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Functionnal analysis (norm in Sobolev space)

  1. Apr 10, 2008 #1

    quasar987

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    [SOLVED] Functionnal analysis (norm in Sobolev space)

    1. The problem statement, all variables and given/known data
    In relations to the problem of finding the eigenvalues of the operator -d²/dt², it can be shown (Lax-Milgram) that given f in L²[0,1], there exists a unique u in [tex]H^1_0[0,1][/tex] such that

    [tex]\int_0^1u'v' = \int_0^1fv \ \ \forall v \in H^1_0[0,1][/tex]

    Consider the map S:L²[0,1]-->[tex]H^1_0[0,1][/tex] defined by S(f)=u.

    In arguing that S in continuous, my book says to take v=u. The equation characterizing u is then

    [tex]\int_0^1(u')^2 = \int_0^1fu [/tex]

    We deduce from this that [tex]||u'||^2_{L^2}\leq ||f||_{L^2}||u||_{L^2}[/tex] (Hölder). And here my book simply says "It results that [tex]||u||_{H^1}\leq C||f||_{L^2}[/tex], for a constant C."

    How does that follow? :confused:


    2. Relevant equations
    [tex]||u||_{H^1}=\sqrt{||u'||_{L^2}^2+||u||_{L^2}^2}[/tex]


    3. The attempt at a solution

    Well, we can use the inequality to write

    [tex]||u||_{H^1}^2=||u'||_{L^2}^2+||u||_{L^2}^2\leq |f||_{L^2}||u||_{L^2}+||u||_{L^2}[/tex], but then what?
     
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  3. Apr 10, 2008 #2

    quasar987

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    That last equation should read

    [tex]||u||_{H^1}^2=||u'||_{L^2}^2+||u||_{L^2}^2\leq ||f||_{L^2}||u||_{L^2}+||u||_{L^2}^2[/tex]
     
  4. Apr 12, 2008 #3

    quasar987

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    Perhaps slightly closer would be to use

    [tex]ab\leq \frac{a^2+b^2}{2}[/tex]

    on Hölder's inequality to obtain instead

    [tex]||u'||^2_{L^2}\leq \frac{||f||_{L^2}^2+||u||_{L^2}^2}{2}[/tex]

    and thus

    [tex]||u||_{H^1}\leq \frac{3}{2}||u||_{L^2}^2+\frac{1}{2}||f||_{L^2}^2[/tex]

    But can we relate ||u||_2 to ||f||_2 ? :grumpy:
     
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