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 $x_e$.

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

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