Radially symmetric heat equation with Gaussian initial conditions

Question

$\rho(\textbf x,t)$ obeys the heat equation i.e. $\rho_t=D \Delta \rho$ with initial conditions $\rho (\textbf x, 0) = \rho_0 e^{-r^2/a^2}$. I have tried separation of variables but this only gives an exponential for time and $\frac{1}{r}\sin{r\sqrt{\lambda/D}}$ for the radial function. Do I just take the radial function to be the initial condition and use the exponential for the time function? The time function I get is $e^{-\lambda t}$. I would normally sum up all of these with the different values of lambda but I don't know how to do this with the Gaussian initial conditions on an unbounded domain. I then tried assuming it was of the form $T(t)\rho_0 e^{-r^2/a^2}$ but this led to T having radial dependence so that doesn't work either. I have read about doing something to do with Fourier transforms but I don't understand it.

Answer 1

What happens with a Gaussian bulge of heat over time? It spreads. Suppose it keeps its Gaussian shape. This suggests looking at solutions of the form $$f(t) e^{-r^2 g(t)}$$ for some functions $f$ and $g$. Take $D=a=\rho_0=1$ to begin with (such constants only clog the computations) and try to solve the equation in this form.