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