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B Functional derivative

  1. Oct 28, 2016 #1

    naima

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    Hi PF
    I try to understand how we get get a Taylor expansion of a non linear functional.
    I found this good paper
    here F maps functions to scalars. F[f] is defined. It has not scalars as arguments. I agree with A13 and A18.
    In another paper (in french) skip to page 9
    the fisrt term is ##\int dx P_0 (x)## and all the terms have one more variable.
    Do you understand the second point of view? we did not start with densities.
     
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  3. Oct 28, 2016 #2

    Krylov

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    Could you explain into some more detail what exactly is confusing you?

    In my experience, "functional derivative" usually means what mathematicians call a Gâteaux derivative. Sometimes it refers to the stronger concept of Fréchet derivative. Using the latter notion it is possible to set up a calculus that looks very much identical to the ordinary multivariable calculus for functions defined on open subsets of ##\mathbb{R}^n##. For this it is necessary to specify the domain of the nonlinear functional (or operator) in question as a suitable open subset of a normed linear space. Once this is done, familiar theorems such as Taylor's hold almost verbatim.

    Now, I know that physicists often do not like to talk about the function spaces that underpin their work, but in this case I believe it really does pay off to break that habit. Unfortunately I cannot read texts that lack rigor (your first link presents itself like that), so for a more rigorous but still gentle reference I would also like to mention "A Primer of Nonlinear Analysis" by Ambrosetti and Prodi.
     
  4. Oct 28, 2016 #3

    stevendaryl

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    I think you mean page 7 of the second paper.

    In the second paper, they seem to be assuming that the functional [itex]F[f][/itex] can be written in the form:

    [itex]F[f] = \int dx F(f(x))[/itex]

    This is a very confusing notation. If you look at the bottom of page 7, you'll see they define:

    [itex]S[x(t)] = \int dt L(x, \dot{x})[/itex]

    which doesn't fit the pattern of [itex]F[f][/itex]. What I think they mean is something like this: They are assuming that the functional [itex]F[f][/itex] can be written in the form:

    [itex]F[f] = \int dx \tilde{F}[/itex]

    Where the expression [itex]\tilde{F}[/itex] inside the integral can involve [itex]f[/itex] and its derivatives evaluated at the point [itex]x[/itex]. Writing this as [itex]F(f(x))[/itex] gives the mistaken impression that it is an ordinary function [itex]F(y)[/itex] evaluated at the point [itex]y=f(x)[/itex].

    But in any case, the two papers seem to agree about the functional derivative: Look at equation A.28 in section A.3 in the first paper, and compare it with equation 1.13 page 6 of the second paper. What's confusing about the second paper is that they seem to be making the distinction between

    1. [itex]\frac{\delta F[f(y)]}{\delta f(x)}[/itex], and
    2. [itex]\frac{\delta F[f]}{\delta f(x)}[/itex]
    They seem to be treating the second expression as the integral of the first expression:

    [itex]\frac{\delta F[f]}{\delta f(x)} = \int dy \frac{\delta F[f(y)]}{\delta f(x)}[/itex]

    To me, this is an extremely confusing convention. And the author is not even consistent about it, because in equation 1.26 on page 8, they write [itex]\frac{\delta S[x(t)]}{\delta x(t)}[/itex], when it would seem like they should be writing [itex]\frac{\delta S[x]}{\delta x(t)}[/itex] (with no argument [itex]t[/itex] on the function [itex]x[/itex] in the expression [itex]S[x][/itex]). It's very confusing, because it's unclear when they are using [itex]x(t)[/itex] to mean a function, and when they are using it to mean a number, the value at point [itex]t[/itex].

    I think it's bad notation, but that the two papers probably mean the same thing.
     
  5. Oct 29, 2016 #4

    naima

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    I found the "trick"
    When you have number a and a normalized function G (such that ##\int dx G(x)= 1## you can write ## a = \int dx aG(x)##
    Here the author expands the functional like in the first paper and then he takes any normalized function G and calls it F(f(x))
     
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