if the solution of a system of ode's blows up, does it mean it's unstable in the Lyapunov sense?

Question

there are many techniques and theorems which let you conclude about the stability of a system without actually solving it.

let's say for example you've got a system : $X'= AX + B, \; X(0)= X_0$

$A \in \mathbb{R^{n \times n}}, \; B \in \mathbb{R^n}$

if $\|X(t)\| \to \infty$ intuitvely it seems that the $\epsilon-\delta$ defintion of stability in the sense of lyapunov doesn't hold.

so does it mean the system is indeed unstable ?

Answer 1

Say $x^{*}$ is an equilibrium point of your system. Then this equilibrium point is stable if for every $\epsilon>0$, there exists a $\delta>0$ such that if $||x(0) - x^{*}||<\delta$, then for all $t>0$, $||x(t)-x^{*}||<\epsilon$.

If in your case $||x(t)|| \to \infty$ no matter what the initial conditions, then you can't satisfy the above condition, so your equilibrium point is by definition unstable.

Or in other words, in your case, there exists an $\epsilon>0$, such that for all $\delta>0$, $||x(0) - x^{*}||<\delta$ and $||x(t)-x^{*}||>\epsilon$.