\documentclass[twoside]{article} \usepackage{amssymb, amsmath} \pagestyle{myheadings} \markboth{\hfil Nonlocal Cauchy problems \hfil EJDE--2002/47} {EJDE--2002/47\hfil Abdelkader Boucherif\hfil} \begin{document} \title{\vspace{-1in}% \parbox{\linewidth}{\footnotesize\noindent {\sc Electronic Journal of Differential Equations}, Vol. {\bf 2002}(2002), No. 47, pp. 1--9. \newline ISSN: 1072-6691. URL: http://ejde.math.swt.edu or http://ejde.math.unt.edu \newline ftp ejde.math.swt.edu (login: ftp)} \vspace{\bigskipamount} \\ % Nonlocal Cauchy problems for first-order multivalued differential equations % \thanks{\emph{Mathematics Subject Classifications:} 34A60, 34G20. \hfil\break\indent {\em Key words:} Cauchy problems, multivalued differential equations, nonlocal condition, \hfil\break\indent topological transversality theorem. \hfil\break\indent \copyright 2002 Southwest Texas State University. \hfil\break\indent Submitted January 27, 2002. Published May 28, 2002.} } \date{} \author{Abdelkader Boucherif} \maketitle \begin{abstract} We prove the existence of solutions for a nonlocal Cauchy problem for a first-order multivalued differential equation. Our approach is based on the topological transversality theory for set-valued maps. \end{abstract} \newtheorem{theorem}{Theorem}[section] \numberwithin{equation}{section} \section{Introduction} In this paper, we investigate the existence of solutions for the nonlocal Cauchy problem \begin{equation} \begin{gathered} x'(t)\in F( t,x(t)) \quad t\in (0,T] \\ x(0)+\sum_{k=1}^{m}a_{k}x( t_{k}) =0 \end{gathered} \label{e1} \end{equation} Here $F:J\times \mathbb{R}\rightarrow {2}^{\mathbb{R}}$ is a set-valued map, $J=[0,T]$, $01$, are completely continuous for $J$ compact \cite{Brezis}. \subsection*{Set-valued Maps} Let $X$ and $Y$ be Banach spaces. A set-valued map $G:X\to 2^{Y}$ is said to be compact if $G(X)=\overline{\cup \{G(x);x\in X\}}$ is compact. $G$ has convex (closed, compact) values if $G(x)$ is convex (closed, compact) for every $x\in X$. $G$ is bounded on bounded subsets of $X$ if $G(B)$ is bounded in $Y$ for every bounded subsets $B$ of $X$. A set-valued map $G$ is upper semicontinuous at $z_{0}\in X$ if for every open set O containing $Gz_{0}$, there exists a neighborhood $\mathcal{M}$ of $z_{0}$ such that $G(\mathcal{M})\subset O$. $G$ is upper semicontinuous on $X$ if it is upper semicontinuous at every point of $X$. If $G$ is nonempty and compact-valued then $G$ is upper semicontinuous if and only if $G$ has a closed graph. The set of all bounded closed convex and nonempty subsets of $X$ is denoted by $bcc(X)$. A set-valued map $G:J\to bcc(X)$ is measurable if for each $x\in X$, the function $t\mapsto \mbox{dist}(x,G(t))$ is measurable on $J$. If $X\subset Y$, $G$ has a fixed point if there exists $x\in X$ such that $x\in Gx$. Also, $|G( x)|=\sup\{ | y| ;\,y\in G(x)\}$. \paragraph{Definition} A multivalued map $F:J\times \mathbb{R}\rightarrow {}2^{\mathbb{R}}$ is said to be $L^{1}$-Carath\'{e}odory if \begin{itemize} \item[(i)] $t\mapsto F(t,y)$ is measurable for each $y\in \mathbb{R}{}$; \item[(ii)] $y\mapsto F(t,y)$ is upper semicontinuous for almost all $t\in J $; \item[(iii)] For each $\sigma >0$, there exists $h_{\sigma }\in L^{1}(J,% \mathbb{R}{}_{+})$ such that \begin{equation*} \Vert F(t,y)\Vert =\sup \{|v|:v\in F(t,y)\}\leq h_{\sigma }(t) \end{equation*} for all $|y|\leq \sigma $ and for almost all $t\in J$. \end{itemize} The set of selectors of $F$ that belong to $L^{1}$ is denoted by \begin{equation*} S_{F( .,y( .) ) }^{1}=\{v\in L^{1}(J,{}):v(t)\in F(t,y(t))\ \hbox{for a.e.}\ t\in J\} \end{equation*} By a solution of (\ref{e1}) we mean an absolutely continuous function $x$ on $J$, such that $x^{\prime}\in L^{1}$ and \begin{equation} \begin{gathered} x'(t)=f(t)\quad \text{a.e. }t\in (0,T] \\ x(0)+\sum_{k=1}^{m}a_{k}x( t_{k}) =0 \end{gathered} \label{e2} \end{equation} where $f\in S_{F( .,x( .) ) }^{1}$. Note that for an $L^{1}$-Carath\'{e}odory multifunction $F:J\times \mathbb{R}% \rightarrow 2^{\mathbb{R}}$ the set $S_{F(.,x(.))}^{1}$ is not empty (see \cite{LaOp}). For more details on set-valued maps we refer to \cite{Deimling}. \subsection*{Topological Transversality Theory for Set-valued Maps} Let $X$ be a Banach space, $C$ a convex subset of $X$ and $U$ an open subset of $C$. $K_{\partial U}(\overline{U},2^{C})$ shall denote the set of all set-valued maps $G:\overline{U}\to 2^{C}$ which are compact, upper semicontinuous with closed convex values and have no fixed points on $\partial U$ (i.e., $u\notin Gu$ for all $u\in \partial U)$. A compact homotopy is a set-valued map $H:[0,1]\times \overline{U}\to 2^{C}$ which is compact, upper semicontinuous with closed convex values. If $u\notin H(\lambda ,u)$ for every $\lambda \in [0,1],u\in \partial U$, $H$ is said to be fixed point free on $\partial U$. Two set-valued maps $F,G\in K_{\partial U}(\overline{U},2^{C})$ are called homotopic in $K_{\partial U}(\overline{U}% ,2^{C})$ if there exists a compact homotopy $H:[0,1]\times \overline{U}\to 2^{C}$ which is fixed point free on $\partial U$ and such that $H(0,\cdot )=F $ and $H(1,\cdot )=G$. $G\in K_{\partial U}(\overline{U},2^{C})$ is called essential if every $F\in K_{\partial U}(\overline{U},2^{C})$ such that $G|_{\partial U}=F|_{\partial U}$, has a fixed point. Otherwise $G$ is called inessential. For more details we refer the reader to \cite{Frigon}. \begin{theorem}[Topological transversality theorem] \label{thm1} Let $F, G$ be two homotopic set-valued maps in $K_{\partial U}( \overline{U}, 2^C)$. Then $F$ is essential if and only if $G$ is essential. \end{theorem} \begin{theorem} \label{thm2} Let $G: \overline{U} \to 2^C$ be the constant set-valued map $G(u) \equiv u_0$. Then, if $u_0 \in U$, $G$ is essential \end{theorem} \begin{theorem}[Nonlinear Alternative] \label{thm3} Let $U$ be an open subset of a convex set $C$, with $0 \in U$. Let $H: [0,1] \times \overline{U} \to 2^C$ be a compact homotopy such that $H_0 \equiv 0$. Then, either \begin{enumerate} \item[(i)] $H(1,\cdot )$ has a fixed point in $\overline{U}$, or \item[(ii)] there exists $u\in \partial U$ and $\lambda \in (0,1)$ such that $u\in H(\lambda ,u)$. \end{enumerate} \end{theorem} \section{Main results} To prove our main results, we assume the following: \begin{enumerate} \item[(H0)] $a_{k}\neq 0$ for each $k=1,2,\dots ,m$ and $\sum_{k=1}^{m}a_{k}+1\neq 0$. \item[(H1)] $F:J\times \mathbb{R}{}\rightarrow bcc(\mathbb{R}),\;(t,x)\mapsto F(t,x)$ is \begin{enumerate} \item[(i)] measurable in $t$, for each $x\in \mathbb{R}{}$ \item[(ii)] upper semicontinuous with respect to $x\in {}$ for a.e. $t\in J$ \end{enumerate} \item[(H2)] $|F(t,x)|\leq \psi (\left| x\right| )$ for a.e. $t\in J$, all $x\in \mathbb{R}{,}$ where $\psi :[0,+\infty )\rightarrow (0,+\infty )$ is continuous nondecreasing and such that $\limsup_{\rho \rightarrow \infty }% \frac{\psi (\rho )}{\rho }=0$. \end{enumerate} Our first result reads as follows. \begin{theorem} \label{thm4} If the assumptions (H0), (H1), and (H2) are satisfied, then the initial-value problem (\ref{e1}) has at least one solution. \end{theorem} \paragraph{Proof} This proof will be given in several steps, and uses some ideas from [6]. \noindent \textbf{Step 1.} Consider the set-valued operator $\Phi :C(J)\rightarrow L^{2}(J)$ defined as \begin{equation*} (\Phi x)(t)=F(t,x(t)). \end{equation*} Note that $\Phi $ is well defined, upper semicontinuous, with convex values and sends bounded subsets of $C(J)$ into bounded subsets of $L^{2}(J)$. In fact, we have \begin{equation*} \Phi x:=\left\{ u:J\rightarrow \mathbb{R\quad }{}\mbox{ measurable;}% \;u(t)\in F(t,x(t))\mbox{ a.e. }t\in J\right\} . \end{equation*} Let $z\in C(J)$. If $u\in \Phi z$ then \begin{equation*} |u(t)|\leq \psi (\left| z(t)\right| )\leq \psi (\Vert z\Vert _{0}). \end{equation*} Hence $\Vert u\Vert _{L^{2}}\leq C_{0}:=\psi (\Vert z\Vert _{0})$. This shows that $\Phi $ is well defined. It is clear that $\Phi $ is convex valued. Now, let $B$ be a bounded subset of $C(J)$. Then, there exists $K>0$ such that $\Vert u\Vert _{0}\leq K$ for $u\in B$. So, for $w\in \Phi u$ we have $\Vert w\Vert _{L^{2}}\leq C_{1}$, where $C_{1}=\psi (K)$. Also, we can argue as in [5, p. 16] to show that $\Phi $ is upper semicontinuous. \noindent \textbf{Step 2}. Let $x$ be a possible solution of (\ref{e1}). Then there exists a positive constant $R^{\ast }$, not depending on $x$, such that \begin{equation*} |x(t)|\leq R^{\ast }\quad \mbox{ for all $t$ in $J$}. \end{equation*} It follows from the definition of solutions of (\ref{e1}) that \begin{equation} \begin{gathered} x'(t)=f(t)\quad \text{a.e. }t\in (0,T] \\ x(0)+\sum_{k=1}^{m}a_{k}x( t_{k}) =0 \end{gathered} \label{e3} \end{equation} where $f\in S_{F(.,x(.))}^{1}$. Simple computations give \begin{equation} x(t)=\big(1+\sum_{k=1}^{m}a_{k}\big)^{-1}\big(-\sum_{k=1}^{m}a_{k}% \int_{0}^{t_{k}}f(s)ds\big)+\int_{0}^{t}f(s)ds \label{e4} \end{equation} Hence \begin{equation*} |x(t)|\leq \big|\big(1+\sum_{k=1}^{m}a_{k}\big)^{-1}\big|\big(% \sum_{k=1}^{m}|a_{k}|\int_{0}^{t_{k}}|f(s)|ds\big)+\int_{0}^{t}|f(s)|ds \end{equation*} Assumption (H2) yields \begin{equation*} |x(t)|\leq \big|\big(1+\sum_{k=1}^{m}a_{k}\big)^{-1}\big|\big(% \sum_{k=1}^{m}|a_{k}|\int_{0}^{t_{k}}\psi (|x(s)|)ds\big)+\int_{0}^{t}\psi (|x(s)|)ds \end{equation*} Let \begin{equation*} R_{0}=\max \left\{ |x(t)|;t\in J\right\} . \end{equation*} Then \begin{equation*} R_{0}\leq \big|\big(1+\sum_{k=1}^{m}a_{k}\big)^{-1}\big|\big(% \sum_{k=1}^{m}|a_{k}|\,t_{k}\psi (R_{0})\big)+T\,\psi (R_{0}) \end{equation*} or \begin{equation*} R_{0}\leq \Big[\big\{\big|\big(1+\sum_{k=1}^{m}a_{k}\big)^{-1}\big|% \sum_{k=1}^{m}|a_{k}|\,t_{k}\big\}+T\,\Big]\psi (R_{0}) \end{equation*} The above inequality implies \begin{equation*} 1\leq \Big(T+\big|(1+\sum_{k=1}^{m}a_{k})^{-1}\big|\sum_{k=1}^{m}|a_{k}|% \,t_{k}\Big)\frac{\psi (R_{0})}{R_{0}} \end{equation*} Now, the condition on $\psi $ in (H2) shows that there exists $R^{\ast }>0$ such that for all $R>R^{\ast }$, \begin{equation*} \Big(T+\big|(1+\sum_{k=1}^{m}a_{k})^{-1}\big|\sum_{k=1}^{m}|a_{k}|\,t_{k}% \Big)\frac{\psi (R)}{R}<1. \end{equation*} Comparing these last two inequalities, we see that $R_{0}\,\leq R^{\ast }$. Consequently, we obtain $\left| x(t)\right| \leq R^{\ast }$ for all $t\in J$. \noindent \textbf{Step 3}. For $0\leq \lambda \leq 1$ consider the one-parameter family of problems \begin{equation} \begin{gathered} x'(t)\in \lambda F( t,x(t)) \quad t\in J, \\ x(0) +\sum_{k=1}^{m}a_{k}x( t_{k}) =0. \end{gathered} \label{e1l} \end{equation} It follows from Step 2 that if $x$ is a solution of (\ref{e1l}) for some $% \lambda \in \lbrack 0,1]$, then \begin{equation*} |x(t)|\leq R^{\ast }\quad \mbox{ for all }t\in J \end{equation*} and $R^{\ast }$ does not depend on $\lambda $. Define $\Phi _{\lambda }:C(J)\rightarrow L^{2}(J)$ as \begin{equation*} (\Phi _{\lambda }x)(t)=\lambda F(t,x(t)). \end{equation*} Step 1 shows that $\Phi _{\lambda }$ is upper semicontinuous, has convex values and sends bounded subsets of $C(J)$ into bounded subsets of $L^{2}(J)$. Let $j:H_{b}^{1}(J)\rightarrow C(J)$ be the completely continuous embedding. The operator $L:H_{b}^{1}(J)\rightarrow L^{2}(J)$, defined by $(Lx)(t)=x^{\prime }(t)$ has a bounded inverse (in fact this follows from the solution of (\ref{e3}) which is given by (\ref{e4})), which we denote by $L^{-1}$. Let $B_{R^{\ast }+1}:=\left\{ x\in C(J);\Vert x\Vert _{0}1 \end{equation*} \end{description} Now, we state our second result. \begin{theorem} \label{thm5} If assumptions (H0), (H1), and (H2') are satisfied, then the initial value problem (\ref{e1}) has at least one solution. \end{theorem} \paragraph{Proof} This proof is similar to the proof of Theorem \ref{thm4}. Let $M_{0}>0 $ be defined by \begin{equation*} \frac{M_{0}}{[\{\left| ( 1+\sum_{k=1}^{m}a_{k}) ^{-1}\right| \sum_{k=1}^{m}|a_k| \,\int_{0}^{t_{k}}p(s)ds\}+\left\| p\right\| _{L^{1}}\,]\psi ( M_{0}) }>1. \end{equation*} Let $U:=\left\{ x\in C(J);\Vert x\Vert _{0}