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

\AtBeginDocument{{\noindent\small
\emph{Electronic Journal of Differential Equations},
Vol. 2013 (2013), No. 60, pp. 1--10.\newline
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu
\newline ftp ejde.math.txstate.edu}
\thanks{\copyright 2013 Texas State University - San Marcos.}
\vspace{9mm}}

\begin{document}
\title[\hfilneg EJDE-2013/60\hfil Nonlocal fractional boundary-value problems]
{Integro-differential equations of fractional order
with  nonlocal fractional boundary conditions associated with
financial asset model}

\author[B. Ahmad, S. K. Ntouyas \hfil EJDE-2013/60\hfilneg]
{Bashir Ahmad, Sotiris K. Ntouyas}  % in alphabetical order

\address{Bashir Ahmad  \newline
Department of Mathematics, Faculty of Science, King Abdulaziz
University, P.O. Box 80203, Jeddah 21589, Saudi Arabia}
\email{bashirahmad\_qau@yahoo.com}

\address{Sotiris K. Ntouyas \newline
Department of Mathematics,  University of
Ioannina, 451 10 Ioannina, Greece}
\email{sntouyas@uoi.gr}

\thanks{Submitted December 6, 2012. Published February 26, 2013.}
\subjclass[2000]{34A08, 34B10, 34B15}
\keywords{Fractional differential equations;
 integral boundary conditions;\hfill\break\indent
 existence; fixed point theorems; financial asset}

\begin{abstract}
 In this article, we discuss the existence of solutions for a
 boundary-value problem of integro-differential equations of
 fractional order with  nonlocal fractional boundary conditions
 by means of some standard tools of fixed point theory.
 Our problem describes a more general form of  fractional stochastic
 dynamic model for financial asset. An illustrative example is also presented.
\end{abstract}

\maketitle
\numberwithin{equation}{section}
\newtheorem{theorem}{Theorem}[section]
\newtheorem{lemma}[theorem]{Lemma}
\newtheorem{remark}[theorem]{Remark}
\allowdisplaybreaks

\section{Formulation and basic result}

Fractional calculus, regarded as a branch of mathematical analysis
dealing with derivatives and integrals of arbitrary order, has
been extensively developed and applied to a variety of problems
appearing in sciences and engineering. It is worthwhile to mention
that this branch of mathematics has played a crucial role in
exploring various characteristics of engineering materials such as
viscoelastic polymers, foams, gels, and animal tissues, and their
engineering and scientific applications. For a recent detailed
survey of the activities involving fractional calculus, we refer a
recent paper by  Machado,  Kiryakova and  Mainardi \cite{mach}.
Some recent work on the topic can be found in 
\cite{Ravi, Agha,  bvp1, BNi, BA5,  ahmad,  Kli,  CuevS, Ford,
Kirane1, Kirane2,  S-T} and references therein.


The underlying dynamics of equity prices following a jump process
or a Levy process provide a basis for modeling of financial
assets.  The CGMY, KoBoL and FMLS are examples of some interesting
financial models involving the dynamics of stock prices. In
\cite{car}, it is shown that the prices of financial derivatives
are expressible in terms of fractional derivative.

In \cite{las}, the author described the dynamics of a financial
asset by the fractional stochastic differential equation of order
$\mu$ (representing the dynamical memory effects in the market
stochastic evolution)  with fractional boundary conditions. In the
present paper, we study a more general model associated with
financial asset. Precisely, we consider the following problem:
\begin{equation}\label{p} 
\begin{gathered}
-D^{\alpha}x(t)= Af(t,  x(t))+BI^{\beta}g(t,  x(t)),
\quad (n-1)<\alpha\leq n, \; t\in [0,1],\\
 D^\delta x(0)=0, \quad D^{\delta+1}x(0)=0, \dots,\;
D^{\delta+(n-2)}x(0)=0, \quad  
D^\delta x(1)=\int_0^\eta D^\delta x(s)ds,
\end{gathered}
\end{equation}
where  $0<\delta\le 1$, $\alpha-\delta >n$, $0<\beta<1$,
$0<\eta<1$, $D^{(\cdot)}$ denotes the Riemann-Lioville fractional
derivative of order $(\cdot)$, $f, g$ are
given continuous function, and $A,B$ are real constants.

We remark that the problem \eqref{p} also arises in
real estate asset securitization modeling \cite{tao}.

By the substitution
$x(t)=I^{\delta}y(t)=D^{-\alpha}y(t)$, the problem \eqref{p} takes
the form
\begin{equation}\label{21}
\begin{gathered}
-D^{\alpha-\delta}y(t)= Af(t,I^{\delta}y(t))+BI^{\beta}g(t, I^{\delta}y(t)), 
\quad t\in [0,1], \\
y(0)=0, \quad y'(0)=0,\dots, y^{(n-2)}(0)=0,\quad  y(1)=\int_0^\eta
y(s)ds.
\end{gathered}
\end{equation}

\begin{lemma}\label{22}
 For any $h \in C(0,1)\cap L(0, 1)$, the unique
solution of the linear fractional boundary-value problem
\begin{equation}\label{11}
\begin{gathered}
-D^{\alpha-\delta}y(t)= h(t), \quad t\in [0,1], \\
y(0)=0, \quad y'(0)=0, \dots , y^{(n-2)}(0)=0, \quad  y(1)=\int_0^\eta
y(s)ds,
\end{gathered}
\end{equation}
is 
$$
y(t)=-I^{\alpha-\delta} h(t)+\frac{(\alpha-\delta)
t^{\alpha-\delta-1}}{\alpha-\delta-\eta^{\alpha-\delta}}
\Big(I^{\alpha-\delta}h(1)-I^{\alpha-\delta+1}h(\eta)\Big),
 $$
where $I^{(\cdot)}(\cdot)$ denotes Riemann-Liouville integral.
\end{lemma}

\begin{proof} 
It is well known that the solutions of fractional
differential equation in \eqref{22} can be written as 
\begin{equation}\label{sol} 
y(t)= -I^{\alpha-\delta}
h(t)+c_1t^{\alpha-\delta-1}+c_2t^{\alpha-\delta-2}+c_3t^{\alpha-\delta-3}
+\dots+c_nt^{\alpha-\delta-n},
\end{equation}
where $c_1, c_2, \dots ,c_n \in \mathbb{R}$ are arbitrary
constants \cite{Kil}. Using the given boundary conditions, we find that 
$c_2=0$, $c_3=0,\dots,c_n=0$ and
$$
c_1=\frac{\alpha-\delta}{\alpha-\delta-\eta^{\alpha-\delta}}
\Big(I^{\alpha-\delta}h(1)-I^{\alpha-\delta+1}h(\eta)\Big).
$$
Substituting these values in \eqref{22} yields
$$
y(t)=-I^{\alpha-\delta} h(t)+\frac{(\alpha-\delta)
t^{\alpha-\delta-1}}{\alpha-\delta-\eta^{\alpha-\delta}}
\Big(I^{\alpha-\delta}h(1)-I^{\alpha-\delta+1}h(\eta)\Big). 
$$
This completes the proof. 
\end{proof}

 Thus, the solution of the linear variant of the
problem \eqref{p} can be written as
\begin{align*}
x(t)& = I^{\delta}y(t)\\&= I^{\delta}\Big[-I^{\alpha-\delta}
h(t)+\frac{(\alpha-\delta)
 t^{\alpha-\delta-1}}{\alpha-\delta-\eta^{\alpha-\delta}}
 \Big(I^{\alpha-\delta}h(1)-I^{\alpha-\delta+1}h(\eta)\Big)\Big]\\
&= -I^{\alpha}h(t)+ \frac{(\alpha-\delta)
 }{\alpha-\delta-\eta^{\alpha-\delta}}
 \Big(I^{\alpha-\delta}h(1)-I^{\alpha-\delta+1}h(\eta)\Big)
\int_0^t\frac{(t-s)^{\delta-1}}{\Gamma(\delta)}s^{\alpha-\delta-1}ds
\\
&= -I^{\alpha}h(t)+ \frac{(\alpha-\delta)
 }{\alpha-\delta-\eta^{\alpha-\delta}}
 \Big(I^{\alpha-\delta}h(1)-I^{\alpha-\delta+1}h(\eta)\Big)\times\\
& \quad\times\Big\{\frac{t^{\alpha-1}}{\Gamma(\delta)}
 \int_0^1(1-\nu)^{\delta-1}\nu^{\alpha-\delta-1}d\nu\Big\},
\end{align*}
where we have used the substitution $s= \nu t$ in the integral of
the last term. Using the relation for Beta function
$B(\cdot,\cdot)$:
$$
B(\beta+1, \alpha)=\int_0^1 (1-u)^{\alpha-1}u^{\beta}du
=\frac{\Gamma(\alpha)\Gamma(\beta+1)}{\Gamma(\alpha+\beta+1)},
$$
we obtain
\begin{equation}\label{vlin}
x(t)= -I^{\alpha}h(t)+ \frac{\Gamma(\alpha-\delta+1)t^{\alpha-1}
 }{(\alpha-\delta-\eta^{\alpha-\delta})\Gamma(\alpha)}
\Big(I^{\alpha-\delta}h(1)-I^{\alpha-\delta+1}h(\eta)\Big).
\end{equation}
The solution of the original nonlinear problem \eqref{p} can be
obtained by replacing $h$ with the right hand side of the
fractional
equation of \eqref{p} in \eqref{vlin}.


Let ${\mathcal{C}}=C([0,1], {\mathbb R})$ denote the Banach space of
all continuous functions from $[0,1] \to {\mathbb R}$ endowed with
the  norm  defined by $\|x\|= \sup \{|x(t)|, t \in [0,1]\}$.

In relation to problem \eqref{p}, we define an operator
$\mathcal{U}: {\mathcal{C}} \to {\mathcal{C}}$ as
\begin{align*}
&(\mathcal{U} x)(t)\\
&= -A\int_0^t \frac{(t-s)^{\alpha-1}}{\Gamma(\alpha)}f(s,
 x(s))ds-B\int_0^t
\frac{(t-s)^{\alpha+\beta-1}}{\Gamma(\alpha+\beta)}g(s,
 x(s))ds\\
&\quad + Q t^{\alpha-1}\Big[A\int_0^1
\frac{(1-s)^{\alpha-\delta-1}}{\Gamma(\alpha-\delta)}f(s,
x(s))ds+B\int_0^1
\frac{(1-s)^{\alpha-\delta+\beta-1}}{\Gamma(\alpha-\delta+\beta)}g(s,
 x(s))ds\\
&\quad - A \int_0^\eta
\frac{(\eta-s)^{\alpha-\delta}}{\Gamma(\alpha-\delta+1)}f(s,
 x(s))ds-B\int_0^\eta
\frac{(\eta-s)^{\alpha-\delta+\beta}}{\Gamma(\alpha-\delta+\beta+1)}g(s,
 x(s))ds\Big],
\end{align*}
where
$$
Q=\frac{\Gamma(\alpha-\delta+1)}{(\alpha-\delta-\eta^{\alpha-\delta})
\Gamma(\alpha)}, \quad \alpha \ne \delta+\eta^{\alpha-\delta}.
$$
For the sake of convenience, we set
\begin{equation}\label{omega}
\begin{aligned}
\Omega
&=\sup_{t \in [0,1]}\Big\{|A|\Big[\frac{t^\alpha}{\Gamma(\alpha+1)}
+|Q|t^{\alpha-1}\Big(\frac{1}{\Gamma(\alpha-\delta+1)}
+\frac{\eta^{\alpha-\delta+1}}{\Gamma(\alpha-\delta+2)}\Big)\Big]
\\ 
 &\quad +|B|\Big[\frac{t^{\alpha+\beta}}{\Gamma(\alpha+\beta+1)}
 +|Q|t^{\alpha-1}\Big(\frac{1}{\Gamma(\alpha-\delta+\beta+1)}
+\frac{\eta^{\alpha-\delta+\beta+1}}{\Gamma(\alpha-\delta+\beta+2)}
\Big)\Big]\Big\}.
\end{aligned}
\end{equation}

\subsection{Existence results via  Banach's fixed point theorem}

\begin{theorem}\label{t1} 
 Assume that $f, g : [0,1]\times \mathbb{R} \to
\mathbb{R}$ are continuous functions satisfying the condition:
\begin{itemize}
\item[(A1)] 
$|f(t,x)-f(t,y)| \le L_1 |x-y|$,
$|g(t,x)-g(t,y)| \le L_2 |x-y|$, for all $t \in [0,1]$, 
$L_1,L_2>0$, $x, y\in \mathbb{R}$.
 \end{itemize}
Then the boundary-value problem \eqref{p} has a unique solution if
 $L  < 1/\Omega$,
where $L=\max\{L_1, L_2\} $ and $\Omega$ is given by
\eqref{omega}.
\end{theorem}

\begin{proof}
Let us define $M=\max\{M_1, M_2\}$, where $M_1, M_2$ are finite numbers 
given by  $\sup_{t \in [0,1]}|f(t,0)|=M_1,\,\sup_{t \in [0,1]}|g(t,0)|=M_2$.
 Selecting $ r \ge \frac{\Omega M}{1-L\Omega}$, we show that
$\mathcal{U} B_r \subset B_r$, where 
$B_r=\{x \in {\mathcal{C}}: \|x\|\le r \}$. 
Using that $|f(s,x(s))\le |f(s,x(s))-f(s,0)|+|f(s,0)|\le L_1r+M_1$, 
$|g(s,x(s))| \le |g(s,x(s))-g(s,0)|+|g(s,0)|\le L_2r+M_2$ for 
$x \in B_r$ and \eqref{omega},  it can easily be shown that
\begin{align*}
&\|(\mathcal{U}x)\|\\
& \le (Lr+M )\sup_{t \in
[0,1]}\Big\{|A|\Big[\frac{t^\alpha}{\Gamma(\alpha+1)}+|Q|t^{\alpha-1}\Big(\frac{1}{\Gamma(\alpha-\delta+1)}
+\frac{\eta^{\alpha-\delta+1}}{\Gamma(\alpha-\delta+2)}\Big)\Big]
\\ 
&\quad +|B|\Big[\frac{t^{\alpha+\beta}}{\Gamma(\alpha+\beta+1)}
 +|Q|t^{\alpha-1}\Big(\frac{1}{\Gamma(\alpha-\delta+\beta+1)}
+\frac{\eta^{\alpha-\delta+\beta+1}}{\Gamma(\alpha-\delta+\beta+2)}
 \Big)\Big]\Big\}
\\
&= (Lr+M)\Omega \le r,
\end{align*}
which implies that $\mathcal{U} B_r \subset B_r$. Now, for
 $x, y \in {\mathcal{C}}$ we obtain
\begin{align*}
&\|\mathcal{U}x-\mathcal{U}y\| \\
& \le   \sup_{t\in
[0,1]}\Big\{|A|\int_0^t
\frac{(t-s)^{\alpha-1}}{\Gamma(\alpha)}|f(s,x(s))-f(s,y(s))|ds\\
 &\quad + |B|\int_0^t
\frac{(t-s)^{\alpha+\beta-1}}{\Gamma(\alpha+\beta)}|g(s,x(s))-g(s,y(s))|ds\\
 &\quad +|Q| t^{\alpha-1}\Big[|A|\int_0^1
\frac{(1-s)^{\alpha-\delta-1}}{\Gamma(\alpha-\delta)}|f(s,x(s))-f(s,y(s))|ds\\
 &\quad +|B|\int_0^1
\frac{(1-s)^{\alpha-\delta+\beta-1}}{\Gamma(\alpha-\delta+\beta)}|g(s,x(s))-g(s,y(s))|ds\\
 &\quad + |A| \int_0^\eta
\frac{(\eta-s)^{\alpha-\delta}}{\Gamma(\alpha-\delta+1)}|f(s,x(s))-f(s,y(s))|ds\\
 &\quad +|B|\int_0^\eta
\frac{(\eta-s)^{\alpha-\delta+\beta}}{\Gamma(\alpha-\delta+\beta+1)}|g(s,x(s))-g(s,y(s))|ds\Big]\Big\}
\\
 &\le L \sup_{t \in
[0,1]}\Big\{|A|\Big[\frac{t^\alpha}{\Gamma(\alpha+1)}+|Q|t^{\alpha-1}\Big(\frac{1}{\Gamma(\alpha-\delta+1)}
+\frac{\eta^{\alpha-\delta+1}}{\Gamma(\alpha-\delta+2)}\Big)\Big]
\\ 
&\quad +|B|\Big[\frac{t^{\alpha+\beta}}{\Gamma(\alpha+\beta+1)}+|Q|t^{\alpha-1}\Big(\frac{1}{\Gamma(\alpha-\delta+\beta+1)}
+\frac{\eta^{\alpha-\delta+\beta+1}}{\Gamma(\alpha-\delta+\beta+2)}\Big)\Big]
\Big\} \\
&\quad\times\|x-y\|\\
&= L \Omega \|x-y\|.
\end{align*}
By the given assumption,   $L<1/\Omega$. Therefore $\mathcal{U}$
is a contraction. Thus, the conclusion of the theorem follows by
the contraction mapping principle (Banach fixed point
theorem).
\end{proof}

 Now we present another variant
of existence-uniqueness result. This result is based on the
H\"older's inequality.

\begin{theorem}\label{t11}
Suppose that the continuous functions $f$ and $g$ satisfy the
following assumptions:
\begin{itemize}
\item [(H1)]  $|f(t,x)-f(t,y)|\le m(t)|x-y|$,
$|g(t,x)-g(t,y)|\le n(t)|x-y|$, for $t\in [0,1]$, 
$x,y\in {\mathbb R}$, and $m,n\in L^{\frac{1}{\gamma}}([0,1], {\mathbb R}^+)$,
$\gamma\in (0,\alpha-\delta-n)$. 

\item[(H2)] $|A|\|m\|Z_1+|B|\|n\|Z_2<1$, where
\begin{align*}
Z_1&= \frac{1}{\Gamma(\alpha)}
\Big(\frac{1-\gamma}{\alpha-\gamma}\Big)^{1-\gamma}
+\frac{|Q|}{\Gamma(\alpha-\delta)}
\Big(\frac{1-\gamma}{\alpha-\delta-\gamma}\Big)^{1-\gamma}
\\
&\quad+\frac{|Q|}{\Gamma(\alpha-\delta+1)}
\Big(\frac{1-\gamma}{\alpha-\delta+1-\gamma}\Big)^{1-\gamma}
 \eta^{\alpha-\delta+1-\gamma},
\end{align*}
\begin{align*}
Z_2&= \frac{1}{\Gamma(\alpha+\beta)}
\Big(\frac{1-\gamma}{\alpha+\beta-\gamma}\Big)^{1-\gamma}
+\frac{|Q|}{\Gamma(\alpha-\delta+\beta)}
\Big(\frac{1-\gamma}{\alpha-\delta+\beta-\gamma}\Big)^{1-\gamma}\\
&\quad +\frac{|Q|}{\Gamma(\alpha-\delta+\beta+1)}
\Big(\frac{1-\gamma}{\alpha-\delta+\beta+1-\gamma}
 \Big)^{1-\gamma}\eta^{\alpha-\delta+\beta+1-\gamma},
\end{align*}
\end{itemize}
and 
$\|\mu\|=\big(\int_0^1|\mu(s)|^{\frac{1}{\gamma}}ds\big)^{\gamma}$,
$\mu=m, n$. Then the boundary value problem \eqref{p} has a unique
solution.
\end{theorem}

\begin{proof}  
 For $x,y\in {\mathbb R}$ and for each $t\in [0,1]$, by H\"older inequality, 
we have
\begin{align*}
&\|\mathcal{U}x-\mathcal{U}y\| \\
&\le \sup_{t\in [0,1]}\Big\{|A|\int_0^t\frac{(t-s)^{\alpha-1}}
{\Gamma(\alpha)}m(s)|x(s)-y(s)|ds\\
&\quad +|B|\int_0^t\frac{(t-s)^{\alpha+\beta-1}}
 {\Gamma(\alpha+\beta)}n(s)|x(s)-y(s)|ds\\
&\quad +|Q|\Big[|A|\int_0^1\frac{(1-s)^{\alpha-\delta-1}}
 {\Gamma(\alpha-\delta)}m(s)|x(s)-y(s)|ds\\
&\quad +|B|\int_0^1\frac{(1-s)^{\alpha-\delta+\beta-1}}
 {\Gamma(\alpha-\delta+\beta)}n(s)|x(s)-y(s)|ds\\
&\quad +|A|\int_0^{\eta}\frac{(\eta-s)^{\alpha-\delta}}
 {\Gamma(\alpha-\delta+1)}m(s)|x(s)-y(s)|ds\\
&\quad +|B|\int_0^{\eta}\frac{(\eta-s)^{\alpha-\delta+\beta}}
 {\Gamma(\alpha-\delta+\beta+1)}n(s)|x(s)-y(s)|ds\Big]\Big\}\\
&\leq  \sup_{t\in [0,1]}\Big\{\frac{|A|\|m\|}{\Gamma(\alpha)}
 \Big(\frac{1-\gamma}{\alpha-\gamma}\Big)^{1-\gamma}t^{\alpha-\gamma}+
\frac{|B|\|n\|}{\Gamma(\alpha+\beta)}
 \Big(\frac{1-\gamma}{\alpha+\beta-\gamma}\Big)^{1-\gamma}
 t^{\alpha+\beta-\gamma}\\
&\quad +|Q|\Big[\frac{|A|\|m\|}{\Gamma(\alpha-\delta)}
 \Big(\frac{1-\gamma}{\alpha-\delta-\gamma}\Big)^{1-\gamma}
 +\frac{|B|\|n\|}{\Gamma(\alpha-\delta+\beta)}
 \Big(\frac{1-\gamma}{\alpha-\delta+\beta-\gamma}\Big)^{1-\gamma} \\
&\quad +\frac{|A|\|m\|}{\Gamma(\alpha-\delta+1)}
 \Big(\frac{1-\gamma}{\alpha-\delta+1-\gamma}\Big)^{1-\gamma}
 \eta^{\alpha-\delta+1-\gamma}\\
&\quad +\frac{|B|\|n\|}{\Gamma(\alpha-\delta+\beta+1)}
 \Big(\frac{1-\gamma}{\alpha-\delta+\beta+1-\gamma}\Big)^{1-\gamma}
 \eta^{\alpha-\delta+\beta+1-\gamma}\Big]\Big\}\|x-y\|\\
&\le |A|\|m\|\Big[\frac{1}{\Gamma(\alpha)}
 \Big(\frac{1-\gamma}{\alpha-\gamma}\Big)^{1-\gamma}
 +\frac{|Q|}{\Gamma(\alpha-\delta)}
 \Big(\frac{1-\gamma}{\alpha-\delta-\gamma}\Big)^{1-\gamma}\\
&\quad +\frac{|Q|}{\Gamma(\alpha-\delta+1)}
 \Big(\frac{1-\gamma}{\alpha-\delta+1-\gamma}\Big)^{1-\gamma}\Big]\|x-y\|\\
&\quad +|B|\|n\|\Big[\frac{1}{\Gamma(\alpha+\beta)}
 \Big(\frac{1-\gamma}{\alpha+\beta-\gamma}\Big)^{1-\gamma}
 +\frac{|Q|}{\Gamma(\alpha-\delta+\beta)}
 \Big(\frac{1-\gamma}{\alpha-\delta+\beta-\gamma}\Big)^{1-\gamma} \\
&\quad +\frac{|Q|}{\Gamma(\alpha-\delta+\beta+1)}
 \Big(\frac{1-\gamma}{\alpha-\delta+\beta+1-\gamma}\Big)^{1-\gamma}
 \eta^{\alpha-\delta+\beta+1-\gamma}\Big]\|x-y\|\\
&= [|A|\|m\|Z_1+|B|\|n\|Z_2]\|x-y\|.
\end{align*}
In view of condition (H2), it follows that  $\mathcal{U}$ is a
contraction mapping. Hence, Banach's fixed point theorem applies
and $\mathcal{U}$ has a unique fixed point which is the unique
solution of problem \eqref{p}. This completes the proof.
\end{proof}

\subsection{Existence result via  Leray-Schauder Alternative}

\begin{lemma}[Nonlinear alternative for single valued maps \cite{GrDu}] 
\label{lls}
Let $E$ be a Banach space, $C$ a closed, convex subset of $E$, $U$ 
an open subset of $C$ and $0\in U$. Suppose that $F:\overline{U}\to C$ 
is a continuous, compact (that is, $F(\overline{U})$ is a relatively 
compact subset of $C$) map. Then either
\begin{itemize}
\item[(i)] $F$ has a fixed point in $\overline{U}$, or 
\item[(ii)] there is a $u\in \partial U$ (the boundary of $U$ in $C$) and
$\lambda\in(0,1)$ with $u=\lambda F(u)$.
\end{itemize}
\end{lemma}

\begin{theorem}\label{tls}
 Assume that $f, g : [0,1]\times \mathbb{R} \to \mathbb{R}$ 
are continuous functions.   Assume that:
\begin{itemize}
 \item[(A3)] There exist   functions   $p_1, p_2 \in
L^1([0,1], {\mathbb R}^+)$, and   nondecreasing functions $\psi_1,
\psi_2: {\mathbb{R}}^+\to { \mathbb{R}}^+$  such that
\[
|f(t,x)|\le p_1(t)\psi_1(\|x\|),\quad
|g(t,x)|\le p_2(t)\psi_2(\|x\|),
\]
for all $(t,x) \in [0,1] \times {\mathbb R}$.

\item[(A4)] There exists a constant $M>0$ such that
$$
\frac{M}{
 |A|\Lambda_1\psi_1(M)\|p_1\|_{L^1}+|B|\Lambda_1\psi_2(M)\|p_2\|_{L^1}}> 1,
$$
where
\begin{gather*}
\Lambda_1= \frac{1}{\Gamma(\alpha+1)}+\frac{|Q|}{\Gamma(\alpha-\delta+1)}
+\frac{|Q|}{\Gamma(\alpha-\delta+2)}, \\
 \Lambda_2= \frac{1}{\Gamma(\alpha+\beta+1)}+\frac{|Q|}{\Gamma(\alpha-\delta+\beta+1)}
+\frac{|Q|}{\Gamma(\alpha-\delta+\beta+2)}.
\end{gather*}
\end{itemize}
 Then the boundary-value problem \eqref{p} has  at least one solution 
on $[0,1]$.
\end{theorem}

\begin{proof} 
Consider the operator   $\mathcal{U}:  \mathcal{C} \to  \mathcal{C}$ with
$ x=\mathcal{U} x$,
 where
\begin{align*}
&(\mathcal{U} x)(t)\\
&=  -A\int_0^t \frac{(t-s)^{\alpha-1}}{\Gamma(\alpha)}f(s,
 x(s))ds-B\int_0^t
\frac{(t-s)^{\alpha+\beta-1}}{\Gamma(\alpha+\beta)}g(s,
 x(s))ds\\
&\quad + Q t^{\alpha-1}\Big[A\int_0^1
\frac{(1-s)^{\alpha-\delta-1}}{\Gamma(\alpha-\delta)}f(s,
x(s))ds+B\int_0^1
\frac{(1-s)^{\alpha-\delta+\beta-1}}{\Gamma(\alpha-\delta+\beta)}g(s,
 x(s))ds\\
&\quad - A \int_0^\eta
\frac{(\eta-s)^{\alpha-\delta}}{\Gamma(\alpha-\delta+1)}f(s,
 x(s))ds-B\int_0^\eta
\frac{(\eta-s)^{\alpha-\delta+\beta}}{\Gamma(\alpha-\delta+\beta+1)}g(s,
 x(s))ds\Big].
\end{align*}
We show that $F$ maps bounded sets into bounded sets in $ C([0,1],
\mathbb{R})$. For a positive number $r$, let 
$B_r = \{x \in C([0,1], \mathbb{R}): \|x\| \le r \}$ be a bounded set in
$C([0,1], \mathbb{R})$. Then
\begin{align*}
&|(\mathcal{U} x)(t)|\\
&\leq  |A|\int_0^t \frac{(t-s)^{\alpha-1}}{\Gamma(\alpha)}p_1(s)\psi_1(\|x\|)ds+|B|\int_0^t
\frac{(t-s)^{\alpha+\beta-1}}{\Gamma(\alpha+\beta)}p_2(s)\psi_2(\|x\|)ds\\
&\quad + |Q| t^{\alpha-1}\Big[|A|\int_0^1
\frac{(1-s)^{\alpha-\delta-1}}{\Gamma(\alpha-\delta)}p_1(s)\psi_1(\|x\|)ds\\
&\quad +|B|\int_0^1
\frac{(1-s)^{\alpha-\delta+\beta-1}}{\Gamma(\alpha-\delta+\beta)}p_2(s)\psi_2(\|x\|)ds\\&\quad +
|A| \int_0^\eta
\frac{(\eta-s)^{\alpha-\delta}}{\Gamma(\alpha-\delta+1)}p_1(s)\psi_1(\|x\|)ds\\
&\quad +|B|\int_0^\eta
\frac{(\eta-s)^{\alpha-\delta+\beta}}{\Gamma(\alpha-\delta+\beta+1)}p_2(s)\psi_2(\|x\|)ds\Big]\\
&\leq
|A|\psi_1(r)\|p_1\|_{L^1}\Big\{\frac{1}{\Gamma(\alpha+1)}+\frac{|Q|}{\Gamma(\alpha-\delta+1)}
+\frac{|Q|}{\Gamma(\alpha-\delta+2)}\Big\}\\
&\quad +
|B|\psi_2(r)\|p_2\|_{L^1}\Big\{\frac{1}{\Gamma(\alpha+\beta+1)}+\frac{|Q|}{\Gamma(\alpha-\delta+\beta+1)}
+\frac{|Q|}{\Gamma(\alpha-\delta+\beta+2)}\Big\}.
\end{align*}
Consequently
\begin{align*}
&\|\mathcal{U} x\|\\&\leq
|A|\psi_1(r)\|p_1\|_{L^1}\Big\{\frac{1}{\Gamma(\alpha+1)}+\frac{|Q|}{\Gamma(\alpha-\delta+1)}
+\frac{|Q|}{\Gamma(\alpha-\delta+2)}\Big\}\\
&\quad +
|B|\psi_2(r)\|p_2\|_{L^1}\Big\{\frac{1}{\Gamma(\alpha+\beta+1)}+\frac{|Q|}{\Gamma(\alpha-\delta+\beta+1)}
+\frac{|Q|}{\Gamma(\alpha-\delta+\beta+2)}\Big\}.
\end{align*}

Next we show that $F$ maps bounded sets into equicontinuous sets
of $ C([0,1], \mathbb{R})$. Let $t_1, t_2 \in [0,1]$ with 
$t_1<t_2$ and  $x \in B_r$, where $B_r$ is a bounded set of 
$C([0,1],\mathbb{R})$. Then we obtain
\begin{align*}
&\|(\mathcal{U} x)(t_2)-(\mathcal{U} x)(t_1)\|\\
&\leq   \Big\|\frac{|A|}{\Gamma(\alpha)}\int_0^{t_1}[(t_2-s)^{\alpha-1}-(t_1-s)^{\alpha-1}]f(s,x(s))ds\\
&\quad + \frac{|A|}{\Gamma(\alpha)}\int_{t_1}^{t_2}(t_2-s)^{\alpha-1}f(s,x(s))ds\\
&\quad + \frac{|B|}{\Gamma(\alpha+\beta)}\int_0^{t_1}[(t_2-s)^{\alpha+\beta-1}-(t_1-s)^{\alpha+\beta-1}]g(s,x(s))ds \\
&\quad + \frac{|B|}{\Gamma(\alpha+\beta)}\int_{t_1}^{t_2}(t_2-s)^{\alpha+\beta-1}g(s,x(s))ds \\
&\quad + |Q| [(t_2)^{\alpha-1}-(t_1)^{\alpha-1}]\Big[|A|\int_0^1
\frac{(1-s)^{\alpha-\delta-1}}{\Gamma(\alpha-\delta)}|f(s,
x(s))|ds\\
&\quad +|B|\int_0^1
\frac{(1-s)^{\alpha-\delta+\beta-1}}{\Gamma(\alpha-\delta+\beta)}|g(s,
 x(s))|ds\\
&\quad + |A| \int_0^\eta
\frac{(\eta-s)^{\alpha-\delta}}{\Gamma(\alpha-\delta+1)}|f(s,
 x(s))|ds+|B|\int_0^\eta
\frac{(\eta-s)^{\alpha-\delta+\beta}}{\Gamma(\alpha-\delta+\beta+1)}|g(s,
 x(s))|ds\Big]\Big\|\\
 &\leq   \Big\|\frac{|A|}{\Gamma(\alpha)}\int_0^{t_1}[(t_2-s)^{\alpha-1}-(t_1-s)^{\alpha-1}]p_1(s)\psi_1(r)ds\\
 &\quad + \frac{|A|}{\Gamma(\alpha)}\int_{t_1}^{t_2}(t_2-s)^{\alpha-1}p_1(s)\psi_1(r)ds\\
&\quad + \frac{|B|}{\Gamma(\alpha+\beta)}\int_0^{t_1}[(t_2-s)^{\alpha+\beta-1}-(t_1-s)^{\alpha+\beta-1}]p_2(s)\psi_2(r)ds \\
&\quad + \frac{|B|}{\Gamma(\alpha+\beta)}\int_{t_1}^{t_2}(t_2-s)^{\alpha+\beta-1}p_2(s)\psi_2(r)ds \\
&\quad + |Q| [(t_2)^{\alpha-1}-(t_1)^{\alpha-1}]\Big[|A|\int_0^1
\frac{(1-s)^{\alpha-\delta-1}}{\Gamma(\alpha-\delta)}p_1(s)\psi_1(r)ds\\
&\quad +|B|\int_0^1
\frac{(1-s)^{\alpha-\delta+\beta-1}}{\Gamma(\alpha-\delta+\beta)}p_2(s)\psi_2(r)ds\\&\quad +
|A| \int_0^\eta
\frac{(\eta-s)^{\alpha-\delta}}{\Gamma(\alpha-\delta+1)}p_1(s)\psi_1(r)ds
+|B|\int_0^\eta
\frac{(\eta-s)^{\alpha-\delta+\beta}}{\Gamma(\alpha-\delta+\beta+1)}
p_2(s)\psi_2(r)ds\Big]\Big\|.
\end{align*}
Obviously the right hand side of the above inequality tends to
zero independently of $x \in B_{r}$ as $t_2- t_1 \to 0$. As
$\mathcal{U}$ satisfies the above   assumptions, therefore it
follows by the Arzel\'a-Ascoli  theorem that
 $\mathcal{U}: C([0,1], \mathbb{R}) \to C([0,1], \mathbb{R})$ is completely
continuous.

The result will follow from the Leray-Schauder nonlinear
alternative (Lemma \ref{lls}) once we have proved the boundendness
of the set of all solutions to equations $x=\lambda \mathcal{U} x$
for $\lambda\in [0,1]$.

Let $x$ be a solution. Then, for $t\in [0,1]$, and using the
computations in proving that $\mathcal{U}$ is bounded,  we have
\begin{align*}
&|x(t)|\\  
&= |\lambda (\mathcal{U} x)(t)| \le
|A|\psi_1(\|x\|)\|p_1\|_{L^1}\Big\{\frac{1}{\Gamma(\alpha+1)}
+\frac{|Q|}{\Gamma(\alpha-\delta+1)}
+\frac{|Q|}{\Gamma(\alpha-\delta+2)}\Big\}\\
&\quad + |B|\psi_2(\|x\|)\|p_2\|_{L^1}\Big\{\frac{1}{\Gamma(\alpha+\beta+1)}
+\frac{|Q|}{\Gamma(\alpha-\delta+\beta+1)}
+\frac{|Q|}{\Gamma(\alpha-\delta+\beta+2)}\Big\}.
\end{align*}
Consequently, 
$$
\frac{\|x\|}{
 |A|\Lambda_1\psi_1(\|x\|)\|p_1\|_{L^1}+|B|\Lambda_1\psi_2(\|x\|)\|p_2\|_{L^1}}
\le 1.
$$
In view of (A4), there exists $M$ such that $\|x\| \ne M$. Let
us set
$$
U = \{x \in  C([0,1], X):\|x\| < M\}.
$$ 
Note that the operator $\mathcal{U}:\overline{U} \to  C([0,1], \mathbb{R})$ is
 continuous and completely continuous. From the choice of $U$,
there is no $x \in \partial U$ such that 
$x =\lambda \mathcal{U}(x)$ for some $\lambda \in (0,1)$. 
Consequently, by the nonlinear alternative of Leray-Schauder type 
(Lemma \ref{lls}), we
deduce that $\mathcal{U}$ has a fixed point $x \in \overline{U}$
which is a solution of the problem \eqref{p}. This completes the
proof.
\end{proof}


\subsection*{Example} 
Consider a boundary-value problem of
integro-differential equations of fractional order with  nonlocal
fractional boundary conditions given by
\begin{equation}\label{exa} 
\begin{gathered}
-D^{5/2}x(t)= Af(t,  x(t))+BI^{\beta}g(t,  x(t)),\quad t\in [0,1],\\ 
D^{1/4} x(0)=0, \quad D^{5/4}x(0)=0, \quad D^{1/4} x(1)=\int_0^\eta
D^{1/4}x(s)ds,
\end{gathered} 
\end{equation}
where $n=3$, $ A=B=1$, $\beta=3/4$, $\eta=2/3$,
$f(t,x)=\frac{3|x|(2+|x|)}{8(1+|x|)}+4t$,
$g(t,x)=\frac{1}{2}\tan^{-1}x+\sin^2 t$. With the given data, we
find that
\[
Q= \frac{\Gamma(\alpha-\delta+1)}{(\alpha-\delta-\eta^{\alpha-\delta})
\Gamma(\alpha)}=1.037485,
\]
and 
\begin{align*}
\Omega&= \sup_{t \in
[0,1]}\Big\{|A|\Big[\frac{t^\alpha}{\Gamma(\alpha+1)}+|Q|t^{\alpha-1}
\Big(\frac{1}{\Gamma(\alpha-\delta+1)}
+\frac{\eta^{\alpha-\delta+1}}{\Gamma(\alpha-\delta+2)}\Big)\Big]
\\
 &\quad +|B|\Big[\frac{t^{\alpha+\beta}}{\Gamma(\alpha+\beta+1)}
 +|Q|t^{\alpha-1}\Big(\frac{1}{\Gamma(\alpha-\delta+\beta+1)}
+\frac{\eta^{\alpha-\delta+\beta+1}}{\Gamma(\alpha-\delta+\beta+2)}\Big)
 \Big]\Big\}\\
&= 1.043555,
\end{align*}
and $L_1=3/4$, $L_2=1/2$ as 
$ |f(t,x)-f(t,y)|\le \frac{3}{4}|x-y|$, 
$|g(t,x)-g(t,y)|\le \frac{1}{2}|x-y|$. 
Clearly $L=\max\{L_1, L_2\}=3/4$ and $L <1/\Omega$. 
Thus all the assumptions of Theorem \ref{t1} are
satisfied. Hence, by the conclusion of Theorem \ref{t1}, the
problem \eqref{exa} has a unique solution.

\subsection*{Acknowledgments} 
The authors gratefully acknowledge
the editor for his constructive comments.

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\end{document}