Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Given the arbitrary linear system of DE's $$x'=A(t)x,$$ with the condition that the spectral bound of $A(t) $ is uniformly bounded by a negative constant, is the trivial solution always stable? All the $(2\times 2)$ matrices I've tried which satisfy the above property yield stable trivial solutions, which seems to suggest this might be the case in general. I can't think of a simple counterexample, so I'm asking if one exists. If there isn't what would be some steps toward proving the statement?

This is indeed homework.

share|cite|improve this question
up vote 2 down vote accepted

You can elongate a vector a bit over a short time using a constant matrix with negative eigenvalues, right? Now just do it and at the very moment it starts to shrink, change the matrix. It is not so easy to come up with an explicit formula (though some periodic systems will do it) but this idea of a counterexample is not hard at all ;).

share|cite|improve this answer

Here is an example.

$$ A(t)=\left( \begin{matrix} -1+\frac32\cos^2 t& 1-\frac32\cos t\sin t\\ -1-\frac32\sin t\cos t &-1+\frac32\sin^2 t \end{matrix} \right) $$

One can check that the eigenvalues of $A(t)$ are $$ \lambda_1=\frac14[-1+\sqrt7 i],\quad \lambda_2=\bar\lambda_1, $$ both of which have negative real parts. But the origin is unstable.

This is an example from the section about Floquet theory in Ordinary Differential Equations and Dynamical Systems by Sideris.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.