Explanation on a proof of the mean-value property for the heat equation. Let $U\subset\mathbb{R}^n$ be an open set, $\Phi$ the fundamental solution of heat equation, $T>0$, $r>0$, $x\in\mathbb{R}^n$ and $t\in\mathbb{R}$. Defines $U_T=U\times(0,T]$ and $$E(x,t;r)=\{(y,s)\in\mathbb{R}^{n+1};\:s\leq t,\;\Phi(x-y,t-s)\geq r^{-n}\}.$$
I would like some explanations about the proof of the following

Theorem (mean-value property for the heat equation): Let $u\in C^2_1(U_T)$ solve the heat equation. Then $$u(x,t)=\frac{1}{4r^n}\iint_{E(x,t;r)}u(y,s)\frac{|x-y|^2}{(t-s)^2}\,dy\,ds.\tag{1}$$ 

Obviously, from $(1)$ follows
$$u(0,0)=\frac{1}{4r^n}\iint_{E(0,0;r)}u(y,s)\frac{|y|^2}{s^2}\,dy\,ds.\tag{2}$$ 
However, the proofs that I saw (for exemple this one, page 4) proves $(2)$. So, my question is: how to prove that $(2)\Rightarrow(1)$?
Thanks.
 A: The reason (2) implies (1) is that the heat equation is invariant under translation in space-time. That is, if $u$ satisfies the PDE, then for any fixed $(x_0,t_0)$ the translated function $\tilde u(x,t)=u(x+x_0,t+t_0)$ also satisfies the PDE. This is easy to check by taking the derivatives of $\tilde u$ with the chain rule. 
So, to prove (1) at the point $(x_0,t_0)$, one can apply (2) to $\tilde u$ defined above, obtaining 
$$\tilde u(0,0)=\frac{1}{4r^n}\iint_{E(0,0;r)}\tilde u(y,s)\frac{|y|^2}{s^2}\,dy\,ds \tag{2} $$
Then return to $u$:
$$\tilde u(x_0,t_0)=\frac{1}{4r^n}\iint_{E(0,0;r)} u(y+x_0,s+t_0)\frac{|y|^2}{s^2}\,dy\,ds  $$
and finally change the variables in the integral: $\tilde y=y+x_0$, $\tilde s=s+t_0$.
You may wonder why do all of this, instead of simply proving (1) directly (which is not very different from proving (2)). But with enough experience, people notice the translation-invariance in their problems right away, and simply say: we can take $(x,t)=(0,0)$ without losing generality, by translating the function. (Or something of the kind. Or say nothing at all, considering it obvious.) 
TeXnical note: the proper way to number displayed formulas here is with \tag{2}, as I did above.
