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Math Help - Partial Derivatives satisfying laplace equation.

  1. #1
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    Partial Derivatives satisfying laplace equation.

    Hi

    I needed help understand the partial derivatives of a solution.

    I have been given the chain rule.

    I have attached my problem.
    Attached Thumbnails Attached Thumbnails Partial Derivatives satisfying laplace equation.-partial-derivates.gif   Partial Derivatives satisfying laplace equation.-chainrule.gif  
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  2. #2
    MHF Contributor FernandoRevilla's Avatar
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    What concrete doubts have you had?



    Fernando Revilla
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  3. #3
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    Please find attached the the bits in red which I don't understand how they appeared and how they were calculated.

    I know that the rest of the terms came from the expansion of the brackets.

    Thanks
    Attached Thumbnails Attached Thumbnails Partial Derivatives satisfying laplace equation.-partial-derivates1.gif  
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  4. #4
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    Remember that each of those derivatives involves a derivative of a product and so, by the product rule, the sum of two terms. That's why, in the top line, you have four terms but in the next eight. Each of the terms you have in red for u_{xx} is the result of applying the derivative to the part of the product that does not already involve a derivative of u.

    The first term in red I wouldn't have written at all. It is from applying the derivative in cos(\theta)\frac{\partial}{\partial r} to the cos(\theta) in \frac{\partial u}{\partial r}cos(\theta): cos(\theta)\frac{\partial cos(\theta)}{\partial r}u_r. But u and r are independent variables: \frac{\partial cos(\theta)}{\partial r}= 0 so that is cos(\theta) (0) u_r= 0. That " u_r(0)" is u_r times 0, not " u_r of 0".

    The second term is red is from applying the derivative in cos(\theta)\frac{\partial}{\partial r} to the \frac{1}{r} in -\frac{\partial u}{\partial r}\frac{sin(\theta)}{r}: \left(\cos(\theta)\frac{\partial u}{\partial \theta}sin(\theta)\right)\frac{\partial}{\partial r}\frac{1}{r}= \left(sin(\theta)cos(\theta)u_\theta\right)\left(-\frac{1}{r^2}\right)= -\frac{sin(\theta)cos(\theta)}{r^2}u_\theta.

    The third is from applying the derivative in -\frac{sin(\theta)}{r}\frac{\partial}{\partial\thet  a} to the cos(\theta) in \frac{\partial u}{\partial\theta}cos(\theta): -\frac{sin(\theta)}{r}\frac{\partial u}{\partial r}\frac{\partial cos(\theta)}{\partial \theta}= -\frac{sin(\theta)}{r}u_r\left(- sin(\theta)\right)= \frac{sin^2(\theta)}{r}u_r

    The fourth is from applying the derivative in -\frac{sin(\theta)}{r}\frac{\partial}{\partial\thet  a} to the sin(\theta) in \frac{\partial u}{\partial \theta}\frac{sin(\theta)}{r}: -\frac{sin(\theta)}{r}\left(-\frac{\partial u}{\partial r}\right)\frac{1}{r}\frac{\partial sin(\theta)}{\partial \theta}= \frac{sin(\theta)cos(\theta)}{r^2}u_r

    In the last, you have encircled two separate terms from two separate derivatives.

    The first, -2\frac{sin(\theta)cos(\theta)}{r^2}u_\theta, is from applying the derivative in sin(\theta)\frac{\partial }{\partial r} to the \frac{1}{r} in \frac{\partial u}{\partial \theta}\frac{cos(\theta)}{r}: sin(\theta)cos(\theta)u_\theta\frac{\partial \frac{1}{r}}{\partial r}= -\frac{sin(\theta)cos(\theta)}{r^2}u_\theta.

    The second, \frac{cos^2(\theta)}{r}u_r, is from applying the derivative in \frac{cos(\theta)}{r}\frac{\partial}{\partial \theta} to the sin(\theta)
    in \frac{\partial u}{\partial r}sin(\theta): \frac{cos(\theta)}{r}\frac{\partial u}{\partial r}\frac{\partial sin(\theta)}{\partial \theta}= \frac{cos(\theta)}{r}u_r (cos(\theta)= \frac{cos^2(\theta)}{r}u_r
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  5. #5
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    thank you for your help. Very much appreciated
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