The equilibrium $x^\ast$ is (Lyapunov) stable iff $$\forall \varepsilon > 0 \; \exists \delta(\varepsilon) : \lvert x(0) – x^\ast \rvert < \delta(\varepsilon) \Rightarrow \forall t \geqslant 0 \; \lvert x(t) – x^\ast \rvert < \varepsilon. $$
And equilibrium $x^\ast$ is asymptotically (Lyapunov) stable iff it's (Lyapunov) stable and every trajectory $x(t)$ from some neighbourhood tends to $x^\ast$ (i.e., $x(t) \rightarrow x^\ast$ when $t \rightarrow +\infty$)
Can someone explain to me why the choice of the word "asymptotic"?
Does it simply describe that the trajectory will converge to $x^\star$ but will not achieve it in finite time? (I am not totally sure if this is true for all dynamical systems)
Best Answer
Assymptotically stable does indeed mean that a solutions assymptotically convergences to the equilibrium point. However this does not mean that it has to happen in infinite time. Namely if $x(\tau)=x^*$ for some finite $\tau$ then it still satisfies the limit that $x(t) \to x^*$ as $t \to \infty$, because $x(t)$ should remain equal to $x^*$ for all $t\geq\tau$.