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

\AtBeginDocument{{\noindent\small
\emph{Electronic Journal of Differential Equations},
Vol. 2016 (2016), No. 202, pp. 1--5.\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-2016/202\hfil Existence and uniqueness of solutions]
{Existence and uniqueness of solutions for two-point
fractional boundary value problems}

\author[R. A. C. Ferreira \hfil EJDE-2016/202\hfilneg]
{Rui A. C. Ferreira}

\address{Rui A. C. Ferreira \newline
Grupo de F\'{i}sica Matem\'{a}tica da Universidade de Lisboa,
Complexo Interdisciplinar,
Av. Prof. Gama Pinto 2,
P-1649-003 Lisboa, Portugal}
\email{raferreira@fc.ul.pt}

\thanks{Submitted June 2, 2016. Published July 27, 2016.}
\subjclass[2010]{34B15, 26A33}
\keywords{Fractional derivative; uniqueness;
 boundary value problem}

\begin{abstract}
 In this note we present an existence and uniqueness of a continuous
 solution for  a fractional boundary-value problem which depends on the
 Riemmann-Liouville operator. We conclude this article by presenting
 an illustrative example.
\end{abstract}

\maketitle
\numberwithin{equation}{section}
\newtheorem{theorem}{Theorem}[section]
\newtheorem{lemma}[theorem]{Lemma}
\newtheorem{proposition}[theorem]{Proposition}
\newtheorem{remark}[theorem]{Remark}
\allowdisplaybreaks

\section{Introduction}

In the book by Kelley and Peterson \cite{Peterson} the following result 
is established:

\begin{theorem}[{\cite[Theorem 7.7]{Peterson}}] \label{thm0}
Assume $f:[a,b]\times\mathbb{R}\to\mathbb{R}$ is continuous and 
satisfies a uniform Lipschitz condition with respect to the second 
variable on $[a,b]\times\mathbb{R}$ with Lipschitz constant $K$; that is,
$$
|f(t,x)-f(t,y)|\leq K|x-y|,
$$
for all $(t,x), (t,y)\in[a,b]\times\mathbb{R}$. 
If
$$
b-a<\frac{2\sqrt{2}}{\sqrt{K}},
$$
then the boundary value problem
\begin{gather*}
y''(t)=-f(t,y(t)),\quad a<t<b,\\
y(a)=A,\ y(b)=B,\quad A,B\in\mathbb{R},
\end{gather*}
has a unique continuous solution.
\end{theorem}

In this work we want to extend the above result by considering a
fractional Riemmann-Liouville derivative (we refer the reader to \cite{Kilbas} 
for the definitions and basic results on fractional calculus)
 instead of the classical operator $y''$, i.e., we prove the existence 
and uniqueness of solutions for the fractional differential boundary value problem
\begin{gather}
{_a}D^\alpha y(t)=-f(t,y(t)),\quad a<t<b,\label{eq0}\\
y(a)=0,\quad y(b)=B,\label{eq1}
\end{gather}
where $1<\alpha\leq 2$.
Existence and uniqueness results for fractional IVPs and BVPs have been 
obtained before in the literature (cf. \cite{Ferreira,Graef} and 
the references cited therein). Nevertheless we believe that our results 
are new and provide useful tools in the study of fractional boundary 
value problems.

\section{Main Results}

We start by writing the boundary value problem \eqref{eq0}--\eqref{eq1} 
in its integral form.

\begin{lemma}\label{lem0}
Suppose that $f$ is a continuous function. A function $y\in C[a,b]$ 
is a solution of \eqref{eq0}--\eqref{eq1} 
if and only if $y$  satisfies the integral equation
\begin{equation*}
y(t)=B\frac{(t-a)^{\alpha-1}}{(b-a)^{\alpha-1}}+\int_a^b G(t,s)f(s,y(s))ds,
\end{equation*}
where
\[
G(t,s)=\frac{1}{\Gamma(\alpha)}
\begin{cases}
\frac{(t-a)^{\alpha-1}}{(b-a)^{\alpha-1}}(b-s)^{\alpha-1}-(t-s)^{\alpha-1},
& a\leq s\leq t\leq b,\\[4pt]
\frac{(t-a)^{\alpha-1}}{(b-a)^{\alpha-1}}(b-s)^{\alpha-1},
& a\leq t\leq s\leq b.
\end{cases}
\]
\end{lemma}

\begin{proof}
The proof is somewhat standard. Nevertheless, for completeness, we provide it here.

It is well known that solving  \eqref{eq0}--\eqref{eq1} is equivalent 
to solving the integral equation
$$
y(t)=c\frac{(t-a)^{\alpha-1}}{\Gamma(\alpha)}
+d\frac{(t-a)^{\alpha-2}}{\Gamma(\alpha-1)}
-\frac{1}{\Gamma(\alpha)}\int_a^t(t-s)^{\alpha-1}f(s,y(s))ds,
$$
where $c$ and $d$ are some real constants. Now, $d=0$ by  
the first boundary condition. On the other hand, $y(b)=B$ implies
$$
B=c\frac{(b-a)^{\alpha-1}}{\Gamma(\alpha)}-\frac{1}{\Gamma(\alpha)}\int_a^b(b-s)^{\alpha-1}f(s,y(s))ds,$$
which after some manipulations yields
$$c=\frac{\Gamma(\alpha)}{(b-a)^{\alpha-1}}
\Big(B+\frac{1}{\Gamma(\alpha)}\int_a^b(b-s)^{\alpha-1}f(s,y(s))ds\Big).
$$
Hence,
\begin{align*}
y(t)
&=\frac{\Gamma(\alpha)}{(b-a)^{\alpha-1}}
\Big(B+\frac{1}{\Gamma(\alpha)}\int_a^b(b-s)^{\alpha-1}f(s,y(s))ds\Big)
\frac{(t-a)^{\alpha-1}}{\Gamma(\alpha)}\\
&\quad -\frac{1}{\Gamma(\alpha)}\int_a^t(t-s)^{\alpha-1}f(s,y(s))ds,\\
&=B\frac{(t-a)^{\alpha-1}}{(b-a)^{\alpha-1}}
 +\frac{1}{\Gamma(\alpha)}\int_a^b(b-s)^{\alpha-1}f(s,y(s))ds
 \frac{(t-a)^{\alpha-1}}{(b-a)^{\alpha-1}}\\
&\quad -\frac{1}{\Gamma(\alpha)}\int_a^t(t-s)^{\alpha-1}f(s,y(s))ds,
\end{align*}
and the proof is complete.
\end{proof}

The next result is essential for proving our main result.

\begin{proposition}\label{prop0}
Let $G$ be the Green function given in Lemma \ref{lem0}. Then
$$
\int_a^b|G(t,s)|ds\leq \frac{1}{\Gamma(\alpha)}
\frac{(\alpha-1)^{\alpha-1}}{\alpha^{\alpha+1}}(b-a)^\alpha.
$$
\end{proposition}

\begin{proof}
It is known \cite[Lemma 2.2]{Ferreira1} that $G(t,s)\geq 0$ for all 
$a\leq t,s\leq b$. Therefore,
\begin{align*}
\int_a^b|G(t,s)|ds
&=\frac{1}{\Gamma(\alpha)}\Big(\int_a^t
\big(\frac{(t-a)^{\alpha-1}}{(b-a)^{\alpha-1}}(b-s)^{\alpha-1}-(t-s)^{\alpha-1}\big)ds
\\
&\quad +\int_t^b\frac{(t-a)^{\alpha-1}}{(b-a)^{\alpha-1}}(b-s)^{\alpha-1}ds\Big)\\
&=\frac{1}{\Gamma(\alpha)}\Big(-\frac{(t-a)^{\alpha-1}}{(b-a)^{\alpha-1}}
 \frac{(b-t)^{\alpha}}{\alpha}+\frac{(t-a)^{\alpha-1}}{(b-a)^{\alpha-1}}
 \frac{(b-a)^{\alpha}}{\alpha}\\
&\quad -\frac{(t-a)^\alpha}{\alpha} 
 +\frac{(t-a)^{\alpha-1}}{(b-a)^{\alpha-1}}\frac{(b-t)^{\alpha}}{\alpha}\Big)\\
&=\frac{1}{\Gamma(\alpha)}\Big((t-a)^{\alpha-1}\frac{b-a}{\alpha}
-\frac{(t-a)^\alpha}{\alpha}\Big)\\
&=\frac{1}{\Gamma(\alpha)}\frac{(t-a)^{\alpha-1}(b-t)}{\alpha}.
\end{align*}
Define $g:[a,b]\to\mathbb{R}$ by
$$
g(t)=\frac{(t-a)^{\alpha-1}(b-t)}{\alpha}.
$$
Differentiating the function $g$ we immediately find that its maximum 
is achieved at the point
$$
t^*=\frac{(\alpha-1)b+a}{\alpha}.
$$
Moreover,
$$
g(t^*)=\frac{(\alpha-1)^{\alpha-1}}{\alpha^{\alpha+1}}(b-a)^\alpha,
$$
which completes the proof.
\end{proof}

\begin{theorem}\label{thm1}
Assume $f:[a,b]\times\mathbb{R}\to\mathbb{R}$ is continuous and satisfies 
a uniform Lipschitz condition with respect to the second variable on 
$[a,b]\times\mathbb{R}$ with Lipschitz constant $K$; that is,
$$
|f(t,x)-f(t,y)|\leq K|x-y|,
$$
for all $(t,x), (t,y)\in[a,b]\times\mathbb{R}$. 
If
\begin{equation}\label{in1}
b-a<\Gamma^{1/\alpha}(\alpha)
\frac{\alpha^{(\alpha+1)/\alpha}}{K^{1/\alpha}(\alpha-1)
^{(\alpha-1)/\alpha}},
\end{equation}
then the boundary-value problem
\begin{gather}
{_a}D^\alpha y(t)=-f(t,y(t)),\quad a<t<b,\label{eq2}\\
y(a)=0,\ y(b)=B,\quad B\in\mathbb{R},\label{eq3}
\end{gather}
has a unique continuous solution.
\end{theorem}

\begin{proof}
Let $\mathcal{B}$ be the Banach space of continuous functions defined 
on $[a,b]$ with the norm 
$$
\|x\|=\max_{t\in[a,b]}|x(t)|.
$$
By Lemma \ref{lem0}, $y\in C[a,b]$ is a solution of  
\eqref{eq2}--\eqref{eq3} if and only if it is a solution of the integral equation
$$
y(t)=B\frac{(t-a)^{\alpha-1}}{(b-a)^{\alpha-1}}+\int_a^b G(t,s)f(s,y(s))ds.
$$
Define the operator $T:\mathcal{B}\to\mathcal{B}$ by
$$
Ty(t)=B\frac{(t-a)^{\alpha-1}}{(b-a)^{\alpha-1}}+\int_a^b G(t,s)f(s,y(s))ds,
$$
for $t\in[a,b]$. We will show that the operator $T$ has a unique fixed point.

Let $x,y\in\mathcal{B}$. Then
\begin{align*}
|Tx(t)-Ty(t)|&\leq \int_a^b |G(t,s)||f(s,x(s))-f(s,y(s))|ds\\
&\leq \int_a^b |G(t,s)|K|x(s)-y(s)|ds\\
&\leq K\int_a^b G(t,s)ds\|x-y\|\\
&\leq K\frac{1}{\Gamma(\alpha)}\frac{(\alpha-1)^{\alpha-1}}
{\alpha^{\alpha+1}}(b-a)^\alpha\|x-y\|,
\end{align*}
where we have used Proposition \ref{prop0}. By \eqref{in1} we conclude that $T$ 
is a contracting mapping on $\mathcal{B}$, and by the Banach contraction 
mapping theorem we get the desired result.
\end{proof}

\begin{remark} \rm
We note that when $\alpha=2$ in Theorem \ref{thm1}, one immediately obtains 
Theorem \ref{thm0} (apart from the restriction $A=0$ ($y(a)=0$),
 which we have to assume in order to consider continuous solutions on $[a,b]$ 
to \eqref{eq2}).
\end{remark}

As an example we consider the initial-value problem
\begin{gather}
{_0}D^{3/2} y(t)=-1-\sin(y(t)),\quad 0<t<1,\label{eq4}\\
y(0)=0,\quad y(1)=0.\label{eq5}
\end{gather}
Here $f(t,y)=-1-\sin(y)$ and, therefore,
$$
|f_y(t,y)|=|\cos(y)|\leq 1=K.
$$
Since $\alpha=3/2$, we have 
$$
\Gamma^{1/\alpha}(\alpha)\frac{\alpha^{(\alpha+1)/\alpha}}
{(\alpha-1)^{(\alpha-1)/\alpha}}
=\frac{3}{4}\pi^{1/3}3^{2/3},
$$
and therefore \eqref{in1} is satisfied.
Now an application of Theorem \ref{thm1} proves that \eqref{eq4}--\eqref{eq5} 
has a unique solution.


\subsection*{Acknowledgements}
The author was supported by the ``Funda\c{c}\~ao para a Ci\^encia e
 a Tecnologia (FCT)" through the program ``Investigador FCT" 
with reference IF/01345/2014.

\begin{thebibliography}{99}

\bibitem{Ferreira} R. A. C. Ferreira;
\emph{A uniqueness result for a fractional differential equation},
 Fract. Calc. Appl. Anal. \textbf{15} (2012), no.~4, 611--615.

\bibitem{Ferreira1} R. A. C. Ferreira;
\emph{A Lyapunov-type inequality for a fractional boundary value problem},
 Fract. Calc. Appl. Anal. \textbf{16} (2013), no.~4, 978--984.

\bibitem{Graef} J. R Graef, L. Kong, Q. Kong, M. Wang;
\emph{Existence and uniqueness of solutions for a fractional boundary value
 problem with Dirichlet boundary condition}, 
Electron. J. Qual. Theory Differ. Equ. \textbf{2013}, No. 55, 11 pp.

\bibitem{Peterson} W. G. Kelley, A. C. Peterson;
\emph{The theory of differential equations}, second edition, 
Universitext, Springer, New York, 2010.

\bibitem{Kilbas} A. A. Kilbas, H. M. Srivastava, J. J. Trujillo;
\emph{Theory and applications of fractional differential equations},
North-Holland Mathematics Studies, 204, Elsevier, Amsterdam, 2006.


\end{thebibliography}


\end{document}