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](https://en.wikipedia.org/wiki/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 $D > 0$)?



Solids with constant surface area during "erosion"
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.


