\documentclass[reqno]{amsart} \AtBeginDocument{{\noindent\small {\em Electronic Journal of Differential Equations}, Vol. 2003(2003), No. 67, pp. 1--12.\newline ISSN: 1072-6691. URL: http://ejde.math.swt.edu or http://ejde.math.unt.edu \newline ftp ejde.math.swt.edu (login: ftp)} \thanks{\copyright 2003 Southwest Texas State University.} \vspace{9mm}} \begin{document} \title[\hfilneg EJDE--2003/67\hfil Impulsive neutral functional differential inclusions] {Impulsive neutral functional differential inclusions with variable times} \author[Mouffak Benchohra \& Abdelghan Ouahabi\hfil EJDE--2003/67\hfilneg] {Mouffak Benchohra \& Abdelghani Ouahab} \address{Laboratoire de Math\'ematiques, Universit\'e de Sidi Bel Abb\`es\\ BP 89, 22000 Sidi Bel Abb\`es, Alg\'erie} \email{benchohra@univ-sba.dz} \date{} \thanks{Submitted November 27, 2002. Published June 13, 2003.} \subjclass[2000]{34A37, 34A60, 34K25} \keywords{Impulsive neutral functional differential inclusions, variable times, \hfill\break\indent condensing map, fixed point} \begin{abstract} In this paper, we study the existence of solutions for first and second order impulsive neutral functional differential inclusions with variable times. Our main tool is a fixed point theorem due to Martelli for condensing multivalued maps. \end{abstract} \maketitle \newtheorem{theorem}{Theorem}[section] \newtheorem{lemma}[theorem]{Lemma} \newtheorem{remark}[theorem]{Remark} \newtheorem{definition}[theorem]{Definition} \numberwithin{equation}{section} \allowdisplaybreaks \section{Introduction} This paper concerns the existence of solutions for initial-value problems for first and second order neutral functional differential inclusions with impulsive effects at variable times. In Section 3, we consider the first order initial-value problem (IVP for short) \begin{gather}\label{e1} \frac{d}{dt}[y(t)-g(t,y_t)]\in F(t,y_{t}), \quad\mbox{a. e. } t\in J=[0,T], \; t\neq \tau_k(y(t)), \; k=1,\ldots,m, \\ \label{e2} y(t^+)=I_{k}(y(t)), \quad t=\tau_k(y(t)), \; k=1,\ldots,m, \\ \label{e3} y(t)=\phi(t), \quad t\in [-r,0], \end{gather} where $F: J\times D\to 2^{\mathbb{R}^{n}}$ is a compact convex valued multivalued map, $ g: J\times D\to\mathbb{R}^n$ is given function, $D=\{\psi:[-r,0]\to\mathbb{R}^n; \psi$ is continuous everywhere except for a finite number of points $\bar{t}$ at which $\psi(\bar{t})$ and $\psi(\bar{t}^+)$ exist and $\psi(\bar{t}^-)=\psi(\bar{t})\}$, $\phi\in D$, $00$, there exists $\phi_{q} \in L^{1}(J,\mathbb{R}_{+})$ such that $$ \|F(t,u)\|=\sup\{|v|: v\in F(t,y)\}\leq\phi_{q}(t) $$ for all $\|u\|_{D}\leq q$ and for almost all $t\in J$. \end{itemize} \end{definition} For more details on multi-valued maps we refer the reader to the books of Deimling \cite{Dei}, Gorniewicz \cite{Gor}, Hu and Papageorgiou \cite{HuPa} and Tolstonogov \cite{Tol}. For a function $y$ defined on $[-r,T]$ we define the set $$ S_{F,y}=\{v\in L^{1}(J,\mathbb{R}^n): \ v(t)\in F(t,y_t) \hbox { for a.e. } t\in J\}, $$ which is known as the set of {\em selection functions}. The following lemmas are crucial in the proof of our main theorem. \begin{lemma}[\cite{LaOp}] \label{l1} Let $I$ be a compact real interval and $X$ be a Banach space. Let $F$ be a multi-valued map satisfying the Carath\'eodory conditions with the set of $L^{1}$-selections $S_{F}$ nonempty, and let $\Gamma$ be a linear continuous mapping from $L^{1}(I, X)$ to $C(I, X)$. Then the operator $$ \Gamma \circ S_{F}:C(I, X)\to CC(C(I, X)), \quad y\mapsto (\Gamma \circ S_{F})(y):=\Gamma(S_{F,y})$$ is a closed graph operator in $C(I, X)\times C(I, X)$. \end{lemma} \begin{lemma}[\cite{Mar}] \label{l2} Let $ N:X\to CC(X)$ be an upper semicontinuous and condensing map. If the set $$ \mathcal{M}:=\{ y\in X: y\in\lambda N(y) \mbox{ for some } 0<\lambda<1 \} $$ is bounded, then $ N$ has a fixed point. \end{lemma} To define the solutions of problems (\ref{e1})-(\ref{e3}) and (\ref{e4})-(\ref{e7}), we shall consider the space \begin{align*} PC&=\big\{y: [0,T]\to\mathbb{R}^n: \hbox{there exist } 0 =t_0 < t_1 < \ldots < t_m 0$ there exists a positive constant $\ell$ such that for each $y\in B_{q}=\{y\in C([-r,T],\mathbb{R}^n): \|y\|_{\infty}\leq q \}$ we have $\|N_1(y)\|\leq \ell$. Let $y\in B_{q}$ and $h\in N_1(y)$ then there exists $v\in S_{F, y}$ such that for each $t\in J$ we have $$ h(t)=\phi(0)+\int_{0}^{t}v(s)ds. $$ Thus, $$ |h(t)|\leq |\phi(0)|+\int_{0}^{t}|v(s)|ds \leq \|\phi\|_{\infty}+\|h_{q}\|_{L^{1}}:=\ell. $$ {\bf Claim 3:} {\em $N_1$ maps bounded sets into equicontinuous sets of $C([-r,T],\mathbb{R}^n)$.} Let $u_{1}, u_{2}\in J$, $u_{1}From Lemma \ref{l1}, it follows that $\Gamma\circ S_{F}$ is a closed graph operator. Since $\big(h_{n}(t)-\phi(0)\big) \in \Gamma( S_{F, y_{n}})$, it follows from Lemma \ref{l1} that for some $v_{*}\in S_{F,v_{*}}$, $$ h_{*}(t)=\phi(0)+\int_{0}^{t}v_{*}(s)ds\,. $$ \noindent{\bf Claim 5:} {\em The following set is bounded,} $$ \mathcal{E}(\mathcal{N}):=\{y\in C([-r,T],\mathbb{R}^n): y\in\lambda\mathcal{N}(y), \hbox{ for some } 0<\lambda<1 \}\,. $$ Let $y\in \mathcal{E}(\mathcal{N})$. Then there exists $v\in S_{F,y}$ such that $y\in\lambda\mathcal{N}(y)$, for some $0<\lambda<1$. Thus, for each $t\in[0,T]$, $$ y(t)=\lambda\big(\phi(0)-g(0,\phi)+g(t,y_{t})+ \int_{0}^{t}v(s)\, ds\big). $$ This implies, by (H2)--(H4), that for each $t\in J$ we have $$ |y(t)|\leq \|\phi\|+d_1\|\phi\|+d_1\|y_t\|+2d_{2}+ \int_{0}^{t}p(s)\psi(\|y_{s}\|)ds. $$ We consider the function $$ \mu(t)=\sup\{|y(s)|: -r\leq s\leq t \}, \ \ 0\leq t\leq T. $$ Let $t^{*}\in [-r,t]$ be such that $\mu(t)=|y(t^{*})|$. If $t^{*}\in J$, by the previous inequality we have for $t\in J$ $$ \mu(t)\leq \|\phi\|+d_1\|\phi\|+d_1\mu(t)+2d_{2}+ \int_{0}^{t}p(s)\psi(\mu(s))ds. $$ Thus $$ \mu(t)\leq\frac{1}{1-d_1}\Big[\|\phi\|+d_1\|\phi\|+2d_{2} +\int_0^{t}p(s)\psi(\mu(s))ds\Big], \quad t\in J. $$ If $t^{*}\in[-r,0]$ then $\mu(t)=\|\phi\|$ and the previous inequality holds. Let us take the right-hand side of the above inequality to be $v(t)$. Then \begin{gather*} c=v(0)=\frac{1}{1-d_1}(\|\phi\|+d_1\|\phi\|+2d_{2}), \\ \mu(t)\leq v(t), \quad t\in J, \\ v'(t)=\frac{1}{1-d_1}p(t)\psi(\mu(t)), \quad t\in J. \end{gather*} Using the nondecreasing character of $\psi$, we obtain $$ v'(t)\leq \frac{1}{1-d_1}p(t)\psi(v(t)). $$ This implies that for each $t\in J$, $$ \int_{v(0)}^{v(t)}\frac{d\gamma}{\psi(\gamma)} \leq\frac{1}{1-d_1}\int_{0}^{T}p(s)ds< \int_{v(0)}^{\infty}\frac{d\gamma}{\psi(\gamma)}. $$ This inequality implies that there exists a constant $K$ such that $v(t)\leq K$, $t\in J$, and hence $\mu(t)\leq K$, $t\in J$. Since for every $t\in [0,T]$, $\|y_{t}\|\leq\mu(t)$, we have $$ \|y\|_{\infty}\leq K'=\max\{\|\phi\|,K\}, $$ where $K'$ depends only $T, d_1, d_2$, and on the functions $p, \phi$ and $\psi$. This shows that $\mathcal{E}(\mathcal{N})$ is bounded. Set $X:=C([-r,T],\mathbb{R}^n)$. As a consequence of Lemma \ref{l2} we deduce that $N$ has a fixed point which is a solution of (\ref{eq1})--(\ref{eq2}). Denote this solution by $y_1$. Define the function $$ r_{k,1}(t)=\tau_{k}(y_1(t))-t\quad \mbox{for } t\geq0. $$ Hypothesis (H1) implies that $r_{k,1}(0)\neq 0$ for $k=1,\ldots,m$. If $r_{k,1}(t)\neq 0$ on $[0,T]$ for $k=1,\ldots,m$; i.e., $$ t\neq \tau_k(y_1(t)) \quad \hbox{on } [0,T]\quad \hbox{for } k=1,\ldots,m, $$ then $y_1$ is a solution of the problem (\ref{e1})-(\ref{e3}). It remains to consider the case when $r_{1,1}(t)=0$ for some $t\in [0,T]$. Now since $r_{1,1}(0)\neq 0$ and $r_{1,1}$ is continuous, there exists $t_1>0$ such that $$ r_{1,1}(t_1)=0, \quad \hbox{and} \quad r_{1,1}(t)\neq 0 \quad \hbox{for all } t\in[0,t_1). $$ Thus by (H1) we have $r_{k,1}(t)\neq 0$ for all $t\in[0,t_1)$, and $k=1,\ldots,m$. \noindent {\bf Step 2:} Consider now the problem \begin{gather}\label{eq3} \frac{d}{dt}[y(t)-g(t,y_t)]\in F(t, y_t), \quad\mbox{a.e. } t\in [t_1,T], \\ \label{eq4} y(t_1^+)=I_1(y_1(t_1)). \end{gather} Transform this problem into a fixed point problem. Consider the operator $N_2: C([t_1,T],\mathbb{R}^n)\to 2^{C([t_1,T],\mathbb{R}^n)}$ defined as $$ N_2(y)=\big\{ h\in C([t_1,T],\mathbb{R}^n): h(t)= I_1(y_1(t_1))-g(t_1,y_{t_1})+g(t,y_t)+\int_{t_{1}}^{t}v(s)ds\big\}, $$ where $v\in S_{F,y}$. As in Step 1 we can show that $N_2$ is completely continuous, and that the following set is bounded, $$ \mathcal{E}(N_2):=\{y\in C([t_1,T],\mathbb{R}^n): y\in\lambda N_2(y), \quad\hbox{for some } 0<\lambda<1 \}\,. $$ Set $X:=C([t_1,T],\mathbb{R}^n)$. As a consequence of Martelli's theorem, we deduce that $N_2$ has a fixed point $y$ which is a solution to problem (\ref{eq3})--(\ref{eq4}). Denote this solution by $y_2$. Define $$ r_{k,2}(t)=\tau_{k}(y_2(t))-t\quad \hbox{for}\ t\geq t_{1}. $$ If $ r_{k,2}(t)\neq 0$ on $(t_1,T]$ for all $k=1,\ldots,m$, then $$ y(t)=\begin{cases} y_{1}(t), & \mbox{if } t\in [0,t_{1}],\\ y_{2}(t), &\mbox{if }t\in (t_{1},T], \end{cases} $$ is a solution of the problem (\ref{e1})--(\ref{e3}). It remains to consider the case when $r_{2,2}(t)=0$, for some $t\in (t_{1},T]$. By (H8), we have \begin{align*} r_{2,2}(t_1^+)&= \tau_{2}(y_{2}(t_{1}^{+}))-t_{1} \\ &= \tau_{2}(I_{1}(y_{1}(t_{1})))-t_{1} \\ &> \tau_{1}(y_{1}(t_{1}))-t_{1} \\ &= r_{1,1}(t_{1})=0. \end{align*} Since $r_{2,2}$ is continuous, there exists $t_2>t_1$ such that $r_{2,2}(t_{2})=0$ and $r_{2,2}(t)\neq 0$ for all $ t\in(t_1,t_2)$. It is clear by (H1) that $$ r_{k,2}(t)\neq 0 \quad \hbox{for all } t\in(t_1,t_2), \; k=2,\ldots,m. $$ Suppose now that there is $\bar s\in (t_1,t_2]$ such that $r_{1,2}(\bar s)=0$. Consider the function $L_1(t)=\tau_1(y_2(t)-g(t,y_t))-t$. >From (H6)-(H8) it follows that \begin{align*} L_{1}(\bar s)&= \tau_{1}(y_2(\bar s)-g(\bar s,y_{\bar s}))-\bar s\\ &\geq \tau_{1}(y_{2}(\bar s))-\bar s \\ &= r_{1,2}(\bar s)=0. \end{align*} Thus the function $L_{1}$ attains a nonnegative maximum at some point $s_1\in(t_1,T]$. Since $$ \frac{d}{dt}[y_2(t)-g(t,y_{2t})]\in F(t,y_{2t}), \quad\mbox{a.e. } t\in (t_{1},T), $$ then there exists $v(\cdot)\in L^1((t_{1},T))$ with $v(t)\in F(t,y_{2t})$, a.e. $t\in (t_{1},T)$ such that $$ \frac{d}{dt}[y_2(t)-g(t,y_{2t})]=v(t), $$ then $$ L'_{1}(s_1)=\tau_1'(y_2(s_1)-g(s_1,y_{2s_1})) \frac{d}{dt}[y_2(s_1)-g(s_1,y_{s_1})]-1=0. $$ Therefore, $$\langle\tau'_1(y_2(s_1)-g(s_1,y_{2s_1})), v(s_1)\rangle=1, $$ which is a contradiction by (H4). \noindent{\bf Step 3:} We continue this process and take into account that $y_{m}:=y\Bigr|_{[t_{m},T]}$ is a solution to the problem \begin{gather}\label{eq5} \frac{d}{dt}[y(t)-g(t,y_{t})]\in F(t,y_{t}), \quad\mbox{a.e. } t\in (t_{m},T),\\ \label{eq13} y(t^+_{m})=I_{m}(y_{m-1}(t_{m})). \end{gather} The solution $y$ of the problem (\ref{e1})-(\ref{e3}) is then defined by $$ y(t)=\begin{cases} y_{1}(t), & \mbox{if } t\in [-r,t_{1}],\\ y_{2}(t), & \mbox{if } t\in (t_{1},t_{2}], \\ \dots \\ y_{m}(t), & \mbox{if } t\in (t_{m},T]. \end{cases} $$ %\end{proof} \section{Second Order Impulsive NFDIs} In this section, we study the initial-value problem (\ref{e4})--(\ref{e7}). \begin{definition}\label{d2} \rm A function $y\in \Omega\cap\cup_{k=0}^{m}AC((t_{k},t_{k+1}),\mathbb{R}^{n})$ is said to be a solution of (\ref{e4})--(\ref{e7}) if there exists $v(t)\in F(t,y_t)$ a.e. $t\in[0,T]$ such that $\frac{d}{dt}[y'(t)-g(t,y_{t})]=v(t)$ a.e. on $J,\ t\neq \tau_k(y(t))$, $k=1,\ldots,m$, $y(t^+)=I_{k}(y(t))$, $t=\tau_k(y(t))$, $k=1,\ldots,m$, $y'(t^{+})=\overline I_{k}(y(t))$, $t=\tau_k(y(t))$ $k=1,\ldots,m$, $ y(t)=\phi(t)$, $t\in[-r,0]$ and $y'(0)=\eta$. \end{definition} For the next theorem we need the following assumptions: \begin{itemize} \item[(A1)] There exist positive constants $\overline d_{k}$ such that $|\overline I_{k}(x)|\leq\overline d_{k}$, $k=1,\ldots,m$ for each $x\in\mathbb{R}^n$ \item[(A2)] There exists a continuous nondecreasing function $\psi:[0,\infty)\to (0,\infty)$ and $p\in L^{1}([0,T],\mathbb{R}_{+})$ such that $\|F(t,u)\|\leq p(t)\psi (\|u\|)$ a.e. $t\in [0,T]$ and each $u\in D$ with $$ \int_{1}^{\infty}\frac{d\gamma}{\gamma+\psi(\gamma)}=\infty, $$ \item[(A3)] For all $(t,\bar s, x)\in [0,T]\times[0,T]\times \mathbb{R}^n$ and all $y_t\in D$ we have, for all $v\in S_{F,y}$, $$ \langle \tau'_k(x),\overline I_k(y(\bar s))-g(\bar s,y_{\bar s})+g(t,y_t) +\int_{\bar s}^{t} v(s)ds\rangle \neq 1\quad \hbox{for } k=1,\ldots,m\,. $$ \item[(A4)] For all $x\in \mathbb{R}^n$, $\tau_k(I_k(x))\leq \tau_k(x)<\tau_{k+1}(I_k(x))$ for $k=1,\ldots,m$. \end{itemize} \begin{theorem} \label{thm4.2} Assume that (H1)-(H3) and (A1)-(A4) are satisfied. Then the IVP (\ref{e4})--(\ref{e7}) has at least one solution. \end{theorem} The proof of this theorem will be given in several steps. \noindent{\bf Step 1:} Consider the problem \begin{gather}\label{eq7} \frac{d}{dt}[y'(t)-g(t,y_t)]\in F(t, y_t), \quad\mbox{a.e. } t\in [0,T],\\ \label{eq8} y(t)=\phi(t),\quad t\in[-r,0], \; y'(0)=\eta\,. \end{gather} Transform the problem into a fixed point problem. Consider the operator $\overline N_1: C([-r,T],\mathbb{R}^n)\to 2^{C([-r,T],\mathbb{R}^n)}$ defined as $\overline N_1(y)=\{h\in C([-r,T],\mathbb{R}^{n})\}$ where $$ h(t)=\begin{cases} \phi(t),& \mbox{if } t\in [-r,0]; \\ \phi(0)+[\eta-g(0,\phi(0))]t +\int^{t}_{0}g(s,y_s)ds +\int_{0}^{t}(t-s)v(s)ds& \mbox{if } t\in [0,T]. \end{cases} $$ As in Theorem \ref{t1}, we can show that $\overline N_1$ is completely continuous. Now we prove only that the following set is bounded, $$ \mathcal{E}(\overline N_1):=\{y\in C([-r,T],\mathbb{R}^n): y\in\lambda \overline N_1(y), \hbox{ for some } 0<\lambda<1 \}\,. $$ Let $y\in \mathcal{E}(\overline N_1)$. Then there exists $v\in S_{F,y}$ such that $y\in\lambda \overline N_1(y)$ for some $0<\lambda<1$. Thus for each $t\in[0,T]$ we have $$ y(t)=\lambda\phi(0)+\lambda[\eta-g(0,\phi(0))]t +\lambda\int^{t}_{0}g(s,y_s)ds +\lambda\int_{0}^{t}(t-s)v(s)ds. $$ This implies, by (H2), (H3), (A1), and (A2), that for each $t\in [0,T]$ we have \begin{align*} |y(t)| &\leq \|\phi\|+T(|\eta|+\|\phi\|d_1+d_2)+\int^{t}_{0}\! d_1{\|y_s\|}ds+Td_{2} +\int_{0}^{t}\! (T-s)p(s)\psi(\|y_s\|)ds \\ &\leq \|\phi\|+T(|\eta|+\|\phi\|d_1+2d_2)+\int^{t}_{0}M(s){\|y_s\|}ds +\int_{0}^{t}M(s)\psi(\|y_s\|)ds, \end{align*} where $M(t)=\max\{d_{1}, (T-t)p(t)\}$. Consider the function $$ \mu(t)=\sup\{|y(s)|:-r\leq s\leq t\}, \quad 0\leq t\leq T. $$ Let $t^*\in [-r,t]$ be such that $\mu(t)=|y(t^*)|$. If $t^*\in J$, by the previous inequality, for $t\in[0,T]$, we have $$ \mu(t) \leq \|\phi\| +T(|\eta|+\|\phi\|d_1+2d_2) +\int^{t}_{0}M(s){\mu(s)}ds +\int^t_0M(s)\psi(\mu(s))ds. $$ Let us denote the right-hand side of the above inequality to be $v(t)$. Then \begin{gather*} v(0)=\|\phi\|+T(|\eta|+\|\phi\|d_1+2d_2), \\ v'(t)=M(t)\mu(t)+M(t)\psi(\mu(t)),\ t\in [0,T]. \end{gather*} Using the nondecreasing character of $\psi$ we obtain, for a.e. $t\in [0,T]$, $$ v'(t)\leq M(t)v(t)+M(t)\psi(v(t)) =M(t)[v(t)+\psi(v(t))]. $$ This implies that, for each $t\in[0,T]$, $$ \int_{v(0)}^{v(t)}\frac{d\gamma}{\gamma+\psi(\gamma)}\leq\int_{0}^{T}M(s)ds< \int_{v(0)}^{\infty}\frac{d\gamma}{\gamma+\psi(\gamma)}. $$ This inequality implies that there exists a constant $b^{*}$ such that $v(t)\leq b^{*}, \ t\in [0,T]$, and hence $\mu(t)\leq b^{*}, \ t\in[0,T] $. Since for every $t\in [0,T], \|y_{t}\|\leq\mu(t)$, we have $$ \|y\|_{\infty}\leq \max\{\|\phi\|,b^{*}\}, $$ where $b^{*}$ depends only $T$ and on the functions $p$ and $\psi$. This shows that $\mathcal{E}( \overline N_1)$ is bounded. Set $X:=C([-r,T],\mathbb{R}^n)$. As a consequence of Martelli's theorem, we deduce that $\overline N_{1}$ has a fixed point $y$ which is a solution to (\ref{eq7})--(\ref{eq8}). Denote this solution by $y_1$. Define the function $$ r_{k,1}(t)=\tau_{k}(y_1(t))-t\quad \hbox{for}\quad t\geq0. $$ Hypothesis (H1) implies that $ r_{k,1}(0)\neq 0$ for $k=1,\ldots,m$. If $r_{k,1}(t)\neq 0$ on $[0,T]$ for $k=1,\ldots,m$; i.e., $$ t\neq \tau_k(y_1(t)) \quad \hbox{on } [0,T] \quad \mbox{and for } k=1,\ldots,m\,. $$ Then $y_1$ is a solution of the problem (\ref{e1})-(\ref{e3}). It remains to consider the case when $r_{1,1}(t)=0$ for some $t\in [0,T]$. Now since $r_{1,1}(0)\neq 0$ and $r_{1,1}$ is continuous, there exists $t_1>0$ such that $$ r_{1,1}(t_1)=0, \quad \mbox{and} \quad r_{1,1}(t)\neq 0 \quad \mbox{for all } t\in[0,t_1). $$ Thus, by (H1) we have $r_{k,1}(t)\neq 0$ for all $t\in[0,t_1)$ and $k=1,\ldots,m$. {\bf Step 2:} Consider now the problem \begin{gather}\label{eq9} \frac{d}{dt}[y'(t)-g(t,y_t)]\in F(t, y_t), \quad\mbox{a.e. } t\in [t_1,T],\\ \label{eq10} y(t_1^+)=I_1(y_1(t_1)), \\ \label{eq11} y'(t_1^+)=\overline I_1(y_1(t_1)). \end{gather} Transform the problem into a fixed point problem. Consider the operator $\overline N_2: C([t_1,T],\mathbb{R}^n)\to 2^{C([t_1,T],\mathbb{R}^n)}$ defined as $\overline N_2(y)=\{ h\in C([t_1,T],\mathbb{R}^n)\}$ where $$ h(t)=I_1(y_1(t_1))+(t-t_1)\overline I_1(y_1(t_1))-(t-t_1)g(t_1,y_{t_1}) +\int^t_{t_1}\! g(s,y_s)ds +\int_{t_{1}}^{t}(t-s)v(s)ds, $$ with $v\in S_{F,y}$. As in Step 1 we can show that $\overline N_2$ is completely continuous, and that the following set is bounded, $$ \mathcal{E}(\overline N_2):=\{y\in C([t_1,T],\mathbb{R}^n): y\in\lambda\overline N_2(y), \hbox{ for some } 0<\lambda<1 \}. $$ Set $X:=C([t_1,T],\mathbb{R}^n)$. As a consequence of Martelli's theorem, we deduce that $N_2$ has a fixed point $y$ which is a solution to problem (\ref{eq9})--(\ref{eq11}). Denote this solution by $y_2$. Define $$ r_{k,2}(t)=\tau_{k}(y_2(t))-t\quad \hbox{for}\ t\geq t_{1}. $$ If $r_{k,2}(t)\neq 0$ on $(t_1,T]$ for all $k=1,\ldots,m$, then $$ y(t)=\begin{cases} y_{1}(t), & \mbox{if } t\in [0,t_{1}],\\ y_{2}(t), & \mbox{if } t\in (t_{1},T], \end{cases} $$ is a solution of the problem (\ref{e4})--(\ref{e7}). It remains to consider the case when $r_{2,2}(t)=0$, for some $t\in (t_{1},T]$. By (A4) we have \begin{align*} r_{2,2}(t_1^+)&= \tau_{2}(y_{2}(t_{1}^{+}))-t_{1} \\ &= \tau_{2}(I_{1}(y_{1}(t_{1}))-t_{1} \\ &> \tau_{1}(y_{1}(t_{1}))-t_{1} \\ &= r_{1,1}(t_{1})=0. \end{align*} Since $r_{2,2}$ is continuous, there exists $t_2>t_1$ such that $r_{2,2}(t_{2})=0$, and $r_{2,2}(t)\neq 0$ for all $t\in(t_1,t_2)$. It is clear by (H1) that $$ r_{k,2}(t)\neq 0 \quad \hbox{for all } t\in(t_1,t_2), \quad k=2,\ldots,m. $$ Suppose now that there is $\bar s\in (t_1,t_2]$ such that $r_{1,2}(\bar s)=0$. From (A4), it follows that \begin{align*} r_{1,2}(t_1^+)&= \tau_{1}(y_{2}(t_{1}^{+}))-t_{1} \\ &= \tau_{1}(I_{1}(y_{1}(t_{1})))-t_{1} \\ &\leq \tau_{1}(y_{1}(t_{1}))-t_{1} \\ &= r_{1,1}(t_{1})=0. \end{align*} Thus, the function $r_{1,2}$ attains a nonnegative maximum at some point $s_1\in(t_1,T]$. Since $$ \frac{d}{dt}[y'_2(t)-g(t, y_{2t})]\in F(t,y_{2t}), \quad\mbox{a.e. } t\in (t_{1},T), $$ there exist $v(\cdot)\in L^1((t_{1},T))$ with $v(t)\in F(t,y_{2t})$, a.e. $t\in (t_{1},T)$ such that $$ y'_2(t)=\overline I_1(y(t_1))-g(t_1,y_{t_1})+g(t,y_{t}) +\int_{t_1}^{t} v(s)ds\,. $$ Then $$ r'_{1,2}(s_1)=\tau_1'(y_2(s_1)) \Big(\overline I_1(y(t_1))-g(t_{1},y_{t_{1}})+g(s_1,y_{s_1}) +\int_{t_1}^{s_{1}} v(s)ds\Big)-1=0. $$ Therefore, $$ \langle\tau'_1(y_2(s_1)), \overline I_1(y(t_1))-g(t_{1},y_{t_{1}}) +g(s_1,y_{s_1}) +\int_{t_1}^{s_{1}} v(s)ds\rangle=1, $$ which is a contradiction by (A3). \noindent{\bf Step 3:} We continue this process by taking into account that $y_{m}:=y\bigr|_{[t_{m},T]}$ is a solution to the problem \begin{gather}\label{eq12} \frac{d}{dt}[y'(t)-g(t,y_{t})]\in F(t,y_{t}), \ a.e.\ t\in (t_{m},T),\\ \label{eq14} y(t^+_{m})=I_{m}(y_{m-1}(t_{m})), \\ \label{eq15} y'(t^+_{m})=\overline I_{m}(y_{m-1}(t_{m})). \end{gather} The solution $y$ of the problem (\ref{e4})-(\ref{e7}) is then defined by $$ y(t)=\begin{cases} y_{1}(t), &\mbox{if } t\in [-r,t_{1}],\\ y_{2}(t), &\mbox{if } t\in (t_{1},t_{2}], \\ \dots \\ y_{m}(t), & \mbox{if } t\in (t_{m},T]\,. \end{cases} $$ %\end{proof} \begin{thebibliography}{00} \bibitem{BaGo} J. 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