I need some help with the following excersice.Find the minimum period and the Floquet multipliers $\bf\lambda_{1},\lambda_{2}$ of the following matrix.

$A(t)=\begin{bmatrix}-1+\frac{3}{2}cos^2(t) & 1-\frac{3}{2}cos(t)sin(t)\\-1-\frac{3}{2}cos(t)sin(t) & -1+\frac{3}{2}sin^2(t)\end{bmatrix}$

It is obvious that the minimum period of A is $\pi$,just by using the double-angle formulas $\cos^2(t)=\frac{1+cos(2t)}{2},sin^2(t)=\frac{1-cos(2t)}{2}$.Furthermore by using the relation $$\prod_{i=1}^2\lambda_{i}=e^{\intop_0^T trace(A(s))ds}$$ we get easily that $\prod_{i=1}^2\lambda_{i}=e^{-\frac{\pi}{2}}$.But now i am stuck!


1 Answer 1


There are two useful facts:

  • $\hat x(t)=e^{t/2}\left(\begin{array}{r}-\cos t\\ \sin t\end{array}\right)$ is a solution of the Markus-Yamabe system;
  • if some nontrivial solution $x(t)$ has the property $x(t+T)=\lambda x(t)$, then $\lambda$ is a Floquet multiplier of the system.

Combining these two facts, we can deduce that $\lambda_1=-e^{\pi/2}$ is a multiplier of the system: $$ \hat x(t+\pi)=e^{(t+\pi)/2}\left(\begin{array}{r}-\cos (t+ \pi)\\ \sin (t+\pi)\end{array}\right) =e^{\pi/2}e^{t/2}\left(\begin{array}{r}\cos t\\ -\sin t\end{array}\right)= -e^{\pi/2}\hat x(t). $$ Another multiplier can be obtained from the equality $\lambda_1 \lambda_2= e^{-\pi/2}$.

  • $\begingroup$ thank you for the correction.As for the solution $\widehat{x(t)}$ you suggest is it out of the blue or it can be derived somehow? $\endgroup$ Commented Dec 5, 2018 at 12:38
  • $\begingroup$ @Perpendicular This solution is a part of the Markus-Yamabe counterexample: it is unbounded and, thus, the origin is not globally asymptotically stable, although the eigenvalues of $A(t)$ do have negative real parts for any $t$. I.e. the system was specially designed to have this solution. There is no general way to find such a solution in any case. $\endgroup$
    – AVK
    Commented Dec 5, 2018 at 14:01
  • 1
    $\begingroup$ For a paper on how to construct such examples, see Josić and Rosenbaum Unstable Solutions of Nonautonomous Linear Differential Equations (however, I am afraid that the paper is behind a paywall). $\endgroup$
    – user539887
    Commented Dec 6, 2018 at 10:06

You must log in to answer this question.

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