Euler-Lagrange equations for a double integral

Question

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.

Answer 1

Let's define $F(x,y):=f(x)-f(y)$ and $G(x,y):=g(x,y)\theta(y-x)$, where $\theta(\cdot)$ is the Heaviside step function. Then, we may rewrite the functional $J$ as $$ J[F]=\int_a^b\int_a^b{\cal L}(x,y,F,F_x)\,dxdy, \tag{1} $$ where $$ {\cal L}(x,y,F,F_x):=\frac{e^{-F(x,y)}}{F_x(x,y)}G(x,y). \tag{2} $$ For this problem, the appropriate form of the Euler-Lagrange equation is given by $$ \frac{\partial \cal{L}}{\partial F}-\frac{\partial}{\partial x}\left(\frac{\partial \cal{L}}{\partial F_x}\right)=0. \tag{3} $$ Explicitly, \begin{align} (3)&\implies -\frac{e^{-F}}{F_x}G-\frac{\partial}{\partial x}\left(-\frac{e^{-F}}{F_x^2}G\right)=0 \\ &\implies -2\frac{e^{-F}}{F_x}G-2\frac{F_{xx}\,e^{-F}}{F_x^3}G+\frac{e^{-F}}{F_x^2}G_x=0 \\ &\implies -2(F_x^2+F_{xx})G+F_xG_x=0. \tag{4} \end{align} Incidentally, it seems that $(4)$ does not have a solution, since $F_x(=f'(x))$ and $F_{xx}(=f''(x))$ depend only on $x$, whereas $G$ is a function of both $x$ and $y$.

Answer 2

Recalling that, in the present case, the functional derivative is taken as $$ \frac{\delta}{\delta f} \equiv \frac{\partial}{\partial f} - \frac{\mathrm{d}}{\mathrm{d}x} \frac{\partial}{\partial f'} \quad\mathrm{with}\quad \frac{\delta f(x)}{\delta f(x')} = \delta(x-x'), $$ where the last $\delta$ denotes the Dirac delta function, one has : $$ \begin{array}{rcl} \displaystyle \frac{\delta J[f]}{\delta f(z)} &=& \displaystyle \int_a^b\mathrm{d}y \int_a^y\mathrm{d}x \left(\frac{e^{f(y)-f(x)}}{f'(x)}(\delta(y-z)-\delta(x-z)) - \frac{e^{f(y)-f(x)}}{f'(x)}(1+2f''(x))\delta(x-z)\right) g(x,y) \\ &=& \displaystyle \int_a^b\mathrm{d}y \int_a^y\mathrm{d}x \left(\frac{e^{f(y)-f(x)}}{f'(x)}(\delta(y-z)-2(1+f''(x))\delta(x-z)\right) g(x,y) \end{array} $$ Then, the Dirac deltas will "kill" some integrals depending on whether $z$ lies within the domains of integration.

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Edit (see comments below) :

The functional differential is given by $$ \delta J[f(x)] = \frac{\partial J}{\partial f}\delta f(x) + \frac{\partial J}{\partial f'}\delta f'(x), $$ hence $$ \frac{\delta J[f]}{\delta f(z)} = \frac{\partial J}{\partial f}\frac{\delta f(x)}{\delta f(z)} + \frac{\partial J}{\partial f'}\frac{\delta f'(z)}{\delta f(z)}, $$ with $$ \frac{\delta f'(x)}{\delta f(z)} = \frac{\delta}{\delta f(z)}\frac{\mathrm{d}f(x)}{\mathrm{d}x} = \frac{\mathrm{d}}{\mathrm{d}x}\frac{\delta f(x)}{\delta f(z)} = \frac{\mathrm{d}}{\mathrm{d}x}\delta'(x-z) \equiv -\delta(x-z)\frac{\mathrm{d}}{\mathrm{d}x} $$ where the last line is due to an integration by parts. In consequence, one gets : $$ \frac{\delta J[f]}{\delta f(z)} = \delta(x-z)\left(\frac{\partial J}{\partial f} - \frac{\mathrm{d}}{\mathrm{d}x}\frac{\partial J}{\partial f'}\right) $$ In the present case, one has $$ \frac{\partial}{\partial f'} \frac{e^{f(y)-f(x)}}{f'(x)} = -\frac{e^{f(y)-f(x)}}{f'(x)^2} $$ and $$ -\frac{e^{f(y)-f(x)}}{f'(x)^2} \frac{\delta f'(x)}{\delta f(z)} \equiv \delta(x-z)\frac{\mathrm{d}}{\mathrm{d}x} \frac{e^{f(y)-f(x)}}{f'(x)^2} = -\delta(x-z)(1+2f''(x))\frac{e^{f(y)-f(x)}}{f'(x)} $$ Note that there is no derivative with respect to $y$ because $f'(y)$ doesn't appear within the functional.

Answer 3

1. 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.

2. Assuming pertinent Dirichlet boundary conditions an infinitesimal variation is of the form $$\begin{align} \delta J[f] ~\stackrel{(1)}{=}~&\iint_{[a,b]^2}\!\mathrm{d}x~\mathrm{d}y~(\delta f(y)-\delta f(x)) \frac{e^{f(y) - f(x)}}{f'(x)} g(x,y)\theta(y\!-\!x)\cr &-\iint_{[a,b]^2}\!\mathrm{d}x~\mathrm{d}y~\delta f^{\prime}(x) \frac{e^{f(y) - f(x)}}{f'(x)^2} g(x,y)\theta(y\!-\!x)\cr ~=~&\int_{[a,b]}\!\mathrm{d}x~\delta f(x) \int_{[a,b]}\!\mathrm{d}y~ \left(\frac{e^{f(x) - f(y)}}{f'(y)} g(y,x)\theta(x\!-\!y) \right.\cr &\qquad -\left. \frac{e^{f(y) - f(x)}}{f'(x)} g(x,y)\theta(y\!-\!x) \right.\cr &\qquad +\left. \frac{d}{dx}\left(\frac{e^{f(y) - f(x)}}{f'(x)^2} g(x,y)\theta(y\!-\!x)\right)\right).\end{align}\tag{2}$$

3. Therefore the functional derivative is $$\begin{align} \frac{\delta J[f]}{\delta f(x)}~\stackrel{(2)}{=}~&\int_{[a,b]}\!\mathrm{d}y~ \left(\frac{e^{f(x) - f(y)}}{f'(y)} g(y,x)\theta(x\!-\!y) \right.\cr &\qquad -\left. \frac{e^{f(y) - f(x)}}{f'(x)} g(x,y)\theta(y\!-\!x) \right.\cr &\qquad +\left. \frac{d}{dx}\left(\frac{e^{f(y) - f(x)}}{f'(x)^2} g(x,y)\theta(y\!-\!x)\right)\right).\end{align}\tag{3}$$