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:

           (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!

  • $\begingroup$ 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.) $\endgroup$ – Kevin Walker Jul 8 '11 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 '11 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. $\endgroup$ – Joseph O'Rourke Jul 8 '11 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?... ... $\endgroup$ – André Henriques Jul 8 '11 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. $\endgroup$ – Joseph O'Rourke Jul 8 '11 at 23:57

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! $\endgroup$ – Joseph O'Rourke Jul 9 '11 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). $\endgroup$ – Andrey Rekalo Jul 9 '11 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 '18 at 21:46

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:


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


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


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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... $\endgroup$ – Joseph O'Rourke Jul 8 '11 at 21:03

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