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

\AtBeginDocument{{\noindent\small
\emph{Electronic Journal of Differential Equations},
Vol. 2012 (2012), No. 30, pp. 1--13.\newline
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu
\newline ftp ejde.math.txstate.edu}
\thanks{\copyright 2012 Texas State University - San Marcos.}
\vspace{9mm}}

\begin{document}
\title[\hfilneg EJDE-2012/30\hfil Higher order viability problem ]
{Higher order viability problem in Banach spaces}

\author[M. Aitalioubrahim, S. Sajid \hfil EJDE-2012/30\hfilneg]
{Myelkebir Aitalioubrahim, Said Sajid}  % in alphabetical order

\address{Myelkebir Aitalioubrahim \hfill\break
University Hassan II-Mohammedia, Laboratory Mathematics,
Cryptography and Mecanics, F.S.T, BP 146, Mohammedia, Morocco}
\email{aitalifr@yahoo.fr}

\address{Said Sajid \hfill\break
University Hassan II-Mohammedia, Laboratory Mathematics,
Cryptography and Mecanics, F.S.T, BP 146, Mohammedia, Morocco}
\email{saidsajid@hotmail.com}

\thanks{Submitted May 30, 2011. Published February 21, 2012.}
\subjclass[2000]{34A60}
\keywords{Differential inclusion; measurability;  selection; viability}

\begin{abstract}
 We show the existence of viable solutions to the differential inclusion
 \begin{gather*}
 x^{(k)}(t) \in F(t,x(t))\\
 x(0)=x_0,\quad x^{(i)}(0)=y^i_0,\quad  i=1,\dots,k-1,\\
 x(t) \in K\quad\text{on } [0,T],
 \end{gather*}
 where $k \geq 1$, $K$ is a closed subset of a  separable  Banach space
 and $F(t,x)$ is an integrable bounded multifunction  with closed values,
 (strongly) measurable in $t$ and Lipschitz  continuous in $x$.
\end{abstract}

\maketitle
\numberwithin{equation}{section}
\newtheorem{theorem}{Theorem}[section]
\newtheorem{lemma}[theorem]{Lemma}
\newtheorem{remark}[theorem]{Remark}
\newtheorem{claim}[theorem]{Claim}
%\newtheorem{definition}[theorem]{Definition}
\allowdisplaybreaks

\section{Introduction}

The aim of this paper is to establish the existence of local
solutions of the higher-order viability problem
\begin{equation} \label{cauchy}
 \begin{gathered}
 x^{(k)}(t) \in F(t,x(t)) \quad\text{a.e on } [0,T]   \\
 x(0)=x_0\in K, x^{(i)}(0)=y_0^{i}\in \Omega_i,\quad  i=1,\dots,k-1, \\
 x(t) \in K \quad\text{on } [0,T].
 \end{gathered}
\end{equation}
 where $K$ is a closed subset of a separable Banach space
$E$, $F:[0,1]\times K \to 2^{E}$ is a measurable
multifunction with respect to the first argument and Lipschitz
continuous with respect to the second argument,
$\Omega_1,\dots,\Omega_{k-1}$ are open subsets of $E$ and
$(x_0, y_0^{1},\dots,y_0^{k-1})$ is given in $ K\times \prod_{i=1}^{k-1}\Omega_i$.

As regards the existence result of such problems, we refer to the
work of Marco and Murillo \cite{marco}, in the case when $F$ is a
convex and compact valued-multifunction in finite-dimensional space.

 First-order viability problems with the non-convex
Carath\'eodory Lipschitzean right-hand side in Banach spaces have been
studied  by Duc Ha \cite{ducha}. The author established a
multi-valued version of Larrieu's work \cite{larrieu}, assuming the
 tangential condition:
$$
   \liminf_{h \to
   0^{+}}\frac{1}{h}d\Big(x+\int_{t}^{t+h}F(s,x)ds,K\Big)=0,
$$
where $K$ is the viability set and $d(.,.)$ denotes
the Hausdorff's excess.

Lupulescu and Necula \cite{lupulescu2} extended the result of Duc Ha
\cite{ ducha} to first-order functional differential inclusions with
the non-convex Carath\'eodory Lipschitzean right-hand side in Banach
space. The authors used the same kind of tangential conditions that
in Duc Ha \cite{ducha}.

Recently, Aitalioubrahim and Sajid \cite{aitali} proved the
existence of viable solution to the following second-order
differential inclusions with the non-convex Carath\'eodory
Lipschitzean right-hand side in Banach space $E$:
\begin{equation} \label{problem1}
 \begin{gathered}
 \ddot{x}(t) \in F(t,x(t),\dot{x}(t))\quad\text{a.e.;}   \\
 (x(0),\dot{x}(0))=(x_0,y_0);  \\
 (x(t),\dot{x}(t))\in K\times \Omega;
 \end{gathered}
\end{equation}
 where $K$ (resp. $\Omega$) is a closed subset (resp. an
open subset) of $E$. The authors introduced the  tangential
condition:
\begin{equation}
   \liminf_{h \to
   0^{+}}\frac{1}{h^2}d\Big(x+hy+\frac{h}{2}\int_{t}^{t+h}F(s,x,y)ds,K\Big)=0.
\label{tangcon}
\end{equation}
In this paper we extend this result to the higher-order case with
the tangential condition:
$$
\liminf_{h \to
   0^{+}}\frac{1}{h^k}d\Big(x+\sum_{i=1}^{k-1}\frac{h^{i}}{i!}y^{i}
   +\frac{h^{k-1}}{k!}\int_{t}^{t+h}F(s,x)ds,K\Big)=0.
$$

\section{Preliminaries and statement of the main result}

Let $E$ be a separable Banach space with the norm $\|.\|$.
 For measurability purpose, $E$ (resp. $U\subset E$) is endowed with the
$\sigma$-algebra $B(E)$ (resp. $B(U)$) of Borel subsets for the strong topology 
and $[0,1]$ is endowed with Lebesgue measure and the $\sigma$-algebra
 of Lebesgue measurable subsets.
 For $x\in E$ and $r>0$ let $B(x,r):=\{y\in E; \| y-x \| <r\}$
be the open ball centered at $x$
with radius $r$ and $\bar B(x,r)$ be its closure and put
$B=B(0,1)$. For $x\in E$ and for nonempty sets $A, B$ of $E$ we
denote $d(x,A):=\inf\{\| y-x
\|; y\in A\}$, $d(A,B):=\sup\{d(x,B); x\in A\}$ and
$H(A,B)=\max\{d(A,B),d(B,A)\}$. 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}.

Let us recall the following Lemmas that will be used in the sequel.
For the proofs, we refer the reader to \cite{qiji}.

\begin{lemma}  \label{lemme1}
Let $\Omega$ be a nonempty set in $E$. Assume that
$F:[a,b]\times \Omega \to 2^{E}$ is a multifunction with
nonempty closed values satisfying:
\begin{itemize}
\item For every $x\in \Omega$, $F(.,x)$ is measurable on $[a,b]$;
\item For every $t\in [a,b]$, $F(t,.)$ is (Hausdorff) continuous on $\Omega$.
\end{itemize}
 Then for any measurable function $x(.): [a,b] \to \Omega$, the multifunction 
 $F(.,x(.))$ is measurable on $[a,b]$.
 \end{lemma}

 \begin{lemma} \label{lemme2}
Let $G: [a,b]\to 2^{E}$ be a measurable multifunction and $y(.): [a,b]\to E$ a
measurable function. Then for any positive measurable function 
$r(.): [a,b]\to \mathbb{R}^{+}$, there exists a measurable selection $g(.)$ of $G$ 
such that for almost all $t\in [a,b]$
\[
\| g(t)-y(t) \| \leq  d\big(y(t),G(t)\big)+r(t).
\]
\end{lemma}

Before stating our main result, for any integer $n\geq 2$, we
recall the tangent set of $n$th order denoted by
$A_K^n(x_0,x_1,\dots,x_{n-1})$ introduced by Marco and Murillo
\cite[Def. 3.1]{marco2} as follows.

For $y \in E$,we say that $y \in A_K^n(x_0,x_1,\dots,x_{n-1})$ if
$$
\liminf_{h \to
   0^{+}}\frac{{n!}}{h^k}d\Big(\sum_{i=0}^{n-1}\frac{h^{i}}{i!}x_{i}
   +\frac{h^n}{n!}y,K\Big)=0.
$$
Let $gr(A_K^n)$ be its graph.

Assume that the following hypotheses hold:
\begin{itemize}
\item[(H1)]   $K$ is a nonempty closed subset in $E$ and for
$i=1,\dots,k-1$, $\Omega_i$  is a nonempty open subset in $E$, such
that $K\times \prod_{i=1}^{k-1}\Omega_i \subset gr(A_K^n)$.
 \item[(H2)] $F:[0,1]\times K \to 2^{
E}$ is a set valued map with nonempty closed values satisfying
\begin{itemize}
\item[(i)] For each $x\in K$, $t\mapsto F(t,x)$ is measurable.
\item[(ii)]There is a function $m\in L^{1}([0,1],\mathbb{R}^{+})$ such that for all $t\in [0,1]$
 and for all $x_{1},x_{2}\in K$
$$
 H(F(t,x_{1}),F(t,x_{2})) \leq m(t)\| x_{1}-x_{2}\|
$$

\item[(iii)] For all  bounded subset $S$ of $K$, there is a function
 $g_{S}\in L^{1}([0,1],\mathbb{R}^{+})$ such that for all $t\in [0,1]$
 and for all $x \in S$
\begin{equation*}
 \| F(t,x) \| := \sup _{z\in F(t,x)} \| z \| \leq g_{S}(t)
\end{equation*}
\end{itemize}

\item[(H3)] (\textbf{Tangential condition})
 For every $(t,x,(y^{1},\dots,y^{k-1})) \in [0,1]\times K\times \prod_{i=1}^{k-1}\Omega_i$,
$$
\liminf_{h\to 0^{+}}\frac{1}{h^k}d\Big(x+\sum_{i=1}^{k-1}\frac{h^{i}}{i!}y^{i}
+\frac{h^{k-1}}{k!}\int_{t}^{t+h}F(s,x)ds,K\Big)=0.
$$
\end{itemize}


\begin{theorem} \label{result}
If assumptions {\rm (H1)--(H3)} are satisfied, then there exist  $T>0$ and
an absolutely continuous function  $x(.):[0,T]\to E$, for
which $x^{(i)}(.):[0,T]\to E$, for all $i=1,\dots,k-1$, is
also absolutely continuous,  such that $x(.)$  is solution of
\eqref{cauchy}.
\end{theorem}

\section{Proof of the main result}



 Let $r>0$ and  $\bar B(y_0^{i},r)\subset \Omega_i$ for $i=1,\dots,k-1$.
 Choose  $g\in L^{1}([0,1],\mathbb{R}^{+})$ such that
 \begin{equation}
 \| F(t,x) \|\leq  g(t)\quad \forall (t,x)\in [0,1]\times (K\cap B(x_0,r)). \label{defdeg}
 \end{equation}
Let $T_{1}>0$ and $T_{2}>0$ be such that
\begin{gather}
\int_0^{T_{1}}m(t)dt <1, \label{relationdem} \\
\int_0^{T_{2}}\Big(g(t)+(k-1)r+1+\sum_{i=1}^{k-1}\| y_0^i\|\Big)dt 
< \frac{r}{2}.\label{defdeg2}
\end{gather}
 For $\varepsilon >0$ there exists $\eta(\varepsilon)>0$ such that
\begin{equation}
\Big|\int_{t_{1}}^{t_2}g(\tau)d\tau\Big| < \varepsilon \;
\quad\text{if }  |t_1-t_2| < \eta(\varepsilon).\label{defdeita}
\end{equation}
  Set
\begin{gather} 
T=\min\{T_1,T_2,1\}, \label{defdegrt}\\
\alpha=\min\Big\{T,\frac{1}{2}\eta(\frac{\varepsilon}{4}),\frac{\varepsilon}{4}\Big\}.
\label{defdealfa}
 \end{gather}
 We will used the following Lemma to prove the main result.

\begin{lemma} \label{construction}
If assumptions {\rm (H1)--(H3)} are satisfied, then for all $\varepsilon>0$ and 
 all $y(.)\in L^{1}([0,T],E)$,
there exists $f\in L^{1}([0,T],E)$, $z(.):[0,T]\to E$
differentiable and a step function $\theta :[0,T]\to [0,T]$
such that
\begin{itemize}
\item $f(t)\in F(t,z(\theta(t)))$ for all $t \in [0,T]$;

\item $\| f(t)-y(t)\|\leq d(y(t),F(t,z(\theta(t)))+ \varepsilon$ for all $t \in [0,T]$;

\item $\big\|z^{(k-1)}(t)-y_0^{k-1}-\int_0^{t}f(\tau)d\tau\big\|\leq
\varepsilon$ for all $t \in [0,T]$.
\end{itemize}
\end{lemma}

\begin{proof} 
Let $\varepsilon>0$ and $y(.)\in L^{1}([0,T],E)$ be fixed.
 For $(0,x_0,(y_0^{1},\dots,y_0^{k-1}))\in [0,T]\times K\times \prod_{i=1}^{k-1}\Omega_i$, 
 by (H3),
$$
\liminf_{h \to
   0^{+}}\frac{1}{h^k}d\Big(x_0+\sum_{i=1}^{k-1}\frac{h^{i}}{i!}y_0^{i}+\frac{h^{k-1}}{k!}\int_0^{h}F(s,x_0)ds,K\Big)=0.
$$
Hence, there exists $0<h\leq\alpha$ such that
$$
d\Big(x_0+\sum_{i=1}^{k-1}\frac{h^{i}}{i!}y_0^{i}+\frac{h^{k-1}}{k!}
\int_0^{h}F(s,x_0)ds,K\Big)<\frac{\alpha h^k}{4k!}.
$$
 Put
 $$
 h_0:=\max\Big\{h\in ]0,\alpha]:
d\Big(x_0+\sum_{i=1}^{k-1}\frac{h^{i}}{i!}y_0^{i}+\frac{h^{k-1}}{k!}\int_0^{h}F(s,x_0)ds,K\Big)<\frac{\alpha
h^k}{4k!}\Big\}.
$$
In view of Lemma \ref{lemme2}, there exists a function
$f_0\in L^{1}([0,h_0],E)$ such that $f_0(t)\in F(t,x_0)$ and
$$
\| f_0(t)-y(t)\|\leq  d(y(t),F(t,x_0))+
\varepsilon,\quad \forall t \in [0,h_0].
$$
Then
$$
d\Big(x_0+\sum_{i=1}^{k-1}\frac{h_0^{i}}{i!}y_0^{i}+\frac{h_0^{k-1}}{k!}\int_0^{h_0}f_0(s)ds,K\Big)<\frac{\alpha
h_0^k}{4k!}.
$$
So, there exists $ x_{1}\in  K $ such that
\begin{align*}
& \frac{k!}{h_0^{k}}\big\|x_{1}-\Big(x_0+\sum_{i=1}^{k-1}
 \frac{h_0^{i}}{i!}y_0^{i}+\frac{h_0^{k-1}}{k!}\int_0^{h_0}f_0(s)ds\Big)\big\|\\
&\leq \frac{k!}{h_0^{k}}d\Big(x_0+\sum_{i=1}^{k-1}\frac{h_0^{i}}{i!}y_0^{i}
 +\frac{h^{k-1}}{k!}\int_0^{h}f_0(s)ds,K\Big)+\frac{\alpha}{4},
\end{align*}
hence
$$
\big\|\frac{x_{1}-x_0-\sum_{i=1}^{k-1}\frac{h_0^{i}}{i!}y_0^{i}}
-\frac{1}{h_0}\int_0^{h_0}f_0(s)ds\big\|< \alpha.
$$
Set
$$
u_0=\frac{x_{1}-x_0-\sum_{i=1}^{k-1}\frac{h_0^{i}}{i!}y_0^{i}}{\frac{h_0^{k}}{k!}},
$$
then
$$
x_{1}=\Big(x_0+\sum_{i=1}^{k-1}\frac{h_0^{i}}{i!}y_0^{i}+\frac{h_0^{k}}{k!}u_0\Big)\in
K, \quad
u_0\in \frac{1}{h_0}\int_0^{h_0}f_0(s)ds +\alpha B.
$$
For $i=1,\dots,k-1$, put
$$
y_1^{i}=\sum_{j=i}^{k-1}\frac{h_0^{j-i}}{(j-i)!}y_0^{j}+\frac{h_0^{k-i}}{(k-i)!}u_0.
$$
Since $f_0(t)\in F(t,x_0)$ for all $t\in [0,h_0]$ and
by \eqref{defdeg}, \eqref{defdeg2} and \eqref{defdealfa}, we have
\begin{align*}
\| x_{1}-x_0\| 
&= \big\| \sum_{i=1}^{k-1}\frac{h_0^{i}}{i!}y_0^{i} +\frac{h_0^{k}}{k!}u_0\big\| \\
&\leq h_0\sum_{i=1}^{k-1}\|y_0^{i}\| + \int_0^{h_0}g(s)ds+h_0\alpha \\
&\leq \int_0^{h_0}\Big(g(s)+1+\sum_{i=1}^{k-1}\|y_0^{i}\|\Big)ds < \frac{r}{2}.
\end{align*}
Then $x_{1}\in  B(x_0,r)$. For $i=1,\dots,k-2$, we have
\begin{align*}
\| y_1^{i}-y_0^{i}\| 
&\leq \sum_{j=i+1}^{k-1}\frac{h_0^{j-i}}{(j-i)!}\|y_0^{j}\|
 +\frac{h_0^{k-i}}{(k-i)!}\| u_0 \|   \\
&\leq h_0\sum_{j=i+1}^{k-1}\|y_0^{j}\|+ \int_0^{h_0}g(s)ds +h_0 \alpha\\
&\leq \int_0^{h_0}\Big(g(s)+1+\sum_{j=i+1}^{k-1}\|y_0^{j}\|\Big)ds <\frac{r}{2},
\end{align*}
and
\begin{align*}
\| y_1^{k-1}-y_0^{k-1}\| &\leq h_0\| u_0 \|   \\
&\leq  \int_0^{h_0}g(s)ds +h_0 \alpha  \\
&\leq \int_0^{h_0}(g(s)+1)ds <\frac{r}{2}.
\end{align*}
Then $y_{1}^{i}\in B(y_0^{i},r)$ for  all $i=1,\dots,k-1$.
 We reiterate this process for constructing sequences
$h_q, t_q, x_q, y_q^1,\dots,y_q^{k-1}, f_q$ and $u_q$
satisfying for some rank $m\geq 1$ the following assertions:
\begin{itemize}
\item[(a)] For all $q\in \{0,\dots,m-1\}$,
$$
h_q:=\max\Big\{h\in ]0,\alpha]:
d\Big(x_q+\sum_{i=1}^{k-1}\frac{h^{i}}{i!}y_q^{i}+\frac{h^{k-1}}{k!}
\int_{t_q}^{t_{q+1}}F(s,x_q)ds,K\Big)<\frac{\alpha h^k}{4k!}\Big\};
$$
\item[(b)] $t_0=0,\,  t_{m-1}< T \leq t_{m}$ with
$t_q=\sum_{j=0}^{q-1}h_{j}$ for all $q\in \{1,\dots,m\}$;

\item[(c)] For all $q\in \{1,\dots,m\}$ and for all $j\in\{1,\dots,k-2\}$
\begin{gather*}
x_q=x_0+\sum_{j=1}^{k-1}\sum_{i=0}^{q-1}\frac{h_i^{j}}{j!}y_i^{j}
+\sum_{i=0}^{q-1}\frac{h_i^{k}}{k!}u_i, \quad
x_q\in K \cap B(x_0,r),
\\
y_{q+1}^{k-1}:= y_q^{k-1}+h_qu_q= y_0^{k-1}+\sum_{i=0}^{q}h_iu_i,\quad
y_q^{k-1}\in B(y_0^{k-1},r),
\\
y_q^{j}=y_0^{j}+\sum_{l=j+1}^{k-1}\sum_{i=0}^{q-1}\frac{h_i^{l-j}}{(l-j)!}y_i^{l}
+\sum_{i=0}^{q-1}\frac{h_i^{k-j}}{(k-j)!}u_i,
\quad y_q^j\in B(y_0^{j},r);
\end{gather*}

\item[(d)] For all $t\in [t_q,t_{q+1}]$ and for all $q\in \{0,\dots,m-1\}$,
\begin{gather*}
u_q\in \frac{1}{h_q}\int_{t_q}^{t_{q+1}}f_q(s)ds +\alpha B,
\quad f_q(t)\in  F(t,x_q),
\\
\| f_q(t)-y(t)\|\leq  d(y(t),F(t,x_q))+
\varepsilon.
\end{gather*}
\end{itemize}


 It is easy to see that for $q=1$ the assertions (a)-(d)
are fulfilled. Let now $q\geq 2$. Assume that (a)-(d) are satisfied
for any $j=1,\dots,q$. If, $T\leq t_{q+1}$, then we take $m=q+1$ and
so the process of iterations is stopped and we get (a)-(d) satisfied
with $t_{m-1}< T \leq t_{m}$. In the other case, i.e, $t_{q+1}< T$,
we define $y_{q+1}^{1},\dots,y_{q+1}^{k-1}$ and $x_{q+1}$ as follows
\begin{gather*}
x_{q+1}:=x_q+\sum_{i=1}^{k-1}\frac{h_q^{i}}{i!}y_q^{i}+\frac{h_q^{k}}{k!}u_q
=\Big(x_0+\sum_{j=1}^{k-1}\sum_{i=0}^{q}\frac{h_i^{j}}{j!}y_i^{j}
+\sum_{i=0}^{q}\frac{h_i^{k}}{k!}u_i\Big)\in K,
\\
y_{q+1}^{k-1}:= y_q^{k-1}+h_qu_q=  y_0^{k-1}+\sum_{i=0}^{q}h_iu_i,
\\
y_{q+1}^{j}:= \sum_{l=j}^{k-1}\frac{h_q^{l-j}}{(l-j)!}y_q^{l}+\frac{h_q^{k-j}}{(k-j)!}u_q=
 y_0^{j}+\sum_{l=j+1}^{k-1}\sum_{i=0}^{q}\frac{h_i^{l-j}}{(l-j)!}y_i^{l}
+\sum_{i=0}^{q}\frac{h_i^{k-j}}{(k-j)!}u_i
\end{gather*}
for $j=1,\dots,k-2$. By \eqref{defdeg}, \eqref{defdeg2} and \eqref{defdealfa}, we have
\begin{align*}
\| x_{q+1}-x_0\| 
&\leq \sum_{j=1}^{k-1}\sum_{i=0}^{q}\frac{h_i^{j}}{j!}\|y_i^{j}\|
+\sum_{i=0}^{q}\frac{h_i^{k}}{k!}\| u_i\|\\
&\leq \sum_{j=1}^{k-1}\sum_{i=0}^{q}h_i(r+\|y_0^{j}\|)+\sum_{i=0}^{q}\| h_iu_i\| \\
&\leq  \sum_{i=0}^{q}h_i \Big((k-1)r + \sum_{j=1}^{k-1}\| y_0^{j}\|\Big) +
 \sum_{i=0}^{q}\Big(\int_{t_i}^{t_{i+1}}\| f_i(t)\| dt + \alpha h _{i}\Big)\\
&\leq \int_0^{t_{q+1}}\Big(g(t) +1+ (k-1)r + \sum_{j=1}^{k-1}\|
y_0^{j}\|\Big)dt < r,
\end{align*}
which ensures that $x_{q+1}\in K\cap B(x_0,r)$. For all $j=1,\dots,k-2$, we have
\begin{align*}
\| y_{q+1}^j-y_0^j\|  
&\leq \sum_{l=j+1}^{k-1}\sum_{i=0}^{q}\frac{h_i^{l-j}}{(l-j)!}\|
y_i^{l}\| + \sum_{i=0}^{q}\frac{h_i^{k-j}}{(k-j)!}\| u_i\| \\
&\leq  \sum_{l=j+1}^{k-1}\sum_{i=0}^{q}h_i(r+\| y_0^{l}\| )+
\sum_{i=0}^{q}\| h_iu_i\|\\
&\leq  \sum_{i=0}^{q}\Big(h_i \Big((k-j-1)r + \sum_{l=j+1}^{k-1}\| y_0^{l}\|\Big) +
 \int_{t_i}^{t_{i+1}}g(t) dt + \alpha h _{i}\Big)\\
&\leq \int_0^{t_{q+1}}\Big(g(t) +1+ (k-j-1)r +  \sum_{l=j+1}^{k-1}\|
y_0^{l}\|\Big)dt  < r
\end{align*}
and
\begin{align*}
\| y_{q+1}^{k-1}-y_0^{k-1}\|  
&\leq \sum_{i=0}^{q}h_i\| u_i\| \\
&\leq  \sum_{i=0}^{q}\Big( \int_{t_i}^{t_{i+1}}g(t) dt + \alpha h _{i}\Big)\\
&\leq \int_0^{t_{q+1}}(g(t) +1)dt  < r,
\end{align*}
which ensures that $y_{q+1}^j\in B(y_0^j,r)$ for all $j=1,\dots,k-1$.

Now, we have to prove that this iterative process is finite; i.e.,
there exists a positive integer $m$ such that $ t_{m-1}< T \le t_{m}$. 
Suppose to the contrary, that is $t_q\le T$, for all $q\ge 1$.
 Then the bounded increasing sequence $\{t_q\}_q$ converges to some $\bar t$ 
such that $\bar t \le T$. Hence for $q>p$,
$$
\|x_q-x_p\|\le \int_{t_p}^{t_q}g(t)dt+ (t_q-t_p)
\Big((k-1)r+1+\sum_{i=1}^{k-1}\| y_0^{i}\| \Big)\to 0 \quad\text{as } q,p\to \infty
$$
and for $j=1,\dots,k-2$,
$$
\|y^j_q-y^j_p\|\le \int_{t_p}^{t_q}g(t)dt+ (t_q-t_p)\Big((k-j-1)r+1
+\sum_{l=j+1}^{k-1}\| y_0^{l}\|\Big)\to 0
\quad\text{as } q,p\to \infty
$$
and
$$
\|y^{k-1}_q-y^{k-1}_p\|\le \int_{t_p}^{t_q}(g(t)+1)dt\to 0 \quad\text{as } q,p\to \infty.
$$
Therefore, the sequences $\{x_q\}_q$ and $\{y_q^j\}_q$, for all
$j=1,\dots,k-1$, are Cauchy sequences and hence, they converge to
some $\bar x\in K$ and $\bar y^j\in \Omega_j$ respectively. Hence,
as $(\overline{t},\bar x,(\bar y^1,\dots,\bar y^{k-1}))\in [
0,T]\times K \times \prod_{i=1}^{k-1}\Omega_i$, by (H3), there
exist $h \in ]0,\alpha]$ and an integer $q_0\ge 1$ such that for
all $q\geq q_0$ and for all $j=1,\dots,k-1$
\begin{gather}
d\Big(\bar x+\sum_{j=1}^{k-1}\frac{h^{j}}{j!}\bar
y^{j}+\frac{h^{k-1}}{k!}\int_{\bar t}^{\bar{t} + h}F(s,\bar
x)ds,K\Big) \leq  \frac{h^k\alpha }{8(k+5)(k!)};  \nonumber \\
\| x_q-\overline{x}\| \leq \frac{h^k\alpha }{8(k+5)(k!)}; \nonumber \\
\| y_q^j-\overline{y}^j\| \leq \frac{h^{k-j}\alpha j!}{8(k+5)(k!)}; \label{1bis}\\
\bar t- t_q  <  \min\{\eta(\frac{ h\alpha}{8(k+5)}),h\}; \nonumber\\
\int_{\bar t}^{\bar{t} + h}m(t)\| x_q-\bar x\|  dt\leq \frac{h\alpha }{8(k+5)}.
\nonumber
\end{gather} 

 Let $q> q_0$ be given. For an arbitrary measurable selection
 $\phi _q$ of $F(t,x_q)$ on $[0,\bar t +h]$, there exists a measurable
selection $\phi$ of $F(t,\bar x)$ on $[0,\bar t +h]$ such that
\begin{equation}
\| \phi _q(t)-\phi (t)\|
\leq  d(\phi _q(t),F(t,\bar x))+\frac{\alpha}{2(k+5)}\leq m(t)\| x_q-\bar x\|
+\frac{\alpha}{8(k+5)}.\label{2bis}
\end{equation}
Relations \eqref{1bis} and \eqref{2bis} imply
\begin{align*}
&d\Big(x_q+\sum_{j=1}^{k-1}\frac{h^{j}}{j!}y_q^{j}+\frac{h^{k-1}}{k!}\int_{t_q}^{t_q
+ h}\phi _q(s)ds,K\Big) \\
&\leq \| x_q-\overline{x }\| +
\sum_{j=1}^{k-1}\frac{h^{j}}{j!}\| y_q^{j}-\overline{y }^{j}\|+
d\Big(\bar x+\sum_{j=1}^{k-1}\frac{h^{j}}{j!}\bar
y^{j}+\frac{h^{k-1}}{k!}\int_{\bar t}^{\bar{t} + h}\phi(s)ds,K\Big)
\\
&\quad +\frac{h^{k-1}}{k!}\int_{t_q}^{\bar t}\|\phi _q(s)\|ds
 +\frac{h^{k-1}}{k!}\int_{\bar t}^{t_q + h}\|\phi _q(s)-\phi(s)\|ds+\frac{h^{k-1}}{k!}\int_{t_q+h}^{\bar t +h}\|\phi (s)\|ds
\\
&\leq  \| x_q-\overline{x }\| + \sum_{j=1}^{k-1}\frac{h^{j}}{j!}\| y_q^{j}-\overline{y }^{j}\|+
\frac{h^{k-1}}{k!}\int_{t_q}^{\bar t}g(s)ds\\
&\quad +  d\Big(\bar x+\sum_{j=1}^{k-1}\frac{h^{j}}{j!}\bar y^{j}
+\frac{h^{k-1}}{k!}\int_{\bar t}^{\bar{t} + h}\phi(s)ds,K\Big)
+\frac{h^{k-1}}{k!}\int_{\bar t}^{\bar t + h}m(s)\| x_q-\bar
x\| ds\\
&\quad +\frac{h^k\alpha}{8(k+5)(k!)}+
\frac{h^{k-1}}{k!}\int_{t_q+h}^{\bar t +h}g(s)ds \\
&\leq \frac{h^k\alpha }{8(k+5)(k!)}+(k-1)\frac{ h^k\alpha }{8(k+5)(k!)}
+\frac{ h^k\alpha }{8(k+5)(k!)}
+\frac{ h^k\alpha }{8(k+5)(k!)}\\&+\frac{ h^k\alpha }{8(k+5)(k!)}
+ \frac{ h^k\alpha }{8(k+5)(k!)}+\frac{ h^k\alpha }{8(k+5)(k!)}
 <\frac{ h^k\alpha }{4k!}.
\end{align*}

 Since $\phi_q$ is an arbitrary measurable selection of
$F(t,x_q)$ on $[0,\bar t +h]$ it follows that
$$
d\Big(x_q+\sum_{j=1}^{k-1}\frac{h^{j}}{j!}y_q^{j}+\frac{h^{k-1}}{k!}\int_{t_q}^{t_q
+ h}F(t,x_q)ds,K\Big)< \frac{ h^k\alpha }{4k!}.
$$
On the other hand, by \eqref{1bis}, we have
$t_{q+1}\le  \overline{t}<t_q+h$ and hence 
$h >t_{q+1}-t_q=h_q$.
Thus, there exists $h > h_q$ (for all $q\ge q_0$) such
that $0<h \leq \alpha$  and
$$
d\Big(x_q+\sum_{i=1}^{k-1}\frac{h^{i}}{i!}y_q^{i}+\frac{h^{k-1}}{k!}\int_{t_q}^{t_q
+ h}F(t,x_q)ds,K\Big) < \frac{ h^k\alpha }{4k!}.
$$
This contradicts the definition of $h_q$. Therefore,
there is an integer $m \ge 1$ such that $t_{m-1}< T \le t_{m}$ and
for which the assertions (a)-(d) are fulfilled.


Now, we take $t_m=T$ and we define the function 
$\theta :[0,T]\to [0,T]$, $z(.):[0,T]\to E$ and
 $f\in L^{1}([0,T],E)$ by setting for all $t\in [t_q,t_{q+1}[$
$$
\theta(t)=t_q,\quad f(t)=f_q(t),\quad
z(t)=x_q+\sum_{i=1}^{k-1}\frac{(t-t_q)^{i}}{i!}y_q^{i}+\frac{(t-t_q)^k}{k!}u_q.
$$

\begin{claim}
For all $q\in \{0,\dots,m\}$ we have
$$
\big\|y_q^{k-1}-y_0^{k-1}-\int_0^{t_q}f(s)ds\big\| \leq \alpha t_q.
$$
\label{claim1}\end{claim}

\begin{proof} 
It is easy to see that for $q=0$ the above
assertion is fulfilled. By induction, assume that
$$
\big\|y_j^{k-1}-y_0^{k-1}-\int_0^{t_j}f(s)ds\big\| \leq \alpha
t_j.
$$
for any $j=1,\dots,q-1$. By (d) we have
\begin{align*}
&\big\|y_q^{k-1}-y_0^{k-1}-\int_0^{t_q}f(s)ds\big\|\\
&=\big\|y_{q-1}^{k-1}-y_0^{k-1}-\int_0^{t_{q-1}}f(s)ds+h_{q-1}u_{q-1}
-\int_{t_{q-1}}^{t_q}f(s)ds\big\|\\
&\leq \big\|y_{q-1}^{k-1}-y_0^{k-1}-\int_0^{t_{q-1}}f(s)ds\big\|+\big\|h_{q-1}u_{q-1}
 -\int_{t_{q-1}}^{t_q}f(s)ds\big\|\\
&\leq \alpha t_{q-1} +\alpha h_{q-1} =\alpha t_{q-1}+\alpha t_q-\alpha t_{q-1}
=\alpha t_q.
\end{align*}
\end{proof}

 Now let $t\in [t_q,t_{q+1}]$, then by Claim \ref{claim1} and the relations (d),
 \eqref{defdeg}, and \eqref{defdealfa}, we have
\begin{align*}
&\big\|z^{(k-1)}(t)-y_0^{k-1}-\int_0^{t}f(s)ds\big\|\\
&=\big\|y_q^{k-1}-y_0^{k-1}-\int_0^{t_q}f(s)ds+(t-t_q)u_q-\int_{t_q}^{t}f(s)ds\big\|\\
&\leq \big\|y_q^{k-1}-y_0^{k-1}-\int_0^{t_q}f(s)ds\big\|+\|h_qu_q\|
 +\int_{t_q}^{t_{q+1}}g(s)ds\\
&\leq  \alpha t_q +2\int_{t_q}^{t_{q+1}}g(s)ds+\alpha h_q \\
&\leq \frac{\varepsilon}{4}+\frac{\varepsilon}{2}+
   \frac{\varepsilon}{4}=\varepsilon.
\end{align*}
The proof of Lemma \ref{construction} is complete.
\end{proof}


\begin{proof}[Proof of the Theorem\ref{result}] 
Let $(\varepsilon_n)_{n=1}^{\infty}$ be a strictly decreasing sequence of
positive scalars such that $\sum_{n=1}^{\infty}\varepsilon_n<\infty$
and $\varepsilon_1<1$. In view of Lemma \ref{construction}, we can
define inductively sequences 
$(f_n)_{n=1}^{\infty}\subset L^{1}([0,T],E)$, $(z_n(.))_{n=1}^{\infty}\subset
\mathcal{C}^k([0,T],E)$ and $(\theta_n)_{n=1}^{\infty}\subset
S([0,T],[0,T])$ ($S([0,T],[0,T])$ is the space of step functions
from $[0,T]$ into $[0,T]$) such that
\begin{itemize}
\item[(1)] $f_n(t)\in F(t,z_n(\theta_n(t)))$ for all $t \in [0,T]$;

\item[(2)] $\| f_{n+1}(t)-f_n(t)\|\leq d(f_n(t),F(t,z_{n+1}(\theta_{n+1}(t))))
+ \varepsilon_{n+1}$ for all $t \in [0,T]$;

\item[(3)] $\big\| z^{(k-1)}_n(t)-y_0^{k-1}-\int_0^{t}f_n(\tau)d\tau \big\|\leq
\varepsilon_n$ for all $t \in [0,T]$.
\end{itemize}
By (1) and (2) we have
\begin{align*}
\| f_{n+1}(t)-f_n(t)\|
&\leq H\Big(F(t,z_n(\theta_n(t))),F(t,z_{n+1}(\theta_{n+1}(t)))\Big)
 +\varepsilon_{n+1}\\
&\leq m(t)\| z_n(\theta_n(t))-z_{n+1}(\theta_{n+1}(t))\| +\varepsilon_{n+1}\\
&\leq m(t)\Big(\|z_n(\theta_n(t))-z_{n}(t) \| + \| z_n(t)-z_{n+1}(t)\| \\
&\quad +\| z_{n+1}(t)-z_{n+1}(\theta_{n+1}(t)) \| \Big)+\varepsilon_{n+1}.
\end{align*}
On the other hand, for $t\in [t_q,t_{q+1}[$ we have
\begin{align*}
\| z_{n}(t)-z_n(\theta_{n}(t))\|
&= \big\| \sum_{i=1}^{k-1}\frac{(t-t_q)^{i}}{i!}y_q^{i}+\frac{(t-t_q)^k}{k!}u_q\big\|\\
&\leq \sum_{i=1}^{k-1}h_q\Big(\| y_q^{i}-y_0^{i}\|+
\| y_0^{i}\|\Big) +\| h_qu_q\| \\
&\leq \frac{\varepsilon_{n}}{4}\Big((k-1)r +\sum_{i=1}^{k-1}\|
y_0^{i}\| \Big)+\alpha +\int_{t_q}^{t_{q+1}}g(s)ds\\
&\leq \frac{\varepsilon_{n}}{4}\Big((k-1)r +\sum_{i=1}^{k-1}\|
y_0^{i}\| \Big)+\frac{\varepsilon_{n}}{4}
+\frac{\varepsilon_{n}}{4}\\
&\leq \frac{\varepsilon_{n}}{4}\Big((k-1)r+ 2 +\sum_{i=1}^{k-1}\|
y_0^{i}\| \Big).
\end{align*}
 Hence
\begin{equation}
\| z_{n}(t)-z_n(\theta_{n}(t))\|\leq
\frac{\varepsilon_{n}}{4}\Big((k-1)r+ 2 +\sum_{i=1}^{k-1}\|
y_0^{i}\| \Big).\label{reldezn}
\end{equation}
 It follows that
\begin{equation}
\begin{split}
&\| f_{n+1}(t)-f_n(t)\|\\
&\leq m(t)\Big(\frac{\varepsilon_{n}}{2}\Big((k-1)r+ 2
+\sum_{i=1}^{k-1}\| y_0^{i}\| \Big)+
\| z_n(.)-z_{n+1}(.)\| _{\infty}\Big)+\varepsilon_{n+1}.
\end{split}\label{reldefn}
\end{equation}
Relations \eqref{reldefn} and \eqref{relationdem} yield
\begin{align*}
&\| z^{(k-1)}_{n+1}(t)-z^{(k-1)}_n(t)\| \\
&\leq \big\|z^{(k-1)}_{n+1}(t)-y_0^{k-1}-\int_0^{t}f_{n+1}(s)ds\big\|
+\big\| z^{(k-1)}_{n}(t)-y_0^{k-1}-\int_0^{t}f_{n}(s)ds\big\|
\\
&\quad +\int_0^t\| f_{n+1}(s)-f_n(s)\| ds
\\
&\leq \varepsilon_{n+1}+\varepsilon_{n}+\int_0^t
m(s)\Big(\frac{\varepsilon_{n}}{2}\Big((k-1)r+ 2
+\sum_{i=1}^{k-1}\| y_0^{i}\| \Big)+ \| z_n(.)-z_{n+1}(.)\|
_{\infty}\Big)ds\\
&\quad +t\varepsilon_{n+1}
\\
&\leq 3\varepsilon_{n}+\| z_n(.)-z_{n+1}(.) \|
_{\infty}\int_0^T m(s)ds\\
&\quad +\frac{\varepsilon_{n}}{2}\Big((k-1)r+ 2
+\sum_{i=1}^{k-1}\| y_0^{i}\| \Big)\int_0^T m(s)ds
\\
&\leq \Big((k-1)r+ 5 +\sum_{i=1}^{k-1}\| y_0^{i}\|
\Big)\varepsilon_{n}+\| z_n(.)-z_{n+1}(.) \| _{\infty}\int_0^T
m(s)ds .
\end{align*}
Since $T\leq 1$ for all $t\in[0,T]$, we have
\begin{align*}
\| z^{(k-2)}_{n+1}(t)-z^{(k-2)}_n(t)\|
&\leq \int_0^t\| z^{(k-1)}_{n+1}(s)-z^{(k-1)}_n(s) \| ds\\
&\leq \Big((k-1)r+ 5 +\sum_{i=1}^{k-1}\| y_0^{i}\|
\Big)\varepsilon_{n}\\
&\quad +\| z_n(.)-z_{n+1}(.) \|_{\infty}\int_0^T m(s)ds.
\end{align*}
Then by the same reasoning, for $j=1,\dots,k-1$,  we obtain
\begin{align*}
\| z^{(j)}_{n+1}(t)-z^{(j)}_n(t)\|
&\leq  \Big((k-1)r+ 5 +\sum_{i=1}^{k-1}\| y_0^{i}\|\Big)\varepsilon_{n}\\
&\quad +\| z_n(.)-z_{n+1}(.) \|_{\infty}\int_0^T m(s)ds
 \end{align*}
and
\begin{equation}
\| z_n(.)-z_{n+1}(.) \| _{\infty}\leq
\frac{\Big((k-1)r+ 5 +\sum_{i=1}^{k-1}\| y_0^{i}\|\Big)\varepsilon_{n}}{1-L}
\label{bis8}
\end{equation}
where $L=\int_0^T m(s)ds$. For  $j=1,\dots,k-1$ we have
\begin{align*}
\| z^{(j)}_{n+1}(.)-z^{(j)}_n(.)\| _{\infty}
&\leq \Big((k-1)r+ 5 +\sum_{i=1}^{k-1}\|
y_0^{i}\|\Big)\varepsilon_{n}+\| z_n(.)-z_{n+1}(.)
\|_{\infty}\\
&\leq \frac{\Big((k-1)r+ 5 +\sum_{i=1}^{k-1}\|
y_0^{i}\|\Big)(2-L)\varepsilon_{n}}{1-L} .
\end{align*}
Therefore, for $n<m$,
$$
\| z_m(.)-z_{n}(.) \| _{\infty}\leq \frac{(k-1)r+ 5
+\sum_{i=1}^{k-1}\| y_0^{i}\|}{1-L}\sum_{i=n}^{m-1}\varepsilon_{i}
$$
and for $j=1,\dots,k-1$,
$$
\| z^{(j)}_m(.)-z^{(j)}_{n}(.) \| _{\infty}\leq \frac{\Big((k-1)r+ 5
+\sum_{i=1}^{k-1}\| y_0^{i}\|\Big)(2-L)}{1-L}\sum_{i=n}^{m-1}\varepsilon_{i}.
$$
Thus the sequences $\{z_n(.)\}_{n=1}^{\infty}$ and
$\{z^{(j)}_n(.)\}_{n=1}^{\infty}$, for $j=1,\dots,k-1$, converge
uniformly on $[0,T]$, namely $x(.)$ and $y_j(.)$ its limits
respectively. Also the relations
$$
z_n(t)=x_0+\int_0^t\dot{z}_n(s)ds
$$
and
$$
z^{(j)}_n(t)=y^j_0+\int_0^tz^{(j+1)}_n(s)ds \quad\text{for }
j=1,\dots,k-2
$$
yield $x(t)=x_0+\int_0^ty_1(s)ds$ and
$$
y_j(t)=y^j_0+\int_0^ty_{j+1}(s)ds \quad\text{for }
j=1,\dots,k-2.
$$
 Thus $\dot{x}(t)=y_1(t)$ and $\dot{y}_j(t)=y_{j+1}(t)$
for all $t\in [0,T]$ and for all $j=1,\dots,k-2$. Hence $x(0)=x_0$ and
$x^{(j)}(0)=y_0^j$ for all $j=1,\dots,k-1$. On the other hand, observe
that $z_{n}(\theta_{n}(t))$ converges uniformly to $x(t)$ on
$[0,T]$. Indeed, for $t\in [t_q,t_{q+1}[$ we have
$$
\|
z_n(\theta_{n}(t))-x(t)\|\leq \|
z_{n}(t)-z_n(\theta_{n}(t))\|+
\| z_{n}(t)-x(t)\|.
$$
  By \eqref{reldezn} and since $(z_{n}(.))$ converges
uniformly to $x(.)$, it follows that
\begin{equation}
z(\theta_{n}(.)) \text{ converges uniformly to $x(.)$ on $[0,T]$.}\label{reldetteta}
\end{equation}
By construction, we have
$z_{n}(\theta_{n}(t))\in K$ for every $t\in [0,T]$ and $K$ is
closed, then $x(t)\in K$ for all $t\in [0,T]$.

 Now we return to relation \eqref{reldefn}. By \eqref{bis8} we have
\begin{align*}
& \| f_{n+1}(t)-f_n(t)\| \\
&\leq  \Big(m(t)\Big(\frac{(k-1)r+ 5 +\sum_{i=1}^{k-1}\| y_0^{i}\|}{1-L}
 +\frac{(k-1)r+ 2 +\sum_{i=1}^{k-1}\| y_0^{i}\|}{2}\Big)+1\Big)\varepsilon_n.
\end{align*}
  This implies (as above) that $\{f_n(t)\}_{i=1}^{n}$ is a
Cauchy sequence and $f_n(t)$ converges to $f(t)$. Further, since
$\| f_n(t)\|\leq  g(t)$, by (3) and Lebesgue's
theorem we have
$$
y_{k-1}(t)=\lim_{n\to \infty }
z^{(k-1)}_n(t)=\lim_{n\to \infty }
\Big(y^{k-1}_0+\int_0^{t}f_n(s)ds\Big)=y^{k-1}_0+\int_0^{t}f(s)ds.
$$
Hence $\dot{y}_{k-1}(t)=f(t)$. Finally, observe that by (1),
\begin{align*}
d(f(t),F(t,x(t)))
&\leq \| f(t)-f_n(t)\| + H\Big(F(t,z_n(\theta_n(t))),F(t,x(t))\Big)\\
&\leq \| f(t)-f_n(t)\| + m(t)\| z_n(\theta_n(t))-x(t)\| .
\end{align*}
Since $f_n(t)$ converges to $f(t)$ and by
\eqref{reldetteta} the last term converges to $0$. So that
$x^{(k)}(t)=\dot{y}_{k-1}(t)=f(t)\in F(t,x(t))$ a.e on $[0,T]$.
The proof is complete.
\end{proof}

\begin{remark} \rm
The tangential condition (H3) provides a sufficient
condition ensuring the existence of solution to \eqref{cauchy}.
However, this condition is not necessary at all. In fact, in the case
$k=2$,  Marco and Murillo \cite[Example 4.1]{marco2} gave a counterexample: 
The multifunction  $F:[0,1]\times \mathbb{R} \to 2^{\mathbb{R}}$ defined as
$$
F(t,x)=[-t^{-a},t^{-a}], \quad 0< t\leq 1 ,  \quad  F(0,x)={0}
$$
with $0<a<(3-\sqrt{3})/2$, satisfies (H2) and
$x(t)=\frac{t^{2-a}}{(1-a)(2-a)}$ is a solution of
\begin{equation} \label{problem1b}
 \begin{gathered}
 \ddot{x}(t) \in F(t,x(t)), \quad t\in [0,1];    \\
 (x(0),\dot{x}(0))=(0,0); \\
 x(t)\in [0,2].
 \end{gathered}
\end{equation}
However (H3) fails on  $[0,2]\times gr(A^{1}_K)$,
because
$$
\frac{1}{h^{2}}d\Big(\frac{h}{2}\int^{t+h}_t
F(s,0)ds,[0,2]\Big)=\frac{{(t+h)}^{1-a}-t^{1-a}}{2(1-a)h}
$$
and
$$
\lim_{h \to 0^{+}}\frac{{(t+h)}^{1-a}-t^{1-a}}{2(1-a)h}
=\begin{cases}
+ \infty &\text{if } t=0 \\
\frac{{t}^{-a}}{2}&\text{if } 0< t\leq 1
\end{cases}
$$
\end{remark}


\begin{remark} \rm
Let $F:[0,1]\times K\times\prod_{i=1}^{k-1}\Omega_i\to 2^{E}$.
 For any $(x_0,y_0^1,y_0^2\dots,y_0^{k-1}) \in K\times\prod_{i=1}^{k-1}\Omega_i$,
 we can prove the existence of
viable solutions of the differential inclusion
\begin{gather*}
 x^{(k)}(t) \in F(t,x(t),x^{(1)}(t),\dots,x^{(k-1)}(t)) \quad\text{a.e. on } [0,T]  \\
 x(0)=x_0\in K, x^{(i)}(0)=y_0^{i}\in \Omega_i, \quad  i=1,\dots,k-1, \\
 x(t) \in K \quad\text{on } [0,T],
 \end{gather*}
by the same technics and the same hypothesis as above
except the condition (H2) part $(ii)$ which must be replaced by:
There is a function $m\in
L^{1}([0,1],\mathbb{R}^{+})$ such that for all $t\in [0,1]$,
  for all $x_{1},x_{2}\in K$ and for all 
$(y_1^1,\dots,y_1^{k-1}),(y_2^1,\dots,y_2^{k-1})\in \prod_{i=1}^{k-1}\Omega_i$,
 $$
 H\Big(F(t,x_{1},y_1^1,\dots,y_1^{k-1}),F(t,x_{2},y_2^1,\dots,y_2^{k-1})\Big) 
\leq m(t)\| y^{k-1}_{1}-y^{k-1}_{2}\|.
 $$
\end{remark}

\subsection*{Acknowledgments}
The authors would like to thank the anonymous referee for his/her careful reading, 
comments and  suggestions.


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\end{document}