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Imagine a drug, a pill that you swallow, which is designed to dissolve in your stomach at a constant rate. It must be shaped such that the surface area remains constant when the volume is "eroded" uniformly over its surface.

Two dimensions

Define "erosion" of a distance $\delta$ from a shape as removing a slither of width $\delta$ around the whole perimeter of the shape. This is a parallel curve, at least initially. (Note that a re-entrant corner will become rounded after erosion. An alternative defintion would have the corner preserved, which makes the question slightly easier but is harder to justify.)

Does a shape exist whose perimeter remains constant after is eroded (at least for erosion of a distance $\delta$, for all $\delta <= D$, for some $D > 0$)?

I have an unsatisfying solution:

An annulus, assuming internal erosion is allowed, because the change of perimeter of the inner and outer circles cancel out until the area is zero. This works mathematically but not for how I posed the question. So I would be interested in solutions that don't have holes.

But for an annulus with a channel of zero width connecting the inner and outer circle, after erosion the perimeter will have been reduced by about twice the width of the channel. So I would conjecture that it can't be done without holes.

(This solution doesn't generalise to spheres in 3D although holes might still help.)

Three dimensions

Define "erosion" of a distance $\delta$ from a solid as shaving a width $\delta$ from the whole surface of the solid.

Does a solid exist whose perimeter remains constant after is eroded (at least for erosion of a distance $\delta$, for all $\delta <= D$, for some $D > 0$)?

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    $\begingroup$ If exists, it should be a solid torus since the integral of Gauss curvature has to be vanish. $\endgroup$ – Anton Petrunin Mar 4 '17 at 18:59
  • $\begingroup$ These may be of interest: Shlomo Sternberg, "Semi-Riemann Geometry and General Relativity," math.harvard.edu/~shlomo/index.html . lightandmatter.com/sr (by me), secs, 3.9 and 9.5.4. $\endgroup$ – Ben Crowell Mar 4 '17 at 22:14
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    $\begingroup$ en.wikipedia.org/wiki/Mathematical_morphology $\endgroup$ – Steve Huntsman Mar 5 '17 at 0:03
  • $\begingroup$ Not an answer to your mathematical question, but if you're seeking an approach to a medical/technical problem, it seems easier to shape the pill into a circular cylinder sheathed in a relatively insoluble (but ultimately digestible) coating on the lateral surface, so the drug dissolves only at the flat ends of the cylinder, or something of that type. $\endgroup$ – Andrew D. Hwang Mar 5 '17 at 17:14
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It seems that such pill exists.

Take a ball and drill a hole through it, so you get a solid torus; we assume it has smooth boundary $\Sigma$.

By Gauss--Bonnet formula, we gave $$\int\limits_\Sigma G=0,$$ where $G$ denotes Gauss curvature.

Denote by $H$ the mean curvature of $\Sigma$; it is mostly very negative in the surface of the hole. It is easy to make a hole such that $$\int\limits_\Sigma H >0,$$ but, if the hole wriggles badly, then $$\int\limits_\Sigma H\approx -\infty.$$

It follows that for a reasonably wriggling hole, we get $$\int\limits_\Sigma G=\int\limits_\Sigma H=0.$$

Let $\Sigma_r$ be the $r$-equidistant surface from $\Sigma$. Then by Weyl's formula $$\textrm{area}\,\Sigma_r=\textrm{area}\,\Sigma+ r\cdot\int\limits_\Sigma H +r^2 \cdot\int\limits_\Sigma G$$ for sufficiently small $r$. Therefore the identities above imply that the area of equidistant surfaces stays constant for a short time.

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    $\begingroup$ Could you (or someone else) please expand on "Weyl's formula"---presumably Weyl’s Tube Formula---and how it leads to your final sentence? I am not questioning you; I just want to understand. $\endgroup$ – Joseph O'Rourke Mar 5 '17 at 0:23
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    $\begingroup$ @JosephO'Rourke The "$\textrm{area}\,\Sigma_r$" looks like equation (4), used in a proof of the Tube Formula, from Curvature Estimation over Smooth Polygonal Meshes using The Half Tube Formula (if so I think the second term needs a coefficient of 2, but it doesn't matter here). $\endgroup$ – Ben C Mar 5 '17 at 19:25
  • $\begingroup$ @AntonPetrunin Thanks, that seems remarkable. $\endgroup$ – Ben C Mar 5 '17 at 21:13
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    $\begingroup$ @BenC about coefficient 2: it depends how you define mean curvature, my definition is $H=k_1+k_2$ and yours $ H=\tfrac12\cdot(k_1+k_2)$. $\endgroup$ – Anton Petrunin Mar 5 '17 at 22:15
  • $\begingroup$ @AntonPetrunin Thanks for the clarification. $\endgroup$ – Ben C Mar 6 '17 at 21:10

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