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Math Help - Leibniz' rule

  1. #1
    Junior Member
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    Leibniz' rule

    hello


    i've been trying for ages and this one question i can't answer, show that:


    u(x,y)=(y/pi) times integral , from - to + infinity, of {f(t)dt/[(x-t)^2 + y^2]}

    satisfies

    'second partial of u w.r.t x' + 'second partial of u w.r.t y' = 0


    i've tried

    du =
    {-(y/pi)times integral, from -to+ inf, of {2(x-t)f(t)dt/[(x-t)^2 + y^2]^2}}dx
    +
    {(1/pi)times integral , from - to + infinity, of {f(t)dt/[(x-t)^2 + y^2]}
    (y/pi)times integral, from -to+ infinity, of {2yf(t)dt/[(x-t)^2 + y^2]^2}}dy

    and then take the second derivatives of the partials and add them but it don't work.

    NOW I AM SURE i am going about this the wrong way, BUT HOPEFULLY some brighter soul out there will see what i am trying to do and any way tell me what is the right way
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  2. #2
    Super Member
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    I'll do the first one (assuming everything sufficient to take the partials):

    \frac{\partial}{\partial x} u=\frac{y}{\pi}\int_{-\infty}^{\infty}\frac{\partial}{\partial x}\left(\frac{f(t)}{(x-t)^2+y^2}\right)dx=-\frac{y}{\pi}\int_{-\infty}^{\infty}\frac{2(x-t)f(t)}{\left[(x-t)^2+y^2\right]^2}dt

    \frac{\partial^2 u}{\partial x^2}=\frac{y}{\pi}\int_{-\infty}^{\infty} \frac{8(x-t)^2f(t)}{((x-t)^2+y^2)^3}-\frac{2f(t)}{((x-t)^2+y^2)^2}dt=\frac{y}{\pi}\int_{-\infty}^{\infty}\frac{[(x-t)^2-2y^2]f(t)}{[(x-t)^2+y^2)^3}dt

    I did check the results in Mathematica and the Laplace expression does equate to zero.
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  3. #3
    Junior Member
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    Quote Originally Posted by shawsend View Post
    I'll do the first one (assuming everything sufficient to take the partials):

    \frac{\partial}{\partial x} u=\frac{y}{\pi}\int_{-\infty}^{\infty}\frac{\partial}{\partial x}\left(\frac{f(t)}{(x-t)^2+y^2}\right)dx=-\frac{y}{\pi}\int_{-\infty}^{\infty}\frac{2(x-t)f(t)}{\left[(x-t)^2+y^2\right]^2}dt

    \frac{\partial^2 u}{\partial x^2}=\frac{y}{\pi}\int_{-\infty}^{\infty} \frac{8(x-t)^2f(t)}{((x-t)^2+y^2)^3}-\frac{2f(t)}{((x-t)^2+y^2)^2}dt=\frac{y}{\pi}\int_{-\infty}^{\infty}\frac{[(x-t)^2-2y^2]f(t)}{[(x-t)^2+y^2)^3}dt

    I did check the results in Mathematica and the Laplace expression does equate to zero.
    yes thanks, i do appreciate this . however i did get that far my self so my issue is with the second one.


    p.s. don't think i am pushing to get the full answer, it simply the case and that's how it is. anyway, i do appreciate your help
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  4. #4
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    Quote Originally Posted by shawsend View Post
    \frac{\partial^2 u}{\partial x^2}=\frac{y}{\pi}\int_{-\infty}^{\infty} \frac{8(x-t)^2f(t)}{((x-t)^2+y^2)^3}-\frac{2f(t)}{((x-t)^2+y^2)^2}dt=\frac{y}{\pi}\int_{-\infty}^{\infty}\frac{[(x-t)^2-2y^2]f(t)}{[(x-t)^2+y^2)^3}dt
    That should be:

    \frac{\partial^2 u}{\partial x^2}=\frac{1}{\pi}\int_{-\infty}^{\infty} \frac{f(t)\left[6y(x-t)^2-2y^3\right]}{((x-t)^2+y^2)^3}dt

    and I think you got the other one by now. It's messy and I had problems with it too but here it is:

    \frac{\partial u}{\partial y}=\frac{y}{\pi}\int_{-\infty}^{\infty}\frac{-2y f(t)}{((x-t)^2+y^2)^2}+\frac{1}{\pi}\int_{-\infty}^{\infty}\frac{f(t)}{(x-t)^2+y^2}

    \frac{\partial^2 u}{\partial y^2}=\frac{y}{\pi}\left[\int_{-\infty}^{\infty} \frac{f(t)\left[(x-t)^2+y^2\right]^2(-2)+8y^2\left[(x-t)^2+y^2\right]}{((x-t)^2+y^2)^4}\right] -\frac{1}{\pi}\int_{-\infty}^{\infty}\frac{4yf(t)}{(x-t)^2+y^2)^2}

    =\frac{1}{\pi}\int_{-\infty}^{\infty} \frac{f(t)\left[-2y^3-2y(x-t)^2+8y^3-\left[4y((x-t)^2+y^2)\right]\right]}{((x-t)^2+y^2)^3}

    =\frac{1}{\pi}\int_{-\infty}^{\infty}\frac{f(t)\left[2y^3-6y(x-t)^2\right]}{((x-t)^2+y^2)^3}

    which cancels the other one.
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