Surface equivalent of catenary curve

Question

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!

Answer 1

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).

Answer 2

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