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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\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)**\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, ~ 00$, $a, b$
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
$\sigma(t)0$ 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
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\endRefs
\enddocument
**