\documentclass[reqno]{amsart}
\usepackage{hyperref}

\AtBeginDocument{{\noindent\small
\emph{Electronic Journal of Differential Equations},
Vol. 2017 (2017), No. 100, pp. 1--18.\newline
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu}
\thanks{\copyright 2017 Texas State University.}
\vspace{8mm}}

\begin{document}
\title[\hfilneg EJDE-2017/100\hfil
BVP with multiple fractional derivatives and integrals]
{Fractional boundary value problems  with multiple orders of
fractional derivatives and integrals}

\author[S. K. Ntouyas, J. Tariboon \hfil EJDE-2017/100\hfilneg]
{Sotiris K. Ntouyas, Jessada Tariboon}

\address{Sotiris K. Ntouyas\newline
Department of Mathematics,
University of Ioannina,
451 10 Ioannina, Greece. \newline
Nonlinear Analysis and Applied
Mathematics (NAAM)-Research Group,
Department of Mathematics, Faculty of Science,
King Abdulaziz University, P.O. Box 80203,
Jeddah 21589, Saudi Arabia}
\email{sntouyas@uoi.gr}

\address{Jessada Tariboon  \newline
Nonlinear Dynamic Analysis Research Center,
Department of  Mathematics, Faculty of Applied Science,
King Mongkut's University of Technology North Bangkok,
Bangkok 10800,  Thailand}
\email{jessada.t@sci.kmutnb.ac.th}

\dedicatory{Communicated by  Mokhtar Kirane}

\thanks{Submitted February 21, 2017. Published April 11, 2017.}
\subjclass[2010]{34A08, 34A12, 34A60}
\keywords{Fractional differential equation; fractional differential inclusion;
\hfill\break\indent boundary value problem; existence; fixed point theorems}

\begin{abstract}
 In this article we study a new class of boundary value problems for
 fractional differential equations and inclusions with multiple orders of
 fractional  derivatives and integrals, in both fractional differential
 equation and  boundary conditions. The Sadovski's  fixed point theorem
 is applied in the single-valued case  while, in multi-valued case,
 the nonlinear alternative for contractive maps is used.
  Some illustrative examples are also included.
\end{abstract}

\maketitle
\numberwithin{equation}{section}
\newtheorem{theorem}{Theorem}[section]
\newtheorem{lemma}[theorem]{Lemma}
\newtheorem{remark}[theorem]{Remark}
\newtheorem{definition}[theorem]{Definition}
\newtheorem{example}[theorem]{Example}
\allowdisplaybreaks

\section{Introduction}

Fractional differential equations have attracted more and more attention in
 recent years, which is partly because of their numerous applications in many
branches of science and engineering including fluid flow, signal and image
processing, fractals theory, control theory, electromagnetic theory, fitting
of experimental data, optics, potential theory, biology, chemistry,
diffusion, and viscoelasticity, etc. For a detailed account of applications
and recent results on initial and boundary value problems of fractional
differential equations, we refer the reader to  a series of books and
papers \cite{f10,fp16,pf4,f2,f11,f5,f3,f8, Kil,f6,f11-a,pod,sam,f33,H4,b9}
 and references cited therein.

In this article, we investigate  boundary value problems which contains multiple
orders of  fractional  derivatives and integrals,  in both  fractional
differential equation  and boundary conditions.  More precisely, we consider
the following boundary value problems which consist from the differential equation
\begin{equation}\label{E}
\left( \lambda D^\alpha + (1-\lambda)D^\beta \right)x(t)=f(t,x(t)),\quad
 t \in (0,T),
\end{equation}
which includes two fractional derivatives, supplemented by boundary conditions with:
\begin{itemize}
\item     two fractional derivatives
\begin{equation}\label{bc1}
 x(0) = 0,\quad
\mu D^{\gamma_1}x(T)+(1-\mu)D^{\gamma_2}x(T) = \gamma_3,
\end{equation}
or
\item     two fractional integrals
\begin{equation}\label{bc2}
 x(0) = 0,\quad
\mu I^{\delta_1}x(T)+(1-\mu)I^{\delta_2}x(T) = \delta_3,
\end{equation}
or
\item   one fractional derivative and one fractional integral
\begin{equation}\label{bc3}
 x(0) = 0,\quad
\mu D^{\gamma_1}x(T)+(1-\mu)I^{\delta_2}x(T) = \gamma_3,
\end{equation}
\end{itemize}
where $D^{\phi}$ is the Riemann-Liouville or Caputo fractional derivative
of order $\phi\in\{\alpha,\beta,\gamma_1,\gamma_2\}$ such that
$1 < \alpha, \beta \leq 2$  and $0<\gamma_1,\gamma_2 < \alpha - \beta $,
$\gamma_3, \delta_3\in \mathbb{R}$,  $I^{\chi}$ is the Riemann-Liouville
fractional integral of order $\chi\in \{\delta_1, \delta_2\}$,
$0 < \lambda\leq 1$, $0\leq\mu\leq 1$ are given constants and
$f: [0,T]\times {\mathbb R}\to {\mathbb R}$ is a continuous function.

Also we consider  the multi-valued  analogues of boundary value problems
above by  studying  the differential inclusion
\begin{equation}\label{E-inc}
 \left( \lambda D^\alpha + (1-\lambda)D^\beta \right)x(t) \in F(t,x(t)),\quad
 t \in (0,T),
\end{equation}
supplemented by boundary conditions \eqref{bc1}-\eqref{bc3}, where
$F: [0,T]\times {\mathbb R}\to \mathcal{P}({\mathbb R})$ is a multivalued
function  ($\mathcal{P}({\mathbb R})$ is the family of all nonempty subsets
of ${\mathbb R}$).

In fact, fractional calculus provide an excellent tool for the description
of memory and hereditary properties of various materials and processes in
mathematical modeling. The fractional differential equation \eqref{E} and
inclusion  \eqref{E-inc} subject to boundary conditions \eqref{bc1}, \eqref{bc2}
 and \eqref{bc3} describe models of physical problems in which often some
parameters have been adjusted to suitable situations. The values of these
parameters can be change the effects of fractional derivatives and integrals.
Especially, in this paper, the linear adjusting or convex combination is used.

Recently in \cite{NNLT} we studied  problem \eqref{E}-\eqref{bc1}  with four
 Riemann-Liouville   fractional derivatives. Existence and uniqueness results
 were proved by using Banach's fixed point theorem, Krasnoselskii's fixed
point theorem and Leray-Schauder's nonlinear alternative.
Similar results for the boundary value problems \eqref{E}-\eqref{bc1} to
\eqref{E}-\eqref{bc3} can be  established also for Caputo fractional
derivatives, with obvious modifications.

In  this article we prove an existence result for the boundary value problem
\eqref{E}-\eqref{bc1}, with four Caputo type fractional derivatives,
 via Sadovski's fixed point theorem and an existence result for the multi-valued
analogue \eqref{E-inc}-\eqref{bc1}, by means of nonlinear alternative for
contractive maps.

This article  is organized as follows. In section 2, we present
the framework in which the boundary value problems \eqref{E}-\eqref{bc1},
\eqref{E}-\eqref{bc2}, \eqref{E}-\eqref{bc3}, are formulated in a fixed point
equation.
Section 3 is devoted to the problem \eqref{E}-\eqref{bc1} and
 Section 4 to the problem \eqref{E-inc}-\eqref{bc1}.
 Illustrative examples are also presented.

\section{Preliminaries}

In this section, we introduce some notation and definitions of
fractional calculus \cite{ Kil,pod} and present preliminary results
needed in our proofs later.

 \begin{definition} \rm
The Riemann-Liouville fractional integral of order $\alpha>0$ of a
function $g: (0,\infty)\to{\mathbb R}$ is defined by
\[
J^{\alpha}g(t)=\int_0^t\frac{(t-s)^{\alpha-1}}{\Gamma(\alpha)}g(s)ds,
\]
provided the right-hand side is point-wise defined on $(0,\infty)$,
where $\Gamma$ is the Gamma function.
\end{definition}

\begin{definition} \rm
The Riemann-Liouville fractional derivative of order $\alpha>0$ of
a continuous function $g: (0,\infty)\to {\mathbb R}$ is defined by
\[
D^{\alpha}g(t)=\frac{1}{\Gamma(n-\alpha)}\Big(\frac{d}{dt}\Big)^n
\int_0^t\frac{g(s)}{(t-s)^{\alpha-n+1}}ds,\quad n-1<\alpha<n,
\]
where $n=[\alpha]+1$, $[\alpha]$ denotes the integer part of real
number $\alpha$, provided the right-hand side is point-wise defined
on $(0,\infty)$.
\end{definition}

\begin{definition} \rm
The Caputo derivative of order $q$ for a function $f:[0,\infty)\to
{\mathbb R}$ can be written as
\[
^cD^q f(t)=
D^q\Big(f(t)-\sum_{k=0}^{n-1}\frac{t^k}{k!}f^{(k)}(0)\Big),\quad
t>0, \; n-1<q<n.
\]
\end{definition}

\begin{remark} \rm
If $f(t)\in C^{n}[0,\infty)$, then
\[
^cD^{q}f(t)= \frac{1}{\Gamma(n-q)}\int_0^t
\frac{f^{(n)}(s)}{(t-s)^{q+1-n}}ds = I^{n-q}f^{(n)}(t),\quad t>0,\; n-1<q<n.
\]
\end{remark}

 \begin{lemma} \label{lem}
For $q >0$, the general solution of the fractional differential
equation $^cD^{q} x(t)=0$ is given by
$$
x(t)=c_0+c_1t+ \ldots +c_{n-1} t^{n-1},
$$
where $c_i \in  \mathbb{R}$, $i=0,1, 2,\ldots, n-1$ ($n=[q]+1$).
\end{lemma}

  In view of Lemma \ref{lem}, it follows that
\begin{equation}\label{e2-lem}
 I^{q}~ {^cD^{q}} x(t)= x(t)+c_0+c_1t+ \ldots +c_{n-1} t^{n-1},
 \end{equation}
for some $c_i \in  \mathbb{R}$, $i=0, 1, 2,\ldots, n-1$ ($n=[q]+1$).

 \begin{lemma}\label{lem-linear}
The boundary value problem
\begin{equation}\label{Problem-linear}
\begin{gathered}
 \left( \lambda D^\alpha + (1-\lambda)D^\beta \right)x(t)=\omega(t),\quad
 t \in (0,T),\\
 x(0) = 0,\quad
\mu D^{\gamma_1}x(T)+(1-\mu)D^{\gamma_2}x(T) = \gamma_3,
\end{gathered}
\end{equation}
is equivalent to the  integral equation
\begin{equation} \label{Pre-operator}
\begin{aligned}
x(t) &=  \frac{\lambda-1}{\lambda \Gamma(\alpha-\beta)}\int_0^t (t-s)^{\alpha-\beta-1}x(s)ds + \frac{1}{\lambda \Gamma(\alpha)}\int_0^t (t-s)^{\alpha-1}\omega(s)ds \\
&\quad+ \frac{t}{\Lambda_1}\Big( \gamma_3
 - \frac{\mu(\lambda-1)}{\lambda \Gamma(\alpha-\beta-\gamma_1)}
 \int_0^T (T-s)^{\alpha-\beta-\gamma_1-1}x(s)ds \\
&\quad - \frac{\mu}{\lambda \Gamma(\alpha-\gamma_1)}
 \int_0^T (T-s)^{\alpha-\gamma_1-1}\omega(s)ds  \\
&\quad - \frac{(1-\mu)(\lambda-1)}{\lambda \Gamma(\alpha-\beta-\gamma_2)}
 \int_0^T (T-s)^{\alpha-\beta-\gamma_2-1}x(s)ds \\
&\quad - \frac{1-\mu}{\lambda \Gamma(\alpha-\gamma_2)}
 \int_0^T (T-s)^{\alpha-\gamma_2-1}\omega(s)ds \Big),\quad t\in J:=[0,T],
\end{aligned}
\end{equation}
where the non zero constant $\Lambda_1$ is defined by
\begin{equation}\label{eq-l-1}
\Lambda_1 = \frac{\mu   T^{1-\gamma_1}}{\Gamma(2-\gamma_1)}
 + \frac{(1-\mu)   T^{1-\gamma_2}}{\Gamma(2-\gamma_2)}.
\end{equation}
\end{lemma}

\begin{proof}
From the first equation of \eqref{Problem-linear}, we have
\begin{equation}\label{Eq2-3}
D^{\alpha}x(t) = \frac{\lambda-1}{\lambda}D^{\beta}x(t)+ \frac{1}{\lambda}\omega(t),
 \quad t \in J.
\end{equation}
Taking the Riemann-Liouville fractional integral of order $\alpha$ to both
sides of \eqref{Eq2-3}, we obtain
\begin{align*}
x(t) &=  \frac{\lambda-1}{\lambda \Gamma(\alpha-\beta)}
 \int_0^t (t-s)^{\alpha-\beta-1}x(s)ds
 + \frac{1}{\lambda \Gamma(\alpha)}\int_0^t (t-s)^{\alpha-1}\omega(s)ds\\
&\quad + C_1  + C_2t,
\end{align*}
for $C_1$, $C_2 \in \mathbb{R}$. The first boundary condition of
\eqref{Problem-linear} implies that $C_1 = 0$. Hence
\begin{equation}\label{Pre-eq-1}
x(t)  = \frac{\lambda-1}{\lambda \Gamma(\alpha-\beta)}
 \int_0^t (t-s)^{\alpha-\beta-1}x(s)ds
  + \frac{1}{\lambda \Gamma(\alpha)}\int_0^t (t-s)^{\alpha-1}\omega(s)ds +C_2t.
\end{equation}
Applying the Caputo fractional derivative of order $\psi\in \{\gamma_1,\gamma_2\}$
such that $0<\psi<\alpha-\beta$ to \eqref{Pre-eq-1}, we have
\begin{align*}
D^{\psi}x(t)
&=  \frac{\lambda-1}{\lambda \Gamma(\alpha-\beta-\psi)}
  \int_0^t (t-s)^{\alpha-\beta-\psi-1}x(s)ds \\
&\quad + \frac{1}{\lambda \Gamma(\alpha-\psi)}
 \int_0^t (t-s)^{\alpha-\psi-1}\omega(s)ds +C_2\frac{1}{\Gamma(2-\psi)}t^{1-\psi} .
\end{align*}
Substituting the values $\psi=\gamma_1$ and $\psi=\gamma_2$ to the above
relation and using the second condition of \eqref{Problem-linear}, we obtain
\begin{align*}
\gamma_3
&= \frac{\mu(\lambda-1)}{\lambda \Gamma(\alpha-\beta-\gamma_1)}\int_0^T (T-s)^{\alpha-\beta-\gamma_1-1}x(s)ds \\
&\quad + \frac{\mu}{\lambda \Gamma(\alpha-\gamma_1)}\int_0^T (T-s)^{\alpha-\gamma_1-1}\omega(s)ds +\frac{\mu   T^{1-\gamma_1}}{\Gamma(2-\gamma_1)}C_2 \\
&\quad + \frac{(1-\mu)(\lambda-1)}{\lambda \Gamma(\alpha-\beta-\gamma_2)}\int_0^T (T-s)^{\alpha-\beta-\gamma_2-1}x(s)ds \\
&\quad + \frac{1-\mu}{\lambda \Gamma(\alpha-\gamma_2)}\int_0^T (T-s)^{\alpha-\gamma_2-1}\omega(s)ds + \frac{(1-\mu)   T^{1-\gamma_2}}{\Gamma(2-\gamma_2)}C_2,
\end{align*}
which leads to
\begin{align*}
C_2 &=  \frac{1}{\Lambda_1}\Big[ \gamma_3
 - \frac{\mu(\lambda-1)}{\lambda \Gamma(\alpha-\beta-\gamma_1)}
 \int_0^T (T-s)^{\alpha-\beta-\gamma_1-1}x(s)ds \\
&\quad - \frac{\mu}{\lambda \Gamma(\alpha-\gamma_1)}
 \int_0^T (T-s)^{\alpha-\gamma_1-1}\omega(s)ds  \\
&\quad - \frac{(1-\mu)(\lambda-1)}{\lambda \Gamma(\alpha-\beta-\gamma_2)}
 \int_0^T (T-s)^{\alpha-\beta-\gamma_2-1}x(s)ds \\
&\quad - \frac{1-\mu}{\lambda \Gamma(\alpha-\gamma_2)}
 \int_0^T (T-s)^{\alpha-\gamma_2-1}\omega(s)ds \Big].
\end{align*}
Substituting the value of the constant $C_2$ in \eqref{Pre-eq-1},
 we deduce the integral equation \eqref{Pre-operator}.
The converse follows by direct computation. This completes the proof.
\end{proof}

The following lemmas concerning the boundary value problems \eqref{E}-\eqref{bc2}
and  \eqref{E}-\eqref{bc3}, are similar to that of Lemma \ref{lem-linear}.
We omit the proofs.

\begin{lemma}\label{lem-linear-2}
The boundary value problem
\begin{equation}\label{Problem-linear-2}
\begin{gathered}
 \left( \lambda D^\alpha + (1-\lambda)D^\beta \right)x(t)=\omega(t),\quad
 t \in (0,T),\\
 x(0) = 0,\quad
\mu I^{\delta_1}x(T)+(1-\mu)I^{\delta_2}x(T) = \delta_3,
\end{gathered}
\end{equation}
is equivalent to the  integral equation
\begin{equation}\label{Pre-operator-2}
\begin{aligned}
x(t)
&=  \frac{\lambda-1}{\lambda \Gamma(\alpha-\beta)}
 \int_0^t (t-s)^{\alpha-\beta-1}x(s)ds
 + \frac{1}{\lambda \Gamma(\alpha)}\int_0^t (t-s)^{\alpha-1}\omega(s)ds \\
&\quad + \frac{t}{\Lambda_2}\Big(\delta_3
 - \frac{\mu(\lambda-1)}{\lambda \Gamma(\delta_1+\alpha-\beta)}
 \int_0^T (T-s)^{\delta_1+\alpha-\beta-1}x(s)ds \\
&\quad- \frac{\mu}{\lambda \Gamma(\delta_1+\alpha)}
 \int_0^T (T-s)^{\delta_1+\alpha-1}\omega(s)ds  \\
&\quad - \frac{(1-\mu)(\lambda-1)}{\lambda \Gamma(\delta_2+\alpha-\beta)}
 \int_0^T (T-s)^{\delta_2+\alpha-\beta-1}x(s)ds \\
&\quad - \frac{1-\mu}{\lambda \Gamma(\delta_2+\alpha)}
 \int_0^T (T-s)^{\delta_2+\alpha-1}\omega(s)ds \Big),\quad t\in J:=[0,T],
\end{aligned}
\end{equation}
where the non-zero constant $\Lambda_2$ is defined by
\begin{equation}\label{eq-l-2}
\Lambda_2 = \frac{\mu   T^{1+\delta_1}}{\Gamma(2+\delta_1)}
+ \frac{(1-\mu)   T^{1+\delta_2}}{\Gamma(2+\delta_2)}.
\end{equation}
\end{lemma}

\begin{lemma}\label{lem-linear-3}
The boundary value problem
\begin{equation}\label{Problem-linear-3}
\begin{gathered}
 \left( \lambda D^\alpha + (1-\lambda)D^\beta \right)x(t)=\omega(t),\quad
 t \in (0,T),\\
 x(0) = 0,\quad
\mu D^{\gamma_1}x(T)+(1-\mu)I^{\delta_2}x(T) = \gamma_3,
\end{gathered}
\end{equation}
is equivalent to the integral equation
\begin{equation}\label{Pre-operator-3}
\begin{aligned}
x(t) &=  \frac{\lambda-1}{\lambda \Gamma(\alpha-\beta)}
 \int_0^t (t-s)^{\alpha-\beta-1}x(s)ds
 + \frac{1}{\lambda \Gamma(\alpha)}\int_0^t (t-s)^{\alpha-1}\omega(s)ds \\
&\quad + \frac{t}{\Lambda_3}\Big( \gamma_3
 - \frac{\mu(\lambda-1)}{\lambda \Gamma(\alpha-\beta-\gamma_1)}
 \int_0^T (T-s)^{\alpha-\beta-\gamma_1-1}x(s)ds \\
&\quad- \frac{\mu}{\lambda \Gamma(\alpha-\gamma_1)}
 \int_0^T (T-s)^{\alpha-\gamma_1-1}\omega(s)ds  \\
&\quad- \frac{(1-\mu)(\lambda-1)}{\lambda \Gamma(\delta_2+\alpha-\beta)}
 \int_0^T (T-s)^{\delta_2+\alpha-\beta-1}x(s)ds \\
&\quad- \frac{1-\mu}{\lambda \Gamma(\delta_2+\alpha)}
 \int_0^T (T-s)^{\delta_2+\alpha-1}\omega(s)ds \Big),\quad t\in J:=[0,T],
\end{aligned}
\end{equation}
where the non zero constant $\Lambda_3$ is defined by
\begin{equation}\label{eq-l-3}
\Lambda_3 = \frac{\mu   T^{1-\gamma_1}}{\Gamma(2-\gamma_1)}
+ \frac{(1-\mu)   T^{1+\delta_2}}{\Gamma(2+\delta_2)}.
\end{equation}
\end{lemma}

\section{Existence  result  for problem \eqref{E}-\eqref{bc1}}

Let $\mathcal{C}:=C([0,T], {\mathbb R})$ denote the Banach space of all
continuous functions from $[0,T]$ into $\mathbb{R}$ with the norm
 $\|x\|= \sup\{|x(t)|$, $t\in [0,T]\}$.

Our   existence result  for the problem \eqref{E}-\eqref{bc1}  is based on
Sadovskii's fixed point theorem. Before proceeding further, let us
recall some auxiliary material.

\begin{definition} \rm
Let $M$ be a   bounded set in metric space $(X, d)$, then
Kuratowskii  measure of noncompactness, $\alpha(M)$ is defined as
$\inf\{\epsilon: M$   covered by a finitely many sets such that
the diameter of each set $\le \epsilon\}$.
\end{definition}

\begin{definition}[\cite{GrDu}] \rm
 Let $\Phi: \mathcal{D}(\Phi)\subseteq X\to  X$ be a bounded and continuous
operator on Banach space $X$. Then $\Phi$ is called a
 condensing map if $\alpha(\Phi(B))<\alpha(B)$ for all bounded sets
$  B\subset \mathcal{D}(\Phi)$, where $\alpha$ denotes
 the Kuratowski measure of noncompactness.
\end{definition}

\begin{lemma}[{\cite[Example 11.7]{Zei}}]   \label{lem-Sa}
 The map $K+C$ is a $k$-set contraction with $0\le k<1$, and thus also
condensing, if
\begin{itemize}
\item[(i)] $K, C: \mathcal{D}\subseteq X\to X$ are operators on
the Banach space $X$;
\item[(ii)] $K$ is $k$-contractive, i.e.,
\[
\|Kx-Ky\|\le k\|x-y\|
\]
for all $x,y\in \mathcal{D}$ and fixed $k\in [0,1)$;
\item[(iii)] $C$ is compact.
\end{itemize}
\end{lemma}

\begin{lemma}[\cite{Sad}] \label{Sad}
 Let $B$ be a convex, bounded and closed subset of a
Banach space $X$ and $\Phi: B\to B$ be a condensing map. Then
$\Phi$ has a fixed point.
\end{lemma}

\begin{theorem}\label{t-Sad}
 Let $f : J\times {\mathbb R} \to {\mathbb R}$ be a
  continuous function. Assume that:
 \begin{itemize}
\item[(H1)]  there exists a function $ \nu \in C(J,{\mathbb R}_{+})$
 such that
$$
|f(t,u)|\leq \nu(t), \quad \text{for  a.e. }  t\in J,  \text{ and each }
 u\in {\mathbb R}.
$$
 \item[(H2)]
\begin{align*}
\Omega_1:=& \frac{T^{\alpha-\beta}|\lambda-1|}
{\lambda\Gamma(\alpha-\beta+1)}+\frac{T^{\alpha-\beta-\gamma_1+1}
\mu|\lambda-1|}{\lambda\Lambda_1\Gamma(\alpha-\beta-\gamma_1+1)}\\
 &+\frac{T^{\alpha-\beta-\gamma_2+1}(1-\mu)|\lambda-1|}
{\lambda\Lambda_1\Gamma(\alpha-\beta-\gamma_2+1)}<1.
\end{align*}
 \end{itemize}
 Then, problem \eqref{E}-\eqref{bc1} has at least one solution on $J$.
\end{theorem}

\begin{proof}
Let $B_r=\{x\in \mathcal{C}: \|x\|\le r\}$  be a closed
bounded and convex subset of $\mathcal{C}$, where $r$ is a fixed constant.
Consider the operator $\mathcal{P}: \mathcal{C}\to \mathcal{C}$ defined by
\begin{equation}\label{e300}
\begin{aligned}
\mathcal{P}x(t)
&=     \frac{\lambda-1}{\lambda \Gamma(\alpha-\beta)}
 \int_0^t (t-s)^{\alpha-\beta-1}x(s)ds
 + \frac{1}{\lambda \Gamma(\alpha)}\int_0^t (t-s)^{\alpha-1}f(s,x(s))ds \\
&\quad +  \frac{t}{\Lambda_1}\Big( \gamma_3
 - \frac{\mu(\lambda-1)}{\lambda \Gamma(\alpha-\beta-\gamma_1)}
 \int_0^T (T-s)^{\alpha-\beta-\gamma_1-1}x(s)ds \\
&\quad - \frac{\mu}{\lambda \Gamma(\alpha-\gamma_1)}
 \int_0^T (T-s)^{\alpha-\gamma_1-1}f(s,x(s))ds \\
&\quad -\frac{(1-\mu)(\lambda-1)}{\lambda \Gamma(\alpha-\beta-\gamma_2)}
 \int_0^T (T-s)^{\alpha-\beta-\gamma_2-1}x(s)ds \\
&\quad - \frac{1-\mu}{\lambda \Gamma(\alpha-\gamma_2)}
 \int_0^T (T-s)^{\alpha-\gamma_2-1}f(s,x(s))ds \Big),  \quad t\in J.
\end{aligned}
\end{equation}

Let us define  $\mathcal{P}_{1},\mathcal{P}_{2}: {\mathcal{C}} \to {\mathcal{C}}$ by
\begin{align*}
(\mathcal{P}_1 x)(t)
&=  \frac{(\lambda-1)}{\lambda\Gamma(\alpha-\beta)}
 \int_0^t (t-s)^{\alpha-\beta-1}x(s)ds \\
&\quad - \frac{t}{\Lambda_1}\Big[\frac{\mu(\lambda-1)}
 {\lambda\Gamma(\alpha-\beta-\gamma_1)}\int_0^T (T-s)^{\alpha-\beta-\gamma_1-1}x(s)ds\\
&\quad + \frac{(1-\mu)(\lambda-1)}{\lambda\Gamma(\alpha-\beta-\gamma_2)}
 \int_0^T (T-s)^{\alpha-\beta-\gamma_2-1}x(s)ds\Big],
\end{align*}
and
\begin{align*}
(\mathcal{P}_2 x)(t)
&= \frac{1}{\lambda\Gamma(\alpha)}\int_0^t (t-s)^{\alpha-1}f(s,x(s))ds \\
&\quad +  \frac{t}{\Lambda_1}\Big[\gamma_3
 - \frac{\mu}{\lambda\Gamma(\alpha-\gamma_1)}
 \int_0^T (T-s)^{\alpha-\gamma_1-1}f(s,x(s))ds\\
&\quad -  \frac{(1-\mu)}{\lambda\Gamma(\alpha-\gamma_2)}
 \int_0^T (T-s)^{\alpha-\gamma_2-1}f(s,x(s))ds \Big].
 \end{align*}

Clearly
\begin{equation}\label{p3}
(\mathcal{P} x)(t)=(\mathcal{P}_1 x)(t)+(\mathcal{P}_2 x)(t), ~~t
\in J.
\end{equation}
Obviously the operator $\mathcal{P}$ has a fixed point  is equivalent to
$\mathcal{P}_1+\mathcal{P}_2$ has one, so it turns to prove that
$\mathcal{P}_1+\mathcal{P}_2$ has a fixed point.  We shall show that the
operators $\mathcal{P}_1$ and $\mathcal{P}_2$ satisfy all conditions of
Lemma \ref{Sad}. The proof will be given in several steps.
\smallskip

\noindent\textbf{Step 1:}  $\mathcal{P} B_r\subset B_r$.
 Let us  select $ r \ge \frac{\|\nu\|\Omega_2+|\gamma_3|T/\Lambda_1}{1-\Omega_1}$
where $\Omega_1$ defined by (H2) and
\begin{equation}\label{Omega-2}
\Omega_2=\frac{T^{\alpha}}{\lambda\Gamma(\alpha+1)}
+\frac{T^{\alpha-\gamma_1+1}\mu}{\lambda\Lambda_1\Gamma(\alpha-\gamma_1+1)}
+\frac{T^{\alpha-\gamma_2+1}(1-\mu)}{\lambda\Lambda_1\Gamma(\alpha-\gamma_2+1)}.
\end{equation}
  For any $x\in B_r$, we have
 \begin{align*}
&\|\mathcal{P}x\| \\
&\leq \sup_{t \in J} \Big|\frac{(\lambda-1)}{\lambda\Gamma(\alpha-\beta)}
 \int_0^t (t-s)^{\alpha-\beta-1}x(s)ds
 + \frac{1}{\lambda\Gamma(\alpha)}\int_0^t (t-s)^{\alpha-1}f(s,x(s))ds \\
&\quad - \frac{t\mu(\lambda-1)}{\lambda\Lambda_1\Gamma(\alpha-\beta-\gamma_1)}
 \int_0^T (T-s)^{\alpha-\beta-\gamma_1-1}x(s)ds\\
&\quad - \frac{t (1-\mu)(\lambda-1)}{\lambda\Lambda_1\Gamma(\alpha-\beta-\gamma_2)}
 \int_0^T (T-s)^{\alpha-\beta-\gamma_2-1}x(s)ds  \\
&\quad + \frac{t}{\Lambda_1}\Big[\gamma_3 - \frac{\mu}{\lambda\Gamma(\alpha-\gamma_1)}
 \int_0^T (T-s)^{\alpha-\beta-\gamma_1-1}f(s,x(s))ds\\
&\quad - \frac{(1-\mu)}{\lambda\Gamma(\alpha-\gamma_2)}
 \int_0^T (T-s)^{\alpha-\gamma_2-1}f(s,x(s))ds \Big] \Big| \\
&\leq \|x\|\Big[\frac{T^{\alpha-\beta}|\lambda-1|}{\lambda\Gamma(\alpha-\beta+1)}
 +\frac{T^{\alpha-\beta-\gamma_1+1}\mu|\lambda-1|}
 {\lambda\Lambda_1\Gamma(\alpha-\beta-\gamma_1+1)}\\
&\quad +\frac{T^{\alpha-\beta-\gamma_2+1}(1-\mu)|\lambda-1|}
 {\lambda\Lambda_1\Gamma(\alpha-\beta-\gamma_2+1)}\Big]
+\frac{|\gamma_3| T}{\Lambda_1}\\
&\quad + \|\nu\| \Big[\frac{T^{\alpha}}{\lambda\Gamma(\alpha+1)}
 +\frac{T^{\alpha-\gamma_1+1}\mu}{\lambda\Lambda_1\Gamma(\alpha-\gamma_1+1)}
 +\frac{T^{\alpha-\gamma_2+1}(1-\mu)}{\lambda\Lambda_1\Gamma(\alpha-\gamma_2+1)}\Big] \\
&\leq r\Omega_1 + \|\nu\| \Omega_2 + \frac{|\gamma_3| T}{\Lambda}  \leq r,
\end{align*}
 which implies that $\mathcal{P}B_r\subset B_r$.
\smallskip

\noindent\textbf{Step 2:} $\mathcal{P}_2$ is compact.
 Observe that the operator $\mathcal{P}_2$ is
uniformly bounded in view of Step 1. Let $t_1,t_2 \in
J$ with $t_1<t_2$ and  $x \in B_{r}$. Then  we obtain
\begin{align*}
&|(\mathcal{P}_2 x)(t_2)-(\mathcal{P}_2 x)(t_1)|\\
&\le   \frac{1}{\lambda\Gamma(\alpha)}
 \Big[\int_0^{t_1} [t_2-s)^{\alpha-1}-(t_1-s)^{\alpha-1}] \nu(s)ds
 + \int_{t_1}^{t_2} (t_1-s)^{\alpha-1} \nu(s)ds \Big]  \\
&\quad +   \frac{|t_2-t_1|}{\Lambda_1}\Big[|\gamma_3|
 + \frac{\mu}{\lambda\Gamma(\alpha-\gamma_1)}
 \int_0^T (T-s)^{\alpha-\beta-\gamma_1-1} \nu(s)ds\\
&\quad + \frac{(1-\mu)}{\lambda\Gamma(\alpha-\gamma_2)}
 \int_0^T (T-s)^{\alpha-\gamma_2-1} \nu(s)ds \Big],
\end{align*}
which is independent of $x$  and tends to zero as
$t_2-t_1\to 0$. Thus, $\mathcal{P}_2$ is equicontinuous.
Hence, by the Arzel\'a-Ascoli  Theorem,
 $\mathcal{P}_2(B_r)$   is a relatively compact set.
\smallskip

\noindent\textbf{Step 3:} $\mathcal{P}_1$ is  $\gamma$-contractive.
 Let $x,y\in B_r$. Then, we have
\begin{align*}
\|\mathcal{P}_1x-\mathcal{P}_1y\|
&\leq \frac{|\lambda-1|}{\lambda\Gamma(\alpha-\beta)}
 \int_0^T (T-s)^{\alpha-\beta-1}|x(s)-y(s)|ds \\
&\quad + \frac{T\mu|\lambda-1|}{\lambda\Lambda_1\Gamma(\alpha-\beta-\gamma_1)}
 \int_0^T (T-s)^{\alpha-\beta-\gamma_1-1}|x(s)-y(s)|ds\\
&\quad + \frac{T(1-\mu)|\lambda-1|}{\lambda\Lambda_1\Gamma(\alpha-\beta-\gamma_2)}
 \int_0^T (T-s)^{\alpha-\beta-\gamma_2-1}|x(s)-y(s)|ds  \\
&\le \Big\{ \frac{T^{\alpha-\beta}|\lambda-1|}{\lambda\Gamma(\alpha-\beta+1)}
 +\frac{T^{\alpha-\beta-\gamma_1+1}\mu|\lambda-1|}
 {\lambda\Lambda_1\Gamma(\alpha-\beta-\gamma_1+1)}\\
&\quad +\frac{T^{\alpha-\beta-\gamma_2+1}(1-\mu)|\lambda-1|}
 {\lambda\Lambda_1\Gamma(\alpha-\beta-\gamma_2+1)}\Big]\Big\}\|x-y\|\\
&= \Omega_1\|x-y\|,
\end{align*}
which   is $\gamma$-contractive, since  $\Omega_1<1$.
\smallskip

\noindent\textbf{Step 4:} $\mathcal{P}$ is condensing.
Since $\mathcal{P}_1$ is continuous,
$\gamma$-contraction  and $\mathcal{P}_2$ is compact, therefore,
by Lemma  \ref{lem-Sa}, $\mathcal{P}: B_r\to B_r$ with
$\mathcal{P}=\mathcal{P}_1+\mathcal{P}_2$ is a condensing map on $B_r$.

From the above four steps, we conclude by Lemma \ref{Sad} that the
map $\mathcal{P}$ has a fixed point which, in turn,  implies that the
problem \eqref{E}-\eqref{bc1} has  a solution.
\end{proof}

Setting two constants
\begin{gather*}
\Omega_3= \frac{|\lambda-1|T^{\alpha-\beta}}{\lambda\Gamma(\alpha-\beta+1)}
 +\frac{\mu|\lambda-1|T^{\delta_1+\alpha-\beta+1}}{\lambda\Lambda_2
\Gamma(\delta_1+\alpha-\beta+1)}
 +\frac{(1-\mu)|\lambda-1|T^{\delta_2+\alpha-\beta+1}}
 {\lambda\Lambda_2\Gamma(\delta_2+\alpha-\beta+1)},\\
\Omega_4= \frac{|\lambda-1|T^{\alpha-\beta}}{\lambda\Gamma(\alpha-\beta+1)}
 +\frac{\mu|\lambda-1|T^{\alpha-\beta-\gamma_1+1}}{\lambda\Lambda_3
\Gamma(\alpha-\beta-\gamma_1+1)}
 +\frac{(1-\mu)|\lambda-1|T^{\delta_2+\alpha-\beta+1}}
 {\lambda\Lambda_3\Gamma(\delta_2+\alpha-\beta+1)}.
\end{gather*}

\begin{theorem}\label{Th-1-2-1}
Let condition {\rm (H1)} of Theorem \ref{t-Sad} be satisfied.
If $\Omega_3<1$, then  problem \eqref{E}-\eqref{bc2} has at least one
solution on $J$.
\end{theorem}

\begin{theorem}\label{Th-1-2-2}
Let condition {\rm (H1)} of Theorem \ref{t-Sad} be satisfied.
If $\Omega_4<1$, then  problem \eqref{E}-\eqref{bc3} has at least one
solution on $J$.
\end{theorem}

\begin{remark} \rm
If $\lambda=1$, then \eqref{E} is reduced to a single order fractional
differential equation and also $\Omega_1=\Omega_3=\Omega_4=0$.
In this case, only condition (H1) can be used for the  existence of solutions
for problems  \eqref{E}-\eqref{bc1}, \eqref{E}-\eqref{bc2} and
\eqref{E}-\eqref{bc3}.
\end{remark}


\begin{example} \rm
Let us consider the following two orders fractional differential equation
with two orders fractional derivative boundary conditions
\begin{gather}
\frac{38}{43}D^{7/4}x(t)+\frac{5}{43}D^{5/4}x(t)
 =\frac{x(t)e^{2t}}{|x(t)|+1}\sin^2x(t)+\frac{2}{3},\quad t\in [0,3/2],\label{EXDE-1}
 \\
x(0)=0,\quad\frac{15}{32}D^{1/3}x\left(\frac{3}{2}\right)
 +\frac{17}{32}D^{1/4}x\left(\frac{3}{2}\right)=\frac{3}{4}.\label{EXBC-1}
\end{gather}
\end{example}

Here $\lambda=38/43$, $\alpha=7/4$, $\beta=5/4$, $\mu=15/32$, $\gamma_1=1/3$,
$\gamma_2=1/4$, $\gamma_3=3/4$, $T=3/2$ and $f(t,x)=(xe^{2t}\sin^2x)/(|x|+1)+(2/3)$.
Observe that $0<\gamma_1,\gamma_2<(1/2)=\alpha-\beta$. It is obvious that
\[
|f(t,x)|\leq e^{2t}+\frac{2}{3}:=v(t),
\]
which satisfies condition (H1) of Theorem \ref{t-Sad}.
In addition, we can find that
\begin{equation*}
\Omega_1=0.3421779589<1.
\end{equation*}
Hence, by Theorem \ref{t-Sad}, the four orders fractional boundary value
problem \eqref{EXDE-1}-\eqref{EXBC-1} has at least one solution on $[0,3/2]$.

\begin{example} \rm
Let us consider the two orders fractional differential equation \eqref{EXDE-1}
subject to two orders fractional boundary conditions
\begin{equation}
x(0)=0,\quad\frac{7}{16}I^{3/4}x\left(\frac{3}{2}\right)
+\frac{9}{16}I^{4/5}x\left(\frac{3}{2}\right)=\frac{1}{6}, \label{EXBC-2}
\end{equation}
and mixed fractional derivative and integral boundary conditions
\begin{equation}
x(0)=0,\quad\frac{13}{28}D^{1/5}x\left(\frac{3}{2}\right)
 +\frac{15}{28}I^{4/3}x\left(\frac{3}{2}\right)=\frac{3}{7}.
\label{EXBC-3}
\end{equation}
\end{example}


\noindent\textbf{Problem I} \eqref{EXDE-1}-\eqref{EXBC-2}.
In this case $\mu=7/16$, $\delta_1=3/4$, $\delta_2=4/5$ and $\delta_3=1/6$.
We can find that $\Lambda_2=1.249160013$ and $\Omega_3=0.4121621065<1$.
Therefore, by applying  Theorem \ref{Th-1-2-1}, the two orders fractional
derivatives and integrals boundary value problem \eqref{EXDE-1}-\eqref{EXBC-2}
 has at least one solution on $[0,3/2]$.

\noindent\textbf{Problem II} \eqref{EXDE-1}-\eqref{EXBC-3}.
In the final case $\mu=13/28$, $\gamma_1=1/5$, $\delta_2=4/3$ and $\gamma_3=3/7$.
We can find that $\Lambda_3=1.186148831$ and $\Omega_4=0.3877544803<1$.
Therefore, by using the conclusion in  Theorem \ref{Th-1-2-2},
the mixed type of fractional derivative and integral boundary value
problem \eqref{EXDE-1}-\eqref{EXBC-3} has at least one solution on $[0,3/2]$.

\section{Existence result  for problem \eqref{E-inc}-\eqref{bc1}}

First of all, we recall some basic concepts for  multi-valued
maps \cite{De, Pa, Sm}.
For a normed space $(X, \|\cdot\|)$, let
$\mathcal{P}_{cl}(X)=\{Y \in {\mathcal{P}}(X) : Y  \text{ is  closed}\}$,
$\mathcal{P}_{b}(X)=\{Y \in {\mathcal{P}}(X) : Y  \text{ is  bounded}\}$,
$\mathcal{P}_{cp}(X)=\{Y \in {\mathcal{P}}(X) : Y  \text{ is  compact}\}$ and
$\mathcal{P}_{cp, c}(X)=\{Y \in {\mathcal{P}}(X) : Y  \text{ is compact and
 convex}\}$.

 A multi-valued map $G : X \to {\mathcal{P}}(X)$:
\begin{itemize}
\item[(i)] is \emph{convex (closed) valued} if $G(x)$ is convex
(closed) for all $x \in X$;

\item[(ii)] is  \emph{bounded} on
bounded sets if $G(\mathbb{B}) = \cup_{x \in \mathbb{B}}G(x)$ is
bounded in $X$ for all $\mathbb{B} \in \mathcal{P}_{b}(X)$  (i.e.
$\sup_{x \in \mathbb{B}}\{\sup \{|y| : y \in G(x)\}\} < \infty)$;

\item[(iii)]  is called \emph{upper semi-continuous (u.s.c.)} on
$X$ if for each $x_0 \in X$, the set $G(x_0)$ is a nonempty closed
subset of $X$, and if for each open set $N$ of $X$ containing
$G(x_0)$, there exists an open neighborhood $\mathcal{N}_0$ of
$x_0$ such that $G(\mathcal{N}_0) \subseteq N$;

\item[(iv)] $G$ is \emph{lower semi-continuous (l.s.c.)} if the set
$\{y \in X : G(y)\cap B \ne \emptyset\}$ is open for any open set $B$ in $E$;

\item[(v)]  is said to be \emph{completely continuous} if
$G(\mathbb{B})$ is relatively compact for every $\mathbb{B} \in
\mathcal{P}_{b}(X)$;

\item[(vi)]  is said to be \emph{measurable} if for every $y \in \mathbb{R}$,
the function
$$
t  \mapsto d(y,G(t)) = \inf\{|y-z|: z \in G(t)\}$$ is measurable;

\item[(vii)] \emph{has a fixed point} if there is $x \in X$ such
that $x \in G(x)$. The fixed point set of the multivalued
operator $G$ will be denoted by ${\it{Fix}} G$.
\end{itemize}


\begin{definition} \rm
A multivalued map $F : J \times \mathbb{R} \to {\mathcal{P}}(\mathbb{R})$,
$J:=[0,T]$, is said to be Carath\'{e}odory if
\begin{itemize}
\item[(i)] $t  \mapsto F(t,x)$ is measurable for each
$x \in \mathbb{R}$;
\item[(ii)] $x  \mapsto F(t,x)$ is upper
semicontinuous for almost all $t\in J$.
 \end{itemize}
Further a Carath\'{e}odory function $F$ is called
$L^1-$Carath\'{e}odory if
\begin{itemize}
\item[(iii)] for each $\alpha > 0$, there exists
$\varphi_{\alpha} \in L^1(J,\mathbb{R}^+)$ such that
$$
\|F (t, x)\| = \sup \{|v| : v \in F (t, x)\} \le \varphi_{\alpha} (t)
$$
for all  $\|x\| \le \alpha$ and for  a.e.
$t \in J$.
\end{itemize}
\end{definition}

Recall that  $\mathcal{C}:=C(J, {\mathbb R})$. For each $x \in \mathcal{C}$,
define the set of selections of $F$ by
$$
S_{F,x} := \{ v \in L^1(J,\mathbb{R}) :
v (t) \in F (t, x(t)) ~\text{for a.e.} ~t \in J\}.
$$

 We define the graph of $G$ to be the set
$Gr(G)=\{(x,y)\in X\times Y, y\in G(x)\}$ and recall two useful results
regarding  closed graphs and upper-semicontinuity.


 \begin{lemma}[{\cite[Proposition 1.2]{De}}] \label{lemusc}
If $G : X \to \mathcal{P}_{cl}(Y)$ is u.s.c., then $Gr(G)$
is a closed subset of $X \times Y$; i.e., for every sequence
$\{x_n\}_{n \in \mathbb{N}} \subset X$ and
$\{y_n\}_{n \in \mathbb{N}} \subset Y$, if when $n \to \infty$, $x_n \to x_*$,
$y_n \to y_*$ and $y_n \in G(x_n)$, then $y_* \in G(x_*)$.
Conversely, if $G$ is completely continuous and has a closed
graph, then it is upper semi-continuous.
\end{lemma}

\begin{lemma}[\cite{LaOp}] \label{l1i}
Let $X$ be a Banach space. Let
$F : J \times \mathbb{R} \to \mathcal{P}_{cp,c}(X)$ be an
$L^1-$ Carath\'{e}odory multivalued map and let $\Theta$ be a
linear continuous mapping from $L^1(J,X)$ to $C(J,X)$.
Then the operator
$$
\Theta \circ S_F : C(J,X) \to \mathcal{P}_{cp,c} (C(J,X)), \quad
 x \mapsto (\Theta \circ S_F) (x) = \Theta( S_{F,x,y})
$$
is a closed graph operator in $C(J,X) \times C(J,X)$.
\end{lemma}

To prove our main result in this section,  we  use  the following
form of  the nonlinear alternative for contractive maps
\cite[Corollary 3.8]{PF}.

\begin{theorem}\label{t21}
Let $X$ be a Banach space, and $D$ a bounded neighborhood of
$0\in X$. Let  $Z_1:X\to \mathcal{P}_{cp,c}(X)$  and
$Z_2:\bar D\to \mathcal{P}_{cp,c}(X)$  two multi-valued operators satisfying
\begin{itemize}
\item[(a)] $Z_1$ is  contraction, and \item [(b)] $Z_2$ is upper semi-continuous
and compact.
\end{itemize}
Then, if $Q=Z_1+Z_2$, either
\begin{itemize}
\item[(i)]  $Q$ has a fixed point in $\bar D$ or
\item[(ii)] there is a point $u\in\partial D$ and $\lambda\in (0,1)$ with
$u\in\lambda Q(u)$.
\end{itemize}
\end{theorem}

\begin{definition}  \rm
A function $x \in C^2(J, \mathbb R)$ is a solution of
 problem  \eqref{E-inc}-\eqref{bc1}  if  $ x(0)=0$,
$\mu D^{\gamma_1}x(T)+(1-\mu)D^{\gamma_2}x(T) = \gamma_3$, and there exists
function  $v  \in L^1(J,\mathbb{R})$ such that $v(t) \in F(t, x(t))$   a.e. on
$J$ and
\begin{equation} \label{Big-F}
\begin{aligned}
 x(t) &=   \frac{\lambda-1}{\lambda \Gamma(\alpha-\beta)}
\int_0^t (t-s)^{\alpha-\beta-1}x(s)ds + \frac{1}{\lambda \Gamma(\alpha)}
 \int_0^t (t-s)^{\alpha-1}v(s)ds \\
&\quad + \frac{t}{\Lambda_1}\Big( \gamma_3
 - \frac{\mu(\lambda-1)}{\lambda \Gamma(\alpha-\beta-\gamma_1)}
 \int_0^T (T-s)^{\alpha-\beta-\gamma_1-1}x(s)ds \\
&\quad - \frac{\mu}{\lambda \Gamma(\alpha-\gamma_1)}
 \int_0^T (T-s)^{\alpha-\gamma_1-1}v(s)ds \\
&\quad - \frac{(1-\mu)(\lambda-1)}{\lambda \Gamma(\alpha-\beta-\gamma_2)}
 \int_0^T (T-s)^{\alpha-\beta-\gamma_2-1}x(s)ds \\
&\quad - \frac{1-\mu}{\lambda \Gamma(\alpha-\gamma_2)}
 \int_0^T (T-s)^{\alpha-\gamma_2-1}v(s)ds \Big), \quad t\in J,
\end{aligned}
\end{equation}
where $ \Lambda_1 \neq 0$ is defined by \eqref{eq-l-1}.
\end{definition}

\begin{theorem}\label{t1i}
 Assume that {\rm (H2)} holds. In addition we assume that:
\begin{itemize}
\item[(H3)] $F :J \times \mathbb{R} \to {\mathcal{P}_{cp,c}}(\mathbb{R})$ is
$L^1$-Carath\'{e}odory;

 \item[(H4)] there exists a continuous  nondecreasing function
$\Phi : [0,\infty) \to (0,\infty)$ and a function $p \in L^1(J,\mathbb{R}^+)$ such
that
$$
\|F(t,x)\|_\mathcal{P}:=\sup\{|y|: y \in F(t,x)\}\le p(t)\Phi(\|x\|) \quad
 \text{for each } (t,x) \in J
\times \mathbb{R};
$$

\item[(H5)] there exists a constant $M>0$ such that
\begin{equation}\label{M-eq}
\frac{(1-\Omega_1)M}{ \Phi(M)\Psi_1+ |\gamma_3| T/\Lambda_1}>1,
 \end{equation}
where
\begin{align*}
\Psi_1&= \frac{1}{\lambda\Gamma(\alpha)}\int_0^T(T-s)^{\alpha-1}p(s)ds
 +\frac{T}{\Lambda_1}\Big[\frac{\mu}{\lambda\Gamma(\alpha-\gamma_1)}
 \int_0^T(T-s)^{\alpha-\gamma_1-1}p(s)ds\\
&\quad +\frac{1-\mu}{\lambda\Gamma(\alpha-\gamma_2)}
 \int_0^T(T-s)^{\alpha-\gamma_2-1}p(s)ds\Big].
\end{align*}
\end{itemize}
Then the boundary value problem \eqref{E-inc}-\eqref{bc1}  has at least one
solution on $J$.
\end{theorem}

\begin{proof}
To transform  problem \eqref{E-inc}-\eqref{bc1}  into a
fixed point problem, we define an operator
$\mathcal{N}: \mathcal{C}\to \mathcal{P}(\mathcal{C})$ by
\begin{align*}
\mathcal{N}(x)=\Big\{&
h \in \mathcal{C}:
h(t) = \frac{\lambda-1}{\lambda \Gamma(\alpha-\beta)}
 \int_0^t (t-s)^{\alpha-\beta-1}x(s)ds  \\
& + \frac{1}{\lambda \Gamma(\alpha)}\int_0^t (t-s)^{\alpha-1}v(s)ds \\
&+  \frac{t}{\Lambda_1}\Big( \gamma_3
 - \frac{\mu(\lambda-1)}{\lambda \Gamma(\alpha-\beta-\gamma_1)}
 \int_0^T (T-s)^{\alpha-\beta-\gamma_1-1}x(s)ds \\
&- \frac{\mu}{\lambda \Gamma(\alpha-\gamma_1)}
 \int_0^T (T-s)^{\alpha-\gamma_1-1}v(s)ds \\
&-\frac{(1-\mu)(\lambda-1)}{\lambda \Gamma(\alpha-\beta-\gamma_2)}
 \int_0^T (T-s)^{\alpha-\beta-\gamma_2-1}x(s)ds \\
&- \frac{1-\mu}{\lambda \Gamma(\alpha-\gamma_2)}
 \int_0^T (T-s)^{\alpha-\gamma_2-1}v(s)ds \Big)
\Big\}
\end{align*}
for $v \in S_{F,x}$.

Next we introduce  the operator  $\mathcal{A}: \mathcal{C}\to   \mathcal{C} $
by
\begin{align*}
\mathcal{A} x(t)
&= \frac{(\lambda-1)}{\lambda\Gamma(\alpha-\beta)}
 \int_0^t (t-s)^{\alpha-\beta-1}x(s)ds \\
&\quad - \frac{t}{\Lambda_1}\Big[\frac{\mu(\lambda-1)}
 {\lambda\Gamma(\alpha-\beta-\gamma_1)}
 \int_0^T (T-s)^{\alpha-\beta-\gamma_1-1}x(s)ds\\
&\quad + \frac{(1-\mu)(\lambda-1)}{\lambda\Gamma(\alpha-\beta-\gamma_2)}
 \int_0^T (T-s)^{\alpha-\beta-\gamma_2-1}x(s)ds\Big],
\end{align*}
and the multi-valued operator
$ \mathcal{B}: \mathcal{C}\to \mathcal{P}(\mathcal{C})$ by
\begin{align*}
\mathcal{B}x(t)=\Big\{& h \in \mathcal{C}:
h(t) =  \frac{1}{\lambda\Gamma(\alpha)}\int_0^t (t-s)^{\alpha-1}v(s)ds \\
& +  \frac{t}{\Lambda_1}\Big[\gamma_3
 - \frac{\mu}{\lambda\Gamma(\alpha-\gamma_1)}
 \int_0^T (T-s)^{\alpha-\gamma_1-1}v(s)ds\\
& -  \frac{(1-\mu)}{\lambda\Gamma(\alpha-\gamma_2)}
 \int_0^T (T-s)^{\alpha-\gamma_2-1}v(s)ds \Big]
\Big\}
\end{align*}
for $v\in S_{F,x}$. Observe that  $\mathcal{N}=\mathcal{A}+\mathcal{B}$. We shall
show that the operators $\mathcal{A}$ and  $\mathcal{B}$ satisfy
all the conditions of Theorem \ref{t21} on $J$.
First, we show that the operators $\mathcal{A}$ and $\mathcal{B}$ define the
 multivalued operators
$\mathcal{A},\mathcal{B}: B_r\to\mathcal{P}_{cp,c}(\mathcal{C})$ where
$B_r = \{x \in \mathcal{C}: \|x\| \le r \}$ is  a bounded
set in $\mathcal{C}$.  First we prove that $\mathcal{B}$ is compact-valued
on $B_r$. Note that the operator $\mathcal{B}$ is equivalent to the composition
$\mathcal{L} \circ S_{F}$, where $\mathcal{L} $ is the continuous linear operator
on $L^1(J, \mathbb{R})$ into $\mathcal{C}$, defined by
 \begin{align*}
&\mathcal{L} (v)(t)\\
&=  \frac{1}{\lambda\Gamma(\alpha)}\int_0^t (t-s)^{\alpha-1}v(s)ds
 +  \frac{t}{\Lambda_1}\Big[\gamma_3 - \frac{\mu}{\lambda\Gamma(\alpha-\gamma_1)}
 \int_0^T (T-s)^{\alpha-\gamma_1-1}v(s)ds\\
&\quad -  \frac{(1-\mu)}{\lambda\Gamma(\alpha-\gamma_2)}
 \int_0^T (T-s)^{\alpha-\gamma_2-1}v(s)ds \Big].
\end{align*}

Suppose that $x\in B_r$ is arbitrary and let $\{v_n\}$ be a sequence in $S_{F,x}$.
Then, by definition of $S_{F,x}$, we have $v_n(t)\in  F(t, x(t))$ for almost
all $t\in J$. Since $F(t, x(t))$ is compact for all $t\in J$, there is a convergent
subsequence of $\{v_{n}(t)\}$ (we denote it by $\{v_{n}(t)\}$ again) that
converges in measure to some $v(t)\in S_{F,x}$ for almost all $t\in J$.
On the other hand, $\mathcal{L} $ is continuous, so
$\mathcal{L} (v_{n})(t)\to \mathcal{L} (v)(t)$ pointwise on $J$.

 To show that the convergence is uniform, we have to show that
$\{\mathcal{L} (v_{n})\}$ is an equi-continuous sequence.
Let $t_1, t_2 \in J$ with $t_1<t_2 $. Then, we have
\begin{align*}
&| \mathcal{L} (v_{n})(t_2)  -  \mathcal{L} (v_{n})(t_1)|\\
&\le  \Big|\frac{1}{\lambda\Gamma(\alpha)}
 \Big[\int_0^{t_2} (t_2-s)^{\alpha-1}v_n(s)ds
  - \int_0^{t_1} (t_1-s)^{\alpha-1}v_n(s)ds \Big] \\
&\quad + \frac{|t_2-t_1|}{\Lambda_1}\Big[|\gamma_3|
 + \frac{\mu}{\lambda\Gamma(\alpha-\gamma_1)}
 \int_0^T (T-s)^{\alpha-\beta-\gamma_1-1}v_n(s)ds\\
&\quad + \frac{(1-\mu)}{\lambda\Gamma(\alpha-\gamma_2)}
 \int_0^T (T-s)^{\alpha-\gamma_2-1}v_n(s)ds \Big]\Big|\\
&\le   \frac{\Phi(r)}{\lambda\Gamma(\alpha)}
 \Big[\int_0^{t_1} [t_2-s)^{\alpha-1}-(t_1-s)^{\alpha-1}] p(s)ds
 + \int_{t_1}^{t_2} (t_1-s)^{\alpha-1} p(s)ds \Big]  \\
&\quad +  \Phi(r)\frac{|t_2-t_1|}{\Lambda_1}\Big[|\gamma_3|
 + \frac{\mu}{\lambda\Gamma(\alpha-\gamma_1)}
 \int_0^T (T-s)^{\alpha-\beta-\gamma_1-1} p(s)ds\\
&\quad + \frac{(1-\mu)}{\lambda\Gamma(\alpha-\gamma_2)}
 \int_0^T (T-s)^{\alpha-\gamma_2-1} p(s)ds \Big].
\end{align*}

We see that the right-hand side of the above inequality tends to zero as
$t_2\to t_1$. Thus, the sequence $\{\mathcal{L} (v_n)\}$ is equi-continuous
and by using the Arzel\'a-Ascoli theorem, we obtain that there is a uniformly
convergent subsequence. So, there is a subsequence of $\{v_{n}\}$
(we denote it again by $\{v_{n}\}$) such that
$\mathcal{L} (v_{n})\to \mathcal{L} (v)$. Note that,
$\mathcal{L} (v) \in \mathcal{L} (S_{F,x})$. Hence,
$\mathcal{B}(x) =\mathcal{L} (S_{F,x})$ is compact for all $x\in B_r$.
So $\mathcal{B}(x)$ is compact.

Now, we show that $\mathcal{B}(x)$ is convex for all $x\in \mathcal{C}$.
 Let $h_1,h_2\in \mathcal{B}(x)$. We select $v_1,v_2\in S_{F,x}$ such that
 \begin{align*}
&h_i(t)\\
&=  \frac{1}{\lambda\Gamma(\alpha)}\int_0^t (t-s)^{\alpha-1}v_i(s)ds
 +  \frac{t}{\Lambda_1}\Big[\gamma_3 - \frac{\mu}{\lambda\Gamma(\alpha-\gamma_1)}
 \int_0^T (T-s)^{\alpha-\gamma_1-1}v_i(s)ds\\
&\quad -  \frac{(1-\mu)}{\lambda\Gamma(\alpha-\gamma_2)}
 \int_0^T (T-s)^{\alpha-\gamma_2-1}v_i(s)ds \Big], \quad i=1,2,
\end{align*}
for almost all $t\in J$. Let $0\leq \theta \leq 1$. Then, we have
 \begin{align*}
&[\theta h_1+(1-\theta )h_2](t)\\
&=  \frac{1}{\lambda\Gamma(\alpha)}\int_0^t (t-s)^{\alpha-1}[\theta v_1(s)
+ (1-\theta)v_2(s)]ds \\
&\quad +  \frac{t}{\Lambda_1}\Big[\gamma_3
 - \frac{\mu}{\lambda\Gamma(\alpha-\gamma_1)}
 \int_0^T (T-s)^{\alpha-\gamma_1-1}[\theta v_1(s) + (1-\theta)v_2(s)]ds\\
&\quad -  \frac{(1-\mu)}{\lambda\Gamma(\alpha-\gamma_2)}
 \int_0^T (T-s)^{\alpha-\gamma_2-1}[\theta v_1(s)
 + (1-\theta)v_2(s)]ds \Big].
\end{align*}


Since $F$ has convex values, so $S_{F,u}$ is convex and
$\theta v_1(s) + (1-\theta)v_2(s)\in S_{F,x}$. Thus
$$
\theta h_1+(1-\theta)h_2\in \mathcal{B}(x).
$$
Consequently, $\mathcal{B}$ is convex-valued. Obviously,  $\mathcal{A}$
is compact and convex-valued.

The rest of the proof consists of several steps and claims.
\smallskip

\noindent\textbf{Step 1:} $\mathcal{A}$  is a
contraction on $\mathcal{C}$.
This was proved in Step 3 of Theorem  \ref{t-Sad}.
\smallskip

\noindent\textbf{Step 2}:
 $\mathcal{B}$ is compact and  upper semi-continuous. This will be
established in several claims.

\noindent{\sc Claim I:} $\mathcal{B}$ maps bounded sets
into bounded sets in $\mathcal{C}$. Let $B_r = \{x \in
\mathcal{C}: \|x\| \le r \}$ be a bounded set in
$\mathcal{C}$. Then, for each $h \in \mathcal{B} (x), x
\in B_r$, there exists  $v \in S_{F,x}$ such that
 \begin{align*}
&h(t)\\
&=  \frac{1}{\lambda\Gamma(\alpha)}\int_0^t (t-s)^{\alpha-1}v(s)ds
 +  \frac{t}{\Lambda_1}\Big[\gamma_3
 - \frac{\mu}{\lambda\Gamma(\alpha-\gamma_1)}
 \int_0^T (T-s)^{\alpha-\gamma_1-1}v(s)ds\\
&\quad -  \frac{(1-\mu)}{\lambda\Gamma(\alpha-\gamma_2)}
 \int_0^T (T-s)^{\alpha-\gamma_2-1}v(s)ds \Big].
\end{align*}
Then, for $t\in J$, we have
\begin{align*}
 |h(t)| &\leq \Phi(r)\frac{1}{\lambda\Gamma(\alpha)}\int_0^T (T-s)^{\alpha-1}p(s)ds\\
&\quad +  \frac{T\Phi(r)}{\Lambda_1}\Big[|\gamma_3|
 + \frac{\mu}{\lambda\Gamma(\alpha-\gamma_1)}
 \int_0^T (T-s)^{\alpha-\gamma_1-1}p(s)ds\\
&\quad +  \frac{(1-\mu)}{\lambda\Gamma(\alpha-\gamma_2)}
 \int_0^T (T-s)^{\alpha-\gamma_2-1}p(s)ds \Big].
\end{align*}
Thus,
\begin{align*}
 \|h\|
&\leq \Phi(r)\frac{1}{\lambda\Gamma(\alpha)}\int_0^T (T-s)^{\alpha-1}p(s)ds\\
&\quad +  \frac{T\Phi(r)}{\Lambda_1}\Big[|\gamma_3|
 + \frac{\mu}{\lambda\Gamma(\alpha-\gamma_1)}
 \int_0^T  (T-s)^{\alpha-\gamma_1-1}p(s)ds\\
&\quad +  \frac{(1-\mu)}{\lambda\Gamma(\alpha-\gamma_2)}
 \int_0^T (T-s)^{\alpha-\gamma_2-1}p(s)ds \Big].
\end{align*}

\noindent{\sc Claim II:} $\mathcal{B}$ maps bounded sets
into equi-continuous sets. Let $t_1, t_2 \in J$ with
$t_1< t_2$ and  $x \in B_{r}$. Then, for each  $h \in \mathcal{B}(x)$, we obtain
\begin{align*}
&|h(t_2)-h(t_1)|\\
&\le  \frac{\Phi(r)}{\lambda\Gamma(\alpha)}
\Big[\int_0^{t_1} [t_2-s)^{\alpha-1}-(t_1-s)^{\alpha-1}] p(s)ds
+ \int_{t_1}^{t_2} (t_1-s)^{\alpha-1} p(s)ds \Big]  \\
&\quad +  \Phi(r)\frac{|t_2-t_1|}{\Lambda_1}\Big[|\gamma_3|
 + \frac{\mu}{\lambda\Gamma(\alpha-\gamma_1)}
 \int_0^T (T-s)^{\alpha-\beta-\gamma_1-1} p(s)ds\\
&\quad + \frac{(1-\mu)}{\lambda\Gamma(\alpha-\gamma_2)}
 \int_0^T (T-s)^{\alpha-\gamma_2-1} p(s)ds \Big].
\end{align*}
Obviously the right-hand side of the above inequality tends to
zero independently of $x \in B_{r}$ as $t_2- t_1 \to 0$.
Therefore it follows by the Ascoli-Arzel\'a theorem that
$\mathcal{B}: \mathcal{C} \to {\mathcal{P}}(\mathcal{C})$ is completely
continuous.

Next we show that $\mathcal{B}$ is an upper semi-continuous multi-valued mapping.
It is knowm by Lemma \ref{lemusc} that $\mathcal{B}$ will be upper
semicontinuous if we establish that it has a closed graph, since already
shown to be completely continuous. Thus we will prove that:

\noindent{\sc Claim III}: $\mathcal{B}$ has a closed graph.
Let $x_n \to x_*, h_n \in \mathcal{B} (x_n)$ and $h_n \to  h_*$. Then
we need to show that $h_* \in  \mathcal{B}(x_*)$. Associated with
$h_n \in \mathcal{B} (x_n)$, there exists $v_n \in S_{F,x_n}$
such that for each $t \in J$,
\begin{align*}
h_n(t)
&=  \frac{1}{\lambda\Gamma(\alpha)}\int_0^t (t-s)^{\alpha-1}v_n(s)ds
+  \frac{t}{\Lambda_1}\Big[\gamma_3 - \frac{\mu}{\lambda\Gamma(\alpha-\gamma_1)}
 \int_0^T (T-s)^{\alpha-\gamma_1-1}v_n(s)ds\\
&\quad -  \frac{(1-\mu)}{\lambda\Gamma(\alpha-\gamma_2)}
 \int_0^T (T-s)^{\alpha-\gamma_2-1}v_n(s)ds \Big].
\end{align*}

Thus it suffices to show that there exists $v_* \in  S_{F,x_*}$
such that for each $t \in J$,
\begin{align*}
h_*(t)
&=  \frac{1}{\lambda\Gamma(\alpha)}\int_0^t (t-s)^{\alpha-1}v_*(s)ds \\
&\quad +  \frac{t}{\Lambda_1}\Big[\gamma_3
 - \frac{\mu}{\lambda\Gamma(\alpha-\gamma_1)}
 \int_0^T (T-s)^{\alpha-\gamma_1-1}v_*(s)ds \\
&\quad -  \frac{(1-\mu)}{\lambda\Gamma(\alpha-\gamma_2)}
 \int_0^T (T-s)^{\alpha-\gamma_2-1}v_*(s)ds \Big].
\end{align*}

 Let us consider the   linear operator
$\Theta : L^1(J, \mathbb{R}) \to \mathcal{C}$ given
by
\begin{align*}
v \mapsto \Theta(v)(t)
&=  \frac{1}{\lambda\Gamma(\alpha)}\int_0^t (t-s)^{\alpha-1}v(s)ds \\
&\quad +  \frac{t}{\Lambda_1}\Big[\gamma_3
 - \frac{\mu}{\lambda\Gamma(\alpha-\gamma_1)}
 \int_0^T (T-s)^{\alpha-\gamma_1-1}v(s)ds\\
&\quad -  \frac{(1-\mu)}{\lambda\Gamma(\alpha-\gamma_2)}
\int_0^T (T-s)^{\alpha-\gamma_2-1}v(s)ds \Big].
\end{align*}
Observe that
 \begin{align*}
\|h_n(t)-h_*(t)\|
&=  \Big\|\frac{1}{\lambda\Gamma(\alpha)}\int_0^t (t-s)^{\alpha-1}(v_n(s)-v_*(s))ds \\
&\quad +  \frac{t}{\Lambda_1}\Big[- \frac{\mu}{\lambda\Gamma(\alpha-\gamma_1)}
 \int_0^T (T-s)^{\alpha-\gamma_1-1}(v_n(s)-v_*(s))ds\\
&\quad -  \frac{(1-\mu)}{\lambda\Gamma(\alpha-\gamma_2)}
 \int_0^T (T-s)^{\alpha-\gamma_2-1}(v_n(s)-v_*(s))ds \Big]\Big\|  \to
0,
\end{align*}
as  $n\to \infty$.

Thus, it follows by Lemma \ref{l1i} that $\Theta \circ S_F$ is a
closed graph operator. Further, we have $h_n(t) \in
\Theta(S_{F,x_n})$. Since  $x_n \to x_*$, we have that
\begin{align*}
&h_*(t)\\
&=  \frac{1}{\lambda\Gamma(\alpha)}\int_0^t (t-s)^{\alpha-1}v_*(s)ds
 +  \frac{t}{\Lambda_1}\Big[\gamma_3 - \frac{\mu}
 {\lambda\Gamma(\alpha-\gamma_1)}\int_0^T (T-s)^{\alpha-\gamma_1-1}v_*(s)ds\\
&\quad -  \frac{(1-\mu)}{\lambda\Gamma(\alpha-\gamma_2)}
 \int_0^T (T-s)^{\alpha-\gamma_2-1}v_*(s)ds \Big],
\end{align*}
for some $v_* \in  S_{F,x_*}$.  Hence $\mathcal{B}$ has a closed
graph (and therefore has closed values). In consequence, the
operator $\mathcal{B}$ is compact valued and upper semi-continuous.

 Thus the operators $\mathcal{A}$ and $\mathcal{B}$ satisfy all the
conditions of Theorem \ref{t21} and hence its conclusion implies
either condition (i) or condition (ii) holds. We show that the
conclusion (ii) is not possible. If $x \in \theta \mathcal{A}
(x)+\theta \mathcal{B}(x)$ for $\theta\in  (0,1)$, then there
exist  $v\in S_{F,x}$     such that
\begin{align*}
 x(t)
&=  \theta \frac{\lambda-1}{\lambda \Gamma(\alpha-\beta)}
 \int_0^t (t-s)^{\alpha-\beta-1}x(s)ds
 + \theta\frac{1}{\lambda \Gamma(\alpha)}\int_0^t (t-s)^{\alpha-1}v(s)ds \\
&\quad +\theta\frac{t}{\Lambda_1}\Big( \gamma_3 - \frac{\mu(\lambda-1)}
 {\lambda \Gamma(\alpha-\beta-\gamma_1)}
 \int_0^T (T-s)^{\alpha-\beta-\gamma_1-1}x(s)ds  \\
&\quad -  \frac{\mu}{\lambda \Gamma(\alpha-\gamma_1)}
 \int_0^T (T-s)^{\alpha-\gamma_1-1}v(s)ds \\
&\quad-  \frac{(1-\mu)(\lambda-1)}{\lambda \Gamma(\alpha-\beta-\gamma_2)}
 \int_0^T (T-s)^{\alpha-\beta-\gamma_2-1}x(s)ds  \\
&\quad-  \frac{1-\mu}{\lambda \Gamma(\alpha-\gamma_2)}
 \int_0^T (T-s)^{\alpha-\gamma_2-1}v(s)ds \Big),  \quad t\in J.
\end{align*}

By our assumptions, we  obtain
\begin{align*}
|x(t)| &\le    \|x\|\Big[\frac{T^{\alpha-\beta}|\lambda-1|}
 {\lambda\Gamma(\alpha-\beta+1)}
 +\frac{T^{\alpha-\beta-\gamma_1+1}\mu|\lambda-1|}
 {\lambda\Lambda_1\Gamma(\alpha-\beta-\gamma_1+1)}\\
&\quad  +\frac{T^{\alpha-\beta-\gamma_2+1}(1-\mu)|\lambda-1|}
 {\lambda\Lambda_1\Gamma(\alpha-\beta-\gamma_2+1)}\Big]
 +\frac{|\gamma_3| T}{\Lambda_1}\\
&\quad  +\Phi(\|x\|)\frac{1}{\lambda\Gamma(\alpha)}
 \int_0^T (T-s)^{\alpha-1}p(s)ds\\
&\quad +  \frac{T\Phi(\|x\|)}{\Lambda_1}
 \Big[\frac{\mu}{\lambda\Gamma(\alpha-\gamma_1)}
 \int_0^T (T-s)^{\alpha-\gamma_1-1}p(s)ds\\
&\quad +  \frac{(1-\mu)}{\lambda\Gamma(\alpha-\gamma_2)}
 \int_0^T (T-s)^{\alpha-\gamma_2-1}p(s)ds \Big].
\end{align*}
Thus
\begin{equation}\label{ineq}
(1-\Omega_1)   \|x\|\le \Phi(\|x\|) \Psi_1+|\gamma_3| T/\Lambda_1.
\end{equation}

If condition (ii) of Theorem \ref{t21} holds, then there exists
$\theta\in  (0, 1)$ and $x\in \partial B_M$ with
$x = \theta \mathcal{N}(x)$. Then, $x$ is a solution of
 \eqref{E-inc}-\eqref{bc1} with
$\|x\| = M$. Now, by the inequality \eqref{ineq}, we obtain
$$
\frac{(1-\Omega_1)M}{ \Phi(M) \Psi_1+|\gamma_3| T/\Lambda_1} \le 1,
$$
 which contradicts  \eqref{M-eq}. Hence, $\mathcal{N}$ has a fixed point in
$J$ by Theorem \ref{t21}, and consequently the problem \eqref{E-inc}-\eqref{bc1}
 has a solution.  This completes the proof.
\end{proof}

In the above results, we define the following two constants
\begin{align*}
\Psi_2&= \frac{1}{\lambda\Gamma(\alpha)}\int_0^T(T-s)^{\alpha-1}p(s)ds
 +\frac{T}{\Lambda_2}\Big[\frac{\mu}{\lambda\Gamma(\delta_1+\alpha)}
 \int_0^T(T-s)^{\delta_1+\alpha-1}p(s)ds\\
&\quad +\frac{1-\mu}{\lambda\Gamma(\delta_2+\alpha)}
 \int_0^T(T-s)^{\delta_2+\alpha-1}p(s)ds\Big],\\
\Psi_3&= \frac{1}{\lambda\Gamma(\alpha)}
 \int_0^T(T-s)^{\alpha-1}p(s)ds+\frac{T}{\Lambda_3}\Big[\frac{\mu}
 {\lambda\Gamma(\alpha-\gamma_1)}
 \int_0^T(T-s)^{\alpha-\gamma_1-1}p(s)ds\\
&\quad +\frac{1-\mu}{\lambda\Gamma(\delta_2+\alpha)}
 \int_0^T(T-s)^{\delta_2+\alpha-1}p(s)ds\Big].
\end{align*}

\begin{theorem}\label{Th-2-2-1}
Let $\Omega_3<1$. Assume that the conditions {\rm (H3), (H4)} are satisfied.
If there exists a positive constant $M$ such that
\begin{equation*}
\frac{(1-\Omega_3)M}{\Phi(M)\Psi_2+|\delta_3|T/\Lambda_2}>1,
\end{equation*}
then  problem \eqref{E-inc}-\eqref{bc2} has at least one solution on $J$.
\end{theorem}

\begin{theorem}\label{Th-2-2-2}
Let $\Omega_4<1$. Suppose that the conditions {\rm (H3), (H4)} are satisfied.
 If there exists a positive constant $M$ such that
\begin{equation*}
\frac{(1-\Omega_4)M}{\Phi(M)\Psi_3+|\gamma_3|T/\Lambda_3}>1,
\end{equation*}
then  problem \eqref{E-inc}-\eqref{bc3} has at least one solution on $J$.
\end{theorem}


\begin{example} \rm
Let us consider the following two order fractional differential inclusion
with two order fractional derivative boundary conditions
\begin{equation}\label{ex1}
\begin{gathered}
  \frac{47}{54}D^{16/9}x(t)+\frac{7}{54}D^{10/9}x(t)\in F(t,x(t)),\quad
 t\in [0,1],\\
 x(0)=0,\quad\frac{9}{23}D^{7/15}x(1)+\frac{14}{23}D^{4/15}x(1)=\frac{1}{12}.
\end{gathered}
\end{equation}
where $F(t,x)$ is the multivalue function
\begin{equation*}
F(t,x)=\Big[\Big(\frac{\sqrt{t}+1}{5}\Big)
\Big(\frac{|x|\sin^2x}{18(1+|x|)}+\frac{1}{4}\Big),
\Big(\sqrt[3]{t}+\frac{1}{4}\Big)\Big(\frac{|x|\sin x}{15}+\frac{1}{2}\Big)\Big].
\end{equation*}
Here $\lambda=47/54$, $\alpha=16/9$, $\beta=10/9$, $\mu=9/23$, $\gamma_1=7/15$,
$\gamma_2=4/15$, $\gamma_3=1/12$, $T=1$. Observe that
$0<\gamma_1,\gamma_2<2/3=\alpha-\beta$. We can find that $\Lambda_1=1.105743248$
and $\Omega_1=0.3147893857$. It is easy to see that
\begin{equation*}
\|F(t,x)\|_{\mathcal{P}}=\sup\{|y|\,:\,y\in F(t,x)\}
\leq\Big(\sqrt[3]{t}+\frac{1}{4}\Big)\Big(\frac{|x|}{15}+\frac{1}{2}\Big).
\end{equation*}
Set   $p(t)=\sqrt[3]{t}+(1/4)$ and $\Phi(x)=(x/15)+(1/2)$.
By direct computation, we have $\Psi_1=1.410896861$. From the given data,
we can prove that there exists a positive constant $M>1.267866938$
satisfying inequality \eqref{M-eq} of Theorem \ref{t1i}. Therefore,
by applying Theorem \ref{t1i}, we deduce that the boundary value
problem \eqref{ex1} has at least one solution on $[0,1]$.
\end{example}

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\end{document}
