Consider the following linear periodic system: \begin{align} \dot{x}=A(t)x \end{align} where $x\in\mathbb{R}^n$ and $A(t)\in\mathbb{R}^{n\times n}$ is piecewise continuous and $A(t)=A(t+T)$ for some positive constant $T$.

Let $\Phi(T,0)$ be the state transition matrix of the system over one period. If $\Phi(T,0)$ is Schur, i.e., the magnitudes of all eigenvalues are smaller than $1$, then the system is asymptotically stable. This follows from the Floquet theory, say https://en.wikipedia.org/wiki/Floquet_theory.

My question is: Whether the converse is true, that is, if the system is asymptotically stable, then $\Phi(T,0)$ must be Schur?

I ended up here:

Since $\Phi(t+T,T)=\Phi(t,0)$ (see Theorem 2.48 of Ordinary Differential Equations with Applications, Chicone Carmen, 1999), the solution of the system satisfies \begin{align} x((k+1)T)&=\Phi((k+1)T,kT)x(kT)\notag\\ &=\Phi(T,0)x(kT) \end{align} which can be viewed as \begin{align} y(k+1)=\bar{A}y(k) \end{align} where $\bar{A}=\Phi(T,0)$.

Update: What I am worried about is that, since the second system is sampled from the first system, it might not capture all the property of the solution of the first system. Thus, the stability of one system might not be equivalent to the stability of the other one.


Yes, in order for this to be asymptotically stable $\Phi(T,0)$ can't have any eigenvalue of absolute value $\ge 1$. Namely, if $\lambda$ is such an eigenvalue, with eigenvector $v$, it is easy to show that $\Phi(T,0)^n \text{Re}(v)$ and $\Phi(T,0)^n \text{Im}(v)$ can't both go to $0$ as $n \to \infty$.

  • $\begingroup$ By 'absolute value', did you mean the magnitude of a complex number? $\endgroup$
    – guluzhu
    Mar 25 '20 at 13:12
  • $\begingroup$ Yes, that's what it means. $\endgroup$ Mar 25 '20 at 13:33

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