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

\AtBeginDocument{{\noindent\small
\emph{Electronic Journal of Differential Equations},
Vol. 2008(2008), No. 146, pp. 1--9.\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 2008 Texas State University - San Marcos.}
\vspace{9mm}}

\begin{document}
\title[\hfilneg EJDE-2008/146\hfil Existence of positive solutions]
{Existence of positive solutions for singular fractional
differential equations}

\author[T. Qiu, Z. Bai \hfil EJDE-2008/146\hfilneg]
{Tingting Qiu, Zhanbing Bai}  % not in alphabetical order

\address{Tingting Qiu \newline
   College of Information Science and Engineering\\
   Shandong University of Science and Technology \\
   Qingdao, 266510, China}
\email{qiutingting19833@163.com}

\address{Zhanbing Bai \newline
   College of Information Science and Engineering\\
   Shandong University of Science and Technology \\
   Qingdao, 266510, China}
\email{zhanbingbai@163.com}

\thanks{Submitted April 4, 2008. Published October 24, 2008.}
\thanks{Supported by grants 10371006 from the National Nature Science 
Foundation of  China, \hfill\break\indent
and 10626033 from the  Mathematics Tianyuan Foundation of China.}
\subjclass[2000]{34B15}
\keywords{Boundary value problem; positive solution; \hfill\break\indent
singular fractional differential equation; fixed-point theorem}

\begin{abstract}
 In this article, we establish the existence of a positive
 solution to a singular boundary-value problem of nonlinear
 fractional differential equation. Our analysis rely on nonlinear
 alternative of Leray-Schauder type and Krasnoselskii's fixed point
 theorem in a cone.
\end{abstract}

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

\section{Introduction}

    Many papers and books on fractional calculus differential
equation have appeared recently. Most of them are devoted to the
solvability of the linear fractional equation in terms of a
special function and to problems of analyticity in the complex
domain(see, for example \cite{LM,YL}). Moreover, Delbosco and Rodino
\cite{DD} considered the existence of a solution for the nonlinear
fractional differential equation $D_{0^{+}}^{\alpha}u=f(t,u)$,
where $0<\alpha<1$, and $f:[0,a]\times \mathbb{R}\to \mathbb{R}, 0<a\leq
+\infty$ is a given function,continuous in $(0,a)\times \mathbb{R}$. They
obtained results for solutions by using the Schauder fixed point
theorem and the Banach contraction principle. Recently,
Zhang \cite{SZ} considered the existence of positive solution for
equation $D_{0^{+}}^{\alpha}u=f(t,u)$, where $0<\alpha<1$, and
$f:[0,1]\times [0,+\infty)\to [0,+\infty)$, is a given
continuous function, by using the sub-and super-solution method.

In this article, we discuss the existence of a positive solution to
boundary-value problems of the nonlinear fractional differential
equation
\begin{equation}
\begin{gathered}
D_{0^+}^{\alpha}u(t)+f(t,u(t))=0,\quad  0<t<1, \\
u(0)=u'(1)=u''(0)=0\,,
\end{gathered}
\end{equation}
 where $ 2<\alpha\leq3$,
 $D_{0^+}^{\alpha}$ is the Caputo's
 differentiation, and $f:(0,1]\times[0,\infty)\to[0,\infty)$ with
 $\lim\limits_{t\to 0^{+}}f(t,\cdot)=+\infty$
(that is $f$ is singular at $t=0$).
We obtain two results about this boundary-value
problem, by using Krasnoselskii's fixed point theorem and nonlinear
alternative of Leray-Schauder type in a cone.

For existence theorems for fractional
differential equation and applications, we refer the reader
to the survey by Kilbas and Trujillo \cite{AA}.
 Concerning the definitions of fractional integral
and derivative and related basic properties, we refer the reader to
Samko, Kilbas, and Marichev \cite{SG} and Delbosco and Rodino
\cite{DD}.

\section{Preliminaries}

    For the convenience of the reader, we present here the
necessary definitions from fractional calculus theory. These
definitions and properties can be found in the literature.

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

\begin{definition} \rm
The Caputo fractional derivative
of order $\alpha>0$ of a continuous function
$f:(0,\infty)\to \mathbb{R}$ is given by
$$
D_{0^+}^{\alpha}f(t)=\frac{1}{\Gamma(n-\alpha)}
\int_{0}^{t}\frac{f^{(n)}(s)}{(t-s)^{\alpha-n+1}}ds,
$$
where $n-1<\alpha\leq n$, provided that the right-hand side is
pointwise defined on $(0,\infty)$.
\end{definition}


\begin{lemma}[\cite{SG2}]
 Let  $n-1<\alpha\leq n$,  $u\in C^{n}[0,1]$. Then
$$
I_{0+}^{\alpha}D_{0+}^{\alpha}u(t)=u(t)-C_{1}-C_{2}t-\dots-C_{n}t^{n-1},
$$
where  $C_{i}\in \mathbb{R}$, $i=1,2,\dots n$.
\end{lemma}

\begin{lemma}[\cite{SG2}]
~~The relation
$$I_{a+}^{\alpha}I_{a+}^{\beta}\varphi=I_{a+}^{\alpha+\beta}\varphi$$
is valid in following case
$$Re\beta >0,~~Re(\alpha+\beta) >0,~~\varphi(x)\in L_{1}(a,b).$$
\end{lemma}
\begin{lemma} Given $f\in C[0,1],$ and
$2<\alpha\leq3,$ the unique solution of
\begin{equation}
\begin{array}{ll}
$$D_{0^+}^{\alpha}u(t)+f(t)=0,$$ & \hbox{$0<t<1,$} \\
$$u(0)=u'(1)=u''(0)=0. $$ & \hbox{} \\
\end{array}
\end{equation}is
$$u(t)=\int_{0}^{1}G(t,s)f(s)ds$$
where
\begin{equation}
G(t,s)=\left\{%
\begin{array}{ll}
    \frac{(\alpha-1)t(1-s)^{\alpha-2}-(t-s)^{\alpha-1}}{\Gamma(\alpha)},  & \hbox{$0\leq s\leq t\leq1$,} \\
   \frac{t(1-s)^{\alpha-2}}{\Gamma(\alpha-1)},  & \hbox{$0\leq t\leq s\leq1$.} \\
\end{array}%
\right.
\end{equation}
\end{lemma}
\begin{proof}We may apply Lemma2.3 to reduce Eq.(2.1) to an
equivalent integral equation
$$u(t)=-I_{0+}^{\alpha}f(t)+C_{1}+C_{2}t+C_{3}t^{2} $$
for some  \ $C_{i}\in R ,\ \ i=1,2,3.$ \ ~By Lemma2.4 we have
\begin{align*}
    u'(t) &=-D^{1}_{0+}I_{0+}^{\alpha}f(t)+C_{2}+2C_{3}t
    =-D^{1}_{0+}I_{0+}^{1}I_{0+}^{\alpha-1}f(t)+C_{2}+2C_{3}t
    \\& =-I_{0+}^{\alpha-1}f(t)+C_{2}+2C_{3}t
\end{align*}
\begin{align*}
u''(t)=-D^{1}_{0+}I_{0+}^{\alpha-1}f(t)+2C_{3}=-D^{1}_{0+}I_{0+}^{1}I_{0+}^{\alpha-2}f(t)+2C_{3}=-I_{0+}^{\alpha-2}f(t)+2C_{3}.
\end{align*}

 From $u(0)=u'(1)=u''(0) = 0 $, \ one has
$$C_{1}=0,
~~~C_{2}=\frac{1}{\Gamma(\alpha-1)}\int_0^1(1-s)^{\alpha-2}f(s)ds,
~~~C_{3}=0.$$ Therefore, \ the unique solution of problem (2.1) is
\begin{align*}
 u(t)& =-\frac{1}{\Gamma(\alpha)}\int_0^t(t-s)^{\alpha-1}f(s)ds+\frac{1}{\Gamma(\alpha-1)}\int_0^1t(1-s)^{\alpha-2}f(s)ds
 \\& =\int_0^t\left[\frac{t(1-s)^{\alpha-2}}{\Gamma(\alpha-1)}-\frac{(t-s)^{\alpha-1}}{\Gamma(\alpha)}\right]f(s)ds+ \int_t^1\frac{t(1-s)^{\alpha-2}}{\Gamma(\alpha-1)}f(s)ds
 \\& =\int_0^1G(t,s)f(s)ds
\end{align*}


For $G(t, s)$, since ~~$2<\alpha\leq3,0\leq s\leq t\leq1$ ~~we can
obtain
$$(\alpha-1)t(1-s)^{\alpha-2}\geq t(1-s)^{\alpha-2}\geq
t(t-s)^{\alpha-2}\geq(t-s)^{\alpha-1}$$ obviously,we get $G(t,
s)>0.$ The proof is complete.
\end{proof}


\begin{lemma}[\cite{MA}]
Let $E$ be a Banach space, $P\subseteq E$
a cone, and $\Omega_1$, $\Omega_2$ are two bounded open balls of
$E$ centered at the origin with
$\overline{\Omega}_1\subset\Omega_2$.  Suppose that
$A:P\cap(\overline{\Omega}_2\setminus\Omega_1)\to P$ is a
completely continuous operator such that either
\begin{itemize}
 \item[(i)] $\|Ax\| \leq \|x\|$, $x\in P\cap\partial\Omega_1$
 and $\|Ax\| \geq \|x\|$,
      $x\in P\cap\partial\Omega_2$, or
 \item[(ii)] $\|Ax\| \geq \|x\|$, $x\in P\cap\partial\Omega_1$
 and $\|Ax\| \leq \|x\|$,
      $x\in P\cap\partial\Omega_2$
\end{itemize}
holds. Then $A$ has a fixed point in
$P\cap(\overline{\Omega}_2\setminus\Omega_1)$.
\end{lemma}


\begin{lemma}[\cite{AG}]
Let $E$ be a Banach space with $C\subseteq E$ closed and convex.
Assume $U$ is a relatively open subset of $C$ with $0\in U$ and
$A:\overline{U}\to C$ is a continuous compact map. Then
either
\begin{itemize}
 \item[(1)] $A$ has a fixed point in  $\overline{U}$; or
  \item[(2)] there exists $u\in\partial U$
  and   $\lambda\in(0,1)$ with   $u=\lambda Au$.
\end{itemize}
\end{lemma}


\section{Main Results}


For our construction, we let $E=C[0,1]$ and $\|u\|=\max\limits_{0
\le t \le 1}|u(t)|$ which is a Banach space. We seek solutions of
(1.1) that lie in the cone
\begin{equation*}
 P=\{u\in E:u(t)\geq0, \; 0\leq t\leq 1\}\,.
\end{equation*}
Define operator $T:P\to P$, by
\begin{equation*}
    Tu(t)=\int_{0}^{1}G(t,s)f(s,u(s))ds\,.
\end{equation*}


\begin{lemma}
Let $0<\sigma<1$, $2<\alpha\leq3$, $F:(0,1]\to \mathbb{R}$ is
continuous and $\lim\limits_{t\to0^{+}}F(t)=\infty$. Suppose that
$t^{\sigma}F(t)$ is continuous function on $[0,1]$.  Then the
function
$$
H(t)=\int_{0}^{t}G(t,s)F(s)ds
$$
is continuous on $[0,1]$.
\end{lemma}

\begin{proof}
By the continuity of $t^{\sigma}F(t)$ and
$H(t)=\int_{0}^{t}G(t,s)s^{-\sigma}s^{\sigma}F(s)ds$ It is easily
to check that $H(0)=0$. The proof is divided into three cases:

\noindent\textbf{Case 1:} $t_{0}=0, \forall t\in (0,1]$.
Since $t^{\sigma}F(t)$ is continuous in $[0,1]$, there exists a
constant $M>0$, such that $\big|t^{\sigma}F(t)\big|\leq M$,
for $t\in[0,1]$. Hence
\begin{align*}
\big|H(t)-H(0)\big|
&=\Big|\int_{0}^{t}\frac{(\alpha-1)t(1-s)^{\alpha-2}-(t-s)^{\alpha-1}}
{\Gamma(\alpha)}s^{-\sigma}s^{\sigma}F(s)ds\\
&\quad +\int_{t}^{1}\frac{t(1-s)^{\alpha-2}}{\Gamma(\alpha-1)}
 s^{-\sigma}s^{\sigma}F(s)ds\Big|\\
& =\Big|\int_{0}^{1}\frac{t(1-s)^{\alpha-2}}{\Gamma(\alpha-1)}
 s^{-\sigma}s^{\sigma}F(s)ds-
\int_{0}^{t}\frac{(t-s)^{\alpha-1}}{\Gamma(\alpha)}s^{-\sigma}
 s^{\sigma}F(s)ds\Big|\\
& \leq \Big|\int_{0}^{1}\frac{t(1-s)^{\alpha-2}}{\Gamma(\alpha-1)}
 s^{-\sigma}s^{\sigma}F(s)ds\Big|
 +\Big|\int_{0}^{t}\frac{(t-s)^{\alpha-1}}{\Gamma(\alpha)}
 s^{-\sigma}s^{\sigma}F(s)ds\Big|\\
& \leq M\int_{0}^{1}\frac{t(1-s)^{\alpha-2}}{\Gamma(\alpha-1)}
 s^{-\sigma}ds+M\int_{0}^{t}\frac{(t-s)^{\alpha-1}}{\Gamma(\alpha)}
 s^{-\sigma}ds\\
&=\frac{Mt}{\Gamma(\alpha-1)}B(1-\sigma,\alpha-1)
 +\frac{M}{\Gamma(\alpha)}t^{\alpha-\sigma}B(1-\sigma,\alpha)\\
& =\frac{\Gamma(1-\sigma)Mt}{\Gamma(\alpha-\sigma)}
 +\frac{\Gamma(1-\sigma)Mt^{\alpha-\sigma}}{\Gamma(1+\alpha-\sigma)}
\to 0  \quad\text{(as $t\to 0$)}
\end{align*}
where $B$ denotes the beta function.

\noindent\textbf{Case 2:} $t_{0}\in(0,1)$, for all $t\in(t_{0},1]$
\begin{align*}
& \big|H(t)-H(t_{0})\big|\\
&=\Big|\int_{0}^{t}\frac{(\alpha-1)t(1-s)^{\alpha-2}-(t-s)^{\alpha-1}}
{\Gamma(\alpha)}s^{-\sigma}s^{\sigma}F(s)ds\\
&\quad +\int_{t}^{1}\frac{t(1-s)^{\alpha-2}}{\Gamma(\alpha-1)}
 s^{-\sigma}s^{\sigma}F(s)ds-\int_{t_{0}}^{1}
 \frac{t_{0}(1-s)^{\alpha-2}}{\Gamma(\alpha-1)}
 s^{-\sigma}s^{\sigma}F(s)ds\\
&\quad -\int_{0}^{t_{0}}\frac{(\alpha-1)t_{0}(1-s)^{\alpha-2}-(t_{0}-s)^{\alpha-1}}
 {\Gamma(\alpha)}s^{-\sigma}s^{\sigma}F(s)ds\Big|\\
& =\Big|\int_{0}^{1}\frac{t(1-s)^{\alpha-2}}{\Gamma(\alpha-1)}
 s^{-\sigma}s^{\sigma}F(s)ds-
\int_{0}^{t}\frac{(t-s)^{\alpha-1}}{\Gamma(\alpha)}s^{-\sigma}s^{\sigma}F(s)ds\\
& \quad -\int_{0}^{1}\frac{t_{0}(1-s)^{\alpha-2}}{\Gamma(\alpha-1)}
 s^{-\sigma}s^{\sigma}F(s)ds
+\int_{0}^{t_{0}}\frac{(t_{0}-s)^{\alpha-1}}{\Gamma(\alpha)}s^{-\sigma}
 s^{\sigma}F(s)ds\Big|\\
& =\Big|\int_{0}^{1}\frac{(t-t_{0})(1-s)^{\alpha-2}}{\Gamma(\alpha-1)}
 s^{-\sigma}s^{\sigma}F(s)ds\\
&\quad -\int_{0}^{t_{0}}\frac{(t-s)^{\alpha-1}-(t_{0}-s)^{\alpha-1}}{\Gamma(\alpha)}
 s^{-\sigma}s^{\sigma}F(s)ds
 -\int_{t_{0}}^{t}\frac{(t-s)^{\alpha-1}}{\Gamma(\alpha)}s^{-\sigma}
 s^{\sigma}F(s)ds\Big|\\
&\leq\frac{M(t-t_{0})}{\Gamma(\alpha-1)}\int_{0}^{1}(1-s)^{\alpha-2}
 s^{-\sigma}ds+\frac{M}{\Gamma(\alpha)}\int_{0}^{t_{0}}
 \Big[(t-s)^{\alpha-1}-(t_{0}-s)^{\alpha-1}\Big]s^{-\sigma}ds\\
&\quad -\frac{M}{\Gamma(\alpha)}\int_{t_{0}}^{t}(t-s)^{\alpha-1}s^{-\sigma}ds\\
&\leq\frac{M(t-t_{0})}{\Gamma(\alpha-1)}B(1-\sigma,\alpha-1)
 +\frac{Mt^{\alpha-\sigma}}{\Gamma(\alpha)}B(1-\sigma,\alpha)
 -\frac{Mt_{0}^{\alpha-\sigma}}{\Gamma(\alpha)}B(1-\sigma,\alpha)\\
&=\frac{\Gamma(1-\sigma)M(t-t_{0})}{\Gamma(\alpha-\sigma)}
 +\frac{\Gamma(1-\sigma)Mt^{\alpha-\sigma}}{\Gamma(1+\alpha-\sigma)}
 -\frac{\Gamma(1-\sigma)Mt_{0}^{\alpha-\sigma}}{\Gamma(1+\alpha-\sigma)}
\to 0 \quad\text{(as $t\to t_{0}$).}
\end{align*}

\noindent\textbf{Case 3:} $t_{0}\in(0,1]$, for all $t\in[0,t_{0})$.
The proof is similar to that of Case 2; we omitted it.
\end{proof}

\begin{lemma}
Let $0<\sigma<1$, $2<\alpha\leq3$,
$f:(0,1]\times[0,+\infty)\to[0,+\infty)$ is continuous and
$\lim\limits_{t\to0^{+}}f(t,\cdot)=+\infty$, $t^{\sigma}f(t,u(t))$
is continuous function on $[0,1]\times[0,+\infty)$, then the
operator $T:P\to P$ is completely continuous.
\end{lemma}

\begin{proof}
For each $u\in P$, let  $Tu(t)=\int_{0}^{1}G(t,s)f(s,u(s))ds$. By
Lemma3.1 and the fact that $f,G(t,s)$ are non-negative, we have
$T:P\to P$.

Let $u_{0}\in P$ and $\|u_{0}\|=C_{0}$, if $u\in P$ and
$\|u-u_{0}\|<1$, then $\|u\|<1+C_{0}=C$. By the continuity of
$t^{\sigma}f(t,u(t))$, we know that $t^{\sigma}f(t,u(t))$ is
uniformly continuous on $[0,1]\times[0,C]$.

Thus for all $\epsilon>0$, there exists $\delta>0(\delta<1)$, such
that $|t^{\sigma}f(t,u_{2})-t^{\sigma}f(t,u_{1})|<\epsilon$, for
all $t\in[0,1]$, and $u_{1},u_{2}\in [0,C]$ with
$|u_{2}-u_{1}|<\delta$. Obviously, if $\|u-u_{0}\|<\delta$, then
$u(t),u_{0}(t)\in[0,C]$ and $\|u(t)-u_{0}(t)\|<\delta$, for all
$t\in[0,1]$. Hence,
\begin{equation}
|t^{\sigma}f(t,u(t))-t^{\sigma}f(t,u_{0}(t))|<\epsilon, \quad
    \mbox{for all t}\in[0,1].
\end{equation}
$u\in P$, with $\|u-u_{0}\|<\delta$. It follows from (3.1) that
\begin{align*}
 \|Tu-Tu_{0}\|&=\max_{0\leq t\leq1}\big|Tu(t)-Tu_{0}(t)\big|\\
 & \leq\max_{0\leq t\leq1}\int_{0}^{1}G(t,s)s^{-\sigma}\big|s^{\sigma}f(s,u(s))
 -s^{\sigma}f(s,u_{0}(s))\big|ds\\
&<\epsilon\int_{0}^{1}G(t,s)s^{-\sigma}ds\\
&=\epsilon\int_{0}^{1}\frac{(\alpha-1)(1-s)^{\alpha-2}-(1-s)^{\alpha-1}}
{\Gamma(\alpha)}s^{-\sigma}ds\\
&\leq\frac{\epsilon}{\Gamma(\alpha-1)}\int_{0}^{1}
(1-s)^{\alpha-2}s^{-\sigma}ds\\
&=\frac{\epsilon}{\Gamma(\alpha-1)}
B(1-\sigma,\alpha-1)=\frac{\Gamma(1-\sigma)\epsilon}{\Gamma(\alpha-\sigma)}.
\end{align*}
By the arbitrariness of $u_{0}$, $T:P\to P$ is continuous.
Let $M\subset P$ be bounded; i.e., there exists a positive constant
$b$ such that $\|u\|\leq b$,  for all $u\in p$.

Since $t^{\sigma}f(t,u)$ is continuous in
$[0,1]\times[0,+\infty)$, let
 $$
 L=\max_{0\leq t\leq1,u\in M}t^{\sigma}f(t,u)+1, \quad  \forall   u  \in M.
$$
Then
\[
\big|Tu(t)\big|
\leq\int_{0}^{1}G(t,s)s^{-\sigma}|s^{\sigma}f(s,u(s))|ds
\leq L\int_{0}^{1}G(1,s)s^{-\sigma}ds
=\frac{\Gamma(1-\sigma)L}{\Gamma(\alpha-\sigma)};
\]
thus
$$
\|Tu\|=\max_{0\leq t\leq1}\big|Tu(t)\big|
\leq\frac{\Gamma(1-\sigma)L}{\Gamma(\alpha-\sigma)}\,.
$$
So, $T(M)$ is equicontinuous.
For  $\epsilon>0$ set
\begin{equation*}
\delta=\min\Big\{\frac{\epsilon}{\frac{\Gamma(1-\alpha)L}{\Gamma(\alpha-\sigma)}
+\frac{\Gamma(1-\alpha)L}{\Gamma(1+\alpha-\sigma)}},
\frac{\epsilon\Gamma(\alpha-\sigma)}{2L\Gamma(1-\sigma)},
\frac{\epsilon}{\frac{\Gamma(1-\alpha)L}{\Gamma(\alpha-\sigma)}
+\frac{\Gamma(1-\alpha)L2^{\alpha}}{\Gamma(1+\alpha-\sigma)}}\Big\}\,.
\end{equation*}
For $u\in M$,  $t_{1}, t_{2}\in[0,1]$,  with
$t_{1}<t_{2}$, for $0<t_{2}- t_{1}<\delta$, we have
\begin{align*}
& \big|Tu(t_{2})-Tu(t_{1})\big|\\
&=\Big|\int_{0}^{1}G(t_{2},s)f(s,u(s))ds-\int_{0}^{1}G(t_{1},s)f(s,u(s))ds\Big|\\
&
=\Big|\int_{0}^{1}\big[G(t_{2},s)-G(t_{1},s)\big]s^{-\sigma}
 s^{\sigma}f(s,u(s))ds\Big|\\
& \leq L\int_{0}^{1}\Big|G(t_{2},s)-G(t_{1},s)\Big|s^{-\sigma}ds\\
& \leq L\Big|\int_{0}^{t_{2}}\frac{(\alpha-1)t_{2}(1-s)^{\alpha-2}-(t_{2}-s)
 ^{\alpha-1}}{\Gamma(\alpha)}s^{-\sigma}ds
 +\int_{t_{2}}^{1}\frac{t_{2}(1-s)^{\alpha-2}}{\Gamma(\alpha-1)}
 s^{-\sigma}ds\\
&\quad -\int_{0}^{t_{1}}\frac{(\alpha-1)t_{1}(1-s)^{\alpha-2}-(t_{1}-s)
 ^{\alpha-1}}{\Gamma(\alpha)}s^{-\sigma}ds-\int_{t_{1}}^{1}
 \frac{t_{1}(1-s)^{\alpha-2}}{\Gamma(\alpha-1)}s^{-\sigma}ds\Big|\\
& \leq L\Big[(t_{2}-t_{1})\int_{0}^{1}\frac{(1-s)^{\alpha-2}}{\Gamma(\alpha-1)}
 s^{-\sigma}ds+\int_{0}^{t_{2}}\frac{(t_{2}-s)^{\alpha-1}}{\Gamma(\alpha)}
 s^{-\sigma}ds\\
&\quad -\int_{0}^{t_{1}}\frac{(t_{1}-s)^{\alpha-1}}{\Gamma(\alpha)}
 s^{-\sigma}ds\Big]\\
&=L\frac{(t_{2}-t_{1})}{\Gamma(\alpha-1)}\int_{0}^{1}s^{-\sigma}(1-s)
 ^{\alpha-2}ds+\frac{L}{\Gamma(\alpha)}\int_{0}^{t_{2}}s^{-\sigma}
 (t_{2}-s)^{\alpha-1}ds\\
&\quad -\frac{L}{\Gamma(\alpha)}\int_{0}^{t_{1}}s^{-\sigma}(t_{1}-s)^{\alpha-1}ds\\
&\leq L\frac{(t_{2}-t_{1})\Gamma(1-\sigma)}{\Gamma(\alpha-\sigma)}
 +\frac{L\Gamma(1-\sigma)}{\Gamma(1+\alpha-\sigma)}(t_{2}^{\alpha-\sigma}
 -t_{1}^{\alpha-\sigma}).
\end{align*}

\noindent\textbf{Case 1:} $t_{1}=0$, $t_{2}< \delta$.
\begin{align*}
\big|Tu(t_{2})-Tu(t_{1})\big|
&=L\frac{t_{2}\Gamma(1-\sigma)}{\Gamma(\alpha-\sigma)}
 +\frac{L\Gamma(1-\sigma)}{\Gamma(1+\alpha-\sigma)}t_{2}^{\alpha-\sigma}\\
& <L\frac{\delta\Gamma(1-\sigma)}{\Gamma(\alpha-\sigma)}
 +\frac{L\Gamma(1-\sigma)}{\Gamma(1+\alpha-\sigma)}\delta<\epsilon\,.
\end{align*}

\noindent\textbf{Case 2:} $\delta\leq t_{1}<t_{2}< 1$.
\begin{align*}
 \big|Tu(t_{2})-Tu(t_{1})\big|
&<L\frac{\delta\Gamma(1-\sigma)}{\Gamma(\alpha-\sigma)}
 +\frac{L\Gamma(1-\sigma)}{\Gamma(1+\alpha-\sigma)}\delta^{\alpha-\sigma}\\
& =\frac{L\delta\Gamma(1-\sigma)
 +L\Gamma(1-\sigma)\delta^{\alpha-\sigma}}{\Gamma(\alpha-\sigma)}\\
&<\frac{2L\delta\Gamma(1-\sigma)}{\Gamma(\alpha-\sigma)}<\epsilon\,.
\end{align*}

\noindent\textbf{Case 3:} $0<t_{1}<\delta$, $t_{2}< 2\delta$.
\[
\big|Tu(t_{2})-Tu(t_{1})\big|
<L\frac{\delta\Gamma(1-\sigma)}{\Gamma(\alpha-\sigma)}
+\frac{L\Gamma(1-\sigma)}{\Gamma(1+\alpha-\sigma)}2^{\alpha}\delta<\epsilon\,.
\]
Therefore, $T(M)$ is equicontinuous. The Arzela-Ascoli theorem
implies that $\overline{T(M)}$ is compact. Thus, the operator
$T:P\to P$ is completely continuous.
\end{proof}

\begin{theorem}
Let $0<\sigma<1$, $2<\alpha\leq3$,
$f:(0,1]\times[0,+\infty)\to[0,+\infty)$ is continuous and
$\lim\limits_{t\to0^{+}}f(t,\cdot)=+\infty$, $t^{\sigma}f(t,y)$ is
continuous function on $[0,1]\times[0,+\infty)$. Assume that there
exist two distinct positive constant $\rho,   \mu   (\rho>\mu)$ such
that
\begin{itemize}

\item[(H1)] $t^{\sigma}f(t,\omega)\leq \rho\frac{\Gamma(\alpha-\sigma)}
{\Gamma(1-\sigma)}$,   for  $(t,\omega)\in[0,1]\times[0,\rho]$;

\item[(H2)] $t^{\sigma}f(t,\omega)\geq \mu\frac{\Gamma(\alpha-\sigma)}
{\Gamma(1-\sigma)}$,   for  $(t,\omega)\in[0,1]\times[0,\mu]$.
\end{itemize}
Then (1.1) has at least one positive solution.
\end{theorem}

\begin{proof}
From Lemma 3.2 we have $T:P\to P$ is completely continuous. We
divide the proof into the following two steps.

\noindent\textbf{Step1:}
 Let $\Omega_{1}=\{u\in P:\|u\|<\frac{\alpha-\sigma-1}{\alpha-\sigma}\mu\}$,
for $u\in K\cap\partial\Omega_{1}$ and all $t\in[0,1]$, we have
$0\leq u(t)\leq\frac{\alpha-\sigma-1}{\alpha-\sigma} \mu$.
It follows from (H2) that
\begin{align*}
Tu(1) &
=\int_{0}^{1}G(1,s)f(s,u(s))ds=\int_{0}^{1}G(1,s)s^{-\sigma}s^{\sigma}f(s,u(s))ds\\
& \geq \mu\frac{\Gamma(\alpha-\sigma)}{\Gamma(1-\sigma)}
 \int_{0}^{1}G(1,s)s^{-\sigma}ds\\
& =\mu\frac{\Gamma(\alpha-\sigma)}{\Gamma(1-\sigma)}
 \Big[\int_{0}^{1}\frac{(\alpha-1)(1-s)^{\alpha-2}-(1-s)^{\alpha-1}}
 {\Gamma(\alpha)}s^{-\sigma}ds\Big]\\
& =\mu\frac{\Gamma(\alpha-\sigma)}{\Gamma(1-\sigma)}
 \Big[\int_{0}^{1}\frac{(1-s)^{\alpha-1}}{\Gamma(\alpha-1)}
 s^{-\sigma}ds-\int_{0}^{1}\frac{(1-s)^{\alpha-1}}{\Gamma(\alpha)}
 s^{-\sigma}ds\Big]\\
& =\mu\frac{\Gamma(\alpha-\sigma)}{\Gamma(1-\sigma)}
 \big[\frac{B(1-\sigma,\alpha-1)}{\Gamma(\alpha-1)}
 -\frac{B(1-\sigma,\alpha-1)}{\Gamma(\alpha)}\big]\\
&\geq\frac{\alpha-\sigma-1}{\alpha-\sigma}\mu=\|u\|\,.
\end{align*}
Hence,
$$
\|Tu\|=\max_{0\leq t\leq1}|Tu(t)|\geq
\frac{\alpha-\sigma-1}{\alpha-\sigma}\mu=\|u\|,
$$
for $u\in P\cap\partial\Omega_{1}$.

\noindent\textbf{Step 2:}
 Let $\Omega_{2}=\{u\in P:\|u\|<\rho\}$,
for $u\in K\cap\partial\Omega_{2}$ and all $t\in[0,1]$, we have
$0\leq u(t)\leq \rho$.
By assumption (H1),
\begin{align*}
 Tu(t)
&=\int_{0}^{1}G(t,s)f(s,u(s))ds\\
&=\int_{0}^{1}G(t,s)s^{-\sigma}s^{\sigma}f(s,u(s))ds\\
& \leq \rho\frac{\Gamma(\alpha-\sigma)}{\Gamma(1-\sigma)}
\Big[\int_{0}^{t}\frac{(\alpha-1)t(1-s)^{\alpha-2}-(t-s)^{\alpha-1}}
{\Gamma(\alpha)}s^{-\sigma}ds \\
&\quad +\int_{t}^{1}\frac{t(1-s)^{\alpha-2}}{\Gamma(\alpha-1)}s^{-\sigma}ds\Big]\\
& \leq \rho\frac{\Gamma(\alpha-\sigma)}{\Gamma(1-\sigma)}
 \Big[\int_{0}^{1}\frac{t(1-s)^{\alpha-2}}{\Gamma(\alpha-1)}s^{-\sigma}ds
-\int_{0}^{t}\frac{(t-s)^{\alpha-1}}{\Gamma(\alpha)}s^{-\sigma}ds\Big]\\
& \leq\rho\frac{\Gamma(\alpha-\sigma)}{\Gamma(1-\sigma)}
 \Big[\frac{t}{\Gamma(\alpha-1)}\int_{0}^{1}s^{-\sigma}(1-s)^{\alpha-2}ds\Big]\\
&\leq\rho\frac{\Gamma(\alpha-\sigma)}{\Gamma(1-\sigma)}
 \frac{B(1-\sigma,\alpha-1)}{\Gamma(\alpha-1)}\\
&=\rho\frac{\Gamma(\alpha-\sigma)}{\Gamma(1-\sigma)}
 \frac{\Gamma(1-\sigma)}{\Gamma(\alpha-\sigma)}=\rho\,.
\end{align*}
So $\|Tu(t)\|\leq\|u\|$,  for $u\in P\cap\partial\Omega_{2}$.
Therefore, by (ii) of Lemma 2.6, we complete the proof.
\end{proof}

\begin{theorem}
Let $0<\sigma<1$, $2<\alpha\leq3$,
$f:(0,1]\times[0,+\infty)\to[0,+\infty)$ is continuous and
$\lim\limits_{t\to0^{+}}f(t,\cdot)=+\infty$, $t^{\sigma}f(t,y)$ is
continuous function on $[0,1]\times[0,+\infty)$.  Suppose the
following conditions are satisfied:
\begin{itemize}
\item[(H3)] there exists a continuous, nondecreasing
function  $\varphi :[0,+\infty)\to(0,\infty)$
with  $t^{\sigma}f(t,\omega)\leq \varphi(\omega)$,  for
$(t,\omega)\in[0,1]\times[0,+\infty)$

\item[(H4)] there exists $r>0$, with
$\frac{r}{\varphi(r)}>\frac{\Gamma(\alpha-\sigma)}{\Gamma(1-\sigma)}$

\end{itemize}
Then (1.1) has one positive solution.
\end{theorem}

\begin{proof}
Let $U=\{u\in P:\|u\|<r\}$, we have $U\subset P$. From Lemma 3.2, we
know $T:\overline{U}\to P$ is completely continuous. If there exists
$u\in\partial U$, $\lambda\in(0,1)$ such that
\begin{equation}
u=\lambda Tu,
\end{equation}
By (H3) and (3.2), for $t\in[0,1]$ we have
\begin{align*}
 u(t)
& =\lambda Tu(t)=\lambda\int_{0}^{1}G(t,s)f(s,u(s))ds\leq\int_{0}^{1}G(t,s)
  s^{-\sigma}s^{\sigma}f(s,u(s))ds \\
& \leq\int_{0}^{1}G(t,s)s^{-\sigma}\varphi(u(s))ds\\
&\leq\varphi(\|u\|)\int_{0}^{1}G(t,s)s^{-\sigma}ds \\
& \leq\varphi(\|u\|)\int_{0}^{1}G(1,s)s^{-\sigma}ds \\
& =\varphi(\|u\|)\int_{0}^{1}\frac{(\alpha-1)t(1-s)^{\alpha-2}-(t-s)
^{\alpha-1}}{\Gamma(\alpha)}s^{-\sigma}ds \\
& \leq\varphi(\|u\|)\frac{tB(1-\sigma,\alpha-1)}{\Gamma(\alpha-1)}\\
&\leq\varphi(\|u\|)\frac{\Gamma(1-\sigma)}{\Gamma(\alpha-\sigma)}\,.
\end{align*}
Consequently,
$\|u\|\leq\varphi(\|u\|)\frac{\Gamma(1-\sigma)}{\Gamma(\alpha-\sigma)}$;
namely,
\[
\frac{\|u\|}{\varphi(\|u\|)}\leq\frac{\Gamma(1-\sigma)}{\Gamma(\alpha-\sigma)}.
\]
Combining (H4) and the above inequality, we have $\|u\|\neq r$,
which is contradiction with $u\in\partial U$. According to Lemma
2.7, $T$ has a fixed point $u\in\overline{U}$, therefore, (1.1) has
a positive solution.
\end{proof}

As an example, consider the fractional differential equation
\begin{equation}
\begin{gathered}
   D_{0^{+}}^{\alpha}u(t)+\frac{(t-\frac{1}{2})^{2}\ln(2+u)}{t^{\sigma}}=0,
   \quad 0<t<1 \\
 u(0)=u'(1)=u''(0)=0,
\end{gathered}
\end{equation}
where $0<\sigma<1$, $2<\alpha\leq3$. Then (3.3) has a positive
solution.

\begin{thebibliography}{00}

\bibitem{ZB} Z. B. Bai;
Positive solutions for boundary value problem
of nonlinear fractional differential equation, \emph{J. Math.
Anal. Appl.} \textbf{311} (2005), 495--505.

\bibitem{LM} L. M. C. M. Campos;
 On the solution of some simple
fractional differential equation, \emph{Int. J. Math. Sci.} \textbf{13}
(1990), 481-496.

\bibitem{DD} D. Delbosco and L. Rodino;
 Existence and uniqueness for a
nonlinear fractional differential equation, \emph{J. Math. Anal.
 Appl.} \textbf{204} (1996), 609--625.

\bibitem{AG} A. Granas, R. B. Guenther, J. W. Lee;
 Some general existence principle in the Caratheodory
theory of nonlinear system,  \emph{J. Math. Pure. Appl. } \textbf{70}
(1991), 153-196.

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

\bibitem{AA} A. A. Kilbas and J. J. Trujillo;
 Differential equations of fractional order: methods, results and problems-I,
\emph{Applicable Analysis } \textbf{78} (2001), 153--192.

\bibitem{MA} M. A. Krasnoselskii;
Positive Solution of Operator Equation,
Noordhoff,Groningen, 1964.

\bibitem{YL} Y. Ling, S. Ding;
 A Class of Analytic Function Defined by Fractive Derivative,
\emph{J. Math. Anal. Appl. } \textbf{186} (1994), 504-513.

\bibitem{KS} K. S. Miller, B. Ross;
An Introduction to the Fractional Calculus and Fractional Differential
Equation, Wiley, New York, 1993.


\bibitem{SG2} S. G. Samko, A. A. Kilbas, O. I. Marichev;
\emph{Fractional Integral and Derivative. Theory and Applications},
Gordon and Breach, 1993.

\bibitem{SZ} S. Q. Zhang;
 The existence of a positive solution for a nonlinear
fractional differential equation, \emph{J. Math. Anal. Appl. } {\bf
252} (2000), 804--812.

\end{thebibliography}

\end{document}