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Help me show the ff inequality

  1. Sep 1, 2006 #1
    If v(0) = v(1) = 0 where v is differentiable over [0,1]

    how do I show the ff 2 inequalities?

    1) [tex]\int_{0}^{1}|v(x)|^{2}dx \leq \frac{1}{2}\int_{0}^{1}|v'(x)|^{2}dx[/tex]

    2) [tex]\int_{0}^{1}|v(x)|^{2}dx \leq \frac{1}{8}\int_{0}^{1}|v'(x)|^{2}dx[/tex]

    I tried using the Cauchy Schwartz inequality...and of course I ended up with the 2 being equal (without 1/2 or 1/8, that is) :(
     
    Last edited: Sep 1, 2006
  2. jcsd
  3. Sep 1, 2006 #2

    quasar987

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    and v is real valued? An indirect hint may be the absolute values. What do you think? If they are not there to give the hint "use a result concerning absolute values", then why are they here? It would have been simpler to just write v(x)² and v'(x)². You've already explored the "cauchy-schwartz path" and that led nowhere. mmmh...

    P.S. what does "ff" means to you?
     
  4. Sep 2, 2006 #3
    ff means following, sorry bout that :)

    is there a theorem or proposition involving absolute values and integrals that I should know about? I'm getting a blank on this...
     
  5. Sep 2, 2006 #4

    quasar987

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    Me too I'm afraid. And the worst is that I'm almost positive I've seen the proof of this before. This "prove its <1/2", then "prove it's actually <1/8" sounds too familiar. But I'm as stumped as you.
     
  6. Sep 2, 2006 #5
    if its any help, the topic we are taking where this question was lifted from is for Poincare Inequalities in Distributions (generalized functions)
     
  7. Sep 2, 2006 #6

    StatusX

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    If you want to show the RHS is at least so big, the idea is to replace it by smaller quantities and show these are still "big enough." Similarly, to show the LHS is at least so small, replace it by larger quantities and show these are still "small enough." (I should really be saying "smaller or equal to", but that gets cumbersome, so I'll leave it out)

    With this in mind, let's say the max of |v(x)| is A, and see if we can bound the LHS and RHS. Clearly the LHS is smaller than A^2. If we pick v(x) as a triangular function with v(1/2)=A, then the integral of the magnitude squared of the derivative is (2A)^2. Can we pick another v(x) with the same max that makes this integral any smaller? It turns out we can't, as I'll let you prove. This directly gives the first inequality. I'll have to think about the second inequality a little more. I'll get back to you.
     
    Last edited: Sep 2, 2006
  8. Sep 2, 2006 #7

    StatusX

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    Two approaches I tried that gave the same result were looking at the Fourier series of v(x) and v'(x) and using calculus of variations. These both suggested that the largest ratio between the two integrals occurs when v(x)=A sin([itex]\pi[/itex]x), and that the factor of 8 can actually be improved to a [itex]\pi^2[/itex]. Maybe you don't need to use these methods (which would be a pain to be completely rigorous about) if you can find another reason why sin is the best. The fact that they tell you to prove it with an 8 instead of a [itex]\pi^2[/itex] tells me that 1) they didn't want to give away that the answer involved trigonometric functions, 2) there is a simpler proof, probably using more boxy, linear shapes, or 3) I'm wrong, and 8 is the best bound.
     
    Last edited: Sep 2, 2006
  9. Sep 3, 2006 #8
    thanks for the help StatusX,

    I was able to find a book which gave the bound as [tex]\frac{1}{\pi^{2}}[/tex], which used Fourier transformation as proof. when I asked my professor about it, she told me that I don't need Fourier transforms to find the proof (which is good, since I am unfamiliar with Fourier transforms.)

    Interestingly, there's a followup question

    --------------------
    let [tex] p = sup \frac{\int_{0}^{1} |v(x)|^{2} dx}{\int_{0}^{1} |v'(x)|^{2} dx}[/tex]
    Show that if k is a constant such that k < 1/p then the following problem admits at most one solution:

    -u"(x) - ku(x) = f(x)
    x an element of (0,1)

    where
    u(0) = u(1) = 0

    and for f = 0, and k not equal to 0 show that the problem above admits a nontrivial explicit solution.

    Deduce that we have [tex]\frac{1}{8} \geq p \geq \frac{1}{\pi^{2}}[/tex]

    -------------------
    so [tex]\frac{1}{\pi^{2}}[/tex] is eventually asked to be found.
     
    Last edited: Sep 3, 2006
  10. Sep 3, 2006 #9

    StatusX

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    What does k refer to in the second problem?
     
  11. Sep 3, 2006 #10
    i edited the entry above.
    it seems the latex image is having problems loading.
     
    Last edited: Sep 3, 2006
  12. Sep 7, 2006 #11
    a question regarding the post I made two posts above:

    again, the question is:
    let [tex] p = sup \frac{\int_{0}^{1} |v(x)|^{2} dx}{\int_{0}^{1} |v'(x)|^{2} dx}[/tex]
    Show that if k is a constant such that k < 1/p then the following problem admits at most one solution:

    -u"(x) - ku(x) = f(x)
    x an element of (0,1)

    where
    u(0) = u(1) = 0

    and for f = 0, and k not equal to 0 show that the problem above admits a nontrivial explicit solution.

    Deduce that we have [tex]\frac{1}{8} \geq p \geq \frac{1}{\pi^{2}}[/tex]

    My question is how exactly do you show that the differential equation above a)has a unique solution for the first condition and b)admits a nontrivial explicit solution for the second condition?

    I have no idea how to prove this. What do I need to do show the proof?

    help please.
    thanks
     
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