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  "body": "\n## Definition\n\nLet $n \\ge 1$, let $\\Omega \\subseteq \\R^n$ be an open set, and let $T \u003e 0$.\nLet $u \\colon \\Omega \\times (0, T) \\to \\R$ have one continuous [[partial derivative|09EGC6F8PT]] in $t$ and two continuous partial derivatives in $x$.\nThe *heat equation* on $\\Omega \\times (0, T)$ is the [[parabolic|1CCTH8X2FT]] equation\n$$\\frac{\\partial u}{\\partial t}(x, t) = \\Delta u(x, t), \\qquad x \\in \\Omega, \\ 0 \u003c t \u003c T.$$\nHere $\\Delta = \\partial_{x_1}^2 + \\cdots + \\partial_{x_n}^2$ is the [[Laplacian|0E0B6RA8K0]] in $x$.\n\n## Remarks\n\nMany authors write $\\partial_t u = \\alpha \\Delta u$ with a constant $\\alpha \u003e 0$, the diffusivity; the change of variable $t \\mapsto \\alpha t$ reduces it to $\\alpha = 1$.\nExample: for $k \u003e 0$, $u(x, t) = e^{-k^2 t} \\sin(k x)$ solves the heat equation on $\\R \\times (0, \\infty)$, since $\\partial_t u = -k^2 u$ and $\\partial_x^2 u = -k^2 u$.\nNon-example: $u(x, t) = e^{k^2 t} \\sin(k x)$ is not a solution, since $\\partial_t u = k^2 u$ while $\\partial_x^2 u = -k^2 u$; it solves the backward heat equation $\\partial_t u = -\\Delta u$.\nThe equation is usually completed by an initial condition $u(x, 0) = g(x)$ and, when $\\Omega \\ne \\R^n$, by boundary conditions on $\\partial \\Omega$.\n",
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