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  "body": "\n## Definition\n\nLet $\\nu \u003e 0$, let $I \\subseteq \\R$ be an open interval, and let $T \u003e 0$.\nLet $\\varphi \\colon I \\times (0, T) \\to (0, \\infty)$ be a positive function with continuous [[partial derivatives|09EGC6F8PT]] $\\varphi_t$, $\\varphi_x$, $\\varphi_{xx}$.\nThe *Cole–Hopf transformation* of $\\varphi$ is the function\n$$u = -2\\nu\\, \\frac{\\varphi_x}{\\varphi} = -2\\nu\\, \\partial_x \\log \\varphi.$$\nIts basic property: if $\\varphi$ solves the [[heat equation|VP2HN86VAG]] $\\varphi_t = \\nu \\varphi_{xx}$, then $u$ solves the viscous [[Burgers' equation|NEGHXN8RE1]] $u_t + u u_x = \\nu u_{xx}$.\n\n## Remarks\n\nProof of the basic property: put $\\psi = \\log \\varphi$, so $u = -2\\nu \\psi_x$.\nThe heat equation for $\\varphi$ becomes $\\psi_t = \\nu \\psi_{xx} + \\nu \\psi_x^2$.\nDifferentiating in $x$ and multiplying by $-2\\nu$ gives $u_t = \\nu u_{xx} - u u_x$.\n\nInverse direction: given initial values $u_0$, take $\\varphi_0(x) = \\exp\\left(-\\frac{1}{2\\nu} \\int_0^x u_0(y)\\, dy\\right)$.\nSolving the heat equation with data $\\varphi_0$ and transforming gives the solution of Burgers' equation with data $u_0$, for bounded continuous $u_0$ on $I = \\R$.\n\nExample: $\\varphi(x, t) = 1 + e^{-k(x - \\nu k t)}$ with $k \u003e 0$ solves $\\varphi_t = \\nu \\varphi_{xx}$, since both sides equal $\\nu k^2 e^{-k(x - \\nu k t)}$.\nIts transformation is $u = \\nu k \\left(1 - \\tanh\\left(\\tfrac{k}{2}(x - \\nu k t)\\right)\\right)$.\nThis is a [[traveling wave|M37M411KP7]] of speed $\\nu k$ from $2\\nu k$ to $0$.\n\nNon-example: $\\varphi(x, t) = \\sin x$ is not allowed on $\\R$, since it is not positive.\nIts transformation $-2\\nu \\cot x$ is undefined at the zeros of $\\sin x$.\n\n## Explanations\n\n### L1 human\n\nThe transformation takes the logarithmic derivative of a solution of the heat equation.\nThis one formula converts a linear equation into a nonlinear one.\nRead backwards, it solves a nonlinear equation with the tools of a linear one.\n",
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