\documentclass[reqno]{amsart}
\usepackage{hyperref}

\AtBeginDocument{{\noindent\small
\emph{Electronic Journal of Differential Equations},
Vol. 2012 (2012), No. 187, pp. 1--26.\newline
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu
\newline ftp ejde.math.txstate.edu}
\thanks{\copyright 2012 Texas State University - San Marcos.}
\vspace{9mm}}

\begin{document}
\title[\hfilneg EJDE-2012/187\hfil Existence of multiple positive solutions]
{Existence of multiple positive solutions for fractional differential 
 inclusions with m-point boundary conditions and two fractional orders}

\author[N. Nyamoradi, M. Javidi \hfil EJDE-2012/187\hfilneg]
{Nemat Nyamoradi, Mohamad Javidi}  % in alphabetical order

\address{Nemat Nyamoradi \newline
Department of Mathematics, Faculty of Sciences\\
Razi University, 67149 Kermanshah, Iran}
\email{nyamoradi@razi.ac.ir, neamat80@yahoo.com}

\address{Mohamad Javidi \newline
Department of Mathematics, Faculty of Sciences\\
Razi University, 67149 Kermanshah, Iran}
\email{mo\_javidi@yahoo.com}

\thanks{Submitted August 28, 2012. Published October 28, 2012.}
\subjclass[2000]{47H10, 26A33, 34A08}
\keywords{Inclusion;  multi point boundary value problem; cone; \hfill\break\indent
fixed point theorem; fractional derivative}

\begin{abstract}
 We study boundary-value problems of nonlinear fractional
 differential equations and inclusions with $m$-point boundary
 conditions. Several results are obtained by using suitable fixed
 point theorems when the right hand side has convex or non
 convex values.
\end{abstract}

\maketitle
\numberwithin{equation}{section}
\newtheorem{theorem}{Theorem}[section]
\newtheorem{lemma}[theorem]{Lemma}
\newtheorem{definition}[theorem]{Definition}
\newtheorem{remark}[theorem]{Remark}
\allowdisplaybreaks 


\section{Introduction}

Fractional calculus is the field of mathematical analysis which
deals with the investigation and applications of integrals and
derivatives of arbitrary order, the fractional calculus may be
considered an old and yet novel topic.

Recently, fractional differential equations have been of great
interest. This is because of both the intensive development of the
theory of fractional calculus itself and its applications in
various sciences, such as physics, mechanics, chemistry,
engineering, etc. For example, for fractional initial value
problems, the existence and multiplicity of solutions were
discussed in \cite{A. Babakhani,D. Delbosco,Trujillo,Trujillo1},
moreover, fractional derivative arises from many physical
processes,such as a charge transport in amorphous semiconductors
\cite{H. Scher}, electrochemistry and material science are also
described by differential equations of fractional order 
\cite{K. Diethelm, L. Gaul, W. G. Glockle, F. Mainardi,R. Metzler}.

The existence of solutions of initial value problems for
fractional order differential equations have been studied in the
literature \cite{R.P. Agarwal, V. Lakshmikantham,
Nia1, Nia2, Nia3, Nia4, Nia5, I. Podlubny, G. Samko} and the references
therein. The study of fractional differential inclusions was
initiated by El-Sayed and Ibrahim \cite{Ibrahim}. Also, recently
several qualitative results for fractional differential inclusions
were obtained in \cite{Benchohra, Cernea, Ntouyas, Ouahab1} and the
references therein.

  Bai and L\"{u} \cite{Z. B. Bai} considered the boundary-value
problem of fractional-order differential equation
\begin{gather*}
D_{0^+}^\alpha u(t) + f (t, u (t))=0,\quad    t \in (0, 1),\\
 u(0)=  u(1)=0,
\end{gather*}
where  $D_{0^+}^\alpha$ is the standard Riemann-Liouville
fractional derivative of order
$1<\alpha \leq 2$ and $f:[0,1]\times [0, \infty)\to [0,\infty)$ is continuous.

 Hussein \cite{Hussein}  considered the
 nonlinear m-point boundary-value problem of fractional
type
\begin{gather*}
    D_{0^+}^\alpha x (t)+ q(t) f (t, x (t))=0,
    \quad   \text{a.e.  on }  [0, 1],\; \alpha \in (n-1,n],\; n\geq 2\\
     x(0)=  x'(0)=x''(0)=\dots=x^{(n-2)} (0)=0,\quad
   x(1)= \sum_{i=1}^{m-2} \xi_i x (\eta_i)
\end{gather*}
where $ 0<\eta_1<\dots<\eta_{m-2}<1$, $\xi_i>0 $ with
$\sum_{i=1}^{m-2} \xi_i \eta_i^{\alpha-1}<1$, $q$ is a
real-valued continuous function and $f$ is a nonlinear Pettis
integrable function.

In the past few decades, many important results relative to
equation \eqref{1} with certain boundary value conditions have
been obtained. we refer the reader to \cite{T. Chen, D. Jiang,
L. Lian, B. Liu, J. Zhang} and the references therein.

 Motivated by the mentioned works,
our purpose in the first part of this paper
is to show the existence and multiplicity of positive solutions
for the boundary-value problem of the fractional differential
equation
\begin{equation}\label{1}
\begin{gathered}
     D_{0^+}^\beta (D_{0^+}^\alpha u) (t) = f (t, u (t)),  \quad   t \in (0, 1),\\
     D_{0^+}^\alpha u(0)= D_{0^+}^\alpha u(1)=0,\quad
     u (0) = 0, \quad u (1) -\sum_{i=1}^{m-2} a_{i}\; u(\xi_{i})=\lambda,
\end{gathered}
\end{equation}
where $D_{0^+}^\alpha$ is the Riemann-Liouville fractional
derivative of order $\alpha$, $m > 2$, $1<\alpha,\beta\leq 2$, 
$2 <\alpha + \beta\leq 4$, $0<\xi_{j1}<\xi_{j2}<\dots<\xi_{j{m-2}}<1$, 
$ a_{i} > 0$ for $i=1,2,\dots,m-2$ and
$\sum_{i=1}^{m-2} a_{i}\xi_{i}^{\alpha-1}<1$, $\lambda> 0$ is a
parameter and $f : [0, 1] \times \mathbb{R} \to \mathbb{R}$ is a given 
continuous function.

In the second part of this paper, we consider a nonlinear
fractional differential inclusion of an arbitrary order with
multi-strip boundary conditions
\begin{equation}\label{110}
     \begin{gathered}
     D_{0^+}^\beta (D_{0^+}^\alpha u) (t) \in F (t, u (t)),
    \quad     t \in (0, 1),\\
              D_{0^+}^\alpha u(0)= D_{0^+}^\alpha u(1)=0,\quad
         u (0) = 0, \quad u (1) -\sum_{i=1}^{m-2} a_{i}\; u(\xi_{i})=\lambda,
\end{gathered}
\end{equation}
where $ D_{0^+}^\alpha$ is the standard Riemann-Liouville
fractional derivative and $F : [0, + \infty) \times \mathbb{R}
\times \mathbb{R} \to \mathcal{P} (\mathbb{R})$ is a
set-valued map.

The aim here is to establish existence results for the problem
\eqref{1}, when the right-hand side is convex as well as nonconvex
valued. In the first result (Theorem \ref{tusc}) we consider the
case when the right hand side has convex values, and prove an
existence result via Nonlinear alternative for Kakutani maps. In
the second result (Theorem \ref{tlsc}), we shall combine the
nonlinear alternative of Leray-Schauder type for single-valued
maps with a selection theorem due to Bressan and Colombo for lower
semicontinuous multivalued maps with nonempty closed and
decomposable values, while in the third result (Theorem
\ref{tlip}), we shall use the fixed point theorem for contraction
multivalued maps due to Covitz and Nadler.


 The rest of the article is organized as follows: in
Section 2, we present some preliminaries that will be used in
Sections 3 and 4. In Section 3, we give the existence of one and
three positive solutions for the problem \eqref{1} by using the
Leray-Schauder Alternative, Leggett-Williams fixed point theorem
and nonlinear contractions. The main result and proof for the
problem \eqref{110} will be given in Section 4. Finally, in
Section 4, an example is given to demonstrate the
application of one our main result.

\section{Preliminaries}

In this section, we present some notation and preliminary lemmas
that will be used in the Sections 3 and 4.

\begin{definition}[\cite{I. Podlubny}] \label{def1} \rm
The Riemann-Liouville fractional integral
operator of order $\alpha> 0$, of function $f \in L^1(\mathbb{R}^+)$ is defined as
\[
I_{0^+}^\alpha f(t) = \frac{1}{\Gamma (\alpha)} \int_0^t (t -
s)^{\alpha - 1} f (s)\,ds,
\]
 where $\Gamma(\cdot)$ is the Euler gamma function.
\end{definition}

\begin{definition}[\cite{A. A. Kilbas}] \label{def2} \rm
The Riemann-Liouville fractional derivative of
order $\alpha > 0$, $n - 1 < \alpha < n$, $n \in \mathbb{N}$ is
defined as
\[
D_{0^+}^\alpha f(t) = \frac{1}{\Gamma (n -
\alpha)}\Big(\frac{d}{dt}\Big)^n \int_0^t (t - s)^{n - \alpha - 1}f (s) ds,
\]
where the function $f (t)$ has
absolutely continuous derivatives up to order $(n - 1)$.
\end{definition}

\begin{lemma}[\cite{A. A. Kilbas}] \label{lemnia1}
 The equality $D_{0^+}^\gamma I_{0^+}^\gamma
f(t) = f(t)$, $\gamma > 0$ holds for $f \in L^1 (0, 1)$.
\end{lemma}

\begin{lemma}[\cite{K. Diethelm1, A. A. Kilbas}] \label{lem1} 
Let $\alpha > 0$ and $u\in C(0,1)\cap L^1(0,1) $. 
Then the differential equation
\begin{gather*}
D_{0^+}^\alpha u (t)= 0
\end{gather*}
has a unique solution
$u(t) = c_1 t^{\alpha - 1} + c_2 t^{\alpha - 2} +\dots + c_n
t^{\alpha - n}$, $c_i \in \mathbb{R}$, $i = 1, \dots, n$, where
$n - 1 < \alpha <n$.
\end{lemma}

\begin{lemma}[(\cite{A. A. Kilbas}] \label{lem2}
 Let $\alpha >0$. Then the following equality holds for $u \in L^1(0, 1)$,
$D_{0^+}^\alpha u \in L^1(0, 1)$;
\[
I_{0^+}^\alpha D_{0^+}^\alpha u (t) = u (t) + c_1 t^{\alpha - 1} +
c_2 t^{\alpha - 2} + \dots + c_n t^{\alpha - n},
\]
$c_i \in \mathbb{R}$, $i =1, \dots, n$, where $n - 1 < \alpha
\leq n$.
\end{lemma}

In the following, we present the Green function of fractional
differential equation boundary value problem.
Let
\begin{equation} \label{c}
 y(t)=-(D_{0^+}^\alpha u) (t),
\end{equation}
then, the problem
\begin{equation*}
      \begin{gathered}
    D_{0^+}^\beta (D_{0^+}^\alpha u) (t) = h (t),
    \quad 1<\beta\leq 2,\;  t \in (0, 1),\\
    D_{0^+}^\alpha u(0)= D_{0^+}^\alpha u(1)=0,
             \end{gathered}
\end{equation*}
where $f\in C[0,1]$, is transformed into the problem
\begin{equation}\label{a1}
      \begin{gathered}
    D_{0^+}^\beta y(t)+ h (t)=0,  \quad 1<\beta\leq 2,\;    t \in (0, 1),\\
    y(0)= y(1)=0,
\end{gathered}
\end{equation}

\begin{lemma}\label{b}
Suppose that $h\in C[0,1]$, then the boundary-value problem
\eqref{a1} has a unique solution
   \begin{equation} \label{a2}
y(t)=\int_0^1 H(t,s) h(s) ds,
   \end{equation}
where
\begin{equation}\label{a3}
 H(t,s)=\begin{cases}
 \frac{t^{\beta-1}(1-s)^{\beta-1}-(t-s)^{\beta-1}}{\Gamma(\beta)},
&0\leq s\leq t\leq 1,\\
\frac{t^{\beta-1}(1-s)^{\beta-1}}{\Gamma(\beta)},
&0\leq t\leq s\leq 1.
\end{cases}
\end{equation}
\end{lemma}

 The proof of the above lemma is similar to that of
\cite[Lemma 2.3]{Z. B. Bai}, so we omit it here.

\begin{lemma}\label{j}
 Let  $\Delta = \sum_{i=1}^{m-2} a_{i} \xi_{i}^{\alpha-1}\neq 1$.  Then,
  for $y \in C[0, 1]$, the boundary-value problem
\begin{equation}\label{j1}
     \begin{gathered}
    D_{0^+}^\alpha u (t) + y(t) = 0,
    \quad     t \in (0, 1), \; 1 < \alpha \leq 2,\\
          u (0) = 0, \quad  u (1) -\sum_{i=1}^{m-2} a_{i}
          u(\xi_{i})=\lambda,
             \end{gathered}
\end{equation}
 has a unique solution
\begin{equation} \label{21}
u(t)= \int_0^1 G (t, s) y(s)\,ds + \frac{t^{\alpha - 1}}{(1
-\Delta)} {\sum_{i=1}^{m-2} a_{i}}\int_0^1 G(\xi_{i}, s) y(s)\,ds +  \frac{\lambda t^{\alpha-1}}{(1-\Delta)},
\end{equation}
where
\begin{equation}\label{H}
 G (t, s) = \begin{cases}
   \frac{ t^{\alpha - 1} (1 - s)^{\alpha -1} - (t - s)^{\alpha - 1}}{\Gamma (\alpha)},
    &   0 \leq s \leq t \leq 1,\\
          \frac{ t^{\alpha - 1} (1 - s)^{\alpha-1}}{\Gamma (\alpha)},
    &   0 \leq t \leq s \leq 1,
 \end{cases}
\end{equation}
\end{lemma}

\begin{proof} 
By applying lemma \eqref{lem1},  equation \eqref{j1} is
 equivalent to the  integral equation
\[
u(t)= - \frac{1} {{\Gamma (\alpha )}}\int_0^t (t - s)^{\alpha  -
1} y(s)ds  -  c_1  t^{\alpha  - 1}-  c_2  t^{\alpha  - 2},
\]
for some arbitrary constants $c_1,c_2\in\mathbb{R}$.
By the boundary condition \eqref{j1}, one can get $c_2=0$ and
\begin{align*}
c_1&=  - \frac{1} {{\Gamma (\alpha )}{(1 - \Delta_j)}}\int_0^1 (1
- s)^{\alpha  - 1} y(s))ds\\
&\quad + \frac{1} {{\Gamma (\alpha )}{(1 - \Delta)}}
{\sum_{i=1}^{m-2} a_{i}}\int_0^{ \xi _{i} }{ {( { \xi _{i}  - s}
)}^{\alpha   - 1}  {y(s)ds}^{} } -{ \frac{ \lambda_j}{1-\Delta}} .
\end{align*}
Then, the unique solution of  \eqref{j1} is given by the
formula
\begin{align*}
&u(t)\\
&= - \frac{1} {{\Gamma (\alpha )}}\int_0^t { {(t -
s)}^{\alpha  - 1} y(s)\,ds}+ \frac{1} {{\Gamma
(\alpha )(1 - \Delta )}}\int_0^1 { t^{\alpha  - 1}
 {(1 - s)}^{\alpha    - 1} y(s)\,ds}\\
&\quad -
\frac{{ t^{\alpha  - 1} }} {{\Gamma (\alpha)(1 - \Delta)}}\sum_{i = 1}^{m - 2}
{ a_{i} } \int_0^{ \xi _{i} } {}
 {(\xi_i - s)}^{\alpha  - 1} y(s)\,ds +  \frac{\lambda  t^{\alpha-1}}{(1-\Delta)}\\
&=  - \frac{1} {{\Gamma (\alpha )}}\int_0^t { {(t -
s)}^{\alpha  - 1}
y(s)\,ds}\\
&\quad +\frac{1}{\Gamma(\alpha)}\int_0^1 {
t^{\alpha  - 1}  {(1 - s)}^{\alpha  - 1}
y(s)\,ds}+ \frac{\Delta} {{\Gamma (\alpha )(1 - \Delta )}}\int_0^1
{ t^{\alpha  - 1}  {(1 -
s)}^{\alpha  - 1} y(s)\,ds}\\
&\quad - \frac{{t^{\alpha  - 1} }} {{\Gamma (\alpha)
(1 - \Delta)}} \sum_{i = 1}^{m - 2} {
a_{i} } \int_0^{ \xi _{i} } {}
{(\xi_i - s)}^{\alpha
- 1} y(s)\,ds + \frac{ \lambda t^{\alpha -
1}}{(1-\Delta)}\\
&= \frac{1}{\Gamma(\alpha)}\Big[ \int_0^t {\left( {
t^{\alpha  - 1}  {(1 - s)}^{\alpha  - 1}  -  {(t - s)}^{\alpha  - 1} } \right)} \,\,y(s)\,ds\, +
\int_t^1 { t^{\alpha  - 1}  {(1 - s)}^{\alpha  - 1} y(s)} \,ds \Big]\\
&\quad + \frac{{ t^{\alpha  - 1} }} {{\Gamma (\alpha)(1 - \Delta)}}\,\,\sum_{i = 1}^{m
- 2} { a_{i} } \int_0^1 {
\xi_{i}^{\alpha - 1}  {( {1 - s})}^{\alpha - 1} y(s)} \,ds\\
&\quad - \frac{{ t^{\alpha  - 1} }} {{\Gamma (\alpha)(1 - \Delta)}} \sum_{i = 1}^{m -
2} { a_{i} } \int_0^{ \xi _{i} }
{}  {(\xi_i - s)}^{\alpha - 1} y(s)\,ds +  \frac{ \lambda
t^{\alpha -
1}}{(1-\Delta_j)}\\
&= \frac{1}{\Gamma(\alpha)}\Big[ \int_0^t {\left( {
t^{\alpha  - 1}  {(1 - s)}^{\alpha  - 1}  -  {(t - s)}^{\alpha  - 1} } \right)} y(s)\,ds\, +
\int_t^1 { t^{\alpha  - 1}  {(1 - s)}^{\alpha  - 1} y(s)} \,ds \Big]\\
&\quad +\frac{{ t^{\alpha  - 1} }} {{\Gamma (\alpha)(1 - \Delta)}}\sum_{i = 1}^{m - 2}
{ a_{ji} } \Big[\int_0^{\xi_{i}}{\Big({
{\xi_{i}}^{\alpha - 1}(1-s)^{\alpha -1}}-{{(\xi_{i}-s)}^{\alpha -
1} }\Big)}y(s) ds
 \\
&\quad +\int_{\xi_{i}}^{1}{{{\xi_{i}}^{\alpha -
1}(1-s)^{\alpha-1} }y(s) ds }\; \Big] +\frac{ \lambda t^{\alpha -
1}}{(1-\Delta)}\\
&= \int_0^1{G(t,s)}y(s) ds +\frac{{ t^{\alpha  -
1} }} {{(1 - \Delta) }}\sum_{i = 1}^{m - 2}
{ a_{i} }\int_0^1{G(\xi_{i},s)}v(s) ds +\frac{
\lambda t^{\alpha - 1}}{(1-\Delta)}.
\end{align*}
Thus, the proof is complete.
\end{proof} 

\begin{lemma}\label{G1}
 The functions $H(t,s)$ and $G(t,s)$ defined by \eqref{a3} and \eqref{H},
 respectively satisfies the following conditions:
\begin{itemize}
\item[(i)] $ G (t, s) \geq 0 $, $G (t, s)\leq G (s, s) \leq
\frac{1}{\Gamma (\alpha)}$ for any $t,s \in [0,1]$;

\item[(ii)] $ H (t, s) \geq 0 $, $H (t, s)\leq H (s, s) \leq
\frac{1}{\Gamma (\alpha)}$ for any $t,s \in [0,1]$;

\item[(iii)]  there
exists a positive function $g \in C(0,1)$ such that $\min_{\gamma
\leq t\leq \delta} G(t,s)\geq g(s) G(s,s), s\in(0,1)$, where $0 <
\gamma < \delta < 1$ and
\begin{equation}\label{D}
g(s)=\begin{cases}
\frac{\delta^{\alpha-1}(1-s)^{\alpha-1}
-(\delta-s)^{\alpha-1}}{s^{\alpha-1}(1-s)^{\alpha-1}} &s\in (0,m_1]\\
(\frac{\gamma}{s})^{\alpha-1} &s\in[m_1,1)
\end{cases}
\end{equation}
with $\gamma  < m_1 < \delta$.
\end{itemize}
\end{lemma}

\begin{proof}
 First, we define
\begin{gather*}
g_1(t,s)  =  \frac{ t^{\alpha - 1} (1 - s)^{\alpha -1} - (t -
s)^{\alpha - 1}}{\Gamma (\alpha)},
    \quad  0 \leq s \leq t \leq 1,\\
g_2(t,s)  =  \frac{ t^{\alpha - 1} (1 - s)^{\alpha -1}}{\Gamma
(\alpha)},     \quad   0 \leq t \leq s \leq 1.
\end{gather*}
One can obtain
\begin{align*}
g_1(t,s) & \geq  \frac{1}{\Gamma (\alpha)} \Big [{t^{\alpha - 1}
(1 - s)^{\alpha -1} - (t -{ts})^{\alpha - 1}} \Big ]\\
& =  \frac{1}{\Gamma (\alpha)} \Big [{ t^{\alpha - 1}} \Big(
{(1-s)^{\alpha-1}}-{(1-s)^{\alpha-1}} \Big) \Big] = 0,
\end{align*}
 On the other hand, it is obvious that
$g_2(t,s)\geq 0,\;0 \leq t \leq s \leq 1$.
Thus, $G(t,s)\geq 0$ for any $t,s\in[0,1]$.

Now, we show that $ G(t,s)\leq G(s,s)$ for any $t,s\in[0,1]$.
Also, we have
\begin{align*}
{\frac{{\partial g_1( {t,s})}} {{\partial t}}}
& = \frac{1}{\Gamma (\alpha)} \Big [ (\alpha-1)t^{\alpha-2}
(1-s)^{\alpha-1}-(\alpha-1)(t-s)^{\alpha-2} \Big ]\\
& =  \frac{1}{\Gamma (\alpha)} (\alpha-1)t^{\alpha-2}\Big [
(1-s)^{\alpha-1}-(1-\frac{s}{t})^{\alpha-2} \Big ]\\
&\leq  \frac{1}{\Gamma (\alpha)}  (\alpha-1)t^{\alpha-2}\Big[
{(1-s)^{\alpha-1}}-{(1-s)^{\alpha-2}}\Big] \leq 0,
\end{align*}
then, $g_1(t,s)$ is non-increasing with respect to $t$ on $[s,1]$,
hence, we obtain 
$$
g_1(t,s)\leq g_1(s,s)\quad \forall  0\leq s\leq t\leq 1.
$$
On the other hand, we obtain
$$
{\frac{{\partial g_2( {t,s})}} {{\partial
t}}}= \frac{1}{\Gamma (\alpha)} \Big [(\alpha-1)t^{\alpha-2}
(1-s)^{\alpha-1}\Big ]\geq 0,
$$
then $g_2(t,s)$ is increasing with respect to $t$ on $[0,s]$,
therefore,
$$
g_2(t,s)\leq g_2(s,s)\quad \forall  0\leq t \leq s\leq 1.
$$
Then, we have
\begin{equation} \label{q}
G(t,s)\leq G(s,s)\quad \forall t,s\in [0,1].
\end{equation}

(ii) Since $g_1(., s)$ is non-increasing and $g_2(.,s)$ is
non-decreasing, for all $ s,t\in [0,1]$, we have
\begin{align*}
\min_{\gamma \leq t\leq \delta}G(t,s)
&= \begin{cases}
\min_{\gamma \leq t\leq \delta}g_1(t,s) &s\in [0,\gamma]\\
\min_{\gamma \leq t\leq \delta}\{g_1(t,s),g_2(t,s)\} &s\in [\gamma,\delta]\\
\min_{\gamma \leq t\leq \delta}g_2(t,s) &s\in [\delta,1]
\end{cases}\\
&= \begin{cases}
\min_{\gamma\leq t\leq\delta}g_1(t,s) &s\in [0,m_1]\\
\min_{\gamma \leq t\leq \delta}{g_2(t,s)} &s\in [m_1,1]
\end{cases}\\
&\geq\begin{cases}
g_1(\delta,s) &s\in [0,m_1]\\
{g_2(\gamma,s)} &s\in [m_1,1]
\end{cases}\\
&= \begin{cases}
{\delta^{\alpha-1}(1-s)^{\alpha-1}-(\delta-s)^{\alpha-1}} &s\in [0,m_1]\\
{\gamma}^{\alpha-1}(1-s)^{\alpha-1}  &s\in[m_1,1]
\end{cases}
\end{align*}
where $\gamma<m_1<\delta$ is the solution of
$$
{\delta^{\alpha-1}(1-m_1)^{\alpha-1}-(\delta-m_1)^{\alpha-1}}
={\gamma}^{\alpha-1}(1-m_1)^{\alpha-1}.
$$
It follows from the monotonicity of $g_1$ and $g_2$ that
$$
\max_{0 \leq t\leq 1} G(t,s) = G (s, s)
= \frac{(s (1 - s))^{\alpha - 1}}{\Gamma (\alpha)} \quad s \in (0,1),
$$
The proof is complete.
\end{proof} 

\begin{remark} \label{rmk1} \rm
If $\gamma \in (0,1/4)$ and $\delta = 1-\gamma$, then 
Lemma \ref{G1} holds.
\end{remark}

In this article, we assume that $\gamma \in (0,1/4)$ and
$\delta = 1-\gamma$.
Now, we consider the system \eqref{1}.
 By applying lemmas \ref{b} and \ref{j}, $u \in C(0, 1)$ is a solution 
of \eqref{1}  if and only if $u \in C[0, 1]$  is a solution of the
 nonlinear integral equation
 \begin{equation} \label{sol}
\begin{split}
  u(t) & =   z (t) + \int_0^1 G (t, s)
\Big(\int_0^1 H(s,r) f (r, u (r)) d r \Big)d s \\
&\quad + \frac{t^{\alpha - 1}}{(1 -\Delta)}\sum_{i=1}^{m-2} a_{i}
\int_0^1 G (\xi_{i}, s) \Big(\int_0^1 H(s,r) f (r, u (r)) d r
\Big)\,ds,
\end{split}
\end{equation}
where $\lambda t^{\alpha-1}/(1-\Delta)$. Let
$z (t) = \frac{\lambda t^{\alpha-1} }{1-\Delta}$ and
$C : = \sup_{t \in I} |z (t)| = \|z\|$.

\section{Main result for the single-valued case}

 Now we are able to present the existence results for problem \eqref{1}.

\subsection{Existence result via Leray-Schauder alternative}

\begin{theorem}[Nonlinear alternative for single valued maps \cite{Granas}]
\label{tls} 
 Let $X$ be a Banach space, $C$ a closed convex subset of $X$, $U$ an open subset of
$D$ and $0 \in U$. Suppose that $T : \overline{U} \to D$
is a continuous, compact (that is, $F (\overline{U})$ is a
relatively compact subsets of $D$) map. Then either
\begin{itemize}
\item[(i)] $T$ has a fixed point in $U$, or

\item[(ii)] there is a $u \in \partial U$ (the boundary of $U$ in $C$)
and $\lambda \in (0, 1)$ with $u = \lambda T(u)$.
\end{itemize}
\end{theorem}


\begin{theorem}\label{tls1}
If the continuous function $f : I \times \mathbb{R} \to
\mathbb{R}$ satisfies:
\begin{itemize}
\item[(B1)] There exist a function $\varphi \in L^1 (I, \mathbb{R}^+)$
and a continuous nondecreasing function $\psi : [0, + \infty)
\to (0, + \infty)$ such that
\begin{gather*}
|f (t, x)| \leq \varphi (t) \psi (\|x\|), \quad  \text{for all } 
(t, x) \in I \times \mathbb{R}.
\end{gather*}

\item[(B2)] There exists a constant $M > 0$ such that
\begin{equation} \label{r}
\begin{split}
&M\Big[ C +  \psi (M) \Big [\int_0^1 G (s, s) \Big(\int_0^1
H(s,r) \varphi (r)  d r \Big)d s \\
&+ \frac{\sum_{i=1}^{m-2}
a_{i}}{(1 -\Delta)} \int_0^1 G (\xi_{i}, s) \Big(\int_0^1 H(s,r)
\varphi (r)  d r \Big)\,ds \Big]\Big]^{-1}   > 1.
\end{split}
\end{equation}
\end{itemize}
Then \eqref{1} has at least one solution on $I$.
\end{theorem}

\begin{proof} 
Consider $F : E \to E$ with $u = F (u)$, where
\begin{align*}
  (T u)(t) 
& =   z (t) + \int_0^1 G (t, s) 
\Big(\int_0^1 H(s,r) f (r, u (r)) d r \Big)d s \\
&\quad + \frac{t^{\alpha - 1}}{(1 -\Delta)}\sum_{i=1}^{m-2} a_{i}
\int_0^1 G (\xi_{i}, s) \Big(\int_0^1 H(s,r) f (r, u (r)) d r\Big)\,ds,
\end{align*}
for all $t \in I$. We show that $T$ maps bounded sets into bounded
sets in $C([0, 1], \mathbb{R})$. For a positive number $r$, let
$B_r = \{u \in C([0, 1], \mathbb{R}) : \|u\| \leq r \}$ be a
bounded set in $C([0, 1], \mathbb{R})$. Then
\begin{align*}
 |(T u)(t) | & \leq   |z (t )| + \psi (\|u\|) 
 \Big [\int_0^1 G (s, s) \Big(\int_0^1 H(s,r) \varphi (r)  d r \Big)d s \\
&\quad + \frac{\sum_{i=1}^{m-2} a_{i}}{(1 -\Delta)}
\int_0^1 G (\xi_{i}, s) \Big(\int_0^1 H(s,r) \varphi (r)  d r
\Big)\,ds \Big ].
\end{align*}
Consequently,
\begin{align*}
 \|T u\| & \leq   C + \psi (r)  \Big [\int_0^1 G (s, s) \Big(\int_0^1 H(s,r) \varphi (r)  d r \Big)d s \\
&\quad + \frac{\sum_{i=1}^{m-2} a_{i}}{(1 -\Delta)}
\int_0^1 G (\xi_{i}, s) \Big(\int_0^1 H(s,r) \varphi (r)  d r
\Big)\,ds \Big ].
\end{align*}

Next we show that $T$ maps bounded sets into equi-continuous 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*}
&| (T u)(t_1) - (T u)(t_2) |  \\
&\leq  \int_0^1 |G(t_1,s)
- G(t_2,s)|  \Big(\int_0^1 H(s,r) f (r, u (r)) d r \Big) ds\\
&\quad +  \dfrac{\sum_{i=1}^{m-2} a_i }{1-\Delta}  |t_1^{\alpha -1}  -
 t_2^{\alpha -1} | \int_0^{1}G_1(\xi_i,s) \Big(\int_0^1 H(s,r) f (r, u (r)) d r \Big) 
  ds\\
&\leq  \int_0^1 |G(t_1,s)
- G(t_2,s)|  \Big(\int_0^1 H(s,r)  \psi (r) \varphi (r)  d r \Big) ds\\
&\quad +  \dfrac{\sum_{i=1}^{m-2} a_i }{1-\Delta}  |t_1^{\alpha -1}  -
 t_2^{\alpha -1} | \int_0^{1}G_1(\xi_i,s) \Big(\int_0^1 H(s,r)  
\psi (r) \varphi (r)  d r \Big) ds
\end{align*}
On the other hand, 
\begin{align*}
&\int_0^1 |G(t_1,s)
- G(t_2,s)|  \Big(\int_0^1 H(s,r)  \psi (r) \varphi (r)  d r \Big) ds  \\
&\leq \Big( \int_0^{t_1}+
\int_{t_1}^{t_2}+\int_{t_2}^{1}\Big) |G(t_1,s)
- G(t_2,s)|  
 \Big(\int_0^1 H(s,r)  \psi (r) \varphi (r)  d r \Big) ds  \\
&\leq \psi (r) \int_0^{t_1} [(t_2^{\alpha -1} -t_1^{\alpha -1}) (1 - s)^{\alpha - 1} +(
(t_2-s)^ {\alpha -1} - (t_1-s)^ {\alpha -1})] \\
&\quad \times\Big(\int_0^1 H(s,r) \varphi (r)  d r \Big)  ds \\
&\quad + \psi (r)\int_{t_1}^{t_2} [(t_2^{\alpha -1}
-t_1^{\alpha -1}) (1 - s)^{\alpha - 1}
+ (t_2-s)^ {\alpha-1} ] \Big(\int_0^1 H(s,r) \varphi (r)  d r \Big) ds \\
 &\quad  + \psi (m) \int_{t_2} ^{1} [(t_2^{\alpha -1}
-t_1^{\alpha -1}) (1 - s)^{\alpha - 1}] \Big(\int_0^1 H(s,r) \varphi (r)  d r \Big) ds \\
& \to  0 \quad \text{uniformly as } t_1 \to t_2.
\end{align*}
Therefore,
\[
| (T u)(t_1) - (T u)(t_2) |  \to  0\quad 
\text{uniformly  as  $t_1 \to t_2$, for $u \in B_r$}.
\]
As $F$ satisfies the above assumptions,  it follows by the
Arzel\'{a}-Ascoli theorem that $T : C([0, 1], \mathbb{R})
\to C([0, 1], \mathbb{R})$ is completely continuous.

The result will follow from the Leray-Schauder nonlinear
alternative (Theorem \ref{tls}) once we have proved the
boundendness of the set of all solutions to equations 
$u = \lambda Fu$ for $\lambda \in [0, 1]$.

Let $u$ be a solution. Then, for $t \in [0, 1]$, and using the
computations in proving that $T$ is bounded, we have
\begin{align*}
 |u (t)| 
= |\lambda (T u)(t) | 
& \leq   |z (t )| + \psi (\|u\|)  \Big [\int_0^1 G (s, s) 
\Big(\int_0^1 H(s,r) \varphi (r)  d r \Big)d s \\
&\quad + \frac{\sum_{i=1}^{m-2} a_{i}}{(1 -\Delta)}
\int_0^1 G (\xi_{i}, s) \Big(\int_0^1 H(s,r) \varphi (r)  d r
\Big)\,ds \Big ].
\end{align*}
Consequently,
\begin{align*}
&\|u\| \Big[C +  \psi (\|u\|) \Big [\int_0^1 G (s, s) 
\Big(\int_0^1 H(s,r) \varphi (r)  d r \Big)d s \\
&+ \frac{\sum_{i=1}^{m-2} a_{i}}{(1 -\Delta)}
\int_0^1 G (\xi_{i}, s) \Big(\int_0^1 H(s,r) \varphi (r)  d r
\Big)\,ds \Big ]\Big]^{-1}   \leq 1.
\end{align*}
In view of (B2), there exists $M$ such that  $\|u\| \ne M$. Let us
set
\[
U : = \{u \in C(I, \mathbb{R}) :  \|u\| < M \}.
\]
Note that the operator $T : \overline{U} \to C([0, 1],
\mathbb{R}$ is continuous and completely continuous. From the
choice of $U$, there is no $u \in \partial U$ such that
 $u = \lambda T (u)$ for some $\lambda \in (0, 1)$. Consequently, by the
nonlinear alternative of Leray-Schauder type (Theorem \ref{tls}),
we deduce that  has a fixed point $u \in U$ which is a solution of
the problem \eqref{1}. This completes the proof.
\end{proof}

\subsection{Existence result via Leggett-Williams fixed point theorem}
  
In this section, we assume that $\gamma \in (0,1/4)$ and $\delta = 1-\gamma$.
To establish the existence of three positive solutions to system
\eqref{1}, we use the following Leggett-Williams fixed
point  theorem.
For the convenience of the reader, we present here the
Leggett-Williams fixed point theorem \cite{R. W. Leggett}. 

Given a cone $K$ in a real Banach space $E$, a map $\alpha$ is
said to be a nonnegative continuous concave (resp. convex)
functional on $K$ provided that $\alpha : K \to [0. +
\infty)$ is continuous and
\begin{gather*}
\alpha (tx + (1-t)y) \geq t \alpha (x) + (1-t) \alpha (y),\\
(\text{resp. } \alpha (tx + (1-t)y) \leq t \alpha (x) + (1-t) \alpha (y)),
\end{gather*}
for all $x,y \in K$ and $t \in [0,1]$.
Let $0 < a < b$ be given and let $\alpha$ be a nonnegative
continuous concave functional on $K$. Define the convex sets $P_r$
and $P(\alpha, a, b)$ by
\begin{gather*}
P_r = \{ x \in K | \|x\| < r\},\\
P(\alpha, a, b) = \{ x \in K |a \leq \alpha (x), \|x\| \leq b\}.
\end{gather*}

\begin{theorem}[Leggett-Williams fixed point theorem] \label{the1}
Let $A : \overline{P_c} \to \overline{P_c}$ be a
completely continuous operator and let $\alpha$ be a nonnegative
continuous concave functional on $K$ such that 
$\alpha (x) \leq \|x\|$ for all $x \in \overline{P_c}$. 
Suppose there exist $0 < a < b < d \leq c$ such that
\begin{itemize}
\item[(A1)] $ \{x \in P(\alpha, b, d) |
\alpha(x) > b \} \neq \emptyset $, and $\alpha (Ax) > b $ for
 $x \in P(\alpha, b, d)$,

\item[(A2)]  $\| Ax \| < a $ for $\| x \| \leq a$, and

\item[(A3)] $\alpha (Ax) > b $ for $x \in P(\alpha, b, c)$ with
 $\| Ax \|> d$.
\end{itemize}
Then $A$ has at least three fixed points $x_1$, $x_2$, and $x_3$
and such that $\|x_1\| < a$, $b < \alpha (x_2)$ and $\|x_3\| > a$,
with $\alpha (x_3) < b$.
\end{theorem}

 Also, we introduce the following notation. Define
\[
\eta=  \min_{\gamma \leq t \leq \delta} \{ g(t) \},\quad 
\sigma = \min \{\eta, \gamma^{\alpha-1}\},
\]
and
\begin{gather*}
\begin{aligned}
M &= \int_0^1 G (s, s) \Big(\int_0^1 H(s,r)  d r \Big)\,ds\\
&\quad +\frac{1}{(1 -\Delta)}\sum_{i=1}^{m-2} a_{i} \int_0^1 G (\xi_{i},
s)  \Big(\int_0^1 H(s,r)  d r \Big)d s, 
\end{aligned} \\
\begin{aligned}
m &= \min_{\gamma \leq t \leq \delta} \Big\{\int_\gamma^\delta G
(t, s) \Big(\int_\gamma^\delta H(s,r)  d r \Big)
\,ds \\
&\quad + \frac{t^{\alpha - 1}}{(1 -\Delta)}\sum_{i=1}^{m-2} a_{i} \int_\gamma^\delta G
(\xi_{i}, s) \Big(\int_\gamma^\delta  H(s,r)  d r \Big)\,ds
\Big\}.
\end{aligned}
\end{gather*}
Note that  $0 < m < M$.
The basic space used in this paper is a real Banach space 
$E = C([0, 1], \mathbb{R})$ with the norm 
$\|u\| : =  \max_{t \in [0,1]}|u(t)|$.
Then, we define a cone $K \subset E$, by
\[
K = \{u \in E : u(t) \geq 0 \; \min_{\gamma\leq t \leq
\delta} (u(t)) \geq \frac{\sigma}{3} \|u \| \},
\]
and an operator $T:E \to E$  by
\begin{equation} \label{t1}
\begin{split}
T (u)(t)  & =   \int_0^1 G (t, s) \Big(\int_0^1 H(s,r) f (r, u (r)) d r \Big)d s \\
&\quad +  \frac{t^{\alpha - 1}}{(1 -\Delta)}\sum_{i=1}^{m-2} a_{i}
\int_0^1 G (\xi_{i}, s) \Big(\int_0^1 H(s,r) f (r, u (r)) d r
\Big)\,ds + \frac{\lambda t^{\alpha-1} }{1-\Delta}.
\end{split}
\end{equation}


\begin{lemma} \label{lem7}
 The operator $T$ is defined from $K$ to $K$; i.e., $T(K) \subseteq K$.
\end{lemma}

\begin{proof} For any $u \in K$,  by Lemma \ref{G1}, 
$T (u)(t) \geq 0$, $t \in [0, 1]$, and it follows from \eqref{t1} that
\begin{equation} \label{a}
\begin{split}
&\| T(u)\|\\
& \leq   \int_0^1 G (s, s)  \Big(\int_0^1 H(s,r) f (r, u (r)) d r \Big)d s \\
&\quad + \frac{1}{(1 -\Delta)}\sum_{i=1}^{m-2} a_{i} \int_0^1 G
(\xi_{i}, s)  \Big(\int_0^1 H(s,r) f (r, u (r)) d
r \Big)\,ds + \frac{\lambda }{(1-\Delta)} \\
&=  \Big(\int_0^\gamma + \int_\gamma^\delta +\int_\delta^1 \Big) \Big(G (s, s)   \Big(\int_0^1 H(s,r) f (r, u (r)) d r \Big)d s \Big)  \\
&\quad +\frac{1}{(1 -\Delta)}\sum_{i=1}^{m-2} a_{i}
\Big(\int_0^\gamma + \int_\gamma^\delta +\int_\delta^1\Big)
\Big(G(\xi_{i}, s)  \Big(\int_0^1 H(s,r) f (r, u (r)) d r \Big)
\,ds \Big)  \\
&\quad  + \frac{ \lambda }{(1-\Delta)} \\
&\leq 3 \Big[\int_\gamma^\delta G (s, s)  \Big(\int_0^1 H(s,r) f (r, u (r)) d r \Big)d s \\
&\quad +\frac{1}{(1 -\Delta)} \sum_{i=1}^{m-2} a_{i}
\int_\gamma^\delta G(\xi_{i}, s)  \Big(\int_0^1 H(s,r) f (r, u
(r)) d r \Big)\,ds + \frac{ \lambda }{3(1-\Delta)}\Big].
\end{split}
\end{equation}
Thus, for any $u \in K$, it follows from Lemma \ref{G1} and
\eqref{t1} that
\begin{align*}
&\min_{\gamma \leq t \leq \delta} T (u)(t)\\
& = \min_{\gamma \leq t \leq \delta}\Big\{ \int_0^1 G (t, s) \Big(\int_0^1 H(s,r) f  (r, u (r)) d r \Big)d s \\
&\quad + \frac{t^{\alpha - 1}}{(1
-\Delta_1)}\sum_{i=1}^{m-2} a_{i} \int_0^1 G (\xi_{i}, s)
\Big(\int_0^1 H(s,r) f (r, u (r)) d
r \Big)\,ds + \frac{\lambda t^{\alpha-1} }{(1-\Delta)}\Big\}\\
&\geq   \int_0^1 g(s) G (s, s)  \Big(\int_0^1 H(s,r) f (r, u (r)) d r \Big)d s \\
&\quad + \frac{\gamma^{\alpha - 1}}{(1
-\Delta_1)}\sum_{i=1}^{m-2} a_{i} \int_0^1 G (\xi_{i}, s)
\Big(\int_0^1 H(s,r) f (r, u (r)) d
r \Big)\,ds + \frac{\lambda \gamma^{\alpha-1} }{3(1-\Delta)}\\
&\geq  \eta  \int_\gamma^\delta  G (s, s)  \Big(\int_0^1 H(s,r) f (r, u (r)) d r \Big)d s \\
&\quad + \frac{\gamma^{\alpha - 1}}{(1
-\Delta_1)}\sum_{i=1}^{m-2} a_{i} \int_\gamma^\delta G (\xi_{i},
s)  \Big(\int_0^1 H(s,r) f (r, u (r)) d
r \Big)\,ds + \frac{\lambda \gamma^{\alpha-1} }{3(1-\Delta)}\\
& \geq \sigma \Big[  \int_\gamma^\delta  G (s, s) \Big(\int_0^1 H(s,r) f (r, u (r)) d r \Big)d s \\
&\quad + \frac{1}{(1 -\Delta)}\sum_{i=1}^{m-2} a_{i}
\int_\gamma^\delta G (\xi_{i}, s)  \Big(\int_0^1 H(s,r) f (r, u
(r)) d
r \Big)\,ds + \frac{\lambda  }{3(1-\Delta)}\Big]\\ \\
&\geq \frac{\sigma}{3} \| T (u)\|.\\
\end{align*}
Therefore, from the above, we conclude that $T (u)(t) \in K$; that
is, $T (K) \subset K$. This completes the proof.
\end{proof}

It is clear that the existence of a positive solution for \eqref{1} is equivalent
to the existence of a nontrivial fixed point of $T$ in $K$.

Next, we define the nonnegative continuous concave functional
on $K$ by
\begin{gather*}
\alpha(u)= \min_{\gamma \leq t \leq \delta}(u(t)).
\end{gather*}
It is obvious that, for each $u \in K$, $\alpha(u) \leq \|u\|$.

Throughout this section, we assume that $p$, $q$, are two positive
numbers satisfying $\frac{1}{p} + \frac{1}{q} \leq 1$.
Now, we can state our main result in this section.

\begin{theorem}\label{the2}
Assume that there exist nonnegative numbers $a, b, c$ such that
 $0 < a < b \leq \frac{\sigma}{3}c$, and  $f (t, u)$, $j=1,2,$ satisfy
the following conditions:
\begin{itemize}
\item[(H1)] $f(t, u) \leq
   \frac{1}{p} \cdot \frac{c}{M}$, for all $t \in [0,1], u  \in [0,c]$;

\item[(H2)] $f (t, u) \leq  \frac{1}{p} \cdot \frac{a}{M}$, for all 
$t \in [0,1], u \in [0,a]$;

\item[(H3)] (i) $f (t, u) > \frac{b}{m}$  for all $t \in [\gamma, \delta]$, 
$u \in [b, \frac{3b}{\sigma}]$. 
\end{itemize}
Then, for
\begin{equation} \label{p}
0 < \lambda < \frac{ c (1-\Delta) }{q },
\end{equation}
problem \eqref{1} has at least three positive solutions $u_1,
u_2, u_3$ such that $\|u_1\| < a$, $b < \min_{\gamma \leq t \leq
\delta}(u_2(t))$, and
$\|u_3\| > a$, with $\min_{\gamma\leq t \leq \delta}(u_3(t)) < b$.
\end{theorem}

\begin{proof} 
First, we show that $T : \overline{P_c} \to \overline{P_c}$ is a 
completely continuous operator. If $u \in \overline{P_c}$, then  $\|u\| \leq c$. 
Then, by applying condition (H1), we have
\begin{align*}
 \|T(u)\| & =  \max_{0 \leq t \leq 1} |T(u)(t)| \\
& = \max_{0 \leq t \leq 1} \Big \{  \int_0^1 G (t, s)  \Big(\int_0^1 H(s,r) f (r, u (r)) d r \Big)d s \\
&\quad +  \frac{t^{\alpha - 1}}{(1 -\Delta)}\sum_{i=1}^{m-2} a_{i}
\int_0^1 G (\xi_{i}, s) \Big(\int_0^1 H(s,r) f (r, u (r)) d
r \Big)\,ds + \frac{\lambda t^{\alpha-1} }{(1-\Delta)} \Big\}\\
& \leq 
 \frac{1}{p} \cdot \frac{c}{M}  \Big\{\int_0^1 G (s, s) \Big(\int_0^1 H(s,r)  d r \Big)\,ds\\
&\quad +  \frac{1}{(1
-\Delta)}\sum_{i=1}^{m-2} a_{i} \int_0^1 G (\xi_{i}, s)  \Big(\int_0^1 H(s,r)  d r \Big)\,ds \Big \}+ \frac{\lambda  }{(1-\Delta)}\\
& \leq  \frac{1}{p} \cdot c +\frac{1}{q} \cdot c  \leq c.
\end{align*}
Therefore, $\|T(u)\| \leq c$, that is, 
$T : \overline{P_c} \to \overline{P_c}$. The operator $T$ is completely
continuous by an application of the Ascoli-Arzela theorem.

 In the same way,  condition (H2) implies that  condition (A2) of
Theorem \ref{the1} is satisfied. We now show that condition (A1)
of Theorem \ref{the1} is satisfied. 
Clearly  $\{u \in P(\alpha, b, \frac{3b}{\sigma}) | \alpha(u) > b\} \neq \emptyset$. 
If $u \in P(\alpha, b, \frac{3b}{\sigma})$, then 
$b \leq u(s) \leq \frac{3b}{\sigma}, s \in [\gamma, \delta]$.

 By condition (H3), we obtain
\begin{align*}
&\alpha (T(u)(t)) \\
&=  \min_{\gamma \leq t \leq \delta} (T(u)(t))\\
&=  \min_{\gamma \leq t \leq \delta}  \Big
\{ \int_0^1 G (t, s) \Big(\int_0^1 H(s,r) f (r, u (r)) d r \Big)d s \\
&\quad + \frac{t^{\alpha - 1}}{(1
-\Delta_1)}\sum_{i=1}^{m-2} a_{i} \int_0^1 G (\xi_{i}, s)
\Big(\int_0^1 H(s,r) f (r, u (r)) d
r \Big)\,ds + \frac{\lambda t^{\alpha-1} }{(1-\Delta)} \Big\}\\
& = \frac{b}{m}  m = b.
\end{align*}
Therefore, condition (A1) of Theorem \ref{the1} is satisfied.

Finally, we show that the condition (A3) of Theorem \ref{the1} is
also satisfied.
If $u \in P(\alpha, b,c)$, and $\|T(u)\| > 3b/\sigma$,
then
\[
 \alpha (T(u)(t))=  \min_{\gamma \leq t \leq \delta}
 (T(u)(t)) \geq \frac{\sigma}{3} \|T(u)\| > b.
\]
Therefore,  condition (A3) of Theorem \ref{the1} is also
satisfied. By Theorem \ref{the1}, there exist three positive
solutions $u_1, u_2, u_3$ such that 
$\|u_1\| < a$, $b < \min_{\gamma \leq t \leq \delta}(u_2(t))$, and $\|u_3\| > a$,
 with $\min_{\gamma \leq t \leq \delta}(u_3(t)) < b$. we have the
conclusion. 
\end{proof}

\subsection{Existence result via nonlinear contractions}

Now we  present the existence results for problem \eqref{1}.

\begin{definition} \label{def3} \rm
 Let $X$ be a Banach space and let $T : X \to X$ be a
mapping. $T$ is said to be a nonlinear contraction if there exists
a continuous nondecrasing function
     $\psi: \mathbb{R}^+ \to \mathbb{R}^+$ such that $\psi (0) = 0$ and 
$\psi (\xi) < \xi$ for all $\xi > 0$ with the property:
\[
\|T(u) - T (v)\| \leq \psi (\|u - v\|), \quad \forall  u, v \in X.
\]
\end{definition}

\begin{lemma}[Boyd and Wong \cite{Boyd}] \label{lembw}
Let $X$ be a Banach space and let $T : X \to X$ be a
nonlinear contraction. Then $T$ has a unique fixed point in $X$.
\end{lemma}

\begin{theorem} \label{thebw}
Assume that $f$ satisfy the  following conditions:
\begin{itemize}
\item[(H4)] $|f(t, u) - f(t, v)| \leq h(t) \frac{|u -v|}{\theta +|u -
v|}$ for all $t \in [0, 1]$, $u, v \geq 0$, where $h : [0, 1]
\to \mathbb{R}^+$ is continuous and
\begin{equation} \label{teta}
\begin{split}
\theta &=  \int_0^1 G (s, s) \Big(\int_0^1 H(s,r) h (r) d r \Big)\,ds\\
 &+  \frac{\sum_{i=1}^{m-2} a_{i}}{(1 -\Delta)} \int_0^1 G
(\xi_{i}, s) \Big(\int_0^1 H(s,r) h (r) d r \Big)\,ds,
\quad t \in [0, 1].
\end{split}\end{equation}
\end{itemize}
\end{theorem}

AUTHORS: IT SEEMS THAT PART OF THE THEOREM IS MISSING  ???

\begin{proof} 
We define the operator $T : (E = C([0, 1], \mathbb{R})) \to E$ by
\begin{align*}
T (u)(t) 
 & =  \int_0^1 G (t, s) \Big(\int_0^1 H(s,r) f (r, u (r)) d r \Big)d s \\
&\quad +  \frac{t^{\alpha - 1}}{(1 -\Delta)}\sum_{i=1}^{m-2} a_{i}
\int_0^1 G (\xi_{i}, s) \Big(\int_0^1 H(s,r) f (r, u (r)) d r
\Big)\,ds + \frac{\lambda t^{\alpha-1} }{1-\Delta},
\end{align*}
for  $t \in [0, 1]$.
Let  $\psi: \mathbb{R}^+ \to \mathbb{R}^+$ a continuous nondecreasing function
such that $\psi (0) = 0$ and $\psi (\xi) < \xi$ for all $\xi > 0$ be defined by
\[
\psi (\xi) = \frac{\theta \xi}{\theta + \xi}, \quad \forall \xi \geq 0.
\]
Let $u, v \in E$. Then
\[
|f(s, u (s)) - f(s, v (s))| \leq \frac{h(s)}{\theta} \psi (\|u -v\|).
\]
Thus
\begin{align*}
|T (u)(t) - T (v)(t)| 
& \leq   \int_0^1 G (s, s) \Big(\int_0^1
H(s,r) h (r) \frac{|u (r) -v (r)|}{\theta +|u (r) -
v (r)|} d r \Big)d s  \\
&\quad +  \frac{\sum_{i=1}^{m-2} a_{i}}{(1 -\Delta)} \int_0^1 G
(\xi_{i}, s) \Big(\int_0^1 H(s,r) \frac{|u (r) -v (r)|}{\theta +|u
(r) - v (r)|} d r \Big)\,ds,
\end{align*}
for all $t \in [0, 1]$. In view of \eqref{teta}, it follows that
$\|T(u) - T (v)\| \leq \psi (\|u - v\|)$ and hence $T$ is a
nonlinear contraction. Thus, by Lemma \ref{lembw}, the operator
$T$ has a unique fixed point in $E$, which in turn is a unique
solution of problem \eqref{1}. 
\end{proof}

\section{Existence results of the multi-valued case}

\subsection{The upper semi-continuous case}

To obtain the complete continuity of existence solutions operator,
the following lemmas and definitions are needed.

Let $(X,d)$ be a metric space with the corresponding norm
$\|\cdot\|$ and let $I = [0, 1]$. Denoted by $\mathcal{L} (I)$ the
$\sigma$-algebra of all Lebesgue measurable subsets of $I$, by
$\mathcal{B} (X)$ the family of all nonempty subsets of $X$ and by
$\mathcal{P} (X)$ the family of all Borel subsets of $X$. If $A
\subset I$ then $\chi_A: I \to \{0, 1\}$ denotes the
characteristic function of $A$. For any subset $A \subset X$ we
denote by $\overline{A}$ the closure of $A$.

Recall that the Pompeiu-Hausdorff distance of the closed subsets
$A,B \subset X$ is defined by
\[
d_H (A, B) = \max \{d^* (A,B), d^* (B, A) \}, \quad
d^* (A, B) = \sup \{d (a, B), a \in A \},
\]
where $d (x, B) = \inf_{y \in B} d (x, y)$. Define
\begin{gather*}
\mathcal{P} (X)  =  \{Y \subset X : \; Y \ne \emptyset \},\\
\mathcal{P}_b (X)  =  \{Y \in \mathcal{P} (X) :  Y  \text{ is bounded} \},\\
\mathcal{P}_{cl} (X)  =  \{Y \in \mathcal{P} (X) :  Y  \text{ is closed} \},\\
\mathcal{P}_{c p} (X)  = \{Y \in \mathcal{P} (X) :  Y  \text{ is compact} \},\\
\mathcal{P}_{c v} (X)  =  \{Y \in \mathcal{P} (X) :  Y \text{ is convex} \}.
\end{gather*}
Also, we denote $C(I, X)$ the Banach space of all continuous
functions $x: I \to X$ endowed with the norm $|x|_c =
\sup_{t \in I} |x(t)|$ and by $L^1(I, X)$ the Banach space of all
(Bochner) integrable functions $x: [0, 1] \to X$ endowed
with the norm $|x|_1 = \int_{I} |x (t)| dt$.

A subset $D \subset L^1(I, X)$ is said to be decomposable if for
any $x, y \in D$ and any subset $A \in \mathcal{L} (I)$ one has $x
\chi_A+ y \chi_B \in D$, where $B= I \setminus A$.

Let $(X, d_1)$ and $(Y, d_2)$ be two metric spaces. If $T : X
\to \mathcal{P} (X)$ a set-valued map, then a point $x \in
X$ is called a fixed point for $T$ if $x \in T(x)$. $T$ is said to
be bounded on bounded sets if $T(B): = \cup_{x \in B} T(x)$ is a
bounded subset of $X$ for all bounded sets $B$ in $X$. $T$ is said
to be compact if $T(B)$ is relatively compact for any bounded sets
$B$ in $X$. $T$ is said to be totally compact if $\overline{T
(X)}$ is a compact subset of $X$. $T$ is said to be upper
semi-continuous if for any open set $D \subset X$, the set $\{x
\in X: T(x) \subset D \}$ is open in $X$. $T$ is called completely
continuous if it is upper semi-continuous and and, for every
bounded subset $A \subset X$, $T(A)$ is relatively compact. It is
well known that a compact set-valued map $T$ with nonempty compact
values is upper semi-continuous if and only if $T$ has a closed
graph.

We define the graph of $T$ to be the set 
$Gr(T) = \{(x, y) \in X \times Y, y = T(x) \}$ 
and recall a useful result regarding
connection between closed graphs and upper semi-continuity.

\begin{lemma}[{\cite[Proposition 1.2]{Deimling}}] \label{lemusc} 
If $T : X \to \mathcal{P}_{cl}(Y)$ is upper
semi-continuous, then $Gr(T)$ 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 $y_n
\in T(x_n)$, then $y_* \in T(x_*)$. Conversely, if $T$ is
completely continuous and has a closed graph, then it is upper
semi-continuous.
\end{lemma}

For  convenience of the reader, we present here the following
nonlinear alternative of Leray-Schauder type and its consequences.

\begin{theorem}[Nonlinear alternative for Kakutani maps \cite{Granas}]
\label{thels} 
Let $X$ be a Banach space, $C$ a closed convex subset of $X$, $U$ 
an open subset of $C$ and $0 \in U$. Suppose that 
$T : \overline{U} \to \mathcal{P}_{cl, cv} (C)$ is a upper semi-continuous 
compact map; here $\mathcal{P}_{cl, cv} (C)$ denotes the family of nonempty,
compact convex subsets of $C$. Then either
\begin{itemize}
\item[(i)] $T$ has a fixed point in $U$, or

\item[(ii)] there is a $u \in \partial U$ and $\lambda \in (0, 1)$ with
$u \in \lambda T(u)$.
\end{itemize}
\end{theorem}

\begin{definition} \label{def4} \rm
The multifunction $T : X \to \mathcal{P} (X)$ is said to
be lower semi-continuous if for any closed subset $C \subset X$,
the subset $\{s \in X:  T(s) \subset C \}$ is closed.
\end{definition}

If $F : I \times \mathbb{R} \times \mathbb{R} \to
\mathcal{P} (\mathbb{R})$ is a set-valued map with compact values
and $x \in C(I, \mathbb{R})$ we define
\[
S_F (x) : = \{f \in L^1 (I, \mathbb{R}) :  f (t) \in F (t, x
(t))  \text{ a.e. }  I \}.
\]
Then $F$ is of lower semi-continuous type if $S_F (\cdot)$ is
lower semi-continuous with closed and decomposable values.

\begin{theorem}[\cite{Bressan}] \label{theb} 
Let $S$ be a separable metric space and $G : S \to
\mathcal{P} (L^1 (I, \mathbb{R}))$ be a lower semi-continuous
set-valued map with closed decomposable values. Then $G$ has a
continuous selection (i.e., there exists a continuous mapping 
$g:S \to L^1 (I, \mathbb{R})$ such that $g(s) \in G(s)$ for
all $s \in S$).
\end{theorem}

\begin{definition} \label{def5} \rm
(i) A set-valued map $G : I \to \mathcal{P} (\mathbb{R})$
with nonempty compact convex values is said to be measurable if
for any $x \in \mathbb{R}$ the function $t \to d(x,G(t))$
is measurable.

(ii) A set-valued map $F : I \times \mathbb{R} \to
\mathcal{P} (\mathbb{R})$ is said to be Carath\'eodory if $t
\to F(t,x)$ is measurable for all $x \in \mathbb{R}$ and
$x \to F(t, x)$ is upper semi-continuous for almost all 
$t \in I$.

(iii)  $F$ is said to be $L^1$-Carath\'eodory if for any $l > 0$
there exists $h_l \in L^1 (I, \mathbb{R})$ such that 
$\sup \{|v|: v  \in F(t,x) \} \leq h_l(t) $ a.e. $I$
for all $x  \in \mathbb{R}$.
\end{definition}


The following results are easily deduced from the
theoretical limit set properties.

\begin{lemma}[{\cite[Lemma 1.1.9]{Aubin}}] \label{lem401} 
Let $\{K_n \}_{n \in \mathbb{N}} \subset K \subset X$ be a
sequence of subsets where $K$ is a compact subset of a separable
Banach space $X$. Then
\[
\overline{\operatorname{co}} (\limsup_{n \to \infty} K_n) =
\cap_{N > 0} \overline{\operatorname{co}} \big(\cup_{n \geq N}K_n\big),
\]
where $\overline{\operatorname{co}} (A)$ refers to the closure of the
convex hull of $A$.
\end{lemma}

\begin{lemma}[{\cite[Lemma 1.4.13]{Aubin}}] \label{lem402} 
Let $X$ and $Y$ be two metric spaces. If $G : X \to
\mathcal{P}_{cp}(Y)$ is upper semi-continuous, then for each 
$x_0 \in X$,
\[
\limsup_{x \to x_0} G (x) = G (x_0).
\]
\end{lemma}

\begin{definition} \label{def6} \rm
Let $X$ be a Banach space. A sequence $\{x_n \}_{n \in \mathbb{N}}
\subset L^1([a, b], X)$ is said to be semi-compact if

(a) it is integrably bounded; i.e., there exists $q \in L^1([a, b],
\mathbb{R}^+)$ such that
\[
|x_n(t)|_E \leq q(t), \quad \text{for a.e. $t \in [a, b]$ 
 and every $n \in \mathbb{N}$},
\]

(b) the image sequence $\{x_n (t) \}_{n―in \mathbb{N}}$ is
relatively compact in $E$ for a.e. $t \in [a, b]$.
\end{definition}

The following important result follows from the Dunford-Pettis
theorem (see \cite[Proposition 4.2.1]{Kamenskii}).

\begin{lemma}\label{lem3.2}
 Every semi-compact sequence $L^1([a, b], X)$ is weakly
compact in the space $L^1([a, b], X)$.
\end{lemma}

When the nonlinearity takes convex values, Mazur's Lemma
may be useful:

\begin{lemma}[{\cite[Theorem 21.4]{Musielak}}] \label{lem3.3}
Let $E$ be a normed space and $\{x_k \}_{k―in \mathbb{N}}
―subset E$ a sequence weakly converging to a limit $x \in E$.
Then there exists a sequence of convex combinations 
$y_m = \sum_{k=1}^m \alpha_{m k} x_k$ with $\alpha_{m k} > 0$ for
 $k = 1, 2, \dots ,m$ and $\sum_{k=1}^m \alpha_{m k} = 1$ which converges
strongly to $x$.
\end{lemma}

\begin{theorem}\label{tusc}
The Carath\'eodory multivalued map $F : I \times \mathbb{R}
\to \mathcal{P} (\mathbb{R})$ has non\-empty, compact,
convex values and satisfies:
\begin{itemize}
\item[(H5)] There exist a continuous nondecreasing function $\psi : [0, +
\infty) \to (0, + \infty)$ and $\varphi \in L^1 (I,
\mathbb{R}^+)$ such that
\[
\|F (t, x)\|_{\mathcal{P}} : = \sup \Big \{|v(t) | : \; v \in F
(t, x) \Big \} \leq \varphi (t) \psi (\|x\|),
\]
for a.e. $t \in I$ and each $x \in \mathbb{R}$.

\item[(H6)] There exists a constant $M > 0$ such that
\begin{equation} \label{r2}
\begin{split}
&M \Big[C +  \psi (M) \Big [\int_0^1 G (s, s) \Big(\int_0^1
H(s,r) \varphi (r)  d r \Big)d s\\
& + \frac{\sum_{i=1}^{m-2}
a_{i}}{(1 -\Delta)} \int_0^1 G (\xi_{i}, s) \Big(\int_0^1 H(s,r)
\varphi (r)  d r \Big)\,ds \Big ] \Big]^{-1}    > 1.
\end{split}
\end{equation}
\end{itemize}
Then  problem \eqref{110} has at least one solution.
\end{theorem}

\begin{proof} Let $X = E$ and consider $M > 0$ as in \eqref{r}. It
is obvious that the existence of solutions to problem \eqref{110}
reduces to the existence of the solutions of the integral
inclusion
\begin{align*}
u (t)  & =   z (t ) + \int_0^1 G (t, s) \Big(\int_0^1 H(s,r) F (r, u (r)) d r \Big)d s \\
&\quad +  \frac{t^{\alpha - 1}}{(1 -\Delta)}\sum_{i=1}^{m-2} a_{i}
\int_0^1 G (\xi_{i}, s) \Big(\int_0^1 H(s,r) F (r, u (r)) d r
\Big)\,ds .
\end{align*}
\begin{equation} \label{3}
\begin{split}
u(t) & \in  z (t ) + \int_0^1 G (t, s) \Big(\int_0^1 H(s,r) F (r, u (r)) d r \Big)d s \\
&\quad +  \frac{t^{\alpha - 1}}{(1 -\Delta)}\sum_{i=1}^{m-2} a_{i}
\int_0^1 G (\xi_{i}, s) \Big(\int_0^1 H(s,r) F (r, u (r)) d r
\Big)\,ds, \quad  t \in I,
\end{split}
\end{equation}
where $G(t, s)$ and $H (t, s)$ defined by \eqref{a3} and
\eqref{H}, respectively. Consider the set-valued map $T : E \to \mathcal{P} (X)$
defined by
\begin{equation} \label{t}
\begin{split}
T(u) &:= \Big \{v \in X ; \; v (t) = z (t ) + \int_0^1 G (t, s) 
\Big(\int_0^1 H(s,r) f (r) d r \Big)d s \\
&\quad + \frac{t^{\alpha - 1}}{(1
-\Delta)}\sum_{i=1}^{m-2} a_{i} \int_0^1 G (\xi_{i}, s)
\Big(\int_0^1 H(s,r) f (r) d r \Big)\,ds,\quad f \in \overline{S_F (u)} \Big \}.
\end{split}
\end{equation}

We show that $T$ satisfies the hypotheses of the Theorem \ref{thels}.

\textbf{Claim 1.} We show that $T(u) \subset X$ is convex for any 
$u \in X$. If $v_1, v_2 \in T(u)$ then there exist 
$f_1, f_2 \in S_F (u)$ such that for any $t \in I$ one has
\begin{equation} \label{3b}
\begin{split}
v_i(t) & =   z (t ) + \int_0^1 G (t, s) \Big(\int_0^1 H(s,r) f_i (r) d r \Big)d s \\
&\quad + \frac{t^{\alpha - 1}}{(1-\Delta)}
\sum_{i=1}^{m-2} a_{i} \int_0^1 G (\xi_{i}, s)
\Big(\int_0^1 H(s,r) f_i (r) d r \Big)\,ds, \quad  i = 1, 2.
\end{split}
\end{equation}
Let $0 \leq \lambda \leq 1$. Then for any $t \in I$ we have
\begin{align*}
&(\lambda v_1 + (1-\lambda) v_2) (t)\\
& =   z (t ) + \int_0^1 G (t, s) \Big(\int_0^1 H(s,r)
  [\lambda f_1 (r) + (1-\lambda) f_2 (r)] d r \Big)d s \\
&\quad +  \frac{t^{\alpha - 1}}{(1 -\Delta)}\sum_{i=1}^{m-2} a_{i}
\int_0^1 G (\xi_{i}, s) \Big(\int_0^1 H(s,r) [\lambda f_1 (r) +
(1-\lambda) f_2 (r)] d r \Big)\,ds.
\end{align*}
The values of $F$ are convex, thus $S_F(u)$ is a convex set and
hence $\lambda v_1 + (1-\lambda) v_2 \in T(u)$.

\textbf{Claim 2.} we show that $T$ is bounded on bounded sets of $X$.
Let $B$ be any bounded subset of $X$. Then there exist $m > 0$
such that $\|u\| \leq m$ for all $u \in B$. If $v \in T (u)$ there
exists $f \in S_F (u)$ such that
\begin{equation} \label{vt}
\begin{split}
v(t) & =   z (t ) + \int_0^1 G (t, s) \Big(\int_0^1 H(s,r) f (r) d r \Big)d s \\
&\quad  + \frac{t^{\alpha - 1}}{(1-\Delta)}\sum_{i=1}^{m-2} a_{i}
\int_0^1 G (\xi_{i}, s) \Big(\int_0^1 H(s,r) f (r) d r \Big)\,ds.
\end{split}
\end{equation}
 One may write for any $t \in I$,
\begin{align*}
 |v (t)|
& \leq   |z (t )| + \int_0^1 G (t, s) \Big(\int_0^1 H(s,r) |f (r)| d r \Big)d s \\
&\quad  + \frac{\sum_{i=1}^{m-2} a_{i} }{(1 -\Delta)}
 \int_0^1 G (\xi_{i}, s)
 \Big(\int_0^1 H(s,r) |f (r)| d r \Big)\,ds\\
& \leq   |z (t )| + \int_0^1 G (s, s)
 \Big(\int_0^1 H(s,r) \varphi (r) \psi (\|u\|) d r \Big)d s \\
&\quad  + \frac{\sum_{i=1}^{m-2} a_{i}}{(1 -\Delta)}
\int_0^1 G (\xi_{i}, s)
 \Big(\int_0^1 H(s,r) \varphi (r) \psi (\|u\|) d r \Big)\,ds\\
& \leq   |z (t )| + \psi (\|u\|)  \Big [\int_0^1 G (s, s)
 \Big(\int_0^1 H(s,r) \varphi (r)  d r \Big)d s \\
&\quad + \frac{\sum_{i=1}^{m-2} a_{i}}{(1 -\Delta)}
\int_0^1 G (\xi_{i}, s) \Big(\int_0^1 H(s,r) \varphi (r)  d r\Big)\,ds \Big ].
\end{align*}
Therefore,
\begin{align*}
 \|v\| & =  \max_{t \in I} |v (t)|\\
 & \leq  C +  \psi (m)  \Big [\int_0^1 G (s, s) \Big(\int_0^1 H(s,r) \varphi (r)  d r
 \Big)d s \\
&\quad +  \frac{\sum_{i=1}^{m-2} a_{i}}{(1 -\Delta)} \int_0^1 G
(\xi_{i}, s) \Big(\int_0^1 H(s,r) \varphi (r)  d r \Big)\,ds \Big],
\end{align*}
for all  $v \in T (u)$; i.e., $T(B)$ is bounded.

\textbf{Claim 3.} We show that $T$ maps bounded sets into
equi-continuous sets. Let $B$ be any bounded subset of $X$ as
before and $v \in T (u)$ for some $u \in B$. Then, there exists 
$f \in S_F (u)$ such that $v(t)$ is defined as \eqref{vt}.
 So, for any $t_1, t_2\in [0, 1]$, without loss of generality we may assume
that $ t_2 > t_1$ and one can get
\begin{align*}
| v(t_1) - v(t_2) |  
&\leq  \int_0^1 |G(t_1,s)
 - G(t_2,s)|  \Big(\int_0^1 H(s,r) f (r) d r \Big) ds\\
&\quad +  \dfrac{\sum_{i=1}^{m-2} a_i }{1-\Delta}  |t_1^{\alpha -1}  -
 t_2^{\alpha -1} | \int_0^{1}G_1(\xi_i,s) \Big(\int_0^1 H(s,r) f (r) d r \Big) ds\\
&\leq  \int_0^1 |G(t_1,s)
- G(t_2,s)|  \Big(\int_0^1 H(s,r)  \psi (m) \varphi (r)  d r \Big) ds\\
&\quad +  \dfrac{\sum_{i=1}^{m-2} a_i }{1-\Delta}  |t_1^{\alpha -1}  -
 t_2^{\alpha -1} | \int_0^{1}G_1(\xi_i,s) \Big(\int_0^1 H(s,r)  
 \psi (m) \varphi (r)  d r \Big) ds
\end{align*}
On the other hand, 
\begin{equation} \label{10}
\begin{split}
& \int_0^1 |G(t_1,s) - G(t_2,s)|  \Big(\int_0^1 H(s,r)  \psi (m) \varphi (r)
  d r \Big) ds  \\
&\leq \Big( \int_0^{t_1}+ \int_{t_1}^{t_2}+\int_{t_2}^{1}\Big) |G(t_1,s)
- G(t_2,s)|  \Big(\int_0^1 H(s,r)  \psi (m) \varphi (r)  d r \Big) ds  \\
&\leq \psi (m) \int_0^{t_1} [(t_2^{\alpha -1} -t_1^{\alpha -1})
 (1 - s)^{\alpha - 1} +( (t_2-s)^ {\alpha -1} - (t_1-s)^ {\alpha -1})]\\
&\quad\times \Big(\int_0^1 H(s,r) \varphi (r)  d r \Big)  ds \\
&\quad  + \psi (m)\int_{t_1}^{t_2} [(t_2^{\alpha -1}
-t_1^{\alpha -1}) (1 - s)^{\alpha - 1}
+ (t_2-s)^ {\alpha-1} ] \Big(\int_0^1 H(s,r) \varphi (r)  d r \Big) ds \\
&\quad   + \psi (m) \int_{t_2} ^{1} [(t_2^{\alpha -1}
-t_1^{\alpha -1}) (1 - s)^{\alpha - 1}] \Big(\int_0^1 H(s,r) \varphi (r)  d r \Big) ds \\
& \to  0 \quad \text{uniformly as }t_1 \to t_2.
\end{split}
\end{equation}
Thus,
\begin{equation} \label{v1}
| v(t_1) - v(t_2) |  \to  0\quad \text{uniformly as }t_1 \to t_2.
\end{equation}
Therefore, $T(B)$ is an equi-continuous set in $X$. As
 satisfies the above Claims 2 and 3, therefore
it follows by the Arzel\'{a}-Ascoli theorem that
 $T : C([0, 1], \mathbb{R}) \to \mathcal{P}(C([0, 1], \mathbb{R}))$ is
completely continuous.

\textbf{Claim 5.} $T$  is upper semi-continuous. To this end, it is
sufficient to show that $T$ has a closed graph.
 Let $v_n \in T(u_n)$ be such that $v_n \to v$ and $u_n \to u$,
as $n \to + \infty$. Then there exists $m > 0$ such that
$\|u_n\| \leq m$. We shall prove that $v \in T(u)$ means that
there exists $f_n \in S_F (u_n)$ such that, for a.e. $t \in I$, we
have
\begin{align*}
v_n(t) & =   z (t ) + \int_0^1 G (t, s) \Big(\int_0^1 H(s,r) f_n (r) d r \Big)d s \\
&\quad  + \frac{t^{\alpha - 1}}{(1-\Delta)}\sum_{i=1}^{m-2} a_{i} 
\int_0^1 G (\xi_{i}, s) \Big(\int_0^1 H(s,r) f_n (r) d r \Big)\,ds.
\end{align*}
The condition (H5) implies that $f_n(t) \in \varphi (t) \psi (m) B_1
(0)$. Then $\{f_n\}_{n―in \mathbb{N}}$ is integrable bounded in
$L^1(I, \mathbb{R})$. Since $F$ has compact values, we deduce that
$\{f_n\}_n$ is semi-compact. By Lemma \ref{lem3.2}, there exists a
subsequence, still denoted $\{f_n\}_{n―in \mathbb{N}}$, which
converges weakly to some limit $f \in L^1(I, \mathbb{R})$.
Moreover, the mapping $\Gamma : L^1(I, \mathbb{R}) \to X =E$ defined by
\begin{align*}
\Gamma (g) (t) 
& =  \int_0^1 G (t, s) \Big(\int_0^1 H(s,r) g (r) d r \Big)d s \\
&\quad + \frac{t^{\alpha - 1}}{(1
-\Delta)}\sum_{i=1}^{m-2} a_{i} \int_0^1 G (\xi_{i}, s)
\Big(\int_0^1 H(s,r) g (r) d r \Big)\,ds
\end{align*}
is a continuous linear operator. Then it remains continuous if
these spaces are endowed with their weak topologies
\cite{Kantorovich, Musielak}. Moreover for a.e. $t \in I$, $u_n(t)$
converges to $u(t)$. Then we have
\begin{align*}
v (t)& =   z (t ) + \int_0^1 G (t, s) \Big(\int_0^1 H(s,r) f (r) d r \Big)d s \\
&\quad + \frac{t^{\alpha - 1}}{(1
-\Delta)}\sum_{i=1}^{m-2} a_{i} \int_0^1 G (\xi_{i}, s)
\Big(\int_0^1 H(s,r) f (r) d r \Big)\,ds.
\end{align*}
It remains to prove that $f \in F (t, u(t))$, a.e. $t \in I$.
Mazur's Lemma \ref{lem3.3} yields the existence of 
$\alpha_i^n \geq 0$, $i = n, \dots, k(n)$ such that 
$\sum_{i=1}^{k (n)} \alpha_i^n = 1$ and the sequence of convex combinations
$g_n(\cdot) = \sum_{i=1}^{k (n)} \alpha_i^n f_i(\cdot)$ converges
strongly to $f$ in $L^1$. Using Lemma \ref{lem401}, we obtain that
\begin{equation} \label{co}
\begin{split}
v (t) & \in  \cap_{n \geq 1} \overline{\{g_n(t) \}}, \quad \text{a.e. }  t \in I   \\
& \subset  \cap_{n \geq 1} \overline{\operatorname{co}} \{f_k (t),
\; k \geq n \}    \\
& \subset  \cap_{n \geq 1} \overline{\operatorname{co}} \Big
\{\cup_{n \geq 1} F (t, u_k (t)) \Big\}    \\
& =  \overline{\operatorname{co}} \Big ( \limsup_{k \to +
\infty} F (t, u_k (t)) \Big).
\end{split}
\end{equation}
The fact that the multivalued function $x \to  F (., x)$
is upper semi-continuous and has compact values, together with
Lemma \ref{lem402}, implies that
\[
\limsup_{n \to + \infty} F (t, u_n (t)) = F (t, u (t)),
\quad  \text{a.e. } t \in I.
\]
This with \eqref{co} yields that
$f(t) \in  \overline{\operatorname{co}}F (t, u (t))$. Finally 
$F (\cdot, \cdot)$ has closed, convex
values; hence $f(t) \in  F (t, u (t))$, a.e. $t \in I$. Thus $v
\in T(u)$, proving that $T$ has a closed graph. Finally, with
Lemma \ref{lemusc} and the compactness of $T$, we conclude that
$T$ is upper semi-continuous.

\textbf{Claim 6.} A priori bounds of solutions. Let $u$ be a solution
of \eqref{110}. Then there exists $f \in L^1(I, \mathbb{R})$ with
$f \in S_F (u)$   such that
\begin{align*}
u (t) & =   z (t ) + \int_0^1 G (t, s) \Big(\int_0^1 H(s,r) f (r) d r \Big)d s \\
&\quad + \frac{t^{\alpha - 1}}{(1
-\Delta)}\sum_{i=1}^{m-2} a_{i} \int_0^1 G (\xi_{i}, s)
\Big(\int_0^1 H(s,r) f (r) d r \Big)\,ds.
\end{align*}
In view of (H5), and
using the computations in the Clime 2 above, for each $t \in I$,
we obtain
\begin{align*}
 |u (t)| & \leq   |z (t )| + \psi (\|u\|)  \Big [\int_0^1 G (s, s) \Big(\int_0^1 H(s,r) \varphi (r)  d r \Big)d s \\
&\quad + \frac{\sum_{i=1}^{m-2} a_{i}}{(1 -\Delta)}
\int_0^1 G (\xi_{i}, s) \Big(\int_0^1 H(s,r) \varphi (r)  d r
\Big)\,ds \Big ].
\end{align*}
Consequently,
\begin{align*}
&\|u\| \Big[C +  \psi (\|u\|) \Big [\int_0^1 G (s, s) 
\Big(\int_0^1 H(s,r) \varphi (r)  d r \Big)d s \\
&+ \frac{\sum_{i=1}^{m-2} a_{i}}{(1 -\Delta)}
\int_0^1 G (\xi_{i}, s) \Big(\int_0^1 H(s,r) \varphi (r)  d r
\Big)\,ds \Big ]\Big]^{-1}  \leq 1.
\end{align*}
In view of (H6), there exists $M$ such that  $\|u\| \ne M$. Let us
set
\[
U : = \{u \in C(I, \mathbb{R}) :   \|u\| < M \}.
\]
Note that the operator $T : \overline{U} \to \mathcal{P}(C([0, 1], \mathbb{R})$ 
is upper semi-continuous and completely continuous. 
From the choice of $U$, there is no $u \in
\partial U$ such that $u = \lambda T (u)$ for some $\lambda \in (0, 1)$. 
Consequently, by the nonlinear alternative of Leray-Schauder type 
(Theorem \ref{thels}), we deduce that
 has a fixed point $u \in U$ which is a solution of
the problem \eqref{110}. This completes the proof.
\end{proof}

\subsection{The lower semi-continuous case}

 Here, we study the case when $T$ is not
necessarily convex valued. Our strategy to deal with this problems
is based on the nonlinear alternative of Leray-Schauder type
together with the selection theorem of Bressan and Colombo
\cite{Bressan} for lower semi-continuous maps with decomposable
values. Consider a Banach space $Y$ and $I = [a, b]$ an interval
of the real line.

\begin{definition} \label{def7} \rm
A subset $A ―subset L^1(I, Y)$ is decomposable if for all $u, v
―in A$ and for every Lebesgue measurable subset $I' \subset I$,
we have $u \chi_{I'} + v \chi_{I \setminus I'} \in A$, where
$\chi_A$ stands for the characteristic function of the set $A$.
\end{definition}

Let $F : I \times \mathbb{R} \to \mathcal{P} (\mathbb{R})$
be a multivalued map with nonempty compact values. Define a
multivalued operator $\mathcal{F} : C([0, + \infty), \mathbb{R})
\to \mathcal{P}(L^1([0, + \infty), \mathbb{R}))$
associated with $F$ as
\begin{equation}\label{f}
\mathcal{F} (u) := \Big \{w \in L^1(I, \mathbb{R})) : w (t) \in
F (t, u (t) ), \;  \text{a.e. }  t \in I \Big \},
\end{equation}
which is called the Nemytskii operator associated with $F$.

\begin{definition} \label{def8} \rm
Let $F : I \times Y  \to \mathcal{P} (Y)$ be a multivalued
map with nonempty compact values. We say that $F$ is of lower
semi-continuous type if its associated Nemytskii operator
$\mathcal{F} : C(I, Y) \to \mathcal{P}(L^1(I, Y))$ defined
by $\mathcal{F} (y) = S_F (y)$ is lower semi-continuous and has
nonempty, closed, and decomposable values.
\end{definition}

\begin{definition} \label{def9} \rm
Let $A$ be a subset of $I \times \mathbb{R}$. $A$ is 
$\mathcal{L}\otimes \mathcal{B} $ measurable if $A$ belongs to the
$\sigma$-algebra generated by all sets of the form $\mathcal{J}
\times \mathcal{D} $, where $\mathcal{J}$ is Lebesgue measurable
in $I$ and $\mathcal{D}$ is Borel measurable in $\mathbb{R}$.
\end{definition}

Next, we state the celebrated selection theorem of Bressan and
Colombo \cite{Bressan}.

\begin{lemma}\label{lemlsc}
Let $X$ be a separable metric space and let $Y$ be a Banach space.
Then every lower semi-continuous multivalued operator $T : X
―to \mathcal{P}_{cl}(L^1(I, Y))$ with nonempty closed
decomposable values has a continuous selection; i.e., there exists
a continuous single-valued function $f : X ―to L^1(I, Y)$
such that $f (x) \in T (x)$ for every $x \in X$.
\end{lemma}

\begin{theorem}\label{tlsc}
Assume that (H5) and the following condition holds:
\begin{itemize}
\item[(H7)] $F : I \times \mathbb{R} \to \mathcal{P}
(\mathbb{R})$ is a nonempty compact-valued multivalued map such
that:
\begin{itemize}
\item[(a)] $(t, x) \to F(t, x)$ is $\mathcal{L} \otimes \mathcal{B}$ measurable,

\item[(b)] $x \to F(t, x )$ is lower semi-continuous
for each $t \in I$.
\end{itemize}
\end{itemize}
Then  boundary value problem \eqref{110} has at least one
solution on $I$.
\end{theorem}

\begin{proof} 
We note first that if (H5) and (H7) are satisfied
then $F$ is of lower semi-continuous type (e.g., \cite{Frignon}).
Then, by Lemma \ref{lemlsc}, there exists a continuous function 
$f : C(I, \mathbb{R}) \to L^1(I, \mathbb{R})$ such that 
$f(u) \in \mathcal{F} (u)$ for all $u \in C(I, \mathbb{R})$.
Consider the problem
\begin{equation}\label{1nia}
      \begin{gathered}
   D_{0^+}^\alpha  u (t)  = f (u (t)), \quad   0< t<+\infty,\\
   D_{0^+}^\alpha u(0)= D_{0^+}^\alpha u(1)=0,\quad
   u (0) = 0, \quad u (1) -\sum_{i=1}^{m-2} a_{i}\; u(\xi_{i})=\lambda,
 \end{gathered}
\end{equation}
in the space $C([0, 1), \mathbb{R})$. It is clear that if $u$ is a
solution of the problem \eqref{1nia}, then $u$ is a solution to
the problem \eqref{110}. In order to transform the problem
\eqref{1nia} into a fixed point problem, we define the operator
as
\begin{align*}
\Theta u (t) & =  \Big \{ z (t ) + \int_0^1 G (t, s) \Big(\int_0^1 H(s,r) f (u(r)) d r \Big)d s \\
&\quad  + \frac{t^{\alpha - 1}}{(1
-\Delta)}\sum_{i=1}^{m-2} a_{i} \int_0^1 G (\xi_{i}, s)
\Big(\int_0^1 H(s,r) f (u(r)) d r \Big)\,ds, \; t \in I \Big \}.
\end{align*}
It can easily be shown that  is continuous and completely continuous. 
The remaining part of the proof is similar to that of Theorem \ref{tusc}. So we
omit it. This completes the proof. 
\end{proof}

\subsection{The Lipschitz case}
 Now we prove the existence of solutions for the problem
\eqref{110} with a nonconvex valued right hand side by applying a
fixed point theorem for multivalued maps due to Covitz and Nadler
\cite{Covitz}.

\begin{definition} \label{def10} \rm
A multivalued operator $N : X \to \mathcal{P}_{cl}(X)$ is
called:
\begin{itemize}
\item[(a)] $\gamma$-Lipschitz if and only if there exists $\gamma > 0$
such that $d_H (N(x), N(y)) \leq d(x, y)$ for each $x, y \in X$;

\item[(b)] a contraction if and only if it is $\gamma$-Lipschitz with
 $\gamma < 1$.
\end{itemize}
\end{definition}

\begin{lemma}[Covitz-Nadler \cite{Covitz}] \label{lemcn}
 Let $(X, d)$ be a complete metric space. If $N : X \to \mathcal{P}_{cl}(X)$ is a
contraction, then $\operatorname{Fix} N \ne \emptyset$.
\end{lemma}


\begin{definition} \label{def11} \rm
A measurable multi-valued function $F : [0, + \infty) \to
\mathcal{P}(X)$ is said to be integrably bounded if there exists a
function $f \in L^1(I, X)$ such that for all $v \in F(t)$, 
$\|v\| \leq f(t)$ for a.e. $t \in [0, + \infty)$.
\end{definition}

\begin{theorem}\label{tlip}
Assume that the following conditions hold:
\begin{itemize}
\item[(H8)] $F : I \times \mathbb{R}  \to \mathcal{P}_{cp}
(\mathbb{R})$ is such that $F(\cdot, x ) : I \to
\mathcal{P}_{cp} (\mathbb{R})$ is measurable for each $x  \in
\mathbb{R}$;

\item[(H9)] There exists $l : I \to [0, + \infty)$ are not
identical zero on any closed subinterval of $I$, and
\[
\int_0^{1} H (r, r) l(r) dr  < + \infty, \quad i = 1, 2,
\]
such that for almost all $t \in [0, + \infty)$,
\begin{equation*}
d_H (F(t, x_1), F(t, x_2)) \leq l (t) |x_1 - x_2|
\end{equation*}
for all $x_1, x_2 \in \mathbb{R}$ with  $d(0, F(t, 0, 0))
\leq l (t)$ for almost all $t \in I$.
\end{itemize}
Then  boundary-value problem \eqref{110} has at least one
solution on $I = [0, 1]$ if
\[
 \int_0^1 G (s, s) \Big(\int_0^1 H(r,r) l (r) d r \Big)d s + \frac{\sum_{i=1}^{m-2} a_{i}}{(1 -\Delta)}
\int_0^1 G (\xi_{i}, s) \Big(\int_0^1 H(r,r)  l (r) d r \Big)\,ds
< 1.
\]
\end{theorem}

\begin{proof}
 We transform  problem \eqref{110} into a fixed
point problem. Consider the set-valued map $T : C [0, 1]
\to \mathcal{P}(C [0, 1], \mathbb{R}))$ defined at the
beginning of the proof of Theorem \ref{tusc}. It is clear that the
fixed points of $T$ are solutions of \eqref{110}.

Note that since the set-valued map $F(\cdot, u(\cdot))$ is
measurable with the measurable selection theorem 
(e.g., \cite[Theorem III.6]{Castaing}) it admits a measurable selection
 $f: I \to \mathbb{R}$. Moreover, since $F$ is integrably
bounded, $f \in L^1([0, 1], \mathbb{R})$. Therefore, $S_F (u) \ne
\emptyset$.

We shall prove that $T$ fulfills the assumptions of Covitz-Nadler
contraction principle (Lemma \ref{lemcn}).

First, we note that since $S_F(u) \ne \emptyset$, $T(u)  \ne
\emptyset$ for any $u \in C([0, + \infty))$.
Second, we prove that $T(u)$ is closed for any $u \in AC^1([0, +
\infty), \mathbb{R})$. Let $\{u_n\}_{n \geq 0} \in T(u)$ such that
$u_n \to u_0$ in $AC^1([0, + \infty), \mathbb{R})$. Then
$u_0 \in AC^1([0, + \infty), \mathbb{R})$ and there exists $f_n
\in S_F(u)$ such that
\begin{align*}
u_n (t)& =   z (t ) + \int_0^1 G (t, s) \Big(\int_0^1 H(s,r) f_n (r) d r \Big)d s \\
&\quad + \frac{t^{\alpha - 1}}{(1-\Delta)}\sum_{i=1}^{m-2} a_{i} 
\int_0^1 G (\xi_{i}, s) \Big(\int_0^1 H(s,r) f_n (r) d r \Big)\,ds.
\end{align*}
Since $F$ has compact values, we may pass onto a subsequence (if
necessary) to obtain that $f_n$ converges to 
$f \in L^1(([0, 1], \mathbb{R}))$ in $L^1(([0, 1], \mathbb{R}))$. In particular, 
$f \in S_F(u)$ and for any $t \in [0, 1]$ we have
\begin{align*}
u_n (t) \to u_0 (t) 
& =   z (t ) + \int_0^1 G (t, s) \Big(\int_0^1 H(s,r) f (r) d r \Big)d s \\
&\quad + \frac{t^{\alpha - 1}}{(1-\Delta)}\sum_{i=1}^{m-2} a_{i} 
\int_0^1 G (\xi_{i}, s) \Big(\int_0^1 H(s,r) f (r) d r \Big)\,ds;
\end{align*}
i.e., $u_0 \in T(u)$ and $T(u)$ is closed.

Next we show that $T$ is a contraction on $C([0, 1], \mathbb{R})$. Let
$u_1, u_2 \in C([0, 1], \mathbb{R})$ and $v_1 \in T (u_1)$. Then
there exist $f_1 \in S_F (u_1)$ such that
\begin{align*}
v_1 (t) & =   z (t ) + \int_0^1 G (t, s) \Big(\int_0^1 H(s,r) f_1 (r) d r \Big)d s \\
&\quad + \frac{t^{\alpha - 1}}{(1
-\Delta)}\sum_{i=1}^{m-2} a_{i} \int_0^1 G (\xi_{i}, s)
\Big(\int_0^1 H(s,r) f_1 (r) d r \Big)\,ds, \quad \; t \in I.
\end{align*}
Consider the set-valued map
\[
H (t) : =  F (t, u_2 (t)) \cap \{u \in \mathbb{R}: |f_1 (t) -
u| \leq l (t) |x_1 - x_2|\}, \quad t \in [0, + \infty).
\]
By (H5), we have
\begin{equation*}
d_H (F(t, x_1), F(t, x_2)) \leq l (t) |x_1 - x_2|,
\end{equation*}
hence $H$ has nonempty closed values. Moreover, since $H$ is
measurable (e.g., \cite[Proposition III.4]{Castaing}), there
exists $f_2$ a measurable selection of $H$. It follows that 
$f_2 \in S_F (u_2)$ and for any $t \in [0, 1]$,
\[
|f_1 (t) - f_2 (t)| \leq l (t) |x_1 - x_2|.
\]
Define
\begin{align*}
v_2 (t)   & =   z (t ) + \int_0^1 G (t, s) \Big(\int_0^1 H(s,r) f_2 (r) d r \Big)d s \\
&\quad + \frac{t^{\alpha - 1}}{(1
-\Delta)}\sum_{i=1}^{m-2} a_{i} \int_0^1 G (\xi_{i}, s)
\Big(\int_0^1 H(s,r) f_2 (r) d r \Big)\,ds, \quad  t \in I,
\end{align*}
and one can obtain
\begin{align*}
  | v_1 (t)  -  v_2 (t) | 
& \leq   \int_0^1 G (s, s) \Big(\int_0^1 H(r,r) |f_1(r)- f_1(r)| d r \Big)d s \\
&\quad + \frac{\sum_{i=1}^{m-2} a_{i}}{(1 -\Delta)}
\int_0^1 G (\xi_{i}, s)
\Big(\int_0^1 H(r,r) |f_1(r)- f_1(r)| d r \Big)\,ds\\
& \leq  \int_0^1 G (s, s) \Big(\int_0^1 H(r,r) l (r) |x_1 (r) - x_2 (r)| d r \Big)d s \\
&\quad + \frac{\sum_{i=1}^{m-2} a_{i}}{(1 -\Delta)}
\int_0^1 G (\xi_{i}, s)
\Big(\int_0^1 H(r,r)  l (r) |x_1 (r) - x_2 (r)| d r \Big)\,ds\\
& \leq   \|x_1 - x_2\| \Big [ \int_0^1 G (s, s) \Big(\int_0^1 H(r,r) l (r) d r \Big)d s \\
&\quad  + \frac{\sum_{i=1}^{m-2} a_{i}}{(1 -\Delta)}
\int_0^1 G (\xi_{i}, s) \Big(\int_0^1 H(r,r)  l (r) d r \Big)\,ds
\Big].
\end{align*}
Therefore,
\begin{align*}
 \|v_1 - v_2 \|
& \leq  \|x_1 - x_2\| \Big [ \int_0^1 G (s, s) \Big(\int_0^1 H(r,r) l (r) d r \Big)d s \\
&\quad  + \frac{\sum_{i=1}^{m-2} a_{i}}{(1 -\Delta)}
\int_0^1 G (\xi_{i}, s) \Big(\int_0^1 H(r,r)  l (r) d r \Big)\,ds
\Big].
\end{align*}
From an analogous reasoning by interchanging the roles of $u_1$
and $u_2$ it follows
\begin{align*}
 d_H (T (u_1), T (u_2))
& \leq  \|x_1 - x_2\| \Big [ \int_0^1 G (s, s) \Big(\int_0^1 H(r,r) l (r) d r \Big)d s \\
&\quad  + \frac{\sum_{i=1}^{m-2} a_{i}}{(1 -\Delta)}
\int_0^1 G (\xi_{i}, s) \Big(\int_0^1 H(r,r)  l (r) d r \Big)\,ds \Big].
\end{align*}
Since $T$ is a contraction, it follows by the Lemma \ref{lemcn}
that $T$ admits a fixed point which is a solution to problem
\eqref{110}.
\end{proof}

\section{An application}

Consider the  singular boundary-value problem
\begin{equation}\label{3c}
\begin{gathered}
  D_{0^+}^{3/2} (D_{0^+}^{3/2} u) (t)=f (t, u (t)) ,\quad  t \in (0, 1),\\
  D_{0^+}^{3/2} u(0)= D_{0^+}^{3/2} u(1)=0,\quad
  u (0) = 0, \quad 
  u (1) -{\frac{7}{4}}u({\frac{1}{16}})-  u({\frac{1}{4}})=\lambda_1.
\end{gathered}
\end{equation}
Here, $ \alpha= \beta=3/2$, $p= 2$, $m=4$,
$‖_{1}=7/4$, $a_{2}=1$, $\xi_{1}=1/16$ and
$\xi_{2}=1/4$. Let
\[
f (t, u) =  \begin{cases} 
\frac{\sqrt{1 - t^2}}{1000} + \frac{1}{400} u^2, &  t \in [0, 1],  \; 0 \leq u \leq 1,\\
\frac{\sqrt{1 - t^2}}{1000} + 5 [u^2 - u] + \frac{1}{400}, &  t \in [0, 1], \; 
 1 < u  < 2,\\
\frac{\sqrt{1 - t^2}}{1000} +2 [ \log_2{u} + 2u] + \frac{1}{400},
&  t \in [0, 1],  \; 2 \leq u \leq 4\\
\frac{\sqrt{1 - t^2}}{1000} + \frac{\sqrt{u}}{2} +19 +
\frac{1}{400}, & t \in [0, 1], \; 4 < u< + \infty,
\end{cases} 
\]
Choose $\gamma = 1/4$ and $\delta= 3/4$.
Then, by direct calculations,
we can obtain that $\eta=0.3780, \sigma= 0.3536$ and
\[
 \Delta =0.9375,\quad  m = 1.0765, \quad M = 2.9786.
\]
Then, by  Choosing $a = 1$, $b = 2$, $c = 1000$, one obtains
\[
  \frac{1}{p }\cdot \frac{a}{M }  = 0.0671,\quad   
\frac{1}{p }\cdot  \frac{c}{M } = 67.1456, \quad
\frac{b}{m }  = 1.8579.
\]
 It is easy to check that $f$ satisfy the conditions
(H1)--(H3).  Then, all conditions of theorem \ref{the2} hold.
Hence, for $0< \lambda < 3.1250$, the system \eqref{3c} has at
least three positive solutions $u_1, u_2, u_3$ such that 
$\|u_1\| < 1, 2 < \min_{\frac{1}{4} \leq t \leq \frac{3}{4}}(u_2(t))$, and
$\|u_3\|> 1$, with
$\min_{\frac{1}{4} \leq t \leq \frac{3}{4}}(u_3(t)) < 2$.

\subsection*{Acknowledgments}
The authors express their sincere gratitude to the anonymous referees 
for their careful reading, and their valuable suggestions for the 
improvement of this article.


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