\documentclass[reqno]{amsart}
\usepackage{hyperref}
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\AtBeginDocument{{\noindent\small
\emph{Electronic Journal of Differential Equations},
Vol. 2016 (2016), No. 296, pp. 1--177.\newline
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu}
\thanks{\copyright 2016 Texas State University.}
\vspace{8mm}}

\begin{document}
\title[\hfilneg EJDE-2016/296\hfil Survey and new results on BVPs for IFDEs]
{Survey and new results on boundary-value problems of singular
fractional differential equations with impulse effects}

\author[Y. Liu \hfil EJDE-2016/296\hfilneg]
{Yuji Liu}

\address{Yuji Liu \newline
Department of Mathematics,
Guangdong University of Finance and Economics,
Guangzhou 510000, China}
\email{liuyuji888@sohu.com}

\thanks{Submitted February 24, 2015. Published November 18, 2016.}
\subjclass[2010]{34A08, 26A33, 39B99, 45G10, 34B37, 34B15, 34B16}
\keywords{Higher order singular fractional differential system;
 impulsive boundary value problem; Riemann-Liouville fractional derivative;
 Caputo fractional derivative; Riemann-Liouville type Hadamard fractional derivative;
 Caputo type Hadamard fractional derivative; fixed point theorem}

\begin{abstract}
 Firstly we prove existence and uniqueness of solutions of Cauchy
 problems of linear fractional differential equations (LFDEs) with two
 variable coefficients involving Caputo fractional derivative, Riemann-Liouville
 derivative, Caputo type Hadamard derivative and Riemann-Liouville type Hadamard
 fractional derivatives with order $q\in [n-1,n)$ by using the iterative method.
 Secondly we obtain exact expressions for piecewise continuous solutions of
 the linear fractional differential equations with a constant coefficient and
 a variable one. These results provide new methods to transform an impulsive
 fractional differential equation (IFDE) to a fractional integral equation (FIE).
 Thirdly, we propose four classes of boundary value problems of singular fractional
 differential equations with impulse effects. Sufficient conditions are given
 for the existence of solutions of these problems. We allow the nonlinearity
 $p(t)f(t,x)$ in fractional differential equations to be singular at $t=0,1$.
 Finally, we point out some incorrect formulas of solutions in cited papers.
 A new Banach space and the compact properties of subsets are proved. By
 establishing a new framework to find the solutions for impulsive fractional
 boundary value problems, the existence of solutions of three classes boundary
 value problems of impulsive fractional differential equations with multi-term
 fractional derivatives are established.
\end{abstract}

\maketitle
\numberwithin{equation}{section}
\newtheorem{theorem}{Theorem}[section]
\newtheorem{lemma}[theorem]{Lemma}
\newtheorem{remark}[theorem]{Remark}
\newtheorem{definition}[theorem]{Definition}
\newtheorem{corollary}[theorem]{Corollary}
\newtheorem{example}[theorem]{Example}
\newtheorem{result}[theorem]{Result}
\allowdisplaybreaks

\tableofcontents

\section{Introduction}

One knows that the fractional derivatives (Riemann-Liouville fractional derivative,
Caputo fractional derivative and Hadamard fractional
 derivative and other type see \cite{Kil}) are actually nonlocal operators
because integrals are nonlocal operators. Moreover, calculating time fractional
derivatives of a function at some time requires all the past history and hence
fractional derivatives can be used for modeling systems with memory.

Fractional order differential equations are generalizations of integer
order differential equations. Using fractional order differential equations
can help us to reduce the errors arising from
the neglected parameters in modeling real life phenomena.
Fractional differential equations have many applications see
\cite[Chapter 10]{8}, and books \cite{Kil, 7, 8, skm}.

In recent years, there have been many results obtained on the existence
and uniqueness of solutions of initial value problems or boundary value
problems for nonlinear fractional differential equations,
see \cite{2, 3, ll11, 11, 13, 14, 15, 18, 20, 21}.


Dynamics of many evolutionary processes from various fields
such as population dynamics, control theory, physics, biology,
and medicine. undergo abrupt changes at certain moments of
time like earthquake, harvesting, shock, and so forth. These
perturbations can be well approximated as instantaneous
change of states or impulses.These processes are modeled by
impulsive differential equations. In 1960, Milman and Myshkis
introduced impulsive differential equations in their paper
\cite{mm1}. Based on their work, several monographs have been
published by many authors like Samoilenko and Perestyuk
\cite{sp}, Lakshmikantham et al. \cite{lbs}, Bainov and Simeonov \cite{bs1, bs3},
Bainov and Covachev \cite{bc}, and Benchohra et al.\ \cite{bhn}.

Fractional differential equations were extended to impulsive fractional
differential equations, since Agarwal and Benchohra
published the first paper on the topic \cite{abs} in 2008.
Since then many authors \cite{1, 4, 22, 6, llz, l3, 9, 12, 15, 17, 18, 19, lll, ll1}
studied the existence or uniqueness of solutions of impulsive initial or
boundary value problems for fractional differential equations. For examples,
impulsive anti-periodic boundary value problems see \cite{an, an1, abs, la, waz},
impulsive periodic boundary value problems see \cite{16, bnr, wz}, impulsive
initial value problems see \cite{bs, dck, m, tb}, two-point, three-point or
multi-point impulsive boundary value problems see \cite{as, gj, 201, 241, wc, zhl},
impulsive boundary value problems on infinite intervals see \cite{zw}.


Feckan and Zhou \cite{fzw} pointed out that the formula of solutions for
impulsive fractional differential equations in \cite{161, 171, 181, 191}
is incorrect and gave their correct formula. In \cite{ 201, 211}, the
authors established a general framework to find the solutions for impulsive
fractional boundary value problems and obtained some sufficient conditions
for the existence of the solutions to a kind of impulsive fractional
differential equations.
In \cite{221}, the authors illustrated their comprehension for the counterexample
in \cite{fzw} and criticized the viewpoint in
\cite{fzw, 201, 211}. Next, in \cite{231}, Feckan et al.\ expanded for the
counterexample in \cite{fzw} and provided further explanations
in the paper.

In a fractional differential equation, there exist two cases concerning the
derivatives: the firs case is $D^\alpha =D_{0^+}^\alpha$, i.e.,
the fractional derivative has a single start point $t=0$.
The other case is $D^\alpha=D_{t_i^+}^\alpha$, i.e., the fractional derivative
has a multiple start points $t=t_i$ $(i\in N[0,m])$.

There have been many authors concerning the existence and uniqueness of solutions
of boundary value problems of impulsive fractional differential equations with
multiple start points $t=t_i$ $(i\in N[0,m])$.

Recently, Wang \cite{wx} consider the second case in which $D^\alpha$ has
 multiple start points, i. e., $D^\alpha=D_{t_i^+}^\alpha$. They studied
the existence and uniqueness of solutions of the following initial value
problem of the impulsive fractional differential equation
\begin{equation}
\begin{gathered}
{}^{C}D_{t_i^+}^\alpha u(t)=f(t,u(t)), \quad t\in (t_i,t_{i+1}],\; i\in \mathbb{N}[0,p],\\
u^{(j)}(0) = u_j,\quad j\in \mathbb{N}[0,n-1],\\
\Delta u^{(j)}(t_i)]=I_{ji}(u(t_i)),\quad i\in \mathbb{N}[1,p],\; j\in \mathbb{N}[0,n-1],
\end{gathered}\label{e1.0.1}
\end{equation}
where $\alpha\in (n-1,n)$ with $n$ being a positive integer,
${}^{C}D_{t_i^+}^\alpha$ represents the standard
Caputo fractional derivatives of order $\alpha$,
$\mathbb{N}[a,b]=\{a,a+1,\dots,b\}$ with $a,b$ being integers,
$0=t_0<t_1<\dots<t_p< t_{p+1}= 1$, $I_{ji}\in C(\mathbb{R},\mathbb{R})$
$(i\in \mathbb{N}[1,p],j\in \mathbb{N}[0,n-1])$, $f:[0,T]\times \mathbb{R}\to \mathbb{R}$ is a continuous
function.

Henderson and Ouahab \cite{ho} studied the existence of solutions of the
following problems
\begin{gather*}
{}^{C}D_{t_i^+}^\alpha u(t)=f(t,u(t)), \quad t\in (t_i,t_{i+1}],\; i\in \mathbb{N}[0,p],\\
u^{(j)}(0) = u_j,\quad j\in \mathbb{N}[0,1],\\
u^{(j)}(t_i)]=I_{ji}(u(t_i)),\quad i\in \mathbb{N}[1,p],\; j\in \mathbb{N}[0,1],
\end{gather*}
and
 \begin{gather*}
{}^{C}D_{t_i^+}^\alpha u(t)=f(t,u(t)), \quad t\in (t_i,t_{i+1}],\; i\in \mathbb{N}[0,p],\\
u^{(j)}(0) = u^{(j)}(b),\quad j\in \mathbb{N}[0,1],\\
u^{(j)}(t_i)]=I_{ji}(u(t_i)),\quad i\in \mathbb{N}[1,p],\; j\in \mathbb{N}[0,1],
\end{gather*}
where $\alpha\in (1,2]$, $b>0$, $0=t_0<t_1<\dots<t_p<t_{p+1}=b$,
$f:[0,b]\times \mathbb{R}\to \mathbb{R} $, $I_{ji}:\mathbb{R}\to \mathbb{R}$ are continuous functions.
Readers should also refer \cite{waz6}.

Zhao and Gong \cite{zg} studied existence of positive solutions of the
 nonlinear impulsive fractional differential equation with generalized
periodic boundary value conditions
\begin{equation}
\begin{gathered}
{}^{C}D_{t_i^+}^q u(t)=f(t,u(t)), \quad t\in (0,T]\setminus\{t_1,\dots,t_p\},\\
\Delta u(t_i)]=I_{i}(u(t_i)),\quad \Delta u'(t_i)]=J_iu(t_i)),\quad i\in \mathbb{N}[1,p],\\
\alpha u(0)-\beta u(1)=0,\quad \alpha u'(0)-\beta u'(1)=0,
\end{gathered}\label{e1.0.2}
\end{equation}
where $q\in (1,2)$, ${}^{C}D_{t_i^+}^q$ represents the standard
Caputo fractional derivatives of order $q$, $\alpha>\beta>0$,
 $0=t_0<t_1<\dots<t_p< t_{p+1}= 1$, $\mathbb{N}[a,b]=\{a,a+1,\dots,b\}$
with $a,b$ being integers,$I_{i},J_i\in C([0,+\infty),[0,+\infty))$
$(i\in \mathbb{N}[1,p]$, $f:[0,1]\times [0,+\infty)\to [0,+\infty)$ is a continuous
function.

Wang, Ahmad and Zhang \cite{waz8} studied the existence and uniqueness of
solutions of the periodic boundary value problems for nonlinear impulsive
fractional differential equation
\begin{equation}
\begin{gathered}
{}^{C}D_{t_i^+}^\alpha u(t)=f(t,u(t)),\quad t\in (0,T]\setminus\{t_1,\dots,t_p\},\\
\Delta u(t_i)=I_{i}(u(t_i)),\quad \Delta u'(t_i)=I_{i}^*(u(t_i)),\quad i\in \mathbb{N}[1,p],\\
u'(0)+(-1)^\theta u(T)=bu(T),\quad u(0)+(-1)^\theta u(T)=0,
\end{gathered} \label{e1.0.3}
\end{equation}
where $\alpha\in (1,2)$, ${}^{C}D_{t_i^+}^\alpha$ represents the standard
Caputo fractional derivatives of order $\alpha$, $\theta=1,2$,
$\mathbb{N}[a,b]=\{a,a+1,\dots,b\}$ with $a,b$ being integers,
$0=t_0<t_1<\dots<t_p< t_{p+1}= T$, $I_{i},I_i^*\in C(\mathbb{R},\mathbb{R})$ $(i\in \mathbb{N}[1,p]$,
 $f:[0,T]\times \mathbb{R}\to \mathbb{R}$ is a continuous function.

Zou and Feng, Li and Shang \cite{fb, ls, zl8} studied the existence of solutions
 of the nonlinear boundary value problem of fractional impulsive differential
equations
\begin{equation}
\begin{gathered}
{}^{C}D_{t_i^+}^\alpha x(t)=w(t)f(t,x(t),x'(t)), \quad
 t\in (0,1]\setminus\{t_1,\dots,t_p\},\\
\Delta x(t_i)=I_{i}(x(t_i)),\quad \Delta x'(t_i)=J_{i}(x(t_i)),\quad i\in \mathbb{N}[1,p],\\
\alpha_1x(0)-\beta_1 u'(0)=g_1(x),\quad \alpha_2x(1)+\beta_2x'(1)=g_2(x),
\end{gathered} \label{e1.0.4}
\end{equation}
where $\alpha\in (1,2)$, ${}^{C}D_{t_i^+}^\alpha$ represents the standard
Caputo fractional derivatives of order $\alpha$,
$\alpha_1,\alpha_2,\beta_1,\beta_2\in \mathbb{R}$ with
$\alpha_1\alpha_2+\alpha_1\beta_2+\alpha_2\beta_1\neq 0$,
$\mathbb{N}[a,b]=\{a,a+1,\dots,b\}$ with $a,b$ being integers,
$0=t_0<t_1<\dots<t_p< t_{p+1}= 1$, $I_{i},J_i\in C(\mathbb{R},\mathbb{R})$
$(i\in \mathbb{N}[1,p]$, $f:[0,T]\times \mathbb{R}^2\to \mathbb{R}$ is continuous,
$w:[0,1]\to [0,+\infty)$ is a continuous function,
 $g_1,g_2:PC(0,1]\to \mathbb{R}$ are two continuous functions.

 Liu and Li \cite{ll1} investigated the existence and uniqueness of solutions
for the nonlinear impulsive fractional differential equations
\begin{equation}
\begin{gathered}
{}^{C}D_{t_i^+}^\alpha u(t)=f(t,u(t),u'(t),u''(t)), \quad
t\in (t_i,t_{i+1}],\; i\in \mathbb{N}[0,p],\\
u(0)=\lambda_1u(T)+\xi_1\int_0^Tq_1(s,u(s),u'(s),u''(s))ds,\\
u'(0)=\lambda_2u'(T)+\xi_2\int_0^Tq_2(s,u(s),u'(s),u''(s))ds,\\
u''(0)=\lambda_3u''(T)+\xi_3\int_0^Tq_3(s,u(s),u'(s),u''(s))ds,\\
\Delta u(t_i)]=A_i(u(t_i)),\quad \Delta u'(t_i)]=B_i(u(t_i)),\quad
\Delta u''(t_i)]=C_i(u(t_i)),
\end{gathered} \label{e1.0.5}
\end{equation}
for $i\in \mathbb{N}[1,p]$,
where $\alpha\in (2,3)$, ${}^{C}D_{t_i^+}^\alpha$ represents the standard
Caputo fractional derivatives of order $\alpha$, $\mathbb{N}[a,b]=\{a,a+1,\dots,b\}$
with $a,b$ being integers, $0=t_0<t_1<\dots<t_p< t_{p+1}= T$,
 $\lambda_i,\xi_i\in \mathbb{R}$ $(i=1,2,3)$ are constants,
$A_i,B_i,C_i\in C(\mathbb{R},\mathbb{R})$ $(i\in\mathbb{N}[1,p]$, $f:[0,T]\times \mathbb{R}^3\to \mathbb{R}$ is continuous.

Recently, in \cite{brt}, to extend the problem for impulsive differential equation
$u''(t)-\lambda u(t)=f(t,u(t)),u(0)=u(T)=0,\Delta u'(t_i)=I_i(u(t)i)),i\mathbb{N}[1,p]$
to impulsive fractional differential equation, the authors studied the
existence and the multiplicity of solutions for the Dirichlet's boundary
value problem for impulsive fractional order differential equation
\begin{equation}
\begin{gathered}
{}^CD_{T^-}^\alpha ({}^CD_{0^+}^\alpha x(t)+a(t)x(t)
=\lambda f(t,x(t)),\quad t\in [0,T],t\neq t_i,\; i\in \mathbb{N}[1,m],\\
\Delta {}^CD_{T^-}^{\alpha-1} ({}^CD_{0^+}^\alpha x(t_i)=\mu I_i(x(t_i^-)),
\quad i\in \mathbb{N}[1,m],\;x(0)=x(T)=0,
\end{gathered}\label{e1.0.6}
\end{equation}
where $\alpha\in (1/2,1]$, $\lambda,\mu>0$ are constants,
 $\mathbb{N}[a,b]=:\{a,a+1,\dots,b]$ with $a\le b$, $0=t_0<t_1<\dots<t_m<t_{m+1}=T$,
$f : [0,T]\times \mathbb{R}\to \mathbb{R}$ is a continuous function,
 $I_i : \mathbb{R}\to \mathbb{R}(i\in \mathbb{N}[1,m])$ are continuous functions, ${}^CD_{0^+}^\alpha$
 (or ${}^CD_{T^-}^\alpha$) is the standard left (or right) Caputo fractional
derivative of order $\alpha$, $a\in C[0,T]$ and there exist constants
$a_1,a_2>0$ such that $a_1\le a(t)\le a_2$ for all $t\in [0,T]$,
$\Delta x|_{t=t_i}=\lim_{t\to t_i^+}x(t)-\lim_{t\to t_i^-}x(t)=x(t_i^+)-x(t_i^-)$
and $x(t_i^+), x(t_i^-)$ represent the right and left limits of $x(t)$ at
$t=t_i$ respectively, $a,b,x_0$ a constant with $a+b\neq 0$. One knows that
the boundary condition $ax(0)+bx(T)=x_0$ becomes $x(0)-x(T)=\frac{x_0}{a}$
when $a+b=0$, that is so called nonhomogeneous periodic type boundary condition.

For impulsive fractional differential equations whose derivatives have single
start points $t=0$, there has been few papers published.
In \cite{re}, authors presented a new method to converting the impulsive
fractional differential equation (with the Caputo fractional derivative) to
an equivalent integral equation and established existence and uniqueness results
for some boundary value problems of impulsive fractional differential
equations involving the Caputo fractional derivatives with single start point.
The existence and uniqueness of solutions of the following initial or
boundary value problems were discussed in \cite{re}:
 \begin{gather*}
{}^{C}D_{0^+}^\alpha x(t)=f(t,x(t)), \quad t\in (0,1]\setminus\{t_1,\dots,t_p\},\\
\Delta x(t_i)]=I_{i}(x(t_i)),\quad \Delta x'(t_i)]=J_{i}(x(t_i)),
\quad i\in \mathbb{N}[1,p],\\
 x(0)=x_0,\quad x'(0)=x_1;
\end{gather*}
\begin{gather*}
{}^{C}D_{0^+}^\alpha x(t)=f(t,x(t)), \quad t\in (0,1]\setminus\{t_1,\dots,t_p\},\\
\Delta x(t_i)]=I_{i}(x(t_i)),\quad \Delta x'(t_i)]=J_{i}(x(t_i)),\;i\in \mathbb{N}[1,p],\\
x(0)+\phi(x)=x_0,\quad x'(0)=x_1;
\end{gather*}
\begin{gather*}
{}^{C}D_{0^+}^\beta x(t)=f(t,x(t)), \quad t\in (0,1]\setminus\{t_1,\dots,t_p\},\\
\Delta x(t_i)]=I_{i}(x(t_i)),\quad i\in \mathbb{N}[1,p],\; ax(0)+bx(1)=0,
\end{gather*}
\begin{gather*}
{}^{C}D_{0^+}^\alpha x(t)=f(t,x(t)), \quad t\in (0,1]\setminus\{t_1,\dots,t_p\},\\
\Delta x(t_i)]=I_{i}(x(t_i)),\quad \Delta x'(t_i)]=J_{i}(x(t_i)),\;i\in \mathbb{N}[1,p],\\
ax(0)-bx'(0)=x_0,\quad cx(1)+dx'(1)=x_1;
\end{gather*}
and
\begin{gather*}
{}^{C}D_{0^+}^\alpha x(t)=f(t,x(t)), \quad t\in (0,1]\setminus\{t_1,\dots,t_p\},\\
\Delta x(t_i)]=I_{i}(x(t_i)),\quad \Delta x'(t_i)]=J_{i}(x(t_i)),\quad i\in \mathbb{N}[1,p],\\
x(0)-ax(\xi)=x(1)-bx(\eta)=0,
\end{gather*}
where $\alpha\in (1,2],\beta\in (0,1]$, $D_{0^+}^*$ is the Caputo fractional
derivative with order $*$ and single start point $t=0$,
$f:[0,1]\times \mathbb{R}\to \mathbb{R}$, $I_i,J_i:\mathbb{R}\to \mathbb{R}$ are continuous functions,
$a,b,c,d,x_0,x_1\in \mathbb{R}$ are constants, $\phi:PC(0,1]\to \mathbb{R}$ is a functional.

We observed that in the above-mentioned work, the authors all require that
the fractional derivatives are the Caputo type derivatives, the nonlinear
term $f$ and the impulse functions are continuous. It is easy to see that
these conditions are very restrictive and difficult to satisfy in applications.
To the author's knowledge, there has been no paper published discussed
the existence of solutions of boundary value problems of impulsive fractional
differential equations involving other fractional derivatives such as
the Riemann-Liouville fractional derivatives, Hadamard fractional derivatives.

In this paper, we study the existence of solutions of four classes of
impulsive boundary value problems of singular fractional differential equations.
The first class is the impulsive Dirichlet type integral boundary value problem
\begin{equation}
\begin{gathered}
{}^{RL}D_{0^+}^{\beta}x(t)-\lambda x(t)=p(t)f(t,x(t)), \quad
\text{a.e., }t\in (t_i,t_{i+1}],\; i\in \mathbb{N}[0,m],\\
\lim_{t\to 0^+}t^{2-\beta }x(t)=\int_0^1\phi(s)G(s,x(s))ds,\quad
x(1)=\int_0^1\psi(s)H(s,x(s))ds,\\
\lim_{t\to t_i^+}(t-t_i)^{2-\beta} x(t)=I(t_i,x(t_i)),\quad
\Delta {}^{RL}D_{0^+}^{\beta-1}x(t_i)=J(t_i,x(t_i)),
\end{gathered}\label{e1.0.7}
\end{equation}
for $i\in \mathbb{N}[1,m]$, where
\begin{itemize}
\item[(1.A1)] $ 1<\beta<2$, $\lambda \in \mathbb{R}$, ${}^{RL}D_{0^+}^{\beta}$
is the Riemann-Liouville
 fractional derivative of order $\beta$,


\item[(1.A2)] $m$ is a positive integer, $0=t_0<t_1<t_2<\dots<t_m<t_{m+1}=1$,
$\mathbb{N}[a,b]=\{a,a+1,a+2,\dots,a+n\}$ with $a,b$ being integers and $a\le b$,


\item[(1.A3)] $\phi,\psi:(0,1)\to \mathbb{R}$ are measurable functions,

\item[(1.A4)] $p:(0,1)\to \mathbb{R}$ is continuous and there exist numbers $k>-1$
and $l\in \max\{-\beta,-2-k,0]$ such that $|p(t)|\le t^k(1-t)^l$ for all
$t\in (0,1)$,

\item[(1.A5)] $f,G,H$ defined on $(0,1]\times \mathbb{R}$ are impulsive
II-Carath\'eodory functions, $I,J:\{t_i:i\in \mathbb{N}[1,m]\}\times \mathbb{R}\to \mathbb{R}$
is a discrete II-Carath\'eodory functions.
\end{itemize}

 The second class is the impulsive mixed type integral boundary value problem
\begin{equation}
\begin{gathered}
{}^CD_{0^+}^{\beta}x(t)-\lambda x(t)=p(t)f(t,x(t)), \quad
\text{a.e., } t\in (t_i,t_{i+1}],\;i\in \mathbb{N}[0,m],\\
\lim_{t\to 0^+}x(t)=\int_0^1\phi(s)G(s,x(s))ds,\quad
x'(1)=\int_0^1\psi(s)H(s,x(s))ds,\\
\Delta x(t_i)=I(t_i,x(t_i)),\quad \Delta x'(t_i)=J(t_i,x(t_i)),\quad i\in \mathbb{N}[1,m],
\end{gathered} \label{e1.0.8}
\end{equation}
where
\begin{itemize}
\item[(1.A6)] $1<\beta<2$, $\lambda \in \mathbb{R}$, ${}^CD_{0^+}^{\beta}$ is
the Caputo fractional derivative of order $\beta$,
$m,t_i,\mathbb{N}[a,b]$ satisfies (1.A2), 
$\phi, \psi:(0,1)\to \mathbb{R}$ satisfy (1.A3),


\item[(1.A7)] $p:(0,1)\to \mathbb{R}$ is continuous and there exist numbers
$k>1-\beta$ and $l\in \max\{-\beta,-\beta-k,0]$ such that
$|p(t)|\le t^k(1-t)^l$ for all $t\in (0,1)$,


\item[(1.A8)] $f,G,H$ defined
 on $(0,1]\times \mathbb{R}$ are impulsive I-Carath\'eodory functions,
$I,J:\{t_i:i\in \mathbb{N}[1,m]\}\times \mathbb{R}\to \mathbb{R}$ are
discrete I-Carath\'eodory functions.
\end{itemize}


We emphasize that much work on fractional boundary value problems involves
either Riemann-Liouville or Caputo type fractional differential equations
see \cite{an2, an3, an4, an1}. Another kind of fractional derivatives that
appears side by side to Riemann-Liouville and Caputo derivatives in the
literature is the fractional derivative due to Hadamard introduced
in 1892 \cite{hhh}, which differs from the preceding ones in the sense that
the kernel of the integral (in the definition of Hadamard derivative)
contains logarithmic function of arbitrary exponent. Recent studies can
be seen in \cite{bkt1, bkt2, bkt3}.

Thirdly we study the following impulsive periodic type integral boundary
value problems of singular fractional differential systems
\begin{equation}
\begin{gathered}
{}^{RLH}D_{1^+}^{\beta}x(t)-\lambda x(t)=p(t)f(t,x(t)),\quad
\text{a.e., } t\in (t_i,t_{i+1}],\; i\in \mathbb{N}[0,m],\\
\lim_{t\to 1^+}(\log t)^{2-\beta }x(t)-x(e)
=\int_1^e \phi(s)G(s,x(s))ds,\\
\lim_{t\to 1^+}{}^{RLH}D_{1^+}^{\beta-1}x(t)-{}^{RLH}D_{1^+}^{\beta-1}x(e)
=\int_1^e \psi(s)H(s,x(s))ds,\\
\lim_{t\to t_i^+}(\log t-\log t_i)^{2-\beta} x(t)=I(t_i,x(t_i)),\quad
\Delta {}^{RLH}D_{1^+}^{\beta-1}x(t_i)=J(t_i,x(t_i)),
\end{gathered} \label{e1.0.9}
\end{equation}
for $i\in \mathbb{N}[1,m]$, where
\begin{itemize}
\item[(1.A9)] $ 1<\beta<2$, $\lambda\in \mathbb{R}$, ${}^{RLH}D_{1^+}^{\beta}$
is the Hadamard fractional derivative of order $\beta$,

\item[(1.A10)] $m$ is a positive integer, $1=t_0<t_1<t_2<\dots<t_m<t_{m+1}=e$,
$\phi,\psi:(1,e)\to \mathbb{R}$ are measurable functions,
 $p:(1,e)\to \mathbb{R}$ is continuous and satisfies 
$|p(t)|\leq (\log t)^k(1-\log t)^l)$ with $k>-1$, $l\leq 0$,
$2+k+l>0$,
$\mathbb{N}[a,b]=\{a,a+1,a+2,\dots,a+n\}$ with $a,b$ being integers and $a\le b$,

\item[(1.A11)] $f,G,H$ defined
 on $(1,e]\times \mathbb{R}$ are impulsive III-Carath\'eodory functions,
$I,J:\{t_i:i\in \mathbb{N}[1,m]\}\times \mathbb{R}\to \mathbb{R}$ are
discrete III-Carath\'eodory functions.
\end{itemize}

Finally we study the following impulsive Neumann type integral boundary value problems
of singular fractional differential systems
\begin{equation}
\begin{gathered}
{}^{CH}D_{1^+}^{\beta}x(t)-\lambda x(t)=p(t)f(t,x(t)), \quad
\text{a.e., } t\in (t_i,t_{i+1}],\quad i\in \mathbb{N}[0,m],\\
(t\frac{d}{dt})x(t)\big|_{t=1} =\int_1^e\phi(s)G(s,x(s))ds,\\
(t\frac{d}{dt})x(t)\big|_{t=e} =\int_1^e \psi(s)H(s,x(s))ds,
\\
\lim_{t\to t_i^+}x(t)-x(t_i)=I(t_i,x(t_i)),\\
\lim_{t\to t_i^+}(t\frac{d}{dt})x(t)-(t\frac{d}{dt})x(t)\big|_{t=t_i}
=J(t_i,x(t_i)),
\end{gathered}\label{e1.0.10}
\end{equation}
for $i\in \mathbb{N}[1,m]$, where
\begin{itemize}
\item[(1.A12)] $ 1<\beta<2$, $\lambda\in \mathbb{R}$, ${}^{CH}D_{1^+}^{\beta}$
is the Caputo type Hadamard fractional derivative of order $\beta$,
$(t\frac{d}{dt})^1x(t)=tx'(t)$,

\item[(1.A13)] $m,t_i,\mathbb{N}[a,b]$ satisfy (1.A10),
 $\phi,\psi:(1,e)\to \mathbb{R}$
are measurable functions, $p:(1,e)\to \mathbb{R}$ is continuous and 
satisfies $|p(t)|\leq (\log t)^k(1-\log t)^l)$ with $k>-1$, $l\leq 0$,
$\beta+k+l>0$,

\item[(1.A14)] $f,G,H$ defined on $(1,e]\times \mathbb{R}$ are
impulsive I-Carath\'eodory functions,
$I,J:\{t_i:i\in \mathbb{N}[1,m]\}\times \mathbb{R}\to \mathbb{R}$ are
discrete I-Carath\'eodory functions.

\end{itemize}

 A function $x:(0,1]\to \mathbb{R}$ is called a solution of BVP \eqref{e1.0.7}
(or of BVP \eqref{e1.0.8}) if
$ x|_{(t_i,t_{i+1}]}(i=0,1,j\in \mathbb{N}[0,m]) $ is continuous,
 the limits below exist
\[
\lim_{t\to t_i^+}(t-t_i)^{2-\beta}x(t),i\in \mathbb{N}[0,m], \quad
\text{(or}\quad
\lim_{t\to t_i^+}x(t)\; i\in \mathbb{N}[0,m])
\]
 and $x$ satisfies \eqref{e1.0.7} (or \eqref{e1.0.8}).

A function $x:(1,e]\to \mathbb{R}$ is called a solution of BVP \eqref{e1.0.9}
 (or of BVP \eqref{e1.0.10})
 if $ x|_{(t_i,t_{i+1}]}(i\in \mathbb{N}[0,m]) $ is continuous,
 the limits below exist
\[
 \lim_{t\to t_i^+}(\log \frac{t}{t_i})^{2-\beta}x(t),i\in \mathbb{N}[0,m], \quad
\text{(or }
\lim_{t\to t_i^+}x(t), i\in \mathbb{N}[0,m])
\]
 and $x$ satisfies \eqref{e1.0.9} (or \eqref{e1.0.10}).

To obtain solutions of a boundary value problem of fractional differential equations,
we firstly define a Banach space $X$, then we transform the boundary value problem
into a integral equation and define a nonlinear operator $T$ on $X$ by using
the integral equation obtained, finally, we prove that $T$ has fixed point in $X$.
The fixed points are just solutions of the boundary value problem.
Three difficulties occur in known papers: one is how to transform the boundary
 value problem into a integral equation; the other one is how to define and
prove a Banach space and the completely continuous property of the nonlinear
operator defined; the third one is to choose a suitable fixed point theorem
and impose suitable growth conditions on functions to get the fixed points of
the operator.

To the best of the authors knowledge, no one has studied the existence of
strong weak or weak solutions of BVPs \eqref{e1.0.7}--\eqref{e1.0.10}.
This paper fills this gap. Another purpose of this paper is to illustrate the
similarity and difference of these three kinds of fractional differential equations.
We obtain results on the existence of at least one solution for
BVPs \eqref{e1.0.7}--\eqref{e1.0.10}.
 For simplicity we only consider the left-sided operators here. The
right-sided operators can be treated similarly. For clarity and brevity,
we restrict our attention to BVPs with one impulse, the difference
between the theory of one or an arbitrary number of impulses is quite similar.


The remainder of this paper is organized as follows:
in Section 2, we present related definitions.
In Section 3 some preliminary results are given (one purpose is to establish
existence and uniqueness of continuous solutions of linear
fractional differential equations (Subsection 3.1),
the second purpose is to get exact expression of piecewise continuous
solutions of the linear fractional differential equations with a constant
coefficient and a variable force term (Subsection 3.2), the third purpose
is to prove preliminary results for establishing existence results of
solutions of \eqref{e1.0.7}--\eqref{e1.0.10}in Subsections 3.3, 3.4, 3.5 and 3.6,
respectively), we transform them into corresponding integral equations and
define completely continuous nonlinear operators.
In Sections 4, the main theorems and their proof
are given (we establish existence results for solutions of BVP
\eqref{e1.0.7}--\eqref{e1.0.10}. In Section 5, we preset applications of theorems
obtained in Subsection 3.2, the solvability of multi-point boundary value
problem, Sturm-Liouville boundary value problem and anti-periodic boundary
value problem for fractional differential equations with impulse effects
are discussed, respectively. In Section 6, some mistakes happened in cited
papers are showed. Corrected expressions of solutions are given.
Finally, in Section 7, we survey some examples
and applications of fractional differential equations in various fields:
population dynamics, control theory,
physics, biology, medicine.

\section{Related definitions}

 For convenience of the readers, we firstly present the
necessary definitions from the fractional calculus theory. These
definitions and results can be found in \cite{Kil, 8, skm}.

Let the Gamma function, Beta function and the classical Mittag-Leffler special
function be
\begin{gather*}
\Gamma(\alpha)=\int_0^{+\infty}x^{\alpha-1}e^{-x}dx,\quad
\mathbf{B}(p,q)=\int_0^1x^{p-1}(1-x)^{q-1}dx,\\
\mathbf{E}_{\delta,\sigma}(x)=\sum_{k=0}^{+\infty}
 \frac{x^k}{\Gamma(\delta k+\sigma)}
\end{gather*}
respectively for $\alpha>0$, $p>0$, $q>0$, $\delta>0$, $\sigma>0$.
 We note that $\mathbf{E}_{\delta,\delta}(x)>0$ for all $x\in \mathbb{R}$ and
$\mathbf{E}_{\delta,\delta}(x)$ is strictly increasing in $x$.
Then for $x>0$ we have
$\mathbf{E}_{\delta,\sigma}(-x)<\mathbf{E}_{\delta,\sigma}(0)
=\frac{1}{\Gamma(\sigma)}<\mathbf{E}_{\delta,\sigma}(x)$.

\begin{definition}[\cite{Kil}] \label{def2.1} \rm
 Let $c\in \mathbb{R}$. The Riemann-Liouville fractional integral
of order $\alpha>0$ of a function $g:(c,\infty)\to \mathbb{R}$ is
$$
I_{c^+}^{\alpha}g(t)=\frac{1}{\Gamma(\alpha)}\int_c^t(t-s)^{\alpha-1}g(s)ds,
$$
provided that the right-hand side exists.
\end{definition}

\begin{definition}[\cite{Kil}] \label{def2.2} \rm
Let $c\in \mathbb{R}$. The Riemann-Liouville fractional derivative
of order $\alpha>0$ of a function $g:(c,+\infty)\to \mathbb{R}$ is
$$
{}^{RL}D_{c^+}^{\alpha}g(t)=\frac{1}{\Gamma(n-\alpha)}\frac{d^{n}}{dt^{n}}
\int_{c}^{t}\frac{g(s)}{(t-s)^{\alpha-n+1}}ds,
$$
where $\alpha< n< \alpha+1$, i.e., $n=\lceil\alpha\rceil$,
 provided that the right-hand side exists.
\end{definition}

\begin{definition}[\cite{Kil}] \label{def2.3} \rm
 Let $c\in \mathbb{R}$. The Caputo fractional derivative
of order $\alpha>0$ of a function $g:(c,+\infty)\to \mathbb{R}$ is
$$
{}^CD_{c^+}^{\alpha}g(t)=\frac{1}{\Gamma(n-\alpha)}\int_{c}^{t}
\frac{g^{(n)}(s)}{(t-s)^{\alpha-n+1}}ds,
$$
where $\alpha< n < \alpha+1$, i.e., $n=\lceil\alpha\rceil$,
provided that the right-hand side exists.
\end{definition}

\begin{definition}[\cite{Kil}] \label{def2.4} \rm
Let $c>0$. The Hadamard fractional integral
of order $\alpha>0$ of a function $g:[c,+\infty)\to \mathbb{R}$ is
$$
{}^HI_{c^+}^{\alpha}g(t)=\frac{1}{\Gamma(\alpha)}
\int_c^t(\log\frac{t}{s})^{\alpha-1}g(s)\frac{ds}{s},
$$
provided that the right-hand side exists.
\end{definition}

\begin{definition}[\cite{Kil}] \label{def2.5} \rm
Let $c>0$. The Hadamard fractional derivative
of order $\alpha>0$ of a function $g:[c,+\infty)\to \mathbb{R}$ is
$$
{}^{RLH}D_{c^+}^{\alpha}g(t)
=\frac{1}{\Gamma(n-\alpha)}(t\frac{d}{dt})^n\int_{c}^{t}
(\log\frac{t}{s})^{n-\alpha-1}g(s)\frac{ds}{s},
$$
where $\alpha< n < \alpha+1$, i.e., $n=\lceil\alpha\rceil$,
provided that the right-hand side exists.
\end{definition}

\begin{definition}[\cite{jab}] \label{def2.6} \rm
Let $c>0$. The Caputo type Hadamard fractional derivative
of order $\alpha>0$ of a function $g:[c,+\infty)\to \mathbb{R}$ is
$$
{}^{CH}D_{c^+}^{\alpha}g(t)
=\frac{1}{\Gamma(n-\alpha)}\int_{c}^{t}(\log\frac{t}{s})^{n-\alpha-1}
(s\frac{d}{ds})^n g(s)\frac{ds}{s},
$$
where $\alpha< n\le \alpha+1$, i.e., $n=\lceil\alpha\rceil$, provided
that the right-hand side exists.
\end{definition}


\begin{definition} \label{def2.7}\rm
We call $F:\cup_{i=0}^m(t_i,t_{i+1})\times \mathbb{R}\to \mathbb{R}$ an
\emph{impulsive I-Carath\'eodory function} if it satisfies
\begin{itemize}
\item[(i)] $t\to F(t, u)$ is measurable on $(t_i,t_{i+1})$ $(i\in \mathbb{N}[0,m])$
for any $u\in \mathbb{R}$,

\item[(ii)] $u\to F(t, u)$ are continuous on $\mathbb{R}$ for almost all $t\in (t_i,t_{i+1})$
$(i\in \mathbb{N}[0,m])$,


\item[(iii)] for each $r>0$ there exists $M_r>0$ such that
$$
 |F(t, u)|\le M_r,t\in (t_i,t_{i+1}),|u|\le r,\quad i\in \mathbb{N}[0,m].
$$
\end{itemize}
\end{definition}

\begin{definition} \label{def2.8}\rm
We call $F:\cup_{i=0}^m(t_i,t_{i+1})\times \mathbb{R}\to \mathbb{R}$ an
\emph{impulsive II-Carath\'eodory function} if it satisfies
\begin{itemize}
\item[(i)] $t\to F(t, (t-t_i)^{\beta-2}u)$ is measurable on
$(t_i,t_{i+1})$ $(i\in\mathbb{N}[0,m])$ for any $u\in \mathbb{R}$,

\item[(ii)] $u\to F(t, (t-t_i)^{\beta-2}u)$ are continuous on $\mathbb{R}$ for almost
all $t\in (t_i,t_{i+1})$ $(i\in \mathbb{N}[0,m])$,


\item[(iii)] for each $r>0$ there exists $M_r>0$ such that
$$
 |F(t, (t-t_i)^{\beta-2}u)|\le M_r,t\in (t_i,t_{i+1}),|u|\le r,\quad i\in \mathbb{N}[0,m].
$$
\end{itemize}
\end{definition}


\begin{definition} \label{def2.9}\rm
We call $F:\cup_{i=0}^m(t_i,t_{i+1})\times \mathbb{R}\to \mathbb{R}$ an
\emph{impulsive III-Carath\'eodory function} if it satisfies
\begin{itemize}
\item[(i)] $t\to F(t, (\log\frac{t}{t_i})^{\beta-2}u)$ is measurable on
$(t_i,t_{i+1})$ $(i\in \mathbb{N}[0,m])$ for any $u\in \mathbb{R}$,

\item[(ii)] $u\to F(t, (\log\frac{t}{t_i})^{\beta-2}u)$ are continuous on
$\mathbb{R}$ for all $t\in (t_i,t_{i+1})$ $(i\in \mathbb{N}[0,m])$,


\item[(iii)] for each $r>0$ there exists $M_r>0$ such that
$$
 |F(t, (\log\frac{t}{t_i})^{\beta-2}u)|\le M_r,t\in (t_i,t_{i+1}),|u|\le r,
\quad i\in \mathbb{N}[0,m].
$$
\end{itemize}
\end{definition}


\begin{definition} \label{def2.10}\rm
 We call $I:\{t_i:i\in \mathbb{N}[1,m]\}\times \mathbb{R}\to \mathbb{R}$ a
\emph{discrete I-Carath\'eodory function} if it satisfies
\begin{itemize}

\item[(i)] $u\to I(t_i, u)$ $(i\in\mathbb{N}[1,m])$ are continuous on $\mathbb{R}$,

\item[(ii)] for each $r>0$ there exists $M_r>0$ such that
$ |I(t_i, u)|\le M_r,|u|\le r$ for $i\in \mathbb{N}[1,m]$.
\end{itemize}
\end{definition}

\begin{definition} \label{def2.11}\rm
We call $I:\{t_i:i\in \mathbb{N}[1,m]\}\times \mathbb{R}\to \mathbb{R}$ a
 \emph{discrete II-Carath\'eodory function} if it satisfies
\begin{itemize}

\item[(i)] $u\to I(t_i, (t_i-t_{i-1})^{\beta-2}u)$ $(i\in\mathbb{N}[1,m])$
 are continuous on $\mathbb{R}$,


\item[(ii)] for each $r>0$ there exists $M_r>0$ such that
$ |I(t_i,(t_i- t_{i-1}^{\beta-2}u)|\le M_r,|u|\le r$ for $i\in \mathbb{N}[1,m]$.
\end{itemize}
\end{definition}


\begin{definition} \label{def2.12}\rm
We call $I:\{t_i:i\in \mathbb{N}[1,m]\}\times \mathbb{R}\to \mathbb{R}$ a
 \emph{discrete III-Carath\'eodory function} if it satisfies
\begin{itemize}

\item[(i)] $u\to I(t_i, (\log {t_i}-\log t_{i-1})^{\beta-n}u)$ $(i\in \mathbb{N}[1,m])$
are continuous on $\mathbb{R}$,

\item[(ii)] for each $r>0$ there exists $M_r>0$ such that $
 |I(t_1, (\log\frac{ {t_i}}{t_{i-1}})^{\beta-n}u)|\le M_r,|u|\le r$ for
$i\in \mathbb{N}[1,m]$.
\end{itemize}
\end{definition}


\begin{definition}[\cite{10}] \label{def2.13}\rm 
Let $E$ and $F$ be Banach spaces.
A operator $T:E\to F$ is called a completely continuous operator
if $T$ is continuous and maps any bounded set into relatively compact set.
\end{definition}

Suppose that $n-1\le \alpha<n$. The following Banach spaces are used:

 Let $a<b$ be constants. $C(a,b]$ denotes the set of continuous
functions on $(a,b]$ with  $\lim_{t\to a^+}x(t)$ existing,
and the norm \[
\|x\|=\sup_{t\in (a,b]}|x(t)|.
\]

 Let $a<b$ be constants. $C_{n-\alpha}(a,b]$ the set of
continuous functions on $(a,b]$ with
$\lim_{t\to a^+}(t-a)^{n-\alpha}x(t)$ existing, the norm
$\|x\|_{n-\alpha}=\sup_{t\in (a,b]}(t-a)^{n-\alpha}|x(t)|$.


 Let $0<a<b$. $LC_{n-\alpha}(a,b]$ denote the set of all continuous
functions on $(a,b]$ with the limit
$\lim_{t\to a^+}(\log\frac{t}{a})^{n-\alpha}x(t)$ existing, and the norm
\[
\|x\|=\sup_{t\in (a,b]}(\log\frac{t}{a})^{n-\alpha}|x(t)|.
\]

For a positive integer  $m$ let $\mathbb{N}[0,m]=\{0,1,2,\dots,m\}$, with
$0=t_0<t_1<\dots<t_m<t_{m+1}=1$. The following Banach spaces are also used
in this paper:
$$
P_mC_{n-\alpha}(0,1]=\{x:(0,1]\to \mathbb{R}:x\big|_{(t_i,t_{i+1}]}
\in C_{n-\alpha}(t_i,t_{i+1}]:i\in \mathbb{N}[0,m]\}
$$
 with the norm
$$
\|x\|=\|x\|_{P_mC_{n-\alpha}}
=\max\Big\{\sup_{t\in (t_i,t_{i+1}]}(t-t_i)^{n-\alpha}|x(t)|:i\in \mathbb{N}[0,m]\Big\}.
$$

$$
P_mC(0,1]=\{x:(0,1]\to \mathbb{R}:x|_{(t_i,t_{i+1}]}\in C(t_i,t_{i+1}]:i\in \mathbb{N}[0,m]
\}
$$
with the norm
$$
\|x\|=\|x\|_{P_mC(0,1]}=\max\Big\{\sup_{t\in (t_i,t_{i+1}]}|x(t)|:
i\in \mathbb{N}[0.m]\Big\}.
$$

For a positive integer $m$ let  $\mathbb{N}[0,m]=\{0,1,2,\dots,m\}$, with
$1=t_0<t_1<\dots<t_m<t_{m+1}=e$. We also use the Banach spaces
\begin{align*}
LP_mC_{n-\alpha}(1,e]
=\Big\{&x:(1,e]\to \mathbb{R}: x\big|_{(t_i,t_{i+1}]}\in C(t_i,t_{i+1}],\; i\in \mathbb{N}[0,m],\\
&\text{and } \lim_{t\to t_i^+}(\log\frac{t}{t_i})^{n-\alpha}x(t)
\text{ exist for }i\in \mathbb{N}[0,m]\Big\}
\end{align*}
with the norm
$$
\|x\|=\|x\|_{LP_mC_{n-\alpha}}
=\max\Big\{\sup_{t\in (t_i,t_{i+1}]}(\log\frac{t}{t_i})^{n-\alpha}|x(t)|,
i\in \mathbb{N}[0,m]\Big\}.
$$

$$
P_mC(1,e]=\big\{x:(1,e]\to \mathbb{R}:x\big|_{(t_i,t_{i+1}]}\in C(t_i,t_{i+1}],\; i\in \mathbb{N}[0.m]
\big\}
$$
with the norm
$$
\|x\|=\|x\|_{P_mC}=\max\Big\{\sup_{t\in (t_i,t_{i+1}]}|x(t)|,i\in \mathbb{N}[0,m]\Big\}.
$$

\section{Preliminaries}

In this section, we present some preliminary results that can be used
in next sections for obtain solutions of \eqref{e1.0.7}--\eqref{e1.0.10}.

\subsection{Basic theory for linear fractional differential equations}

Lakshmikantham et al.\ \cite{lv1, lv2, lv3, lv4} investigated the basic theory
of initial value problems for fractional differential equations involving
Riemann-Liouville differential operators of order $q\in (0,1)$.
The existence and uniqueness of solutions of the following initial value
problems of fractional differential equations were discussed under the
assumption that $f\in C_r[0,1]$. We will establish existence and uniqueness
results for these problems under more weaker assumptions see (3.A1)--(3.A4) below.

Suppose that $n-1< \alpha<n$ and $\eta_j\in \mathbb{R}(j\in N[0,n-1])$,
$F,A:(0,1)\to \mathbb{R}$ and $B,G:(1,e)\to \mathbb{R}$ are continuous functions.
We consider the following four classes of initial value problems of non-homogeneous
linear fractional differential equations:
\begin{equation}
\begin{gathered}
{}^{C}D_{0^+}^{\alpha}{x}(t)=A(t){x}(t)+{F}(t),\quad\text{a.e. } t\in (0,1),\\
\lim_{t\to 0^+}{x}^{(j)}(t)={\eta}_j,\quad j\in \mathbb{N}[0,n-1],
\end{gathered}\label{e3.1.1}
\end{equation}

\begin{equation}
\begin{gathered}
{}^{RL}D_{0^+}^{\alpha}{x}(t)=A(t){x}(t)+{F}(t),\quad\text{a.e. }t\in (0,1),\\
\lim_{t\to 0^+}t^{n-\alpha}{x}(t)=\frac{{\eta}_n}{\Gamma(\alpha-n+1)},\\
\lim_{t\to 0^+}{}^{RL}D_{0^+}^{\alpha-j}x(t)=\eta_j,\quad j\in \mathbb{N}[1,n-1],
\end{gathered}\label{e3.1.2}
\end{equation}

\begin{equation} \begin{gathered}
{}^{RLH}D_{0^+}^{\alpha}{x}(t)=B(t){x}(t)+{G}(t),\quad\text{a.e. }t\in (1,e),\\
\lim_{t\to 1^+}(\log t)^{n-\alpha}{x}(t)=\frac{{\eta}_n}{\Gamma(\alpha-n+1)},\\
\lim_{t\to 1^+}{}^{RLH}D_{1^+}^{\alpha-j}x(t)=\eta_j, \quad j\in \mathbb{N}[1,n-1],
\end{gathered} \label{e3.1.3}
\end{equation}

\begin{equation} \begin{gathered}
{}^{CH}D_{0^+}^{\alpha}{x}(t)=B(t){x}(t)+{G}(t),\quad\text{a.e. }t\in (1,e),\\
\lim_{t\to 1^+}(t\frac{d}{dt})^{j}x(t)={\eta}_j,j\in \mathbb{N}[0,n-1].
\end{gathered}\label{e3.1.4}
\end{equation}
where $(t\frac{d}{dt})^jx(t)=t\frac{d(t\frac{d}{dt})^{j-1}x(t)}{dt}$
for $j=2,3,\dots$.

To obtain solutions of \eqref{e3.1.1}, we need the following assumptions:
\begin{itemize}
\item[(3.A1)] there exist constants $k_i>-\alpha+n-1$, $l_i\le 0$ with
$l_i>\max\{-\alpha,-\alpha-k_i\}(i=1,2)$, $M_A\ge 0$ and
$M_F\ge 0$ such that $|A(t)|\le M_At^{k_1}(1-t)^{l_1}$ and
$|F(t)|\le M_Ft^{k_2}(1-t)^{l_2}$ for all $t\in (0,1)$.
\end{itemize}
Choose the Picard function sequence as
\begin{gather*}
\phi_0(t)=\sum_{j=0}^{n-1}\frac{\eta_jt^j}{j!},\quad t\in [0,1],\\
\phi_i(t)=\sum_{j=0}^{n-1}\frac{\eta_jt^j}{j!}
+\int_0^t\frac{(t-s)^{\alpha-1}}{\Gamma(\alpha)}
[A(s)\phi_{i-1}(s)+F(s)]ds,\quad t\in (0,1],\; i=1,2,\dots.
\end{gather*}


\noindent\textbf{Claim 1.} $\phi_i\in C[0,1]$.

\begin{proof}
One sees $\phi_0\in C[0,1]$. Then
\begin{align*}
&|\int_0^t\frac{(t-s)^{\alpha-1}}{\Gamma(\alpha)}[A(s)\phi_0(s) +F(s)]ds| \\
&\le \int_0^t\frac{(t-s)^{\alpha-1}}{\Gamma(\alpha)}
[M_A|\phi_0(s)|s^{k_1}(1-s)^{l_1}+M_Fs^{k_2}(1-s)^{l_2}]ds\\
&\le M_A\|\phi_0\|\int_0^t\frac{(t-s)^{\alpha+l_1-1}}{\Gamma(\alpha)}
s^{k_1}ds+M_F\int_0^t\frac{(t-s)^{\alpha+l_2-1}}{\Gamma(\alpha)}s^{k_2}ds\\
&=M_A\|\phi_0\|t^{\alpha+k_1+l_1}\int_0^1
 \frac{(1-w)^{\alpha+l_1-1}}{\Gamma(\alpha)}w^{k_1}dw \\
&\quad +M_Ft^{\alpha+k_2+l_2}
\int_0^1\frac{(1-w)^{\alpha+l_2-1}}{\Gamma(\alpha)}w^{k_2}dw\\
&=M_A\|\phi_0\|t^{\alpha+k_1+l_1}
 \frac{\mathbf{B}(\alpha+l_1,k_1+1)}{\Gamma(\alpha)}
 +M_Ft^{\alpha+k_2+l_2}\frac{\mathbf{B}(\alpha+l_2,k_2+1)}{\Gamma(\alpha)} \\
&\to 0\quad \text{as }t\to 0^+.
\end{align*}
It follows that $\phi_1$ is continuous on $(0,1]$ and
 $\lim_{t\to 0^+}\phi_1(t)$ exists. So $\phi_1\in C[0,1]$.
By mathematical induction, we can prove that $\phi_i\in C[0,1]$.
\end{proof}

\noindent\textbf{Claim 2.} $\{\phi_i\}$ is convergent uniformly on $[0,1]$.

\begin{proof} For $t\in [0,1]$ we have
\begin{align*}
&|\phi_1(t)-\phi_0(t)| \\
&=\big|\int_0^t\frac{(t-s)^{\alpha-1}}{\Gamma(\alpha)}[A(s)\phi_0(s)+F(s)]ds\big|\\
&\le M_A\|\phi_0\| \int_0^t\frac{(t-s)^{\alpha-1}}{\Gamma(\alpha)}
s^{k_1}(1-s)^{l_1}ds +M_F\int_0^t\frac{(t-s)^{\alpha-1}}
{\Gamma(\alpha)}s^{k_2}(1-s)^{l_2}ds\\
&\le M_A\|\phi_0\|\int_0^t\frac{(t-s)^{\alpha+l_1-1}}
 {\Gamma(\alpha)}s^{k_1}ds+M_F\int_0^t\frac{(t-s)^{\alpha+l_2-1}}
 {\Gamma(\alpha)}s^{k_2}ds\\
&=M_A\|\phi_0\| t^{\alpha+k_1+l_1}\frac{\mathbf{B}(\alpha+l_1,k_1+1)}
 {\Gamma(\alpha)}+M_F t^{\alpha+k_2+l_2}
 \frac{\mathbf{B}(\alpha+l_2,k_2+1)}{\Gamma(\alpha)}.
\end{align*}
So
\begin{align*}
&|\phi_2(t)-\phi_1(t)| \\
&=|\int_0^t\frac{(t-s)^{\alpha-1}}{\Gamma(\alpha)}A(s)[\phi_1(s)-\phi_0(s)]ds|\\
&\le \int_0^t\frac{(t-s)^{\alpha-1}}{\Gamma(\alpha)}M_As^{k_1}
 (1-s)^{l_1}(M_A\|\phi_0\| s^{\alpha+k_1+l_1}
 \frac{\mathbf{B}(\alpha+l_1,k_1+1)}{\Gamma(\alpha)} \\
&\quad +M_F s^{\alpha+k_2+l_2}
 \frac{\mathbf{B}(\alpha+l_2,k_2+1)}{\Gamma(\alpha)})ds\\
& \le \|\phi_0\| M_A^2\int_0^t\frac{(t-s)^{\alpha+l_1-1}}
 {\Gamma(\alpha)}s^{\alpha+2k_1+l_1}\frac{\mathbf{B}(\alpha+l_1,k_1+1)}
 {\Gamma(\alpha)}ds\\
&\quad +M_AM_F\int_0^t\frac{(t-s)^{\alpha+l_1-1}}{\Gamma(\alpha)}
 s^{\alpha+k_1+k_2+l_2}\frac{\mathbf{B}(\alpha+l_2,k_2+1)}{\Gamma(\alpha)}ds\\
&=\|\phi_0\| M_A^2t^{2\alpha+2k_1+2l_1}\frac{\mathbf{B}(\alpha+l_1,k_1+1)}
 {\Gamma(\alpha)}\frac{\mathbf{B}(\alpha+l_1,\alpha+2k_1+l_1+1)}{\Gamma(\alpha)} \\
&\quad +M_AM_Ft^{2\alpha+k_1+k_2+l_1+l_2}\frac{\mathbf{B}(\alpha+l_2,k_2+1)}
 {\Gamma(\alpha)}\frac{\mathbf{B}(\alpha+l_1,\alpha+k_1+k_2+l_2+1)}{\Gamma(\alpha)}.
\end{align*}
Now suppose that
\begin{align*}
&|\phi_j(t)-\phi_{j-1}(t)|\\
&\le \|\phi_0\| M_A^jt^{j\alpha+jk_1+jl_1}\prod_{i=0}^{j-1}
 \frac{\mathbf{B}(\alpha+l_1,i\alpha+(i+1)k_1+il_1+1)}{\Gamma(\alpha)}\\
&\quad +M_A^{j-1}M_Ft^{j\alpha+(j-1)k_1+k_2+(j-1)l_1+l_2}
 \frac{\mathbf{B}(\alpha+l_2,k_2+1)}{\Gamma(\alpha)} \\
&\quad\times \prod_{i=1}^{j-1}\frac{\mathbf{B}
(\alpha+l_1,i\alpha+ik_1+k_2+(i-1)l_1+l_2+1)}
{\Gamma(\alpha)}.
\end{align*}
Then we have
\begin{align*}
&|\phi_{j+1}(t)-\phi_j(t)| \\
&=\big|\int_0^t\frac{(t-s)^{\alpha-1}}{\Gamma(\alpha)}A(s)[\phi_j(s)
 -\phi_{j-1}(s)]ds\big|\\
&\le \int_0^t\frac{(t-s)^{\alpha-1}}{\Gamma(\alpha)}M_A
\Big(\|\phi_0\| M_A^js^{j\alpha+jk_1+jl_1}\prod_{i=0}^{j-1}
 \frac{\mathbf{B}(\alpha+l_1,i\alpha+(i+1)k_1+il_1+1)}{\Gamma(\alpha)} \\
&\quad +M_A^{j-1}M_Fs^{j\alpha+(j-1)k_1+k_2+(j-1)l_1+l_2}
\frac{\mathbf{B}(\alpha+l_2,k_2+1)}{\Gamma(\alpha)} \\
&\quad\times \prod_{i=1}^{j-1}\frac{\mathbf{B}
 (\alpha+l_1,i\alpha+ik_1+k_2+(i-1)l_1+l_2+1)}{\Gamma(\alpha)})s^{k_1}(1-s)^{l_1}ds \\
&\le \int_0^t\frac{(t-s)^{\alpha+l_1-1}}{\Gamma(\alpha)}
 M_A\Big(\|\phi_0\| M_A^js^{j\alpha+jk_1+jl_1}
 \prod_{i=0}^{j-1}\frac{\mathbf{B}(\alpha+l_1,i\alpha+(i+1)k_1+il_1+1)}
 {\Gamma(\alpha)}\\
&\quad +M_A^{j-1}M_Fs^{j\alpha+(j-1)k_1+k_2+(j-1)l_1+l_2}
\frac{\mathbf{B}(\alpha+l_2,k_2+1)}{\Gamma(\alpha)} \\
&\quad\times \prod_{i=1}^{j-1}\frac{\mathbf{B}(\alpha+l_1,i\alpha+ik_1+k_2+(i-1)l_1+l_2+1)}
{\Gamma(\alpha)}\Big)s^{k_1}ds\\
&\le \|\phi_0\|M_A^{j+1}t^{(j+1)\alpha+(j+1)k_1+(j+1)l_1}
\prod_{i=0}^{j}\frac{\mathbf{B}(\alpha+l_1,i\alpha+(i+1)k_1+il_1+1)}
{\Gamma(\alpha)}\\
&\quad +M_A^jM_Ft^{(j+1)\alpha+jk_1+k_2+jl_1+l_2}\frac{\mathbf{B}
 (\alpha+l_2,k_2+1)}{\Gamma(\alpha)} \\
&\quad\times \prod_{i=1}^{j}
 \frac{\mathbf{B}(\alpha+l_1,i\alpha+ik_1+k_2+(i-1)l_1+l_2+1)}{\Gamma(\alpha)} .
\end{align*}
Using mathematical induction, for every $i=1,2,\dots$ we obtain
\begin{align*}
&|\phi_{i+1}(t)-\phi_{i}(t)| \\
&\le \|\phi_0\|M_A^{i+1}t^{(i+1)\alpha+(i+1)k_1+(i+1)l_1}
\prod_{j=0}^i\frac{\mathbf{B}(\alpha+l_1,j\alpha+(j+1)k_1+jl_1+1)}
 {\Gamma(\alpha)}\\
&\quad +M_A^iM_Ft^{(i+1)\alpha+ik_1+k_2+il_1+l_2}
\frac{\mathbf{B}(\alpha+l_2,k_2+1)}{\Gamma(\alpha)} \\
&\quad\times \prod_{j=1}^i
\frac{\mathbf{B}(\alpha+l_1,j\alpha+jk_1+k_2+(j-1)l_1+l_2+1)}{\Gamma(\alpha)}\\
&\le\|\phi_0\| M_A^{i+1}\prod_{j=0}^i
\frac{\mathbf{B}(\alpha+l_1,j\alpha+(j+1)k_1+jl_1+1)}{\Gamma(\alpha)}\\
&\quad +M_A^iM_F\frac{\mathbf{B}(\alpha+l_2,k_2+1)}{\Gamma(\alpha)}
\prod_{j=1}^i\frac{\mathbf{B}(\alpha+l_1,j\alpha+jk_1+k_2+(j-1)l_1+l_2+1)}
{\Gamma(\alpha)},
\end{align*}
for $t\in [0,1]$. Consider
\begin{gather*}
\sum_{i=1}^{+\infty}u_i=\sum_{i=1}^{+\infty}\|\phi_0\| M_A^{i+1}
\prod_{i=0}^i\frac{\mathbf{B}(\alpha+l_1,i\alpha+(i+1)k_1+il_1+1)}
{\Gamma(\alpha)},\\
\begin{aligned}
\sum_{i=1}^{+\infty}v_i
&=\sum_{i=1}^{+\infty}M_A^iM_F
\frac{\mathbf{B}(\alpha+l_2,k_2+1)}{\Gamma(\alpha)} \\
&\quad\times  \prod_{j=1}^i\frac{\mathbf{B}(\alpha+l_1,j\alpha+jk_1+k_2+(j-1)l_1+l_2+1)}
{\Gamma(\alpha)}.
\end{aligned}
\end{gather*}
One sees that for sufficiently large $n$ with $\delta\in (0,\frac{1}{2})$,
\begin{align*}
\frac{u_{i+1}}{u_i}
&=M_A\frac{\mathbf{B}(\alpha+l_1,(i+1)\alpha+(i+1)k_1+(i+1)l_1)}{\Gamma(\alpha)} \\
&=M_A\int_0^1(1-x)^{\alpha+l_1-1}x^{(i+1)\alpha+(i+1)k_1+(i+1)l_1}dx\\
&\le M_A\int_0^\delta(1-x)^{\alpha+l_1-1}x^{(i+1)\alpha+(i+1)k_1+(i+1)l_1}dx
 +M_A\int_\delta^1(1-x)^{\alpha+l_1-1}dx \\
&\le M_A\int_0^\delta(1-x)^{\alpha+l_1-1}dx\delta^{(i+1)\alpha+(i+1)k_1+(i+1)l_1}
+\frac{M_A}{\alpha+l_1}\delta^{\alpha+l_1}\\
&\le \frac{M_A}{\alpha+l_1}\delta^{(i+1)\alpha+(i+1)k_1+(i+1)l_1}
+\frac{M_A}{\alpha+l_1}\delta^{\alpha+l_1}.
\end{align*}
It is easy to see that for any $\epsilon>0$ there exists $\delta\in (0,\frac{1}{2})$
such that $\frac{M_A}{\alpha+l_1}\delta^{\alpha+l_1}<\frac{\epsilon}{2}$.
For this $\delta$, there exists an integer $N>0$ sufficiently large such that
\[
\frac{M_A}{\alpha+l_1}\delta^{(i+1)\alpha+(i+1)k_1+(i+1)l_1}<\frac{\epsilon}{2}
\]
 for all $i>N$. So $0<\frac{u_{i+1}}{u_i}<\frac{\epsilon}{2}
+\frac{\epsilon}{2}=\epsilon$ for all $i>N$. It follows that
$\lim_{i\to +\infty} u_{i+1}/u_i=0$.
Then $\sum_{i=1}^{+\infty} u_i$ converges. Similarly we obtain
$\sum_{i=1}^{+\infty} v_i$ converges. Hence
$$
\phi_0(t)+[\phi_1(t)-\phi_0(t)]+[\phi_2(t)-\phi_1(t)]+\dots
+[\phi_i(t)-\phi_{i-1}(t)]+\dots,\quad t\in [0,1]
$$
is uniformly convergent. Then $\{\phi_i(t)\}$ is convergent uniformly on $[0,1]$.
\end{proof}


\noindent\textbf{Claim 3.}
 $\phi(t)=\lim_{i\to +\infty}\phi_i(t)$ defined on $[0,1]$ is a unique continuous
solution of the integral equation
\begin{equation}
x(t)=\sum_{j=0}^{n-1}\frac{\eta_j}{j!}+\frac{1}{\Gamma(\alpha)}
\int_0^t(t-s)^{\alpha-1}[A(s)x(s)+F(s)]ds,\quad t\in [0,1]. \label{e3.1.5}
\end{equation}


\begin{proof}
From $\phi(t)=\lim_{i\to +\infty}\phi_i(t)$ and the uniformly convergence, we
 see that $\phi(t)$ is continuous on $[0,1]$. From
\begin{align*}
&\Big|\int_0^t\frac{(t-s)^{\alpha-1}}{\Gamma(\alpha)}[A(s)\phi_{p-1}(s)+F(s)]ds
-\int_0^t\frac{(t-s)^{\alpha-1}}{\Gamma(\alpha)}[A(s) \phi_{q-1}(s)+F(s)]ds\Big|\\
&\le M_A\|\phi_{p-1}-\phi_{q-1}\|\int_0^t
 \frac{(t-s)^{\alpha-1}}{\Gamma(\alpha)}s^{k_1}(1-s)^{l_1}ds\\
&\le M_A\|\phi_{p-1}-\phi_{q-1}\|t^{\alpha+k_1+l_1}
 \frac{\mathbf{B}(\alpha+l_1,k_1+1)}{\Gamma(\alpha)}\\
&\le M_A\|\phi_{p-1}-\phi_{q-1}\|
 \frac{\mathbf{B}(\alpha+l_1,k_1+1)}{\Gamma(\alpha)}\to 0\quad
 \text{as }p,q\to +\infty,
\end{align*}
we have
\begin{align*}
\phi(t)
&=\lim_{i\to+ \infty}\phi_i(t)\\
&=\lim_{i\to +\infty}\Big[\sum_{j=0}^{n-1}\frac{\eta_jt^j}{j!}
+\int_0^t\frac{(t-s)^{\alpha-1}}{\Gamma(\alpha)}[A(s)\phi_{i-1}(s)+F(s)]ds\Big]\\
&=\sum_{j=0}^{n-1}\frac{\eta_jt^j}{j!}
 +\lim_{i\to +\infty}\int_0^t\frac{(t-s)^{\alpha-1}}
 {\Gamma(\alpha)}[A(s)\phi_{i-1}(s)+F(s)]ds\\
&=\sum_{j=0}^{n-1}\frac{\eta_jt^j}{j!}
+\int_0^t\frac{(t-s)^{\alpha-1}}{\Gamma(\alpha)}
\big[A(s)\lim_{i\to +\infty}\phi_{i-1}(s)+F(s)\big]ds\\
&=\sum_{j=0}^{n-1}\frac{\eta_jt^j}{j!}
 +\int_0^t\frac{(t-s)^{\alpha-1}}{\Gamma(\alpha)}[A(s)\phi(s)+F(s)]ds.
\end{align*}
Then $\phi$ is a continuous solution of \eqref{e3.1.5} defined on $[0,1]$.

Suppose that $\psi$ defined on $[0,1]$ is also a solution of \eqref{e3.1.5}. Then
\[
\psi(t)=\sum_{j=0}^{n-1}\frac{\eta_j}{j!}
+\int_0^t\frac{(t-s)^{\alpha-1}}{\Gamma(\alpha)}[A(s)\psi(s)+F(s)]ds,
\quad t\in (0,1].
\]
We need to prove that $\phi(t)\equiv\psi(t)$ on $[0,1]$.
Then
\begin{align*}
&|\psi(t)-\phi_0(t)| \\
&=|\int_0^t\frac{(t-s)^{\alpha-1}}{\Gamma(\alpha)}|A(s)\psi_0(s)+F(s)|ds|\\
&\le \|\phi_0\|M_A t^{\alpha+k_1+l_1}\frac{\mathbf{B}(\alpha+l_1,k_1+1)}
 {\Gamma(\alpha)}+M_F t^{\alpha+k_2+l_2}
 \frac{\mathbf{B}(\alpha+l_2,k_2+1)}{\Gamma(\alpha)}.
\end{align*}
Furthermore,
\begin{align*}
&|\psi(t)-\phi_1(t)| \\
&=|\int_0^t\frac{(t-s)^{\alpha-1}}{\Gamma(\alpha)}A(s) [\psi(s)-\phi_0(s)]ds|\\
&\le\|\phi_0\|M_A^2t^{2\alpha+2k_1+2l_1}\frac{\mathbf{B}(\alpha+l_1,k_1+1)}
 {\Gamma(\alpha)}\frac{\mathbf{B}(\alpha+l_1,\alpha+2k_1+l_1+1)}{\Gamma(\alpha)}\\
&\quad +M_AM_Ft^{2\alpha+k_1+k_2+l_1+l_2}\frac{\mathbf{B}(\alpha+l_2,k_2+1)}
{\Gamma(\alpha)}\frac{\mathbf{B}(\alpha+l_1,\alpha+k_1+k_2+l_2+1)}{\Gamma(\alpha)}.
\end{align*}
Now suppose that
\begin{align*}
|\psi(t)-\phi_{j-1}(t)|
&\le \|\phi_0\|M_A^jt^{j\alpha+jk_1+jl_1}
 \prod_{i=0}^{j-1}\frac{\mathbf{B}(\alpha+l_1,i\alpha+(i+1)k_1+il_1+1)}
 {\Gamma(\alpha)}\\
&\quad +M_A^{j-1}M_Ft^{j\alpha+(j-1)k_1+k_2+(j-1)l_1+l_2}
\frac{\mathbf{B}(\alpha+l_2,k_2+1)}{\Gamma(\alpha)} \\
&\quad\times \prod_{i=1}^{j-1}\frac{\mathbf{B}
 (\alpha+l_1,i\alpha+ik_1+k_2+(i-1)l_1+l_2+1)}{\Gamma(\alpha)}.
\end{align*}
Then
\begin{align*}
&|\psi(t)-\phi_{j}(t)| \\
&=\Big|\int_0^t\frac{(t-s)^{\alpha-1}}{\Gamma(\alpha)}A(s)
[\psi(s)-\phi_{j-1}(s)]ds\Big|\\
& \le \|\phi_0\| M_A^{j+1}t^{(j+1)\alpha+(j+1)k_1+(j+1)l_1}
 \prod_{i=0}^{j}\frac{\mathbf{B}(\alpha+l_1,i\alpha+(i+1)k_1+il_1+1)}
 {\Gamma(\alpha)}\\
&\quad +M_A^jM_Ft^{(j+1)\alpha+jk_1+k_2+jl_1+l_2}
 \frac{\mathbf{B}(\alpha+l_2,k_2+1)}{\Gamma(\alpha)} \\
&\quad\times \prod_{i=1}^{j}\frac{\mathbf{B}(\alpha+l_1,i\alpha+ik_1+k_2
 +(i-1)l_1+l_2+1)}{\Gamma(\alpha)} .
\end{align*}
Hence,
\begin{align*}
&|\psi(t)-\phi_{i}(t)| \\
&\le \|\phi_0\| M_A^{i+1}t^{(i+1)\alpha+(i+1)k_1+(i+1)l_1}
\prod_{j=0}^i\frac{\mathbf{B}(\alpha+l_1,j\alpha+(j+1)k_1+jl_1+1)}{\Gamma(\alpha)}\\
&\quad+M_A^iM_Ft^{(i+1)\alpha+ik_1+k_2+il_1+l_2}
 \frac{\mathbf{B}(\alpha+l_2,k_2+1)}{\Gamma(\alpha)} \\
&\quad\times \prod_{j=1}^i\frac{\mathbf{B}(\alpha+l_1,j\alpha+jk_1+k_2+(j-1)l_1+l_2+1)}
 {\Gamma(\alpha)}\\
&\le\|\phi_0\| M_A^{i+1}\prod_{j=0}^i
 \frac{\mathbf{B}(\alpha+l_1,j\alpha+(j+1)k_1+jl_1+1)}{\Gamma(\alpha)}\\
&\quad +M_A^iM_F\frac{\mathbf{B}(\alpha+l_2,k_2+1)}{\Gamma(\alpha)}
\prod_{j=1}^i\frac{\mathbf{B}(\alpha+l_1,j\alpha+jk_1+k_2+(j-1)l_1+l_2+1)}
{\Gamma(\alpha)}
\end{align*}
for $i=1,2,\dots$. Similarly we have
\begin{gather*}
\lim_{i\to +\infty}\|\phi_0\| M_A^{i+1}\prod_{j=0}^i
\frac{\mathbf{B}(\alpha+l_1,j\alpha+(j+1)k_1+jl_1+1)}{\Gamma(\alpha)}=0,\\
\begin{aligned}
&\lim_{i\to +\infty}M_A^iM_F\frac{\mathbf{B}(\alpha+l_2,k_2+1)}{\Gamma(\alpha)} \\
&\prod_{j=1}^i\frac{\mathbf{B}(\alpha+l_1,j\alpha+jk_1+k_2+(j-1)l_1+l_2+1)}
{\Gamma(\alpha)}=0.
\end{aligned}
\end{gather*}
Then $\lim_{i\to +\infty}\phi_i(t)=\psi(t)$ uniformly on $[0,1]$.
Then $\phi(t)\equiv\psi(t)$.
Then \eqref{e3.1.5} has a unique solution $\phi$. The proof is complete.
\end{proof}


\begin{theorem} \label{thm3.1.1}
 Suppose that {\rm (3.A1)} holds. Then ${x}$ is a solution of IVP \eqref{e3.1.1}
if and only if $x$ is a solution of the integral equation \eqref{e3.1.5}.
\end{theorem}


\begin{proof}
 Suppose that $x$ is a solution of \eqref{e3.1.1}. Then $\lim_{t\to 0^+}x(t)=\eta$
and $\|x\|=r<+\infty$. From (3.A1), we have for $t\in (0,1)$
\begin{align*}
&\Big|\int_0^t\frac{(t-s)^{\alpha-n}}{\Gamma(\alpha-n+1)}[A(s)x(s)+F(s)]ds\Big| \\
&\le \|x\|\int_0^t\frac{(t-s)^{\alpha-n}}{\Gamma(\alpha-n+1)}|A(s)|ds
+\int_0^t\frac{(t-s)^{\alpha-n}}{\Gamma(\alpha-n+1)}|F(s)|ds\\
&\le \int_0^t\frac{(t-s)^{\alpha-n}}{\Gamma(\alpha-n+1)}[M_Ars^{k_1}(1-s)^{l_1}
+M_Fs^{k_2}(1-s)^{l_2}]ds\\
&\le \int_0^t\frac{(t-s)^{\alpha-n}}{\Gamma(\alpha-n+1)}[M_Ars^{k_1}(1-t)^{l_1}
+M_Fs^{k_2}(1-t)^{l_2}]ds \quad \text{by }\frac{s}{t}=w\\
&=M_Ar(1-t)^{l_1}t^{\alpha+k_1-n+1}\int_0^1
 \frac{(1-w)^{\alpha-n}}{\Gamma(\alpha-n+1)}w^{k_1}dw \\
&\quad +M_F(1-t)^{l_2}t^{\alpha+k_2-n+1}
\int_0^1\frac{(1-w)^{\alpha-n}}{\Gamma(\alpha-n+1)}w^{k_2}dw\\
&=M_Ar(1-t)^{l_1}t^{\alpha+k_1-n+1}
 \frac{\mathbf{B}(\alpha-n+1,k_1+1)}{\Gamma(\alpha)} \\
&\quad  +M_F(1-t)^{l_2}t^{\alpha+k_2-n+1}\frac{\mathbf{B}
(\alpha-n+1,k_2+1)}{\Gamma(\alpha)}.
\end{align*}
So $t\to \int_0^t\frac{(t-s)^{\alpha-n}}{\Gamma(\alpha-n+1)}[A(s)x(s)+F(s)]ds$
 is defined on $(0,1)$. $k_i>-\alpha+n-1$ implies that
\begin{equation}
\lim_{t\to 0^+}\int_0^t\frac{(t-s)^{\alpha-n}}{\Gamma(\alpha-n+1)}A(s)x(s)ds
=\lim_{t\to 0^+}\int_0^t\frac{(t-s)^{\alpha-n}}{\Gamma(\alpha-n+1)}F(s)ds=0.
 \label{e3.1.6}
\end{equation}
 Furthermore, for $t_1,t_2\in (0,1)$ with $t_1<t_2$ we have
\begin{align*}
&\Big|\int_0^{t_1}\frac{(t_1-s)^{\alpha-1}}{\Gamma(\alpha)}[A(s)x(s)
+F(s)]ds-\int_0^{t_2}\frac{(t_2
-s)^{\alpha-1}}{\Gamma(\alpha)}[A(s)x(s)+F(s)]ds\Big|\\
&\le \int_{t_1}^{t_2}\frac{(t_2-s)^{\alpha-1}}{\Gamma(\alpha)}|A(s)x(s)+F(s)|ds \\
&\quad +\int_0^{t_1}\frac{|(t_1-s)^{\alpha-1}-(t_2-s)^{\alpha-1}|}
{\Gamma(\alpha)}|A(s)x(s)+F(s)|ds\\
&\le M_Ar\Big[\int_{t_1}^{t_2}\frac{(t_2-s)^{\alpha-1}}{\Gamma(\alpha)}
 s^{k_1}(1-s)^{l_1}ds \\
&\quad +\int_0^{t_1}\frac{(t_1-s)^{\alpha-1}-(t_2-s)^{\alpha-1}}
{\Gamma(\alpha)}s^{k_1}(1-s)^{l_1}ds\Big]\\
&\quad +M_F\Big[\int_{t_1}^{t_2}\frac{(t_2-s)^{\alpha-1}}{\Gamma(\alpha)}
 s^{k_2}(1-s)^{l_2}ds \\
&\quad +\int_0^{t_1}\frac{(t_1-s)^{\alpha-1}-(t_2-s)^{\alpha-1}}
{\Gamma(\alpha)}s^{k_2}(1-s)^{l_2}ds\Big]\\
&\le M_Ar\Big[\int_{t_1}^{t_2}\frac{(t_2-s)^{\alpha+l_1-1}}
 {\Gamma(\alpha)}s^{k_1}ds \\
&\quad +\int_0^{t_1}\frac{(t_1-s)^{\alpha-1}-(t_2-s)^{\alpha-1}}
 {\Gamma(\alpha)}s^{k_1}(t_2-s)^{l_1}ds\Big]\\
&\quad +M_F\Big[\int_{t_1}^{t_2}\frac{(t_2-s)^{\alpha+l_2-1}}
 {\Gamma(\alpha)}s^{k_2}ds \\
&\quad +\int_0^{t_1}\frac{(t_1-s)^{\alpha-1}-(t_2-s)^{\alpha-1}}
 {\Gamma(\alpha)}s^{k_2}(t_2-s)^{l_2}ds\Big]\\
&=M_Ar\Big[t_2^{\alpha+k_1+l_1}\int_{\frac{t_1}{t_2}}^1
 \frac{(1-w)^{\alpha+l_1-1}}{\Gamma(\alpha)}w^{k_1}dw \\
&\quad +\int_0^{t_1}\frac{(t_1-s)^{\alpha+l_1-1}}
 {\Gamma(\alpha)}s^{k_1}ds-\int_0^{t_1}\frac{(t_2-s)^{\alpha+l_1-1}}
 {\Gamma(\alpha)}s^{k_1}ds\Big]\\
&\quad +M_F\Big[t_2^{\alpha+k_2+l_2}\int_{\frac{t_1}{t_2}}^1
 \frac{(1-w)^{\alpha+l_2-1}}{\Gamma(\alpha)}w^{k_2}dw \\
&\quad +\int_0^{t_1}\frac{(t_1-s)^{\alpha+l_2-1}}{\Gamma(\alpha)}s^{k_2}ds
-\int_0^{t_1}\frac{(t_2-s)^{\alpha+l_2-1}}{\Gamma(\alpha)}s^{k_2}ds\Big]\\
&=M_Ar\Big[t_2^{\alpha+k_1+l_1}\int_{\frac{t_1}{t_2}}^1
 \frac{(1-w)^{\alpha+l_1-1}}{\Gamma(\alpha)}w^{k_1}dw
 +t_1^{\alpha+k_1+l_1}\int_0^1 \frac{(1-w)^{\alpha+l_1-1}}
 {\Gamma(\alpha)}w^{k_1}dw \\
&\quad -t_2^{\alpha+k_1+l_1} \int_0^{\frac{t_1}{t_2}}\frac{(1-w)^{\alpha+l_1-1}}
 {\Gamma(\alpha)}w^{k_1}dw\Big]\\
&\quad +M_F\Big[t_2^{\alpha+k_2+l_2}\int_{\frac{t_1}{t_2}}^1
\frac{(1-w)^{\alpha+l_2-1}}{\Gamma(\alpha)}w^{k_2}dw \\
&\quad +t_1^{\alpha+k_2+l_2}\int_0^1
 \frac{(1-w)^{\alpha+l_2-1}}{\Gamma(\alpha)}w^{k_2}dw-t_2^{\alpha+k_2+l_2}
 \int_0^{\frac{t_1}{t_2}}\frac{(1-w)^{\alpha+l_2-1}}
 {\Gamma(\alpha)}w^{k_2}dw\Big]\\
&=M_Ar\Big[t_2^{\alpha+k_1+l_1}\int_{\frac{t_1}{t_2}}^1
 \frac{(1-w)^{\alpha+l_1-1}}{\Gamma(\alpha)}w^{k_1}dw \\
&\quad +|t_1^{\alpha+k_1+l_1}-t_2^{\alpha+k_1+l_1}
 |\frac{\mathbf{B}(\alpha+l_1,k_1+1)}{\Gamma(\alpha)}-t_2^{\alpha+k_1+l_1}
 \int_{\frac{t_1}{t_2}}^1\frac{(1-w)^{\alpha+l_1-1}} {\Gamma(\alpha)}w^{k_1}dw\Big]\\
&\quad +M_F\Big[t_2^{\alpha+k_2+l_2}\int_{\frac{t_1}{t_2}}^1
\frac{(1-w)^{\alpha+l_2-1}}{\Gamma(\alpha)}w^{k_2}dw \\
&\quad +|t_1^{\alpha+k_2+l_2}-t_2^{\alpha+k_2+l_2}|
 \frac{\mathbf{B}(\alpha+l_2,k_2+1)}{\Gamma(\alpha)}-t_2^{\alpha+k_2+l_2}
 \int_{\frac{t_1}{t_2}}^1\frac{(1-w)^{\alpha+l_2-1}}{\Gamma(\alpha)}w^{k_2}dw\Big]\\
&\to 0 \quad \text{as }t_1\to t_2.
\end{align*}
So $t\to \int_0^t\frac{(t-s)^{\alpha-1}}{\Gamma(\alpha)}[A(s)x(s)+F(s)]ds$
is continuous on $(0,1]$, by defining
$$
\int_0^t\frac{(t-s)^{\alpha-1}}{\Gamma(\alpha)}[A(s)x(s)+F(s)]ds\Big|_{t=0}
=\lim_{t\to 0^+}\int_0^t\frac{(t-s)^{\alpha-1}}{\Gamma(\alpha)}[A(s)x(s)+F(s)]ds.
 $$
We have $I_{0^+}^\alpha {}^CD_{0^+}^\alpha x(t)=I_{0^+}^\alpha [A(t)x(t)+F(t)]$. So
\begin{align*}
&\int_0^t\frac{(t-s)^{\alpha-1}}{\Gamma(\alpha)}[A(s)x(s)+F(s)]ds \\
&=I_{0^+}^\alpha [A(t)x(t)+F(t)]=I_{0^+}^\alpha {}^CD_{0^+}^\alpha x(t)\\
&=\int_0^t\frac{(t-s)^{\alpha-1}}{\Gamma(\alpha)}
\big(\frac{1}{\Gamma(n-\alpha)}\int_0^s(s-w)^{-\alpha}x^{(n)}(w)dw\big)ds
\\ 
&\quad  \text{(interchange the order of integration)}\\
&=\frac{1}{\Gamma(\alpha)\Gamma(n-\alpha)}\int_0^t\int_w^t(t-s)^{\alpha-1}
(s-w)^{n-\alpha-1}ds x^{(n)}(w)dw\quad\text{using  }\frac{s-w}{t-w}=u\\
&=\frac{1}{\Gamma(\alpha)\Gamma(n-\alpha)}\int_0^t(t-u)^{n-1}
\int_0^1(1-u)^{\alpha-1}u^{n-\alpha-1}du x^{(n)}(w)dw \\
&\quad \text{(using $\mathbf{B}(\alpha,1-\alpha)
 =\frac{\Gamma(\alpha)\Gamma(1-\alpha)}{\Gamma(1)}$)}\\
&=-\frac{1}{(n-1)!}\int_0^t(t-u)^{n-1}x^{(n)}(w)dw \\
&=\frac{1}{(n-1)!}\Big[(t-u)^{n-1}x^{(n-1)}(w)
\big|_0^t+(n-1)\int_0^t(t-u)^{n-2}x^{(n-1)}(w)dw\Big]\\
&=-\frac{\eta_{n-1}}{(n-1)!}+\frac{1}{(n-2)!}\int_0^t(t-u)^{n-2}x^{(n-1)}(w)dw\\
&=\dots \\
&=-\sum_{j=1}^{n-1}\frac{\eta_{j}}{j!}+\int_0^tx'(s)ds
=x(t)-\sum_{j=0}^{n-1}\frac{\eta_j}{j!}.
\end{align*}
Then $x\in C(0,1]$ is a solution of \eqref{e3.1.5}.

On the other hand, if $x$ is a solution of \eqref{e3.1.5}. From
 Cases 1, 2 and 3, we have $x\in C(0,1]$ and
$\lim_{t\to 0^+}{x}^{(j)}(t)={\eta}_j(j\in N[0,n-1])$.
 So $x\in C[0,1]$. Furthermore, from \eqref{e3.1.6} we have
\begin{align*}
&{}^CD_{0^+}^\alpha x(t) \\
&=\frac{1}{\Gamma(n-\alpha)}\int_0^t(t-s)^{n-\alpha-1}x^{(n)}(s)ds\\
&=\frac{1}{\Gamma(n-\alpha)}\int_0^t(t-s)^{n-\alpha-1}
\Big(\sum_{j=0}^{n-1}\frac{\eta_js^j}{j!} \\
&\quad +\int_0^s\frac{(s-w)^{\alpha-1}}{\Gamma(\alpha)}
[A(w)x(w)+F(w)]dw\Big)^{(n)}ds \\
&=\frac{1}{\Gamma(n-\alpha)}\int_0^t(t-s)^{n-\alpha-1}
\Big(\int_0^s\frac{(s-w)^{\alpha-1}}{\Gamma(\alpha)}[A(w)x(w)+F(w)]dw\Big)^{(n)}ds\\
&=\frac{1}{\Gamma(n-\alpha)}\int_0^t(t-s)^{n-\alpha-1}
\frac{1}{\Gamma(\alpha-n+1)} \\
&\quad\times \Big(\int_0^s(s-w)^{\alpha-n}[A(w)x(w)+F(w)]dw\Big)'ds\\
&=\frac{1}{\Gamma(\alpha-n+1)}\frac{1}{\Gamma(n+1-\alpha)}
\Big[\int_0^t(t-s)^{n-\alpha} \\
&\quad\times \Big(\int_0^s(s-w)^{\alpha-n}[A(w)x(w)+F(w)]dw\Big)'ds\Big]'\\
&=\frac{1}{\Gamma(\alpha-n+1)}\frac{1}{\Gamma(n+1-\alpha)}
\Big[(t-s)^{n-\alpha}\int_0^s(s-w)^{\alpha-n}[A(w)x(w)+F(w)]dw|_0^t \\
&\quad +(n-\alpha)\int_0^t(t-s)^{n-\alpha-1}\int_0^s(s-w)^{\alpha-n}[A(w)x(w)+F(w)]
\,dw\,ds\Big]'\\
&=\frac{1}{\Gamma(\alpha-n+1)}\frac{1}{\Gamma(n-\alpha)}
\Big[\int_0^t(t-s)^{n-\alpha-1} \\
&\quad\times \int_0^s(s-w)^{\alpha-n}
[A(w)x(w)+F(w)]\,dw\,ds\Big]'\quad \text{by \eqref{e3.1.6}}\\
&=\frac{1}{\Gamma(\alpha-n+1)}\frac{1}{\Gamma(n-\alpha)}
\frac{1}{\Gamma(\alpha)}\Big[\int_0^t\int_{w}^t(t-s)^{n-\alpha-1}(s-w)^{\alpha-n}ds \\
&\quad\times [A(w)x(w)+F(w)] dw\Big]'\\
&\quad \text{by changing the order of integration}\\
&=\frac{1}{\Gamma(\alpha-n+1)}\frac{1}{\Gamma(n-\alpha)}
\Big[\int_0^t\int_0^1(1-u)^{n-\alpha-1}u^{\alpha-n}du[A(w)x(w)+F(w)]dw\Big]' \\
&\quad\text{(because $\frac{s-w}{t-w}=u$)}\\
&=\Big[\int_0^t[A(w)x(w)+F(w)]dw\Big]'
\end{align*}
by using $\mathbf{B}(n-\alpha,\alpha-n+1)=\Gamma(n-\alpha)\Gamma(\alpha-n+1)
=A(t)x(t)+F(t)$ in the last equality.
So $x\in C[0,1]$ is a solution of \eqref{e3.1.1}. The proof is complete.
\end{proof}


\begin{theorem} \label{thm3.1.2}
 Suppose that {\rm (3.A1)} holds. Then \eqref{e3.1.1} has a unique solution.
If there exists constants $k_2>-\alpha+n-1$, $l_2\le 0$ with
$l_2>\max\{-\alpha,-\alpha-k_2\}$, $M_F\ge 0$ such that
 $|F(t)|\le M_Ft^{k_2}(1-t)^{l_2}$ for all $t\in (0,1)$, then the following
problem
\begin{equation}
 \begin{gathered}
{}^{C}D_{0^+}^{\alpha}{x}(t)=\lambda {x}(t)+{F}(t),\quad\text{a.e., }
t\in (0,1],\\
\lim_{t\to 0^+}{x}^{(j)}(t)={\eta}_j,\quad j\in \mathbb{N}[0,n-1]
\end{gathered}\label{e3.1.7}
\end{equation}
has a unique solution
\begin{equation}
x(t)=\sum_{j=0}^{n-1}\eta_j\mathbf{E}_{\alpha,j+1}(\lambda t^\alpha)t^j
+\int_0^t(t-s)^{\alpha-1}\mathbf{E}_{\alpha,\alpha}(\lambda(t-s)^\alpha)F(s)ds,
\quad t\in (0,1].\label{e3.1.8}
\end{equation}
\end{theorem}


\begin{proof} (i) From Claims 1, 2 and 3, Theorem \ref{thm3.1.1} implies
that \eqref{e3.1.1} has a unique solution.

 (ii) From the assumption and $A(t)\equiv\lambda$, it is easy to see that (3.A1)
holds with $k_1=l_1=0$ and $k_2$, $l_2$ mentioned. Thus \eqref{e3.1.7} has
a unique solution. From the Picard function sequence we have
\begin{align*}
&\phi_i(t) \\
&=\sum_{j=0}^{n-1}\frac{\eta_jt^j}{j!}
 +\lambda\int_0^t\frac{(t-s)^{\alpha-1}}{\Gamma(\alpha)}\phi_{i-1}(s)ds
 +\int_0^t\frac{(t-s)^{\alpha-1}}{\Gamma(\alpha)}F(s)ds\\
&=\sum_{j=0}^{n-1}\frac{\eta_jt^j}{j!}+\lambda\int_0^t
 \frac{(t-s)^{\alpha-1}}{\Gamma(\alpha)}
\Big(\sum_{j=0}^{n-1}\frac{\eta_js^j}{j!}+\lambda\int_0^s
 \frac{(s-u)^{\alpha-1}}{\Gamma(\alpha)} \phi_{i-2}(u)du \\
&\quad +\int_0^s\frac{(s-u)^{\alpha-1}}{\Gamma(\alpha)}F(u)du\Big)ds
 +\int_0^t\frac{(t-s)^{\alpha-1}}{\Gamma(\alpha)}F(s)ds\\
&=\sum_{j=0}^{n-1}\frac{\eta_jt^j}{j!}+\lambda\sum_{j=0}^{n-1}
 \frac{\eta_j}{\Gamma(\alpha)j!}\int_0^t(t-s)^{\alpha-1}s^jds \\
&\quad +\lambda^2\int_0^t\frac{(t-s)^{\alpha-1}}{\Gamma(\alpha)}
 \int_0^s\frac{(s-u)^{\alpha-1}}{\Gamma(\alpha)}\phi_{i-2}(u)\,du\,ds\\
&\quad +\lambda\int_0^t\frac{(t-s)^{\alpha-1}}{\Gamma(\alpha)}
 \int_0^s\frac{(s-u)^{\alpha-1}}{\Gamma(\alpha)}F(u)\,du\,ds
+\int_0^t\frac{(t-s)^{\alpha-1}}{\Gamma(\alpha)}F(s)ds\\
&=\sum_{j=0}^{n-1}\frac{\eta_jt^j}{j!}+\lambda\sum_{j=0}^{n-1}
 \frac{\eta_j}{\Gamma(\alpha)j!}t^{\alpha+j}\int_0^1(1-w)^{\alpha-1}w^jdw \\
&\quad +\frac{\lambda^2}{\Gamma(\alpha)^2}\int_0^t\int_u^t(t-s)^{\alpha-1}
 (s-u)^{\alpha-1}ds\phi_{i-2}(u)du\\
&\quad +\frac{\lambda}{\Gamma(\alpha)^2}\int_0^t\int_u^t(t-s)^{\alpha-1}
 (s-u)^{\alpha-1}dsF(u)du \\
&\quad +\int_0^t\frac{(t-s)^{\alpha-1}}{\Gamma(\alpha)}F(s)ds\\
&=\sum_{j=0}^{n-1}\frac{\eta_jt^j}{j!}+\sum_{j=0}^{n-1}
 \frac{\lambda\eta_jt^{\alpha+j}}{\Gamma(\alpha+j+1)}
 +\frac{\lambda^2}{\Gamma(2\alpha)}\int_0^t(t-u)^{2\alpha-1}\phi_{i-2}(u)du\\
&\quad +\frac{\lambda}{\Gamma(2\alpha)}\int_0^t(t-u)^{2\alpha-1}F(u)du
+\int_0^t\frac{(t-s)^{\alpha-1}}{\Gamma(\alpha)}F(s)ds\\
&=\sum_{j=0}^{n-1}\eta_jt^j(\frac{1}{\Gamma(0\alpha+j+1)}
 +\frac{\lambda t^{\alpha}}{\Gamma(\alpha+j+1)})
+\frac{\lambda^2}{\Gamma(2\alpha)}\int_0^t(t-u)^{2\alpha-1}\phi_{i-2}(u)du\\
&\quad +\int_0^t(t-s)^{\alpha-1}(\frac{\lambda(t-s)^\alpha}{\Gamma(2\alpha)}
 +\frac{1}{\Gamma(\alpha)})F(s)ds\\
&=\dots\\
&=\sum_{j=0}^{n-1}\eta_jt^j(\sum_{v=0}^{i-1}
 \frac{\lambda^vt^{\alpha v}}{\Gamma(v\alpha+j+1)})
+\frac{\lambda^i}{\Gamma(i\alpha)}\int_0^t(t-u)^{i\alpha-1}\phi_0(u)du\\
&\quad +\int_0^t(t-s)^{\alpha-1}(\sum_{v=0}^i
 \frac{\lambda^v(t-s)^{\alpha v}}{\Gamma((v+1)\alpha)})F(s)ds\\
&=\sum_{j=0}^{n-1}\eta_jt^j(\sum_{v=0}^{i-1}
 \frac{\lambda^vt^{\alpha v}}{\Gamma(v\alpha+j+1)})
+\frac{\lambda^i}{\Gamma(i\alpha+j+1)}\sum_{j=0}^{n-1}\eta_jt^{\alpha n+j} \\
&\quad +\int_0^t(t-s)^{\alpha-1}(\sum_{v=0}^{i-1}
 \frac{\lambda^v(t-s)^{\alpha v}}{\Gamma((v+1)\alpha)})F(s)ds \\
&=\sum_{j=0}^{n-1}\eta_jt^j\Big(\sum_{v=0}^i
 \frac{\lambda^vt^{\alpha v}}{\Gamma(v\alpha+j+1)}\Big)
+\int_0^t(t-s)^{\alpha-1}\Big(\sum_{v=0}^{i-1}
 \frac{\lambda^v(t-s)^{\alpha v}}{\Gamma((v+1)\alpha)}\Big)F(s)ds\\
&\to \sum_{j=0}^{n-1}\eta_jt^j\mathbf{E}_{\alpha,j+1}
 (\lambda t^\alpha)+\int_0^t(t-s)^{\alpha-1}\mathbf{E}_{\alpha,\alpha}
(\lambda(t-s)^\alpha)F(s)ds,
\end{align*}
as $m\to +\infty$.
Then $x(t)=\lim_{i\to +\infty}\phi_i(t)$ is a unique solution of \eqref{e3.1.7}.
So $x$ satisfies \eqref{e3.1.8}. The proof is complete.
\end{proof}


To obtain solutions of \eqref{e3.1.2}, we need the following assumption:
\begin{itemize}
\item[(3.A2)] There exist constants $k_i>-1$, $l_i\le 0$ with
$l_1>\max\{-\alpha,-\alpha-k_1\}$, $l_2>\max\{-\alpha,-n-k_2\}$, $M_A\ge 0$
and $M_F\ge 0$ such that $|A(t)|\le M_At^{k_1}(1-t)^{l_1}$ and
$|F(t)|\le M_Ft^{k_2}(1-t)^{l_2}$ for all $t\in (0,1)$.
\end{itemize}
We choose Picard function sequence as
\begin{gather*}
\phi_0(t)=\sum_{v=1}^n\frac{\eta_v}{\Gamma(\alpha-v+1)}t^{\alpha-v},\quad
t\in (0,1],\\
\phi_i(t)=\sum_{v=1}^n\frac{\eta_v}{\Gamma(\alpha-v+1)}t^{\alpha-v}
+\int_0^t\frac{(t-s)^{\alpha-1}}{\Gamma(\alpha)}[A(s)\phi_{i-1}(s)
+F(s)]ds,
\end{gather*}
for $t\in (0,1]$, $i=1,2,\dots$.
\smallskip

\noindent\textbf{Claim 1.} $\phi_i\in C_{n-\alpha}[0,1]$.

\begin{proof}
It is easy to see that $\phi_0\in C_{n-\alpha}[0,1]$. We have
\begin{align*}
& t^{n-\alpha}\Big|\int_0^t\frac{(t-s)^{\alpha-1}}{\Gamma(\alpha)}[A(s)
\phi_0(s)+F(s)]ds\Big|\\
& \le t^{n-\alpha}\int_0^t\frac{(t-s)^{\alpha-1}}
 {\Gamma(\alpha)}s^{k_1}(1-s)^{l_1}s^{\alpha-n}[s^{n-\alpha}|\phi_{n-1}(s)|]ds \\
&\quad + M_Ft^{n-\alpha}\int_0^t\frac{(t-s)^{\alpha-1}}
 {\Gamma(\alpha)}s^{k_1}(1-s)^{l_1}\\
& \le t^{n-\alpha}\|\phi_0\|\int_0^t\frac{(t-s)^{\alpha+l_1-1}}
 {\Gamma(\alpha)}s^{\alpha+k_1-n}ds
 + M_Ft^{n-\alpha}\int_0^t\frac{(t-s)^{\alpha+l_2-1}}{\Gamma(\alpha)}s^{k_2}ds\\
& =\|\phi_0\|t^{\alpha+k_1+l_1}\frac{\mathbf{B}(\alpha+l_1,\alpha+k_1-n+1)}
 {\Gamma(\alpha)} +M_Ft^{n+k_2+l_2}\frac{\mathbf{B}(\alpha+l_2,k_2+1)}
 {\Gamma(\alpha)}.
\end{align*}
Then $t\to \int_0^t\frac{(t-s)^{\alpha-1}}{\Gamma(\alpha)}[A(s)\phi_{n-1}(s)
+F(s)]ds$ is convergent on $(0,1]$ and
\[
\lim_{t\to 0^+}t^{n-\alpha}\int_0^t\frac{(t-s)^{\alpha-1}}{\Gamma(\alpha)}
[A(s)\phi_{n-1}(s)+F(s)]ds=0.
\]
 We see that $\phi_1\in C_{n-\alpha}[0,1]$. By mathematical induction, we can
prove that $\phi_n\in C_{n-\alpha}[0,1]$.
\end{proof}
\smallskip

\noindent\textbf{Claim 2.}
 $\{t\to t^{n-\alpha}\phi_i(t)\}$ converges uniformly on $[0,1]$.


\begin{proof} As in Case 1, for $t\in [0,1]$ we have
\begin{align*}
&t^{n-\alpha}|\phi_1(t)-\phi_0(t)| \\
&=\Big|\int_0^t\frac{(t-s)^{\alpha-1}}{\Gamma(\alpha)}[A(s)\phi_0(s)+F(s)]ds\Big|\\
&\le\|\phi_0\|t^{\alpha+k_1+l_1}\frac{\mathbf{B}(\alpha+l_1,\alpha+k_1-n+1)}
{\Gamma(\alpha)} +t^{n+k_2+l_2}\frac{\mathbf{B}(\alpha+l_2,k_2+1)}{\Gamma(\alpha)}.
\end{align*}
So
\begin{align*}
&t^{n-\alpha}|\phi_2(t)-\phi_1(t)| \\
&=|\int_0^t\frac{(t-s)^{\alpha-1}}{\Gamma(\alpha)}A(s)[\phi_1(s)-\phi_0(s)]ds|\\
&\le t^{n-\alpha}\int_0^t\frac{(t-s)^{\alpha-1}}{\Gamma(\alpha)}M_As^{k_1}
(1-s)^{l_1}s^{\alpha-n} \\
&\quad\times \Big(\|\phi_0\|s^{\alpha+k_1+l_1}
\frac{\mathbf{B}(\alpha+l_1,\alpha+k_1-n+1)}{\Gamma(\alpha)}
+s^{n+k_2+l_2}\frac{\mathbf{B}(\alpha+l_2,k_2+1)}{\Gamma(\alpha)}\Big)ds\\
&\le M_A\|\phi_0\|t^{n-\alpha}\int_0^t
\frac{(t-s)^{\alpha+l_1-1}}{\Gamma(\alpha)}s^{2\alpha-n+2k_1+l_1}ds
 \frac{\mathbf{B}(\alpha+l_1,\alpha+k_1-n+1)}{\Gamma(\alpha)} \\
&\quad +M_AM_Ft^{n-\alpha}\int_0^t\frac{(t-s)^{\alpha+l_1-1}}
{\Gamma(\alpha)}s^{\alpha+k_1+k_2+l_2}ds \
\frac{\mathbf{B}(\alpha+l_2,k_2+1)}{\Gamma(\alpha)}\\
&=M_A\|\phi_0\|t^{2\alpha+2k_1+2l_1}
\frac{\mathbf{B}(\alpha+l_1,2\alpha-n+2k_1+l_1+1)}
{\Gamma(\alpha)}\frac{\mathbf{B}(\alpha+l_1,\alpha+k_1-n+1)}{\Gamma(\alpha)} \\
&\quad +M_AM_Ft^{\alpha+n+k_1+l_1+k_2+l_2}\frac{\mathbf{B}
(\alpha+l_1,\alpha+k_1+k_2+l_2+1)}{\Gamma(\alpha)}
\frac{\mathbf{B}(\alpha+l_2,k_2+1)}{\Gamma(\alpha)}.
\end{align*}
Furthermore,
\begin{align*}
& t^{n-\alpha}|\phi_3(t)-\phi_2(t)| \\
&=\Big|\int_0^t\frac{(t-s)^{\alpha-1}}{\Gamma(\alpha)}A(s)[\phi_2(s)
 -\phi_1(s)]ds\Big|\\
&\le t^{n-\alpha}\int_0^t
\frac{(t-s)^{\alpha-1}}{\Gamma(\alpha)}M_As^{k_1}(1-s)^{l_1}s^{\alpha-n}
\Big(M_A\|\phi_0\|s^{2\alpha+2k_1+2l_1} \\
&\quad \times \frac{\mathbf{B}(\alpha+l_1,2\alpha-n+2k_1+l_1+1)}{\Gamma(\alpha)}
\frac{\mathbf{B}(\alpha+l_1,\alpha+k_1-n+1)}{\Gamma(\alpha)} \\
&\quad +M_AM_Fs^{\alpha+n+k_1+l_1+k_2+l_2}
\frac{\mathbf{B}(\alpha+l_1,\alpha+k_1+k_2+l_2+1)}{\Gamma(\alpha)}
 \frac{\mathbf{B}(\alpha+l_2,k_2+1)}{\Gamma(\alpha)}\Big)ds\\
&\le M_A^2\|\phi_0\|t^{n-\alpha}\int_0^t
 \frac{(t-s)^{\alpha+l_1-1}}{\Gamma(\alpha)}s^{3\alpha-n+3k_1+2l_1}ds \\
&\quad\times  \frac{\mathbf{B}(\alpha+l_1,2\alpha-n+2k_1+l_1+1)}{\Gamma(\alpha)}
 \frac{\mathbf{B}(\alpha+l_1,\alpha+k_1-n+1)}{\Gamma(\alpha)}\\
&\quad +M_A^2M_Ft^{n-\alpha}\int_0^t\frac{(t-s)^{\alpha+l_1-1}}
{\Gamma(\alpha)}s^{2\alpha+2k_1+l_1+k_2+l_2}ds \\
&\quad\times \frac{\mathbf{B}(\alpha+l_1,\alpha+k_1+k_2+l_2+1)}{\Gamma(\alpha)}
\frac{\mathbf{B}(\alpha+l_2,k_2+1)}{\Gamma(\alpha)}\\
&=M_A^2\|\phi_0\|t^{3\alpha+3k_1+3l_1}
 \frac{\mathbf{B}(\alpha+l_1,3\alpha-n+3k_1+2l_1+1)}{\Gamma(\alpha)} \\
&\quad\times  \frac{\mathbf{B}(\alpha+l_1,2\alpha-n+2k_1+l_1+1)}{\Gamma(\alpha)}
 \frac{\mathbf{B}(\alpha+l_1,\alpha+k_1-n+1)}{\Gamma(\alpha)}\\
&\quad +M_A^2M_Ft^{2\alpha+n+2k_1+2l_1+k_2+l_2}
 \frac{\mathbf{B}(\alpha+l_1,2\alpha+2k_1+l_1+k_2+l_2+1)}{\Gamma(\alpha)} \\
&\quad\times \frac{\mathbf{B}(\alpha+l_1,\alpha+k_1+k_2+l_2+1)}{\Gamma(\alpha)}
 \frac{\mathbf{B}(\alpha+l_2,k_2+1)}{\Gamma(\alpha)}.
\end{align*}
Similarly by the mathematical induction, for every $i=1,2,\dots$ we obtain
\begin{align*}
&t^{n-\alpha}|\phi_i(t)-\phi_{i-1}(t)|\\
&\le M_A^i\|\phi_0\|t^{i\alpha+ik_1+il_1}
 \frac{\mathbf{B}(\alpha+l_1,\alpha+k_1-n+1)}{\Gamma(\alpha)} \\
&\quad\times \prod_{j=0}^{i-2}\frac{\mathbf{B}(\alpha+l_1,(j+1)\alpha+(j+1)k_1+jl_1-n+1)}
 {\Gamma(\alpha)}\\
&\quad +M_A^{m-1}M_Ft^{(i-1)\alpha+n+(i-1)k_1+(i-1)l_1+k_2+l_2}
 \frac{\mathbf{B}(\alpha+l_2,k_2+1)}{\Gamma(\alpha)} \\
&\quad\times \prod_{j=0}^{i-2}\frac{\mathbf{B}(\alpha+l_1,(j+1)\alpha
 +(j+1)k_1+jl_1+k_2+l_2+1)}{\Gamma(\alpha)}\\
&\le M_A^i\|\phi_0\|\frac{\mathbf{B}(\alpha+l_1,\alpha+k_1-n+1)}{\Gamma(\alpha)} \\
&\quad\times  \prod_{j=0}^{i-2}\frac{\mathbf{B}(\alpha+l_1,(j+1)\alpha+(j+1)k_1+jl_1-n+1)}
 {\Gamma(\alpha)}\\
&\quad +M_A^{i-1}M_F\frac{\mathbf{B}(\alpha+l_2,k_2+1)}{\Gamma(\alpha)} \\
&\quad\times \prod_{j=0}^{i-2}\frac{\mathbf{B}(\alpha+l_1,(j+1)\alpha+(j+1)k_1+jl_1+k_2+l_2+1)}
{\Gamma(\alpha)}, \quad t\in [0,1].
\end{align*}
Similarly we can prove that both
\begin{align*}
\sum_{i=1}^{+\infty}u_i&=\sum_{i=1}^{+\infty} M_A^i\|\phi_0\|
\frac{\mathbf{B}(\alpha+l_1,\alpha+k_1-n+1)}{\Gamma(\alpha)} \\
&\quad\times \prod_{j=0}^{i-2}\frac{\mathbf{B}(\alpha+l_1,(j+1)\alpha+(j+1)k_1+jl_1-n+1)}
{\Gamma(\alpha)},\\
\sum_{i=1}^{+\infty}v_i&=\sum_{i=1}^{+\infty}M_A^{i-1}M_F
 \frac{\mathbf{B}(\alpha+l_2,k_2+1)}{\Gamma(\alpha)}\\
&\quad\times \prod_{j=0}^{i-2}\frac{\mathbf{B}(\alpha+l_1,(j+1)\alpha+(j+1)k_1+jl_1+k_2+l_2+1)}
{\Gamma(\alpha)}
\end{align*}
are convergent. Hence,
$$
t^{n-\alpha}\phi_0(t)+t^{n-\alpha}[\phi_1(t)-\phi_0(t)]+t^{n-\alpha}
[\phi_2(t)-\phi_1(t)]+\dots+t^{n-\alpha}[\phi_i(t)-\phi_{i-1}(t)]+\dots,
$$
for $t\in [0,1]$, is uniformly convergent. Then $\{t\to t^{n-\alpha}\phi_i(t)\}$
is convergent uniformly on $(0,1]$.
\end{proof}
\smallskip


\noindent\textbf{Claim 3.}
 $\phi(t)=t^{\alpha-n}\lim_{i\to +\infty}t^{n-\alpha}\phi_i(t)$
defined on $(0,1]$ is a unique continuous solution of the integral equation
\begin{equation}
{x}(t)=\sum_{v=1}^n\frac{\eta_v}{\Gamma(\alpha-v+1)}t^{\alpha-v}
+\int_0^t\frac{(t-s)^{\alpha-1}}{\Gamma(\alpha)}[A(s)x(s)+F(s)]ds, \quad t\in (0,1].
\label{e3.1.9}
\end{equation}

\begin{proof}
By $\lim_{i\to +\infty}t^{n-\alpha}\phi_i(t)=t^{n-\alpha}\phi(t)$ and the
uniformly convergence, we see $\phi(t)$ is continuous on $(0,1]$. From
\begin{align*}
&t^{n-\alpha}\Big|\int_0^t\frac{(t-s)^{\alpha-1}}{\Gamma(\alpha)}[A(s)
\phi_{p-1}(s)+F(s)]ds \\
&\-\int_0^t\frac{(t-s)^{\alpha-1}}{\Gamma(\alpha)}[A(s)
\phi_{q-1}(s)+F(s)]ds\Big|\\
& \le M_A\|\phi_{p-1}-\phi_{q-1}\|t^{n-\alpha}\int_0^t\frac{(t-s)^{\alpha-1}}
 {\Gamma(\alpha)}s^{k_1}(1-s)^{l_1}s^{\alpha-n}ds\\
&\le M_A\|\phi_{p-1}-\phi_{q-1}\|t^{n-\alpha}\int_0^t
 \frac{(t-s)^{\alpha+l_1-1}}{\Gamma(\alpha)}s^{\alpha+k_1-n}ds\\
&\le M_A\|\phi_{p-1}-\phi_{q-1}\|t^{\alpha+k_1+l_1}
 \frac{\mathbf{B}(\alpha+l_1,\alpha+k_1-n+1)}{\Gamma(\alpha)}\\
&\le M_A\|\phi_{p-1}-\phi_{q-1}\|
 \frac{\mathbf{B}(\alpha+l_1,\alpha+k_1-n+1)}{\Gamma(\alpha)} \\
&\to 0 \quad \text{uniformly as }p,q\to +\infty,
\end{align*}
we know that
\begin{align*}
\phi(t)&=t^{\alpha-n}\lim_{i\to \infty}t^{n-\alpha}\phi_i(t)\\
&=t^{\alpha-n}\lim_{i\to +\infty}\Big[t^{n-\alpha}\sum_{v=1}^n
\frac{\eta_v}{\Gamma(\alpha-v+1)}t^{\alpha-v} \\
&\quad +t^{n-\alpha}\int_0^t\frac{(t-s)^{\alpha-1}}
{\Gamma(\alpha)}[A(s)\phi_{i-1}(s)+F(s)]ds\Big]\\
&=\sum_{v=1}^n\frac{\eta_v}{\Gamma(\alpha-v+1)}t^{\alpha-v}
+\lim_{i\to +\infty}\int_0^t\frac{(t-s)^{\alpha-1}}
{\Gamma(\alpha)}[A(s)\phi_{i-1}(s)+F(s)]ds\\
&=\sum_{v=1}^n\frac{\eta_v}{\Gamma(\alpha-v+1)}t^{\alpha-v}
+\int_0^t\frac{(t-s)^{\alpha-1}}{\Gamma(\alpha)} [A(s)\phi(s)+F(s)]ds.
\end{align*}
Then $\phi$ is a continuous solution of \eqref{e3.1.9} defined on $(0,1]$.

Suppose that $\psi$ defined on $(0,1]$ is also a solution of \eqref{e3.1.9}. Then
$$
\psi(t)=\sum_{v=1}^n\frac{\eta_v}{\Gamma(\alpha-v+1)}t^{\alpha-v}
+\int_0^t\frac{(t-s)^{\alpha-1}}{\Gamma(\alpha)}[A(s)\psi(s)+F(s)]ds, \quad
t\in [0,1].
$$
We need to prove that $\phi(t)\equiv\psi(t)$ on $(0,1]$.
Then
\begin{align*}
&t^{n-\alpha}|\psi(t)-\phi_0(t)| \\
&=t^{n-\alpha}\big|\int_0^t\frac{(t-s)^{\alpha-1}}{\Gamma(\alpha)}
|A(s)\psi(s)+F(s)|ds\big|\\
&\le \|\psi\|t^{\alpha+k_1+l_1}
\frac{\mathbf{B}(\alpha+l_1,\alpha+k_1-n+1)}{\Gamma(\alpha)}
+t^{n+k_2+l_2}\frac{\mathbf{B}(\alpha+l_2,k_2+1)}{\Gamma(\alpha)}.
\end{align*}
Furthermore, we have
\begin{align*}
&t^{n-\alpha}|\psi(t)-\phi_1(t)| \\
&=t^{n-\alpha}\big|\int_0^t\frac{(t-s)^{\alpha-1}}{\Gamma(\alpha)}A(s)
[\psi(s)-\phi_0(s)]ds\big|\\
&\le M_A\|\phi_0\|t^{2\alpha+2k_1+2l_1}
\frac{\mathbf{B}(\alpha+l_1,2\alpha-n+2k_1+l_1+1)}{\Gamma(\alpha)}
\frac{\mathbf{B}(\alpha+l_1,\alpha+k_1-n+1)}{\Gamma(\alpha)} \\
&\quad +M_AM_Ft^{\alpha+n+k_1+l_1+k_2+l_2}
 \frac{\mathbf{B}(\alpha+l_1,\alpha+k_1+k_2+l_2+1)}{\Gamma(\alpha)}
 \frac{\mathbf{B}(\alpha+l_2,k_2+1)}{\Gamma(\alpha)}.
\end{align*}
Using mathematical induction, we have
\begin{align*}
& t^{n-\alpha}|\psi(t)-\phi_{i-1}(t)|\\
&=t^{n-\alpha}\big|\int_0^t\frac{(t-s)^{\alpha-1}}{\Gamma(\alpha)}
A(s)[\psi(s)-\phi_{i-2}(s)]ds\big|\\
&\le M_A^i\|\phi_0\|t^{i\alpha+ik_1+il_1}
 \frac{\mathbf{B}(\alpha+l_1,\alpha+k_1-n+1)}{\Gamma(\alpha)}\\
&\quad\times \prod_{j=0}^{i-2}\frac{\mathbf{B}(\alpha+l_1,(j+1)\alpha+(j+1)k_1+jl_1-n+1)}
 {\Gamma(\alpha)}\\
&\quad +M_A^{m-1}M_Ft^{(i-1)\alpha+n+(i-1)k_1+(i-1)l_1+k_2+l_2}
 \frac{\mathbf{B}(\alpha+l_2,k_2+1)}{\Gamma(\alpha)} \\
&\quad\times \prod_{j=0}^{i-2}\frac{\mathbf{B}(\alpha+l_1,(j+1)\alpha+(j+1)k_1+jl_1+k_2+l_2+1)}
 {\Gamma(\alpha)}\\
&\le M_A^i\|\phi_0\|\frac{\mathbf{B}(\alpha+l_1,\alpha+k_1-n+1)}{\Gamma(\alpha)}\\
&\quad\times \prod_{j=0}^{i-2}\frac{\mathbf{B}(\alpha+l_1,(j+1)\alpha+(j+1)k_1+jl_1-n+1)}
 {\Gamma(\alpha)}\\
&\quad +M_A^{i-1}M_F\frac{\mathbf{B}(\alpha+l_2,k_2+1)}{\Gamma(\alpha)} \\
&\quad\times \prod_{j=0}^{i-2}\frac{\mathbf{B}(\alpha+l_1,(j+1)\alpha+(j+1)k_1+jl_1+k_2+l_2+1)}
 {\Gamma(\alpha)},t\in [0,1].
\end{align*}
Hence,
\begin{align*}
t^{n-\alpha}|\psi(t)-\phi_{i-1}(t)|
&\le M_A^i\|\phi_0\|\frac{\mathbf{B}(\alpha+l_1,\alpha+k_1-n+1)}
{\Gamma(\alpha)} \\
&\quad\times \prod_{j=0}^{i-2}
\frac{\mathbf{B}(\alpha+l_1,(j+1)\alpha+(j+1)k_1+jl_1-n+1)}{\Gamma(\alpha)}\\
&\quad +M_A^{i-1}M_F\frac{\mathbf{B}(\alpha+l_2,k_2+1)}{\Gamma(\alpha)} \\
&\quad\times \prod_{j=0}^{i-2}\frac{\mathbf{B}(\alpha+l_1,(j+1)\alpha+(j+1)k_1+jl_1+k_2+l_2+1)}
{\Gamma(\alpha)},
\end{align*}
 for $i=1,2,\dots$.
Similarly we have $\lim_{i\to +\infty}t^{n-\alpha}\phi_i(t)=t^{n-\alpha}\psi(t)$
uniformly on $(0,1]$. Then $\phi(t)\equiv\psi(t)$ on $(0,1]$.
Then \eqref{e3.1.9} has a unique solution $\phi$. The proof is complete.
\end{proof}


\begin{theorem} \label{thm3.1.3}
 Suppose that {\rm (3.A2)} holds. Then ${x}\in C_{n-\alpha}(0,1]$ is a solution
of IVP \eqref{e3.1.2} if and only if $x\in C_{n-\alpha}(0,1]$ is a solution of
the integral equation \eqref{e3.1.9}.
\end{theorem}

\begin{proof}
Suppose that $x\in C_{n-\alpha}(0,1]$ is a solution of \eqref{e3.1.2}.
Then $t\to t^{n-\alpha}x(t)$is continuous on $(0,1]$ by defining
$t^{n-\alpha}x(t)|_{t=0}=\lim_{t\to 0^+}t^{n-\alpha}x(t)$ and $\|x\|=r<+\infty$.
So from $\frac{w}{s}=u$, we obtain
\begin{equation}
\begin{aligned}
&\lim_{s\to 0^+}\int_0^s(s-w)^{n-\alpha-1}x(w)dw \\
&=\lim_{s\to 0^+}\int_0^s(s-w)^{n-\alpha-1}w^{\alpha-n}w^{n-\alpha}x(w)dw\\
&=\lim_{s\to 0^+}\xi^{n-\alpha}x(\xi)\int_0^s(s-w)^{n-\alpha-1}w^{\alpha-n}dw \\
&\quad\text{ by mean value theorem  with }\xi \in (0,s)\\
&=\lim_{s\to 0^+}\xi^{n-\alpha}x(\xi)\int_0^1(1-u)^{n-\alpha-1}u^{\alpha-n}du\\
&=\frac{\eta_n}{\Gamma(\alpha-n+1)} \mathbf{B}(n-\alpha,\alpha-n+1).
\end{aligned}\label{e3.1.10}
\end{equation}
 From (3.A2), we have similarly to Case 1 that
\begin{align*}
&t^{n-\alpha}\big|\int_0^t\frac{(t-s)^{\alpha-1}}{\Gamma(\alpha)}[A(s)x(s)
+F(s)]ds\big|\\
&=t^{n-\alpha}|\int_0^t\frac{(t-s)^{\alpha-1}}{\Gamma(\alpha)}
[A(s)s^{\alpha-n}s^{n-\alpha}x(s)+F(s)]ds|\\
& \le t^{n-\alpha}\int_0^t\frac{(t-s)^{\alpha-1}}{\Gamma(\alpha)}
 [M_Ars^{\alpha-n}s^{k_1}(1-s)^{l_1}+M_Fs^{k_2}(1-s)^{l_2}]ds\\
&\le rM_At^{\alpha+k_1+l_1}\frac{\mathbf{B}(\alpha+l_1,\alpha+k_1-n+1)}
{\Gamma(\alpha)} +M_Ft^{n+k_2+l_2}\frac{\mathbf{B}(\alpha+l_2,k_2+1)}
{\Gamma(\alpha)}.
\end{align*}
So $t\to t^{n-\alpha}\int_0^t\frac{(t-s)^{\alpha-1}}{\Gamma(\alpha)}
[A(s)x(s)+F(s)]ds$ is defined on $(0,1]$ and
\begin{equation}
\lim_{t\to 0^+}t^{n-\alpha}\int_0^t\frac{(t-s)^{\alpha-1}}{\Gamma(\alpha)}
[A(s)x(s)+F(s)]ds=0. \label{e3.1.11}
\end{equation}
 Furthermore, we have similarly to Theorem \ref{thm3.1.1} that
$t\to \int_0^t\frac{(t-s)^{\alpha-1}}{\Gamma(\alpha)}[A(s)x(s)+F(s)]ds$
is continuous on $(0,1]$. So
$t\to t^{n-\alpha}\int_0^t\frac{(t-s)^{\alpha-1}}{\Gamma(\alpha)}[A(s)x(s)+F(s)]ds$
 is continuous on $[0,1]$ by defining
\begin{align*}
&t^{n-\alpha}\int_0^t\frac{(t-s)^{\alpha-1}}{\Gamma(\alpha)}[A(s)x(s)+F(s)]ds
\Big|_{t=0} \\
&=\lim_{t\to 0^+}t^{n-\alpha}\int_0^t\frac{(t-s)^{\alpha-1}}
 {\Gamma(\alpha)}[A(s)x(s)+F(s)]ds.
\end{align*}
 We have $I_{0^+}^\alpha {}^{RL}D_{0^+}^\alpha x(t)=I_{0^+}^\alpha [A(t)x(t)+F(t)]$.
So
\begin{align*}
&\int_0^t\frac{(t-s)^{\alpha-1}}{\Gamma(\alpha)}[A(s)x(s)+F(s)]ds \\
&=I_{0^+}^\alpha [A(t)x(t)+F(t)]=I_{0^+}^\alpha {}^{RL}D_{0^+}^\alpha x(t)\\
&=\int_0^t\frac{(t-s)^{\alpha-1}}{\Gamma(\alpha)}
 \Big[\frac{1}{\Gamma(n-\alpha)}(\int_0^s(s-w)^{n-\alpha-1}x(w)dw)^{(n)}\Big]ds\\
&=\frac{1}{\Gamma(\alpha)}\int_0^t(t-s)^{\alpha-1}d
\Big(\int_0^s\frac{(s-w)^{n-\alpha-1}}{\Gamma(n-\alpha)}x(w)dw\Big)^{(n-1)}\\
&=\frac{1}{\Gamma(\alpha)}\int_0^t(t-s)^{\alpha-1}
d\big({}^{RL}D_{0^+}^{\alpha-1}x(s)\big)\\
&=\frac{1}{\Gamma(\alpha)} (t-s)^{\alpha-1}({}^{RL}D_{0^+}^{\alpha-1}x(s))\big|_0^t
+\frac{1}{\Gamma(n-\alpha)\Gamma(\alpha-1)}
\int_0^t(t-s)^{\alpha-2} \\
&\quad\times \Big(\int_0^s(s-w)^{n-\alpha-1}x(w)dw\Big)^{(n-1)}ds\\
&=\frac{1}{\Gamma(n-\alpha)\Gamma(\alpha-1)}\int_0^t(t-s)^{\alpha-2}
\Big(\int_0^s(s-w)^{n-\alpha-1}x(w)dw\Big)^{(n-1)}ds
-\frac{\eta_1}{\Gamma(\alpha)}t^{\alpha-1}\\
&=\frac{1}{\Gamma(n-\alpha)\Gamma(\alpha-2)}\int_0^t(t-s)^{\alpha-3}
\Big(\int_0^s(s-w)^{n-\alpha-1}x(w)dw\Big)^{(n-2)}ds \\
&\quad -\frac{\eta_1}{\Gamma(\alpha)}t^{\alpha-1}
-\frac{\eta_2}{\Gamma(\alpha-1)}t^{\alpha-2}\\
&=\dots \\
&=\frac{1}{\Gamma(n-\alpha)\Gamma(\alpha-(n-1))}
\int_0^t(t-s)^{\alpha-n}(\int_0^s(s-w)^{n-\alpha-1}x(w)dw)'ds \\
&\quad -\sum_{v=1}^{n-1}\frac{\eta_v}{\Gamma(\alpha-v+1)}t^{\alpha-v}\\
&=\frac{1}{\Gamma(n-\alpha)\Gamma(\alpha-(n-2))}
\Big[\int_0^t(t-s)^{\alpha-n+1}(\int_0^s (s-w)^{n-\alpha-1}x(w)dw)'ds\Big]'\\
&\quad -\sum_{v=1}^{n-1}\frac{\eta_v}{\Gamma(\alpha-v+1)}t^{\alpha-v} \\
&=\frac{1}{\Gamma(n-\alpha)\Gamma(\alpha-(n-2))}
\Big[(t-s)^{\alpha-n+1}\Big(\int_0^s(s-w)^{n-\alpha-1}x(w)dw\Big)\Big|_0^t\\
&\quad +(\alpha-n+1)\int_0^t(t-s)^{\alpha-n}
 \int_0^s(s-w)^{n-\alpha-1}x(w)\,dw\,ds\Big]' \\
&\quad -\sum_{v=1}^{n-1}\frac{\eta_v}{\Gamma(\alpha-v+1)}t^{\alpha-v}
 \quad \text{(using \eqref{e3.1.10})} \\
&=\frac{1}{\Gamma(n-\alpha)\Gamma(\alpha-(n-1))}
\Big[\int_0^t\int_u^t(t-s)^{\alpha-n}(s-w)^{n-\alpha-1}dsx(w)dw\Big]' \\
&\quad -\frac{\eta_n}{\Gamma(\alpha-n+1)} t^{\alpha-n}-\sum_{v=1}^{n-1}
\frac{\eta_v}{\Gamma(\alpha-v+1)}t^{\alpha-v}\\
&=\frac{1}{\Gamma(n-\alpha)\Gamma(\alpha-(n-1))}
\Big[\int_0^t\int_0^1(1-w)^{\alpha-n}w^{n-\alpha-1}dwx(w)dw\Big]' \\
&\quad -\sum_{v=1}^{n}\frac{\eta_v}{\Gamma(\alpha-v+1)}t^{\alpha-v} t^{\alpha-n} \\
&=x(t)-\sum_{v=1}^{n}\frac{\eta_v}{\Gamma(\alpha-v+1)}t^{\alpha-v}.
\end{align*}
Then $x\in C_{n-\alpha}(0,1]$ is a solution of \eqref{e3.1.9}.


On the other hand, if $x\in C_{n-\alpha}(0,1]$ is a solution of \eqref{e3.1.9}.
Then \eqref{e3.1.10} implies
$\lim_{t\to 0^+}t^{n-\alpha}{x}(t)=\frac{{\eta}_n}{\Gamma(\alpha-n+1)}$.
Furthermore, we have
\begin{align*}
{}^{RL}D_{0^+}^\alpha x(t)
&=\frac{1}{\Gamma(n-\alpha)}\Big(\int_0^t(t-s)^{n-\alpha-1}x(s)ds\Big)^{(n)}\\
&=\frac{1}{\Gamma(n-\alpha)}\Big(\int_0^t(t-s)^{n-\alpha-1}
\Big(\sum_{v=1}^n\frac{\eta_v}{\Gamma(\alpha-v+1)}s^{\alpha-v} \\
&\quad +\int_0^s\frac{(s-u)^{\alpha-1}}{\Gamma(\alpha)}[A(u)x(u)+F(u)]du\Big)
ds\Big)^{(n)}\\
&=\frac{1}{\Gamma(n-\alpha)}\Big(\sum_{v=1}^n\frac{\eta_v}{\Gamma(\alpha-v+1)}
 \int_0^t(t-s)^{n-\alpha-1}s^{\alpha-v}ds \\
&\quad +\int_0^t(t-s)^{n-\alpha-1}\int_0^s\frac{(s-u)^{\alpha-1}}{\Gamma(\alpha)}
 [A(u)x(u)+F(u)]\,du\,ds\Big)^{(n)}\\
&=\frac{1}{\Gamma(n-\alpha)}\Big(\sum_{v=1}^n
 \frac{\eta_v}{\Gamma(\alpha-v+1)}t^{n-v}\int_0^1(1-w)^{n-\alpha-1}w
^{\alpha-v}dw \\
&\quad +\int_0^t\int_u^t(t-s)^{n-\alpha-1}\frac{(s-u)^{\alpha-1}}{\Gamma(\alpha)}
ds[A(u)x(u)+F(u)]du\Big)^{(n)}\\
&=\frac{1}{\Gamma(n-\alpha)}\Big(\sum_{v=1}^n\frac{\eta_v}{\Gamma(\alpha-v+1)}t^{n-v}\int_0^1(1-w)^{n-\alpha-1}w
^{\alpha-v}dw\\
&\quad +\int_0^t(t-u)^{n-1}\int_0^1(1-w)^{n-\alpha-1}
\frac{w^{\alpha-1}}{\Gamma(\alpha)}dw[A(u)x(u)+F(u)]du\Big)^{(n)} \\
&=A(t)x(t)+F(t).
\end{align*}
So $x\in C_{n-\alpha}(0,1]$ is a solution of IVP\eqref{e3.1.2}.
The proof is complete.
\end{proof}

\begin{theorem} \label{thm3.1.4}
 Suppose that {\rm (3.A2)} holds. Then \eqref{e3.1.2} has a unique solution.
If $A(t)\equiv\lambda$ and there exists constants $k_2>-1$, $l_2\le 0$ with
$l_2>\max\{-\alpha,-n-k_2\}$ and $M_F\ge 0$ such that
$|F(t)|\le M_Ft^{k_2}(1-t)^{l_2}$ for all $t\in (0,1)$, then the
problem
\begin{equation} \begin{gathered}
{}^{RL}D_{0^+}^{\alpha}{x}(t)=\lambda {x}(t)+{F}(t),\quad\text{a.e. }
t\in (0,1],\\
\lim_{t\to 0^+}t^{n-\alpha}{x}(t)=\frac{{\eta}_n}{\Gamma(\alpha-n+1)},
\\
\lim_{t\to 0^+}{}^{RL}D_{0^+}^{\alpha-j}x(t)=\eta_j,\quad j\in \mathbb{N}[1,n-1]
\end{gathered}\label{e3.1.12}
\end{equation}
has a unique solution
\begin{equation}
x(t)=\sum_{v=1}^n\eta_vt^{\alpha-v}\mathbf{E}_{\alpha,\alpha-v+1}
(\lambda t^\alpha)+\int_0^t(t-s)^{\alpha-1}\mathbf{E}_{\alpha,\alpha}
(\lambda(t-s)^\alpha)F(s)ds, \label{e3.1.13}
\end{equation}
for $t\in (0,1]$.
\end{theorem}

\begin{proof}
(i) From Claims 1, 2 and 3, and Theorem \ref{thm3.1.3}, we see that \eqref{e3.1.2}
has a unique solution.

 (ii) From the assumption and $A(t)\equiv\lambda$, one sees that (3.A2) holds
with $k_1=l_1=0$ and $k_2,l_2$ mentioned. Thus \eqref{e3.1.12} has a unique solution.
From the Picard function sequence we have
\begin{align*}
&\phi_i(t) \\
&=\sum_{v=1}^n\frac{\eta_v}{\Gamma(\alpha-v+1)}t^{\alpha-v}
+\lambda\int_0^t\frac{(t-s)^{\alpha-1}}{\Gamma(\alpha)}
\phi_{i-1}(s)ds+\int_0^t\frac{(t-s)^{\alpha-1}}{\Gamma(\alpha)}F(s)ds\\
&=\sum_{v=1}^n\frac{\eta_v}{\Gamma(\alpha-v+1)}t^{\alpha-v}
 +\lambda\int_0^t\frac{(t-s)^{\alpha-1}}{\Gamma(\alpha)}
\Big(\sum_{v=1}^n\frac{\eta_v}{\Gamma(\alpha-v+1)}s^{\alpha-v} \\
&\quad +\lambda\int_0^s\frac{(s-u)^{\alpha-1}}{\Gamma(\alpha)} \phi_{i-2}(u)du
 +\int_0^s\frac{(s-u)^{\alpha-1}}{\Gamma(\alpha)}F(u)du\Big)ds \\
&\quad +\int_0^t\frac{(t-s)^{\alpha-1}}{\Gamma(\alpha)}F(s)ds\\
&=\sum_{v=1}^n\frac{\eta_v}{\Gamma(\alpha-v+1)}t^{\alpha-v}+
\lambda\sum_{v=1}^n\frac{\eta_v}{\Gamma(\alpha-v+1)}
\int_0^t\frac{(t-s)^{\alpha-1}}{\Gamma(\alpha)}s^{\alpha-v}ds \\
&\quad +\lambda^2\int_0^t\int_u^t\frac{(t-s)^{\alpha-1}}{\Gamma(\alpha)}
\frac{(s-u)^{\alpha-1}}{\Gamma(\alpha)} ds\phi_{i-2}(u)du\\
&\quad +\lambda\int_0^t\int_u^t\frac{(t-s)^{\alpha-1}}{\Gamma(\alpha)}
 \frac{(s-u)^{\alpha-1}}{\Gamma(\alpha)}dsF(u)du
+\int_0^t\frac{(t-s)^{\alpha-1}}{\Gamma(\alpha)}F(s)ds \\
&=\sum_{v=1}^n\frac{\eta_v}{\Gamma(\alpha-v+1)}t^{\alpha-v}+
\lambda\sum_{v=1}^n\frac{\eta_v}{\Gamma(2\alpha-v+1)}t^{2\alpha-v}\\
&\quad +\lambda^2\int_0^t\frac{(t-u)^{2\alpha-1}}{\Gamma(2\alpha)}\phi_{i-2}(u)du
 +\lambda\int_0^t\frac{(t-u)^{2\alpha-1}}{\Gamma(2\alpha)}F(u)du \\
&\quad +\int_0^t\frac{(t-s)^{\alpha-1}}{\Gamma(\alpha)}F(s)ds\\
&=\sum_{v=1}^n\eta_vt^{\alpha-v}(\frac{1}{\Gamma(\alpha-v+1)}
 + \frac{\lambda t^{\alpha}}{\Gamma(2\alpha-v+1)})
 +\lambda^2\int_0^t\frac{(t-u)^{2\alpha-1}}{\Gamma(2\alpha)}\phi_{i-2}(u)du\\
&\quad +\int_0^t(t-s)^{\alpha-1}(\frac{\lambda(t-s)^{\alpha}}{\Gamma(2\alpha)}
 +\frac{1}{\Gamma(\alpha)})F(s)ds\\
&=\dots\\
&=\sum_{v=1}^n\eta_vt^{\alpha-v}\Big(\sum_{j=0}^{i-1}
 \frac{\lambda^jt^{\alpha j}}{\Gamma(j\alpha+\alpha-v+1)}\Big)
 +\lambda^i\int_0^t\frac{(t-u)^{i\alpha-1}}{\Gamma(m\alpha)}\phi_0(u)du \\
&\quad +\int_0^t(t-s)^{\alpha-1}
 \Big(\sum_{j=0}^{m-1}\frac{\lambda^j(t-s)^{\alpha j}}{\Gamma((j+1)\alpha)}\Big)
 F(s)ds\\
&=\sum_{v=1}^n\eta_vt^{\alpha-v}(\sum_{j=0}^i
 \frac{\lambda^jt^{\alpha j}}{\Gamma(j\alpha+\alpha-v+1)})
+\int_0^t(t-s)^{\alpha-1}\Big(\sum_{j=0}^{i-1}
 \frac{\lambda^j(t-s)^{\alpha j}}{\Gamma((j+1)\alpha)}\Big)F(s)ds\\
&\to \sum_{v=1}^n\eta_vt^{\alpha-v}\mathbf{E}_{\alpha,\alpha-v+1}
 (\lambda t^\alpha)
 +\int_0^t(t-s)^{\alpha-1}\mathbf{E}_{\alpha,\alpha}(\lambda(t-s)^\alpha)F(s)ds.
\end{align*}
Then we obtain $x(t)=\lim_{i\to +\infty}\phi_i(t)$ is a unique solution
of \eqref{e3.1.12}. Then $x$ satisfies \eqref{e3.1.13}.
The proof is complete.
\end{proof}

To obtain solutions of \eqref{e3.1.3}, we need the following assumption:
\begin{itemize}
\item[(3.A3)] There exist constants $k_i>-1$, $l_i\le 0$ with
$l_1>\max\{-\alpha,-\alpha-k_1\}$, $l_2>\max\{-\alpha,-n-k_2\}$,
$M_B\ge 0$ and $M_G\ge 0$ such that $|B(t)|\le M_B(\log t)^{k_1}(1-\log t)^{l_1}$
and $|G(t)|\le M_G(\log t)^{k_2}(1-\log t)^{l_2}$ for all $t\in (1,e)$.
\end{itemize}
We choose Picard function sequence as
\begin{gather*}
\phi_0(t)=\sum_{v=1}^n\frac{\eta_v}{\Gamma(\alpha-v+1)}(\log t)^{\alpha-v},
\quad t\in (1,e],\\
\phi_i(t)=\sum_{v=1}^n\frac{\eta_v}{\Gamma(\alpha-v+1)}(\log t)^{\alpha-v}
+\frac{1}{\Gamma(\alpha)}\int_1^t(\log\frac{t}{s})^{\alpha-1}
[B(s)\phi_{i-1}(s)+G(s)]\frac{ds}{s},\\
 t\in (1,e],\; i=1,2,\dots.
\end{gather*}
\smallskip

\noindent\textbf{Claim 1.} $\phi_i\in LC_{n-\alpha}(1,e]$.

\begin{proof}
We have $\phi_0\in LC_{n-\alpha}[1,e]$ and
\begin{align*}
&(\log t)^{n-\alpha}\Big|\int_1^t(\log \frac{t}{s})^{\alpha-1}
[B(s)\phi_0(s)+G(s)]\frac{ds}{s}\Big|\\
&\le (\log t)^{n-\alpha}\int_1^t(\log \frac{t}{s})^{\alpha-1}
\Big[M_B\|\phi_0\|(\log s)^{\alpha-n}(\log s)^{k_1}(1-\log s)^{l_1}\\
&\quad +M_G(\log s)^{k_2}(1-\log s)^{l_2}\Big]\frac{ds}{s}\\
&\le (\log t)^{n-\alpha}M_B\|\phi_0\|\int_1^t(\log \frac{t}{s})^{\alpha+l_1-1}
(\log s)^{\alpha+k_1-n}\frac{ds}{s} \\
&\quad +(\log t)^{n-\alpha}M_G \int_1^t(\log \frac{t}{s})^{\alpha+l_2-1}
(\log s)^{k_2}\frac{ds}{s} \\
&=M_B\|\phi_0\|(\log t)^{\alpha+k_1+l_1}\mathbf{B}(\alpha+l_1,\alpha+k_1-n+1)\\
&\quad +M_G(\log t)^{n+k_1+l_1}\mathbf{B}(\alpha+l_2,k_2+1)\to 0\quad
\text{as }t\to 0^+,
\end{align*}
we know that
$t\to \frac{1}{\Gamma(\alpha)}\int_1^t(\log\frac{t}{s})^{\alpha-1}
[B(s)\phi_0(s)+G(s)]\frac{ds}{s}$ is continuous on $(1,e]$ and
$\lim_{t\to 0^+}(\log t)^{n-\alpha}\phi_1(t)$ exists.
Then $\phi_1\in LC_{n-\alpha}[1,e]$. By mathematical induction,
we can show $\phi_i\in LC_{n-\alpha}[1,e]$.
\end{proof}
\smallskip

\noindent\textbf{Claim 2.}
 $\{t\to (\log t)^{n-\alpha}\phi_i(t)\}$ is convergent uniformly on $[1,e]$.

\begin{proof} As above, for $t\in [1,e]$ we have
\begin{align*}
&(\log t)^{n-\alpha}|\phi_1(t)-\phi_0(t)| \\
&=\frac{1}{\Gamma(\alpha)}(\log t)^{n-\alpha}|\int_1^t
(\log\frac{t}{s})^{\alpha-1}[B(s)\phi_0(s)+G(s)]\frac{ds}{s}|\\
&\le \frac{1}{\Gamma(\alpha)}(\log t)^{n-\alpha}
 \int_1^t(\log\frac{t}{s})^{\alpha-1}\Big[\|\phi_0\|M_B(\log s)^{\alpha-n+k_1}
 (1-\log s)^{l_1} \\
&\quad +M_G(\log s)^{k_2}(1-\log s)^{l_2}\Big]\frac{ds}{s}\\
&\le M_B\|\phi_0\|(\log t)^{\alpha+k_1+l_1}\frac{\mathbf{B}
 (\alpha+l_1,\alpha+k_1-n+1)}{\Gamma(\alpha)} \\
&\quad +M_G(\log t)^{n+k_2+l_2}
 \frac{\mathbf{B}(\alpha+l_2,k_2+1)}{\Gamma(\alpha)}.
\end{align*}
So
\begin{align*}
&(\log t)^{n-\alpha}|\phi_2(t)-\phi_1(t)|\\
&=\frac{1}{\Gamma(\alpha)}(\log t)^{n-\alpha}
\big|\int_1^t(\log \frac{t}{s})^{\alpha-1}B(s)[\phi_1(s)-\phi_0(s)]\frac{ds}{s}\big|\\
& \le \frac{1}{\Gamma(\alpha)}(\log t)^{n-\alpha}
 \int_1^t(\log \frac{t}{s})^{\alpha+l_1-1}M_B(\log s)^{k_1+\alpha-n} \\
&\quad\times \Big(M_B\|\phi_0\|(\log s)^{\alpha+k_1+l_1}
\frac{\mathbf{B}(\alpha+l_1,\alpha-n+k_1+1)} {\Gamma(\alpha)} \\
&\quad +M_G(\log s)^{n+k_2+l_2}
 \frac{\mathbf{B}(\alpha+l_2,k_2+1)}{\Gamma(\alpha)}\Big)
\frac{ds}{s}\\
& \le \|\phi_0\|M_B^2(\log t)^{2\alpha+2k_1+2l_1}
\frac{\mathbf{B}(\alpha+l_1,2\alpha+2k_1+l_1-n+1)}{\Gamma(\alpha)} \\
&\quad\times \frac{\mathbf{B}(\alpha+l_1,\alpha+k_1-n+1)}{\Gamma(\alpha)}\\
&\quad +M_BM_G(\log t)^{\alpha+n+k_1+l_1+k_2+l_2}
 \frac{\mathbf{B}(\alpha+l_1,\alpha+k_1+k_2+l_2+1)}{\Gamma(\alpha)} \\
&\quad\times  \frac{\mathbf{B}(\alpha+l_2,k_2+1)}{\Gamma(\alpha)}.
\end{align*}
Then
\begin{align*}
&(\log t)^{n-\alpha}|\phi_3(t)-\phi_2(t)| \\
&=\frac{1}{\Gamma(\alpha)}(\log t)^{n-\alpha}
\big|\int_1^t(\log \frac{t}{s})^{\alpha-1}B(s)[\phi_2(s)-\phi_1(s)]ds\big|\\
&\le \frac{1}{\Gamma(\alpha)}(\log t)^{n-\alpha}\
int_1^t(\log \frac{t}{s})^{\alpha+l_1-1}M_B(\log s)^{k_1+\alpha-n}\\
&\quad\times \Big(\|\phi_0\|M_B^2(\log s)^{\alpha+n+2k_1+2l_1}
\frac{\mathbf{B}(\alpha+l_1,\alpha+2k_1+l_1+1)}{\Gamma(\alpha)} \\
&\quad\times \frac{\mathbf{B}(\alpha+l_1,\alpha+k_1-n+1)}{\Gamma(\alpha)}
 +M_BM_G(\log s)^{2n+k_1+l_1+k_2+l_2} \\
&\quad\times \frac{\mathbf{B}(\alpha+l_1,n+k_1+k_2+l_2+1)}{\Gamma(\alpha)}
\frac{\mathbf{B}(\alpha+l_2,k_2+1)}{\Gamma(\alpha)}\Big)\frac{ds}{s}\\
& \le \|\phi_0\|M_B^3(\log t)^{3\alpha+3k_1+3l_1}
 \frac{\mathbf{B}(\alpha+l_1,3\alpha+3k_1+2l_1-n+1)}{\Gamma(\alpha)} \\
&\quad\times \frac{\mathbf{B}(\alpha+l_1,2\alpha+2k_1+l_1-n+1)}{\Gamma(\alpha)}
 \frac{\mathbf{B}(\alpha+l_1,\alpha+k_1-n+1)}{\Gamma(\alpha)}\\
&\quad +M_B^2M_G(\log t)^{2\alpha+n+2k_1+2l_1+k_2+l_2}
 \frac{\mathbf{B}(\alpha+l_1,2\alpha+2k_1+l_1+k_2+l_2+1)}{\Gamma(\alpha)}\\
&\quad\times \frac{\mathbf{B}(\alpha+l_1,\alpha+k_1+k_2+l_2+1)}{\Gamma(\alpha)}
 \frac{\mathbf{B}(\alpha+l_2,k_2+1)}{\Gamma(\alpha)}.
\end{align*}
Furthermore,
\begin{align*}
&(\log t)^{n-\alpha}|\phi_4(t)-\phi_3(t)| \\
&=\frac{1}{\Gamma(\alpha)}(\log t)^{n-\alpha}
\big|\int_1^t(\log \frac{t}{s})^{\alpha-1}B(s)[\phi_3(s)-\phi_2(s)]ds\big|\\
&\le \frac{(\log t)^{n-\alpha}}{\Gamma(\alpha)}
\int_1^t(\log \frac{t}{s})^{\alpha+l_1-1}M_B(\log s)^{k_1+\alpha-n}
 \Big(\|\phi_0\|M_B^3(\log s)^{\alpha+2n+3k_1+3l_1} \\
&\quad \times\frac{\mathbf{B}(\alpha+l_1,\alpha+n+3k_1+2l_1+1)}{\Gamma(\alpha)}
\frac{\mathbf{B}(\alpha+l_1,\alpha+2k_1+l_1+1)}{\Gamma(\alpha)} \\
&\quad\times \frac{\mathbf{B}(\alpha+l_1,\alpha+k_1-n+1)}{\Gamma(\alpha)}
 +M_B^2M_G(\log s)^{3n+2k_1+2l_1+k_2+l_2} \\
&\quad \times \frac{\mathbf{B}(\alpha+l_1,2n+2k_1+l_1+k_2+l_2+1)}{\Gamma(\alpha)} \\
&\quad\times \frac{\mathbf{B}(\alpha+l_1,n+k_1+k_2+l_2+1)}{\Gamma(\alpha)}
\frac{\mathbf{B}(\alpha+l_2,k_2+1)}{\Gamma(\alpha)}\Big)\frac{ds}{s}\\
&\le \|\phi_0\|M_B^4(\log t)^{4\alpha+4k_1+4l_1}
 \frac{\mathbf{B}(\alpha+l_1,4\alpha+4k_1+3l_1-n+1)}{\Gamma(\alpha)} \\
&\quad\times \frac{\mathbf{B}(\alpha+l_1,3\alpha+3k_1+2l_1-n+1)}{\Gamma(\alpha)}
\frac{\mathbf{B}(\alpha+l_1,2\alpha+2k_1+l_1-n+1)}{\Gamma(\alpha)} \\
&\quad\times \frac{\mathbf{B}(\alpha+l_1,\alpha+k_1-n+1)}{\Gamma(\alpha)}
 +M_B^3M_G(\log t)^{3\alpha+n+3k_1+3l_1+k_2+l_2} \\
&\quad \times \frac{\mathbf{B}(\alpha+l_1,3\alpha+3k_1+2l_1+k_2+l_2+1)}
{\Gamma(\alpha)}
\frac{\mathbf{B}(\alpha+l_1,2\alpha+2k_1+l_1+k_2+l_2+1)}{\Gamma(\alpha)} \\
&\quad \times \frac{\mathbf{B}(\alpha+l_1,\alpha+k_1+k_2+l_2+1)}{\Gamma(\alpha)}
 \frac{\mathbf{B}(\alpha+l_2,k_2+1)}{\Gamma(\alpha)}.
\end{align*}
Similarly by mathematical induction, for every
$i=1,2,\dots$ we obtain
\begin{align*}
&(\log t)^{n-\alpha}|\phi_i(t)-\phi_{i-1}(t)|\\
&\le \|\phi_0\|M_B^i(\log t)^{i\alpha +ik_1+il_1}
\frac{\mathbf{B}(\alpha+l_1,\alpha+k_1-n+1)}{\Gamma(\alpha)} \\
&\quad \times \prod_{j=1}^{i-1}\frac{\mathbf{B}(\alpha+l_1,(j+1)\alpha+(j+1)k_1+jl_1-n+1)}
{\Gamma(\alpha)}\\
&\quad +M_B^{i-1}M_G(\log t)^{(i-1)\alpha+n+(i-1)k_1+(i-1)l_1+k_2+l_2}
 \frac{\mathbf{B}(\alpha+l_2,k_2+1)}{\Gamma(\alpha)} \\
&\quad\times \prod_{j=1}^{i-1}\frac{\mathbf{B}
(\alpha+l_1,jn+jk_1+(j-1)l_1+k_2+l_2+1)}{\Gamma(\alpha)}\\
&\le \|\phi_0\|M_B^i\frac{\mathbf{B}(\alpha+l_1,\alpha+k_1-n+1)}{\Gamma(\alpha)}\\
&\quad \times \prod_{j=1}^{i-1}\frac{\mathbf{B}(\alpha+l_1,(j+1)\alpha+(j+1)k_1+jl_1-n+1)}
{\Gamma(\alpha)} \\
&\quad +M_B^{i-1}M_G\frac{\mathbf{B}(\alpha+l_2,k_2+1)}{\Gamma(\alpha)}\\
&\quad \times \prod_{j=1}^{i-1}\frac{\mathbf{B}(\alpha+l_1,jn+jk_1+(j-1)l_1+k_2+l_2+1)}
{\Gamma(\alpha)},\quad t\in (1,e].
\end{align*}
Similarly we can prove that both
\begin{align*}
\sum_{i=1}^{+\infty}u_i
&=\sum_{i=1}^{+\infty} \|\phi_0\|M_B^i
\frac{\mathbf{B}(\alpha+l_1,\alpha+k_1-n+1)}{\Gamma(\alpha)}\\
&\quad \times \prod_{j=1}^{i-1}\frac{\mathbf{B}(\alpha+l_1,(j+1)\alpha+(j+1)k_1+jl_1-n+1)}
{\Gamma(\alpha)},
\\
\sum_{i=1}^{+\infty}v_i
&=\sum_{i=1}^{+\infty}M_B^{m-1}M_G\frac{\mathbf{B}(\alpha+l_2,k_2+1)}{\Gamma(\alpha)}\\
&\quad \times \prod_{j=1}^{i-1}\frac{\mathbf{B}(\alpha+l_1,jn+jk_1+(j-1)l_1+k_2+l_2+1)}
{\Gamma(\alpha)}
\end{align*}
converge. Hence,
$$
(\log t)^{n-\alpha}\phi_0(t)+(\log t)^{n-\alpha}[\phi_1(t)-\phi_0(t)]+\dots
+(\log t)^{n-\alpha}[\phi_i(t)-\phi_{i-1}(t)]+\dots,
$$
for $t\in (1,e]$
converges uniformly. Then $\{t\to (\log t)^{n-\alpha}\phi_i(t)\}$
converges uniformly on $[1,e]$.
\end{proof}
\smallskip

\noindent\textbf{Claim 3.}
 $\phi(t)=(\log t)^{\alpha-n}\lim_{i\to +\infty}(\log t)^{n-\alpha}\phi_i(t)$
defined on $[1,e]$ is a unique continuous solution of the integral equation
\begin{equation}
{x}(t)=\sum_{v=1}^n\frac{\eta_v}{\Gamma(\alpha-v+1)}(\log t)^{\alpha-v}
+\frac{1}{\Gamma(\alpha)}\int_1^t(\log \frac{t}{s})^{\alpha-1}[B(s)x(s)+G(s)]
\frac{ds}{s},
\label{e3.1.14}
\end{equation}
for $t\in (1,e]$.

\begin{proof}
From $\lim_{i\to +\infty}(\log t)^{n-\alpha}\phi_i(t)=(\log t)^{n-\alpha}\phi(t)$
and the uniformly convergence, we see that $\phi(t)$ is continuous on $[1,e]$. From
\begin{align*}
&(\log t)^{n-\alpha}|\int_1^t(\log \frac{t}{s})^{\alpha-1}[A(s)\phi_{p-1}(s)
+F(s)]\frac{ds}{s}\\
&-\int_1^t(\log \frac{t}{s})^{\alpha-1}[B(s)
\phi_{q-1}(s)+G(s)]\frac{ds}{s}|\\
&\le M_B\|\phi_{p-1}-\phi_{q-1}\|(\log t)^{n-\alpha}
\int_1^t(\log\frac{t}{s})^{\alpha+l_1-1}(\log s)^{k_1}(\log s)^{\alpha-n}
\frac{ds}{s}\\
&\le M_B\|\phi_{p-1}-\phi_{q-1}\|(\log t)^{n-\alpha}
\int_1^t(\log\frac{t}{s})^{\alpha+l_1-1}(\log s)^{\alpha+k_1-n}\frac{ds}{s}\\
&\le M_B\|\phi_{p-1}-\phi_{q-1}\|(\log t)^{n+k_1+l_1}\mathbf{B}
 (\alpha+l_1,\alpha+k_1)\\
&\le M_B\|\phi_{p-1}-\phi_{q-1}\|\mathbf{B}(\alpha+l_1,\alpha+k_1-n+1) \\
&\to 0\quad \text{uniformly as }p,q\to +\infty,
\end{align*}
we know that
\begin{align*}
\phi(t)
&=(\log t)^{\alpha-n}\lim_{i\to +\infty}(\log t)^{n-\alpha}\phi_i(t)\\
&=\lim_{i\to +\infty}\Big[\sum_{v=1}^n\frac{\eta_v}{\Gamma(\alpha-v+1)}
(\log t)^{\alpha-v} \\
&\quad + \frac{1}{\Gamma(\alpha)}\int_1^t(\log \frac{t}{s})^{\alpha-1}
[B(s)\phi_{i-1}(s)+G(s)]\frac{ds}{s}\Big]\\
&=\sum_{v=1}^n\frac{\eta_v}{\Gamma(\alpha-v+1)}(\log t)^{\alpha-v}
+\lim_{i\to +\infty}\int_1^t(\log \frac{t}{s})^{\alpha-1}
[B(s)\phi_{i-1}(s)+G(s)]\frac{ds}{s}\\
&=\sum_{v=1}^n\frac{\eta_v}{\Gamma(\alpha-v+1)}(\log t)^{\alpha-v}
+\frac{1}{\Gamma(\alpha)}\int_1^t(\log \frac{t}{s})^{\alpha-1}
\left[B(s)\phi(s)+G(s)\right]\frac{ds}{s}.
\end{align*}
Then $\phi$ is a continuous solution of \eqref{e3.1.14} defined on $(1,e]$.

Suppose that $\psi$ defined on $(1,e]$ is also a solution of \eqref{e3.1.14}.
Then
$$
\psi(t)=\sum_{v=1}^n\frac{\eta_v}{\Gamma(\alpha-v+1)}(\log t)^{\alpha-v}
+\frac{1}{\Gamma(\alpha)}\int_1^t(\log \frac{t}{s})^{\alpha-1}
[B(s)\psi(s)+G(s)]\frac{ds}{s},
$$
for $t\in (1,e]$.
We need to prove that $\phi(t)\equiv\psi(t)$ on $(1,e]$.
Then
\begin{align*}
&(\log t)^{n-\alpha}|\psi(t)-\phi_0(t)| \\
&=(\log t)^{n-\alpha}\Big|\int_1^t(\log \frac{t}{s})^{\alpha-1}|B(s)
\psi(s)+G(s)|\frac{ds}{s}\Big|\\
&\le M_B\|\phi_0\|(\log t)^{\alpha+k_1+l_1}
\frac{\mathbf{B}(\alpha+l_1,\alpha+k_1-n+1)}{\Gamma(\alpha)} \\
&\quad +M_G(\log t)^{n+k_2+l_2}\frac{\mathbf{B}(\alpha+l_2,k_2+1)}{\Gamma(\alpha)}.
\end{align*}
Furthermore,
\begin{align*}
&(\log t)^{n-\alpha}|\psi(t)-\phi_1(t)| \\
&=(\log t)^{n-\alpha}\frac{1}{\Gamma(\alpha)}
\Big|\int_1^t(\log \frac{t}{s})^{\alpha-1}B(s)
[\psi(s)-\phi_0(s)]\frac{ds}{s}\Big|\\
&\le \|\phi_0\|M_B^2(\log t)^{2\alpha+2k_1+2l_1}
\frac{\mathbf{B}(\alpha+l_1,2\alpha+2k_1+l_1-n+1)}{\Gamma(\alpha)} \\
&\quad\times \frac{\mathbf{B}(\alpha+l_1,\alpha+k_1-n+1)}{\Gamma(\alpha)}
 +M_BM_G(\log t)^{\alpha+n+k_1+l_1+k_2+l_2} \\
&\quad\times \frac{\mathbf{B}(\alpha+l_1,\alpha+k_1+k_2+l_2+1)}{\Gamma(\alpha)}
\frac{\mathbf{B}(\alpha+l_2,k_2+1)}{\Gamma(\alpha)}.
\end{align*}
By mathematical induction, we obtain
\begin{align*}
&(\log t)^{1-\alpha}|\psi(t)-\phi_{i-1}(t)| \\
&=(\log t)^{n-\alpha}\frac{1}{\Gamma(\alpha)}
\big|\int_1^t(\log \frac{t}{s})^{\alpha-1}B(s)[\psi(s)-\phi_{i-1}(s)]ds\big|\\
&\le \|\phi_0\|M_B^i(\log t)^{i\alpha +ik_1+il_1}
\frac{\mathbf{B}(\alpha+l_1,\alpha+k_1-n+1)}{\Gamma(\alpha)} \\
&\quad\times \prod_{j=1}^{i-1} \frac{\mathbf{B}(\alpha+l_1,(j+1)\alpha+(j+1)k_1+jl_1-n+1)}
 {\Gamma(\alpha)}\\
&\quad +M_B^{i-1}M_G(\log t)^{(i-1)\alpha+n+(i-1)k_1+(i-1)l_1+k_2+l_2}
\frac{\mathbf{B}(\alpha+l_2,k_2+1)}{\Gamma(\alpha)} \\
&\quad\times \prod_{j=1}^{i-1}
 \frac{\mathbf{B}(\alpha+l_1,jn+jk_1+(j-1)l_1+k_2+l_2+1)}{\Gamma(\alpha)}\\
&\le \|\phi_0\|M_B^i\frac{\mathbf{B}(\alpha+l_1,\alpha+k_1-n+1)}{\Gamma(\alpha)}\\
&\quad\times  \prod_{j=1}^{i-1}
\frac{\mathbf{B}(\alpha+l_1,(j+1)\alpha+(j+1)k_1+jl_1-n+1)}
{\Gamma(\alpha)}\\
&\quad +M_B^{i-1}M_G\frac{\mathbf{B}(\alpha+l_2,k_2+1)}{\Gamma(\alpha)}
\prod_{j=1}^{i-1}\frac{\mathbf{B}(\alpha+l_1,jn+jk_1+(j-1)l_1+k_2+l_2+1)}
{\Gamma(\alpha)},
\end{align*}
for $t\in (1,e]$, $i=1,2,\dots$.
Similarly we have
$\lim_{i\to +\infty}(\log t)^{n-\alpha}\phi_i(t)=(\log t)^{n-\alpha}\psi(t)$
 uniformly on $(1,e]$. Then $\phi(t)\equiv\psi(t)$ on $(1,e]$.
Then \eqref{e3.1.14} has a unique solution $\phi$. The proof is complete.
\end{proof}

\begin{theorem} \label{thm3.1.5}
 Suppose that {\rm (3.A3)} holds. Then ${x}$ is a solution of IVP \eqref{e3.1.3}
if and only if $x\in LC_{n-\alpha}(1,e]$ is a solution of the integral equation
\eqref{e3.1.14}.
\end{theorem}

\begin{proof}
Suppose that $x$ is a solution of \eqref{e3.1.3}.
 Then $t\to (\log t)^{n-\alpha}x(t)$is continuous on $(1,e]$ by defining
$(\log t)^{n-\alpha}x(t)|_{t=1}=\lim_{t\to 1^+}(\log t)^{n-\alpha}x(t)$
and $\|x\|=r<+\infty$. So
\begin{align*}
&\lim_{s\to 1^+}\int_1^s(\log \frac{s}{w})^{n-\alpha-1}x(w)\frac{dw}{w} \\
&=\lim_{s\to 1^+}\int_1^s(\log \frac{s}{w})^{n-\alpha-1}(\log w)^{\alpha-n}
(\log w)^{n-\alpha}x(w)\frac{dw}{w}\\
&=\lim_{s\to 1^+}(\log \xi)^{n-\alpha}x(\xi)\int_1^s
(\log \frac{s}{w})^{n-\alpha-1}(\log w)^{\alpha-n}\frac{dw}{w} \\
&\quad \text{(by the mean value theorem with $\xi \in (1,s)$)}\\
&=\lim_{s\to 1^+}(\log \xi)^{n-\alpha}x(\xi)\int_0^1(1-u)^{n-\alpha-1}
u^{\alpha-n}du\quad \text{(because $\frac{\log w}{\log s}=u$)}\\
&=\frac{\eta_n}{\Gamma(\alpha-n+1)} \mathbf{B}(n-\alpha,\alpha-n+1).
\end{align*}
and for $v\in N[1,n-1]$ we have
\begin{align*}
\lim_{t\to 1^+}
&(s\frac{d}{ds})^{n-v}\Big(\int_1^s(\log \frac{s}{w})^{n-\alpha-1}x(w)
 \frac{dw}{w}\Big)\\
&=\Gamma(n-v-(\alpha-v))\lim_{t\to 1^+} {}^{RLH}D_{0^+}^{\alpha-v}x(t)
=\Gamma(n-\alpha)\eta_v.
\end{align*}
 From (3.A3), we have
\begin{align*}
&(\log t)^{n-\alpha}\big|\int_1^t(\log \frac{t}{s})^{\alpha-1}
[B(s)x(s)+G(s)]\frac{ds}{s}\big|\\
&\le (\log t)^{n-\alpha}\int_1^t(\log \frac{t}{s})^{\alpha-1}
[M_Br(\log s)^{\alpha-n}(\log s)^{k_1}(1-\log s)^{l_1} \\
&\quad +M_G(\log s)^{k_2}(1-\log s)^{l_2}]\frac{ds}{s}\\
&\le (\log t)^{n-\alpha}M_Br\int_1^t(\log \frac{t}{s})^{\alpha+l_1-1}
 (\log s)^{\alpha+k_1-n}\frac{ds}{s} \\
&\quad +(\log t)^{n-\alpha}M_G
\int_1^t(\log \frac{t}{s})^{\alpha+l_2-1}(\log s)^{k_2}\frac{ds}{s}\\
&=M_Br(\log t)^{\alpha+k_1+l_1}\mathbf{B}(\alpha+l_1,k_1+\alpha)
+M_G(\log t)^{n+k_1+l_1}\mathbf{B}(\alpha+l_2,k_2+1).
\end{align*}
So $t\to (\log t)^{n-\alpha}\int_1^t(\log \frac{t}{s})^{\alpha-1}
[B(s)x(s)+G(s)]\frac{ds}{s}$ is defined on $(1,e]$ and
\begin{equation}
\lim_{t\to 1^+}(\log t)^{n-\alpha}\int_1^t(\log \frac{t}{s})^{\alpha-1}
[B(s)x(s)+G(s)]\frac{ds}{s}=0.
 \label{e3.1.15}
\end{equation}
 Furthermore, similarly to Theorem \ref{thm3.1.1} we have
$t\to \int_1^t(\log \frac{t}{s})^{\alpha-1}[B(s)x(s)+G(s)]\frac{ds}{s}$
is continuous on $(1,e]$. So
 $t\to (\log t)^{n-\alpha}\int_1^t(\log \frac{t}{s})^{\alpha-1}
[B(s)x(s)+G(s)]\frac{ds}{s}$ is continuous on $[1,e]$ by defining
\begin{equation}
 (\log t)^{n-\alpha}\int_1^t(\log \frac{t}{s})^{\alpha-1}
[B(s)x(s)+G(s)]\frac{ds}{s}|_{t=1}=0. \label{e3.1.16}
\end{equation}
 We have ${}^HI_{1^+}^\alpha {}^{RLH}D_{1^+}^\alpha x(t)
={}^HI_{1^+}^\alpha [B(t)x(t)+G(t)]$. So
\begin{align*}
&\frac{1}{\Gamma(\alpha)}\int_1^t(\log \frac{t}{s})^{\alpha-1}
[B(s)x(s)+G(s)]\frac{ds}{s}\\
&={}^HI_{1^+}^\alpha [B(t)x(t)+G(t)]
={}^HI_{1^+}^\alpha {}^{RLH}D_{1^+}^\alpha x(t)\\
&=\frac{1}{\Gamma(\alpha)}\frac{1}{\Gamma(n-\alpha)}\int_1^t
(\log \frac{t}{s})^{\alpha-1}(s\frac{d}{ds})^n
\Big(\int_1^s(\log \frac{s}{w})^{n-\alpha-1}x(w)\frac{dw}{w}\Big)\frac{ds}{s}\\
&=\frac{1}{\Gamma(\alpha)}\frac{1}{\Gamma(n-\alpha)}\int_1^t
(\log \frac{t}{s})^{\alpha-1}d\Big[ (s\frac{d}{ds})^{n-1}
\Big(\int_1^s(\log \frac{s}{w})^{n-\alpha-1}x(w)\frac{dw}{w}\Big)\Big]\\
&=\frac{1}{\Gamma(\alpha)}\frac{1}{\Gamma(n-\alpha)}
\Big[ (\log \frac{t}{s})^{\alpha-1}
(s\frac{d}{ds})^{n-1}\Big(\int_1^s(\log \frac{s}{w})^{n-\alpha-1}x(w)\frac{dw}{w}\Big)
\Big|_1^t \\
&\quad +(\alpha-1)\int_1^t(\log \frac{t}{s})^{\alpha-2}(s\frac{d}{ds})^{n-1}
\Big(\int_1^s(\log \frac{s}{w})^{n-\alpha-1}x(w)\frac{dw}{w}\Big)\frac{ds}{s}\Big]\\
&=-\frac{1}{\Gamma(\alpha)}\frac{1}{\Gamma(n-\alpha)}
 (\log t)^{\alpha-1}\lim_{t\to 1^+}
(s\frac{d}{ds})^{n-1}(\int_1^s(\log \frac{s}{w})^{n-\alpha-1}x(w)\frac{dw}{w}) \\
&\quad +\frac{1}{\Gamma(\alpha-1)}\frac{1}{\Gamma(n-\alpha)}
\int_1^t(\log \frac{t}{s})^{\alpha-2}(s\frac{d}{ds})^{n-1}
\Big(\int_1^s(\log \frac{s}{w})^{n-\alpha-1}x(w)\frac{dw}{w}\Big)\frac{ds}{s}\\
&=-\frac{\eta_1 }{\Gamma(\alpha)}(\log t)^{\alpha-1}
+\frac{1}{\Gamma(\alpha-1)}\frac{1}{\Gamma(n-\alpha)}
\int_1^t(\log \frac{t}{s})^{\alpha-2}(s\frac{d}{ds})^{n-1} \\
&\quad\times \Big(\int_1^s(\log \frac{s}{w})^{n-\alpha-1}x(w)
\frac{dw}{w}\Big)\frac{ds}{s}\\
&=\dots \\
&=-\sum_{v=1}^{n-1}\frac{\eta_v}{\Gamma(\alpha-v+1)}(\log t)^{\alpha-v}
 +\frac{1}{\Gamma(\alpha-n+1)}\frac{1}{\Gamma(n-\alpha)}\int_1^t
(\log \frac{t}{s})^{\alpha-n} \\
&\quad\times\Big(\int_1^s(\log \frac{s}{w})^{-\alpha}x(w)
 \frac{dw}{w}\Big)'ds\\
&=-\sum_{v=1}^{n-1}\frac{\eta_v}{\Gamma(\alpha-v+1)}
 (\log t)^{\alpha-v}+\frac{1}{\Gamma(\alpha-n+2)}\frac{1}{\Gamma(n-\alpha)}t\\
&\quad\times \Big[\int_1^t(\log \frac{t}{s})^{\alpha-n+1}
\Big(\int_1^s(\log \frac{s}{w})^{-\alpha}x(w)\frac{dw}{w}Big)'ds\Big]'\\
&=-\sum_{v=1}^{n-1}\frac{\eta_v}{\Gamma(\alpha-v+1)}(\log t)^{\alpha-v}
+\frac{1}{\Gamma(n-\alpha)\Gamma(\alpha-n+2)}t \\
&\quad\times \Big[ (\log \frac{t}{s})^{\alpha-n+1}\int_1^s(\log \frac{s}{w})^{n-\alpha-1}x(w)
\frac{dw}{w}|_1^t
\\
&\quad + (\alpha-n+1)\int_1^t(\log \frac{t}{s})^{\alpha-n}\int_1^s
(\log \frac{s}{w})^{n-\alpha-1}x(w)\frac{dw}{w}\frac{ds}{s}\Big]' \\
&=-\sum_{v=1}^{n-1}\frac{\eta_v}{\Gamma(\alpha-v+1)}
(\log t)^{\alpha-v}+\frac{1}{\Gamma(n-\alpha)\Gamma(\alpha-n+2)}t \\
&\quad\times \Big[\lim_{t\to 1^+}(\log t)^{\alpha-n+1}\int_1^s
(\log \frac{s}{w})^{n-\alpha-1}x(w)\frac{dw}{w}\\
&\quad + (\alpha-n+1)\int_1^t\int_u^t(\log \frac{t}{s})^{\alpha-n}
(\log \frac{s}{w})^{n-\alpha-1}\frac{ds}{s}x(w)\frac{dw}{w}\Big]' \\
&=-\sum_{v=1}^{n-1}\frac{\eta_v}{\Gamma(\alpha-v+1)}(\log t)^{\alpha-v}
+\frac{1}{\Gamma(n-\alpha)\Gamma(\alpha-n+2)}t \\
&\quad\times \Big[\frac{\eta_n}{\Gamma(\alpha-n+1)} \mathbf{B}(n-\alpha,\alpha-n+1)
(\log t)^{\alpha-n+1} \\
&\quad + (\alpha-n+1)\int_1^t\int_0^1(1-u)^{\alpha-n}u^{n-\alpha-1}dux(w)
\frac{dw}{w}\Big]' \\
&=x(t)-\sum_{v=1}^{n}\frac{\eta_v}{\Gamma(\alpha-v+1)}(\log t)^{\alpha-v}.
\end{align*}
Then $x\in LC_{n-\alpha}(1,e]$ is a solution of \eqref{e3.1.14}.

On the other hand, if $x$ is a solution of \eqref{e3.1.14}, Cases 1, 2, 3 and
\eqref{e3.1.15} imply
$\lim_{t\to 1^+}(\log t)^{n-\alpha}{x}(t)=\frac{{\eta}_n}{\Gamma(\alpha-n+1)}$.
Then $x\in LC_{n-\alpha}(1,e]$. Furthermore, by Definition \ref{def2.5} we have
\begin{align*}
&{}^{RLH}D_{1^+}^\alpha x(t) \\
&=\frac{1}{\Gamma(n-\alpha)}(t\frac{d}{dt})^n
\Big(\int_1^t(\log\frac{t}{s})^{n-\alpha-1}x(s)\frac{ds}{s}\Big)\\
&=\frac{1}{\Gamma(n-\alpha)}(t\frac{d}{dt})^n
\Big[\int_1^t(\log\frac{t}{s})^{n-\alpha-1}
\Big(\sum_{v=1}^n\frac{\eta_v}{\Gamma(\alpha-v+1)}(\log s)^{\alpha-v} \\
&\quad +\frac{1}{\Gamma(\alpha)}\int_1^s(\log\frac{s}{w})^{\alpha-1}
[A(w)x(w)+F(w)]\frac{dw}{w}\Big)\frac{ds}{s}\Big]\\
&=\frac{1}{\Gamma(n-\alpha)}\sum_{v=1}^n\frac{\eta_v}{\Gamma(\alpha-v+1)}
(t\frac{d}{dt})^n\int_1^t(\log\frac{t}{s})^{n-\alpha-1}
(\log s)^{\alpha-v}\frac{ds}{s}\\
&\quad +\frac{1}{\Gamma(\alpha)}\frac{1}{\Gamma(n-\alpha)}(t\frac{d}{dt})^n
\int_1^t(\log\frac{t}{s})^{n-\alpha-1}\int_1^s(\log\frac{s}{w})^{\alpha-1}
[B(w)x(w)+G(w)]\frac{dw}{w}\frac{ds}{s}\\
&=\frac{1}{\Gamma(n-\alpha)}\sum_{v=1}^n\frac{\eta_v}{\Gamma(\alpha-v+1)}
(t\frac{d}{dt})^n(\log t)^{n-v}\int_0^1(1-w)^{n-\alpha-1} w^{\alpha-v}dw\\
&\quad +\frac{1}{\Gamma(\alpha)}\frac{1}{\Gamma(n-\alpha)}(t\frac{d}{dt})^n
\int_1^t(\log\frac{t}{s})^{n-\alpha-1}\int_1^s(\log\frac{s}{w})^{\alpha-1}
[B(w)x(w)+G(w)]\frac{dw}{w}\frac{ds}{s}\\
&=\frac{1}{\Gamma(\alpha)}\frac{1}{\Gamma(n-\alpha)}(t\frac{d}{dt})^n
\int_1^t\int_u^t(\log\frac{t}{s})^{n-\alpha-1}(\log\frac{s}{w})^{\alpha-1}
\frac{ds}{s}[B(w)x(w)+G(w)]\frac{dw}{w}\\
&=\frac{1}{\Gamma(\alpha)}\frac{1}{\Gamma(n-\alpha)}(t\frac{d}{dt})^n
\int_1^t(\log\frac{t}{w})^{n-1}\int_0^1(1-w)^{n-\alpha-1}w^{\alpha-1}dw \\
&\quad\times [B(w)x(w)+G(w)]\frac{dw}{w}\\
&=\frac{1}{\Gamma(n)}(t\frac{d}{dt})^n
\int_1^t(\log\frac{t}{w})^{n-1}[B(w)x(w)+G(w)]\frac{dw}{w}\\
&=B(t)x(t)+G(t).
\end{align*}
So $x\in LC_{n-\alpha}(1,e]$ is a solution of IVP\eqref{e3.1.3}.
The proof is complete.
\end{proof}

\begin{theorem} \label{thm3.1.6}
Suppose that {\rm (3.A3)} holds. Then \eqref{e3.1.14} has a unique solution.
If $B(t)\equiv\lambda$ and there exists constants $k_2>-1$, $l_2\le 0$
with $l_2>\max\{-\alpha,-n-k_2\}$ and $M_G\ge 0$ such that
$|G(t)|\le M_Gt^{k_2}(1-t)^{l_2}$ for all $t\in (1,e)$, then following problem
\begin{equation}
\begin{gathered}
{}^{RLH}D_{1^+}^{\alpha}{x}(t)=\lambda {x}(t)+{G}(t),\quad
 \text{a.e., t}\in (1,e],\\
\lim_{t\to 1^+}(\log t)^{n-\alpha}{x}(t)=\frac{{\eta}_n}{\Gamma(\alpha-n+1)},\\
\lim_{t\to 1^+}{}^{RLH}D_{1^+}^{\alpha-j}x(t)=\eta_j, \quad j\in \mathbb{N}[1,n-1]
\end{gathered}\label{e3.1.17}
\end{equation}
has a unique solution
\begin{equation}
\begin{aligned}
x(t)&=\sum_{v=1}^n\eta_v(\log t)^{\alpha-v}
\mathbf{E}_{\alpha,\alpha-v+1}(\lambda (\log t)^{\alpha}) \\
&\quad +\int_1^t(\log \frac{t}{u})^{\alpha-1}\mathbf{E}_{\alpha,\alpha}
(\lambda (\log \frac{t}{u})^{\alpha})G(s)\frac{ds}{s},\quad t\in (1,e].
\end{aligned}\label{e3.1.18}
\end{equation}
\end{theorem}

\begin{proof} (i)
From Claims 1, 2 and 3, \eqref{e3.1.14} has a unique solution.

 (ii) From the assumption and $B(t)\equiv\lambda$, one sees that (3.A3) holds
with $k_1=l_1=0$ and $k_2,l_2$ mentioned in assumption. Thus \eqref{e3.1.17}
has a unique solution. From the Picard function sequence we obtain
\begin{align*}
&\phi_i(t)\\
&=\sum_{v=1}^n\frac{\eta_v}{\Gamma(\alpha-v+1)}(\log t)^{\alpha-v}
+\lambda\frac{1}{\Gamma(\alpha)}\int_1^t
(\log \frac{t}{s})^{\alpha-1}\phi_{i-1}(s)\frac{ds}{s} \\
&\quad +\frac{1}{\Gamma(\alpha)}\int_1^t(\log \frac{t}{s})^{\alpha-1}G(s)
\frac{ds}{s}\\
&=\sum_{v=1}^n\frac{\eta_v}{\Gamma(\alpha-v+1)}(\log t)^{\alpha-v}
 +\lambda\sum_{v=1}^n\frac{\eta_v}{\Gamma(\alpha-v+1)}
 \frac{1}{\Gamma(\alpha)}\int_1^t(\log \frac{t}{s})^{\alpha-1}(\log s)^{\alpha-v}
\frac{ds}{s}\\
&\quad +\lambda^2\frac{1}{\Gamma(\alpha)}\int_1^t(\log \frac{t}{s})^{\alpha-1}
\frac{1}{\Gamma(\alpha)}\int_1^s(\log \frac{s}{u})^{\alpha-1}\phi_{i-2}(u)
\frac{du}{u}\frac{ds}{s}\\
&\quad +\frac{1}{\Gamma(\alpha)}\lambda\frac{1}{\Gamma(\alpha)}
\int_1^t(\log \frac{t}{s})^{\alpha-1}\int_1^s(\log \frac{s}{u})^{\alpha-1}G(u)
\frac{du}{u}\frac{ds}{s}+\frac{1}{\Gamma(\alpha)}\int_1^t
(\log \frac{t}{s})^{\alpha-1}G(s)\frac{ds}{s}\\
&=\sum_{v=1}^n\frac{\eta_v}{\Gamma(\alpha-v+1)}(\log t)^{\alpha-v}
 +\lambda\sum_{v=1}^n\frac{\eta_v}{\Gamma(\alpha-v+1)}
\frac{1}{\Gamma(\alpha)}(\log t)^{2\alpha-v} \\
&\quad\times \int_0^1(1-w)^{\alpha-1}w^{\alpha-v}dw
 +\lambda^2\frac{1}{\Gamma(\alpha)}\int_1^t
\int_u^t(\log \frac{t}{s})^{\alpha-1}\frac{1}{\Gamma(\alpha)}
(\log \frac{s}{u})^{\alpha-1}\frac{ds}{s}\phi_{i-2}(u)\frac{du}{u}\\
&\quad +\frac{1}{\Gamma(\alpha)}\lambda\frac{1}{\Gamma(\alpha)}
\int_1^t\int_u^t(\log \frac{t}{s})^{\alpha-1}(\log \frac{s}{u})^{\alpha-1}
\frac{ds}{s}G(u)\frac{du}{u}+\frac{1}{\Gamma(\alpha)}
\int_1^t(\log \frac{t}{s})^{\alpha-1}G(s)\frac{ds}{s}\\
&=\sum_{v=1}^n\frac{\eta_v}{\Gamma(\alpha-v+1)}(\log t)^{\alpha-v}
 +\lambda\sum_{v=1}^n\frac{\eta_v}{\Gamma(2\alpha-v+1)}(\log t)^{2\alpha-v} \\
&\quad  +\frac{\lambda^2 }{\Gamma(2\alpha)}\int_1^t
 (\log \frac{t}{u})^{2\alpha-1}\phi_{i-2}(u)\frac{du}{u}\\
&\quad +\frac{\lambda }{\Gamma(2\alpha)}\int_1^t
 (\log \frac{t}{u})^{2\alpha-1}G(u)\frac{du}{u}
 +\frac{1}{\Gamma(\alpha)}\int_1^t(\log \frac{t}{s})^{\alpha-1}G(s)\frac{ds}{s}\\
&=\sum_{v=1}^n\eta_v(\log t)^{\alpha-v}\Big(\frac{1}{\Gamma(\alpha-v+1)}
 +\frac{\lambda(\log t)^{\alpha}}{\Gamma(2\alpha-v+1)}\Big)
 +\frac{\lambda^2 }{\Gamma(2\alpha)}\int_1^t
 (\log \frac{t}{u})^{2\alpha-1}\phi_{i-2}(u)\frac{du}{u}\\
&\quad +\int_1^t(\log \frac{t}{u})^{\alpha-1}\Big(\frac{\lambda }{\Gamma(2\alpha)}
(\log \frac{t}{u})^{\alpha}+\frac{1}{\Gamma(\alpha)}\Big)G(s)\frac{ds}{s}\\
&=\dots\\
&=\sum_{v=1}^n\eta_v(\log t)^{\alpha-v}\Big(\sum_{j=0}^{i-1}
 \frac{\lambda^j(\log t)^{j\alpha}}{\Gamma(j\alpha+\alpha-v+1)}\Big)
 +\frac{\lambda^i }{\Gamma(i\alpha)}\int_1^t
 (\log \frac{t}{u})^{i\alpha-1}\phi_0(u)\frac{du}{u}\\
&\quad +\int_1^t(\log \frac{t}{u})^{\alpha-1}\Big(\sum_{j=0}^{i-1}
 \frac{\lambda^j }{\Gamma((j+1)\alpha)}
 (\log \frac{t}{u})^{j\alpha}\Big)G(s)\frac{ds}{s}\\
&=\sum_{v=1}^n\eta_v(\log t)^{\alpha-v}
\Big(\sum_{j=0}^i\frac{\lambda^j(\log t)^{j\alpha}}{\Gamma(j\alpha+\alpha-v+1)}\Big)\\
&\quad +\int_1^t(\log \frac{t}{u})^{\alpha-1}
\Big(\sum_{j=0}^{i-1}\frac{\lambda^j }{\Gamma((j+1)\alpha)}
(\log \frac{t}{u})^{j\alpha}\Big)G(s)\frac{ds}{s}\\
&\to \sum_{v=1}^n\eta_v(\log t)^{\alpha-v}\mathbf{E}_{\alpha,\alpha-v+1}
(\lambda (\log t)^{\alpha})+\int_1^t(\log \frac{t}{u})^{\alpha-1}
\mathbf{E}_{\alpha,\alpha}(\lambda (\log \frac{t}{u})^{\alpha})G(s)\frac{ds}{s}.
\end{align*}
Then $x(t)=\lim_{i\to +\infty}\phi_i(t)$ is the unique solution of \eqref{e3.1.17}.
 $x$ is just as in \eqref{e3.1.18}. The proof is complete.
\end{proof}

To obtain solutions of \eqref{e3.1.4}, we need the following assumption:
\begin{itemize}
\item[(3.A4)] there exist constants $k_i>-\alpha+n-1$, $l_i\le 0$ with
$l_i>\max\{-\alpha,-\alpha-k_i\}$, $M_B\ge 0$ and $M_G\ge 0$ such that
$|B(t)|\le M_B(\log t)^{k_1}(1-\log t)^{l_1}$ and
$|G(t)|\le M_G(\log t)^{k_2}(1-\log t)^{l_2}$ for all $t\in (1,e)$.
\end{itemize}
We choose the Picard function sequence as
\begin{gather*}
\phi_0(t)=\sum_{j=0}^{n-1}\frac{\eta_j}{j!}(\log t)^j ,\quad t\in (1,e],\\
\phi_i(t)=\sum_{j=0}^{n-1}\frac{\eta_j}{j!}(\log t)^j
+\frac{1}{\Gamma(\alpha)}\int_1^t(\log \frac{t}{s})^{\alpha-1}
[B(s)\phi_{i-1}(s)+G(s)]\frac{ds}{s},
\end{gather*}
for $t\in (1,e]$, $i=1,2,\dots$.
\smallskip

\noindent\textbf{Claim 1.} $\phi_i\in C(1,e]$.

\begin{proof}
On sees that $\phi_0\in C[1,e]$.
From
\begin{align*}
&\big|\int_1^t(\log \frac{t}{s})^{\alpha-1}[B(s)\phi_0(s)+G(s)]\frac{ds}{s}\big|\\
&\le \int_1^t(\log \frac{t}{s})^{\alpha-1}
[M_B\|\phi_0\|(\log s)^{k_1}(1-\log s)^{l_1} \\
&\quad +M_G(\log s)^{k_2}(1-\log s)^{l_2}]\frac{ds}{s}\\
&\le M_B\|\phi_0\|\int_1^t(\log \frac{t}{s})^{\alpha-1}
 (\log s)^{k_1}(1-\log s)^{l_1}\frac{ds}{s} \\
&\quad +M_G\int_1^t(\log \frac{t}{s})^{\alpha-1}(\log s)^{k_2}(1-\log s)^{l_2}
\frac{ds}{s}\\
&=M_B\|\phi_0\|(\log t)^{\alpha+k_1+l_1}\mathbf{B}(\alpha+l_1,k_1+1)
 +M_G(\log t)^{\alpha+k_2+l_2}\mathbf{B}(\alpha+l_2,k_2+1) \\
&\to 0\quad \text{as }t\to 1^+,
\end{align*}
we obtain that $\lim_{t\to 1^+}\phi_1(s)$ exists and $\phi_1$ is continuous
on $(1,e]$. Then $\phi_1\in C[1,e]$. By mathematical induction, we see that
$\phi_i\in C[1,e]$.
\end{proof}
\smallskip

\noindent\textbf{Claim 2.}
 $\phi_i$  converges uniformly on $[1,e]$.

\begin{proof}
For $t\in [1,e]$ we have
\begin{align*}
&|\phi_1(t)-\phi_0(t)| \\
&=\big|\frac{1}{\Gamma(\alpha)}\int_1^t(\log\frac{t}{s})^{\alpha-1}
[B(s)\phi_0(s)+G(s)]\frac{ds}{s}\big|\\
&\le \frac{1}{\Gamma(\alpha)}\int_1^t(\log\frac{t}{s})^{\alpha-1}
[M_B\|\phi_0\|(\log s)^{k_1}(1-\log s)^{l_1}+M_G(\log s)^{k_2}
(1-\log s)^{l_2}]\frac{ds}{s}\\
& \le \|\phi_0\|M_B\frac{1}{\Gamma(\alpha)}\int_1^t(\log\frac{t}{s})^{\alpha+l_1-1}
(\log s)^{k_1}\frac{ds}{s}+M_G\frac{1}{\Gamma(\alpha)}
\int_1^t(\log\frac{t}{s})^{\alpha+l_2-1}(\log s)^{k_2}\frac{ds}{s}\\
&=\|\phi_0\|M_B(\log t)^{\alpha+k_1+l_1}
 \frac{\mathbf{B}(\alpha+l_1,k_1+1)}{\Gamma(\alpha)} \\
&\quad +M_G(\log t)^{\alpha+k_2+l_2}
\frac{\mathbf{B}(\alpha+l_2,k_2+1)}{\Gamma(\alpha)}.
\end{align*}
So
\begin{align*}
&|\phi_2(t)-\phi_1(t)| \\
&=\big|\frac{1}{\Gamma(\alpha)}\int_1^t(\log\frac{t}{s})^{\alpha-1}B(s)
[\phi_1(s)-\phi_0(s)]\frac{ds}{s}\big|\\
&\le \frac{1}{\Gamma(\alpha)}\int_1^t(\log\frac{t}{s})^{\alpha-1}
M_B(\log s)^{k_1}(1-\log s)^{l_1} \\
&\quad\times \Big(\|\phi_0\|M_B(\log s)^{\alpha+k_1+l_1}
\frac{\mathbf{B}(\alpha+l_1,k_1+1)}{\Gamma(\alpha)} \\
&\quad +M_G(\log s)^{\alpha+k_2+l_2}
\frac{\mathbf{B}(\alpha+l_2,k_2+1)}{\Gamma(\alpha)}\Big)\frac{ds}{s}\\
&\le \frac{1}{\Gamma(\alpha)}\|\phi_0\|M_B^2\int_1^t(\log\frac{t}{s})^{\alpha+l_1-1}
(\log s)^{\alpha+2k_1+l_1}
\frac{\mathbf{B}(\alpha+l_1,k_1+1)}{\Gamma(\alpha)}\frac{ds}{s}\\
\\
&\quad +M_BM_G\le \frac{1}{\Gamma(\alpha)}
 \int_1^t(\log\frac{t}{s})^{\alpha+l_1-1}(\log s)^{\alpha+k_1+k_2+l_2}
 \frac{\mathbf{B}(\alpha+l_2,k_2+1)}{\Gamma(\alpha)}\frac{ds}{s}\\
&=\|\phi_0\|M_B^2(\log t)^{2k_1+2l_1+2}
 \frac{\mathbf{B}(\alpha+l_1,k_1+1)}{\Gamma(\alpha)}
 \frac{\mathbf{B}(\alpha+l_1,\alpha+2k_1+l_1+1)}{\Gamma(\alpha)}\\
&\quad +M_BM_G(\log t)^{2\alpha+k_1+k_2+l_1+l_2}
 \frac{\mathbf{B}(\alpha+l_2,k_2+1)}{\Gamma(\alpha)} \\
&\quad\times  \frac{\mathbf{B}(\alpha+l_1,\alpha+k_1+k_2+l_2+1)}{\Gamma(\alpha)},
\end{align*}
and
\begin{align*}
&|\phi_3(t)-\phi_2(t)| \\
&=\big|\frac{1}{\Gamma(\alpha)}\int_1^t(\log\frac{t}{s})^{\alpha-1}B(s)
[\phi_2(s)-\phi_1(s)]\frac{ds}{s}\big|\\
&\le \frac{1}{\Gamma(\alpha)}\int_1^t(\log\frac{t}{s})^{\alpha-1}
 M_B(\log s)^{k_1}(1-\log s)^{l_1}\\
&\quad\times\Big(\|\phi_0\|M_B^2(\log s)^{2k_1+2l_1+2}
 \frac{\mathbf{B}(\alpha+l_1,k_1+1)}{\Gamma(\alpha)}
 \frac{\mathbf{B}(\alpha+l_1,\alpha+2k_1+l_1+1)}{\Gamma(\alpha)} \\
&\quad +M_BM_G(\log s)^{2\alpha+k_1+k_2+l_1+l_2}
\frac{\mathbf{B}(\alpha+l_2,k_2+1)}{\Gamma(\alpha)} \\
&\quad\times \frac{\mathbf{B}(\alpha+l_1,\alpha+k_1+k_2+l_2+1)}{\Gamma(\alpha)}\Big)
\frac{ds}{s}\\
& \le \|\phi_0\|M_B^3\frac{1}{\Gamma(\alpha)}\int_1^t
 (\log\frac{t}{s})^{\alpha+l_1-1}(\log )s^{2\alpha+3k_1+2l_1}
 \frac{\mathbf{B}(\alpha+l_1,k_1+1)}{\Gamma(\alpha)} \\
&\quad\times  \frac{\mathbf{B}(\alpha+l_1,\alpha+2k_1+l_1+1)}{\Gamma(\alpha)}\frac{ds}{s}\\
&\quad +M_B^2M_G\frac{1}{\Gamma(\alpha)}
\int_1^t(\log\frac{t}{s})^{\alpha+l_1-1}(\log s)^{2\alpha+2k_1+k_2+l_1+l_2} \\
&\quad\times \frac{\mathbf{B}(\alpha+l_2,k_2+1)}{\Gamma(\alpha)}
 \frac{\mathbf{B}(\alpha+l_1,\alpha+k_1+k_2+l_2+1)}{\Gamma(\alpha)}
 \frac{ds}{s} \\
&=\|\phi_0\|M_B^3(\log t)^{3\alpha+3k_1+3l_1}
 \frac{\mathbf{B}(\alpha+l_1,k_1+1)}{\Gamma(\alpha)}
 \frac{\mathbf{B}(\alpha+l_1,\alpha+2k_1+l_1+1)}{\Gamma(\alpha)} \\
&\quad\times \frac{\mathbf{B}(\alpha+l_1,2\alpha+3k_1+2l_1+1)}{\Gamma(\alpha)}\\
&\quad +M_B^2M_G(\log t)^{3\alpha+2k_1+k_2+2l_1+l_2}
 \frac{\mathbf{B}(\alpha+l_2,k_2+1)}{\Gamma(\alpha)}
 \frac{\mathbf{B}(\alpha+l_1,\alpha+k_1+k_2+l_2+1)}{\Gamma(\alpha)}\\
&\quad\times  \frac{\mathbf{B}(\alpha+l_1,2\alpha+2k_1+k_2+l_1+l_2+1)}{\Gamma(\alpha)}.
\end{align*}
\begin{align*}
&|\phi_4(t)-\phi_3(t)| \\
&=\big|\frac{1}{\Gamma(\alpha)}\int_1^t(\log\frac{t}{s})^{\alpha-1}B(s)
[\phi_3(s)-\phi_2(s)]\frac{ds}{s}\big| \\
&\quad +\frac{\|\phi_0\|M_B^4}{\Gamma(\alpha)}\int_1^t(\log\frac{t}{s})^{\alpha+l_1-1}
(\log s)^{3\alpha+4k_1+3l_1}
 \frac{\mathbf{B}(\alpha+l_1,k_1+1)}{\Gamma(\alpha)} \\
&\quad\times \frac{\mathbf{B}(\alpha+l_1,\alpha+2k_1+l_1+1)}{\Gamma(\alpha)}
 \frac{\mathbf{B}(\alpha+l_1,2\alpha+3k_1+2l_1+1)}{\Gamma(\alpha)}\frac{ds}{s}\\
&\quad +\frac{M_B^3M_G}{\Gamma(\alpha)}\int_1^t(\log\frac{t}{s})^{\alpha+l_1-1}
(\log s)^{3\alpha+3k_1+k_2+2l_1+l_2}\\
&\quad \times \frac{\mathbf{B}(\alpha+l_2,k_2+1)}{\Gamma(\alpha)}
 \frac{\mathbf{B}(\alpha+l_1,\alpha+k_1+k_2+l_2+1)}{\Gamma(\alpha)} \\
&\quad\times \frac{\mathbf{B}(\alpha+l_1,2\alpha+2k_1+k_2+l_1+l_2+1)}{\Gamma(\alpha)}
 \frac{ds}{s}\\
&\le \|\phi_0\|M_B^4(\log t)^{4\alpha+4k_1+4l_1}
 \frac{\mathbf{B}(\alpha+l_1,k_1+1)}{\Gamma(\alpha)}
 \frac{\mathbf{B}(\alpha+l_1,\alpha+2k_1+l_1+1)}{\Gamma(\alpha)} \\
&\quad\times \frac{\mathbf{B}(\alpha+l_1,2\alpha+3k_1+2l_1+1)}{\Gamma(\alpha)}
 \frac{ds}{s}\frac{\mathbf{B}(\alpha+l_1,3\alpha+4k_1+3l_1+1)}{\Gamma(\alpha)}\\
&\quad +M_B^3M_G(\log t)^{4\alpha+3k_1+k_2+3l_1+l_2}
 \frac{\mathbf{B}(\alpha+l_2,k_2+1)}{\Gamma(\alpha)} \\
&\quad\times \frac{\mathbf{B}(\alpha+l_1,\alpha+k_1+k_2+l_2+1)}{\Gamma(\alpha)}
 \frac{\mathbf{B}(\alpha+l_1,2\alpha+2k_1+k_2+l_1+l_2+1)}{\Gamma(\alpha)}\\
&\quad\times \frac{\mathbf{B}(\alpha+l_1,3\alpha+3k_1+k_2+2l_1+l_2+1)}
 {\Gamma(\alpha)}.
\end{align*}
Similarly by mathematical induction, for every $i=1,2,\dots$ we obtain
\begin{align*}
&|\phi_i(t)-\phi_{i-1}(t)|\\
&\le \|\phi_0\|M_B^i(\log t)^{i\alpha+ik_1+il_1}
\prod_{j=0}^{i-1}\frac{\mathbf{B}(\alpha+l_1,j\alpha+(j+1)k_1+jl_1+1}{\Gamma(\alpha)}\\
&\quad +M_B^{i-1}M_G(\log t)^{i\alpha+(i-1)k_1+k_2+(i-1)l_1+l_2}
 \frac{\mathbf{B}(\alpha+l_2,k_2+1)}{\Gamma(\alpha)} \\
&\quad\times \prod_{j=1}^{i-1}\frac{\mathbf{B}(\alpha+l,j\alpha+jk_1+k_2+(j-1)l_1+l_2+1)}
 {\Gamma(\alpha)}\\
&\le \|\phi_0\|M_B^i\prod_{j=0}^{i-1}
 \frac{\mathbf{B}(\alpha+l_1,j\alpha+(j+1)k_1+jl_1+1}{\Gamma(\alpha)}\\
&\quad +M_B^{i-1}M_G\frac{\mathbf{B}(\alpha+l_2,k_2+1)}{\Gamma(\alpha)}
 \prod_{j=1}^{i-1}\frac{\mathbf{B}(\alpha+l,j\alpha+jk_1+k_2+(j-1)l_1+l_2+1)}
{\Gamma(\alpha)},
\end{align*}
for $t\in [1,e]$.
Similarly we can prove that both
\begin{align*}
\sum_{i=1}^{+\infty}u_i
&=\sum_{i=1}^{+\infty} M_B^{i-1}M_G
\frac{\mathbf{B}(\alpha+l_2,k_2+1)}{\Gamma(\alpha)} \\
&\quad\times \prod_{j=1}^{i-1}\frac{\mathbf{B}(\alpha+l,j\alpha+jk_1+k_2+(j-1)l_1+l_2+1)}
{\Gamma(\alpha)}, \\
\sum_{i=1}^{+\infty}v_i
&=\sum_{i=1}^{+\infty}M_B^{i-1}M_G
\frac{\mathbf{B}(\alpha+l_2,k_2+1)}{\Gamma(\alpha)} \\
&\quad\times  \prod_{j=1}^{i-1}\frac{\mathbf{B}(\alpha+l,j\alpha+jk_1+k_2+(j-1)l_1+l_2+1)}
{\Gamma(\alpha)}
\end{align*}
are convergent. Hence,
$$
\phi_0(t)+[\phi_1(t)-\phi_0(t)]+[\phi_2(t)-\phi_1(t)]+\dots
+[\phi_i(t)-\phi_{i-1}(t)]+\dots,\quad t\in [1,e]
$$
is uniformly convergent. Then $\{\phi_i(t)\}$ is convergent uniformly on $[1,e]$.
\end{proof}
\smallskip

\noindent\textbf{Claim 3.}
 $\phi(t)=\lim_{i\to +\infty}\phi_i(t)$ defined on $(1,e]$ is a unique continuous
solution of the integral equation
$$
{x}(t)=\sum_{j=0}^{n-1}\frac{\eta_j}{j!}(\log t)^j
+\frac{1}{\Gamma(\alpha)}\int_1^t(\log \frac{t}{s})^{\alpha-1}
[B(s)x(s)+G(s)]\frac{ds}{s}.
\label{e3.1.19}
$$


\begin{proof}
From $\lim_{i\to +\infty}\phi_i(t)=\phi(t)$ and the uniformly convergence,
we see that $\phi(t)$ is continuous on $[1,e]$. From
\begin{align*}
&\big|\int_1^t(\log\frac{t}{s})^{\alpha-1}[B(s)\phi_{p-1}(s)+G(s)]\frac{ds}{s}
-\int_1^t(\log\frac{t}{s})^{\alpha-1}
[B(s) \phi_{q-1}(s)+G(s)]\frac{ds}{s}\big|\\
&\le M_B\|\phi_{p-1}-\phi_{q-1}\|\int_1^t(\log\frac{t}{s})^{\alpha-1}
(\log s)^{k_1}(1-\log s)^{l_1}\frac{ds}{s}\\
& \le M_B\|\phi_{p-1}-\phi_{q-1}\|\int_1^t(\log\frac{t}{s})^{\alpha+l_1-1}
(\log s)^{k_1}\frac{ds}{s}\\
& \le M_B\|\phi_{p-1}-\phi_{q-1}\|(\log t)^{\alpha+k_1+l_1}
 \frac{\mathbf{B}(\alpha+l_1,k_1+1)}{\Gamma(\alpha)}\\
&\le M_B\|\phi_{p-1}-\phi_{q-1}\|\frac{\mathbf{B}(\alpha+l_1,k_1+1)}{\Gamma(\alpha)}\\
&\to 0 \quad \text{uniformly as } p,q\to +\infty,
\end{align*}
we know that
\begin{align*}
\phi(t)
&=\lim_{i\to \infty}\phi_i(t) \\
&=\lim_{i\to +\infty}\Big[\sum_{j=0}^{n-1}\frac{\eta_j}{j!}(\log t)^j
+\frac{1}{\Gamma(\alpha)}\int_1^t(\log \frac{t}{s})^{\alpha-1}
[B(s)\phi_{i-1}(s)+G(s)]\frac{ds}{s}\Big]\\
&=\sum_{j=0}^{n-1}\frac{\eta_j}{j!}(\log t)^j+\frac{1}{\Gamma(\alpha)}
\int_1^t(\log \frac{t}{s})^{\alpha-1}[B(s)\phi(s)+G(s)]\frac{ds}{s}.
\end{align*}
Then $\phi$ is a continuous solution of \eqref{e3.1.19} defined on $(1,e]$.

Suppose that $\psi$ defined on $(1,e]$ is also a solution of \eqref{e3.1.19}. Then
\[
\psi(t) =\sum_{j=0}^{n-1}\frac{\eta_j}{j!}(\log t)^j
+\frac{1}{\Gamma(\alpha)}\int_1^t(\log \frac{t}{s})^{\alpha-1}[B(s)
\phi(s)+G(s)]\frac{ds}{s},t\in (1,e].
\]
We need to prove that $\phi(t)\equiv\psi(t)$ on $(0,1]$.
Now we have
\begin{align*}
& |\psi(t)-\phi_0(t)| \\
&=\frac{1}{\Gamma(\alpha)}|\int_1^t(\log \frac{t}{s})^{\alpha-1}
[B(s)\phi_0(s)+G(s)]\frac{ds}{s}|\\
&\le \|\phi_0\|M_B(\log t)^{\alpha+k_1+l_1}
\frac{\mathbf{B}(\alpha+l_1,k_1+1)}{\Gamma(\alpha)}
+M_G(\log t)^{\alpha+k_2+l_2}\frac{\mathbf{B}(\alpha+l_2,k_2+1)}{\Gamma(\alpha)}.
\end{align*}
Furthermore,
\begin{align*}
& |\psi(t)-\phi_1(t)| \\
&=\frac{1}{\Gamma(\alpha)}|\int_1^t(\log \frac{t}{s})^{\alpha-1}
B(s)[\psi(s)-\phi_0(s)]\frac{ds}{s}|\\
&\le \|\phi_0\|M_B^2(\log t)^{2k_1+2l_1+2}
\frac{\mathbf{B}(\alpha+l_1,k_1+1)}{\Gamma(\alpha)}
\frac{\mathbf{B}(\alpha+l_1,\alpha+2k_1+l_1+1)}{\Gamma(\alpha)}\\
&\quad +M_BM_G(\log t)^{2\alpha+k_1+k_2+l_1+l_2}
 \frac{\mathbf{B}(\alpha+l_2,k_2+1)}{\Gamma(\alpha)}
 \frac{\mathbf{B}(\alpha+l_1,\alpha+k_1+k_2+l_2+1)}{\Gamma(\alpha)}.
\end{align*}
By mathematical induction, we obtain
\begin{align*}
&|\psi(t)-\phi_{i-1}(t)| \\
&=\frac{1}{\Gamma(\alpha)}|\int_1^t(\log \frac{t}{s})^{\alpha-1}B(s)
[\psi(s)-\phi_{m-2}(s)]\frac{ds}{s}|\\
&\le \|\phi_0\|M_B^i(\log t)^{i\alpha+ik_1+il_1}
\prod_{j=0}^{i-1}\frac{\mathbf{B}(\alpha+l_1,j\alpha+(j+1)k_1+jl_1+1}{\Gamma(\alpha)}\\
&\quad +M_B^{i-1}M_G(\log t)^{i\alpha+(i-1)k_1+k_2+(i-1)l_1+l_2}
 \frac{\mathbf{B}(\alpha+l_2,k_2+1)}{\Gamma(\alpha)} \\
&\quad\times \prod_{j=1}^{i-1}
 \frac{\mathbf{B}(\alpha+l,j\alpha+jk_1+k_2+(j-1)l_1+l_2+1)}{\Gamma(\alpha)}\\
&\le \|\phi_0\|M_B^i\prod_{j=0}^{i-1}
 \frac{\mathbf{B}(\alpha+l_1,j\alpha+(j+1)k_1+jl_1+1}{\Gamma(\alpha)}\\
&\quad +M_B^{i-1}M_G\frac{\mathbf{B}(\alpha+l_2,k_2+1)}{\Gamma(\alpha)}
\prod_{j=1}^{i-1}\frac{\mathbf{B}(\alpha+l,j\alpha+jk_1+k_2+(j-1)l_1+l_2+1)}
{\Gamma(\alpha)},
\end{align*}
for $t\in [1,e]$.
Similarly we have $\lim_{i\to +\infty}\phi_i(t)=\psi(t)$ uniformly on $[1,e]$.
Then $\phi(t)\equiv\psi(t)$ on $(1,e]$.
Then \eqref{e3.1.19} has a unique solution $\phi$. The proof is complete.
\end{proof}


\begin{theorem} \label{thm3.1.7}
 Suppose that {\rm (3.A4)} holds. Then ${x}\in C(1,e]$ is a solution of IVP
\eqref{e3.1.4} if and only if $x\in C(1,e]$ is a solution of the integral
equation \eqref{e3.1.19}.
\end{theorem}

\begin{proof}
Suppose that $x\in C(1,e]$ is a solution of \eqref{e3.1.4}. Then $t\to x(t)$
is continuous on $[1,e]$ by defining $x(t)|_{t=1}=\lim_{t\to 1^+}x(t)$ and
$\|x\|=r<+\infty$. One can see that
\begin{align*}
&\int_1^t(\log \frac{t}{s})^{\alpha-1}(\log s)^{k_1}(1-\log s)^{l_1}\frac{ds}{s}\\
&\le \int_1^t(\log \frac{t}{s})^{\alpha+l_1-1}(\log s)^{k_1}
 \frac{ds}{s}
\quad \text{(because $\frac{\log s}{\log t}=u$)}\\
&=(\log t)^{\alpha+k_1+l_1}\int_0^1(1-u)^{\alpha+l_1-1}u^{k_1}du\\
&\le (\log t)^{\alpha+k_1+l_1}\int_0^1(1-u)^{\alpha+l_1-1}u^{k_1}du\\
&=(\log t)^{\alpha+k_1+l_1}\mathbf{B}(\alpha+l_1,k_1+1).
\end{align*}
From (3.A4), for $t\in (1,e]$ we have
\begin{align*}
&\big|\int_1^t(\log \frac{t}{s})^{\alpha-1}[B(s)x(s)+G(s)]\frac{ds}{s}\big|\\
&\le \int_1^t(\log \frac{t}{s})^{\alpha-1}[M_Br(\log s)^{k_1}(1-\log s)^{l_1}
+M_G(\log s)^{k_2}(1-\log s)^{l_2}]\frac{ds}{s}\\
&\le M_Br\int_1^t(\log \frac{t}{s})^{\alpha-1}(\log s)^{k_1}(1-\log s)^{l_1}
\frac{ds}{s}\\
&\quad +M_G\int_1^t(\log \frac{t}{s})^{\alpha-1}(\log s)^{k_2}(1-\log s)^{l_2}
 \frac{ds}{s}\\
&=M_Br(\log t)^{\alpha+k_1+l_1}\mathbf{B}(\alpha+l_1,k_1+1)
+M_G(\log t)^{\alpha+k_2+l_2}\mathbf{B}(\alpha+l_2,k_2+1).
\end{align*}
So $t\to \int_1^t(\log \frac{t}{s})^{\alpha-1}[B(s)x(s)+G(s)]\frac{ds}{s}$
is defined on $(1,e]$ and
\begin{equation}
\lim_{t\to 1^+}\int_1^t(\log \frac{t}{s})^{\alpha-1}[B(s)x(s)+G(s)]\frac{ds}{s}=0.
 \label{e3.1.20}
\end{equation}
 Furthermore, similarly to Theorem \ref{thm3.1.1} we have
$t\to \int_1^t(\log \frac{t}{s})^{\alpha-1}[B(s)x(s)+G(s)]\frac{ds}{s}$
is continuous on $(1,e]$. So
$t\to \int_1^t(\log \frac{t}{s})^{\alpha-1}[B(s)x(s)+G(s)]\frac{ds}{s}$
is continuous on $[1,e]$ by defining
\begin{equation}
 \int_1^t(\log \frac{t}{s})^{\alpha-1}[B(s)x(s)+G(s)]\frac{ds}{s}\big|_{t=1}=0.
 \label{e3.1.21}
\end{equation}
 One sees that
\[
\int_w^t(\log\frac{t}{s})^{\alpha-1}(\log\frac{s}{w})^{-\alpha}\frac{ds}{s}
 =\int_0^1(1-u)^{\alpha-1}u^{-\alpha}du =\Gamma(1-\alpha)\Gamma(\alpha),
\]
because $\frac{\log s-\log w}{\log t-\log w}=u$.
By Definition \ref{def2.6} and
${}^HI_{1^+}^\alpha {}^{CH}D_{1^+}^\alpha x(t)={}^HI_{1^+}^\alpha [B(t)x(t)+G(t)]$
 We have
\begin{align*}
&\frac{1}{\Gamma(\alpha)}\int_1^t(\log\frac{t}{s})^{\alpha-1}
[A(s)x(s)+F(s)]\frac{ds}{s} \\
&={}^HI_{1^+}^\alpha [B(t)x(t)+G(t)]\\
&={}^HI_{1^+}^\alpha {}^{CH}D_{1^+}^\alpha x(t)\\
&=\frac{1}{\Gamma(\alpha)}\int_1^t(\log\frac{t}{s})^{\alpha-1}
\Big[\frac{1}{\Gamma(n-\alpha)}
\int_1^s(\log\frac{s}{w})^{n-\alpha-1}(w\frac{d}{dw})^nx(w)\frac{dw}{w}\Big]
\frac{ds}{s}\\
&=\frac{1}{\Gamma(\alpha)}\frac{1}{\Gamma(n-\alpha)}
 \int_1^t\int_u^t(\log\frac{t}{s})^{\alpha-1}
(\log\frac{s}{w})^{n-\alpha-1}\frac{ds}{s}(w\frac{d}{dw})^nx(w)\frac{dw}{w}\\
&=\frac{1}{\Gamma(\alpha)}\frac{1}{\Gamma(n-\alpha)}
 \int_1^t(\log \frac{t}{w})^{n-1}\int_0^1(1-u)^{\alpha-1}
u^{n-\alpha-1}du(w\frac{d}{dw})^nx(w)\frac{dw}{w}\\
&=\frac{1}{(n-1)!}\int_1^t(\log \frac{t}{w})^{n-1}(w\frac{d}{dw})^nx(w)\frac{dw}{w}\\
&=\int_1^t(\log \frac{t}{w})^{n-1}d\big[(w\frac{d}{dw})^{n-1}x(w)\big]\\
&=\frac{1}{(n-1)!} (\log \frac{t}{w})^{n-1}
\big[(w\frac{d}{dw})^{n-1}x(w)\big]|_1^t \\
&\quad +\frac{1}{(n-2)!}\int_1^t(\log \frac{t}{w})^{n-2}
\big[(w\frac{d}{dw})^{n-1}x(w)\big]\frac{dw}{w}\\
&=-\frac{\eta_{n-1}}{(n-1)!}(\log t)^{n-1}+\frac{1}{(n-2)!}
\int_1^t(\log \frac{t}{w})^{n-2}\big[(w\frac{d}{dw})^{n-1}x(w)\big]\frac{dw}{w}\\
&=\dots \\
&=-\sum_{j=1}^{n-2}\frac{\eta_{n-j}}{(n-j)!}(\log t)^{n-j}+\int_1^tx'(w)dw \\
&=x(t)-\sum_{j=0}^{n-1}\frac{\eta_{j}}{j!}(\log t)^{j}.
\end{align*}
Then $x\in C(1,e]$ is a solution of \eqref{e3.1.19}.


On the other hand, if $x\in C(1,e]$ is a solution of \eqref{e3.1.19},
then \eqref{e3.1.20} implies $\lim_{t\to 1^+}(t)={\eta}_0$. Furthermore,
 for $t\in (1,e)$ we have
\begin{align*}
&\big|\int_1^t(\log \frac{t}{s})^{\alpha-n}[B(s)x(s)+G(s)]\frac{ds}{s}\big|\\
&\le \int_1^t(\log \frac{t}{s})^{\alpha-n}[M_Br(\log s)^{k_1}(1-\log s)^{l_1}
+M_G(\log s)^{k_2}(1-\log s)^{l_2}]\frac{ds}{s}\\
&\le M_Br\int_1^t(\log \frac{t}{s})^{\alpha-n}(\log s)^{k_1}(1-\log t)^{l_1}
 \frac{ds}{s} \\
&\quad +M_G\int_1^t(\log \frac{t}{s})^{\alpha-n}(\log s)^{k_2}
(1-\log t)^{l_2}\frac{ds}{s}\\
&=M_Br(1-\log t)^{l_1}\int_1^t(\log \frac{t}{s})^{\alpha-n}(\log s)^{k_1}
 \frac{ds}{s} \\
&\quad +M_G(1-\log t)^{l_2}\int_1^t(\log \frac{t}{s})^{\alpha-n}(\log s)^{k_2}
 \frac{ds}{s}\\
&=M_Br(1-\log t)^{l_1}(\log t)^{\alpha-n+k_1+1}\mathbf{B}(\alpha-n+1,k_1+1)\\
&\quad +M_G(1-\log t)^{l_2}(\log t)^{\alpha-n+k_2+1}\mathbf{B}(\alpha-n+1,k_2+1)
 \to 0\quad \text{ as }t\to 1^+.
\end{align*}
Then
\begin{align*}
&{}^{CH}D_{1^+}^\alpha x(t) \\
&=\frac{1}{\Gamma(n-\alpha)}\int_1^t(\log \frac{t}{s})^{n-\alpha-1}
(s\frac{d}{ds})^nx(s)\frac{ds}{s}\\
&=\frac{1}{\Gamma(n-\alpha)}\int_1^t(\log \frac{t}{s})^{n-\alpha-1}
 (s\frac{d}{ds})^n\Big(\sum_{j=0}^{n-1}\frac{\eta_j}{j!}(\log s)^j \\
&\quad +\frac{1}{\Gamma(\alpha)}\int_1^s(\log \frac{s}{u})^{\alpha-1}[B(u)x(u)+G(u)]
 \frac{du}{u}\Big)\frac{ds}{s}\\
&=\frac{1}{\Gamma(n-\alpha)}\int_1^t(\log \frac{t}{s})^{n-\alpha-1}
 (s\frac{d}{ds})^n\sum_{j=0}^{n-1}\frac{\eta_j}{j!}(\log s)^j\frac{ds}{s}\\
&\quad +\frac{1}{\Gamma(\alpha)}\frac{1}{\Gamma(n-\alpha)}
 \int_1^t(\log \frac{t}{s})^{n-\alpha-1}(s\frac{d}{ds})^n
 \Big(\int_1^s(\log \frac{s}{u})^{\alpha-1}[B(u)x(u)+G(u)]\frac{du}{u}\Big)
 \frac{ds}{s}\\
&=\frac{1}{\Gamma(\alpha)}\frac{1}{\Gamma(n-\alpha)}
 \int_1^t(\log \frac{t}{s})^{n-\alpha-1}(s\frac{d}{ds})^n
\Big(\int_1^s(\log \frac{s}{u})^{\alpha-1}[B(u)x(u)+G(u)]\frac{du}{u}\Big)
 \frac{ds}{s}\\
&=\frac{1}{\Gamma(\alpha)}\frac{1}{\Gamma(n-\alpha)}
 \int_1^t(\log \frac{t}{s})^{n-\alpha-1}(s\frac{d}{ds})^{n-1}
 \Big(\int_1^s(\log \frac{s}{u})^{\alpha-2}[B(u)x(u)+G(u)]\frac{du}{u}\Big)
 \frac{ds}{s}\\
&=\dots \\
&=\frac{1}{\Gamma(\alpha)}\frac{1}{\Gamma(n-\alpha)}
 \int_1^t(\log \frac{t}{s})^{n-\alpha-1}
\Big(\int_1^s(\log \frac{s}{u})^{\alpha-n}[B(u)x(u)+G(u)]\frac{du}{u}\Big)'ds\\
&=\frac{1}{\Gamma(\alpha)}\frac{1}{\Gamma(n-\alpha+1)}t
\Big[\int_1^t(\log \frac{t}{s})^{n-\alpha}(\int_1^s
 (\log \frac{s}{u})^{\alpha-1}[B(u)x(u)+G(u)]\frac{du}{u})'ds\Big]'\\
&=\frac{1}{\Gamma(\alpha)}\frac{1}{\Gamma(n-\alpha+1)}t
 \Big[ (\log \frac{t}{s})^{n-\alpha}\int_1^s(\log \frac{s}{w})^{\alpha-n}
 [B(w)x(w)+G(w)]\frac{dw}{w}\big|_1^t \\
&\quad +(n-\alpha)\frac{1}{s}\int_1^t(\log \frac{t}{s})^{n-\alpha-1}
 \int_1^s(\log \frac{s}{w})^{\alpha-n}[B(w)x(w)+G(w)]\frac{dw}{w}ds\Big]'\\
&=\frac{1}{\Gamma(\alpha)}\frac{1}{\Gamma(n-\alpha)}t
\Big[\int_1^t\int_u^t(\log \frac{t}{s})^{n-\alpha-1}(\log
\frac{s}{w})^{\alpha-n}\frac{ds}{s}[B(w)x(w)+G(w)]\frac{dw}{w}\Big]'\\
&=\frac{1}{\Gamma(\alpha)}\frac{1}{\Gamma(n-\alpha)}t
\Big[\int_1^t\int_u^t(\log \frac{t}{s})^{n-\alpha-1}(\log \frac{s}{w})^{\alpha-n}
\frac{ds}{s}[B(w)x(w)+G(w)]\frac{dw}{w}\Big]'\\
&=\frac{1}{\Gamma(\alpha)}\frac{1}{\Gamma(n-\alpha)}t
\Big[\int_1^t\int_0^1(1-u)^{n-\alpha-1}u^{\alpha-n}du[B(w)x(w)+G(w)]\frac{dw}{w}\Big]'\\
&=B(t)x(t)+G(t).
\end{align*}
So $x\in C(1,e]$ is a solution of \eqref{e3.1.4}. The proof is complete.
\end{proof}

\begin{theorem} \label{thm3.1.8}
 Suppose that {\rm (3.A4)} holds. Then \eqref{e3.1.4} has a unique solution.
If $B(t)\equiv\lambda$ and there exists constants $k_2>-\alpha+n-1$,
$l_2\le 0$ with $l_2>\max\{-\alpha,-\alpha-k_2\}$ and $M_G\ge 0$ such that
$|G(t)|\le M_Gt^{k_2}(1-t)^{l_2}$ for all $t\in (1,e)$,
then the problem
\begin{equation}
\begin{gathered}
{}^{CH}D_{0^+}^{\alpha}{x}(t)=\lambda {x}(t)+{G}(t),\quad
 \text{a.e. t}\in (1,e],\\
\lim_{t\to 1^+}(t\frac{d}{dt})^j{x}(t)={\eta}_j,\quad j\in \mathbb{N}[0,n-1]
\end{gathered}\label{e3.1.22}
\end{equation}
has a unique solution
\begin{equation}
\begin{aligned}
x(t)&= \sum_{j=0}^{n-1}\eta_j(\log t)^j\mathbf{E}_{\alpha,j+1}
 (\lambda (\log t)^\alpha) \\
&\quad +\int_1^t(\log\frac{t}{s})^{\alpha-1}
\mathbf{E}_{\alpha,\alpha}(\lambda (\log t-\log s)^\alpha)G(s)\frac{ds}{s},
\quad t\in (1,e].
\end{aligned} \label{e3.1.23}
\end{equation}
\end{theorem}

\begin{proof} From Claims 1, 2 and 3, Theorem \ref{thm3.1.7}, \eqref{e3.1.4} has a
unique solution. From the assumption and $A(t)\equiv\lambda$, one sees that
 (3.A4) holds with $k_1=l_1=0$ and $k_2,l_2$ mentioned. Thus \eqref{e3.1.22}
has a unique solution. From the Picard function sequence we obtain
\begin{align*}
&\phi_i(t) \\
&=\sum_{j=0}^{n-1}\frac{\eta_j}{j!}(\log t)^j
+\lambda\frac{1}{\Gamma(\alpha)}\int_1^t(\log\frac{t}{s})^{\alpha-1}
\phi_{i-1}(s)\frac{ds}{s}
+\frac{1}{\Gamma(\alpha)}\int_1^t(\log\frac{t}{s})^{\alpha-1}G(s)\frac{ds}{s} \\
&=\sum_{j=0}^{n-1}\frac{\eta_j}{j!}(\log t)^j
 +\frac{\lambda }{\Gamma(\alpha)}\sum_{j=0}^{n-1}
 \frac{\eta_j}{j!}\int_1^t(\log\frac{t}{s})^{\alpha-1}
(\log s)^j \frac{ds}{s}\\
&\quad +\frac{\lambda }{\Gamma(\alpha)}
\frac{\lambda }{\Gamma(\alpha)}\int_1^t(\log\frac{t}{s})^{\alpha-1}
 \int_1^s(\log\frac{s}{u})^{\alpha-1} \phi_{i-2}(u)\frac{du}{u}\frac{ds}{s}\\
&\quad +\frac{1}{\Gamma(\alpha)}\frac{\lambda }{\Gamma(\alpha)}
 \int_1^t(\log\frac{t}{s})^{\alpha-1}\int_1^s(\log\frac{s}{u})^{\alpha-1}G(u)
 \frac{du}{u} \frac{ds}{s}+\frac{1}{\Gamma(\alpha)}
 \int_1^t(\log\frac{t}{s})^{\alpha-1}G(s)\frac{ds}{s} \\
&=\sum_{j=0}^{n-1}\frac{\eta_j}{j!}(\log t)^j
 +\frac{\lambda }{\Gamma(\alpha)}\sum_{j=0}^{n-1}
 \frac{\eta_j}{j!}(\log t)^{\alpha+j}\int_0^1(1-w)^{\alpha-1}w^j dw\\
&\quad +\frac{\lambda }{\Gamma(\alpha)}\frac{\lambda }{\Gamma(\alpha)}
 \int_1^t\int_u^t(\log\frac{t}{s})^{\alpha-1}(\log\frac{s}{u})^{\alpha-1}\frac{ds}{s}
 \phi_{i-2}(u)\frac{du}{u}\\
&\quad +\frac{1}{\Gamma(\alpha)}\frac{\lambda }{\Gamma(\alpha)}
 \int_1^t\int_u^t(\log\frac{t}{s})^{\alpha-1}(\log\frac{s}{u})^{\alpha-1}
 \frac{ds}{s}G(u)\frac{du}{u}
+\frac{1}{\Gamma(\alpha)}\int_1^t(\log\frac{t}{s})^{\alpha-1}G(s)\frac{ds}{s} \\
&=\sum_{j=0}^{n-1}\frac{\eta_j}{\Gamma(j+1)}(\log t)^j
 +\sum_{j=0}^{n-1}\frac{\lambda\eta_j}{\Gamma(\alpha+j+1)}(\log t)^{\alpha+j} \\
&\quad +\frac{\lambda }{\Gamma(\alpha)}\frac{\lambda }{\Gamma(\alpha)}
 \int_1^t(\log\frac{t}{u})^{2\alpha-1}\int_0^1(1-w)^{\alpha-1}w^{\alpha-1}dw
 \phi_{i-2}(u)\frac{du}{u}\\
&\quad +\frac{1}{\Gamma(\alpha)}\frac{\lambda }{\Gamma(\alpha)}
 \int_1^t(\log\frac{t}{u})^{2\alpha-1}
 \int_0^1(1-w)^{\alpha-1}w^{\alpha-1}dwG(u)\frac{du}{u} \\
&\quad +\frac{1}{\Gamma(\alpha)}\int_1^t(\log\frac{t}{s})^{\alpha-1}G(s)\frac{ds}{s}\\
&=\sum_{j=0}^{n-1}\eta_j(\log t)^j\big(\frac{1}{\Gamma(j+1)}
 +\frac{\lambda(\log t)^{\alpha}}{\Gamma(\alpha+j+1)}\big)
 +\frac{\lambda^2 }{\Gamma(2\alpha)}\int_1^t
 (\log\frac{t}{u})^{2\alpha-1}\phi_{i-2}(u)\frac{du}{u}\\
&\quad +\int_1^t(\log\frac{t}{s})^{\alpha-1}\big(
 \frac{\lambda (\log t\log s)^\alpha }{\Gamma(2\alpha)}+\frac{1}{\Gamma(\alpha)}\big)
 G(s)\frac{ds}{s} \\
&=\dots\\
&=\sum_{j=0}^{n-1}\eta_j(\log t)^j(\sum_{v=0}^{i-1}
 \frac{\lambda^v(\log t)^{v\alpha}}{\Gamma(v\alpha+j+1)})
 +\frac{\lambda^i }{\Gamma(i\alpha)}\int_1^t(\log\frac{t}{u})^{m\alpha-1}
 \phi_0(u)\frac{du}{u}\\
&\quad +\int_1^t(\log\frac{t}{s})^{\alpha-1}(\sum_{v=0}^{i-1}
 \frac{\lambda^v (\log t\log s)^{v\alpha } }{\Gamma((v+1)\alpha)})G(s)\frac{ds}{s}\\
&=\sum_{j=0}^{n-1}\eta_j(\log t)^j
 \Big(\sum_{v=0}^i\frac{\lambda^v(\log t)^{v\alpha}}{\Gamma(v\alpha+j+1)}\Big)
+\int_1^t(\log\frac{t}{s})^{\alpha-1}
 \Big(\sum_{v=0}^{i-1}\frac{\lambda^v (\log t\log s)^{v\alpha } }{\Gamma((v+1)\alpha)}
 \Big)G(s)\frac{ds}{s} \\
&\to \sum_{j=0}^{n-1}\eta_j(\log t)^j\mathbf{E}_{\alpha,j+1}
 (\lambda (\log t)^\alpha)
+\int_1^t(\log\frac{t}{s})^{\alpha-1}\mathbf{E}_{\alpha,\alpha}
 (\lambda (\log t-\log s)^\alpha)G(s)\frac{ds}{s}.
\end{align*}
Then $x(t)=\lim_{i\to+\infty}\phi_i(t)$ is the unique solution of \eqref{e3.1.22}.
$x$ is just as in \eqref{e3.1.23}. The proof is complete.
\end{proof}

We list the following two fixed point theorems which will be used in Section 4.

\begin{theorem}[Schaefer's fixed point theorem \cite{10}] \label{thm3.1.9}
 Let $E$ be a Banach spaces and $T : E\to E$ be a completely continuous operator. If
the set $E(T)=\{x=\theta (Tx):\text{ for some }\theta\in [0,1),x\in E\}$
is bounded, then $T$ has at least a fixed point in $E$.
\end{theorem}


\begin{theorem}[\cite{sss8}] \label{thm3.1.10}
 Let $X$ be a Banach space. Assume that $\Omega$ is an open bounded subset
of $X$ with $0\in \Omega$ and let $T :X\to X$ be a completely continuous
operator such that
$\|Tx\|\le \|x\|$ for all $x\in \partial \Omega$. Then $T$ has a fixed point
in $\Omega$.
\end{theorem}

\subsection{Exact piecewise continuous solutions of LFDEs}

In this section, we present exact piecewise continuous solutions of the
linear fractional differential equations (LFDEs)
\begin{gather}
{}^{C}D_{0^+}^{\alpha}{x}(t)=\lambda {x}(t)+{F}(t),\quad
\text{a.e., }t\in (t_i,t_{i+1}],\; i\in \mathbb{N}[0,m], \label{e3.2.1} \\
{}^{RL}D_{0^+}^{\alpha}{x}(t)=\lambda {x}(t)+{F}(t),\quad
\text{a.e., }t\in (t_i,t_{i+1}],\; i\in \mathbb{N}[0,m], \label{e3.2.2} \\
{}^{RLH}D_{0^+}^{\alpha}{x}(t)=\lambda {x}(t)+{G}(t),\quad
\text{a.e., }t\in (s_i,s_{i+1}],\; i\in \mathbb{N}[0,m], \label{e3.2.3} \\
{}^{CH}D_{0^+}^{\alpha}{x}(t)=\lambda {x}(t)+{G}(t),\quad
\text{a.e., }t\in (s_i,s_{i+1}],\;i\in \mathbb{N}[0,m],\label{e3.2.4}
\end{gather}
where $n-1< \alpha<n$, $\lambda\in \mathbb{R}$, 
$0=s_0<t_1<\dots<s_m<s_{m+1}=1$
in \eqref{e3.2.1} and \eqref{e3.2.2} and $1=t_0<t_1<\dots<t_m<t_{m+1}=e$
in \eqref{e3.2.3} and \eqref{e3.2.4}.

We say that $x:(0,1]\to \mathbb{R}$ is a piecewise solution of \eqref{e3.2.1}
(or \eqref{e3.2.2} if $x\in P_mC(0,1]$ (or $P_mC_{n-\alpha}(0,1]$
and satisfies \eqref{e3.2.1} or \eqref{e3.2.2}. We say that $x:(1,e]\to \mathbb{R}$
is a piecewise continuous solutions of \eqref{e3.2.3} (or \eqref{e3.2.4})
if $x\in LP_mC_{n-\alpha}(1,e]$, (or $LP_mC(1,e]$) and $x$ satisfies
all equations in \eqref{e3.2.3} (or \eqref{e3.2.4}).

\begin{theorem} \label{thm3.2.1}
Suppose that $F$ is continuous on $(0,1)$ and there exist constants $k>-\alpha+n-1$
 and $l\in (-\alpha,-\alpha-k,0]$ such that $|F(t)|\le t^k(1-t)^l$ for all
$t\in (0,1)$. Then $x$ is a piecewise solution of \eqref{e3.2.1} if and only
if there exist constants $c_{iv} \in \mathbb{R}(i\in \mathbb{N}[0,m]$, $v\in \mathbb{N}[0,n-1])$ such that
\begin{equation}
\begin{aligned}
x(t)&=\sum_{\sigma=0}^i\sum_{v=0}^{n-1}c_{\sigma v}\mathbf{E}_{\alpha,v+1}
 (\lambda (t-t_\sigma)^\alpha)(t-t_\sigma)^v\\
&\quad +\int_0^t(t-s)^{\alpha-1}\mathbf{E}_{\alpha,\alpha}
 (\lambda(t-s)^\alpha)F(s)ds, \quad t\in (t_i,t_{i+1}],\; i\in N[0,m].
\end{aligned} \label{e3.2.5}
\end{equation}
\end{theorem}


\begin{proof}
Firstly, for $t\in (t_i,t_{i+1}]$ we have
\begin{align*}
&\big|\int_0^t(t-s)^{\alpha-1}\mathbf{E}_{\alpha,\alpha}
(\lambda (t-s)^\alpha)F(s)ds\big| \\
&\le \int_0^t(t-s)^{\alpha-1}\mathbf{E}_{\alpha,\alpha}
 (\lambda (t-s)^\alpha)|F(s)|ds \\
& \le \int_0^t(t-s)^{\alpha-1}\mathbf{E}_{\alpha,\alpha}
 (\lambda (t-s)^\alpha)s^k(1-s)^lds\\
&= \sum_{j=0}^{+\infty}\frac{\lambda^j}{\Gamma ((j+1)\alpha)}
 \int_0^t(t-s)^{\alpha-1}(t-s)^{\alpha j}s^k(1-s)^lds\\
&\le\sum_{j=0}^{+\infty}\frac{\lambda^j}{\Gamma ((j+1)\alpha)}
 \int_0^t(t-s)^{\alpha+l-1}(t-s)^{\alpha j}s^kds\\
&= \sum_{j=0}^{+\infty}\frac{\lambda^j}{\Gamma ((j+1)\alpha)}
 t^{\alpha+\alpha j+k+l}\int_0^1(1-w)^{\alpha+\alpha j+l-1}w^kdw \\
&\le \sum_{j=0}^{+\infty}\frac{\lambda^j t^{\alpha j}}{\Gamma
 ((j+1)\alpha)}t^{\alpha+k+l}\int_0^1(1-w)^{\alpha+l-1}w^kdw\\
&=t^{\alpha+k+l}\mathbf{E}_{\alpha,\alpha}(\lambda t^\alpha)\mathbf{B}(\alpha+l,k+1).
\end{align*}
Then $\int_0^t(t-s)^{\alpha-1}\mathbf{E}_{\alpha,\alpha}(\lambda (t-s)^\alpha)F(s)ds$
 is convergent and is continuous on $[0,1]$.
If $x$ is a solution of \eqref{e3.2.5}, then we know that
 $\lim_{t\to t_i^+}x(t)$ $(i\in N_0)$ exist and $x\in P_mC(0,1]$.
From
\begin{align*}
\big|\int_0^s\frac{\lambda^\tau (s-u)^{\tau\alpha+\alpha-n}}
{\Gamma((\tau+1)\alpha-n+1)}F(u)du|
&\le \int_0^s\frac{\lambda^\tau (s-u)^{\tau\alpha+\alpha-n+l}}
 {\Gamma((\tau+1)\alpha-n+1)}u^kdu\\
&=s^{\tau\alpha+\alpha-n+k+l+1}
 \int_0^1\frac{\lambda^\tau (1-w)^{\tau\alpha+\alpha-n+l}}
 {\Gamma((\tau+1)\alpha-n+1)}w^kdw \\
&\le s^{\alpha-n+k+l+1}\int_0^1 (1-w)^{\alpha-n+l}w^kdw\\
&=s^{\alpha-n+k+l+1}\frac{\lambda^\tau }{\Gamma((\tau+1)\alpha-n+1)},
\end{align*}
we know that
$$
\lim_{s\to 0^+}\sum_{\tau=0}^{+\infty}
\int_0^s\frac{\lambda^\tau (s-u)^{\tau\alpha+\alpha-n}}
{\Gamma((\tau+1)\alpha-n+1)}F(u)du=0.
$$
Now we prove that $x$ satisfies differential equation in \eqref{e3.2.1}.
 In fact, for $t\in (t_0,t_1]$ we have by Theorem \ref{thm3.1.2} that
${}^CD_{0^+}^\alpha x(t)=\lambda x(t)+F(t)$. For $t\in (t_i,t_{i+1}](i\in N[1,m])$,
 we have by Definition \ref{def2.3} that
\begin{align*}
&{}^CD_{0^+}^\alpha x(t) \\
&=\frac{1}{\Gamma (n-\alpha)}\int_0^t(t-s)^{n-\alpha-1}x^{(n)}(s)ds\\
&=\frac{1}{\Gamma (n-\alpha)}
\Big[\sum_{j=0}^{i-1}\int_{t_j}^{t_{j+1}}(t-s)^{n-\alpha-1}x^{(n)}(s)ds
+\int_{t_i}^t(t-s)^{n-\alpha-1}x^{(n)}(s)ds\Big] \\
&=\frac{1}{\Gamma (n-\alpha)}
\Big[\sum_{j=0}^{i-1}\int_{t_j}^{t_{j+1}}(t-s)^{n-\alpha-1}
\Big(\sum_{\sigma=0}^j\sum_{v=0}^{n-1}c_{\sigma v}
\mathbf{E}_{\alpha,v+1}(\lambda (s-t_\sigma)^\alpha)(s-t_\sigma)^v \\
&\quad +\int_0^s(s-u)^{\alpha-1}\mathbf{E}_{\alpha,\alpha}
 (\lambda(s-u)^\alpha)F(u)du \Big)^{(n)}ds \\
&\quad +\int_{t_i}^t(t-s)^{n-\alpha-1}
\Big(\sum_{\sigma=0}^i\sum_{v=0}^{n-1}c_{\sigma v}\mathbf{E}_{\alpha,v+1}
(\lambda (s-t_\sigma)^\alpha)(s-t_\sigma)^v \\
&\quad +\int_0^s(s-u)^{\alpha-1}\mathbf{E}_{\alpha,\alpha}
(\lambda(s-u)^\alpha)F(u)du)^{(n)}ds\Big]\\
&=\frac{1}{\Gamma (n-\alpha)}
\Big[\sum_{j=0}^{i-1}\int_{t_j}^{t_{j+1}}(t-s)^{n-\alpha-1}
\Big(\sum_{\sigma=0}^j\sum_{v=0}^{n-1}c_{\sigma v}
 \frac{(s-t_\sigma)^{v}}{\Gamma(v+1)} \\
&\quad +\sum_{\sigma=0}^j\sum_{v=0}^{n-1}c_{\sigma v}
\sum_{\tau=1}^{+\infty}\frac{\lambda^\tau(s-t_\sigma)^{\tau\alpha+v}}
{\Gamma(\tau \alpha +v+1)}\Big)^{(n)}ds
\\
&\quad +\int_{t_i}^t(t-s)^{n-\alpha-1}\Big(\sum_{\sigma=0}^i\sum_{v=0}^{n-1}c_{\sigma v}
\frac{(s-t_\sigma)^{v}}{\Gamma(v+1)}+\sum_{\sigma=1}^i\sum_{v=0}^{n-1}c_{\sigma v}\sum_{\tau=0}^{+\infty}
\frac{\lambda^\tau(s-t_\sigma)^{\tau\alpha+v}}{\Gamma(\tau \alpha +v+1)}
\Big)^{(n)}ds \\
&\quad +\int_0^t(t-s)^{n-\alpha-1}Big(\sum_{\tau=0}^{+\infty}
 \int_0^s\frac{\lambda^\tau (s-u)^{\tau\alpha+\alpha-1}}
 {\Gamma((\tau+1)\alpha)}F(u)du\Big)^{(n)}ds\Big]\\
&=\frac{1}{\Gamma (n-\alpha)}
\Big[\sum_{j=0}^{i-1}\int_{t_j}^{t_{j+1}}(t-s)^{n-\alpha-1}
\Big(\sum_{\sigma=0}^j\sum_{v=0}^{n-1}c_{\sigma v}
\sum_{\tau=1}^{+\infty}\frac{\lambda^\tau(s-t_\sigma)^{\tau\alpha+v}}
{\Gamma(\tau \alpha +v+1)}\Big)^{(n)}ds \\
&\quad +\int_{t_i}^t(t-s)^{n-\alpha-1}\Big(\sum_{\sigma=0}^i\sum_{v=0}^{n-1}c_{\sigma v}\sum_{\tau=0}^{+\infty}
\frac{\lambda^\tau(s-t_\sigma)^{\tau\alpha+v}}{\Gamma(\tau \alpha +v+1)}
\Big)^{(n)}ds \\
&\quad +\int_0^t(t-s)^{n-\alpha-1}
\Big(\sum_{\tau=0}^{+\infty}\int_0^s\frac{\lambda^\tau (s-u)^{\tau\alpha+\alpha-n}}
{\Gamma((\tau+1)\alpha-n+1)}F(u)du\Big)'ds\Big]\\
&=\frac{1}{\Gamma (n-\alpha)}\sum_{j=0}^{i-1}\sum_{\sigma=0}^j
\sum_{v=0}^{n-1}c_{\sigma v}\sum_{\tau=1}^{+\infty}
\frac{\lambda^\tau}{\Gamma(\tau \alpha +v+1-n)} \\
&\quad \times \int_{t_j}^{t_{j+1}}(t-s)^{n-\alpha-1}
(s-t_\sigma)^{\tau\alpha+v-n}ds\\
&\quad +\frac{1}{\Gamma (n-\alpha)}\sum_{\sigma=0}^i
 \sum_{v=0}^{n-1}c_{\sigma v}\sum_{\tau=0}^{+\infty}
\frac{\lambda^\tau}{\Gamma(\tau \alpha +v+1-n)} \\
&\quad\times  \int_{t_i}^t(t-s)^{n-\alpha-1}(s-t_\sigma)^{\tau\alpha+v-n}ds\\
&\quad +\frac{1}{\Gamma (n-\alpha+1)}
\Big[\int_0^t(t-s)^{n-\alpha}(\sum_{\tau=0}^{+\infty}\int_0^s
\frac{\lambda^\tau (s-u)^{\tau\alpha+\alpha-n}}{\Gamma((\tau+1)\alpha-n+1)}F(u)du)'ds
\Big]'\\
&=\frac{1}{\Gamma (n-\alpha)}\sum_{j=0}^{i-1}\sum_{\sigma=0}^j\sum_{v=0}^{n-1}
c_{\sigma v}\sum_{\tau=1}^{+\infty}\frac{\lambda^\tau}{\Gamma(\tau \alpha +v+1-n)}
(t-t_\sigma)^{\tau\alpha-\alpha+v} \\
&\quad\times \int_{\frac{t_j-t_\sigma}{t-t_\sigma}}
 ^{\frac{t_{j+1}-t_\sigma}{t-t_\sigma}}
(1-w)^{n-\alpha-1} w^{\tau\alpha+v-n}dw\\
&\quad +\frac{1}{\Gamma (n-\alpha)}\sum_{\sigma=0}^i
 \sum_{v=0}^{n-1}c_{\sigma v}\sum_{\tau=0}^{+\infty}
\frac{\lambda^\tau}{\Gamma(\tau \alpha +v+1-n)}(t-t_\sigma)^{\tau\alpha-\alpha+v} \\
&\quad\times \int_{\frac{t_i-t_\sigma}{t-t_\sigma}}^1(1-w)^{n-\alpha-1}w^{\tau\alpha+v-n}dw \\
&\quad +\frac{1}{\Gamma (n-\alpha+1)}
\Big[ (t-s)^{n-\alpha}(\sum_{\tau=0}^{+\infty}\int_0^s
\frac{\lambda^\tau (s-u)^{\tau\alpha+\alpha-n}}{\Gamma((\tau+1)\alpha-n+1)}
F(u)du)|_0^t \\
&\quad +(n-\alpha)\int_0^t(t-s)^{n-\alpha-1}(\sum_{\tau=0}^{+\infty}
\int_0^s\frac{\lambda^\tau (s-u)^{\tau\alpha+\alpha-n}}{\Gamma((\tau+1)\alpha-n+1)}
F(u)du)ds\Big]'\\
&=\frac{1}{\Gamma (n-\alpha)}\sum_{\sigma=0}^{i-1}
\sum_{j=\sigma}^{i-1}\sum_{v=0}^{n-1}c_{\sigma v}\sum_{\tau=1}^{+\infty}
\frac{\lambda^\tau}{\Gamma(\tau \alpha +v+1-n)}(t-t_\sigma)^{\tau\alpha-\alpha+v} \\
&\quad \times \int_{\frac{t_j-t_\sigma}{t-t_\sigma}}^{\frac{t_{j+1}-t_\sigma}{t-t_\sigma}}
(1-w)^{n-\alpha-1} w^{\tau\alpha+v-n}dw\\
&\quad +\frac{1}{\Gamma (n-\alpha)}\sum_{\sigma=0}^i\sum_{v=0}^{n-1}
c_{\sigma v}\sum_{\tau=0}^{+\infty}
\frac{\lambda^\tau}{\Gamma(\tau \alpha +v+1-n)}(t-t_\sigma)^{\tau\alpha-\alpha+v} \\
&\quad \times \int_{\frac{t_i-t_\sigma}{t-t_\sigma}}^1(1-w)^{n-\alpha-1}
w^{\tau\alpha+v-n}dw \\
&\quad +\frac{1}{\Gamma (n-\alpha)}
\Big[\sum_{\tau=0}^{+\infty}\int_0^t\int_u^t(t-s)^{n-\alpha-1}
\frac{\lambda^\tau (s-u)^{\tau\alpha+\alpha-n}}{\Gamma((\tau+1)\alpha-n+1)}dsF(u)du
\Big]'\\
&=\frac{1}{\Gamma (n-\alpha)}\sum_{\sigma=0}^{i-1}
\sum_{v=0}^{n-1}c_{\sigma v}\sum_{\tau=1}^{+\infty}
\frac{\lambda^\tau}{\Gamma(\tau \alpha +v+1-n)}
(t-t_\sigma)^{\tau\alpha-\alpha+v} \\
&\quad\times \int_0^{\frac{t_{i}-t_\sigma}{t-t_\sigma}}(1-w)^{n-\alpha-1}
w^{\tau\alpha+v-n}dw\\
&\quad +\frac{1}{\Gamma (n-\alpha)}\sum_{\sigma=0}^i
\sum_{v=0}^{n-1}c_{\sigma v}\sum_{\tau=0}^{+\infty}
\frac{\lambda^\tau}{\Gamma(\tau \alpha +v+1-n)}
(t-t_\sigma)^{\tau\alpha-\alpha+v} \\
&\quad \int_{\frac{t_i-t_\sigma}{t-t_\sigma}}^1(1-w)^{n-\alpha-1}w^{\tau\alpha+v-n}dw\\
&\quad +\frac{1}{\Gamma (n-\alpha)}\Big[\sum_{\tau=0}^{+\infty}
\int_0^t(t-u)^{\tau\alpha}\int_0^1(1-w)^{n-\alpha-1}\frac{\lambda^\tau
w^{\tau\alpha+\alpha-n}}{\Gamma((\tau+1)\alpha-n+1)}dwF(u)du\Big]' \\
&=\sum_{\sigma=0}^i\sum_{v=0}^{n-1}c_{\sigma v}\sum_{\tau=1}^{+\infty}
\frac{\lambda^\tau}{\Gamma((\tau -1)\alpha +v+1)}(t-t_\sigma)^{\tau\alpha-\alpha+v} \\
&\quad +\Big[\sum_{\tau=0}^{+\infty}\int_0^t(t-u)^{\tau\alpha}
\frac{\lambda^\tau }{\Gamma(\tau\alpha+1)}F(u)du\Big]'\\
&=\lambda x(t)+F(t).
\end{align*}
We have shown that $x$ satisfies \eqref{e3.2.1} if $x$ satisfies \eqref{e3.2.5}.

Now, we suppose that $x$ is a solution of \eqref{e3.2.1}. We will prove that
$x$ satisfies \eqref{e3.2.5} by mathematical induction. Since $x$ is continuous
on $(t_i,t_{i+1}]$ and the limit $\lim_{t\to t_i^+}x(t)$ $(i\in N_0)$ exists,
it follows that $x\in P_mC(0,1]$. For $t\in (t_0,t_1]$, we know from 
Theorem \ref{thm3.1.2}
that there exists $c_{0v}\in \mathbb{R}$ such that
$$
x(t)=\sum_{v=0}^{n-1}c_{0v}\mathbf{E}_{\alpha,1}
(\lambda t^\alpha)t^v+\int_0^t(t-s)^{\alpha-1}
\mathbf{E}_{\alpha,\alpha}(\lambda (t-s)^\alpha)F(s)ds, \quad t\in (t_0,t_1].
$$
Then \eqref{e3.2.5} holds for $i=0$. We suppose that \eqref{e3.2.5} holds for
all $i=0,1,\dots,j\le m-1$.
We derive the expression of $x$ on $(t_{j+1},t_{j+2}]$. Suppose that
\begin{equation}
\begin{aligned}
x(t)&=\Phi(t)+\sum_{\sigma=0}^j\sum_{v=0}^{n-1}c_{\sigma v}
\mathbf{E}_{\alpha,v+1}(\lambda (t-t_\sigma)^\alpha)(t-t_\sigma)^v\\
&\quad +\int_0^t(t-s)^{\alpha-1}\mathbf{E}_{\alpha,\alpha}
(\lambda (t-s)^\alpha)F(s)ds, \quad t\in (t_{j+1},t_{j+2}].
\end{aligned} \label{e3.2.6}
\end{equation}
By ${}^CD_{0^+}^\alpha x(t)-\lambda x(t)=f(t),t\in (t_{j+1},t_{j+2}]$, we obtain
\begin{align*}
&F(t)+\lambda x(t) \\
&={}^CD_{0^+}^\alpha x(t)
 =\frac{1}{\Gamma (n-\alpha)}\int_0^t(t-s)^{n-\alpha-1}x^{(n)}(s)ds\\
&=\sum_{\rho=0}^j\int_{t_\rho}^{t_{\rho+1}}
\frac{(t-s)^{n-\alpha-1}}{\Gamma (n-\alpha)}
\Big(\sum_{\sigma=0}^\rho\sum_{v=0}^{n-1}c_{\sigma v}
\mathbf{E}_{\alpha,v+1}(\lambda (s-t_\sigma)^\alpha)(s-t_\sigma)^v \\
&\quad +\int_0^s(s-u)^{\alpha-1}\mathbf{E}_{\alpha,\alpha}
(\lambda (s-u)^\alpha)F(u)du\Big)^{(n)}ds\\
&\quad +\int_{t_{j+1}}^t\frac{(t-s)^{n-\alpha-1}}{\Gamma (n-\alpha)}
(\Phi(s)+\sum_{\sigma=0}^j\sum_{v=0}^{n-1}c_{\sigma v}
\mathbf{E}_{\alpha,v+1}(\lambda (s-t_\sigma)^\alpha)(s-t_\sigma)^v \\
&\quad +\int_0^s(s-u)^{\alpha-1}\mathbf{E}_{\alpha,\alpha}
 (\lambda (s-u)^\alpha)F(u)du)^{(n)}ds \\
&={}^CD_{t_{j+1}^+}^\alpha \Phi(t)+\frac{1}{\Gamma (n-\alpha)}
 \sum_{\rho=0}^j\int_{t_\rho}^{t_{\rho+1}}(t-s)^{n-\alpha-1} \\
&\quad\times \Big(\sum_{\sigma=0}^\rho\sum_{v=0}^{n-1}c_{\sigma v}\mathbf{E}_{\alpha,v+1}
 (\lambda (s-t_\sigma)^\alpha)(s-t_\sigma)^v\Big)^{(n)}ds\\
&\quad +\frac{1}{\Gamma (n-\alpha)}\int_{t_{j+1}}^t(t-s)^{n-\alpha-1}
\Big(\sum_{\sigma=0}^j\sum_{v=0}^{n-1}c_{\sigma v}\mathbf{E}_{\alpha,v+1}
(\lambda (s-t_\sigma)^\alpha)(s-t_\sigma)^v\Big)^{(n)}ds\\
&\quad +\frac{1}{\Gamma (n-\alpha)}\int_0^{t}(t-s)^{n-\alpha-1}
(\int_0^s(s-u)^{\alpha-1}\mathbf{E}_{\alpha,\alpha}(\lambda (s-u)^\alpha)
F(u)du)^{(n)}ds\\
&={}^CD_{t_{j+1}^+}^\alpha \Phi(t)+\frac{1}{\Gamma (n-\alpha)}
\sum_{\rho=0}^j\int_{t_\rho}^{t_{\rho+1}}(t-s)^{n-\alpha-1} \\
&\quad\times \Big(\sum_{\sigma=0}^\rho\sum_{v=0}^{n-1}c_{\sigma v}
\sum_{\tau=0}^{+\infty}\frac{\lambda^\tau(s-t_\sigma)^{\tau\alpha+v}}
{\Gamma(\tau \alpha+v+1)}\Big)^{(n)}ds\\
&\quad +\frac{1}{\Gamma (n-\alpha)}\int_{t_{j+1}}^t(t-s)^{n-\alpha-1}
\Big(\sum_{\sigma=0}^j\sum_{v=0}^{n-1}c_{\sigma v}\sum_{\tau=0}^{+\infty}
\frac{\lambda^\tau(s-t_\sigma)^{\tau\alpha+v}}{\Gamma(\tau \alpha+v+1)}\Big)^{(n)}ds\\
&\quad +\frac{1}{\Gamma (n-\alpha)}\int_0^{t}(t-s)^{n-\alpha-1}
\Big(\int_0^s(s-u)^{\alpha-1}\sum_{\tau=0}^{+\infty}
 \frac{\lambda^\tau(s-u)^{\tau\alpha}}{\Gamma((\tau +1)\alpha)}F(u)du\Big)^{(n)}ds\\
&={}^CD_{t_{j+1}^+}^\alpha \Phi(t)+\frac{1}{\Gamma (n-\alpha)}
 \sum_{\rho=0}^j\int_{t_\rho}^{t_{\rho+1}}(t-s)^{n-\alpha-1} \\
&\quad\times  \Big(\sum_{\sigma=0}^\rho\sum_{v=0}^{n-1}c_{\sigma v}
 \sum_{\tau=1}^{+\infty}
\frac{\lambda^\tau(s-t_\sigma)^{\tau\alpha+v-n}}{\Gamma(\tau \alpha+v+1-n)}\Big)ds\\
&\quad +\frac{1}{\Gamma (n-\alpha)}\int_{t_{j+1}}^t(t-s)^{n-\alpha-1}
\Big(\sum_{\sigma=0}^j\sum_{v=0}^{n-1}c_{\sigma v}\sum_{\tau=1}^{+\infty}
\frac{\lambda^\tau(s-t_\sigma)^{\tau\alpha+v-n}}{\Gamma(\tau \alpha+v+1-n)}\Big)ds\\
&\quad +\frac{1}{\Gamma (n-\alpha)}\int_0^{t}(t-s)^{n-\alpha-1}
\Big(\sum_{\tau=0}^{+\infty}\int_0^s
\frac{\lambda^\tau(s-u)^{\tau\alpha+\alpha-n}}{\Gamma((\tau +1)\alpha-n+1)}F(u)du
\Big)'ds\\
&={}^CD_{t_{j+1}^+}^\alpha \Phi(t)+\frac{1}{\Gamma (n-\alpha)}
\sum_{\rho=0}^j\sum_{\sigma=0}^\rho\sum_{v=0}^{n-1}c_{\sigma v}\sum_{\tau=1}^{+\infty}
\int_{t_\rho}^{t_{\rho+1}}(t-s)^{n-\alpha-1} \\
&\quad\times \frac{\lambda^\tau(s-t_\sigma)^{\tau\alpha+v-n}}
 {\Gamma(\tau \alpha+v+1-n)}ds\\
&\quad +\frac{1}{\Gamma (n-\alpha)}\sum_{\sigma=0}^j\sum_{v=0}^{n-1}
c_{\sigma v}\sum_{\tau=1}^{+\infty}\int_{t_{j+1}}^t(t-s)^{n-\alpha-1}
\frac{\lambda^\tau(s-t_\sigma)^{\tau\alpha+v-n}}{\Gamma(\tau \alpha+v+1-n)}ds \\
&\quad +\frac{1}{\Gamma (n-\alpha+1)}\Big[\int_0^{t}(t-s)^{n-\alpha}
(\sum_{\tau=0}^{+\infty}\int_0^s
\frac{\lambda^\tau(s-u)^{\tau\alpha+\alpha-n}}{\Gamma((\tau +1)\alpha-n+1)}
F(u)du)'ds\Big]'.
\end{align*}
By a similar computation, we obtain
$$
F(t)+\lambda x(t)=F(t)+\lambda x(t)+{}^cD_{t_{j+1}^+}^\alpha \Phi(t)
-\lambda \Phi(t).
$$
It follows that ${}^CD_{t_{j+1}^+}^{\alpha} \Phi(t)-\lambda \Phi(t)=0$ for all
$t\in (t_{j+1},t_{j+2}]$. By Theorem \ref{thm3.1.2}, we know that there exists
$c_{j+1v}\in \mathbb{R}(v\in \mathbb{N}[0,n-1])$ such that
$\Phi(t)=\sum_{v=0}^{n-1}c_{j+1v}\mathbf{E}_{\alpha,v+1}
(\lambda (t-t_{j+1})^\alpha)(t-t_{j+1})^v$ for $t\in (t_{j+1},t_{j+2}]$.
Substituting $\Phi$ into \eqref{e3.2.6}, we obtain that \eqref{e3.2.5} holds
for $i=j+1$. Now suppose that \eqref{e3.2.5} holds for all $j\in N_0$.
By the mathematical induction, we know that $x$ satisfies \eqref{e3.2.5}
and $x|_{(t_i,t_{i+1}]}$ is continuous and $\lim_{t\to t_i^+}x(t)$ exists.
The proof is complete.
\end{proof}

\begin{theorem} \label{thm3.2.2}
 Suppose that $F$ is continuous on $(0,1)$ and there exist constants $k>-1$
and $l\in (-\alpha,-n-k,0]$ such that $|F(t)|\le t^k(1-t)^l$ for all $t\in (0,1)$.
Then $x$ is a solution of \eqref{e3.2.2} if and only if there exist constants
$c_{\sigma v}\in \mathbb{R}(\sigma\in \mathbb{N}[0,m],v\in \mathbb{N}[1,n])$ such that
\begin{equation}
\begin{aligned}
x(t)&=\sum_{\sigma=0}^i\sum_{v=1}^{n}c_{\sigma v}\mathbf{E}_{\alpha,\alpha-v+1}
(\lambda (t-t_\sigma)^\alpha)(t-t_\sigma)^{\alpha-v}\\
&\quad +\int_0^t(t-s)^{\alpha-1}\mathbf{E}_{\alpha,\alpha}
(\lambda (t-s)^\alpha)F(s)ds, \quad t\in (t_i,t_{i+1}],\; i\in \mathbb{N}[0,m].
\end{aligned} \label{e3.2.7}
\end{equation}
\end{theorem}


\begin{proof}
For $t\in (t_j,t_{j+1}](j\in \mathbb{N}_0^m)$, similarly to the beginning of the
proof of Theorem \ref{thm3.2.1} we know that
\begin{align*}
&t^{n-\alpha}|\int_0^t(t-s)^{\alpha-1}\mathbf{E}_{\alpha,\alpha}
(\lambda (t-s)^\alpha)F(s)ds| \\
&\le \int_0^t(t-s)^{\alpha-1}\mathbf{E}_{\alpha,\alpha}(\lambda (t-s)^\alpha)
 |F(s)|ds\\
&\le t^{n-\alpha}\int_0^t(t-s)^{\alpha-1}\mathbf{E}_{\alpha,\alpha}
 (\lambda (t-s)^\alpha)s^k(1-s)^lds\\
&= t^{n-\alpha}\sum_{j=0}^{+\infty}\frac{\lambda^j}{\Gamma ((j+1)\alpha)}
\int_0^t(t-s)^{\alpha-1}(t-s)^{\alpha j}s^k(1-s)^lds\\
&\le t^{n-\alpha}\sum_{j=0}^{+\infty}\frac{\lambda^j}{\Gamma ((j+1)\alpha)}
 \int_0^t(t-s)^{\alpha+l-1}(t-s)^{\alpha j}s^kds\\
&= t^{n-\alpha}\sum_{j=0}^{+\infty}\frac{\lambda^j}{\Gamma ((j+1)\alpha)}
 t^{\alpha+\alpha j+k+l}\int_0^1(1-w)^{\alpha+\alpha j+l-1}w^kdw \\
& \le t^{n-\alpha}\sum_{j=0}^{+\infty}
\frac{\lambda^j t^{\alpha j}}{\Gamma ((j+1)\alpha)}t^{\alpha+k+l}
\int_0^1(1-w)^{\alpha+l-1}w^kdw \\
&=t^{n+k+l}\mathbf{E}_{\alpha,\alpha}
(\lambda t^\alpha)\mathbf{B}(\alpha+l,k+1).
\end{align*}
So $t^{n-\alpha}\int_0^t(t-s)^{\alpha-1}\mathbf{E}_{\alpha,\alpha}
(\lambda (t-s)^\alpha)F(s)ds$ is convergent and is continuous on $[0,1]$.

% Page 58

If $x$ is a solution of \eqref{e3.2.7}, we have $x\in P_mC_{1-\alpha}(0,1]$.
It follows for $t\in (t_i,t_{i+1}]$ and from Definition \ref{def2.2} that
\begin{align*}
&{}^{RL}D_{0^+}^\alpha x(t) \\
&=\frac{1}{\Gamma (n-\alpha )}\Big[\int_0^t(t-s)^{n-\alpha-1}x(s)ds\Big]^{(n)}\\
&=\frac{1}{\Gamma (n-\alpha )}\Big[\sum_{j=0}^{i-1}
\int_{t_j}^{t_{j+1}}(t-s)^{n-\alpha-1} \\
&\quad\times \sum_{\sigma=0}^j\sum_{v=1}^{n}c_{\sigma v}
\mathbf{E}_{\alpha,\alpha-v+1}(\lambda (s-t_\sigma)^\alpha)
(s-t_\sigma)^{\alpha-v}ds\Big]^{(n)}\\
&\quad +\frac{1}{\Gamma (n-\alpha )}
\Big[\int_{t_i}^t(t-s)^{n-\alpha-1}\sum_{\sigma=0}^i\sum_{v=1}^{n}c_{\sigma v}
\mathbf{E}_{\alpha,\alpha-v+1}(\lambda (s-t_\sigma)^\alpha)(s-t_\sigma)^{\alpha-v}
\Big]^{(n)}\\
&\quad +\frac{1}{\Gamma(n-\alpha)}\Big[\int_0^{t}(t-s)^{n-\alpha-1}
\int_0^s(t-u)^{\alpha-1}\mathbf{E}_{\alpha,\alpha}(\lambda (s-u)^\alpha)F(u)\,du\,ds
\Big]^{(n)}\\
&=\frac{1}{\Gamma (n-\alpha )}\Big[\sum_{j=0}^{i-1}
 \int_{t_j}^{t_{j+1}}(t-s)^{n-\alpha-1}\sum_{\sigma=0}^j\sum_{v=1}^{n}
 c_{\sigma v}\sum_{\tau=0}^{+\infty}
 \frac{\lambda^\tau(s-t_\sigma)^{\tau\alpha+\alpha-v}}{\Gamma(\tau \alpha+\alpha-v+1)}
 ds\Big]^{(n)}\\
&\quad +\frac{1}{\Gamma (n-\alpha )}
\Big[\int_{t_i}^t(t-s)^{n-\alpha-1}\sum_{\sigma=0}^i\sum_{v=1}^{n}
c_{\sigma v}\sum_{\tau=0}^{+\infty}
\frac{\lambda^\tau(s-t_\sigma)^{\tau\alpha+\alpha-v}}{\Gamma(\tau \alpha+\alpha-v+1)}
ds\Big]^{(n)}\\
&\quad +\frac{1}{\Gamma (n-\alpha )}
\Big[\int_0^t(t-s)^{n-\alpha-1}\int_0^s(s-u)^{\alpha-1}\sum_{\tau=0}^{+\infty}
\frac{\lambda^\tau(s-u)^{\tau\alpha}}{\Gamma((\tau +1) \alpha)}F(u)\,du\,ds\Big]^{(n)}\\
&=\frac{1}{\Gamma (n-\alpha )}
\Big[\sum_{j=0}^{i-1}\sum_{\sigma=0}^j\sum_{v=1}^{n}c_{\sigma v}
\sum_{\tau=0}^{+\infty} \\
&\quad\times \int_{t_j}^{t_{j+1}}(t-s)^{n-\alpha-1}
\frac{\lambda^\tau(s-t_\sigma)^{\tau\alpha+\alpha-v}}{\Gamma(\tau \alpha+\alpha-v+1)}
ds\Big]^{(n)}\\
&\quad +\frac{1}{\Gamma (n-\alpha )}
\Big[\sum_{\sigma=0}^i\sum_{v=1}^{n}c_{\sigma v}\sum_{\tau=0}^{+\infty} \\
&\quad\times \int_{t_i}^t(t-s)^{n-\alpha-1}\frac{\lambda^\tau(s-t_\sigma)^{\tau\alpha+\alpha-v}}
{\Gamma(\tau \alpha+\alpha-v+1)}ds\Big]^{(n)}\\
&\quad +\frac{1}{\Gamma (n-\alpha )}
\Big[\sum_{\tau=0}^{+\infty}\int_0^t\int_u^t(t-s)^{n-\alpha-1}(s-u)^{\alpha-1}
\frac{\lambda^\tau(s-u)^{\tau\alpha}}{\Gamma((\tau +1) \alpha)}dsF(u)du\Big]^{(n)}\\
&=\frac{1}{\Gamma (n-\alpha )}\Big[\sum_{j=0}^{i-1}\sum_{\sigma=0}^j
\sum_{v=1}^{n}c_{\sigma v}\sum_{\tau=0}^{+\infty}(t-t_\sigma)^{\tau \alpha+n-v} \\
&\quad\times \int_{\frac{t_j-t_\sigma}{t-t_\sigma}}^{\frac{t_{j+1}-t_\sigma}{t-t_\sigma}}
(1-w)^{n-\alpha-1}
\frac{\lambda^\tau w^{\tau\alpha+\alpha-v}}{\Gamma(\tau \alpha+\alpha-v+1)}dw
\Big]^{(n)}\\
&\quad +\frac{1}{\Gamma (n-\alpha )}
\Big[\sum_{\sigma=0}^i\sum_{v=1}^{n}c_{\sigma v}\sum_{\tau=0}^{+\infty}
(t-t_\sigma)^{\tau\alpha+n-v} \\
&\quad\times \int_{\frac{t_i-t_\sigma}{t-t_\sigma}}^1
(1-w)^{n-\alpha-1}\frac{\lambda^\tau w^{\tau\alpha+\alpha-v}}
{\Gamma(\tau \alpha+\alpha-v+1)}dw\Big]^{(n)}\\
&\quad +\frac{1}{\Gamma (n-\alpha )}
\Big[\sum_{\tau=0}^{+\infty}\int_0^t(t-u)^{\tau\alpha+n-1}
\int_0^1(1-w)^{n-\alpha-1}\frac{\lambda^\tau w^{\tau\alpha+\alpha-1}}
{\Gamma((\tau +1) \alpha)}dwF(u)du\Big]^{(n)}\\
&=\frac{1}{\Gamma (n-\alpha )}
\Big[\sum_{\sigma=0}^{i-1}\sum_{j=\sigma}^{i-1}\sum_{v=1}^{n}c_{\sigma v}
\sum_{\tau=0}^{+\infty}(t-t_\sigma)^{\tau \alpha+n-v} \\
&\quad\times  \int_{\frac{t_j-t_\sigma}{t-t_\sigma}}^{\frac{t_{j+1}-t_\sigma}{t-t_\sigma}}
(1-w)^{n-\alpha-1}\frac{\lambda^\tau w^{\tau\alpha+\alpha-v}}
{\Gamma(\tau \alpha+\alpha-v+1)}dw\Big]^{(n)}\\
&\quad +\frac{1}{\Gamma (n-\alpha )}
\Big[\sum_{\sigma=0}^i\sum_{v=1}^{n}c_{\sigma v}\sum_{\tau=0}^{+\infty}
(t-t_\sigma)^{\tau\alpha+n-v} \\
&\quad\times \int_{\frac{t_i-t_\sigma}{t-t_\sigma}}^1
(1-w)^{n-\alpha-1}\frac{\lambda^\tau w^{\tau\alpha+\alpha-v}}
{\Gamma(\tau \alpha+\alpha-v+1)}dw\Big]^{(n)}\\
&\quad +\Big[\sum_{\tau=0}^{+\infty}\int_0^t(t-u)^{\tau\alpha+n-1}
\frac{\lambda^\tau }{\Gamma(\tau\alpha+n)}F(u)du\Big]^{(n)}\\
&=\frac{1}{\Gamma (n-\alpha )}
\Big[\sum_{\sigma=0}^i\sum_{v=1}^{n}c_{\sigma v}\sum_{\tau=0}^{+\infty}
(t-t_\sigma)^{\tau \alpha+n-v} \\
&\quad\times \int_0^1 (1-w)^{n-\alpha-1}
\frac{\lambda^\tau w^{\tau\alpha+\alpha-v}}{\Gamma(\tau \alpha+\alpha-v+1)}dw
\Big]^{(n)} \\
&\quad +\Big[\sum_{\tau=0}^{+\infty}\int_0^t(t-u)^{\tau\alpha+n-1}
\frac{\lambda^\tau }{\Gamma(\tau\alpha+n)}F(u)du\Big]^{(n)}\\
&=\Big[\sum_{\sigma=0}^i\sum_{v=1}^{n}c_{\sigma v}
 \sum_{\tau=0}^{+\infty}(t-t_\sigma)^{\tau \alpha+n-v}\frac{\lambda^\tau }
 {\Gamma(\tau \alpha+n-v+1)}\Big]^{(n)} \\
&\quad +\Big[\sum_{\tau=0}^{+\infty}
 \int_0^t(t-u)^{\tau\alpha+n-1}
\frac{\lambda^\tau }{\Gamma(\tau\alpha+n)}F(u)du\Big]^{(n)}\\
&=\Big[\sum_{\sigma=0}^i\sum_{v=1}^{n}c_{\sigma v}\sum_{\tau=1}^{+\infty}
 (t-t_\sigma)^{\tau \alpha-v}\frac{\lambda^\tau }{\Gamma(\tau \alpha-v+1)}\Big]\\
&\quad +\Big[\sum_{\tau=1}^{+\infty}\int_0^t(t-u)^{\tau\alpha-1}
\frac{\lambda^\tau }{\Gamma(\tau\alpha)}F(u)du\Big]^{(n)}+F(t) \\
&=\lambda x(t)+F(t),t\in (t_i,t_{i+1}].
\end{align*}
It follows that $x$ is a solution of \eqref{e3.2.2}.

Now we prove that if $x$ is a solution of \eqref{e3.2.2}, then $x$ satisfies
\eqref{e3.2.7} and $x\in P_mC_{1-\alpha}(0,1]$ by mathematical induction.
By Theorem \ref{thm3.1.4}, we know that there exists a constant $c_{0v}\in \mathbb{R}(v\in N[1,n]$
such that
$$
x(t)=\sum_{v=1}^nc_{0v}t^{\alpha-v}\mathbf{E}_{\alpha,\alpha-v+1}
(\lambda t^\alpha)+\int_0^t(t-s)^{\alpha -1}\mathbf{E}_{\alpha ,\alpha}
(\lambda(t-s)^\alpha)F(s)ds,\quad t\in (t_0,t_1].
$$
Hence, \eqref{e3.2.7} holds for $i=0$. Assume that \eqref{e3.2.7} holds
for $i=0,1,2,\dots,j\le m-1$, we will prove that \eqref{e3.2.7} holds for $i=j+1$.
Suppose that
\begin{align*}
x(t)&=\Phi(t)+\sum_{\sigma=0}^j\sum_{v=1}^{n}c_{\sigma v}
\mathbf{E}_{\alpha,\alpha-v+1}(\lambda (t-t_\sigma)^\alpha)(t-t_\sigma)^{\alpha-v}\\
&\quad +\int_0^t(t-s)^{\alpha-1}\mathbf{E}_{\alpha,\alpha}
(\lambda (t-s)^\alpha)F(s)ds, \quad t\in (t_{j+1},t_{j+2}].
\end{align*}
Then for $t\in (t_{j+1},t_{j+2}]$ we have
\begin{align*}
&F(t)+\lambda x(t) \\
&={}^{RL}D_{0^+}^\alpha x(t)
=\frac{1}{\Gamma (n-\alpha)}
\Big[\sum_{\rho=0}^{j}\int_{t_\rho}^{t_{\rho+1}}(t-s)^{-\alpha}x(s)ds
+\int_{t_{j+1}}^t(t-s)^{-\alpha}x(s)ds\Big]^{(n)}\\
&=\frac{1}{\Gamma (n-\alpha)}
\Big[\sum_{\rho=0}^{j}\int_{t_\rho}^{t_{\rho+1}}(t-s)^{-\alpha}
\Big(\sum_{\sigma=0}^\rho
\sum_{v=1}^{n}c_{\sigma v}\mathbf{E}_{\alpha,\alpha-v+1}
(\lambda (s-t_\sigma)^\alpha)(s-t_\sigma)^{\alpha-v} \\
&\quad +\int_0^s(s-u)^{\alpha-1}\mathbf{E}_{\alpha,\alpha}
(\lambda (s-u)^\alpha)F(u)du\Big)ds\Big]^{(n)}\\
&\quad +\frac{1}{\Gamma (n-\alpha)}
\Big[\int_{t_{j+1}}^t(t-s)^{-\alpha}\Big(\Phi(s)
+\sum_{\sigma=0}^j\sum_{v=1}^{n}c_{\sigma v}\mathbf{E}_{\alpha,\alpha-v+1}
(\lambda (s-t_\sigma)^\alpha)(s-t_\sigma)^{\alpha-v} \\
&\quad +\int_0^s(s-u)^{\alpha-1}\mathbf{E}_{\alpha,\alpha}
(\lambda (s-u)^\alpha)F(u)du\Big)ds\Big]^{(n)} \\
&=\frac{1}{\Gamma (n-\alpha)}
\Big[\sum_{\rho=0}^{j}\int_{t_\rho}^{t_{\rho+1}}(t-s)^{-\alpha}\sum_{\sigma=0}^\rho
\sum_{v=1}^{n}c_{\sigma v}\mathbf{E}_{\alpha,\alpha-v+1}(\lambda (s-t_\sigma)^\alpha)
(s-t_\sigma)^{\alpha-v}ds \Big]^{(n)} \\
&\quad +\frac{1}{\Gamma (n-\alpha)}
\Big[\int_{t_{j+1}}^t(t-s)^{-\alpha}\Phi(s) \\
&\quad +\sum_{\sigma=0}^j
\sum_{v=1}^{n}c_{\sigma v}\mathbf{E}_{\alpha,\alpha-v+1}
(\lambda (s-t_\sigma)^\alpha)(s-t_\sigma)^{\alpha-v}ds\Big]^{(n)}\\
&\quad +\frac{1}{\Gamma (n-\alpha)}
\Big[\int_0^t(t-s)^{-\alpha}\int_0^s(s-u)^{\alpha-1}
\mathbf{E}_{\alpha,\alpha}(\lambda (s-u)^\alpha)F(u)\,du\,ds\Big]^{(n)}.
\end{align*}
As in the above discussion, we obtain
$$
F(t)+\lambda x(t)={}^{RL}D_{0^+}^\alpha x(t)
=F(t)+\lambda x(t)+{}^{RL}D_{t_{j+1}^+}^\alpha \Phi(t)-\lambda \Phi(t).
$$
So ${}^{RL}D_{t_{j+1}^+}^\alpha \Phi(t)-\lambda \Phi(t)=0$ on $(t_{j+1},t_{j+2}]$.
Then Theorem \ref{thm3.1.4} implies that there exists a constant $c_{j+1v}\in \mathbb{R}$
such that $\Phi(t)=\sum_{v=1}^nc_{j+1v}(t-t_{i+1})^{\alpha-v}
\mathbf{E}_{\alpha,\alpha-v+1}(\lambda (t-t_{i+1})^\alpha)$ on
$(t_{j+1},t_{j+2}]$. Hence,
\begin{align*}
x(t)&=\sum_{\rho=0}^{j+1}\sum_{v=1}^nc_{\rho v}(t-t_\rho)^{\alpha-v}
\mathbf{E}_{\alpha,\alpha-v+1}(\lambda (t-t_\rho)^\alpha) \\
&\quad +\int_1^t(t-s)^{\alpha-1}\mathbf{E}_{\alpha,\alpha}
 (\lambda (t-s)^\alpha)f(s)ds,\quad t\in (t_{j+1},t_{j+2}].
\end{align*}
By mathematical induction, we know that \eqref{e3.2.7} holds for $j\in \mathbb{N}[0,m]$.
The proof is complete.
\end{proof}

\begin{theorem} \label{thm3.2.3}
Suppose that $G$ is continuous on $(1,e)$ and there exist constants $k>-1$
and $l\in (-\alpha,-n-k,0]$ such that $|G(t)|\le (\log t)^k(1-\log t)^l$
for all $t\in (1,e)$. Then $x$ is a solution of \eqref{e3.2.3} if and only if
there exist constants $c_{jv}\in \mathbb{R}(j\in \mathbb{N}[0,m]$, $v\in \mathbb{N}[1,n])$ such that
\begin{equation}
\begin{aligned}
x(t)&=\sum_{j=0}^i\sum_{v=1}^{n}c_{jv}(\log\frac{ t}{t_j})^{\alpha-v}
\mathbf{E}_{\alpha,\alpha-v+1}\big(\lambda (\log \frac{t}{t_j})^\alpha\big)\\
&\quad +\int_1^t(\log \frac{t}{s})^{\alpha-1}\mathbf{E}_{\alpha,\alpha}
(\lambda(\log \frac{t}{s})^\alpha)G(s)\frac{ds}{s},\quad t\in (t_i,t_{i+1}],\;
i\in \mathbb{N}[0,m].
\end{aligned}\label{e3.2.8}
\end{equation}
\end{theorem}

\begin{proof}
For $t\in (t_i,t_{i+1}](i\in \mathbb{N}[0,m])$, similarly to the beginning of the
proof of Theorem \ref{thm3.2.2} we know that
\begin{align*}
&(\log t)^{n-\alpha} \big|\int_1^t(\log \frac{t}{s})^{\alpha-1}
\mathbf{E}_{\alpha,\alpha}(\lambda(\log \frac{t}{s})^\alpha)G(s)\frac{ds}{s}\big|\\
&\le (\log t)^{n-\alpha}\int_1^t(\log \frac{t}{s})^{\alpha-1}
 \mathbf{E}_{\alpha,\alpha}(\lambda(\log \frac{t}{s})^\alpha)(\log s)^k(1-\log s)^l
 \frac{ds}{s}\\
&\le (\log t)^{n-\alpha}\sum_{\iota=0}^{+\infty}
\frac{\lambda^\iota}{\Gamma(\alpha(\iota+1))}
\int_1^t(\log \frac{t}{s})^{\alpha \iota+\alpha+l-1}(\log s)^k
\frac{ds}{s} \\
&\quad\quad\text{(because $\frac{\log s}{\log t}=w$)}\\
&= (\log t)^{n-\alpha}\sum_{\iota=0}^{+\infty}
 \frac{\lambda^\iota}{\Gamma(\alpha(\iota+1))}(\log t)^{\alpha \iota+\alpha+k+l}
 \int_0^1(1-w)^{\alpha \iota+\alpha+l-1}w^kdw\\
&\le (\log t)^{n+k+l} \mathbf{E}_{\alpha,\alpha}
 (\lambda(\log \frac{t}{s})^\alpha)\int_0^1(1-w)^{\alpha+l-1}w^kdw\\
&= (\log t)^{n+k+l} \mathbf{E}_{\alpha,\alpha}
 \big(\lambda(\log \frac{t}{s})^\alpha\big)\mathbf{B}(\alpha+l,k+1).
\end{align*}
So $\int_1^t(\log \frac{t}{s})^{\alpha-1}\mathbf{E}_{\alpha,\alpha}
\big(\lambda(\log \frac{t}{s})^\alpha\big)G(s)\frac{ds}{s}$ is convergent
for all $t\in (1,e]$ and
$$
\lim_{t\to 1^+}(\log t)^{n-\alpha}\int_1^t(\log \frac{t}{s})^{\alpha-1}
\mathbf{E}_{\alpha,\alpha}\big(\lambda(\log \frac{t}{s})^\alpha\big)G(s)
\frac{ds}{s}
$$
exists.

If $x$ is a solution of \eqref{e3.2.8}, we have $x\in LP_mC_{n-\alpha}(1,e]$.
By using Definition \ref{def2.5}, it follows for $t\in (t_i,t_{i+1}]$ that
\begin{align*}
&{}^{RLH}D_{1^+}^\alpha x(t) \\
&=\frac{1}{\Gamma (n-\alpha )}(t\frac{d}{dt})^n
\Big[\int_1^t(\log\frac{t}{s})^{n-\alpha-1}x(s)\frac{ds}{s}\Big]\\
&=\frac{1}{\Gamma (n-\alpha )}(t\frac{d}{dt})^n
\Big[\sum_{\sigma=0}^{i-1}\int_{t_\sigma}^{t_{\sigma+1}}
(\log\frac{t}{s})^{n-\alpha-1} \\
&\quad\times \Big(\sum_{j=0}^\sigma\sum_{v=1}^{n}c_{jv}(\log\frac{s}{t_j})^{\alpha-v}
\mathbf{E}_{\alpha,\alpha-v+1}(\lambda (\log \frac{s}{t_j})^\alpha) \\
&\quad +\int_1^s(\log \frac{s}{u})^{\alpha-1}\mathbf{E}_{\alpha,\alpha}
(\lambda(\log \frac{s}{u})^\alpha)G(u)\frac{du}{u}\Big)\frac{ds}{s}\\
&\quad +\int_{t_{i}}^t(\log\frac{t}{s})^{n-\alpha-1}
\Big(\sum_{j=0}^i\sum_{v=1}^{n}c_{jv}(\log\frac{ s}{t_j})^{\alpha-v}
\mathbf{E}_{\alpha,\alpha-v+1}(\lambda (\log \frac{s}{t_j})^\alpha) \\
&\quad +\int_1^s(\log \frac{s}{u})^{\alpha-1}\mathbf{E}_{\alpha,\alpha}
(\lambda(\log \frac{s}{u})^\alpha)G(u)\frac{du}{u}\Big)\frac{ds}{s}\Big]\\
&=\frac{1}{\Gamma (n-\alpha )}(t\frac{d}{dt})^n
\Big[\sum_{\sigma=0}^{i-1}\sum_{j=0}^\sigma\sum_{v=1}^{n}c_{jv}
\int_{t_\sigma}^{t_{\sigma+1}}(\log\frac{t}{s})^{n-\alpha-1}
(\log\frac{s}{t_j})^{\alpha-v} \\
&\quad\times \mathbf{E}_{\alpha,\alpha-v+1}
(\lambda (\log \frac{s}{t_j})^\alpha)\frac{ds}{s}\Big]\\
&\quad +(t\frac{d}{dt})^n\Big[\sum_{j=0}^i\sum_{v=1}^{n}c_{jv}
 \int_{t_{i}}^t(\log\frac{t}{s})^{n-\alpha-1}
 (\log\frac{ s}{t_j})^{\alpha-v}\mathbf{E}_{\alpha,\alpha-v+1}
 (\lambda (\log \frac{s}{t_j})^\alpha)\frac{ds}{s}\Big]\\
&\quad +\frac{1}{\Gamma (n-\alpha )}(t\frac{d}{dt})^n
\Big[\int_1^t(\log\frac{t}{s})^{n-\alpha-1}
\int_1^s(\log \frac{s}{u})^{\alpha-1}\mathbf{E}_{\alpha,\alpha}
(\lambda(\log \frac{s}{u})^\alpha)G(u)\frac{du}{u}\frac{ds}{s}\Big].
\end{align*}
One sees that
\begin{align*}
&\int_{t_{i}}^t(\log\frac{t}{s})^{n-\alpha-1}
(\log\frac{ s}{t_v})^{\alpha-v}\mathbf{E}_{\alpha,\alpha-v+1}
(\lambda (\log \frac{s}{t_v})^\alpha)\frac{ds}{s}\\
&=\sum_{\kappa=0}^{+\infty}\frac{\lambda^\kappa}{\Gamma(\kappa\alpha+\alpha-v+1))}
\int_{t_{i}}^t(\log\frac{t}{s})^{n-\alpha-1}
(\log\frac{ s}{t_v})^{\alpha\kappa+\alpha-v}\frac{ds}{s}\\
&\quad \text{(because $\frac{\log s-\log t_v}{\log t-\log t_v}=w$)}\\
&=\sum_{\kappa=0}^{+\infty}\frac{\lambda^\kappa}{\Gamma(\kappa\alpha+\alpha-v+1)}
(\log\frac{t}{t_v})^{\alpha\kappa+n-v}
\int_{\frac{\log t_{i}-\log t_v}{\log t-\log t_v}}^1(1-w)^{n-\alpha-1}
w^{\alpha\kappa+\alpha-v}dw
\end{align*}
and
\begin{align*}
&\int_{t_i}^{t_{i+1}}(\log\frac{t}{s})^{n-\alpha-1}
(\log\frac{ s}{t_v})^{\alpha-v}\mathbf{E}_{\alpha,\alpha-v+1}
\big(\lambda (\log \frac{s}{t_v})^\alpha\big)\frac{ds}{s}\\
&=\sum_{\kappa=0}^{+\infty}\frac{\lambda^\kappa}
{\Gamma(\kappa\alpha+\alpha-v+1)}(\log\frac{t}{t_v})^{\alpha\kappa+n-v}
\int_{\frac{\log t_{i}-\log t_v}{\log t-\log t_v}}
^{\frac{\log t_{i+1}-\log t_v}{\log t-\log t_v}} (1-w)^{n-\alpha-1}
w^{\alpha \kappa+\alpha-v}dw.
\end{align*}
Similarly,
\begin{align*}
&\int_1^t(\log\frac{t}{s})^{n-\alpha-1}\int_1^s
(\log \frac{s}{u})^{\alpha-1}\mathbf{E}_{\alpha,\alpha}
(\lambda(\log \frac{s}{u})^\alpha)G(u)\frac{du}{u}\frac{ds}{s}\\
&=\int_1^t\int_u^t(\log\frac{t}{s})^{n-\alpha-1}
 (\log \frac{s}{u})^{\alpha-1}\mathbf{E}_{\alpha,\alpha}
 (\lambda(\log \frac{s}{u})^\alpha)\frac{ds}{s}G(u)\frac{du}{u}\\
&=\int_1^t\int_u^t(\log\frac{t}{s})^{n-\alpha-1}
 (\log \frac{s}{u})^{\alpha-1}\sum_{\kappa=0}^{+\infty}
 \frac{\lambda^\kappa (\log \frac{s}{u})^{\kappa\alpha}}{\Gamma((\kappa+1)\alpha)}
 \frac{ds}{s}G(u)\frac{du}{u}\\
&=\sum_{\kappa=0}^{+\infty}\int_1^t(\log\frac{t}{u})^{\alpha\kappa+n-1}
 \frac{\lambda^\kappa}{\Gamma(\alpha(\kappa+1))}
\int_0^1(1-w)^{n-\alpha-1} w^{\alpha \kappa+\alpha-1}dwG(u)\frac{du}{u}.
\end{align*}
So
\begin{align*}
&{}^{RLH}D_{1^+}^\alpha x(t)\\
&=\frac{1}{\Gamma (n-\alpha )}(t\frac{d}{dt})^n
\Big[\sum_{\sigma=0}^{i-1}\sum_{j=0}^\sigma\sum_{v=1}^{n}c_{jv}
\sum_{\kappa=0}^{+\infty}
\frac{\lambda^\kappa(\log\frac{t}{t_j})^{\alpha\kappa+n-v}}
{\Gamma(\kappa\alpha+\alpha-v+1)} \\
&\quad\times \int_{\frac{\log t_{\sigma}-\log t_j}{\log t-\log t_j}}
^{\frac{\log t_{\sigma+1}-\log t_j}{\log t-\log t_j}}
(1-w)^{n-\alpha-1}w^{\alpha \kappa+\alpha-v}dw\Big]\\
&\quad +(t\frac{d}{dt})^n\Big[\sum_{j=0}^i\sum_{v=1}^{n}c_{jv}
\sum_{\kappa=0}^{+\infty} \frac{\lambda^\kappa}{\Gamma(\kappa\alpha+\alpha-v+1)}
(\log\frac{t}{t_j})^{\alpha\kappa+n-v} \\
&\quad\times \int_{\frac{\log t_{i}-\log t_j}{\log t-\log t_j}}^1(1-w)^{n-\alpha-1}
w^{\alpha\kappa+\alpha-v}dw\Big]\\
&\quad +\frac{1}{\Gamma (n-\alpha )}(t\frac{d}{dt})^n
\Big[\sum_{\kappa=0}^{+\infty}\int_1^t(\log\frac{t}{u})^{\alpha\kappa+n-1}
 \frac{\lambda^\kappa}{\Gamma(\alpha(\kappa+1))} \\
&\quad\times \int_0^1(1-w)^{n-\alpha-1} w^{\alpha \kappa+\alpha-1}dwG(u)
\frac{du}{u}\Big]\\
&=\frac{1}{\Gamma (n-\alpha )}(t\frac{d}{dt})^n
 \Big[\sum_{j=0}^{i-1}\sum_{v=1}^{n}c_{jv}
\sum_{\kappa=0}^{+\infty}\frac{\lambda^\kappa(\log\frac{t}{t_j})^{\alpha\kappa+n-v}}
 {\Gamma(\kappa\alpha+\alpha-v+1)} \\
&\quad\times \int_0^{\frac{\log t_{i}-\log t_j}{\log t-\log t_j}} (1-w)^{n-\alpha-1}
w^{\alpha \kappa+\alpha-v}dw\Big]\\
&\quad +(t\frac{d}{dt})^n\Big[\sum_{j=0}^i\sum_{v=1}^{n}c_{jv}\sum_{\kappa=0}^{+\infty}
\frac{\lambda^\kappa}{\Gamma(\kappa\alpha+\alpha-v+1)}
(\log\frac{t}{t_j})^{\alpha\kappa+n-v} \\
&\quad\times \int_{\frac{\log t_{i}-\log t_j}{\log t-\log t_j}}^1
(1-w)^{n-\alpha-1}w^{\alpha\kappa+\alpha-v}dw\Big]\\
&\quad +\frac{1}{\Gamma (n-\alpha )}(t\frac{d}{dt})^n
\Big[\sum_{\kappa=0}^{+\infty}\int_1^t(\log\frac{t}{u})^{\alpha\kappa+n-1}
 \frac{\lambda^\kappa}{\Gamma(\alpha(\kappa+1))} \\
&\quad\times \int_0^1(1-w)^{n-\alpha-1} w^{\alpha \kappa+\alpha-1}dwG(u)
 \frac{du}{u}\Big]\\
&=(t\frac{d}{dt})^n\Big[\sum_{j=0}^i\sum_{v=1}^{n}c_{jv}
\sum_{\kappa=0}^{+\infty}\frac{\lambda^\kappa(\log
\frac{t}{t_j})^{\alpha\kappa+n-v}}{\Gamma(\kappa\alpha+n-v+1)}\Big] \\
&\quad +(t\frac{d}{dt})^n\Big[\sum_{\kappa=0}^{+\infty}
\int_1^t(\log\frac{t}{u})^{\alpha\kappa+n-1}
\frac{\lambda^\kappa}{\Gamma(\alpha\kappa+n)} G(u)\frac{du}{u}\Big]\\
&=(t\frac{d}{dt})^{n-1}\Big[\sum_{j=0}^i\sum_{v=1}^{n}c_{jv}
\sum_{\kappa=0}^{+\infty}\frac{\lambda^\kappa
(\log\frac{t}{t_j})^{\alpha\kappa+n-v-1}}{\Gamma(\kappa\alpha+n-v)}\Big]\\
&\quad +(t\frac{d}{dt})^{n-1}\Big[\sum_{\kappa=0}^{+\infty}
\int_1^t(\log\frac{t}{u})^{\alpha\kappa+n-2}
\frac{\lambda^\kappa}{\Gamma(\alpha\kappa+n-1)}
G(u)\frac{du}{u}\Big]\\
&=\dots \\
&=\sum_{j=0}^i\sum_{v=1}^{n}c_{jv}
\sum_{\kappa=0}^{+\infty}
\frac{\lambda^\kappa(\log\frac{t}{t_j})^{\alpha\kappa-v}}{\Gamma(\kappa\alpha-v+1)}
+\sum_{\kappa=0}^{+\infty}\int_1^t(\log\frac{t}{u})^{\alpha\kappa-1}
\frac{\lambda^\kappa}{\Gamma(\alpha\kappa)}
G(u)\frac{du}{u}\\
&=\lambda x(t)+F(t),t\in (t_i,t_{i+1}].
\end{align*}
It follows that $x$ is a solution of \eqref{e3.2.3}.

Now we prove that if $x$ is a solution of \eqref{e3.2.3}, then $x$
satisfies \eqref{e3.2.8} and $x\in LP_mC_{n-\alpha}(1,e]$ by mathematical
induction. By Theorem \ref{thm3.1.6}, we know that there exists a constant
$c_{0v}\in \mathbb{R}(v\in N[1,n])$ such that
\[
x(t)=\sum_{v=0}^nc_{0v}(\log t)^{\alpha-v}\mathbf{E}_{\alpha,\alpha}
(\lambda (\log t)^\alpha)+\int_1^t(\log\frac{t}{s})^{\alpha -1}
\mathbf{E}_{\alpha ,\alpha}(\lambda(\frac{t}{s})^\alpha)F(s)ds,
\]
for $t\in (t_0,t_1]$.
Hence \eqref{e3.2.8} holds for $i=0$. Assume that \eqref{e3.2.8} holds for
$i=0,1,2,\dots,j\le m-1$, we will prove that \eqref{e3.2.8} holds for $i=j+1$.
Suppose that
\begin{align*}
x(t)&=\Phi(t)+\sum_{\sigma=0}^j\sum_{v=1}^nc_{\sigma v}
(\log\frac{ t}{t_\sigma})^{\alpha-v}\mathbf{E}_{\alpha,\alpha-v+1}
\big(\lambda (\log \frac{t}{t_\sigma})^\alpha\big)\\
&\quad +\int_1^t(\log \frac{t}{s})^{\alpha-1}\mathbf{E}_{\alpha,\alpha}
\big(\lambda(\log \frac{t}{s})^\alpha\big)G(s)\frac{ds}{s},
\quad t\in (t_{j+1},t_{j+2}].
\end{align*}
Then for $t\in (t_{j+1},t_{j+2}]$ we have
\begin{align*}
&F(t)+\lambda x(t) \\
&={}^{RLH}D_{1^+}^\alpha x(t) \\
&=\frac{1}{\Gamma (n-\alpha)}(t\frac{d}{dt})^n
\Big[\sum_{\rho=0}^{j}\int_{t_\rho}^{t_{\rho+1}}
(\log\frac{t}{s})^{n-\alpha-1}x(s)\frac{ds}{s}
+\int_{t_{j+1}}^t(\log\frac{t}{s})^{n-\alpha-1}x(s)\frac{ds}{s}\Big]\\
&=\frac{1}{\Gamma (n-\alpha)}(t\frac{d}{dt})^n
\Big[\sum_{\rho=0}^{j}\int_{t_\rho}^{t_{\rho+1}}
(\log\frac{t}{s})^{n-\alpha-1} \\
&\quad\times \Big(\sum_{\sigma=0}^\rho\sum_{v=1}^nc_{\sigma v}
(\log\frac{ s}{t_\sigma})^{\alpha-v}\mathbf{E}_{\alpha,\alpha-v+1}
\big(\lambda (\log \frac{s}{t_\sigma})^\alpha\big) \\
&\quad +\int_1^s(\log \frac{s}{u})^{\alpha-1}
\mathbf{E}_{\alpha,\alpha}(\lambda(\log \frac{s}{u})^\alpha)G(u)
\frac{du}{u}\Big)\frac{ds}{s} \\
&\quad +\int_{t_{j+1}}^t(\log\frac{t}{s})^{n-\alpha-1}(\Phi(s)
 +\sum_{\sigma=0}^{j}\sum_{v=1}^nc_{\sigma v}(\log\frac{ s}{t_\sigma})^{\alpha-v}
\mathbf{E}_{\alpha,\alpha-v+1}(\lambda (\log \frac{s}{t_\sigma})^\alpha) \\
&\quad +\int_1^s(\log \frac{s}{u})^{\alpha-1}\mathbf{E}_{\alpha,\alpha}
(\lambda(\log \frac{s}{u})^\alpha)G(u)\frac{du}{u})\frac{ds}{s}\Big] \\
&={}^{RLH}D_{t_{j+1}^+}^\alpha \Phi(t)
 +\frac{1}{\Gamma (n-\alpha)}(t\frac{d}{dt})^n
\Big[\sum_{\rho=0}^{j}\int_{t_\rho}^{t_{\rho+1}}
(\log\frac{t}{s})^{n-\alpha-1} \\
&\quad\times \sum_{\sigma=0}^\rho\sum_{v=1}^n
c_{\sigma v}(\log\frac{ s}{t_\sigma})^{\alpha-v}
 \mathbf{E}_{\alpha,\alpha-v+1}
\big(\lambda (\log \frac{s}{t_\sigma})^\alpha\big)\Big]
 +\frac{1}{\Gamma (n-\alpha)}(t\frac{d}{dt})^n \\
&\quad\times \Big[\int_{t_{j+1}}^t(\log\frac{t}{s})^{n-\alpha-1}\sum_{\sigma=0}^{j}
\sum_{v=1}^nc_{\sigma v}(\log\frac{ s}{t_\sigma})^{\alpha-v}
\mathbf{E}_{\alpha,\alpha-v+1}(\lambda (\log \frac{s}{t_\sigma})^\alpha)\Big]\\
&\quad +\frac{1}{\Gamma (n-\alpha)}(t\frac{d}{dt})^n
\Big[\int_1^t(\log\frac{t}{s})^{n-\alpha-1}\int_1^s(\log \frac{s}{u})^{\alpha-1}
\mathbf{E}_{\alpha,\alpha}(\lambda(\log \frac{s}{u})^\alpha)
G(u)\frac{du}{u}\frac{ds}{s}\Big].
\end{align*}
As above, we can obtain
$$
F(t)+\lambda x(t)={}^{RLH}D_{1^+}^\alpha x(t)
=F(t)+\lambda x(t)+{}^{RLH}D_{t_{j+1}^+}^\alpha \Phi(t)-\lambda \Phi(t).
$$
So ${}^{RLH}D_{t_{j+1}^+}^\alpha \Phi(t)-\lambda \Phi(t)=0$ on
$(t_{j+1},t_{j+2}]$. Then Theorem \ref{thm3.1.6} implies that there exists a
constant $c_{j+1v}\in \mathbb{R}(v\in N[1,n])$ such that
\[
\Phi(t)=\sum_{v=1}^nc_{j+1v}(\log\frac{t}{t_{j+1}})^{\alpha-v}
\mathbf{E}_{\alpha,\alpha}(\lambda (\frac{t}{t_{j+1}})^\alpha)
\]
 on $(t_{j+1},t_{j+2}]$. Hence
\begin{align*}
x(t)&=\sum_{\rho=0}^{j+1}\sum_{v=1}^{n}c_{\rho v}
(\log\frac{ t}{t_\rho})^{\alpha-v}\mathbf{E}_{\alpha,\alpha-v+1}
(\lambda (\log \frac{t}{t_\rho})^\alpha)\\
&\quad +\int_1^t(\log \frac{t}{s})^{\alpha-1}\mathbf{E}_{\alpha,\alpha}
(\lambda(\log \frac{t}{s})^\alpha)G(s)\frac{ds}{s},\quad t\in (t_{j+1},t_{j+2}].
\end{align*}
By mathematical induction, we know that \eqref{e3.2.8} holds for $j\in N_0$.
The proof is complete.
\end{proof}

\begin{theorem} \label{thm3.2.4}
Suppose that $G$ is continuous on $(1,e)$ and there exist constants
$k>-\alpha+n-1$ and $l\in (-\alpha,-\alpha+k,0]$ such that
$|G(t)|\le (\log t)^k(1-\log t)^l$ for all $t\in (1,e)$.
Then $x$ is a piecewise solution of \eqref{e3.2.4} if and only if there exist
constants $c_{jv}\in \mathbb{R}(j\in N[0,m],v\in N[1,n])$ such that
\begin{equation}
\begin{aligned}
x(t)&=\sum_{\rho=0}^j\sum_{v=0}^{n-1}c_{\rho v}(\log\frac{t}{t_\rho})^v
 \mathbf{E}_{\alpha,v+1}(\lambda (\log \frac{t}{t_\rho})^\alpha)\\
&\quad +\int_1^t(\log\frac{t}{s})^{\alpha-1}
\mathbf{E}_{\alpha,\alpha}(\lambda(\log\frac{t}{s})^\alpha)
G(s)\frac{ds}{s}, \quad
t\in (t_j,t_{j+1}],\; j\in N[0,m].
\end{aligned}\label{e3.2.9}
\end{equation}
\end{theorem}

\begin{proof}
For $t\in (t_j,t_{j+1}](j\in N[0,m])$, similarly to the beginning of the
 proof of Theorem \ref{thm3.2.3} we know that
\begin{align*}
&\big|\int_1^t(\log \frac{t}{s})^{\alpha-1}
 \mathbf{E}_{\alpha,\alpha}(\lambda(\log \frac{t}{s})^\alpha)G(s)\frac{ds}{s}\big|
\\
&\le (\log t)^{\alpha+k+l} \mathbf{E}_{\alpha,\alpha}
(\lambda(\log \frac{t}{s})^\alpha)\mathbf{B}(\alpha+l,k+1).
\end{align*}
So $\int_1^t(\log \frac{t}{s})^{\alpha-1}\mathbf{E}_{\alpha,\alpha}
(\lambda(\log \frac{t}{s})^\alpha)G(s)\frac{ds}{s}$ is convergent for all
$t\in (1,e]$ and
$$
\lim_{t\to 1^+}\int_1^t(\log \frac{t}{s})^{\alpha-1}
\mathbf{E}_{\alpha,\alpha}(\lambda(\log \frac{t}{s})^\alpha)G(s)
\frac{ds}{s}
$$
exists.
 We have
\begin{align*}
&\sum_{\tau=0}^{+\infty}
\frac{\lambda^\tau}{\Gamma(\alpha(\tau+1)-n+1)}
\int_1^s(\log\frac{s}{u})^{\tau\alpha+\alpha-n}
|G(u)|\frac{du}{u}\\
&\le \sum_{\tau=0}^{+\infty}
\frac{\lambda^\tau}{\Gamma(\alpha(\tau+1)-n+1)}
\int_1^s(\log\frac{s}{u})^{\tau\alpha+\alpha-n}
(\log \frac{e}{u})^l(\log u)^k\frac{du}{u}\\
& \le \sum_{\tau=0}^{+\infty}
\frac{\lambda^\tau}{\Gamma(\alpha(\tau+1)-n+1)}
\int_1^s(\log\frac{s}{u})^{\tau\alpha+\alpha+l-n} (\log u)^k\frac{du}{u}\\
&=\sum_{\tau=0}^{+\infty}\frac{\lambda^\tau}{\Gamma(\alpha(\tau+1)-n+1)}
(\log s)^{\tau\alpha+\alpha+k+l-n+1}\int_0^1(1-w)^{\alpha\tau+\alpha+l-n}w^kdw\\
&\le (\log s)^{\alpha+k+l-n+1}\mathbf{E}_{\alpha,\alpha-v+1}
(\lambda (\log s)^\alpha)\int_0^1(1-w)^{\alpha+l-n}w^kdw\\
&\le \mathbf{E}_{\alpha,\alpha-v+1}(\lambda (\log s)^\alpha)\mathbf{B}
(\alpha+l-n+1,k+1).
\end{align*}
So
$$
\lim_{s\to 1^+}\sum_{\tau=0}^{+\infty}
\frac{\lambda^\tau}{\Gamma(\alpha(\tau+1)-n+1)}
\int_1^s(\log\frac{s}{u}) ^{\tau\alpha+\alpha-n}
|G(u)|\frac{du}{u}=0.
$$
If $x$ is a solution of \eqref{e3.2.9}, we have $x\in LP_mC(1,e]$.
By using Definition \ref{def2.6}, it follows for $t\in (t_i,t_{i+1}]$ that
\begin{align*}
&{}^{CH}D_{1^+}^\alpha x(t) \\
&=\frac{1}{\Gamma (n-\alpha )}
 \int_1^t(\log\frac{t}{s})^{n-\alpha-1}(s\frac{d}{ds})^nx(s)\frac{ds}{s}\\
&=\frac{1}{\Gamma (n-\alpha )}
 \Big[\sum_{\sigma=0}^{i-1}\int_{t_\sigma}^{t_{\sigma+1}}
 (\log\frac{t}{s})^{n-\alpha-1}(s\frac{d}{ds})^n
 \Big(\sum_{\rho=0}^\sigma\sum_{v=0}^{n-1}c_{\rho v}(\log\frac{s}{t_\rho})^v
 \mathbf{E}_{\alpha,v+1}(\lambda (\log \frac{s}{t_\rho})^\alpha) \\
&\quad +\int_1^s(\log\frac{s}{u})^{\alpha-1}
 \mathbf{E}_{\alpha,\alpha}(\lambda(\log\frac{s}{u})^\alpha)
G(u)\frac{du}{u}\Big)\frac{ds}{s} \\
&\quad +\int_{t_{i}}^t(\log\frac{t}{s})^{n-\alpha-1}(s\frac{d}{ds})^n
\Big(\sum_{\rho=0}^i\sum_{v=0}^{n-1}c_{\rho v}(\log\frac{s}{t_\rho})^v
 \mathbf{E}_{\alpha,v+1}(\lambda (\log \frac{s}{t_\rho})^\alpha) \\
&\quad +\int_1^s(\log\frac{s}{u})^{\alpha-1}\mathbf{E}_{\alpha,\alpha}
 (\lambda(\log\frac{s}{u})^\alpha)
G(u)\frac{du}{u}\Big)\frac{ds}{s}\Big]\\
&=\frac{1}{\Gamma (n-\alpha )}\sum_{\sigma=0}^{i-1}
 \sum_{\rho=0}^\sigma\sum_{v=0}^{n-1}c_{\rho v}
 \int_{t_\sigma}^{t_{\sigma+1}}(\log\frac{t}{s})^{n-\alpha-1}(s\frac{d}{ds})^n
 (\log\frac{s}{t_\rho})^v \mathbf{E}_{\alpha,v+1}
 \big(\lambda (\log \frac{s}{t_\rho})^\alpha\big)\frac{ds}{s}\\
&\quad +\frac{1}{\Gamma (n-\alpha )}\sum_{\rho=0}^i\sum_{v=0}^{n-1}
 c_{\rho v}\int_{t_{i}}^t(\log\frac{t}{s})^{n-\alpha-1}
 (s\frac{d}{ds})^n(\log\frac{s}{t_\rho})^v \mathbf{E}_{\alpha,v+1}
 (\lambda (\log \frac{s}{t_\rho})^\alpha)\frac{ds}{s}\\
&\quad\times \frac{1}{\Gamma (n-\alpha )}\int_1^t
(\log\frac{t}{s})^{n-\alpha-1}(s\frac{d}{ds})^n\int_1^s(\log\frac{s}{u})
^{\alpha-1}\mathbf{E}_{\alpha,\alpha}(\lambda(\log\frac{s}{u})^\alpha)
G(u)\frac{du}{u}\frac{ds}{s}.
\end{align*}
One sees that
\begin{align*}
&\int_{t_{i}}^t(\log\frac{t}{s})^{n-\alpha-1}(s\frac{d}{ds})^n
\big((\log\frac{s}{t_\rho})^v\mathbf{E}_{\alpha,v+1}
\big(\lambda (\log \frac{s}{t_v})^\alpha\big)\Big)\frac{ds}{s}\\
&=\int_{t_{i}}^t(\log\frac{t}{s})^{n-\alpha-1}(s\frac{d}{ds})^n
\Big(\sum_{\tau=0}^{+\infty}\frac{\lambda^\tau
(\log\frac{s}{t_\rho})^{\tau\alpha+v}}{\Gamma(\tau \alpha+v+1)}\Big)\frac{ds}{s}\\
&=\int_{t_{i}}^t(\log\frac{t}{s})^{n-\alpha-1}
\Big(\sum_{\tau=1}^{+\infty}\frac{\lambda^\tau
(\log\frac{s}{t_\rho})^{\tau\alpha+v-n}}{\Gamma(\tau \alpha+v-n+1)}\Big)\frac{ds}{s}\\
&=\sum_{\tau=1}^{+\infty}\frac{\lambda^\tau}{\Gamma(\tau \alpha+v-n+1)}
(\log\frac{t}{t_\rho})^{\alpha(\tau-1)+v}
\int_{\frac{\log t_{i}-\log t_\rho}{\log t-\log t_\rho}}^1(1-w)^{n-\alpha-1}w
^{\tau\alpha+v-n}dw
\end{align*}
and
\begin{align*}
&\int_{t_\sigma}^{t_{\sigma+1}}(\log\frac{t}{s})^{n-\alpha-1}(s\frac{d}{ds})^n
(\log\frac{s}{t_\rho})^v \mathbf{E}_{\alpha,v+1}
\big(\lambda (\log \frac{s}{t_\rho})^\alpha\big)\frac{ds}{s}\\
&=\sum_{\tau=1}^{+\infty}\frac{\lambda^\tau}{\Gamma(\tau \alpha+v-n+1)}
(\log\frac{t}{t_\rho})^{\alpha(\tau-1)+v}
\int_{\frac{\log t_{\sigma}-\log t_\rho}{\log t-\log t_\rho}}^{\frac{\log t_{\sigma+1}
-\log t_\rho}{\log t-\log t_\rho}} (1-w)^{n-\alpha-1}w^{\alpha \tau+v-n}dw.
\end{align*}
Similarly,
\begin{align*}
&\int_1^t(\log\frac{t}{s})^{n-\alpha-1}(s\frac{d}{ds})^n\int_1^s(\log\frac{s}{u})
^{\alpha-1}\mathbf{E}_{\alpha,\alpha}(\lambda(\log\frac{s}{u})^\alpha)
G(u)\frac{du}{u}\frac{ds}{s}\\
&=\int_1^t(\log\frac{t}{s})^{n-\alpha-1}(s\frac{d}{ds})^n
\sum_{\tau=0}^{+\infty}\frac{\lambda^\tau}{\Gamma(\alpha(\tau+1))}
\int_1^s(\log\frac{s}{u})
^{\tau\alpha+\alpha-1}
G(u)\frac{du}{u}\frac{ds}{s}\\
&=\int_1^t(\log\frac{t}{s})^{n-\alpha-1}(s\frac{d}{ds})\sum_{\tau=0}^{+\infty}
\frac{\lambda^\tau}{\Gamma(\alpha(\tau+1)-n+1)}
\int_1^s(\log\frac{s}{u})
^{\tau\alpha+\alpha-n}
G(u)\frac{du}{u}\frac{ds}{s}\\
&=\frac{t}{n-\alpha}\Big[\int_1^t(\log\frac{t}{s})^{n-\alpha}(\sum_{\tau=0}^{+\infty}
\frac{\lambda^\tau}{\Gamma(\alpha(\tau+1)-n+1)}
\int_1^s(\log\frac{s}{u})
^{\tau\alpha+\alpha-n} G(u)\frac{du}{u})'ds\Big]'\\
&=\frac{t}{n-\alpha}\Big[ (\log\frac{t}{s})^{n-\alpha}(\sum_{\tau=0}^{+\infty}
\frac{\lambda^\tau}{\Gamma(\alpha(\tau+1)-n+1)}
\int_1^s(\log\frac{s}{u})
^{\tau\alpha+\alpha-n} G(u)\frac{du}{u})|_1^t \\
&\quad +(n-\alpha)\int_1^t(\log\frac{t}{s})^{n-\alpha-1}\sum_{\tau=0}^{+\infty}
\frac{\lambda^\tau}{\Gamma(\alpha(\tau+1)-n+1)}
\int_1^s(\log\frac{s}{u})^{\tau\alpha+\alpha-n}
G(u)\frac{du}{u}\frac{ds}{s}\Big]'\\
&=t\Big[\int_1^t(\log\frac{t}{s})^{n-\alpha-1}\sum_{\tau=0}^{+\infty}
\frac{\lambda^\tau}{\Gamma(\alpha(\tau+1)-n+1)}
\int_1^s(\log\frac{s}{u})^{\tau\alpha+\alpha-n}
G(u)\frac{du}{u}\frac{ds}{s}\Big]'\\
&=t\Big[\sum_{\tau=0}^{+\infty}\frac{\lambda^\tau}{\Gamma(\alpha(\tau+1)-n+1)}
\int_1^t\int_u^t(\log\frac{t}{s})^{n-\alpha-1}
(\log\frac{s}{u})^{\tau\alpha+\alpha-n}\frac{ds}{s} G(u)\frac{du}{u}\Big]'\\
&=t\left[\sum_{\tau=0}^{+\infty}\frac{\lambda^\tau}{\Gamma(\alpha(\tau+1)-n+1)}
\int_1^t(\log\frac{t}{u})^{\alpha\tau}
\int_0^1(1-w)^{n-\alpha-1}w^{\tau\alpha+\alpha-n}dwG(u)\frac{du}{u}\right]'\\
&=G(t)\frac{\mathbf{B}(n-\alpha,\alpha-n+1)}{\Gamma(\alpha-n+1)}
+\sum_{\tau=1}^{+\infty}\frac{(\alpha\tau)\lambda^\tau\mathbf{B}
(n-\alpha,\alpha\tau+\alpha-n+1)}{\Gamma(\alpha(\tau+1)-n+1)} \\
&\quad\times \int_1^t(\log\frac{t}{u})^{\alpha\tau-1}G(u)\frac{du}{u}.
\end{align*}
So
\begin{align*}
&{}^{CH}D_{1^+}^\alpha x(t) \\
&=\frac{1}{\Gamma (n-\alpha )}\sum_{\sigma=0}^{i-1}\sum_{\rho=0}^\sigma
\sum_{v=0}^{n-1}c_{\rho v}\sum_{\tau=1}^{+\infty}\frac{\lambda^\tau}
{\Gamma(\tau \alpha+v-n+1)}(\log\frac{t}{t_\rho})^{\alpha(\tau-1)+v} \\
&\quad \times \int_{\frac{\log t_{\sigma}
 -\log t_\rho}{\log t-\log t_\rho}}^{\frac{\log t_{\sigma+1}
 -\log t_\rho}{\log t-\log t_\rho}} (1-w)^{n-\alpha-1}w^{\alpha \tau+v-n}dw\\
&\quad +\frac{1}{\Gamma (n-\alpha )}\sum_{\rho=0}^i\sum_{v=0}^{n-1}c_{\rho v}
\sum_{\tau=1}^{+\infty}\frac{\lambda^\tau}{\Gamma(\tau \alpha+v-n+1)}
(\log\frac{t}{t_\rho})^{\alpha(\tau-1)+v} \\
&\quad\times\int_{\frac{\log t_{i}-\log t_\rho}{\log t-\log t_\rho}}^1(1-w)^{n-\alpha-1}w
^{\tau\alpha+v-n}dw\\
&\quad +G(t)+\sum_{\tau=1}^{+\infty}\frac{(\alpha\tau)
\lambda^\tau}{\Gamma(\alpha\tau+1)}
\int_1^t(\log\frac{t}{u})^{\alpha\tau-1}G(u)\frac{du}{u}\\
&=\frac{1}{\Gamma (n-\alpha )}\sum_{\rho=0}^{i-1}
 \sum_{v=0}^{n-1}c_{\rho v}\sum_{\tau=1}^{+\infty}
 \frac{\lambda^\tau}{\Gamma(\tau \alpha+v-n+1)}
 (\log\frac{t}{t_\rho})^{\alpha(\tau-1)+v} \\
&\quad\times\sum_{\sigma=\rho}^{i-1}
\int_{\frac{\log t_{\sigma}-\log t_\rho}
 {\log t-\log t_\rho}}^{\frac{\log t_{\sigma+1}-\log t_\rho}
{\log t-\log t_\rho}} (1-w)^{n-\alpha-1}w^{\alpha \tau+v-n}dw\\
&\quad +\frac{1}{\Gamma (n-\alpha )}\sum_{\rho=0}^i\sum_{v=0}^{n-1}c_{\rho v}
\sum_{\tau=1}^{+\infty}\frac{\lambda^\tau}{\Gamma(\tau \alpha+v-n+1)}
(\log\frac{t}{t_\rho})^{\alpha(\tau-1)+v} \\
&\quad\times\int_{\frac{\log t_{i}-\log t_\rho}{\log t-\log t_\rho}}^1(1-w)^{n-\alpha-1}w
^{\tau\alpha+v-n}dw\\
&\quad +G(t)+\sum_{\tau=1}^{+\infty}\frac{(\alpha\tau)\lambda^\tau}
{\Gamma(\alpha\tau+1)} \int_1^t(\log\frac{t}{u})^{\alpha\tau-1}G(u)\frac{du}{u}\\
&=\frac{1}{\Gamma (n-\alpha )}\sum_{\rho=0}^{i-1}\sum_{v=0}^{n-1}c_{\rho v}
\sum_{\tau=1}^{+\infty}\frac{\lambda^\tau}{\Gamma(\tau \alpha+v-n+1)}
(\log\frac{t}{t_\rho})^{\alpha(\tau-1)+v} \\
&\quad\times\int_0^{\frac{\log t_{i+1}-\log t_\rho}{\log t-\log t_\rho}}
(1-w)^{n-\alpha-1}w^{\alpha \tau+v-n}dw\\
&\quad +\frac{1}{\Gamma (n-\alpha )}\sum_{\rho=0}^i\sum_{v=0}^{n-1}c_{\rho v}
\sum_{\tau=1}^{+\infty}\frac{\lambda^\tau}{\Gamma(\tau \alpha+v-n+1)}
(\log\frac{t}{t_\rho})^{\alpha(\tau-1)+v} \\
&\quad\times \int_{\frac{\log t_{i}-\log t_\rho}
{\log t-\log t_\rho}}^1(1-w)^{n-\alpha-1}w^{\tau\alpha+v-n}dw\\
&\quad  +G(t) \\
&\quad +\sum_{\tau=1}^{+\infty}\frac{(\alpha\tau)\lambda^\tau}
{\Gamma(\alpha\tau+1)}
\int_1^t(\log\frac{t}{u})^{\alpha\tau-1}G(u)\frac{du}{u} \\
&=\frac{1}{\Gamma (n-\alpha )}\sum_{\rho=0}^i\sum_{v=0}^{n-1}c_{\rho v}
\sum_{\tau=1}^{+\infty}\frac{\lambda^\tau}{\Gamma(\tau \alpha+v-n+1)}
(\log\frac{t}{t_\rho})^{\alpha(\tau-1)+v} \\
&\quad\times\int_0^1 (1-w)^{n-\alpha-1}w^{\alpha \tau+v-n}dw
 +G(t)\\
&\quad +\sum_{\tau=1}^{+\infty}\frac{(\alpha\tau)
 \lambda^\tau}{\Gamma(\alpha\tau+1)}
\int_1^t(\log\frac{t}{u})^{\alpha\tau-1}G(u)\frac{du}{u}\\
&=\sum_{\rho=0}^i\sum_{v=0}^{n-1}c_{\rho v}\sum_{\tau=1}^{+\infty}
\frac{\lambda^\tau}{\Gamma(\tau \alpha-\alpha+v+1)}
 (\log\frac{t}{t_\rho})^{\alpha(\tau-1)+v}+
G(t)\\
&\quad +\sum_{\tau=1}^{+\infty}\frac{(\alpha\tau)\lambda^\tau}{\Gamma(\alpha\tau+1)}
\int_1^t(\log\frac{t}{u})^{\alpha\tau-1}G(u)\frac{du}{u}\\
&=\lambda x(t)+G(t),t\in (t_i,t_{i+1}].
\end{align*}
It follows that $x$ is a solution of \eqref{e3.2.4}.

Now we prove that if $x$ is a solution of \eqref{e3.2.4}, then $x$ satisfies
 \eqref{e3.2.9} and $x\in LP_mC(1,e]$ by mathematical induction.
 By Theorem \ref{thm3.1.8}, we know that there exists a constant
$c_{0v}\in \mathbb{R}(v\in N[0,n-1])$ such that
\[
x(t)=\sum_{v=0}^{n-1}c_{0v}(\log t)^v\mathbf{E}_{\alpha,v+1}
(\lambda (\log t)^\alpha)+\int_1^t(\log\frac{t}{s})^{\alpha -1}
\mathbf{E}_{\alpha ,\alpha}(\lambda(\frac{t}{s})^\alpha)F(s)ds,
\]
for $t\in (t_0,t_1]$.
Hence \eqref{e3.2.9} holds for $j=0$. Assume that \eqref{e3.2.9} holds for
$j=0,1,2,\dots,i\le m-1$, we will prove that \eqref{e3.2.9} holds for $j=i+1$.
 Suppose that
\begin{align*}
x(t)&=\Phi(t)+\sum_{\sigma=0}^i\sum_{v=0}^{n-1}c_{\sigma v}
(\log\frac{t}{t_\sigma})^v
\mathbf{E}_{\alpha,v+1}\big(\lambda (\log \frac{t}{t_\sigma})^\alpha\big)\\
&\quad +\int_1^t(\log \frac{t}{s})^{\alpha-1}
 \mathbf{E}_{\alpha,\alpha}(\lambda(\log \frac{t}{s})^\alpha)G(s)\frac{ds}{s},
\quad t\in (t_{i+1},t_{i+2}].
\end{align*}
Then for $t\in (t_{i+1},t_{i+2}]$ we have
\begin{align*}
&F(t)+\lambda x(t) \\
&={}^{CH}D_{1^+}^\alpha x(t)\\
&=\frac{1}{\Gamma (n-\alpha)}\Big[\sum_{j=0}^i\int_{t_j}^{t_{j+1}}
(\log\frac{t}{s})^{n-\alpha-1}(s\frac{d}{ds})^nx(s)\frac{ds}{s} \\
&\quad +\int_{t_{i+1}}^t(\log\frac{t}{s})^{n-\alpha-1}
 (s\frac{d}{ds})^nx(s)\frac{ds}{s}\Big]\\
&=\frac{1}{\Gamma (n-\alpha)}\sum_{j=0}^i\int_{t_j}^{t_{j+1}}
(\log\frac{t}{s})^{n-\alpha-1}(s\frac{d}{ds})^n
\Big(\sum_{\sigma=0}^j\sum_{v=0}^{n-1}c_{\sigma v}(\log\frac{s}{t_\sigma})^v
\mathbf{E}_{\alpha,v+1}\big(\lambda (\log \frac{s}{t_\sigma})^\alpha\big)\\
&\quad +\int_1^s(\log \frac{s}{u})^{\alpha-1}
\mathbf{E}_{\alpha,\alpha}(\lambda(\log \frac{s}{u})^\alpha)G(u)
\frac{du}{u}\Big)\frac{ds}{s}
 +\frac{1}{\Gamma (n-\alpha)}\int_{t_{i+1}}^t(\log\frac{t}{s})^{n-\alpha-1} \\
&\quad\times (s\frac{d}{ds})^n\Big(\Phi(s)
+\sum_{\sigma=0}^i\sum_{v=0}^{n-1}c_{\sigma v}(\log\frac{s}{t_\sigma})^v
\mathbf{E}_{\alpha,v+1}(\lambda (\log \frac{s}{t_\sigma})^\alpha)\\
&\quad +\int_1^s(\log \frac{s}{u})^{\alpha-1}
\mathbf{E}_{\alpha,\alpha}(\lambda(\log \frac{s}{u})^\alpha)
G(u)\frac{du}{u}\Big)\frac{ds}{s}\\
&={}^{CH}D_{t_{i+1}^+}^\alpha \Phi(t)+\frac{1}{\Gamma (n-\alpha)}
\sum_{j=0}^i\int_{t_j}^{t_{j+1}}(\log\frac{t}{s})^{n-\alpha-1}(s\frac{d}{ds})^n \\
&\quad\times \sum_{\sigma=0}^j\sum_{v=0}^{n-1}c_{\sigma v}(\log\frac{s}{t_\sigma})^v
\mathbf{E}_{\alpha,v+1}(\lambda (\log \frac{s}{t_\sigma})^\alpha)\frac{ds}{s} \\
&\quad +\frac{1}{\Gamma (n-\alpha)}\int_{t_{i+1}}^t(\log\frac{t}{s})^{n-\alpha-1}
 (s\frac{d}{ds})^n\sum_{\sigma=0}^i\sum_{v=0}^{n-1}c_{\sigma v}
 (\log\frac{s}{t_\sigma})^v \mathbf{E}_{\alpha,v+1}
\big(\lambda (\log \frac{s}{t_\sigma})^\alpha\big)\frac{ds}{s} \\
&\quad +\frac{1}{\Gamma (n-\alpha)}\int_1^t(\log\frac{t}{s})^{n-\alpha-1}
(s\frac{d}{ds})^n\int_1^s(\log \frac{s}{u})^{\alpha-1}
\mathbf{E}_{\alpha,\alpha}(\lambda(\log \frac{s}{u})^\alpha)G(u)\frac{du}{u}\frac{ds}{s}.
\end{align*}
as above discussion we obtain
$$
F(t)+\lambda x(t)={}^{CH}D_{1^+}^\alpha x(t)
=F(t)+\lambda x(t)+{}^{CH}D_{t_{i+1}^+}^\alpha \Phi(t)-\lambda \Phi(t).
$$
So
${}^{CH}D_{t_{i+1}^+}^\alpha \Phi(t)-\lambda \Phi(t)=0$ for $t\in (t_{i+1},t_{i+2}]$.
 Then Theorem \ref{thm3.1.8} implies that there exists a constant
$c_{i+1v}\in \mathbb{R}(v\in N[0,n-1])$ such that
$$
 \Phi(t)=\sum_{v=0}^{n-1}c_{i+1v}(\log\frac{t}{t_{i+1}})^v
\mathbf{E}_{\alpha,v+1}(\lambda (\frac{t}{t_{i+1}})^\alpha),\quad
t\in (t_{i+1},t_{i+2}].
 $$
 Hence
\begin{align*}
x(t)&=\sum_{j=0}^{i+1}\sum_{v=0}^{n-1}c_{jv}(\log\frac{t}{t_j})^v
\mathbf{E}_{\alpha,v+1}\big(\lambda (\log \frac{t}{t_j})^\alpha\big) \\
&\quad +\int_1^t(\log \frac{t}{s})^{\alpha-1}\mathbf{E}_{\alpha,\alpha}
(\lambda(\log \frac{t}{s})^\alpha)G(s)\frac{ds}{s},t\in (t_{i+1},t_{i+2}].
\end{align*}
By mathematical induction, we know that \eqref{e3.2.9} holds for $j\in N_0$.
 The proof is complete.
\end{proof}

\subsection{Preliminaries for BVP \eqref{e1.0.7}}

In this section, we present some preliminary results that can be used in
next sections for obtaining solutions to \eqref{e1.0.7}.

\begin{lemma} \label{lem3.3.1}
 Suppose that $\sigma:(0,1)\to \mathbb{R}$ is continuous and satisfies that there exist
numbers $k>-1$ and $\max\{-\beta,-2-k\}<l\le 0$ such that
$|\sigma(t)|\le t^k(1-t)^l$ for all $t\in (0,1)$. The $x$ is a solutions of
\begin{equation}
\begin{gathered}
{}^{RL}D_{0^+}^{\beta}x(t)-\lambda x(t)=\sigma(t), \quad
t\in (t_i,t_{i+1}],\; i\in \mathbb{N}[0,m],\\
\lim_{t\to 0^+}t^{2-\beta }x(t)=a,\quad x(1)=b,\\
\lim_{t\to t_i^+}(t-t_i)^{2-\beta} x(t)=I_i,\quad i\in \mathbb{N}[1,m],\\
\Delta {}^{RL}D_{0^+}^{\beta-1}x(t_i)=J_i,\quad i\in \mathbb{N}[1,m],
\end{gathered} \label{e3.3.1}
\end{equation}
if and only if $x\in P_1C_{1-\alpha}(0,1]$ and
\begin{equation}
\begin{aligned}
x(t)
&= \frac{t^{\beta-1}\mathbf{E}_{\beta,\beta}(\lambda t^\beta)}
{\mathbf{E}_{\beta,\beta}(\lambda)}\Big[b-a\mathbf{E}_{\beta,\beta-1}(\lambda)
-\int_0^1(1-s)^{\beta-1}\mathbf{E}_{\beta,\beta}(\lambda(1-s)^\beta)\sigma(s)ds \\
&\quad -\sum_{\sigma=1}^m((1-t_\sigma)^{\beta-1}
\mathbf{E}_{\beta,\beta}(\lambda (1-t_\sigma)^\beta)J_\sigma
+(1-t_\sigma)^{\beta-2}\mathbf{E}_{\beta,\beta-1}
(\lambda (1-t_\sigma)^\beta)I_\sigma)\Big]\\
&\quad +at^{\beta-2}\mathbf{E}_{\beta,\beta-1}(\lambda t^\alpha)\\
&\quad +\sum_{\sigma=1}^i\big[(t-t_\sigma)^{\beta-1}\mathbf{E}_{\beta,\beta}
(\lambda (t-t_\sigma)^\beta)J_\sigma+(t-t_\sigma)^{\beta-2}
\mathbf{E}_{\beta,\beta-1}(\lambda (t-t_\sigma)^\beta)I_\sigma\big]\\
&\quad +\int_0^t(t-s)^{\beta-1}\mathbf{E}_{\beta,\beta}
(\lambda(t-s)^\beta)\sigma(s)ds,\quad t\in (t_i,t_{i+1}],\; i\in \mathbb{N}[0,m].
\end{aligned} \label{e3.3.2}
\end{equation}
\end{lemma}

\begin{proof}
Let $x$ be a solution of \eqref{e3.3.1}.
By Theorem \ref{thm3.2.2}, we know that there exist numbers
$c_{\sigma 0},c_{\sigma 1}\in \mathbb{R}(\sigma\in \mathbb{N}[1,n])$ such that
\begin{equation}
\begin{aligned}
x(t)&=\sum_{\sigma=0}^i\big[c_{\sigma 1}(t-t_\sigma)^{\beta-1}
\mathbf{E}_{\beta,\beta}(\lambda (t-t_\sigma)^\beta)
+c_{\sigma 2}(t-t_\sigma)^{\beta-2}\mathbf{E}_{\beta,\beta-1}
(\lambda (t-t_\sigma)^\beta)\big]\\
&\quad +\int_0^t(t-s)^{\beta-1}\mathbf{E}_{\beta,\beta}(\lambda(t-s)^\beta)
\sigma(s)ds, \quad t\in (t_i,t_{i+1}],\; i\in \mathbb{N}[0,m].
\end{aligned}\label{e3.3.3}
\end{equation}
One has
\begin{gather*}
\begin{aligned}
&{}^{RL}D_{0^+}^{\beta-1}[(t-t_\sigma)^{\beta-1}\mathbf{E}_{\beta,\beta}
(\lambda (t-t_\sigma)^\beta)]\\
&=\frac{1}{\Gamma(2-\beta)}\Big[\sum_{\tau=0}^{+\infty}
\frac{\lambda^\tau}{\Gamma((\tau+1)\beta)}
\int_0^t(t-s)^{1-\beta}(s-t_\sigma)^{\tau\beta+\beta-1}ds\Big]'\\
&=\frac{1}{\Gamma(2-\beta)}\Big[\sum_{\tau=0}^{+\infty}
\frac{\lambda^\tau}{\Gamma((\tau+1)\beta)}
(t-t_\sigma)^{\tau\beta+1}\int_0^1(1-w)^{1-\beta}w^{\tau\beta+\beta-1}dw\Big]'\\
&=\mathbf{E}_{\beta,1}(\lambda(t-t_\sigma)^\beta),
\end{aligned} \\
{}^{RL}D_{0^+}^{\beta-1}\left[(t-t_\sigma)^{\beta-2}
\mathbf{E}_{\beta,\beta-1}(\lambda (t-t_\sigma)^\beta)\right]
=\lambda (t-t_\sigma)^\beta\mathbf{E}_{\beta,\beta+1}(\lambda(t-t_\sigma)^\beta),
\\
{}^{RL}D_{0^+}^{\beta-1}\left[(t-s)^{\beta-1}\mathbf{E}_{\beta,\beta}
(\lambda(t-s)^\beta)\right]=\mathbf{E}_{\beta,1}(\lambda(t-s)^\beta).
\end{gather*}
It follows that
\begin{equation}
\begin{aligned}
&{}^{RL}D_{0^+}^{\beta-1}x(t) \\
&=\sum_{\sigma=0}^i\big[c_{\sigma 1}\mathbf{E}_{\beta,1}
(\lambda(t-t_\sigma)^\beta)+c_{\sigma 2}\lambda
(t-t_\sigma)^\beta\mathbf{E}_{\beta,\beta+1}(\lambda(t-t_\sigma)^\beta)\big]\\
&\quad +\int_0^t\mathbf{E}_{\beta,1}(\lambda(t-s)^\beta)\sigma(s)ds, \quad
t\in (t_i,t_{i+1}],\; i\in \mathbb{N}[0,m].
\end{aligned} \label{e3.3.4}
\end{equation}
It follows from the boundary conditions and the impulse assumption in \eqref{e3.3.1}
that $c_{02}=a$, $c_{\sigma2}=I_\sigma(\sigma\in \mathbb{N}[1,m])$,
$c_{\sigma1}=J_\sigma(\sigma\in \mathbb{N}[1,m])$ and
\begin{align*}
&\sum_{\sigma=0}^m\left[c_{\sigma 1}(1-t_\sigma)^{\beta-1}
\mathbf{E}_{\beta,\beta}(\lambda (1-t_\sigma)^\beta)
+c_{\sigma 2}(1-t_\sigma)^{\beta-2}\mathbf{E}_{\beta,\beta-1}
(\lambda (1-t_\sigma)^\beta)\right]\\
&\quad +\int_0^1(1-s)^{\beta-1}\mathbf{E}_{\beta,\beta}
(\lambda(1-s)^\beta)\sigma(s)ds=b.
\end{align*}
Then
\begin{equation}
\begin{aligned}
c_{0,1}&=\frac{1}{\mathbf{E}_{\beta,\beta}(\lambda)}
\Big[b-a\mathbf{E}_{\beta,\beta-1}(\lambda)-\int_0^1(1-s)^{\beta-1}
\mathbf{E}_{\beta,\beta}(\lambda(1-s)^\beta)\sigma(s)ds \\
& -\sum_{\sigma=1}^m((1-t_\sigma)^{\beta-1}\mathbf{E}_{\beta,\beta}
(\lambda (1-t_\sigma)^\beta)J_\sigma \\
&\quad +(1-t_\sigma)^{\beta-2}
\mathbf{E}_{\beta,\beta-1}(\lambda (1-t_\sigma)^\beta)I_\sigma)\Big].
\end{aligned} \label{e3.3.5}
\end{equation}
Substituting $c_{\sigma1},c_{\sigma2}(\sigma\in N[0,m])$ into \eqref{e3.2.3},
we obtain \eqref{e3.3.2} obviously.


On the other hand, if $x$ satisfies \eqref{e3.3.2},
 then $x|_{(t_i,t_{i+1}]}(i\in \mathbb{N}[0,m])$ are continuous and the limits
$\lim_{t\to t_i^+}(t-t_i)^{2-\beta}x(t)$ $(i\in\mathbb{N}[0,m])$ exist.
So $x\in P_mC_{2-\beta}(0,1]$.
Using \eqref{e3.3.5} and $c_{02}=a$, $c_{\sigma2}=I_\sigma(\sigma\in N[1,m])$,
 $c_{\sigma1}=J_\sigma(\sigma\in \mathbb{N}[1,m])$, we rewrite $x$ by
\begin{align*}
x(t)&=\sum_{\sigma=0}^i\left[c_{\sigma 1}(t-t_\sigma)^{\beta-1}
 \mathbf{E}_{\beta,\beta}(\lambda (t-t_\sigma)^\beta)
 +c_{\sigma2}(t-t_\sigma)^{\beta-2}
 \mathbf{E}_{\beta,\beta-1}(\lambda (t-t_\sigma)^\beta)\right]\\
&\quad +\int_0^t(t-s)^{\beta-1}\mathbf{E}_{\beta,\beta}(\lambda(t-s)^\alpha)
\sigma(s)ds,\quad t\in (t_i,t_{i+1}],\; i\in \mathbb{N}[0,m].
\end{align*}
Since $\sigma$ is continuous on $(0,1)$ and $|\sigma(t)|\le t^k(1-t)^l$,
one can show easily that $x$ is continuous on $(t_i,t_{i+1}](i=0,1)$ and
using the method at the beginning of the proof of this lemma, we know that
both limits $\lim_{t\to t_i^+}(t-t_i)^{2-\beta}x(t)$ $(i\in\mathbb{N}[0,m])$ exist.
So $x\in P_mC_{2-\beta}(0,1]$. Furthermore, by direct computation, we have $x(1)=b$,
and $ \lim_{t\to 0^+}t^{1-\beta} x(t)=a$. One have from Theorem \ref{thm3.2.2} easily
for $t\in (t_0,t_1]$ that $D_{0^+}^{\beta}x(t)=\lambda x(t)+\sigma(t)$ and
for $t\in (t_j,t_{j+1}]$ that
\begin{align*}
&{}^{RL}D_{0^+}^\beta x(t) \\
&=\frac{1}{\Gamma(2-\beta)}\Big[\int_0^t(t-s)^{1-\beta}x(s)ds\Big]'' \\
&=\frac{1}{\Gamma(2-\beta)}\Big[\sum_{i=0}^{j-1}
 \int_{t_i}^{t_{i+1}}(t-s)^{1-\beta}\Big(\sum_{\sigma=0}^i\sum_{v=1}^2
c_{\sigma v}(s-t_\sigma)^{\beta-v}\mathbf{E}_{\beta,\beta-v+1}
 (\lambda (s-t_\sigma)^\beta) \\
&\quad +\int_0^s(s-u)^{\beta-1}\mathbf{E}_{\beta,\beta}
(\lambda(s-u)^\beta)\sigma(u)du\Big)ds\Big]'' \\
&\quad +\frac{1}{\Gamma(2-\beta)}\Big[\int_{t_j}^t(t-s)^{1-\beta}
\Big(\sum_{\sigma=0}^j\sum_{v=1}^2c_{\sigma v}(s-t_\sigma)^{\beta-v}
 \mathbf{E}_{\beta,\beta-v+1}(\lambda (s-t_\sigma)^\beta) \\
&\quad +\int_0^s(s-u)^{\beta-1}\mathbf{E}_{\beta,\beta}
(\lambda(s-u)^\beta)\sigma(u)du\Big)ds\Big]''\\
&=\frac{1}{\Gamma(2-\beta)}
\Big[\sum_{i=0}^{j-1}\int_{t_i}^{t_{i+1}}(t-s)^{1-\beta}\sum_{\sigma=0}^i
\sum_{v=1}^2c_{\sigma v}(s-t_\sigma)^{\beta-v}\mathbf{E}_{\beta,\beta-v+1}
(\lambda (s-t_\sigma)^\beta)ds\Big]''\\
&\quad +\frac{1}{\Gamma(2-\beta)}\Big[\int_{t_j}^t(t-s)^{1-\beta}
\sum_{\sigma=0}^j\sum_{v=1}^2c_{\sigma v}(s-t_\sigma)^{\beta-v}
\mathbf{E}_{\beta,\beta-v+1}(\lambda (s-t_\sigma)^\beta)ds\Big]''\\
&\quad +\frac{1}{\Gamma(2-\beta)}\Big[\int_0^t(t-s)^{1-\beta}
 \int_0^s(s-u)^{\beta-1}\mathbf{E}_{\beta,\beta}(\lambda(s-u)^\beta)
\sigma(u)\,du\,ds\Big]''\\
&=\frac{1}{\Gamma(2-\beta)}
\Big[\sum_{i=0}^{j-1}\sum_{\sigma=0}^i\sum_{v=1}^2c_{\sigma v}
\sum_{\tau=0}^{+\infty}\frac{\lambda^\tau}{\Gamma((\tau+1)\beta-v+1)} \\
&\quad\times \int_{t_i}^{t_{i+1}}(t-s)^{1-\beta}(s-t_\sigma)^{\tau\beta+\beta-v}ds\Big]''\\
&\quad +\frac{1}{\Gamma(2-\beta)}\Big[\sum_{\sigma=0}^j
\sum_{v=1}^2c_{\sigma v}\sum_{\tau=0}^{+\infty}\frac{\lambda^\tau}
{\Gamma((\tau+1)\beta-v+1)}\int_{t_j}^t(t-s)^{1-\beta}
(s-t_\sigma)^{\tau\beta+\beta-v}ds\Big]''\\
&\quad +\frac{1}{\Gamma(2-\beta)}
\Big[\sum_{\tau=0}^{+\infty}\frac{\lambda^\tau}{\Gamma((\tau+1)\beta)}
\int_0^t\int_u^t(t-s)^{1-\beta}(s-u)^{\tau\beta+\beta-1}ds\sigma(u)du\Big]''\\
&=\frac{1}{\Gamma(2-\beta)}\Big[\sum_{\sigma=0}^{j-1}
 \sum_{i=\sigma}^{j-1}\sum_{v=1}^2c_{\sigma v}
 \sum_{\tau=0}^{+\infty}\frac{\lambda^\tau}{\Gamma((\tau+1)\beta-v+1)}
 (t-t_\sigma)^{\tau\beta-v+2} \\
&\quad\times \int_{\frac{t_i-t_\sigma}{t-t_\sigma}}^{\frac{t_{i+1}-t_\sigma}{t-t_\sigma}}
 (1-w)^{1-\beta}w^{\tau\beta+\beta-v}dw\Big]''\\
&\quad +\frac{1}{\Gamma(2-\beta)}
 \Big[\sum_{\sigma=0}^j\sum_{v=1}^2c_{\sigma v}\sum_{\tau=0}^{+\infty}
 \frac{\lambda^\tau}{\Gamma((\tau+1)\beta-v+1)}(t-t_\sigma)^{\tau\beta-v+2}\\
&\quad\times \int_{\frac{t_j-t_\sigma}{t-t_\sigma}}^1(1-w)^{1-\beta}
 w^{\tau\beta+\beta-v}dw\Big]''\\
&\quad +\frac{1}{\Gamma(2-\beta)}
 \Big[\sum_{\tau=0}^{+\infty}\frac{\lambda^\tau}{\Gamma((\tau+1)\beta)}
 \int_0^t(t-u)^{\tau\beta+1}\int_0^1(1-w)^{1-\beta}
 w^{\tau\beta+\beta-1}dw\sigma(u)du\Big]'' \\
&=\frac{1}{\Gamma(2-\beta)}\Big[\sum_{\sigma=0}^{j-1}\sum_{v=1}^2c_{\sigma v}
 \sum_{\tau=0}^{+\infty}\frac{\lambda^\tau}{\Gamma((\tau+1)\beta-v+1)}
 (t-t_\sigma)^{\tau\beta-v+2} \\
&\quad\times \int_0^{\frac{t_{j}-t_\sigma}{t-t_\sigma}}
 (1-w)^{1-\beta}w^{\tau\beta+\beta-v}dw\Big]''\\
&\quad +\frac{1}{\Gamma(2-\beta)}
 \Big[\sum_{\sigma=0}^j\sum_{v=1}^2c_{\sigma v}\sum_{\tau=0}^{+\infty}
 \frac{\lambda^\tau}{\Gamma((\tau+1)\beta-v+1)}(t-t_\sigma)^{\tau\beta-v+2}\\
&\quad\times \int_{\frac{t_j-t_\sigma}{t-t_\sigma}}^1(1-w)^{1-\beta}
 w^{\tau\beta+\beta-v}dw\Big]''
 \Big[\sum_{\tau=0}^{+\infty}\frac{\lambda^\tau}{\Gamma(\tau\beta+2)}
\int_0^t(t-u)^{\tau\beta+1}\sigma(u)du\Big]''\\
&=\frac{1}{\Gamma(2-\beta)}\Big[\sum_{\sigma=0}^{j}\sum_{v=1}^2c_{\sigma v}
 \sum_{\tau=0}^{+\infty}\frac{\lambda^\tau}{\Gamma((\tau+1)\beta-v+1)}
 (t-t_\sigma)^{\tau\beta-v+2} \\
&\quad \int_0^1 (1-w)^{1-\beta}w^{\tau\beta+\beta-v}dw\Big]''
 +\sigma(t)+\sum_{\tau=1}^{+\infty}\frac{\lambda^\tau}{\Gamma(\tau\beta)}
\int_0^t(t-u)^{\tau\beta-1}\sigma(u)du\\
&=\Big[\sum_{\sigma=0}^{j}\sum_{v=1}^2c_{\sigma v}
 \sum_{\tau=0}^{+\infty}\frac{\lambda^\tau}{\Gamma(\tau\beta-v+3)}
(t-t_\sigma)^{\tau\beta-v+2}\Big]''
+\sigma(t)+\sum_{\tau=1}^{+\infty}\frac{\lambda^\tau}{\Gamma(\tau\beta)} \\
&\quad \times \int_0^t(t-u)^{\tau\beta-1}\sigma(u)du \\
&=\sum_{\sigma=0}^{j}\sum_{v=1}^2c_{\sigma v}
 \sum_{\tau=1}^{+\infty}\frac{\lambda^\tau}{\Gamma(\tau\beta-v+1)}
(t-t_\sigma)^{\tau\beta-v} +\sigma(t) \\
&\quad +\sum_{\tau=1}^{+\infty}\frac{\lambda^\tau}{\Gamma(\tau\beta)}
\int_0^t(t-u)^{\tau\beta-1}\sigma(u)du\\
&=\lambda x(t)+\sigma(t).
\end{align*}
So $x$ is a solution of \eqref{e3.3.1}. The proof is complete.
\end{proof}

 Define the nonlinear operator $T$ on $P_mC_{2-\beta}(0,1]$ for
$x\in P_mC_{2-\beta}(0,1]$ by
\begin{align*}
(Tx)(t)
&= \frac{t^{\beta-1}\mathbf{E}_{\beta,\beta}
(\lambda t^\beta)}{\mathbf{E}_{\beta,\beta}(\lambda)}
\Big[\int_0^1\psi(s)H(s,x(s))ds-\mathbf{E}_{\beta,\beta-1}
(\lambda)\int_0^1\phi(s)G(s,x(s))ds \\
&\quad -\int_0^1(1-s)^{\beta-1}\mathbf{E}_{\beta,\beta}
 (\lambda(1-s)^\beta)p(s)f(s,x(s))ds \\
&\quad -\sum_{\sigma=1}^m((1-t_\sigma)^{\beta-1}\mathbf{E}_{\beta,\beta}
(\lambda (1-t_\sigma)^\beta)J(t_\sigma,x(t_\sigma)) \\
&\quad +(1-t_\sigma)^{\beta-2}\mathbf{E}_{\beta,\beta-1}
(\lambda (1-t_\sigma)^\beta)I(t_\sigma,x(t_\sigma)))\Big] \\
&\quad +t^{\beta-2}\mathbf{E}_{\beta,\beta-1}(\lambda t^\alpha)
 \int_0^1\phi(s)G(s,x(s))ds\\
&\quad +\sum_{\sigma=1}^i\Big[(t-t_\sigma)^{\beta-1}
\mathbf{E}_{\beta,\beta}(\lambda (t-t_\sigma)^\beta)J(t_\sigma,x(t_\sigma)) \\
&\quad +(t-t_\sigma)^{\beta-2}\mathbf{E}_{\beta,\beta-1}
 (\lambda (t-t_\sigma)^\beta)I(t_\sigma,x(t_\sigma))\Big]\\
&\quad +\int_0^t(t-s)^{\beta-1}\mathbf{E}_{\beta,\beta}
 (\lambda(t-s)^\beta)p(s)f(s,x(s))ds,
\end{align*}
for $t\in (t_i,t_{i+1}]$, $i\in \mathbb{N}[0,m]$.

\begin{lemma} \label{lem3.3.2}
 Suppose that {\rm (1.A1)--(1.A5)} hold, and that $f, G,H$ are
impulsive \\ II-Carath\'eodory functions, $I,J$
discrete II-Carath\'odory functions. Then
$T:P_mC_{2-\beta}(0,1]\to P_mC_{2-\beta}(0,1]$ is well defined and is
completely continuous, $x\in P_mC_{2-\beta}(0,1]$ is a solution of
\eqref{e1.0.7} if and only if $x\in P_mC_{2-\beta}(0,1]$ is a fixed point of $T$.
\end{lemma}


\begin{proof}
\textbf{Step (i)} $T:P_mC_{2-\beta}(0,1]\to P_mC_{2-\beta}(0,1]$
is well defined.
It comes from the method in Theorem \ref{thm3.2.2} that $Tx|_{(t_i,t_{i+1}]}$ $(i\in \mathbb{N}[0,m])$
are continuous and the limits $\lim_{t\to t_i^+}(t-t_i)^{2-\beta}(Tx)(t)$
$(i\in\mathbb{N}[0,m])$ exist. We see from Lemma \ref{lem3.3.1} that
$x\in P_mC_{2-\beta}(0,1]$ is a solution of BVP\eqref{e1.0.7} if and only if
$x\in P_mC_{2-\beta}(0,1]$ is a fixed point of $T$ in $P_mC_{2-\beta}(0,1]$.
\smallskip

\noindent\textbf{Step (ii)} $T$ is continuous.
Let $x_n\in P_mC_{2-\beta}(0,1]$ with $x_n\to x_0$ as $n\to +\infty$.
We can show that $Tx_n\to Tx_0$ as $n\to +\infty$ by using the dominant
convergence theorem. We refer the readers to the papers \cite{sg2, wxl, rk}.
\smallskip

\noindent\textbf{Step (iii)}
 $T$ is compact, i.e., $T(\overline{\Omega})$ is relatively
compact for every bounded subset $\Omega\subset P_1C_{1-\alpha}(0,1]$.
Let $\Omega$ be a bounded open nonempty subset of $P_mC_{2-\beta}(0,1]$. We have
\begin{equation}
\|x\|=\max\{\sup_{t\in (t_i,t_{i+1}]}(t-t_i)^{2-\beta}|x(t)|:i\in \mathbb{N}[0,m]\}
\le r<+\infty,\label{e3.3.6}
\end{equation}
for $(x,y)\in \overline{\Omega}$.
Since $f,G,H$ are impulsive II-Carath\'eodory functions, $I,J$ are
discrete II-Carath\'eodory functions, then there exists constants
$M_f,M_I,M_J,M_G,M_H\ge 0$ such that
\begin{equation}
\begin{gathered}
|f(t,x(t)))|
=|f(t,(t-t_i)^{\alpha-2}(t-t_i)^{2-\alpha}x(t))|
\le M_f,\quad t\in (t_i,t_{i+1}],\; i\in \mathbb{N}[0,m],\\
|G(t,x(t))|\le M_G,t\in (t_i,t_{i+1}],\quad i\in \mathbb{N}[0,m],\\
|H(t,x(t))|\le M_H,t\in (t_i,t_{i+1}],\quad i\in \mathbb{N}[0,m],\\
|I(t_i,x(t_i))|=|I(t_i,(t_i-t_{i-1})^{\beta-2}(t_i-t_{i-1})^{2-\beta}x(t_i))|
\le M_I,\quad i\in \mathbb{N}[1,m],\\
|J(t_i,x(t_i))|=|I(t_i,(t_i-t_{i-1})^{\beta-2}(t_i-t_{i-1})^{2-\beta}x(t_i))|
\le M_J,\quad i\in \mathbb{N}[1,m].
\end{gathered}\label{e3.3.7}
\end{equation}
This step is done in three sub-steps:
\smallskip

\noindent\textbf{Sub-step (iii1)}
 $T(\overline{\Omega})$ is uniformly bounded.
Using \eqref{e3.3.2} and \eqref{e3.3.7}, we have for $t\in (t_i,t_{i+1}]$ that
\begin{align*}
&(t-t_i)^{2-\beta}|(Tx)(t)| \\
&\le \frac{(t-t_i)^{2-\beta}t^{\beta-1}\mathbf{E}_{\beta,\beta}
(\lambda t^\beta)}{\mathbf{E}_{\beta,\beta}(\lambda)}
\Big[\|\psi\|_1M_H+\mathbf{E}_{\beta,\beta-1}(\lambda)\|\phi\|_1M_G \\
&\quad +\int_0^1(1-s)^{\beta-1}\mathbf{E}_{\beta,\beta}(\lambda(1-s)^\beta)
s^k(1-s)^ldsM_f \\
&\quad +\sum_{\sigma=1}^m((1-t_\sigma)^{\beta-1}\mathbf{E}_{\beta,\beta}
 (\lambda (1-t_\sigma)^\beta)M_J
 +(1-t_\sigma)^{\beta-2}\mathbf{E}_{\beta,\beta-1}
 (\lambda (1-t_\sigma)^\beta)M_I)\Big]\\
&\quad +(t-t_i)^{2-\beta}t^{\beta-2}\mathbf{E}_{\beta,\beta-1}
 (\lambda t^\alpha)\|\phi\|_1M_G\\
&\quad +(t-t_i)^{2-\beta}\sum_{\sigma=1}^i
\Big[(t-t_\sigma)^{\beta-1}\mathbf{E}_{\beta,\beta}(\lambda (t-t_\sigma)^\beta)M_J\\
&\quad +(t-t_\sigma)^{\beta-2}\mathbf{E}_{\beta,\beta-1}(\lambda (t-t_\sigma)^\beta)M_I
\Big] \\
&\quad +\int_0^t(t-s)^{\beta-1}\mathbf{E}_{\beta,\beta}(\lambda(t-s)^\beta)
s^k(1-s)^ldsM_f\\
&\le \frac{\mathbf{E}_{\beta,\beta}(|\lambda|)}
 {\mathbf{E}_{\beta,\beta}(\lambda)}
\Big[\|\psi\|_1M_H+\mathbf{E}_{\beta,\beta-1}(\lambda)\|\phi\|_1M_G \\
&\quad +\sum_{\tau=0}^{+\infty}\frac{\lambda^\tau}{\Gamma((\tau+1)\beta)}
 \int_0^1(1-s)^{\tau\beta+\beta+l-1}s^kdsM_f  \\
&\quad +\sum_{\sigma=1}^m
 (\mathbf{E}_{\beta,\beta}(|\lambda|)M_J
 +(1-t_\sigma)^{\beta-2}\mathbf{E}_{\beta,\beta-1}(|\lambda|)M_I)\Big]\\
&\quad +\mathbf{E}_{\beta,\beta-1}(|\lambda|)\|\phi\|_1M_G
+\sum_{\sigma=1}^m\Big[\mathbf{E}_{\beta,\beta}(|\lambda|)M_J
+\mathbf{E}_{\beta,\beta-1}(|\lambda|)M_I\Big] \\
&\quad +\sum_{\tau=0}^{+\infty}\frac{\lambda^\tau}{\Gamma((\tau+1)\beta)}
\int_0^t(t-s)^{\tau\beta+\beta+l-1}s^kdsM_f\\
&\le \Big(\frac{\mathbf{E}_{\beta,\beta}(|\lambda|)
 \mathbf{E}_{\beta,\beta-1}(\lambda)\|\phi\|_1}{\mathbf{E}_{\beta,\beta}(\lambda)}
+\mathbf{E}_{\beta,\beta-1}(|\lambda|)\|\phi\|_1\Big)M_G
 +\frac{\mathbf{E}_{\beta,\beta}(|\lambda|)\|\psi\|_1}{\mathbf{E}_{\beta,\beta}
 (\lambda)}M_H\\
&\quad +\Big(\frac{\mathbf{E}_{\beta,\beta}(|\lambda|)\mathbf{E}_{\beta,\beta-1}
(|\lambda|)}{\mathbf{E}_{\beta,\beta}(\lambda)}\sum_{\sigma=1}^m
 (1-t_\sigma)^{\beta-2}+m\mathbf{E}_{\beta,\beta-1}(|\lambda|)\Big)M_I\\
&\quad +\Big(\frac{m\mathbf{E}_{\beta,\beta}(|\lambda|)\mathbf{E}_{\beta,\beta}
 (|\lambda|)}{\mathbf{E}_{\beta,\beta}(\lambda)}+m\mathbf{E}_{\beta,\beta}
 (|\lambda|)\Big)M_J \\
&\quad +\Big(\frac{\mathbf{E}_{\beta,\beta}(|\lambda|)}
{\mathbf{E}_{\beta,\beta}(\lambda)}\mathbf{E}_{\beta,\beta}(|\lambda|)
+\mathbf{E}_{\beta,\beta}(|\lambda|)\Big)\mathbf{B}(\beta+l,k+1)M_f.
\end{align*}
From the above discussion, we obtain
\begin{equation}
\begin{aligned}
\|Tx\|
&\le \Big(\frac{\mathbf{E}_{\beta,\beta}(|\lambda|)\mathbf{E}_{\beta,\beta-1}(\lambda)\|\phi\|_1}{\mathbf{E}_{\beta,\beta}(\lambda)}
+\mathbf{E}_{\beta,\beta-1}(|\lambda|)\|\phi\|_1\Big)M_G
 +\frac{\mathbf{E}_{\beta,\beta}(|\lambda|)\|\psi\|_1}{\mathbf{E}_{\beta,\beta}
 (\lambda)}M_H\\
&\quad +\Big(\frac{\mathbf{E}_{\beta,\beta}(|\lambda|)
 \mathbf{E}_{\beta,\beta-1}(|\lambda|)}{\mathbf{E}_{\beta,\beta}
 (\lambda)}\sum_{\sigma=1}^m(1-t_\sigma)^{\beta-2}
 +m\mathbf{E}_{\beta,\beta-1}(|\lambda|)\Big)M_I \\
&\quad +\Big(\frac{m\mathbf{E}_{\beta,\beta}(|\lambda|)\mathbf{E}_{\beta,\beta}
(|\lambda|)}{\mathbf{E}_{\beta,\beta}(\lambda)}+m\mathbf{E}_{\beta,\beta}(|\lambda|)
\Big)M_J \\
&\quad +(\frac{\mathbf{E}_{\beta,\beta}(|\lambda|)}{\mathbf{E}_{\beta,\beta}(\lambda)}
\mathbf{E}_{\beta,\beta}(|\lambda|)
+\mathbf{E}_{\beta,\beta}(|\lambda|))\mathbf{B}(\beta+l,k+1)M_f.
\end{aligned} \label{e3.3.8}
\end{equation}
 From the above discussion, $T(\overline{\Omega})$ is uniformly bounded.
\smallskip

\noindent\textbf{Sub-step (iii2)}
 Prove that $t\to (t-t_i)^{2-\beta}T(\overline{\Omega})$ is equi-continuous
 on $(t_i,t_{i+1}](i\in \mathbb{N}[0,m])$.
Let
$$
(t-t_i)^{2-\beta}\overline{(Tx)}(t)
= \begin{cases}
(t-t_i)^{2-\beta}(Tx)(t), & t\in (t_i,t_{i+1}],\\
\lim_{t\to t_i^+}(t-t_i)^{2-\beta}(Tx)(t), & t=t_i.
\end{cases}
$$
Then $t\to (t-t_i)^{2-\beta}\overline{(Tx)}(t)$ is continuous on $[t_i,t_{i+1}]$.
 Let $s_2\le s_1$ and $s_1,s_2\in [t_i,t_{i+1}]$. By Ascoli-CArzela theorem on
the closed interval, we have
$$
\big|(s_1-t_i)^{2-\beta}\overline{{(Tx)}}(s_1)
-(s_2-t_i)^{2-\beta}\overline{{(Tx)}}(s_2)\big|\to 0 \quad\text{uniformly as }
s_1\to s_2.
$$
Then $t\to (t-t_i)^{2-\beta}T(\overline{\Omega})$ is equi-continuous on
$(t_i,t_{i+1}](i\in N[0,m])$. So $T(\overline{\Omega})$ is relatively compact.
Then $T$ is completely continuous.
The proof is complete.
\end{proof}

\subsection{Preliminaries for BVP \eqref{e1.0.8}}
In this section, we present some preliminary results that can be used in
next sections for get solutions of \eqref{e1.0.8}.

\begin{lemma} \label{lem3.4.1}
 Suppose that $\sigma:(0,1)\to \mathbb{R}$ is continuous and satisfies that there exist
 numbers $k>1-\beta$ and $l\le 0$ with $l>\max\{-\beta,-\beta-k\}$ such that
$|\sigma(t)|\le t^k(1-t)^l$ for all $t\in (0,1)$. The $x$ is a solutions of
\begin{equation}
\begin{gathered}
{}^CD_{0^+}^{\beta}x(t)-\lambda x(t)=\sigma(t), \quad t\in (t_i,t_{i+1}],\;
i\in \mathbb{N}[0,m],\\
x(0)=a,\quad x'(1)=b,\\
\Delta x(t_i)=I_i,\quad \Delta x'(t_i)=J_i,\quad i\in \mathbb{N}[1,m]
\end{gathered} \label{e3.4.1}
\end{equation}
if and only if
\begin{equation}
\begin{aligned}
x(t)&=a+\frac{t}{\mathbf{E}_{\beta,1}(\lambda)}
\Big[b-\lambda \mathbf{E}_{\beta,\beta}(\lambda)a-\sum_{\sigma=1}^m
\Big(\lambda \mathbf{E}_{\beta,\beta}(\lambda (1-t_\sigma)^\beta)
(1-t_\sigma)^{\beta-1}I_\sigma \\
&\quad +\mathbf{E}_{\beta,1}(\lambda(1-t_\sigma)^\beta)J_\sigma\Big)
-\int_0^1(1-s)^{\beta-1}\mathbf{E}_{\beta,\beta-1}
(\lambda (1-s)^\beta)\sigma(s)ds\Big]\\
&\quad +\sum_{j=1}^i[\mathbf{E}_{\beta,1}(\lambda (t-t_\sigma)^\beta)
I_\sigma+(t-t_\sigma)\mathbf{E}_{\beta,2}(\lambda (t-t_\sigma)^\beta)J_\sigma]\\
&\quad +\int_0^t(t-s)^{\beta-1}\mathbf{E}_{\beta,\beta}
 (\lambda (t-s)^\beta)\sigma(s)ds, \quad t\in (t_i,t_{i+1}],\; i\in \mathbb{N}[0,m]
\end{aligned} \label{e3.4.2}
\end{equation}
\end{lemma}

\begin{proof}
Let $x$ be a solution of \eqref{e3.4.1}. We know by Theorem \ref{thm3.2.1} that there exist
numbers $c_{\sigma 0},c_{\sigma 1}\in \mathbb{R}(\sigma\in \mathbb{N}[0,m])$ such that
\begin{equation}
\begin{aligned}
x(t)&=\sum_{\sigma=0}^i\big[c_{\sigma0}\mathbf{E}_{\beta,1}
 (\lambda (t-t_\sigma)^\beta)+c_{\sigma1}(t-t_\sigma)
 \mathbf{E}_{\beta,2}(\lambda (t-t_\sigma)^\beta)\big]\\
&\quad +\int_0^t(t-s)^{\beta-1}\mathbf{E}_{\beta,\beta}
 (\lambda(t-s)^\beta)\sigma(s)ds, \quad t\in (t_i,t_{i+1}],\; i\in \mathbb{N}[0,m].
\end{aligned} \label{e3.4.3}
\end{equation}
It is easy to show that
\begin{gather*}
\left[\mathbf{E}_{\beta,1}(\lambda(t-t_\sigma)^\beta)\right]'
=\lambda \mathbf{E}_{\beta,\beta}(\lambda(t-t_\sigma)^\beta)(t-t_\sigma)^{\beta-1},\\
\left[(t-t_\sigma)\mathbf{E}_{\beta,2}(\lambda(t-t_\sigma)^\beta)\right]'
=\mathbf{E}_{\beta,1}(\lambda(t-t_\sigma)^\beta),\\
\left[(t-s)^{\beta-1}\mathbf{E}_{\beta,\beta}(\lambda(t-s)^\beta)\right]'
=(t-s)^{\beta-2}\mathbf{E}_{\beta,\beta-1}(\lambda(t-s)^\beta).
\end{gather*}
So
\begin{equation}
\begin{aligned}
x'(t)&=\sum_{\sigma=0}^i\left[\lambda c_{\sigma0}
\mathbf{E}_{\beta,\beta}(\lambda (t-t_\sigma)^\beta)(t-t_\sigma)^{\beta-1}
+c_{\sigma1}\mathbf{E}_{\beta,1}(\lambda (t-t_\sigma)^\beta)\right]\\
&+\int_0^t(t-s)^{\beta-2}\mathbf{E}_{\beta,\beta-1}(\lambda(t-s)^\beta)
\sigma(s)ds,\quad t\in (t_i,t_{i+1}],\; i\in \mathbb{N}[0,m].
\end{aligned} \label{e3.4.4}
\end{equation}
Note $\mathbf{E}_{\beta,1}(0)=1$, $\mathbf{E}_{\beta,2}(0)=1$ and
$\mathbf{E}_{\beta,\beta}(0)=\frac{1}{\Gamma(\beta)}$ and
\begin{align*}
&\big|\int_0^t(t-s)^{\beta-1}\mathbf{E}_{\beta,\beta}(\lambda(t-s)^\beta)
 \sigma(s)ds\big|\\
&\le \int_0^t(t-s)^{\beta-1}\mathbf{E}_{\beta,\beta}(\lambda(t-s)^\beta)
 s^k(1-s)^lds\\
&\le \sum_{\tau=0}^{+\infty}\frac{\lambda^\tau}{\Gamma((\tau+1)\beta)}
 \int_0^t(t-s)^{\tau\beta+\beta+l-1}s^kds\\
&=\sum_{\tau=0}^{+\infty}\frac{\lambda^\tau t^{\tau\beta+\beta+k+l}}
 {\Gamma((\tau+1)\beta)}\int_0^1(1-w)^{\tau\beta+\beta+l-1}w^kdw\\
&\le t^{\beta+k+l}\mathbf{E}_{\beta,\beta}(\lambda t^\beta)\mathbf{B}
 (\beta+l,k+1)\to 0\quad \text{as }t\to 0^+.
\end{align*}
It follows from \eqref{e3.4.3}, \eqref{e3.4.4}, the boundary conditions
and the impulse assumption in \eqref{e3.4.1} that
$c_{00}=a$, $c_{\sigma0}=I_\sigma$, $c_{\sigma1}=J_\sigma$,
$\sigma\in \mathbb{N}[1,m]$, and
\begin{align*}
&\sum_{\sigma=0}^m\big[\lambda c_{\sigma0}\mathbf{E}_{\beta,\beta}
(\lambda(1-t_\sigma)^\beta)(1-t_\sigma)^{\beta-1}
+c_{\sigma1}\mathbf{E}_{\beta,1}(\lambda(1-t_\sigma)^\beta)\big]\\
& +\int_0^1(1-s)^{\beta-2}\mathbf{E}_{\beta,\beta-1}(\lambda(1-s)^\beta)
\sigma(s)ds=b.
\end{align*}
Then
\begin{equation}
\begin{aligned}
c_{01}&=\frac{1}{\mathbf{E}_{\beta,1}(\lambda )}
\Big[b-\int_0^1(1-s)^{\beta-2}\mathbf{E}_{\beta,\beta-1}
(\lambda(1-s)^\beta)\sigma(s)ds \\
&\quad\times \sum_{\sigma=1}^m(\lambda \mathbf{E}_{\beta,\beta}
(\lambda(1-t_\sigma)^\beta)(1-t_\sigma)^{\beta-1}I_\sigma
+\mathbf{E}_{\beta,1}(\lambda(1-t_\sigma)^\beta)J_\sigma)
-\lambda \mathbf{E}_{\beta,\beta}(\lambda)a\Big].
\end{aligned}\label{e3.4.5}
\end{equation}
Substituting $c_{\sigma0},c_{\sigma1}(\sigma\in N[0,m])$ into \eqref{e3.4.3},
 we obtain \eqref{e3.3.2}.

On the other hand, if $x$ satisfies \eqref{e3.4.2}, then
$x|_{(t_i,t_{i+1}]}(i\in \mathbb{N}[0,m])$ are continuous and the limits
 $\lim_{t\to t_i^+}x(t)$ $(i\in\mathbb{N}[0,m])$ exist. So $x\in P_mC(0,1]$.
Using \eqref{e3.4.5} and $c_{00}=a$, $c_{\sigma0}=I_\sigma$, $c_{\sigma1}=J_\sigma$,
$\sigma\in \mathbb{N}[1,m]$, we rewrite $x$ by
\begin{align*}
x(t)&=\sum_{\sigma=0}^i\left[c_{\sigma0}\mathbf{E}_{\beta,1}
 (\lambda(t-t_\sigma)^\beta)+c_{\sigma1}(t-t_\sigma)
 \mathbf{E}_{\beta,2}(\lambda(t-t_\sigma)^\beta)\right]\\
&\quad +\int_0^t(t-s)^{\beta-1}\mathbf{E}_{\beta,\beta}
(\lambda(t-s)^\beta)\sigma(s)ds,\quad t\in (t_i,t_{i+1}],\; i\in \mathbb{N}[0,1].
\end{align*}
Since $\sigma$ is continuous on $(0,1)$ and $|\sigma(t)|\le t^k(1-t)^l$,
one can show easily that $x$ is continuous on $(t_i,t_{i+1}](i=0,1)$
 and using the method at the beginning of the proof of this lemma,
we know that the limits $\lim_{t\to t_i^+}x(t)$ $(i\in\mathbb{N}[0,m])$ exist.
So $x\in P_mC(0,1]$. Next, by direct computation, we have $x(0)=a,x'(1)=b$,
$\lim_{t\to t_i^+}x(t)-x(t_i)=I_i$ and $\lim_{t\to t_i^+}x(t)-x(t_i)=J_i$.
Furthermore, we have
\begin{align*}
&\Big|\sum_{\tau=0}^{+\infty}
\frac{\lambda^\tau}{\Gamma((\tau+1)\beta-1)}
\int_0^s(s-u)^{\tau\beta+\beta-2}\sigma(u)du\Big|\\
&\le \sum_{\tau=0}^{+\infty} \frac{\lambda^\tau}{\Gamma((\tau+1)\beta-1)}
\int_0^s(s-u)^{\tau\beta+\beta-2}u^k(1-u)^ldu\\
&\le \sum_{\tau=0}^{+\infty} \frac{\lambda^\tau}{\Gamma((\tau+1)\beta-1)}
\int_0^s(s-u)^{\tau\beta+\beta+l-2}u^kdu\\
&=\sum_{\tau=0}^{+\infty} \frac{\lambda^\tau}{\Gamma((\tau+1)\beta-1)}
s^{\tau\beta+\beta+k+l-1}\int_0^1(1-w)^{\tau\beta+\beta+l-2}w^kdw\\
&\le \sum_{\tau=0}^{+\infty} \frac{\lambda^\tau s^{\tau\beta}}
{\Gamma((\tau+1)\beta-1)}s^{\beta+k+l-1}\int_0^1(1-w)^{\beta+l-2}w^kdw\\
&=\mathbf{E}_{\alpha,\beta-1}(\lambda s^\beta)s^{\beta+k+l-1}\mathbf{B}
(\beta+l-1,k+1)\to 0\text{ as }s\to 0^+.
\end{align*}
From Theorem \ref{thm3.2.1} for $t\in (t_0,t_1]$ we have easily that
${}^CD_{0^+}^{\beta}x(t)=\lambda x(t)+\sigma(t)$, and for $t\in (t_i,t_{i+1}]$ that
\begin{align*}
&{}^CD_{0^+}^\beta x(t) \\
&=\frac{1}{\Gamma(2-\beta)}\int_0^t(t-s)^{1-\beta}x''(s)ds\\
&=\frac{1}{\Gamma(2-\beta)}\sum_{j=0}^{i-1}\int_{t_j}^{t_{j+1}}
(t-s)^{1-\beta}\Big(\sum_{\sigma=0}^j\sum_{v=0}^1c_{\sigma v}(s-t_\sigma)^v
\mathbf{E}_{\beta,v+1}(\lambda(s-t_\sigma)^\beta) \\
&\quad +\int_0^s(s-u)^{\beta-1}\mathbf{E}_{\beta,\beta}(\lambda(s-u)^\beta)
\sigma(u)du\Big)''ds\\
&\quad+\frac{1}{\Gamma(2-\beta)}\int_{t_i}^t(t-s)^{1-\beta}
\Big(\sum_{\sigma=0}^i\sum_{v=0}^1c_{\sigma v}(s-t_\sigma)^v
\mathbf{E}_{\beta,v+1}(\lambda(s-t_\sigma)^\beta) \\
&\quad +\int_0^s(s-u)^{\beta-1}\mathbf{E}_{\beta,\beta}
 (\lambda(s-u)^\beta)\sigma(u)du\Big)''ds\\
&=\frac{1}{\Gamma(2-\beta)}\sum_{j=0}^{i-1}\int_{t_j}^{t_{j+1}}
 (t-s)^{1-\beta}(\sum_{\sigma=0}^j\sum_{v=0}^1c_{\sigma v}\sum_{\tau=0}^{+\infty}
\frac{\lambda^\tau}{\Gamma(\tau\beta+v+1)}(s-t_\sigma)^{\beta\tau+v})''ds\\
&\quad +\frac{1}{\Gamma(2-\beta)}\int_{t_i}^t(t-s)^{1-\beta}
Big(\sum_{\sigma=0}^i\sum_{v=0}^1c_{\sigma v}\sum_{\tau=0}^{+\infty}
\frac{\lambda^\tau}{\Gamma(\tau\beta+v+1)}(s-t_\sigma)^{\beta\tau+v}\Big)''ds\\
&\quad +\frac{1}{\Gamma(2-\beta)}\int_0^t(t-s)^{1-\beta}
\Big(\sum_{\tau=0}^{+\infty}
\frac{\lambda^\tau}{\Gamma((\tau+1)\beta)}\int_0^s(s-u)^{\tau\beta+\beta-1}
\sigma(u)du\Big)''ds\\
&=\frac{1}{\Gamma(2-\beta)}\sum_{j=0}^{i-1}\int_{t_j}^{t_{j+1}}
(t-s)^{1-\beta}\Big(\sum_{\sigma=0}^j\sum_{v=0}^1c_{\sigma v}\sum_{\tau=1}^{+\infty}
\frac{\lambda^\tau}{\Gamma(\tau\beta+v-1)}(s-t_\sigma)^{\beta\tau+v-2}\Big)ds\\
&\quad +\frac{1}{\Gamma(2-\beta)}\int_{t_i}^t(t-s)^{1-\beta}\Big(\sum_{\sigma=0}^i\sum_{v=0}^1c_{\sigma v}\sum_{\tau=1}^{+\infty}
\frac{\lambda^\tau}{\Gamma(\tau\beta+v-1)}(s-t_\sigma)^{\beta\tau+v-2}\Big)ds\\
&\quad +\frac{1}{\Gamma(2-\beta)}\int_0^t(t-s)^{1-\beta}(\sum_{\tau=0}^{+\infty}
\frac{\lambda^\tau}{\Gamma((\tau+1)\beta-1)}\int_0^s(s-u)^{\tau\beta+\beta-2}
 \sigma(u)du)'ds \\
&=\frac{1}{\Gamma(2-\beta)}\sum_{j=0}^{i-1}\sum_{\sigma=0}^j
 \sum_{v=0}^1c_{\sigma v}\sum_{\tau=1}^{+\infty}
\frac{\lambda^\tau}{\Gamma(\tau\beta+v-1)}\int_{t_j}^{t_{j+1}}(t-s)^{1-\beta}
(s-t_\sigma)^{\beta\tau+v-2}ds\\
&\quad +\frac{1}{\Gamma(2-\beta)}\sum_{\sigma=0}^i\sum_{v=0}^1c_{\sigma v}
\sum_{\tau=1}^{+\infty} \frac{\lambda^\tau}{\Gamma(\tau\beta+v-1)}
\int_{t_i}^t(t-s)^{1-\beta}(s-t_\sigma)^{\beta\tau+v-2}ds\\
&\quad +\frac{1}{\Gamma(3-\beta)}\Big[ (t-s)^{2-\beta}(\sum_{\tau=0}^{+\infty}
\frac{\lambda^\tau}{\Gamma((\tau+1)\beta-1)}\int_0^s(s-u)^{\tau\beta+\beta-2}
\sigma(u)du)|_0^t \\
&\quad +((2-\beta)\int_0^t(t-s)^{1-\beta}(\sum_{\tau=0}^{+\infty}
\frac{\lambda^\tau}{\Gamma((\tau+1)\beta-1)}
\int_0^s(s-u)^{\tau\beta+\beta-2}\sigma(u)du)ds\Big]'\\
&=\frac{1}{\Gamma(2-\beta)}\sum_{j=0}^{i-1}\sum_{\sigma=0}^j\sum_{v=0}^1
c_{\sigma v}\sum_{\tau=1}^{+\infty}
\frac{\lambda^\tau}{\Gamma(\tau\beta+v-1)}(t-t_\sigma)^{\beta\tau-\beta+v} \\
&\quad\times \int_{\frac{t_j-t_\sigma}{t-t_\sigma}}^{\frac{t_{j+1}-t_\sigma}{t-t_\sigma}
}(1-w)^{1-\beta}w^{\beta\tau+v-2}dw\\
&\quad +\frac{1}{\Gamma(2-\beta)}\sum_{\sigma=0}^i\sum_{v=0}^1c_{\sigma v}
\sum_{\tau=1}^{+\infty} \frac{\lambda^\tau}{\Gamma(\tau\beta+v-1)}
(t-t_\sigma)^{\beta\tau-\beta+v} \\
&\quad\times \int_{\frac{t_i-t_\sigma}{t-t_\sigma}}^1(1-w)^{1-\beta}w^{\beta\tau+v-2}dw\\
&\quad +\frac{1}{\Gamma(2-\beta)}\Big[\sum_{\tau=0}^{+\infty}
\frac{\lambda^\tau}{\Gamma((\tau+1)\beta-1)} 
 \int_0^t\int_u^t(t-s)^{1-\beta}
(s-u)^{\tau\beta+\beta-2}ds\sigma(u)du\Big]'\\
&=\frac{1}{\Gamma(2-\beta)}\sum_{\sigma=0}^{i-1}
\sum_{j=\sigma}^{i-1}\sum_{v=0}^1c_{\sigma v}\sum_{\tau=1}^{+\infty}
\frac{\lambda^\tau}{\Gamma(\tau\beta+v-1)}(t-t_\sigma)^{\beta\tau-\beta+v}\\
&\quad\times \int_{\frac{t_j-t_\sigma}{t-t_\sigma}}^{\frac{t_{j+1}-t_\sigma}{t-t_\sigma}
}(1-w)^{1-\beta}w^{\beta\tau+v-2}dw \\
&\quad +\frac{1}{\Gamma(2-\beta)}\sum_{\sigma=0}^i\sum_{v=0}^1
 c_{\sigma v}\sum_{\tau=1}^{+\infty}
\frac{\lambda^\tau}{\Gamma(\tau\beta+v-1)}(t-t_\sigma)^{\beta\tau-\beta+v}\\
&\quad\times \int_{\frac{t_i-t_\sigma}{t-t_\sigma}}^1(1-w)^{1-\beta}
 w^{\beta\tau+v-2}dw
+\Big[\sum_{\tau=0}^{+\infty}
\frac{\lambda^\tau}{\Gamma(\tau\beta+1)}\int_0^t(t-u)^{\tau\beta}\sigma(u)du\Big]'\\
&=\frac{1}{\Gamma(2-\beta)}\sum_{\sigma=0}^{i-1}\sum_{v=0}^1
 c_{\sigma v}\sum_{\tau=1}^{+\infty}
\frac{\lambda^\tau}{\Gamma(\tau\beta+v-1)}(t-t_\sigma)^{\beta\tau-\beta+v} \\
&\quad\times \int_0^{\frac{t_{i}-t_\sigma}{t-t_\sigma}}(1-w)^{1-\beta}w^{\beta\tau+v-2}dw\\
&\quad +\frac{1}{\Gamma(2-\beta)}\sum_{\sigma=0}^i\sum_{v=0}^1
 c_{\sigma v}\sum_{\tau=1}^{+\infty}
\frac{\lambda^\tau}{\Gamma(\tau\beta+v-1)}(t-t_\sigma)^{\beta\tau-\beta+v} \\
&\quad\times \int_{\frac{t_i-t_\sigma}{t-t_\sigma}}^1
 (1-w)^{1-\beta}w^{\beta\tau+v-2}dw
 +\Big[\sum_{\tau=0}^{+\infty}\frac{\lambda^\tau}{\Gamma(\tau\beta+1)}
 \int_0^t(t-u)^{\tau\beta}\sigma(u)du\Big]'\\
&=\frac{1}{\Gamma(2-\beta)}\sum_{\sigma=0}^i\sum_{v=0}^1c_{\sigma v}
\sum_{\tau=1}^{+\infty}
\frac{\lambda^\tau}{\Gamma(\tau\beta+v-1)}(t-t_\sigma)^{\beta\tau-\beta+v} \\
&\quad\times \int_0^1 (1-w)^{1-\beta}w^{\beta\tau+v-2}dw
 +\sigma(t)+\sum_{\tau=1}^{+\infty}
\frac{\lambda^\tau}{\Gamma(\tau\beta)}\int_0^t(t-u)^{\tau\beta-1}\sigma(u)du\\
&=\sum_{\sigma=0}^i\sum_{v=0}^1c_{\sigma v}\sum_{\tau=1}^{+\infty}
\frac{\lambda^\tau}{\Gamma((\tau-1)\beta+v+1)}(t-t_\sigma)^{\beta\tau-\beta+v}
+\sigma(t) \\
&\quad +\sum_{\tau=1}^{+\infty} \frac{\lambda^\tau}{\Gamma(\tau\beta)}
\int_0^t(t-u)^{\tau\beta-1}\sigma(u)du \\
&=\lambda x(t)+\sigma(t).
\end{align*}
So $x$ is a solution of \eqref{e3.4.1}. The proof is complete.
\end{proof}


 Define the nonlinear operator $Q$ on $P_mC(0,1]$ by $Qx$ for $x\in P_mC(0,1]$ with
\begin{align*}
&(Qx)(t) \\
&= \int_0^1\phi(s)G(s,x(s))ds+\frac{t}{\mathbf{E}_{\beta,1}(\lambda)}
\Big[\int_0^1\psi(s)H(s,x(s))ds \\
&\quad -\lambda \mathbf{E}_{\beta,\beta}(\lambda)\int_0^1\phi(s)G(s,x(s))ds\\
&\quad -\sum_{\sigma=1}^m(\lambda \mathbf{E}_{\beta,\beta}
(\lambda (1-t_\sigma)^\beta)(1-t_\sigma)^{\beta-1}I(t_\sigma,x(t_\sigma))
 +\mathbf{E}_{\beta,1}(\lambda(1-t_\sigma)^\beta)J(t_\sigma,x(t_\sigma)))\\
&\quad -\int_0^1(1-s)^{\beta-1}\mathbf{E}_{\beta,\beta-1}(\lambda (1-s)^\beta)
p(s)f(s,x(s))ds\Big]\\
&\quad +\sum_{j=1}^i\big[\mathbf{E}_{\beta,1}(\lambda (t-t_\sigma)^\beta)
 I(t_\sigma,x(t_\sigma))+(t-t_\sigma)\mathbf{E}_{\beta,2}
 (\lambda (t-t_\sigma)^\beta)J(t_\sigma,x(t_\sigma))\big] \\
&\quad +\int_0^t(t-s)^{\beta-1}\mathbf{E}_{\beta,\beta}(\lambda (t-s)^\beta)p(s)
f(s,x(s))ds,\quad t\in (t_i,t_{i+1}],\; i\in \mathbb{N}[0,m]
\end{align*}

\begin{lemma} \label{lem3.4.2}
Suppose that {(1.A2)--(1.A3), (1.A6)--(1.A8)} in the introduction hold, 
and $f,G,H$ are impulsive I-Carath\'eodory functions, $I,J$ discrete I-Carath\'odory functions.
Then $Q:P_mC(0,1]\to P_mC(0,1]$ is well defined and is completely continuous,
$x\in P_mC(0,1]$ is a solution of BVP\eqref{e1.0.8} if and only if
$x\in P_mC(0,1]$ is a fixed point of $Q$.
\end{lemma}

 The proof of the above lemma is similar to that of Lemma \ref{lem3.3.2} and is omitted.


\subsection{Preliminaries for BVP \eqref{e1.0.9}}
In this section, we present some preliminary results that can be used in next
sections for get solutions of \eqref{e1.0.9}. Denote
$$
\Delta =\lambda \mathbf{E}_{\beta,\beta}(\lambda)^2
-\Big(\frac{1}{\Gamma(\beta-1)}
-\mathbf{E}_{\beta,\beta-1}(\lambda)\Big)\Big(\frac{1}{\Gamma(\beta)}
-\mathbf{E}_{\beta,\beta}(\lambda)\Big).
$$

\begin{lemma} \label{lem3.5.1}
 Suppose that $\Delta\neq 0$, $\sigma:(0,1)\to \mathbb{R}$ is continuous and
that there exist numbers $k>-1$ and $\max\{-\beta,-2-k\}<l\le 0$ such that
$|\sigma(t)|\le (\log t)^k(1-\log t)^l$ for all $t\in (1,e)$.
Then $x$ is a solution of
\begin{equation}
\begin{gathered}
{}^{RLH}D_{1^+}^{\beta}x(t)-\lambda x(t)
=\sigma(t), \quad t\in (t_i,t_{i+1}],\; i\in \mathbb{N}[0,m],\\
\lim_{t\to 1^+}(\log t)^{2-\beta }x(t)-x(e)=a,\\
\lim_{t\to 1^+}{}^{RLH}D_{1^+}^{\beta-1}x(t)-{}^{RLH}D_{1^+}^{\beta-1}x(e)=b,\\
\lim_{t\to t_i^+}(\log t-\log t_i)^{2-\beta} x(t)=I_i,i\in \mathbb{N}[1,m],\\
\Delta {}^{RL}D_{1^+}^{\beta-1}x(t_i)=J_i,i\in \mathbb{N}[1,m],
\end{gathered}\label{e3.5.1}
\end{equation}
if and only if $x\in LP_mC_{2-\alpha}(1,e]$ and
\begin{equation}
\begin{aligned}
x(t)&=M_1(\log t)^{\beta-1}\mathbf{E}_{\beta,\beta}\big(\lambda(\log t)^\beta\big)
-M_2(\log t)^{\beta-2}\mathbf{E}_{\beta,\beta-1}\big(\lambda(\log t)^\beta\big)\\
&\quad +\sum_{\sigma=1}^i(\log\frac{t}{t_\sigma})^{\beta-1}
\mathbf{E}_{\beta,\beta}\big(\lambda(\log \frac{t}{t_\sigma})^\beta\big)
\Gamma(\beta)J_\sigma\\
&\quad +\sum_{\sigma=1}^i(\log\frac{t}{t_\sigma})^{\beta-2}
\mathbf{E}_{\beta,\beta-1}\big(\lambda(\log \frac{t}{t_\sigma})^\beta\big)
\Gamma(\beta-1)I_\sigma\\
&\quad +\int_1^t(\log\frac{t}{s})^{\beta-1}\mathbf{E}_{\beta,\beta}
\big(\lambda(\log \frac{t}{s})^\beta\big)\sigma(s)\frac{ds}{s},\quad
t\in (t_i,t_{i+1}],\; i\in \mathbb{N}[0,m].
\end{aligned} \label{e3.5.2}
\end{equation}
where
\begin{align*}
M_1&=\frac{\lambda\mathbf{E}_{\beta,\beta}(\lambda)}{\Delta}a
+\frac{\frac{1}{\Gamma(\beta-1)}-\mathbf{E}_{\beta,\beta-1}(\lambda)}{\Delta}b\\
&\quad +\sum_{\sigma=1}^m\Big(\frac{\frac{1}{\Gamma(\beta-1)}
-\mathbf{E}_{\beta,\beta-1}(\lambda)}{\Delta}\mathbf{E}_{\beta,\beta}
\big(\lambda(\log\frac{e}{t_\sigma})^\beta\big) \\
&\quad +\frac{\lambda\mathbf{E}_{\beta,\beta}(\lambda)}{\Delta}
(\log\frac{e}{t_\sigma})^{\beta-1}\mathbf{E}_{\beta,\beta}
\big(\lambda(\log \frac{e}{t_\sigma})^\beta\big)\Big)\Gamma(\beta)J_\sigma\\
&\quad +\sum_{\sigma=1}^m\Big(\lambda\frac{\frac{1}{\Gamma(\beta-1)}
-\mathbf{E}_{\beta,\beta-1}(\lambda)}{\Delta}(\log\frac{e}{t_\sigma})^{\beta-1}
\mathbf{E}_{\beta,\beta}\big(\lambda(\log\frac{e}{t_\sigma})^\beta\big) \\
&\quad +\frac{\lambda\mathbf{E}_{\beta,\beta}(\lambda)}{\Delta}
 (\log\frac{e}{t_\sigma})^{\beta-2}\mathbf{E}_{\beta,\beta-1}
\big(\lambda(\log \frac{e}{t_\sigma})^\beta\big)\Big)\Gamma(\beta-1)I_\sigma\\
&\quad +\int_1^e \Big(\lambda\frac{\frac{1}{\Gamma(\beta-1)}
 -\mathbf{E}_{\beta,\beta-1}(\lambda)}{\Delta}\mathbf{E}_{\beta,1}
\big(\lambda(\log \frac{e}{s})^{\beta}\big) \\
&\quad +\frac{\lambda\mathbf{E}_{\beta,\beta}(\lambda)}{\Delta}
(\log \frac{e}{s})^{\beta-1}\mathbf{E}_{\beta,\beta}
\big(\lambda(\log \frac{e}{s})^\beta\big)\Big)\sigma(s)\frac{ds}{s},
\end{align*}
and
\begin{align*}
M_2
&=\frac{\frac{1}{\Gamma(\beta)}-\mathbf{E}_{\beta,\beta}
(\lambda)}{\Delta}a+\frac{\mathbf{E}_{\beta,\beta}(\lambda)}{\Delta}b
+\sum_{\sigma=1}^m\Big(\frac{\frac{1}{\Gamma(\beta)}-\mathbf{E}_{\beta,\beta}
(\lambda)}{\Delta}(\log\frac{e}{t_\sigma})^{\beta-1} \\
&\quad\times \mathbf{E}_{\beta,\beta}\big(\lambda(\log \frac{e}{t_\sigma})^\beta\big)
 +\frac{\mathbf{E}_{\beta,\beta}(\lambda)}{\Delta}\mathbf{E}_{\beta,\beta}
\big(\lambda(\log\frac{e}{t_\sigma})^\beta\big)\Big)
\Gamma(\beta)J_\sigma \\
&\quad +\sum_{\sigma=1}^m\Big(\frac{\frac{1}{\Gamma(\beta)}
-\mathbf{E}_{\beta,\beta}(\lambda)}{\Delta}(\log\frac{e}{t_\sigma})^{\beta-2}
\mathbf{E}_{\beta,\beta-1}\big(\lambda(\log \frac{e}{t_\sigma})^\beta\big) \\
&\quad +\frac{\lambda\mathbf{E}_{\beta,\beta}(\lambda)}{\Delta}
(\log\frac{e}{t_\sigma})^{\beta-1}\mathbf{E}_{\beta,\beta}
\big(\lambda(\log\frac{e}{t_\sigma})^\beta\big)\Big)\Gamma(\beta-1)I_\sigma\\
&\quad +\int_1^e\Big(\frac{\frac{1}{\Gamma(\beta)}
 -\mathbf{E}_{\beta,\beta}(\lambda)}{\Delta}(\log \frac{e}{s})^{\beta-1}
\mathbf{E}_{\beta,\beta}\big(\lambda(\log \frac{e}{s})^\beta\big) \\
&\quad +\lambda\frac{\mathbf{E}_{\beta,\beta}(9(\lambda)}{\Delta}\mathbf{E}_{\beta,1}
\big(\lambda(\log \frac{e}{s})^{\beta}\big)\Big)\sigma(s)\frac{ds}{s}.
\end{align*}
\end{lemma}

\begin{proof}
Let $x$ be a solution of \eqref{e3.5.1}.
We know from Theorem \ref{thm3.1.6} that there exist numbers
$c_{\sigma 1},c_{\sigma 2}\in \mathbb{R}$ such that
\begin{equation}
\begin{aligned}
x(t)&=\sum_{\sigma=0}^i(c_{\sigma 1}(\log\frac{t}{t_\sigma})^{\beta-1}
\mathbf{E}_{\beta,\beta}\big(\lambda(\log \frac{t}{t_\sigma})^\beta\big) \\
&\quad +c_{\sigma 2}(\log\frac{t}{t_\sigma})^{\beta-2}\mathbf{E}_{\beta,\beta-1}
\big(\lambda(\log \frac{t}{t_\sigma})^\beta\big))\\
&\quad +\int_1^t(\log \frac{t}{s})^{\beta-1}
\mathbf{E}_{\beta,\beta}\big(\lambda(\log \frac{t}{s})^\beta\big)\sigma(s)
 \frac{ds}{s},
\end{aligned} \label{e3.5.3}
\end{equation}
for $t\in (t_i,t_{i+1}]$, $i\in \mathbb{N}[0,m]$.
From
\begin{align*}
&{}^{RLH}D_{1^+}^{\beta-1}(\log\frac{t}{t_\sigma})^{\beta-v}
 \mathbf{E}_{\beta,\beta-v+1}\big(\lambda(\log \frac{t}{t_\sigma})^\beta\big)\\
&=\frac{1}{\Gamma(2-\beta)}(t\frac{d}{dt})\int_1^t(\log \frac{t}{s})^{1-\beta}
(\log\frac{s}{t_\sigma})^{\beta-v}
\mathbf{E}_{\beta,\beta-v+1}\big(\lambda(\log \frac{s}{t_\sigma})^\beta\big)\frac{ds}{s}\\
&=\frac{1}{\Gamma(2-\beta)}(t\frac{d}{dt})\sum_{\tau=0}^{+\infty}
\frac{\lambda^\tau}{\Gamma(\tau\beta+\beta-v+1)}\int_1^t
(\log \frac{t}{s})^{1-\beta}(\log\frac{s}{t_\sigma})^{\tau\beta+\beta-v}\frac{ds}{s}\\
&=\frac{1}{\Gamma(2-\beta)}(t\frac{d}{dt})\sum_{\tau=0}^{+\infty}
\frac{\lambda^\tau}{\Gamma(\tau\beta+\beta-v+1)}
 (\log \frac{t}{t_\sigma})^{\tau\beta-v+2} \\
&\quad\times \int_0^1(1-w)^{1-\beta}w^{\tau\beta+\beta-v}dw\\
&=(t\frac{d}{dt})\sum_{\tau=0}^{+\infty}
\frac{\lambda^\tau}{\Gamma(\tau\beta-v+3)}
(\log \frac{t}{t_\sigma})^{\tau\beta-v+2}\\
&=\begin{cases}
1+\sum_{\tau=1}^{+\infty}
\frac{\lambda^\tau}{\Gamma(\tau\beta)}(\log \frac{t}{t_\sigma})^{\tau\beta}
=\mathbf{E}_{\beta,\beta}(\lambda(\log\frac{t}{t_\sigma})^\beta), &v=1,\\
\sum_{\tau=1}^{+\infty}
\frac{\lambda^\tau}{\Gamma(\tau\beta)}(\log \frac{t}{t_\sigma})^{\tau\beta-1}
=\lambda\mathbf{E}_{\beta,\beta}(\lambda(\log\frac{t}{t_\sigma})^\beta)
(\log\frac{t}{t_\sigma})^{\beta-1}, &v=2,
\end{cases}
\end{align*}
and
\begin{align*}
&{}^{RLH}D_{1^+}^{\beta-1}(\log \frac{t}{s})^{\beta-1}
\mathbf{E}_{\beta,\beta}\big(\lambda(\log \frac{t}{s})^\beta\big)\\
&=\frac{1}{\Gamma(2-\beta)}(t\frac{d}{dt})\int_1^t
 (\log \frac{t}{u})^{1-\beta}(\log\frac{u}{s})^{\beta-1}
 \mathbf{E}_{\beta,\beta}(\lambda(\log \frac{u}{s})^\beta)\frac{du}{u}\\
&=\frac{1}{\Gamma(2-\beta)}(t\frac{d}{dt})
 \sum_{\tau=0}^{+\infty}\frac{\lambda^\tau}{\Gamma((\tau+1)\beta)}
 \int_1^t(\log \frac{t}{u})^{1-\beta}(\log\frac{u}{s})^{\tau\beta+\beta-1}
 \frac{du}{u}\\
&=\frac{1}{\Gamma(2-\beta)}(t\frac{d}{dt})\sum_{\tau=0}^{+\infty}
 \frac{\lambda^\tau}{\Gamma((\tau+1)\beta)}
 (\log \frac{t}{s})^{\tau\beta+1}\int_0^1(1-w)^{1-\beta}w^{\tau\beta+\beta-1}dw\\
&=(t\frac{d}{dt})\sum_{\tau=0}^{+\infty}\frac{\lambda^\tau}{\Gamma(\tau\beta+2)}
(\log \frac{t}{s})^{\tau\beta+1} \\
&=\sum_{\tau=0}^{+\infty}\frac{\lambda^\tau}{\Gamma(\tau\beta+1)}
(\log \frac{t}{s})^{\tau\beta}
=\mathbf{E}_{\beta,1}(\lambda(\log \frac{t}{s})^{\beta}),
\end{align*}
we have
\begin{equation}
\begin{aligned}
&{}^{RLH}D_{1^+}^{\beta-1}x(t) \\
&=\sum_{\sigma=0}^i\Big[c_{\sigma 1}\mathbf{E}_{\beta,\beta}
(\lambda(\log\frac{t}{t_\sigma})^\beta)+\lambda c_{\sigma 2}
\mathbf{E}_{\beta,\beta}(\lambda(\log\frac{t}{t_\sigma})^\beta)
(\log\frac{t}{t_\sigma})^{\beta-1}\Big]\\
&\quad +\lambda\int_1^t \mathbf{E}_{\beta,1}
\big(\lambda(\log \frac{t}{s})^{\beta}\big)\sigma(s)\frac{ds}{s}, \quad
t\in (t_i,t_{i+1}],\; i\in \mathbb{N}[0,m].
\end{aligned} \label{e3.5.4}
\end{equation}
One sees that
\begin{align*}
&(\log t)^{2-\beta}|\int_1^t(\log \frac{t}{s})^{\beta-1}
\mathbf{E}_{\beta,\beta}\big(\lambda(\log \frac{t}{s})^\beta\big)\sigma(s)\frac{ds}{s}|\\
& \le (\log t)^{2-\beta}\int_1^t(\log \frac{t}{s})^{\beta-1}
 \mathbf{E}_{\beta,\beta}\big(\lambda(\log \frac{t}{s})^\beta\big)
 (\log s)^k(\log \frac{e}{s})^l\frac{ds}{s} \\
&\le (\log t)^{2-\beta}\sum_{\tau=0}^{+\infty}
 \frac{\lambda^\tau}{\Gamma((\tau+1)\beta)}\int_1^t
 (\log \frac{t}{s})^{\tau\beta+\beta+l-1}(\log s)^k\frac{ds}{s}\\
&=(\log t)^{2-\beta}\sum_{\tau=0}^{+\infty}
 \frac{\lambda^\tau}{\Gamma((\tau+1)\beta)}(\log t)^{\tau\beta+\beta+k+l}
 \int_0^1(1-w)^{\tau\beta+\beta+l-1}w^kdw\\
&\le (\log t)^{2-\beta}\sum_{\tau=0}^{+\infty}
\frac{\lambda^\tau}{\Gamma((\tau+1)\beta)}(\log t)^{\tau\beta+\beta+k+l}
 \int_0^1(1-w)^{\beta+l-1}w^kdw\\
&= (\log t)^{2+k+l}\mathbf{E}_{\beta,\beta}(\lambda (\log t)^\beta)
 \mathbf{B}(\beta+l,k+1)\to 0\text{ as }t\to 1^+.
\end{align*}
Similarly we have
\begin{align*}
&\big|\int_1^t \mathbf{E}_{\beta,1}\big(\lambda(\log \frac{t}{s})^{\beta}\big)
\sigma(s)\frac{ds}{s}\big| \\
&\le \int_1^t \mathbf{E}_{\beta,1}\big(\lambda(\log \frac{t}{s})^{\beta}\big)
 (\log s)^k(\log \frac{e}{s})^l\frac{ds}{s}\\
&\le \sum_{\tau=0}^{+\infty}\frac{\lambda^\tau}{\Gamma(\tau\beta+1)}
 \int_1^t (\log \frac{t}{s})^{\tau\beta+l}(\log s)^k\frac{ds}{s} \\
&=\sum_{\tau=0}^{+\infty}\frac{\lambda^\tau}{\Gamma(\tau\beta+1)}
(\log t)^{\tau\beta+k+l+1}\int_0^1 (1-w)^{\tau\beta+l}w^kdw\\
&\le \sum_{\tau=0}^{+\infty}\frac{\lambda^\tau}{\Gamma(\tau\beta+1)}
 (\log t)^{\tau\beta+k+l+1}\int_0^1 (1-w)^{l}w^kdw\\
&=(\log t)^{k+l+1}\mathbf{E}_{\beta,1}\big(\lambda(\log t)^\beta\big)
\mathbf{B}(l+1,k+1)
\to 0 \quad \text{as }t\to 1^+.
\end{align*}
 It follows from \eqref{e3.5.3}, \eqref{e3.5.4}, the boundary conditions
and the impulse assumption in \eqref{e3.5.1} that
\begin{align*}
&\frac{1}{\Gamma(\beta-1)}c_{02}-\Big[\sum_{\sigma=0}^m
\Big(c_{\sigma 1}(\log\frac{e}{t_\sigma})^{\beta-1}
\mathbf{E}_{\beta,\beta}\big(\lambda(\log \frac{e}{t_\sigma})^\beta\big) \\
&+c_{\sigma 2}(\log\frac{e}{t_\sigma})^{\beta-2}\mathbf{E}_{\beta,\beta-1}
 \big(\lambda(\log \frac{e}{t_\sigma})^\beta\big)\Big)
 +\int_1^e(\log \frac{e}{s})^{\beta-1}
 \mathbf{E}_{\beta,\beta}\big(\lambda(\log \frac{e}{s})^\beta\big)\sigma(s)
 \frac{ds}{s}\Big]=a,
\end{align*}
\begin{align*}
&\frac{1}{\Gamma(\beta)}c_{01}-\Big[\sum_{\sigma=0}^m\Big(c_{\sigma 1}
\mathbf{E}_{\beta,\beta}\big(\lambda(\log\frac{e}{t_\sigma})^\beta\big) \\
&+\lambda c_{\sigma 2}\mathbf{E}_{\beta,\beta}
\big(\lambda(\log\frac{e}{t_\sigma})^\beta\big)
 (\log\frac{e}{t_\sigma})^{\beta-1}\Big)
 +\lambda\int_1^e \mathbf{E}_{\beta,1}
\big(\lambda(\log \frac{e}{s})^{\beta}\big)\sigma(s)\frac{ds}{s}\Big]=b
\end{align*}
and $c_{i2}=\Gamma(\beta-1)I_i(i\in \mathbb{N}[1,m])$,
$c_{i1}=\Gamma(\beta)J_i(i\in \mathbb{N}[1,m])$.
Then
\begin{align*}
c_{01}&=\frac{\lambda\mathbf{E}_{\beta,\beta}(\lambda)}{\Delta}a
 +\frac{\frac{1}{\Gamma(\beta-1)}-\mathbf{E}_{\beta,\beta-1}(\lambda)}{\Delta}b\\
&\quad +\sum_{\sigma=1}^m\Big[\frac{\frac{1}{\Gamma(\beta-1)}-\mathbf{E}_{\beta,\beta-1}
(\lambda)}{\Delta}\mathbf{E}_{\beta,\beta}
\big(\lambda(\log\frac{e}{t_\sigma})^\beta\big) \\
&\quad +\frac{\lambda\mathbf{E}_{\beta,\beta}(\lambda)}{\Delta}
(\log\frac{e}{t_\sigma})^{\beta-1}\mathbf{E}_{\beta,\beta}
\big(\lambda(\log \frac{e}{t_\sigma})^\beta\big)\Big]\Gamma(\beta)J_\sigma\\
&\quad +\sum_{\sigma=1}^m\Big[\lambda\frac{\frac{1}{\Gamma(\beta-1)}
 -\mathbf{E}_{\beta,\beta-1}(\lambda)}{\Delta}
 (\log\frac{e}{t_\sigma})^{\beta-1}\mathbf{E}_{\beta,\beta}
 \big(\lambda(\log\frac{e}{t_\sigma})^\beta\big) \\
&\quad +\frac{\lambda\mathbf{E}_{\beta,\beta}(\lambda)}{\Delta}
 (\log\frac{e}{t_\sigma})^{\beta-2}\mathbf{E}_{\beta,\beta-1}
 \big(\lambda(\log \frac{e}{t_\sigma})^\beta\big)\Big]\Gamma(\beta-1)I_\sigma \\
&\quad +\int_1^e \Big[\lambda\frac{\frac{1}{\Gamma(\beta-1)}
 -\mathbf{E}_{\beta,\beta-1}(\lambda)}{\Delta}\mathbf{E}_{\beta,1}
 \big(\lambda(\log \frac{e}{s})^{\beta}\big) \\
&\quad +\frac{\lambda\mathbf{E}_{\beta,\beta}(\lambda)}{\Delta}
 (\log \frac{e}{s})^{\beta-1}\mathbf{E}_{\beta,\beta}
 \big(\lambda(\log \frac{e}{s})^\beta\big)\Big]\sigma(s)\frac{ds}{s}=:M_1,
\end{align*}
\begin{equation}
\begin{aligned}
c_{02}
&=-\frac{\frac{1}{\Gamma(\beta)}-\mathbf{E}_{\beta,\beta}(\lambda)}{\Delta}a
 -\frac{\mathbf{E}_{\beta,\beta}(\lambda)}{\Delta}b\\
&\quad -\sum_{\sigma=1}^m\Big[\frac{\frac{1}{\Gamma(\beta)}
 -\mathbf{E}_{\beta,\beta}(\lambda)}{\Delta}(\log\frac{e}{t_\sigma})^{\beta-1}
 \mathbf{E}_{\beta,\beta}\big(\lambda(\log \frac{e}{t_\sigma})^\beta\big) \\
&\quad +\frac{\mathbf{E}_{\beta,\beta}(\lambda)}{\Delta}
 \mathbf{E}_{\beta,\beta}\big(\lambda(\log\frac{e}{t_\sigma})^\beta\big)\Big]
 \Gamma(\beta)J_\sigma\\
&\quad -\sum_{\sigma=1}^m\Big[\frac{\frac{1}{\Gamma(\beta)}
 -\mathbf{E}_{\beta,\beta}(\lambda)}{\Delta}(\log\frac{e}{t_\sigma})^{\beta-2}
 \mathbf{E}_{\beta,\beta-1}\big(\lambda(\log \frac{e}{t_\sigma})^\beta\big) \\
&\quad +\frac{\lambda\mathbf{E}_{\beta,\beta}(\lambda)}{\Delta}
 (\log\frac{e}{t_\sigma})^{\beta-1}\mathbf{E}_{\beta,\beta}
 \big(\lambda(\log\frac{e}{t_\sigma})^\beta\big)\Big]\Gamma(\beta-1)I_\sigma\\
&\quad -\int_1^e\Big[\frac{\frac{1}{\Gamma(\beta)}-\mathbf{E}_{\beta,\beta}
 (\lambda)}{\Delta}(\log \frac{e}{s})^{\beta-1}
 \mathbf{E}_{\beta,\beta}\big(\lambda(\log \frac{e}{s})^\beta\big) \\
&\quad +\lambda\frac{\mathbf{E}_{\beta,\beta}(\lambda)}{\Delta}
\mathbf{E}_{\beta,1}\big(\lambda(\log \frac{e}{s})^{\beta}\big)\Big]
 \sigma(s)\frac{ds}{s}=:M_2.
\end{aligned} \label{e3.5.5}
\end{equation}
Substituting $c_{\sigma v}(\sigma\in N[0,m],v\in N[1,2]$ into
 \eqref{e3.5.3} and \eqref{e3.5.4}, we obtain \eqref{e3.5.2}.

On the other hand, if $x$ satisfies \eqref{e3.5.2}, then
$x|_{(t_i,t_{i+1}]}$ $(i\in \mathbb{N}[0,m])$ are continuous and the limits
$\lim_{t\to t_i^+}(\log t-\log t_i)^{2-\beta}x(t)$ $(i\in\mathbb{N}[0,m]$ exist.
So $x\in LP_mC_{2-\beta}(1,e]$.
Furthermore, by direct computation, we have
$\lim_{t\to 1^+}(\log t)^{2-\beta }x(t)-x(e)=a$,
$\lim_{t\to 1^+}{}^{RLH}D_{1^+}^{\beta-1}x(t)-{}^{RLH}D_{1^+}^{\beta-1}x(e)=b$,
$\lim_{t\to t_i^+}(\log t-\log t_i)^{2-\beta} x(t)=I_i$, $i\in \mathbb{N}[1,m]$ and
 $\Delta {}^{RL}D_{1^+}^{\beta-1}x(t_i)=J_i$, $i\in \mathbb{N}[1,m]$.


Using \eqref{e3.5.5} and $c_{i2}=\Gamma(\beta-1)I_i(i\in \mathbb{N}[1,m])$,
$c_{i1}=\Gamma(\beta)J_i(i\in \mathbb{N}[1,m])$, we rewrite $x$ by \eqref{e3.5.3}.
From Theorem \ref{thm3.2.3}, for $t\in (t_0,t_1]$ easily one has
${}^{RLH}D_{0^+}^{\beta}x(t)=\lambda x(t)+\sigma(t)$ for $t\in (t_0,t_1]$.
 For $t\in (t_j,t_{j+1}]$, by Definition \ref{def2.5} we have
\begin{align*}
&{}^{RLH}D_{1^+}^\beta x(t) \\
&=\frac{1}{\Gamma(2-\beta)}(t\frac{d}{dt})^2\int_1^t(\log \frac{t}{s})^{1-\beta}x(s)
 \frac{ds}{s}\\
&=\frac{1}{\Gamma(2-\beta)}(t\frac{d}{dt})^2
 \Big[\sum_{i=0}^{j-1}\int_{t_i}^{t_{i+1}}(\log \frac{t}{s})^{1-\beta}
 \Big(\sum_{\sigma=0}^i\sum_{v=1}^2c_{\sigma v}
 (\log\frac{s}{t_\sigma})^{\beta-v} \\
&\quad\times \mathbf{E}_{\beta,\beta-v+1}
\big(\lambda(\log \frac{s}{t_\sigma})^\beta\big)
 +\int_1^s(\log \frac{s}{u})^{\beta-1}
\mathbf{E}_{\beta,\beta}\big(\lambda(\log \frac{s}{u})^\beta\big)
\sigma(u)\frac{du}{u}\Big)\frac{ds}{s}\\
&\quad +\int_{t_j}^t(\log \frac{t}{s})^{1-\beta}
\Big(\sum_{\sigma=0}^j\sum_{v=1}^2c_{\sigma v}(\log\frac{s}{t_\sigma})^{\beta-v}
\mathbf{E}_{\beta,\beta-v+1}\big(\lambda(\log \frac{s}{t_\sigma})^\beta\big) \\
&\quad +\int_1^s(\log \frac{s}{u})^{\beta-1}\mathbf{E}_{\beta,\beta}
\big(\lambda(\log \frac{s}{u})^\beta\big)\sigma(u)\frac{du}{u}\Big)\frac{ds}{s}\Big]\\
&=\frac{1}{\Gamma(2-\beta)}(t\frac{d}{dt})^2
 \Big[\sum_{i=0}^{j-1}\int_{t_i}^{t_{i+1}}(\log \frac{t}{s})^{1-\beta}
 \sum_{\sigma=0}^i\sum_{v=1}^2c_{\sigma v}(\log\frac{s}{t_\sigma})^{\beta-v} \\
&\quad\times \mathbf{E}_{\beta,\beta-v+1}
 \big(\lambda(\log \frac{s}{t_\sigma})^\beta\big) \frac{ds}{s}\Big]
 +\frac{1}{\Gamma(2-\beta)}(t\frac{d}{dt})^2 \\
&\quad\times \Big[\int_{t_j}^t(\log \frac{t}{s})^{1-\beta}
\sum_{\sigma=0}^j\sum_{v=1}^2c_{\sigma v}(\log\frac{s}{t_\sigma})^{\beta-v}
\mathbf{E}_{\beta,\beta-v+1}\big(\lambda(\log \frac{s}{t_\sigma})^\beta\big)
\frac{ds}{s}\Big] \\
&\quad +\frac{1}{\Gamma(2-\beta)}(t\frac{d}{dt})^2
\Big[\int_1^t(\log \frac{t}{s})^{1-\beta}\int_1^s(\log \frac{s}{u})^{\beta-1}
\mathbf{E}_{\beta,\beta}\big(\lambda(\log \frac{s}{u})^\beta\big)
\sigma(u)\frac{du}{u}\frac{ds}{s}\Big]
\end{align*}
By using the method in the proof of Theorem \ref{thm3.2.3}, we have
$$
{}^{RLH}D_{1^+}^\beta x(t)=\frac{1}{\Gamma(2-\beta)}
(t\frac{d}{dt})^2\int_1^t(\log \frac{t}{s})^{1-\beta}x(s)\frac{ds}{s}
=\lambda x(t)+\sigma(t).
$$
So $x$ is a solution of \eqref{e3.5.1}. The proof is complete.
\end{proof}

 Define the nonlinear operator $R$ on $LP_mC_{2-\beta}(1,e]$
for $x\in LP_mC_{2-\alpha}(1,e]$ by
\begin{align*}
(Rx)(t)&=
M_{1x}(\log t)^{\beta-1}\mathbf{E}_{\beta,\beta}\big(\lambda(\log t)^\beta\big)
-M_{2x}(\log t)^{\beta-2}\mathbf{E}_{\beta,\beta-1}\big(\lambda(\log t)^\beta\big)\\
&\quad +\sum_{\sigma=1}^i\Big[(\log\frac{t}{t_\sigma})^{\beta-1}
\mathbf{E}_{\beta,\beta}\big(\lambda(\log \frac{t}{t_\sigma})^\beta\big)
\Gamma(\beta)J(t_\sigma,x(t_\sigma)) \\
&\quad +(\log\frac{t}{t_\sigma})^{\beta-2}
\mathbf{E}_{\beta,\beta-1}\big(\lambda(\log \frac{t}{t_\sigma})^\beta\big)
\Gamma(\beta-1)I(t_\sigma,x(t_\sigma))\Big]\\
&\quad +\int_1^t(\log\frac{t}{s})^{\beta-1}\mathbf{E}_{\beta,\beta}
\big(\lambda(\log \frac{t}{s})^\beta\big)p(s)f(s,x(s))
\frac{ds}{s},
\end{align*}
for  $t\in (t_i,t_{i+1}]$ and $i\in \mathbb{N}[0,m]$,
where
\begin{align*}
M_{1x}
&=\frac{\lambda\mathbf{E}_{\beta,\beta}(\lambda)}{\Delta}
 \int_0^1\phi(s)G(s,x(s))ds
 +\frac{\frac{1}{\Gamma(\beta-1)}-\mathbf{E}_{\beta,\beta-1}(\lambda)}{\Delta}
 \int_0^1\psi(s)H(s,x(s))ds\\
&\quad +\sum_{\sigma=1}^m(\frac{\frac{1}{\Gamma(\beta-1)}-\mathbf{E}_{\beta,\beta-1}
 (\lambda)}{\Delta}\mathbf{E}_{\beta,\beta}
 \big(\lambda(\log\frac{e}{t_\sigma})^\beta\big) \\
&\quad +\frac{\lambda\mathbf{E}_{\beta,\beta}(\lambda)}{\Delta}
 (\log\frac{e}{t_\sigma})^{\beta-1}\mathbf{E}_{\beta,\beta}
 \big(\lambda(\log \frac{e}{t_\sigma})^\beta\big))
 \Gamma(\beta)J(t_\sigma,x(t_\sigma))\\
&\quad +\sum_{\sigma=1}^m(\lambda\frac{\frac{1}{\Gamma(\beta-1)}
 -\mathbf{E}_{\beta,\beta-1}(\lambda)}{\Delta}
 (\log\frac{e}{t_\sigma})^{\beta-1}\mathbf{E}_{\beta,\beta}
 \big(\lambda(\log\frac{e}{t_\sigma})^\beta\big) \\
&\quad +\frac{\lambda\mathbf{E}_{\beta,\beta}(\lambda)}{\Delta}
(\log\frac{e}{t_\sigma})^{\beta-2}\mathbf{E}_{\beta,\beta-1}
\big(\lambda(\log \frac{e}{t_\sigma})^\beta\big))
 \Gamma(\beta-1)I(t_\sigma,x(t_\sigma))\\
&\quad +\int_1^e (\lambda\frac{\frac{1}{\Gamma(\beta-1)}
 -\mathbf{E}_{\beta,\beta-1}(\lambda)}{\Delta}\mathbf{E}_{\beta,1}
 \big(\lambda(\log \frac{e}{s})^{\beta}\big) \\
&\quad +\frac{\lambda\mathbf{E}_{\beta,\beta}(\lambda)}{\Delta}
 (\log \frac{e}{s})^{\beta-1}\mathbf{E}_{\beta,\beta}
\big(\lambda(\log \frac{e}{s})^\beta\big))p(s)f(s,x(s))\frac{ds}{s}
\end{align*}
and
\begin{align*}
M_{2x}&=\frac{\frac{1}{\Gamma(\beta)}-\mathbf{E}_{\beta,\beta}
(\lambda)}{\Delta}\int_0^1\phi(s)G(s,x(s))ds
 +\frac{\mathbf{E}_{\beta,\beta}(\lambda)}{\Delta}\int_0^1\psi(s)H(s,x(s))ds\\
&\quad +\sum_{\sigma=1}^m(\frac{\frac{1}{\Gamma(\beta)}
-\mathbf{E}_{\beta,\beta}(\lambda)}{\Delta}(\log\frac{e}{t_\sigma})^{\beta-1}
 \mathbf{E}_{\beta,\beta}\big(\lambda(\log \frac{e}{t_\sigma})^\beta\big) \\
&\quad +\frac{\mathbf{E}_{\beta,\beta}(\lambda)}{\Delta}
 \mathbf{E}_{\beta,\beta}\big(\lambda(\log\frac{e}{t_\sigma})^\beta\big))
 \Gamma(\beta)J(t_\sigma,x(t_\sigma))\\
&\quad +\sum_{\sigma=1}^m(\frac{\frac{1}{\Gamma(\beta)}
 -\mathbf{E}_{\beta,\beta}(\lambda)}{\Delta}(\log\frac{e}{t_\sigma})^{\beta-2}
 \mathbf{E}_{\beta,\beta-1}\big(\lambda(\log \frac{e}{t_\sigma})^\beta\big) \\
&\quad +\frac{\lambda\mathbf{E}_{\beta,\beta}(\lambda)}{\Delta}
 (\log\frac{e}{t_\sigma})^{\beta-1}\mathbf{E}_{\beta,\beta}
 \big(\lambda(\log\frac{e}{t_\sigma})^\beta\big))\Gamma(\beta-1)
 I(t_\sigma,x(t_\sigma))\\
&\quad +\int_1^e(\frac{\frac{1}{\Gamma(\beta)}-\mathbf{E}_{\beta,\beta}(\lambda)}
 {\Delta}(\log \frac{e}{s})^{\beta-1}\mathbf{E}_{\beta,\beta}
 \big(\lambda(\log \frac{e}{s})^\beta\big) \\
&\quad +\lambda\frac{\mathbf{E}_{\beta,\beta}(\lambda)}{\Delta}
 \mathbf{E}_{\beta,1}\big(\lambda(\log \frac{e}{s})^{\beta}\big))p(s)f(s,x(s))
 \frac{ds}{s}.
\end{align*}

\begin{lemma} \label{lem3.5.2}
 Suppose that {\rm (d), (i), (j), (k)} in the intoruction hold,
$\Delta\neq 0$, and $f,G,H$ are impulsive III-Carath\'eodory functions,
$I,J$ are discrete III-Carath\'odory functions.
Then $R:LP_mC_{2-\beta}(1,e]\to LP_mC_{2-\alpha}(1,e]$ is well defined and
is completely continuous, $x$ is a solution of BVP\eqref{e1.0.9} if and only
if $x$ is a fixed point of $R$ in $LP_mC_{2-\alpha}(1,e]$.
\end{lemma}

The proof of the above lemma is similar to that of Lemma \ref{lem3.3.2} and is omitted.

\subsection{Preliminaries for BVP \eqref{e1.0.10}}

In this section, we present some preliminary results that can be used
in next sections for obtain solutions of \eqref{e1.0.10}.

\begin{lemma} \label{lem3.6.1}
 Suppose that $\lambda\neq 0$, $\sigma:(0,1)\to \mathbb{R}$ is continuous and satisfies
that there exist numbers $k>1-\beta$ and $l\le 0$ with $l>\max\{-\beta,-\beta-k\}$
such that $|\sigma(t)|\le (\log t)^k(1-\log t)^l$ for all $t\in (0,1)$.
The $x$ is a solutions of
\begin{equation}
\begin{gathered}
{}^{CH}D_{1^+}^{\beta}x(t)-\lambda x(t)=\sigma(t), \quad t\in (t_i,t_{i+1}],\;
i\in \mathbb{N}[0,m],\\
(t\frac{d}{dt})x(t)\big|_{t=1}=a,\quad
 (t\frac{d}{dt})x(t)\big|_{t=e}=b,\quad
\lim_{t\to t_i^+}x(t)-x(t_i)=I_i,\\
\lim_{t\to t_i^+}(t\frac{d}{dt})x(t)- (t\frac{d}{dt})x(t)\big|_{t=t_i}=J_i,
\quad i\in \mathbb{N}[1,m],
\end{gathered}\label{e3.6.1}
\end{equation}
if and only if $x\in LP_mC(1,e]$ and
\begin{equation}
\begin{aligned}
x(t)
&=\frac{\mathbf{E}_{\beta,1}\big(\lambda(\log t)^\beta\big)}
{\lambda\mathbf{E}_{\beta,\beta}(\lambda)}b
+\big[(\log t)\mathbf{E}_{\beta,2}(\lambda (\log t)^\beta)
 -\frac{\mathbf{E}_{\beta,1}\big(\lambda(\log t)^\beta\big)}
 {\lambda\mathbf{E}_{\beta,\beta}(\lambda)} \mathbf{E}_{\beta,1}(\lambda )\big]a\\
&\quad -\frac{\mathbf{E}_{\beta,1}
 \big(\lambda(\log t)^\beta\big)}{\lambda\mathbf{E}_{\beta,\beta}(\lambda)}
 \sum_{\sigma=1}^m\Big(\lambda(\log \frac{e}{t_\sigma})^{\beta-1}
 \mathbf{E}_{\beta,\beta}\big(\lambda(\log \frac{e}{t_\sigma})^\beta\big)
 I_{\sigma} \\
&\quad +\mathbf{E}_{\beta,1}(\lambda (\log \frac{e}{t_\sigma})^\beta)J_{\sigma}\Big)\\
&\quad -\frac{\mathbf{E}_{\beta,1}\big(\lambda(\log t)^\beta\big)}
 {\lambda\mathbf{E}_{\beta,\beta}(\lambda)}
 \int_1^e(\log\frac{ e}{s})^{\beta-2}\mathbf{E}_{\beta,\beta-1}
 \big(\lambda (\log\frac{e}{s})^\beta\big) \sigma(s)\frac{ds}{s}\\
&\quad + \sum_{\sigma=1}^i\Big(\mathbf{E}_{\beta,1}
\big(\lambda (\log\frac{ t}{t_\sigma})^\beta\big)I_\sigma+(\log\frac{ t}{t_\sigma})
\mathbf{E}_{\beta,2}\big(\lambda (\log\frac{ t}{t_\sigma})^\beta\big)J_\sigma\Big)\\
&\quad +\int_1^t(\log\frac{t}{s})^{\beta-1}
 \mathbf{E}_{\beta,\beta}\big(\lambda(\log\frac{t}{s})^\beta\big)
 \sigma(s)\frac{ds}{s},
\end{aligned} \label{e3.6.2}
\end{equation}
for $t\in (t_i,t_{i+1}]$ and $i\in \mathbb{N}[0,m]$.
\end{lemma}


\begin{proof}
Let $x$ be a solution of \eqref{e3.6.1}.
We know by Theorem \ref{thm3.2.4} that there exist numbers
$c_{\sigma 0},c_{\sigma1}\in \mathbb{R}(\sigma\in \mathbb{N}[0,n-1])$ such that
\begin{equation}
\begin{aligned}
x(t)&=\sum_{\sigma=0}^i(c_{\sigma0}\mathbf{E}_{\beta,1}
\big(\lambda (\log\frac{ t}{t_\sigma})^\beta\big)
+c_{\sigma 1}(\log\frac{ t}{t_\sigma})\mathbf{E}_{\beta,2}
\big(\lambda (\log\frac{ t}{t_\sigma})^\beta\big))\\
&\quad +\int_1^t(\log\frac{t}{s})^{\beta-1}
 \mathbf{E}_{\beta,\beta}\big(\lambda(\log\frac{t}{s})^\beta\big)
\sigma(s)\frac{ds}{s},
\end{aligned} \label{e3.6.3}
\end{equation}
for $t\in (t_i,t_{i+1}]$ and $i\in \mathbb{N}[0,m]$.
One has
\begin{align*}
(t\frac{d}{dt})[\mathbf{E}_{\beta,1}(\lambda (\log t-\log t_\sigma)^\beta)]
&=(t\frac{d}{dt})\Big[\sum_{\tau=0}^{+\infty}\frac{\lambda^\tau
(\log t-\log t_\sigma)^{\tau\beta}}{\Gamma(\tau\beta+1)}\Big]\\
&=\sum_{\tau=1}^{+\infty}\frac{\lambda^\tau (\log t-\log t_\sigma)^{\tau\beta-1}}
 {\Gamma(\tau\beta)} \\
&=\lambda(\log \frac{t}{t_\sigma})^{\beta-1}\mathbf{E}_{\beta,\beta}
\big(\lambda(\log \frac{t}{t_\sigma})^\beta\big),
\end{align*}
\begin{align*}
(t\frac{d}{dt})\big[(\log t-\log t_\sigma)\mathbf{E}_{\beta,2}
\big(\lambda (\log t-\log t_\sigma)^\beta\big)\big]'
&=\Big[\sum_{\tau=0}^{+\infty}\frac{\lambda^\tau
 (\log t-\log t_\sigma)^{\tau\beta+1}}{\Gamma(\beta\tau+2)}\Big]\\
&=\sum_{\tau=0}^{+\infty}\frac{\lambda^\tau(\log t-\log t_\sigma)^{\tau\beta}}
{\Gamma(\beta\tau+1)} \\
&=\mathbf{E}_{\beta,1}(\lambda (\log \frac{t}{t_\sigma})^\beta),
\end{align*}
\begin{align*}
&(t\frac{d}{dt})\big[(\log\frac{t}{s})^{\beta-1}
 \mathbf{E}_{\beta,\beta}\big(\lambda(\log\frac{t}{s})^\beta\big)\big] \\
&=(t\frac{d}{dt})\Big[\sum_{\tau=0}^{+\infty}
\frac{\lambda^\tau(\log t-\log s)^{\tau\beta+\beta-1}}{\Gamma((\tau+1)\beta)}\Big]\\
&=\sum_{\tau=0}^{+\infty}\frac{\lambda^\tau(\log t-\log s)^{\tau\beta+\beta-2}}
 {\Gamma((\tau+1)\beta-1)} \\
&=(\log\frac{ t}{s})^{\beta-2}\mathbf{E}_{\beta,\beta-1}
 (\lambda (\log\frac{t}{s})^\beta).
\end{align*}
It follows that
\begin{equation}
\begin{aligned}
(t\frac{d}{dt})x(t)
&=\sum_{\sigma=0}^i\Big(\lambda(\log \frac{t}{t_\sigma})^{\beta-1}
\mathbf{E}_{\beta,\beta}\big(\lambda(\log \frac{t}{t_\sigma})^\beta\big)
c_{\sigma0}+\mathbf{E}_{\beta,1}(\lambda (\log \frac{t}{t_\sigma})^\beta)
c_{\sigma 1}\Big)\\
&\quad +\int_1^t(\log\frac{ t}{s})^{\beta-2}
 \mathbf{E}_{\beta,\beta-1}(\lambda (\log\frac{t}{s})^\beta)
\sigma(s)\frac{ds}{s},
\end{aligned} \label{e3.6.4}
\end{equation}
for $t\in (t_i,t_{i+1}]$ and $i\in N[0,m]$.
From \eqref{e3.6.3}, \eqref{e3.6.4}, the boundary conditions
 and the impulse assumption in \eqref{e3.6.1} it follows that $c_{01}=a$, and
\begin{gather*}
\begin{aligned}
&\sum_{\sigma=0}^m\Big(\lambda(\log \frac{e}{t_\sigma})^{\beta-1}
 \mathbf{E}_{\beta,\beta}\big(\lambda(\log \frac{e}{t_\sigma})^\beta\big)
c_{\sigma0}+\mathbf{E}_{\beta,1}(\lambda (\log \frac{e}{t_\sigma})^\beta)
 c_{\sigma 1}\Big)\\
&\quad +\int_1^e(\log\frac{ e}{s})^{\beta-2}\mathbf{E}_{\beta,\beta-1}
\big(\lambda (\log\frac{e}{s})^\beta\big) \sigma(s)\frac{ds}{s}=b,
\end{aligned}\\
c_{\sigma0}=I_\sigma,\quad c_{\sigma1}=J_\sigma,\quad \sigma\in \mathbb{N}[1,m].
\end{gather*}
Then
\begin{equation}
\begin{aligned}
c_{00}
&=\frac{1}{\lambda\mathbf{E}_{\beta,\beta}(\lambda)}
\Big[b-\mathbf{E}_{\beta,1}(\lambda )a-\sum_{\sigma=1}^m
\Big(\lambda(\log \frac{e}{t_\sigma})^{\beta-1}\mathbf{E}_{\beta,\beta}
\big(\lambda(\log \frac{e}{t_\sigma})^\beta\big)I_{\sigma} \\
&\quad +\mathbf{E}_{\beta,1}(\lambda (\log \frac{e}{t_\sigma})^\beta)J_{\sigma}\Big)
 -\int_1^e(\log\frac{ e}{s})^{\beta-2}\mathbf{E}_{\beta,\beta-1}
\big(\lambda (\log\frac{e}{s})^\beta\big)\sigma(s)\frac{ds}{s}\Big].
\end{aligned}\label{e3.6.5}
\end{equation}
Substituting $c_{\sigma0},c_{\sigma1}(\sigma\in N[0,m])$ into \eqref{e3.6.3}
and \eqref{e3.6.4}, we obtain \eqref{e3.6.2}.

On the other hand, if $x$ satisfies \eqref{e3.6.2}, then
$x|_{(t_i,t_{i+1}]}(i\in \mathbb{N}[0,m])$ are continuous and the limits
 $\lim_{t\to t_i^+}x(t)$. So $x\in LP_mC(1,e]$. Using \eqref{e3.6.5},
 $c_{01}=a$ and $c_{\sigma0}=I_\sigma$, $c_{\sigma1}=J_\sigma$, $\sigma\in \mathbb{N}[1,m]$,
we rewrite $x$ by \eqref{e3.6.3}.
 One have from Theorem \ref{thm3.2.4} easily for $t\in (t_0,t_1]$ that
${}^{CH}D_{1^+}^{\beta}x(t)=\lambda x(t)+\sigma(t)$ and for $t\in (t_i,t_{i+1}]$
similarly to the proof of Theorem \ref{thm3.2.4} that
\begin{align*}
&{}^{CH}D_{1^+}^\beta x(t) \\
&=\frac{1}{\Gamma(2-\beta)}\int_1^t(\log \frac{t}{s})^{1-\alpha}
 (s\frac{d}{ds})^2x(s)\frac{ds}{s}\\
&=\frac{1}{\Gamma(2-\beta)}\sum_{i=0}^{j-1}\int_{t_i}^{t_{i+1}}
(\log \frac{t}{s})^{1-\alpha}(s\frac{d}{ds})^2
\Big[\sum_{\sigma=0}^i\Big(c_{\sigma0}\mathbf{E}_{\beta,1}\big(\lambda (\log\frac{s}{t_\sigma})^\beta\big) \\
&\quad +c_{\sigma 1}(\log\frac{ s}{t_\sigma})
\mathbf{E}_{\beta,2}\big(\lambda (\log\frac{ s}{t_\sigma})^\beta\big)\Big) \\
&\quad +\int_1^s(\log\frac{s}{u})^{\beta-1}\mathbf{E}_{\beta,\beta}
\big(\lambda(\log\frac{s}{u})^\beta\big)\sigma(u)\frac{du}{u}\Big]\frac{ds}{s}\\
&\quad +\frac{1}{\Gamma(2-\beta)}\int_{t_j}^{t}
 (\log \frac{t}{s})^{1-\alpha}(s\frac{d}{ds})^2
\Big[\sum_{\sigma=0}^j\Big(c_{\sigma0}\mathbf{E}_{\beta,1}
 \big(\lambda (\log\frac{s}{t_\sigma})^\beta\big) \\
&\quad +c_{\sigma 1}
 (\log\frac{ s}{t_\sigma})\mathbf{E}_{\beta,2}
 \big(\lambda (\log\frac{ s}{t_\sigma})^\beta\big)\Big) \\
&\quad +\int_1^s(\log\frac{s}{u})^{\beta-1}
 \mathbf{E}_{\beta,\beta}\big(\lambda(\log\frac{s}{u})^\beta\big)
\sigma(u)\frac{du}{u}\Big]\frac{ds}{s}\\
&=\frac{1}{\Gamma(2-\beta)}\sum_{i=0}^{j-1}\int_{t_i}^{t_{i+1}}
(\log \frac{t}{s})^{1-\alpha}(s\frac{d}{ds})^2\sum_{\sigma=0}^i
\Big[c_{\sigma0}\mathbf{E}_{\beta,1}\big(\lambda (\log\frac{s}{t_\sigma})^\beta\big) \\
&\quad +c_{\sigma 1}(\log\frac{ s}{t_\sigma})\mathbf{E}_{\beta,2}
\big(\lambda (\log\frac{ s}{t_\sigma})^\beta\big)\Big]\frac{ds}{s}\\
&\quad +\frac{1}{\Gamma(2-\beta)}\int_{t_j}^{t}
(\log \frac{t}{s})^{1-\alpha}(s\frac{d}{ds})^2
\sum_{\sigma=0}^j\Big[c_{\sigma0}\mathbf{E}_{\beta,1}
\big(\lambda (\log\frac{s}{t_\sigma})^\beta\big) \\
&\quad +c_{\sigma 1}(\log\frac{ s}{t_\sigma})\mathbf{E}_{\beta,2}
 \big(\lambda (\log\frac{ s}{t_\sigma})^\beta\big)\Big]\frac{ds}{s} \\
&\quad +\frac{1}{\Gamma(2-\beta)}\int_1^{t}
 (\log \frac{t}{s})^{1-\alpha}(s\frac{d}{ds})^2
 \int_1^s(\log\frac{s}{u})^{\beta-1}
 \mathbf{E}_{\beta,\beta}\big(\lambda(\log\frac{s}{u})^\beta\big)
\sigma(u)\frac{du}{u}\frac{ds}{s} \\
&=\lambda x(t)+\sigma(t).
\end{align*}
So $x$ is a solution of \eqref{e3.6.1}. The proof is complete.
\end{proof}

 Define the nonlinear operator $J$ on $LP_mC(1,e]$ by $(Jx)$ by
\begin{align*}
&(Jx)(t) \\
&=\frac{\mathbf{E}_{\beta,1}\big(\lambda(\log t)^\beta\big)}
{\lambda\mathbf{E}_{\beta,\beta}(\lambda)}\int_0^1\psi(s)H(s,x(s))ds\\
&\quad +\Big[(\log t)\mathbf{E}_{\beta,2}
\big(\lambda (\log t)^\beta\big)
-\frac{\mathbf{E}_{\beta,1}\big(\lambda(\log t)^\beta\big)}
{\lambda\mathbf{E}_{\beta,\beta}(\lambda)}
\mathbf{E}_{\beta,1}(\lambda )\Big]\int_0^1\phi(s)G(s,x(s))ds\\
&\quad -\frac{\mathbf{E}_{\beta,1}\big(\lambda(\log t)^\beta\big)}
{\lambda\mathbf{E}_{\beta,\beta}(\lambda)}\sum_{\sigma=1}^m
(\lambda(\log \frac{e}{t_\sigma})^{\beta-1}\mathbf{E}_{\beta,\beta}
\big(\lambda(\log \frac{e}{t_\sigma})^\beta\big)I(t_{\sigma},x(t_\sigma)) \\
&\quad +\mathbf{E}_{\beta,1}(\lambda (\log \frac{e}{t_\sigma})^\beta)
J(t_{\sigma},x(t_\sigma)))\\
&\quad -\frac{\mathbf{E}_{\beta,1}\big(\lambda(\log t)^\beta\big)}
{\lambda\mathbf{E}_{\beta,\beta}(\lambda)}
\int_1^e(\log\frac{ e}{s})^{\beta-2}\mathbf{E}_{\beta,\beta-1}
\big(\lambda (\log\frac{e}{s})^\beta\big) p(s)f(s,x(s))\frac{ds}{s}\\
&\quad +\sum_{\sigma=1}^i(\mathbf{E}_{\beta,1}
 \big(\lambda (\log\frac{ t}{t_\sigma})^\beta\big)
 I(t_\sigma,x(t_\sigma))+(\log\frac{ t}{t_\sigma})
 \mathbf{E}_{\beta,2}\big(\lambda (\log\frac{ t}{t_\sigma})^\beta\big)
 J(t_\sigma,x(t_\sigma)))\\
&\quad +\int_1^t(\log\frac{t}{s})^{\beta-1}\mathbf{E}_{\beta,\beta}
\big(\lambda(\log\frac{t}{s})^\beta\big)
p(s)f(s,x(s))\frac{ds}{s},\quad t\in (t_i,t_{i+1}],\;i\in \mathbb{N}[0,m].
\end{align*}

\begin{lemma} \label{lem3.6.2}
Suppose that {\rm (1.A7), (1.A12), (1.A13, (1.A14)} hold,
$\lambda\neq 0$, and $f,G,H$ are impulsive I-Carath\'eodory functions, and
$I,J$ are discrete I-Carath\'odory functions.
Then $R:LP_mC(1,e]\to LP_mC(1,e]$ is well defined and is completely continuous,
$x$ is a solution of BVP \eqref{e1.0.10} if and only if $x$ is a fixed point
of $J$ in $LP_mC(1,e]$.
\end{lemma}

The proof of the above lemma is similar to that of the proof of
Lemma \ref{lem3.3.2} and is omitted.


\section{Solvability of BVPs \eqref{e1.0.7}--\eqref{e1.0.10}}

Now, we prove that main theorems in this article by using the Schaefer's
fixed point theorem, i.e., \cite[Lemma 3.1.9]{10}.

\begin{theorem} \label{thm4.1}
Suppose that {\rm (1.A1)--(1.A5)} are satisfied and
\begin{itemize}
\item[(4.A1)] There exist nondecreasing functions
$M_f,M_g,M_h,M_I,M_J$  from $[0,+\infty)$ to $[0,+\infty)$ such that
\begin{gather*}
|f(t,(t-t_i)^{\beta-2}x)|\le M_f(|x|),\quad t\in (t_i,t_{i+1}],\;i\in \mathbb{N}[0,m],x\in \mathbb{R},\\
|G(t,(t-t_i)^{\beta-2}x)|\le M_G(|x|),\quad t\in (t_i,t_{i1}],\;i\in \mathbb{N}[0,m],x\in \mathbb{R},\\
|H(t,(t-t_i)^{\beta-2}x)|\le M_H(|x|),\quad t\in (t_i,t_{i1}],\;i\in \mathbb{N}[0,m],x\in \mathbb{R},\\
|I(t_i,(t_i-t_{i-1})^{\beta-2}x)|\le M_I(|x|),\quad i\in N[1,m],\;x\in \mathbb{R},\\
|J(t_i,(t_i-t_{i-1})^{\beta-2}x)|\le M_J(|x|),\quad i\in N[1,m],\; x\in \mathbb{R}.
\end{gather*}
\end{itemize}
Then \eqref{e1.0.7} has at least one solution if there exists a $r_0>0$
such that
\begin{equation}
\begin{aligned}
&\big[\frac{\mathbf{E}_{\beta,\beta}(|\lambda|)
 \mathbf{E}_{\beta,\beta-1}(\lambda)\|\phi\|_1}
 {\mathbf{E}_{\beta,\beta}(\lambda)}+\mathbf{E}_{\beta,\beta-1}
 (|\lambda|)\|\phi\|_1\Big]M_G(r_0) \\
&+\frac{\mathbf{E}_{\beta,\beta}(|\lambda|)\|\psi\|_1}
 {\mathbf{E}_{\beta,\beta}(\lambda)}M_H(r_0)
+[\frac{m\mathbf{E}_{\beta,\beta}(|\lambda|)^2}{\mathbf{E}_{\beta,\beta}(\lambda)}
 +m\mathbf{E}_{\beta,\beta}(|\lambda|)]M_J(r_0) \\
&+\big[\frac{\mathbf{E}_{\beta,\beta}(|\lambda|)
\mathbf{E}_{\beta,\beta-1}(|\lambda|)}
{\mathbf{E}_{\beta,\beta}(\lambda)(1-t_\sigma)^{2-\beta}}
+m\mathbf{E}_{\beta,\beta-1}(|\lambda|)\big]M_I(r_0)\\
&+\big[\frac{\mathbf{E}_{\beta,\beta}(|\lambda|)
 \mathbf{E}_{\beta,\beta}(|\lambda|)\mathbf{B}(\beta+l,k+1)}
 {\mathbf{E}_{\beta,\beta}(\lambda)}+\mathbf{E}_{\beta,\beta}
 (\lambda|)\mathbf{B}(\beta+l,k+1)\big]M_f(r_0) \\
&<r_0.
\end{aligned}\label{e4.0.1}
\end{equation}
\end{theorem}

\begin{proof}
From Lemmas \ref{lem3.3.1} and \ref{lem3.3.2}, and the definition of $T$, 
it follows that $x\in P_mC_{2-\beta}(0,1]$
is a solution of \eqref{e1.0.7} if and only if $x\in P_mC_{2-\beta}(0,1]$
is a fixed point of $T$ in $P_mC_{2-\beta}(0,1]$.
 Lemma \ref{lem3.3.2} implies that $T$ is a completely continuous operator.
 From (4.A1), we have for $x\in P_1C(0,1]$ that
\begin{gather*}
\begin{aligned}
|f(t,x(t))|&=|f(t,(t-t_i)^{\beta-2}(t-t_i)^{2-\beta}x(t))|\\
&\le M_f(|(t-t_i)^{2-\beta}x(t)|) \\
&\le M_f(\|x\|),\quad t\in (t_i,t_{i+1}],\;i\in \mathbb{N}[0,m],
\end{aligned}\\
|G(t,x(t))|\le M_G(\|x\|),\quad t\in (t_i,t_{i+1}],\;i\in \mathbb{N}[0,m],\\
|H(t,x(t))|\le M_H(\|x\|),\quad t\in (t_i,t_{i+1}],\;i\in \mathbb{N}[0,m],\\
\begin{aligned}
|I(t_i,x(t_i))|
&=|I(t_i,(t_i-t_{i-1})^{\beta-2}(t_i-t_{i-1})^{2-\beta}x(t_i))|\\
&\le M_I((t_i-t_{i-1})^{2-\beta}|x(t_i)|) \\
&\le M_I(\|x\|),\quad i\in \mathbb{N}[1,m],
\end{aligned}\\
|J(t_i,x(t_i))|\le M_J(\|x\|), \quad i\in \mathbb{N}[1,m].
\end{gather*}
We consider the set
$\Omega=\{x\in P_mC_{2-\beta}(0,1]:x=\lambda (Tx),\text{ for some }
\lambda \in [0,1]\}$. For $x\in \Omega$ and $t\in (t_i,t_{i+1}]$ we have
\begin{align*}
&(t-t_i)^{2-\beta}|(Tx)(t)| \\
&\le \frac{(t-t_i)^{2-\beta}t^{\beta-1}\mathbf{E}_{\beta,\beta}
(\lambda t^\beta)}{\mathbf{E}_{\beta,\beta}(\lambda)}
\Big[\|\psi\|_1M_H(\|x\|)+\mathbf{E}_{\beta,\beta-1}(\lambda)\|\phi\|_1M_G(\|x\|) \\
&\quad +\sum_{\tau=0}^{+\infty}\frac{\lambda^\tau}{\Gamma((\tau+1)\beta)}
\int_0^1(1-s)^{\tau\beta+\beta-1}s^k(1-s)^ldsM_f(\|x\|) \\
&\quad +\sum_{\sigma=1}^m\Big(
 (1-t_\sigma)^{\beta-1}\mathbf{E}_{\beta,\beta}(\lambda (1-t_\sigma)^\beta)M_J(\|x\|)\\
&\quad +(1-t_\sigma)^{\beta-2}\mathbf{E}_{\beta,\beta-1}
 (\lambda (1-t_\sigma)^\beta)M_I(\|x\|)\Big)\Big]\\
&\quad +(t-t_i)^{2-\beta}t^{\beta-2}\mathbf{E}_{\beta,\beta-1}(|\lambda|)
 \|\phi\|_1M_G(\|x\|)\\
&\quad +(t-t_i)^{2-\beta}\sum_{\sigma=1}^i
\Big[(t-t_\sigma)^{\beta-1}\mathbf{E}_{\beta,\beta}
(\lambda (t-t_\sigma)^\beta)M_J(\|x\|) \\
&\quad +(t-t_\sigma)^{\beta-2}
 \mathbf{E}_{\beta,\beta-1}(\lambda (t-t_\sigma)^\beta)M_I(\|x\|)\Big]\\
&\quad +(t-t_i)^{2-\beta}\sum_{\tau=0}^{+\infty}
 \frac{\lambda^\tau}{\Gamma((\tau+1)\beta)}
\int_0^t(t-s)^{\tau\beta+\beta-1}s^k(1-s)^ldsM_f(\|x\|) \\
&\le\frac{\mathbf{E}_{\beta,\beta}(|\lambda|)}{\mathbf{E}_{\beta,\beta}
 (\lambda)}\Big[\|\psi\|_1M_H(\|x\|)+\mathbf{E}_{\beta,\beta-1}
(\lambda)\|\phi\|_1M_G(\|x\|) \\
&\quad +\sum_{\tau=0}^{+\infty}\frac{\lambda^\tau}{\Gamma((\tau+1)\beta)}
\int_0^1(1-s)^{\beta+l-1}s^kdsM_f(\|x\|)
 +m\mathbf{E}_{\beta,\beta}(|\lambda|)M_J(\|x\|) \\
&\quad +(1-t_\sigma)^{\beta-2}\mathbf{E}_{\beta,\beta-1}(|\lambda|)M_I(\|x\|)\Big]
+\mathbf{E}_{\beta,\beta-1}(|\lambda|)\|\phi\|_1M_G(\|x\|)\\
&\quad +\sum_{\sigma=1}^i
\big[\mathbf{E}_{\beta,\beta}(|\lambda|)M_J(\|x\|)
+\mathbf{E}_{\beta,\beta-1}(|\lambda|)M_I(\|x\|)\big]\\
&\quad +(t-t_i)^{2-\beta}\sum_{\tau=0}^{+\infty}
 \frac{\lambda^\tau t^{\tau\beta+\beta+k+l}}{\Gamma((\tau+1)\beta)}
 \int_0^1(1-w)^{\tau\beta+\beta+l-1}w^kdwM_f(\|x\|)\\
&\le\frac{\mathbf{E}_{\beta,\beta}(|\lambda|)\|\psi\|_1}
 {\mathbf{E}_{\beta,\beta}(\lambda)}M_H(\|x\|)
 +\frac{\mathbf{E}_{\beta,\beta}(|\lambda|)\mathbf{E}_{\beta,\beta-1}
 (\lambda)\|\phi\|_1}{\mathbf{E}_{\beta,\beta}(\lambda)}M_G(\|x\|)\\
&\quad +\frac{\mathbf{E}_{\beta,\beta}(|\lambda|)
 \mathbf{E}_{\beta,\beta}(|\lambda|)\mathbf{B}(\beta+l,k+1)}
 {\mathbf{E}_{\beta,\beta}(\lambda)}M_f(\|x\|)
 +\frac{m\mathbf{E}_{\beta,\beta}(|\lambda|)^2}
 {\mathbf{E}_{\beta,\beta}(\lambda)}M_J(\|x\|)\\
&\quad +\frac{\mathbf{E}_{\beta,\beta}(|\lambda|)\mathbf{E}_{\beta,\beta-1}
 (|\lambda|)}{\mathbf{E}_{\beta,\beta}(\lambda)(1-t_\sigma)^{2-\beta}}M_I(\|x\|)
 +\mathbf{E}_{\beta,\beta-1}(|\lambda|)\|\phi\|_1M_G(\|x\|)\\
&\quad +m\mathbf{E}_{\beta,\beta}(|\lambda|)M_J(\|x\|)
 +m\mathbf{E}_{\beta,\beta-1}(|\lambda|)M_I(\|x\|) \\
&\quad +\mathbf{E}_{\beta,\beta}(\lambda|)\mathbf{B}(\beta+l,k+1)M_f(\|x\|)\\
&=\big[\frac{\mathbf{E}_{\beta,\beta}(|\lambda|)
 \mathbf{E}_{\beta,\beta-1}(\lambda)\|\phi\|_1}
 {\mathbf{E}_{\beta,\beta}(\lambda)}
 +\mathbf{E}_{\beta,\beta-1}(|\lambda|)\|\phi\|_1\big]M_G(\|x\|) \\
&\quad +\frac{\mathbf{E}_{\beta,\beta}(|\lambda|)
 \|\psi\|_1}{\mathbf{E}_{\beta,\beta}(\lambda)}M_H(\|x\|)
+\big[\frac{m\mathbf{E}_{\beta,\beta}(|\lambda|)^2}{\mathbf{E}_{\beta,\beta}
 (\lambda)}+m\mathbf{E}_{\beta,\beta}(|\lambda|)\big]M_J(\|x\|) \\
&\quad +\big[\frac{\mathbf{E}_{\beta,\beta}(|\lambda|)
\mathbf{E}_{\beta,\beta-1}(|\lambda|)}
{\mathbf{E}_{\beta,\beta}(\lambda)(1-t_\sigma)^{2-\beta}}
+m\mathbf{E}_{\beta,\beta-1}(|\lambda|)\big]M_I(\|x\|)\\
&\quad +\big[\frac{\mathbf{E}_{\beta,\beta}(|\lambda|)
\mathbf{E}_{\beta,\beta}(|\lambda|)\mathbf{B}(\beta+l,k+1)}
{\mathbf{E}_{\beta,\beta}(\lambda)}+\mathbf{E}_{\beta,\beta}(\lambda|)
\mathbf{B}(\beta+l,k+1)\big]M_f(\|x\|).
\end{align*}
It follows that
\begin{align*}
\|x\|
&=\lambda \|Tx\|\le \|Tx\| \\
&\le \big[\frac{\mathbf{E}_{\beta,\beta}(|\lambda|)
\mathbf{E}_{\beta,\beta-1}(\lambda)\|\phi\|_1}
{\mathbf{E}_{\beta,\beta}(\lambda)}+\mathbf{E}_{\beta,\beta-1}(|\lambda|)
\|\phi\|_1\big]M_G(\|x\|) \\
&\quad +\frac{\mathbf{E}_{\beta,\beta}(|\lambda|)\|\psi\|_1}
 {\mathbf{E}_{\beta,\beta}(\lambda)}M_H(\|x\|)\\
&\quad +\big[\frac{m\mathbf{E}_{\beta,\beta}(|\lambda|)^2}
 {\mathbf{E}_{\beta,\beta}(\lambda)}+m\mathbf{E}_{\beta,\beta}(|\lambda|)
 \big]M_J(\|x\|)
+\big[\frac{\mathbf{E}_{\beta,\beta}(|\lambda|)
 \mathbf{E}_{\beta,\beta-1}(|\lambda|)}{\mathbf{E}_{\beta,\beta}
 (\lambda)(1-t_\sigma)^{2-\beta}} \\
&\quad +m\mathbf{E}_{\beta,\beta-1}(|\lambda|)\big]M_I(\|x\|)
 +\big[\frac{\mathbf{E}_{\beta,\beta}(|\lambda|)
 \mathbf{E}_{\beta,\beta}(|\lambda|)\mathbf{B}(\beta+l,k+1)}
 {\mathbf{E}_{\beta,\beta}(\lambda)} \\
&\quad +\mathbf{E}_{\beta,\beta}(\lambda|)
 \mathbf{B}(\beta+l,k+1)\big]M_f(\|x\|).
\end{align*}
From \eqref{e4.0.1}, we choose $ \Omega=\{x\in P_1C_{2-\beta}(0,1]:\|x\|\le r_0\}$.
For $x\in \Omega$, we obtain $x\neq \lambda (Tx)$ for any $\lambda\in [0,1]$
and $x\in \partial\Omega$. In fact, if $x=\lambda (Tx)$ for some
$\lambda\in [0,1]$ and $x\in \partial\Omega$, then
\begin{align*}
r_0&=\|x\| \\
&\le \big[\frac{\mathbf{E}_{\beta,\beta}(|\lambda|)
 \mathbf{E}_{\beta,\beta-1}(\lambda)\|\phi\|_1}{\mathbf{E}_{\beta,\beta}
(\lambda)}+\mathbf{E}_{\beta,\beta-1}(|\lambda|)\|\phi\|_1\big]
M_G(r_0) \\
&\quad +\frac{\mathbf{E}_{\beta,\beta}(|\lambda|)\|\psi\|_1}
 {\mathbf{E}_{\beta,\beta}(\lambda)}M_H(r_0)
 +\big[\frac{m\mathbf{E}_{\beta,\beta}(|\lambda|)^2}
 {\mathbf{E}_{\beta,\beta}(\lambda)}
 +m\mathbf{E}_{\beta,\beta}(|\lambda|) \big]M_J(r_0) \\
&\quad +\big[\frac{\mathbf{E}_{\beta,\beta}(|\lambda|)
 \mathbf{E}_{\beta,\beta-1}(|\lambda|)}{\mathbf{E}_{\beta,\beta}(\lambda)
 (1-t_\sigma)^{2-\beta}}
 +m\mathbf{E}_{\beta,\beta-1}(|\lambda|)\big]M_I(r_0)\\
&\quad +\big[\frac{\mathbf{E}_{\beta,\beta}(|\lambda|)
 \mathbf{E}_{\beta,\beta}(|\lambda|)\mathbf{B}(\beta+l,k+1)}
 {\mathbf{E}_{\beta,\beta}(\lambda)}+\mathbf{E}_{\beta,\beta}
 (\lambda|)\mathbf{B}(\beta+l,k+1)\big]M_f(r_0)<r_0,
\end{align*}
which is a contradiction.
 As a consequence of Schaefer's fixed point theorem, we deduce that $T$
has a fixed point which is a solution of the problem
BVP\eqref{e1.0.7}. The proof is complete.
\end{proof}

\begin{corollary} \label{coro4.1} 
Suppose that {\rm (1.A1)--(1.A5)}
and {\rm (4.A1)} hold.
Then \eqref{e1.0.7} has at least one solution if
\begin{align*}
&\inf_{r\in (0,+\infty)}\frac{1}{r}
\Big[(\frac{\mathbf{E}_{\beta,\beta}(|\lambda|)
\mathbf{E}_{\beta,\beta-1}(\lambda)\|\phi\|_1}
{\mathbf{E}_{\beta,\beta}(\lambda)}+\mathbf{E}_{\beta,\beta-1}
(|\lambda|)\|\phi\|_1)M_G(r)\\
&+\frac{\mathbf{E}_{\beta,\beta}(|\lambda|)\|\psi\|_1}
{\mathbf{E}_{\beta,\beta}(\lambda)}M_H(r)
 +\big[\frac{m\mathbf{E}_{\beta,\beta}(|\lambda|)^2}
 {\mathbf{E}_{\beta,\beta}(\lambda)}
 +m\mathbf{E}_{\beta,\beta}(|\lambda|)\big]M_J(r) \\
&+\big[\frac{\mathbf{E}_{\beta,\beta}(|\lambda|)
 \mathbf{E}_{\beta,\beta-1}(|\lambda|)}
 {\mathbf{E}_{\beta,\beta}(\lambda)(1-t_\sigma)^{2-\beta}}
 +m\mathbf{E}_{\beta,\beta-1}(|\lambda|)\big]M_I(r)\\
&+\big[\frac{\mathbf{E}_{\beta,\beta}(|\lambda|)
\mathbf{E}_{\beta,\beta}(|\lambda|)\mathbf{B}(\beta+l,k+1)}
{\mathbf{E}_{\beta,\beta}(\lambda)}
+\mathbf{E}_{\beta,\beta}(\lambda|)\mathbf{B}(\beta+l,k+1)\big]M_f(r)\Big]<1.
\end{align*}
\end{corollary}

\begin{proof}
Form the assumption, we know that there exists $r_0>0$ such that \eqref{e4.0.1}
 holds. By Theorem \ref{thm4.1}, \eqref{e1.0.7} has at least one solution.
The proof is omitted.
\end{proof}

\begin{theorem} \label{thm4.2}
 Suppose that {\rm (1.A2), (1.A3) (1.A6)--(1.A8)} hold, and
\begin{itemize}
\item[(4.A2)] there exist nondecreasing functions
$M_f,M_g,M_h,M_I,M_J:[0,+\infty)\to [0,+\infty)$ such that
\begin{gather*}
|f(t,x)|\le M_f(|x|),\quad t\in (0,1),\; x\in \mathbb{R},\\
|G(t,x)|\le M_G(|x|),\quad t\in (0,1),\; x\in \mathbb{R},\\
|H(t,x)|\le M_H(|x|),\quad t\in (0,1),\; x\in \mathbb{R},\\
|I(t_i,x)|\le M_I(|x|),\quad i\in \mathbb{N}[1,m],\; x\in \mathbb{R},\\
|J(t_i,x)|\le M_J(|x|),\quad i\in \mathbb{N}[1,m],\; x\in \mathbb{R}.
\end{gather*}
\end{itemize}
Then \eqref{e1.0.8} has at least one solution if there exists $r_0>0$ such that
\begin{equation}
\begin{aligned}
&\big[\|\phi\|_1+\frac{|\lambda |\mathbf{E}_{\beta,\beta}
 (\lambda)\|\phi\|_1}{\mathbf{E}_{\beta,1}(\lambda)}\big]M_G(r_0)
+\frac{\|\psi\|_1}{\mathbf{E}_{\beta,1}(\lambda)}M_H(r_0)\\
&+\big[\frac{m|\lambda| \mathbf{E}_{\beta,\beta}(|\lambda|)}
 {\mathbf{E}_{\beta,1}(\lambda)}+m\mathbf{E}_{\beta,1}(|\lambda|)\big]M_I(r_0)
+\big[m\mathbf{E}_{\beta,2}(|\lambda|) \\
&+\frac{m\mathbf{E}_{\beta,1}(|\lambda|)}{\mathbf{E}_{\beta,1}(\lambda)}
 \big]M_J(r_0)
 +\big[\frac{\mathbf{E}_{\beta,\beta-1}(|\lambda|)
 \mathbf{B}(\beta+l,k+1)}{\mathbf{E}_{\beta,1}(\lambda)} \\
&+\mathbf{E}_{\beta,\beta}(|\lambda|)\mathbf{B}(\beta+l,k+1)\big]M_f(r_0)< r_0.
\end{aligned}\label{e4.0.2}
\end{equation}
\end{theorem}

\begin{proof}
 From Lemmas \ref{lem3.4.1} and \ref{lem3.4.2}, and the definition of $Q$, 
it follows that $x\in P_mC(0,1]$
is a solution of \eqref{e1.0.8} if and only if $x\in P_mC(0,1]$ is a fixed
point of $Q$. Lemma \ref{lem3.4.2} implies that $Q$ is a completely continuous operator.
From (4.A2), for $x\in P_mC(0,1]$ we have
\begin{gather*}
|f(t,x(t))|\le M_f(|x(t)|)\le M_f(\|x\|),\quad t\in (0,1),\\
|G(t,x(t))|\le M_G(\|x\|),t\in (0,1),\\
|H(t,x(t))|\le M_H(\|x\|),t\in (0,1),\\
|I(t_i,x(t_i))|\le M_I(\|x\|),i\in \mathbb{N}[1,m],\\
|J(t_i,x(t_i))|\le M_J(\|x\|),i\in \mathbb{N}[1,m].
\end{gather*}
We consider the set
$\Omega=\{x\in P_mC(0,1]:x=\lambda (Tx),\text{ for some }\lambda \in [0,1]\}$.
For $x\in \Omega$, and $t\in (t_i,t_{i+1}]$ we have
\begin{align*}
&|(Qx)(t)| \\
&\le \|\phi\|_1M_G(\|x\|)+\frac{1}{\mathbf{E}_{\beta,1}(\lambda)}
\Big[\|\psi\|_1M_H(\|x\|)+|\lambda |\mathbf{E}_{\beta,\beta}(\lambda)\|
\phi\|_1M_G(\|x\|) \\
&\quad +\sum_{\sigma=1}^m(|\lambda| \mathbf{E}_{\beta,\beta}(|\lambda|)
M_I(\|x\|)+\mathbf{E}_{\beta,1}(|\lambda|)M_J(\|x\|)) \\
&\quad +\sum_{\tau=0}^{+\infty}
\frac{\lambda^\tau}{\Gamma(\tau\beta+\beta-1)}
\int_0^1(1-s)^{\tau\beta+\beta+l-1}s^kdsM_f(\|x\|)\Big] \\
&\quad +\sum_{j=1}^i[\mathbf{E}_{\beta,1}(|\lambda|)M_I(\|x\|)
+\mathbf{E}_{\beta,2}(|\lambda|)M_J(\|x\|)]\\
&\quad +\sum_{\tau=0}^{+\infty}\frac{\lambda^\tau}{\Gamma((\tau+1)\beta)}
\int_0^t(t-s)^{\tau\beta+\beta-1}s^k(1-s)^ldsM_f(\|x\|)\\
&\le \|\phi\|_1M_G(\|x\|)+\frac{1}{\mathbf{E}_{\beta,1}(\lambda)}
\Big[\|\psi\|_1M_H(\|x\|)+|\lambda |\mathbf{E}_{\beta,\beta}(\lambda)
\|\phi\|_1M_G(\|x\|) \\
&\quad +m|\lambda| \mathbf{E}_{\beta,\beta}(|\lambda|)M_I(\|x\|)
+m\mathbf{E}_{\beta,1}(|\lambda|)M_J(\|x\|)\\
&\quad +\sum_{\tau=0}^{+\infty}\frac{\lambda^\tau}
 {\Gamma(\tau\beta+\beta-1)}\int_0^1(1-s)^{\beta+l-1}s^kdsM_f(\|x\|)\Big]\\
&\quad +m\mathbf{E}_{\beta,1}(|\lambda|)M_I(\|x\|)
 +m\mathbf{E}_{\beta,2}(|\lambda|)M_J(\|x\|) \\
&\quad+\sum_{\tau=0}^{+\infty}\frac{\lambda^\tau t^{\tau\beta+\beta+k+l}}
 {\Gamma((\tau+1)\beta)}\int_0^1(1-w)^{\tau\beta+\beta+l-1}w^kdwM_f(\|x\|) \\
& \le [\|\phi\|_1+\frac{|\lambda |\mathbf{E}_{\beta,\beta}(\lambda)
 \|\phi\|_1}{\mathbf{E}_{\beta,1}(\lambda)}]
 M_G(\|x\|)+\frac{\|\psi\|_1}{\mathbf{E}_{\beta,1}(\lambda)}M_H(\|x\|)
 +[\frac{m|\lambda| \mathbf{E}_{\beta,\beta}(|\lambda|)}{\mathbf{E}_{\beta,1}
 (\lambda)} \\
&\quad +m\mathbf{E}_{\beta,1}(|\lambda|)]M_I(\|x\|)
 +[m\mathbf{E}_{\beta,2}(|\lambda|)
 +\frac{m\mathbf{E}_{\beta,1}(|\lambda|)}{\mathbf{E}_{\beta,1}(\lambda)}]M_J(\|x\|)\\
&\quad +\big[\frac{\mathbf{E}_{\beta,\beta-1}(|\lambda|)
\mathbf{B}(\beta+l,k+1)}{\mathbf{E}_{\beta,1}(\lambda)}
+\mathbf{E}_{\beta,\beta}(|\lambda|)\mathbf{B}(\beta+l,k+1)\big]M_f(\|x\|).
\end{align*}
It follows that
\begin{align*}
\|x\|&=\lambda \|Tx\|\le \|Tx\| \\
&\le \big[\|\phi\|_1+\frac{|\lambda |\mathbf{E}_{\beta,\beta}(\lambda)
 \|\phi\|_1}{\mathbf{E}_{\beta,1}(\lambda)}\big] M_G(\|x\|)
+\frac{\|\psi\|_1}{\mathbf{E}_{\beta,1}(\lambda)}M_H(\|x\|)\\
&\quad +\big[\frac{m|\lambda| \mathbf{E}_{\beta,\beta}(|\lambda|)}
 {\mathbf{E}_{\beta,1}(\lambda)}+m\mathbf{E}_{\beta,1}(|\lambda|)\big]M_I(\|x\|)\\
&\quad +\big[m\mathbf{E}_{\beta,2}(|\lambda|)+\frac{m\mathbf{E}_{\beta,1}
 (|\lambda|)}{\mathbf{E}_{\beta,1}(\lambda)}\big]M_J(\|x\|)
 +\big[\frac{\mathbf{E}_{\beta,\beta-1}(|\lambda|)
 \mathbf{B}(\beta+l,k+1)}{\mathbf{E}_{\beta,1}(\lambda)} \\
&\quad +\mathbf{E}_{\beta,\beta}(|\lambda|)\mathbf{B}(\beta+l,k+1)\big]M_f(\|x\|).
\end{align*}
From \eqref{e4.0.2}, we choose $ \Omega=\{x\in P_mC(0,1]:\|x\|\le r_0\}$.
For $x\in \Omega$, we obtain $x\neq \lambda (Tx)$ for any
$\lambda\in [0,1]$ and $x\in \partial\Omega$.

 As a consequence of Schaefer's fixed point theorem, we deduce that $Q$ has
a fixed point which is a solution of problem \eqref{e1.0.8}.
 The proof is complete.
\end{proof}


\begin{theorem} \label{thm4.3}
Suppose that {\rm (1.A4), (1.A4) (1.A9)--(1.A11)} hold,
$\Delta \neq 0$, and
\begin{itemize}
\item[(4.A3)] there exist nondecreasing functions
$M_f,M_g,M_h,M_I,M_J:[0,+\infty)\to [0,+\infty)$ such that
\begin{gather*}
|f(t,(\log \frac{t}{t_i})^{\beta-2}x)|
\le M_f(|x|),\quad t\in (t_i,t_{i+1}],\;i\in \mathbb{N}[0,m],x\in \mathbb{R},\\
|G(t,(\log \frac{t}{t_i})^{\beta-2}x)|
\le M_f(|x|),\quad t\in (t_i,t_{i+1}],\;i\in \mathbb{N}[0,m],x\in \mathbb{R},\\
|H(t,(\log \frac{t}{t_i})^{\beta-2}x)|
\le M_f(|x|),\quad t\in (t_i,t_{i+1}],\;i\in \mathbb{N}[0,m],x\in \mathbb{R},\\
|I(t_i,(\log \frac{t_i}{t_{i-1}})^{\beta-2}x)|
\le M_I(|x|),\quad i\in \mathbb{N}[1,m],\; x\in \mathbb{R},\\
|J(t_i,(\log \frac{t_i}{t_{i-1}})^{\beta-2}x)|
\le M_I(|x|),\quad i\in \mathbb{N}[1,m],\; x\in \mathbb{R}.
\end{gather*}
\end{itemize}
Then \eqref{e1.0.9} has at least one solution if there exists a constant
$r_0>0$ such that
\begin{equation}
B_{1G}M_G(r_0)+B_{2H}M_H(r_0)+B_{3J}M_J(r_0)+B_{4I}M_I(r_0)+B_{5f}M_f(r_0)< r_0,
\label{e4.0.3}
\end{equation}
where
\begin{gather*}
B_{1G}=\frac{|\lambda|\mathbf{E}_{\beta,\beta}(\lambda)
\mathbf{E}_{\beta,\beta}(|\lambda|)\|\phi\|_1}{|\Delta|}
+\frac{|\frac{1}{\Gamma(\beta)}-\mathbf{E}_{\beta,\beta}(\lambda)
|\mathbf{E}_{\beta,\beta-1}(|\lambda|)\|\phi\|_1}{|\Delta|},
\\
B_{2H}=\frac{\mathbf{E}_{\beta,\beta}(\lambda)\mathbf{E}_{\beta,\beta-1}
(|\lambda|)\|\psi\|_1}{|\Delta|}+\frac{|\frac{1}{\Gamma(\beta-1)}
-\mathbf{E}_{\beta,\beta-1}(\lambda)|\mathbf{E}_{\beta,\beta}(|\lambda|)
\|\psi\|_1}{|\Delta|},\\
\begin{aligned}
B_{3J}&=\frac{m\Gamma(\beta)|\frac{1}{\Gamma(\beta-1)}
 -\mathbf{E}_{\beta,\beta-1}(\lambda)|\mathbf{E}_{\beta,\beta}
 (|\lambda|)^2}{|\Delta|}
 +\frac{m|\lambda|\Gamma(\beta)\mathbf{E}_{\beta,\beta}(\lambda)
 \mathbf{E}_{\beta,\beta}(|\lambda|)^2}{|\Delta|}\\
&\quad +\frac{m\Gamma(\beta)|\frac{1}{\Gamma(\beta)}
 -\mathbf{E}_{\beta,\beta}(\lambda)|\mathbf{E}_{\beta,\beta}(|\lambda|)
 \mathbf{E}_{\beta,\beta-1}(|\lambda|)}{|\Delta|} \\
&\quad +\frac{m\Gamma(\beta)\mathbf{E}_{\beta,\beta}(\lambda)\mathbf{E}_{\beta,\beta}
 (|\lambda|)\mathbf{E}_{\beta,\beta-1}(|\lambda|)}{|\Delta|}
 +m\mathbf{E}_{\beta,\beta}(|\lambda|)\Gamma(\beta),
\end{aligned}\\
\begin{aligned}
B_{4I}
&=\frac{m|\lambda|\Gamma(\beta-1)|\frac{1}{\Gamma(\beta-1)}
 -\mathbf{E}_{\beta,\beta-1}(\lambda)|\mathbf{E}_{\beta,\beta}
 (|\lambda|)^2}{|\Delta|} \\
&\quad +\frac{|\lambda|\Gamma(\beta-1)\mathbf{E}_{\beta,\beta}(\lambda)
 \mathbf{E}_{\beta,\beta-1}(|\lambda|)\mathbf{E}_{\beta,\beta}
 (|\lambda|)}{|\Delta|}
 \sum_{\sigma=1}^m(\log\frac{e}{t_\sigma})^{\beta-2}\\
&\quad +\Big(\Gamma(\beta-1)\mathbf{E}_{\beta,\beta-1}(|\lambda|)
 (|\frac{1}{\Gamma(\beta)}-\mathbf{E}_{\beta,\beta}(\lambda)
 |\mathbf{E}_{\beta,\beta-1}(|\lambda|)
 \sum_{\sigma=1}^m(\log\frac{e}{t_\sigma})^{\beta-2} \\
&\quad  +m|\lambda| \mathbf{E}_{\beta,\beta}(\lambda)\mathbf{E}_{\beta,\beta}
(|\lambda|) )\Big)/ |\Delta|
+m\mathbf{E}_{\beta,\beta-1}(|\lambda|)\Gamma(\beta-1),
\end{aligned}\\
\begin{aligned}
B_{5f}
&=\frac{|\frac{1}{\Gamma(\beta-1)}-\mathbf{E}_{\beta,\beta-1}
(\lambda)|\big[|\lambda\|\mathbf{E}_{\beta,1}(|\lambda|)
 \mathbf{E}_{\beta,\beta}(|\lambda|)+
\mathbf{E}_{\beta,\beta}(|\lambda|)\mathbf{E}_{\beta,\beta-1}(|\lambda|)\big]}
{|\Delta|} \\
&\quad\times \mathbf{B}(l+1,k+1)
 +\frac{|\lambda|\mathbf{E}_{\beta,\beta}(\lambda)
[\mathbf{E}_{\beta,\beta}(|\lambda|)^2+\mathbf{E}_{\beta,1}
(|\lambda|)\mathbf{E}_{\beta,\beta-1}(|\lambda|)]}{|\Delta|} \\
&\quad\times \mathbf{B}(l+1,k+1)
+\mathbf{E}_{\beta,\beta}(|\lambda|)\mathbf{B}(\beta+l,k+1).
\end{aligned}
\end{gather*}
\end{theorem}


\begin{proof}
From Lemmas \ref{lem3.5.1} and \ref{lem3.5.2}, and the definition of $R$,
$x\in LP_mC_{2-\beta}(1,e]$ is a solution of BVP\eqref{e1.0.9} if and only if
 $x\in LP_mC_{2-\beta}(1,e]$ is a fixed point of $R$.
 Lemma \ref{lem3.4.2} implies that $R$ is a completely continuous operator.

From (4.A3), for $x\in LP_mC_{2-\beta}(1,e]$ we have
\begin{gather*}
\begin{aligned}
|f(t,x(t))|
&=\Big|f\big(t,(\log\frac{t}{t_i})^{\beta-2}(\log\frac{t}{t_i})^{2-\beta}x(t)
 \big)\Big|\\
&\le M_f\Big(|(\log\frac{t}{t_i})^{2-\beta}x(t)|\Big) \\
&\le M_f(\|x\|),\quad t\in (t_i,t_{i+1}],\;i\in \mathbb{N}[0,m],
\end{aligned} \\
|G(t,x(t))|\le M_G(\|x\|),\quad t\in (t_i,t_{i+1}],\;i\in \mathbb{N}[0,m],\\
|H(t,x(t))|\le M_H(\|x\|),\quad t\in (t_i,t_{i+1}],\;i\in \mathbb{N}[0,m],\\
\begin{aligned}
|I(t_i,x(t_i))| & =|I(t_i,(\log\frac{t_i}{t_{i-1}})^{\beta-2}
(\log\frac{t_i}{t_{i-1}})^{2-\beta}x(t))|\\
&\le M_f(|(\log\frac{t_i}{t_{i-1}})^{2-\beta}x(t)|)\le M_G(\|x\|),i\in \mathbb{N}[1,m],
\end{aligned} \\
|I(t_i,x(t_i))|\le M_H(\|x\|),i\in \mathbb{N}[1,m].
\end{gather*}
We consider the set
$\Omega=\{x\in LP_mC_{2-\beta}(1,e]:x=\lambda (Rx),\text{ for some }
\lambda \in [0,1]\}$. For $x\in \Omega$ and $t\in (t_i,t_{i+1}]$ we have
\begin{align*}
&(\log\frac{t}{t_i})^{2-\beta}|(Rx)(t)| \\
&\le \mathbf{E}_{\beta,\beta}(|\lambda|)
\Big[\frac{|\lambda|\mathbf{E}_{\beta,\beta}(\lambda)}{|\Delta|}
\|\phi\|_1M_G(\|x\|)+\frac{|\frac{1}{\Gamma(\beta-1)}
 -\mathbf{E}_{\beta,\beta-1}(\lambda)|}{|\Delta|}\|\psi\|_1M_H(\|x) \\
&\quad +\Gamma(\beta)\sum_{\sigma=1}^m(\frac{|\frac{1}{\Gamma(\beta-1)}
 -\mathbf{E}_{\beta,\beta-1}(\lambda)|}{|\Delta|}\mathbf{E}_{\beta,\beta}
 (|\lambda|)+\frac{\lambda\mathbf{E}_{\beta,\beta}(\lambda)}{|\Delta|}
 \mathbf{E}_{\beta,\beta}(|\lambda|))M_J(\|x\|) \\
&\quad +\Gamma(\beta-1)\sum_{\sigma=1}^m
\Big(|\lambda|\frac{|\frac{1}{\Gamma(\beta-1)}-\mathbf{E}_{\beta,\beta-1}
(\lambda)|}{|\Delta|}\mathbf{E}_{\beta,\beta}(|\lambda|) \\
&\quad +\frac{\lambda\mathbf{E}_{\beta,\beta}(\lambda)}{|\Delta|}
(\log\frac{e}{t_\sigma})^{\beta-2}\mathbf{E}_{\beta,\beta-1}
 (|\lambda|)\Big)M_I(\|x\|)\\
&\quad +\Big(|\lambda|\frac{|\frac{1}{\Gamma(\beta-1)}
 -\mathbf{E}_{\beta,\beta-1}(\lambda)|}{|\Delta|}\mathbf{E}_{\beta,1}(|\lambda|)
+\frac{|\lambda|\mathbf{E}_{\beta,\beta}(\lambda)}{|\Delta|}
 \mathbf{E}_{\beta,\beta}(|\lambda|)\Big) \\
&\quad\times \int_1^e (\log s)^k(1-\log s)^l\frac{ds}{s}M_f(\|x\|)\Big]\\
&\quad +\mathbf{E}_{\beta,\beta-1}(|\lambda|)
\Big[\frac{|\frac{1}{\Gamma(\beta)}-\mathbf{E}_{\beta,\beta}(\lambda)|}
{|\Delta|}\|\phi\|_1M_G(\|x\|)+\frac{\mathbf{E}_{\beta,\beta}(\lambda)}
{|\Delta|}\|\psi\|_1M_H(\|x\|)
\\
&\quad +\Gamma(\beta)\sum_{\sigma=1}^m\Big(\frac{|\frac{1}{\Gamma(\beta)}
-\mathbf{E}_{\beta,\beta}(\lambda)|}{|\Delta|}\mathbf{E}_{\beta,\beta}
 (|\lambda|)+\frac{\mathbf{E}_{\beta,\beta}(\lambda)}{|\Delta|}
 \mathbf{E}_{\beta,\beta}(|\lambda|)\Big)M_J(\|x\|)\\
&\quad +\Gamma(\beta-1)\sum_{\sigma=1}^m\Big(\frac{|\frac{1}{\Gamma(\beta)}
 -\mathbf{E}_{\beta,\beta}(\lambda)|}{|\Delta|}(\log\frac{e}{t_\sigma})^{\beta-2}
 \mathbf{E}_{\beta,\beta-1}(|\lambda|) \\
&\quad +\frac{|\lambda|\mathbf{E}_{\beta,\beta}
 (\lambda)}{|\Delta|}\mathbf{E}_{\beta,\beta}(|\lambda|)\Big)M_I(\|x\|)\\
&\quad +\Big(\frac{|\frac{1}{\Gamma(\beta)}-\mathbf{E}_{\beta,\beta}(\lambda)|}{|\Delta|}
 \mathbf{E}_{\beta,\beta}(|\lambda|)+|\lambda|\frac{\mathbf{E}_{\beta,\beta}
(\lambda)}{|\Delta|}\mathbf{E}_{\beta,1}(|\lambda|)\Big)
\\
&\quad \times \int_1^e(\log s)^k (1-\log s)^l\frac{ds}{s}M_f(\|x\|)\Big]\\
&\quad +\sum_{\sigma=1}^i
\big[\mathbf{E}_{\beta,\beta}(|\lambda|)\Gamma(\beta)M_J(\|x\|)
+\mathbf{E}_{\beta,\beta-1}(|\lambda|)\Gamma(\beta-1)M_I(\|x\|)\big]\\
&\quad +(\log\frac{t}{t_i})^{2-\beta}\sum_{\tau=0}^{+\infty}
\frac{\lambda^\tau}{\Gamma((\tau+1)\beta)}\int_1^t
(\log\frac{t}{s})^{\tau\beta+\beta+l-1}(\log s)^k\frac{ds}{s}M_f(\|x\|)\\
&\le \frac{|\lambda|\mathbf{E}_{\beta,\beta}(\lambda)
\mathbf{E}_{\beta,\beta}(|\lambda|)\|\phi\|_1}{|\Delta|}M_G(\|x\|)
+\frac{|\frac{1}{\Gamma(\beta-1)}-\mathbf{E}_{\beta,\beta-1}(\lambda)
|\mathbf{E}_{\beta,\beta}(|\lambda|)\|\psi\|_1}{|\Delta|}M_H(\|x))\\
&\quad +m\Gamma(\beta)\Big(\frac{|\frac{1}{\Gamma(\beta-1)}
-\mathbf{E}_{\beta,\beta-1}(\lambda)|\mathbf{E}_{\beta,\beta}
(|\lambda|)^2}{|\Delta|}+\frac{\lambda\mathbf{E}_{\beta,\beta}(\lambda)
\mathbf{E}_{\beta,\beta}(|\lambda|)^2}{|\Delta|}\Big)M_J(\|x\|)\\
&\quad +\Gamma(\beta-1)\Big(\frac{m|\lambda\|\frac{1}{\Gamma(\beta-1)}
-\mathbf{E}_{\beta,\beta-1}(\lambda)|\mathbf{E}_{\beta,\beta}
(|\lambda|)^2}{|\Delta|} \\
&\quad +\frac{\lambda\mathbf{E}_{\beta,\beta}
(\lambda)\mathbf{E}_{\beta,\beta-1}(|\lambda|)\mathbf{E}_{\beta,\beta}
(|\lambda|)}{|\Delta|}\sum_{\sigma=1}^m(\log\frac{e}{t_\sigma})^{\beta-2}
\Big)M_I(\|x\|)\\
&\quad +\Big(\frac{|\lambda\|\frac{1}{\Gamma(\beta-1)}-\mathbf{E}_{\beta,\beta-1}
(\lambda)|\mathbf{E}_{\beta,1}(|\lambda|)\mathbf{E}_{\beta,\beta}
(|\lambda|)}{|\Delta|} \\
&\quad +\frac{|\lambda|\mathbf{E}_{\beta,\beta}(\lambda)\mathbf{E}_{\beta,\beta}
(|\lambda|)^2}{|\Delta|}\Big)\mathbf{B}(l+1,k+1)M_f(\|x\|)\\
&\quad +\frac{|\frac{1}{\Gamma(\beta)}-\mathbf{E}_{\beta,\beta}(\lambda)|
\mathbf{E}_{\beta,\beta-1}(|\lambda|)\|\phi\|_1}{|\Delta|}M_G(\|x\|)
+\frac{\mathbf{E}_{\beta,\beta}(\lambda)\mathbf{E}_{\beta,\beta-1}
(|\lambda|)\|\psi\|_1}{|\Delta|}M_H(\|x\|)\\
&\quad +m\Gamma(\beta)\Big(\frac{|\frac{1}{\Gamma(\beta)}-\mathbf{E}_{\beta,\beta}
(\lambda)|\mathbf{E}_{\beta,\beta}(|\lambda|)\mathbf{E}_{\beta,\beta-1}
(|\lambda|)}{|\Delta|} \\
&\quad +\frac{\mathbf{E}_{\beta,\beta}(\lambda)
\mathbf{E}_{\beta,\beta}(|\lambda|)\mathbf{E}_{\beta,\beta-1}
(|\lambda|)}{|\Delta|}\Big)M_J(\|x\|)\\
&\quad +\Gamma(\beta-1)\Big(\frac{|\frac{1}{\Gamma(\beta)}-\mathbf{E}_{\beta,\beta}
(\lambda)|\mathbf{E}_{\beta,\beta-1}(|\lambda|)^2}{|\Delta|}\sum_{\sigma=1}^m
(\log\frac{e}{t_\sigma})^{\beta-2}\\
&\quad +m\frac{|\lambda|\mathbf{E}_{\beta,\beta}
(\lambda)\mathbf{E}_{\beta,\beta}(|\lambda|)\mathbf{E}_{\beta,\beta-1}
(|\lambda|)}{|\Delta|}\Big)M_I(\|x\|)\\
&\quad +\Big(\frac{|\frac{1}{\Gamma(\beta)}-\mathbf{E}_{\beta,\beta}(\lambda)|
\mathbf{E}_{\beta,\beta}(|\lambda|)\mathbf{E}_{\beta,\beta-1}(|\lambda|)}
{|\Delta|}+\frac{|\lambda|\mathbf{E}_{\beta,\beta}(\lambda)
\mathbf{E}_{\beta,1}(|\lambda|)\mathbf{E}_{\beta,\beta-1}(|\lambda|)}
{|\Delta|}\Big) \\
&\quad\times \mathbf{B}(l+1,k+1)M_f(\|x\|)
 +m\mathbf{E}_{\beta,\beta}(|\lambda|)\Gamma(\beta)M_J(\|x\|)\\
&\quad +m\mathbf{E}_{\beta,\beta-1}(|\lambda|)\Gamma(\beta-1)M_I(\|x\|)
 +\mathbf{E}_{\beta,\beta}(|\lambda|)\mathbf{B}(\beta+l,k+1)M_f(\|x\|) \\
&=\Big[\frac{|\lambda|\mathbf{E}_{\beta,\beta}(\lambda)
\mathbf{E}_{\beta,\beta}(|\lambda|)\|\phi\|_1}{|\Delta|}
+\frac{|\frac{1}{\Gamma(\beta)}-\mathbf{E}_{\beta,\beta}(\lambda)|
\mathbf{E}_{\beta,\beta-1}(|\lambda|)\|\phi\|_1}{|\Delta|}\Big]M_G(\|x\|)\\
&\quad +\Big[\frac{\mathbf{E}_{\beta,\beta}(\lambda)\mathbf{E}_{\beta,\beta-1}
(|\lambda|)\|\psi\|_1}{|\Delta|}+\frac{|\frac{1}{\Gamma(\beta-1)}
-\mathbf{E}_{\beta,\beta-1}(\lambda)|\mathbf{E}_{\beta,\beta}(|\lambda|)
\|\psi\|_1}{|\Delta|}\Big]M_H(\|x))\\
&\quad +\Big[\frac{m\Gamma(\beta)|\frac{1}{\Gamma(\beta-1)}
-\mathbf{E}_{\beta,\beta-1}(\lambda)|\mathbf{E}_{\beta,\beta}
(|\lambda|)^2}{|\Delta|}+\frac{m|\lambda|\Gamma(\beta)
\mathbf{E}_{\beta,\beta}(\lambda)\mathbf{E}_{\beta,\beta}(|\lambda|)^2}{|\Delta|} \\
&\quad +\frac{m\Gamma(\beta)|\frac{1}{\Gamma(\beta)}
 -\mathbf{E}_{\beta,\beta}(\lambda)
|\mathbf{E}_{\beta,\beta}(|\lambda|)\mathbf{E}_{\beta,\beta-1}(|\lambda|)}{|\Delta|}\\
&\quad +\frac{m\Gamma(\beta)\mathbf{E}_{\beta,\beta}(\lambda)
\mathbf{E}_{\beta,\beta}(|\lambda|)\mathbf{E}_{\beta,\beta-1}(|\lambda|)}
{|\Delta|}
 +m\mathbf{E}_{\beta,\beta}(|\lambda|)\Gamma(\beta)\Big]M_J(\|x\|)\\
&\quad +\Big[\frac{m|\lambda|\Gamma(\beta-1)|\frac{1}{\Gamma(\beta-1)}
-\mathbf{E}_{\beta,\beta-1}(\lambda)|\mathbf{E}_{\beta,\beta}(|\lambda|)^2}
{|\Delta|} \\
&\quad +\frac{|\lambda|\Gamma(\beta-1)\mathbf{E}_{\beta,\beta}
(\lambda)\mathbf{E}_{\beta,\beta-1}(|\lambda|)
\mathbf{E}_{\beta,\beta}(|\lambda|)}{|\Delta|}\sum_{\sigma=1}^m
(\log\frac{e}{t_\sigma})^{\beta-2} \\
&\quad +\Big(\Gamma(\beta-1)\mathbf{E}_{\beta,\beta-1}(|\lambda|)
(|\frac{1}{\Gamma(\beta)}-\mathbf{E}_{\beta,\beta}(\lambda)
|\mathbf{E}_{\beta,\beta-1}(|\lambda|)\sum_{\sigma=1}^m
(\log\frac{e}{t_\sigma})^{\beta-2} \\
&\quad +m|\lambda|\mathbf{E}_{\beta,\beta}(\lambda)\mathbf{E}_{\beta,\beta}(|\lambda|)
)\Big)/|\Delta|
 +m\mathbf{E}_{\beta,\beta-1}(|\lambda|)\Gamma(\beta-1)\Big]M_I(\|x\|)\\
&\quad +\Big[\frac{|\frac{1}{\Gamma(\beta-1)}-\mathbf{E}_{\beta,\beta-1}(\lambda)|
\big[|\lambda\|\mathbf{E}_{\beta,1}(|\lambda|)\mathbf{E}_{\beta,\beta}(|\lambda|)+
\mathbf{E}_{\beta,\beta}(|\lambda|)\mathbf{E}_{\beta,\beta-1}(|\lambda|)\big]}
{|\Delta|} \\
&\quad\times \mathbf{B}(l+1,k+1)
 +\frac{|\lambda|\mathbf{E}_{\beta,\beta}(\lambda)
[\mathbf{E}_{\beta,\beta}(|\lambda|)^2+\mathbf{E}_{\beta,1}(|\lambda|)
\mathbf{E}_{\beta,\beta-1}(|\lambda|)]}{|\Delta|}
\mathbf{B}(l+1,k+1) \\
&\quad +\mathbf{E}_{\beta,\beta}(|\lambda|)\mathbf{B}
(\beta+l,k+1)\Big]M_f(\|x\|)\\
&=B_{1G}M_G(\|x\|)+B_{2H}M_H(\|x\|)+B_{3J}M_J(\|x\|)
+B_{4I}M_I(\|x\|)+B_{5f}M_f(\|x\|).
\end{align*}
It follows that
\begin{align*}
\|x\|&=\lambda \|Rx\|\le \|Rx\| \\
&\le B_{1G}M_G(\|x\|)+B_{2H}M_H(\|x\|)+B_{3J}M_J(\|x\|)
 +B_{4I}M_I(\|x\|)+B_{5f}M_f(\|x\|).
\end{align*}
From \eqref{e4.0.3}, we choose $ \Omega=\{x\in LP_mC_{2-\beta}(1,e]:\|x\|\le r_0\}$.
For $x\in \partial \Omega$, we obtain $x\neq \lambda (Rx)$ for any
$\lambda\in [0,1]$. In fact, if there exists $x\in \partial \Omega$ such that
$x=\lambda (Rx)$ for some $\lambda\in [0,1]$.
Then $r_0=\|x\|=\lambda\|Rx\|\le \|Rx\|
\le B_{1G}M_G(r_0)+B_{2H}M_H(r_0)+B_{3J}M_J(r_0)+B_{4I}M_I(r_0)+B_{5f}M_f(r_0)<r_0$,
 which is a contradiction.

 As a consequence of Schaefer's fixed point theorem, we deduce that $R$ has a
fixed point which is a solution of problem \eqref{e1.0.9}.
 The proof is complete.
\end{proof}


\begin{theorem} \label{thm4.4}
 Suppose that {\rm (1.A7), (1.A12)--(1.A14)} hold, $\lambda\neq 0$, and
\begin{itemize}
\item[(4.A4)] there exist nondecreasing functions
$M_f,M_g,M_h,M_I,M_J:[0,+\infty)\to [0,+\infty)$ such that
\begin{gather*}
|f(t,x)|\le M_f(|x|),\quad t\in (t_i,t_{i+1}],\;i\in \mathbb{N}[0,m],x\in \mathbb{R},\\
|G(t,x)|\le M_f(|x|),\quad t\in (t_i,t_{i+1}],\;i\in \mathbb{N}[0,m],x\in \mathbb{R},\\
|H(t,x)|\le M_f(|x|),\quad t\in (t_i,t_{i+1}],\;i\in \mathbb{N}[0,m],x\in \mathbb{R},\\
|I(t_i,x)|\le M_I(|x|),\quad i\in \mathbb{N}[1,m],\; x\in \mathbb{R},\\
|J(t_i,x)|\le M_J(|x|),\quad i\in \mathbb{N}[1,m],\; x\in \mathbb{R}.
\end{gather*}
\end{itemize}
Then BVP \eqref{e1.0.10} has at least one solution if there exists a constant
 $r_0>0$ such that
\begin{equation}
\begin{aligned}
&\frac{\mathbf{E}_{\beta,1}(|\lambda|)\|\psi\|_1}
 {\lambda\mathbf{E}_{\beta,\beta}(\lambda)}M_H(r_0)
+\big[\mathbf{E}_{\beta,2}(|\lambda|)+\frac{\mathbf{E}_{\beta,1}
 (|\lambda|)}{|\lambda|\mathbf{E}_{\beta,\beta}(\lambda)}
\mathbf{E}_{\beta,1}(|\lambda |)\big]\|\phi\|_1M_G(r_0)\\
&+\big[\frac{\mathbf{E}_{\beta,1}(|\lambda|)}
 {|\lambda|\mathbf{E}_{\beta,\beta}(\lambda)}m|\lambda|
 \mathbf{E}_{\beta,\beta}(|\lambda|)+
m\mathbf{E}_{\beta,1}(|\lambda|)\big]M_I(r_0)
+[m\mathbf{E}_{\beta,1}(|\lambda|) \\
&+m\mathbf{E}_{\beta,2}(|\lambda|)]M_J(r_0)
 +\big[\frac{\mathbf{E}_{\beta,1}(|\lambda|)}
{|\lambda|\mathbf{E}_{\beta,\beta}(\lambda)}\mathbf{E}_{\beta,\beta}
(|\lambda|)\mathbf{B}(\beta+l-1,k+1) \\
&+\mathbf{E}_{\beta,\beta}(|\lambda|)\mathbf{B}(\beta+l,k+1)\big]M_f(r_0)
<r_0.
\end{aligned} \label{e4.0.4}
\end{equation}
\end{theorem}


\begin{proof}
From Lemmas \ref{lem3.6.1} and \ref{lem3.6.2}, and the definition of $J$,
it follows that  $x\in LP_mC(1,e]$
is a solution of \eqref{e1.0.10} if and only if $x\in LP_mC(1,e]$ is a fixed
point of $R$. Lemma \ref{lem3.6.2} implies that $J$ is a completely continuous operator.

From (4.A4), for $x\in LP_mC(1,e]$ we have
\begin{gather*}
|f(t,x(t))|\le M_f(\|x\|),\quad t\in (t_i,t_{i+1}],\;i\in \mathbb{N}[0,m],\\
|G(t,x(t))|\le M_f(\|x\|),\quad t\in (t_i,t_{i+1}],\;i\in \mathbb{N}[0,m],\\
|H(t,x(t))|\le M_f(\|x\|),\quad t\in (t_i,t_{i+1}],\;i\in \mathbb{N}[0,m],\\
|I(t_i,x(t_i))|\le M_I(\|x\|),\quad i\in \mathbb{N}[1,m],\\
|J(t_i,x(t_i))|\le M_J(\|x\|),\quad i\in \mathbb{N}[1,m].
\end{gather*}
We consider the set
$\Omega=\{x\in LP_mC(1,e]:x=\lambda (Jx),\text{ for some }\lambda \in [0,1]\}$.
 For $x\in \Omega$ and $t\in (t_i,t_{i+1}]$, we have
\begin{align*}
&|(Jx)(t)| \\
&\le \frac{\mathbf{E}_{\beta,1}(|\lambda|)\|\psi\|_1}
{\lambda\mathbf{E}_{\beta,\beta}(\lambda)}M_H(\|x\|)
+\big[\mathbf{E}_{\beta,2}(|\lambda|)
+\frac{\mathbf{E}_{\beta,1}(|\lambda|)}{|\lambda|\mathbf{E}_{\beta,\beta}(\lambda)}
\mathbf{E}_{\beta,1}(|\lambda |)\big]\|\phi\|_1M_G(\|x\|)\\
&\quad +\big[\frac{\mathbf{E}_{\beta,1}(|\lambda|)}
{|\lambda|\mathbf{E}_{\beta,\beta}(\lambda)}m|\lambda|
 \mathbf{E}_{\beta,\beta}(|\lambda|)+
m\mathbf{E}_{\beta,1}(|\lambda|)\big]M_I(\|x\|) \\
&\quad +[m\mathbf{E}_{\beta,1}(|\lambda|)+m\mathbf{E}_{\beta,2}
 (|\lambda|)]M_J(\|x\|)
 +\Big[\frac{\mathbf{E}_{\beta,1}(|\lambda|)}
{|\lambda|\mathbf{E}_{\beta,\beta}(\lambda)}
\mathbf{E}_{\beta,\beta}(|\lambda|)\mathbf{B}(\beta+l-1,k+1) \\
&\quad +\mathbf{E}_{\beta,\beta}(|\lambda|)\mathbf{B}(\beta+l,k+1)\Big]M_f(\|x\|).
\end{align*}
It follows that
\begin{align*}
\|x\|
&=\lambda \|Rx\|\le \|Rx\| \\
&\le \frac{\mathbf{E}_{\beta,1}(|\lambda|)\|\psi\|_1}
 {\lambda\mathbf{E}_{\beta,\beta}(\lambda)}M_H(\|x\|)
 +\big[\mathbf{E}_{\beta,2}(|\lambda|)
 +\frac{\mathbf{E}_{\beta,1}(|\lambda|)}
 {|\lambda|\mathbf{E}_{\beta,\beta}(\lambda)}
\mathbf{E}_{\beta,1}(|\lambda |)\big]\|\phi\|_1M_G(\|x\|)\\
&\quad +\big[\frac{\mathbf{E}_{\beta,1}(|\lambda|)}
{|\lambda|\mathbf{E}_{\beta,\beta}(\lambda)}m|\lambda|
 \mathbf{E}_{\beta,\beta}(|\lambda|)
+ m\mathbf{E}_{\beta,1}(|\lambda|)\big]M_I(\|x\|) \\
&\quad +[m\mathbf{E}_{\beta,1}(|\lambda|)+m\mathbf{E}_{\beta,2}(|\lambda|)]M_J(\|x\|)
 +\Big[\frac{\mathbf{E}_{\beta,1}(|\lambda|)}
 {|\lambda|\mathbf{E}_{\beta,\beta}(\lambda)}
 \mathbf{E}_{\beta,\beta}(|\lambda|) \\
&\quad\times \mathbf{B}(\beta+l-1,k+1)
 +\mathbf{E}_{\beta,\beta}(|\lambda|)\mathbf{B}(\beta+l,k+1)\Big]M_f(\|x\|).
\end{align*}
From \eqref{e4.0.4}, we choose $ \Omega=\{x\in LP_mC(1,e]:\|x\|\le r_0\}$.
For $x\in \partial \Omega$, we obtain $x\neq \lambda (Jx)$ for any
 $\lambda\in [0,1]$. In fact, if there exists $x\in \partial \Omega$
such that $x=\lambda (Jx)$ for some $\lambda\in [0,1]$. Then
\begin{align*}
r_0&=\|x\|=\lambda\|Jx\|\le \|Jx\| \\
&\le \frac{\mathbf{E}_{\beta,1}(|\lambda|)\|\psi\|_1}
 {\lambda\mathbf{E}_{\beta,\beta}(\lambda)}M_H(r_0)
+[\mathbf{E}_{\beta,2}(|\lambda|)
 +\frac{\mathbf{E}_{\beta,1}(|\lambda|)}{|\lambda|\mathbf{E}_{\beta,\beta}(\lambda)}
\mathbf{E}_{\beta,1}(|\lambda |)]\|\phi\|_1M_G(r_0) \\
&\quad +\big[\frac{\mathbf{E}_{\beta,1}(|\lambda|)}
 {|\lambda|\mathbf{E}_{\beta,\beta}(\lambda)}m|\lambda|
 \mathbf{E}_{\beta,\beta}(|\lambda|)
 + m\mathbf{E}_{\beta,1}(|\lambda|)\big]M_I(r_0)
+[m\mathbf{E}_{\beta,1}(|\lambda|) \\
&\quad +m\mathbf{E}_{\beta,2}(|\lambda|)]M_J(r_0)
+\big[\frac{\mathbf{E}_{\beta,1}(|\lambda|)}
{|\lambda|\mathbf{E}_{\beta,\beta}(\lambda)}\mathbf{E}_{\beta,\beta}
(|\lambda|)\mathbf{B}(\beta+l-1,k+1) \\
&\quad +\mathbf{E}_{\beta,\beta}(|\lambda|)\mathbf{B}(\beta+l,k+1)\big]M_f(r_0)<r_0,
\end{align*}
 which is a contradiction.

As a consequence of Schaefer's fixed point theorem, we deduce that $J$ has
a fixed point which is a solution of \eqref{e1.0.10}. The proof of is
complete.
\end{proof}

\section{Applications of main results}

In this section, we present firstly applications of the results obtained in
Section 3.2. We also point out some mistakes occurred in cited papers.
Finally we establish sufficient conditions for the existence of solutions
of three classes of boundary value problems of impulsive fractional
differential equations.

Applying the results obtained in Section 3.2, choose $\lambda=0$, by
Theorems \ref{thm3.2.1}--\ref{thm3.2.4}, we obtain the exact piecewise
continuous solutions of the following fractional differential equations
 (see Corollaries \ref{coro5.1}, \ref{coro5.2},  \ref{coro5.3} and \ref{coro5.4} 
below)
\begin{gather}
{}^{C}D_{0^+}^{\alpha}{x}(t)={F}(t),\quad t\in (t_i,t_{i+1}],\;i\in \mathbb{N}[0,m],
\label{e5.0.1} \\
{}^{RL}D_{0^+}^{\alpha}{x}(t)={F}(t),\quad t\in (t_i,t_{i+1}],\;i\in \mathbb{N}[0,m],
\label{e5.0.2} \\
{}^{RLH}D_{0^+}^{\alpha}{x}(t)={G}(t),\quad t\in (s_i,s_{i+1}],\; i\in \mathbb{N}[0,m],
\label{e5.0.3} \\
{}^{CH}D_{0^+}^{\alpha}{x}(t)={G}(t),\quad t\in (s_i,s_{i+1}],\; i\in \mathbb{N}[0,m],
\label{e5.0.4}
\end{gather}
where $n-1\le \alpha<n$, $0=t_0<t_1<\dots<t_m<t_{m+1}=1$ in \eqref{e5.0.1}
and \eqref{e5.0.2} and $1=t_0<t_1<\dots<t_m<t_{m+1}=e$ in \eqref{e5.0.3}
and \eqref{e5.0.4}.

\begin{corollary} \label{coro5.1}
Suppose that $F$ is continuous on $(0,1)$ and there exist constants $k>-\alpha+n-1$
and $l\in (-\alpha,-\alpha-k,0]$ such that $|F(t)|\le t^k(1-t)^l$ for all
$t\in (0,1)$. Then $x$ is a piecewise solution of \eqref{e5.0.1} if and only
if there exist constants $c_{iv}(i\in N[0,m],v\in \mathbb{N}[0,n-1])\in \mathbb{R}$ such that
\begin{equation}
x(t)=\sum_{\sigma=0}^i\sum_{v=0}^{n-1}\frac{c_{\sigma v}}
{\Gamma(v+1)}(t-t_\sigma)^v
+\frac{1}{\Gamma(\alpha)}\int_0^t(t-s)^{\alpha-1}F(s)ds,
\label{e5.0.5}
\end{equation}
for $t\in (t_i,t_{i+1}]$ and $i\in \mathbb{N}[0,m]$.
\end{corollary}

The above corollary follows from Theorem \ref{thm3.2.1}, with $\lambda=0$.

\begin{remark} \label{rmk5.1}\rm 
We note that $(t-t_\sigma)^v=\sum_{\tau=0}^v(-1)^{\tau}
\binom{v}{\tau}t_\sigma^{\tau}t^{v-\tau}$.
By Corollary \ref{coro5.1}, we can transform \eqref{e5.0.5} into
\begin{equation}
x(t)=\sum_{v=0}^{n-1}\frac{d_{i v}}{\Gamma(v+1)}t^v
+\frac{1}{\Gamma(\alpha)}\int_0^t(t-s)^{\alpha-1}F(s)ds,
\label{e5.0.6}
\end{equation}
for $t\in (t_i,t_{i+1}]$ and $i\in \mathbb{N}[0,m]$.
Here $d_{i v}$ are constants. Certainly \eqref{e5.0.6} can be transformed
into
\begin{equation}
x(t)=\sum_{v=0}^{n-1}\frac{e_{i v}}{\Gamma(v+1)}(t-t_i)^v
+\frac{1}{\Gamma(\alpha)}\int_0^t(t-s)^{\alpha-1}F(s)ds,
\label{e5.0.7}
\end{equation}
for $t\in (t_i,t_{i+1}]$ and $i\in \mathbb{N}[0,m]$.
\end{remark}

%\begin{remark} \label{rmk5.2}\rm
 In \cite{ll1}, the authors actually used \eqref{e5.0.6} to construct the
nonlinear operator for getting solutions of \eqref{e1.0.5}, but they did
 not prove the equivalence between the boundary value problem and the integral
equation, see \cite[Lemma 2.6]{ll1}.
%\end{remark}

%\begin{remark} \label{rmk5.3}\rm
Rehman and Eloe \cite{re} proved that
\eqref{e5.0.1} is equivalent to \eqref{e5.0.6} under the assumption that
$F\in PC[0,1]$. So Corollary generalizes \cite[Lemma 2.4]{re}.
Furthermore, it is difficult to convert a BVP for impulsive fractional
differential equation to a integral equation by using \eqref{e5.0.6},
 while it is easy to do this job by using \eqref{e5.0.5}. See the examples
in Section 6.
%\end{remark}


\begin{corollary} \label{coro5.2}
Suppose that $F$ is continuous on $(0,1)$ and there exist constants $k>-1$
and $l\in (-\alpha,-n-k,0]$ such that $|F(t)|\le t^k(1-t)^l$ for all $t\in (0,1)$.
Then $x$ is a solution of \eqref{e5.0.2} if and only if there exist constants
$c_{\sigma v}(\sigma\in \mathbb{N}[0,m],v\in \mathbb{N}[1,n])\in \mathbb{R}$ such that
\begin{equation}
x(t)=\sum_{\sigma=0}^i\sum_{v=1}^{n}\frac{c_{\sigma v}}{\Gamma(\alpha-v+1)}
(t-t_\sigma)^{\alpha-v} +\frac{1}{\Gamma(\alpha)}\int_0^t(t-s)^{\alpha-1}F(s)ds,
\label{e5.0.8}
\end{equation}
for $t\in (t_i,t_{i+1}]$, $i\in \mathbb{N}[0,m]$.
\end{corollary}

\begin{remark} \label{rmk5.4}\rm
 When $\alpha$ is not an integer, \eqref{e5.0.8} is not equivalent to the
 equation
$$
x(t)=\sum_{v=1}^{n}\frac{c_{i v}}{\Gamma(\alpha-v+1)}(t-t_i)^{\alpha-v}
+\frac{1}{\Gamma(\alpha)}\int_0^t(t-s)^{\alpha-1}F(s)ds,
$$
for $t\in (t_i,t_{i+1}]$, $i\in \mathbb{N}[0,m]$.
\end{remark}

\begin{corollary} \label{coro5.3}
Suppose that $G$ is continuous on $(1,e)$ and there exist constants $k>-1$
and $l\in (-\alpha,-n-k,0]$ such that $|G(t)|\le (\log t)^k(1-\log t)^l$
for all $t\in (1,e)$. Then $x$ is a solution of \eqref{e5.0.3} if and only
if there exist constants $c_{jv}\in \mathbb{R}(j\in \mathbb{N}[0,m],v\in \mathbb{N}[1,n])$ such that
\begin{equation}
x(t)=\sum_{j=0}^i\sum_{v=1}^{n}\frac{c_{jv}}{\Gamma(\alpha-v+1)}
(\log\frac{ t}{t_j})^{\alpha-v}
+\frac{1}{\Gamma(\alpha)}\int_1^t(\log \frac{t}{s})^{\alpha-1}G(s)
\frac{ds}{s},\label{e5.0.9}
\end{equation}
for $t\in (t_i,t_{i+1}]$, $i\in \mathbb{N}[0,m]$.
\end{corollary}

%\begin{remark} \label{rmk5.5} \rm
In \cite{7}, the same result was proved under the assumption that $G\in L(1,e)$.
So Corollary \ref{coro5.3} improve the results in \cite{7}.
%\end{remark}

\begin{corollary} \label{coro5.4}
Suppose that $G$ is continuous on $(1,e)$ and there exist constants
$k>-\alpha+n-1$ and $l\in (-\alpha,-\alpha+k,0]$ such that
$|G(t)|\le (\log t)^k(1-\log t)^l$ for all $t\in (1,e)$.
Then $x$ is a piecewise solution of \eqref{e5.0.4} if and only if there exist
constants $c_{jv}\in \mathbb{R}(j\in \mathbb{N}[0,m],i\in \mathbb{N}[0,n-1])$ such that
\begin{equation}
x(t)=\sum_{\rho=0}^j\sum_{v=0}^{n-1}\frac{c_{\rho v}}{\Gamma(v+1)}
(\log\frac{t}{t_\rho})^v
+\frac{1}{\Gamma(\alpha)}\int_1^t(\log\frac{t}{s})^{\alpha-1}
G(s)\frac{ds}{s},
\label{e5.0.10}
\end{equation}
for $t\in (t_j,t_{j+1}]$, $j\in \mathbb{N}[0,m]$.
\end{corollary}

We now construct a Banach space $X$ and prove the compact criterion for
subsets of $X$. $X$ will be used in next three subsections. Choose the set
of functions
\begin{align*}
X=\Big\{&x: x\big|_{(t_i,t_{i+1}]},\;D_{0^+}^\beta x|_{(t_i,t_{i+1}]}
\text{ are continuous, }i\in \mathbb{N}[0,m],\;
\lim_{t\to t_i^+}(t-t_i)^{2-\alpha}x(t),\\
 &\lim_{t\to t_i^+}(t-t_i)^{2+\beta-\alpha}D_{0^+}^\beta x(t)\text{ exist, }
i\in \mathbb{N}[0,m] \Big\}
\end{align*}
For $x\in X$ define the norm
\[
\|x\|=\max\big\{\sup_{t\in (t_i,t_{i+1}]}(t-t_i)^{2-\alpha}|x(t)|,\;
\sup_{t\in (t_i,t_{i+1}]}(t-t_i)^{2+\beta-\alpha}|D_{0^+}^\beta x(t)|:
i\in \mathbb{N}[0,m]\big\}.
\]

\begin{lemma} \label{claim5.1}
$X$ is a Banach space with the norm defined above.
\end{lemma}

\begin{proof}
It is easy to see that $X$ is a normed linear space. Let $\{x_u\}$ be a
Cauchy sequence in $X$. Then $ \|x_u-x_v\|\to 0$, $u,v \to +\infty$.
It follows that
\begin{equation}
\lim_{t\to t_i^+}(t-t_i)^{2-\alpha}x_u(t),\quad
\lim_{t\to t_i^+}(t-t_i)^{2+\beta-\alpha}D_{0^+}^\beta x_u(t)\quad \text{exist for }
i\in \mathbb{N}[0,m]. \label{e5.0.11}
\end{equation}
Let
\begin{gather*}
\overline{x}_{u,i}(t)=\begin{cases}
\lim_{t\to t_i^+}(t-t_i)^{2-\alpha}x_u(t), & t=t_i,\\
(t-t_i)^{2-\alpha}x_u(t), & t\in (t_i,t_{i+1}],
\end{cases} \\
\overline{Dx}_{u,i}(t)=\begin{cases}
\lim_{t\to t_i^+}(t-t_i)^{2+\beta-\alpha}D_{0^+}^\beta x_u(t), &t=t_i,\\
(t-t_i)^{2+\beta\alpha}D_{0^+}^\beta x_u(t), & t\in (t_i,t_{i+1}].
\end{cases}
\end{gather*}
Then both $\overline{x}_{u,i}$ and $\overline{Dx}_{u,i}$ are continuous on
$[t_i,t_{i+1}]$. Hence there exist two continuous function $x_{0,i},y_{0,i}$
defined on $[t_i,t_{i+1}]$ such that
$\max_{t\in [t_i,t_{i+1}]}|\overline{x}_{u,i}(t)-x_{0,i}(t)|\to 0$ as $u\to +\infty$
and $\max_{t\in [t_i,t_{i+1}]}|\overline{Dx}_{u,i}(t)-y_{0,i}(t)|\to 0$
as $u\to +\infty$.

Denote $x_0(t)=(t-t_i)^{\alpha-2}x_{0,i}(t)$ and
$y_0(t)=(t-t_i)^{\alpha-\beta-2}y_{0,i}(t)$ for $t\in (t_i,t_{i+1}]$.
One sees that $x_0$ and $y_0$ defined on $(0,1]$ such that
\begin{gather*}
\lim_{t\to t_i^+}(t-t_i)^{2-\alpha}x_0(t),\quad
\lim_{t\to t_i^+}(t-t_i)^{2+\beta-\alpha}D_{0^+}^\beta y_0(t)\quad \text{exist for }
i\in \mathbb{N}[0,m],\\
\lim_{u\to +\infty}x_u(t)=x_0(t),\quad
\lim_{u\to +\infty}D_{0^+}^\beta x_u(t)=y_0(t),\quad t\in (0,1],\\
\lim_{u\to +\infty}\sup_{t\to (t_i,t_{i+1}]}(t-t_i)^{2-\alpha}|x_u(t)-x_0(t)|=0,
\quad i\in \mathbb{N}[0,m],\\
\lim_{u\to +\infty}\sup_{t\in (t_i,t_{i+1}]}(t-t_i)^{2+\beta-\alpha}|D_{0^+}^\beta
 x_u(t)-y_0(t)|=0,\quad i\in \mathbb{N}[0,m].
\end{gather*}
Now, denote $D_{0^+}^\beta x_u(t)=y_u(t)$ for $t\in (0,1]$.
 Then by Theorem \ref{thm3.2.2} 
(with $\lambda=0$, $F(t)=y_n(t)$,  $n=1$)
there exist numbers $c_{u,\sigma}(\sigma\in \mathbb{N}[0,m])$
such that $x_u(t)=I_{0^+}^\beta y_u(t)+\sum_{\sigma=0}^ic_{u,\sigma}
(t-t_\sigma)^{\beta-1}$ for $t\in (t_i,t_{i+1}]$, $i\in \mathbb{N}[0,m]$.
So for $t\in (t_i,t_{i+1}]$, we have
\begin{align*}
&|x_u(t)-\sum_{\sigma=0}^ic_{u,\sigma} (t-t_\sigma)^{\beta-1}
 - I_{0^+}^\beta (t-t_\sigma)^{\alpha-\beta-2}y_0(t)| \\
&=|I_{0^+}^\beta y_u(t)- I_{0^+}^\beta (t-t_\sigma)^{\alpha-\beta-2}y_0(t)|\\
&=|I_{0^+}^\beta D_{0^+}^\beta x_u(t)- I_{0^+}^\beta (t-t_\sigma)^{\alpha-\beta-2}
 y_0(t)| \\
&=|\int_0^t\frac{(t-s)^{\beta-1}}{\Gamma(\beta)}[D_{0^+}^\beta x_u(s)
 -(s-t_\sigma)^{\alpha-\beta-2}y_0(s)]ds|\\
&\le \int_0^t\frac{(t-s)^{\beta-1}}{\Gamma(\beta)}|D_{0^+}^\beta
 x_u(s)-(s-t_\sigma)^{\alpha-\beta-2}y_0(s)|ds\\
&=\sum_{\sigma=0}^{i-1}\int_{t_\sigma}^{t_{\sigma+1}}
 \frac{(t-s)^{\beta-1}}{\Gamma(\beta)}|D_{0^+}^\beta x_u(s)-y_0(s)|ds \\
&\quad +\int_{t_i}^{t}\frac{(t-s)^{\beta-1}}{\Gamma(\beta)}|D_{0^+}^\beta x_u(s)
 -(s-t_\sigma)^{\alpha-\beta-2}y_0(s)|ds\\
&=\sum_{\sigma=0}^{i-1}\int_{t_\sigma}^{t_{\sigma+1}}
 \frac{(t-s)^{\beta-1}}{\Gamma(\beta)}|D_{0^+}^\beta x_u(s)
 -(s-t_\sigma)^{\alpha-\beta-2}y_{0,\sigma}(s)|ds\\
&\quad +\int_{t_i}^{t}\frac{(t-s)^{\beta-1}}{\Gamma(\beta)}|D_{0^+}^\beta x_u(s)
 -(s-t_i)^{\alpha-\beta-2}y_{0,i}(s)|ds\\
&=\sum_{\sigma=0}^{i-1}\int_{t_\sigma}^{t_{\sigma+1}}
 \frac{(t-s)^{\beta-1}}{\Gamma(\beta)}(s-t_\sigma)^{\alpha-\beta-2}
 |(s-t_\sigma)^{2+\beta-\alpha}D_{0^+}^\beta x_u(s)-y_{0,\sigma}(s)|ds\\
&\quad +\int_{t_i}^{t}\frac{(t-s)^{\beta-1}}{\Gamma(\beta)}(s-t_i)^{\alpha-\beta-2}
 |(s-t_\sigma)^{2+\beta-\alpha}D_{0^+}^\beta x_u(s)-y_{0,i}(s)|ds\\
&\le \sum_{\sigma=0}^{i-1}\int_{t_\sigma}^{t_{\sigma+1}}
 \frac{(t-s)^{\beta-1}}{\Gamma(\beta)}(s-t_\sigma)^{\alpha-\beta-2}
ds\sup_{t\in (t_\sigma,t_{\sigma+1}]}|(t-t_\sigma)^{2+\beta
 -\alpha}D_{0^+}^\beta x_u(t)-y_{0,\sigma}(t)|\\
&\quad +\int_{t_i}^{t}\frac{(t-s)^{\beta-1}}{\Gamma(\beta)}(s-t_i)^{\alpha-\beta-2}ds
 \sup_{t\in(t_i,t_{i+1}]}|(t-t_\sigma)^{2+\beta-\alpha}D_{0^+}^\beta x_u(t)
 -y_{0,i}(t)| \\
&=\sum_{\sigma=0}^{i-1}(t-t_\sigma)^{\alpha-2}\int_0^{\frac{t_{\sigma+1}
 -t_\sigma}{t-t_\sigma}}\frac{(1-w)^{\beta-1}}{\Gamma(\beta)}w^{\alpha-\beta-2}dw \\
&\quad\times \sup_{t\in (t_\sigma,t_{\sigma+1}]}|(t-t_\sigma)^{2+\beta-\alpha}
 D_{0^+}^\beta x_u(t)-y_{0,\sigma}(t)| \\
&\quad +\int_0^1 \frac{(1-w)^{\beta-1}}{\Gamma(\beta)}
 w^{\alpha-\beta-2}dw\sup_{t\in(t_i,t_{i+1}]}|(t
-t_\sigma)^{2+\beta-\alpha}D_{0^+}^\beta x_u(t)-y_{0,i}(t)|\\
&\le \sum_{\sigma=0}^{i-1}(t-t_\sigma)^{\alpha-2}\int_0^1
 \frac{(1-w)^{\beta-1}}{\Gamma(\beta)}w^{\alpha-\beta-2}dw \\
&\quad\times \sup_{t\in (t_\sigma,t_{\sigma+1}]}|(t-t_\sigma)^{2+\beta-\alpha}D_{0^+}^\beta
 x_u(t)-y_{0,\sigma}(t)|\\
&\quad +\int_0^1 \frac{(1-w)^{\beta-1}}{\Gamma(\beta)}
 w^{\alpha-\beta-2}dw\sup_{t\in(t_i,t_{i+1}]}|(t -t_\sigma)^{2+\beta-\alpha}
 D_{0^+}^\beta x_u(t)-y_{0,i}(t)| \\
&\to 0\quad \text{as }u\to +\infty.
\end{align*}
It follows that
$$
\lim_{u\to +\infty}\big[x_u(t)-\sum_{\sigma=0}^ic_{u,\sigma} (t-t_\sigma)^{\beta-1}
\big]= I_{0^+}^\beta (t-t_\sigma)^{\alpha-\beta-2}y_0(t).
$$
We have
\[
x_0(t)-\sum_{\sigma=0}^ic_{0,\sigma} (t-t_\sigma)^{\beta-1}
= I_{0^+}^\beta (t-t_\sigma)^{\alpha-\beta-2}y_0(t),\quad
\]
for $t\in (t_i,t_{i+1}]$, $i\in \mathbb{N}[0,m]$.
Then for $t\in (t_i,t_{i+1}]$, we have
\begin{align*}
&(t-t_\sigma)^{\alpha-\beta-2}y_0(t) \\
&=D_{0^+}^\beta I_{0^+}^\beta (t-t_\sigma)^{\alpha-\beta-2}y_0(t)\\
&=\frac{1}{\Gamma(1-\beta)}
\Big[\sum_{j=0}^{i-1}\int_{t_j}^{t_{j+1}}(t-s)^{-\beta}\Big(x_0(s)
-\sum_{\sigma=0}^j c_{0,\sigma} (s-t_\sigma)^{\beta-1}\Big)ds\Big]'\\
&\quad +\frac{1}{\Gamma(1-\beta)}
 \Big[\int_{t_i}^{t}(t-s)^{-\beta}\Big(x_0(s)-\sum_{\sigma=0}^i c_{0,\sigma}
(s-t_\sigma)^{\beta-1}\Big)ds\Big]'\\
&=\frac{1}{\Gamma(\beta)}\Big[\int_0^t(t-s)^{-\beta}x_0(s)ds\Big]' \\
&\quad -\frac{1}{\Gamma(1-\beta)}
\Big[\sum_{j=0}^{i-1}\int_{t_j}^{t_{j+1}}(t-s)^{-\beta}\sum_{\sigma=0}^j
c_{0,\sigma} (s-t_\sigma)^{\beta-1}ds\Big]'\\
&\quad -\frac{1}{\Gamma(1-\beta)}
 \Big[\int_{t_i}^{t}(t-s)^{-\beta}\sum_{\sigma=0}^i c_{0,\sigma}
 (s-t_\sigma)^{\beta-1}ds\Big]'\\
&=D_{0^+}^\beta x_0(t)-\frac{1}{\Gamma(1-\beta)}
\Big[\sum_{j=0}^{i-1}\sum_{\sigma=0}^j c_{0,\sigma}
\int_{\frac{t_j-t_\sigma}{t-t_\sigma}}^{\frac{t_{j+1}-t_\sigma}
 {t-t_\sigma}}(1-w)^{-\beta}w^{\beta-1}dw\Big]'\\
&\quad -\frac{1}{\Gamma(1-\beta)}\Big[\sum_{\sigma=0}^i
 c_{0,\sigma}\int_{\frac{t_i-t_\sigma}{t-t_\sigma}}^1 (1-w)^{-\beta} w^{\beta-1}dw
 \Big]'\\
&=D_{0^+}^\beta x_0(t)-\frac{1}{\Gamma(1-\beta)}
\Big[\sum_{\sigma=0}^{i-1}c_{0,\sigma}\sum_{j=\sigma}^{i-1}
\int_{\frac{t_j-t_\sigma}{t-t_\sigma}}^{\frac{t_{j+1}
 -t_\sigma}{t-t_\sigma}}(1-w)^{-\beta}w^{\beta-1}dw\Big]'\\
&\quad -\frac{1}{\Gamma(1-\beta)}\Big[\sum_{\sigma=0}^i
 c_{0,\sigma}\int_{\frac{t_i-t_\sigma}{t-t_\sigma}}^1
 (1-w)^{-\beta} w^{\beta-1}dw\Big]'\\
&=D_{0^+}^\beta x_0(t)-\frac{1}{\Gamma(1-\beta)}
\Big[\sum_{\sigma=0}^ic_{0,\sigma}\int_0^1 (1-w)^{-\beta}
 w^{\beta-1}dw\Big]' \\
&=D_{0^+}^\beta x_0(t).
\end{align*}
Then $(t-t_\sigma)^{\alpha-\beta-2}y_0(t)=D_{0^+}^\beta x_0(t)$ for
$t\in (t_i,t_{i+1}]$, $i\in \mathbb{N}[0,m]$. From above discussion, $X$ is a
Banach space.
\end{proof}

\begin{lemma} \label{claim5.2}
 Let $M$ be a subset of $X$. Then $M$ is relatively compact if and only
if the following conditions are satisfied:
\begin{itemize}
\item[(i)] both $\{t\to (t-t_i)^{2-\alpha}x(t):x\in M\}$ and
$\{t\to (t-t_i)^{2+\beta-\alpha}D_{0^+}^\beta x(t):x\in M\}$ are uniformly bounded,

\item[(ii)] both $\{t\to (t-t_i)^{2-\alpha}x(t):x\in M\}$ and
$\{t\to (t-t_i)^{2+\beta-\alpha}D_{0^+}^\beta x(t):x\in M\}$
are equi-continuous in $(t_i,t_{i+1}](i\in N[0,m])$.
\end{itemize}
\end{lemma}

\begin{proof} $"\Leftarrow"$. From lemma \ref{claim5.1}, we know $X$ is a Banach space.
To prove that the subset $M$ is relatively compact in $X$, we only
need to show $M$ is totally bounded in $X$, that is for all $\epsilon>0$, $M$
has a finite $\epsilon$-net.

For any given $\epsilon>0$, by (i) and (ii), there exist a constants
$A\ge 0,\delta>0$ and a $t_{s_0}$, such that
\begin{gather*}
|(u_1-t_i)^{2-\alpha}x(u_1)-(u_2-t_i)^{2-\alpha}x(u_2)|
\le \frac{\epsilon}{3},\\
\text{for } t_i<u_1,u_2\le t_{i+1},\; |u_1-u_2|<\delta,\; x\in M,\\
|(u_1-t_i)^{2+\beta-\alpha}D_{0^+}^\beta x(u_1)
-(u_2-t_i)^{2+\beta-\alpha}D_{0^+}^\beta x(u_2)|<\frac{\epsilon}{3}, \\
\text{for } t_i<u_1,u_2\le t_{i+1},\; |u_1-u_2|<\delta,\; x\in M,\\
(t-t_i)^{2-\alpha}|x(t)|,(t-t_i)^{2+\beta-\alpha}|D_{0^+}^\beta x(t)|\le A,\\
\text{for }t\in (t_i,t_{i+1}],\; i\in N[0,m],\; x\in M.
\end{gather*}
 Define
\begin{align*}
X|_{(t_i,t_{i+1}]}=\Big\{&x: x\big|_{(t_i,t_{i+1}]},D_{0^+}^\beta x
\text{ is continuous on }(t_i,t_{i+1}] \text{ and }\\
&\lim_{t\to t_i^+}(t-t_i)^{2-\alpha}x(t),\;
 \lim_{t\to t_i^+}(t-t_i)^{2+\beta-\alpha}D_{0^+}^\beta x(t)
\Big\}.
\end{align*}
 For $x\in X|_{(t_i,t_{i+1}]}$, define
$$
\|x\|_{i}=\max\Big\{\sup_{t\in (t_i,t_{i+1}]}(t-t_i)^{2-\alpha}|x(t)|,\;
\sup_{t\in (t_i,t_{i+1}]}(t-t_i)^{2+\beta-\alpha}|D_{0^+}^\beta x(t)|\Big\}.
$$
Similarly to Lemma \ref{claim5.1}, we can prove that $X_{(t_i,t_{i+1}]}$
is a Banach space.
Let $M|_{(t_i,t_{i+1}]}=\{t\to x(t),t\in (t_i,t_{i+1}]:x\in M\}$.
Then $M|_{(t_i,t_{i+1}]}$ is a subset of $X|_{(t_i,t_{i+1}]}$.
By \textbf{(i)} and \textbf{(ii)}, and Ascoli-Arzela theorem, we can know that
$M|_{(t_i,t_{i+1}]}$ is relatively compact. Thus, there exist
$x_{i1}, x_{i2},\dots, x_{ik_i}\in M|_{(t_i,t_{i+1}]}$ such that
$\{U_{x_{i1}}, U_{x_{i2}},\dots, U_{x_{ik_i}}\}$ is a finite $\epsilon$-net of
$M|_{(t_i,t_{i+1}]}$.

Denote $x_{l_0l_1l_2\dots l_m}(t)=x_{il_i}(t)$, $t\in (t_i,t_{i+1}]$,
$i\in \mathbb{N}[0,m]$, $l_i\in \mathbb{N}[1,k_{l_i}]$.
For any $x\in M$, we have $x|_{(t_i,t_{i+1} ]}\in M|_{(t_i,t_{i+1}]}$,
so there exists $l_i\in \mathbb{N}[1,k_i]$ such that
\begin{align*}
\|x|_{(t_i,t_{i+1} ]}-x_{il_i}\|_{i}
&=\max\Big\{\sup_{t\in (t_i,t_{i+1}]}(t-t_i)^{2-\alpha}|x(t)-x_{il_i}(t)|,\\
&\quad \sup_{t\in (t_i,t_{i+1}]}(t-t_i)^{2+\beta-\alpha}|D_{0^+}^\beta x(t)
 -D_{0^+}^\beta x_{il_i}(t)|\} \le \epsilon.
\end{align*}
Then, for $x\in M$,
\begin{align*}
&\|x-x_{l_0l_1l_2\dots l_m}\|_X \\
&= \max\Big\{\sup_{t\in (t_i,t_{i+1}]}(t-t_i)^{2-\alpha}|x(t)-x_{l_0l_1l_2\dots l_m}(t)|, \\
&\quad \sup_{t\in (t_i,t_{i+1}]}(t-t_i)^{2+\beta-\alpha}|D_{0^+}^\beta x(t)
 -D_{0^+}^\beta x_{l_0l_1l_2\dots l_m}(t)|:i\in \mathbb{N}[0,m]\Big\} \\
& \le\max\big\{\sup_{t\in (t_i,t_{i+1}]}(t-t_i)^{2-\alpha}|x(t) -x_{il_i}t)|, \\
&\quad \sup_{t\in (t_i,t_{i+1}]}(t-t_i)^{2+\beta-\alpha}
 |D_{0^+}^\beta x(t)-D_{0^+}^\beta x_{il_i}(t)|:i\in \mathbb{N}[0,m]\Big\}\\
&<\epsilon.
\end{align*}
So, for any $\epsilon>0$, $M$ has a finite $\epsilon$-net
$\{U_{x_{l_0l_1l_2\dots l_m}}:l_i\in N[1,k_i],i\in \mathbb{N}[0,m]\}$;
 that is, $M$ is totally bounded in $X$. Hence, $M$ is relatively compact in $X$.

$\Rightarrow$ Assume that $M$ is relatively compact, then for any $\epsilon>0$,
there exists a finite $\epsilon$-net of $M$. Let the finite $\epsilon$-net be
$\{U_{x_1}, U_{x_2},\dots,U_{x_k}\}$ with $x_i\subset M$. Then for any $x\in M$,
there exists $U_{x_i}$ such that $x\in U_{x_i}$ and
$$
\|x\|\le \|x-x_i\|+\|x_i\|\le \epsilon+\max\{\|x_i\|:i\in \mathbb{N}[1,k]\}.
$$
It follows that $M$ is uniformly bounded. Then (i) holds.

 Furthermore, let $x\in M$, then there exists $x_i$ such that $\|x-x_i\|<\epsilon$.
Since $\lim_{t\to t_j^+} (t-t_i)^{2-\alpha}x_i(t)$ exists and $x_i$ is continuous
on $(t_j,t_{j+1}]$, then there exists $\delta>0$ such that
$u_1,u_2\in (t_j,t_{j+1}]$ with $|u_1-u_2|<\delta$ implies that
$|(u_1-t_j)^{2-\alpha}x_i(u_1)-(u_2-t_j)^{2-\alpha}x_i(u_2)|<\epsilon$.
Then we have for $u_1,u_2\ge (t_{j},t_{j+1}]$ with $|u_1-u_2|<\delta$
\begin{align*}
&|(u_1-t_j)^{2-\alpha}x(u_1)-(u_2-t_j)^{2-\alpha}x(u_2)|\\
& \le |(u_1-t_j)^{2-\alpha}x(u_1)-(u_1-t_j)^{2-\alpha}x_i(u_1)|
 +\big|(u_1-t_j)^{2-\alpha}x_i(u_1) \\
&\quad -(u_2-t_j)^{2-\alpha}x_i(u_2)\big|
 +|(u_2-t_j)^{2-\alpha}x_i(u_2)-(u_2-t_j)^{2-\alpha}x(u_2)|\\
&< 3\epsilon,\quad x\in M.
 \end{align*}
 Similarly we have $t\to (t-t_j)^{2+\beta-\alpha}D_{0^+}^\beta x(t)$ is
equi-continuous on $(t_j,t_{j+1}]$.
Thus (iii) is valid. Similarly we can prove that (ii) holds.
Consequently, the claim is proved.
\end{proof}

\subsection{Impulsive multi-point boundary value problems}

In \cite{zzw}, the authors studied the impulsive boundary value problem
\begin{equation}
\begin{gathered}
{}^{RL}D_{0^+}^\alpha u(t)=f(t,v(t),{}^{RL}D_{0^+}^p v(t)),\\
{}^{RL}D_{0^+}^\beta v(t)=g(t,u(t),{}^{RL}D_{0^+}^q u(t)),\quad t\in (0,1),\\
\Delta u(t_i)=A_i(v(t_i),{}^{RL}D_{0^+}^p v(t_i))),\\
\Delta {}^{RL}D_{0^+}^q u(t_i))=B_i(v(t_i),{}^{RL}D_{0^+}^p v(t_i))), \quad
 i\in \mathbb{N}[1,k],\\
\Delta v(t_i)=C_i(u(t_i),{}^{RL}D_{0^+}^q u(t_i))),\quad
\Delta {}^{RL}D_{0^+}^p v(t_i))=D_i(u(t_i),\\
 {}^{RL}D_{0^+}^q u(t_i))), \quad
 i\in \mathbb{N}[1,k],\\
{}^{RL}D_{0^+}^{\alpha-1}u(0)=\sum_{i=1}^ma_i{}^{RL}D_{0^+}^{\alpha-1}u(\xi_i),
\quad u(1)=\sum_{i=1}^mb_i\eta^{2-\alpha}u(\eta_i),\\
{}^{RL}D_{0^+}^{\beta-1}v(0)=\sum_{i=1}^mc_i{}^{RL}D_{0^+}^{\beta-1}v(\zeta_i),\quad
v(1)=\sum_{i=1}^md_i\eta^{2-\beta}u(\theta_i),
\end{gathered} \label{e5.1.1}
\end{equation}
where
\begin{itemize}
\item[(i)] $\alpha,\beta\in (1,2)$, $p\in (0,\beta-1]$, $q\in (0,\alpha-1]$,
 $\{t_i:i\in \mathbb{N}[1,k]\}$, $\{\xi_i:i\in \mathbb{N}[1,m]\}$, $\{\eta_i:i\in \mathbb{N}[1,m]\}$,
$\{\zeta_i:i\in \mathbb{N}[1,m]\}$, $\{\theta_i:i\in \mathbb{N}[1,m]\}\subset(0,1)$
are increasing sequences,


\item[(ii)] $f,g:[0,1]\times \mathbb{R}^2\to \mathbb{R}$ are Carath\'eodory functions,

\item[(iii)] $A_i,B_i,C_i,D_i:R^2\to \mathbb{R}$ are continuous functions,
$k,m$ are positive integers, $a_i,b_i,c_i,d_i$ are fixed constants,
 $i\in \mathbb{N}[1,m]$.

\end{itemize}
Zhang et al. claimed that the following assumptions
\begin{equation}
\sum_{i=1}^ma_i=\sum_{i=1}^mb_i=\sum_{i=1}^mc_i=\sum_{i=1}^md_i=1,\quad
\sum_{i=1}^mb_i\eta_i=\sum_{i=1}^md_i\eta_i=1 \label{e5.1.2}
\end{equation}
make \eqref{e5.1.1} be a resonant problem and tried to establish existence
result for solutions of BVP \eqref{e5.0.1} by using the coincidence
degree theory due to Mawhin \cite{10}. However, some mistakes occurred
in \cite[(2.10)--(2.14)]{zzw}. There, it means that 
\begin{equation}
{}^{RL}D_{0^+}^\alpha u(t)=z_1(t),\quad
\Delta u(t_i)=\delta_i,\quad
\Delta {}^{RL}D_{0^+}^q u(t_i)=\omega_i, \quad i\in \mathbb{N}[1,m]
\label{e5.1.3}
\end{equation}
imply that
\begin{align*}
u(t)&=\frac{1}{\Gamma(\alpha)}\int_0^t(t-s)^{\alpha-1}z_1(s)ds
+\big(h_1+\frac{\Gamma(\alpha-q)}{\Gamma(\alpha)}
 \sum_{t_i<t}\omega_it_i^{q+1-\alpha} \Big)t^{\alpha-1}\\
&\quad +\Big(h_2+\sum_{t_i<t}\delta_it_i^{2-\alpha}-\frac{\Gamma(\alpha-q)}{\Gamma(\alpha)}\sum_{t_i<t}\omega_i
t_i^{q+2-\alpha}\Big)t^{\alpha-2},\quad h_1,h_2\in \mathbb{R}.
\end{align*}
This claim is false because of the following two items:

(i) In fact, when $t\in (0,t_1]$, this claim is correct. When
$t\in (t_i,t_{i+1}]$, $i\ge 1$, we have by Definition \ref{def2.2}
\begin{align*}
&{}^{RL}D_{0^+}^\alpha u(t) \\
&=\frac{1}{\Gamma(2-\alpha)}\Big(\int_0^t(t-s)^{1-\alpha}u(s)ds\Big)''\\
&=\frac{1}{\Gamma(2-\alpha)}
\Big(\sum_{j=0}^{i-1}\int_{t_j}^{t_{j+1}}(t-s)^{1-\alpha}u(s)ds\Big)''
+\frac{1}{\Gamma(2-\alpha)}\Big(\int_{t_i}^t(t-s)^{1-\alpha}u(s)ds\Big)''\\
&=\frac{1}{\Gamma(2-\alpha)}
\Big[\sum_{j=0}^{i-1}\int_{t_j}^{t_{j+1}}
(t-s)^{1-\alpha}\Big(\frac{1}{\Gamma(\alpha)}\int_0^s(s-u)^{\alpha-1}z_1(u)du\\
&\quad +(h_1+\frac{\Gamma(\alpha-q)}{\Gamma(\alpha)}
\sum_{\tau=1}^j\omega_\tau t_\tau^{q+1-\alpha})s^{\alpha-1} \\
&\quad +(h_2+\sum_{\tau=1}^j\delta_\tau t_\tau^{2-\alpha}
 -\frac{\Gamma(\alpha-q)}{\Gamma(\alpha)}\sum_{\tau=1}^j\omega_\tau
 t_\tau^{q+2-\alpha})s^{\alpha-2}\Big)ds\Big]''\\
&\quad +\frac{1}{\Gamma(2-\alpha)}
\Big[\int_{t_i}^t(t-s)^{1-\alpha}\Big(\frac{1}{\Gamma(\alpha)}
\int_0^s(s-u)^{\alpha-1}z_1(u)du \\
&\quad +\Big(h_1+\frac{\Gamma(\alpha-q)}{\Gamma(\alpha)}
\sum_{\tau=1}^i\omega_\tau t_\tau^{q+1-\alpha}\Big)s^{\alpha-1} \\
&\quad +\Big(h_2+\sum_{\tau=1}^j\delta_\tau t_\tau^{2-\alpha}
-\frac{\Gamma(\alpha-q)}{\Gamma(\alpha)}\sum_{\tau=1}^i\omega_\tau
t_\tau^{q+2-\alpha}\Big)s^{\alpha-2}\Big)ds\Big]''\\
&=\frac{1}{\Gamma(2-\alpha)}
 \Big[\sum_{j=0}^{i-1}\int_{t_j}^{t_{j+1}}
(t-s)^{1-\alpha}\Big(\frac{1}{\Gamma(\alpha)}\int_0^s(s-u)^{\alpha-1}z_1(u)du \\
&\quad +\Big(h_1+\frac{\Gamma(\alpha-q)}{\Gamma(\alpha)}
\sum_{\tau=1}^j\omega_\tau t_\tau^{q+1-\alpha}
\Big)s^{\alpha-1}\Big)ds \Big]''\\
&\quad +\frac{1}{\Gamma(2-\alpha)}
\Big[\int_{t_i}^t(t-s)^{1-\alpha} \Big(\frac{1}{\Gamma(\alpha)}
\int_0^s(s-u)^{\alpha-1}z_1(u)du \\
&\quad +\Big(h_1+\frac{\Gamma(\alpha-q)}{\Gamma(\alpha)}
\sum_{\tau=1}^i\omega_\tau t_\tau^{q+1-\alpha}\Big)s^{\alpha-1}\Big)ds \Big]'' \\
&\quad +\frac{1}{\Gamma(2-\alpha)}
\Big[\int_0^t(t-s)^{1-\alpha}\Big(h_2+\sum_{\tau=1}^j\delta_\tau t_\tau^{2-\alpha}
-\frac{\Gamma(\alpha-q)}{\Gamma(\alpha)}\sum_{\tau=1}^i\omega_\tau
t_\tau^{q+2-\alpha}\Big)s^{\alpha-2}ds\Big]''\\
&\neq z_1(t),\quad \text{by direct computation}.
\end{align*}

 (ii) Problem \eqref{e5.1.3} is unsuitable proposed. By Corollary \ref{coro5.2},
one sees the piecewise continuous solutions of ${}^{RL}D_{0^+}^\alpha u(t)=z_1(t)$
are given by
\[
u(t)=\sum_{\sigma=0}^i(\frac{c_{\sigma 1}}{\Gamma(\alpha)}
(t-t_\sigma)^{\alpha-1}+\frac{c_{\sigma 2}}{\Gamma(\alpha-1)}
(t-t_\sigma)^{\alpha-2})
+\frac{1}{\Gamma(\alpha)}\int_0^t(t-s)^{\alpha-1}z_1(s)ds,
\]
for $t\in (t_i,t_{i+1}]$, $i\in \mathbb{N}[0,m]$.
It is easy to see that $u$ may be discontinuous at $t=t_i$ if $c_{i2}\neq 0$.
So $\Delta u(t_i)=\delta_i$ is unsuitable. Since the expression of $u$
on $(t_i,t_{i+1}],i\in N[0,m]$ are different from each other, then the
resonant conditions \eqref{e5.1.2} may be false.

To show the readers a correct result, now, we consider the boundary value problem
(for ease expression, we consider the one of a impulsive fractional differential
equation, not a fractional differential system):
\begin{equation}
\begin{gathered}
{}^{RL}D_{0^+}^\alpha u(t)=p(t)f(t,u(t),{}^{RL}D_{0^+}^\beta u(t)),
\quad t\in (t_i,t_{i+1}],\;i\in \mathbb{N}[0,m],\\
\lim_{t\to t_i^+}(t-t_i)^{2-\alpha}u(t)=I(t_i,u(t_i),{}^{RL}D_{0^+}^\beta u(t_i)),
\quad i\in \mathbb{N}[1,m],\\
\Delta {}^{RL}D_{0^+}^{\alpha-1}u(t_i))=J(t_i,u(t_i),{}^{RL}D_{0^+}^\beta u(t_i)),
\quad i\in \mathbb{N}[1,m],\\
{}^{RL}D_{0^+}^{\alpha-1}u(0)=\sum_{i=1}^ma_i{}^{RL}D_{0^+}^{\alpha-1}u(\xi_i),\quad
u(1)=\sum_{i=1}^mb_iu(\eta_i),
\end{gathered} \label{e5.1.4}
\end{equation}
where:

(5.A1) $\alpha\in (1,2)$, $\beta\in (0,\alpha-1]$,
$\{t_i:i\in \mathbb{N}[1,m]\}$, $\{\xi_i:i\in \mathbb{N}[1,m]\},\{\eta_i:i\in \mathbb{N}[1,m]\}$
 are increasing sequences with $\xi_i,\eta_i\in (t_i,t_{i+1}]$,
$i\in \mathbb{N}[0,m]$, $a_i,b_i\in \mathbb{R}$ are fixed constants, $m$ is a positive integer,

(5.A2) $f:(0,1)\times \mathbb{R}^2\to \mathbb{R}$ satisfies the following items:
$t\to f(t,(t-t_i)^{\alpha-2}u,(t-t_i)^{\alpha-\beta-2}v)$
is measurable on $(0,1)$ for each $(u,v)\in \mathbb{R}^2$,
$(u,v)\to f(t,(t-t_i)^{\alpha-2}u,(t-t_i)^{\alpha-\beta-2}v)$
is continuous on $\mathbb{R}^2$ for all $t\in (t_i,t_{i+1}]$, $i\in \mathbb{N}[0,m]$,
for each $r>0$ there exists a constant $M_r\ge 0$ such that
$|f(t,(t-t_i)^{\alpha-2}u,(t-t_i)^{\alpha-\beta-2}v)|\le M_r$ holds for all
$t\in (t_i,t_{i+1}]$, $i\in \mathbb{N}[0,m]$ and $|u|,|v|\le r$;

(5.A3) $I,J:\{t_i:i\in \mathbb{N}[1,m]\}\times \mathbb{R}^2\to \mathbb{R}$ satisfies the following items:
$(u,v)\to I\text{ or }J(t_i,(t_i-t_{i-1})^{\alpha-2}u,
(t_i-t_{i-1})^{\alpha-\beta-2}v)$ is continuous on $\mathbb{R}^2$ for all $i\in \mathbb{N}[1,m]$,
for each $r>0$ there exists a constant $M_r\ge 0$ such that
$|I\text{ or }J(t_i,(t_i-t_{i-1})^{\alpha-2}u,(t_i-t_{i-1})^{\alpha-\beta-2}v)|
\le M_r$ holds for all $i\in \mathbb{N}[1,m]$ and $|u|,|v|\le r$;

(5.A4) $p:(0,1)\to \mathbb{R}$ satisfies that there exist number $k>-1$,
$l\in (\max\{-\alpha,-2-k\},0]$ and $\beta-\alpha<l\le 0$ such that
$|p(t)|\le t^k(1-t)^l$ for all $t\in (0,1)$.

The homogeneous problem of \eqref{e5.1.4} is as follows:
\begin{equation} \begin{gathered}
{}^{RL}D_{0^+}^\alpha u(t)=0,\quad t\in (t_i,t_{i+1}],\;i\in \mathbb{N}[0,m],\\
\lim_{t\to t_i^+}(t-t_i)^{2-\alpha}u(t)=\Delta {}^{RL}D_{0^+}^{\alpha-1}u(t_i))=0,
\quad i\in \mathbb{N}[1,m],\\
{}^{RL}D_{0^+}^{\alpha-1}u(0)=\sum_{i=1}^ma_i{}^{RL}D_{0^+}^{\alpha-1}u(\xi_i),\quad
u(1)=\sum_{i=1}^mb_iu(\eta_i),
\end{gathered}\label{e5.1.5}
\end{equation}
By Corollary \ref{coro5.2}, ${}^{RL}D_{0^+}^\alpha u(t)=0$ with
$t\in (t_i,t_{i+1}]$, $i\in \mathbb{N}[0,m]$ implies that there exist constants
$c_{\sigma 1},c_{\sigma2}\in \mathbb{R}$ such that
\begin{equation}
u(t)=\sum_{\sigma=0}^i(c_{\sigma1}(t-t_\sigma)^{\alpha-1}
+c_{\sigma2}(t-t_\sigma)^{\alpha-2}),\quad t\in (t_i,t_{i+1}],\;i\in \mathbb{N}[0,m].
\label{e5.1.6}
\end{equation}
 So Definition \ref{def2.2} implies for $t\in (t_j,t_{j+1}]$, $i\in \mathbb{N}[0,m]$ that
\begin{align*}
&{}^{RL}D_{0^+}^{\alpha-1}u(t)\\
&=\frac{1}{\Gamma(2-\alpha)}\Big[\int_0^t(t-s)^{1-\alpha}u(s)ds\Big]'\\
&=\frac{1}{\Gamma(2-\alpha)}
 \Big[\sum_{i=0}^{j-1}\int_{t_i}^{t_{i+1}}(t-s)^{1-\alpha}
\sum_{\sigma=0}^i(c_{\sigma1}(s-t_\sigma)^{\alpha-1}
 +c_{\sigma2}(s-t_\sigma)^{\alpha-2})ds\Big]'\\
&\quad +\frac{1}{\Gamma(2-\alpha)}
 \Big[\int_{t_j}^{t}(t-s)^{1-\alpha}
\sum_{\sigma=0}^j(c_{\sigma1}(s-t_\sigma)^{\alpha-1}
+c_{\sigma2}(s-t_\sigma)^{\alpha-2})ds\Big]'\\
&=\frac{1}{\Gamma(2-\alpha)}\Big[\sum_{i=0}^{j-1}
 \sum_{\sigma=0}^i(c_{\sigma1}\int_{t_i}^{t_{i+1}}(t-s)^{1-\alpha}
(s-t_\sigma)^{\alpha-1}ds \\
&\quad +c_{\sigma2}\int_{t_i}^{t_{i+1}}(t-s)^{1-\alpha}
 (s-t_\sigma)^{\alpha-2}ds)\Big]'\\
&\quad +\frac{1}{\Gamma(2-\alpha)}
 \Big[\sum_{\sigma=0}^j(c_{\sigma1}\int_{t_j}^{t}(t-s)^{1-\alpha}
(s-t_\sigma)^{\alpha-1}ds \\
&\quad +c_{\sigma2}\int_{t_j}^{t}(t-s)^{1-\alpha}
 (s-t_\sigma)^{\alpha-2}ds)\Big]'\\
&=\frac{1}{\Gamma(2-\alpha)}
 \Big[\sum_{i=0}^{j-1}\sum_{\sigma=0}^i(c_{\sigma1}(t-t_\sigma)
 \int_{\frac{t_i-t_\sigma}{t-t_\sigma}}^{\frac{t_{i+1}
 -t_\sigma}{t-t_\sigma}}(1-w)^{1-\alpha} w^{\alpha-1}dw \\
&\quad +c_{\sigma2}\int_{\frac{t_i-t_\sigma}{t-t_\sigma}}^{\frac{t_{i+1}
 -t_\sigma}{t-t_\sigma}}(1-w)^{1-\alpha}w^{\alpha-2}dw)\Big]'\\
&\quad +\frac{1}{\Gamma(2-\alpha)}\Big[\sum_{\sigma=0}^j
 (c_{\sigma1}(t-t_\sigma)\int_{\frac{t_j-t_\sigma}{t-t_\sigma}}^1 (1-w)^{1-\alpha}
 w^{\alpha-1}dw \\
&\quad +c_{\sigma2}\int_{\frac{t_j-t_\sigma}{t-t_\sigma}}^1
 (1-w)^{1-\alpha}w^{\alpha-2}dw)\Big]' \\
&=\frac{1}{\Gamma(2-\alpha)}\Big[\sum_{\sigma=0}^{j-1}
 \sum_{i=\sigma}^{j-1}(c_{\sigma1}(t-t_\sigma)
 \int_{\frac{t_i-t_\sigma}{t-t_\sigma}}^{\frac{t_{i+1}
 -t_\sigma}{t-t_\sigma}}(1-w)^{1-\alpha} w^{\alpha-1}dw \\
&\quad +c_{\sigma2} \int_{\frac{t_i-t_\sigma}{t-t_\sigma}}^{\frac{t_{i+1}-t_\sigma}
 {t-t_\sigma}}(1-w)^{1-\alpha}w^{\alpha-2}dw)\Big]'\\
&\quad +\frac{1}{\Gamma(2-\alpha)}
 \Big[\sum_{\sigma=0}^j\Big(c_{\sigma1}(t-t_\sigma)
 \int_{\frac{t_j-t_\sigma}{t-t_\sigma}}^1 (1-w)^{1-\alpha}w^{\alpha-1}dw \\
&\quad +c_{\sigma2} \int_{\frac{t_j-t_\sigma}{t-t_\sigma}}^1 (1-w)^{1-\alpha}w^{\alpha-2}dw\Big)\Big]'\\
&=\frac{1}{\Gamma(2-\alpha)}\Big[\sum_{\sigma=0}^{j}
\Big(c_{\sigma1}(t-t_\sigma)\int_0^1 (1-w)^{1-\alpha} w^{\alpha-1}dw \\
&\quad +c_{\sigma2}\int_0^1 (1-w)^{1-\alpha}w^{\alpha-2}dw\Big)\Big]' \\
&=\Gamma(\alpha)\sum_{\sigma=0}^jc_{\sigma1}.
\end{align*}
It follows that
\begin{equation}
{}^{RL}D_{0^+}^{\alpha-1}u(t)=\Gamma(\alpha)\sum_{\sigma=0}^ic_{\sigma1},
\quad t\in (t_i,t_{i+1}],\;i\in \mathbb{N}[0,m].\label{e5.1.7}
\end{equation}
Then the other equations in \eqref{e5.1.5} with \eqref{e5.1.6} and
\eqref{e5.1.7} imply
\begin{equation}
\begin{gathered}
c_{i2}=0,\quad c_{i1}=0,i\in \mathbb{N}[1,m],\\
\Gamma(\alpha )c_{01}=\sum_{i=1}^ma_i\sum_{\sigma=0}^i\Gamma(\alpha)c_{\sigma1},\\
\begin{aligned}
&\sum_{\sigma=0}^{m}(c_{\sigma1}(1-t_\sigma)^{\alpha-1}
 +c_{\sigma2}(1-t_\sigma)^{\alpha-2})\\
&=\sum_{i=1}^mb_i\sum_{\sigma=0}^i(c_{\sigma1}(\eta_i-t_\sigma)^{\alpha-1}
 +c_{\sigma2}(\eta_i-t_\sigma)^{\alpha-2}).
\end{aligned}
\end{gathered} \label{e5.1.8}
\end{equation}
Thus \eqref{e5.1.8} implies that $c_{\sigma1}=0$ and $ c_{\sigma2}=0$ for all
$\sigma\in \mathbb{N}[1,m]$. So by \eqref{e5.1.8} we have
\begin{equation}
c_{01}=c_{01}\sum_{i=1}^ma_i,\quad
c_{01}(1-\sum_{i=1}^mb_i\eta_i^{\alpha-1})
=c_{02}\Big[\sum_{i=1}^mb_i\eta_i^{\alpha-2}-1\Big]=0.
\label{e5.1.9}
\end{equation}
From \eqref{e5.1.9} it is easy to see the following:

\noindent Case (i): \eqref{e5.1.5} has a unique trivial solution $u(t)=0$ if and
 only if
\begin{equation}
\sum_{i=1}^ma_i\neq 1,\quad \sum_{i=1}^mb_i\eta_i^{\alpha-2}\neq 1.
\label{e5.1.10}
\end{equation}

\noindent Case (ii): \eqref{e5.1.5} has a group of nontrivial solutions
$u(t)=c_{02}t^{\alpha-2},c_{02}\in \mathbb{R}$ if and only if
\begin{equation}
\sum_{i=1}^ma_i\neq 1,\quad \sum_{i=1}^mb_i\eta_i^{\alpha-2}=1. \label{e5.1.11}
\end{equation}

\noindent Case (iii): \eqref{e5.1.5} has a group of nontrivial solutions
\[
u(t)=c_{01}t^{\alpha-1}+c_{01}\frac{1-\sum_{i=1}^mb_i\eta_i^{\alpha-1}}
{\sum_{i=1}^mb_i\eta_i^{\alpha-2}-1}t^{\alpha-2}, \quad c_{01},
c_{02}\in \mathbb{R}
\]
if and only if
\begin{equation}
\sum_{i=1}^ma_i=1,\quad \sum_{i=1}^mb_i\eta_i^{\alpha-1}\neq 1,\quad
\sum_{i=1}^mb_i\eta_i^{\alpha-2}\neq 1.
\label{e5.1.12}
\end{equation}

\noindent Case (iv): \eqref{e5.1.5} has a group of nontrivial solutions
$u(t)=c_{02}t^{\alpha-2}$, $c_{02}\in \mathbb{R}$ if and only if
\begin{equation}
\sum_{i=1}^ma_i=1,\quad \sum_{i=1}^mb_i\eta_i^{\alpha-1}\neq 1,\quad
\sum_{i=1}^mb_i\eta_i^{\alpha-2}=1.
\label{e5.1.13}
\end{equation}

\noindent Case (v): \eqref{e5.1.5} has a group of nontrivial solutions
$u(t)=c_{01}t^{\alpha-1}$, $c_{01}\in \mathbb{R}$
if and only if
\begin{equation}
\sum_{i=1}^ma_i=1,\quad \sum_{i=1}^mb_i\eta_i^{\alpha-1}=1,\quad
\sum_{i=1}^mb_i\eta_i^{\alpha-2}=1. \label{e5.1.14}
\end{equation}


% Claim5.1
 If \eqref{e5.1.10} holds, then BVP \eqref{e5.1.4} is a un-resonant problem.
While \eqref{e5.1.11} or \eqref{e5.1.12} or \eqref{e5.1.13} or \eqref{e5.1.14}
implies that BVP\eqref{e5.1.4} is a resonant problem.
Concerning Case (i), we establish an existence result for solutions of
\eqref{e5.1.4}. Similar results can be established for other cases.
The readers should try it.


\begin{lemma} \label{lem5.1.1}
 Suppose that \eqref{e5.1.10} holds, denote $\Delta_1=1-\sum_{i=1}^ma_i$ and
$\Delta_2=1-\sum_{i=1}^mb_i\eta_i^{\alpha-2}$,
$\Delta _3= \sum_{i=1}^mb_i\eta_i^{\alpha-1}-1$, $\sigma$ is continuous on
$(0,1)$ and there exist $k>-1$ and $l\in (\max\{-\alpha,-2-k\},0]$ such that
$|\sigma(t)|\le t^k(1-t)^l$ for all $t\in (0,1)$. Then $x$ is a solution of
the problem
\begin{equation}
\begin{gathered}
{}^{RL}D_{0^+}^\alpha x(t)=\sigma(t),\quad t\in (t_i,t_{i+1}],\;i\in \mathbb{N}[0,m],\\
\lim_{t\to t_i^+}(t-t_i)^{2-\alpha}x(t)=I_i, \quad i\in \mathbb{N}[1,m],\\
\Delta {}^{RL}D_{0^+}^{\alpha-1}x(t_i))=J_i, \quad \;i\in \mathbb{N}[1,m],\\
{}^{RL}D_{0^+}^{\alpha-1}x(0)=\sum_{i=1}^ma_i{}^{RL}D_{0^+}^{\alpha-1}x(\xi_i),\quad
x(1)=\sum_{i=1}^mb_ix(\eta_i)
\end{gathered} \label{e5.1.15}
\end{equation}
if and only if
\begin{equation}
\begin{aligned}
&x(t) \\
&=\frac{1}{\Gamma(\alpha)}
\Big[\frac{t^{\alpha-1}}{\Delta_1}\sum_{\sigma=1}^m\sum_{i=\sigma}^m a_i
+\frac{\Delta_3}{\Delta_1}\frac{t^{\alpha-2}}{\Delta_2}
\sum_{\sigma=1}^m\sum_{i=\sigma}^m a_i+\frac{t^{\alpha-2}}{\Delta_2}
\sum_{\sigma=1}^m\sum_{i=\sigma}^m
b_i(\eta_i-t_\sigma)^{\alpha-1}\Big]J_\sigma\\
&\quad +\frac{t^{\alpha-2}}{\Delta_2}\sum_{\sigma=1}^m
 \sum_{i=\sigma}^mb_i(\eta_i-t_\sigma)^{\alpha-2}I_\sigma+\frac{1}{\Gamma(\alpha)}
\sum_{\sigma=1}^i(t-t_\sigma)^{\alpha-1}J_\sigma
 +\sum_{\sigma=1}^i(t-t_\sigma)^{\alpha-2}I_\sigma\\
&\quad +\frac{1}{\Gamma(\alpha)}\frac{t^{\alpha-2}}{\Delta_2}
 \frac{\Delta_3}{\Delta_1}\sum_{i=1}^ma_i
\int_0^{\xi_i}\sigma(s)ds+\frac{t^{\alpha-2}}{\Delta_2}\sum_{i=1}^mb_i
\int_0^{\eta_i}\frac{(\eta_i-s)^{\alpha-1}}{\Gamma(\alpha)}\sigma(s)ds \\
&\quad -\frac{t^{\alpha-2}}{\Delta_2}
\int_0^1 \frac{(1-s)^{\alpha-1}}{\Gamma(\alpha)}\sigma(s)ds
 +\frac{1}{\Gamma(\alpha)}
\frac{t^{\alpha-1}}{\Delta_1}\sum_{i=1}^ma_i
\int_0^{\xi_i}\sigma(s)ds \\
&\quad +\int_0^t\frac{(t-s)^{\alpha-1}}{\Gamma(\alpha)}
 \sigma(s)ds,\quad t\in (t_i,t_{i+1}],\;i\in \mathbb{N}[0,m].
\end{aligned} \label{e5.1.16}
\end{equation}
\end{lemma}

\begin{proof}
By Theorem \ref{thm3.2.2}, we have ${}^{RL}D_{0^+}^\alpha x(t)=\sigma(t)$,
$t\in (t_i,t_{i+1}]$, $i\in \mathbb{N}[0,m]$ if and only if there exit numbers
$c_{\sigma1},c_{\sigma2}$, $i\in \mathbb{N}[0,m]$ such that
\begin{equation}
x(t)=\sum_{\sigma=0}^i[c_{\sigma 1}(t-t_\sigma)^{\alpha-1}
+c_{\sigma2}(t-t_\sigma)^{\alpha-2}]
+\int_0^t\frac{(t-s)^{\alpha-1}}{\Gamma(\alpha)} \sigma(s)ds,
\label{e5.1.17}
\end{equation}
for $t\in (t_i,t_{i+1}]$, $i\in \mathbb{N}[0,m]$.
Then
\begin{equation}
{}^{RL}D_{0^+}^{\alpha-1}x(t)=\Gamma(\alpha)
\sum_{\sigma=0}^ic_{\sigma1}+\int_0^t\sigma(s)ds,\quad
t\in (t_i,t_{i+1}],\;i\in \mathbb{N}[0,m].
\label{e5.1.18}
\end{equation}
Furthermore, by direct computations we have
\begin{equation}
\begin{aligned}
&{}^{RL}D_{0^+}^{\beta}x(t) \\
&=\frac{1}{\Gamma(1-\beta)}\Big[\int_0^t(t-s)^{-\beta}x(s)ds\Big]'\\
&=\frac{1}{\Gamma(1-\beta)}\Big[\sum_{j=0}^{i-1}\int_{t_j}^{t_{j+1}}(t-s)^{-\beta}
x(s)ds\Big]'
+\frac{1}{\Gamma(1-\beta)}\Big[\int_{t_i}^{t}(t-s)^{-\beta} x(s)ds\Big]'\\
&=\sum_{\sigma=0}^i(c_{\sigma1}\frac{\Gamma(\alpha)}
 {\Gamma(\alpha-\beta+1)}(t-t_\sigma)^{\alpha-\beta-1}
 +c_{\sigma2}\frac{\Gamma(\alpha-1)}{\Gamma(\alpha-\beta-1)}
(t-t_\sigma)^{\alpha-\beta-2})\\
&\quad +\int_0^t\frac{(t-s)^{\alpha-\beta-1}}{\Gamma(\alpha-\beta)}
 \sigma(s)ds,\quad t\in (t_i,t_{i+1}],\; i\in N[0,m].
\end{aligned} \label{e5.1.19}
\end{equation}
From $\lim_{t\to t_i^+}(t-t_i)^{2-\alpha}x(t)=I_i$, we have
$c_{\sigma2}=I_\sigma$ for all $\sigma\in \mathbb{N}[1,m]$. From
$\Delta {}^{RL}D_{0^+}^{\alpha-1}x(t_i))=J_i,\;i\in \mathbb{N}[1,m]$, one has
 $c_{\sigma1}=\frac{J_\sigma}{\Gamma(\alpha)}$ for all $\sigma\in \mathbb{N}[1,m]$.
 From the boundary conditions in \eqref{e5.1.14}, we obtain
\begin{align*}
&\Gamma(\alpha )c_{01}=\sum_{i=1}^ma_i
\Big[\Gamma(\alpha)\sum_{\sigma=0}^ic_{\sigma1}+\int_0^{\xi_i}\sigma(s)ds\Big],\\
&\sum_{\sigma=0}^m [c_{\sigma 1}(1-t_\sigma)^{\alpha-1}
 +c_{\sigma2}(1-t_\sigma)^{\alpha-2}]
 +\int_0^1\frac{(1-s)^{\alpha-1}}{\Gamma(\alpha)}\sigma(s)ds\\
&=\sum_{i=1}^mb_i\Big[\sum_{\sigma=0}^i
 [c_{\sigma 1}(\eta_i-t_\sigma)^{\alpha-1}+c_{\sigma2}(\eta_i-t_\sigma)^{\alpha-2}
 ]+\int_0^{\eta_i}\frac{(\eta_i-s)^{\alpha-1}}{\Gamma(\alpha)}\sigma(s)ds\Big].
\end{align*}
It follows that
\begin{gather*}
\begin{aligned}
c_{01}&=\frac{1}{\Gamma(\alpha)}\frac{1}{\Delta_1}
\Big[\sum_{i=1}^ma_i\sum_{\sigma=1}^iJ_{\sigma}+\sum_{i=1}^ma_i
\int_0^{\xi_i}\sigma(s)ds\Big] \\
&=\frac{1}{\Gamma(\alpha)}\frac{1}{\Delta_1}
\sum_{\sigma=1}^m\sum_{i=\sigma}^ma_iJ_{\sigma}+\frac{1}{\Gamma(\alpha)}\frac{1}{\Delta_1}\sum_{i=1}^ma_i
\int_0^{\xi_i}\sigma(s)ds,
\end{aligned} \\
\begin{aligned}
c_{02}&=\frac{1}{\Delta_2}
\Big[\frac{1}{\Gamma(\alpha)}\frac{\Delta_3}{\Delta_1}
\sum_{\sigma=1}^m\sum_{i=\sigma}^ma_iJ_{\sigma}
 +\frac{1}{\Gamma(\alpha)}\sum_{i=1}^m b_i\sum_{\sigma=1}^i
 (\eta_i-t_\sigma)^{\alpha-1}J_\sigma \\
&\quad +\sum_{i=1}^mb_i\sum_{\sigma=1}^i(\eta_i-t_\sigma)^{\alpha-2}I_\sigma
 +\frac{1}{\Gamma(\alpha)}\frac{\Delta_3}{\Delta_1}\sum_{i=1}^ma_i
\int_0^{\xi_i}\sigma(s)ds \\
&\quad +\sum_{i=1}^mb_i \int_0^{\eta_i}\frac{(\eta_i-s)^{\alpha-1}}{\Gamma(\alpha)}
\sigma(s)ds-\int_0^1 \frac{(1-s)^{\alpha-1}}{\Gamma(\alpha)}\sigma(s)ds\Big]\\
&=\frac{1}{\Delta_2}\Big[\frac{1}{\Gamma(\alpha)}\frac{\Delta_3}{\Delta_1}
\sum_{\sigma=1}^m\sum_{i=\sigma}^ma_iJ_{\sigma}
 +\frac{1}{\Gamma(\alpha)}\sum_{\sigma=1}^m\sum_{i=\sigma}^m
b_i(\eta_i-t_\sigma)^{\alpha-1}J_\sigma \\
&\quad +\sum_{\sigma=1}^m\sum_{i=\sigma}^mb_i(\eta_i-t_\sigma)^{\alpha-2}I_\sigma
 +\frac{1}{\Gamma(\alpha)}\frac{\Delta_3}{\Delta_1}\sum_{i=1}^ma_i
\int_0^{\xi_i}\sigma(s)ds \\
&\quad +\sum_{i=1}^mb_i
\int_0^{\eta_i}\frac{(\eta_i-s)^{\alpha-1}}{\Gamma(\alpha)}\sigma(s)ds
-\int_0^1 \frac{(1-s)^{\alpha-1}}{\Gamma(\alpha)}\sigma(s)ds\Big].
\end{aligned}
\end{gather*}
Substituting $c_{\sigma1},c_{\sigma2}$ into \eqref{e5.1.17}, we obtain
\eqref{e5.1.16} by changing the order of the terms. It is easy to show
from \eqref{e5.1.17} and \eqref{e5.1.19} that $x\in X$.
The proof is compete.
\end{proof}

To simplify notation, let
 $H_x(t)=H(t,x(t),{}^{RL}D_{0^+}^\beta x(t))$ for functions
 $H:(0,1)\times \mathbb{R}^2\to \mathbb{R}$ and $x:(0,1]\to \mathbb{R}$.
Define the operator $(Tx)(t)$ for $x\in X$ by
\begin{align*}
&(Tx)(t) \\
&=\frac{1}{\Gamma(\alpha)}
\Big[ \frac{t^{\alpha-1}}{\Delta_1}\sum_{\sigma=1}^m
\sum_{i=\sigma}^ma_i+\frac{\Delta_3}{\Delta_1}\frac{t^{\alpha-2}}{\Delta_2}
\sum_{\sigma=1}^m\sum_{i=\sigma}^ma_i+\frac{t^{\alpha-2}}{\Delta_2}
\sum_{\sigma=1}^m\sum_{i=\sigma}^m
b_i(\eta_i-t_\sigma)^{\alpha-1}\Big]J_x(t_\sigma)\\
&\quad +\frac{t^{\alpha-2}}{\Delta_2}\sum_{\sigma=1}^m
\sum_{i=\sigma}^mb_i(\eta_i-t_\sigma)^{\alpha-2}I_x(t_\sigma)
+\frac{1}{\Gamma(\alpha)}\sum_{\sigma=1}^i(t-t_\sigma)^{\alpha-1}J_x(t_\sigma) \\
&\quad +\sum_{\sigma=1}^i (t-t_\sigma)^{\alpha-2}I_x(t_\sigma)
+\frac{1}{\Gamma(\alpha)}\frac{t^{\alpha-2}}{\Delta_2}
 \frac{\Delta_3}{\Delta_1}\sum_{i=1}^m a_i \int_0^{\xi_i}p(s)f_x(s)ds \\
&\quad +\frac{t^{\alpha-2}}{\Delta_2}\sum_{i=1}^mb_i
\int_0^{\eta_i}\frac{(\eta_i-s)^{\alpha-1}}{\Gamma(\alpha)}p(s)f_x(s)ds
-\frac{t^{\alpha-2}}{\Delta_2}
\int_0^1 \frac{(1-s)^{\alpha-1}}{\Gamma(\alpha)}p(s)f_x(s)ds \\
&\quad +\frac{1}{\Gamma(\alpha)}\frac{t^{\alpha-1}}{\Delta_1}\sum_{i=1}^ma_i
\int_0^{\xi_i}p(s)f_x(s)ds
+\int_0^t\frac{(t-s)^{\alpha-1}}{\Gamma(\alpha)}p(s)f_x(s)ds,
\end{align*}
for $t\in (t_i,t_{i+1}]$ and $i\in \mathbb{N}[0,m]$.

\begin{theorem} \label{thm5.1.1}
 Suppose that \eqref{e5.1.10}, {\rm (5.A1)--(5.A4)} hold and
\begin{itemize}
\item[(H1)] there exist non-decreasing functions
$M_f,M_I,M_J:[0,+\infty)\times [0,+\infty)\to [0,+\infty)$ such that
\begin{gather*}
|f(t,(t-t_i)^{\alpha-2}u,(t-t_i)^{\alpha-\beta-2}v)|
\le M_f(|u|,|v|),\\
\text{for } t\in (t_i,t_{i+1}],\;i\in \mathbb{N}[0,m],u,v\in \mathbb{R},\\
|I(t_i,(t_i-t_{i-1})^{\alpha-2}u,(t_i-t_{i-1})^{\alpha-\beta-2}v)|
\le M_I(|u|,|v|),\quad i\in \mathbb{N}[1,m],u,v\in \mathbb{R},\\
|J(t_i,(t_i-t_{i-1})^{\alpha-2}u,(t_i-t_{i-1})^{\alpha-\beta-2}v)|
\le M_J(|u|,|v|),\quad i\in \mathbb{N}[1,m],u,v\in \mathbb{R}
\end{gather*}
\end{itemize}
Then \eqref{e5.1.4} has at least one solution if there is $r_0>0$ such
that $A_1M_J(r_0,r_0)+A_2M_I(r_0,r_0)+A_3M_f(r_0,r_0)\le r_0$,
where
\begin{gather*}
\begin{aligned}
A_1&=\max\Big\{\frac{1}{\Gamma(\alpha)}\Big[\Big(
\frac{1}{|\Delta_1|}+\frac{|\Delta_3|}{|\Delta_1|}\frac{1}{|\Delta_2|}\Big)
\sum_{i=1}^mi|a_i|+\frac{1}{|\Delta_2|}
\sum_{i=1}^mi|b_i|+m\Big], \\
&\quad \Big(\frac{1}{\Gamma(\alpha-\beta+1)|\Delta_1|}
+\frac{\Gamma(\alpha-1)}{\Gamma(\alpha-\beta-1)}
 \frac{1}{\Gamma(\alpha)|\Delta_2|}\frac{|\Delta_3|}{|\Delta_1|}\Big)
\sum_{i=1}^mi|a_i| \\
&\quad +\frac{\Gamma(\alpha-1)}{\Gamma(\alpha-\beta-1)}
\frac{1}{\Gamma(\alpha)|\Delta_2|}\sum_{i=1}^m i|b_i|
+\frac{m}{\Gamma(\alpha-\beta+1)}\Big\};
\end{aligned}\\
\begin{aligned}
A_2&=\max\Big\{\frac{1}{|\Delta_2|}\sum_{\sigma=1}^m
 \sum_{i=\sigma}^m|b_i|(\eta_i-t_\sigma)^{\alpha-2}+m, \\
&\quad \frac{\Gamma(\alpha-1)}{\Gamma(\alpha-\beta-1)}\frac{1}{|\Delta_2|}
\sum_{\sigma=1}^m\sum_{i=\sigma}^m|b_i|(\eta_i-t_\sigma)^{\alpha-2}+
\frac{m\Gamma(\alpha-1)}{\Gamma(\alpha-\beta-1)}\Big\};
\end{aligned} \\
\begin{aligned}
A_3&=\max\Big\{(\frac{\mathbf{B}(l+1,k+1)}{\Gamma(\alpha)|\Delta_2|}
 \frac{|\Delta_3|}{|\Delta_1|}
 +\frac{\mathbf{B}(l+1,k+1)}{\Gamma(\alpha)|\Delta_1|})\sum_{i=1}^m|a_i| \\
&\quad +\frac{\mathbf{B}(\alpha+l,k+1)}{\Gamma(\alpha)|\Delta_2|}\sum_{i=1}^m|b_i|
+\frac{\mathbf{B}(\alpha+l,k+1)}{\Gamma(\alpha)|\Delta_2|}
+\frac{\mathbf{B}(l+1,k+1)}{\Gamma(\alpha)}, \\
&\quad  \frac{\mathbf{B}(l+1,k+1)}{\Gamma(\alpha)}\frac{\Gamma(\alpha-1)}
 {\Gamma(\alpha-\beta-1)}\frac{1}{|\Delta_2|}
 \frac{|\Delta_3|}{|\Delta_1|} \sum_{i=1}^m|a_i| \\
&\quad +\frac{\Gamma(\alpha-1)}{\Gamma(\alpha-\beta-1)}
 \frac{1}{|\Delta_2|}\sum_{i=1}^m|b_i|
 \frac{\mathbf{B}(\alpha+l,k+1)}{\Gamma(\alpha)}
 +\frac{\mathbf{B}(\alpha+l,k+1)}{\Gamma(\alpha)} \\
&\quad +\frac{\mathbf{B}(l+1,k+1)}{\Gamma(\alpha-\beta+1)}\frac{1}{\Delta_1}
+\frac{\mathbf{B}(\alpha-\beta+l,k+1)}{\Gamma(\alpha-\beta)}\Big\}.
\end{aligned}
\end{gather*}
\end{theorem}

\begin{proof}
Let $X$ be defined above. By Lemma \ref{lem5.1.1}, we know that $x$
is a solution of \eqref{e5.1.4} if and only if $x$ is a fixed point of $T$.
By a standard proof, we can see that $T:X\to X$ is a completely continuous operator.
From (H1), for $x\in X$ we have
\begin{gather*}
|f_x(t)|=|f(t,x(t),{}^{RL}D_{0^+}^\beta x(t))|
\le M_f(\|x\|,\|x\|),\quad t\in (t_i,t_{i+1}],\;i\in \mathbb{N}[0,m],\\
|I(t_i,x(t_i),{}^{RL}D_{0^+}^\beta x(t_i))|\le M_I(\|x\|,\|x\|),
\quad i\in \mathbb{N}[1,m],\; x\in \mathbb{R},\\
|J(t_i,x(t_i),{}^{RL}D_{0^+}^\beta x(t_i))|\le M_J(\|x\|,\|x\|),\quad
i\in \mathbb{N}[1,m],\; x\in \mathbb{R}.
\end{gather*}
We consider the set
$\Omega=\{x\in X:x=\lambda (Tx),\text{ for some }\lambda \in [0,1)\}$.
 For $x\in \Omega$, by the definition of $T$ for $t\in (t_i,t_{i+1}]$ we have
\begin{align*}
&{}^{RL}D_{0^+}^\beta (Tx)(t) \\
&=\frac{1}{\Gamma(\alpha-\beta+1)}\frac{t^{\alpha-\beta-1}}{\Delta_1}
 \sum_{\sigma=1}^m\sum_{i=\sigma}^ma_iJ_{\sigma}
+\frac{1}{\Gamma(\alpha)}\frac{\Gamma(\alpha-1)}{\Gamma(\alpha-\beta-1)}
 \frac{t^{\alpha-\beta-2}}{\Delta_2}\frac{\Delta_3}{\Delta_1}
\sum_{\sigma=1}^m\sum_{i=\sigma}^ma_iJ_{\sigma}\\
&\quad +\frac{1}{\Gamma(\alpha)}\frac{\Gamma(\alpha-1)}{\Gamma(\alpha-\beta-1)}
\frac{t^{\alpha-\beta-2}}{\Delta_2}\sum_{\sigma=1}^m\sum_{i=\sigma}^m
b_i(\eta_i-t_\sigma)^{\alpha-1}J_\sigma \\
&\quad +\frac{\Gamma(\alpha-1)}{\Gamma(\alpha-\beta-1)}
 \frac{t^{\alpha-\beta-2}}{\Delta_2}
\sum_{\sigma=1}^m\sum_{i=\sigma}^mb_i(\eta_i-t_\sigma)^{\alpha-2}
I_\sigma \\
&\quad +\sum_{\sigma=1}^i(\frac{1}{\Gamma(\alpha-\beta+1)}
 (t-t_\sigma)^{\alpha-\beta-1}J_\sigma
 + \frac{\Gamma(\alpha-1)}{\Gamma(\alpha-\beta-1)}
 (t-t_\sigma)^{\alpha-\beta-2}I_\sigma)\\
&\quad +\frac{1}{\Gamma(\alpha)}\frac{\Gamma(\alpha-1)}
 {\Gamma(\alpha-\beta-1)}\frac{t^{\alpha-\beta-2}}{\Delta_2}
 \frac{\Delta_3}{\Delta_1}\sum_{i=1}^ma_i \int_0^{\xi_i}\sigma(s)ds \\
&\quad +\frac{\Gamma(\alpha-1)}
 {\Gamma(\alpha-\beta-1)}\frac{t^{\alpha-\beta-2}}{\Delta_2}\sum_{i=1}^mb_i
 \int_0^{\eta_i}\frac{(\eta_i-s)^{\alpha-1}}{\Gamma(\alpha)}\sigma(s)ds\\
&\quad -\int_0^1 \frac{(1-s)^{\alpha-1}}{\Gamma(\alpha)}\sigma(s)ds
 +\frac{1}{\Gamma(\alpha-\beta+1)}\frac{t^{\alpha-\beta-1}}{\Delta_1}\sum_{i=1}^ma_i
 \int_0^{\xi_i}\sigma(s)ds\\
&\quad +\int_0^t\frac{(t-s)^{\alpha-\beta-1}}{\Gamma(\alpha-\beta)}\sigma(s)ds,
\quad t\in (t_i,t_{i+1}],\;i\in \mathbb{N}[0,m].
\end{align*}
So
\begin{align*}
&(t-t_i)^{2- \alpha}|(Tx)(t)|\\
&\le \frac{1}{\Gamma(\alpha)}\Big[\frac{1}{|\Delta_1|}
 \sum_{\sigma=1}^m\sum_{i=\sigma}^m|a_i|
 +\frac{|\Delta_3|}{|\Delta_1|}\frac{1}{|\Delta_2|}
 \sum_{\sigma=1}^m\sum_{i=\sigma}^m|a_i| \\
&\quad +\frac{1}{|\Delta_2|} \sum_{\sigma=1}^m\sum_{i=\sigma}^m
 |b_i|(\eta_i-t_\sigma)^{\alpha-1}\Big]M_J(\|x\|,\|x\|)\\
&\quad +\frac{1}{|\Delta_2|}\sum_{\sigma=1}^m
 \sum_{i=\sigma}^m|b_i|(\eta_i-t_\sigma)^{\alpha-2}M_I(\|x\|,\|x\|) \\
&\quad +\frac{1}{\Gamma(\alpha)} \sum_{\sigma=1}^iM_J(\|x\|,\|x\|)
 +\sum_{\sigma=1}^iM_I(\|x\|,\|x\|)\\
&\quad +\frac{1}{\Gamma(\alpha)}\frac{1}{|\Delta_2|}
 \frac{|\Delta_3|}{|\Delta_1|}\sum_{i=1}^m|a_i|
 \int_0^{\xi_i}s^{k}(1-s)^ldsM_f(\|x\|,\|x\|) \\
&\quad +\frac{1}{|\Delta_2|}\sum_{i=1}^m|b_i|\int_0^{\eta_i}
 \frac{(\eta_i-s)^{\alpha-1}}{\Gamma(\alpha)}s^k(1-s)^ldsM_f(\|x\|,\|x\|)\\
&\quad +\frac{1}{|\Delta_2|} \int_0^1 \frac{(1-s)^{\alpha-1}}
 {\Gamma(\alpha)}s^k(1-s)^ldsM_f(\|x\|,\|x\|) \\
&\quad +\frac{1}{\Gamma(\alpha)}
 \frac{1}{|\Delta_1|}\sum_{i=1}^m|a_i|
 \int_0^{\xi_i}s^k(1-s)^ldsM_f(\|x\|,\|x\|)\\
&\quad +(t-t_i)^{2-\alpha}\int_0^t\frac{(t-s)^{\alpha-1}}
 {\Gamma(\alpha)}s^k(1-s)^ldsM_f(\|x\|,\|x\|) \\
&\le \frac{1}{\Gamma(\alpha)}\Big[\Big(
\frac{1}{|\Delta_1|}+\frac{|\Delta_3|}{|\Delta_1|}\frac{1}{|\Delta_2|}\Big)
\sum_{i=1}^mi|a_i|+\frac{1}{|\Delta_2|}
\sum_{i=1}^mi|b_i|+m\Big]M_J(\|x\|,\|x\|)\\
&\quad +\Big[\frac{1}{|\Delta_2|}\sum_{\sigma=1}^m
\sum_{i=\sigma}^m|b_i|(\eta_i-t_\sigma)^{\alpha-2}
+m\Big]M_I(\|x\|,\|x\|)\\
&\quad +\Big[(\frac{\mathbf{B}(l+1,k+1)}{\Gamma(\alpha)|\Delta_2|}
 \frac{|\Delta_3|}{|\Delta_1|}
 +\frac{\mathbf{B}(l+1,k+1)}{\Gamma(\alpha)|\Delta_1|})\sum_{i=1}^m|a_i|
+\frac{\mathbf{B}(\alpha+l,k+1)}{\Gamma(\alpha)|\Delta_2|}\sum_{i=1}^m|b_i| \\
&\quad +\frac{\mathbf{B}(\alpha+l,k+1)}{\Gamma(\alpha)|\Delta_2|}
+\frac{\mathbf{B}(l+1,k+1)}{\Gamma(\alpha)}\Big]M_f(\|x\|,\|x\|).
\end{align*}
Furthermore,
\begin{align*}
&(t-t_\sigma)^{2+\beta-\alpha}|{}^{RL}D_{0^+}^\beta (Tx)(t)|\\
&\le \frac{1}{\Gamma(\alpha-\beta+1)}\frac{1}{|\Delta_1|}
 \sum_{\sigma=1}^m\sum_{i=\sigma}^m|a_i|M_J(\|x\|,\|x\|) \\
&\quad +\frac{1}{\Gamma(\alpha)}
 \frac{\Gamma(\alpha-1)}{\Gamma(\alpha-\beta-1)}
 \frac{1}{|\Delta_2|}\frac{|\Delta_3|}{|\Delta_1|}
 \sum_{\sigma=1}^m\sum_{i=\sigma}^m|a_i|M_J(\|x\|,\|x\|)\\
&\quad +\frac{1}{\Gamma(\alpha)}\frac{\Gamma(\alpha-1)}{\Gamma(\alpha-\beta-1)}
\frac{1}{|\Delta_2|}\sum_{\sigma=1}^m\sum_{i=\sigma}^m
|b_i|M_J(\|x\|,\|x\|) \\
&\quad +\frac{\Gamma(\alpha-1)}{\Gamma(\alpha-\beta-1)}\frac{1}{|\Delta_2|}
\sum_{\sigma=1}^m\sum_{i=\sigma}^m|b_i|(\eta_i-t_\sigma)^{\alpha-2}M_I(\|x\|,\|x\|)\\
&\quad +\sum_{\sigma=1}^i\Big(\frac{1}{\Gamma(\alpha-\beta+1)}M_{J}(\|x\|,\|x\|)+
\frac{\Gamma(\alpha-1)}{\Gamma(\alpha-\beta-1)}
M_I(\|x\|,\|x\|)\Big) \\
&\quad +\frac{1}{\Gamma(\alpha)}\frac{\Gamma(\alpha-1)}
{\Gamma(\alpha-\beta-1)}\frac{1}{|\Delta_2|}
\frac{|\Delta_3|}{|\Delta_1|}\sum_{i=1}^m|a_i|
\int_0^{\xi_i}s^k(1-s)^ldsM_f(\|x\|,\|x\|)\\
&\quad +\frac{\Gamma(\alpha-1)}{\Gamma(\alpha-\beta-1)}
 \frac{1}{|\Delta_2|}\sum_{i=1}^m|b_i|
\int_0^{\eta_i}\frac{(\eta_i-s)^{\alpha-1}}{\Gamma(\alpha)}s^k(1-s)^lds
 M_f(\|x\|,\|x\|) \\
&\quad +\int_0^1 \frac{(1-s)^{\alpha-1}}{\Gamma(\alpha)}s^k(1-s)^lds
 M_f(\|x\|,\|x\|) \\
&\quad +\frac{1}{\Gamma(\alpha-\beta+1)}\frac{1}{\Delta_1}\sum_{i=1}^ma_i
\int_0^{\xi_i}s^k(1-s)^ldsM_f(\|x\|,\|x\|)\\
&\quad +(t-t_i)^{2+\beta-\alpha}\int_0^t\frac{(t-s)^{\alpha-\beta-1}}
{\Gamma(\alpha-\beta)}s^k(1-s)^ldsM_f(\|x\|,\|x\|)\\
&\le \Big[\Big(\frac{1}{\Gamma(\alpha-\beta+1)|\Delta_1|}
+\frac{\Gamma(\alpha-1)}{\Gamma(\alpha-\beta-1)}
 \frac{1}{\Gamma(\alpha)|\Delta_2|}\frac{|\Delta_3|}{|\Delta_1|}\Big)
\sum_{i=1}^mi|a_i| \\
&\quad +\frac{\Gamma(\alpha-1)}{\Gamma(\alpha-\beta-1)}
\frac{1}{\Gamma(\alpha)|\Delta_2|}\sum_{i=1}^mi|b_i|
 +\frac{m}{\Gamma(\alpha-\beta+1)}\Big]M_{J}(\|x\|,\|x\|)\\
&\quad +\Big[\frac{\Gamma(\alpha-1)}{\Gamma(\alpha-\beta-1)}\frac{1}{|\Delta_2|}
\sum_{\sigma=1}^m\sum_{i=\sigma}^m|b_i|(\eta_i-t_\sigma)^{\alpha-2}+
\frac{m\Gamma(\alpha-1)}{\Gamma(\alpha-\beta-1)}\Big]
M_I(\|x\|,\|x\|)\\
&\quad +\Big[\frac{\mathbf{B}(l+1,k+1)}{\Gamma(\alpha)}
 \frac{\Gamma(\alpha-1)}{\Gamma(\alpha-\beta-1)}\frac{1}{|\Delta_2|}
 \frac{|\Delta_3|}{|\Delta_1|} \sum_{i=1}^m|a_i| \\
&\quad +\frac{\Gamma(\alpha-1)}{\Gamma(\alpha-\beta-1)}\frac{1}{|\Delta_2|}
 \sum_{i=1}^m|b_i| \frac{\mathbf{B}(\alpha+l,k+1)}{\Gamma(\alpha)}
 +\frac{\mathbf{B}(\alpha+l,k+1)}{\Gamma(\alpha)}\\
&\quad +\frac{\mathbf{B}(l+1,k+1)}{\Gamma(\alpha-\beta+1)}\frac{1}{\Delta_1}
+\frac{\mathbf{B}(\alpha-\beta+l,k+1)}{\Gamma(\alpha-\beta)}
\Big]M_f(\|x\|,\|x\|).
\end{align*}
It follows that
\[
\|Tx\|\le A_1M_J(\|x\|,\|x\|)+A_2M_I(\|x\|,\|x\|) +A_3M_f(\|x\|,\|x\|).
\]
From the assumption, we choose $ \Omega=\{x\in X:\|x\|\le r_0\}$.
For $x\in \partial \Omega$, we obtain $x\neq \lambda (Tx)$ for any
$\lambda\in [0,1]$. In fact, if there exists $x\in \partial \Omega$ such that
 $x=\lambda (Tx)$ for some $\lambda\in [0,1]$. Then
\[
r_0=\|x\|=\lambda\|Tx\|<\|Tx\|
\le A_1M_J(r_0,r_0)+A_2M_I(r_0,r_0)+A_3M_f(r_0,r_0)\le r_0,
\]
which is a contradiction.

 As a consequence of Schaefer's fixed point theorem, we deduce that
$T$ has a fixed point which is a solution of problem
\eqref{e5.1.4}. The proof is complete.
 \end{proof}

\begin{theorem} \label{thm5.1.2}
 Suppose that \eqref{e5.1.10}, {\rm (5.A1)-(5.A4), (H1)} hold and
\[
\lim_{r\to 0^+}\frac{ M_f(r,r)}{r}
=\lim_{r\to 0^+}\frac{M_I(r,r)}{r}=\lim_{r\to 0^+}\frac{ M_J(r,r)}{r}=0
\]
 or
\[
\lim_{r\to +\infty}\frac{ M_f(r,r)}{r}
=\lim_{r\to +\infty}\frac{M_I(r,r)}{r}
=\lim_{r\to +\infty}\frac{ M_J(r,r)}{r}=0.
\]
Then \eqref{e5.1.4} has at least one solution.
\end{theorem}

\begin{proof}
Let $X$ be defined above. By Lemma \ref{lem5.1.1}, we know that $x$ is a solution of
\eqref{e5.1.4} if and only if $x$ is a fixed point of $T$.
By a standard proof, we can see that $T:X\to X$ is a completely continuous
operator.

From (H1), as in the proof of Theorem \ref{thm5.1.1}, we have
\[
\|Tx\|\le A_1M_J(\|x\|,\|x\|)+A_2M_I(\|x\|,\|x\|) +A_3M_f(\|x\|,\|x\|).
\]
Choose $\delta_0>0$ such that $(A_1+A_2+A_3)\delta_0\le 1$.
By the assumption, we know that there exist a constant $M>0$ such that
$$
 M_f(r,r)\le \delta_0r,\quad
 M_I(r,r)\le \delta_0r,\quad
 M_J(r,r)\le \delta_0r,\quad
 r\in [0,M] \text{ or }r\in [M,+\infty).
$$
We choose $ \Omega=\{x\in X:\|x\|< M\}$.
 Then $\Omega$ is an open bounded subset of $X$ and $0\in \Omega$.
For $x\in \partial \Omega$, we have $\|x\|=M$. Thus
\begin{align*}
\|Tx\|&\le A_1M_J(\|x\|,\|x\|)+A_2M_I(\|x\|,\|x\|)
+A_3M_f(\|x\|,\|x\|)\\
&\le A_1\delta_0M+A_2\delta_0M+A_3\delta_0M\le M=\|x\|.
\end{align*}
 As a consequence of Theorem \ref{thm3.1.10}, we deduce that $T$ has a fixed point
which is a solution of \eqref{e5.1.4}. The proof is complete.
\end{proof}

\subsection{Impulsive Sturm-Liouville boundary value problems}

Zhang and Feng \cite{zf} studied the Sturm-Liouville boundary value problem
of impulsive fractional differential equation
\begin{equation}
\begin{gathered}
{}^{C}D_{0^+}^{q} x(t)=\omega(t)f(t,x(t),x'(t)),\quad
t\in (t_i,t_{i+1}],\;i\in \mathbb{N}[0,m],\\
\Delta x(t_i)=I_i(x(t_i)), \quad i\in \mathbb{N}[1,m],\\
\Delta x'(t_i)=J_i(x(t_i)),\quad i\in \mathbb{N}[1,m],\\
\alpha_1x(0)-\beta_1x'(0)=\alpha_2x(1)+\beta_2x'(1)=0,
\end{gathered} \label{e5.2.1}
\end{equation}
where $q\in (1,2]$, ${}^{C}D_{0^+}^{q}u$ is the Caputo fractional derivative,
 $\omega:[0,1]\to [0,+\infty)$, $f:[0,1]\times \mathbb{R}^2\to \mathbb{R}$,
$I_i,J_i:\mathbb{R}\to \mathbb{R}$ are continuous functions,
$0=t_0<t_1<\dots<t_m<t_{m+1}=1$,
 $\alpha_1\alpha_2+\alpha_1\beta_2+\alpha_2\beta_1\neq 0$.
The existence and uniqueness of solutions of BVP\eqref{e5.2.1} were
established under the assumptions that $f$, $I_i,J_i$ are bounded functions.
The following theorem was proved in \cite{zf}.

\begin{theorem}[\cite{zf}] \label{thmZF}
 Suppose that $\sigma\in C[0,1]$, $I_i,J_i$ are continuous, and
$\eta=\alpha_1\alpha_2+\alpha_1\beta_2+\alpha_1\beta_1\neq 0$.
The solution of the problem
\begin{equation}
\begin{gathered}
{}^{C}D_{0^+}^{q} x(t)=\sigma(t),\quad t\in (t_i,t_{i+1}],\;i\in \mathbb{N}[0,m],\\
\Delta x(t_i)=I_i(x(t_i)),\quad i\in \mathbb{N}[1,m],\\
\Delta x'(t_i)=J_i(x(t_i)),\quad i\in \mathbb{N}[1,m],\\
\alpha_1x(0)-\beta_1x'(0)=\alpha_2x(1)+\beta_2x'(1)=0,
\end{gathered} \label{eI}
\end{equation}
can be expressed as
\begin{equation}
\begin{aligned}
x(t)&=\int_{t_k}^t\frac{(t-s)^{q-1}}{\Gamma(q)}\sigma(s)ds
 +\sum_{i=1}^{m+1}G'_{1s}(t,t_i)\int_{t_{i-1}}^{t_i}
 \frac{(t_i-s)^{q-1}}{\Gamma(q)}\sigma(s)ds\\
&-\sum_{i=1}^nG_1(t,t_i)\int_{t_{i-1}}^{t_i}
 \frac{(t_i-s)^{q-2}}{\Gamma(q-1)}\sigma(s)ds+\sum_{i=1}^nG_{1s}'(t,t_i)I_i(x(t_i))\\
&-\sum_{i=1}^nG_1(t,t_i)J_i(x(t_i)),\quad t\in (t_k,t_{k+1}],\; k\in \mathbb{N}[0,m],
\end{aligned} \label{eII}
\end{equation}
where
$$
G_1(t,s)=-\frac{1}{\eta} \begin{cases}
(\beta_1+\alpha_1t)(\alpha_2(1-s)+\beta_2), & t\le s,\\
(\beta_1+\alpha_1s)(\alpha_2(1-t)+\beta_2), & s\le t.
\end{cases}
$$
\end{theorem}

This result is wrong. In fact, suppose that $u$ is a solution of \eqref{eI}.
By Theorem \ref{thm3.2.1} (with $\lambda=0$), we know from
${}^{C}D_{0^+}^{q} x(t)=\sigma(t)$, $t\in (t_i,t_{i+1}],\;i\in \mathbb{N}[0,m]$
that there exist constants $c_{\sigma v}\in \mathbb{R}$ such that
$$
x(t)=\sum_{\sigma=0}^ic_{\sigma }+\sum_{\sigma=0}^id_{\sigma}(t-t_\sigma)
+\int_0^t\frac{(t-s)^{\alpha-1}}{\Gamma(\alpha)}\sigma(s)ds,
\quad t\in (t_i,t_{i+1}],\; i\in N[0,m].
$$
Then
$$
x'(t)=\sum_{\sigma=0}^id_{\sigma}+\int_0^t\frac{(t-s)^{\alpha-2}}
{\Gamma(\alpha-1)}\sigma(s)ds,\quad t\in (t_i,t_{i+1}],\; i\in N[0,m].
$$
From $\Delta x(t_i)=I_i(x(t_i)),\Delta x'(t_i)=J_i(x(t_i)),i\in \mathbb{N}[1,m]$,
we obtain $c_i=I_i(x(t_i))$ and $d_i=J_i(x(t_i)),i\in \mathbb{N}[1,m]$.
By $\alpha_1x(0)-\beta_1x'(0)=\alpha_2x(1)+\beta_2x'(1)=0$, we have
\begin{gather*}
\alpha_1c_0-\beta_1d_0=0,\\
\begin{aligned}
&\alpha_2\Big[\sum_{\sigma=0}^mc_{\sigma }+\sum_{\sigma=0}^md_{\sigma}(1-t_\sigma)
+\int_0^1\frac{(1-s)^{\alpha-1}}{\Gamma(\alpha)}\sigma(s)ds\Big] \\
&+\beta_2\Big[\sum_{\sigma=0}^md_{\sigma}
 +\int_0^1\frac{(1-s)^{\alpha-2}}{\Gamma(\alpha-1)}\sigma(s)ds\Big]=0.
\end{aligned}
\end{gather*}
It follows that
\begin{gather*}
\alpha_1c_0-\beta_1d_0=0, \\
\begin{aligned}
\alpha_2c_0+(\alpha_2+\beta_2)d_0
&=-\alpha_2\sum_{\sigma=1}^mI_\sigma(x(t_\sigma))
 -\sum_{\sigma=1}^m[\alpha_2(1-t_\sigma)+\beta_2]J_\sigma(x(t_\sigma))\\
&\quad -\int_0^1\big[\frac{\alpha_2(1-s)^{\alpha-1}}{\Gamma(\alpha)}
+\frac{\beta_2(1-s)^{\alpha-2}}{\Gamma(\alpha-1)}\big]\sigma(s)ds.
\end{aligned}
\end{gather*}
Then
\begin{gather*}
\begin{aligned}
c_0&=\beta_1\Big(-\alpha_2\sum_{\sigma=1}^mI_\sigma(x(t_\sigma))
-\sum_{\sigma=1}^m[\alpha_2(1-t_\sigma)+\beta_2]
J_\sigma(x(t_\sigma)) \\
&\quad -\int_0^1[\frac{\alpha_2(1-s)^{\alpha-1}}
{\Gamma(\alpha)}-\frac{\beta_2(1-s)^{\alpha-2}}{\Gamma(\alpha-1)}
]\sigma(s)ds\Big)/\eta,
\end{aligned}\\
\begin{aligned}
d_0&=\alpha_1\Big(-\alpha_2\sum_{\sigma=1}^mI_\sigma(x(t_\sigma))
-\sum_{\sigma=1}^m[\alpha_2(1-t_\sigma)+\beta_2]
J_\sigma(x(t_\sigma)) \\
&\quad -\int_0^1[\frac{\alpha_2(1-s)^{\alpha-1}}{\Gamma(\alpha)}
+\frac{\beta_2(1-s)^{\alpha-2}}{\Gamma(\alpha-1)}]\sigma(s)ds\Big)/\eta.
\end{aligned}
\end{gather*}
Hence
\begin{align*}
x(t)&=\beta_1\Big(-\alpha_2\sum_{\sigma=1}^mI_\sigma(x(t_\sigma))
-\sum_{\sigma=1}^m[\alpha_2(1-t_\sigma)+\beta_2]
J_\sigma(x(t_\sigma)) \\
&\quad -\int_0^1[\frac{\alpha_2(1-s)^{\alpha-1}}
{\Gamma(\alpha)}-\frac{\beta_2(1-s)^{\alpha-2}}{\Gamma(\alpha-1)}]\sigma(s)ds
\Big)/\eta
\\
&\quad +\alpha_1\Big(-\alpha_2\sum_{\sigma=1}^mI_\sigma(x(t_\sigma))
-\sum_{\sigma=1}^m[\alpha_2(1-t_\sigma)+\beta_2] J_\sigma(x(t_\sigma)) \\
&\quad -\int_0^1[\frac{\alpha_2(1-s)^{\alpha-1}}{\Gamma(\alpha)}
 +\frac{\beta_2(1-s)^{\alpha-2}}{\Gamma(\alpha-1)}]\sigma(s)ds\Big)t/\eta\\
&\quad\times \sum_{\sigma=1}^iI_\sigma(x(t_\sigma))
 +\sum_{\sigma=1}^iJ_\sigma(x(t_\sigma))(t-t_\sigma)
+\int_0^t\frac{(t-s)^{\alpha-1}}{\Gamma(\alpha)}\sigma(s)ds,
\end{align*}
for $t\in (t_i,t_{i+1}]$ and $i\in N[0,m]$.
This is the correct expression of solutions of \eqref{eI}.
This result shows that \cite[Theorem ZF]{zf} is wrong.

\begin{theorem} \label{thm5.2.0}
 Suppose that $\sigma\in C[0,1]$, $I_i,J_i$ are continuous, and
\begin{equation}
\zeta=\alpha_1\Big[\alpha_2\sum_{\tau=0}^{m-1}(t_{\tau+1}
-t_{\tau})+[\alpha_2+\beta_2](1-t_m)\Big]+\alpha_2\beta_1\neq 0.
\end{equation}
Then the solution of BVP for fractional differential equation with
multiple starting points $t_i$,
\begin{equation}
\begin{gathered}
{}^{C}D_{t_i^+}^{q} x(t)=\sigma(t),\quad t\in (t_i,t_{i+1}],\;i\in \mathbb{N}[0,m],\\
\Delta x(t_i)=I_i(x(t_i)),\quad i\in \mathbb{N}[1,m],\\
\Delta x'(t_i)=J_i(x(t_i)),\quad i\in \mathbb{N}[1,m],\\
\alpha_1x(0)-\beta_1x'(0)=\alpha_2x(1)+\beta_2x'(1)=0,
\end{gathered} \label{eIII}
\end{equation}
can be expressed as
\begin{equation}
x(t)= \begin{cases}
c_0+d_0t+\int_0^t\frac{(t-s)^{q-1}}{\Gamma(q)}\sigma(s)ds,\quad t\in (0,t_1],\\[4pt]
c_0+\sum_{\tau=0}^{i-1}(t_{\tau+1}-t_{\tau})d_0
+\sum_{j=1}^{i-1}\sum_{\tau=j}^{i-1}(t_{\tau+1}-t_{\tau})J_j(x(t_j)) \\
+\sum_{\tau=1}^iI_\tau(x(t_\tau))
+\sum_{\tau=0}^{i-1}\int_{t_{\tau}}^{t_{\tau+1}}
 \frac{(t_{\tau+1}-s)^{\alpha-1}}{\Gamma(\alpha)}\sigma(s)ds \\
+ \sum_{j=0}^{i-2}\sum_{\tau=j+1}^{i-1}(t_{\tau+1}-t_{\tau})
\int_{t_{j}}^{t_{j+1}} \frac{(t_{j+1}-s)^{\alpha-2}}{\Gamma(\alpha-1)}\sigma(s)ds\\
+\big[d_0+\sum_{j=0}^{i-1}\int_{t_{j}}^{t_{j+1}}\frac{(t_{j+1}-s)^{\alpha-2}}
 {\Gamma(\alpha-1)}\sigma(s)ds+\sum_{j=1}^iJ_j(x(t_j))\big](t-t_i)\\
+\int_{t_i}^t\frac{(t-s)^{q-1}}{\Gamma(q)}\sigma(s)ds, \quad
 t\in (t_i,t_{i+1}],\; i\in \mathbb{N}[1,m].
\end{cases}
\label{eIV}
\end{equation}
Here $c_0,d_0$ are defined by
\begin{align*}
d_0&=-\frac{\alpha_1}{\zeta}\Big[\alpha_2\sum_{j=1}^{m-1}
 \sum_{\tau=j}^{m-1}(t_{\tau+1}-t_{\tau})J_j(x(t_j))
 + \sum_{j=1}^{m}J_j(x(t_j))+\alpha_2\sum_{\tau=1}^{m}I_\tau(x(t_\tau)) \\
&\quad +\alpha_2\sum_{\tau=0}^{m-1}\int_{t_{\tau}}^{t_{\tau+1}}
 \frac{(t_{\tau+1}-s)^{\alpha-1}}{\Gamma(\alpha)}\sigma(s)ds \\
&\quad + \alpha_2\sum_{j=0}^{m-2}\sum_{\tau=j+1}^{m-1}(t_{\tau+1}-t_{\tau})
 \int_{t_{j}}^{t_{j+1}}
\frac{(t_{j+1}-s)^{\alpha-2}}{\Gamma(\alpha-1)}\sigma(s)ds\\
&\quad +[\alpha_2+\beta_2](1-t_m)\sum_{j=0}^{m-1}\int_{t_{j}}^{t_{j+1}}
 \frac{(t_{j+1}-s)^{\alpha-2}}{\Gamma(\alpha-1)}\sigma(s)ds\\
&\quad +\alpha_2\int_{t_m}^1\frac{(1-s)^{\alpha-1}}{\Gamma(\alpha)}\sigma(s)ds
+\beta_2\int_{t_m}^1\frac{(1-s)^{\alpha-2}}{\Gamma(\alpha-1)}\sigma(s)ds\Big],
\end{align*}
\begin{align*}
c_0&=-\frac{\beta_1}{\zeta}\Big[\alpha_2\sum_{j=1}^{m-1}
\sum_{\tau=j}^{m-1}(t_{\tau+1}-t_{\tau})J_j(x(t_j))+
\sum_{j=1}^{m}J_j(x(t_j))+\alpha_2\sum_{\tau=1}^{m}I_\tau(x(t_\tau)) \\
&\quad +\alpha_2\sum_{\tau=0}^{m-1}\int_{t_{\tau}}^{t_{\tau+1}}
 \frac{(t_{\tau+1}-s)^{\alpha-1}}{\Gamma(\alpha)}\sigma(s)ds \\
&\quad +\alpha_2\sum_{j=0}^{m-2}\sum_{\tau=j+1}^{m-1}(t_{\tau+1}-t_{\tau})
\int_{t_{j}}^{t_{j+1}}
\frac{(t_{j+1}-s)^{\alpha-2}}{\Gamma(\alpha-1)}\sigma(s)ds\\
&\quad +[\alpha_2+\beta_2](1-t_m)\sum_{j=0}^{m-1}
 \int_{t_{j}}^{t_{j+1}}\frac{(t_{j+1}-s)^{\alpha-2}}{\Gamma(\alpha-1)}\sigma(s)ds\\
&\quad +\alpha_2\int_{t_m}^1\frac{(1-s)^{\alpha-1}}{\Gamma(\alpha)}\sigma(s)ds
+\beta_2\int_{t_m}^1\frac{(1-s)^{\alpha-2}}{\Gamma(\alpha-1)}\sigma(s)ds\Big].
\end{align*}
\end{theorem}

\begin{proof}
Suppose that $x$ is a solution of \eqref{eIII}. From Theorem \ref{thm3.2.1} and
${}^{C}D_{t_i^+}^{q} x(t)=\sigma(t),t\in (t_i,t_{i+1}]$,
there exist constants $c_i,d_i\in \mathbb{R}$ such that
$$
x(t)=c_{i}+d_{i}(t-t_i)
+\int_{t_i}^t\frac{(t-s)^{\alpha-1}}{\Gamma(\alpha)}\sigma(s)ds,\quad
t\in (t_i,t_{i+1}],\; i\in N[0,m].
$$
Then
$$
x'(t)=d_{i}+\int_{t_i}^t\frac{(t-s)^{\alpha-2}}{\Gamma(\alpha-1)}\sigma(s)ds,
\quad t\in (t_i,t_{i+1}],\; i\in N[0,m].
$$
By $\Delta x(t_i)=I_i(x(t_i)),\Delta x'(t_i)=J_i(x(t_i)),i\in \mathbb{N}[1,m]$, we obtain
\begin{gather*}
c_i-\Big[c_{i-1}+d_{i-1}(t_i-t_{i-1})
+\int_{t_{i-1}}^{t_i}\frac{(t_i-s)^{\alpha-1}}{\Gamma(\alpha)}\sigma(s)ds\Big]
=I_i(x(t_i)),\\
d_i-\Big[d_{i-1}+\int_{t_{i-1}}^{t_i}\frac{(t_i-s)^{\alpha-2}}{\Gamma(\alpha-1)}
\sigma(s)ds\Big]=J_i(x(t_i)), \quad i\in \mathbb{N}[1,m].
 \end{gather*}
 It follows that
 $$
d_i=d_0+\sum_{j=0}^{i-1}\int_{t_{j}}^{t_{j+1}}
 \frac{(t_{j+1}-s)^{\alpha-2}}{\Gamma(\alpha-1)}\sigma(s)ds
+\sum_{j=1}^iJ_j(x(t_j)), \quad i\in \mathbb{N}[1,m].
 $$
 Then
\begin{align*}
c_i 
&=c_{i-1}+(t_i-t_{i-1})d_0
+(t_i-t_{i-1})\sum_{j=1}^{i-1}J_j(x(t_j))+I_i(x(t_i))\\
&\quad +\int_{t_{i-1}}^{t_i}\frac{(t_i-s)^{\alpha-1}}
 {\Gamma(\alpha)}\sigma(s)ds+(t_i-t_{i-1})\sum_{j=0}^{i-2}\int_{t_{j}}^{t_{j+1}}
\frac{(t_{j+1}-s)^{\alpha-2}}{\Gamma(\alpha-1)}\sigma(s)ds\\
&=c_0+\sum_{\tau=0}^{i-1}(t_{\tau+1}-t_{\tau})d_0
+\sum_{\tau=0}^{i-1}(t_{\tau+1}-t_{\tau})
 \sum_{j=1}^{\tau}J_j(x(t_j))+\sum_{\tau=1}^iI_\tau(x(t_\tau))\\
&\quad +\sum_{\tau=0}^{i-1}\int_{t_{\tau}}^{t_{\tau+1}}
 \frac{(t_{\tau+1}-s)^{\alpha-1}}{\Gamma(\alpha)}\sigma(s)ds \\
&\quad +\sum_{\tau=0}^{i-1}(t_{\tau+1}-t_{\tau})
\sum_{j=0}^{\tau-1}\int_{t_{j}}^{t_{j+1}}
\frac{(t_{j+1}-s)^{\alpha-2}}{\Gamma(\alpha-1)}\sigma(s)ds\\
&=c_0+\sum_{\tau=0}^{i-1}(t_{\tau+1}-t_{\tau})d_0
+\sum_{j=1}^{i-1}\sum_{\tau=j}^{i-1}(t_{\tau+1}-t_{\tau})J_j(x(t_j))
+\sum_{\tau=1}^iI_\tau(x(t_\tau))\\
&\quad +\sum_{\tau=0}^{i-1}\int_{t_{\tau}}^{t_{\tau+1}}
 \frac{(t_{\tau+1}-s)^{\alpha-1}}{\Gamma(\alpha)}\sigma(s)ds \\
&\quad + \sum_{j=0}^{i-2}\sum_{\tau=j+1}^{i-1}(t_{\tau+1}-t_{\tau}) \int_{t_{j}}^{t_{j+1}}
\frac{(t_{j+1}-s)^{\alpha-2}}{\Gamma(\alpha-1)}\sigma(s)ds,\quad i\in \mathbb{N}[1,m].
 \end{align*}
By $\alpha_1x(0)-\beta_1x'(0)=\alpha_2x(1)+\beta_2x'(1)=0$, we have
\begin{gather*}
\alpha_1c_0-\beta_1d_0=0,\\
\begin{aligned}
&\alpha_2\Big[c_{m}+d_{m}(1-t_m)
+\int_{t_m}^1\frac{(1-s)^{\alpha-1}}{\Gamma(\alpha)}\sigma(s)ds\Big]\\
&+\beta_2\Big[d_{m}+\int_{t_m}^1\frac{(1-s)^{\alpha-2}}
{\Gamma(\alpha-1)}\sigma(s)ds\Big]=0.
\end{aligned}
\end{gather*}
It follows that
\begin{gather*}
\alpha_1c_0-\beta_1d_0=0,\\
\begin{aligned}
&\alpha_2 c_0+\Big[\alpha_2\sum_{\tau=0}^{m-1}(t_{\tau+1}-t_{\tau})
+[\alpha_2+\beta_2](1-t_m)\Big]d_0\\
&+\alpha_2\sum_{j=1}^{m-1}\sum_{\tau=j}^{m-1}(t_{\tau+1}-t_{\tau})J_j(x(t_j))+
\sum_{j=1}^{m}J_j(x(t_j))+\alpha_2\sum_{\tau=1}^{m}I_\tau(x(t_\tau))\\
&+\alpha_2\sum_{\tau=0}^{m-1}\int_{t_{\tau}}^{t_{\tau+1}}
 \frac{(t_{\tau+1}-s)^{\alpha-1}}{\Gamma(\alpha)}\sigma(s)ds \\
&+\alpha_2\sum_{j=0}^{m-2}\sum_{\tau=j+1}^{m-1}(t_{\tau+1}-t_{\tau})
 \int_{t_{j}}^{t_{j+1}}
 \frac{(t_{j+1}-s)^{\alpha-2}}{\Gamma(\alpha-1)}\sigma(s)ds\\
&\quad +[\alpha_2+\beta_2](1-t_m)\sum_{j=0}^{m-1}\int_{t_{j}}^{t_{j+1}}
 \frac{(t_{j+1}-s)^{\alpha-2}}{\Gamma(\alpha-1)}\sigma(s)ds\\
&\quad +\alpha_2\int_{t_m}^1\frac{(1-s)^{\alpha-1}}{\Gamma(\alpha)}\sigma(s)ds
+\beta_2\int_{t_m}^1\frac{(1-s)^{\alpha-2}}{\Gamma(\alpha-1)}\sigma(s)ds=0.
\end{aligned}
\end{gather*}
Hence
\begin{align*}
d_0&=-\frac{\alpha_1}{\zeta}\Big[\alpha_2\sum_{j=1}^{m-1}
 \sum_{\tau=j}^{m-1}(t_{\tau+1}-t_{\tau})J_j(x(t_j))
 +\sum_{j=1}^{m}J_j(x(t_j))+\alpha_2\sum_{\tau=1}^{m}I_\tau(x(t_\tau)) \\
&\quad +\alpha_2\sum_{\tau=0}^{m-1}\int_{t_{\tau}}^{t_{\tau+1}}
 \frac{(t_{\tau+1}-s)^{\alpha-1}}{\Gamma(\alpha)}\sigma(s)ds \\
&\quad +\alpha_2\sum_{j=0}^{m-2}\sum_{\tau=j+1}^{m-1}(t_{\tau+1}-t_{\tau})
\int_{t_{j}}^{t_{j+1}}\frac{(t_{j+1}-s)^{\alpha-2}}{\Gamma(\alpha-1)}\sigma(s)ds\\
&\quad +[\alpha_2+\beta_2](1-t_m)\sum_{j=0}^{m-1}\int_{t_{j}}^{t_{j+1}}
\frac{(t_{j+1}-s)^{\alpha-2}}{\Gamma(\alpha-1)}\sigma(s)ds\\
&\quad +\alpha_2\int_{t_m}^1\frac{(1-s)^{\alpha-1}}{\Gamma(\alpha)}\sigma(s)ds
+\beta_2\int_{t_m}^1\frac{(1-s)^{\alpha-2}}{\Gamma(\alpha-1)}\sigma(s)ds\Big],
\end{align*}
\begin{align*}
c_0&=-\frac{\beta_1}{\zeta}\Big[\alpha_2\sum_{j=1}^{m-1}
 \sum_{\tau=j}^{m-1}(t_{\tau+1}-t_{\tau})J_j(x(t_j))
 +\sum_{j=1}^{m}J_j(x(t_j))+\alpha_2\sum_{\tau=1}^{m}I_\tau(x(t_\tau)) \\
&\quad +\alpha_2\sum_{\tau=0}^{m-1}\int_{t_{\tau}}^{t_{\tau+1}}
 \frac{(t_{\tau+1}-s)^{\alpha-1}}{\Gamma(\alpha)}\sigma(s)ds \\
&\quad +\alpha_2\sum_{j=0}^{m-2}\sum_{\tau=j+1}^{m-1}(t_{\tau+1}-t_{\tau})
\int_{t_{j}}^{t_{j+1}}
\frac{(t_{j+1}-s)^{\alpha-2}}{\Gamma(\alpha-1)}\sigma(s)ds\\
&\quad +[\alpha_2+\beta_2](1-t_m)\sum_{j=0}^{m-1}
 \int_{t_{j}}^{t_{j+1}}\frac{(t_{j+1}-s)^{\alpha-2}}
 {\Gamma(\alpha-1)}\sigma(s)ds\\
&\quad +\alpha_2\int_{t_m}^1\frac{(1-s)^{\alpha-1}}{\Gamma(\alpha)}\sigma(s)ds
+\beta_2\int_{t_m}^1\frac{(1-s)^{\alpha-2}}{\Gamma(\alpha-1)}\sigma(s)ds\Big].
\end{align*}
It follows that
$$
x(t)= \begin{cases}
c_0+d_0t+\int_0^t\frac{(t-s)^{q-1}}{\Gamma(q)}\sigma(s)ds, \quad t\in (0,t_1],\\[4pt]
c_0+\sum_{\tau=0}^{i-1}(t_{\tau+1}-t_{\tau})d_0
+\sum_{j=1}^{i-1}\sum_{\tau=j}^{i-1}(t_{\tau+1}-t_{\tau})J_j(x(t_j)) \\
+\sum_{\tau=1}^iI_\tau(x(t_\tau))
+\sum_{\tau=0}^{i-1}\int_{t_{\tau}}^{t_{\tau+1}}
 \frac{(t_{\tau+1}-s)^{\alpha-1}}{\Gamma(\alpha)}\sigma(s)ds \\
+\sum_{j=0}^{i-2}\sum_{\tau=j+1}^{i-1}(t_{\tau+1}-t_{\tau})
\int_{t_{j}}^{t_{j+1}}
\frac{(t_{j+1}-s)^{\alpha-2}}{\Gamma(\alpha-1)}\sigma(s)ds\\
+\Big[d_0+\sum_{j=0}^{i-1}\int_{t_{j}}^{t_{j+1}}\frac{(t_{j+1}-s)^{\alpha-2}}
{\Gamma(\alpha-1)}\sigma(s)ds+\sum_{j=1}^iJ_j(x(t_j))\Big](t-t_i)\\
+\int_{t_i}^t\frac{(t-s)^{q-1}}{\Gamma(q)}\sigma(s)ds,\quad
t\in (t_i,t_{i+1}],\; i\in \mathbb{N}[1,m].
\end{cases}
$$
This is \eqref{eIV}. From Theorem \ref{thm5.2.0}, Result \ref{res6.7.1} is wrong even
$D_{0^+}^q$ with a single starting point in \eqref{eI} is replaced by
 $D_{t_i^+}^q$ with multiple starting points $\{t_i\}$.
\end{proof}


Karaca and Tokmak \cite{kt} studied the kind of impulsive Sturm-Liouville boundary
value problem
\begin{equation}
\begin{gathered}
[\phi_p({}^{C}D_{0^+}^{\beta} x(t)]'=f(t,x(t)),\quad
 t\in (t_i,t_{i+1}],\;i\in \mathbb{N}[0,m],\\
\Delta x(t_i)=I_i(x(t_i)),\quad i\in \mathbb{N}[1,m],\\
\Delta x'(t_i)=J_i(x(t_i)),\quad i\in \mathbb{N}[1,m],\\
\alpha_1x(0)-\beta_1x'(0)=\alpha_2x(1)+\beta_2x'(1)=0,
\end{gathered} \label{e5.2.2}
\end{equation}
where $\beta\in (0,1]$, ${}^{C}D_{0^+}^{\beta}u$ is the Caputo fractional derivative,
 $f:[0,1]\times \mathbb{R}\to \mathbb{R}$, $I_i,J_i:\mathbb{R}\to \mathbb{R}$ are continuous functions,
$0=t_0<t_1<\dots<t_m<t_{m+1}=1$,
 $\alpha_1\alpha_2+\alpha_1\beta_2+\alpha_2\beta_1\neq 0$.
The existence and uniqueness of solutions of \eqref{e5.2.2} were established
under the assumptions that $f$, $I_i,J_i$ are bounded functions.

In \cite{zl8, zlz}, the authors studied the existence and uniqueness of solution
for the boundary value problems for the semilinear impulsive fractional
integro-differential equations:
\begin{gather*}
{}^{C}D^{q} x(t)=\lambda x(t)+f(t,x(t),(Kx)(t),(Hx)(t)),\quad
t\in (t_i,t_{i+1}],\;i\in \mathbb{N}[0,m],\\
\Delta x(t_i)=I_i(x(t_i)),\quad i\in \mathbb{N}[1,m],\\
\Delta x'(t_i)=J_i(x(t_i)),\quad i\in \mathbb{N}[1,m],\\
\alpha_1x(0)-\beta_1x'(0)=x_0,\quad \alpha_2x(1)+\beta_2x'(1)=x_1,
\end{gather*}
where $q\in (1,2]$, $\lambda\ge 0$, $\alpha_1\ge 0$, $\beta_1>0$,
$\alpha_2\ge 0$, $\beta_2>0$,
$\Gamma=\alpha_1\alpha_2+\alpha_1\beta_2+\alpha_2\beta_1\neq 0$, and
$x_0,x_1\in \mathbb{R}$, $f:[0,1]\times \mathbb{R}^3\to \mathbb{R}$,
 $I_i,J_i:\mathbb{R}\to \mathbb{R}$ are continuous functions,
$0=t_0<t_1<\dots<t_m<t_{m+1}=1$, and $H,K$ are integral operators with
integral kernels
$$
(Kx)(t)=\int_0^tk(t,s)x(s)ds,\quad
(Hx)(t)=\int_0^1h(t,s)x(s)ds,\quad t\in [0,1],
$$
where $k:D_0\to \mathbb{R}$ and $h:D\to \mathbb{R}$ satisfies
$\sup_{t\in [0,1]}\int_0^tk(t,s)|ds<+\infty$ and
$ \sup_{t\in [0,1]}\int_0^1h(t,s)|ds<+\infty$.
 The existence and uniqueness of solutions were established under the assumptions
that $f$, $I_i,J_i$ are bounded functions and satisfies the Lipschitz conditions.
 Some examples were presented in \cite{zl8, zlz}.
These examples are unsuitable. Since \cite[Lemma 2.3]{zl8}) and
\cite[Lemma 2.3]{zlz})
are from \cite{ww8} in which the derivative is the Caputo derivative with
 multiple start point $t_i(i\in N_0)$, see \cite[Lemma 2.2]{ww8} and \cite{wx}.
So ${}^{C}D_{0^+}^{q} $ in examples in \cite{zl8, zlz} should be replaced
by the one with multiple start point $t_i(i\in N_0)$.
But The derivative in Examples in \cite{zl8, zlz} is the Riemann-Liouville
derivative with single start point $t=0$.

There has been no papers concerning with the solvability of Sturm-Liouville
boundary value problems of impulsive fractional differential equations
involving the other fractional derivatives such as the Riemman-Liouville
fractional derivatives.
Now we consider the problem
\begin{equation}
\begin{gathered}
{}^{RL}D_{0^+}^{\alpha} x(t)=p(t)f(t,x(t),{}^{RL}D_{0^+}^{\beta} x(t)),\quad
t\in (t_i,t_{i+1}],\;i\in \mathbb{N}[0,m],\\
\lim_{t\to t_i^+}(t-t_i)^{2-\alpha}x(t)=I(t_i,x(t_i),{}^{RL}D_{0^+}^{\beta} x(t_i)),
\quad i\in \mathbb{N}[1,m],\\
\Delta {}^{RL}D_{0^+}^{\alpha-1}x(t_i)=J(t_i,x(t_i),{}^{RL}D_{0^+}^{\beta}
x(t_i)),\quad i\in \mathbb{N}[1,m],\\
\alpha_1\lim_{t\to 0^+}t^{2-\alpha}x(t)-\beta_1{}^{RL}D_{0^+}^{\alpha-1} x(0)
=\alpha_2x(1)+\beta_2{}^{RL}D_{0^+}^{\alpha-1} x (1)=0,
\end{gathered}
\label{e5.2.3}
\end{equation}
where $\alpha\in (1,2]$, $\beta\in (0,\alpha-1]$, ${}^{RL}D_{0^+}^{*}u$
is the Riemann-Liouville fractional derivative of order $*$,
$p:(0,1)\to \mathbb{R}$ satisfies (5.A4) in \eqref{e5.1.4},
$f:(0,1)\times \mathbb{R}^2\to \mathbb{R}$ satisfies (5.A2) in  \eqref{e5.1.4},
and $I,J:\{t_i:i\in \mathbb{N}[1,m]\times \mathbb{R}^2\to \mathbb{R}$ 
in \eqref{e5.1.4} satisfies (5.A3) , $0=t_0<t_1<\dots<t_m<t_{m+1}=1$,
$\alpha_1\alpha_2+\alpha_1\beta_2+\alpha_2\beta_1\neq 0$.

\begin{lemma} \label{lem5.2.1}
 Suppose that $\Theta=\alpha_1\alpha_2+\Gamma(\alpha)\alpha_1\beta_2
+\Gamma(\alpha)\alpha_2\beta_1\neq 0$, $\sigma$ is continuous on $(0,1)$
and there exist $k>-1$ and $l\in (\max\{-\alpha,-2-k\},0]$ such that
$|\sigma(t)|\le t^k(1-t)^l$ for all $t\in (0,1)$. Then $x$ is a solution
of the problem
\begin{equation}
\begin{gathered}
{}^{RL}D_{0^+}^{\alpha} x(t)=\sigma(t),\quad t\in (t_i,t_{i+1}],\;i\in \mathbb{N}[0,m],\\
\lim_{t\to t_i^+}(t-t_i)^{2-\alpha}x(t)=I_i, \quad i\in \mathbb{N}[1,m],\\
\Delta {}^{RL}D_{0^+}^{\alpha-1}x(t_i)=J_i, \quad i\in \mathbb{N}[1,m],\\
\alpha_1\lim_{t\to 0^+}t^{2-\alpha}x(t)-\beta_1{}^{RL}D_{0^+}^{\alpha-1}
x(0)=\alpha_2x(1)+\beta_2{}^{RL}D_{0^+}^{\alpha-1} x (1)=0,
\end{gathered} \label{e5.2.4}
\end{equation}
if and only if
\begin{equation}
\begin{aligned}
&x(t) \\
&=\frac{\alpha_1}{\Theta}\Big[\sum_{\sigma=1}^m
 \frac{(1-t_\sigma)^{\alpha-1}}{\Gamma(\alpha)}J_\sigma
 +\sum_{\sigma=1}^m(1-t_\sigma)^{\alpha-2}I_\sigma
 +\beta_2\int_0^1\sigma(s)ds \\
&\quad +\alpha_2\int_0^1\frac{(1-s)^{\alpha-1}}
 {\Gamma(\alpha)}\sigma(s)ds\Big]t^{\alpha-1}
 -\frac{\Gamma(\alpha)\beta_1}{\Theta}
 \Big[\sum_{\sigma=1}^m\frac{(1-t_\sigma)^{\alpha-1}}{\Gamma(\alpha)}
 J_\sigma \\
&\quad +\sum_{\sigma=1}^m(1-t_\sigma)^{\alpha-2}I_\sigma
 +\beta_2\int_0^1\sigma(s)ds+\alpha_2\int_0^1
 \frac{(1-s)^{\alpha-1}}{\Gamma(\alpha)}\sigma(s)ds\Big]t^{\alpha-2}\\
&\quad +\sum_{\sigma=1}^i\frac{(1-t_\sigma)^{\alpha-1}}{\Gamma(\alpha)}
 J_\sigma+\sum_{\sigma=1}^i(1-t_\sigma)^{\alpha-2}I_\sigma
+\int_0^t\frac{(t-s)^{\alpha-1}}{\Gamma(\alpha)}\sigma(s)ds,
\end{aligned}\label{e5.2.5}
\end{equation}
for $t\in (t_i,t_{i+1}]$ and $i\in N[0,m]$.
\end{lemma}

\begin{proof}
By Theorem \ref{thm3.2.2}, we have ${}^{RL}D_{0^+}^\alpha x(t)=\sigma(t)$,
$t\in (t_i,t_{i+1}]$, $i\in \mathbb{N}[0,m]$ if and only if there exit numbers
 $c_{\sigma1},c_{\sigma2},i\in \mathbb{N}[0,m]$ such that
\begin{equation}
x(t)=\sum_{\sigma=0}^i[c_{\sigma 1}(t-t_\sigma)^{\alpha-1}
+c_{\sigma2}(t-t_\sigma)^{\alpha-2}]
+\int_0^t\frac{(t-s)^{\alpha-1}}{\Gamma(\alpha)}\sigma(s)ds,
\label{e5.2.6}
\end{equation}
for $t\in (t_i,t_{i+1}]$ and $i\in \mathbb{N}[0,m]$.
Then
\begin{equation}
{}^{RL}D_{0^+}^{\alpha-1}x(t)=\Gamma(\alpha)\sum_{\sigma=0}^ic_{\sigma1}
+\int_0^t\sigma(s)ds,\quad t\in (t_i,t_{i+1}],\;i\in \mathbb{N}[0,m].
\label{e5.2.7}
\end{equation}
Furthermore, by direct computations we have
\begin{equation}
\begin{aligned}
&{}^{RL}D_{0^+}^{\beta}x(t) \\
&=\frac{1}{\Gamma(1-\beta)}\Big[\int_0^t(t-s)^{-\beta}x(s)ds\Big]'\\
&=\frac{1}{\Gamma(1-\beta)}\Big[\sum_{j=0}^{i-1}\int_{t_j}^{t_{j+1}}(t-s)^{-\beta}
x(s)ds\Big]'+\frac{1}{\Gamma(1-\beta)}\Big[\int_{t_i}^{t}(t-s)^{-\beta}
x(s)ds\Big]'\\
&=\sum_{\sigma=0}^i\Big(c_{\sigma1}\frac{\Gamma(\alpha)}
{\Gamma(\alpha-\beta+1)}(t-t_\sigma)^{\alpha-\beta-1}
 +c_{\sigma2}\frac{\Gamma(\alpha-1)}{\Gamma(\alpha-\beta-1)}
(t-t_\sigma)^{\alpha-\beta-2}\Big)\\
&\quad +\int_0^t\frac{(t-s)^{\alpha-\beta-1}}{\Gamma(\alpha-\beta)}\sigma(s)ds,
\quad t\in (t_i,t_{i+1}],\;i\in \mathbb{N}[0,m].
\end{aligned} \label{e5.2.8}
\end{equation}
From $\lim_{t\to t_i^+}(t-t_i)^{2-\alpha}x(t)=I_i$, we have
$c_{\sigma2}=I_\sigma$ for all $\sigma\in \mathbb{N}[1,m]$. From
$\Delta {}^{RL}D_{0^+}^{\alpha-1}x(t_i))=J_i,\;i\in \mathbb{N}[1,m]$, one has
$c_{\sigma1}=\frac{J_\sigma}{\Gamma(\alpha)}$ for all $\sigma\in \mathbb{N}[1,m]$.
 From the boundary conditions in \eqref{e5.2.4}, we obtain
\begin{gather*}
\alpha_1c_{02}-\Gamma(\alpha)\beta_1c_{01}=0,\\
\begin{aligned}
&\alpha_2\Big[\sum_{\sigma=0}^m(c_{\sigma 1}(1-t_\sigma)^{\alpha-1}
+c_{\sigma2}(1-t_\sigma)^{\alpha-2})
+\int_0^1\frac{(1-s)^{\alpha-1}}{\Gamma(\alpha)}\sigma(s)ds\Big]\\
& +\beta_2\Big[\Gamma(\alpha)\sum_{\sigma=0}^mc_{\sigma1}
+\int_0^1\sigma(s)ds\Big]=0.
\end{aligned}
\end{gather*}
It follows that
\begin{gather*}
\begin{aligned}
c_{01}&=\frac{\alpha_1}{\Theta}\Big[\sum_{\sigma=1}^m
 \frac{(1-t_\sigma)^{\alpha-1}}{\Gamma(\alpha)}J_\sigma
 +\sum_{\sigma=1}^m(1-t_\sigma)^{\alpha-2}I_\sigma \\
&\quad +\beta_2\int_0^1\sigma(s)ds+\alpha_2
 \int_0^1\frac{(1-s)^{\alpha-1}}{\Gamma(\alpha)}\sigma(s)ds\Big],
\end{aligned}\\
\begin{aligned}
c_{02}&=-\frac{\Gamma(\alpha)\beta_1}{\Theta}
 \Big[\sum_{\sigma=1}^m\frac{(1-t_\sigma)^{\alpha-1}}{\Gamma(\alpha)}
J_\sigma+\sum_{\sigma=1}^m(1-t_\sigma)^{\alpha-2}I_\sigma \\
&\quad +\beta_2\int_0^1\sigma(s)ds+\alpha_2\int_0^1\frac{(1-s)^{\alpha-1}}
 {\Gamma(\alpha)}\sigma(s)ds\Big].
\end{aligned}
\end{gather*}
Substituting $c_{\sigma1},c_{\sigma2}$ into \eqref{e5.2.6}, we obtain
\eqref{e5.2.5} by changing the order of the terms.
It is easy to show from \eqref{e5.2.5} and \eqref{e5.2.8} that $x\in X$.
The proof is compete.
\end{proof}

To abbreviate notation, let $H_x(t)=H(t,x(t),{}^{RL}D_{0^+}^\beta x(t))$
for functions $H:(0,1)\times \mathbb{R}^2\to \mathbb{R}$ and $x:(0,]\to \mathbb{R}$.
Define the operator $(Tx)(t)$ for $x\in X$ by
\begin{align*}
(Tx)(t)&=\frac{\alpha_1}{\Theta}\Big[\sum_{\sigma=1}^m
 \frac{(1-t_\sigma)^{\alpha-1}}{\Gamma(\alpha)}J_x(t_\sigma)
+\sum_{\sigma=1}^m(1-t_\sigma)^{\alpha-2}I_x(t_\sigma) \\
&\quad +\beta_2\int_0^1p(s)f_x(s)ds+\alpha_2
 \int_0^1\frac{(1-s)^{\alpha-1}}{\Gamma(\alpha)}p(s)f_x(s)ds\Big]t^{\alpha-1}\\
&\quad -\frac{\Gamma(\alpha)\beta_1}{\Theta}
 \Big[\sum_{\sigma=1}^m\frac{(1-t_\sigma)^{\alpha-1}}{\Gamma(\alpha)}J_x(t_\sigma)
+\sum_{\sigma=1}^m(1-t_\sigma)^{\alpha-2}I_x(t_\sigma) \\
&\quad +\beta_2\int_0^1p(s)f_x(s)ds+\alpha_2
 \int_0^1\frac{(1-s)^{\alpha-1}}{\Gamma(\alpha)}p(s)f_x(s)ds\Big]t^{\alpha-2}\\
&\quad +\sum_{\sigma=1}^i\frac{(1-t_\sigma)^{\alpha-1}}{\Gamma(\alpha)}
 J_x(t_\sigma)(t-t_\sigma)^{\alpha-1}
+\sum_{\sigma=1}^i(1-t_\sigma)^{\alpha-2}I_x(t_\sigma)(t-t_\sigma)^{\alpha-2}\\
&\quad +\int_0^t\frac{(t-s)^{\alpha-1}}{\Gamma(\alpha)}p(s)f_x(s)ds,
\quad t\in (t_i,t_{i+1}],\; i\in N[0,m].
\end{align*}

\begin{theorem} \label{thm5.2.1}
 Suppose that $\Theta\neq 0$ and {\rm (H1)} in Theorem \ref{thm5.1.1} holds.
Then \eqref{e5.2.3} has at least one solution if there exists $r_0>0$
such that
\[
A_1M_J(r_0,r_0)+A_2M_I(r_0,r_0)+A_3M_f(r_0,r_0)\le r_0,
\]
where
\begin{align*}
A_1&=\max\Big\{\frac{|\alpha_1|}{|\Theta|}\sum_{\sigma=1}^m
\frac{(1-t_\sigma)^{\alpha-1}}{\Gamma(\alpha)}
+\sum_{\sigma=1}^m\frac{(1-t_\sigma)^{\alpha-1}}{\Gamma(\alpha)}
+\frac{\Gamma(\alpha)|\beta_1|}{|\Theta|}\sum_{\sigma=1}^m
\frac{(1-t_\sigma)^{\alpha-1}}{\Gamma(\alpha)}, \\
&\quad \frac{|\alpha_1|}{|\Theta|}\frac{\Gamma(\alpha)}{\Gamma(\alpha-\beta)}
\sum_{\sigma=1}^m\frac{(1-t_\sigma)^{\alpha-1}}{\Gamma(\alpha)}
+\sum_{\sigma=1}^m\frac{(1-t_\sigma)^{\alpha-1}}{\Gamma(\alpha)}
\frac{\Gamma(\alpha)}{\Gamma(\alpha-\beta)} \\
&\quad +\frac{\Gamma(\alpha)|\beta_1|}{|\Theta|}
 \frac{\Gamma(\alpha-1)}{\Gamma(\alpha-\beta-1)}
\sum_{\sigma=1}^m\frac{(1-t_\sigma)^{\alpha-1}}{\Gamma(\alpha)}\Big\},
\end{align*}
and
\begin{align*}
A_2&=\max\Big\{\frac{\Gamma(\alpha)|\beta_1|}{|\Theta|}
 \sum_{\sigma=1}^m(1-t_\sigma)^{\alpha-2}
+\sum_{\sigma=1}^m(1-t_\sigma)^{\alpha-2}
+\frac{|\alpha_1|}{|\Theta|}\sum_{\sigma=1}^m(1-t_\sigma)^{\alpha-2}, \\
&\quad \frac{\Gamma(\alpha)|\beta_1|}{|\Theta|}
 \frac{\Gamma(\alpha-1)}{\Gamma(\alpha-\beta-1)}
\sum_{\sigma=1}^m(1-t_\sigma)^{\alpha-2}
+\sum_{\sigma=1}^m(1-t_\sigma)^{\alpha-2}\frac{\Gamma(\alpha-1)}
 {\Gamma(\alpha-\beta-1)} \\
&\quad +\frac{|\alpha_1|}{|\Theta|}\frac{\Gamma(\alpha)}{\Gamma(\alpha-\beta)}
\sum_{\sigma=1}^m(1-t_\sigma)^{\alpha-2}\},
\end{align*}
\begin{align*}
A_3&=\max\Big\{\frac{|\alpha_1\|\beta_2|\mathbf{B}(l+1,k+1)}{|\Theta|}
 +\frac{|\alpha_1\|\alpha_2|}{|\Theta|}\frac{\mathbf{B}(\alpha+l,k+1)}
 {\Gamma(\alpha)} \\
&\quad+\frac{\Gamma(\alpha)|\beta_1\|\beta_2|\mathbf{B}(l+1,k+1)}{|\Theta|}
 +\frac{\Gamma(\alpha)|\beta_1\|\alpha_2|}{|\Theta|}
\frac{\mathbf{B}(\alpha+l,k+1)}{\Gamma(\alpha)} \\
&\quad+\frac{\mathbf{B}(\alpha+l,k+1)}{\Gamma(\alpha)} ,\;
 \frac{|\alpha_1\|\beta_2|\mathbf{B}(l+1,k+1)}{|\Theta|}
 \frac{\Gamma(\alpha)}{\Gamma(\alpha-\beta)} \\
&\quad +\frac{|\alpha_1\|\alpha_2|}{|\Theta|}
 \frac{\Gamma(\alpha)}{\Gamma(\alpha-\beta)}
\frac{\mathbf{B}(\alpha+l,k+1)}{\Gamma(\alpha)} \\
&\quad +\frac{\Gamma(\alpha)|\beta_1\|\beta_2|
\mathbf{B}(l+1,k+1)}{|\Theta|}\frac{\Gamma(\alpha-1)}{\Gamma(\alpha-\beta-1)}
 \\
&\quad +\frac{\Gamma(\alpha)|\beta_1\|\alpha_2|}{|\Theta|}
 \frac{\Gamma(\alpha-1)}{\Gamma(\alpha-\beta-1)}
 \frac{\mathbf{B}(\alpha+l,k+1)}{\Gamma(\alpha)}
+\frac{\mathbf{B}(\alpha-\beta+l,k+1)}{\Gamma(\alpha-\beta)}\Big\}.
\end{align*}
\end{theorem}

\begin{proof}
Let $X$ be defined above. By Lemma \ref{lem5.2.1}, we know that $x$ is a solution of
\eqref{e5.2.3} if and only if $x$ is a fixed point of $T$.
By a standard proof, we can see that $T:X\to X$ is a completely
continuous operator.

From (H1), for $x\in X$ we have
\begin{gather*}
|f_x(t)|=|f(t,x(t),{}^{RL}D_{0^+}^\beta x(t))|\le M_f(\|x\|,\|x\|),\quad
t\in (t_i,t_{i+1}],\;i\in \mathbb{N}[0,m],\\
|I(t_i,x(t_i),{}^{RL}D_{0^+}^\beta x(t_i))|\le M_I(\|x\|,\|x\|),\quad
 i\in \mathbb{N}[1,m],\; x\in \mathbb{R},\\
|J(t_i,x(t_i),{}^{RL}D_{0^+}^\beta x(t_i))|\le M_J(\|x\|,\|x\|),\quad
i\in \mathbb{N}[1,m],\; x\in \mathbb{R}.
\end{gather*}
We consider the set
$\Omega=\{x\in X:x=\lambda (Tx),\text{ for some }\lambda \in [0,1)\}$.
 For $x\in \Omega$, we have by definition of $T$ for $t\in (t_i,t_{i+1}]$ that
\begin{align*}
&{}^{RL}D_{0^+}^\beta (Tx)(t) \\
&=\frac{\alpha_1}{\Theta}\Big[\sum_{\sigma=1}^m\frac{(1-t_\sigma)^{\alpha-1}}
{\Gamma(\alpha)}J_x(t_\sigma) +\sum_{\sigma=1}^m(1-t_\sigma)^{\alpha-2}I_x(t_\sigma) \\
&\quad +\beta_2\int_0^1p(s)f_x(s)ds+\alpha_2\int_0^1\frac{(1-s)^{\alpha-1}}
{\Gamma(\alpha)}p(s)f_x(s)ds\Big]
\frac{\Gamma(\alpha)}{\Gamma(\alpha-\beta)}t^{\alpha-\beta-1}\\
&\quad -\frac{\Gamma(\alpha)\beta_1}{\Theta}
 \Big[\sum_{\sigma=1}^m\frac{(1-t_\sigma)^{\alpha-1}}{\Gamma(\alpha)}J_x(t_\sigma)
+\sum_{\sigma=1}^m(1-t_\sigma)^{\alpha-2}I_x(t_\sigma) \\
&\quad +\beta_2\int_0^1p(s)f_x(s)ds
 +\alpha_2\int_0^1\frac{(1-s)^{\alpha-1}}{\Gamma(\alpha)}p(s)f_x(s)ds\Big]
\frac{\Gamma(\alpha-1)}{\Gamma(\alpha-\beta-1)}t^{\alpha-\beta -2}\\
&\quad +\sum_{\sigma=1}^i\frac{(1-t_\sigma)^{\alpha-1}}{\Gamma(\alpha)}
 J_x(t_\sigma)\frac{\Gamma(\alpha)}{\Gamma(\alpha-\beta)}(t-t_\sigma)^{\alpha-1} \\
&\quad +\sum_{\sigma=1}^i(1-t_\sigma)^{\alpha-2}I_x(t_\sigma)
 \frac{\Gamma(\alpha-1)}{\Gamma(\alpha-\beta-1)}(t-t_\sigma)^{\alpha-2}\\
&\quad +\int_0^t\frac{(t-s)^{\alpha-\beta-1}}{\Gamma(\alpha-\beta)}p(s)f_x(s)ds,
\quad t\in (t_i,t_{i+1}],\;i\in \mathbb{N}[0,m].
\end{align*}
So
\begin{align*}
&(t-t_i)^{2-\alpha}|(Tx)(t)| \\
&\le \frac{|\alpha_1|}{|\Theta|}
 \Big[\sum_{\sigma=1}^m\frac{(1-t_\sigma)^{\alpha-1}}{\Gamma(\alpha)}M_J(\|x\|,\|x\|)
+\sum_{\sigma=1}^m(1-t_\sigma)^{\alpha-2}M_I(\|x\|,\|x\|) \\
&\quad +|\beta_2|\int_0^1s^k(1-s)^ldsM_f(\|x\|,\|x\|) \\
&\quad +|\alpha_2|\int_0^1\frac{(1-s)^{\alpha-1}}{\Gamma(\alpha)}s^k(1-s)^lds
M_f(\|x\|,\|x\|)\Big]\\
&\quad +\frac{\Gamma(\alpha)|\beta_1|}{|\Theta|}
\Big[\sum_{\sigma=1}^m\frac{(1-t_\sigma)^{\alpha-1}}
{\Gamma(\alpha)}M_J(\|x\|,\|x\|)
+\sum_{\sigma=1}^m(1-t_\sigma)^{\alpha-2}M_I(\|x\|,\|x\|) \\
&\quad +|\beta_2|\int_0^1s^k(1-s)^ldsM_f(\|x\|,\|x\|) \\
&\quad +|\alpha_2|\int_0^1\frac{(1-s)^{\alpha-1}}{\Gamma(\alpha)}s^k(1-s)^lds
 M_f(\|x\|,\|x\|)\Big]\\
&\quad
+\sum_{\sigma=1}^m\frac{(1-t_\sigma)^{\alpha-1}}{\Gamma(\alpha)}M_J(\|x\|,\|x\|)
+\sum_{\sigma=1}^m(1-t_\sigma)^{\alpha-2}M_I(\|x\|,\|x\|)\\
&\quad +(t-t_i)^{2-\alpha}\int_0^t\frac{(t-s)^{\alpha-1}}{\Gamma(\alpha)}
 s^k(1-s)^ldsM_f(\|x\|,\|x\|)\\
&\le \Big[\frac{|\alpha_1|}{|\Theta|}\sum_{\sigma=1}^m
 \frac{(1-t_\sigma)^{\alpha-1}}{\Gamma(\alpha)}
+\sum_{\sigma=1}^m\frac{(1-t_\sigma)^{\alpha-1}}{\Gamma(\alpha)} \\
&\quad +\frac{\Gamma(\alpha)|\beta_1|}{|\Theta|}\sum_{\sigma=1}^m
 \frac{(1-t_\sigma)^{\alpha-1}}{\Gamma(\alpha)}\Big]M_J(\|x\|,\|x\|)
 +\Big[\frac{\Gamma(\alpha)|\beta_1|}{|\Theta|}\sum_{\sigma=1}^m
 (1-t_\sigma)^{\alpha-2} \\
&\quad +\sum_{\sigma=1}^m(1-t_\sigma)^{\alpha-2}
+\frac{|\alpha_1|}{|\Theta|}\sum_{\sigma=1}^m(1-t_\sigma)^{\alpha-2}
 \Big]M_I(\|x\|,\|x\|)\\
&\quad +\Big[\frac{|\alpha_1\|\beta_2|\mathbf{B}(l+1,k+1)}{|\Theta|}
 +\frac{|\alpha_1\|\alpha_2|}{|\Theta|}\frac{\mathbf{B}(\alpha+l,k+1)}
 {\Gamma(\alpha)} \\
&\quad +\frac{\Gamma(\alpha)|\beta_1\|\beta_2|\mathbf{B}(l+1,k+1)}{|\Theta|}
 +\frac{\Gamma(\alpha)|\beta_1\|\alpha_2|}{|\Theta|}
\frac{\mathbf{B}(\alpha+l,k+1)}{\Gamma(\alpha)} \\
&\quad +\frac{\mathbf{B}(\alpha+l,k+1)}{\Gamma(\alpha)}\Big]M_f(\|x\|,\|x\|).
\end{align*}
Furthermore,
\begin{align*}
&(t-t_\sigma)^{2+\beta-\alpha}|{}^{RL}D_{0^+}^\beta (Tx)(t)| \\
&\le \Big[\frac{|\alpha_1|}{|\Theta|}\frac{\Gamma(\alpha)}{\Gamma(\alpha-\beta)}
\sum_{\sigma=1}^m\frac{(1-t_\sigma)^{\alpha-1}}{\Gamma(\alpha)}
+\sum_{\sigma=1}^m\frac{(1-t_\sigma)^{\alpha-1}}{\Gamma(\alpha)}
\frac{\Gamma(\alpha)}{\Gamma(\alpha-\beta)} \\
&\quad +\frac{\Gamma(\alpha)|\beta_1|}{|\Theta|}\frac{\Gamma(\alpha-1)}
{\Gamma(\alpha-\beta-1)}
\sum_{\sigma=1}^m\frac{(1-t_\sigma)^{\alpha-1}}{\Gamma(\alpha)}\Big]
 M_J(\|x\|,\|x\|)\\
&\quad +\Big[\frac{\Gamma(\alpha)|\beta_1|}{|\Theta|}
 \frac{\Gamma(\alpha-1)}{\Gamma(\alpha-\beta-1)}
\sum_{\sigma=1}^m(1-t_\sigma)^{\alpha-2}
+\sum_{\sigma=1}^m(1-t_\sigma)^{\alpha-2}
\frac{\Gamma(\alpha-1)}{\Gamma(\alpha-\beta-1)} \\
&\quad +\frac{|\alpha_1|}{|\Theta|}\frac{\Gamma(\alpha)}{\Gamma(\alpha-\beta)}
\sum_{\sigma=1}^m(1-t_\sigma)^{\alpha-2}\Big]M_I(\|x\|,\|x\|) \\
&\quad +\Big[\frac{|\alpha_1\|\beta_2|\mathbf{B}(l+1,k+1)}{|\Theta|}
 \frac{\Gamma(\alpha)}{\Gamma(\alpha-\beta)}
+\frac{|\alpha_1\|\alpha_2|}{|\Theta|}\frac{\Gamma(\alpha)}{\Gamma(\alpha-\beta)}
\frac{\mathbf{B}(\alpha+l,k+1)}{\Gamma(\alpha)}\\
&\quad + \frac{\Gamma(\alpha)|\beta_1\|\beta_2|
\mathbf{B}(l+1,k+1)}{|\Theta|}\frac{\Gamma(\alpha-1)}{\Gamma(\alpha-\beta-1)} \\
&\quad +\frac{\Gamma(\alpha)|\beta_1\|\alpha_2|}{|\Theta|}
 \frac{\Gamma(\alpha-1)}{\Gamma(\alpha-\beta-1)}
 \frac{\mathbf{B}(\alpha+l,k+1)}{\Gamma(\alpha)}
+\frac{\mathbf{B}(\alpha-\beta+l,k+1)}{\Gamma(\alpha-\beta)}\Big]M_f(\|x\|,\|x\|).
\end{align*}
It follows that
$$
\|Tx\|\le A_1M_J(\|x\|,\|x\|)+A_2M_I(\|x\|,\|x\|)+A_3M_f(\|x\|,\|x\|).
$$
From the assumption, we choose $ \Omega=\{x\in X:\|x\|\le r_0\}$.
For $x\in \partial \Omega$, we obtain $x\neq \lambda (Tx)$ for any
$\lambda\in [0,1]$. In fact, if there exists $x\in \partial \Omega$ such that
$x=\lambda (Tx)$ for some $\lambda\in [0,1]$. Then
\[
r_0=\|x\|=\lambda\|Tx\|< \|Tx\|\le A_1M_J(r_0,r_0)+A_2M_I(r_0,r_0)
+A_3M_f(r_0,r_0)\le r_0,
\]
which is a contradiction.

As a consequence of Schaefer's fixed point theorem, we deduce that $T$
has a fixed point which is a solution of \eqref{e5.2.3}.
 The proof is complete.
\end{proof}


\begin{theorem} \label{thm5.2.2}
Suppose that $\Theta\neq 0$, {\rm (1.A1)--(1.A4)}
and {\rm (H1)} in Theorem \ref{thm5.1.1} hold and
\[
\lim_{r\to 0^+}\frac{ M_f(r,r)}{r}=\lim_{r\to 0^+}\frac{M_I(r,r)}{r}
=\lim_{r\to 0^+}\frac{ M_J(r,r)}{r}=0
\]
or
\[
\lim_{r\to +\infty}\frac{ M_f(r,r)}{r}=\lim_{r\to +\infty}\frac{M_I(r,r)}{r}
=\lim_{r\to +\infty}\frac{ M_J(r,r)}{r}=0.
\]
Then BVP \eqref{e5.2.3} has at least one solution.
\end{theorem}

The proof of the above theorem is similar to that of Theorem \ref{thm5.1.2}
 and is omitted.

\begin{example} \label{examp5.2.1}\rm
Consider the problem
\begin{equation}
\begin{gathered}
\begin{aligned}
{}^{RL}D_{0^+}^{3/2} x(t)
&=t^{-\frac{1}{8}}(1-t)^{-\frac{1}{8}}
\Big[\overline{A}_1+\overline{B}_1((t-t_i)^{1/2}x(t))^\sigma \\
&\quad +\overline{C}_1((t-t_i)^{5/8}{}^{RL}D_{0^+}^{1/4}
 x(t))^\sigma\Big],\quad
 t\in (t_i,t_{i+1}],\;i\in \mathbb{N}[0,m],
\end{aligned}\\
\begin{aligned}
\lim_{t\to t_i^+}(t-t_i)^{1/2}x(t)
&=\overline{A}_2+\overline{B}_2((t_i-t_{i-1})^{1/2}x(t_i))^\sigma \\
&\quad +\overline{C}_2((t_i-t_{i-1})^{5/8}{}^{RL}D_{0^+}^{1/4}
 x(t_i))^\sigma,
\end{aligned} \\
\begin{aligned}
\Delta {}^{RL}D_{0^+}^{1/2}x(t_i)
&=\overline{A}_3 +\overline{B}_3((t_i-t_{i-1})^{1/2}x(t_i))^\sigma \\
&\quad +\overline{C}_3((t_i-t_{i-1})^{5/8}{}^{RL}D_{0^+}^{1/4} x(t_i))^\sigma,
\end{aligned}\\
\lim_{t\to 0^+}t^{1/2}x(t)-{}^{RL}D_{0^+}^{1/2} x(0)
=x(1)+{}^{RL}D_{0^+}^{1/2} x (1)=0,
\end{gathered} \label{e5.2.9}
\end{equation}
where $A_1,B_i,C_i\in \mathbb{R}(i\in \mathbb{N}[1,3],\sigma\ge 0$ are constants,
$0=t_0<t_1<\dots<t_m<t_{m+1}=1$. Then \eqref{e5.2.9} has at least one solution if
one of the following items holds:
\begin{itemize}
\item[(i)] $\sigma\in [0,1]$ or

\item[(ii)] $\sigma=1$ with $23.5780m|\overline{B}_1|+23.5780m|\overline{C}_1|
+14.2468m|\overline{B}_2|+14.2468m|\overline{C}_2|\\ +34.6784m|\overline{B}_3|
+34.6784m|\overline{C}_3|<1$
or

\item[(iii)] $\sigma>1$ with
$[23.5780m|\overline{A}_1|+14.2468m|\overline{A}_2|+34.6784m|
\overline{A}_3|]^{\sigma-1} [23.5780m|\overline{B}_1| \\\
+23.5780m| \overline{C}_1|+14.2468m|\overline{B}_2|+14.2468m|\overline{C}_2|
+A_3|\overline{B}_3|+34.6784m|\overline{C}_3|]
\le \frac{(\sigma-1)^{\sigma-1}}{\sigma^\sigma}$.
\end{itemize}
\end{example}

\begin{proof} Corresponding to \eqref{e5.2.3}, we have $\alpha=\frac{3}{2}$,
$\beta=\frac{1}{4}$, $\alpha_1=\beta_1=\alpha_2=\beta_1=1$,
 $0=t_0<t_1<\dots<t_m<t_{m+1}=1$, $p(t)=t^{-\frac{1}{8}}(1-t)^{-\frac{1}{8}}$
 with $k=l=-1/8$, and
\begin{gather*}
f(t,u,v)=\overline{A}_1+\overline{B}_1((t-t_i)^{1/2}u)^\sigma
 +\overline{C}_1((t-t_i)^{5/8}v)^\sigma,\quad
t\in (t_i,t_{i+1}], \; i\in \mathbb{N}_0^m,\\
I(t_i,u,v)=\overline{A}_2+\overline{B}_2((t_i-t_{i-1})^{1/2}u)^\sigma
 +\overline{C}_2((t_i-t_{i-1})^{5/8}v)^\sigma,\quad i\in \mathbb{N}_1^m,\\
J(t_i,u,v)=\overline{A}_3+\overline{B}_3((t_i-t_{i-1})^{1/2}u)^\sigma
 +\overline{C}_3((t_i-t_{i-1})^{5/8}v)^\sigma,\quad i\in \mathbb{N}_1^m.
\end{gather*}
We see that
\begin{gather*}
|f(t,(t-t_i)^{\alpha-2}u,(t-t_i)^{\alpha-\beta-2}v)|
\le M_f(|u|,|v|)=|\overline{A}_1|+|\overline{B}_1|u^\sigma
+|\overline{C}_1|v^\sigma,\\
 t\in (t_i,t_{i+1}],\;i\in \mathbb{N}[0,m],u,v\in \mathbb{R},\\
|I(t_i,(t_i-t_{i-1})^{\alpha-2}u,(t_i-t_{i-1})^{\alpha-\beta-2}v)|
\le M_I(|u|,|v|)=|\overline{A}_2|+|\overline{B}_2|u^\sigma
+|\overline{C}_2|v^\sigma, \\
i\in \mathbb{N}[1,m],u,v\in \mathbb{R},\\
|J(t_i,(t_i-t_{i-1})^{\alpha-2}u,(t_i-t_{i-1})^{\alpha-\beta-2}v)|
\le M_J(|u|,|v|)=|\overline{A}_3|+|\overline{B}_3|u^\sigma
+|\overline{C}_3|v^\sigma, \\
i\in \mathbb{N}[1,m],\; u,v\in \mathbb{R}.
\end{gather*}
By direct computations, we obtain
$\Theta=1+2\Gamma(3/2)$,
and by using Mathlab 7.0 that
\begin{align*}
A_1&=\max\Big\{\frac{1}{1+2\Gamma(3/2)}\sum_{\sigma=1}^m
 \frac{(1-t_\sigma)^{1/2}}{\Gamma(3/2)}
+\sum_{\sigma=1}^m\frac{(1-t_\sigma)^{1/2}}{\Gamma(3/2)} \\
&\quad +\frac{\Gamma(3/2)}{1+2\Gamma(3/2)}\sum_{\sigma=1}^m
 \frac{(1-t_\sigma)^{1/2}}{\Gamma(3/2)},
 \frac{1}{1+2\Gamma(3/2)}\frac{\Gamma(3/2)}{\Gamma(11/8)}
\sum_{\sigma=1}^m\frac{(1-t_\sigma)^{1/2}}{\Gamma(3/2)} \\
&\quad +\sum_{\sigma=1}^i\frac{(1-t_\sigma)^{1/2}}{\Gamma(3/2)}
\frac{\Gamma(3/2)}{\Gamma(11/8)}
 +\frac{\Gamma(3/2)}{1+2\Gamma(3/2)}\frac{\Gamma(1/2)}{\Gamma(3/8)}
\sum_{\sigma=1}^m\frac{(1-t_\sigma)^{1/2}}{\Gamma(3/2)}\Big\}\\
& \le m\max\{\frac{2+3\Gamma(3/2)}{(1+2\Gamma(3/2))\Gamma(3/2)},
\frac{(2+2\Gamma(3/2))\Gamma(3/8)+\Gamma(11/8)\Gamma(1/2)}
{(1+2\Gamma(3/2))\Gamma(11/8)\Gamma(3/8)}\Big\} \\
&<23.5780m,
\end{align*}
\begin{align*}
A_2&=\max\Big\{\frac{\Gamma(3/2)}{1+2\Gamma(3/2)}
\sum_{\sigma=1}^m(1-t_\sigma)^{-1/2}
+\sum_{\sigma=1}^m(1-t_\sigma)^{-1/2} \\
&\quad +\frac{1}{1+2\Gamma(3/2)}\sum_{\sigma=1}^m(1-t_\sigma)^{-1/2},
 \frac{\Gamma(3/2)}{1+2\Gamma(3/2)}\frac{\Gamma(1/2)}{\Gamma(3/8)}
\sum_{\sigma=1}^m(1-t_\sigma)^{-1/2} \\
&\quad +\sum_{\sigma=1}^i(1-t_\sigma)^{-1/2}\frac{\Gamma(1/2)}{\Gamma(3/8)}
 +\frac{1}{1+2\Gamma(3/2)}\frac{\Gamma(3/2)}{\Gamma(11/8)}
\sum_{\sigma=1}^m(1-t_\sigma)^{-1/2}\Big\}\\
&\le \frac{m}{\sqrt{1-t_m}}\max
\Big\{\frac{2+3\Gamma(3/2)}{1+2\Gamma(3/2)},
\frac{\Gamma(3/2)}{1+2\Gamma(3/2)}\frac{\Gamma(1/2)}{\Gamma(3/8)}
+\frac{\Gamma(1/2)}{\Gamma(3/8)} \\
&\quad +\frac{1}{1+2\Gamma(3/2)}\frac{\Gamma(3/2)}{\Gamma(11/8)}
\Big\}
<14.2468m,
\end{align*}
and
\begin{align*}
A_3&=\max\Big\{\frac{(1+\Gamma(3/2))\mathbf{B}(7/8,7/8)}
{1+2\Gamma(3/2)}+\frac{2+3\Gamma(3/2)}{1+2\Gamma(3/2)}
\frac{\mathbf{B}(11/8,7/8)}{\Gamma(3/2)} ,\\
&\quad \frac{\mathbf{B}(7/8,7/8)}{1+2\Gamma(3/2)}\frac{\Gamma(3/2)}{\Gamma(11/8)}
 +\frac{1}{1+2\Gamma(3/2)}\frac{\Gamma(3/2)}{\Gamma(11/8)}
\frac{\mathbf{B}(11/8,7/8)}{\Gamma(3/2)} \\
&\quad +\frac{\Gamma(3/2)\mathbf{B}(7/8,7/8)}{1+2\Gamma(3/2)}
\frac{\Gamma(1/2)}{\Gamma(3/8)}
 +\frac{\Gamma(3/2)}{1+2\Gamma(3/2)}
 \frac{\Gamma(1/2)}{\Gamma(3/8)}\frac{\mathbf{B}(11/8,7/8)}{\Gamma(3/2)} \\
&\quad +\frac{\mathbf{B}(5/4,7/8)}{\Gamma(11/8)}\Big\}
<34.6784m.
\end{align*}


By Theorem \ref{thm5.2.1}, BVP \eqref{e5.2.9} has at least one solution if
\begin{equation}
\begin{aligned}
&A_1|\overline{A}_1|+A_2|\overline{A}_2|+A_3|\overline{A}_3|
+[A_1|\overline{B}_1|+A_1|\overline{C}_1| \\
&+A_2|\overline{B}_2|+A_2|\overline{C}_2|
+A_3|\overline{B}_3|+A_3|\overline{C}_3|]r^\sigma\le r
\end{aligned} \label{e5.2.10}
\end{equation}
has a positive solution $r_0$.

It is easy to see that
$\sigma\in [0,1]$ or
\[
\sigma=1\text{ with }A_1|\overline{B}_1|+A_1|\overline{C}_1|
+A_2|\overline{B}_2|+A_2|\overline{C}_2|
+A_3|\overline{B}_3|+A_3|\overline{C}_3|<1
\]
or $\sigma>1$ with
\begin{align*}
&[A_1|\overline{A}_1|+A_2|\overline{A}_2|+A_3|\overline{A}_3|]^{\sigma-1}
[A_1|\overline{B}_1|+A_1|\overline{C}_1|+A_2|\overline{B}_2| \\\
&+A_2|\overline{C}_2| +A_3|\overline{B}_3|+A_3|\overline{C}_3|]
\le \frac{(\sigma-1)^{\sigma-1}}{\sigma^\sigma}
\end{align*}
implies that \eqref{e5.2.10} holds. Hence \eqref{e5.2.9} has at least one
solution if (i) or (ii) or (iii) holds. The proof is complete.
\end{proof}

\subsection{Impulsive anti-periodic boundary value problems}

The solvability of anti-periodic boundary value problems of impulsive fractional
differential equations involving the Caputo fractional derivatives with
 multiple start points were studied by many authors, see \cite{an1, ll1, wx, ww8}
and the references therein. In \cite{re}, authors presented a new method to
converting the impulsive fractional differential equation (with the Caputo
fractional derivative) to an equivalent integral equation and established
existence and uniqueness results for some boundary value problems of impulsive
fractional differential equations involving the Caputo fractional derivatives.
There has been no paper concerning the solvability of anti-periodic boundary
value problems of impulsive fractional differential equations involving other
 fractional derivatives with single start point.

Now we consider the problem
\begin{equation}
\begin{gathered}
{}^{RL}D_{0^+}^{\alpha} x(t)=p(t)f(t,x(t),{}^{RL}D_{0^+}^{\beta} x(t)),\quad
t\in (t_i,t_{i+1}],\;i\in \mathbb{N}[0,m],\\
\Delta x(t_i)=I(t_i,x(t_i),{}^{RL}D_{0^+}^{\beta} x(t_i)),\quad i\in \mathbb{N}[1,m],\\
\Delta {}^{RL}D_{0^+}^{\alpha-1}x(t_i)
=J(t_i,x(t_i),{}^{RL}D_{0^+}^{\beta} x(t_i)),\quad i\in \mathbb{N}[1,m],\\
\lim_{t\to 0^+}t^{2-\alpha}x(t)+ x(1)
=\lim_{t\to 0^+}t^{2+\beta-\alpha}{}^{RL}D_{0^+}^{\alpha-1}x(t)
+{}^{RL}D_{0^+}^{\alpha-1} x (1)=0,
\end{gathered}\label{e5.3.1}
\end{equation}
where $\alpha\in (1,2]$, $\beta\in (0,\alpha-1]$, ${}^{RL}D_{0^+}^{*}u$
is the Riemann-Liouville fractional derivative of order $*$,
$p:(0,1)\to \mathbb{R}$ in \eqref{e5.1.4} satisfies (5.A4),
$f:(0,1)\times \mathbb{R}^2\to \mathbb{R}$  in \eqref{e5.1.4} satisfies (5.A2), and
$I,J:\{t_i:i\in \mathbb{N}[1,m]\}\times \mathbb{R}^2\to \mathbb{R}$ 
in \eqref{e5.1.4} satisfy  (5.A3),
$0=t_0<t_1<\dots<t_m<t_{m+1}=1$.

\begin{lemma} \label{lem5.3.1} 
Suppose that $\sigma$ is continuous on
$(0,1)$ and there exist $k>-1$ and $l\in (\max\{-\alpha,-2-k\},0]$ such that
$|\sigma(t)|\le t^k(1-t)^l$ for all $t\in (0,1)$. Then $x$ is a solution
of the problem
\begin{equation}
\begin{gathered}
{}^{RL}D_{0^+}^{\alpha} x(t)=\sigma(t),\quad t\in (t_i,t_{i+1}],\;i\in \mathbb{N}[0,m],\\
\Delta x(t_i)=I_i,\quad
\Delta {}^{RL}D_{0^+}^{\alpha-1}x(t_i)=J_i, \quad i\in \mathbb{N}[1,m],\\
\lim_{t\to 0^+}t^{2-\alpha}x(t)+ x(1)
=\lim_{t\to 0^+}t^{2+\beta-\alpha}{}^{RL}D_{0^+}^{\alpha-1}x(t)
+{}^{RL}D_{0^+}^{\alpha-1} x (1)=0,
\end{gathered} \label{e5.3.2}
\end{equation}
if and only if
\begin{equation}
\begin{aligned}
x(t)&=-\frac{1}{2\Gamma(\alpha)}
\Big[\sum_{\sigma=1}^mJ_\sigma+\int_0^1\sigma(s)ds\Big]t^{\alpha-1}
-\frac{1}{2}\Big[
\sum_{\sigma=1}^m\frac{(1-t_\sigma)^{\alpha-1}}{2\Gamma(\alpha)}J_\sigma \\
&\quad +\sum_{\sigma=1}^m(1-t_\sigma)^{\alpha-2}I_\sigma
 -\frac{1}{2\Gamma(\alpha)}\int_0^1\sigma(s)ds \\
&\quad +\int_0^1\frac{(1-s)^{\alpha-1}}{\Gamma(\alpha)}\sigma(s)ds\Big]
t^{\alpha-2}
 +\sum_{\sigma=1}^i\frac{(1-t_\sigma)^{\alpha-1}}{\Gamma(\alpha)}
 J_\sigma(t-t_\sigma)^{\alpha-1} \\
&\quad +\sum_{\sigma=1}^i(1-t_\sigma)^{\alpha-2}I_\sigma(t-t_\sigma)^{\alpha-2}
 +\int_0^t\frac{(t-s)^{\alpha-1}}{\Gamma(\alpha)}\sigma(s)ds,
\end{aligned} \label{e5.3.3}
\end{equation}
for $t\in (t_i,t_{i+1}]$, $i\in \mathbb{N}[0,m]$.
\end{lemma}

\begin{proof}
By Theorem \ref{thm3.2.2}, we have ${}^{RL}D_{0^+}^\alpha x(t)=\sigma(t)$,
$t\in (t_i,t_{i+1}]$, $i\in \mathbb{N}[0,m]$ if and only if there exit numbers
 $c_{\sigma1},c_{\sigma2},i\in \mathbb{N}[0,m]$ such that
\begin{equation}
x(t)=\sum_{\sigma=0}^i[c_{\sigma 1}(t-t_\sigma)^{\alpha-1}
+c_{\sigma2}(t-t_\sigma)^{\alpha-2}]
+\int_0^t\frac{(t-s)^{\alpha-1}}{\Gamma(\alpha)}\sigma(s)ds,
\label{e5.3.4}
\end{equation}
for $t\in (t_i,t_{i+1}]$, $i\in N[0,m]$.
Then
\begin{equation}
{}^{RL}D_{0^+}^{\alpha-1}x(t)=\Gamma(\alpha)\sum_{\sigma=0}^ic_{\sigma1}
+\int_0^t\sigma(s)ds,\quad t\in (t_i,t_{i+1}],\;i\in \mathbb{N}[0,m].
\label{e5.3.5}
\end{equation}
Furthermore, by direct computations we have
\begin{equation}
\begin{aligned}
&{}^{RL}D_{0^+}^{\beta}x(t) \\
&=\frac{1}{\Gamma(1-\beta)}\Big[\int_0^t(t-s)^{-\beta}x(s)ds\Big]'\\
&=\frac{1}{\Gamma(1-\beta)}\Big[\sum_{j=0}^{i-1}\int_{t_j}^{t_{j+1}}(t-s)^{-\beta}
x(s)ds\Big]'+\frac{1}{\Gamma(1-\beta)}\Big[\int_{t_i}^{t}(t-s)^{-\beta}
x(s)ds\Big]'\\
&=\sum_{\sigma=0}^i\Big(c_{\sigma1}\frac{\Gamma(\alpha)}
{\Gamma(\alpha-\beta+1)}(t-t_\sigma)^{\alpha-\beta-1}
+c_{\sigma2}\frac{\Gamma(\alpha-1)}{\Gamma(\alpha-\beta-1)}
(t-t_\sigma)^{\alpha-\beta-2}\Big)\\
&\quad +\int_0^t\frac{(t-s)^{\alpha-\beta-1}}{\Gamma(\alpha-\beta)}
\sigma(s)ds,\quad t\in (t_i,t_{i+1}],\;i\in \mathbb{N}[0,m].
\end{aligned}\label{e5.3.6}
\end{equation}
From $\lim_{t\to t_i^+}(t-t_i)^{2-\alpha}x(t)=I_i$, we have
$c_{\sigma2}=I_\sigma$ for all $\sigma\in \mathbb{N}[1,m]$.
 From $\Delta {}^{RL}D_{0^+}^{\alpha-1}x(t_i))=J_i,\;i\in \mathbb{N}[1,m]$, one has
$c_{\sigma1}=\frac{J_\sigma}{\Gamma(\alpha)}$ for all $\sigma\in \mathbb{N}[1,m]$.
From the boundary conditions in \eqref{e5.2.4}, we obtain
\begin{gather*}
c_{02}+\sum_{\sigma=0}^m [c_{\sigma 1}(1-t_\sigma)^{\alpha-1}
+c_{\sigma2}(1-t_\sigma)^{\alpha-2}]
+\int_0^1\frac{(1-s)^{\alpha-1}}{\Gamma(\alpha)}\sigma(s)ds=0,\\
\Gamma(\alpha)c_{01}+\Gamma(\alpha)\sum_{\sigma=0}^mc_{\sigma1}
+\int_0^1\sigma(s)ds=0.
\end{gather*}
It follows that
\begin{gather*}
c_{01}=-\frac{1}{2\Gamma(\alpha)}
\Big[\sum_{\sigma=1}^mJ_\sigma+\int_0^1\sigma(s)ds\Big],\\
\begin{aligned}
c_{02}&=-\frac{1}{2}\Big[\sum_{\sigma=1}^m
\frac{(1-t_\sigma)^{\alpha-1}}{2\Gamma(\alpha)}J_\sigma
+\sum_{\sigma=1}^m(1-t_\sigma)^{\alpha-2}I_\sigma \\
&\quad -\frac{1}{2\Gamma(\alpha)}\int_0^1\sigma(s)ds
+\int_0^1\frac{(1-s)^{\alpha-1}}{\Gamma(\alpha)}\sigma(s)ds\Big].
\end{aligned}
\end{gather*}
Substituting $c_{\sigma1},c_{\sigma2}$ into \eqref{e5.3.4},
we obtain \eqref{e5.3.3} by changing the order of the terms.
It is easy to show from \eqref{e5.3.4} and \eqref{e5.3.6} that $x\in X$.
The proof is compete.
\end{proof}

To abbreviate notation let $H_x(t)=H(t,x(t),{}^{RL}D_{0^+}^\beta x(t))$
for functions $H:(0,1)\times \mathbb{R}^2\to \mathbb{R}$ and $x:(0,]\to \mathbb{R}$.
Define the operator $(Tx)(t)$ for $x\in X$ by
\begin{align*}
(Tx)(t)&=-\frac{1}{2\Gamma(\alpha)}
\Big[\sum_{\sigma=1}^mJ_x(t_\sigma)+\int_0^1p(s)f_x(s)ds\Big]t^{\alpha-1}
-\frac{1}{2}\Big[\sum_{\sigma=1}^m\frac{(1-t_\sigma)^{\alpha-1}}
{2\Gamma(\alpha)}J_x(t_\sigma) \\
&\quad +\sum_{\sigma=1}^m(1-t_\sigma)^{\alpha-2}I_x(t_\sigma)
-\frac{1}{2\Gamma(\alpha)}\int_0^1p(s)f_x(s)ds \\
&\quad +\int_0^1\frac{(1-s)^{\alpha-1}}{\Gamma(\alpha)}p(s)f_x(s)ds\Big]t^{\alpha-2}
 +\sum_{\sigma=1}^i\frac{(1-t_\sigma)^{\alpha-1}}{\Gamma(\alpha)}
 J_x(t_\sigma)(t-t_\sigma)^{\alpha-1} \\
&\quad +\sum_{\sigma=1}^i(1-t_\sigma)^{\alpha-2}I_x(t_\sigma)(t-t_\sigma)^{\alpha-2}
 +\int_0^t\frac{(t-s)^{\alpha-1}}{\Gamma(\alpha)}p(s)f_x(s)ds,
\end{align*}
for $t\in (t_i,t_{i+1}]$ and $i\in N[0,m]$.

\begin{theorem} \label{thm5.3.1}
Suppose that {\rm (H1)} in Theorem \ref{thm5.1.1} holds.
Then \eqref{e5.3.1} has at least one solution if there is $r_0>0$
such that $A_1M_J(r_0,r_0)+A_2M_I(r_0,r_0)+A_3M_f(r_0,r_0)\le r_0$,
where
\begin{gather*}
\begin{aligned}
A_1&=\max\Big\{\frac{m}{2\Gamma(\alpha)}
 +\sum_{\sigma=1}^m\frac{(1-t_\sigma)^{\alpha-1}}{\Gamma(\alpha)}
 +\sum_{\sigma=1}^m\frac{(1-t_\sigma)^{\alpha-1}}{4\Gamma(\alpha)},
 \frac{m}{2\Gamma(\alpha-\beta)} \\
&\quad +\sum_{\sigma=1}^m \frac{(1-t_\sigma)^{\alpha-1}}{\Gamma(\alpha)}
 +\frac{\Gamma(\alpha-1)}{2\Gamma(\alpha-\beta-1)}
\sum_{\sigma=1}^m\frac{(1-t_\sigma)^{\alpha-1}}{2\Gamma(\alpha)}\Big\},\\
\end{aligned} \\
\begin{aligned}
A_2&=\max\Big\{\sum_{\sigma=1}^m\frac{(1-t_\sigma)^{\alpha-2}}{2}
+\sum_{\sigma=1}^m(1-t_\sigma)^{\alpha-2}, \;
\frac{\Gamma(\alpha)}{\Gamma(\alpha-\beta)} \\
&\quad +\sum_{\sigma=1}^m(1-t_\sigma)^{\alpha-2}
 \frac{\Gamma(\alpha-1)}{\Gamma(\alpha-\beta-1)}
 +\frac{\Gamma(\alpha-1)}{2\Gamma(\alpha-\beta-1)}
\sum_{\sigma=1}^m(1-t_\sigma)^{\alpha-2}\Big\},
\end{aligned}\\
\begin{aligned}
A_3&=\max\Big\{\frac{\mathbf{B}(l+1,k+1)}{4\Gamma(\alpha)}
+ \frac{\mathbf{B}(\alpha+l,k+1)}{2\Gamma(\alpha)}
+\frac{\mathbf{B}(l+1,k+1)}{2\Gamma(\alpha)} \\
&\quad +\frac{\mathbf{B}(\alpha+l,k+1)}{\Gamma(\alpha)},\;
 \frac{\mathbf{B}(l+1,k+1)}{2\Gamma(\alpha-\beta)}
+\frac{\Gamma(\alpha-1)}{2\Gamma(\alpha-\beta-1)}
\frac{\mathbf{B}(l+1,k+1)}{2\Gamma(\alpha)} \\
&\quad +\frac{\Gamma(\alpha-1)}{2\Gamma(\alpha-\beta-1)}
\frac{\mathbf{B}(\alpha+l,k+1)}{\Gamma(\alpha)}
+\frac{\mathbf{B}(\alpha-\beta+l,k+1)}{\Gamma(\alpha-\beta)}\Big\}.
\end{aligned}
\end{gather*}
\end{theorem}

\begin{proof}
Let $X$ be defined above. By Lemma \ref{lem5.3.1}, we know that $x$ is a solution
of \eqref{e5.3.1} if and only if $x$ is a fixed point of $T$.
By a standard proof, we can see that $T:X\to X$ is a completely
continuous operator.

From (H1), for $x\in X$ we have
\begin{gather*}
|f_x(t)|=|f(t,x(t),{}^{RL}D_{0^+}^\beta x(t))|
\le M_f(\|x\|,\|x\|),\quad t\in (t_i,t_{i+1}],\;i\in \mathbb{N}[0,m],\\
|I(t_i,x(t_i),{}^{RL}D_{0^+}^\beta x(t_i))|\le M_I(\|x\|,\|x\|),\quad
i\in \mathbb{N}[1,m],\; x\in \mathbb{R},\\
|J(t_i,x(t_i),{}^{RL}D_{0^+}^\beta x(t_i))|\le M_J(\|x\|,\|x\|),\quad
i\in \mathbb{N}[1,m],\; x\in \mathbb{R}.
\end{gather*}
We consider the set
$\Omega=\{x\in X:x=\lambda (Tx),\text{ for some }\lambda \in [0,1)\}$.
For $x\in \Omega$, we have by definition of $T$ for $t\in (t_i,t_{i+1}]$ that
\begin{align*}
&{}^{RL}D_{0^+}^\beta (Tx)(t) \\
&=-\frac{1}{2\Gamma(\alpha)}\Big[\sum_{\sigma=1}^mJ_x(t_\sigma)
+\int_0^1p(s)f_x(s)ds\Big]\frac{\Gamma(\alpha)}{\Gamma(\alpha-\beta)}
t^{\alpha-\beta-1}\\
&\quad -\frac{1}{2}\Big[\sum_{\sigma=1}^m
 \frac{(1-t_\sigma)^{\alpha-1}}{2\Gamma(\alpha)}J_x(t_\sigma)
+\sum_{\sigma=1}^m(1-t_\sigma)^{\alpha-2}I_x(t_\sigma)
 -\frac{1}{2\Gamma(\alpha)}\int_0^1p(s)f_x(s)ds \\
&\quad +\int_0^1\frac{(1-s)^{\alpha-1}}{\Gamma(\alpha)}p(s)f_x(s)ds\Big]
\frac{\Gamma(\alpha-1)}{\Gamma(\alpha-\beta-1)}t^{\alpha-\beta-2}\\
&\quad +\sum_{\sigma=1}^i\frac{(1-t_\sigma)^{\alpha-1}}{\Gamma(\alpha)}J_x(t_\sigma)
\frac{\Gamma(\alpha)}{\Gamma(\alpha-\beta)}(t-t_\sigma)^{\alpha-\beta-1} \\
&\quad +\sum_{\sigma=1}^i(1-t_\sigma)^{\alpha-2}I_x(t_\sigma)
 \frac{\Gamma(\alpha-1)}{\Gamma(\alpha-\beta-1)}(t-t_\sigma)^{\alpha-\beta-2} \\
&\quad +\int_0^t\frac{(t-s)^{\alpha-\beta-1}}{\Gamma(\alpha-\beta)}p(s)f_x(s)ds
,\quad t\in (t_i,t_{i+1}],\;i\in \mathbb{N}[0,m].
\end{align*}
So
\begin{align*}
&(t-t_i)^{2-\alpha}|(Tx)(t)| \\
&\le \frac{1}{2\Gamma(\alpha)}
\big[mM_J(\|x\|,\|x\|)+\int_0^1s^k(1-s)^ldsM_f(\|x\|,\|x\|)\big]\\
&\quad +\frac{1}{2}\Big[\sum_{\sigma=1}^m
 \frac{(1-t_\sigma)^{\alpha-1}}{2\Gamma(\alpha)}M_J(\|x\|,\|x\|)
 + \sum_{\sigma=1}^m(1-t_\sigma)^{\alpha-2}M_I(\|x\|,\|x\|) \\
&\quad +\frac{1}{2\Gamma(\alpha)}\int_0^1s^k(1-s)^ldsM_f(\|x\|,\|x\|) \\
&\quad +\int_0^1\frac{(1-s)^{\alpha-1}}{\Gamma(\alpha)}s^k(1-s)^lds
M_f(\|x\|,\|x\|)\Big]\\
&\quad +\sum_{\sigma=1}^i\frac{(1-t_\sigma)^{\alpha-1}}{\Gamma(\alpha)}
M_J(\|x\|,\|x\|) +\sum_{\sigma=1}^i(1-t_\sigma)^{\alpha-2}M_I(\|x\|,\|x\|)\\
&\quad +(t-t_i)^{2-\alpha}\int_0^t\frac{(t-s)^{\alpha-1}}{\Gamma(\alpha)}
s^k(1-s)^ldsM_f(\|x\|,\|x\|)\\
&\le \Big[\frac{m}{2\Gamma(\alpha)}+\sum_{\sigma=1}^m
 \frac{(1-t_\sigma)^{\alpha-1}}{\Gamma(\alpha)}
+\sum_{\sigma=1}^m\frac{(1-t_\sigma)^{\alpha-1}}{4\Gamma(\alpha)}
 \Big]M_J(\|x\|,\|x\|)\\
&\quad + \big[\sum_{\sigma=1}^m\frac{(1-t_\sigma)^{\alpha-2}}{2}
+\sum_{\sigma=1}^m(1-t_\sigma)^{\alpha-2}\big]M_I(\|x\|,\|x\|)\\
&\quad + \big[\frac{\mathbf{B}(l+1,k+1)}{4\Gamma(\alpha)}+
\frac{\mathbf{B}(\alpha+l,k+1)}{2\Gamma(\alpha)}
+\frac{\mathbf{B}(l+1,k+1)}{2\Gamma(\alpha)} \\
&\quad +\frac{\mathbf{B}(\alpha+l,k+1)}{\Gamma(\alpha)}\big]M_f(\|x\|,\|x\|).
\end{align*}
Furthermore,
\begin{align*}
&(t-t_\sigma)^{2+\beta-\alpha}|{}^{RL}D_{0^+}^\beta (Tx)(t)| \\
&\le\Big[ \frac{m}{2\Gamma(\alpha-\beta)}+\sum_{\sigma=1}^m
 \frac{(1-t_\sigma)^{\alpha-1}}{\Gamma(\alpha)}
+\frac{\Gamma(\alpha-1)}{2\Gamma(\alpha-\beta-1)}
\sum_{\sigma=1}^m\frac{(1-t_\sigma)^{\alpha-1}}{2\Gamma(\alpha)}\Big]
M_J(\|x\|,\|x\|)\\
&\quad + \Big[\frac{\Gamma(\alpha)}{\Gamma(\alpha-\beta)}
+\sum_{\sigma=1}^m(1-t_\sigma)^{\alpha-2}
 \frac{\Gamma(\alpha-1)}{\Gamma(\alpha-\beta-1)}
+\frac{\Gamma(\alpha-1)}{2\Gamma(\alpha-\beta-1)}
\sum_{\sigma=1}^m(1-t_\sigma)^{\alpha-2}\Big] \\
&\quad\times M_I(\|x\|,\|x\|)
 +\Big[\frac{\mathbf{B}(l+1,k+1)}{2\Gamma(\alpha-\beta)}
 +\frac{\Gamma(\alpha-1)}{2\Gamma(\alpha-\beta-1)}
\frac{\mathbf{B}(l+1,k+1)}{2\Gamma(\alpha)} \\
&\quad +\frac{\Gamma(\alpha-1)}{2\Gamma(\alpha-\beta-1)}
\frac{\mathbf{B}(\alpha+l,k+1)}{\Gamma(\alpha)}
+\frac{\mathbf{B}(\alpha-\beta+l,k+1)}{\Gamma(\alpha-\beta)}\Big]M_f(\|x\|,\|x\|).
\end{align*}
It follows that
$$
\|Tx\|\le A_1M_J(\|x\|,\|x\|)+A_2M_I(\|x\|,\|x\|)
+A_3M_f(\|x\|,\|x\|).
$$
From the assumption, we choose $ \Omega=\{x\in X:\|x\|\le r_0\}$.
For $x\in \partial \Omega$, we obtain $x\neq \lambda (Tx)$ for any
 $\lambda\in [0,1]$. In fact, if there exists $x\in \partial \Omega$ such that
$x=\lambda (Tx)$ for some $\lambda\in [0,1]$. Then
\[
r_0=\|x\|=\lambda\|Tx\|\le \|Tx\|\le A_1M_J(r_0,r_0)+A_2M_I(r_0,r_0)
+A_3M_f(r_0,r_0)<r_0,
\]
which is a contradiction.

 As a consequence of Schaefer's fixed point theorem, we deduce that $T$
has a fixed point which is a solution of \eqref{e5.3.1}. The proof is
complete.
\end{proof}


\begin{theorem} \label{thm5.3.2}
Suppose that {\rm (H1)} in Theorem \ref{thm5.1.1} holds and
\[
\lim_{r\to 0^+}\frac{ M_f(r,r)}{r}=\lim_{r\to 0^+}\frac{M_I(r,r)}{r}
=\lim_{r\to 0^+}\frac{ M_J(r,r)}{r}=0
\]
 or
\[
\lim_{r\to +\infty}\frac{ M_f(r,r)}{r}
=\lim_{r\to +\infty}\frac{M_I(r,r)}{r}=\lim_{r\to +\infty}\frac{ M_J(r,r)}{r}=0.
\]

Then \eqref{e5.3.1} has at least one solution.
\end{theorem}

\section{Comments on some published articles}

In some recently published articles, the existence and uniqueness of solutions
of initial or boundary value problems for impulsive fractional differential
equations have been studied, see \cite{wl, wl2, wl3, z, z9, zw, zw1, l9, 241}.
However, we find that some results in such papers are wrong from a mathematical
point of view. To avoiding misleading readers, in this section we
make some comments on these papers.

\subsection{Corrected results from \cite{241}}

 In \cite{241}, the authors studied the solvability of the initial value problems
for the impulsive fractional differential equation
\begin{equation}
\begin{gathered}
{}_0D_{t}^qx(t)=f(t,x(t)),\quad t\in [0,T],\; t\neq t_1,t_2,\dots,t_m,\\
\Delta x(t_k)=I_k(x(t_k^-)),\quad k=1,2,\dots,m,\; x(0)=x_0,
\end{gathered}\label{e6.1.1}
\end{equation}
where $q\in (0,1)$, $x_0\in \mathbb{R}$, ${}_0D_t^q$ is the is the Caputo fractional
derivative in interval $[0,t]$, $f:[0,T]\times\times \mathbb{R}\mapsto \mathbb{R}$
is an appropriate continuous function, $I_k:\mathbb{R}\mapsto \mathbb{R}(k=1,2,\dots,m)$
are continuous functions, $0=t_0<t_1<\dots<t_m<t_{m+1}=T$,
$\Delta x(t_k)=x(t_k^+)-x(t_k^-)=\lim_{\epsilon\to 0^+}x(t_k+\epsilon)
-\lim_{\epsilon\to 0^+}x(t_k-\epsilon)$. We find that the main result
(\cite[Theorem 2.1]{241} with its long proof) is as follows:

\begin{result} \label{res6.1.1} \rm
System \eqref{e6.1.1} is equivalent to the integral equation
\begin{equation}
x(t)=\begin{cases}
x_0+\frac{1}{\Gamma(q)}\int_0^t(t-s)^{q-1}f(s,x(s))ds, \quad t\in (t_0,t_1],\\
x_0+\sum_{k=1}^nI_k(x(t_k^-))+\frac{1}{\Gamma(q)}\int_0^t(t-s)^{q-1}f(s,x(s))ds \\
+h\sum_{k=1}^n\Big[I_k(x(t_k^-))\Big(\int_0^{t_k}(t_k-s)^{q-1}f(s,x(s))ds \\
+\int_{t_k}^t(t-s)^{q-1}f(s,x(s))ds-\int_0^{t}(t-s)^{q-1}f(s,x(s))ds
\Big)\Big]/\Gamma(q),\\
\quad t\in (t_n,t_{n+1}],\; n=1,2,\dots,m
\end{cases}
\label{e6.1.2}
\end{equation}
provided that the integral in \eqref{e6.1.2} exists, where $q\in (0,1),h\in \mathbb{R}$
are constants.
\end{result}

However, this result is in-correct.
In fact, the following example was given \cite[Example 1]{241}:
\begin{equation}
{}_0D_t^{1/4}x(t)=t,\quad t\in [0,2]\setminus\{1\},\quad
x(1)-x(1^-)=I(x(1^-)),\quad x(0)=0.\label{e6.1.3}
\end{equation}
It was claimed that the general solution of \eqref{e6.1.3} is
\begin{equation}
x(t)=\begin{cases}
\frac{16}{5\Gamma(1/4)}t^{5/4}, & t\in (0,1],\\
I_1(x(1^-))+\frac{16}{5\Gamma(1/4)}t^{5/4}\\
+\frac{4hI_1(x(1^-))}
{5\Gamma(1/4)}[4+(t-1)^{1/4}(4t+1)-4t^{5/4}],
& t\in (1,2],
\end{cases}\label{e6.1.4}
\end{equation}
where $h$ is a constant. Let $x$ be defined by \eqref{e6.1.4}.
One notes (by \cite[Definition 2.2]{241}) for $t\in (1,2]$ that
\begin{align*}
{}_0D_t^{1/4}x(t)
&=\frac{1}{\Gamma(1-1/4)}\int_0^t(t-s)^{-1/4}x'(s)ds\\
&=\frac{1}{\Gamma(3/4)}\int_0^1(t-s)^{-1/4}x'(s)ds
+\frac{1}{\Gamma(3/4)}\int_1^t(t-s)^{-1/4}x'(s)ds\\
&=\frac{1}{\Gamma(3/4)}\int_0^1(t-s)^{-1/4}
\frac{16}{5\Gamma(1/4)}[s^{5/4}]'ds\\
&\quad +\frac{1}{\Gamma(3/4)}\int_1^t(t-s)^{-1/4}
\Big[I_1(x(1^-))+\frac{16}{5\Gamma(1/4)}s^{5/4} \\
&\quad +\frac{4hI_1(x(1^-))}{5\Gamma(1/4)}(4+(s-1)^{1/4}(4s+1)-4s^{5/4})\Big]'ds.
\end{align*}
When $w=s/t$, we obtain
\begin{align*}
{}_0D_t^{1/4}x(t)
&=\frac{1}{\Gamma(3/4)}\frac{4}{\Gamma(1/4)}\int_0^t(t-s)^{-1/4}s^{1/4}ds \\
&\quad + \frac{1}{\Gamma(3/4)}\int_1^t(t-s)^{-1/4}
\Big[\frac{4hI_1(x(1^-))}{5\Gamma(1/4)}
\Big(4(s-1)^{1/4}  +s(s-1)^{-3/4}\\
&\quad -\frac{1}{4}(s-1)^{-3/4}-5s^{1/4}\Big)\Big] ds \\
&=\frac{1}{\Gamma(3/4)}\frac{4}{\Gamma(1/4)}t\int_0^1(1-w)^{-1/4}w^{1/4}dw \\
&\quad +\frac{1}{\Gamma(3/4)}\int_1^t(t-s)^{-1/4}
\Big[\frac{4hI_1(x(1^-))}{5\Gamma(1/4)}(4(s-1)^{1/4}
 +s(s-1)^{-3/4} \\
&\quad -\frac{1}{4}(s-1)^{-3/4}-5s^{1/4})\Big] ds
\neq t.
\end{align*}
This shows us that $x$ does not satisfy ${}_0D_t^{1/4}x(t)=t$ on $(1,2]$.
In fact, we have
\begin{gather*}
\begin{aligned}
&\frac{1}{\Gamma(1/4)}\int_1^t(t-s)^{\frac{1}{4}-1}sds \\
&=\frac{1}{\Gamma(1/4)}\int_1^t(t-s)^{\frac{1}{4}-1}(s-1) ds
 +\frac{1}{\Gamma(1/4)}\int_1^t(t-s)^{\frac{1}{4}-1}ds\\
&=\frac{1}{\Gamma(1/4)}(t-1)^{5/4}\int_0^1(1-w)^{\frac{1}{4}-1}wdw
+\frac{4}{\Gamma(1/4)}(t-1)^{1/4} \end{aligned} \\
\text{(using $\frac{s-1}{t-1}=w$)}, \\
\begin{aligned}
\mathbf{B}(1/4,2)
&=\frac{\Gamma(1/4)\Gamma(2)}{\Gamma(9/4)}\\
&=\frac{1}{\Gamma(9/4)}(t-1)^{5/4}+\frac{4}{\Gamma(1/4)}(t-1)^{1/4}\\
&=\frac{4}{5\Gamma(1/4)}(t-1)^{1/4}(4t+1).
\end{aligned}
\end{gather*}
This mistake comes from the following incorrect equality used in \cite{241}:
$$
\frac{1}{\Gamma(1/4)}\int_1^t(t-s)^{\frac{1}{4}-1}sds
=\frac{4}{5\Gamma(1/4)}(t-1)^{1/4}(4t+1).
$$
The correct formula of solution of \eqref{e6.1.3} is
$$
x(t)=\begin{cases}
\frac{16}{5\Gamma(1/4)}t^{5/4}, & t\in (0,1],\\
I_1(x(1^-))+\frac{16}{5\Gamma(1/4)}t^{5/4}, &t\in (1,2].
\end{cases}
$$
We consider the more general problem
\begin{equation}
\begin{gathered}
{}^cD_{0^+}^\alpha x(t)-\lambda x(t)
=f(t,x(t)), \quad t\in (t_i,t_{i+1}],i\in \mathbb{N}_0^m,\\
\Delta x(t_i)=I(t_i,x(t_i^-)), \quad x(0)=x_0,
\end{gathered} \label{e6.1.5}
\end{equation}
where $\alpha\in (0,1)$, $\lambda\in \mathbb{R}$, ${}^cD_{0^+}^*$ is
the Caputo fractional derivative of order $*$, $f:[0,1]\times \mathbb{R}\mapsto \mathbb{R}$,
$I:\{t_i:i\in \mathbb{N}_1^m\}\times \mathbb{R}\mapsto\mathbb{R}$ are continuous functions,
$0=t_0<t_1<\dots<t_m<t_{m+1}=1$, $x_0,\overline{y}_0\in \mathbb{R}$.


\begin{theorem} \label{thm6.1.1}
$x$ is a solution of \eqref{e6.1.5} if and only if
\begin{equation}
\begin{aligned}
x(t)&=x_0\mathbf{E}_{\alpha-\beta,1}(\lambda t^{\alpha-\beta})
 -\lambda x_0t^{\alpha-\beta}\mathbf{E}_{\alpha-\beta,\alpha-\beta+1}
 (\lambda t^{\alpha-\beta})\\
&\quad +\sum_{j=1}^iI(t_j,x(t_j))\mathbf{E}_{\alpha-\beta,1}
 (\lambda (t-t_j)^{\alpha-\beta}) \\
&\quad -\lambda\sum_{j=1}^iI(t_j,x(t_j))
 (t-t_j)^{\alpha-\beta}\mathbf{E}_{\alpha-\beta,\alpha
 -\beta+1}(\lambda(t-t_j)^{\alpha-\beta})\\
&\quad +\int_0^t(t-s)^{\alpha-1}\mathbf{E}_{\alpha-\beta,\alpha}
(\lambda(t-s)^{\alpha-\beta})f(s,x(s))ds,\
\end{aligned} \label{e6.1.6}
\end{equation}
for $t\in (t_i,t_{i+1}]$ and $i\in \mathbb{N}_0^m$.
\end{theorem}

\begin{proof}
By Theorem \ref{thm3.2.1} (with $\alpha=1)$, we know that $x$ is a solution of
\eqref{e6.1.5} if and only if there exist constants $d_j\in \mathbb{R}(j\in \mathbb{N}_0^m)$
such that
\begin{equation}
\begin{aligned}
x(t)
&=\sum_{j=0}^id_j\mathbf{E}_{\alpha-\beta,1}(\lambda (t-t_j)^{\alpha-\beta})
 -\lambda\sum_{j=0}^id_j(t-t_j)^{\alpha-\beta} \\
&\quad\times  \mathbf{E}_{\alpha-\beta,\alpha-\beta+1}
 (\lambda(t-t_j)^{\alpha-\beta}) \\
&\quad +\int_0^t(t-s)^{\alpha-1}\mathbf{E}_{\alpha-\beta,\alpha}
 (\lambda(t-s)^{\alpha-\beta})f(s,x(s))ds,
\end{aligned}\label{e6.1.7}
\end{equation}
$t\in (t_i,t_{i+1}]$ and $i\in \mathbb{N}_0^m$.
From $x(0)=x_0$, we obtain $d_0=x_0$. By $\Delta x(t_i)=I(t_i,x(t_i^-))$,
we have $d_i=I(t_i,x(t_i^-))$ $(i\in\mathbb{N}_1^m)$. Substituting $d_i$ into
\eqref{e6.1.7}, we obtain \eqref{e6.1.6}. The proof is complete.
\end{proof}

\begin{remark} \label{rmk6.1.1}\rm 
Let $\lambda=0$. Then Theorem \ref{thm6.1.1} implies that
\begin{gather*}
{}^cD_{0^+}^\alpha x(t)=f(t,x(t)),\quad t\in (t_i,t_{i+1}], \; i\in \mathbb{N}_0^m,\\
\Delta x(t_i)=I(t_i,x(t_i^-)),x(0)=x_0,
\end{gather*}
is equivalent to
$$
x(t)=x_0+\sum_{j=1}^iI(t_j,x(t_j))
 +\int_0^t\frac{(t-s)^{\alpha-1}}{\Gamma(\alpha)}f(s,x(s))ds,\quad
t\in (t_i,t_{i+1}],\; i\in \mathbb{N}_0^m.
$$
It is easy to see that Result \ref{res6.1.1}
(the equivalent integral equation \eqref{e6.1.1})
is wrong. Theorem \ref{thm6.1.1} (with $\lambda=0$) is a corrected version of the
result.
\end{remark}


\begin{remark} \label{rmk6.1.2}\rm
 Similar to Theorem \ref{thm6.1.1}, we can establish equivalent integral equations
for the following problems:
\begin{gather*}
{}^{c}D_{0^+}^\alpha x(t)-\lambda x(t)=f(t,x(t)),\quad
 t\in (0,1]\setminus\{t_1,\dots,t_p\},\\
\Delta x(t_i)=I_{i}(x(t_i)),\quad \Delta x'(t_i)=J_{i}(x(t_i)),\quad i\in \mathbb{N}_1^p,\\
x(0)=x_0,\quad x'(0)=x_1;
\end{gather*}

\begin{gather*}
{}^{c}D_{0^+}^\alpha x(t)-\lambda x(t)=f(t,x(t)),\quad
 t\in (0,1]\setminus\{t_1,\dots,t_p\},\\
\Delta x(t_i)=I_{i}(x(t_i)),\quad \Delta x'(t_i)=J_{i}(x(t_i)),\quad
i\in \mathbb{N}_1^p,\\
x(0)+\phi(x)=x_0,\quad x'(0)=x_1;
\end{gather*}

\begin{gather*}
{}^{c}D_{0^+}^\beta x(t)-\lambda x(t)=f(t,x(t)), \quad
 t\in (0,1]\setminus\{t_1,\dots,t_p\}, \\
\Delta x(t_i)=I_{i}(x(t_i)),\quad i\in \mathbb{N}_1^p,\\
ax(0)+bx(1)=0;
\end{gather*}


\begin{gather*}
{}^{c}D_{0^+}^\alpha x(t)-\lambda x(t)=f(t,x(t)), \quad
 t\in (0,1]\setminus\{t_1,\dots,t_p\},\\
\Delta x(t_i)=I_{i}(x(t_i)),\quad \Delta x'(t_i)=J_{i}(x(t_i)),\;i\in \mathbb{N}_1^p,\\
ax(0)-bx'(0)=x_0,\quad cx(1)+dx'(1)=x_1;
\end{gather*}
and
\begin{gather*}
{}^{c}D_{0^+}^\alpha x(t)-\lambda x(t)=f(t,x(t)), \quad
 t\in (0,1]\setminus\{t_1,\dots,t_p\},\\
\Delta x(t_i)=I_{i}(x(t_i)),\quad \Delta x'(t_i)=J_{i}(x(t_i)),\quad i\in \mathbb{N}_1^p,\\
x(0)-ax(\xi)=x(1)-bx(\eta)=0.
\end{gather*}
These problems are generalized forms of the problems discussed in \cite{re}.
\end{remark}

\subsection{Corrected results from \cite{z}}

In a recent article, Zhang \cite{z} studied the solvability of the boundary
value problems for the impulsive differential equations with fractional derivative
\begin{equation}
\begin{gathered}
{}_0D_{t}^qy(t)=f(t,y(t)), \quad t\in [0,T],\; t\neq t_k,\;
t\neq \overline{t}_l,\; k=1,2,\dots,m,\; l=1,2,\dots,p,\\
\Delta y(t_k)=I_k(y(t_k^-)),\quad k=1,2,\dots,m,\quad
\Delta y'(\overline{t}_l)=\overline{I}_l(y(\overline{t}_l^-)),\quad l=1,2,\dots,p,\\
y(0)=y_0,\quad y'(0)=\overline{y}_0,
\end{gathered} \label{e6.2.1}
\end{equation}
and its special case
\begin{equation}
\begin{gathered}
{}_0D_{t}^qy(t)=f(t,y(t)), \quad t\in [0,T],\; t\neq t_k,\; k=1,2,\dots,m,\\
\Delta y(t_k)=I_k(y(t_k^-)),\quad \Delta y'({t}_k)=\overline{I}_k(y({t}_k^-)),\;
k=1,2,\dots,m, \\
y(0)=y_0,\quad y'(0)=\overline{y}_0,
\end{gathered} \label{e6.2.2}
\end{equation}
where $q\in (1,2)$, $y_0,\overline{y}_0\in \mathbb{R}$, ${}_0D_t^q$ is the Caputo
fractional derivative in interval $[0,t]$, $f:[0,T]\times\times \mathbb{R}\mapsto \mathbb{R}$
is an appropriate continuous function,
$I_k,\overline{I}_l:\mathbb{R}\mapsto \mathbb{R}(k=1,2,\dots,m,l=1,2,\dots,p)$ are continuous
functions, $0=t_0<t_1<\dots<t_m<t_{m+1}=T$,
$0=\overline{t}_0<\overline{t}_1<\dots<\overline{t}_p<\overline{t}_{p+1}=T$,
$\Delta y(t_k)=y(t_k^+)-y(t_k^-)
=\lim_{\epsilon\to 0^+}y(t_k+\epsilon)-\lim_{\epsilon\to 0^+}y(t_k-\epsilon)$,
$\Delta y'(t_k)=y'(t_k^+)-y'(t_k^-)=\lim_{\epsilon\to 0^+}y'(t_k+\epsilon)
-\lim_{\epsilon\to 0^+}y'(t_k-\epsilon)$.
The main theorem (\cite[Theorem 2.1]{z}) claims that

\begin{result} \label{res6.2.1}\rm
System \eqref{e6.2.1} is equivalent to the integral equation
\[
y(t)=\begin{cases}
y_0+\overline{y}_0t+\int_0^t\frac{(t-s)^{q-1}}{\Gamma(q)}f(s,y(s))ds,
\quad t\in J_0',
\\[4pt]
y_0+\overline{y}_0t+\sum_{1\le k\le n_1}I_k(y(t_k^-))
+\sum_{1\le l\le n_2}(t-\overline{t}_l)\overline{I}_l(y(\overline{t}_l^-)) \\
+\int_0^t\frac{(t-s)^{q-1}}{\Gamma(q)}f(s,y(s))ds
+\xi\sum_{1\le k\le n_1}\Big[{I}_k(y({t}_k^-)) \\
\times \Big(\int_0^{{t}_k}({t}_k-s)^{q-1}f(s,y(s))ds
+\int_{{t}_k}^t(t-s)^{q-1}f(s,y(s))ds \\
-\int_0^t(t-s)^{q-1}f(s,y(s))ds\Big)\Big] / \Gamma(q)\\
+\zeta\sum_{1\le l\le n_2}\Big[\overline{I}_l(y(\overline{t}_l^-))
\Big(\int_0^{\overline{t}_l}(\overline{t}_l-s)^{q-1}f(s,y(s))ds \\
+\int_{\overline{t}_l}^t(t-s)^{q-1}f(s,y(s))ds
-\int_0^t(t-s)^{q-1}f(s,y(s))ds\Big)\Big]/\Gamma(q)
\\
+\frac{\xi}{\Gamma(q-1)}\sum_{1\le k\le n_1}
\Big[(t-{t}_k){I}_k(y({t}_k^-))\int_0^{{t}_k}({t}_k-s)^{q-2}f(s,y(s))ds\Big]
\\
+\frac{\zeta}{\Gamma(q-1)}\sum_{1\le l\le n_2}
\Big[(t-\overline{t}_l)\overline{I}_l(y(\overline{t}_l^-))
\int_0^{\overline{t}_l}(\overline{t}_l-s)^{q-2}f(s,y(s))ds\Big], \\
\quad t\in J_n',\; n=1,2,\dots,\Delta,
\end{cases}
\]
provided that the integral exists, where $q\in(1,2)$, $\xi,\zeta\in\mathbb{R}$
are two constants,
$$
\{t_1,t_2,\dots,t_m,\overline{t}_1,\overline{t}_2,\dots,\overline{t}_p\}
=\{t_1',t_2'\dots,t_\Delta\}
$$
with $0=t_0'<t_1'<t_2'<\dots<t_{\Delta}<t_{\Delta+1}=T$,
$J'_k=(t_k',t_{k+1}']$ $(k=0,1,2,\dots,\Delta)$.
\end{result}

The following statement was also claimed \cite[Corollary 2.4]{z}.

\begin{result} \label{res6.2.2} \rm
System \eqref{e6.2.2} is equivalent to the integral equation
\[
y(t)=\begin{cases}
y_0+\overline{y}_0t+\int_0^t\frac{(t-s)^{q-1}}{\Gamma(q)}f(s,y(s))ds, \quad
t\in [0,t_1],
\\[4pt]
y_0+\overline{y}_0t+\sum_{1\le i\le k}I_k(y(t_k^-))
+\sum_{1\le i\le k}(t-\overline{t}_i)\overline{I}_i(y(\overline{t}_i^-)) \\
+\int_0^t\frac{(t-s)^{q-1}}{\Gamma(q)}f(s,y(s))ds
+\sum_{1\le i\le k}\Big[(\xi{I}_i(y({t}_i^-)) \\
+\zeta\overline{I}_i(y(t_i^-)))
\Big(\int_0^{{t}_i}({t}_i-s)^{q-1}f(s,y(s))ds \\
+\int_{{t}_i}^t(t-s)^{q-1}f(s,y(s))ds-\int_0^t(t
-s)^{q-1}f(s,y(s))ds\Big)\Big]/\Gamma(q)
\\
+\frac{\sum_{1\le i\le k}\Big[(\xi{I}_i(y({t}_i^-))+\zeta\overline{I}_i(y(t_i^-)))(t-t_i)\int_0^{{t}_i}({t}_i-s)^{q-2}
f(s,y(s))ds\Big]/\Gamma(q-1)},\\
\quad t\in (t_k,t_{k+1}],\; k=1,2,\dots,m.
\end{cases}
\]
\end{result}
We find that Results \ref{res6.2.1} and \ref{res6.2.2} are also incorrect.

It is easy to see that \eqref{e6.2.1} and \eqref{e6.2.2} can be generalized
by the IVP
\begin{equation}
\begin{gathered}
{}^cD_{0^+}^\alpha y(t)-\lambda y(t)=f(t,y(t)),\quad t\in (t_i,t_{i+1}],\;
i\in \mathbb{N}_0^m,\\
\Delta y(t_i)=I(t_i,y(t_i^-)),\Delta y'({t}_i)
=\overline{I}(t_i,y({t}_i^-)),\quad i\in \mathbb{N}_1^m,\\
y(0)=y_0,\quad y'(0)=\overline{y}_0,
\end{gathered} \label{e6.2.3}
\end{equation}
where $\alpha\in (1,2)$, ${}^cD_{0^+}^*$ is the Caputo fractional derivative
of order $*$, $f:[0,1]\times \mathbb{R}\mapsto \mathbb{R}$,
$I,J:\{t_i:i\in \mathbb{N}_1^m\}\times \mathbb{R}\mapsto\mathbb{R}$ are continuous functions,
$\lambda\in \mathbb{R}$, $0=t_0<t_1<\dots<t_m<t_{m+1}=1$, $y_0,\overline{y}_0\in \mathbb{R}$.
One sees that BVP\eqref{e6.2.3} generalizes BVP\eqref{e6.2.1} and
\eqref{e6.2.2}. We now establish existence results for \eqref{e6.2.3}.


\begin{theorem} \label{thm6.2.1}
A function $y$ is a solution of \eqref{e6.2.3} if and only if
\begin{equation}
\begin{aligned}
y(t)&=y_0\mathbf{E}_{\alpha,1}(\lambda t^{\alpha})
+\overline{y}_0t\mathbf{E}_{\alpha,2}(\lambda t^{\alpha})
+\sum_{j=1}^iI_{j}(t_j,y(t_j))\mathbf{E}_{\alpha,1}(\lambda(t-t_j)^{\alpha}) \\
&\quad +\sum_{j=1}^i\overline{I}_j(t_j,y(t_j))(t-t_j)
 \mathbf{E}_{\alpha,2}(\lambda(t-t_j)^{\alpha})\\
&\quad + \int_0^t(t-s)^{\alpha-1}\mathbf{E}_{\alpha,\alpha}
(\lambda (t-s)^{\alpha})f(s,y(s))ds, \quad
t\in (t_i,t_{i+1}],\; i\in \mathbb{N}_0^m.
\end{aligned}\label{e6.2.4}
\end{equation}
\end{theorem}

\begin{proof}
From Theorem \ref{thm3.2.1} (with $n=2$), $y$ is a solution of \eqref{e6.2.3} if and only
if there exist constants $c_j,d_j\in \mathbb{R}(j\in \mathbb{N}_0^m)$ such that
\begin{equation}
\begin{aligned}
y(t)&=\sum_{j=0}^ic_j\mathbf{E}_{\alpha,1}(\lambda(t-t_j)^{\alpha})
+\sum_{j=0}^id_j(t-t_j)\mathbf{E}_{\alpha,2}(\lambda(t-t_j)^{\alpha})\\
&\quad + \int_0^t(t-s)^{\alpha-1}\mathbf{E}_{\alpha,\alpha}
 (\lambda(t-s)^{\alpha})f(s,y(s))ds,
\end{aligned}\label{e6.2.5}
\end{equation}
for  $t\in (t_i,t_{i+1}]$ and $i\in \mathbb{N}_0^m$.
By direct computations we obtain
\begin{equation}
\begin{aligned}
y'(t)&=\lambda\sum_{j=0}^ic_j(t-t_j)^{\alpha-1}
\mathbf{E}_{\alpha,\alpha}(\lambda(t-t_j)^{\alpha})
+\sum_{j=0}^id_j\mathbf{E}_{\alpha-1,1}(\lambda(t-t_j)^{\alpha-1})\\
&\quad + \int_0^t(t-s)^{\alpha-2}\mathbf{E}_{\alpha,\alpha-1}
(\lambda(t-s)^{\alpha})f(s,y(s))ds,
\end{aligned} \label{e6.2.6}
\end{equation}
for $t\in (t_i,t_{i+1}]$ and $i\in \mathbb{N}_0^m$.
From $y(0)=y_0,y'(0)=\overline{y}_0$, we obtain $c_0=y_0$ and
$d_0=\overline{y}_0$.
From $\Delta y(t_i)=I(t_i,y(t_i^-)),\Delta y'({t}_i)=\overline{I}(t_i,y({t}_i^-))$,
we obtain $c_i=I(t_i,y(t_i^-))$ and $d_i=\overline{I}(t_i,y({t}_i^-))$ for all
$i\in \mathbb{N}_1^m$.
Substituting $c_i,d_i$ into \eqref{e6.2.5}, we obtain \eqref{e6.2.4}.
The proof is complete.
\end{proof}

\begin{remark} \label{rmk6.2.1} \rm
Let $\lambda=0$. By \eqref{e6.2.4}, we obtain that the IVP
\begin{gather*}
{}^cD_{0^+}^\alpha y(t)=f(t,y(t)),\quad t\in (t_i,t_{i+1}],\; i\in \mathbb{N}_0^m,\\
\Delta y(t_i)=I(t_i,y(t_i^-)),\quad \Delta y'({t}_i)=\overline{I}(t_i,y({t}_i^-)),
\; i\in \mathbb{N}_1^m,\\
y(0)=y_0,\quad y'(0)=\overline{y}_0,
\end{gather*}
is equivalent to the integral equation
\begin{align*}
y(t)
&=y_0+\overline{y}_0t+\sum_{j=1}^iI_{j}(t_j,y(t_j))
+\sum_{j=1}^i\overline{I}_j(t_j,y(t_j))(t-t_j)\\
&\quad + \int_0^t\frac{(t-s)^{\alpha-1}}{\Gamma(\alpha)}f(s,y(s))ds,
\quad t\in (t_i,t_{i+1}],\; i\in \mathbb{N}_0^m.
\end{align*}
It is easy to see that Results \ref{res6.2.1} and \ref{res6.2.2}
 (obtained in \cite{z}) are wrong.
\end{remark}

\subsection{Corrected results from \cite{z9}}
Zhang \cite{z9} considered the initial value problem for fractional
differential equation with Caputo-Hadamard fractional derivative and
impulsive effect,
\begin{equation}
\begin{gathered}
{}^{ch}D_{a^+}^qx(t)=f(t,x(t)),\quad t\in (a,T],\; t\neq t_k, \; k=1,2,\dots,m,\\
\Delta x(t_k)=\Delta_k(x(t_k^-)),\quad k=1,2,\dots,m,\; x(a)=x_a,
\end{gathered} \label{e6.3.1}
\end{equation}
where $q\in \mathbb{C}$ and $\mathbb{R}(q)\in (0,1)$, ${}^{ch}D_{a^+}^q$
denotes the left-sided Caputo-Hadamard fractional
derivative of order $q$ with the low limit $a (>0)$,
 $ a=t_0<t_1<\dots< t_m < t_{m+1}= T$, $f: (a,T]\times\mathbb{C}\mapsto \mathbb{C}$
is an appropriate continuous function,
$x(t^+_k ) = \lim_{\epsilon\to0^+}x(t_k+\epsilon)$ and
$x(t^-_ k) =\lim_{\epsilon\to0^-}x(t_k+\epsilon)$ represent the right and
left limits of $x(t)$ at $t = t_k$, respectively.
The following result was claimed \cite[Theorem 3.2]{z9}.

\begin{result} \label{res6.3.1}\rm
System \eqref{e6.3.1} is equivalent to the fractional integral equation
\[
x(t)=\begin{cases}
x_a+\int_a^t\frac{(\ln t-\ln s)^{q-1}}{\Gamma(q)}f(s,x(s))\frac{ds}{s}, \quad
t\in (a,t_1],\\[4pt]
x_a+\sum_{i=1}^k\Delta_i(x(t_i^-))+\int_a^t
\frac{(\ln t-\ln s)^{q-1}}{\Gamma(q)}f(s,x(s))\frac{ds}{s}\\
+\sum_{i=1}^k\frac{h\Delta_i(x(t_i^-))}{\Gamma(q)}
\Big[\int_a^{t_i}\frac{(\ln t_i-\ln s)^{q-1}}{\Gamma(q)}f(s,x(s))\frac{ds}{s} \\
 +\int_{t_i}^t\frac{(\ln t-\ln s)^{q-1}}{\Gamma(q)}
f(s,x(s))\frac{ds}{s}-\int_a^t\frac{(\ln t-\ln s)^{q-1}}{\Gamma(q)}f(s,x(s))
\frac{ds}{s}\Big], \\
\quad t\in (t_i,t_{i+1}],\; i=1,2,\dots,m,
\end{cases}
\]
where $h$ is a constant. We find that this result is wrong.
We consider the more general problem
\begin{equation}
\begin{gathered}
{}^{ch}D_{0^+}^q x(t)-\lambda x(t)=f(t,x(t)),\quad t\in (t_i,t_{i+1}],\; i\in \mathbb{N}_0^m,\\
\Delta x(t_i)=x(t_i^+)-x(t_i^-)=I(t_i,x(t_i)),\quad i\in \mathbb{N}_1^m,\; x(1)=x_0,
\end{gathered} \label{e6.3.2}
\end{equation}
where $q\in (0,1)$, $p\in (0,q)$, ${}^{ch}D_{0^+}^*$ is the Caputo-Hadamard
type fractional derivative of order $*$, $\lambda\in \mathbb{R}$,
$f:[1,e]\times \mathbb{R}\mapsto \mathbb{R}$, $I:\{t_i:i\in \mathbb{N}_1^m\}\mapsto \mathbb{R}$
are continuous functions, $1=t_0<t_1<t_2<\dots<t_m<t_{m+1}=e$, $x_0\in \mathbb{R}$.
One sees that \eqref{e6.3.2} is a generalization of \eqref{e6.3.1}
 ($a=1$ and $\lambda=0$).
\end{result}


\begin{theorem} \label{thm13.1}
 BVP \eqref{e6.3.2} is equivalent to the integral equation
\begin{equation}
\begin{aligned}
x(t)&=x_0\mathbf{E}_{q,1}(\lambda (\ln t)^{q})
+\sum_{j=1}^iI(t_j,x(t_j))\mathbf{E}_{q,1}(\lambda(\ln t-\ln t_j)^{q})\\
&\quad +\int_0^t(\ln t-\ln s)^{q-1}\mathbf{E}_{q,q}
 (\lambda(\ln t-\ln s)^{q})f(s,x(s))\frac{ds}{s},
\end{aligned} \label{e6.3.3}
\end{equation}
for $t\in (t_i,t_{i+1}]$ and $i\in \mathbb{N}_0^m$.
\end{theorem}

\begin{proof}
 By Theorem \ref{thm3.2.4}, $x$ is a solution of BVP\eqref{e6.3.2} if and only if there
exist constants $d_j\in \mathbb{R}(j\in \mathbb{N}_0^m)$ such that
\begin{equation}
\begin{aligned}
x(t)&=\sum_{j=0}^id_j\mathbf{E}_{q,1}(\lambda(\ln t-\ln t_j)^{q})\\
&\quad +\int_0^t(\ln t-\ln s)^{q-1}\mathbf{E}_{q,q}
(\lambda(\ln t-\ln s)^{q})f(s,x(s))\frac{ds}{s},
\end{aligned}\label{e6.3.4}
\end{equation}
for $t\in (t_i,t_{i+1}]$ and $i\in \mathbb{N}_0^m$.
From $\Delta x(t_i)=I(t_i,x(t_i))$, we obtain $d_i=I(t_i,x(t_i))$ $(i\in\mathbb{N}_1^m)$.
 From $x(1)=x_0$, we obtain $d_0=x_0$.
Substituting $d_j$ into \eqref{e6.3.4}, we obtain \eqref{e6.3.3}.
The proof is complete.
\end{proof}

\begin{remark} \label{rmk6.3.1}\rm
 From \eqref{e6.3.3}, letting $\lambda=0$, we obtain that the system
 \begin{gather*}
{}^{ch}D_{0^+}^q x(t)=f(t,x(t)),\quad t\in (t_i,t_{i+1}],\; i\in \mathbb{N}_0^m,\\
\Delta x(t_i)=I(t_i,x(t_i)),\quad i\in \mathbb{N}_1^m,\;x(1)=x_0
\end{gather*}
 is equivalent to
$$
x(t)=x_0+\sum_{j=1}^iI(t_j,x(t_j))
+\int_0^t\frac{(\ln t-\ln s)^{q-1}}{\Gamma(q)}f(s,x(s))\frac{ds}{s},
\quad t\in (t_i,t_{i+1}],\; i\in \mathbb{N}_0^m.
$$
Then we know that Result \ref{res6.3.1} obtained in \cite{z9} is wrong.
\end{remark}

\subsection{Corrected results from \cite{wl2, wl3}}

In \cite{wl2, wl3}, the authors studied the existence of solutions of the
 anti-periodic boundary value problems for impulsive fractional differential
equations,
\begin{gather*}
D_{0^+}^q x(t)+\lambda(t) x(t)=f(t,x(t)),\quad
t\in [0,1]\setminus\{t_1,t_2,\dots,t_m\},\\
I_{0^+}^\alpha x(t_i^+)-I_{0^+}^\alpha (t_i^-)=J_i(x(t_i)),\quad
i=1,2,\dots,m,\; t^{1-q}x(t)|_{t=0}+t^{1-q}x(t)|_{t=1}=0,
\end{gather*}
and
\begin{gather*}
{}^cD_{0^+}^q x(t)+\lambda((t) x(t)=f(t,x(t)),\quad
t\in [0,1]\setminus\{t_1,t_2,\dots,t_m\},\\
I_{0^+}^\alpha x(t_i^+)-I_{0^+}^\alpha (t_i^-)=J_i(x(t_i)),\quad
i=1,2,\dots,m,\; x(0)+x(1)=0,
\end{gather*}
where $q,\alpha\in (0,1)$, ${}^cD_{0^+}^q$ is the Caputo fractional derivative,
$D_{0^+}^q$ is the Riemann-Liouville fractional derivative, $I_{0^+}^\alpha$
is the Riemann-Liouville fractional integral,
$0 = t_0 < t_1 < \dots< t_m < t_{m+1} = 1$,
$\lambda\in C^0([0, 1], R)$ satisfies $\lambda_0 =:\max_{t\in [0,1]}\lambda(t)> 0$,
$J_k : \mathbb{R} \mapsto \mathbb{R}$ is continuous, $f$ is a given piecewise continuous function.
The main results in \cite{wl3} are based upon the following two results:

\begin{result}[{\cite[Lemma2.7]{wl3}}] \label{res6.4.1} \rm
Suppose that $q,\alpha\in (0,1)$. Then $x$ is a solution of
\begin{equation}
\begin{gathered}
D_{0^+}^q x(t)+\lambda_0 x(t)=f(t,x(t)),\quad
t\in [0,1]\setminus\{t_1,t_2,\dots,t_m\},\\
I_{0^+}^\alpha x(t_i^+)-I_{0^+}^\alpha (t_i^-)=J_i(x(t_i)),\quad
i=1,2,\dots,m,\; t^{1-q}x(t)|_{t=0}+t^{1-q}x(t)|_{t=1}=0
\end{gathered} \label{e6.4.1}
\end{equation}
if and only if $x$ is a fixed point of the operator
$T_q:PC_q(0,1]\mapsto PC_q(0,1]$, where $T_q$ is defined by
\begin{align*}
(T_qx)(t)&=\frac{\Gamma(q)t^{q-1}\mathbf{E}_{q,q}
(-\lambda_0t^{q})}{1+\Gamma(q)\mathbf{E}_{q,q}(-\lambda_0)}
\Big[\sum_{i=1}^m\frac{J_i(x(t_i))}{\Gamma(q)t_i^{\alpha+q-1}
\mathbf{E}_{q,q+\alpha}(-\lambda_0t_i^q)} \\
&\quad -\int_0^1(1-s)^{q-1}\mathbf{E}_{q,q}(-\lambda_0(1-s)^q)f(s,x(s))ds\Big]\\
&-t^{q-1}\mathbf{E}_{q,q}(-\lambda_0t^q)
 \sum_{t\le t_i<1}\frac{J_i(x(t_i))}{t_i^{\alpha+q-1}
 \mathbf{E}_{q,q+\alpha}(-\lambda_0t_i^q)}\\
&\quad +\int_0^t(t-s)^{q-1}\mathbf{E}_{q,q}(-\lambda_0(t-s)^q)f(s,x(s))ds.
\end{align*}
\end{result}

\begin{result}[{cite[Lemma 2.8]{wl3}}] \label{res6.4.2} \rm
Suppose that $q,\alpha\in (0,1)$. Then $x$ is a solution of
\begin{equation}
\begin{gathered}
{}^cD_{0^+}^q x(t)+\lambda_0 x(t)=f(t,x(t)),\quad
t\in [0,1]\setminus\{t_1,t_2,\dots,t_m\},\\
I_{0^+}^\alpha x(t_i^+)-I_{0^+}^\alpha (t_i^-)=J_i(x(t_i)),\quad
i=1,2,\dots,m,\; x(0)+x(1)=0
\end{gathered} \label{e6.4.2}
\end{equation}
if and only if $x$ is a fixed point of the operator $T:PC(0,1]\mapsto PC(0,1]$,
where $T$ is defined by
\begin{align*}
(Tx)(t)&=\frac{\mathbf{E}_{q,1}(-\lambda_0t^{q})}{1+\mathbf{E}_{q,1}(-\lambda_0)}
\Big[\sum_{i=1}^m\frac{J_i(x(t_i))}{t_i^{\alpha}\mathbf{E}_{q,1+\alpha}
 (-\lambda_0t_i^q)} \\
&\quad -\int_0^1(1-s)^{q-1}\mathbf{E}_{q,q}
 (-\lambda_0(1-s)^q)f(s,x(s))ds\Big]\\
&\quad -\mathbf{E}_{q,1}(-\lambda_0t^q)\sum_{t\le t_i<1}
 \frac{J_i(x(t_i))}{t_i^{\alpha}\mathbf{E}_{q,1+\alpha}(-\lambda_0t_i^q)} \\
&\quad +\int_0^t(t-s)^{q-1}\mathbf{E}_{q,q}(-\lambda_0(t-s)^q)f(s,x(s))ds.
\end{align*}
\end{result}

We consider the problem
\begin{equation}
\begin{gathered}
{}^cD_{0^+}^\alpha x(t)+\lambda x(t)=f(t,x(t)),\quad t\in (t_i,t_{i+1}],\;
i\in \mathbb{N}_0^m,\\
\Delta x(t_i)=x(t_i^+)-x(t_i^-)=I(t_i,x(t)i)),\quad i\in \mathbb{N}_1^m,\; x(0)+x(1)=0,
\end{gathered} \label{e6.4.3}
\end{equation}
where $\lambda\in \mathbb{R}$, $f:[0,1]\times\mathbb{R}\mapsto \mathbb{R}$,
$I:\{t_i:i\in \mathbb{N}_1^m\}\times \mathbb{R}\mapsto \mathbb{R}$ are continuous functions,
${}^cD_{0^+}^\alpha$ is the Caputo fractional derivative with the order
$\alpha\in (0,1)$, $0=t_0<t_1<t_2<\dots<t_m<t_{m+1}=1$.

By Theorem \ref{thm3.2.1}, we know $x$ is a solution of \eqref{e6.4.3} if and only if
there exists constants $c_i(i\in \mathbb{N}_0^m)$ such that
\begin{equation}
x(t)=\sum_{v=0}^jc_v\mathbf{E}_{\alpha,1} (-\lambda (t-t_v)^\alpha)
+\int_0^t(t-s)^{\alpha-1}\mathbf{E}_{\alpha,\alpha}(-\lambda (t-s)^\alpha)f(s,x(s))ds,
\label{e6.4.4}
\end{equation}
for $t\in (t_j,t_{j+1}]$, $j\in \mathbb{N}_0^m$.
From $\Delta x(t_i)=I(t_i,x(t)i))$, we have $c_i=I(t_i,x(t)i))$
$(i\in\mathbb{N}_1^m)$. By $x(0)+x(1)=0$, we have
\[
c_0+\sum_{v=0}^mc_v\mathbf{E}_{\alpha,1}(-\lambda (1-t_v)^\alpha)
+\int_0^1(1-s)^{\alpha-1}\mathbf{E}_{\alpha,\alpha}(-\lambda (t-s)^\alpha)
f(s,x(s))ds=0.
\]
It follows that
\begin{align*}
c_0&=-\frac{1}{1+\mathbf{E}_{\alpha,1}(-\lambda)}
\Big[\sum_{v=1}^mI(t_v,x(t_v))\mathbf{E}_{\alpha,1}(-\lambda (1-t_v)^\alpha) \\
&\quad +\int_0^1(1-s)^{\alpha-1}\mathbf{E}_{\alpha,\alpha}
 (-\lambda (t-s)^\alpha)f(s,x(s))ds\Big].
\end{align*}
Substituting $c_v$ into \eqref{e6.4.4}, we obtain
\begin{equation}
\begin{aligned}
x(t)&=-\frac{\mathbf{E}_{\alpha,1}(\lambda t^\alpha) }
{1+\mathbf{E}_{\alpha,1}(-\lambda)}
\Big[\sum_{v=1}^mI(t_v,x(t_v))\mathbf{E}_{\alpha,1}(-\lambda (1-t_v)^\alpha) \\
&\quad +\int_0^1(1-s)^{\alpha-1}\mathbf{E}_{\alpha,\alpha}
 (-\lambda (t-s)^\alpha)f(s,x(s))ds\Big]\\
&\quad +\sum_{v=1}^jI(t_v,x(t_v))\mathbf{E}_{\alpha,1}(-\lambda (t-t_v)^\alpha) \\
&\quad +\int_0^t(t-s)^{\alpha-1}\mathbf{E}_{\alpha,\alpha}
 (-\lambda (t-s)^\alpha)f(s,x(s))ds,
\end{aligned}\label{e6.4.5}
\end{equation}
for $t\in (t_j,t_{j+1}]$ and $j\in \mathbb{N}_0^m$.
From \eqref{e6.4.5}, Result \ref{res6.4.2} is wrong.

We consider problem \eqref{e6.4.1}. By Theorem \ref{thm3.2.2}, we know $x$ is a solution of
\eqref{e6.4.1} if and only if there exists constants $c_i(i\in \mathbb{N}_0^m)$ such that
\begin{equation}
\begin{aligned}
x(t)&=\sum_{v=0}^jc_v(t-t_v)^{q-1}\mathbf{E}_{q,q}(-\lambda_0 (t-t_v)^q)\\
&\quad +\int_0^t(t-s)^{q-1}\mathbf{E}_{q,q}(-\lambda_0 (t-s)^q)f(s,x(s))ds,
\end{aligned}\label{e6.4.6}
\end{equation}
for  $t\in (t_j,t_{j+1}]$ and $j\in \mathbb{N}_0^m$.
By Definition \ref{def2.1} and direct computations, we have
\begin{equation}
\begin{aligned}
I_{0^+}^\alpha x(t)
&=\sum_{v=0}^jc_v(t-t_v)^{q+\alpha-1}\mathbf{E}_{q,q+\alpha}(-\lambda_0 (t-t_v)^q)\\
&\quad +\int_0^t(t-s)^{q+\alpha-1}\mathbf{E}_{q,q+\alpha}
 (-\lambda_0 (t-s)^q)f(s,x(s))ds, 
\end{aligned}\label{e6.4.6b}
\end{equation}
for $t\in (t_j,t_{j+1}]$ and $j\in \mathbb{N}_0^m$.
Using $t^{1-q}x(t)|_{t=0}+t^{1-q}x(t)|_{t=1}=0$, we obtain
\begin{align*}
&\frac{c_0}{\Gamma(q)}+\sum_{v=0}^mc_v(1-t_v)^{q-1}\mathbf{E}_{q,q}
(-\lambda_0 (1-t_v)^q) \\
&+\int_0^1(1-s)^{q-1}\mathbf{E}_{q,q}(-\lambda_0 (1-s)^q)f(s,x(s))ds=0.
\end{align*}
\smallskip

\noindent\textbf{Case 1: $\alpha+q<1$.}
 From \eqref{e6.4.6} we know that the existence of $I_{0^+}^\alpha x(t_i^+)$
implies $c_i=0(i\in \mathbb{N}[1,m])$. So
\[
I_{0^+}^\alpha x(t)=c_0t^{q+\alpha-1}\mathbf{E}_{q,q+\alpha}
(-\lambda_0 t^q)+\int_0^t(t-s)^{q+\alpha-1}
\mathbf{E}_{q,q+\alpha}(-\lambda_0 (t-s)^q)f(s,x(s))ds,
\]
for $t\in (t_j,t_{j+1}]$, $j\in \mathbb{N}_0^m$.
Then $I_{0^+}^\alpha x(t_i^+)-I_{0^+}^\alpha (t_i^-)=J_i(x(t_i))$
implies that $J_i(x(t_i))=0(i\in \mathbb{N}[1,m]$
 So this impulse model is unsuitable.
\smallskip

\noindent\textbf{Case 2: $\alpha+q=1$.}
 From $I_{0^+}^\alpha x(t_i^+)-I_{0^+}^\alpha (t_i^-)=J_i(x(t_i))$ and
\eqref{e6.4.6}, we obtain
$$
c_i-
\sum_{v=0}^{i-1}c_v\mathbf{E}_{q,q+\alpha}(-\lambda_0 (t_i-t_v)^q)=J_i(x(t_i)),
\quad i\in \mathbb{N}_1^m.
$$
\smallskip

\noindent\textbf{Case 3: $\alpha+q>1$.}
From $I_{0^+}^\alpha x(t_i^+)-I_{0^+}^\alpha (t_i^-)=J_i(x(t_i))$ and \eqref{e6.4.6},
we obtain
$$
-\sum_{v=0}^{i-1}(t_i-t_v)^{\alpha+q-1}c_v\mathbf{E}_{q,q+\alpha}
(-\lambda_0 (t_i-t_v)^q)=J_i(x(t_i)),
\quad i\in \mathbb{N}_1^m.
$$

We know that Result \ref{res6.4.1} is wrong.


\subsection{Corrected results from \cite{wz}}

In \cite{wz}, authors established the existence of solutions for a class of nonlinear
impulsive Hadamard fractional differential equations with initial condition of the
form
\begin{equation}
\begin{gathered}
{}^{rh}D_{1^+}^\alpha x(t)=f(t,x(t)),\quad t\in (1,e]\setminus\{t_1,t_2,\dots,t_m\},\\
\Delta^*x(t_i)={}^h\mathfrak{J}_{1^+}^{1-\alpha} x(t_i^+)-{}^h
\mathfrak{J}_{1^+}^{1-\alpha}x(t_i^-)=p_i, \quad i=1,2,\dots,m,\\
{}^h\mathfrak{J}_{1^+}^{1-\alpha}u(1)=u_0,
\end{gathered} \label{e6.5.1}
\end{equation}
where ${}^{rh}D_{1^+}^\alpha$ is the left-side Riemann-Liouville type
Hadamard fderivative of order $\alpha\in (0,1)$ with the starting point
$1$ and ${}^h\mathfrak{J}_{1^+}^{1-\alpha}$ denotes left-side Hadamard
fractional integral of order $1-\alpha$, $1=t_0<t_1<\dots<t_m<t_{m+1}=e$,
$u_0,p_i\in \mathbb{R}(i=1,2,\dots,m)$, $f:[1,e]\times\mathbb{R}\mapsto \mathbb{R}$ is a continuous
function. It was claimed the following result \cite[Lemma 2.9, p. 87]{wz}:

\begin{result} \label{res6.5.1}\rm
 Let $f:[1,e]\times \mathbb{R}\mapsto \mathbb{R}$ and $t\mapsto (\ln t)^{1-\alpha}f(t,u)$
are continuous functions. Then $x$ is a solution of the fractional integral
equation
$$
x(t)= \begin{cases}
\frac{u_0}{\Gamma(\alpha)}(\ln t)^{\alpha-1}
+\int_1^t\frac{(\ln t-\ln s)^{\alpha-1}}{\Gamma(\alpha)}f(s,x(s))
\frac{ds}{s},\quad t\in (1,t_1],\\[4pt]
\frac{u_0}{\Gamma(\alpha)}(\ln t)^{\alpha-1}
+\sum_{j=1}^i\frac{p_j}{\Gamma(\alpha)}(\ln t)^{\alpha-1}
+\int_1^t\frac{(\ln t-\ln s)^{\alpha-1}}{\Gamma(\alpha)}f(s,x(s))\frac{ds}{s},\\
\quad t\in (t_i,t_{i+1}],i=1,2,\dots,m
\end{cases}
$$
if and only if $x$ is a solution of IVP \eqref{e6.5.1}.
\end{result}

We note that Result \ref{res6.5.1} is also wrong.
Now we consider the initial value problem for impulsive fractional differential
equation involving the Riemann-Liouville type Hadamard fractional derivatives
\begin{equation}
\begin{gathered}
{}^{rh}D_{1^+}^\alpha x(t)-\lambda {}^{rh}D_{1^+}^\beta x(t)=f(t,x(t)),
\quad t\in (t_i,t_{i+1}],\; i\in \mathbb{N}_0^m,\\
\Delta^*x(t_i)=I(t_i,{}^h\mathfrak{J}_{1^+}^{1-\alpha}x(t_i)),\quad i\in \mathbb{N}_1^m,\\
{}^h\mathfrak{J}_{1^+}^{1-\alpha}u(1)=u_0,
\end{gathered} \label{e6.5.2}
\end{equation}
where $\alpha\in (0,1)$ and $\beta\in (0,\beta)$, ${}^{rh}D_{1^+}^*$ is the
left-side Riemann-Liouville type Hadamard fderivative of order
$*\in (0,1)$ with the starting point $1$ and ${}^h\mathfrak{J}_{1^+}^{1-\alpha}$
denotes left-side Hadamard fractional integral of order $1-\alpha$, $u_0\in \mathbb{R}$,
$1=t_0<t_1<\dots<t_m<t_{m+1}=e$, $f:[1,e]\times\mathbb{R}\mapsto \mathbb{R}$ is a
V-Carath\'eodory function, $I:\{t_i:i\in \mathbb{N}_1^m\}\times \mathbb{R}\mapsto \mathbb{R}$ is a
discrete V-Carath\'eodory function,
\[
\Delta^*x(t_i)={}^h\mathfrak{J}_{1^+}^{1-\alpha} x(t_i^+)-{}^h
\mathfrak{J}_{1^+}^{1-\alpha}x(t_i^-).
\]
It is easy to see that \eqref{e6.5.2} generalizes \eqref{e6.5.1}
 ($\lambda=0$ and $I(t_i,u)=p_i$).

\begin{theorem} \label{thm6.5.1}
 Suppose that $\alpha\in (0,1)$. Then $x$ is a solution of \eqref{e6.5.2}
if and only $x$ is the solution of the integral equation
\begin{equation}
\begin{aligned}
x(t)&=u_0(\ln t)^{\alpha-1}
\mathbf{E}_{\alpha-\beta,\alpha}(\lambda (\ln t)^{\alpha-\beta}) \\
&\quad +\sum_{j=1}^iI(t_j,x(t_j))(\ln t-\ln t_j)^{\alpha-1}
\mathbf{E}_{\alpha-\beta,\alpha}(\lambda (\ln t-\ln t_j)^{\alpha-\beta})\\
&\quad +\int_1^t(\ln t-\ln s)^{\alpha-1}\mathbf{E}_{\alpha-\beta,\alpha}
(\lambda (\ln t-\ln s)^{\alpha-\beta})f(s,x(s))\frac{ds}{s},
\end{aligned}\label{e6.5.3}
\end{equation}
for $t\in (t_i,t_{i+1}]$, $i\in \mathbb{N}_0^m$.
\end{theorem} 

\begin{proof}
We can prove that $x$ is a piecewise continuous solution of
\[
{}^{rh}D_{1^+}^\alpha x(t)-\lambda {}^{rh}D_{1^+}^\beta x(t)=f(t,x(t)), \quad
t\in (t_i,t_{i+1}],\;  i\in \mathbb{N}_0^m.
\]
if and only if there exist constants $d_j\in \mathbb{R}(j\in \mathbb{N}_0^m)$ such that
\begin{equation}
\begin{aligned}
x(t)&=\sum_{j=0}^id_j(\ln t-\ln t_j)^{\alpha-1}
\mathbf{E}_{\alpha-\beta,\alpha}(\lambda (\ln t-\ln t_j)^{\alpha-\beta})\\
&\quad +\int_1^t(\ln t-\ln s)^{\alpha-1}\mathbf{E}_{\alpha-\beta,\alpha}
(\lambda (\ln t-\ln s)^{\alpha-\beta})f(s,x(s))\frac{ds}{s},
\end{aligned} \label{e15.3}
\end{equation}
for $t\in (t_i,t_{i+1}]$ and $i\in \mathbb{N}_0^m$.
For $t\in (t_i,t_{i+1}]$,  using Definition \ref{def2.1}-(ii) we have
\begin{align*}
&{}^h\mathfrak{J}_{1^+}^{1-\alpha}x(t) \\
&=\int_1^t\frac{(\ln t-\ln s)^{-\alpha}}{\Gamma(1-\alpha)}x(s)\frac{ds}{s} \\
&=\Big(\sum_{\tau=0}^{i-1}\int_{t_{\tau}}^{t_{\tau+1}}
 (\ln t-\ln s)^{-\alpha}x(s)\frac{ds}{s}+\int_{t_i}^t
(\ln t-\ln s)^{-\alpha}x(s)\frac{ds}{s}\Big)\Gamma(1-\alpha)
\\
&=\Big(\sum_{\tau=0}^{i-1}\int_{t_{\tau}}^{t_{\tau+1}}
(\ln t-\ln s)^{-\alpha}
\Big[\sum_{j=0}^\tau d_j(\ln s-\ln t_j)^{\alpha-1}
\mathbf{E}_{\alpha-\beta,\alpha}\big(\lambda (\ln s-\ln t_j)^{\alpha-\beta}\big) \\
&\quad +\int_1^s(\ln s-\ln u)^{\alpha-1}\mathbf{E}_{\alpha-\beta,\alpha}
(\lambda (\ln s-\ln u)^{\alpha-\beta})f(u,x(u))\frac{du}{u}\Big]
\frac{ds}{s}\Big)/\Gamma(1-\alpha)
\\
&\quad +\Big(\int_{t_i}^t(\ln t-\ln s)^{-\alpha}
\Big[\sum_{j=0}^id_j(\ln s-\ln t_j)^{\alpha-1}
\mathbf{E}_{\alpha-\beta,\alpha}\big(\lambda (\ln s-\ln t_j)^{\alpha-\beta}\big) \\
&\quad +\int_1^s(\ln s-\ln u)^{\alpha-1}\mathbf{E}_{\alpha-\beta,\alpha}
(\lambda (\ln s-\ln u)^{\alpha-\beta})f(u,x(u))\frac{du}{u}\Big]
\frac{ds}{s}\Big)/\Gamma(1-\alpha)
\\
&=\Big(\sum_{\tau=0}^{i-1}\int_{t_{\tau}}^{t_{\tau+1}}(\ln t-\ln s)^{-\alpha}
\Big[\sum_{j=0}^\tau d_j(\ln s-\ln t_j)^{\alpha-1} \\
&\quad\times \mathbf{E}_{\alpha-\beta,\alpha}
\big(\lambda (\ln s-\ln t_j)^{\alpha-\beta}\big)\Big]
\frac{ds}{s}\Big)/\Gamma(1-\alpha)
\\
&\quad +\Big(\int_{t_i}^t(\ln t-\ln s)^{-\alpha}
\Big[\sum_{j=0}^id_j(\ln s-\ln t_j)^{\alpha-1} \\
&\quad\times \mathbf{E}_{\alpha-\beta,\alpha}\big(\lambda (\ln s-\ln t_j)^{\alpha-\beta}\big)\Big]
\frac{ds}{s}\Big)/\Gamma(1-\alpha)\\
\\
&\quad +\Big(\int_1^t(\ln t-\ln s)^{-\alpha}
\Big[\int_1^s(\ln s-\ln u)^{\alpha-1}\mathbf{E}_{\alpha-\beta,\alpha}
(\lambda (\ln s-\ln u)^{\alpha-\beta}) \\
&\quad\times f(u,x(u))\frac{du}{u}\Big]
\frac{ds}{s}\Big)/\Gamma(1-\alpha)\\
&\quad \text{(by changing the order of the sum and the integral)}\\
&=\Big(\sum_{j=0}^{i-1}\sum_{\tau=j}^{i-1} d_j
 \int_{t_{\tau}}^{t_{\tau+1}}(\ln t-\ln s)^{-\alpha}(\ln s-\ln t_j)^{\alpha-1} \\
&\quad\times \mathbf{E}_{\alpha-\beta,\alpha}\big(\lambda (\ln s-\ln t_j)^{\alpha-\beta}\big)
\frac{ds}{s}\Big)/\Gamma(1-\alpha)
\\
&\quad +\Big(\sum_{j=0}^id_j\int_{t_i}^t(\ln t-\ln s)^{-\alpha}
(\ln s-\ln t_j)^{\alpha-1} \\
&\quad\times \mathbf{E}_{\alpha-\beta,\alpha}\big(\lambda (\ln s-\ln t_j)^{\alpha-\beta}\big)
\frac{ds}{s}\Big)/\Gamma(1-\alpha)
\\
&\quad +\Big(\int_1^t\int_u^t(\ln t-\ln s)^{-\alpha}(\ln s-\ln u)^{\alpha-1} \\
&\quad\times \mathbf{E}_{\alpha-\beta,\alpha}(\lambda (\ln s-\ln u)^{\alpha-\beta})
\frac{ds}{s}f(u,x(u))\frac{du}{u}\Big)/\Gamma(1-\alpha)
\\
&=\Big(\sum_{j=0}^id_j\int_{t_{j}}^{t}(\ln t-\ln s)^{-\alpha}
(\ln s-\ln t_j)^{\alpha-1} \\
&\quad\times \mathbf{E}_{\alpha-\beta,\alpha}\big(\lambda (\ln s-\ln t_j)^{\alpha-\beta}\big)
\frac{ds}{s}\Big)/\Gamma(1-\alpha)
\\
&\quad +\Big(\int_1^t\int_u^t(\ln t-\ln s)^{-\alpha}(\ln s-\ln u)^{\alpha-1} \\
&\quad\times \mathbf{E}_{\alpha-\beta,\alpha}(\lambda (\ln s-\ln u)^{\alpha-\beta})
\frac{ds}{s}f(u,x(u))\frac{du}{u}\Big)/\Gamma(1-\alpha)
\\
&=\Big(\sum_{j=0}^id_j\int_{t_{j}}^{t}(\ln t-\ln s)^{-\alpha}
(\ln s-\ln t_j)^{\alpha-1} \\
&\quad\times \sum_{\chi=0}^\infty
 \frac{\lambda^\chi}{\Gamma(\chi(\alpha-\beta)+\alpha)}
(\ln s-\ln t_j)^{\chi(\alpha-\beta)}\frac{ds}{s}\Big)/\Gamma(1-\alpha) \\
&\quad \text{(using $\frac{\ln s-\ln t_j}{\ln t-\ln t_j}=w$)}
\\
&\quad + \Big(\int_1^t\int_u^t(\ln t-\ln s)^{-\alpha}(\ln s-\ln u)^{\alpha-1} \\
&\quad\times \sum_{\chi=0}^\infty\frac{\lambda^\chi}
{\Gamma(\chi(\alpha-\beta)+\alpha)}
(\ln s-\ln u)^{\chi(\alpha-\beta)}\frac{ds}{s}f(u,x(u))\frac{du}{u}
\Big)/\Gamma(1-\alpha) \\
&\quad \text{(using $\frac{\ln s-\ln u}{\ln t-\ln s}=w$)}
\\
&=\Big(\sum_{j=0}^id_j\sum_{\chi=0}^\infty\frac{\lambda^\chi}
{\Gamma(\chi(\alpha-\beta)+\alpha)}(\ln t-\ln t_j)^{\chi(\alpha-\beta)} \\
&\quad\times \int_0^1 (1-w)^{-\alpha}w^{\alpha-1+\chi(\alpha-\beta)}dw
\Big)/ \Gamma(1-\alpha)
\\
&\quad +\Big(\sum_{\chi=0}^\infty\frac{\lambda^\chi}
{\Gamma(\chi(\alpha-\beta)+\alpha)}\int_1^t
(\ln t-\ln u)^{\chi(\alpha-\beta)} \\
&\quad\times \int_0^1(1-w)^{-\alpha}
w^{\alpha-1+\chi(\alpha-\beta)}dwf(u,x(u))\frac{du}{u}\Big)/\Gamma(1-\alpha)
\\
&=\sum_{j=0}^id_j\sum_{\chi=0}^\infty\frac{\lambda^\chi}
{\Gamma(\chi(\alpha-\beta)+1)}(\ln t-\ln t_j)^{\chi(\alpha-\beta)} \\
&\quad +\sum_{\chi=0}^\infty\frac{\lambda^\chi}{\Gamma(\chi(\alpha-\beta)+1)}
\int_1^t(\ln t-\ln u)^{\chi(\alpha-\beta)}f(u,x(u))\frac{du}{u}
\\
&=\sum_{j=0}^id_j\mathbf{E}_{\alpha-\beta,1}(\lambda(\ln t-\ln t_j)^{\alpha-\beta})
+\int_1^t\mathbf{E}_{\alpha-\beta,1}(\lambda(\ln t-\ln s)^{\alpha-\beta})
f(u,x(u))\frac{du}{u}.
\end{align*}
Then
\begin{equation}
\begin{aligned}
{}^h\mathfrak{J}_{1^+}^{1-\alpha}x(t)
&=\sum_{j=0}^id_j\mathbf{E}_{\alpha-\beta,1}(\lambda(\ln t-\ln t_j)^{\alpha-\beta})\\
&\quad +\int_1^t\mathbf{E}_{\alpha-\beta,1}(\lambda(\ln t-\ln s)^{\alpha-\beta})
 f(u,x(u))\frac{du}{u},
\end{aligned} \label{e15.4}
\end{equation}
for $t\in (t_i,t_{i+1}]$ and $i\in \mathbb{N}_0^m$.
Using the above inequality,
 $\Delta^*x(t_i)=I(t_i,{}^h\mathfrak{J}_{1^+}^{1-\alpha}x(t_i))$, $i\in \mathbb{N}_1^m$,
and ${}^h\mathfrak{J}_{1^+}^{1-\alpha}u(1)=u_0$, we obtain
 $d_i=I(t_i,{}^h\mathfrak{J}_{1^+}^{1-\alpha}x(t_i))$, $i\in \mathbb{N}_1^m$ and $d_0=u_0$.
Substituting $d_i$ into \eqref{e15.3}, we obtain \eqref{e15.3}.
\end{proof}

Now we show that Result \ref{res6.5.1} is wrong.
Note that $\mathbf{E}_{\alpha-\beta,\alpha}(0)=\frac{1}{\Gamma(\alpha)}$.
Let $A=0$ and $I(t_i,x)=p_i$. We know from Theorem \ref{thm6.5.1} that $x$ is a solution
of \eqref{e6.5.1} if and only if
\[
x(t)=\frac{u_0}{\Gamma(\alpha)}(\ln t)^{\alpha-1}
+\sum_{j=1}^i\frac{p_j}{\Gamma(\alpha)}(\ln t-\ln t_j)^{\alpha-1}
+\int_1^t\frac{(\ln t-\ln s)^{\alpha-1}}{\Gamma(\alpha)}f(s,x(s))
\frac{ds}{s},
\]
for $t\in (t_i,t_{i+1}]$ and $i\in \mathbb{N}_0^m$.
This shows that Result \ref{res6.5.1} is wrong.

\subsection{Corrected results from \cite{l9,zw,zw1}}

The following BVP was studied in \cite{zw1}:
\begin{equation}
\begin{gathered}
{}^cD_{0^+}^\gamma x(t)+ax(t)=f(t,x(t),y(t)),\quad
t\in [0,1]\setminus\{t_1,\dots,t_m\},\\
{}^cD_{0^+}^\gamma y(t)+by(t)=g(t,x(t),y(t)),\quad
t\in [0,1]\setminus\{t_1,\dots,t_m\},\\
x(0)=-\sum_{i=1}^m\alpha_i x(\tau_i),\quad
y(0)=-\sum_{i=1}^m\beta_iy(\tau_i),\\
\Delta x(t_i)=I_i(x(t_i)),\quad
\Delta y(t_i)=J_i(y(t_i)),\quad i=1,2,\dots,m,
\end{gathered} \label{e6.6.1}
\end{equation}
where $\gamma\in (0,1)$, $a,b>0$, $0=t_0<t_1<\dots<t_m<t_{m+1}=1$,
$\tau_i\in (t_i,t_{i+1})$, $\alpha_i,\beta_i\in \mathbb{R}$ with
$1+\sum_{i=1}^m\alpha_i\neq 0$ and $1+\sum_{i=1}^m\beta_i\neq 0$,
 $\Delta x(t_i)=\lim_{t\to t_i^+}x(t)-\lim_{t\to t_i^-}x(t)$ and
 $\Delta y(t_i)=\lim_{t\to t_i^+}y(t)-\lim_{t\to t_i^-}y(t)$,
${}^cD_{0^+}^\gamma$ is the Caputo type fractional derivative of
order $\gamma$ with starting point $0$, $I_i,J_i:\mathbb{R}\mapsto \mathbb{R}$ are continuous,
$f,g$ are jointly continuous functions. The following claim was made:

\begin{result}[\cite{zw1}] \label{res6.6.1}
 BVP \eqref{e6.6.1} is equivalent to the integral system
\begin{gather*}
\begin{aligned}
x(t)&=-\alpha\sum_{i=1}^m\alpha_i
\Big[\sum_{0<t_i<\tau_i}\mathbf{E}_{\gamma,1}(-at_i^\gamma)I_i(x(t_i)) \\
&\quad + \int_0^{\tau_i}(\tau_i-s)^{\gamma-1}\mathbf{E}_{\gamma,\gamma}
(-a(\tau_i-s)^\gamma)f(s,x(s),y(s))ds\Big]\\
&\quad +\sum_{0<t_i<t}\mathbf{E}_{\gamma,1}(-at^\gamma)I_i(x(t_i)) \\
&\quad +\int_0^t(t-s)^{\gamma-1}\mathbf{E}_{\gamma,\gamma}(-a(t-s)^\gamma)
f(s,x(s),y(s))ds,
\end{aligned}\\
\begin{aligned}
y(t)&=-\beta\sum_{i=1}^m\beta_i\Big[\sum_{0<t_i<\tau_i}\mathbf{E}_{\gamma,1}
(-bt_i^\gamma)J_i(y(t_i)) \\
&\quad + \int_0^{\tau_i}(\tau_i-s)^{\gamma-1}\mathbf{E}_{\gamma,\gamma}
(-b(\tau_i-s)^\gamma)g(s,x(s),y(s))ds\Big]\\
&\quad +\sum_{0<t_i<t}\mathbf{E}_{\gamma,1}(-bt^\gamma)J_i(y(t_i)) \\
&\quad +\int_0^t(t-s)^{\gamma-1}\mathbf{E}_{\gamma,\gamma}(-b(t-s)^\gamma)g(s,x(s),y(s))ds,
\end{aligned}
\end{gather*}
where $\alpha=[1+\sum_{i=1}^m\alpha_i\mathbf{E}_{\gamma,1}(-a\tau_i^\gamma)]^{-1}$
and $\beta=[1+\sum_{i=1}^m\beta_i\mathbf{E}_{\gamma,1}(-b\tau_i^\gamma)]^{-1}$.
\end{result}

\begin{theorem} \label{thm6.6.1}
BVP \eqref{e6.6.1} is equivalent to the integral system
\begin{equation}
\begin{gathered}
\begin{aligned}
x(t)&=-\alpha\sum_{i=1}^m\alpha_i\Big[\sum_{0<t_i<\tau_i}
\mathbf{E}_{\gamma,1}(-at_i^\gamma)I_i(x(t_i)) \\
&\quad + \int_0^{\tau_i}(\tau_i-s)^{\gamma-1}\mathbf{E}_{\gamma,\gamma}
(-a(\tau_i-s)^\gamma)f(s,x(s),y(s))ds\Big]\\
&\quad +\sum_{0<t_i<t}\mathbf{E}_{\gamma,1}(-at^\gamma)I_i(x(t_i)) \\
&\quad +\int_0^t(t-s)^{\gamma-1}\mathbf{E}_{\gamma,\gamma}(-a(t-s)^\gamma)
f(s,x(s),y(s))ds,
\end{aligned} \\
\begin{aligned}
y(t)&=-\beta\sum_{i=1}^m\beta_i\Big[\sum_{0<t_i<\tau_i}
 \mathbf{E}_{\gamma,1}(-bt_i^\gamma)J_i(y(t_i)) \\
&\quad + \int_0^{\tau_i}(\tau_i-s)^{\gamma-1}\mathbf{E}_{\gamma,\gamma}
(-b(\tau_i-s)^\gamma)g(s,x(s),y(s))ds\Big]\\
&\quad +\sum_{0<t_i<t}\mathbf{E}_{\gamma,1}(-bt^\gamma)J_i(y(t_i)) \\
&\quad +\int_0^t(t-s)^{\gamma-1}\mathbf{E}_{\gamma,\gamma}(-b(t-s)^\gamma)g(s,x(s),y(s))ds,
\end{aligned}
\end{gathered}\label{e6.6.2}
\end{equation}
\end{theorem}

\begin{proof}
Suppose that $(x,y)$ is a solution of \eqref{e6.6.1}.
 By Theorem \ref{thm3.2.4} (choose $\alpha=\gamma\in (0,1)$ in \eqref{e3.2.4}),
we know that there exist constants $c_i,d_i\in \mathbb{R}(i\in \mathbb{N}[0,m])$ such that
\begin{gather*}
x(t)=\sum_{j=0}^ic_j\mathbf{E}_{\gamma,1}(-a(\log t-\log t_j)^\gamma)
+\int_1^t(\log t-\log s)^{\gamma-1}f(s,x(s),y(s))\frac{ds}{s},\\
 t\in (t_i,t_{i+1}],\;i\in \mathbb{N}[0,m],\\
y(t)=\sum_{j=0}^id_j\mathbf{E}_{\gamma,1}(-b(\log t-\log t_j)^\gamma)
+\int_1^t(\log t-\log s)^{\gamma-1}g(s,x(s),y(s))\frac{ds}{s},\\
 t\in (t_i,t_{i+1}],\;i\in \mathbb{N}[0,m].
\end{gather*}
By $\Delta x(t_i)=I_i(x(t_i))$, $\Delta y(t_i)=J_i(y(t_i))$, $i\in \mathbb{N}[1,m]$,
we obtain
$c_i=I_i(x(t_i)),\;d_i=J_i(y(t_i))$, $i\in \mathbb{N}[1,m]$.
By $x(0)=-\sum_{i=1}^m\alpha_i x(\tau_i)$, we obtain
\begin{align*}
& c_0+\sum_{i=1}^m\alpha_i\Big[\sum_{j=0}^ic_j\mathbf{E}_{\gamma,1}
(-a(\log \tau_i-\log t_j)^\gamma) \\
&+\int_1^{\tau_i}(\log \tau_i-\log s)^{\gamma-1}
f(s,x(s),y(s))\frac{ds}{s}\Big]=0.
\end{align*}
It follows that
\begin{align*}
c_0&=-\Big(\sum_{i=1}^m\alpha_i\Big[\sum_{j=1}^iI_j(x(t_j))
\mathbf{E}_{\gamma,1}(-a(\log \tau_i-\log t_j)^\gamma) \\
&\quad +\int_1^{\tau_i}(\log \tau_i-\log s)^{\gamma-1}f(s,x(s),y(s))
\frac{ds}{s}\Big]\Big)
/\Big(1+\sum_{i=1}^m\alpha_i\mathbf{E}_{\gamma,1}(-a(\log \tau_i))\Big).
\end{align*}
Hence
\begin{align*}
x(t)&=-\Big(\sum_{i=1}^m\alpha_i\Big[\sum_{j=1}^iI_j(x(t_j))
\mathbf{E}_{\gamma,1}(-a(\log \tau_i-\log t_j)^\gamma) \\
&\quad +\int_1^{\tau_i}(\log \tau_i-\log s)^{\gamma-1}f(s,x(s),y(s))
\frac{ds}{s}\Big]\Big) \mathbf{E}_{\gamma,1}(-a(\log t)^\gamma) \\
&\quad \div \Big(1+\sum_{i=1}^m\alpha_i\mathbf{E}_{\gamma,1}
(-a(\log \tau_i))\Big) \\
&\quad +\sum_{j=1}^iI_j(x(t_j))\mathbf{E}_{\gamma,1}
 (-a(\log t-\log t_j)^\gamma) \\
&\quad +\int_1^t(\log t-\log s)^{\gamma-1}f(s,x(s),y(s))\frac{ds}{s},
\quad  t\in (t_i,t_{i+1}],\;i\in \mathbb{N}[0,m].
\end{align*}
Similarly we obtain
\begin{align*}
y(t)&=-\Big(\sum_{i=1}^m\beta_i
\Big[\sum_{j=1}^iJ_j(y(t_j))\mathbf{E}_{\gamma,1}
(-b(\log \tau_i-\log t_j)^\gamma) \\
&\quad +\int_1^{\tau_i}
(\log \tau_i-\log s)^{\gamma-1}g(s,x(s),y(s))\frac{ds}{s}\Big]\Big)
\mathbf{E}_{\gamma,1}(-b(\log t)^\gamma) \\
&\quad\div \Big(1+\sum_{i=1}^m\beta_i\mathbf{E}_{\gamma,1}(-b(\log \tau_i))\Big)\\
&\quad +\sum_{j=1}^iJ_j(y(t_j))\mathbf{E}_{\gamma,1}(-b(\log t-\log t_j)^\gamma) \\
&\quad +\int_1^t(\log t-\log s)^{\gamma-1}g(s,x(s),y(s))\frac{ds}{s},\quad
 t\in (t_i,t_{i+1}],\;i\in \mathbb{N}[0,m].
\end{align*}
On the other hand, if $(x,y)$ satisfies \eqref{e6.6.2}, we can prove that
$(x,y)$ is a solution of \eqref{e6.6.1}. The proof is complete.
\end{proof}

 From Theorem \ref{thm6.6.1}, we know the Result \ref{res6.6.1}
 claimed in \cite{zw1} is wrong.

In \cite{zw}, authors studied the fractional impulsive boundary value problem
 on infinite intervals
\begin{equation}
\begin{gathered}
D_{0^+}^{\alpha}u(t)+f(t,u(t))=0,\quad t\in (0,+\infty),\; t\neq t_k,k=1,2,\dots,m, \\
u(t_k^+)-u(t_k^-)=-I_k(u(t_k)),\quad k=1,2,\dots,m,\\
u(0)=0,\quad D_{0^+}^{\alpha-1}u(+\infty)=0,
\end{gathered} \label{e6.6.3}
\end{equation}
where $\alpha\in (1,2]$, $D_{0^+}^*$ is the Riemann-Liouville
 fractional derivatives of orders $*>0$, $t_0=0$, $1<t_1<\dots<t_m<+\infty$,
$u(t_k^+)=\lim_{t\to t_k^+}u(t)$ and $u(t_k^-)=\lim_{t\to t_k^-}u(t)$,
$D_{0^+}^{\alpha-1}u(+\infty)=\lim_{t\to +\infty}D_{0^+}^{\alpha-1}u(t)$,
$(t,u)\to f(t,(1+t^\alpha)u)$ is nonnegative, continuous on
$[0,+\infty)\times [0,+\infty)$ and $u\to I_k(u)$ is nonnegative, continuous
and bounded. Existence, uniqueness and computational method of unbounded
positive solutions of \eqref{e6.6.3} were established.
\cite[Lemma 3.1]{zw} claimed the following result:

 \begin{result} \label{res6.6.2}\rm
Let $y\in C^0[0,\infty)$ with $\int_0^\infty y(t)dt$ convergent,
$\alpha\in (1,2)$. If $u$ is a solution of
\begin{equation}
u(t)=\int_0^\infty G(t,s)y(s)ds+\sum_{i=1}^mW_i(t,u(t_i)), \label{e6.6.4}
\end{equation}
where
\begin{gather*}
 G(t,s)= \begin{cases}
 \frac{t^{\alpha-1}-(t-s)^{\alpha-1}}{\Gamma(\alpha)}, & 0\le s\le t<\infty,\\
\frac{t^{\alpha-1}}{\Gamma(\alpha)}, & 0\le t\le s<\infty,
\end{cases} \\
 W_i(t,u(t_i))= \begin{cases}
 \frac{I_i(u(t_i))t^{\alpha-1}}{t_i^{\alpha-1}-t_i^{\alpha-2}},& 0\le t\le t_i,\\
 \frac{I_i(u(t_i))t^{\alpha-2}}{t_i^{\alpha-1}-t_i^{\alpha-2}},
& t_i<t<\infty\end{cases}
\end{gather*}
then $u$ is a solution of
\begin{equation}
\begin{gathered}
D_{0^+}^{\alpha}u(t)+y(t)=0,\quad t\in (0,+\infty),\; t\neq t_k, \; k=1,2,\dots,m, \\
u(t_k^+)-u(t_k^-)=-I_k(u(t_k)),\quad k=1,2,\dots,m,\\
u(0)=0,\quad  D_{0^+}^{\alpha-1}u(+\infty)=0.
\end{gathered} \label{e6.6.5}
\end{equation}
\end{result}

We find that this result is wrong. In fact, \eqref{e6.6.4} can be re-written as
\begin{align*}
 u(t)&=-\int_0^t\frac{(t-s)^{\alpha-1}}{\Gamma(\alpha)}y(s)ds
 +\frac{t^{\alpha-1}}{\Gamma(\alpha)}\int_0^\infty y(s)ds\\
&\quad +\sum_{i=k+1}^m\frac{I_i(u(t_i))t^{\alpha-1}}{t_i^{\alpha-1}
 -t_i^{\alpha-2}}+\sum_{i=1}^k\frac{I_i(u(t_i))t^{\alpha-2}}{t_i^{\alpha-1}
 -t_i^{\alpha-2}}, \quad t\in (t_k,t_{k+1}], \; k=0,1,2,\dots.
 \end{align*}
Hence  for $t\in (t_k,t_{k+1}](k=1,2,\dots,m)$ we have
 \begin{align*}
&D_{0^+}^\alpha u(t) \\
&=\frac{1}{\Gamma(2-\alpha)}\Big[\int_0^t(t-s)^{1-\alpha}u(s)ds\Big]'' \\
&=\Big[\sum_{\nu=0}^{k-1}\int_{t_\nu}^{t_{\nu+1}}(t-s)^{1-\alpha}u(s)ds
+\int_{t_k}^t(t-s)^{1-\alpha}u(s)ds\Big]''/ \Gamma(2-\alpha) \\
&=\Big[\sum_{\nu=0}^{k-1}\int_{t_\nu}^{t_{\nu+1}}(t-s)^{1-\alpha}
 (-\int_0^s\frac{(s-u)^{\alpha-1}}{\Gamma(\alpha)}y(u)du
 +\frac{s^{\alpha-1}}{\Gamma(\alpha)}\int_0^\infty y(u)du \\
&\quad +\sum_{i=\nu+1}^m\frac{I_i(u(t_i))s^{\alpha-1}}{t_i^{\alpha-1}
 -t_i^{\alpha-2}}+\sum_{i=1}^\nu\frac{I_i(u(t_i))s^{\alpha-2}}{t_i^{\alpha-1}
 -t_i^{\alpha-2}})ds\Big]''/ \Gamma(2-\alpha) \\
&\quad  +\Big[\int_{t_k}^t(t-s)^{1-\alpha}
(-\int_0^s\frac{(s-u)^{\alpha-1}}{\Gamma(\alpha)}y(u)du+\frac{s
 ^{\alpha-1}}{\Gamma(\alpha)}\int_0^\infty y(u)du \\
&\quad +\sum_{i=k+1}^m\frac{I_i(u(t_i))s^{\alpha-1}}{t_i^{\alpha-1}-t_i^{\alpha-2}}
+\sum_{i=1}^k\frac{I_i(u(t_i))s^{\alpha-2}}{t_i^{\alpha-1}-t_i^{\alpha-2}})ds
\Big]''/\Gamma(2-\alpha) \\
& =\Big[ \sum_{\nu=0}^{k-1}\sum_{i=\nu+1}^m
 \frac{I_i(u(t_i))}{t_i^{\alpha-1}-t_i^{\alpha-2}}
 \int_{t_\nu}^{t_{\nu+1}}(t-s)^{1-\alpha}s^{\alpha-1}ds \\
&\quad  +\sum_{\nu=0}^{k-1}\sum_{i=1}^\nu\frac{I_i(u(t_i))}{t_i^{\alpha-1}-t_i^{\alpha-2}}
 \int_{t_\nu}^{t_{\nu+1}}(t-s)^{1-\alpha}s^{\alpha-2}ds\Big]''
/\Gamma(2-\alpha) \\
&\quad  +\Big[\sum_{i=k+1}^m\frac{I_i(u(t_i))}{t_i^{\alpha-1}-t_i^{\alpha-2}}
\int_{t_k}^t(t-s)^{1-\alpha}s^{\alpha-1}ds \\
&\quad +\sum_{i=1}^k\frac{I_i(u(t_i))}{t_i^{\alpha-1}-t_i^{\alpha-2}}
\int_{t_k}^t(t-s)^{1-\alpha}s^{\alpha-2}
ds\Big]''/\Gamma(2-\alpha) \\
&\quad  +\Big[-\int_0^t(t-s)^{1-\alpha}
 \int_0^s\frac{(s-u)^{\alpha-1}}{\Gamma(\alpha)}y(u)\,du\,ds \\
&\quad +\int_0^t(t-s)^{1-\alpha}\frac{s
 ^{\alpha-1}}{\Gamma(\alpha)}\int_0^\infty y(u)\,du\,ds\Big]''
/\Gamma(2-\alpha) \\
&\quad \text{(by changing the order of the sum and the integral, we obtain)}\\
&=\Big[ \sum_{i=1}^{k}\sum_{\nu=0}^{i-1}\frac{I_i(u(t_i))}{t_i^{\alpha-1}-t_i^{\alpha-2}}
\int_{t_\nu}^{t_{\nu+1}}(t-s)^{1-\alpha}s^{\alpha-1}ds
\Big]'' / \Gamma(2-\alpha) \\
&\quad  +\Big[\sum_{i=k+1}^{m}\sum_{\nu=0}^{k-1}\frac{I_i(u(t_i))}{t_i^{\alpha-1}
 -t_i^{\alpha-2}}\int_{t_\nu}^{t_{\nu+1}}(t-s)^{1-\alpha}s^{\alpha-1}ds \\
&\quad  +\sum_{i=1}^{k-1}\sum_{\nu=i}^{k-1}\frac{I_i(u(t_i))}
 {t_i^{\alpha-1}-t_i^{\alpha-2}}\int_{t_\nu}^{t_{\nu+1}}(t-s)^{1-\alpha}s^{\alpha-2}
ds\Big]''/\Gamma(2-\alpha) \\
&\quad \Big[\sum_{i=k+1}^m\frac{I_i(u(t_i))}{t_i^{\alpha-1}-t_i^{\alpha-2}}
 \int_{t_k}^t(t-s)^{1-\alpha}s^{\alpha-1}ds \\
&\quad +\sum_{i=1}^k\frac{I_i(u(t_i))}{t_i^{\alpha-1}-t_i^{\alpha-2}}
 \int_{t_k}^t(t-s)^{1-\alpha}s^{\alpha-2}
ds\Big]''/\Gamma(2-\alpha)  \\
&\quad  +\Big[-\int_0^t\int_u^t(t-s)^{1-\alpha}
\frac{(s-u)^{\alpha-1}}{\Gamma(\alpha)}dsy(u)du \\
&\quad +\int_0^\infty\int_0^t(t-s)^{1-\alpha}\frac{s
 ^{\alpha-1}}{\Gamma(\alpha)}ds y(u)du\Big]''
/\Gamma(2-\alpha) \\
&\quad \text{(using that $\frac{s-u}{t-u}=w,\frac{s}{t}=w$)}  \\
& =\Big[ \sum_{i=1}^{k}\frac{I_i(u(t_i))}{t_i^{\alpha-1}-t_i^{\alpha-2}}
 \int_0^{t_{i}}(t-s)^{1-\alpha}s^{\alpha-1}ds\Big]''/ \Gamma(2-\alpha) \\
&\quad  +\Big[\sum_{i=k+1}^m\frac{I_i(u(t_i))}{t_i^{\alpha-1}-t_i^{\alpha-2}}
\int_0^t(t-s)^{1-\alpha}s^{\alpha-1}ds \\
&\quad +\sum_{i=1}^k\frac{I_i(u(t_i))}{t_i^{\alpha-1}-t_i^{\alpha-2}}
\int_{t_i}^t(t-s)^{1-\alpha}s^{\alpha-2}
ds\Big]''/ \Gamma(2-\alpha)\\
&\quad  +\Big[-\int_0^t(t-u)\int_0^1(1-w)^{1-\alpha}\frac{w^{\alpha-1}}
 {\Gamma(\alpha)}dwy(u)du \\
&\quad +\int_0^\infty t\int_0^1(1-w)^{1-\alpha}
\frac{w ^{\alpha-1}}{\Gamma(\alpha)}dw y(u)du\Big]''/ \Gamma(2-\alpha)\\
&\quad \text{(using that $\mathbf{B}(p,q)=\int_0^1x^{p-1}(1-x)^{q-1}dx
=\frac{\Gamma(p)\Gamma(q)}{\Gamma(p+q)}$)}  \\
&=y(t)+\Big[
 \sum_{i=1}^{k}\frac{I_i(u(t_i))}{t_i^{\alpha-1}-t_i^{\alpha-2}}t
\int_0^{\frac{t_{i}}{t}}(1-w)^{1-\alpha}w^{\alpha-1}dw
\Big]'' / \Gamma(2-\alpha)  \\
&\quad +\Big[\sum_{i=k+1}^m\frac{I_i(u(t_i))}{t_i^{\alpha-1}-t_i^{\alpha-2}}t
 \int_0^1(1-w)^{1-\alpha}w^{\alpha-1}dw \\
&\quad +\sum_{i=1}^k\frac{I_i(u(t_i))}{t_i^{\alpha-1}-t_i^{\alpha-2}}t
\int_{\frac{t_i}{t}}^1(1-w)^{1-\alpha}w^{\alpha-2}
dw\Big]''/ \Gamma(2-\alpha)\\
&=y(t)+\Big[ \sum_{i=1}^{k}\frac{I_i(u(t_i))}{t_i^{\alpha-1}
 -t_i^{\alpha-2}}t\int_0^{\frac{t_{i}}{t}}(1-w)^{1-\alpha}w^{\alpha-1}dw
\Big]''/ \Gamma(2-\alpha) \\
&\quad -\Big[\sum_{i=1}^k\frac{I_i(u(t_i))}{t_i^{\alpha-1}-t_i^{\alpha-2}}
 t\int_0^{\frac{t_i}{t}}(1-w)^{1-\alpha}w^{\alpha-2}
dw\Big]''/\Gamma(2-\alpha)\\
&=y(t)+\Big[ \sum_{i=1}^{k}\frac{I_i(u(t_i))}{t_i^{\alpha-1}
 -t_i^{\alpha-2}}t\int_0^{\frac{t_{i}}{t}}(1-w)^{1-\alpha}[w^{\alpha-1}
 -w^{\alpha-2}]dw\Big]''/ \Gamma(2-\alpha) \\
&\neq y(t), \quad t\in (t_k,t_{k+1}],\; k=1,2,\dots,m.
\end{align*}
Thus $u$ given by \eqref{e6.6.4} is not a solution of \eqref{e6.6.3}.
Hence Result \ref{res6.6.2} is wrong.

To correct the results in \cite{zw}, we consider the BVP
\begin{equation}
\begin{gathered}
D_{0^+}^{\alpha}u(t)+f(t,u(t))=0,\quad t\in (0,+\infty),\;
t\neq t_k, \; k\in \mathbb{N}[0,\infty), \\
\Delta D_{0^+}^{2-\alpha}u(t_k)=J(t_k,u(t_k)),\quad
\Delta I_{0^+}^{2-\alpha}u(t_k)=I(t_k,u(t_k)),\quad k\in \mathbb{N}[1,\infty),\\
I_{0^+}^{2-\alpha}u(0)=0,\quad D_{0^+}^{\alpha-1}u(+\infty)=0,
\end{gathered} \label{e6.6.6}
\end{equation}
where $\alpha\in (1,2]$, $D_{0^+}^*$ is the Riemann-Liouville
 fractional derivatives of orders $*>0$, $I_{0^+}^*$ is the Riemann-Liouville
 fractional integral of order $*>0$, $t_0=0$, $1<t_1<\dots<t_m<\dots<+\infty$,
\begin{gather*}
\Delta D_{0^+}^{\alpha-1}u(t_k)
=\lim_{\epsilon\to 0^+} D_{0^+}^{\alpha-1}u(t_k+\epsilon)-\lim_{\epsilon\to 0^-}
D_{0^+}^{\alpha-1}u(t_k+\epsilon),  \\
\Delta I_{0^+}^{2-\alpha}u(t_k)
=\lim_{\epsilon\to 0^+} I_{0^+}^{2-\alpha}u(t_k+\epsilon)
-\lim_{\epsilon\to 0^-} I_{0^+}^{2-\alpha}u(t_k+\epsilon), \\
D_{0^+}^{\alpha-1}u(+\infty)=\lim_{t\to +\infty}D_{0^+}^{\alpha-1}u(t),
\end{gather*}
and $f,I,J$ satisfy some suitable assumptions.

\begin{theorem} \label{thm6.6.2}
A function $u$ is a solution of \eqref{e6.6.6} if and only if
 \eqref{e6.6.6}.
\end{theorem}

\begin{proof}
 Suppose that $u$ is a solution of \eqref{e6.6.6}.
By Theorem \ref{thm3.2.2} (with $\lambda =0$, $\alpha\in (1,2)$,  there exist constants
 $c_i,d_i\in \mathbb{R}$ such that
\begin{align*}
u(t)&=\sum_{j=0}^i\frac{c_{j}}{\Gamma(\alpha)}(t-t_j)^{\alpha-1}
 +\sum_{j=0}^i\frac{d_{j}}{\Gamma(\alpha-1)}(t-t_j)^{\alpha-2}\\
&\quad +\frac{1}{\Gamma(\alpha)}\int_1^t(t-s)^{\alpha-1}f(s,u(s))ds,
\quad t\in (t_i,t_{i+1}],\; i\in \mathbb{N}[0,\infty).
\end{align*}
By direct computations, we obtain
\begin{gather*}
I_{0^+}^{2-\alpha}u(t)=\sum_{j=0}^ic_{j}(t-t_j)+\sum_{j=0}^id_{j}
+\int_1^t(t-s)f(s,u(s))ds, \\
t\in (t_i,t_{i+1}],\; i\in \mathbb{N}[0,\infty),\\
D_{0^+}^{\alpha-1}u(t)=\sum_{j=0}^ic_{j}+\int_1^tf(s,u(s))ds, \quad
t\in (t_i,t_{i+1}],\; i\in \mathbb{N}[0,\infty).
\end{gather*}
From $\Delta D_{0^+}^{2-\alpha}u(t_k)=J(t_k,u(t_k))$,
$\Delta I_{0^+}^{2-\alpha}u(t_k)=I(t_k,u(t_k))$, $k\in \mathbb{N}[1,\infty)$,
we know that $c_k=J(t_k,u(t_k))$ and $d_k=I(t_k,u(t_k))$, $k\in \mathbb{N}[1,\infty)$.
By $I_{0^+}^{2-\alpha}u(0)=0$, $D_{0^+}^{\alpha-1}u(+\infty)=0$, we know that
$d_0=u_0$ and $c_0=u_2$.
Hence we obtain
\begin{align*}
u(t)&=u_2t^{\alpha-1}+u_1t^{\alpha-2}
 +\sum_{j=1}^i\frac{J(t_j,u(t_j))}{\Gamma(\alpha)}(t-t_j)^{\alpha-1}
 +\sum_{j=1}^i\frac{I(t_j,u(t_j))}{\Gamma(\alpha-1)}(t-t_j)^{\alpha-2}\\
&\quad +\frac{1}{\Gamma(\alpha)}\int_1^t(t-s)^{\alpha-1}f(s,u(s))ds,
\quad t\in (t_i,t_{i+1}],;i\in \mathbb{N}[0,\infty).
\end{align*}
On the other hand, if $u$ satisfies above integral equation, we can prove that
$u$ satisfies \eqref{e6.6.6}. The proof is complete.
\end{proof}

Liu \cite{l9} studied the existence of positive solutions of the
 boundary value problem of fractional impulsive differential equation
\begin{equation}
\begin{gathered}
D_{0^+}^\alpha x(t)+f(t,x(t)),\quad t\in (0,1),\quad t\neq t_i,\;
i\in\mathbb{N}[1,m], \\
x(0)=x(1)=0,\quad x(t_i^+)-x(t_i^-)=c_ix(t_i^-),\quad i\in\mathbb{N}[1,m],
\end{gathered}\label{e6.6.7}
\end{equation}
where $\alpha\in (1,2]$, $0=t_0<t_1<\dots<t_{m}<t_{m+1}=1$, $D_{0^+}^\alpha$
is the standard Riemann-Liouville fractional derivative,
$c_i\in (0,\frac{1}{2})$, $f:[0,1]\times \mathbb{R}^+\to \mathbb{R}^+$ is a continuous function.
\cite[Lemma 3.1]{l9} claimed that:

\begin{result} \label{res6.6.3}\rm
 If $u\in PC([0,1])$ is a fixed point of the operator $A$ defined by
\begin{equation}
Ax(t)=\int_0^1G(t,s)f(s,x(s))ds
+t^{\alpha-1}\sum_{t<t_k<1}\frac{c_k}{1-c_k}t_k^{1-\alpha}x(t_k),
\label{e6.6.8}
\end{equation}
for all $x\in PC([0,1])$,
then $u$ is a solution of\eqref{e6.6.7} ($x$ is continuous at each point $t\neq t_i$,
right continuous at $t_i$, the left limit $\lim_{t\to t_i^-}x(t)$ is finite
and satisfies \eqref{e6.6.7}), where
$$
G(t,s)=\frac{1}{\Gamma(\alpha)}
 \begin{cases}
[t(1-s)]^{\alpha -1}-(t-s)^{\alpha-1}, & 0\le s\le t\le 1,\\
[t(1-s)]^{\alpha -1}, & 0\le t\le s\le 1.
\end{cases}
$$
\end{result}

We find that Result \ref{res6.6.3} is wrong. In fact, if $u$ is a fixed point of $A$,
the we obtain
$$
x(t)=\int_0^1G(t,s)f(s,x(s))ds+t^{\alpha-1}
\sum_{t<t_k<1}\frac{c_k}{1-c_k}t_k^{1-\alpha}x(t_k).
$$
This is re-written as
\begin{align*}
x(t)&=-\int_0^t\frac{(t-s)^{\alpha-1}}{\Gamma(\alpha)}f(s,x(s))ds
 +t^{\alpha-1}\Big[\int_0^1\frac{(1-s)^{\alpha -1}}{\Gamma(\alpha)}
 f(s,x(s))ds \\
&\quad +\sum_{i=1}^m\frac{c_i}{1-c_i}t_i^{1-\alpha}x(t_i)\Big]
-t^{\alpha-1}\sum_{i=1}^{k-1}\frac{c_i}{1-c_i}t_i^{1-\alpha}x(t_i),
\end{align*}
for $ t\in [t_{k-1},t_{k})$ and $k=1,\dots, m+1$.
One can easily verify that
$D_{0^+}^\alpha x(t)\neq f(t,x(t))$, $t\in (t_i,t_{i+1}]$, $i=1,\dots,m$
similarly to above discussion. We can improve \eqref{e6.6.7} and correct
Result \ref{res6.6.3}. We omit the details.

\subsection{Corrected results from \cite{zz9,zal,zscld}}

In \cite{zscld}, the authors studied the  impulsive system
with Hadamard fractional derivative:
\begin{equation}
\begin{gathered}
{}_HD_{a^+}^qu(t)=f(t,u(t)), \quad t\in (a,T],\; t\neq t_i,
\overline{t}_j,\; i\in\mathbb{N}[1,m],\; j\in\mathbb{N}[1,n],
\\
\Delta {}_HI_{a^+}^{2-q}u(t_i)={}_HI_{a^+}^{2-q}u(t_i^+)-{}_HI_{a^+}^{2-q}u(t_i^-)
=\Delta _i(u(t_i^-)),\quad i\in\mathbb{N}[1,m],
\\
\Delta {}_HD_{a^+}^{q-1}u(\overline{t}_j)
={}_HD_{a^+}^{q-1}u(\overline{t}_j^+)-{}_HD_{a^+}^{q-1}u(\overline{t}_j^-)
=\overline{\Delta} _j(u(\overline{t}_j^-)),\quad j\in\mathbb{N}[1,n],
\\
{}_HI_{a^+}^{2-q}u(a)=u_1,\quad {}_HD_{a^+}^{q-1}u(a)=u_2,
\end{gathered}\label{e6.7.1}
\end{equation}
and its special case
\begin{equation}
\begin{gathered}
{}_HD_{a^+}^qu(t)=f(t,u(t)),\quad t\in (a,T],\; t\neq t_i,\;
i\in\mathbb{N}[1,m],
\\
\Delta {}_HI_{a^+}^{2-q}u(t_i)={}_HI_{a^+}^{2-q}u(t_i^+)-{}_HI_{a^+}^{2-q}u(t_i^-)
=\Delta _i(u(t_i^-)), \quad i\in\mathbb{N}[1,m],
\\
\Delta {}_HD_{a^+}^{q-1}u(t_j)={}_HD_{a^+}^{q-1}u(t_j^+)-{}_HD_{a^+}^{q-1}u(t_j^-)
=\overline{\Delta} _j(u(t_j^-)),\quad j\in\mathbb{N}[1,n],
\\
{}_HI_{a^+}^{2-q}u(a)=u_1,\quad {}_HD_{a^+}^{q-1}u(a)=u_2,
\end{gathered}\label{e6.7.2}
\end{equation}
where $a>0$, ${}_HD^q_{a^+}$ denotes left-sided Riemann-Liouville type Hadamard
fractional derivative of order $q\in (1,2)$ with the starting point $a$, $f:
[a,T]\times \mathbb{R}\to\mathbb{R}$ is an appropriate continuous function,
 $a=t_0<t_1<\dots<t_m<t_{m+1}=T$, and $a=\overline{t}_0<\overline{t}_1<\dots<\overline{t}_m<\overline{t}_{n+1}=T$, ${}_HI_{a^+}^*$ denotes the left-sided Hadamard fractional
integral of order $*>0$, and
\[
{}_HI_{a^+}^{2-q}u(t_i^+)=\lim_{\epsilon\to 0^+}{}_HI_{a^+}^{2-q}u(t_i+\epsilon),\quad
{}_HI_{a^+}^{2-q}u(t_i^-)=\lim_{\epsilon\to 0^+}{}_HI_{a^+}^{2-q}u(t_i-\epsilon)
\]
 represent the right and left limits of ${}_HI_{a^+}^{2-q}u(t)$ at $t=t_i$,
respectively.
The derivatives ${}_HD_{a^+}^{q-1}u(t_i^+)$ and ${}_HD_{a^+}^{q-1}u(t_i^-)$
have a similar meaning for ${}_HD_{a^+}^{q-1}u(t)$.
\cite[Theorem 3.4 and Corollary 3.5]{zscld} are as follows:

\begin{theorem}[\cite{zscld}] \label{thm3.4}
Let $q\in (1,2)$, $\overline{\lambda},\overline{h}$ be constants. Then system
\eqref{e6.7.1} is equivalent to the fractional integral equation
\begin{align*}
u(t)&= \frac{u_1}{\Gamma(q)}\Big(\int_a^t\frac{ds}{s}\Big)^{q-1}
+\frac{u_2}{\Gamma(q-1)}\Big(\int_a^t\frac{ds}{s}\Big)^{q-2} \\
&\quad +\frac{1}{\Gamma(q)} \int_a^t(\log\frac{t}{s})^{q-1}f(s,u(s))\frac{ds}{s}, \quad
 t\in (1,t_1];\\
u(t)&=\frac{u_1}{\Gamma(q)}\Big(\int_a^t\frac{ds}{s}\Big)^{q-1}
 +\frac{u_2}{\Gamma(q-1)}\Big(\int_a^t\frac{ds}{s}\Big)^{q-2}
 +\frac{1}{\Gamma(q)} \int_a^t(\log \frac{t}{s})^{q-1}f(s,u(s))\frac{ds}{s}\\
&\quad +\sum_{i=1}^{k_0}\frac{\Delta_i(u(t_i^-))}{\Gamma(q-1)}
\Big(\int_{t_i}^t\frac{ds}{s}\Big)^{q-2}
 +\sum_{j=1}^{k_1}\frac{\overline{\Delta}_j(u(\overline{t}_j^-))}
 {\Gamma(q)}\Big(\int_{\overline{t}_j}^t\frac{ds}{s}\Big)^{q-1}\\
&\quad -\overline{\lambda}\sum_{i=1}^{k_0}\Delta_i(u(t_i^-))
 \Big[\frac{u_1}{\Gamma(q)}\Big(\int_{a}^t\frac{ds}{s}\Big)^{q-1}
 +\frac{u_2}{\Gamma(q-1)}\Big(\int_{a}^t\frac{ds}{s}\Big)^{q-2} \\
&\quad  +\frac{1}{\Gamma(q)} \int_a^t(\log \frac{t}{s})^{q-1}f(s,u(s))
 \frac{ds}{s} \\
&\quad  -\frac{u_1+\int_a^{t_i}f(s,u(s))\frac{ds}{s}}{\Gamma(q)}
 \Big(\int_{t_i}^t\frac{ds}{s}\Big)^{q-1} \\
&\quad -\frac{u_1\log \frac{t_i}{a}+u_2+\int_a^{t_i}
 \log \frac{t_i}{s}f(s,u(s))\frac{ds}{s}}{\Gamma(q-1)}
 \Big(\int_{t_i}^t\frac{ds}{s}\Big)^{q-2} \\
&\quad  -\frac{1}{\Gamma(q)} \int_{t_i}^t(\log \frac{t}{s})^{q-1}
 f(s,u(s))\frac{ds}{s}\Big]\\
&\quad -\overline{h}\sum_{j=1}^{k_1}\overline{\Delta}_j(u(t_j^-))
 \Big[\frac{u_1}{\Gamma(q)}(\int_{a}^t\frac{ds}{s})^{q-1}
+\frac{u_2}{\Gamma(q-1)}(\int_{a}^t\frac{ds}{s})^{q-2} \\
&\quad +\frac{1}{\Gamma(q)} \int_a^t(\log \frac{t}{s})^{q-1}f(s,u(s))
 \frac{ds}{s} \\
&\quad  -\frac{u_1+\int_a^{\overline{t}_j}f(s,u(s))
 \frac{ds}{s}}{\Gamma(q)}
 \Big(\int_{\overline{t}_j}^t\frac{ds}{s}\Big)^{q-1} \\
&\quad -\frac{u_1\log \frac{\overline{t}_j}{a}+u_2+\int_a^{\overline{t}_j}
\log \frac{\overline{t}_j}{s}f(s,u(s))\frac{ds}{s}}{\Gamma(q-1)}
\Big(\int_{\overline{t}_j}^t\frac{ds}{s}\Big)^{q-2} \\
&\quad  -\frac{1}{\Gamma(q)} \int_{\overline{t}_j}^t
 (\log \frac{t}{s})^{q-1}f(s,u(s))\frac{ds}{s}\Big],
\end{align*}  % \label{e6.7.3}
for $t\in (t_k',t_{k+1}']$ and $k=1,2,\dots,\Omega$,
where $a,t_1,\dots,t_m,\overline{t}_1,\dots,\overline{t}_n,T$
are queued to $a=t'_0<t_1'<t_2'<\dots<t_{\Omega}'<t_{\Omega}'=T$ so
that $\{t_1,t_2,\dots,t_m,\overline{t_1},\dots,\overline{t}_n\}
=\{t_1',t_2',\dots,t_\Omega'\}$.
\end{theorem}

\begin{corollary}[\cite{zscld}] \label{coro3.5}
 Let $q\in (1,2)$, $\overline{\lambda},\overline{h}$ be constants. Then system
\eqref{e6.7.2} is equivalent to the fractional integral equation
\begin{equation}
u(t)= \begin{cases}
\frac{u_1}{\Gamma(q)}\Big(\int_a^t\frac{ds}{s}\Big)^{q-1}
+\frac{u_2}{\Gamma(q-1)}\Big(\int_a^t\frac{ds}{s}\Big)^{q-2}
+\frac{1}{\Gamma(q)} \int_a^t(\log \frac{t}{s})^{q-1}f(s,u(s))
\frac{ds}{s},\\
\quad t\in (1,t_1]; \\[4pt]
\frac{u_1}{\Gamma(q)}\Big(\int_a^t\frac{ds}{s}\Big)^{q-1}
+\frac{u_2}{\Gamma(q-1)}\Big(\int_a^t\frac{ds}{s}\Big)^{q-2}
+\frac{1}{\Gamma(q)} \int_a^t(\log \frac{t}{s})^{q-1}f(s,u(s))\frac{ds}{s}
\\
+\sum_{i=1}^{k}\Big[\frac{\Delta_i(u(t_i^-))}{\Gamma(q-1)}
\Big(\int_{t_i}^t\frac{ds}{s}\Big)^{q-2}
 + \frac{\overline{\Delta}_i(u({t}_i^-))}{\Gamma(q)}
\Big(\int_{{t}_i}^t\frac{ds}{s}\Big)^{q-1}\Big]
\\
-\sum_{i=1}^{k_0}[\overline{\lambda}\Delta_i(u(t_i^-))
+\overline{h}\overline{\Delta}_j(u(t_j^-))]
\Big[\frac{u_1}{\Gamma(q)}(\int_{a}^t\frac{ds}{s})^{q-1}
 +\frac{u_2}{\Gamma(q-1)}(\int_{a}^t\frac{ds}{s})^{q-2}
\\
+\frac{1}{\Gamma(q)} \int_a^t(\log \frac{t}{s})^{q-1}f(s,u(s))\frac{ds}{s}
-\frac{u_1+\int_a^{t_i}f(s,u(s))\frac{ds}{s}}{\Gamma(q)}
\Big(\int_{t_i}^t\frac{ds}{s}\Big)^{q-1} \\
-\frac{u_1\log \frac{t_i}{a}+u_2
 +\int_a^{t_i}\log \frac{t_i}{s}f(s,u(s))\frac{ds}{s}}{\Gamma(q-1)}
\Big(\int_{t_i}^t\frac{ds}{s}\Big)^{q-2}
\\
 -\frac{1}{\Gamma(q)} \int_{t_i}^t(\log \frac{t}{s})^{q-1}f(s,u(s))
\frac{ds}{s}\Big], \\
\quad t\in (t_k,t_{k+1}],\; k=1,2,\dots,m.
\end{cases}  \label{e6.7.4}
\end{equation}
\end{corollary}

\begin{theorem} \label{thm6.7.1}
A function $x$ is a solution of \eqref{e6.7.2} if and only if $x$ satisfies
the integral equation
\begin{align*}
x(t)&=\frac{u_2}{\Gamma(q)}(\log t)^{q-1}
 +\frac{u_1}{\Gamma(q-1)}(\log t)^{q-2}
 +\sum_{j=1}^i\frac{\overline{\Delta} _j(u(t_j^-))}{\Gamma(q)}
 (\log\frac{ t}{t_j})^{q-1} \\
&\quad +\sum_{j=1}^i\frac{\Delta _j(u(t_j^-))}{\Gamma(q-1)}
 (\log\frac{ t}{t_j})^{q-2}
 +\frac{1}{\Gamma(q)}\int_1^t(\log \frac{t}{s})^{q-1}f(s,x(s))\frac{ds}{s},
\end{align*}
for $t\in (t_i,t_{i+1}]$, $i\in \mathbb{N}[0,m]$.
\end{theorem}

\begin{proof}
Suppose that $x$ is a solution of BVP\eqref{e6.7.2} with $a=1$ and $T=e$.
 By Theorem \ref{thm3.2.3} (with $\lambda=0$), we know from
${}_HD_{a^+}^qx(t)=f(t,x(t))$ and $q\in (1,2)$ that there exist constants
$c_i,d_i\in \mathbb{R}(i\in \mathbb{N}[0,m]$ such that
\begin{equation}
\begin{aligned}
x(t)&=\sum_{j=0}^i\frac{c_{j}}{\Gamma(q)}(\log\frac{ t}{t_j})^{q-1}
+\sum_{j=0}^i\frac{d_{j}}{\Gamma(q-1)}(\log\frac{ t}{t_j})^{q-2}\\
&\quad +\frac{1}{\Gamma(q)}\int_1^t(\log \frac{t}{s})^{q-1}f(s,x(s))\frac{ds}{s},
\quad t\in (t_i,t_{i+1}],\;i\in \mathbb{N}[0,m].
\end{aligned} \label{e6.7.5}
\end{equation}
Then  by Definitions  \ref{def2.4} and  \ref{def2.5},  for $t\in (t_i,t_{i+1}]$, we have
\begin{align*}
&{}_HI_{1^+}^{2-q}x(t) \\
&=\frac{1}{\Gamma(2-q)}\int_1^t(\log \frac{t}{s})^{1-q}x(s)\frac{ds}{s} \\
&=\Big(\sum_{\tau=0}^{i-1}\int_{t_{\tau}}^{t_{\tau+1}}
 (\log \frac{t}{s})^{1-q}x(s)\frac{ds}{s}+\int_{t_i}^t
 (\log \frac{t}{s})^{1-q}x(s)\frac{ds}{s}\Big)/ \Gamma(2-q)
\\
&=\Big(\sum_{\tau=0}^{i-1}\int_{t_{\tau}}^{t_{\tau+1}}
(\log \frac{t}{s})^{1-q}
\Big[\sum_{j=0}^\tau\frac{c_{j}}{\Gamma(q)} (\log\frac{ s}{t_j})^{q-1} \\
&\quad +\sum_{j=0}^\tau\frac{d_{j}}{\Gamma(q-1)}
(\log\frac{ s}{t_j})^{q-2}+\frac{1}{\Gamma(q)}\int_1^s
(\log \frac{s}{u})^{q-1}f(u,x(u))\frac{du}{u}\Big]
\frac{ds}{s}\Big) /\Gamma(2-q)
\\
&\quad +\Big(\int_{t_i}^t(\log \frac{t}{s})^{1-q}
\Big[\sum_{j=0}^i\frac{c_{j}}{\Gamma(q)}(\log\frac{ s}{t_j})^{q-1}
+\sum_{j=0}^i\frac{d_{j}}{\Gamma(q-1)}(\log\frac{ s}{t_j})^{q-2} \\
&\quad +\frac{1}{\Gamma(q)}\int_1^s(\log \frac{s}{u})^{q-1}f(u,x(u))
\frac{du}{u}\Big]
\frac{ds}{s}\Big)/ \Gamma(2-q)
\\
&=\Big(\sum_{\tau=0}^{i-1}\sum_{j=0}^\tau\frac{c_{j}}{\Gamma(q)}
\int_{t_{\tau}}^{t_{\tau+1}}(\log \frac{t}{s})^{1-q}(\log\frac{ s}{t_j})^{q-1}
\frac{ds}{s} \\
&\quad +\sum_{\tau=0}^{i-1}\sum_{j=0}^\tau\frac{d_{j}}{\Gamma(q-1)}
\int_{t_{\tau}}^{t_{\tau+1}}(\log \frac{t}{s})^{1-q}
(\log\frac{ s}{t_j})^{q-2}\frac{ds}{s}\Big) /\Gamma(2-q) \\
\\
&\quad +\Big(\int_{t_i}^t(\log \frac{t}{s})^{1-q}
\sum_{j=0}^i\frac{c_{j}}{\Gamma(q)}(\log\frac{ s}{t_j})^{q-1}
\frac{ds}{s} \\
&\quad +\int_{t_i}^t(\log \frac{t}{s})^{1-q}\sum_{j=0}^i
\frac{d_{j}}{\Gamma(q-1)}(\log\frac{ s}{t_j})^{q-2}\frac{ds}{s}\Big)/
 \Gamma(2-q)\\
&\quad +\Big(\frac{1}{\Gamma(q)}\int_1^t(\log \frac{t}{s})^{1-q}
\int_1^s(\log \frac{s}{u})^{q-1}f(u,x(u))\frac{du}{u}
\frac{ds}{s}\Big)/ \Gamma(2-q) \\
&\quad \text{(changing the order of the sums and integral)}\\
&=\Big(\sum_{j=0}^{i-1}\sum_{\tau=j}^{i-1}\frac{c_{j}}{\Gamma(q)}
 \int_{t_{\tau}}^{t_{\tau+1}}(\log \frac{t}{s})^{1-q}(\log\frac{ s}{t_j})^{q-1}
 \frac{ds}{s} \\
&\quad +\sum_{j=0}^{i-1}\sum_{\tau=j}^{i-1}\frac{d_{j}}{\Gamma(q-1)}
 \int_{t_{\tau}}^{t_{\tau+1}}(\log \frac{t}{s})^{1-q}(\log\frac{ s}{t_j})^{q-2}
 \frac{ds}{s}\Big)/ \Gamma(2-q)\\
&\quad +\Big(\sum_{j=0}^i\frac{c_{j}}{\Gamma(q)}\int_{t_i}^t
(\log \frac{t}{s})^{1-q}(\log\frac{ s}{t_j})^{q-1}\frac{ds}{s} \\
&\quad +\sum_{j=0}^i\frac{d_{j}}{\Gamma(q-1)}\int_{t_i}^t
(\log \frac{t}{s})^{1-q}(\log\frac{ s}{t_j})^{q-2}\frac{ds}{s}\Big)/
 \Gamma(2-q)\\
&\quad +\Big(\frac{1}{\Gamma(q)}\int_1^t\int_u^t(\log \frac{t}{s})^{1-q}
(\log \frac{s}{u})^{q-1}\frac{ds}{s}f(u,x(u))\frac{du}{u}\Big)/ \Gamma(2-q)\\
&=\Big(\sum_{j=0}^i\frac{c_{j}}{\Gamma(q)}\int_{t_j}^t
 (\log \frac{t}{s})^{1-q}(\log\frac{ s}{t_j})^{q-1}\frac{ds}{s} \\
&\quad +\sum_{j=0}^i\frac{d_{j}}{\Gamma(q-1)}\int_{t_j}^t(\log \frac{t}{s})^{1-q}
 (\log\frac{ s}{t_j})^{q-2}\frac{ds}{s}\Big)/\Gamma(2-q)\\
&\quad +\Big(\frac{1}{\Gamma(q)}\int_1^t\int_u^t(\log \frac{t}{s})^{1-q}
(\log \frac{s}{u})^{q-1}\frac{ds}{s}f(u,x(u))\frac{du}{u}\Big)
/ \Gamma(2-q) \\
&\quad \text{using }\frac{\log s-\log t_j}{\log t-\log t_j}=w,
\frac{\log s-\log u}{\log t-\log u}=w\\
&=\Big(\sum_{j=0}^i\frac{c_{j}}{\Gamma(q)}(\log \frac{t}{t_j})
\int_0^1(1-w)^{1-q}w^{q-1}dw \\
&\quad  +\sum_{j=0}^i\frac{d_{j}}{\Gamma(q-1)}\int_0^1(1-w)^{1-q}w^{q-2}dw\Big)
/\Gamma(2-q)\\
&\quad +\Big(\frac{1}{\Gamma(q)}\int_1^t(\log \frac{t}{u})\int_0^1(1-w)^{1-q}
w^{q-1}dwf(u,x(u))\frac{du}{u}\Big)/\Gamma(2-q) \\
&=\sum_{j=0}^ic_{j}(\log\frac{ t}{t_j})+\sum_{j=0}^id_{j}
+\int_1^t(\log \frac{t}{u})f(u,x(u))\frac{du}{u}.
\end{align*}
and similarly
\begin{align*}
&{}_HD_{1^+}^{q-1}x(t) \\
&=\Big(\sum_{\tau=0}^{i-1}(t\frac{d}{dt})
 \int_1^t(\log\frac{t}{s})^{1-q}x(s)\frac{ds}{s}
+(t\frac{d}{dt})\int_{t_i}^t(\log\frac{t}{s})^{1-q}x(s)
 \frac{ds}{s}\Big)/ \Gamma(2-q) \\
&=\Big(\sum_{\tau=0}^{i-1}(t\frac{d}{dt})\int_1^t(\log\frac{t}{s})^{1-q}
\Big[\sum_{j=0}^\tau\frac{c_{j}}{\Gamma(q)}(\log\frac{ s}{t_j})^{q-1} \\
&\quad +\sum_{j=0}^\tau\frac{d_{j}}{\Gamma(q-1)}(\log\frac{ s}{t_j})^{q-2}
+\frac{1}{\Gamma(q)}\int_1^s(\log \frac{s}{u})^{q-1}f(u,x(u))
\frac{du}{u}\Big]\frac{ds}{s}\Big)/\Gamma(2-q) \\
&\quad +\Big((t\frac{d}{dt})\int_{t_i}^t(\log\frac{t}{s})^{1-q}
\Big[\sum_{j=0}^i\frac{c_{j}}{\Gamma(q)}(\log\frac{ s}{t_j})^{q-1} \\
&\quad +\sum_{j=0}^i\frac{d_{j}}{\Gamma(q-1)}(\log\frac{ s}{t_j})^{q-2}
+\frac{1}{\Gamma(q)}\int_1^s(\log \frac{s}{u})^{q-1}f(u,x(u))
 \frac{du}{u}\Big]\frac{ds}{s}\Big)/ \Gamma(2-q)\\
&=\Big(\sum_{\tau=0}^{i-1}(t\frac{d}{dt})\int_1^t(\log\frac{t}{s})^{1-q}
\sum_{j=0}^\tau\frac{c_{j}}{\Gamma(q)}(\log\frac{ s}{t_j})^{q-1}\frac{ds}{s} \\
&\quad +\sum_{\tau=0}^{i-1}(t\frac{d}{dt})\int_1^t(\log\frac{t}{s})^{1-q}
\sum_{j=0}^\tau\frac{d_{j}}{\Gamma(q-1)}(\log\frac{ s}{t_j})^{q-2}\frac{ds}{s}\Big)
/\Gamma(2-q)\\
&\quad +\Big((t\frac{d}{dt})\int_{t_i}^t(\log\frac{t}{s})^{1-q}
\sum_{j=0}^i\frac{c_{j}}{\Gamma(q)}(\log\frac{ s}{t_j})^{q-1}\frac{ds}{s} \\
&\quad +(t\frac{d}{dt})\int_{t_i}^t(\log\frac{t}{s})^{1-q}
\sum_{j=0}^i\frac{d_{j}}{\Gamma(q-1)}(\log\frac{ s}{t_j})^{q-2}
 \frac{ds}{s}\Big)/ \Gamma(2-q)\\
&\quad +\Big(\frac{1}{\Gamma(q)}(t\frac{d}{dt})\int_1^t(\log\frac{t}{s})^{1-q}
\int_1^s(\log \frac{s}{u})^{q-1}f(u,x(u))\frac{du}{u}\frac{ds}{s}
\Big)/\Gamma(2-q)\\
&=\Big(\sum_{j=0}^i\frac{c_{j}}{\Gamma(q)}(t\frac{d}{dt})
\int_{t_j}^t(\log\frac{t}{s})^{1-q}
(\log\frac{ s}{t_j})^{q-1}\frac{ds}{s} \\
&\quad +\sum_{j=0}^i\frac{d_{j}}{\Gamma(q-1)}(t\frac{d}{dt})
\int_{t_j}^t(\log\frac{t}{s})^{1-q}
(\log\frac{ s}{t_j})^{q-2}\frac{ds}{s}\Big)/ \Gamma(2-q)\\
&\quad +\Big(\frac{1}{\Gamma(q)}(t\frac{d}{dt})
 \int_1^t\int_u^t(\log\frac{t}{s})^{1-q}
(\log \frac{s}{u})^{q-1}\frac{ds}{s}f(u,x(u))\frac{du}{u}\Big)/ \Gamma(2-q)\\
&=\Big(\sum_{j=0}^i\frac{c_{j}}{\Gamma(q)}(t\frac{d}{dt})
(\log\frac{t}{t_j})\int_0^1 (1-w)^{1-q}w^{q-1}dw \\
&\quad +\sum_{j=0}^i \frac{d_{j}}{\Gamma(q-1)}(t\frac{d}{dt})\int_0^1 (1-w)^{1-q}
w^{q-2}dw\Big)/ \Gamma(2-q)\\
&\quad +\Big(\frac{1}{\Gamma(q)}(t\frac{d}{dt})\int_1^t(\log \frac{t}{u})\int_0^1(1-w)^{1-q}
w^{q-1}dwf(u,x(u))\frac{du}{u}\Big)/ \Gamma(2-q) \\
&= \sum_{j=0}^ic_{j}+\int_1^tf(s,x(s))\frac{ds}{s}.
\end{align*}
It follows that
\begin{equation}
\begin{gathered}
{}_HI_{1^+}^{2-\alpha}x(t)
= \sum_{j=0}^ic_{j}(\log\frac{ t}{t_j})
 +\sum_{j=0}^id_{j}+\int_1^t(\log \frac{t}{u})f(u,x(u))\frac{du}{u},\\
 t\in (t_i,t_{i+1}],\;i\in \mathbb{N}[0,m],\\
{}_HD_{1^+}^{\alpha-1}x(t)=\sum_{j=0}^ic_{j}+\int_1^tf(s,x(s))\frac{ds}{s},
\quad t\in (t_i,t_{i+1}],\;i\in \mathbb{N}[0,m].
\end{gathered}\label{e6.7.6}
\end{equation}
From $\Delta {}_HI_{1^+}^{2-q}u(t_i)=\Delta _i(u(t_i^-))$, $i\in \mathbb{N}[1,m] $
and \eqref{e6.7.6}, we obtain $d_i=\Delta _i(u(t_i^-))$, $i\in \mathbb{N}[1,m] $.
From  $\Delta {}_HD_{1^+}^{q-1}u(t_j)=\overline{\Delta} _j(u(t_j^-))$,
$j\in \mathbb{N}[1,m]$ and \eqref{e6.7.6}, we obtain
 $c_j=\overline{\Delta} _j(u(t_j^-))$, $j\in \mathbb{N}[1,m]$.
From  ${}_HI_{1^+}^{2-q}u(1)=u_1,\;{}_HD_{1^+}^{q-1}u(1)=u_2$ and \eqref{e6.7.6},
we obtain $d_0=u_1,c_0=u_2$.
Substituting $c_i,d_i$ into \eqref{e6.7.5}, we obtain
\begin{equation}
\begin{aligned}
x(t)&=\frac{u_2}{\Gamma(q)}(\log t)^{q-1}
+\frac{u_1}{\Gamma(q-1)}(\log t)^{q-2}
+\sum_{j=1}^i\frac{\overline{\Delta} _j(u(t_j^-))}{\Gamma(q)}
 (\log\frac{ t}{t_j})^{q-1} \\
&\quad +\sum_{j=1}^i\frac{\Delta _j(u(t_j^-))}{\Gamma(q-1)}(\log\frac{ t}{t_j})^{q-2}
+\frac{1}{\Gamma(q)}\int_1^t(\log \frac{t}{s})^{q-1}f(s,x(s))\frac{ds}{s},
\end{aligned} \label{e6.7.8}
\end{equation}
for $t\in (t_i,t_{i+1}]$, $i\in \mathbb{N}[0,m]$.
On the other hand, if $x$ satisfies \eqref{e6.7.8}, we can prove that $x$
is a solution of \eqref{e6.7.2} by direct computation.
The proof is complete.
\end{proof}

\begin{example}[\cite{zscld}] \label{examp6.7.1}  \rm
 Consider the  problem
\begin{equation}
\begin{gathered}
{}_HD_{1^+}^{3/2}x(t)=\ln t, \quad t\in (1,3],\; t\neq 2,\\
\Delta {}_HI_{1^+}^{1/2}u(2)={}_HI_{1^+}^{1/2}u(2^+)-{}_HI_{1^+}^{1/2}u(2^-)
=\delta,\\
\Delta {}_HD_{1^+}^{1/2}u(2)={}_HD_{1^+}^{1/2}u(2^+)-{}_HD_{1^+}^{1/2}u(2^-)
=\overline{\delta},\\
{}_HI_{1^+}^{1/2}u(1)=u_1,\quad {}_HD_{1^+}^{1/2}u(1)=u_2.
\end{gathered}\label{e6.7.9}
\end{equation}
By \cite[Theorem 3.4]{zscld}, its solution is
$$
x(t)= \begin{cases}
\frac{u_1}{\Gamma(3/2)}\Big(\int_1^t\frac{ds}{s}\Big)^{3/2-1}
+\frac{u_2}{\Gamma(3/2-1)}\Big(\int_1^t\frac{ds}{s}\Big)^{3/2-2} \\
+\frac{1}{\Gamma(3/2)} \int_1^t(\log \frac{t}{s})^{3/2-1}
\log s\frac{ds}{s},\\
\quad t\in (1,2];
\\[4pt]
\frac{u_1}{\Gamma(3/2)}\Big(\int_1^t\frac{ds}{s}\Big)^{3/2-1}
+\frac{u_2}{\Gamma(3/2-1)}\Big(\int_1^t\frac{ds}{s}\Big)^{3/2-2} \\
+\frac{1}{\Gamma(3/2)} \int_1^t(\log \frac{t}{s})^{3/2-1}\log s\frac{ds}{s}
+\frac{\delta}{\Gamma(3/2-1)}\Big(\int_2^t\frac{ds}{s}\Big)^{3/2-2} \\
+\frac{\overline{\delta}}{\Gamma(3/2)}
\Big(\int_2^t\frac{ds}{s}\Big)^{3/2-1}
-[\overline{\lambda}\delta+\overline{h}\overline{\delta}]
\Big[\frac{u_1}{\Gamma(3/2)}\Big(\int_1^t\frac{ds}{s}\Big)^{3/2-1} \\
+\frac{u_2}{\Gamma(3/2-1)}\Big(\int_1^t\frac{ds}{s}\Big)^{3/2-2}
+\frac{1}{\Gamma(3/2)} \int_1^t(\log \frac{t}{s})^{3/2-1}\log s\frac{ds}{s}\\
-\frac{u_1+\int_1^{2}\log s\frac{ds}{s}}{\Gamma(3/2)}(\int_{2}^t
\frac{ds}{s})^{3/2-1}
-\frac{u_1\log 2+u_2+\int_1^{2}\log \frac{2}{s}\log s\frac{ds}{s}}{\Gamma(3/2-1)}
\Big(\int_{2}^t\frac{ds}{s}\Big)^{3/2-2} \\
 -\frac{1}{\Gamma(3/2)} \int_{2}^t(\log \frac{t}{s})^{3/2-1}\log s\frac{ds}{s}\Big],\\
\quad t\in (2,3],
\end{cases}
$$
where $\overline{\lambda},\overline{h}$ are constants.
It is easy to see that
\[
\int_1^t(\log \frac{t}{s})^{3/2-1}\log s\frac{ds}{s}
=(\log t)^{5/2}\int_0^1(1-w)^{3/2-1}wdw
=(\log t)^{5/2}\mathbf{B}(3/2,2).
\]
Then
\begin{align*}
x(t)&= \begin{cases}
\frac{u_1}{\Gamma(3/2)}(\log t)^{1/2}+
\frac{u_2}{\Gamma(1/2)}(\log t)^{-1/2}
+\frac{1}{\Gamma(7/2)} (\log t)^{5/2},\\
\quad t\in (1,2];\\[4pt]
\frac{u_1}{\Gamma(3/2)}(\log t)^{1/2}+
\frac{u_2}{\Gamma(1/2)}(\log t)^{-1/2}
+\frac{1}{\Gamma(7/2)} (\log t)^{5/2}
+\frac{\delta}{\Gamma(1/2)}(\log \frac{t}{2})^{-1/2} \\
+\frac{\overline{\delta}}{\Gamma(3/2)}(\log \frac{t}{2})^{1/2}
-[\overline{\lambda}\delta+\overline{h}\overline{\delta}]
\Big[\frac{u_1}{\Gamma(3/2)}(\log t)^{1/2}+\frac{u_2}{\Gamma(1/2)}(\log t)^{-1/2}
\\
+\frac{1}{\Gamma(7/2)} (\log t)^{5/2}-\frac{u_1+\frac{1}{2}(\log 2)^2}{\Gamma(3/2)}
(\log \frac{t}{2})^{1/2} \\
-\frac{u_1\log 2+u_2+\frac{1}{6}(\log 2)^3}{\Gamma(1/2)}
(\log\frac{t}{2})^{-1/2}
-\frac{1}{\Gamma(7/2)}(\log t)^{5/2}\Big], \\
\quad t\in (2,3]
\end{cases} \\
&= \begin{cases}
\frac{u_1}{\Gamma(3/2)}(\log t)^{1/2}+
\frac{u_2}{\Gamma(1/2)}(\log t)^{-1/2}
+\frac{1}{\Gamma(7/2)} (\log t)^{5/2}, \quad t\in (1,2];
\\[4pt]
\frac{u_1+\overline{\delta}+[\overline{\lambda}\delta+\overline{h}
 \overline{\delta}]u_1}{\Gamma(3/2)}(\log t)^{1/2}
+ \frac{u_2+\delta+[\overline{\lambda}\delta+\overline{h}
 \overline{\delta}]u_2}{\Gamma(1/2)}(\log t)^{-1/2}
+\frac{1}{\Gamma(7/2)} (\log t)^{5/2}
\\
-\frac{[\overline{\lambda}\delta+\overline{h}\overline{\delta}]
 (u_1+\frac{1}{2}(\log 2)^2)}{\Gamma(3/2)}(\log \frac{t}{2})^{1/2}
-\frac{[\overline{\lambda}\delta+\overline{h}\overline{\delta}]
 (u_1\log 2+u_2+\frac{1}{6}(\log 2)^3)}{\Gamma(1/2)}(\log\frac{t}{2})^{-1/2}, \\
\quad t\in (2,3].
\end{cases}
\end{align*}
\end{example} 

First, this example shows us that \eqref{e6.7.9} has infinitely many solutions
since $\overline{\lambda},\overline{h}\in \mathbb{R}$ are two variables.
Second, we can obtain
$$
{}_HI_{1^+}^{1/2}x(t)
=\begin{cases}
u_1(\log t)+ u_2+\frac{1}{\Gamma(4)} (\log t)^3, & t\in (1,2];
\\
(u_1+\overline{\delta}+[\overline{\lambda}\delta+\overline{h}
 \overline{\delta}]u_1)(\log t)+ (u_2+\delta+[\overline{\lambda}\delta
 +\overline{h}\overline{\delta}]u_2)  \\
+\frac{1}{\Gamma(4)} (\log t)^3
-([\overline{\lambda}\delta+\overline{h}\overline{\delta}]
 (u_1+\frac{1}{2}(\log 2)^2))(\log \frac{t}{2}) \\
 -([\overline{\lambda}\delta+\overline{h}\overline{\delta}]
 (u_1\log 2+u_2+\frac{1}{6}(\log 2)^3)), 
& t\in (2,3].
\end{cases}
$$
It is easy to see that
\[
\Delta {}_HI_{1^+}^{1/2}u(2)={}_HI_{1^+}^{\frac{1}{2}
}u(2^+)-{}_HI_{1^+}^{1/2}u(2^-)\neq \delta.
\]
 Hence Corollary \ref{coro3.5} is wrong.
In fact, the correct expression for the solution of \eqref{e6.7.9} is
\begin{align*}
x(t)&=\frac{u_2}{\Gamma(3/2)}(\log t)^{3/2-1}
+\frac{u_1}{\Gamma(3/2-1)}(\log t)^{3/2-2}
+\sum_{j=1}^i\frac{\overline{\delta}}{\Gamma(3/2)}(\log\frac{ t}{t_j})^{3/2-1} \\
&\quad +\sum_{j=1}^i\frac{\delta }{\Gamma(3/2-1)}(\log\frac{ t}{t_j})^{3/2-2}
 +\frac{1}{\Gamma(3/2)}\int_1^t(\log \frac{t}{s})^{3/2-1}f(s,x(s))\frac{ds}{s},
\end{align*}
for $t\in (t_i,t_{i+1}]$ and $i\in \mathbb{N}[0,1]$.
Our result is easy to understand.


\begin{remark} \label{rmk6.7.1}\rm
 A similar initial value problem for impulsive fractional
differential equation involving the Riemann-Liouville fractional
derivatives were studied in \cite{zal}.
Similarly we remark that \cite[Theorem 3.5, page 920]{zal}) and
\cite[Corollary 3.6, page 927]{zal}) are wrong.
 We omit the details.
\end{remark}

Zhang \cite{zz9} studied a class of higher-order nonlinear Riemann-Liouville
fractional differential equations with Riemann-Stieltjes integral boundary value
conditions and impulses as follows:
\begin{equation}
\begin{gathered}
-D_{0^+}^\alpha u(t)=\lambda a(t)f(t,u(t)),\quad
t\in (0,1)\setminus\{t_1,t_2,\dots,t_m\},\\
\Delta u(t_k)=I_k(u(t_k)),\quad k=1,2,\dots,m,\\
u(0)=u'(0)=\dots=u^{(n-2)}(0)=0, \quad u'(1)=\int_0^1u(s)dH(s),
\end{gathered} \label{e6.7.10}
\end{equation}
where $D_{0^+}^\alpha$ is the standard Riemann-Liouville fractional derivative
of order $n-1<\alpha\le n$, $n\ge 3$. The number $n$ is the smallest integer
 greater than or equal to $\alpha$. The impulsive point sequence $\{t_k\}_{k=1}^m$
satisfies $0=t_0<t_1<\dots<t_m<t_{m+1}=1$,
$\Delta u(t_k) =\lim_{\epsilon\to 0^+}u(t_k+\epsilon)-\lim_{\epsilon\to 0^-}
u(t_k+\epsilon)$, $\lambda>0$ is a parameter,
$f\in C([0,1]\times [0,+\infty),[0,+\infty))$,
$a\in C((0,1),[0,+\infty))$, $I_k\in C([0,+\infty),[0,+\infty))$,
the integral $\int_0^1u(s)dH(s)$ is the Riemann-Stieltjes integral with
$H : [0,1]\to \mathbb{R}$. \cite[Lemma 2.4]{zz9} claimed the following result:

\begin{result} \label{res6.7.1}\rm
 Suppose that $H:[0,1]\to \mathbb{R}$ is a function of bounded variation
 $\delta=\int_0^1s^{\alpha-1}dH(s)\not\alpha-1$, $h\in C[0,1]$.
Then the unique solution of
\begin{equation}
\begin{gathered}
-D_{0^+}^\alpha u(t)=h(t),\quad t\in (0,1)\setminus\{t_1,t_2,\dots,t_m\},\\
\Delta u(t_k)=I_k(u(t_k)),\quad k\in\mathbb{N}[1,m],\\
u(0)=u'(0)=\dots=u^{(n-2)}(0)=0, \quad u'(1)=\int_0^1u(s)dH(s),
\end{gathered} \label{e6.7.11}
\end{equation}
is
\begin{equation}
u(t)=\int_0^1G(t,s)h(s)ds+t^{\alpha-1}\sum_{t\le t_k<1}t_k^{1-\alpha}I_k(u(t_k)),
\quad t\in (0,1], \label{e6.7.12}
\end{equation}
where $G(t,s)=G_1(t,s)+G_2(t,s)$ with
\begin{gather*}
G_1(t,s)=\begin{cases}
\frac{t^{\alpha-1}(1-s)^{\alpha-2}-(t-s)^{\alpha-1}}{\Gamma(\alpha)},
& 0\le s\le t\le 1,\\
\frac{t^{\alpha-1}(1-s)^{\alpha-2}}{\Gamma(\alpha)}, & 0\le t\le s\le 1,
\end{cases} \\
 G_2(t,s)=\frac{t^{\alpha-1}}{\alpha-1-\delta}\int_0^1G_1(\tau,s)dH(\tau).
\end{gather*}
\end{result}

This result is wrong. In fact, \eqref{e6.7.12} can be re-written  as
\begin{align*}
u(t)&=-\int_0^t\frac{(t-s)^{\alpha-1}}{\Gamma(\alpha)}h(s)ds
+t^{\alpha-1}\int_0^1\frac{(1-s)^{\alpha-2}}{\Gamma(\alpha)}h(s)ds \\
&\quad +\frac{t^{\alpha-1}}{\alpha-1-\delta}
 \int_0^1\int_0^1G_1(\tau,s)dH(\tau)h(s)ds \\
&\quad +t^{\alpha-1}\Big[\sum_{k=1}^mt_k^{1-\alpha}I_k(u(t_k))
-\sum_{j=1}^{k-1}t_j^{1-\alpha}I_j(u(t_j))\Big]\\
&=:-\int_0^t\frac{(t-s)^{\alpha-1}}{\Gamma(\alpha)}h(s)ds
 +A_kt^{\alpha-1},\quad t\in (t_{k-1},t_k],\; k\in \mathbb{N}[1,m].
\end{align*}
By Definition \ref{def2.2}, for $\alpha\in (n-1,n)$, and  $t\in (t_i,t_{i+1}]$ 
we have 
\begin{align*}
&D_{0^+}^\alpha x(t) \\
&=\Big[\int_0^t(t-s)^{n-\alpha-1}x(s)ds\Big]^{(n)} / \Gamma(n-\alpha) \\
&=\Big[\sum_{\tau=0}^{i-1}\int_{t_\tau}^{t_{\tau+1}}(t-s)^{n-\alpha-1}x(s)ds
+\int_{t_i}^t(t-s)^{n-\alpha-1}x(s)ds\Big]^{(n)} / \Gamma(n-\alpha) \\
&=\Big[\sum_{\tau=0}^{i-1}\int_{t_\tau}^{t_{\tau+1}}(t-s)^{n-\alpha-1}
\Big(-\int_0^s\frac{(s-u)^{\alpha-1}}{\Gamma(\alpha)}h(u)du+A_{\tau+1}
s^{\alpha-1}\Big)ds \Big]^{(n)} /\Gamma(n-\alpha)\\
&\quad +\Big[\int_{t_i}^t(t-s)^{n-\alpha-1}
\Big(-\int_0^s\frac{(s-u)^{\alpha-1}}{\Gamma(\alpha)}h(u)du
+A_{i+1}s^{\alpha-1}\Big)ds\Big]^{(n)}/ \Gamma(n-\alpha)\\
&=\Big[\sum_{\tau=0}^{i-1}A_{\tau+1}\int_{t_\tau}^{t_{\tau+1}}(t-s)^{n-\alpha-1}
\sum_{\tau=0}^{i-1}\int_{t_\tau}^{t_{\tau+1}}(t-s)^{n-\alpha-1}s^{\alpha-1}ds
\Big]^{(n)} / \Gamma(n-\alpha)
\\
&\quad +\Big[-\int_0^t(t-s)^{n-\alpha-1}
\int_0^s\frac{(s-u)^{\alpha-1}}{\Gamma(\alpha)}h(u)\,du\,ds\\
&\quad +A_{i+1}\int_{t_i}^t(t-s)^{n-\alpha-1}s^{\alpha-1}ds\Big]^{(n)}/
\Gamma(n-\alpha)\\
&=\Big[\sum_{\tau=0}^{i-1}A_{\tau+1}\int_{t_\tau}^{t_{\tau+1}}(t-s)^{n-\alpha-1}
s^{\alpha-1}ds\Big]^{(n)} {\Gamma(n-\alpha)}\\
&\quad +\Big[-\int_0^t\int_u^t(t-s)^{n-\alpha-1}
\frac{(s-u)^{\alpha-1}}{\Gamma(\alpha)}dsh(u)du \\
&\quad +A_{i+1}\int_{t_i}^t(t-s)^{n-\alpha-1}s^{\alpha-1}ds\Big]^{(n)}/ \Gamma(n-\alpha)\\
&=\Big[\sum_{\tau=0}^{i-1}A_{\tau+1}t^{n-1}
 \int_{\frac{t_\tau}{t}}^{\frac{t_{\tau+1}}{t}}(1-w)^{n-\alpha-1}
w^{\alpha-1}dw \Big]^{(n)} / \Gamma(n-\alpha)
\\
&\quad +\Big[-\int_0^t(t-u)^{n-1}\int_0^1(1-w)^{n-\alpha-1}
\frac{w^{\alpha-1}}{\Gamma(\alpha)}dwh(u)du \\
&\quad +A_{i+1}t^{n-1}\int_{\frac{t_i}{t}}^1(1-w)^{n-\alpha-1}w^{\alpha-1}dw
\Big]^{(n)}/ \Gamma(n-\alpha)
=h(t)
\end{align*}
if and only if $A_1=A_2=\dots=A_{i+1}$ if and only if
\[
I_1(u(t_1))=I_2(u(t_2))=I_3(u(t_3))=\dots=I_m(u(t_m))=0.
\]
Hence the impulse functions are not suitable. Then Result \ref{res6.7.1} is wrong.

We consider the  improved problem 
\begin{equation}
\begin{gathered}
-D_{0^+}^\alpha u(t)=h(t),\quad t\in (t_i,t_{i+1}],\;i\in \mathbb{N}[0,m],\\
\Delta I_{0^+}^{n-\alpha} u(t_k)=I_{nk},\quad \Delta D_{0^+}^{\alpha-i}
u(t_k)=I_{ik},\; k\in \mathbb{N}[1,m], \; i\in \mathbb{N}[1,n-1]\\
I_{0^+}^{n-\alpha}u(0)=D_{0^+}^{\alpha-{n-1}}u(0)=\dots=D_{0^+}^{n-2}u(0)=0, \\
 D_{0^+}^{\alpha-(n-1)}u(1)=\int_0^1I_{0^+}^{n-\alpha}u(s)dH(s),
\end{gathered} \label{e6.7.13}
\end{equation}

\begin{theorem} \label{thm6.7.2}
 Suppose that
\[
\delta=\frac{1}{\Gamma(n-1)}-\sum_{\tau=0}^m\frac{t_{\tau+1}^{n}
-t_{\tau}^{n}}{\Gamma(n+1)}\neq 0.
\]
 Then $u$ is a solution of \eqref{e6.7.13} if and only if $u$ satisfies the
integral equation
\begin{equation}
\begin{aligned}
u(t)&=\frac{t^{\alpha-1}}{\Gamma(\alpha)}\frac{1}{\delta}
\Big[ \int_0^1\frac{(t-s)^{n-2}}{\Gamma(i)}h(s)ds
-\sum_{\sigma=1}^m\sum_{v=1}^{n-1}
\frac{I_{v\sigma }}{\Gamma(n-v)}(1-t_\sigma)^{n-1-v} \\
&\quad +\int_0^1\frac{(t-s)^{n-2}}{\Gamma(n-1)}h(s)ds \\
&\quad +\sum_{\sigma=1}^m\sum_{\tau=\sigma}^m\sum_{v=1}^{n}\frac{I
_{v\sigma }[(t_{\tau+1}-t_\sigma)^{n-v+1}-(t_{\tau}-t_\sigma)^{n-v+1}]}
{\Gamma(n-v+2)} \\
&\quad -\int_0^1 (\int_u^1\frac{(s-u)^{n-1}}{\Gamma(n)}dH(s)) h(u)du\Big]
 +\sum_{\sigma=1}^i\sum_{v=1}^{n}\frac{I_{v\sigma }}{\Gamma(\alpha-v+1)}
 (t-t_\sigma)^{\alpha-v} \\
&\quad -\int_0^t\frac{(t-s)^{\alpha-1}}{\Gamma(\alpha)}h(s)ds,
\quad t\in (t_i,t_{i+1}],\;i\in \mathbb{N}[0,m].
\end{aligned}\label{e6.7.14}
\end{equation}
\end{theorem}

\begin{proof}
Suppose that $u$ is a solution of \eqref{e6.7.13}. By Theorem \ref{thm3.2.2}
(with $\lambda=0$), there exist constants
 $c_{\sigma v}\in \mathbb{R}(\sigma\in \mathbb{N}[0,m]$, $v\in \mathbb{N}[1,n])$ such that
\begin{equation}
u(t)=\sum_{\sigma=0}^i\sum_{v=1}^{n}\frac{c_{\sigma v}}{\Gamma(\alpha-v+1)}
(t-t_\sigma)^{\alpha-v}-\int_0^t\frac{(t-s)^{\alpha-1}}{\Gamma(\alpha)}h(s)ds,
\label{e6.7.15}
\end{equation}
for $t\in (t_i,t_{i+1}]$ and $i\in \mathbb{N}[0,m]$.
Then we obtain
\begin{equation}
I_{0^+}^{n-\alpha}u(t)
=\sum_{\sigma=0}^i\sum_{v=1}^{n}
\frac{c_{\sigma v}}{\Gamma(n-v+1)}(t-t_\sigma)^{n-v}
-\int_0^t\frac{(t-s)^{n-1}}{\Gamma(n)}h(s)ds, \label{e6.7.16}
\end{equation}
for $t\in (t_i,t_{i+1}]$, $i\in \mathbb{N}[0,m]$, 
and
\begin{equation}
D_{0^+}^{\alpha-i}u(t)=\sum_{\sigma=0}^\tau\sum_{v=1}^i
\frac{c_{\sigma v}}{\Gamma(i-v+1)}(t-t_\sigma)^{i-v}
-\int_0^t\frac{(t-s)^{i-1}}{\Gamma(i)}h(s)ds, \label{e6.7.17}
\end{equation}
for $t\in (t_\tau,t_{\tau+1}]$, $\tau\in \mathbb{N}[0,m]$, $i\in \mathbb{N}[1,n-1]$.

From $\Delta I_{0^+}^{n-\alpha} u(t_k)=I_{nk}$ and \eqref{e6.7.16},
 we obtain $c_{kn}=I_{nk}(k\in \mathbb{N}[1,m])$.
From $\Delta D_{0^+}^{\alpha-i} u(t_k)=I_{ik}$ and \eqref{e6.7.17}
and \eqref{e6.7.16}, we obtain $c_{ki}=I_{ik}(k\in \mathbb{N}[1,m],i\in \mathbb{N}[1,n-1])$.
From $I_{0^+}^{n-\alpha}u(0)=0$ and \eqref{e6.7.16}, we obtain $c_{0n}=0$.
From  $D_{0^+}^{\alpha-{n-1}}u(0)=\dots=D_{0^+}^{n-2}u(0)=0$ and \eqref{e6.7.17},
we obtain $c_{0i}=0(i\in \mathbb{N}[2,n-1])$.
From $D_{0^+}^{\alpha-(n-1)}u(1)=\int_0^1I_{0^+}^{n-\alpha}u(s)dH(s)$ and
\eqref{e6.7.17}, we obtain
\begin{align*}
&\sum_{\sigma=0}^m\sum_{v=1}^{n-1}
\frac{c_{\sigma v}}{\Gamma(n-v)}(1-t_\sigma)^{n-1-v}
 -\int_0^1\frac{(t-s)^{n-2}}{\Gamma(n-1)}h(s)ds\\
&=\sum_{\tau=0}^m\int_{t_\tau}^{t_{\tau+1}}(\sum_{\sigma=0}^\tau
\sum_{v=1}^{n}\frac{c_{\sigma v}}{\Gamma(n-v+1)}(s-t_\sigma)^{n-v}
-\int_0^s\frac{(s-u)^{n-1}}{\Gamma(n)}h(u)du)dH(s).
\end{align*}
It follows that
\begin{align*}
c_{01}&=\frac{1}{\frac{1}{\Gamma(n-1)}
-\sum_{\tau=0}^m\frac{t_{\tau+1}^{n}-t_{\tau}^{n}}{\Gamma(n+1)}}
\Big[\int_0^1\frac{(t-s)^{n-2}}{\Gamma(i)}h(s)ds \\
&\quad -\sum_{\sigma=1}^m\sum_{v=1}^{n-1}
\frac{I_{v\sigma }}{\Gamma(n-v)}(1-t_\sigma)^{n-1-v}
+\int_0^1\frac{(t-s)^{n-2}}{\Gamma(n-1)}h(s)ds \\
&\quad  +\sum_{\sigma=1}^m\sum_{\tau=\sigma}^m\sum_{v=1}^{n}\frac{I
_{v\sigma }
\Big[(t_{\tau+1}-t_\sigma)^{n-v+1}-(t_{\tau}-t_\sigma)^{n-v+1}\Big]}
{\Gamma(n-v+2)} \\
&\quad -\int_0^1 (\int_u^1\frac{(s-u)^{n-1}}{\Gamma(n)}dH(s))
h(u)du\Big].
\end{align*}
Substituting $c_{\sigma v}$ into \eqref{e6.7.15}, we obtain
\begin{align*}
u(t)&=\frac{t^{\alpha-1}}{\Gamma(\alpha)}\frac{1}{\delta}
\Big[\int_0^1\frac{(t-s)^{n-2}}{\Gamma(i)}h(s)ds-\sum_{\sigma=1}^m\sum_{v=1}^{n-1}
\frac{I_{v\sigma }}{\Gamma(n-v)}(1-t_\sigma)^{n-1-v}\\
&\quad +\int_0^1\frac{(t-s)^{n-2}}{\Gamma(n-1)}h(s)ds \\
&\quad +\sum_{\sigma=1}^m\sum_{\tau=\sigma}^m\sum_{v=1}^{n}\frac{I
_{v\sigma }[(t_{\tau+1}-t_\sigma)^{n-v+1}-(t_{\tau}-t_\sigma)^{n-v+1}]}
{\Gamma(n-v+2)} \\
&\quad -\int_0^1 (\int_u^1\frac{(s-u)^{n-1}}{\Gamma(n)}dH(s))
h(u)du\Big]
+\sum_{\sigma=1}^i\sum_{v=1}^{n}\frac{I_{v\sigma }}{\Gamma(\alpha-v+1)}
(t-t_\sigma)^{\alpha-v} \\
&\quad -\int_0^t\frac{(t-s)^{\alpha-1}}{\Gamma(\alpha)}h(s)ds,
\quad t\in (t_i,t_{i+1}],\;i\in \mathbb{N}[0,m].
\end{align*}
This is \eqref{e6.7.14}.
On the other hand, if $u$ satisfies \eqref{e6.7.14}, we can prove that
 $u$ is a solution of \eqref{e6.7.13} by direct computation similar
to the one in the proof of Theorem \ref{thm3.2.2}.
\end{proof} 

\section{Applications of impulsive fractional differential equations}

It is generally known that integer-order derivatives and integrals have clear
physical and geometric interpretations. In \cite{8}, it is shown that
geometric interpretation of fractional integration is ``Shadows on the walls''
and its Physical interpretation is ``Shadows of the past''.
 Geometric and physical interpretation of fractional integration and fractional
 differentiation were introduced in \cite{p10}. The physical meaning of initial
value problems for fractional differential equations was expressed in \cite{g9, hp9}.

It was shown that some fractional
operators describe in a better way some complex physical phenomena, especially when
dealing with memory processes or viscoelastic and viscoplastic materials \cite{aln}.
 Well known references about the application of fractional operators in rheology
modeling are \cite{bt1, bt2}.
One of the most important advantage of fractional order models in comparison with
integer order ones is that fractional integrals and derivatives are a powerful
tool for the description of memory and hereditary properties of some materials.
Notice that integer order derivatives are local operators, but the fractional
order derivative of a function in a point depends on the past values of such function.
This features motivated the successful
use of fractional calculus in population dynamics, control theory, physics, biology,
medicine and so forth see \cite{abm, aee, cdf, r} and \cite[Chap. 10]{8}.

A fractional differential equations (FDEs)-based theory involving 1- and
2-term equations was developed to predict the nonlinear survival and
growth curves of food-borne pathogens. It is interesting to note that the
solution of 1-term FDE leads to the Weibull model. Two-term FDE was
more successful in describing the complex shapes of microbial survival
and growth curves as compared to the linear and Weibull models \cite{ktsmb}.
The Schr\"odinger equation control the dynamical behaviour of quantum
particles. In \cite{ac}, F. B. Adda and J. Cresson, considered to study of
$\alpha$-differential equations and discussed a fundamental problem concerning
the Schr\"odinger equation in the framework of Nottale's scale relativity
theory.

Impulsive fractional differential equations represent a real framework for
mathematical modeling to real world problems. Significant progress has been
made in the theory of impulsive fractional differential equations
\cite{abs, bs}. Xu et al.\ in their paper \cite{xhl} have described an
impulsive delay fishing model. In \cite{maf}, the authors introduced the
fractional impulsive logistic model.
Fractional impulsive neural networks, fractional impulsive biological models,
 Lasota-Wazewska models, Lotka-Volterra models. Kolmogorov-type models and
fractional impulsive models in economics were introduce in recent book \cite{ss}.

It is well known that $x'(t)=t$ has solutions $x(t)=c+\frac{1}{2}t^2$ which is
continuous on $\mathbb{R}$, where $c\in \mathbb{R}$. Let $0=t_0<t_1<t_2<\dots<t_m<t_{m+1}=1$
be fixed points on $\mathbb{R}$. It also has piecewise continuous solutions
$x(t)=\sum_{j=0}^ic_i+\frac{1}{2}t^2$, where $c_i\in \mathbb{R}(i\in \mathbb{N}[0,m]$.

\begin{example} \label{examp7.1}\rm
 We consider the fractional differential equation ${}^CD_{0^+}^\alpha x(t)=t$
with $\alpha\in (0,1)$. It has continuous solutions
\begin{align*}
x(t)&=c+\frac{1}{\Gamma(\alpha)}\int_0^t(t-s)^{\alpha-1}sds\\
&=c+\frac{1}{\Gamma(\alpha)}t^{\alpha+1}
\int_0^1(1-w)^{\alpha-1}wdw \\
&=c+\frac{t^{\alpha+1}}{\Gamma(\alpha+2)},\quad t\ge 0,\; c\in \mathbb{R}.
\end{align*}
By Theorem \ref{thm3.2.1}, it also has piecewise continuous solutions
\[
x(t)=\sum_{j=0}^ic_j+\frac{1}{\Gamma(\alpha)}
\int_0^t(t-s)^{\alpha-1}sds
=\sum_{j=0}^ic_j+\frac{t^{\alpha+1}}{\Gamma(\alpha+2)},\quad
\] 
for $t\in (t_i,t_{i+1}]$, $c_i\in \mathbb{R}$ and $i\in \mathbb{N}_0^m$.
\end{example} 

\begin{example} \label{examp7.2}\rm
 We consider the fractional differential equation ${}^{RL}D_{0^+}^\alpha x(t)=t$
with $\alpha\in (0,1)$. It has continuous solutions
\[
x(t)=ct^{\alpha-1}+\frac{1}{\Gamma(\alpha)}\int_0^t(t-s)^{\alpha-1}sds
=ct^{\alpha-1}+\frac{t^{\alpha+1}}{\Gamma(\alpha+2)},\quad t>0,\; c\in \mathbb{R}.
\]
By Theorem \ref{thm3.2.2}, it also has piecewise continuous solutions
\[
x(t)=\sum_{j=0}^ic_j(t-t_j)^{\alpha-1}
+\frac{1}{\Gamma(\alpha)}\int_0^t(t-s)^{\alpha-1}sds
=\sum_{j=0}^ic_j(t-t_j)^{\alpha-1}+\frac{t^{\alpha+1}}{\Gamma(\alpha+2)},
\]
for $t\in (t_i,t_{i+1}]$ and $i\in \mathbb{N}[0,m]$.
Here the $c_i\in \mathbb{R}$ are constants.
\end{example}

\begin{example} \label{examp7.3}\rm
 We consider the fractional differential equation
${}^{CH}D_{1^+}^\alpha x(t)=\log t$ with $\alpha\in (0,1)$.
It has continuous solutions
$$
x(t)=c+\frac{1}{\Gamma(\alpha)}\int_1^t(\log t-\log s)^{\alpha-1}
\log s\frac{ds}{s}=c+\frac{(\log t)^{\alpha+1}}{\Gamma(\alpha+2)},\quad
t\ge 1,c\in \mathbb{R}.
$$
Let $1=s_0<s_1<\dots<s_m<s_{m+1}=e$ be fixed. By Theorem \ref{thm3.2.4},
it also has piecewise continuous solutions
\[
x(t)=\sum_{j=0}^ic_j+\frac{1}{\Gamma(\alpha)}\int_1^t(\log t-\log s)^{\alpha-1}
\log s\frac{ds}{s} =\sum_{j=0}^ic_j+\frac{(\log t)^{\alpha+1}}{\Gamma(\alpha+2)},
\]
for $t\in (s_i,s_{i+1}]$, $c_i\in \mathbb{R}$ and $i\in \mathbb{N}[0,m]$.
\end{example}

\begin{example} \label{examp7.4}\rm
 We consider the fractional differential equation
${}^{RLH}D_{1^+}^\alpha x(t)=\log t$ with $\alpha\in (0,1)$.
It has continuous solutions
\[
x(t)=c(\log t)^{\alpha-1}+\frac{1}{\Gamma(\alpha)}
\int_1^t(\log t-\log s)^{\alpha-1}\log s\frac{ds}{s}
=c(\log t)^{\alpha-1}+\frac{(\log t)^{\alpha+1}}{\Gamma(\alpha+2)},
\]
for $t>1$ and $c\in \mathbb{R}$.
Let $1=s_0<s_1<\dots<s_m<s_{m+1}=e$ be fixed. By Theorem \ref{thm3.2.3},
it also has piecewise continuous solutions
\begin{align*}
x(t)&=\sum_{j=0}^ic_j(\log \frac{t}{t_j})^{\alpha-1}
 +\frac{1}{\Gamma(\alpha)}\int_1^t(\log t-\log s)^{\alpha-1}\log s\frac{ds}{s}\\
&=\sum_{j=0}^ic_j(\log \frac{t}{t_j})^{\alpha-1}
 +\frac{(\log t)^{\alpha+1}}{\Gamma(\alpha+2)},\quad
t\in (s_i,s_{i+1}],\; c_i\in \mathbb{R}, \; i\in \mathbb{N}[0,m].
\end{align*}
\end{example} 

A typical application of the Logistic equation
\begin{equation}
u'(t)=\rho u(t)(1-u(t)),\label{e7.0.1}
\end{equation}
 is a common model of population
growth. Let $u(t)$ represents the population size and $t$ represents the time where
the constant $\rho>0$ defines the growth rate.
Another application of Logistic curve is in medicine, where the Logistic
differential equation is used to model the growth of tumors. This application
can be considered an extension of the above mentioned use in the framework
of ecology. Denoting with $u(t)$ the size of the tumor at time $t$.

The fractional order Logistic model
\begin{equation}
D_{*^+}^\alpha u(t)=\rho u(t)(1-u(t)),\quad t\ge 0,\; \alpha\in (0,1),\label{e7.0.2}
\end{equation}
can be obtained by applying the fractional derivative
operator on the Logistic equation. The model is initially published by Pierre
Verhulst in 1838 \cite{c, msk}. In \eqref{e7.0.2}, $D_{0^+}^\alpha $
denotes the fractional derivative. One sees that the exact solutions of
\eqref{e7.0.1} is $u(t)=\frac{u_0}{(1-u_0)e^{-\rho t}+u_0}$.
While it is difficult to solve \eqref{e7.0.2}. When
$D_{*^+}u(t)={}^CD_{0^+}^\alpha u(t)$, by using the Picard iterative method,
we can get its iterative solutions:
\begin{gather*}
\phi_0(t)=u_0,\\
\phi_1(t)=u_0+\rho\int_0^t\frac{(t-s)^{\alpha-1}}{\Gamma(\alpha)}u_0(1-u_0)ds=u_0
+\frac{\rho u_0(1-u_0)}{\Gamma(\alpha+1)}t^{\alpha},\\
\begin{aligned}
\phi_2(t)&=u_0+\rho\int_0^t\frac{(t-s)^{\alpha-1}}{\Gamma(\alpha)}\phi_1(s)
(1-\phi_1(s))ds \\
&=u_0+\frac{\rho u_0}{\Gamma(\alpha+1)}t^{\alpha}
+\frac{\rho^2 u_0(1-u_0)}{\Gamma(2\alpha+1)}t^{2\alpha},
\end{aligned} \\
\phi_i(t)=u_0+\rho\int_0^t\frac{(t-s)^{\alpha-1}}{\Gamma(\alpha)}
\phi_{i-1}(s)(1-\phi_{i-1}(s))ds,\quad \dots.
\end{gather*}

\begin{remark} \label{rmk7.1}\rm
We see that \eqref{e1.0.7} with $\lambda=0$ can be re-written as
\begin{equation}
\begin{gathered}
{}^{RL}D_{0^+}^{\beta}x(t)=p(t)f(t,x(t)), \quad t\in (t_i,t_{i+1}],\;i\in \mathbb{N}[0,m],\\
\lim_{t\to 0^+}t^{2-\beta }x(t)=\int_0^1\phi(s)G(s,x(s))ds,\quad
x(1)=\int_0^1\psi(s)H(s,x(s))ds,\\
\lim_{t\to t_i^+}(t-t_i)^{2-\beta}[ x(t)-x(t_i)]=I(t_i,x(t_i)),\quad
\Delta {}^{RL}D_{0^+}^{\beta-1}x(t_i)=J(t_i,x(t_i)),
\end{gathered} \label{e7.0.3}
\end{equation}
for $i\in \mathbb{N}[1,m]$.
When $\lambda =0$, BVP \eqref{e1.0.8} becomes
\begin{equation}
\begin{gathered}
{}^CD_{0^+}^{\beta}x(t)=p(t)f(t,x(t)), \quad t\in (t_i,t_{i+1}],\;i\in \mathbb{N}[0,m],\\
\lim_{t\to 0^+}x(t)=\int_0^1\phi(s)G(s,x(s))ds,\quad
x'(1)=\int_0^1\psi(s)H(s,x(s))ds,\\
\Delta x(t_i)=I(t_i,x(t_i)),\;\;\Delta x'(t_i)=J(t_i,x(t_i)),\quad i\in \mathbb{N}[1,m].
\end{gathered} \label{e7.0.4}
\end{equation}
If $\beta\to 2$, we obtain both that \eqref{e7.0.3} and \eqref{e7.0.4} become
the following Dirichlet type BVP for second order impulsive differential equation
\begin{equation}
\begin{gathered}
x''(t)=p(t)f(t,x(t)), \quad t\in (t_i,t_{i+1}],\;i\in \mathbb{N}[0,m],\\
x(0)=\int_0^1\phi(s)G(s,x(s))ds,\quad ;x(1)=\int_0^1\psi(s)H(s,x(s))ds,\\
\Delta x(t_i)=I(t_i,x(t_i)),\quad
\Delta {}^{RL}D_{0^+}^{\beta-1}x(t_i)=J(t_i,x(t_i)),\quad i\in \mathbb{N}[1,m].
\end{gathered} \label{e7.0.5}
\end{equation}
From the equivalent integral equation \eqref{e3.3.2} (Lemma \ref{lem3.3.1})
of BVP \eqref{e1.0.7} with $\lambda =0$, we know that the equivalent integral
equation of \eqref{e7.0.3} become the integral equation
\begin{equation}
\begin{aligned}
x(t)&= t\Big[\int_0^1\psi(s)H(s,x(s))ds-\int_0^1\phi(s)G(s,x(s))ds
-\int_0^1(1-s)\sigma(s)ds \\
& -\sum_{\sigma=1}^m(
(1-t_\sigma)J(t_\sigma,x(t_\sigma))+I(t_\sigma,x(t_\sigma)))\Big]
+\int_0^1\phi(s)G(s,x(s))ds\\
&\quad +\sum_{\sigma=1}^i[(t-t_\sigma)J(t_\sigma,x(t_\sigma))
+I(t_\sigma,x(t_\sigma))]
+\int_0^t(t-s)\sigma(s)ds,
\end{aligned} \label{e7.0.6}
\end{equation}
for $t\in (t_i,t_{i+1}]$ and $i\in \mathbb{N}[0,m]$.
From the equivalent integral equation \eqref{e3.4.2} (Lemma \ref{lem3.4.1})
of BVP\eqref{e1.0.8} with $\lambda =0$, we know that the equivalent integral
equation of \eqref{e7.0.4} becomes the integral equation \eqref{e7.0.6}
when $\beta\to 2$.
\end{remark} 

When $G(t,x)=H(t,x)=0$, \eqref{e7.0.6} is the so called Dirichlat boundary
value problem for second order impulsive differential equation.
Its equivalent integral equation becomes
\begin{align*}
x(t)&= t\Big[-\int_0^1(1-s)\sigma(s)ds -\sum_{\sigma=1}^m(
(1-t_\sigma)J(t_\sigma,x(t_\sigma))+I(t_\sigma,x(t_\sigma)))\Big]\\
&\quad +\sum_{\sigma=1}^i [(t-t_\sigma)J(t_\sigma,x(t_\sigma))
 +I(t_\sigma,x(t_\sigma))\big]
+\int_0^t(t-s)\sigma(s)ds,
\end{align*}
for $t\in (t_i,t_{i+1}]$ and $i\in \mathbb{N}[0,m]$.
This result was established in \cite{zlj} when $I(t,x)=0$.
The solvability of Dirichlet boundary value problems \eqref{e7.0.5}
 or its special cases were studied in \cite{lj, lcz, no, zml}.

\subsection*{Acknowledgments}
This work was supported by the National Natural Science
Foundation of China (No. 11401111), the Natural Science
Foundation of Guangdong province (No. S2011010001900)
and the Foundation for High-level talents in Guangdong Higher Education Project.

 The author would like to thank the
anonymous referees and the editors for their careful reading and some useful
comments on improving the presentation of this paper.


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\section{Addendum posted February 13, 2017}

 In response to comments from readers, the author wants to correct 
some typos and other mistakes in the original article.
More precise proofs and extension of the current results will be presented
in a future article.


Page 6, (1.A4): $l\in \max\{-\beta,-2-k,0]$ 
\textcolor{blue}{should be replaced by}
$l\in (\max\{-\beta+1,-2-k\},0]$.

Page 6, (1.A7): $k>1-\beta$ should be  replaced by $k>-1$; \\
$l\in \max\{-\beta,-\beta-k,0]$ \textcolor{blue}{should be replaced by}
$l\in (\max\{-\beta+1,-\beta-k+1\},0]$.

Page 7, (1.A10): $l\leq 0$, $2+k+l>0$ \textcolor{blue}{should be replaced by}  
$l\in (\max\{-\beta+1,-2-k\},0]$.

Page 8, (1.A13): $l\leq 0$, $\beta+k+l>0$ \textcolor{blue}{should be replaced by}
$l\in (\max\{-\beta+1,-\beta-k+1\},0]$.

Page 9, Definitions 2.1:
 \textcolor{blue}{this definition should be replaced by}
\\
\textbf{Definition 2.1} (\cite[page 69]{Kil})
 Let $-\infty<a<b<+\infty$. The Riemann-Liouville fractional integrals
$I_{a^+}^\alpha g$ and $I_{b^-}^\alpha g$
of order $\alpha\in \mathbb{C}$ with $(\operatorname{Re}(\alpha)> 0)$ 
are defined by
\begin{gather*}
I_{a^+}^{\alpha}g(t)=\frac{1}{\Gamma(\alpha)}\int_a^t(t-s)^{\alpha-1}g(s)ds,t>a,\\
I_{b^-}^\alpha g(t)=\frac{1}{\Gamma(\alpha)}\int_t^b(s-t)^{\alpha-1}g(s)ds,t<b
\end{gather*}
respectively. These integrals are called he left side and the right side
fractional integrals.

Page 9, Definition 2.2: \textcolor{blue}{this definition should
 be replaced by} \\
\textbf{Definition 2.2} (\cite[page 70]{Kil})
 Let $-\infty<a<b<+\infty$.  The Riemann-Liouville fractional derivatives
$D_{a^+}^\alpha g$ and $D_{b^-}^\alpha g$
of order $\alpha\in \mathbb{C}$ with  with $\operatorname{Re}(\alpha)\ge 0$ 
are defined by
\begin{gather*}
{}^{RL}D_{a^+}^{\alpha}g(t)=(\frac{d}{dt})^nI_{a^+}^{n-\alpha}g(t)
=\frac{1}{\Gamma(n-\alpha)}\frac{d^{n}}{dt^{n}}
\int_{a}^{t}\frac{g(s)}{(t-s)^{\alpha-n+1}}ds,t>a,\\
{}^{RL}D_{b^-}^{\alpha}g(t)==(-\frac{d}{dt})^nI_{b^-}^{n-\alpha}g(t)
=\frac{1}{\Gamma(n-\alpha)}\frac{d^{n}}{dt^{n}}
\int_{t}^{b}\frac{g(s)}{(s-t)^{\alpha-n+1}}ds,t<b
\end{gather*}
where $n=[\operatorname{Re}(\alpha)]+1$. In particular, 
when $\alpha=n\in \mathbb{N}$,
then $D_{a^+}^0g(t)=D_{b^-}^0g(t)=g(t)$ and
$D_{a^+}^ng(t)=g^{(n)}(t),D_{b^-}^ng(t)=(-1)^ng^{(n)}(t)$, where
$g^{(n)}(t)$ is the usual derivative of $g(t)$ of order $n$.


Page 9, Definition 2.3: 
\textcolor{blue}{this definition should be replaced by} \\
\textbf{Definition 2.3} \cite[page 91]{Kil}
Let $-\infty<a<b<+\infty$. The Caputo fractional derivatives 
${}^CD_{a^+}^\alpha g$ and ${}^CD_{b^-}^\alpha g$
of order $\alpha\in \mathbb{C}$ with  with $\operatorname{Re}(\alpha)\ge 0$
are defined via the fractional integrals by
\begin{gather*}
{}^CD_{a^+}^{\alpha}g(t)
=\int_a^t\frac{(t-s)^{n-\alpha-1}}{\Gamma(n-\alpha)}
g^{(n)}(s)ds,\quad t>a,\\
{}^CD_{b^-}^{\alpha}g(t)
=\int_t^b\frac{(s-t)^{n-\alpha-1}}{\Gamma(n-\alpha)}g^{(n)}(s)ds,\quad t<b
\end{gather*}
respectively, where $n=[\operatorname{Re}(\alpha)]+1$ for 
$\alpha \not\in \mathbb{N}$ and
$n=\alpha$ for $\alpha\in \mathbb{N}$. These derivatives are called left side
and right side Caputo fractional derivatives of order $\alpha$.

Page 10, Definition 2.4: \textcolor{blue}{this definition should be replaced by} \\
\textbf{Definition 2.4} \cite[page 110]{Kil}
 Let $0<a<b<+\infty$. The left side and the right side Hadamard fractional 
integrals ${}^HI_{a^+}^\alpha g$ and ${}^HI_{b^-}^\alpha g$ of order
 $\alpha\in \mathbb{C}(\operatorname{Re}(\alpha)>0)$ are defined by
\begin{gather*}
{}^HI_{a^+}^{\alpha}g(t)=\frac{1}{\Gamma(\alpha)}
\int_a^t(\log\frac{t}{s})^{\alpha-1}g(s)\frac{ds}{s},\quad t>a,\\
{}^HI_{b^-}^{\alpha}g(t)=\frac{1}{\Gamma(\alpha)}
 \int_t^b(\log\frac{s}{t})^{\alpha-1}g(s)\frac{ds}{s},\quad t<b
\end{gather*}
respectively.

Page 10, Definition 2.5: \textcolor{blue}{this definition should be replaced by} \\
\textbf{Definition 2.5} \cite[page 111]{Kil}
 Let $0<a<b<+\infty$. The left side and the right side Hadamard fractional
derivatives ${}^{RLH}D_{a^+}^\alpha g$ and ${}^{RRH}D_{b^-}^\alpha g$
of order $\alpha\in \mathbb{C}$ with $\operatorname{Re}(\alpha)\ge 0$ are 
defined by
\begin{gather*}
{}^{RLH}D_{a^+}^{\alpha}g(t)=\frac{1}{\Gamma(n-\alpha)}(t\frac{d}{dt})^n
\int_{a}^{t}(\log\frac{t}{s})^{n-\alpha-1}g(s)\frac{ds}{s},\quad t>a,\\
{}^{RRH}D_{b^-}^{\alpha}g(t)=\frac{1}{\Gamma(n-\alpha)}(-t\frac{d}{dt})^n
\int_{t}^{b}(\log\frac{s}{t})^{n-\alpha-1}g(s)\frac{ds}{s},\quad t<b
\end{gather*}
respectively, where $n=[\operatorname{Re}(\alpha)]+1$.

Page 10, Definition 2.6: \textcolor{blue}{this definition should be replaced by} \\
\textbf{Definition 2.6} \cite{jab}
 Let $0<a<b<+\infty$. The left side and right side Caputo type Hadamard fractional
derivatives ${}^{CH}D_{a^+}^\alpha g$ and ${}^{CH}D_{b^-}^\alpha g$ of order
$\alpha\in \mathbb{C}$ with $\operatorname{Re}(\alpha)\ge 0$ are defined by
\begin{gather*}
{}^{CH}D_{a^+}^{\alpha}g(t)=\int_a^t\frac{(\log t-\log s)^{n-\alpha-1}}
{\Gamma(n-\alpha)}(s\frac{d}{ds})^ng(s)\frac{ds}{s},\quad t>a,\\
{}^{CH}D_{b^-}^{\alpha}g(t)=\int_t^b\frac{(\log s-\log t)^{n-\alpha-1}}
{\Gamma(n-\alpha)}(s\frac{d}{ds})^ng(s)\frac{ds}{s},\quad t<b
\end{gather*}
respectively, where $n=[\operatorname{Re}(\alpha)]+1$ for 
$\alpha \not\in \mathbb{N}$
and $n=\alpha$ for $\alpha\in \mathbb{N}$.

Page 12, line 10: $j\in N[0,n-1]$ \textcolor{blue}{should be replaced by}
$j\in \mathbb{N}[0,n-1]$.

Page 12, equation (3.3)  \textcolor{blue}{should be replaced by}
\begin{equation}
\begin{gathered}
{}^{RLH}D_{1^+}^{\alpha }{x}(t)=B(t){x}(t)+{G}(t),\quad \text{a.e. }t\in (1,e),\\
\lim_{t\to 1^+}(\log t)^{n-\alpha}{x}(t)=\frac{{\eta}_n}{\Gamma(\alpha-n+1)},\\
\lim_{t\to 1^+}{}^{RLH}D_{1^+}^{\alpha-j}x(t)=\eta_j,\quad j\in \mathbb{N}[1,n-1],
\end{gathered}\tag{3.3}
\end{equation}

Page 12, equation (3.4) \textcolor{blue}{should be replaced by}
\begin{equation}
\begin{gathered}
{}^{CH}D_{1^+}^{\alpha }{x}(t)=B(t){x}(t)+{G}(t),\quad\text{a.e. }t\in (1,e),\\
\lim_{t\to 1^+}(t\frac{d}{dt})^{j}x(t)={\eta}_j,\quad j\in \mathbb{N}[0,n-1].
\end{gathered} \tag{3.4}
\end{equation}

Page 12, (3.A1):  \textcolor{blue}{the assumptions should be replaced by}
\begin{itemize}
\item[(3.A1)] there exist constants $k_i>-\alpha+n-1$, $l_i\le 0$ with 
$l_i>\max\{-\alpha+n-1,-\alpha-k_i+n-1\}(i=1,2)$, 
$M_A\ge 0$ and $M_F\ge 0$ such that $|A(t)|\le M_At^{k_1}(1-t)^{l_1}$ and 
$|F(t)|\le M_Ft^{k_2}(1-t)^{l_2}$ for all $t\in (0,1)$
\end{itemize}

Page 16, line 3: $\frac{\eta_j}{j!}$" \textcolor{blue}{should be replaced by}
$\frac{\eta_j}{j!}t^j$.


Page 20 line -5: From Cases 1, 2 and 3 \textcolor{blue}{should be replaced by}  
From Claims 1, 2 and 3.

Page 20, line -4: $j\in N[0,n-1]$ \textcolor{blue}{should be replaced by} 
$j\in \mathbb{N}[0,n-1]$.


Page 52, equation (3.25)  \textcolor{blue}{should be replaced by}
\begin{equation}
{}^{RLH}D_{1^+}^{\alpha }{x}(t)=\lambda {x}(t)+{G}(t),\quad
\text{a.e. }t\in (s_i,s_{i+1}],\; i\in \mathbb{N}[0,m], \tag{3.25}\\
\end{equation}

Page 52, equation (3.26) \textcolor{blue}{should be replaced by}
\begin{equation}
{}^{CH}D_{1^+}^{\alpha }{x}(t)=\lambda {x}(t)+{G}(t),\quad\text{a.e. }
t\in (s_i,s_{i+1}],\; i\in \mathbb{N}[0,m],\tag{3.26}
\end{equation}

Page 52, line 8: where $n-1<\alpha < n$, dots (3.25) and (3.26).
\textcolor{blue}{should be replaced by}
where $0=t_0<t_1<\dots<t_{m}<t_{m+1}=1$ in (3.23) and (3.24),
$1=s_0<s_1<\dots<s_m<s_{m+1}=e$ in (3.25) and (3.26).


Page 74, line -1: $P_1C_{1-\alpha}(0,1]$ \textcolor{blue}{should be replaced by}
$P_mC_{2-\beta}(0,1]$.


Page 75, line 3: for $(x,y)\in \overline{\Omega}$ 
\textcolor{blue}{should be replaced by}
$x\in \overline{\Omega}$.

Page 75, line 12: Using (3.33) \textcolor{blue}{should be replaced by}
Using the definition of $T$.


Page 76 line -12:
 Ascoli-CArzela 
\textcolor{blue}{should be replaced by}
Ascoli-Arzela 

Page 101: equations (5.1)--(5.4) \textcolor{blue}{should be replaced by}
\begin{gather}
{}^{C}D_{0^+}^{\alpha }{x}(t)={G}(t),\quad\text{a.e. }t\in (t_i,t_{i+1}],\;
i\in \mathbb{N}[0,m], \tag{5.1} \\
{}^{RL}D_{0^+}^{\alpha }{x}(t)={G}(t),\quad\text{a.e. }
t\in (t_i,t_{i+1}],\; i\in \mathbb{N}[0,m], \tag{5.2}\\
{}^{RLH}D_{1^+}^{\alpha }{x}(t)={G}(t),\quad\text{a.e. }
t\in (s_i,s_{i+1}],\; i\in \mathbb{N}[0,m], \tag{5.3} \\
{}^{CH}D_{1^+}^{\alpha }{x}(t)={G}(t),\quad\text{a.e. }
t\in (s_i,s_{i+1}],\; i\in \mathbb{N}[0,m]. \tag{5.4}
\end{gather}

Page 151: equation (6.31) \textcolor{blue}{should be replaced by}
\begin{align*}
x(t)
&=\sum_{j=1}^iI_j(x(t_j))\mathbf{E}_{\gamma,1}(-a(\log t-\log t_j)^\gamma) \\
&\quad -\Big[\Big[\sum_{i=1}^m\alpha_i\Big[\sum_{j=1}^iI_j(x(t_j))
\mathbf{E}_{\gamma,1}(-a(\log \tau_i-\log t_j)^\gamma)  \\
&\quad +\int_1^{\tau_i}(\log \tau_i-\log s)^{\gamma-1}f(s,x(s),y(s))
\frac{ds}{s}\Big]\Big] \nonumber \\
&\quad\div \Big[1+\sum_{i=1}^m\alpha_i\mathbf{E}_{\gamma,1}
(-a(\log \tau_i))\Big] \Big]\mathbf{E}_{\gamma,1}(-a(\log t)^\gamma) \\
&\quad +\int_1^t(\log t-\log s)^{\gamma-1}f(s,x(s),y(s))\frac{ds}{s},\quad
t\in (t_i,t_{i+1}],\; i\in \mathbb{N}[0,m].
\end{align*} 
\begin{equation}
\begin{aligned}
y(t)
&=\sum_{j=1}^iJ_j(y(t_j))\mathbf{E}_{\gamma,1}(-b(\log t-\log t_j)^\gamma)\\
&\quad -\Big[\Big[\sum_{i=1}^m\beta_i\Big[\sum_{j=1}^iJ_j(y(t_j))
\mathbf{E}_{\gamma,1}(-b(\log \tau_i-\log t_j)^\gamma) \\
&\quad +\int_1^{\tau_i}(\log \tau_i-\log s)^{\gamma-1}g(s,x(s),y(s))
\frac{ds}{s}\Big]\Big] \\
&\quad\div 
\Big[1+\sum_{i=1}^m\beta_i\mathbf{E}_{\gamma,1}(-b(\log \tau_i))\Big] \Big]
\mathbf{E}_{\gamma,1}(-b(\log t)^\gamma)\\ 
&\quad +\int_1^t(\log t-\log s)^{\gamma-1}g(s,x(s),y(s))\frac{ds}{s},\quad
t\in (t_i,t_{i+1}],\; i\in \mathbb{N}[0,m].
\end{aligned} \tag{6.31}
\end{equation}

Page 170, line 1:
$D_{0^+}^\alpha$ \textcolor{blue}{should be replaced by}
$D_{*^+}^\alpha$.

Page 170, equation (7.5): ``;" \textcolor{blue}{should be deleted}.

Page 171, line 3:  \textcolor{blue}{label} (7.6) 
\textcolor{blue}{should be replaced by label}  (7.5).


End of addendum



\end{document}