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

\AtBeginDocument{{\noindent\small
\emph{Electronic Journal of Differential Equations},
Vol. 2018 (2018), No. 87, pp. 1--16.\newline
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu}
\thanks{\copyright 2018 Texas State University.}
\vspace{7mm}}

\begin{document}
\title[\hfilneg EJDE-2018/87\hfil 
 Multi-term fractional-order boundary-value problems]
{Multi-term fractional-order boundary-value problems with nonlocal integral
boundary conditions}

\author[A. Alsaedi, N. Alghamdi, R. P. Agarwal, S. K. Ntouyas, B. Ahmad 
 \hfil EJDE-2018/87\hfilneg]
{Ahmed Alsaedi, Najla Alghamdi, Ravi P. Agarwal, \\
Sotiris K. Ntouyas, Bashir Ahmad}

\address{Ahmed Alsaedi \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{aalsaedi@hotmail.com}

\address{Najla Alghamdi \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.\newline
Department of Mathematics, Faculty of Science, 
University of Jeddah, P.O. Box 80327, Jeddah 21589, Saudi Arabia}
\email{njl-ghamdi@hotmail.com}

\address{Ravi P. Agarwal \newline
Department of Mathematics,
Texas A\&M University,
Kingsville, TX 78363-8202, USA. \newline
Distinguished University Professor of Mathematics,
Florida Institute of Technology,
150 West University Boulevard, Melbourne, FL 32901, USA}
\email{Ravi.Agarwal@tamuk.edu}

\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{Bashir Ahmad \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{bashirahmad\_qau@yahoo.com}

\dedicatory{Communicated by Vicentiu D. Radulescu}

\thanks{Submitted February 25, 2018. Published April 10, 2018.}
\subjclass[2010]{34A08, 34B10}
\keywords{Caputo fractional derivative; integral boundary condition;
\hfill\break\indent multi-term fractional differential equations;
 existence; fixed point}

\begin{abstract}
 In this article, we introduce a class of multi-term fractional-order
 boundary-value problems involving nonlocal integral boundary conditions.
 Existence results for the given problem are obtained by means of standard
 tools of fixed point theory. The results are illustrated with the
 aid of examples and make a useful contribution to the existing literature
 on the topic.
\end{abstract}

\maketitle
\numberwithin{equation}{section}
\newtheorem{theorem}{Theorem}[section]
\newtheorem{lemma}[theorem]{Lemma}
\newtheorem{property}[theorem]{Property}
\newtheorem{remark}[theorem]{Remark}
\newtheorem{definition}[theorem]{Definition}
\newtheorem{example}[theorem]{Example}
\allowdisplaybreaks

\section{Introduction}

Fractional differential equations arise in the mathematical modeling of many
engineering and scientific disciplines such as biophysics, bio-engineering,
virology, control theory, signal and image processing, blood flow phenomena,
 etc. A huge amount of mathematically and physically interesting works
published in recent years, including
several excellent monographs, clearly reflects the overwhelming interest
in the topic. For details we refer the reader the texts
\cite{Dieth,Kil,Mil,Pod,Sam}
and references cited therein.

Nonlocal boundary-value problems of fractional-order differential equations
and inclusions have received significant attention. One can witness a great
deal of work on the topic involving different kinds of boundary conditions
in the literature, for example, see \cite{Bas2,Bas1,f3, fp1,Rod, NTS}
and the references cited therein.

 There is another class of differential equations containing more than one
fractional-order differential operators. Such equations appear in the modeling
of the motion of a rigid plate immersed in a Newtonian fluid.
Other typical examples include Bagley-Torvik \cite{Tov} and Basset
equation \cite{Main}. Some recent results on multi-term fractional
differential equations can be found in the articles
\cite{Ravi,Bas,Bash-IJAA,f1,Li,f4}.

In this article, we introduce and investigate the following nonlinear
multi-term fractional order boundary value problem with nonlocal integral
conditions:
\begin{gather}\label{a1}
 (p_2{}^cD^{\delta+2}+p_1{}^cD^{\delta+1}+p_0{}^cD^\delta)x(t)=f(t,x(t)), \quad
 0<\delta<1, \; 0<t<1, \\
\label{a2}
 x(0)=0, \quad x(\xi)=0, \quad x(1)= \lambda \int_0^\sigma x(s) ds, \quad
 0<\sigma<\xi<1, \; \lambda \in \mathbb{R},
\end{gather}
where $^cD^\delta$ denote the Caputo fractional derivative of order
$\delta$, $f:[0,1]\times \mathbb{R} \to \mathbb{R}$ is a given
continuous functions, and $p_j, j=0,1,2$ are real constants.

Existence results for problem \eqref{a1}-\eqref{a2} are obtained with the
help of Krasnoselskii fixed point theorem and Leray-Schauder nonlinear alternative,
while the uniqueness result is proved via Banach contraction mapping principle.
These results are presented in Section 3. Some preliminary concepts and lemmas
are given in Section 2. The obtained results are well illustrated by examples.

\section{Preliminary concepts and basic result}

We begin this section with some definitions \cite{Kil}.

\begin{definition} \label{def2.1} \rm
The Riemann-Liouville fractional integral of order $\tau>0$ for a function
 $h: [0, 1] \to \mathbb{R}$ with $h \in L(0, 1)$ is defined by
 \begin{equation}
I^{\tau}h(u)=\int_0^u\frac{(u-v)^{\tau-1}}{\Gamma(\tau)}h(v)dv, \quad\text{for a.e, }
 u \in [0, 1],
\end{equation}
 where $\Gamma$ is the Gamma function.
\end{definition}


\begin{definition} \label{def2.2} \rm
The Caputo derivative of order $\tau \in (n-1, n)$ for a function
$h:[0,1]\to {\mathbb R}$ with
 $h\in C^{n}[0,1]$ is defined by
$$
^cD^{\tau} h (u)= \frac{1}{\Gamma(n-\tau)}\int_0^u
\frac{h^{(n)}(v)}{(u-v)^{\tau+1-n}}dv
= I^{n-\tau}h^{(n)}(u),\quad u>0.
$$
\end{definition}

\begin{property}\label{property1}
With the given notations, the following equality holds:
\begin{equation} \label{prop}
 I^{\tau} ({^cD^{\tau} h (u)})=h (u)-c_0-c_1u-\cdots-c_{n-1}u^{n-1},\quad
 u>0, \; n-1<\tau <n,
\end{equation}
where $c_i ~(i=1,\ldots, n-1)$ are arbitrary constants.
\end{property}

To define the solution for problem \eqref{a1}-\eqref{a2}, we consider
its linear variant in the following lemma.

\begin{lemma}\label{lemma1}
Let $p_0,p_1,p_2$ be positive constants such that ${p_1}^2-4p_0p_2>0$ and
$y \in C(0, 1)\cap L(0,1)$. Then the solution of the linear multi-term
fractional differential equation
\begin{equation}\label{a3}
 (p_2{}^cD^{\delta+2}+p_1{}^cD^{\delta+1}+p_0{}^cD^\delta)x(t)=y(t), \quad
 0<\delta<1, \; 0<t<1,
\end{equation}
supplemented with the boundary conditions \eqref{a2} is given by
\begin{equation} \label{a4}
\begin{aligned}
&x(t)\\
&=\frac{1}{p_2(m_2-m_1)}\Big\{\int_0^t\int_0^s \Phi(t)
 \frac{(s-u)^{\delta-1}}{\Gamma(\delta)}y(u)\,du\,ds\\
&\quad+\rho_1(t)\int_0^\xi \int_0^s \Phi(\xi)
 \frac{(s-u)^{\delta-1}}{\Gamma(\delta)}y(u)\,du\,ds \\
&\quad  +\rho_2(t)\Big [\int_0^1 \int_0^s \Phi(1)\frac{(s-u)^{\delta-1}}
{\Gamma(\delta)}y(u)\,du\,ds\\
&\quad -\lambda \int_0^\sigma \int_0^s \Big(\frac{(e^{m_2(\sigma-s)}-1)}{m_2}
 -\frac{(e^{m_1(\sigma-s)}-1)}{m_1}\Big)
 \frac{(s-u)^{\delta-1}}{\Gamma(\delta)}y(u)\,du\,ds\Big] \Big\},
\end{aligned}
\end{equation}
where
\begin{equation}\label{eq-A-1}
\begin{gathered}
 \Phi(\kappa)=e^{m_2(\kappa-s)}-e^{m_1(\kappa-s)} \quad \kappa=t,1,\xi\,,\\
 m_1=\frac{-p_1-\sqrt{p_1^2-4p_0p_2}}{2p_2}, \quad
 m_2=\frac{-p_1+\sqrt{p_1^2-4p_0p_2}}{2p_2},\\
  \rho_1(t)=\frac{\omega_4\varrho_1(t)-\omega_3\varrho_2(t)}{\mu_1}, \quad
 \rho_2(t)=\frac{\omega_1\varrho_2(t)-\omega_2\varrho_1(t)}{\mu_1},\\
 \varrho_1(t)=\frac{m_1(1-e^{m_2t})-m_2(1-e^{m_1t})}{m_1m_2}, \\
 \varrho_2(t)=p_2(m_2-m_1)(e^{m_2t}-e^{m_1t}),\\
 \mu_1=\omega_1\omega_4-\omega_2\omega_3\neq0,\quad
 \omega_1=\frac{m_2(1-e^{m_1\xi})-m_1(1-e^{m_2\xi})}{m_1m_2}, \\
 \omega_2=p_2(m_2-m_1)(e^{m_1\xi}-e^{m_2\xi}),\\
\begin{aligned}
 \omega_3&=\Big(m_2\big(1-e^{m_1}-\lambda \sigma+\lambda/m_1 (e^{m_1\sigma}-1)\big)\\
&\quad  -m_1\big(1-e^{m_2}-\lambda \sigma+\lambda/m_2 (e^{m_2 \sigma}-1)\big)\Big)
/(m_1m_2),
\end{aligned}\\
\begin{aligned}
\omega_4&=p_2(m_2-m_1)\Big((e^{m_1}+\lambda/m_1(1-e^{m_1 \sigma}))\\
&\quad -(e^{m_2}+\lambda/m_2(1-e^{m_2 \sigma}))\Big).
\end{aligned}
 \end{gathered}
\end{equation}
\end{lemma}


\begin{proof}
Applying the operator $I^\delta$ on \eqref{a3} and using Property \eqref{property1},
we get
\begin{equation}\label{a6}
 (p_2D^2+p_1D+p_0)x(t)=\int_0^t \frac{(t-s)^{\delta-1}}{\Gamma(\delta)}y(s)ds+c_1,
\end{equation}
where $c_1$ is an arbitrary constant. By the method of variation of parameters,
the solution of \eqref{a6} can be written as
\begin{equation} \label{b1}
\begin{aligned}
 x(t)
&= c_2 e^{m_1t}+c_3 e^{m_2t}- \frac{1}{p_2(m_2-m_1)}
 \int_0^t e^{m_1(t-s)}\Big(\int_0^s\frac{(s-u)^{\delta-1}}{\Gamma(\delta)}
 y(u)du+c_1\Big)ds \\
&\quad +\frac{1}{p_2(m_2-m_1)}\int_0^t e^{m_2(t-s)}
\Big(\int_0^s\frac{(s-u)^{\delta-1}}{\Gamma(\delta)}y(u)du+c_1\Big)ds,
\end{aligned}
 \end{equation}
where $m_1$ and $m_2$ are given by \eqref{eq-A-1}. Using $x(0)=0$ in \eqref{b1},
we get
 \begin{equation}\label{b2}
\begin{aligned}
 x(t)
&= c_1\Big[\frac{m_2(1-e^{m_1t})-m_1(1-e^{m_2t})}{p_2m_1m_2(m_2-m_1)}\Big]
 +c_2\Big(e^{m_1t}-e^{m_2t}\Big) \\
&\quad - \frac{1}{p_2(m_2-m_1)}\Big[\int_0^t\Big(e^{m_1(t-s)}-e^{m_2(t-s)}\Big)
\Big(\int_0^s\frac{(s-u)^{\delta-1}}{\Gamma(\delta)}y(u)du\Big)ds\Big],
\end{aligned}
 \end{equation}
which together with the conditions $x(\xi)=0$ and
$x(1)= \lambda\int_0^\sigma x(s) ds$ yields the following system of equations
in the unknown constants $c_1$ and $c_2$:
\begin{gather}
 c_1\omega_1+c_2\omega_2 = V_1,\label{s1}\\
 c_1\omega_3+c_2\omega_4 = V_2 \label{s2}.
 \end{gather}
where
 \begin{gather*}
 V_1 =\int_0^\xi \int_0^s \Phi(\xi) \frac{(s-u)^{\delta-1}}{\Gamma(\delta)}y(u)\,du\,ds,
 \\
\begin{aligned}
V_2 &= \int_0^1 \int_0^s \Phi(1) \frac{(s-u)^{\delta-1}}{\Gamma(\delta)}y(u)\,du\,ds \\
 &\quad - \lambda \int_0^\sigma \int_0^s \Big[\frac{(e^{m_1(\sigma-s)}-1)}{m_1}
 -\frac{(e^{m_2(\sigma-s)}-1)}{m_2}\Big]\frac{(s-u)^{\delta-1}}{\Gamma(\delta)}y(u)
\,du\,ds.
\end{aligned}
 \end{gather*}

Solving  system \eqref{s1}-\eqref{s2} and using \eqref{eq-A-1}, we find that
\begin{equation*}
c_1 = \frac{V_1\omega_4-V_2\omega_2}{\mu_1},\quad
c_2 = \frac{V_2\omega_1-V_1\omega_3}{\mu_1}.
\end{equation*}

Substituting the value of $c_1$ and $c_2$ in \eqref{b2}, we obtain the
solution \eqref{a4}. The converse of the lemma follows by direct computation.
This completes the proof.
\end{proof}

\begin{remark}\label{rmk2.6} \rm
(i) When ${p_1}^2-4p_0p_2=0$ the solution of \eqref{a3} equipped
with  condition \eqref{a2} is
 \begin{equation}\label{a6-1}
\begin{aligned}
  x(t)&= \frac{1}{p_2}\Big\{\int_0^t\int_0^s \Psi(t)
 \frac{(s-u)^{\delta-1}}{\Gamma(\delta)}y(u)\,du\,ds\\
 &\quad  +\chi_1(t) \int_0^{\xi} \int_0^s \Psi(\xi)
\frac{(s-u)^{\delta-1}}{\Gamma(\delta)}y(u)\,du\,ds\\
 &\quad  +\chi_2(t)\Big[\int_0^1 \int_0^s \Psi(1)
\frac{(s-u)^{\delta-1}}{\Gamma(\delta)}y(u)\,du\,ds\\
 &\quad  -\lambda\int_0^\sigma\Big(\frac{m(\sigma-s)e^{m(\sigma-s)}
-e^{m(\sigma-s)}+1}{m^2}\Big)\frac{(s-u)^{\delta-1}}{\Gamma(\delta)}y(u)\,du\,ds
\Big]\Big\},
\end{aligned}
\end{equation}
where
\begin{equation}\label{eq-A-2}
\begin{gathered}
 \Psi(\kappa)=(\kappa-s) e^{m(\kappa-s)} \quad \kappa=t,1,\xi\,,\\
 m=\frac{-p_1}{2p_2},\\
  \chi_1(t)=\frac{\varpi_3v_2(t)-\varpi_4v_1(t)}{\mu_2},\quad
\chi_2(t)=\frac{\varpi_2v_1(t)-\varpi_1v_2(t)}{\mu_2},\\
 v_1(t)=\frac{mt e^{mt}-e^{mt}+1}{m^2},\quad v_2(t)=p_2 t e^{mt},\\
\varpi_1=\frac{m\xi e^{m\xi}-e^{m\xi}+1}{m^2},\quad
\varpi_2=p_2 \xi e^{m\xi},\\
 \varpi_3=\frac{m^2 e^m-m e^m+m-m\sigma e^{m\sigma}+2e^{m\sigma}-2-m\sigma}{m^3},\\
\varpi_4=p_2\frac{m^2 e^m-\lambda m\sigma e^{m\sigma}
 +\lambda e^{m\sigma}-\lambda}{m^2},\\
\mu_2=\varpi_1\varpi_4-\varpi_2\varpi_3\neq0;
 \end{gathered}
\end{equation}

(ii) When ${p_1}^2-4p_0p_2<0$ the solution of \eqref{a3} equipped with
 condition \eqref{a2} is
\begin{align*} %\label{a7}
x(t)&= \frac{1}{p_2 b}\Big\{\int_0^t\int_0^s \Omega(t)
 \frac{(s-u)^{\delta-1}}{\Gamma(\delta)}y(u)\,du\,ds \\
&\quad +\tau_1(t) \int_0^{\xi} \int_0^s \Omega(\xi)
 \frac{(s-u)^{\delta-1}}{\Gamma(\delta)}y(u)\,du\,ds  \\
 &\quad  +\tau_2(t)\Big[\int_0^1 \int_0^s \Omega(1)
 \frac{(s-u)^{\delta-1}}{\Gamma(\delta)}y(u)\,du\,ds \\
&\quad -\frac{\lambda}{a^2+b^2}\int_0^\sigma\Big(b+be^{-a(\sigma-s)}
 \cos{b(\sigma-s)} \nonumber\\
 &\quad -a e^{-a(\sigma-s)}\sin{b(\sigma-s)}\Big)\frac{(s-u)^{\delta-1}}
{\Gamma(\delta)}y(u)\,du\,ds\Big]\Big\}, 
\end{align*}
where
\begin{equation}\label{eq-A-3}
\begin{gathered}
\Omega(\kappa)= e^{-a(\kappa-s)}\sin{b(\kappa-s)} \quad \kappa=t,1,\xi\,,\\
m_{1,2}=-a\pm b i,\quad  a=\frac{p_1}{2p_2},\quad
 b=\frac{\sqrt{4p_0p_2-{p_1}^2}}{2p_2},\\
\tau_1(t)=\frac{q_3\nu_2(t)-q_4\nu_1(t)}{\mu_3},\quad
\tau_2(t)=\frac{q_2\nu_1(t)-q_1\nu_2(t)}{\mu_3},\\
\nu_1(t)=\frac{b+be^{-at}\cos{bt}-a e^{-at}\sin{bt}}{a^2+b^2},\quad
\nu_2(t)=p_2b e^{-at}\sin{bt}\\
q_1=\frac{b-be^{-a\xi}\cos{b\xi}-a e^{-a\xi}\sin{b\xi}}{a^2+b^2},\quad
q_2=p_2b e^{-a\xi}\sin{b\xi},\\
\begin{aligned}
q_3&=\frac{1}{a^2+b^2}\Big[b-be^{-a}\cos{b}-a e^{-a}\sin{b}-b\lambda\sigma
 +\frac{b\lambda}{a^2+b^2}(a-ae^{-a\sigma}\cos{b\sigma}\\
 &\quad +b e^{-a\sigma}\sin{b\sigma})+\frac{a\lambda}{a^2+b^2}
 (b-be^{-a\sigma}\cos{b\sigma}-a e^{-a\sigma}\sin{b\sigma})\Big],
\end{aligned}\\
q_4=p_2 b\Big[e^{-a}\sin{b}-\frac{\lambda}{a^2+b^2}
 (b-be^{-a\sigma}\cos{b\sigma}-a e^{-a\sigma}\sin{b\sigma})\Big],\\
\mu_3=q_1q_4-q_2q_3\neq 0.
 \end{gathered}
\end{equation}
\end{remark}

\section{Existence and uniqueness results}

Denote by $\mathcal{C}=C([0,1],\mathbb{R})$ the Banach space of all continuous
functions from $[0,1]\to \mathbb{R}$ endowed with the norm defined by
$\|x\|=\sup{\{|x(t)|:t \in[0,1]\}}$.
 By Lemma \ref{lemma1}, we can transform problem \eqref{a1}-\eqref{a2}
into a fixed point problem as follows:

 (i) For ${p_1}^2-4p_0p_2>0$, we introduce an operator
$\mathcal{J}: \mathcal{C} \to \mathcal{C}$ given by
 \begin{equation}\label{J1}
\begin{aligned}
(\mathcal{J}x)(t)
&=  \frac{1}{p_2(m_2-m_1)}\Big\{\int_0^t\int_0^s \Phi(t)\frac{(s-u)^{\delta-1}}
 {\Gamma(\delta)}f(u,x(u)) \,du\,ds\\
&\quad  +\rho_1(t)\int_0^\xi \int_0^s \Phi(\xi)\frac{(s-u)^{\delta-1}}
 {\Gamma(\delta)}f(u,x(u))\,du\,ds\\
&\quad  +\rho_2(t)\Big [\int_0^1 \int_0^s \Phi(1)\frac{(s-u)^{\delta-1}}
 {\Gamma(\delta)}f(u,x(u))\,du\,ds\\
&\quad  -\lambda \int_0^\sigma \int_0^s
 \Big(\frac{(e^{m_2(\sigma-s)}-1)}{m_2}-\frac{(e^{m_1(\sigma-s)}-1)}{m_1}\Big)\\
&\quad\times  \frac{(s-u)^{\delta-1}}{\Gamma(\delta)}f(u,x(u))\,du\,ds\Big] \Big\},
\end{aligned}
 \end{equation}
such that
 \begin{equation}\label{J}
 x=\mathcal{J}x.
\end{equation}

(ii) For ${p_1}^2-4p_0p_2=0$, we have an operator equation
 \begin{equation}\label{H}
 x=\mathcal{H}x,
 \end{equation}
where the operator $\mathcal{H}: \mathcal{C} \to \mathcal{C}$ is defined by
\begin{equation}\label{H1}
\begin{aligned}
&(\mathcal{H}x)(t)\\
&= \frac{1}{p_2}\Big\{\int_0^t\int_0^s \Psi(t)\frac{(s-u)^{\delta-1}}
 {\Gamma(\delta)}f(u,x(u))\,du\,ds\\
 &\quad  +\chi_1(t) \int_0^{\xi} \int_0^s \Psi(\xi)
\frac{(s-u)^{\delta-1}}{\Gamma(\delta)}f(u,x(u))\,du\,ds\\
 &\quad  +\chi_2(t)\Big[\int_0^1 \int_0^s \Psi(1)
\frac{(s-u)^{\delta-1}}{\Gamma(\delta)}f(u,x(u)) \,du\,ds\\
 &\quad  -\lambda\int_0^\sigma\Big(\frac{m(\sigma-s)e^{m(\sigma-s)}
-e^{m(\sigma-s)}+1}{m^2}\Big)\frac{(s-u)^{\delta-1}}{\Gamma(\delta)}
 f(u,x(u))\,du\,ds\Big]\Big\}.
\end{aligned}
\end{equation}

(iii) For ${p_1}^2-4p_0p_2<0$, we have the fixed point problem:
 \begin{equation}\label{K-a}
 x=\mathcal{K}x,
 \end{equation}
where the operator $\mathcal{K}: \mathcal{C} \to \mathcal{C}$ is defined by
 \begin{equation}\label{K1}
\begin{aligned}
 (\mathcal{K}x)(t)
&=  \frac{1}{p_2 b}\Big\{\int_0^t\int_0^s \Omega(t)
\frac{(s-u)^{\delta-1}}{\Gamma(\delta)}f(u,x(u))\,du\,ds \\
 &\quad  +\tau_1(t) \int_0^{\xi} \int_0^s \Omega(\xi)
\frac{(s-u)^{\delta-1}}{\Gamma(\delta)}f(u,x(u))\,du\,ds \\
 &\quad  +\tau_2(t)\Big[\int_0^1 \int_0^s \Omega(1)
\frac{(s-u)^{\delta-1}}{\Gamma(\delta)}f(u,x(u))\,du\,ds\\
 &\quad  -\frac{\lambda}{a^2+b^2}\int_0^\sigma
\Big(b+be^{-a(\sigma-s)}\cos{b(\sigma-s)}
 a e^{-a(\sigma-s)}\sin{b(\sigma-s)}\Big) \\
&\quad\times \frac{(s-u)^{\delta-1}}{\Gamma(\delta)}f(u,x(u))\,du\,ds\Big]\Big\}.
\end{aligned}
\end{equation}
 Now we set
 \begin{equation}\label{max}
 \begin{gathered}
\widehat{\rho}_1=\max_{t\in[0,1]}{|\rho_1(t)|},\quad
\widehat{\rho}_2=\max_{t\in[0,1]}{|\rho_2(t)|}, \\
\varepsilon = \max_{t\in[0,1]}\Big|m_2(1-e^{m_1t})-m_1(1-e^{m_2t})\Big|,\\
\begin{aligned}
\alpha 
&=\frac{1}{p_2m_1m_2(m_2-m_1)\Gamma(\delta+1)}
 \Big\{ \varepsilon +\xi^\delta \widehat{\rho}_1(m_2(1-e^{m_1\xi})
 - m_1(1-e^{m_2\xi}))\\
 &\quad +\widehat{\rho}_2\Big[(m_2(1-e^{m_1})-m_1(1-e^{m_2}))\\
 &\quad +\frac{\sigma^\delta |\lambda|}{m_1m_2}(m^2_1(m_2\sigma-e^{m_2\sigma}+1)
 -m^2_2(m_1\sigma-e^{m_1\sigma}+1))\Big]\Big\},
\end{aligned}\\
\alpha_1=\alpha-\frac{ \varepsilon}{p_2m_1m_2(m_2-m_1)\Gamma(\delta+1)};
\end{gathered}
 \end{equation}
also we set
\begin{equation}\label{max1}
\begin{gathered}
\widehat{\chi}_1=\max_{t\in[0,1]}{|\chi_1(t)|},\quad
\widehat{\chi}_2=\max_{t\in[0,1]}{|\chi_2(t)|},\\
\begin{aligned}
 \beta &=\frac{1}{p_2 m^2 \Gamma(\delta+1)}
 \Big\{ (1 +\widehat{\chi}_2)|me^m-e^m+1| 
 +\xi^\delta\widehat{\chi}_1|m\xi e^{m\xi} -e^{m\xi}+1|\\
 &\quad +\frac{|\lambda| \sigma^\delta \widehat{\chi}_2 }{|m|}
 |2(e^{m\sigma}-1)-m\sigma(e^{m\sigma}+1)|\Big\},
\end{aligned}\\
 \beta_1=\beta-\frac{|me^m-e^m+1 |}{p_2m^2\Gamma(\delta+1)};
 \end{gathered}
 \end{equation}
 and
 \begin{equation}\label{max2}
\begin{gathered}
 \widehat{\tau}_1=\max_{t\in[0,1]}{|\tau_1(t)|}, \quad
\widehat{\tau}_2=\max_{t\in[0,1]}{|\tau_2(t)|}\\
\begin{aligned}
 \gamma &=\frac{1}{p_2 b(a^2+b^2) \Gamma(\delta+1)}
\Big\{ (1 +\widehat{\tau}_2)|b-be^{-a}\cos b-ae^{-a}\sin b|\\
 &\quad +\xi^\delta\widehat{\tau}_1|b-be^{-a \xi}\cos {b\xi}
 -ae^{-a \xi}\sin{b\xi}|+\frac{|\lambda| \sigma^\delta \widehat{\tau}_2}
 {a^2+b^2} \\
&\quad\times \big|2ab-(a^2+b^2)b\sigma
  -2abe^{-a\sigma}\cos{b\sigma}+(a^2-b^2)e^{-a\sigma}\sin{b\sigma}\big| \Big\},
\end{aligned}\\
\gamma_1=\gamma-\frac{|b-be^{-a}\cos b-a e^{-a}\sin b |}{p_2 b(a^2+b^2)
 \Gamma(\delta+1)}.
 \end{gathered}
 \end{equation}
Now we discus the existence and uniqueness of solutions for the problem
\eqref{a1}-\eqref{a2} by using the standard fixed point theorems.
We give the details for the case where ${p_1}^2-4p_0p_2>0$, while the details
for other two cases ${p_1}^2-4p_0p_2=0$ and ${p_1}^2-4p_0p_2<0$ can be
completed in a similar manner.

 Our first result is based on Krasnoselskii's fixed point theorem,
which is stated below.

 \begin{theorem}[\cite{K}]\label{l1}
Let $Y$ be a bounded, closed, convex, and nonempty subset of a Banach space $X$.
Let $A_1,A_2$ be the operators such that
$(i)$ $A_1x+A_2y \in M$ whenever  $x,y \in Y$;
$(ii)$ $A_1$ is compact and continuous; and
$(iii)$ $A_2$ is a contraction mapping. Then there exists $w\in Y $ such that
 $w=A_1w+A_2w. $
 \end{theorem}

\begin{theorem}\label{th1}
 Let $f:[0,1]\times \mathbb{R}\to \mathbb{R}$ be a continuous
function satisfying the conditions:
 \begin{itemize}
 \item[(A1)] $|f(t,x)-f(t,y)|\le \ell|x-y|$ for all $t\in [0,1]$, $x, y \in\mathbb{R}$,
 $\ell>0$;

 \item[(A2)] $|f(t,x)|\leq \theta(t)$, for all
$(t,x)\in[0,1]\times \mathbb{R}$ and $\theta\in C([0,1],\mathbb{R}^+)$.
\end{itemize}
Then problem \eqref{a1}-\eqref{a2} with $p_1^2-4p_0p_2>0,$ has at least one
solution on $[0,1]$ if
 \begin{equation}\label{J4}
 \ell \alpha_1 < 1,
 \end{equation}
where $\alpha_1$ is given by \eqref{max}.
\end{theorem}

\begin{proof}
 Setting $\sup_{t\in [0,1]}|\theta(t)|=\|\theta\|$, we can fix
\begin{equation}\label{c2}
\begin{aligned}
 r &\geq  \frac{\|\theta\|}{p_2m_1m_2(m_2-m_1)\Gamma(\delta+1)}
\Big\{ \varepsilon +\xi^\delta \widehat{\rho}_1(m_2(1-e^{m_1\xi})
 +m_1(1-e^{m_2\xi}))\\
&\quad  +\widehat{\rho}_2\Big[(m_2(1-e^{m_1})-m_1(1-e^{m_2}))
 +\frac{\sigma^\delta |\lambda|}{m_1m_2}(m^2_1(m_2\sigma-e^{m_2\sigma}+1)\\
 &\quad -m^2_2(m_1\sigma-e^{m_1\sigma}+1))\Big]\Big\},
\end{aligned}
 \end{equation}
and consider $ B_{r}=\{ x \in \mathcal{C} : \|x\|\leq r\} $.
 Introduce the operators $\mathcal{J}_1$ and $\mathcal{J}_2$ defined on $B_{r}$
 as follows:
 \begin{gather}\label{c3}
 (\mathcal{J}_1 x)(t)=\frac{1}{p_2(m_2-m_1)}\int_0^t \int_0^s \Phi(t)
\frac{(s-u)^{\delta-1}}{\Gamma(\delta)}f(u, x(u))\,du\,ds, \\
\label{c4}
\begin{aligned}
 (\mathcal{J}_2 x)(t)
&=  \frac{1}{p_2(m_2-m_1)}\Big\{\rho_1(t)\int_0^\xi \int_0^s
 \Phi(\xi)\frac{(s-u)^{\delta-1}}{\Gamma(\delta)}f(u, x(u))\,du\,ds\\
 &\quad  +\rho_2(t)\Big [\int_0^1 \int_0^s \Phi(1)\frac{(s-u)^{\delta-1}}
{\Gamma(\delta)}f(u, x(u))\,du\,ds\\
 &\quad  -\lambda \int_0^\sigma \int_0^s
 \Big(\frac{(e^{m_2(\sigma-s)}-1)}{m_2} -\frac{(e^{m_1(\sigma-s)}-1)}{m_1}\Big)\\
&\quad\times \frac{(s-u)^{\delta-1}}{\Gamma(\delta)}f(u, x(u))\,du\,ds\Big] \Big\}.
\end{aligned}
 \end{gather}
 Observe that $\mathcal{J}=\mathcal{J}_1+\mathcal{J}_2$. For $x,y \in B_{r}$,
we have
 \begin{align*}
 &\|\mathcal{J}_1x+\mathcal{J}_2y\|\\ &=  \sup_{t\in[0,1]} |(\mathcal{J}_1x)(t)+(\mathcal{J}_2y)(t)|\\
 &\leq \frac{1}{p_2(m_2-m_1)}\sup_{t\in[0,1]}
 \Big\{\int_0^t\int_0^s \Phi(t)\frac{(s-u)^{\delta-1}}{\Gamma(\delta)}
 |f(u, x(u))|\,du\,ds\\
 &\quad  +|\rho_1(t)|\int_0^\xi \int_0^s \Phi(\xi)
 \frac{(s-u)^{\delta-1}}{\Gamma(\delta)}|f(u, y(u))|\,du\,ds\\
 &\quad  +|\rho_2(t)|\Big [\int_0^1 \int_0^s \Phi(1)
 \frac{(s-u)^{\delta-1}}{\Gamma(\delta)}|f(u, y(u))|\,du\,ds\\
&\quad  +|\lambda| \int_0^\sigma \int_0^s \Big(\frac{(e^{m_2(\sigma-s)}-1)}{m_2}
 -\frac{(e^{m_1(\sigma-s)}-1)}{m_1}\Big) \\
&\quad\times \frac{(s-u)^{\delta-1}}{\Gamma(\delta)}|f(u, y(u))|\,du\,ds\Big] \Big\}\\
&\leq \frac{\|\theta\|}{p_2(m_2-m_1)\Gamma{(\delta+1)}}\sup_{t\in[0,1]}
 \Big\{ t^\delta \int_0^t\Big(e^{m_2(t-s)}-e^{m_1(t-s)}\Big)ds\\
&\quad  + \xi^ {\delta} |\rho_1(t)|\int_0^\xi \Big(e^{m_2(\xi-s)}
 -e^{m_1(\xi-s)}\Big)ds\\
&\quad  + |\rho_2(t)|\Big[\int_0^1\Big(e^{m_2(1-s)}-e^{m_1(1-s)}\Big)ds\\
&\quad  +|\lambda| \sigma^\delta \int_0^\sigma\Big(\frac{(e^{m_2(\sigma-s)}-1)}{m_2}
 -\frac{(e^{m_1(\sigma-s)}-1)}{m_1}\Big)ds \Big] \Big\}\\
&\leq \frac{\|\theta\|}{p_2m_1m_2(m_2-m_1)\Gamma(\delta+1)}
 \Big\{ \varepsilon +\widehat{\rho}_1\xi^\delta(m_2(1-e^{m_1\xi})-m_1(1-e^{m_2\xi})) \\
 &\quad  +\widehat{\rho}_2[(m_2(1-e^{m_1})-m_1(1-e^{m_2}))\\
 &\quad  +\frac{\sigma^\delta |\lambda|}{m_1m_2}
 (m^2_1(m_2\sigma-e^{m_2\sigma}+1)-m^2_2(m_1\sigma-e^{m_1\sigma}+1))]\Big\}
 \leq r,
\end{align*}
 where we used \eqref{c2}.
 Thus $ \mathcal{J}_1x+\mathcal{J}_2y\in B_{r}$. Using the assumption (A1)
together with \eqref{J4}, we show that $\mathcal{J}_2$ is a contraction as follows:
 \begin{align*}
 &\|\mathcal{J}_2x-\mathcal{J}_2y\|\\ &=  \sup_{t\in[0,1]} |(\mathcal{J}_2x)(t)
 -(\mathcal{J}_2y)(t)|\\
&\leq \frac{1}{p_2(m_2-m_1)}\sup_{t\in[0,1]}
 \Big\{ |\rho_1(t)|\int_0^\xi \int_0^s \Phi(\xi)\frac{(s-u)^{\delta-1}}
 {\Gamma(\delta)} \\
&\quad\times |f(u, x(u))-f(u, y(u))|\,du\,ds\\
&\quad  +|\rho_2(t)|\Big[\int_0^1 \int_0^s \Phi(1)\frac{(s-u)^{\delta-1}}
 {\Gamma(\delta)}|f(u, x(u))-f(u, y(u))|\,du\,ds\\
&\quad  +|\lambda| \int_0^\sigma \int_0^s \Big(\frac{(e^{m_2(\sigma-s)}-1)}{m_2}
 -\frac{(e^{m_1(\sigma-s)}-1)}{m_1}\Big)
 \frac{(s-u)^{\delta-1}}{\Gamma(\delta)} \\
&\quad\times |f(u, x(u))-f(u, y(u))|\,du\,ds\Big] \Big\}\\
&\leq \frac{\ell}{p_2(m_2-m_1)}\sup_{t\in[0,1]}
 \Big\{\xi^\delta|\rho_1(t)|\int_0^\xi\Big(e^{m_2(\xi-s)}-e^{m_1(\xi-s)}\Big)ds\\
&\quad  +|\rho_2(t)|\Big [\int_0^1 \int_0^s \Big(e^{m_2(1-s)}-e^{m_1(1-s)}\Big)ds\\
&\quad  +|\lambda| \sigma^\delta \int_0^\sigma
 \Big(\frac{(e^{m_2(\sigma-s)}-1)}{m_2}-\frac{(e^{m_1(\sigma-s)}-1)}{m_1}\Big)ds \Big] \Big\} \|x-y\|\\
&\leq \frac{\ell}{p_2m_1m_2(m_2-m_1)\Gamma(\delta+1)}
 \Big\{\widehat{\rho}_1\xi^\delta(m_2(1-e^{m_1\xi})-m_1(1-e^{m_2\xi}))\\
&\quad  +\widehat{\rho}_2[(m_2(1-e^{m_1})-m_1(1-e^{m_2}))
 +\frac{\sigma^\delta |\lambda|}{m_1m_2}(m^2_1(m_2\sigma-e^{m_2\sigma}+1)\\
&\quad  -m^2_2(m_1\sigma-e^{m_1\sigma}+1))]\Big\}\|x-y\|\\
&= \ell \alpha_1 \|x-y\|.
 \end{align*}
Note that continuity of $f$ implies that the operator $\mathcal{J}_1$
is continuous. Also, $\mathcal{J}_1$ is uniformly bounded on $B_{r}$ as
 \begin{equation*}
 \|\mathcal{J}_1x\|= \sup_{t\in[0,1]} |(\mathcal{J}_1x)(t)|
\leq \frac{\|\theta\|\varepsilon}{p_2m_1m_2(m_2-m_1)\Gamma{(\delta+1)}}.
 \end{equation*}

Now we prove  compactness of operator $\mathcal{J}_1$. We define
 $\sup_{(t,x)\in{[0,1]\times B_{r}}}|f(t,x)|=\overline{f}$.
 Thus, for $0<t_1<t_2<1$, we have
\begin{align*}
 &|(\mathcal{J}_1x)(t_2)-(\mathcal{J}_1x)(t_1)|\\
&= \frac{1}{p_2(m_2-m_1)}\Big|\int_0^{t_1}
 \int_0^s \Big[\Phi(t_2)-\Phi(t_1)\Big]\frac{(s-u)^{\delta-1}}{\Gamma(\delta)}
 f(u, x(u))du ds \\
&\quad  + \int_{t_1}^{t_2} \int_0^s \Phi(t_2)
 \frac{(s-u)^{\delta-1}}{\Gamma(\delta)}f(u, x(u))du ds \Big| \\
&\leq \frac{\overline{f}}{p_2m_1m_2(m_2-m_1)\Gamma{(\delta+1)}}
 \Big\{\Big(t_1^\delta-t_2^\delta\Big)
 \Big(m_1(1-e^{m_2(t_2-t_1)}) \\
&\quad -m_2(1-e^{m_1(t_2-t_1)})\Big)
  + t_1^\delta\Big(m_1(e^{m_2t_2}-e^{m_2t_1})-m_2(e^{m_1t_2}
 -e^{m_1t_1})\Big)\Big\} \\
&\to 0,\quad \text{as } t_1\to t_2,
 \end{align*}
and is independent of $x$. Thus, $\mathcal{J}_1$ is relatively compact on $B_{r}$.
Hence, by the Arzel\'a-Ascoli Theorem, $\mathcal{J}_1$ is compact on $B_{r}$.
 Thus all the assumption of Theorem \eqref{l1} are satisfied.
So by the conclusion of Theorem \ref{l1}, the problem \eqref{a1}-\eqref{a2}
 has at least one solution $[0,1]$. The proof is complete.
\end{proof}

\begin{remark} \label{rmk3.3} \rm
In the above theorem we can interchange the roles of the operators $\mathcal{J}_1$
and $\mathcal{J}_2$ to obtain a second result by replacing \eqref{J4}
by the following condition:
$$
\frac {\ell \varepsilon}{p_2m_1m_2(m_2-m_1)\Gamma(\delta+1)}<1.
$$
\end{remark}

Now we apply Banach's contraction mapping principle to prove existence and
uniqueness of solutions for the problem \eqref{a1}-\eqref{a2}.

\begin{theorem}\label{th2}
 Assume that $f:[0,1]\times \mathbb{R }\to \mathbb{R}$ is a continuous function
such that {\rm (A1)} is satisfied.
 Then there exists a unique solution for the problem \eqref{a1}-\eqref{a2}
on $[0, 1]$ if $\ell<1/\alpha,$ where $\alpha$ is given by \eqref{max}.
 \end{theorem}

 \begin{proof}
Let us define $\sup_{t\in[0,1]}{|f(t,0)|=M}$ and select
$ \bar{r}\geq\frac{\alpha M}{1-\ell \alpha }$ to show that
$\mathcal{J}B_{\bar{r}} \subset B_{\bar{r}}$, where
$B_{\bar{r}}=\{x\in\mathcal{C}:\|x\|\leq \bar{r}\}$ and ${\mathcal J}$
is defined by \eqref{J1}.
 Using the condition (A1), we have
\begin{equation}\label{c1}
\begin{aligned}
 |f(t,x)| &=  |f(t,x)-f(t,0)+f(t,0)|\leq|f(t,x)-f(t,0)|+|f(x,0)| \\
 &\leq  \ell \|x\| + M \leq \ell \bar{r}+M.
\end{aligned}
 \end{equation}
Then, for $x\in B_{\bar{r}}$, we obtain
 \begin{align*}
&\|\mathcal{J}(x)\|\\
&=  \sup_{t\in[0,1]}|\mathcal{J}(x)(t)| \\
&\leq \frac{1}{p_2(m_2-m_1)}\sup_{t\in[0,1]}
 \Big\{\int_0^t\int_0^s \Phi(t)\frac{(s-u)^{\delta-1}}
 {\Gamma(\delta)}|f(u, x(u))|\,du\,ds\\
 &\quad  +|\rho_1(t)|\int_0^\xi \int_0^s \Phi(\xi)
 \frac{(s-u)^{\delta-1}}{\Gamma(\delta)}|f(u, x(u))|\,du\,ds\\
 &\quad  +|\rho_2(t)|\Big [\int_0^1 \int_0^s \Phi(1)
\frac{(s-u)^{\delta-1}}{\Gamma(\delta)}|f(u, x(u))|du ds\\
 &\quad  +|\lambda| \int_0^\sigma \int_0^s \Big(\frac{(e^{m_2(\sigma-s)}-1)}{m_2}
-\frac{(e^{m_1(\sigma-s)}-1)}{m_1}\Big)\frac{(s-u)^{\delta-1}}{\Gamma(\delta)}
 |f(u, y(u))|\,du\,ds\Big] \Big\}\\
& \leq \frac{(\ell \bar{r}+M)}{p_2(m_2-m_1)}\sup_{t\in[0,1]}
 \Big\{ \int_0^t\Big(e^{m_2(t-s)}-e^{m_1(t-s)}\Big)
 \frac{s^\delta}{\Gamma{(\delta+1)}}ds\\
&\quad  + |\rho_1(t)|\int_0^\xi \Big(e^{m_2(\xi-s)}-e^{m_1(\xi-s)}\Big)
 \frac{s^\delta}{\Gamma{(\delta+1)}}ds\\
&\quad + |\rho_2(t)|\Big[\int_0^1\Big(e^{m_2(1-s)}-e^{m_1(1-s)}\Big)
 \frac{s^\delta}{\Gamma{(\delta+1)}} ds\\
&\quad  +|\lambda| \int_0^\sigma\Big(\frac{(e^{m_2(\sigma-s)}-1)}{m_2}
 -\frac{(e^{m_1(\sigma-s)}-1)}{m_1}\Big)\frac{s^\delta}{\Gamma{(\delta+1)}}ds \Big]
 \Big\}\\
& \leq  \frac{(\ell \bar{r}+M)}{p_2m_1m_2(m_2-m_1)\Gamma(\delta+1)}
 \Big\{ \varepsilon +\widehat{\rho}_1\xi^\delta(m_2(1-e^{m_1\xi})-m_1(1-e^{m_2\xi})) \\
&\quad  +\widehat{\rho}_2[(m_2(1-e^{m_1})-m_1(1-e^{m_2}))\\
&\quad  +\frac{\sigma^\delta |\lambda|}{m_1m_2}
 (m^2_1(m_2\sigma-e^{m_2\sigma}+1)-m^2_2(m_1\sigma-e^{m_1\sigma}+1))]\Big\}\\
&=  (\ell \bar{r}+M) \alpha \leq \bar{r},
 \end{align*}
which clearly shows that $\mathcal{J}x\in B_{\bar{r}}$ for any $x\in B_{\bar{r}}$.
Thus $\mathcal{J}B_{\bar{r}}\subset B_{\bar{r}}$.
 Now, for $x,y\in \mathcal{C}$ and for each $t\in [0,1]$, we have
 \begin{align*}
 &\|(\mathcal{J}x)-(\mathcal{J}y)\| \\
&\leq  \frac{1}{p_2(m_2-m_1)}\sup_{t\in[0,1]}\Big\{\int_0^t\int_0^s \Phi(t)
 \frac{(s-u)^{\delta-1}}{\Gamma(\delta)}|f(u, x(u))-f(u,y(u))|\,du\,ds\\
 &\quad  +|\rho_1(t)|\int_0^\xi \int_0^s \Phi(\xi)\frac{(s-u)^{\delta-1}}
 {\Gamma(\delta)}|f(u, x(u))-f(u,y(u))|\,du\,ds\\
 &\quad  +|\rho_2(t)|\Big [\int_0^1 \int_0^s \Phi(1)\frac{(s-u)^{\delta-1}}
 {\Gamma(\delta)}  |f(u, x(u))-f(u,y(u))|\,du\,ds\\
 &\quad  +|\lambda| \int_0^\sigma \int_0^s \Big(\frac{(e^{m_2(\sigma-s)}-1)}{m_2}
-\frac{(e^{m_1(\sigma-s)}-1)}{m_1}\Big)\frac{(s-u)^{\delta-1}}{\Gamma(\delta)}\\
&\quad\times  |f(u, x(u))-f(u,y(u))|\,du\,ds\Big] \Big\}\\
&\leq  \frac{\ell}{p_2(m_2-m_1)}\sup_{t\in[0,1]}\Big\{ \int_0^t\Big(e^{m_2(t-s)}
 -e^{m_1(t-s)}\Big) \frac{s^\delta}{\Gamma{(\delta+1)}}ds\\
 &\quad  + |\rho_1(t)|\int_0^\xi \Big(e^{m_2(\xi-s)}-e^{m_1(\xi-s)}\Big)
  \frac{s^\delta}{\Gamma{(\delta+1)}}ds\\
&\quad + |\rho_2(t)|
 \Big[\int_0^1\Big(e^{m_2(1-s)}-e^{m_1(1-s)}\Big)
 \frac{s^\delta}{\Gamma{(\delta+1)}} ds\\
&\quad  +|\lambda| \int_0^\sigma\Big(\frac{(e^{m_2(\sigma-s)}-1)}{m_2}
 -\frac{(e^{m_1(\sigma-s)}-1)}{m_1}\Big)\frac{s^\delta}{\Gamma{(\delta+1)}}ds \Big]
 \Big\}\|x-y\|\\
&\leq  \frac {\ell}{p_2m_1m_2(m_2-m_1)\Gamma(\delta+1)}
 \Big\{ \varepsilon +\widehat{\rho}_1\xi^\delta(m_2(1-e^{m_1\xi})
 -m_1(1-e^{m_2\xi})) \\
&\quad  +\widehat{\rho}_2[(m_2(1-e^{m_1})-m_1(1-e^{m_2}))\\
&\quad  +\frac{\sigma^\delta |\lambda|}{m_1m_2}(m^2_1(m_2\sigma-e^{m_2\sigma}+1)
 -m^2_2(m_1\sigma-e^{m_1\sigma}+1))]\Big\}\|x-y\|\\
&= \ell \alpha \|x-y\|,
 \end{align*}
 where $\alpha$ is given by \eqref{max} and depends only on the parameters
involved in the problem. In view of the condition $\ell<1/\alpha$,
it follows that $\mathcal{J}$ is a contraction. Thus, by the contraction
 mapping principle (Banach fixed point theorem), the problem \eqref{a1}
and \eqref{a2} has a unique solution on $[0,1]$. This completes the proof.
\end{proof}

 The next existence result is based on Leray-Schauder nonlinear alternative.

\begin{theorem}[Nonlinear alternative for single valued maps \cite{GrDu}]
\label{th3}
Let $C$ be a closed, convex subset of a Banach space $E$ and $U$ be an
 open subset of $C$ with $0\in U$. Suppose that
$F:\overline{U}\to C$ is a continuous, compact (that is, $F(\overline{U})$
is a relatively compact subset of $C$) map. Then either $(i)$ $F$ has a fixed point
in $\overline{U},$ or $(ii)$ there is a $u\in \partial U$
(the boundary of $U$ in $C$) and
$\epsilon\in(0,1)$ such that $u=\epsilon Fu$.
\end{theorem}

\begin{theorem}  \label{LSA}
Let $f: [0,1]\times \mathbb{R} \to \mathbb{R}$ be a continuous function
satisfying the conditions:
\begin{itemize}
\item[(A3)] There exist a function $g\in C([0,1],{\mathbb{R}}^{+})$,
 and a nondecreasing function $\psi :{\mathbb{R}}^{+}\to {\ \mathbb{R}}^{+}$
 such that $| f(t,y)| \leq g(t)\psi (\| y\|)$ for all
$(t,y)\in [0,1]\times {\mathbb{R}}$;

\item[(A4)] There exists a constant $K>0$ such that
\begin{equation*}
\frac{K}{ \|g\|\psi(K)\alpha}>1.
\end{equation*}
\end{itemize}
 Then the problem \eqref{a1}-\eqref{a2} has at least one solution
on $[0,1]$.
\end{theorem}

\begin{proof}
 Consider the operator $\mathcal{J}: \mathcal{C} \to \mathcal{C}$ defined by
\eqref{J1}. We show that $\mathcal{J}$ maps bounded sets into bounded sets in
$\mathcal{C}= C([0,1], \mathbb{R})$. For a positive number $\zeta$, let
 ${\mathcal B}_\zeta = \{x \in \mathcal{C}: \|x\| \le \zeta \}$ be a
 bounded set in $\mathcal{C}$. Then we have
\begin{align*}
&\|\mathcal{J}(x)\|\\
&=  \sup_{t\in[0,1]}|\mathcal{J}(x)(t)| \\
 &\leq \frac{1}{p_2(m_2-m_1)} \sup_{t\in[0,1]}
 \Big\{\int_0^t \int_0^s \Phi(t)
 \frac{(s-u)^{\delta-1}}{\Gamma(\delta)}|f(u, x(u))|\,du\,ds\\
 &\quad  +|\rho_1|(t)\int_0^\xi \int_0^s \Phi(\xi)
 \frac{(s-u)^{\delta-1}}{\Gamma(\delta)}|f(u, x(u))|\,du\,ds\\
 &\quad  +|\rho_2(t)|\Big [\int_0^1 \int_0^s \Phi(1)
 \frac{(s-u)^{\delta-1}}{\Gamma(\delta)}|f(u, x(u))|\,du\,ds\\
 &\quad  +|\lambda| \int_0^\sigma \int_0^s
 \Big(\frac{(e^{m_2(\sigma-s)}-1)}{m_2}-\frac{(e^{m_1(\sigma-s)}-1)}{m_1}\Big)
 \frac{(s-u)^{\delta-1}}{\Gamma(\delta)}|f(u, x(u))|\,du\,ds\Big] \Big\}\\
 & \leq \frac{\|g\|\psi(\zeta)}{p_2(m_2-m_1)} \sup_{t\in[0,1]}
 \Big\{ \int_0^t\Big(e^{m_2(t-s)}-e^{m_1(t-s)}\Big)
 \frac{s^\delta}{\Gamma{(\delta+1)}}ds\\
 &\quad  + |\rho_1(t)|\int_0^\xi \Big(e^{m_2(\xi-s)}-e^{m_1(\xi-s)}\Big)
 \frac{s^\delta}{\Gamma{(\delta+1)}}ds\\
&\quad  + |\rho_2(t)|\Big[\int_0^1\Big(e^{m_2(1-s)}-e^{m_1(1-s)}\Big)
 \frac{s^\delta}{\Gamma{(\delta+1)}} ds\\
 &\quad  +|\lambda| \int_0^\sigma\Big(\frac{(e^{m_2(\sigma-s)}-1)}{m_2}
 -\frac{(e^{m_1(\sigma-s)}-1)}{m_1}\Big) \frac{s^\delta}{\Gamma{(\delta+1)}}ds \Big]
  \Big\}\\
& \leq  \frac{\|g\|\psi(\zeta)}{p_2m_1m_2(m_2-m_1)\Gamma(\delta+1)}
 \big\{\varepsilon +\widehat{\rho}_1\xi^\delta(m_2(1-e^{m_1\xi})-m_1(1-e^{m_2\xi})) \\
 &\quad  +\widehat{\rho}_2[(m_2(1-e^{m_1})-m_1(1-e^{m_2}))\\
 &\quad  +\frac{\sigma^\delta |\lambda|}{m_1m_2}
 (m^2_1(m_2\sigma-e^{m_2\sigma}+1)-m^2_2(m_1\sigma-e^{m_1\sigma}+1))]\big\},
 \end{align*}
which yields
 \begin{align*}
 \|{\mathcal J}x\|
&\le  \frac{\|g\|\psi(\zeta)}{p_2m_1m_2(m_2-m_1)\Gamma(\delta+1)}
 \Big\{\varepsilon +\widehat{\rho}_1\xi^\delta(m_2(1-e^{m_1\xi})-m_1(1-e^{m_2\xi}))\\
&\quad  +\widehat{\rho}_2[(m_2(1-e^{m_1})-m_1(1-e^{m_2}))\\
&\quad  +\frac{\sigma^\delta |\lambda|}{m_1m_2}(m^2_1(m_2\sigma-e^{m_2\sigma}+1)
 -m^2_2(m_1\sigma-e^{m_1\sigma}+1))]\Big\}.
\end{align*}

 Next we show that ${\mathcal J}$ maps bounded sets into equicontinuous sets
of $\mathcal{C}$. Let $t_1, t_2 \in [0,1]$ with $t_1< t_2$
and $y \in {\mathcal B}_\zeta,$ where ${\mathcal B}_\zeta$ is a bounded set of
$\mathcal{C}$. Then we obtain
\begin{align*}
 &|(\mathcal{J}x)(t_2)-(\mathcal{J}x)(t_1)|\\
 &\le \frac{1}{p_2(m_2-m_1)}\Big\{\Big|\int_0^{t_1} \int_0^s
\Big[\Phi(t_2)-\Phi(t_1)\Big]\frac{(s-u)^{\delta-1}}{\Gamma(\delta)}f(u, x(u))
 \,du\,ds\\
&\quad  + \int_{t_1}^{t_2} \int_0^s \Phi(t_2)
 \frac{(s-u)^{\delta-1}}{\Gamma(\delta)}f(u, x(u))\,du\,ds \Big|\\
&\quad   +|\rho_1(t_2)-\rho_1(t_1)|\int_0^\xi \int_0^s \Phi(\xi)
 \frac{(s-u)^{\delta-1}}{\Gamma(\delta)}|f(u, y(u))|\,du\,ds\\
 &\quad  + |\rho_2(t_2)-\rho_2(t_1)|\Big[\int_0^1\int_0^s \Phi(1)
 \frac{(s-u)^{\delta-1}}{\Gamma(\delta)}|f(u, y(u))|\,du\,ds\\
&\quad  +|\lambda|\int_0^\sigma\int_0^s \Big(\frac{(e^{m_2(\sigma-s)}-1)}{m_2}
 -\frac{(e^{m_1(\sigma-s)}-1)}{m_1}\Big)\frac{(s-u)^{\delta-1}}{\Gamma(\delta)}
 |f(u, y(u))|\,du\,ds\Big]\Big\}\\
&\leq \frac{\overline{f}}{p_2m_1m_2(m_2-m_1)\Gamma{(\delta+1)}}
 \Big\{\Big(t_1^\delta-t_2^\delta\Big)\Big(m_1(1-e^{m_2(t_2-t_1)}) \\
&\quad -m_2(1-e^{m_1(t_2-t_1)})\Big)
  + t_1^\delta\Big(m_1(e^{m_2t_2}-e^{m_2t_1})-m_2(e^{m_1t_2}-e^{m_1t_1})\Big)\\
&\quad   +|\rho_1(t_2)-\rho_1(t_1)|\xi^\delta(m_2(1-e^{m_1\xi})
 -m_1(1-e^{m_2\xi}))\\
&\quad  +|\rho_2(t_2)-\rho_2(t_1)|[(m_2(1-e^{m_1})-m_1(1-e^{m_2}))\\
&\quad  +\frac{\sigma^\delta |\lambda|}{m_1m_2}(m^2_1(m_2\sigma-e^{m_2\sigma}+1)
 -m^2_2(m_1\sigma-e^{m_1\sigma}+1))]\Big\},
 \end{align*}
which tends to zero independently of $x \in {\mathcal B}_\zeta$ as
$t_2- t_1 \to 0$. As ${\mathcal J}$
satisfies the above assumptions, therefore it follows by the
Arzel\'a-Ascoli theorem that ${\mathcal J}: \mathcal{C} \to
\mathcal{C}$ is completely continuous.

The result will follow from the Leray-Schauder nonlinear
alternative once it is shown that the set of all solutions
to the equation $x=\vartheta {\mathcal J}x$ is bounded for
$\vartheta \in [ 0,1]. $ For that,
let $x$ be a solution of $x=\vartheta {\mathcal J}x$ for
$\vartheta \in [0,1]$. Then, for $t\in [0,1]$,  we have
\begin{align*}
&|x(t)| \\
&=  |\vartheta {\mathcal J}x(t)| \\
 &\leq \frac{1}{p_2(m_2-m_1)} \sup_{t\in[0,1]}
\Big\{\int_0^t \int_0^s \Phi(t)\frac{(s-u)^{\delta-1}}{\Gamma(\delta)}
|f(u, x(u))|\,du\,ds\\
 &\quad  +|\rho_1|(t)\int_0^\xi \int_0^s \Phi(\xi)
\frac{(s-u)^{\delta-1}}{\Gamma(\delta)}|f(u, x(u))|\,du\,ds\\
 &\quad  +|\rho_2(t)|\Big [\int_0^1 \int_0^s \Phi(1)
\frac{(s-u)^{\delta-1}}{\Gamma(\delta)}|f(u, x(u))|\,du\,ds\\
 &\quad  +|\lambda| \int_0^\sigma \int_0^s
\Big(\frac{(e^{m_2(\sigma-s)}-1)}{m_2}-\frac{(e^{m_1(\sigma-s)}-1)}{m_1}\Big)
\frac{(s-u)^{\delta-1}}{\Gamma(\delta)}|f(u, x(u))|\,du\,ds\Big] \Big\}\\
 & \leq \frac{\|g\|\psi(\|x\|)}{p_2(m_2-m_1)} \sup_{t\in[0,1]}
\Big\{ \int_0^t\Big(e^{m_2(t-s)}-e^{m_1(t-s)}\Big) \frac{s^\delta}
{\Gamma{(\delta+1)}}ds\\
 &\quad  + |\rho_1(t)|\int_0^\xi \Big(e^{m_2(\xi-s)}-e^{m_1(\xi-s)}\Big)
 \frac{s^\delta}{\Gamma{(\delta+1)}}ds\\&\quad  + |\rho_2(t)|
\Big[\int_0^1\Big(e^{m_2(1-s)}-e^{m_1(1-s)}\Big)\frac{s^\delta}{\Gamma{(\delta+1)}}
 ds\\
&\quad  +|\lambda| \int_0^\sigma\Big(\frac{(e^{m_2(\sigma-s)}-1)}{m_2}
-\frac{(e^{m_1(\sigma-s)}-1)}{m_1}\Big) \frac{s^\delta}{\Gamma{(\delta+1)}}ds \Big]
 \Big\}\\
& \leq  \frac{\|g\|\psi(\|x\|)}{p_2m_1m_2(m_2-m_1)\Gamma(\delta+1)}
\Big\{\varepsilon +\widehat{\rho}_1\xi^\delta(m_2(1-e^{m_1\xi})-m_1(1-e^{m_2\xi})) \\
&\quad  +\widehat{\rho}_2[(m_2(1-e^{m_1})-m_1(1-e^{m_2}))\\
&\quad  +\frac{\sigma^\delta |\lambda|}{m_1m_2}
 (m^2_1(m_2\sigma-e^{m_2\sigma}+1)-m^2_2(m_1\sigma-e^{m_1\sigma}+1))]\Big\}\\
&= \|g\|\psi(\|x\|) \alpha,
\end{align*}
which implies
\begin{equation*}
\frac{\| x\| }{ \|g\|\psi(\|x\|) \alpha}\leq 1.
\end{equation*}
In view of (A4), there is no solution $x$ such that $\|x\| \neq K$.
 Let us set
\begin{equation*}
U=\{x\in {\mathcal C}:\|x\| <K\}.
\end{equation*}
The operator ${\mathcal J}:\overline{U}\to {\mathcal C}$
 is continuous and completely continuous.
From the choice of $U$, there is no $u\in \partial U$ such that
$u=\vartheta{\mathcal J}(u)$ for some $\vartheta \in (0,1)$. Consequently,
by the nonlinear alternative of Leray-Schauder type \cite{GrDu}, we deduce
that $\mathcal{J}$
 has a fixed point $u\in \overline{U}$ which is a solution of the
problem \eqref{a1}-\eqref{a2}. This completes the proof.
\end{proof}

\begin{example}\label{example1} \rm
Consider the boundary-value problem
\begin{gather}\label{ex1}
 (^cD^{5/2}+3{}^cD^{3/2}+2{}^cD^{1/2})x(t)
=\frac{A}{\sqrt{t^2+49}}\Big(\cos x + \tan^{-1}t\Big), \quad  0<t<1,\\
\label{co1}
 x(0)=0, \quad x(1/3)=0,\quad  x(1)=\int_0^{1/5}x(s)ds.
\end{gather}
Here, $\delta=1/2$, $\sigma=3/5$, $\xi=1/3$, $p_2=1$, $p_1=3$, $p_0=2$, $\lambda=1$,
 $A$ is a positive constant and
\begin{equation*}
 f(t,x)=\frac{A}{\sqrt{t^2+49}}\Big(\cos x + \tan^{-1}t\Big).
\end{equation*}
Clearly the constants $p_2, p_1,$ and $p_0$ satisfy the condition of Lemma
\ref{lemma1}, and
\begin{equation*}
 |f(t,x)-f(t,y)|\leq A |x-y|/7,
\end{equation*}
where $\ell=A/7$. Using the given values, we find $\alpha\approx 0.44269$
and $\alpha_1\approx 0.21725$,
It is easy to check that $ |f(t,x)|\leq \frac{A(2+\pi)}{2\sqrt{t^2+49}}=\theta(t)$
and $\ell \alpha_1<1$ when $A<32.22094$. As all the condition of Theorem
\ref{th1} are satisfied the problem \eqref{ex1}-\eqref{co1} has at least one
solution on $[0,1]$. On the other hand, $\ell \alpha <1$ whenever $A<15.81242$
and thus there exists a unique solution for the problem \eqref{ex1}-\eqref{co1}
 on $[0,1]$ by Theorem \ref{th2}.
\end{example}


\begin{example}\label{example2} \rm
Consider the boundary-value problem
\begin{gather}\label{ex2}
 (^cD^{5/2}+3{}^cD^{3/2}+2{}^cD^{1/2})x(t)
=\frac{1}{4\pi}\sin {(2\pi x)}+\frac{|x|^2}{1+|x|^2}, \quad 0<t<1,\\
\label{co2}
 x(0)=0, \quad  x(1/3)=0,\quad x(1)=\int_0^{1/5}x(s)ds.
\end{gather}
Here, $\delta=1/2$, $\sigma=3/5$, $\xi=1/3$, $p^2_1-4p_2p_0=1>0$, $\lambda=1$, and
\begin{equation*}
 f(t,x)=\frac{1}{4\pi}\sin {(2\pi x)}+\frac{|x|^2}{1+|x|^2}.
\end{equation*}
Clearly
\begin{equation*}
 |f(t,x)|\leq |\frac{1}{4\pi}\sin {(2\pi x)}+\frac{|x|^2}{1+|x|^2}|
\leq \frac{1}{2}\|x\|+1,
\end{equation*}
where $g(t)=1$, $\psi(\|x\|)=\frac{1}{2}\|x\|+1$.

Then by using the condition (A4),
we find that $K> 0.56853$ (we have used $\alpha = 0.44269$).
Thus, the conclusion of Theorem \ref{LSA} applies to problem
\eqref{ex2}-\eqref{co2}.
\end{example}

\subsection*{Acknowledgements}
 This project was funded by the Deanship of Scientific Research (DSR), 
King Abdulaziz University, Jeddah, Saudi Arabia under grant no. 
(KEP-PhD-11-130-39). The authors, therefore, acknowledge with 
gratitude the DSR technical and financial support. 

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\section*{Addendum posted by the editor on May 2, 2018}

A reader informed us that the second part of Lemma 2.4 is incorrect:
\begin{quote}
``The converse of the lemma follows by direct computation''
is not valid since solutions of (2.4) are found in the space
$C[0,1]$ and it has to be shown that such a solution of (2.4)
is $(2+\delta)$-Caputo differentiable for all $t \in(0,1)$ (or almost all).

The authors should (probably) use the alternative
definition of  Caputo differential operator as given in  K. Diethelm,
The analysis of fractional differential equations.
Lecture Notes in Mathematics, 2004. Springer-Verlag, Berlin, 2010.
\end{quote}

The fifth author tried to prove the part needed, but instead decided
to  write 
``The converse of Lemma 2.4 remains an open problem under
the current definition of fractional derivative''
\medskip

End of addendum.

\end{document}