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\markboth{\hfil Nonlinear fractional differential equation \hfil }
{\hfil B. Tellab \&  K. Haouam \hfil}
\begin{document}
\title{\textbf{{Existence and uniqueness of solutions for  nonlinear hyperbolic fractional differential equation with integral boundary conditions
}}}


\author{\textbf{Brahim \textsc{Tellab*}} ~ and  ~
\textbf{Kamel \textsc{Haouam**}}}

\date{}

\maketitle

\begin{center}
{\scriptsize {* Departement of Mathematics University Mentouri , } \\[0pt]
Constantine 1, Algeria. \\[0pt]
}

{\scriptsize {** Mathematics and Informatics dept, LAMIS Laboratory, Tebessa
University,} \\[0pt]
12000 Tebessa, Algeria. ~ \\[0pt]
Mail : brahimtel@yahoo.fr (B. Tellab)\\[0pt]
haouam@yahoo.fr (K. Haouam) }
\end{center}

\begin{abstract}
%\bigskip

In this paper, we investigate the existence and uniqueness of solution for second order nonlinear fractional differential equation with integral boundary conditions. Our result is an application of the Banach contraction principle and the Krasnoselskii fixed point theorem.
\medskip

\noindent {\bf Keywords:} fractional derivatives, contraction principle, fixed point theorem, integral equation.

\smallskip

\end{abstract}

\maketitle









\section{Introduction}\label{sec:1}

\medskip

Fractional differential equations have been of great interest and attracted many researchers in recent years, this is due to the development of the above cited concept. It have found applications in several different disciplines as physics, engineering, economics, electrochemistry, electromagnetism etc. (See \cite{b03, b04, b09, b13, b20}).\\
Such equations have recently proved to be valuable tools in modelling of many phenomena. ( See papers \cite{b02,b07,b12, b14, b16,   b17} ).\\
\\In \cite{b10}, Benchohra and Ouaar discussed the existence of solutions to the boundary value problem:


\begin{eqnarray}
\label{equ1}
^CD^{\alpha}y(t)=f(t,y(t)),\hspace{0.2cm} t\in J=[0,T],\hspace{0.4cm}\alpha\in(0,1],
\end{eqnarray}
\begin{eqnarray}
\label{equ2}
y(0)+\mu\int_{0}^{T}y(s)ds=y(T),
\end{eqnarray}
 $^CD^{\alpha}$ is the Caputo fractional derivative,  $f:J\times\R\rightarrow\R$ is continuous function and $\mu\in\R^{\ast}.$\\
In \cite{b08}, Sotiris K. Ntouyas investigated the existence and uniqueness of solutions of the following problem:
\begin{eqnarray}
\label{equ3}
^CD^{q}x(t)=f(t,x(t)),\hspace{0.2cm} 0<t<1,\hspace{0.2cm}0<q\leq1,
\end{eqnarray}
\begin{eqnarray}
\label{equ4}
x(0)=\alpha I^{p}x(\eta),\hspace{0.2cm}0<\eta<1,
\end{eqnarray}
$^CD^{q}$ denotes always the Caputo fractional derivative of order $q,$  $f:[0,1]\times\R\rightarrow\R$ is continuous function, $\alpha\in\R$ such that $\alpha\neq\Gamma(p+1)/\eta^{p},$ $\Gamma$ is the Euler funtion and $I^{p},$ 0<p<1 is the  Riemann-Liouville fractional integral of order $p.$\\
In this paper, we consider the following nonlinear fractional differential equation with integral boundary conditions:

\begin{eqnarray}
\label{equ5}
^CD^{\alpha}y(t)=f(t,y(t)),\hspace{0.2cm} t\in J=[0,1],
\end{eqnarray}
\begin{eqnarray}
\label{equ6}
y(0)=\int_{0}^{1}y(s)ds
\end{eqnarray}
\begin{eqnarray}
\label{equ7}
y(1)=\frac{1}{\Gamma(\beta)}\int_{0}^{1}(1-s)^{\beta-1}y(s)ds
\end{eqnarray}
where $^CD^{\alpha}$ is the Caputo fractional derivative of order $\alpha,$ $1<\alpha\leq2,$ $0<\beta\leq1$ and $f:[0,1]\times\R\rightarrow\R$ is continuous function.
\section{preliminaries}\label{sec:2}

Now, we present some basic definitions and lemmas of fractional calculus which will be used in our theorems \cite{b01, b04, b18}.
%%%%  definition 1 %%%%
\begin{definition}\label{Def1}
For a differentiable function $h:[0,+\infty)\rightarrow\R,$ the Caputo derivative of fractional order $\alpha$ is defined by
$$
^CD^{\alpha}h(t)=\frac{1}{\Gamma(n-\alpha)}\int_{0}^{t}(t-s)^{n-\alpha-1}h^{(n)}(s)ds,\hspace{0,3cm}n-1<\alpha<n,n=[\alpha]+1,
$$
where $[\alpha]$ denotes the integer part of $\alpha$ and $\Gamma$ is the gamma function.
\end{definition}
%%%%  definition 2 %%%%
\begin{definition}
  The Riemann-Liouville fractional integral of order $\alpha$ is given by
$$
I^{\alpha}h(t)=\frac{1}{\Gamma(\alpha)}\int_{0}^{t}(t-s)^{\alpha-1}h(s)ds,
$$
where $h:\R^{+}\rightarrow\R$ is a Lebesgue measurable function, provided that the integral exists.
\end{definition}
\begin{lemma}\label{lemma1}
\cite{b19} Let $\alpha>0,$ then the differential equation
$$
^CD^{\alpha}h(t)=0
$$
has solutions
$$
h(t)=c_{0}+c_{1}t+c_{2}t^{2}+...+c_{n-1}t^{n-1},
$$
$c_{i}\in\R,$ $i=0,1,2,...,n-1,$ $n=[\alpha]+1.$

\end{lemma}
\begin{lemma}\label{lemma2}
\cite{b19} Let $\alpha>0,$ then
$$
I^{\alpha}{^CD^{\alpha}}h(t)=h(t)+c_{0}+c_{1}t+c_{2}t^{2}+...+c_{n-1}t^{n-1},
$$
where $c_{i}\in\R,$ $i=0,1,2,...,n-1,$ $n=[\alpha]+1.$
\end{lemma}
\begin{lemma}\label{lemma3}
Let $1<\alpha\leq2$ and let $h:J\times\R\rightarrow\R$ be a given continuous function. Then, the boundary-value problem
\begin{eqnarray}
\label{equ8}
^CD^{\alpha}y(t)=h(t),\hspace{0.2cm} t\in J
\end{eqnarray}
\begin{eqnarray}
\label{equ9}
y(0)=\int_{0}^{1}y(s)ds
\end{eqnarray}
\begin{eqnarray}
\label{equ10}
y(1)=\frac{1}{\Gamma(\beta)}\int_{0}^{1}(1-s)^{\beta-1}y(s)ds
\end{eqnarray}
has a unique solution defined by:
\begin{eqnarray}
&&y(t)=\frac{1}{\Gamma(\alpha)}\int_{0}^{t}(t-s)^{\alpha-1}h(s)ds+\nonumber\\
&&\int_{0}^{1}\bigg[\frac{1}{\gamma_{1}\Gamma(\alpha)\Gamma(\beta)}\int_{s}^{1}(1-r)^{\beta-1}(r-s)^{\alpha-1}dr-
\frac{(1-s)^{\alpha-1}}{\gamma_{1}\Gamma(\alpha)}+\bigg(\frac{2\gamma_{2}}{\gamma_{1}\alpha\Gamma(\alpha)}-\frac{2t}
{\alpha\Gamma(\alpha)}
\bigg)(1-s)^{\alpha}\bigg]h(s)ds\nonumber
\end{eqnarray}
where
$$
\gamma_{1}=1-\frac{1}{\Gamma(\beta+1)},\hspace{0.4cm}\gamma_{2}=1-\frac{1}{\Gamma(\beta+2)}.
$$
\end{lemma}
%\begin{proof}
\subsubsection*{ \bf Proof of Lemma~\ref{lemma3}}


Applying lemma \ref{lemma2}, we can reduce the problem (\ref{equ8})-(\ref{equ10})to an equivalent integral equation
\begin{eqnarray}
\label{equ11}
y(t)&=&I_{0}^{\alpha}h(t)+c_{0}+c_{1}t\nonumber\\
&=&\frac{1}{\Gamma(\alpha)}\int_{0}^{t}(t-s)^{\alpha-1}h(s)ds+c_{0}+c_{1}t,
\end{eqnarray}
for some constants $c_{0},c_{1}\in\R.$ By integrating and using Fubini's theorem, we obtain:
\begin{eqnarray}
\label{equ12}
\int_{0}^{1}y(s)ds=\int_{0}^{1}\frac{(1-\tau)^{\alpha}}{\alpha\Gamma(\alpha)}h(\tau)d\tau+c_{0}+\frac{c_{1}}{2}.
\end{eqnarray}
Applying (\ref{equ9}), we find:
$$
y(0)=c_{0},
$$
and with (\ref{equ12}), we arrive to
$$
y(0)=\int_{0}^{1}\frac{(1-\tau)^{\alpha}}{\alpha\Gamma(\alpha)}h(\tau)d\tau+c_{0}+\frac{c_{1}}{2},
$$
then,
\begin{eqnarray}
\label{equ13}
c_{1}=-2\int_{0}^{1}\frac{(1-\tau)^{\alpha}}{\alpha\Gamma(\alpha)}h(\tau)d\tau.
\end{eqnarray}
Now, we operate (\ref{equ10}) and (\ref{equ11}), we get
\begin{eqnarray}
\frac{1}{\Gamma(\beta)}\int_{0}^{1}(1-s)^{\beta-1}y(s)ds&=&\frac{1}{\Gamma(\alpha)\Gamma(\beta)}\int_{0}^{1}\int_{0}^{s}
(1-s)^{\beta-1}(s-r)^{\alpha-1}h(r)drds\nonumber\\
&+&\frac{c_{0}}{\Gamma(\beta)}\int_{0}^{1}(1-s)^{\beta-1}ds+\frac{c_{1}}{\Gamma(\beta)}\int_{0}^{1}(1-s)^{\beta-1}sds,\nonumber
\end{eqnarray}
that is to say:
\begin{eqnarray}
\frac{1}{\Gamma(\alpha)}\int_{0}^{1}(1-s)^{\alpha-1}h(s)ds+c_{0}+c_{1}&=&
\frac{1}{\Gamma(\alpha)\Gamma(\beta)}\int_{0}^{1}\int_{0}^{s}
(1-s)^{\beta-1}(s-r)^{\alpha-1}h(r)drds\nonumber\\
&+&\frac{c_{0}}{\Gamma(\beta)}\int_{0}^{1}(1-s)^{\beta-1}ds+\frac{c_{1}}{\Gamma(\beta)}\int_{0}^{1}(1-s)^{\beta-1}sds,\nonumber
\end{eqnarray}
after the simplifications, we obtain:
\begin{eqnarray}
\frac{1}{\Gamma(\alpha)}\int_{0}^{1}(1-s)^{\alpha-1}h(s)ds+c_{0}+c_{1}&=&
\frac{1}{\Gamma(\alpha)\Gamma(\beta)}\int_{0}^{1}\int_{s}^{1}
(1-r)^{\beta-1}(r-s)^{\alpha-1}h(s)drds\nonumber\\
&+&\frac{c_{0}}{\Gamma(\beta+1)}+\frac{c_{1}}{\Gamma(\beta+2)},\nonumber
\end{eqnarray}
which may be written,
\begin{eqnarray}
\label{equ14}
\bigg(1-\frac{1}{\Gamma(\beta+1)}\bigg)c_{0}+\bigg(1-\frac{1}{\Gamma(\beta+2)}\bigg)c_{1}&=&\frac{1}{\Gamma(\alpha)\Gamma(\beta)}
\int_{0}^{1}\int_{s}^{1}(1-r)^{\beta-1}(r-s)^{\alpha-1}h(s)drds\nonumber\\
&-&\frac{1}{\Gamma(\alpha)}\int_{0}^{1}(1-s)^{\alpha-1}h(s)ds.
\end{eqnarray}
In putting  $\gamma_{1}=1-\frac{1}{\Gamma(\beta+1)},\hspace{0.4cm}\gamma_{2}=1-\frac{1}{\Gamma(\beta+2)},$ (\ref{equ14}) becomes:
\begin{eqnarray}
\gamma_{1}c_{0}+\gamma_{2}c_{1}=\frac{1}{\Gamma(\alpha)\Gamma(\beta)}
\int_{0}^{1}\int_{s}^{1}(1-r)^{\beta-1}(r-s)^{\alpha-1}h(s)drds
-\frac{1}{\Gamma(\alpha)}\int_{0}^{1}(1-s)^{\alpha-1}h(s)ds.\nonumber
\end{eqnarray}
Using (\ref{equ13}), we obtain:
\begin{eqnarray}
\label{equ15}
c_{0}&=&\frac{1}{\gamma_{1}\Gamma(\alpha)\Gamma(\beta)}\int_{0}^{1}\int_{s}^{1}(1-r)^{\beta-1}(r-s)^{\alpha-1}h(s)drds
-\frac{1}{\gamma_{1}\Gamma(\alpha)}\int_{0}^{1}(1-s)^{\alpha-1}h(s)ds\nonumber\\
&+&\frac{2\gamma_{2}}{\gamma_{1}\alpha\Gamma(\alpha)}\int_{0}^{1}(1-s)^{\alpha}h(s)ds
\end{eqnarray}
a combination of (\ref{equ11}),(\ref{equ13}) and (\ref{equ15}) leads us to:
\begin{eqnarray}
y(t)&=&\frac{1}{\Gamma(\alpha)}\int_{0}^{t}(t-s)^{\alpha-1}h(s)ds+\frac{1}{\gamma_{1}\Gamma(\alpha)\Gamma(\beta)}\int_{0}^{1}
\int_{s}^{1}(1-r)^{\beta-1}(r-s)^{\alpha-1}h(s)drds\nonumber\\
&-&\frac{1}{\gamma_{1}\Gamma(\alpha)}\int_{0}^{1}(1-s)^{\alpha-1}h(s)ds+\bigg[\frac{2\gamma_{2}}{\gamma_{1}\alpha\Gamma(\alpha)}
-\frac{2t}{\alpha\Gamma(\alpha)}\bigg]\int_{0}^{1}(1-s)^{\alpha}h(s)ds,\nonumber
\end{eqnarray}
i.e,
\begin{eqnarray}
&&y(t)=\frac{1}{\Gamma(\alpha)}\int_{0}^{t}(t-s)^{\alpha-1}h(s)ds+\nonumber\\
&&\int_{0}^{1}\bigg[\frac{1}{\gamma_{1}\Gamma(\alpha)\Gamma(\beta)}\int_{s}^{1}(1-r)^{\beta-1}(r-s)^{\alpha-1}dr-
\frac{(1-s)^{\alpha-1}}{\gamma_{1}\Gamma(\alpha)}+\bigg(\frac{2\gamma_{2}}{\gamma_{1}\alpha\Gamma(\alpha)}-\frac{2t}
{\alpha\Gamma(\alpha)}
\bigg)(1-s)^{\alpha}\bigg]h(s)ds\nonumber
\end{eqnarray}
{\flushright$\blacksquare$\\}
%\end{proof}
\section{Existence and uniqueness result}\label{sec:3}

\begin{theorem}\label{Th1}{\textbf{( fixed point theorem of Banach)}}\cite{b06}
Let $X$ be a  Banach space and $T:X\longrightarrow X$ a contracting mapping. Then $T$ has a unique fixed point i.e.
$$
\exists! x\in X: \hspace{0.1cm}Tx=x.
$$
\end{theorem}
Our first result of existence is based on theorem of Banach contracting application.
\begin{theorem}\label{Th2}
Suppose that the function $f:[0,1]\times\R\rightarrow\R$ is continuous and there is a constant $L>0$ such that:\\
$(H_{1}): |f(t,x)-f(t,y)|\leq L|x-y|,\hspace{0.2cm}t\in[0,1],\hspace{0.2cm}x,y\in\R.$\\








If $LA<1,$ then the boundary value problem (\ref{equ5})-(\ref{equ7}) has a unique solution, where
\begin{eqnarray}
\label{equ16}
A=\frac{1}{\Gamma(\alpha+1)}+\frac{\mathfrak{B(\beta,\alpha)}}{|\gamma_{1}|(\alpha+\beta)\Gamma(\alpha)\Gamma(\beta)}+
\frac{1}{|\gamma_{1}|\Gamma(\alpha+1)}+\frac{2|\gamma_{2}|}{|\gamma_{1}|\Gamma(\alpha+2)}+\frac{2}{\Gamma(\alpha+2)}.
\end{eqnarray}
\end{theorem}
%\textbf{Proof:}\\
\subsubsection*{ \bf Proof of Theorem~\ref{Th2}}
We define the operator $F,$ by:
\begin{eqnarray}
\label{equ17}
&&(Fy)(t)=\frac{1}{\Gamma(\alpha)}\int_{0}^{t}(t-s)^{\alpha-1}f(s,y(s))ds+
\int_{0}^{1}\bigg[\frac{1}{\gamma_{1}\Gamma(\alpha)\Gamma(\beta)}\int_{s}^{1}(1-r)^{\beta-1}(r-s)^{\alpha-1}dr\nonumber\\&&-
\frac{(1-s)^{\alpha-1}}{\gamma_{1}\Gamma(\alpha)}+
\bigg(\frac{2\gamma_{2}}{\gamma_{1}\alpha\Gamma(\alpha)}-\frac{2t}
{\alpha\Gamma(\alpha)}
\bigg)(1-s)^{\alpha}\bigg]f(s,y(s))ds,\hspace{0.2cm}t\in[0,1].
\end{eqnarray}
When we put $\begin{displaystyle}\sup_{t\in[0,1]}\end{displaystyle}|f(t,0)|=M$ we show that $FB_{\rho}\subset B_{\rho},$ where
$B_{\rho}=\{y\in C([0,1],\R):\hspace{0.1cm}\|y\|\leq \rho$\} and $\rho\geq\frac{MA}{1-LA}.$\\
\\
Let $y\in B_{\rho},$ $t\in [0,1],$ this leads to write:
\begin{eqnarray}
&&\|(Fy)(t)\|\nonumber\\
&\leq&\begin{displaystyle}\sup_{t\in[0,1]}\end{displaystyle}\bigg\{
\frac{1}{\Gamma(\alpha)}\int_{0}^{t}(t-s)^{\alpha-1}f(s,y(s))ds+\frac{1}{|\gamma_{1}|\Gamma(\alpha)\Gamma(\beta)}\int_{0}^{1}
\int_{s}^{1}(1-r)^{\beta-1}(r-s)^{\alpha-1}f(s,y(s))drds\nonumber\\
&+&\frac{1}{|\gamma_{1}|\Gamma(\alpha)}\int_{0}^{1}(1-s)^{\alpha-1}f(s,y(s)) ds+\frac{2|\gamma_{2}|}{|\gamma_{1}|\alpha\Gamma(\alpha)}
\int_{0}^{1}(1-s)^{\alpha}f(s,y(s)) ds \nonumber\\
&+&\frac{2t}{\alpha\Gamma(\alpha)}\int_{0}^{1}(1-s)^{\alpha}f(s,y(s))ds\bigg\}\nonumber
\end{eqnarray}
\begin{eqnarray}
&\leq&\begin{displaystyle}\sup_{t\in[0,1]}\end{displaystyle}\bigg\{
\frac{1}{\Gamma(\alpha)}\int_{0}^{t}(t-s)^{\alpha-1}\bigg(|f(s,y(s))-f(s,0)|+|f(s,0)|\bigg)ds\nonumber\\
&+&\frac{1}{|\gamma_{1}|\Gamma(\alpha)\Gamma(\beta)}\int_{0}^{1}\int_{s}^{1}(1-r)^{\beta-1}(r-s)^{\alpha-1}
\bigg(|f(s,y(s))-f(s,0)|+|f(s,0)|\bigg)drds\nonumber
\end{eqnarray}
\begin{eqnarray}
\label{equ18}
&+&\frac{1}{|\gamma_{1}|\Gamma(\alpha)}\int_{0}^{1}(1-s)^{\alpha-1}\bigg(|f(s,y(s))-f(s,0)|+|f(s,0)|\bigg)ds\nonumber\\
&+&\frac{2|\gamma_{2}|}{|\gamma_{1}|\alpha\Gamma(\alpha)}\int_{0}^{1}(1-s)^{\alpha}
\bigg(|f(s,y(s))-f(s,0)|+|f(s,0)|\bigg)ds\nonumber\\
&+&\frac{2}{\alpha\Gamma(\alpha)}\int_{0}^{1}(1-s)^{\alpha}\bigg(|f(s,y(s))-f(s,0)|+|f(s,0)|\bigg)ds\bigg\}.
\end{eqnarray}
To put $u=\frac{r-s}{1-r},$ i.e, $1-r=(1-u)(1-s),$ $dr=(1-s)du$ we obtain:
\begin{eqnarray}
\label{equ19}
\int_{0}^{1}\int_{s}^{1}(1-r)^{\beta-1}(r-s)^{\alpha-1}drds&=&\int_{0}^{1}(1-s)^{\alpha+\beta-1}ds \int_{0}^{1}(1-u)^{\beta-1}u^{\alpha-1}du\nonumber\\
&=&\frac{\mathfrak{B(\beta,\alpha)}}{\alpha+\beta}.
\end{eqnarray}
By substitution in (\ref{equ18}), and after simplification, we get
\begin{eqnarray}
\label{equ20}
\|(Fy)(t)\|&\leq&(L\rho+M)\bigg\{\frac{1}{\Gamma(\alpha+1)}+\frac{\mathfrak{B(\beta,\alpha)}}
{|\gamma_{1}|(\alpha+\beta)\Gamma(\alpha)\Gamma(\beta)}+\frac{1}{|\gamma_{1}|\Gamma(\alpha+1)}+
\frac{2|\gamma_{2}|}{|\gamma_{1}|\Gamma(\alpha+2)}\nonumber\\
&&+\frac{2}{\Gamma(\alpha+2)}\bigg\}\nonumber\\
\\
&\leq&(L\rho+M)A\leq\rho,
\end{eqnarray}
which means that $FB_{\rho}\subset B_{\rho}.$\\
\\
Now, suppose that $x,y\in C([0,1],\R)$ and $t\in[0,1],$ then we have:
\begin{eqnarray}
\|(Fx)-(Fy)\|
&\leq&\begin{displaystyle}\sup_{t\in[0,1]}\end{displaystyle}\bigg\{
\frac{1}{\Gamma(\alpha)}\int_{0}^{t}(t-s)^{\alpha-1}|f(s,x(s))-f(s,y(s))|ds\nonumber\\
&+&\frac{1}{|\gamma_{1}|\Gamma(\alpha)\Gamma(\beta)}\int_{0}^{1}\int_{s}^{1}(1-r)^{\beta-1}(r-s)^{\alpha-1}
|f(s,x(s))-f(s,y(s))|drds\nonumber\\
&+&\frac{1}{|\gamma_{1}|\Gamma(\alpha)}\int_{0}^{1}(1-s)^{\alpha-1}|f(s,x(s))-f(s,y(s))|ds\nonumber\\
&+&\frac{2|\gamma_{2}|}{|\gamma_{1}|\alpha\Gamma(\alpha)}\int_{0}^{1}(1-s)^{\alpha}
|f(s,x(s))-f(s,y(s))|ds\nonumber\\
&+&\frac{2}{\alpha\Gamma(\alpha)}\int_{0}^{1}(1-s)^{\alpha}|f(s,x(s))-f(s,y(s))|ds\bigg\}\nonumber
\end{eqnarray}
\begin{eqnarray}
&\leq&L\|x-y\|\bigg\{\frac{1}{\Gamma(\alpha+1)}+\frac{\mathfrak{B(\beta,\alpha)}}
{|\gamma_{1}|(\alpha+\beta)\Gamma(\alpha)\Gamma(\beta)}+\frac{1}{|\gamma_{1}|\Gamma(\alpha+1)}+
\frac{2|\gamma_{2}|}{|\gamma_{1}|\Gamma(\alpha+2)}\nonumber\\
&&+\frac{2}{\Gamma(\alpha+2)}\bigg\}\nonumber\\
&=&LA\|x-y\|.\nonumber
\end{eqnarray}
Since we assumed $0<LA<1,$ then $F$ is a contraction. Using the principle of Banach fixed point, we deduce that the problem (\ref{equ5})-(\ref{equ7}) has a unique solution.
{\flushright$\blacksquare$\\}




%\underline{\textbf{Example:}}\\


\begin{example}\label{Example}
Consider the following boundary value problem:
\begin{eqnarray}
\label{equ21}
^CD^{\frac{3}{2}}y(t)=\frac{1}{t^{2}+4}\times\frac{|x|}{1+|x|}+t\cos^{2}t,\hspace{0.2cm} t\in [0,1]
\end{eqnarray}
\begin{eqnarray}
\label{equ22}
y(0)=\int_{0}^{1}y(s)ds
\end{eqnarray}
\begin{eqnarray}
\label{equ23}
y(1)=\frac{1}{\Gamma(\frac{1}{2})}\int_{0}^{1}(1-s)^{-\frac{1}{2}}y(s)ds.
\end{eqnarray}
In this example, $\alpha=\frac{3}{2},$ $\beta=\frac{1}{2}$ et $f(t,x)=\frac{1}{t^{2}+4}\times\frac{|x|}{1+|x|}+t\cos^{2}t.$\\
We have:
\begin{eqnarray}
|f(t,x)-f(t,y)|&=&\frac{1}{t^{2}+4}\times\frac{||x|-|y||}{(1+|x|)(1+|y|)}\nonumber\\
&\leq&\frac{1}{4}|x-y|,\nonumber
\end{eqnarray}
then, $L=\frac{1}{4}.$\\
\\
 By a simple calculation, we find: $LA\simeq 0.3715...<1,$ and by theorem \ref{Th1} we deduce that the problem (\ref{equ21})-(\ref{equ23}) has a unique solution.\\
\end{example}
\medskip


\begin{theorem}\label{Th3}{\textbf{( Arzel\`{a}-Ascoli's theorem)}}\cite{b05}
Let $A\subset C(K,\R^{n}),$  $(K=[a,b]\subset\R).$\\ $A$ is relatively compact (i.e, $\bar{A}$ is compact) if and only if:\\
1.\hspace{0.2cm} $A$ is uniformly bounded.\\
2.\hspace{0.2cm} $A$ is equicontinuous.\\
$\bullet$ Recall that a function $f$ is uniformly bounded in $A$ if there exists a constant $M>0$ such that:
$$
\|f\|=\sup_{x\in K}|f(x)|\leq M,\hspace{0.3cm}\forall \hspace{0.05cm} f\in A
$$
\end{theorem}
\begin{theorem}\label{Th4}{\textbf{(Fixed point theorem of Krasnoselskii )}}\cite{b15}
Let $F$ a non-empty set, closed and convex in a Banach space $X.$ $T_{1}$ and $T_{2}$
are two applications of $F$ in $X$ such that:\\
1.\hspace{0.2cm}$T_{1}(x)+T_{2}(y)\in F,$\hspace{0.2cm}$\forall x,y\in F,$\\
2.\hspace{0.2cm}$T_{1}$ is a contraction,\\
3.\hspace{0.2cm} $T_{2}$ is compact and continuous.\\
Then, $T_{1}+T_{2}$ has a fixed point in $F,$ i.e, there exits $x\in F$ such that $T_{1}(x)+T_{2}(x)=x.$
\end{theorem}
The second existence result is obtained by using the fixed point theorem of Krasnoselskii.
\begin{theorem}\label{Th5}
Let $f:[0,1]\times\R\rightarrow\R$ be a continuous fonctions which satisfies the conditions $(H_{1})$ and
$(H_{2}):|f(t,x)|\leq \mu(t)$,\hspace{0.2cm} $\forall (t,x)\in[0,1]$ et $\mu\in C([0,1],\R^{+}).$
Suppose that
\begin{eqnarray}
\label{equ24}
L\bigg\{\frac{\mathfrak{B(\beta,\alpha)}}
{|\gamma_{1}|(\alpha+\beta)\Gamma(\alpha)\Gamma(\beta)}+\frac{1}{|\gamma_{1}|\Gamma(\alpha+1)}+
\frac{2|\gamma_{2}|}{|\gamma_{1}|\Gamma(\alpha+2)}+\frac{2}{\Gamma(\alpha+2)}\bigg\}<1.
\end{eqnarray}
Then, the value problem (\ref{equ5})-(\ref{equ7}) has a unique solution.
\end{theorem}
%\textbf{Proof:}\\
\subsubsection*{ \bf Proof of Theorem~\ref{Th5}}
Let $\begin{displaystyle}\sup_{t\in[0,1]}|\mu(t)|=\|\mu(t)\|.\end{displaystyle}$\\ We set $\rho^{\ast}\geq\|\mu\|\bigg\{\frac{1}{\Gamma(\alpha+1)}+\frac{\mathfrak{B(\beta,\alpha)}}
{|\gamma_{1}|(\alpha+\beta)\Gamma(\alpha)\Gamma(\beta)}+\frac{1}{|\gamma_{1}|\Gamma(\alpha+1)}+
\frac{2|\gamma_{2}|}{|\gamma_{1}|\Gamma(\alpha+2)}+\frac{2}{\Gamma(\alpha+2)}\bigg\}$ and we consider the set
$B_{\rho^{\ast}}=\{y\in C([0,1],\R):\hspace{0.1cm}\|y\|\leq \rho^{\ast}\}.$
We define two operators  $P$ and $Q$ on $B_{\rho^{\ast}}$ by:
\begin{eqnarray}
(Py)(t)&=&\frac{1}{\Gamma(\alpha)}\int_{0}^{t}(t-s)^{\alpha-1}f(s,y(s))ds, \hspace{0.2cm} t\in[0,1]\nonumber
\end{eqnarray}
\begin{eqnarray}
(Qy)(t)&=&\frac{1}{\gamma_{1}\Gamma(\alpha)\Gamma(\beta)}\int_{0}^{1}
\int_{s}^{1}(1-r)^{\beta-1}(r-s)^{\alpha-1}f(s,y(s))drds\nonumber\\
&+&\frac{1}{\gamma_{1}\Gamma(\alpha)}\int_{0}^{1}(1-s)^{\alpha-1}f(s,y(s)) ds
+\frac{2\gamma_{2}}{\gamma_{1}\Gamma(\alpha+1)}
\int_{0}^{1}(1-s)^{\alpha}f(s,y(s)) ds \nonumber\\
 &+&\frac{2t}{\Gamma(\alpha+1)}\int_{0}^{1}(1-s)^{\alpha}f(s,y(s))ds, \hspace{0.2cm} t\in[0,1]\nonumber
\end{eqnarray}
Let $x,y\in B_{\rho^{\ast}},$ we write:
\begin{eqnarray}
\|Px+Qy\|&\leq& \frac{\|\mu\|}{\Gamma(\alpha)}\int_{0}^{t}(t-s)^{\alpha-1}ds+\frac{\|\mu\|}
{|\gamma_{1}|\Gamma(\alpha)\Gamma(\beta)}\int_{0}^{1}\int_{s}^{1}(1-r)^{\beta-1}(r-s)^{\alpha-1}drds\nonumber\\
&+&\frac{\|\mu\|}{|\gamma_{1}|\Gamma(\alpha)}\int_{0}^{1}(1-s)^{\alpha-1} ds+\frac{2|\gamma_{2}|\|\mu\|}
{|\gamma_{1}|\Gamma(\alpha+1)}\int_{0}^{1}(1-s)^{\alpha} ds \nonumber\\
&+&\frac{2\|\mu\|}{\Gamma(\alpha+1)}\int_{0}^{1}(1-s)^{\alpha}ds\nonumber\\
&\leq&\|\mu\|\bigg\{\frac{1}{\Gamma(\alpha+1)}+\frac{\mathfrak{B(\beta,\alpha)}}
{|\gamma_{1}|(\alpha+\beta)\Gamma(\alpha)\Gamma(\beta)}+\frac{1}{|\gamma_{1}|\Gamma(\alpha+1)}+
\frac{2|\gamma_{2}|}{|\gamma_{1}|\Gamma(\alpha+2)}+\frac{2}{\Gamma(\alpha+2)}\bigg\}\nonumber\\
&\leq&\rho^{\ast}.\nonumber
\end{eqnarray}
Then, $Px+Qy\in B_{\rho^{\ast}},$ we have:
$$
\|Q(x)-Qy\|\leq L\|x-y\|\bigg\{\frac{\mathfrak{B(\beta,\alpha)}}
{|\gamma_{1}|(\alpha+\beta)\Gamma(\alpha)\Gamma(\beta)}+\frac{1}{|\gamma_{1}|\Gamma(\alpha+1)}+
\frac{2|\gamma_{2}|}{|\gamma_{1}|\Gamma(\alpha+2)}+\frac{2}{\Gamma(\alpha+2)}\bigg\}
$$
By exploiting (\ref{equ24}), we deduce that $Q$ is a contraction.\\
\\
According to the definition of the operator $P,$ we deduce that the continuity of $f$ implies that of $P.$ In addition, we have:
\begin{eqnarray}
\label{equ25}
\|Px\|&\leq&\|\mu\|\int_{0}^{t}\frac{(t-s)^{\alpha-1}}{\Gamma(\alpha)}ds\nonumber\\
&\leq&\frac{\|\mu\|}{\Gamma(\alpha+1)},
\end{eqnarray}
which implies that $P$ is uniformly bounded.\\
\\
Now, we show that  $P$ is compact. We have:
\begin{eqnarray}
(Py)(t_{1})-(Py)(t_{2})&=&\frac{1}{\Gamma(\alpha)}\int_{0}^{t_{1}}(t_{1}-s)^{\alpha-1}f(s,y(s))ds-
\frac{1}{\Gamma(\alpha)}\int_{0}^{t_{2}}(t_{2}-s)^{\alpha-1}f(s,y(s))ds\nonumber\\
&=&\frac{1}{\Gamma(\alpha)}\bigg(\int_{0}^{t_{1}}(t_{1}-s)^{\alpha-1}f(s,y(s))ds-
\int_{0}^{t_{1}}(t_{2}-s)^{\alpha-1}f(s,y(s))ds\nonumber\\
&-&\int_{t_{1}}^{t_{2}}(t_{2}-s)^{\alpha-1}f(s,y(s))ds\bigg)\nonumber
\end{eqnarray}
Taking into account the condition $(H_{1}),$
we set $f^{\ast}=\begin{displaystyle}\sup_{(t,x)\in[0,1]\times B_{\rho^{\ast}}}|f(t,x)|\end{displaystyle}.$ \\Then we can write:

\begin{eqnarray}
|(Py)(t_{1})-(Py)(t_{2})|&=&\frac{1}{\Gamma(\alpha)}\bigg|\int_{0}^{t_{1}}\bigg[(t_{2}-s)^{\alpha-1}-(t_{1}-s)^{\alpha-1}\bigg]
f(s,y(s))ds\nonumber\\
&&+\int_{t_{1}}^{t_{2}}(t_{2}-s)^{\alpha-1}f(s,y(s))ds \bigg|\nonumber\\
&\leq&\frac{f^{\ast}}{\Gamma(\alpha)}\bigg|\int_{0}^{t_{1}}\bigg[(t_{2}-s)^{\alpha-1}-(t_{1}-s)^{\alpha-1}\bigg]ds+
\int_{t_{1}}^{t_{2}}(t_{2}-s)^{\alpha-1}ds \bigg|,\nonumber
\end{eqnarray}
a simple calculation leads us to:
\begin{eqnarray}
\label{equ26}
|(Py)(t_{1})-(Py)(t_{2})|&\leq&\frac{f^{\ast}}{\Gamma(\alpha+1)}|t_{2}^{\alpha}-t_{1}^{\alpha}|.
\end{eqnarray}
The second member of (\ref{equ26}) is independent of $y$ and tends to zero when $t_{2}-t_{1}\rightarrow 0,$ so $P$ is equicontinuous. Using the Arzel\`{a}-Ascoli theorem, we deduce that  $P$ is compact in $ B_{\rho^{\ast}}.$ Thus all the assumptions of the fixed point theorem of Krasnoselskii are satisfied. Which implies that the boundary value problem (\ref{equ5})-(\ref{equ7}) has a unique solution on  $[0,1].$
\newpage
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%\noindent\textsc{B.Tellab } \\
%Department of Mathematics and Informatics, Ghardaia University \\
 %47000 Ghardaia, Algeria  \\
%e-mail:   brahimtel@yahoo.fr \smallskip

%\noindent\textsc{K.Houam }\\
%Department of Mathematics and Informatics, University of Tebessa \\
% 12000 Tebessa, Algeria\\
%e-mail: haouam@yahoo.fr\\
\end{document} 