From what I've read, Lyapunov functions provide a very nice mechanism for verifying stability of continuous-time linear systems (non-linear as well, but that's not my concern at the moment). For example, for the system $x' = -x$, a Lyapunov function is $V(x) = \frac{1}{2}x^2$. It is easy to see that
- $V(0) = 0$
- $V(x) > 0$ for $x \neq 0$
- $V(x)' = \frac{dV(x)}{dx}\frac{dx}{dt} = -x^2 < 0$ for $x\neq 0$
which is sufficient to establish global asymptotic stability. However, what happens if I change this to a zero-order hold system with a non-constant sample period that is at most $\Delta$? In other words, at least every $\Delta$ time, $v$ is set to $-\frac{1}{\Delta}x$ and the system evolves according to $x' = v, v' = 0$. I think this system is still globally asymptotically stable. Is there a nice way to prove this, analogous to Lyapunov functions for continuous-time systems? Is there some way of saying that the zero-hold system is "close enough" to the continuous-time equivalent that the stability results still hold?