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Math Help - Sobolev spaces, weak derivative

  1. #1
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    Sobolev spaces, weak derivative

    Hi,

    in which sobolev spaces H^{1,p}(\Omega), \ 1\leq p\leq \infty, are the functions

    a) u(x)=x^\frac{3}{2}\sin\left ( \tfrac{1}{x} \right ) with  \Omega=(0,1)

    b)  u(x)= \left |\ln x  \right |^{-1} with  \Omega=\left ( 0, \frac{1}{2} \right ) ?

    So, do I have to find out in which spaces L^p(\Omega) the functions above has a first weak derivative?

    And if yes: How can I find it out?

    Thanks in advance,
    Alexander
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  2. #2
    Super Member girdav's Avatar
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    Re: Sobolev spaces, weak derivative

    If the function is differentiable in the usual sense, then it's also weakly differentiable and these two derivatives are the same.
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  3. #3
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    Re: Sobolev spaces, weak derivative

    Quote Originally Posted by AlexanderW View Post
    So, do I have to find out in which spaces L^p(\Omega) the functions above has a first weak derivative?
    Alexander
    Yes, so as girdav noted, you have to check for which p do the derivatives lie in L^p.

    For the first one, notice that u'(x)=\frac{3}{2}x^{1/2} \sin(1/x) - x^{-1/2}\cos(1/x) so that u'\in L^p(0,1) iff f(x)=x^{-1/2}\cos(1/x) \in L^p(0,1). Now |f(x)|\leq x^{-1/2} so f\in L^q for all 1\leq q<2, and \int_0^1 f^2(x)dx=\infty (this is easy to see: change x=1/y and bound below by a harmonic series). So that u\in W^{1,p}(0,1) iff 1\leq p<2.

    The second one is easier: u'(x)=\frac{\ln^2(x)}{x}\geq 0 and \int _0^{1/2} u'(x)=\infty so that u doesn't belong to any Sobolev space W^{1,p} for 1\leq p \leq \infty.
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  4. #4
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    Re: Sobolev spaces, weak derivative

    Thank you very much for your helpful answers!

    Bye,
    Alexander
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  5. #5
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    Re: Sobolev spaces, weak derivative

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