The effect of a constant heat source depends on dimension

by greenturtle3141, Mar 4, 2025, 9:51 PM

Reading Difficulty: 3-6/5

Prerequisites: You should probably know what the heat equation is, and hopefully you know how to solve it...

It is a well-known fact among those that have studied PDEs that the nature of waves depends wildly on the dimension of space. Particularly, it depends on the parity of the dimension $d$ : If we take space to be $\mathbb{R}^d$ where $d \geq 3$ is odd, then solutions to the wave equation
$\partial_t^2 u - \Delta u = 0$ behave the way we're used to: They travel in a direction and don't go backwards because they feel like it. In other words, if you yell at someone for a second, they'll only hear you for about a second because your yell will travel past them. However, if $d$ is even, then is not true: If you yell at someone for a second, then your yell will keep traveling in all directions for some reason, and so they'll hear you forever! (But your yell will get quieter at an exponential rate.)

For some bizarre reason, there is a similar fascinating phenomenon about the heat equation that I haven't yet come across. In fact, I've been unable to find a source for this.

A Thought Experiment

Go deep into the woods on a cold night and build a fire somewhere. Let's assume that the fire neither spreads nor dies out, so that it is constant in time as a heat source. Do you expect the distribution of heat to approach some stable distribution? Or will it keep accumulating and make the woods infinitely hot over time?

As someone that has seen a fire before, I'm pretty sure the latter doesn't happen. I mean, does a fireplace make a room infinitely hot? I thought not. This setting is the natural physical interpretation for the following PDE in one dimension:
$\begin{cases}\partial_tu - \partial_x^2 u = 1_{(0,1)}\\ u\,|_{t=0} = 0\end{cases}$ Here the ``fire" is built on the interval $(0,1)$ and persists as a heat source. If God is good, surely $\lim_{t \to \infty} u(\cdot,t)$ should converge pointwise to something.

To my horror, it turns out that $\lim_{t \to \infty} u(\cdot,t) \equiv +\infty$ , meaning that the fire makes space infinitely hot. Huh?

It turns out that the reason why this limit defies our intuition is because of dimension.

Theorem: Given a constant heat source on $\mathbb{R}^d$ , the heat distribution approaches a stable distribution for $d \geq 3$ . But for $d = 1,2$ , heat accumulates infinitely in the sense that space gets arbitrarily hot.

An Explanation

Let us begin with a slightly scuffed argument for this principle by considering a "delta" heat source: That is, if $u$ solves
$\begin{cases}\partial_t u - \Delta u = \delta_0 \\ u\,|_{t=0} = 0\end{cases}$ in the sense of distributions, where $\delta_0$ is the Dirac delta mass at $x = 0$ , then $u(\cdot,t)$ converges pointwise iff $d \geq 3$ and explodes to $+\infty$ for $d=1,2$ . This is not too hard. By Duhamel's principle we have the following explicit form for $u$ (where $\star$ is convolution and $\Phi$ is the heat kernel):
$u(x,t) = \int_0^t (\Phi(\cdot,t-s) \star \delta_0)(x)\,ds$ $= \int_0^t \Phi(x,t-s)\,ds = \int_0^t \Phi(x,s)\,ds$ $= \int_0^t \frac{1}{(4\pi s)^{d/2}}e^{\frac{-|x|^2}{4s}}\,ds$ We are interested in what happens when we take $t = +\infty$ . Morally speaking (the constants and $x$ are irrelevant) this reduces to studying the convergence of the integral $\int_0^\infty \frac{1}{s^{d/2}}e^{-1/s}\,ds$ , which in turn (from $u = 1/s$ ) reduces to the convergence of $\int_0^\infty \frac{1}{u^{2-\frac{d}{2}}}e^{-u}\,du$ , which is $< \infty$ iff $2 - \frac{d}{2} < 1$ , i.e. $d > 2$ .

This is pretty much a complete proof, but if we wish to be more precise, the integral at the end is
$\int_0^\infty \frac{1}{(4\pi )^{d/2}s^{2-d/2}}e^{\frac{-|x|^2s}{4}}\,ds = \frac{|x|^{2-d}}{2^{6-d}\pi^{d/2}}\begin{cases}\Gamma(d/2-1), & d/2 - 1 > 0 \\ +\infty, & \text{otherwise}\end{cases}.$ (Note that when $d/2 -1 < 0$ , $\Gamma(d/2-1)$ technically has a value, but strictly speaking this is the analytic continuation of the integral as the integral diverges in this case.)

Hence if we light a fire somewhere in a space of at least three dimensions then the heat should approach a distribution with polynomial decay on the order $\frac{1}{|x|^{d-2}}$ , which is not too surprising. In fact, this form makes some intuitive sense in that for large dimensions, heat should have more directions to disperse in, which matches with the fact that the exponent here increases with $d$ . For the low dimensions $d=1,2$ , heat does not have enough directions to spread out in to offset the speed at which a constant heat source increases the heat in a region, so the heat will grow infinitely.

The langugage used in this intuitive picture suggests a wildly different approach.

The Probabilist's View

The theorem we've proven here bears an uncanny resemblance to a theorem in probability: A symmetric random walk on the lattice $\mathbb{Z}^d$ will visit the origin infinitely often almost surely when $d = 1,2$ , but not for $d \geq 3$ . Since we can imagine that heat proliferates as a random walk with infinitesimally small steps, we may expect an interesting connection between these two results. Indeed, we can "use" the probabilistic result to prove our result for the heat equation. Though, I believe there is some unavoidable reliance on the integral $\int_0^\infty \frac{1}{s^{d/2}}e^{-1/s}\,ds$ . If you can somehow dodge this, do let me know. The main tool we require is the theorem which establishes the connection between probability and the heat equation with a forcing term.

Theorem (Brownian motion solves the heat equation; forcing term variant): Let $\phi \in C_0(\mathbb{R}^d)$ and let $W_t$ be a Brownian motion on $\mathbb{R}^d$ started at $x$ . Take
$u(x,t) := \mathbb{E}[\phi(W_t)] + \mathbb{E}\int_0^t f(W_s)\,ds.$ Then $u$ solves the PDE
$\begin{cases}\partial_t u - \frac12\Delta u = f \\ u\,|_{t=0} = \phi\end{cases}.$

This makes a lot of intuitive sense: You start out with $\phi$ heat, and then move along and collect heat according to $f$ .

Proof. By Ito's lemma applied to $\phi$ ,
$\phi(W_t) = \phi(x) + \int_0^t \nabla \phi(W_s) \cdot dW_s + \frac12\int_0^t \Delta \phi(W_s)\,ds.$ Take the expectation of both sides. Since $W_t$ is a continuous local martingale, so is $\int_0^t \nabla \phi(W_s) \cdot dW_s$ , so this has zero mean. Thus
$\mathbb{E}[\phi(W_t)] = \phi(x) + \frac12\int_0^t \Delta \mathbb{E}[\phi(W_s)]\,ds \qquad (1).$ On the other hand, the same reasoning applied to $f$ gives
$\mathbb{E}[f(W_t)] = f(x) + \frac12\int_0^t \Delta \mathbb{E}[f(W_s)]\,ds \qquad (2).$ Adding the time derivative of $(1)$ to $(2)$ gives
$\partial_t\mathbb{E}[\phi(W_t)] + \mathbb{E}[f(W_t)] = \frac12\Delta \mathbb{E}[\phi(W_t)] + \frac12\int_0^t \Delta \mathbb{E}[f(W_s)]\,ds + f(x).$ The LHS is $\partial_t u$ , and the term $\frac12\Delta \mathbb{E}[\phi(W_t)] + \frac12\int_0^t \Delta \mathbb{E}[f(W_s)]\,ds$ is equal to $\frac12\Delta u$ , so the above equality is $\partial t u = \frac12\Delta u + f$ . It's clear that $u(x,0) = \phi$ from the definition of $u$ so this completes the proof. $\square$

The second tool we need is the continuous version of the transience/recurrence theorem for discrete random walks.

Theorem (Transience and Recurrence of Brownian motion): Let $W_t$ be a Brownian motion on $\mathbb{R}^d$ started at some $x$ . For $d = 1,2$ , $W_t$ will visit the unit ball infinitely often with probability 1. For $d \geq 3$ , there is a positive probability that $W_t$ never visits the unit ball. (Quantitatively, the probability of ever visiting the ball $B(x,r)$ is $\frac{1}{|x/r|^{d-2}}$ for $|x| > r$ .)

Proof

It would also be nice if there were a proof for this that involves appealing directly to the discrete version of this theorem (and using the fact that random walks on increasingly finer lattices converge to a Brownian motion in distribution), but I find it unclear how this can be done without a more quantiative bound on the probabilities involved in the discrete case.

Anyways, with these two results the connection is clearer and a probabilistic argument can be made. Let $f$ be a heat source somewhere on $\mathbb{R}^d$ --- say, an indicator on some open set $U$ , for simplicity. If $d = 1,2$ , take two balls $B_1 \Subset B_2 \Subset U$ . Then any Brownian motion will visit $B_1$ infinitely often with probability 1, and with each such visit we expect $W_t$ to spend at least $\varepsilon > 0$ time inside $B_2$ , hence accumulating at least $\varepsilon$ heat in expectation. Thus $u(x,t) = \int_0^t \mathbb{E} f(W_s)\,ds \xrightarrow{t \to \infty} \infty\varepsilon = \infty$ for every $x \in \mathbb{R}^d$ .

If instead $d \geq 3$ , we instead have that (for each $x$ , where we recall $W_0 = x$ ) the random variable $\int_0^\infty f(W_s)\,ds$ is finite almost surely, as with probability 1 it will only visit $U$ finitely many times, with each visit accumulating only a finite amount of heat. But sadly, this is not quite enough to imply that the expectation is finite, and it's also not possible to prove that this random variable has an upper bound (it doesn't!). Unfortunately this means we have to be more delicate to complete this weird probabilistic painting, and here is where I believe we are forced back into considering the integral $\int_0^\infty \frac{1}{s^{d/2}}e^{-1/s}\,ds$ . Indeed,
$u(x,\infty) = \mathbb{E}\int_0^\infty f(W_s)\,ds = \int_0^\infty \mathbb{P}(W_s \in U)\,ds$ $= \int_0^\infty \int_{x + U} \frac{c_1}{s^{d/2}}e^{\frac{-|y|^2}{c_2s}}\,dy\,ds$ for some irrelevant constants $c_1$ and $c_2$ . Interchanging the integrals again will bring this inevitable integral to light and show that $u(x,\infty) < \infty$ .

I had a few ideas for trying to get this argument to work without this integral, relying as much as possible on the "random walk" result, but none of them seemed very feasible, and certainly not more elegant than what I've shown. If you have ideas for alternate approaches I'd be delighted to hear it.

Probability hurts my brain, let's go back to spamming integrals.

I'll close with generalizing the initial result, with $f$ not necessarily being an indicator.

Theorem: Let $f \geq 0$ everywhere, compactly supported, with $f > 0$ over a set of positive measure. Let $u$ solve the heat equation with $f$ as a heat source, constant in time. Then $u$ explodes for $d = 1,2$ , and approaches a finite stable distribution a.e. for $d \geq 3$ .

Proof. We have
$u(x,t) = \int_0^t (\Phi(\cdot,t-s) \star f)(x)\,ds = \int_0^t \int_{\mathbb{R}^d} \frac{1}{(4\pi(t-s))^{d/2}}e^{\frac{|x-y|^2}{4(t-s)}}f(y)\,dy\,ds$ $= \int_{\mathbb{R}^d} f(y)\int_0^t \frac{1}{(4\pi(t-s))^{d/2}}e^{\frac{|x-y|^2}{4(t-s)}}\,ds\,dy$ $= \int_{\mathbb{R}^d} f(y)\int_0^t \frac{1}{(4\pi s)^{d/2}}e^{\frac{|x-y|^2}{4s}}\,ds\,dy.$ If $d = 1,2$ then the inner integral explodes as $t \to \infty$ for every $(x,y)$ , and so the outer integral explodes over a positive-measure subset where $f > 0$ . (Note that funny things happen if we didn't take $f$ to be signed... better to not think about it. And interchanging all the integrals and limits is 100% safe because of this restriction on $f$ , so...)

If $d \geq 3$ , then, well, taking $t = +\infty$ and running through the computation again gives
$= C_d\int_{\mathbb{R}^d} \frac{f(y)}{|x-y|^{d-2}}\,dy,$ for a constant $C_d < \infty$ depending on $d$ that I could not care less about, and this is $<\infty$ since we assume $f$ is compactly supported (and because $d-2 < d$ is always true). The compact support condition can certainly be weakened here but I'm going to call the fire brigade if you manage to get your hands on an unbounded heat source. More importantly, the above is a convolution of $f$ with a (multiple of) the fundamental solution to the Laplacian, so we see that the stable distribution that we approach is the (possibly distributional) solution to the problem $-\Delta u = f$ . This makes a lot of sense: If $u$ were constant in time then the time derivative term goes away leaving us with $\partial_t u - \Delta u = -\Delta u = f$ . That's cool.

This post has been edited 2 times. Last edited by greenturtle3141, Mar 4, 2025, 9:57 PM

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The effect of a constant heat source depends on dimension

by greenturtle3141, Mar 4, 2025, 9:51 PM

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