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Math Help - Does the function u match the differential equation

  1. #1
    Junior Member
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    Does the function u match the differential equation

    Problem statement: If u=\frac{1}{r} where r=\sqrt{(x-a)^2+(y-b)^2+(z-c)^2}, for r\ne0 prove that \bigtriangleup u \equiv \frac{\partial^2 u}{\partial x^2}+\frac{\partial^2 u}{\partial y^2}+\frac{\partial^2 u}{\partial z^2}=0

    This is my approach: \frac{\partial u}{\partial x}=\frac{\partial u}{\partial r}\frac{dr}{dx}=-\frac{x-a}{r^2\sqrt{(x-a)^2+(y-b)^2+(z-c)^2}}
    \frac{\partial^2 u}{\partial x^2}=\frac{\partial(\frac{\partial u}{\partial x})}{\partial x}=\frac{\partial(\frac{\partial u}{\partial x})}{\partial r}\frac{dr}{dx}=\frac{2(x-a)^2}{r^3[(x-a)^2+(y-b)^2+(z-c)^2]}

    \frac{\partial^2 u}{\partial y^2}=\frac{\partial(\frac{\partial  u}{\partial y})}{\partial y}=\frac{\partial(\frac{\partial u}{\partial y})}{\partial  r}\frac{dr}{dy}=\frac{2(y-b)^2}{r^3[(x-a)^2+(y-b)^2+(z-c)^2]}

    \frac{\partial^2 u}{\partial z^2}=\frac{\partial(\frac{\partial   u}{\partial z})}{\partial z}=\frac{\partial(\frac{\partial u}{\partial z})}{\partial   r}\frac{dr}{dz}=\frac{2(z-c)^2}{r^3[(x-a)^2+(y-b)^2+(z-c)^2]}


    So \frac{\partial^2 u}{\partial x^2}+\frac{\partial^2 u}{\partial y^2}+\frac{\partial^2 u}{\partial z^2}=\frac{2(x-a)^2+2(y-b)^2+2(z-c)^2}{r^3[(x-a)^2+(y-b)^2+(z-c)^2]}=\frac{2}{r}\ne0

    It seems I made an major error somewhere or I have conceptual problems.
    Help me!

    Thank you
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  2. #2
    Super Member
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    Athens, OH, USA
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    Re: Does the function u match the differential equation

    Yes, you made an error in the calculation of the 2nd partials. Here's the situation: you have a function r(x,y,z) and a function u=f(r) (in your case 1/r). So you want the partials of u w.r.t. x , y and z. The first partial ux=f'(r(x,y,z) times rx is used correctly. But the second partial is:

    u_{xx}=f^{\prime\prime}(r(x,y,z)r_x\cdot r_x+f^{\prime}(r(x,y,z)r_{xx}, using the product rule.

    If you do this (replacing occurrences of r) you get:

    u_{xx}={3(x-a)^2\over r^5}-{1\over r^3}

    Similarly for the other 2nd partials. It is then easy to see that Laplace's equation is satisfied.
    Thanks from patzer
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  3. #3
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    Re: Does the function u match the differential equation

    Thank you John
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