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Question on Sobolev Spaces

  1. May 28, 2010 #1
    Hello there,

    I am learning the first rudiments of functional analysis and have a couple of queries.
    1) A book says that the space W (1,p) is defined as a subset of functions u(x) of Lp (so far, good) such that the integral, for any smooth epsilon (x), of u(x) times the derivative of epsilon is equal to MINUS epsion times the derivative of u(x). I have to admit I do not get it.
    2) Wikpedia reports the example of a non - continuous W (1,1) function on the unit ball in R3, namely u(x) = 1 / abs(x). Also it adds that "space Wk,p(Ω) will contain only continuous functions, but for which k this is already true depends both on p and on the dimension<...>Intuitively, the blow-up of f at 0 "counts for less" when n is large since the unit ball is "smaller" in higher dimensions". Why should the unti ball be "smaller" in higher dimensions?

    Many thanks for your help

    Muzialis
     
  2. jcsd
  3. May 28, 2010 #2
    So it's {u in Lp such that [tex]\int \frac{d\epsilon}{dx} u dx = -\int \epsilon \frac{du}{dx} dx[/tex]} ?
    If so, that basically means that u dies at infinity (and negative infinity).

    Probably to achieve results that you might derive for Schwarz spaces where you need to integrate by parts a bajillion times.

    Just trying to decode what you're saying first x.x
    I did some stuff on distributions and sobolev spaces, but it wasn't very rigorous.

    Don't understand what "already true" means. I guess it means the result holds for smaller k. Not sure about the unit ball thing.
     
    Last edited: May 28, 2010
  4. May 28, 2010 #3
    Hi,

    many thanks for your reply.

    Yes you decoded correctly what I meant, although I forgot to mention that integration is carried over a finite domain Omega.

    Does this clarify?

    I simply do not understand how this could hold. A caveman's counterexample, u(x) = x, eps(x) = x^2, they are both regolar enough to fit in the definition, belonging to W 1,p

    But the equality would then look as

    integral over the domain of (x^2) vs. intergral over the domain of (2 x ^2).

    Many thanks

    Best Regards
    Muzialis
     
  5. May 28, 2010 #4
    (edited)
    Uhm. Start on the LHS.
    [tex]\int \frac{d\epsilon}{dx}u dx= [\epsilon u] - \int \epsilon \frac{du}{dx}{dx} [/tex]
    I'm taking Omega as an open subset of R.
    epsilon is arbitrary, so we NEED u to vanish on the boundary of Omega, for it to be in the Sobolev space surely?

    Only functions that disappear on the boundary are allowed to be in the Sobolev space. I think your caveman example shows that u=u(x)=x isn't allowed.

    Let me check the Wiki article, and then a textbook.
     
  6. May 28, 2010 #5
    Sorry, can we verify the conditions on epsilons?
    Smooth (C-infinity)? Compact support?
    (also can I ask like, are you in university, or doing this for kicks or...?)
     
    Last edited: May 28, 2010
  7. May 28, 2010 #6
    Hi Jarbarrrrr,

    first of all thank you, and of course you can ask, I am a PhD student in computation mechanics but I am deepning a bit the maths for myself.

    For rigorousness, let me quote the definition I found in toto,

    "The Sobolev space W 1,p (Omega) is defined as the set of functions u belonging to Lp (Omega) such that there exist functions du / dxi belonging to Lp(omega) with the property

    Integral over mega (u(x)*(deps/dxi(x))dx= - Integral over omega (eps(x)*du/dxi(x))dx

    for any compactly supported, smooth function eps.
    This definition can also be stated by saying that the distributational partila derivatives of u belong to Lp(Omega)"

    (PS: I write Lp for Lebesgue spaces and Wp forsobolev spaces, where the "p" should be superscripts)

    The vanishing condition on the frontier of the domain is something I have not heard as part of the definition of Sobolev spaces, which explains my doubts.

    Many thanks

    Muzialis
     
  8. May 28, 2010 #7
    Dear Jerbearrrrr,

    it is "only" 2.26 AM here in the UK but I think we made it.

    I overlooked the fact that as the epsilon function is compactly supported on omega, it has to vanish on the boundary. I did not give much weight to the compactly supported bit, my apologies.

    Still, many thanks for your help

    Muzialis
     
  9. May 28, 2010 #8
    It's always in the assumptions haha.
    You're welcome, glad you sorted it out.

    I would have overlooked the compact support thing too, but there are PDEs in my exams...which are next week. I might need your help as I finish my revision actually x.x

    I probably won't continue into further study though (graduating this year).
     
  10. May 29, 2010 #9
    Best of luck for your exams then!
     
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