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\markboth{\hfil Periodic and almost periodic solutions \hfil EJDE--2000/24}
{EJDE--2000/24\hfil E. Hanebaly \& B. Marzouki \hfil}
\begin{document}
\title{\vspace{-1in}\parbox{\linewidth}{\footnotesize\noindent
{\sc  Electronic Journal of Differential Equations},
Vol.~{\bf 2000}(2000), No.~24, pp.~1--16. \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} \\
%
 Periodic and almost periodic solutions for multi-valued differential
 equations in Banach spaces
\thanks{ {\em Mathematics Subject Classifications:} 34A60, 34C25, 34C27, 47H10.
\hfil\break\indent
{\em Key words and phrases:} Multi-valued differential equation, Hyper-accretive,
 \hfil\break\indent
Almost periodicity, Banach space.
\hfil\break\indent
\copyright 2000 Southwest Texas State University  and University of
North Texas. \hfil\break\indent
Submitted July 16, 1999. Published March 30, 2000.} }
\date{}
%
\author{ E. Hanebaly \& B. Marzouki }
\maketitle

\begin{abstract}
 It is known that for $\omega$-periodic differential
 equations of monotonous type, in uniformly convex Banach spaces,
 the existence of a bounded solution on ${\mathbb R}^+$ is equivalent to the
 existence of an $\omega$-periodic solution (see Haraux \cite{B.H} and
 Hanebaly \cite{H1,H4}). It is also known that if the Banach space
 is strictly convex and the equation is almost periodic and of
 monotonous type, then the existence of a continuous solution with
 a precompact range is equivalent to the existence of an almost
 periodic solution (see Hanebaly \cite{H2} ).
 In this note we want to generalize the results above for multi-valued
 differential equations.
\end{abstract}

\newtheorem{theorem}{Theorem}
\newtheorem{lemma}{Lemma}

\section{Preliminaries}

Let $X$ and $Y$ be Banach spaces, and $2^Y$ denote the collection of
subsets of $Y$. For a multi-valued map $F: X\to 2^Y$ we define the following
conditions:

 $F$ is  {\bf upper semi-continuous} (u.s.c.) in
$X$ if for every  $x_0$ in $X$ and every open set $G\subset Y$
with $Fx_0\subset G$ there exists a neighborhood $U$ of $x_0$ such
that $Fx_0\subset G$ for all $x\in U$.
In practice $F$ is u.s.c. at $x_0$ means that
$Fx\subset Fx_0 + B_\varepsilon (0)$ for all $x$ sufficiently close to
$x_0$ and for $\varepsilon$ sufficiently small.

 $F$ is {\bf bounding} if it maps bounded
subsets of $X$ into bounded subsets of $Y$.

 $F$ is {\bf dissipative} if  $X=Y$ and
$$\langle Fx-Fy,x-y \rangle_- \leq 0 \quad
 \forall x\in X , \forall y\in X\,.$$
This implies that for all $x_1\in Fx$ and all $y_1\in Fy$,
$$\langle x_1-y_1,x-y \rangle_-\leq 0\,,$$
where the lower semi-inner product on $X$ introduced by Lumer \cite{L}
is defined as
$$\langle x,y \rangle_-=\|y\|\lim_{h\to 0^-}{{\|y+hx\|-\|y\|}\over h}\,.
$$


$F$ is {\bf accretive} if
$\langle Fx-Fy,x-y \rangle_+ \geq 0 $ where the upper semi-inner product on $X$
is defined as
$$ \langle x,y \rangle_+=\|y\|\lim_{h\to 0^+}{{\|y+hx\|-\|y\|}\over h}\,.$$

We denote by $\rightharpoonup$ the convergence for the weak topology
$\sigma (X,X^*)$.
Recall that $x : J\subset {\mathbb R}\to X$ is said
to be absolutely continuous (a.c. for short) if for each
$\varepsilon > 0$ there is $\delta > 0$ such that
$\sum\|x(t_i)-x(s_i)\|\leq \varepsilon$ whenever the finitely many
intervals $[s_i,t_i] \subset J$ do not overlap and $\sum |
t_i-s_i|\leq \delta$. In particular every Lipschitzean map
is a.c. When $X$ is of finite dimension it is known that $x$ is
a.c. if and only if $x$ is differentiable almost everywhere (a.e.
for short) and $x'\in L^1(J,X)$, but if $X$ is of infinite
dimension and $X$ is not reflexive, then an a.c. function need not
be differentiable at any point (see e.g Deimling \cite{D} p.138).

By a solution of the Cauchy problem
\begin{equation}
x'\in F(t,x)\,;\quad x(t_0)=x_0
\end{equation}
in some interval $I$ (with $t_0\in I$),  we mean a  continuous function on $I$,
a.c. in every compact subset of $I$, differentiable a.e., and that
satisfies (1)  a.e. on $I$.

The collection non-empty compact convex subsets of $X$ will be denoted
by $CV(X)$.

\section{Boundedness and periodicity of solutions}

We begin by giving a result concerning the existence of a global
solutions. Let $(X,\|.\|)$ be a real reflexive Banach space.
Consider the multi-valued Cauchy problem
\begin{eqnarray}
&x'(t) \in   F(t,x(t))& \label{2}\\
&x(0)  = x_0\,,& \label{3}
\end{eqnarray}
where $F:{\mathbb R}^+\times X \to CV(X)$ is u.s.c. and
bounding.

\begin{theorem} \label{thm1}
If for all $(t,x,y)\in {\mathbb R}^+\times X\times X$, $\langle
F(t,x)-F(t,y),x-y \rangle_-\leq 0$,  then the Cauchy problem
(2)-(3) has a unique solution defined on ${\mathbb R}^+$.
\end{theorem}

\paragraph{Remark.}
This theorem is well known for the inclusion of type $$x'\in -Ax+f(t)$$
where A is a multi-valued maximal monotone operator on a Hilbert space and
$f$ is a uni-valued map (see Brezis \cite{B}).

\paragraph{Proof of Theorem 1.}
Since $F$ is u.s.c. with convex values, by the approximate selection
theorem (see Cellina \cite{A.C}) for each $n\geq 0$ there exists a
locally lipschitzean map  $f_n:{\mathbb R}^+\times X \to X$ such that
$$
f_n(t,x)\in F({\mathbb R}^+\times X\cap B_{1/n}(t,x))+B_{1\over
n}(0)  \quad \forall (t,x)\in {\mathbb R}^+\times X\,,$$
where  $B_{1/n}(t,x)$  is a ball in  ${\mathbb R}^+\times X$
and  $ B_{1/n}(0)$  is a ball in $X$.  Since $F$ is
u.s.c. at $(t,x)$, for each $\varepsilon > 0$ there exists $\delta > 0$
such that
$$   F({\mathbb R}^+\times X\cap B_\delta (t,x))
\subset F(t,x)+B_\varepsilon (0)\,.$$
Then for $n$ large we can choose
$\delta$ such that $ B_{1/n}(t,x)\subset B_\delta
(t,x)$ and
 $$ F({\mathbb R}^+\times X\cap B_{1/n}(t,x))\subset F(t,x)+
B_\varepsilon (0)\,.
$$
Consequently, for $\varepsilon =1/m$ with $m\geq n$ we obtain
$$f_n(t,x)\in F(t,x)+B_{1/n} (0)+B_{1/n}(0)\subset F(t,x)+B_{2/n}(0)\,.
$$
Now we show that for any $a > 0$ the uni-valued Cauchy
problem
\begin{eqnarray}
&x'(t) = f_n(t,x(t))& \label{4} \\
&x(0) = x_0 &\label{5}
\end{eqnarray}
has a unique solution $x_n$ on $[0,a]$ and the sequence $x_n$
converges uniformly to the solution of the Cauchy problem (2)-(3).

 Consider $f_n$ from $[0,a]\times X$ to $X$, then $f_n$
satisfies\\
i) $f_n$ is continuous and locally lipschitzean with
respect to $x$.\\
ii) $\langle f_n(t,x)-f_n(t,y),x-y \rangle_-\leq
{4\over n}\|x-y\|$.\\
For proving ii), we take  $f_n(t,x)\in F(t,x)+B_{2/n}(0)$  and
$f_n(t,y)\in F(t,y)+B_{2/n}(0)$,  so  that $f_n(t,x)=a+\alpha_n$
and  $f_n(t,y)=b+\beta_n$  with  $a\in F(t,x),
b\in F(t,y)$  and  $\alpha_n,\beta_n \in B_{2/n}(0)$. Then
\begin{eqnarray*}
\langle f_n(t,x)-f_n(t,y),x-y \rangle_-&=&\langle
a+\alpha_n-b-\beta_n,x-y \rangle_- \\ & \leq &  \langle a-b,x-y
\rangle_-+\langle \alpha_n-\beta_n,x-y \rangle_+ \\
                    & \leq &  \langle \alpha_n-\beta_n,x-y \rangle_- \\
                    & \leq &  \|\alpha_n-\beta_n\|\|x-y\|\\
                    & \leq & {4\over n}\|x-y\|.
\end{eqnarray*}

It is well known that by i) the uni-valued Cauchy problem (4)-(5) has a
unique local solution $x_n$, and that by ii) this solution can
be extended on $[0,a]$. This statement is proven by the standard procedure
of bounding the derivative of $x_n$.

Taking $y=0$ in ii) , we obtain
$$\langle f_n(t,x)-f_n(t,0),x \rangle_-\leq {4\over n}\|x\|.$$
Therefore,
\begin{eqnarray*}
\langle x'_n(t),x_n(t)\rangle_-&=&\langle
f_n(t,x_n(t))-f_n(t,0)+f_n(t,0),x_n(t)\rangle_-\\
&\leq &\langle
f_n(t,x_n(t))-f_n(t,0),x_n(t)\rangle_-+\langle
f_n(t,0),x_n(t)\rangle_+\\ &\leq &{4\over
n}\|x_n(t)\|+\|f_n(t,0)\|\|x_n(t)\|\\
&\leq &( 1+\sup\limits_{t\in
[0,a]}\|f_n(t,0)\| )\|x_n(t)\|.
\end{eqnarray*}
We deduce that (see appendix II)
$$D^-\|x_n(t)\|\leq
1+\sup\limits_{t\in [0,a]}\|f_n(t,0)\|=k_n $$ with $k_n$  a
constant which does not depend on $t$. This follows because there is
$t^n_0 \in [0,a]$ such that
$$\sup\limits_{t\in [0,a]}\|f_n(t,0)\|=\|f_n(t^n_0,0)\|.
$$
consequently, we have a sequence $x_n \in C([0,a],X)$ that satisfies
\begin{equation}
x'_n(t)\in F(t,x_n(t))+B_{2/n}(0)\,. \label{6}
\end{equation}
Next we show that $x_n$ is a Cauchy sequence. Let
$\Phi_{n,m} (t)=\|x_n(t)-x_m(t)\|$.
Then  $\Phi_{n,m} (0)=0$  and using the same
technique as for proving ii) we deduce that
\begin{eqnarray*}
\Phi_{n,m} (t)D^-\Phi_{n,m} (t)& = & \langle
x'_n(t)-x'_m(t),x_n(t)-x_m(t)\rangle_-\\ & \leq &  ({2\over
n}+{2\over m})\Phi_{n,m} (t)\,.
\end{eqnarray*}
Therefore,  $\Phi_{n,m} (t) \leq ({2\over n}+{2\over
m})a$ and then $$\sup\limits_{t\in [0,a]}\|x_n(t)-x_m(t)\|
\to 0 \quad \mbox{as }  n,m\to +\infty$$
Let $x$ be the limit
of $x_n$. Then we have in particular  $x(0)=x_0$, now we
have to show that $x$ is a.e. differentiable and satisfies
$$x'(t)\in F(t,x(t)) \quad \mbox{a.e. in } [0,a].$$
Since $F$ is u.s.c. and  $x_n\to x$  uniformly on $[0,a]$,
we deduce that for $n$ large,
$$F(t,x_n(t))\subset F(t,x(t)) +
B_1(0)\,.$$
Since $F$ is bounding, by (6) we have
$\|x'_n(t)\|\leq c$  uniformly on $[0,a]$ for some $c > 0$.

Put  $J=[0,a]$,  then we have  $x'_n\in L^\infty
(J,X)\subset L^2(J,X)$. Since $L^2(J,X)$ is reflexive (because
$X$ is reflexive), there is a subsequence
(which we denote by the same symbol)
such that  $x'_n\rightharpoonup y\in L^2(J,X)$  so
\begin{eqnarray*}
x_n(t)&=&x_0+\int_{0}^{t}x'_n(s)\,ds
= x_0+\int_{J}\chi_{[0,t]}(s)x'_n(s)\,ds \\
 &\rightharpoonup&  x_0+\int_{J}\chi_{[0,t]}(s)y(s)\,ds =
x_0+\int_{0}^{t}y(s)ds\,.
\end{eqnarray*}
Since $x_n(t)\to x(t)$, it follows that $x_n(t)\rightharpoonup
x(t)$. Consequently $$x(t)=x_0+\int_{0}^{t}y(s)ds \quad \mbox
{and} \quad x'(t)=y(t) \quad \mbox{ a.e. in}  J.$$ We deduce that
$x'_n \rightharpoonup x'$  in $L^2(J,X)$  for the weak topology
$\sigma(L^2(J,X),L^2 (J,X^*)).$  Let $\varepsilon > 0$ and put
$$A_\varepsilon =\big \{ z\in L^2(J,X)  :  z(t)\in F(t,x(t)) +
\overline{B}_\varepsilon(0) \mbox{ a.e. } \big\}$$ Then
$A_\varepsilon$ is nonempty (because $x_n(t)\to x(t) $ and $F$ is
u.s.c., so $x'_n\in A_\varepsilon$ for n large), $A_\varepsilon$
is closed and convex, hence $A_\varepsilon$ is weakly closed.
Since $x'_n\in A_\varepsilon$ and  $x'_n\rightharpoonup x'$
 we deduce that
 $$x'(t)\in \overline{F(t,x(t))}=F(t,x(t))
\mbox{ a.e.}$$
 So $x$ is a solution of the Cauchy problem (2)-(3).
Since $a > 0$ is arbitrary we deduce that the sequence $x_n$
converges in the Banach space  $C({\mathbb R}^+,X)$ equipped  with the
topology of uniform convergence in compact subsets of
${\mathbb R}^+$.

That $x$ is unique follows from the dissipativeness of $F$.
Indeed let $x$ and $y$ be two solutions of the Cauchy problem (2)-(3),
then we have
$$\langle x'(t) - y'(t),x(t) - y(t) \rangle_-
\leq 0 \quad \mbox{and}\quad  {{1}\over 2}D^-\|x(t) - y(t)\|^2 \leq
0\,.$$ Hence the map $t \mapsto \|x(t) - y(t)\|^2$  is
non increasing, and consequently
\begin{equation} \label{7}
\|x(t) - y(t)\| \leq \|x(0) - y(0)\|.
\end{equation}
Now we present a result that gives us the
relationship between the existence of bounded solution  and
the existence of an $\omega$-periodic solution of (2) when $F$ is
$\omega$-periodic. Observe that under the hypothesis of Theorem 1
the condition: There exists a positive $R$ such that
$$<F(t,x),x>_-\leq 0 \quad\mbox{for }  \|x\| > R$$
ensures the existence of a bounded solution on $[0,+\infty[$
(see Browder \cite{Br} and Hanebaly \cite{H2}).

\begin{theorem} Under the hypothesis of Theorem 1,
assuming that $X$ is uniformly convex, and
$F(t+\omega,x)=F(t,x)$ ($\omega > 0$), the equation
(2) has an $\omega$-periodic solution if and only if it has a
bounded solution on $[0,+\infty[$.
\end{theorem}

\paragraph{Proof.} The necessity condition is obvious
because a continuous periodic map is bounded. Conversely we
consider the Poincar\'e map  $P : X \to X$
defined by  $Px_0 = x(\omega)$  where $x_0$ is given in
$X$ and $x$ is a solution of (2) which satisfies  $x(0) =
x_0$. The map P is well defined because of the uniqueness of solutions
for the Cauchy problem (2)-(3).  Now let $x$ be the
solution of (2) which is bounded on $[0,+\infty [$ and put
\begin{eqnarray*}
x_1 &=& Px_0 = x(\omega)\\
x_2 &=& Px_1 = x(2\omega)\\ &\vdots& \\
x_n &=& Px_{n-1} = x(n\omega)
\end{eqnarray*}
Note that the solution $x$ is bounded, so the sequence $x_n$ is bounded,
and that $P$ is non-expansive. Indeed, let $y$ and $z$ be two solutions
of (2) such that $y(0) = y_0$ and $z(0) = z_0$  so by dissipativeness of
$F$ and the inequality (7) we have
 $$\|y(t) - z(t)\| \leq \|y(0) - z(0)\| = \|y_0 - z_0\|. $$
Taking $t=\omega$ we deduce that
$$\|Py_0 - Pz_0\| \leq \|y_0 - z_0\|.$$
So by the
Browder-Petryshyn's fixed point theorem (see Petryshyn \cite{B.P}),
 $P$ has a fixed  point. So there is a solution $\widetilde{x}$ of
(2) which satisfies  $\widetilde{x}(0) =
\widetilde{x}(\omega)$  and $\widetilde{x}$ is
$\omega$-periodic. Indeed, put   $\widetilde{y}(t) =
\widetilde{x}(t+\omega)$  then $$\widetilde{y'}(t) =
\widetilde{x'}(t+\omega)\in F(t+\omega,\widetilde{x}(t+\omega)) =
F(t,\widetilde{y}(t))\,.$$
Now since $\widetilde{y}(0) = \widetilde{x}(\omega) = \widetilde{x}(0)$,
  by (7) we deduce that
$$\widetilde{x}(t) = \widetilde{y}(t) = \widetilde{x}(t+\omega)$$
hence $\widetilde{x}$ is $\omega$-periodic.\hfill$\diamondsuit$

\paragraph{Remark.}  Let $x$ be an $\omega$-periodic solution of (2),
if $y$ is another $\omega$-periodic solution (respectively an
$T$-periodic solution with $\frac{\omega}{T}\notin {\mathbb Q}$)
then $\|x(t)-y(t)\|$ is constant for all $t\in {\mathbb R}^+$.
From the dissipativeness of $F$ it follows that the map $t\mapsto
\|x(t)-y(t)\|$  is decreasing. Since it is continuous and periodic
(respectively almost-periodic) we conclude that it is constant.

\paragraph{Example.} Consider $({\mathbb R}^n,\|.\|)$ with $\|.\|$ the
Euclidean norm and $\langle .,. \rangle$ the associated inner
product. We consider the differential equation
$$x'+x\|x\|^{\alpha}+\beta \mathop{\rm sgn}(x) = f(t)$$
where $\alpha \geq 0$, $\beta \geq 0$,
$f :{\mathbb R}^+ \to {\mathbb R}^n$ is continuous and $\omega$-periodic,
and
$$ \mathop{\rm sgn}(x) = \left \{
\begin{array}{ll} \frac{x}{\|x\|} &\mbox {if  $x\neq 0$} \\[5pt]
 \overline{B}(0,1) &\mbox{if  $x = 0$}
\end{array}\right.
$$
Then the above equation becomes  $x'\in F(t,x)$ where
 $F(t,x)=f(t)-x\|x\|^{\alpha}-\beta \mathop{\rm sgn}(x)$  is a
bounding multi-valued map with compact and convex values.
To conclude that the inclusion has an $\omega$-periodic
solution, we have to prove the following lemma.

\begin{lemma}  1) F is upper semi-continuous on ${\mathbb R}^+
\times {\mathbb R}^n$.\\
 2) There exist a positive  $c_{\alpha}$ and $r_{\alpha}\geq 2$ such that
for all $(t,x)\in {\mathbb R}^+ \times {\mathbb R}^n $,
$$\langle F(t,x)-F(t,y),x-y \rangle\leq
-c_{\alpha}\|x-y\|^{r_{\alpha}}\,.$$ In particular $F$
is dissipative with respect to $x$ \\
3) Every solution of the inclusion  $x'\in F(t,x)$  is bounded.
\end{lemma}

\paragraph{Proof of 1)} We have to show that for every closed
$A\subset {\mathbb R}^n$, the set
$$F^{-1}(A)=\{ (t,x)\in {\mathbb R}^+\times {\mathbb R}^n  :
 F(t,x)\cap A\neq \emptyset \}$$
is closed in ${\mathbb R}^+\times{\mathbb R}^n$.
Let   $(t_n,x_n)\in {\mathbb R}^+\times {\mathbb R}^n$ be such that
  $(t_n,x_n)\to (t,x)$  and  $F(t_n,x_n)\cap A\neq\emptyset$.
 We have to show that  $F(t,x)\cap A\neq\emptyset.$ Let
$y_n\in F(t_n,x_n)\cap A$,  then
$y_n=f(t_n)-x_n\|x_n\|^{\alpha}-\beta\gamma_n$  with
$\|\gamma_n\|\leq 1$, $\gamma_n$ has a subsequence (which we
denote by the same) such that  $\gamma_n\to \gamma$
with  $(\|\gamma \|\leq 1)$, so
$$y_n=f(t_n)-x_n\|x_n\|^{\alpha}-\beta\gamma_n\to
y:=f(t)-x\|x\|^{\alpha}-\beta\gamma\in F(t,x)\cap A.$$
Hence $F(t,x)\cap A\neq \emptyset$  and F is upper semi-continuous on
 ${\mathbb R}^+\times {\mathbb R}^n$.

\paragraph{Proof of 2)}  It is easy to see that  for all $x,y\in
{\mathbb R}^n$,
$\langle \mathop{\rm sgn}(x)-\mathop{\rm sgn}(y),x-y \rangle\geq
0$. Now let  $x,y\in {\mathbb R}^n$,  then
\begin{eqnarray*}
\lefteqn{\langle x\|x\|^{\alpha}-y\|y\|^{\alpha},x-y \rangle} \\
&=&\langle x\|x\|^{\alpha}-y\|x\|^{\alpha} +x\|y\|^{\alpha}-
  y\|y\|^{\alpha} +y\|x\|^{\alpha}-x\|y\|^{\alpha},x-y \rangle\\
&=&\|x-y\|^2(\|x\|^{\alpha}+\|y\|^{\alpha})+\langle
y\|x\|^{\alpha}-x\|y\|^{\alpha},x-y \rangle\\
&=&\frac{1}{2}\|x-y\|^2(\|x\|^{\alpha}+\|y\|^{\alpha})
+\frac{1}{2}\langle (x+y)\|x\|^{\alpha}-(x+y)\|y\|^{\alpha},x-y\rangle\\
&=& \frac{1}{2}\|x-y\|^2(\|x\|^{\alpha}+\|y\|^{\alpha})+
\frac{1}{2}(\|x\|^{\alpha}-\|y\|^{\alpha})(\|x\|^{2}-\|y\|^{2})\\
&\geq& \frac{1}{2}\|x-y\|^2(\|x\|^{\alpha}+\|y\|^{\alpha})
\end{eqnarray*}
The last inequality comes from the fact that the map
$\varphi(t)=t^{\alpha}$ is increasing on ${\mathbb R}^+$, so
$(\|x\|^{\alpha}-\|y\|^{\alpha})(\|x\|-\|y\|)\geq 0$. Hence for
$\alpha =0$,
$$\langle x\|x\|^{\alpha}-y\|y\|^{\alpha},x-y \rangle\geq \|x-y\|^2\,.$$
If $0<\alpha \leq 1$ then
$\|x\|^{\alpha}+\|y\|^{\alpha}\geq (\|x\|+\|y\|)^{\alpha}\geq
\|x-y\|^{\alpha}$,  (because the map
$\varphi(t)=1+t^{\alpha}-(1+t)^{\alpha}$  is positive on
${\mathbb R}^+$), so
$$\langle x\|x\|^{\alpha}-y\|y\|^{\alpha},x-y
\rangle\geq \frac{1}{2}\|x-y\|^{\alpha +2}
$$
If $\alpha \geq 1$ then the map  $\varphi(t)=t^{\alpha}$  is convex on
${\mathbb R}^+$, so
$$\|x\|^{\alpha}+\|y\|^{\alpha}\geq \frac{1}{2^{\alpha
-1}}(\|x\|+\|y\|)^{\alpha}\geq \frac{1}{2^{\alpha
-1}}\|x-y\|^{\alpha}\,.$$
Hence
$$\langle x\|x\|^{\alpha}-y\|y\|^{\alpha},x-y \rangle\geq
\frac{1}{2^\alpha}\|x-y\|^{\alpha +2}$$

\paragraph{Proof of 3)}  From 2) we deduce that
$$\langle F(t,x)-F(t,0),x \rangle\leq
-c_{\alpha}\|x\|^{r_{\alpha}}\,,$$
where
$$\begin{array}{ll}
 c_\alpha =1  \mbox{ and } r_\alpha =2
&\mbox { if  $\alpha = 0$} \\
 c_\alpha = 1/2 \mbox{ and } r_\alpha =\alpha +2
& \mbox { if  $0<\alpha\leq 1$} \\
 c_\alpha =1/ 2^\alpha \mbox{ and }  r_\alpha =\alpha +2
& \mbox { if  $\alpha\geq 1$}
\end{array}
$$
Let $x$ be a solution of  $x'\in F(t,x)$,  and
let $a\in F(t,0)$. Then  $a=f(t)-\beta\gamma$,
$(\|\gamma\|\leq 1)$,  and we have
\begin{eqnarray*}
\langle x'(t),x(t) \rangle&=&\langle x'(t)-a+a,x(t) \rangle\\
&=&\langle x'(t)-a,x(t) \rangle+\langle a,x(t) \rangle\\ &\leq&
-c_{\alpha}\|x(t)\|^{r_{\alpha}} + ( M+\beta )\|x(t)\|\,,
\end{eqnarray*}
where  $M=\sup\limits_{t\in {\mathbb R}}\|f(t)\|$. Therefore,
$$\frac{d}{2dt}\|x(t)\|^2\leq 0  \quad\mbox{for }
\|x(t)\|\geq \big( \frac{M+\beta}{c_{\alpha}} \big
)^{1/(r_{\alpha}-1)}\,.$$
Consequently
$$\sup\limits_{t\in {\mathbb R}}\|x(t)\|\leq \max \bigl [ \|x(0)\|,
 (\frac{M+\beta}{c_{\alpha}} )^{1/(r_{\alpha}-1)} \bigr ]$$
because the map  $t\mapsto \|x(t)\|^2$  is
decreasing outside
$B\big(0,(\frac{M+\beta}{c_{\alpha}})^{1/(r_{\alpha}-1)}\big )$.

\section {Almost periodic solutions}

 Let $(E,\|.\|)$ be a uniformly convex Banach space with $E^*$
uniformly convex. We consider the problem
\begin{equation} \label{8}
x'(t) \in -Ax(t)+f(t)\,,
\end{equation}
where  $f:{\mathbb R}\to E$  is a continuous
almost periodic function (see appendix I for the definition of
almost periodicity) and  $A:E\to 2^E\setminus
\emptyset$  is a hyper-accretive multi-valued map which means
that  for all $\lambda > 0$,  Im$(I+\lambda A)=E$  and
 $ \langle Ax-Ay , x-y \rangle_+ \geq 0$ for all $x,y\in E$.

\begin{theorem} Problem (8) has a solution on
$[t_0,+\infty[$ ($t_0\in {\mathbb R}$), which is uniformly continuous
with precompact range if and only if it has a weak almost periodic
solution.
\end{theorem}

\paragraph{Remark.}
Since a continuous almost periodic map is uniformly continuous with
precompact range, it is convenient to relate the existence of
a solution to that of uniformly continuous with the precompact range.

\paragraph{Proof of Theorem 3.}
The proof will be divided into four steps.

\paragraph{Step 1.} The Cauchy problem
\begin{eqnarray}
&x'(t) \in   -Ax(t) +f(t) &  \label{9}\\
& x(t_0)  = x_0& \label{10}
\end{eqnarray}
has a unique weak solution on $ [t_0,+\infty[$. (weak solution
means that there are sequences $x_n$ and $f_n$ where $x_n$ is a
strong  solution and $x_n\to x$ uniformly in every compact subset
J of $[t_0,+\infty[$ and  $f_n \to f$ in $L^1(J,E)$ ). Indeed,
Since $E$ and $E^*$ are uniformly convex, the Cauchy  problem
\begin{eqnarray*}
&x'(t) \in  -Ax(t)&   \\
&x(t_0)  = x_0&
\end{eqnarray*}
has a unique strong solution on $[t_0,+\infty[$ (see Deimling
\cite{D}). Since $f$ is almost periodic, $f \in L^1(J,E)$
for every compact $J\subset [t_0,+\infty[$, with $t_0\in J$, so
there is a sequence $f_n$ of stairs functions which converges
uniformly to $f$, hence  $f_n\to f$ in $L^1(J,E)$. On
the other hand for every $f_n$ there is $x_n$ such that
\begin{eqnarray*}
&x_n'(t) \in  -Ax_n(t) +f_n(t) & \\
&x_n(t_0)  =  x_0\,.&
\end{eqnarray*}
Because if $g$ is a stair function defined on
$a=b_0<b_1<...<b_p=T$ ( $T > a$ )  by $g(t)=y_i$ on
$[b_{i-1} , b_i[$ the Cauchy problem
\begin{eqnarray*}
&x'(t) \in -Ax(t) + g(t)&  \\
&x(t_0)  = x_0 &
\end{eqnarray*}
has also a unique strong solution $x$ defined by
$x(t)=S_i(t-b_{i-1}).x(b_{i-1})$  for $t\in [b_{i-1} , b_i]$ and
$x(t_0)=x_0$ where $S_i(t)$ is the semigroup generated by the
hyper-accretive operator $-( A - y_i )$.

Let us show that $(x_n)$ is a Cauchy sequence in the
Banach space $C([t_0,+\infty[,E)$ equipped with the  topology of
uniformly convergence in compact subsets.
 Since $-A$ is dissipative, we have
$$\langle x_n'(t)-f_n(t)-x'_p(t)+f_p(t) , x_n(t)-x_p(t)
 \rangle_-\leq 0\,,$$
$E^*$ is uniformly convex, $\langle .,. \rangle_-= \langle
.,. \rangle _+$, and $\langle .,. \rangle_-$ is linear on the
first argument. Then
\begin{eqnarray*}
\lefteqn{ \frac{d^-}{2dt} \|x_n(t)-x_p(t)\|^2}\\
&=&\langle x'_n(t)-x'_p(t) , x_n(t)-x_p(t) \rangle_-  \\
& = &\langle x'_n(t)-f_n(t)-x'_p(t)+f_p(t)+f_n(t)-f_p(t) ,
x_n(t) -x_p(t) \rangle_- \\
& = & \langle x'_n(t)-f_n(t)-x'_p(t)+f_p(t) , x_n(t)-x_p(t) \rangle_- \\
&&+  \langle f_n(t)-f_p(t) , x_n(t)-x_p(t) \rangle_-\\
& \leq & \langle f_n(t)-f_p(t) , x_n(t)-x_p(t) \rangle_-  \\
&\leq  & \|f_n(t)-f_p(t)\|\|x_n(t)-x_p(t)\|
\end{eqnarray*}
Hence
\begin{eqnarray*}
\|x_n(t)-x_p(t)\|&\leq&
\|x_n(t_0)-x_p(t_0)\|+\int_{t_0} ^{t} \|f_n(s)-f_p(s)\|ds \\
&=& \int_{t_0}^{t}\|f_n(s)-f_p(s)\|ds \to 0 \quad \mbox{as }
n,p\to +\infty\,.
\end{eqnarray*}
Without loss of generality, we can assume that the
Cauchy problem (9)-(10) has a strong solution. Let
$x:[t_0,+\infty[ \to E$  be the uniformly
continuous solution of the Cauchy problem (9)-(10) with
$x([t_0,+\infty [)$  precompact. Since  $f$ is almost periodic,
there is  $t_n\to +\infty$  such that  $f(t+t_n)\to
f(t)$  uniformly on ${\mathbb R}$ (see appendix I). Consider the
sequences of translated functions
$$x_n(t)=x(t+t_n) \quad\mbox{and}\quad  f_n(t)=f(t+t_n)$$
which are defined on the real
interval $[a,+\infty[$ when $n\geq n(a)$. Since
$x([t_0,+\infty[)$  is precompact, we deduce that  $\{
x_n(t), t\geq a, n\geq n(a) \}$ is also precompact.
On the other hand  that $\{ x_n ,  n\geq n(a) \}$  is
equi-continuous follows from the following lemma which is easy to
proof.

\begin{lemma} Let E be a Banach space, $J\subset {\mathbb R}$ be
an interval and ${\cal M}$ a bounded subset of the Banach space
$C_b(J,E)$ of continuous bounded functions. Then ${\cal M}$ is
uniformly equi-continuous if and only if the mapping
$(\psi,t) \mapsto \psi(t)$  of  ${\cal M}\times
J\subset C_b(J,E)\times {\mathbb R}$  into E is uniformly continuous
on ${\cal M}\times J$.
\end{lemma}

Now applying  Ascoli's theorem in the intervals $[-N,N]$,
$N=1,2,\dots$ and using the diagonal procedure (see Zaidman
\cite{Z}) it is possible to find a subsequence which converges
uniformly in every compact subset $J$ of ${\mathbb R}$. But $f_n$
is almost periodic, so $f_n\to f$ in  $L^1(J,E)$. Therefore, we
obtain a weak solution $x^*$ of (8) defined on ${\mathbb R}$ which
is uniformly  continuous with range contained in the closure of
$x([t_0,+\infty [)$,  hence with precompact range.

\paragraph{Step 2.} Put  $K_0=\overline {Co} ( x^*({\mathbb R}) )$,
so that $K_0$ is a compact convex subset of $E$.
Let $$\Omega =\big \{ x: {\mathbb R}\to E | x({\mathbb R})\subset K_0
\big \}$$
with $x$ a uniformly continuous solution of (8)  and
 $ J:\Omega \to {\mathbb R}^+$  defined by
$Jx=\sup\limits_{t\in {\mathbb R}} \|x(t)\|$.
Put  $\mu=\inf\limits_{x\in \Omega }Jx$,  so there is $x_n\in \Omega $
such that  $J(x_n)\to \mu $. By Lemma 2 and Ascoli's
theorem there is a subsequence of $x_n$ which converges uniformly
in every compact subset of ${\mathbb R}$, let $\widetilde {x}$ be this
limit, then $\widetilde {x}\in \Omega $ and $J\widetilde
{x}=\mu$.

\paragraph{Step 3.} We show that $\widetilde{x}$ is
unique. Assume that there are $x_1$ and $x_2$ in $\Omega$ such
that  $Jx_1=Jx_2=\mu $.  Since $f$ is almost periodic,
there is  $t_n\to -\infty$  such that  $f(t+t_n)\to f(t)$
 uniformly on ${\mathbb R}$. By Ascoli's theorem, we can extract
from $t_n$ a subsequence (which we denote by the same symbol) such that
$x_1(t+t_n)$ and $x_2(t+t_n)$ converge  uniformly in every compact
subset of ${\mathbb R}$. Let
$$y_1=\lim  x_1(t+t_n)\quad  \mbox{and} \quad
  y_2=\lim x_2(t+t_n\,.)$$
Then $y_1$ and $y_2$ are weak solutions of (8), and
$y_1,y_2\in \Omega$  with $J(y_1)=J(y_2)=\mu$. Now since
$$x'_1(t+t_n)\in -Ax_1(t+t_n)+f(t+t_n)$$
 and
$$x'_2(t+t_n)\in -Ax_2(t+t_n)+f(t+t_n)$$
and $-A$ is dissipative, we deduce that
$$\langle x'_1(t+t_n)-x'_2(t+t_n),x_1(t+t_n)-x_2(t+t_n)
\rangle_-\leq 0$$ So
$$\frac{d^-}{2dt}\|x_1(t+t_n)-x_2(t+t_n)\|^2\leq 0$$ Consequently
the map $t\longmapsto \|x_1(t+t_n)-x_2(t+t_n)\|$  is
non increasing. Since  $x_i({\mathbb R})\subset K_0$ for
i=1,2, we deduce that
\begin{eqnarray}
\|y_1(t)-y_2(t)\|&=&\lim\limits_{n\to
+\infty}\|x_1(t+t_n)-x_2(t+t_n)\| \nonumber\\
&=&\lim\limits_{\tau\to -\infty} \|x_1(\tau )-x_2(\tau )\| \label{11}\\
&=&\sup\limits_{t\in {\mathbb R}}\|x_1(t)-x_2(t)\| \nonumber\\
&=&\mbox{a constant} \nonumber
\end{eqnarray}
To continue, we need the following lemma.

\begin{lemma} Let $E$ be a strictly convex Banach space,
$C$ a closed convex subset of $E$. Let  $T:C\to C$ be a
non expansive map and $x_0,y_0$ in $C$ such that
$$\|Tx_0-Ty_0\|=\|x_0-y_0\|\,.$$
Then $$T({{x_0+y_0}\over 2})={{Tx_0+Ty_0}\over 2}.$$
\end{lemma}

Let the operator  $T_t:E\to E$ be defined by
 by  $T_tx(0)=x(t)$ where $x(.)$ is a weak solution
of (8). Then $T_ty_1(0)=y_1(t)$  and
$T_ty_2(0)=y_2(t)$ where $y_1$ and $y_2$  are in $\Omega$. By
(11),
$$\|T_ty_1(0)-T_ty_2(0)\|=
\|y_1(t)-y_2(t)\|=\|y_1(0)-y_2(0)\|\,.$$
So that by Lemma 3,
$$T_t({{y_1(0)+y_2(0)}\over 2})= {{T_ty_1(0)+y_2(0)}\over 2}=
{{y_1(t)+y_2(t)}\over 2}$$ and
$y(t):=\frac{y_1(t)+y_2(t)}{2}$   is also a solution of (8)
satisfying  $y(0)=\frac{y_1(0)+y_2(0)}{ 2}$. Since
$K_0$ is convex, $y({\mathbb R})\subset K_0$ and $y\in \Omega$.
We have $$Jy_1=Jy_2=\mu$$ So,  $\mu
=\inf\limits_{x\in \Omega }Jx$  and
$\frac{y_1+y_2}{2}\in \Omega$.  We deduce that
$$\mu \leq J({{y_1+y_2)}\over 2}) \leq {{Jy_1}\over 2}+{{Jy_2}\over 2}
=\mu $$
and consequently $Jy=\mu$. Since
$J(\frac{y_1+y_2}{2})=\frac{Jy_1}{ 2}+\frac{Jy_2}{ 2}$  we
have
$$\sup\limits_{t\in {\mathbb R}}\big\|{{y_1(t)+y_2(t)}\over 2}\big\|=
\frac{1}{2} \sup\limits_{t\in {\mathbb R}}\|y_1(t)\|+ \frac{1}{2}
\sup\limits_{t\in {\mathbb R}}\|y_2(t)\|$$
So there is  $s_n \in {\mathbb R}$ such that
\begin{eqnarray*}
\mu -\frac{1}{n} &<& \big\|{{y_1(s_n)+y_2(s_n)}\over 2}\big\| \\
&\leq& \frac{\|y_1(s_n)\|}{2} + \frac{\|y_2(s_n)\|}{2}\\
&\leq& \mu
\end{eqnarray*}
and since  $y_1(s_n)\in K_0;  y_2(s_n)\in K_0$
there is a subsequence (which we denote by the same symbol)
such that  $ y_1(s_n)\to l_1$  and  $y_2(s_n)\to l_2$. Then
$$\big\|{{l_1+l_2}\over 2}\big\|={{\|l_1\|}\over 2}+{{\|l_2\|}\over2}=
\mu\,.$$
On the other hand
$\|y_i(s_n)\|\leq \mu$ implies $\|l_i\| \leq \mu$ and
${{\|l_1\|}\over 2}+{{\|l_2\|}\over2}=\mu$ implies
$ \|l_i\|\geq \mu$ for $i=1,2$.
Hence  $\|l_1\|=\|l_2\|=\mu$.
Since the norm of $E$ is strictly convex, we deduce that
$l_1=l_2$  and consequently
\begin{eqnarray*}
\|l_1-l_2\|&=&\|y_1(t)-y_2(t)\| \\
&=& \lim\limits_{\tau \to -\infty}\|x_1(\tau) - x_2(\tau)\| \\
&=& \|x_1(-\infty) - x_2(-\infty)\| \\
&=& \sup\limits_{t\in {\mathbb R}}\|x_1(t)-x_2(t)\|
\end{eqnarray*}
So  $x_1(t)=x_2(t)$  for every $t\in {\mathbb R}$.

\paragraph{Remark.}  In the case of a Hilbert space, by the
parallelogram formula and by (11), we deduce directly
that  $x_1(t)=x_2(t)$  for all $t\in {\mathbb R}$.

\paragraph{Step 4.} Finally we show that
$\widetilde {x}$ the unique element of $\Omega $ which satisfies
 $J\widetilde {x}=\inf_{x\in \Omega}Jx$  is
almost periodic. For this purpose, we use the $2^{nd}$ Bochner's
characterization of almost periodicity (see appendix I). Let
$t_n$ and $s_n$ be two real sequences, then by  Ascoli's theorem
there is a subsequence of $t_n$ (which we denote by the same symbol)
such that  $\widetilde {x}(t+t_n) \to y(t)$  uniformly
in every compact subset of ${\mathbb R}$. Then $y(.)$ is a  weak solution of
\begin{equation} \label{12}
x'\in -Ax+g(t)\,,
\end{equation}  where  $g(t)=\lim f(t+t_n)$. Now consider
$\widetilde {x}(t+t_n+s_n)$ and  $y(t+s_n)$,  then by Ascoli's theorem
we can extract from $t_n$ and $s_n$ sub-sequences such that
$\widetilde {x}(t+t_n+s_n)\to z_1(t)$  and  $y(t+s_n)\to z_2(y)$,
 but  $f(t+t_n+s_n)$ and $g(t+s_n)$ have the same
limit which we denote by $h(t)$. Then $z_1(.)$ and $z_2(.)$ are weak
solutions of
\begin{equation} \label{13}
x'\in -Ax+h(t)
\end{equation}
so $\mu=J_f(K_0)=J_h(K_0)\leq Jz_i \quad i=1,2$ where
$$J_f(K_0)=\inf \big \{Jx :  x   \mbox {is a weak solution of
(8)}, x({\mathbb R})\subset K_0 \big \}$$ and $$J_h(K_0)=\inf \big
\{Jx :  x  \mbox {is a weak solution of (13)},x({\mathbb
R})\subset K_0 \big \}\,.$$ We have  $\mu =Jz_1=Jz_2$,  but the
equation (13) has the same property as the equation (8) because
the map $h(.)$ is almost periodic. Therefore, there is a unique
solution which satisfies $$J_h(K_0)=\inf \big \{Ju :   u  \mbox {
is a weak solution of (13)}, u({\mathbb R})\subset K_0 \big \}.$$
Consequently $z_1=z_2$. Also  $\widetilde {x}(t+t_n+s_n)$  and
 $y(t+s_n)$  have the same limit, hence $\widetilde{x}$
is almost  periodic.\hfill$\diamondsuit$

\paragraph{Example.} Let $E=({\mathbb R}^n,\|.\|)$ with the Euclidean
norm $\|.\|$, and let  $\varphi (x)=\|x\|$. Consider
$$Ax=\partial \varphi (x) + kx$$
where  $k>0$ and  $\partial \varphi$ is the sub-differential of $\varphi$.
Since $\varphi $ is continuous and convex,
$$\overbrace{\mathop{\rm Dom}(\varphi)}^{\circ} \subset
\mathop{\rm Dom}(\partial \varphi)
 \quad\mbox{so}\quad \mathop{\rm Dom}(A)={\mathbb R}^n.$$
The problem
$$x'\in -Ax+f(t)\,,$$
with  $f:{\mathbb R}\to {\mathbb R}^n$
continuous and almost  periodic, has a strong solution defined on
 $[t_0,+\infty [$  ( $t_0\in {\mathbb R}$ )  (see Brezis
\cite{B}). Now since $0\in \partial \varphi (0)$  we
have $$\langle f(t)-kx-x'(t) , x(t) \rangle\geq 0\,.$$
Therefore,
\begin{eqnarray*}
\langle x'(t),x(t) \rangle & \leq & \langle f(t),x(t) \rangle
-k\|x(t)\|^2
\\ & \leq & ( M-k\|x(t)\| )\|x(t)\|
\end{eqnarray*}
where   $M=\sup\limits_{t\in {\mathbb R}}\|f(t)\|$. We deduce
that $$D^-\|x(t)\|\leq M$$
and $$\frac{d}{2dt}\|x(t)\|^2\leq
0 \quad\mbox{for }  \|x(t)\|\geq \frac{M}{k}.$$
The first
inequality shows that $x$ is lipschitzean, hence uniformly
continuous and the second one shows that the map
$t\longmapsto \|x(t)\|$ is non increasing outside of the ball
$B(0,\frac{M}{k})$. Consequently
$$\|x(t)\|\leq \sup(\|x(t_0)\|,\frac{M}{k} ) \quad \forall t\geq t_0.$$
So that the problem $x'\in -Ax+f(t)$   has a uniformly continuous
solution which is bounded, hence with precompact range, so it has
an almost periodic solution. \hfill$\diamondsuit$

\subsection*{Appendix I}
Let $E$ be a real Banach space, a map  $f: {\mathbb R} \to E$  is
said to be almost periodic if for each $\varepsilon > 0$ there
exists $l_\varepsilon$ such that for all $a\in {\mathbb R}$ there
exists $\tau\in [a,a+l_\varepsilon]$ such that $$
\|f(t+\tau)-f(t)\|\leq \varepsilon  \quad \forall t\in {\mathbb
R}.$$

If $f$ is almost periodic then there exist $t_n\to +\infty$
 and  $s_n\to -\infty$  such that
$f(t+t_n)\to f(t)$  and  $f(t+s_n)\to f(t)$
uniformly on ${\mathbb R}$. In practice, we use the following
Bochner's characterizations of almost periodicity (Yoshisawa \cite {Y}).

\paragraph{First characterization.} $f\in C({\mathbb R},E)$ is
almost periodic if and only if from every real  sequence $t'_n$
one can extract a subsequence $t_n$ such that $\lim f(t+t_n)$
exists uniformly on the real line, furthermore the limit is also
almost periodic.

\paragraph{Second characterization.} $f\in C({\mathbb R},E)$ is
almost periodic if and only if for every pair of real sequences
$h'_n$ and $k'_n$ there are sub-sequences $h_n$ and $k_n$ such that
$f(t+h_n)$ has a pointwise limit $g(t)$ on ${\mathbb R}$, and
$f(t+h_n+k_n)$ and $g(t+k_n)$ have a same limit $h(t)$ on
${\mathbb R}$, and $h$ is also almost periodic.

\subsection*{Appendix II}
 Let E be a real Banach space and
$x : [a,b]\subset {\mathbb R}\to E$  differentiable, and put
$\Phi(t)=\|x(t)\|$. Then
$$\Phi(t)D^-\Phi(t)=<x'(t),x(t)>_-$$ where
$$D^-\Phi(t)=\limsup\limits_{h\to 0^-}\frac{\Phi(t+h)-\Phi(t)}{h}$$
is the upper Dini's derivative of $\Phi$ (see e.g Deimling \cite{D}).

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\noindent{\sc E. Hanebaly } \\
Universit\'e Mohammed V Facult\'e des  Sciences\\
D\'epartement de Math\'ematiques, Rabat, Maroc.\\
e-mail: hanebaly@fsr.ac.ma \smallskip

\noindent{\sc Brahim Marzouki} \\
Universit\'e Mohammed I Facult\'e des Sciences\\
D\'epartement de Math\'ematiques,  Oujda, Maroc.\\
e-mail: marzouki@sciences.univ-oujda.ac.ma

\end{document}