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

\AtBeginDocument{{\noindent\small
\emph{Electronic Journal of Differential Equations},
Vol. 2012 (2012), No. 54, pp. 1--10.\newline
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu
\newline ftp ejde.math.txstate.edu}
\thanks{\copyright 2012 Texas State University - San Marcos.}
\vspace{9mm}}

\begin{document}
\title[\hfilneg EJDE-2012/54\hfil Existence of solutions]
{Existence of solutions for multi-point nonlinear differential 
 equations of fractional orders with integral boundary conditions}

\author[G. Wang, W. Liu, C. Ren \hfil EJDE-2012/54\hfilneg]
{Gang Wang, Wenbin Liu, Can Ren} 

\address{Gang Wang \newline
Department of mathematics,  University of Mining and Technology,
 Xuzhou 221008, China}
\email{wangg0824@163.com}

\address{Wenbin Liu \newline
Department of mathematics,  University of Mining and Technology,
 Xuzhou 221008, China}
\email{wblium@163.com}

\address{Can Ren \newline
Department of mathematics,  University of Mining and Technology,
 Xuzhou 221008, China}
\email{rencan0502@163.com}

\thanks{Submitted November 27, 2011. Published April 5, 2012.}
\subjclass[2000]{34B15}
\keywords{Fractional differential equation;  boundary value problem;  
\hfill\break\indent fixed point theorem;
 existence and uniqueness}

\begin{abstract}
 In this article, we study the multi-point boundary-value problem
 of nonlinear fractional differential equation
 \begin{gather*}
 D^\alpha_{0+}u(t)=f(t,u(t)),\quad 1<\alpha\leq 2,\; t\in[0,T],\; T>0,\\
 I_{0+}^{2-\alpha}u(t)|_{t=0}=0,\quad
 D_{0+}^{\alpha-2}u(T)=\sum_{i=1}^ma_i I_{0+}^{\alpha-1}u(\xi_i),
 \end{gather*}
 where $D_{0^+}^\alpha$ and $I_{0^+}^\alpha$ are the standard Riemann-Liouville
 fractional derivative and fractional integral respectively.
 Some existence and uniqueness results are obtained by applying some standard
 fixed point principles. Several examples are given to illustrate the results.
\end{abstract}

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

\section{Introduction}

The study of fractional differential equations ranges from the theoretical aspects 
of existence and uniqueness of solutions to the analytic and numerical methods 
for finding solutions. Fractional differential equations appear naturally in a 
number of fields such as physics, polymer rheology, regular variation in 
thermodynamics, biophysics,blood flow phenomena, aerodynamics, electro-dynamics of
complex medium, viscoelasticity, Bode¡¯s analysis of feedback amplifiers, 
capacitor theory, electrical circuits, electron-analytical chemistry, biology, 
control theory, fitting of experimental data, etc.
 An excellent account in the study of fractional differential equations can be 
found in \cite{k1,l1,s1,s2}. 
Boundary  value problems for fractional differential equations have been 
discussed in \cite{a1,b1,d1,h1,l2,s4,w1,w2}.
 
Integral boundary conditions have various applications in applied fields such as blood
flow problems, chemical engineering, thermo-elasticity, underground water flow, population
dynamics, and so forth. For a detailed description of the integral boundary conditions, we
refer the reader to a recent paper \cite{a6}. 
For more details of nonlocal and integral boundary conditions, 
see \cite{a7,b2,c1} and references therein.

Ahmada and Nieto \cite{a1}  considered the  anti-periodic fractional
 boundary value problem given
\begin{gather*}
 ^cD^qu(t)=f(t,u(t)),\quad 1<\alpha\leq 2,\\
 u(0)=-u(T),\quad ^cD^pu(0)=^cD^pu(T),
\end{gather*}
where $^cD^q$ is the standard Caputo fractional derivative.
Using of some existence and uniqueness results are obtained by 
applying some standard fixed point principles.

Ahmada and  Nieto \cite{a3}  considered the  fractional integro-differential equation 
with integral boundary conditions
\begin{gather*}
^cD^qx(t)=f(t,x(t),(\chi x)(t)),\quad 1<q\leq 2,\;t\in (0,1),\\
\alpha x(0)+\beta x'(0)=\int_0^1q_1(x(s))ds,\quad
\alpha x(1)+\beta x'(1)=\int_0^1q_2(x(s))ds,
\end{gather*}
where $^cD^q$ is the standard Caputo fractional derivative,
$$
(\chi x)(t)=\int_0^t\gamma(t,s)x(s)ds.
$$
Some existence and uniqueness results are obtained by applying 
standard fixed point principles.

In this paper, we investigate the existence and uniqueness of solutions 
for the fractional boundary-value problem
\begin{gather} \label{e1.1}
D^\alpha_{0+}u(t)=f(t,u(t)),\quad 1<\alpha\leq 2,\; t\in[0,T],\;T>0, \\
 \label{e1.2}
I_{0+}^{2-\alpha}u(t)|_{t=0}=0,\quad
D_{0+}^{\alpha-2}u(T)=\sum_{i=1}^ma_i I_{0+}^{\alpha-1}u(\xi_i),
\end{gather}
where $0<\xi_i<T$, $T>0$, $a_i\in \mathbb{R}$, $m\geq 2$, $D_{0^+}^\alpha$ and
$I_{0^+}^\alpha$ are the standard Riemann-Liouville fractional derivative 
and fractional integral respectively, $f:[0,T]\times\mathbb{R}\to\mathbb{R}$
is continuous.

\section{Preliminaries}

For the convenience of the reader, we present here some necessary basic knowledge 
and definitions for fractional calculus theory, that can be found in the 
recent literature.

\begin{definition} \label{def2.1} \rm
The fractional integral of order $\alpha>0$ of a function $y:(0,\infty)\to R$ is given by
$$
I_{0+}^\alpha y(t)=\frac{1}{\Gamma(\alpha)}\int_0^t(t-s)^{\alpha-1}y(s)ds,
$$
provided the right side is pointwise defined on $(0,\infty)$, 
where $\Gamma(\cdot)$ is the Gamma function.
\end{definition}

\begin{definition} \label{def2.2} \rm
The fractional derivative  of order $\alpha>0$ of a function $y:(0,\infty)\to R$ is given by
$$
D_{0+}^\alpha y(t)=\frac{1}{\Gamma(n-\alpha)}(\frac{d}{dt})^n
\int_0^t\frac{y(s)}{(t-s)^{\alpha-n+1}}ds,
$$
where $n=[\alpha]+1$, provided the right side is pointwise defined on $(0,\infty)$.
\end{definition}

\begin{lemma} \label{lem2.1}
Let $\alpha>0$ and $u\in C(0,1)\cap L^1(0,1)$.Then fractional differential equation
$D^\alpha_{0+}u(t)=0$
has
$$
u(t)=c_1t^{\alpha-1}+c_2t^{\alpha-2}+\dots +c_Nt^{\alpha-N},\quad
c_i\in \mathbb{R},\; N=[\alpha]+1,
$$
as unique solution.
\end{lemma}

\begin{lemma} \label{lem2.2}
Assume that $u\in C(0,1)\cap L^1(0,1)$ with a fractional derivative of order $\alpha>0$ 
that belongs to $C(0,1)\cap L^1(0,1)$. Then 
$$
I_{0+}^\alpha D_{0+}^\alpha u(t)=u(t)+c_1t^{\alpha-1}+c_2t^{\alpha-2}+\dots 
+c_Nt^{\alpha-N},
$$
for some $c_i\in \mathbb{R},i=1,2,\dots ,N$, where $N$ is the smallest integer grater
than or equal to $\alpha$.
\end{lemma}

\begin{definition} \label{def2.3} \rm
For $n\in N$, we denote by $AC^n[0,1]$ the space of functions $u(t)$ which have 
continuous derivatives up to order $n-1$ on $[0,1]$ such that $u^{(n-1)}(t)$ 
is absolutely continuous:
$AC^n[0,1]$ =$\{u|[0,1]\to R$ and $(D^{(n-1)})u(t)$ is absolutely continuous in $[0,1]\}$.
\end{definition}

\begin{lemma}[\cite{k1}] \label{lem2.3}
 Let $\alpha>0$, $n=[\alpha]+1$. Assume that $u\in L^1(0,1)$ with a fractional 
integration of order $n-\alpha$ that belongs to $AC^n[0,1]$. Then the equality
$$
(I_{0+}^{\alpha}D_{0+}^{\alpha}u)(t)=u(t)
-\sum_{i=1}^n\frac{((I_{0+}^{n-\alpha}u)(t))^{n-i}|_{t=0}}{\Gamma(\alpha-i+1)}t^{\alpha-i}
$$
holds almost everywhere on $[0, 1]$.
\end{lemma}

\begin{lemma}[\cite{k1}]  \label{lem2.4}
\begin{itemize}
\item[(i)]  Let $k\in N,\alpha > 0$. If $D_{a+}^{\alpha}y(t)$ and
 $(D_{a+}^{\alpha+k}y)(t)$ exist, then
$$
(D^kD_{a+}^{\alpha})y(t)=(D_{a+}^{\alpha+k}y)(t);
$$

\item[(ii)] If $\alpha>0, \beta > 0, \alpha + \beta > 1$, then
$$
(I_{a+}^\alpha I_{a+}^{\alpha})y(t)=(I_{a+}^{\alpha+\beta}y)(t)
$$
satisfies at any point on $[a,b]$ for $y\in L_p(a,b)$ and $1\leq p \leq \infty$;

\item[(iii)] Let $\alpha > 0$ and $y\in C[a,b]$. Then
$(D_{a+}^\alpha I_{a+}^{\alpha})y(t)=y(t)$
holds on $[a,b]$;

\item[(iv)] Note that for $\lambda > -1, \lambda \neq \alpha-1,\alpha-2,\dots ,\alpha-n$,
 we have
\begin{gather*}
D^\alpha t^\lambda=\frac{\Gamma(\lambda+1)}{\Gamma(\lambda-\alpha+1)}t^{\lambda-\alpha},\\
D^\alpha t^{\alpha-i}=0,i=1,2,\dots ,n
\end{gather*}
\end{itemize}
\end{lemma}

\begin{lemma}  \label{lem2.5}
For any $y(t)\in C[0,1]$, the linear fractional boundary-value
 problem
\begin{equation} \label{e2.1}
  \begin{gathered}
   D^\alpha_{0+}u(t)=y(t),\quad 1<\alpha\leq 2,\; t\in[0,T],\\
    I_{0+}^{2-\alpha}u(t)|_{t=0}=0,\quad
 D_{0+}^{\alpha-2}u(T)=\sum_{i=1}^ma_i I_{0+}^{\alpha-1}u(\xi_i),
  \end{gathered}
\end{equation}
 has unique solution 
\begin{equation} \label{e2.2}
\begin{split}
u(t)&= \int_0^t\frac{(t-s)^{\alpha-1}}{\Gamma(\alpha)}y(s)ds\\
&\quad+\frac{t^{\alpha-1}}{\Gamma(\alpha)(T-A)}
\Big[\frac{\sum_{i=1}^ma_i}{\Gamma(2\alpha-1)}\int_0^{\xi_i}(\xi_i-s)^{2\alpha-2}y(s)ds
-\int_0^T(T-s)y(s)ds\Big],
\end{split}
\end{equation}
where $A=\sum_{i=1}^ma_i\xi_i^{2\alpha-2}/\Gamma(2\alpha-1)$ and $ T\neq A $.
\end{lemma}

\begin{proof}
By Lemma \ref{lem2.2}. the solution of \eqref{e2.1} can be written as
$$
u(t)=c_1t^{\alpha-1}+c_2t^{\alpha-2}
+\frac{1}{\Gamma(\alpha)}\int^t_0(t-s)^{\alpha-1}y(s)ds.
$$
From $I_{0+}^{2-\alpha}u(t)|_{t=0}=0$, and by Lemmas \ref{lem2.3} and \ref{lem2.4},
we know that $c_2=0$, and
\begin{gather*}
D_{0+}^{\alpha-2}u(t)=c_1t\Gamma(\alpha)+I_{0+}^2y(t),\\
I_{0+}^{\alpha-1}u(t)=c_1\frac{\Gamma(\alpha)}{\Gamma(2\alpha-1)}t^{2\alpha-2}
+I_{0+}^{\alpha-1}I_{0+}^\alpha y(t),
\end{gather*}
from $D_{0+}^{\alpha-2}u(T)=\sum_{i=1}^ma_i I_{0+}^{\alpha-1}u(\xi_i)$, we have
$$
c_1=\frac{1}{\Gamma(\alpha)(T-A)}
\Big[\frac{\sum_{i=1}^ma_i}{\Gamma(2\alpha-1)}\int_0^{\xi_i}(\xi_i-s)
^{2\alpha-2}y(s)ds -\int_0^T(T-s)y(s)ds\Big],
$$
where $A=\sum_{i=1}^ma_i\xi_i^{2\alpha-2}/\Gamma(2\alpha-1)$ and $ T\neq A$, so
\begin{align*}
u(t)
&= \int_0^t\frac{(t-s)^{\alpha-1}}{\Gamma(\alpha)}y(s)ds\\
&\quad \frac{t^{\alpha-1}}{\Gamma(\alpha)(T-A)}
\Big[\frac{\sum_{i=1}^ma_i}{\Gamma(2\alpha-1)}
 \int_0^{\xi_i}(\xi_i-s)^{2\alpha-2}y(s)ds
-\int_0^T(T-s)y(s)ds\Big].
\end{align*}
The proof is complete.
\end{proof}


\section{Existence and uniqueness of solutions}

Let $E =C([0,T],R)$ denote the Banach space of all continuous functions 
from $[0,T]\to R$ endowed with the norm
defined by $\|x\|=sup\{|x(t)|,t\in[0,T]\}$.
Now we state some known fixed point theorems which are needed to prove the
 existence of solutions for \eqref{e1.1}--\eqref{e1.2}.

\begin{theorem}[\cite{s3}] \label{thm3.1}
Let $X$ be a Banach space. Assume that $T:X\to X$ is a completely continuous 
operator and the set $V=\{u\in X |u=\mu Tu,0<\mu< 1\}$ is bounded. 
Then $T$ has a fixed point in $X$.
\end{theorem}

\begin{theorem}\cite{s3} \label{thm3.2}
Let $X$ be a Banach space. Assume that $\Omega$ is an open bounded subset of $X$ with $\theta \in \Omega $and let $T:\bar{\Omega}\to X$
be a completely continuous operator such that
$$\|Tu\|\leq \|u\|, \forall u\in \partial \Omega.$$
Then $T$ has a fixed point in $ \bar{\Omega}$.
\end{theorem}



We define, in relation to \eqref{e2.2}, an operator $P:E\to E$, as
\begin{equation} \label{e3.1}
\begin{split}
(Pu)(t)
&= \int_0^t\frac{(t-s)^{\alpha-1}}{\Gamma(\alpha)}f(t,u(s))ds\\
&\quad +\frac{t^{\alpha-1}}{\Gamma(\alpha)(T-A)}
\Big(\frac{\sum_{i=1}^ma_i }{\Gamma(2\alpha-1)}
 \int_{0}^{\xi_i}(\xi_i-s)^{2\alpha-2}f(t,u(s))ds\\
&\quad -\int_{0}^{T}(T-s)f(t,u(s))ds\Big).
\end{split}
\end{equation}
Observe that  this equation has a solution if and only if the operator $P$
 has a fixed point.

\begin{theorem} \label{thm3.3}
Assume that there exists a positive constant $L_1$ such that 
$|f(t,u)|\leq L_1$ for $t\in[0,T],u\in E$. Then  \eqref{e1.1}-\eqref{e1.2}
 has at least one solution.
\end{theorem}


\begin{proof}
We show, as a first step, that the operator $P$ is completely continuous. 
Clearly, continuity of the operator $P$ follows from the continuity of $f$.
 Let $\Omega\subset E$ be bounded. Then, $\forall u\in \Omega $ together 
with the assumption $|f(t,u)|\leq L_1$, we obtain
\begin{align*}
(Pu)(t)
&\leq \int_0^t\frac{(t-s)^{\alpha-1}}{\Gamma(\alpha)}|f(t,u(s))|ds\\
&+\frac{t^{\alpha-1}}{\Gamma(\alpha)|T-A|}
\Big(\frac{\sum_{i=1}^ma_i }{\Gamma(2\alpha-1)}
 \int_{0}^{\xi_i}(\xi_i-s)^{2\alpha-2}|f(t,u(s))|ds\\
&\quad -\int_{0}^{T}(T-s)|f(t,u(s))|ds\Big)\\
&\leq  L_1\Big[\int_0^t\frac{(t-s)^{\alpha-1}}{\Gamma(\alpha)}ds\\
&\quad +\frac{t^{\alpha-1}}{\Gamma(\alpha)|T-A|}
\Big(\frac{\sum_{i=1}^ma_i }{\Gamma(2\alpha-1)}
 \int_{0}^{\xi_i}(\xi_i-s)^{2\alpha-2}ds
-\int_{0}^{T}(T-s)ds\Big)\Big]\\
&\leq  L_1\Big[\frac{T^{\alpha}}{\Gamma(\alpha+1)}
 +\frac{T^{\alpha-1}}{\Gamma(\alpha)|T-A|}
\Big(\frac{\sum_{i=1}^ma_i \xi^{2\alpha-1}}{\Gamma(2\alpha)}
 -\frac{T^2}{2}\Big)\Big],
\end{align*}
which implies 
$$
\|Pu\|\leq L_1\Big[\frac{T^{\alpha}}{\Gamma(\alpha+1)}
+\frac{T^{\alpha-1}}{\Gamma(\alpha)|T-A|}
\Big(\frac{\sum_{i=1}^ma_i \xi^{2\alpha-1}}{\Gamma(2\alpha)}-\frac{T^2}{2}
\Big)\Big]<\infty.
$$
Hence, $T(\Omega)$ is uniformly bounded.

For any $t_1, t_2\in[0, T], u\in \Omega$, we have
\begin{align*}
&|(Pu)(t_1)-(Pu)(t_2)|\\
&=\Big|\int_0^{t_1}\frac{(t_1-s)^{\alpha-1}}{\Gamma(\alpha)}f(s,u(s))ds\\
&\quad +\frac{t_1^{\alpha-1}}{\Gamma(\alpha)(T-A)}
 \Big(\frac{\sum_{i=1}^ma_i }{\Gamma(2\alpha-1)}
 \int_{0}^{\xi_i}(\xi_i-s)^{2\alpha-2}f(s,u(s))ds\\
&\quad -\int_{0}^{T}(T-s)^f(s,u(s))ds\Big)
 -\int_0^{t_2}\frac{(t_2-s)^{\alpha-1}}{\Gamma(\alpha)}f(s,u(s))ds
 -\frac{t_2^{\alpha-1}}{\Gamma(\alpha)(T-A)}\\
&\quad\times\Big(\frac{\sum_{i=1}^ma_i }{\Gamma(2\alpha-1)}
\int_{0}^{\xi_i}(\xi_i-s)^{2\alpha-2}f(s,u(s))ds
-\int_{0}^{T}(T-s)f(s,u(s))ds\Big)\Big|
\\
&\leq L_1\Big|\int_0^{t_1}\frac{(t_1-s)^{\alpha-1}
 -(t_2-s)^{\alpha-1}}{\Gamma(\alpha)}ds\\
&\quad  +\frac{t_1^{\alpha-1}-t_2^{\alpha-1}}{\Gamma(\alpha)(T-A)}
 \Big(\frac{\sum_{i=1}^ma_i }{\Gamma(2\alpha-1)}\int_{0}^{\xi_i}
 (\xi_i-s)^{2\alpha-2}ds \\
&\quad -\int_{0}^{T}(T-s)ds\Big)-
\int_{t_1}^{t_2}\frac{(t_2-s)^{\alpha-1}}{\Gamma(\alpha)}ds\Big|
\\
&\leq L_1\Big[\Big|\int_0^{t_1}\frac{(t_1-s)^{\alpha-1}
 -(t_2-s)^{\alpha-1}}{\Gamma(\alpha)}ds-
\int_{t_1}^{t_2}\frac{(t_2-s)^{\alpha-1}}{\Gamma(\alpha)}ds\Big|\\
&\quad +\Big|\frac{t_1^{\alpha-1}-t_2^{\alpha-1}}{\Gamma(\alpha)(T-A)}
\Big(\frac{\sum_{i=1}^ma_i }{\Gamma(2\alpha-1)}
 \int_{0}^{\xi_i}(\xi_i-s)^{2\alpha-2}ds-
\int_{0}^{T}(T-s)ds\Big)\Big|\Big]\\
& \to 0 \quad\text{as }  t_1\to t_2.
\end{align*}
Thus, by the Arzela-Ascoli theorem, $P(\Omega)$ is equicontinuous. 
Consequently, the operator $P$ is compact.

Next, we consider the set
$V=\{u\in E:u=\mu Pu,0<\mu<1\}$,
and show that it is bounded. 
Let $u\in V;$ then $u=\mu Pu,0 <\mu< 1$. For any $t\in [0,T]$, we have
\begin{align*}
u(t)&=  \int_0^t\frac{(t-s)^{\alpha-1}}{\Gamma(\alpha)}f(t,u(s))ds\\
&\quad +\frac{t^{\alpha-1}}{\Gamma(\alpha)(T-A)}
\Big(\frac{\sum_{i=1}^ma_i }{\Gamma(2\alpha-1)}
 \int_{0}^{\xi_i}(\xi_i-s)^{2\alpha-2}f(t,u(s))ds\\
&\quad -\int_{0}^{T}(T-s)f(t,u(s))ds\Big),
\end{align*}
and
\begin{align*}
|u(t)|
&=\mu|Pu|\\
&\leq\int_0^t\frac{(t-s)^{\alpha-1}}{\Gamma(\alpha)}|f(t,u(s))|ds\\
&\quad +\frac{t^{\alpha-1}}{\Gamma(\alpha)(T-A)}
 \Big(\frac{\sum_{i=1}^ma_i }{\Gamma(2\alpha-1)}
 \int_{0}^{\xi_i}(\xi_i-s)^{2\alpha-2}|f(t,u(s))|ds\\
&\quad -\int_{0}^{T}(T-s)|f(t,u(s))|ds\Big)
\\
&\leq L_1\Big[\int_0^t\frac{(t-s)^{\alpha-1}}{\Gamma(\alpha)}ds\\
&\quad +\frac{t^{\alpha-1}}{\Gamma(\alpha)|T-A|}
\Big(\frac{\sum_{i=1}^ma_i }{\Gamma(2\alpha-1)}
 \int_{0}^{\xi_i}(\xi_i-s)^{2\alpha-2}ds
 -\int_{0}^{T}(T-s)ds\Big)\Big]\\
&\leq \max_{t\in[0,T]}\Big\{L_1\Big[\frac{|t^{\alpha}|}{\Gamma(\alpha+1)}
+\frac{|t^{\alpha-1}|}{\Gamma(\alpha)|T-A|}
\Big(\frac{\sum_{i=1}^ma_i \xi^{2\alpha-1}}{\Gamma(2\alpha)}-\frac{T^2}{2}\Big)\Big]
\Big\}
=M.
\end{align*}
Thus, $\|u\|\leq M$. So, the set $V$ is bounded. Thus, by the conclusion of
Theorem \ref{thm3.1}, the operator $P$ has at
least one fixed point, which implies that \eqref{e1.1}-\eqref{e1.2}
 has at least one solution.
\end{proof}

\begin{theorem} \label{thm3.4}
Let $lim_{x\to 0}\frac{f(t,x)}{x}=0$. Then  \eqref{e1.1}-\eqref{e1.2} has at least 
one solution.
\end{theorem}

\begin{proof}
Since $\lim_{x\to 0}\frac{f(t,x)}{x}=0$, there exists a constant $r>0$ such 
that $|f(t,x)|\leq \varepsilon|x|$ for $0<|x|<r$, where
$\varepsilon>0$ is such that
\begin{equation} \label{e3.2}
\max_{t\in[0,T]}\Big\{\frac{|t^{\alpha}|}{\Gamma(\alpha+1)}
+\frac{|t^{\alpha-1}|}{\Gamma(\alpha)|T-A|}
\Big(\frac{\sum_{i=1}^ma_i \xi^{2\alpha-1}}{\Gamma(2\alpha)}-\frac{T^2}{2}\Big)\Big\}
\varepsilon\leq 1,
\end{equation}
Define $\Omega_1=\{x\in E:\|x\|<r\}$ and take $x\in E$ such that $\|x\|=r$;
 that is, $x\in \Omega_1$. As before, it can be shown that $T$ is
completely continuous and
$$
|(Tx)(t)|\leq max_{t\in[0,T]}\Big\{\frac{|t^{\alpha}|}{\Gamma(\alpha+1)}
+\frac{|t^{\alpha-1}|}{\Gamma(\alpha)|T-A|}
\Big(\frac{\sum_{i=1}^ma_i \xi^{2\alpha-1}}{\Gamma(2\alpha)}-\frac{T^2}{2}\Big)\Big\}
\varepsilon\|x\| ,
$$
which, in view of \eqref{e3.2}, yields $\|Tx\|\leq\|x\|, x\in \partial\Omega_1$. 
Therefore, by Theorem \ref{thm3.2}, the operator $T$ has at least one fixed point,
which in turn implies that  \eqref{e1.1}-\eqref{e1.2} has at least one solution.
\end{proof}

For the next theorem we use the following two assumptions:
\begin{itemize}
\item[(H1)]
there exist positive functions $L$ , such that
$$
|f(t,x)-f(t,y)|\leq L|x-y|,\quad\forall  t\in [0,T],x,y\in \mathbb{R},
$$
\item[(H2)] The function $L$ satisfies
$$
2L\leq \big[\frac{T^\alpha}{\Gamma(\alpha+1)}
+\frac{T^{\alpha-1}}{\Gamma(\alpha)|T-A|}
\Big(\frac{\sum_{i=1}^ma_i\xi_i^{2\alpha-1}}{\Gamma(2\alpha)}
-\frac{T^2}{2}\Big)\Big]^{-1}\,.
$$
\end{itemize}

\begin{theorem} \label{thm3.5}
Assume thatUnder assumptions {\rm (H1), (H2)}, 
Problem \eqref{e1.1}--\eqref{e1.2}) has a unique solution in $C[0,T]$.
\end{theorem}

\begin{proof}
Let us set $\sup_{t\in [0,T]}|f(t,0)| = M_1$, and choose
$$
r\geq 2M_1\Big[\frac{T^\alpha}{\Gamma(\alpha+1)}
+\frac{T^{\alpha-1}}{\Gamma(\alpha)|T-A|}
\Big(\frac{\sum_{i=1}^ma_i\xi_i^{2\alpha-1}}{\Gamma(2\alpha)}
-\frac{T^2}{2}\Big)\Big]
$$
Then we show that $PBr\subset Br$, where $Br =\{u \in E :\|u\|\leq r\}$.
For $u\in Br$, we have
\begin{align*}
&\|(Pu)(t)\|\\
&= \sup_{t\in [0,T]}\Big|\int_0^t\frac{(t-s)^{\alpha-1}}{\Gamma(\alpha)}f(s,u(s))ds\\
&\quad +\frac{t^{\alpha-1}}{\Gamma(\alpha)(T-A)}
\Big(\frac{\sum_{i=1}^ma_i }{\Gamma(2\alpha-1)}
 \int_{0}^{\xi_i}(\xi_i-s)^{2\alpha-2}f(s,u(s))ds\\
&\quad -\int_{0}^{T}(T-s)f(s,u(s))ds\Big)\Big|
\\
&\leq  \sup_{t\in [0,T]}\Big[ \int_0^t\frac{(t-s)^{\alpha-1}}{\Gamma(\alpha)}|f(s,u(s)|ds\\
&\quad +\frac{t^{\alpha-1}}{\Gamma(\alpha)(T-A)}
\Big(\frac{\sum_{i=1}^ma_i }{\Gamma(2\alpha-1)}
 \int_{0}^{\xi_i}(\xi_i-s)^{2\alpha-2}|f(s,u(s)|ds\\
&\quad -\int_{0}^{T}(T-s)|f(s,u(s))|ds\Big)\Big]
\\
&\leq  \sup_{t\in [0,T]}\Big[\int_0^t\frac{(t-s)^{\alpha-1}}{\Gamma(\alpha)}
(|f(s,u(s)-f(s,0)|+|f(s,0)|)ds\\
&\quad +\frac{t^{\alpha-1}}{\Gamma(\alpha)|T-A|}
 \Big(\frac{\sum_{i=1}^ma_i }{\Gamma(2\alpha-1)}
 \int_{0}^{\xi_i}(\xi_i-s)^{2\alpha-2}
 (|f(s,u(s)-f(s,0)|+|f(s,0)|)ds\\
&\quad -\int_{0}^{T}(T-s)(|f(s,u(s)-f(s,0)|
 +|f(s,0)|)ds\Big)\Big]
\\
&\leq  \sup_{t\in [0,T]}\Big[(Lr+M_1)\Big(\int_0^t\frac{(t-s)^{\alpha-1}}{\Gamma(\alpha)}ds\\
&\quad +\frac{t^{\alpha-1}}{\Gamma(\alpha)|T-A|}\Big(\frac{\sum_{i=1}^ma_i }{\Gamma(2\alpha-1)}
\int_{0}^{\xi_i}(\xi_i-s)^{2\alpha-2}ds
-\int_{0}^{T}(T-s)ds\Big)\Big)\Big]
\\
&\leq  (Lr+M_1)\Big[\frac{T^\alpha}{\Gamma(\alpha+1)}
+\frac{T^{\alpha-1}}{\Gamma(\alpha)|T-A|}
 \Big(\frac{\sum_{i=1}^ma_i\xi_i^{2\alpha-1}}{\Gamma(2\alpha)}
-\frac{T^2}{2}\Big)\Big]\leq r
\end{align*}
Taking the maximum over the interval $[0,T]$, we obtain $\|(Pu)(t)\|\leq r$.

In view of (H1), for every $t\in [0,T]$, we have
\begin{align*}
&\|(Px)(t)-(Py)(t)\|\\
&= \sup_{t\in [0,T]}\Big|\int_0^t\frac{(t-s)^{\alpha-1}}{\Gamma(\alpha)}
 (f(t,x)-f(t,y)ds\\
&\quad +\frac{t^{\alpha-1}}{\Gamma(\alpha)(T-A)}
\Big(\frac{\sum_{i=1}^ma_i }{\Gamma(2\alpha-1)}
 \int_{0}^{\xi_i}(\xi_i-s)^{2\alpha-2} (f(t,x)-f(t,y)ds\\
&\quad -\int_{0}^{T}(T-s)(f(t,x)-f(t,y)ds\Big)\Big|
\\
&\leq  \sup_{t\in [0,T]}\Big[\int_0^t\frac{(t-s)^{\alpha-1}}{\Gamma(\alpha)}|
 (f(t,x)-f(t,y)|ds\\
&\quad +\frac{t^{\alpha-1}}{\Gamma(\alpha)|T-A|}
\Big(\frac{\sum_{i=1}^ma_i }{\Gamma(2\alpha-1)}\int_{0}^{\xi_i}(\xi_i-s)^{2\alpha-2}
 |(f(t,x)-f(t,y)|ds \\
&\quad -\int_{0}^{T}(T-s)|(f(t,x)-f(t,y)|ds\Big)\Big]
\\
&\leq  \sup_{t\in [0,T]}\Big[L\|x-y\|\Big(\int_0^t\frac{(t-s)
 ^{\alpha-1}}{\Gamma(\alpha)}ds 
\\
&\quad +\frac{t^{\alpha-1}}{\Gamma(\alpha)|T-A|}
\Big(\frac{\sum_{i=1}^ma_i }{\Gamma(2\alpha-1)}
 \int_{0}^{\xi_i}(\xi_i-s)^{2\alpha-2}ds
 -\int_{0}^{T}(T-s)ds\Big)\Big)\Big]
\\
&\leq  L\|x-y\|\Big[\frac{T^\alpha}{\Gamma(\alpha+1)}
+\frac{T^{\alpha-1}}{\Gamma(\alpha)|T-A|}
\Big(\frac{\sum_{i=1}^ma_i\xi_i^{2\alpha-1}}{\Gamma(2\alpha)}
-\frac{T^2}{2}\Big)\Big]
= A\|x-y\|,
\end{align*}
where
$$
A=L\Big[\frac{T^\alpha}{\Gamma(\alpha+1)}
+\frac{T^{\alpha-1}}{\Gamma(\alpha)|T-A|}
\Big(\frac{\sum_{i=1}^ma_i\xi_i^{2\alpha-1}}{\Gamma(2\alpha)}
-\frac{T^2}{2}\Big)\Big],
$$
which depends only on the parameters involved in the problem. 
As $ A<1$, $T$ is therefore a contraction. Thus, 
the conclusion of the theorem follows by the contraction mapping principle 
(the Banach fixed point theorem).
\end{proof}

\begin{example} \label{examp3.1} \rm
Consider the following three-point nonlinear differential equations 
\begin{gather} \label{e3.3}
D^{3/2}_{0+}u(t)=f(t,u(t)),\quad 0<t<1,\\
 \label{e3.4}
I_{0+}^{2-\alpha}u(t)|_{t=0}=0,\quad
D_{0+}^{\alpha-2}u(T)=\sum_{i=1}^ma_i I_{0+}^{\alpha-1}u(\xi_i),
\end{gather}
where $f(t,u)=e^{-2sin^2(u(t))}[3+5\sin (2t)+4ln(5+2\cos^2 (u(t)))]/(2+\cos t)$,
$a_1=4$, $a_2=2$, $\xi_1=1/2$, $\xi_2=1/4$, $T=1$ we have
$A=\sum_{i=1}^ma_i\xi_i^{2\alpha-2}/\Gamma(2\alpha-1)=5/2\neq T=1$.

Clearly $L_1=4+2ln7$, and the hypothesis of Theorem \ref{thm3.3} holds. 
Therefore, the conclusion of Theorem 3.3 applies to \eqref{e3.3}--\eqref{e3.4}.
 Then, there exists at least one solution.
\end{example}

\begin{example} \label{examp3.2} \rm
Consider the problem
\begin{gather} \label{e3.5}
D^{3/2}_{0+}u(t)=f(t,u(t)),\quad 0<t<1, \\
\label{e3.6}
I_{0+}^{2-\alpha}u(t)|_{t=0}=0,\quad
D_{0+}^{\alpha-2}u(T) =\sum_{i=1}^ma_i I_{0+}^{\alpha-1}u(\xi_i),
\end{gather}
where $f(t,u)=(8+2u^3(t))^{1/3}+(2t-1)(2u-2\sin (u(t)))-2$,
$a_1=1/2$, $a_2=1/3$, $\xi_1=1/3$, $\xi_2=1/4$, $T=2$ we have
$A=\sum_{i=1}^ma_i\xi_i^{2\alpha-2}/\Gamma(2\alpha-1)=1/4\neq T=2$.
Clearly $\lim_{u\to0}\frac{f(t,u)}{u}=0$. It can easily be verified that all 
the assumptions of Theorem \ref{thm3.4} hold. Consequently,  \eqref{e3.5}-\eqref{e3.6}
 has at least one solution.
\end{example}

\begin{example} \label{examp3.3} \rm
Consider the  three-point nonlinear differential equation
\begin{gather} \label{e3.7}
D^{3/2}_{0+}u(t)+f(t,u(t))=0,\quad 0<t<1, \\
 \label{e3.8}
I_{0+}^{2-\alpha}u(t)|_{t=0}=0,\quad D_{0+}^{\alpha-2}u(T)
=\sum_{i=1}^ma_i I_{0+}^{\alpha-1}u(\xi_i),
\end{gather}
where $f(t,u)=\frac{1}{(2t+8)^2}\frac{8\|u\|}{1+\|u\|}, a_1=2,a_2=3,\xi_1=1/2,\xi_2=1/3,T=2$ we have
$A=\sum_{i=1}^ma_i\xi_i^{2\alpha-2}/\Gamma(2\alpha-1)=1\neq T=2$.
Clearly, $L=1/8$ as
$$
|f(t,u)-f(t,v)|\leq 1/8\|u-v\|.
$$
Further,
$$
L\Big[\frac{T^\alpha}{\Gamma(\alpha+1)}
+\frac{T^{\alpha-1}}{\Gamma(\alpha)|T-A|}
\Big(\frac{\sum_{i=1}^ma_i\xi_i^{2\alpha-1}}{\Gamma(2\alpha)}
-\frac{T^2}{2}\Big)\Big]\approx 0.3<1.
$$
Thus, all the assumptions of Theorem \ref{thm3.5} are satisfied. Hence,
 \eqref{e3.7}-\eqref{e3.8} has a unique solution on $[0, 1]$.
\end{example}

\subsection*{Acknowledgements}
The authors want to thank the anonymous referee for his or her valuable comments
 and suggestions.
This study was supported by grants 10771212 from 
the NNSF of China, and 2010LKSX09 from 
the Fundamental Research Funds for the Central Universities.


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