# mean value theorem for heat equation

Hi I am looking at the proof of mean value property for heat equations (evans chapter 2 theorem 3)

Again, I got through every step except the very last line of the proof $$\phi(r)=\lim_{t\to 0}\Phi(t)=u(0,0)(\cdots)\cdots$$

I just don't understand why is $u(0,0)$, why not $u(y,s)$, assuming we meant to plug that back to the definition of heat ball...

I look through many notes, for instance p11 http://math.uchicago.edu/~may/REU2014/REUPapers/Ji.pdf

None explains why $u(0,0)$.

• No. U is just assumed to solved the heat eqn – math101 Jun 17 '15 at 3:16

By Evans, in the final lines, we have:

$$\phi (r)=\lim_{t\to 0}\phi(t)=\lim_{t\to 0} \frac{1}{t^n}\int\int_{E(t)}u(y,s)\frac{|y|^2}{s^2}dyds.$$

Changing of variables, we have

$$\phi (r)=\lim_{t\to 0} \int\int_{E(1)}u(ty,t^2s)\frac{|y|^2}{s^2}dyds,$$

So,

$$\phi (r)=u(0,0)\lim_{t\to 0} \int\int_{E(1)}\frac{|y|^2}{s^2}dyds,$$

• Does your $r$ mean $t$ – math101 Jun 17 '15 at 3:19
• Thanks for that, and the prompt editing – math101 Jun 17 '15 at 3:27
• I fixed, thanks. So, you just need to do the change of variable. – Irddo Jun 17 '15 at 3:27