# Find the Floquet multipliersfor the Markus & Yamabe system

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!

• $$\hat x(t)=e^{t/2}\left(\begin{array}{r}-\cos t\\ \sin t\end{array}\right)$$ 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 $$\lambda_1 \lambda_2= e^{-\pi/2}$$.
• 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? Commented Dec 5, 2018 at 12:38
• @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.