2
$\begingroup$

Consider the differential equation

$$ m \ddot{x} + k \dot{x} = - W_t x $$

where

  • $m$ and $k$ are nonnegative.
  • $x_t \in \mathbb{R}^n$
  • $W_t$ is a matrix that satisfies $$ \alpha I \succeq W_t \succeq \beta I > 0 ~~\mbox{ for all } t$$

My question is: can we conclude that $x_t$ bounded?

If $W_t= W$ did not depend on $t$, this would be trivial ($x^T W x + m \dot{x}^2$ would decrease along trajectores). However, I'm not sure if the statement remains true with the time dependence. On an intuitive level, this describes the motion of a particle with a time-varying force which always pushes it a direction roughly opposite its position (since $x^T (-Wx) < 0$).

$\endgroup$

1 Answer 1

1
$\begingroup$

This does not follow. We can find a counterexample in dimension $n=1$ (and let's also set $m=1$). If we write $x=ye^{-kt/2}$, then $y$ solves $$ -y'' - W(t) y = -\frac{k^2}{4} y , $$ and I want to interpret this as a 1D Schrödinger equation at energy $E=-k^2/4$. For a periodic potential, the spectrum has band structure and the Lyapunov exponent is positive in the gaps; in other words, we have (modulo a periodic factor) exponentially increasing solutions in those gaps. The restriction that $-\alpha\le -W \le -\beta$ is not an issue; typically, there are gaps at arbitrarily large energies.

We can now just choose such a $W$ (with a suitable constant added, so that $E=-k^2/4$ will lie in a gap), and we obtain a solution $|y|\gtrsim e^{\gamma t}$ (where the periodic factor is not close to zero). Here, the Lyapunov exponent $\gamma$ is not affected by $k$, so if $k$ was small enough compared to $\gamma$, then $x$ will also have exponential increase on average.

In fact, I could do this more carefully (using facts about the location of gaps), and it then follows that such a counterexample is possible for any given bounds $-\alpha<-\beta$ as long as I'm allowed to make $k>0$ small in comparison.

$\endgroup$
2
  • $\begingroup$ Thanks. Could I ask if you know of any conditions on the RHS that allow it to be time-varying but guarantee the boundedness of the resulting trajectory? $\endgroup$
    – Michael S.
    Commented Aug 22, 2017 at 18:28
  • $\begingroup$ @MichaelS.: If $W$ approaches a limit $W_0$ as $t\to\infty$, then you're in much better shape and there are perturbative results. For example, if $W-W_0\in L^1$ (or $BV$ would work too), then the solutions will look like the ones of the unperturbed equation ($W\equiv W_0$) asymptotically, and thus are bounded. If $W$ varies over an interval, then I don't think there can be any very general positive results. $\endgroup$ Commented Aug 22, 2017 at 22:40

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .