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Math Help - Vector calculus identities mess

  1. #1
    Member Mollier's Avatar
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    Vector calculus identities mess

    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.

     \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}

    The author turns the left side of the equation into: -\nabla\cdot(p\vec{e} + \rho\vec{u}(\vec{u}\cdot\vec{e})) + \rho\vec{b}\cdot\vec{e}

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


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

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

    I am not sure what to do with is. \rho is a scalar valued function, but so is I think \vec{u}\cdot\vec{e}. 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.
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  2. #2
    MHF Contributor FernandoRevilla's Avatar
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    Re: Vector calculus identities mess

    Quote Originally Posted by Mollier View Post
    Next I need to find  \nabla\cdot (\rho\vec{u}(\vec{u}\cdot\vec{e}))
    We have \nabla\cdot(f\vec{F})=\textrm{div}(f\vec{F})=f\; \textrm{div}\vec{F}+\vec{F}\cdot\nabla f=f(\nabla\cdot \vec{F})+\vec{F}\cdot \nabla f . In our case, f=\rho (\vec{u}\cdot\vec{e}) and \vec{F}=\vec{u} .
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  3. #3
    Member Mollier's Avatar
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    Re: Vector calculus identities mess

    Hi,

    I tried that but ran into a few problems while trying to make the result look identical to the textbook.

     \nabla\cdot(\rho(\vec{u}\cdot\vec{e})\vec{u}) = \rho((\vec{u}\cdot\vec{e})(\nabla\cdot(\vec{u}))) + \vec{u}\cdot\nabla(\rho(\vec{u}\cdot\vec{e}))

    I play a bit with the last term and get,

    \vec{u}\cdot\nabla(\rho(\vec{u}\cdot\vec{e})) = \vec{u}\cdot [\rho\nabla(\vec{u}\cdot\vec{e})+(\vec{u}\cdot\vec{  e})\nabla(\rho)]

    by using the product rule for gradients. Now I know that there is an identity that tells us how to take the gradient of a vector dot product. I have here a dot product between a vector valued functon ( \vec{u}) and a fixed vector ( \vec{e}). If I treat both as vectors I get,

    \rho\nabla(\vec{u}\cdot\vec{e}) = \rho[(\vec{u}\cdot\nabla)\vec{e}+(\vec{e}\cdot\nabla ) \vec{u}+\vec{u}\times (\nabla\times\vec{e})+\vec{e}\times(\nabla \times \vec{u})]

    The second and third term are zero leaving me with

    \rho\nabla(\vec{u}\cdot\vec{e}) = \rho[(\vec{u}\cdot\nabla)\vec{e}+\vec{e}\times(\nabla \times \vec{u})]

    To get what I want, I need to let the last term be zero. Then,

    \rho\nabla(\vec{u}\cdot\vec{e})\cdot\vec{u} = \rho(\vec{u}\cdot\nabla)\vec{u}\cdot\vec{e}

    This looks like one of the terms in the textbook (see my first post).
    I really hope I am missing something because this is a mess Thanks!
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