shape formed by a stiff string with ends pinched together Suppose I have a string of length $L$ with a bending energy given by 
$$E=\frac{1}{2}\epsilon \int_0^L ds\, (\mathbf{R}''(s))^2 $$
Let's say I form a bight with it by pinching the ends together, similar to this but with the string only meeting at the ends:

In other words, $\mathbf{R}(0)=\mathbf{R}(L)$, and $\mathbf{R}'(0)=-\mathbf{R}'(L)$. 
What is the shape that will be formed and what is its energy?
 A: Oh, I like this. Hopefully I can avoid stupid math mistakes. (No warranty is made, express or implied, that I will not make stupid math mistakes. Read at your own risk)
We can attack this problem with calculus of variations. Specifically, we can map this to a problem in Lagrangian mechanics using a Lagrangian proportional to $\mathcal{L}=\ddot{\mathbf{R}}$. The Euler-Lagrange equations with this extra derivative are given by:
$\frac{\partial\mathcal{L}}{\partial q}-\frac{d}{ds}\frac{\partial\mathcal{L}}{\partial{\dot{\mathbf{R}}}}+\frac{d^2}{ds^2}\frac{\partial\mathcal{L}}{\partial\ddot{\mathbf{R}}}=0$
Note I'm playing a little fast and loose with partial derivatives with respect to vectors: if you write out the full form of $\mathbf{R}$ and do partial derivatives with respect to each variable it gives you the same thing.
From force of habit, I'm using the dots to denote the partial derivative with respect to s here. The first two terms are zero, and we quickly see that we are left with:
$\frac{d^4\mathbf{R}}{ds^4}=0$
So then we have to solve this differential equation with the given initial conditions.
First, pick the origin and the axes- the simplest choice is that $\mathbf{R}(0)=\mathbf{R}(L)=0$ and $\mathbf{R}^\prime(0)=-\mathbf{R}^\prime(L)=\hat{x}$, where $\hat{x}$ is a unit vector in the x direction.
We can now break the vector equation into separate component equations (I'm going to assume that the string is in a plane, but you can trivially add a third dimension):
$\mathbf{R}=x(s)\hat{x}+y(s)\hat{y}$
$\frac{d^4 x}{ds^4}=\frac{d^4 y}{ds^4}=0$
with the constraint that $\left(\frac{dx}{ds}\right)^2+\left(\frac{dy}{ds}\right)^2=1$ so that $s$ is the distance along the string.
At this point you can just plug it in to a numerical ODE solver, or know what kind of shape this makes if you have a bigger geometry hat than I do. Sadly, Mathematica is acting up for me right now, so I'll try to edit this in later. Hope this gives you a starting point though!
EDIT: This is totally wrong. I'll fix it later when I get a chance. 
A: This turns out to be more involved than I though. The problem is that the boundary conditions I gave can't be matched by a stationary solution. You need a source term to force the ends together.
The solution with a constant source is given in this paper. In brief, if we define a curvature $\kappa$ so that $\kappa^2=(\mathbf{R}'')^2$, then 
$$-\kappa'' - \kappa (\frac{1}{2} \kappa^2-\mu) = S $$
where $\mu$ is a Lagrange multiplier enforcing the length constraint on the string and $S$ is the source term. The solutions when parameters are scaled so $S=-1$ are
$$\kappa(s) = c_1 + \frac{c_2}{c_3+\mathrm{cn}(c_4\,s - 2K(c_5), c_5)}$$
where cn is the Jacobi elliptic function and $K$ is the complete elliptic integral of the first kind. The $c$ constants are given in the paper.
This doesn't exactly solve the original question, but I now think the solution must be related to how exactly we physically describe "pinching" the ends, so the question is not detailed enough.
