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

\AtBeginDocument{{\noindent\small
\emph{Electronic Journal of Differential Equations},
Vol. 2018 (2018), No. 105, pp. 1--24.\newline
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu}
\thanks{\copyright 2018 Texas State University.}
\vspace{8mm}}

\begin{document}
\title[\hfilneg EJDE-2018/105\hfil $p$-Laplacian fractional differential equations]
{Existence of solutions to nonlinear $p$-Laplacian fractional differential equations
with higher-order derivative terms}

\author[Y.-H. Su, Y. Yun, D. Wang, W. Hu \hfil EJDE-2018/105\hfilneg]
{You-Hui Su, Yongzhen Yun, Dongdong Wang, Weimin Hu}

\address{You-Hui Su (corresponding author) \newline
School of Mathematics and Physics,
Xuzhou University of Technology,
Xuzhou,  Jiangsu 221018, China}
\email{suyh02@163.com, suyouhui@xzit.edu.cn, fax 86-0516-85608333}

\address{Yongzhen Yun \newline
College of Science, Hohai University,
Nanjing, Jiangsu 211100, China.\newline
School of Mathematics and Physics,
Xuzhou University of Technology,
Xuzhou, Jiangsu 221018, China}
\email{yongzhen0614@163.com}

\address{Dongdong Wang \newline
School of Mathematics and Physics,
Xuzhou University of Technology,
Xuzhou, Jiangsu 221018, China}
\email{wdd@xzit.edu.cn}

\address{Weimin Hu \newline
School of Mathematics and Statistic,
Yili Normal University,
Yining, Xinjiang 835000, China}
\email{hwm680702@163.com}

\dedicatory{Communicated by Goong Chen}

\thanks{Submitted June 19, 2016. Published May 7, 2018.}
\subjclass[2010]{34B25, 34B18, 35G30}
\keywords{Fractional differential equation; Green's function;
\hfill\break\indent $p$-Laplacian operator; upper and lower solution method}

\begin{abstract}
 In this article, we discuss the existence of positive solution to a
 nonlinear $p$-Laplacian fractional differential equation
 whose nonlinearity contains a higher-order derivative
 \begin{gather*}
 D_{0^+}^{\beta}\phi_p\big(D_{0^+}^{\alpha}u(t)\big)
 +f\big(t,u(t),u'(t),\dots,u^{(n-2)}(t)\big)=0,\quad t\in ( 0,1 ),\\
 u(0)=u'(0)=\dots=u^{(n-2)}(0)=0,\\
 u^{(n-2)}(1)=au^{(n-2)}(\xi)=0,\quad D_{0^+}^{\alpha}u(0)=D_{0^+}^{\alpha}u(1)=0,
 \end{gather*}
 where ${n-1}<\alpha \leq n$, $n\geq 2$, $1<\beta \leq 2$, $0<\xi <1$,
 $0\leq a\leq 1$ and $0\leq a\xi ^{\alpha-n}\leq 1$,
 $\phi_{p}(s)=|s|^{p-2}s$, $p>1$, $\phi_{p}^{-1}=\phi_q$,
 $\frac{1}{p}+\frac{1}{q}=1$. $D_{0^+}^{\alpha}$, $D_{0^+}^{\beta}$
 are the standard Riemann-Liouville fractional derivatives, and
 $f\in C((0,1)\times[0,+\infty)^{n-1},[0,+\infty))$.
 The Green's function of the fractional differential equation mentioned above
 and its relevant properties are presented, and some novel results on the
 existence of positive solution   are established  by using the mixed monotone
 fixed point theorem and  the upper and lower solution method. The interesting
 of this paper is  that the  nonlinearity involves the higher-order derivative,
 and also, two examples are given in this paper to illustrate our main results
 from the perspective of application.
\end{abstract}

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

\section{Introduction}

In the past decades, there has been a growing interest in the
study of the  fractional differential equations due to the intensive
development of the fractional calculus theory itself and its applications
in various sciences such as engineering,  control theory,
blood flow phenomena, bode analysis of feedback amplifiers, electro-analytical
chemistry, and aerodynamics, etc.,  for details,
 see \cite{exp4,exp5,exp1,exp2,exp3,pa6s} and references therein.
For example, in studying a transfer process in porous material,
Mehaute \cite{Mehaute} discussed the following fractional differential
equations
$$
_0D_t^{1/d-1}J(t)=LX(t),
$$
where $J(t)$ is the macroscopic flow across the material interface, $X(t)$
is the local driving force, $L$ is a constant, and $d$ is
the fractal dimension of the material.
In the meantime, the existence theory of  solutions to the fractional
boundary-value problems  has attracted the attention of many researchers
quite recently,  see
\cite{bai, pa5, pa9,Kirane1,Kirane2,pxhd3,pxhd1,su,sujmaa,sum,zhou}
and their references.

We find that $p$-Laplacian differential equation has been widely applied in
analyzing mechanics, physics, dynamic systems and other related
fields of mathematical modeling. Hence, there have been many published
papers which are devoted to the existence of solutions  to  the  differential
equations with $p$-Laplacian operator,
see \cite{pa12,pa7, pxh02,pa28,  pxh01, pa24,sufeng,  pa26, paa07, pa30}
and their references. For example, in studying the turbulent flow in a porous medium,
Leibenson introduced the $p$-Laplacian equation in \cite{Leibenson} as follows
$$
(\phi_p(x'(t)))'=f(t,x(t),x'(t)),
$$
where $\phi_{p}(s)$ is $p$-Laplacian
operator, i.e., $\phi_{p}(s)=| s | ^{p-2}s$ for $p>1$
and $(\phi_{p})^{-1}=\phi_q,$ and $1/p+1/q=1$.

So,  based on the above illustration, it is of significance to make the
study of  the nonlinear $p$-Laplacian fractional differential equation.
In order to better explore the existence of positive solution to
 the nonlinear $p$-Laplacian fractional differential equation, here we
briefly review some related results in the
existing literature \cite{paa07,paa01,liu,pa004}.

Tian and Li \cite{paa07}   investigated the existence of positive solution
to the following  fractional differential equations with
$p$-Laplacian operator
\begin{equation}
\begin{gathered}
D_{0^+}^{\alpha}\phi_p\big(D_{0^+}^{\beta}u(t)\big)+f\big(t,u(t)\big)=0,\quad t\in ( 0,1 ),\\
u(0)=0,D_{0^+}^{\gamma}u(1)=\lambda D_{0^+}^{\gamma}u(\xi)=0,D_{0^+}^{\beta}u(0)=0,
\end{gathered}  \label{eqsu1}
\end{equation}
where $\phi_{p}(s)=|s|^{p-2}s$, $p>1$, $\phi_{p}^{-1}=\phi_q$,
$\frac{1}{p}+\frac{1}{q}=1$, $\alpha,\beta,\gamma\in \mathbb R$,
$0<\alpha<1$, $1<\beta\leq2$, $0<\gamma\leq1$
and $1+\gamma\leq\beta$, $0<\xi<1$, $\lambda\in[0,+\infty)$ and
$\lambda\xi^{\beta-\gamma-1}<1$. $D_{0^+}^{\alpha}$, $D_{0^+}^{\beta}$
are the standard Riemann-Liouville
fractional derivatives, and $f\in C([0,1]\times[0,+\infty),[0,+\infty))$.
The  existence results on positive solution to fractional differential
equations \eqref{eqsu1} are obtained by using some fixed point theorems in a  cone.

There are very few publications  concerning the existence
of positive solutions to fractional differential
equations with nonlinear terms involving the derivative \cite{pa004,paa01,liu}.
Cheng et al.\ \cite{pa004} investigated the positive solutions to the
following fractional differential equations  whose nonlinearity contains
the one-order  derivative as the form
\begin{equation}
\begin{gathered}
D_{0^+}^{\alpha}u(t)+f\big(t,u(t),u'(t)\big)=0,\quad t\in ( 0,1 ),\; n-1<\alpha\leq n,\\
u^{(i)}(0)=0,\; i=0,1,2,\dots,n-2,\quad[D_{0^+}^{\beta}u(t)]_{t=1}=0,\;
2\leq\beta\leq n-2,
\end{gathered} \label{eqsu2}
\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.
Some sufficient conditions  for the existence of positive solutions to
boundary-value problem \eqref{eqsu2} are established by using
fixed-point theorem.


It is notable that the nonlinear term $f(t,u(t))$ in equation \eqref{eqsu1}
does not involve the derivative.  In
 \cite{pa004,paa01,liu}, attention was mainly focused on the existence of
fractional differential equations with nonlinear terms involving the
first-order derivative and the $p$-Laplacian operator is not involved.
Apparently, the nonlinear term which is to be studied in this paper
$$
f(t,u(t),u'(t),\dots,u^{(n-2)}(t)), n=1,2,\dots,
$$
contains the higher-derivative, and we believe the study in this paper is
theoretically and practically significant because it will represent a more
general case.
 Naturally, it is interesting and necessary to study the existence
of positive solutions to $p$-Laplacian fractional differential
equations with nonlinear terms involving the higher-derivative.

In this paper, we mainly study the
existence of positive solutions to the following $p$-Laplacian  fractional
differential equations with nonlinear terms involving
the higher-derivative:
\begin{equation}
\begin{gathered}
D_{0^+}^{\beta}\phi_p(D_{0^+}^{\alpha}u(t))+f\big(t,u(t),u'(t),\dots,u^{(n-2)}(t)\big)=0,\quad
 t\in ( 0,1 ),\\
u(0)=u'(0)=\dots=u^{(n-2)}(0)=0,\\
u^{(n-2)}(1)=au^{(n-2)}(\xi)=0,\quad D_{0^+}^{\alpha}u(0)=D_{0^+}^{\alpha}u(1)=0,
\end{gathered} \label{FBVP1}
\end{equation}
where ${n-1}<\alpha \leq n$, $n\geq 2$, $1<\beta \leq 2$, $0<\xi <1$,
 $0\leq a\leq 1$ and $0\leq a\xi ^{\alpha-n}\leq 1$, $\phi_{p}(s)=|s|^{p-2}s$, $p>1$,
$\phi_{p}^{-1}=\phi_q$, $\frac{1}{p}+\frac{1}{q}=1$. $D_{0^+}^{\alpha}$,
$D_{0^+}^{\beta}$ are the standard Riemann-Liouville fractional derivatives,
and $f\in C((0,1)\times[0,+\infty)^{n-1},[0,+\infty))$.

The Green's function of the  boundary-value
problem \eqref{FBVP1} and the relevant properties are to be presented later,
and because of the nonlinear terms involving the higher-derivative in fractional
differential equations \eqref{FBVP1}, it's very difficult or even impossible
to obtain the existence of  positive solution of it
by using some fixed point theorem in a cone, such as  nonlinear alternative
of Leray-Schauder type and  Krasnosel'skii's fixed point theorem, and it is
the same for the methods listed in \cite{pa004,paa01,liu,paa07}.
The reason for that is the  nonlinearity  is in a  high dimensional space and
is not controlled  in a cone   because  of the  nonlinear terms involving the
higher-derivative, and so we establish some novel results on the existence of
positive solution by using the mixed monotone fixed point theorem and
the upper and lower solution method.

The first special feature and innovative contribution of our work is that we
present in this paper the Green's function of the differential equation and
its relevant properties, which is very difficult because the differential equation
relates to the standard Riemann-Liouville fractional derivatives and
$p$-Laplacian operator.
The second special feature and innovative contribution of our work is that
the nonlinearity involves the higher-order derivative, which is also not
so easy for the  nonlinearity is not controlled in a cone  because  of the
nonlinear terms involving the higher-derivative. Therefore, we try to deal
with this problem by using a new method which is different from many other
works \cite{pa004,paa01,liu,paa07}.
In addition, two examples are also given in this paper to illustrate our
 main results from the viewpoint of applications.

The structure of our paper is as follows. Section 1 is the introduction of
the paper. In Section 2,  some necessary definitions and lemmas which are
cited in our paper are presented.  In Section 3, we construct an
equivalent fractional differential equation.  The Green's function of
the equivalent fractional differential equation is constructed, and its
properties are presented in Section 4.
The existence  results on  unique positive solution  to the fractional
boundary-value problem \eqref{FBVP1} are obtained in Section 5.
The existence  theorem of at least  single  positive solution to the
fractional boundary-value problem \eqref{FBVP1} is proved in
Section 6. In Section 7, we give two examples to
illustrate our main results.

\section{Preliminaries}

To prove our main results, in this section we present some basic definitions
and  technical lemmas which can help us to better understand our main results
and proofs. For the basic terminologies,  we refer the
reader to references \cite{exp5,paa01,noor,mix1}.


\begin{definition}[\cite{paa01}] \rm
The Riemann-Liouville fractional integral of
order $\alpha> 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{paa01}] \rm
The Riemann-Liouville fractional derivative of order $\alpha > 0 $ of a
function $y: (0,\infty)\to\mathbb R$ is given by
$$
D_{0^+}^{\alpha}y(t)=\frac{1}{\Gamma(n-\alpha)}
(\frac{d}{dt})^{n}\int_{0}^{t}(t-s)^{n-\alpha-1}y(s)ds,
$$
where $n$ is the smallest integer greater than or equal to $\alpha$,
provided that the right side is pointwise defined on $(0,\infty)$.
\end{definition}

Let $P$ be a normal cone of a Banach space $E$,
and $e\in P$ with $\|e\| \leq 1$, $e\neq \theta$ ($\theta$ is zero element of $E$).
Define
\[
Q_e=\{x\in P:\text{ there exist constants $m,M>0$ such that }
me\leq x \leq Me\}.
\]
\begin{definition}[\cite{mix1}] \rm
Let $T$ be a operator satisfies $T:Q_{e}\times Q_{e}\to Q_{e}$. $T$ is said
to be mixed monotone if $T(x,y)$ is nondecreasing in $x$
and nonincreasing in $y$, i.e., if
$x_1\leq x_{2}$ $(x_1,x_{2}\in Q_{e})$ implies $T(x_1,y)\leq T(x_{2},y)$
for any $y\in Q_{e}$, and $y_1\geq y_{2}~(y_1,y_{2}\in Q_{e})$ implies
$T(x,y_1)\leq T(x,y_{2})$ for any $x\in Q_{e}$. Element $x^{*}\in Q_{e}$ is
called a fixed point of $T$ if $T(x^{*},x^{*})=x^{*}$.
\end{definition}

Next we give some Lemmas which are used in  our main  results.

\begin{lemma}[\cite{paa01}] \label{Lemma 1}
The equality $I_{0^+}^{\gamma}I_{0^+}^{\delta}y(t)=I_{0^+}^{\gamma +\delta}y(t)$,
$\gamma >0$, $\delta>0$ holds for $y\in C(0,1)\cap L(0,1)$.
\end{lemma}

\begin{lemma}[\cite{paa01}]\label{Lemma 2}
The equality $D_{0^+}^{\gamma}I_{0^+}^{\gamma}y(t)=y(t)$, $\gamma >0$ holds
for $y\in C(0,1)\cap L(0,1)$.
\end{lemma}

\begin{lemma}[\cite{paa01}]
Assume that $u\in C(0,1)\cap L(0,1)$ with a fractional derivative of $\alpha >0$
that belongs to $C(0,1)\cap L(0,1)$. Then
the fractional differential equation
$$ D_{0^+}^{\alpha}y(t)=0,
$$
has a unique solution
$y(t)=C_1t^{\alpha -1}+C_{2}t^{\alpha -2}+\dots +C_{n}t^{\alpha -n}$, where
$C_{i}\in R$, $i=1,2,\dots,n$, $n$ is the smallest integer greater than or
equal to $\alpha$.
\end{lemma}

\begin{lemma}[\cite{paa01}] \label{Lemma 3}
Assume that $u\in C(0,1)\cap L(0,1)$ with a fractional derivative of $\alpha >0$
that belongs to $C(0,1)\cap L(0,1)$. Then
$$
I_{0^+}^{\alpha}D_{0^+}^{\alpha}y(t)=y(t)+C_1t^{\alpha -1}+C_{2}t^{\alpha -2}
+\dots +C_{n}t^{\alpha -n},
$$
for some $C_{i}\in R$, $i=1,2,\dots,n$, where $n$ is the smallest integer
greater than or equal to $\alpha$.
\end{lemma}

\begin{lemma}[\cite{mix1}] \label{Lemma 8}
Assume that the operator $T:Q_{e}\times Q_{e}\to Q_{e}$ is mixed monotone
operator and there exists a constant $\delta ~(0<\delta<1)$, such that
$$
T(tx,\frac{1}{t}y)\geq t^{\delta}T(x,y), x,y\in Q_{e},0< t<1.
$$
Then the operator $T$ has a unique fixed point $x^{*}(x^{*}\in Q_{e})$.
\end{lemma}

\section{Equivalence of fractional differential equation}

In this section, we  construct   an  equivalent fractional differential equation,
and prove that search for  the solution of
fractional differential equation \eqref{FBVP1} is equivalent  to finding
the solution of  it.

\begin{lemma} \label{Lemma 7}
Let $u(t)=I_{0^+}^{n-2}v(t)$, $v\in C[0,1]$, then  the fractional boundary-value problem {\rm\eqref{FBVP1}} is equivalent  to  the following
fractional differential equation
\begin{equation}
\begin{gathered}
D_{0^+}^{\beta}\phi_{p}(D_{0^+}^{\alpha-n+2}v(t))
+f(t,I_{0^+}^{n-2}v(t),I_{0^+}^{n-3}v(t),\dots,I_{0^+}^{1}v(t),v(t))=0,
\\ t\in( 0,1 )\\
v(0)=0,\quad v(1)=av(\xi),\quad D_{0^+}^{\alpha-n+2}v(0)=D_{0^+}^{\alpha-n+2}v(1)=0,
\end{gathered}  \label{FBVP5}
\end{equation}
where ${n-1}<\alpha \leq n$, $n\geq 2$, $1<\beta \leq 2$, $0<\xi <1$,
$0\leq a\leq 1$ and $0\leq a\xi ^{\alpha-n}\leq 1$, $\phi_{p}=|s|^{p-2}s$,
$p>1$, $\phi_{p}^{-1}=\phi_q$, $\frac{1}{p}+\frac{1}{q}=1$ and
$f\in C((0,1)\times[0,+\infty)^{n-1},[0,+\infty))$. Moreover,
 $v\in C([0,1],[0,+\infty))$ is a positive solution of the differential
equation \eqref{FBVP5} means that $u(t)=I_{0^+}^{n-2}v(t)$ is a positive
solution of the differential equation \eqref{FBVP1}.
\end{lemma}

\begin{proof}
Let $u(t)=I_{0^+}^{n-2}v(t)$, it follow from the definition of
Riemann-Liouville fractional derivative, Lemma \ref{Lemma 1} and
Lemma \ref{Lemma 2} that
\begin{gather*}
D_{0^+}^{n}I_{0^+}^{n-\alpha}u(t)=D_{0^+}^{n}I_{0^+}^{n-\alpha}I_{0^+}^{n-2}v(t)
=D_{0^+}^{n}I_{0^+}^{2n-\alpha-2}v(t)=D_{0^+}^{\alpha-n+2}v(t),\\
u'(t)=D_{0^+}^{1}I_{0^+}^{n-2}v(t)=D_{0^+}^{1}I_{0^+}^{1}I_{0^+}^{n-3}v(t)
 =I_{0^+}^{n-3}v(t), \\
u''(t)=D_{0^+}^{2}I_{0^+}^{n-2}v(t)=D_{0^+}^{2}I_{0^+}^{2}I_{0^+}^{n-4}v(t)
  =I_{0^+}^{n-4}v(t), \\
\dots \\
u^{(n-3)}(t)=D_{0^+}^{n-3}I_{0^+}^{n-2}v(t)
 =D_{0^+}^{n-3}I_{0^+}^{n-3}I_{0^+}^{1}v(t)=I_{0^+}^{1}v(t),\\
u^{(n-2)}(t)=D_{0^+}^{n-2}I_{0^+}^{n-2}v(t)=v(t).
\end{gather*}
Therefore
\begin{gather*}
D_{0^+}^{\beta}(\phi_{p}D_{0^+}^{\alpha-n+2}v(t))
 +f(t,I_{0^+}^{n-2}v(t),I_{0^+}^{n-3}v(t),\dots,I_{0^+}^{1}v(t),v(t))=0, \\
v(0)=u^{n-2}(0)=u^{n-3}(0)=\dots =u(0)=0, \\
v(1)=av(\xi),D_{0^+}^{\alpha-n+2}v(0)=D_{0^+}^{\alpha-n+2}v(1)=0.
\end{gather*}
From above discussions,  let $u(t)=I_{0^+}^{n-2}v(t)$, then the differential
equation \eqref{FBVP1} is equivalent  to the differential equation \eqref{FBVP5}.

Now, let $v\in C([0,1],[0,+\infty))$ is a positive solution of  the differential
equation \eqref{FBVP5}. Then
\begin{align*}
D_{0^+}^{\beta}\phi_p(D_{0^+}^{\alpha-n+2}v(t))
&= D_{0^+}^{\beta}\phi_p(D_{0^+}^{n}I_{0^+}^{2n-\alpha-2}v(t))\\
&= D_{0^+}^{\beta}\phi_p(D_{0^+}^{n}I_{0^+}^{n-\alpha}I_{0^+}^{n-2}v(t))\\
&= D_{0^+}^{\beta}\phi_p(D_{0^+}^{\alpha}I_{0^+}^{n-2}v(t))\\
&= D_{0^+}^{\beta}\phi_p(D_{0^+}^{\alpha}u(t)),
\end{align*}
and
$$
f(t,I_{0^+}^{n-2}v(t),I_{0^+}^{n-3}v(t),\dots,I_{0^+}^{1}v(t),v(t))
=f(t,u(t),u'(t),\dots,u^{(n-2)}(t)),
$$
which implies that $u(t)=I_{0^+}^{n-2}v(t)$ is a positive solution of
 the differential equation \eqref{FBVP1}.
The proof is complete.
\end{proof}

\section{Properties of Green's Function}

In this section,  we obtain the Green's function of  fractional boundary-value
problem \eqref{FBVP5}   and its some properties.

\begin{lemma} \label{Lemma 4}
Assume that $y\in C[0,1]$ and $n-1<\alpha \leq n$, then  the following fractional
 boundary-value problem
\begin{equation}
\begin{gathered}
D_{0^+}^{\alpha-n+2}v(t)+y(t)=0,\quad t\in( 0,1 ),\\
v(0)=0,\quad v(1)=av(\xi),
\end{gathered}  \label{FBVP2}
\end{equation}
has a unique solution
$$
v(t)=\int_0^{1}G(t,s)y(s)d(s),
$$
where
\begin{equation}\label{G(t)}
\begin{aligned}
&G(t,s) \\
&=\begin{cases}
\frac{[t(1-s)]^{\alpha-n+1}-a[t(\xi-s)]^{\alpha-n+1}
-(1-a\xi^{\alpha-n+1})(t-s)^{\alpha-n+1}}{(1-a\xi^{\alpha-n+1})
\Gamma(\alpha-n+2)},
 &0\leq s \leq t \leq 1, s\leq\xi,
\\
\frac{[t(1-s)]^{\alpha-n+1}-(1-a\xi^{\alpha-n+1})(t-s)^{\alpha-n+1}}
{(1-a\xi^{\alpha-n+1})\Gamma(\alpha-n+2)},
 &0< \xi\leq s \leq t \leq 1,
\\
\frac{[t(1-s)]^{\alpha-n+1}-a[t(\xi-s)]^{\alpha-n+1}}
 {(1-a\xi^{\alpha-n+1})\Gamma(\alpha-n+2)},
& 0\leq t\leq s \leq \xi < 1,
\\
\frac{[t(1-s)]^{\alpha-n+1}}{(1-a\xi^{\alpha-n+1})\Gamma(\alpha-n+2)},
& 0\leq t\leq s \leq 1,\xi\leq s.
\end{cases}
\end{aligned}
\end{equation}
\end{lemma}

\begin{lemma} \label{Lemma 001}
Let $n-1<\alpha\leq n,0<\xi<1,0\leq a \leq 1$. If $y(t)\in C[0,1]$ and
$y(t)\geq 0$ hold, then the fractional differential equation {\rm\eqref{FBVP2}}
has a unique solution $v(t)\geq 0,~t\in[0,1]$.
\end{lemma}


\begin{lemma} \label{Lemma 5}
Assume that $y\in C[0,1]$ and $n-1<\alpha \leq n$, $0<\xi<1$, $1<\beta \leq 2$,
$0\leq a \leq1$, then the following fractional differential equation
\begin{equation}
\begin{gathered}
D_{0^+}^{\beta}\phi_{p}(D_{0^+}^{\alpha-n+2}v(t))=y(t),\quad t\in( 0,1 ),\\
v(0)=0,v(1)=av(\xi),D_{0^+}^{\alpha-n+2}v(0)=D_{0^+}^{\alpha-n+2}v(1)=0,
\end{gathered}\label{FBVP3}
\end{equation}
has a unique solution
$$
v(t)=\int_0^{1}G(t,s)\phi_q\Big(\int_0^{1}H(s,\tau)y(\tau)d\tau\Big)ds,
$$
where
\begin{equation}\label{H(s)}
H(s,\tau)=\begin{cases}
\frac{s^{\beta-1}(1-\tau)^{\beta-1}-(s-\tau)^{\beta-1}}{\Gamma(\beta)},
& 0\leq \tau \leq s\leq1,\\
\frac{s^{\beta-1}(1-\tau)^{\beta-1}}{\Gamma(\beta)},
& 0\leq s\leq \tau \leq 1,
\end{cases}
\end{equation} and
$G(t,s)$ is defined in Lemma \rm{\ref{Lemma 4}}.
\end{lemma}

\begin{proof}
It follows from  Lemma \ref{Lemma 3}  that
$$
\phi_{p}(D_{0^+}^{\alpha-n+2}v(t))
=I_{0^+}^{\beta}y(t)+C_1t^{\beta -1}+C_{2}t^{\beta -2}, \quad t\in(0,1),
$$
where $C_1,C_{2}\in \mathbb R$.
According to the  boundary condition $D_{0^+}^{\alpha-n+2}v(0)=0$ and
$D_{0^+}^{\alpha-n+2}v(1)=0,$  one has
$$
C_1=-I_{0^+}^{\beta}y(t)|_{t=1}
=-\frac{1}{\Gamma(\beta)}\int_0^{1}(1-\tau)^{\beta-1}y(\tau)d\tau,~C_{2}=0,
$$
this implies
\begin{align*}
\phi_{p}(D_{0^+}^{\alpha-n+2}v(t))
&= I_{0^+}^{\beta}y(t)-t^{\beta -1}I_{0^+}^{\beta}y(1)\\
&=\frac{1}{\Gamma(\beta)}\int_0^{t}(t-\tau)^{\beta-1}y(\tau)d\tau
 -\frac{t^{\beta-1}}{\Gamma(\beta)}\int_0^{1}(1-\tau)^{\beta-1}y(\tau)d\tau\\
&=-\int_0^{1}H(t,\tau)y(\tau)d\tau,
\end{align*}
i.e.,
$$
D_{0^+}^{\alpha-n+2}v(t)+\phi_q\Big(\int_0^{1}H(t,\tau)y(\tau)d\tau\Big)=0.
$$
Therefore, the fractional boundary-value problems \eqref{FBVP3} is equivalent
to the following fractional boundary-value problems
\begin{equation}
\begin{gathered}
D_{0^+}^{\alpha-n+2}v(t)+\phi_q\Big(\int_0^{1}H(t,\tau)y(\tau)d\tau\Big)=0,\quad
t\in( 0,1 ),\\
v(0)=0, \quad v(1)=av(\xi).
\end{gathered} \label{FBVP4}
\end{equation}
It follows from Lemma \ref{Lemma 4} that  the fractional boundary-value
problems \eqref{FBVP4} exists a unique solution
$$
v(t)=\int_0^{1}G(t,s)\phi_q\Big(\int_0^{1}H(s,\tau)y(\tau)d\tau\Big)ds.
$$
The proof is complete.
\end{proof}

\begin{lemma} \label{Lemma 6}
Assume that  $0\leq a\xi^{\alpha-n}\leq 1$ holds, there exist  the functions
$G(t,s)$ and $H(t,s)$ be defined by \eqref{G(t)} and \eqref{H(s)} such that
\item[(i)]  $G(t,s)$ and $H(t,s)$ are  continuous functions on $[0,1]\times [0,1]$;

\item[(ii)] $G(t,s)\leq \frac{t^{\alpha-n+1}}{\Gamma(\alpha-n+2)}$ for
$(t,s)\in[0,1]\times [0,1]$, $H(t,s)\leq \frac{t^{\beta-1}}{\Gamma(\beta)}$
for $(t,s)\in[0,1]\times [0,1]$;

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

\item[(iv)]  $G(t,s)\leq G(s,s)$ for $(t,s)\in[0,1]\times [0,1]$,
$H(t,s)\leq H(s,s )$ for $(t,s)\in[0,1]\times [0,1]$;

\item[(v)] there exist positive functions $\gamma(s) \in C[0,1]$ and
 $\rho(s) \in C[0,1]$ such that
 $$
\min_{t\in [\xi,1]}G(t,s)\geq \gamma(s) \max_{t\in [0,1]}G(t,s)
=\gamma(s) G(s,s)\quad\text{for }0< s <1,
$$
and
$$
\min_{t\in [\xi,1]}H(t,s)\geq \rho(s) \max_{t\in [0,1]}H(t,s)
=\rho(s) H(s,s)\quad\text{for }0< s <1.
$$
\end{lemma}

\begin{proof}
From the definition of $G(t,s)$ and $H(t,s)$, it is easy to check that (i)
and (ii) be satisfied.
 We shall prove that (iii) holds, set
\begin{gather*}
g_1(t,s)=[t(1-s)]^{\alpha-n+1}-a[t(\xi-s)]^{\alpha-n+1}
-(1-a\xi^{\alpha-n+1})(t-s)^{\alpha-n+1}, \\
\text{for }0\leq s \leq t \leq 1, s\leq\xi; \\
g_{2}(t,s)=[t(1-s)]^{\alpha-n+1}-(1-a\xi^{\alpha-n+1})(t-s)^{\alpha-n+1},\\
\text{for }0< \xi\leq s \leq t \leq 1; \\
g_{3}(t,s)=[t(1-s)]^{\alpha-n+1}-a[t(\xi-s)]^{\alpha-n+1},\quad
0\leq t\leq s \leq \xi < 1;;\\
g_{4}(t,s)=[t(1-s)]^{\alpha-n+1},\quad 0\leq t\leq s \leq 1,\xi\leq s.
\end{gather*}
To prove that (iii) is true, we  need to show that $g_{i}\geq0$ for $i=1,2,3,4$.

(1) If $t<\xi$, since $0\leq a\xi^{\alpha-n+1} \leq 1$ and $0<\xi<1$, we have
\begin{align*}
&g_1(t,s) \\
&= (t-ts)^{\alpha -n+1}-a(t\xi -ts)^{\alpha -n+1}-(t-s)^{\alpha -n+1}
 +a\xi^{\alpha -n+1}(t-s)^{\alpha -n+1}\\
&= [(t-ts)^{\alpha -n+1}-(t-s)^{\alpha -n+1}]
 -a\xi^{\alpha -n+1}
 \big[\big(t-\frac{ts}{\xi}\big)^{\alpha-n+1}-(t-s)^{\alpha-n+1}\big]\\
&\geq (t-ts)^{\alpha-n+1}-\big(t-\frac{ts}{\xi}\big)^{\alpha-n+1}
\geq 0.
\end{align*}
Moreover, if $t\geq \xi$,  then
\begin{align*}
g_1(t,s)
&= t^{\alpha-n+1}\big[(1-s)^{\alpha-n+1}-a(\xi-s)^{\alpha-n+1}
 -(1-a\xi^{\alpha-n+1})(1-\frac{s}{t})^{\alpha-n+1}\big]\\
&= t^{\alpha-n+1}\big\{[(1-s)^{\alpha-n+1}-(1-\frac{s}{t})^{\alpha-n+1}]\\
&\quad  +a\xi^{\alpha-n+1}
[(1-\frac{s}{t})^{\alpha-n+1}-(1-\frac{s}{\xi})^{\alpha-n+1}]\big\}
\geq 0.
\end{align*}

(2) If $0< \xi \leq s \leq t \leq 1$, according to $0\leq a\xi^{\alpha-n}\leq 1$,
there is
\begin{align*}
g_{2}(t,s)
&\geq (1-s)^{\alpha-n+1}t^{\alpha-n+1}-(t-s)^{\alpha-n+1}\\
&=  t^{\alpha-n+1}[(1-s)^{\alpha-n+1}-(1-\frac{s}{t})^{\alpha-n+1}]
\geq 0.
\end{align*}

(3) If $0\leq t \leq s \leq \xi <1$, we obtain
\begin{align*}
g_{3}(t,s)
&= t^{\alpha-n+1}\big[(1-s)^{\alpha-n+1}-a\xi^{\alpha-n+1}
 (1-\frac{s}{\xi})^{\alpha-n+1}\big]\\
&\geq t^{\alpha-n+1}[(1-s)^{\alpha-n+1}-(1-\frac{s}{\xi})^{\alpha-n+1}]\\
&\geq 0, \quad \text{for }0 \leq a\xi^{\alpha-n} \leq 1.
\end{align*}

(4) It is obvious that $g_{4}(t,s)\geq 0$ for $0 \leq t \leq s \leq 1$,
$\xi \leq s$.

Similarly,  $H(t,s)\geq 0$  for $t,s \in (0,1)$.
From above discussions, we conclude that $G(t,s)\geq 0$ and $H(t,s)\geq 0$
for any $t,s \in (0,1)$. So property (iii) holds.

Now we prove that (iv) holds.
Firstly, we check that $g_1(t,s)$ and $g_{2}(t,s)$ are nonincreasing with
respect to $t\in[s,1]$.
\begin{align*}
\frac{\partial{g_1(t,s)}}{\partial(t)}
&= (\alpha-n+1)t^{\alpha-n}(1-s)^{\alpha-n+1}
 -a(\xi-s)^{\alpha-n+1}(\alpha-n+1)t^{\alpha-n}\\
&\quad -(1-a\xi^{\alpha-n+1})(\alpha-n+1)(t-s)^{\alpha-n}\\
&= (\alpha-n+1)[t^{\alpha-n}(1-s)^{\alpha-n+1}-a(\xi-s)^{\alpha-n+1}t^{\alpha-n}\\
&\quad -(1-a\xi^{\alpha-n+1})(t-s)^{\alpha-n}]\\
&= (\alpha-n+1)t^{\alpha-n}[(1-s)^{\alpha-n+1}-a\xi^{\alpha-n+1}
 (1-\frac{s}{\xi})^{\alpha-n+1} \\
&\quad -(1-a\xi^{\alpha-n+1})(1-\frac{s}{t})^{\alpha-n}]\\
&\leq (\alpha-n+1)t^{\alpha-n}\Big[(1-s)^{\alpha-n+1}
 -a\xi^{\alpha-n+1}(1-\frac{s}{\xi})(1-s)^{\alpha-n+1}\\
&\quad -(1-a\xi^{\alpha-n+1})(1-s)^{\alpha-n}\Big]\\
&= (\alpha-n+1)t^{\alpha-n}(1-s)^{\alpha-n}[(1-s)
 -a\xi^{\alpha-n+1}(1-\frac{s}{\xi})-(1-a\xi^{\alpha-n+1})]\\
&= (\alpha-n+1)t^{\alpha-n}(1-s)^{\alpha-n}[s(a\xi^{\alpha-n}-1)]\\
&\leq 0,
\end{align*}
and
\begin{align*}
\frac{\partial{g_{2}(t,s)}}{\partial(t)}
&= (\alpha-n+1)t^{\alpha-n}(1-s)^{\alpha-n+1}-(1-a\xi^{\alpha-n+1})
 (\alpha-n+1)(t-s)^{\alpha-n}\\
&= (\alpha-n+1)[t^{\alpha-n}(1-s)^{\alpha-n+1}-(1-a\xi^{\alpha-n+1})
 (t-s)^{\alpha-n}]\\
&= (\alpha-n+1)t^{\alpha-n}[(1-s)^{\alpha-n+1}
 -(1-\frac{s}{t})^{\alpha-n}(1-a\xi^{\alpha-n+1})]\\
&\leq (\alpha-n+1)t^{\alpha-n}[(1-s)^{\alpha-n+1}
 -(1-s)^{\alpha-n}(1-a\xi^{\alpha-n+1})]\\
&= (\alpha-n+1)t^{\alpha-n}(1-s)^{\alpha-n}[(1-s)-(1-a\xi^{\alpha-n+1})]\\
&= (\alpha-n+1)t^{\alpha-n}(1-s)^{\alpha-n}(a\xi^{\alpha-n+1}-s)\\
&\leq (\alpha-n+1)t^{\alpha-n}(1-s)^{\alpha-n}(a\xi^{\alpha-n+1}-\xi)\\
&= (\alpha-n+1)t^{\alpha-n}(1-s)^{\alpha-n}\xi(a\xi^{\alpha-n}-1)\\
&\leq  0.
\end{align*}
Then,  $g_1(t,s)$ and $g_{2}(t,s)$ is non-increasing with respect to $t\in[s,1]$.

Secondly, we  show that $g_{3}(t,s)$ and $g_{4}(t,s)$ are nondecreasing with
respect to $t\in[0,s]$.
\begin{align*}
\frac{\partial{g_{3}(t,s)}}{\partial(t)}
&= (\alpha-n+1)t^{\alpha-n}(1-s)^{\alpha-n+1}
 -a(\xi-s)^{\alpha-n+1}(\alpha-n+1)t^{\alpha-n}\\
&= (\alpha-n+1)t^{\alpha-n}[(1-s)^{\alpha-n+1}-a(\xi-s)^{\alpha-n+1}]\\
&\geq (\alpha-n+1)t^{\alpha-n}[(1-s)^{\alpha-n+1}-a(1-s)^{\alpha-n+1}]\\
&= (\alpha-n+1)t^{\alpha-n}(1-s)^{\alpha-n}(1-a)
\geq 0,
\end{align*}
which implies that $g_{3}(t,s)$ is nondecreasing with respect to $t$ on $[0,s]$.
It is obvious that $g_{4}(t,s)$ is nondecreasing with respect to $t$ on $[0,s]$.
Therefore,
\begin{gather*}
G(t,s)\leq G(s,s)\quad\text{for }0\leq s \leq t \leq 1, \\
G(t,s)\leq G(s,s)\quad\text{for }0\leq t \leq s \leq 1.
\end{gather*}
In conclusion
$$
G(t,s)\leq G(s,s)\quad \text{for }(t,s)\in[0,1]\times [0,1].
$$

Thirdly, setting
\begin{gather*}
h_1(t,s)=t^{\beta-1}(1-s)^{\beta-1}-(t-s)^{\beta-1},\quad 0\leq s \leq t \leq 1,\\
h_{2}(t,s)=t^{\beta-1}(1-s)^{\beta-1}, \quad 0\leq t \leq s \leq 1,
\end{gather*}
we have
\begin{align*}
\frac{\partial{h_1(t,s)}}{\partial(t)}
&= (\beta-1)t^{\beta-2}(1-s)^{\beta-1}-(\beta-1)(t-s)^{\beta-2}\\
&= (\beta-1)[t^{\beta-2}(1-s)^{\beta-1}-(t-s)^{\beta-2}]\\
&= (\beta-1)t^{\beta-2}[(1-s)^{\beta-1}-(1-\frac{s}{t})^{\beta-2}]
\leq 0,
\end{align*}
which means that $h_1(t,s)$ is nonincreasing with respect to $t$ for
$0\leq s \leq t \leq 1$.
It is easily to see that $h_{2}(t,s)$ is nondecreasing with respect to $t$
for $0\leq t \leq s\leq 1$.
Thus $$H(t,s)\leq H(s,s)\text{for }0\leq s \leq t \leq 1$$ and
$$
H(t,s)\leq H(s,s)\quad \text{for } 0\leq t \leq s \leq 1.
$$
From  the above discussion,
$$
H(t,s)\leq H(s,s)\quad \text{for }(t,s)\in[0,1]\times [0,1].
$$
So property (iv) holds.


Let's now show that (v) is true.
First,  $g_1(t,s)$, $g_{2}(t,s)$ are nonincreasing with respect to $t\in[s,1]$,
and $g_{3}(t,s)$, $g_{4}(t,s)$ are nondecreasing with respect to $t\in[0,s]$,
so there is
\begin{align*}
\min_{\xi\leq t \leq 1}G(t,s)
&=\begin{cases}
\min_{\xi\leq t \leq 1}\{g_1(t,s),g_{3}(t,s)\}, & 0\leq s <\xi,\\
\min_{\xi\leq t \leq 1}\{g_{2}(t,s),g_{4}(t,s)\}, & \xi \leq s < 1,
\end{cases} \\
&=\begin{cases}
g_1(t,s), & 0\leq s <\xi,\\
\lambda_1(s), &\xi \leq s < 1,
\end{cases}
\end{align*}
where $\lambda_1(s)=\min\{g_{2}(1,s),g_{4}(\xi,s)\}$,
$\lambda_1(s)\in C(\xi,1)$ and $\lambda_1(s)>0$.
Let
\[
\gamma(s)=\begin{cases}
\frac{g_1(t,s)}{G(s,s)}, & 0\leq s <\xi,\\
\frac{\lambda_1(s)}{G(s,s)}, &\xi \leq s < 1,
\end{cases}
\]
where
\[
G(s,s)=\begin{cases}
\frac{[s(1-s)]^{\alpha-n+1}-a[s(\xi-s)]^{\alpha-n+1}}{(1-a\xi^{\alpha-n+1})
\Gamma(\alpha-n+2)}, & 0\leq s <\xi,\\
\frac{[s(1-s)]^{\alpha-n+1}}{(1-a\xi^{\alpha-n+1})\Gamma(\alpha-n+2)},
 & \xi \leq s < 1,
\end{cases}
\]
 From  above discussions,
$$
\min_{t\in [\xi,1]}G(t,s)\geq \gamma(s) \max_{t\in [0,1]}G(t,s)
=\gamma(s) G(s,s),\quad  0< s <1.
$$
Second,  $h_1(t,s)$ is nonincreasing with respect to $t$ on $[s,1]$,  and
$h_{2}(t,s)$ is nondecreasing with respect to $t$ on $[0,s]$, one has
\begin{align*}
\min_{\xi\leq t \leq 1}H(t,s)
&=\begin{cases}
\min_{\xi\leq t \leq 1}\{h_1(t,s),h_{2}(t,s)\}, & 0\leq s <\xi,\\
\min_{\xi\leq t \leq 1}\{h_1(t,s),h_{2}(t,s)\}, & \xi \leq s < 1,
\end{cases}\\
&=\begin{cases}
h_1(t,s), & 0\leq s <\xi,\\
\lambda_{2}(s), & \xi \leq s < 1,
\end{cases}
\end{align*}
where $\lambda_{2}(s)=\min\{h_1(1,s),h_{2}(\xi,s)\}$,
$\lambda_{2}(s)\in C(\xi,1)$ and $\lambda_{2}(s)>0$.
Let
\[
\rho(s)=\begin{cases}
\frac{h_1(t,s)}{H(s,s)}, & 0\leq s <\xi,\\
\frac{\lambda_{2}(s)}{H(s,s)}, & \xi \leq s < 1,
\end{cases}
\]
where $H(s,s)=\frac{1}{\Gamma(\beta)}[s(1-s)]^{\beta-1}$.
Therefore, we obtain that
$$
\min_{s\in [\xi,1]}H(t,s)\geq \rho(s) \max_{t\in [0,1]}H(t,s)
=\rho(s) H(s,s), \quad 0< s <1.
$$
The proof is complete.
\end{proof}


\section{Existence of a unique positive solution}

In this section,  we discuss the existence of a unique positive solution to
fractional boundary-value problem \eqref{FBVP1} by using the mixed
monotone fixed point theorem.
We need the following assumptions:
\begin{itemize}
\item[(H1)] Let $f(t,x_1,x_2,\dots,x_{n-1})
 =g(t,x_1,x_2,\dots,x_{n-1})+h(t,x_1,x_2,\dots,x_{n-1})$, where
$g:(0,1)\times [0,+\infty)\times \mathbb R^{n-1}\to [0,+\infty)$
and $h:(0,1)\times[0,+\infty)\times (\mathbb R/{0})^{n-2}\to [0,+\infty)$
are continuous;

\item[(H2)] $g(t,x_1,x_2,\dots,x_{n-1})$ is nondecreasing in $t$ and
$x_i$ and $h(t,x_1,x_2,\dots,x_{n-1})$ is nonincreasing in $t$ and $x_i$,
where $t,x_i\in(0,1)\times[0,+\infty)\times (\mathbb R/{0})^{n-2},i=1,2,\dots,n-1$;

\item[(H3)] There exists a constant $b\in (0,1)$ such that
\begin{gather*}
g(t,kx_1,kx_2,\dots,kx_{n-1})
 \geq  k^{b}g(t,x_1,x_2,\dots,x_{n-1}),\quad k\in (0,1), \\
h(t,k^{-1}x_1,k^{-1}x_2,\dots,k^{-1}x_{n-1})
 \geq  k^{b}h(t,x_1,x_2,\dots,x_{n-1}),k\in (0,1).
\end{gather*}
 where $x_i>0$ and $i=1,2,\dots,n-1$;

\item[(H4)] $k^r:[0,1]\to [0,+\infty)$ is continuous and
$\int_0^{1}\phi_q(s^{-b(\alpha-1)})ds<+\infty$, where $0\leq r <1$.
\end{itemize}

Let us denote $E_1=C(0,1)$ equipped with the norm $||v||=\sup\limits_{t\in[0,1]}|v(t)|$, then $E_1$ is a Banach space. Let $P$ be a normal cone
of $E_1$ defined by
$$
P=\big\{v\in E_1: v(t)\geq 0,t\in[0,1]\big\}.
$$
Define
$$
Q_e=\Big\{v\in P : \frac{1}{M}e(t)\leq v(t)\leq Me(t),t\in [0,1]\Big\},
$$
where $e(t)=t^{\alpha-n+1}$,
$M$ is a positive constant defined by
\begin{align*}
M&>\min \Big\{1,
\Big[\frac{(g(1,1,\dots,1))^{q-1}\int_0^{1}\phi_q(s^{\beta-1})ds}
{\Gamma(\alpha-n+1)(\Gamma(\beta))^{q-1}} \\
&\quad +\frac{(\zeta^{-b}h(0,1,1,\dots,1))^{q-1}\int_0^{1}\phi_q(s^{\beta-1}
\int_0^{1}\tau^{-b(\alpha-1)}d\tau)ds}{\Gamma(\alpha-n+1)(\Gamma(\beta))^{q-1}}
 \Big]^{\frac{1}{b(1-q)}},\\
&\quad\Big[(h(1,1,1,\dots,1))^{q-1}\int_\xi^{1}\gamma(s)G(s,s)\phi_q
 \Big(\int_\xi^{1}\rho(\tau)H(\tau,\tau)\tau^{b(\alpha-1)}\Big)d\tau)ds\\
&\quad+(h(1,1,1,\dots,1))^{q-1}\int_\xi^{1}\gamma(s)G(s,s)\phi_q
 \Big(\int_\xi^{1}\rho(\tau)H(\tau,\tau)d\tau\Big)ds
\Big]^{\frac{1}{b(1-q)}}\Big\},
\end{align*}
where
$0<\zeta<\min\big\{1,\frac{\Gamma(\alpha-n+2)}{\Gamma(\alpha)},
\frac{\Gamma(\alpha-n+2)}{\Gamma(\alpha-1)},\dots,
\frac{\Gamma(\alpha-n+2)}{\Gamma(\alpha)-n+3}\big\}$.
It is easy to obtain that $e\in P$ and $\| e \| =1$,
$e\neq \theta$.
The operator $T$ is defined by
\begin{align*}
T(v,w)(t)&=\int_0^{1}G(t,s)\phi_q\Big(\int_0^{1}H(s,\tau)
(g(\tau,I_{0^+}^{n-2}v(\tau),
,\dots,I_{0^+}^{1}v(\tau),v(\tau)) \\
&\quad +h(\tau,I_{0^+}^{n-2}w(\tau),
,\dots,I_{0^+}^{1}w(\tau),w(\tau))d\tau\Big)ds,\quad t\in (0,1).
\end{align*}

\begin{theorem} \label{theorem1}
Assume that {\rm (H1)--(H4)} are satisfied. Then the fractional boundary-value problem \eqref{FBVP1} has a unique positive solution.
\end{theorem}

\begin{proof}
By the definition of the operator $T$ and its
properties, it suffices to show that all conditions of
 Lemma \ref{Lemma 8} are satisfied with respect to $t$.

Firstly.  we show that $T:Q_{e}\times Q_{e}\to Q_{e}$.
Let $x_i=1$, Assumption (H3) implies that
$$
g(t,k,k,\dots,k)\geq k^{b}g(t,1,1,\dots,1),\quad k\in (0,1).
$$
Set $\bar{x}:=x_1=x_2=\dots=x_{n-1}$, and $k=\frac{1}{\bar{x}}$, $\bar{x}>1$,
one has
$$
g(t,\bar{x},\dots,\bar{x})\leq \bar{x}^{b}g(t,1,\dots,1),~\bar{x}>1.
$$
Similarly, from (H3), for $x_i>0$, if we let $k^{-1}x_i=y_i$, $i=1,2,\dots,n-1$,
then
$$
h(t,y_1,\dots,y_{n-1})\geq k^{b}h(t,ky_1,\dots,ky_{n-1}),
\quad k\in (0,1),~y_i>0,~i=1,2,\dots,n-1.
$$
Now, let $y_i=1,i=1,2,\dots,n-1$, we obtain
$$
h(t,1,\dots,1)\geq k^{b}h(t,k,\dots,k),\quad k\in (0,1).
$$
From the above discussion, we have
\begin{gather*}
h(t,k^{-1},\dots,k^{-1})\geq k^{b}h(t,1,\dots,1), \\
h(t,ky_1,\dots,ky_{n-1})\leq k^{-b}h(t,y_1,\dots,y_{n-1}), \\
h(t,k,\dots,k)\leq k^{-b}h(t,1,\dots,1),
\end{gather*}
where $k\in (0,1)$, $y_i>0$, $i=1,2,\dots,n-1$.

Since $v\in Q_e$ and the monotonicity of Riemann-Liouville fractional
integral $I_{0^+}^\delta$, we obtain that
$$
I_{0^+}^{n-2}v(t)>0,\; I_{0^+}^{n-3}v(t)>0,\;\dots,\;
I_{0^+}^{1}v(t)>0,v(t)>0,
$$
\begin{align*}
&g\Big(t,I_{0^+}^{n-2}v(t),I_{0^+}^{n-3}v(t),\dots,I_{0^+}^{1}v(t),v(t)\Big) \\
&\leq g\Big(t,I_{0^+}^{n-2}Me(t),I_{0^+}^{n-3}Me(t),\dots,I_{0^+}^{1}Me(t),Me(t)\Big)\\
&\leq g\Big(t,I_{0^+}^{n-2}M,I_{0^+}^{n-3}M,\dots,I_{0^+}^{1}M,M\Big)\\
&= g\Big(t,\frac{M}{(n-2)!}t^{n-2},\frac{M}{(n-3)!}t^{n-3},\dots,Mt,M\Big)\\
&\leq g(t,M,M,\dots,M,M)\\
&\leq M^{b}g(t,1,1,\dots,1,1)\\
&\leq M^{b}g(1,1,1,\dots,1,1),
\end{align*}
and
\begin{align*}
&h\Big(t,I_{0^+}^{n-2}w(t),I_{0^+}^{n-3}w(t),\dots,I_{0^+}^{1}w(t),w(t)\Big)\\
&\leq h\Big(t,I_{0^+}^{n-2}\frac{1}{M}e(t),I_{0^+}^{n-3}
 \frac{1}{M}e(t),\dots,I_{0^+}^{1}\frac{1}{M}e(t),\frac{1}{M}e(t)\Big)\\
&= h\Big(t,\frac{\Gamma(\alpha-n+2)}{M\Gamma(\alpha)}t^{\alpha -1},
 \frac{\Gamma(\alpha-n+2)}{M\Gamma(\alpha-1)}t^{\alpha -2},\dots, \\
&\quad \frac{\Gamma(\alpha-n+2)}{M\Gamma(\alpha-n+3)}t^{\alpha -n+2},
\frac{1}{M}t^{\alpha -n+1}\Big)\\
&\leq h\Big(t,\frac{\zeta}{M}t^{\alpha -1},\frac{\zeta}{M}t^{\alpha -1},\dots,
 \frac{\zeta}{M}t^{\alpha -n+3},\frac{\zeta}{M}t^{\alpha -n+2}\Big)\\
&\leq h\Big(t,\frac{\zeta}{M}t^{\alpha -1},\frac{\zeta}{M}t^{\alpha -1},\dots,
\frac{\zeta}{M}t^{\alpha -n+4},\frac{\zeta}{M}t^{\alpha -n+3}\Big)\\
&\leq \dots\\
&\leq h\Big(t,\frac{\zeta}{M}t^{\alpha -1},\frac{\zeta}{M}t^{\alpha -1},\dots,
 \frac{\zeta}{M}t^{\alpha -1},\frac{\zeta}{M}t^{\alpha -1}\Big)\\
&\leq (\frac{\zeta}{M})^{-b}t^{-b(\alpha-1)}h(t,1,1,\dots,1,1)\\
&\leq (\frac{\zeta}{M})^{-b}t^{-b(\alpha-1)}h(0,1,1,\dots,1,1),
\end{align*}
where
\[
0<\zeta<\min\Big\{1,\frac{\Gamma(\alpha-n+2)}{\Gamma(\alpha)},
 \frac{\Gamma(\alpha-n+2)}{\Gamma(\alpha-1)},\dots,
\frac{\Gamma(\alpha-n+2)}{\Gamma(\alpha)-n+3}\Big\},
\]
for $0< \frac{\zeta}{M}t^{\alpha-1}<1$.
We also obtain
\begin{align*}
&g\Big(t,I_{0^+}^{n-2}v(t),I_{0^+}^{n-3}v(t),\dots,I_{0^+}^{1}v(t),v(t)\Big)\\
&\geq g\Big(t,I_{0^+}^{n-2}\frac{1}{M}e(t),I_{0^+}^{n-3}\frac{1}{M}e(t),\dots,
 I_{0^+}^{1}\frac{1}{M}e(t),\frac{1}{M}e(t)\Big)\\
&= g\Big(t,\frac{\Gamma(\alpha-n+2)}{M\Gamma(\alpha)}t^{\alpha -1},
 \frac{\Gamma(\alpha-n+2)}{M\Gamma(\alpha-1)}t^{\alpha -2},\dots, \\
&\quad \frac{\Gamma(\alpha-n+2)}{M\Gamma(\alpha-n+3)}t^{\alpha -n+2},
 \frac{1}{M}t^{\alpha -n+1}\Big)\\
&\geq g\Big(t,\frac{\zeta}{M}t^{\alpha -1},\frac{\zeta}{M}t^{\alpha -1},\dots,
 \frac{\zeta}{M}t^{\alpha -n+3},\frac{\zeta}{M}t^{\alpha -n+2}\Big)\\
&\geq g\Big(t,\frac{\zeta}{M}t^{\alpha -1},\frac{\zeta}{M}t^{\alpha -1},\dots,
 \frac{\zeta}{M}t^{\alpha -n+4},\frac{\zeta}{M}t^{\alpha -n+3}\Big)\\
&\geq \dots\\
&\geq g\Big(t,\frac{\zeta}{M}t^{\alpha -1},\frac{\zeta}{M}t^{\alpha -1},\dots,
 \frac{\zeta}{M}t^{\alpha -1},\frac{\zeta}{M}t^{\alpha -1}\Big)\\
&\geq (\frac{\zeta}{M})^{-b}t^{b(\alpha-1)}g(t,1,1,\dots,1,1)\\
&\geq (\frac{\zeta}{M})^{-b}t^{b(\alpha-1)}g(0,1,1,\dots,1,1),
\end{align*}
where $0< \frac{\zeta}{M}t^{\alpha-1}<1$,
and
\begin{align*}
& h\Big(t,I_{0^+}^{n-2}w(t),I_{0^+}^{n-3}w(t),\dots,I_{0^+}^{1}w(t),w(t)\Big)\\
&\geq h\Big(t,I_{0^+}^{n-2}Me(t),I_{0^+}^{n-3}Me(t),\dots,I_{0^+}^{1}Me(t),Me(t)\Big)\\
&\geq h\Big(t,I_{0^+}^{n-2}M,I_{0^+}^{n-3}M,\dots,I_{0^+}^{1}M,M\Big)\\
&= h\Big(t,\frac{M}{(n-2)!}t^{n-2},\frac{M}{(n-3)!}t^{n-3},\dots,Mt,M\Big)\\
&\geq h(t,M,M,\dots,M,M)\\
&\geq M^{-b}h(t,1,1,\dots,1,1)\\
&\geq M^{-b}h(1,1,1,\dots,1,1).
\end{align*}
From the above  and Lemma \ref{Lemma 6}  it follows that
$T(v,w)\in C([0,1],[0,+\infty))$. Then
\begin{align*}
&\int_0^{1}G(t,s)\phi_q\Big(\int_0^{1}H(s,\tau)g(\tau,I_{0^+}^{n-2}v(\tau),
 \dots,I_{0^+}^{1}v(\tau),v(\tau))d\tau\Big)ds\\
&\leq \frac{t^{\alpha-n+1}}{\Gamma(\alpha-n+2)}\int_0^{1}
\phi_q\Big( \frac{s^{\beta-1}}{\Gamma(\beta)}\int_0^{1}
 g(\tau,I_{0^+}^{n-2}v(\tau),\dots,I_{0^+}^{1}v(\tau),v(\tau))d\tau\Big)ds\\
&\leq \frac{t^{\alpha-n+1}}{\Gamma(\alpha-n+2)}\int_0^{1}\phi_q
 \Big(\frac{s^{\beta-1}M^{b}g(1,1,\dots,1)}{\Gamma(\beta)}\Big)ds\\
&\leq \frac{t^{\alpha-n+1}(M^{b}g(1,1,\dots,1))^{q-1}}
 {\Gamma(\alpha-n+1)(\Gamma(\beta))^{q-1}}
\int_0^{1}\phi_q(s^{\beta-1})ds,
\end{align*}
and
\begin{align*}
& \int_0^{1}G(t,s)\phi_q\Big(\int_0^{1}H(s,\tau)h(\tau,I_{0^+}^{n-2}w(\tau),
 \dots,I_{0^+}^{1}w(\tau),w(\tau))d\tau\Big)ds\\
&\leq \frac{t^{\alpha-n+1}}{\Gamma(\alpha-n+2)}\int_0^{1}
\phi_q\Big( \frac{s^{\beta-1}}{\Gamma(\beta)}\int_0^{1}
 h(\tau,I_{0^+}^{n-2}w(\tau),\dots,I_{0^+}^{1}w(\tau),w(\tau))d\tau\Big)ds\\
&\leq \frac{t^{\alpha-n+1}}{\Gamma(\alpha-n+2)}\int_0^{1}
\phi_q\Big( \frac{s^{\beta-1}}{\Gamma(\beta)}\int_0^{1}
(\frac{\zeta}{M})^{-b}\tau^{-b(\alpha-1)}h(0,1,1,\dots,1)d\tau\Big)ds\\
&\leq \frac{t^{\alpha-n+1}(\zeta^{-b}M^{b}h(0,1,1,\dots,1))^{q-1}}
{\Gamma(\alpha-n+1)(\Gamma(\beta))^{q-1}}
\int_0^{1}\phi_q\Big(s^{\beta-1}\int_0^{1}\tau^{-b(\alpha-1)}d\tau\Big)ds.
\end{align*}
Then
$$
T(v,w)(t)\leq Mt^{\alpha-n+1}=Me(t),\quad t\in(0,1).
$$
From the inequalities
\begin{align*}
&\int_0^{1}G(t,s)\phi_q\Big(\int_0^{1}H(s,\tau)g(\tau,I_{0^+}^{n-2}v(\tau),
 \dots,I_{0^+}^{1}v(\tau),v(\tau))d\tau\Big)ds\\
&\geq \int_\xi^{1}G(t,s)\phi_q
 \Big(\int_\xi^{1}H(s,\tau)g(\tau,I_{0^+}^{n-2}v(\tau),\dots,
 I_{0^+}^{1}v(\tau),v(\tau))d\tau\Big)ds\\
&\geq \int_\xi^{1}\gamma(s)G(s,s)\phi_q
 \Big(\int_\xi^{1}\rho(\tau)H(\tau,\tau)g(\tau,I_{0^+}^{n-2}v(\tau),\dots,
 I_{0^+}^{1}v(\tau), v(\tau))d\tau\Big)ds\\
&\geq \int_\xi^{1}\gamma(s)G(s,s)\phi_q\Big(\int_\xi^{1}\rho(\tau)
 H(\tau,\tau)(\frac{\zeta}{M})^{b}\tau^{b(\alpha-1)} g(0,1,1,\dots,1)d\tau\Big)ds\\
&= \big(\zeta^{b}M^{-b}g(0,1,1,\dots,1)\big)^{q-1}
 \int_\xi^{1}\gamma(s)G(s,s)\phi_q\Big(\int_\xi^{1}\rho(\tau)H(\tau,\tau)
 \tau^{b(\alpha-1)} d\tau\Big) ds\\
&\geq t^{\alpha-n+1}\big(\zeta^{b}M^{-b}g(0,1,1,\dots,1)\big)^{q-1} \\
&\quad\times \int_\xi^{1}\gamma(s)G(s,s)\phi_q
 \Big(\int_\xi^{1}\rho(\tau)H(\tau,\tau)\tau^{b(\alpha-1)}d\tau\Big)ds.
\end{align*}
and
\begin{align*}
&\int_0^{1}G(t,s)\phi_q(\int_0^{1}H(s,\tau)
 h\Big(\tau,I_{0^+}^{n-2}w(\tau),\dots,I_{0^+}^{1}w(\tau),w(\tau)\Big)d\tau)ds\\
&\geq \int_\xi^{1}G(t,s)\phi_q
 \Big(\int_\xi^{1}H(s,\tau)h(\tau,I_{0^+}^{n-2}w(\tau),\dots,
 I_{0^+}^{1}w(\tau),w(\tau))d\tau\Big)ds\\
&\geq \int_\xi^{1}\gamma(s)G(s,s)\phi_q
 \Big(\int_\xi^{1}\rho(\tau)H(\tau,\tau)h(\tau,I_{0^+}^{n-2}w(\tau),\dots,
 I_{0^+}^{1}w(\tau), w(\tau))d\tau\Big)ds\\
&\geq \int_\xi^{1}\gamma(s)G(s,s)\phi_q\Big(\int_\xi^{1}\rho(\tau)
 H(\tau,\tau)M^{-b}h(1,1,1,\dots,1)d\tau\Big)ds\\
&\geq \big(M^{-b}h(1,1,1,\dots,1)\big)^{q-1}\int_\xi^{1}\gamma(s)G(s,s)\phi_q(\int_\xi^{1}\rho(\tau)H(\tau,\tau)d\tau)ds\\
&\geq t^{\alpha-n+1}\big(M^{-b}h(1,1,1,\dots,1)\big)^{q-1}
\int_\xi^{1}\gamma(s)G(s,s)\phi_q\Big(\int_\xi^{1}\rho(\tau)
H(\tau,\tau)d\tau\Big)ds,
\end{align*}
we deduce that
$$
T(v,w)(t)\geq \frac{1}{M}t^{\alpha-n+1}=\frac{1}{M}e(t),\quad t\in(0,1).
$$
Therefore, we concluded that $T:Q_{e}\times Q_{e}\to Q_{e}$.

Secondly, we prove that $T:Q_{e}\times Q_{e}\to Q_{e}$ is a mixed monotone operator.
Let $v_1,v_2 \in Q_e$ and $v_1\leq v_2$, we obtain
\begin{align*}
&\int_0^{1}G(t,s)\phi_q\Big(\int_0^{1}H(s,\tau)g(\tau,I_{0^+}^{n-2}v_1(\tau),
 \dots,I_{0^+}^{1}v_1(\tau),v_1(\tau))d\tau\Big)ds\\
&\leq \int_0^{1}G(t,s)\phi_q\Big(\int_0^{1}H(s,\tau)g(\tau,I_{0^+}^{n-2}v_2(\tau),
\dots,I_{0^+}^{1}v_2(\tau),v_2(\tau))d\tau\Big)ds,
\end{align*}
i.e.,
\begin{equation}\label{equTv1w}
T(v_1,w)(t)\leq T(v_2,w)(t),\quad w\in Q_e.
\end{equation}
Thus $T(v,w)(t)$ is nondecreasing in $v$ for any $w\in Q_e$.

Let $w_1,w_2 \in Q_e$ and $w_1\geq w_2$. Then
\begin{align*}
&\int_0^{1}G(t,s)\phi_q(\int_0^{1}H(s,\tau)h(\tau,I_{0^+}^{n-2}w_1(\tau),
 \dots,I_{0^+}^{1}w_1(\tau),w_1(\tau))d\tau)ds\\
&\leq \int_0^{1}G(t,s)\phi_q(\int_0^{1}H(s,\tau)h(\tau,I_{0^+}^{n-2}w_2(\tau),
\dots,I_{0^+}^{1}w_2(\tau),w_2(\tau))d\tau)ds,
\end{align*}
i.e.,
\begin{equation}\label{equTvw1}
T(v,w_1)(t)\leq T(v,w_2)(t),\quad w\in Q_e.
\end{equation}
Therefore $T(v,w)(t)$ is nonincreasing in $w$ for any $v\in Q_e$.
Consequently, according to \eqref{equTv1w} and \eqref{equTvw1}, we conclude
that the operator $T:Q_{e}\times Q_{e}\to Q_{e}$ is a mixed monotone operator.

Finally, we show that the operator $T$ has a fixed point.
If $v,w\in Q_e$, it follows from (H3) that
\begin{align*}
&\int_0^{1}G(t,s)\phi_q
\Big(\int_0^{1}H(s,\tau)
 g\Big(\tau,I_{0^+}^{n-2}tv(\tau),\dots,I_{0^+}^{1}tv(\tau),tv(\tau)\Big)d\tau\Big)ds\\
&= \int_0^{1}G(t,s)\phi_q\Big(\int_0^{1}H(s,\tau)
 g\Big(\tau,t I_{0^+}^{n-2}v(\tau),\dots,t I_{0^+}^{1}v(\tau),tv(\tau)\Big)d\tau\Big)ds\\
&\geq \int_0^{1}G(t,s)\phi_q\Big(\int_0^{1}H(s,\tau)t^{b}
 g\Big(\tau,I_{0^+}^{n-2}v(\tau),\dots,I_{0^+}^{1}v(\tau),v(\tau)\Big)d\tau\Big)ds\\
&\geq t^{b}\int_0^{1}G(t,s)\phi_q\Big(\int_0^{1}H(s,\tau)
 g\Big(\tau,I_{0^+}^{n-2}v(\tau),\dots,I_{0^+}^{1}v(\tau),v(\tau)\Big)d\tau\Big)ds,
\end{align*}
and
\begin{align*}
& \int_0^{1}G(t,s)\phi_q
 \Big(\int_0^{1}H(s,\tau)h(\tau,I_{0^+}^{n-2}t^{-1}w(\tau),\dots,I_{0^+}^{1}t^{-1}w(\tau),
t^{-1}w(\tau))d\tau\Big)ds\\
&= \int_0^{1}G(t,s)\phi_q
 \Big(\int_0^{1}H(s,\tau)h(\tau,t^{-1} I_{0^+}^{n-2}w(\tau),\dots,
 t^{-1} I_{0^+}^{1}w(\tau),t^{-1}w(\tau))d\tau\Big)ds\\
&\geq \int_0^{1}G(t,s)\phi_q
 \Big(\int_0^{1}H(s,\tau)t^{b}h(\tau,I_{0^+}^{n-2}w(\tau),\dots,
 I_{0^+}^{1}w(\tau),w(\tau))d\tau\Big)ds\\
&\geq t^{b}\int_0^{1}G(t,s)\phi_q
 \Big(\int_0^{1}H(s,\tau)h(\tau,I_{0^+}^{n-2}w(\tau),\dots,I_{0^+}^{1}w(\tau),
 w(\tau))d\tau\Big)ds,
\end{align*}
we obtain
$$
T\Big(tx,\frac{1}{t}y\Big)\geq t^{b}T(x,y), \quad x,y\in Q_{e},\; t\in(0,1),\; b\in(0,1).
$$
Therefore, from Lemma \ref{Lemma 8} it follows that the operator $T$
has a fixed point. That is to say, the fractional differential equation
\eqref{FBVP5} has a unique positive solution $v(t)$, $v\in Q_e$.
By Lemma \ref{Lemma 7}, we know that the fractional boundary-value problem
\eqref{FBVP1} has a unique positive solution $u(t)$, such that
\begin{align*}
\frac{\Gamma(\alpha-n+2)}{M\Gamma(\alpha)}t^{\alpha-1}
&=\frac{1}{M}I_{0^+}^{n-2}e(t) \leq u(t) \\
&\leq MI_{0^+}^{n-2}e(t)=\frac{M\Gamma(\alpha-n+2)}{\Gamma(\alpha)}
 t^{\alpha-1},\quad t\in(0,1).
\end{align*}
The proof is complete.
\end{proof}

Now we introduce the following assumptions:
\begin{itemize}
\item[(H5)] $f(t,x_1,x_2,\dots,x_{n-1})=g(t,x_1,x_2,\dots,x_{n-1})\times
 h(t,x_1,x_2,\dots,x_{n-1})$,
where $g:(0,1)\times [0,+\infty)\times \mathbb R^{n-1}\to [0,+\infty)$ and
$h:(0,1)\times[0,+\infty)\times
(\mathbb R/{0})^{n-2}\to [0,+\infty)$ are continuous;

\item[(H6)] For  $x_i>0$, $i=1,2,\dots,n-1$, there exist constants
 $b_1,b_2>0,0<b_1+b_2<1$, such that
\begin{gather*}
 g(t,kx_1,kx_2,\dots,kx_{n-1})\geq k^{b_1}g(t,x_1,x_2,\dots,x_{n-1}),k\in (0,1),\\
\\
h(t,k^{-1}x_1,k^{-1}x_2,\dots,k^{-1}x_{n-1})\geq k^{b_2}h(t,x_1,x_2,\dots,x_{n-1}),
 k\in (0,1);
\end{gather*}

\end{itemize}

\begin{corollary} \label{corollary1}
Assume that {\rm(H2), (H4), (H5)} and {\rm(H6)} are satisfied, then the fractional boundary-value
problem \eqref{FBVP1} has a unique positive solution.
\end{corollary}

The proof is done in the same way as the proof of Theorem \ref{theorem1};
we omit it.

\section{Existence of at least one positive solution}

In this section, we show the existence of at least one positive solution to
the fractional boundary-value problem \eqref{FBVP1} by using the upper
and lower solution method.

Let $E_2=\{v:v(t)\in C^{2}[0,1]\text{~and~}\phi_p(D_{0^+}^{\alpha-n+2}v(t))\in C^{2}[0,1]\}$ denote the Banach space endowed with the norm
$\mid\mid v\mid\mid=\max\limits_{t\in[0,1]}\{\sup|v(t)|,\sup|\phi_p(D_{0^+}^{\alpha-n+2}v(t)|\}$.


The operator $F$ is defined by
$$
(Fv)(t)=\int_0^{1}G(t,s)\phi_q
 \Big(\int_0^{1}H(s,\tau)f(\tau,I_{0^+}^{n-2}v(\tau),\dots,
 I_{0^+}^{1}v(\tau),v(\tau))d\tau\Big)ds,
$$
for $t\in(0,1)$.
Now we have two definitions on  the lower and
upper solutions of the fractional differential
equation \eqref{FBVP5}.

\begin{definition} \label{def4}\rm
A function $m(t)$ is called a lower solution of the fractional differential
equation \eqref{FBVP5}, if $m(t)\in E_2$,
and $m(t)$ satisfies
\begin{equation}\label{lower1}
\begin{gathered}
D_{0^+}^{\beta}(\phi_{p}D_{0^+}^{\alpha-n+2}m(t))\geq f(t,I_{0^+}^{n-2}m(t),
I_{0^+}^{n-3}m(t),\dots,I_{0^+}^{1}m(t),m(t)), \quad t\in(0,1),\\
m(0)\leq 0,m(1)\leq am(\xi),D_{0^+}^{\alpha-n+2}m(0)\geq D_{0^+}^{\alpha-n+2}m(1).
\end{gathered}
\end{equation}
\end{definition}

\begin{definition} \label{def5} \rm
A function $n(t)$ is called an upper solution of the fractional differential
equation \eqref{FBVP5}, if $n(t)\in E_2$,
and $n(t)$ satisfies
\begin{equation}\label{upper1}
\begin{gathered}
D_{0^+}^{\beta}(\phi_{p}D_{0^+}^{\alpha-n+2}n(t))
\leq f(t,I_{0^+}^{n-2}n(t),I_{0^+}^{n-3}n(t),\dots,I_{0^+}^{1}n(t),n(t)),\quad
t\in(0,1),\\
n(0)\geq 0,n(1)\geq an(\xi),D_{0^+}^{\alpha-n+2}n(0)\leq D_{0^+}^{\alpha-n+2}n(1).
\end{gathered}
\end{equation}
\end{definition}

We introduce the following assumptions:
\begin{itemize}

\item[(H9)] $f(t,x_1,x_2,\dots,x_{n-1})\in C((0,1)
 \times(0,+\infty)^{n-1},[0,+\infty))$ is nonincressing relative to $x_i$, ,
and there exists a constant $L_1>0$, such that $\mid f(t,x_1,x_2,\dots,x_{n-1})\mid \leq L_1$,
where $x_i>0$, $i=1,2,\dots,n-1$;

\item[(H10)] For any constant $\mu>0$, we have
\[
0<\int_0^{1}H(t,t)
f(t,I_{0^+}^{n-2}\mu t^{\alpha-n+1},\dots,I_{0^+}^{1}
\mu t^{\alpha-n+1},\mu t^{\alpha-n+1})dt<+\infty;
\]

\item[(H11)] There exists a continuous function $p(t),t\in[0,1]$, such that
\begin{gather*}
\int_0^{1}G(t,s)\phi_q\Big(\int_0^{1}H(s,\tau)f(\tau,I_{0^+}^{n-2}p(\tau),\dots,
I_{0^+}^{1}p(\tau),p(\tau))d\tau\Big)ds=q(t)\geq p(t), \\
\int_0^{1}G(t,s)\phi_q\Big(\int_0^{1}H(s,\tau)f(\tau,I_{0^+}^{n-2}q(\tau),
\dots,I_{0^+}^{1}q(\tau),q(\tau))d\tau\Big)ds\geq p(t).
\end{gather*}
\end{itemize}

\begin{theorem} \label{theorem3}
If {\rm(H9)--(H11)} are satisfied, then
 problem \eqref{FBVP1} has at least one
positive solution.
\end{theorem}

\begin{proof}
We  divide our proof into four steps.
\smallskip

\noindent\textbf{Step 1.}
Let $M_1:=\max\limits_{[0,1]\times[0,1]}G(t,s), M_2:=\max\limits_{[0,1]\times[0,1]}H(t,s).$
Set $\Omega_1=\{v\in E_2:||v||\leq M_1\phi_q(M_2L_1)\}$,
we prove that $F(\Omega_1)\subset \Omega_1$.

For any $v\in \Omega_1$,
we obtain that
\begin{align*}
(Fv)(t)
&= \int_0^{1}G(t,s)\phi_q\Big(\int_0^{1}H(s,\tau)f(\tau,I_{0^+}^{n-2}v(\tau),
 \dots,I_{0^+}^{1}v(\tau),v(\tau))d\tau\Big)ds\\
&\leq M_1\phi_q(M_2L_1).
\end{align*}
Consequently  $F(\Omega_1)\subset \Omega_1$.

By computations, we have
\begin{equation}
\begin{gathered}
\int_0^{1}G(t,s)\phi_{p}\Big(\int_0^{1}H(s,\tau)(Fv)(t)d\tau\Big)ds
=f\Big(t,I_{0^+}^{n-2}v(t),\dots,I_{0^+}^{1}v(t),v(t)\Big), \\
\text{for  }t\in(0,1),\\
(Fv)(0)=0,\quad (Fv)(1)=a(Fv)(\xi),\\\
D_{0^+}^{\alpha-n+2}(Fv)(0) =D_{0^+}^{\alpha-n+2}(Fv)(1)=0.
\end{gathered} \label{eqt2}
\end{equation}
\smallskip

\noindent\textbf{Step 2.} Set $m(t)=Fq(t),n(t)=Fp(t)$, in this step, we
prove that $m(t),n(t)$ are lower and upper solutions of the fractional differential
equation \eqref{FBVP5}, respectively.
From the assumptions (H9) and (H11), we obtain
\begin{equation}\label{eqt3}
p(t)\leq q(t)=Fp(t),\quad Fq(t)\leq q(t)=Fp(t),\quad t\in[0,1],
\end{equation}
this means that $m(t)\leq n(t)$. Since  $F(\Omega_1)\subset \Omega_1$, there is
$m(t),n(t)\in \Omega_1$.  According to \eqref{lower1}, \eqref{upper1}, we have
\begin{equation}
\begin{gathered}
\begin{aligned}
&D_{0^+}^{\beta}\phi_{p}\Big(D_{0^+}^{\alpha-n+2}m(t)\Big)
-f\Big(t,I_{0^+}^{n-2}m(t),I_{0^+}^{n-3}m(t),\dots,I_{0^+}^{1}m(t),m(t)\Big) \\
&\leq D_{0^+}^{\beta}\phi_{p}\Big(D_{0^+}^{\alpha-n+2}(Fq)(t)\Big)
-f\Big(t,I_{0^+}^{n-2}q(t),I_{0^+}^{n-3}q(t),\dots,I_{0^+}^{1}q(t),q(t)\Big)=0,
\end{aligned}\\
m(0)=0,\quad m(1)=am(\xi),\quad D_{0^+}^{\alpha-n+2}m(0)=D_{0^+}^{\alpha-n+2}m(1)=0.
\end{gathered}
\end{equation}
and
\begin{equation}
\begin{gathered}
\begin{aligned}
&D_{0^+}^{\beta}\phi_{p}\Big(D_{0^+}^{\alpha-n+2}n(t)\Big)
-f\Big(t,I_{0^+}^{n-2}n(t),I_{0^+}^{n-3}n(t),\dots,I_{0^+}^{1}n(t),n(t)\Big) \\
&\geq D_{0^+}^{\beta}\phi_{p}\big(D_{0^+}^{\alpha-n+2}(Fp)(t)\big)
-f\Big(t,I_{0^+}^{n-2}p(t),I_{0^+}^{n-3}p(t),\dots,I_{0^+}^{1}p(t),p(t)\Big)=0,
\end{aligned}\\
n(0)=0,\quad n(1)=an(\xi),\quad D_{0^+}^{\alpha-n+2}n(0)=D_{0^+}^{\alpha-n+2}n(1)=0.
\end{gathered}
\end{equation}
Hence, $m(t)$ and $n(t)$ are lower and upper solutions of the fractional differential
equation \eqref{FBVP5}, respectively.
\smallskip

\noindent\textbf{Step 3.}
Let
\begin{equation}
\begin{aligned}
&g_1\Big(t,I_{0^+}^{n-2}v(t),I_{0^+}^{n-3}v(t),\dots,I_{0^+}^{1}v(t),v(t)\Big) \\
&=\begin{cases}
f\big(t,I_{0^+}^{n-2}m(t),I_{0^+}^{n-3}m(t),\dots,I_{0^+}^{1}m(t),m(t)\big),
 &v(t)<m(t),\\
f\big(t,I_{0^+}^{n-2}v(t),I_{0^+}^{n-3}v(t),\dots,I_{0^+}^{1}v(t),v(t)\big),
 &m(t)\leq v(t)\leq n(t),\\
f\big(t,I_{0^+}^{n-2}n(t),I_{0^+}^{n-3}n(t),\dots,I_{0^+}^{1}n(t),n(t)\big),
 &v(t)>n(t).
\end{cases}
\end{aligned}
\end{equation}

Consider the  fractional differential equation
\begin{equation}
\begin{gathered}
D_{0^+}^{\beta}\phi_{p}\big(D_{0^+}^{\alpha-n+2}n(t)\big)
+g_1\big(t,I_{0^+}^{n-2}v(t),I_{0^+}^{n-3}v(t),\dots,I_{0^+}^{1}v(t),v(t)\big)=0,\\
 0<t<1,\\
v(0)=0,\quad v(1)=av(\xi),\quad D_{0^+}^{\alpha-n+2}v(0)=D_{0^+}^{\alpha-n+2}v(1)=0.
\end{gathered} \label{FBVP6}
\end{equation}

Set
$\Omega_2=\{v\in E_2:||v||\leq M_1\phi_q(M_2L_2)\}$, then $\Omega_2$
is a closed, bounder and convex set, where
$$
L_2:=\sup_{^{t\in [0, 1 ],v\in \Omega_2}}| g_1(t,I_{0^+}^{n-2}v(t),
\dots,I_{0^+}^{1}v(t),v(t))|+1.
$$


The operator $A:\Omega_2\to E_2$ is defined by
$$
Av(t)=\int_0^{1}G(t,s)\phi_q\Big(\int_0^{1}H(s,\tau)g_1\big(\tau,I_{0^+}^{n-2}v(\tau),
\dots,I_{0^+}^{1}v(\tau),v(\tau)\big)d\tau\Big)ds,
$$
where $G(t,s)$ and $H(s,\tau)$ are defined in Lemma \ref{Lemma 4} and Lemma
\ref{Lemma 5}. From Lemma  \ref{Lemma 001}, if $v\in \Omega_2$,  we have $Av(t)\geq 0$,
and the fixed point of the operator $A$ is the solution of the fractional
differential equation \eqref{FBVP6}.


Now, we show that $A$ is a completely continuous operator.
Let $v\in \Omega_2$, it follow from  the nonnegative and continuity of $G(t,s),H(t,s)$
and Lemma \ref{Lemma 6} that $A:\Omega_2\to \Omega_2$ is continuous.

For any $v\in \Omega_2$,
\begin{align*}
&| (Av)(t)| \\
&=\bigg|\int_0^{1}G(t,s)\phi_q\Big(\int_0^{1}H(s,\tau)g_1(\tau,I_{0^+}^{n-2}v(\tau),
 \dots,I_{0^+}^{1}v(\tau),v(\tau))d\tau\Big)ds\bigg|\\
&\leq \int_0^{1}G(t,s)\phi_q\Big(\int_0^{1}H(s,\tau)| g_1(\tau,I_{0^+}^{n-2}v(\tau),
\dots,I_{0^+}^{1}v(\tau),v(\tau))| d\tau\Big)ds\\
&\leq L_2^{q-1}\int_0^{1}G(t,s)\phi_q\Big(\int_0^{1}H(s,\tau)d\tau\Big)ds\\
&\leq L_2^{q-1}\int_0^{1}G(s,s)\phi_q\Big(\int_0^{1}H(\tau,\tau)d\tau\Big)ds\\
&< +\infty ,
\end{align*}
which means that $A$ is uniformly bounded.

The function $G(t,s)$ is a continuous function for $t,s\in[0,1]\times [0,1]$; then
it is uniformly continuous for $t,s\in[0,1]\times [0,1]$. Hence for fixed
$s\in[0,1]$ and any $\varepsilon >0$, there exists a constant $\delta>0$, such that
$$
G(t_1,s)-G(t_2,s)<\frac{\varepsilon}{L_2^{q-1}\phi_q(\int_0^{1}H(\tau,\tau)d\tau)},
$$
where $t_1,t_2\in[0,1]$ and $| t_1-t_2|<\delta$.
Therefore
\begin{align*}
&| Av(t_2)-Av(t_1)| \\
&\leq \int_0^{1}| G(t_2,s)-G(t_1,s)|\phi_q
 \Big(\int_0^{1}H(s,\tau)g_1(\tau,I_{0^+}^{n-2}v(\tau),\dots,I_{0^+}^{1}v(\tau),
 v(\tau))d\tau\Big)ds\\
&\leq L_2^{q-1}\int_0^{1}| G(t_2,s)-G(t_1,s)|\phi_q(\int_0^{1}H(\tau,\tau)d\tau)ds\\
&\leq L_2^{q-1}\phi_q(\int_0^{1}H(\tau,\tau)d\tau)\int_0^{1}| G(t_2,s)-G(t_1,s)| ds\\
&<  \varepsilon \quad \text{for any } v\in \Omega_2,
\end{align*}
which implies that $A$ is equicontinuous.
It follows from the Arzela-Ascoli theorem  that the operator $A:\Omega_2\to \Omega_2$
is completely continuous.

Similar to Step 1, we have $A(\Omega_2)\subset \Omega_2$.
According to the
Schauder's fixed point theorem, the operator $A$ has a fixed point, that is to say,
the  fractional differential equation \eqref{FBVP6} has a positive solution.
\smallskip

\noindent\textbf{Step 4.} We prove that the fractional differential
equation \eqref{FBVP1} has at least one positive solution.
Suppose that $d(t)$ is a solution of \eqref{FBVP6}, then
$$
d(0)=0,\quad d(1)=ad(\xi),\quad D_{0^+}^{\alpha-n+2}d(0)=D_{0^+}^{\alpha-n+2}d(1)=0.
$$
From (H9), we know that
\begin{align*}
&f\Big(t,I_{0^+}^{n-2}n(t),I_{0^+}^{n-3}n(t),\dots,I_{0^+}^{1}n(t),n(t)\Big)\\
&\leq g_1\Big(t,I_{0^+}^{n-2}d(t),I_{0^+}^{n-3}d(t),\dots,I_{0^+}^{1}d(t),d(t)\Big)\\
&\leq f(t,I_{0^+}^{n-2}m(t),I_{0^+}^{n-3}m(t),\dots,I_{0^+}^{1}m(t),m(t)),
\quad t\in [0,1].
\end{align*}
According to (H11) and \eqref{eqt3},
\begin{align*}
&f\Big(t,I_{0^+}^{n-2}q(t),I_{0^+}^{n-3}q(t),\dots,I_{0^+}^{1}q(t),q(t)\Big)\\
&\leq g_1\Big(t,I_{0^+}^{n-2}d(t),I_{0^+}^{n-3}d(t),\dots,I_{0^+}^{1}d(t),d(t)\Big)\\
&\leq f\Big(t,I_{0^+}^{n-2}p(t),I_{0^+}^{n-3}p(t),\dots,I_{0^+}^{1}p(t),p(t)\Big),
\quad t\in [0,1].
\end{align*}
It follows from $p(t)\in \Omega_2$ and \eqref{eqt2}  that
\begin{align*}
D_{0^+}^{\beta}\phi_{p}\big(D_{0^+}^{\alpha-n+2}n(t)\big)
&= D_{0^+}^{\beta}\phi_{p}\big(D_{0^+}^{\alpha-n+2}(Fp)(t)\big)\\
&= f\big(t,I_{0^+}^{n-2}p(t),I_{0^+}^{n-3}p(t),\dots,I_{0^+}^{1}p(t),p(t)\big),
\quad t\in [0,1].
\end{align*}
From the above discussions, we obtain
\begin{align*}
&D_{0^+}^{\beta}\phi_{p}\big(D_{0^+}^{\alpha-n+2}n(t)\big)
 -D_{0^+}^{\beta}\phi_{p}\big(D_{0^+}^{\alpha-n+2}d(t)\big)\\
&= f\Big(t,I_{0^+}^{n-2}p(t),I_{0^+}^{n-3}p(t),\dots,I_{0^+}^{1}p(t),p(t)\Big)\\
&\quad -g_1\Big(t,I_{0^+}^{n-2}d(t),I_{0^+}^{n-3}d(t),\dots,I_{0^+}^{1}d(t),d(t)\Big)\\
&\geq 0,\quad t\in [0,1].
\end{align*}
and
\begin{gather*}
(n-d)(0)=0,\quad (n-d)(1)=a(n-d)(\xi),\\
D_{0^+}^{\alpha-n+2}(n-d)(0)=D_{0^+}^{\alpha-n+2}(n-d)(1)=0.
\end{gather*}

If we let $z(t)=\phi_{p}(D_{0^+}^{\alpha-n+2}n(t))-\phi_{p}(D_{0^+}^{\alpha-n+2}d(t))$,
then  $z(0)=z(1)=0$. By Lemma \ref{Lemma 001}, we have $z(t)\leq 0$. Hence,
$$
\phi_{p}(D_{0^+}^{\alpha-n+2}n(t))
\leq \phi_{p}(D_{0^+}^{\alpha-n+2}d(t)),\quad t\in [0,1].
$$
Since  $\phi_{p}$ is monotone increasing,
$$
D_{0^+}^{\alpha-n+2}n(t)\leq D_{0^+}^{\alpha-n+2}d(t),
$$
that is $$
D_{0^+}^{\alpha-n+2}(n-d)(t)\leq0.
$$
By Lemma \ref{Lemma 001}, we have $(n-d)(t)\geq 0$. Thus we conclude that
 $n(t)\geq d(t),t\in [0,1]$.

In the same way, it is easy to prove that $m(t)\leq d(t),t\in [0,1]$.
Hence $d(t)$ is a positive solution of the boundary-value problem \eqref{FBVP5},
that is,  Lemma \ref{Lemma 7} imolies that $u(t)=I_{0^+}^{n-2}d(t)$
is a positive solution of boundary-value problem \eqref{FBVP1}.
The proof is complete.
\end{proof}

\section{Examples}

In this section, we give two simple examples to illustrate our main
results.

\begin{example} \label{examp1} \rm
We consider the boundary-value problem
\begin{equation}
\begin{gathered}
\begin{aligned}
&D_{0^+}^{4/3}\phi_{p}(D_{0^+}^{7/2}u(t))+t(u^{1/4}(t)+(u')^{1/2}(t)+(u'')^{3/4}(t)) \\
&+t^{-1/2}(u^{-1/4}(t)+(u')^{-1/2}(t)+(u'')^{-3/4}(t))=0, \quad 0<t<1,
\end{aligned}\\
u(0)=u'(0)=u''(0)=0,u''(1)=au''(\xi),D_{0^+}^{7/2}u(0)=D_{0^+}^{7/2}u(1)=0.
\end{gathered} \label{BVP5}
\end{equation}
\end{example}

\begin{proof}
Set
\begin{gather*}
g(t,u(t),u'(t),u''(t))=t(u^{1/4}(t)+(u')^{1/2}(t)+(u'')^{3/4}(t)),\\
h(t,u(t),u'(t),u''(t))=t^{-1/2}(u^{-1/4}(t)+(u')^{-1/2}(t)+(u'')^{-3/4}(t)).
\end{gather*}
where $0<t<1$  and $u(t) >0$, $u'(t) >0$, $u''(t) >0$.
Obviously,
$g(t,u(t),u'(t),u''(t))$ is nondecreasing relative to $t$ and $u(t),u'(t),u''(t)$, and
$h(t,u(t),u'(t),u''(t))$ is nonincreasing relative to $t$ and $u(t),u'(t),u''(t)$.

By $k\in(0,1)$ and $u(t) >0,u'(t) >0,u''(t) >0$, we have
\begin{align*}
g(t,ku(t),ku'(t),ku''(t))
&= t(k^{1/4}u^{1/4}(t)+k^{1/2}(u')^{1/2}(t)+k^{3/4}(u'')^{3/4}(t))\\
&\geq tk^{3/4}(u^{1/4}(t)+(u')^{1/2}(t)+(u'')^{3/4}(t))\\
&= k^{3/4}g(t,u(t),u'(t),u''(t)),
\end{align*}
and
\begin{align*}
&h(t,k^{-1}u(t),k^{-1}u'(t),k^{-1}u''(t))\\
&= t^{-1/2}(k^{1/4}u^{-1/4}(t)+k^{1/2}(u')^{-1/2}(t)+k^{3/4}(u'')^{-3/4}(t))\\
&\leq t^{-1/2}k^{3/4}(u^{-1/4}(t)+(u')^{-1/2}(t)+(u'')^{-3/4}(t))\\
&= k^{3/4}h(t,u(t),u'(t),u''(t)).
\end{align*}
From above discussion, the assumptions (H1)--(H4) are satisfied.
It follows from Theorem \ref{theorem1}  that the boundary-value problem
\eqref{BVP5} has a unique positive solution. The proof is complete.
\end{proof}

Next we give  an example which is difficult to obtain the
existence of positive solution to  the fractional
boundary-value problem \eqref{BVP6} by using  Theorem \ref{theorem1}.

\begin{example} \rm
We consider the  boundary-value problem
\begin{equation}
\begin{gathered}
D_{0^+}^{7/3}\phi_{p}\big(D_{0^+}^{7/2}u(t)\big)
=t^2+\frac{u(t)}{2}+\frac{u'(t)}{3}+\frac{u''(t)}{4},\quad 0<t<1,\\
u(0)=u'(0)=u''(0)=0,\quad u''(1)=\frac{1}{2}u''(\frac{1}{3}),\quad
D_{0^+}^{7/2}u(0)=D_{0^+}^{7/2}u(1)=0.
\end{gathered}  \label{BVP6}
\end{equation}
\end{example}

\begin{proof} The function
$$D_{0^+}^{7/3}\phi_{p}\big(D_{0^+}^{7/2}u(t)\big)
=t^2+\frac{u(t)}{2}+\frac{u'(t)}{3}+\frac{u''(t)}{4}$$
is changed into the form of
$$D_{0^+}^{7/3}\phi_{p}\big(D_{0^+}^{7/2}u(t)\big)
-t^2-\frac{u(t)}{2}-\frac{u'(t)}{3}-\frac{u''(t)}{4}=0.$$
Let $u(t)=I_{0^+}^{2}v(t)$, we have
\begin{equation}
\begin{gathered}
D_{0^+}^{7/3}\phi_{p}\big(D_{0^+}^{3/2}v(t)\big)
-t^2-\frac{1}{2}I_{0^+}^{2}v(t)-\frac{1}{3}I_{0^+}^{1}v(t)-\frac{1}{4}v(t)=0,\quad
 0<t<1,\\
v(0)=0,v(1)=\frac{1}{2}v(\frac{1}{3}),D_{0^+}^{3/2}v(0)=D_{0^+}^{3/2}v(1)=0,
\end{gathered} \label{BVP7}
\end{equation}
then the hypotheses (H9)  are satisfied.
For any $\mu>0$, we obtain
\begin{align*}
&\int_0^{1}H(\tau,\tau)f\big(\tau,I_{0^+}^{2}\mu\tau^{1/2},I_{0^+}^{1}
 \mu\tau^{1/2},\mu\tau^{1/2}\big)d\tau\\
&= \int_0^{1}H(\tau,\tau)\big(t^2+\frac{1}{2}I_{0^+}^{2}\mu\tau^{1/2}
+\frac{1}{3}I_{0^+}^{1}\mu\tau^{1/2}+\frac{1}{4}\mu\tau^{1/2}\big)d\tau
<+\infty,
\end{align*}
which implies that (H10) holds.
\begin{align*}
f\Big(t,\frac{1}{2}I_{0^+}^{2}v(t),\frac{1}{3}I_{0^+}^{1}v(t),\frac{1}{4}v(t)\Big)
&= t^2+\frac{1}{2}I_{0^+}^{2}v(t)+\frac{1}{3}I_{0^+}^{1}v(t)+\frac{1}{4}v(t)\\
&\leq r^{-1/2}\Big(t^2+\frac{1}{2}I_{0^+}^{2}v(t)+\frac{1}{3}I_{0^+}^{1}v(t)
 +\frac{1}{4}v(t)\Big)\\
&= r^{-1/2}f\Big(t,\frac{1}{2}I_{0^+}^{2}v(t),\frac{1}{3}I_{0^+}^{1}v(t),
 \frac{1}{4}v(t)\Big).
\end{align*}
Let $a(t)=t^{1/2}$, we obtain
$$ b(t):=Ta(t)=\int_0^{1}G(t,s)\phi_{p}
(\int_0^{1}H(\tau,\tau)f(\tau,I_{0^+}^{2}\mu\tau^{1/2},I_{0^+}^{1}
\mu\tau^{1/2},\mu\tau^{1/2})d\tau)ds\in \Omega_1,
$$
and
$Tb(t)=T^2a(t)\in \Omega_1$.
Hence, there exist two positive numbers $\mu_1,\mu_2$, such that
$$
Ta(t)\geq \mu_1a(t),T^{2}a(t)\geq \mu_{2}a(t).
$$

Let $0<\mu_0\leq \min\{1,\mu_1,\mu_2^{7/3}\}$, by the monotonicity of $T$, we obtain
$$
T(\mu_{0}a(t))\geq Ta(t)\geq \mu_1a(t)\geq \mu_{0}a(t),
$$
and
$$
T^{2}(\mu_{0}a(t))\geq \mu_{0}^{1/4}T^{2}a(t)
\geq \mu_{0}^{1/4}\mu_{2}a(t)\geq \mu_{0}a(t).
$$
If we take $p(t)=\mu_{0}t^{1/2}$, then
\begin{align*}
q(t)&= Tp(t)\\
&= \int_0^{1}G(t,s)\phi_{p}
\Big(\int_0^{1}H(\tau,\tau)f(\tau,I_{0^+}^{2}\mu\tau^{1/2},
 I_{0^+}^{1}\mu\tau^{1/2},\mu\tau^{1/2})d\tau\Big)ds\\
&\geq \mu_{0}t^{1/2}=p(t),
\end{align*}
and
\begin{align*}
Tq(t)
&= T^{2}p(t)\\
&= \int_0^{1}G(t,s)\phi_{p}
\Big(\int_0^{1}H(\tau,\tau)f(\tau,T(I_{0^+}^{2}\mu\tau^{1/2}),T(I_{0^+}^{1}
 \mu\tau^{1/2}),T(\mu\tau^{1/2}))d\tau\Big)ds\\
&\geq \mu_{0}t^{1/2}=p(t).
\end{align*}
Therefore,  condition (H11) holds. By Theorem \ref{theorem3}, the fractional
boundary-value problem \eqref{BVP7} has at least
one positive solution, that is to say, it follows from Lemma \ref{Lemma 7}
that the fractional boundary-value problem \eqref{BVP6} has at least
one positive solution. The proof is complete.
\end{proof}

\subsection*{Acknowledgments}
This work was  partially supported by Natural Science Foundation of China
(No. 11361047, 11501560), by the Natural Science Foundation of Jiangsu
 Province (No. BK20151160), by the Six Talent Peaks Project of Jiangsu
 Province (2013-JY-003) and by the 333 High-Level Talents Training Program
 of Jiangsu Province(BRA2016275).



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\end{document}