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I can't understand two things regarding the following example:

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1- Why the minimal surface $S$ will not minimize the area among all surfaces with boundary the two circles if $S_0 < S$? I don't understand this phrase at all and I don't know why the inequality must hold?

2- What is the meaning of the last paragraph and how to show that it holds (a proof)?

A clear simple explanation would be much appreciated.

PS - Source is the book Elementary Differential Geometry by A. N. Pressley.

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    $\begingroup$ If $a>0$ is big enough, one has $1+ e^{-2a} < 2a$, which is the same as $\cosh^2 a < a+ \sinh a\cosh a$. And this inequality is the same as saying that the area of the catenoid $S$ (bounded by the two circles $\mathcal C^\pm$) is bigger than the area of the two disks. That is, the catenoid does not minimize area. Indeed, there is no reason why a minimal surfaces minimizes area, as this example shows. $\endgroup$ – user99914 Oct 31 '15 at 2:47
  • $\begingroup$ Can "the two disks" be considered as another surface with boundaries "the two circles"? I thought that in evaluation of minimal surfaces in differential geometry we always consider one connected surface "ended" to one or many boundaries. $\endgroup$ – L.G. Oct 31 '15 at 3:19
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    $\begingroup$ Yes, that two disk are another surfaces with the same boundary $\mathcal C^\pm$ as the catenoid. $\endgroup$ – user99914 Oct 31 '15 at 3:34
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The exposition is somewhat unclear because they use $a$ twice in two different contexts. For the sake of this discussion, consider only the function $$x = \cosh z, \quad -\infty < z < \infty,$$ from which the minimal surface is formed by the surface of revolution $r(z) = \cosh z$ rotated about the $z$-axis. All subsequent use of the letter $a$ is in the context of some positive constant that can be arbitrarily chosen.

The whole catenoid, of course, is an infinite surface. But if we select a portion of the catenoid satisfying $|z| \le a$, then we get what might seem to be a surface $S$ whose area minimizes the surface area of any surface that "spans" the boundary. But this is not true: the surface area of the catenoid between the planes $z = a$ and $z = -a$ is easily found even by elementary methods, as it is a surface of revolution defined by the function $r(z)$: $$|S| = \int_{z=-a}^a 2 \pi r(z) \sqrt{1 + \left(\frac{dr}{dz}\right)^2} \, dz = 4\pi \int_{z=0}^a \cosh^2 z \, dz = 4\pi \left[\frac{z}{2} + \frac{\sinh 2z}{4}\right]_{z=0}^a = \pi (2a + \sinh 2a).$$ This is equivalent to the area described in the text. But the combined area of the two disks is just $$|S_0| = 2(\pi \cosh^2 a),$$ because their common radius is $r(a) = r(-a) = \cosh a$. Now we want to see which quantity is smaller for a given $a > 0$: to this end, we set $$2\pi \cosh^2 a < \pi(2a + \sinh 2a),$$ and expanding this into exponential form, and simplifying, we get $$1+e^{-2a} < 2a,$$ as claimed. This means that whenever $a$ satisfies this inequality, the disks will have less area than the catenoid. Our intuition should tell us that this happens for "sufficiently large" $a$: indeed, numerically solving this inequality gives $a > a_0 \approx 0.639232\ldots$. This is because when $a$ gets large, the catenoid becomes much more "disk-like" at the boundary, whereas if $a$ is small, the catenoid is "cylinder-like". But simply having two disks is going to clearly win out over two deformed funnel-like shapes that have been made to join at their centers.

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  • $\begingroup$ Considering definition of minimal surfaces in differential geometry, would $S_0$ be another surface to evaluate and compare? I think $S$ means a surface 'between' the two-circle boundaries. Also why in case $a<a_0$, i.e. approaching a cylinder-like, the catenoid will not have minimum value area? Thank you. $\endgroup$ – L.G. Oct 31 '15 at 3:29
  • $\begingroup$ Well, for a given boundary consisting of the two circles $C^+$ and $C^-$, yes, the surface $S_0$ is a minimal surface in that it obviously has mean curvature of $0$. As for the second part of your comment, no: when $0 < a < a_0$, the catenoid has SMALLER area than the two disks, as should be evident by the animation. When $a$ is small, the catenoid is cylinder-like, and the disks spanning the boundary clearly have greater area. When $a = a_0$, the catenoid will have surface area equivalent to the two disks. $\endgroup$ – heropup Oct 31 '15 at 3:38

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