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

\AtBeginDocument{{\noindent\small
\emph{Electronic Journal of Differential Equations},
Vol. 2009(2009), No. 26, pp. 1--15.\newline
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu
\newline ftp ejde.math.txstate.edu}
\thanks{\copyright 2009 Texas State University - San Marcos.}
\vspace{9mm}}

\begin{document}
\title[\hfilneg EJDE-2009/26\hfil Solutions to boundary-value problems]
{Solutions to boundary-value problems for
 nonlinear differential equations of\\ fractional order}

\author[X. Su, S. Zhang\hfil EJDE-2009/26\hfilneg]
{Xinwei Su, Shuqin Zhang}  % in alphabetical order

\address{Department of Mathematics, China University
of Mining and Technology, Beijing, 100083, China}
\email[Xinwei Su]{kuangdasuxinwei@163.com}
\email[Shuqin Zhang]{zhangshuqin@tsinghua.org.cn}

\thanks{Submitted June 25, 2008. Published February 3, 2009.}
\subjclass[2000]{34B05, 26A33}
\keywords{Boundary value problem; fractional derivative;
fixed-point theorem; \hfill\break\indent
Green's function; existence and uniqueness; continuous dependence}

\begin{abstract}
 we discuss the existence, uniqueness
 and continuous dependence of solutions for a boundary value
 problem of nonlinear fractional differential equation.
\end{abstract}

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

\section{Introduction}

Fractional differential equations have gained considerable
popularity and importance during the past three decades or so, due
mainly to their varied applications in many fields of science and
engineering. Analysis of fractional differential equations has
been carried out by various authors. As for the research in
solutions and also many real applications for factional
differential equations, we refer to the book by Kilbas, Srivastava
and Trujillo \cite{k1} and references therein. Boundary-value
problems for fractional differential equations have been discussed
in \cite{a1,b1,g1,n1,s1,z1,z2}. Bai and L\"u \cite{b1} used
fixed-point theorems on a cone to obtain the existence and
multiplicity of positive solutions for a Dirichlet-type problem of
the nonlinear fractional differential equation
\begin{gather*}
D_{0^{+}}^{\alpha}u(t)+f(t,u(t))=0,\quad 0<t<1,\; 1<\alpha\leq 2,\\
u(0)=u(1)=0,
\end{gather*}
where $f:[0,1]\times[0,\infty)\to[0,\infty)$ is continuous
and $D_{0^{+}}^{\alpha}$ is the fractional derivative in the sense
of Riemann-Liouville. However, as mentioned in \cite{z1}, the
Riemann-Liouville fractional derivative is not suitable for
nonzero boundary values. Therefore, Zhang \cite{z1} investigated the
existence and multiplicity of positive solutions for the problem
\begin{gather*}
\mathbf{D}_{0^{+}}^{\alpha}u(t)=f(t,u(t)),\quad 0<t<1, \; 1<\alpha\leq 2,\\
u(0)+u'(0)=0,\quad u(1)+u'(1)=0,
\end{gather*}
with the Caputo's fractional derivative
$\mathbf{D}_{0^{+}}^{\alpha}$ and a nonnegative continuous
function $f$ on $[0,1]\times [0,\infty)$. The existence of
solutions for the nonlinear fractional differential equation
\begin{gather*}
{_0^CD_{t}^{\delta}}u(t)=g(t,u(t)),\quad 0<t<1,\; 1<\delta<2, \\
u(0)=\alpha\neq0,\quad u(1)=\beta\neq0
\end{gather*}
has been discussed using the  Laplace transform method in
\cite{z2}, where ${_0^CD_{t}^{\delta}}$ denotes the Caputo's
fractional derivative and $g:[0,1]\times \mathbb{R}\to \mathbb{R}$
is a given continuous function. By means of Schauder fixed-point
theorem, Su \cite{s1} proved an existence result for the problem
\begin{gather*}
D^{\alpha}u(t)=f(t,v(t),D^{\mu}v(t)), \quad 0<t<1, \\
D^{\beta}v(t)=g(t,u(t),D^{\nu}u(t)), \quad  0<t<1, \\
u(0)=u(1)=v(0)=v(1)=0,
\end{gather*}
where $1<\alpha$, $\beta<2$, $\mu,\nu>0$, $\alpha-\nu\geq 1$,
$\beta-\mu\geq 1$, $f,g:[0,1]\times\mathbb{R} \times \mathbb{R}\to\mathbb{R}$ are
given functions and $D$ is the standard Riemann-Liouville
differentiation.

Motivated by the previous results, we present in this paper
analysis of a boundary value problem for the fractional
differential equation involving more general boundary conditions
and a nonlinear term dependent on the fractional derivative of the
unknown function
\begin{equation}
\begin{gathered}
{^CD_{0^{+}}^{\alpha}}u(t)=f(t,u(t),{^CD_{0^{+}}^{\beta}}u(t)), \ \ 0<t<1, \\
a_{1}u(0)-a_{2}u'(0)=A,b_{1}u(1)+b_{2}u'(1)=B,
\end{gathered}
\label{e1}
\end{equation}
where $1<\alpha\leq 2$, $0<\beta\leq 1$,
$a_{i},b_{i}\geq 0$, $i=1,2$, $a_{1}b_{1}+a_{1}b_{2} +a_{2}b_{1}>0$,
${^CD_{0^{+}}^{\alpha}}$ and ${^CD_{0^{+}}^{\beta}}$ are the
Caputo's fractional derivatives and $f:[0,1]\times\mathbb{R}\times
\mathbb{R}\to\mathbb{R}$ is continuous. We impose a growth condition on the
function $f$ to prove an existence result  for \eqref{e1}.
For $f$ Lipschitz in the second and third variables,
the uniqueness of solution and the solution's dependence on the
order $\alpha$ of the differential operator, the boundary values
$A$ and $B$, and the nonlinear term  $f$ are also discussed.

Throughout this work, we denote by $I_{0^{+}}^{\alpha}$ and
$D_{0^{+}}^{\alpha}$ the Riemann-Liouville fractional integral and
derivative respectively. The definitions and some properties of
fractional integrals and fractional derivatives of different types
can be found in \cite{k1,p1}. In order to proceed, we recall some
fundamental facts of fractional calculus theory.

\begin{remark} \label{rmk1.1} \rm
If $\alpha=n$ is an integer, the Riemann-Liouville fractional
derivative  of order $\alpha$ is the usual derivative of order
$n$. The following properties are well known:
$I_{0^{+}}^{\alpha}I_{0^{+}}^{\beta}f(t)=I_{0^{+}}^{\alpha+\beta}f(t)$,
$D_{0^{+}}^{\alpha}I_{0^{+}}^{\alpha}f(t)=f(t), \alpha>0, \beta>0,
f\in L^{1}(0,1)$; $I_{0^{+}}^{\alpha}: C[0,1]\to C[0,1]$,
$\alpha>0$.
\end{remark}

\begin{remark} \label{rmk1.2} \rm
For $\alpha=n$, the Caputo's fractional derivative of order
$\alpha$ becomes a conventional $n$-th derivative. The Caputo's
fractional derivative is defined in \cite{k1} as follows:
$^{C}D_{0^{+}}^{\alpha}f(t)
=D_{0^{+}}^{\alpha}(f(t)-\sum_{k=0}^{n-1}\frac{f^{(k)}(0^{+})}{k!}t^{k})$,
provided that the right-side derivative exists. In particular,
${^{C}D_{0^{+}}^{\alpha}}C=0$ for any constant $C\in \mathbb{R}$,
$\alpha>0$. Moreover, we can derive the following useful
properties from \cite[Lemmas 2.21 and 2.22]{k1}:
${^{C}D_{0^{+}}^{\alpha}}I_{0^{+}}^{\alpha}f(t)=f(t), \alpha>0,
f(t)\in C[0,1];
I_{0^{+}}^{\alpha}{^{C}D_{0^{+}}^{\alpha}}f(t)=f(t)-f(0),
0<\alpha\leq1, f(t)\in C[0,1]$.
\end{remark}

Similar composition relation below between
$I_{0^{+}}^{\alpha}$ and ${^{C}D_{0^{+}}^{\alpha}}$ can be found
in \cite[Lemma 2.3]{z1}, but the author did not point out the space
to which $u(t)$ belongs. Besides, the subscript $n$ of the
coefficient $c_{n}$ is wrong.

\begin{lemma} \label{lem1.1}
Assume that $u(t)\in C(0,1)\cap L^{1}(0,1)$ with a derivative of
order $n$ that belongs to $C(0,1)\cap L^{1}(0,1)$. Then
$$
I_{0^{+}}^{\alpha}{^{C}D_{0^{+}}^{\alpha}}u(t)
=u(t)+c_{0}+c_{1}t+c_{2}t^{2}+\dots+c_{n-1}t^{n-1}
$$
for some $c_{i}\in \mathbb{R}$, $i=0,1,2,\dots,n-1$, where $n$ is the
smallest integer greater than or equal to $\alpha$.
\end{lemma}

We can also prove this lemma using \cite[Lemma 2.2]{b1} and
Remark \ref{rmk1.1}.
This proof is  obvious and we omit it here.

\section{Existence and uniqueness results}

In this section, we first impose a growth condition on $f$ which
allows us to establish an existence result of solution, and then
utilize the Lipschitz condition on $f$ to prove a uniqueness
theorem for the problem \eqref{e1}. Our approaches are based on the
fixed-point theorems due to Schauder and Banach.

Let $I=[0,1]$ and $C(I)$ be the space of all continuous real
functions defined on $I$. Define the space $X=\{u(t)\mid u(t)\in
C(I)\mbox{ and } {^{C}D_{0^{+}}^{\beta}}u(t)\in C(I), 0<\beta\leq
1\}$ endowed with the norm
$\|u\|=\max_{t\in I}|u(t)|+\max_{t\in I}|{^{C}D_{0^{+}}^{\beta}}u(t)|$.
Then
by the method in \cite[Lemma 3.2]{s1} and Remark \ref{rmk1.2} we can know that
$(X,\|\cdot\|)$ is a Banach space.

Now we present the Green's function for boundary value problem of
fractional differential equation.

\begin{lemma} \label{lem2.1}
Let $1<\alpha\leq2$. Assume that $g:[0,1]\to \mathbb{R}$ is a
continuous function. Then the unique solution of
\begin{gather*}
^{C}D_{0^{+}}^{\alpha}u(t)=g(t),\ \ 0<t<1, \\
a_{1}u(0)-a_{2}u'(0)=0,b_{1}u(1)+b_{2}u'(1)=0,
\end{gather*}
is $u(t)=\int_{0}^{1}G(t,s)g(s)\mathrm{d}s$, where
$$
G(t,s)=\begin{cases}
\frac{1}{\Gamma(\alpha)}[(t-s)^{\alpha-1}-\frac{a_{2}b_{1}}{l}(1-s)^{\alpha-1}
-\frac{a_{1}b_{1}}{l}(1-s)^{\alpha-1}t]\\
+\frac{1}{\Gamma(\alpha-1)}[-\frac{a_{2}b_{2}}{l}(1-s)^{\alpha-2}
-\frac{a_{1}b_{2}}{l}(1-s)^{\alpha-2}t],
& s\leq t, \\[4pt]
\frac{1}{\Gamma(\alpha)}[-\frac{a_{2}b_{1}}{l}(1-s)^{\alpha-1}
-\frac{a_{1}b_{1}}{l}(1-s)^{\alpha-1}t]\\
+\frac{1}{\Gamma(\alpha-1)}[-\frac{a_{2}b_{2}}{l}(1-s)^{\alpha-2}
-\frac{a_{1}b_{2}}{l}(1-s)^{\alpha-2}t], & t\leq s,
\end{cases}
$$
here $l=a_{1}b_{1}+a_{1}b_{2}+a_{2}b_{1}$.
\end{lemma}

This lemma can be proved using Lemma \ref{lem1.1} and Remark
\ref{rmk1.1}. For details, we refer the reader to \cite[Lemma
3.1]{z1}.

Similarly, we can obtain the solution for the boundary-value
problem with homogeneous equation and nonhomogeneous boundary
conditions.

\begin{lemma} \label{lem2.2}
Let $1<\alpha\leq2$. Then the unique solution of
\begin{gather*}
^{C}D_{0^{+}}^{\alpha}u(t)=0,\quad 0<t<1, \\
a_{1}u(0)-a_{2}u'(0)=A,\quad b_{1}u(1)+b_{2}u'(1)=B,
\end{gather*}
is
$u(t)=\frac{(b_{1}+b_{2})A+a_{2}B}{l}+\frac{a_{1}B-b_{1}A}{l}t$.
\end{lemma}

In the following discussion, we denote
\[
\varphi(t):=\frac{(b_{1}+b_{2})A+a_{2}B}{l}+\frac{a_{1}B-b_{1}A}{l}t,
\]
and use the assumption
\begin{itemize}
\item[(H)] $1<\alpha\leq 2$, $0<\beta\leq 1$, $a_{i},b_{i}\geq 0$,
$i=1,2$, $a_{1}b_{1}+a_{1}b_{2} +a_{2}b_{1}>0$,
$f:[0,1]\times\mathbb{R}\times \mathbb{R}\to\mathbb{R}$ is a given
continuous function.
\end{itemize}

\begin{lemma} \label{lem2.3}
Assume that {\rm (H)} holds. Then  \eqref{e1} is
equivalent to the nonlinear integral equation
\begin{equation}
u(t)=\int_{0}^{1}G(t,s)f(s,u(s),{^{C}D_{0^{+}}^{\beta}}u(s))\mathrm{d}s
+\varphi(t).
\label{e2}
\end{equation}
In other words, every solution of \eqref{e1} is also a solution of
\eqref{e2} and vice versa.
\end{lemma}

\begin{proof}
Let $u\in X$ be a solution of \eqref{e1}, applying the method used to
prove Lemma \ref{lem2.1}, we can obtain that $u$ is a solution of \eqref{e2}.

Conversely, let $u\in X$ be a solution of \eqref{e2}. We denote the
right-hand side of the equation \eqref{e2} by $w(t)$; i.e.,
\begin{align*}
w(t)&= I_{0^{+}}^{\alpha}f(t,u(t),{^{C}D_{0^{+}}^{\beta}}u(t))
+\frac{-a_{2}b_{1}-a_{1}b_{1}t}{l}I_{0^{+}}^{\alpha}f(1,u(1),{^{C}D_{0^{+}}^{\beta}}u(1))\\
&\quad +\frac{-a_{2}b_{2}-a_{1}b_{2}t}{l}I_{0^{+}}^{\alpha-1}f(1,u(1),{^{C}D_{0^{+}}^{\beta}}u(1))
+\varphi(t).
\end{align*}
Using Remarks \ref{rmk1.1} and \ref{rmk1.2}, we have
\begin{align*}
w'(t)
&= D_{0^{+}}^{1}I_{0^{+}}^{1}I_{0^{+}}^{\alpha-1}f(t,u(t),{^{C}D_{0^{+}}^{\beta}}u(t))
-\frac{a_{1}b_{1}}{l}I_{0^{+}}^{\alpha}f(1,u(1),{^{C}D_{0^{+}}^{\beta}}u(1))\\
&\quad
-\frac{a_{1}b_{2}}{l}I_{0^{+}}^{\alpha-1}f(1,u(1),{^{C}D_{0^{+}}^{\beta}}u(1))
+\frac{a_{1}B-b_{1}A}{l}\\
&= I_{0^{+}}^{\alpha-1}f(t,u(t),{^{C}D_{0^{+}}^{\beta}}u(t))-
\frac{a_{1}b_{1}}{l}I_{0^{+}}^{\alpha}f(1,u(1),{^{C}D_{0^{+}}^{\beta}}u(1))\\
&\quad
-\frac{a_{1}b_{2}}{l}I_{0^{+}}^{\alpha-1}f(1,u(1),{^{C}D_{0^{+}}^{\beta}}u(1))
+\frac{a_{1}B-b_{1}A}{l},
\end{align*}
\begin{align*}
{^{C}D_{0^{+}}^{\alpha}}w(t)&= D_{0^{+}}^{\alpha}(w(t)-w(0^{+})-w'(0^{+})t)
=D_{0^{+}}^{\alpha}I_{0^{+}}^{\alpha}f(t,u(t),{^{C}D_{0^{+}}^{\beta}}u(t))\\
&= f(t,u(t),{^{C}D_{0^{+}}^{\beta}}u(t)),
\end{align*}
namely, ${^{C}D_{0^{+}}^{\alpha}}u(t)=f(t,u(t),{^{C}D_{0^{+}}^{\beta}}u(t))$.
One can verify easily that
$a_{1}u(0)-a_{2}u'(0)=A,b_{1}u(1)+b_{2}u'(1)=B$. Therefore, $u$
is a solution of \eqref{e1}, which completes the proof.
\end{proof}

Lemma \ref{lem2.3} indicates that the solution of the problem \eqref{e1} coincides
with the fixed point of the operator $T$ defined as
$$
Tu(t)=\int_{0}^{1}G(t,s)f(s,u(s),{^{C}D_{0^{+}}^{\beta}}u(s))\mathrm{d}s
+\varphi(t).
$$

Now, we give the main results of this section.

\begin{theorem} \label{thm2.1}
Let the assumption {\rm (H)} be satisfied. Suppose further
that
\begin{itemize}
\item[(H1)]
\[
\lim_{(|x|+|y|)\to \infty}\frac{\max_{t\in I}|f(t,x,y)|}{|x|+|y|}
<\frac{l\Gamma(\alpha)\Gamma(2-\beta)}
{(2l+a_{2}b_{2})\Gamma(2-\beta)+2l}=:K.
\]
\end{itemize}
Then there exists at least one solution $u(t)$ to the
boundary-value problem {\rm\eqref{e1}}.
\end{theorem}

\begin{proof}
For any $t\in I$, we find
\begin{align*}
&\int_{0}^{1}|G(t,s)|\mathrm{d}s\\
&\leq \frac{1}{\Gamma(\alpha)}\Big(\int_{0}^{t}(t-s)^{\alpha-1}\mathrm{d}s
 + \frac{a_{2}b_{1}}{l}\int_{0}^{1}(1-s)^{\alpha-1}\mathrm{d}s
 + \frac{a_{1}b_{1}t}{l}\int_{0}^{1}(1-s)^{\alpha-1}\mathrm{d}s\Big) \\
&\quad +\frac{1}{\Gamma(\alpha-1)}\Big(\frac{a_{2}b_{2}}{l}
  \int_{0}^{1}(1-s)^{\alpha-2}\mathrm{d}s
 +\frac{a_{1}b_{2}t}{l}\int_{0}^{1}(1-s)^{\alpha-2}\mathrm{d}s\Big)\\
&= \frac{1}{\Gamma(\alpha+1)}\Big(t^{\alpha}
 +\frac{a_{2}b_{1}+a_{1}b_{1}t}{l}\Big)
 +\frac{1}{\Gamma(\alpha)}\frac{a_{2}b_{2}+a_{1}b_{2}t}{l}\\
&\leq \frac{1}{\Gamma(\alpha)}\Big(1+\frac{a_{2}b_{1}+a_{1}b_{1}}{l}
 +\frac{a_{2}b_{2}+a_{1}b_{2}}{l}\Big)\\
&=\frac{2l+a_{2}b_{2}}{l\Gamma(\alpha)}
\end{align*}
and
\begin{align*}
&\int_{0}^{1}|G'_{t}(t,s)|\mathrm{d}s\\
&\leq \frac{1}{\Gamma(\alpha-1)}\int_{0}^{t}(t-s)^{\alpha-2}\mathrm{d}s
 +\frac{a_{1}b_{1}}{l\Gamma(\alpha)}\int_{0}^{1}(1-s)^{\alpha-1}\mathrm{d}s
 +\frac{a_{1}b_{2}}{l\Gamma(\alpha-1)}\int_{0}^{1}(1-s)^{\alpha-2}\mathrm{d}s\\
&= \frac{t^{\alpha-1}}{\Gamma(\alpha)}+\frac{a_{1}b_{1}}{l\Gamma(\alpha+1)}
+\frac{a_{1}b_{2}}{l\Gamma(\alpha)} \\
&\leq\frac{1}{\Gamma(\alpha)}\Big(1+\frac{a_{1}b_{1}}{l}+
\frac{a_{1}b_{2}}{l}\Big)\leq\frac{2}{\Gamma(\alpha)}.
\end{align*}
Therefore, $|G(t,\cdot)|$ and $|G'_{t}(t,\cdot)|$ are integrable
for any $t\in I$.

Denote $h(x,y)=\max_{t\in I}|f(t,x,y)|$ and Choose
$\varepsilon=\frac{1}{2}(K -\lim_{(|x|+|y|)\to
\infty}\frac{h(x,y)}{|x|+|y|})$. It follows from the condition
 (H1) that there exists a constant $d_{1}>0$ such
that $h(x,y)\leq (K-\varepsilon)(|x|+|y|)$ for $|x|+|y|\geq
d_{1}$. Let $M=\max\{h(x,y): |x|+|y|\leq d_{1}\}$ and choose
$d_{2}>d_{1}$ such that $M/d_{2}\leq K-\varepsilon$. Then we get
$h(x,y)\leq (K-\varepsilon)d_{2}, |x|+|y|\leq d_{2}$. Therefore,
$h(x,y)\leq (K-\varepsilon)c$, $|x|+|y|\leq c$ for any $c\geq
d_{2}$.

Let $k_{1}=\max_{t\in I} |\varphi(t)|$,
$k_{2}=\max_{t\in I}|\varphi'(t)|$,
$k=\max\{k_{1},k_{2}/\Gamma(2-\beta)\}$,
$d_{3}=2Kk/\varepsilon$ and
$d=\max\{d_{2}, d_{3}\}$. Define
$$
U=\{u(t): u(t)\in X,\|u(t)\|\leq d,\; t\in I\}.
$$
Then $U$ is a convex, closed and bounded subset of $X$. Moreover,
for any $u\in U$, $h(u(t),{^{C}D_{0^{+}}^{\beta}}u(t))\leq (K-\varepsilon)d$.

Now we prove that the operator $T$ maps $U$ to itself. For any
$u\in U$, we can get
\begin{align*}
|Tu(t)|
&\leq |\varphi(t)|+\int_{0}^{1}|G(t,s)h(u(s),{^{C}D_{0^{+}}^{\beta}}u(s)
)|\mathrm{d}s \\
&\leq k_{1}+d(K-\varepsilon)\int_{0}^{1}|G(t,s)|\mathrm{d}s \\
&\leq k_{1}+d(K-\varepsilon)\frac{2l+a_{2}b_{2}}{l\Gamma(\alpha)},
\end{align*}
\begin{align*}
&|{^{C}D_{0^{+}}^{\beta}}(Tu)(t)|\\
&= \big|\frac{1}{\Gamma(1-\beta)}
\int_{0}^{t}(t-s)^{-\beta}(Tu)'(s)\mathrm{d}s\big| \\
&\leq \frac{1}{\Gamma(1-\beta)}\int_{0}^{t}(t-s)^{-\beta}
\Big(\int_{0}^{1}|G'_{s}(s,\tau)
f(\tau,u(\tau),{^{C}D_{0^{+}}^{\beta}}u(\tau))|\mathrm{d}\tau
+|\varphi'(s)|\Big)\mathrm{d}s \\
&\leq \frac{1}{\Gamma(1-\beta)}\int_{0}^{t}(t-s)^{-\beta}
\Big(|\varphi'(s)|+\int_{0}^{1}|G'_{s}(s,\tau)h(u(\tau)
,{^{C}D_{0^{+}}^{\beta}}u(\tau))|\mathrm{d}\tau\Big)
\mathrm{d}s\\
&\leq \frac{k_{2}}{\Gamma(1-\beta)}\int_{0}^{t}(t-s)^{-\beta}\mathrm{d}s
 +d(K-\varepsilon) \frac{1}{\Gamma(1-\beta)}\int_{0}^{t}(t-s)^{-\beta}
 \Big(\int_{0}^{1}|G'_{s}(s,\tau)|\mathrm{d}\tau \Big)\mathrm{d}s \\
&\leq \frac{k_{2}}{\Gamma(2-\beta)}+d(K-\varepsilon)
 \frac{2}{\Gamma(2-\beta)\Gamma(\alpha)}
\end{align*}
for $0<\beta<1$, and
\begin{align*}
|(Tu)'(t)|
&= \big| \int_{0}^{1}G'_{t}(t,s)
f(s,u(s),{^{C}D_{0^{+}}^{\beta}}u(s))\mathrm{d}s+\varphi'(t)\big| \\
&\leq |\varphi'(t)|+
\int_{0}^{1}|G'_{t}(t,s)h(u(s),{^{C}D_{0^{+}}^{\beta}}u(s))|\mathrm{d}s
\\
&\leq k_{2}+d(K-\varepsilon) \int_{0}^{1}|G'_{t}(t,s)|\mathrm{d}s
\leq k_{2}+d(K-\varepsilon) \frac{2}{\Gamma(\alpha)}
\end{align*}
for $\beta=1$. Hence,
\[
\|Tu\|\leq 2k+d(K-\varepsilon)\frac{(2l+a_{2}b_{2})\Gamma(2-\beta)+2l}
{l\Gamma(\alpha)\Gamma(2-\beta)} \leq
d\frac{\varepsilon}{K}+d(K-\varepsilon)\frac{1}{K}=d.
\]
Note also that
\begin{align*}
Tu(t)&= I_{0^{+}}^{\alpha}f(t,u(t),{^{C}D_{0^{+}}^{\beta}}u(t))
+\frac{-a_{2}b_{1}-a_{1}b_{1}t}{l}I_{0^{+}}^{\alpha}f(1,u(1),{^{C}D_{0^{+}}^{\beta}}u(1))\\
&\quad +\frac{-a_{2}b_{2}-a_{1}b_{2}t}{l}I_{0^{+}}^{\alpha-1}f(1,u(1),{^{C}D_{0^{+}}^{\beta}}u(1))
+\varphi(t),
\end{align*}
\begin{align*}
(Tu)'(t)&= I_{0^{+}}^{\alpha-1}f(t,u(t),{^{C}D_{0^{+}}^{\beta}}u(t))-
\frac{a_{1}b_{1}}{l}I_{0^{+}}^{\alpha}f(1,u(1),{^{C}D_{0^{+}}^{\beta}}u(1))\\
&\quad
-\frac{a_{1}b_{2}}{l}I_{0^{+}}^{\alpha-1}f(1,u(1),{^{C}D_{0^{+}}^{\beta}}u(1))
+\frac{a_{1}B-b_{1}A}{l},
\end{align*}
\begin{align*}
{^{C}D_{0^{+}}^{\beta}}(Tu)(t)&= I_{0^{+}}^{1-\beta}(Tu)'(t)\\
&= I_{0^{+}}^{\alpha-\beta}f(t,u(t),{^{C}D_{0^{+}}^{\beta}}u(t))-
\frac{a_{1}b_{1}}{l}I_{0^{+}}^{\alpha-\beta+1}f(1,u(1),{^{C}D_{0^{+}}^{\beta}}u(1)) \\
&\quad
-\frac{a_{1}b_{2}}{l}I_{0^{+}}^{\alpha-\beta}f(1,u(1),{^{C}D_{0^{+}}^{\beta}}u(1))
+I_{0^{+}}^{1-\beta}\frac{a_{1}B-b_{1}A}{l}.
\end{align*}
It is easy to see $Tu(t), {^{C}D_{0^{+}}^{\beta}}(Tu)(t)\in C(I)$.
Therefore, $T:U\to U$.

Claim: $T$ is a continuous operator. In fact, for $u_{n}$, $n=0,1,2,\dots$
and $u\in U$ such that $\lim_{n\to \infty}\|u_{n}-u\|\to 0$, we have
\begin{align*}
&|Tu_{n}(t)-Tu(t)|\\
& =\big| \int_{0}^{1}G(t,s)\left(
f(s,u_{n}(s),{^{C}D_{0^{+}}^{\beta}}u_{n}(s))
-f(s,u(s),{^{C}D_{0^{+}}^{\beta}}u(s))\right) \mathrm{d}s \big| \\
&\leq \max_{t\in I}|f(t,u_{n}(t),{^{C}D_{0^{+}}^{\beta}}u_{n}(t))
-f(t,u(t),{^{C}D_{0^{+}}^{\beta}}u(t))|
\int_{0}^{1}|G(t,s)|\mathrm{d}s \\
&\leq \frac{2l+a_{2}b_{2}}{l\Gamma(\alpha)}\max_{t\in I}|
f(t,u_{n}(t),{^{C}D_{0^{+}}^{\beta}}u_{n}(t))
-f(t,u(t),{^{C}D_{0^{+}}^{\beta}}u(t))|,
\end{align*}
\begin{align*}
&|{^{C}D_{0^{+}}^{\beta}}(Tu_{n})(t)-{^{C}D_{0^{+}}^{\beta}}(Tu)(t)|\\
&=\big| \frac{1}{\Gamma(1-\beta)}
\int_{0}^{t}(t-s)^{-\beta}((Tu_{n})'(s)-(Tu)'(s))\mathrm{d}s\big| \\
&\leq \frac{1}{\Gamma(1-\beta)}\int_{0}^{t}(t-s)^{-\beta}
 \Big(\int_{0}^{1}\big| G'_{s}(s,\tau) \Big(
f(\tau,u_{n}(\tau),{^{C}D_{0^{+}}^{\beta}}u_{n}(\tau))\\
&\quad -f(\tau,u(\tau),{^{C}D_{0^{+}}^{\beta}}u(\tau))\Big)\big|\mathrm{d}\tau\Big)
\mathrm{d}s \\
&\leq \max_{t\in I} |f(t,u_{n}(t),{^{C}D_{0^{+}}^{\beta}}u_{n}(t))
-f(t,u(t),{^{C}D_{0^{+}}^{\beta}}u(t))| \\
&\quad\times \frac{1}{\Gamma(1-\beta)}\int_{0}^{t}(t-s)^{-\beta}
\Big( \int_{0}^{1}|G'_{s}(s,\tau)|\mathrm{d}\tau\Big)\mathrm{d}s \\
&\leq \max_{t\in I}| f(t,u_{n}(t),{^{C}D_{0^{+}}^{\beta}}u_{n}(t))
-f(t,u(t),{^{C}D_{0^{+}}^{\beta}}u(t))|
 \frac{2}{\Gamma(1-\beta)\Gamma(\alpha)}\int_{0}^{t}(t-s)^{-\beta}
\mathrm{d}s \\
&\leq \frac{2}{\Gamma(2-\beta)\Gamma(\alpha)}\max_{t\in I}|
f(t,u_{n}(t),{^{C}D_{0^{+}}^{\beta}}u_{n}(t))
-f(t,u(t),{^{C}D_{0^{+}}^{\beta}}u(t))|
\end{align*}
for $0<\beta<1$, and
\begin{align*}
&|(Tu_{n})'(t)-(Tu)'(t)| \\
&=\big| \int_{0}^{1}G'_{t}(t,s)
\left(f(s,u_{n}(s),{^{C}D_{0^{+}}^{\beta}}u_{n}(s))
-f(s,u(s),{^{C}D_{0^{+}}^{\beta}}u(s))\right) \mathrm{d}s \big| \\
&\leq \max_{t\in I}|f(t,u_{n}(t),{^{C}D_{0^{+}}^{\beta}}u_{n}(t))
-f(t,u(t),{^{C}D_{0^{+}}^{\beta}}u(t))|
\int_{0}^{1}|G'_{t}(t,s)|\mathrm{d}s \\
&\leq \frac{2}{\Gamma(\alpha)}\max_{t\in
I}|f(t,u_{n}(t),{^{C}D_{0^{+}}^{\beta}}u_{n}(t))
-f(t,u(t),{^{C}D_{0^{+}}^{\beta}}u(t))|
\end{align*}
for $\beta=1$. Then in view of the uniform continuity of the
function $f$ on $I\times [-d,d]\times [-d,d]$, we obtain that $T$
is continuous.
\par
The last step is to prove that $T$ is a completely continuous
operator. Let $t, \tau\in I$ be such that $t<\tau$ and
$N=\max_{t\in I,u\in
U}|f(t,u(t),{^{C}D_{0^{+}}^{\beta}}u(t))|+1$. Then we have
\begin{align*}
&|Tu(t)-Tu(\tau)|\\
&=\Big|
\int_{0}^{1}(G(t,s)-G(\tau,s))f(s,u(s),{^{C}D_{0^{+}}^{\beta}}u(s))\mathrm{d}s+\varphi(t)
-\varphi(\tau)\Big| \\
&\leq N\Big(
\int_{0}^{t}|G(t,s)-G(\tau,s)|\mathrm{d}s+\int_{t}^{\tau}
|G(t,s)-G(\tau,s)|\mathrm{d}s \\
&\quad +\int_{\tau}^{1}|G(t,s)-G(\tau,s)|\mathrm{d}s\Big)
+|\varphi(t)-\varphi(\tau)|\\
&\leq N\Big[\int_{0}^{t}\Big(\frac{(\tau-s)^{\alpha-1}
 -(t-s)^{\alpha-1}}{\Gamma(\alpha)}
+(\tau-t)\Big(\frac{a_{1}b_{1}}{l}\frac{(1-s)^{\alpha-1}}{\Gamma(\alpha)}\\
&\quad + \frac{a_{1}b_{2}}{l}\frac{(1-s)^{\alpha-2}}{\Gamma(\alpha-1)}
 \Big)\Big)\mathrm{d}s\\
&\quad +\int_{t}^{\tau}\Big(\frac{(\tau-s)^{\alpha-1}}{\Gamma(\alpha)}
+(\tau-t)\Big(\frac{a_{1}b_{1}}{l}\frac{(1-s)^{\alpha-1}}{\Gamma(\alpha)}
+\frac{a_{1}b_{2}}{l}\frac{(1-s)^{\alpha-2}}{\Gamma(\alpha-1)}\Big)\Big)
 \mathrm{d}s\\
&\quad +\int_{\tau}^{1}(\tau-t)\Big(\frac{a_{1}b_{1}}{l}
 \frac{(1-s)^{\alpha-1}}{\Gamma(\alpha)}
+\frac{a_{1}b_{2}}{l}\frac{(1-s)^{\alpha-2}}{\Gamma(\alpha-1)}\Big)
\mathrm{d}s\Big]+|\varphi(t)-\varphi(\tau)|\\
&= N\Big[\int_{0}^{\tau}\frac{(\tau-s)^{\alpha-1}}{\Gamma(\alpha)}\mathrm{d}s
-\int_{0}^{t}\frac{(t-s)^{\alpha-1}}{\Gamma(\alpha)}\mathrm{d}s
+(\tau-t)\Big(\frac{a_{1}b_{1}}{l}\int_{0}^{1}
 \frac{(1-s)^{\alpha-1}}{\Gamma(\alpha)}\mathrm{d}s \\
&\quad +\frac{a_{1}b_{2}}{l}\int_{0}^{1}\frac{(1-s)^{\alpha-2}}{\Gamma(\alpha-1)}
\mathrm{d}s \Big)\Big]
+(\tau-t)\frac{|a_{1}B-b_{1}A|}{l}\\
&= N\Big[\frac{\tau^{\alpha}-t^{\alpha}}{\Gamma(\alpha+1)}
+(\tau-t)\Big(\frac{a_{1}b_{1}}{l\Gamma(\alpha+1)}
+\frac{a_{1}b_{2}}{l\Gamma(\alpha)}\Big)\Big]
+(\tau-t)\frac{|a_{1}B-b_{1}A|}{l},
\end{align*}
\begin{align*}
&|{^{C}D_{0^{+}}^{\beta}}(Tu)(t)-{^{C}D_{0^{+}}^{\beta}}(Tu)(\tau)| \\
&= \frac{1}{\Gamma(1-\beta)}\big|
\int_{0}^{t}(t-s)^{-\beta}\Big(\int_{0}^{1}G'_{s}(s,\theta)
f(\theta,u(\theta),{^{C}D_{0^{+}}^{\beta}}u(\theta))\mathrm{d}\theta
 +\varphi'(s)\Big) \mathrm{d}s  \\
&\quad  -\int_{0}^{\tau}(\tau-s)^{-\beta}\Big(
\int_{0}^{1}G'_{s}(s,\theta)
f(\theta,u(\theta),{^{C}D_{0^{+}}^{\beta}}u(\theta))\mathrm{d}\theta
+\varphi'(s)\Big) \mathrm{d}s \big|
\\
&\leq \frac{1}{\Gamma(1-\beta)}\big|
\int_{0}^{t}(t-s)^{-\beta}\Big(\int_{0}^{1}G'_{s}(s,\theta)
f(\theta,u(\theta),{^{C}D_{0^{+}}^{\beta}}u(\theta))\mathrm{d}\theta\Big)
\mathrm{d}s  \\
&\quad  - \int_{0}^{t}(\tau-s)^{-\beta}\Big(
\int_{0}^{1}G'_{s}(s,\theta)
f(\theta,u(\theta),{^{C}D_{0^{+}}^{\beta}}u(\theta))\mathrm{d}\theta\Big)
\mathrm{d}s \big| \\
&\quad +\frac{1}{\Gamma(1-\beta)}\big|
\int_{0}^{t}(\tau-s)^{-\beta}\Big( \int_{0}^{1}G'_{s}(s,\theta)
f(\theta,u(\theta),{^{C}D_{0^{+}}^{\beta}}u(\theta))\mathrm{d}\theta\Big)
\mathrm{d}s \\
&\quad -\int_{0}^{\tau}(\tau-s)^{-\beta}\Big(
\int_{0}^{1}G'_{s}(s,\theta)
f(\theta,u(\theta),{^{C}D_{0^{+}}^{\beta}}u(\theta))\mathrm{d}\theta\Big)
\mathrm{d}s \big|\\
&\quad +\frac{1}{\Gamma(1-\beta)}\big|
\int_{0}^{t}(t-s)^{-\beta}\varphi'(s)\mathrm{d}s-
\int_{0}^{t}(\tau-s)^{-\beta}\varphi'(s)\mathrm{d}s\big|\\
&\quad +\frac{1}{\Gamma(1-\beta)}\big|\int_{0}^{t}(\tau-s)^{-\beta}\varphi'(s)\mathrm{d}s-
\int_{0}^{\tau}(\tau-s)^{-\beta}\varphi'(s)\mathrm{d}s\big|\\
&\leq \frac{2N}{\Gamma(1-\beta)\Gamma(\alpha)} \int_{0}^{t}(
(t-s)^{-\beta}-(\tau-s)^{-\beta}) \mathrm{d}s
+\frac{2N}{\Gamma(1-\beta)\Gamma(\alpha)}
\int_{t}^{\tau}(\tau-s)^{-\beta}\mathrm{d}s \\
&\quad +\frac{|a_{1}B-b_{1}A|}{l\Gamma(1-\beta)}\int_{0}^{t}(
(t-s)^{-\beta}-(\tau-s)^{-\beta})
\mathrm{d}s+\frac{|a_{1}B-b_{1}A|}{l\Gamma(1-\beta)}
\int_{t}^{\tau}(\tau-s)^{-\beta}\mathrm{d}s \\
&\leq \Big(\frac{2N}{\Gamma(2-\beta)\Gamma(\alpha)}
+\frac{|a_{1}B-b_{1}A|}{l\Gamma(2-\beta)}\Big)(
\tau^{1-\beta}-t^{1-\beta}+2(\tau-t)^{1-\beta})
\end{align*}
for $0<\beta<1$, and
\begin{align*}
&|(Tu)'(t)-(Tu)'(\tau)| \\
&=\big| \int_{0}^{1}G'_{t}(t,s)
f(s,u(s),{^{C}D_{0^{+}}^{\beta}}u(s))\mathrm{d}s+\varphi'(t) \\
&\quad -\int_{0}^{1}G'_{\tau}(\tau,s)
f(s,u(s),{^{C}D_{0^{+}}^{\beta}}u(s))\mathrm{d}s-\varphi'(\tau) \big| \\
&\leq \frac{N}{\Gamma(\alpha-1)}\Big( \int_{0}^{t}(
(t-s)^{\alpha-2} -(\tau-s)^{\alpha-2})\mathrm{d}s
+\int_{t}^{\tau}(\tau-s)^{\alpha-2}\mathrm{d}s \Big) \\
&\leq \frac{N}{\Gamma(\alpha)}(\tau^{\alpha-1}-t^{\alpha-1}
 +2(\tau-t)^{\alpha-1})
\end{align*}
for $\beta=1$.

Now, using the fact that the functions
$\tau^{\alpha}-t^{\alpha},\tau^{\alpha-1}-t^{\alpha-1}$ and
$\tau^{1-\beta}-t^{1-\beta}$  are uniformly continuous on the
interval $I$, we conclude that $TU$ is an equicontinuous set.
Obviously it is uniformly bounded since $TU\subseteq U$. Thus, $T$
is completely continuous. The Schauder fixed-point theorem asserts
the existence of solution in $U$ for the problem \eqref{e1} and the
theorem is proved.
\end{proof}

The following corollary is obvious.

\begin{corollary} \label{coro2.1}
Let the assumption {\rm (H)} be satisfied. Suppose further
that there exist two nonnegative functions $a(t),b(t)\in
C[0,1]$ such that $|f(t,x,y)|\leq a(t)|x|^{\rho}+b(t)|y|^{\theta}$,
where $0<\rho, \theta<1$. Then there exists at least one solution
for the boundary value problem {\rm\eqref{e1}}.
\end{corollary}

\begin{example} \label{exa2.1} \rm
Consider the problem
\begin{gather*}
^CD_{0^{+}}^{3/2}u=(t-\frac{1}{2})^{3}(u(t)
 +{^CD_{0^{+}}^{1/2}}u(t)), \quad 0<t<1,\\
a_{1}u(0)-a_{2}u'(0)=A,b_{1}u(1)+b_{2}u'(1)=B.
\end{gather*}
Using $\Gamma(1/2)=\sqrt{\pi}$, a simple
computation shows $K =(l\pi)/(2(2l+a_{2}b_{2})\sqrt{\pi}+8l)$.
Since
$|f(t,x,y)|=|t-\frac{1}{2}|^{3}|x+y|\leq\frac{|x|+|y|}{8}$,
\[
\lim_{(|x|+|y|)\to\infty}\frac{\frac{|x|+|y|}{8}}{|x|+|y|}=\frac{1}{8},
\]
then, if $1/8<(l\pi)/(2(2l+a_{2}b_{2})\sqrt{\pi}+8l)$
(for example, we choose $a_{1}=b_{2}=1,a_{2}=b_{1}=0$, then
$K\approx0.2082>1/8$), Theorem \ref{thm2.1} ensures the existence
of solution for this problem.
\end{example}

\begin{theorem} \label{thm2.2}
Let the assumption {\rm (H)} be satisfied. Furthermore, let
the function $f$ fulfill a Lipschitz condition with respect to the
second and third variables; i.e., $|f(t,x,y) -f(t,u,v)|\leq
L(|x-u|+|v-y|)$ with a Lipschitz constant $L$ such that $0<L<K$,
where $K$ is as the same as that in Theorem  \ref{thm2.1}. Then the
boundary value problem \eqref{e1} has a unique solution $u(t)\in X$.
\end{theorem}

\begin{proof}
We have shown in Theorem \ref{thm2.1} that
$Tu(t),{^{C}D_{0^{+}}^{\beta}}(Tu)(t)\in C(I)$; i.e., $T:X\to X$.
To apply the Banach fixed-point theorem, we need to verify that
$T$ is a contraction mapping. For any $u,v\in X$, we have
\begin{align*}
&|Tu(t)-Tv(t)|\\
& =\big| \int_{0}^{1}G(t,s)\left(
f(s,u(s),{^{C}D_{0^{+}}^{\beta}}u(s))
-f(s,v(s),{^{C}D_{0^{+}}^{\beta}}v(s))\right) \mathrm{d}s \big| \\
&\leq L\|u-v\|\int_{0}^{1}|G(t,s)|\mathrm{d}s\\
&\leq\frac{(2l+a_{2}b_{2})L}{l\Gamma(\alpha)}\|u-v\|,
\end{align*}
\begin{align*}
&|{^{C}D_{0^{+}}^{\beta}}(Tu)(t)-{^{C}D_{0^{+}}^{\beta}}(Tv)(t)|\\
&=\big| \frac{1}{\Gamma(1-\beta)}
\int_{0}^{t}(t-s)^{-\beta}((Tu)'(s)-(Tv)'(s))\mathrm{d}s\big| \\
&\leq \frac{1}{\Gamma(1-\beta)}\int_{0}^{t}(t-s)^{-\beta}
\Big(\int_{0}^{1}\big| G'_{s}(s,\tau) \big(
f(\tau,u(\tau),{^{C}D_{0^{+}}^{\beta}}u(\tau))\\
&\quad -f(\tau,v(\tau),{^{C}D_{0^{+}}^{\beta}}v(\tau))\big)
 \big|\mathrm{d}\tau\Big)\mathrm{d}s \\
&\leq  L\|u-v\|\frac{1}{\Gamma(1-\beta)}\int_{0}^{t}(t-s)^{-\beta}
\Big(\int_{0}^{1}|G'_{s}(s,\tau)|\mathrm{d}\tau\Big)\mathrm{d}s \\
&\leq  \frac{2L}{\Gamma(2-\beta)\Gamma(\alpha)}\|u-v\|
\end{align*}
for $0<\beta<1$, and
\begin{align*}
&|(Tu)'(t)-(Tv)'(t)| \\
&=\big| \int_{0}^{1}G'_{t}(t,s)
\left(f(s,u(s),{^{C}D_{0^{+}}^{\beta}}u(s))
-f(s,v(s),{^{C}D_{0^{+}}^{\beta}}v(s))\right) \mathrm{d}s \big| \\
&\leq L\|u-v\|\int_{0}^{1}|G'_{t}(t,s)|\mathrm{d}s
\leq\frac{2L}{\Gamma(\alpha)}\|u-v\|
\end{align*}
for $\beta=1$. Thus, $\|Tu-Tv\|\leq
L(\frac{2l+a_{2}b_{2}}{l\Gamma(\alpha)}
+\frac{2}{\Gamma(2-\beta)\Gamma(\alpha)})\|u-v\|
=\frac{L}{K}\|u-v\|$. Hence, the Banach fixed-point theorem yields
that $T$ has a unique fixed point which is the unique solution of
the problem \eqref{e1}. The proof is therefore complete.
\end{proof}

\section{Dependence on the parameters}

The present section is devoted to the study of the dependence of
solution on the parameters $\alpha$, $A$ and $B$, and $f$ for the
problem \eqref{e1}, provided that the function $f(t,x,y)$ is Lipschitz
with respect to $x$ and $y$.

\begin{theorem} \label{thm3.1}
Suppose that the conditions of Theorem \ref{thm2.2} hold. Let
$u_{1}(t), u_{2}(t)$ be the solutions, respectively, of the
problems {\rm \eqref{e1}} and
\begin{equation}
\begin{gathered}
{^CD_{0^{+}}^{\alpha-\varepsilon}}u(t)= f(t,u(t),{^CD_{0^{+}}^{\beta}}u(t)),
\quad 0<t<1,\\
a_{1}u(0)-a_{2}u'(0)=A, \quad b_{1}u(1)+b_{2}u'(1)=B,
\end{gathered}
\label{e3}
\end{equation}
where $1<\alpha-\varepsilon<\alpha\leq 2$. Then
$\|u_{1}-u_{2}\|=\textsl{O}(\varepsilon)$.
\end{theorem}

\begin{proof}
Let $G_{1}(t,s)=G(t,s)$ and
 $$
G_{2}(t,s)=\begin{cases}
\frac{1}{\Gamma(\alpha-\varepsilon)}[(t-s)^{\alpha-\varepsilon-1}-\frac{a_{2}b_{1}}{l}(1-s)^{\alpha-\varepsilon-1}
-\frac{a_{1}b_{1}}{l}(1-s)^{\alpha-\varepsilon-1}t]\\
+\frac{1}{\Gamma(\alpha-\varepsilon-1)}[-\frac{a_{2}b_{2}}{l}(1-s)^{\alpha-\varepsilon-2}
-\frac{a_{1}b_{2}}{l}(1-s)^{\alpha-\varepsilon-2}t],
& s\leq t, \\[4pt]
\frac{1}{\Gamma(\alpha-\varepsilon)}[-\frac{a_{2}b_{1}}{l}(1-s)^{\alpha-\varepsilon-1}
-\frac{a_{1}b_{1}}{l}(1-s)^{\alpha-\varepsilon-1}t]\\
+\frac{1}{\Gamma(\alpha-\varepsilon-1)}[-\frac{a_{2}b_{2}}{l}(1-s)^{\alpha-\varepsilon-2}
-\frac{a_{1}b_{2}}{l}(1-s)^{\alpha-\varepsilon-2}t], & t\leq s,
\end{cases}
$$
be the Green's function of \eqref{e3}. Then
\begin{gather*}
u_{1}(t)=\int_{0}^{1}G_{1}(t,s)f(t,u_{1}(s),{^{C}D_{0^{+}}^{\beta}}u_{1}(s))
  \mathrm{d}s+\varphi(t),\\
u_{2}(t)=\int_{0}^{1}G_{2}(t,s)f(t,u_{2}(s),{^{C}D_{0^{+}}^{\beta}}u_{2}(s))
  \mathrm{d}s+\varphi(t).
\end{gather*}
First we show that
\begin{equation}
\int_{0}^{1}|G_{1}(t,s)-G_{2}(t,s)|\mathrm{d}s=\textsl{O}(\varepsilon),\quad
\int_{0}^{1}|G'_{1t}(t,s)-G'_{2t}(t,s)|\mathrm{d}s=\textsl{O}(\varepsilon).
\label{e4}
\end{equation}
Observing that
\begin{align*}
\int_{0}^{1}|G_{1}(t,s)-G_{2}(t,s)|\mathrm{d}s
&\leq\int_{0}^{t}\big| \frac{(t-s)^{\alpha-1}}{\Gamma(\alpha)}
-\frac{(t-s)^{\alpha-\varepsilon-1}}{\Gamma(\alpha-\varepsilon)}\big|
 \mathrm{d}s\\
&\quad +(\frac{a_{2}b_{1}}{l}+\frac{a_{1}b_{1}t}{l})
\int_{0}^{1}\big|\frac{(1-s)^{\alpha-1}}{\Gamma(\alpha)}
-\frac{(1-s)^{\alpha-\varepsilon-1}}{\Gamma(\alpha-\varepsilon)}\big|
\mathrm{d}s \\
&\quad +(\frac{a_{2}b_{2}}{l}+\frac{a_{1}b_{2}t}{l})
\int_{0}^{1}\big|\frac{(1-s)^{\alpha-2}}{\Gamma(\alpha-1)}
-\frac{(1-s)^{\alpha-\varepsilon-2}}{\Gamma(\alpha-\varepsilon-1)}\big|
\mathrm{d}s
\end{align*}
and
\begin{align*}
&\int_{0}^{1}|G'_{1t}(t,s)-G'_{2t}(t,s)|\mathrm{d}s\\
&\leq\int_{0}^{t}\big| \frac{(t-s)^{\alpha-2}}{\Gamma(\alpha-1)}
 -\frac{(t-s)^{\alpha-\varepsilon-2}}{\Gamma(\alpha-\varepsilon-1)}\big|
 \mathrm{d}s
+\frac{a_{1}b_{1}}{l}\int_{0}^{1}\big|
\frac{(1-s)^{\alpha-1}}{\Gamma(\alpha)}
-\frac{(1-s)^{\alpha-\varepsilon-1}}{\Gamma(\alpha-\varepsilon)}\big|
\mathrm{d}s \\
&\quad +\frac{a_{1}b_{2}}{l}\int_{0}^{1}\big|
\frac{(1-s)^{\alpha-2}}{\Gamma(\alpha-1)}-\frac{(1-s)^{\alpha-\varepsilon-2}}
{\Gamma(\alpha-\varepsilon-1)}\big|\mathrm{d}s,
\end{align*}
without loss of generality, we only estimate one of the integrals
in the right-hand side of the two inequalities above.
\begin{align*}
&\int_{0}^{t}\big|\frac{(t-s)^{\alpha-1}}{\Gamma(\alpha)}-
 \frac{(t-s)^{\alpha-\varepsilon-1}}{\Gamma(\alpha-\varepsilon)}\big|
\mathrm{d}s \\
&\leq \int_{0}^{t}\big|\frac{(t-s)^{\alpha-1}}{\Gamma(\alpha)}-
\frac{(t-s)^{\alpha-\varepsilon-1}}{\Gamma(\alpha)}\big|\mathrm{d}s
+\int_{0}^{t}\big|\frac{(t-s)^{\alpha-\varepsilon-1}}{\Gamma(\alpha)}
-\frac{(t-s)^{\alpha-\varepsilon-1}}{\Gamma(\alpha-\varepsilon)}\big|
\mathrm{d}s\\
&= \frac{1}{\Gamma(\alpha)}\int_{0}^{t}\big|x^{\alpha-1}-
x^{\alpha-\varepsilon-1}\big|\mathrm{d}x+
\big|\frac{1}{\Gamma(\alpha)}-\frac{1}{\Gamma(\alpha-\varepsilon)}\big|
\int_{0}^{t}(t-s)^{\alpha-\varepsilon-1}\mathrm{d}s\\
&\leq \frac{1}{\Gamma(\alpha)}\big(\frac{1}{\alpha-\varepsilon}
-\frac{1}{\alpha}\big)
+\big|\frac{1}{\Gamma(\alpha)}-\frac{1}{\Gamma(\alpha-\varepsilon)}\big|
\frac{1}{\alpha-\varepsilon}\\
&= \varepsilon\Big(\frac{1}{\alpha(\alpha-\varepsilon)\Gamma(\alpha)}
+\frac{|\Gamma'(\alpha-\varepsilon+\theta\varepsilon)|}{(\alpha
-\varepsilon)\Gamma(\alpha)\Gamma(\alpha-\varepsilon)}\Big),
\end{align*}
for some $\theta$ such that $0<\theta<1$. So we arrive at the
relations in \eqref{e4}.
Furthermore,
\begin{align*}
&|u_{1}(t)-u_{2}(t)|\\
&= \big|\int_{0}^{1}G_{1}(t,s)f(t,u_{1}(s),{^{C}D_{0^{+}}^{\beta}}u_{1}(s))
 \mathrm{d}s
-\int_{0}^{1}G_{2}(t,s)f(t,u_{2}(s),{^{C}D_{0^{+}}^{\beta}}u_{2}(s))
\mathrm{d}s\big| \\
&\leq \int_{0}^{1}\big|G_{1}(t,s)(f(t,u_{1}(s),{^{C}D_{0^{+}}^{\beta}}
 u_{1}(s))\mathrm{d}s
-f(t,u_{2}(s),{^{C}D_{0^{+}}^{\beta}}u_{2}(s))\big|\mathrm{d}s\\
&\quad +\int_{0}^{1}\big|G_{1}(t,s)f(t,u_{2}(s),{^{C}D_{0^{+}}^{\beta}}
 u_{2}(s))\mathrm{d}s
-G_{2}(t,s)f(t,u_{2}(s),{^{C}D_{0^{+}}^{\beta}}u_{2}(s))\big|\mathrm{d}s\\
&\leq L\|u_{1}-u_{2}\|\int_{0}^{1}|G_{1}(t,s)|\mathrm{d}s+|\|f\||\int_{0}^{1}|G_{1}(t,s)-G_{2}(t,s)|\mathrm{d}s\\
&\leq \frac{(2l+a_{2}b_{2})L}{l\Gamma(\alpha)}\|u_{1}-u_{2}\|+|\|f\||\int_{0}^{1}|G_{1}(t,s)-G_{2}(t,s)|\mathrm{d}s,
\end{align*}
where
$|\|f\||=\sup_{0<\varepsilon<\alpha-1}\{\max_{t\in
I}|f(t,u_{2}(t),{^{C}D_{0^{+}}^{\beta}}u_{2}(t))|\}$.
\begin{align*}
&|{^{C}D_{0^{+}}^{\beta}}u_{1}(t)-{^{C}D_{0^{+}}^{\beta}}u_{2}(t)|\\
&\leq\frac{1}{\Gamma(1-\beta)}\int_{0}^{t}(t-s)^{-\beta}
 \Big(\int_{0}^{1}\big| G'_{1s}(s,\tau)
 f(\tau,u_{1}(\tau),{^{C}D_{0^{+}}^{\beta}}u_{1}(\tau)) \\
&\quad -  G'_{2s}(s,\tau)f(\tau,u_{2}(\tau),{^{C}D_{0^{+}}^{\beta}}u_{2}
 (\tau))\big|\mathrm{d}\tau\Big)\mathrm{d}s \\
&\leq \frac{1}{\Gamma(1-\beta)}\int_{0}^{t}(t-s)^{-\beta}
\Big(\int_{0}^{1}\big| G'_{1s}(s,\tau)
f(\tau,u_{1}(\tau),{^{C}D_{0^{+}}^{\beta}}u_{1}(\tau)) \\
&\quad- G'_{1s}(s,\tau)f(\tau,u_{2}(\tau),{^{C}D_{0^{+}}^{\beta}}u_{2}
(\tau))\big|\mathrm{d}\tau\\
&\quad +\int_{0}^{1}\!\big| G'_{1s}(s,\tau)
f(\tau,u_{2}(\tau),{^{C}D_{0^{+}}^{\beta}}u_{2}(\tau))
-G'_{2s}(s,\tau)f(\tau,u_{2}(\tau),{^{C}D_{0^{+}}^{\beta}}u_{2}(\tau))\big|
\mathrm{d}\tau\Big)\mathrm{d}s
\\
&\leq L\|u_{1}-u_{2}\|\frac{1}{\Gamma(1-\beta)}\int_{0}^{t}(t-s)^{-\beta}
\Big(\int_{0}^{1}|G'_{1s}(s,\tau)|\mathrm{d}\tau\Big)\mathrm{d}s \\
&\quad +|\|f\||\frac{1}{\Gamma(1-\beta)}\int_{0}^{t}(t-s)^{-\beta}
\Big(\int_{0}^{1}|G'_{1s}(s,\tau)-G'_{2s}(s,\tau)|\mathrm{d}\tau\Big)
 \mathrm{d}s \\
&\leq  \frac{2L}{\Gamma(2-\beta)\Gamma(\alpha)}\|u_{1}-v_{2}\|\\
&\quad +|\|f\||\frac{1}{\Gamma(1-\beta)}\int_{0}^{t}(t-s)^{-\beta}
\Big(\int_{0}^{1}|G'_{1s}(s,\tau)-G'_{2s}(s,\tau)|\mathrm{d}\tau\Big)
\mathrm{d}s
\end{align*}
for $0<\beta<1$, and
\begin{align*}
|u'_{1}(t)-u'_{2}(t)|
&\leq L\|u_{1}-u_{2}\|\int_{0}^{1}|G'_{1t}(t,s)|\mathrm{d}s +|\|f\||
\int_{0}^{1}|G'_{1t}(t,s)-G'_{2t}(t,s)|\mathrm{d}s \\
&\leq \frac{2L}{\Gamma(\alpha)}\|u_{1}-u_{2}\|+|\|f\||
\int_{0}^{1}|G'_{1t}(t,s)-G'_{2t}(t,s)|\mathrm{d}s
\end{align*}
for $\beta=1$. It follows that
\begin{align*}
&\|u_{1}-u_{2}\|\\
&\leq  \frac{1}{1-L/K}
\Big[|\|f\||\frac{1}{\Gamma(1-\beta)}\int_{0}^{t}(t-s)^{-\beta}
\Big(\int_{0}^{1} |G'_{1s}(s,\tau)-G'_{2s}(s,\tau)|\mathrm{d}\tau\Big)
\mathrm{d}s \\
&\quad + |\|f\||\int_{0}^{1}|G_{1}(t,s)-G_{2}(t,s)|\mathrm{d}s\Big]
\end{align*}
for $0<\beta<1$ and
\begin{align*}
&\|u_{1}-u_{2}\|\\
&\leq \frac{1}{1-L/K}
\Big(|\|f\||\int_{0}^{1}|G'_{1t}(t,s)-G'_{2t}(t,s)|\mathrm{d}s
+|\|f\||\int_{0}^{1}|G_{1}(t,s)-G_{2}(t,s)|\mathrm{d}s\Big)
\end{align*}
for $\beta=1$. Thus, in accordance with \eqref{e4}, we obtain
$\|u_{1}-u_{2}\|=\textsl{O}(\varepsilon)$, which completes the
proof.
\end{proof}

\begin{theorem} \label{thm3.2}
Assume  the conditions of Theorem  \ref{thm2.2} are valid. Let
$u_{1}(t), u_{2}(t)$ be the solutions, respectively, of the
problems
\begin{gather*}
{^CD_{0^{+}}^{\alpha}}u(t)= f(t,u(t),{^CD_{0^{+}}^{\beta}}u(t)), \quad
 0<t<1,\\
a_{1}u(0)-a_{2}u'(0)=A,b_{1}u(1)+b_{2}u'(1)=B,
\end{gather*}
and
\begin{gather*}
{^CD_{0^{+}}^{\alpha}}u(t)= f(t,u(t),{^CD_{0^{+}}^{\beta}}u(t)), \quad
 0<t<1,\\
a_{1}u(0)-a_{2}u'(0)=A+\varepsilon_{1},\quad
b_{1}u(1)+b_{2}u'(1)=B+\varepsilon_{2},
\end{gather*}
Then $\|u_{1}-u_{2}\|=\textsl{O}(\max\{\varepsilon_{1},
\varepsilon_{2}\})$.
\end{theorem}

\begin{proof}
Let
\begin{gather*}
\varphi_{1}(t)=\frac{(b_{1}+b_{2})A+a_{2}B}{l}+\frac{a_{1}B-b_{1}A}{l}t,\\
\varphi_{2}(t)=\frac{(b_{1}+b_{2})(A+\varepsilon_{1})+a_{2}(B+\varepsilon_{2})}{l}
+\frac{a_{1}(B+\varepsilon_{2})-b_{1}(A+\varepsilon_{1})}{l}t.
\end{gather*}
Then
\begin{gather*}
u_{1}(t)=\int_{0}^{1}G(t,s)
f(s,u_{1}(s),{^{C}D_{0^{+}}^{\beta}}u_{1}(s))\mathrm{d}s+\varphi_{1}(t),\\
u_{2}(t)=\int_{0}^{1}G(t,s)
f(s,u_{2}(s),{^{C}D_{0^{+}}^{\beta}}u_{2}(s))\mathrm{d}s+\varphi_{2}(t).
\end{gather*}
So we obtain
\begin{align*}
|u_{1}(t)-u_{2}(t)|
&\leq L\|u_{1}-u_{2}\|\int_{0}^{1}|G(t,s)|\mathrm{d}s+|\varphi_{1}(t)
 -\varphi_{2}(t)|\\
&\leq \frac{L(2l+a_{2}b_{2})}{l\Gamma(\alpha)}\|u_{1}-u_{2}\|
+\frac{(b_{1}+b_{2})\varepsilon_{1}+a_{2}\varepsilon_{2}}{l}+
\frac{|a_{1}\varepsilon_{2}-b_{1}\varepsilon_{1}|}{l},
\end{align*}
\begin{align*}
&|{^{C}D_{0^{+}}^{\beta}}u_{1}(t)-{^{C}D_{0^{+}}^{\beta}}u_{2}(t)|\\
&\leq L\|u_{1}-u_{2}\|\frac{1}{\Gamma(1-\beta)}\int_{0}^{t}(t-s)^{-\beta}
\Big(\int_{0}^{1}|G'_{s}(s,\tau)|\mathrm{d}\tau\Big)\mathrm{d}s\\
&\quad +\frac{1}{\Gamma(1-\beta)}\int_{0}^{t}(t-s)^{-\beta}|\varphi'_{1}(s)
-\varphi'_{2}(s)|\mathrm{d}s \\
&\leq
\frac{2L}{\Gamma(2-\beta)\Gamma(\alpha)}\|u_{1}-u_{2}\|+\frac{1}{\Gamma(2-\beta)}
\frac{|a_{1}\varepsilon_{2}-b_{1}\varepsilon_{1}|}{l}
\end{align*}
for $0<\beta<1$, and
\begin{align*}
|u'_{1}(t)-u'_{2}(t)|
&\leq L\|u_{1}-u_{2}\|\int_{0}^{1}|G'_{t}(t,s)|\mathrm{d}s+|\varphi'_{1}(t)-\varphi'_{2}(t)|\\
&\leq \frac{2L}{\Gamma(\alpha)}\|u_{1}-u_{2}\|
+\frac{|a_{1}\varepsilon_{2}-b_{1}\varepsilon_{1}|}{l}
\end{align*}
for $\beta=1$. Thus,
\begin{align*}
&\|u_{1}-u_{2}\|\\
&\leq \frac{1}{1-L/K}
\Big(\frac{(b_{1}+b_{2})\varepsilon_{1}+a_{2}\varepsilon_{2}}{l}+
\frac{|a_{1}\varepsilon_{2}-b_{1}\varepsilon_{1}|}{l}+\frac{1}{\Gamma(2-\beta)}
\frac{|a_{1}\varepsilon_{2}-b_{1}\varepsilon_{1}|}{l}\Big)\\
&\leq \frac{\max\{\varepsilon_{1}, \varepsilon_{2}\}}{1-L/K}
\Big(\frac{b_{1}+b_{2}+a_{2}}{l}+
\frac{a_{1}+b_{1}}{l}+\frac{1}{\Gamma(2-\beta)}
\frac{a_{1}+b_{1}}{l}\Big).
\end{align*}
Therefore, the conclusion of the theorem follows.
\end{proof}

\begin{theorem} \label{thm3.3}
Suppose  the conditions of Theorem \ref{thm2.2} are satisfied.
Let $u_{1}(t), u_{2}(t)$ be the solutions, respectively, of the
problems
\begin{gather*}
{^CD_{0^{+}}^{\alpha}}u(t)= f(t,u(t),{^CD_{0^{+}}^{\beta}}u(t)), \ \ 0<t<1,\\
a_{1}u(0)-a_{2}u'(0)=A,b_{1}u(1)+b_{2}u'(1)=B,
\end{gather*}
and
\begin{gather*}
{^CD_{0^{+}}^{\alpha}}u(t)= f(t,u(t),{^CD_{0^{+}}^{\beta}}u(t))+\varepsilon, \ \ 0<t<1,\\
a_{1}u(0)-a_{2}u'(0)=A, b_{1}u(1)+b_{2}u'(1)=B.
\end{gather*}
Then $\|u_{1}-u_{2}\|=\textsl{O}(\varepsilon)$.
\end{theorem}

\begin{proof}
Note that $u_{1}(t)=\int_{0}^{1}G(t,s)
f(s,u_{1}(s),{^{C}D_{0^{+}}^{\beta}}u_{1}(s))\mathrm{d}s+\varphi(t)$,\\
$u_{2}(t)=\int_{0}^{1}G(t,s)
(f(s,u_{2}(s),{^{C}D_{0^{+}}^{\beta}}u_{2}(s))+\varepsilon)\mathrm{d}s+\varphi(t)$.
Thus,
\begin{align*}
|u_{1}(t)-u_{2}(t)|
&\leq L\|u_{1}-u_{2}\|\int_{0}^{1}|G(t,s)|\mathrm{d}s+\varepsilon\int_{0}^{1}|G(t,s)|\mathrm{d}s\\
&\leq  \frac{L(2l+a_{2}b_{2})}{l\Gamma(\alpha)}\|u_{1}-u_{2}\|
+\frac{\varepsilon(2l+a_{2}b_{2})}{l\Gamma(\alpha)},
\end{align*}
\begin{align*}
&|{^{C}D_{0^{+}}^{\beta}}u_{1}(t)-{^{C}D_{0^{+}}^{\beta}}u_{2}(t)|\\
&\leq L\|u_{1}-u_{2}\|\frac{1}{\Gamma(1-\beta)}\int_{0}^{t}(t-s)^{-\beta}
\Big(\int_{0}^{1}|G'_{s}(s,\tau)|\mathrm{d}\tau\Big)\mathrm{d}s\\
&\quad +\frac{\varepsilon}{\Gamma(1-\beta)}\int_{0}^{t}(t-s)^{-\beta}
\Big(\int_{0}^{1}|G'_{s}(s,\tau)|\mathrm{d}\tau\Big)\mathrm{d}s\\
&\leq \frac{2L}{\Gamma(2-\beta)\Gamma(\alpha)}\|u_{1}-u_{2}\|
 +\frac{2\varepsilon}{\Gamma(2-\beta)\Gamma(\alpha)}
\end{align*}
for $0<\beta<1$, and
\begin{align*}
|u'_{1}(t)-u'_{2}(t)|
&\leq L\|u_{1}-u_{2}\|\int_{0}^{1}|G'_{t}(t,s)|\mathrm{d}s
+\varepsilon\int_{0}^{1}|G'_{t}(t,s)|\mathrm{d}s\\
&\leq \frac{2L}{\Gamma(\alpha)}\|u_{1}-u_{2}\|
+\frac{2\varepsilon}{\Gamma(\alpha)}
\end{align*}
for $\beta=1$. Then,
$$
\|u_{1}-u_{2}\|\leq \frac{\varepsilon}{1-L/K}
\Big(\frac{2l+a_{2}b_{2}}{l\Gamma(\alpha)}
+\frac{2}{\Gamma(2-\beta)\Gamma(\alpha)}\Big),
$$
and we get the desired result.
\end{proof}

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\end{document}