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  "body": "\n## Statement\n\nLet $U$ 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 $x' = f(x)$.\nLet $W \\subseteq U$ be an open neighbourhood of $x^*$ and let $V : W \\to \\mathbb{R}$ be a [[Lyapunov function|04P7B5XC22]] for $x^*$.\n(1) Then $x^*$ is [[Lyapunov stable|092S6VBECQ]].\n(2) If $V$ is a strict Lyapunov function, then $x^*$ is [[asymptotically stable|05Y1JAM3Y5]].\n(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.\n\n## Remarks\n\nProof idea for (1): choose $\\varepsilon$ so small that the sphere $S$ of radius $\\varepsilon$ around $x^*$ lies in $W$. Let $m \u003e 0$ be the minimum of $V$ on $S$. A solution that starts where $V \u003c m$ keeps $V \u003c m$, so it cannot reach $S$.\nFor the [[damped harmonic oscillator|0RNXVVTJTJ]] $x' = y$, $y' = -x - c y$ with $c \u003e 0$, 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|0659J5X7B5]] gives asymptotic stability.\nThe method needs no solution formula. The difficulty is to find a suitable $V$.\n\n## History\n\nThe method is named after Aleksandr Lyapunov, who introduced it in his study of the stability of motion.\n\n## Explanations\n\n### L1 human\n\nIf some energy-like quantity decreases along all motions near an equilibrium, the motions stay near it.\nIf the quantity decreases strictly, the motions must also run down to the equilibrium.\n",
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