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  "body": "\n## Definition\n\nLet $D \\subseteq \\R^k$, let $f \\colon D \\to D$ be [[continuous|0X3WGX95FE]], and consider the difference equation $x_{n+1} = f(x_n)$.\nLet $x^* \\in D$ be an [[equilibrium|D62PEKHRV5]], so $f(x^*) = x^*$.\nLet $G \\subseteq D$ be a [[neighbourhood|0ZCY0Z80B6]] of $x^*$ relative to $D$.\nA *Lyapunov function* for the equation on $G$ is a continuous function $V \\colon G \\to \\R$ such that:\n\n1. $V(x^*) = 0$.\n2. $V(x) \u003e 0$ for every $x \\in G$ with $x \\ne x^*$.\n3. $\\Delta V(x) = V(f(x)) - V(x) \\le 0$ for every $x \\in G$ with $f(x) \\in G$.\n\nThe function $V$ is a *strict* Lyapunov function if moreover $\\Delta V(x) \u003c 0$ for every $x \\ne x^*$ in $G$ with $f(x) \\in G$.\n\n## Remarks\n\nExample: for $f(x) = x / (1 + x^2)$ on $\\R$ with $x^* = 0$, take $V(x) = x^2$.\nThen $V(0) = 0$, $V(x) \u003e 0$ for $x \\ne 0$, and $\\Delta V(x) = x^2 \\left( (1 + x^2)^{-2} - 1 \\right) \u003c 0$ for $x \\ne 0$, so $V$ is a strict Lyapunov function.\n\nNon-example: for $f(x) = 2x$ and $V(x) = x^2$, clauses 1 and 2 hold but $\\Delta V(x) = 3x^2 \u003e 0$ for $x \\ne 0$, so clause 3 fails.\n\nThe quantity $\\Delta V$ is the forward difference of $V$ along solutions, the analogue of the derivative $\\dot V$ along solutions of a differential equation.\nIf a Lyapunov function exists, the equilibrium is stable; if a strict one exists, it is asymptotically stable.\nFor a linear map $f(x) = Ax$ one often uses a quadratic form $V(x) = x^T P x$ with $P$ positive definite and $A^T P A - P$ negative definite.\n",
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