I am considering the following system of ODE:
$$x'=y, y'=x-x^3$$
where (0,0) is a critical point of it.
By considering its locally linear system, I found that this critical point is an unstable saddle point. And I believe that this conclusion is true since I had verified it by plotting the phase portrait.
However, when I consider the following Lyapunov Function, $$V(x,y)=\frac{1}{4}(u^2-1)^2+\frac{1}{2}v^2$$ upon computation, the derivative of V by the system is constant zero, so it is negative semidefinite. Would this fact show that (0,0) is a stable critical point?
Thank you very much for reading the post.
While it is true that solutions of the ODE follow the level curves of $V$, the point $(0,0)$ is also a saddle point of $$ V=\frac12y^2+\frac14-\frac12x^2+\frac14x^4, $$ thus your claim is wrong/not applicable.