
$\|f(t,u) \| \leq g(t,\|u\| )$  for every
$t\geq t_{0}$ and $u\in \mathbb{R }^{n}$,
where $g(t,v) $ is nonnegative and nondecreasing in $v$, for each
$t\geq t_{0}$, and
\[
\|I_{k}(t_{k}) \| \leq J_{k}( t_{k}), \quad\text{for }
t=t_{k}, \quad k=1,\dots m.
\]
If
\eqref{13}-\eqref{15} has a positive solution $v(t) $ in
$\mathcal{PC}_{1-\alpha }( [t_{0},+\infty [,\mathbb{R}) $, then
 \eqref{1}-\eqref{3} has at least a solution in
$\mathcal{PC}_{1-\alpha }([t_{0},+\infty [,\mathbb{R}^{n}) $ such
that $\|u\| _{1-\alpha }\leq |v|_{1-\alpha }$, for each
$u_{0}\in \mathbb{R}^{n}$ satisfying $\|u_{0}\| \leq v_{0}$.
\end{theorem}

\begin{proof}
 To apply Brouwer's theorem we use the Banach
space $\mathcal{PC}_{1-\alpha }([t_{0},+\infty [,\mathbb{R} ^{n})
$, $0<\alpha \leq 1$ , which we equip with the norm
\[
\|u\| _{1-\alpha }=\sup_{t\in [ t_{0},+\infty [\backslash \{
t_{k}\}_{k=0,1,\dots m}} ( t-t_{0}) ^{2-\alpha
}\prod_{i=1}^m (t-t_{i}) ^{2-\alpha }\|u(t)\| .
\]
Next, define a subset of $\mathcal{PC}_{1-\alpha }([ t_{0},+\infty
[,\mathbb{R}^{n})  $ by
\begin{align*}
\mathcal{V}_{1-\alpha }=\big\{& u\in \mathcal{PC}_{1-\alpha
}([t_{0},+\infty [,\mathbb{R}^{n}) :\|u\| _{1-\alpha }\leq
|v|_{1-\alpha };\\
&\text{where $v(t)$ is a positive solution of \eqref{13}-\eqref{15}}
\big\}.
\end{align*}
It is not difficult to verify that $\mathcal{V}_{1-\alpha }$ is a closed,
convex, and bounded subsect of
$\mathcal{PC}_{1-\alpha }([ t_{0},+\infty [, \mathbb{R}^{n}) $.

The operator $A$ defined by (\ref{6b}) is continuous, and so, it
remains to prove that $A(\mathcal{V}_{1-\alpha }) \subset
\mathcal{V} _{1-\alpha }$. For each $u\in \mathcal{V}_{1-\alpha
}$, by (\ref{6b}), we have
\begin{align*}
\|Au(t) \| &\leq \prod_{i=1}^m (t-t_{i})
^{\alpha +1}\frac{\|u_{0}\| }{\Gamma (\alpha ) }+\frac{1}{\Gamma
(\alpha ) } \int_{t_0}^t(t-s) ^{\alpha -1}\|
f(s,u(s) ) \| \,ds \\
&\quad +(t-t_{0}) ^{\alpha +1}\prod_{i=1,i\neq k}^m
(t-t_{i}) ^{\alpha +1}\frac{1}{\Gamma (\alpha )
}\sum_{t_0<t_k\leq t} \|I_{k}( t_{k}) \| \,.
\end{align*}
 From the assumptions, we obtain
\begin{equation} \label{17}
\begin{aligned}
\|Au(t) \| &\leq \frac{v_{0}}{\Gamma (\alpha ) }(t-t_{0})
^{\alpha -1}+\frac{1}{\Gamma (\alpha ) }\int_{t_0}^t( t-s)
^{\alpha -1}g(s,v(s) ) \,ds   \\
&\quad +\frac{1}{\Gamma (\alpha ) }\sum_{t_0<t_k\leq t}
(t-t_{k}) ^{\alpha -1}J_{k}(t_{k})
\end{aligned}
\end{equation}
It follows from (\ref{16}) that
\[
\|Au(t) \| \leq v(t) ,\quad t\geq t_{0};
\]
and since
\begin{gather*}
\lim_{t\to t_0^+} (t-t_{0}) ^{1-\alpha }\|Au(t) \|
\leq \lim_{t\to t_0^+} (t-t_{0}) ^{1-\alpha }v(t) , \\
\lim_{t\to t_K^+} (t-t_{k}) ^{1-\alpha}\|Au(t) \|
\leq \lim_{t\to t_K^+} (t-t_{k}) ^{1-\alpha }v(t) ;\quad
 k=1,\dots ,m,
\end{gather*}
it follows that
\[
\|Au\| _{1-\alpha }\leq |v|_{1-\alpha }.
\]
Hence, $A(\mathcal{V}_{1-\alpha }) \subset \mathcal{V}_{1-\alpha
}$. Hence, as all the requirements of Brouwer's fixed-point
theorem are satisfied, then $A$ has a fixed point in
$\mathcal{V}_{1-\alpha }$ which is the solution of
\eqref{1}-\eqref{3} such that $\|u\| _{1-\alpha }\leq
|v|_{1-\alpha }$.
\end{proof}

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\end{thebibliography}


\section*{Corrigendum posted on August 31, 2010.}

First, we apologize for the misprints in the original article.
Now we correct those misprints and present a new proof of the
global existence result, using Schauder's fixed-point theorem instead
of Brouwer's theorem.

- Page 2, line 6: Replace ``For this reason we prefer to use
Caputo's definition which gives better results than those of
Riemann-Liouville'' by 
``For this reason, and despite our use of
the Riemann-Liouville derivative, many authors prefer to use
Caputo's definition''

- Page 2, line 10:
Replace $D^{\alpha }u(t) $  by  $^{c}D^{\alpha }u(t)$

- Page 2, in the second condition of (2.3):
Replace $u(t)$ by $u(t^{+})$

- Page 2, line 7 from the bottom:
Replace $(t_0^{+})$ by $u(t_0^{+})$

- Page 3, Eq. (2.6): Replace 
$\lim\limits_{t\to t_k^{+}} ( t-t_k)^\alpha u(t)$
   by
$\lim\limits_{t\to t_k^{+}} ( t-t_k) ^{1-\alpha }u(t)$

- Page 3 line 1: Insert the sentence:\\
The notation $\mathcal{PC}([t_0,t_0+\tau ])$
stands for $\mathcal{PC}([ t_0,t_0+\tau ], \mathbb{R}^{n})$
throughout this article.

- Page 6, last line: Replace $u(t)$  by $u(t^{+})$

- Page 7:  Delete the entire Theorem 4.1.

- Page 7, in the second condition of (4.3):
Replace $v(t)$ by $v(t^{+})$


- Page 7, line 5 from the bottom: Replace
 ``theorem 3.1'' by
``Lemma 2.1''

- Page 7 after Eq. (4.4), insert the paragraph:\\
 Since, $\mathcal{PC}_{1-\alpha }([ t_0,+\infty [ ,\mathbb{R}^{n})$
is not a Banach space,  we introduce the Banach space
\begin{align*}
&\mathcal{PC}_{1-\alpha }^{b}([t_0,+\infty [ ,\mathbb{R}^{n})\\
&=\{ u\in \mathcal{PC}_{1-\alpha }([t_0,+\infty [ ,\mathbb{R}^{n}):
\sup_{t\in J^{\ast }} \prod_{i=0}^m
(t-t_{i})^{2-\alpha }\| u(t)\|<+\infty \},
\end{align*}
where $J^{\ast }=[ t_0,+\infty [\backslash \{ t_k\} _{k=0,\dots,m}$.
This space is endowed with the  norm $\|\cdot\|_{\alpha-1}$ defined on
page 8, which is still valid for $n=1$.
Therefore, it is clear that the solution of (4.1)-(4.3)
satisfies (4.4) in
$\mathcal{PC}_{1-\alpha }^{b}([t_0,+\infty [,\mathbb{R}_{+})$.

- Page 7, line 2 from the bottom:
Replace $\mathcal{PC}_{1-\alpha }([ t_0,+\infty [ ,\mathbb{R})$
by \\
$\mathcal{PC}_{1-\alpha }^{b}([ t_0,+\infty [ ,\mathbb{R}_{+})$

- Page 8,  line 4:
Replace $\mathcal{PC}_{1-\alpha }([t_0,+\infty [ ,\mathbb{R})$
 by $\mathcal{PC}_{1-\alpha }^{b}([ t_0,+\infty [ ,\mathbb{R}_{+})$,

- Page 8, line 5: Replace $\mathcal{PC}_{1-\alpha }([ t_0,+\infty
 [ ,\mathbb{R}^{n})$ by
$\mathcal{PC}_{1-\alpha }^{b}([ t_0,+\infty[ ,\mathbb{R}^{n})$

- Page 8: Replace the proof of Theorem 4.2 by the following proof.

\begin{proof}[Proof of Theorem 4.2]
 To apply Schauder's theorem we have to establish that the
operator $A$, defined by (3.1), is completely continuous.
To prove that claim we define the set
\begin{align*}
\mathcal{V}_{1-\alpha }
&=\Big\{ u\in \mathcal{PC}_{1-\alpha }^{b}
([t_0,+\infty [,\mathbb{R}^{n}):D^{\alpha -1}u(t_0)=u_0,\\
&\quad \| u(t)\| \leq v(t),\; t\neq t_k,\; k=0,\dots,m, \\
&\quad \sup_{t\in J^{\ast }} \prod_{i=0}^m (t-t_{i})^{2-\alpha }
\| u(t)\| \leq \sup_{t\in J^{\ast }} \prod_{i=0}^m
(t-t_{i})^{2-\alpha }v(t),\\
&\quad v(t)\text{ being a positive solution of
(4.1)-(4.3) in }\mathcal{PC}_{1-\alpha }^{b}([t_0,+\infty
[,\mathbb{R}_{+})\Big\}.
\end{align*}

It is not difficult to verify that $\mathcal{V}_{1-\alpha }$
is not empty, closed, convex and bounded in
$\mathcal{PC}_{1-\alpha }^{b}([t_0,+\infty [,\mathbb{R}^{n})$.
The operator $A$  is continuous, and so for each
$u\in \mathcal{V}_{1-\alpha }$, we have for each
$t\neq t_k$, $k=0,\dots,m$,
\begin{align*}
\| Au(t)\|
&\leq (t-t_0) ^{\alpha -1}\frac{\| u_0\| }{\Gamma (\alpha )
}+\frac{1}{\Gamma (\alpha )}\int_{t_0}^t
 (t-s)^{\alpha -1}\| f(s,u(s))\| ds \\
&\quad +\frac{1}{\Gamma (\alpha )}\sum_{t_0<t_k\leq t}
(t-t_k)^{\alpha -1}\| I_k(t_k)\| .
\end{align*}
 From the assumptions of theorem 4.2, we obtain
\[
\| Au(t)\| \leq v(t),\quad t\neq t_k,\; k=0,\dots,m,
\]
and since
\[
\lim_{t\to t_k^{+}} (t-t_k)^{1-\alpha}\| Au(t)\|
\leq \lim_{t\to t_k^{+}} (t-t_k)^{1-\alpha }v(t),\quad
k=0,\dots,m,
\]
we have
\[
\sup_{t\in J^{\ast }} \prod_{i=0}^m (t-t_{i})^{2-\alpha }\| Au(t)\|
\leq \sup_{t\in J^{\ast }} \prod_{i=0}^m (t-t_{i})^{2-\alpha }v(t).
\]
Hence, $A(\mathcal{V}_{1-\alpha })\subset \mathcal{V}_{1-\alpha}$.
The elements of $A\mathcal{V}_{1-\alpha }$ are uniformly
bounded in $\mathcal{PC}_{1-\alpha }^{b}([t_0,+\infty [,R^{n})$
because
\[
\prod_{i=0}^m (t-t_{i})^{2-\alpha}\| Au(t)\| \leq
\prod_{i=0}^m (t-t_{i})^{2-\alpha }v(t)<\infty .
\]
Next, we prove that the elements in the set $A\mathcal{V}_{1-\alpha }$ are
equicontinuous in the space $\mathcal{PC}_{1-\alpha }^{b}([t_0,+\infty [,R^{n})$.
To do this, we show that the derivative of
$u_{1}(t)$ (defined in section 3) is uniformly bounded
in $\mathcal{PC}_{1-\alpha }^{b}([t_0,+\infty [,R^{n})$.
Note that
\begin{align*}
\| u_{1}'(t)\|  &\leq
\frac{(1-\alpha )}{\Gamma (\alpha )}\Big(v_0(
t-t_0)^{\alpha -2}+\underset{t_0<t_k\leq t}{\sum }(
t-t_k)^{\alpha -2}J_k(t_k)\Big)\\
&\leq \frac{(1-\alpha )}{\Gamma (\alpha )}v(
t);\quad t\neq t_k,\; k=0,\dots,m,
\end{align*}
and since $v(t)$ is a solution of (4.1)-(4.3) in
$\mathcal{PC}_{1-\alpha }^{b}([t_0,+\infty [,\mathbb{R}_{+})$,
we have
\[
\| u_{1}'(t)\| _{1-\alpha }\leq
| v(t)| _{1-\alpha }\frac{(1-\alpha
)}{\Gamma (\alpha )}<+\infty .
\]

On the other hand, regarding the set $Bu(t)$ defined in section 3,
the uniform convergence in $\mathcal{PC}_{1-\alpha }^{b}([
t_0,+\infty [,\mathbb{R}_{+})$ is equivalent to the uniform
convergence in $\mathcal{PC}_{1-\alpha }^{b}([t_0,T_{p}]
,R^{n})$, for each $T_{p}$, with
$[t_0,t_{m}] \subset [t_0,T_{p}] $ and
$\lim\limits_{p\to \infty }T_{p}=\infty $.
Thus, for each $s_{1}$ and $s_{2}$ different from $t_k$,
$k=0,\dots,m$, satisfying $t_0<s_{1}<$ $s_{2}<T_{p}$, we have
\begin{align*}
&\| Bu(s_{2})-Bu(s_{1})\|\\
& \leq \frac{1}{\Gamma (\alpha )}\Big[
 \int_{t_0}^{s_2} (s_{2}-s)^{\alpha -1}
 -\int_{t_0}^{s_1} (s_{1}-s)^{\alpha -1}\Big]
 \| f(s,u(s))\| ds \\
&\leq \frac{1}{\Gamma (\alpha )}\Big[
\int_{t_0}^{s_1} (s_{2}-s)^{\alpha -1}-(s_{1}-s) ^{\alpha -1}
+ \int_{s_1}^{s_2} (s_{2}-s)^{\alpha -1}\Big] g(s,v(s))ds,
\end{align*}
and since $g(s,v(s))$ is continuous and positive,
we  obtain
\[
\| Bu(s_{2})-Bu(s_{1})\| \leq
\frac{G}{\Gamma (\alpha )}[2(s_{2}-s_{1})
^{\alpha }+| (s_{2}-t_0)^{\alpha }-(s_{1}-t_0)^{\alpha }| ],
\]
where 
\[
G=\sup_{t\in [ t_0,+\infty [} g(t,v(t)).
\]
Clearly, the right hand side tends to zero as $s_{1}\to s_{2}$.
We infer that
\[
\| Bu(s_{2})-Bu(s_{1})\|
_{1-\alpha }\to 0,\quad\text{as }s_{1}\to s_{2}.
\]
Therefore, $\overline{A\mathcal{V}_{1-\alpha }}$ is compact and
$A$  is completely continuous. We conclude by Schauder's theorem
that $A$ has at least one fixed point in $\mathcal{V}_{1-\alpha }$
which is the solution to the given problem (2.1)-(2.3).
Furthermore, it satisfies the estimate
$\|u\| _{1-\alpha }\leq |v| _{1-\alpha }$. The proof  is complete.
\end{proof}


End of the corrigendum.

\end{document}