Why is there a delta for locally stable in the sense of lyapunov?

Question

Why does this definition (screenshotted above) for locally stable i.s.L. require a $\delta$ term? Why not just say, something is locally stable i.s.L. if "for any ball of size $\epsilon$ around the fixed point, all trajectories initialized within the $\epsilon$ ball stay inside the $\epsilon$ ball"?

Can someone provide an example where introducing this $\delta$ ball is significant?

Answer 1

This is because even if a system is Lyapunov stable it doesn't mean that $\|x(0) - x^*\| \leq \|x(t) - x^*\|\ \forall\,t \geq 0$.

For example consider a simple mass spring system with unit mass and spring constant $k>0$

$$ \ddot{x} = -k\,x. $$

For this system one can use the following Lyapunov function to also show Lyapunov stability

$$ V = \dot{x}^2 + k\,x^2, $$

since $\dot{V} = 0$. Namely, for $\mathbf{x}(t) = \begin{bmatrix}x(t) & \dot{x}(t)\end{bmatrix}^\top$ and $\|\mathbf{x}(0)\| = \delta$ it could mean that $x(0) = \delta$ and $\dot{x}(0) = 0$ or $x(0) = 0$ and $\dot{x}(0) = \delta$. However, since the proposed Lyapunov function should stay constant it also follows that the trajectory starting at $x(0) = \delta$ and $\dot{x}(0) = 0$ can reach $x(t) = 0$ and $\dot{x}(t) = \delta\,\sqrt{k}$ and thus $\|\mathbf{x}(t)\| = k\,\delta$. Similarly, for $x(0) = 0$ and $\dot{x}(0) = \delta$ is follows that the system could reach $\|\mathbf{x}(t)\| = \delta/k$. Therefore, for all initial conditions satisfying $\|\mathbf{x}(0)\| < \delta$ one can only ensure that all their trajectories are bounded by $\|\mathbf{x}(t)\| < k\,\delta$ when $k\geq1$ or $\|\mathbf{x}(t)\| < \delta/k$ when $k\leq1$. It can be noted that many initial conditions would yield trajectories that would remain inside the ball with radius $\delta$ but some initial conditions yield trajectories that are only bounded by the ball with radius $\epsilon$ with $\epsilon = k\,\delta$ when $k\geq1$ or $\epsilon = \delta/k$ when $k\leq1$.