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

\AtBeginDocument{{\noindent\small
\emph{Electronic Journal of Differential Equations},
Vol. 2012 (2012), No. 215, pp. 1--27.\newline
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu
\newline ftp ejde.math.txstate.edu}
\thanks{\copyright 2012 Texas State University - San Marcos.}
\vspace{9mm}}

\begin{document}
\title[\hfilneg EJDE-2012/215\hfil Positive solutions]
{Positive solutions of fractional differential equations with derivative terms}

\author[C. Cheng, Z. Feng,  Y. Su \hfil EJDE-2012/215\hfilneg]
{Cuiping Cheng, Zhaosheng Feng, Youhui Su}  % in alphabetical order

\address{Cuiping Cheng \newline
Department of Applied Mathematics, 
Hangzhou Dianzi University, Hangzhou 310018, China}
\email{chengcp0611@163.com}

\address{Zhaosheng Feng \newline
Department of Mathematics, 
University of Texas--Pan American, Edinburg, TX 78539, USA}
\email{zsfeng@utpa.edu; fax: (956) 665-5091}

\address{Youhui Su \newline
Department of Mathematics, 
Xuzhou Institute of Technology, Xuzhou 221116, China\newline
Department of Mathematics, Indiana University,
Bloomington, IN 47405, USA}
\email{youhsu@indiana.edu}


\thanks{Submitted August 15, 2012. Published November 29, 2012.}
\subjclass[2000]{34A08, 34B18, 34K37}
\keywords{Positive solution; equicontinuity; 
 fractional differential equation; \hfill\break\indent fixed point theorem;
 Carath\'eodory  type condition}

\begin{abstract}
 In this article, we are concerned with the existence of positive
 solutions for nonlinear fractional differential equation
 whose nonlinearity contains the first-order derivative,
 \begin{gather*}
 D_{0^+}^{\alpha}u(t)+f(t,u(t),u'(t))=0,\quad t\in (0,1),\;
 n-1<\alpha\leq n,\\
 u^{(i)}(0)=0, \quad  i=0,1,2,\dots,n-2,\\
 [D_{0^+}^{\beta}u(t)]_{t=1}=0, \quad 2\leq\beta\leq n-2,
 \end{gather*}
 where $n>4 $ $ (n\in\mathbb{N})$, $D_{0^+}^{\alpha}$ is the
 standard Riemann-Liouville fractional derivative of order $\alpha$
 and $f(t,u,u'):[0,1] \times [0,\infty)\times(-\infty,+\infty)
 \to [0,\infty)$ satisfies the Carath\'eodory  type condition.
 Sufficient conditions are obtained for the existence of
 at least one or two positive solutions by using  the nonlinear
 alternative of the Leray-Schauder type and Krasnosel'skii's fixed
 point theorem. In addition, several other sufficient
 conditions are established for the existence of at least triple,
 $n$  or  $2n-1$ positive solutions. Two examples
 are given to illustrate our theoretical results.
\end{abstract}

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

\section{Introduction}

Fractional calculus is a generalization of ordinary
differentiation and integration to arbitrary non-integer order.
Although the notion of fractional derivative dates back to the
time when Leibnitz and Newton invented differential calculus and
the tools of fractional calculus have been available and
applicable to various fields of study, the qualitative and
quantitative investigation of the theory of fractional
differential equations has been started recently
\cite{pa1,pa2,ag}. In the past two decades, we have seen that
differential equations involving Riemann-Liouville differential
operators of fractional order arise in many engineering and
scientific disciplines as the mathematical modelling of systems
and processes in the fields of physics, chemistry, aerodynamics,
electrodynamics of complex medium, polymer rheology, etc
\cite{pa3,pa4,pa5,pa6,pa7}. In consequence, the subject of
fractional differential equations is gaining diverse and
continuous attention. For more details of some recent theoretical
results on fractional differential equations and their
applications, we refer the reader to 
\cite{aga,p2,Suat,p6,1a,zhou,su,lin} and the references therein.

In order to better understand the background of our fractional
differential system, let us briefly review some related studies on the topic.
 Xu et al \cite{xu} considered the equation
\begin{equation}\label{eq1}
\begin{gathered}
D_{0^+}^{\alpha}u(t)+f(t,u(t))=0,\quad t\in ( 0,1 ), \\
u(0)=u'(0)=u(1)=u'(1)=0,
\end{gathered}
\end{equation}
where
$D_{0^+}^{\alpha}$ is the standard Riemann-Liouville fractional
derivative of order $\alpha$  ($2<\alpha\leq 3$) and
$f: [0,\,1]\times[0,\infty)\to[0,\infty)$ is continuous.
The existence of multiple positive
solutions to system \eqref{eq1} is established by applying the fixed
point theorems.

  Goodrich \cite{10} considered  \eqref{eq1}
subject to the boundary conditions
\begin{gather*}
u^{(i)}(0)=0,\quad i=0,1,2,\dots,n-2,\\
[D_{0^+}^{\beta}u(t)]_{t=1}=0, \quad 2\leq\beta\leq n-2,
\end{gather*}
where $n-1<\alpha\leq n$ and $n>4$ $ (n\in {\mathbb{N}})$.
The existence of one positive solution was explored.

It is notable that the nonlinear term $f(t,u(t))$ in \eqref{eq1}
does not involve the derivative.
Apparently, the nonlinear term $f(t,u(t),u'(t))$ containing the derivative is
a more general case, and the study of such fractional
differential equations is of significance theoretically and
practically \cite{pa6}. So far, to the best of our knowledge, it appears that
there is only  a few articles  concerning the existence
of positive solutions to fractional differential
equations with nonlinear terms involving the derivative \cite{zhang}.
In the present article, we restrict our attention to this problem
in some respects.
More precisely, we consider the problem
\begin{equation}
\begin{gathered}
D_{0^+}^{\alpha}u(t)+f(t,u(t),u'(t))=0,\quad t\in
( 0,1 ),\; n-1<\alpha\leq n,\\
u^{(i)}(0)=0, \quad
i=0,1,2,\dots,n-2,\\
{}[D_{0^+}^{\beta}u(t)]_{t=1}=0,
\quad  2\leq\beta\leq n-2,
\end{gathered}\label{equ1}
\end{equation}
where $u^{(i)}$ represents the $i$th derivative of
$u$, $n>4$ $ (n\in\mathbb{N})$, $D_{0^+}^{\alpha}$ is the standard
Riemann-Liouville fractional derivative of order
$n-1<\alpha\leq n$  and $f(t,u,u'):[0,1] \times [0,\infty
)\times(-\infty,+\infty) \to [0,\infty)$ satisfies
 Carath\'eodory type conditions.
Our goal is to establish the existence  of
at least one or two positive solutions by using the nonlinear
alternative of Leray-Schauder type and the Krasnosel'skii's fixed
point theorem, and find sufficient conditions of the existence
of at least $n$  or  $2n-1$ distinct  positive solutions by  means
of the Leggett-Williams fixed point theorem, the generalized
Avery-Henderson fixed point theorem as well as the Avery-Peterson
fixed point theorem.

The rest of the paper is organized as follows. In Section 2, we presents
some basic definitions and several fixed point
theorems. In Section 3, we state properties of the associated Green's function.
In Section 4, we discuss the completely continuous operator of fractional
differential \eqref{equ1}. In Section 5,
by using the nonlinear alternative of Leray-Schauder
type and the Krasnosel'skii's fixed point theorem, some new
sufficient conditions of the existence of at least \textit{one} or
\textit{two} positive solutions of fractional differential
\eqref{equ1} are obtained. In Section 6, the
existence criteria for at least three  or arbitrary $n$
or $2n-1$ positive solutions of fractional differential \eqref{equ1} are established. In Section 7, we present
two examples.


In this study, we assume that
$f(t,u_1,u_2): [0,1] \times [ 0,\infty )\times(-\infty,+\infty) \to [
0,\infty )$ satisfies the following conditions of
Carath\'eodory type:
\begin{itemize}

\item[(S1)] $f(t,u_1,u_2)$ is Lebesgue measurable with respect to
$t$ on $[0,1]$;

\item[(S2)] for a.e. $t\in[0,1]$, $f(t,.,.)$ is continuous on
$[0,1] \times [0,\infty )\times(-\infty,+\infty) $.
\end{itemize}


\section{Preliminaries}

In this section,  we present some basic definitions
and the fixed point theorems which help us to better understand
discussions presented in next a few sections.

\begin{definition}[\cite{pa1}]\rm
 The Riemann-Liouville fractional integral of order $a > 0 $ of a function
 $y:(0,\,\infty)\to\mathbb{R}$  is given by
$$
I_{0^+}^{\alpha}y(t)
=\frac{1}{\Gamma(\alpha)}\int_{0}^{t}(t-s)^{\alpha-1}y(s)ds,
$$
provided that the right side is pointwise defined on
$(0,\infty)$.
\end{definition}

\begin{definition}[\cite{pa1}] \rm
The Riemann-Liouville fractional derivative of order $a > 0 $ of a
function $y: (0,\infty)\to\mathbb{R}$ is given by
$$
D_{0^+}^{\alpha}y(t)=\frac{1}{\Gamma(n-\alpha)}(\frac{d}{dt})^{n}
\int_{0}^{t}(t-s)^{n-\alpha-1}y(s)ds,
$$
provided that the right side is pointwise defined
on $(0,\infty)$, where $n=[\alpha]+1$.
\end{definition}

The following are two fixed point theorems. The former one is the
so-called nonlinear alternative of Leray-Schauder
type and the latter one is the Krasnosel'skii's fixed point
theorem \cite{kr, a}.

\begin{lemma}\label{Lemma2.4}
Let $X$ be a Banach space with $C\subset X$ being closed and convex.
Assume that $U$ is a relatively
open subset of $C$ with $0\in U$  and
$A:\overline{U}\to C$ is a completely continuous operator,  then either
\begin{itemize}
\item[(i)] $A$  has a fixed point in $\overline{U}$, or

\item[(ii)] there exists $u\in \partial U$ and
$\gamma_1^*\in(0,1)$ with $u=\gamma_1^* A u$.
\end{itemize}
\end{lemma}


\begin{lemma}\label{Lemma2.3} Let $P$ be a cone in a Banach
space $E$. Assume $\Omega _1$ and $\Omega _2$ are open subsets of $E$
with $0\in \Omega _1$ and $\overline{\Omega}_1\subset \Omega _2$. If
$A:P\cap (\overline{\Omega}_2\backslash \Omega _1)\to P $
is a completely continuous operator such that either
\begin{itemize}
\item[(i)] $ \| Ax\| \leq \| x\|$ for all $x\in P\cap
\partial \Omega _1$, and $\| Ax\| \geq \|x\|$ for all
$x\in P\cap \partial \Omega _2$, or

\item[(ii)] $ \| Ax\| \geq \| x\|$ for all
$x\in P\cap \partial \Omega _1$  and
$\| Ax\| \leq \| x\|$ for all $x\in P\cap \partial \Omega _2$.
\end{itemize}
Then $A$ has a fixed point in
$P\cap (\overline{\Omega_2}\backslash \Omega _1)$.
\end{lemma}

Define that $P_c = \{u \in P : \|u\|< c\}$ and
$P(q, b, d) = \{u\in P : b \leq q(u),\,\|u\|\leq d\}$,
where the map $q$ is a nonnegative continuous concave functional on  $P$.
The following are two fixed-point theorems due to Leggett and Williams
\cite{p9}.

\begin{lemma} \label{lw}
Suppose that $A : \overline{P}_c \to \overline{P}_c$ is completely
continuous and there exists a concave positive functional $q$ on $P$ such that
$q(u) \leq \|u\|$ for $u \in \overline{P}_c$. Suppose that there
exist constants $0 < a < b < d \leq c$ such that
\begin{itemize}
\item[(i)] $\{u\in P(q, b, d) : q(u)
> b\} \neq \emptyset $ and $q(Tu) > b$ if $u \in P(q, b, d)$;

\item[(ii)] $\|Tu\| < a$ if $u \in P_a$;

\item[(iii)] $q(Tu) > b$ for $u \in P(q, b, c)$ with $\|Tu\| > d$.
\end{itemize}
Then $A$ has at least three fixed points $u_1, u_2$ and $u_3$ such
that
$$
\|u_1\| < a,\quad b < q(u_2) \text{ and }u_3 > a \text{ with }q(u_3) < b.
$$
\end{lemma}

For each $d>0$, let  $ P(\gamma,d)=\left\{ x\in P:\gamma (x)<d\right\}$,
 where  $\gamma$ is a nonnegative continuous functional on a cone $P$
of a real Banach space $E$.


\begin{lemma}\label{Lemma2.11}
Let $P$ be a cone in a real Banach space $E$. Let $\alpha,\beta $
and $\gamma $ be increasing,
nonnegative continuous functionals on $P$ such that for some $c>0$
and $H>0$, $ \gamma (x)\leq \beta (x)\leq \alpha (x)$  and
$\| x\| \leq H\gamma (x) $ for all $x\in \overline{P}(\gamma ,c)$.
Suppose that there exist  positive
numbers $a$ and $b$ with $a<b<c$, and $A: \overline{P}(\gamma
,c)\to P$ is a completely continuous operator  such that:
\begin{itemize}
\item[(i)] $\gamma (Ax)<c$ for all $x\in \partial P(\gamma ,c)$;

\item[(ii)] $\beta (Ax)>b$ for all $x\in \partial P(\beta ,b)$;

\item[(iii)] $P(\alpha ,a)\neq \emptyset $ and $\alpha (Ax)<a$ for
$x\in \partial P(\alpha ,a)$.
\end{itemize}
Then $A$ has at least three fixed points $x_1$, $x_2$ and $x_3$
belonging to $ \overline{P}(\gamma ,c)$ such that
\[
0\leq \alpha (x_1)<a<\alpha (x_2)\text{ with }\beta
(x_2)<b<\beta(x_3)\text{ and }\gamma (x_3)<c.
\]
\end{lemma}

Let $\beta$ and $\phi$ be nonnegative continuous convex
functionals on $P$, $\lambda$ be a nonnegative continuous concave
functional on $P$
and $\varphi$ be a nonnegative continuous functional
on $P$. We define the following convex sets:
\begin{gather*}
P(\phi ,\lambda ,b,d)=\{ x\in P:\, b\leq \lambda (x),\, \phi
(x)\leq d\},\\
P(\phi,\beta,\lambda,b,c,d)=\{ x\in
P:\, b\leq \lambda (x),\, \beta (x)\leq c,\, \phi (x)\leq d\},
\end{gather*}
and
$$
R(\phi,\varphi,a,d)=\{ x\in P:\, a\leq \varphi (x),\ \phi (x)\leq d\}.
$$
We are ready to recall the Avery-Peterson fixed point theorem \cite{ap}.

\begin{lemma}\label{Lemma2.7}
Let $P$ be a cone in a real Banach space
$E$, and $\beta$, $\phi$,  $\lambda$ and $\varphi $ be defined as
the above. Moreover,  $\varphi $  satisfies
$\varphi(\lambda' x)\leq \lambda' \varphi(x)$ for $0\leq\lambda'\leq1$
such that for some positive numbers $h$ and $d$,
\begin{equation}
\lambda (x)\leq \varphi (x),\quad \| x\|\leq h\phi (x)\label{hui}
\end{equation}
holds for all $x\in \overline{P(\phi ,d)}$. Suppose that $A$:
$\overline{P(\phi ,d)}\to \overline{P(\phi ,d)} $ is
completely continuous and there exist positive real numbers
$a,b,c$, with $a<b$ such that:
\begin{itemize}
\item[(i)] $\{ x\in P(\phi ,\beta ,\lambda ,b,c,d):\lambda
(x)>b\} \neq \emptyset $ and $\lambda (A (x))>b$ for
 $x$ in the set $ P(\phi ,\beta ,\lambda ,b,c,d)$;


\item[(ii)] $\lambda (A (x))>b$ for $x\in P(\phi ,\lambda ,b,d)$
with $\beta (A(x))>c$;

\item[(iii)] $0\notin R(\phi,\varphi,a,d)$ and  $\lambda (A(x))<a$
for all $x\in R(\phi,\varphi,a,d)$ with $\varphi (x)=a$.
\end{itemize}
Then $A$ has at least three fixed points
$x_1,x_2,x_3\in \overline{P(\phi ,d)}$ such that
\begin{equation*}
\phi(x_i)\leq d \text{  for i=1,\,2,\,3,  }\ b<\lambda (x_1),
\quad a<\varphi (x_2)\text{ and }  \lambda (x_2)<b\text{ with }
\varphi(x_3)<a.\end{equation*}
\end{lemma}


\section{Properties of Green's function}

In this section, we present some properties of  the Green's function
which will be used in our discussions. Note that using a similar discussion
as to proofs of Theorems 3.1 and 3.2 in \cite{10}, we have the following lemmas.

\begin{lemma}\label{23}
  Assume that $y(t)\in L[0,1]$, then
the   fractional differential equation
\begin{equation}\label{eq3-1}
\begin{gathered}
D_{0^+}^{\alpha}u(t)+y(t)=0,\quad  t\in (0,1 ),\; n-1<\alpha\leq n,\\
u^{(i)}(0)=0, \quad i=0,1,2,\dots,n-2,\\
[D_{0^+}^{\beta}u(t)]_{t=1}=0,\quad  2\leq\beta\leq n-2,
\end{gathered}
\end{equation}
has the unique solution
$$
u(t)=\int_0^1G(t,s)y(s)ds,
$$
where
\begin{equation}
G(t,s)=\begin{cases}
\frac{t^{\alpha-1}(1-s)^{\alpha-\beta-1}-(t-s)^{\alpha-1}}{\Gamma(\alpha)},
&0\leq s \leq t \leq 1, \\
\frac{t^{\alpha-1}(1-s)^{\alpha-\beta-1}}{\Gamma(\alpha)},
&0\leq t \leq s \leq 1,
\end{cases}\label{2a}
\end{equation}
is the Green's function of problem \eqref{eq3-1} with $n>4$.
\end{lemma}

\begin{lemma}\label{lem2} Let $G(t,s)$ be  given as \eqref{2a},
then:
\begin{itemize}
\item[(i)]  $G(t,s)$ is a continuous function on the unit square
$[0,1]\times [0,1]$;

\item[(ii)] $G(t,s)\geq 0$ for $(t,s)\in[0,1]\times [0,1]$;

\item[(iii)]  $\max_{t\in [0,1]}G(t,s)=G(1,s)$ for each $s\in [0,1]$;

\item[(iv)] there exists a constant $\gamma\in(0,1)$ such that
   $$
\min_{t\in [1/2,1]}G(t,s)\geq \gamma\max_{t\in [0,1]}G(t,s)
=\gamma G(1,s),
$$
where
\begin{equation}\label{3a}
\gamma=\min\Big\{\frac{(1/2)^{\alpha-\beta-1}}{2^{\beta}-1},\,
(1/2)^{\alpha-1}\Big\}.
\end{equation}
\end{itemize}
\end{lemma}

\begin{lemma}\label{lem3}
Let $G(t,s)$ be  given as {\rm\eqref{2a}}, then:
\begin{itemize}
\item[(i)]  $\frac{\partial G(t,s)}{\partial t}$ is a continuous
function on the unit square $[0,1]\times [0,1]$;

\item[(ii)] $\frac{\partial G(t,s)}{\partial t}\geq 0$ for
$(t,s)\in[0,1]\times [0,1]$;

\item[(iii)]  $\max_{t\in [0,1]}\frac{\partial G(t,s)}{\partial t}
= \frac{\partial G(1,s)}{\partial t}$   for each $s\in  [0,1]$;

\item[(iv)]   $\max_{t\in[0,1]}\frac{\partial
G(1,s)}{\partial t}\leq (\alpha-1)\max_{t\in
[0,1]}G(t,s)=(\alpha-1)G(1,s)$ for each $s\in
  [0,1]$.
\end{itemize}
   \end{lemma}

\begin{proof}[Proof of Lemma \ref{lem3}]
Note that
\begin{equation}
\frac{\partial G(t,s)}{\partial t}
=\begin{cases}
(\alpha-1)\frac{t^{\alpha-2}(1-s)^{\alpha-\beta-1}
 -(t-s)^{\alpha-2}}{\Gamma(\alpha)},& 0\leq s \leq t \leq 1, \\
(\alpha-1)\frac{t^{\alpha-2}(1-s)^{\alpha-\beta-1}}{\Gamma(\alpha)},
& 0\leq t \leq s \leq 1.
\end{cases} \label{qu3}
\end{equation}
 Let
\begin{gather*}
G_1(t,s)=(\alpha-1)\frac{t^{\alpha-2}(1-s)^{\alpha-\beta-1}
-(t-s)^{\alpha-2}}{\Gamma(\alpha)},\quad 0\leq s \leq t \leq 1;\\
G_2(t,s)=(\alpha-1)\frac{t^{\alpha-2}(1-s)^{\alpha-\beta-1}}{\Gamma(\alpha)},
\quad 0\leq t \leq s \leq 1.
\end{gather*}
It is easy to see that property (i) holds since $G_1$
and $G_2$ are continuous on their  domains and  $G_1(s,s)=G_2(s,s)$.

When $0\leq s \leq t \leq 1$, since $\alpha-2\geq \alpha-\beta-1$,
we have
\begin{align*}
\frac{\partial G(t,s)}{\partial t}
&= (\alpha-1)\frac{t^{\alpha-2}(1-s)^{\alpha-\beta-1}-(t-s)^{\alpha-2}}
{\Gamma(\alpha)}\\
&\geq  (\alpha-1)t^{\alpha-2}\frac{(1-s)^{\alpha-\beta-1}
 -(1-\frac{s}{t})^{\alpha-2}}{\Gamma(\alpha)}
\\
&\geq
(\alpha-1)t^{\alpha-2}\frac{(1-s)^{\alpha-\beta-1}
-(1-s)^{\alpha-2}}{\Gamma(\alpha)}
\geq  0.
\end{align*}
Hence, property (ii) is  true in view of
$G_2(t,s)\geq 0$ for $0\leq t \leq s \leq 1$.

When $0\leq s \leq t \leq 1$, since $\alpha-2\geq \alpha-\beta-1$,
we have
\begin{align*}
\frac{\partial G_1(t,s)}{\partial t}
&= (\alpha-1)(\alpha-3)\frac{t^{\alpha-3}(1-s)^{\alpha-\beta-1}
-(t-s)^{\alpha-3}}{\Gamma(\alpha)}
\\
 &\geq
(\alpha-1)(\alpha-2)t^{\alpha-3}\frac{(1-s)^{\alpha-\beta-1}
-(1-\frac{s}{t})^{\alpha-3}}{\Gamma(\alpha)}
\\
&\geq
(\alpha-1)(\alpha-2)t^{\alpha-3}\frac{(1-s)^{\alpha-\beta-1}
-(1-s)^{\alpha-3}}{\Gamma(\alpha)}
\geq  0.
\end{align*}
Similarly, when $0\leq t \leq s \leq 1$, we have
$$
\frac{\partial G_2(t,s)}{\partial t}
=(\alpha-1)(\alpha-2)\frac{t^{\alpha-3}(1-s)^{\alpha-\beta-1}}{\Gamma(\alpha)}
\geq 0.
$$
This implies that
$\frac{\partial G(t,s)}{\partial t}$ is increasing on its domain.
Moreover,
\begin{equation}\label{qu3b}
\max_{t\in[0,1]}\frac{\partial
G(t,s)}{\partial t}=\frac{\partial G(1,s)}{\partial t}\quad \text{for
each }\ s\in [0,1].
\end{equation}
So property (iii) holds.

When $0\leq s \leq 1$, we have
\begin{align*}
\max_{t\in[0,1]}\frac{\partial
G(1,s)}{\partial t}&= \frac{\partial G(1,s)}{\partial t}\\
&= (\alpha-1)\frac{(1-s)^{\alpha-\beta-1}-(1-s)^{\alpha-2}}{\Gamma(\alpha)}\\
&\leq
(\alpha-1)\frac{(1-s)^{\alpha-\beta-1}-(1-s)^{\alpha-1}}{\Gamma(\alpha)}\\
&= (\alpha-1)\max_{t\in [0,1]}G(t,s)\\
&=  (\alpha-1)G(1,s).
\end{align*}
This implies that property (iv) holds.
\end{proof}

\begin{lemma}\label{lem4}
Let $G(t,s)$ be  given as \eqref{2a}, then there
exists a constant $\gamma\in(0,1)$ such that
   $$
\min_{t\in [1/2,1]}\frac{\partial G(t,s)}{\partial t}\geq
   \gamma^*\max_{t\in [0,1]}\frac{\partial G(t,s)}{\partial t}
=\gamma^* \frac{\partial G(1,s)}{\partial t}.
$$
   \end{lemma}

\begin{proof}[Proof of Lemma \ref{lem4}]
It follows from the proof of Lemma \ref{lem3} that
\begin{align*}
\min_{t\in [1/2,1]}\frac{\partial
G(t,s)}{\partial t}
&=\begin{cases}
G_1(\frac{1}{2},s), & s\in(0,\frac{1}{2}] \\
G_2(\frac{1}{2},s), & s\in[\frac{1}{2},1)
\end{cases} \\
&= \begin{cases}
(\alpha-1)\frac{(1/2)^{\alpha-2}(1-s)^{\alpha-\beta-1}
-(\frac{1}{2}-s)^{\alpha-2}}{\Gamma(\alpha)},
 & s\in(0,\frac{1}{2}],  \\
(\alpha-1)\frac{(1/2)^{\alpha-2}(1-s)^{\alpha-\beta-1}}{\Gamma(\alpha)},
 & s\in[\frac{1}{2},1).
\end{cases}
\end{align*}
L'Hopital's  rule applies:
\begin{equation}\label{12}
\begin{split}
& \lim_{s\to 0+}\frac{(\frac{1}{2})^{\alpha-2}(1-s)^{\alpha-\beta-1}
 -(\frac{1}{2}-s)^{\alpha-2}}{(1-s)^{\alpha-\beta-1}-(1-s)^{\alpha-2}}\\
& = \lim_{s\to
0+}\frac{(\frac{1}{2})^{\alpha-2}(1-s)^{\alpha-\beta-1}-(\frac{1}{2}-s)
^{\alpha-2}}{(1-s)^{\alpha-\beta-1}[1-(1-s)^{\beta-1}]}\\
&=  \lim_{s\to
0+}\frac{-(\alpha-\beta-1)(\frac{1}{2})^{\alpha-2}(1-s)^{\alpha-\beta-2}+(\alpha-2)(\frac{1}{2}-s)
^{\alpha-3}}{-(\alpha-\beta-1)(1-s)^{\alpha-\beta-2}+(\alpha-2)(1-s)^{\alpha-3}}\\
& = \frac{(\frac{1}{2})^{\alpha-2}(\alpha+\beta-3)}{\beta-1}
> 0.
\end{split}
\end{equation}

When $0<s\leq 1/2$, a straightforward calculation gives
\begin{align*}
\frac{(1/2)^{\alpha-2}(1-s)^{\alpha-\beta-1}
 -(\frac{1}{2}-s)^{\alpha-2}}{(1-s)^{\alpha-\beta-1}-(1-s)^{\alpha-2}}
& = \frac{(\frac{1}{2})^{\alpha-2}(1-s)^{\alpha-\beta-1}-(\frac{1}{2}-s)
^{\alpha-2}}{(1-s)^{\alpha-\beta-1}[1-(1-s)^{\beta-1}]} \\
&\geq
\frac{(1/2)^{\alpha-\beta-2}}{2^{\beta-1}-1}.
\end{align*}
Similarly, when $\frac{1}{2} \leq s \leq 1 $, we have
\begin{equation*}
\frac{(\frac{1}{2})^{\alpha-2}}{1-(1-s)^{\beta}}\geq
(\frac{1}{2})^{\alpha-2}.
\end{equation*}
Define
$$
\overline{\gamma}(s):=
\begin{cases}
\frac{(1/2)^{\alpha-2}(1-s)^{\alpha-\beta-1}
-(\frac{1}{2}-s)^{\alpha-2}}{(1-s)^{\alpha-\beta-1}-(1-s)^{\alpha-2}},
&s\in(0,\frac{1}{2}],  \\
\frac{(1/2)^{\alpha-2}}{1-(1-s)^{\beta}}, &s\in[1/2,1],
\end{cases}
$$
where $\overline{\gamma}(0)=\lim_{s\to 0+}\overline{\gamma}(s)>0$ due to
\eqref{12}. Let
$$
\gamma^*=\Big\{\frac{(\frac{1}{2})^{\alpha-\beta-2}}{2^{\beta-1}-1},\,
 (\frac{1}{2})^{\alpha-2}\Big\}.
$$
It is obvious that $0<\gamma^* <1$. Consequently, we have
\[
\min_{t\in [1/2,1]}\frac{\partial
G(t,s)}{\partial t}= \overline{\gamma}(s)\max_{t\in
[0,1]}\frac{\partial G(t,s)}{\partial t}
\geq \gamma^*\max_{t\in [0,1]}\frac{\partial G(t,s)}{\partial t}
 =  \gamma^* \frac{\partial G(1,s)}{\partial t}.
\]
\end{proof}


\section{Completely Continuous  Operator}

In this section, we construct the completely continuous  operator for
our system, and show that finding the solution of
fractional differential \eqref{equ1} is equivalent  to finding
the fixed points of the associated  completely continuous  operator.

Let us denote  $E_1=C^1([0,1],\,\mathbb{R})$. Then
$E_1$ is a Banach space endowed with norm
$$
\| u\| =\max\{\|u\|_1,\ \|u\|_{2}\},
$$
where
$$
\|u\|_1=\sup_{t\in [0,1] } |u(t)|,\quad
\|u\|_{2}=\sup_{t\in [0,1] } | u'(t)|.
$$
The cone $P_1\subset E_1$ is defined by
\[
P_1=\big\{u\in E_1 :   u(0)=0 \text{ and }u(t)\geq 0\text { for }t\in [0,1]
\big\}.
\]
Assume that
$y(t)=f(t,u(t),u'(t))$, then it follows from  Lemma
\ref{23} that the solutions of  fractional differential \eqref{equ1}
are the corresponding  fixed points of the  operator
 $A: E_1\to E_1$, which is defined by
\begin{equation}
Au=\int_0^1G(t,s)f(s,u(s),u'(s))\,ds.\label{su11}
\end{equation}

\begin{lemma}\label{Lem213}
Suppose that  conditions {\rm (S1)} and {\rm (S2)} hold.
For  $t\in [0,1]$ and all $(u_1,u_2)\in [0,+\infty)\times(-\infty,+\infty)$,
we assume that there exist two nonnegative real-value functions
$a_1,\, a_2\in L[0,1]$ such that
\begin{equation}
f(t,u_1,u_2)\leq a_1(t)+a_2(t)\max_{t\in[0,1]}u_1(t),\label{qu1}
\end{equation}
or
\begin{equation}
f(t,u_1,u_2)\leq a_1(t)+a_2(t)\max_{t\in[0,1]}|u_2(t)|.\label{qu2}
\end{equation}
Then the operator $A:P_1 \to P_1$ is completely continuous.
\end{lemma}

\begin{proof}[Proof of Lemma \ref{Lem213}]
Firstly, we show that $A$: $P_1 \to P_1$ is  continuous.

Let $u\in P_1$. It is obvious that $Au(0)=0$ because of
$G(0,s)=0$. Suppose that  $\{u_n\}_{n=1}^\infty\subset
\overline{P}_1$ and $u_n(t)$ converges to $u(t)$
uniformly on $[0,1]$ as $n \to \infty$; that is,
   $$
\lim_{n\to\infty}\|u_n-u\|=0.
$$
So we have
$$
\lim_{n\to\infty}\|u_n-u\|_1=0 \text{ and }
   \lim_{n\to\infty}\|u_n-u\|_2=0,
$$
which implies that
 $$
\lim_{n\to\infty}u_n(t)=u(t) \text{ and }
   \lim_{n\to\infty}u'_n(t)=u'(t),\quad  t\in[0,1].
$$
It follows from (S1) that
   $$\lim_{n\to \infty}
   f(t,u_n(t),u_n'(t))=f(t,u(t),u'(t)),\quad
 t\in[0,1],
$$
 which gives
\begin{equation}
|Au_n(t)-Au(t)|
\leq \big|\int_0^1G(1,s)(f(s,u_n(s),u_n'(s))
-f(s,u(s),u'(s)))\,ds\big| \to 0
\label{lianxu1}
\end{equation}
as $n\to \infty$,
and
\begin{equation} \label{lianxu2}
\begin{split}
&|A'u_n(t)-A'u(t)| \\
&= |\int_0^1\frac{\partial G(t,s)}{\partial
t}(f(s,u_n(s),u_n'(s))
-f(s,u(s),u'(s)))\,ds| \\
&\leq \int_0^1\frac{\partial G(1,s)}{\partial t}\big|f(s,u_n(s),u_n'(s))
-f(s,u(s),u'(s))\big|d s\to 0,\quad \text{as }n\to \infty.
\end{split}
\end{equation}
By  \eqref{lianxu1} and  \eqref{lianxu2}, we have
$$
\|(Au_n)(t)-(Au)(t)\|\to 0 \quad \text{as } n \to \infty,
$$
 which means that $A$ is continuous.

Secondly, we show that $A$ maps bounded sets into
bounded sets in $P_1$. It suffices to show that for any
$\eta>0$, there is a positive constant $l>0$  such that for each
$u\in B_\eta=\{u\in P_1:\|u\|\leq \eta\}$, we have $\|Au\|\leq l$.
Let
$$
l=(\alpha-1)(\int_0^1a_1(t)G(1,s)d
s+\eta\int_0^1a_2(t)G(1,s)d s)>0.
$$
Using \eqref{qu1} yields
\begin{align*}
|Au(t)|&= \big|\int_0^1G(t,s)f(s,u(s),u'(s)) ds \big|
\\
&\leq
\int_0^1(a_1(t)+a_2(t)\max_{t\in[0,1]}u(t))G(1,s)d s\\
&\leq
\int_0^1a_1(t)G(1,s)d s+\int_0^1a_2(t)G(1,s)d s\|u(t)\|\\
&\leq
\int_0^1a_1(t)G(1,s)d s+\eta\int_0^1a_2(t)G(1,s)d s
<l.
\end{align*}
Using \eqref{qu2} yields
\begin{align*}
|Au(t)|&= \big|\int_0^1G(t,s)f(s,u(s),u'(s)) ds \big|
\\
&\leq \int_0^1(a_1(t)+a_2(t)\max_{t\in[0,1]}|u'(t)|)G(1,s)d
s\\
&\leq  \int_0^1a_1(t)G(1,s)d s+\int_0^1a_2(t)G(1,s)d
s\|u(t)\|\\
&\leq  \int_0^1a_1(t)G(1,s)d s+\eta\int_0^1a_2(t)G(1,s)d s
<l.
\end{align*}
In view of Lemma \ref{lem3},  we have
\begin{align*}
|A'u(t)|
&= \big|\int_0^1\frac{\partial G(t,s)}{\partial
t}f(s,u(s),u'(s)) ds \big|
\\
&\leq (\alpha-1) \int_0^1 G(1,s)f(s,u(s),u'(s))d s\\
&\leq (\alpha-1)(\int_0^1a_1(t)G(1,s)d s+\eta\int_0^1a_2(t)G(1,s)d
s)
<l.
\end{align*}
Hence, we have $\|Au\|\leq l$.

Thirdly, we consider that $A$ maps bounded sets into
equicontinuous  sets of $P_1$.
It follows from  Lemma \ref{qu3} that
$\frac{\partial G(t,s)}{\partial t}$ is continuous in $[0,1]\times [0,1]$. In
addition, $G(t,s)$ is continuous in $[0,1]\times [0,1]$, then
$\frac{\partial G(t,s)}{\partial t}$ and  $G(t,s)$ are uniformly
continuous in $[0,1]\times [0,1]$. Take $t_1,\,t_2\in [0,1]$.
For any $\varepsilon>0$,
 there exists $\delta>0$, whenever $|t_1-t_2|<\delta , $ we have
 $$
|G(t_2,s)-G(t_1,s)|<\frac{\varepsilon}{1+a_1(s)+\eta
 a_2(s)},
$$
and
 $$
\big|\frac{\partial
G(t_2,s)}{\partial t}-\frac{\partial G(t_1,s)}{\partial
t}\big|<\frac{\varepsilon}{1+a_1(s)+\eta
 a_2(s)}.
$$
For convenience, we assume $t_1<t_2$. For any $u\in B_\eta$,
according to \eqref{qu1} and \eqref{qu2}, we obtain
\begin{align*}
|Au(t_2)-Au(t_1)|
&= \big|\int_0^1(G(t_2,s)-G(t_1,s))f(s,u(s),u'(s))\,ds\big|
\\
&\leq  \int_0^1|G(t_2,s)-G(t_1,s)|[a_1(s)+\eta a_2(s)]
\,ds
< \varepsilon,
\end{align*}
and
\begin{align*}
|A'u(t_2)-A'u(t_1)|
&= \big|\int_0^1[\frac{\partial
G(t_2,s)}{\partial t}-\frac{\partial G(t_1,s)}{\partial
t}]f(s,u(s),u'(s))\,ds\big|
\\
&\leq  \int_0^1\big|\frac{\partial
G(t_2,s)}{\partial t}-\frac{\partial G(t_1,s)}{\partial
t}\big|[a_1(s)+\eta a_2(s)]\,ds
< \varepsilon.
\end{align*}
Consequently,
\begin{equation*}
\|Au(t_2)-Au(t_1)\|<\varepsilon,
\end{equation*}
which implies that the family of functions $\{Au:u\in B_\eta\}$ is
equicontinuous. By  the Arzela-Ascoli theorem,
we conclude that the operator
$A: P_1\to P_1$ is completely continuous.
\end{proof}

\begin{remark}\label{rem}\rm
If $f(t,u_1,u_2):[0,1] \times [0,\infty )\times(-\infty,+\infty) \to
[0,\infty )$ is continuous, we can see that
$A:P_1\to P_1$ is completely continuous by using a similar
argument as the above.
 \end{remark}

Since $u(t)=u(0)+\int_0^tu'(s)d s$, it leads to
\begin{equation*}
\max_{t\in[0,1]}u(t)=u(0)+\max_{t\in[0,1]}\int_0^tu'(s)d
s \leq\max_{t\in[0,1]}|u'(s)|.
\end{equation*}
That is,

\begin{lemma}\label{Lem21} If $u\in P_1$, then
$ \max_{t\in[0,1]}u(t)\leq  \max_{t\in[0,1]} |u'(t)|$.
 \end{lemma}

\section{Existence of one or two solutions}

 In this section,  we discuss the existence of single or twin  positive
 solutions to problem \eqref{equ1} by the nonlinear alternative of Leray-Schauder
 type and the Krasnosel'skii's fixed point theorem, respectively.

\begin{theorem}\label{3.1}
 Assume that all assumptions of Lemma \ref{Lem213}    hold, and
$$
(\alpha-1)\int_0^{1}G(1,s)a_2(s)\,ds<1,
$$
Then \eqref{equ1} has at least one positive solution.
\end{theorem}

\begin{proof}
Let
$U=\{u\in P_1:\|u\|<r\}$,
where
$$
r=\frac{(\alpha-1)\int_0^1G(1,s)a_1(s)ds}{1-(\alpha-1)
\int_0^1G(1,s)a_2(s)ds}>0.
$$
It is easily seen that the operator $A:\overline{U}\to P_1$
defined by \eqref{su11} is completely continuous.

Assume that there exist $u\in P_1$ and $\gamma_1^*\in (0,1)$ such
that $u=\gamma_1^* Au$. The we find that
\begin{align*}
u(t) &=  \gamma_1^* Au\\
&=  \gamma_1^*\int_0^1G(t,s)f(s,u(s),u'(s))\,ds
\\
&\leq  \gamma_1^*\int_0^1G(1,s)[a_1(s)+a_2(s)\|u\|]\,ds
\\
&\leq  (\alpha-1)\gamma_1^*[\int_0^1G(1,s)a_1(s)d
s+\|u\|\int_0^1G(1,s)a_2(s)\,ds],
\end{align*}
and
\begin{align*}
|u'(t)| &=  |\gamma_1^* A'u(t)|\\
&=  |\gamma_1^*\int_0^1\frac{\partial G(t,s)}{\partial
t}f(s,u(s),u'(s))\,ds|
\\
&\leq  (\alpha-1)
\gamma_1^*\int_0^1G(1,s)[a_1(s)+a_2(s)\|u\|]\,ds
\\
&\leq  (\alpha-1)\gamma_1^*[\int_0^1G(1,s)a_1(s)d
s+\|u\|\int_0^1G(1,s)a_2(s)\,ds].
\end{align*}
Thus, we have
\[
\|u\| < (\alpha-1)\gamma_1^*[\int_0^1G(1,s)a_1(s)d
s+r\int_0^1G(1,s)a_2(s)\,ds]
=  \gamma_1^*r,
\]
which means that $u\notin \partial U$ and $\|u\|\neq r$.

By  Lemma \ref{Lemma2.4}, we conclude that \eqref{equ1}
 has at least one positive solution.
\end{proof}

\begin{remark} \rm
In Theorem \ref{3.1}, ``All assumptions
of Lemma \ref{Lem213} hold"  can be replaced by
``$f(t,u_1,u_2):[0,1] \times [0,\infty)\times(-\infty,+\infty) \to [0,\infty
)$ is continuous".
\end{remark}

\begin{theorem} \label{Theorem3.2}
Assume that all assumptions of Lemma
\ref{Lem213}   and  the following conditions hold:
\begin{itemize}
\item[(i)] there exists a constant $p>0$ such that
$f(t,u_1,u_2)\leq p\Lambda _1$ for
$(t,u_1,u_2)\in[0,1]\times[0,p]\times[-p,p]$, where
$ \Lambda _1= ((\alpha-1)\int_0^{1}G(1,s)\,ds) ^{-1}$;

\item[(ii)] there exists a constant $q>0$ such that
$f(t,u_1,u_2)\geq q\Lambda _2$ for
$(t,u_1,u_2)\in[1/2,1]\times [
0,q]\times[-q,q] $, where $ \Lambda
_2=(\gamma \int_{1/2}^{1}G(1,s)\,ds ) ^{-1}$,
and $p\neq q$.
\end{itemize}
Then problem  \eqref{equ1} has at least one positive
  solution $u$ such that $\|u\|$ lies in between $p$
and $q$.
\end{theorem}

\begin{proof}
 Without loss of generality, we assume that $p<q$.
Let
$$
\Omega _{p}=\{ u\in E_1:\| u\| <p\}.
$$
For any $u\in P_1\cap \partial \Omega _{p}$, we see that
$$
\max_{t\in[0,1]}|u'(t)|\leq \|u\|<p \quad \text{and}\quad
\max_{t\in[0,1]}u(t)\leq\|u\|<p.
$$
It follows from Lemma \ref{lem3} and condition (i) that
\begin{equation} \label{feng}
\begin{aligned}
&\| Au\|\\
&=  \max\{\|Au\|_1, \|Au\|_{2}\}\\
&=  \max\Big\{\max_{t\in
[0,1]}\int_0^1G(t,s)f(s,u(s),u'(s))\,ds,\max_{t\in
[0,1]}\int_0^1\frac{\partial G(t,s)}{\partial
t}f(s,u(s),u'(s))\,ds\Big\}
\\
&=  (\alpha-1)\int_0^1G(1,s)f(s,u(s),u'(s)) ds
\\
&< (\alpha-1)p\Lambda_1\int_0^1G(1,s)\,ds,
\end{aligned}
\end{equation}
which implies that
\begin{equation}
\| Au\| \leq \| u\| \quad \text{for }u\in P_1\cap \partial \Omega _{p}.
 \label{4.1j}
\end{equation}

We define
$$
\Omega _{q}=\{ u\in E_1:\| u\|<q\}
$$
for arbitrary  $u\in P_1\cap \partial \Omega _{q}$,
and find
$$ \max_{t\in[0,1]}|u'(t)|\leq \|u\|<q \quad \text{and}\quad
\max_{t\in[0,1]}u(t)\leq\|u\|<q.
$$
On the other hand, it follows from Lemma \ref{lem2}
and condition (ii) that
\begin{align*}
\| Au\|
 &\geq  \max_{t\in [0,1]}\int_0^1G(t,s)f(s,u(s),u'(s))\,ds \\
&\geq  \int_0^\frac{1}{2}G(t,s)f(s,u(s),u'(s))\,ds
+\int_{1/2}^1G(t,s)f(s,u(s),u'(s))\,ds \\
&\geq
\int_{1/2}^1G(t,s)f(s,u(s),u'(s))\,ds \\
&\geq
\int_{1/2}^1\min_{t\in[1/2,1]}G(t,s)f(s,u(s),u'(s))
\,ds \\
&\geq  \int_{1/2}^1\gamma G(1,s)q\Lambda_2\,ds\\
&\geq  \gamma q\Lambda_2 \int_{1/2}^1G(1,s)\,ds,
\end{align*}
which implies that
\begin{equation}
\| Au\| \geq \| u\| \quad \text{for }u\in P_1\cap
\partial \Omega _{q}.  \label{4.2j}
\end{equation}

In view of \eqref{4.1j} and (\eqref {4.2j}, it
follows from Lemma \ref{Lemma2.3}, that problem
  \eqref{equ1} has a positive  solution $u$ in
$P_1\cap (\overline{\Omega }_{q}\backslash \Omega _{p})$.
 \end{proof}


For $u_1,u_2\in P_1$, we denote
 \begin{gather*}
 f^0 =  \lim_{(u_1,u_2)\to (0,0)}\ \sup_{t\in[0,1]}\frac{f(t,u_1,u_2)}{|u_2|},\\
f_\infty =  \lim_{u_1+|u_2|\to \infty} \ \inf_{t\in[0,1]}
\frac{f(t,u_1,u_2)}{u_1+|u_2|},\\
f_0 =  \lim_{(u_1,u_2)\to (0,0)}  \inf_{t\in[0,1]}
\frac{f(t,u_1,u_2)}{u_1+|u_2|},\\
f^\infty=\lim_{(u_1,u_2)\to (\infty,\infty)}
\sup_{t\in[0,1]}\frac{f(t,u_1,u_2)}{|u_2|}.
\end{gather*}
Now, we have the following two theorems.

\begin{theorem} \label{Theorem3.41}
Assume that all assumptions  of Lemma
\ref{Lem213} are satisfied, and
$f^0\in [0, \Lambda _1 ) $ and
$f_\infty \in (\Lambda_2,\infty)\cup\{\infty\}$. Then
\eqref{equ1} has at least one positive solution.
\end{theorem}

\begin{proof}
According to the assumption $f^0<\Lambda _1$ and Lemma \ref{Lem21},
there exists a sufficiently small $p>0$ such that
\begin{equation*}
f(t,u,u')\leq  \Lambda _1|u'|\leq \Lambda _1p~~\text{ for
}(t,u,u')\in[0,1]\times [0,p]\times[-p,p],
\end{equation*}
which implies that the condition (i) of Theorem \ref{Theorem3.2}
is true. That is, if we let
$$
\Omega _{p}=\{ u\in E_1:\|u\| <p\} ,
$$
then \eqref{4.1j} is satisfied.
It follows from $f_\infty > \Lambda _2 $ that there exists an
$H>2p$ satisfying
\begin{equation}
 f(t,u,u')\geq \Lambda _2(u+|u'|)\geq\Lambda _2\|u\|,\label{11}
\end{equation}
where $t\in[0,1]$  and $u+|u'|\geq H$.
Set  \begin{equation*}
\Omega _{H}=\{ u\in E_1: u+|u'|<H\} ,
\end{equation*}
then we see that $\overline{\Omega} _{p}\subset\Omega _{H}$.

For any $u\in P_1\cap \partial \Omega _{H}$, we have
 $u+|u'|=H$. Using \eqref{11} gives
\begin{align*}
\| Au\| &\geq  \max_{t\in [0,1]}\int_0^1G(t,s)f(s,u(s),u'(s))\,ds \\
&\geq \int_{1/2}^1G(t,s)f(s,u(s),u'(s))\,ds \\
&\geq \int_{1/2}^1\min_{t\in[1/2,1]}G(t,s)f(s,u(s),u'(s))
\,ds \\
&\geq  \int_{1/2}^1\gamma G(1,s)\Lambda_2 \|u\|d s\\
&\geq  \gamma \Lambda_2 \int_{1/2}^1G(1,s)\,ds\|u\|
=\| u\|.
\end{align*}
Consequently, by  Lemma \ref{Lemma2.3}, we conclude that
  \eqref{equ1} has a positive  solution $u$ in
 $P_1\cap (\overline{\Omega }_{H}\backslash \Omega _{p})$.
\end{proof}

\begin{theorem} \label{Theorem3.4}
Assume that all assumptions of Lemma \ref{Lem213} hold, and
that $f_0\in (  \Lambda _2 ,\infty ) \cup\{\infty\}$ and
$f^\infty \in [0,\Lambda _1)$. Then \eqref{equ1} has at
least one positive solution.
\end{theorem}

\begin{proof}
It follows from $f_0>\Lambda _2$ that there exists a
sufficiently small $q>0$ such that
\begin{align*}
f(t,u,u')
 &\geq  \Lambda _2(u+|u'|)\\
&\geq  \Lambda _2\max \{\|u\|_1,\|u\|_{2}\} \quad
\text{for }(t,u,u')\in[0,1]\times [0,q]\times[-q,q].
\end{align*}
When $(t,u,u')\in [1/2,1]\times[0,q ]\times[-q,q]$,
we obtain
\[
f(t,u,u') \geq  \Lambda _2\max \{\|u\|_1,\|u\|_{2}\}
=  \Lambda _2q,
\]
which implies that the condition (ii) in Theorem \ref{Theorem3.2}
is satisfied. Take
$$
\Omega _{q}=\{ u\in E_1:\| u\| <q\}.
$$
Then, inequality \eqref{4.2j} holds.

Let $\varepsilon_1= \Lambda _1-f^\infty\ (>0)$. Since
$f^\infty<\Lambda _1$, there exists a $p_1$  $ (>q)$  such
that
\begin{equation}
f(t,u,u')\leq (\varepsilon_1+f^\infty)|u'|=
 \Lambda _1|u'|, \label{3.1a}
\end{equation}
where $(t,u,u')\in [0,1]\times[p_1,\infty)
\times(-\infty, -p_1]\cup[p_1,+\infty )$.
Note that
$$
f\in C([0,1]\times[0,\infty )\times(-\infty,\infty ),[0,\infty )).
$$
So there exists a $C_4>0$ satisfying
\begin{equation}
f(t,u,u')\leq C_{4}\text{ for }(t,u,u')\in
[0,1]\times[0,p_1]\times[-p_1,p_1].\label{3.1a1}
\end{equation}
According to \eqref{3.1a} and \eqref{3.1a1}, we have
$$
f(t,u,u')\leq \max\{C_4,  \Lambda _1|u'|\} \quad
\text{for }(t,u,u')\in [0,1]\times[0,\infty )\times(-\infty,
\infty ).
$$
Let
$$
p_2^*>\max\{C_4/\Lambda _1,2q\},
$$
and
$$ \Omega _{p_2}=\{ u\in E_1:\|u\| <p_2^*\}.
$$
If $u\in P_1\cap \partial \Omega_{p_2}$, by \eqref{feng},
  one has $\|u\| =p_2^*$ and
\begin{align*}
\| Au\|
&\leq  (\alpha-1)\int_0^1G(1,s)f(s,u(s),u'(s))\,ds
\\
&\leq  (\alpha-1)\int_0^1G(1,s) \max\left\{C_4,  \Lambda _1|u'|\right\}\,ds \\
&\leq  (\alpha-1)\Lambda_1p_2^*\int_0^1G(1,s)\,ds
=  \|u\|.
\end{align*}
This implies our desired result.
\end{proof}


Next, we deal with  the existence of at least two distinct positive
solutions to problem \eqref{equ1}.


\begin{theorem} \label{Theorem3.7}
Assume that all assumptions of Lemma
\ref{Lem213} hold. Moreover,  suppose that $f_0=\infty $
and $f_\infty =\infty$, and the condition (i) in
Theorem \ref{Theorem3.2} is satisfied. Then problem
\eqref{equ1} has at least two distinct positive solutions
$u_1$, $u_2\in P_1$.
\end{theorem}

\begin{proof}
 In view of $f_0=\infty$, there exists an $H_1$  such that $0<H_1<p$ and
\begin{equation}
f(t,u,u')\geq m(u+|u'|)\geq m \|u\|\quad \text{for }(t,u,u')\in[0,1]\times(0,
H_1]\times[-H_1,H_1],\label{kn11}
\end{equation}
 where $m$ is given by
\begin{equation}
\gamma m\int_{1/2}^{1} G(1,s)d s\geq  1.\label{kn1}
\end{equation}
Take
$$
\Omega _{H_1}=\{ u\in E_1:\| u\| <H_1\} .
$$
If $u\in \Omega _{H_1}$ with $\| u\| =H_1$, it means that
   \begin{equation*}
\max_{t\in[0,1]}u(t)\leq\max_{t\in[0,1]}|u'(t)|
\leq \|u||=H_1\text{ for }t\in[0,1]
.\end{equation*}
It follows from  \eqref{kn11} and  \eqref{kn1} that
\begin{align*}
\| Au\|
&\geq \int_{1/2}^1\min_{t\in[1/2,1]}G(t,s)f(s,u(s),u'(s))
\,ds \\
&\geq  \int_{1/2}^1\gamma G(1,s)m \|u\|d s\\
&\geq  \gamma m \int_{1/2}^1G(1,s)\,ds\|u\|,
%\label{sd}
\end{align*}
which implies that
$$
\| Au\| \geq \| u\|\quad \text{ for } u\in P_1\cap
\partial \Omega _{H_1}.
$$
Let
$$
\Omega _{p}=\{ u\in E_1:\| u\| <p\}.
$$
Then, we obtain that \eqref{4.1j} holds by using the
condition \item[(i)] of Theorem \ref{Theorem3.2}.
According to Lemma \ref{Lemma2.3},  problem
  \eqref{equ1} has a positive  solution $u_1$ in
 $P_1\cap (\overline{\Omega }_{p}\backslash \Omega _{H_1})$.

It follows from   $f_\infty =\infty $ that there exists an
$H_2>4p$ such that
\begin{equation}
f(t,u,u')\geq k(u+|u'|)\geq k \|u\|,\label{sa1}
\end{equation}
where $t\in[0,1]$ and $u+|u'|\geq H_2$. Moreover, $k$
satisfies
$$
k\gamma \int_{1/2}^{1}G(1,s)d r\geq 1.
$$
Let
\begin{equation*}
\Omega_{H_2}=\{ u\in \Omega _{H}:u+|u'|<H_2\} ,
\end{equation*}
then we see that $\overline{\Omega} _{p}\subset\Omega _{H_2}$.

For  any $u\in P_1\cap \partial \Omega _{H_2}$, we have
$u+|u'|=H_2$. According to
 \eqref{sa1}, we deduce that
\begin{align*}
\| Au\| &\geq
\int_{1/2}^1\min_{t\in[1/2,1]}G(t,s)f(s,u(s),u'(s))
\,ds \\
&\geq  \int_{1/2}^1\gamma G(1,s)k \|u\|d s\\
&\geq  \gamma k
\int_{1/2}^1G(1,s)\,ds\|u\|
\geq  \| u\|.% \label{sc}
\end{align*}
 Thus,  it follows from (i) of Lemma
\ref{Lemma2.3} that   problem \eqref{equ1} has at least a
single positive   solution   $u_2$ in
$P_1\cap (\overline{\Omega }_{H_2}\backslash \Omega _{p})$ with
$$
p\leq \| u_2\| \text{ and }u_2+|u_2'|\leq H_2.
$$
It is easily seen that $u_1$ and $u_2$ are distinct.
\end{proof}

By a closely similar way, we can obtain the following result.

\begin{theorem} \label{Theorem3.8}
 Assume that all assumptions of Lemma
\ref{Lem213} hold. Moreover, suppose that $f^0=0 $ and
$f^\infty =0$, and the condition (ii) in Theorem
\ref{Theorem3.2} is satisfied, then problem \eqref{equ1}
has at least two distinct positive
 solutions $u_1$, $u_2\in P_1$.
\end{theorem}

\section{Existence of triple or multiple solutions}

We have obtained  some  existence results of at least one or two
distinct positive solutions to  fractional differential \eqref{equ1}
in the preceding section. In this section, we will
further discuss the existence of at least $3$, $n$  or $2n-1$
positive solutions to fractional differential \eqref{equ1} by using
different fixed point theorems.

For the notational convenience, we define
\[
M =\int_{1/2}^1G(1,s)d s, \quad
N =\gamma\int_{1/2}^1G(1,s)\,ds,\quad
L=(\alpha-1)\int_0^1G(1,s)\,ds.
\]

\subsection{Existence of Three  Solutions}

In this subsection, we investigate the existence of at least
three distinct positive  solutions of \eqref{equ1}.


\begin{theorem} \label{Theorem4.1'}
Let $a, b$ and $c$ be constants such that
$0 <a < b <d\leq c$ and $bL<cN$. In addition,
 if  all assumptions of Lemma
\ref{Lem213} hold and $f(t,u_1,u_2)$ satisfies
the following conditions:
\begin{itemize}
\item[(i)] $f(t,u_1,u_2) < \frac{a}{L}$ for
$(t,u_1,u_2)\in [0, 1]\times [0, a]\times [-a, a]$;


\item[(ii)] $f(t,u_1,u_2)
> \frac{b}{N}$ for  $(t,u_1,u_2) \in [\frac{1 }{2} , 1]
\times [b, d]\times [-c, c]$;

\item[(iii)] $f(t,u_1,u_2)\leq \frac{c}{L}$ for
$(t,u_1,u_2) \in [0, 1]\times [0, c]\times [-c, c]$.
\end{itemize}
Then \eqref{equ1} has at least three positive
solutions $u_1, u_2, u_3 \in P_1$  such that
\begin{equation}\label{eeq1}
0< \|u_1\| < a,\quad
b <\inf_{t\in[1/2,1]}u_2,\quad
a< u_3 \quad \text{with }
\inf_{t\in [1/2,1]}u_3 < b.
\end{equation}
\end{theorem}

\begin{proof}
By the definition of the completely continuous operator $A$
and by Lemma \ref{lw}, we consider all conditions of Lemma
\ref{lw} with respect to $A$.
Let
$$
q(u) =\inf_{t\in [1/2,1]}u(t)\quad
\text{for }u\in \overline{P}_1,
$$
then $q(u)$ is a nonnegative continuous
concave function and satisfies
$$
q(u) \leq \|u\|\text{ for } u \in P_{1_{c}}=\{u\in P_1:\|u\|\leq c\}.
$$
Since
$$
u\in[0,c] \text{ and } u'\in[-c,c] \text{ for }u \in
P_{1_{c}},
$$
according to condition (iii) and \eqref{feng},  we
have
\begin{equation}
\begin{split}
\|A u\|
&\leq (\alpha-1)\int_0^1G(1,s)f(s,u(s),u'(s))\,ds\\
&\leq  (\alpha-1) \int_0^1 G(1,s)\frac{c}{L}\,ds
\leq c,
\end{split}\label{4.2s}
\end{equation}
which implies $A : \overline{P}_{1_{c}} \to \overline{P}_{1_{c}}$.
When $u \in P_{1_{a}}=\{u\in P_1:\|u\|\leq a\}$, it
implies that
$$
u\in[0,a]\text{ and  }u'\in[-a,a].
$$
 We observe that the conditions (ii) of Lemma \ref{lw} is true.

Let $d$ be a fixed constant such that $b < d \leq c$, then we have
$q(d)= d > b $ and $\|d\| = d$. This means that
$$
d\in P_1(q, b, d)=\{u\in P_1 : b \leq q(u),\|u\|\leq d\}.
$$
For any $u \in P_1(q, b,d)$, we obtain
$$
\|u\| \leq d\text{  and }q(u) = \inf_{t\in [1/2,1]}u \geq b,
$$
which implies
$$
u\in [b,d ] \text{ and } u'\in[-d,d] \quad\text{for }
t\in [1/2,1].
$$
Hence,
\begin{align*}
q(Au)&= \inf_{t\in [1/2,1]}Au\\
&= \int_{1/2}^1\inf_{t\in[1/2,1]}G(t,s)f(s,u(s),u'(s))\,ds \\
&> \int_{1/2}^1\gamma G(1,s)\frac{b}{N}\,ds
> b,\end{align*}
which means that condition (i) of Lemma \ref{lw} holds.

For any $u \in P_1(q, b, c)$ with $\|Au\|> d$, it gives
 $\|u\|\leq c$ and $\inf_{t\in[1/2,1 ]}u\geq b $.
By using the same argument as the above,
we see that $q(Au) > b$. This implies that the condition (iii)
 of Lemma \ref{lw} is fulfilled.

Consequently, all conditions of  Lemma \ref{lw} are verified. That
is, problem\eqref{equ1} has at least three distinct
solutions distributed as \eqref{eeq1}.
\end{proof}

\begin{corollary} \label{Corollary5.1s*}
Assume that all assumptions of Lemma
\ref{Lem213} hold. If the condition (iii) in Theorem
\ref{Theorem4.1'} is replaced by
\begin{itemize}
\item[(iii')] $f^\infty=\lim_{(u_1,u_2)\to
(\infty,\infty)}\sup_{t\in[0,1]}\frac{f(t,u_1,u_2)}{|u_2|}\leq
\frac 1{L}$ for
$u_1,u_2\in P_1$,
\end{itemize}
then \eqref{eeq1} in Theorem  \ref{Theorem4.1'} also holds.
\end{corollary}

\begin{proof}
 We only need to prove that the condition (iii')
implies the condition (iii) in Theorem \ref{Theorem4.1'}. That
is, assume that (iii') holds, then there exists a
number $ c^*\geq  d^*$ such that
$$
 f(t,u,u')\leq \frac{c^*}{L} \text{ for }(t,u,u')\in[0,1]\times
[0,c^*]\times[-c^*,c^*].
$$
Conversely, we suppose that  for any $c^*\geq d^*$, there exists
$$
(u_{c},u'_{c})\in [0,c^*]\times[-c^*,c^*]
$$
such that
$$
f(t,u_{c},u'_{c})> \frac{c^*}{L}\quad \text{for }t\in[0,1] .
$$
Take
$$
c_n^*> d^*\quad  (n=1,2,\dots )\text{ with } c_n^*\to \infty.
$$
Then, there exists
$$
(u_n,u'_n)\in [0,c_n^*]\times[-c_n^*,c_n^*]
$$
such that
\begin{gather}
f(t,u_n,u'_n)> \frac{c_n^*}{L} \quad \text{for }t\in[0,1],\label{5.1}\\
\lim_{n\to \infty }f(t,u_n,u_n')=\infty \quad \text{for }
t\in[0,1]. \label{5.2}
\end{gather}

Since  condition (iii') holds,  there is a $\tau >0$ such that
\begin{equation}
f(t,u,u')\leq\frac {|u'|}{L} \quad \text{for }(t,u,u')\in[0,1]
\times[\tau,\infty)\times(-\infty,\tau]\cup
[\tau,\infty).  \label{5.3}
\end{equation}
Thus,
 $$
|u'_n(t)|\leq \tau\quad \text{for } t\in[0,1],
$$
which implies $u_n(t)\leq \tau \text{ for }t\in[0,1]$.
 Otherwise, if
   $$
|u'_n(t)|>\tau \text{  and } u_n(t)>\tau  \quad\text{ for }t\in[0,1],
$$
  it follows from  \eqref{5.3} that
\begin{equation*}
f(t,u_n,u'_n)\leq \frac{|u'_n|}{L} \leq \frac{c_n^*}{L} \quad
\text{for } t\in[0,1],
 \end{equation*}
which contradicts inequality \eqref{5.1}.

Let
$$
W=\max_{(t,u,u')\in [0,1 ]\times [0,\tau]\times [-\tau,\tau ]}f(t,u,u'),
$$
so we have
$$
f(t,u_n,u'_n)\leq W \ (n=1,2,\dots ).
$$
This yields a contradiction with formula
\eqref{5.2}. The proof is complete.
\end{proof}


\subsection{Existence of $n$ solutions}

In this subsection,  the existence criteria for at least
three  or an arbitrary number $n$  positive solutions to
fractional differential \eqref{equ1} are obtained by
using the generalized  Avery-Henderson fixed point theorem.

 We define the nonnegative, increasing,
continuous functionals $\gamma_1 $, $\beta_1 $ and $\alpha_1 $ by
\begin{gather*}
\gamma_1 (u)= \beta_1(u)=\max\{\inf_{t\in [1/2,1]}u,\, \inf_{t\in [1/2,1 ]}|u'|\}
\quad \text{for } u\in P_1,\\
\alpha_1 (u)=\max\{\sup_{t\in [0,1 ]}u,\ \sup_{t\in [0,1 ]}|u'|\}
\quad \text{for } u\in P_1,
\end{gather*}
so we have
$$
\gamma_1 (u)=\beta_1 (u)\leq \alpha_1 (u) \quad \text{for each } u\in P_1.
$$
Since
$u(t)=\int_0^1G(t,s)y(s)\,ds$, in view of Lemma \ref{lem2},  we deduce
\begin{align*}
\inf_{t\in[1/2,1]}u(t) 
= \min_{t\in [1/2,1]}u(t)
&=\int_0^1\min_{t\in [1/2,1]}G(t,s)y(s)\,ds\\
&\geq  \gamma \max_{t\in [0,1]} \int_0^1G(t,s)y(s)\,ds\\
&=  \gamma\|u\|_1. %\label{1}
\end{align*}
 According to Lemma \ref{lem4},
\begin{align*}
\inf_{t\in [1/2,1]}|u'(t)|
&= \int_0^1\min_{t\in [1/2,1]}\frac{\partial G(t,s)}{\partial t}y(s)\,ds\\
&\geq  \gamma^* \max_{t\in [0,1]}\int_0^1
 \frac{\partial G(t,s)}{\partial t}y(s)\,ds\\
&=  \gamma^*\|u\|_2.%\label{1}
\end{align*}
Hence, we have
$$
\| u\| \leq \max\{\frac{1}{\gamma},\frac{1}{\gamma^*}\}\gamma _1(u)
=\frac{1}{\gamma}\gamma _1(u)\quad \text{for all }u\in P_1.
$$


\begin{theorem}\label{Theorem4.1a}
Assume that there exist real numbers $a',b',c'$ with $a'<b'<c'$
such that $0<Lb'<a'N. $ In addition, if
all assumptions of Lemma \ref{Lem213} hold and
$f(t,u_1,u_2)$ satisfies the following conditions:
\begin{itemize}
\item[(i)] $f(t,u_1,u_2)<\frac{c'}{L}$ for
$(t,u_1,u_2)\in [1/2,1]\times
[0,\frac{1}{\gamma}c']\times[-\frac{1}{\gamma
}c',\frac{1}{\gamma }c']$;

\item[(ii)] $f(t,u_1,u_2)>\frac{b'}{N}$ for
$(t,u_1,u_2)\in [1/2,1]\times
[0,\frac{1}{\gamma}b']]\times[-\frac{1}{\gamma
}b',\frac{1}{\gamma }b']$;

\item[(iii)] $f(t,u_1,u_2)<\frac{a'}{L}$ for
$(t,u_1,u_2)\in [0 ,1]\times
[0,a']\times[-a',a']$.
\end{itemize}
Then \eqref{equ1} has at least three distinct
positive solutions $ u_1 , u_2 ,u_3\in \overline{P_1}(\gamma_1,c')$
such that
\begin{equation} \label{eeq2}
\begin{gathered}
0< \|u_1\|<a'<\|u_2\|,\\
\max\{\inf_{t\in [1/2,1]}u_2,\, \inf_{t\in [1/2,1]}|u_2'|\}<b'
<\max\{\inf_{t\in [1/2,1]}u_3,\, \inf_{t\in [1/2,1]}|u_3'|\}<c'.
\end{gathered}
\end{equation}
\end{theorem}

\begin{proof}
We only need to check whether all conditions of Lemma
\ref{Lemma2.11} are fulfilled with respect to the operator $A$.
 By using a
similar way as to the proof of inequality \eqref{4.2s}, we can see  that
$$
A: \overline{P}_1(\gamma_1,c)\to \overline{P}_1.
$$
For arbitrary $u\in \partial P_1(\gamma_1 ,c')$, one has
\begin{gather*}
\gamma_1 (u)=\max\left\{\inf_{t\in [1/2,1]}u,\, \inf_{t\in [1/2,1]}|u'|\right\}=c',
\\
\| u\| \leq \frac{1}{\gamma}\gamma_1 (u)=\frac{1}{\gamma }c'.
\end{gather*}
This implies that
\begin{gather*}
0\leq u\leq \frac{1}{\gamma }c',\quad t\in [1/2,1],\\
-\frac{1}{\gamma}c'\leq u'\leq \frac{1}{\gamma}c',\quad t\in[1/2,1].
\end{gather*}
According to the condition (i) and $\alpha>3$, we have
\begin{align*}
\gamma_1 (A u)
&=  \max\big\{\inf_{t\in [1/2,1]}Au,\inf_{t\in [1/2,1]}|Au'|\big\}\\
&<(\alpha-1)\int_0^1G(1,s)f(s,u(s),u'(s))d s \\
&\leq  (\alpha-1)\int_0^1G(1,s) \frac{c'}{L}d s
< c'.
\end{align*}
We see that $\gamma_1 (Au)<c'$ for $u\in \partial P_1(\gamma_1 ,c')$.

For any $u\in \partial P_1(\beta_1 ,b')$, we have
\begin{gather*}
\beta_1 (u)=\max\big\{\inf_{t\in [1/2,1]}u,\ \inf_{t\in [1/2,1]}|u'|\big\}
=b',\\
\| u\| \leq \frac{1}{\gamma }\beta_1
(u)=\frac{1}{\gamma }\gamma_1 (u)=\frac{1}{\gamma
}b'.
\end{gather*}
This implies
\begin{gather*}
0\leq u\leq \frac{1}{\gamma}b',\quad t\in [1/2,1],\\
-\frac{1}{\gamma }b'\leq u'\leq \frac{1}{\gamma }b',\quad t\in
[1/2,1].
\end{gather*}
Using condition (ii), we obtain
\begin{align*}
\beta_1 (A u)
&=  \max\big\{\inf_{t\in [1/2,1]}Au,\inf_{t\in [1/2,1]}|Au'|\big\}\\
&> \int_{1/2}^1\min_{t\in[1/2,1]}G(t,s)f(s,u(s),u'(s)) \,ds \\
&\geq  \int_{1/2}^1\gamma G(1,s)\frac{b'}{N}\,ds\\
&\geq \frac{b'}{N} \gamma \int_{1/2}^1G(1,s)\,ds
=  b'.
\end{align*}
So we have $\beta_1(Au)>b'$ for $u\in \partial P_1(\beta_1 ,b')$.

We now show that $P_1 (\alpha_1 ,a')\neq \emptyset $ and
$\alpha_1 (Au)<a'$ for arbitrary  $u\in \partial P_1(\alpha_1 ,a')$.
Since  $\frac{a'}2\in P_1(\alpha_1 ,a')$, for $u\in \partial P_1(\alpha_1 ,a')$
we have
$$
\alpha_1 (u)=\max\{\sup_{t\in [0,1 ]}u ,\ \sup_{t\in [0,1 ]}|u'|\}=a',
$$
which gives
$$
0\leq u\leq a'\text{ and } -a'\leq u'\leq a'\quad \text{for }t\in [0,1].
$$
It follow from assumption (iii) and \eqref{feng} that
\begin{align*}
\alpha_1 (Au)
&=  \max\big\{\sup_{t\in [0,1]}Au, \sup_{t\in [0,1]}|A'u|\big\}\\
&=  (\alpha-1)\int_0^1G(1,s)f(s,u(s),u'(s))\,ds \\
&< (\alpha-1)\int_0^1G(1,s)\frac{a'}{L}\,ds \\
&< (\alpha-1)\frac{a'}{L} \int_0^1G(1,s)\,ds
= a'.
\end{align*}

All conditions in Lemma \ref{Lemma2.11} are satisfied.
From  (S1) and (S2), we know that solutions of \eqref{equ1}  do not
vanish identically on any closed subinterval of $[0,1]$.
 Consequently, \eqref{equ1}  has at least three distinct  positive solutions
$u_1$, $u_2$ and $u_3$ belonging to $\overline{P}(\gamma_1 ,c')$
distributed as \eqref{eeq2}.
\end{proof}

The following  result is regarded as a corollary of Theorem \ref{Theorem4.1a}.

\begin{corollary} \label{Corollary4.2}
Assume that all assumptions of Lemma \ref{Lem213} hold and
 $f$ satisfies the following conditions:
\begin{itemize}
\item[(i)] $f^0=0$ and $f^\infty =0$;

\item[(ii)] there exists a constant $c_{0}>0$ such that
 $$
f(t,u_1,u_2)>\frac{\gamma c_{0} }{N} \quad
\text{for }(t,u_1,u_2)\in[1/2,1]\times[0,c_0]\times[-c_0,c_0].
$$
\end{itemize}
Then \eqref{equ1} has
at least three distinct positive solutions.
\end{corollary}

\begin{proof}
Let $b'=\gamma c_{0}$. It follows from the condition (ii) that
\begin{align*}
f(t,u_1,u_2)>\frac{b' }{N} \quad \text{for }(t,u_1,u_2)\in[1/2,1]\times[
0,\frac{b'}{\gamma}]\times[-\frac{1}{\gamma
}b',\frac{1}{\gamma }b'],
\end{align*}
which implies that the condition (ii) of Theorem \ref{Theorem4.1a}
holds.

We choose a sufficiently small $\varepsilon_5>0$  such that
\begin{equation}
\varepsilon_5L=\varepsilon_5(\alpha-1)
\Big(\int_0^1G(1,s)\,ds\Big) <1. \label{ans}
\end{equation}
In view of $f^0=0 $, there exists a sufficiently small $k_1>0$  such that
\begin{equation}
f(t,u_1,u_2)\leq \varepsilon_5 |u_2|\quad \text{for }
(t,u_1,u_2)\in[0,1]\times [0,k_1]\times [-k_1,k_1].
\label{ans2}
\end{equation}
Without loss of generality, let $ k_1=a'<b'$.
 Because of $\max_{t\in[0,1]}|u_2|\leq a'$, we have
$$
\max_{t\in[0,1]}u_1\leq \max_{t\in[0,1]}|u_2|\leq k_1.
$$
It follows from \eqref{ans} and \eqref{ans2} that
\begin{equation*}
f(t,u_1,u_2)\leq \varepsilon_5|u_2|\leq \varepsilon_5a'<
\frac{a'}{L} \text{ for }(t,u_1,u_2)\in[0,1]\times [0,a']\times
[-a',a'],
\end{equation*}
which implies that  condition (iii) of Theorem \ref{Theorem4.1a} holds.

Choose $\varepsilon_6$ sufficiently small such that
\begin{equation*}
\frac{\varepsilon_6}{\gamma}L=(\alpha-1)\frac{\varepsilon_6}{\gamma}
\Big(\int_{1/2}^1G(1,s)ds\Big) <1.
\end{equation*}
By using the continuity of $f$,  there exists a constant $C^*$ such
that
\begin{equation}
f(t,u_1,u_2)\leq C^* \text{ for }(t,u_1,u_2)\in[0,1]\times
[0,  c'/\gamma]\times[-c',c'].\label{55}
\end{equation}
 Since $f^\infty =0 $, there exists a sufficiently large $k_2>LC^*$
such that
\[
f(t,u_1,u_2)\leq \varepsilon_6 |u_2| \quad\text{ for}
(t,u_1,u_2)\in[0,1]\times [k_2,+\infty)\times [k_2,+\infty)\cup
(-\infty,k_2].
\]
Without loss of generality, let $k_2>b'/\gamma$ and $c'=k_2$. We find
$$
(t,u_1,u_2)\in[0,1]\times [0, c'/\gamma]\times[- c'/\gamma ,-c']
\cup[c', c'/\gamma],
$$
and
\begin{equation}
f(t,u_1,u_2)\leq \varepsilon_6 |u_2|\leq
\varepsilon_6\frac{c'}{\gamma}<\frac{ c'}{L}.\label{56}
\end{equation}
Moreover, in view of \eqref{55}, one has
\begin{equation}
f(t,u_1,u_2)\leq C^*<\frac{k_2}{L}=\frac{c'}{L}\quad
\text{for }(t,u_1,u_2)\in[0,1]\times [0, \frac{1}{\gamma }
c']\times[- c',c'].\label{57}
\end{equation}
From \eqref{56} and \eqref{57}, we see  that condition (i) of
Theorem \ref{Theorem4.1a} is fulfilled. Hence,
\eqref{equ1} has at least three distinct positive solutions
according to  Theorem \ref{Theorem4.1a}.
 \end{proof}

According to Theorem \ref{Theorem4.1a},    we can prove that the
existence for multiple positive solutions to the \eqref{equ1}
when conditions (i), (ii) and (iii) are modified
appropriately on $f$.

\begin{theorem}\label{Theorem4.1a1} If there exist constant numbers
$ a_i',\,b_i'$ and $c_i'$ such that $ 0<a_1'<b_1'<c_1'<\dots<a_n'<b_{n}'<c_n'$
 together with
\begin{equation}
0<Lb_1'<a_1'N<Lb_2'<a_2'N<\dots<Lb_n'<a_n'N,\quad n\in\mathbb{N},\label{7}
\end{equation}
 where $i=1,2,\dots,n$.
 In addition, if all assumptions of Lemma
\ref{Lem213} hold and the function $f$ satisfies:
\begin{itemize}
\item[(i)] $f(t,u_1,u_2)<\frac{c_i'}{L}$ for
$(t,u_1,u_2)\in [1/2,1]\times [0,\frac{1}{\gamma}c_i']\times[-\frac{1}{\gamma
}c_i',\frac{1}{\gamma }c_i']$;

\item[(ii)] $f(t,u_1,u_2)>\frac{b_i'}{N}$ for
$(t,u_1,u_2)\in [1/2,1]\times
[0,\frac{1}{\gamma}b_i']\times[-\frac{1}{\gamma
}b_i',\frac{1}{\gamma }b_i']$;

\item[(iii)] $f(t,u_1,u_2)<\frac{a_i'}{L}$ for
$(t,u_1,u_2)\in [1/2,1]\times
[0,a_i']\times[-a_i',a_i']$.
\end{itemize}
Then \eqref{equ1}  has at least $n$ distinct
positive solutions.
\end{theorem}

\begin{proof} (By Mathematical Induction)
 If $n=1$, from the condition (iii), we have
$$
A: \overline{P}_{a_1'}\to P_{a_1'}\subset \overline{P}_{a_1'}.
$$
It follows from the Schauder fixed point theorem that
$A$ has at least one fixed point $u_{01}\in \overline{P}_{a_1'}$.

 If $i=2$, we let $a'=a'_1,\,b'=b'_1$ and $c'=c'_1$. By using Theorem
\ref{Theorem4.1a},  problem \eqref{equ1} has at least
three distinct positive solutions $u_{11}$, $u_{12}$ and $u_{13}$
such that
\begin{gather*}
0< \|u_{11}\|<a_1'<\|u_{12}\|,\\
\max\Big\{\inf_{t\in[1/2,1]}u_{12},\, 
\inf_{t\in [1/2,1]}|u_{12}'|\Big\}
<b_1' <\max\Big\{\inf_{t\in [1/2,1]}u_{13},\, 
\inf_{t\in [1/2,1]}|u_{13}'|\Big\}<c_1',
\end{gather*}
which implies that problem \eqref{equ1} has at least $2$
 distinct positive solutions.

Assume that problem \eqref{equ1} has at least $k-1$
distinct positive solutions  when $n=k-1$. We denote by $u_i$
again. It follows from  the solution position and local
 properties that
\begin{equation}
0<\max\Big\{\inf_{t\in
[1/2,1]}u_i,\ \inf_{t\in [1/2,1]}|u_i'|\Big\}<c_{k-1}',\quad
 i=1,2,\dots,k-1, \text{ where }c_0'=a_1'.
\label{4.3}
\end{equation}

When $n=k$, we let $a'=a'_k$, $b'=b'_k$ and $c'=c'_{k}$.
According to Theorem \ref{Theorem4.1a},  there exist at least three positive
solutions $u_{k1}$, $u_{k2}$ and $u_{k3}$  such that
\begin{equation} \label{buden}
\begin{gathered}
0< \|u_{k1}\|<a_k'<\|u_{k2}\|,\\
\max\Big\{\inf_{t\in [1/2,1]}u_{k2},\ \inf_{t\in
[1/2,1]}|u_{k2}'|\Big\}<b_k'<\max\Big\{\inf_{t\in
[1/2,1]}u_{k3},\ \inf_{t\in [1/2,1]}|u_{k3}'|\Big\}<c_k'.
\end{gathered}
\end{equation}
Combining \eqref{4.3} and  \eqref{buden} gives
$$
\max\left\{\inf_{t\in
[1/2,1]}u_i,\ \inf_{t\in [1/2,1]}|u_i'|\right\}
<c_{k-1}'<b_k'<\max\left\{\inf_{t\in
[1/2,1]}u_{k3},\ \inf_{t\in [1/2,1]}|u_{k3}'|\right\}.
$$
This implies
$$
u_i\neq u_{k3},\quad  i=1,2,\dots,k-1.
$$
Therefore, \eqref{equ1} has
at least $n$ distinct positive solutions.
\end{proof}

By Lemma \ref{Lemma2.11} and Theorem
\ref{Theorem4.1a}, we can obtain the following results:

\begin{theorem}\label{Theorem4.1as}
Assume that there exist positive numbers $a',b',c'$ with  $ a'<b'<c'$
such that $c'L<b'N. $ In addition, if
all assumptions of Lemma \ref{Lem213} hold and
$f(t,u_1,u_2)$ satisfies the following conditions:
\begin{itemize}
\item[(i)] $f(t,u_1,u_2)>\frac{c'}{N}$ for
$(t,u_1,u_2)\in [1/2,1]\times
[0,\frac{1}{\gamma}c']\times
[-\frac{1}{\gamma}c',\frac{1}{\gamma}c']$;

\item[(ii)] $f(t,u_1,u_2)<\frac{b'}{L}$ for
$(t,u_1,u_2)\in [1/2,1]\times
[0,\frac{1}{\gamma}b']\times
[-\frac{1}{\gamma}b',\frac{1}{\gamma}b']$;

\item[(iii)] $f(t,u_1,u_2)>\frac{a'}{N}$ for
$(t,u_1,u_2)\in [0 ,1]\times [0,a']\times
[-a',a']$.
\end{itemize}
Then \eqref{equ1}  has at least three distinct
positive solutions $ u_1, u_2 ,u_3\in \overline{P_1}(\gamma_1,c')$
such that
\begin{gather*}
0\leq \|u_1\|<a'<\|u_2\|,\\
\max\Big\{\inf_{t\in
[1/2,1]}u_{2},\, \inf_{t\in [1/2,1]}|u_{2}'|\Big\}<b'
<\max\Big\{\inf_{t\in [1/2,1]}u_{3},\, \inf_{t\in [1/2,1]}|u_{3}'|\Big\}<c'.
\end{gather*}
\end{theorem}


\begin{corollary}\label{Corollary4.1}
Assume that all assumptions of Lemma
\ref{Lem213} hold and $f$ satisfies conditions
\begin{itemize}
\item[(i)] $f_0=\infty $ and $f_\infty =\infty $;

\item[(ii)] there exists $c_{0}>0$ such that $
f(t,u_1,u_2)<\frac{\gamma}{M}c_0 \text{ for }(t,u_1,u_2)\in
[1/2,1]\times[0, c_{0}]\times[- c_0,
c_{0}]$.
\end{itemize}
Then  \eqref{equ1} has at least three distinct
positive solutions.
\end{corollary}

\begin{theorem}\label{Theorem4.1as1}
Assume that all assumptions of Lemma \ref{Lem213} hold and  there are
positive numbers $a_i',b_i',c_i'$  such that
$a_1'<b_1'<c_1'<\dots<a_{n}'<b_{n}'<c_{n}'$ together with
\[
0<c_1'L<Nb_1'<c_2'L<Nb_2'<\dots<c_n'L<Nb_n',\quad n\in\mathbb{N},
\]
where $i=1,2,\dots,n$. In addition, $f(t,u_1,u_2)$ satisfies the following
conditions:
\begin{itemize}
\item[(i)] $f(t,u_1,u_2)>\frac{c_i'}{N}$ for
$(t,u_1,u_2)\in [1/2,1]\times
[0,\frac{1}{\gamma}c_i']\times
[-\frac{1}{\gamma}c_i',\frac{1}{\gamma}c_i']$;

\item[(ii)] $f(t,u_1,u_2)<\frac{b_i'}{L}$ for
$(t,u_1,u_2)\in [1/2,1]\times
[0,\frac{1}{\gamma}b_i']\times
[-\frac{1}{\gamma}b_i',\frac{1}{\gamma}b_i']$;

\item[(iii)] $f(t,u_1,u_2)>\frac{a_i'}{N}$ for
$(t,u_1,u_2)\in [0 ,1]\times [0,a_i']\times
[-a_i',a_i']$.
\end{itemize}
Then \eqref{equ1} has at least $n$ distinct positive solutions.
\end{theorem}

\subsection{Existence of $2n-1$ solutions}

In this subsection, we are concerned with the existence of at least three
 or $2n-1$ positive solutions to \eqref{equ1}.

Define the nonnegative continuous convex functionals $\phi$ and
$\beta$,  concave functional $\lambda$ and functional $\varphi$
 on $P_1$ by
\begin{gather*}
\phi(u) =\max\Big\{\sup_{t\in [0,1]}u,\ \sup_{t\in [0,1]}|u'|\Big\},\\
 \beta(u)=\varphi (u)=\sup_{t\in [1/2,1]}u,\quad
\lambda (u) =\inf_{t\in [\frac{1}{2},1]}u.
\end{gather*}

\begin{theorem}\label{anmao}
Assume that all assumptions of Lemma
\ref{Lem213} hold and there exist constants
$a^{*},b^{*},d^{*}$ such that $0<a^{*}<b^{*}<\frac{N}{L}d^{*}$. In
addition,   $f(t,u_1,u_2)$ satisfies the following
conditions:
\begin{itemize}
\item[(i)] $f(t,u_1,u_2)\leq\frac{d^{*}}{L}$ for all
$(t,u_1,u_2)\in[0,1]\times  [0, d^{*}]\times [-d^{*}, d^{*}] $;

\item[(ii)] $f(t,u_1,u_2)>\frac{b^{*}}{N}$
 for all
$(t,u_1,u_2)\in[1/2,1]\times
 [b^*, d^{*}]\times
 [-d^{*}, d^{*}] $;

\item[(iii)] $f(t,u_1,u_2)<\frac {a^{*}}{M}$
 for all $(t,u_1,u_2)\in[1/2,1]\times [0,a^*]\times
  [-d^{*}, d^{*}]$.
\end{itemize}
Then  \eqref{equ1}  has at least three distinct
positive solutions $u_1$, $u_2$, $u_3$ such that
\[
\|x_i\|\leq d^{*} \text{ for i=1,2,3,  }\quad
b^{*}<\inf_{t\in [1/2,1]}u_1, \quad
a^{*}<\sup_{t\in [\frac{1}{2},1]}u_2, \quad
\inf_{t\in [\frac{1}{2},1]}u_2<b^{*}
\]
with $\sup_{t\in [\frac{1}{2},1]}u_3<a^{*}$.
\end{theorem}

\begin{proof}
It suffices to show that all conditions of
Lemma \ref{Lemma2.7} hold with respect to the completely continuous
operator $A$.
For arbitrary  $u\in P_1$, we have $\lambda (u)=\varphi (u)$
 and $\|u\|=\phi(u)$. This implies that inequality \eqref{hui}
in Lemma \ref{Lemma2.7} holds.

For any  $u\in\overline{P_1(\phi ,d^*)}$, from
$\phi(u)=\|u\|\leq d^*$ and the assumption (i),  we have
\begin{align*}
\| Au\| &\leq  (\alpha-1)\int_0^1G(1,s)f(s,u(s),u'(s))\,ds \\
&\leq  (\alpha-1) \frac{d^*}{L}\int_0^1G(1,s)\,ds
 = d^*.
\end{align*}
This means that $A:\overline{P_1(\phi,d^*)}\to\overline{P_1(\phi ,d^*)}$.

It remains to show that assumptions (i)-(iii) of Lemma \ref{Lemma2.7}
are fulfilled with respect to the operator $A$.

Let $u\equiv kb^*, $ where $k=L/N$. It is obvious that
$k>1$, $u=kb^*>b^*$ and $\beta(u)=kb^*$. We see that
  $b^*<\frac{N}{L}d^*$ and $\phi(u)=kb^*<d^*$. So we have
$$
\{ u\in P_1(\phi ,\beta ,\lambda ,b^*,kb^*,d^*):\lambda
(x)>b^*\} \neq \emptyset .
$$
For any $u\in P_1(\phi ,\beta ,\lambda,b^*,kb^*,d^*)$,
 we obtain $b^*\leq u\leq d^*$ and $-d^*\leq u'\leq d^*$
for all $t\in [1/2,1]$. It follows from
the assumption (ii) that
\begin{align*}
\lambda( Au)
&= \int_{1/2}^1\inf_{t\in[1/2,1]}G(t,s)f(s,u(s),u'(s))\,ds \\
&\geq  \int_{1/2}^1\gamma G(1,s)\frac{b^*}{N}\,ds
=  b^*,
\end{align*}
which implies that the assumption (i) of Lemma \ref{Lemma2.7} is
satisfied.

For any $u\in P_1(\phi,\lambda, b^*, d^*)$ with
$\beta(Au)>kb^*$, we have $b^*\leq u\leq d^*$ and
$-d^*\leq u'\leq d^*$  for $t\in[1/2,1]$. So we have
\begin{align*}
\lambda( Au) & =
\int_{1/2}^1\inf_{t\in[1/2,1]}G(t,s)f(s,u(s),u'(s))\,ds \\
&\geq  \int_{1/2}^1\gamma G(1,s)\frac{b^*}{N}\,ds
=  b^*.
\end{align*}
This  implies that assumption  (ii) of Lemma \ref{Lemma2.7} is
fulfilled.

Since $\varphi(0)=0<a^*$, we have
$0\notin R(\phi,\varphi,a^*,d^*)$. If
$$
u\in R(\phi,\varphi,a^*,d^*)\quad \text{with }
\varphi (u) =\sup_{t\in [1/2,1]}u=a^*,
$$
it reduces to
$$
0\leq u\leq a^*\text{  and } -d^*\leq u'\leq d^* \quad \text{for all }
t\in [1/2,1].
$$
A straightforward calculation gives
\begin{align*}
\lambda(Au) &=  \inf_{t\in[1/2,1]} \int_{1/2}^1G(t,s)f(s,u(s),u'(s))
\,ds  \\
&\leq  \int_{1/2}^1G(t,s)f(s,u(s),u'(s))\,ds
  \\
&< \int_{1/2}^1G(1,s)\frac{a^*}{M}d s
 = a^* .
\end{align*}
It is easy to see the assumption  (iii) of Lemma \ref{Lemma2.7} is
fulfilled too. The proof is complete.
\end{proof}


\begin{corollary}\label{Corollary5.1s}
Assume that all assumptions of Lemma
\ref{Lem213} hold and the condition {\rm (i)} in Theorem
\ref{anmao} is replaced by {\rm (i')}, then the conclusion
of Theorem \ref{anmao} also holds.
\end{corollary}

Similar to the proof of Theorem \ref{Theorem4.1a1} by mathematical
induction, we have

\begin{theorem}\label{Theorem5.2s}
Assume that all assumptions of Lemma \ref{Lem213}
 hold and there exist constants
$a_i^{*}, b_i^{*}$ and $d_i^{*}$ such that
$$
0<a_1^{*}<b_1^{*}<\frac{N}{L}d_1^{*}
<a_2^{*}<b_2^{*}<\frac{N}{L}d_2^{*}<a_3^{*} <\dots<a_n^{*},\quad
n\in\mathbb{N},
$$ where $i=1,2,\dots,n$.
In addition,   $f$ satisfies the following
conditions:
\begin{itemize}
\item[(i)] $f(t,u_1,u_2)\leq\frac{d_i^{*}}{L}$ for all
$(t,u_1,u_2)\in[0,1]\times
 [0, d_i^{*}]\times  [-d_i^{*}, d_i^{*}] $;

\item[(ii)] $f(t,u_1,u_2)>\frac{b_i^{*}}{N}$
 for all $(t,u_1,u_2)\in[0,1]\times
 [b_i^*, d_i^{*}]\times
 [-d_i^{*}, d_i^{*}] $;

\item[(iii)] $f(t,u_1,u_2)<\frac {a_i^{*}}{M}$
 for all $(t,u_1,u_2)\in[1/2,1]\times
 [0,a_i^*]\times  [-d_i^{*}, d_i^{*}]$.
\end{itemize}
Then \eqref{equ1}has at least $2n-1$ positive solutions.
\end{theorem}


\section{Examples}

In this section, we give two simple examples to illustrate our theoretical
results. In Example \ref{examp1}, it shows the difference between
two cases, which indicates our theorems presented in
Sections 5 and 6 are complementary.

\begin{example}\label{examp1} \rm
Consider the equation
\begin{equation}
\begin{gathered}
D_{0^+}^{\alpha}u(t)+f(t,u(t),u'(t))=0,\quad  t\in ( 0,1 ), n-1<\alpha\leq n,\\
u^{(i)}(0)=0, \quad i=0,1,2,\dots,n-2,\\
[D_{0^+}^{\beta}u(t)]_{t=1}=0,\quad 1\leq\beta\leq n-2.
\end{gathered} \label{6.1}
\end{equation}
\end{example}

Case 1: when $f(t,u,u')$ takes the form as
\[
f(t,u,u')=\frac{t}{4(\alpha-1)G(1,t)}(u(t)+u'(t))\quad
\text{for } (t,u,u')\in[0,1] \times [
0,\infty )\times(-\infty,+\infty).
\]
It is easy to see that $a_1(t)=0$ and
$a_2(t)=\frac{t}{2(\alpha-1)G(1,t)}$. Moreover, we see that
$$
(\alpha-1)\int_0^{1}G(1,s)a_2(s)\,ds=\int_0^{1}\frac{s}{2}\,ds<1.
$$
By means of Theorem \ref{3.1}, we find that
the fractional differential equation \eqref{6.1} has at least
one positive solution. However, it is difficult for us to obtain
the existence of at least one positive solution to  the fractional
differential equation \eqref{6.1} by using theorems of the
super-linearity and sub-linearity.

Case 2: when $f(t,u,u')$ takes the form as
\[
f(t,u,u')=\begin{cases}   -e^{u'+1} & \text{for } u'\in (-\infty,-1),\\
u'^3  &\text{for } u'\in [-1,1],\\
e^{u'-1} &\text{for } u'\in (1,+\infty).
\end{cases}
\]
Since $f$  is continuous, we know that the operator $A$ is completely continuous.
It is easy to check that $f_0=0$ and $f_\infty=\infty$. By means
of Theorem \ref{Theorem3.41}, we conclude  that fractional
differential equation \eqref{6.1} has at least one positive
solution.  But it is difficult for us to know the existence of
 positive solution to  the fractional
differential equation \eqref{6.1} if we use  Theorem \ref{3.1}.

\begin{example} \label{examp2}\rm
Consider the equation
\begin{equation}
\begin{gathered}
D_{0^+}^{\frac{7}{2}}u(t)+f(t,u(t),u'(t))=0,\quad t\in ( 0,1),\\
u(0)=u^{(1)}(0)=0,\\
[D_{0^+}^{2}u(t)]_{t=1}=0,
\end{gathered} \label{6.2}
\end{equation}
where
\[
f(t,u,u')=\begin{cases}
 2\Gamma(\frac{7}{2}) &\text{for }u\in[0,1],\\
\Gamma(\frac{7}{2}) (57u-55) &\text{for } u\in(1,2),\\
59\Gamma(\frac{7}{2}) &\text{for } u\in[2,+\infty).
\end{cases}
\]
\end{example}

Since $\alpha=7/2$  and $\beta=2$, a straightforward calculation gives
\begin{align*}
N &=  \gamma\int_{1/2}^1G(1,s)\,ds \\
&\approx \frac{1}{\Gamma(\frac{7}{2})}\Big(0.1768\int_{1/2}^1(1-s)
^{1/2}ds-0.1768\int_{1/2}^1(1-s)^{5/2}ds\Big)\\
&\approx \frac{1}{\Gamma(\frac{7}{2})}3.7207\times 10^{-2},
\end{align*}
and
$$
L\approx\frac{1}{\Gamma(\frac{7}{2})}0.3910.
$$
Taking $a=1$, $b=2$, $d=80$ and $c=100$, we see that
$0<a<b<d<\frac{N}{L}c$ and
\begin{gather*}
f(t,u,u')<\frac{a}{L}=\frac {\Gamma(\frac{7}{2})}{ 0.3910}\approx2.625
\Gamma(\frac{7}{2}) \quad \text{for } (t,u,u')\in[0,1]\times[0,1]\times[-1,1],
\\
f(t,u,u')>\frac{b}{N}=\frac {2\Gamma(\frac{7}{2})}{3.7207\times
10^{-2}}\approx 53.753 \Gamma(\frac{7}{2})\\
 \text{for }(t,u,u')\in[0,1]\times[2,80]\times[-80,80],
\\
f(t,u_1,u_2)<\frac{c}{L}=\frac{100\Gamma(\frac{7}{2})}{ 0.17778}
\approx 262.5 \Gamma(\frac{7}{2}) \\
\text{for }(t,u,u')\in[0,1]\times[0,100]\times[-100,100].
\end{gather*}
It follows from Theorem \ref{Theorem4.1'} that the
fractional differential equation \eqref{6.2} has at least
three distinct positive solutions such that
$$
0< \|u_1\| < 1,\quad
2 < \inf_{t\in[1/2,1]}u_2,\quad
1< u_3 \quad \text{with } \inf_{t\in [1/2,1]}u_3 < 2.
$$

\subsection*{Acknowledgments}
This work is supported by  Natural Science Foundation of China (No.11126091),
and Jiangsu Overseas Research and Training Program for University
Prominent Young Faculty.


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\end{document}