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\def\rightheadline{EJDE--1997/17\hfil Stability of a linear oscillator 
\hfil\folio}
\def\leftheadline{\folio\hfil A. O. Ignatyev
 \hfil EJDE--1997/17}

\def\pretitle{\vbox{\eightrm\noindent\baselineskip 9pt %
 Electronic Journal of Differential Equations,
Vol.\ {\eightbf 1997}(1997), No.\ 17, pp.\ 1--6.\hfil\break
ISSN: 1072-6691. URL: http://ejde.math.swt.edu or http://ejde.math.unt.edu
\hfill\break 
ftp (login: ftp) 147.26.103.110 or 129.120.3.113\bigskip} }

\topmatter
\title
Stability of a linear oscillator with variable parameters
\endtitle

\thanks \noindent
{\it 1991 Mathematics Subject Classifications:} 34D20, 70J25.\hfil\break
{\it Key words and phrases:} differential equation, stable solution, Lyapunov function.
\hfil\break
\copyright 1997 Southwest Texas State University  and
University of North Texas.\hfil\break
Submitted June 20, 1997. Published October 29, 1997. 
\endthanks
\author A. O. Ignatyev   \endauthor
\address
Alexander O. Ignatyev \hfil\break 
Institute for Applied Mathematics and Mechanics\hfil\break
R.Luxemburg Street 74, Donetsk-340114, Ukraine
\endaddress
\email ignat\@iamm.ac.donetsk.ua
\endemail

\abstract
A criterion  of  asymptotic stability for a
linear oscillator with variable parameters is obtained. It is
shown  that this  criterion is  close  to a  necessary  and
sufficient conditions of asymptotic stability. An instability
theorem is  proved, and a mechanical example is considered.
\endabstract
\endtopmatter

\document
\head 1. Introduction \endhead

Consider an oscillator described  by  the  following
differential equation
$$\ddot x+f(t)\dot x+g(t)x=0\,, \tag1$$
where  the damping and rigidity coefficients
$f(t)$  and  $g(t)$  are continuous and bounded functions of the
time $t$.
Most of  the  theories  examining  a stability problem of the
zero solution are  based  on  the  Lyapunov  stability  and
instability theorems  and  the corresponding Lyapunov function is assumed as
an energy-type function
     $$V=\frac{1}{2}c_1(t)\dot x^2+\frac{1}{2}c_2(t)x^2\,,$$
where $c_1(t),  c_2(t)$ are time variable functions. In [6], A. P. Merkin
 considered  the  case  $c_1(t)=c_2(t)=1$  and  stability
conditions were obtained only for constant
$f$ and $g$.  An extension was done
in [15] for periodic functions $f(t)$ and $g(t)$.
By means of a Lyapunov function which is a  quadratic
form with respect to $x$ and $\dot x$,  V. M. Starzhinsky [10]
 (assuming that $0<l\le  f(t)\le  L, \  0<m\le  g(t)\le  M$)
obtained sufficient conditions of asymptotical stability for
the solution
     $$x=0, \quad \dot x=0 \tag2 $$
of equation (1). They are written as restrictions to the constants
$l,L,m,M$.
We note
that for a linear $n$-dimensional  system $\dot x=A
(t)x$, the  problem  of  asymptotical  stability
 has been considered by many  of  authors  [1-5,
11-14],
but the obtained conditions on the
elements of  the  matrix  $A(t)$ are rather restrictive.

In  this  paper  sufficient
asymptotic stability conditions  of  the  solution  (2)  are
obtained which are close to necessary and sufficient
conditions of   stability.

     \head 2. Criterion of the asymptotic stability \endhead
     We suppose that $g(t)$ is continuously differentiable
and that the inequalities
     $$|f(t)|<M_1,\quad |g(t)|<M_2,\quad |\dot g(t)|<M_3 \tag3$$
     hold for $t\in R_+=[0;\infty ).$

     \proclaim{Theorem 1} If the conditions
$$g(t)>\alpha_1>0,\qquad   p(t)=\frac{1}{2}\frac{\dot
g(t)}{g(t)}+f(t)>\alpha_2>0 \tag4$$
     are fulfilled,  then the  solution  (2)  of the differential
equation (1) is uniformly asymptotically stable.
     \endproclaim
\demo{Proof}
\define\be{\beta}
Let us consider the function
$$V_1=\frac12\left(x^2+2\be \frac{x\dot x}{\sqrt{g(t)}}+
     \frac{\dot x^2}{g(t)}\right)\quad \ ~(\be =const).$$
Its time  derivative along the solutions of equation (1) has the
form
$$
\dot                                          V_1=\frac1{\sqrt
{g(t)}}\left(\left(-\frac{p(t)}{\sqrt{g(t)}}+\be\right)\dot x^
2-\be p(t)x\dot x-\be g(t)x^2\right)
$$
If we take $\be>0$ sufficiently small,  then $V_1$ is positive
definite ($V_1>0$)  and  $\dot V_1$ is negative definite.
Carrying out  conditions $V_1>0, \dot V_1
<0$, we can take
\define\a{\alpha}
$$0<\be<\min\left\{1, \ \frac{\a_2}{2\sqrt{M_2}}, \
     \frac{8\a_1^3\a_2}{(M_3+2\a_1M_1)^2\sqrt{M_2}}\right\}.$$
\par
     Thus all   conditions   of  Lyapunov  theorem  [6, 9]  are
fulfilled and the zero solution of equation (1) is
uniformly asymptotically stable.
\enddemo
     \proclaim{Corollary 1} If conditions (4)  are  fulfilled,
then there exist positive numbers $B,\alpha $ such, that for
$t>t_0\ge 0$ inequalities
$$|x(t)<B\exp{[-\alpha (t-t_0)]},~~\quad  |\dot x(t)|<B\exp{[-\alpha(t
-t_0)]}$$
     hold.
     \endproclaim
\demo{Proof} The  results  of  Theorem  1  and [8] imply
 Corollary 1.
     \enddemo
     \remark{Remark 1}   Inequalities   (4)   are   sufficient
conditions for asymptotical stability of the trivial solution.
They are  close  to  necessary  and  sufficient  ones  in  the
following sense. If conditions (4) are  changed to one of
the next
     $$g(t)>\a_1>0, \quad p(t)<-\a_2<0, \tag5$$
     $$g(t)<-\a_1<0, \quad p(t)>\a_2>0, \tag6$$
     $$g(t)<-\a_1<0, \quad p(t)<-\a_2<0, \tag7$$
     then (2) is unstable.
     \endremark
     \demo{Proof} Let (5) be fulfilled.  Let  us  take  $\be<0$
with $|\be|$ so small  that $V_1>0, \ \dot V_1>0$. This proves
the instability of the zero solution. If one of the conditions
(6), (7) is true, then consider the Lyapunov function
$$V_2=\frac12\left(x^2+2\be \frac{x\dot x}{\sqrt {-g(t)}}+
     \frac{\dot x^2}{g(t)}\right),$$
whose time derivative along the solutions of equation (1) has the form
$$\dot V_2=\frac1{\sqrt            {-g(t)}}\left(\left(\frac{p(t)}{\sqrt
{-g(t)}}+\be\right)\dot x^2-\be p(t)x\dot x-\be g(t)x^2\right).$$
     \par
     Choosing  $|\be|$
small enough,  one can make the function $\dot V_2$  of  fixed
sign (in the case (6) we suppose  $\be>0$,  in  the  case  (7)
$\be<0$). But $V_2$ changes its sign. Thus  according to [9], the
trivial solution of (1) is unstable.
\enddemo
\remark{Remark 2}  If $f(t)$ and $g(t)$ are constants, then
the conditions (4) amount to the usual Routh-Hurwitz criterion.
\endremark


\head3. Instability of the zero solution\endhead


     Now let us obtain instability conditions. Noting $\dot
x=y$, we get the system
     $$\dot x=y,\qquad  \dot y=-g(t)x-f(t)y \tag8$$
which is equivalent  to  equation  (1).  It  has  the  trivial
solution
     $$x=0,\quad y=0\tag9$$

     \proclaim{Theorem 2} The solution (9) of the  system  (8)
is unstable  if there exists some $t_0$ such that, for each $t>t_0$, one
of the following conditions
     $$D(t)=\frac 14f^2(t)+g(t)\le 0,\tag10$$
$$D(t)>0,\quad  4f(t)D(t)+\frac12\dot f(t)f(t)+\dot g(t)-
     (\dot f(t)+f^2(t)+4D(t))\sqrt {D(t)}<0,\tag11$$
$$D(t)>0,\quad  4f(t)D(t)+\frac12\dot f(t)f(t)+\dot g(t)+
     (\dot f(t)+f^2(t)+4D(t))\sqrt {D(t)}<0\tag12$$
holds.
     \endproclaim
     \demo{Proof}
     Let  $\epsilon$  be an arbitrary positive number.  We
shall show that, for any sufficiently small~ $\delta>0$,~  there
exists some ~$x_0, y_0$ ~with
     $$|x_0|<\delta,\qquad |y_0|<\delta \tag13$$
and some   ~$T>0$ ~such that, for $t=t_0+T$, the trajectory
$x(t), y(t)\quad
(x(t_0)=x_0, y(t_0)=y_0)$   reaches   the
boundary of the domain
     $$|x|<\epsilon,\quad |y|<\epsilon\tag14$$

     Consider the  function $V=xy$.  Its time derivative along
the solutions of (8) has the form
     $$\dot V=y^2-f(t)xy-g(t)x^2.$$

     Take $x_0>0, ~y_0>0$ satisfying (13) and such that
$$\dot V(t_0,x_0,y_0)=y_0^2-f(t_0)x_0y_0-g(t_0)x_0^2>0.$$
     Consider the trajectory $x(t),  y(t)$ of (8) with initial
data $x(t_0)=x_0,  y(t_0)=y_0.$  Without loss of generality we
can assume         $D(t_0)<0$.         Let         $[t_0; t_1]$,
$[t_2; t_3]$,..., $[t_{2n}; t_{2n+1}]$,... be segments    on    which
condition (10) holds and $(t_1; t_2)$,
$(t_3; t_4)$,...,$(t_{2n-1}; t_{2n}),...$ be  the intervals on which
inequalities (11) or (12) are valid.  As $\dot V\ge 0$ on  $[t_0; t_1],$
 the  trajectory is staying in the domain
$xy\ge x_0y_0$ on this segment.  Now let us consider $x(t),  y(t)$ when  $t\in
(t_1; t_2)$. On this interval $\dot V$ changes its sign.  $\dot
V=0$ if
     $$y=(\frac12f+\sqrt D)x\tag15$$
or
$$y=(\frac12f-\sqrt D)x\tag16$$
     and $\dot V>0$ if
     $$y>(\frac12f+\sqrt D)x\tag17$$
or
  $$y<(\frac12f-\sqrt D)x.\tag18$$
     Let $t_*\in (t_1; t_2)$ be such moment of time, that
     $y(t_*)=(\frac12f(t_*)+\sqrt {D(t_*)})x(t_*)$,  i.e.  the
point of the trajectory belongs to the straight line (15) when
$t=t_*$.

     We shall show that $x(t),  y(t)$ satisfy  the  inequality
(17) if   $t\in   (t_*;t_*+\Delta  t)$  and  $\Delta  t>0$  is
sufficiently small.  To this end, we write $\ddot V$
 under the condition $\dot V=0$:
     $$  \left. \ddot V\right|_{(15)}=-(4f(t)D(t)+\frac12\dot f(t)f(t)+\dot g(t)+
     (\dot f(t)+f^2(t)+4D(t))\sqrt{D(t)})x^2$$

     Taking into account conditions (12), we obtain $\ddot V>0$
under $\dot V=0$,  i.e.  the trajectory belongs to the  domain
$\dot V>0$  when  $t\in (t_*;t_*+\Delta t)$.

If $t_*'\in[t_1;t_2]$ is such moment of time, that
$y(t_*')=(\frac12f(t_*')-\sqrt{D(t_*')})x(t_*')$ (i.e. the point
of the trajectory belongs to the straight line (16) when $t=t_
*'$), then, using conditions (11), we obtain that $x(t), y(t)$
satisfy the inequality  (18)  for  $t\in(t_*';t_*'+\Delta  t)$
where $\Delta t>0$ is sufficiently small.
Thus it is proved
the trajectory lies in the domain $\dot  V\ge  0$  when  $t\in
[t_1;t_2].$

     One can  show  analogously  that  the point $x(t),  y(t)$
belongs to the set $\dot V\ge 0$ when $t\in [t_n;t_{n+1}]$ 
$(n=3, 4,... )$.
It means that for the trajectory the inequality $\dot V(x(t),y
(t))\ge 0$  holds  for  every  $t\ge  t_0$.  But from the last
inequality it follows, that $x(t)y(t)\ge x_0y_0$ for every $t
\ge t_0$.

Let us  show  that  the  boundary  of  (14) is reached for the
finite interval of time. Consider on the plane $x,y$ the domain
$$\Omega=\{x,y: ~xy\ge     x_0y_0,     ~    0<x<\epsilon,    ~
0<y<\epsilon\}$$

We shall estimate the time for which  the  trajectory  $x(t),y(t)$  can
stay in  $\Omega$.  In  this  domain  the  inequality  $y\ge
\epsilon^{-1}x_0y_0$ holds, so from the first of the equations
(8) we  obtain $x(t)\ge x_0+\epsilon^{-1}x_0y_0(t-t_0)$.  This
relation implies that the interval  of  time,  for  which  the
trajectory lies in $\Omega$, can be estimated by the number $T
=\epsilon(\epsilon-x_0)x_0^{-1}y_0^{-1}$. The trajectory cannot
leave $\Omega$  intersecting the hyperbola $xy=x_0y_0$,  hence
one of the inequalities (14) is  broken.  This  completes  the
proof.
\enddemo
\example{Example}
N. N. Moiseyev [7]  wrote down a differential equation of small
plane oscillations  of  a   rocket,   whose   centre of mass   moves
rectilinearly with constant velocity. It has a form (1), where
$f(t)=ae^{-\alpha t}$,  $g(t)=be^{-\alpha t}$; $\alpha>0$,  $a, b$
are positive constants, $x$ is an angle of attack. The author obtained
sufficient conditions for stability of the zero solution  and showed that
in the  case  of  plane  small  oscillations of a rocket, these
conditions are not fulfilled. But he did not prove instability of
the small oscillations.

Let us apply Theorem 2 in order to  prove instability  of  the  solution  (2).
 Actually,   there   exists   $t_0>0$  such  that
inequalities (12) hold for $t\ge t_0$. This proves instability
of the small oscillations.
\endexample
\proclaim{Theorem 3} If in  equation (1) the functions $f(t), g(t)$
are vanishing, i.e.
$$\lim_{t\to\infty}f(t)=0,\quad\lim_{t\to\infty}g(t)=0,$$
then the equilibrium (2) cannot be uniformly stable.
\endproclaim
\demo
{Proof} Consider a system of differential equations (8) which has
the trivial solution (9). Let us take arbitrary $\epsilon>0$. We
shall show that for every $\delta>0$, there exist $x_0, y_0$,
satisfying  (13)  and some $t_0\ge 0$ such  that  the
trajectory $x(t), y(t)$, where $x(t_0)=x_0,~ y(t_0)=y_0$, leaves
the domain (14) with time increasing. Denote
$$
\sigma(t)=\frac12|f(t)|+\frac12\sqrt{f^2(t)+4|g(t)|}
$$

The functions $f(t), g(t)$ are vanishing, hence $\sigma(t)$ is also
vanishing. Let  us  choose  such   $t_0>0$   that
$\sigma(t)<x_0y_0\epsilon^{-2}$ holds for $t\ge t_0$.
Using the auxiliary function $V=xy$, the  trajectory  $x(t),  y(t)$  will  be
disposed in the domain $\dot V>0$ for $t\ge t_0$. Then, as it
follows from Theorem 2 proof, there exists such time moment
$t>t_0$, under which the trajectory leaves the domain (14). The
proof is complete.
\enddemo
\Refs

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\by M.S.P. Eastham
\book The Asymptotic Solution of Linear Differential Systems:
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\publ Oxford Science Publications
\publaddr Oxford
\yr 1989
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\paper Globally   Analytic  Simplification  and  the  Levinson
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\ref \no 4
\by N. Levinson
\paper The asymptotic nature of solutions of  linear  ordinary
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\vol 96
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\by A.P. Merkin
\book
Stability of Motion
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\book Asymptotic Methods of Nonlinear Mechanics
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\ref\no 8
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\endref
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\by V.M. Starzhinsky
\paper Sufficient conditions of stability of the mechanical system
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\vol 16 \yr 1952 \pages 369--374
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\bysame
\paper Asymptotic Behavior of Solutions of Asymptotically
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\paper Estimates  of  solutions  of  a  system  of   perturbed
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\ref\no 14
\by Van Dan-Chzhy
\paper On  the stability of the zero solution of the system of
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\ref\no 15
\by
V.A. Yakubovich and V.M. Starzhinsky
\book Linear differential equations with periodic coefficients
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\publ ``Nauka'' \publaddr Moscow \yr 1972  \lang Russian
\endref

\endRefs

\enddocument