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

\AtBeginDocument{{\noindent\small
\emph{Electronic Journal of Differential Equations},
Vol. 2016 (2016), No. 108, pp. 1--15.\newline
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu}
\thanks{\copyright 2016 Texas State University.}
\vspace{9mm}}

\begin{document}
\title[\hfilneg EJDE-2016/108\hfil Superlinear singular fractional problems]
{Superlinear singular fractional boundary-value problems}

\author[I. Bachar,  H. M\^{a}agli \hfil EJDE-2016/108\hfilneg]
{Imed Bachar, Habib M\^{a}agli}

\address{Imed Bachar \newline
King Saud University, College of Science, Mathematics
Department, P.O. Box 2455, Riyadh 11451, Saudi Arabia}
\email{abachar@ksu.edu.sa}

\address{Habib M\^{a}agli \newline
King Abdulaziz University, Rabigh Campus,
College of Sciences and Arts, Department of Mathematics,
P.O. Box 344, Rabigh 21911, Saudi Arabia.\newline
D\'{e}partement de Math\'{e}matiques,
Facult\'{e} des Sciences de Tunis,
Campus Universitaire, 2092 Tunis, Tunisia}
\email{abobaker@kau.edu.sa, habib.maagli@fst.rnu.tn}

\thanks{Submitted  February 8, 2016. Published April 26, 2016.}
\subjclass[2010]{34A08, 34B15, 34B18, 34B27}
\keywords{Fractional differential equation; positive solution; Green's
function; \hfill\break\indent perturbation arguments}

\begin{abstract}
 In this article, we study the  superlinear fractional
 boundary-value problem
 \begin{gather*}
 D^{\alpha }u(x) =u(x)g(x,u(x)),\quad 0<x<1, \\
 u(0)=0,\quad \lim_{x\to0^{+}} D^{\alpha -3}u(x)=0,\quad
 \lim_{x\to0^{+}} D^{\alpha -2}u(x)=\xi ,\quad u''(1)=\zeta ,
 \end{gather*}
 where $3<\alpha \leq 4$, $D^{\alpha }$ is the Riemann-Liouville
 fractional derivative and $\xi ,\zeta \geq 0$ are such that $\xi +\zeta >0$.
 The function $g(x,u)\in C((0,1)\times [ 0,\infty ),[0,\infty)) $
 that may be singular at $x=0$ and $x=1$ is required to satisfy
 convenient hypotheses to be stated later.

 By means of a perturbation argument, we establish the existence,
 uniqueness and global asymptotic behavior of a positive continuous solution
 to the above problem.An example is given to illustrate our main results.
\end{abstract}

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


\section{Introduction}

Fractional differential equations have been of great interest
recently. Many phenomena in viscoelasticity, porous structures,
fluid flows, electrical networks can be modeled by these fractional
boundary-value problems (see, for instance, \cite{H,KST,MR,P,SKM}
and references therein) for discussions of various applications.

Fractional boundary-value problems of the form
\begin{equation}
D^{\alpha }u(x) +f(x,u(x))=0,\quad 0<x<1,\; 3<\alpha \leq 4,  \label{e1.1}
\end{equation}
subject to various boundary-value conditions have been considered by many
authors, see for example, \cite{AOS,BMR, BMTZ,CHS, GMS, LZ, MMZ, MRS, MRe,
PXZ, R, XCW,ZYY, ZSHL} and the references therein.

 Here $D^{\alpha }$ is the Riemann-Liouville fractional derivative
of order $\alpha$ ($3<\alpha \leq 4$) defined by \cite{KST,P,SKM},
\begin{equation*}
D^{\alpha }u(x)=\begin{cases}
(\frac{d}{dx}) ^{4}I^{4-\alpha }u(x), & \text{if }3<\alpha <4 \\[4pt]
(\frac{d}{dx}) ^{4}u(x), & \text{if }\alpha =4,
\end{cases}
\end{equation*}
 where for $\beta >0$,
\begin{equation*}
I^{\beta }u(x)=\frac{1}{\Gamma (\beta ) }\int_0^{x}(
x-y) ^{\beta -1}u(y) \,dy.
\end{equation*}
Liang and Zhang  \cite{LZ} established the existence of
positive solutions to problem \eqref{e1.1} subject to
\begin{equation}
u(0)=u'(0)=u''(0)=u''(1)=0, \label{e1.2}
\end{equation}
where $f(x,u)\in C([0,1]\times [ 0,\infty ),[0,\infty)) $ is nondecreasing 
with respect to $u$,
\[
f\Big(t,\frac{1}{\Gamma (\alpha )}\big(\frac{t^{\alpha -1}}{
\alpha -2}-\frac{t^{\alpha }}{\alpha }\big) \Big) \neq 0
\]
 for $t\in (0,1) $ and there exists a positive constant $\gamma <1$ such that 
$f$ is $ \gamma $-concave with respect to $u$, that is, for all 
$\lambda \in [0,1]$,
\begin{equation*}
\lambda ^{\gamma }f(x,u)\leq f(x,\lambda u).
\end{equation*}
Their approach is based on lower and upper solution method.

 Recently Zhai et al \cite{ZYY}, by means of fixed point theorem
for a sum operator proved the existence and uniqueness of a positive
solution to problem \eqref{e1.1}-\eqref{e1.2} with 
$f(x,u)=\varphi (x,u)+\psi (x,u)$, where 
$\varphi ,\psi \in C([0,1]\times [0,\infty ),[0,\infty )) $ increasing 
with respect to the second
variable. The function $\varphi $ is $\gamma $-concave with respect to $u$
for some $\gamma \in (0,1)$, $\varphi \geq \delta _0\psi $ for some
positive constant $\delta _0$, $\psi (x,0)\neq 0$ and
 $\psi (x,\lambda u)\geq \lambda \psi (x,u)$ for $\lambda \in (0,1)$.

 In this article, we consider the  superlinear fractional problem
\begin{equation}
\begin{gathered}
D^{\alpha }u(x) -u(x)g(x,u(x))=0,\quad 0<x<1, \\
u(0)=0,\quad \lim_{x\to0^{+}} D^{\alpha -3}u(x)=0,\quad 
\lim_{x\to0^{+}} D^{\alpha -2}u(x)=\xi ,\quad u''(1)=\zeta ,
\end{gathered} \label{e1.3}
\end{equation}
where $3<\alpha \leq 4$ and $\xi ,\zeta \geq 0$ with $\xi +\zeta >0 $.

The function $g(x,u)\in C((0,1)\times [ 0,\infty ),[0,\infty )) $, which may 
be singular at $x=0$ and $x=1$ is required
to satisfy some convenient hypotheses to be stated later.
 We emphasize that the condition $\xi +\zeta >0$ on the boundary
data is essential to obtain positive solutions.

 To simplify our statements, we use the following notation:

(i) $\mathcal{B}^{+}((0,1)) $ denotes the
set of nonnegative measurable functions on $(0,1)$.

(ii) $C(X)$ (resp. $C^{+}(X)$) denotes the set of continuous (resp.
nonnegative continuous) functions on a metric space $X$.

(iii) We denote by $G(x,y)$ the Green's function of the operator 
$u\to -D^{\alpha }u$,
with boundary conditions 
\[
u(0)=\lim_{x\to0^{+}} D^{\alpha -3}u(x)=\lim_{x\to0^{+}} D^{\alpha -2}u(x)=u''(1)=0.
\]

(iv) For $\alpha \in (3,4]$, we let
\begin{equation}
\mathcal{J}_{\alpha }=\{p\in \mathcal{B}^{+}((0,1)) 
:\int_0^{1}t^{\alpha -1}(1-t)^{\alpha -3}p(t)dt<\infty \}.
\label{e1.4}
\end{equation}

(v) For $p\in \mathcal{B}^{+}((0,1)) $, we denote
\begin{equation}
\tau _p:=\sup_{x,y\in (0,1)}\int_0^{1}\frac{G(x,t)G(t,y)}{G(x,y)}p(t)dt.
 \label{e1.5}
\end{equation}
and we will prove that if $p\in \mathcal{J}_{\alpha }$, then
 $\tau_p<\infty$.

(vi) For $3<\alpha \leq 4$ and $\xi ,\zeta \geq 0$ with $\xi +\zeta >0$,
we define the function $h$ on $[0,1]$
by
\begin{equation}
\begin{aligned}
h(x) &= \frac{\xi }{\Gamma (\alpha )}x^{\alpha -2}(\alpha -1-(\alpha -3)x)+
\frac{\zeta }{(\alpha -1)(\alpha -2)}x^{\alpha -1}  \\
&= h_1(x)+h_{2}(x).
\end{aligned}  \label{e1.6}
\end{equation}
It is easy to show that $h$ is the unique solution of the problem
\begin{equation}
\begin{gathered}
D^{\alpha }u(x)=0,\quad 0<x<1, \\
u(0)=0,\quad \lim_{x\to0^{+}} D^{\alpha -3}u(x)=0,\quad 
\lim_{x\to0^{+}} D^{\alpha -2}u(x)=\xi ,\quad
u''(1)=\zeta .
\end{gathered}  \label{e1.7}
\end{equation}
 Note also that, there exists a constant $M>0$, such that
\begin{equation}
\frac{1}{M}\phi (x)\leq h(x)\leq M\phi (x),\text{ for all }x\in [ 0,1]
\label{e1.8}
\end{equation}
where
\begin{equation*}
\phi (x)=\begin{cases}
x^{\alpha -1}, & \text{if }\xi =0, \\
x^{\alpha -2}, & \text{if }\xi >0.
\end{cases}
\end{equation*}
To state our main results, we require a combination of the
following conditions.
\begin{itemize}
\item[(H1)] $g:(0,1)\times [ 0,\infty )\to [ 0,\infty )$, continuous,

\item[(H2)] There exists a function $p\in
C((0,1))\mathcal{\cap J}_{\alpha }$ with $\tau _p\leq \frac{1}{2}$ such
that, for all $x\in (0,1)$, the map $y\to y(p(x) -g(x,yh(x) ) ) $ is
nondecreasing on $[ 0,1]$.

\item[(H3)]  For all $x\in (0,1)$, the function $y\to yg(x,y) $ is
nondecreasing on $[0,\infty )$.
\end{itemize}
Using a perturbation method, we establish the following result.

\begin{theorem}\label{thm1.1}
Under assumptions {\rm (H1)--(H2)},
problem \eqref{e1.3} admits a solution $u\in C([0,1])$ such that, for
all $x\in [ 0,1] $,
\begin{equation}
c_0h(x)\leq u(x) \leq h(x),  \label{e1.9}
\end{equation}
where $c_0\in [ 0,1]$.
Furthermore, if assumption (H3) is also
fulfilled, then this solution is unique.
\end{theorem}

\begin{corollary} \label{coro1.2}
Let $\psi \in C^{1}([0,\infty ))$, $\psi \geq 0$ such that
the map $y\to \varphi (y)=y\psi (y)$ is nondecreasing on $[0,\infty)$. 
Let $q\in C^{+}((0,1)) $ such that the function 
$x\to \widetilde{q}(x):=q(x)\max_{0\leq t\leq h(x) } \varphi '(t)\in 
\mathcal{J}_{\alpha }$. Then for $\lambda \in [ 0,\frac{1}{2\tau _{\widetilde{q}}})$, 
the problem
\begin{gather*}
D^{\alpha }u(x)=\lambda q(x)u(x)\psi (u(x)),\quad  x\in (0,1),\\
u(0)=0,\quad \lim_{x\to0^{+}} D^{\alpha -3}u(x)=0,\quad 
\lim_{x\to0^{+}} D^{\alpha -2}u(x)=\xi ,\quad
u''(1)=\zeta ,
\end{gather*}
admits a unique positive solution $u\in C([0,1])$ such that
\begin{equation*}
(1-\lambda \tau _{\widetilde{q}})h(x)\leq u(x) \leq h(x),\quad 
\text{for all }x\in [ 0,1].
\end{equation*}
\end{corollary}

Our paper is organized as follows. In section 2, we give the explicit
expression of the Green's function $G(x,y)$ and we establish some sharp
estimates on it. In section 3, first for a convenient nonnegative given
function $p$, we construct the Green's function $\mathcal{H}(
x,y) $ of the operator $u\to -D^{\alpha }u+pu$, with boundary
conditions $u(0)=\lim_{x\to0^{+}} D^{\alpha -3}u(x)=
\lim_{x\to0^{+}} D^{\alpha -2}u(x)=u''(1)=0 $ and we derive some 
of its properties. In particular, we prove the following statements:

(i) There exists a constant $c\in (0,1]$ such that for 
$(x,y) \in [ 0,1]\times [ 0,1]$,
\begin{equation*}
cG(x,y) \leq \mathcal{H}(x,y) \leq G(x,y) .
\end{equation*}

(ii) The  equation holds
\begin{equation*}
U\psi =U_p\psi +U_p(pU\psi ) =U_p\psi +U(pU_p\psi) ,\quad
\text{for all }\psi \in \mathcal{B}^{+}((0,1)) .
\end{equation*}
 where the kernels $U$ and $U_p$ are defined on $\mathcal{B}^{+}((0,1)) $ by
\begin{equation}
U\psi (x) :=\int_0^{1}G(x,y) \psi (y)dy,\quad 
U_p\psi (x) :=\int_0^{1}\mathcal{H}(x,y) \psi
(y)dy,\text{\ }x\in [ 0,1].  \label{e1.10}
\end{equation}
 By exploiting these properties, we prove our main results.

\section{On the Green function}

We recall the following known properties.

\begin{lemma}[\cite{KST,P,SKM}] \label{lem2.4}
 Let $\alpha \in (3,4)$ and $u\in C((0,1) ) \cap L^{1}((0,1) )$. Then we have
\begin{itemize}
\item[(i)] For $\ 0<\gamma <\alpha $, $\ D^{\gamma }I^{\alpha
}u=I^{\alpha -\gamma }u$ and $\ D^{\alpha }I^{\alpha }u=u$.

\item[(ii)]  $D^{\alpha }u(x)=0$ if and only if $u(x)=c_1x^{\alpha
-1}+c_{2}x^{\alpha -2}+c_{3}x^{\alpha -3}+c_{4}x^{\alpha -4}$,
 where $c_{i}\in \mathbb{R}$,
 for $i\in \{1,2,3,4\}$.

\item[(iii)] Assume that $D^{\alpha }u\in C((0,1)) \cap L^{1}((0,1) ) $, then
\begin{equation*}
I^{\alpha }D^{\alpha }u(x) =u(x)+c_1x^{\alpha
-1}+c_{2}x^{\alpha -2}+c_{3}x^{\alpha -3}+c_{4}x^{\alpha -4},
\end{equation*}
 where $c_{i}\in \mathbb{R}$, for $i\in \{1,2,3,4\}$.
\end{itemize}
\end{lemma}

 Next we give the explicit expression of the Green's function $G(x,y) $.

\begin{lemma} \label{lem2.5}
Let $\alpha \in (3,4]$  and $\psi \in C^{+}([0,1])$. Then the problem
\begin{equation}
\begin{gathered}
-D^{\alpha }u(x) =\psi (x),\quad 0<x<1, \\
u(0)=0,\quad \lim_{x\to0^{+}} D^{\alpha -3}u(x)=0,\quad 
\lim_{x\to0^{+}} D^{\alpha -2}u(x)=0,\quad u''(1)=0,
\end{gathered}  \label{e2.1}
\end{equation}
 has a unique nonnegative solution
\begin{equation}
u(x)=\int_0^{1}G(x,y) \psi (y) dy,  \label{e2.2}
\end{equation}
 where for $ x,y\in [ 0,1] $,
\begin{equation}
G(x,y) =\frac{1}{\Gamma (\alpha ) }
\begin{cases}
x^{\alpha -1}(1-y)^{\alpha -3}-(x-y)^{\alpha -1}, & 0\leq y\leq x\leq
1; \\
x^{\alpha -1}(1-y)^{\alpha -3}, & 0\leq x\leq y\leq 1.
\end{cases}  \label{e2.G}
\end{equation}
\end{lemma}

\begin{proof}
Since $\psi \in C([0,1])$,  by Lemma \ref{lem2.4}, we have
\begin{equation*}
u(x)=c_1x^{\alpha -1}+c_{2}x^{\alpha -2}+c_{3}x^{\alpha -3}+c_{4}x^{\alpha
-4}-I^{\alpha }\psi (x).
\end{equation*}
Using the fact that $u(0)=0$, $\lim_{x\to0^{+}}
D^{\alpha -3}u(x)=0$ and $\lim_{x\to0^{+}} D^{\alpha
-2}u(x)=0$, $u''(1)=0$, we obtain $c_{2}=c_{3}=c_{4}=0$ and 
$c_1=\frac{1}{\Gamma (\alpha ) }\int_0^{1}(1-y)^{\alpha -3}\psi (y) dy$.
Then, the unique solution of \eqref{e2.1} is 
\begin{align*}
u(x) &= \frac{1}{\Gamma (\alpha ) }\int_0^{1}x^{\alpha
-1}(1-y) ^{\alpha -3}\psi (y) dy-\frac{1}{\Gamma
(\alpha ) }\int_0^{x}(x-y) ^{\alpha -1}\psi (y) dy \\
&= \int_0^{1}G(x,y) \psi (y) dy.
\end{align*}
This completes the proof.
\end{proof}

\begin{proposition} \label{prop2.6}
The Green function $G(x,y)$ in Lemma \ref{lem2.5} has the
following properties:
\begin{itemize}
\item[(i)] For $y\in [ 0,1]$, the function $x\to G(
x,y) $ belongs to $C^{2}([0,1])$.

\item[(ii)] For $x,y\in [ 0,1]$,
\begin{equation*}
\frac{1}{\Gamma (\alpha ) }H_0(x,y)\leq G(x,y)
\leq \frac{2(\alpha -1)}{\Gamma (\alpha ) }H_0(x,y),
\end{equation*}
where $H_0(x,y)=x^{\alpha -2}(1-y) ^{\alpha -3}\min
(x,y)$.

\item[(iii)] For $x,y\in [ 0,1]$,
\begin{equation*}
\frac{1}{\Gamma (\alpha ) }x^{\alpha -1}y(1-y)
^{\alpha -3}\leq G(x,y) \leq \frac{2(\alpha -1)}{\Gamma (
\alpha ) }x^{\alpha -2}y(1-y) ^{\alpha -3}.
\end{equation*}

\item[(iv)] For $x\in (0,1]$ and $y\in [ 0,1)$,
\begin{equation*}
\frac{(\alpha -1)}{\Gamma (\alpha ) }H(x,y)\leq \frac{\partial }{
\partial x}G(x,y) \leq \frac{(\alpha -1)(\alpha -2)}{\Gamma
(\alpha ) }H(x,y),
\end{equation*}
 where $H(x,y)=x^{\alpha -3}(1-y) ^{\alpha -3}\min
(x,y)$.

\item[(v)] For $x\in (0,1]$ and $y\in [ 0,1)$,
\begin{equation*}
\frac{(\alpha -1)(\alpha -2)(\alpha -3)}{\Gamma (\alpha ) }
\widetilde{H}(x,y)\leq \frac{\partial ^{2}}{\partial x^{2}}G(
x,y) \leq \frac{(\alpha -1)(\alpha -2)}{\Gamma (\alpha ) }
\widetilde{H}(x,y),
\end{equation*}
 where $\widetilde{H}(x,y)=x^{\alpha -4}(1-y) ^{\alpha -4}\min (x,y)(1-\max (x,y))$.
\end{itemize}
\end{proposition}

\begin{proof}
(i) From Lemma \ref{lem2.5}, for $x,y\in [ 0,1]$, we have
\begin{align*}
G(x,y) &= \frac{1}{\Gamma (\alpha ) }
\begin{cases}
x^{\alpha -1}(1-y)^{\alpha -3}-(x-y)^{\alpha -1}, & 0\leq y\leq x\leq
1; \\
x^{\alpha -1}(1-y)^{\alpha -3}, & 0\leq x\leq y\leq 1,
\end{cases} \\
&= \frac{1}{\Gamma (\alpha ) }\Big[ x^{\alpha -1}(1-y)^{\alpha
-3}-(\max (x-y,0)) ^{\alpha -1}\Big] .
\end{align*}
Since $\alpha >3$, if follows that the function $x\to
(\max (x-y,0)) ^{\alpha -1}$ belongs to $C^{2}([0,1])$. This
implies the result.

(ii) Observe that for $a,b>0$ and $c,y\in [ 0,1]$, we have
\begin{equation}
\min (1,\frac{b}{a})(1-cy^{a})\leq 1-cy^{b}\leq \max (1,\frac{b}{a}
)(1-cy^{a}).  \label{e2.3}
\end{equation}
 Now, since for $x,y\in [ 0,1]$, we have
\begin{equation*}
G(x,y) =\frac{1}{\Gamma (\alpha ) }x^{\alpha
-1}(1-y)^{\alpha -3}\Big[ 1-(1-y)^{2}(\frac{\max (x-y,0)}{x(1-y)}
) ^{\alpha -1}\Big] ,
\end{equation*}
and $\frac{\max (x-y,0)}{x(1-y)}\in [ 0,1]$, for $x\in (0,1]$
and $y\in [ 0,1)$, then the required result follows from \eqref{e2.3}
with $b=\alpha -1$, $a=1$ and $c=(1-y)^{2}$.

(iii) The inequalities follows from (i) and the fact that
\begin{equation*}
xy\leq \min (x,y)\leq y,\text{ for }x,y\in [ 0,1].
\end{equation*}

(iv) Since for $x,y\in [ 0,1]$,
\begin{align*}
\frac{\partial }{\partial x}G(x,y) 
&= \frac{\alpha -1}{\Gamma(\alpha ) }
\begin{cases}
x^{\alpha -2}(1-y)^{\alpha -3}-(x-y)^{\alpha -2}, & 0\leq y\leq x\leq
1; \\
x^{\alpha -2}(1-y)^{\alpha -3}, & 0\leq x\leq y\leq 1,
\end{cases} \\
&= \frac{\alpha -1}{\Gamma (\alpha ) }x^{\alpha -2}(1-y)^{\alpha
-3}\Big[ 1-(1-y)\Big(\frac{\max (x-y,0)}{x(1-y)}\Big) ^{\alpha -2}
\Big] ,
\end{align*}
 the required result follows from \eqref{e2.3} with $b=\alpha -2$, $a=1$ and $c=(1-y)$.

(v) Since for $x\in (0,1]$ and $y\in [ 0,1)$,
\begin{align*}
\frac{\partial ^{2}}{\partial x^{2}}G(x,y) 
&= \frac{(\alpha -1) (\alpha -2) }{\Gamma (\alpha ) }
\begin{cases}
x^{\alpha -3}(1-y)^{\alpha -3}-(x-y)^{\alpha -3}, & 0\leq y\leq x\leq
1; \\
x^{\alpha -3}(1-y)^{\alpha -3}, & 0\leq x\leq y\leq 1,
\end{cases}
 \\
&= \frac{(\alpha -1) (\alpha -2) }{\Gamma (
\alpha ) }x^{\alpha -3}(1-y)^{\alpha -3}[ 1-(\frac{\max
(x-y,0)}{x(1-y)}) ^{\alpha -3}] ,
\end{align*}
 the required result follows again from $\eqref{e2.3}$ with 
$b=\alpha -3$, $a=1$ and $c=1$.
This completes the proof.
\end{proof}

 From Proposition \ref{prop2.6} (iii), we deduce the following result.

\begin{corollary} \label{coro2.7}
Let $\psi \in \mathcal{B}^{+}((0,1)) $, then
\begin{equation*}
U\psi \in C([0,1])\Longleftrightarrow \int_0^{1}y(1-y)
^{\alpha -3}\psi (y) dy<\infty .
\end{equation*}
\end{corollary}

\begin{proposition}\label{prop2.8}
Let $3<\alpha <4$ and $\psi \in C((0,1))$. Assume that the
function $y\to y(1-y) ^{\alpha -3}\psi (y)\in
C((0,1))\cap L^{1}((0,1)$, then $U\psi $ is the unique solution in $C([0,1])$
of
\begin{equation}
\begin{gathered}
-D^{\alpha }u(x)=\psi (x),\quad  0<x<1, \\
u(0)=0,\quad \lim_{x\to0^{+}} D^{\alpha -3}u(x)=0,\quad 
\lim_{x\to0^{+}} D^{\alpha -2}u(x)=0,\quad u''(1)=0.
\end{gathered}  \label{e2.4}
\end{equation}
\end{proposition}

\begin{proof}
From Corollary \ref{coro2.7}, we deduce that the function 
$U\psi \in C([0,1])$. This implies that
 $I^{4-\alpha }(U| \psi | ) $ is finite on $[0,1]$.
Hence, we obtain
\begin{align*}
I^{4-\alpha }(U\psi ) (x) 
&= \frac{1}{\Gamma (4-\alpha )}
\int_0^{x}(x-y)^{3-\alpha }U\psi (y)dy \\
&= \frac{1}{\Gamma (4-\alpha )}\int_0^{1}\Big(
\int_0^{x}(x-y)^{3-\alpha }G(y,z)dy\Big) \psi (z)dz \\
&= \int_0^{1}\mathcal{K}(x,z)\psi (z)dz,
\end{align*}
where 
\[
\mathcal{K}(x,z):=\frac{1}{\Gamma (4-\alpha )}
\int_0^{x}(x-y)^{3-\alpha }G(y,z)dy.
\]

Next we will express explicitly $\mathcal{K}(x,z)$.
Using $\eqref{e2.G}$, we obtain
\begin{align*}
&\mathcal{K}(x,z) \\
&= \frac{(1-z)^{\alpha -3}}{\Gamma (4-\alpha )\Gamma (\alpha )}
\int_0^{x}(x-y)^{3-\alpha }y^{\alpha -1}dy \\
&\quad -\frac{1}{\Gamma (4-\alpha )\Gamma (\alpha )}\int_0^{x}(x-y)^{3-\alpha
}(\max (y-z,0))^{\alpha -1}dy \\
&= \frac{1}{6}x^{3}(1-z)^{\alpha -3}-\frac{1}{\Gamma (4-\alpha )\Gamma
(\alpha )}\int_0^{x}(x-y)^{3-\alpha }(\max (y-z,0))^{\alpha -1}dy
\end{align*}
 If $z\leq x$, then we have
\begin{equation}
\begin{aligned}
\int_0^{x}(x-y)^{3-\alpha }((y-z)^{+})^{\alpha -1}dy
&= \int_{z}^{x}(x-y)^{3-\alpha }(y-z)^{\alpha -1}dy   \\
&= \frac{\Gamma (\alpha )\Gamma (4-\alpha )}{6}(x-z)^{3}.
\end{aligned}  \label{e2.6}
\end{equation}
On the other hand, if $x\leq z$ and $y\in (0,x)$, we have
\begin{equation}
\int_0^{x}(x-y)^{3-\alpha }(\max (y-z,0))^{\alpha -1}dy=0.  \label{e2.7}
\end{equation}
From \eqref{e2.6} and \eqref{e2.7}, we obtain
\begin{equation*}
\mathcal{K}(x,z)=\frac{1}{6}x^{3}(1-z)^{\alpha -3}-\frac{1}{6}(\max
(x-z,0))^{3}.
\end{equation*}
Hence for $x\in (0,1)$, we have
\begin{align*}
6I^{4-\alpha }(U\psi ) (x) &= 6\int_0^{1}\mathcal{K}(x,z)\psi
(z)dz \\
&= x^{3}\int_0^{x}[(1-z)^{\alpha -3}-1]\psi (z)dz+3x^{2}\int_0^{x}z\psi
(z)dz \\
&\quad -3x\int_0^{x}z^{2}\psi (z)dz+\int_0^{x}z^{3}\psi (z)dz 
 +x^{3}\int_{x}^{1}(1-z)^{\alpha -3}\psi (z)dz \\
&:=J_1(x)+J_{2}(x)+J_{3}(x)+J_{4}(x)+J_{5}(x).
\end{align*}
 We claim that
\begin{equation*}
D^{\alpha }(U\psi ) (x):=\frac{d^{4}}{dx^{4}}(I^{4-\alpha
}(U\psi ) )(x)=-\psi (x),\text{ for }x\in (0,1).
\end{equation*}
Indeed, since the function $z\mapsto z\psi (z)$ is continuous
and integrable in a neighborhood of $0$ and the function 
$z\mapsto(1-z)^{\alpha -3}\psi (z)$ is continuous and integrable in 
a neighborhood of $1$, we deduce that $J_{2}(x),J_{3}(x),J_{4}(x)$ and 
$J_{5}(x)$ are differentiable.

 On the other hand, since $(1-z)^{\alpha -3}-1=O(z)$ near $0$, it
follows that $J_1(x)$ is differentiable. So, we have
\begin{align*}
\frac{d}{dx}(6I^{4-\alpha }(U\psi ) )(x)
&= 3x^{2}\int_0^{x}[(1-z)^{\alpha -3}-1]\psi (z)dz+6x\int_0^{x}z\psi
(z)dz \\
&\quad -3x\int_0^{x}z^{2}\psi (z)dz+3x^{2}\int_{x}^{1}(1-z)^{\alpha -3}\psi
(z)dz, \\
&= K_1(x)+K_{2}(x)+K_{3}(x)+K_{4}(x).
\end{align*}
Similarly, we obtain
\begin{equation*}
\frac{d^{4}}{dx^{4}}(I^{4-\alpha }(U\psi ) )(x)=-\psi (x),\quad 
\text{for }x\in (0,1).
\end{equation*}

It remains to verify the boundary conditions. Since $U\psi \in
C([0,1])$,  we deduce that $U\psi (0)=0$.
On the other hand, clearly we have 
\[
\lim_{x\to 0^{+}}K_1(x)=\lim_{x\to0^{+}} K_{2}(x)
=\lim_{x\to 0^{+}} K_{3}(x)=0
\]
 and by \cite[Lemma 2.2]{MT}, we have 
$\lim_{x\to0^{+}} K_{4}(x)=0$.
 Now, since $D^{\alpha -3}(U\psi ) (x)=\frac{d}{dx}
(I^{4-\alpha }(U\psi ) )(x)$, we deduce that
\begin{equation*}
\lim_{x\to0^{+}} D^{\alpha -3}(U\psi ) (x)=0.
\end{equation*}
Similarly, we show that $\lim_{x\to0^{+}}D^{\alpha -2}(U\psi ) (x)=0$, 
by using the fact that
\begin{equation*}
D^{\alpha -2}(U\psi ) (x)=\frac{d^{2}}{dx^{2}}(I^{4-\alpha}(U\psi ) )(x).
\end{equation*}

Let $\eta >0$. By Proposition \ref{prop2.6} (v), there exists a
constant $c>0$, such that for $x\in (\eta ,1]$ and $y\in (0,1)$, we have
\begin{equation*}
\big| \frac{\partial ^{2}}{\partial x^{2}}G(x,y)\big| 
\leq c\eta ^{\alpha -4}y(1-y) ^{\alpha -4}(1-\max (x,y))
\leq c\eta ^{\alpha -4}y(1-y) ^{\alpha -3}.
\end{equation*}
So by the Lebesgue theorem, we deduce that $(U\psi )''(1)=0$.

 Finally, the uniqueness follows immediately from Lemma \ref{lem2.4}.
 The proof is complete.
\end{proof}

Same properties in Proposition \ref{prop2.8} remain true for $\alpha =4$.

\begin{proposition}\label{prop2.9}
For each $x,t,y\in (0,1)$, we have
\begin{equation}
\frac{G(x,t)G(t,y)}{G(x,y)}\leq \frac{4(\alpha -1)^{2}}{\Gamma (\alpha )}
t^{\alpha -1}(1-t)^{\alpha -3}.  \label{e2.8}
\end{equation}
\end{proposition}

\begin{proof}
 Using Proposition \ref{prop2.6} (ii), for each $x,t,y\in (0,1)$,
we have
\begin{equation*}
\frac{G(x,t)G(t,y)}{G(x,y)}\leq \frac{4(\alpha -1)^{2}}{\Gamma (\alpha )}
t^{\alpha -2}(1-t)^{\alpha -3}\frac{\min (x,t)\min (t,y)}{\min (x,y)}.
\end{equation*}
 So the result follows from the fact that
\begin{equation*}
\frac{\min (x,t)\min (t,y)}{\min (x,y)}\leq t.
\end{equation*}
This completes the proof.
\end{proof}

\begin{proposition} \label{prop2.10}
Let $p\in \mathcal{J}_{\alpha }$. We have:
(i)
\begin{equation}
\tau _p\leq \frac{4(\alpha -1)^{2}}{\Gamma (\alpha )}\int
_0^{1}t^{\alpha -1}(1-t)^{\alpha -3}p(t)dt<\infty ,  \label{e2.9}
\end{equation}
where $\tau _p$ is given by \eqref{e1.5}.

(ii)
\begin{equation}
U(ph)(x)\leq \tau _ph(x),\text{ for }x\in [ 0,1].  \label{e2.10}
\end{equation}
\end{proposition}

\begin{proof}
Let $p\in \mathcal{J}_{\alpha }$.
(i) Using \eqref{e1.5} and \eqref{e2.8}, we obtain \eqref{e2.9}.
(ii) Since $h=h_1+h_{2}$, we need to prove \eqref{e2.10} for $
h_1$ and $h_{2}$.

 To this end, observe that for each $x,y\in (0,1]$, we have 
$\lim_{z\to 1} \frac{G(y,z)}{G(x,z)}=\frac{h_{2}(y)}{h_{2}(x)}$.
Therefore, by applying Fatou lemma and \eqref{e1.5}, we obtain
\begin{align*}
\frac{1}{h_{2}(x)}U(ph_{2})(x) 
&= \int_0^{1}G(x,y)\frac{h_{2}(y)}{h_{2}(x)} p(y)dy \\
&\leq \liminf_{z\to 1}\int_0^{1}G(x,y)\frac{G(y,z)}{
G(x,z)}p(y)dy\leq \tau _p.
\end{align*}
Similarly, we prove $U(ph_1)(x)\leq \tau _ph_1(x)$, by
observing that $\lim_{z\to 0} \frac{G(y,z)}{G(x,z)}=
\frac{h_1(y)}{h_1(x)}$.
This completes the proof.
\end{proof}

\section{Proofs of main results}

\subsection{On the Green's function of the perturbed operator.}
In this subsection, our goal is to determine the positive solution to the
 linear fractional problem
\begin{equation}
\begin{gathered}
-D^{\alpha }u(x)+p(x)u(x)=\psi (x),\quad 0<x<1, \\
u(0)=\lim_{x\to0^{+}} D^{\alpha -3}u(x)
=\lim_{x\to 0^{+}} D^{\alpha -2}u(x)=u''(1)=0.
\end{gathered} \label{e3.p}
\end{equation}
To this end, we need to construct the Green's function to the
homogeneous problem associated with \eqref{e3.p}.

 Let $p\in \mathcal{J}_{\alpha }$. For $(x,y) \in
[ 0,1]\times [ 0,1]$, put $G_0(x,y)=G(x,y)$ and
\begin{equation}
G_{n}(x,y)=\int_0^{1}G(x,t)G_{n-1}(t,y)p(t)dt,\quad n\geq 1.
\label{e3.2}
\end{equation}
Let $\mathcal{H}:[0,1]\times [ 0,1]\to \mathbb{R}$,
be defined by
\begin{equation}
\mathcal{H}(x,y) =\sum_{n=0}^{\infty }
(-1) ^{n}G_{n}(x,y),  \label{e3.1}
\end{equation}
provided that the series converges.

\begin{lemma}\label{lem3.1}
Let $p\in \mathcal{J}_{\alpha }$ with $\tau _p<1$, then for
all $(x,y) \in [ 0,1]\times [ 0,1]$, we have
\begin{itemize}
\item[(i)] $G_{n}(x,y)\leq \tau _p^{n}G(x,y)$ for each $n\in \mathbb{N}$.
So, $\mathcal{H}(x,y) $ is well defined in $
[0,1]\times [ 0,1]$.

\item[(ii)] For each $n\in \mathbb{N}$,
\begin{equation}
L_{n}x^{\alpha -1}y(1-y) ^{\alpha -3}\leq G_{n}(x,y)\leq
R_{n}x^{\alpha -2}y(1-y) ^{\alpha -3},  \label{e3.3}
\end{equation}
where
\begin{gather*}
L_{n}=\frac{1}{(\Gamma (\alpha ))^{n+1}}\Big(\int_0^{1}t^{
\alpha }(1-t)^{\alpha -3}p(t)dt\Big)^{n},\\
R_{n}=(\frac{2\alpha -2}{\Gamma
(\alpha )})^{n+1}\Big(\int_0^{1}t^{\alpha -1}(1-t)^{\alpha
-3}p(t)dt\Big)^{n}.
\end{gather*}

\item[(iii)] $G_{n+1}(x,y)=\int_0^{1}G_{n}(x,t)G(t,y)p(t)dt$ for each
$n\in \mathbb{N}$.

\item[(iv)] $\int_0^{1}\mathcal{H}(x,t) G(t,y)p(t)dt=\int_0^{1}G(x,t) 
\mathcal{H}(t,y)p(t)dt$.
\end{itemize}
\end{lemma}

\begin{proof}
(i) Obviously the inequality is valid for $n=0$.
Assume that $G_{n}(x,y)\leq \tau _p^{n}G(x,y)$, then by using 
\eqref{e3.2} and \eqref{e1.5}, we obtain
\begin{equation*}
G_{n+1}(x,y)\leq \tau _p^{n}\int_0^{1}G(x,t)G(t,y)p(t)dt
\leq \tau _p^{n+1}G(x,y).
\end{equation*}
 So, $\mathcal{H}(x,y) $ is well defined in $[0,1]\times [ 0,1]$.

(ii) The inequality in \eqref{e3.3}, follows by induction and
Proposition \ref{prop2.6} (iii).

\item[(iii)] We will proceed by induction. Obviously the equality is
valid for $n=0$.
Assume that
\begin{equation}
G_{n}(x,y)=\int_0^{1}G_{n-1}(x,t)G(t,y)p(t)dt.  \label{e3.4}
\end{equation}
Then by using \eqref{e3.2} and the Fubini-Tonelli theorem, we
obtain
\begin{align*}
G_{n+1}(x,y) 
&= \int_0^{1}G(x,t)\Big(\int_0^{1}G_{n-1}(t,z)G(z,y)p(z)dz\Big) p(t)dt \\
&= \int_0^{1}\Big(\int_0^{1}G(x,t)G_{n-1}(t,z)p(t)dt\Big) G(z,y)p(z)dz \\
&= \int_0^{1}G_{n}(x,z)G(z,y)p(z)dz.
\end{align*}
(iv) Let $n\in \mathbb{N}$ and $x,t,y\in [ 0,1]$. From
Lemma \ref{lem3.1} (i) we deduce that
\begin{equation*}
0\leq G_{n}(x,t)G(t,y)p(t)\leq 
\tau _p^{n}G(x,t)G(t,y)p(t).
\end{equation*}
So, the series $\sum_{n\geq 0} \int_0^{1}G_{n}(x,t)G(t,y)p(t)dt$ is convergent.
By the dominated convergence theorem and Lemma \ref{lem3.1} (iii),
we deduce that
\begin{align*}
\int_0^{1}\mathcal{H}(x,t) G(t,y)p(t)dt 
&= \sum_{n=0}^{\infty } \int_0^{1}(-1)
^{n}G_{n}(x,t)G(t,y)p(t)dt \\
&= \sum_{n=0}^{\infty } \int_0^{1}(-1) ^{n}G(x,t)G_{n}(t,y)p(t)dt \\
&= \int_0^{1}G(x,t) \mathcal{H}(t,y)p(t)dt.
\end{align*}
\end{proof}

\begin{proposition}\label{prop3.2}
For $p\in \mathcal{J}_{\alpha }$ with $\tau _p<1$, the
function $(x,y)\to \mathcal{H}(x,y) $ belongs to $C([0,1]\times [ 0,1]) $.
\end{proposition}

\begin{proof}
The function $(x,y)\to G_{n}(x,y)\in C([0,1]\times [0,1]) $, for all 
$n\in \mathbb{N}$.
Clearly $G_0=G\in C([0,1]\times [0,1]) $,

 Assume that the function $(x,y)\to G_{n-1}(x,y)\in
C([0,1]\times [ 0,1])$.
 Using Lemma \ref{lem3.1} (i) and Proposition \ref{prop2.6} (iii),
we obtain
\begin{align*}
G(x,t)G_{n-1}(t,y)p(t) 
&\leq \tau _p^{n-1}G(x,t)G(t,y)p(t) \\
&\leq 4(\frac{\alpha -1}{\Gamma (\alpha )})^{2}t^{\alpha -1}(1-t)^{\alpha
-3}p(t).
\end{align*}
Therefore by \eqref{e3.2} and the dominated convergence theorem,
we deduce that the function $(x,y)\to G_{n}(x,y)\in C([0,1]\times
[ 0,1])$.

 On the other hand, from Lemma \ref{lem3.1} (i) and Proposition 
\ref{prop2.6} (iii), we have
\begin{equation*}
G_{n}(x,y)\leq \tau _p^{n}G(x,y)\leq \frac{2(\alpha -1)}{\Gamma (\alpha )}
\tau _p^{n}.
\end{equation*}
So, the series $\sum_{n\geq 0} (-1)^{n}G_{n}(x,y)$ is uniformly convergent 
on $[0,1]\times [ 0,1]$ and
therefore the function $(x,y)\to \mathcal{H}(x,y) $
belongs to $C([0,1]\times [ 0,1]) $.
\end{proof}

\begin{lemma} \label{lem3.3}
Let $p\in \mathcal{J}_{\alpha }$ such that $\tau _p\leq 1/2$. 
On $[0,1]\times [ 0,1]$, one has
\begin{equation}
(1-\tau _p) G(x,y) \leq \mathcal{H}(x,y) \leq G(x,y) .  \label{e3.5}
\end{equation}
\end{lemma}

\begin{proof}
By using Lemma \ref{lem3.1} (i), we obtain
\begin{equation}
| \mathcal{H}(x,y) | \leq \sum_{n=0}^{\infty } (\tau _p) ^{n}G(x,y) =
\frac{1}{1-\tau _p}G(x,y) .  \label{e3.6}
\end{equation}
On the other hand, we have
\begin{equation}
\mathcal{H}(x,y) =G(x,y) -\sum_{n=0}^{\infty } (-1) ^{n}G_{n+1}(x,y).  \label{e3.7}
\end{equation}
Since the series $\sum_{n\geq 0}\int_0^{1}G(x,z)G_{n}(z,y)p(z)dz$ 
is convergent,  we deduce by \eqref{e3.7}
 and \eqref{e3.2} that
\begin{align*}
\mathcal{H}(x,y) 
&= G(x,y) -\sum_{n=0}^{\infty } (-1) ^{n}\int_0^{1}G(x,z)G_{n}(z,y)p(z)dz \\
&= G(x,y) -\int_0^{1}G(x,z)\Big(\sum_{n=0}^{\infty } (-1) ^{n}G_{n}(z,y)\Big)p(z)dz;
\end{align*}
 that is,
\begin{equation}
\mathcal{H}(x,y) =G(x,y) -U(p\mathcal{H}(.,y) ) (x) .  \label{e3.8}
\end{equation}
 Using \eqref{e3.6} and Lemma \ref{lem3.1} (i), we obtain
\begin{equation*}
U(p\mathcal{H}(.,y) ) (x) 
\leq \frac{1}{ 1-\tau _p}U(pG(.,y) ) (x) 
=\frac{1}{1-\tau _p}G_1(x,y)\leq \frac{\tau _p}{1-\tau _p}G(x,y) .
\end{equation*}
So, by \eqref{e3.8}, we obtain
\begin{equation*}
\mathcal{H}(x,y) \geq G(x,y) -\frac{\tau _p}{
1-\tau _p}G(x,y) =\frac{1-2\tau _p}{1-\tau _p}G(x,y) \geq 0.
\end{equation*}
So, $\mathcal{H}(x,y) \leq G(x,y) $ and by
\eqref{e3.8} and Lemma \ref{lem3.1} (i), we obtain
\begin{equation*}
\mathcal{H}(x,y) \geq G(x,y) -U(pG(.,y) ) (x) \geq (1-\tau _p) G(x,y) .
\end{equation*}
\end{proof}

\begin{corollary}\label{coro3.4}
Let $p\in \mathcal{J}_{\alpha }$ with $\tau _p\leq \frac{1}{2}$ and
 $\psi \in \mathcal{B}^{+}((0,1)) $.
Then
\begin{equation*}
U_p\psi \in C([0,1])\Longleftrightarrow \int_0^{1}y(1-y)
^{\alpha -3}\psi (y) dy<\infty .
\end{equation*}
\end{corollary}

\begin{proof}
The assertion follows from Proposition \ref{prop3.2}, \eqref{e3.5} and
Proposition \ref{prop2.6} (iii).
\end{proof}

\begin{lemma}\label{lem3.5}
Let $p\in \mathcal{J}_{\alpha }$ with $\tau _p\leq \frac{1}{2}$ and 
$\psi \in \mathcal{B}^{+}((0,1)) $. Then we have
\begin{equation}
U\psi =U_p\psi +U_p(pU\psi ) =U_p\psi +U(pU_p\psi) .  \label{e3.9}
\end{equation}
In particular, if $U(p\psi )<\infty $, then
\begin{equation}
(I-U_p(p.) )(I+U(p.) )\psi =(I+U(p.) )(I-U_p(p.) )\psi =\psi .  \label{e3.10}
\end{equation}
Here $U(p.) (\psi ):=U(p\psi ) $.
\end{lemma}

\begin{proof}
From \eqref{e3.8}, we have
\begin{equation*}
G(x,y) =\mathcal{H}(x,y) +U(p\mathcal{H}(\cdot,y) ) (x) ,\quad
\text{for }(x,y)\in [ 0,1]\times [ 0,1].
\end{equation*}
 So by the Fubini-Tonelli theorem we deduce that
\begin{align*}
U\psi (x) 
&= \int_0^{1}(\mathcal{H}(x,y) +U(p
\mathcal{H}(.,y) ) (x) ) \psi (y)dy \\
&= U_p\psi (x)+U(pU_p\psi ) (x).
\end{align*}
Now, using Lemma \ref{lem3.1} (iv) and again the Fubini theorem,
we obtain
\begin{equation*}
\int_0^{1}\int_0^{1}\mathcal{H}(x,t) G(t,y)p(t)\psi (y)\,dt\,dy
=\int_0^{1}\int_0^{1}G(x,t) \mathcal{H}(t,y)p(t)\psi (y)\,dt\,dy;
\end{equation*}
that is,
\begin{equation*}
U_p(pU\psi ) (x)=U(pU_p\psi ) (x).
\end{equation*}
Therefore
\begin{equation*}
U\psi =U_p\psi +U(pU_p\psi ) =U_p\psi +U_p(pU\psi
) (x).
\end{equation*}
\end{proof}

\begin{proposition}\label{prop3.6}
Let $\psi \in \mathcal{B}^{+}((0,1)) $ such that 
$y\mapsto y(1-y)^{\alpha -3}\psi (y)$ belongs to
$C((0,1))\mathcal{\cap }L^{1}((0,1))$ and 
$p\in C((0,1))\mathcal{\cap J}_{\alpha }$ with 
$\tau_p\leq \frac{1}{2}$. Then $u=U_p\psi $ is the unique nonnegative
solution in $C([0,1])$ to problem \eqref{e3.p} satisfying
\begin{equation}
(1-\tau _p) U\psi \leq u\leq U\psi .  \label{e3.11}
\end{equation}
\end{proposition}

\begin{proof}
By Corollary \ref{coro3.4}, we conclude that the function 
$x\to p(x)U_p\psi (x) \in C((0,1)) $.

 On the other hand, from \eqref{e3.9} and Proposition \ref{prop2.6}
(iii), we have that there exists $m\geq 0$ such that
\begin{equation}
U_p\psi (x)\leq U\psi (x)\leq \frac{2(\alpha -1)}{\Gamma (\alpha )}
\int_0^{1}x^{\alpha -2}y(1-y)^{\alpha -3}\psi (y)dy\equiv mx^{\alpha -2}.
\label{e3.12}
\end{equation}
Therefore
\begin{equation*}
\int_0^{1}y(1-y)^{\alpha -3}p(y)U_p\psi (y)dy\leq m\int_0^{1}y^{\alpha
-1}(1-y)^{\alpha -3}p(y)dy<\infty .
\end{equation*}
By applying Proposition \ref{prop2.8}, the function $u=U_p\psi
=U\psi -U(pU_p\psi ) $ satisfies the equation
\begin{gather*}
D^{\alpha }u(x)=-\psi (x)+p(x)u(x),\quad x\in (0,1), \\
u(0)=\lim_{x\to0^{+}} D^{\alpha -3}u(x)
=\lim_{x\to 0^{+}} D^{\alpha -2}u(x)=u''(1)=0.
\end{gather*}
Integrating the inequalities \eqref{e3.5}, we obtain \eqref{e3.11}.

 Next, we prove the uniqueness. Let $w\in C^{+}([0,1])$ be another
solution to problem $\eqref{e3.p}$ satisfying $w\leq U\psi $.
 Since by \eqref{e3.11} and \eqref{e3.12} the function 
$y\to y(1-y)^{\alpha -3}p(y)w(y)\in C((0,1))\mathcal{\cap }
L^{1}((0,1))$,  by Proposition \ref{prop2.8} the function $\widetilde{w}
:=w+U(pw)$ satisfies
\begin{gather*}
D^{\alpha }\widetilde{w}(x)+\psi (x)=0,\quad x\in (0,1), \\
\widetilde{w}(0)=\lim_{x\to0^{+}}  D^{\alpha -3}\widetilde{w}(x)
=\lim_{x\to 0^{+}} D^{\alpha -2}\widetilde{w}(x)
=\widetilde{w}''(1)=0.
\end{gather*}
 From Proposition \ref{prop2.8} we deduce that
\begin{equation*}
\widetilde{w}:=w+U(pw)=U\psi .
\end{equation*}
Therefore,
\begin{equation*}
(I+U(p.))((w-u)^{+}) =(I+U(p.))((w-u)^{-}) ,
\end{equation*}
 where $(w-u)^{+}=\max (w-u,0)$ and $(w-u)^{-}=\max (u-w,0)$.

 Since $| w(y)-u(y)| \leq 2U\psi (y)\leq 2my^{\alpha -2}$,
 we deduce by Proposition \ref{prop2.6} (ii) that
\begin{align*}
U(p| w-u| )(x) 
&\leq \frac{4m(\alpha -1)}{\Gamma (\alpha
)}\int_0^{1}y^{\alpha -2}(1-y)^{\alpha -3}\min (x,y)p(y)dy \\
&\leq \frac{4m(\alpha -1)}{\Gamma (\alpha )}\int_0^{1}y^{\alpha
-1}(1-y)^{\alpha -3}p(y)dy<\infty .
\end{align*}
So by \eqref{e3.10}, we obtain that $u=w$.
\end{proof}

\subsection{Proofs of main results}


\begin{proof}[Proof of Theorem \ref{thm1.1}]
Let $3<\alpha \leq 4$ and $\xi ,\zeta \geq 0$ with $\xi +\zeta >0$.
We recall that
\begin{equation*}
h(x):=\frac{\xi }{\Gamma (\alpha )}x^{\alpha -2}(\alpha -1-(\alpha -3)x)+
\frac{\zeta }{(\alpha -1)(\alpha -2)}x^{\alpha -1},\quad \text{for }x\in [0,1].
\end{equation*}
Let $p\in C((0,1))\mathcal{\cap J}_{\alpha }$ with 
$\tau _p\leq \frac{1}{2}$ such that assumption (H2) is satisfied.
Let
\begin{equation*}
F:=\{ u\in \mathcal{B}^{+}((0,1)) :(1-\tau_p) h\leq u\leq h\} .
\end{equation*}
Consider the operator $A$ defined on $F$ by
\begin{equation*}
Au=h-U_p(ph) +U_p((p-g(.,u) ) u).
\end{equation*}
By \eqref{e3.9} and \eqref{e2.10} we have
\begin{equation}
U_p(ph)\leq U(ph) \leq \tau _ph\leq h,  \label{e3.13}
\end{equation}
\noindent and by (H2), we obtain
\begin{equation}
0\leq g(.,u)\leq p\quad \text{for all }u\in F.  \label{e3.14}
\end{equation}
Next, we prove that $A(F)\subset F$.
Form \eqref{e3.14} and \eqref{e3.13}, we obtain
\begin{gather*}
Au\leq h-U_p(ph) +U_p(pu)\leq h, \\
Au\geq h-U_p(ph) \geq (1-\tau _p) h.
\end{gather*}
 Since the map $y\mapsto y(p(x) -g(x,yh(x) ) ) $ is nondecreasing on $[ 0,1]$, 
for $x\in (0,1)$, the operator $A$ becomes nondecreasing on $F$.

 Define the sequence $\{ u_{k}\} $ by 
$u_0=(1-\tau _p) h$ and $u_{k+1}=Au_{k}$ for $k\in \mathbb{N}$. 
Since $F$ is invariant under $A$, we have $u_1=Au_0\geq $ $u_0$ and by the
monotonicity of $A$, we obtain
\begin{equation*}
(1-\tau _p) h=u_0\leq u_1\leq \dots \leq u_{k}\leq u_{k+1}\leq h.
\end{equation*}
Therefore, the sequence $\left\{ u_{k}\right\} $ converges to a
function $u\in F$ satisfying
\begin{equation*}
u=(I-U_p(p.) ) h+U_p((p-g(.,u)) u).
\end{equation*}
Namely
\begin{equation}
(I-U_p(p.) ) u=(I-U_p(p.)) h-U_p(ug(.,u) ) .  \label{e3.14a}
\end{equation}
 Now, since $U(pu) \leq U(ph) \leq h<\infty$,  by applying the operator
 $(I+U(p.) ) $ on \eqref{e3.14a} and using \eqref{e3.9} and \eqref{e3.10},
 we conclude that $u$ satisfies
\begin{equation}
u=h-U(ug(.,u) ) .  \label{e3.15}
\end{equation}
We claim that $u$ is a solution of $\eqref{e1.3}$.
From \eqref{e3.14} and \eqref{e1.8}, we have
\begin{equation}
u(y)g(y,u(y)) \leq p(y)h(y)\leq Mp(y)\phi (y)\leq My^{\alpha
-2}p(y).  \label{e3.16}
\end{equation}
This implies that $\int_0^{1}y(1-y)^{\alpha -3}u(y)g(y,u(y)) dy<\infty $.
 Hence from Corollary \ref{coro2.7},  we deduce that the function 
$x\mapsto U(ug(.,u) ) (x)\in C([0,1])$ and from
\eqref{e3.15}, we conclude that $u\in C([0,1])$.

Using (H1) and \eqref{e3.16}, we obtain that the function 
$y\mapsto y(1-y)^{\alpha -3}u(y)g(y,u(y))$ belongs to
$C((0,1)) \mathcal{\cap }L^{1}((0,1)))$, which implies by 
Proposition \ref{prop2.8}
that $u$ is a solution of \eqref{e1.3}.

 Now assume further that condition (H3) is
satisfied. Let $v\in C([0,1])$ be another nonnegative solution to problem 
\eqref{e1.3} satisfying \eqref{e1.9}.
As above, we have
\begin{equation*}
0\leq v(y)g(y,v(y)) \leq p(y)h(y)\leq My^{\alpha -2}p(y).
\end{equation*}
So the function $y\to y(1-y)^{\alpha -3}v(y)g(
y,v(y)) \in C((0,1))\mathcal{\cap }L^{1}((0,1)))$.
Put $\widetilde{v}:=v+U(vg(.,v) ) $. By
Proposition \ref{prop2.8} we have
\begin{gather*}
D^{\alpha }\widetilde{v}(x)=0,\quad 0<x<1, \\
u(0)=0,\quad \lim_{x\to0^{+}} D^{\alpha -3}u(x)=0,\quad 
\lim_{x\to0^{+}} D^{\alpha -2}u(x)=\xi ,\quad
u''(1)=\zeta .
\end{gather*}
 Hence
\begin{equation*}
\widetilde{v}:=v+U(vg(.,v) ) =h.
\end{equation*}
 That is,
\begin{equation}
v=h-U(vg(.,v) ) .  \label{e3.17}
\end{equation}
 For $z\in (0,1)$, we let
\begin{equation*}
\varrho (z)=\begin{cases}
\frac{v(z)g(z,v(z)) -u(z)g(z,u(z)) }{v(z)-u(z)}, &
\text{if }v(z)\neq u(z), \\
0, & \text{if }v(z)=u(z).
\end{cases}
\end{equation*}
Note that, from (H3), we have $\varrho \in \mathcal{B}^{+}((0,1)) $.
Using \eqref{e3.15} and \eqref{e3.17} we deduce
\begin{equation*}
(I+U(\varrho .))((v-u)^{+}) =(I+U(\varrho .))(
(v-u)^{-}) ,
\end{equation*}
where $(v-u)^{+}=\max (v-u,0)$ and $(v-u)^{-}=\max (u-v,0)$.
Since $\varrho \leq p$, we deduce by \eqref{e2.10}, that
\begin{equation*}
U(\varrho | v-u| )\leq 2U(ph)\leq 2\tau _ph<\infty .
\end{equation*}
Hence $u=v$ by \eqref{e3.10}. This ends the proof.
\end{proof}

\begin{proof}[Proof of Corollary \ref{coro1.2}]
The conclusion follows from Theorem \ref{thm1.1} with $g(x,y)
=\lambda q(x)\psi (y)$
and $p(x):=\lambda \widetilde{q}(x)$.
\end{proof}

\begin{example} \label{examp3.7} \rm
Let $3<\alpha \leq 4$ and  $\xi ,\zeta \geq 0$ with $\xi +\zeta >0$.
 Let $ r\geq 0$, $\nu \geq 0$ and $q\in C^{+}((0,1)) $
such that
\begin{equation*}
\int_0^{1}t^{(\alpha -1)+(\alpha -2)(r+\nu )}(1-t)^{\alpha
-3}q(t)dt<\infty .
\end{equation*}
Let $\varphi (s)=s^{r+1}\log (1+s^{\nu })$ and 
$\widetilde{q} (y):=q(y)\max_{0\leq t\leq h(y) } \varphi '(t)$. 
Since $\widetilde{q}\in \mathcal{J}_{\alpha }$, then for 
$\lambda \in [ 0,\frac{1}{2\tau _{\widetilde{q}}})$, the problem
\begin{gather*}
D^{\alpha }u(x)=\lambda q(x)u^{r+1}(x)\log (1+u^{\nu }(x)),\quad 0<x<1, \\
u(0)=0,\quad \lim_{x\to0^{+}} D^{\alpha -3}u(x)=0,\quad 
\lim_{x\to0^{+}} D^{\alpha -2}u(x)=\xi ,\quad
u''(1)=\zeta ,
\end{gather*}
admits a unique positive solution $u\in C([0,1])$ satisfying
\begin{equation*}
(1-\lambda \tau _{\widetilde{q}})h(x)\leq u(x) \leq h(x),\quad
\text{for all }0\leq x\leq 1.
\end{equation*}
\end{example}

\subsection*{Acknowledgments}
 The authors would like to extend their
sincere appreciation to the Deanship of Scientific Research at King Saud
University for its funding this Research group No RG-1435-043.

\begin{thebibliography}{99}

\bibitem{AOS} R. P. Agarwal, D. O'Regan, S. Stan\u{e}k;
\emph{Positive solutions for Dirichlet problems of singular nonlinear 
fractional differential equations}, J. Math. Anal. Appl., 371 (2010), 57-68.

\bibitem{BMR} I. Bachar, M. M\^{a}agli, V. D. R\u{a}dulescu;
\emph{Fractional Navier boundary value problems},
 Bound Value Probl 2016:79 (2016)  DOI 10.1186/s13661-016-0586-7

\bibitem{BMTZ} I. Bachar, M. M\^{a}agli, F. Toumi, Z. Zine el Abidine;
\emph{Existence and Global Asymptotic Behavior of Positive Solutions for Sublinear
and Superlinear Fractional Boundary Value Problems}, Chinese Annals of
Mathematics, Series B, 37 (2016), no. 1, 1-28.

\bibitem{CHS} J. Caballero Mena, J. Harjani, K. Sadarangani;
\emph{Existence and uniqueness of positive and nondecreasing solutions for a class 
of singular fractional boundary value problems}. Bound Value 
Probl 2009:421310 (2009) DOI 10.1155/2009/421310.

\bibitem{GMS} J. Giacomoni, P. Kumar Mishra, K. Sreenadh;
\emph{ Fractional elliptic equations with critical exponential nonlinearity}, 
Adv. Nonlinear Anal., 5 (2016), 57-74.

\bibitem{H} R. Hilfer;
\emph{Applications of Fractional Calculus in Physics}, World
Scientific, Singapore, 2000.

\bibitem{KST} A. Kilbas, H. Srivastava, J. Trujillo;
\emph{Theory and Applications of Fractional Differential Equations}, 
in: North-Holland Mathematics studies, Vol. 204, Elsevier, Amsterdam, 2006.

\bibitem{LZ} S. Liang, J. Zhang;
\emph{Positive solutions for boundary value problems of nonlinear fractional 
differential equation}, Nonlinear Anal., 71
(2009), 5545--5550.

\bibitem{MT} R. Ma, C. C. Tisdell;
\emph{Positive solutions of singular sublinear
fourth order boundary value problems}, Appl. Anal., 84 (2005), no. 12,
1199--1220.

\bibitem{MMZ} H. M\^{a}agli, N. Mhadhebi, N. Zeddini;
\emph{Existence and Estimates of Positive Solutions for Some Singular 
Fractional Boundary Value Problems}, Abstract and Applied Analysis 
Vol. 2014 (2014), Article ID 120781, 7 pages.

\bibitem{MR} K. Miller, B. Ross;
\emph{An introduction to the Fractional Calculus
and Fractional Differential Equations}, Wiley, New York, 1993.

\bibitem{MRS} G. Molica Bisci, V. D. R\u{a}dulescu, R. Servadei;
\emph{Variational Methods for Nonlocal Fractional Problems}, 
Encyclopedia of Mathematics and its Applications, Vol. 162, 
Cambridge University Press, Cambridge, 2016.

\bibitem{MRe} G. Molica Bisci, D. Repov\v{s};
\emph{Existence and localization of solutions for nonlocal fractional equations}, 
Asymptot. Anal. 90 (2014), no. 3-4, 367--378.

\bibitem{P} I. Podlubny;
\emph{Fractional Differential Equations}, in:
Mathematics in Sciences and Engineering, vol. 198, Academic Press, San
Diego, 1999.

\bibitem{PXZ} P. Pucci, M. Xiang, B. Zhang;
\emph{Existence and multiplicity of entire solutions for fractional p-Kirchhof 
equations}, Adv. Nonlinear Anal., 5 (2016), 27-55.

\bibitem{R} M. Ro\c{s}iu;
\emph{Trajectory structure near critical points}. An.
Univ. Craiova Ser. Mat. Inform. 25 (1998), 35-44.

\bibitem{SKM} S. Samko, A. Kilbas, O. Marichev;
\emph{Fractional Integrals and derivative}. Theory and applications. 
Gordon and Breach, Yverdon, 1993.

\bibitem{T} V. Tarasov;
\emph{Fractional Dynamics: Applications of Fractional
Calculus to Dynamics of Particles, Fields and Media}, Springer-Verlag, New
York, 2011.

\bibitem{XCW} Z. Xiangkui, C. Chengwen, G. Weigao;
\emph{Existence and nonexistence results for a class of fractional boundary 
value problems}, J Appl Math Comput, 41 (2013), 17-31.

\bibitem{ZYY} C. Zhai, W. Yan, C. Yang;
\emph{A sum operator method for the existence and uniqueness of positive 
solutions to Riemann-Liouville fractional differential equation boundary 
value problems}, Commun Nonlinear Sci Numer Simul, 18 (2013), 858-866.

\bibitem{ZSHL} Y. Zhao, S. Sun, Z. Ha, Q. Li;
\emph{The existence of multiple positive solutions for boundary value problems
 of nonlinear fractional differential equations}, Commun Nonlinear Sci Numer Simul, 
16 (2011), 2086-2097.

\end{thebibliography}

\end{document}