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  "body": "\n## Definition\n\nLet $\\Omega \\subseteq \\R^n$ be open and let $\\varepsilon \u003e 0$.\nLet $W(u) = \\frac{1}{4}(1 - u^2)^2$, so that $W'(u) = u^3 - u$.\nThe *Allen–Cahn equation* is the [[reaction–diffusion equation|TAAHWE917M]] for $u \\colon \\Omega \\times (0, T) \\to \\R$:\n$$u_t = \\Delta u - \\frac{1}{\\varepsilon^2} W'(u) = \\Delta u - \\frac{1}{\\varepsilon^2}(u^3 - u),$$\nwhere $\\Delta$ is the [[Laplacian|0E0B6RA8K0]] in $x$.\n\n## Remarks\n\nExample: with $n = 1$, the function $u(x) = \\tanh\\!\\big(x/(\\sqrt{2}\\,\\varepsilon)\\big)$ is a stationary solution.\nIndeed, with $s = x/(\\sqrt{2}\\,\\varepsilon)$ one finds $u'' = -\\varepsilon^{-2}\\operatorname{sech}^2(s)\\tanh(s) = \\varepsilon^{-2}(u^3 - u)$.\nNon-example: the Cahn–Hilliard equation $u_t = \\Delta(u^3 - u - \\varepsilon^2 \\Delta u)$ is of fourth order, so it is not of this form.\nThe equation is the $L^2$ gradient flow of the energy\n$$E(u) = \\int_\\Omega \\Big( \\frac{1}{2}\\abs{\\nabla u}^2 + \\frac{1}{\\varepsilon^2} W(u) \\Big)\\, dx,$$\nso $E$ does not increase along smooth solutions with Neumann boundary conditions.\nSome authors scale the equation as $u_t = \\varepsilon^2 \\Delta u - W'(u)$ instead.\n",
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