Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Extrema of functionals

  1. Mar 12, 2013 #1
    Hello guys,

    Recently I came across a definition to which I'd never given much thought. I was reading through Gelfand and Fomin's "Calculus of variations" and I read the part about weak and strong extrema, and I really can't manage to wrap my head around these definitions. They can be found in the wiki article (basically a copy-paste of the book):

    http://en.wikipedia.org/wiki/Calculus_of_variations#Extrema

    My immediate thought is that these definitions must be inverted. I mean, if you have:

    [tex] \| f-f_0 \|_1 < \delta [/tex]

    For certain f's, then, a fortiori you have:

    [tex]\| f-f_0\|_0 < \delta [/tex]

    For all these f's. So this means that the extremum defined with the 1st order norm should be stronger!

    At some point I found a definition that was the exact opposite of the one given in wiki (in some google book), but I can't find it anymore. Maybe I should just sleep on it, but I'd still like your input. While we're at it, can you recommend a good, modern book on the calculus of variations? I find that most books on the subject used as references are pretty dated and often not very clear.
     
    Last edited by a moderator: Mar 12, 2013
  2. jcsd
  3. Mar 12, 2013 #2

    mathman

    User Avatar
    Science Advisor
    Gold Member

    The Wiki article looks wrong. As you noted, it seems to have weak and strong reversed.
     
  4. Mar 15, 2013 #3
    They aren't reversed. Strong is called strong because it allows a wider class of variations. To be a weak extremum you only have to check differentiable variations.

    It would be nice to hear some book recommendations. I love Gelfand Fomin, but a more advanced presentation that assumes more background analysis would be great.
     
  5. Mar 15, 2013 #4
    I reread the definition again just to clarify to myself whether strong implies weak. And it does. The norms are only used to restrict the class of variations. So the "stronger norm" is more restrictive on the allowed variations. Thus it produces a weaker form of extremum.
     
  6. Jul 8, 2015 #5
    well it is kind of late but yet..

    I would say that extrema in "0-norm" are stronger.

    Consider a functional ##J## defined in ##\mathcal{C}[a,b]## (that is the space of all continuous functions from ##[a,b]## to ##\mathcal{R}##) then if ##J## has an extremum at ##y_0 \in \mathcal{C}## certanly it also has an extremum at ##y_0 \in \mathcal{D}_1## if it is defined on ##\mathcal{D}_1##.
    I dont know if i was clear enough. I hope so..
     
  7. Jul 9, 2015 #6
    i forgot to say that ##\mathcal{D}_1## is the set of all smooth functions ##y:[a,b] \rightarrow \mathcal{R}##.
    Then obviously ##\mathcal{D}_1 \subset \mathcal{C}##. Let ##y_0 \in \mathcal{D}_1## (##\Rightarrow y_0 \in \mathcal{C}##). Then if ##J## has an extremum at ##y_0## in ##\mathcal{C}##, then it also has an extremum at ##y_0## in ##\mathcal{D}_1## (since ##\mathcal{D}_1## is a subset of ##\mathcal{C}##).
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Extrema of functionals
  1. Extrema of function (Replies: 1)

Loading...