\documentclass[reqno]{amsart}
\usepackage{hyperref}

\AtBeginDocument{{\noindent\small
\emph{Electronic Journal of Differential Equations},
Vol. 2012 (2012), No. 234, pp. 1--11.\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-2012/234\hfil Existence and uniqueness of positive solutions]
{Existence and uniqueness of positive solutions to higher-order
nonlinear fractional differential equation with integral boundary
conditions}

\author[C. Zhou \hfil EJDE-2012/234\hfilneg]
{Chenxing Zhou}  % in alphabetical order

\address{Chenxing Zhou \newline
College of Mathematics, 
Changchun Normal University, 
Changchun 130032, Jilin, China}
\email{mathcxzhou@163.com}

\thanks{Submitted September 19, 2012. Published December 21, 2012.}
\subjclass[2000]{26A33, 34B18, 34B27}
\keywords{Partially ordered sets; fixed-point theorem; positive solution}

\begin{abstract}
 In this article, we consider the nonlinear fractional order
 three-point boundary-value problem
 \begin{gather*}
 D_{0+}^{\alpha} u(t) + f(t,u(t))=0, \quad 0 < t < 1,\\
 u(0) = u'(0) = \dots = u^{(n-2)}(0)=0, \quad u^{(n-2)}(1) =
 \int_0^\eta u(s)ds,
 \end{gather*}
 where $D_{0+}^{\alpha}$ is the standard Riemann-Liouville fractional
 derivative, $n-1 < \alpha \leq n$, $n \geq 3$. By using a fixed-point
 theorem  in partially ordered sets,  we obtain sufficient conditions
 for the existence and uniqueness of positive and nondecreasing
 solutions to  the above boundary value problem.
\end{abstract}

\maketitle
\numberwithin{equation}{section}
\newtheorem{theorem}{Theorem}[section]
\newtheorem{lemma}[theorem]{Lemma}
%\newtheorem{corollary}[theorem]{Corollary}
\newtheorem{definition}[theorem]{Definition}
%\newtheorem{example}[theorem]{Example}
\newtheorem{remark}[theorem]{Remark}
\allowdisplaybreaks


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\newcommand{\oild}{\omega_{I(L)}(\delta)}
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\newcommand{\om}{\omega}
\newcommand{\de}{\delta}
\newcommand{\De}{\Delta}
%\newcommand{\vs}{\vspace}
%\newcommand{\un}{\underline}

\section{Introduction}

 Fractional differential equations
arise in many engineering and scientific disciplines as the
mathematical models of systems and processes in the fields of
physics, chemistry, aerodynamics, electrodynamics of complex medium,
polymer rheology, Bode's analysis of feedback amplifiers, capacitor
theory, electrical circuits, electron-analytical chemistry, biology,
control theory, fitting of experimental data, and so on, and
involves derivatives of fractional order. Fractional derivatives
provide an excellent tool for the description of memory and
hereditary properties of various materials and processes. This is
the main advantage of fractional differential equations in
comparison with classical integer-order models. For an extensive
collection of such results, we refer the readers to the monographs
by Samko et al \cite{s1}, Podlubny \cite{p1} and Kilbas et al
\cite{k1}. For the basic theory and recent development of the
subject, we refer a text by Lakshmikantham \cite{la1}. For more
details and examples, see \cite{b1,bai1,be1,e1,e2,l1,l2,la1,z2} and
the references therein. However, the theory of boundary value
problems for nonlinear fractional differential equations is still in
the initial stages and many aspects of this theory need to be
explored.

Zhang \cite{z1}  considered the  singular
fractional differential equation
\begin{gather*}
 D_{0+}^{\alpha} u(t) +  a(t)f(t,u(t),u'(t),\dots,u^{(n-2)}(t))=0, \quad
0 < t < 1, \\
u(0) =  u'(0)= \dots = u^{(n-2)}(0) = u^{(n-2)}(1) = 0,
\end{gather*}
where $D_{0+}^{\alpha}$ is the standard Riemann-Liouville fractional
derivative of order $n-1 < \alpha \leq n$, $n \geq 2$. They used a
fixed-point theorem for the mixed monotone operator to show the
existence of positive solutions for the above fractional boundary
value problem.  But the uniqueness is not treated.

 In \cite{f1}, the authors obtained the existence and
multiplicity of positive solutions for a class of higher-order
nonlinear fractional differential equations with integral boundary
conditions by applying Krasnoselskii's fixed-point theorem in cones.
But the uniqueness is also not treated.

 On the other hand, the study of the existence of solutions
of multi-point boundary value problems for linear second-order
ordinary differential equations was initiated by Il'in and Moiseev
\cite{i1}. Then Gupta \cite{g1} studied three-point boundary value
problems for nonlinear second-order ordinary differential equations.
Since then, nonlinear second-order three-point boundary value
problems have also been studied by several authors. We refer the
reader to \cite{g2,h2,li2,ma1,ma2} and the references therein.
However, all these papers are concerned with problems with
three-point boundary condition restrictions on the slope of the
solutions and the solutions themselves, for example,
\begin{gather*}
 u(0) = 0, \quad \alpha u(\eta) = u(1);\\
 u(0) = \beta u(\eta), \quad \alpha u(\eta) = u(1);\\
 u'(0) = 0, \quad \alpha u(\eta) = u(1);\\
 u(0) - \beta u'(0) = 0, \quad \alpha u(\eta) = u(1);\\
 \alpha u(0) - \beta u'(0) = 0, \quad u'(\eta) + u'(1) = 0; etc.
\end{gather*}

 In this article, we study the  higher-order
three-point boundary-value problem of fractional differential
equation.
\begin{gather}
\label{e1.1} 
D_{0+}^{\alpha} u(t) + f(t,u(t))=0, \quad 0 < t < 1,\\
\label{e1.2} u(0) = u'(0) = \dots = u^{(n-2)}(0)=0, \quad
u^{(n-2)}(1) = \int_0^\eta u(s)ds,
\end{gather}
where $D_{0+}^{\alpha}$ is the standard Riemann-Liouville fractional
derivative.  $n-1 < \alpha \leq n, n \geq 3$, $0 < \eta^\alpha <
\alpha(\alpha-1)(\alpha-2)\dots(\alpha-n+2)$. We will prove the
existence and uniqueness of a positive and nondecreasing solution
for the boundary value problems \eqref{e1.1}-\eqref{e1.2} by using a
fixed point theorem in partially ordered sets. 

 We note that the new three-point boundary conditions are
related to the area under the curve of solutions $u(t)$ from $t = 0$
to $t = \eta$. Existence of fixed point in partially ordered sets
has been considered recently in \cite{c1,h1,n1,n2,o1}. This work is
motivated by papers \cite{c1,li1}.

\section{Some definitions and fixed point theorems}

 The following definitions and lemmas will be
used for proving our the main results.

\begin{definition} \label{def2.1}\rm
Let $(E,  \|\cdot\|)$ be a real Banach space. A nonempty, closed,
convex set $P \subset E$ is said to be a cone
provided the following are satisfied:
\begin{itemize}
\item[(a)] if $y \in P$ and $\lambda\ \geq 0$, then $\lambda y \in P$; 
\item[(b)] if $y \in P$ and $-y \in P$, then $y = 0$.
\end{itemize}
 If $P \subset E$ is a cone, we denote the order induced by
$P$ on $E$ by $\leq$, that is, $x \leq y$  if and only if  $y - x
\in P$.
\end{definition}

\begin{definition}[\cite{p1}] \label{def2.2}\rm 
The integral
\[
I_{0+}^sf(x)= \frac
1{\Gamma(s)}\int_0^x\frac{f(t)}{(x-t)^{1-s}}dt,\quad x>0,
\]
 where $s>0$, is called Riemann-Liouville fractional integral
of order $s$ and $\Gamma(s)$ is the Euler gamma function defined by
\[
\Gamma(s) = \int_0^{+\infty} t^{s-1}e^{-t}dt,\ s > 0.
\]
\end{definition}


\begin{definition}[\cite{k1}] \label{def2.3}\rm   
For a function $f(x)$ given in the interval $[0,\infty)$, the
expression
\[
D_{0+}^sf(x)=\frac 1{\Gamma(n-s)}(\frac d{dx})^n
\int_0^x\frac{f(t)}{(x-t)^{s-n+1}}dt,
 \]
 where $ n=[s]+1, [s]$
denotes the integer part of number $s$, is called the
Riemann-Liouville fractional derivative of order $s$.
\end{definition}

 The following two lemmas can be found in \cite{b2,k1} which
are crucial in finding an integral representation of fractional
boundary value problem \eqref{e1.1} and \eqref{e1.2}.

\begin{lemma}[\cite{b2,k1}]\label{lemma2.1}  
Let $\alpha>0$ and $u \in C(0, 1)\cap L(0, 1)$.  Then  the
fractional differential equation
\[
{D}_{0+}^\alpha u(t)=0
\]
has  
$$
u(t)= c_1t^{\alpha-1}+c_2t^{\alpha-2}+\dots+c_nt^{\alpha-n},\quad
c_i\in \mathbb{R},\ i=0,1,\dots,n,\ n=[\alpha]+1
$$ 
as unique solution.
\end{lemma}

\begin{lemma}[\cite{b2,k1}] \label{lemma2.2}  
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
\[
I_{0+}^\alpha {D}_{0+}^\alpha
u(t)=u(t)+c_1t^{\alpha-1}+c_2t^{\alpha-2}+\dots+c_nt^{\alpha-n},
\]
for some $c_i\in \mathbb{R}$, $i=0,1,\dots,n$, $n=[\alpha]+1$.
\end{lemma}

  The following  fixed-point  theorems  in partially ordered sets
are fundamental and important to the proofs of our main results.

 \begin{theorem}[\cite{h1}]\label{the2.1}
 Let $(E, \leq)$ be a  partially ordered set and suppose that 
there exists a metric $d$
 in $E$ such that $(E, d)$ is a complete metric space. Assume that $E$ 
satisfies the following  condition:
\begin{equation} \label{e2.1}
\parbox{8cm}{
if $\{x_n\}$ is a nondecreasing sequence in $E$ such that
$x_n\to x$,
then $x_n \leq x$ for all $n\in \mathbb{N}$.}
\end{equation}
Let $T: E \to E$ be nondecreasing mapping such that
\[
d(Tx, Ty) \leq d(x, y) - \psi(d(x, y)), \quad for\ x \geq y,
\]
where $\psi: [0, +\infty) \to [0, +\infty)$ is a continuous
and nondecreasing function such that $\psi$ is positive in
 $(0, +\infty)$, $\psi(0) = 0$ and
$\lim_ {t \to \infty}\psi(t) = \infty$. If there exists
$x_0 \in E$ with $x_0 \leq T(x_0)$, then $T$ has a fixed point.
\end{theorem}

If we consider that $(E, \leq)$ satisfies the  condition
\begin{equation}\label{e2.2}
\text{for $ x, y \in E$ there exists $z \in E$ which
is comparable to $x$ and $y$},
\end{equation}
then we have the following result.

 \begin{theorem}[\cite{n1}]\label{the2.2} 
Adding condition \eqref{e2.2} to the hypotheses of Theorem \ref{the2.1}, we
 obtain uniqueness of the fixed point.
\end{theorem}

\section{Related lemmas} 

The basic space used in this paper is $E = C[0, 1]$. 
Then $E$ is a real Banach space with the norm 
$\|u\| = \max_{0 \leq t\leq 1}|u(t)|$. Note that this
space can be equipped with a partial order given by
\[
x, y \in C[0, 1],\quad x \leq y \Leftrightarrow x(t) \leq y(t), \quad
t \in [0, 1].
\]
In \cite{n1} it is proved that $(C[0, 1], \leq)$ with the classic
metric given by
\[
d(x, y) = \sup_{0 \leq t \leq 1} \{|x(t) - y(t)|\}
\]
satisfied condition \eqref{e2.1} of Theorem \ref{the2.1}. Moreover,
for $x, y \in C[0, 1]$ as the function $\max\{x, y\} \in C[0, 1]$,
$(C[0, 1], \leq)$ satisfies condition \eqref{e2.2}.

\begin{lemma}\label{lemma3.1}  
Let $0 < \eta^\alpha < \alpha(\alpha-1)(\alpha-2)\dots(\alpha-n+2)$. 
If $h \in C[0, 1]$, then the boundary-value problem
\begin{gather}
\label{e3.1} D_{0+}^{\alpha} u(t) + h(t) =0, \quad
0 < t < 1, \quad n-1 < \alpha \leq n,\\
\label{e3.2} u(0) = u'(0) = \dots = u^{(n-2)}(0)=0, \quad
u^{(n-2)}(1) = \int_0^\eta u(s)ds,
\end{gather}
has a unique solution
\begin{equation}\label{e3.3}
u(t) = \int_0^1 G(t,s)h(s)ds,
\end{equation}
where
\begin{gather}\label{e3.4}
G(t,s) = G_1(t, s) + G_2(t, s), \\
\label{e3.5}
G_1(t,s) = \frac{1}{\Gamma(\alpha)}\begin{cases}
t^{\alpha-1}(1-s)^{\alpha-n+1}-(t-s)^{\alpha-1}, & 0 \leq s \leq
t \leq 1, \\
t^{\alpha-1}(1-s)^{\alpha-n+1}, & 0 \leq t \leq s \leq 1,
\end{cases}
\\
\label{e3.6}
G_2(t,s) = \frac{\alpha t^{\alpha-1}
}{\alpha(\alpha-1)(\alpha-2)\dots(\alpha-n+2)-\eta^\alpha}\int_0^\eta
G_1(t,s)dt.
\end{gather}
\end{lemma}

\begin{proof}
By Lemma \ref{lemma2.2}, the solution of \eqref{e3.1} can be written
as
\[
u(t) = c_1 t^{\alpha-1} + c_2 t^{\alpha-2} +
\dots + c_n t^{\alpha-n} - \int_0^t
\frac{(t-s)^{\alpha-1}}{\Gamma(\alpha)}h(s)ds.
\]
From \eqref{e3.2}, we know that $c_2 = c_3 = \dots =c_{n} = 0$ and
\begin{align*}
u^{(n-2)}(t)
&= c_1(\alpha-1)(\alpha-2)\dots(\alpha-n+2) t^{\alpha-n+1}\\
&\quad - (\alpha-1)(\alpha-2)\dots(\alpha-n+2)\int_0^t
\frac{(t-s)^{\alpha-n+1}}{\Gamma(\alpha)}h(s)ds.
\end{align*}
Thus, together with $u^{(n-2)}(1) = \int_0^\eta u(s)ds$, we have
\[
c_1 = \frac{1 }{(\alpha-1)(\alpha-2)\dots(\alpha-n+2)}\int_0^\eta u(s)ds+
\int_0^1 \frac{(1-s)^{\alpha-n+1}}{\Gamma(\alpha)}h(s)ds.
\]
Therefore, the unique solution of boundary value problem
\eqref{e3.1}-\eqref{e3.2}  is
\begin{equation} \label{e3.7}
\begin{split}
u(t) &=  - \int_0^t \frac{(t-s)^{\alpha-1}}{\Gamma(\alpha)}h(s)ds\\
&\quad + \frac{t^{\alpha-1}
}{(\alpha-1)(\alpha-2)\dots(\alpha-n+2)}\int_0^\eta
u(s)ds+
t^{\alpha-1}\int_0^1 \frac{(1-s)^{\alpha-n+1}}{\Gamma(\alpha)}h(s)ds\\
&=  - \int_0^t
\frac{(t-s)^{\alpha-1}}{\Gamma(\alpha)}h(s)ds   + \frac{t^{\alpha-1}
}{(\alpha-1)(\alpha-2)\dots(\alpha-n+2)}\int_0^\eta
u(s)dx\\
&\quad + \int_0^t \frac{t^{\alpha-1} (1-s)^{\alpha-n+1}}{\Gamma(\alpha)}h(s)ds+ \int_t^1 \frac{t^{\alpha-1} (1-s)^{\alpha-n+1}}{\Gamma(\alpha)}h(s)ds\\
&=  \frac{1}{\Gamma(\alpha)}
\int_0^t(t^{\alpha-1}(1-s)^{\alpha-n+1}-(t-s)^{\alpha-1})h(s)ds \\
&\quad +\frac{1}{\Gamma(\alpha)}\int_t^1t^{\alpha-1}(1-s)^{\alpha-n+1}h(s)ds\\
&\quad + \frac{t^{\alpha-1}
}{(\alpha-1)(\alpha-2)\dots(\alpha-n+2)}\int_0^\eta
u(s)ds\\
&= \int_0^1 G_1(t,s)h(s)ds + \frac{t^{\alpha-1}
}{(\alpha-1)(\alpha-2)\dots(\alpha-n+2)}\int_0^\eta
u(s)ds,
\end{split}
\end{equation}
where $G_1(t, s)$ is defined by \eqref{e3.5}.
 From \eqref{e3.7}, we have
\[
\int_0^\eta u(t)dt = \int_0^\eta\int_0^1 G_1(t,s)h(s)\,ds\,dt +
\frac{\eta^{\alpha}
}{\alpha(\alpha-1)(\alpha-2)\dots(\alpha-n+2)}\int_0^\eta
u(s)ds.
\]
It follows that
\begin{equation} \label{e3.8}
\int_0^\eta u(t)dt =
\frac{\alpha(\alpha-1)(\alpha-2)\dots(\alpha-n+2)
}{\alpha(\alpha-1)(\alpha-2)\dots(\alpha-n+2)-\eta^\alpha}\int_0^\eta\int_0^1
G_1(t,s)h(s)\,ds\,dt.
\end{equation}
Substituting \eqref{e3.8} into \eqref{e3.7}, we obtain
\begin{align*}
u(t) &=  \int_0^1 G_1(t,s)h(s)ds \\
&\quad + \frac{\alpha t^{\alpha-1}}{\alpha(\alpha-1)(\alpha-2)\dots(\alpha-n+2)-\eta^\alpha}\int_0^\eta\int_0^1
G_1(t,s)h(s)\,ds\,dt\\
&=  \int_0^1 G_1(t,s)h(s)ds + \int_0^1 G_2(t,s)h(s)ds\\
&=   \int_0^1 G(t,s)h(s)ds,
\end{align*}
where $G(t, s)$, $G_1(t, s)$ and $G_2(t, s)$ are defined by
\eqref{e3.4}, \eqref{e3.5}, \eqref{e3.6}, respectively. The proof is
complete.
\end{proof}

\begin{lemma}\label{lemma3.2}
The function $G_1(t,s)$ defined by \eqref{e3.5} satisfies
\begin{itemize}
\item[(i)]  $G_1$ is a  continuous function and $G_1(t,s)\geq 0$ 
for $(t,s)\in [0,1]\times[0,1]$;
\item[(ii)] 
\[ \sup_{t\in [0,1]}\int_0^1G_1(t,s)ds =
\frac{n-2}{(\alpha-n+2)\Gamma(\alpha+1)}.
\]
\end{itemize}
\end{lemma}

\begin{proof}
(i) The continuity of $G_1$ is easily checked. On the other hand,
for $0 \leq t \leq s \leq 1$ it is obvious that
\[
G_1(t,s) = \frac{t^{\alpha-1}(1-s)^{\alpha-n+1}}{\Gamma(\alpha)}
\geq 0.
\]
In the case $0 \leq s \leq t \leq 1$ $(s \neq 1)$, we have
\begin{align*}
G_1(t,s) &=
\frac{1}{\Gamma(\alpha)}\left[\frac{t^{\alpha-1}(1-s)^{\alpha-1}}{(1-s)^{n-2}}
- (t-s)^{\alpha-1}\right]\\
&\geq \frac{1}{\Gamma(\alpha)}\left[t^{\alpha-1}(1-s)^{\alpha-1} -
(t-s)^{\alpha-1}\right]\\
&=  \frac{1}{\Gamma(\alpha)}\left[(t-ts)^{\alpha-1} -
(t-s)^{\alpha-1}\right]
\geq 0.
\end{align*}
Moreover, as $G_1(t,1) = 0$, then we conclude that $G_1(t,s) \geq 0$
for all $(t,s) \in [0,1]\times[0,1]$.

(ii)  Since
\begin{align*}
\int_0^1G_1(t,s)ds &=
\int_0^tG_1(t,s)ds + \int_t^1G_1(t,s)ds\\
&=   \frac{1}{\Gamma(\alpha)}
\int_0^t(t^{\alpha-1}(1-s)^{\alpha-n+1}-(t-s)^{\alpha-1})ds \\
&\quad +
\frac{1}{\Gamma(\alpha)}\int_t^1t^{\alpha-1}(1-s)^{\alpha-n+1}ds \\
&=
\frac{1}{\Gamma(\alpha)}\Big(\frac{t^{\alpha-1}}{\alpha-n+2}
-\frac{1}{\alpha}t^\alpha\Big).
\end{align*}
On the other hand, let
\[
\rho(t) = \int_0^1G_1(t,s)ds =
\frac{1}{\Gamma(\alpha)}
\Big(\frac{t^{\alpha-1}}{\alpha-n+2}-\frac{1}{\alpha}t^\alpha\Big),
\]
then, as $n \geq 3$, we have
\[
\rho'(t) =
\frac{1}{\Gamma(\alpha)}\Big(\frac{\alpha-1}{\alpha-n+2}t^{\alpha-2}
-t^{\alpha-1}\Big)> 0, \quad \text{for } t >0,
\]
the function $\rho(t)$ is strictly increasing and, consequently,
\begin{align*}
\sup_{t\in [0,1]}\rho(t)
&=  \sup_{t\in [0,1]}\int_0^1G_1(t,s)ds
= \rho(1) = \frac{1}{\Gamma(\alpha)}
\Big(\frac{1}{\alpha-n+2}-\frac{1}{\alpha}\Big)\\
&=   \frac{n-2}{\alpha(\alpha-n+2)\Gamma(\alpha)} =
\frac{n-2}{(\alpha-n+2)\Gamma(\alpha+1)}.
\end{align*}
The proof is complete.
\end{proof}

\begin{remark}\label{remark3.1} \rm
Obviously, by Lemma \ref{lemma3.1} and \ref{lemma3.2}, we have $u(t)
\geq 0$ if $h(t) \geq 0$ on $t \in [0, 1]$.
\end{remark}

\begin{lemma}\label{lemma3.3}
$G_1(t,s)$ is strictly increasing in the first variable.
\end{lemma}

\begin{proof} For $s$ fixed, we let
\begin{gather*}
g_1(t) = \frac{1}{\Gamma(\alpha)}\left(t^{\alpha -
1}(1-s)^{\alpha-n+1} - (t-s)^{\alpha-1}\right) \quad \text{for } s \leq t,\\
g_2(t)= \frac{1}{\Gamma(\alpha)}t^{\alpha-1}(1-s)^{\alpha-n+1}
\quad \text{for } t \leq s.
\end{gather*}
It is easy to check that $g_1(t)$ is strictly increasing on $[s, 1]$
and $g_2(t)$ is strictly increasing on $[0, s]$. Then we have the
following cases:

Case 1: $t_1, t_2 \leq s$ and $ t_1 < t_2$. In this case, we
have $g_2(t_1) < g_2(t_2)$, i.e. $G_1(t_1,s) < G_2(t_2, s)$.

Case 2: $s \leq t_1, t_2$ and $t_1 < t_2$. In this case, we
have $g_1(t_1) < g_1(t_2)$, i.e. $G_1(t_1,s) < G_1(t_2, s)$.

Case 3: $t_1 \leq s \leq  t_2$ and $t_1 < t_2$. In this case, we
have $g_2(t_1) \leq g_2(s) = g_1(s) \leq g_1(t_2) $. We claim that
$g_2(t_1)<g_1(t_2)$. In fact, if $g_2(t_1)=g_1(t_2)$, then $g_2(t_1)
= g_2(s) = g_1(s) = g_1(t_2)$, from the monotone of $g_1$ and $g_2$,
we have $t_1 = s = t_2$, which contradicts with $t_1 < t_2$. This
fact implies that $G_1(t_1,s) < G_1(t_2, s)$. The proof is complete.
\end{proof}

\begin{remark}\label{remark3.2} \rm
Obviously, by Lemma \ref{lemma3.3}, we have
\begin{equation} \label{e3.9}
\int_0^1 G_2(t, s)ds \leq
\frac{\eta(n-2)}{\Gamma(\alpha)[\alpha(\alpha-1)(\alpha-2)
\dots(\alpha-n+2)-\eta^{\alpha}](\alpha-n+2)}.
\end{equation}
\end{remark}

\begin{proof}
In fact, from Lemma \ref{lemma3.3} and \eqref{e3.6}, we have
\begin{align*}
G_2(t, s) \leq G_2(1, s) 
&=  \frac{\alpha\eta
G_1(1,s)}{\alpha(\alpha-1)(\alpha-2)\dots(\alpha-n+2)-\eta^\alpha}\\
&=   \frac{\alpha\eta
((1-s)^{\alpha-n+1}-(1-s)^{\alpha-1})}{\Gamma(\alpha)
[\alpha(\alpha-1)(\alpha-2)\dots(\alpha-n+2)-\eta^\alpha]}.
\end{align*}
Thus,
\begin{align*}
\int_0^1G_2(t,s)ds 
&\leq \frac{\alpha\eta\int_0^1
((1-s)^{\alpha-n+1}-(1-s)^{\alpha-1})ds
}{\Gamma(\alpha)[\alpha(\alpha-1)(\alpha-2)\dots(\alpha-n+2)
 -\eta^\alpha]}\\
&=\frac{\eta(n-2)}{\Gamma(\alpha) [\alpha(\alpha-1)(\alpha-2)\dots(\alpha-n+2)
 -\eta^{\alpha}](\alpha-n+2)},
\end{align*}
for $s, t \in [0, 1]\times [0, 1]$. 
\end{proof}

\section{Main Result} 

  For notational convenience, we denote 
\begin{align*}
L &:= \frac{n-2}{(\alpha-n+2)\Gamma(\alpha+1)}\\
&\quad +\frac{\eta(n-2)}{\Gamma(\alpha)\left[\alpha(\alpha-1)
(\alpha-2)\dots(\alpha-n+2)-\eta^{\alpha}\right](\alpha-n+2)}
> 0.
\end{align*}
The main result of this paper is the following.

\begin{theorem}\label{the3.1} 
The boundary value problem \eqref{e1.1}-\eqref{e1.2} has
a unique positive and strictly increasing solution $u(t)$ if the
following conditions are satisfied:
\begin{itemize}
\item[(i)] $f: [0, 1]\times[0, +\infty) \to [0,
+\infty)$ is continuous and nondecreasing respect to the second
variable and $f(t,u(t)) \not\equiv 0$ for $t \in Z \subset [0, 1]$
with $\mu(Z) > 0\ (\mu$ denotes the Lebesgue measure);

\item[(ii)] There exists $0 < \lambda  < \frac{1}{L}$ such that for $u, v \in [0, +\infty)$ with $u \geq v$ and
$t \in [0, 1]$
\[
 f(t,u)-f(t,v) \leq  \lambda \ln(u-v+1).
\]
\end{itemize}
\end{theorem}

\begin{proof}
Consider the cone
\[
K = \{u \in C[0, 1]: u(t) \geq 0\}.
\]
As $K$ is a closed set of $C[0, 1]$, $K$ is a complete metric space
with the distance given by $d(u, v) = \sup_{t\in [0,
1]}|u(t)-v(t)|$.
 Now, we consider the operator $T$ defined by
\begin{align*}
 Tu(t) &=  \int_0^1 G(t,s)f(s,u(s))ds,
\end{align*}
where $G(t, s)$ is defined by \eqref{e3.4}.
By Lemma \ref{lemma3.2} and condition (i), we have that $T(K) \subset K$.

 We now show that all the conditions of Theorem \ref{the2.1}
and  Theorem \ref{the2.2} are satisfied.

 Firstly, by condition (i), for $u, v \in K$ and $u \geq v$,
we have
\[
 Tu(t) =  \int_0^1 G(t,s)f(s,u(s))ds
\geq  \int_0^1 G(t,s)f(s,v(s))ds  = Tv(t).
\]
This proves that $T$ is a nondecreasing operator.

 On the other hand, for $u \geq v$ and by condition (ii) we
have
\begin{align*}
d(Tu, Tv) 
&=   \sup_{0 \leq t \leq 1} \left|(Tu)(t) - (Tv)(t)\right|
= \sup_{0 \leq t \leq 1} \left((Tu)(t) - (Tv)(t)\right)\\
&=  \sup_{0 \leq t \leq 1}\int_0^1G(t,s)(f(s,u(s))-f(s,v(s)))ds  \\
&\leq  \sup_{0 \leq t \leq
1}\int_0^1G(t,s)\lambda\cdot\ln(u(s)-v(s)+1)ds.
\end{align*}
Since the function $h(x) = \ln(x+1)$ is nondecreasing, by Lemma
\ref{lemma3.2} (ii), Remark \ref{remark3.2} and condition (ii),  we
have
\begin{align*}
&d(Tu, Tv)\\
&\leq  \lambda\ln(\|u-v\|+1)\Big(\sup_{0 \leq t \leq
1}\int_0^1G_1(t,s)ds + \sup_{0 \leq t \leq
1}\int_0^1 G_2(t, s)ds\Big)\\
&\leq \lambda\ln(\|u-v\|+1)\cdot L\\
&\leq  \|u - v\| - (\|u - v\| - \ln(\|u-v\|+1)).
\end{align*}
Let $\psi(x) = x - \ln(x+1)$. Obviously $\psi: [0, +\infty)
\to [0, +\infty)$ is continuous, nondecreasing, positive in
$(0, +\infty)$, $\psi(0) = 0$ and $\lim_{x \to
+\infty}\psi(x) = +\infty$. Thus, for $u \geq v$, we have
\[
d(Tu, Tv) \leq  d(u, v) - \psi(d(u, v)).
\]
As $G(t,s)\geq 0$ and $f \geq 0$, $(T0)(t) = \int_0^1G(t,s)f(s,0)ds
\geq 0$ and by Theorem \ref{the2.1} we know that problem
\eqref{e1.1}-\eqref{e1.2} has at least one nonnegative solution. As
$(K, \leq)$ satisfies condition \eqref{e2.2}, thus, Theorem
\ref{the2.2} implies that uniqueness of the solution. 

 Finally, we will prove that this solution $u(t)$ is strictly
increasing function. As $u(0)=\int_0^1G(0,s)f(s,u(s))ds$ and
$G(0,s)=0$ we have $u(0)=0$.

 Moreover, if we take $t_1,t_2\in[0,1]$ with $t_1<t_2$, we
can consider the following cases.

Case 1: $t_1 =0$, in this case, $u(t_1) = 0$ and, as $u(t) \geq 0$,
suppose that $u(t_2) = 0$. Then
\[
 0 = u(t_2) = \int_0^1G(t_2,s)f(s,u(s))ds.
\]
This implies that
\[
G(t_2,s)\cdot f(s,u(s)) = 0, \quad \text{a.e. }(s)
\]
and as $G(t_2,s) \neq 0$ a.e.$(s)$ we get $f(s,u(s)) = 0$
a.e. $(s)$.
 On the other hand, $f$ is nondecreasing respect to the
second variable, then we have
\[
f(s,0) \leq f(s,u(s)) = 0, \quad \text{a.e. }(s)
\]
which contradicts the condition (i) $f(t,0) \neq 0$ for $t \in Z
\subset [0, 1] (\mu(Z) \neq 0)$. Thus $u(t_1) = 0 < u(t_2)$.

Case 2: $0 < t_1$. In this case, let us take $t_2,t_1\in [0,1]$ with
$t_1<t_2$, then
\begin{align*}
u(t_2)-u(t_1) 
&=  (Tu)(t_2)-(Tu)(t_1) \\
&=  \int_0^1 (G(t_2,s)-G(t_1,s))f(s,u(s))ds.
\end{align*}
Taking into account Lemma \ref{lemma3.3} and the fact that $f \geq
0$, we get $u(t_2) - u(t_1) \geq 0$.

 Suppose that $u(t_2)=u(t_1)$ then
\[
\int_0^1 (G(t_2,s)-G(t_1,s))f(s,u(s))ds = 0
\]
and this implies
\[
(G(t_2,s)-G(t_1,s))f(s,u(s)) = 0 \quad\text{a.e. }(s).
\]
Again, Lemma \ref{lemma3.3} gives us
\[
f(s,u(s)) = 0 \quad\text{a.e. }(s)
\]
and using the same reasoning as above we have that this contradicts
condition (i) $f(t,0) \neq 0$ for $t \in Z \subset [0, 1] (\mu(Z)
\neq 0)$. Thus $u(t_1) = 0 < u(t_2)$. The proof is complete.
\end{proof}

\section{Example}

The fractional boundary-value problem
\begin{equation}
\begin{gathered}
 D_{0+}^{7/2} u(t) + (t^2+1)\ln(2+u(t))=0, \quad 0 < t < 1, \\
u(0) = u'(0) = u''(0)=0, \quad u''(1) =  \int_0^{1/2} u(s)ds.
\end{gathered}
\end{equation}
has a unique and strictly increasing solution.


 In this case, $n=4$, $\alpha = 7/2$, $\eta= 1/2$,  
$f(t, u) = (t^2+1)\ln(2+u(t))$ for $(t, u) \in [0, 1]\times[0, \infty)$.
 Note that $f$ is a continuous function and
$f(t,u) \neq 0$ for $t \in [0, 1]$.  Moreover, $f$ is nondecreasing
respect to the second variable since $\frac{\partial{f}}{\partial u}
= \frac{1}{u+2}(t^2+1)> 0$. On the other hand, for $u \geq v$ and $t
\in [0, 1]$, we have
\begin{align*}
f(t,u)-f(t,v) 
&=   (t^2+1)\ln(2+u)-(t^2+1)\ln(2+v) 
= (t^2+1)\ln\Big(\frac{2+u}{2+v}\Big) \\
&=  (t^2+1)\ln\Big(\frac{2+v+u-v}{2+v}\Big)
 = (t^2+1)\ln\Big(1+\frac{u-v}{2+v}\Big)\\
&\leq  (t^2+1)\ln(1+(u-v)) \leq 2\ln(1+u-v).
 \end{align*}
In this case, $\lambda = 2$. By simple computation, we have 
$0 < \lambda < 1/L$. Thus, Theorem \ref{the3.1} implies that
boundary value problem \eqref{e1.1}-\eqref{e1.2} has a unique and
strictly increasing solution.


\subsection*{Acknowledgment} 
The authors are supported by Research Foundation for
Chunmiao project of Department of Education of Jilin Province, China
(Ji Jiao Ke He Zi [2013] No. 252).


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\end{document}

