Catenoid is a minimal surface

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.