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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!

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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 $s\in (0,L^{-2}T)$.

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