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Confused on partial differentiation.

  1. Apr 26, 2010 #1
    In the process of deriving .[tex]\nabla^2u(x,y)[/tex]. in polar coordinates I am confuse how to work through the steps. The first step from the book is to find [tex]\frac{\partial u}{\partial x} [/tex]

    Standard partial differentiation is at follow:
    [tex]\frac{\partial u}{\partial x} = \frac{\partial u}{\partial r} \frac{\partial r}{\partial x}\;+\; \frac{\partial u}{\partial \theta}\frac{\partial \theta}{\partial x} [/tex]

    If we use the same method of partial differentiation to continue to find .[tex]\frac{\partial^2 u}{\partial x^2}[/tex]. we should perform:

    [tex]\frac{\partial^2 u}{\partial x^2}= \frac{\partial}{\partial r}(\frac{\partial u}{\partial x}) \;\; \frac{\partial r}{\partial x} \;+\; \frac{\partial}{\partial \theta}(\frac{\partial u}{\partial x}) \;\; \frac{\partial \theta}{\partial x} [/tex]

    [tex]\Rightarrow \frac{\partial^2 u}{\partial x^2}= \frac{\partial}{\partial r}[\frac{\partial u}{\partial r} \frac{\partial r}{\partial x}\;+\; \frac{\partial u}{\partial \theta}\frac{\partial \theta}{\partial x}] \frac{\partial r}{\partial x} \;+\; \frac{\partial}{\partial \theta}[\frac{\partial u}{\partial r} \frac{\partial r}{\partial x}\;+\; \frac{\partial u}{\partial \theta}\frac{\partial \theta}{\partial x}] \frac{\partial \theta}{\partial x} [/tex]




    But the book did this differently:

    [tex]\frac{\partial^2 u}{\partial x^2}= \frac{\partial}{\partial x}[\frac{\partial u}{\partial r} \frac{\partial r}{\partial x}\;+\; \frac{\partial u}{\partial \theta}\frac{\partial \theta}{\partial x}] [/tex]

    Then apply product rule:

    [tex]\frac{\partial^2 u}{\partial x^2}= [\frac{\partial r}{\partial x}\frac{\partial}{\partial x}(\frac{\partial u}{\partial r}) + \frac{\partial u}{\partial r}\frac{\partial^2r}{\partial x^2}]\;+\; [\frac{\partial \theta}{\partial x}\frac{\partial}{\partial x}(\frac{\partial u}{\partial \theta}) + \frac{\partial u}{\partial \theta}\frac{\partial^2 \theta}{\partial x^2}][/tex]

    Obvious the two don't give the same answer. Why don't the book need to perform the second derivative like the first one like what I did?
     
  2. jcsd
  3. Apr 26, 2010 #2
    I am confused on what you did. When I calculated the second partial, I got the same answer as your book. The way your book did it is the easiest way I can see. By definition:

    [tex]\frac{\partial^2 u}{\partial x^2} = \frac{\partial}{\partial x}\left( \frac{\partial u}{\partial x} \right)[/tex]

    Can you explain this statement more:

    [tex]\frac{\partial^2 u}{\partial x^2}= \frac{\partial}{\partial r}\left(\frac{\partial u}{\partial x}\right) \frac{\partial r}{\partial x} + \frac{\partial}{\partial \theta}\left(\frac{\partial u}{\partial x}\right) \frac{\partial \theta}{\partial x}[/tex]

    I am just not seeing the transition...
     
    Last edited: Apr 26, 2010
  4. Apr 26, 2010 #3
    Thanks so much for the reply. I am sure I am wrong and the book is right....so are you. I am just looking at the first partial differentiation:

    [tex]\frac{\partial u}{\partial x} = \frac{\partial u}{\partial r} \frac{\partial r}{\partial x}\;+\; \frac{\partial u}{\partial \theta}\frac{\partial \theta}{\partial x} [/tex]

    You notice even though it is .[tex]\frac{\partial u}{\partial x}[/tex]. , you still see differentiation of theta in the second part. All I did is treating [tex]\frac{\partial u}{\partial x}[/tex] as u of the first differentiation and so there is a term of differentiation by theta.

    I am just following the partial differentiation rule by the book that if u is function of .[tex]r \;and\; \theta[/tex]. and both are function of x, then [tex]\frac{\partial u}{\partial x}[/tex]. has to have terms differentiation by both .[tex]r \;and\; \theta[/tex]

    Tell me why it does not work the same way as the first differentiation?

    Thanks

    Alan
     
  5. Apr 27, 2010 #4
    Oh I see what you did there. That substitution is a bit of a risky move. When you substitute the first partial of u into the equation, you seem to be ignoring the product rule.

    The function u(r(x),θ(x)) is easy to differentiate with respect to x and find the formula you have, just by applying the chain rule. But now your partial ux is a completely new function, which includes products of functions (the other partials), therefore you have to apply the product rule.
     
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