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

\AtBeginDocument{{\noindent\small
\emph{Electronic Journal of Differential Equations},
Vol. 2015 (2015), No. 302, pp. 1--11.\newline
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu
\newline ftp ejde.math.txstate.edu}
\thanks{\copyright 2015 Texas State University.}
\vspace{9mm}}

\begin{document}
\title[\hfilneg EJDE-2015/302\hfil Asymptotic behavior]
{Asymptotic behavior for second-order differential equations
with nonlinear slowly time-decaying damping
and integrable source}

\author[M. Balti \hfil EJDE-2015/302\hfilneg]
{Mounir Balti}

\address{Mounir Balti \newline
Universit\'e de Carthage,
Institut Pr\'eparatoire aux Etudes Scientifiques et Techniques,
B.P. 51, 2070 La Marsa, Tunisia.\newline
Facult\'e des sciences de Tunis, Laboratoire EDP-LR03ES04
Universit\'e de Tunis El Manar, Tunis, Tunisia}
\email{mounir.balti@gmail.com}

\thanks{Submitted November 24, 2015. Published December 11, 2015.}
\subjclass[2010]{34A12, 34A34, 34A40, 34D05}
\keywords{Dissipative dynamical system; asymptotic behavior;
\hfill\break\indent nonautonomous asymptotically small dissipation; 
 gradient system}

\begin{abstract}
 In this article we establish convergence to the equilibrium of all
 global and bounded solutions of a gradient-like system of second-order
 with slow dissipation. Also we estimate the rate of convergence.
\end{abstract}

\maketitle
\numberwithin{equation}{section}
\newtheorem{theorem}{Theorem}[section]
\newtheorem{lemma}[theorem]{Lemma}
\newtheorem{remark}[theorem]{Remark}
\allowdisplaybreaks

\section{Introduction and main results}

In this article we study the asymptotic behaviour of global and bounded
solutions of the following gradient like system
\begin{equation}  \label{equa pr}
\begin{gathered}
\ddot{x}(t) +a( t) \| \dot{x}(t)\|^{\alpha }\dot{x}(t) +\nabla \Phi (x
( t)) =g( t) \\
x( 0) =x_0\in \mathbb{R}^N,\quad \dot{x}( 0) =x_1\in \mathbb{R}^N
\end{gathered}
\end{equation}
where
$N\in \mathbb{N}^{\ast }$, $\alpha \geq0$, 
$\Phi \in W_{\rm loc}^{2,\infty }(\mathbb{R}^N,\mathbb{R})$,
$g\in L^1({\mathbb{R}}_+,{\mathbb{R}}^N)$,
$a\in L^\infty({\mathbb{R}}_+)$, $a\geq 0$.
We denote by $S$ the set of critical points of $\Phi$:
\[
S=\{ x\in \mathbb{R}^N:\nabla \Phi (x) =0\}.
\]
Recently, Haraux and Jendoubi \cite{HaJe13} studied the asymptotic behavior
of global solutions to the nonlinear differential equation
\begin{equation}  \label{EquHJ}
\ddot{x}( t) +a(t)\dot{x}( t) +\nabla \Phi (x( t) ) =0.
\end{equation}
They prove among other things that if 
$a(t)\geq \frac{c}{(1+t)^{\beta }}$ with $\beta\geq 0$ small enough and 
$S=\arg\min \Phi$
then the solution converge as $t$ goes to infinity to $S$. Moreover, they
proved that if the potential $\Phi$ satisfies an adapted uniform Lojasiewicz
gradient inequality then the solution converge to some point $b\in S$.
The purpose of this paper is to generalize the results obtained by the
authors of \cite{HaJe13} to the equation \eqref{equa pr}.

Before stating the results of this paper, recall that equation \eqref{EquHJ}
with $a(t)=1$ has been studied by several authors. When $\Phi$ is analytic,
Haraux and Jendoubi \cite{HaJe98} 
(see also \cite{AABR02,Ch03,ChHaJe09, HaJe01})  proved convergence 
to equilibrium of all global and bounded
solutions. Now when the potential $\Phi$ is assumed to be convex and still
in the case where $a(t)=1$, Attouch et al \cite{AtGoRe00}  proved a
similar convergence result.

Equation \eqref{EquHJ} in the case where $a(t)$ tends to 0 was initiated by
Cabot et al \cite{CabotEnglerGadat} in the case where the potential $\Phi$
is convex (see also \cite{CabotFrankel,JeMa14}).

The main results of this paper read as follows.

\begin{theorem}\label{Omegalimit}
 Let $\Phi \in W_{\rm loc}^{2,\infty }( \mathbb{R}^N,
\mathbb{R})$, $a\in L^\infty(\mathbb{R}_{+})$ be a positive function,
and $x\in W_{\rm loc}^{2,1 }\cap L^\infty ( \mathbb{R}_{+},\mathbb{R}^N ) $ be
a solution of \eqref{equa pr}. Assume that
\begin{itemize}
\item[(H1)] $S=\arg \min \Phi$.

\item[(H2)] There exists $\delta >0$, $d>0$ such that 
$\| g(t) \| \leqslant \frac{d}{( 1+t) ^{1+\delta }}$. 

\item[(H3)] There exists $\beta \in ]0,1[,\ c>0$  for all $t\geqslant 0$,
$a( t) \geqslant \frac{c}{( 1+t) ^{\beta }}$.
\end{itemize}
Then
\begin{equation}  \label{Vitessetend0}
\lim_{t\to +\infty} \| \dot{x}( t) \| +\operatorname{dist}(x( t) ,S) =0.
\end{equation}
\end{theorem}

\begin{theorem} \label{ThePrincipal} 
Let $\Phi \in W_{\rm loc}^{2,\infty }(\mathbb{R}^N,\mathbb{R})$, 
$a\in L^{\infty }(\mathbb{R}_{+})$ be a positive function. 
Let $x\in W_{\rm loc}^{2,1}\cap W^{1,\infty }(\mathbb{R}_{+},\mathbb{R}^N)$ 
a solution of \eqref{equa pr}.
Assume {\rm (H1), (H2)} and that 
\begin{itemize}
\item[(H4)]  There exists $\theta \in ]0,\frac{1}{2}]$ for all 
$b\in S\exists \sigma _{b}>0$, $\exists C_{b}>0$ for all $x\in B(b,\sigma _{b})$,
$\| \nabla \Phi ( x) \| \geqslant C_{b}|\Phi ( x) -\Phi ( b) | ^{1-\theta }$.

\item[(H5)] There exists $c>0$, $\exists \beta \geq 0$:
 $\alpha +\beta \in ]0,\inf ( \frac{\theta }{1-\theta };\delta ) [$ and 
$a(t) \geq c/( 1+t) ^{\beta }$ for all $t\geq 0$.
\end{itemize}
Then there exists $b^{\ast }\in S$, $T>0$ and $M>0$ such that for every 
$t>T$
\begin{equation*}
\| x( t) -b^{\ast }\| \leqslant Mt^{-\lambda }
\end{equation*}
where
\begin{equation*}
\lambda =\inf \Big( \big[ \frac{\theta -( \alpha +\beta )
( 1-\theta ) }{( 1-\theta ) ( \alpha +2) -1}
\big] ,\big[ \frac{\delta -( \alpha +\beta ) }{( \alpha
+1) }\big] \Big) .
\end{equation*}
\end{theorem}

\begin{remark} \label{rmk1.3} \rm
(1) If $g=0$ and $\alpha=0$, we recover a result previously obtained by
Haraux and Jendoubi, see \cite[Theorem 2.3]{HaJe13}.
(2) If $\beta=0$, we recover a result obtained by Ben Hassen and Chergui, see
\cite[Theorem 1.6]{BenHassenChergui}.
\end{remark}

\begin{remark} \label{rmk1.4} \rm
Assumption {\rm (H4)} is satisfied if one of the following two conditions is
satisfied 
\begin{itemize}
\item $F$ is a polynomial \cite{Kurd05}, or
\item $F$ is analytic and $S$ is compact \cite{Chergui08}.
\end{itemize}
\end{remark}

\begin{remark} \label{rmk1.5} \rm
Let us observe that the condition $\alpha+\beta<\delta$ in (H5) is
necessary. Here is an example of a nonconvergent solution of the following
scalar equation
\begin{equation}  \label{equCE}
\ddot{x}(t) + \vert \dot{x}(t)\vert^\alpha\frac{\dot{x}(t)}{(1+t)^\beta}
 =g(t).
\end{equation}
Let $x(t)=\cos(\ln(1 + t))$ be a solution of \eqref{equCE}. Then we can
easily see that $g$ satisfies assumption (H2) with $\delta=\alpha+\beta$
and that $x$ is a non convergent solution of \eqref{equCE} with $\Phi=0$.
Note also that in this case assumption (H4) holds true with $\theta=1/2$.
\end{remark}

\begin{remark} \label{rmk1.6} \rm
The hypothesis that $\alpha+\beta<\frac{\theta}{1-\theta}$ in (H5) is in
some sense optimal.  Haraux \cite{Ha90} gives an example of a function $f$
such that the $\omega-$limit set of a global and bounded solution of the
following equation
\begin{equation*}
\ddot{x}(t) + \vert \dot{x}(t)\vert\dot{x}(t) +f(x)=0
\end{equation*}
is equal to an interval and then this solution does not converge. The
nonlinearity can be chosen such that its primitive satisfies assumption 
$(H4)$ with $\theta=1/2$.
\end{remark}

\begin{remark} \label{rmk1.7} \rm
Theorems \ref{Omegalimit} and \ref{ThePrincipal} remain true if the
dissipation term $a( t) \| \dot{x}(t)\|^{\alpha }\dot{x}(t)$ in the
equation \ref{equa pr} is replaced by $a(t)\gamma(\dot{x}(t))$ where 
$\gamma: \mathbb{R}^N \to \mathbb{R}^N$ is a continuous
function satisfying
\[
  \langle \gamma(v),v \rangle \geq \rho_1 
\| v \|^{\alpha+2}, \quad 
\| \gamma(v) \| \leq \rho_2 \| v \| ^{\alpha+1} \quad \forall v \in \mathbb{R}^N,
\]
with $0<\rho_1< \rho_2<\infty$ and $\alpha$ is as in Theorems 
\ref{Omegalimit} and \ref{ThePrincipal}.
\end{remark}

\section{Proof of Theorem \protect\ref{Omegalimit}}

 We define the two functions
\begin{gather}
E( t) =\frac{1}{2}\| \dot{x}( t) \| ^2+\Phi ( x( t) ) -\min \Phi, \nonumber\\
K( t) =E( t) +\int_t^{+\infty }\frac{M}{(1+s)^{\frac{-\beta
+( 1+\delta ) ( \alpha +2) }{\alpha +1}}}ds  \label{6-1}
\end{gather}
where
\begin{equation*}
M=\big( \frac{c}{2}\big)^{-\frac{1}{\alpha +1}}d^{\frac{\alpha +2}{\alpha
+1}},
\end{equation*}
$c$ is as in (H6) and $d$ is as in (H2). Note that by hypotheses 
(H4)--(H6), we have 
$\frac{-\beta +( 1+\delta ) ( \alpha +2) }{\alpha +1}>1$ and $K$ is well defined. 
Now by differentiating $E$
we obtain
\begin{equation*}
E'( t) =-a\| \dot{x}( t) \|^{2+\alpha }+\langle g( t) ;\dot{x}( t)\rangle.
\end{equation*}
By the Cauchy-Schwarz inequality we obtain
\begin{equation*}
E'( t) \leq -\frac{c}{( 1+t) ^{\beta }}
\| \dot{x}( t) \| ^{2+\alpha }+\Big( \frac{c}{
2( 1+t) ^{\beta }}\Big) ^{\frac{1}{\alpha +2}}\| \dot{x}
( t) \| \Big( \frac{c}{2( 1+t) ^{\beta }}
\Big) ^{-\frac{1}{\alpha +2}}\| g( t) \| .
\end{equation*}
Thanks to Young's inequality we obtain 
\begin{align*}
E'( t)
 &\leq -\frac{c}{2( 1+t) ^{\beta }}
\| \dot{x}( t) \| ^{2+\alpha }+\Big( \frac{c}{
2( 1+t) ^{\beta }}\Big) ^{-\frac{1}{\alpha +1}}\| g(
t)\|^{\frac{\alpha +2}{\alpha +1}} \\
&\leq -\frac{c}{2( 1+t) ^{\beta }}\| \dot{x}(
t) \| ^{2+\alpha }+\frac{M}{( 1+t) ^{\frac{-\beta
+( 1+\delta ) ( \alpha +2) }{\alpha +1}}}.
\end{align*}
Now by differentiating $K$ we obtain
\begin{equation*}
K'( t)  \leq -\frac{c}{2( 1+t) ^{\beta }}
\| \dot{x}( t) \| ^{2+\alpha }.
\end{equation*}
So $K$ is a decreasing and positive function. Hence
\begin{equation}
\dot{x}\in L^{\infty }({\mathbb{R}}_+,{\mathbb{R}}^N)  \label{3}
\end{equation}
and there exists $l\in \mathbb{R}_{+}$ such that
\begin{equation*}
\lim_{t\to +\infty} K( t) =\lim_{t\to +\infty } E( t) =l.
\end{equation*}
We define the function
\begin{equation*}
{\mathcal{E}}(t) =E(t) +\int_t^{+\infty }\langle g(s),\dot{x}(s) \rangle ds.
\end{equation*}
Using \eqref{3} and (H2) which implies that $g\in L^1$, we see that 
$ \lim_{t\to\infty } {\mathcal{E}}( t) = \lim_{t\to+\infty }E(t)$ and 
${\mathcal{E}}'( t) =-a ( t )\| \dot{x}(t) \| ^{2+\alpha }$. Then we obtain
\begin{equation}  \label{1}
\int_0^\infty a ( t ) \| \dot{x}(t)\| ^{2+\alpha }\,dt<\infty.
\end{equation}
Let $r>0$ and assume that there exists $\varepsilon >0$ and $t_0>0$ such
that for all $t\geqslant t_0$
\begin{equation*}
\int_t^{t+r}\| \dot{x}( s) \| ^{2+\alpha }ds\geq \varepsilon.
\end{equation*}
Then
\begin{equation*}
\int_t^{t+r}a( s) \| \dot{
x}( s) \| ^{2+\alpha }ds\geq \frac{\varepsilon c}{(
1+t+r) ^{\beta }}\quad \forall t\geqslant t_0\,.
\end{equation*}
It follows that
\begin{align*}
\int_{t_0}^{+\infty }a( s) \| \dot{x}( s)\| ^{2+\alpha }ds
&\geq \sum_{n=0}^{+\infty
}\int_{t_0+nr}^{t_0+( n+1) r}a( s) \| \dot{x}( s) \| ^{2\ast \alpha }ds \\
&\geq \sum_{n=0}^{+\infty }\frac{\varepsilon c}{(1+t_0+( n+1) r) ^{\beta }}
=\infty
\end{align*}
which contradicts \eqref{1}. Then for every $n\in \mathbb{N}^{\ast }$, there
exists $t_{n}\geq n$ such that
\begin{equation*}
\int_{t_{n}}^{t_{n}+r}\| \dot{x}( t) \|
^{2+\alpha }dt\leq \frac{1}{n}.
\end{equation*}
Hence there exists a real sequence $( t_{n}) _{n}$ such that $
 \lim_{n\to\infty }t_{n}=\infty $ and
\begin{equation}
\lim_{n\to +\infty } \int_{t_{n}}^{t_{n}+r}\| \dot{x}( t) \| ^{2+\alpha }dt=0.  
\label{4}
\end{equation}
By \eqref{3}, $x$ and $\dot{x}$ are bounded, and then by the equation
\eqref{equa pr}, $\ddot{x}$ is bounded. Hence $\dot{x}$ is Lipschitz
continuous. Thanks to the Cauchy-Schwarz inequality we obtain for all 
$t\in [ t_{n},t_{n}+r]$
\begin{align*}
| \| \dot{x}( t_{n}+t) \| ^{2+\alpha
}-\| \dot{x}( t_{n}) \| ^{2+\alpha }|
&= ( 2+\alpha ) | \int_{t_{n}}^{t_{n}+t}\| \dot{x
}( s) \| ^{\alpha }<\dot{x}( s) ,\ddot{x}
( s) >ds| \\
&\leq  ( 2+\alpha ) ( \int_{t_{n}}^{t_{n}+t}\| \dot{x
}( s) \| ^{\alpha +1}\| \ddot{x}(s)
\| ds) \\
&\leqslant  ( 2+\alpha ) \| \dot{x}\| _{\infty}^{\alpha/2}\|
\ddot{x}\| _{\infty
}(\int_{t_{n}}^{t_{n}+r}\| \dot{x}( s) \| ^{\frac{\alpha }{2}+1}ds) \\
&\leq  ( 2+\alpha ) \| \dot{x}\| _{\infty}^{\alpha/2}
\| \ddot{x}\| _{\infty }\sqrt{r}
(\int_{t_{n}}^{t_{n}+r}\| \dot{x}( s) \|
^{\alpha+2}ds) ^{1/2}.
\end{align*}
Then from \eqref{4} we obtain 
\begin{equation}  \label{5}
\lim_{n\to\infty }\sup_{s\in [ 0,r] }\| \dot{x}(t_{n}+s) \| =0.
\end{equation}
Since $x$ is a bounded function and $\nabla \Phi $ is a Lipschitz continuous
function on every bounded domain, then there exists $\lambda >0$ such that
for all $( t,s) \in \mathbb{R}_{+}^2$
\[
\| \nabla \Phi ( x( t) ) -\nabla \Phi (
x( s) ) \| \leq \lambda \| x(t) -x( s) \| .
\]
Then
\begin{equation}  \label{estgradPhi}
\begin{aligned}
\big\| r\nabla \Phi ( x( t_{n})
)-\int_{t_{n}}^{t_{n}+r}\nabla \Phi ( x( s))\,ds\big\| 
&\leq \int_{t_{n}}^{t_{n}+r}\| \nabla \Phi
( x( t_{n}) ) -\nabla \Phi ( x( s)) \| ds   \\
&\leq \lambda \int_{t_{n}}^{t_{n}+r}\| x( t_{n})
-x( s) \| ds   \\
&\leq \lambda \int_{t_{n}}^{t_{n}+r}\int_{t_{n}}^{s}\| \dot{x}
( u) \| du\,ds   \\
&\leq \lambda r^2\sup_{s\in [ 0,r] } \| \dot{x}( t_{n}+s) \| .
\end{aligned}
\end{equation}
Since
\[
\int_{t_{n}}^{t_{n}+r}\nabla \Phi ( x( s) )ds
=-\int_{t_{n}}^{t_{n}+r}\ddot{x}( t) dt
 -\int_{t_{n}}^{t_{n}+r}a( t) \| \dot{x}( t)\| ^{\alpha }\dot{x}( t)dt
+\int_{t_{n}}^{t_{n}+r}g( t) dt,
\]
then
\begin{equation}
\lim_{n\to +\infty} \int_{t_{n}}^{t_{n}+r}\nabla \Phi
( x( s) ) ds=0.  \label{6}
\end{equation}
Combining \eqref{5}, \eqref{estgradPhi} and \eqref{6} yields
\begin{equation}
\lim_{n\to +\infty} \| \nabla \Phi ( x(t_{n}) ) \| =0.  \label{7}
\end{equation}
Hence
\begin{equation*}
l=\lim_{t\to +\infty} E( t) =\underset{
n\to +\infty }{\lim }E( t_{n}) =\lim_{n\to +\infty} \Phi ( x( t_{n}) ) -\min \Phi .
\end{equation*}

Since $( x( t_{n}) ) _{n}$ is a bounded sequence, 
we can extract a subsequence, still denoted by $( x( t_{n})
) _{n}$ such that $ \lim_{n\to\infty }x( t_{n}) =a$.
From \eqref{7} we obtain
\begin{equation*}
\lim_{n\to +\infty} \nabla \Phi ( x( t_{n}) ) =0=\nabla \Phi ( a) .
\end{equation*}
Then $a\in S$. By (H1), $S=\arg \min \Phi $, and then it follows that
$\lim_{t\to\infty }E( t) =0$,
so $ \lim_{t\to\infty }\| \dot{x}( t)
\| =0$ and $ \lim_{t\to\infty }\Phi ( x(
t) ) =\min \Phi $.
Assume that
\begin{equation*}
\lim_{t\to\infty } \operatorname{dist}( x( t) ,S) \neq 0.
\end{equation*}
Then there exists $\varepsilon >0$ and $t_{n} \to  \infty $ such
that
\begin{equation}
d( x( t_{n}) ,S) \geqslant \varepsilon.  \label{8}
\end{equation}
Therefore, we can extract a subsequence still denoted by 
$( t_{n}) $ such that
\begin{equation*}
\lim_{n\to +\infty} x( t_{ n}) =a.
\end{equation*}
Then $  \lim_{n\to \infty}\Phi ( x( t_{n })) =\Phi ( a) =\min \Phi$ that is 
$a\in S$, which contradicts \eqref{8}.

\section{Proof of theorem \ref{ThePrincipal}}

To prove Theorem \ref{ThePrincipal}, we need some lemmas. We begin
with the following lemma proved by Alvarez et al \cite{AABR02}, here we
give a slightly different proof.

\begin{lemma} \label{LojUniforme} 
Under hypothesis {\rm (H4)}, let $\Gamma $ be a compact
subset of ${\mathbb{R}}^N$ such that
\begin{equation*}
\exists K\in {\mathbb{R}}:  \forall a\in \Gamma\ \Phi (a)=K .
\end{equation*}
Then there exist $\sigma, C>0$ and $\theta \in (0,1/2)$ such that
\begin{equation}
[ d(u,\Gamma )\leq \sigma \Rightarrow \| \nabla \Phi (u)\| \geq C|\Phi
(u)-K|^{1-\theta }] .  \label{2}
\end{equation}
\end{lemma}

\begin{proof}
Using  (H4),  there exists $\theta \in ]0,1/2]$ 
such that for all $a\in \Gamma $ there exists $C_{a}>0$, $\sigma _{a}>0$ such that
\begin{equation}
\| \nabla \Phi (u)\| \geq C_{a}|\Phi (u)-\Phi (a)|^{1-\theta }\quad
\forall u\in B( a,\sigma _{a}) .  \label{LojIneq}
\end{equation}
Since $\Gamma $ is compact, then there exists $( a_1,\dots ,a_{n})
\in \Gamma ^{n}$ such that
\begin{equation*}
\Gamma \subset ( \cup_{i=1}^{n}B( a_{i},\frac{\sigma
_{a_{i}}}{2}) ) .
\end{equation*}
We choose $\sigma =\inf \sigma _{a_{i}}/2$ and $C=\inf C_{a_{i}}$.
Let $u\in {{{\mathbb{R}}}}^N$ such that $d(u,\Gamma )\leq \sigma $. Then
there exists $a\in \Gamma $ such that $d(u,a)\leq \sigma $ and 
$i\in \{1,2,\dots ,n\} $ such that 
$a\in B( a_{i},\frac{\sigma _{a_{i}}}{2}) $. Hence we obtain
 $d(u,a_{i})\leq \sigma _{a_{i}}$. From
\eqref{LojIneq} we obtain
\[
\| \nabla \Phi (u)\| 
\geq C_{a}|\Phi (u)-\Phi (a)|^{1-\theta } 
\geq C|\Phi (u)-K|^{1-\theta }.
\]
\end{proof}

\begin{lemma} \label{PrinComp} 
Let $f$ and $g:\mathbb{R}_{+}\to \mathbb{R}_{+}$ be
two continuously differentiable functions, 
$h:\mathbb{R}_{+}^2\to \mathbb{R}$ be 
continuously differentiable function, and $T\geq 0$ be such that for all 
$t\geq T$,
\begin{gather*}
f'( t) +h( t,f( t) ) \leq g'(t) +h( t,g( t) ) , \\
f( T) \leq g( T).
\end{gather*}
Then for all $t\geq T$,
$f( t) \leq g( t)$.
\end{lemma}

\begin{proof}
Let $k:\mathbb{R}_{+}\to \mathbb{R}$ be a function such that
\[
k'( t) =\begin{cases}
\frac{h( t,g( t) ) -h( t,f( t) ) }{g( t) -f( t) } &\text{if }g( t) \neq
f( t) \\[4pt]
\partial _2h( t,f( t) ) &\text{if }g( t) =f( t)
\end{cases}
\]
and
$\phi :\mathbb{R}_{+}\to \mathbb{R};t\longmapsto e^{k( t)}( g-f) ( t)$.
Then for all $t\geq T$,
\[
\phi '( t) =[ ( g-f) '(t) +k'( t) ( g-f) ( t)] e^{k( t) }\geq 0.
\]
So $\phi $ is an increasing function in $[T,+\infty [ $. Finally we
see that for all $t\in $ $[T,+\infty [ $ we obtain
$f( t) \leq g( t)$.
\end{proof}

\begin{lemma}\label{IneqDiff} 
Let $H\in W_{\rm loc}^{1,1}(\mathbb{R}_{+},\mathbb{R}_{+})$.
Assume that there exist constants $k_1>0$, $k_2\geq 0$, $T>0$, 
$\mu >1>\beta $ and $\gamma >\beta >0$ such that for almost every $t\geq T$ we
have
\begin{equation*}
H'( t) +\frac{k_1}{(1+t)^{\beta }}( H(t))
^{\mu }\leq \frac{k_2}{( 1+t)^{\gamma }}.
\end{equation*}
Then there exists $M>0$ such that for all $t\geq T$,
\begin{equation*}
H( t) \leq \frac{M}{( 1+t) ^{c}}
\end{equation*}
where
\begin{equation*}
c=\inf \Big( \frac{\gamma -\beta }{\mu },\frac{1-\beta }{\mu -1}\Big) .
\end{equation*}
\end{lemma}

\begin{proof}
Let $M>0$ such that $k_1M^{\mu }-cM>k_2$ and $M>H( T) (
1+T) ^{c}$. We define the function $\phi:{\mathbb{R}}_+\to \mathbb{R}$ by
\begin{equation*}
\phi ( t) =\frac{M}{( 1+t) ^{c}}.
\end{equation*}
Hence for all $t\geq T$, we have
\begin{align*}
\phi'( t) +\frac{k_1}{( 1+t)^{\beta }}(\phi ( t) )^{\mu }
&=\frac{k_1M^{\mu }}{(
1+t) ^{\beta +c\mu }}( 1-\frac{cM^{1-\mu }}{k_1( 1+t)
^{1-\beta +c( 1-\mu ) }}) \\
&\geq \frac{k_1M^{\mu }}{( 1+t) ^{\beta +c\mu }}( 1-\frac{
cM^{1-\mu }}{k_1}) \\
&\geq \frac{k_2}{( 1+t)^{\gamma }} \\
&\geq H'( t) +\frac{k_1}{( 1+t) ^{\beta }}( H( t) ) ^{\mu }.
\end{align*}
Since $\phi ( T) \geq H( T) $, thanks to Lemma \ref{PrinComp},  
for all $t\geq T$, we obtain
\[
H( t) \leq \phi ( t) =\frac{M}{( 1+t)^{c}}.
\]
\end{proof}

\begin{proof}[Proof of Theorem \ref{ThePrincipal}]
 Let $\varepsilon >0$.
We define the function
\begin{equation}
H(t)={\mathcal{E}}( t) +\frac{\varepsilon \| \nabla \Phi
( x( t) ) \| ^{\alpha }}{( 1+t)
^{\beta }}\langle \nabla \Phi (x(t)),\dot{x}(t)\rangle+\frac{\varepsilon }{2}
\int_t^{\infty }\frac{\| \nabla \Phi (x(s))\| ^{\alpha }\|
g(s)\| ^2}{(1+s)^{\beta }}\,ds.  \label{9}
\end{equation}
By differentiating $H$ we obtain
\begin{align*}
H'( t) 
&= -a( t) \| \dot{x}(t) \| ^{\alpha +2}-\frac{\varepsilon \beta }{(
1+t) ^{\beta +1}}\| \nabla \Phi ( x( t)
) \| ^{\alpha }\langle \nabla \Phi ( x(
t) ) ,\dot{x}( t) \rangle \\
&\quad +\frac{\varepsilon }{( 1+t) ^{\beta }}\| \nabla \Phi
( x( t) ) \| ^{\alpha }\langle \nabla
^2\Phi ( x( t) ) \dot{x}( t) ,\dot{x}
( t) \rangle \\
&\quad +\frac{\varepsilon }{( 1+t) ^{\beta }}\alpha \| \nabla
\Phi ( x( t) ) \| ^{\alpha -2}\langle
\nabla ^2\Phi ( x( t) ) \dot{x}( t)
,\nabla \Phi ( x( t) ) \rangle \langle
\nabla \Phi ( x( t) ) ,\dot{x}( t)\rangle \\
&\quad -\frac{\varepsilon a( t) }{( 1+t) ^{\beta }}
\| \dot{x}( t) \| ^{\alpha }\| \nabla
\Phi ( x( t) ) \| ^{\alpha }\langle
\nabla \Phi ( x( t) ) ,\dot{x}( t)
\rangle -\frac{\varepsilon }{( 1+t) ^{\beta }}\|
\nabla \Phi ( x) \| ^{2+\alpha } \\
&\quad +\frac{\varepsilon }{( 1+t) ^{\beta }}\| \nabla \Phi
( x( t) ) \| ^{\alpha }\langle \nabla
\Phi ( x( t) ) ,g( t) \rangle -\frac{
\varepsilon }{2}\frac{\| \nabla \Phi (x(t))\| ^{\alpha }\|
g(t)\| ^2}{(1+t)^{\beta }}.
\end{align*}
By the Cauchy-Schwarz inequality and by setting 
$M_1=\|\nabla^2\Phi ( x) \| _{\infty }$ and 
$M_2=\| a\| _{\infty }$ we obtain
\begin{align*}
H'(t) 
&\leq -a( t) \| \dot{x}( t)
\| ^{\alpha +2}-\frac{\varepsilon }{( 1+t) ^{\beta }}
\| \nabla \Phi ( x) \| ^{2+\alpha }
+\frac{\varepsilon \beta }{( 1+t) ^{\beta +1}}\| \dot{x}(
t) \| \| \nabla \Phi ( x( t) )\| ^{\alpha +1} \\
&\quad +\frac{\varepsilon M_1( \alpha +1) }{( 1+t)
^{\beta }}\| \dot{x}( t) \| ^2\|
\nabla \Phi ( x( t) ) \| ^{\alpha } 
 +\frac{\varepsilon M_2}{( 1+t) ^{\beta }}\| \dot{x}
( t) \| ^{\alpha +1}\| \nabla \Phi (
x( t) ) \| ^{\alpha +1} \\
&\quad +\frac{\varepsilon }{( 1+t) ^{\beta }}\| \nabla \Phi
( x( t) ) \| ^{\alpha }\| \nabla \Phi
(x(t))\| \| g(t)\| -\frac{\varepsilon }{2}\frac{\| \nabla \Phi
(x(t))\| ^{\alpha }\| g(t)\| ^2}{(1+t)^{\beta }}.
\end{align*}
By Young's inequality, there exist $C_1,C_2>0$ such that
\begin{align*}
H'(t) &\leq -a( t) \| \dot{x}( t)
\| ^{2+\alpha }-\frac{\varepsilon }{2( 1+t) ^{\beta }}
\| \nabla \Phi ( x( t) ) \|^{2+\alpha
} \\
&\quad +\frac{\varepsilon \beta }{( 1+t) ^{\beta +1}}( \|
\dot{x}( t) \| ^{\alpha +2}+\| \nabla \Phi
( x( t) ) \| ^{\alpha +2}) \\
&\quad +\frac{\varepsilon }{( 1+t) ^{\beta }}( C_1\|
\dot{x}( t) \| ^{\alpha +2}+\frac{1}{8}\| \nabla
\Phi ( x( t) ) \| ^{\alpha +2}) \\
&\quad +\frac{\varepsilon }{( 1+t) ^{\beta }}( C_2\|
\dot{x}( t) \| ^{( \alpha +2) ( \alpha
+1) }+\frac{1}{8}\| \nabla \Phi ( x( t)
) \| ^{\alpha +2}) .
\end{align*}
By using \eqref{Vitessetend0}, there exists $T>0$ such that
\begin{equation}  \label{vitesseinf1}
\| \dot{x}(t)\| <1\quad \forall t\geq T.
\end{equation}
Then we obtain that for all $t\geq T$ 
(with $T$  large enough so that $(1/((1+T)^{\beta)}\leq 1/8$),
\begin{equation*}
H'( t) \leqslant \Big( \frac{-c+\varepsilon (\beta
+C_1+C_2)}{( 1+t) ^{\beta }}\Big) \| \dot{x}(t) \| ^{2+\alpha }
-\frac{\varepsilon }{8(1+t)^{\beta }}\| \nabla \Phi ( x( t) ) \|^{2+\alpha }.
\end{equation*}
By choosing $\varepsilon $ small enough, we obtain that for all $t\geqslant T$,
\begin{equation}  \label{7-2}
H'( t) \leqslant -\frac{\varepsilon }{8( 1+t)
^{\beta }}( \| \dot{x}( t) \| ^{2+\alpha
}+\| \nabla \Phi ( x( t) ) \|^{2+\alpha }) .
\end{equation}
So $H$ is nonincreasing on $[T,\infty )$ and
$\lim_{t\to \infty }H(t)=0$.
From \eqref{9} together with the Cauchy-Schwarz inequality we obtain for all 
$t>T$,
\begin{align*}
&[ H(t)] ^{(1-\theta )(\alpha +2)} \\
&\leq \Big[ {\mathcal{E}}( t) +\frac{\varepsilon \|
\nabla \Phi ( x( t) ) \| ^{\alpha +1}}{(
1+t) ^{\beta }}\| \dot{x}(t)\| +\frac{\varepsilon }{2}
\int_t^{\infty }\frac{\| \nabla \Phi (x(s))\| ^{\alpha }\|
g(s)\| ^2}{(1+s)^{\beta }}\,ds\Big] ^{(1-\theta )(\alpha +2)}
\end{align*}
By using the inequality $\big(\sum_{i=1}^{5} a_i\big)^\lambda \leq
5^\lambda \sum_{i=1}^5 a_i^\lambda$ for $a_i$ nonnegative for all $i$ and 
$0\leq\lambda\leq2$, we obtain that for all $t\geq T$,
\begin{equation} \label{EstiH}
\begin{aligned}
{}[ H(t)]^{(1-\theta )(\alpha +2)}  
&\leq C_3\big[ \frac{1}{2}\| \dot{x}( t) \|^2
\big] ^{(1-\theta )(\alpha +2)}+C_3[ \Phi ( x( t)
) -\min \Phi ] ^{(1-\theta )(\alpha +2)}   \\
&\quad +C_3\Big[ \int_t^{+\infty }\langle g( s) ,\dot{x}( s) \rangle\,ds
\Big] ^{(1-\theta )(\alpha +2)}\\
&\quad +C_3\Big[ \frac{\varepsilon \|
\nabla \Phi ( x( t) ) \| ^{\alpha +1}}{(
1+t) ^{\beta }}\| \dot{x}(t)\|\Big] ^{(1-\theta )(\alpha +2)}\\
&\quad + C_3\Big[ \frac{\varepsilon }{2}\int_t^{\infty }\frac{\| \nabla \Phi
(x(s))\| ^{\alpha }\| g(s)\| ^2}{(1+s)^{\beta }}\,ds\Big]
^{(1-\theta )(\alpha +2)}.
\end{aligned}
\end{equation}
where $C_3=5^{(1-\theta )(\alpha +2)}$. By using \eqref{vitesseinf1} and
since $2(1-\theta )(\alpha +2)\geq \alpha +2$ we have
\begin{equation}
[ \| \dot{x}(t)\| ^2]^{(1-\theta )(\alpha +2)}\leq \| \dot{x
}(t)\| ^{\alpha +2}  \label{EstimaVitesse}
\end{equation}
Now by using  (H4) and Lemma \ref{LojUniforme} we obtain that for all
$t\geq T$,
\begin{equation}
[ \Phi ( x( t) ) -\min \Phi ] ^{(1-\theta
)(\alpha +2)}\leq \| \nabla \Phi (x(t))\| ^{\alpha +2}.
\label{EstimaPhi}
\end{equation}
Young's inequality yields
\begin{equation} \label{Estimag}
\begin{aligned}
&\Big| \int_t^{\infty }\langle g(s),\dot{x}(s)\rangle\,ds\Big|^{(1-\theta )
 (\alpha +2)}\\
&\leq K({\rho })\Big( \int_t^{+\infty }\|
g(s)\| ^{\frac{\alpha +2}{\alpha +1}}(1+t)^{\frac{\beta }{\alpha +1}
}\,ds\Big) ^{(\alpha +2)(1-\theta )} \\
&\quad + \rho ( \int_t^{+\infty }\frac{\| \dot{x}(s)\| ^{\alpha +2}}{
(1+t)^{\beta }}\,ds) ^{(1-\theta )(\alpha +2)} ,
\end{aligned}
\end{equation}
where $\rho $ is a small positive constant which will be fixed in the sequel.
Using (H2) we obtain
\begin{equation}
\Big( \int_t^{+\infty }\| g(s)\| ^{\frac{\alpha +2}{\alpha +1}
}(1+t)^{\frac{\beta }{\alpha +1}}\,ds\Big) ^{(\alpha +2)(1-\theta )}
\leq C_{4}(1+t)^{-\chi},  \label{Estimag+1}
\end{equation}
where 
\[
\chi =\frac{1+\alpha \delta +2\delta -\beta }{1+\alpha }(\alpha +2)(1-\theta ).
\]
 Once again, by applying Young's inequality and
using the fact that $(1-\theta )(\alpha +2)\geq 1$, we obtain that for all 
$t\geq T$,
\begin{equation} \label{Equationphivitx}
\begin{aligned}
&C_3\Big[ \frac{\varepsilon \| \nabla \Phi ( x( t)
) \| ^{\alpha +1}}{( 1+t) ^{\beta }}\| \dot{x}
(t)\| \Big] ^{(1-\theta )(\alpha +2)} \\
&\leq C_5[ \frac{\|\nabla \Phi (x(t))\| ^{\alpha +2}
 +\| \dot{x}(t)\| ^{\alpha +2}}{(1+t)^{\beta }}] ^{(1-\theta )(\alpha +2)}
 \\
&\leq C_5(\| \nabla \Phi (x(t))\| ^{\alpha +2}+\| \dot{x}
(t)\| ^{\alpha +2}).
\end{aligned}
\end{equation}
Now, since $x$ is bounded and by (H2), we obtain
\begin{equation}
C_3\Big[ \frac{\varepsilon }{2}\int_t^{\infty }\frac{\| \nabla \Phi
(x(s))\| ^{\alpha }\| g(s)\| ^2}{(1+s)^{\beta }}\,ds\Big]
^{(1-\theta )(\alpha +2)}\leq C_{6}(1+t)^{-\eta },  \label{EstimaPhig}
\end{equation}
where $\eta =(1+2\delta +\beta )(1-\theta )(\alpha +2)$.

By combining \eqref{EstiH}, \eqref{EstimaVitesse}, \eqref{EstimaPhi},
\eqref{Estimag}, \eqref{Estimag+1}, \eqref{Equationphivitx} and
\eqref{EstimaPhig} we obtain
\begin{equation} \label{estimHavantFinal}
\begin{aligned}
&[ H(t)] ^{(1-\theta )(\alpha +2)} \\
&\leq C_{7}(\| \dot{x}
(t)\| ^{\alpha +2}+\| \nabla \Phi (x(t))\| ^{\alpha +2}) \\
&\quad +C_{8}(1+t)^{-\chi }+C_{9}(1+t)^{-\eta }
+\rho \Big(\int_t^{+\infty }\frac{\| \dot{x}(s)\| ^{\alpha +2}}
{(1+t)^{\beta }} \,ds\Big) ^{(1-\theta )(\alpha +2)}   \\
&\leq C_{7}(\| \dot{x}(t)\| ^{\alpha +2}+\| \nabla \Phi
(x(t))\| ^{\alpha +2}) \\
&\quad +C_{10}(1+t)^{-\chi }+\rho \Big( \int_t^{+\infty }\frac{\|
\dot{x}(s)\| ^{\alpha +2}}{(1+t)^{\beta }}\,ds\Big) ^{(1-\theta)(\alpha +2)} ,
\end{aligned}
\end{equation}
where we use the fact that $\eta >\chi $ in the last inequality.
On the other hand, by integrating \eqref{7-2} over $(t,\infty )$, we obtain
\[
\Big( \int_t^{\infty }\frac{\| \dot{x}(s)\| ^{\alpha +2}}{
(1+t)^{\beta }}\,ds\Big) ^{(1-\theta )(\alpha +2)}
\leq \Big( \frac{8}{\varepsilon }H(t)\Big) ^{(1-\theta )(\alpha +2)}.
\]
Now by choosing $\rho $ in \eqref{Estimag} such that 
$\rho (8/\varepsilon)^{(1-\theta )(\alpha +2)}<1/2$, estimate
\eqref{estimHavantFinal} becomes
\begin{equation}
[ H(t)]^{(1-\theta )(\alpha +2)}\leq C_{11}(\| \dot{x}(t)\|
^{\alpha +2}+\| \nabla \Phi (x(t))\| ^{\alpha +2})+C_{12}(1+t)^{-\chi}.
\end{equation}
Now, by combining \eqref{7-2} with the above inequality, we obtain that for all 
$t\geq T$
\begin{align*}
-H'( t) &\geq \frac{\varepsilon }{8( 1+t)
^{\beta }}( \| \dot{x}( t) \| ^{2+\alpha}+\| \nabla \Phi ( x( t) ) \|
^{2+\alpha }) \\
&\geq C_{13}\frac{[H(t)]^{(1-\theta )(\alpha +2)}}{(1+t)^{\beta }}
-C_{14}(1+t)^{-\chi -\beta }.
\end{align*}
We finally obtain the differential inequality
\[
H'( t) +\frac{C_{13}}{( 1+t) ^{\beta }}[
H( t) ] ^{( 2+\alpha ) ( 1-\theta )
}\leqslant \frac{C_{14}}{( 1+t) ^{^{\chi +\beta }}}.
\]
By using Lemma \ref{IneqDiff}, there exists $M>0$ such that for all $t\geq T$,
\begin{equation*}
H( t) \leq \frac{M}{( 1+t) ^{\nu }}
\end{equation*}
where
\begin{align*}
\nu &=\inf \Big( \frac{\chi }{( 2+\alpha ) ( 1-\theta) },
 \frac{1-\beta }{( 2+\alpha ) ( 1-\theta ) -1}\Big) \\
& = \inf \Big( \delta +\frac{1+\delta -\beta }{1+\alpha },\frac{1-\beta }{
( 2+\alpha ) ( 1-\theta ) -1}\Big) .
\end{align*}
Once again from \eqref{7-2},
\[
\frac{\varepsilon }{8( 1+t) ^{\beta }}\| \dot{x}(t) \| ^{2+\alpha }\leq -H'( t) .
\]
Then for all $t>T$,
\[
\int_t^{2t}\frac{\varepsilon }{8( 1+s) ^{\beta }}\|
\dot{x}( s) \| ^{2+\alpha }ds 
\leq H( t) \leq \frac{M}{( 1+t)^{\nu }}.
\]
H\"older's inequality yields
\begin{align*}
\int_t^{2t}\| \dot{x}( s) \| ds 
&\leq t^{\frac{1+\alpha }{2+\alpha }}\Big( \int_t^{2t}\| \dot{x}
(s) \| ^{2+\alpha }ds\Big) ^{\frac{1}{2+\alpha }} \\
&\leq t^{\frac{1+\alpha }{2+\alpha }}\Big( \frac{8( 1+2t)
^{\beta }}{\varepsilon }\int_t^{2t}\frac{\varepsilon }{8( 1+s)
^{\beta }}\| \dot{x}( s) \| ^{2+\alpha
}ds\Big) ^{\frac{1}{2+\alpha }} \\
&\leq t^{\frac{1+\alpha }{2+\alpha }}\Big( \frac{8(
1+2t)^{\beta }}{\varepsilon }\frac{M}{( 1+t)^{\nu }}
\Big)^{\frac{1}{2+\alpha }} 
\leq \frac{C_{15}}{t^{\lambda }},
\end{align*}
where
\begin{align*}
\lambda 
&=\frac{\nu }{2+\alpha }-\frac{\alpha +1+\beta }{2+\alpha } \\
&=\inf \Big( \big[ \frac{\theta -( \alpha +\beta ) (
1-\theta ) }{( 1-\theta ) ( \alpha +2) -1}\big]
,\big[ \frac{\delta -( \alpha +\beta ) }{( \alpha +1)
}\big] \Big)>0 .
\end{align*}
Then
\begin{align*}
\int_t^{+\infty }\| \dot{x}( s) \| ds 
&\leq \sum_{n=0}^{+\infty }\int_{2^{n}t}^{2^{n+1}t}\| \dot{x}
( s) \| ds \\
&\leq \sum_{n=0}^{+\infty }\frac{C_{15}}{2^{n\lambda }t^{\lambda }}\\
&\leq \frac{C_{15}}{t^{\lambda }( 1-2^{-\lambda }) }
\end{align*}
and the result follows since
\[
\| x(t) -x(\tau)\| \leq \int_t^\tau \| \dot{x}
(s)\|\, ds \leq \int_t^{\infty } \| \dot{x}(s)\| ds.
\]
\end{proof}


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\end{document}