In this Lagrangian (from the paper: https://arxiv.org/abs/1302.0192 – page 4), $\eta, \mu, \nu, \& \lambda$ are lagrange multipliers.
My question is: why do they include $\nu$ and $\lambda$ inside the integral but the others not? In other words, when do we use varying lagrange multipliers?
In the same paper they use this theorem of calculus of variation:
\begin{equation}
\mathcal{L'}(X_e)(\bar{X}) = \frac{d}{d\epsilon}L(X_e + \epsilon \bar{X})|_{\epsilon=0} = 0
\end{equation}
What is $\bar{X}$? What do this theorem differ from differentiation?
Best Answer
Presumably the mulipliers are inside the integral because they wish to enforce $x'=\cos \theta$ at all points $s$. To do this they need a separate Lagrange multiplier $\nu(s)$ at each point $s$.
If they wrote $\nu\int_0^D (x'-\cos \theta) ds$, with only one $\nu$, they would only get $\int_0^D x' ds= \int_0^D \cos\theta ds$, which is not at all the same thing.
Look at problem 3 in this homework set for an example. In that problem the lagrenace multiplier for point with coordinate $s$ is the tension in the cable at that point.