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

\AtBeginDocument{{\noindent\small
\emph{Electronic Journal of Differential Equations},
Vol. 2015 (2015), No. 286, pp. 1--12.\newline
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu
\newline ftp ejde.math.txstate.edu}
\thanks{\copyright 2015 Texas State University.}
\vspace{9mm}}

\begin{document}
\title[\hfilneg EJDE-2015/286\hfil Solution of fractional differential equations]
{Solution of fractional differential equations via coupled fixed point}

\author[H. Afshari, S. Kalantari, E. Karapinar \hfil EJDE-2015/286\hfilneg]
{Hojjat Afshari, Sabileh Kalantari, Erdal Karapinar}

\address{Hojjat Afshari \newline
Faculty of Basic Sciences,
Bonab University, Bonab, Iran}
\email{hojat.afshari@yahoo.com}

\address{Sabileh Kalantari \newline
Faculty of Basic Sciences,
Bonab University, Bonab, Iran}
\email{kalantari.math@gmail.com}

\address{Erdal Karapinar  \newline
Atilim University, Department of Mathematics,
06836, \.Incek, Ankara, Turkey.\newline
Nonlinear Analysis and Applied Mathematics (NAAM) 
Research Group, King Abdulaziz University, 
21589, Jeddah, Saudi Arabia}
\email{erdalkarapinar@yahoo.com}

\thanks{Submitted June 25, 2015. Published November 16, 2015.}
\subjclass[2010]{47H10, 54H25}
\keywords{Fixed point; mixed monotone operator;  normal cone; 
\hfill\break\indent sub-homogeneous operator}

\begin{abstract}
 In this article, we investigate the existence and uniqueness of a solution
 for the fractional differential equation by introducing some new coupled
 fixed point theorems for the class of mixed monotone operators with 
 perturbations in the context of partially ordered complete metric space.
\end{abstract}

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

\section{Introduction and preliminaries}

In the previous decade, one of the most attractive research subject is 
to investigate the existence and uniqueness of a fixed point of certain 
operator in the setting of complete metric space endowed with a partial order
(see e.g. \cite{AEO1}-\cite{Zhao} and related reference therein).
Recently, CB. Zhai \cite{Zhai2} proved some results on a class of mixed monotone
operators with perturbations.
The aim of this article is to propose a method for  the existence and uniqueness 
of a solution of certain  fractional differential equations
by following the paper by Zhai \cite{Zhai2}. 
For this purpose,  we shall consider some
coupled fixed point theorems for a class of mixed monotone operators
with perturbations on ordered Banach spaces with the different conditions 
that was introduced by Zhai \cite{Zhai2}.
On the other hand, our result are finer than the results of Zhai \cite{Zhai2} 
since we obtain the existence and uniqueness of coupled fixed points without
assuming continuity of compactness of the operator.

For the sake of completeness of the paper, we present here some 
basic definitions, notations and known results.

Suppose $(E,\| \cdot \|)$ is a Banach space which is
partially ordered by a cone $P\subseteq E$, that is, $x\leq y$ if
and only if $y-x\in P$. If $x\neq y$, then we denote $x<y$ or $x>y$.
We denote the zero element of $E$ by $\theta$. Recall that a
non-empty closed convex set $P\subset E$ is a cone if it satisfies
(i)  $x\in P,~\lambda\geq 0 \Longrightarrow \lambda x\in P$; (ii)
$x\in P,~-x\in P \Longrightarrow x=\theta$. A cone $P$ is called
normal if there exists a constant $N>0$  such that $\theta\leq x\leq
y$ implies $\| x \| \leq N \| y \|$.
Also we define the order interval $[x_1,x_2]=\{x\in E|x_1\leq x\leq
x_2\}$ for all $x_1,x_2\in E$. We say that and operator
$A:E\to E$ is increasing whenever $x\leq y$ implies $Ax\leq
Ay$.  For all $x,y\in E$, the notation $x\sim y$ means that there exist 
$\lambda>0$ and $\mu>0$, such that $\lambda x\leq y\leq \mu x$. 
Clearly, $\sim$ is an equivalent relation.  Given $e>\theta$, we 
denote by $P_e$ the set $P_e=\{x\in E|x\sim e\}$. It is easy to see that 
$P_e\subset P$ is convex and $\lambda P_e= P_e$ for all $\lambda>0$. 
If $P\neq \phi$ and $e\in P$, it is clear that $P_e=P$.

\begin{definition}[\cite{Guo1,Guo2}] \rm
$A:P\times P\to P$ is said to be a mixed monotone operator
if $A(x,y)$ is increasing in $x$ and decreasing in $y$, i.e., 
$u_i,v_i$ $(i=1,2)\in P$, $u_1\leq u_2$,
$v_1\geq v_2$ imply $A(u_1,v_1)\leq A(u_2,v_2)$. The element 
$x\in P$ is called a fixed point of $A$ if $A(x,x)=x$.
\end{definition}

The following conditions were was assumed in \cite{Zhai1}: 
\begin{itemize}
\item[(A1)] there exists $h\in P$ with $h\neq \theta$ such that $A(h,h)\in P_h$,
\item[(A2)] for any $u,v\in P$ and $t\in (0,1)$, there exists $\varphi(t)\in(t,1]$  
such that
\begin{equation} \label{t1}
A(tu,t^{-1}v)\geq {\varphi(t)}A(u,v).
\end{equation}
\end{itemize}

\begin{lemma}[\cite{Zhai1}] \label{nnnt2}
Assume that {\rm (A1), (A2)} hold. 
Then $A:P_h \times P_h \to P_h$; and there exist $u_0, v_0\in P_h$ and 
$r\in(0,1)$ such that
$$
rv_0\leq u_0<v_0,\quad  u_0\leq A(u_0,v_0)\leq A(v_0,u_0)\leq v_0.
$$
\end{lemma}

\begin{definition}[\cite{Zhai2}] \label{t9} \rm
An operator $A:P\to P$ is said to be sub-homogeneous if it satisfies 
$$
A(tx)\geq tA(x),\quad \forall t\in (0,1),\; x\in P.
$$
\end{definition}

The following result can be found in Zhai and Zhang \cite{Zhai1}.

\begin{theorem}[\cite{Zhai1}] \label{t11}
Let $P$ be a normal cone in $E$. Assume that $T:P\times P\to P$ is a mixed 
monotone operator and satisfies:
\begin{itemize}
\item[(A3)] there exists $h\in P$ with $h\neq \theta$ such that $T(h,h)\in P_h$;
\item[(A4)] for any $u,v\in P$ and $t\in (0,1)$, there exists $\varphi(t)\in(t,1]$ 
such that
\begin{equation} \label{nnt1}
T(tu,t^{-1}v)\geq \varphi(t)T(u,v).
\end{equation}
\end{itemize}
Then
\begin{itemize}
\item[(1)] $T:P_h\times P_h\to P_h$;
\item[(2)] there exist $u_0,v_0\in P_h$ and $r\in (0,1)$ such that
$rv_0\leq u_0<v_0,~u_0\leq T(u_0,v_0)\leq T(v_0,u_0)\leq v_0$;

\item[(3)] $T$ has a unique fixed point $x^*$ in $P_h$;

\item[(4)] for any initial values $x_0,y_0\in P_h$, constructing successively 
the sequences 
$$
x_n=T(x_{n-1},y_{n-1}),\quad 
y_n=T(y_{n-1},x_{n-1}),\quad n=1,2\dots,
$$
we have $x_n\to x^*$ and  $y_n\to x^*$ as $n\to \infty$.
\end{itemize}
\end{theorem}

\section{Main result}

In this section, we state and prove our main results. 
First, we consider the mixed monotone operator $A:P\times P \to P$. 
The following conditions will be assumed:
\begin{itemize}
\item[(A5)] there exists $h\in P$ with $h\neq \theta$ such that $A(h,h)\in P_h$,
\item[(A6)] for any $u,v\in P$ and $s,t\in (0,1)$ such that $s\leq t$, 
there exists 
$\varphi(t)\in ({t}^2,1]$ and $\varphi$ is decreasing such that
\begin{equation} \label{nt1}
A(tu,t^{-1}v)+A(tu,s^{-1}v)\geq 2\frac{\varphi(t)}{t}A(u,v).
\end{equation}
\end{itemize}

\begin{lemma}\label{t2}
Assume {\rm (A5), (A6)} hold. Then $A:P_h \times P_h \to P_h$; and there 
exist $u_0, v_0\in P_h$ and $r\in(0,1)$ such that
$$
rv_0\leq u_0<v_0, u_0\leq A(u_0,v_0)\leq A(v_0,u_0)\leq v_0.
$$
\end{lemma}

\begin{proof}
For $s\leq t$ from condition (A6) we obtain
\begin{equation}\label{t3}
A(t^{-1}x,ty)\leq \frac{1}{2\frac{\varphi(t)}{t}}(A(x,y)+A(x,\frac{t}{s}y)),
\quad \forall s,t\in (0,1),\; x,y\in P.
\end{equation}
For any $u,v\in P_h$, there exist $\mu_1,\mu_2\in (0,1)$, such that
$$
\mu_1h\leq u\leq \frac{1}{\mu_1}h,~\mu_2h\leq v\leq \frac{1}{\mu_2}h.
$$
Let $\mu=\min\{\mu_2, \mu_1\}$. Then $\mu\in (0,1)$. From \eqref{t3} 
and the mixed monotone properties of operator $A$ and regarding $0<\mu$, 
there exists $0<\mu'<\mu$ such that
\begin{align*}
A(u,v)
&\leq A(\frac{1}{\mu_1}h,{\mu_2}h)
 \leq A(\frac{1}{\mu}h,{\mu}h)\\
&\leq \frac{1}{2(\frac{\varphi(\mu)}{\mu})}(A(h,h)+A(h,\frac{\mu}{\mu'}h))\\
&\leq\frac{1}{2(\frac{\varphi(\mu)}{\mu})}(A(h,h)+A(h,h))\\
&=\frac{1}{(\frac{\varphi(\mu)}{\mu})}(A(h,h))\leq \frac{1}{\varphi(\mu)}A(h,h).
\end{align*}
Regarding the inequality
$$
A(u,v)\geq A({\mu_1}h,\frac{1}{\mu_2}h)\geq A({\mu}h,\frac{1}{\mu}h),
$$
we derive that
\begin{align*}
2A(u,v)
&\geq A({\mu}h,\frac{1}{\mu}h)+A({\mu}h,\frac{1}{\mu}h)\\
&\geq 2(\frac{\varphi(\mu)}{\mu})(A(h,h)+A(h,h))\\
&=2(\frac{\varphi(\mu)}{\mu})(A(h,h))\geq 2\varphi(\mu)A(h,h).
\end{align*}
Tehrefore, we obtain
$$
A(u,v)\geq \varphi(\mu)A(h,h).
$$
It follows from $A(h,h)\in P_h$ that $A(u,v)\in P_h$.
 Hence we have $A :P_h\times P_h \to P_h$. Since $A(h,h)\in P_h$, 
we can choose a sufficiently small number $t_0\in (0, 1)$ such that
\begin{equation}\label{t4}
t_0h\leq A(h,h)\leq \frac{1}{t_0}h.
\end{equation}
For $k>2$ we have
\begin{equation}\label{t5}
{t_0^k}h\leq A(h,h)\leq  \frac{1}{t_0^k}h\,.
\end{equation}
Put $u_0={t_0}^kh$ and $v_0=\frac{1}{t_0^k}h$. Evidently, $u_0,v_0\in P_h$ 
and $u_0 = {t_0}^{2k}v_0<v_0$.
Take any $r\in (0,{t_0}^{2k}]$, then $r\in (0, 1)$ and $u_0\geq rv_0.$
By the mixed monotone properties of $A$, we have $A(u_0,v_0)\leq A(v_0,u_0)$. 
Because $t_0\in(0,1)$, then there exists $s_0 \in (0,1)$ such that 
$0 <s_0 \leq t_0$. Further, combining condition (A2) with
 \eqref{t4}, and since $s_0\leq t_0$ we have
\begin{align*}
A(u_0,v_0)
&=A({t_0^k}h,\frac{1}{t_0^k}h)\\
&\geq 2(\frac{\varphi(t_0^k)}{t_0^k})A(h,h)-A(h,h)\\
&\geq(2(\frac{\varphi(t_0)}{t_0^2})-1)A(h,h)>A(h,h)\\
&\geq t_0^kh=u_0,
\end{align*}
and
\begin{align*}
A(v_0,u_0)
&=A(\frac{1}{t_0^k}h,{t_0^k}h)\\
&\leq \frac{1}{2(\frac{\varphi(t_0^k)}{t_0^k})}(A(h,h)+A(h,\frac{t_0^k}{s_0^k}h))\\
&\leq \frac{1}{2(\frac{\varphi(t_0^k)}{t_0^k})}2A(h,h)\\
&\leq A(h,h)\leq \frac{1}{t_0^k}h=v_0.
\end{align*}
Consequently, we have $u_0\leq A(u_0,v_0)\leq A(v_0,u_0)\leq v_0$.
\end{proof}

\begin{corollary}
If in \eqref{nt1} put $s=t$ then we obtain \eqref{nnt1}. 
Consequently the Lemma \ref{t2} yields the Lemma \ref{nnnt2}.
\end{corollary}

\begin{theorem}\label{a}
Suppose that $P$ is a normal cone of $E$, and 
{\rm (A5), (A6)} hold. Then operator $A$ has a unique fixed point $x^*$ 
in $P_h$. Moreover, for any initial $x_0,y_0\in P_h$, constructing successively 
the sequences
\[
x_n = A(x_{n-1}, y_{n-1}), \quad y_n = A(y_{n-1}, x_{n-1}) \quad n = 1, 2,\ldots,
\]
we have $\| x_n-x^*\|\to 0$,$\| y_n-x^*\|\to 0$ as $n\to \infty$.
\end{theorem}

\begin{proof}
From Lemma \ref{t2}, there exist $u_0,v_0\in P_h$ and $r\in (0, 1)$ such that
 $$
rv_0\leq u_0<v_0, u_0\leq A(u_0,v_0)\leq A(v_0,u_0)\leq v_0.
$$
Construct recursively the sequences
$$
u_n= A(u_{n-1}, v_{n-1}),\quad  v_n = A(v_{n-1}, u_{n-1}),
\quad n=1,2,\ldots.
$$
Evidently $u_1\leq v_1$. By the mixed monotone properties of $A$, we obtain
\[
u_n\leq v_n~ ,~n = 1, 2,\dots .
\] 
It also follows from Lemma \ref{t2} and the mixed monotone properties of $A$ that
\begin{equation}\label{t6}
u_0\leq u_1\leq \ldots\leq u_n\leq \ldots \leq v_n\leq \ldots\leq v_1\leq v_0.
\end{equation}
Note that $u_0\geq rv_0$. We can get $u_n\geq u_0\geq rv_0\geq rv_n,~n = 1, 2,\dots$. 
Let
\[
t_n = \sup\{t>0|u_n\geq tv_n\},\quad
s_n=\sup\{s>0|u_n\geq sv_n\}, \quad
s_n \leq t_n\; n = 1, 2,\ldots.
\]
Thus we have $u_n\geq t_nv_n,u_n\geq s_nv_n,n=1,2,\ldots$, 
then $u_n\geq t_nv_n\geq s_nv_n$, also 
$u_{n+1}\geq u_n\geq t_nv_n\geq t_nv_{n+1}\geq s_nv_{n+1},n=1,2,\ldots$. 
Therefore, $t_{n+1}\geq t_n,~i.e., {t_n}$ is increasing with 
${t_n}\subset(0, 1]$. Suppose $t_n\to t^*$ as $n\to \infty$, then $t^*=1$. 
Otherwise, $0 <t^*<1$. Then from condition (A2) and $t_n\leq t^*$, we have
$A(u_n,v_n)\geq A(t_nv_n,\frac{1}{t_n}u_n)$ and 
$A(u_n,v_n)\geq A(t_nv_n,\frac{1}{s_n}u_n)$,
so
\begin{align*}
u_{n+1}
&=A(u_n,v_n)\\
&\geq \frac{1}{2}(A(t_nv_n,\frac{1}{t_n}u_n)+A(t_nv_n,\frac{1}{s_n}u_n))\\
&\geq \frac{\varphi(t_n)}{t_n}A(v_n,u_n) \geq \frac{\varphi(t^*)}{t_n}A(v_n,u_n)\\
&=\frac{\varphi(t^*)}{t_n}v_{n+1}.
\end{align*}
By the definition of $t_n,~t_{n+1}\geq \frac{\varphi(t^*)}{t_n}$. 
Let $n\to \infty$, we obtain ${t^*}^2\geq \varphi(t^*)>{t^*}^2$, 
which is a contradiction. Thus, $\lim_{n\to\infty}t_n=1$. 
For any natural number $p$ we have
\begin{gather*}
\theta\leq u_{n+p}-u_n\leq v_n-u_n\leq v_n-t_nv_n=(1-t_n)v_n\leq (1-t_n)v_0,\\
\theta\leq v_n-v_{n+p}\leq v_n-u_n\leq (1-t_n)v_0.
\end{gather*}
Since the cone $P$ is normal, we have
\begin{gather*}
\| u_{n+p}-u_n \| \leq M(1-t_n) \| v_0 \|\to 0,\\
\|  v_n-v_{n+p} \| \leq M(1-t_n) \| v_0 \|\to 0,
\end{gather*}
as $n\to \infty$,
where $M$ is the normality constant of $P$. 
So we can claim that ${u_n}$ and ${v_n}$ are Cauchy sequences. 
Since $E$ is complete, there exist $u^*,v^*$ such that
 $u_n\to u^*,v_n\to v^*,$ as $n\to \infty.$ By \eqref{t6}, we know that 
$u_n\leq u^*\leq v^*\leq v_n$ with $u^*,v^*\in P_h$ and
\[
\theta\leq v^*-u^*\leq v_n-u_n\leq (1-t_n)v_0.
\]
Further
\[
\| v^*-u^* \|\leq M(1-t_n)\| v_0\| \to 0 \quad (n\to\infty),
\]
and thus $u^*=v^*$. Let $x^*:=u^*=v^*$ and then we obtain
$$
u_{n+1}=A(u_n,v_n)\leq A(x^*,x^*)\leq A(v_n,u_n)=v_{n+1}.
$$
Let $n\to\infty$, then we obtain $x^*=A(x^*,x^*)$. 
That is, $x^*$ is a fixed point of $A$ in $P_h$.
Next we shall prove that $x^*$ is the unique fixed point of 
$A$ in $P_h$. In fact, suppose $\bar x$ is a fixed point of $A$ in $P_h$. 
Since $x^*,\bar x\in P_h$, there exists positive numbers 
$\bar\mu_1,\bar\mu_2,\bar\lambda_1,\bar\lambda_2>0$ such that
$$
\bar\mu_1h\leq x^*\leq \bar\lambda_1,\quad 
\bar\mu_2h\leq \bar x\leq \bar\lambda_2h.
$$
Then we obtain
$$
\bar x\leq \bar\lambda_2h=\frac{\bar\lambda_2}{\bar\mu_1}\bar\mu_1h
\leq \frac{\bar\lambda_2}{\bar\mu_1}x^*,\quad
\bar x\geq \bar\mu_2h=\frac{\bar\mu_2}{\bar\lambda_1}\bar\lambda_1h
\geq \frac{\bar\mu_2}{\bar\lambda_1}x^*.
$$
Let $e_1=\sup\{t>0|tx^*\leq \bar x\leq t^{-1}x^*\}$. 
Evidently, $0<e_1\leq 1,e_1x^*\leq \bar x\leq \frac{1}{e_1}x^*$. 
Next we prove $e_1=1$. If $0<e_1<1$, then
$\bar x=A(\bar x,\bar x)\geq A(e_1x^*,\frac{1}{e_1}x^*)$, then
\begin{align*}
2A(\bar x,\bar x)
&\geq 2A(e_1x^*,\frac{1}{e_1}x^*)=A(e_1x^*,\frac{1}{e_1}x^*)
 +A(e_1x^*,\frac{1}{e_1}x^*)\\
&\geq 2(\frac{\varphi(e_1)}{e_1})A(x^*,x^*).
\end{align*}
So we have
\[
A(\bar x,\bar x)
\geq (\frac{\varphi(e_1)}{e_1})A(x^*,x^*)\geq \frac{\varphi(e_1)}{e_1}A(x^*,x^*)
\geq \varphi(e_1)A(x^*,x^*).
\]
Since $\varphi(e_1)>e_1$, this contradicts the definition of $e_1$. 
Hence  $e_1=1$, and we obtain $\bar x=x^*$. Therefore, $A$ has a unique 
fixed point $x^*$ in $P_h$. Note that $[u_0,v_0]\subset P_h$, then we know 
that $x^*$ is the unique fixed point of $A$ in
 $[u_0,v_0]$.

 Now we construct the sequences recursively as follows:
 $$
x_n=A(x_{n-1},y_{n-1}), \quad
y_n=A(y_{n-1},x_{n-1}),\quad n=1,2,\ldots,
$$
  for any initial points $x_0,y_0\in P_h$. Since $x_0,y_0\in P_h$ we can 
choose small numbers $e_2,e_3\in (0,1)$ such that
 $$
 e_2h\leq x_0\leq \frac{1}{e_2}h,\quad 
e_3h\leq y_0\leq \frac{1}{e_3}h.
 $$
 Let $e^*=\min\{e_2,e_3\}$. Then $e^*\in (0,1)$ and
 $$
 e^*h\leq x_0,\quad y_0\leq \frac{1}{e^*}h.
 $$
 We can choose a sufficiently large positive integer $m$ such that
 $$
 \big[\frac{\varphi(e^*)}{e^*}\big]^m\geq \frac{1}{e^*}.
 $$
Put $\bar u_0={e^*}^mh,\bar v_0=\frac{1}{{e^*}^m}h$, it easy to see 
that $\bar u_0,\bar v_0\in P_h$ and $\bar u_0<x_0,y_0<\bar v_0$. Let
$$
 \bar u_n=A(\bar u_{n-1},\bar v_{n-1}),\quad
\bar v_n=A(\bar v_{n-1},\bar u_{n-1}),\quad  n=1,2,\ldots.
 $$
Analogously, it follows that there exists $y^*\in P_h$ such that 
$A(y^*,y^*)=y^*$ and \\
$\lim_{n\to \infty}\bar u_n=\lim_{n\to \infty}\bar v_n=y^*$. 
By the uniqueness of fixed point of operator $A$ in $P_h$. We get $x^*=y^*$ 
and by induction $\bar u_n\leq x_n,y_n\leq \bar v_n,~n=1,2,\ldots$. 
Since cone $P$ is normal we have $\lim_{n\to \infty} x_n=\lim_{n\to \infty} y_n=x^*$.
\end{proof}

\begin{theorem} \label{thm2.4}
Let $\alpha\in (0,1)$, $A:P\times P\to P$ be a mixed monotone operator 
satisfying
\begin{equation}\label{t8}
A(tx,t^{-1}y)+A(tx,s^{-1}y)\geq 2t^{2\alpha-1} A(x,y),
\quad s,t\in(0,1),\; s \leq t \; x,y\in P.
\end{equation}
Suppose that $B:P\to P$ is an in increasing sub-homogeneous operator. 
Assume also that
\begin{itemize}
\item[(i)] there is $h_0\in P_h$ such that $A(h_0,h_0)\in P_h$ and $Bh_0\in P_h$;
\item[(ii)] there exists a constant $\delta_0>0$ such that 
$A(x,y)\geq \delta_0Bx$ for all $x,y\in P$.
\end{itemize}
Then
\begin{itemize}
\item[(1)] $A:P_h\times P_h\to P_h,B:P_h\to P_h$;

\item[(2)] there exist $u_0,v_0\in P_h$ and $r\in(0,1)$ such that 
$$
rv_0\leq u_0<v_0,\quad u_0\leq A(u_0,v_0)+Bu_0\leq A(v_0,u_0)+Bv_0\leq v_0;
$$

\item[(3)] the operator $A(x,x)+Bx=x$ has a unique solution $x^*$ in $P_h$;

\item[(4)] for any initial values $x_0,y_0\in P_h$, constructing successively 
the sequences
$$
x_n = A(x_{n-1}, y_{n-1})+Bx_{n-1}, \quad
y_n = A(y_{n-1}, x_{n-1})+By_{n-1},\quad n=1,2,\ldots,
$$
we have $x_n\to x^*$ and $y_n\to x^*$ as $n\to\infty$.
\end{itemize}
\end{theorem}

\begin{proof}
Notice that from \eqref{t8} and Definition \ref{t9}, we have 
\begin{equation}\label{t10}
A(\frac{1}{t}x,ty)\leq \frac{1}{2t^{2\alpha-1}}(A(x,y)+A(x,\frac{t}{s}y)),
\end{equation}
and $B(\frac{1}{t}x)\leq \frac{1}{t}Bx$ for $s,t\in (0,1),x,y\in P$ and $s\leq t$.

Since $A(h_0,h_0),Bh_0\in P_h$, there exist constants 
$\lambda_1,\lambda_2,\nu_1,\nu_2>0$ such that 
$$
\lambda_1h\leq A(h_0,h_0)\leq \lambda_2h,\quad \nu_1h\leq Bh_0\leq \nu_2h\,.
$$
Also from $h_0\in P_h$, there exists a constant $t_0\in (0,1)$ such that 
$t_0h\leq h_0\leq\frac{1}{t_0}h$, and let $s_0\in(0,1)$ such that 
$s_0\leq t_0$,  then we have
$$
s_0h\leq t_0h\leq h_0\leq \frac{1}{t_0}h\leq \frac{1}{s_0}h.
$$
From $s_0\leq t_0$, \eqref{t8}, \eqref{t10} and the mixed monotone properties 
of operator $A$, we have
$$
A(h,h)\geq A(t_0h_0,\frac{1}{t_0}h_0),\quad
A(h,h)\geq A(t_0h_0,\frac{1}{s_0}h_0).
$$
So we have
$$
2A(h,h)\geq 2t_0^{2\alpha-1} A(h_0,h_0)\,.
$$
By combining the inequalities above, we have
\[
A(h,h)\geq t_0^{2\alpha-1} A(h_0,h_0)
\geq t_0^{2\alpha} A(h_0,h_0)\geq \lambda_1 t_0^{2\alpha} h,
\]
and
\begin{align*}
A(h,h)&\leq  A(\frac{1}{t_0}h_0,t_0h_0)
 \leq \frac{1}{2t_0^{2\alpha-1}}(A(h_0,h_0)+A(h_0,\frac{t_0}{s_0}h_0))\\
&\leq \frac{1}{t_0^{2\alpha}}A(h_0,h_0)\leq \frac{\lambda_2}{t_0^\alpha}h.
\end{align*}
Noting that $\frac{\lambda_2}{t_0^{2\alpha}},\lambda_1 t_0^2\alpha>0$, 
we can get $A(h,h)\in P_h$.  By Definition \ref{t9} and the monotone 
property of operator $B$, we have 
$$
Bh\leq B(\frac{1}{t_0}h_0)\leq \frac{1}{t_0}Bh_0\leq \frac{\nu_2}{t_0}h,\quad
Bh\geq B(t_0h_0)\geq t_0Bh_0\geq\nu_1t_0h.
$$
Next we show $B:P_h\to P_h$. For any $x\in P_h$, we can choose a sufficiently 
small number $\mu\in (0,1)$ such that 
\[
\mu h\leq x\leq \frac{1}{\mu}h.
\]
Consequently, 
$$
Bx\leq B(\frac{1}{\mu}h)\leq \frac{1}{\mu}\frac{\nu_2}{t_0}h,\quad
Bx\geq B(\mu h)\geq \mu t_0\nu_1h.
$$
Evidently, we have $\frac{\nu_2}{\mu t_0},\mu t_0 \nu_1>0$. Thus $Bx\in P_h$; 
that is, $B:P_h\to P_h$. So the conclusion $(1)$ holds.
Now we define an operator $T=A+B$ by $T(x,y)=A(x,y)+Bx$. Then 
$T:P\times P\to P$ is a mixed monotone operator and $T(h,h)\in P_h$. 
In the following we show that there exists $\varphi(t)\in (t,1]$ 
with respect to $s,t\in (0,1),s\leq t$ such that
$$
T(tx,t^{-1}y)+T(tx,s^{-1}y)\geq 2(\frac{\varphi(t)}{t})A(x,y),\quad
\forall,x,y\in P.
$$
Consider the function 
\[
f(t)=\frac{t^{2\beta-1}-t}{t^{2\alpha-1}-t^{2\beta-1}},
\]
for $t\in (0,1)$, where $\beta\in (\alpha,1)$.
It is easy to prove that $f$ is increasing in $(0,1)$ and
$$ 
\lim_{t\to 0^+}f(t)=0,\quad \lim_{t\to1^{-}}f(t)=\frac{1-\beta}{\beta-\alpha}.
$$
Further, fixing $t\in(0,1)$, we have
$$
\lim_{\beta\to1^{-}}f(t)=\lim_{\beta\to1^{-}}
\frac{t^{2\beta-1}-t}{t^{2\alpha-1}-t^{2\beta-1}}=0.
$$
So there exists $\beta_0(t)\in (0,1)$ with respect to $t$ such that
$$
\frac{t^{2{\beta_0(t)}-1}-t}{t^{2\alpha-1}-t^{2{\beta_0(t)}-1}}\leq \delta_0,
\quad t\in(0,1).
$$
Hence we have
$$
A(x,y)\geq \delta_0Bx
\geq \frac{t^{2\beta_0(t)-1}-t}{t^{2\alpha-1}-t^{2\beta_0(t)-1}}Bx,\quad
\forall t\in (0,1),~x,y\in P.
$$
Then we obtain
$$
t^{2\alpha-1 }A(x,y)+tBx\geq t^{2\beta_0(t)-1}[A(x,y)+Bx],\quad 
\forall t\in (0,1),\; x,y\in P.
$$
Consequently, for any $t\in(0,1)$ and $x,y\in P$,
\begin{align*}
T(tx,t^{-1}y)+T(tx,s^{-1}y)
&=A(tx,t^{-1}y)+B(tx)+A(tx,s^{-1}y)+B(tx)\\
&\geq 2t^{2\alpha-1} A(x,y)+2tBx\\
&\geq 2t^{2\beta_0(t)-1}( A(x,y)+Bx)\\
&=2t^{2\beta_0(t)-1}T(x,y).
\end{align*}
Let $\varphi(t)=t^{2\beta_0(t)},~~t\in (0,1)$. Then $\varphi(t)\in (t^{2},1]$ and
$$
T(tx,t^{-1}y)+T(tx,s^{-1}y)\geq 2(\frac{\varphi(t)}{t})A(x,y),
$$
 for any $s,t\in (0,1)$ and $x,y\in P$.
Hence the condition (A2) in Lemma \ref{t2} is satisfied. 
By Lemma  \ref{t2} we conclude that: 
(a) there exist $u_0,v_0\in P_h$ and $r\in (0,1)$ such that 
$rv_0\leq u_0<v_0,u_0\leq T(u_0,v_0)\leq T(v_0,u_0)\leq v_0$; 
(b) $T$ has a unique fixed point $x^*$ in $P_h$; 
(c) for any initial values $x_0,y_0\in P_h$, constructing successively 
the sequences
$$
x_n=T(x_{n-1},y_{n-1}),\quad  y_n=T(y_{n-1},x_{n-1}),\quad n=1,2,\dots, 
$$
we have $x_n\to x^*$ and $y_n\to x^*$ as $n\to \infty$. That is, 
 conclusions (2)--(4) hold.
\end{proof}

\begin{corollary}
Let $\alpha\in (0,1),~A:P\times P\to P$ is a mixed monotone operator. 
Assume \eqref{t8} holds and there is $h_0>\theta$ such that $A(h_0,h_0)\in P_h$.
Then
\begin{itemize}
\item[(1)] $A:P_h\times P_h\to P_h$;

\item[(2)] there exist $u_0,v_0\in P_h$ and $r\in(0,1)$ such that 
$$
rv_0\leq u_0<v_0,\quad
u_0\leq A(u_0,v_0)\leq A(v_0,u_0)\leq v_0;
$$

\item[(3)] the operator $A(x,x)=x$ has a unique solution $x^*$ in $P_h$;

\item[(4)] for any initial values $x_0,y_0\in P_h$, constructing successively 
the sequences
$$
x_n = A(x_{n-1}, y_{n-1}), \quad y_n = A(y_{n-1}, x_{n-1}),\quad n=1,2,\ldots,
$$
we have $x_n\to x^*$ and $y_n\to x^*$ as $n\to\infty$.
\end{itemize}
\end{corollary}

\section{Solution to fractional differential equations}

In this section, we shall propose a method for showing the existence and uniqueness 
of a solution for the fractional differential equation
\begin{equation}\label{21}
\begin{gathered}
\frac{D^{\alpha}}{Dt}u(s,t)+f(s,t,u(s,t),v(s,t))=0, \\
0<\epsilon<T,\quad T\geq 1, \quad t\in [\epsilon,T],\quad
0<\alpha<1,\quad s\in [a,b]
\end{gathered}
\end{equation}
subject to the condition
\begin{equation}\label{22}
u(s,\zeta)=u(s,T),\quad (s,\zeta)\in [a,b]\times(\epsilon,t),
\end{equation}
where $D^{\alpha}$ is the Riemann-Liouville fractional derivative of 
order $\alpha$. We will suppose that $a,b\in (0,\infty),~a<b$. Let
$$
E=C({[a,b]}\times{[\epsilon,T]}).
$$
Consider the Banach space of  continuous functions on 
${[a,b]}\times{[\epsilon,T]}$ with sup norm and set
$$
P=\{y\in C({[a,b]}\times{[\epsilon,T]}):
\min_{(s,t)\in [a,b]\times[\epsilon,T]}y(s,t)\geq 0\}.
$$
Then $P$ is a normal cone.

\begin{lemma} \label{23}
Let $(s,t)\in [a,b]\times[\epsilon,T],(s,\zeta)\in [a,b]\times(\epsilon,t)$ 
and $0<\alpha<1$. Then the problem
$$
\frac{D^{\alpha}}{Dt}u(s,t)+f(s,t,u(s,t),v(s,t))=0
$$
with the boundary value condition
 $u(s,\zeta)=u(s,T)$ has a solution $u_0$ if and only if  $u_0$ is a 
solution of the  fractional integral equation
$$
u(s,t)=\int_{\epsilon}^TG(t,\xi)f(s,\xi,u(s,\xi),v(s,t))d{\xi},
$$
where
\[
G(t,\xi)=\begin{cases}
\frac{t^{\alpha-1}(\zeta-\xi)^{\alpha-1}
 -t^{\alpha-1}(T-\xi)^{\alpha-1}}{({\zeta}^{\alpha-1}-T^
{\alpha-1})\Gamma(\alpha)}-\frac{(t-\xi)^{\alpha-1}}{\Gamma(\alpha)},
 &\epsilon\leq \xi\leq \zeta\leq t\leq T,\\[4pt]
\frac{-t^{\alpha-1}-(T-\xi)^{\alpha-1}}{({\zeta}^{\alpha-1}-T^
{\alpha-1})\Gamma(\alpha)}-\frac{(t-\xi)^{\alpha-1}}{\Gamma(\alpha)},
&\epsilon\leq \zeta\leq \xi\leq t\leq T,\\[4pt]
\frac{-t^{\alpha-1}(T-\xi)^{\alpha-1}}{({\zeta}^{\alpha-1}-T^
{\alpha-1})\Gamma(\alpha)},&\epsilon\leq \zeta\leq t\leq \xi\leq T.
\end{cases}
\]
\end{lemma}

\begin{proof}
From $\frac{D^{\alpha}}{Dt}u(s,t)+f(s,t,u(s,t),v(s,t))=0$ and the boundary 
condition, it is easy to see that 
$u(s,t)-c_1t^{\alpha-1}=-I_{\epsilon}^{\alpha}f(s,t,u(s,t),v(s,t))$. 
By the definition of a fractional integral, we obtain
\begin{gather*}
u(s,t)=c_1t^{\alpha-1}-\int_{\epsilon}^{\zeta}
 \frac{(t-\xi)^{\alpha-1}}{\Gamma(\alpha)}
f(s,\xi,u(s,\xi),v(s,\xi))d\xi,
\\
u(s,\zeta) =c_1T^{\alpha-1}-\int_{\epsilon}^{\zeta}
 \frac{(\zeta-\xi)^{\alpha-1}}{\Gamma(\alpha)}
f(s,\xi,u(s,\xi),v(s,\xi))d\xi,
\\
u(s,T)=c_1T^{\alpha-1}-\int_{\epsilon}^{T}\frac{(t-\xi)^{\alpha-1}}{\Gamma(\alpha)}
f(s,\xi,u(s,\xi),v(s,\xi))d\xi.
\end{gather*}
Since $u(s,\zeta)=u(s,T)$, we obtain
\begin{align*}
c_1&=\frac{1}{{\zeta}^{\alpha-1}-T^{\alpha-1}}
 \int_{\epsilon}^{\zeta}\frac{(\zeta-\xi)^{\alpha-1}}
{\Gamma(\alpha)}f(s,\xi,u(s,\xi),v(s,\xi))d\xi\\
&\quad -\frac{1}{{\zeta}^{\alpha-1}-T^{\alpha-1}}
 \int_{\epsilon}^{T}\frac{(T-\xi)^{\alpha-1}}
{\Gamma(\alpha)}f(s,\xi,u(s,\xi),v(s,\xi))d\xi.
\end{align*}
Hence
\begin{align*}
u(s,t)
&=\frac{t^{\alpha-1}}{{\zeta}^{\alpha-1}-T^{\alpha-1}}\int_{\epsilon}^{\zeta}
\frac{(\zeta-\xi)^{\alpha-1}}{\Gamma(\alpha)}f(s,\xi,u(s,\xi),v(s,\xi))d\xi\\
&\quad -\frac{t^{\alpha-1}}{{\zeta}^{\alpha-1}-T^{\alpha-1}}
 \int_{\epsilon}^{T}\frac{(T-\xi)^{\alpha-1}}
{\Gamma(\alpha)}f(s,\xi,u(s,\xi),v(s,\xi))d\xi\\
&\quad -\int_{\epsilon}^{t}\frac{(t-\xi)^{\alpha-1}}{\Gamma(\alpha)}
 f(s,\xi,u(s,\xi),v(s,\xi) )d\xi\\
&=\int_{\epsilon}^{T}G(t,\xi)f(s,\xi,u(s,\xi),v(s,\xi))d\xi.
\end{align*}
This completes the proof.
\end{proof}

\begin{theorem}\label{0}
Let $0<\epsilon<T$  and let
$f(s,t,u(s,t),v(s,t))$ be function in the space 
$C([a,b],[\epsilon,T],[0,\infty],[0,\infty])$,
that is increasing in $u$, decreasing in $v$, with positive values.
Also assume that for any $u,v\in P$ and $c,c'\in (0,1)$ with  $c'\leq c$,
there exists $\varphi(c)\in ({c}^2,1]$ and $\varphi$ is decreasing such that
\begin{align*} %\label{nt1}
&\int_{\epsilon}^TG(t,\xi)f(s,\xi,cu(s,\xi),c^{-1}v(s,\xi))d{\xi}
 +\int_{\epsilon}^TG(t,\xi)f(s,\xi,cu(s,\xi),{c'}^{-1}v(s,\xi))d{\xi}\\
&\geq 2\frac{\varphi(c)}{c}\int_{\epsilon}^TG(t,\xi)f(s,\xi,u(s,\xi),
v(s,\xi))d{\xi},
\end{align*}
and $f(s,t,u(s,t),v(s,t))=0$, whenever $G(s,t)<0$. Also assume that there
 exist $M_1>0$, $M_2>0$ and $h\neq \theta\in P$ such that
$$
M_1h(t)\leq\int_{\epsilon}^TG(t,\xi)f(s,\xi,u(s,\xi),v(s,\xi))d{\xi
}\leq M_2h(t),
$$
for all $t\in [\epsilon,T]$, where $G(t,s)$ is the green function defined in 
Lemma \ref{23}.
Then  problem \eqref{21} with the boundary value condition \eqref{22} 
has unique solution $u^*$.
\end{theorem}

\begin{proof}
By using Lemma \eqref{23}, the problem is equivalent to the integral equation
\[
u(s,t)=\int_{\epsilon}^TG(t,\xi)f(s,\xi,u(s,\xi),v(s,\xi))d{\xi},
\]
where
\[ 
G(t,\xi)=\begin{cases}
\frac{t^{\alpha-1}(\eta-\xi)^{\alpha-1}-t^{\alpha-1}
 (T-\xi)^{\alpha-1}}{({\eta}^{\alpha-1}-T^
{\alpha-1})\Gamma(\alpha)}-\frac{(t-\xi)^{\alpha-1}}{\Gamma(\alpha)}
 &\epsilon\leq \xi\leq \eta\leq t\leq T\\[4pt]
\frac{-t^{\alpha-1}-(T-\xi)^{\alpha-1}}{({\eta}^{\alpha-1}-T^
{\alpha-1})\Gamma(\alpha)}-\frac{(t-\xi)^{\alpha-1}}{\Gamma(\alpha)}
 &\epsilon\leq \eta\leq \xi\leq t\leq T\\[4pt]
\frac{-t^{\alpha-1}(T-\xi)^{\alpha-1}}{({\eta}^{\alpha-1}-T^
{\alpha-1})\Gamma(\alpha)}&\epsilon\leq \eta\leq t\leq \xi\leq T
\end{cases}
\]
Define the operator $A:P\times P\to E$ by 
\[
 A(u(s,t),v(s,t))=\int_{\epsilon}^TG(t,\xi)f(s,\xi,u(s,\xi),v(s,\xi))d{\xi}.
\]
 Then $u$ is solution for the problem if and only if $u=A(u,u)$. 
It is easy to see to check that the operator $A$ is increasing in $u$ 
and decreasing in $v$ on $P$. By assumptions of theorem we have;
\begin{itemize}
\item[(A7)] there exists $h\in P$ with $h\neq \theta$ such that 
$$
M_1h(t)\leq\int_{\epsilon}^TG(t,\xi)f(s,\xi,u(s,\xi),v(s,\xi))d{\xi}\leq M_2h(t),
$$
thus
$A(h,h)\in P_h$,

\item[(A8)] for any $u,v\in P$ and $c,c'\in (0,1)$ such that $c'\leq c$ , 
there exists $\varphi(c)\in ({c}^2,1]$ and $\varphi$ is decreasing such that
$$
A(cu,c^{-1}v)+A(cu,{c'}^{-1}v)\geq 2\frac{\varphi(c)}{c}A(u,v).
$$
\end{itemize}
Now by using theorem \eqref{a}, the operator $A$ has a unique fixed point 
$u^*$ in $P_h$. Therefore the boundary value problem \eqref{21} has 
unique solution $u^*$.
\end{proof}


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\end{document}