Let $U\subset\mathbb
Theorem (mean-value property for the heat equation): Let $u\in C^2_1(U_T)$ solve the heat equation. Then $$u(x,t)=\frac\iint_u(y,s)\frac<|x-y|^2>\,dy\,ds.\tag$$
Obviously, from $(1)$ follows $$u(0,0)=\frac<1>\iint_u(y,s)\frac<|y|^2>\,dy\,ds.\tag$$ 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.1>
$\begingroup$ Is there something preventing you from following the exact same proof as in your reference for arbitrary $x,t$? $\endgroup$
Commented Oct 6, 2013 at 19:10The 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 $$\tilde u(0,0)=\frac\iint_\tilde u(y,s)\frac<|y|^2>\,dy\,ds \tag $$ Then return to $u$: $$\tilde u(x_0,t_0)=\frac\iint_ u(y+x_0,s+t_0)\frac<|y|^2>\,dy\,ds $$ 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 , as I did above.