# What physical information does the mean value property of heat equation convey?

I'm reading through the Evans' book on PDE, the chapter on heat equation. The definitions are the same as here.

I see that mean value property of heat equation is useful for proving maximum principle and various uniqueness results, but I'm curious are there any physical interpretations of definition of heat ball and mean value property itself?

It seems to me that definition of heat ball $E(\mathbf x, t; r)$ should have something to do with an idea of bringing a point source to the point $\mathbf x$ at the time $t$, but the fact that we are looking only at the past (points $(y, s)$ of the space-time with $s \leq t$) confuses me.

Can someone help?

• – Christian Blatter Dec 20 '14 at 13:52
• @ChristianBlatter I'm aware of intuitive interpretation of heat equation, but nonetheless I can't figure out how to explain in simple words what these ''heat balls'' are... Maybe I'm missing something obvious. – ante.ceperic Dec 20 '14 at 19:26

Your textbook says the fundamental solution is $\Phi(x,t) = \frac{1}{(4\pi t)^{n/2}}e^{-\frac{|x|^2}{2t}}$ then we plug into the definition of the energy ball:

\begin{eqnarray} E(x,t,r) &=& \left\{ (y,s): s \leq t \text{ and } \frac{1}{(4\pi (t-s))^{n/2}}e^{-\frac{1}{2}\frac{|x-y|^2}{t-s}} \geq \frac{1}{r^2}\right\} \\ &=& \bigcup_{s \leq t} \left\{ (y,s): |x-y|^2 \leq -2(t-s)\log \frac{(4\pi (t-s))^{n/2}}{r^2}\right\} \end{eqnarray}

So the energy ball lives in (plain-old Galilean) space-time $\mathbb{R}^n \times \mathbb{R}$ and it's fibered by Euclidean balls. This is something like a Euclidean light-cone in physics. ### Stochastic Point of View

One can show the heat equation can be solved by random walk. I believe the formula is $f(x) = \mathbb{E}[f(B_\tau)]$ where $\tau$ is the hitting time of a Brownian motion with $B_0 = x$ hitting the boundary $B_\tau \in \partial V$.

One can imagine heat diffusing by way of a Brownian motion. See Greg Lawler Heat Equation and Random Walk.

Actually come to think of it, you solve the heat equation by convolving the initial solution $u(t=0, x)$ with the heat kernel $\frac{1}{\sqrt{t}}e^{-x^2/t}$. Convolving with the heat kernel is as if diffusing the original solution via Brownian motion.

• for $t \ll 1$ this is a point distribution $\frac{1}{\sqrt{2\pi / t}}e^{-x^2/t} \to \delta(t)$
• for $t \gg 1$ this is uniform across space $\frac{1}{\sqrt{2\pi/ t}}e^{-x^2/t} \to \frac{1}{\sqrt{t}}$.

For linear PDE, that convolution can be thought of as just the Minkowski sum, or the theory of "wavefronts"as mentioned in the book of Hormander: Analysis of Linear Partial Differential Operators I or Tao's blog Computing Convolutions of Measures.

The mean value property is rather intuitive in the stochastic point of view:

$$\Delta^2 f \approx \frac{f(x+h,y)+ f(x,y+h)+f(x-h,y)+f(x,y-h) }{4} = \mathbf{E}f\big((x,y) + (\Delta x, \Delta y)\big)$$

where $(\Delta x, \Delta y)= (\pm 1, \pm 1)$ each with probability $\mathbb{P}=\frac{1}{4}$. Or in a very geometric way the Laplacian is just the average of the values of $f$ over a circle:

$$\Delta^2f = \frac{1}{2\pi} \oint f\bigg((x,y) + \epsilon(\cos \theta, \sin \theta)\bigg)d\theta = f(x)$$

Mean Value Property

The physical process behind heat transfer is Brownian motion of particles. Brownian motion is a type of martingale. Martingales are random variables that satisfy the mean value property (as such, they model harmonic functions.)

Intuitively speaking, heat is the random motion of particles. These particles have no reason to pick a favorite direction, so therefore as an aggregate they travel in a harmonic fashion.

Heat Ball

Unlike the name suggests, I don't think of a heat ball as a spatial object. It's more like the information available to a particle, based on the particles it could have interacted with in the past.

• pre-kidney's last remark adequately answers the question. The heat ball, much like all characteristic objects of these equations (like the plain ball for the Laplace equation), are not spatial objects. They are characteristic objects stemming from the symmetries of the equation; in the context of mean value theorems they are best viewed as the right 'shape of information distribution'. The shape of actual spatial observables will depend heavily on 1) the manifold where the equation is being solved and 2) the initial-boundary conditions imposed. – guest Jan 1 '15 at 6:12
• One more remark: john mangual is also correct in drawing the parallel (more than parallel actually) with the light cone, as a similar case of 'shape of distribution of information'. – guest Jan 1 '15 at 6:15