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

\AtBeginDocument{{\noindent\small 
{\em Electronic Journal of Differential Equations},
Vol. 2001(2001), No. 12, pp. 1--9.\newline
ISSN: 1072-6691. URL: http://ejde.math.swt.edu or http://ejde.math.unt.edu
\newline ftp ejde.math.swt.edu \quad ejde.math.unt.edu (login: ftp)}
\thanks{\copyright 2001 Southwest Texas State University.} 
\vspace{1cm}}

\begin{document} 

\title[\hfilneg EJDE--2001/12\hfil Asymptotic behavior of solutions ]
{Asymptotic behavior of the solutions of a class of second order
 differential systems } 

\author[ Svetoslav Ivanov Nenov \hfil EJDE--2001/12\hfilneg]
{ Svetoslav Ivanov Nenov }

\address{ Svetoslav Ivanov Nenov \hfill\break
Department of Mathematics, University of Chemical Technology 
and Metallurgy; 8, Kliment Ohridsky blvd., Sofia 1756, Bulgaria }

\email{ s.nenov@uctm.edu, s\_nenov@hotmail.com } 

\date{}
\thanks{Submitted April 4, 2000; December 28, 2000. Published January 30, 2001.}
\subjclass{34D05, 34D10, 34E05}
 \keywords{asymptotic behaviour, gradient systems, T. Wazewski's theorem} 

\newtheorem{thm}{Theorem}      
\newtheorem{lem}{Lemma}
\newtheorem{defn}{Definition} 
\newtheorem{example}{Example}

\begin{abstract}
 In the present paper it is proved that for any solution $x_1(t)$ of the 
 system
 $M \ddot x + \dot x = f(t,x)$, for which 
 $\lim\limits_{t\to\infty}\|\dot x_1(t)\|=0$,
 there exists a solution $x_2(t)$ of the system $\dot x = f(t,x)$ such that
 $\lim\limits_{t\to\infty}\|x_1(t)-x_2(t)\|=0$. 
 Some generalizations of this result are also presented.
 The case $f(t,x)=-\nabla U(x)$ has been investigated explicitly.
\end{abstract}

\maketitle

\section{Statements and main results}
We consider the following two $n$-dimensional systems
\begin{gather}\label{eq:1}
M \ddot x + \dot x = f(t,x)
\intertext{and}
\label{eq:2}
\dot x = f(t,x),
\end{gather}
where $M \geq 0$ is a constant; ``$\cdot$" denotes differentiation with respect to 
$t$;  $f:\mathbb{R}_+ \times \Omega \to  \mathbb{R}^n$;
$\mathbb{R}_+ \equiv [0,\infty)$; $\Omega$ is a domain in $\mathbb{R}^n$; 
$\mathbb{R}^n$ is the $n$-dimensional Euclidean
space with Euclidean scalar product $\langle\cdot ,\cdot\rangle$
and corresponding norm $\|\cdot\|$.

    Let $(x_{11}, x_{12})\in\Omega\times\mathbb{R}^n$ and $x_2 \in \Omega$ 
    be fixed initial points and
$t_0\in\mathbb{R}_+$ be a fixed initial moment;
$x_1 (t; t_0, x_{11}, x_{12})$ and $x_2 (t;t_0,x_2)$ denote the solutions of the
systems \eqref{eq:1} and \eqref{eq:2} with initial conditions
\begin{gather}\label{ini_for_eq:1}
x_1 (t_0;t_0,x_{11},x_{12}) = x_{11}, \quad 
\dot x_1 (t_0;t_0,x_{11},x_{12}) = x_{12}
\intertext{and}
\label{ini_for_eq:2}
x_2 (t_0;t_0,x_2) = x_2,
\end{gather}
respectively.

We introduce the following hypotheses (H1):\\
\begin{enumerate}
\item[(H1.1)] $\Omega$ is a bounded domain in $\mathbb{R}^n$;
    $f\in {\bf C} (\mathbb{R}_+\times\Omega, \mathbb{R}^n)$.

\item[(H1.2)] The function $f$ is Lipschitz with respect to the second 
  argument with  Lipschitz constant $L \geq 0$.
  
\item[(H1.3)] For arbitrary initial conditions 
$(t_0,x_{11}, x_{12})\in\mathbb{R}_+\times\Omega\times\mathbb{R}^n$,
  $(t_0,x_2) \in \mathbb{R}_+\times\Omega$, the Cauchy problems \eqref{eq:1},
    \eqref{ini_for_eq:1} and \eqref{eq:2}, \eqref{ini_for_eq:2}
    have unique solutions $x_1(t;t_0,x_{11},x_{12})$ and $x_2(t;t_0,x_2)$, 
    respectively,
    defined on $t$-interval $\mathbb{R}_+$. Moreover
    $$
    \{x_1(t;t_0,x_{11},x_{12}) : t \in \mathbb{R}_+\} \subset \Omega,\quad
    \{x_2(t;t_0,x_2) : t \in \mathbb{R}_+\} \subset \Omega .
    $$
\end{enumerate}

 In this article,  the following theorem contains one of the basic results.

\begin{thm}\label{main-theorem_1}
    Assume the following conditions hold:
\begin{enumerate}
\item The hypothesis (H1) holds.
\item $(t_0, x_{11},x_{12}) \in  \mathbb{R}_+\times\Omega\times\mathbb{R}^n$ 
is the fixed initial condition for \eqref{eq:1}, \eqref{ini_for_eq:1}.
\item 
\begin{equation}\label{thm_1:condition-1}
\lim\limits_{t\to\infty} \|\dot x_1 (t;t_0,x_{11}, x_{12})\| = 0\,.
\end{equation}

\end{enumerate}

    Then there exists at least one initial condition $x_2\in\Omega$ for the system
\eqref{eq:2} such that
\begin{equation}\label{thm_1:condition-2a}
\lim\limits_{t\to\infty} \|x_1(t;t_0,x_{11}, x_{12}) - x_2(t;t_0,x_2) \| = 0.
\end{equation}
\end{thm}

    Theorem \ref{main-theorem_1} is proved in subsection \ref{proof_of_Theorem_1}.

\begin{example} \rm
    Let us consider the differential equations:
\begin{gather}\label{ex-1:eq1}
\ddot x + \dot x = t,
\end{gather}
and
\begin{gather}\label{ex-1:eq2}
\dot x = t.
\end{gather}
    An immediate integration of these equations yields:
$$
x_1 (t;0,x_{11},x_{12}) = 1+x_{12}+x_{11}-t+\frac{t^2}{2} - e^{-t} (1+x_{12})
$$
and
$$
x_2(t;0,x_2) = x_2 + \frac{t^2}{2}.
$$
    It is not difficult to check that for any three points
$x_{11},x_{12},x_2 \in \mathbb{R}$ we have
$$
\lim\limits_{t\to\infty}|x_1 (t;0,x_{11},x_{12}) - x_2(t;0,x_2)| \not = 0.
$$
Moreover, in this example, for any initial conditions $x_{11},x_{12}\in\mathbb{R}$
of the problem \eqref{ex-1:eq1}, \eqref{ini_for_eq:1} we have
$$
\lim\limits_{t\to\infty}|\dot x_1 (t;0,x_{11},x_{12})| =
\lim\limits_{t\to\infty}|-1+t+e^{-t}(1+x_{12})| = \infty.
$$

    From the above equality it follows that \eqref{thm_1:condition-1}
is not true and, as we have shown, equality \eqref{thm_1:condition-2a} is 
not valid.
Thus \eqref{thm_1:condition-1} is an essential condition.
\end{example}

    The following result is derived similarly to the proof of Theorem
\ref{main-theorem_1} (see subsection \ref{proof_of_Theorem_2}).

\begin{thm} \label{main-theorem_2}
    Assume the following conditions are fulfilled:
\begin{enumerate}
\item The hypothesis (H1) holds.
\item $(t_0, x_{11},x_{12}) \in  \mathbb{R}_+\times\Omega\times\mathbb{R}^n$ is
 a fixed initial condition for  \eqref{eq:1}, \eqref{ini_for_eq:1}.
\item \begin{equation}\label{thm_2:condition-1}
\lim\limits_{t\to\infty} \|\ddot x_1 (t;t_0,x_{11}, x_{12})\| = 0.
\end{equation}
\end{enumerate}
Then there exists at least one initial condition $x_2\in\Omega$ of 
system \eqref{eq:2} such that the equality \eqref{thm_1:condition-2a} 
is valid.
\end{thm}


\section{An application: Second order gradient systems}

In the present section we shall discuss some  asymptotic properties  of 
the solutions of the following system
\begin{gather}\label{II sys}
M \ddot x + \dot x = -\nabla U(x),
\end{gather}
where $M\geq 0$ is a constant; $U\in {\bf C^1} (\Omega, \mathbb{R})$; $\Omega$ is a domain in
$\mathbb{R}^n$ such that any solution of \eqref{II sys} starting in $\Omega$ remains
in $\Omega$.

    First, let us write \eqref{II sys} as a first order system in $\mathbb{R}^{2n}$:
\begin{gather}\label{II sys-1}
\dot x = y,\quad \dot y = M^{-1}\left(-y-\nabla U(x)\right).
\end{gather}
    Setting
$$
L(x,y) = \frac{M}{2} y^2 + U(x),
$$
it is not difficult to see that
\begin{gather}\label{for U -1}
L^\prime (x,y) = M y \dot y + \nabla U(x) y = (M\dot y + \nabla U(x)) y = - y^2 \leq 0,
\end{gather}
for any $(x,y)\in\Omega\times\mathbb{R}^{n}$. Therefore, if $U(x) \geq 0$,
$x\in\Omega$ then $L(x,y)$ is a Liapunov function (i.e. a continuous non-negative
function which satisfies locally a Lipschitz condition) for \eqref{II sys-1}.

    Let $\mathcal{M}=\{(x,y):L^\prime(x,y) = 0\}=\{(x,0):x\in\Omega\}$ and let $\mathcal{M}_1$
be the union of all points $(x_0,y_0)$ of all orbits $(x(t;x_0,y_0),y(t;x_0,y_0))$
such that
$$
\{(x(t;x_0,y_0),y(t;x_0,y_0)):t\in\mathbb{R}\}\subset \mathcal{M}.
$$
\begin{thm}\label{LaSalle type thm}
    Suppose that $U(x) \geq 0$ for all $x\in\Omega$, and $\lim\limits_{\|x\|\to\infty} U(x) = \infty$.

    Then:
\begin{enumerate}
\item All solutions of \eqref{II sys-1} are bounded.
\item Every solution of \eqref{II sys-1} approaches $\mathcal{M}_1$ as $t\to\infty$, i.e.
for any $(x_0,y_0)\in\Omega\times\mathbb{R}^n$ we have
$$
(x(t;x_0,y_0),y(t;x_0,y_0)) \to \mathcal{M}_1,\quad \text{as}\quad t\to\infty.
$$
\item For any $(x_0,y_0)\in\Omega\times\mathbb{R}^n$, 
$$
\lim\limits_{t\to\infty} \dot x(t;x_0,y_0)=\lim\limits_{t\to\infty} 
y(t;x_0,y_0)= 0\,.
$$
\end{enumerate}
\end{thm}

\begin{proof}
    Obviously, $\lim_{\|x\|^2+\|y\|^2\to\infty}L(x,y)=\infty$. Then
the first statement of Theorem \ref{LaSalle type thm} follows from
\cite[Theorem 10.1]{Taro}.

    The second statement follows immediately from results in \cite{LaSalle},
see also \cite[Theorem 14.4, Theorem 14.7]{Taro}.

    The third statement follows from second one and implication $\mathcal{M}_1 \subset \mathcal{M} = \{ (x,0) : x \in \Omega \}$.
\end{proof}


    The following result follows immediately from
Theorem \ref{main-theorem_1} and Theorem \ref{LaSalle type thm}.
\begin{thm}\label{main-theorem_3}
    Let the following conditions hold true:
\begin{enumerate}
\item $\Omega$ is a domain in $\mathbb{R}^n$; $U\in {\bf C^1} (\Omega, \mathbb{R}^n)$; the
hypothesis (H 1.3) is valid, where $f(t,x)=-\nabla U(x)$.
\item $(t_0,x_{11},x_{12})\in\mathbb{R}_+\times\Omega\times\mathbb{R}^n$
is a fixed initial condition for the initial-value problem \eqref{II sys}, \eqref{ini_for_eq:1}.
\end{enumerate}

    Then there exists at least one initial condition $x_2\in\Omega$ for the system
$\dot x = -\nabla U(x)$ such that
\begin{equation}\label{thm_1:condition-2c}
\lim\limits_{t\to\infty} \|x_1(t;t_0,x_{11}, x_{12}) - x_2(t;t_0,x_2) \| = 0.
\end{equation}
\end{thm}

    It is not difficult to derive some properties of the solutions of
systems \eqref{II sys} or \eqref{II sys-1}: the $\omega$-limit set of a
solution of system \eqref{II sys-1} consists of critical points only;
if there are two critical points in the $\omega$-limit set then there are
infinitely many critical points in the same $\omega$-limit set; there
is no non-trivial periodic solutions of \eqref{II sys-1}, etc. The
proofs of these facts follow from Theorem \ref{main-theorem_3} and results
in \cite[Chapter 1, \S 1]{PalisandMelo}.

\section{A Topological Principle}

    In the present Section we shall deduce the Topological Principle in the
theory of autonomous dynamical systems or the so-called T.~Wazewski's Theorem.
The Topological Principle is related to the initial value-problem
\begin{equation}\label{eq:topol_principle}
\dot x=f(t,x),\quad x(t_0)=x_0,
\end{equation}
where $f\in {\bf C}({\mathcal{E}},{  R}^n)$; ${\mathcal{E}}$ is an open $(t,x)$-set in
$\mathbb{R} \times \mathbb{R}^n$; $(t_0,x_0)\in \mathcal{E}$. Let $\mathcal{E}_0$ be a non-empty open subset
in ${\mathcal{E}}$.

    We recall the following definitions.
\begin{defn}\label{defn:egress_points}
    The point $(t_0,x_0)\in {\mathcal{E}}\cap \partial {\mathcal{E}}_0$ is said to be:
\begin{enumerate}
\item  {\it an egress point} of ${\mathcal{E}}_0$ with respect to the system \eqref{eq:topol_principle}
if, for every solution $x(t;t_0,x_0)$ of \eqref{eq:topol_principle} there exists
$\theta > 0$ such that $\{(t,x(t;t_0,x_0)):t\in [t_0-\theta,t_0)\}\subset {\mathcal{E}}_0$.
\item  {\it an strict egress point} of ${\mathcal{E}}_0$ with respect to the system \eqref{eq:topol_principle}
if, $(t_0,x_0)$ is an egress point of ${\mathcal{E}}_0$ and
$\{(t,x(t;t_0,x_0)):t\in [t_0,t_0+\theta ]\}\subset {\mathcal{E}}\setminus {\mathcal{E}}_0$
for sufficiently small $\theta >0$. See Figure \ref{figure:Topological_principle}.
\end{enumerate}
\end{defn}

 In the following, ${\mathcal{E}}_0^e$ (${\mathcal{E}}_0^{se}$) denotes the 
 set of all egress (strict egress) points of ${\mathcal{E}}_0$. 
 It is clear that ${\mathcal{E}}_0^{se} \subset {\mathcal{E}}_0^e$.

\begin{defn}\label{defn:UV-set_definition}
    The open subset ${\mathcal{E}}_0$ in ${\mathcal{E}}$ is said to be an
 open $[U,V]$-subset in ${\mathcal{E}}$ with respect to the system
\eqref{eq:topol_principle} if:
\begin{enumerate}
\item     There exist integers $p,q\geq 1$ and continuous functions
$U_j:{\mathcal{E}}\to \mathbb{R}$, $j=1,\dots ,p$ and  $V_k:{\mathcal{E}}\to \mathbb{R}$,
$k=1,\dots ,q$ such that
$$
{\mathcal{E}}_0=\left\{ (t,x):U_j(t,x)<0\ \text{ and }\ V_k(t,x)<0,\ 1\leq j\leq p,\ 1\leq k\leq q\right\} .
$$
\item   If for any two indexes $\alpha = 1,\dots , p$ and 
$\beta = 1,\dots , q$ we denote
\begin{eqnarray*}
{\mathcal{U}}_\alpha &=& \big\{ (t,x): U_\alpha (t,x)=0,\; U_j(t,x)\leq 0 
\text{ and } V_k(t,x) < 0, \\
&&1\leq j\leq p,\; j\neq \alpha,\; 1\leq k\leq q \big\} , 
\\[3pt]
 {\mathcal{V}}_\beta &=&\big\{ (t,x): U_j(t,x) < 0,\; V_\beta (t,x)=0
\text{ and } V_k(t,x)\leq 0, \\
&&1\leq j\leq p,\; 1\leq k\leq q,\; k\not=\beta\big\} ,
\end{eqnarray*}
then the trajectory derivatives
$$
    U_\alpha ^{\prime }(t_0,x_0)=\frac{d U_\alpha (t,x(t;t_0,x_0))}{dt}\!\!\mid_{t=t_0},
$$
$$
    V_\beta ^{\prime }(t_0,x_0)=\frac{d V_\beta (t,x(t;t_0,x_0))}{dt}\!\!\mid _{t=t_0}
$$
exist, satisfying the inequalities
$$
    U_\alpha ^{\prime }(t_0,x_0)>0,\ \text{ for any point }\ (t_0,x_0)\in {\mathcal{U} }_\alpha ,
$$
$$
    V_\beta ^{\prime }(t_0,x_0)<0,\ \text{ for any point }\ (t_0,x_0)\in {\mathcal{V}}_\beta ,
$$
along all solutions of \eqref{eq:topol_principle} through $(t_0,x_0)$.
\end{enumerate}
\end{defn}

\begin{figure}[t]
%\includegraphics{nenovg2.eps}
\epsfxsize=10cm
\epsffile{nenovg2.eps}
\caption{Egress points and strict egress points
\label{figure:Topological_principle}}
\end{figure}

    The theorem (T. Wazewski's Theorem) is known also as the Topological
 Principle in the theory of autonomous dynamical systems.
\begin{thm}\label{thm:topological_principle}
   Assume  the following conditions:
\begin{enumerate}
\item  ${\mathcal{E}}$ is an open $(t,x)$-set in $\mathbb{R}\times\mathbb{R}^n$; $f\in {\bf C}({\mathcal{E}},R^n)$.
\item  The initial-value problem \eqref{eq:topol_principle} has a unique
solution through every point of ${\mathcal{E}}$, and these solutions depend continuously
on initial values.
\item  ${\mathcal{E}}_0$ is an open subset in ${\mathcal{E}}$.
\item  All egress points of the set ${\mathcal{E}}_0$ are strict egress points, i.e.
${\mathcal{E}}_0^e={\mathcal{E}}_0^{se}$.
\item  $\mathcal{W}$ is a non-empty subset in ${\mathcal{E}}_0\cup {\mathcal{E}}_0^e$ such that
$\mathcal{W}\cap {\mathcal{E}}_0^e$ is a retract of ${\mathcal{E}}_0^e$, but is not a retract of $\mathcal{W}$.
\end{enumerate}
    Then there exists at least one point $(t_0,x_0)\in \mathcal{W}\cap {\mathcal{E}}_0$ such
that $x(t;t_0,x_0)\in {\mathcal{E}}_0$ for any $t$ in the right-maximal interval of
existence of $x(t;t_0,x_0)$.
\end{thm}

    An useful tool for checking the validity of condition 4 of 
Theorem \ref{thm:topological_principle} is the following lemma.

\begin{lem}\label{lemma:help_for_topological_principle}
    Assume the following conditions:
\begin{enumerate}
\item  The conditions 1 and 2 of Theorem \ref{thm:topological_principle} 
hold.
\item  ${\mathcal{E}}_0$ is an open $[U,V]$-subset of ${\mathcal{E}}$ with respect to the 
system \eqref{eq:topol_principle}.
\end{enumerate}
    Then
$$
{\mathcal{E}}_0^e={\mathcal{E}}_0^{se}=\bigcup\limits_{\alpha =1}^p{\mathcal{U}}_\alpha \
\setminus \ \bigcup\limits_{\beta =1}^q{\mathcal{V}}_\beta,
$$
where ${\mathcal{U}}_\alpha$ and ${\mathcal{V}}_\beta$ are the sets introduced in
Definition \ref{defn:UV-set_definition}.
\end{lem}

    One may find the circumstantial explanation and proofs of all results
in this Section in \cite[Chapter X, \S 3]{Hartman}.


\section{Proofs}

\subsection{Proof of Theorem \ref{main-theorem_1}}
\label{proof_of_Theorem_1}
    Let $(x_{11},x_{12}) \in \Omega\times\mathbb{R}^n$ be a fixed initial condition
for the system \eqref{eq:1}. For simplification of notations we suppose
that $t_0 = 0$. Further, we shall use the notation $x_1(t) = x_1(t;0,x_{11},x_{12})$.

    We set
$$
g:\mathbb{R}_+\times\mathbb{R}_+\to\mathbb{R},\quad g(t,u) = - \frac{6}{5} L u^2 - \frac{6}{5} L M \sqrt[3]{u} \|\dot x_1(t)\|.
$$

For any initial condition $u_0\in\mathbb{R}_+$, the differential inequality
\begin{gather}\label{proof:ineq-1}
2 u \dot u < g(t,u)
\end{gather}
has a solution $u(t;u_0)$ for which:
\begin{gather}\label{proof:for_ineq-1a}
u(0;u_0)=u_0,
\end{gather}
and
\begin{gather}\label{proof:for_ineq-1c}
\lim\limits_{t\to\infty} u(t;u_0) = 0.
\end{gather}
To prove these facts, it is sufficient to see that the initial-value
problem
\begin{gather}\label{proof:eq-1_for_ineq-1}
2 u \dot u = -2 L u^2 - \frac{6}{5} L M \sqrt[3]{u} \|\dot x_1(t)\|, \quad u(0; u_0) = u_0
\end{gather}
has solution for which \eqref{proof:for_ineq-1a} and \eqref{proof:for_ineq-1c} hold true. The mentioned solution is
$$
u(t,u_0) = e^{-Lt} \left(u_0^{\frac{5}{3}} - LM\int\limits_{0}^{t} e^{\frac{5Ls}{3}}\|\dot x_1(s)\|\,ds\right)^{\frac{3}{5}}.
$$
Below, we shall use the notation $u(t) = u(t;u_0)$.

    We set:
$$
\begin{array}{ll}
 U:\mathbb{R}_+\times \Omega\to \mathbb{R} ,&\quad
 U(t,x)=\|M\dot x_1(t) + x_1(t) - x \|^\frac{6}{5} - u^2(t);\\[2pt]
 V: \mathbb{R}_+ \to \mathbb{R}_-\equiv (-\infty,0],&\quad  V(t)= - t;
\end{array}
$$
$$
\begin{array}{ll}
 \mathcal{U}=\left\{ (t,x) \in \mathbb{R}_+ \times \Omega : U(t,x)=0\ \text{ and }V(t) < 0\right\}; &\\[2pt]
 {\mathcal{V}}=\left\{ (t,x) \in \mathbb{R}_+ \times \Omega :U(t,x) < 0\ \text{ and }V(t)=0\right\};  &\\[2pt]
{\mathcal{E}}_0=\left\{ (t,x) \in \mathbb{R}_+ \times \Omega :U(t,x)<0\ \text{ and }V(t)<0\right\}; & \quad  {\mathcal{E}} = \mathbb{R}_+\times \Omega .
\end{array}
$$

    Our goal is to show that there exists at least one initial condition
$\xi_2 \in \Omega$ and initial moment $\tau > 0$ for the problem
\eqref{eq:2}, \eqref{ini_for_eq:2} such that
\begin{gather}\label{proof:subset_1}
\{(t,x_2 (t;\tau,\xi_2)) : t > \tau\} \subset {\mathcal{E}}_0.
\end{gather}

    First, we shall prove if
\begin{gather}\label{proof:for_U_prime}
(t_*,x_*) \in \mathcal{U}, \text{ then } U^\prime (t_*,x_*)>0,
\end{gather}
where $^\prime$ denotes the derivative of function $U(t,x)$ along the trajectories
of system \eqref{eq:2}, i.e. $U^\prime (t_*,x_*) = \frac{d}{dt} U(t,x_2(t;t_*,x_*))|_{t=t_*}$.

    Let $(t_*,x_*) \in \mathcal{U}$ be a fixed point. We set
$$
m: \mathbb{R}_+ \to \mathbb{R}_+ ,\quad m(t) = \|M\dot x_1(t) + x_1(t) - x_2(t;t_*,x_*)\|^2.
$$
The definition of the set $\mathcal{U}$ implies
\begin{eqnarray*}\label{proof:help_1}
0 = U(t_*,x_*) & =& \|M\dot x_1(t_*) + x_1(t_*) - x_*\|^{\frac{6}{5}}
 - u^2 (t_*)  \\
&=& \|M\dot x_1(t_*) + x_1(t_*) - x_2(t_*;t_*,x_*)\|^{\frac{6}{5}}
 - u^2 (t_*) \\
&=& m^{\frac{3}{5}}(t_*) - u^2 (t_*) \nonumber
\end{eqnarray*}
or $m^{\frac{3}{5}}(t_*) = u^2(t_*)$. On the other hand, 
if $h_0 > 0$ is sufficiently small and if
$h \in (-h_0,h_0)$ then
\begin{eqnarray*}\label{proof:help_2}
m(t_*+h)&=& m(t_*) + 2 h 
\Big\langle M\dot x_1(t_*) + x_1(t_*) - x_2(t_*;t_*,x_*) ,\\
&& M\ddot x_1(t_*)+ \dot x_1(t_*) - \dot x_2(t_*;t_*,x_*)\Big\rangle 
+ \varepsilon_1 (h), \nonumber
\end{eqnarray*}
where $\varepsilon_1 :(-h_0,h_0)\to\mathbb{R}$ and
\begin{gather}\label{proof:limit_of_epsilon}
\lim\limits_{h\to 0}\frac{\varepsilon_1 (h)}{h} = 0.
\end{gather}
The equalities
$M \ddot x_1(t) + \dot x_1(t) = f(t,x_1(t))$  and 
$\dot x_2(t;t_*,x_*) = f(t,x_2(t;t_*,x_*))$
imply
\begin{multline}\label{proof:help_2aa}
\|M\ddot x_1(t) + \dot x_1(t) - \dot x_2(t;t_*,x_*)\|    =
                \|f(t,x_1(t)) - f(t,x_2(t;t_*,x_*))\| \leq\\
    \leq L \|x_1(t) - x_2(t;t_*,x_*)\|
    = L\|M\dot x_1(t) + x_1(t) - x_2(t;t_*,x_*) - M\dot x_1(t)\|\leq\\
    \leq L \sqrt{m(t)} + LM\|\dot x_1(t)\|.
\end{multline}
From \eqref{proof:help_2} and \eqref{proof:help_2aa} at $t=t_*$ we obtain
\begin{gather}\label{proof:help_3}
m(t_*+h)\leq m(t_*) + 2 |h| \left(L m(t_*) + LM \sqrt{m(t_*)} \|\dot x_1(t_*)\|\right) + \varepsilon_1 (h).
\end{gather}

    The formula \eqref{proof:help_3} yields
\begin{gather}\label{proof:help_4}
\begin{array}{lll}
\frac{m(t_*+h)-m(t_*)}{h} &\leq 2 L m(t_*) + 2LM \sqrt{m(t_*)} \|\dot x_1(t_*)\| + \frac{\varepsilon_1 (h)}{h}, &\text{ for } h>0,\\[2pt]
\frac{m(t_*+h)-m(t_*)}{h} &\geq -2 L m(t_*) - 2LM \sqrt{m(t_*)} \|\dot x_1(t_*)\| + \frac{\varepsilon_1 (h)}{h},&\text{ for } h<0.
\end{array}
\end{gather}

    From the definition of the function $m(t)$ it follows that $m(t)$ is ${\bf C^1}$-smooth.
Letting $h\to\pm 0$ in the inequalities \eqref{proof:help_4} and using \eqref{proof:limit_of_epsilon}
we obtain the following estimates for the derivative of function $m(t)$ at $t=t_*$
\begin{gather}\label{proof:help_5}
-2L m(t_*) - 2LM \sqrt{m(t_*)}\|\dot x_1(t_*)\|\leq \dot m(t_*) \leq 2L m(t_*) + 2LM \sqrt{m(t_*)}\|\dot x_1(t_*)\|.
\end{gather}

    Therefore, from definitions of functions $U(t,x)$, $u(t)$,
\eqref{proof:help_1} and left hand-side of \eqref{proof:help_5} it follows that
$$
\begin{array}{rl}
U^\prime (t_*,x_*) =& \frac{d}{dt} \left(m^{\frac{3}{5}}(t_*) - u^2(t_*)
\right) =
    \frac{3}{5} m^{-\frac{2}{5}}(t_*) \dot m(t_*) - 2 u \dot u(t_*) \\[2pt]
    \geq &  \frac{3}{5} m^{-\frac{2}{5}}(t_*) \left(-2L m(t_*)
 - 2LM\sqrt{m(t_*)}\|\dot x_1(t_*)\|\right) -2u(t_*) \dot u(t_*) \\[2pt]
    =&  -\frac{6}{5}L m^{\frac{3}{5}}(t_*) - \frac{6}{5}LM
 m^{\frac{1}{10}}(t_*) -2u(t_*) \dot u(t_*) \\[2pt]
    =&  -\frac{6}{5}L u^{2}(t_*) - \frac{6}{5}LM u^{\frac{1}{3}}(t_*)
     -2u(t_*) \dot u(t_*) \\[2pt]
    =&g(t_*,u(t_*))- 2  u(t_*)\dot u(t_*) > 0.
\end{array}
$$
The last inequality prove the implication \eqref{proof:for_U_prime}.
Immediately, the definition of function $V(t)$ yields
\begin{gather}\label{proof:for_V_prime}
\text{if }(t_*,x_*) \in {\mathcal{V}}, \text{ then } V^\prime (t_*,x_*) = -1 < 0.
\end{gather}

    From \eqref{proof:for_U_prime} and \eqref{proof:for_V_prime} it follows,
${\mathcal{E}}_0$ is an open $[U,V]$-subset in ${\mathcal{E}}$ with respect to the system \eqref{eq:2}.
Therefore, using Lemma \ref{lemma:help_for_topological_principle} we conclude
\begin{gather}\label{proof:for_E0-help-1}
{\mathcal{E}}_0^e = {\mathcal{E}}_0^{se} = \mathcal{U}\setminus {\mathcal{V}} = \mathcal{U}.
\end{gather}

    Now, from the definitions of the sets $\mathcal{U}$, ${\mathcal{V}}$ and equality
\eqref{proof:for_E0-help-1} it is not difficult to conclude (see. Figure \ref{figure:101})
\begin{gather}\label{proof:for_€0-help-2}
{\mathcal{E}}_0^e = \left\{ (t,x) \in \mathbb{R} \times \Omega : t>0 \text{ and } \|\phi(t) - x\| = u(t)\right\},
\end{gather}
where $\phi(t) = M \dot x_1(t) + x_1(t)$.

\begin{figure}[t]
%\includegraphics{nenovg1.eps}
\epsfxsize=10cm
\epsffile{nenovg1.eps}
\label{figure:101}\caption{}
\end{figure}

    Let $\tau > 0$ be a fixed number. Setting
$$
{\mathcal{W}} = \left\{ (t,x) \in \mathbb{R}_+ \times \Omega : t = \tau \text{ and } \|\phi(\tau) - x\| \leq u(\tau)\right\} \subset {\mathcal{E}}_0 \cup {\mathcal{E}}_0^e.
$$
we obtain that ${\mathcal{W}}$ is a ball in $\mathbb{R}^n$,  and
\begin{gather}\label{proof:for_W-help_2}
{\mathcal{W}} \cap {\mathcal{E}}_0^e = \left\{ (t,x) \in \mathbb{R}_+ \times \Omega : t = \tau \text{ and } \|\phi(\tau) - x\| = u(\tau)\right\}.
\end{gather}

    Obviously, the boundary $\partial {\mathcal{W}}$ of
the set ${\mathcal{W}}$ is not a retract of ${\mathcal{W}}$, i.e. the set ${\mathcal{W}}\cap {\mathcal{E}}_0^e$ is
not a retract of ${\mathcal{W}}$. We shall show that ${\mathcal{W}} \cap {\mathcal{E}}_0^e$ is a retract of
${\mathcal{E}}_0^e$. For this purpose we introduce the map
$$
\pi : {\mathcal{E}}_0^e \to  \mathbb{R}^{1+n},\quad \pi(t,x) = (\tau,\pi_2 (t,x)),
$$
where
$$
\pi_2(t,x) = \phi(\tau) + (x - \phi(t))\frac{u(\tau)}{u(t)}.
$$

    Obviously $\pi$ is a continuous map. Moreover, if
$(\widetilde t, \widetilde x) \in {\mathcal{E}}_0^e$, then
$$
\|\phi(\widetilde t) - \widetilde x\| = u(\widetilde t).
$$
That is why
$$
\|\phi(\tau) - \pi_2 (\widetilde t,\widetilde x)\| = \|\phi (\widetilde t) - \widetilde x\| \frac{u(\tau)}{u(\widetilde t)} = u(\tau),
$$
or $\pi :{\mathcal{E}}_0^e \to {\mathcal{W}} \cap {\mathcal{E}}_0^e$. For
$(\tau, \widetilde x) \in {\mathcal{W}} \cap {\mathcal{E}}_0^e$, we have
$$
\pi (\tau, \widetilde x) = (\tau, \pi_2(\tau,\widetilde x)) = \left(\tau, \phi(\tau) + (\widetilde x - \phi(\tau))\right) = (\tau,\widetilde x).
$$
Therefore, $\pi$ is a retraction.

    From the Wazewski's Theorem (see Theorem \ref{thm:topological_principle})
it follows that there exists at least one point $(\tau,\xi_2) \in {\mathcal{W}} \cap {\mathcal{E}}_0^e$,
such that \eqref{proof:subset_1} holds true.

    The definition of set ${\mathcal{E}}_0$ yields
\begin{gather}\label{proof:the_end}
\|M\dot x_1(t) + x_1(t) - x_2(t;\tau,\xi_2)\| < u^{\frac{5}{3}}(t)\text{ for } t>\tau.
\end{gather}
From \eqref{proof:the_end} and \eqref{proof:for_ineq-1c}
we conclude that
\begin{gather}\label{proof:the_end-1}
\lim\limits_{t\to\infty} \|M\dot x_1(t) + x_1(t) - x_2(t;\tau,\xi_2)\| \leq
\lim\limits_{t\to\infty} u^{\frac{5}{3}}(t) = 0.
\end{gather}
Therefore, \eqref{proof:the_end-1} and \eqref{thm_1:condition-1} imply
\begin{multline*}
\lim\limits_{t\to\infty} \|x_1(t) - x_2(t;\tau,\xi_2)\| =
\lim\limits_{t\to\infty} \|M\dot x_1(t) + x_1(t) - x_2(t;\tau,\xi_2) - M\dot x_1(t)\| \leq\\[2pt]
\leq\lim\limits_{t\to\infty} \|M\dot x_1(t) + x_1(t) - x_2(t;\tau,\xi_2)\| +
      M \lim\limits_{t\to\infty} \|\dot x_1(t)\| = 0.
\end{multline*}

    To complete the proof of Theorem \ref{main-theorem_1} it is enough
to set $x_2 = x_2 (0;\tau,\xi_2)$.


\subsection{Proof of Theorem \ref{main-theorem_2}}
\label{proof_of_Theorem_2}

    The proof of the Theorem \ref{main-theorem_2} is analogous to the
proof of Theorem \ref{main-theorem_1}. We shall present only
the appropriate settings:
$$
g:\mathbb{R}_+\times\mathbb{R} \to  \mathbb{R},\quad g(t,u) = - \frac{6}{5}L u^2 - \frac{6}{5} L M u^{\frac{1}{3}}\|\ddot x_1(t)\|
$$
and
$$
 U:\mathbb{R}_+\times \Omega\to \mathbb{R},\quad  U(t,x)=\|x_1(t) - x \|^{\frac{6}{5}} - u^2(t).
$$

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\end{document}