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  "body": "\n## Definition\n\nLet $f \\colon \\R \\times \\R \\to \\R$ be [[smooth|0V33BNKZ9S]], and consider the family of scalar equations $x' = f(x, \\mu)$ with parameter $\\mu$.\nAssume that $f$ is [[odd|0FP31SX5EA]] in $x$: $f(-x, \\mu) = -f(x, \\mu)$ for all $x$ and $\\mu$.\nThen $f(0, \\mu) = 0$, so $0$ is an [[equilibrium point|0G4X36AE0X]] for every $\\mu$.\nThe family has a *pitchfork bifurcation* at $(x, \\mu) = (0, 0)$ if:\n\n1. $f_x(0, 0) = 0$;\n2. $f_{x\\mu}(0, 0) \\ne 0$;\n3. $f_{xxx}(0, 0) \\ne 0$.\n\n## Remarks\n\nUnder these conditions, besides $x = 0$, the equilibria near $(0, 0)$ form a curve $\\mu = m(x)$ with $m$ even, $m(0) = 0$ and $m''(0) = -f_{xxx}(0, 0) / (3 f_{x\\mu}(0, 0))$.\nSo on one side of $\\mu = 0$ there is a symmetric pair of new equilibria, and the stability of $0$ changes at $\\mu = 0$.\nThe bifurcation is supercritical when the new equilibria are stable and subcritical when they are unstable.\n\nExample: $f(x, \\mu) = \\mu x - x^3$.\nIt is odd in $x$, $f_x(0, 0) = 0$, $f_{x\\mu} = 1$ and $f_{xxx} = -6$.\nThe new equilibria are $\\pm\\sqrt{\\mu}$ for $\\mu \u003e 0$, and they are stable since $f_x(\\pm\\sqrt{\\mu}, \\mu) = -2\\mu \u003c 0$.\nThe equation $x' = \\mu x + x^3$ gives the subcritical case.\n\nNon-example: $f(x, \\mu) = \\mu x - x^2$ is not odd in $x$, so the symmetry hypothesis fails; it gives a transcritical bifurcation.\nWithout the symmetry, a small perturbation such as $\\mu x - x^3 + \\varepsilon$ breaks the pitchfork into a saddle-node and a separate branch.\n",
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    "name": "Pitchfork bifurcation",
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  "name": "Pitchfork bifurcation",
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