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Math Help - Is this function harmonic?

  1. #1
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    Is this function harmonic?[SOLVED]

    u=\frac{1}{r} and r=\sqrt{x^2+y^2} Calculate the Laplacian L(u) and give your anser in terms of r. Use dr/dx=d/x etc.

    \frac{du}{dx}=\frac{du}{dr}=\frac{-1}{r^2}\frac{dr}{dx}=\frac{-x}{r^3}

    now I'm stuck calculating the second order derivative
    Last edited by superdude; October 8th 2009 at 03:21 PM. Reason: solved
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  2. #2
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    Quote Originally Posted by superdude View Post
    u=\frac{1}{r} and r=\sqrt{x^2+y^2} Calculate the Laplacian L(u) and give your anser in terms of r. Use dr/dx=d/x etc.

    \frac{du}{dx}=\frac{du}{dr}=\frac{-1}{r^2}\frac{dr}{dx}=\frac{-x}{r^3}

    now I'm stuck calculating the second order derivative
    If u_x = r_x u_r then  <br />
u_{xx} = r_x^2 u_{rr} + r_{xx} u_r<br />
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  3. #3
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    Quote Originally Posted by Danny View Post
    If u_x = r_x u_r then  <br />
u_{xx} = r_x^2 u_{rr} + r_{xx} u_r<br />
    could you please elaborate? where does this come from? Does r_x^2 mean square r before taking it's derivative with respect to x?

    This is what I've got but I know it's wrong:
    \frac{\partial^2 u}{\partial x^2}=\frac{\partial}{\partial x}(\frac{-x}{r^3}) = \frac{3x}{r^45}\times\frac{x}{r}=\frac{3x^2}{r^5}
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  4. #4
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    Quote Originally Posted by Danny View Post
    If u_x = r_x u_r then  <br />
u_{xx} = r_x^2 u_{rr} + r_{xx} u_r<br />
    If u_x = r_x u_r then we have the operator \frac{\partial}{\partial x} = r_x \frac{\partial}{\partial x}


    Therefore  <br />
\frac{\partial}{\partial x} u_x = \frac{\partial}{\partial x} \left( r_x u_r\right) = \frac{\partial}{\partial x} \left(r_x \right)u_r + r_x \frac{\partial}{\partial x} \left( u_r\right) <br />

     <br />
= r_{xx}u_r + r_x r_x \frac{\partial}{\partial r} \left( u_r\right) \left(\text{from the operator above}\right)

     <br />
= \left(r_x\right)^2 u_{rr} + r_{xx} u_r.<br />
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  5. #5
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    got it
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