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

\AtBeginDocument{{\noindent\small
\emph{Electronic Journal of Differential Equations},
Vol. 2010(2010), No. 85, pp. 1--9.\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-2010/85\hfil Second order differential inclusions]
{Second-order differential inclusions with Lipschitz right-hand
sides}

\author[D. Azzam-Laouir, F. Bounama\hfil EJDE-2010/85\hfilneg]
{Dalila Azzam-Laouir, Fatiha Bounama}  % in alphabetical order

\address{Dalila Azzam-Laouir \newline
Laboratoire de Math\'ematiques Pures et Appliqu\'ees,
D\'epartement de Math\'ematiques, 
Universit\'e de Jijel, Alg\'erie}
\email{laouir.dalila@gmail.com}

\address{Fatiha Bounama \newline
Laboratoire de Math\'ematiques Pures et Appliqu\'ees,
D\'epartement de Math\'ematiques, 
Universit\'e de Jijel, Alg\'erie}
\email{bounamaf@yahoo.fr}

\thanks{Submitted March 29, 2010. Published June 18, 2010.}
\subjclass[2000]{34A60, 34B15, 47H10}
\keywords{Differential inclusion; Lipschitz multifunction}

\begin{abstract}
 We study the existence of solutions of a three-point boundary-value
 problem for a second-order differential inclusion,
 \begin{gather*}
 \ddot u(t)\in F(t,u(t),\dot u(t)),\quad\text{a.e. }t\in [0,1],\\
 u(0)=0, \quad u(\theta)=u(1).
 \end{gather*}
 Here $F$ is a set-valued mapping from $[0,1]\times E\times E$ to
 $E$ with nonempty closed values satisfying  a standard Lipschitz
 condition, and $E$ is a separable Banach space.
\end{abstract}

\maketitle
\numberwithin{equation}{section}
\newtheorem{theorem}{Theorem}[section]
\newtheorem{proposition}[theorem]{Proposition}
\newtheorem{corollary}[theorem]{Corollary}

\section{Introduction}

We study the existence of solutions to the second-order
differential inclusion
\begin{equation} \label{PF}
\begin{gathered}
 \ddot u(t)\in F(t,u(t),\dot u(t)),\quad \text{a.e. }t\in [0,1],\\
 u(0)=0, \quad u(\theta)=u(1),
 \end{gathered}
\end{equation}
where $F:[0,1]\times E\times E \to E$ is a nonempty closed
valued multifunction and $\theta$ is a given number in $[0,1]$.
Existence of solutions for \eqref{PF} has been investigated
by many authors \cite{ACT,AL,ALT,IG}
under the assumption that $F$ is a convex bounded-valued
multifunction upper semicontinuous on $E\times E$ and
integrably compact.

The aim of this article is to provide existence of solutions
for \eqref{PF} under the standard Lipschitz condition for
the multifunction $F$, when it is nonconvex.

After some preliminaries in section 3, we present our
main result which is the existence of
$\mathbf{W}^{2,1}_{E}([0,1])$-solutions for \eqref{PF}.
We suppose that $F$ is  a closed valued multifunction satisfying
the  Lipschitz condition
$$
 \mathcal{H}(F(t,x_1,y_1),F(t,x_2,y_2))
 \leq k_1(t)\| x_1-y_1\|+k_2(t)\| x_2-y_2\|
$$
where $\mathcal{H}(\cdot,\cdot)$ stands for the Hausdorff distance.

For  first-order differential inclusions satisfying the
standard Lipschitz condition we refer  the reader to
\cite{CV, F,G,W} and the  references therein.

 \section{Notation and preliminaries}

 In this article, $(E, \|\cdot\|)$ is a separable Banach
space and $E'$ is its topological dual, $\overline{\mathbf{B}}_E$ is
the closed unit  ball of $E$,
$\mathcal{L}([0,1])$ is the $\sigma$-algebra of Lebesgue-measurable
sets of $[0,1]$, $\lambda=dt$ is the Lebesgue measure on $[0,1]$,
and $\mathcal{B}(E)$ is the $\sigma$-algebra of Borel subsets of
$E$. By $ L^{1}_{E}([0,1])$, we denote the space of all
Lebesgue-Bochner integrable E-valued mappings defined on $[0,1]$.

Let $\mathbf{C}_{E}([0,1])$ be the Banach space of all continuous
mappings $u:[0,1] \to  E$, endowed with the supremum norm, and let
$\mathbf{C}^{1}_{E}([0,1])$ be the Banach space of all continuous
mappings $u : [0,1] \to E$ with continuous derivative,
equipped with the norm
$$
\| u\|_{\mathbf{C}^{1}}=\max\{\max_{t\in [0,1]} \| u(t)\|, \max_{t\in
[0,1]}\| \dot u(t)\|\}.
$$
Recall that a mapping $v:[0,1]\to E$ is said to be scalarly derivable
when there exists some mapping
$\dot v:[0,1]\to E$ (called the weak derivative of $v$) such that,
for every $x'\in E'$, the scalar function
$\langle x',v(\cdot)\rangle$ is derivable and its derivative is equal to
$\langle x',\dot v(\cdot)\rangle$. The weak derivative
$\ddot v$ of $\dot v$
when it exists is the weak second derivative.

By $\mathbf{W}^{2,1}_{E}([0,1])$ we denote the space
of all continuous mappings $u\in \mathbf{C}_{E}([0,1])$ such that
their first usual derivatives are continuous and scalarly
derivable and such that
$\ddot u\in  L^{1}_{E}([0,1])$.

 For closed subsets $A$
and $B$ of $E$, the Hausdorff distance between $A$ and $B$ is
defined by
$$
\mathcal{H}(A,B)=\sup(e(A,B)), e(B,A))
$$
where
$$
e(A,B) =\sup_{a\in A}d(a,B)=\sup_{a\in A}(\inf_{b\in B}\| a-b\|)
$$
stands for the excess of $A$ over $B$.

\section{Existence results under Lipschitz condition}

We begin with a proposition  that
summarizes some properties of some Green type function
(see \cite{Az,ACT,H,IG}). It will  use it in the study of our
 boundary value problems.

\begin{proposition} \label{prop3.1}
 Let $E$ be a separable Banach space and let $G:[0,1]\times [0,1]\to
\mathbb{R}$ be the function defined by
\[
G(t,s) =\begin{cases}
    -s &\text{if } 0\leq s\leq t,\\
    -t &\text{if } t<s\leq \theta, \\
    t(s-1)/(1-\theta)&\text{if }\theta<s\leq 1,
\end{cases}
\]
if $0\leq t<\theta$, and
\[
G(t,s) =\begin{cases}
-s &\text{if } 0\leq s< \theta,\\
    (\theta(s-t)+s(t-1))/(1-\theta)&\text{if } \theta\leq s\leq t, \\
    t(s-1)/(1-\theta)&\text{if }t<s\leq 1,
\end{cases}
\]
if $\theta\leq t\leq 1$.
Then the following assertions hold.
\begin{itemize}
\item[(1)] If $u\in \mathbf{W}^{2,1}_{E}([0,1])$ with $u(0)=0$
and $u(\theta)=u(1)$, then
$$
u(t)=\int_{0}^{1} G(t,s)\ddot u(s)ds,\quad \forall t\in [0,1].
$$

\item[(2)] $G(\cdot, s)$ is derivable on $[0,1]$ for every
$s\in [0,1]$, and its derivative is given by
\[
\frac{\partial G}{\partial t}(t,s)
=\begin{cases}
0 &\text{if } 0\leq s\leq t,\\
    -1 &\text{if } t<s\leq \theta, \\
    (s-1)/(1-\theta)&\text{if }\theta<s\leq 1,
\end{cases}
\]
if $0\leq t< \theta$, and
\[
\frac{\partial G}{\partial t}(t,s)
=\begin{cases}
0 &\text{if } 0\leq s< \theta,\\
    (s-\theta)/(1-\theta)&\text{if }  \theta\leq s\leq t, \\
    (s-1)/(1-\theta)&\text{if } t<s\leq 1,
\end{cases}
\]
if $\theta\leq t\leq 1$.

\item[(3)] $G(\cdot,\cdot)$, and
$\frac{\partial G}{\partial t}(\cdot,\cdot)$ satisfy
\begin{equation}
\sup _{t,s\in [0,1]} \vert G(t,s)\vert\leq 1,\quad
\sup _{t,s\in [0,1]} \vert \frac{\partial G}{\partial
t}(t,s)\vert\leq 1.\label{e3.1}
\end{equation}

\item[(4)] For $f\in  L^{1}_{E}([0,1])$ and for
the mapping $u_f: [0,1]\to E$ defined by
\begin{equation}
u_f(t)=\int_{0}^{1} G(t,s)f(s)ds, \quad \forall t\in [0,1],\label{e3.2}
\end{equation}
one has $u_f(0)=0$ and $u_f(\theta)=u_f(1)$.
Furthermore, the mapping $u_f$ is derivable, and its derivative
$\dot u_f$ satisfies
\begin{equation}
 \lim_{h\to 0}\frac{u_f(t+h)-u_f(t)}{h}=
\dot u_f(t)=\int_{0}^{1} \frac{\partial G}{\partial
t}(t,s)f(s)ds\label{e3.3}
\end{equation}
for all $t\in [0,1]$. Consequently, $\dot u_f$ is a continuous
mapping from $[0,1]$ into the space $E$.

\item[(5)] The mapping $\dot u_f$ is scalarly derivable; that is,
there exists a mapping $\ddot u_f:[0,1]\to E$ such that,
for every $x'\in E'$, the scalar function
$\langle x',\dot u_f(\cdot)\rangle$ is
derivable with $\frac{d}{dt}\langle x',\dot u_f(t)\rangle=\langle
x',\ddot u_f(t)\rangle$; furthermore
\begin{equation}
\ddot u_f=f\quad \text{a.e. } \text{on } [0,1].\label{e3.4}
\end{equation}
\end{itemize}
 \end{proposition}

Let us mention a useful consequence of Proposition \ref{prop3.1}.

\begin{proposition} \label{prop3.2}
Let $E$ be a separable Banach space and let $f:[0,1]\to E$
be a continuous mapping
(respectively a mapping in $ L^{1}_{E}([0,1]))$.
Then the mapping
$$
u_f(t)=\int_{0}^{1} G(t,s)f(s)ds,\quad \forall t\in [0,1]
$$
is the unique $\mathbf{C}^{2}_{E}([0,1])$-solution (respectively
$\mathbf{W}^{2,1}_{E}([0,1])$-solution) to the differential equation
\begin{gather*}
    \ddot u(t)=f(t),\quad \forall t\in [0,1],\\
    u(0)=0,\quad  u(\theta)=u(1).
\end{gather*}
\end{proposition}

Now we are able to state and prove our main result. The approach
below used some techniques and arguments from \cite{ACT,F,G}.

\begin{theorem} \label{thm3.1}
Let $E$ be a separable Banach space and let $F:[0,1]\times E\times
E\to E$ be a measurable multifunction with nonempty
closed values. Let $g\in  L^1_E([0,1])$ and let
$u_g:[0,1]\to E$ be the mapping defined by
$$
u_g(t)=\int_0^1G(t,s)g(s)ds,\quad \forall t\in [0,1].
$$
Assume that for some $r\in ]0,+\infty]$ and
$$
\mathbf{X}_r=\{(t,x,y)\in [0,1]\times E\times E:
\| x-u_g(t)\|<r;\;\| y-\dot u_g(t)\|<r\},
$$
the following conditions hold:
\begin{itemize}
\item[(i)] there exist two functions $k_1,k_2\in
 L^1_{\mathbb{R}}([0,1])$ with $k_1(t)\geq 0$ and
$k_2(t)\geq 0$ satisfying
$\| k_1+k_2\|_{ L^1_{\mathbb{R}}}<1$ such that
$$
\mathcal{H}(F(t,x_1,y_1),F(t,x_2,y_2))\leq k_1(t)\| x_1-x_2\|+k_2(t)\|
y_1-y_2\|
$$
for all $(t,x_1,y_1),\,(t,x_2,y_2)\in \mathbf{X}_r$;

\item[(ii)]  there is $\eta\in  L^1_{\mathbb{R}}([0,1])$
satisfying $\| \eta\|_{ L^1_{\mathbb{R}}}<[1-\|
k_1+k_2\|_{ L^1_{\mathbb{R}}}]r$, such that
$$
d(g(t),F(t,u_g(t),\dot u_g(t)))\leq \eta(t),\quad \forall t\in [0,1].
$$
\end{itemize}
Then the differential inclusion \eqref{PF} has at least one solution
$u\in \mathbf{W}^{2,1}_E([0,1])$, with
$$
\| u(t)\|\leq r+\| g(t)\|,\quad
\| \dot u(t)\|\leq r+\|g(t)\|,\quad \forall t\in [0,1].
$$
\end{theorem}

\begin{proof} \textit{Step 1}.
Since $\| k_1+k_2\|_{ L^1_{\mathbb{R}}}<1$ and
$\|\eta\|_{ L^1_{\mathbb{R}}}<[1-\|
k_1+k_2\|_{ L^1_{\mathbb{R}}}]r$ we may choose some real
number $\alpha>0$ satisfying
\begin{equation}
(1+\alpha)\|k_1+k_2\|_{ L^1_{\mathbb{R}}}<1,\quad
(1+\alpha)\|\eta\|_{ L^1_{\mathbb{R}}}<[1-(1+\alpha)\|
k_1+k_2\|_{ L^1_{\mathbb{R}}}]r .\label{e3.5}
\end{equation}
We will define a sequence of mappings $f_n$, $n\in \mathbb{N}$, of
$ L^1_E([0,1])$ such that  the following
conditions are fulfilled (see \eqref{e3.2} for the definition of $u_f$).
\begin{gather}
f_n\in  L^1_E([0,1]), \quad
f_{n}(t)\in F(t,u_{f_{n-1}}(t),\dot u_{f_{n-1}}(t)),\quad
\text{a.e. }t\in [0,1];\label{e3.6}
\\
\| f_{n}(t)-f_{n-1}(t)\|\leq(1+\alpha)d(f_{n-1}(t),F(t,u_{f_{n-1}}(t),
\dot u_{f_{n-1}}(t))),\quad \forall t\in[0,1];\label{e3.7}
\\
\mathop{\rm gph}(u_{f_n}(\cdot),\dot u_{f_n}(\cdot))
=\{(u_{f_n}(t),\dot u_{f_n}(t)):\;t\in[0,1]\}
\subset \mathbf{X}_r.\label{e3.8}
\end{gather}
We put $f_0=g$ and $u_{f_0}(t)=\int_0^1G(t,s)f_0(s)ds=u_g(t)$,
for all $t\in [0,1]$. Let us consider the
multifunction $H_0:[0,1]\to E$ defined by
$$
H_0(t)=\{v\in F(t,u_{f_0}(t),\dot u_{f_0}(t)):\| v-f_0(t)\|
\leq(1+\alpha)d(f_0(t),F(t,u_{f_0}(t),\dot u_{f_0}(t)))\}.
$$
Observe first that $H_0(t)\neq \emptyset$ for any $t\in [0,1]$.

Since $F(\cdot,u_{f_0}(\cdot),\dot u_{f_0}(\cdot))$ is measurable,
the multifunction $H_0$ is also measurable with nonempty closed
values. In view of the existence theorem of measurable selections
(see \cite{CV}), there is a measurable mapping
$f_1:[0,1]\to E$ such that $f_1(t)\in
H_0(t)$, for all $t\in [0,1]$. This yields, for all $t\in [0,1]$,
$f_1(t)\in F(t,u_{f_0}(t),\dot u_{f_0}(t))$ and
$\|f_1(t)-f_0(t)\|\leq(1+\alpha)d(f_0(t),F(t,u_{f_0}(t),\dot
u_{f_0}(t)))$, and hence according to the assumption
(ii),
$$
\| f_1(t)-f_0(t)\|\leq (1+\alpha)\eta(t).
$$
So, we have
\begin{equation}
\| f_1(t)\| \leq \| f_1(t)-f_0(t)\| +\| f_0(t)\|
\leq(1+\alpha)\eta(t) +\| f_0(t)\|\label{e3.9}.
\end{equation}
Since $\eta\in  L^1_{\mathbb{R}}([0,1])$ and
$f_0\in  L^1_{E}([0,1])$, the last inequality shows that
$f_1\in  L^1_{E}([0,1])$. Then we define the mapping
$u_{f_1}:[0,1]\to E$ by
$$
u_{f_1}(t)=\int_0^1 G(t,s)f_1(s)ds,\quad\forall t\in [0,1],
$$
 and by relation \eqref{e3.3} in Proposition \ref{prop3.1}
$$
\dot u_{f_1}(t)=\int_0^1 \frac{\partial G}{\partial t}(t,s)f_1(s)ds,
\quad \forall t\in [0,1].
$$
On the other hand,
\begin{align*}
\| u_{f_1}(t)-u_{f_0}(t)\|
&=\|\int_0^1 G(t,s)(f_1(s)-f_0(s))ds\|\\
&\leq \int_0^1 \| f_1(s)-f_0(s)\| ds\\
&\leq (1+\alpha)\int_0^1 d(f_0(s),F(t,u_{f_0}(s),\dot u_{f_0}(s)))ds\\
&\leq  (1+\alpha)\|\eta\|_{ L^1_{\mathbb{R}}}\\
&<[1-(1+\alpha)\| k_1+k_2\|_{ L^1_{E}}]r<r,
\end{align*}
the first inequality being due to \eqref{e3.1} and the fourth one to
\eqref{e3.5}. Similarly we  have
$$
\| \dot u_{f_1}(t)-\dot u_{f_0}(t)\|
=\|\int_0^1 \frac{\partial G}{\partial t}(t,s)(f_1(s)-f_0(s))ds\|
\leq \int_0^1 \| f_1(s)-f_0(s)\| ds <r.
$$
This shows that $\mathop{\rm gph}(u_{f_1}(\cdot),\dot u_{f_1}(\cdot))
\subset \mathbf{X}_r$.

Suppose that $f_i$ and $u_{f_i}$ have been defined on $[0,1]$
satisfying \eqref{e3.6}, \eqref{e3.7} and \eqref{e3.8} for
$i=0,1,\dots,n$. Let us
consider the multifunction $H_n:[0,1]\to E$ defined by
\begin{align*}
H_n(t)=\big\{&v\in F(t,u_{f_n}(t),\dot u_{f_n}(t)):\|
v-f_n(t)\|\leq(1+\alpha)d(f_n(t),\\
& F(t,u_{f_n}(t),\dot u_{f_n}(t))\big\}.
\end{align*}
Observe first that $H_n(t)\neq \emptyset$ for
any $t\in [0,1]$.

 Since $F(\cdot,u_{f_n}(\cdot),\dot u_{f_n}(\cdot))$ is measurable,
the multifunction $H_n$ is also measurable with nonempty closed values.
As above, in view of the
existence theorem of measurable selections (see \cite{CV}), there is
a measurable mapping $f_{n+1}:[0,1]\to E$ such that
$f_{n+1}(t)\in H_n(t)$, for all $t\in [0,1]$. This yields for all
$t\in [0,1]$, $f_{n+1}(t)\in  F(t,u_{f_n}(t),\dot u_{f_n}(t))$ and
$\|f_{n+1}(t)-f_n(t)\|\leq(1+\alpha)d(f_n(t),F(t,u_{f_n}(t),\dot
u_{f_n}(t))$. The second inequality implies
\begin{equation}
\begin{aligned}
&\| f_{n+1}(t)-f_n(t)\|\\
& \leq (1+\alpha)d(f_n(t),F(t,u_{f_n}(t),\dot u_{f_n}(t)))\\
&\leq (1+\alpha)\mathcal{H}(F(t,u_{f_{n-1}}(t),\dot u_{f_{n-1}}(t)),
 F(t,u_{f_n}(t),\dot u_{f_n}(t)))\\
&\leq  (1+\alpha) [k_1(t)\|
u_{f_{n}}(t)-u_{f_{n-1}}(t)\|+k_2(t)\| \dot u_{f_{n}}(t)-\dot
u_{f_{n-1}}(t)\|],
\end{aligned} \label{e3.10}
\end{equation}
where the last inequality follows from  assumption (i).
However, the mapping $f_{n-1}$ and $f_n$ being integrable by the
induction assumption, we have on the one hand
\begin{align*}
\| u_{f_n}(t)-u_{f_{n-1}}(t)\|
&= \|\int_0^1 G(t,s)f_n(s)ds-\int_0^1 G(t,s)f_{n-1}(s)ds\|\\
&\leq \int_0^1 \vert G(t,s)\vert\| f_n(s)-f_{n-1}(s)\| ds\\
&\leq \| f_n-f_{n-1} \|_{ L^1_{E}},
\end{align*}
where the last inequality follows from the first inequality in
\eqref{e3.1}. On the other hand using the second
inequality in \eqref{e3.6}, we may write
\begin{align*}
\| \dot u_{f_n}(t)-\dot u_{f_{n-1}}(t)\|
&= \|\int_0^1
\frac{\partial G}{\partial t}(t,s)f_n(s)ds-\int_0^1
\frac{\partial G}{\partial t}(t,s)f_{n-1}(s)ds\|\\
&\leq \int_0^1 \vert \frac{\partial G}{\partial t}(t,s)
\vert\| f_n(s)-f_{n-1}(s)\| ds\\
&\leq \| f_n-f_{n-1}\|_{ L^1_{E}}.
\end{align*}
Combining  those last inequalities and \eqref{e3.10}, we obtain
\begin{equation}
\| f_{n+1}(t)-f_n(t)\|
\leq(1+\alpha)(k_1(t)+k_2(t))\| f_n-f_{n-1}\|_{ L^1_{E}}.\label{e3.11}
\end{equation}
Since $k_1,\,k_2\in  L^1_{\mathbb{R}}([0,1])$ and
$f_n, f_{n-1}\in  L^1_{E}([0,1])$, we see that
$f_{n+1}\in  L^1_{E}([0,1])$. We may then integrate \eqref{e3.11},
\begin{align*}
\int_0^1 \| f_{n+1}(t)-f_n(t)\| dt
&\leq (1+\alpha)\int_0^1(k_1(t)+k_2(t))\| f_n-f_{n-1}\|_{ L^1_{E}}dt\\
&= (1+\alpha)\| k_1+k_2\|_{ L^1_{\mathbb{R}}}\|
f_n-f_{n-1}\|_{ L^1_{E}};
\end{align*}
that is,
\begin{equation}
\| f_{n+1}-f_{n}\|_{ L^1_{E}}
\leq (1+\alpha)\| k_1+k_2\|_{ L^1_{\mathbb{R}}}
\| f_n-f_{n-1}\|_{ L^1_{E}}. \label{e3.12}
\end{equation}
 Taking \eqref{e3.2} into account, we define the mapping
$u_{f_{n+1}}:[0,1]\to E$ by
$$
u_{f_{n+1}}(t)=\int_0^1 G(t,s)f_{n+1}(s)ds,\quad
\forall t\in [0,1],
$$
and  relation \eqref{e3.3} in Proposition \ref{prop3.1} says that
$u_{f_{n+1}}$ is derivable with
$$
\dot u_{f_{n+1}}(t)=\int_0^1 \frac{\partial G}{\partial t}
(t,s)f_{n+1}(s)ds,\quad \forall t\in [0,1].
$$
Next, let us prove that the graph of
$(u_{f_{n+1}}(\cdot),\dot u_{f_{n+1}}(\cdot))$
is contained in $\mathbf{X}_r$. Setting
$\gamma=(1+\alpha)\| k_1+k_2\|_{ L^1_{\mathbb{R}}}$
and using successively relation \eqref{e3.12}, we obtain
\begin{equation}
\| f_{n+1}-f_n\|_{ L^1_{E}}\leq \gamma^n\|
f_1-f_0\|\leq \gamma^n(1+\alpha)\|
\eta\|_{ L^1_{\mathbb{R}}}\label{e3.13}
\end{equation}
with $\gamma<1$, the last inequality being due to \eqref{e3.9}.
 On the other hand, since
\begin{gather*}
\| u_{f_{n+1}}(t)-u_{f_n}(t)\|\leq \| f_{n+1}-f_n\|_{ L^1_{E}},\\
\| \dot u_{f_{n+1}}(t)-\dot u_{f_n}(t)\|\leq \| f_{n+1}-f_n\|_{ L^1_{E}},
\end{gather*}
\eqref{e3.13} yields
\begin{equation}
\| u_{f_{n+1}}-u_{f_n}\|_{\mathbf{C}^1}
\leq\| f_{n+1}-f_n\|_{ L^1_{E}}\leq \gamma^n(1+\alpha)\|\eta\|_
{ L^1_{\mathbb{R}}}.\label{e3.14}
\end{equation}
Writing,
\begin{align*}
\| u_{f_{n+1}}(t)-u_{f_0}(t)\|
&\leq \|  u_{f_{n+1}}(t)-u_{f_n}(t)\| + \| u_{f_n}(t)-u_{f_0}(t)\|\\
&\leq \gamma^n (1+\alpha)\|\eta
\|_{ L^1_{\mathbb{R}}}+\| u_{f_n}(t)-u_{f_0}(t)\|,
\end{align*}
and using successively this relation, we obtain thanks to the
second inequality of \eqref{e3.11},
\begin{equation}
\| u_{f_{n+1}}(t)-u_{f_0}(t)\|\leq(\sum_{p=0}^n\gamma^p)(1+\alpha)\|\eta \|_{ L^1_{\mathbb{R}}}
\leq \frac{1}{1-\gamma}(1+\alpha)\|\eta
\|_{ L^1_{\mathbb{R}}} <r.\label{e3.15}
\end{equation}
Using again \eqref{e3.14} to write
\begin{align*}
\| \dot u_{f_{n+1}}(t)-\dot u_{f_0}(t)\|
&\leq \|  \dot u_{f_{n+1}}(t)-\dot u_{f_n}(t)\| + \| \dot u_{f_n}(t)-\dot u_{f_0}(t)\|\\
&\leq \gamma^n (1+\alpha)\|\eta
\|_{ L^1_{\mathbb{R}}}+\| \dot u_{f_n}(t)-\dot
u_{f_0}(t)\|,
\end{align*}
we obtain, in a similar way,
\begin{equation}
 \| \dot u_{f_{n+1}}(t)-\dot u_{f_0}(t)\| <r.\label{e3.16}
 \end{equation}
Consequently the sequences $(f_n)$ and $(u_{f_n})$ are well defined
satisfying \eqref{e3.6}, \eqref{e3.7} and \eqref{e3.8}.

\textit{Step 2}. By  \eqref{e3.13} we see that $(f_n)$ is a Cauchy
sequence in $ L^1_{E}([0,1])$, hence it converges to some mapping
$f\in  L^1_{E}([0,1])$. In the same way \eqref{e3.14} shows that
$(u_{f_n})$ is a Cauchy sequence in $\mathbf{C}^1_{E}([0,1])$,
consequently it converges to some mapping $w\in
\mathbf{C}^1_{E}([0,1])$. Observe that
\begin{align*}
\| u_{f_n}(t)-u_f(t)\|
&= \|\int_0^1 G(t,s)f_n(s)ds-\int_0^1 G(t,s)f(s)ds\|\\
&\leq \int_0^1\| f_n(s)-f(s)\| ds=\| f_n-f\|_{ L^1_{E}},
\end{align*}
and
\begin{align*}
\| \dot u_{f_n}(t)-\dot u_f(t)\|
&= \|\int_0^1 \frac{\partial G}{\partial t}(t,s)f_n(s)ds-\int_0^1
\frac{\partial G}{\partial t}(t,s)f(s)ds\|\\
&\leq \int_0^1\| f_n(s)-f(s)\| ds=\| f_n-f\|_{ L^1_{E}},
\end{align*}
which, according to the strong convergence in
$ L^1_{E}([0,1])$ of $(f_n)$ to the mapping $f$ means that
$(u_{f_n})$ converges in
$(\mathbf{C}^1_{E}([0,1]),\|\cdot\|_{\mathbf{C}^1})$ to $u_f$.
Thus we get $w=u_f$, and by Proposition \ref{prop3.1}
(relations \eqref{e3.2},
\eqref{e3.3} and \eqref{e3.4}) we have $\ddot u_f=f$, with
$u_f(0)=0$, $u_f(\theta)=u_f(1)$.

Let us prove now that $u_f$ is a solution of the problem
\eqref{PF}. For this purpose, let us prove that, for each
$t\in [0,1]$, the graph of the multifunction $(x,y)\mapsto
F(t,x,y)$ is closed relatively to $\mathbf{X}_r(t)\times E$ where
$$
\mathbf{X}_r(t)=\{(x,y)\in E\times E:\;(t,x,y)\in
\mathbf{X}_r\}.
$$
 Let $(x_n,y_n,v_n)_n$ be a sequence
in $\mathop{\rm gph}(F(t,\cdot,\cdot))$ converging to
$(x,y,v)\in \mathbf{X}_r(t)\times E$.
For each integer $n$, $v_n\in
F(t,x_n,y_n)$, and hence
\begin{align*}
d(v,F(t,x,y))
&\leq \| v-v_n\| + d(v_n,F(t,x,y))\\
&\leq \| v-v_n\|+\mathcal{H}(F(t,x_n,y_n),F(t,x,y))\\
&\leq \| v-v_n\| +k_1(t)\| x_n-x\| +k_2(t)\|
y_n-y\|.
\end{align*}
Since the last member goes to $0$ as $n$ tends to $+\infty$, this
says that $v\in F(t,x,y)$ according to the closedness of this set.
Consequently the graph of $F(t,\cdot,\cdot)$ is closed relatively to
$\mathbf{X}_r(t)\times E$. Since $(f_n)$ converges to $f$ strongly
in $ L^1_{E}([0,1])$, by extracting a subsequence we may
suppose that $(f_n)$ converges to $f$ almost everywhere on $[0,1]$.
As $f_{n+1}(t)\in F(t,u_{f_n}(t),\dot u_{f_n}(t))$ and as
$(u_{f_n})$ converges to $u_f$ in $\mathbf{C}^1_{E}([0,1])$ and
$(t,u_{f_n}(t),\dot u_{f_n}(t))$,
$(t,u_{f}(t),\dot u_{f}(t))\in \mathbf{X}_r$, we
conclude that $f(t)\in F(t,u_f(t),\dot u_f(t))$, a.e., equivalently
$\ddot u_f(t)\in F(t,u_f(t),\dot u_f(t))$, a.e., with
$u_f(0)=0;\,u_f(\theta)=u_f(1)$. Furthermore, the relations
\eqref{e3.15}
and \eqref{e3.16} show that
$$
\| u_f(t)\|\leq r+\|g(t)\|,\;\;\| \dot u_f(t)\|
\leq r+\| g(t)\|,\quad \forall t\in [0,1].
$$
This completes the proof of our theorem.
\end{proof}

 The following corollary translates the above result in
a more amenable way.

\begin{corollary} \label{coro3.4}
Let $E$ be a separable Banach space and  $F:[0,1]\times E\times
E\to E$ be a measurable multifunction with nonempty
closed values such that
\begin{itemize}
 \item[(i)] there exist two functions
$k_1,\,k_2\in  L^1_{\mathbb{R}}([0,1])$ with $k_1(t)\geq 0$
and $k_2(t)\geq 0$ satisfying
$\|k_1+k_2\|_{ L^1_{\mathbb{R}}}<1$ such that
$$
\mathcal{H}(F(t,x_1,y_1),F(t,x_2,y_2))
\leq k_1(t)\| x_1-x_2\|+k_2(t)\| y_1-y_2\|
$$
for all $(t,x_1,y_1),\,(t,x_2,y_2)\in [0,1]\times E\times E$;

\item[(ii)]  the function $t\mapsto d(0,F(t,0,0))$ is
integrable.
\end{itemize}
Then the differential inclusion \eqref{PF} has at least a
solution $u\in \mathbf{W}^{2,1}_E([0,1])$.
\end{corollary}

\begin{proof}
Taking $g\equiv 0$ and $r=+\infty$, we see in
Theorem \ref{thm3.1} that $\mathbf{X}_r=[0,1]\times E\times E$. Further
putting $\eta(t)=d(0, F(t,0,0))$, the function $\eta$ is integrable
and satisfies the assumption (ii) of Theorem \ref{thm3.1}. We may
then conclude that the corollary is a consequence of
Theorem \ref{thm3.1}.
\end{proof}

\subsection*{Acknowledgements}
The authors would like to thank Professor L. Thibault
for his important comments and suggestions.

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\end{document}