Minimal surface between two parallel circle of same radius : why is it a surface of revolution?

Question

I would like to find a minimal surface between two parallel circle of same radius (i.e. they are coaxial). I in fact just need to know that it's a revolution surface to conclude that it will be a catenoid. So let $\Sigma$ such a surface. How can I show that $\Sigma$ must be a revolution surface ?

Answer 1

On re-reading your question I suspect you are looking for a deeper answer to that I am offering, but having typed it I might as well post it.

Let the radius of the surface be $r(l)$ at a distance $l$ from one end. Then the area of an annulus will be $2\pi r \sqrt{dr^2+dl^2}=2\pi r\sqrt{1+r'^2}dl$. This is now in the form of a standard Euler-Lagrange minimization problem. As there is no explicit $l$ dependence the Beltrami identity can be used so the EL equations reduce to $r \sqrt{1+r'^2}-\frac{r r'^2}{\sqrt{1+r'^2}}=const$, which simplifies to $r=C \sqrt{1+r'^2}$. This is easily solvable to get a catenary.

Answer 2

The solution need not unconditionally be axisymmetric i.e. be a surface of revolution. It is just one among a countless set of possibilities. It is just that the surface of revolution by virtue of its symmetery is easier to compute. And included in text-books. You are free to hold two circles in either hand, dip them in soap solution and take one out offsetting an axis of one circle relative to other, when you see a non-axisymmetric oblique film that involves elliptic / hyper-elliptic integrals for its full differential desription. Not one example of the unsymmetric cases is given in the text-books that I so far came across. You see it so frequently in books that you perhaps started believing that it should be a surface of revolution. What has been done for convenience has lead you perhaps to a wrong conclusion and further towards belief induced by repetition.

The offset that is stably maintainable is given by Goldschmitt conditions mentioned in Cyril Isenberg's book cited above. The more the offset $d$, shorter is the film $h$ formed. It is tougher to solve, numeric/experimental methods are more suitable to look at the shapes of zero mean curvature $H=0$. That page does not appear in the pdf, but I shall sketch it from my memory.. $h$ should be correctly labelled as vertical dimension..

For easier symmetric case when you assume an axisymmetric surface of revolution at start and finally arrive at a catenoid's DE, we conclude that the start assumption is indeed correct. It is in-built into formulation of axi-symmetry. It can be calculated with variational calculation easily.

Area of soap film in cylindrical axisymmetric cordinates $(r,z)$ is

$$ \int 2 \pi r \sqrt{1+r^{'2} }) dz $$

Using Euler-Lagrage on functional $ r \sqrt{1+r^{'2} }$ and when there is no $z$ term explicitly, Beltrami solution

$$ r \sqrt{1+r^{'2} }) -r^{'} \cdot r \cdot \frac {r^{'}}{ \sqrt{1+r^{'2} }}= const $$

leads to differential equation of a rotationally symmetric catenoid where the constant is properly interpreted:

$$ \dfrac{r} {\sqrt{1+r^{'2}} } = r_{min}. $$

Answer 3

The following information was provided by Brian C. White. The key publication here is

Uniqueness, symmetry, and embeddedness of minimal surfaces, by Richard M. Schoen, in J. Differential Geom. 18(4): 791-809 (1983). DOI: 10.4310/jdg/1214438183

Here on page 791 Schoen writes:

In particular this shows that if $\Gamma_1$ and $\Gamma_2$ are circles situated so that the line joining their centers is perpendicular to the planes in which they lie, then $M$ is a surface of rotation, hence a catenoid.

That's a consequence of the Alexandrov reflection argument. Thus the minimal surface is given by the solution of an ODE, from which it follows that it is a portion of a catenoid.

However, that catenoidal surface is not necessarily unique. As a preliminary, consider the family of catenary curves $y=a\cosh \frac{a}{z}$. Note that these pairwise intersect. It is these points of intersection that correspond to boundary circles after we rotate the catenary around the $z$-axis to obtain the catenoid as the corresponding surface of revolution.

Let $C$ be the standard catenoid. Dilate $C$ about the origin to get another catenoid $C'$. Then $C$ and $C'$ will intersect in a pair a circles. That pair of circles bounds a minimal annulus contained in $C$ and also a minimal annulus contained in $C'$.

In general, one can show that for any pair of horizontal circles centered on the $z$-axis, the number of minimal annuli they bound is 0,1, or 2.

Suppose we fix the radii of the two circles, and vary the distance $d$ between the centers of the two circles. If $d$ is large, they won't any connected minimal surface. For a certain critical value of $d$, they bound exactly one connected minimal surface. For $0<d<$ that critical, value, the circles will bound exactly two connected minimal surfaces. In all 3 cases, the surfaces in question are rotationally symmetric and therefore portions of catenoids.