# The lyapunov function of gradient system

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.

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

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!
• 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.
• Thanks. And if $f$ is not real analytic, it remains to be prove in another way.