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

A FEM, Space embedding doubt

  1. Apr 15, 2016 #1
    Hi. I'm studying Finite Elements Method, I was readding a paper written by Danielle Boffi and in a part dedicated to the approximation of eigenvalues in mixed form, it's about approximating eigenvalues in the Hilbert Spaces [tex]\Phi[/tex] and [tex]\Xi[/tex]
    Then it says:
    "If we suppose that there exists Hilbert Spaces [tex]H_{\Phi}[/tex] and [tex]H_{\Xi}[/tex] such that the following dense and continuous embeddings hold in a compatible way
    [tex]\Phi \subset H_{\Phi} \simeq H_{\Phi} ' \subset \Phi'[/tex]
    [tex]\Xi \subset H_{\Xi} \simeq H_{\Xi} ' \subset \Xi'[/tex]
    Here [tex]\subset[/tex] means dense and continously embedded.

    I'm not sure why it does that, I have read more than one time that they work with a space identified with it's dual space, as they're Hilbert Spaces I know you can do it using Riesz Representation theorem, but I don't exacly see why they're doing this.

    I didn't know why they don't simply identify [tex]\Phi[/tex] with [tex]\Phi'[/tex].

    I have found an advice against identifying a space which is not [tex]L^{2}[/tex] with it's dual, because otherwise in constructions like this, if [tex]H_{\Phi}=L^{2}[/tex] and [tex]\Phi=H^{1}[/tex] we would end up identifying the four spaces [tex]\Phi \equiv H_{\Phi} \equiv H_{\Phi} ' \equiv \Phi'[/tex] and it would mean that you're identifying a function with it's laplacian which is (and here I'm quoting a book) ""the beggining of the end""

    I'm new in this of variational formulations and I don't have a strong background on partial differential equations, I have more background on functional and real analysis, I'm studying this subject to make my Licenciatura's thesis in this but I'm really lost sometimes, this is as clear as I could make the question so feel free to ask me to clarify something if I wasn't clear enough.

    PD: I don't know how to write in latex without making a new line.
    Last edited: Apr 15, 2016
  2. jcsd
  3. Apr 20, 2016 #2
    Thanks for the post! This is an automated courtesy bump. Sorry you aren't generating responses at the moment. Do you have any further information, come to any new conclusions or is it possible to reword the post?
  4. May 2, 2016 #3
    Hi there.
    does not make sense. i do not think i am negligent with my work in saying a function can't be equal to it's derivative. unless it's e^x. albeit Danielle boffi might have a different finding.

    This comes from a guy who would enjoys going to a Lollapalooza wearing a rippling red scirt. This site helps me with your problem http://math.stackexchange.com/questions/644879/function-is-equal-to-its-own-derivative .
    Last edited by a moderator: May 2, 2016
  5. May 3, 2016 #4
    When working with PDEs and variational methods it is not advised to identify [itex]H^1_0[/itex] with its dual [itex]H^{-1}[/itex], because you lose some subtleties. The space [itex]L^2[/itex] is sometimes called the pivot space and is sort of in the middle in terms of regularity, i.e. functions in [itex]L^2[/itex] are first derivatives of functions in [itex]H^1_0[/itex] and the elements of [itex]H^{-1}[/itex] are first derivatives of functions in [itex]L^2[/itex]. We can use this to characterize the dual space [itex]H^{-1}[/itex] in the following way:

    For every [itex]\mathcal{l}\in H^{-1}[/itex] there exist [itex]v_0,v_1,\ldots,v_d[/itex] such that
    [tex]\langle\mathcal{l},u\rangle_{H^{-1},H^1_0} = (v_o,u)_{L^2} + \sum_{i=1}^d (v_i, u_{x_i})_{L^2}[/tex]
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted