Doubt about Lyapunov's theorem proof

Question

Given the autonomous system $\dot x=f(x)$ and an equilibrium point $\bar x$, we know that it is stable if
$\exists\phi:U_0\to \mathbb R$, $\phi\in\mathcal C^1(U_0;\mathbb R)$, with $ U_0$ open nbh of $\bar x$, such that $$ \phi (x)>\phi(\bar x), \forall x\in U_0\setminus\{\bar x\}\\(\mathcal L_f\phi)(x)\le 0,\forall x\in U_0. $$ My question is: why do we need during the proof of the theorem the hypotesis that $\bar x$ is a $\textbf {strict minimum}$ for $\phi$?

Answer 1

You need this because the proof is somewhat indirect. You show that $t\mapsto \phi(x(t))$ is a falling function. As it is bound below, it will have a limit. But this chain of arguments does not tell anything constructive about this limit. Neither if it is the minimum $\bar \phi$, nor that there is a corresponding limit in $x(t)$.

Next you need some property that $\phi\circ x$ can not have an equilibrium except at the minimum, that is $\cal L_f\phi$ has to be sufficiently negative. But this alone does not tell you that $x(t)$ itself converges. To show that it is helpful if $\phi^{-1}(\bar \phi)$ only contains one point, that is, that $\bar \phi$ is a strict minimum.