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

\AtBeginDocument{{\noindent\small
\emph{Electronic Journal of Differential Equations},
Vol. 2016 (2016), No. 207, pp. 1--14.\newline
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu}
\thanks{\copyright 2016 Texas State University.}
\vspace{8mm}}

\begin{document}
\title[\hfilneg EJDE-2016/207\hfil Existence and continuation of solutions]
{Existence and continuation of solutions for Caputo type fractional
differential equations}

\author[C. P. Li, S. Sarwar \hfil EJDE-2016/207\hfilneg]
{Changpin Li, Shahzad Sarwar}

\address{Changpin Li \newline
Department of Mathematics, Shanghai University,
Shanghai 200444, China}
\email{lcp@shu.edu.cn}

\address{Shahzad Sarwar \newline
Department of Mathematics, Shanghai University,
Shanghai 200444, China}
\email{shahzadppn@gmail.com}

\thanks{Submitted  April 11, 2016. Published August 1, 2016.}
\subjclass[2010]{34A08, 35A01}
\keywords{Fractional differential equation; Caputo derivative;
\hfill\break\indent local solution; continuation theorem; global solution}

\begin{abstract}
 In this article, we consider a fractional differential equation
 (FDE) with Caputo derivative and study the existence and continuation of
 its solution. Firstly, we prove a theorem on the existence of local solutions.
 Then we  extend the continuation theorems for ODEs to those FDEs. Also
 several global existence results for FDE are obtained.
\end{abstract}

\maketitle
\numberwithin{equation}{section}
\newtheorem{theorem}{Theorem}[section]
\newtheorem{lemma}[theorem]{Lemma}
\newtheorem{remark}[theorem]{Remark}
\newtheorem{definition}[theorem]{Definition}
\allowdisplaybreaks


\section{Introduction}

Recently, fractional differential equations (FDEs) have been the center of
attention of many studies and played a vital role due to emergence in
various applications and exact description of nonlinear phenomena. It has
been found that models using mathematical tools from fractional calculus can
describe various phenomena such as viscoelasticity, electrochemistry,
control, porous media, and many other branches of sciences \cite{13,16,18,34}.
However, the development of existence and uniqueness of solution of FDEs
are very slow. Some contributions about existence of solution of FDEs can be
found in \cite{16,17,22,29}.

Many authors
\cite{1,3,6,9,7,11,12,19,20,26,30,31,32,33,36,39,37}, studied the
existence-uniqueness of solution for FDEs on the finite interval $[0,T]$.
But few researchers \cite{2,4,5,21} present results about the
global existence-uniqueness of solution FDEs on the half axis $[0,+\infty )$.
As far as we know, we cannot find  directly the  existence of global
solution of FDEs by using the results from local existence because, yet
continuation theorems for FDEs have not been derived. Recently, Kou, et al.
\cite{40} found the existence and continuation theorems for
Riemann-Liouville type FDEs. Motivated by that work, a natural question is,
do there also exist local existence, continuation theorems and global
existence for Caputo type FDEs? In this paper, we give an active answer.

In this article, we consider the fractional order initial value problems
(IVPs) of the  form
\begin{equation}
\begin{gathered}
_{C}D_{0,t}^{\alpha }\ x(t)=f(t,x),\quad  0<\alpha <1,\; t\in (0,+\infty ), \\
x(t)|_{t=0}=x_0,\quad  x\in \mathbb{R}.
\end{gathered}   \label{1}
\end{equation}
To ensure the existence of a unique solution to \eqref{1} we always
assume that $f$ satisfies Lipschitz condition with respect to the second
variable, that is, $| f(t,x_1)-f(t,x_2))
| \leq L| x_1-x_2| $, where $L>0$.

For the system of equations
\begin{equation}
\begin{gathered}
_{C}D_{0,t}^{\alpha } x_1(t)=f_1(t,x_1,x_2,\dots,x_n),  \quad 0<\alpha
<1,\; t\in (0,+\infty ), \\
_{C}D_{0,t}^{\alpha } x_2(t)=f_2(t,x_1,x_2,\dots,x_n),
\quad x\in\mathbb{R}^{n}, \\
\cdots   \\
_{C}D_{0,t}^{\alpha } x_n(t)=f_n(t,x_1,x_2,\dots,x_n),  \\
x_i(t)|_{t=0}=x_0, \quad  i=1,2,\dots,n,
\end{gathered}   \label{1a}
\end{equation}
we assume that $f_n(t,x_1,x_2,\dots,x_n)$ satisfy the Lipschitzian
conditions,
\begin{equation*}
| f_{k}(t,x_1,x_2,\dots,x_n)-f_{k}(t,\tilde{x}_1,\tilde{x}
_2,\dots,\tilde{x}_n)| \leq \sum\limits_{k=1}^{n}L_{k}|
x_{k}-\tilde{x}_{k}| ,
\end{equation*}
($L_{k}>0$, $k=1,2,\dots,n$),
where $_{C}D_{0,t}^{\alpha }$ is the Caputo derivative,
$f:\mathbb{R}^{+}\times\mathbb{R}\to\mathbb{R}$ in
the IVP \eqref{1} and $f_i:\mathbb{R}^{+}\times\mathbb{R}^{n}\to\mathbb{R}^{n}$
in  IVP \eqref{1a} have weak singularities with respect to $t$
respectively. In this paper, we establish the local existence for IVP
(\eqref{1} and IVP \eqref{1a}. Then we extend the continuation theorems for ODEs
to those of FDEs. Furthermore, we present global existence of solutions for
IVP \eqref{1}.

The rest of this article is organized as follows:
In Section 2, we introduce
some basic definitions and previously known results that will be used in our
main results. A new local existence theorem for IVP \eqref{1} is given in
Section 3. In Section 4 we present two new continuation theorems for IVP
\eqref{1} which are generalization of the continuation theorems for ODEs.
Concluding remarks and comments are included in the last section.

\section{Preliminaries}

In this section, we introduce some basic definitions and lemmas
\cite{17,22,23,25,29,41} from the theory of fractional calculus which are used
later.
Let $C[a,b]$ be the Bannach space of all continuous functions mapping $[a,b]$
into $\mathbb{R}$ where the norm $\| x\| _{[a,b]}=\max_{t\in [a,b]}| x(t)| $

\begin{definition} \label{def2.1} \rm
The Riemann-Liouville integral of function $f(t)$
with order $\alpha >0$ is defined as
\begin{equation*}
_{RL}D_{0,t}^{-\alpha }f(t)=\frac{1}{\Gamma (\alpha )}\int_0^{t}(t-s)^{
\alpha -1}f(s)ds,\quad t>0.
\end{equation*}
\end{definition}

\begin{definition} \label{def2.2} \rm
The Riemann-Liouville derivative of function $f(t)$
with order $\alpha >0$ is defined as
\begin{equation*}
_{RL}D_{0,t}^{\alpha }f(t)=\frac{1}{\Gamma (n-\alpha )}\frac{d^{n}}{dt^{n}}
\int_0^{t}(t-s)^{n-\alpha -1}f(s)ds,\ t>0,
\end{equation*}
where $n-1<\alpha <n\in \mathbb{Z}^{+}$.
\end{definition}

\begin{definition} \label{def2.3} \rm
The Caputo derivative of function $f(t)$ with order $\alpha >0$ is defined as
\begin{equation*}
_{C}D_{0,t}^{\alpha }f(t)=\frac{1}{\Gamma (n-\alpha )}\int_0^{t}(t-s)^{
\alpha -1}f^{(n)}(s)ds,\ t>0,
\end{equation*}
where $n-1<\alpha <n\in \mathbb{Z}^{+}$.
\end{definition}

\begin{lemma} \label{lem2.1}
Suppose that $f(t,x)$ is a continuous function. Then the
initial value problem \eqref{1} is equivalent to the nonlinear Volterra
integral equation of the second kind
\begin{equation}
x(t)=x_0+\frac{1}{\Gamma (\alpha )}\int_0^{t}(t-s)^{\alpha
-1}f(s,x(s))ds.  \label{2}
\end{equation}
In other words, every solution of the Volterra integral equation \eqref{2} is
also the solution of our original IVP \eqref{1} and vise versa.
\end{lemma}

\begin{lemma} \label{lem2.2}
Let $M$ be a subset of $C[0,T]$. Them $M$ is precompact
if and only if the following conditions  hold:
\begin{enumerate}
\item $\{x(t):x\in M\}$ is uniformly bounded,

\item $\{x(t):x\in M\}$ is equicontinuous on $[0,T]$.
\end{enumerate}
\end{lemma}

\begin{lemma}[Schauder fixed point theorem] \label{lem2.3 }
Let $U$ be a closed bounded convex subset of Bannach space $X$. Suppose that
$T:U\to U$ is completely continuous. Then $T$ has a fixed point in $U$.
\end{lemma}

\section{Local existence theorems}

In this section, we study the  existence of local solutions for
\eqref{1}. Suppose that $f(t,x)$ in \eqref{1} and $f_i(t,x_i)$,
$i=1,2,\dots,n$ in  \eqref{1a} have some weak singularity with respect to
$t$ respectively. By applying Schauder fixed point theorem, a new local
existence theorem is obtained. For this, we make the following hypothesis
for our discussion.
\begin{itemize}
\item[(H1)] Let $f:R^{+}\times R\to R$ in \eqref{1} be a continuous
function then there exists a constant $0\leq \delta <1$ such that $
(Ax)(t)=t^{\delta }f(t,x)$ is a continuous bounded map from $C[0,T]$ into $
C[0,T]$ where $T$ is positive.

\item[(H2)] Let $f_i:R^{+}\times R^{n}\to R$ in \eqref{1a} be
continuous functions then there exist constants $0\leq \delta _i<1$,
such that $(A_ix_i)(t)=t^{\delta
_i}f_i(t,x_1,x_2,\dots,x_n)$, $i=1,2,\dots,n$ are continuous bounded
maps from $C[0,T]$ into $C[0,T]$ where $T$ is positive.
\end{itemize}

\begin{theorem} \label{thm3.1}
Suppose that condition {\rm (H1)} is satisfied. Then IVP \eqref{1} has at least one
solution $x\in C[0,h]$ for some $(T\geq )$ $h>0$.
\end{theorem}

\begin{proof} Let
\begin{equation*}
E=\{ x\in C[0,T]:\| x-x_0\| _{C[0,T]}=\sup_{0\leq
t\leq T}| x-x_0| \leq b\} ,
\end{equation*}
where $b>0$ is a constant. Since operator $A$ is bounded then there exists a
constant $M>0$ such that
\begin{equation*}
\sup \{ | (Ax)(t)| :t\in [ 0,T],x\in E\} \leq M.
\end{equation*}
Again let
\begin{equation*}
D_{h}=\big\{ x:x\in C[0,h],\ \sup_{0\leq t\leq h}| x-x_0| \leq b\big\} ,
\end{equation*}
where $h=\min \{ (\frac{b\Gamma (\alpha +1-\delta )}{M\ \Gamma
(1-\alpha )}) ^{\frac{1}{\alpha -\delta }},T\}$, $\alpha >\delta$.

It is clear that $D_{h}\subseteq C[0,h]$ is nonempty, bounded closed and
convex subset. Note that $h\leq T$, define an operator $B$ as follows
\begin{equation}
(Bx)(t)=x_0+\frac{1}{\Gamma (\alpha )}\int_0^{t}(t-s)^{\alpha
-1}f(s,x(s))ds,\quad t\in [ 0,h].  \label{3}
\end{equation}
By \eqref{3}, for any $x\in C[0,h]$ we have
\[
| (Bx)(t)-x_0|  \leq \frac{M}{\Gamma (\alpha )}
\int_0^{t}(t-s)^{\alpha -1}s^{-\delta }ds
\leq \frac{M\Gamma (1-\alpha )}{\Gamma (\alpha +1-\delta )}h^{\alpha
-\delta }\leq b,
\]
which shows that $BD_{h}\subset D_{h}$.

Next we show that $B$ is continuous. Let $x_n,\ x\in D_{h}$ such that
$\| x_n-x\| _{C[0,h]}\to 0$ as $n\to +\infty$. In the continuity of $A$
we have $\| Ax_n-Ax\|_{[0,h]}\to 0$ as $n\to +\infty $. Now
\begin{align*}
&| (Bx_n)(t)-(Bx)(t)|\\
&= | \frac{1}{\Gamma (\alpha )}\int_0^{t}(t-s)^{\alpha -1}f(s,x_n(s))ds
-\frac{1}{\Gamma (\alpha )}\int_0^{t}(t-s)^{\alpha -1}f(s,x(s))ds|  \\
&\leq \frac{1}{\Gamma (\alpha )}\int_0^{t}(t-s)^{\alpha -1}|
f(s,x_n(s))-f(s,x(s))| ds \\
&\leq \frac{1}{\Gamma (\alpha )}\int_0^{t}(t-s)^{\alpha -1}s^{-\delta
}| (Ax_n)(s)-(Ax)(s)| ds \\
&\leq \frac{1}{\Gamma (\alpha )}\int_0^{t}(t-s)^{\alpha -1}s^{-\delta
}ds\| (Ax_n)(s)-(Ax)(s)\| _{[0,h]}.
\end{align*}
We have
\begin{equation*}
\| (Bx_n)(s)-(Bx)(s)\| _{[0,h]}\leq \frac{\Gamma
(1-\alpha )}{\Gamma (\alpha +1-\delta )}h^{\alpha -\delta }\|
(Ax_n)(s)-(Ax)(s)\| _{[0,h]}.
\end{equation*}
Then $\| (Bx_n)(s)-(Bx)(s)\| _{[0,h]}\to 0$ as
$n\to +\infty $. Thus $B$ is continuous.

Furthermore, we prove that operator $BD_{h}$ is continuous.
Let $x\in D_{h}$ and $0\leq t_1\leq t_2\leq h$. For any $\epsilon >0$, note that
\begin{equation*}
\frac{1}{\Gamma (\alpha )}\int_0^{t}(t-s)^{\alpha -1}s^{-\delta }ds
=\frac{\Gamma (1-\alpha )}{\Gamma (\alpha +1-\delta )}t^{\alpha -\delta
}\to 0,\ \text{as }t\to 0^{+},
\end{equation*}
where $0\leq \delta <1$. There exists a $\tilde{\delta}>0$ such that for
$t\in [ 0,h]$,
\begin{equation*}
\frac{2M}{\Gamma (\alpha )}\int_0^{t}(t-s)^{\alpha -1}s^{-\delta}ds<\epsilon
\end{equation*}
holds.
In this case, for $t_1,t_2\in [ 0,\tilde{\delta}]$ one has
\begin{equation}
\begin{aligned}
&\big| \frac{1}{\Gamma (\alpha )}\int_0^{t_1}(t_1-s)^{\alpha
-1}f(s,x(s))ds-\frac{1}{\Gamma (\alpha )}\int_0^{t_2}(t_2-s)^{\alpha
-1}f(s,x(s))ds\big|    \\
&\leq \frac{M}{\Gamma (\alpha )}\int_0^{t_1}(t_1-s)^{\alpha
-1}s^{-\delta }ds+\frac{M}{\Gamma (\alpha )}\int_0^{t_2}(t_2-s)^{
\alpha -1}s^{-\delta }ds<\epsilon .
\end{aligned} \label{4}
\end{equation}
In this case for $t_1,t_2\in [ \frac{\tilde{\delta}}{2},h]$ one gets
\begin{equation}
\begin{aligned}
&| (Bx)(t_1)-(Bx)(t_2)| \\
&= \Big| \frac{1}{\Gamma
(\alpha )}\int_0^{t_1}(t_1-s)^{\alpha -1}f(s,x(s))ds
-\frac{1}{\Gamma (\alpha )}\int_0^{t_2}(t_2-s)^{\alpha -1}f(s,x(s))ds\Big|
\\
&\leq \big| \frac{1}{\Gamma (\alpha )}\int_0^{t_1}[
(t_1-s)^{\alpha -1}-(t_2-s)^{\alpha -1}] f(s,x(s))ds\big|
 \\
&\quad +\big| \frac{1}{\Gamma (\alpha )}\int_{t_1}^{t_2}(t_2-s)^{
\alpha -1}f(s,x(s))ds\big| .
\end{aligned} \label{5}
\end{equation}
Now, from the first term on the right hand side of \eqref{5} one has
\begin{equation}
\begin{aligned}
&| \frac{1}{\Gamma (\alpha )}\int_0^{t_1}[
(t_1-s)^{\alpha -1}-(t_2-s)^{\alpha -1}] f(s,x(s))ds|
 \\
&\leq \frac{M}{\Gamma (\alpha )}\int_0^{t_1}| [
(t_1-s)^{\alpha -1}-(t_2-s)^{\alpha -1}] s^{-\delta }| ds
 \\
&\leq \frac{M}{\Gamma (\alpha )}\int_0^{\tilde{\delta}/2}
| [ (t_1-s)^{\alpha -1}-(t_2-s)^{\alpha -1}]
s^{-\delta }| ds   \\
&\quad +\frac{M(\frac{\tilde{\delta}}{2})^{-\delta }}{\Gamma (\alpha )}\int_{
\frac{\tilde{\delta}}{2}}^{t_1}| [ (t_1-s)^{\alpha
-1}-(t_2-s)^{\alpha -1}] | ds   \\
&\leq \frac{2M}{\Gamma (\alpha )}\int_0^{\frac{\delta _1}{2}}\big(
\frac{\tilde{\delta}}{2}-s\big) ^{\alpha -1}s^{-\delta }ds+\frac{M(\frac{
\tilde{\delta}}{2})^{-\delta }}{\Gamma (\alpha )}[ (t_2-t_1)^{
\alpha }  \\
&\quad +\big(t_1-\frac{\tilde{\delta}}{2}\big) ^{\alpha }
-\big(t_2-\frac{\tilde{\delta}}{2}\big) ^{\alpha }]   \\
&\leq \epsilon +\frac{M(\frac{\tilde{\delta}}{2})^{-\delta }}{\Gamma
(\alpha )}[ (t_2-t_1)^{\alpha }+\big(t_1-\frac{\tilde{\delta}}{2
}\big) ^{\alpha }-\big(t_2-\frac{\tilde{\delta}}{2}\big) ^{\alpha }] .
\end{aligned}  \label{6}
\end{equation}
Next from the second term on the right hand side of \eqref{5}, one has
\begin{equation}
\begin{aligned}
\big| \frac{1}{\Gamma (\alpha )}\int_{t_1}^{t_2}(t_2-s)^{\alpha
-1}f(s,x(s))ds\big|
&\leq \frac{M(\frac{\delta _1}{2})^{-\delta }}{
\Gamma (\alpha )}\int_{t_1}^{t_2}(t_2-s)^{\alpha -1}ds   \\
&\leq \frac{M(\frac{\delta _1}{2})^{-\delta }}{\Gamma (\alpha +1)}
(t_2-t_1)^{\alpha }.
\end{aligned}  \label{7}
\end{equation}
From the above discussion, there exists a
$(\frac{\tilde{\delta}}{2}>)\tilde{\delta}_1>0$  such that for
$t_1,t_2\in [ \frac{\tilde{\delta}}{2},h]$
and $| t_1-t_2| <\tilde{\delta}_1$,
\begin{equation}
| (Bx)(t_1)-(Bx)(t_2)| <2\epsilon .  \label{8}
\end{equation}

It follows from \eqref{4} and \eqref{8} that $\{(Bx)(t):x\in D_{h}\}$ is
equicontinuous. It is also clear that $\{(Bx)(t):x\in D_{h}\}$ is uniformly
bounded due to $BD_{h}\subset D_{h}$. So $BD_{h}$ is precompact. Therefore
$B $ is completely continuous. By Schauder fixed point theorem and Lemma \ref{lem2.1},
IVP \eqref{1} has a local solution. The proof is thus completed.
\end{proof}

\begin{theorem} \label{thm3.2}
 Suppose that condition {\rm (H2)} is satisfied. Then IVP
\eqref{1a} has at least one solution $x_i\in C[0,h]$ for some $(T\geq )$
$h>0$.
\end{theorem}

\begin{proof} Let
\begin{equation*}
E=\big\{ x_i\in C[0,T]:\| x_i-x_0\|
_{C[0,T]}=\sup_{0\leq t\leq T}| x_i-x_0| \leq
b_i,\ i=1,2,\dots,n\big\} ,
\end{equation*}
where $b_i>0$, $i=1,2,\dots,n$ are constants. Since the operators
$A_i$, $i=1,2,\dots,n$ are bounded then there exist constants
$M_i>0$, $i=1,2,\dots,n$
such that
\begin{equation*}
\sup \{ | (A_ix_i)(t)| :t\in [ 0,T],\
x_i\in E\} \leq M_i,\quad i=1,2,\dots,n.
\end{equation*}
Again let
\begin{equation*}
D_{ih}=\{ x_i:x_i\in C[0,h],\, \sup_{0\leq t\leq h}|
x_i-x_0| \leq b_i,\; i=1,2,\dots,n\} ,
\end{equation*}
where
\begin{align*}
h=\min \big\{& \big(\frac{b_1\Gamma (\alpha +1-\delta _1)}{
M_1\ \Gamma (1-\alpha )}\big) ^{\frac{1}{\alpha -\delta _1}},
\big(\frac{b_2\Gamma (\alpha +1-\delta _2)}{M_2\ \Gamma (1-\alpha )}\big)
^{\frac{1}{\alpha -\delta _2}}, \dots, \\
&\big(\frac{b_n\Gamma (\alpha
+1-\delta _n)}{M_n\ \Gamma (1-\alpha )}\big) ^{\frac{1}{\alpha -\delta
_n}},T\big\},
\end{align*}
$\alpha >\delta _i$, $i=1,2,\dots,n$.

It is clear that $D_{ih}\subseteq C[0,h]$ are nonempty, bounded closed,
and convex subsets. Note that $h\leq T$, define operators $B_i$ as follows
\begin{equation}
\begin{gathered}
(B_1x_1)(t)=x_0+\frac{1}{\Gamma (\alpha )}\int_0^{t}(t-s)^{\alpha
-1}f_1(s,x_1(s),x_2(s),\dots,x_n(s))ds,  \\
(B_2x_2)(t)=x_0+\frac{1}{\Gamma (\alpha )}\int_0^{t}(t-s)^{\alpha
-1}f_2(s,x_1(s),x_2(s),\dots,x_n(s))ds,   \\
\cdots   \\
(B_nx_n)(t)=x_0+\frac{1}{\Gamma (\alpha )}\int_0^{t}(t-s)^{\alpha
-1}f_n(s,x_1(s),x_2(s),\dots,x_n(s))ds,
\end{gathered}  \label{a}
\end{equation}
for $t\in [ 0,h]$.
By \eqref{a}, for any $x_i\in C[0,h]$ we have
\begin{gather*}
| (B_1x_1)(t)-x_0| \leq \frac{M_1}{\Gamma (\alpha
)}\int_0^{t}(t-s)^{\alpha -1}s^{-\delta _1}ds, \\
| (B_2x_2)(t)-x_0| \leq \frac{M_2}{\Gamma (\alpha
)}\int_0^{t}(t-s)^{\alpha -1}s^{-\delta _2}ds, \\
\cdots \\
| (B_nx_n)(t)-x_0| \leq \frac{M_n}{\Gamma (\alpha
)}\int_0^{t}(t-s)^{\alpha -1}s^{-\delta _n}ds,
\end{gather*}
and
\begin{gather*}
| (B_1x_1)(t)-x_0| \leq \frac{M_1\Gamma (1-\alpha
)}{\Gamma (\alpha +1-\delta _1)}h^{\alpha -\delta _1}\leq b_1, \\
| (B_2x_2)(t)-x_0| \leq \frac{M_2\Gamma (1-\alpha
)}{\Gamma (\alpha +1-\delta _2)}h^{\alpha -\delta _2}\leq b_2, \\
\cdots \\
| (B_nx_n)(t)-x_0| \leq \frac{M_n\Gamma (1-\alpha
)}{\Gamma (\alpha +1-\delta _1)}h^{\alpha -\delta _n}\leq b_n,
\end{gather*}
which shows that, $B_iD_{ih}\subset D_{ih}$, $i=1,2,\dots,n$.

Next we show that operators $B_i$ are continuous.
Let $x_{m}$, $x_i\in D_{ih}$, $m>n$, $i=1,2,\dots,n$ such that
$\| x_{m}-x_i\| _{C[0,h]}\to 0$ as $m\to +\infty $. In view of
continuity of operators $A_i$ we have $\|A_ix_{m}-A_ix_i\| _{[0,h]}\to 0$ as
$m\to+\infty $. Now
\begin{align*}
&| (B_ix_{m})(t)-(B_ix_i)(t)| \\
&= \big| \frac{1}{\Gamma (\alpha )}\int_0^{t}(t-s)^{\alpha -1}f_i(s,x_{m}(s))ds
-\frac{1}{\Gamma (\alpha )}\int_0^{t}(t-s)^{\alpha -1}f_i(s,x_i(s))ds\big|
\\
&\leq \frac{1}{\Gamma (\alpha )}\int_0^{t}(t-s)^{\alpha -1}|
f_i(s,x_{m}(s))-f_i(s,x_i(s))| ds \\
&\leq \frac{1}{\Gamma (\alpha )}\int_0^{t}(t-s)^{\alpha -1}s^{-\delta
_i}| (A_ix_{m})(s)-(A_ix_i)(s)| ds \\
&\leq \frac{1}{\Gamma (\alpha )}\int_0^{t}(t-s)^{\alpha -1}s^{-\delta
_i}ds\| (A_ix_{m})(s)-(A_ix_i)(s)\| _{[0,h]}.
\end{align*}
We have
\begin{equation*}
\| (B_ix_{m})(s)-(B_ix_i)(s)\| _{[0,h]}\leq \frac{
\Gamma (1-\alpha )}{\Gamma (\alpha +1-\delta _i)}h^{\alpha -\delta
_i}\| (A_ix_{m})(s)-(A_ix_i)(s)\| _{[0,h]}.
\end{equation*}
Then $\| (B_ix_{m})(s)-(B_ix_i)(s)\|_{[0,h]}\to 0$ as $m\to +\infty $.
Thus $B_i$ are continuous.
Furthermore, we prove that operators $B_iD_{ih}$ are continuous.
Let $x_i\in D_{ih}$ and $0\leq t_1\leq t_2\leq h$. For any
$\epsilon >0$, note that
\begin{equation*}
\frac{1}{\Gamma (\alpha )}\int_0^{t}(t-s)^{\alpha -1}s^{-\delta _i}ds
=\frac{\Gamma (1-\alpha )}{\Gamma (\alpha +1-\delta _i)}t^{\alpha -\delta
_i}\to 0,\quad \text{as }t\to 0^{+},
\end{equation*}
where $0\leq \delta _i<1$. There exists $\tilde{\delta}_i>0$
such that for $t\in [ 0,h]$,
\begin{equation*}
\frac{2M_i}{\Gamma (\alpha )}\int_0^{t}(t-s)^{\alpha -1}s^{-\delta_i}ds<\epsilon .
\end{equation*}
In this case, for $t_1,t_2\in [ 0,\tilde{\delta}_i]$,
one has
\begin{equation}
\begin{aligned}
&\big| \frac{1}{\Gamma (\alpha )}\int_0^{t_1}(t_1-s)^{\alpha
-1}f_i(s,x_i(s))ds-\frac{1}{\Gamma (\alpha )}\int_0^{t_2}(t_2-s)^{
\alpha -1}f_i(s,x_i(s))ds\big|   \\
&\leq \frac{M_i}{\Gamma (\alpha )}\int_0^{t_1}(t_1-s)^{\alpha
-1}s^{-\delta _i}ds+\frac{M_i}{\Gamma (\alpha )}
\int_0^{t_2}(t_2-s)^{\alpha -1}s^{-\delta _i}ds<\epsilon .
\end{aligned} \label{4a}
\end{equation}
In this case, for $t_1,t_2\in [ \frac{\tilde{\delta}_i}{2},h]$, one gets
\begin{equation}
\begin{aligned}
&| (B_ix_i)(t_1)-(B_ix_i)(t_2)| \\
&= \big|\frac{1}{\Gamma (\alpha )}\int_0^{t_1}(t_1-s)^{\alpha
-1}f_i(s,x_i(s))ds
-\frac{1}{\Gamma (\alpha )}\int_0^{t_2}(t_2-s)^{
\alpha -1}f_i(s,x_i(s))ds\big|   \\
&\leq \big| \frac{1}{\Gamma (\alpha )}\int_0^{t_1}[
(t_1-s)^{\alpha -1}-(t_2-s)^{\alpha -1}] f_i(s,x_i(s))ds\big|\\
&\quad +\big| \frac{1}{\Gamma (\alpha )}\int_{t_1}^{t_2}(t_2-s)^{
\alpha -1}f_i(s,x_i(s))ds\big|.
\end{aligned}   \label{5a}
\end{equation}
Now, from the first term on the right hand side of \eqref{5a} one has
\begin{equation}
\begin{aligned}
&\big| \frac{1}{\Gamma (\alpha )}\int_0^{t_1}[
(t_1-s)^{\alpha -1}-(t_2-s)^{\alpha -1}] f_i(s,x_i(s))ds\big|
   \\
&\leq \frac{M_i}{\Gamma (\alpha )}\int_0^{t_1}| [
(t_1-s)^{\alpha -1}-(t_2-s)^{\alpha -1}] s^{-\delta
_i}| ds   \\
&\leq \frac{M_i}{\Gamma (\alpha )}\int_0^{\frac{\tilde{\delta}_i}{2}
}| [ (t_1-s)^{\alpha -1}-(t_2-s)^{\alpha -1}]
s^{-\delta _i}| ds   \\
&\quad +\frac{M_i(\frac{\tilde{\delta}_i}{2})^{-\delta _i}}{\Gamma (\alpha )
}\int_{\frac{\tilde{\delta}_i}{2}}^{t_1}| [
(t_1-s)^{\alpha -1}-(t_2-s)^{\alpha -1}] | ds   \\
&\leq \frac{2M_i}{\Gamma (\alpha )}\int_0^{\tilde{\delta}_i/2}
\big(\frac{\tilde{\delta}_i}{2}-s\big) ^{\alpha -1}s^{-\delta _i}ds
+ \frac{M_i(\frac{\tilde{\delta}_i}{2})^{-\delta _i}}{\Gamma (\alpha )}
\Big[ (t_2-t_1)^{\alpha }   \\
&\quad +\big(t_1-\frac{\tilde{\delta}_i}{2}\big) ^{\alpha }
-\big( t_2-\frac{\tilde{\delta}_i}{2}\big) ^{\alpha }\Big]    \\
&\leq \epsilon +\frac{M_i(\frac{\tilde{\delta}_i}{2})^{-\delta _i}}{
\Gamma (\alpha )}\big[ (t_2-t_1)^{\alpha }+\big(t_1-\frac{\tilde{
\delta}_i}{2}\big) ^{\alpha }-\big(t_2-\frac{\tilde{\delta}_i}{2}
\big) ^{\alpha }\big] .
\end{aligned}
\end{equation}
Next from the second term on the right hand side of \eqref{5}, one has
\begin{align*}
\big| \frac{1}{\Gamma (\alpha )}\int_{t_1}^{t_2}(t_2-s)^{\alpha
-1}f_i(s,x_i(s))ds\big|
&\leq \frac{M_i(\frac{\tilde{\delta}_i}{2})^{-\delta _i}}{\Gamma (\alpha )}
\int_{t_1}^{t_2}(t_2-s)^{\alpha -1}ds \\
&\leq \frac{M_i(\frac{\tilde{\delta}_i}{2})^{-\delta _i}}{\Gamma
(\alpha +1)}(t_2-t_1)^{\alpha }.
\end{align*}
So from the above discussion, there exist $(\frac{\tilde{\delta}_i}{2}>)$
$\lambda >0$ such that for $t_1,t_2\in [ \frac{\tilde{\delta}_i}{2
},h]$ and $| t_1-t_2| <\lambda $,
\begin{equation}
| (B_ix_i)(t_1)-(B_ix_i)(t_2)| <2\epsilon .
\label{6a}
\end{equation}
It follows from \eqref{4} and \eqref{8} that
$\{(B_ix_i)(t):x_i\in D_{ih}\}$ are equicontinuous. It is also clear
that $\{(B_ix_i)(t):x_i\in D_{ih}\}$ are uniformly bounded due to
$B_iD_{ih}\subset D_{ih}$. So $B_iD_{ih}$ are precompact. Therefore
operators $B_i$ are completely continuous. By Schauder fixed point theorem
and Lemma \ref{lem2.1}, IVP \eqref{1a} has a local solution. The proof is thus
completed.
\end{proof}

\section{Continuation theorems}

In this section, we study the continuation of solution for IVP \eqref{1}.
 The basic techniques may be applied to system \eqref{1a}, so we omit
the detail here or leave to the interested readers. We extend the
continuation theorem for ODEs to Caputo type FDEs. Initially, we give the
following definition.

\begin{definition}[\cite{40}] \label{def4.1} \rm
Let $x(t)$ on $(0,\beta )$ and $\tilde{x}(t)$ on $(0,\tilde{\beta})$
both are the solutions of \eqref{1}. If
$\beta <\tilde{\beta}$ and $x(t)=\tilde{x}(t)$ for $t\in (0,\beta )$, we say
that $\tilde{x}(t)$ can be continued to $(0,\tilde{\beta})$.
A solution $x(t)$ is noncontinuable if it has no continuation.
The existing interval of noncontinuable solution $x(t)$ is called the maximum
 existing interval of $x(t)$.
\end{definition}

\begin{theorem} \label{thm4.1}
Assume that condition {\rm (H1)} is satisfied. Then
$x=x(t)$, $t\in (0,\beta )$ is noncontinuable if and if only for some
$\eta \in (0,\frac{\beta }{2})$ and any bounded closed subset
$S\subset [\eta ,+\infty )\times\mathbb{R}$ there exists a
$t^{\ast }\in [ \eta ,\beta )$ such that $(t^{\ast},x(t^{\ast }))\notin S$.
\end{theorem}

\begin{proof}
The proof of this theorem is given in two steps.
Suppose that there exists a compact subset
$S\subset [ \eta,+\infty )\times\mathbb{R}$ such that
$\{ (t,x(t)):t\in [ \eta ,\beta )\} \subset S$.
The compactness of $S$ implies $\beta <+\infty $. By (H1) there exists a $K>0$
such that $\sup_{(t,x)\in S}| f(t,x)| \leq K$.
\smallskip

\noindent\textbf{Step 1.}
 We show that $\lim_{t\to \beta ^{-}}x(t)$ exists.
Let
\[
J(t)=\int_0^{\eta }(t-s)^{\alpha -1}s^{-\delta }ds,\quad t\in [2\eta ,\beta].
\]
We can easily see that $J(t)$ is uniformly continuous on
$[2\eta ,\beta ]$. For all $t_1,t_2\in [ 2\eta ,\beta ),\ t_1<t_2$ we
have
\begin{align*}
&| x(t_1)-x(t_2)| \\
&= \big| \frac{1}{\Gamma
(\alpha )}\int_0^{t_1}(t_1-s)^{\alpha -1}f(s,x(s))ds-\frac{1}{\Gamma
(\alpha )}\int_0^{t_2}(t_2-s)^{\alpha -1}f(s,x(s))ds\big| \\
&\leq \big| \frac{1}{\Gamma (\alpha )}\int_0^{\eta } [
(t_1-s)^{\alpha -1}-(t_2-s)^{\alpha -1}] s^{-\delta
}(Ax)(s)ds\big| \\
&\quad +| \frac{1}{\Gamma (\alpha )}\int_{\eta }^{t_1}[
(t_1-s)^{\alpha -1}-(t_2-s)^{\alpha -1}] f(s,x(s))ds\big| \\
&\quad +| \frac{1}{\Gamma (\alpha )}\int_{t_1}^{t_2}(t_2-s)^{
\alpha -1}f(s,x(s))ds| \\
&\leq \frac{\| Ax\| _{[0,\eta ]}}{\Gamma (\alpha )}
\int_0^{\eta }\left[ (t_1-s)^{\alpha -1}-(t_2-s)^{\alpha -1}\right]
s^{-\delta }ds \\
&\quad +\frac{K}{\Gamma (\alpha )}\int_{\eta }^{t_1}[ (t_1-s)^{\alpha
-1}-(t_2-s)^{\alpha -1}] ds
+\frac{K}{\Gamma (\alpha )} \int_{t_1}^{t_2}(t_2-s)^{\alpha -1}ds \\
&\leq | J(t_1)-J(t_2)| \frac{\|
Ax\| _{[0,\eta ]}}{\Gamma (\alpha )}+\frac{K}{\Gamma (\alpha )}
[ 2(t_2-t_1)^{\alpha }+(t_1-\eta )^{\alpha }-(t_2-\eta
)^{\alpha }] .
\end{align*}
From the continuity of $J(t)$ and Cauchy convergence
criterion, it follows that $\lim_{t\to \beta ^{-}}x(t)=x^{\ast }$.
\smallskip

\noindent\textbf{Step 2.}
 Now we show that $x(t)$ is continuable. Since $S$ is a
closed subset, we have $(\beta ,x^{\ast })\in S$.
Define $x(\beta )=x^{\ast }$. Then $x(t)\in C[0,\beta ]$, we define
operator $D$ as follows
\begin{equation*}
(Dy)(t)=x_1+\frac{1}{\Gamma (\alpha )}\int_{\beta }^{t}(t-s)^{\alpha
-1}f(s,y(s))ds,
\end{equation*}
where
\begin{equation*}
x_1=x_0+\frac{1}{\Gamma (\alpha )}\int_0^{\beta }(t-s)^{\alpha
-1}f(s,y(s))ds,\ \ y\in C[\beta ,\beta +1],\quad  t\in [ \beta ,\beta+1].
\end{equation*}
Let
\begin{equation*}
E_{b}=\{ (t,y):\beta \leq t\leq \beta +1,| y| \leq
\max_{\beta \leq t\leq \beta +1}| x_1(t)| +b\}.
\end{equation*}
In view of the continuation of $f$ on $E_{b}$, denote
$M=\max_{(t,y)\in E_{b}}| f(t,y)| $. Again let
\begin{equation*}
E_{h}=\{ y\in C[\beta ,\beta +1]:\max_{t\in [ \beta ,\beta
+h]}| y(t)-x_1(t)| \leq b,y(\beta )=x_1(\beta)\} ,
\end{equation*}
where $h=\min \big\{ 1,(\frac{\Gamma (\alpha +1)b}{M})
^{\frac{1}{\alpha }}\big\}$. We can claim that $D$ is completely
continuous on $E_{b}$. Set $\{y_n\}\subseteq C[\beta ,\beta +h]$,
$\| y_n-y\| _{[\beta ,\beta +h]}\to 0$ as $n\to +\infty $. Then we have
\begin{align*}
| (Dy_n)(t)-(Dy)(t)|
&= \big| \frac{1}{\Gamma (\alpha )}\int_{\beta }^{t}(t-s)^{\alpha -1}
[ f(s,y_n(s))-f(s,y(s))] ds\big| \\
&\leq \frac{h^{\alpha }}{\Gamma (\alpha +1)}\|
f(s,y_n(s))-f(s,y(s))\| _{[\beta ,\beta +h]}.
\end{align*}
By the continuity of $f$ we have
$\|f(s,y_n(s))-f(s,y(s))\| _{[\beta ,\beta +h]}\to 0$ as
$n\to +\infty $. Therefore,
$\| (Dy_n)(t)-(Dy)(t)\| _{_{[\beta ,\beta+h]}}\to 0$ as $n\to +\infty $,
which implies that operator $D$ is continuous.

Secondly, we prove that $DE_{h}$ is equicontinuous. For any
$y\in E_{h}$ we have $(Dy)(\beta )=x_1(\beta )$ and
\begin{align*}
| (Dy)(t)-x_1|
&= \big| \frac{1}{\Gamma (\alpha )}
 \int_{\beta }^{t}(t-s)^{\alpha -1}f(s,y(s))ds\big| \\
&\leq \frac{M(t-\beta )^{\alpha }}{\Gamma (\alpha +1)}\leq \frac{Mh^{\alpha
}}{\Gamma (\alpha +1)}\leq b.
\end{align*}
Thus $DE_{h}\subset E_{h}$. Set
$I(t)=\frac{1}{\Gamma (\alpha )} \int_0^{\beta }(t-s)^{\alpha -1}f(s,x(s))ds$.
We know that $I(t)$ is continuous on $[\beta ,\beta +1]$.
For all $y\in E_{h}$, $\beta \leq t_1\leq t_2\leq \beta +h$, we have
\begin{equation} \label{9}
\begin{aligned}
&| (Dy)(t_1)-(Dy)(t_2)| \\
& \leq | \frac{1}{\Gamma (\alpha )}\int_0^{\beta }\left[ (t_1-s)^{\alpha
-1}-(t_2-s)^{\alpha -1}\right] f(s,y(s))ds|   \\
&\quad +\frac{1}{\Gamma (\alpha )}| \int_{\beta }^{t_1}\left[
(t_1-s)^{\alpha -1}-(t_2-s)^{\alpha -1}\right] f(s,y(s))ds|
 \\
&\quad +\frac{1}{\Gamma (\alpha )}|
\int_{t_1}^{t_2}(t_2-s)^{\alpha -1}f(s,y(s))ds|   \\
& \leq | I(t_1)-I(t_2)| +\frac{M}{\Gamma (\alpha +1)}
[ 2(t_2-t_1)^{\alpha } +(t_1-\beta )^{\alpha }-(t_2-\beta )^{\alpha }] .
\end{aligned}
\end{equation}
In view of the uniform continuity of $I(t)$ on $[\beta ,\beta +h]$ and
\eqref{9}, we conclude that $\{ (Dy)(t):y\in E_{h}\} $ is
equicontinuous. Therefore $D$ is completely continuous. By Schauder fixed
point theorem, operator $D$ has a fixed point $\tilde{x}(t)\in E_{h}$, i.e.,
\begin{equation} \label{10}
\begin{aligned}
\tilde{x}(t)
&= x_1+\frac{1}{\Gamma (\alpha )}\int_{\beta}^{t}
 (t-s)^{\alpha -1}f(s,\tilde{x}(s))ds,   \\
&= x_0+\frac{1}{\Gamma (\alpha )}\int_0^{t}(t-s)^{\alpha -1}f(s,\tilde{x}
(s))ds,\quad t\in [ \beta ,\beta +h],
\end{aligned}
 \end{equation}
where
\[
\tilde{x}(t)=\begin{cases}
x(t), & t\in (0,\beta ] \\
\tilde{x}(t), & t\in [ \beta ,\beta +h]
\end{cases}
\]
It follows that $\tilde{x}(t)\in C[0,\beta +h]$ and
\begin{equation}
\tilde{x}(t)=x_0+\frac{1}{\Gamma (\alpha )}\int_0^{t}(t-s)^{\alpha
-1}f(s,\tilde{x}(s))ds.  \label{11}
\end{equation}
Therefore, according to Lemma \ref{lem2.1}, $\tilde{x}(t)$ is a solution of
\eqref{1} on $(0,\beta +h]$. This yields a contradiction (since $x(t)$ is
noncontinuable). The proof is thus complete.
\end{proof}

\begin{remark} \label{rmk4.1} \rm
Theorem \ref{thm4.1} is generalization of \cite[Theorem C]{10},
which is the continuation theorem for the ODE. To see this
\eqref{1} is reduced to an ODE if we set $\alpha =1$.
\end{remark}

Now we present another continuation theorem, which is more
convenient for applications.

\begin{theorem}[[Continuation Theorem II] \label{thm4.2}
Suppose that condition {\rm (H1)} is satisfied. Then $x=x(t)$, $t\in (0,\beta )$
is noncontinuable if and only if
\begin{equation}
\lim_{t\to \beta ^{-}}\sup |K(t)|=+\infty ,  \label{12}
\end{equation}
where $K(t)=(t,x(t))$, $\| K(t)\| =(x^{2}(t)+t^{2})^{\frac{1}{2}}$.
\end{theorem}

\begin{proof}
We prove this theorem by contradiction. Suppose that \eqref{12}
 is not true. Then there exist a sequence $\{t_n\}$ and a positive
constant $L>0$ such that $t_n<t_{n+1},\ n\in \mathbb{N}$,
\begin{equation}
\lim_{n\to \infty }t_n=\beta ,\quad |K(t_n)|\leq L,\quad
\text{i.e., }(x^{2}(t_n)+t_n^{2})\leq L^{2} \label{13}
\end{equation}
Since $\{x(t_n)\}$ is a bounded convergent sub-sequence, one can let
\begin{equation}
\lim_{n\to \infty }x(t_n)=x^{\ast }.  \label{14}
\end{equation}
Now we show that, for any given $\varepsilon >0$ there exists
$T\in (0,\beta )$, such that $|x(t)-x^{\ast }|<\varepsilon$, $t\in (T,\beta )$,
i.e.,
\begin{equation}
\lim_{t\to \beta ^{-}}x(t)=x^{\ast }.  \label{15}
\end{equation}
For sufficiently small $\tau >0$, let
\begin{equation*}
E_1=\big\{ (t,x):t\in [ \tau ,\beta ],|x|\leq \sup_{t\in [
\tau ,\beta )}|x(t)|\big\} .
\end{equation*}
Since $f$ is continuous on $E_1$, we can denote
$K=\max_{(t,y)\in E_1}|f(t,y)|$. It follows from \eqref{13} and \eqref{14}
that there exists $n_0$ such that $t_{n_0}>\tau $ and for $n\geq n_0$ we have
\begin{equation*}
|x(t_n)-x^{\ast }|\leq \frac{\varepsilon }{2}.
\end{equation*}
If \eqref{15} is not true, then for $n\geq n_0$, there exists
$\lambda _n\in (t_n,\beta )$ such that
$|x(\lambda _n)-x^{\ast }| \geq \varepsilon $ and
$|x(t)-x^{\ast }|<\varepsilon$, $t\in (t_n,\lambda_n)$. Thus
\begin{align*}
\varepsilon
&\leq |x(\lambda _n)-x^{\ast }| \\
&\leq |x(t_n)-x^{\ast }|+|x(\lambda _n)-x(t_n)| \\
&\leq \frac{\varepsilon }{2}+\big| \frac{1}{\Gamma (\alpha )}
\int_0^{t_n}(t_n-s)^{\alpha -1}f(s,x(s))ds-\frac{1}{\Gamma (\alpha )}
\int_0^{\lambda _n}(\lambda _n-s)^{\alpha -1}f(s,x(s))ds\big| \\
&\leq \frac{\varepsilon }{2}+\frac{1}{\Gamma (\alpha )}\big|
\int_0^{\tau }[ (t_n-s)^{\alpha -1}-(\lambda _n-s)^{\alpha -1}
] f(s,x(s))ds\big| \\
&\quad +\frac{1}{\Gamma (\alpha )}| \int_{\tau }^{t_n}[
(t_n-s)^{\alpha -1}-(\lambda _n-s)^{\alpha -1}] f(s,x(s))ds\big| \\
&\quad +\big| \frac{1}{\Gamma (\alpha )}\int_{t_n}^{\lambda _n}(\lambda
_n-s)^{\alpha -1}f(s,x(s))ds\big| \\
&\leq \frac{\varepsilon }{2}+\frac{\| Ax\| _{[0,\tau ]}}{
\Gamma (\alpha )}| I(t_n)-I(\lambda _n)| +\frac{M}{
\Gamma (\alpha +1)}[ 2(\lambda _n-t_n)^{\alpha } \\
&\quad +(t_n-\tau)^{\alpha }-(\lambda _n-\tau )^{\alpha }] .
\end{align*}
In view of continuity of $I(t)$ on $[t_{n_0},\beta ]$, for sufficiently
large $n\geq n_0$, we have
\begin{equation*}
\varepsilon \leq |x(\lambda _n)-x^{\ast }|<\frac{\varepsilon }{2}+\frac{
\varepsilon }{2}=\varepsilon .
\end{equation*}
This implies the contradiction that $\lim_{t\to \beta ^{-}}x(t)$ exists.
By the similar argument to the proof of Theorem \ref{thm4.1}, we
can find a continuation of $x(t)$. The proof is ended.
\end{proof}

\begin{remark} \label{rmk4.2} \rm
 If $f$ in  \eqref{1} satisfies the global
Lipschitz condition with the second variable, then its solution globally
exists and it is unique.
\end{remark}

\section{Global existence theorems}

In this section, we study the existence of a  global solution for
\eqref{1} which is based on the previously  results. The basic
techniques may be applied to system \eqref{1a}, so we omit the details here,
and leave them for the interested readers.
Applying Theorem \ref{thm4.2}, in a straight way we acquire the following
conclusion about the existence of global solution of \eqref{1}.

\begin{theorem} \label{thm5.1}
 Suppose that condition {\rm (H1)} is satisfied. Let $x(t)$ be a solution
of \eqref{1} on $(0,\beta )$. If $x(t)$ is bounded on $[\tau ,\beta )$ for some
$\tau >0$, then $\beta =+\infty$.
\end{theorem}

Continuing our discussion, we firstly present the
following lemma, which is useful in our analysis.

\begin{lemma}[\cite{15,35}] \label{lem5.1}
Let $v:[0,b]\to [0,+\infty )$ be a real function, and $w(\cdot )$
be a nonnegative, locally integrable function on $[0,b]$.
Suppose that there exist $a>0$ and $0<\alpha <1$ such that
\begin{equation*}
v(t)\leq w(t)+a\int_0^{t}\frac{v(s)}{(t-s)^{\alpha }}ds.
\end{equation*}
Then there exists a constant $k=k(\alpha )$ such that for $t\in [ 0,b]$, we have
\begin{equation*}
v(t)\leq w(t)+ka\int_0^{t}\frac{w(s)}{(t-s)^{\alpha }}ds.
\end{equation*}
\end{lemma}

\begin{theorem} \label{thm5.2}
Suppose that condition {\rm (H1)} is satisfied and
there exist three non-negative continuous functions
$l(t)$, $m(t)$, $p(t):[0,+\infty )\to [0,+\infty )$ such that
$|f(t,x)|\leq l(t)m(|x|)+p(t)$, where $m(r)\leq r$ for $r\geq 0$.
Then \eqref{1} has one solution in $C[0,+\infty )$.
\end{theorem}

\begin{proof}
The  existence of a local solution $x(t)$ of \eqref{1} can be
concluded by Theorem \ref{thm3.1}. By Lemma \ref{lem2.1}, $x(t)$ satisfies the
integral equation
\begin{equation*}
x(t)=x_0+\frac{1}{\Gamma (\alpha )}\int_0^{t}(t-s)^{\alpha
-1}f(s,x(s))ds.
\end{equation*}
Suppose that the maximum existing interval of $x(t)$ is
$[0,\beta )$ $(\beta<+\infty )$. Then
\begin{align*}
| x(t)|
&= \big| x_0+\frac{1}{\Gamma (\alpha )}
\int_0^{t}(t-s)^{\alpha -1}f(s,x(s))ds\big| \\
&\leq x_0+\frac{1}{\Gamma (\alpha )}\int_0^{t}(t-s)^{\alpha -1}(
l(s)m(|x|)+p(s)) ds \\
&\leq x_0+\frac{\| l\| _{[0,\beta ]}}{\Gamma (\alpha )}
\int_0^{t}(t-s)^{\alpha -1}(m(|x|) ds+\frac{1}{\Gamma (\alpha )
}\int_0^{t}(t-s)^{\alpha -1}p(s)ds.
\end{align*}
We take $v(t)=| x(t)| ,\ w(t)=x_0+\frac{1}{\Gamma
(\alpha )}\int_0^{t}(t-s)^{\alpha -1}p(s)ds$,
$a=\frac{\|l\| _{[0,\beta ]}}{\Gamma (\alpha )}$. By Lemma \ref{lem5.1}, we know that
$v(t)=| x(t)| $ is bounded on $[0,\beta )$. Thus for any
$\tau \in (0,\beta )$, $x(t)$ is bounded on $[\tau ,\beta )$.
By theorem \ref{thm5.1}, IVP \eqref{1} has a solution $x(t)$ on $(0,+\infty )$.
\end{proof}

The following result guarantees the existence and uniqueness of
global solution of \eqref{1} on $\mathbb{R}^{+}$.

\begin{theorem} \label{thm5.3}
Suppose that  {\rm (H1)} is satisfied and
there exists a non-negative continuous function $l(t)$ defined on
$[0,\infty )$ such that $|f(t,x)-f(t,y)|\ \leq l(t)|x-y|$. Then  \eqref{1} has a
unique solution in $C[0,+\infty )$.
\end{theorem}

The  existence of a global solution can be obtained by using the same
arguments as above. From the Lipschitz-type condition and Lemma \ref{lem5.1}, we can
conclude the uniqueness of global solution. The proof is omitted here.

\subsection*{Conclusion}

In this article, we  obtained a new local existence theorem for
Caputo type general FDE which has a certain singularity.
Then we derived two continuation theorems which have been never studied before.
Next we established global existence theorems for the FDEs.

\subsection*{Acknowledgments}
The present work was partially supported by the
National Natural Science Foundation of China under grant no. 11372170.

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