Catenoid

A catenoid is the surface of revolution obtained by rotating a catenary about its directrix. One conformal parametrization is

An axisymmetric smooth connected soap film spanning equal radius-$R$ rings at heights $\pm H$ has stationary profile $r=k^{-1}\cosh (kz)$, $k>0$. The boundary equation is $R/H=\cosh u/u$, $u=kH$. Its unique minimum occurs at $u\tanh u=1$, because the derivative numerator is $u\sinh u-\cosh u$. If $A$ is this root, the minimum is $\cosh A/A=\sinh A$. There are no catenoids below that ratio, one at equality, and two above it. Existence of stationary connected films does not assert global minimality against pinching or disconnected competitors.

For the standard catenoid, the two principal curvatures are opposite, so its mean curvature vanishes. Its Gaussian curvature is $K=-{\mathrm{sech}}^{4}u<0$, which also shows that no neighbourhood is locally isometric to a plane, cylinder, or regular part of a cone, all of which have zero Gaussian curvature.

For the catenoid profile $y(x)=E\cosh (x/E)$ spanning fixed-endpoint rings, the second variation of $F[y]=2\pi \int y\sqrt{1+y^{\prime 2}}dx$ isThis refers to axisymmetric graph variations, not all possible changes in surface topology. Strict positivity for every nonzero admissible variation gives a strict local minimum in the $C^1$ topology: the positive lowest Dirichlet eigenvalue of the regular Sturm-Liouville problem gives coercivity, and sufficiently small coefficient perturbations preserve it. The zero variation must be excluded from a strict-positivity hypothesis.

A catenoid has zero mean curvature, but need not minimize area for its fixed boundary rings. For the profile $y(x)=\cosh x$ on $[-2,2]$, take an endpoint-zero variation $\eta (x)=\cosh x-\cosh 2$. The second variation of the rotational area isTo derive the integral, differentiate $y\sqrt{1+y^{\prime 2}}$ twice, then integrate its mixed term by parts; the endpoint term vanishes. To evaluate it, substitute ${\eta }^{\prime 2}-{\eta }^2=-1-C^2+2C\cosh x$, using the antiderivatives of $\mathrm{sech}x$ and ${\mathrm{sech}}^{2}x$. The result is negative because $\arctan (\sinh 2)<\pi /2$ and $(C+C^{-1})\tanh 2>\pi$.

Small positive and negative perturbations remain positive and preserve the boundary values. The first variation vanishes, while the negative second variation gives nearby smaller areas. Thus being a minimal surface in the zero-mean-curvature sense is a stationarity condition, not a guarantee of area minimization for arbitrary boundary separations.