\documentclass[reqno]{amsart}
\usepackage{hyperref}

\AtBeginDocument{{\noindent\small
\emph{Electronic Journal of Differential Equations},
Vol. 2012 (2012), No. 70, pp. 1--8.\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{8mm}}

\begin{document}
\title[\hfilneg EJDE-2012/70\hfil Uniqueness of positive solutions]
{Uniqueness of positive solutions for a fractional differential equation
 via a fixed point theorem of a sum operator}

\author[C. Yang, C. Zhai\hfil EJDE-2012/70\hfilneg]
{Chen Yang, Chengbo Zhai}  % in alphabetical order

\address{Chen Yang \newline
 Department of Mathematics,  
 Business College of Shanxi University, Taiyuan  030031 Shanxi,  China}
\email{yangchen0809@126.com}

\address{Chengbo Zhai \newline
 School of Mathematical Sciences, 
 Shanxi University, Taiyuan  030006  Shanxi, China}
\email{cbzhai@sxu.edu.cn, cbzhai215@sohu.com}

\thanks{Submitted January 27, 2012. Published May 7, 2012.}
\thanks{Supported by  grant 2010021002-1 from the Youth Science
 Foundation of Shanxi Province}
\subjclass[2000]{34B18}
\keywords{Riemann-Liouville fractional derivative; positive solution;
\hfill\break\indent 
fractional differential equation; existence and uniqueness; 
fixed point theorem}

\begin{abstract}
 In this work, we study the existence and uniqueness of positive
 solutions for nonlinear fractional differential equation
 boundary-value problems. Our analysis relies on a fixed point
 theorem of a sum operator. Our results guarantee  the existence
 of a  unique  positive solution, and can be applied for constructing
 an iterative scheme for obtaining the solution.
\end{abstract}

\maketitle
\numberwithin{equation}{section}
\newtheorem{theorem}{Theorem}[section]
\newtheorem{lemma}[theorem]{Lemma}
\newtheorem{remark}[theorem]{Remark}
\newtheorem{definition}[theorem]{Definition}
\newtheorem{example}[theorem]{Example}
\allowdisplaybreaks

\section{Introduction}

Fractional differential equations arise in many fields, such
as physics, mechanics, chemistry, economics, engineering and
biological sciences,etc; see \cite{d1,g1,g2,k2,m1,m2,m3,o1,p1,r1} 
for example. In the recent years,
there has been a significant development in ordinary and partial
differential equations involving fractional derivatives, see the
monographs of  Miller and Ross \cite{m3},
 Podlubny \cite{p1}, Kilbas et al \cite{k2},
and the articles \cite{b1,b2,j1,k1,k3,l1,l2,x1,y1,w1,z2,z3,z4,z5}
and the references therein. In these papers,
many authors have investigated the existence of positive solutions
for nonlinear fractional differential equation boundary value
problems. On the other hand, the uniqueness  of positive solutions
for nonlinear fractional differential equation boundary value
problems has been studied by  some authors, see \cite{y1,z3,z4,z5}
for example.

By means of a fixed point theorem for mixed monotone
operators, Xu, Jiang and Yuan \cite{x1}
considered the existence and the
uniqueness of positive solutions for the following
\begin{equation}
\begin{gathered}
D^\alpha_{0+} u(t)=f(t,u(t)),\quad 0<t<1,\; 3<\alpha\leq4,\\
u(0)=u(1)=u'(0)=u'(1)=0,
\end{gathered}
\label{e1.1}
\end{equation}
where $f(t,u)=q(t)[g(u)+h(u)]$, $g:[0,+\infty)\to[0,+\infty)$
is continuous and nondecreasing, $h:(0,+\infty)\to(0,+\infty)$ is
continuous and nonincreasing, and  $q\in C((0,1),(0,+\infty))$ satisfies
$$
\int^1_0 s^{2-\eta(2-\alpha)}(1-s)^{\alpha-2-2\eta}q(s)ds<+\infty,\quad
\eta\in (0,1).
$$
By a similar method,  Zhang  \cite{z3} studied a unique positive
solution for the singular boundary value problem
\begin{equation}
\begin{gathered}
D^\alpha_{0+} u(t)+q(t)f(t,u,u',\dots, u^{(n-2)})=0,\quad
0<t<1,\; n-1<\alpha\leq n,\; n\geq 2,\\
u(0)=u'(0)=\dots=u^{(n-2)}(0)=u^{(n-2)}(1)=0,
\end{gathered} \label{e1.2}
\end{equation}
where $f=g+h$ and $g,h$ have different monotone properties.

By means of a fixed point theorem for $u_0$ concave
operators, Yang and Chen \cite{y1} investigated the existence and  uniqueness
of positive solutions for the following boundary value problem
\begin{equation}
\begin{gathered}
D^\alpha_{0+} u(t)+f(t,u,u',\dots, u^{(n-2)})=0,\quad 0<t<1,\; n-1<\alpha\leq n,\; n\geq 2,\\
u(0)=u'(0)=\dots=u^{(n-2)}(0)=u^{(n-2)}(1)=0,
\end{gathered}\label{e1.3}
\end{equation}
where $f\in C([0,1]\times[0,+\infty)\times
R^{n-2}\to[0,+\infty))$, $f(t,y_1,y_2,\dots, y_{n-1}) $
is increasing for $y_i \geq 0, i=1,2,\dots n-1$, and
$ f\not \equiv 0$.

Different from the works mentioned above, we will use a fixed point
theorem for a sum operator  to show the existence and  uniqueness of
positive solutions for the following  fractional equation boundary
value problem
\begin{equation}
\begin{gathered}
D^\alpha_{0+} u(t)+f(t,u(t))+g(t,u(t))=0,\quad 0<t<1,\; 1<\alpha\leq 2,\\
u(0)=u(1)=0.
\end{gathered}\label{e1.4}
\end{equation}
Moreover, we can construct a sequence for approximating the unique
solution. It must be pointed out that the method used in this
article can be applied to \eqref{e1.1}-\eqref{e1.3}.

\section{Preliminaries}

 For the convenience of the reader, we present here some
definitions, lemmas and basic results that will be used in the proof
of our theorem.

\begin{definition}[{\cite[Definiton 2.1]{s1}}] \label{def2.1} \rm
The integral
$$
I^\alpha_{0+}f(x)=\frac 1{\Gamma(\alpha)} \int^x_0 \frac {f(t)}
{(x-t)^{1-\alpha}} dt,\quad x>0
 $$
is called the Riemann-Liouville fractional integral of order $\alpha$, where $\alpha>0$ and $\Gamma(\alpha)$ denotes the gamma function.
\end{definition}

\begin{definition}[{\cite[page 36-37]{s1}}]  \label{def2.2} \rm
For a function $f(x)$ given in the interval  $[0,\infty)$, the expression
$$
D^\alpha_{0+}f(x)=\frac 1{\Gamma(n-\alpha)}
\big(\frac d {dx}\big)^n \int^x_0 \frac {f(t)}
{(x-t)^{\alpha-n+1}} dt,
 $$
where $n=[\alpha]+1,[\alpha]$ denotes the
integer part of number $\alpha$, is called the Riemann-Liouville
fractional derivative   of order $\alpha$.
\end{definition}

\begin{lemma}[\cite{b1}] \label{lem2.3}
 Given $y \in C[0,1]$ and $1<\alpha\leq 2$,
the  boundary-value problem
\begin{equation}
\begin{gathered}
D^\alpha_{0+} u(t)+y(t)=0,\quad 0<t<1,\\
u(0)=u(1)=0,
\end{gathered} \label{e2.1}
\end{equation}
has a unique solution
$$
u(t)=\int^1_0 G(t,s)y(s)ds,
$$
where
$$
G(t,s)=\frac 1 {\Gamma(\alpha)}
 \begin{cases}
 [t(1-s)]^{\alpha-1}-(t-s)^{\alpha-1},& 0\leq s\leq t\leq 1,\\
 {[t(1-s)]}^{\alpha-1}, &  0\leq t \leq s\leq 1,
 \end{cases}
$$
which is the Green function for this  boundary-value problem.
\end{lemma}

In \cite{j1}, the authors obtained the following result.

\begin{lemma} \label{lem2.4}
Let $1<\alpha\leq 2$. Then the Green function
$G(t,s)$ in Lemma \ref{lem2.3}  satisfies
$$
\frac {\alpha-1}
{\Gamma(\alpha)}h(t)(1-s)^{\alpha-1}s\leq G(t,s)\leq \frac 1
{\Gamma(\alpha)}h(t)(1-s)^{\alpha-2} \quad \text{for } t,s \in[0,1],
$$
where $h(t)=t^{\alpha-1}(1-t)$, $t\in [0,1]$.
\end{lemma}

In the sequel, we present some basic concepts in ordered Banach
spaces for completeness and a fixed-point theorem which we will be
used later. For convenience of readers, we suggest that one refer to
\cite{g3,z1} for details.

Suppose that $(E,\|\cdot\|)$ is a real Banach space which is
partially ordered by a cone $P\subset E$; i.e.,
$ x\leq y $ if and only if $y-x\in P$.
If $x\leq y$ and $x\neq y$, then we denote $x<y$ or
$y>x$. By $\theta$ we denote the zero element of $E$. Recall that a
non-empty closed convex set  $P\subset E$ is a cone if it satisfies
(i) $x\in P,\lambda\geq 0$ implies $\lambda x\in P$;
(ii) $x\in P$ and $-x\in P$ imply $x=\theta$.

Putting $\mathring{P}=\{x\in P:  x\text{ is an interior point of }P\}$,
a cone $P$ is said to be solid if  $\mathring{P}$ is
non-empty. Moreover, $P$ is called normal if there exists a constant
$N>0$ such that, for all $x,y\in E$,
$\theta\leq x\leq y$ implies $\|x\|\leq N\|y\|;$  in this case $N$
is called the normality constant of $P$.
 We say that an operator $A:E\to E$ is
increasing if $x\leq y$ 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 equivalence relation.  Given
 $h>\theta$ (i.e., $h\geq \theta$ and $h\neq \theta$), we denote by
 $P_h$ the set $P_h=\{x\in E|\ x\sim h\}$.
It is easy to see that $P_h\subset P$.

\begin{definition} \label{def2.5} \rm
Let $D=P$ or $D=\mathring{P}$ and $\beta$ be a
real number with $0\leq \beta<1$. An operator $A:P\to P$ is
said to be $\beta$-concave if it satisfies
$$
A(tx)\geq t^\beta Ax, \forall\ t\in (0,1),\ x\in D.
$$
Notice that the definition of a $\beta$-concave operator mentioned
above is different from that in \cite{g3}, because we need not
require the cone to be solid in general.
\end{definition}

\begin{definition} \label{def2.6}\rm
An operator $A:E\to E$ is said to be homogeneous if it satisfies
$$
A(\lambda x)=\lambda Ax, \quad \forall \lambda>0,\; x\in E.
$$
An operator $A:P\to P$ is said to be sub-homogeneous if it
satisfies
$$
A(tx)\geq t Ax, \forall\ t\in (0,1),\ x\in P.
$$
\end{definition}

In a recent paper Zhai and Anderson \cite{z1} considered the
sum operator equation
$$
Ax+Bx+Cx=x,
$$
where $A$ is an increasing $\beta$-concave operator,
$B$ is an increasing sub-homogeneous
operator and $C$ is a homogeneous operator. They established the
existence and uniqueness of positive solutions for the above
equation, and when $C$ is a null operator, they present the
following interesting result.

\begin{lemma} \label{lem2.7}
Let $P$ be a normal cone in a real Banach space
$E$, $A:P\to P$ be an increasing $\beta$-concave operator
and $B:P\to P$ be an increasing sub-homogeneous operator.
Assume that
\begin{itemize}
\item[(i)] there is $h>\theta$ such that $Ah\in P_h$ and $Bh\in P_h;$

\item[(ii)] there exists a constant $\delta_0>0$ such that
$Ax\geq \delta_0Bx$, for all $x\in P$.
\end{itemize}
Then  the operator equation $Ax+Bx=x$ has a unique solution $x^*$ in
$P_h$. Moreover, constructing successively the sequence
$y_n=Ay_{n-1}+By_{n-1}, n=1,2,\dots$ for any initial value
$y_0\in P_h$, we have $y_n\to x^*$ as $n\to \infty$.
\end{lemma}

\section{Existence and uniqueness of positive solutions for \eqref{e1.4}}

In this section, we apply Lemma  \ref{lem2.7} to study problem \eqref{e1.4}
and we obtain a new result on the existence and uniqueness of
positive solutions. The method used here is new to the literature
and so is the existence and uniqueness result to the fractional
differential equations.

In our considerations we will work in the Banach space
 $C[0,1]=\{x: [0,1]\to  \mathbf{R} \text{ is continuous}\}$ with the
standard norm $\|x\| = \sup\{|x(t)|:t\in [0,1]\}$. Note that this
space can be equipped with a partial order given by
$$
x,y \in C[0,1], x \leq y \Leftrightarrow x(t)\leq y(t) \text{ for }
 t\in [0,1].
$$
Set $ P=\{x\in C[0,1]| x(t)\geq 0,\; t\in [0,1]\}$, the
standard cone. It is clear that $P$ is a normal cone in $C[0,1]$ and
the normality constant is 1.  Our main result is summarized in the
following theorem using the following assumptions:
\begin{itemize}
\item[(H1)] $f,g:[0,1]\times [0,+\infty)\to [0,+\infty)$ are
continuous and increasing respect to the second argument,
$g(t,0)\not\equiv 0$;

\item[(H2)]   $g(t,\lambda x)\geq \lambda
g(t,x)$ for  $\lambda\in (0,1)$, $t\in [0,1]$, $x\in [0,+\infty)$, and
there exists a constant $\beta\in (0,1)$ such that
 $f(t,\lambda x)\geq \lambda^\beta f(t,x)$, for all
$t\in [0,1]$, $\lambda\in (0,1)$, $x\in [0,+\infty)$;

\item[(H3)]   there exists a constant $\delta_0>0$ such that
 $f(t,x)\geq \delta_0g(t,x)$, $t\in [0,1]$, $x\geq 0$.
\end{itemize}

\begin{theorem} \label{thm3.1}
Under assumptions {\rm (H1)--(H3)},  problem \eqref{e1.4} has a unique positive solution $u^*$ in
$P_h$, where $h(t)=t^{\alpha-1}(1-t)$, $t\in [0,1]$.
 Moreover, for any initial value  $u_0\in P_h$,  the sequence
$$
u_{n+1}(t)=\int^1_0 G(t,s)f(s,u_n(s))ds+\int^1_0 G(t,s)g(s,u_n(s))ds,\quad n=0,1,2,\dots
$$
satisfies $u_n(t)\to u^*(t)$  as $n\to \infty$.
\end{theorem}

\begin{proof}
From \cite{b1}, problem \eqref{e1.4} has the
integral formulation
$$
u(t)=\int_0^1G(t,s)[f(s,u(s))+g(s,u(s))]ds,
$$
where $G(t,s)$ is given as in Lemma  \ref{lem2.3}.
Define two operators $A:P\to E$ and $ B:P\to E$ by
$$
Au(t)=\int_0^1G(t,s)f(s,u(s)))ds,\quad
Bu(t)=\int_0^1G(t,s)g(s,u(s))ds.
$$
It is easy to prove that $u$ is the solution of  \eqref{e1.4} if
and only if $u=Au+Bu$. From (H1) and Lemma  \ref{lem2.4}, we know   that
$A:P\to P$ and $B:P\to P$. In the sequel we check
that $A,B$ satisfy all assumptions of Lemma  \ref{lem2.7}.

Firstly, we prove that $A,B$ are two increasing operators. In fact,
by (H1) and Lemma  \ref{lem2.4}, for $u,v\in P$ with $u\geq v$, we know
that $u(t)\geq v(t)$, $t\in [0,1]$ and obtain
$$
Au(t)=\int_0^1G(t,s)f(s,u(s))ds\geq
\int_0^1G(t,s)f(s,v(s))ds=Av(t).
$$
 That is, $Au\geq Av$. Similarly, $Bu\geq Bv$.
Next we show that $A$ is a $\beta$-concave operator and $B$ is a
sub-homogeneous operator. In fact, for any $\lambda \in (0,1)$ and
$u\in P$, by $(H_2)$ we obtain
\[
A(\lambda u)(t)=\int^1_0G(t,s)f(s,\lambda u(s)) ds \geq
\lambda^\beta\int^1_0G(t,s)f(s,u(s)) ds=\lambda^\beta Au(t).
\]
That is, $A(\lambda u)\geq \lambda^\beta Au $ for
 $ \lambda \in (0,1)$, $u\in P$. So the operator $A$ is a $\beta$-concave
operator.
Also, for any $\lambda \in (0,1)$ and $u\in P$, from (H2) we know
that
$$
B(\lambda u)(t)=\int_0^1G(t,s)g(s,\lambda u(s))ds\geq \lambda\int_0^1G(t,s)g(s,u(s))ds=\lambda
Bu(t);
$$
that is, $B(\lambda u)\geq \lambda Bu$ for
$ \lambda \in (0,1),\ u\in P$. So the operator $B$ is sub-homogeneous.

Now we show that $Ah\in P_h$ and $Bh\in P_h$. Let
$h_{\rm max}=\max\{h(t)=t^{\alpha-1}(1-t):t\in [0,1]\}$.
Then $h_{\rm max}>0$. From (H1) and Lemma  \ref{lem2.4},
\begin{gather*}
Ah(t)=\int_0^1G(t,s)f(s,h(s))ds\leq\frac
1{\Gamma(\alpha)}h(t)\int_0^1(1-s)^{\alpha-2}f(s,h_{\rm max})ds ,
\\
Ah(t)=\int_0^1G(t,s)f(s,h(s))ds\geq \frac
{\alpha-1}{\Gamma(\alpha)}h(t)\int_0^1s(1-s)^{\alpha-1}f(s,0)ds.
\end{gather*}
From (H1) and (H3), we have
$$
f(s,h_{\rm max})\geq f(s,0)\geq \delta_0g(s,0)\geq 0.
$$
Since $g(t,0)\not\equiv 0$, we can obtain
$$
\int_0^1f(s,h_{\rm max})ds\geq \int^1_0f(s,0)ds\geq \delta_0\int^1_0g(s,0)ds> 0,
$$
and in consequence,
\begin{gather*}
l_1:=\frac {\alpha-1}{\Gamma(\alpha)}\int_0^1s(1-s)^{\alpha-1}f(s,0)ds>0,\\
l_2:=\frac 1{\Gamma(\alpha)}\int_0^1(1-s)^{\alpha-2}f(s,h_{\rm max})ds>0.
\end{gather*}
So $l_1h(t)\leq Ah(t)\leq l_2h(t)$, $t\in [0,1]$;  and hence we have
$Ah\in P_h$.

Similarly,
$$
\frac{\alpha-1}{\Gamma(\alpha)}h(t)\int_0^1s(1-s)^{\alpha-1}g(s,0)ds\leq
Bh(t)\leq\frac 1{\Gamma(\alpha)}h(t)\int_0^1(1-s)^{\alpha-2}g(s,h_{\rm max})ds,
$$
from $g(t,0)\not\equiv 0$, we easily prove  $Bh\in P_h$. Hence the
condition (i) of Lemma  \ref{lem2.7} is satisfied.

In the following we show the condition (ii) of Lemma  \ref{lem2.7} is
satisfied. For $u\in P$, from  (H3),
$$
Au(t)=\int_0^1G(t,s)f(s,u(s))ds\geq
\delta_0\int^1_0G(t,s)g(s,u(s))ds =\delta_0Bu(t).
$$
 Then we get $Au\geq \delta_0Bu, u\in P$. Finally, an application of 
Lemma  \ref{lem2.7}
 implies: the  operator equation $Ax+Bx=x$ has a unique solution $u^*$ in $P_h$.
Moreover, constructing successively the sequence
$y_n=Ay_{n-1}+By_{n-1}, n=1,2,\dots$ for any initial value
 $y_0\in P_h$, we have $y_n\to u^*$ as $n\to \infty$. That
is, problem \eqref{e1.4} has a unique positive solution $u^*$ in $P_h$.
Moreover, for any initial value  $u_0\in P_h$, constructing
successively the sequence
$$
u_{n+1}(t)=\int^1_0 G(t,s)f(s,u_n(s))ds
+\int^1_0 G(t,s)g(s,u_n(s))ds,\quad n=0,1,2,\dots,
$$
we have $u_n(t)\to u^*(t)$  as $n\to \infty$.
\end{proof}


\begin{remark} \label{rmk3.1} \rm
A simple example that illustrates Theorem \ref{thm3.1} is as follows:
let $f(t,x)\equiv 2\delta$,
$g(t,x)\equiv 1, \delta>0$.
Then the conditions (H1)--(H3) are satisfied and
\eqref{e1.4} has a unique solution $u(t)=(2\delta+1)\int^1_0G(t,s)ds$,
$t\in [0,1]$. Evidently,
\begin{gather*}
u(t)\geq \frac{(2\delta+1)(\alpha-1)}{\Gamma(\alpha)}
\int^1_0s(1-s)^{\alpha-1}ds\cdot h(t)
=\frac{(2\delta+1)(\alpha-1)}{\alpha(\alpha+1)\Gamma(\alpha)}\cdot h(t),
\\
u(t)\leq  \frac {2\delta+1}{\Gamma(\alpha)}
\int^1_0(1-s)^{\alpha-2}ds\cdot h(t)=\frac
 {2\delta+1}{(\alpha-1)\Gamma(\alpha)}\cdot h(t), \quad t\in [0,1].
\end{gather*}
So the unique solution $u$ is a positive solution and satisfies 
$u\in P_h=P_{t^{\alpha-1}(1-t)}$.
\end{remark}

\begin{example} \label{examp3.1} \rm
 \begin{equation}
\begin{gathered}
D^{\frac 32}_{0+}u(t)+ u^{1/2}(t)+\frac {u(t)}{1+u(t)}q(t)+t^2+a=0,\quad
0<t<1,\\
u(0)=u(1)=0,
\end{gathered} \label{e3.1}
\end{equation}
where $a>0$ is a constant, $q:[0,1]\to [0,+\infty)$ is
continuous with $q\not\equiv 0$.

In this example, we have $\alpha=3/2$. Take $0<b<a$ and let
\begin{gather*}
f(t,x)=x^{1/2}+t^2+b,\quad
g(t,x)=\frac x{1+x}q(t)+a-b,\\
\beta=\frac 12,\quad q_{\rm max}=\max\{q(t): t\in [0,1]\}.
\end{gather*}
Obviously, $q_{\rm max}>0$; $ f,g:[0,1]\times [0,+\infty)\to [0,+\infty)$
are continuous and increasing respect to the second argument,
$g(t,0)=a-b> 0$. Besides, for  $\lambda\in (0,1)$, $t\in [0,1]$, 
$x\in [0,+\infty)$, we have
\begin{gather*}
g(t,\lambda x)=\frac {\lambda x}{1+\lambda x}q(t)+a-b\geq
\frac {\lambda x}{1+ x}q(t)+\lambda(a-b)=\lambda g(t,x),\\
f(t,\lambda x)=\lambda^{1/2}x^{1/2}+t^2+b\geq
\lambda^{1/2}(x^{1/2}+t^2+b)= \lambda^\beta f(t,x).
\end{gather*}
Moreover, if we take $\delta_0\in (0,\frac b{q_{\rm max}+a-b}]$, then
we obtain
\begin{align*}
f(t,x)&=x^{1/2}+t^2+b\geq b=\frac b{q_{\rm max}+a-b}
(q_{\rm max}+a-b)\\
&\geq \delta_0 [\frac x{1+x}q(t)+a-b]
=\delta_0 g(t,x).
\end{align*} 
Hence all the conditions of Theorem \ref{thm3.1} are satisfied. 
This implies that \eqref{e3.1} has a unique positive solution 
 in $P_h$, where $h(t)=t^{\alpha-1}(1-t)$, $t\in [0,1]$.
\end{example}

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