Euler-Lagrange equation for a given functional I have a certain functional, very similar to the following:
$\mathcal{L}=\int \limits_{\Omega} \left\{- b(x)\ c(x)\ + \int \limits_{\Omega '} x x' [c(x)-c(x')] \ \mathrm{d}x'\right\} \mathrm{d}x =\int \limits_{\Omega} \Lambda \ \mathrm{d}x$
where for brevity:
$\Lambda=- b(x)\ c(x)\ + \int \limits_{\Omega '} x x' [c(x)-c(x')] \ \mathrm{d}x'$
Here the functional depends on a function, $c(x)$, evaluated at different points of the domain $\Omega$, namely $x$ and $x'$. I am trying to evaluate the first variation of $\mathcal{L}$:
$\delta \mathcal{L}=0$
Am I right when I consider that $\mathcal{L}$ depends on $(x,c(x))$? Or should I consider also $(x',c(x'))$? I am a bit confused about this.
 A: Suppose $\Omega$ and $\Omega'$ are such that $\mathcal{L}$ is well-defined. Provided that
$$
\int_{\Omega'}xx'\left[c(x)-c(x')\right]{\rm d}x'=xc(x)\int_{\Omega'}x'{\rm d}x'-x\int_{\Omega'}x'c(x'){\rm d}x',
$$
the original functional reads
$$
\mathcal{L}=-\int_{\Omega}b(x)c(x){\rm d}x+\int_{\Omega}xc(x){\rm d}x\int_{\Omega'}x'{\rm d}x'-\int_{\Omega}x{\rm d}x\int_{\Omega'}x'c(x'){\rm d}x'.
$$
Let 
$$
\alpha=\int_{\Omega}x{\rm d}x\quad\text{and}\quad\alpha'=\int_{\Omega'}x'{\rm d}x',
$$
which are independent from $c=c(x)$. Then the functional reads
$$
\mathcal{L}=-\int_{\Omega}b(x)c(x){\rm d}x+\int_{\Omega}\alpha'xc(x){\rm d}x-\int_{\Omega'}\alpha x'c(x'){\rm d}x'.
$$
Let $\Omega,\Omega'\subset\mathbb{R}^d$. The functional can be written as
$$
\mathcal{L}=\int_{\mathbb{R}^d}\left[\left(\alpha'x-b(x)\right)c(x)\mathbb{1}_{\Omega}(x)-\alpha xc(x)\mathbb{1}_{\Omega'}(x)\right]{\rm d}x.
$$
Thus it suffices to consider
\begin{align}
\Lambda&=\left(\alpha'x-b(x)\right)c(x)\mathbb{1}_{\Omega}(x)-\alpha xc(x)\mathbb{1}_{\Omega'}(x)\\
&=\left[\left(\alpha'x-b(x)\right)\mathbb{1}_{\Omega}(x)-\alpha x\mathbb{1}_{\Omega'}(x)\right]c(x).
\end{align}
