\documentclass[reqno]{amsart} \usepackage{hyperref} \AtBeginDocument{{\noindent\small \emph{Electronic Journal of Differential Equations}, Vol. 2010(2010), No. 79, pp. 1--12.\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-2010/79\hfil Urysohn integral inclusion] {Controllability of Urysohn integral inclusions of Volterra type} \author[T. S. Angell, R. K. George, J. P. Sharma \hfil EJDE-2010/79\hfilneg] {Thomas S. Angell, Raju K. George, Jaita Pankaj Sharma} % in alphabetical order \address{Thomas S. Angell \newline Department of Mathematics, University of Delaware, Newark, DE 19716, USA} \email{angell@math.udel.edu} \address{Raju K. George \newline Department of Applied Mathematics, Faculty of Tech. \& Engg., M.S. University of Baroda, Vadodara 390001, India} \email{raju\_k\_george@yahoo.com} \address{Jaita Pankaj Sharma \newline Department of Applied Mathematics, Faculty of Tech. \& Engg., M.S. University of Baroda, Vadodara 390001, India} \email{jaita\_sharma@yahoo.co.uk} \thanks{Submitted March 5, 2010. Published June 15, 2010.} \subjclass[2000]{93B05, 93C10} \keywords{Controllability; Urysohn operator; delay systems; \hfill\break\indent set-valued mappings; Kakutani's Fixed-point theorem} \begin{abstract} The aim of this paper is to study the controllability of a system described by an integral inclusion of Urysohn type with delay. In our approach we reduce the controllability problem of the nonlinear system into solvability problem of another integral inclusion. The solvability of this integral inclusion is subsequently established by imposing suitable standard boundedness, convexity and semicontinuity conditions on the set-valued mapping defining the integral inclusion, and by employing Bohnenblust-Karlin extension of Kakutani's fixed point theorem for set-valued mappings. \end{abstract} \maketitle \numberwithin{equation}{section} \newtheorem{theorem}{Theorem}[section] \newtheorem{example}[theorem]{Example} \newtheorem{definition}[theorem]{Definition} \allowdisplaybreaks \section{Introduction} In recent years a number of papers appeared in the literature concerning integral inclusions, in particular inclusions of Hammerstein type and Urysohn type; see Rangimchannov \cite{RAN}, Gaidarov \cite{GAI}, Angel \cite{AN1}. This type of inclusions have been used to model many thermostatic devices; see Glashoff and Sperckels \cite{GLAS1}, \cite{GLAS2}. Here we consider the nonlinear control system described by the following Urysohn integral inclusion on the time interval $[0,T]$, $T>0$ \begin{equation}\label{e1} x(t)\in (Hx)(t)+\int_0^t g(t,s,x_s)F(s,x_s)ds + \int_0^t K(t,s)u(s)ds. \end{equation} where, for each $t \in [0,T]$ the state x(t) is in $\mathbb{R}^n$ and the control $u(t)\in \mathbb{R}^m$. For any given real number $00}cl\cup\{F(\overline{t},\phi), ||\phi -\overline{\phi}||\leq\delta\} \] where $\phi \in L^{\infty}([-r,0];\mathbb{R}^n)$. Since the intersection of closed set is closed, so each of $F(\overline{t},\overline{\phi})$ is closed; \item[(c)] there exist a measurable set-valued function $P:[0,T]\to E^{1}$, a constant $M > 0$, and for each $\epsilon > 0$, a function $\psi_\epsilon \in {L}^{1}([0,T];{\mathbb{R}^{n}})$, with $\psi_{\epsilon}(t)>0$, such that, for given $x \in {L}^\infty([-r,T];{\mathbb{R}^n})$ and selection $\xi(t) \in F(t,x_t)$, there exists a selection $\eta(t) \in P(t)$, with \begin{enumerate} \item $\int_0^T \eta(t)dt \leq M$ \item $|\xi(t)| \leq \psi_\epsilon(t)+\epsilon \eta(t)$ \end{enumerate} \end{itemize} \item[(K)] for each $(t,s) \in [0,T]\times [0,T]$, $(t,s)\to K(t,s)$ is continuous with $\|K(t,s)\| \leq k(t,s)$ for $k(t,s)\in L^{2}([0,T]\times [0,T])$ \end{itemize} Here the conditions (H)(a) and (H)(b) are used to establish the complete continuity of the Urysohn operator (see Krasnoselskii, \cite{KRA}), and the condition (F)(c) is used for proving equi - absolute integrability (see Ioffe [6]) condition of the set of selections. Here we will use operator theory in the analysis of controllability (Joshi and George \cite{JG}). So some basic definition regarding control operator are as follows: \begin{definition} \rm The control operator $C: L^2([0,t];\mathbb{R}^m) \to \mathbb{R}^n$ of \eqref{e1}-\eqref{e2} be defined by \begin{equation} Cu=\int_0^T K(T, \tau)u(\tau) d\tau\,. \end{equation} \end{definition} \begin{definition} \rm A bounded linear operator $S:{\mathbb{R}^n} \to L^2([0,t];{\mathbb{R}^m})$ is said to be steering operator for the associated linear system \begin{equation} \label{e3} \begin{gathered} x(t)= \int_0^t K(t,s)u(s) ds\\ x(0) = 0 \end{gathered} \end{equation} if $CS=I$, where $I$ is the identity operator on ${\mathbb{R}^n}$ \end{definition} \begin{definition} \rm An $m\times n$ matrix function P(t) with entries in $L^2([0,T];{\mathbb{R}^m})$ is said to be a steering function for \eqref{e3}on [0,T] if \[ \int_0^T K(T,s)P(s) ds =I \] \end{definition} We note that if the linear system \eqref{e3} is controllable then there exists a steering function P(t), Russel \cite{RU}. \section{Controllability and feed-back formulation} For studying the controllability of \eqref{e1}-\eqref{e2}, we assume that the corresponding linear system \eqref{e3} is controllable and let $P(t)$ be a steering function for it. Now the nonlinear system \eqref{e1}-\eqref{e2} is controllable on $[0,T]$ if and only if there exists a control u which steers a given initial state $\phi(0)$ of the system to a desired final state $x_1$. That is, there exists a control function u such that \[ x_1=x(T)=\phi(0)+\int_0^T h(T,s,x_s) ds +\int_0^T g(T,s,x_s) v(s) ds \] \begin{equation} \label{e4}+ \int_0^T K(T,s)u(s) ds, \end{equation} for any selection $v \in L^{1}([0,T];\mathbb{R}^{n})$ satisfying the inclusion $v(t)\in F(t,x_t)$ almost everywhere on $[0,T]$. Let us define a control $u(t)$ by \begin{equation} \label{e5} u(t)=P(t)[x_1-\phi(0)-\int_0^T h(T,s,x_s) ds -\int_0^T g(T,s,x_s) v(s) ds ], \end{equation} where $x(.)$ satisfies the nonlinear system \eqref{e1}-\eqref{e2}. Now substituting this control u(t) into the nonlinear integral equation \eqref{e1}-\eqref{e2}, we get \begin{equation} \label{e6} \begin{aligned} x(t)&=\phi(0)+\int_0^T h(t,s,x_s) ds+\int_0^t g(t,s,x_s) v(s) ds +\int_0^t K(t,s)P(s) \\ &\quad \times \Big[x_1 -\phi(0) -\int_0^T h(T,\tau,x_\tau)d\tau -\int_0^T g(T,\tau,x_\tau) v(\tau) d\tau \Big]ds. \end{aligned} \end{equation} If this equation is solvable then $x(t)$ satisfies $x(0)=\phi(0)$ and $x(T)=x_1$. This implies that \eqref{e1}-\eqref{e2} is controllable with a control $u$ given by \eqref{e5}. Hence the controllability of the nonlinear integral inclusion system \eqref{e1}-\eqref{e2} is equivalent to the solvability of the integral equation \eqref{e6} with suitable selection $v(t) \in F(t,x_t)$. \section{Solvability of nonlinear integral equation} We apply fixed point theorem for establishing solvability of the nonlinear integral equation \eqref{e6}. We now recast the integral equation \eqref{e6} with a selection $v$ as a set-valued mapping and apply fixed point theorem for a set-valued mapping. We introduce two set-valued mappings $\Phi$ and $\Psi$ whose domain $S$ is defined by \begin{equation} \label{e7} S =\{ x \in {L}^\infty([-r,T];{\mathbb{R}^n}): x|_{[-r,0]}=\phi,\; x|_{[0,T]} \in C([0,T];{\mathbb{R}^n})\} \end{equation} The maps $\Phi : S \to L^{1}([0,T];\mathbb{R}^n)$ and $\Psi : S \to S$, are defined by \begin{gather} \label{e8} \Phi(x)=\{v \in {L^{1}}([0,T];{\mathbb{R}^n}) | v(t) \in F(t,x_t), a.e., on [0,T]\}, \\ \label{e9} \begin{aligned} \Psi(x)=\Big\{& z \in S : z(t)=(Hx)(t) +\int_0^t g(t,s,x_s)v(s)ds\\ &+\int_0^t K(t,s)P(s)\Big[x_1-\phi(0) -\int_0^T h(T,\tau,x_\tau) d\tau \\ & -\int_0^T g(T,\tau,x_\tau)v(\tau)d\tau\Big] ds,\; z|_{[-r,0]}=\phi,\; v \in \Phi(x) \Big\} \end{aligned} \end{gather} We will use the following Bohnenblust-Karlin extension of KaKutani's fixed point theorem for set-valued mappings. \begin{theorem}[Bohnenblust-Karlin \cite{BOH}] \label{tBOH} Let $\Sigma$ be a non-empty, closed convex subsets of a Banach space ${\mathcal{B}}$. If $\Gamma: \Sigma \to 2^{\Sigma}$ is such that \begin{itemize} \item[(a)] $\Gamma(a)$ is non-empty and convex for each $a \in \Sigma$, \item[(b)] the graph of $\Gamma$, $\mathcal{G}(\Gamma) \subset \Sigma \times \Sigma $, is closed, \item[(c)] $\cup$ $\{\Gamma(a): a \in \Sigma \}$ is contained in a sequentially compact set $\mathcal{F} \in {B},$\\then the map $\Gamma$ has a fixed point, that is, there exists a $\sigma_0 \in \Sigma$ such that $\sigma_0 \in \Gamma(\sigma_0)$. \end{itemize} \end{theorem} We will apply this theorem to the map $\Psi$ defined on the closed convex set $S \subset L^{\infty} ([-r,T];\mathbb{R}^n)$. In order to apply Theorem \ref{tBOH}, we need to prove that the set $\Psi(S)$ is relatively sequentially compact. This property in turn, depends on the weak relative compactness of $\Phi(S)$ in ${L}^1 ([0,T];\mathbb{R}^n)$. \begin{theorem}[Angel \cite{AN1}] \label{tRCPhi} The set $\Phi(S)$ defined by the relation \eqref{e8} is an equi-absolutely integrable set and is weakly compact in $L^{1}([0,T] ;\mathbb{R}^n)$. \end{theorem} We have the following theorem on the relative compactness of the set $\Psi(s)$. \begin{theorem}\label{tRCPsi} Under the hypotheses {\rm (H), (G), (F), (K)}, for each $x \in S$, $\Psi(x)$ is a non-empty and the set $\Psi(S)$ defined by the relation \eqref{e9} is a relatively sequentially compact subset of $L^\infty ([0,T]; \mathbb{R}^n)$ \end{theorem} \begin{proof} First we shall show that $\psi(S)\neq \emptyset$ for all $x\in S$. For a given $x \in S $ we have $ \phi(x) \neq \emptyset $ (Angel \cite{AN1}). Hence choosing $v \in \phi(x)$ we define \begin{align*} y(t)& = \phi(0)+\int_0^T h(t,s,x_s) ds +\int_0^t g(t,s,x_s) v(s) ds \\ &\quad + \int_0^t K(t,s)P(s) \Big\{x_1-\phi(0) -\int_0^T h(T,\tau,x_\tau) d\tau -\int_0^T g(T,\tau,x_\tau)v(\tau)d\tau\Big\}ds \end{align*} For any $t',t''\in[0,T]$, we have \begin{align*} &| y(t')-y(t'')|\\ & = \Big|\int_0^T (h(t',s,x_s)-h(t'',s,x_s))ds +\Big(\int_0^{t'} g(t',s,x_s) v(s) ds\\ &\quad - \int_0^{t''}g(t'',s,x_s) v(s) ds \Big) + \Big\{\int_0^{t'} K(t',s)P(s)ds - \int_0^{t''} K(t'',s)P(s)ds\Big\}\\ &\quad\times\Big[x_1-\phi(0)-\int_0^T h(T,\tau,x_\tau) d\tau - \int_0^T g(T,\tau,x_\tau)v(\tau)d\tau\Big]\Big| \\ &\leq \int_0^T{\sup_{\phi \in L^{\infty}}} |h(t',s,\phi) -h(t'',s,\phi)|ds + \int_0^{t'} |g(t',s,x_s) - g(t'',s,x_s)||v(s)| ds\\ &\quad + \int_{t'}^{t''}|g(t'',s,x_s)|| v(s)| ds\\ &\quad + \Big\{\int_0^{t'}|K(t',s)-K(t'',s)||P(s)|ds +\int_{t'}^{t''} |K(t'',s)||P(s)|ds\Big\}\\ &\quad\times\Big\{|x_1|+|\phi(0)| + \int_0^T |h(T,\tau,x_\tau)|d\tau + \int_0^T|g(T,\tau,x_\tau)| |v(\tau)|d\tau\Big\}ds \\ & = I_1 + I_2 + I_3 + I_4. \end{align*} By (H)(c), there exists $\delta_1 > 0$ such that \[ I_1 = \int_0^T \sup_{\phi \in L^{\infty}} |h(t',s,\phi) -h(t'',s,\phi)|ds < \frac{\epsilon}{5}\qquad if\;\; |t'-t''| < \delta_{1} \] Using the condition (F)(c) for $\delta_{2}> 0$ \begin{align*} I_2 & = \int_0^{t'}|g(t',s,x_s)-g(t'',s,x_s)| |v(s)|ds\\ & \leq \int_0^{t'} \sup_{\phi \in L^{\infty}} |g(t',s,\phi)-g(t'',s,\phi) |\psi_{\delta_2}(s)+\delta_2 \eta(s)|ds\\ & = \int_{0}^{t'} \sup_{\phi \in L^{\infty}}|g(t',s,\phi) -g(t'',s,\phi)| \psi_{\delta_1}(s)ds \\ &\quad +\delta_2 \int_{0}^{t'} \sup_{\phi \in L^{\infty}}|g(t',s,\phi)-g(t'',s,\phi) | \eta(s) ds \\ & = I_{21} + I_{22} \end{align*} Since $g$ is a bounded function, taking its bound as $M_{g}$ and $\psi_{\delta_2} \in L^1([0,T];\mathbb{R}^n)$. We apply the Lesbesgue dominated convergence theorem. Therefore for a small $\delta_3 >0$, \begin{align*} &\lim_{t'\to t''}\int_{0}^{t'} \sup_{\phi \in L^{\infty}} |g(t',s,\phi) -g(t'',s,\phi)|\psi_{\delta_2}(s)ds\\ &= \int_{0}^{t'} \lim_{t'\to t''} \sup_{\phi \in L^{\infty}}|g(t',s,\phi)-g(t'',s,\phi)|\psi_{\delta_2}(s)ds\,. \end{align*} Using (G)(c), \begin{align*} I_{21} & = \int_{0}^{t'} \sup_{\phi \in L^{\infty}}|g(t',s,\phi)-g(t'',s,\phi) |\psi_{\delta_2}(s)ds\\ & \leq \frac{\epsilon}{5k_1}\int_0^{t'}\psi_{\delta_2}(s)ds\quad\text{for } \int_{0}^{t'}\psi_{\delta_2}(s)ds = k_1 \leq \infty\\ & \leq \frac{\epsilon}{5k_1}k_1\\ & = \frac{\epsilon}{5}\quad \text{if } |t'-t''| \leq \delta_3\,. \end{align*} \begin{align*} I_{22} & = \delta_2 \int_{0}^{t'} \sup_{\phi \in L^{\infty}} |g(t',s,\phi)-g(t'',s,\phi)|\eta(s)ds\\ & \leq \delta_2 2 M_g \int_0^{t'}\eta(s)ds\\ & \leq \delta_1 2 M_g M \\ & \leq \frac{\epsilon}{10 M_g M} 2 M_g M\quad \text{(taking $\delta_1 \leq \frac{\epsilon}{10 M_g M}$)} \\ & = \frac{\epsilon}{5}\,. \end{align*} \begin{align*} I_{3} & = \int_{t'}^{t''}|g(t'',s,x_s)|| v(s)| ds\\ & \leq (M_g\int_{t'}^{t''}| v(s)| ds)\\ & \leq M_g \Big(\int_{t'}^{t''}|v(s)|ds\Big)^{1/2} \Big(\int_{t'}^{t''}ds\Big)^{1/2}\\ & \leq M_g (k_1+M)^{1/2} (t''-t')^{1/2} \quad \text{for } t'' -t' \leq \delta_4\\ & \leq M_g (k_1+M)^{1/2} \delta_4 \\ & \leq \frac{\epsilon}{5}\quad \text{taking }\delta_4 \leq \frac{\epsilon}{5M_g(M+K)^{1/2}} \end{align*} The function $h$ satisfies the condition (H), so for any given $t\in[0,T]$, there exists a finite $b=b(\hat{t})$ such that \[ \int_0^T h(\hat{t},s,x_s) ds \leq \int_0^T {\sup_{\phi \in L^{\infty}}}|h(\hat{t},s,\phi)ds \leq b(\hat{t})\,. \] \begin{align*} I_{4} & = \Big\{|x_1|+|\phi(0)|+\int_0^T |h(T,\tau,x_\tau)|d\tau +\int_0^T |g(T,\tau,x_\tau)||v(\tau)|d\tau\Big\}\\ &\quad\times \Big(\int_{0}^{t'}|K(t',s)-K(t'',s)||P(s)|ds + \int_{t'}^{t''} |K(t'',s)||P(s)|ds\Big) \\ & \leq \big\{|x_1|+|\phi(0)| + b(T)+M_g(M+k_1)\big\} \\ & \quad\times\Big(P \int_0^{t'}|K(t',s)-K(t'',s)|ds + KP \int_{t'}^{t''} ds\Big)\\ & \quad\text{where $K$ and $P$ are bounds of $K(t,s)$ and $P(s)$}\\ & \leq R(P t'\frac{\epsilon}{10 P R t'}+ KP (t'-t''))\\ & \leq R(\frac{\epsilon}{10R}+ KP \delta_5) \quad \text{(if $|t'-t''|\leq \delta_5$ and taking $\delta_5 \leq \frac{\epsilon}{10KPR}$)}\\ & \leq R(\frac{\epsilon}{10R} + KP\frac{\epsilon}{10RKP})\\ & = \frac{\epsilon}{5} \end{align*} Thus continuity of $y$, follows by choosing $\delta \leq \min({\delta_1},\delta_2,\delta_3,\delta_4,\delta_5)$ and so the piecewise continuous function $z$ is defined by \[ z(t)=\begin{cases}We prefer JPG f \phi(t) &\text{if } -r\leq t\leq 0 \\ y(t) &\text{if } 0 \leq t \leq T \end{cases} \] lies in $S$. Hence $\Psi(x)\neq \emptyset $. Here the elements of $\Psi(S)$ in the interval $[0,T]$ form an equicontinuos family. Hence Relative sequential compactness will now follow from the equiboundedness of $\Psi(S)$, since then any sequence in $\Psi(S)$ say, $\{z_k \}$, restricted to $[0,T]$, will have a uniformly convergent subsequence by the Arzela - Ascoli theorem. Now to show that $\Psi(S)$ is equibounded, let us consider $y\in\Psi(S)$ on $[0,T]$, for a given $t_{0}\in [0,T]$, \begin{align*} y(t_0) & = \phi(0)+\int_0^T h(t_0,s,x_s) ds+\int_0^{t_0}g(t_0,s,x_s) v(s) ds +\int_0^{t_0}K(t_{0},s)P(s)\\ & \quad\times\Big\{x_1-\phi(0)-\int_0^T h(T,\tau,x_\tau) d\tau -\int_0^T g(T,\tau,x_\tau)v(\tau) d\tau\Big\}ds \end{align*} and \begin{align*} &|y(t_0)|\\ & \leq |\phi(0)|+\int_0^T |h(t_0,s,x_s)|ds +\int_0^{t_0}|g(t_0,s,x_s)|| v(s)| ds \\ &\quad +\int_0^{t_0}|K(t_{0},s)||P(s)| \Big|x_1-\phi(0)-\int_0^T h(T,\tau,x_\tau)d\tau -\int_0^T g(T,\tau,x_\tau)v(\tau)d\tau\Big|ds\\ & \leq |\phi(0)|+\int_0^T |h(t_0,s,x_s)| ds + M_g(M+k_1)\\ &\quad + (KP t_0)\Big\{|x_1|+|\phi(0)| +\int_0^T |h(T,\tau,x_\tau)| d\tau +M_g(M+k_1)\Big\}\\ & \leq |\phi(0)|+b(t_0) + M_g(M+k_1)+(KP t_0) \big\{|x_1|+|\phi(0)|+h(T)+M_g(M+k_1)\big\}\\ & < \infty \end{align*} Hence $y$ is bounded uniformly on $[0,T]$. It follows that the set $\Psi(S)$ is relatively sequentially compact, since the initial function $\phi$ is fixed and the restrictions of elements of $S$ to $[0,T]$ are continuous. \end{proof} \begin{theorem}\label{tCONV} The set $\Psi(x)$ is convex for each $x \in S$. \end{theorem} \begin{proof} Let $y^{(1)},y^{(2)}\in \Psi(x)$. Then there exists $v^{(i)}(t)\in F(t,x_t)$, $i=1,2$, such that \begin{align*} &y^{(i)}(t)\\ &=\phi(0)+\int_0^T h(t,s,x_s) ds+\int_0^t g(t,s,x_s) v^{(i)}(s) ds\\ &\quad +\int_0^t K(s,t)P(s) \Big[x_1-\phi(0)-\int_0^T h(T,\tau,x_\tau) d\tau -\int_0^T g(T,\tau,x_\tau) v^{(i)}(\tau)d\tau\Big] ds. \end{align*} Thus, for $0<\lambda<1$, \begin{align*} &\lambda y^{(1)}(t)+(1-\lambda)y^{(2)}(t)\\ &=\phi(0)+\int_0^T h(t,s,x_s) ds +\int_0^t g(t,s,x_s) (\lambda v^{(1)}(s)+(1-\lambda)v^{(2)}(s)) ds\\ &\quad + \int_0^t K(s,t)P(s) \Big\{x_1-\phi(0)-\int_0^T h(T,\tau,x_\tau) d\tau\\ &\quad -\int_0^T g(T,\tau,x_\tau) (\lambda v^{(1)}(\tau) +(1-\lambda)v^{(2)}(\tau))d\tau\Big\}ds. \end{align*} By the convexity of $F(t,x_t)$ we have $(\lambda v^{(1)}(t)+(1-\lambda)v^{(2)}(t))\in F(t,x_t)$. And hence $\Psi(x)$ is convex. \end{proof} Now we prove that $\mathcal{G}$$(\Psi)$ is closed. For proving this, we use the following theorems, which were used in \cite{AN2} and modified by Cesari \cite{CE}. \begin{theorem}\label{tclos} Let $I=[0,T]$, consider the set-valued mapping $F:I\times{L}^{\infty}\to 2^{E^{n}}$, and assume that $F$ satisfies the conditions (F)(a) and (F)(b) with respect to $\phi$. Let $\xi,\xi_{k},x,x_{k}$ be functions measurable on I, $x,x_k$ bounded, and let $\xi,\xi_{k}\in {L}^{1}(I;\mathbb{R}^{n})$. Then if $\xi_{k}(t)\in F(t,x_t)$ a.e. in $I$ and $\xi_{k}\to\xi$ weakly in ${L}^{1}(I;\mathbb{R}^n)$, while $x \to x_{k}$ uniformly on $I$, then $\xi(t)\in F(t,x_t)$ in $I$. \end{theorem} We now use Theorem \ref{tclos} to show that the graph of the map $\Psi$, defined by the relation \eqref{e9}, has a closed graph. \begin{theorem}\label{tPsiclos} Under the assumption {\rm (H), (G), (F), (K)} the map $\Psi:S\to 2^{S}$ has a closed graph. That is, \{${(x,y) \in {S \times S} : y\in \Psi(x)}$\}is closed. \end{theorem} \begin{proof} Let $\{x_k , y_k\}$ be a sequence of functions such that $y_k \in \Psi(x_k)$ which converges to a limit point $(x,y)$ of $\mathcal{G}(\Psi)$. Thus, $x_k \to x$ and $y_k \to y$ uniformly on $[0,T]$. By definition of $\Psi$ there exists a sequence ${v_k}$, with $v_k \in \Phi(x_k)$, such that \begin{align*} y_k(t) & = \phi(0)+\int_0^T h(t,s,x_{k_s})ds +\int_0^t g(t,s,x_{k_s})v_k(s)ds\\ &\quad +\int_0^t K(t,s)P(s) \Big[x_1-\phi(0) -\int_0^T h(T,\tau,x_{k_\tau}) d\tau\\ &\quad -\int_0^T g(T,\tau,x_{k_\tau}) v_k(\tau)d\tau\Big]ds \end{align*} Without loss of generality we may assume that $v_k \to v$ weakly in $L^{1}([0,T];\mathbb{R}^n)$ and $v(s)\in F(s,x_s)$. We wish to show y satisfies the equation \begin{align*} y(t) &= \phi(0)+\int_0^T h(t,s,x_s)ds+\int_0^t g(t,s,x_s) v(s)ds\\ &\quad +\int_0^t K(t,s)P(s) \Big[x_1-\phi(0)-\int_0^T h(T,\tau,x_\tau) d\tau -\int_0^T g(T,\tau,x_\tau) v(\tau)d\tau\Big] ds \end{align*} Now considering that \begin{align*} &\Big|y(t)-\phi(0)-\int_0^T h(t,s,x_s)ds-\int_0^t g(t,s,x_s) v(s)ds\\ &-\int_0^t K(t,s)P(s) \Big[x_1-\phi(0)-\int_0^T h(T,\tau,x_\tau) d\tau -\int_0^T g(T,\tau,x_\tau) v(\tau)d\tau\Big]ds\Big|\\ &=\Big|y(t)-y_k(t)+y_k(t)-\phi(0)-\int_0^T h(t,s,x_s)ds -\int_0^t g(t,s,x_s) v(s)ds\\ &\quad -\int_0^t K(t,s)P(s) \Big[x_1-\phi(0)-\int_0^T h(T,\tau,x_\tau)d\tau -\int_0^T g(T,\tau,x_\tau)v(\tau)d\tau\Big]ds\Big|\\ &=\Big|y(t)-y_k(t)+\phi(0)+\int_0^T h(t,s,x_{k_s})ds +\int_0^t g(t,s,x_{k_s})v_k(s)ds\\ &\quad +\int_0^t K(t,s)P(s) \Big[x_1-\phi(0)-\int_0^T h(T,\tau,x_{k_\tau}) d\tau -\int_0^T g(T,\tau,x_{k_\tau}) v_k(\tau)d\tau\Big]ds\\ &\quad -\phi(0)-\int_0^T h(t,s,x_s)ds-\int_0^t g(t,s,x_s) v(s)ds\\ &\quad -\int_0^t K(t,s)P(s) \Big[x_1-\phi(0)-\int_0^T h(T,\tau,x_\tau)d\tau -\int_0^T g(T,\tau,x_\tau)v(\tau)d\tau\Big]ds\Big| \\ &\leq |y(t)-y_k(t)|+\int_0^T |h(t,s,x_{k_s})- h(t,s,x_s)|ds\\ &\quad +\int_0^t |g(t,s,x_{k_s}) v_k(s)-g(t,s,x_s) v(s)|ds + \int_0^t \Big|K(t,s)P(s)\\ &\quad\times \Big[\int_0^T (h(T,\tau,x_\tau)-h(T,\tau,x_{k_\tau}))d\tau\\ &\quad + \int_0^T g(T,\tau,x_\tau)v(\tau) -g(T,\tau,x_{k_\tau})v_k(\tau)d\tau\Big]ds\Big| \\ &\leq |y(t)-y_k(t)|+\int_0^T |h(t,s,x_{k_s})- h(t,s,x_s)|ds + \int_0^t |g(t,s,x_s)||v_k(s)-v(s)|ds\\ &\quad +\int_0^t|g(t,s,x_{k_s}) - g(t,s,x_s)||v_k(s)|ds + \int_0^t \big|K(t,s)P(s)|\\ &\quad\times \Big|\int_0^T (h(T,\tau,x_\tau)-h(T,\tau,x_{k_\tau}))d\tau\\ &\quad + \int_0^T g(T,\tau,x_\tau)v(\tau) -g(T,\tau,x_{k_\tau})v_k(\tau)d\tau\Big|ds \end{align*} Here we need to show that the relation holds pointwise so for a fixed $t_0$ we consider each term separately. \[ |y(t_0)-y_{k}(t_0)|\leq \frac{\epsilon}{5} \] since $y_k \to y$ uniformly. From (H) each element of the sequence of functions $s\to |h(t_{0},s,x_k)|$ k=1,2,\dots is bounded above by the integrable function $s\to \sup |h(t_0),s,\phi|$. Since $x_k \to x$ uniformly we have from (H) that $h(t_{0},s,x_{k_s})\to h(t_0,s,x_s)$ pointwise a.e. in $[0,T]$ and so \[ \lim_{k \to \infty} \int_0^T h(t_{0},s,x_{k_s})ds = \int_0^T h(t_0,s,x_s)ds\,. \] Also, \[ \int_0^{t_{0}} |g(t,s,x_s)|| v_{k}(s)- v(s)|ds \leq \frac{\epsilon}{5} \] Applying Egorov's theorem and condition (G), \[ \int_0^{t_{0}} |g(t_0,s,x_{k_s}) - g(t_0,s,x_s)||v_k(s)|ds \] can be made less than $\epsilon/5$ Using the continuity and boundedness of $K, P$ and the conditions (H),(G) and (F) for the following terms, we obtain \begin{align*} &\int_0^t |K(t,s)P(s)|\Big|\int_0^T (h(T,\tau,x_\tau) - h(T,\tau,x_{k_\tau}))d\tau \\ &+ \int_0^T g(T,\tau,x_\tau) v(\tau) - g(T,\tau,x_{k_\tau})v_k(\tau)d\tau\Big|ds \leq \frac{\epsilon}{5} \end{align*} Hence for a given $\epsilon>0$, \begin{align*} &|y(t) - \phi(0)-\int_0^T h(t,s,x_s)ds+\int_0^t g(t,s,x_s) v(s)ds\\ &- \int_0^t K(t,s)P(s) \Big[x_1-\phi(0)-\int_0^T h(T,\tau,x_\tau) d\tau -\int_0^T g(T,\tau,x_\tau) v(\tau)d\tau\Big] ds| \leq \epsilon \end{align*} Hence, $(x,y)\in \mathcal{G}$ and the graph of $\Psi$ is closed. \end{proof} With this theorem all of the hypothesis of the fixed point theorem are satisfied. And now we consider the main controllability theorem. \section{The Main Result} \begin{theorem}\label{tM} Under assumption {\rm (H)(b)}, the nonlinear system described by the integral inclusion \eqref{e4} is controllable. \end{theorem} \begin{proof} We have proved in Theorem \ref{tRCPhi}, Theorem \ref{tRCPsi}, Theorem \ref{tCONV} and Theorem \ref{tPsiclos} that under the assumptions (H)(b) the map $\psi:S\to 2^{S}$ satisfies all the hypotheses of the Bohnenblust-Karlin extension of KaKutani's fixed point theorem. Hence $\Psi$ has a fixed point in $S$. Let $x\in S$ be the fixed point of the mapping $\psi$ defined by the relation \eqref{e9} that is $x\in \psi(x)$. Therefore, for a selection $v\in\phi(x)$ such that $v(t)\in F(t,x_t)$ a.e, we have \begin{align*} &x(t)\\ &=\phi(0)+\int_0^T h(t,s,x_s) ds+\int_0^t g(t,s,x_s) v(s) ds\\ &\quad +\int_0^t K(s,t)B(s)P(s) \Big[x_1-\phi(0)-\int_0^T h(T,\tau,x_\tau) d\tau -\int_0^T g(T,\tau,x_\tau) v(\tau) d\tau\Big] ds. \end{align*} Obviously $x(0)= x_0$ and $x(T)=x_1$. Hence the system is controllable. \end{proof} We conclude this section with an example similar to one presented by Angel \cite{AN1}, which illustrate our result. \begin{example} \rm Let us consider the integral inclusion \[ x(t)\in \int_0^1[\frac{\sin(s^2)\sin(t^2)}{3+\arctan x(s)}+\frac{\cos(sx(s))}{3\sqrt{1+t}}F(s,x(s))]ds +\int_{0}^{t}e^{t-s}u(s)ds \] where $r=0$, $m=n=1$, and $F:[0,T]\times R\to 2^R$ is the set-valued map defined by \[ F(t,x) = \begin{cases} u \quad\text{with } |u|\leq t+|x| &\text{if $t\neq 0$ and }-1-\frac{1}{\sqrt{t}}\leq x \leq 1+\frac{1}{\sqrt{t}}, , \\ u \quad \text{with }|u|\leq |x|, &\text{if } t=0, \\ 0 &\text{otherwise}. \end{cases} \] \end{example} Here $F$ has a closed graph and convex values, also the growth condition (F)(c) is satisfied for the set-valued map $F$, since any selection $\xi(t)\in F(t,x(t))$ satisfies \[ |\xi(t)|\leq t+|x(t)|\leq t+1+\frac{1}{\sqrt(t)} \] it follows that $F(t,x(t)$ is integrably bounded. 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