\documentclass[reqno]{amsart}
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\AtBeginDocument{{\noindent\small
\emph{Electronic Journal of Differential Equations},
Vol. 2010(2010), No. 104, pp. 1--5.\newline
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu
\newline ftp ejde.math.txstate.edu}
\thanks{\copyright 2010 Texas State University - San Marcos.}
\vspace{9mm}}

\begin{document}
\title[\hfilneg EJDE-2010/104\hfil Neumann boundary-value problems]
{Neumann boundary-value problems for differential
inclusions in Banach spaces}

\author[M. Aitalioubrahim\hfil EJDE-2010/104\hfilneg]
{Myelkebir Aitalioubrahim}

\address{Myelkebir Aitalioubrahim \newline
High school Ibn Khaldoune, commune Bouznika, Morocco}
\email{aitalifr@hotmail.com}


\thanks{Submitted May 3, 2010. Published August 2, 2010.}
\subjclass[2000]{34A60, 34K10}
\keywords{Boundary-value problems; differential
inclusion; \hfill\break\indent fixed point; measurability}

\begin{abstract}
 In this article, a fixed point theorem is used to investigate
 the existence of solutions for differential inclusions,
 with Neumann boundary conditions, in Banach spaces.
\end{abstract}

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


\section{Introduction}

The aim of this article is to establish the existence of
solutions to the  boundary-value problem
\begin{equation} \label{cauchy}
\begin{gathered}
 \ddot{x}(t) \in  F(t,x(t),\dot{x}(t)), \quad \text{a.e. on } [0,1]\\
\dot{x}(0)=r,\quad \dot{x}(1)=s,
\end{gathered}
\end{equation}
where $F$ is a closed multifunction, measurable in the
first argument and Lipschitz continuous in the second argument; and
 $r,s$ are in a Banach space $E$.

Neumann boundary-value problems have received the attention of many
authors. For review of results on Neumann boundary-value problems
for differential equations, we refer the reader to the papers by
Boucherif and Al-malki \cite{boucherif}, Wang, Cui and Zhang
\cite{wang}, Mawhin and Ruiz \cite{mawhin}, Guennoun
\cite{guennoun}, Granas, Guenther and Lee \cite{granas} and the
references therein. The techniques involved, in  most of these
works, are based on the upper and lower solution method, the
topological degree, and the topological transversality theorem.

In the literature there are a few papers dealing with the existence of
solutions for Neumann boundary-value problems for differential
inclusions. Our main purpose, in this paper, is to obtain the
existence of solutions to \eqref{cauchy}, in infinite
dimensional space, and in the case when the multifunction $F$ is
nonconvex. We used the fixed point theorem introduced by Covitz and
Nadler for contraction multi-valued maps.

 The paper is organized as follows. In Section 2 we
recall some preliminary facts that we need in the sequel while in
Section 3, we prove our main result.

\section{Preliminaries and statement of the main results}

Let $E$ be a real separable Banach space with the norm $\|\cdot\|$.
 We denote by $\mathcal{C}([0,1],E)$ the Banach space of continuous
functions from $[0,1]$ to $E$ equipped with the norm $\| x(\cdot)\|
_{\infty}:=\sup \big\{\| x(t)\| ; t\in [0,1]\big\}$. For $x\in E$
and for nonempty sets $A, B$ of $E$ we denote $d(x,A)=\inf\{d(x,y);
y\in A\}$, $e(A,B):=\sup\{d(x,B); x\in A\}$ and
$H(A,B):=\max\{e(A,B),e(B,A)\}$. A multifunction is said to be
measurable if its graph is measurable. For more detail on
measurability theory, we refer the reader to the book of
Castaing-Valadier \cite{castaing}.

For the sake of completeness, we recall the following results that
will be used in the sequel.

\begin{definition} \label{def2.1} \rm
Let $T: E \to 2^{E}$ be a multi function with
closed values
\begin{itemize}
\item $T$ is $k$-Lipschitz if
$ H\big(T(x),T(y)\big)\leq k\|x-y\|$  for each $x, y\in E$.

\item $T$ is a contraction if  it is
    $k$-Lipschitz with $k<1$.

\item $T$ has a fixed point if there exists $x\in E$ such that
    $x\in T(x)$.
\end{itemize}
\end{definition}

\begin{lemma}[\cite{covitz}]  \label{lemme1}
If $T:E \to 2^{E}$ is a contraction with nonempty closed
values, then it has a fixed point.
\end{lemma}

\begin{lemma}[\cite{qiji}]  \label{lemme2}
Assume that $F:[a,b]\times E \to 2^{E}$ is a multi  function
with nonempty closed values satisfying:
\begin{itemize}
\item  For every $x\in E$, $F(.,x)$ is measurable on
$[a,b]$;
\item For every $t\in [a,b]$, $F(t,.)$ is (Hausdorff)
continuous on  $E$.
 \end{itemize}
Then for any measurable function $x(\cdot): [a,b] \to E$,
 the multi function $F(.,x(\cdot))$ is measurable on $[a,b]$.
\end{lemma}

\begin{definition} \label{def2.2} \rm
 A measurable multi-valued function $F:[0,1] \to 2^{E}$ is
 said to be integrably bounded if there exists a function $h\in
 L^{1}([0,1],E)$ such that for all $v\in F(t)$, $\|v\|\leq h(t)$ for
 almost every $t\in [0,1]$.
 \end{definition}

\begin{definition} \label{def2.3} \rm
A function $x(\cdot):[0,1]\to E$ is said to be a solution of
\eqref{cauchy} if $x(\cdot)$ is absolutely continuous on $[0,1]$ and
satisfies \eqref{cauchy}.
\end{definition}

Let $G(t,s)$ be the Green function associated with the problem
\begin{gather*}
-\ddot{y}(t)+y(t)=0,\quad t\in[0,1],\\
\dot{y}(0)=\dot{y}(1)=0,
\end{gather*}
(see \cite{green1,green2}) which is explicitly given by
$$
G(t,s)=\begin{cases}
\frac{\cosh(1-s)\cosh(t)}{\sinh(1)}, & 0\leq t\leq s\leq 1,\\[3pt]
 \frac{\cosh(1-t)\cosh(s)}{\sinh(1)}, & 0\leq s\leq t\leq 1,
\end{cases}
$$
where $\cosh(x)=(e^x+e^{-x})/2$ and
$\sinh(x)=(e^x-e^{-x})/2$. Obviously, $G(t,s)$ is continuous
 on $[0,1]\times [0,1]$ and
$0\leq G(t,s)\leq \cosh^2(1)/ \sinh(1) =\lambda$, for each $t,s\in
[0,1]$.
We shall prove the following theorem.

\begin{theorem}\label{theoreme}
Let $F:[0,1]\times E\times E\to 2^{ E}$ be a set-valued map with
nonempty closed values satisfying
\begin{itemize}
\item[(i)] For each $(x,y)\in E\times E$, $t\mapsto F(t,x,y)$ is
measurable and integrably bounded;

\item[(ii)] There exists a function $m(\cdot)\in
L^{1}([0,1],\mathbb{R}^{+})$ such that for all $t\in [0,1]$
 and for all $x_{1}, x_{2},y\in E$
 $$
 H\big(F(t,x_{1},y),F(t,x_{2},y)\big) \leq m(t)\|x_{1}-x_{2}\|.
 $$
 \end{itemize}
Then, if $\int_0^1(1+m(s))ds<1/\lambda$, for all $r,s\in E$,
Problem \eqref{cauchy} has at least one solution on
 $[0,1]$.
 \end{theorem}


\section{Proof of the main result}

 Let $r,s$ be in $E$. We introduce first the function
 $\rho:[0,1]\to E$ defined by
$$
 \rho(t)= \frac{1}{2}(s-r)t^2+rt,\quad \forall t\in [0,1],
$$
and the multifunction
 $H:[0,1]\times\mathcal{C}([0,1],E)\to 2^E$ defined by
\begin{equation}
H(t,y(\cdot))=y(t)-F(t,y(t)+\rho(t),\dot{y}(t)+\dot{\rho}(t))+(s-r)
\label{relg}
\end{equation}
for all $(t,y(\cdot))\in[0,1]\times\mathcal{C}([0,1],E)$.
Consider the problem:
\begin{equation} \label{Axcauchy}
  \begin{gathered}
 -\ddot{y}(t)+y(t) \in  H(t,y(\cdot)), \quad \text{a.e. on } [0,1] \\
\dot{y}(0)=0,\quad \dot{y}(1)=0.
\end{gathered}
\end{equation}

We should point out that the function $y(\cdot)$
is a solution of \eqref{Axcauchy}, if and only if the function
$x(t)= y(t)+\rho(t)$ is a solution of \eqref{cauchy},
for all $t\in [0,1]$.

Next, by Lemma \ref{lemme2}, for $y(\cdot)\in \mathcal{C}([0,1],E)$,
$F(\cdot,y(\cdot),\dot{y}(\cdot))$ is closed and measurable,
then it has a measurable selection which, by hypothesis (i),
 belongs to $L^{1}([0,1],E)$. Thus the set
$$
S_{F,y(\cdot)}:=\Big\{f\in L^{1}([0,1],E): f(t)\in
F(t,y(t),\dot{y}(t)) \text{ for a.e. } t\in [0,1]\Big\}
$$
is nonempty. Let us transform  problem \eqref{Axcauchy}
into a fixed point problem. Consider the multi-valued map,
$$
T:\mathcal{C}([0,1],E)\to 2^{\mathcal{C}([0,1],E)}
$$
defined as follows, for $y(\cdot)\in \mathcal{C}([0,1],E)$,
$$
T(y(\cdot))=\Big\{z(\cdot)\in \mathcal{C}([0,1],E):
z(t)=\int_{0}^{1}G(t,s)h(s)ds,\forall t\in [0,1], h\in
S_{H,y(\cdot)}\Big\},
$$
where
$$
S_{H,y(\cdot)}:=\Big\{h\in L^{1}([0,1],E): h(t)\in H(t,y(\cdot))\;
\text{for a.e.}\; t\in [0,1]\Big\}.
$$

We shall show that $T$ satisfies the assumptions of Lemma
\ref{lemme1}. The proof will be given in two steps:

\paragraph{Step 1: $T$ has non-empty closed-values.}
 Indeed, let $(y_p(\cdot))_{p\geq 0}\in T(y(\cdot))$ such
that $(y_p(\cdot))_{p\geq 0}$ converges to $\bar y(\cdot)$ in
$\mathcal{C}([0,1],E)$. Then $\bar y(\cdot)\in \mathcal{C}([0,1],E)$
and for each $t\in [0,1]$,
$$
y_p(t)\in \int_{0}^{1}G(t,s)H(s,y(\cdot))ds,
$$
where
$ \int_{0}^{1}G(t,s)H(s,y(\cdot))ds$
is the Aumann's integral of $G(t,.)H(.,y)$, which is defined as
$$
 \int_{0}^{1}G(t,s)H(s,y(\cdot))ds
 =\Big\{\int_{0}^{1}G(t,s)h(s)ds,\;h\in S_{H,y(\cdot)}\Big\}.
$$
Using the fact that the set-valued map $F$ is closed
and from \eqref{relg}, we conclude that the set
\begin{gather*}
\int_{0}^{1}G(t,s)H(s,y(\cdot))ds
\end{gather*}
is closed for all $t\in [0,1]$. Then
$$
\bar y(t)\in \int_{0}^{1}G(t,s)H(s,y(\cdot))ds.
$$
 So, there exists $h\in S_{H,y(\cdot)}$ such that
$$
\bar y(t)=\int_{0}^{1}G(t,s)h(s)ds.
$$
 Hence $\bar y(\cdot)\in T(y(\cdot))$. So $T(y(\cdot))$ is closed for
each $y(\cdot)\in \mathcal{C}([0,1],E)$.



\paragraph{Step 2: $T$ is a contraction.}
 Indeed, let $y_1(\cdot), y_2(\cdot)\in \mathcal{C}([0,1],E)$
and consider $z_1(\cdot)\in T(y_1(\cdot))$. Then there
exists $h_1\in S_{H,y_1(\cdot)}$ such that
$$
z_1(t)=\int_{0}^{1}G(t,s)h_1(s)ds,\quad \forall t\in [0,1].
$$
Using \eqref{relg}, there exists $f_1\in S_{F,y_1(\cdot)}$ such
that
 $$
 h_1(t)=y_1(t)-f_1(t)+(s-r),\quad \forall t\in [0,1].
 $$
On the other hand, let $\varepsilon>0$ and consider
the valued map $U_{\varepsilon}:[0,1]\to 2^{E}$, given by
$$
U_{\varepsilon}(t)=\big\{x\in E:\|f_1(t)-x\|\leq m(t)\|y_1(t)-y_2(t)\|
+\varepsilon\big\}.
$$
We claim that $U_{\varepsilon}(t)$ is nonempty, for each
$t\in [0,1]$.
Indeed, let $t\in [0,1]$, we have
\[
H\big(F(t,y_1(t),\dot{y}_1(t)),F(t,y_2(t),\dot{y}_2(t))\big)
\leq  m(t)\|y_1(t)-y_2(t)\|.
\]
Hence, there exists $x\in F(t,y_2(t),\dot{y}_2(t))$, such that
\[
\|f_1(t)-x\| \leq  m(t)\|y_1(t)-y_2(t)\|+\varepsilon.
\]
 By \cite[Theorem III.40]{castaing}, the multifunction
\begin{equation}
V:t\to U_{\varepsilon}(t)\cap F(t,y_2(t),\dot{y}_2(t))\quad
\text{ is measurable.} \label{measura}
\end{equation}
Then there exists a measurable selection for $V$ denoted
$f_2$ such that, for all $t\in [0,1]$,
$$
f_2(t)\in F(t,y_2(t),\dot{y}_2(t))
$$
and
$$
\|f_1(t)-f_2(t)\|\:\leq\: m(t)\|y_1(t)-y_2(t)\|+\varepsilon.
$$
Now,  for all $t\in[0,1]$, set $h_2(t)=y_2(t)-f_2(t)+(s-r)$ and
$$
z_2(t)=\int_{0}^{1}G(t,s)h_2(s)ds.
$$
 We have
\begin{align*}
\|z_1(t)-z_2(t)\|
& \leq \int_{0}^{1}\|G(t,s)\|\|h_1(s)-h_2(s)\|ds\\
& \leq \lambda\int_0^1\|y_1(s)-y_2(s)\|ds
 +\lambda\int_0^1\|f_1(s)-f_2(s)\|ds \\
& \leq \lambda\int_0^1\|y_1(s)-y_2(s)\|ds
 +\lambda\int_{0}^{1}m(s)\|y_1(s)-y_2(s)\|ds+\lambda\varepsilon\\
&\leq \lambda\|y_1(\cdot)-y_2(\cdot)\|_\infty\int_{0}^{1}(1+m(s))ds
 +\lambda\varepsilon.
\end{align*}
So, we conclude that
\[
\|z_1(\cdot)-z_2(\cdot)\|_{\infty}
\leq \lambda\|y_1(\cdot)-y_2(\cdot)\|_\infty\int_{0}^{1}(1+m(s))ds
+\lambda\varepsilon.
\]
 By an analogous relation, obtained by interchanging the
roles of $y_1(\cdot)$ and $y_2(\cdot)$, it follows that
\[
H\big(T(y_1(\cdot)),T(y_2(\cdot))\big)
\leq \lambda\|y_1(\cdot)-y_2(\cdot)\|_\infty\int_{0}^{1}(1+m(s))ds
 +\lambda\varepsilon.
\]
By letting $\varepsilon\to 0$, we obtain
\[
H\big(T(y_1(\cdot)),T(y_2(\cdot))\big)
\leq \lambda\|y_1(\cdot)-y_2(\cdot)\|_\infty\int_{0}^{1}(1+m(s))ds.
\]
Consequently, if $\int_0^1(1+m(s))ds<\frac{1}{\lambda}$, $T$
is a contraction. By Lemma \ref{lemme1}, $T$ has a fixed point which
is a solution of \eqref{Axcauchy}. The proof is complete.

\subsection*{Acknowledgements}
The author would like to thank the anonymous referee for his/her
careful and thorough reading of the paper.


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\end{thebibliography}

\end{document}