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A catenary curve is the shape taken by an idealized hanging chain or rope under the influence of gravity. It has the equation $y= a \cosh (x/a)$. My question is:

What is the shape taken by an idealized, thin two-dimensional sheet, pinned on a plane parallel to the ground, under the influence of gravity?

The answer surely depends on how it is pinned to the plane, the boundary conditions. Natural options are:

  • A disk sheet fixed to a circle.
  • A square sheet fixed to a square.
  • A square sheet pinned at its four corners.

The middle option above would look something like this when inverted:


           CatDome
           (Image by Tim Tyler at hexdome.com.)


I don't think any of these shapes is a catenoid, which is the surface of revolution formed by a catenary curve. Is there a simple analytic description of any of these surfaces, analogous to the $\cosh$ equation for the catenary curve? I have been unsuccessful in finding anything but simulations of solutions of the differential equations.

This question arose in imagining a higher-dimensional version of the property that an inverted catenary supports smooth rides of a square-wheeled bicycle (explored in this MO question). Thanks for pointers!

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  • $\begingroup$ How is the surface allowed to deform in the interior? Area preserving? Conformal? Preserving the volume of the region is incloses? (Like a soap film, but with gravity taken into account.) $\endgroup$ Jul 8, 2011 at 19:18
  • $\begingroup$ Keivn, good point. For example, a rotated catenary surface is quite simply not isometric to a flat disc. So we might, for instance, be asking about a rubber sheet in the shape of a disc, glued down along a circular boundary, and allowed to sag in the middle under gravity. The elastic energy less resembles the mean curvature operator in favor of the ordinary Laplacian. $\endgroup$
    – Will Jagy
    Jul 8, 2011 at 19:28
  • $\begingroup$ @Kevin,Will: Good questions! I had imagined a thin bedsheet, or loose chain mail. In the above image, the hexagon edge lengths are fixed, so it is akin to chain mail. $\endgroup$ Jul 8, 2011 at 20:34
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    $\begingroup$ Dear Joseph. AS mentioned by others, your problem is somewhat ill-posed... Are you trying to minimize the energy of a surface or of a parametrized surface? Are you fixing the area of the surface? Does the energy contain only a potential term, or does it also contain a term having to do with the derivative of the parametrizing map? For the last question "a square sheet pinned at its four corners": is the length of the free boundary fixed?... ... $\endgroup$ Jul 8, 2011 at 22:52
  • $\begingroup$ @André: Point well-taken! I guess the model in the image is closest: tesselate the surface with identical polygons, each composed of rigid links joined at universal joints. Maintain the length of each link, allowing rotation at the joints. This is analogous to a hanging chain composed of rigid links connected at rotatable joints. It would approximate woven cloth of a certain constitution. $\endgroup$ Jul 8, 2011 at 23:57

2 Answers 2

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A model equation for an inextensible, flexible, heavy surface in a gravitational field was deduced by Poisson Lagrange and later the problem was also studied by Poisson (see the references in the linked papers below). The equilibrium condition for a hanging heavy surface of constant mass density reads $$\sqrt{1+|\nabla u|^2}\ \nabla\cdot{}\frac{\nabla u}{\sqrt{1+|\nabla u|^2}}=\frac{1}{u+\lambda},\qquad x\in\Omega\subset\mathbb R^2,\qquad\qquad(1)$$ where $u=u(x)$ is the vertical displacement and $\lambda\in\mathbb R$ is an arbitrary constant (a Lagrange multiplier). (1) is the Euler equation of the variational integral $$I(u)=\int_{\Omega}u\sqrt{1+|\nabla u|^2}dx,$$ which can be interpreted as the vertical coordinate of the center of gravity of the surface $$\mbox{graph}(u)=\{(x,u(x)):\ x\in\Omega\}\subset\mathbb R^2\times\mathbb R.$$

Equation (1) is to be supplemented with the requirement that the surface has a prescribed area $A$ $$\qquad\qquad\qquad\qquad\qquad\int_{\Omega}\sqrt{1+|\nabla u|^2}dx=A,\qquad\qquad\qquad\qquad\qquad\qquad\quad\quad(2) $$ and the Dirichlet boundary condition describing the curve from which the surface is being suspended $$\qquad\qquad\qquad\qquad\qquad\qquad\qquad\quad\left.u\right|_{\partial \Omega}=g.\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\quad(3) $$ One can check formally that a solution to (1)-(3) provides a graph of a heavy surface of prescribed area and boundary with the lowest center of gravity, so this is a precise 2D analogue of the classical catenary problem.

It is known that problem (1)-(3) has no classical solutions for the values of area $A$ outside of some bounded interval $[A_{\min},A_{\max}]$. Moreover, the corresponding variational problem has no global solutions for all $A\in\mathbb R$. A short survey of some old and relatively new results concerning well-posedness of (1)-(3) and its multidimensional analogues can be found in the paper by Dierkes and Huisken, "The N-dimensional analogue of the catenary: Prescribed area", in J. Jost (ed) Calculus of Variations and Geometric Analysis, Int. Press (1996), pp. 1-13.

Addendum. Here is a more recent survey by Dierkes: "Singular Minimal Surfaces" (in Geometric Analysis and Nonlinear Partial Differential Equations, Springer (2003), pp. 177-194).

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  • $\begingroup$ @Andrey: That link is broken but you must mean this: "The $n$-dimensional analogue of the catenary: Existence and nonexistence," U. Dierkes and G. Huisken, Pacific J. Math., Volume 141, Number 1 (1990), 47-54. projecteuclid.org/… This seems exactly what I sought---Thanks so much! $\endgroup$ Jul 9, 2011 at 0:16
  • $\begingroup$ @Joseph: Thank you, I've replaced the broken link with a reference. Actually, I referred to a different, more recent publication with a similar title (please see the corrected answer). $\endgroup$ Jul 9, 2011 at 12:40
  • $\begingroup$ I believe the humble cone is a solution to this (with circular boundary), at least if I did my math correctly, which I'm not sure I did. $\endgroup$
    – Tally
    Sep 18, 2018 at 21:46
  • $\begingroup$ From where did you get equation (1)? Why does $u$ ecist in the denominator of the right hand side? $\endgroup$ Oct 31, 2020 at 1:11
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The thing that comes to mind is the capillary surface including gravity. See the note by Finn, available free as a pdf, as a reference at the end of:

http://en.wikipedia.org/wiki/Capillary_surface

Hmmm, maybe not. Your surface would not have a large flat region in the middle...

A rotated catenary surface is quite simply not isometric to a flat disc. So we might, for instance, be asking about a rubber sheet, glued down along a boundary, and allowed to sag in the middle under gravity. The elastic energy less resembles the mean curvature operator in favor of the ordinary Laplacian

http://en.wikipedia.org/wiki/Elastic_energy

It appears you are looking for the biharmonic equation, as the force of gravity vector field will be considered constant and divergence free, so the displacement $u$ satisfies $\Delta^2 u = 0.$ See

http://en.wikipedia.org/wiki/Linear_elasticity#Elastostatics

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  • $\begingroup$ Thanks, Will! Your last link especially seems quite useful. But it's leading me to suspect that there may be no cleanly expressible solutions to the differential equations... $\endgroup$ Jul 8, 2011 at 21:03
  • $\begingroup$ While a water drop may not be it, a soap film would probably obtain the sought after shape, even though inverted. The rubber sheet is probably not a good model as it gets a lower surface density on places where it is stretched out, where it will also need to take a greater amount of force. $\endgroup$ Oct 31, 2020 at 0:58

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