\documentclass[reqno]{amsart} \usepackage{hyperref} \AtBeginDocument{{\noindent\small \emph{Electronic Journal of Differential Equations}, Vol. 2008(2008), No. 62, pp. 1--11.\newline ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu \newline ftp ejde.math.txstate.edu (login: ftp)} \thanks{\copyright 2008 Texas State University - San Marcos.} \vspace{9mm}} \begin{document} \title[\hfilneg EJDE-2008/62\hfil Boundary value problems] {On boundary-value problems for higher-order differential inclusions} \author[M. Aitalioubrahim, S. Sajid\hfil EJDE-2008/62\hfilneg] {Myelkebir Aitalioubrahim, Said Sajid} % in alphabetical order \address{Myelkebir Aitalioubrahim \newline University Hassan II-Mohammedia, U.F.R Mathematics and applications, F.S.T, BP 146, Mohammedia, Morocco} \email{aitalifr@yahoo.fr} \address{Said Sajid \newline University Hassan II-Mohammedia, U.F.R Mathematics and applications, F.S.T, BP 146, Mohammedia, Morocco} \email{saidsajid@hotmail.com} \thanks{Submitted March 14, 2007. Published April 22, 2008.} \subjclass[2000]{34A60, 34B10, 34B15} \keywords{Boundary value problems; contraction; measurability; multifunction} \begin{abstract} We show the existence of solutions to boundary-value problems for higher-order differential inclusion $x^{(n)}(t) \in F(t,x(t))$, where $F(.,.)$ is a closed multifunction, measurable in $t$ and Lipschitz continuous in $x$. We use the fixed point theorem introduced by Covitz and Nadler for contraction multivalued maps. \end{abstract} \maketitle \numberwithin{equation}{section} \newtheorem{theorem}{Theorem}[section] \newtheorem{definition}[theorem]{Definition} \newtheorem{proposition}[theorem]{Proposition} \newtheorem{remark}[theorem]{Remark} \newtheorem{lemma}[theorem]{Lemma} \allowdisplaybreaks \section{Introduction} The aim of this paper is to establish the existence of solutions of the following higher-order boundary-value problems: \begin{itemize} \item For $n \geq 2$ \begin{equation}\label{cauchy1} \begin{gathered} x^{(n)}(t) \in F(t,x(t)) \quad \mbox{a.e. on } [0,1]; \\ x^{(i)}(0)=0, \quad 0\leq i \leq n-2; \\ x(\eta)=x(1). \end{gathered} \end{equation} \item For $n \geq 2$ \begin{equation}\label{cauchy4} \begin{gathered} x^{(n)}(t) \in F(t,x(t))\quad \mbox{a.e. on } [0,1]; \\ x(0)=x'(\eta);\quad x(1)=x(\tau). \end{gathered} \end{equation} \item For $n \geq 4$ \begin{equation}\label{cauchy2} \begin{gathered} x^{(n)}(t) \in F(t,x(t))\quad \mbox{ a.e. on } [0,1]; \\ x^{(i)}(0)=x^{(i+1)}(\eta),\quad 2\leq i \leq n-2; \\ x(0)=x'(\eta);\quad x(1)=x(\tau). \end{gathered} \end{equation} \item For $n \geq 2$ \begin{equation}\label{cauchy3} \begin{gathered} x^{(n)}(t) \in F(t,x(t))\quad \mbox{ a.e. on } [0,1]; \\ x^{(i)}(0)=x^{(i+1)}(\eta),\quad 0\leq i \leq n-2. \end{gathered} \end{equation} \end{itemize} where $F:[0,1]\times \mathbb{R}\to 2^{\mathbb{R}}$ is a closed multivalued map, measurable with respect to the first argument and Lipschitz with respect to the second argument, and $(\eta,\tau)\in ]0,1[^{2}$. Three and four-point boundary-value problems for second-order differential inclusions was initiated by Benchohra and Ntouyas, see \cite{benchohra1,benchohra2,benchohra3}. The authors investigate the existence of solutions on compact intervals for the problems \eqref{cauchy1} and \eqref{cauchy4} in the particular case $n=2$. In order to obtain solutions of \eqref{cauchy1} and \eqref{cauchy4} when $F$ is not necessarily convex values, Benchohra and Ntouyas (see \cite{benchohra3}) reduce the existence of solutions to the search for fixed points of a suitable multivalued map on the Banach space $\mathcal{C}([0,1],\mathbb{R})$. Indeed, they used the fixed point theorem for contraction multivalued maps, due to Covitz and Nadler \cite{covitz}. In this paper, we give an extension of the Benchohra and Ntouyas's result \cite{benchohra3} to the $n-$order non-convex boundary-value problems and we prove the existence of solutions for \eqref{cauchy2} and \eqref{cauchy3}. We shall adopt the technique used by Benchohra and Ntouyas in the previous paper. \section{Preliminaries and statement of the main results} Let $(E,d)$ be a complete metric space. We denote by $\mathcal{C}([0,1],E)$ the Banach space of continuous functions from $[0,1]$ to $E$ with the norm $\| x(.)\| _{\infty}:=\sup \big\{\| x(t)\| ; t\in [0,1]\big\}$, where $\|\cdot\|$ is the norm of $E$. 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 and Valadier \cite{castaing}. \begin{definition} \rm Let $T: E \to 2^{E}$ be a multifunction with closed values. \begin{enumerate} \item $T$ is $k$-Lipschitz if and only if $$ H\big(T(x),T(y)\big)\:\leq \:kd(x,y),\quad \mbox{for each } x, y\in E. $$ \item $T$ is a contraction if and only 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{enumerate} \end{definition} Let us recall the following results that will be used in the sequel. \begin{lemma}\cite{covitz} If $T:E \to 2^{E}$ is a contraction with nonempty closed values, then it has a fixed point. \label{lemme1}\end{lemma} \begin{lemma}\cite{qiji} Assume that $F:[a,b]\times \mathbb{R} \to 2^{\mathbb{R}}$ is a multifunction with nonempty closed values satisfying: \begin{itemize} \item For every $x\in \mathbb{R}$, $F(.,x)$ is measurable on $[a,b];$ \item For every $t\in [a,b]$, $F(t,.)$ is (Hausdorff) continuous on $\mathbb{R}$. \end{itemize} Then for any measurable function $x(.): [a,b] \to \mathbb{R}$, the multifunction $F(.,x(.))$ is measurable on $[a,b]$. \label{lemme2}\end{lemma} \begin{definition} \rm A function $x(.):[0,1]\to \mathbb{R}$ is said to be a solution of \eqref{cauchy1} (resp. \eqref{cauchy4}, \eqref{cauchy2}, \eqref{cauchy3}) if $x(.)$ is $(n-1)$-times differentiable, $x^{(n-1)}(.)$ is absolutely continuous and $x(.)$ satisfies the conditions of \eqref{cauchy1} (resp. \eqref{cauchy4}, \eqref{cauchy2}, \eqref{cauchy3}). \end{definition} Let $\eta\in \mathbb{R}$ and $n\in \mathbb{N}\setminus\{0,1\}$. Define a sequence of functions $(\varphi_p(.))_{2\leq p\leq n}$ by: For all $t\in[0,1]$ \begin{gather*} \varphi_2(t)=1;\\ \varphi_3(t)= t+\varphi_{2}(\eta);\\ \varphi_p(t)= \frac{t^{p-2}}{(p-2)!} +\sum_{k=3}^{p-1}\varphi_{k-1}(\eta)\frac{t^{p-k}}{(p-k)!}+\varphi_{p-1}(\eta). \end{gather*} We remark that \begin{itemize} \item[(a)]For $t\in [0,1]$ and $k\in \{0,\dots ,n-2\}$, $\varphi_n^{(k)}(t) = \varphi_{n-k}(t)$; \item[(b)]For $k\in\{0,\dots ,n-3\}$, $\varphi_{n-k}(0) = \varphi_{n-k-1}(\eta)$; \item[(c)]For $k\in\{0,\dots ,n-2\}$ the function $\varphi_{n}^{(k)}(.)$ is increasing. \end{itemize} \subsection*{Assumptions} We will use the following hypotheses: \begin{itemize} \item[(H1)] $F:[0,1]\times \mathbb{R} \to 2^{ \mathbb{R}}$ is a multivalued map with nonempty closed values satisfying \begin{itemize} \item[(i)] For each $x\in \mathbb{R}$, $t\mapsto F(t,x)$ is measurable; \item[(ii)] There exists a function $m(.)\in L^{1}([0,1],\mathbb{R}^{+})$ such that for all $t\in [0,1]$ and for all $x_{1}, x_{2}\in \mathbb{R}$, $$ H\big(F(t,x_{1}),F(t,x_{2})\big) \:\leq \:m(t)| x_{1}-x_{2}|. $$ \end{itemize} \item[(H2)] For $\eta\in]0,1[$, $$ \frac{1}{(n-1)!}\Big(L(1)+ \frac{L(\eta)+L(1)}{1-\eta^{n-1}}\Big)\:<\:1 $$ where $L(t)=\int_0^t m(s)ds$ for all $t\in [0,1];$ \item[(H3)] For $(\eta,\tau)\in ]0,1[^2$, $$ \frac{(3-\tau)L(1)+2L(\tau)}{(1-\tau)(n-1)!}+ \sum_{k=0}^{n-2}\frac{L(\eta)}{(1-\tau)k!} \big[(3-\tau)\varphi_n^{(k)}(1)+ 2\varphi_n^{(k)}(\tau)\big]\:<\:1; $$ \item[(H4)] For $\eta\in]0,1[$, $$ \frac{L(1)}{(n-1)!}+ L(\eta)\sum_{k=0}^{n-2} \frac{\varphi_n^{(k)}(1)}{k!}\:<\:1. $$ \end{itemize} \subsection*{Main results} We shall prove the following results. \begin{theorem}\label{theoreme1} If assumptions {\rm (H1)} and {\rm (H2)} are satisfied, then problem \eqref{cauchy1} has at least one solution on $[0,1]$. \end{theorem} \begin{theorem}\label{theoreme2} If assumptions {\rm (H1)} and {\rm (H3)} are satisfied, then problems \eqref{cauchy4} and \eqref{cauchy2} have at least one solution on $[0,1]$. \end{theorem} \begin{theorem}\label{theoreme3} If assumptions {\rm (H1)} and {\rm (H4)} are satisfied, then problem \eqref{cauchy3} has at least one solution on $[0,1]$. \end{theorem} \section{Proof of the main results} \subsection*{Proof of Theorem \ref{theoreme1}} For $y(.)\in \mathcal{C}([0,1],\mathbb{R})$, set $$ S_{F,y(.)}:=\big\{g\in L^{1}([0,1],\mathbb{R}): g(t)\in F(t,y(t))\; \mbox{for a.e.}\; t\in [0,1]\big\}. $$ By Lemma \ref{lemme2}, for $y(.)\in \mathcal{C}([0,1],\mathbb{R})$, $F(.,y(.))$ is closed and measurable, then it has a selection. Thus $S_{F,y(.)}$ is nonempty. Let us transform the problem into a fixed point problem. Consider the multivalued map $T:\mathcal{C}([0,1],\mathbb{R})\to 2^{\mathcal{C}([0,1],\mathbb{R})}$ defined as follows: for $y(.)\in L^{1}([0,1],\mathbb{R})$, $T(y(.))$ is the set of all $z(.)\in \mathcal{C}([0,1],\mathbb{R})$, such that \begin{align*} z(t)&=\int_0^t\frac{(t-s)^{n-1}}{(n-1)!}g(s)ds + \frac{t^{n-1}}{1-\eta^{n-1}}\int_0^\eta\frac{(\eta-s)^{n-1}}{(n-1)!}g(s)ds\\ &\quad -\frac{t^{n-1}}{1-\eta^{n-1}}\int_0^1\frac{(1-s)^{n-1}}{(n-1)!}g(s)ds, \end{align*} where $g\in S_{F,y(.)}$. We shall show that $T$ satisfies the assumptions of Lemma \ref{lemme1}. The proof will be given in two steps: \noindent{\bf Step 1:} \textit{$T$ has non-empty closed values.} Indeed, let $(y_p(.))_{p\geq 0}\in T(y(.))$ converges to $\bar y(.)$ in $\mathcal{C}([0,1],\mathbb{R})$. Then $\bar y(.)\in \mathcal{C}([0,1],\mathbb{R})$ and for each $t\in [0,1]$, \begin{align*} y_p(t)&\in \int_0^t\frac{(t-s)^{n-1}}{(n-1)!}F(s,y(s))ds + \frac{t^{n-1}}{1-\eta^{n-1}} \int_0^\eta\frac{(\eta-s)^{n-1}}{(n-1)!}F(s,y(s))ds\\ &\quad -\frac{t^{n-1}}{1-\eta^{n-1}}\int_0^1\frac{(1-s)^{n-1}}{(n-1)!}F(s,y(s))ds. \end{align*} Since the sets \begin{gather*} \int_0^t\frac{(t-s)^{n-1}}{(n-1)!}F(s,y(s))ds\,,\quad \frac{t^{n-1}}{1-\eta^{n-1}} \int_0^\eta\frac{(\eta-s)^{n-1}}{(n-1)!}F(s,y(s))ds\,,\\ \frac{t^{n-1}}{1-\eta^{n-1}}\int_0^1 \frac{(1-s)^{n-1}}{(n-1)!}F(s,y(s))ds \end{gather*} are closed for all $t\in [0,1]$, we have \begin{align*} \bar y(t)&\in \int_0^t\frac{(t-s)^{n-1}}{(n-1)!}F(s,y(s))ds + \frac{t^{n-1}}{1-\eta^{n-1}} \int_0^\eta\frac{(\eta-s)^{n-1}}{(n-1)!}F(s,y(s))ds\\ &\quad -\frac{t^{n-1}}{1-\eta^{n-1}} \int_0^1\frac{(1-s)^{n-1}}{(n-1)!}F(s,y(s))ds. \end{align*} Then $\bar y(.)\in T(y(.))$. So $T(y(.))$ is closed for each $y(.)\in \mathcal{C}([0,1],\mathbb{R})$. \noindent{\bf Step 2:} \textit{$T$ is a contraction.} Indeed, let $y_1(.), y_2(.)\in \mathcal{C}([0,1],\mathbb{R})$ and $z_1(.)\in T(y_1(.))$. Then \begin{align*} z_1(t)&=\int_0^t\frac{(t-s)^{n-1}}{(n-1)!}g_1(s)ds + \frac{t^{n-1}}{1-\eta^{n-1}} \int_0^\eta\frac{(\eta-s)^{n-1}}{(n-1)!}g_1(s)ds\\ &\quad -\frac{t^{n-1}}{1-\eta^{n-1}}\int_0^1 \frac{(1-s)^{n-1}}{(n-1)!}g_1(s)ds, \end{align*} where $g_1\in S_{F,y_1(.)}$. Consider the multivalued map $U:[0,1]\to 2^{\mathbb{R}}$, defined by $$ U(t)=\big\{x\in \mathbb{R}:|g_1(t)-x|\:\leq \:m(t)|y_1(t)-y_2(t)|\big\}. $$ For each $t\in [0,1]$, $U(t)$ is nonempty. Indeed, let $t\in [0,1]$, from (H1) we have \[ H\big(F(t,y_1(t)),F(t,y_2(t))\big)\leq m(t)|y_1(t)-y_2(t)|. \] Hence, there exists $x\in F(t,y_2(t))$, such that \[ |g_1(t)-x|\leq m(t)|y_1(t)-y_2(t)|. \] By \cite[Proposition III.4]{castaing}, the multifunction \begin{equation} V:t\to U(t)\cap F(t,y_2(t)) \label{measura} \end{equation} is measurable. Then there exists a measurable selection of $V$ denoted $g_2$ such that $$ g_2(t)\in F(t,y_2(t))\quad\mbox{and}\quad |g_1(t)-g_2(t)|\leq m(t)|y_1(t)-y_2(t)|,\quad \mbox{for each } t\in [0,1]. $$ Now, for $t\in[0,1]$ set \begin{align*} z_2(t)&=\int_0^t\frac{(t-s)^{n-1}}{(n-1)!}g_2(s)ds + \frac{t^{n-1}}{1-\eta^{n-1}} \int_0^\eta\frac{(\eta-s)^{n-1}}{(n-1)!}g_2(s)ds\\ &\quad -\frac{t^{n-1}}{1-\eta^{n-1}}\int_0^1\frac{(1-s)^{n-1}}{(n-1)!}g_2(s)ds. \end{align*} Then \begin{align*} |z_1(t)-z_2(t)| & \leq \int_0^t\frac{(t-s)^{n-1}}{(n-1)!}|g_1(t)-g_2(s)|ds\\ &\quad + \frac{t^{n-1}}{1-\eta^{n-1}} \int_0^\eta\frac{(\eta-s)^{n-1}}{(n-1)!}|g_1(s)-g_2(s)|ds\\ &\quad +\frac{t^{n-1}}{1-\eta^{n-1}}\int_0^1\frac{(1-s)^{n-1}}{(n-1)!} |g_1(s)-g_2(s)|ds\\ &\leq \int_0^t\frac{(t-s)^{n-1}}{(n-1)!}m(s)|y_1(s)-y_2(s)|ds\\ &\quad + \frac{t^{n-1}}{1-\eta^{n-1}} \int_0^\eta\frac{(\eta-s)^{n-1}}{(n-1)!}m(s)|y_1(s)-y_2(s)|ds\\ &\quad +\frac{t^{n-1}}{1-\eta^{n-1}}\int_0^1\frac{(1-s)^{n-1}}{(n-1)!} m(s)|y_1(s)-y_2(s)|ds\\ &\leq \frac{1}{(n-1)!}\|y_1(.)-y_2(.)\|_{\infty}\int_0^1m(s)ds \\ &\quad + \frac{1}{(1-\eta^{n-1})(n-1)!} \|y_1(.)-y_2(.)\|_{\infty}\int_0^\eta m(s)ds\\ &\quad +\frac{1}{(1-\eta^{n-1})(n-1)!}\|y_1(.)-y_2(.)\|_{\infty}\int_0^1 m(s)ds\\ &\leq \frac{1}{(n-1)!}\Big(L(1)+ \frac{L(\eta)+L(1)}{1-\eta^{n-1}}\Big)\|y_1(.)-y_2(.)\|_{\infty}. \end{align*} So, we conclude that \[ \|z_1(.)-z_2(.)\|_{\infty}\leq \frac{1}{(n-1)!}\Big(L(1)+ \frac{L(\eta)+L(1)}{1-\eta^{n-1}}\Big)\|y_1(.)-y_2(.)\|_{\infty}. \] By the analogous relation, obtained by interchanging the roles of $y_1(.)$ and $y_2(.)$, it follows that \[ H\big(T(y_1(.)),T(y_2(.))\big)\leq \frac{1}{(n-1)!}\Big(L(1)+ \frac{L(\eta)+L(1)}{1-\eta^{n-1}}\Big)\|y_1(.)-y_2(.)\|_{\infty}. \] Consequently, $T$ is a contraction. Hence, by Lemma \ref{lemme1}, $T$ has a fixed point $y(.)$. \begin{proposition} $y(.)$ is a solution of \eqref{cauchy1}. \end{proposition} \begin{proof} We have \begin{align*} y(t)&=\int_0^t\frac{(t-s)^{n-1}}{(n-1)!}g(s)ds + \frac{t^{n-1}}{1-\eta^{n-1}} \int_0^\eta\frac{(\eta-s)^{n-1}}{(n-1)!}g(s)ds\\ &\quad -\frac{t^{n-1}}{1-\eta^{n-1}}\int_0^1\frac{(1-s)^{n-1}}{(n-1)!}g(s)ds, \end{align*} where $g\in S_{F,y(.)}$. Then \begin{align*} y(\eta)&=\int_0^\eta\frac{(\eta-s)^{n-1}}{(n-1)!}g(s)ds + \frac{\eta^{n-1}}{1-\eta^{n-1}} \int_0^\eta\frac{(\eta-s)^{n-1}}{(n-1)!}g(s)ds\\ &\quad -\frac{\eta^{n-1}}{1-\eta^{n-1}}\int_0^1\frac{(1-s)^{n-1}}{(n-1)!}g(s)ds\\ &= \frac{1}{1-\eta^{n-1}} \int_0^\eta\frac{(\eta-s)^{n-1}}{(n-1)!}g(s)ds-\frac{\eta^{n-1}}{1-\eta^{n-1}}\int_0^1\frac{(1-s)^{n-1}}{(n-1)!}g(s)ds \end{align*} and \begin{align*} y(1)&=\int_0^1\frac{(1-s)^{n-1}}{(n-1)!}g(s)ds + \frac{1}{1-\eta^{n-1}} \int_0^\eta\frac{(\eta-s)^{n-1}}{(n-1)!}g(s)ds\\ &\quad -\frac{1}{1-\eta^{n-1}}\int_0^1\frac{(1-s)^{n-1}}{(n-1)!}g(s)ds\\ &= \frac{1}{1-\eta^{n-1}} \int_0^\eta\frac{(\eta-s)^{n-1}}{(n-1)!}g(s)ds-\frac{\eta^{n-1}}{1-\eta^{n-1}}\int_0^1\frac{(1-s)^{n-1}}{(n-1)!}g(s)ds, \end{align*} hence $y(1)=y(\eta)$. On the other hand, for $0\leq i \leq n-2$, we have \begin{align*} y^{(i)}(t)&=\int_0^t\frac{(t-s)^{n-i-1}}{(n-i-1)!}g(s)ds + \frac{(n-1)\dots (n-i)t^{n-i-1}}{1-\eta^{n-1}} \int_0^\eta\frac{(\eta-s)^{n-1}}{(n-1)!}g(s)ds\\ &\quad -\frac{(n-1)\dots (n-i)t^{n-i-1}}{1-\eta^{n-1}} \int_0^1\frac{(1-s)^{n-1}}{(n-1)!}g(s)ds, \end{align*} hence $y^{(i)}(0)=0$. Finally, it is clear that $y^{(n)}(t)=g(t)$, so $y^{(n)}(t)\in F(t,y(t))$. \end{proof} \subsection*{Proof of Theorem \ref{theoreme2}} We transform the problem into a fixed point problem. For $t\in[0,1]$, set \[ \psi_n^g(t)=\int_0^t\frac{(t-s)^{n-1}}{(n-1)!}g(s)ds + \sum_{k=0}^{n-2}\varphi_{n}^{(k)}(t)\int_0^\eta \frac{(\eta-s)^{k}}{k!}g(s)ds, \] where $g\in S_{F,y(.)}$. Consider the multivalued map, $T:\mathcal{C}([0,1],\mathbb{R})\to 2^{\mathcal{C}([0,1],\mathbb{R})}$ defined as follows: for $y(.)\in \mathcal{C}([0,1],\mathbb{R})$, $$ T(y(.)):=\big\{z(.)\in \mathcal{C}([0,1],\mathbb{R}): z(t)=\psi_n^g(t)+ \frac{1+t}{1-\tau}\big(\psi_n^g(\tau)-\psi_n^g(1)\big) \big\}. $$ We shall show that $T$ satisfies the assumptions of Lemma \ref{lemme1}. The proof will be given in two steps: \noindent{\bf Step 1:} \textit{$T$ has non-empty closed values.} Indeed, let $(y_p(.))_{p\geq 0}\in T(y(.))$ converges to $\bar y(.)$ in $\mathcal{C}([0,1],\mathbb{R})$. Then $\bar y(.)\in \mathcal{C}([0,1],\mathbb{R})$ and for each $t\in [0,1]$, \begin{align*} y_p(t)&\in \int_0^t\frac{(t-s)^{n-1}}{(n-1)!}F(s,y(s))ds + \sum_{k=0}^{n-2}\varphi_{n}^{(k)}(t)\int_0^\eta \frac{(\eta-s)^{k}}{k!}F(s,y(s))ds\\ &\quad+\frac{1+t}{1-\tau}\Big[\int_0^\tau\frac{(\tau-s)^{n-1}}{(n-1)!} F(s,y(s))ds\\ &\quad+ \sum_{k=0}^{n-2}\varphi_{n}^{(k)}(\tau)\int_0^\eta \frac{(\eta-s)^{k}}{k!}F(s,y(s))ds\\ &\quad -\int_0^1\frac{(1-s)^{n-1}}{(n-1)!} F(s,y(s))ds - \sum_{k=0}^{n-2}\varphi_{n}^{(k)}(1)\int_0^\eta \frac{(\eta-s)^{k}}{k!}F(s,y(s))ds\Big]. \end{align*} Since the set \begin{gather*} \int_0^t \frac{(t-s)^{k}}{k!}F(s,y(s))ds \end{gather*} is closed for all $t\in [0,1]$ and $0\leq k\leq n-1$, we have \begin{align*} \bar y(t)&\in \int_0^t\frac{(t-s)^{n-1}}{(n-1)!}F(s,y(s))ds + \sum_{k=0}^{n-2}\varphi_{n}^{(k)}(t)\int_0^\eta \frac{(\eta-s)^{k}}{k!}F(s,y(s))ds\\ &\quad +\frac{1+t}{1-\tau}\Big[\int_0^\tau\frac{(\tau-s)^{n-1}}{(n-1)!}F(s,y(s))ds \\ &\quad + \sum_{k=0}^{n-2}\varphi_{n}^{(k)}(\tau)\int_0^\eta \frac{(\eta-s)^{k}}{k!}F(s,y(s))ds\\ &\quad -\int_0^1\frac{(1-s)^{n-1}}{(n-1)!}F(s,y(s))ds - \sum_{k=0}^{n-2}\varphi_{n}^{(k)}(1)\int_0^\eta \frac{(\eta-s)^{k}}{k!}F(s,y(s))ds\Big]. \end{align*} Then $\bar y(.)\in T(y(.))$. So $T(y(.))$ is closed for each $y(.)\in \mathcal{C}([0,1],\mathbb{R})$. \noindent{\bf Step 2:} \textit{$T$ is a contraction.} Indeed, let $y_1(.), y_2(.)\in \mathcal{C}([0,1],\mathbb{R})$ and $z_1(.)\in T(y_1(.))$. Then \begin{align*} z_1(t)=\psi_n^{g_1}(t)+ \frac{1+t}{1-\tau}\big(\psi_n^{g_1}(\tau)-\psi_n^{g_1}(1)\big), \end{align*} where $g_1\in S_{F,y_1(.)}$. By (\ref{measura}), there exists $g_2$ such that $$ g_2(t)\in F(t,y_2(t))\quad\mbox{and}\quad |g_1(t)-g_2(t)|\leq m(t)|y_1(t)-y_2(t)|,\quad\mbox{for each }t\in [0,1]. $$ Now, set for all $t\in[0,1]$, \[ z_2(t)=\psi_n^{g_2}(t)+ \frac{1+t}{1-\tau}\big(\psi_n^{g_2}(\tau)-\psi_n^{g_2}(1)\big). \] On the other hand, we have \begin{align*} |\psi_n^{g_2}(t)-\psi_n^{g_1}(t)| & \leq \int_0^t\frac{(t-s)^{n-1}}{(n-1)!}|g_1(s)-g_2(s)|ds\\ & \quad + \sum_{k=0}^{n-2}\varphi_n^{(k)}(t) \int_0^\eta\frac{(\eta-s)^{k}}{k!}|g_1(s)-g_2(s)|ds\\ &\leq \frac{1}{(n-1)!}\int_0^t m(s)|y_1(s)-y_2(s)|ds\\ &\quad + \sum_{k=0}^{n-2}\varphi_n^{(k)}(t) \frac{1}{k!}\int_0^\eta m(s)|y_1(s)-y_2(s)|ds\\ &\leq \frac{1}{(n-1)!}\|y_1(.)-y_2(.)\|_\infty\int_0^t m(s)ds\\ &\quad + \sum_{k=0}^{n-2}\varphi_n^{(k)}(t) \frac{1}{k!}\|y_1(.)-y_2(.)\|_\infty\int_0^\eta m(s)ds\\ &\leq \Big(\frac{L(1)}{(n-1)!}+ L(\eta)\sum_{k=0}^{n-2} \frac{\varphi_n^{(k)}(1)}{k!}\Big)\|y_1(.)-y_2(.)\|_\infty. \end{align*} Then, by (c) \begin{align*} |z_2(t)-z_1(t)| &\leq |\psi_n^{g_2}(t)-\psi_n^{g_1}(t)| +\frac{1+t}{1-\tau}\Big[|\psi_n^{g_2}(\tau)-\psi_n^{g_1}(\tau)| +|\psi_n^{g_2}(1)-\psi_n^{g_1}(1)|\\ & \leq \Big(\frac{L(1)}{(n-1)!}+ L(\eta)\sum_{k=0}^{n-2} \frac{\varphi_n^{(k)}(1)}{k!}\Big)\|y_1(.)-y_2(.)\|_\infty \\ &\quad +\frac{2}{1-\tau}\Big[\Big(\frac{L(\tau)}{(n-1)!}+ L(\eta)\sum_{k=0}^{n-2} \frac{\varphi_n^{(k)}(\tau)}{k!}\Big)\|y_1(.)-y_2(.)\|_\infty\\ &\quad +\Big(\frac{L(1)}{(n-1)!}+ L(\eta)\sum_{k=0}^{n-2} \frac{\varphi_n^{(k)}(1)}{k!}\Big)\|y_1(.)-y_2(.)\|_\infty\Big]\\ &\leq \Big[\frac{(3-\tau)L(1)+2L(\tau)}{(1-\tau)(n-1)!}\\ &\quad + \sum_{k=0}^{n-2}\frac{L(\eta)}{(1-\tau)k!} \big[(3-\tau)\varphi_n^{(k)}(1)+ 2\varphi_n^{(k)}(\tau)\big]\Big]\|y_1(.)-y_2(.)\|_\infty. \end{align*} By the analogous relation, obtained by interchanging the roles of $y_1(.)$ and $y_2(.)$, it follows that \begin{align*} H\big(T(y_1(.)),T(y_2(.))\big) &\leq \Big[\frac{(3-\tau)L(1)+2L(\tau)}{(1-\tau)(n-1)!} + \sum_{k=0}^{n-2}\frac{L(\eta)}{(1-\tau)k!} \big[(3-\tau)\varphi_n^{(k)}(1)\\ &\quad + 2\varphi_n^{(k)}(\tau)\big]\Big]\|y_1(.)-y_2(.)\|_\infty. \end{align*} Consequently, $T$ is a contraction. Thus, by Lemma \ref{lemme1}, $T$ has a fixed point $y(.)$. \begin{proposition} \label{prop3.2} $y(.)$ is a solution of \eqref{cauchy4} and \eqref{cauchy2}. \end{proposition} \begin{proof} We have \[ y(t)=\psi_n^{g}(t)+\frac{1+t}{1-\tau}\big(\psi_n^{g}(\tau)-\psi_n^{g}(1)\big), \] where $g\in S_{F,y(.)}$. Then \[ y(1)=\psi_n^{g}(1)+ \frac{2}{1-\tau}\big(\psi_n^{g}(\tau)-\psi_n^{g}(1)\big) =\frac{-1-\tau}{1-\tau}\psi_n^{g}(1)+\frac{2}{1-\tau}\psi_n^{g}(\tau) \] and \[ y(\tau)=\psi_n^{g}(\tau)+ \frac{1+\tau}{1-\tau}\big(\psi_n^{g}(\tau)-\psi_n^{g}(1)\big) =\frac{-1-\tau}{1-\tau}\psi_n^{g}(1)+\frac{2}{1-\tau}\psi_n^{g}(\tau), \] hence $y(1)=y(\tau)$. On the other hand, for $0\leq i\leq n-2$ and $t\in[0,1]$, we have \begin{align*} [\psi_n^g]^{(i)}(t)&=\int_0^t\frac{(t-s)^{n-i-1}}{(n-i-1)!}g(s)ds + \sum_{k=0}^{n-2}\varphi_{n}^{(k+i)}(t)\int_0^\eta \frac{(\eta-s)^{k}}{k!}g(s)ds\\ &=\int_0^t\frac{(t-s)^{n-i-1}}{(n-i-1)!}g(s)ds + \sum_{l=i}^{n+i-2}\varphi_{n}^{(l)}(t)\int_0^\eta \frac{(\eta-s)^{l-i}}{(l-i)!}g(s)ds\\ &=\int_0^t\frac{(t-s)^{n-i-1}}{(n-i-1)!}g(s)ds + \sum_{l=i}^{n-2}\varphi_{n}^{(l)}(t)\int_0^\eta \frac{(\eta-s)^{l-i}}{(l-i)!}g(s)ds. \end{align*} Then, by (a) and (b) \begin{align*} [\psi_n^g]^{(i)}(0) &=\int_0^\eta\frac{(\eta-s)^{n-i-2}}{(n-i-2)!}g(s)ds + \sum_{l=i}^{n-3}\varphi_{n}^{(l)}(0)\int_0^\eta \frac{(\eta-s)^{l-i}}{(l-i)!}g(s)ds\\ &=\int_0^\eta\frac{(\eta-s)^{n-i-2}}{(n-i-2)!}g(s)ds + \sum_{l=i}^{n-3}\varphi_{n-l-1}(\eta)\int_0^\eta \frac{(\eta-s)^{l-i}}{(l-i)!}g(s)ds \end{align*} and by (a) \begin{align*} [\psi_n^g]^{(i+1)}(\eta) &=\int_0^\eta\frac{(\eta-s)^{n-i-2}}{(n-i-2)!}g(s)ds + \sum_{l=i}^{n-3}\varphi_{n}^{(l+1)}(\eta)\int_0^\eta \frac{(\eta-s)^{l-i}}{(l-i)!}g(s)ds\\ &=\int_0^\eta\frac{(\eta-s)^{n-i-2}}{(n-i-2)!}g(s)ds + \sum_{l=i}^{n-3}\varphi_{n-l-1}(\eta)\int_0^\eta \frac{(\eta-s)^{l-i}}{(l-i)!}g(s)ds, \end{align*} consequently \begin{equation} [\psi_n^g]^{(i+1)}(\eta)=[\psi_n^g]^{(i)}(0),\label{relationp} \end{equation} which implies that $y(0)=y'(\eta)$ and $y^{(i)}(0)=y^{(i+1)}(\eta)$ for $2\leq i\leq n-2$ whenever if $n\geq 4$. Finally, it is clear that $y^{(n)}(t)=g(t)$, hence $y^{(n)}(t)\in F(t,y(t))$. \end{proof} \begin{proof}[Proof of Theorem \ref{theoreme3}] Consider the multivalued map $T:\mathcal{C}([0,1],\mathbb{R})\to 2^{\mathcal{C}([0,1],\mathbb{R})}$ defined as follows: for $y(.)\in \mathcal{C}([0,1],\mathbb{R})$, $$ T(y(.)):=\big\{z(.)\in \mathcal{C}([0,1],\mathbb{R}): z(t)=\psi_n^g(t)\big\}. $$ We shall show that $T$ satisfies the assumptions of Lemma \ref{lemme1}. The proof will be given in two steps: \noindent{\bf Step 1:} \textit{$T$ has non-empty closed values.} Indeed, let $(y_p(.))_{p\geq 0}\in T(y(.))$ converges to $\bar y(.)$ in $\mathcal{C}([0,1],\mathbb{R})$. Then $\bar y(.)\in \mathcal{C}([0,1],\mathbb{R})$ and for each $t\in [0,1]$, \begin{align*} y_p(t)&\in \int_0^t\frac{(t-s)^{n-1}}{(n-1)!}F(s,y(s))ds + \sum_{k=0}^{n-2}\varphi_{n}^{(k)}(t)\int_0^\eta \frac{(\eta-s)^{k}}{k!}F(s,y(s))ds. \end{align*} Since the set \[ \int_0^t \frac{(t-s)^{k}}{k!}F(s,y(s))ds \] is closed for all $t\in [0,1]$ and $0\leq k\leq n-1$, we have \[ \bar y(t)\in \int_0^t\frac{(t-s)^{n-1}}{(n-1)!}F(s,y(s))ds + \sum_{k=0}^{n-2}\varphi_{n}^{(k)}(t)\int_0^\eta \frac{(\eta-s)^{k}}{k!}F(s,y(s))ds. \] Then $\bar y(.)\in T(y(.))$. So $T(y(.))$ is closed for each $y(.)\in \mathcal{C}([0,1],\mathbb{R})$. \noindent {\bf Step 2:} \textit{$T$ is a contraction.} Indeed, let $y_1(.), y_2(.)\in \mathcal{C}([0,1],\mathbb{R})$ and $z_1(.)\in T(y_1(.))$. Then \[ z_1(t)=\psi_n^{g_1}(t), \] where $g_1\in S_{F,y_1(.)}$. By (\ref{measura}), there exists $g_2$ such that $$ g_2(t)\in F(t,y_2(t))\quad \mbox{and}\quad |g_1(t)-g_2(t)|\leq m(t)|y_1(t)-y_2(t)|,\quad \mbox{for each }t\in [0,1]. $$ Now, for $t\in[0,1]$, we set $z_2(t)=\psi_n^{g_2}(t)$. On the other hand, we have \begin{align*} |\psi_n^{g_2}(t)-\psi_n^{g_1}(t)| & \leq \int_0^t\frac{(t-s)^{n-1}}{(n-1)!}|g_1(s)-g_2(s)|ds\\ &\quad + \sum_{k=0}^{n-2}\varphi_n^{(k)}(t) \int_0^\eta\frac{(\eta-s)^{k}}{k!}|g_1(s)-g_2(s)|ds\\ &\leq \frac{1}{(n-1)!}\int_0^t m(s)|y_1(s)-y_2(s)|ds\\ &\quad + \sum_{k=0}^{n-2}\varphi_n^{(k)}(t) \frac{1}{k!}\int_0^\eta m(s)|y_1(s)-y_2(s)|ds\\ &\leq \frac{1}{(n-1)!}\|y_1(.)-y_2(.)\|_\infty\int_0^t m(s)ds\\ &\quad + \sum_{k=0}^{n-2}\varphi_n^{(k)}(t) \frac{1}{k!}\|y_1(.)-y_2(.)\|_\infty\int_0^\eta m(s)ds\\ &\leq \Big(\frac{L(t)}{(n-1)!}+ L(\eta)\sum_{k=0}^{n-2} \frac{\varphi_n^{(k)}(t)}{k!}\Big)\|y_1(.)-y_2(.)\|_\infty. \end{align*} Then, by (c) \[ |z_2(t)-z_1(t)|\leq \Big(\frac{L(1)}{(n-1)!}+ L(\eta)\sum_{k=0}^{n-2} \frac{\varphi_n^{(k)}(1)}{k!}\Big)\|y_1(.)-y_2(.)\|_\infty. \] By the analogous relation, obtained by interchanging the roles of $y_1(.)$ and $y_2(.)$, it follows that \[ H\big(T(y_1(.)),T(y_2(.))\big) \leq \Big(\frac{L(1)}{(n-1)!}+ L(\eta)\sum_{k=0}^{n-2} \frac{\varphi_n^{(k)}(1)}{k!}\Big)\|y_1(.)-y_2(.)\|_\infty. \] Consequently, $T$ is a contraction. Hence, by Lemma \ref{lemme1}, $T$ has a fixed point $y(.)$. \end{proof} \begin{proposition} $y(.)$ is a solution of \eqref{cauchy3}. \end{proposition} \begin{proof} By (\ref{relationp}), we have $y^{(i)}(0)=y^{(i+1)}(\eta),$ for $0\leq i\leq n-2$. Since $y^{(n)}(t)=g(t)$, we have $y^{(n)}(t)\in F(t,y(t))$. \end{proof} \begin{thebibliography}{0} \bibitem[1]{boucherif} A. Boucherif and S. M. Bouguima; \emph{Nonlinear second order ordinary differential equations with nonlocal boundary conditions}, Commu. Appl. Nonl. Anal., 5(2), (1998), 73-85. \bibitem[2]{castaing} C. Castaing and M. Valadier; \emph{Convex Analysis and Measurable Multifunctions}, Lecture Notes in Mathematics 580, Springer-Verlag, Berlin-Heidelberg-New York, (1977). \bibitem[3]{covitz} H. Covitz and S. B. Jr. Nadler; \emph{Multivalued contraction mappings in generalized metric spaces}, Israel J. Math., 8, (1970), 5-11. \bibitem[4]{benchohra1} M. Benchohra and S. K. 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