Let $K_0$ and $K_1$ be two bounded, disjoint convex sets in $R^n,n\geq 3$, and $u$ the equibrium potential, that is the harmonic functon in $R^n\backslash\{ K_0\cup K_1\}$ such that $u$ has boundary values $1$ on $K_1\cup K_2$ and $u(\infty)= 0$. Denote $$r_j=r_j(K_0,K_1)= \int_{\partial K_j}\frac{\partial u}{\partial n}ds,\quad j=0,1,$$ where $n$ is the inner normal and $ds$ is the surface area element.

Is it true that each $r_j$ decreases if we move $K_0$ and $K_1$ closer to each other?

More precisely: Let $f:R^n\to R^n$ be a distance decreasing homeomorphism, whose restriction on each $K_j$ is an isometry onto $f(K_j)$. Is it true that $r_j(f(K_0),f(K_1))\leq r_j(K_0,K_1)$ for each $j$?

Same question can be asked with more than two sets, but there are counterexamples with many balls $K_j$, even of equal radii: this is called a "Faraday cage".

Comments. This problem originates in the attempts of biologists to explain why certain animals (like armadillos) group close together when they sleep. In this model, $K_j$ are the animals with body temperature $1$, while the medum has temperature $0$. Then $u$ is the time independent temperature near the bodies, and $r_j$ is the rate of heat loss. Presumably moving closer together desceases the rates of heat loss $r_j$. It is easy to prove that the TOTAL heat loss of the group $r_1+r_2$ decreases when we move animals $K_1$ and $K_2$ closer together, see http://www.math.purdue.edu/~eremenko/dvi/armadillo.pdf and references there. But each individual animal $K_j$ feels only his own $r_j$, not the sum. Also some condition of convexity type is indeed needed here; think of a kangaroo putting her child in the bag. This decreases the heat loss by the child but (slightly) increases the cooling rate of the mother.

The problem is unsolved even when $K_0$ and $K_1$ are two balls of different radii. In this case one can obtain $r_k$ as infinite series of functions of the distance, but these series are alternate and it is unclear how to show that they are monotone.

EDIT. By "Faraday cage" I mean the following. Take a spherical shell of the form $1-r\leq |x|\leq 1$, $r$ is small, and break it into little convex pieces of size approximately $r$, with some space between pieces. These pieces are $n$ animals. The temperature they create is approximately the same as the shell itself would create, which is almost constant $1$ inside, and almost $1/|x|$ outside. Consider a new animal of the same size outside. It slightly decreases the heat loss by some animals on the shell close to it. Now move the same animal inside. It will have no effect at all on the shell animals.

This is the same principle as kangaroo bag, I just broke the bag into many small convex pieces. More careful argument shows that the animals can be balls of the same radius here. (Faraday's own cage does the same with electric field which satisfies the same equation).

  • $\begingroup$ Can one show that neither $r_j$ increase as the bodies move closer? One could imagine that a really cold armadillo seeks shelter close to a hot one, and for a moment, the hot armadillo experience a pair of cold feet against its back... but as you mention in the notes, I see both are assumed to have uniform surface temperature 1, so the cold feet scenario is not part of this model. $\endgroup$ – Per Alexandersson Sep 5 '15 at 15:45
  • $\begingroup$ I was convinced that animals group together because so they form a body with the same volume and lesser surface area, and heat dispertion is somehow proportional to the area of the interface between "body" and "environment"... $\endgroup$ – Qfwfq Sep 5 '15 at 19:07
  • $\begingroup$ @Qfwfq: The precise statement is in my question under "It is easy to prove...". The point is that this minimizes the heat loss of the whole group of animals. And I am asking whether this minimizes the heat loss of each individual animal. $\endgroup$ – Alexandre Eremenko Sep 5 '15 at 20:05
  • $\begingroup$ @per Alexandersson: the sum of $r_j$ decreases. See the linked file. All animals have the same body temperature $1$ in this model. $\endgroup$ – Alexandre Eremenko Sep 5 '15 at 20:17
  • 1
    $\begingroup$ @Eckhard: I do not have a good reference but I added to the question an explanation what I mean. In the Wikipedia article explanation is non-mathematical. $\endgroup$ – Alexandre Eremenko Sep 6 '15 at 14:14

The answer is negative.

Assume that $K_{0}$ and $K_{1}$ are two disjoint balls. If $K_{0}$ is at least twice as big as $K_{1}$, then $K_{0}$ should keep some nonzero distance in order to minimize his heat loss while $K_{1}$ should try to be as close as possible to $K_{0}$.

To be more precise let $r_{0}$ and $r_{1}$denote the radii of the balls $K_{0}$ and $K_{1}$. Let \begin{align*} Q_{j}(d) = \int_{\partial K_{j}} \frac{\partial u}{\partial n} ds\quad \text{for} \quad j=0,1 \end{align*} denote the heat loss as a function of distance between the balls. Then $Q_{0}(d)$ is not monotonically increasing provided that $\frac{r_{0}}{r_{1}} > \ell$ where $\ell\approx 1.95$ is the positive solution of the equation \begin{align*} 2(1+x^{3})\left(\gamma+\psi\left(\frac{1}{1+x}\right) \right)+x^{2}+x+2(x^{2}-x)\psi'\left(\frac{1}{1+x}\right)=0. \end{align*} In the above notation $\gamma$ is the Euler--Mascheroni constant, and $\psi$ is the digamma function. I have uploaded a preprint on arXiv about this problem.

In the following figure the graph of $Q_{0}(d)$ is given when $r_{0}=20$, $r_{1}=1$, $0\leq d \leq 80$ and the temperature of the balls equals $1$. The heat loss $Q_{0}(d)$ of the big ball. $r_{0}=20, r_{1}=1$, $0\leq d \leq 80$

Actually numerical computations show that if the sizes of the balls are comparable, namely, $\frac{1}{\ell} \leq \frac{r_{0}}{r_{1}} \leq \ell$ then the individual heat loss $Q_{j}$, $j=0,1$ is monotonically increasing, however, I was not able to prove this.

Also based on numerical computations if $\frac{r_{0}}{r_{1}} > \ell$ then the nonzero distance that the big ball has to keep approximately equal to $r_{0}$. The heat loss of the small ball is always monotonically increasing.


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