Cole–Hopf transformation

Remarks

Proof of the basic property: put $\psi = \log \varphi$, so $u = -2\nu \psi_x$. The heat equation for $\varphi$ becomes $\psi_t = \nu \psi_{xx} + \nu \psi_x^2$. Differentiating in $x$ and multiplying by $-2\nu$ gives $u_t = \nu u_{xx} - u u_x$.

Inverse direction: given initial values $u_0$, take $\varphi_0(x) = \exp\left(-\frac{1}{2\nu} \int_0^x u_0(y)\, dy\right)$. Solving 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 $I = \R$.

Example: $\varphi(x, t) = 1 + e^{-k(x - \nu k t)}$ with $k > 0$ solves $\varphi_t = \nu \varphi_{xx}$, since both sides equal $\nu k^2 e^{-k(x - \nu k t)}$. Its transformation is $u = \nu k \left(1 - \tanh\left(\tfrac{k}{2}(x - \nu k t)\right)\right)$. This is a traveling wave of speed $\nu k$ from $2\nu k$ to $0$.

Non-example: $\varphi(x, t) = \sin x$ is not allowed on $\R$, since it is not positive. Its transformation $-2\nu \cot x$ is undefined at the zeros of $\sin x$.