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 November 3rd, 2018, 09:17 PM #1 Newbie   Joined: Nov 2013 Posts: 20 Thanks: 0 Calculate the derivative of a functional. How to calculate the $J'\left(y\right)$ of $J\left(y\right) = \int_0^1 g\left( y\left( x \right) \right) \operatorname dx$ ? It will be useful if can provide the derivation procss.
 November 5th, 2018, 06:53 AM #2 Banned Camp   Joined: Mar 2015 From: New Jersey Posts: 1,720 Thanks: 126 $\displaystyle J(y)=\int_{0}^{1}g(y)dx$ $\displaystyle J(y+\Delta y)=\int_{0}^{1}g(y+\Delta y)dx$ $\displaystyle J(y+\Delta y)-J(y)=\int_{0}^{1}g'(y)\Delta ydx$ Now you have to think through is it ok to divide by $\displaystyle \Delta y$ EDIT: Let $\displaystyle \Delta y$ remain constant throughout the integration. Thanks from rubis Last edited by zylo; November 5th, 2018 at 07:07 AM.
November 5th, 2018, 11:34 PM   #3
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Quote:
 Originally Posted by zylo $\displaystyle J(y)=\int_{0}^{1}g(y)dx$ $\displaystyle J(y+\Delta y)=\int_{0}^{1}g(y+\Delta y)dx$ $\displaystyle J(y+\Delta y)-J(y)=\int_{0}^{1}g'(y)\Delta ydx$ Now you have to think through is it ok to divide by $\displaystyle \Delta y$ EDIT: Let $\displaystyle \Delta y$ remain constant throughout the integration.
So $\triangle y$ has nothing to do with $\triangle x$ or something like that?

 November 6th, 2018, 02:17 AM #4 Global Moderator   Joined: Dec 2006 Posts: 20,933 Thanks: 2207 Have you calculated $J'(y)$ for some particular $y$ and $g$? If so, what did you get?
November 6th, 2018, 06:17 AM   #5
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 Originally Posted by skipjack Have you calculated $J'(y)$ for some particular $y$ and $g$? If so, what did you get?
When calculating the derivative, you specify a $\displaystyle \Delta y$, it's not a variable.

There's nothing to compare against if you take a definite integral. You can see it with an indefinite integral:

Let $\displaystyle J(y)=\int g(y(x))dx$
$\displaystyle y(x)=ax$
$\displaystyle g(y)=\sin y$
$\displaystyle J(y)=\int \sin axdx=-\frac{1}{a}\cos ax=-\frac{1}{a}\cos y$
$\displaystyle J'(y)= \frac{1}{a}\sin y$

$\displaystyle J'(y)=\int g'(y)dx=\int \cos y dx=\int \cos axdx=\frac{1}{a}\sin ax=\frac{1}{a}\sin y$

Last edited by skipjack; November 6th, 2018 at 12:27 PM.

 November 6th, 2018, 12:41 PM #6 Global Moderator   Joined: Dec 2006 Posts: 20,933 Thanks: 2207 So, zylo, are you claiming that the original problem, using a definite integral, (a) doesn't make sense, or (b) makes sense, but can't be answered by your method, or (c) makes sense, but always has an answer that doesn't depend on the choice of g, or (d) makes sense, but always has an answer that doesn't depend on the choice of y?
 November 7th, 2018, 03:55 AM #7 Banned Camp   Joined: Mar 2015 From: New Jersey Posts: 1,720 Thanks: 126 None of the above. I gave an answer that makes sense in the context of the question. EDIT: Elaboration. You can think of it as a special case of Calculus of Variations. Instead of taking an increment of y as an arbitrary close function, you take the case of a constant increment. That's indicated by the J'(y) notation. Last edited by zylo; November 7th, 2018 at 04:11 AM.
 November 7th, 2018, 11:26 AM #8 Global Moderator   Joined: Dec 2006 Posts: 20,933 Thanks: 2207 The original question is about a definite integral, but you haven't answered it for such an integral, either in the general case or for some particular choice of $g$ and $y$. Stating that there's nothing to compare against if you take a definite integral, doesn't make it clear whether you mean that the original question, using a definite integral, makes sense or can be answered by your method. Your method doesn't include a comparison, so "nothing to compare against" is referring to some comparison that you never specify.
 November 7th, 2018, 11:36 AM #9 Senior Member   Joined: Sep 2016 From: USA Posts: 635 Thanks: 401 Math Focus: Dynamical systems, analytic function theory, numerics As usual, zylo has "solved" the problem by just making up nonsense at will and throwing around notation that doesn't make any sense. For starters, a derivative is a linear functional and while I have no clue what half of the manipulations he has done are supposed to mean, it's obvious that his solution is not a linear functional so it isn't even the correct type of object. At this point I consider him a lost cause, but I wanted to caution you that everything he has written should be ignored as it is complete nonsense. That out of the way, it would help to know in what context this question is being asked. Usually this material is first seen in a course on functional analysis or calculus of variations. If you are in a course such as this, then start by explaining what you have already done/thought about and where you are stuck. If this problem is arising in some other context and you haven't had a course in functional analysis, then you just need to do a bit of reading. The derivative in this context may seem awkward, since the functional you have written is defined on some function space (e.g. continuous functions which map from $\mathbb{R}$ into $[0,1]$). The derivative is thus a linear functional defined on an infinite dimensional vector space. You should google "functional derivative" to get started and feel free to ask additional questions as you go. But they should be specific and you should include any work that you have done. The wikipedia page on this is fairly easy to follow as well: https://en.wikipedia.org/wiki/Functional_derivative Last edited by skipjack; November 7th, 2018 at 11:57 AM.
November 7th, 2018, 12:55 PM   #10
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Quote:
 Originally Posted by zylo You can think of it as a special case of Calculus of Variations. Instead of taking an increment of y as an arbitrary close function, you take the case of a constant increment. That's indicated by the J'(y) notation.
Quote:
 Originally Posted by zylo Example: Let $\displaystyle J(y)=\int g(y(x))dx$ $\displaystyle y(x)=ax$ $\displaystyle g(y)=\sin y$ $\displaystyle J(y)=\int \sin axdx=-\frac{1}{a}\cos ax=-\frac{1}{a}\cos y$ $\displaystyle J'(y)= \frac{1}{a}\sin y$ $\displaystyle J'(y)=\int g'(y)dx=\int \cos y dx=\int \cos axdx=\frac{1}{a}\sin ax=\frac{1}{a}\sin y$
$\displaystyle J'(y)=\int_{0}^{1} g'(y)dx=\frac{1}{a}\sin ax\biggr\rvert_{0}^{1}=\frac{1}{a}\sin a$

I think OP is a good introductory example, if properly understood, of general concept of calculus of variations, as opposed to just referencing wiki articles without understanding. It took me a while to see the connection as a special case. See above.

Other solutions to OP. with explanations and derivations, are welcome.

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