\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 $$\label{cauchy} \begin{gathered} x^{(k)}(t) \in F(t,x(t)) \quad\text{a.e on } [0,T] \\ x(0)=x_0\in K, x^{(i)}(0)=y_0^{i}\in \Omega_i,\quad i=1,\dots,k-1, \\ x(t) \in K \quad\text{on } [0,T]. \end{gathered}$$ 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 $E$: $$\label{problem1} \begin{gathered} \ddot{x}(t) \in F(t,x(t),\dot{x}(t))\quad\text{a.e.;} \\ (x(0),\dot{x}(0))=(x_0,y_0); \\ (x(t),\dot{x}(t))\in K\times \Omega; \end{gathered}$$ where $K$ (resp. $\Omega$) is a closed subset (resp. an open subset) of $E$. The authors introduced the tangential condition: $$\liminf_{h \to 0^{+}}\frac{1}{h^2}d\Big(x+hy+\frac{h}{2}\int_{t}^{t+h}F(s,x,y)ds,K\Big)=0. \label{tangcon}$$ 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 $B(x,r):=\{y\in E; \| y-x \| 0$ 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 $$\| F(t,x) \|\leq g(t)\quad \forall (t,x)\in [0,1]\times (K\cap B(x_0,r)). \label{defdeg}$$ 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 $$\Big|\int_{t_{1}}^{t_2}g(\tau)d\tau\Big| < \varepsilon \; \quad\text{if } |t_1-t_2| < \eta(\varepsilon).\label{defdeita}$$ 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 $0p$, $$\|x_q-x_p\|\le \int_{t_p}^{t_q}g(t)dt+ (t_q-t_p) \Big((k-1)r+1+\sum_{i=1}^{k-1}\| y_0^{i}\| \Big)\to 0 \quad\text{as } q,p\to \infty$$ 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 $$\| \phi _q(t)-\phi (t)\| \leq d(\phi _q(t),F(t,\bar x))+\frac{\alpha}{2(k+5)}\leq m(t)\| x_q-\bar x\| +\frac{\alpha}{8(k+5)}.\label{2bis}$$ 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 $t_{q+1}\le \overline{t}t_{q+1}-t_q=h_q$. Thus, there exists $h > h_q$ (for all $q\ge q_0$) such that \$0