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  "body": "\n## Definition\n\nLet $U$ be an open subset of $\\R^n$, let $f \\colon U \\to \\R^n$ be locally Lipschitz, and consider the [[autonomous system|07BNRFQ0PJ]] $x' = f(x)$.\nLet $x_0 \\in U$ be an [[equilibrium point|0G4X36AE0X]].\nA solution $\\gamma \\colon \\R \\to U$ is homoclinic to $x_0$ if it is not constant and\n$$\\lim_{t \\to -\\infty} \\gamma(t) = x_0 = \\lim_{t \\to +\\infty} \\gamma(t).$$\nA homoclinic orbit is the [[trajectory|07TZ2Q4C1M]] $\\gamma(\\R)$ of such a solution.\n\n## Remarks\n\nA homoclinic orbit lies in the intersection of the stable and the unstable set of $x_0$, and it does not contain $x_0$.\nExample: for $x' = y$, $y' = x - x^2$, the function $\\gamma(t) = (x(t), x'(t))$ with $x(t) = \\frac{3}{2} \\operatorname{sech}^2(t/2)$ is a solution.\nIndeed $x'' = \\frac{3}{2} s - \\frac{9}{4} s^2 = x - x^2$ with $s = \\operatorname{sech}^2(t/2)$, and $x(t) \\to 0$, $x'(t) \\to 0$ as $t \\to \\pm\\infty$, so $\\gamma$ is homoclinic to the origin.\nIts trajectory is the part of the curve $y^2 = x^2 - \\frac{2}{3} x^3$ with $x \u003e 0$.\nNon-example: for the pendulum $x' = y$, $y' = -\\sin x$, the solution with $x(t) \\to -\\pi$ as $t \\to -\\infty$ and $x(t) \\to \\pi$ as $t \\to +\\infty$ joins two different equilibria in $\\R^2$.\nSo it is a heteroclinic orbit, not a homoclinic one.\nA periodic orbit is also not homoclinic, since it does not tend to an equilibrium.\n\n## Explanations\n\n### L1 human\n\nA homoclinic orbit is a path that leaves an equilibrium and comes back to the same one, taking infinite time in both directions.\nIn the plane it looks like a loop pinned at a saddle point.\n",
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