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

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?

• Since it is a different ball with a different radius. Jun 2, 2020 at 19:45
• @useless-machine Why do you need a 2nd ball? Why does my definition with a single ball not make sense? Jun 3, 2020 at 7:08

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 $$k>0$$
$$\ddot{x} = -k\,x.$$
$$V = \dot{x}^2 + k\,x^2,$$
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 $$k\leq1$$.