Consider a functional
$$
J[f] = \int_{a}^{b} \int_{a}^{y} \frac{e^{f(y) – f(x)}}{f'(x)} g(x,y) \, dx \,dy,
$$
where the constants $a$, $b$ and the function $g$ are given. I am looking for a function $f$ that minimizes this functional. I have tried to use the Euler-Lagrange equation, but I am unsure how to correctly apply it in this case. The difficulties are:
- There are two integrals, and the function of interest is nested within the inner one.
- The upper limit of the inner integral is not constant.
- Both $f(x)$ and $f(y)$ are present within the functional.
I know that there exist various versions of the Euler-Lagrange equations, and would be very grateful if someone could point out which one could be used in this case.
Best Answer
OP's functional reads $$ J[f]~=~\iint_{[a,b]^2}\!\mathrm{d}x~\mathrm{d}y~\frac{e^{f(y) - f(x)}}{f'(x)} g(x,y)\theta(y\!-\!x),\tag{1}$$ where $\theta$ denotes the Heaviside step function.
Assuming pertinent Dirichlet boundary conditions an infinitesimal variation is of the form $$\begin{align} \delta J[f] ~\stackrel{(1)}{=}~&\iint_{[a,b]^2}\!\mathrm{d}x~\mathrm{d}y~(\delta f(y)-\delta f(x)) \frac{e^{f(y) - f(x)}}{f'(x)} g(x,y)\theta(y\!-\!x)\cr &-\iint_{[a,b]^2}\!\mathrm{d}x~\mathrm{d}y~\delta f^{\prime}(x) \frac{e^{f(y) - f(x)}}{f'(x)^2} g(x,y)\theta(y\!-\!x)\cr ~=~&\int_{[a,b]}\!\mathrm{d}x~\delta f(x) \int_{[a,b]}\!\mathrm{d}y~ \left(\frac{e^{f(x) - f(y)}}{f'(y)} g(y,x)\theta(x\!-\!y) \right.\cr &\qquad -\left. \frac{e^{f(y) - f(x)}}{f'(x)} g(x,y)\theta(y\!-\!x) \right.\cr &\qquad +\left. \frac{d}{dx}\left(\frac{e^{f(y) - f(x)}}{f'(x)^2} g(x,y)\theta(y\!-\!x)\right)\right).\end{align}\tag{2}$$
Therefore the functional derivative is $$\begin{align} \frac{\delta J[f]}{\delta f(x)}~\stackrel{(2)}{=}~&\int_{[a,b]}\!\mathrm{d}y~ \left(\frac{e^{f(x) - f(y)}}{f'(y)} g(y,x)\theta(x\!-\!y) \right.\cr &\qquad -\left. \frac{e^{f(y) - f(x)}}{f'(x)} g(x,y)\theta(y\!-\!x) \right.\cr &\qquad +\left. \frac{d}{dx}\left(\frac{e^{f(y) - f(x)}}{f'(x)^2} g(x,y)\theta(y\!-\!x)\right)\right).\end{align}\tag{3}$$