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Math Help - Derivatives, Integrals, Absolute Values Oh My!

  1. #1
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    Derivatives, Integrals, and Absolute Values, Oh My!

    This is making me feel quite stupid right now, but I have the following integral that is messing with my head
    \frac{1}{2\gamma} \int_{-\infty}^{\infty}\! \frac{d^2}{dx^2}\left[e^{-\gamma |x - y|}\right] q(y)\, dy
    Where q:\mathbb{R}\rightarrow\mathbb{R} is continuous with compact support.
    My first bit of instinct is to split the integral, because I need to know how to take that derivative. So there are two cases x < y and x>y, but how do I translate this into the limits of the integral? It seems that I would split it like so
    \frac{1}{2\gamma} \left(\int_{-\infty}^{x}\! \frac{d^2}{dx^2}\left[e^{-\gamma |x - y|}\right] q(y)\, dy + \int_{x}^{\infty}\! \frac{d^2}{dx^2}\left[e^{-\gamma |x - y|}\right] q(y)\, dy\right).
    =\frac{1}{2\gamma} \left(\int_{-\infty}^{x}\! \frac{d^2}{dx^2}\left[e^{-\gamma (x - y)}\right] q(y)\, dy + \int_{x}^{\infty}\! \frac{d^2}{dx^2}\left[e^{-\gamma (y - x)}\right] q(y)\, dy\right)
    =\frac{1}{2\gamma} \left(\int_{-\infty}^{x}\! \gamma^2e^{-\gamma (x - y)} q(y)\, dy + \int_{x}^{\infty}\! \gamma^2 e^{-\gamma (y - x)} q(y)\, dy\right)
    =\frac{\gamma}{2} \left(\int_{-\infty}^{x}\! e^{-\gamma (x - y)} q(y)\, dy + \int_{x}^{\infty}\! e^{-\gamma (y - x)} q(y)\, dy\right)

    Does this seem correct?
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  2. #2
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    Looks good to me so far.
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  3. #3
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    I was hoping it didn't. I guess I will post the whole problem.

    Show that
    u(x) = \frac{1}{2\gamma}\int_{-\infty}^{\infty} e^{-\gamma |x - y|}q(y)\, dy
    satisfies
    -u''(x) + \gamma^2 u(x) = q(x).
    What I end up getting in the end is  0 = q(x) instead of q(x) = q(x).

    Now my question is miscategorized. lol
    Last edited by lvleph; July 26th 2010 at 08:46 AM.
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  4. #4
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    Well, show what you get when you plug in the derivative and the original function u(x) into your DE.
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  5. #5
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    -u'' + \gamma^2 u = -\frac{d^2}{dx^2}\left(\frac{1}{2\gamma}\int_{-\infty}^{\infty}\! e^{-\gamma|x-y|} q(y)\, dy\right) + \gamma^2\left(\frac{1}{2\gamma}\int_{-\infty}^{\infty}\! e^{-\gamma|x-y|} q(y)\, dy\right)
    = -\frac{1}{2\gamma}\int_{-\infty}^{\infty}\! \frac{d^2}{dx^2}\left[e^{-\gamma|x-y|}\right] q(y)\, dy+ \frac{\gamma}{2}\int_{-\infty}^{\infty}\! e^{-\gamma|x-y|} q(y)\, dy
    = -\frac{1}{2\gamma}\left(\int_{-\infty}^{x}\! \frac{d^2}{dx^2}\left[e^{-\gamma(x-y)}\right] q(y)\, dy + \int_{x}^{\infty}\! \frac{d^2}{dx^2}\left[e^{-\gamma(y-x)}\right] q(y)\, dy\right) + \frac{\gamma}{2}\left(\int_{-\infty}^{x}\! e^{-\gamma(x-y)} q(y)\, dy + \int_{x}^{\infty}\! e^{-\gamma(y-x)} q(y)\, dy\right)
    = -\frac{1}{2\gamma}\left(\int_{-\infty}^{x}\! \frac{d^2}{dx^2}\left[e^{-\gamma(x-y)}\right] q(y)\, dy + \int_{x}^{\infty}\! \frac{d^2}{dx^2}\left[e^{-\gamma(y-x)}\right] q(y)\, dy\right) + \frac{\gamma}{2}\left(\int_{-\infty}^{x}\! e^{-\gamma(x-y)} q(y)\, dy + \int_{x}^{\infty}\! e^{-\gamma(y-x)} q(y)\, dy\right)
    = -\frac{1}{2\gamma}\left(\int_{-\infty}^{x}\! \gamma^2 e^{-\gamma(x-y)} q(y)\, dy + \int_{x}^{\infty}\! \gamma^2 e^{-\gamma(y-x)} q(y)\, dy\right) + \frac{\gamma}{2}\left(\int_{-\infty}^{x}\! e^{-\gamma(x-y)} q(y)\, dy + \int_{x}^{\infty}\! e^{-\gamma(y-x)} q(y)\, dy\right)
    = -\frac{\gamma}{2}\left(\int_{-\infty}^{x}\! e^{-\gamma(x-y)} q(y)\, dy + \int_{x}^{\infty}\! e^{-\gamma(y-x)} q(y)\, dy\right) + \frac{\gamma}{2}\left(\int_{-\infty}^{x}\! e^{-\gamma(x-y)} q(y)\, dy + \int_{x}^{\infty}\! e^{-\gamma(y-x)} q(y)\, dy\right) = 0
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  6. #6
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    Where did the minus sign out front come from? It's not in post # 3.

    Incidentally, I'm thinking that a Dirac delta function is going to show up any minute. I can feel it in my bones.
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  7. #7
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    It is now. That was a typo. It is suppose to be -u''(x) + \gamma^2 u(x) = q(x).

    EDIT: Maybe the test had a typo? Hmmm.
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  8. #8
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    Find the Green's function. That is, solve

    -G''+\gamma^{2}G=\delta(x).

    I'll bet you anything that the Green's function for this DE is \frac{1}{2\gamma}\,e^{-\gamma|x-y|}.

    At that point, I think you're pretty much done, since you can write the solution as

    u=\int_{-\infty}^{\infty}G(x,y)g(y)\,dy.

    I'll bet that works for you.
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  9. #9
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    That might be. It does look like a Green's Function.
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  10. #10
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    Whoops. I think the Green's function satisfies

    -G''+\gamma^{2}G=\delta(x-y).

    Try that.
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  11. #11
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    I have to come back to this. I have to get through one more test, before the end of the day and I have spent too much time on this problem. Thank you.
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  12. #12
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    Ha, I started the next test and it has the same thing attached to a problem with Green's Functions. So I guess that is what I need to do.
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