Lyapunov's stability theorem

Statement

Let $U$ be an open subset of $\mathbb{R}^n$, let $f : U \to \mathbb{R}^n$ be locally Lipschitz, and let $x^* \in U$ be an equilibrium point of $x' = f(x)$. Let $W \subseteq U$ be an open neighbourhood of $x^*$ and let $V : W \to \mathbb{R}$ be a Lyapunov function for $x^*$. (1) Then $x^*$ is Lyapunov stable. (2) If $V$ is a strict Lyapunov function, then $x^*$ is asymptotically stable. (3) If $W = U = \mathbb{R}^n$, $V$ is strict, and $V(x) \to \infty$ as $\|x\| \to \infty$, then $x^*$ is globally asymptotically stable.

Remarks

Proof idea for (1): choose $\varepsilon$ so small that the sphere $S$ of radius $\varepsilon$ around $x^*$ lies in $W$. Let $m > 0$ be the minimum of $V$ on $S$. A solution that starts where $V < m$ keeps $V < m$, so it cannot reach $S$. For the damped harmonic oscillator $x' = y$, $y' = -x - c y$ with $c > 0$, the energy $V = \frac{1}{2}(x^2 + y^2)$ has $\dot{V} = -c y^2 \le 0$. This gives stability, and LaSalle's invariance principle gives asymptotic stability. The method needs no solution formula. The difficulty is to find a suitable $V$.

History

The method is named after Aleksandr Lyapunov, who introduced it in his study of the stability of motion.

Explanations

L1 human

If some energy-like quantity decreases along all motions near an equilibrium, the motions stay near it. If the quantity decreases strictly, the motions must also run down to the equilibrium.