# Uniform stability of equilibrium

I have been given the following definition for uniform stability: the equilibrium state $$x_e$$ is uniformly stable, if for any $$\epsilon > 0$$ there is a $$\delta > 0$$ such that

$$\|x(0)-x_e\|<\delta~~\Rightarrow ~~\|x(t)-x_e\|<\epsilon, ~~\forall t\geq 0$$

In my opinion this definition does not have anything to do with stability. Imagine a system with $$x(t)$$ going to infinity and $$x(t) \geq 10^{100} ~~\forall t \geq 0$$ and $$x_e = 0$$.

Then the system would be uniformly stable following the above definition, as for any $$\epsilon$$ I can use $$\delta = 10^{100}$$ and that would fulfill the implication. The reason is that $$x(0) - x_e$$ is never smaller than $$10^{100}$$ so the left side of the implication is always false and therefore the implication always true.

What am I missing here?

I think you are missing a quantifier about the curves $$x(t)$$. Something like: for all $$\epsilon>0$$, there exists a $$\delta>0$$ such that for all integral curves $$x:\mathbb{R}\to\mathbb{R}$$,

$$\|x(0)-x_e\|<\delta~~\Rightarrow (\forall t\geq 0,\|x(t)-x_e\|<\epsilon).$$

Then this precisly says that, if you start near enough you equilibrium, you stay arbitrarily close to it the whole time.

Your definition of uniform stability is strange.

Usual definition:

Uniform stability: for each $$\varepsilon>0$$, there is $$\delta=\delta(\mathbf{\epsilon})>0$$, $$\textbf{independent of}$$ $$\mathbf{t_0}$$, such that $$\|x(t_0)-x_0(t_0)\|<\delta~~\Rightarrow ~~\|x(t)-x_0(t)\|<\epsilon, ~~\forall t\geq t_0\geq0$$

The assumption is that the initial value $$x(0)$$ can be chosen arbitrarily in some neighbourhood of the equilibrium point $$x_e$$.

• Could you elaborate this further? How does it avoid simply setting $\delta = \min \{ x(t) \}$ ? – NightRain23 Nov 18 '19 at 17:32