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

\AtBeginDocument{{\noindent\small
{\em Electronic Journal of Differential Equations},
Vol. 2006(2006), No. 36, 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  (login: ftp)}
\thanks{\copyright 2006 Texas State University - San Marcos.}
\vspace{9mm}}

\begin{document}
\title[\hfilneg EJDE-2006/36\hfil
Positive solutions for boundary-value problems]
{Positive solutions for boundary-value problems of nonlinear
fractional differential equations}
\author[S. Zhang\hfil EJDE-2006/36\hfilneg]
{Shuqin Zhang} 

\address{Shuqin Zhang \hfill\break
Department of Mathematics, University of Mining
and Technology, Beijing, 100083 China}
\email{zhangshuqin@tsinghua.org.cn}

\date{}
\thanks{Submitted November 30, 2005. Published March 21, 2006.}
\subjclass[2000]{34B15}
\keywords{Caputo's fractional derivative;
 fractional differential equation; \hfill\break\indent
 boundary-value problem; positive solution; fractional Green's
 function; fixed-point theorem}

\begin{abstract}
 In this paper, we  consider the existence and multiplicity
 of positive solutions for the nonlinear fractional differential
 equation boundary-value problem
 \begin{gather*}
 \mathbf{D}_{0+}^\alpha u(t)=f(t,u(t)),\quad 0<t<1\\
 u(0)+u'(0)=0,\quad   u(1)+u'(1)=0
 \end{gather*}
 where $1<\alpha\leq 2$ is a real number, and
 $\mathbf{D}_{0+}^\alpha$ is the Caputo's fractional derivative,
 and $f:[0,1]\times[0,+\infty)\to [0,+\infty)$ is
 continuous. By means of a fixed-point theorem on cones, some
 existence and multiplicity results of positive solutions are
 obtained.
\end{abstract}

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

\section{Introduction}

Fractional calculus has played a significant role in
engineering, science, economy, and other fields. Many papers and
books on fractional calculus, fractional differential equations
have appeared recently, (see \cite{a,e,h,j}).
As cited in \cite{d} ``There have appeared
lots of works, in which fractional derivatives are used for a
better description considered material properties, mathematical
modelling base on enhanced rheological models naturally leads to
differential equations of fractional order-and to the necessity of
the formulation of initial conditions to such equations. Applied
problems require definitions of fractional derivatives allowing
the utilization of physically in interpretable initial conditions,
which contain $f(a),f'(a)$, etc". In fact, there has the same
requirements for boundary conditions. Caputo's fractional derivative
exactly satisfies these demands. Here, we  consider the
existence and multiplicity of positive solutions of nonlinear
fractional differential equation boundary-value problem involving
 Caputo's derivative.
\begin{equation} \label{e1}
\begin{gathered}
\mathbf{D}_{0+}^\alpha u(t)=f(t,u(t)),\quad 0<t<1\\
u(0)+u'(0)=0, \quad u(1)+u'(1)=0
\end{gathered}
\end{equation}
where $1<\alpha\leq 2$ is a real number and
$\mathbf{D}_{0+}^\alpha$ is the Caputo's fractional derivative,
and $f:[0,1]\times[0,+\infty)\to [0,+\infty)$ is
continuous. As far as we known, there has few papers which deal
with the boundary-value problem for nonlinear fractional
differential equation.

In \cite{e}, the authors consider the existence and multiplicity of
positive solutions of nonlinear fractional differential equation
boundary-value problem
\begin{equation} \label{e2}
\begin{gathered}
D_{0+}^\alpha u(t)+f(t,u(t))=0,\quad 0<t<1\\
 u(0)=u(1)=0
\end{gathered}
\end{equation}
where $1<\alpha\leq 2$ is a real number. $D_{0+}^\alpha$ is the
standard Riemann-Liouville fractional derivative, and
$f:[0,1]\times[0,+\infty)\to [0,+\infty)$ is
continuous. Due to the reasons cited above, when conditions of
\eqref{e2} are not zero boundary value, the Riemann-Liouville
fractional derivative $D_{0+}^\alpha$ is not suitable. Therefore,
in the sense of practicable demand, we investigate boundary-value
problem \eqref{e1} involving  the Caputo's fractional
derivative.

In this paper, analogy with boundary-value problem for
differential equations of integer order, we firstly derive the
corresponding Green' function-named by fractional Green' function.
Consequently problem \eqref{e1} is reduced to a equivalent Fredholm
integral equation of the second kind. Finally, using some
fixed-point theorems, the existence and multiplicity of positive
solutions are obtained.

\section{Preliminaries}

For completeness, in this section, we will demonstrate and
study the definitions and some fundamental facts of Caputo's
derivatives of
fractional order which can been founded in \cite{b}.
\smallskip

\noindent \textbf{Definition.}
\cite[(2.138)]{b} Caputo's derivative for a function
$f:[0,\infty)\to R$  can been written as
\begin{equation} \label{e3}
\mathbf{D}_{0+}^sf(x)=\frac
1{\Gamma(n-s)}\int_0^x\frac{f^{n}(t)dt}{(x-t)^{s+1-n}},\quad
n=[s]+1
\end{equation}
where $[s]$ denotes the integer part of real number $s$.

\begin{remark} \label{rmk2.1} \rm
Under natural conditions on the function $f(x)$, for $s\to n$
Caputo's derivative becomes a conventional $n$-th derivative of
the function $f(x)$. See \cite[79]{b}
\end{remark}

\noindent\textbf{Definition.} \cite[Definition 2.1]{a}
The integral
$$
I_{0+}^sf(x)= \frac
1{\Gamma(s)}\int_0^x\frac{f(t)}{(x-t)^{1-s}}dt,\quad x>0
$$
 where $s>0$, is called Riemann-Liouville fractional integral
of order $s$. \smallskip

\noindent\textbf{Definition.}  \cite[page 36-37]{a}
For a function $f(x)$ given in the interval $[0,\infty)$, the expression
$$
D_{0+}^sf(x)=\frac 1{\Gamma(n-s)}(\frac d{dx})^n
\int_0^x\frac{f(t)}{(x-t)^{s-n+1}}dt
$$
 where $ n=[s]+1, [s]$
denotes the integer part of number $s$, is called the
Riemann-Liouville fractional derivative of order $s$.

 As examples, for $\mu>-1$, we have
\begin{gather*}
\mathbf{D}_{0+}^\alpha x^\mu=\mu(\mu-1)\dots (\mu-n+1)
\frac{\Gamma(1+\mu-n)}{\Gamma(1+\mu-\alpha)}x^{\mu-\alpha}
\\
D_{0+}^\alpha
x^\mu=\frac{\Gamma(1+\mu-n)}{\Gamma(1+\mu-\alpha)}x^{\mu-\alpha}
\end{gather*}
where $n=[\alpha]+1$.

From the definition of Caputo's derivative and Remark \ref{rmk2.1}, we
can obtain the statement.

\begin{lemma} \label{lem2.1}
Let $\alpha>0$, then the differential equation
\[
\mathbf{D}_{0+}^\alpha u(t)=0
\]
has solutions $u(t)=c_0+c_1t+c_2t^2+\dots+c_nt^{n-1}$,
$c_i\in \mathbb{R}$, $i=0,1,\dots,n$,
$n=[\alpha]+1$.
\end{lemma}

 From the lemma above, we deduce the following statement.

\begin{lemma} \label{lem2.2}
 Let $\alpha>0$, then
\begin{align*}
I_{0+}^\alpha \mathbf{D}_{0+}^\alpha
u(t)=u(t)+c_0+c_1t+c_2t^2+\dots+c_nt^{n-1}
\end{align*}
for some $c_i\in \mathbb{R}$, $i=0,1,\dots,n$, $n=[\alpha]+1$.
\end{lemma}

The following theorems will play major role in our next
analysis.

\begin{lemma}[\cite{g}] \label{lem2.3}
Let $X$ be a Banach space, and let
$P\subset X$ be a cone in $X$. Assume $\Omega_1, \Omega_2$ are
open subsets of $X$ with
$0\in \Omega_1\subset {\overline{\Omega}_1}\subset{\Omega _2}$,
and let $S:P\to P$ be a completely continuous operator such that, either
\begin{enumerate}
\item $\|Sw\|\leq \|w\|$, $w\in P\cap {\partial \Omega_1}$,
$ \|Sw\|\geq \|w\|$, $w\in P\cap {\partial \Omega_2}$, or
\item $\|Sw\|\geq \|w\|$, $w\in P\cap {\partial \Omega_1}$,
$\|Sw\|\leq \|w\|$ $w\in P\cap {\partial \Omega_2}$
\end{enumerate}
Then $S$ has a fixed point in
$P\cap {\overline{\Omega}_2\backslash{\Omega_1}}$.
\end{lemma}


\noindent\textbf{Definition} %2.4
A map $\delta$ is said to be a nonnegative
continuous concave functional on $K$ if
$\delta : K\to [0,+\infty)$ is continuous and
$$
\delta(tx+(1-t)y)\geq t\delta(x)+(1-t)\delta(y)
$$
for all $x, y\in K$ and $0\leq t\leq 1$. And let
$$
K(\delta ,a,b)=\{u\in K|a\leq \delta (u),\|u\|\leq b\}
$$

\begin{lemma}[\cite{f}] \label{lem2.4}
Let $K$ be a cone and $K_c=\{y\in K|\|y\|\leq c\}$, and
$A:\overline{K_c}\to \overline{K_c}$
be completely continuous and $\alpha $ be a nonnegative continuous
concave function on $K$ such that $\alpha(y)\leq\|y\|$, for all
$y\in\overline{K_c}$. Suppose there exist $0<a<b<d\leq c$ such
that
\begin{itemize}
\item[(C1)] $\{y\in K(\alpha,b,d\}|\alpha(y)>b\}\neq\emptyset$ and
$\alpha(Ay)>b$, for all $y\in K\{\alpha,b,d\}$,

\item[(C2)] $\|Ay\|<a$, for $\|y\|\leq a$, and

\item[(C3)] $\alpha(Ay)>b$, for $y\in K\{\alpha,b,c\}$ with
$\|Ay\|>d$.
\end{itemize}
Then $A$ has at least three fixed points $y_1,y_2,y_3$ satisfying
$$
\|y_1\|<a,\quad b<\alpha(y_2),\quad\mbox{and}\quad
\|y_3\|>a\quad \mbox{with}\quad \alpha(y_3)<b
$$
\end{lemma}

\section{Main Results}

In this section, we consider the existence and multiplicity
of positive solutions of problem \eqref{e1} by means of the
Lemmas \ref{lem2.3}
and \ref{lem2.4}. First of all, we  find the Green's
function for boundary-value problem \eqref{e1}.

\begin{lemma} \label{lem3.1}
 Let $h(t)\in{C[0,1]}$ be a given function,
then the boundary-value problem
\begin{equation} \label{e4}
 \begin {gathered}
\mathbf{D}_{0+}^\alpha u(t)=h(t),\quad 0<t<1\\
u(0)+u'(0)=0,\quad u(1)+u'(1)=0
\end{gathered}
\end{equation}
has a unique solution
\begin{equation} \label{e5}
u(t)=\int_0^1G(t,s)h(s)ds
\end{equation}
where
\begin{equation} \label{e6}
G(t,s)= \begin {cases}
 \frac{(1-s)^{\alpha-1}(1-t)+(t-s)^{\alpha-1}}{\Gamma(\alpha)}+
 \frac{(1-s)^{\alpha-2}(1-t)}{\Gamma(\alpha-1)}, & s\leq t\\
  \frac{(1-s)^{\alpha-1}(1-t)}{\Gamma(\alpha)}+
 \frac{(1-s)^{\alpha-2}(1-t)}{\Gamma(\alpha-1)}, & t\leq s
\end{cases}
\end{equation}
Here $G(t,s)$ is called the Green's function of boundary-value
 problem \eqref{e4}.
\end{lemma}

\begin{proof} By the Lemma \ref{lem2.2}, we can reduce the equation
of problem \eqref{e4} to an equivalent integral equation
\[
u(t)=I_{0+}^\alpha h(t)-c_1-c_2t=\frac
1{\Gamma(\alpha)}\int_0^t(t-s)^{\alpha-1}h(s)ds-c_1-c_2t
\]
for some constants $c_1, c_2\in \mathbb{R}$. On the other hand, by
relations $D_{0+}^\alpha I_{0+}^\alpha u(t)=u(t)$ and
$I_{0+}^\alpha I_{0+}^\beta u(t)=I_{0+}^{\alpha+\beta}u(t)$, for
$\alpha,\beta>0$, $u\in L(0,1)$ (see \cite{a}), we have
\[
u'(t)=\frac 1{\Gamma(\alpha-1)}\int_0^t(t-s)^{\alpha-2}h(s)ds-c_2
\]
 As boundary conditions for problem \eqref{e4}, we have
\begin{gather*}
 -c_1-c_2=0\\
 -c_1-2c_2=-I_{0+}^\alpha h(1)-I_{0+}^{\alpha-1}h(1);
\end{gather*}
that is,
\begin{gather*}
 c_1=-I_{0+}^\alpha h(1)-I_{0+}^{\alpha-1}h(1)\\
 c_2=I_{0+}^\alpha h(1)+I_{0+}^{\alpha-1}h(1)
\end{gather*}
Therefore, the unique solution of \eqref{e4} is
\begin{align*}
&u(t)=\frac 1{\Gamma(\alpha)}\int_0^t(t-s)^{\alpha-1}h(s)ds+
\frac{1}{\Gamma(\alpha)}\int_0^1(1-s)^{\alpha
-1}h(s)ds\\
&\quad +\frac{1}{\Gamma(\alpha-1)}\int_0^1(1-s)^{\alpha
-2}h(s)ds-\frac{t}{\Gamma(\alpha)}\int_0^1(1-s)^{\alpha
-1}h(s)ds\\
&\quad -\frac{t}{\Gamma(\alpha-1)}\int_0^1(1-s)^{\alpha
-2}h(s)ds\\
&= \int_0^t(\frac{(1-s)^{\alpha-1}(1-t)+(t-s)^{\alpha-1}}{\Gamma(\alpha)}+
 \frac{(1-s)^{\alpha-2}(1-t)}{\Gamma(\alpha-1)})h(s)ds\\
&\quad +\int_t^1 (\frac{(1-s)^{\alpha-1}(1-t)}{\Gamma(\alpha)}+
 \frac{(1-s)^{\alpha-2}(1-t)}{\Gamma(\alpha-1)})h(s)ds\\
&=\int_0^1G(t,s)h(s)ds
\end{align*}
which completes the proof.
\end{proof}

\begin{lemma} \label{lem3.2}
 Let $h(t)\in{C[0,1]}$ be a given function,
 then function $G(t,s)$ defined by \eqref{e6} has the
following properties:
\begin{itemize}
\item[(R1)]
 $G(t,s)\in C([0,1]\times [0,1))$, and $G(t,s)>0$ for $t,s\in (0,1)$;

\item[(R2)] There exists a positive function $\gamma\in C(0,1)$ such
that
\begin{equation} \label{e7}
 \begin {gathered}
 \min_{1/4\leq t\leq 3/4}G(t,s)\geq \gamma(s)M(s), \quad\ s\in(0,1)
\\
\max_{0\leq t\leq 1}G(t,s)\leq M(s),
\end{gathered}
\end{equation}
where
\begin{equation} \label{e8}
M(s)=\frac{2(1-s)^{\alpha-1}}{\Gamma(\alpha)}+
\frac{(1-s)^{\alpha-2}}{\Gamma(\alpha-1)},\quad s\in [0,1)
 \end{equation}
\end{itemize}
\end{lemma}

\begin{proof} From the expression of $G(t,s)$, it is
obvious that $G(t,s)\in C([0,1]\times[0,1))$ and $G(t,s)\geq 0$
for $s, t\in (0,1)$. Next, we will prove (R2). From the
definition of $G(t,s)$, we can known that, for given $s\in (0,1)$,
$G(t,s)$ is decreasing with respect to $t$ for $t\leq s$, we let
\begin{gather*}
g_1(t,s)=\frac{(1-t)(1-s)^{\alpha
-1}+(t-s)^{\alpha-1}}{\Gamma(\alpha)}+\frac{(1-t)(1-s)^{\alpha-2}}{
\Gamma(\alpha-1)},\quad  s\leq t
\\
g_2(t,s)=\frac{(1-t)(1-s)^{\alpha
-1}}{\Gamma(\alpha)}+\frac{(1-t)(1-s)^{\alpha-2}}{
\Gamma(\alpha-1)},\quad t\leq s
\end{gather*}
 That is, $g_1(t,s)$ is a continuous function for
$\frac 14\leq t\leq \frac 34$,
 and $g_2(t,s)$ is decreasing with
respect to $t$. Hence, we have
\begin{gather*}
g_1(t,s)\geq \frac{ (1-s)^{\alpha}}{4\Gamma(\alpha)}+ \frac{
(1-s)^{\alpha-2}}{4\Gamma(\alpha-1)},\quad \mbox{for $1/4\leq t\leq 3/4$}
\\
\max_{0\leq t\leq
1}g_1(t,s)\leq\frac{2(1-s)^{\alpha-1}}{\Gamma(\alpha)}+
\frac{(1-s)^{\alpha-2}}{\Gamma(\alpha-1)}
\\
\min_{1/4\leq t\leq 3/4}g_2(t,s)=
 g_2(\frac 34,s)=\frac{(1-s)^{\alpha-1}}{4\Gamma(\alpha)}+
\frac{(1-s)^{\alpha-2}}{4\Gamma(\alpha-1)}
\\
\begin{aligned}
\max_{0\leq t\leq 1}g_2(t,s)=
 g_2(0,s)&=\frac{(1-s)^{\alpha-1}}{\Gamma(\alpha)}+
\frac{(1-s)^{\alpha-2}}{\Gamma(\alpha-1)}\\
&<\frac{2(1-s)^{\alpha-1}}{\Gamma(\alpha)}+
\frac{(1-s)^{\alpha-2}}{\Gamma(\alpha-1)}
\end{aligned}
\end{gather*}
Thus, we have
\begin{gather} \label{e9}
\min_{1/4\leq t\leq 3/4}G(t,s)\geq
m(s)=\frac{(1-s)^{\alpha-1}}{4\Gamma(\alpha)}+
\frac{(1-s)^{\alpha-2}}{4\Gamma(\alpha-1)},\quad s\in[0,1)
\\
 \label{e10}
\max_{0\leq t\leq 1}G(t,s)\leq
M(s)=\frac{2(1-s)^{\alpha-1}}{\Gamma(\alpha)}+
\frac{(1-s)^{\alpha-2}}{\Gamma(\alpha-1)},\quad s\in [0,1)
 \end{gather}
 Let
\begin{equation} \label{e11}
\gamma(s)= m(s)/M(s)=\frac 14
\frac{(1-s)^{\alpha-1}+(\alpha-1)(1-s)^{\alpha-2}}{2(1-s)^{\alpha-1}+
(\alpha-1)(1-s)^{\alpha-2}},\quad s\in (0,1)
\end{equation}
It is obviously that $\gamma(s)\in C((0,1), (0,+\infty))$.
The proof is completed.
\end{proof}

\begin{remark} \label{rmk3.1} \rm
 From the definition of function $\gamma(s)$,
 we see that $\gamma(s)\geq \frac 18$.
\end{remark}

Let $E=C[0,1]$ be endowed with the ordering $u\leq v$ if
$u(t)\leq v(t)$ for all $t\in[0,1]$, and the maximum norm,
$\|u\|=\max_{0\leq t\leq 1}|u(t)|$, Define the cone $K\subset E$
by
\[
K=\{u\in E|u(t)\geq 0, \min_{1/4\leq t\leq 3/4}\geq \frac
18\|u\|\}
\]
and the nonnegative continuous concave functional $\varphi$ on the
cone $K$ by
\[
\varphi(u)=\min_{1/4\leq t\leq 3/4}|u(t)|
\]

\begin{lemma} \label{lem3.3}
 Assume that $f(t,u)$ is continuous on
$[0,1]\times [0,\infty)$. A function $u\in K$ is a solution of
boundary-value problem \eqref{e1} if and only if it is a solution of
the integral equation \eqref{e5}.
\end{lemma}

\begin{proof}
Let $u\in K$ be a solution of boundary-value problem \eqref{e1}.
Applying the operator $I_{0+}^\alpha$ to both sides
of equation of problem \eqref{e1}, we have
\[
u(t)=c_1+c_2t+I_{0+}^\alpha f(t,u(t))
\]
for some $c_1,c_2\in \mathbb{R}$. By the same methods as obtaining the
Green's function of problem \eqref{e1} (Lemma \ref{lem3.1}), by boundary value
conditions of problem \eqref{e1}, we can calculate out constants $c_1$
and $c_2$, so
\[
u(t)=\int_0^1G(t,s)f(s,u(s))ds
\]
From  Lemma \ref{lem3.2} and Remark \ref{rmk3.1}, we  obtain
that $\int_0^1 G(t,s) f(s,u(s))ds \in K$. Hence, $u$ is also a
solution of integral equation \eqref{e5}.

Let $u\in K$ be a solution of integral equation \eqref{e5}. If we
denote the right-hand side of integral equation \eqref{e5} by $w(t)$,
then, applying Caputo's fractional operator to both sides of
integral equation \eqref{e5}, by the Definition of function $G(t,s)$,
since
\begin{align*}
w(t)&=\int_0^1G(t,s)f(s,u(s))ds\\
&=\frac 1{\Gamma(\alpha)}\int_0^t(t-s)^{\alpha-1}f(s,u(s))ds+
\frac{1}{\Gamma(\alpha)}\int_0^1(1-s)^{\alpha
-1}f(s,u(s))ds\\
&\quad +\frac{1}{\Gamma(\alpha-1)}\int_0^1(1-s)^{\alpha
-2}f(s,u(s))ds-\frac{t}{\Gamma(\alpha)}\int_0^1(1-s)^{\alpha
-1}f(s,u(s))ds\\
&\quad -\frac{t}{\Gamma(\alpha-1)}\int_0^1(1-s)^{\alpha -2}f(s,u(s))ds\,.
\end{align*}
Therefore,
\begin{align*}
w'(t)&=\frac d{dt}I_{0+}^\alpha f(t,u(t))-I_{0+}^\alpha
f(1,u(1))-I_{0+}^{\alpha-1} f(1,u(1))\\
&= D_{0+}^1I_{0+}^\alpha f(t,u(t))-I_{0+}^\alpha
f(1,u(1))-I_{0+}^{\alpha-1} f(1,u(1))\\
&= D_{0+}^1I_{0+}^1I_{0+}^{\alpha-1} f(t,u(t))-I_{0+}^\alpha
f(1,u(1))-I_{0+}^{\alpha-1} f(1,u(1))\\
&= I_{0+}^{\alpha-1} f(t,u(t))-I_{0+}^\alpha
f(1,u(1))-I_{0+}^{\alpha-1} f(1,u(1))
\end{align*}
and
\begin{gather*}
w''(t)=D_{0+}^1I_{0+}^{\alpha-1}
f(t,u(t))=D_{0+}^{2-\alpha}f(t,u(t))
\\
\mathbf{D}_{0+}^\alpha w(t)=I_{0+}^{2-\alpha}w''(t)
=I_{0+}^{2-\alpha}D_{0+}^{2-\alpha}f(t,u(t))
=f(t,u(t))
\end{gather*}
 here, use the relation
$I_{0+}^sI_{0+}^tg(t)=I_{0+}^{s+t}g(t)$,
$D_{0+}^sI_{0+}^sg(t)=g(t)$, $s>0$, $t>0$, $g\in L(0,1)$ and
$I_{0+}^s D_{0+}^s g(t)=g(t)$,  $s>0$, $g\in C[0,1]$
(see \cite{a}),
where $D_{0+}^s$ is Riemann-Liouville fractional derivative.
That is, $\mathbf{D}_{0+}^\alpha u(t)=f(t,u(t))$. On the other
hand, one has
\begin{gather*}
u(0) =\int_0^1(\frac{(1-s)^{\alpha-1}}{\Gamma(\alpha)}+
\frac{(1-s)^{\alpha-2}}{\Gamma(\alpha-1)})f(s,u(s))ds
\\
u'(0) =-\int_0^1(\frac{(1-s)^{\alpha-1}}{\Gamma(\alpha)}+
\frac{(1-s)^{\alpha-2}}{\Gamma(\alpha-1)})f(s,u(s))ds
\\
u(1) =\int_0^1\frac{(1-s)^{\alpha-1}}{\Gamma(\alpha)}f(s,u(s))ds
\\
u'(1)
=-\int_0^1\frac{(1-s)^{\alpha-1}}{\Gamma(\alpha)}f(s,u(s))ds\,.
\end{gather*}
We obtain
\[
u(0)+u'(0)=0,  u(1)+u'(1)=0
\]
which implies that $u\in K$ is a solution of problem \eqref{e1}.
\end{proof}

 \begin{lemma} \label{lem3.4}
 Assume that $f(t,u)$ is
continuous on $[0,1]\times [0,\infty)$, and define the operator
$T:K\to E$ by
\[
Tu(t)=\int_0^1G(t,s)f(s,u(s))ds
\]
Then $T:K\to K$ is completely continuous.
\end{lemma}

\begin{proof}
Firstly, we prove that $T:K\to K$.  In
view of the expression of $G(t,s)$, it is clear that,
$Tu(t)\geq 0$, $t\in [0,1]$, $Tu(t)$ is continuous for $u\in K$.
And that, for $u\in K$, by means of the Lemma \ref{lem3.1}
 and Remark \ref{rmk3.1},
we have
\[
\min_{1/4\leq t\leq 3/4}Tu(t)=\min_{1/4\leq t\leq
3/4}\int_0^1G(t,s)f(s,u(s))ds\geq\frac 18\int_0^1M(s)f(s,u(s))ds
\]
On the other hand,
\begin{align*}
\|Tu\|=\max_{0\leq t\leq 1}|Tu(t)|\leq\int_0^1M(s)f(s,u(s))ds\,.
\end{align*}
Thus, we obtain
\[
\min_{1/4\leq t\leq 3/4}Tu(t)\geq\frac 18\|Tu\|
\]
 which implies $T:K\to K$.

Let $P\subset K$ be bounded, i.e. there exists a positive constant
$L>0$ such that $\|u\|\leq L$, for all $u\in P$. Let
$M=\max_{0\leq t\leq 1, 0\leq u\leq L}|f(t,u)|+1$, then for $u\in
P$, from the Lemma \ref{lem3.1}, one has
\[
|Tu(t)|\leq \int_0^1|G(t,s)f(t,u(s))|ds\leq M\int_0^1M(s)ds
\]
Hence, $T(P)$ is bounded.
For all $\varepsilon>0$, each $u\in P$, $t_1, t_2\in [0,1]$,
$t_1<t_2$, let $$\eta=\min\{\frac 12,
\frac{\Gamma(\alpha)\varepsilon}{12M},\frac{\Gamma(1+\alpha)\varepsilon}{8M}\}$$
we will prove that $|Tu(t_2)-Tu(t_1)|<\varepsilon$, when
$t_2-t_1<\eta$. One has
\begin{align*}
&|Tu(t_2)-Tu(t_1)|\\
&=|\int_0^1G(t_2,s)f(s,u(s))ds-\int_0^1G(t_1,s)f(s,u(s))ds|\\
&\leq \int_0^{t_1}|(G(t_2,s)-G(t_1,s))f(s,u(s))|ds+\int_{t_2}^1|(G(t_2,s)
-G(t_1,s)) f(s,u(s))|ds\\
&\quad +\int_{t_1}^{t_2}|(G(t_2,s)-G(t_1,s))f(s,u(s))|ds
\\
&\leq M(\int_0^{t_1}|(G(t_2,s)-G(t_1,s))|ds+\int_{t_2}^1|(G(t_2,s)
-G(t_1,s))| ds
\\
&\quad +\int_{t_1}^{t_2}|(G(t_2,s)-G(t_1,s))|ds)\\
&= M(\int_0^{t_1}(\frac{(t_2-t_1)(1-s)^{\alpha-1}+((t_2-s)^{\alpha-1}-
(t_1-s)^{\alpha-1})}{\Gamma(\alpha)}\\
&\quad + \frac{(t_2-t_1)(1-s)^{\alpha-2}}{\Gamma(\alpha-1)})ds\\
&\quad +\int_{t_2}^1(\frac{(t_2-t_1)(1-s)^{\alpha-1}}{\Gamma(\alpha)}+
 \frac{(t_2-t_1)(1-s)^{\alpha-2}}{\Gamma(\alpha-1)})ds
 \\
 &\quad +\int_{t_1}^{t_2}(
\frac{(t_2-t_1)(1-s)^{\alpha-1}+(t_2-s)^{\alpha-1}}{\Gamma(\alpha)}+
 \frac{(t_2-t_1)(1-s)^{\alpha-2}}{\Gamma(\alpha-1)})ds)
\\
&\leq M(\int_0^{t_1}(\frac{\eta+((t_2-s)^{\alpha-1}-
(t_1-s)^{\alpha-1})}{\Gamma(\alpha)}+
 \frac{\eta(1-s)^{\alpha-2}}{\Gamma(\alpha-1)})ds\\
&\quad +\int_{t_2}^1(\frac{\eta}{\Gamma(\alpha)}+
 \frac{\eta(1-s)^{\alpha-2}}{\Gamma(\alpha-1)})ds
 +\int_{t_1}^{t_2}(
\frac{\eta+(t_2-s)^{\alpha-1}}{\Gamma(\alpha)}+
 \frac{\eta(1-s)^{\alpha-2}}{\Gamma(\alpha-1)})ds)
 \\
&\leq M(\frac{2\eta}{\Gamma(\alpha)}+\frac{t_2^{\alpha}-
t_1^{\alpha}}{\Gamma(1+\alpha)}+\frac{2\eta}{\Gamma(\alpha)}+
\frac{2\eta}{\Gamma(\alpha)}+\frac{2\eta^\alpha}{\Gamma(1+\alpha)})
\\
&= M(\frac{6\eta}{\Gamma(\alpha)}+\frac{2\eta^\alpha+(t_2^{\alpha}-
t_1^{\alpha})}{\Gamma(1+\alpha)})
\\
&< M(\frac{6\eta}{\Gamma(\alpha)}+\frac{2\eta+(t_2^{\alpha}-
t_1^{\alpha})}{\Gamma(1+\alpha)})
\end{align*}
 In order to estimate $t_2^\alpha-t_1^\alpha$, we can apply a method
 used in \cite{e}; that is, for $\eta\leq t_1<t_2\leq 1$, by means of mean value
 theorem of differentiation, we have
\[
t_2^\alpha-t_1^\alpha\leq \alpha(t_2-t_1)<\alpha \eta\leq 2\eta
\]
for $0\leq t_1<\eta, t_2<2\eta$, we have
\[
t_2^\alpha-t_1^\alpha\leq t_2^\alpha<(2\eta)^\alpha\leq 2\eta\,.
\]
while for $0\leq t_1<t_2\leq\eta$, there has
\[
t_2^\alpha-t_1^\alpha\leq t_2^\alpha<\eta^\alpha< 2\eta\,.
\]
Thus, we obtain
\[
|Tu(t_2)-Tu(t_1)|<\frac{6M\eta}{\Gamma(\alpha)}+\frac{4M\eta}{\Gamma(1+\alpha)}<\frac\varepsilon
2+\frac\varepsilon 2=\varepsilon
\]
By means of the Arzela-Ascoli theorem,
$T:K\to K$ is completely continuous.
\end{proof}

\begin{theorem} \label{thm3.1}
 Assume that $f(t,u)$ is continuous on
$[0,1]\times [0,\infty)$,  and satisfies one of the following
conditions
\begin{itemize}
\item[(H1)] There exist $0<\mu_1,\nu_1\leq 1$ such that
 $$
\lim_{u\to \infty}\frac{f(t,u(t))}{u^{\mu_1}}=0,\quad
\lim_{u\to 0}\frac{f(t,u(t))}{u^{\nu_1}}=\infty
$$
for all $t\in [0,1]$
\item[(H1')] There exist $\mu_2,\nu_2\geq 1$ such that
 $$\lim_{u\to \infty}\frac{f(t,u(t))}{u^{\mu_2}}=\infty,\quad
 \lim_{u\to 0}\frac{f(t,u(t))}{u^{\nu_2}}=0
$$
 for all $t\in [0,1]$.
\end{itemize}
Then problem \eqref{e1} has one positive solution.
\end{theorem}

\begin{proof} By the Lemma \ref{lem3.3}, we know that we only need
 to consider existence of fixed point of operator  $T$ in $K$.
 It follows from the Lemma \ref{lem3.4} that
 $T: K\to K$ is a completely continuous operator.
Assume that (H1) holds, then there exist $N_1>0,
N_2>0$, such that for all $0<\varepsilon<(2\int_0^1M(s)ds)^{-1}$  and
$\rho>\frac {64}{\int_{1/4}^{3/4}M(s)ds}>0$.
Then
\begin{gather*}
f(t,u(t))\leq \varepsilon u^{\mu_1}, \quad \mbox{for $t\in [0,1],u\geq N_1$}\\
f(t,u(t))>\rho u^{\nu_1}, \quad \mbox{for $t\in [0,1],0\leq u\leq N_2$}
\end{gather*}
So we have
 $$
f(t,u(t))\leq \varepsilon u^{\mu_1}+c,
\quad \mbox{for $t\in[0,1],u\in [0,+\infty)$}
$$
 where
$$
c=\max_{0\leq t\leq 1  ,0\leq u\leq N_1}|f(t,u(t))|+1
$$
 Let
$$ \Omega_1=\{u\in K; \|u\|<R_1\}
$$
 where $R_1>\{1,2c\int_0^1M(s)ds\}$. For
$u\in  \partial{\Omega_1}$, from the Lemma \ref{lem3.2}, we have
\begin{align*}
|T u(t)|&=\int_0^1G(t,s)f(s,u(s))ds\\
&\leq\int_0^1M(s)(\varepsilon |u|^{\mu_1}+c)ds\\
&\leq \varepsilon R_1^{\mu_1}\int_0^1M(s)+c\int_0^1M(s)ds\\
&\leq \frac {R_1}2+\frac {R_1}2=R_1\,,
\end{align*}
$\|Tu\|\leq R_1=\|u\|$.
Let
$$
\Omega_2=\{u\in K; \|u\|<R_2\}
$$
 where $0<R_2<\{1,N_2\}$, then for $u\in  \partial{ \Omega_2}$,
we obtain
\begin{align*}
|T u(t)|&=|\int_0^1G(t,s)f(s,u(s))ds|\\
&\geq\int_{1/4}^{3/4}G(t,s)f(s,u(s))ds\\
&>\frac \rho 8\int_{1/4}^{3/4}M(s)u(s)^{\nu_1} ds\\
&\geq \frac \rho {64}\int_{1/4}^{3/4}M(s)\|u\|^{\nu_1} ds\\
&=\frac \rho {64}\int_{1/4}^{3/4}M(s)R_2R_2^{\nu_1-1} ds\\
&\geq \frac \rho {64}\int_{1/4}^{3/4}M(s)R_2 ds\\
&> R_2=\|u\|
\end{align*}
 so $\|Tu\|\geq R_2=\|u\|$.
 Then Lemma \ref{lem2.3} implies that operator $T$ has one fixed point
$u^*(t)\in  {\overline {\Omega_1}\backslash {\Omega_2}}$.
Then $u^*(t)$ is one  positive solution of problem \eqref{e1}.

 For condition (H1'), we can obtain the result in a similarly way.
Now, we  give a briefly description. Assume that (H1') holds,
thus, there exist $M_1>0, M_2>0$, such that for all
$0< \varepsilon<(\int_0^1M(s)ds)^{-1}$  and
$\lambda>(\frac{\int_{1/4}^{3/4}M(s)ds}{64})^{-1}>0$,
 we have have
\begin{gather*}
f(t,u(t))>\lambda u^{\mu_2}, \quad \mbox{for $t\in [0,1],u\geq M_1$}
\\
f(t,u(t))\leq\varepsilon u^{\nu_2}, \quad
\mbox{for $t\in [0,1],0\leq u\leq M_2$}
\end{gather*}
Let
$$
 \Omega_1=\{u\in K; \|u\|<R_1\},\quad \Omega_2=\{u\in K; \|u\|<R_2\}
$$
 where $R_1>\{1,8M_1\}$, $0<R_2<\{1,M_2\}$.  Then for
$u\in  \partial{\Omega_1}$, for $\frac 14\leq t\leq \frac 34$,
one has $ u(t)\geq \min_{1/4\leq t\leq 3/4}u(t)\geq \frac 18
 \|u\|=\frac {R_1}8>M_1$. thus, from the Lemma \ref{lem3.2}, we have
 \begin{align*}
|T u(t)|&\geq\int_{1/4}^{3/4}G(t,s)f(s,u(s))ds\\
&\geq \frac \lambda {64}\int_{1/4}^{3/4}M(s)\|u\|^{\mu_2} ds\\
&\geq \frac \lambda {64}\int_{1/4}^{3/4}M(s)\|u\| ds\\
&> R_1=\|u\|
\end{align*}
 for $u\in \partial{ \Omega_2}$, we obtain
\[
|T u(t)|\leq\int_0^1M(s)\varepsilon \|u\|^{\nu_2}ds
\leq \varepsilon R_2\int_0^1M(s)ds
\leq R_2
\]
Thence, the Lemma \ref{lem2.3} implies that operator $T$ has one fixed
point $u^*(t)\in
 {\overline {\Omega_1}\backslash {\Omega_2}}$, then $u^*(t)$ is a
 positive solution of problem \eqref{e1}.
\end{proof}

 Let
 \[
 M=(\int_0^1M(s)ds)^{-1},\quad
 N=(\int_{1/4}^{3/4}\gamma(s)M(s)ds)^{-1}
\]

\begin{theorem} \label{thm3.2}
Assume that $f(t,u)$ is continuous on $[0,1]\times [0,\infty)$,
and there exist constants $0<b<c$ such that:
\begin{itemize}
\item[(H2)] There exists $r\geq c$ such that
$ f(t,u(t))<Mu$ for all $(t,u)\in [0,1]  \times [0,r]$;

\item[(H3)] $f(t,u)\geq Nb$, for all
$(t,u)\in  [\frac 14,\frac 34]\times [b,c]$.
\end{itemize}
  Then problem \eqref{e1} has at least three positive solutions
 $u_1, u_2, u_3$ with
 \begin{gather*}
 \|u_1\|<a,\quad b<\min_{1/4\leq t\leq  3/4}|u_2(t)|
\\
 a<\|u_3\|,\quad \min_{1/4\leq t\leq  3/4}|u_3(t)|<b
 \end{gather*}
\end{theorem}

\begin{proof} We will apply the Lemma \ref{lem2.4} to prove this
 result. Next, we show that all conditions of the Lemma \ref{lem2.4}
 are satisfied.  By the Lemma \ref{lem3.3}, we know that we only need
 to consider existence of fixed point of operator  $T$ in $K$.
 It follows from the Lemma \ref{lem3.4} that
 $T: K\to K$ is a completely continuous operator.\\
By (H2), there exist $r$, such that
 $$
f(t,u(t))<M u, \quad \mbox{for $t\in [0,1],0\leq u\leq r$}
$$
Let $0<a<b<c\leq r$, then if $u\in \overline{K}_c$
($\overline{K}_c=\{u\in K|\|u\|\leq c\}$,$K_c= \{u\in K|\|u\|<
c\}$), we can obtain
\begin{align*}
\|Tu\|=\max_{0\leq t\leq 1}|\int_0^1G(t,s)f(s,u(s))ds|<
M\int_0^1M(s)\|u\| ds=\|u\|\leq c
\end{align*}
Hence, combining with the Lemma \ref{lem3.4}, we know that
$T:\overline{K}_c\to \overline{K}_c$ is completely continuous. In
the same way, let $0<a<c$, then if $u\in \overline{K}_a$, we can
also obtain that $\|Tu\|<a$ which satisfies the condition (C2) of
the Lemma \ref{lem2.4}. Now, we  check condition (C1) of the Lemma
\ref{lem2.4}. Let $u(t)=\frac {b+c}2$, $0\leq t\leq 1$. It is
obvious that $u(t)=\frac {b+c}2\in{K(\delta,b,c)}$,
$\delta(u)=\frac {b+c}2>b$, thus, $\{u\in
K(\delta,b,c)|\delta(u)>b\}\neq\emptyset$. Thence, if $u\in
K(\delta,b,c)$, then $b\leq u(t)\leq c$ for $1/4\leq t\leq 3/4$,
by assumption $(H_3)$, we have $f(t,u)\geq Nb$, for $\frac 14\leq
t\leq\frac 34$,  so, by the Lemma \ref{lem3.2}, there has
\[
\delta(Tu)=\min_{1/4\leq t\leq 3/4}|Tu(t)|>
\int_{1/4}^{3/4}\gamma(s)M(s)Nbds=b
\]
By Lemma \ref{lem2.4}, problem \eqref{e1} has at least three positive
solutions  $u_1, u_2, u_3$ with
the required conditions; which completes the proof.
\end{proof}

 \begin{thebibliography}{99}

\bibitem{d} O. P. Agrawal, \emph{Formulation of Euler-Larange
equations for fractional variational problems}, J. Math. Anal.
Appl. 272, 368-379, 2002.

 \bibitem{c} D. Delbosco and L.Rodino, \emph{Existence
and Uniqueness for a Nonlinear Fractional Differential Equation},
J. Math. Appl, 204, 609-625, 1996.

 \bibitem{g} Krasnosel'skii M. A.,
\emph{Positive solutions of operator equations},   Noordhoff
Gronigen, Netherland, 1964.

\bibitem{f} R. W. Leggett, L. R. Williams, \emph{Multiple positive
 solutions of nonlinear operators on ordered Banach spaces},
 Indiana Univ.Math.J.28(1979) 673-688.

\bibitem{b} I. Podlubny, \emph{Fractional Differential
equations, Mathematics in Science and Engineering}, vol, 198,
Academic Press, New Tork/Londin/Toronto, 1999.

\bibitem{a} S. G. Samko, A. A. Kilbas, O. I. Marichev, \emph{
Fractional Integral And Derivatives (Theorey and Applications)}.
Gordon and Breach, Switzerland, 1993.

 \bibitem{e} Zhanbing Bai, Haishen L\"{u}, \emph{Positive
 solutions for boundary-value problem of nonlinear fractional
 differential equation},  J.
Math. Anal. Appl. 311, 495-505, 2005.

\bibitem{h} Shu-qin Zhang,\emph{The Existence of a Positive
Solution for a Nonlinear Fractional Differential Equation}, J.
Math. Anal. Appl. 252, 804-812, 2000.

\bibitem{j} Shu-qin Zhang, \emph{Existence of Positive
Solution for some class of Nonlinear Fractional Differential
Equations}, J. Math. Anal. Appl. 278, 1, 136-148, 2003.

 \end{thebibliography}


\section*{Addendum posted on November 9, 2009.}

The definition  $n=[\alpha]+1$  in Lemmas \ref{lem2.1} and \ref{lem2.2}
is incomplete. It should be
\[
n=\begin{cases}
[\alpha]+1 &\text{if } n \not\in \{0,1,2,\dots\}\\
\alpha  &\text{if } n \in \{0,1,2,\dots\}\,.
\end{cases}
\]
The author wants to thank Yige Zhao, Shurong Sun, Zhenlai Han, and
Meng Zhang (at the University of Jinan) for pointing out 
this misprint.

\end{document}