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Properties of mixed partial derivatives

  1. Aug 27, 2013 #1
    Hi, I am sort of hung up with a particular step in a derivation, and this has caused me to ponder a few properties of partial derivatives. As a result, I believe I may be correct for the wrong reasons. For this example, the starting term is

    ([itex]\frac{\partial}{\partial x}[/itex][itex]\frac{\partial v}{\partial t}[/itex]-[itex]\frac{\partial}{\partial y}[/itex][itex]\frac{\partial u}{\partial t}[/itex])

    I want to go from the above term to

    [itex]\frac{\partial}{\partial t}[/itex] ([itex]\frac{\partial v}{\partial x}[/itex] - [itex]\frac{\partial u}{\partial y}[/itex])

    I am a little confused how this is done. I am not sure if you can "factor" out the [itex]\frac{\partial}{\partial t}[/itex] or not. I thought about simply rearranging the partials, but I don't think I can assume the function is smooth or symmetric. Any help or insight you can provide will be appreciated.

    Thanks,
    wefoust
     
  2. jcsd
  3. Aug 27, 2013 #2

    cristo

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    Well, partial derivatives commute, so

    [tex]\frac{\partial}{\partial x} \Big(\frac{\partial v}{\partial t}\Big)=\frac{\partial^2v}{\partial x \partial t}=\frac{\partial}{\partial t}\Big(\frac{\partial v }{\partial x}\Big) [/tex]

    from there you can factor out the time derivative.
     
  4. Aug 27, 2013 #3

    tiny-tim

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  5. Aug 27, 2013 #4
    Doesn't that require the continuity of the second partial derivatives in order to be true in general? It seems to me that OP explicitly stated that not enough is known about the function to guarantee that?
     
  6. Aug 27, 2013 #5
    Hi guys,
    Thank you very much for you input so far! I've looked up some more info based on the links and terms you all have used, and I feel that I am getting descent understanding about the property. It seems that assuming all partials commute is a bit of an overstatement. In most cases the partials should commute, but there do exists special cases, as when the 2nd order is not defined or continuous (this holds for both ways as in Fxy or Fyx).

    Again, thanks for the help, and let me know if you come across something more convincing.
     
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