\documentclass[twoside]{article} \usepackage{amssymb, amsmath, graphicx } \pagestyle{myheadings} \markboth{\hfil Cauchy problem for derivors \hfil EJDE--2001/32} {EJDE--2001/32\hfil J-F Couchouron, C. Dellacherie, \& M. Grandcolas \hfil} \begin{document} \title{\vspace{-1in}\parbox{\linewidth}{\footnotesize\noindent {\sc Electronic Journal of Differential Equations}, Vol. {\bf 2001}(2001), No. 32, pp. 1--19. \newline ISSN: 1072-6691. URL: http://ejde.math.swt.edu or http://ejde.math.unt.edu \newline ftp ejde.math.swt.edu \quad ftp ejde.math.unt.edu (login: ftp)} \vspace{\bigskipamount} \\ % Cauchy problem for derivors in finite dimension % \thanks{ {\em Mathematics Subject Classifications:} 34A12, 34A40, 34A45, 34D05. \hfil\break\indent {\em Key words:} derivor, quasimonotone operator, accretive operator, Cauchy problem, \hfil\break\indent uniqueness condition. \hfil\break\indent \copyright 2001 Southwest Texas State University. \hfil\break\indent Submitted December 4, 2000. Published May 8, 2001.} } \date{} % \author{ Jean-Fran\c{c}ois Couchouron, Claude Dellacherie, \\ \& Michel Grandcolas} \maketitle \begin{abstract} In this paper we study the uniqueness of solutions to ordinary differential equations which fail to satisfy both accretivity condition and the uniqueness condition of Nagumo, Osgood and Kamke. The evolution systems considered here are governed by a continuous operators $A$ defined on $\mathbb{R}^N$ such that $A$ is a derivor; i.e., $-A$ is quasi-monotone with respect to $(\mathbb{R}^{+})^N$. \end{abstract} \section{Introduction} For $T>0$, we study the Cauchy Problem (CP) $$ \begin{gathered} \dot u(t) +Au(t)=f(t),\quad t\in [0,T]\cr u(0)=u_{0}, \end{gathered} \eqno(1.1) $$ where $A$ is a continuous operator on $ \mathbb{R}^N$ and $f$ belongs to $L^{1}([0,T]:\mathbb{R}^N)$. We require in addition that $A$ be a derivor on $ \mathbb{R}^N$ (or equivalently that $-A$ be quasi-monotone with respect to the cone $ (\mathbb{R^{+}})^N$) and has an additional order property (see Assumption \textrm{H1T} in Section 2). The existence of local solutions of (1.1) is proved by standard arguments (see \cite{w1} and Lemma 4.2). For instance, in the continuous case, this local existence comes from the Peano's Theorem. So the problem is essentially to prove the uniqueness of a local solution and the existence of global solutions. An important remark is that the identity operator minus the limit of infinitesimal generators of increasing semigroups is a derivor on the domains of the operators (see remark 2.1.d). The aim of this paper consists of giving a special converse of this previous property. General studies of evolution problems governed by derivors can be found in \cite{b2,d1,d2,w1} (for existence of extremal solutions of differential inclusions in $ \mathbb{R}^N$) and in \cite{h2} for the behavior of the flow (stability, etc.) in the regular case: $A$ is $C^{1}$. This work establishes uniqueness for the Cauchy Problem and complements previous studies. Let us point out that derivors often occur in the theory of production processes in Economics (for cooperative systems, see \cite{d3,h3}), in Chemistry \cite{h1}, and in Biology \cite{h1}. Our uniqueness result given in the sequel applies to these situations. Notice also that the additional order property; namely, existence of uniform ascents, (see Definition 2.2) has obvious interpretations in applications and may be considered as a special extension of the submarkovian property (see remark 2.2.3) and \cite{b1}). Nevertheless the notion of uniform ascents is a new concept built from the concept of progressions in \cite{d3}. This ascent notion which extends the usual submarkovian property seems to lead naturally to the maximum principle worked in \cite{d3}. Finally, we emphasize that the ascent notion is the key to obtain a suitable increasing resolvent (see Proposition 2.4 and Theorem 3.2). In this paper, the operator $A$ does not satisfy either uniqueness conditions such as those given by Nagumo, Osgood and Kamke \cite{c2,k1,k2} nor accretivity conditions, even in a generalized sense as in \cite{c2,c4,k2}. We will exhibit in Section 5 a simple example of operator on $\mathbb{R}^{2}$ which satisfies all our conditions and none of the uniqueness conditions quoted above. Consequently our framework is not included in the submarkovian case, since a continuous submarkovian derivor is accretive in ($\mathbb{R}^N,\|\cdot \|_{\infty}$). Moreover based in our analysis, it appear that a simple natural-order property can replace a classical Lipschitz condition about uniqueness in the Cauchy Problem. Uniqueness and order-preserving dependence with respect to the initial value $u_{0}$ are stated in Theorem 3.1. In the case $f=0$, Theorem 3.2 guarantees the existence of a global solution and a special form of the Crandall-Ligget exponential formula \cite[p. 319]{c5} involving suitable selections of the multi-valued operators $(I+\lambda A)^{-1}$ (while in \cite{d1} $(I+\lambda A)^{-1}$ is single valued and Lipschitz). This paper is organized as follows. Section 2 is devoted to general definitions and preliminaries. The main results are stated in Section 3, while the proofs are given in the next section. Section 5 gives an example in $ \mathbb{R}^{2}$ which demonstrates the need for Theorems 3.1 and 3.2. Some remarks about the asymptotic behavior follow in Section 6. \section{Generalities} We supply $\mathbb{R}^N$ with the usual partial order relation $u\leq v$ if $u^i\leq v^i$ for all $i=1,\dots ,N$, where $u^i$ is the $i$-th component. The vector in $\mathbb{R}^N$ whose components are $C,\dots,C$ is denoted by $C$. The symbol $\|\cdot \|$ stands for any norm in $\mathbb{R}^N$. The symbol $\mathbb{N}^{*}$ denotes the set of integers greater than zero. \paragraph {Definition 2.1} We say that the map $A$ is a {\bf derivor} on $\mathbb{R}^N$ if it satisfies the condition \begin{enumerate} \item[(i)] For each $(u,v)\in (\mathbb{R}^N)^{2}$ and each $i\in \{1,\dots ,N\}$ $$(u\leq v \hbox{ and } u^i=v^i) \quad \hbox{implies that}\quad A^iu\geq A^iv\eqno (2.1)$$ \end{enumerate} We say that the map $A$ is a {\bf moderate derivor} (resp. a {\bf strong derivor}) if, in addition to (i), it satisfies \begin{enumerate} \item[(ii)] For each $u \in \mathbb{R}^N$, there exist $u_{1},u_{2} \in \mathbb{R}^N$ (resp. two sequences $(u_{k})_{k}\to +\infty$, $(v_{k})_{k}\to -\infty$) such that $u_{1}\leq u \leq u_{2}$ and $Au_{1}\leq 0 \leq Au_{2}$ (see \cite{c2}). (resp. $ \lim_{k \to +\infty} Au_{k}=+\infty$ and $ \lim_{k \to +\infty} Av_{k}=-\infty$). \end{enumerate} The previous notation $ \lim_{k \to +\infty} w_{k}=+\infty$ in $\mathbb{R}^N$ may be interpreted to mean that $ \lim_{k \to +\infty} w_{k}^{j}=+\infty$ in $\mathbb{R}^N$ for each $j \in \{1,\dots ,N\}$. The derivor notion coincides with the notion of quasimonotone operator on $\mathbb{R}^N$, except the sign (see \cite{b2}, \cite[p. 91]{d1}, \cite{k1}). In these references, $-A$ is quasimonotone with respect to $(\mathbb{R^{+}})^N$ if $$(u\leq v \hbox{ and } x^{*}(u)=x^{*}(v)) \quad\hbox{implies that}\quad x^{*}(-Au)\leq x^{*}(-Av) \eqno (2.2)$$ for any linear positive form $x^{*}$ on $\mathbb{R}^N$. Hence, $A$ is a derivor, because if $x^{*}$ is a linear positive form on $ \mathbb{R}^N$, $x^{*}$ is a linear combination with positive coefficients of coordinate forms on $\mathbb{R}^N$. \paragraph{Remark 2.1} a) Condition (i) in the definition of derivor is automatically fulfilled for any operator $A$ from $\mathbb{R}$ to $\mathbb{R}$, but it is not in the case of Condition (ii). A special case where (ii) holds for an operator $A$ from $\mathbb{R}$ to $\mathbb{R}$ is the case where $A$ is a non-decreasing operator such that there is $v\in \mathbb{R}$ satisfying $Av=0$. \\ b) When $A$ is a linear derivor, the reader can check that Condition (ii) is equivalent to: there is $u\geq 1$ satisfying $Au\geq 0$. \\ c) An equivalent form of definition 2.1.(i) is: $ A^i $ is decreasing with respect to $ x^{j}$ for each $i\not= j$ with $i,j \in \{1,\dots ,N\}$ (see \cite{h3}). \\ d) If $P$ is an increasing operator on $\mathbb{R}^N$, then $A=I-P$ is a derivor. Therefore, when $(P_{t})$ is an increasing semi-group on $\mathbb{R}^N$, then $A_{t}=\frac{I-P_{t}}{t}$ with $t>0$ is a derivor and so is $A_{0}$, defined by $A_{0}u=\lim_{t\downarrow 0}A_{t}u$ (on the domain where this limit exists). \subsection*{Ascents} We denote by $V_{K}(u_{0})$ the set of compact neighborhoods of $u_{0}$. \paragraph{Definition 2.2} We say that a derivor $A$ has a (strict) {\bf uniform ascent} at $u_{0}$ if there are $V \in V_{K}(u_{0})$ and a sequence $(v_{k})$ in $\mathbb{R}^N$ convergent to 0 such that $(v_{k}^i)_{k \in \mathbb{N}^{*}}$ is strictly decreasing for all $i=1,\dots ,N$ and $$\min_{i\in \{1,\dots ,N\}}(A^i(u+v_{k})-A^iu)>0\eqno (2.3)$$ for each $k\in \mathbb{N}^{*}$ and each $u \in V$. \paragraph{Remark 2.2} 1) In terms of production operator ($A^iu$ is the production of the $i$-th input of the product $u$), the uniform ascent property at $u_{0}$ means that in a neighborhood of $u_{0}$ it is possible to increase the level of production by means of small uniform augmentations around $u_{0}$. \\ 2) The notion of uniform ascent plays a crucial part in this work. In our opinion, this concept is new, but it was inspired from the progression notion carried out in \cite{d3}. \\ 3) The uniform ascent property may be connected to the submarkovian property namely, $$A(u+C)-Au \geq 0 \eqno (2.4)$$ for all $u\in \mathbb{R}^N$ and all $C\in \mathbb{R}^{+}$. Notice that a submarkovian derivor in $\mathbb{R}^N$ is accretive in ($\mathbb{R}^N,\|\cdot \|_{\infty}$); the verification of this claim is left to the reader. \\ 4) The following dual notion of uniform ascent at $u_{0}$ provides again the results of Section 3: There are $V \in V_{K}(u_{0})$ and a strictly increasing sequence $(v_{k})$ in $\mathbb{R}^N$ convergent to 0 such that $(v_{k}^i)_{k \in \mathbb{N}^{*}}$ is strictly increasing for all $i=1,\dots ,N$ and $$\sup_{i\in \{1,\dots ,N\}}(A^i(u+v_{k})-A^iu)<0$$ for each $k\in \mathbb{N}^{*}$ and each $u \in V$. \\ 5) In the case $A=I-P$ with an increasing operator $P$, (see Remark 2.1.d)), Definition 2.2 means that the required sequence $(v_{k})$ satisfies $$P^i(u+v_{k})-P^i(u)0$ are single-valued contractions. But in our case $(I+\lambda A)^{-1}$ is a priori multi valued. Nevertheless it is possible to define suitable selectors $J_{\lambda}$ of $(I+\lambda A)^{-1}$ as claimed in the following lemma. \paragraph{Lemma 2.4} {\sl Let $A$ be a moderate continuous derivor on $\mathbb{R}^N$. Let $u$ in $\mathbb{R}^N$, $\lambda \in \mathbb{R}^{+}$ and $v$ a solution of $$ \begin{gathered} v\leq u\cr Av \leq 0.\end{gathered}\eqno (2.7)$$ Then the system $$\begin{gathered}w\geq v\cr (I+\lambda A)w \geq u\end{gathered}\eqno (2.8)$$ has a smallest solution denoted by $J_{\lambda,v}u$. Moreover we have $$J_{\lambda,v}u\in (I+\lambda A)^{-1}(u).\eqno (2.9)$$ } \paragraph{Proof.} According to (ii) in definition 2.1, Systems (2.7) and (2.8) have solutions. Let $v$ be a solution of (2.7). Since $B=I+\lambda A$ is a continuous derivor, the existence of the smallest solution $J_{\lambda,v}u$ of (2.8) is guaranteed by the Theorem 2.4. It remains to prove $$(I+\lambda A)J_{\lambda,v}u=u.\eqno (2.10)$$ Since the constraints are optimal in (2.8), we have for each $i\in \{1,\dots ,N\}$, $((I+\lambda A)J_{\lambda,v}u)^i=u^i$ or $(J_{\lambda,v}u)^i=v^i$. Thus we have to prove $((I+\lambda A)J_{\lambda,v}u)^i=u^i$ when $(J_{\lambda,v}u)^i=v^i$. So assume $(J_{\lambda,v}u)^i=v^i$ for some $i\in \{1,\dots ,N\}$, Relation (2.1) and $J_{\lambda,v}u\geq v$ yield $$(AJ_{\lambda,v}u)^i\leq (Av)^i.\eqno (2.11)$$ Now (2.11) and (2.7) provide $$ (J_{\lambda,v}u)^i+\lambda (AJ_{\lambda,v}u)^i =v^i+\lambda (AJ_{\lambda,v}u)^i \leq u^i+\lambda (AJ_{\lambda,v}u)^i \leq u^i. $$ Therefore, $((I+\lambda A)J_{\lambda,v}u)^i\leq u^i$. But from (2.8) we have $((I+\lambda A)J_{\lambda,v}u)^i\geq u^i$. Finally (2.10) is proved. \hfill$\square$\medskip In the same way, let $v$ be a solution of $$\begin{gathered} v\geq u\cr Av \geq 0.\end{gathered}\eqno (2.12)$$ Then the system $$\begin{gathered} w\leq v\cr (I+\lambda A)w \leq u\end{gathered}\eqno (2.13)$$ has a largest solution $w={\tilde J}_{\lambda,v}u$. Moreover ${\tilde J}_{\lambda,v}u$ satisfies again (2.9). Set $J_{\lambda}u=J_{\lambda,v}u$ (resp. $J_{\lambda}u={\tilde J}_{\lambda,v}u$) for an arbitrary $v$ satisfying (2.7) (resp. (2.12)). Let us notice that $J_{\lambda}$ is defined on $D_{v}=\{u \in \mathbb{R}^N, u\geq v\}$ (resp. $D_{v}=\{u \in \mathbb{R}^N, u\leq v\}$). The family of selectors $(J_{\lambda})_{\lambda\geq 0}$ of $(I+\lambda A)^{-1}$ is said to be the {\bf resolvent} of $A$. \paragraph{Definition 2.4} For $u$ given, the notation ${\overline u}$ (resp. $ {\hat u}$ ) stands for the largest solution of (2.7) (resp. the smallest solution of (2.12)). \smallskip Thanks to Theorem 2.4, such extremal elements ${\overline u}$ and $ {\hat u}$ exist. Furthermore we have clearly $$u\leq v \Longrightarrow ({\overline u}\leq {\overline v} \quad\mbox{and}\quad {\hat u}\leq {\hat v})\eqno (2.14) $$ and $${\overline {\overline u}}={\overline u}\quad\mbox{and}\quad {\hat {\hat u}}={\hat u}.\eqno (2.15)$$ The resolvent operators satisfy the following properties. \paragraph{Proposition 2.4} {\sl For a given $u\in \mathbb{R}^N$, let $v,v'\in \mathbb{R}^N$ satisfying (2.7) and $w,w' \in \mathbb{R}^N$ satisfying (2.12). Then \begin{enumerate} \item[(a)] The map $J_{\lambda}$ is single-valued and increasing on $D_{v}$. \item[(b)] We have $$v\leq J_{\lambda}u \leq w\eqno (2.16)$$ In particular $${\overline u}\leq J_{\lambda}u \leq {\hat u}\eqno (2.17)$$ \item[(c)] If $Au\geq 0$ (resp. $Au\leq 0$), then $AJ_{\lambda}u\geq 0$ (resp. $AJ_{\lambda}u\leq 0$) for each $\lambda \geq 0$. \item[(d)] If $Au\geq 0$, then $\lambda \to J_{\lambda}u$ is decreasing on $\mathbb{R}^N$ (and increasing if $Au\leq 0$). \item[(e)] We have $J_{\lambda,v}u\leq {\tilde J}_{\lambda,w}u$. In particular $J_{\lambda,{\overline u}}u\leq {\tilde J}_{\lambda,{\hat u}}u$. \item[(f)] $J_{\lambda,{\overline u}}{\overline u}={\overline u}$ and ${\tilde J}_{\lambda,{\hat u}}{\hat u}={\hat u}$. \item[(g)] If $v\leq v'$ and $w\leq w'$, then $J_{\lambda,v}u\leq J_{\lambda,v'}u$ and ${\tilde J}_{\lambda,w}u\leq {\tilde J}_{\lambda,w'}u$. \end{enumerate} } %end proposition \paragraph{Proof.} We prove only results (a),(b),(c),(d) in the case $J_{\lambda}=J_{\lambda,v}$.\\ (a) Let $u\geq w$. Then $J_{\lambda,v}u$ satisfies $$\displaylines{J_{\lambda,v}u\geq v\cr (I+\lambda A)J_{\lambda,v}u \geq u \geq w.}$$ Hence we get (a) from minimality of $J_{\lambda,v}w$ for the previous system. \\ (b) Inequality $J_{\lambda,v}u\geq v$ is required in the definition of $J_{\lambda,v}$. Since $w$ satisfies $$\displaylines{w\geq v\cr (I+\lambda A)w \geq w\geq u,}$$ we get (b) from minimality of $J_{\lambda,v}u$ in the previous system. \\ (c) Let $Au\geq 0$ and $\lambda \geq 0$. We have $$\displaylines{(I+\lambda A)u\geq u\cr u \geq {v}}$$ so $J_{\lambda,v}u \leq u$. Hence $u=J_{\lambda,v}u +\lambda AJ_{\lambda,v}u \leq u+\lambda AJ_{\lambda,v}u$ and so $AJ_{\lambda}u \geq 0$. \\ (d) Let $0\leq \lambda \leq \mu$. Then $u=(I+\lambda A)J_{\lambda,v}u \leq (I+\mu A)J_{\lambda,v}u$. Since we have $J_{\lambda,v}u \geq v$, from minimality of $J_{\mu}u$ for these two constraints, it comes $J_{\mu,v}u \leq J_{\lambda,v}u$. The proof is similar when $Au\leq 0$. \\ (e) Since $Av\leq 0$ and $Aw\geq 0$, from (c) it follows $AJ_{\lambda,v}u\geq 0$ and $A{\tilde J}_{\lambda,w}u\leq 0$. Hence (e) results from $J_{\lambda,v}u+\lambda AJ_{\lambda,v}u=u={\tilde J}_{\lambda,w}u+\lambda A{\tilde J}_{\lambda,w}u$. \\ Properties (f) and (g) result immediately from the definitions. \hfill$\square$ \subsection*{Solution of (1.1)} We recall that a (local) \textbf{strong solution} of (1.1) is a continuous function $u$ defined on $[0,\theta) \subset [0,T],\theta >0$ such that $u(t)=u_{0}+\int_{0}^{t}(-Au(\tau)+f(\tau))d\tau$ for $t\in [0,\theta)$. In the sequel we only look for (local) strong solutions of (1.1). A {\bf maximal} (resp. {\bf minimal}) {\bf solution } of (1.1) is the strong solution $u=S_{A,f}^{\text{max}}(t)u_{0}$ (resp. $u=S_{A,f}^{\text{min}}(t)u_{0}$) of (1.1) defined as follows: \\ (i) The interval of definition $[0,\theta)$ of $S_{A,f}^{\text{max}}(.)u_{0}$ (resp. $S_{A,f}^{\text{min}}(.)u_{0}$) is maximal on $[0,T]$, i.e. there is no solution $v \not= u$, such that $v=u$ on $[0,\theta]$. \\ (ii) For each solution $v$ of (1.1) on $[0,T_{1})\subset [0,T]$, we have $v(t)\leq S_{A,f}^{\text{max}}(t)u_{0}$ (resp. $v(t)\geq S_{A,f}^{\text{min}}(t)u_{0}$) on $[0,\inf (\theta, T_{1}))$. \section{Main results} For the following results, we assume the hypothesis \textrm{H1T} defined in Section 2. \paragraph{Theorem 3.1} {\sl The problem $CP(A,f,u_{0})$ has a unique local solution denoted by $S_{A,f}(t)u_{0}$ (or $S_{A}(t)u_{0}$ if $f=0$) and defined on a maximal interval $[0,T_{\max}) \subset [0,T]$. Moreover if $u_{0}\leq u_{1}$ in $\mathbb{R}^N$ and if $f\leq g$ in $L^{1}([0,T],\mathbb{R}^N)$ then $S_{A,f}(t)u_{0}\leq S_{A,g}(t)u_{1}$ on the common domain of existence of these two solutions.} \medskip The next result concerns the autonomous case, for which we have global solutions. \paragraph{Theorem 3.2} {\sl Assume that $f\equiv 0$. Then $S_{A}(.)u_{0}$ is defined on the whole interval $[0,T]$ and $$S_{A}(t)u_{0}=\lim_{n\to +\infty}J_{{t/n},n}(u_{0}),\eqno (3.1)$$ for $t\in [0,T]$, where $J_{\lambda}=J_{\lambda,{\overline u}_{0}}$ is as defined in Section 2.} This is an exponential Crandall-Liggett's type formula, but here $(I+{\lambda}A)^{-1}$ is a priori multi-valued. In the non-autonomous case $f \not\equiv 0$, it is possible to exhibit a formula as (3.1) which gives the solution of (1.1) as a limit of a discrete scheme. But such a formula is more complicated than (3.1) and thus, is not of a particular interest. When $f \in L^{\infty}([0,T], \mathbb{R}^N)$, from Theorem 3.1 and Theorem 3.2, we can deduce that $CP(A,f)$ has solution on $[0,T]$ if $A$ is a strong continuous derivor (see def. 2.1). Unfortunately, we do not know what happens in the general case $f \in L^{1}([0,T], \mathbb{R}^N)$ without extra assumptions. \section{Proofs} The proof of Theorem 3.1 follows immediately from the three lemmas below. \paragraph{Lemma 4.1} {\sl Let $A$ be a continuous derivor. Let $V$ be an element of $V_{K}(u_{0})$. Then the operator B defined by $$B(v):=\inf_{w\in V}[A(w+v)-A(w)]\eqno (4.1)$$ is a continuous derivor.} \paragraph{Proof.} 1.) Let us show that $B$ is a derivor on $\mathbb{R}^N$. If $u\leq v$ and $u^i=v^i$ for some $i\in \{1,\dots ,N\}$, we have $u+w\leq v+w$ and $(u+w)^i=(v+w)^i$ for each $w\in V$. Since A is a derivor, it follows $A^i(u+w)-A^iw\geq A^i(v+w)-A^iw$. Thus $$\inf_{w\in V}(A^i(u+w)-A^iw)\geq \inf_{w\in V}(A^i(v+w)-A^iw).$$ So $B^iu\geq B^iv$ for $u\leq v$ and $u^i=v^i$. \\ 2.) At this stage we will show that $B$ is continuous on $\mathbb{R}^N$. According to (4.1), $B$ is clearly upper semi-continuous (see \cite[pp. 132-137]{c1}). So it is enough to prove that for each $i\in \{1,\dots ,N\}$, $B^i$ is lower semi-continuous on $\mathbb{R}^N$. Fix $i\in \{1,\dots ,N\}$. For each $u\in \mathbb{R}^N$, thanks to the compactness of $V$, there exists $\chi(u)$ (which depends on $i$) in $V$ satisfying $$B^i u=A^i( u+{\chi(u)})-A^i({\chi(u)})\eqno (4.2)$$ We have to prove now that $B^i$ is lower semi-continuous, that is $(B^i)^{-1}(]-\infty,\alpha])$ is closed for all $\alpha \in \mathbb{R}$. In this goal, consider $\alpha \in \mathbb{R}$ and a sequence $(u_{k})_{k \in \mathbb{N}^{*}}$ of elements of $\mathbb{R}^N$ such that $\lim u_{k}=u_{\infty}$ and $B^i(u_{k})\leq \alpha$. It suffices to prove $B^i(u_{\infty})\leq \alpha$. By contradiction, let us suppose $B^i(u_{\infty})> \alpha$. Without loss of generality, thanks to the compactness of $V$, we can suppose $$\lim_{k\to +\infty}{\chi(u_{k})}=v_{\infty}\in V.$$ Equation (4.1) yields $$\alpha 0$ such that $v^i(\tau)\leq u_{n}^i(\tau)$ for $\tau \in [t_{0},t_{0}+\eta]\subset [0,\tilde T \wedge T_{n})$. Finally $E$ is open in $[0,\tilde T \wedge T_{n})$ and thus $E=[0,\tilde T \wedge T_{n})$. \\ 2) We have $u_{n+1}\leq u_{n}$ on $[0,T_{n+1} \wedge T_{n})$. Indeed the proof is the same as 1) if we replace $v$ by $u_{n+1}$ and $-Au_{n}(\tau)+Av(\tau)$ by $-Au_{n}(\tau)+Au_{n+1}(\tau)$. \\ 3) We have $\tilde T \wedge T_{n} \geq \tilde T \wedge T_{1}$. Indeed, from parts 1) and 2), for each $n\in \mathbb{N^{*}}$ we have $$v\leq u_{n+1} \leq u_{n} \leq u_{1}\eqno (4.7)$$ on the common interval of existence of these solutions. Then the extension principle of solutions implies $\tilde T \wedge T_{n+1}\geq \tilde T \wedge T_{n}$ since a bounded solution is extendable. \\ 4) The sequence $(u_{n})$ converges uniformly to $u_{\infty}$ on each compact sub-interval of $[0,\tilde T \wedge T_{1})$ thanks to (4.7) and the Lebesgue's Dominated Convergence Theorem. Furthermore $u_{\infty}$ is solution of $CP(A,f,u_{0},\tilde T \wedge T_{1})$ on $[0,\tilde T \wedge T_{1})$. Moreover, clearly $u_{\infty}$ is the maximal solution of $CP(A,f,u_{0},\tilde T \wedge T_{1})$ (see Section 2). Let $F$ be the set of $S\in [0,T]$ such that $u_{\infty}$ is extendable into a continuous function on $[0,S)$ which is the maximal solution of $CP(A,f,u_{0},S)$. One has $\tilde T \wedge T_{1}\in F$. By considering $S_{\infty}=\sup F$, we obtain a maximal extension of $u_{\infty}$ as a local solution of $CP(A,f,u_{0},T)$ which is by construction the maximal solution of $CP(A,f,u_{0},T)$. \hfill$\square$\smallskip The next lemma makes use of the ascent assumption. \paragraph{Lemma 4.3} {\sl With the notation in Lemma 4.2, if \textrm{H1T} holds, we have $$S_{A,f}^{\text{min}}(t)u_{0}=S_{A,f}^{\text{max}}(t)u_{0}$$ on $[0,T^{1}\wedge T^{2}) =[0,T^{1})$.} \paragraph{Proof.} Thanks to Lemma 4.2(a), (1.1) has a maximal solution $S_{A,f}^{\text{max}}(t)u_{0}$ defined on a sub-interval $[0,T^{1})$ of $[0,T]$ and a minimal solution $u(t)=S_{A,f}^{\text{min}}(t)u_{0}$ defined on a sub-interval $[0,T^{2})$ of $[0,T]$. Set $T_{3}=T^{1}\wedge T^{2}$ and $$w(t):=S_{A,f}^{\text{max}}(t)u_{0}-S_{A,f}^{\text{min}}(t)u_{0}\eqno (4.8)$$ for $t\in [0,T_{3})$. We have to prove $w=0$ on $[0,T_{3})$, that is $E=[0,T_{3})$ where $E=\{ t\in [0,T_{3}), w(\tau)=0,\forall \tau \in [0,t]\}$. Since $E=w^{-1}(0)$ is closed in $[0,T_{3})$ ($w$ being continuous), it just remains to show that $E$ is open to the right. Let $t_{0} \in E, t_{0}< T_{3}$. We have to prove that there exists $h>0$ such that $w=0$ on $[t_{0},t_{0}+h]$. Eventually, by changing $w$ into $w(t_{0}+.)$ and $f$ into $f(t_{0}+.)$, we will suppose $t_{0}=0$. Let $V \in V_{K}(u_{0})$ and $B$ as in (4.1), in view of the continuity of $u$ at 0, there exists $T_{4}\in ]0,T_{3}[$ such that, for each $t\in [0,T_{4}]$, $u(t)\in V$, hence $w$ satisfies a.e.: $$\begin{gathered} w'(t)=-(A(u(t)+w(t))-Au(t))\leq -Bw(t) \cr w(0)=0,\end{gathered}\eqno (4.9)$$ a.e. $t\in [0,T_{4}]$. By using Lemma 4.2 (b) with $B$ instead of $A$, we have $$w(t)\leq S_{B}^{\text{max}}(t)(0)\eqno (4.10)$$ for each $t\in [0,T_{4}\wedge T_{5}]$, where $[0,T_{5}]$ is the maximal interval of existence of $S_{B}^{\text{max}}(t)(0)$. The function $x(t)=S_{B}^{\text{max}}(t)(0)$ satisfies $$\begin{gathered} x'(t)= -Bx(t)\cr x(0)=0.\end{gathered}\eqno (4.11)$$ Let $(v_{k})_{k\in \mathbb{N^{*}}}$ be a sequence which defines a uniform ascent at the point $u_{0}$ for the operator $A$ on the set$ V$ (see section 2). $$B^i(v_{k})=A^i(v_{k}+{\hat v_{k}}(i))-A^i({\hat v_{k}}(i))>0 \eqno (4.12)$$ for $k\in \mathbb{N^{*}}$ and $i \in \{1,\dots ,N\}$ where ${\hat v_{k}}(i)$ is a vector minimizing $ v \to A^i(v_{k}+ v)-A^i(v)$ on $V$. Let $k\in \mathbb{N}$ be fixed, then due to Lemma 4.2(b) there exists $s_{k}>0$ such that $s_{k}\leq T_{4}\wedge T_{5}$ and $$S_{B}^{\text{max}}(t)(0)\leq S_{B}^{\text{max}}(t)(v_{k})\eqno (4.13)$$ for each $t \in [0,s_{k}]$. Equation (4.12) and the continuity of $B$ give the existence of $t_{k}>0$ and $t_{k}\leq s_{k}$ such that: $$B(S_{B}^{\text{max}}(t)(v_{k}))\geq 0$$ for $t\in [0,t_{k}]$. Thus $t \to S_{B}^{\text{max}}(t)(v_{k})$ is decreasing on $[0,t_{k}]$. Consequently, from (4.10) and (4.13), it results $$w(t)\leq S_{B}^{\text{max}}(t)(v_{k})\leq S_{B}^{\text{max}}(0)(v_{k})=v_{k}\eqno (4.14)$$ for each $t\in [0,t_{k}]$. In particular, we have $w(t_{k})\leq v_{k} $. If we put $y(t)=w(t_{k}+t)$, we get $$\begin{gathered} y'(t)\leq -By(t) \cr y(0)=w(t_{k})\leq v_{k}\end{gathered}\eqno (4.15)$$ for a.e. $t\in [0,t_{k}]$. Hence, according to (4.14) and (4.15), one has $$w(t_{k}+t)\leq S_{B}^{\text{max}}(t)(v_{k})\leq v_{k}$$ for $t\in [0,t_{k}]$. So $w(t)\leq v_{k}$ for $t\in [0,2t_{k}\wedge T_{4}]$. Whence by induction, we get $$0\leq w(t)\leq v_{k}\eqno (4.16)$$ for $t\in [0,T_{4}]$. Since (4.16) is valid for each $k\in \mathbb{N^{*}}$ and $\lim v_{k}=0$, it follows $w(t)=0$ for each $t\in [0,T_{4}]$. Hence for $h=T_{4}>0$, we have $[0,h] \subset E$ which completes the proof. \hfill$\square$ \subsection*{Proof of Theorem 3.2} In this subsection, we assume that $A$ satisfies \textrm{H1T}, and $f \equiv 0$ on $[0,T]$. First, let us recall some basic facts about the discretization (1.1) in the Theory of Nonlinear Semigroups. It is known \cite{c5} that a strong solution of (1.1) is a mild solution, i.e. a continuous function which is a uniform limit of Euler's implicit discrete schemes. Such discrete schemes are defined as follows. Let $\epsilon >0$ be fixed. Then an $\epsilon$-discretization on $[0,T]$ of $\dot u +Au=0$ on $[0,T]$ consists of a partition $0=t_{0}\leq t_{1}\leq \dots \leq t_{n}$ of the interval $[0,t_{n}]$ and a finite sequence $(f_{1},f_{2},\dots ,f_{n})$ in $\mathbb{R}^N$ such that \\ (a) $t_{i}-t_{i-1}<\epsilon$ for $i=1,\dots ,n$ and $T-\epsilon0$ there is an $\epsilon$-approximate solution $v$ of $CP(A,0,u_{0})$ on $[0,T]$ such that $\| v(t)-u(t) \| \leq \epsilon$ for $t$ in the domain of $v$. Now, for $n\in \mathbb{N^{*}}$, let $J=J_{T/n, {\overline u_{0}}}$, and define the function $u_{n}$ by $u_{n}(0)=u_{0}$ and $u_{n}(t)=J^i(u_{0})$ for $(i-1)T/n 0$ there exists $N_{\epsilon}\in \mathbb{N}$ and $\eta_{\epsilon}>0$ such that ($n\geq N_{\epsilon}$ and $\vert t-s \vert \leq \eta_{\epsilon}$) implies $\| u_{n}(t)-u_{n}(s) \|_{\infty}\leq \epsilon$. Indeed, relations (4.17) lead to $$ u_{n}(t_{j}^{n})-u_{n}(t_{i}^{n})= -\int_{t_{i}^{n}}^{t_{j}^{n}}Au_{n}(t)dt.\eqno(4.21) $$ Using (4.18), Relation (4.21) yields $$\| u_{n}(t)-u_{n}(s) \| \leq M(\vert t-s \vert +2\frac{T}{n}), \eqno (4.22)$$ where $M=\sup_{{\overline u}_{0}\leq w \leq {\hat u_{0}}} \| Aw \|$. Consequently (see \cite[p. 260]{d4}) the sequence $(u_{n})$ is relatively compact in the Banach space ${\cal B}([0,T],\mathbb{R}^N, \| \hskip 0.2cm \|_{\infty})$ of bounded functions on $[0,T]$ with values in $\mathbb{R}^N$. So there exists a subsequence $(u_{n_{k}})$ converging to a continuous function $u_{\infty}$ which is a mild solution of $CP(A,0,u_{0})$. Then, passing to the limit in (4.21) (or from \cite[p. 314]{c5}), we see that $u_{\infty}$ is a strong (even a classical) solution of $CP(A,0,u_{0})$ on $[0,T]$. From Theorem 3.1, it results $$u_{\infty}=S_{A}(.)u_{0}\eqno (4.23)$$ on $[0,T]$. Thus (4.20) follows from (4.23) and (4.18) on $[0,T]$. Then, taking $T=t$, (4.23) yields $$S_{A}(t)u_{0}=\lim_{n\to +\infty}J_{t/n}^{n} (u_{0}), $$ where $J_{t/n}=J_{{t/n},{\overline u_{0}}}$. The proof is complete. \section{An example in $\mathbb{R}^{2}$} Let $A_{0}$ be the operator defined on $\mathbb{R}^{2}$ by $$A_{0}\begin{pmatrix} x\cr y\end{pmatrix} :=\begin{pmatrix} x+(x-2y)^{1/3}\cr y+(2y-x)^{1/5}\end{pmatrix} \eqno (5.1)$$ \paragraph{Lemma 5.1} {\sl The operator $A_{0}$ satisfies \textrm{H1T} and \textrm{H2T} for all $T>0$.} \medskip The proof is left to the reader. In particular, the relation $$A_{0}\begin{pmatrix} x\cr y\end{pmatrix} +\begin{pmatrix} 2t\cr t\end{pmatrix} =A_{0}\begin{pmatrix} x\cr y\end{pmatrix} +\begin{pmatrix} 2t\cr t\end{pmatrix}\eqno (5.2)$$ for $t\in \mathbb{R}^{+}$, provides uniform ascents at each point. The sublinearity at infinity implies \textrm{H2T}. Therefore we can apply the results of Section 3 to the operator $A_{0}$ for any $T>0$. Hence $CP(A_{0},f,u_{0},+\infty)$ has a unique global solution, on $[0,+\infty[$. Now, our task is to prove that no condition of Nagumo-Osgood-Kamke and no accretivity condition (even in a generalized sense) can be applied to obtain the uniqueness of solutions of $CP(A_{0},f,u_{0})$. \subsection*{Generalized accretivity conditions} Let $\|\cdot \|_{p}$, $p\in [1,+\infty]$, be the classical $l_{p}$-norm in $\mathbb{R}^{2}$. As usual (see \cite{c5,d1}), we set $$[u,v]=\lim_{\lambda \downarrow 0} \frac{\| u+\lambda v \|-\| u \|}{\lambda}\eqno (5.3) $$ for $u,v\in \mathbb{R}^{2}$. For $p\in [1,+\infty]$, the notation $[u,v]_{p}, p\in [1,+\infty]$ means $[u,v]$, with $\| \cdot \|_{p}$ instead of $\| \cdot \|$ in (5.3). In the sequel, $\phi$ stands for a continuous function $\phi:\mathbb{R}\to \mathbb{R}^{+}$ satisfying the following condition ${\cal U}$: For each $T_{0}$, the function $x\equiv 0$ is the unique positive solution on $[0,T_{0}]$ of $$\displaylines{\dot x(t)=\phi(x(t))\cr x(0)=0. }$$ \paragraph{Definition 5.2} We will say that an operator $B$ defined on $\mathbb{R}^{2}$ is {\bf $\phi$-accretive } in $(\mathbb{R}^{2},\| \cdot \|)$ if $$-[u-v,Bu-Bv]\leq \phi(\| u-v \|)\eqno (5.4)$$ for all $u,v\in \mathbb{R}^{2}$. We will say that $B$ satisfies a {\bf $\phi$-Osgood condition} if $$\| Bu-Bv \| \leq \phi(\| u-v \|)$$ for all $u,v\in \mathbb{R}^{2}$. \paragraph{Remark} a) The condition $B+\omega I$ is accretive $(\omega \geq 0)$ means $B$ is $\phi$-accretive with $\phi(x)=\omega x$. General studies of $\phi$-accretive conditions can be found in \cite{c4,k2}. \\ b) A $\phi$-{\bf Osgood condition} is a particular case of $\phi$-accretivity. \paragraph{Lemma 5.2} {\sl Let $p\in [1,+\infty]$. Then, there is no $\phi$, such that $A_{0}$ is $\phi$-accretive in ($\mathbb{R}^{2},\|\cdot \|_{p}$). Moreover, there are no $\phi$ and no norm $\|\cdot \|$ such that $A_{0}$ satisfies a $\phi$-Osgood condition in ($\mathbb{R}^{2},\|\cdot\|$).} \paragraph{Proof.} a) Suppose first $p=+\infty$. By contradiction, suppose that $A_{0}$ is $\phi$-accretive in ($\mathbb{R}^{2},\| \cdot \|_{\infty}$) for some $\phi$. Let $x\in [0,1[$. A direct computation yields $A_{0}\begin{pmatrix}0\cr 0\end{pmatrix} =\begin{pmatrix}0\cr 0\end{pmatrix}$ and $$[\begin{pmatrix}x\cr x-\frac 12 x^{2}\end{pmatrix}, A\begin{pmatrix} x\cr x-\frac 12 x^{2}\end{pmatrix}]_{\infty} =x^{1/3}(x^{2/3}+(-1+x)^{1/3}).\eqno (5.5) $$ So, thanks to the $\phi$-accretivity, (5.5) implies $$\frac 12 x^{1/3}\leq \phi(x),\eqno (5.6)$$ for $x\geq 0$ sufficiently small. Set $$z(t)=H^{-1}(t)\,,\quad H(\sigma) =\int_{0}^{\sigma}\frac{d\xi}{\phi(\xi)}.\eqno (5.7)$$ From (5.6), $H$ is defined for $\sigma\geq 0$ sufficiently small and $$z(t)>0\eqno (5.8)$$ on some interval $]0,T_{0}]$ with $T_{0}>0$. By using (5.7), a straightforward computation gives $z'(t)=\phi (z(t))$ and $z(0)=0$. Then ${\cal U}$ provides $$z\equiv 0 \quad\mbox{on } [0,T_{0}].\eqno (5.9) $$ Hence there is a contradiction between (5.8) and (5.9). \noindent b) Suppose now $p\in [1,+\infty[$. By contradiction again, suppose that $A_{0}$ is $\phi$-accretive in ($\mathbb{R}^{2},\| \cdot \|_{p}$). In this case, for $x\in [0,1]$, by setting $$u=\begin{pmatrix} x\cr \frac 12 x-\frac 12 x^{2} \end{pmatrix} $$ a direct computation gives $$[u, A_{0}u]_{p} =\left(1+\frac{x^{p-\frac 13}-(\frac{x-x^2}{2})^{p-1}x^{2/5}} {\| u\|_{p}^{p}}\right)\| u \|_{p}.\eqno (5.10)$$ According to (5.10), the reader can check that the $\phi$-accretivity property implies $\phi(\| u \|_{p})\geq -[u,A_{0}u]_{p}\geq \frac{1}{2^{p}+1}x^{2/5}$ for $x\in [0,1]$ sufficiently small. Then we can deduce that for some $x_{0}\in ]0,1]$, there is $C>0$ (for instance $C=\frac{e^{-1/5}}{2(2^{p}+1)}$), such that $$C\| u \|_{p}^{2/5}\leq \phi(\| u \|_{p}) $$ for all $x\in [0,x_{0}]$. Finally, there exists $\xi_{0}> 0$ such that $\phi(\xi)\geq C\xi^{2/5}$ for $\xi \in [0,\xi_{0}]$. Now, as in step a), using the function $H$ defined in (5.7), we can easily derive a contradiction. \noindent c) Let $\| \cdot \|$ be a norm in $\mathbb{R}^{2}$ and suppose that $A_{0}$ satisfies a $\phi$-Osgood condition in $(\mathbb{R}^{2}, \| \cdot \|)$. Then, by taking $u=\begin{pmatrix} 0\cr x\end{pmatrix}$, in the $\phi$-Osgood property we obtain $\phi(\xi)\geq c{\xi^{1/5}}$ for a constant $c>0$, $\xi_{1}>0$ and all $\xi \in [0,\xi_{1}]$. So we can conclude as before and the lemma is proved. \hfill$\square$ \begin{figure} \begin{center} \includegraphics[width=0.7\textwidth]{fig1.eps} \end{center} \caption{Flow relative to $A_{0}$} \end{figure} \section{Asymptotic behavior} Figure 5.1 motivates the following remarks about asymptotic behavior of solutions of (1.1). Hypothesis \textbf{H3} stands for following three conditions\begin{itemize} \item $f\equiv 0$ \item The assumption \textrm{H2T} holds for all $T>0$ \item $A$ is a continuous derivor on $ \mathbb{R}^N$. \end{itemize} We do not assume the uniqueness of solutions of $CP(A,0,u_{0},+\infty)$. We set $A^{+}=\{u;Au\geq 0\}$ and $A^{-}=\{u;Au\leq 0\}$. \paragraph{Definition 6.1} A derivor $A$ is \textit{absorbent} if $ u_{0}\in A^{+}$ (resp. $ u_{0}\in A^{-}$) implies $u(t)\in A^{+}$ (resp. $ u(t)\in A^{-}$) for all $t\geq 0$ and each solution $u(.)$ of the autonomous problem $CP(A,0,u_{0},+\infty)$. We say that A is $u_{\infty}$-absorbent if B defined by $Bu=Au-Au_{\infty}$ is absorbent. \paragraph{Proposition 6.2} {\sl Assume \textrm{H3}. Let $u_{0},v_{0},w_{0}$ be in $\mathbb{R}^N$ such that $Av_{0}\leq 0$, $Aw_{0}\geq 0$, $v_{0}\leq w_{0}$ and $v_{0} \leq u_{0} \leq w_{0}$. Suppose $A$ is a continuous $u_{\infty}$-absorbent derivor on $\mathbb{R}^N$ such that the equation $Av=0$ has a unique solution $u_{\infty}$ in $[v_{0},w_{0}]$. Then every solution $u$ of $CP(A,0,u_{0},+\infty)$ satisfies $$\lim_{t\to +\infty}u(t)=u_{\infty} $$} %\end proposition \paragraph{Proof.} It is sufficient to prove the result for $S_{A}^{\text{max}}(t)w_{0}$ and $S_{A}^{\text{min}}(t)v_{0}$ since from Lemma 4.2 such extremal solutions exist and satisfy $S_{A}^{\text{min}}(t)v_{0}\leq u(t) \leq S_{A}^{\text{max}}(t)w_{0}$, $t \in [0,+\infty [$. If $w(t)=S_{A}^{\text{max}}(t)w_{0}$ we have $$w(t)-w_{0}=-\int_{0}^{t}Aw(x)dx. \eqno (5.11)$$ Consequently $t\to w(t)$ is decreasing because from the absorbent property $w'(t)=-Aw(t)\leq 0$. In an analogous way $v(t)=S_{A}^{\text{min}}(t)u$ is increasing because $v'(t)=-Av(t)\leq 0$ for each $t\in [0,+\infty [$. Hence we get $$v_{0}\leq v(t) \leq w(t)\leq w_{0}.$$ Then $l_{1}=\lim_{t\to +\infty}w(t)$ and $l_{2}=\lim_{t\to +\infty}v(t)$ exist in $\mathbb{R}^N$. Hence, according to (5.11), $\int_{0}^{+\infty}Av(\tau)d\tau$ and $\int_{0}^{+\infty}Aw(\tau)d\tau$ converge. Since $\lim_{t\to +\infty}Aw(t)=Al_{1}$ and $\lim_{t\to +\infty}Av(t)=Al_{2}$, we have necessarily $Al_{1}=Al_{2}=0$. So by hypothesis $\lim_{t\to +\infty}w(t)=\lim_{t\to +\infty}v(t)=u_{\infty}$. \paragraph{Corollary 6.3} {\sl For the operator $A_{0}$ introduced in (5.1), we have $$\lim_{t\to \infty}S_{A_{0}}(t)(u_{0})= \begin{pmatrix} 0\cr 0\end{pmatrix}.$$} \paragraph{Proof.} We can show that $A_{0}u=\begin{pmatrix} 0\cr 0\end{pmatrix}$ holds if and only if $u=\begin{pmatrix} 0\cr 0\end{pmatrix}$. Moreover $A_{0}$ is absorbent. Indeed, with the notation of Lemma 4.4, let $u_{0}\in A_{0}^{+}$ (resp. $A_{0}^{-}$) and $u_{n}(t)=J_{T/n}^i(u_{0})$ for $(i-1)T/ n