Let $n \in \mathbb{N}$ with $n\geq 1$, let $D \subset \mathbb{R}^n$ be an open neighbourhood of $0$, and let $f: D \to \mathbb{R}^n$ be locally Lipschitz with $f(0)=0$. For each $x_0 \in D$, let $T_{\max}(x_0) \in (0,\infty]$ and let $x(\cdot;x_0): [0,T_{\max}(x_0)) \to D$ denote the unique maximal forward solution of the initial value problem

\begin{align*} \dot{x}(t)=f(x(t)), \qquad x(0)=x_0. \end{align*}

Let $V \in C^1(D;\mathbb{R})$ satisfy $V(0)=0$ and $V(x)>0$ for every $x \in D \setminus \{0\}$. Define the orbital derivative $\dot V: D \to \mathbb{R}$ by

\begin{align*} \dot V(x)=\nabla V(x)\cdot f(x). \end{align*}

Assume that there exists $r>0$ such that $\overline{B}(0,r)\subset D$ and $\dot V(x)\leq 0$ for every $x \in B(0,r)$. Then the equilibrium $0$ is Lyapunov stable: for every $\varepsilon \in (0,r]$ there exists $\delta>0$ such that, for every $x_0 \in B(0,\delta)$, one has $T_{\max}(x_0)=\infty$ and $x(t;x_0)\in B(0,\varepsilon)$ for every $t\geq 0$.

If, in addition, there exists $r_a \in (0,r]$ such that $\dot V(x)<0$ for every $x \in B(0,r_a)\setminus\{0\}$, then the equilibrium $0$ is locally asymptotically stable: it is Lyapunov stable and there exists $\delta_a>0$ such that, for every $x_0 \in B(0,\delta_a)$, one has $T_{\max}(x_0)=\infty$ and

\begin{align*} \lim_{t\to\infty} x(t;x_0)=0. \end{align*}