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  "body": "\n## Definition\n\nLet $U \\subseteq \\R^n$ be open, let $f \\colon U \\to \\R^n$ be [[continuously differentiable|0VW90BBSJ4]], and let $\\varphi$ be the [[flow|02903SSA48]] of $x' = f(x)$.\nAn [[equilibrium point|0G4X36AE0X]] $x^*$ is *exponentially stable* if there are constants $\\delta \u003e 0$, $C \\ge 1$ and $\\alpha \u003e 0$ such that for every $x_0 \\in U$ with $\\norm{x_0 - x^*} \u003c \\delta$:\n\n1. the solution $\\varphi_t(x_0)$ exists for all $t \\ge 0$;\n2. $\\norm{\\varphi_t(x_0) - x^*} \\le C e^{-\\alpha t} \\norm{x_0 - x^*}$ for all $t \\ge 0$.\n\n## Remarks\n\nAn exponentially stable equilibrium is asymptotically stable.\nIf $f$ is of class $C^2$, then $x^*$ is exponentially stable if and only if every eigenvalue of $Df(x^*)$ has negative real part.\n\nExample: $x' = -x$ on $\\R$ with $x^* = 0$.\nThe solution is $x_0 e^{-t}$, so clause 2 holds with $C = 1$ and $\\alpha = 1$, for any $\\delta$.\n\nNon-example: $x' = -x^3$ on $\\R$ with $x^* = 0$.\nThe solution is $x(t) = x_0 / \\sqrt{1 + 2 x_0^2 t}$, since $\\frac{d}{dt} x^{-2} = 2$.\nIt tends to $0$, but $x(t) e^{\\alpha t} \\to \\infty$ for each $\\alpha \u003e 0$ and $x_0 \\ne 0$, so clause 2 fails.\n\nFor a time-dependent system $x' = f(t, x)$ with $f(t, 0) = 0$, the origin is uniformly exponentially stable if $\\norm{x(t)} \\le C e^{-\\alpha (t - t_0)} \\norm{x(t_0)}$ for all $t \\ge t_0$, all initial times $t_0$ and all $\\norm{x(t_0)} \u003c \\delta$.\n",
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