First of all, apologies if this isn't quite in the right section.(adsbygoogle = window.adsbygoogle || []).push({});

I've been studying functionals, in particular pertaining to variational calculus. My query relates to defining a functional as an integral over some interval [itex]x\in [a,b][/itex] in the following manner [tex]I[y]= \int_{a}^{b} F\left(x, y(x), y'(x)\right)dx [/tex]

Clearly from this we see that [itex]I[/itex] is not dependent on [itex]x[/itex], but instead it depends only on the function [itex]y(x)[/itex]. [itex]I[/itex] is a functional and as such it defines a mapping from the set of all functions [itex]y(x)[/itex] satisfying [itex]y(a)=0=y(b)[/itex] to [itex]\mathbb{R}[/itex].

My question really, is why the integrand a function of the set of curves [itex]y(x)[/itex] (as defined above) and their derivatives [itex]y'(x)[/itex] (I've kept it to first-order for simplicity, but I know that in general it can be dependent on higher orders)?

Is this because, as [itex]I[/itex] is depends on every single value that [itex]y(x)[/itex] takes in the interval [itex]x\in [a, b][/itex], and not just its value at a single point, we must consider how [itex]y(x)[/itex] changes (i.e. we must consider it's derivatives) over this interval as we integrate over it. Thus, this implies that the integrand should be a function of the curve and it's rate of change?

Please could someone let me know if my thinking is correct, and if not, provide an explanation.

Thanks for your time.

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# A question on functionals

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