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Applying Divergence Theorem

  1. May 15, 2012 #1
    1. The problem statement, all variables and given/known data

    attachment.php?attachmentid=47293&stc=1&d=1337106295.png

    2. Relevant equations

    So I have that [itex] v \otimes n = \left( \begin{array}{ccc}
    v_{1}n_{1} & v_{1}n_{2} & v_{1}n_{3} \\
    v_{2}n_{1} & v_{2}n_{2} & v_{2}n_{3} \\
    v_{3}n_{1} & v_{3}n_{2} & v_{3}n_{3} \end{array} \right)
    [/itex]

    3. The attempt at a solution

    I've tried applying the Divergence theorem for Tensors:
    [itex]
    \int_{\partial B} ( v \otimes n )n dA = \int_{B} \nabla \cdot ( v \otimes n ) dV
    [/itex]

    But that doesn't lead anywhere particularly useful. I thought it might be worth noting that [itex]\nabla \cdot ( v \otimes n ) = \frac{dv_{1}}{dx_{1}}n_{1}+\frac{dv_{2}}{dx_{2}}n_{2} + \frac{dv_{3}}{dx_{3}}n_{3}[/itex] but I can't seem to get anywhere near [itex]\nabla v[/itex]

    And this problem isn't homework, it's just an optional exercise, but it's frustrated me for a while and I figured I should get some pointers.
     

    Attached Files:

    Last edited: May 15, 2012
  2. jcsd
  3. May 15, 2012 #2
    No it's not! I think you will find (assuming n is constant) that [itex] \nabla \cdot ( v \otimes n ) = n \cdot (\nabla v) + n (\nabla \cdot v) [/itex]

    Especially note that it is a vector and not a scalar
     
  4. May 15, 2012 #3
    Isn't [itex]\nabla \cdot v \otimes n = (n_{1}(\frac{dv_{1}}{dx_{1}}+\frac{dv_{2}}{dx_{2}}+\frac{dv_{3}}{dx_{3}}), n_{2}(\frac{dv_{1}}{dx_{1}}+\frac{dv_{2}}{dx_{2}}+\frac{dv_{3}}{dx_{3}}), n_{3}(\frac{dv_{1}}{dx_{1}}+\frac{dv_{2}}{dx_{2}}+\frac{dv_{3}}{dx_{3}}))[/itex]

    Which is [itex]n \cdot \nabla v[/itex]?

    This would make sense since it gives that

    [itex]\int_{\partial B} ( v \otimes n )n dA = \int_{B} (\nabla v) ndV[/itex]

    And then since n is just so constant vector, the result follows.

    Is that right? Or am I missing something?
     
  5. May 16, 2012 #4
    Of course not; you can't just get rid of the terms that give you trouble... This is what you need to show, but the formula I gave above is still correct.
     
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