The Isoperimetric problem for domains constrained to lie between two parallel planes

Question

It is well known that for a given volume $V$, a sphere is the shape that minimizes the surface area. I am interested in the same problem under the constraint that the shape must lie between the planes $x=-a$ and $x=a$, i.e. all its points $(x,y,z)$ must have $-a \leq x \leq a$. Obviously, if $V\leq\frac{4}{3}\pi(2a)^3$ then the shape is still a sphere since the sphere with the right volume fits entirely between the planes. However if the volume becomes larger, the sphere can not fit between the planes and it will have to be flattened. My question is: what is this shape? This question is about the 3 dimensional case.

From here I'll describe what I have found out so far.

Since the problem is completely symmetric in the $y,z$ directions, the solution can be represented as the rotation of the graph of a function $r(x)$ around the $x$ axis. Then the volume of this function rotated around the $x$ axis will be:

$$V=\int_{-a}^a\pi r(x)^2 dx$$

and the area will be:

$$A = \pi r(-a)^2 + \pi r(a)^2 + \int_{-a}^a2\pi r(x)\sqrt{1+r'(x)^2}dx$$

(the first two terms are for the disk shaped sides pressed against the plane, and the last term is for the surface of revolution between the planes)

Numerical minimization by discretizing $r$ into a piecewise linear function has produced these results:

The final 3D shapes are these 2D shapes rotated around the horizontal axis. From left to right the distance $a$ is decreasing, thus constraining the shapes more and more. The solution appears to be of the form $r(x) = \sqrt{a^2-x^2}+b$, but on closer inspection it seems that this is not the case. The top and bottom are not exact half circles. For the almost spherical case, and for the very elongated case (with $a$ small), they are very close to half circles, but in the intermediate case there is quite a bit of difference.

By the method of calculus of variations I have derived a differential equation that $r$ should satisfy:

$$\sqrt{1+r'^2} + \lambda r = \frac{d}{dx} \frac{rr'}{\sqrt{1+r'^2}}$$

Where $\lambda$ is some real number dependent on the parameters of the problem ($V$ and $a$). And indeed, the function $r(x) = \sqrt{a^2-x^2}$ satisfies this equation as expected (indicating that with no constraint the minimal surface is a sphere), but $r(x) = \sqrt{a^2-x^2}+b$ unfortunately does not for $b\neq0$.

It would of course be awesome to have a complete solution, but I would also be very happy with asymptotic special cases, particularly the case where the sphere is only a little compressed (so the case where $V$ is only a little greater than $\frac{4}{3}\pi(2a)^3$, i.e. the green shape in the picture).

Thanks for your help!

Answer 1

It seems in your definition of $A$ you are counting the area of the caps (if they exist).
This should probably lead to surfaces which are CMC and have certain contact angle at the two planes which explains your solutions. I also don't think your intuition is wrong but that the solutions should look like $$ r(x)=\sqrt{(a+b)^2-x^2} $$ for $b\geq 0$ (unless somehow Delanauy surfaces are better which I would find surprising -- this should in any case be a simple, though necessary, comparison argument). This is consistent with them being pieces of spheres.

As a (somewhat lengthy) aside, the more standard way to consider this sort of problem is to not count the area of the "caps". In this case the question has been completely answered (at least for $\mathbb{R}^3$). According to page 6 of this nice survey by Ritore and Ros the relevant people are Athanassenas, Vogel, and Pedrosa and Ritore.

In this case the answer is either an appropriate region of the sphere or of the cylinder. Essentially, your conditions are of free boundary type and so the surfaces all have to meet the two planes orthogonally (as pointed out by Will). This means that you actually have a hemi-sphere up to some critical volume where there is a cylinder with the same area and enclosing the same volume and for larger volumes the cylinder is optimal.

According to the survey article this continues to hold in $\mathbb{R}^{n+1}$ for $n\leq 7$. For $n\geq 9$ this is no longer true and there are certain types of rotationally symmetric CMC surfaces (called unduloids) which are optimal for some intermediary values of the volume (i.e. small volumes are still hemi-spheres and large volumes are still cylinders). The argument for this is actually quite beautiful. Essentially the hemisphere large enough to touch one plane tangentially actually has less area than the corresponding cylinder. The regularity of the problem then implies that there has to be a better surface which must be an unduloid.

Surprisingly, this question seems to be open for $n=8$.

Answer 2

You have not been very specific about boundary conditions. Still, the constant mean curvature surfaces (they are not called minimal if the mean curvature is nonzero) of revolution in $\mathbb R^3$ are the circular cylinder and the Delaunay surfaces, http://en.wikipedia.org/wiki/Constant-mean-curvature_surface

Well, I am not done, but let me at least suggest that the free boundary problem you appear to be describing ought to have the surface meeting the planes orthogonally. Furthermore, I think that the actual minimum surface area for a prescribed volume will simply be a circular cylinder of soap film. It is easy enough to prove that the minimizing surface cannot have a pinched waist and meet the planes obtusely, as would the catenoid. More work to be done.

Anyway, see Is there a complete classification of constant mean curvature surfaces? for a beginning.