Heat equation

Remarks

Many authors write $\partial_t u = \alpha \Delta u$ with a constant $\alpha > 0$, the diffusivity; the change of variable $t \mapsto \alpha t$ reduces it to $\alpha = 1$. Example: for $k > 0$, $u(x, t) = e^{-k^2 t} \sin(k x)$ solves the heat equation on $\R \times (0, \infty)$, since $\partial_t u = -k^2 u$ and $\partial_x^2 u = -k^2 u$. Non-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 $\partial_t u = -\Delta u$. The equation is usually completed by an initial condition $u(x, 0) = g(x)$ and, when $\Omega \ne \R^n$, by boundary conditions on $\partial \Omega$.