Summary. I am studying a variational problem with Dirichlet b.c. and an integral equality constraint on an interval $[0,L]$, in which determining $L$ is part of the problem. I have derived extremality conditions in two different ways, and they give me different equations to determine the Lagrange multiplier for the equality constraint.
I have verified numerically that the equations I get from the two derivations result in the same extremal and the same value for the Lagrange multiplier. But it is not clear to me why the two expressions for the Lagrange multiplier would be equal at the extremal. Is there a direct way to see this, perhaps a physical explanation?—One that is more informative than "because the formulas say so" :)
Setup. I want to study extremals of the following variational problem: Among all plane curves connecting the origin to some point on the line $y=1$ such that the curve is horizontal at the starting point, and vertical at the endpoint, which ones have extremal bending energy?
We can formalize this problem by using the turning angle $\alpha:[0,L]\to\mathbb{R}$ of the arc-length parametrized curve as the primary variable. Then we are looking for extremals of the functional $\int_0^L \alpha'^2$ subject to $1=\int_0^L \sin\alpha$, and the Dirichlet b.c. $\alpha(0)=0$ and $\alpha(L)=\pi/2$. The Lagrangian of the problem is $\int_0^L (\alpha'^2+\lambda\sin\alpha)$. Denote $f=\alpha'^2+\lambda \sin\alpha$.
Derivation 1. The Euler–Lagrange equation is easily seen to be $\alpha''=\frac{1}{2} \lambda \cos \alpha$. The transversality condition for a problem of the given form is $0=f-\alpha' f_{\alpha'}$ at the endpoint $L$ of the interval. This equation is equivalent to $\lambda=(\alpha'(L))^2$. The E–L equation, the equality constraint, and the transversality condition suffice to determine a solution numerically. This gives approximately $L=2.187$, $\lambda=1.436$, and an energy $\int_0^L \alpha'^2=1.436$.
Derivation 2. Another way of deriving extremality conditions is by mapping the interval $[0,L]$ to $[0,1]$, and accounting for the length change with an additional unknown $L$. To do this, we use an auxiliary primary variable $\beta:[0,1]\to\mathbb{R}$, and set $\alpha(s):=\beta(s/L)$. Then, the Lagrangian can be represented as $\int_0^1 (\frac{1}{L}\beta'^2+\lambda L \sin \beta)$. We can set the partial derivative of this integral wrt $L$ to zero. The resulting equation simplifies to $\lambda=\frac{1}{L}\int_0^1 \beta'^2=\int_0^L \alpha'^2$, which replaces the transversality condition from Derivation 1. The E–L equation is the same as for Derivation 1.
Question. The numerical experiment suggested that $(\alpha'(L))^2=\int_0^L \alpha'^2$ may hold at extremals, and the two different derivations prove that this is actually true. Is there a direct way to see why? Without going through two separate formalizations of the problem? Is there an explanation of why the total energy would be equal to the energy density at the endpoint?