Show that if $u(x,t)$ satisfies $u_t = k \Delta u$ in a bounded region G, then for any $L>0$, $u(Lx, L^2t)$ solves the same equation for $x \in L^{-1}G$, where $L^{-1}G$ is the set of points $L^{-1}y$ for $y \in G$. Here $k$ is a positive constant.
I am considering $L$ to be some form of a linear operator. In terms of sets, this statement makes sense, but I'm not quite too sure how to definitively show it.
Thanks in advance!
To be sure, you need to make difference between new and old variables, between new and old unkowns. Namely, let $u_t(x,t)=k\Delta_x u(x,t)$ for $x\in G$ and $t\in (0,T)$. Consider a function $\widetilde{u}(y,s)\overset{\rm def}{=}u(Ly,L^2s)$. It is clear that $\widetilde{u}_s(y,s)=k\Delta_y \widetilde{u}_s(y,s)$ for $Ly=x\in G$ and $L^2s=t\in (0,T)$, i.e., for $y\in L^{-1}G$ and $s\in (0,L^{-2}T)$.