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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.

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  • $\begingroup$ you can define $c(e,x') = d(x')$ to avoid the ambiguity, where $e$ is some constant. Now you have $b,c,d$ to consider. $\endgroup$ – L.F. Cavenaghi May 10 '18 at 15:21
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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}

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