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:

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