# Solids with constant surface area during "erosion"

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 $D > 0$)?

• If exists, it should be a solid torus since the integral of Gauss curvature has to be vanish. Commented Mar 4, 2017 at 18:59
• 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.
– user21349
Commented Mar 4, 2017 at 22:14
• en.wikipedia.org/wiki/Mathematical_morphology Commented Mar 5, 2017 at 0:03
• 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. Commented Mar 5, 2017 at 17:14

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.

• 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. Commented Mar 5, 2017 at 0:23
• @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). Commented Mar 5, 2017 at 19:25
• @AntonPetrunin Thanks, that seems remarkable. Commented Mar 5, 2017 at 21:13
• @BenC about coefficient 2: it depends how you define mean curvature, my definition is $H=k_1+k_2$ and yours $H=\tfrac12\cdot(k_1+k_2)$. Commented Mar 5, 2017 at 22:15
• @AntonPetrunin Thanks for the clarification. Commented Mar 6, 2017 at 21:10