# How to get the height of a constrained, infinite elastic surface?

Suppose I have an infinite elastic sheet, constrained by several poles of height 1. The height of the surface would be 1 at the coordinates of each of the poles, and converge to 0 as the distance from the poles goes to infinity.

How can I calculate the height of the surface as a function of coordinates, given the mass density, elasticity coefficient, and the coordinates of the poles?

• You may find it useful to explore the literature on the infinite plate subject to one or more perpendicular point loads. It seems like an appropriate solution would minimize the sum of the total bending strain energy and the total gravitational potential energy. It the sheet is prevented from slipping at the tops of multiple poles, then stretching strain energy would also come into play. With only a single pole and negligible bending stiffness, I believe the solution would be trivial—perfectly vertical draping around the pole to minimize the gravitational potential energy. – Chemomechanics May 9 at 23:04
• I think the problem may be ill-posed without knowing the radii of the poles. Consider the limiting case of a single "point pole" at $\vec{r} = 0$, holding up a sheet of negligible mass. If the displacement is "small" in some appropriate sense, then the height of the sheet will approximately obey Laplace's equation. But there are no solutions to Laplace's equation that approach a finite value as $r \to 0$ and $r \to \infty$. So you can't have a solution that approaches the height of the pole if the height of the pole is finite. – Michael Seifert May 13 at 18:42
• @MichaelSeifert: why do you think that Laplace equation should hold near the singularity? Born-Infeld electrostatics, for example, has Laplace equation as its weak field limit and a regular behaviour near point sources. – A.V.S. May 14 at 7:40
• A similar solution holds for a rope with mass density. It can be found in almost any book on variation calculus with a catenary curve as the solution. I am not saying that this will be the solution to your problem but you may find the technique useful, and it may have been solved. In theory you should be able to set up Newton's eqn for a small piece of the sheet and require a balancing of the forces throughout the sheet. In the limit as the pieces -->0 you will recover a PDE for the shape. – ggcg May 15 at 18:49
• Are you sure you can construct a bound system with given parameters? If the poles are of finite length, $\lim\limits_{r\to{\infty}} h(r) = -\infty$, and I cannot see a scenario where $\lim\limits_{r\to{\infty}} h(r)$ would be finite. – acarturk May 17 at 22:18

The short answer is: such surface does not (always) exist.

## The model

Let's assume for simplicity that this elastic sheet is a "membrane" that can support tension but not bending. It's energy $$E$$ is then proportional to the stretch it undergoes $$E = \alpha\int u_x^2 + u_y^2$$ where $$u$$ is the (vertical) displacement of the membrane from its reference configuration, $$(u_x,u_y)$$ is the gradient of $$u$$, and $$\alpha$$ is equivalent to an elasticity modulus. I have assumed that the sheet is isotropic and that the displacements are small in some sense. Let's also forget about gravity for a second. Minimizing $$E$$ leads to the Poisson's (Laplace?) equation $$u_{xx}+u_{yy} = 0\quad\text{or}\quad u_{rr}+\frac{u_r}{r}+\frac{u_{\theta\theta}}{r^2}=0$$ in polar coordinates; here too, index means derivative not component.

## A (non)solution

Suppose we are using a single pole placed at $$r=0$$. The surface is expected to be axisymmetric with $$u=u(r)$$ being independent of $$\theta$$. Poisson's equation can be integrated into $$u = a\ln(r)+b.$$ From here, it is clear that there are no solutions satisfying both $$u(0)=1$$ and $$u(\infty)=0$$. In fact, satisfying either is enough to make the solution trivial. Adding gravity only worsens our predicament; it adds a quadratic term to $$u$$ too soft to appease the singularity at $$0$$ and too wild to appease the divergence at $$\infty$$.

## Workaround 1

To eliminate both problems at $$0$$ and at $$\infty$$, we could assume that the pole has a finite radius $$a$$ and that the membrane has a finite radius $$b$$ in which case the solution is $$u = \frac{\ln(r/b)}{\ln(a/b)}.$$ With gravity, we add an $$r^2$$ contribution and adjust the integration constants. Last, with multiple poles, solutions can be linearly combined. In that case, axisymmetry (of elementary solutions) no longer holds and we need to solve the "full" equation; which I believe shouldn't be too hard as long as the inner and outer boundaries are concentric circles.

## Workaround 2

Taking into account bending energy has a regularizing effect since it involves higher order derivatives. This is expected to remove the singularity at $$0$$ but I do not see why it would enforce a decay towards $$0$$ at $$\infty$$. In any case, the simplest bending energy would be proportional to the mean curvature $$u_{xx}+u_{yy}$$ squared. All in all, the total energy one would minimize would look like $$E = \alpha\int(u_x^2 + u_y^2) + \beta\int(u_{xx}+u_{yy})^2 - \int\rho g u.$$ The resulting PDE is of order 4. For axisymmetric loading and in the absence of membrane energy now (i.e., $$\alpha=0$$), analytical solutions are known and are linear combinations of the terms $$r^4,\quad \ln r,\quad r^2,\quad r^2(2\ln r -1),\quad 1.$$ The singularity at $$0$$ can then vanish by omitting the $$\ln r$$ term from the solution (i.e., by enforcing ad-hoc boundary conditions) but the divergence at $$\infty$$ cannot.

• The first "workaround" does not satisfy the condition of being finite when $r\rightarrow \infty$. – nicoguaro May 23 at 21:36
• Yes, which why the workaround is to suppose that the membrane has a finite radius. – Hussein May 23 at 22:38
• I understand now, thanks. – nicoguaro May 23 at 22:40