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

\AtBeginDocument{{\noindent\small
\emph{Electronic Journal of Differential Equations},
Vol. 2012 (2012), No. 100, pp. 1--13.\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-2012/100\hfil Method of upper and lower solutions]
{Method of upper and lower solutions for fractional differential
 equations}

\author[L. Lin, X. Liu,  H. Fang\hfil EJDE-2012/100\hfilneg]
{Legang Lin, Xiping Liu, Haiqin Fang}  % in alphabetical order

\address{Legang Lin \newline
College of Science, University of Shanghai for Science and Technology \\
Shanghai 200093, China}
\email{llg870830@163.com}

\address{Xiping Liu \newline
College of Science, University of Shanghai for Science and Technology \\
Shanghai 200093, China}
\email{xipingliu@163.com}

\address{Haiqin Fang \newline
College of Science, University of Shanghai for Science and Technology \\
Shanghai 200093, China}
\email{244307094@qq.com}

\thanks{Submitted May 4, 2012. Published June 12, 2012.}
\thanks{Supported by grants 10ZZ93 from the Foundation of the Innovation
Program of \hfill\break\indent
Shanghai Municipal Education Commission, and 11171220
from the National Natural \hfill\break\indent
Science Foundation of  China}
\subjclass[2000]{34B15, 26A33}
\keywords{Fractional differential equations; boundary-value problems;
\hfill\break\indent upper and lower solutions; monotone iterative algorithm}

\begin{abstract}
 In this paper, we show the existence and uniqueness of solutions
 for the  boundary-value problems of fractional differential equations,
 using the upper and lower solutions method and monotone iterative algorithm. 
 An example is  also included to illustrate our results.
\end{abstract}

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

\section{Introduction}

 In this article, we show the existence and uniqueness of
solutions for the boundary value problems of the fractional
differential equation
\begin{equation}\label{eq1.1}
   \begin{gathered}
      D^\delta u(t)-Mu(t)=f(t,u(t)),\quad t\in J,\; 0<\delta<1,\\
      u(0)=ru(T),
   \end{gathered}
\end{equation}
where $J=[0,T]$, $0<T<+\infty$, 
$f\in C(J\times \mathbb{R},\mathbb{R})$, $M\geq0$,
 $0<r<\frac {1}{E_{\delta,1}(MT^\delta)}$. 
$E_{n_1,n_2}(z)=\sum^{\infty}_{j=0}\frac {z^j}{\Gamma(jn_1+n_2)}$,
 $n_1, n_2>0$
 is the Mittag-Leffler function 
 (see \cite{kdtafde,ahj}). $D^\delta$ is the Caputo fractional derivative of
 order $\delta$
 (see \cite{kdtafde,ahj}); that is,
\begin{equation*}
    D^\delta u(t)=\frac {1}{\Gamma(1-\delta)}\int_0^{t}
    (t-s)^{-\delta}u'(s)\mathrm{d}s.
\end{equation*}
The Riemann-Liouville fractional integral operator of order
 $\delta$ is defined by
$$
  I^{\delta}u(t)=\frac{1}{\Gamma(\delta)}\int_0^{t}
  (t-s)^{\delta-1}u(s)\mathrm{d}s.
$$

Fractional differential equations are thought of an important
research branch of fractional calculus, to which much attention has
been paid. They arise in the models of many phenomena in various
fields of science and engineering as a valuable tool. Indeed, we can
find numerous applications in physics, chemistry, biology, etc. (See
\cite{kdf,rh}). Hence, some meaningful results on this kind of
problems have been obtained, (see \cite{vlsjv,aksm,szhang,rwimd,
sljllac,L-J8,Jia,LXP,szrhmeama}).

The theory of upper and lower solutions is known to be an effective
method to deal with the boundary-value problems of the ordinary
differential equations and functional differential equations, (see
\cite{coster,lfjmlxp,lfjsjm,Yang,jnr,djnwz,tjankowski,nyazidi,wys}).
Li et al. \cite{lfjmlxp,lfjsjm} found a direct method which is quite
simple and practical to prove the existence and uniqueness of
solutions of the second-order three-point boundary-value problem and
gave some examples to illustrate the effectiveness of the result. In
\cite{jnr}, using the upper and lower solutions method, the authors
considered the periodic boundary-value problems for functional
differential equations. In \cite{wys}, the authors presented the
existence of extreme solutions of the boundary-value problem for a
class of first-order functional equations with a nonlinear boundary
condition
\begin{gather*}
      u'(t)=f(t,u(t),u(\theta(t))),\\
      g(u(0))=ru(T),
\end{gather*}
by the method of upper and lower solutions and monotone iterative
techniques. One interesting thing is that the method is also
appropriate for fractional differential equations, (see
\cite{szxs,barrettjh}). In \cite{barrettjh}, Barrett proved the
existence and uniqueness of solutions for the  following initial
value problems
\begin{gather*}
      (D^{\delta}_{a+}u)(t)-Mu(t)=f(t),\quad (n-1<\delta<n),\\
      (D^{\delta-k}_{a+}u)(a+)=b_k,\quad b_{k}\in C(k=1,\dots,n).
\end{gather*}
Zhang and Su \cite{szxs} used the method of upper and lower
solutions to study the existence for a linear fractional
differential equation with nonlinear boundary condition
\begin{equation}\label{eq1.2}
   \begin{gathered}
      D^\delta u(t)-du(t)=h(t),\quad t\in J,\; 0<\delta<1,\\
      g(u(0))=u(T),
   \end{gathered}
\end{equation}
where $d\geq0$, $h\in C^1[0,T ]$ is a given function, $D^{\delta}$
is a regularized fractional derivative (the Caputo derivative) of
order $0<\delta<1$, and there exists a constant 
$\frac{dT^{\delta}}{\Gamma(\delta)-dT^{\delta}}< r_0 < 1$ such that
 $r_0<g'(s)< 1$ for $s\in \mathbb{R}$. 
They presented an existence theorem for the
boundary-value problem \eqref{eq1.2}.

The boundary-value problem \eqref{eq1.1} which we study is different
from the problem \eqref{eq1.2}. We study a nonlinear fractional
differential equation, and $0<r<\frac {1}{E_{\delta,1}(MT^\delta)}$.
Since $E_{\delta,1}(MT^\delta)=\sum^{\infty}_{j=0}\frac
{MT^j}{\Gamma(j\delta+1)}\geq \frac{1}{\Gamma(1)}=1$, which implies
$r\leq 1$. In fact, in the boundary-value problem \eqref{eq1.1},
$g(s)=\frac{1}{r}s$, and $g'(s)=\frac{1}{r}\geq 1$, the condition
$0<r_0<g'(s)< 1$ in \eqref{eq1.2} is not satisfied. The purpose of
this paper is to prove the existence and uniqueness of solutions to
the boundary-value problem \eqref{eq1.1} by the method of upper and
lower solutions and monotone iterative techniques. We not only
present the existence and the uniqueness theorem, but also present
the iterative sequence for solving the solution and its error
estimate formula under the condition of unique solution.

This article is organized as follows. In section 2, we introduce the
basic properties, and some comparison principles are studied. In
section 3, we consider the existence and uniqueness of the solution
of a linear problem associated with \eqref{eq1.1}. In section 4, we
obtain the extreme solution of \eqref{eq1.1} and prove that there
exists a unique solution of \eqref{eq1.1} by using the method of
upper and lower solutions and monotone iterative technique. In
section 5, we give an example to illustrate the results which have
been proved.


\section{Preliminaries and comparison principle}

 In this article, we use the following conditions:
\begin{itemize}
\item[(H0)] $0\leq\frac {MT^\delta}{\Gamma(\delta+1)}<1$,
$c(t)=1-\cos{\frac {\pi t}{2T}}$, $t\in J$.
\end{itemize}
The followings fundamental properties for fractional
differential equations, which are necessary for our analysis.

\begin{lemma}[{\cite[Example 4.9]{ahj}}]  \label{lem2.1}
The linear initial value problem
\begin{equation}\label{eq2.0}
   \begin{gathered}
      D^\delta u(t)-Mu(t)=x(t),\quad t\in J,\\
      u(0)=u_0,
   \end{gathered}
\end{equation}
where $M$ is a constant, has a unique solution 
$$
  u(t)=u_0E_{\delta,1}(Mt^\delta)
     +\int_0^{t}(t-s)^{\delta-1}E_{\delta,\delta}(M(t-s)^\delta)x(s)\mathrm{d}s.
$$
\end{lemma}

In particular, when $M=0$, the initial problem \eqref{eq2.0}
 has the solution
$$
  u(t)=u_0+\frac {1}{\Gamma(\delta)}\int_0^{t}(t-s)^{\delta-1}x(s)\mathrm{d}s.
$$

\begin{lemma}[{\cite[Lemma 2.22]{ahj}}]  \label{lem2.1.1}
 If $u\in C(J)$ and $0<\alpha<1$, then $I^{\alpha}D^{\alpha} u(t)=u(t)-u(0)$.
\end{lemma}

To investigate the boundary-value problem \eqref{eq1.1}, for
 $x\in C(J)$, we consider the boundary-value problem
\begin{equation}\label{eq2.1}
   \begin{gathered}
      D^\delta u(t)-Mu(t)=x(t),\quad t\in J,\\
      u(0)=ru(T).
   \end{gathered}
\end{equation}


\begin{definition} \label{def2.1} \rm
Let $\alpha \in C^{1}(J)$. We say that $\alpha $ is a
lower solution of the boundary-value problem \eqref{eq2.1}, if
$$
  D^{\delta}\alpha(t)-M\alpha(t)\leq x(t)-a_{\alpha}(t),\ t\in J,
$$
where
\begin{equation}\label{eq2.3}
   a_{\alpha}(t)=
      \begin{cases}
         0, & r\alpha(T)\geq \alpha(0),\\
         \frac {1}{r}(D^{\delta}c(t)-Mc(t))(\alpha(0)-r\alpha(T)),
         & r\alpha(T)<\alpha(0).
      \end{cases}
\end{equation}
Let $\beta\in C^{1}(J)$. We say that $\beta$ is an upper solution of
the boundary-value problem \eqref{eq2.1}, if
$$
  D^{\delta}\beta(t)-M\beta(t)\geq x(t)+b_{\beta}(t),\ t\in J,
$$
where
\begin{equation}\label{eq2.4}
   b_{\beta}(t)=
      \begin{cases}
         0, & r\beta(T)\leq \beta(0),\\
         \frac {1}{r}(D^{\delta}c(t)-Mc(t))(r\beta(T)-\beta(0)), &
         r\beta(T)>\beta(0),
       \end{cases}
\end{equation}
where $c(t)$ is defined in (H0).
\end{definition}

The following comparison principle will play a very important role
in our main results.

\begin{lemma} \label{lem2.2}
Let {\rm (H0)} hold. Assume that $u\in C^{1}(J)$ and satisfies
\begin{equation}\label{eq2.5}
   \begin{gathered}
      D^\delta u(t)-Mu(t)\leq 0, \\
      u(0)\leq0.
   \end{gathered}
\end{equation}
Then $u(t)\leq0$ for $t\in J$.
\end{lemma}

 \begin{proof} Suppose this is false, then
there exist $t_1, t_2\in J$ such that $u(t)\leq 0$ for 
$0\leq t\leq t_1$ and $u(t)>0$ for $t_1<t\leq t_2$. Let
\begin{equation}\label{eq2.6}
    u(t_0)=\max_{t\in [t_1,t_2]}u(t)>0.
\end{equation}
By \eqref{eq2.5} and Lemma \ref{lem2.1.1}, we can obtain 
\begin{gather*}
  I^{\delta}D^{\delta}u(t)-I^{\delta}Mu(t)\leq0, \\
  u(t)-u(0)-I^{\delta}Mu(t)\leq0, \\
  u(t)-I^{\delta}Mu(t)\leq0.
\end{gather*}
So
\begin{equation*} %\label{eq2.7}
   u(t_0)- MI^{\delta}u(t_0)\leq 0.
\end{equation*}
Since
\begin{align*} %\label{eq2.8}
I^{\delta}u(t_0)
& = \frac {1}{\Gamma(\delta)}\int_0^{t_1}(t_0-s)^{\delta-1}u(s)\mathrm{d}s
        +\frac {1}{\Gamma(\delta)}\int_{t_1}^{t_0}(t_0-s)^{\delta-1}u(s)\mathrm{d}s\\
      & \leq \frac {1}{\Gamma(\delta)}\int_{t_1}^{t_0}(t_0-s)^{\delta-1}u(s)\mathrm{d}s\\
      & \leq \frac {1}{\Gamma(\delta)}\int_{t_1}^{t_0}(t_0-s)^{\delta-1}u(t_0)\mathrm{d}s\\
      & \leq \frac {u(t_0)}{\delta\Gamma(\delta)}(t_0-t_1)^\delta\\
      & \leq \frac {u(t_0)T^\delta}{\Gamma(\delta+1)},
   \end{align*}
we have
$$
  u(t_0)-\frac {Mu(t_0)T^\delta}{\Gamma(\delta+1)}\leq u(t_0)-
  MI^{\delta}u(t_0)\leq 0;
$$
that is,
$$
  u(t_0)(1-\frac {MT^\delta}{\Gamma(\delta+1)})\leq 0.
$$
It follows from (H0), that $1-\frac {MT^\delta}{\Gamma(\delta+1)}>0$, which
contradicts $u(t_0)>0$. This proves that $u(t)\leq0$ on
$J$.
The proof is complete.
\end{proof}

\begin{lemma} \label{lem2.3}
Suppose {\rm (H0)} holds, and  $u\in C^{1}(J)$ satisfies
\begin{equation}\label{eq2.9}
   \begin{gathered}
      D^\delta u(t)-Mu(t)\leq -a_{\alpha}(t),\quad t\in J,\\
      u(0)\leq0.
   \end{gathered}
\end{equation}
Then $u(t)\leq0$ for $t\in J$.
\end{lemma}

\begin{proof}
We consider the following two cases.

Case 1. $r\alpha(T)\geq \alpha(0)$.
We have $a_{\alpha}(t)=0$ if $r\alpha(T)\geq \alpha(0)$. By Lemma
\ref{lem2.2}, we can obtain $u(t)\leq 0$ for $t\in J$.

Case 2. $r\alpha(T)<\alpha(0)$.
When $r\alpha(T)<\alpha(0)$,
 $a_{\alpha}(t)=\frac{1}{r}(D^{\delta}c(t)-Mc(t))(\alpha(0)-r\alpha(T))$.

Let $v(t)=u(t)+\frac {1}{r}c(t)(\alpha(0)-r\alpha(T))$. Obviously,
$v(t)\geq u(t)$.
Since $c(t)=1-\cos{\frac {\pi t}{2T}}\geq0$ for $t\in J$, we can get
that $v(t)\geq u(t)$ for all $t\in J$. Follows from \eqref{eq2.9} we
have 
\begin{align*}
&D^\delta v(t)-Mv(t)= D^\delta u(t)-Mu(t)+\frac{1}{r}(D^{\delta}c(t)
-Mc(t))(\alpha(0)  -r\alpha(T))\\
&= D^\delta u(t)-Mu(t)+a_{\alpha}(t)\leq 0,
\end{align*}
and
$$
  v(0)=u(0)+c(0)(\alpha(0)-r\alpha(T))=u(0)\leq0.
$$
In view of Lemma \ref{lem2.2},  $v(t)\leq0$ for $t\in J$, which
implies that $u(t)\leq 0$.
This completes the proof.
\end{proof}

In a similar way, we can get the following lemma.

\begin{lemma} \label{lem2.4}
Suppose that  {\rm (H0)} holds, $u\in C^{1}(J)$ and satisfies
\begin{equation}\label{eq2.10}
   \begin{gathered}
      D^\delta u(t)-Mu(t)\leq -b_{\beta}(t),\quad t\in J,\\
      u(0)\leq0.
   \end{gathered}
\end{equation}
Then $u(t)\leq0$, for $t\in J$.
\end{lemma}


\begin{lemma}[{\cite[Theorem 4.1]{kdtafde}}] \label{lem2.6}
 Consider the two-parameter Mittag-Leffler function $E_{n_1,n_2}$ for 
some $n_1, n_2>0$. The power series defining $E_{n_1,n_2}(z)$ is convergent 
for all $z\in \mathbb{R}$.
\end{lemma}

\begin{lemma}[{\cite[Lemma 2.5]{lfjmlxp}}] \label{lem2.7}
Let $E$ be a partially ordered Banach space, $\{x_n\}\subset E$
a monotone sequence and relatively compact set, then $\{x_n\}$ is convergent.
\end{lemma}

\begin{lemma}[{\cite[Lemma 2.6]{lfjmlxp}}] \label{lem2.8}
 Let $E$ be a partially ordered Banach space, $x_n\leq y_n\ (n=1,2,3\dots)_0$,
if $x_n\to x^*,\ y_n\to y^*_0$, we have $x^*\leq y^*$.
\end{lemma}


\section{Existence and uniqueness of solutions for the Linear problems}

 In this section, we present the existence and uniqueness
theorems of solutions of the boundary-value problem \eqref{eq2.1}
based on the method of  the upper and the lower solutions.

\begin{lemma} \label{lem3.1} 
Let {\rm (H0)} hold. Assume that there
exist upper and lower solutions $\beta, \alpha\in C^{1}(J)$ of the
boundary-value problem \eqref{eq2.1} such that
$\alpha(t)\leq\beta(t)$ on $J$. Then the boundary-value problem
\eqref{eq2.1} has a unique solution $u$. Moreover, 
$\alpha \leq u \leq \beta$ on $J$, respectively.
\end{lemma}

\begin{proof} 
(1) We can prove that the boundary-value problem
\eqref{eq2.1} has a unique solution.
Let
$$
p(t)=\begin{cases}
  r\alpha(t), & r\alpha(T)\geq \alpha(0),\\
  r\alpha(t)+c(t)(\alpha(0)-r\alpha(T)), & r\alpha(T)<\alpha(0),
\end{cases}
$$
and
$$
q(t)=
\begin{cases}
r\beta(t), & r\beta(T)\leq \beta(0),\\
r\beta(t)-c(t)(r\beta(T)-\beta(0)), & r\beta(T)>\beta(0).
\end{cases}
$$
It is obvious that $p(0)=r\alpha(0)$ and $q(0)=r\beta(0)$.

If $r\alpha(T)\geq \alpha(0)$, then
$$
  p(T)=r\alpha(T)\geq \alpha(0)=\frac {p(0)}{r}.
$$
If $r\alpha(T)<\alpha(0)$, then
$$
  p(T)=r\alpha(T)-r\alpha(T)+\alpha(0)=\alpha(0)=\frac {p(0)}{r}.
$$
Thus, $rp(T)\geq p(0)$.

Analogously, we can get $rq(T)\leq q(0)$.
Therefore,
\begin{gather}\label{eq3.1}
   p(0)=r\alpha(0),\quad rp(T)\geq p(0), \\
\label{eq3.2}
   q(0)=r\beta(0),\quad rq(T)\leq q(0).
\end{gather}
If $r\alpha(T)\geq \alpha(0)$, we  obtain
$$
  D^{\delta}p(t)-Mp(t)=r(D^{\delta}\alpha(t)-M\alpha(t))\leq rx(t),\quad
 t\in   J,
$$
and if $r\alpha(T)<\alpha(0)$, for $t\in J$, we have
\begin{align*}
D^\delta p(t)-Mp(t)
& = r(D^\delta\alpha(t)-M\alpha(t))+(D^{\delta}c(t)-Mc(t))(\alpha(0)
     -r\alpha(T))\\
& \leq rx(t)-ra_{\alpha}(t)+ra_{\alpha}(t)=rx(t).
\end{align*}
 Hence, we obtain
\begin{equation}\label{eq3.3}
   D^{\delta}p(t)-Mp(t)\leq rx(t),\quad t\in J.
\end{equation}
Similarly, we can show that
\begin{equation}\label{eq3.4}
   D^{\delta}q(t)-Mq(t)\geq rx(t),\quad t\in J.
\end{equation}

Let $y(t)=p(t)-q(t),\ t\in J$. It follows that
\begin{gather*}
  D^\delta y(t)-My(t)\leq 0,\\
      y(0)=p(0)-q(0)\leq 0.
\end{gather*}
From \eqref{eq3.1}, \eqref{eq3.2}, \eqref{eq3.3} and \eqref{eq3.4}.
By Lemma \ref{lem2.2}, we have 
\begin{equation}\label{eq3.5}
  p(t)\leq q(t)\quad  \text{for } t\in J.
\end{equation}
For each $\lambda\in \mathbb{R}$, we consider the  initial
problem
\begin{equation}\label{eq3.6}
   \begin{gathered}
      D^{\delta}u(t)-Mu(t)=x(t),\\
      u(0)=\lambda.
   \end{gathered}
\end{equation}
According to Lemma \ref{lem2.1}, the initial problem \eqref{eq3.6}
has a unique solution
\begin{equation}\label{eq3.7}
   u(t,\lambda)=\lambda E_{\delta,1}(Mt^\delta)+\int_0^{t}(t-s)^{\delta-1}E_{\delta,
   \delta}(M(t-s)^\delta)x(s)\mathrm{d}s,\ t\in J.
\end{equation}
It is easy to see $u(t,\lambda)$ is continuous on $\lambda\in \mathbb{R}$.

Let $z(t)=p(t)-ru(t,\lambda)$, where $u(t,\lambda)$ is the solution
of \eqref{eq3.6}.
If $p(T)\leq\lambda\leq q(T)$, then
\begin{gather*}
     D^{\delta}z(t)-Mz(t)\leq rx(t)-rx(t)=0,\\
     z(0)=p(0)-ru(0,\lambda)\leq rp(T)-r\lambda\leq0.
\end{gather*}
From Lemma \ref{lem2.2}, we  obtain that $z(t)\leq0$ for $t\in
J$, so $p(T)\leq ru(T,\lambda)$.

Using the same method, we can get $ru(T,\lambda)\leq q(T)$.
Hence,
$$
  p(T)\leq ru(T,\lambda)\leq q(T)\quad \text{for } \lambda
  \in[p(T),q(T)].
$$
Let $g(\lambda)=ru(T,\lambda)-\lambda$ for $\lambda\in\mathbb{R}$,
then $g'(\lambda)=rE_{\delta,1}(MT^\delta)-1<0$, and $g$ is
strictly decreasing. We see that the equation $g(\lambda)=0$ has at
most one solution on $\mathbb{R}$.

Since $g(q(T))g(p(T))=(ru(T,q(T))-q(T))(ru(T,p(T))-p(T))\leq0$ and
$g(\lambda)$ is continuous for $\lambda\in\mathbb{R}$, we can show
that the equation $g(\lambda)=0$ has a unique solution
$\lambda_0\in\mathbb{R}$ and $p(T)\leq\lambda_0\leq q(T)$ with
$ru(T,\lambda_0)=\lambda_0=u(0)$.
Hence,
$$
  u(t,\lambda_0)=\lambda_0 E_{\delta,1}(Mt^\delta)
 +\int_0^{t}(t-s)^{\delta-1}E_{\delta,
   \delta}(M(t-s)^\delta)x(s)\mathrm{d}s,\quad t\in J
$$ 
is the unique solution of the boundary-value problem
\eqref{eq2.1}.


(2) We can prove that the solution $u(t,\lambda_0)$ satisfies
$\alpha(t)\leq u(t,\lambda_0)\leq\beta(t)$ for $t\in J$.
Let $h(t)=\alpha(t)-u(t,\lambda_0)$. If $r\alpha(T)\geq \alpha(0)$,
we know $a_{\alpha}(t)=0$. By \eqref{eq3.2}, we have
\begin{gather*}
     D^{\delta}h(t)-Mh(t)=D^{\delta}\alpha(t)-M\alpha(t)-x(t)\leq0,\\
     h(0)=\alpha(0)-u(0,\lambda_0)\leq p(T)-\lambda_0\leq0.
\end{gather*}
In view of Lemma \ref{lem2.2}, we have 
$h(t)=\alpha(t)-u(t,\lambda_0)\leq0$ for $t\in J$.

If $r\alpha(T)<\alpha(0)$, then $a_{\alpha}(t)=\frac
{1}{r}(D^{\delta}c(t)-Mc(t))(\alpha(0)-r\alpha(T))$. By
\eqref{eq3.2}, it is easy to see that
\begin{gather*}
     D^{\delta}h(t)-Mh(t)=D^{\delta}\alpha(t)-M\alpha(t)-x(t)\leq -a_{\alpha}(t),\\
     h(0)=\alpha(0)-u(0,\lambda_0)\leq p(T)-\lambda_0\leq0.
\end{gather*}
As a consequence of Lemma \ref{lem2.3}, we obtain that $h(t)\leq0$,
for $t\in J$, this is
 $u(t,\lambda_0)\geq \alpha(t)$.

In a similar way, we can obtain that $\beta(t)\geq u(t,\lambda_0)$
for $t\in J$.
Therefore, the unique solution $u(t,\lambda_0)$ of the boundary-value 
problem satisfies $\alpha(t)\leq u(t,\lambda_0)\leq\beta(t)$
for $t\in J$.
The proof is complete.
\end{proof}

\section{Main results}

\begin{definition} \label{def4.1} \rm
Let $\beta_0, \alpha_0 \in C^{1}(J)$. We say that $\beta_0,
\alpha_0$ are the upper solution and the lower solution of the
boundary-value problem \eqref{eq1.1}, respectively, if
$$
  D^{\delta}\alpha_0(t)-M\alpha_0(t)\leq f(t,\alpha_0(t))-
  a_{\alpha_0}(t),\ t\in J,
$$
and
$$
  D^{\delta}\beta_0(t)-M\beta_0(t)\geq
  f(t,\beta_0(t))+b_{\beta_0}(t),\ t\in J,
$$
 where
$a_{\alpha_0}(t), b_{\beta_0}(t)$ are defined in \eqref{eq2.3},
\eqref{eq2.4}, respectively.
\end{definition}

Let $E=C(J)$ with $\|x\|=\max_{t\in J}|x(t)|$ for $x\in E$.
Then $E$ is a Banach space.

\begin{theorem}\label{Th4.1}
Suppose {\rm (H0)} holds and there exist $\beta_0, \alpha_0
\in C^{1}(J)$ such that $\beta_0, \alpha_0$ are upper and lower
solutions of the boundary-value problem \eqref{eq1.1} with
$\alpha_0(t)\leq\beta_0(t)$
 for  $t\in J$, respectively, and $f$ satisfies
\begin{itemize}
\item[(H1)] $f(t,x_1)\leq f(t,x_2)$ for any $x_1,\ x_2\in \mathbb{R}$ with
$x_1\leq x_2$ and $t\in J$.
\end{itemize}
Then the boundary-value problem
\eqref{eq1.1} has a minimal solution $ \alpha^{*}$ and a maximal
solution $\beta^{*}$ on $[\alpha_0,\beta_0]=\{u\in C(J)|\;
\alpha_0(t)\leq u(t)\leq\beta_0(t),\;t\in(J)\}$. Moreover, the
monotone iterative sequences defined by
\begin{align*}
  \alpha_n(t)&=\frac {rE_{\delta,1}(Mt^\delta)}{1-rE_{\delta,
     1}(MT^\delta)}\int_0^{T}(T-s)^{\delta-1}E_{\delta,\delta}
     (M(T-s)^\delta)f(s,\alpha_{n-1}(s))\mathrm{d}s\\
    &\quad+ \int_0^{t}(t-s)^{\delta-1}E_{\delta,\delta}(M(t-s)^\delta)
   f(s,\alpha_{n-1}(s))\mathrm{d}s,
\end{align*}
and
\begin{align*}
  \beta_n(t)&=\frac {rE_{\delta,1}(Mt^\delta)}{1-rE_{\delta,
     1}(MT^\delta)}\int_0^{T}(T-s)^{\delta-1}E_{\delta,\delta}
     (M(T-s)^\delta)f(s,\beta_{n-1}(s))\mathrm{d}s\\
    &\quad+ \int_0^{t}(t-s)^{\delta-1}E_{\delta,\delta}(M(t-s)^\delta)
   f(s,\beta_{n-1}(s))\mathrm{d}s,
\end{align*}
converge uniformly on $J$ to $\alpha^{*}$ and $\beta^{*}$,
respectively. Namely, for $t\in J$,
 $\{\alpha_n(t)\}$, $\{\beta_n(t)\}$ with
$$
  \lim_{n\to\infty}\alpha_n(t)=\alpha^{*}(t),\
  \lim_{n\to\infty}\beta_n(t)=\beta^{*}(t).
$$
\end{theorem}

\begin{proof} We divide the proof into five parts.

(1) We denote $D=[\alpha_0,\beta_0]$. For any $\varphi\in D$, we
consider first the boundary-value problem
\begin{equation}\label{eq4.1}
   \begin{gathered}
      D^{\delta}u(t)-Mu(t)=f(t,\varphi(t)),\\
      u(0)=ru(T).
   \end{gathered}
\end{equation}
Since $\beta_0(t), \alpha_0(t)$ are upper and lower solutions
of the boundary-value problem \eqref{eq1.1}, by (H1), for $t\in J$,
we have
$$
  D^{\delta} \alpha_0(t)-M\alpha_0(t)\leq f(t,\alpha_0(t))-a_{\alpha_0}(t)
  \leq f(t,\varphi(t))-a_{\alpha_0}(t),
$$
and
$$
  D^{\delta} \beta_0(t)-M\beta_0(t)\geq f(t,\beta_0(t))+b_{\beta_0}(t)
  \geq f(t,\varphi(t))+b_{\beta_0}(t).
$$
Therefore, $\beta_0(t), \alpha_0(t)$ are also the upper and
lower solutions of the boundary-value problem \eqref{eq4.1}. In view
of Lemma 3.1, the boundary-value problem \eqref{eq4.1} has the
following unique solution $u$ with $u\in D$.
$$
  u(t)=u(0)E_{\delta,1}(Mt^\delta)+\int_0^{t}(t-s)^{\delta-1}E_{\delta,
  \delta}(M(t-s)^\delta)f(s,\varphi(s))\mathrm{d}s,\quad t\in J.
$$
Because $u(0)=ru(T)$, we can easily obtain that
$$
  u(0)=\frac {r}{1-rE_{\delta,1}(MT^\delta)}\int_0^{T}(T-s)^{\delta-1}
  E_{\delta, \delta}(M(T-s)^\delta)f(s,\varphi(s))\mathrm{d}s.
$$
We define an operator $A$: $D\to E$ by
\begin{align*}
     A\varphi(t) 
&= \frac {rE_{\delta,1}(Mt^\delta)}{1-rE_{\delta,1}(MT^\delta)}
     \int_0^{T}(T-s)^{\delta-1}E_{\delta,\delta}(M(T-s)^\delta)
 f(s,\varphi(s))\mathrm{d}s\\
&\quad + \int_0^{t}(t-s)^{\delta-1}E_{\delta,\delta}(M(t-s)^\delta)
 f(s,\varphi(s))\mathrm{d}s.
\end{align*}
By Lemma \ref{lem3.1} we can show that $\alpha_0(t)\leq
A\varphi(t)\leq\beta_0(t)$, for $t\in J$.
Hence, we obtain
\begin{equation}\label{eq4.3}
  A\alpha_0\geq\alpha_0,\quad A\beta_0\leq\beta_0.
\end{equation}


(2) We prove that $A$ is completely continuous.
Since $D$ is bounded, there exists a constant $M_0>0$ such that
$\|\varphi\|\leq M_0$ for $\varphi\in D$. Because $f$ is continuous,
there exists a constant $M_1>0$ such that $\max_{s\in
J}|f(s,\varphi(s))|\leq M_1$ for $s\in J$. We have that
\begin{align*}
 |A\varphi(t)|
   & \leq \frac {rE_{\delta,1}(MT^\delta)}{1-rE_{\delta,1}(MT^\delta)}M_1
     E_{\delta,\delta}(MT^\delta)\int_0^{T}(T-s)^{\delta-1}\mathrm{d}s\\
 &\quad  +M_1E_{\delta,\delta}(MT^\delta)\int_0^{t}(t-s)^{\delta-1}\mathrm{d}s\\
   & \leq \frac {M_1T^{\delta}E_{\delta,
     \delta}(MT^\delta)}{\delta(1-rE_{\delta,1}(MT^\delta))}.
\end{align*}
So $A(D)$ is uniformly bounded.

Let $F(t)=E_{\delta,1}(Mt^\delta)$ for $t\in J$,
$G(t,s)=E_{\delta,\delta}(M(t-s)^\delta)$ for
$(t,s)\in\Omega=\{(t,s)|\;t\in J,\;0\leq s\leq t\}$. As a
consequence of Lemma \ref{lem2.6}, we have $F$ and $G$ are uniformly
continuous on $J$ and $\Omega$, respectively.

For any $\varepsilon>0$, there exists $\delta_1>0$, for any $t_1,t_2\in J$
 with $t_2>t_1$, whenever $|t_2-t_1|<\delta_1$, we obtain
$$
  |F(t_2)-F(t_1)|<\frac {\varepsilon\delta(1-rE_{\delta,1}
  (MT^\delta))}{3rM_1T^{\delta}E_{\delta,\delta}(MT^\delta)},
$$
moreover, if $(t_1,s)$, $(t_2,s)\in\Omega$, then
$$
  |G(t_2,s)-G(t_1,s)|<\frac {\varepsilon\delta}{6M_1T^{\delta}}.
$$
We take $0<\delta_0\leq \min\{\delta_1, \delta_2, (\frac
{\varepsilon\delta}{6M_1E_{\delta, \delta}(MT^\delta)})^{\frac
{1}{\delta}}\}$.
Thus, as $t_1, t_2\in J$ with $t_2>t_1$, whenever
$|t_2-t_1|<\delta_0$ and $\varphi\in D$, we have
\begin{align*}
 &|A\varphi(t_2)-A\varphi(t_1)|\\
&\leq \Big|\frac {r\int_0^{T}(T-s)^{\delta-1}E_{\delta,\delta}(M(T-s)^\delta)
     f(s,\varphi(s))\mathrm{d}s}{1-rE_{\delta,1}(MT^\delta)}\cdot\big(F(t_2)
     -F(t_1)\big)\Big|\\
&\quad +\Big|\int_{t_1}^{t_2}(t_2-s)^{\delta-1}E_{\delta,
     \delta}(M(t_2-s)^\delta)f(s,\varphi(s))\mathrm{d}s\Big|\\
&\quad +\Big|\int_0^{t_1}\left((t_2-s)^{\delta-1}E_{\delta,
     \delta}(M(t_2-s)^\delta)-(t_1-s)^{\delta-1}E_{\delta,
     \delta}(M(t_1-s)^\delta)\right)f(s,\varphi(s))\mathrm{d}s\Big|\\
&\leq \frac {\varepsilon}{3}+\frac {M_1E_{\delta,
     \delta}(MT^\delta)}{\delta}\cdot|t_2-t_1|^{\delta}\\
&\quad +M_1\int_0^{t_1}(t_2-s)^{\delta-1}|E_{\delta,
     \delta}(M(t_2-s)^\delta)-E_{\delta,
     \delta}(M(t_1-s)^\delta)|\mathrm{d}s\\
&\quad +M_1\int_0^{t_1}\left((t_1-s)^{\delta-1}-(t_2-s)^{\delta-1}\right)E_{\delta,
     \delta}(M(t_1-s)^\delta)\mathrm{d}s.
\end{align*}
Since
\begin{align*}
& \int_0^{t_1}(t_2-s)^{\delta-1}|E_{\delta,\delta}(M(t_2-s)^\delta)-E_{\delta,
    \delta}(M(t_1-s)^\delta)|\mathrm{d}s\\
&=    \int_0^{t_1}(t_2-s)^{\delta-1}|G(t_2,s)-G(t_1,s)|\mathrm{d}s\\
\leq  & \frac{T^{\delta}}{\delta}\cdot\frac{\varepsilon\delta}{6M_1T^{\delta}}
    =\frac{\varepsilon}{6M_1},
\end{align*}
and
\begin{align*}
 & \int_0^{t_1}\left((t_1-s)^{\delta-1}-(t_2-s)^{\delta-1}\right)E_{\delta,
     \delta}(M(t_1-s)^\delta)\mathrm{d}s\\
 &\leq E_{\delta,\delta}(MT^\delta)\int_0^{t_1}\left((t_1-s)^{\delta-1}
     -(t_2-s)^{\delta-1}\right)\mathrm{d}s\\
 &\leq \frac{E_{\delta,\delta}(MT^\delta)}{\delta}\left((t_2-t_1)^{\delta}
    -(t_2^{\delta}-t_1^{\delta})\right)\\
 &\leq\frac{E_{\delta,\delta}(MT^\delta)}{\delta}(t_2-t_1)^{\delta}
   \leq\frac{\varepsilon}{6M_1}.
\end{align*}
Therefore, as $t_1, t_2\in J$ with $t_2>t_1$, whenever
$|t_2-t_1|<\delta_0$ and $\varphi\in D$, we can show that
$$
  |A\varphi(t_2)-A\varphi(t_1)|<\varepsilon.
$$
We prove $A$ is equi-continuous.
By Arzela-Ascoli theorem, we know that $A(D)$ is relatively compact.
We can easily show that $A$ is continuous since $f$ is
continuous.
Hence, $A$ is completely continuous.

(3) $A$ is an increasing operator on $J$.
For $\alpha_0\leq \omega_1\leq \omega_2\leq \beta_0$, $t\in
J$, let
$$
  B(s)=f(s,\omega_2(s))-f(s,\omega_1(s)),\ s\in J.
$$
From (H1), we have $B(s)\geq0$ and
\begin{align*}
A\omega_2(t)-A\omega_1(t)
& = \frac {rE_{\delta,1}(Mt^\delta)}{1-rE_{\delta,1}
     (MT^\delta)}\int_0^{T}(T-s)^{\delta-1}E_{\delta,\delta}
     (M(T-s)^\delta)B(s)\mathrm{d}s\\
&\quad + \int_0^{t}(t-s)^{\delta-1}E_{\delta,\delta}
     (M(t-s)^\delta)B(s)\mathrm{d}s\geq0.
\end{align*}
Thus, $A$ is an increasing operator.

(4) Let $\alpha_n=A \alpha_{n-1},\ {\beta_n=A \beta_{n-1}}$, for
$n=1,2,\dots$.
By \eqref{eq4.3}, we get monotone iterative sequences
$$
  \alpha_1\leq\alpha_2\leq\dots\leq\alpha_n\leq\dots\leq\beta_n
  \leq\dots\leq\beta_2\leq\beta_1.
$$
 As $\{\alpha_n\},\ \{\beta_n\}\subset A(D)$, we get
that $\{\alpha_n\}$ and $\{\beta_n\}$ are monotone sequences and
relatively compact set respectively. In view of  Lemma \ref{lem2.7},
we can obtain that there  exist $\alpha^{*},\ \beta^{*}\in C(J)$
such that
$$
  \lim_{n\to\infty}\alpha_n(t)=\alpha^{*}(t),\quad
  \lim_{n\to\infty}\beta_n(t)=\beta^{*}(t).
$$
By the continuity of $A$, we have $\alpha^{*}=A\alpha^{*}$,
$\beta^{*}=A\beta^{*}$. So $\alpha^{*}, \beta^{*}$ are the fixed
points of $A$.

It is clear that $u$ is a solution of the boundary-value problem
\eqref{eq1.1} if and only if $u$ is a fixed point of $A$. Hence,
$\alpha^{*}(t),\beta^{*}(t)$ are solutions of the boundary-value
 problem \eqref{eq1.1}.

(5) We prove that $\alpha^{*}$, $\beta^{*}$ are the minimal solution
and the maximal solution of the boundary-value problem
\eqref{eq1.1}, respectively.

Assume $u\in[\alpha_0,\beta_0]$ is a solution of the boundary-value 
problem \eqref{eq1.1}. We can easily obtain that
$A\alpha_0(t)\leq Au(t)\leq A\beta_0(t)$ by the fact that $A$ is
increasing in $[\alpha_0,\beta_0]$. That is $\alpha_1(t)\leq
u(t) \leq \beta_1(t)$. Doing this repeatedly, we have
$\alpha_n(t)\leq u(t)\leq\beta_n(t)$, for $n=1,2,\dots$. From
Lemma \ref{lem2.8}, we obtain that $\alpha^{*}(t)\leq u(t)\leq
\beta^{*}(t)$, as $n\to\infty$.

Therefore, $\alpha^{*}$, $\beta^{*}$ are the minimal solution and
the maximal solution of the boundary-value problem \eqref{eq1.1},
respectively.
The proof is complete.
\end{proof}

\begin{theorem}\label{Th4.2}
 Suppose the conditions of Theorem \ref{Th4.1} hold.
 There exists a constant
 $\gamma\in\big[0,\frac {\delta(1-rE_{\delta,1}(MT^\delta))}{T^{\delta}
 E_{\delta,\delta}(MT^\delta)}\big)$ and $f$ satisfies
\begin{itemize}
\item[(H2)] $f(t,x_2)-f(t,x_1)\leq \gamma (x_2-x_1)$ for any $x_1, x_2\in
\mathbb{R}$ and $x_1\leq x_2$.
\end{itemize}
 Then the boundary-value problem \eqref{eq1.1} has a unique
solution $u^*$ on $[\alpha_0,\beta_0]$. Moreover, for each
$u_0\in [\alpha_0,\beta_0]$, the iterative sequence
\begin{align*}
     u_n(t) 
&= \frac {rE_{\delta,1}(Mt^\delta)}{1-rE_{\delta,1}(MT^\delta)}
     \int_0^{T}(T-s)^{\delta-1}E_{\delta,\delta}(M(T-s)^\delta)
     f(s,u_{n-1}(s))\mathrm{d}s\\
&\quad + \int_0^{t}(t-s)^{\delta-1}E_{\delta,\delta}(M(t-s)^\delta)
     f(s,u_{n-1}(s))\mathrm{d}s,\; n=1,2,\dots
\end{align*}
converges uniformly to $u^*$ on $J$. Its error estimate is
$$
\|u_n-u^*\|\leq \rho^{n}\|\beta_0-\alpha_0\|,\ n=1,2,3\dots,
$$
where
$$
\rho=\frac {\gamma T^{\delta}E_{\delta,\delta}(MT^\delta)}
{\delta(1-rE_{\delta,1}(MT^\delta))}.
$$
\end{theorem}

\begin{proof} 
For $\alpha_0\leq \omega_1\leq \omega_2\leq \beta_0$, $t\in J$, from 
(H2) we have
\begin{align*}
 0&\leq A\omega_2(t)-A\omega_1(t)\\
  & \leq \frac {rE_{\delta,1}(MT^\delta)}{1-rE_{\delta,1}
    (MT^\delta)}\int_0^{T}(T-s)^{\delta-1}E_{\delta,\delta}
    (M(T-s)^\delta)(f(s,\omega_2(s))-f(s,\omega_1(s)))\mathrm{d}s\\
  & \quad + \int_0^{t}(t-s)^{\delta-1}E_{\delta,\delta}(M(t-s)^\delta)
    (f(s,\omega_2(s))-f(s,\omega_1(s)))\mathrm{d}s\\
  & \leq \Big(\frac {T^{\delta}E_{\delta,1}(MT^\delta)E_{\delta,\delta}
    (MT^\delta)}{\delta(\frac {1}{r}-E_{\delta,1}(MT^\delta))}
    +\frac
    {T^\delta}{\delta}E_{\delta,\delta}(MT^\delta)\Big)\gamma
    \|\omega_2(t)-\omega_1(t)\|\\
  & = \frac {\gamma T^{\delta}E_{\delta,\delta}(MT^\delta)}{\delta(1-r
    E_{\delta,1}(MT^\delta))}\|\omega_2(t)-\omega_1(t)\|.
\end{align*}
Hence,
$$
  \|A\omega_2-A\omega_1\|\leq\rho\|\omega_2-\omega_1\|.
$$
 It is easy to obtain that
\begin{equation*}
\|\beta_n-\alpha_n\|=\|A\beta_{n-1}-A\alpha_{n-1}\|
                        \leq \rho\|\beta_{n-1}-\alpha_{n-1}\| \leq \dots
                        \leq \rho^{n}\|\beta_0-\alpha_0\|.
\end{equation*}
From $0\leq \gamma <\frac{\delta(1-rE_{\delta,1}(MT^\delta))}
{T^{\delta}E_{\delta,\delta}(MT^\delta)}$,
 we have $0\leq \rho<1$,  and
$$
  \|\beta_n-\alpha_n\|\to0,\quad n\to\infty.
$$
It follows from Theorem \ref{Th4.1}, there exists a unique
$u^{*}\in[\alpha_0,\beta_0]$ such that
 $\alpha_n\to u^{*},\ \beta_n\to u^{*}$, when
 $n\to\infty$. As $\alpha_n\leq u^{*}\leq\beta_n$,
 by the monotonicity of $A$ we have $A\alpha_n\leq u^{*}\leq A\beta_n$.
When $n\to\infty$, we have that $u^{*}=Au^{*}$, $u^{*}$ is a
fixed point of $A$. Hence, $u^{*}$ is the unique solution of the
boundary-value problem \eqref{eq1.1}.

For each $u_0\in [\alpha_0,\beta_0]$, about the iterative
sequence $u_n=Au_{n-1}$, we have $\alpha_n\leq u_n\leq \beta_n$,
and
$$
  \|u_n-u^*\|\leq\|\beta_n-\alpha_n\|\leq \rho^{n}\|\beta_0-\alpha_0\|.
$$
This completes  the proof.
\end{proof}

\section{Example}
 Consider the boundary-value problem
\begin{equation}\label{eq5.1}
   \begin{gathered}
      D^{1/2} u(t)=\frac {1}{2}+\frac {\cos t\arctan u(t)}
      {\pi (t+6)^2},\quad 0\leq t\leq1,\\
      3u(0)=u(1).
   \end{gathered}
\end{equation}
Let $f(t,x)=\frac {1}{2}+\frac {\cos t \arctan x}{\pi (t+6)^2}$.
Clearly, $T=1$, $\delta =\frac {1}{2}$, $r=\frac {1}{3}$, $M=0$ and
$0\leq f(t,x)\leq 1$. We take $\gamma=\frac{1}{36\sqrt{\pi}}$, then
$E_{\delta,\delta}(MT^\delta)=\frac{1}{\sqrt{\pi}}$,
$E_{\delta,1}(MT^\delta)=1$, and
$$
  \rho=\frac {\gamma T^{\delta}E_{\delta,\delta}(MT^\delta)}
{\delta(1-rE_{\delta,1}(MT^\delta))}
  =\frac{1}{12}<1.
$$
Obviously, $\alpha_0(t)=0$ is a lower solution of the boundary-value
 problem \eqref{eq5.1}.
Let $\beta_0(t)=\frac{5}{8\sqrt{\pi}}(1+2\sqrt{t})$,
 we have $\frac {1}{3}\beta_0(1)=\beta_0(0)$, then $b_{\beta_0}(t)=0$.
\begin{equation}
  D^{1/2}\beta_0(t)  =\frac{5}{8}  \geq f(t,\beta_0(t)).
\end{equation}
So $\beta_0$ is an upper solution of the boundary-value problem
\eqref{eq5.1}.

It is easy to verify that the assumptions of Theorem \ref{Th4.2} are
satisfied. Hence, the boundary-value problem \eqref{eq5.1} has a
unique solution $u^*(t)$ with $0\leq u^*(t)\leq
\frac{5}{8\sqrt{\pi}}(1+2\sqrt{t})$ on $[0,1]$.

For each $0\leq u_0(t)\leq \frac{5}{8\sqrt{\pi}}(1+2\sqrt{t})$, let
the iterative sequence $u_n=Au_{n-1}$, we have
\begin{align*}
     u_n(t) =
    \frac {1}{2\sqrt{\pi}}
     \int_0^{1}(1-s)^{-\frac{1}{2}}f(s,u_{n-1}(s))\mathrm{d}s
    + \frac {1}{\sqrt{\pi}}\int_0^{t}(t-s)^{-\frac{1}{2}}
    f(s,u_{n-1}(s))\mathrm{d}s
\end{align*}
converges uniformly to $u^*$ on $J$. Its error estimate is
$$
  \|u_n-u^*\|\leq \frac{15}{8\sqrt{\pi}}\cdot\big(\frac{1}{12}\big)^{n}.
$$
We take $u_0=0$. If $n=2$, its error is not more than 0.00734622; if
$n=3$, its error is not more than 0.000612185; if $n=4$, its error
is not more than 0.0000510154.

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1269-1274.

\end{thebibliography}

\end{document}