\documentclass[reqno]{amsart}
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\AtBeginDocument{{\noindent\small
Seventh Mississippi State - UAB Conference on Differential Equations and
Computational Simulations,
{\em Electronic Journal of Differential Equations},
Conf. 17 (2009), pp. 33--38.\newline
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu
\newline ftp ejde.math.txstate.edu}
\thanks{\copyright 2009 Texas State University - San Marcos.}
\vspace{9mm}}
\begin{document} \setcounter{page}{33}
\title[\hfilneg EJDE-2009/Conf/17/\hfil
Second-order differential equations]
{Second-order differential equations with asymptotically small
dissipation and piecewise flat potentials}
\author[A. Cabot, H. Engler, S. Gadat \hfil EJDE/Conf/17 \hfilneg]
{Alexandre Cabot, Hans Engler, S\'ebastien Gadat} % in alphabetical order
\address{Alexandre Cabot \newline
D\'epartement de Math\'ematiques, Universit\'e Montpellier II, CC 051\\
Place Eug\`ene Bataillon, 34095 Montpellier Cedex 5, France}
\email{acabot@math.univ-montp2.fr}
\address{Hans Engler \newline
Department of Mathematics, Georgetown University\\
Box 571233\\ Washington, DC 20057, USA}
\email{engler@georgetown.edu}
\address{S\'ebastien Gadat \newline
Institut de Math\'ematiques de Toulouse, Universit\'e Paul Sabatier\\
118, Route de Narbonne 31062 Toulouse Cedex 9, France}
\email{Sebastien.Gadat@math.ups-tlse.fr}
\dedicatory{Pour Alban, n\'e le 27 mars 2008}
\thanks{Published April 15, 2009.}
\subjclass[2000]{34G20, 34A12, 34D05}
\keywords{Differential equation; dissipative dynamical system;
\hfill\break\indent
vanishing damping; asymptotic behavior}
\begin{abstract}
We investigate the asymptotic properties as
$t\to \infty$ of the differential equation
$$
\ddot{x}(t)+a(t)\dot{x}(t)+ \nabla G(x(t))=0, \quad t\geq 0
$$
where $x(\cdot)$ is $\mathbb{R}$-valued, the map
$a:\mathbb{R}_+\to \mathbb{R}_+$ is non increasing, and
$G:\mathbb{R} \to \mathbb{R}$ is a potential with locally Lipschitz
continuous derivative. We identify conditions on the function $a(\cdot)$
that guarantee or exclude the convergence of solutions of this problem
to points in $\mathop{\rm argmin} G$, in the case where $G$ is
convex and $\mathop{\rm argmin} G$ is an interval. The condition
$$
\int_0^{\infty} e^{-\int_0^t a(s)\, ds}dt<\infty
$$
is known to be necessary for convergence of trajectories. We give a
slightly stronger condition that is sufficient.
\end{abstract}
\maketitle
\numberwithin{equation}{section}
\newtheorem{theorem}{Theorem}[section]
\newtheorem{claim}[theorem]{Claim}
\newtheorem{remark}[theorem]{Remark}
\section{Introduction}
In this note, we study the differential equation
\begin{equation}
\ddot{x}(t)+a(t)\dot{x}(t)+ \nabla G(x(t))=0, \quad t\geq 0 \label{eqcS}
\end{equation}
where $x(\cdot)$ is $\mathbb{R}$-valued, the map
$G:\mathbb{R} \to \mathbb{R}$ is at least of class $\mathcal{C}^1$,
and $a:\mathbb{R}_+\to \mathbb{R}_+$ is a non increasing function.
In a previous paper \cite{CabEngGad},
we studied this differential equation in a finite- or infinite-dimensional
Hilbert space $\mathcal{H}$. We are interested in the case where $a(t) \to0$ as
$t \to\infty$. Broadly speaking, convergence of solutions can be expected
if $a(t)\to 0$ sufficiently slowly. One of the questions left open
in that paper was whether solutions converge to a limit if the property
\begin{equation} \label{eq.exp.int.infinite}
\int_0^\infty e^{-\int_0^t a(s) ds} dt = \infty
\end{equation}
does \emph{not} hold and if $\mathop{\rm argmin} G$ consists of more than just one point.
In this note, we give a positive answer to this question,
in the one dimensional case.
\section{Preliminary Facts}\label{se.preli}
Throughout this paper, we will denote
by $G: \mathbb{R} \to \mathbb{R}$ a $\mathcal{C}^1$ function for which the derivative $G'$ is Lipschitz continuous, uniformly on bounded sets.
The function $a:\mathbb{R}_+ \to \mathbb{R}_+$ will always be assumed to be continuous and non-increasing. We also define the energy
\begin{equation*}
\mathcal{E}(t) = G(x(t)) + \frac12 |\dot{x}(t)|^2 \,.
\end{equation*}
Here are some basic results for solutions of \eqref{eqcS} from \cite{CabEngGad}.
For any $(x_0, x_1) \in \mathbb{R}^2$, the problem \eqref{eqcS} has a unique
solution $x(\cdot) \in \mathcal{C}^2([0,T),\mathbb{R})$ satisfying
$x(0) = x_0, \dot{x}(0) = x_1$ on some maximal time interval
$[0,T) \subset [0,\infty)$.
For every $t\in [0,T)$, the energy identity holds
$$\frac{d}{dt} \mathcal{E}(t) = -a(t)|\dot{x}(t)|^2.$$
If in addition $G$ is bounded from below, then
\begin{equation}\label{energy3}
\int_0^T a(t)|\dot{x}(t)|^2 dt < \infty \, ,
\end{equation}
and the solution exists for all $T > 0$.
If also $G(\xi) \to \infty$ as $|\xi| \to \infty$
(i.e. if $G$ is \emph{coercive}), then all solutions to \eqref{eqcS}
remain bounded together with their first and second derivatives
for all $t > 0$. The bound
depends only on the initial data.
If a solution $x$ to \eqref{eqcS} converges toward some $\overline{x}\in \mathbb{R}$,
then $\lim_{t\to \infty}\dot x(t)=\lim_{t\to \infty}\ddot x(t)=0$
and $G'(\overline{x})=0$.
If $\int_0^{\infty} a(s)\, ds<\infty$ and if $\inf G > - \infty$,
then solutions $x(\cdot)$ of \eqref{eqcS} for which
$(x(0),\dot{x}(0))\not \in \mathop{\rm argmin} G\times \{0\}$ cannot converge to a
point in $\mathop{\rm argmin} G$.
For the remainder of this note we shall assume that
$\mathop{\rm argmin} G \ne \emptyset$. Without loss of generality, we may assume
that $\min_\mathbb{R} G = 0$ and $G(0) = 0$. If for some $\rho \in \mathbb{R}_+$ and $z\in \mathop{\rm argmin} G$
$$\forall x\in \mathbb{R}, \quad G(x)-G(z)\leq \rho\, G'(x)(x-z)$$
then it is possible to show that any solution $x$ to the differential
equation \eqref{eqcS} satisfies
$$\int_0^{\infty}a(t) \, \mathcal{E}(t)\, dt <\infty.$$
Since $t \mapsto \mathcal{E}(t)$ is decreasing, this estimate implies that
$\mathcal{E}(t) \to \min G = 0$ as $t\to \infty$, provided that
$\int_0^{\infty}a(t) \, dt=\infty$. If now
$\mathop{\rm argmin} G = \{\overline{x}\}$ is a singleton, then trajectories must
converge to $\overline{x}$ under fairly weak additional conditions.
The reader is referred to \cite{CabEngGad} for details.
\section{Convex potentials with non-unique minima}\label{ConvTraj}
In this section, we investigate the convergence of the trajectories of
\eqref{eqcS} when $\mathop{\rm argmin} G$ is \emph{not} a singleton. While the previous
discussion shows that $\int_0^\infty a(s) ds = \infty$ is a necessary
condition for trajectories to converge to a point in $\mathop{\rm argmin} G$, this
condition is clearly not sufficient, as the particular case $G\equiv 0$ shows.
In this case, the solution is given by
$$
x(t)= x(0)+\dot x(0)\,\int_{0}^{t}e^{-\int_{0}^s a(u)\, du}ds
$$
and the solution $x$ converges if and only if
\eqref{eq.exp.int.infinite} does not hold. Therefore it is natural to ask
whether for a general potential $G$, the trajectory $x$ is convergent if this
condition does not hold. The potential $G$ is assumed to have all the
properties listed in the previous section. A general result of non-convergence
of the trajectories under the condition \eqref{eq.exp.int.infinite} is shown in
\cite{CabEngGad}. There, we assume that $G$ is coercive, $\inf_{\mathbb{R}} G = 0$,
$\mathop{\rm argmin} G = [\alpha,\beta]$ for some $\alpha < \beta$, and that $G$ is
non-increasing on $(-\infty,\alpha]$ and non-decreasing on $[\beta,\infty)$. It
is also assumed that $a$ satisfies condition \eqref{eq.exp.int.infinite}. Then
either a solution satisfies $(x(0),\dot{x}(0))\in [\alpha,\beta]\times
\{0\}$, or else the $\omega$- limit set $\omega(x_0,\dot x_0)$ contains
$[\alpha,\beta]$ and hence the trajectory $x$ does not converge.
We now ask if the converse assertion is true: do the trajectories
$x$ of \eqref{eqcS} converge if \eqref{eq.exp.int.infinite} does not hold?
We give a positive answer when the map $a$ satisfies the following stronger
condition
\begin{equation}\label{cond-theta}
\int_0^\infty e^{-\theta\, \int_0^s a(u)\, du} ds<\infty,
\end{equation}
for some $\theta\in (0,1)$.
\begin{theorem}\label{pr.conv_dim1}
Let $G:\mathbb{R}\to \mathbb{R}$ be a convex function of class $\mathcal{C}^1$ such that $G'$ is
Lipschitz continuous on the bounded sets of $\mathbb{R}$. Assume that
$\mathop{\rm argmin} G=[\alpha,\beta]$ with $\alpha < \beta$ and that there exists $\delta>0$ such that
$$
\forall \xi\in (-\infty,\alpha], \quad G'(\xi)\leq 2\, \delta\, (\xi-\alpha)
\quad \mbox{and} \quad
\forall \xi\in [\beta, \infty), \quad G'(\xi)\geq 2\, \delta\, (\xi-\beta).
$$
Let $a:\mathbb{R}_+\to \mathbb{R}_+$ be a differentiable non increasing
map such that $\lim_{t\to\infty}a(t)= 0$ and such that
condition \eqref{cond-theta} holds for some positive $\theta< 1$.
Then, for any solution $x$ to the differential equation \eqref{eqcS},
$\lim_{t\to\infty} x(t)$ exists.
\end{theorem}
\begin{proof} We may assume without loss of generality that $\alpha = 0, \beta
= 1$. The conditions on $G$ imply that it is coercive, hence
$\lim_{t\to\infty}\mathcal{E}(t) = 0$ and $|x(t)|\le M$ for some $M>0$, for all
$t\in \mathbb{R}_+$.
Define the set $\mathcal{T}=\{t\geq 0\,|\,\dot x(t)=0\}$. We shall show
that either $\mathcal{T} = [0,\infty)$ or $\mathcal{T}$ is a finite set. Assume first
that $\mathcal{T}$ has an accumulation point $t^*$. Then $\dot x(t^*)=0$ and
$\ddot x(t^*)=0$ by Rolle's Theorem. Since then
$\dot x(t^*)=\ddot x(t^*)=G'(x(t^*))=0$, $x(\cdot)$ must be constant
by forward and backward uniqueness, $\mathcal{T} = [0,\infty)$, and clearly the
limit exists. Therefore we may now assume that $\mathcal{T}$ is discrete.
If $\mathcal{T}$ is a finite set, then $\dot x$ does not change sign for
sufficiently large $t$, and the trajectory $x$ has a limit.
It remains to consider the case $\mathcal{T}=\{t_n\,| \, n\in \mathbb{N}\}$,
where the $t_n$ are increasing and tend to $\infty$. We want to show
that this is impossible.
Observe that at each $t_n$, $\dot x$ must change its sign and
$G'(x(t_n)) \ne 0$, since otherwise also $\ddot x(t_n) = 0$ and we
would again have a stationary solution. Without loss of generality,
we can assume that $\dot x(0) < 0, \, x(0) < 0$ and therefore $x(t_0)< 0$.
Since $G'(x(t_0))<0$, equation \eqref{eqcS} shows that $\ddot x(t_0)>0$,
hence the map $\dot x$ is positive on $(t_0, t_1)$, $x(t_1) > 1$, $\dot x$
is negative on $(t_1,t_2)$, and so on.
The argument so far shows that $G'(x(t))$ vanishes on a union of infinitely many disjoint closed intervals,
$$
\{t \, | \, 0 \le x(t) \le 1\} = \cup_{k\ge 0} [u_{2k},u_{2k+1}]
$$
where $0 < t_0 < u_0$ and $u_{2k-1} < t_k < u_{2k}$ for $k = 1, \, 2, \dots$.
Let us observe that, for every $k\in \mathbb{N}$,
$$
1=|x(u_{2k+1})-x(u_{2k})|=\int_{u_{2k}}^{u_{2k+1}}|\dot x (t)|\, dt\leq
|u_{2k+1}-u_{2k}|\, \max_{t\ge u_{2k}}|\dot x (t)|.
$$
Since $\lim_{t\to \infty}\dot x (t)=0$, we deduce that
$\lim_{k\to \infty} |u_{2k+1}-u_{2k}|=\infty$.
We next observe that for $u_{2k} \le t \le u_{2k+1}$ the function
$v = \dot x$ satisfies $\dot v(t) + a(t)v(t) = 0$ and hence
\begin{equation}\label{expres.dotx}
\forall t\in [u_{2k},u_{2k+1}], \quad
\dot x(t) = \dot x(u_{2k}) e^{-\int_{u_{2k}}^t a(\tau) d \tau} \,.
\end{equation}
\begin{claim}\label{clm_1}
There is a constant $\gamma$ such that $u_{2k+2} - u_{2k+1} \le \gamma$
for all $k\in \mathbb{N}$.
\end{claim}
To show this claim, fix $k\in \mathbb{N}$ and assume that
$t\in [u_{2k+1},u_{2k+2}]$. Assume for now that $k$ is odd and
thus $x(t)\le 0$. Define the quantity
$A(t)=\exp\left(\frac{1}{2}\int_0^t a(s)\, ds\right)$ and set $y(t)= A(t)\, x(t)$.
Then $y$ is the solution of the differential equation
\begin{equation}\label{diff_eq_y}
\ddot y(t)+ A(t) \, G'\left( \frac{y(t)}{A(t)}\right)
-\left(\frac{a^2(t)}{4}+ \frac{\dot a (t)}{2}\right) y(t)=0,
\end{equation}
and satisfies $y(u_{2k+1}) = y(u_{2k+2})=0$ and
$\dot y(u_{2k+1})=A(u_{2k+1})\, \dot x(u_{2k+1}) < 0$.
Since the map $a$ converges to $0$, we can choose $k$ large enough so that
$a(t)<2\,\sqrt{\delta}$ for every $t\in [u_{2k+1},u_{2k+2}]$. On the other hand, the
assumption on $G'$ shows that, for every $t\in [u_{2k+1},u_{2k+2}]$,
$$
A(t) \, G'\left( \frac{y(t)}{A(t)}\right)\leq 2\, \delta\, y(t).
$$
Recalling finally that $\dot a (t)\leq 0$ for every $t\geq 0$, we deduce from
(\ref{diff_eq_y}) that
$$
\forall t\in [u_{2k+1},u_{2k+2}], \quad \ddot y(t)+\delta\, y(t)\geq 0.
$$
The unique solution $z$ of the differential equation
$\ddot z(t)+\delta\, z(t)=0$
with the same initial conditions as $y$ has the first zero larger than
$u_{2k+1}$ at
$u_{2k+1}+\frac{\pi}{\sqrt{\delta}}$. By a standard comparison argument,
we deduce that $y$ vanishes before $z$ does, hence
$$
u_{2k+2}\leq u_{2k+1}+ \gamma, \quad \gamma = \frac{\pi}{\sqrt{\delta}} \,.
$$
The same argument applies if $k$ is even. This proves the claim.
\begin{claim}\label{clm_2}
There is a $k_0\in \mathbb{N}$ such that for $k \ge k_0$
$$
|\dot x(u_{2k+2})|\leq |\dot x(u_{2k})|\, e^{-\theta \int_{u_{2k}}^{u_{2k+2}} a(s)\, ds}.
$$
where $\theta$ is as in \eqref{cond-theta}.
\end{claim}
To prove this, pick $k_0$ so large that for all $k \ge k_0$,
$$
(1-\theta)(u_{2k+2} - u_{2k}) \ge \gamma\theta \,.
$$
This is possible since $u_{2k+2} - u_{2k} \to \infty$ as $k \to \infty$.
Since $a$ is non-increasing, this implies that
\begin{align*}
\theta \int_{u_{2k+1}}^{u_{2k+2}} a(\tau) d \tau
&\leq \gamma \theta a(u_{2k+1}) \\
&\le (1 - \theta)(u_{2k+1} - u_{2k})a(u_{2k+1}) \\
&\leq (1-\theta) \int_{u_{2k}}^{u_{2k+1}} a(\tau) d \tau
\end{align*}
and hence
$$
\theta \int_{u_{2k}}^{u_{2k+2}}a(\tau) d \tau \le \int_{u_{2k}}^{u_{2k+1}}
a(\tau) d\tau \,.
$$
Then for $k \ge k_0$,
\begin{align*}
|\dot x(u_{2k+2})|
&\leq |\dot x(u_{2k+1})|
= |\dot x(u_{2k})| e^{-\int_{u_{2k}}^{u_{2k+1}} a(s)\, ds} \\
&\leq |\dot x(u_{2k})| e^{-\theta \int_{u_{2k}}^{u_{2k+2}} a(s)\, ds}
\end{align*}
proving the claim.
\begin{claim}\label{clm_3}
If the set $\mathcal{T}$ is unbounded, there must exist a constant $C$,
depending on $\mathcal{T}$ and on $x(0), \dot x(0)$ such that for all $t \ge 0$
\begin{equation}
\label{majo_dotx_all}
|\dot x(t)|\leq C\, e^{-\theta \int_0^{t} a(s)\, ds}.
\end{equation}
\end{claim}
By making sure that $C$ is sufficiently large, we only have to prove
the estimate for $t \ge u_{2k_0}$. First assume that
$u_{2k} \le t \le u_{2k+1}$ for some $k$. Then from \eqref{expres.dotx}
$$
|\dot x(t)|\leq |\dot x(u_{2k})| \, e^{-\int_{u_{2k}}^t a(s)\, ds}
\le |\dot x(u_{2k})| \, e^{-\theta \int_{u_{2k}}^t a(s)\, ds} \,.
$$
Using induction, we deduce from Claim \ref{clm_2} that
$$
|\dot x(t)|\leq |\dot x(u_{2k_0})| \, e^{-\theta \int_{u_{2k_0}}^t a(s)\, ds}
= C_1\, e^{-\theta \int_{0}^t a(s)\, ds}
$$
with $C_1 = |\dot x(u_{2k_0})| \, e^{\theta \int_0^{u_{2k_0}} a(s)\, ds}$.
Next consider the case where $u_{2k+1} < t \le u_{2k+2}$ for some $k$. Then
$$
|\dot x(t)|\leq |\dot x(u_{2k+1})| \le C_1\, e^{-\theta \int_{0}^{u_{2k+1}} a(s)
\, ds}
\le C_1 e^{\theta \int_{u_{2k+1}}^{u_{2k+2}}a(\tau) d \tau} \,
e^{-\theta \int_0^t a(s)\, ds} \,.
$$
Due to Claim \ref{clm_1},
$e^{\theta\,\int_{u_{2k+1}}^{u_{2k+2}}a(\tau) d \tau} \le C_2$ for all $k$,
for some constant $C_2$. Estimate (\ref{majo_dotx_all}) now follows for
$t \ge u_{2k_0}$ with $C = C_1 C_2$. By enlarging $C$ further,
the estimate follows for all $t\geq 0$.
Let us now conclude the proof of the theorem. From assumption
\eqref{cond-theta} and estimate (\ref{majo_dotx_all}), we derive that
$\dot x \in L^1(0,\infty)$. Hence $\lim_{t \to \infty} x(t)$ exists,
contradicting the initial assumption.
Therefore $\lim_{t \to \infty} x(t)$ exists after all, and the
theorem has been proved.
\end{proof}
\begin{remark}\label{re.power_neg}
Note that the map $t\mapsto \frac{c}{t+1}$ with $c>1$ satisfies
condition \eqref{cond-theta} for every $\theta\in (\frac{1}{c},1)$.
In fact, if merely $a(t)\geq \frac{c}{t+1}$
for $t$ large enough for some $c>1$, then condition \eqref{cond-theta}
is satisfied. Consider next the family of maps
$a:\mathbb{R}_+\to \mathbb{R}_+$ defined by
$$
a(t)= \frac{1}{t+1}+ \frac{d}{(t+1)\,\ln (t+2)},
$$
for some $d>0$. It is immediate to check that condition
\eqref{eq.exp.int.infinite}
holds if and only if $d\in (0, 1]$. Thus non-stationary trajectories
of \eqref{eqcS} do not converge when
$d\in (0, 1]$. But condition \eqref{cond-theta} is never satisfied,
for any $\theta\in (0,1)$ and $d>0$, and the convergence of trajectories
remains an open question. Thus there remains a ``logarithmic''
gap between the criteria for existence and non-existence of limits.
\end{remark}
We conclude with some remarks on convergence results in dimension $n>1$.
It is possible to extend the non-convergence result given at the beginning
of this section to the case where the differential equation is given in
a Hilbert space $\mathcal{H}$, see \cite{CabEngGad}. However, it is not clear
how to prove that $\lim_{t \to \infty} x(t)$ exists, in a general
Hilbert space $\mathcal{H}$ and for the case where $G$ is convex and $\mathop{\rm argmin} G$
is not a singleton.
Since in this case $|\dot x(t)| \le \sqrt{2 \mathcal{E}(t)}$, it appears
natural to derive convergence results from suitable estimates for $\mathcal{E}(t)$.
In \cite{CabEngGad}, we give conditions that imply $\mathcal{E}(t) \le D a(t)$
for all $t$, for some constant $D>0$. However, since we must also assume
that $\int_0^\infty a(s) ds = \infty$, these estimates are not strong
enough to guarantee the convergence of trajectories.
One could try to extend the proof of Theorem \ref{pr.conv_dim1}. Set
$a_1(t) = a(t) \cdot \chi_{S}(x(t))$, where $\chi_S$ is the characteristic
function of $S= \mathop{\rm argmin} G$, then
$\frac{d}{dt}\mathcal{E}(t) \le - 2a_1(t)\mathcal{E}(t)$, and hence $\mathcal{E}(t) \le \mathcal{E}(0)
e^{-2\int_0^t a_1(s) ds}$. If the function
$t \mapsto e^{-\int_0^t a_1(s) ds}$ can be shown to be in $L^1(0,\infty)$,
it would follow that $|\dot{x}|$ is
integrable, implying the convergence of trajectories. This works in the
one-dimensional case since the behavior of trajectories is quite simple.
However, if $\dim \mathcal{H} > 1$, it is difficult to satisfy this property, since
trajectories corresponding to \eqref{eqcS} can be expected to behave like
trajectories of a billiard problem in $S = \mathop{\rm argmin} G$ for large times.
When the map $a$ is constant and positive, it is established in
\cite{Alv,AttGouRed} that
the trajectories of \eqref{eqcS} are weakly convergent if the potential
$G:\mathcal{H} \to \mathbb{R}$ is convex
and $\mathop{\rm argmin} G \neq \emptyset$, in an arbitrary Hilbert space $\mathcal{H}$.
The key ingredient of the proof is the Opial lemma \cite{Opi}, which allows the authors of these papers to prove convergence even if $|\dot{x}(\cdot)|$ is only in $L^2(0,\infty)$ and not in $L^1(0,\infty)$. However, if e.g. $a(t) = \frac{c}{t+1}$, then Opial's lemma requires that we show $\int_0^\infty (t+1) |\dot x(t)|^2 \, d t < \infty $, while (\ref{energy3}) implies only
$\int_0^\infty \frac{1}{t+1} |\dot x(t)|^2 \, d t < \infty $.
Hence there remains a gap if arguments similar to those
in \cite{Alv} or \cite{AttGouRed} are to be used.
It is unclear how this gap can be closed.
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\end{document}