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Vector calculus identities navigation

  1. Jan 11, 2012 #1
    1. The problem statement, all variables and given/known data
    I'm reading in a fluid dynamics book and in it the author shortens an equation using identities my rusty vector calculus brain cannot reproduce.


    2. Relevant equations
    [tex] \vec{e} \cdot \frac{\partial}{\partial t}(\rho \vec{u}) =
    -\nabla\cdot (\rho\vec{u})\cdot\vec{e} - \rho(\vec{u}\cdot\nabla)\vec{u}\cdot\vec{e}
    - (\nabla p)\cdot\vec{e} + \rho\vec{b}\cdot\vec{e}
    [/tex]

    The author turns the left side of the equation into:

    [tex]-\nabla\cdot(p\vec{e} + \rho\vec{u}(\vec{u}\cdot\vec{e})) + \rho\vec{b}\cdot\vec{e} [/tex]

    Just to be clear;
    [tex]\vec{u}[/tex] is a vector valued function,
    [tex]\vec{e}[/tex] is a fixed vector,
    [tex]\rho, p[/tex] are scalar valued functions.

    3. The attempt at a solution

    The first part is fine:
    [tex]\nabla\cdot (p\vec{e}) = \nabla(p)\cdot\vec{e} + p(\nabla\cdot\vec{e})[/tex]
    The divergence of a fixed vector is zero and so,
    [tex]\nabla\cdot (p\vec{e}) = \nabla(p)\cdot\vec{e}[/tex]

    Next I need to find
    [tex] \nabla\cdot (\rho\vec{u}(\vec{u}\cdot\vec{e})) [/tex]

    I am not sure what to do with is. [tex]\rho[/tex] is a scalar valued function, but so is I think [tex]\vec{u}\cdot\vec{e}[/tex]. I know of the product rule between a scalar and a vector valued function, but what happens when there are two scalar valued functions?

    Any suggestions are welcome, thanks.
     
  2. jcsd
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