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

\AtBeginDocument{{\noindent\small
\emph{Electronic Journal of Differential Equations},
Vol. 2012 (2012), No. 88, pp. 1--12.\newline
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu
\newline ftp ejde.math.txstate.edu}
\thanks{\copyright 2012 Texas State University - San Marcos.}
\vspace{9mm}}

\begin{document}
\title[\hfilneg EJDE-2012/88\hfil Existence of three non-negative solutions]
{Existence of three non-negative solutions for a three-point
 boundary-value problem of nonlinear fractional differential equations}

\author[H. Li, X. Kong, C. Yu \hfil EJDE-2012/88\hfilneg]
{Haitao Li, Xiangshan Kong, Changtian Yu}  % in alphabetical order

\address{Haitao Li  \newline
 School of Control Science and Engineering,
Shandong University, Jinan 250061,  China}
\email{haitaoli09@gmail.com}

\address{Xiangshan Kong \newline
 Basic Science Department, Qingdao Binhai University,
Qingdao 266555, China}
\email{kong\_xiangshan@126.com}

\address{Changtian Yu \newline
 Department of Mathematics, Shandong Normal University,
Jinan 250014, China}
\email{yuchangtian1986@163.com}


\thanks{Submitted February 14, 2012. Published June 4, 2012.}
\thanks{Supported by grants G61174036 from the National Natural Science Foundation
of China, \hfill\break\indent and yzc10064 from the Graduate Independent Innovation
Foundation of Shandong University}
\subjclass[2000]{34B10, 34A08}
\keywords{Non-negative solutions; Riemann-Liouville fractional derivative;
\hfill\break\indent three-point boundary-value problems;
 Leggett-Williams fixed point theorem}

\begin{abstract}
 This article concerns the existence of three non-negative solutions for
 two kinds of three-point boundary-value problems of nonlinear fractional
 differential equations, where the fractional derivative is taken in the
 Riemann-Liouville sense.
 Using Leggett-Williams fixed point theorem, we  present some existence criteria 
 and then illustrate our results with examples.
\end{abstract}

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

\section{Introduction}

With the development of fractional calculus and its  applications
\cite{non1995,old1974,sam1993} in mathematics, technology, biology, chemical
process etc., increasing attention has been paid to the study of fractional
differential equations including
the existence  of solutions to fractional differential equations
\cite{ahm2009, bai2010, bal2010, li2010, shi2009, wei2010, xu2012-1,xu2012-2,yang2012,
zhang2006,zhao2011},
the stability analysis of fractional differential equations
 \cite{deng2010,khu2001,pet2000}, and so on.
As a fundamental issue of the theory of fractional differential equations,
 the existence of (positive) solutions for kinds of boundary-value problems
(BVPs) of fractional differential equations has been studied recently
by many scholars, and lots of excellent results have been obtained
for both two-point BVPs and nonlocal BVPs by means of fixed point
index theory \cite{bai2010}, fixed point theorems \cite{ahm2009,xu2012-2,zhang2006},
mixed monotone method \cite{xu2012-1}, upper and lower solutions technique
\cite{shi2009}, and so on.

Xu et al \cite{xu2012-1,xu2012-2}  investigated the  fractional
differential equation
\begin{equation}
\label{eq1.1} D_{0+}^{\alpha}u(t)+f(t,u(t))=0,\ t\in(0,1),
\end{equation}
subject to the following two kinds of three-point boundary conditions:
\begin{equation} \label{eq1.2}
u(0)=0,\ D_{0+}^{\beta}u(1)=m_1D_{0+}^{\beta}u(\xi),
\end{equation}
and
\begin{equation}\label{eq1.3}
 u(0)=0,\ u(1)=m_2u(\xi),
\end{equation}
respectively, where $1<\alpha<2$, $0<\beta\leq1$, $\alpha-\beta-1\geq0$,
$0\leq m_1\leq1$, $0<m_2,\ \xi<1$, and $D_{0+}^{\alpha}$ is the
standard Riemann-Liouville derivative. They obtained some interesting
positive properties of Green's functions for \eqref{eq1.1}
with \eqref{eq1.2} and  \eqref{eq1.1} with \eqref{eq1.3},
 respectively, and presented some existence criteria of positive
solutions by using some fixed point theorems and the mixed monotone method.

It should be noted that there are fewer results on the existence of
multiple (triple) solutions for nonlocal BVPs of fractional differential
equations such as  \eqref{eq1.1} with \eqref{eq1.2} and  \eqref{eq1.1} with
\eqref{eq1.3}. As a
result, the purpose of this paper is to establish the existence results
for triple non-negative solutions of \eqref{eq1.1} with \eqref{eq1.2} and
\eqref{eq1.1} with \eqref{eq1.3} by virtue of Leggett-Williams fixed point
theorem and enrich this academic area.

Throughout this paper, we assume that the nonlinearity
$f: [0,1]\times[0,+\infty)\to[0,+\infty)$ is continuous
in \eqref{eq1.1}. Moreover, let $E=C[0,1]$ with the norm
$\|x\|=\max_{t\in[0,1]}|x(t)|$. Then, $E$ is a Banach space.
 We will consider the existence of non-negative solutions for \eqref{eq1.1}
with \eqref{eq1.2} and  \eqref{eq1.1} with \eqref{eq1.3} in $E$.

The rest of this work is organized as follows. Section 2 contains
some preliminaries on the standard Riemann-Liouville
derivative and the properties of Green's functions for \eqref{eq1.1}
 with \eqref{eq1.2} and  \eqref{eq1.1} with \eqref{eq1.3}.
Section 3 investigates the existence of triple non-negative solutions for
 \eqref{eq1.1} with \eqref{eq1.2} and  \eqref{eq1.1} with \eqref{eq1.3},
respectively, and presents the main results of this paper.
In Section 4, two illustrative examples are worked out to support
our obtained results.

\section{Preliminaries}

In this section, we give some necessary preliminaries on the
Riemann-Liouville derivative and the properties of Green's functions
for  \eqref{eq1.1} with \eqref{eq1.2} and \eqref{eq1.1} with \eqref{eq1.3},
which will be used in the sequel.

We first recall some well known results about Riemann-Liouville derivative.
 For details, please refer to \cite{old1974,non1995,sam1993} and the
references therein.

\begin{definition}[\cite{sam1993}] \label{def2.1} \rm
 The Riemann-Liouville fractional
integral of order $\alpha>0$ of a function $y:(0,\infty)\to \mathbb{R}$ is given by
\begin{equation}\label{eq2.1}
I_{0+}^{\alpha}y(t)=\frac{1}{\Gamma(\alpha)}\int_0^{t}(t-s)^{\alpha-1}y(s)ds,
\end{equation}
provided the right side is pointwise defined on $(0,\infty)$.
\end{definition}

\begin{definition}[\cite{sam1993}] \label{def2.2} \rm
The Riemann-Liouville fractional
derivative of order $\alpha>0$ of a continuous function
$y:(0,\infty)\to \mathbb{R}$ is given by
\begin{equation}\label{eq2.2}
 D_{0+}^{\alpha}y(t)=\frac{1}{\Gamma(n-\alpha)}(\frac{d}{dt})^{n}\int_0^{t}
\frac{y(s)}{(t-s)^{\alpha-n+1}}ds,
\end{equation}
where $n=[\alpha]+1$, $[\alpha]$ denotes the integer part of
$\alpha$, provided that the right side is pointwise defined on
$(0,\infty)$.
\end{definition}

One can easily obtain the following properties from the definition
of Riemann-Liouville derivative.

\begin{proposition}[\cite{sam1993}] \label{prop2.1}
 Let $\alpha>0$, if we assume $u\in C(0,1)\cap L(0,1)$, then, the
fractional differential equation $D_{0+}^{\alpha}u(t)=0$ has
$u(t)=C_1t^{\alpha-1}+C_2t^{\alpha-2}+\dots+C_{N}t^{\alpha-N}$,
$C_{i}\in \mathbb{R}$, $i=1, 2, \dots, N$ as unique solution, where
$N$ is the smallest integer greater than or equal to $\alpha$.
\end{proposition}

\begin{proposition}[\cite{sam1993}] \label{prop2.2} \rm
 Assume that $u\in C(0,1)\cap L(0,1)$ with a fractional derivative of order
$\alpha>0$ that belongs to $C(0,1)\cap L(0,1)$. Then,
\begin{equation}\label{eq2.3}
I_{0+}^{\alpha}D_{0+}^{\alpha}u(t)=u(t)+C_1t^{\alpha-1}
+C_2t^{\alpha-2}+\dots+C_{N}t^{\alpha-N},
\end{equation}
for some $C_{i}\in \mathbb{R}$, $i=1, 2, \dots, N$, where $N$ is the smallest
integer greater than or equal to $\alpha$.
\end{proposition}

In the following, we present some important properties of Green's functions
for \eqref{eq1.1} with \eqref{eq1.2} and  \eqref{eq1.1} with \eqref{eq1.3},
which have been proved in \cite{bai2010,li2010, xu2012-1,xu2012-2}.

\begin{lemma}[\cite{xu2012-2}] \label{lemma2.1}
$x(t)\in E$ is a solution to \eqref{eq1.1} with \eqref{eq1.2}, if and only if
$x(t)=T_1x(t)$, where
\begin{equation}\label{eq2.4}
T_1x(t)=\int_0^{1}G_1(t,s)f(s,x(s))ds,
\end{equation}
\begin{equation}\label{eq2.5}
G_1(t,s)=\begin{cases}
\frac{Dt^{\alpha-1}(1-s)^{\alpha-\beta-1}-m_1Dt^{\alpha-1}(\xi-s)^{\alpha-\beta-1}
 -(t-s)^{\alpha-1}}{\Gamma(\alpha)},
 &  0\leq s\leq t\leq 1,\; s\leq \xi,\\
\frac{Dt^{\alpha-1}(1-s)^{\alpha-\beta-1}-(t-s)^{\alpha-1}}{\Gamma(\alpha)},
 & 0<\xi \leq s\leq t\leq 1,\\
\frac{Dt^{\alpha-1}(1-s)^{\alpha-\beta-1}-m_1
 Dt^{\alpha-1}(\xi-s)^{\alpha-\beta-1}}{\Gamma(\alpha)},
&  0\leq t\leq s\leq \xi<1, \\
\frac{Dt^{\alpha-1}(1-s)^{\alpha-\beta-1}}{\Gamma(\alpha)},
& 0\leq t\leq s\leq 1,\; \xi\leq s,
\end{cases}
\end{equation}
and $D=(1-m_1\xi^{\alpha-\beta-1})^{-1}$.
\end{lemma}

\begin{lemma}[\cite{xu2012-2}] \label{lemma2.2}
 The Green's function $G_1(t,s)$ given
in \eqref{eq2.5} satisfies
\begin{equation} \label{eq2.6}
 \frac{\beta
t^{\alpha-1}s(1-s)^{\alpha-\beta-1}}{\Gamma(\alpha)}\leq G_1(t,s)\leq
\frac{Dt^{\alpha-1}(1-s)^{\alpha-\beta-1}}{\Gamma(\alpha)},\quad
 \forall\ t, s\in[0,1].
\end{equation}
\end{lemma}

\begin{lemma}[\cite{xu2012-1}] \label{lemma2.3}
 $x(t)\in E$ is a solution to \eqref{eq1.1} with \eqref{eq1.3},
if and only if $x(t)=T_2x(t)$, where
\begin{equation} \label{eq2.7}
T_2x(t)=\int_0^{1}G_2(t,s)f(s,x(s))ds,
\end{equation}
and
\begin{equation} \label{eq2.8}
G_2(t,s)=\begin{cases}
\frac{[t(1-s)]^{\alpha-1}-m_2t^{\alpha-1}(\xi-s)^{\alpha-1}
 -(t-s)^{\alpha-1}(1-m_2\xi^{\alpha-1})}{(1-m_2\xi^{\alpha-1})
\Gamma(\alpha)},
&  0\leq s\leq t\leq 1,\; s\leq \xi,\\
\frac{[t(1-s)]^{\alpha-1}-(t-s)^{\alpha-1}(1-m_2\xi^{\alpha-1})}{(1-m_2
 \xi^{\alpha-1})\Gamma(\alpha)},
&  0<\xi \leq s\leq t\leq 1,\\
\frac{[t(1-s)]^{\alpha-1}-m_2t^{\alpha-1}(\xi-s)^{\alpha-1}}{(1-m_2
 \xi^{\alpha-1})\Gamma(\alpha)},
&  0\leq t\leq s\leq \xi<1, \\
\frac{[t(1-s)]^{\alpha-1}}{(1-m_2\xi^{\alpha-1})\Gamma(\alpha)},
& 0\leq t\leq s\leq 1,\; \xi\leq s.
\end{cases}
\end{equation}
\end{lemma}

\begin{lemma}[\cite{xu2012-1}] \label{lemma2.4}
The Green's function $G_2(t,s)$ given
in \eqref{eq2.8} satisfies
\begin{equation}\label{eq2.9}
\frac{M_0 t^{\alpha-1}s(1-s)^{\alpha-1}}{(1-m_2\xi^{\alpha-1})
 \Gamma(\alpha)}\leq G_2(t,s)\leq
\frac{t^{\alpha-1}(1-s)^{\alpha-1}}{(1-m_2\xi^{\alpha-1})\Gamma(\alpha)},
\quad \forall\ t,\; s\in[0,1],
\end{equation}
where $0<M_0=\min\{1-m_2\xi^{\alpha-1},
m_2\xi^{\alpha-2}(1-\xi)(\alpha-1), m_2\xi^{\alpha-1}\}<1$.
\end{lemma}

Finally, we recall the Leggett-Williams fixed point theorem.
Let $E=(E,\|\cdot\|)$ be a Banach space and $P\subset E$ be a cone on $E$.
 A continuous mapping $\omega: P\to [0,+\infty)$ is said to be
a concave non-negative continuous functional on $P$, if $\omega$
satisfies $\omega(\lambda x+(1-\lambda) y)\geq\lambda\omega(x)+(1-\lambda)\omega(y)$
for all $x,y\in P$ and $\lambda\in [0,1]$.

Let $a, b, d>0$ be constants. Define $P_{d}=\{x\in P: \|x\|<d\}$,
$\overline{P_{d}}=\{x\in P: \|x\|\leq d\}$ and
$P(\omega, a, b)=\{x\in P: \omega(x)\geq a, \|x\|\leq b\}$.

\begin{lemma}[\cite{guo1988}] \label{lemma2.5}
Let $E=(E,\|\cdot\|)$ be a Banach space, $P\subset E$ be a cone of $E$ and
 $c>0$ be a constant. Suppose there exists a concave non-negative
 continuous functional $\omega$ on $P$ with $\omega(x)\leq\|x\|$ for
all $x\in\overline{P_{c}}$. Let $T: \overline{P_{c}}\to\overline{P_{c}}$
be a completely continuous operator. Assume that there are numbers $a$, $b$
and $d$ with $0<d<a<b\leq c$, such that
\begin{itemize}
\item[(i)] $\{x\in P(\omega, a, b): \omega(x)>a\}\neq\emptyset$ and
$\omega(Tx)>a$ for all $x\in P(\omega, a, b)$;

\item[(ii)] $\|Tx\|<d$ for all $x\in \overline{P_{d}}$;

\item[(iii)]
 $\omega(Tx)>a$ for all $x\in P(\omega, a, c)$ with $\|Tx\|>b$.
\end{itemize}
Then, $T$ has at least three fixed points $x_1$, $x_2$ and
$x_3$ in $\overline{P_{c}}$. Furthermore, $x_1\in P_{a}$;
 $x_2\in\{x\in P(\omega, a, c): \omega(x)>a\}$;
$x_3\in \overline{P_{c}}\setminus(P(\omega, b, c)\cup \overline{P_{a}})$.
\end{lemma}

\section{Main Results}

In this section, we investigate the existence of triple non-negative solutions
for \eqref{eq1.1} with \eqref{eq1.2} and \eqref{eq1.1} with \eqref{eq1.3},
and present some existence criteria.
Denote
\begin{gather*}
\Phi_1=\frac{\Gamma(\alpha)(\alpha-\beta)}{D},\quad
\Phi_2=\alpha(1-m_2\xi^{\alpha-1})\Gamma(\alpha),
\\
\Psi_1=\frac{\Gamma(\alpha)(\alpha-\beta)(\alpha-\beta+1)}{\beta
\xi^{\alpha-1}(1-\xi)^{\alpha-\beta}(\alpha \xi-\beta \xi+1)},\quad
\Psi_2=\frac{\alpha(\alpha+1)(1-m_2\xi^{\alpha-1})\Gamma(\alpha)}{M_0
\xi^{\alpha-1}(1-\xi)^{\alpha}(\alpha \xi+1)},
\\
\Pi_1=\frac{\Gamma(\alpha)(\alpha-\beta)(\alpha-\beta+1)}{\beta
\xi^{\alpha-1}}, \quad
\Pi_2=\frac{\alpha(\alpha+1)(1-m_2\xi^{\alpha-1})\Gamma(\alpha)}{M_0
\xi^{\alpha-1}},\\
\overline{f_0}=\limsup_{x\to 0^{+}}\sup_{t\in [0,1]}\frac{f(t,x)}{x},
\end{gather*}
where $D$ and $M_0$ are given in Lemmas \ref{lemma2.1} and \ref{lemma2.4},
 respectively.

To use Lemma \ref{lemma2.5}, we define a cone
$P=\{x\in E: x(t)\geq0, \forall t\in [0,1]\}$
and a functional $\omega : P\to [0, +\infty)$ by
\begin{equation}\label{eq3.1}
\omega(x)=\min_{t\in [\xi,1]}x(t),
\end{equation}
then, one can easily see that $\omega$ is a concave non-negative continuous
functional on $P$, and satisfies $\omega(x)\leq\|x\|$ for all $x\in P$.

We first consider \eqref{eq1.1} with \eqref{eq1.2} and obtain the following result.

\begin{theorem} \label{th3.1}
Consider \eqref{eq1.1} with \eqref{eq1.2}. Assume that
there exist two constants $a$ and $c$ with $0<a<\min\{c,\frac{\Phi_1}{\Psi_1}c\}$,
such that
\begin{itemize}
\item[(H1)] $f(t,x)\leq\Phi_1c$,
for all $(t,x)\in [0,1]\times [0,c]$;

\item[(H2)] there exists a constant
$\eta$ with $0\leq \eta <\Phi_1$,
such that $\overline{f_0}=\eta$;

\item[(H3)] $f(t,x)>\Psi_1a$,
for all $(t,x)\in [\xi,1]\times [a,c]$.
\end{itemize}
Then \eqref{eq1.1} with \eqref{eq1.2} has at least three non-negative solutions.
\end{theorem}

\begin{proof}  Let us divide the proof into 4 steps.

Step1. From (H1) and Lemma \ref{lemma2.2}, for all $x\in \overline{P_{c}}$, we have
\begin{align*}
\|T_1x\|
&= \max_{t\in[0,1]}\int_0^{1}G_1(t,s)f(s,x(s))ds\\
&\leq \Phi_1c\max_{t\in [0,1]}\int_0^{1}G_1(t,s)ds\\
&\leq \Phi_1c\max_{t\in [0,1]}\int_0^{1}\frac{Dt^{\alpha-1}(1-s)
^{\alpha-\beta-1}}{\Gamma(\alpha)}ds \\
&\leq c(\alpha-\beta)\int_0^{1}(1-s)^{\alpha-\beta-1}ds =c.
\end{align*}
Thus, $T_1:\overline{P_{c}}\to \overline{P_{c}}$.

Now, let us show that $T_1:\overline{P_{c}}\to \overline{P_{c}}$ is completely
continuous.
Let $x_{n}, x_0\in \overline{P_{c}}$ with
$\|x_{n}-x_0\|\to0$ as $n\to+\infty$. Then,
\begin{align*}
\|T_1x_{n}-T_1x_0\|
&= \max_{t\in [0,1]}|\int_0^{1}G_1(t,s)f(s,x_{n}(s))ds
-\int_0^{1}G_1(t,s)f(s,x_0(s))ds|\\
&\leq \max_{t\in [0,1]}\int_0^{1}G_1(t,s)|f(s,x_{n}(s))-f(s,x_0(s))|ds\\
&\leq \frac{D}{\Gamma(\alpha)}\int_0^{1}(1-s)^{\alpha-\beta-1}
|f(s,x_{n}(s))-f(s,x_0(s))|ds \to0,
\end{align*}
as $n\to +\infty$.
Hence, $T_1:\overline{P_{c}}\to \overline{P_{c}}$ is continuous.

In addition, for any $t_1,\ t_2\in[0, 1]$ and $x\in \overline{P_{c}}$,
we have
\begin{equation} \label{eq3.2}
|(T_1x)(t_1)-(T_1x)(t_2)|\leq\Phi_1c\int_0^{1}|G_1(t_1,s)-G_1(t_2,s)|ds.
\end{equation}
Since $G_1(t, s)$ is uniformly continuous on $(t, s)\in[0,1]\times[0, 1]$,
it is easy to see that $|(T_1x)(t_1)-(T_1x)(t_2)|\to0$ as $|t_1-t_2|\to0$.
Moreover, $T_1(\overline{P_{c}})$ is bounded. Thus, the Arzela-Ascoli theorem
guarantees that $T_1:\overline{P_{c}}\to \overline{P_{c}}$ is compact.
Therefore, $T_1:\overline{P_{c}}\to \overline{P_{c}}$ is completely
continuous.

Step2. Choose a constant $b\in(a,c]$. Let
$x_0(t)=\frac{a+b}{2},\ \forall\ t\in [0,1]$. Then,
$\omega(x_0)=\frac{a+b}{2}>a$ and $\|x_0\|=\frac{a+b}{2}<b$.
Thus, $x_0\in\{x\in P(\omega, a, b):\omega(x)>a\}\neq \emptyset$.

Now, let us prove that $\omega(T_1x)>a$ holds for all
$x\in P(\omega, a, b)$. In fact, $x\in P(\omega, a, b)$ implies that
$a\leq x(t)\leq b$, for all $t\in [\xi,1]$.
 One can obtain from (H3) and Lemma \ref{lemma2.2} that
\begin{align*}
 \omega(T_1x)&= \min_{t\in [\xi,1]}(T_1x)(t)=\min_{t\in
[\xi,1]}\int_0^{1}G_1(t,s)f(s,x(s))ds\\
&>\Psi_1a
\min_{t\in [\xi,1]}\int_{\xi}^{1}G_1(t,s)ds\\
&\geq \Psi_1a \min_{t\in [\xi,1]}\int_{\xi}^{1}\frac{\beta
t^{\alpha-1}s(1-s)^{\alpha-\beta-1}}{\Gamma(\alpha)}ds\\
&\geq \Psi_1a\frac{\beta\xi^{\alpha-1}}{\Gamma(\alpha)}
\int_{\xi}^{1}s(1-s)^{\alpha-\beta-1}ds=a.
\end{align*}
Hence, the condition (i) of Lemma \ref{lemma2.5} holds.

Step3. It is easy to see from (H2) that for all $t\in [0,1]$,
$\forall\ 0<\varepsilon\leq \frac{\alpha-\beta}{D}\Gamma(\alpha)-\eta$,
there exists $\delta>0$, such that for $0\leq x<\delta$, we have
\begin{equation} \label{eq3.3}
f(t,x)<(\eta+\varepsilon)x.
\end{equation}
Let $0<d<\min\{\delta, a\}$. Now, we prove that $\|T_1x\|<d$ for all
$x\in \overline{P_{d}}=\{x\in P:\|x\|\leq d\}$.
As a matter of fact, for all $x\in \overline{P_{d}}$, one can see
that
\begin{align*}
 \|T_1x\|&= \max_{t\in [0,1]}\int_0^{1}G_1(t,s)f(s,x(s))ds\\
&< (\eta+\varepsilon)\|x\|\max_{t\in [0,1]}\int_0^{1}G_1(t,s)ds\\
&\leq (\eta+\varepsilon)\|x\|\max_{t\in [0,1]}
\int_0^{1}\frac{Dt^{\alpha-1}(1-s)^{\alpha-\beta-1}}{\Gamma(\alpha)}ds
\\
&\leq (\eta+\varepsilon)\|x\|\frac{D}{\Gamma(\alpha)}
\int_0^{1}(1-s)^{\alpha-\beta-1}ds
\leq \|x\|\leq d
\end{align*}
Thus, $\|T_1x\|<d$, for all $x\in \overline{P_{d}}$.

Step4. Let us prove that $\omega(T_1x)>a$ holds for all $x\in
P(\omega, a, c)$ with $\|T_1x\|>b$.
For $x\in P(\omega, a, c)$ with $\|T_1x\|>b$, we have
 $a\leq x(t)\leq c$, for all $t\in [\xi,1]$. From (H3) and
 Lemma \ref{lemma2.2}, one can see that
\begin{align*}
\omega(T_1x)
&= \min_{t\in [\xi,1]}(T_1x)(t)\\
&= \min_{t\in[\xi,1]}\int_0^{1}G_1(t,s)f(s,x(s))ds\\
&\geq \min_{t\in [\xi,1]}\int_0^{1}\frac{\beta
t^{\alpha-1}s(1-s)^{\alpha-\beta-1}}{\Gamma(\alpha)}f(s,x(s))ds\\
&\geq \frac{\beta \xi^{\alpha-1}}{\Gamma(\alpha)}
 \int_{\xi}^{1}s(1-s)^{\alpha-\beta-1}f(s,x(s))ds\\
&> \frac{\beta \xi^{\alpha-1}}{\Gamma(\alpha)}\Psi_1
 a \int_{\xi}^{1}s(1-s)^{\alpha-\beta-1}ds=a.
\end{align*}
 Therefore, the condition (iii) of Lemma \ref{lemma2.5} is satisfied.

To summing up, all conditions of Lemma \ref{lemma2.5} hold;
therefore, BVP \eqref{eq1.1} with \eqref{eq1.2} has at least three
non-negative solutions.
\end{proof}


Next, we study \eqref{eq1.1} with \eqref{eq1.3} and establish a sufficient
condition for the existence of triple non-negative solutions of
 \eqref{eq1.1} with \eqref{eq1.3}.

\begin{theorem} \label{th3.2}
Consider  \eqref{eq1.1} with \eqref{eq1.3}. Assume that
there exist two constants $a$ and $c$ with $0<a<\min\{c,\frac{\Phi_2}{\Psi_2}c\}$,
such that
\begin{itemize}
\item[(H4)] $f(t,x)\leq\Phi_2c$ for all $(t,x)\in [0,1]\times [0,c]$;

\item[(H5)] there exists a constant
$\mu$ with $0\leq \mu <\Phi_2$,
such that $\overline{f_0}=\mu$;

\item[(H6)] $f(t,x)>\Psi_2a$
 for all $(t,x)\in [\xi,1]\times [a,c]$.
\end{itemize}
Then \eqref{eq1.1} with \eqref{eq1.3} has at least three non-negative solutions.
\end{theorem}

\begin{proof}
By (H4) and Lemma \ref{lemma2.4}, for all $x\in \overline{P_{c}}$, we have
\begin{align*}
\|T_2x\|
&= \max_{t\in [0,1]}\int_0^{1}G_2(t,s)f(s,x(s))ds\\
&\leq \Phi_2c\max_{t\in [0,1]}\int_0^{1}G_2(t,s)ds \\
&\leq \Phi_2c\max_{t\in [0,1]}\int_0^{1}\frac{t^{\alpha-1}
(1-s)^{\alpha-1}}{(1-m_2\xi^{\alpha-1})\Gamma(\alpha)}ds \\
&\leq \Phi_2c\frac{1}{(1-m_2\xi^{\alpha-1})\Gamma(\alpha)}\int_0^{1}(1-s)^{\alpha-1}ds
=c.
\end{align*}
Thus, $T_2:\overline{P_{c}}\to \overline{P_{c}}$.
Similar to the proof of Theorem \ref{th3.1}, it is easy to see that
$T_2:\overline{P_{c}}\to \overline{P_{c}}$ is completely
continuous.

Choose a constant $b\in(a,c]$. Denote by
$x_0(t)=\frac{a+b}{2}$ for all$ t\in [0,1]$. Then,
$\omega(x_0)=\frac{a+b}{2}>a$ and $\|x_0\|=\frac{a+b}{2}<b$.
Thus, $x_0\in\{x\in P(\omega, a, b):\omega(x)>a\}\neq \emptyset$.

Next, let us prove that $\omega(T_2x)>a$ holds for all
$x\in P(\omega, a, b)$. In fact, $x\in P(\omega, a, b)$ implies that
$a\leq x(t)\leq b$ for all $t\in [\xi,1]$. One can obtain from
(H6) and Lemma \ref{lemma2.4} that
 \begin{align*}
\omega(T_2x)
&= \min_{t\in [\xi,1]}(T_2x)(t)=\min_{t\in [\xi,1]}\int_0^{1}G_2(t,s)f(s,x(s))ds\\
&> \Psi_2a \min_{t\in [\xi,1]}\int_{\xi}^{1}G_2(t,s)ds\\
&\geq \Psi_2a \min_{t\in [\xi,1]}\int_{\xi}^{1}
 \frac{M_0}{(1-m_2\xi^{\alpha-1})\Gamma(\alpha)}
t^{\alpha-1}s(1-s)^{\alpha-1}ds\\
&\geq \Psi_2a\frac{M_0\xi^{\alpha-1}}{(1-m_2\xi^{\alpha-1})\Gamma(\alpha)}
\int_{\xi}^{1}s(1-s)^{\alpha-1}ds=a.
\end{align*}
Hence, the condition (i) of Lemma \ref{lemma2.5} holds.

Next, we prove that the condition (ii) of Lemma \ref{lemma2.5}
holds.
(H5) implies that for all $t\in [0,1]$,
and all $0<\varepsilon\leq \Phi_2-\mu$, there exists $\delta>0$,
such that for $0\leq x<\delta$, we have
\begin{equation}\label{eq3.4}
f(t,x)<(\mu+\varepsilon)x.
\end{equation}
Let $0<d<\min\{\delta, a\}$. Now, we prove that $\|T_2x\|<d$ for all
$x\in \overline{P_{d}}=\{x\in P:\|x\|\leq d\}$.

For all $x\in \overline{P_{d}}$, one can see
that
\begin{align*}
\|T_2x\|
&= \max_{t\in [0,1]}\int_0^{1}G_2(t,s)f(s,x(s))ds\\
&< (\mu+\varepsilon)\|x\|\max_{t\in [0,1]}\int_0^{1}G_2(t,s)ds\\
&\leq (\mu+\varepsilon)\|x\|\max_{t\in[0,1]}\int_0^{1}
\frac{t^{\alpha-1}(1-s)^{\alpha-1}}{(1-m_2\xi^{\alpha-1})\Gamma(\alpha)}ds
\\
&\leq (\mu+\varepsilon)\|x\|\frac{1}{(1-m_2\xi^{\alpha-1})\Gamma(\alpha)}
\int_0^{1}(1-s)^{\alpha-1}ds \\
&\leq  \|x\|\leq d.
\end{align*}
Thus, $\|T_2x\|<d$ for all $x\in \overline{P_{d}}$.

Finally, let us prove that $\omega(T_2x)>a$ holds for all $x\in
P(\omega, a, c)$ with $\|T_2x\|>b$.
For $x\in P(\omega, a, c)$ with $\|T_2x\|>b$, we have
 $a\leq x(t)\leq c,\ \forall\ t\in [\xi,1]$. From (H6) and
Lemma \ref{lemma2.4}, it is easy to see
\begin{align*}
\omega(T_2x)
&= \min_{t\in [\xi,1]}(T_2x)(t)\\
&= \min_{t\in[\xi,1]}\int_0^{1}G_2(t,s)f(s,x(s))ds\\
&\geq \min_{t\in [\xi,1]}\int_0^{1}\frac{M_0}{(1-m_2\xi^{\alpha-1})\Gamma(\alpha)}
t^{\alpha-1}s(1-s)^{\alpha-1}f(s,x(s))ds \\
&\geq \frac{M_0}{(1-m_2\xi^{\alpha-1})\Gamma(\alpha)}
\xi^{\alpha-1}\int_{\xi}^{1}s(1-s)^{\alpha-1}f(s,x(s))ds\\
&> \frac{M_0}{(1-m_2\xi^{\alpha-1})\Gamma(\alpha)}
\xi^{\alpha-1}\Psi_2a\int_{\xi}^{1}s(1-s)^{\alpha-1}ds=a.
\end{align*}
Therefore, condition (iii) of Lemma \ref{lemma2.5} is satisfied.

Hence, all conditions of Lemma \ref{lemma2.5} hold; therefore,
 \eqref{eq1.1} with \eqref{eq1.3} has at least three non-negative solutions.
\end{proof}

\begin{remark}\label{rem3.1}  \rm
It is noted that in the proof of Theorems \ref{th3.1} and \ref{th3.2},
the condition $\|T_{i}x\|>b$, $i=1,2$ is not applied in Step 4.
This is because  (H3) or (H6) is sufficient for the proof,
 which makes (H3) or (H6) strong.
\end{remark}

In the following, to apply the condition $\|T_{i}x\|>b$, $i=1,2$
in the proof and relax (H3) or (H6), we study \eqref{eq1.1} with \eqref{eq1.2}
and  \eqref{eq1.1} with \eqref{eq1.3} by constructing the following two cones:
\begin{gather}\label{eq3.5}
 P_1=\{x\in E: x(t)\geq0,\,\forall t\in [0,1];\
x(t)\geq \gamma_1\|x\|,\, \forall t\in[\xi,1]\},\\
P_2=\{x\in E: x(t)\geq0,\, \forall t\in [0,1];\,
x(t)\geq \gamma_2\|x\|,\, \forall t\in [\xi,1]\},
\end{gather}
where $0<\gamma_i<\min\{1, \frac{\Phi_{i}}{\Pi_{i}},
\frac{\Psi_{i}}{\Pi_{i}}\},\ i=1,2$. In this case,
$\omega : P_{i}\to [0, +\infty)$, $i=1,2$.

We first consider \eqref{eq1.1} with \eqref{eq1.2} by using the cone $P_1$,
and obtain the following result.

\begin{theorem} \label{th3.3}
Consider  \eqref{eq1.1} with \eqref{eq1.2}. Assume that there exist constants
$a$, $b$, $c$ and $d$ with $0<\frac{\gamma_1\Pi_1}{\Phi_1}c<d<a$ and
$0<\frac{\gamma_1\Pi_1}{\Psi_1}c\leq a<\gamma_1b<b\leq c$, such that
\begin{itemize}
\item[(H1')] $\gamma_1\Pi_1c\leq f(t,x)\leq \Phi_1 c$ for all
 $(t,x)\in [0,1]\times [0,c]$;

\item[(H2')] $f(t,x)<\Phi_1d$ for all $(t,x)\in [0,1]\times
[0,d]$;

\item[(H3')] $f(t,x)>\Psi_1a$ for all $(t,x)\in [\xi,1]\times
[a,b]$.
\end{itemize}
Then, \eqref{eq1.1} with \eqref{eq1.2} has at least three non-negative solutions.
\end{theorem}

\begin{proof}
 Let us divide the proof into 4 steps.

Step 1. By (H1') and Lemma
\ref{lemma2.2}, for any $x\in\overline{P_{c}}=\{x\in P_1: \|x\|\leq c\}$ we have
\begin{align*}
\|T_1x\|
&= \max_{t\in [0,1]}\int_0^{1}G_1(t,s)f(s,x(s))ds\\
&\leq \Phi_1 c\max_{t\in [0,1]}\int_0^{1}\frac{Dt^{\alpha-1}(1-s)
 ^{\alpha-\beta-1}}{\Gamma(\alpha)}ds \\
&\leq \frac{D}{(\alpha-\beta)\Gamma(\alpha)}\Phi_1 c=c,
\end{align*}
and
\begin{align*}
\min_{t\in [\xi,1]}(T_1x)(t)
&= \min_{t\in [\xi,1]}\int_0^{1}G_1(t,s)f(s,x(s))ds\\
&\geq \gamma_1\Pi_1c \min_{t\in [\xi,1]}\int_0^{1}G_1(t,s)ds\\
&\geq \gamma_1\Pi_1c \min_{t\in [\xi,1]}\int_0^{1}\frac{\beta
t^{\alpha-1}s(1-s)^{\alpha-\beta-1}}{\Gamma(\alpha)}ds\\
&\geq \gamma_1\Pi_1c\frac{1}{\Pi_1}
=\gamma_1c\geq\gamma_1\|T_1x\|.
\end{align*}
Thus, $T_1:\overline{P_{c}}\to \overline{P_{c}}$.

Next, let us show that $T_1:\overline{P_{c}}\to \overline{P_{c}}$ is completely
continuous.
Let $x_{n},\ x_0\in \overline{P_{c}}$ with $\|x_{n}-x_0\|\to0$ as $n\to+\infty$.
 Then
\begin{align*}
 \|T_1x_{n}-T_1x_0\|
&= \max_{t\in [0,1]}|\int_0^{1}G_1(t,s)f(s,x_{n}(s))ds
 -\int_0^{1}G_1(t,s)f(s,x_0(s))ds|\\
&\leq \max_{t\in [0,1]}\int_0^{1}G_1(t,s)|f(s,x_{n}(s))-f(s,x_0(s))|ds\\
&\leq \frac{D}{\Gamma(\alpha)}\int_0^{1}(1-s)^{\alpha-\beta-1}
|f(s,x_{n}(s))-f(s,x_0(s))|ds \to0,
\end{align*}
as $n\to+\infty$.
Hence, $T_1:\overline{P_{c}}\to \overline{P_{c}}$ is continuous.

In addition, for any $t_1, t_2\in[0, 1]$ and
 $x\in \overline{P_{c}}$, we have
\begin{equation} \label{eq3.2b}
|(T_1x)(t_1)-(T_1x)(t_2)|\leq \Phi_1 c\int_0^{1}|G_1(t_1,s)-G_1(t_2,s)|ds.
\end{equation}
Since $G_1(t, s)$ is uniformly continuous on
$(t, s)\in[0,1]\times[0, 1]$, it is easy to see that
$|(T_1x)(t_1)-(T_1x)(t_2)|\to0$ as
$|t_1-t_2|\to0$. Moreover, $T_1(\overline{P_{c}})$ is
bounded. Thus, the Arzela-Ascoli theorem guarantees that
$T_1:\overline{P_{c}}\to \overline{P_{c}}$ is compact.
Therefore, $T_1:\overline{P_{c}}\to \overline{P_{c}}$ is completely
continuous.

Step 2. Let $x_0(t)=\frac{a+b}{2}$ for all $t\in [0,1]$. Then
$\omega(x_0)=\frac{a+b}{2}>a$ and $\|x_0\|=\frac{a+b}{2}<b$.
Thus, $x_0\in\{x\in P(\omega, a, b):\omega(x)>a\}\neq \emptyset$.

Now, let us prove that $\omega(T_1x)>a$ holds for all
$x\in P(\omega, a,b)$. In fact, $x\in P(\omega, a, b)$ implies that
$a\leq x(t)\leq b$ for all $t\in [\xi,1]$. One can obtain from (H3')
and Lemma \ref{lemma2.2} that
\begin{align*}
\omega(T_1x)
&= \min_{t\in [\xi,1]}(T_1x)(t)=\min_{t\in [\xi,1]}\int_0^{1}G_1(t,s)f(s,x(s))ds\\
&> \Psi_1a \min_{t\in [\xi,1]}\int_{\xi}^{1}G_1(t,s)ds\\
&\geq \Psi_1a \min_{t\in [\xi,1]}\int_{\xi}^{1}\frac{\beta
t^{\alpha-1}s(1-s)^{\alpha-\beta-1}}{\Gamma(\alpha)}ds\\
&\geq \Psi_1a\frac{\beta\xi^{\alpha-1}}{\Gamma(\alpha)}
\int_{\xi}^{1}s(1-s)^{\alpha-\beta-1}ds=a.
\end{align*}
Hence,  condition (i) of Lemma \ref{lemma2.5} holds.

Step 3. It is easy to see from (H2') that for all $x\in \overline{P_{d}}$, we have
\begin{align*}
\|T_1x\|&= \max_{t\in [0,1]}\int_0^{1}G_1(t,s)f(s,x(s))ds\\
&< \Phi_1d\max_{t\in [0,1]}\int_0^{1}G_1(t,s)ds \\
&\leq \Phi_1d\max_{t\in [0,1]}\int_0^{1}\frac{Dt^{\alpha-1}
(1-s)^{\alpha-\beta-1}}{\Gamma(\alpha)}ds \\
&\leq \Phi_1d\frac{D}{\Gamma(\alpha)}\int_0^{1}(1-s)^{\alpha-\beta-1}ds=d.
\end{align*}
Thus, $\|T_1x\|<d$, for all $x\in \overline{P_{d}}$.

Step 4. Let us prove that $\omega(T_1x)>a$ holds for all
$x\in P(\omega, a, c)$ with $\|T_1x\|>b$.
For $x\in P(\omega, a, c)$ with $\|T_1x\|>b$, we have
$a \leq x(t)\leq c$ for all $t\in [\xi,1]$. Then, one can see that
$$
\omega(T_1x)=\min_{t\in [\xi,1]}(T_1x)(t)
\geq \gamma_1\|T_1x\|>\gamma_1b \geq a.
$$
Therefore, condition (iii) of Lemma \ref{lemma2.5} is satisfied.
By Lemma \ref{lemma2.5}, \eqref{eq1.1} with \eqref{eq1.2} has at
least three non-negative solutions.
\end{proof}


Next, we study \eqref{eq1.1} with \eqref{eq1.3} in the cone $P_2$
and establish the following result.

\begin{theorem} \label{th3.4}
Consider \eqref{eq1.1} with \eqref{eq1.3}. Assume that there exist constants
 $a$, $b$, $c$ and $d$ with $0<\frac{\gamma_2\Pi_2}{\Phi_2}c<d<a$ and
$0<\frac{\gamma_2\Pi_2}{\Psi_2}c\leq a<\gamma_2b<b\leq c$, such that
\begin{itemize}
\item[(H4')] $\gamma_2\Pi_2c\leq f(t,x)\leq \Phi_2 c$ for all
$ (t,x)\in [0,1]\times [0,c]$;

\item[(H5')] $f(t,x)<\Phi_2d$ for all $(t,x)\in [0,1]\times [0,d]$;

\item[(H6')] $f(t,x)>\Psi_2a$ for all $(t,x)\in [\xi,1]\times [a,b]$.
\end{itemize}
Then \eqref{eq1.1} with \eqref{eq1.3} has at least three non-negative solutions.
\end{theorem}

 The proof of the above theorem is similar to that of Theorem \ref{th3.3};
thus we omit it.



\begin{remark} \label{rem3.2}\rm
 From the proof of Theorems \ref{th3.1}-\ref{th3.4}, one can see that at
least two of the three non-negative solutions are positive.
\end{remark}

\begin{remark} \label{rem3.3} \rm
Comparing Theorem \ref{th3.1} and Theorem \ref{th3.3}, one can see that
 (H1) and (H2) are weaker than (H1') and (H2'), while (H3) is stronger than (H3').
Similarly, for Theorem \ref{th3.2} and Theorem \ref{th3.4}:
 (H4) and (H5) are weaker than (H4') and (H5'), while (H6) is stronger than (H6').
\end{remark}

\section{Examples}

In this section, we give two illustrative examples to support our new results.

\begin{example} \label{ex4.1}\rm
 Consider the  fractional order three-point BVP
\begin{equation}\label{eq4.1}
\begin{gathered}
D_{0+}^{3/2}u(t)+f(t,u(t))=0,\quad t\in(0,1),\\
u(0)=0,\quad D_{0+}^{1/2}u(1)=\frac{1}{2}D_{0+}^{1/2}u(\frac{1}{2}),
\end{gathered}
\end{equation}
where
\begin{equation}\label{eq4.2}
 f(t,x)=\begin{cases}
 \frac{1}{5}(1+t)x,&  0\leq t\leq 1,\, 0\leq x\leq 0.5,\\
\frac{99}{5}(1+t)x-\frac{49}{5}(1+t), &  0\leq t\leq 1,\, 0.5 <x<1,\\
10(1+t), & 0\leq t\leq 1,\, x\geq 1.
\end{cases}
\end{equation}
A simple calculation shows that $\Phi_1\approx0.4431$, $\Psi_1\approx6.6843$.
Set $a=1$ and $c=50$, then one can see that
$$
f(t,x)\geq15>\Psi_1a,\quad \forall (t,x)\in[\frac{1}{2},1]\times[1,50],
$$
and
$$
f(t,x)\leq20\leq\Phi_1c,\quad \forall (t,x)\in[0,1]\times[0,50].
$$
Thus, (H1) and (H3) hold.
Since
$\overline{f_0}=\overline{\lim}_{x\to 0^{+}}\sup_{t\in [0,1]}\frac{f(t,x)}{x}
=0.4<\Phi_1$, we conclude that (H2) holds.
By Theorem \ref{th3.1},  \eqref{eq4.1} has at least three non-negative solutions.
\end{example}


\begin{example} \label{ex4.2} \rm
 Consider the fractional order three-point BVP
\begin{equation} \label{eq4.3}
\begin{gathered}
D_{0+}^{3/2}u(t)+f(t,u(t))=0,\quad t\in(0,1),\\
u(0)=0,\quad u(1)=\frac{1}{2}u(\frac{1}{2}),
\end{gathered}
\end{equation}
where
\begin{equation} \label{eq4.4}
f(t,x)=\begin{cases}
 \frac{1}{4}(1+t)x, &  0\leq t\leq 1,\, 0\leq x\leq 0.05,\\
\frac{239}{4}(1+t)x-\frac{119}{40}(1+t), &  0\leq t\leq 1,\ 0.05 <x<0.1,\\
3(1+t),& 0\leq t\leq 1,\; x\geq 0.1.
\end{cases}
\end{equation}
One can calculate that $\Phi_2\approx0.8593$, $\Psi_2\approx27.7778$.
Set $a=0.1$ and $c=10$, then it is easy to see that
$$
f(t,x)\geq\frac{9}{2}>\Psi_2a,\quad \forall (t,x)\in[\frac{1}{2},1]\times[0.1,10],
$$
and
$$
f(t,x)\leq6\leq\Phi_2c,\quad \forall (t,x)\in[0,1]\times[0,10].
$$
Thus, (H4) and (H6) are satisfied.
A straightforward computation implies that
 $\overline{f_0}=\limsup_{x\to 0^{+}}\sup_{t\in [0,1]}\frac{f(t,x)}{x}=0.5<\Phi_2$,
and thus (H5) holds.
By Theorem \ref{th3.2}, \eqref{eq4.3} has at least three non-negative solutions.
\end{example}

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\end{document}