I am learning Mean value property (MVP) of the heat equation. MVP of Laplace equation was relatively easy to understand I think it is because of the spherical symmetry. But I am not able to appreciate the MVP of heat equation. It's not very easy to imagine the "heat ball" in the following theorem from a note:
Here are questions:
- How do I define a heat ball?
- How does it actually look like?
The "heat ball" is defined as it is in the note you cited which is bases on Evans's Partial Differential Equations Chapter 2.3.
For fixed $x\in{\bf R}^n$, $t\in{\bf R}$ and $r>0$, we define $$ E(x,t;r)=\left\{(y,s)\in {\bf R}^{n+1}\bigg|\; s\leqslant t,\ \dfrac{1}{(4\pi(t-s))^{n/2}}\exp\left({-\dfrac{|x-y|^2}{4(t-s)}}\right)\geqslant\frac{1}{r^n}\right\}. $$
The Wikipedia article Mean-value property for the heat equation also gives a similar definition.
Note that in the definition, one should replace $s\leqslant t$ with $s<t$. To get some ideas of what such "ball" would look like, consider $n=1$ and $$ E(0,0;1)=\left\{(y,s)\in {\bf R}^{2}\bigg|\; s<0,\ \dfrac{1}{(4\pi(-s))^{1/2}}\exp\left({-\dfrac{|-y|^2}{4(-s)}}\right)\geqslant 1\right\}\\ =\left\{(y,s)\in{\bf R}^2\bigg|\; 0<-s\leqslant\frac{1}{4\pi}, y^2\leq 2s\log(-4\pi s)\right\} $$ To get $-s\leqslant\frac{1}{4\pi}$ one can simply observe that $\exp\left(-\dfrac{|-y|^2}{4(-s)}\right)\leqslant 1$ for $s<0$ and thus $$ \sqrt{-4\pi s}\leqslant 1. $$ On the other hand, taking the logarithm of $\dfrac{1}{(4\pi(-s))^{1/2}}\exp\left({-\dfrac{|-y|^2}{4(-s)}}\right)\geqslant 1$ gives $$ y^2\leq 2s\log(-4\pi s). $$
The boundary of the heat ball is like this:
There is an illustration on page 53 of PDE by Evans. Nothing mysterious, just an ellipsoid-like shape with the "center" $(x,t)$ located at the center on the top boundary (not in the interior, as for elliptic PDE).
The definition is in the book you are reading, formula (23).
