# Is the Lyapunov stability definition ambiguous?

I understand the conceptual idea of Lyapunov stability.

My question is about the formal definition.

Wikipedia's definition is very much in line with others I found:

An equilibrium, $$x_{e}$$, is said to be Lyapunov stable, if, for every $$\epsilon >0$$, there exists a $$\delta >0$$ such that, if $$\|x(0)-x_{e}\|<\delta$$, then for every $$t\geq 0$$ we have $$\|x(t)-x_{e}\|<\epsilon$$.

The confusing part for me is the phrase "for every $$\epsilon >0$$".

Assume $$\|x(0)-x_{e}\|>0$$ and that some $$\delta$$ exists as required. If I choose $$\epsilon = \frac{|x(0)-x_{e}|}2$$ then $$x(0)$$ falls outside the circle ($$\|x(t)-x_{e}\|<\epsilon$$). By this interpretation no fixed point can be Lyapunov stable. And this is how I know I am in trouble :-).

This makes me wonder why this definition does not draw some relation between $$\epsilon$$ and $$x(0)$$. For example, if it includes $$\epsilon > \|x(0)-x_{e}\|$$ it would aid in my understanding and remove the argument I just provided.

Alternatively, I could wonder why not use the phrase "for some $$\epsilon$$ where $$0 < \epsilon < \infty$$" instead of "for every $$\epsilon >0$$".

• In plain English, it's saying "for any desired accuracy, you can find a tolerance that would guarantee it in perpetuity". Commented Oct 31, 2021 at 8:55
• You probably come from a different field, the problem that you are struggling with is one that maths students usually encounter first with the epsilon delta definition of continuity or the definition of a limit of a sequence (which has an $\epsilon$ and an $N$). Maybe it will help you to review those. Commented Oct 31, 2021 at 11:45
• (fixed "assume" sentence) Commented Nov 1, 2021 at 13:51
• @CarstenS You are correct - I am new to this field. I see the two problems as similar, and the solution is also similar. Thank you. Commented Nov 1, 2021 at 13:59

You're not interpreting the logic correctly.

It doesn't say “there exists a single positive number $$\delta$$ which works for all positive numbers $$\epsilon$$”, it says “for any positive number $$\epsilon$$, you can find a corresponding positive number $$\delta$$ which works for that particular number $$\epsilon$$”. If you change $$\epsilon$$ to a smaller number, you will in general have to make $$\delta$$ smaller too. But the point is that no matter how small an $$\epsilon$$ you take, you can find a $$\delta$$ that works (and that's why it has to say “for every $$\epsilon$$”, not just “for some $$\epsilon$$”).

This is just as for the $$\epsilon$$-$$\delta$$ definition of limits.

• Exactly. Quantifiers do not commute in general. For every person there exists a father, but there doesn't exist a person which is father of everyone (at least I hope; and lets exclude spiritual fatherhood like the pope....) Commented Oct 31, 2021 at 9:19
• I see now that the key mistake I made was to ignore the role of $x(0)$ in the expressions. By deciding on a $\delta$ that is less than $x(0)$ I am forced to choose another value for $\epsilon$. (meta: it is difficult to decide which of the two answers is more correct). Commented Nov 1, 2021 at 14:09

$$ϵ$$ is primary, $$δ$$ is secondary depending on it.

As you have chosen $$x(0)$$, you already know the radius $$δ$$ in $$\|x(0)-x_e\|<δ$$ and thus also the $$ϵ$$ it is based upon. So there remains no freedom to now alter $$ϵ$$ to some other value.

Or in other words, if you now select a new $$ϵ$$, then the stability property results in a new $$δ$$, for which the selected $$x(0)$$ is likely no longer admissible.

• Thank you. The correspondence lies in the value of $x(0)$. This value is used in the clause that contains $\epsilon$ and the one that contains $\delta$. When I chose an $\epsilon$ that excluded the $x(0)$, I also messed up the value of $\delta$. Now I understand the definition. Commented Oct 30, 2021 at 10:33