I have recently come across Lyapunov's stability theorem, which states:
Let $0$ be an equilibrium of a differential autonomous system $X'=f(X)$ where $f$ is smooth.
Suppose $V$ is a Lyapunov function of the system in some neighborhood of $0$; that is, $V(X(t))$ is non-increasing on $t$ for every solution $X$ of the system, and furthermore suppose that $V$ is definite positive.
Then $0$ is a stable equilibrium.
Sadly, all the proofs I have seen feel rather obscure to me.
Do you know of any good, intuitive proof of the theorem? (or would you be so kind to outline one yourself?)
Recall $0$ is a stable equilibrium if for any $\epsilon>0$ there exists $\delta>0$ such that $\|X(t_1)\| < \epsilon$ for all $X$ satisfying $\|X(t_0)\| < \delta$ for some $t_0<t_1.$
Given $\epsilon>0,$ by continuity of $V$ there exists $\delta>0$ such that $V(x)<\min_{\|x'\|=\epsilon}V(x')$ for $\|x\|<\delta.$ Assume $\|X(t_0)\|<\delta.$ For any $t>t_0$ we have $V(X(t))\leq V(X(t_0))$ and hence $\|X(t)\|\neq\epsilon.$ By the intermediate value theorem, $\|X(t)\|$ will never exceed $\epsilon$ with $t\geq t_0.$