Typically the Euler-Lagrange equations are defined for the functional


$$ J[u] = \int\_a^b L(x,u,u') dx. $$


However, I was wondering if anyone knows if they can be solved when the expression involves the inverse of $u$? The way my problem is formulated, it is simplest to write as


$$ J[u] = \int\_{u^{-1}(a)}^{u^{-1}(b)} L(x,u,u') dx, $$


but it could equivalently be written 


$$ J[u] = \int\_{u(a)}^{u(b)}L(x,u,u^{-1},u') dx $$


I had a bit of a play with it and I have a feeling it is not generally solvable, but I'm really just playing -- not an expert in this area.


Cheers!


EDIT: 


To make it more clear why this situation might arise, consider a distribution with pdf $g(x,\vec{\theta})$, but for which the variates are measured with some uncertainty (i.e. the "observed" distribution is a convolution of the original with the uncertainty distribution of the variates). Whether we can directly calculate the convolution or not, since it preserves number density (between different limits), it can be parameterised as a function $\hat{u}(x)$. 


Now consider a hierarchical likelihood model which aims to recover the original underlying distribution. Such a likelihood may look like this:


$$ \ln\mathcal{L} = \sum \ln g(x\_i) + \ln P(x\_i|\hat{x\_i})$$


where the $x$ are (hyper-)parameters and the $\hat{x}$ are the observed variates including uncertainty, and $P$ is the uncertainty distribution. Let's call the conversion from estimated to observed $u$, i.e. $\hat{x} = u(x)$ (so that if $u$ = $\hat{u}$ we recover the initial distribution). In the continuous limit, the likelihood should look like


$$ \ln \mathcal{L} = \int\_{u^{-1}(x\_0)}^{u^{-1}(x\_1)} \frac{g(\hat{u}^{-1}(u(x)),\tilde{\vec{\theta}})}{d\hat{u}/d\tilde{x}} \left[ \ln g(m,\vec{\theta}) + \ln P(m|u(m))\right] dx $$


(I think I have all my jacobian adjustments etc. right here -- oh, and tilde represents original input parameters). Thus you can see the inverse in the limits. If instead I defined $u$ the other way around, so that $x = u(\hat{x})$, then the limits have $u$ but the integrand itself has $u^{-1}$.


Hope that makes sense!



Calculus of variations when functional involves inverse of the function
Probably, the best thing to do would be to write $x = f(u)$ and then use
$$
\int\_{u^{-1}(a)}^{u^{-1}(b)} L(x,u,u') dx = \int\_a^b L\left(f(u),u,\frac{1}{f'(u)}\right)f'(u)\ du
= \int\_a^b M\left(u,f(u),f'(u)\right)\ du
$$
and now compute the Euler-Lagrange equation for $M$, which will give you the differential equation you want to solve for the function $f$.


