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Let us consider the functional $J$ defined over smooth functions $$ J(f) := \int_0^1 \int_0^x f(y) \cos(f(x)-f(y)) dydx $$ My question is, can I make sense of the functional derivate $\delta J (f, \eta) = \lim_{\epsilon \to 0} \left( \frac{d}{d\epsilon} J(f+\epsilon \eta)\right)$ and if so, is it possible to avoid to integrate the variation $\eta(y)$ over $dy$? Or is it possible to separate the $\eta$'s from the $f$'s such that the functional derivative $\delta J (f, \eta) = \int_0^1 F(f,x) G(\eta,x) dx$. For some functionals $F, G$.

I am trying to find critical points on some larger functional and this term for me is hard to get.

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  • $\begingroup$ I think the inner integral can be evaluated to $ \int_0^x f'(y) \cos(f(x)-f(y) dy = -\sin(f(x)-f(y)) + \sin(f(x)-f(0)) $, which might help. $\endgroup$
    – daw
    Commented Jan 20, 2020 at 6:46
  • $\begingroup$ Oh, I just realized I mistyped the derivative. I am sorry. $\endgroup$
    – MatixCubix
    Commented Jan 21, 2020 at 1:50
  • $\begingroup$ On could also do it with a derivative, but one would obtain integrating by parts $\int_0^1 ( f(x)-f(0) ) \cos( f(x)-f(0) ) dx - \int_0^1 \int_0^x f(y) f'(y) \sin(f(x)-f(y)) dy dx $. So one has to deal with an even more complicated integral. $\endgroup$
    – MatixCubix
    Commented Jan 21, 2020 at 4:41

1 Answer 1

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The first variation can be written as $$ J'(f,h) = \int_0^1 \int_0^x h(y) \cos(f(x)-f(y))dydx -\int_0^1 \int_0^x f(y) \sin(f(x)-f(y))(h(x)-h(y)dydx\\ = \int_0^1 \int_y^1 \cos(f(x)-f(y))dx\ h(y) dy -\int_0^1 \int_0^x f(y) \sin(f(x)-f(y))dy\ h(x)dx\\ +\int_0^1 \int_y^1 f(y) \sin(f(x)-f(y))dx\ h(y) dy, $$ which is in turn equal to $$ J'(f,h) = \int_0^1\left(\int_y^1 \cos(f(x)-f(y))dx - \int_0^y f(x) \sin(f(y)-f(x))dx +\int_y^1 f(y) \sin(f(x)-f(y))dx \right) h(y)dy. $$ I am afraid, that there is no convenient way to get rid of the double integrals.

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