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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.

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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 ;).

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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.

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