Why the derivative is negative definite?

Question

Why the derivative is negative definite? Shouldn't be negative semi definite? It only depends on the $x$.

Answer 1

You're right, this argument is faulty. The standard Lyapunov theorem does not give asymptotic stability in this case.

However, there's a stronger theorem by LaSalle that you can use. Since the set where $\dot V=0$, namely the line $x=0$, contains no complete trajectories except the equilibrium $(0,0)$ (which you see from $\dot x = y - 0$ when $x=0$, so that the vector field is transversal to the line $x=0$ except at the origin), you indeed get asymptotic stability. But there is that extra condition that one needs to check.