I recently started studying The Mathod of Lyapunov, and I thought I understood how to distinguish a positive definite from a positive semi-definite functions, or negative from semi-definite functions. However, in the exercises of the book, I was given a function
$$V(x,y)=x^{2}$$
My first thought was that this function is a positive definite $V>0$ when $(x,y)\neq (0,0)$. However, the solution mentioned that this function is positive semi-definite $V\geq 0$ when $(x,y)\neq (0,0)$ can someone please explain to me how is this even possible for values of $x$ when $x \neq 0$?
I think the answer is simple:
$V_1(x)=\frac{1}{2}x^2$ is positive definite and $V_2(x,y)=\frac{1}{2}x^2$ is positive semi-definitive. Because $V_1$ might tell us the system converges to nowhere except $x=0$, but $V_2$ could lead to a invariant set $\{x=0,y\in\mathbb{R}\}$.
Actaully, any lower bounded smooth functions can be selected as Lyapunov candidates, such as $V_3(x,y)=\frac{1}{2}x^2+1$ (it is positive everywhere) or even $V_4(x,y)=\frac{1}{2}x^2-1$. Their names are not important.