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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 $\epsilon >0$".

I am clearly missing something. Please help me to eliminate the ambiguity I read in this definition.

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  • $\begingroup$ In plain English, it's saying "for any desired accuracy, you can find a tolerance that would guarantee it in perpetuity". $\endgroup$
    – user541686
    Commented Oct 31, 2021 at 8:55
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    $\begingroup$ 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. $\endgroup$
    – Carsten S
    Commented Oct 31, 2021 at 11:45
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    $\begingroup$ (fixed "assume" sentence) $\endgroup$
    – Willem
    Commented Nov 1, 2021 at 13:51
  • $\begingroup$ @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. $\endgroup$
    – Willem
    Commented Nov 1, 2021 at 13:59

2 Answers 2

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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.

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    $\begingroup$ 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....) $\endgroup$
    – lalala
    Commented Oct 31, 2021 at 9:19
  • $\begingroup$ 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). $\endgroup$
    – Willem
    Commented Nov 1, 2021 at 14:09
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$ϵ$ 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.

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  • $\begingroup$ 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. $\endgroup$
    – Willem
    Commented Oct 30, 2021 at 10:33

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