0
$\begingroup$

I seem to be stuck at formally showing something that intuitively seems to be true. I have a linear non-autonomous system of the form

$$ \dot{x} = A(t)x $$

where $A(t):\mathbb{R}\to \mathbb{R}^{n\times n}$ is a uniformly bounded square upper block triangular matrix, such that its $m$ diagonal blocks $\{B_i\}_{i=0}^m$ are strictly negative definite (SND) for all $t\ge0$, and $\lim_{t\to\infty} B_i(t)$ is bounded from above by a SND matrix (i.e. there is an SND matrix $B_0$ such that $B_i(t)-B_0$ is also SND). By strictly negative definite I mean a matrix such that $y^TB_i y<0$ for all nonzero $y$.

Q: Are Lyapunov exponents of such a system all negative? If so, how to show this?

Now, I know that if the whole $A(t)$ is negative definite, the Lyapunov exponents (LE) of the system are all negative. I also know from theorem 5.1 that if $A(t)$ is triangular with negative elements on the diagonal, its LE's are also all negative (the time average of a negative function is negative). I can't figure out how to couple these two statements to arrive at a result for a block-triangular matrix. don't see

$\endgroup$
  • $\begingroup$ In dimension 1, $A(t)=-1/(1+t)$ has Lyapunov exponent $-1/(1+t)$, which is negative for every nonnegative $t$, but the Lyapunov exponent $\lim\limits_{t\to+\infty}t^{-1}\log|x(t)|$ is $0$, not negative. $\endgroup$ – Did Feb 8 '15 at 19:15
  • $\begingroup$ @Did looks like your function is 0 as $t\to\infty$, which is a limiting case I'm not considering. I'm not sure how your comment helps answering the question $\endgroup$ – dmytro Feb 8 '15 at 19:24
  • $\begingroup$ Your "which is a limiting case I'm not considering" seems to show that your question is not complete. What do you mean exactly by negative ($\lt0$ or $\leqslant0$) and which cases exactly, limiting or not, do you wish to exclude? $\endgroup$ – Did Feb 8 '15 at 19:29
  • $\begingroup$ @Did ok, I see. I tried to add some more detail to the question now. Any thoughts? $\endgroup$ – dmytro Feb 8 '15 at 19:42
  • $\begingroup$ The uniform bound is new (and changes completely the question). Picking some negative definite $B$ such that $A(t)\leqslant B$ for every $t$ large enough and comparing $x(t)$ with the solutions $z$ of $\dot z(t)=B\cdot z(t)$ should yield the result. $\endgroup$ – Did Feb 8 '15 at 19:49

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.