Continuous-time Switched Linear Systems

Question

Suppose we have a set of matrices $\Sigma$ and a switching rule $\sigma: \mathbb{R} \to \Sigma$. It is known in the discrete case that if the joint spectral radius, $\rho(\Sigma)$, is less than one, then the following so called discrete-time switched linear dynamical system is convergent

$$ x_{t + 1} = \sigma(t)x_t. $$

What is known about the continuous case? Does there exist a condition on the joint spectral radius of $\Sigma$ that would imply that the following dynamical system is globally asymptotically stable?

$$ \frac{dx}{dt} = (\sigma(t)-I)x $$

Answer 1

For continuous-time linear switched systems, there exists an equivalent notion of the JSR, it is the "least measure value" of the set of matrices. It is proven that the system is exp. stable iff this quantity is negative. For a formal definition and properties of this notion, refer to section 2.3.3 in the book of Sun&Ge : Stability Theory of Switched Dynamical Systems(2011). In general, the condition $\rho(\Sigma)<1$ does not tell about the stability in the continuous-time case. For more results on the stability equivalence between the continuous-time and discrete-time cases, refer to section 3.5 in Jacques Theys thesis: Joint Spectral Radius: theory and approximations (2009).

Answer 2

Assuming I understand the question correctly, the answer is no.

Let $\sigma(t) = \begin{bmatrix}0 & e^t \\ 0 & 0 \end{bmatrix}$. That is, $\Sigma = \{ \sigma(t) | t \in \mathbb{R}\}$, and $\rho(\Sigma) = 0$

However, the system $\dot{x}=(\sigma(t)-I) x$ is not asymptotically stable, starting from $x(0)=(0,v)$, we have $x(t) = (v, e^{-t}v)$.