\documentclass[reqno]{amsart}
\usepackage{graphicx}  % to input a PostScript figure.

\AtBeginDocument{{\noindent\small
{\em Electronic Journal of Differential Equations},
Vol. 2003(2003), No. 70, pp. 1--12.\newline
ISSN: 1072-6691. URL: http://ejde.math.swt.edu or http://ejde.math.unt.edu
\newline ftp ejde.math.swt.edu  (login: ftp)}
\thanks{\copyright 2003 Southwest Texas State University.}
\vspace{9mm}}

\begin{document}

\title[\hfilneg EJDE--2003/70\hfil Lienard's equation with quadratic viscous feedback]
{Energy decay estimates for Lienard's equation with quadratic viscous feedback}


\author[A. Y. Khapalov \& P. Nag\hfil EJDE--2003/70\hfilneg]
{Alexander Y. Khapalov \& Parthasarathi Nag }

\address{Alexander Y. Khapalov \newline
Department of Mathematics, Washington State University,\newline
Pullman, WA 99164-3113, USA}
\email{khapala@wsu.edu}

\address{Parthasarathi Nag \newline
Department of Mathematics, Washington State University,\newline
Pullman, WA 99164-3113, USA}
\email{pnag@math.wsu.edu}

\date{}
\thanks{Submitted April 9, 2003. Published June 21, 2003.}
\subjclass[2000]{93D05, 93D15, 93D20}
\keywords{Bilinear systems, stabilization, quadratic feedback, \hfill\break\indent
 energy decay estimate}

\begin{abstract}
 This article concerns the stabilization for a well-known
 Lienard's system of ordinary differential equations modelling oscillatory
 phenomena. It is known that such a system is asymptotically stable
 when a linear viscous (motion-activated) damping with constant gain is
 engaged. However, in many applications it seems more realistic that the
 aforementioned gain is not constant and does depend on the deviation from
 equilibrium. In this article, we consider a (nonlinear) gain, introduced in
 \cite{b1}, which is proportional to the square of such deviation and
 derive an explicit energy decay estimate for solutions of the corresponding
 ``damped'' Lienard's system. We also discuss the place of our result in the
 framework of stabilization of so-called critical bilinear systems.
\end{abstract}

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

\section{Introduction}

\subsection*{Motivation and main results}
This article concerns the stabilization of a single oscillatory motion
(or, more generally, of an oscillatory phenomenon) by means of suitable damping.
Without loss of generality (that is, one might need to perform a routine change of variables first) such
motion is described by a two dimensional system of ordinary differential equations of Lienard's type like
\begin{equation}
\dot{x} = \begin{pmatrix} 0 &-1 \\ 1 &0\end{pmatrix} x + u Bx ,
\label{S}
\end{equation}
where $ x = (x_1, x_2)$ and $u B x$ models a damping device of structure described
by matrix $B$ with gain $u = u (x_1, x_2)$.

A system like \eqref{S} is widely used to model various kinds of
periodic oscillatory phenomena, for example, in electrical or civil engineering (see, e.g.,
\cite{j1,p1} and the references therein). In particular, if we assume that \eqref{S} models a motion
of a particle,
then $x_1 (t)$ would describe the deviation (position) of this particle with respect to equilibrium
$x_1 = 0$ at time $t$, while $-x_2 (t)$ would describe its velocity.

One of the most typical damping devices used in applications is so-called the
``{\em viscous (or motion-activated)'' damping} which, in the framework of model
 \eqref{S}, is associated with the matrix
$$
B = \begin{pmatrix} 0 & 0 \\ 0 & 1\end{pmatrix}
$$
and a {\em negative} gain $ u$, in which case system \eqref{S} looks as follows:
\begin{gather} \label{e1.1a}
\begin{gathered}
\dot{x}_1 = -x_{2}, \quad t > 0,\\
\dot{x}_2 = x_{1} + u x_{2},
\end{gathered}   \\
x_{1}(0) = x_{10}, x_{2}(0) = x_{20}.
\label{e1.1b}
\end{gather}
Note that the damping term in \eqref{e1.1a} is active only when the ``velocity'' $ x_{2}(t) $
is not zero and it ``acts'' in the direction which is opposite of the ``motion of the particle'' at
any given time $ t$.

It is classical result that system \eqref{e1.1a}--\eqref{e1.1b} is asymptotically
stable when a linear viscous damping with a negative constant gain is engaged,
that is, when $ u $ in \eqref{e1.1a} is a negative number.
(Note that in this case one can obtain the explicit formula for solutions.)
However, it seems more realistic in many applications that the aforementioned gain
$ u $ should depend on deviation from equilibrium. In this respect we refer
the reader to Jurdjevic and Quinn \cite{j1} who considered the 
{\em quadratic} gain
\begin{equation} \label{e1.1c}
u(x_1,x_2) = -x_1^2,
\end{equation}
in which case the gain is proportional to the square of the magnitude of deviation.
It was shown in \cite{j1} that system \eqref{e1.1a}--\eqref{e1.1c}
is asymptotically stable. On the other hand,
the method of \cite{j1} does not provide any explicit estimates for the energy decay of the
corresponding solutions, which evaluate the effectiveness of nonlinear feedback
\eqref{e1.1c} and thus are of principal importance in applications.

The main goal of this paper is to derive such an {\em explicit energy decay estimate} for system
\eqref{e1.1a}--\eqref{e1.1c}. It is clear from the start that it cannot be of
exponential type, because system \eqref{e1.1a}--\eqref{e1.1c} is
critical (see the discussion in the next section for details).
 We have the following result.


\begin{theorem} \label{thm1.1}
There exist positive constants $\beta$, $\gamma$, and
$\bar{t} = \bar{t}(\parallel (x_{10},x_{20}) \parallel_{R^{2}})$ such that the
energy
$$
E(t) = \frac{x_{1}^{2}(t) + x_{2}^{2}(t)}{2}
$$
of system \eqref{e1.1a}--\eqref{e1.1c} satisfies
\begin{equation} \label{e1.2}
E(t) \leq \frac{E(0)}{\beta + \gamma E(0)t}, \quad t > \bar{t}.
\end{equation}
\end{theorem}

The values of the constants $\beta$ and $\gamma$ are given in
the proof of this theorem (see \eqref{e2.69} below).

\subsection*{Problem background: Critical bilinear systems}

System \eqref{e1.1a}--\eqref{e1.1c} (or \eqref{S}) is an important principal
case of planar, so-called critical bilinear systems (BLS), which are of traditional
interest in the context of asymptotic stabilization
(see, e.g., \cite{b1} for more detail). Let us remind the reader
that a planar BLS of the form
\begin{equation} \label{e1.3}
\dot{x} = A x + u(x_1,x_2) Bx , \ x(0) = x_0,
\end{equation}
is called {\em critical} if the given matrices $ A $ and $ B $ satisfy the
following two conditions:
\begin{itemize}
\item For each $u^{\star} \in R $ at least one eigenvalue of $A + u^{\star}B$
has nonnegative real part.
\item There exists a $u_0 \in R $ for which all the real parts of the eigenvalues
of $A + u_0 B$ are non positive.
\end{itemize}
Note that the latter condition implies that system \eqref{e1.3} is stable when
the constant feedback $ u (x_1,x_2) = u_0$ is engaged, while the former condition
implies that \eqref{e1.3} is not asymptotically stabilizable by any constant
feedback.

It is well-known (e.g., \cite{b1}) that, by the standard linear change of
coordinates and time-rescaling, all the matrices $ A $ satisfying the above-cited
two conditions can be transformed either into the matrix
$$ \begin{pmatrix} 0 &-1 \\ 1 &0\end{pmatrix}
$$
as our matrix $ A $ in \eqref{S} or in \eqref{e1.1a}--\eqref{e1.1c}, associated
with a periodic oscillatory motion, or into one of
the following three ``simpler'' matrices:
$$
\begin{pmatrix} 0 &0 \\ 0 &0\end{pmatrix}, \quad
\begin{pmatrix} 0 &0 \\ 0 &-1\end{pmatrix}, \quad
 \begin{pmatrix} 0 &1 \\ 0 &0\end{pmatrix} .
 $$
We refer the reader to \cite{a1,b1,h1,k1} for various positive and
negative results on stabilization of planar critical BLS by means of constant,
linear, quadratic and piecewise
constant feedback laws. However, it seems that all the available results do
not provide any explicit energy decay rates.

\begin{figure}[ht]
\begin{center}
\includegraphics[width=0.7\textwidth]{fig1.eps}
\end{center}
\caption{Trajectory for system \eqref{e1.1a}--\eqref{e1.1c}}
\end{figure}

\section{Proof of Theorem \ref{thm1.1}}

Our plan for this section as follows:
First of all, as we are concerned with the large time behavior of
\eqref{e1.1a}--\eqref{e1.1c},
without loss of generality (see Step 8 for the general case), we can assume that
\begin{equation} \label{e2.1}
\{(x_{10},x_{20}) \} \subset B = \{(\xi,\eta) | \xi^{2} + \eta^{2}
\leq \frac{1}{2} \}.
\end{equation}
Select any angle $\theta_0 \in (0,\frac{\pi}{4})$ and split the ball $B$
into eight sectors $I$, $I,'$ $II$, $II,'$ $III$, $III'$ and $IV,$$IV'$ as shown on Figure 1, so that the central angles $\angle{A_1OA_8}$, $\angle{A_2OA_3}$, $\angle{A_4OA_5}$ and $\angle{A_6OA_7}$ are all equal to $2 \theta_0$. Again,
without loss of generality, we may assume that the initial point $(x_{10},x_{20})$ is located at point
$A_1$, also denoted by $q_0$ on Figure 1 (see Step 8 otherwise), where respectively the
solid line starting from $q_0$ describes the trajectory $(x_{1}(t),x_{2}(t))$ of system \eqref{e1.1a}--\eqref{e1.1c} as it
crosses all the eight aforementioned sectors.

Our plan is to derive the energy decay estimate for the solution
to \eqref{e1.1a}--\eqref{e1.1c} starting from $q_0$ by evaluating its energy
subsequently in each of the above mentioned sectors $I-IV'$.
Then we extend this result to the general case.

\begin{remark} \label{rmk2.1} \rm
Below we use the notation  $x_1 = x_1(t)$, $x_2 = x_2(t)$, keeping in mind that
$(x_1,x_2) = (x_1(t),x_2(t))$ represent the solution to
\eqref{e1.1a}--\eqref{e1.1c} at hand.
\end{remark}

\noindent{\bf Step 1:} Denote by $t_{p_{1}}$ the time required for the trajectory
of \eqref{e1.1a}--\eqref{e1.1c} starting from
$q_0 = (x_{1}(0),x_{2}(0)) = (x_{10},x_{20})$ to reach point $p_{1}$.
We have (see Figure 1):
\begin{equation} \label{e2.2}
x_1(t) \geq \sqrt{2E(t_{p_{1}})}\sin(\theta_0), \quad
x_2(t) \geq \sqrt{2E(0)}\sin(\theta_0)
\end{equation}
and hence
\begin{equation} \label{e2.3}
x_{1}^{2}(t)x_{2}^{2}(t) \geq 4E(0)E(t_{p_{1}})\sin^{4}(\theta_0) .
\end{equation}
Multiplying equations \eqref{e1.1a}--\eqref{e1.1c} respectively by $x_1$ and
$x_2$ and adding them, we obtain:
\begin{equation} \label{e2.4}
x_1(t) \dot{x_1}(t) + x_2(t) \dot{x_2}(t) = \frac{dE(t)}{dt}
= \frac{1}{2} \frac{d\parallel x(t)\parallel_{R^{2}}^{2}}{dt}
= - x_{1}^{2}(t)x_{2}^{2}(t).
\end{equation}
Integrating this equation over $(0,t_{p_{1}})$ yields:
\begin{equation} \label{e2.5}
E(t_{p_{1}}) = \frac{x_{1}^{2}(t_{p_{1}}) + x_{2}^{2}(t_{p_{1}})}{2}
= - \int_{0}^{t_{p_{1}}}x_{1}^{2}x_{2}^{2}dt + E(0).
\end{equation}
Applying \eqref{e2.3} \eqref{e2.5}, we obtain
\begin{equation} \label{e2.6}
E(t_{p_{1}}) \leq E(0) - 4E(0)E(t_{p_{1}})\sin^{4}(\theta_0)t_{p_{1}},
\end{equation}
which yields:
\begin{equation} \label{e2.7}
E(t_{p_{1}}) \leq \frac{E(0)}{1 + 4E(0)\delta^{2}t_{p_{1}}},
\end{equation}
where $\delta = \sin^{2}(\theta_0)$.
Thus, to estimate $E(t_{p_{1}})$, we need to evaluate $t_{p_{1}}$. This evaluation is accomplished
in steps 2-4 below, based on the ``visualization technique,''
\cite{k1} which makes use
of the phase-portrait of system \eqref{e1.1a}--\eqref{e1.1c}.
Namely, to evaluate $t_{p_{1}}$, we intend to derive first explicit estimates both (a) for the length of
arc of the trajectory of \eqref{e1.1a}--\eqref{e1.1c} lying in sector $I$ and (b) for the
speed at which the point describing the position of system \eqref{e1.1a}--\eqref{e1.1c} at time $t$ moves along this arc. These
estimates, obtained in terms of the energy function, will provide us with the bounds for the duration of
the time-interval required by the system to traverse sector I, that is, for $t_{p_{1}}$. It turns out
that the derived bounds for $t_{p_{1}}$, do not depend on the energy of the system in sector $I$.

\begin{remark} \label{rmk2.2} \rm
 From \eqref{e2.4} we conclude that the distance $\parallel x(t) \parallel_{R^{2}}$
monotonically decreases as $t \rightarrow \infty$. This in turn implies that the
trajectory initiated at $q_0$ remains in $B$ for all $t > 0$.
In other words, system \eqref{e1.1a}--\eqref{e1.1c} is stable
(see also \cite{j1}).
\end{remark}

\noindent {\bf Step 2: Estimate for $\sqrt{\dot{x_{1}}(t)^{2} + \dot{x_{2}}(t)^{2}}$
in sector $I$.}
In this step we obtain an estimate for the value of the speed
$\sqrt{\dot{x_{1}}^2(t) + \dot{x_{2}}^2(t)}$ at which the point $(x_1(t),x_2(t))$
 moves in sector $I$.
Since $(x_1,x_2) \in B$, in sector $I$ we have
\begin{equation} \label{e2.8}
0 < x_{1} \leq \frac{1}{\sqrt{2}} ,\quad
0 < x_{2} \leq \frac{1}{\sqrt{2}} \quad\mbox{and}\quad
x_{1}x_{2} \leq \frac{1}{2} .
\end{equation}
Hence, using \eqref{e1.1a} and applying \eqref{e2.8},
\begin{equation} \label{e2.9}
\dot{x_{1}} = - x_{2} , \quad
\dot{x_{2}} = x_{1} - x_{1}^{2}x_{2} = x_{1}(1 - x_{1}x_{2}) \geq
\frac{x_{1}}{2}.
\end{equation}
Therefore, from \eqref{e2.9} we derive:
\begin{equation} \label{e2.10}
\dot{x_{1}}^{2} + \dot{x_{2}}^{2} \geq \frac{x_{1}^{2}}{4}+ x_{2}^{2}
 \geq \frac{x_{1}^{2} + x_{2}^{2}}{4} = \frac{E(t)}{2}.
\end{equation}
Since $\dot{E}(t) \leq 0$, in view of \eqref{e2.10}, we have the following
estimates:
\begin{equation} \label{e2.11}
\sqrt{\dot{x_{1}}^{2}(t) + \dot{x_{2}}^{2}(t)}
\geq \sqrt{\frac{E(t)}{2}}
\geq \frac{\sqrt{E(t_{p_{1}})}}{\sqrt{2}} ,\quad t \in [0,t_{p_{1}}].
\end{equation}
Applying the triangular inequality to the second equation in \eqref{e1.1a} and
using \eqref{e2.8}, we obtain:
\begin{equation} \label{e2.12}
|\dot{x_{2}}(t)| = |x_{1}(t) - x_{1}^{2}(t)x_{2}(t)|
\leq |x_{1}(t)| + x_{1}^{2}(t)|x_{2}(t)| \leq |x_{1}(t)| + \frac{|x_{1}(t)|}{2}
= \frac{3|x_{1}(t)|}{2}.
\end{equation}

\begin{remark} \label{rmk2.3} \rm
 Note that inequalities \eqref{e2.10}--\eqref{e2.12}  hold
 everywhere in $B$.
\end{remark}

Using (2.12) and the first equation in \eqref{e1.1a}, we further derive that
for $t \in [0,t_{p_{1}}]$,
\begin{equation} \label{e2.13}
\sqrt{\dot{x_{1}}^{2}(t) + \dot{x_{2}}^{2}(t)}
\leq \sqrt{\frac{9x_{1}^{2}(t)}{4} + x_{2}^{2}(t)}
\leq \sqrt{\frac{9(x_{1}^{2}(t) + x_{2}^{2}(t))}{4}}
\leq \frac{3 \sqrt{E(0)}}{\sqrt{2}}\,.
\end{equation}
Combining equations \eqref{e2.11}  and \eqref{e2.13} yields
\begin{equation} \label{e2.14}
\frac{\sqrt{E(t_{p_{1}})}}{\sqrt{2}} \leq \sqrt{\dot{x_{1}}^{2}(t)
+ \dot{x_{2}}^{2}(t)}  \leq  \frac{3 \sqrt{E(0)}}{\sqrt{2}} , \quad
t \in [0,t_{p_{1}}].
\end{equation}
These estimates will be used  to evaluate $t_{p_{1}}$ in step 4 below. \smallskip

\noindent {\sf Step 3: Estimate for $|q_0p_{1}|$.}
In this step we obtain an estimate for the length of the curve $q_0p_{1}$
 connecting the points $q_0$ and $p_1$ (see Figure 1).
It follows from (2.8) and (2.9) that the curve
$q_0p_{1}$ is rising as $t$ increases in $[0,t_{p_{1}}]$. Hence, we can describe it geometrically as a
graph of some monotone decreasing function
$x_2 = x_2(x_1)$, $x_1 \in [\sqrt{2E(t_{p_{1}})}\sin(\theta_0),\sqrt{2E(0)}
\cos(\theta_0)]$ and
\begin{equation} \label{e2.15}
|q_0p_{1}| = \int_{\sqrt{2E(t_{p_{1}})}\sin(\theta_0)}^{\sqrt{2E(0)}\cos(\theta_0)}
\sqrt{1 + (\frac{dx_{2}}{dx_{1}}})^{2}dx_{1}.
\end{equation}
Using the first equation of \eqref{e1.1a} and the fact that $\dot{x_{2}}(t)$
is strictly positive on the interval
$t \in [0,t_{p_{1}}]$, we have from (2.12):
\begin{equation} \label{e2.16}
\dot{x_{1}} = -x_{2}(t) , \dot{x_{2}} \leq \frac{3x_{1}(t)}{2}.
\end{equation}
Note that in sector $I$, $x_{1}(t)$ and $x_2(t)$ are both positive, while
$\dot{x_{1}}(t)$ is strictly negative.
Hence,
\begin{equation} \label{e2.17}
\frac{\dot{x_{2}}}{-\dot{x_{1}}} \leq \frac{3x_{1}(t)}{2 x_{2}(t)},
t \in [0,t_{p_{1}}].
\end{equation}
Therefore, \cite[pp.105]{r1}, we conclude that
\begin{equation} \label{e2.18}
|\frac{dx_{2}}{dx_{1}}| < \frac{3x_{1}(t)}{2 x_{2}(t)}, t \in [0,t_{p_{1}}] .
\end{equation}
Thus,
\begin{equation} \label{e2.19}
1 + (\frac{dx_{2}}{dx_{1}})^{2} \leq \frac{9(x_{1}^2(t)
+ x_{2}^2(t))}{4x_{2}^2(t)} = \frac{9 E(t)}{2 x_{2}^2(t)}, \quad
t \in [0,t_{p_{1}}].
\end{equation}
Using (2.19) and the fact that $E(t)$ is a decreasing function of $t$, we further
obtain that
\begin{equation} \label{e2.20}
1 + (\frac{dx_{2}}{dx_{1}})^{2} \leq \frac{9E(0)}{2x_{2}(t)^{2}} \quad
\mbox{or} \quad
\sqrt{1 + (\frac{dx_{2}}{dx_{1}})^{2}} \leq \frac{3\sqrt{E(0)}}{\sqrt{2}x_{2}(t)},
\quad t \in [0,t_{p_{1}}].
\end{equation}
Since $\dot{x_{2}}(t)$ is strictly positive,
\begin{equation} \label{e2.21}
\sqrt{2E(0)}\sin(\theta_0) \leq x_{2}(t)
 \leq \sqrt{2E(t_{p_{1}})}\cos(\theta_0)
 \leq \sqrt{2E(t_{p_{1}})}, \quad t \in [0,t_{p_{1}}].
\end{equation}
Applying (2.21) to (2.20) gives
\begin{equation} \label{e2.22}
\sqrt{1 + (\frac{dx_{2}}{dx_{1}})^{2}} \leq
\frac{3}{2\sin(\theta_0)},\quad t \in [0,t_{p_{1}}] \,.
\end{equation}

\begin{remark} \label{rmk2.4} \rm
Note that, in view of Remark \ref{rmk2.1}, inequality (2.22) also holds in
sector $I'$, namely in the portion that lies in the first quadrant.
\end{remark}

Combining (2.15) with (2.22) yields:
\begin{equation} \label{e2.23}
\begin{aligned}
|q_0p_{1}| &= \int_{\sqrt{2E(t_{p_{1}})}\sin(\theta_0)}^{\sqrt{2E(0)}\cos(\theta_0)}
\sqrt{1+(\frac{dx_{2}}{dx_{1}})^{2}}dx_{1} \\
&\leq \frac{3(\sqrt{2E(0)}\cos(\theta_0)
- \sqrt{2E(t_{p_{1}})}\sin(\theta_0))}{2 \sin(\theta_0)}.
\end{aligned}
\end{equation}
Moreover, since $E(0) > E(t_{p_{1}})$ and $\cos(\theta_0) \geq \sin(\theta_0)$ for
any $\theta_0 \in [0,\frac{\pi}{4}]$, further from (2.23), we obtain
\begin{equation} \label{e2.24}
|q_0p_{1}| = \int_{\sqrt{2E(t_{p_{1}})}\sin(\theta_0)}^{\sqrt{2E(0)}\cos(\theta_0)}
\sqrt{1+(\frac{dx_{2}}{dx_{1}})^{2}}dx_{1}
\ \leq \frac{3\sqrt{E(0)}}{\sqrt{2}\sin(\theta_0)}.
\end{equation}
We shall now evaluate the length of the curve $q_0p_{1}$ from below.
Once again, (2.9) yields
\begin{equation} \label{e2.25}
\frac{\dot{x_2}}{-\dot{x_1}}
= |\frac{dx_{2}}{dx_{1}}| \geq \frac{x_{1}(t)}{2x_{2}(t)}, t
\in [0 , t_{p_{1}}].
\end{equation}
Therefore,
$$
1 + (\frac{dx_{2}}{dx_{1}})^{2} \geq \frac{4 x_{1}^2(t) + x_{2}^2(t)}{4x_{2}(t)^{2}}
\geq \frac{x_{1}^2(t) + x_{2}^2(t)}{4x_{2}(t)^{2}},\quad t \in [0 , t_{p_{1}}].
$$
Applying (2.24) to this inequality, we obtain
\begin{equation} \label{e2.26}
\sqrt{1 + (\frac{dx_{2}}{dx_{1}})^{2}} \geq \frac{\sqrt{E(t_{p_{1}})}}{\sqrt{2}x_{2}(t)},
\quad t \in [0 , t_{p_{1}}].
\end{equation}
Combining (2.21) with (2.26) yields
\begin{equation} \label{e2.27}
\sqrt{1 + (\frac{dx_{2}}{dx_{1}})^{2}} \geq \frac{1}{2},\quad t
\in [0 , t_{p_{1}}].
\end{equation}
Applying (2.27) to (2.15), we obtain the following estimate for the length of the
curve $q_0p_{1}$ from below:
\begin{equation} \label{e2.28}
\begin{aligned}
|q_0p_{1}| &= \int_{\sqrt{2E(t_{p_{1}})}\sin(\theta_0)}^{\sqrt{2E(0)}\cos(\theta_0)}
\sqrt{1+(\frac{dx_{2}}{dx_{1}})^{2}}dx_{1} \\
&\geq \frac{(\sqrt{2E(0)}\cos(\theta_0) - \sqrt{2E(t_{p_{1}})}\sin(\theta_0))}{2}
\geq \frac{K}{\sqrt{2}} \sqrt{E(0)} ,
\end{aligned}
\end{equation}
where $K = \cos(\theta_0) - \sin(\theta_0)$. Combining (2.24) and (2.28), we finally obtain:
\begin{equation} \label{e2.29}
 \frac{3\sqrt{E(0)}}{\sqrt{2}\sin(\theta_0)} \geq |q_0p_{1}|
 \geq \frac{K}{\sqrt{2}} \sqrt{E(0)}.
\end{equation}

\noindent{\bf Step 4: Estimate for $t_{p_{1}}$.}
In this step we shall derive an estimate for $t_{p_{1}}$, i.e., the time
required for the trajectory of \eqref{e1.1a}--\eqref{e1.1c}, initiated at $q_0$,
to reach the point $p_1$.
We begin by computing the lower bound estimate for $t_{p_{1}}$.
Making use of (2.14)and (2.29), we obtain that
\begin{equation} \label{e2.30}
\frac{K}{\sqrt{2}} \sqrt{E(0)} \leq |q_0p_{1}|
= \int_{0}^{t_{p_{1}}}\sqrt{\dot{x_{1}}^2(t) + \dot{x_{2}}^2(t)}dt
\leq  \frac{3 \sqrt{E(0)}}{\sqrt{2}}t_{p_{1}},
\end{equation}
which in turn yields:
\begin{equation} \label{e2.31}
\frac{K}{3} \leq t_{p_{1}}.
\end{equation}
Analogously, it follows from (2.14) and (2.29) that
\begin{equation} \label{e2.32}
\frac{\sqrt{E(t_{p_{1}})}}{\sqrt{2}}t_{p_{1}} \leq |q_0p_{1}|
= \int_{0}^{t_{p_{1}}}\sqrt{\dot{x_{1}}^2(t) + \dot{x_{2}}^2(t)}dt
 \leq \frac{3\sqrt{E(0)}}{\sqrt{2}\sin(\theta_0)},
\end{equation}
which implies
\begin{equation} \label{e2.33}
t_{p_{1}} \leq \frac{3\sqrt{E(0)}}{\sin(\theta_0) \sqrt{E(t_{p_{1}})}}.
 \end{equation}
Combining (2.31) and (2.33) yields
\begin{equation} \label{e2.34}
 \frac{K}{3} \leq t_{p_{1}} \leq \frac{3\sqrt{E(0)}}{\sin(\theta_0)
 \sqrt{E(t_{p_{1}})}}.
\end{equation}
Since $\dot{x_{2}}(t)$ is strictly positive in sector $I$, we have
$x_{2}(t_{p_{1}}) \geq x_{2}(0)$. Hence,
$$
E(t_{p_{1}}) = \frac{x^{2}_1(t_{p_{1}}) + x^{2}_2(t_{p_{1}})}{2}
 \geq \frac{x^{2}_1(t_{p_{1}})}{2} \geq \frac{x_{2}^{2}(0)}{2}
 = E(0)\sin^{2}(\theta_0)
 $$
(see Figure 1), which implies that
\begin{equation} \label{e2.35}
\sqrt{\frac{E(0)}{E(t_{p_{1}})}} \leq \frac{1}{\sin(\theta_0)} .
\end{equation}
Applying inequality (2.35) to (2.34) yields:
\begin{equation} \label{e2.36}
 \frac{K}{3} \leq t_{p_{1}} \leq \frac{3}{\sin^{2}(\theta_0)} .
\end{equation}

\noindent{\bf Step 5: Estimate for the time taken by the trajectory
of \eqref{e1.1a}--\eqref{e1.1c} to pass sectors $I$ and $I'$.}
In this step we shall derive an estimate for $t_{q_{1}}$, i.e., for the time
required by the trajectory of \eqref{e1.1a}--\eqref{e1.1c}, initiated at $q_0$,
to reach point $q_1$ (see Figure 1).
Split the trajectory in sector $I'$ into two arcs $p_{1}r_{1}$ and $r_{1}q_{1}$
(see Figure 1).
Denote the time required by the trajectory of \eqref{e1.1a}--\eqref{e1.1c}
to connect the points $p_1$ and $r_1$ by $t_{p_{1}r_{1}}$ and the time required
to connect the points $r_1$ and $q_1$ by $t_{r_{1}q_{1}}$.

\noindent{\bf Step 5.1: Estimate for time $t_{p_{1}r_{1}}$.}
The arc $p_1r_1$ can be described geometrically as a graph of function
$x_2 = x_2(x_1)$, $x_1 \in [0,\sqrt{2E(t_{p_{1}})}\sin(\theta_0)]$.
Then, applying the inequality (2.22)
(see Remarks \ref{rmk2.1} and \ref{rmk2.4}) to (2.37) in
sector $I'$, we obtain
\begin{equation} \label{e2.37}
|p_{1}r_{1}| = \int_{0}^{\sqrt{2E(t_{p_{1}})}\sin(\theta_0)}
\sqrt{1 + (\frac{dx_{2}}{dx_{1}}})^{2}dx_{1}
\leq \frac{3 \sqrt{2 E(t_{p_{1}})}\sin(\theta_0)}{2 \sin(\theta_0)}
 \leq \frac{3 \sqrt{E(t_{p_{1}})}}{\sqrt{2}}.
\end{equation}
Since $\dot{E}(t) < 0$ for any time $t \in [t_{p_{1}},t_{p_{1}} + t_{p_{1}r_{1}}]$,
we have from (2.10),
\begin{equation} \label{e2.38}
\sqrt{\dot{x_{1}}^{2}(t) + \dot{x_{2}}^{2}(t)} \geq \sqrt{\frac{x_{1}^{2}(t)
+ x_{2}^{2}(t)}{4}}
= \frac{\sqrt{E(t)}}{\sqrt{2}}
\geq \frac{\sqrt{E(t_{p_{1}}+t_{p_{1}r_{1}})}}{\sqrt{2}}.
\end{equation}
Hence, in view of (2.37),
\begin{equation} \label{e2.39}
\frac{\sqrt{E(t_{p_{1}}+t_{p_{1}r_{1}})}}{\sqrt{2}}t_{p_{1}r_{1}}
\leq |p_1r_{1}| = \int_{t_{p_{1}}}^{t_{p_{1}}+t_{p_{1}r_{1}}}
\sqrt{\dot{x_{1}}^2(t) + \dot{x_{2}}^2(t)}dt
\leq 3\frac{\sqrt{E(t_{p_{1}})}}{\sqrt{2}},
\end{equation}
which yields
\begin{equation} \label{e2.40}
t_{p_{1}r_{1}} \leq \frac{3 \sqrt{E(t_{p_{1}})}}{\sqrt{E(t_{p_{1}}+t_{p_{1}r_{1}})}}.
\end{equation}
Since $\dot{x_{2}}(t)$ is positive, we have
$x_{2}(t_{p_{1}}+t_{p_{1}r_{1}}) \geq x_{2}(t_{p_{1}})$. Hence,
$$
E(t_{p_{1}}+t_{p_{1}r_{1}}) = \frac{x^{2}_2(t_{p_{1}}+t_{p_{1}r_{1}})}{2}
\geq \frac{x_{2}^{2}(t_{p_{1}})}{2} = E(t_{p_{1}})\cos^{2}(\theta_0),
$$
which implies
\begin{equation} \label{e2.41}
\sqrt{\frac{E(t_{p_{1}})}{E(t_{p_{1}}+t_{p_{1}r_{1}})}}
\leq \frac{1}{\cos(\theta_0)}.
\end{equation}
Finally, combining (2.40) and (2.41), we obtain the following estimate for
$t_{p_{1}r_{1}}$,
\begin{equation} \label{e2.42}
t_{p_{1}r_{1}} \leq \frac{3}{\cos(\theta_0)}.
\end{equation}
{\bf Step 5.2: Estimate for the time $t_{r_{1}q_{1}}$.}
Once again, the arc $r_1q_1$ can be described geometrically as the graph of a
function $x_2 = x_2(x_1)$, with
$x_1$ in the interval
 $[-\sqrt{2E(t_{p_{1}}+t_{p_{1}r_{1}}+t_{r_{1}q_{1}})}\sin(\theta_0),0]$.
Then
\begin{equation} \label{e2.43}
|r_{1}q_{1}| = \int_{-\sqrt{2E(t_{p_{1}}+t_{p_{1}r_{1}}
+t_{r_{1}q_{1}})}\sin(\theta_0)}^{0}
\sqrt{1 + (\frac{dx_{2}}{dx_{1}}})^{2}dx_{1}.
\end{equation}
Note that $x_{1}(t)$, $\dot{x_{1}}(t)$, and $\dot{x_{2}}(t)$ are strictly negative
on the interval
$t \in [t_{p_{1}}+t_{p_{1}r_{1}},t_{p_{1}}+t_{p_{1}r_{1}}+t_{r_{1}q_{1}}]$,
where $x_{2}(t)$ is positive.
Since it follows from \eqref{e1.1a}--\eqref{e1.1c}
and (2.12) (see Remark \ref{rmk2.3}) that
$$
-\dot{x_2} \leq -\frac{3x_1}{2} \quad\mbox{and}\quad  \dot{x_1} = -x_2,
$$
we have
\begin{equation} \label{e2.44}
0 \leq \frac{\dot{x_{2}}}{\dot{x_{1}}} \leq \frac{3x_{1}(t)}{-2 x_{2}(t)},
\quad  t \in [t_{p_{1}}+t_{p_{1}r_{1}},t_{p_{1}}+t_{p_{1}r_{1}}+t_{r_{1}q_{1}}].
\end{equation}
Therefore,
\begin{equation} \label{e2.45}
1 + (\frac{dx_{2}}{dx_{1}})^{2} \leq \frac{9(x_{1}^2(t) + x_{2}^2(t))}{4x_{2}^2(t)} 
\leq \frac{9 E(t)}{2 x_{2}^2(t)},
\quad t \in [t_{p_{1}}+t_{p_{1}r_{1}},t_{p_{1}}+t_{p_{1}r_{1}}+t_{r_{1}q_{1}}].
\end{equation}
Using (2.45) and the fact that $E(t)$ is a decreasing function of $t$, we obtain
\begin{equation} \label{e2.46}
\sqrt{1 + (\frac{dx_{2}}{dx_{1}})^{2}} \leq
\frac{3\sqrt{E(t_{p_{1}}+t_{p_{1}r_{1}})}}{\sqrt{2}x_{2}(t)}.
\end{equation}
Since $\dot{x_{2}}(t)$ is strictly negative, for $t$ in
$[t_{p_{1}}+t_{p_{1}r_{1}},t_{p_{1}}+t_{p_{1}r_{1}}+t_{r_{1}q_{1}}]$,
\begin{equation} \label{e2.47}
\sqrt{2E(t_{p_{1}}+t_{p_{1}r_{1}}+t_{r_{1}q_{1}})}\cos(\theta_0) \leq x_{2}(t)
\leq \sqrt{2E(t_{p_{1}}+t_{p_{1}r_{1}})}.
\end{equation}
Applying (2.47) to (2.46) gives
\begin{equation} \label{e2.48}
\sqrt{1 + (\frac{dx_{2}}{dx_{1}})^{2}} \leq
\frac{3\sqrt{E(t_{p_{1}}+t_{p_{1}r_{1}})}}{\sqrt{2}\sqrt{2E(t_{p_{1}}
+t_{p_{1}r_{1}}+ t_{r_{1}q_{1}})}\cos(\theta_0) },
\end{equation}
for $t \in [t_{p_{1}}+t_{p_{1}r_{1}},t_{p_{1}}+t_{p_{1}r_{1}}+t_{r_{1}q_{1}}]$.
Thus applying inequality (2.48) to (2.43), we obtain an upper estimate for the length of the
arc $r_{1}q_{1}$:
\begin{equation} \label{e2.49}
\begin{aligned}
|r_{1}q_{1}| &= \int_{-\sqrt{2E(t_{p_{1}}+t_{p_{1}r_{1}}+t_{r_{1}q_{1}})}
\sin(\theta_0)}^{0} \sqrt{1 + (\frac{dx_{2}}{dx_{1}}})^{2}dx_{1}\\
&\leq \frac{3 \sqrt{2E(t_{p_{1}}+t_{p_{1}r_{1}})}\tan(\theta_0)}{\sqrt{2}} .
\end{aligned}
\end{equation}
Since $\dot{E}(t) < 0$ for any
$t \in [t_{p_{1}}\ + t_{p_{1}r_{1}},t_{p_{1}} + t_{p_{1}r_{1}} + t_{r_{1}q_{1}}]$, making
use of (2.10) (see Remark \ref{rmk2.3}), we derive that
\begin{equation} \label{e2.50}
\sqrt{\dot{x_{1}}^{2}(t) + \dot{x_{2}}^{2}(t)}
\geq \sqrt{\frac{x_{1}^{2}(t) + x_{2}^{2}(t)}{4}}
\geq \frac{\sqrt{E(t_{p_{1}}+t_{p_{1}r_{1}}+t_{r_{1}q_{1}})}}{\sqrt{2}}.
\end{equation}
Using  inequalities (2.49) and (2.50), we obtain
\begin{equation} \label{e2.51}
\begin{aligned}
\frac{\sqrt{E(t_{p_{1}}+t_{p_{1}r_{1}}+t_{r_{1}q_{1}})}}{\sqrt{2}}t_{r_{1}q_{1}}
&\leq |r_1q_{1}| = \int_{t_{p_{1}}+t_{p_{1}r_{1}}}^{t_{p_{1}}+t_{p_{1}r_{1}}+t_{r_{1}q_{1}}}
\sqrt{\dot{x_{1}}^2(t) + \dot{x_{2}}^2(t)}dt\\
&\leq \frac{3 \sqrt{E(t_{p_{1}}+t_{p_{1}r_{1}})}\tan(\theta_0)}{\sqrt{2}}.
\end{aligned}
\end{equation}
which, in turn, provides
\begin{equation} \label{e2.52}
t_{r_{1}q_{1}} \leq \frac{3\tan(\theta_0)\sqrt{E(t_{p_{1}}+t_{p_{1}r_{1}})}}
{\sqrt{E(t_{p_{1}}+t_{p_{1}r_{1}}+t_{r_{1}q_{1}})}}.
\end{equation}
Once again, integrating (2.4) over $(t_{p_{1}}\ + t_{p_{1}r_{1}},
t_{p_{1}} + t_{p_{1}r_{1}} + t_{r_{1}q_{1}})$ yields (compare with (2.6))
\begin{equation} \label{e2.53}
E(t_{p_{1}}+t_{p_{1}r_{1}}) = E(t_{p_{1}}+t_{p_{1}r_{1}}+t_{r_{1}q_{1}})
+ \int_{t_{p_{1}}+t_{p_{1}r_{1}}}^{t_{p_{1}}+t_{p_{1}r_{1}}
+t_{r_{1}q_{1}}}x_{1}^{2}x_{2}^{2}dt.
\end{equation}
Observe that
$$ x_{2}^{2}(t) \leq \frac{1}{2} \quad\mbox{and} \quad
 0 \leq x_{1}^{2}(t)
 \leq 2E(t_{p_{1}}+t_{p_{1}r_{1}}+t_{r_{1}q_{1}}) \sin^{2}(\theta_0)
$$
for any
$t \in [t_{p_{1}}\ + t_{p_{1}r_{1}},t_{p_{1}} + t_{p_{1}r_{1}} + t_{r_{1}q_{1}}]$.
Hence, it follows from (2.53) that
\begin{equation} \label{e2.54}
E(t_{p_{1}}+t_{p_{1}r_{1}}) \leq E(t_{p_{1}}+t_{p_{1}r_{1}}+t_{r_{1}q_{1}})(1
+ t_{r_{1}q_{1}}\sin^{2}(\theta_0)).
\end{equation}
Hence,
\begin{equation} \label{e2.55}
\sqrt{\frac{E(t_{p_{1}}+t_{p_{1}r_{1}})}{ E(t_{p_{1}}
+t_{p_{1}r_{1}}+t_{r_{1}q_{1}})}}
\ \leq \sqrt{1 + \sin^{2}(\theta_0)t_{r_{1}q_{1}}} .
\end{equation}
Applying inequality (2.55) to (2.52) yields:
\begin{equation} \label{e2.56}
0 < t_{r_{1}q_{1}} \leq 3 \tan(\theta_0)
 \sqrt{1 + \sin^{2}(\theta_0)t_{r_{1}q_{1}}},
\end{equation}
which gives the following estimate for $t_{r_{1}q_{1}}$:
\begin{equation} \label{e2.57}
0 \leq t_{r_{1}q_{1}} \leq C
= \frac{9 \tan^{2}(\theta_0) \sin^{2}(\theta_0) + 3\tan(\theta_0)
\sqrt{9\tan^{2}(\theta_0)\sin^{4}(\theta_0)+4}}{2} .
 \end{equation}

\noindent{\bf Step 5.3: Estimate for time $t_{q_{1}}$.}
Combining (2.36), (2.42) and (2.57) yields
\begin{equation} \label{e2.58}
\frac{K}{3} \leq t_{q_{1}} = \ t_{p_{1}} + t_{p_{1}r_{1}} + t_{r_{1}q_{1}}
 \leq \frac{3}{\sin^{2}(\theta_0)} + \frac{3}{\cos(\theta_0)} + C.
\end{equation}
{\bf Step 6: Estimates for $E(t_{p_{1}})$ and $E(t_{q_{1}})$.}
Substituting the lower bound for $t_{p_{1}}$ from
(2.36) into (2.7) and recalling that since $E(t)$ is decreasing in time,
we obtain from (2.7) that
\begin{equation} \label{e2.59}
E(t_{q_{1}}) \leq E(t_{p_{1}}) \leq \frac{E(0)}{1 + 4E(0)\delta^{2}\frac{K}{3}}.
\end{equation}

\noindent{\bf Step 7: Estimate for $E(t)$ for any time $t > 0$.}
It is not hard to see that in the next two sectors $II$ and $II'$
one can apply the same strategy as described in the above steps 1-6, namely, leading to
estimate (2.59). Denote the time required by the trajectory of \eqref{e1.1a}--\eqref{e1.1c} to reach points $p_2$ and $q_2$
from $q_0$ respectively by
$t_{p_{2}}$ and $t_{q_{2}}$. Then, similarly as in (2.59) and (2.36), (2.58)
(see Remark \ref{rmk2.3}), we obtain:
\begin{equation} \label{e2.60}
E(t_{q_{2}}) \leq \ E(t_{p_{2}}) \leq \frac{E(t_{q_{1}})}{1 + 4E(t_{q_{1}})\delta^{2}
(t_{p_{2}} - t_{q_{1}})}
\leq \frac{E(t_{q_{1}})}{1 + 4E(t_{q_{1}})\delta^{2}\frac{K}{3}},
\end{equation}
where, analogously to (2.58), we have:
$$
\frac{K}{3} \leq t_{p_{2}} - t_{q_{1}} \leq \ t_{q_{2}} - t_{q_{1}}
\ \leq \frac{3}{\sin^{2}(\theta_0)} + \frac{3}{\cos(\theta_0)} + C.
$$

\begin{remark} \label{rmk2.5} \rm
Consider the function
$f(x) = \frac{x}{1 + \kappa x}$,  with $\kappa$  a positive constant which is
associated with the right hand sides in (2.59), (2.60), considered as functions
of energy. Since $f'(x) > 0$ for all $x > 0$, $f(x)$ is a monotonically increasing
function.
\end{remark}

Since $E(t_{q_{1}}) \leq E(t_{p_{1}})$ and (2.59) holds, Remark \ref{rmk2.5} implies
\begin{equation} \label{e2.61}
E(t_{q_{2}}) \leq \frac{\frac{E(0)}{ 1 + 4E(0)\delta^{2}\frac{K}{3}}}
{1 + 4\frac{E(0)}{ 1 + 4E(0)\delta^{2}\frac{K}{3}}\delta^{2}\frac{K}{3}}
= \frac{E(0)}{1 + \frac{2K}{3}4E(0)\delta^{2}} .
\end{equation}
Denote the following subsequent intersecting points of the trajectory at hand
with the lines $OA_6, OA_7, \dots$ by $p_3,q_3, \ldots, p_n,q_n, \dots$.
Analogously, we have
\begin{equation} \label{e2.62}
E(t_{q_{n}}) \leq E(t_{p_{n}})
\leq \frac{E(0)}{1 + \frac{nK}{3}4E(0)\delta^{2}}, \quad\forall
n = 1,\dots .
\end{equation}
Consider any $t > 0$. Then there exists a positive integer $n$ such that
\begin{equation} \label{e2.63}
t_{q_{n-1}} \leq t \leq t_{q_{n}} \leq \frac{3n}
{\sin^{2}(\theta_0)} + \frac{3n}{\cos(\theta_0)} + nC,
\end{equation}
where, thus,
\begin{equation} \label{e2.64}
\frac{t}{\frac{3}{\sin^{2}(\theta_0)} + \frac{3}{\cos(\theta_0)} + C}
\leq n.
\end{equation}
Note that, since $t > t_{q_{n-1}}$ and $E(t)$ decreases in time, (2.62) also
implies that
\begin{equation} \label{e2.65}
E(t) \leq E(t_{q_{n-1}}) \leq \frac{E(0)}{1 + \frac{(n-1)K}{3}4E(0)\delta^{2}}.
\end{equation}
Combined with (2.63), this gives the following energy decay estimate for any
$t > 0$ in the case when the initial state $(x_{10},x_{20})$ lies in $B$ on
the line $OA_{1}$ as shown on Figure 1:
\begin{equation} \label{e2.66}
E(t) \leq \frac{E(0)}{1 - \frac{4KE(0)\delta^{2}}{3}
+ \frac{4KE(0)\delta^{2}}{3}(\frac{3}{\sin^{2}(\theta_0)}
+ \frac{3}{\cos(\theta_0)} + C)^{-1}t}.
\end{equation}
{\bf Step 8: The general case.} Consider any initial point $(x_{10},x_{20})$.
Since \eqref{e1.1a}--\eqref{e1.1c} has an
isolated equilibrium point $(0,0)$, by Poincare-Bendixson theorem (see \cite{p1}),
the trajectory to \eqref{e1.1a}--\eqref{e1.1c} starting from $(x_{10},x_{20})$
after some finite time will enter the ball $B$ and intersect the
line $OA_1$. Denote the time of this intersection by $t_0$.
Then, by the scheme discussed in steps 1-7, we
have
\begin{equation} \label{e2.67}
E(t) \leq \frac{E(t_0)}{1 - \frac{4KE(t_0)\delta^{2}}{3} + \alpha E(t_0)(t-t_0)},
\quad t \geq t_0,
\end{equation}
where
$$
\alpha = \frac{4K\sin^{2}(\theta_0)\cos(\theta_0)
\sin^{4}(\theta_0)}{3C\sin^{2}(\theta_0)\cos(\theta_0) + 9\sin^{2}(\theta_0)
+ 9\cos(\theta_0)}.
$$
Using (2.1) (i.e., $E(t_0) \leq \frac{1}{4}$) and the argument of Remark \ref{rmk2.5},
we derive from (2.67)
that
\begin{equation} \label{e2.68}
\begin{aligned}
E(t) &\leq \frac{E(t_0)}{1 - \frac{4E(t_0)K\delta^{2}}{3} + \alpha E(t_0)(t-t_0)}\\
&\leq \frac{E(t_0)}{1 - \frac{K\delta^{2}}{3} + \alpha E(t_0)(t-t_0)}\\
&\leq  \frac{E(0)}{1 - \frac{K\delta^{2}}{3} + \alpha E(0)(1-\frac{t_0}{t})t}
\quad t > t_0.
\end{aligned}
\end{equation}
Choose $\bar{t} = 2 t_0$ and denote
\begin{equation} \label{e2.69}
\beta = 1 - \frac{K\sin^{4}(\theta_0)}{3},\quad
\gamma = \frac{\alpha}{2}.
\end{equation}
Recall that $\delta = \sin^{2}(\theta_0)$ and $\theta_0$ is any number
in $(0,\frac{\pi}{4})$. \smallskip

Then (2.68) implies (1.2). This concludes the proof of Theorem \ref{thm1.1}.


\subsection*{Acknowledgement} 
This work was supported in part by grant DMS-0204037 from the NSF.

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\end{document}