Calculus of variations with inverse of derivative composited within?

Question

Suppose $s \in S \subset \mathbb{R}^+$ with c.d.f. $F(s)$. A real-valued function $\beta(s) \leq s$ is determined by a first order condition $\gamma'(s - \beta(s)) = g(F(s)) \in \mathbb{R}$, where $\gamma$ is a real valued penalty function. I need to find the $\gamma$ which $$ \min \int_{S} (s - \beta(s))^2 dF(s) $$ subject to a constraint $$ \int_{S} \gamma (s - \beta(s)) dF(s) \leq K \in \mathbb{R}. $$ The Lagrangian with the first order condition plugged in, and change of variable, is $$ \mathcal{L}=\int_0^1 [\gamma'^{-1} (g(x))]^2 + \lambda \gamma(\gamma'^{-1}(g(x))) dx $$ I am not looking for an analytical solution, but any guidance about which direction I should look into? I would like to obtain a differential equation about $\gamma$ so that I can quantitatively analyze the function $\gamma$.

Answer 1

I'll start with the final expression for your Langrangian, assuming that $g,\gamma, \gamma'$ are all invertible functions and well defined in the respective intervals that the compositions define.

By making the substitution $u=\gamma'^{-1}(g(x))$ the Lagrangian takes the following form:

$\mathcal{L}=\int^{\gamma'^{-1}(g(1))}_{\gamma'^{-1}(g(0))}\Big(\frac{\gamma''(u)}{g'(g^{-1}(\gamma'(u)))}(\lambda\gamma(u)+u^2)\Big)du$

which now has the form $\mathcal{L}=\int{duF(\gamma(u),\gamma'(u), \gamma''(u),u)}$ and can be tackled using standard variational techniques. The Euler-Lagrange equation should read:

$\frac{d^2}{du^2}(\frac{\partial F}{\partial\gamma''})-\frac{d}{du}(\frac{\partial F}{\partial\gamma'})+\frac{\partial F}{\partial \gamma}=0$.