How do Lagrange multipliers work for functionals? (Motivation: the catenary)

Question

I'm fairly new to the calculus of variations, so please bear with me. This question is very similar to this other question posted six months ago that has as yet not received any answers. If a good discussion is had here I will be sure to send the findings over to the previous asker. With that out of the way, here is my question:

I want to derive the equation of the catenary using the Euler-Lagrange equation. I know that this has been done many times, but I want to do it myself. Below is a diagram:

Let the path of the wire between the two points be described by some function $y$. It is natural to want to minimize the potential energy of the wire. Assuming the wire has a constant density $\rho$ this boils down to $$U[y]=\rho g\int_{x_1}^{x_2}y~\mathrm{d}x$$ But of course, there are three constraints: $$y(x_1)=y_1~;~y(x_2)=y_2$$ $$\int_{x_1}^{x_2}\sqrt{1+{y'}^2}~\mathrm{d}x=L$$ The problem is I don't really know how to apply these constraints properly. I tried using the Euler-Lagrange equation but this only brought me to $$\rho g(x_2-x_1)=0\implies x_1=x_2$$ Which is obviously nonsense. I looked at Wikiversity's derivation of the catenary and they mention using a Lagrange multiplier. But, the only context in which I'm familiar with using Lagrange multipliers in is when we want to extremize some function $f:\mathbb{R}^n\to\mathbb{R}$ subject to some constraint $g(\mathbf{x})=C$. My main question is: how do we use Lagrange multipliers for extremizing functionals? And why do the same methods work as to extremizing functions?

Thanks for your time,

-K.defaoite

Answer 1

The problem with your formulation is that the integration along the solution curve should be

$$ \mathcal{L}=\rho g\int_{x_1}^{x_2}y \sqrt{1+y'^2}dx+\lambda\left(\int_{x_1}^{x_2}\sqrt{1+y'^2}dx-L\right) $$

You are using in the first integral $dx$ instead of $ds$.

NOTE

The potential energy is given by

$$ \rho g \int_{x_1}^{x_2}y ds $$

and the length is given by

$$ L = \int_{x_1}^{x_2}ds $$

so the lagrangian is

$$ \mathcal{L}=\rho g\int_{x_1}^{x_2}y ds + \lambda\left(\int_{x_1}^{x_2}ds-L\right) $$

or

$$ \mathcal{L}=\int_{x_1}^{x_2}\left(\left(\rho g y+\lambda\right)\sqrt{1+y'^2}-\frac{\lambda L}{(x_2-x_1)}\right)dx $$

because $ds = \sqrt{dx^2+dy^2} = \sqrt{1+y'^2}dx$