On Lyapunov's method

Question

I been reading the standard proof of Lyapunov's method (https://en.wikipedia.org/wiki/Lyapunov_stability)

for example page $7$ here,

https://ocw.mit.edu/courses/electrical-engineering-and-computer-science/6-241j-dynamic-systems-and-control-spring-2011/readings/MIT6_241JS11_chap13.pdf

I can't see why and where we really need that the function $V(x)\ \ge0$. Could someone explain this to me?

Answer 1

Some geometric ideas about Lyapunov's method.

Given a dynamical system

$$ \dot x = f(x,t) $$

Suppose that $V(0) = 0$ and $V(x) > 0 $ for $x \ne 0$ then there exist a closed level curve/surface around $x=0$ indexed by a suitable $\epsilon > 0$ or $V(x) = \epsilon$

Now $\dot V(x) = \nabla V(x)\dot x = \nabla V(x)f(x,t) \le 0$

tell us that the directional derivative for $V(x)$ in the dynamic system's direction is negative. This show us that the dynamic system's path is entering the level curve/surface associated to $V(x) = \epsilon$

So it is essential that around $x=0$ assumed as the equilibrium point, the existence of a closed curve/surface. This is guaranteed by $V(x) = \epsilon > 0$

I hope this helps.