I am curious if the "catenary problem" of finding the shape of a hanging chain can be solved without the implicit assumption that the chain never "doubles back" on itself so that there is a unique vertical displacement $y$ for a given horizontal displacement $x$.
To this end, I consider curves $\mathbf{r}(s)=(x(s),y(s))$ parameterized by some $s \in [s_o,s']$. It seems we can express the potential energy functional $V[\mathbf{r}(s)]$ to be minimized via $$ V[\mathbf{r}(s)] = \int_{s_o}^{s'} ds \rho y(s) $$ where $\rho$ is the linear mass density of the chain. This functional represents the true potential energy only for those chain trajectories satisfying \begin{equation} \left|\frac{d}{ds}{\mathbf{r}}(s)\right| = 1 \tag{a}\label{a} \end{equation} for all $s$.
This constraint is however not a constraint to which we can directly associate a Lagrange multiplier and solve the Euler-Lagrange equations. What we require instead is a functional mapping the entire trajectory $\mathbf{r}(s)$ to the reals. It seems we can accomplish this by defining the following constraint $g[\mathbf{r}(s)]$: $$ g[\mathbf{r}(s)] = \int_{s_o}^{s'}ds \left(\left|\frac{d}{ds}{\mathbf{r}}(s)\right| - 1\right)^2 \tag{b}\label{b} $$ It seems this constraint is "correct" in the sense that the curves represented by the trajectories $g^{-1}(0)$, namely the curves of length $\Delta s = s'- s_o$, are precisely the curves over which we want to minimize the potential energy, and for these curves we are also sure that $V[\cdot]$ yields the accurate potential energy.
After trying to solve the Euler-Lagrange equations I only seem to obtain equations like $0=0$. I suspect that I have "cheated" in some sense by constructing \eqref{b} by integrating the square of \eqref{a}. Is there some additional condition that constraints must obey in order to apply the method of Lagrange multipliers?