\documentclass[reqno]{amsart}
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\AtBeginDocument{{\noindent\small
\emph{Electronic Journal of Differential Equations},
Vol. 2010(2010), No. 78, 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}
\thanks{\copyright 2010 Texas State University - San Marcos.}
\vspace{9mm}}
\begin{document}
\title[\hfilneg EJDE-2010/78\hfil Lyapunov stability of closed sets]
{Lyapunov stability of closed sets in impulsive semidynamical systems}
\author[E. M. Bonotto, N. G. Grulha\hfil EJDE-2010/78\hfilneg]
{Everaldo M. Bonotto, Nivaldo G. Grulha Jr.} % in alphabetical order
\address{Everaldo M. Bonotto \newline
Instituto de Ci\^encias Matem\'aticas e de Computa\c{c}\~ao,
Universidade de S\~ao Paulo-Campus de S\~ao Carlos, Caixa Postal
668, 13560-970 S\~ao Carlos SP, Brazil}
\email{ebonotto@icmc.usp.br}
\address{Nivaldo G. Grulha Jr. \newline
Instituto de Ci\^encias Matem\'aticas e de
Computa\c{c}\~ao, Universidade de S\~ao Paulo-Campus de S\~ao
Carlos, Caixa Postal 668, 13560-970 S\~ao Carlos SP, Brazil}
\email{njunior@icmc.usp.br}
\thanks{Submitted September 8, 2009. Published June 8, 2010.}
\thanks{The first author was supported by grant 2008/03680-4
from Fapesp, and the second author \hfill\break\indent
was supported by grant 03/13929-6 from Fapesp.}
\subjclass[2000]{34A37, 37B25, 54H20}
\keywords{Impulsive semidynamical systems; stability;
Lyapunov functionals}
\begin{abstract}
In this article, we consider impulsive semidynamical systems,
defined in a metric space, with impulse effects at variable times.
Converse-type theorems are included in our results giving necessary
and sufficient conditions for various types of stability of
closed subsets of the metric space.
These results are achieved by means of Lyapunov functionals
which indicate how the solutions behave when entering a
``stable'' set.
\end{abstract}
\maketitle
\numberwithin{equation}{section}
\newtheorem{theorem}{Theorem}[section]
\newtheorem{lemma}[theorem]{Lemma}
\newtheorem{proposition}[theorem]{Proposition}
\newtheorem{corollary}[theorem]{Corollary}
\newtheorem{example}[theorem]{Example}
\newtheorem{remark}[theorem]{Remark}
\newtheorem{definition}[theorem]{Definition}
\section{Introduction}
Impulsive semidynamical systems present interesting and important
phenomena such as ``beating'', ``dying", ``merging",
``noncontinuation of solutions", etc. These systems present a more
complex structure than the non-impulsive systems because of their
irregularity. In recent years, the theory of such systems has been
studied and developed intensively. See for instance
\cite{Bonotto}-\cite{Kaul2}.
Lyapunov stability theory has been studied by several authors in
investigations of continuous dynamical systems and impulsive
dynamical systems. The majority of these papers in the impulsive
case deal with systems with impulse effects at pre-assigned times.
In \cite{Kaul1}, the author considers a more general case where
the impulsive semidynamical system admits impulse effects at
variable times. He considers an impulsive semidynamical system
$(\Omega, \widetilde{\pi})$, where $\Omega \subset X$ is an
open set in a metric space $X$ and the continuous impulsive
function $I$ is defined from $\partial \Omega$ to $X$
$(\partial\Omega$ is the boundary of $\Omega$ in $ X)$. He
introduces a continuous Lyapunov function in $(\Omega,
\widetilde{\pi})$ denoted by $V: \overline{G} \to
\mathbb{R}$, where $G \subset \Omega$ is a positively invariant
closed set and $\overline{G}$ denotes the closure of a set $G$ in
$X$. The derivative of the function $V$ is defined by
$$
\dot{V}(x) = \lim_{t \to 0^{+}}\frac{V(\widetilde{\pi}(x, t))
- V(x)}{t}
$$
and the set $E$ by $\{x \in G: \dot{V}(x) = 0\}$. Considering $A \subset E$ as
being the largest invariant set under $\widetilde{\pi}$, Kaul
proved that $A$ is asymptotically stable, provided $A \subset int
G$ and $V(A) = a$ for some $a \in \mathbb{R}$. The converse result
is given as follows: if $A$ is asymptotically stable, then there
exists a positively invariant set $G$ in $\Omega$ containing $A$
which admits a Lyapunov function $V:G \to \mathbb{R}_{+}$
satisfying some properties. We can observe that the set $A$ does
not contain points of $\partial \Omega$ where the discontinuities
of the impulsive system occur.
In the present paper, we extend the results from \cite{Bhatia} for
the impulsive case. In \cite{Bhatia} the authors present several
results which give necessary and sufficient conditions for the
stability of closed sets in non-impulsive dynamical systems
defined in a metric space. We consider impulsive semidynamical
systems of type $(X, \pi; M , I)$ subject to impulse
action which varies in time, where $X$ is a metric space,
$( X, \pi)$ is a semidynamical system, $ M$ is a
non-empty closed subset of $X$ that denotes the impulsive set
and $I: M \to X$ is the impulse function. We give
necessary and sufficient conditions for various types of stability
of closed sets of $X$. In other words, we establish necessary
and sufficient conditions so that the solutions of the impulsive
system become ``stable'' in some sense after entering a closed
subset of $X$. Converse-type results are included in the main
theorems. In contrast to the paper \cite{Kaul1}, the set which we
prove to be ``stable" can contain points of $M$ where the
discontinuities of the impulsive system occur. %Since the structure
In the first part of this article, we present the basis of the theory
of impulsive semidynamical systems. We present basic definitions
and notations and then we discuss the continuity of a function
which describes the times of reaching the impulsive set. We also
present additional useful definitions.
The second part of the paper concerns the main results. We
introduce two new concepts of stability of sets in impulsive
semidynamical systems and we relate these concepts of stability to
other known concepts. We give necessary and sufficient conditions
for the various types of stability of closed sets of $X$. We
prove that there exists a functional which plays the role of a
Lyapunov functional indicating how the solutions behave when
entering a ``stable'' closed set provided this set is ``stable''
and we also state the reciprocal of this fact. In addition, we
show that this Lyapunov functional is continuous when the
impulsive set is contained in the closed set. Finally we present
two examples to show how the theory can be employed.
\section{Preliminaries}
In this section we present the basic definitions and notation of the
theory of impulsive semidynamical systems. We also include some
fundamental results which are necessary for understanding the basis
of the theory.
\subsection{Basic definitions and terminology}
Let $X$ be a metric space and $\mathbb{R}_{+}$
be the set of non-negative real numbers. The triple $( X,
\pi, \mathbb{R}_{+})$ is called a \emph{semidynamical system},
if the function $\pi : X \times
\mathbb{R}_{+}\to X$ is continuous with $\pi
(x,0)=x$ and $\pi (\pi (x,t),s)= \pi (x, t+s), $ for all $x \in
X$ and $t,s\in \mathbb {R_{+}}$. We denote such system by
$( X, \pi, \mathbb{R}_{+})$ or simply $( X, \pi)$.
When $\mathbb{R}_{+}$ is replaced by $\mathbb{R}$ in the
definition above, the triple $( X, \pi, \mathbb{R})$ is a
\emph{dynamical system}. For every $x \in X$, we consider the
continuous function $\pi_{x}: \mathbb{R}_{+} \to X$ given by
$\pi _{x}(t) = \pi (x,t)$ and we call it the
\emph{motion} of $x$.
Let $(X, \pi)$ be a semidynamical system. Given $x \in X$,
the \emph{positive orbit} of $x$ is given by $\text {C}^{+}
(x) = \{\pi (x,t) : t \in \mathbb{R}_{+}\}$ which we also
denote by $\pi ^{+} (x)$. For $t \geq 0$ and $x \in X$, we
define $F(x,t)= \{y \in X: \pi (y, t) = x \}$
and, for $\Delta \subset [0, +\infty)$ and $D \subset X$, we define
$$
F(D, \Delta) = \cup \{F(x, t): x \in D
\text{ and } t \in \Delta \}.
$$
Then a point $x \in X$ is
called an \emph{initial point}, if $F(x, t) = \emptyset$
for all $t>0$.
Now we define semidynamical systems with impulse action. An
\emph{impulsive semidynamical system} $( X, \pi; M,
I)$ consists of a semidynamical system, $( X, \pi)$, a
non-empty closed subset $M$ of $X$ such that for every $x
\in M$, there exists $\varepsilon_x >0$ such that
\[
F(x, (0, \varepsilon_x))\cap M = \emptyset
\quad \text{and} \quad \pi (x, (0,
\varepsilon_x))\cap M = \emptyset,
\]
and a continuous function $I : M \to X$ whose
action we explain below in the description of the impulsive
trajectory of an impulsive semidynamical system. The points of
$M$ are isolated in every trajectory of system $( X,
\pi)$. The set $M$ is called the \emph{impulsive set}, the
function $I$ is called \emph{impulse function} and we write
$N = I(M)$. We also define
$$
M^{+}(x)= (\pi ^{+}(x)\cap M)\setminus \{x\}.
$$
Another property of the impulsive set $M$ is that $M$ is a
meager set in $X$ as shown by the next lemma.
\begin{lemma} \label{lem2.0}
Let $(X, \pi; M, I)$ be an impulsive semidynamical
system. The impulsive set $M$ is a meager set in $X$.
\end{lemma}
\begin{proof}
The proof is immediate because the points of $M$ are isolated in
every trajectory of the system
$( X, \pi)$. Therefore $\mathop{\rm int}(\overline{ M}) = \emptyset$ in
$X$ and the result follows.
\end{proof}
Given an impulsive semidynamical systems $(X, \pi; M,I)$ and
$x\in X$ such that $ M^{+}(x) \neq \emptyset$,
it is always possible to find a smallest number $s$ such that the
trajectory $\pi _{x}(t)$ for $0 < t < s$ does not intercept
the set $M$. This result is stated next and a proof of it can
be found in \cite{Bonotto}.
\begin{lemma} \label{lemma2.1} Let $(X, \pi; M, I)$ be
an impulsive semidynamical system. Then for every $x \in X$,
there is a positive number $s$, $0 < s\leq +\infty$, such that
$\pi (x, t) \notin M$, whenever $0 < t < s$, and $\pi (x,
s) \in M$ if $ M^{+}(x) \neq \emptyset$.
\end{lemma}
Let $(X, \pi; M, I)$ be an impulsive semidynamical
system and $x \in X$. By means of Lemma \ref{lemma2.1}, it is
possible to define a function
$\phi : X \to (0,+\infty]$ in the following manner
\[
\phi(x)=\begin{cases}
s, & \text{if $\pi (x, s) \in M$ and
$\pi(x, t) \notin M$ for $0 < t < s$,} \\
+\infty, & \text{if } M^{+}(x) = \emptyset .
\end{cases}
\]
This means that $\phi (x)$ is the least positive time for which the
trajectory of $x$ meets $M$. Thus for each $x \in X$, we
call $\pi (x, \phi (x))$ the \emph{impulsive point} of $x$.
The \emph{impulsive trajectory} of $x$ in $(X, \pi; M,
I)$ is an $ X-$valued function $\widetilde{\pi } _{x}$ defined on
the subset $[0,s)$ of $\mathbb{R}_{+}$ $(s$ may be $+\infty)$. The
description of such trajectory follows inductively as described in
the following lines.
If $ M^{+} (x) = \emptyset$, then $\widetilde{\pi}_{x}(t) =
\pi (x,t)$, for all $t \in \mathbb{R}_{+}$, and $\phi (x) = +
\infty$. However if $ M^{+} (x) \neq \emptyset$, it follows
from Lemma \ref{lemma2.1} that there is a smallest positive number
$s_{0}$ such that $\pi (x, s_{0}) = x_{1} \in M$
and $\pi (x, t) \notin M$, for $0< t 0$, if there exists a closed set $L\subset X$ such that
\begin{itemize}
\item[(a)] $F(L, \lambda) = S$;
\item[(b)] $F(L, [0, 2 \lambda])$ is a neighborhood of $x$;
\item[(c)] $F(L, \mu) \cap F(L,\nu) = \emptyset$, for
$0 \leq \mu < \nu \leq 2 \lambda$.
\end{itemize}
The set $F(L,[0, 2 \lambda])$ is called a \emph{tube} or
a \emph{$\lambda$-tube} and the set $L$ is called a \emph{bar}.
Let $( X, \pi)$ be a semidynamical system. We now present the
conditions TC and STC for a tube.
Any tube $F(L,[0,2 \lambda])$ given by a section $S$
through $x \in X$ such that
$S \subset M \cap F(L, [0,2 \lambda])$ is called \emph{TC-tube} on $x$.
We say that a point $x\in M$ fulfills the \emph{Tube Condition}
and we write (TC), if there exists a TC-tube
$F(L, [0, 2 \lambda])$ through $x$.
In particular, if $S = M \cap F(L, [0, 2 \lambda])$ we
have a \emph{STC-tube} on $x$ and we say that a point $x \in M$
fulfills the \emph{Strong Tube Condition} (we write (STC)), if there
exists a STC-tube $F(L, [0, 2 \lambda])$ through $x$.
The following theorem concerns the continuity of $\phi$ which is
accomplished outside $M$ for $M$ satisfying the condition
TC. See \cite{Ciesielski1}, Theorem 3.8.
\begin{theorem}\label{theorem2.1}
Consider an impulsive semidynamical system
$(X, \pi; M, I)$. Assume that no initial point in
$( X, \pi)$ belongs to the impulsive set $M$ and that each
element of $M$ satisfies the condition $(TC)$. Then $\phi$
is continuous at $x$ if and only if $x \notin M$.
\end{theorem}
\begin{remark}
Suppose the conditions of Theorem \ref{theorem2.1} are true.
Although the function $\widetilde{\pi}$ is not continuous, by the
continuity of the impulse function $I: M \to I(M)$ and
function $\phi$, we can obtain the following result: Suppose
$x \in X \setminus M$. Given $\varepsilon > 0$, for each
$k =0,1,2,\dots$. and $t \in [0, \phi(x_k^{+})]$, there is a
$\delta_{k}> 0$ such that
$\rho (\pi(x_{k}^{+}, t), \pi(y_{k}^{+}, t)) <\varepsilon$
whenever $\rho(y_{k}^{+}, x_{k}^{+}) < \delta_k$
$(\rho$ is a metric in $X$ and $x_0^+ = x)$. This result is
applied in the proofs of the main theorems of the next section.
\end{remark}
\subsection{Additional definitions}
Let us consider a metric space $X$ with metric
$\rho$. By $B(x, \delta )$ we mean the open ball with center at
$x \in X$ and ratio $\delta$. Let $ B(A, \delta) = \{ x \in
X : \rho_{A}(x) < \delta \}$ and $B[A, \delta] = \{ x \in
X : \rho_{A}(x) \leq \delta \}$, where $\rho_{A}(x) = \inf
\{\rho(x, y) : y \in A\}$. Throughout this paper, we use the
notation $\partial A$, $\mathop{\rm int}(A)$ and $\overline{A}$ to denote
respectively the boundary, interior and closure of $A$ in $X$.
In what follows, $(X, \pi; M, I)$ is an impulsive
semidynamical system and $x \in X$.
We define the \emph{prolongation set} of $x$ in $(X, \pi; M, I)$ by
\[
\widetilde{D}^{+}(x) = \{y \in X :
\widetilde{\pi}(x_{n}, t_{n}) \stackrel{n \to
+\infty}{\to} y, \text{ for some } x_{n}
\stackrel{n \to +\infty}{\to} x
\text{ and } t_{n} \in [0, +\infty) \}.
\]
For a set $A \subset X$ we consider $\widetilde{D}^{+}(A)
= \cup \{\widetilde{D}^{+}(x): x \in A\}$.
If $\widetilde{\pi}^{+}(A) \subset A$, we say that $A$ is
\emph{$\widetilde{\pi}$-invariant}.
A point $x \in X$ is called \emph{stationary} or \emph{rest
point} with respect to $\widetilde{\pi}$, if $\widetilde{\pi} (x,
t) = x$ for all $t \geq 0$, it is a \emph{periodic point} with
respect to $\widetilde{\pi}$, if $\widetilde{\pi} (x, t) = x$ for
some $t > 0$ and $x$ is not stationary, and it is a \emph{regular
point} if it is neither a rest point nor a periodic point.
Let $A \subset X$. If for every $\varepsilon >0$ and every $x
\in A$, there is $\delta =\delta (x, \varepsilon) >0$ such that
$\widetilde{\pi} (B(x, \delta), [0, +\infty)) \subset
B(A, \varepsilon)$, then $A$ is called
$\widetilde{\pi}$-stable. The set $A$ is orbitally
$\widetilde{\pi}$-stable if for every neighborhood $U$ of $A$,
there is a positively $\widetilde{\pi}-$invariant neighborhood $V$
of $A$, $V \subset U$. If for all $x \in A$ and all $y \notin A$,
there exist a neighborhood $V$ of $x$ and a neighborhood $W$ of
$y$ such that $W \cap \widetilde{\pi} (V, [0, +\infty)) =
\emptyset$, we say that $A$ is $\widetilde{\pi}$-stable according
to Bhatia-Hajek \cite{Bhatia}. We define the set
\begin{align*}
\widetilde{P}^{+}_{W}(A)
= \big\{&x \in X: \text{ for every neighborhood }
U \text{ of } A, \text{ there is a sequence } \\
&\{t_n\} \subset \mathbb{R}_{+}, t_n \stackrel{n \to
+\infty}{\to} +\infty \text{ such that }
\widetilde{\pi}(x, t_n) \in U\}.
\end{align*}
The set
$\widetilde{P}^{+}_{W}(A)$ is called
\emph{region of weak attraction} of $A$ with respect to
$\widetilde{\pi}$. If $x \in \widetilde{P}^{+}_{W}(A)$, then we say that
$x$ is \emph{$\widetilde{\pi}$-weakly attracted} to $A$. A
subset $A \subset X$ is called a \emph{weak
$\widetilde{\pi}$-attractor}, if
$\widetilde{\mathrm{P}}^{+}_{W} (A)$ is a
neighborhood of $A$. A set $A \subset X$ is called
\emph{asymptotically $\widetilde{\pi}$-stable}, if it is both a weak
$\widetilde{\pi}$-attractor and orbitally $\widetilde{\pi}$-stable.
For results concerning the stability and invariancy of sets in an
impulsive system, the reader may want to consult \cite{Bonotto},
\cite{Bonotto1}, \cite{Ciesielski2} and \cite{Kaul1}.
\section{Main Results}
We divide this section into two parts. The first
part concerns the relations among some concepts of stability. In
the second part, we discuss Lyapunov stability of closed sets in
impulsive semidynamical systems where the results give necessary
and sufficient conditions for various types of stability of closed
sets.
Let $(X, \pi; M, I)$ be an impulsive semidynamical system
where $X$ is a metric space. We assume the following additional
hypotheses:
$\bullet$ No initial point in $(X, \pi)$ belongs to the
impulsive set $M$, that is, given $x \in M$ there are $y \in
X$ and $t \in \mathbb{R}_{+}$ such that $\pi (y, t) = x$.
$\bullet$ Each element of $M$ satisfies the condition $
(STC)$ (\emph{consequently, $\phi$ is continuous on $ X
\setminus M$}).
$\bullet$ $ M \cap \rm I(M) = \emptyset$.
$\bullet$ For all $x \in X$ and for all $k \geq 1$, $x_k^+$ is
defined and $ M^+ (x^+_k) \neq \emptyset$, that is, the
trajectory of $x \in X$ has infinitely many discontinuities.
Consequently, $\phi (x) < +\infty$ for all $x \in X$.
\subsection{Stability}
We introduce two new concepts of stability for impulsive
semidynamical systems. Then, we relate these new concepts to known
ones.
\begin{definition}\label{def1.1} \rm
Let $(X, \pi; M, I)$
be an impulsive semidynamical system. A set $A \subset X$ is
said to be
\begin{itemize}
\item[(a)] equi $\widetilde{\pi}$-stable, if for each $x \notin A$,
there is a $\delta = \delta (x) >0$ such that $$x \notin
\overline{\widetilde{\pi}(B(A, \delta), [0, +\infty))}.$$
\item[(b)] uniformly $\widetilde{\pi}$-stable, if for each
$\varepsilon >0$, there is a $\delta = \delta (\varepsilon) >0$
such that
$$
\widetilde{\pi}(B(A, \delta),[0, +\infty)) \subset
B(A, \varepsilon).
$$
\end{itemize}
\end{definition}
The next result deals with the equivalence between equi
$\widetilde{\pi}$-stability and uniform
$\widetilde{\pi}$-stability of a compact subset $A \subset X$.
This result is also valid when we replace the condition of equi
$\widetilde{\pi}$-stability by $\widetilde{\pi}$-stability. The
proof is similar to the continuous case, see \cite{Bhatia}.
\begin{theorem}\label{theo1.1}
Let $(X, \pi; M, I)$ be an impulsive semidynamical
system, $X$ is locally compact and $A \subset X$ is
compact. Then, $A$ is equi $\widetilde{\pi}$-stable if and
only if $A$ is uniformly $\widetilde{\pi}$-stable. Replacing
the hypotheses equi $\widetilde{\pi}-$stability by
$\widetilde{\pi}-$stability, the result remains true.
\end{theorem}
\begin{remark}\label{rem1.1} \rm
If the set $A \subset X$ is closed but not compact, then the
sufficiency of the theorem does not necessarily hold. Indeed,
consider the discontinuous flow shown in Figure \ref{fig1}, where
$\widehat{M} = \{(-1, x_2): x_2 \in \mathbb{R}\}$,
$\widehat{N} = \{(2, x_2): x_2 \in
\mathbb{R}\}$, $p = (x_1', 0) \in \mathbb{R}^{2}$
and the impulsive function $I_1: \widehat{M} \to \widehat{N}$
is given by
$I_1 (-1, x_2) = (2, x_2')$ such that
$x_{2}' < x_{2}'' < x_{2}$, where $x_{2}''$ is such that
for some unique $\lambda >0$,
$\pi ((2, x_{2}'), \lambda) = (x_1', x_{2}'')$.
\end{remark}
\begin{figure}[ht] \label{fig1}
\begin{center}
\setlength{\unitlength}{1mm}
\begin{picture}(0,0)(0,0)
\includegraphics[width=0.8\textwidth]{fig1}
\end{picture}
\begin{picture}(98,49)(0,0)
\put(51,45){$x_2$}
\put(41,40){$\widehat{M}$}
\put(60,40){$\widehat{N}$}
\put(68,10){$x_1'$}
\put(96,10){$x_1$}
\end{picture}
\end{center}
\caption{Discontinuous flow}
\end{figure}
Note that the trajectories for $x_1> x_1'$ are straight
lines parallel to the axis $0x_1$. This discontinuous
flow has the property that for all $x \in \mathbb{R}^{2}$,
$\lim _{t \to -\infty} \widetilde{\pi} (x,t) = 0$. Now,
consider the sets
\[
M = \widehat{M}
\cup _{n=1}^{+\infty} \{(x_1' + 3n, x_2): x_2
\in \mathbb{R}\},
\]
\[
\mathrm{N} = \widehat{N}\cup _{n=1}^{+\infty} \{(x_1' + 3n + 1,
x_2): x_2 \in \mathbb{R}\}
\]
and define the function $I: M \to N$ as follows
\[
I(-1, x_2) = I_1(-1, x_2) \quad \text{for all } x_2 \in \mathbb{R}
\]
and
\[
I(x_1' + 3n, x_2) = (x_1' + 3n + 1, x_2) \quad
\text{for all } x_2 \in \mathbb{R} \text{ and } n = 1,2,\dots.
\]
Then, the impulsive semidynamical system
$(\mathbb{R}^{2}, \pi; M, I)$ has infinitely many discontinuities.
Let $A = \{(x_1, x_2) \in \mathbb{R}^{2}: x_2 = 0\}$. Clearly $A$ is
$\widetilde{\pi}$-stable, but it is not uniformly
$\widetilde{\pi}$-stable.
The next result shows the equivalence between the orbital
stability and the uniform stability in impulsive semidynamical
systems. The proof is similar to the continuous case, see
\cite{Bhatia}.
\begin{theorem}\label{theo1.2}
Let $(X, \pi; M, I)$ be an impulsive semidynamical
system. Assume that $X$ is locally compact and $A \subset X$ is compact.
Then $A$ is orbitally $\widetilde{\pi}$-stable if
and only if $A$ is uniformly $\widetilde{\pi}$-stable.
\end{theorem}
By \cite[Theorem 4.1]{Ciesielski2} and Theorems \ref{theo1.1}
and \ref{theo1.2} above, we have the following result which relates
various concepts of stability.
\begin{theorem}\label{THEO3.5}
Let $(X, \pi; M, I)$ be an impulsive semidynamical
system. Assume that $X$ is locally compact and $A$ is a compact
subset of $X$. Then the following conditions are equivalent:
\begin{itemize}
\item[(a)] $A$ is $\widetilde{\pi}$-stable.
\item[(b)] $A$ is orbitally $\widetilde{\pi}$-stable.
\item[(c)] $A$ is $\widetilde{\pi}$-stable in the sense of Bhatia and
Hajek.
\item[(d)] $A$ is uniformly $\widetilde{\pi}$-stable.
\item[(e)] $A$ is equi $\widetilde{\pi}$-stable.
\item[(f)] $\widetilde{\mathrm{D}}^{+}(A)=A$.
\end{itemize}
\end{theorem}
The $\widetilde{\pi}$-stability of a closed subset $A$ of $X$
implies that $I( M) \subset A$, for $ M
\subset A$, as shown by the next lemma.
\begin{lemma} \label{lem1.1}
Let $(X, \pi; M, I)$ be an impulsive semidynamical
system and $A \subset X$ be closed. If $A$ is
$\widetilde{\pi}$-stable and $ M \subset A$, then
$I( M) \subset A$.
\end{lemma}
\begin{proof}
Given $x \in A$ and $\varepsilon >0$, there is a
$\delta = \delta (x, \varepsilon)>0$ such
that
$$
\widetilde{\pi} (B(x, \delta), [0, +\infty)) \subset
B(A, \varepsilon).
$$
Since $\varepsilon$ is arbitrary, we have
$\widetilde{\pi}^{+}(x) \subset \overline{A} = A$. Therefore,
$I(M) \subset A$ provided $ M \subset A$.
\end{proof}
\subsection{Lyapunov Stability}
In this section, we shall present the results that concern the
Lyapunov stability of certain closed sets of $X$. These results
are achieved by means of functionals which play the role of a
Lyapunov functional indicating how the solutions behave when
entering a ``stable'' set. The results give necessary and sufficient
conditions for the various types of stability of closed sets of $X$. We start by presenting a result on $\widetilde{\pi}$-stability.
\begin{theorem}\label{Theorem3.4}
Let $(X, \pi; M, I)$ be an impulsive semidynamical
system and $A \subset X$ be closed.
\begin{itemize}
\item[(1)] If there exists a functional $\psi : X \to
\mathbb{R}_{+}$ with the following properties:
\begin{itemize}
\item[(a)] $\psi$ is continuous in $X \setminus
( M \setminus A)$.
\item[(b)] For every $\varepsilon >0$,
there is a $\delta >0$ such that $\psi (x) \geq \delta$ whenever
$\rho (x, A) \geq \varepsilon$ and $x \notin M$, and for
any sequence $\{w_n\}_{n \geq 1} \subset X$ such that $w_n
\stackrel{n \to +\infty}{\to} x \in A$ implies
$\psi (w_n) \stackrel{n \to +\infty}{\to} 0$.
\item[(c)] $\psi (\pi(x, t)) \leq \psi (x)$ if $x \in
X \setminus M$ and $0 \leq t \leq \phi(x)$, and
$\psi (I(x)) \leq \psi (x)$ if $x \in M$.
\end{itemize}
Then $A$ is $\widetilde{\pi}$-stable.
\item[(2)] Reciprocally, if $A$ is $\widetilde{\pi}$-stable, then
there is a functional $\psi : X \to \mathbb{R}_{+}$
satisfying conditions a), b) and c) above.
\end{itemize}
\end{theorem}
\begin{proof}
Let us prove the necessary condition. Given $\varepsilon >0$
and $x \in A$, set $\mu = \inf \{\psi (w): w \notin
M \text{ and } \rho (w, A) \geq
\frac{\varepsilon}{2}\}$. Note that $\mu
>0$, because by item (b), there is a $\delta >0$ such that $\psi (a)
\geq \delta$ whenever $\rho (a, A) \geq
\frac{\varepsilon}{2}$ and $a \notin M$. We have
two cases to consider: when $x \in int (A)$ and when $x \in
\partial A$.
First, suppose $x \in int (A)$. Then, by the second part of item
(b) and by the continuity of $\psi$ in $A$, there is a $\delta _1
>0$ such that
\begin{equation}\label{eq1}
\psi (y) < \mu \quad \text{for all } y \in B(x, \delta
_1) \subset A.
\end{equation}
We suppose by contradiction that
$\widetilde{\pi}(B(x, \delta_1), [0, +\infty))$ is not contained
in $B(A, \varepsilon)$. Thus, there are $z \in B(x, \delta _1)$
and $t_1 \in (0, +\infty)$ such that
\begin{equation}\label{eq11}
\widetilde{\pi} (z, t_1) \notin B(A, \varepsilon).
\end{equation}
Note that $\widetilde{\pi} (z, t_1) \notin M$ because $M
\cap I(M) = \emptyset$. By equation \eqref{eq11}, $\rho
(\widetilde{\pi} (z, t_1), A) \geq \varepsilon$ and this
implies
\begin{equation}\label{eq2}
\psi(\widetilde{\pi} (z, t_1)) \geq \inf \big\{\psi (w): w
\notin M \; \text{and} \; \rho (w, A) \geq
\frac{\varepsilon}{2} \big\} = \mu .
\end{equation}
We have two cases to consider: when $z \in M$ and when $z \notin
M$. First suppose that $z \notin M$. Note that as $z \in
B(x, \delta _1)$, then $\psi (z) < \mu$ by \eqref{eq1}. Hence for
$0 \leq t < \phi (z)$, we have
\[
\psi (\widetilde{\pi} (z, t)) = \psi (\pi (z, t))
\stackrel{(c)}{\leq} \psi (z) < \mu .
\]
If $t = \phi (z)$ and remembering from the definition of
$\widetilde{\pi}$ that $z_1 = \pi (z, \phi (z))$, then
\begin{equation}\label{eq2.1}
\psi (\widetilde{\pi} (z, t)) = \psi (\widetilde{\pi} (z, \phi
(z))) = \psi (I(z_1)) \stackrel{(c)}{\leq} \psi (z_1) = \psi
(\pi (z, \phi (z))) \stackrel{(c)}{\leq} \psi (z) < \mu .
\end{equation}
Now if $\phi (z) < t < \phi(z) + \phi (z_1^{+})$, then
\[
\psi (\widetilde{\pi} (z, t)) = \psi (\pi (z_1^{+}, t - \phi
(z))) \stackrel{(c)}{\leq} \psi (z_1^{+})
= \psi (\widetilde{\pi} (z, \phi (z))) \stackrel{\eqref{eq2.1}}{<} \mu .
\]
Repeating this argument, we get $\psi(\widetilde{\pi} (z, t)) <
\mu$ for all $t \geq 0$. In particular for $t = t_1$,
$\psi(\widetilde{\pi} (z, t_1)) < \mu$ which is a contradiction
by \eqref{eq2}. Hence, $\widetilde{\pi}(B(x, \delta _1), [0, +\infty)) \subset B(A, \varepsilon)$. Now, suppose $z \in
M$. Take $\nu >0$, $\nu < t_1$, such that $\widetilde{\pi}(z, \nu) = \pi (z, \nu) \in B(x, \delta _{1}) \setminus M$. By the same argument used above for $z \notin M$, we get
$\psi(\widetilde{\pi} (\widetilde{\pi}(z, \nu), t)) < \mu$
for all $t \geq 0$. In particular for $t = t_1 - \nu$,
$\psi(\widetilde{\pi}(z, t_1)) = \psi(\widetilde{\pi}
(\widetilde{\pi}(z, \nu), t_1 - \nu)) < \mu$ which is a
contradiction by \eqref{eq2}. Therefore, we get again
$\widetilde{\pi}(B(x, \delta _1), [0, +\infty)) \subset
B(A, \varepsilon)$.
Now we assume that $x \in \partial A$. Since $\psi$ is continuous
in $X \setminus ( M \setminus A)$, $M$ is a
meager set in $X$ and by the second part of item (b), there is
a $\delta _2>0$, $\delta_2 < \varepsilon$, such that $\psi (y) <
\mu$ for all $y \in B(x, \delta _2) \setminus M$. Supposing
that $\widetilde{\pi}(B(x, \delta _2), [0, +\infty))$ is
not contained in $B(A, \varepsilon)$, there are $z \in B(x,
\delta _2)$ and $t_2 \in (0, +\infty)$ such that
$\widetilde{\pi} (z, t_2) \notin B(A, \varepsilon)$. Thus
$\rho (\widetilde{\pi} (z, t_2), A) \geq \varepsilon$,
$\widetilde{\pi} (z, t_2) \notin M$ because $M \cap
I(M) = \emptyset$ and therefore
\begin{equation}\label{eq3}
\psi(\widetilde{\pi} (z, t_2)) \geq \inf \big\{\psi (w): w
\notin M \text{ and } \rho (w, A) \geq
\frac{\varepsilon}{2} \big\} = \mu .
\end{equation}
If $z \in B(x, \delta _2) \setminus M$, then it can be
shown that $\psi (\widetilde{\pi} (z, t)) < \mu$ for all $t
\geq 0$ as we did before. Hence, $\psi(\widetilde{\pi} (z,
t_2)) < \mu$ which is a contradiction by \eqref{eq3}. Also, if $z
\in B(x, \delta _2) \cap M$, then $z$ is an initial point
for the impulsive system and there is a time $\tau > 0$ such that
$\widetilde{\pi}(z, (0, \tau)) = \pi (z, (0, \tau))
\subset B(x, \delta _2)\setminus M$. Taking $t^{*}$, $0 <
t^{*} < \tau$. By the previous case, $\psi (\widetilde{\pi} (\pi
(z, t^{*}), t)) < \mu$ for all $t\geq0$. As a result, $\psi
(\widetilde{\pi} (z, t_2)) = \psi (\widetilde{\pi} (\pi (z,
t^{*}), t_2 - t^{*})) < \mu$ and this is a contradiction by
\eqref{eq3}. Therefore, $\widetilde{\pi}(B(x, \delta _2),
[0, +\infty)) \subset B(A, \varepsilon)$.
Consequently, $A$ is $\widetilde{\pi}$-stable.
Let us prove the sufficient condition. Define the function $\psi :
X \to \mathbb{R}_{+}$ by
$$
\psi(x) = \begin{cases}
\sup_{k \geq 0} \Big(\sup _{0 \leq t \leq \phi (x_k^{+})} \frac{\rho
(\pi(x_{k}^{+}, t), A)}{1 + \rho (\pi(x_{k}^{+},
t), A)}\Big), & \text{if } x \in X \setminus M, \\
\psi (I(x)), & \text{if } x \in M,
\end{cases}
$$
where $x_{0}^{+} = x$.
We shall verify that $\psi$ satisfies conditions (a), (b) and (c).
(a) Take $x \in X \setminus M$. Since $\{x\}$ is compact and
$M$ is closed, there is an $\eta >0$, such that $B(x, \eta)
\cap M = \emptyset$. Given a sequence $\{w_n\}_{n \geq 1}
\subset X$ such that $w_n \stackrel{n \to
+\infty}{\to} x$, there is an integer $n_{0} >0$ such
that $w_n \in B(x, \eta)$ for $n > n_{0}$. Since $I$ is a
continuous function and $\phi$ is continuous on
$ X \setminus M$ we have
\[
(w_n)_{1}^{+} = I (\pi (w_n, \phi (w_n))) \stackrel{n
\to +\infty}{\to} I (\pi (x,
\phi(x))) = x_{1}^{+}.
\]
Note that $x_{1}^{+} \notin M$ because $M \cap I(M) =
\emptyset$. But $\{x_{1}^{+}\}$ is compact and $M$ is closed,
then there is an $\eta _1 > 0$ such that $B(x_{1}^{+}, \eta _1)
\cap M = \emptyset$. As $(w_n)_{1}^{+} \stackrel{n \to
+\infty}{\to} x_{1}^{+}$, there is an integer
$n^{1}_{0}>0$ such that $(w_n)_{1}^{+} \in B(x_{1}^{+}, \eta _{1})$
for $n > n^{1}_{0}$.
By the continuity of $\phi$ on $ X \setminus M$ we have
\[
\phi((w_n)_{1}^{+}) \stackrel{n
\to +\infty}{\to} \phi(x_{1}^{+}).
\]
Then
\[
\sup _{0 \leq t \leq \phi ((w_n)_{1}^{+})}
\frac{\rho (\pi((w_n)_{1}^{+}, t), A)}{1 + \rho
(\pi((w_n)_{1}^{+}, t), A)} \stackrel{n \to
+\infty}{\to} \sup _{0 \leq t \leq \phi (x_1^+)}
\frac{\rho (\pi(x_{1}^{+}, t), A)}{1 + \rho (\pi(x_{1}^{+}, t), A)}.
\]
Analogously, since $(w_n)_{1}^{+} \stackrel{n \to
+\infty}{\to} x_{1}^{+}$ and $\phi((w_n)_{1}^{+})
\stackrel{n \to +\infty}{\to}
\phi(x_{1}^{+})$, it follows
\[
(w_n)_{2}^{+} = I (\pi ((w_n)_{1}^{+}, \phi
((w_n)_{1}^{+}))) \stackrel{n \to
+\infty}{\to} I (\pi (x_{1}^{+},
\phi(x_{1}^{+}))) = x_{2}^{+}
\]
and
\[
\sup _{0 \leq t \leq \phi ((w_n)_{2}^{+})}
\frac{\rho (\pi((w_n)_{2}^{+}, t), A)}{1 + \rho
(\pi((w_n)_{2}^{+}, t), A)} \stackrel{n \to
+\infty}{\to} \sup _{0 \leq t \leq \phi (x_2^+)}
\frac{\rho (\pi(x_{2}^{+}, t), A)}{1 + \rho (\pi(x_{2}^{+}, t), A)}.
\]
We can continue with this process because $x_{k}^{+} \notin M$
for all $k \geq 1$. Thus, for each integer $k \geq 1$ we obtain
\[
(w_n)_{k}^{+} = I (\pi ((w_n)_{k-1}^{+}, \phi
((w_n)_{k-1}^{+}))) \stackrel{n \to
+\infty}{\to} I (\pi (x_{k-1}^{+},
\phi(x_{k-1}^{+}))) = x_{k}^{+}
\]
and
\[
\sup _{0 \leq t \leq \phi
((w_n)_{k}^{+})} \frac{\rho (\pi((w_n)_{k}^{+}, t), A)}{1 +
\rho (\pi((w_n)_{k}^{+}, t), A)} \stackrel{n
\to +\infty}{\to} \sup
_{0 \leq t \leq \phi (x_k^+)} \frac{\rho (\pi(x_{k}^{+}, t), A)}{1 + \rho (\pi(x_{k}^{+}, t), A)}.
\]
Therefore,
\begin{align*}
&\sup_{k \geq 0}\Big(\sup _{0 \leq t \leq \phi
((w_n)_{k}^{+})} \frac{\rho (\pi((w_n)_{k}^{+}, t), A)}{1 +
\rho (\pi((w_n)_{k}^{+}, t), A)}\Big)\\
& \stackrel{n
\to +\infty}{\to} \sup_{k \geq 0}\Big(\sup
_{0 \leq t \leq \phi (x_k^+)} \frac{\rho (\pi(x_{k}^{+}, t), A)}
{1 + \rho (\pi(x_{k}^{+}, t), A)}\Big).
\end{align*}
In conclusion, $\psi (w_n) \stackrel{n \to
+\infty}{\to} \psi(x)$ and $\psi$ is continuous on
$ X \setminus M$.
Since we want to prove that $\psi$ is continuous on
$X\setminus (M \setminus A)$, it is enough to
show that $\psi$ is continuous on $A \cap M$. Assume that $x
\in A \cap M$. Since $A$ is $\widetilde{\pi}$-stable, given
$\varepsilon >0$, there is a $\delta = \delta (x,
\varepsilon)>0$ such that $\widetilde{\pi} (B(x, \delta),
[0, +\infty)) \subset B(A, \varepsilon)$. Since
$\varepsilon$ is arbitrary, we have $\widetilde{\pi}^{+}(x)
\subset \overline{A} = A$. Thus, $\pi (x_k^{+}, t) \subset A$
for all $0 \leq t \leq \phi (x_k^{+})$, $k \geq 0$ $(x_0^{+} =
x)$. Consequently $\rho (\pi (x_k^{+}, t), A) = 0$ for all
$0 \leq t \leq \phi (x_k^{+})$, $k \geq 0$. Hence $\psi (x) = \psi
(I(x)) = \psi (x_1^{+}) = 0$.
Considering $\widetilde{\pi} (B(x, \delta), [0, +\infty))
\subset B(A, \varepsilon)$ from the $\widetilde{\pi}$-stability
of $\widetilde{\pi}$, if $\{z_n\}_{n \geq 1}$ is a sequence in
$X$ such that $z_n \stackrel{n \to
+\infty}{\to} x$, then there is a positive integer
$n_0 > 0$ such that $z_n \in B(x, \delta)$ for $n > n_0$.
Consequently, $\widetilde{\pi}(z_n, [0, +\infty)) \subset
B(A, \varepsilon)$ for all $n > n_0$, that is, $\pi
((z_n)_k^{+}, t)) \subset B(A, \varepsilon)$ for $0 \leq t
\leq \phi ((z_n)_k^{+})$, $k = 0, 1,2,\dots$. and $n > n_0$
$((z_n)_0^{+} = z_n)$. Then $\psi (z_n) < \varepsilon$ for all $n
> n_0$, that is, $\psi (z_n) \stackrel{n \to
+\infty}{\to} 0 = \psi (x)$. Therefore, $\psi$ is
continuous on $X\setminus ( M \setminus A)$.
(b) Consider $x \in X \setminus M$. Given $\varepsilon > 0$,
let $\delta = \frac{\varepsilon}{1 + \varepsilon}$. Thus, if $\rho
(x, A)\geq \varepsilon$ then $\frac{\rho (x, A)}{1+ \rho (x, A)} \geq \delta$. Therefore, $\psi (x) \geq \delta$.
For the second part of item (b), let us assume that $x \in A$. If $x
\notin M$, as $M$ is closed and $\{x\}$ is compact, there is
a $\delta >0$ such that $B(x, \delta) \cap M = \emptyset$. Thus, if
$\{w_n\}_{n \geq 1}$ is any sequence in $X$ such that $w_n
\stackrel{n \to +\infty}{\to} x$, there exists
a positive integer $N >0$ such that $w_n \in B(x, \delta)$ for
$n > N$, by continuity of $\psi$ in $X\setminus
( M \setminus A)$,
\[
\psi (w_n) \stackrel{n \to +\infty}{\to} \psi
(x).
\]
Now, suppose $x \in M$. First of all, we should note that if
$\{z_n\}_{n \geq 1} \subset X \setminus M$ and $z_n \stackrel{n
\to +\infty}{\to} x$, then the continuity of
$\psi$ in $X\setminus ( M \setminus A)$ implies
\[
\psi (z_n) \stackrel{n \to +\infty}{\to} \psi
(x).
\]
Also, if $\{z_n\}_{n \geq 1} \subset M$ and $z_n \stackrel{n
\to +\infty}{\to} x$, since the impulsive
operator $I$ is continuous, we have
\[
I(z_n) \stackrel{n \to +\infty}{\to} I(x).
\]
Thus, since $I(z_n) \notin M$ for all $n \in
\mathbb{N}$, $I(x) \notin M$ and $\psi$ is continuous
in $X\setminus ( M \setminus A)$, it follows that
\[
\psi (I(z_n)) \stackrel{n \to
+\infty}{\to} \psi (I(x)),
\]
then by the definition of $\psi$,
\[
\psi (z_n) \stackrel{n \to +\infty}{\to} \psi
(x).
\]
Consequently, if $\{\overline{x}_n\}_{n \geq 1} \subset X$ is
any sequence such that $\overline{x}_n \to x$ as
$n \to +\infty$, then $\psi (\overline{x}_n) \to \psi (x)$ as
$n \to +\infty$. Since
$x \in A$, we have $\psi (x) = 0$ (as shown above) and therefore,
$\psi (\overline{x}_n)\to 0$ as $n \to +\infty$.
(c) Let $x \in X \setminus M$ and $0 \leq s < \phi (x)$. Let $y
= \pi (x, s)$. Since $0 \leq s < \phi (x)$, we have $y_{0}^{+}
= \pi (x, s)$ and $y_{k}^{+} = x_{k}^{+}$ for all integers $k
\geq 1$. Then
\begin{align*}
\psi (\pi (x, s))
&= \psi(y) = \sup_{k \geq 0}
\Big(\sup _{0 \leq t \leq \phi (y_{k}^{+})} \frac{\rho
(\pi(y_{k}^{+}, t), A)}{1 + \rho (\pi(y_{k}^{+}, t), t),
A)}\Big) \\
& \leq \sup_{k \geq 0} \Big(\sup _{0 \leq t \leq \phi
(x_{k}^{+})} \frac{\rho (\pi(x_{k}^{+}, t), A)}{1 + \rho
(\pi(x_{k}^{+}, t), A)}\Big) = \psi (x).
\end{align*}
Now, we shall prove that $\psi (\pi (x, \phi (x))) \leq \psi
(x)$. Since $\pi (x, \phi (x))= x_1 \in M$, we have
\begin{align*}
\psi (\pi (x, \phi (x)))
&= \psi(x_1) = \psi (I(x_1))
= \psi (x_{1}^{+}) \\
&= \sup_{k \geq 1} \Big(\sup _{0 \leq t \leq \phi
(x_{k}^{+})} \frac{\rho (\pi(x_{k}^{+}, t), A)}{1 + \rho
(\pi(x_{k}^{+}, t), A)}\Big) \\
&\leq \sup_{k \geq 0} \Big(\sup _{0 \leq t \leq \phi (x_{k}^{+})}
\frac{\rho (\pi(x_{k}^{+}, t), A)}{1 + \rho (\pi(x_{k}^{+}, t), A)}
\Big) = \psi (x).
\end{align*}
Consequently, $\psi (\pi (x, t)) \leq \psi (x)$ for $0 \leq t
\leq \phi (x)$. Now we will prove the second part of c). Let $x
\in M$. Then by the definition of $\psi$, $\psi
(I(x)) = \psi (x)$ and the theorem is proved.
\end{proof}
The next result is a corollary of Theorem \ref{Theorem3.4}. It
says that if $ M \subset A$, then we get the continuity of
the function $\psi$.
\begin{corollary}\label{Corollary3.1}
Let $(X, \pi; M, I)$ be an impulsive semidynamical
system. A closed subset $A \subset X$ such that $ M
\subset A$ is $\widetilde{\pi}$-stable if and only if there exists a
functional $\psi : X \to \mathbb{R}_{+}$, with the
following properties:
\begin{itemize}
\item[(a)] $\psi$ is continuous.
\item[(b)] For every $\varepsilon >0$, there is a $\delta >0$ such
that $\psi (x) \geq \delta$ whenever $\rho (x, A) \geq
\varepsilon$, and for any sequence $\{x_n\}_{n \geq 1} \subset X$
such that $x_n \stackrel{n \to +\infty}{\to}
x \in A$ implies $\psi (x_n) \stackrel{n \to +\infty}{\to} 0$.
\item[(c)] $\psi (\pi(x, t)) \leq \psi (x)$ if $x \in X \setminus M$
and $t \geq 0$, and, $\psi (I(x)) \leq \psi (x)$ if $x \in M$.
\end{itemize}
\end{corollary}
Theorem \ref{Theorem1.4} below deals with the equi
$\widetilde{\pi}$-stability of a closed set of $X$.
\begin{theorem}\label{Theorem1.4}
Let $(X, \pi; M, I)$ be an impulsive semidynamical
system, $A \subset X$ be closed and
$I( M\setminus A) \subset (X\setminus A)\setminus M$.
\begin{itemize}
\item[(1)] If there exists a functional $\psi : X \to
\mathbb{R}_{+}$ with the following properties:
\begin{itemize}
\item[(a)] $\psi$ is continuous in $X\setminus
( M\setminus A)$.
\item[(b)] $\psi (x) = 0$ for $x \in A$,
$\psi (x) >0$ for $x \notin A \cup M$.
\item[(c)] for every
$\varepsilon > 0$, there is a $\delta > 0$ such that $\psi (x)
\leq \varepsilon$ whenever $\rho(x, A) \leq \delta$.
\item[(d)] $\psi (\pi(x, t)) \leq \psi (x)$ for $x \in
X \setminus M$ and $0 \leq t \leq \phi (x)$,
and $\psi (I(x)) \leq \psi (x)$ for $x \in M$.
\end{itemize}
Then $A$ is equi $\widetilde{\pi}$-stable.
\item[(2)] Reciprocally, if $A$ is equi $\widetilde{\pi}$-stable, then
there is a functional $\psi : X \to \mathbb{R}_{+}$
satisfying conditions (a), (b), (c) and (d) above.
\end{itemize}
\end{theorem}
\begin{proof} (1) Suppose $x \notin A$. We have two cases to
consider: when $x \in M$ and otherwise $x \notin M$.
Suppose $x \notin M$. Set $\rho (x, A) = \varepsilon
>0$. Since $x \notin M$, then $\psi (x) > 0$. Let $\psi (x) =
\mu$. The condition (c) says that there is an $\eta > 0$ such that
\begin{equation}\label{eq7}
\psi (y) \leq \frac{\mu}{2}
\end{equation}
whenever $\rho(y,A) \leq \eta$.
Let $\delta < \min \{\eta, \varepsilon\}$. We assert that $x
\notin \overline{\widetilde{\pi}(B(A, \delta), [0,
+\infty))}$. Indeed. Suppose the contrary. Then there are
sequences $\{y_n\}_{n \geq 1} \subseteq B(A, \delta)$ and
$\{T_n\}_{n \geq 1} \subseteq \, [0, +\infty)$ such that
\[
\widetilde{\pi} (y_n, T_n) \stackrel{n \to
+\infty}{\to} x.
\]
Since $x \notin M$, $\{x\}$ is compact and $M$ is closed,
there is a $\varrho>0$ such that $B(x, \varrho) \cap M = \emptyset$.
Moreover, there is an integer $n_{0}>0$ such that
$\widetilde{\pi} (y_n, T_n) \in B(x, \varrho)$ for all $n > n_0$.
Since $\psi$ is a continuous function on
$X\setminus( M\setminus A)$, there exists an integer $n_1 > n_0$ such
that
\[
|\psi (\widetilde{\pi}(y_n, T_n))- \psi(x)| < \frac{\mu}{3}
\]
for all $n \geq n_1$. As $\psi(x) = \mu$, we have
\begin{equation}\label{eq8}
\frac{2\mu}{3} < \psi (\widetilde{\pi}(y_{n_1}, T_{n_1})) <
\frac{4\mu}{3}.
\end{equation}
Now note that $\psi(\widetilde{\pi} (w, t)) \leq \psi (w)$ for
all $t \geq 0$ and $w \in X \setminus M$. In fact, given
$w \in X\setminus M$ we have $\psi (\pi(w, t)) \leq \psi
(w)$ for $0 \leq t \leq \phi (w)$ and $\psi (w_{1}^{+}) = \psi
(I(w_1)) \leq \psi (w_1) = \psi (\pi (w, \phi (w)))
\leq \psi (w)$. If $\phi (w) < t < \phi (w) + \phi (w_1^{+})$, it
follows that
\[
\psi (\widetilde{\pi} (w, t)) = \psi (\pi (w_1^{+}, t - \phi
(w))) \leq \psi (w_1^{+}) = \psi (\widetilde{\pi} (w, \phi
(w))) \leq \psi (w).
\]
For $t = \phi (w) + \phi (w_1^{+})$,
\[
\psi (\widetilde{\pi} (x, t)) = \psi (w_{2}^{+}) = \psi
(I(w_{2})) \leq \psi (w_2) = \psi (\pi (w_{1}^{+},
\phi (w_1^{+}))) \leq \psi (w_{1}^{+}) \leq \psi (w),
\]
and so on. Thus, $\psi(\widetilde{\pi} (w, t)) \leq \psi (w)$
for all $t \geq 0$ and $w \in X \setminus M$. Using this
fact and $\eqref{eq7}$ we have
\[
\psi (\widetilde{\pi}(y_{n_1}, T_{n_1})) \leq \psi (y_{n_1})
\leq \frac{\mu}{2}
\]
which contradicts \eqref{eq8}. Hence, $x \notin
\overline{\widetilde{\pi}(B(A, \delta), [0, +\infty))}$.
Now we assume that $x \in M$. Suppose $x \in
\overline{\widetilde{\pi}(B(A, \delta), [0, +\infty))}$
for every $\delta > 0$. Then, there are sequences
$\{w_n^{\delta}\}_{n \geq 1} \subset B(A, \delta)$ and
$\{t_n^{\delta}\}_{n \geq 1} \subset [0, +\infty)$ such that
\[
\widetilde{\pi}(w_n^{\delta}, t_n^{\delta}) \stackrel{n
\to +\infty}{\to} x,
\]
for each $\delta >0$. Since each element of $M$ satisfies the
condition STC, we have two cases to consider.
Case 1: There are countably many elements of
$\{\widetilde{\pi}(w_n^{\delta}, t_n^{\delta})\}_{n \geq 1}$,
denoted by $\{\widetilde{\pi}(w_{n_k}^{\delta}, t_{n_k}^{\delta})\}_{k
\geq 1}$, such that
\[
\pi(\widetilde{\pi}(w_{n_k}^{\delta}, t_{n_k}^{\delta}),
\phi (\widetilde{\pi}(w_{n_k}^{\delta}, t_{n_k}^{\delta})))
\stackrel{k \to +\infty}{\to} x.
\]
In this case, since $\pi(\widetilde{\pi}(w_{n_k}^{\delta},
t_{n_k}^{\delta}), \phi (\widetilde{\pi}(w_{n_k}^{\delta},
t_{n_k}^{\delta}))) \in M$ and $I$ is continuous, we have
\[
\widetilde{\pi}(\widetilde{\pi}(w_{n_k}^{\delta},
t_{n_k}^{\delta}), \phi (\widetilde{\pi}(w_{n_k}^{\delta},
t_{n_k}^{\delta}))) \stackrel{k \to
+\infty}{\to} I(x);
\]
that is,
\[
\widetilde{\pi}(w_{n_k}^{\delta}, t_{n_k}^{\delta} + \phi
(\widetilde{\pi}(w_{n_k}^{\delta}, t_{n_k}^{\delta})))
\stackrel{k \to +\infty}{\to} I(x).
\]
But this is a contradiction, because $I(x) \notin A \cup M$
and by the previous case, there is a $\overline{\delta}>0$ such that
$I(x) \notin \overline{\widetilde{\pi}(B(A,
\overline{\delta}), [0, +\infty))}$.
Case 2: There are countably many elements of
$\{\widetilde{\pi}(w_n^{\delta}, t_n^{\delta})\}_{n \geq 1}$,
denoted by $\{\widetilde{\pi}(w_{n_k}^{\delta}, t_{n_k}^{\delta})\}_{k
\geq 1}$, such that $\phi (\widetilde{\pi}(w_{n_k}^{\delta},
t_{n_k}^{\delta})) \stackrel{k \to
+\infty}{\to} \phi (x)$. In this case,
\[
\pi (\widetilde{\pi}(w_{n_k}^{\delta}, t_{n_k}^{\delta}), t)
\stackrel{k \to +\infty}{\to} \pi (x, t),
\]
for all $0 \leq t < \phi (x)$. Let $0 < t_0 < \phi (x)$. Thus,
$\pi (x, t_0) \notin M$ and $\pi
(\widetilde{\pi}(w_{n_k}^{\delta}, t_{n_k}^{\delta}), t_0) =
\widetilde{\pi} (w_{n_k}^{\delta}, t_{n_k}^{\delta} + t_0 )$.
This is a contradiction.
Therefore, if $x \in M \setminus A$, there is a
$\delta >0$ such that $x \notin \overline{\widetilde{\pi}(B(A, \delta), [0,
+\infty))}$ and $A$ is equi $\widetilde{\pi}$-stable.
(2) Consider the function $\psi (x)$ defined in Theorem
\ref{Theorem3.4}. The result follows similarly as in Theorem
\ref{Theorem3.4} .
\end{proof}
We have the following corollary where we obtain the continuity of
the function $\psi$, provided $ M \subset A$.
\begin{corollary}\label{Corollary3.2}
Let $(X, \pi; M, I)$ be an impulsive semidynamical
system. A closed subset $A \subset X$ such that $ M
\subset A$ is equi $\widetilde{\pi}$-stable if and only if there
exists a functional $\psi : X \to \mathbb{R}_{+}$,
with the following properties:
\begin{itemize}
\item[(a)] $\psi$ is continuous.
\item[(b)] $\psi (x) = 0$ for
$x \in A$, $\psi (x) >0$ for $x \notin A$.
\item[(c)] for
every $\varepsilon > 0$, there is a $\delta > 0$ such that
$\psi (x) \leq \varepsilon$ whenever $\rho(x, A) \leq
\delta$.
\item[d)] $\psi (\pi(x, t)) \leq \psi (x)$ for
$x \in X \setminus M$ and $t \geq 0$, and $\psi
(I(x)) \leq \psi (x)$ for $x \in M$.
\end{itemize}
\end{corollary}
Lemma \ref{Lemma3.2} will be necessary to prove Theorem
\ref{theo1.5}.
\begin{lemma}\label{Lemma3.2}
Let $(X, \pi; M, I)$ be an impulsive semidynamical
system and $A \subset X$ be closed. Let $\psi : X
\to \mathbb{R}_{+}$ be a continuous function on
$X\setminus ( M \setminus A)$ satisfying:
\begin{itemize}
\item[(1)] for every $\varepsilon >0$, there is a $\delta >0$ such
that $\psi (x) \geq \delta$ whenever $\rho (x, A) \geq
\varepsilon$ and $x \notin M$.
\item[(2)] for every $\varepsilon >0$, there is a $\delta
>0$ such that $\psi (x) \leq \varepsilon$ whenever $\rho (x, A)
\leq \delta$.
\end{itemize}
Suppose there is a $\widetilde{\delta} >0$ such that $\psi
(\widetilde{\pi} (w, t)) \leq \psi (w)$ for all $t \geq0$ and
$w \in \overline{B(A, \widetilde{\delta})}\setminus M$.
Then, there is a $\overline{\delta}
>0$, $0 < \overline{\delta} \leq \widetilde{\delta}$,
such that $\widetilde{\pi} (B(A, \overline{\delta}),
[0, +\infty)) \subset B(A,
\widetilde{\delta})$.
\end{lemma}
\begin{proof}
We shall suppose that for each $\delta _n = \frac{\widetilde{\delta}}{n}
>0$, $n \in \mathbb{N}$, there are $w_n \in B\big(A,
\widetilde{\delta}/n \big)$ and $t_{0}^{n} \in (0,
+\infty)$ such that
\[
\widetilde{\pi} (w_n, t_{0}^{n}) \notin B\Big(A,
\widetilde{\delta} \Big).
\]
Take $\mu = \inf \big\{\psi (x): x \notin M \;
\text{and} \; \rho (x, A) \geq \widetilde{\delta}\big\}$.
Note that $\mu >0$ by item (1).
By item (2), there is an $\eta >0$, $\eta <
\widetilde{\delta}$, such that $\psi (y) \leq \frac{\mu}{2}$
whenever $\rho (y, A) \leq \eta$. Note that, there is a
positive integer $n_{k_0}$ such that $B(A, \delta _{n_{k_0}})
\subset B(A, \eta)$. Since $w_{n_{k_0}} \in B(A, \delta
_{n_{k_0}})$ we have $\psi (w_{n_{k_0}}) \leq \frac{\mu}{2}$. We
have two cases to consider: when $w_{n_{k_0}} \in M$ and
otherwise $w_{n_{k_0}} \notin M$.
First, suppose $w_{n_{k_0}} \notin M$. Then, $w_{n_{k_0}} \in
\overline{B(A, \widetilde{\delta})}\setminus M$. Thus,
$\psi (\widetilde{\pi} (w_{n_{k_0}}, t_{0}^{n_{k_0}})) \leq
\psi (w_{n_{k_0}}) \leq \mu/2$, which leads to a contradiction
because $\rho (\widetilde{\pi}(w_{n_{k_0}},
t_{0}^{n_{k_{0}}}), A) \geq \widetilde{\delta}$ and
$\widetilde{\pi}(w_{n_{k_0}}, t_{0}^{n_{k_{0}}}) \notin M$
$( M \cap I(M) =\emptyset)$.
Now we assume that $w_{n_{k_0}} \in M$. Then there is an $0<
\epsilon _{k_0} 0$, $0 < \overline{\delta} \leq \widetilde{\delta}$, such that
$\widetilde{\pi} (B(A, \overline{\delta}), [0, +\infty))
\subset B(A, \widetilde{\delta})$.
\end{proof}
For the case of uniformly $\widetilde{\pi}$-stability, we have the
following result.
\begin{theorem}\label{theo1.5}
Let $(X, \pi; M, I)$ be an impulsive semidynamical
system and $A \subset X$ be closed.
\begin{itemize}
\item[(1)] If there exists a functional $\psi : X \to
\mathbb{R}_{+}$ with the following properties:
\begin{itemize}
\item[(a)] $\psi$ is continuous in $X\setminus ( M\setminus A)$.
\item[(b)] for every $\varepsilon >0$, there is a $\delta >0$
such that $\psi (x) \geq \delta$ whenever
$\rho (x, A) \geq \varepsilon$ and $x\notin M$.
\item[(c)] for every $\varepsilon >0$, there is a $\delta >0$ such
that $\psi (x) \leq \varepsilon$ wherever $\rho (x, A) \leq
\delta$.
\item[(d)] $\psi (\pi(x, t)) \leq \psi (x)$ if $x \in
X\setminus M$ and $0 \leq t \leq \phi (x)$,
and $\psi (I(x)) \leq \psi (x)$ if $x \in M$.
\end{itemize}
Then $A$ is uniformly $\widetilde{\pi}$-stable.
\item[(2)] Reciprocally, if $A$ is uniformly $\widetilde{\pi}$-stable,
then
there is a functional $\psi : X \to \mathbb{R}_{+}$
satisfying conditions (a), (b), (c) and (d) above.
\end{itemize}
\end{theorem}
\begin{proof}
(1) Note that $\psi
(\widetilde{\pi} (x, t)) \leq \psi (x)$ for all $t \geq0$ and
$x \in X \setminus M$ $($the proof is the same as in Theorem
\ref{Theorem1.4} item (1). Given $\varepsilon > 0$, in particular
we have $\psi (\widetilde{\pi} (x, t)) \leq \psi (x)$ for all
$t \geq0$ and $x \in \overline{B(A, \varepsilon)}\setminus M$. By Lemma \ref{Lemma3.2} there exists a $\delta >0$ such that
$\widetilde{\pi}(B(A, \delta), [0, +\infty)) \subset B(A, \varepsilon)$. Therefore, $A$ is uniformly
$\widetilde{\pi}$-stable.
To prove condition (2), note that $A$ is uniformly
$\widetilde{\pi}$-stable, then $A$ is $\widetilde{\pi}$-stable.
Thus, the proof follows as in Theorem \ref{Theorem3.4}.
\end{proof}
\begin{corollary}\label{coro1.5}
Let $(X, \pi; M, I)$ be an impulsive semidynamical
system. A closed subset $A \subset X$ such that $ M
\subset A$ is uniformly $\widetilde{\pi}$-stable if and only if
there exists a functional $\psi : X \to
\mathbb{R}_{+}$ with the following properties:
\begin{itemize}
\item[(a)] $\psi$ is continuous.
\item[(b)] for every $\varepsilon >0$, there is a $\delta >0$ such that $\psi (x) \geq \delta$ whenever $\rho (x, A) \geq \varepsilon$.
\item[(c)] for every $\varepsilon >0$, there is a $\delta >0$ such
that $\psi (x) \leq \varepsilon$ wherever $\rho (x, A) \leq
\delta$.
\item[(d)] $\psi (\pi(x, t)) \leq \psi (x)$ if $x \in X \setminus M$ and $t \geq 0$, and, $\psi (I(x)) \leq
\psi (x)$ if $x \in M$.
\end{itemize}
\end{corollary}
Now, we present the result concerning asymptotically
$\widetilde{\pi}$-stability. In this case we consider the
stability of a compact set.
\begin{theorem}\label{theo1.6}
Let $(X, \pi; M, I)$ be an impulsive semidynamical
system, $X$ be locally compact and $A \subset X$ be
compact.
\begin{itemize}
\item[(1)] If there exists a functional $\psi : X \to
\mathbb{R}_{+}$ with the following properties:
\begin{itemize}
\item[(a)] $\psi$ is continuous in $X\setminus
( M\setminus A)$.
\item[(b)] for every
$\varepsilon >0$, there is a $\delta >0$ such that
$\psi (x) \leq \varepsilon$ wherever $\rho (x, A)
\leq \delta$.
\item[(c)] for every $\varepsilon >0$,
there is a $\delta >0$ such that $\psi (x) \geq
\delta$ whenever $\rho (x, A) \geq \varepsilon$ and
$x \notin M$.
\item[(d)] $\psi (\pi(x, t)) \leq
\psi (x)$ if $x \in X\setminus M$ and $0 \leq
t \leq \phi(x)$, and $\psi (I(x)) \leq \psi
(x)$ if $x \in M$.
\item[(e)] there is a $\delta
>0$ such that if $x \in B(A, \delta)\setminus A$,
then $\psi (\widetilde{\pi} (x, t)) \to 0$
as $t \to + \infty$.
\end{itemize}
Then $A$ is asymptotically $\widetilde{\pi}$-stable.
\item[(2)]
Reciprocally, if $A$ is asymptotically $\widetilde{\pi}$-stable,
then there is a functional $\psi : X \to
\mathbb{R}_{+}$ satisfying conditions (a), (b), (c) , (d) and (e)
above.
\end{itemize}
\end{theorem}
\begin{proof}
(1) By Theorem \ref{theo1.5}, $A$ is uniformly
$\widetilde{\pi}$-stable, and by Theorem \ref{theo1.2} $A$ is
orbitally $\widetilde{\pi}$-stable. The condition e) says that
$B(A, \delta) \subset \widetilde{P}^{+}_{W}(A)$, then $A$ is a
weak $\widetilde{\pi}-$attractor. Hence, $A$ is asymptotically
$\widetilde{\pi}$-stable.
(2) Clearly the functional $\psi$ given by
$$
\psi(x) = \begin{cases}
\sup_{k \geq 0} \Big(\sup _{0 \leq t \leq \phi (x_k^{+})} \frac{\rho
(\pi(x_{k}^{+}, t), A)}{1 + \rho (\pi(x_{k}^{+},
t), A)}\Big), & \text{if } x \in X \setminus M, \\
\psi (I(x)), & \text{if } x \in M,
\end{cases}
$$
where $x_{0}^{+} = x$, satisfies the conditions of the theorem.
\end{proof}
\begin{corollary}\label{cor1.6}
Let $(X, \pi; M, I)$ be an impulsive semidynamical
system, $X$ be locally compact, $A \subset X$ be compact
and $ M \subset A$. Then, $A$ is asymptotically
$\widetilde{\pi}$-stable if and only if there exists a functional
$\psi : X \to \mathbb{R}_{+}$, with the following
properties:
\begin{itemize}
\item[(a)] $\psi$ is continuous.
\item[(b)] for every $\varepsilon
>0$, there is a $\delta >0$ such that $\psi (x) \leq \varepsilon$
wherever $\rho (x, A) \leq \delta$.
\item[(c)] for every
$\varepsilon >0$, there is a $\delta >0$ such that $\psi (x) \geq
\delta$ whenever $\rho (x, A) \geq \varepsilon$.
\item[(d)] $\psi (\pi(x, t)) \leq \psi (x)$ if $x \in
X\setminus M$ and $t \geq 0$, and, $\psi (I(x)) \leq \psi
(x)$ if $x \in M$.
\item[(e)] there is a $\delta >0$ such
that if $x \in B(A, \delta)\setminus A$, then $\psi
(\widetilde{\pi} (x, t)) \to 0$ as $t \to + \infty$.
\end{itemize}
\end{corollary}
\subsection{Examples}
We apply the results above to two examples presented in
\cite{Ciesielski2}.
\begin{example} \label{exa3.1} \rm
Let $X = \{(x, y) \in
\mathbb{R}^{2}: 1 \leq x^2 + y^2 \leq 4\}$. Consider the planar
dynamical system
\begin{gather*}
\dot{x} = -y, \\
\dot{y} = x.
\end{gather*}
Let $A_1 = \{(x, y) \in \mathbb{R}^{2}: x^2 + y^2 = 1\}$,
$A_2 = \{(x, y) \in \mathbb{R}^{2}: x^2 + y^2 = 4\}$ and
$A= A_1 \cup A_2$. Set
$ M = \{(x, y) \in \mathbb{R}^{2}: x = 0, -2 \leq y \leq -1\}$
and consider the impulsive
function $I$ given by $I((0, y)) = (0, y + 3)$
for $y \in [-2, -1]$. Now define the function
$$
\psi(x, y) = \begin{cases}
0, &\text{if } \sqrt{x^2 + y^2} = 1, \\
\frac{\sqrt{x^2 + y^2} - 1}{\sqrt{x^2 + y^2}},
& \text{if } 1 < \sqrt{x^2 + y^2} \leq \frac{3}{2}, \\
\frac{2 - \sqrt{x^2 + y^2}}{3 - \sqrt{x^2 + y^2}},
& \text{if } \frac{3}{2} < \sqrt{x^2 + y^2} < 2,\\
0, &\text{if } \sqrt{x^2 + y^2} = 2.
\end{cases}
$$
Let us show that the set $A$ is $\widetilde{\pi}$-stable. In order
to do this, we are going to show that the three conditions of
item (1) from Theorem \ref{Theorem3.4} are satisfied. In fact:
(a) $\psi$ is continuous in $X$ and in particular in
$X \setminus ( M \setminus A)$.
(b) Given $\epsilon >0$, let $(x, y) \in X$ be such that
$\rho ((x, y), A) \geq \varepsilon$ and $(x, y) \notin M$.
Taking $\delta = \frac{\epsilon}{1 + \epsilon}$, we have two
cases to consider:
$\bullet$ If $\rho ((x, y), A) \geq \varepsilon$ and $1 <
\sqrt{x^2 + y^2} \leq \frac{3}{2}$, then
\[
\frac{\sqrt{x^2 + y^2} - 1}{\sqrt{x^2 + y^2}}
\geq \frac{\epsilon}{1 + \epsilon} = \delta.
\]
$\bullet$ If $\rho ((x, y), A)
\geq \varepsilon$ and $\frac{3}{2} < \sqrt{x^2 + y^2} < 2$,
then
\[
\frac{2 - \sqrt{x^2 + y^2}}{3 - \sqrt{x^2 + y^2}}
\geq \frac{\epsilon}{1 + \epsilon} = \delta.
\]
Therefore, given $\epsilon > 0$ there is a $\delta > 0$ such that
$\psi (x, y) \geq \delta$ whenever $\rho ((x, y), A)
\geq \varepsilon$ and $(x, y) \notin M$. On the other
hand, for any sequence $\{(w_n, k_n)\}_{n \geq 1} \subset
X$ such that $(w_n, y_n) \stackrel{n \to
+\infty}{\to} (x, y) \in A$, we have
$\psi ((w_n, k_n)) \stackrel{n \to +\infty}{\to} 0$,
because $w_n^2 + k_n^2 \stackrel{n \to
+\infty}{\to} 1$ if $(x, y) \in A_1$ and $w_n^2 +
k_n^2 \stackrel{n \to +\infty}{\to} 4$ if $(x, y) \in A_2$.
(c) If $1 < \sqrt{x^2 + y^2} \leq \frac{3}{2}$, we have
\[
\dot{\psi}(x, y) = \frac{-xy}{(x^2 + y^2)\sqrt{x^2 + y^2}}
+ \frac{xy}{(x^2 + y^2)\sqrt{x^2 + y^2}} = 0
\]
and if $\frac{3}{2} < \sqrt{x^2 + y^2} < 2$,
\[
\dot{\psi}(x, y) = \frac{xy}{(x^2 + y^2)[3 - \sqrt{x^2 + y^2}]^{2}}
+ \frac{-xy}{(x^2 + y^2)[3 - \sqrt{x^2 + y^2}]^{2}} = 0.
\]
Therefore, $\dot{\psi}(x, y) = 0$ for all $(x, y) \in X$. Then
$\psi (\pi((x, y), t)) = \psi ((x, y))$ if $(x, y) \in X
\setminus M$ and $0 \leq t \leq \phi((x, y))$. Furthermore, $\psi
(I(x, y)) = \psi (x, y)$ if $(x, y) \in M$. By Theorem
\ref{Theorem3.4}, $A$ is $\widetilde{\pi}$-stable.
\end{example}
\begin{example} \label{exa3.2} \rm
Consider the space $X = \mathbb{R}^{2}\times \{0, 1\}$ and the
dynamical system
\begin{equation}\label{eq1b}
\begin{gathered}
\dot{x} = -x, \\
\dot{y} = -y,
\end{gathered}
\end{equation}
on $\mathbb{R}^{2}\times \{0\}$ and $\mathbb{R}^{2}\times \{1\}$,
independently. Now let $ M_0 = \{(x, y, z) \in
\mathbb{R}^{3}: x^2 + y^2 = 1, z = 0\}$,
$ M_1 = \{(x, y, z) \in \mathbb{R}^{3}: x^2 + y^2 = 1/4, z =
1\}$ and $M = M_0 \cup M_1$. We define $I(x, y,
0) = (x, y, 1)$ for $(x, y, 0) \in M_0$ and
$I(x, y, 1) = (x, y, 0)$ for $(x, y, 1)
\in M_1$. Take $A_0 = \{(x, y) \in \mathbb{R}^2: x^2 +
y^2 \leq 1\} \times \{0\}$,
$A_1 = \{(x, y) \in \mathbb{R}^2: x^2 + y^2 \leq 1\} \times \{1\}$
and $A = A_0 \cup A_1$. We claim that the set $A$ is
$\widetilde{\pi}-$asymptotically stable.
In fact. Consider the function
$$
\psi(x, y, z) = \begin{cases}
\frac{\sqrt{x^2 + y^2} - 1}{\sqrt{x^2 + y^2}} ,
&\text{if $\sqrt{x^2 + y^2} > 1$ and $z \in \{0, 1\}$},\\
0, &\text{if $\sqrt{x^2 + y^2} \leq 1$ and $z \in \{0, 1\}$}.
\end{cases}
$$
We are going to use Corollary \ref{cor1.6} to prove it, because $
M \subset A$. Let us start by verifying the five conditions of
Corollary \ref{cor1.6}:
(a) It is easy to see that the function $\psi$ is continuous on
$X$.
(b) Given $\epsilon > 0$, $\epsilon < 1$, if $\rho ((x, y,
0), A_0) < \frac{\epsilon}{1 - \epsilon}$ then
\[
\sqrt{x^2 + y^2} -1 \leq \frac{\epsilon}{1 - \epsilon} \Rightarrow 1 - \frac{1}{\sqrt{x^2 + y^2}} \leq \epsilon.
\]
Thus, $\psi (x, y, 0) < \epsilon$. Analogously, if $\rho
((x, y, 1), A_1) < \frac{\epsilon}{1 - \epsilon}$ then
$\psi (x, y, 1) < \epsilon$. Hence, given $\epsilon > 0$,
there is $\delta > 0$ such that $\psi (x, y, z) \leq
\epsilon$ whenever $\rho ((x, y, z), A) \leq \delta$.
(c) Given $\epsilon > 0$, there is a $\delta = \frac{\epsilon}{1 +
\epsilon}$ such that
\[
\psi (x, y, z) \geq \frac{\epsilon}{1 + \epsilon}:= \delta,
\]
whenever $\rho ((x, y, z), A) \geq \epsilon$.
(d) Consider the two flows $\varphi_1((x_0, y_0, 0), t) =
(x(x_0, t), y(y_0, t), 0)$ and \\
$\varphi_2((x_0, y_0, 1), t) = (x(x_0, t), y(y_0, t), 1)$ such that
$(x(t), y(t)) = (x(x_0, t), y(y_0, t))$ satisfies
system \eqref{eq1b} and $(x(0), y(0)) = (x_0, y_0)$. Let
$z_0 = (x_0, y_0, 0)$ and $w_0 = (x_0, y_0, 1)$.
If $\sqrt{x^2 + y^2} > 1$, we have
\[
\dot{\psi}(\varphi_1(z_0, t)) = \frac{\partial \psi}{\partial x}\dot{x}(x_0, t) +
\frac{\partial \psi}{\partial y}\dot{y}(y_0, t) + \frac{\partial \psi}{\partial z}\dot{z}(z_0, t) =
- \frac{1}{\sqrt{x^2(x_0, t) + y^2(y_0, t)}} < 0
\]
and
\[
\dot{\psi}(\varphi_2(w_0, t)) = - \frac{x^2(x_0, t) + y^2(y_0, t)}{\sqrt{x^2(x_0, t)
+ y^2(y_0, t)}\left[\frac{1}{2} + \sqrt{x^2(x_0, t) + y^2(y_0, t)}\right]^{2}} < 0,
\]
for $0 \leq t < \phi(z_0)$. Hence, $\dot{\psi}(\varphi_1(z_0,
t)) \leq 0$ and $\dot{\psi}(\varphi_2(w_0, t)) \leq 0$ whenever
$(x_0, y_0) \in \mathbb{R}^{2}$ and $t \geq 0$. Then,
$\psi(\varphi_1(z_0, t)) \leq \psi (z_0)$ and $\psi
(\varphi_2(w_0, t)) \leq \psi (w_0)$ whenever $(x_0, y_0)
\in \mathbb{R}^{2}$ and $t \geq 0$.
Since $\psi (x, y, z) = 0$ for each $(x, y, z) \in A$
and $I(x, y, z) \subset \rm A$ for $(x, y, z)
\in M$, we have $\psi (I(x, y, z)) = \psi (x, y, z) = 0$ if $(x, y, z) \in M$.
(e) Let $\varphi_1$ and $\varphi_2$ be flows in
$\mathbb{R}^{2}\times\{0\}$ and $\mathbb{R}^{2}\times\{1\}$
respectively. Now, consider $\widetilde{\pi}$ an impulsive flow in
$X$ obtained from $\varphi_1$ and $\varphi_2$. By Corollary
\ref{Corollary3.1}, $A$ is $\widetilde{\pi}$-stable and by Theorem
\ref{THEO3.5}, $A$ is uniformly $\widetilde{\pi}$-stable. Then,
given $\epsilon
> 0$, there is a $\delta > 0$ such that
\[
\widetilde{\pi} (B(A, \delta)) \subset B(A, \epsilon).
\]
Let $(x_0, y_0, 0) \in B(A_1, \delta) \setminus A_1$.
Since
\[
\frac{d}{dt}\psi (\varphi_1((x_0, y_0, 0), t)) \leq 0
\]
for $t \geq0$, the limit $\lim_{t \to
+\infty} \psi (\varphi_1((x_0, y_0, 0), t))$ exists.
Suppose $\lim_{t \to +\infty} \psi
(\varphi_1((x_0, y_0, 0), t)): = \ell > 0$.
Define $K = \{(x, y, z) \in \mathbb{R}^{2}\times\{0\}:
\rho ((x, y, z), A_1) \leq \epsilon \text{ and }
\psi(x, y, z) \geq \ell\}$. It is clear that $K$ is
compact. Note $\varphi_1((x_0, y_0, 0), t) \in K$ for all
$t \geq 0$. Now, define
\[
\eta := \min \{-\dot{\psi}(w): w \in K\}.
\]
Since $A$ is not contained in $K$, then $\eta > 0$. Thus
\[
-\dot{\psi} (\varphi_1((x_0, y_0, 0), t)) \geq \eta
\]
for all $t \geq 0$. Then, integrating the inequality above from
$0$ to $t$, we have
\[
\psi (\varphi_1((x_0, y_0, 0), t)) \leq \psi(x_0, y_0, 0) - \eta t,
\]
for all $t \geq 0$, which is a contradiction since $\psi$ is
positive. Therefore, $\ell = 0$.
If $(x_0, y_0, 0) \in B(A_2, \delta) \setminus A_2$, the result
follows analogously. By Corollary \ref{cor1.6}, $A$ is
$\tilde{\pi}$-asymptotically stable.
\end{example}
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