\documentclass[reqno]{amsart}
\usepackage{hyperref}

\AtBeginDocument{{\noindent\small
\emph{Electronic Journal of Differential Equations},
Vol. 2007(2007), No. 145, pp. 1--18.\newline
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu
\newline ftp ejde.math.txstate.edu  (login: ftp)}
\thanks{\copyright 2007 Texas State University - San Marcos.}
\vspace{9mm}}

\begin{document}
\title[\hfilneg EJDE-2007/145\hfil Semi-flows and regularized evolution]
{Regularization for evolution equations in Hilbert spaces
involving monotone operators via the semi-flows method}

\author[G. L. Karakostas, K. G. Palaska\hfil EJDE-2007/145\hfilneg]
{George L. Karakostas, Konstantina G. Palaska}

\address{George L. Karakostas \newline
 Department of Mathematics, University of Ioannina,
 451 10 Ioannina, Greece}
\email{gkarako@uoi.gr}

\address{Konstantina G. Palaska \newline
Department of Mathematics, University of Ioannina,
 451 10 Ioannina, Greece}

\thanks{Submitted September 10, 2007. Published October 30, 2007.}
\subjclass[2000]{39B42, 34D45, 47H05, 47A52, 70G60}
\keywords{Hilbert spaces; monotone operators; regularization; evolutions;
\hfill\break\indent semi-flows; limiting equations;
full limiting functions; convergence}

\begin{abstract}
 In a Hilbert space $H$ consider the equation
 $$
 \frac{d}{dt}x(t)+T(t)x(t)+\alpha(t)x(t)=f(t),\quad t\geq 0,
 $$
 where the family of operators $T(t)$, $t\geq 0$ converges in a
 certain sense to a monotone operator $S$, the function $\alpha$
 vanishes at infinity and the function $f$ converges to a point
 $h$. In this paper we provide sufficient conditions that
 guarantee the fact that full limiting functions of any solution of
 the equation are points of the orthogonality set
 $\mathcal{O}(h;S)$ of $S$ at $h$, namely the set of all $x\in H$
 such that $\langle Sx-h, x-z\rangle=0$, for all $z\in S^{-1}(h)$.
 If the set $\mathcal{O}(h;S)$ is a singleton, then the original
 solution converges to a solution of the algebraic equation $Sz=h$.
 Our problem is faced by using the semi-flow theory and it extends
 to various directions the works \cite{a1,a12}.
\end{abstract}

\maketitle
\numberwithin{equation}{section}
\newtheorem{theorem}{Theorem}[section]
\newtheorem{corollary}[theorem]{Corollary}
\newtheorem{lemma}[theorem]{Lemma}
\newtheorem{definition}[theorem]{Definition}
\newtheorem{remark}[theorem]{Remark}

\section{Introduction}

Let $H$ be a (real) Hilbert space with inner product
$\langle\cdot,\cdot\rangle$ and norm $\|\cdot\|$ and let $S:H\to
H$ be a given operator. The operator $S$ is said to be
{\em monotone}, if it satisfies
$$
\langle Sx_1-Sx_2,
x_1-x_2\rangle\geq 0, \quad{\rm{for \quad all}}\quad x_1,
x_2\in H.
$$
The operator $S$ is called {\em strictly monotone}, if the above
inequality is strict for all $x_1\neq x_2$.

\begin{definition} \label{def1.1} \rm
Let $S:H\to H$ be a monotone operator and let $h\in\mathcal{R}(S)$,
the range of $S$. We define the orthogonality of $S$ at $h$ to be the set
$$
\mathcal{O}(h;S):=\{x\in H:\langle Sx-h,
x-z\rangle=0, \; z\in S^{-1}(h)\}.
$$
\end{definition}

It is clear that, in general,
\begin{equation} \label{e1.1}
S^{-1}(h)\subseteq
\mathcal{O}(h;S);
\end{equation}
and if, for instance, $S$ is a strictly monotone operator (hence
$S^{-1}(h)$ is a singleton), then this relation holds as equality.
Some illustrative examples will be given later.

In this paper
we investigate when the orthogonality set attracts all solutions
of a class of so called $(S,h)$-admissible differential equations
in the space $H$. Special role in our approach plays the
minimizer of $S$, namely a point $x^*$ such that
$$
\|x^*\| = \inf\{\|x\|: x\in S^{-1}(h)\}.
$$
In the literature one can find a great number of works dealing
with continuous, or discrete paths (i.e. iterates generating sequences)
converging strongly to a point of $S^{-1}(h)$, see, e.g.,  [2 - 11, 13, 16,
17, 21 - 30, 32 - 34, 40, 42, 46, 47, 51, 54, 57, 59, 60].
Applications of operators and problems generating the situation
can be found elsewhere, see, e.g, [20, 56].

Moreover there are approaches of the problem where a point of
$S^{-1}(h)$ is approximated by the solutions of a suitably
formulated ordinary differential equation related to the pair
$(S,h)$.  We term it $(S,h)$-admissible equations. More precisely,
consider  the evolution equation in $H$ of the form
\begin{equation} \label{e1.2}
\frac{d}{dt}x(t)+T(t)x(t)+\alpha(t)x(t)=f(t), \quad t\geq 0,
\end{equation}
 with arbitrary initial value, where $\alpha(t)$,
$t \geq  t_0,$ is a convex, positive, decreasing  real valued function. In case
the family of operators $T(t)$ approaches $S$ in
a certain sense, the perturbation $f(t)$ approaches $h$ and the function
$\alpha$ satisfies at least the conditions
\begin{equation} \label{e1.3}
\lim_{t\to+\infty}\alpha(t)=0,\quad
\int_{t_0}^{+\infty}\alpha(t)dt=+\infty, \quad
\lim_{t\to+\infty}\frac{\alpha'(t)}{\alpha^2(t)}=0,
\end{equation}
Alber [1] and Alber-Ryazantseva [12] followed methods of
regularization in essentially ill posed problems [48, p.180] have
shown that any solution of \eqref{e1.2} converges strongly to the
point $x^*$.

Our results are proved by setting rather mild conditions on the
coefficients and especially to the scalar function $\alpha$
appeared in \eqref{e1.2}, which are weaker than \eqref{e1.3}.
Indeed, we need only the function $\alpha$ vanished at $+\infty$
and being decreasing and satisfying an integral-like condition
weaker than the two last conditions in \eqref{e1.3}, see the remark
following Lemma \ref{lem3.10} below. However,  we show convergence to the
orthogonality set, while in [12] it is proved convergence to the
minimizer.

 Here let us consider for a moment a point $z$ in
$S^{-1}(h)$, i.e. a point that satisfies the equation
$$
Sz=h.
$$
Then the constant function $u(t):=z$, $t\in\mathbb{R}$ satisfies the
ordinary differential equation
\begin{equation} \label{e1.4}
\dot{u}+Su(t)=h.
\end{equation}

In this work we put the problem of approximation in a new frame by
considering the autonomous ordinary differential equation
\eqref{e1.4} as the full limiting equation of \eqref{e1.2},
generated by a semi-flow with phase-space the cartesian product of
four spaces. According to this semi-flow each point of this space
consists of translations of the solution $x$ and of the
coefficients $T, \alpha, f$, which appear in equation
\eqref{e1.2}. These translations produce semi-dynamical systems
similar to the Bebutov dynamical system (see, e.g., [55] p. 81) and
it is widely used elsewhere [15, 36 - 38, 50, 52, 53]. For the
discrete version of this semi-dynamical system one can consult
e.g. [39].

The limiting properties of the solutions of \eqref{e1.2} are
described by the solutions of \eqref{e1.4}. For instance, if the
latter admits  solutions, which are constant functions, then all solutions of
\eqref{e1.2} converge, as the time increases, to such a constant
function.

 In the references stated above it was also shown how to
use the semi-flow theory to study the asymptotic behavior of
solutions of nonautonomous ordinary differential equations,
Volterra integral equations and more general operator equations in
the the space of continuous functions. Applications to stability
theory can be found e.g. in [38, 52]. For more applications,
details, and references, see e.g. [50, 52], [45].

   Next we shall recall some facts from the limiting equations theory,
as it is developed in [14,  36, 37. 52, 53], [45, Appendix], etc.
We shall not follow the lines of any of these papers, but instead
we prefer to combine ideas from them to establish our semi-flow.
The basic idea is to take into consideration the continuous
changes of the coefficients in the equation as the time progresses
and then to get asymptotic properties of the solutions via the
behavior of the solutions of the full limiting equations.

The major difficulty in the semi-flow approach is to define the
appropriate topology in the phase space. First we will construct
the  shifting family of the original evolution equation in the
sense of nonautonomous differential equation  [45, 53] and then,
following the classical method of limiting equations, we will
investigate the convergence of the trajectories to their
$\omega$-limit sets.  The most general facts for dynamical systems
generated by Volterra-integral operators, or abstract operators,
(see, e.g. [50, 36], etc.]) are not used here.

 \section{Some auxiliary facts}

Before presenting our main results we shall recall some auxiliary
facts about the geometric properties of Hilbert spaces.   To begin
with we first give illustrative examples for the orthogonality of
an operator in Hilbert spaces.

\subsection*{Examples} (a) Let for any $x$
in the plane $Rx$ be the point obtained by rotating $x$
counterclockwise by a right angle. Define $Sx:=x-Rx$ and observe
that $S0=0$ and $S$ is a monotone operator having orthogonality at
any $h$ the singleton $S^{-1}(h)$.

(b) In $H:=l^2$ define the operator
$$
B:(x_1, x_2, x_3,x_4,\dots)\to (x_2, -x_1, x_3,x_4,\dots).
$$
Then $B$ is an isometry and so the linear operator $Sx:=x-Bx$ is
monotone. Observe that
\begin{gather*}
S^{-1}(0)=\{0\},\\
\mathcal{O}(0;S)=\{(0,0,x_2,x_3,\dots ):
\sum_{i=2}^{+\infty}x_i^2<+\infty\}.
\end{gather*}

(c) Let $C$ be a convex closed subset of a Hilbert space $H$ and let
$Sx$ be the  point of the minimum distance of $x$ from $C:$
$$
Sx:=\{z\in C:\|z-x\|=\inf\{\|y-x\|: y\in C\}\}.
$$
It is well known that $S$ defines an
operator with domain $H$, which, as it is shown in [35, p. 42],
is monotone.

 We claim that it holds
\begin{equation} \label{e2.1}
\mathcal{O}(h;S)=S^{-1}(h),
\end{equation}
for all $h\in C$. Before we prove this fact, we shall shift
the point $h$ to the origin as follows: Take any point $h\in C$.
 Defining the set
$$
C':=\{u=x-h:\quad x\in C\}
$$
and the operator
$$
S':H\to C':u\to S(u+h)-h,
$$
it is clear that $0\in C'$ and it holds $Sx=h$, if and only if $S'x'=0$,
where $x':=x-h$. In this case we have
\begin{align*}
0&=S'x'=Sx-h\\
&=\{z\in C:\|z-x\|=\inf\{\|y-x\|: y\in C\}\}-h\\
&=\{u:=z-h\in C':\|z-x\|=\inf\{\|y-x\|: y\in C\}\}\\
&=\{u:=z-h\in C':\|u-(x-h)\|=\inf\{\|v-(x-h)\|: v\in C'\}\}\\
&=\{u\in C':\|u-x'\|=\inf\{\|v-x'\|: v\in C'\}\}.
\end{align*}
This shows that $S'x'$ is the point at the minimum distance of $x'$ from $C'$. \par
Now it is clear that relation \eqref{e2.1} holds if and only if
\begin{equation} \label{e2.2}
\mathcal{O}(0;S')=S'^{-1}(0).
\end{equation}
 Because of \eqref{e1.1}, in order to show (2.2) it is enough to prove that
\begin{equation} \label{e2.3}
\mathcal{O}(0;S')\subseteq S'^{-1}(0).
\end{equation}

 Take a point $z\in C'$  with $S'z=0$ and consider any
$u\in\mathcal{O}(0;S')$, namely a point such that
\begin{equation} \label{e2.4}
\langle S'u, u-z\rangle =0.
\end{equation}
 Then, on one hand, we have
$$
\|S'u-u\|^2\leq \|u\|^2
$$
(notice that $0\in C'$), which gives
\begin{equation} \label{e2.5}
\|S'u\|^2- 2\langle S'u, u\rangle\leq 0
\end{equation}
 and on the other hand
$$
\|z\|^2\leq \|S'u-z\|^2,
$$
since $0$ and  $S'u$ are points of the set  $C'$. The latter implies that
\begin{equation} \label{e2.6}
0\leq \|S'u\|^2- 2\langle S'u, z\rangle,
\end{equation}
Relations
\eqref{e2.4}--\eqref{e2.6} imply that
$$
\|S'u\|^2=2\langle S'u,z\rangle,
$$
from which, we  get
$$
\|S'u-z\|=\|z\|.
$$ The latter, clearly, says that $S'u=S'z=0$ and so our claim is proved.
\medskip

 Now we present some well known notions which have been introduced
in connection with nonlinear
problems in functional analysis (see, e.g., \cite{b2,b3,b4,k6,m2,v1}).

 An operator $S$ is {\em hemicontinuous at a certain point $x_0$}, if
for any sequence $(x_n)$  converging to $x_0$ along a line, the
sequence $(Sx_n)$ converges weakly to $Sx_0$. That is, $\langle
S(x_0+t_nx)-Sx_0, y\rangle\to 0$, as $t_n\to 0,$ for all $x, y\in
H$. An operator $S$ is {\em demicontinuous at a certain point
$x_0$}, if for any sequence $(x_n)$  converging to  $x_0$, the
sequence $(Sx_n)$ converges weakly to $Sx_0$. This means that, if
$\lim x_n=x_0$, then for each $u\in H$ it holds
$$
\lim\langle Sx_n-Sx_0, u\rangle=0.
$$
 A consequence of this fact is the
following: Let $x:I\to H$ be a continuous function, where $I$ is
any subset of the real line. Then, for each $u\in H$, the function
$$
t\to\langle Sx(t), u\rangle
$$
is continuous.

An operator $S$
is  {\em bounded}, if it maps bounded sets into bounded sets.
An operator $S$ is {\em maximal monotone,} if it has no proper
monotone extensions.

\begin{lemma}[{\cite[p. 48]{j1}}] \label{lem2.1}
 If $S : H\to H$  is a monotone hemicontinuous  operator, then $S$ is maximal
monotone.
\end{lemma}

 Actually we need an operator $S$ which is at least demicontinuous.
Then it is hemicontinuous and so by
the previous lemma it is maximal monotone. The latter property
guarantees special properties for the
operator $S$ as in the following result: 

\begin{lemma}[{\cite[p. 55]{j1}}] \label{lem2.2}
 If $S: H\to H$ be a monotone hemicontinuous operator, then the range
${\mathcal{R}}(I+S)$ of the operator $I+S$ is the entire space $H$.
(Here $I$ stands for the identity on
$H$.)
\end{lemma}

Let $\theta>0$ be given. If $S: H\to H$ is a hemicontinuous operator,
then, clearly, $\frac{1}{\theta} S$ does so. Thus, from the previous
lemma, it follows that  the range of the operator $\theta I+S$ is the
entire space $H$. For our purpose we need the following well
known result (stated for general set-relations) as \cite[Lemma 1]{b7}.
See, also, \cite{r1}.

\begin{lemma} \label{lem2.3}
If $S$ is a maximal monotone operator, then for each $\theta>0$,
there exists a unique $y_{\theta}\in H$ for which
$$
0=\theta y_{\theta}+Sy_{\theta}.
$$
Also, if $0\in{\mathcal{R}}(S)$, then the strong limit
$$
\lim_{\theta\to 0^+}y_{\theta}=x^*
$$
exists and is the point of $A:=S^{-1}(0)$ closest to 0.
\end{lemma}

 The first part of the lemma was given in \cite{m2}, but for continuous
monotone operators.\par 
 Now assume that $S : H\to H$ is a demicontinuous operator and
let $h\in\mathcal{R}(S)$. Define the set
\begin{equation} \label{e2.7}
A: =S^{-1}(h).
\end{equation}
The following result which characterizes the minimizer of the set $A$
is known, but for completeness of this work we
shall give the proof here.

\begin{lemma} \label{lem2.4}
Under the conditions above, the set $A$ is convex, (strongly) closed and
there is a unique point $x^*$  in $A$ such
that
$$
\|x^*\| = \min\{\|x\|: x \in A\}.
$$
\end{lemma}

\begin{proof} First we claim that a point
$x$ is an element of the set $A$ if and only if the inequality
\begin{equation} \label{e2.8}
\langle Sz-h, z-x\rangle \geq 0
\end{equation}
holds for all $z\in H$. Indeed, if $Sx=h$,
then, for all $z\in H,$ we have
$$
\langle Sz-h, z-x\rangle =\langle Sz-Sx, z-x\rangle \geq 0,
$$
by the monotonicity of $S$.

 The inverse: Assume that $x$ is a point in $H$ satisfying \eqref{e2.8} for
all $z$. Let $s\in(0,1)$ and define  $w:=h-Sx$. Consider the
point $z:=x+sw$ and observe that \eqref{e2.8} gives
$$
\langle S(x+sw)-h, w\rangle \geq 0.
$$
Letting $s\to 0$, by the hemicontinuity of the
operator $S$, we get $\langle Sx-h, w\rangle\geq 0$. Thus it
follows that
$$
-\|Sx-h\|^2=\langle Sx-h, h-Sx\rangle=\langle Sx-h, w\rangle\geq 0,
$$
which implies that $Sx=h$. Hence $x\in A$.
 By using relation \eqref{e2.8} we can easily show the convexity of $A$.

We show the closedeness of the set $A$. Indeed, let $(x_n)$ be a
sequence in $A$ converging (strongly) to a point $x\in H$.
If $x$ does not belong to $A$, according to our claim above,
there is a $z\in H$ such that
$$
\langle Sz-h, z-x\rangle \quad <-\varepsilon,
$$
for some $\varepsilon>0$. From this relation we get $\|Sz-h\|>0$ and moreover, for all indices $n$ 
\begin{equation}
\begin{aligned}
\varepsilon & < - \langle Sz-h, z-x\rangle
=-\langle Sz-h, z-x_n\rangle - \langle Sz-h, x_n-x\rangle\\
&=-\langle Sz-Sx_n, z-x_n\rangle - \langle Sz-h, x_n-x\rangle\\
&\leq - \langle Sz-h, x_n-x\rangle\leq \|Sz-h\|\|x_n-x\|,
\end{aligned}
\end{equation}
contrary to the fact that $x_n\to x$. Therefore the set $A$ is closed.

 Finally, we show uniqueness. To this end, assume that there are two points $x*\neq y*$ such that
$$
\|x^*\|= \|y^*\|=\min\{\|x\|: x \in A\}.
$$
Here we proceed as in the proof of the main claim in Example (a).
 If $x*=0$, then, clearly, $y^*=0$, too, which is false.
So, assume that $x^*\neq 0$.
Consider the set
$$
A_0: =\{\frac{x}{\|x^*\|}: x \in A\}.
$$
Then the points
$$
x_0^*:=\frac{x^*}{\|x^*\|}, \quad y_0^*: =\frac{y^*}{\|x^*\|}
$$
are different and satisfy
$$
\|x_0^*\|=\|y_0^*\|=1.
$$
Since $H$ is a Hilbert space, it is strictly convex
\cite[p. 31]{c8}, thus by standard properties of such spaces
\cite[p. 41]{c8} we get
$$
\|x_0^*+y_0^*\|<2.
$$
Since $A_0$ is a convex set, we have $\frac{x_0^*+y_0^*}{2}\in A_0$.
Hence, by the minimality of $x_0^*,$  we get
$$
1=\|x_0^*\|\leq\|\frac{x_0^*+y_0^*}{2}\|\leq\frac{\|x_0^*+y_0^*\|}{2}<1,
$$
a contradiction.
The proof of lemma is complete.
\end{proof}

\begin{remark} \label{rmk2.5} \rm
If a monotone  operator $S$ is demicontinuous, then, it is hemicontinuous and, from Lemma
\ref{lem2.1},
$S$ is maximal monotone.  Thus from Lemma \ref{lem2.3} it follows that
the operator equation
\begin{equation} \label{e2.10}
Sx+\lambda x = h
\end{equation}
admits a solution $x_{\lambda}$ which approaches $x^{*}$ as $\lambda$
tends to zero.
\end{remark}

 To proceed we make the following assumptions:
\begin{itemize}
\item[(H1)]
$\alpha: [0,+\infty)\to (0,+\infty)$ is
a decreasing continuous function such
that
$\lim_{t\to+\infty}\alpha(t)=0$.

\item[(H2)]  $f:[0,+\infty)\to H$ is a continuous function such that
$\lim_{t\to+\infty}f(t)=h$.

\end{itemize}
Using Remark \ref{rmk2.5}, it follows that there is a unique point
$z(t)$ in $H$ satisfying  the relation
\begin{equation} \label{e2.11}
Sz(t)+\alpha(t)z(t)=h,
\end{equation}
for  $t\geq 0$.

\begin{lemma} \label{lem2.6}
The function $t\mapsto z(t)$ satisfying equation in \eqref{e2.11} is continuous
on $[0,+\infty)$ and it converges to $x^*$
as $t$ tends to $+\infty$. In particular, $z$ is bounded.
\end{lemma}

\begin{proof}
For $t_0$ fixed and any $t\geq 0$, we obtain from \eqref{e2.11} that
$$
Sz(t_0)-Sz(t)=\alpha(t)[z(t)-z(t_0)]+[\alpha(t)-\alpha(t_0)]z(t_0).
$$
Multiplying both sides with $z(t)-z(t_0)$ and using monotonicity of
the operator $S$ we get
$$
\|z(t)-z(t_0)\|\leq\frac{|\alpha(t)-\alpha(t_0)|}{\alpha(t)}\|z(t_0)\|,
$$
which, clearly, implies the continuity of
$z$ at $t_0$. The fact that $z(t)$ converges to $x^*$ follows from
Remark \ref{rmk2.5}.
\end{proof}

\begin{lemma} \label{lem2.7}
Let $f, \alpha$ be functions satisfying the assumptions {\rm (H2)} and
{\rm (H1)} and moreover
\begin{equation} \label{e2.12}
\sup_{t\geq 0}\frac{\|f(t)-h\|}{\alpha(t)}<+\infty.
\end{equation}
If $z_{\alpha}(t)$ is the (unique) point satisfying the algebraic equation
\begin{equation} \label{e2.13}
Sz_{\alpha}(t)+\alpha(t)z_{\alpha}(t)=f(t),
\end{equation}
for all $t\geq 0$, then the function
$ t\mapsto z_{\alpha}(t)$ is
continuous and bounded.
\end{lemma}

\begin{proof} Continuity
follows as that in Lemma \ref{lem2.6}. To show boundedness, we subtract
relation \eqref{e2.11} from \eqref{e2.13} and obtain
$$
[Sz_{\alpha}(t)-Sz(t)]+\alpha(t)[z_{\alpha}(t)-z(t)]=f(t)-h.
$$
Therefore we have
$$
\langle Sz_{\alpha}(t)-Sz(t),z_{\alpha}(t)-z(t)\rangle
+\alpha(t)\|z_{\alpha}(t)-z(t)\|^2=\langle
f(t)-h,z_{\alpha}(t)-z(t)\rangle,
$$
which, because of the
monotonicity of the operator $S$, gives
$$
\alpha(t)\|z_{\alpha}(t)-z(t)\|^2\leq\|f(t)-h\|\|z_{\alpha}(t)-z(t)\|.
$$
Let $U$ be the set of all $t$ such that $z_{\alpha}(t)=z(t)$.
Then, because of \eqref{e2.12} there is a positive real
number $M$  such that
$$
\|z_{\alpha}(t)-z(t)\|\leq M,
$$
for all $t\notin U$ we obtain. This and boundedness of $z$ proves the result.
\end{proof}

\section{Some semi-flow facts}

 Let us denote by $\mathcal{M}$ the set of all demicontinuous operators
with domain the Hilbert space $H$ and range in  $H$.
 We endow the set $\mathcal{M}$ by the following convergence structure:
A sequence $(S_n)$ converges
to a certain $S$, and write it as $S_n\to S$, if it holds
$$
\lim\|S_nz-Sz\|=0,
$$
uniformly for $z$ in bounded sets.

 To unify things we use the symbol $E$ to denote either the
real line $\mathbb{R}$, or the Hilbert space $H$, or the space
$\mathcal{M}$. Then we shall denote by $C(\mathbb{R},E)$ and by
$C(\mathbb{R}^+,E)$ the sets of all continuous functions defined on
the real line  and on $[0,+\infty)$ respectively and with
values in $E$. We furnish these sets with the topology of uniform
convergence on compact intervals.

First we shall present briefly the shifting semi-flow on $C(\mathbb{R}^+,E)$
in a formal way, although it was previously used implicitly by many
authors (see, e.g., \cite{k1,m3} and the references therein). This is
actually the Bebutov dynamical system (see, e.g., \cite[p. 81]{s4}),
where the functions are restricted on $\mathbb{R}^+$.

 We shall assume that the reader is familiar with all basic notions on
dynamical systems (namely,
$\omega$-limit set,
$\alpha$-limit set, invariance, compactness, stationary points, etc).
The classical book \cite{s4} provides a
good basis of the theory of dynamical systems on metric spaces.

For any $u\in C(\mathbb{R}^+,E)$ and  $t\geq 0$ we define the
translation of $u$ by $t$ to be the function $s\to u(t+s)$,
$ s\in \mathbb{R}^+$. Then consider the mapping
$$
p_E(t,u)(\cdot)= u(t+\cdot): [0,+\infty)\times C(\mathbb{R}^+,E)
\to C(\mathbb{R}^+,E)
$$
and we show the following result.

\begin{lemma} \label{lem3.1}
Under the hypotheses above the mapping  $p_E$ defines a semi-flow
with phase-space the set $C(\mathbb{R}^+,E)$.
\end{lemma}

\begin{proof}
First we see that it holds
$$
p_E(0,u)(\cdot)=u(0+\cdot)=u(\cdot).
$$
Also, the semi-group property holds, since we have
$$
p_E(t,p_E(r,u)(\cdot))(s)=p_E(r,u)(t+s)=u(r+t+s)=p_E(t+r,u)(s).
$$
Finally, to show continuity of
$p_E(t,u)$ with respect to the pair $(t,u)$ we let $(t_n,u_n)$ be
a sequence in $[0,+\infty)\times C(\mathbb{R}^+,E)$ converging to
$(t_0, u_0)$. This means that $(t_n)$ converges to $t_0,$ thus the
set $N: =\{t_0, t_n: n=1, 2, \dots\}$ is compact and the sequence
$(u_n)$ converges to the function $u_0$ uniformly on compact
intervals.

 Now, for any compact interval $I$ of $\mathbb{R}^+$
and any $s\in I$, it holds
\begin{align*}
&\|p_E(t_n,u_n)(s)-p_E(t_0,u_0)(s)\|\\
&=\|u_n(t_n+s)-u_0(t_0+s)\| \\
&\leq\|u_n(t_n+s)-u_0(t_n+s)\|+\|u_0(t_n+s)-u_0(t_0+s)\|\\
&\leq\sup_{\theta\in N+I}\|u_n(\theta)-u_0(\theta)\|
  +\|u_0(t_n+s)-u_0(t_0+s)\|.
\end{align*}
Here $N+I$ is the (compact) set $\{r+t:r\in N, \; t\in I\}$.
Hence continuity of $p_E$ follows from the
previous relation, since both quantities in the right side tent to
zero as $n\to+\infty$.
\end{proof}

  It is clear that a point $u$ is a rest (or, stationary) point with
respect the semi-flow, if and only if the function
$u$ is constant. A motion is called {\em compact} if its trajectory
lies in a compact set.

\begin{lemma} \label{lem3.2}
 Assume that $u$ is a uniformly continuous function in $C(\mathbb{R}^+,E)$
with relatively compact range in case $E=H$ and with
bounded range in case $E=\mathbb{R}$. Then the $\omega$-limit set of $u$
is nonempty, compact,
connected and invariant.
\end{lemma}

\begin{proof}
Take any sequence $(t_n)$ in $\mathbb{R}^+$, a compact set $I$ of
positive reals and consider the family
$$
\mathcal{G}:=\{u(t_n+s): s\in I, \quad n=1,2,\dots \}.
$$
Then, applying a generalization of the Ascoli's theorem
(see, e.g., \cite[p. 290]{w1}, we see that  $\mathcal{G}$
is relatively compact. This implies that
the $\omega$-limit set is nonempty. The other properties follow
as for the Bebutov dynamical system exhibited in
\cite[pp. 28-32]{s4}.
\end{proof}

Usually, a function $u$ satisfying the assumptions of the previous lemma
as well as its motion, are called compact. Notice that Langrange stable
motions in \cite{s4} correspond to compact motions in our case.

\begin{corollary} \label{coro3.1}
Let $u\in C(\mathbb{R}^+,E)$ be such that the limit
$\lim_{t\to+\infty}u(t)$ exists in $E$. Then
the trajectory of the point $u$ is compact.
\end{corollary}

 \begin{proof} The result follows from the previous lemma and the fact
that the existence of the limit and the continuity of
$u$ implies  uniform continuity on $\mathbb{R}^+$, as well as the
relative compactness of the range of $u$.
\end{proof}

\begin{definition} \label{def3.3} \rm
A function $\bar{u}\in C(\mathbb{R}^+,E)$ is said to be a limiting
function of $u$,  if $\bar{u}\in\omega(u)$.
\end{definition}

For any  $u\in C(\mathbb{R}^+,E)$ and any limiting function
$\bar{u}$ of $u$,  there is a full orbit $Q$ through $\bar{u}$,
which lies in $\omega(u)$. Thus a flow is generated in
$\omega(u)$. The latter is the {\em invariance property} in dynamical
 systems \cite{l2}. To prove the above fact
one can use the following lemma adopted to our case,
which is known in the theory of semi-flows and its
proof can be succeeded by using a Cantor type diagonalization argument.

\begin{lemma} \label{lem3.4}
Let $p_E(t,u)$ be a (positively) compact motion with $\omega$-limit
set $\omega(u)$. Then for each
$\bar{u}\in \omega(u)$, there is a continuous function $Q:
\mathbb{R}\to\omega(u)$, which satisfies the following properties:
\par a) $Q(0) =\bar{u},$ and \par b) $p_E(t,Q(s)) = Q(t + s),$ for
all $s\in\mathbb{R}$ and $t\geq 0$.
\end{lemma}

In the sequel we  say that $Q$ is a {\it full limiting orbit} of $u$,
if $Q$ is a full orbit through a point $\bar{u}\in\omega(u)$.

\begin{definition} \label{def3.5} \rm
A function $\hat{u}\in C(\mathbb{R}, E)$ is said to be a full limiting function
of $u$, if it can be written in the form $\hat{u}(t) = Q(t)(0)$,
for a certain full limiting orbit $Q$ of $u$.
\end{definition}

Let $C_{un}(\mathbb{R}^+,E)$ be the set of all uniformly continuous
functions of $C(\mathbb{R}^+,E)$ with relatively compact range. (If
$E=\mathbb{R}$, then boundedness is sufficient for relative
compactness.) Then the set of all full limiting functions of any
$u\in C_{un}(\mathbb{R}^+,E)$ is a nonempty, compact, connected and
invariant subset of $C_{un}(\mathbb{R}^+,E)$. Indeed, the set of full
limiting functions of $u$ is the $\omega$-limit set of any
continuation $z \in C_{un}(\mathbb{R},E)$ of $u$ (for instance, define
$z(t): = u(0)$, if $t < 0$, and $z(t): = u(t),$ if $t \geq 0$) with
respect to the Bebutov dynamical system (see, e.g. \cite[p. 81]{s4},
and therefore it has the above properties. Notice that in \cite{s4} the above properties are referred to metric spaces and are proved by using sequential approach. This approach con be used even in our case, where $E$ is the space $H$ or the space $\mathcal{M}.$ 

 In the sequel we shall use the subspace
${{C}}_{\rm mon}({\mathbb{R}^+},\mathcal{M})$ of all functions
$T\in{{C}}({\mathbb{R}^+},\mathcal{M})$ such that
$$
\langle T(t)z-T(t)y,z-y\rangle \geq0,
$$
for all nonnegative reals $t\geq 0$. It is not hard to see that the space
${{C}}_{\rm mon}({\mathbb{R}^+},\mathcal{M})$ is a closed subset of
$C({\mathbb{R}^+},\mathcal{M})$.

Now consider the differential equation \eqref{e1.2}, where
$T\in{\mathcal{C}}_{\rm mon}({\mathbb{R}},\mathcal{M})$,
$\alpha\in C(\mathbb{R}^+,\mathbb{R}^+)$ and $f\in C(\mathbb{R},H)$.

\begin{definition} \label{def3.6} \rm
An equation of the form \eqref{e1.2} is called
$(S,h)$-admissible if any bounded solution of \eqref{e1.2} has
range in a compact set, the coefficients $\alpha, f$ satisfy the
conditions (H1), (H2) and  the function  $T$ satisfies
the following condition:
\begin{itemize}
\item[(H3)]  The function $T(t)$ converges
to $S$ as $t\to+\infty$ with respect to the convergence structure
of $\mathcal{M}$.
\end{itemize}
\end{definition}

 \begin{lemma} \label{lem3.7}
If $x,y$ are two solutions of \eqref{e1.2}, then they have the same domain,
$\mathcal{D}$, say. Also if
$x(0)=y(0)$, then $x(t)=y(t)$, for all $t$ in
$\mathcal{D}$.
\end{lemma}

\begin{proof}
Assume that $x, y$ are two solutions of \eqref{e1.2} defined on an
interval of the form $[0,t_0)$. Then, for all $t\in[0,t_0)$, we have
$$
x'(t)-y'(t)+T(t)x(t)-T(t)y(t)+\alpha(t)(x(t)-y(t))=0.
$$
Multiplying both sides by the factor $x(t)-y(t)$ we get
\begin{equation} \label{e3.1}
\langle x'(t)-y'(t),x(t)-y(t)\rangle \leq
-\alpha(t) \langle x(t)-y(t),x(t)-y(t)\rangle,
\end{equation}
 for all $t\in[0,t_0)$, or
$$
\frac{1}{2}\frac{d}{dt}\|x(t)-y(t)\|^2\leq-\alpha(t)\|x(t)-y(t)\|^2,
$$
from which we get
\begin{equation} \label{e3.2}
\|x(t)-y(t)\|\leq \|x(0)-y(0)\|\exp\Big(\int_0^t\alpha(s)ds\Big), \quad
t\in [0,t_0).
\end{equation}
This relation implies that the
solutions $x,y$  have the same interval of domain and moreover, if
$x(0)=y(0)$, then $x(t)=y(t)$, for all $t$, hence the uniqueness.
\end{proof}

\begin{lemma} \label{lem3.8}
Consider the function $T\in{{C}}_{\rm mon}({\mathbb{R}},\mathcal{M})$
satisfying assumption
{\rm (H3)} and let us assume that $\alpha$ satisfies assumption $(H1)$. Moreover suppose that
\begin{equation} \label{e3.3}
\sup_{t\geq 0}\frac{\|T(t)z-Sz\|}{\alpha(t)}<+\infty
\end{equation}
holds uniformly for $z$ in bounded sets.
Then  the (unique) solution $y_{\alpha}$ of the functional equation
\begin{equation} \label{e3.4}
T(t)y_{\alpha}(t)+\alpha(t) y_{\alpha}(t)=f(t),\quad t\geq 0,
\end{equation}
is defined for all $t\geq 0$ and it is bounded.
\end{lemma}

\begin{proof}
  Let $z_{\alpha}(t)$ be the solution of the algebraic equation
\begin{equation} \label{e3.5}
Sz_{\alpha}(t)+\alpha(t)z_{\alpha}(t)=f(t).
\end{equation}
Subtracting relation
\eqref{e3.5} from \eqref{e3.4} we have
$$
[T(t)y_{\alpha}(t)-T(t)z_{\alpha}(t)]+[T(t)z_{\alpha}(t)-Sz_{\alpha}(t)]
+\alpha(t)[y_{\alpha}(t)-z_{\alpha}(t)]=0
$$
and so
$$
\langle T(t)z_{\alpha}(t)-Sz_{\alpha}(t),y_{\alpha}(t)-z_{\alpha}(t)\rangle
+\alpha(t)\|y_{\alpha}(t)-z_{\alpha}(t)\|^2\leq 0,
$$
because of the monotonicity of the operator $T(t)$. Thus we have
$$
\alpha(t)\|y_{\alpha}(t)-z_{\alpha}(t)\|^2\leq\|T(t)z_{\alpha}(t)
-Sz_{\alpha}(t)\|\|y_{\alpha}(t)-z_{\alpha}(t)\|.
$$
By using  the fact that the set $\{z_{\alpha}(t):t\geq 0\}$ is bounded
(Lemma \ref{lem2.7}), from \eqref{e3.3}
the result follows.
\end{proof}

\begin{lemma} \label{lem3.9}
Under the conditions  \eqref{e2.12} and \eqref{e3.3}, for each
$\tau>0$ any solution of the differential equation
\begin{equation} \label{e3.6}
\frac{d}{dt}z(t;\tau)+T(t)z(t;\tau)+\alpha(\tau)z(t;\tau)=f(t)
\end{equation}
is defined on the entire interval $[0,+\infty)$ and it is bounded uniformly
for all $\tau\geq 0$.
\end{lemma}

 \begin{proof}
Fix a $\tau\geq 0$ and let $y_{\alpha}(\tau)$ be the solution of the
algebraic equation \eqref{e3.4} for
$t=\tau$. From \eqref{e3.4} and \eqref{e3.6} we obtain
\begin{align*}
&z'(t;\tau)+[T(t)z(t;\tau)-T(t)y_{\alpha}(\tau)]\\
&+[T(t)y_{\alpha}(\tau)-T(\tau)y_{\alpha}(\tau)]
+\alpha(\tau)[z(t;\tau)-y_{\alpha}(\tau)]\\
&=f(t)-f(\tau).
\end{align*}
Multiply both sides with the factor $z(t;\tau)-y_{\alpha}(\tau)$ and
use the monotonicity of the operator $T(t)$ to obtain
\begin{align*}
&\langle z'(t;\tau), z(t;\tau)-y_{\alpha}(\tau)\rangle
 +\langle T(t)y_{\alpha}(\tau)-T(\tau)y_{\alpha}(\tau), z(t;\tau)
 -y_{\alpha}(\tau)\rangle\\
& +\alpha(\tau)\|z(t;\tau)-y_{\alpha}(\tau)\|^2\\
&\leq \langle f(t)-f(\tau), z(t;\tau)-y_{\alpha}(\tau)\rangle.
\end{align*}
From this relation we obtain
\begin{equation} \label{e3.7}
\begin{aligned}
\frac{1}{2}\frac{d}{dt}\|z(t;\tau)-y_{\alpha}(\tau)\|^2
&\leq \|T(t)y_{\alpha}(\tau)-T(\tau)y_{\alpha}(\tau)\|
   \|z(t;\tau)-y_{\alpha}(\tau)\|\\
&\quad -\alpha(\tau)\|z(t;\tau)-y_{\alpha}(\tau)\|^2\\
&\quad + \|f(t)-f(\tau)\|\|z(t;\tau)-y_{\alpha}(\tau)\|.
\end{aligned}
\end{equation}
Therefore for all $t$ for which (3.7) holds, the (nonnegative)
function $\psi(t;\tau):=\|z(t;\tau)-y_{\alpha}(\tau)\|$ satisfies
the inequality
\begin{equation} \label{e3.8}
\psi'(t;\tau)\leq-\alpha(\tau)\psi(t;\tau)+\|T(t)y_{\alpha}(\tau)
-T(\tau)y_{\alpha}(\tau)\|+ \|f(t)-f(\tau)\|.
\end{equation}
 Thus by using standard comparison arguments for differential inequalities
(see, e.g.  \cite[p. 15]{l1}) we get that $\psi(t;\tau)\leq r(t;\tau)$, for
all $t\geq 0$, for which the solution $r(t;\tau)$ of the
differential equation
$$
r'(t;\tau)=-\alpha(\tau)r(t;\tau)+\|T(t)y_{\alpha}(\tau)
-T(\tau)y_{\alpha}(\tau)\|+ \|f(t)-f(\tau)\|
$$
with $r(0;\tau)=\psi(0;\tau)$ exists. But it is clear that
$r(\cdot;\tau)$ is defined
up to $+\infty$. This proves boundedness for each fixed $\tau$.

 To prove the existence of a uniform bound assume that for any
positive integer $n$ there are $\tau_n>0$ and $t_n\geq \tau_n$
such that $\psi(t_n;\tau_n)\geq n$. Without loss of generality we
can assume that $\psi'(t_n;\tau_n)\geq 0$, for any $n$. Then from
\eqref{e3.8} we get
$$
n\alpha(\tau_n)\leq\|T(t_n)y_{\alpha}(\tau_n)-T(\tau_n)y_{\alpha}(\tau_n)\|
+\|f(t_n)-f(\tau_n)\|
$$
and so
\begin{align*}
n&\leq \frac{\|T(t_n)y_{\alpha}(\tau_n)-Sy_{\alpha}(\tau_n)\|}{\alpha(t_n)}
  + \frac{\|f(t_n)-h\|}{\alpha(t_n)}\\
&\quad+\frac{\|T(\tau_n)y_{\alpha}(\tau_n)
  -Sy_{\alpha}(\tau_n)\|}{\alpha(\tau_n)}
  +\frac{\|f(\tau_n)-h\|}{\alpha(\tau_n)},
\end{align*}
contrary to the assumptions on $T$ and $f$. The proof is complete.
\end{proof}

\begin{lemma} \label{lem3.10}
Assume that $S$ is a bounded operator and \eqref{e2.12},  \eqref{e3.3}
 keep in force. If the condition
\begin{equation} \label{e3.9}
\sup_{t\geq 0}\int_{\sigma}^{t}[\alpha(s)-\alpha(t)]
\exp\Big(-\int_{s}^{t}\alpha(u)du\Big)ds<+\infty,
\end{equation}
 holds, then any solution
of the differential equation
\begin{equation} \label{e3.10}
x'+T(t)x(t)+\alpha(t)x(t)=f(t)
\end{equation}
is defined on the whole interval $[0,+\infty)$, it is bounded and
uniformly continuous.
\end{lemma}

It is easy to see that conditions \eqref{e1.3} plus
convexity of the function $\alpha$ imply \eqref{e3.9}. On the other hand
there are functions satisfying \eqref{e3.9} but not \eqref{e1.3}. For
instance, $\alpha(t):=(t+1)^{-1}$, $t\geq 0$ is such a
function.

\begin{proof}[Proof of Lemma \ref{lem3.10}]
 Let $x$ be a solution of \eqref{e3.10}. First we shall prove that $x$ is bounded, and it has domain of
definition the entire interval $[0,+\infty)$.
 Let $\tau$ be any positive real in the domain of the solution. Consider the
solution $z(t;\tau)$ of the differential equation \eqref{e3.6}. Then we have
\begin{equation}\nonumber
\begin{aligned}
&x'(t)-z'(t;\tau)+[T(t)x(t)-T(t)z(t;\tau)]\\
&+\alpha(t)[x(t)-z(t;\tau)]+[\alpha(t)-\alpha(\tau)]z(t;\tau)
=0.
\end{aligned}
\end{equation}
\par Take any $t\in [0,\tau]$, multiply both sides with the quantity
$x(t)-z(t;\tau)$ and use monotonicity of $T(t)$. Then we obtain
$$
\frac{1}{2}\frac{d}{dt} \|x(t)-z(t;\tau)\|^2
\leq-\alpha(t)\|x(t)-z(t;\tau)\|^2
 +[\alpha(t)-\alpha(\tau)]\|z(t;\tau)\|\|x(t)-z(t;\tau)\|.
$$
Therefore the function
$\phi(t;\tau):=\|x(t)-z(t;\tau)\|$ satisfies the differential inequality
\begin{equation} \label{e3.11}
\frac{d}{dt}\phi(t;\tau)\leq -\alpha(t)\phi(t;\tau)+[\alpha(t)
-\alpha(\tau)]m,
\end{equation}
where $m$ is a bound (guaranteed from Lemma \ref{lem3.9}) of the set
$\{z(t;\tau): t\geq 0\}$. Integrate the
differential inequality \eqref{e3.11} from $\sigma$ to $t (\leq\tau)$ to get
\[
\phi(t;\tau)\leq
\phi(\sigma;\tau)\exp\Big(-\int_{\sigma}^t\alpha(s)ds\Big)
+m\int_{\sigma}^t[\alpha(s)-\alpha(\tau)]
\exp\Big(-\int_{s}^t\alpha(u)du\Big)ds.
\]
Thus it follows that
\begin{equation} \label{e3.12}
\begin{aligned}
\phi(\tau;\tau)&\leq
\phi(\sigma;\tau)\exp\Big(-\int_{\sigma}^{\tau}\alpha(s)ds\Big)\\
&\quad +m\int_{\sigma}^{\tau}[\alpha(s)-\alpha(\tau)]
\exp\Big(-\int_{s}^{\tau}\alpha(u)du\Big)ds.
\end{aligned}
\end{equation}
Now, our claim is implied from \eqref{e3.12} and the fact that the set
$\{\phi(\sigma;\tau):\tau\geq 0\}$ is bounded, because
of Lemma \ref{lem3.9}.

To show uniform continuity it is, clearly, enough to show that $x'(t)$
is bounded. This follows from
the fact that the set $\{x(t):t\geq 0\}$ is bounded, the operator
$S$ maps bounded sets into bounded sets and it
holds
$$
\|x'(t)\|\leq\|T(t)x(t)-Sx(t)\|+\|Sx(t)\|+\alpha(t)\|x(t)\|+\|f(t)\|.
$$
\end{proof}

\section{The main semi-flow}

In the sequel we shall assume that $S$ is a bounded demicontinuous
monotone operator. \par 
We shall define the semi-flow
for $(S,h)$-admissible equations and whose the (unique) limiting
equation is of the form \eqref{e1.4}. To be more clear consider
the product
$$
{\mathcal{E}}:={{C}}({\bf{R}}^+,H)
\times{{C}}({\bf{R}}^+,{{M}})\times{{C}}({\bf{R}}^+,
{\bf{R}}^+)\times{{C}}({\bf{R}}^+,H)
$$
endowed with the product topology.

 Let $\mathcal{U}$ be the subset of
${\mathcal{E}}$ with the following characteristic:

\begin{quote}
{\em Property} (P): Given any point $(x, T,\alpha, f)$ of
${\mathcal{U}}$ the function $x$ is a solution of the ordinary
differential equation \eqref{e1.2} with coefficients $T, \alpha,
f$ and equation \eqref{e1.2} is $(S,h)$-admissible.
\end{quote}

Observe that the mapping
$$
{\mathcal{\pi}}(t, (x, T, \alpha, f)):=(p_H(t,x), p_{\mathcal{M}}(t,T),
p_{\mathbb{R}}(t,\alpha), p_H(t,f))
$$
leaves the set $\mathcal{U}$ invariant. Indeed, given
$(x, T, \alpha, f)\in\mathcal{U}$ and any $t\geq 0$ we
have $$x'(t+s)+T(t+s)x(t+s)+\alpha(t+s)x(t+s)=f(t+s),$$ for all $s\geq 0$.
Clearly, this shows the invariance.

 Now we prove the following result:

\begin{theorem} \label{thm4.1}
Any trajectory in $\mathcal{U}$ is a closed subset of $\mathcal{E}$.
 Moreover, Property {\rm{(P)}} holds for all points in the closure
of any trajectory.
\end{theorem}

\begin{proof}
Let $(p_H(t_n,x), p_{\mathcal{M}}(t_n,T),
p_{\mathbb{R}}(t_n,\alpha), p_H(t_n,f))$ be a sequence of points in the
trajectory of some
$(x, T,\alpha, f)$ converging to a certain point $(x_0, T_0, \alpha_0, f_0)$.
Then from the $(S,h)$-admissibility, it follows that
equation
$$
x'(t_n+s)+T(t_n+s)x(t_n+s)+\alpha(t_n+s)x(t_n+s)=f(t_n+s)
$$
is equivalent to the integral equation
\begin{equation} \label{e4.1}
\begin{aligned}
p_H(t_n,x)(t)
&=p_H(t_n,x)(0)-\int_0^t[p_{\mathcal{M}}(t_n,T)(s)p_H(t_n,x)(s)\\
&\quad +p_{\mathbb{R}}(t_n,\alpha)(s)p_H(t_n,x)(s)-p_H(t_n,f)(s)]ds,
\end{aligned}
\end{equation}
 where the latter is meant with respect to the weak sense.
By using the convergence of the original sequence,
the demicontinuity of the operator $T(t)$ and applying the
classical Lebesgue Dominated Theorem we conclude
that the point
 $(x_0, T_0, \alpha_0, f_0)$ lies in ${\mathcal{U}}$.
\end{proof}

Next, we show that property (P) holds for any point
$(x_0, T_0, \alpha_0, f_0)$
in the closure of the trajectory. If $(x_0, T_0, \alpha_0,f_0)$ lies
in (a finite portion of) the trajectory, this fact follows from
the previous arguments.
 So, assume that $(x_0, T_0, \alpha_0, f_0)$ lies in the
$\omega$-limit set of $(x, T, \alpha, f)$. It is clear that
such a point is of the form $(\bar{x}, S,0,h)$,
 where $\bar{x}$ is in the $\omega$-limit set of $x$
with respect to the semi-flow $p_H(\cdot,\cdot)$. Hence, there is a
 sequence of positive
reals $(t_n)$ converging to $+\infty$ and such that
$$
p_H(t_n,x)(s)=x(t_n+s)\to \bar{x}(s),
$$
uniformly for all $s$ is compact intervals.
By using $(S,h)$-admissibility we take (weak) limits in \eqref{e4.1} and
conclude that
\begin{equation} \label{e4.2}
\bar{x}(t)=\bar{x}(0)-\int_0^t(S\bar{x}(s)-h)ds,
\end{equation}
in the weak sense, from which the result follows, since
$S\bar{x}(s)$ is demicontinuous.

 The previous arguments imply the following useful result.

\begin{corollary} \label{coro4.1}
The mapping ${\mathcal{\pi}}$ defines a semi-flow with phase-space 
${\mathcal{U}}$. Moreover any $(S,h)$-admissible equation
\eqref{e1.2} has limiting equations consisting of equations of the
form
\begin{equation} \label{e4.3}
v'(t)+Sv(t)=h,
\end{equation}
 where $v$ is a point of the $\omega$-limit set of $x$ with respect
to the semi-flow $p_H(\cdot,\cdot)$.
\end{corollary}

Now we are ready to give our convergence result.

\begin{theorem} \label{thm4.2}
Assume that $S: H\to H$ is a demicontinuous bounded monotone
operator and let \eqref{e1.2} be a $(S,h)$-admissible ordinary
differential equation. Assume, also, that $\alpha$ satisfies
relation \eqref{e3.9} and conditions \eqref{e2.12} and \eqref{e3.3} keep in force. Then
any solution $x$ of \eqref{e1.2} is defined up to $+\infty$ and
any full limiting function $\hat{x}$ of $x$ satisfies
$$
\alpha(\hat{x})\cup\omega(\hat{x})\subseteq \mathcal{O}(S;h),
$$
with respect to the flow $p_H(\cdot,\cdot)$.
Moreover, if the set $\mathcal{O}(S;h)$ is a singleton, then any
solution of \eqref{e1.2} converges to the unique point of
$S^{-1}(h)$.
\end{theorem}

\begin{proof}
Let $x$ be a solution of equation \eqref{e1.2}. By Lemma \ref{lem3.10}, $x$ is bounded and uniformly continuous. Hence, by
using the $(S,h)$-admissibility the point $x$ is compact with
respect to the semi-flow $p_H(\cdot,\cdot)$.
 Take any full limiting function $\hat{x}$ of $x$. Then  we have
\begin{equation} \label{e4.4}
\hat{x}'(t)+S\hat{x}(t)=h.
\end{equation}
Let $z$ be any point such that $Sz=h$. Hence from (4.4) we get
$$
\hat{x}'(t)+S\hat{x}(t)-Sz=0,
$$
which implies that
$$
\langle\hat{x}'(t), \hat{x}(t)-z\rangle+\langle S\hat{x}(t)-Sz,
 \hat{x}(t)-z\rangle=0,
$$
or
\begin{equation} \label{e4.5}
\frac{1}{2}\frac{d}{dt}\|\hat{x}(t)-z\|^2+\langle S\hat{x}(t)-Sz,
\hat{x}(t)-z\rangle=0,
\end{equation}
Since $S$ is monotone and the solution $\hat{x}$ is bounded,
from \eqref{e4.5}, we conclude that the function
$\|\hat{x}(t)-z\|$ is nonnegative and decreasing, thus the limits
$$
l_-:=\lim_{t\to-\infty}\|\hat{x}(t)-z\|,\quad
l_+:=\lim_{t\to+\infty}\|\hat{x}(t)-z\|
$$
exist as real numbers. Integrate \eqref{e4.5}
from $t$ to $t+\tau$ $(\tau\in\bf{R})$ and get
\begin{equation} \label{e4.6}
\begin{aligned}
\frac{1}{2}\|\hat{x}(t+\tau)-z\|^2&=\frac{1}{2}\|\hat{x}(t)-z\|^2
-\int_t^{t+\tau}\langle S\hat{x}(s)-Sz,\hat{x}(s)-z\rangle\\
&=\frac{1}{2}\|\hat{x}(t)-z\|^2-\int_0^{\tau}\langle
S\hat{x}(t+s)-Sz,\hat{x}(t+s)-z\rangle ds,
\end{aligned}
\end{equation}
Take any point $u$ in the $\alpha$-limit set of $\hat{x}$.
Then
$$
u(s)=\lim\hat{x}(t_n+s),
$$
uniformly for all $s$ in compact intervals of the real line and for
some sequence $(t_n)$ converging to $-\infty$.
Then from \eqref{e4.6} we obtain
$$
\frac{1}{2}l_-^2=\frac{1}{2}l_-^2-\int_0^{\tau}\langle
Su(s)-Sz,u(s)-z\rangle ds,
$$
which, since $\tau$ is arbitrary, implies that
$$
\langle Su(s)-Sz,u(s)-z\rangle=0,
$$
 for all $s$. This shows that $u$ lies in the orthogonality of $S$ at
$h$. Similarly we show that any function in the $\omega$-limit
set of $\hat{x}$ is an element of the  set $\mathcal{O}(h; S).$ .

 If the orthogonality set $\mathcal{O}(S;h)$ is a singleton, $\{w\}$,
say, then, certainly, $S^{-1}(h)=\{w\}$. Therefore, from
the previous arguments, we have
$$
l_-=\lim_{t\to-\infty}\|\hat{x}(t)-z\|=\|w-z\|
=\lim_{t\to+\infty}\|\hat{x}(t)-z\|=l_+.
$$
Setting $\tau=-2t$ in \eqref{e4.6} and  taking the limits as $t$
tends to $+\infty$ we get
$$
\int_{-\infty}^{+\infty}\langle Su(s)-Sz,u(s)-z\rangle ds=0.
$$
Since the integrand is nonnegative, we conclude that
$$
\langle Su(s)-Sz,u(s)-z\rangle=0,
$$
which implies that $u(s)$ lies in the
orthogonality of $S$ at $h$, hence it is a constant equal to $w$.
The proof is complete.
\end{proof}


\subsection*{Acknowledgements}
The author would like to express many thanks to Yakov Alber for his
kindness to provide me with pre-prints of his work. Also, special
thanks are expressed to the referee for his/her useful remarks.

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\section*{Addendum posted on September 15, 2008.}

The authors wish to correct some misprints and to clarify the 
uniform boundedness  (in $\tau$) of the function 
$r(t,\tau)$ considered in Lemma 3.10 of  this article.
\begin{itemize}
 
\item Page 7:  In (H1) add: $\sup_{t\geq 0}t\alpha(t)=:N<+\infty$
and in (2.12) add:  
$$
\sup_{t\geq0}\frac{\|f(t)-h\|}{\alpha(t)}=:\delta<+\infty.
$$

\item Page 10, line $-10$: Instead of $\exp(\int_0^t\dots)$ write $\exp(-\int_0^t\dots)$
 
\item Page 10, line $-5$:  
$\epsilon(B):=\sup_{z\in B}\sup_{t\geq0}\frac{\|T(t)z-Sz\|}{\alpha(t)}<+\infty$
 holds for any bounded $B\subset\mathcal{M}$.
 
\item Page 11 line 10:  Let $Y$ be the range of $y_{\alpha}$.

\item Page 11:  In equation (3.6) add the initial value:  
$z(0,\tau)=y_{\alpha}(\tau)$.

\item Page 11, line 14: In  Lemma 3.10 add the assumption:   
$$
P: = \sup_{u\in Y}\sup_{t\geq 0}\int_0^t\big(\|T(s)u-Su\|
+\|f(s)-h\|\big)e^{-(t-s)\alpha(t)}ds<+\infty.
$$

\item Page 11, line $-2$: This proves that $\psi(\cdot,\tau)$ is defined up 
to $+\infty$.


\item Page 12, line 1: Replace the paragraph with the following: 
To prove the uniform boundedness,   it is enough to prove the same fact 
for $r(t,\tau)$. To this end  fix any $\tau\geq 0$.
Integrating 
\begin{align*}
&[r'(s;\tau)+\alpha(\tau)r(s;\tau)]e^{s\alpha(\tau)}\\
&=[\|T(s)y_{\alpha}(\tau)-T(\tau)y_{\alpha}(\tau)\|
+ \|f(s)-f(\tau)\|]e^{s\alpha(\tau)}
\end{align*}
on the interval $[0,\tau]$  we get 
$r(\tau;\tau)\leq P+\epsilon(Y)+\delta$.
 Now, if there is  $s\leq\tau$ with 
$r(s;\tau)=R(\tau):=\sup_{t\geq 0}r(t;\tau)$,
we obtain  $0=r(0;\tau)\leq r(s;\tau)\leq r(s;\tau)e^{s\alpha(\tau)}
\leq r(\tau;\tau)e^{N}\leq (P+\epsilon(Y)+\delta)e^{N}$,  
while, if for all $t\leq\tau$ it happens 
$R(\tau)>r(t;\tau)$, then there is a sequence $t_n>\tau$ such that 
$r'(t_n;\tau)\geq 0$ and $r(t_n;\tau)\to R(\tau)$.
 Thus $\alpha(\tau)r(t_n;\tau)\leq \|T(t_n)y_{\alpha}(\tau)
-T(\tau)y_{\alpha}(\tau)\|+ \|f(t_n)-f(\tau)\|$ and therefore  
$R(\tau)\leq \max\{2\epsilon(Y)+2\delta,(P+\epsilon(Y)+\delta)e^{N}\}$.
 This completes the proof.
 
\item Page 12, line $-3$: Replace the set $\{z(t;\tau): t\geq 0\}$ by 
$\{z(t;\tau): t\geq 0,\, \tau\geq 0\}$.

\item Page 15, lines $-15,-16,-18$: Replace $u(s)$ with ${\hat{x}}(s)$.

\end{itemize}
\end{document}