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Given a dynamical system

$$\frac{dx}{dt}=-\nabla f(x)$$

which $x=0$ is the only equilibrium point, i.e. $-\nabla f(x)|_{x=0}=0$.

I am reading this tutorial, and it states:

$f(x)$ is a lyapunov function such that $x=0$ is a locally asymptotic stable equilibrium point.

I am confusing about this statement because from my understanding, to be a lyapunov function, the function should satisfy the following three conditions. However I can only see the condition (3) is satisfied and have no idea how condition (1) (2) are satisfied. Could someone help me to understand? Thanks a lot!

(1) $f(x)|_{x=0}=0$

(2) $f(x)|_{x\neq 0}>0$

(3) $\frac{df(x)}{dt}|_{x\neq 0}<0$

Note:

Condition (3) holds, since $\frac{df(x)}{dt}|_{x\neq 0}=\nabla f(x)\cdot (-\nabla f(x))|_{x\neq 0}=-\|\nabla f(x)\|^2_{x\neq 0}<0$. But I have no idea about condition (1) and (2).

I think maybe I have a misunderstanding of this point, maybe this would be correct:

$f(x)$ is not neccessary to be a lyapunov function. If in addition, $f(x)$ meets the conditions (1) (2), then it is a lyapunov funciton.

Could someone help me to clarify this point? Thanks a lot for any suggestion.

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    $\begingroup$ Please provide a full reference. Eventually, all links will be broken. $\endgroup$ May 11, 2022 at 19:46

1 Answer 1

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I think you need to read a bit further in the tutorial to clarify this point. At the beginning of the tutorial, the author suggests that $f(x)$ is a candidate for a Lyapunov functional. Note that the function $f(x)$ is not unique - any $f(x)+c$ would also be a potential for the vector field $\nabla f(x)$. Therefore, you can just pick $f_2 (x)=f(x)-f(0)$ as your potential function, and it will automatically satisfy $f_2 (0)=0$. This is to say that condition 1 can basically be automatically satisfied. The author addresses your point about condition 2 in Theorem 2.3, in which they state that $f_2 (x)$ will be a Lyapunov function when $0$ is a strict local minimum of $f$. This will satisfy your criteria 2, at least locally. I hope this helps!

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    $\begingroup$ Thank you! I got it! maybe here is the proof: from the paper [On the stable equilibrium points of gradient systems], under the assumption that $f(x)$ is real analytic, local minimality is necessary and sufficient for stability. It gives: since $x=0$ is locally asymptotically stable, thus it is locally stable, thus we have $f(x)$ attain local minimum at $x=0$. Then, conditions (1) and (2) are satisfied. $\endgroup$
    – M.K
    May 12, 2022 at 16:24
  • $\begingroup$ Looks right to me. $\endgroup$
    – mwalth
    May 12, 2022 at 16:52
  • $\begingroup$ Thanks. And if $f$ is not real analytic, it remains to be prove in another way. $\endgroup$
    – M.K
    May 12, 2022 at 16:54

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