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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:

enter image description here

Here are questions:

  • How do I define a heat ball?
  • How does it actually look like?
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    @abatkai : I have added the reference . Thank you2012-06-22

2 Answers 2

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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. 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:

enter image description here

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    I have just noticed that I'm responding to a comment about one year old.2017-08-11
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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).

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    @Theorem For each fixed value of time variable $s$, you get a two-dimensional slice which is a circle. The radius of the circle depends on $s$: it drops to zero when $s=t$ and when $s$ is much smaller than $t$.2012-06-22