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First a little background. Mircowaves do not heat uniformly. To help overcome this, your 'food' is rotated, however this is not usually sufficient to produce a totally uniform heating. Informally, the question is when can we find a way of moving our 'food' in order to heat it uniformly throughout?

Let $f : \mathbb{R}^n \to R$ be our heat function. Let $I^n = [-0.5,0.5] \times \cdots \times [-0.5,0.5]$ be the unit n-dimensional cube centered at the origin, this will be our 'food'. Let $\gamma : [0,1] \to \mathbb{R}^n \times SO(n)$ be a map specifying a path along which to translate and rotate $I^n$. If $x \in I^n$ then let $h(x)$ denote the total 'heat absorbed' by $x$ as it travels along $\gamma$. Note if $\gamma(t) = (\gamma_1(t), \gamma_2(t))$ then $h(x) = \int_0^1 f(\gamma_2(t)(x) + \gamma_1(t)) dt$.

We will call a curve $\gamma$ uniformly heated iff $\forall x,y \in I^n$, $h(x) = h(y)$.

How sufficiently 'nice' must our heat function $f$ be in order to guarantee that there exists a uniformly heated curve? Do these requirements change if we consider a different 'food' to heat? For example, if we heat $I^m \times 0^{n-m}$ in $\mathbb{R}^n$.

Note that in $\mathbb{R}^1$, as $SO(1) = 1$, if $f$ is a strictly monotonic function then there cannot exist any uniformly heated curves as (assuming wlog $f$ is increasing) $h(-0.5) < h(0.5)$.

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Did you actualy mean $h(x) = \int_0^1 f(\gamma_2(t)(x) + \gamma_1(t)) dt$? –  fedja May 9 '10 at 23:16
This is a great title for a question. –  Neel Krishnaswami May 10 '10 at 21:27
This might just be my favourite MO Q. –  Spencer May 10 '10 at 21:29
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1 Answer

up vote 18 down vote accepted

You can uniformly cook the cube if $f$ is harmonic, i.e. $\Delta f=0$. Note that, if e.g. $\Delta f>0$ everywhere, then the center of the cube will always receive less heat than the average over a sphere with the same center. Thus if $\Delta f$ happens to have constant sign, it must be zero.

To achieve uniform cooking in the harmonic case, let the center of the cube stay at the origin and rotate the cube using a Peano-like curve $\gamma:[0,1]\to SO(n)$ such that the push-forward measure $\gamma_*m$ (where $m$ is the Lebesgue measure on $[0,1]$) is the normalized Haar measure on $SO(n)$. Then the heat received by a point $x\in I^n$ is the average of $f$ over the sphere of radius $r=|x|$ centered at the origin. Since $f$ is harmonic, this average value equals $f(0)$.

To construct such a curve $\gamma$, follow the standard procedure for the Peano curve: partition $SO(n)$ into reasonable sets (connected and with piecewise smooth boundaries) and visit all of them by a continuous path; this path is the first approximation. Then subdivide the partition and change the path so that it visits all sub-parts but do not make new intersections with boundaries of old parts. And so on. At each step, choose the parametrization so that the time spent in each piece equals its Haar volume. Let the diameters go to zero, then the paths will converge to a Peano-like curve $\gamma$. The push-forward measure $\gamma_*m$ coincides with the Haar measure on all elements of the partitions and hence on all Borel sets.

Remark. A similar trick works if $f$ has compact support (more precisely, its support should be separated away from the microwave walls by distance at least 1). Just move around so as to realize a Haar measure on the relevant subset of the translation group rather than $SO(n)$.

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Remark: the harmonic function construction obviously won't work in the 1 dimensional case, since SO(1) does not act transitively on "S^0". But the compact support construction works well for 1 dimension. –  Willie Wong May 11 '10 at 8:48
Yes, $f$ being harmonic (or having compact support) is certainly sufficient however is it necessary? –  Mark Bell May 11 '10 at 12:32
Uh... of course not? You just need f to be harmonic (heck, even constant) on an open set that can strictly contain, say, the ball of radius 1. Or if $f$ has compact support, then modification of $f$ outside the set $supp(f) + B_1$ will not change this property. For any $f$ with a solution $\gamma$, you can of course modify $f$ outside the image of $I$ under the transport by $\gamma$ and still have a new function with the requisite property. Since the property you are looking for is strongly local, any global characterization of the function can be made as bad as possible. –  Willie Wong May 11 '10 at 13:42
Another example is any periodic function. I don't think there is a sensible "if and only if" condition with this formulation. If you require something stronger - e.g. a cube of any size must be cookable within any cubical region of size 100 times that of the cube - then being harmonic is necessary (at least if $f$ is $C^2$). –  Sergei Ivanov May 11 '10 at 14:32
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