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A problem from finite element book

  1. Mar 25, 2013 #1
    ..though I figure it's sort of an analysis type problem.
    ∫wvdx=0 (int from 0 to 1) for all v in V. w is continuous on [0,1]. What it means to be in V: v in V satisfies being continuous on [0,1], v(0)=v(1)=0, and derivatives of v are piecewise continuous .
    Problem is:
    Show that w(x)=0 for x in [0,1].

    I have spent hours with this. The book I'm looking at describes a couple problems:
    for u in V, u is a solution to -u''=f(x), u is also a solution to (u',v')=(f,v) and also a solution to
    1/2(u',u')-(f,u) less than or equal to 1/2(v',v')-(f,v).

    I have tried in many combinations to use this information, and this is the only given info in the chapter.

    Help.
     
  2. jcsd
  3. Mar 25, 2013 #2

    AlephZero

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    Suppose ##w(x) \ne 0## somewhere in [0,1].

    Now invent a function ##v(x)## such that ##\int_0^1 w(x)v(x)\,dx \ne 0##.

    Hint: think how you could make ## w(x)v(x) \ge 0## everywhere in [0, 1].
     
  4. Mar 25, 2013 #3
    Maybe you could elaborate a bit as my proof skills are not strong.
    Youre saying assume w does not equal 0. But then why would we consider a single arbitrary v, when this condition must hold for all v?
    Are you saying that, if we can find a v where the condition doesnt hold, then we have a contradiction?
     
  5. Mar 25, 2013 #4
    In other words, is finding an arbitrary v that contradicts the given conditions really something that leads to the conclusion w must equal 0?
     
  6. Mar 25, 2013 #5

    AlephZero

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    The theorem says that if ##\int_0^1 wv\,dx = 0## for every function ##v##, then ##w## must be 0.

    Yes, I'm suggesting you prove this by contradition. Suppose ##w \ne 0## and the integral = 0 for every possible function ##v##. But you can show that for any given continuous nonzero function ##w##, you can find a function ##v## that makes the integral non-zero, so it is impossible that ##w \ne 0##.
     
  7. Mar 25, 2013 #6
    thank you sir
     
  8. Mar 25, 2013 #7
    so, wait: Please confirm this is a valid write up:
    Suppose w does not equal 0. Then choose v=sin(x^2-x) (a negative function on [0,1] I just randomly came up with which satisfies being 0 at x=0 and x=1 and is continuous, with a continuous derivative, hence all that satisfies v in space V). Then if w<0 on [0,1], the integral is positive. If w>0 on [0,1], the integral is negative. This is a contradiction, so w must be 0.
    Would this be a 'proper' proof?
     
  9. Mar 27, 2013 #8
    Is that a valid approach? I lack analysis skills.
     
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