The above is taken from Evans book. I can not understand how this scaling indicates that the $\frac{r^2}{t}\;\;$ is invariant. I would appreciate if somebody could explain to me this part with more details. Thanks in advance!
EDIT: I thought also to do this: I chose $λ^2 t\;=\; 1$ and so I found $λ$ to be equal $\frac{1}{\sqrt t}$. Then by substitution to $u$ I concluded $u(λx,λ^2 t)=u(\frac{x}{\sqrt t} ,1)=v(\frac{x}{\sqrt t})$ Does this somehow imply that $v(\frac{{\vert x \vert}^2}{ t})$ ?
Note that if $u(x,t)$ is a solution of
$$u_t = \Delta u$$
then $w(x,t) = u(\lambda x, \lambda^2 t)$ also solves the heat equation:
$$w_t(x,t) = \lambda^2 u_t(\lambda x, \lambda^2 t) = \lambda^2 \Delta u(\lambda x, \lambda^2 t) = \Delta w(x,t).$$
This scaling $x\to \lambda x$, $t\to \lambda^2 t$ is thus an invariance of the heat equation. In other words this is a symmetry of the PDE. Then one observes that by taking $y=\lambda x$ and $s = \lambda^2 t$ we have $\frac{|y|^2}{t} = \frac{|x|^2}{t}$ is independent of $\lambda$. This independence from $\lambda$ is important because the variable $\xi = \frac{|x|^2}{t}$ takes advantage of the symmetry property of the heat equation. In fact, $\frac{r^2}{t}$ is not an invariant as you ask; rather the $\lambda$-independence leads us to conjecture a solution of the heat equation of the form $v(\xi)=v\left(\frac{|x|^2}{t}\right).$
I admit that this is not the most satisfying answer as it still requires some divine inspiration! Perhaps someone has a more rigorous way to "guess" $v(\xi)$...