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

\AtBeginDocument{{\noindent\small
\emph{Electronic Journal of Differential Equations},
Vol. 2015 (2015), No. 166, pp. 1--8.\newline
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu
\newline ftp ejde.math.txstate.edu}
\thanks{\copyright 2015 Texas State University - San Marcos.}
\vspace{9mm}}

\begin{document}
\title[\hfilneg EJDE-2015/166\hfil Existence of local  and global solutions]
{Existence of local and global solutions for Hadamard fractional
differential equations}

\author[M. Li, J. Wang \hfil EJDE-2015/166\hfilneg]
{Mengmeng Li, Jinrong Wang}

\address{Mengmeng Li  \newline
Department of Mathematics,
Guizhou University, Guiyang, Guizhou 550025, China}
\email{Lmm0424@126.com}

\address{Jinrong Wang (corresponding author)\newline
Department of Mathematics,
Guizhou University, Guiyang, Guizhou 550025, China}
\email{sci.jrwang@gzu.edu.cn}

\thanks{Submitted February 2, 2015. Published June 17, 2015.}
\subjclass[2010]{26A33, 34A12}
\keywords{Hadamard fractional differential equation; local solution;
\hfill\break\indent global solution}

\begin{abstract}
 In this article, we study a class of Hadamard fractional
 differential equations and give  sufficient conditions
 on the existence of local and global of solutions.
\end{abstract}

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

\section{Introduction}

Let $0\leq\gamma<1$, $1<a<T$ and $\mathbb{G}$ be an open set in
$\mathbb{R}$. Denote a Banach space by
$C_{\gamma,\ln}[a,T]:=\{\mu(x):(\ln\frac{x}{a})^\gamma \mu(x) \in C[a,T] \}$ 
endowed with the norm 
$ \|\mu\|_{C_{\gamma,\ln}}=\|(\ln \frac{x}{a})^\gamma \mu(x)\|_C$.
In this article, we study the existence of local and global 
solutions to the Hadamard type fractional differential equation
\begin{equation}\label{2-1}
\begin{gathered}
{}_{H}D^{\alpha}_{a,x}y(x)=f(x,y(x)),\quad 0<\alpha<1,\; x\in \mathbb{J},\\
{}_{H}D^{\alpha-1}_{a,x}y(a^{+})=c,\quad c\in \mathbb{R},
\end{gathered}
\end{equation}
where  $\mathbb{J}=[a,a+h]$, $h>0$  or $[a+\infty)$ and the
symbol $_{H}D^{\alpha}_{a,x}y(x)$ is defined by
\[
{}_{H}D_{a,x}^{\alpha}y(x)
=\frac{1}{\Gamma(1-\alpha)}\big(x\frac{d}{dx}\big) 
\int_{a}^{x} (\ln \frac{x}{\tau})^{-\alpha}y(\tau)\frac{d\tau}{\tau}.
\]
We use the notation
${}_{H}D_{a,x}^{\alpha-1}y(a^+)=\lim_{x\to a^+} 
\mathcal{J}_{a,x}^{\alpha-1}y(x)$ and
\[
\mathcal{J}_{a,x}^{\alpha-1}y(x)
=\frac{1}{\Gamma(\alpha)} \int_1^x (\ln
\frac{x}{t})^{\alpha-1} y(t))\frac{dt}{t}\,.
\]
Following \cite[Theorem 3.28]{Kilbas}, the solution $y\in
C_{1-\gamma,\ln}[a,a+h]$ of \eqref{2-1} satisfies
\begin{equation}\label{2-2}
y(x)=y_0(x)+\frac{1}{\Gamma(\alpha)}\int_{a}^{x} (\ln
\frac{x}{\tau})^{\alpha-1}f(\tau,y(\tau))\frac{d\tau}{\tau},\quad
x\in(a,a+h]
\end{equation}
where $y_0(x)=\frac{c}{\Gamma(\alpha)}(\ln
\frac{x}{a})^{\alpha-1}$, if $f:(a,a+h]\times \mathbb{G}\to
\mathbb{R}$ and $f(x,y)\in C_{\gamma,\ln}[a,a+h]$ for any $y\in
\mathbb{G}$.

Inspired by the work in \cite{Kilbas,Lin,TMNA-WZM}, we examine other
explicit sufficient conditions on the nonlinear term $f$ to
guarantee the local existence of solutions in
$C_{\gamma,\ln}[a,a+h]$ and global existence of solutions in
$C_{\gamma,\ln}[a,+\infty)$.

\section{Main results}

The following equality will be used in the sequel.

\begin{lemma}[{\cite[p.296]{PBM-book}}] \label{lem4}
Let $\alpha,\beta,\gamma,p>0$, then
\[
\int_{0}^{x}(x^{\alpha}-s^{\alpha})^{p(\beta-1)}s^{p(\gamma-1)}ds
=\frac{x^{\theta}}{\alpha}B\big[\frac{p(\gamma-1)+1}{\alpha},
p(\beta-1)+1\big],\quad x>0,
\]
where $\theta=p[\alpha(\beta-1)+\gamma-1]+1$ and
$B[\xi,\eta]=\int_{0}^{1}s^{\xi-1}(1-s)^{\eta-1}ds$.
\end{lemma}

Let $\mathbb{B}=\{y\in \mathbb{R}: \|
y-y_{0}(x)\|_{C_{\gamma,\ln}}\leq b\}$ where $b$ will be chosen
latter. Define $\mathbb{D}=\{ (x,y)\in R\times R: x\in
\mathbb{J},\, y\in \mathbb{B} \}$. We assume that
$f:\mathbb{D}\to \mathbb{R}$ satisfies the following conditions:
\begin{itemize}
\item[(H1)] $f(x,y)$ is Lebesgue measurable with regard to $x$ on
$\mathbb{J}$ and $f(x,y)$ is continuous with respect to $y$ on
$\mathbb{B}$.


\item[(H2)] there exists  $m(\cdot)\in L^{q}(\mathbb{J})$, $q>1$ such
that $ |f(x,y)|\leq m(x), $ for arbitrary $x\in \mathbb{J}$, $y\in
\mathbb{B}$.

\end{itemize}
Now we  use Picard iterative approach to derive the existence of a local
solutions to \eqref{2-1}.

\begin{theorem}\label{thm1}
Assume that {\rm (H1)--(H2)} hold for $\mathbb{J}=[a,a+h]$ and
$p,q,\alpha$ satisfy $p(\alpha-1)+1>0,\frac{1}{p}+\frac{1}{q}=1$.
Then \eqref{2-1} has a solution in $C_{\gamma,\ln}[a,a+h]$ for some
$h>0$.
\end{theorem}

\begin{proof} 
To achieve our aim, we divide our proof into three steps.
\smallskip

\noindent\textbf{Step 1.}
 Linking our assumptions and using H\"{o}lder inequality via
$p(\alpha-1)+1>0$ and $\frac{1}{p}+\frac{1}{q}=1$, one can obtain
\begin{equation}\label{2-3}
\int_{a}^{x}|(\ln x-\ln
\tau)^{\alpha-1}f(\tau,y(\tau))|\frac{dt}{\tau}
\leq\sqrt[1/p]{\frac{a^{1-p}h^{p(\alpha-1)+1}}{p(\alpha-1)+1}}\|
m(\cdot)\|_{L^{q}[a,a+h]},
\end{equation}
where we use basic inequalities: $\ln u-\ln v\leq u-v$ for $u\geq
v>1$ and \cite[Lemma 2.2]{MWWK},
\begin{equation}\label{1-3}
\int_{a}^{x}(\ln x-\ln \tau)^{p(\alpha-1)}\tau^{-p}d\tau
\leq\frac{a^{1-p}(\ln x-\ln a)^{p(\alpha-1)+1}}{p(\alpha-1)+1}.
\end{equation}
This  proves that $(\ln x-\ln \tau)f(\tau,y(\tau))$ is Lebesgue
integrable with respect to $\tau\in[a,x]$ for arbitrary $x$ on
$\mathbb{J}$, provided that $y(\tau)$ is Lebesgue measurable on the
interval $[a,a+h]$.
\smallskip

\noindent\textbf{Step 2.}
 For a given $M>0$, there exists a $h'>0$ satisfying
\begin{equation}\label{2-4}
\int_{a}^{a+h'}m^{q}(\tau)d\tau\leq M^{q},
\end{equation}
whenever
$h=\min\{h',T,[\frac{b\Gamma(\alpha)(p(\alpha-1)+1)}{Ma^{p-1}(\ln
T-\ln a)^{\gamma}}]^{\frac{1}{p(\alpha-1)+1}}\}$.
For $\delta$ to be chosen latter, define
\[
y_{n}(x)=\begin{cases}
0, \quad\text{if } a\leq x<a+\delta,\; 0<\delta<\frac{h}{n},\\[4pt]
\frac{c}{\Gamma(\alpha)}(\ln\frac{x}{a})^{\alpha-1}, 
\quad \text{if }  a+\delta\leq x< a+\frac{h}{n}  ,\\[4pt]
\frac{c}{\Gamma(\alpha)}(\ln\frac{x}{a})^{\alpha-1}
+\frac{1}{\Gamma(\alpha)}\int_{a}^{x-\frac{h}{n}}
(\ln\frac{x}{\tau})^{\alpha-1}f(\tau,y_{n}(\tau))
\frac{d\tau}{\tau},\\
\quad\text{if } a+\frac{h}{n}\leq x\leq a+h.
\end{cases}
\]
We show that $y_{n}(x)$ is continuous on $[a,a+\frac{2h}{n}]$ for
all $n$.
\smallskip

\noindent\textbf{Case 1.}
For $a<a+\delta\leq x_{1}<a+\frac{h}{n}<x_{2}\leq a+h$,
\begin{align*}
|y_{n}(x_{2})-y_{n}(x_{1})|
&\leq \frac{|c|}{\Gamma(\alpha)}
 \big|(\ln\frac{x_{2}}{a})^{\alpha-1}-(\ln\frac{x_{1}}{a})^{\alpha-1}\big| \\
&\quad +\frac{1}{\Gamma(\alpha)}\int_{a}^{x_{2}
 -\frac{h}{n}}(\ln\frac{x_{2}}{\tau})^{\alpha-1}|f(\tau,y_{n}(\tau))
  |\frac{d\tau}{\tau} \\
&:= I_{1}+I_{2}.
\end{align*}

For each $\varepsilon>0$, there exists
$0<\delta_{1}<[\frac{a\varepsilon\Gamma(\alpha)
\delta^{2-\alpha}} {2|c|(1-\alpha)}]$ such that for all
$x_{2}-x_{1}\leq\delta_{1}$ and for all $n$, we  derive that
\begin{align*}
I_{1}
&=\frac{|c|}{\Gamma(\alpha)}|\chi(\frac{x_2}{a})
-\chi(\frac{x_1}{a})|
\leq\frac{|c|}{\Gamma(\alpha)}
|\chi'(\xi)||\frac{x_{2}}{a}-\frac{x_{1}}{a}|,~\xi\in(\frac{x_1}{a},\frac{x_2}{a})\\
&\leq \frac{|c|(1-\alpha)(x_{2}-x_{1})}{\Gamma(\alpha)(a+\delta)(\ln
(a+\delta)-\ln a)^{2-\alpha}}\\
&<\frac{|c|(1-\alpha)(x_{2}-x_{1})}{\Gamma(\alpha)a(\ln
(a+\delta)-\ln a)^{2-\alpha}}<\varepsilon/2,
\end{align*}
where $\chi(x)=(\ln x-\ln a)^{\alpha-1}$ and
$\chi'(x)=(\alpha-1)(\ln x-\ln a)^{\alpha-2}\frac{a}{x}$.

For any $\varepsilon>0$, there exists
$0<\delta_{2}<[\frac{\varepsilon^{p}\Gamma^{p}(\alpha)
(p(\alpha-1)+1)}{2^{p}a^{1-p}
M^{p}}]^{\frac{1}{p(\alpha-1)+1}}$ such that for all
$x_{2}-\frac{h}{n}-a\leq x_{2}-x_{1}\leq\delta_{2}$ and for all $n$,
we use \eqref{1-3} and \eqref{2-4} to obtain
\begin{align*}
I_{2}
&=\frac{1}{\Gamma(\alpha)}\int_{a}^{x_{2}-\frac{h}{n}}
 (\ln\frac{x_{2}}{\tau})^{\alpha-1}|f(\tau,y_{n}(\tau))|\frac{d\tau}{\tau}\\
&\leq \frac{1}{\Gamma(\alpha)}\int_{a}^{x_{2}-\frac{h}{n}}(\ln\frac{x_{2}}{\tau})^{\alpha-1}m(\tau)\frac{d\tau}{\tau} \\
&\leq \frac{M}{\Gamma(\alpha)}\big[\frac{a^{1-p}(\ln
(x_{2}-\frac{h}{n})-\ln a)^{p(\alpha-1)+1}}
{p(\alpha-1)+1}\big]^{1/p}\\
&\leq\frac{M}{\Gamma(\alpha)}
 \big[\frac{a^{1-p}(x_{2}-\frac{h}{n}-a)^{p(\alpha-1)+1}}{p(\alpha-1)+1}
  \big]^{1/p}
<\varepsilon/2,
\end{align*}
where we use that $ \ln u-\ln v\leq u-v$, $u>v>1$ again in the
last inequality.

From above, we can choose
$\bar{\delta}=$min$\{\delta_{1},\delta_{2},h/n\}$ such that for all
$x_{2}-x_{1}\leq\bar{\delta}$ and for all $n$, such that
$|y_{n}(x_{2})-y_{n}(x_{1})|\leq I_{1}+I_{2}<\varepsilon$.
\smallskip

\noindent\textbf{Case 2.}
 For $a+\frac{h}{n}\leq x_{1}<x_{2}\leq a+\frac{2h}{n}$. One has
\begin{align*}
&|y_{n}(x_{2})-y_{n}(x_{1})|\\
&\leq S_{1}+\frac{1}{\Gamma(\alpha)}\int_{a}^{x_{1}
 -\frac{h}{n}}\Big((\ln\frac{x_{2}}{\tau})^{\alpha-1}
 -(\ln\frac{x_{1}}{\tau})^{\alpha-1}\Big)
|f(\tau,y_{n}(\tau))|\frac{d\tau}{\tau} \\
&\quad +\frac{1}{\Gamma(\alpha)}\int_{x_{1}-\frac{h}{n}}^{x_{2}-\frac{h}{n}}(\ln\frac{x_{2}}{\tau})^{\alpha-1}|f(\tau,y_{n}(\tau))|\frac{d\tau}{\tau}\\
&:=S_{1}+S_{2}+S_{3}.
\end{align*}
For each $\varepsilon>0$, there exists
$0<\bar{\delta_{1}}<\left[\frac{a\varepsilon\Gamma(\alpha)
(\frac{h}{n})^{2-\alpha}} {3|c|(1-\alpha)}\right]$ such that for all
$x_{2}-x_{1}\leq\bar{\delta_{1}}$, we have
\[
S_{1} \leq\frac{|c|(1-\alpha)(x_{2}-x_{1})}{\Gamma(\alpha)a(\ln
(a+\frac{h}{n})-\ln a)^{2-\alpha}}<\varepsilon/3.
\]

For each $\varepsilon>0$, there exists
$0<\bar{\delta_{2}}<\left[\frac{\varepsilon^{p}\Gamma(\alpha)^{p}
(p(\alpha-1)+1)}{3^{p}M^{p}a^{1-p}}\right]^{\frac{1}{p(\alpha-1)+1}}$
such that for all $x_{2}-x_{1}\leq\bar{\delta_{2}}$, by using the
similar estimation methods of $I_2$ we have
\begin{align*}
S_{2}&=\frac{1}{\Gamma(\alpha)}\int_{a}^{x_{1}-\frac{h}{n}}
\Big((\ln\frac{x_{2}}{\tau})^{\alpha-1}-
(\ln\frac{x_{1}}{\tau})^{\alpha-1}\Big)
|f(\tau,y_n(\tau))|\frac{d\tau}{\tau} \\
&\leq \frac{1}{\Gamma(\alpha)}\int_{a}^{x_{1}-\frac{h}{n}}
\Big((\ln\frac{x_{2}}{\tau})^{\alpha-1}-
(\ln\frac{x_{1}}{\tau})^{\alpha-1}\Big)
m(\tau)\frac{d\tau}{\tau} \\
&\leq \frac{M}{\Gamma(\alpha)}\Big[\Big(\int_{a}^{x_{1}-\frac{h}{n}}(\ln\frac{x_{2}}{\tau})^{p(\alpha-1)}
\tau^{-p}d\tau\Big)^{1/p}
-\Big(\int_{a}^{x_{1}-\frac{h}{n}}(\ln\frac{x_{1}}{\tau})^{p(\alpha-1)}\tau^{-p}d\tau
\Big)^{1/p}\Big] \\
&\leq \frac{M}{\Gamma(\alpha)}
\Big[\frac{(\ln x_{2}-\ln x_{1})^{p(\alpha-1)+1}}{a^{p-1}(p(\alpha-1)+1)}
 \Big]^{1/p}\\
&\leq \frac{M}{\Gamma(\alpha)}
\Big[\frac{(x_{2}-x_{1})^{p(\alpha-1)+1}}{a^{p-1}(p(\alpha-1)+1)}\Big]^{1/p}<
\varepsilon/3.
\end{align*}

For each $\varepsilon>0$, there exists
$0<\bar{\delta_{3}}<[\frac{\varepsilon^{p}\Gamma(\alpha)^{p}(p(\alpha-1)+1)}
{3^{p}M^{p}a^{1-p}}]^{\frac{1}{p(\alpha-1)+1}}$, such that for
all $x_{2}-x_{1}\leq\bar{\delta_{3}}$, by using the similar
estimation methods of $I_2$ we have
\begin{align*}
S_{3}&= \frac{1}{\Gamma(\alpha)}\int_{x_{1}-\frac{h}{n}}^{x_{2}
 -\frac{h}{n}}(\ln\frac{x_{2}}{\tau})^{\alpha-1}|f(\tau,y_n(\tau))
  |\frac{d\tau}{\tau}\\
&\leq\frac{1}{\Gamma(\alpha)}\int_{x_{1}
 -\frac{h}{n}}^{x_{2}-\frac{h}{n}}(\ln\frac{x_{2}}{\tau})^{\alpha-1}m(\tau)
 \frac{d\tau}{\tau} \\
&\leq \frac{M}{\Gamma(\alpha)}
 \big[\frac{a^{1-p}(x_{2}-x_{1})^{p(\alpha-1)+1}}{p(\alpha-1)+1}\big]^{1/p}
 <\varepsilon/3.
\end{align*}

From the above, we choose $\bar{\bar{\delta}}=\min\{\bar{\delta_{1}},
\bar{\delta_{2}},\bar{\delta_{3}},h/n\}$ such that
$x_{2}-x_{1}\leq\bar{\bar{\delta}}$, then 
$|y_{n}(x_{2})-y_{n}(x_{1})|\leq S_{1}+S_{2}+S_{3}<\varepsilon$.
Therefore, we choose $\delta=\min
\{\bar{\delta},\bar{\bar{\delta}}\}$ will lead to $y_{n}(x)$ is
continuous with regard to $x$ on $[a,a+\frac{2h}{n}]$ for all
positive integers $n$. Note that $(\ln x-\ln a)^{\gamma}$ is
continuous function, so $y_{n}(x)(\ln x-\ln a)^{\gamma}$ is also
continuous.

Nevertheless, for all $x\in[a+\delta,a+\frac{h}{n}]$, one has $
|y_{n}(x)-\frac{c}{\Gamma(\alpha)}(\ln\frac{x}{a})^{\alpha-1}|=0,
$ and for all $x\in[a+\frac{h}{n},a+h]$, using H\"{o}lder
inequality again,
\begin{equation}\label{b-value}
\begin{aligned}
&(\ln\frac{x}{a})^{\gamma}|y_{n}(x)
 -\frac{c}{\Gamma(\alpha)}(\ln\frac{x}{a})^{\alpha-1}|\\
&\leq \frac{1}{\Gamma(\alpha)}\int_{a}^{x-\frac{h}{n}}
 (\ln\frac{x}{a})^{\gamma}(\ln\frac{x}{\tau})^{\alpha-1}
  |f(\tau,y_n(\tau))|\frac{d\tau}{\tau} \\
&\leq \frac{(\ln (a+h)-\ln a)^{\gamma}a^{1-p}Mh^{p(\alpha-1)+1}}
 {\Gamma(\alpha)(p(\alpha-1)+1)}\leq b,
\end{aligned}
\end{equation}
which implies that $(x,y_{n}(x))\in \mathbb{D}$ for all $n$.
Therefore, $\{y_{n}(x)\}_{n=1}^{\infty}$ defined on $[a,a+h]$ is
equicontinuous and uniformly bounded.
\smallskip

\noindent\textbf{Step 3.}
 By using Arzel\`{o}-Ascoli lemma and Step 2, there must
exist $\{y_{n_{k}}(x)\}_{k=1}^{\infty}:=\{y_{k}(x)\}_{k=1}^{\infty}$
contained in $\{{y_{n}(x)\}_{n=1}^{\infty}}$, such that
$\{{y_{k}(x)\}_{k=1}^{\infty}}$ is uniformly convergent to $y(x)$
which is continuous with regard to $x$ on $[a,a+h]$. Now we only
need to prove that this limit function $y(x)$ is a solution of
\eqref{2-2}.

For each $\varepsilon>0$, there exists $K_{1}>0$, such that for all
$k>K_{1}$, and $x\in[a,a+h]$, we have
\begin{equation}\label{2-5}
|f(x,y_{k}(x))-f(x,y(x))|<\frac{\Gamma(\alpha+1)\varepsilon}{2(\ln (a+h)-\ln a)^{\gamma}h^{\alpha}}.
\end{equation}
Note that
\begin{align*}
&(\ln x-\ln a)^{\gamma}|y_{k}(x)-y(x)|\\
&\leq \frac{1}{\Gamma(\alpha)}\int_{a}^{x}(\ln\frac{x}{a})^{\gamma}(\ln
\frac{x}{\tau})^{\alpha-1}|f(\tau,y_{k}(\tau))-f(\tau,y(\tau))
|\frac{d\tau}{\tau}\\
&\quad +\frac{1}{\Gamma(\alpha)}
\int_{x-\frac{h}{k}}^{x}(\ln\frac{x}{a})^{\gamma}(\ln \frac{x}{\tau})^{\alpha-1}|f(\tau,y_{k}(\tau))|\frac{d\tau}{\tau} \\
&:= S_{4}+S_{5}.
\end{align*}
Using \eqref{2-5} one obtains,
\begin{align*}
S_{4}&=\frac{1}{\Gamma(\alpha)}\int_{a}^{x}(\ln\frac{x}{a})^{\gamma}
 (\ln \frac{x}{\tau})^{\alpha-1}
|f(\tau,y_{k}(\tau))-f(\tau,y(\tau))|\frac{d\tau}{\tau} \\
&\leq \frac{(\ln (a+h)-\ln a)^{\gamma}}{\Gamma(\alpha)}
 \int_{x}^{a}(\ln \frac{x}{\tau})^{\alpha-1}|f(\tau,y_{k}(\tau))-f(\tau,y(\tau))|d(\ln x-\ln \tau) \\
&\leq \frac{(\ln (a+h)-\ln
a)^{\gamma}h^{\alpha}}{a\Gamma(\alpha)}\frac{\Gamma(\alpha+1)\varepsilon}{2(\ln
(a+h)-\ln a)^{\gamma}h^{\alpha}}<\varepsilon/2.
\end{align*}
Also there exists 
\[
0<K_{2}=h\big[\frac{a^{p-1}\varepsilon^{p}\Gamma(\alpha)^{p}
(p(\alpha-1)+1)}{2^{p}M^{p}(\ln
(a+h)-\ln a)^{p\gamma}}\big] ^{\frac{-1}{p(\alpha-1)+1}},
\] 
such that for all $k>K_{2}$,
\begin{align*}
S_{5}&=\frac{1}{\Gamma(\alpha)}\int_{x-\frac{h}{k}}^{x}
(\ln\frac{x}{a})^{\gamma}(\ln
\frac{x}{\tau})^{\alpha-1}
|f(\tau,y_{k}(\tau))|\frac{d\tau}{\tau} \\
&\leq \frac{(\ln (a+h)-\ln a)^{\gamma}M}{\Gamma(\alpha)}
\Big(\int_{x-\frac{h}{k}}^{x}(\ln \frac{x}{\tau})^{p(\alpha-1)}\tau^{-p}dt\Big)^{1/p} 
 \\
&\leq \frac{(\ln (a+h)-\ln a)^{\gamma}M}{\Gamma(\alpha)}
\Big(\frac{(\frac{h}{k})^{p(\alpha-1)+1}}{a^{p-1}(p(\alpha-1)+1)}\Big)^{1/p}
 <\varepsilon/2.
\end{align*}
Hence, taking $K=\max\{K_{1},K_{2}\}$ and for all $k>K$, one arrives
at $\|y_{k}(x)-y(x)\|_{C_{\gamma,\ln}}<\varepsilon$. Consequently,
$y(x)$ satisfies \eqref{2-2} which means that there at least exists
a solution of \eqref{2-1}.
\end{proof}
Next, we give an existence and uniqueness theorem, using the assumption
\begin{itemize}
\item[(H3)]  there exists a $\mu(\cdot)\in L^{q}(\mathbb{J})$,
$\frac{1}{q}=1-\frac{1}{p}$, $p>1$ such that $|f(x,y)-f(x,z)|\leq
\mu(x)|y-z| $ for $x\in \mathbb{J}$ and $y,z\in \mathbb{B}$,
\end{itemize}

\begin{theorem}\label{thm2}
Let $0\leq\gamma\leq \min\{\alpha-1+\frac{1}{p},\frac{1}{2p}\}$,
$p>1$.  Assume that {\rm (H1)--(H3)} are satisfied for
$\mathbb{J}=[a,a+h]$. Then \eqref{2-1} has a unique solution in
$C_{\gamma,\ln}[a,a+h]$ for some $h>0$.
\end{theorem}

\begin{proof} 
There exits  $h^{*}>0$, such that for all $x\in[a,a+h^{*}]$,
\begin{equation}\label{ineq-2}
\int_{a}^{x}\mu^{q}(\tau)d\tau\leq\int_{a}^{a+h^{*}}\mu^{q}(\tau)d\tau<g^{q}.
\end{equation}

Let $\Psi_{h}=\{y\in C_{\gamma,\ln}[a,a+h]:
\|y(x)-y_{0}(x)\|_{C_{\gamma,\ln}}\leq b,~x\in[a,a+h]\}$, where $h$
is the smaller one between $h^{*}$ and $h$ obtained in Theorem
\ref{thm1}. Note $\Psi_{h}$ endowed with $\|\cdot\|_{C_{r\ln}}$ is a
Banach space.
Define
\[
\psi(y)=y_{0}(x)+\frac{1}{\Gamma(\alpha)}\int_{a}^{x}(\ln
\frac{x}{\tau})^{\alpha-1}
f(\tau,y(\tau))\frac{d\tau}{\tau},\quad \forall x\in[a,a+h].
\]

Firstly, assume that $a\leq x_{1}<x_{2}\leq a+h$, then according to
the proof in Step 2 of Theorem \ref{thm1}, we obtain that
$\psi(y)$ is continuous with regard to $x$ on $[a,a+h]$. Note
that $(\ln x-\ln a)^{\gamma}$ is continuous function, so
$\psi(y)(\ln x-\ln a)^{\gamma}$ is also continuous. Secondly, it
following \eqref{b-value} that $\psi(y)\in \Psi_{h}$ with
$y\in \Psi_{h}$. Thirdly, the condition $0\leq\gamma\leq
\min\{\alpha-1+\frac{1}{p},\frac{1}{2p}\}$, $p>1$ implies that
$p(\alpha-1-\gamma)+1>0$, $0\leq\gamma<\frac{1}{2p}$.
 For any $y_{1},y_{2}\in \Psi_{h}$,
then using H\"{o}lder inequality, \eqref{ineq-2} and Lemma
\ref{lem4}, we have
\begin{align*}
&(\ln x-\ln a)^{\gamma}|\psi(y_{2})-\psi (y_{1})|\\
&\leq \frac{(\ln x-\ln
a)^{\gamma}}{\Gamma(\alpha)}\int_{a}^{x}(\ln
\frac{x}{\tau})^{\alpha-1}\mu(\tau)
|y_{1}(\tau)-y_{2}(\tau)|\frac{d\tau}{\tau} \\
&\leq \frac{(\ln (a+h)-\ln a)^{\gamma}}{\Gamma(\alpha)}\int_{a}^{x}(\ln
\frac{x}{\tau})^{\alpha-1} (\ln \frac{\tau}{a}
)^{-\gamma}\mu(\tau)\frac{d\tau}{\tau} \| y_{1}-y_{2}\|_{C_{r\ln}}\\
&\leq \frac{(\ln (a+h)-\ln a)^{\gamma}ga^{1-p}}{\Gamma(\alpha)}
\Big(\int_{0}^{\ln x-\ln a}(\ln\frac{x}{a}-t)^{p(\alpha-1)}
t^{-p\gamma}dt\Big)^{1/p} \| y_{1}-y_{2}\|_{C_{\gamma\ln}}\\
&\leq \frac{(\ln (a+h)-\ln a)^{\gamma}ga^{1-p}}{\Gamma(\alpha)}
\Big((\ln\frac{x}{a})^{p(\alpha-1-\gamma)+1} B[1-p\gamma,p(\alpha-1)+1]\Big)^{1/p}\\
&\quad\times \| y_{1}-y_{2}\|_{C_{\gamma\ln}}\\
&\leq a^{1-p}\frac{h^{\gamma}g}{\Gamma(\alpha)}(h^{p(\alpha-1-\gamma)+1}B[\frac{1}{2},p(\alpha-1)+1])^{1/p}
\| y_{1}-y_{2}\|_{C_{\gamma\ln}}\\
&=a^{1-p}\Big[\frac{h^{\alpha-1+\frac{1}{p}}g}{\Gamma(\alpha)}
\Big(B[\frac{1}{2},p(\alpha-1)+1]\Big)^{1/p}\Big]
\| y_{1}-y_{2}\|_{C_{\gamma\ln}},
\end{align*}
where we use that
\begin{gather*}
\gamma<\frac{1}{2p}\Rightarrow1-p\gamma>\frac{1}{2}
\Rightarrow t^{1-pr}\leq t^{\frac{1}{2}}~(0\leq t\leq1),\\
\begin{aligned}
B[1-p\gamma,p(\alpha-1)+1] 
&=\int_{0}^{1}t^{-p\gamma}(1-t)^{p(\alpha-1)}dt \\
&\leq\int_{0}^{1}t^{-\frac{1}{2}}(1-t)^{p(\alpha-1)}dt
=B[\frac{1}{2},p(\alpha-1)+1].
\end{aligned}
\end{gather*}
Obviously, one can choose
\[
h\leq\Big(\frac{\Gamma(\alpha)}{g(B[\frac{1}{2},p(\alpha-1)+1])^{1/p}}
\Big)^{\frac{p}{p(\alpha-1)+1}},
\]
then
$$
\frac{h^{\alpha-1+\frac{1}{p}}g}{\Gamma(\alpha)}
\Big(B[\frac{1}{2},p(\alpha-1)+1]\Big)^{1/p}\leq 1.
$$
Therefore,
\[
\|\psi(y_{2})-\psi (y_{1})\|_{C_{\gamma\ln}}
\leq a^{1-p}\| y_{1}-y_{2}\|_{C_{\gamma\ln}}.
\]
Obviously, $a^{1-p}<1$ due to $a,p>1$, applying the Banach
Contractive Mapping Principle, one concludes that there exists a
unique $y^{*}(x)\in \Psi_{h}$, such that \eqref{2-2}. The proof is
compete.
\end{proof}

Next, we give the existence of a  global solution, using the assumption
\begin{itemize}
\item[(H2')] there exist $\omega,\nu>0$ such that 
$ |f(x,y)|\leq \omega+\nu|y| $ for $x\in (a,+\infty)$ and $y\in \mathbb{R}$.
\end{itemize}

\begin{theorem}\label{thm3}
Assume that {\rm (H1), (H2'), (H3)} hold for
$\mathbb{J}=(a,+\infty)$. Further, choose $\gamma= 1-\alpha\leq
\min\{\alpha-1+\frac{1}{p},\frac{1}{2p}\}$, $p>1$. 
Then \eqref{2-1} has a unique solution in $C_{\gamma,\ln}[a,+\infty)$.
\end{theorem}

\begin{proof}
It follows (H2') that $f$ is locally bounded in the domain
$\mathbb{D}$. By Theorem \ref{thm2}, \eqref{2-1} has a unique
solution in $C_{\gamma,\ln}[a,a+h]$. Next, we present proof by
contradiction. Assume that the solution $y(x)$ admits a maximal
existence interval, denoted by $(a,T)\subset (a,+\infty)$. To
achieve our aim, it is sufficient to verify that
$\|y\|_{C_{\gamma,\ln}}$ is bounded. In fact,
\begin{align*}
&(\ln x-\ln a)^{\gamma}|y(x)|\\
&\leq \frac{|c|}{\Gamma(\alpha)}+\frac{1}{\Gamma(\alpha)}\int_{a}^{x}(\ln\frac{x}{a})^{\gamma}
(\ln \frac{x}{\tau})^{\alpha-1}|f(\tau,y(\tau))|\frac{d\tau}{\tau} \\
&\leq \frac{|c|}{\Gamma(\alpha)}+\frac{1}{\Gamma(\alpha)}\int_{a}^{x}(\ln
\frac{x}{a})^{\gamma}
(\ln \frac{x}{\tau})^{\alpha-1}(\omega+\nu|y(\tau)|)\frac{d\tau}{\tau} \\
&\leq \frac{|c|}{\Gamma(\alpha)}+\frac{\omega(\ln T-\ln a)^{1-\alpha}(T-a)^{\alpha}}{\Gamma(a+1)}\\
&\quad +\frac{\nu}{\Gamma(\alpha)}\int_{a}^{x}(\ln
\frac{x}{\tau})^{\alpha-1}(\ln x-\ln a)^{\gamma}|y(\tau)|\frac{d\tau}{\tau}\\
&\leq \frac{|c|}{\Gamma(\alpha)}+\frac{\omega(T-a)}{\Gamma(\alpha+1)}
+\frac{\nu}{\Gamma(\alpha)}\int_{a}^{x}(\ln \frac{x}{\tau})^{\alpha-1}
(\ln x-\ln a)^{\gamma}|y(\tau)|\frac{d\tau}{\tau}.
\end{align*}

By applying the generalized Gronwall inequality from
\cite[Corollary 3.4]{TMNA-WZM}, one can conclude that there exists
$l:=\mathbb{E}_{\alpha}(\nu (\ln T)^\alpha)>0$
($\mathbb{E}_{\alpha}$ denotes Mittag-Leffler function) such that
\[
(\ln x-\ln a)^{\gamma}|y(x)|
\leq l(\frac{|c|}{\Gamma(\alpha)}
 +\frac{\omega(T-a)}{\Gamma(\alpha+1)}):=\rho<+\infty.
\]
This implies that $\|y\|_{C_{\gamma,\ln}}< b$ on $[a,T)$ when $b$ is
chosen as a larger number than
\begin{equation}\label{b-chose}
b=\rho+\frac{|c|}{\Gamma(\alpha)}.
\end{equation} 
This contradicts the assumption that $(a,T)$ is the maximal existence interval. 
The proof is complete.
\end{proof}

To finish this article, we give an example that illustrates our theoretical results.
Consider
\begin{equation}\label{ex-2}
\begin{gathered}
{}_{H}D^{3/4}_{e,x}y(x)=x^{2}+\frac{4|y|}{1+|y|}\sin x,\quad
x\in \mathbb{J}=(e,e^{2}]\mbox{ or } (e,+\infty),\\
_{H}D^{-1/4}_{e,x}y(e^{+})=1,
\end{gathered}
\end{equation}
where $\alpha=3/4$, $T=e^2$, $\gamma=1/4$,
$a=e$, $c=1$, and $p=q=2$.

Define $f(x,y)=x^{2}+\frac{4|y|}{1+|y|}\sin x$, $\mu(x)=4$ and
$\omega=e^{2}+4$ and $\nu=0$. Thus $|f(x,y)-f(x,z)|\leq\mu(x) |y-z|$
and $|f(x,y)|\leq \omega$. Then $l:=\mathbb{E}_{\alpha}(0)=1$ 
(see \cite[Lemma 2]{Wang-EPJ}) and
$b=\frac{2}{\Gamma(3/4)}+\frac{(e^{2}+4)(e^2-e)}{\Gamma(7/4)}$
(see \eqref{b-chose}).

Let $h'=h^{*}=e$. Set
$M^{2}=\int_{e}^{2e}(e^{4}+4)^2dx=e(e^{4}+4)^2$ (see \eqref{2-4})
and $g^{2}=\int_{e}^{2e}16dx=16e$ (see \eqref{ineq-2}). Moreover,
one can find $\gamma=1-\alpha= \min\{\alpha-1+\frac{1}{p},\frac{1}{2p}\}$.

$\bullet$ According to Theorem \ref{thm2}, \eqref{ex-2} admits a
unique solution $y\in C_{\frac{1}{4},\ln}(e,e+h]$ where
\begin{align*}
h&=\min\Big\{h',T,\Big[\frac{b\Gamma(\alpha)(p(\alpha-1)+1)}{Ma^{p-1}}
  \Big]^{\frac{1}{p(\alpha-1)+1}}, \\
&\qquad \Big(\frac{\Gamma(\alpha)}{g(B[\frac{1}{2},p(\alpha-1)+1])^{1/p}}
 \Big)^{\frac{p}{p(\alpha-1)+1}}\Big\}\\
\\
&=\Big\{e,e^2, \Big[\frac{(\frac{2}{\Gamma(3/4)}
 +\frac{(e^{2}+4)(e^2-e)}{\Gamma(7/4)})\Gamma(3/4)}
  {2\sqrt{e}(e^4+e)e}\Big]^{2},
 \Big[\frac{\Gamma(3/4)}{4\sqrt{e}\sqrt{B[\frac{1}{2},
  \frac{1}{2}]}}\Big]^4\Big\}.
\end{align*}

$\bullet$ According to Theorem \ref{thm3}, \eqref{ex-2} has a unique
solution $y\in C_{\frac{1}{4},\ln}(e,+\infty)$.

\subsection*{Acknowledgments}
This work is supported by the National Natural Science Foundation of
China (11201091) and by the Outstanding Scientific and Technological
Innovation Talent Award of Education Department of Guizhou Province
([2014]240).

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\end{document}