I have a system $V(x)$, in $R^2$, and I've calculated that
$V(x) \geq 0$ for all $x$ not equal to zero and that $V(0,0) = 0$
I've also calculated that $V'(x) \leq 0$
Since $V'(x)$ is NSD and not ND I have deduced that the system is globally asymptotically stable.
Is there any flaw in my analysis? Or is it correct?
(I've also calculated that as |V(x)| goes to infinity as ||x|| goes to infinity)
My system is
$x' = -y-x^3$
$y' = x^5$
My Lyapunov function is $V(x) = x^6+3y^2$. I'm trying to see if it's globally asymptotically stable or not but I'm not sure if my answer is correct.
You have the system
$$ \begin{align} \dot{x} &= -x^3 - y \\ \dot{y} &= x^5 \end{align} $$
And the Lyapunov function
$$ \begin{align} V(x) &= x^6 + 3 y^2 \\ \dot{V}(x) &= -6 x^8 \end{align} $$
So your Lyapunov function is positive definite (PD), means it is $V(x) > 0$ for all $x \neq 0$, not $\geq$ like you wrote in the question. This is important because otherwise it would be only positive semi definite (and therefore not valid).
The derivative $\dot{V}(x) \leq 0$ for $x \neq 0$, negative semi definite (NSD) so the trajectories are bounded but we don't know yet if the system is also asymptotically stable.
We can use LaSalle for it: $\dot{V}(x) = 0 \iff x = 0$ but then $\dot{x} = -y$. If $y \neq 0$ you can't keep $\dot{V}(x) = 0$.
Conclusion: The system is globally asymptotically stable.
In general if we only know $V(x)$ is PD and $\dot{V}(x)$ is NSD then we don't know if the system is asymptotically stable without more investigation (like LaSalle).