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Thread: Vector Differentiation

  1. #1
    Member kjchauhan's Avatar
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    Vector Differentiation

    Please help me to solve this problem:

    Prove that:

    $\displaystyle \nabla^2 f(r) = f''(r) + \frac{2}{r}f'(r)$

    Where $\displaystyle r^2=x^2+y^2+z^2$

    Thanks in advance..
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  2. #2
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    Hint:

    $\displaystyle \displaystyle \frac{\partial f(r)}{\partial x} = f'(r) \frac{\partial r}{\partial x} = f'(r)\frac{x}{r} $

    $\displaystyle \displaystyle \frac{\partial^2 f(r)}{\partial^2 x} = \frac{\partial^2 f'(r)\frac{x}{r}}{\partial x} = f''(r)\frac{x^2}{r^2} + f'\frac{r^2 - x^2}{r^3}$.
    Last edited by snowtea; Jan 10th 2011 at 06:51 AM.
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  3. #3
    Member kjchauhan's Avatar
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    Quote Originally Posted by snowtea View Post
    Hint:

    $\displaystyle \displaystyle \frac{\partial f(r)}{\partial x} = f'(r) \frac{\partial r}{\partial x} = f'(r)\frac{x}{r} $

    $\displaystyle \displaystyle \frac{\partial^2 f(r)}{\partial^2 x} = \frac{\partial^2 f'(r)\frac{x}{r}}{\partial x} = f''(r)\frac{x^2}{r} + f'\frac{r^2 - x^2}{r^3}$.
    Thanks for Hint..
    This is correct but I couldn't find the right hand side ..
    According to your hint, the first term of the right will be $\displaystyle rf''(r)$..
    So, is it correct??

    Ok, well, got it..

    Thank you very much..
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  4. #4
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    Quote Originally Posted by kjchauhan View Post
    Thanks for Hint..
    This is correct but I couldn't find the right hand side ..
    According to your hint, the first term of the right will be $\displaystyle rf''(r)$..
    So, is it correct??
    I made a typo. It should be $\displaystyle \frac{x^2}{r^2}$. It is now fixed in the previous post.
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