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Functional diffential-integral equations

  1. Nov 2, 2009 #1
    I'm reading Quantum Field Theory Of Point Particles And Strings, by Brian Hatfield, chapter 9 called Functional Calculus. But he seems to assume some famiality with the subject. I'm intriqued by his notation. He uses notation for functional derivatives almost as if it were ordinary derivatives; he uses the notation:

    [tex]\[
    \frac{{\delta F[a]}}{{\delta a(x)}}
    \][/tex], [tex] \[
    \frac{{\delta ^2 G[a]}}{{\delta a(y)^2 }}
    \][/tex], and [tex] \[
    \int {\Delta a\,\,e^{ - F[a]} }
    \]
    [/tex].

    I'm looking for a more complete development of these ideas using this kind of notation. I wonder if it is developed enough to solve for Functionals F[a], like differential equations solve for functions. It makes me wonder if, say, the functional Lagrangian of physics can be derived on first principles using these methods. Any guidance would be appreciated. Thanks.
     
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  3. Nov 2, 2009 #2

    alxm

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    Yeah, no, you can't really 'solve' functionals the same way you would a differential equation.
    (Oh the elusive Exact Density Functional! Hear the lament of the quantum chemist! 45 years since we http://www-theor.ch.cam.ac.uk/people/ross/thesis/node32.html" you existed and no one yet knows what you are! :cry:)

    Anyway, http://julian.tau.ac.il/~bqs/functionals.pdf" [Broken] seems like an okay introduction - or you could check out some Mathematical Methods for Physics books (e.g. Arfken) that have chapters on functional analysis, or one of the many dedicated books on the topic.
     
    Last edited by a moderator: May 4, 2017
  4. Nov 2, 2009 #3
    I wonder if there is any proof of this statement. Or are you speaking from your own experience? Could it be because there is no way to specify boundary conditions for functionals as there is for functions? Maybe not in general, but perhaps in special circumstances. Or maybe it is sufficient to show that some functional does obey the Functional differential equation. Could uniqueness be addressed by showing that a slight variation of the functional is not a solution?

    In the bibliography, Brian Hatfield references F. A. Berezine, The Method of Second Quantization, 1966. I looked this up on Amazon.com; they want $530 a copy for this book. Does it really have this special information in it?

    What exactly would be the topic that I would look up for this subject? I looked in some of my variational calculus books. They don't seem to use this notation and don't get into functional integration (path integrals) very much.
     
    Last edited by a moderator: May 4, 2017
  5. Nov 4, 2009 #4

    haushofer

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    I was also looking for this kind of stuff. For instance whenever it comes to obtain Taylor-expansions of functionals. :)
     
  6. Nov 4, 2009 #5
    It seems there is some similarity between functional differential-integral calculus and that of operators on a Hilbert space. You can have infinite dimensional vectors in Hilbert space used to describe functions. And the operators that act on these vectors are derivatives and integrals, much like functionals that act on functions. I wonder how close the relationship is. Perhaps operator math on Hilbert space simply uses a different notation, where, perhaps, the operators are limited to being linear. There are operators defined in terms of other operators. And if I remember correctly, there are taylor expansions and power series of operators. Perhaps these techniques transfers to functional math.
     
    Last edited: Nov 4, 2009
  7. Nov 5, 2009 #6

    haushofer

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    I guess they do, because in field theory people are quite used to use these functional Taylor expressions. For instance, when you perform a coordinate transformation and look at the functional and coordinate change of the action. I tried to find a rigorous justification for this, but somehow few people take the effort to explain this in detail. Maybe I should try it for myself :)
     
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