Picking a Lyapunov function that is dependent on $\dot{x}$

Question

I have a system $\dot{x}=f(x)$. Is it a good idea if I pick a Lyapunov function V that is dependent on $\dot{x}$. So that $V=V(x,\dot{x})$? One of the conditions for stability is that the origin is stable, that is $x=0$ implies $V=0$. But this won't be the case if V is dependent on $\dot{x}$ as well. Am I missing something or is it just a bad idea to pick a V dependent on $\dot{x}$.

Answer 1

As @whpowell96 said you can always rewrite $V(x,\dot{x})=V(x,f(x))=\tilde{V}(x)$. Your assumption that $x=0$ must imply $V=0$ is not correct in the most general case. If we have $\dot{x}=f(x)$ with the equilibrium point $x_\text{eq}$ then $V(x=x_\text{eq})=0$. In most cases we transform the system such that the equilibirum point is in the origin.

By the converse theorem of Lyapunov, the existence of an asymptotically stable equilibrium implies that there exists a Lyapunov function $V(x)$ that can be used to prove the asymptotical stability of the equilibrium point. Hence, it is not necessary to use $V(x,\dot{x})$. But you could try to use it as a construction method for Lyapunov functions. If your equilibrium point is $x=0$. Then $V(x,\dot{x})=V(x,f(x))$ might help you to construct a Lyapunov function candidate. But I never saw it actively used to construction. But there is Krasovski's method which is using the Jacobian of $f(x)$.

Answer 2

It depends I would say. Take for example the system

$$ \dot{x} = -x \tag{1} $$

Then you can easily choose as Lyapunov function

$$ V(\dot{x}) = \frac{1}{2} \dot{x}^2 = \frac{1}{2} (-x)^2 = \frac{1}{2} x^2 $$

The time derivative is given by

$$ \dot{V}(\dot{x}) = x \dot{x} = -x^2 $$

which is negative definite and so the system $(1)$ is globally stable. So, it can work to choose $V$ as a function of $\dot{x}$ (which, of course results then in $V$ being a function of $x$).

However, this only worked in this trivial example "by accident". It might be worth a try if other methods fail, but I am not aware of a method that efficiently uses a direct dependence of $V$ on $\dot{x}$.