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Laplacian term in Navier-Stokes equation

  1. Sep 9, 2014 #1
    I am trying to derive part of the navier-stokes equations. Consider the following link:

    http://www.gps.caltech.edu/~cdp/Desktop/Navier-Stokes Eqn.pdf

    Equation 1, without the lambda term, is given in vector form in Equation 3 as [itex]\eta\nabla^2\mathbf{u}[/itex]. However, when I try to get this from Eq. 1, I get [itex]2\eta\nabla^2\mathbf{u}[/itex]. I am getting [itex]2\eta[/itex] because

    [itex]\eta\left(\frac{\partial u_i}{\partial x_j}+\frac{\partial u_j}{\partial x_i}\right)=\eta(\nabla \mathbf{u}+\nabla \mathbf{u}^T)=2\eta\nabla \mathbf{u}[/itex]
    and then the taking the divergence gives [itex]\nabla\cdot 2\eta\nabla\mathbf{u}=2\eta\nabla^2\mathbf{u}[/itex] for constant viscosity [itex]\eta[/itex]

    I suspect that my second step in the first line is wrong. But I don't get it.

    Thanks in advance.
     
  2. jcsd
  3. Sep 9, 2014 #2

    olivermsun

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    You might want to keep the divergence and the terms in the parentheses together to remind yourself that you only sum over the j subscript (which is repeated).

    You should notice that the derivatives of the second term in the parentheses add up to the divergence of u.
     
  4. Sep 9, 2014 #3
    I don't see a repeated subscript. Do you mean that [itex]\nabla\mathbf{u}^T=\nabla\mathbf{u}[/itex]? That is in fact my assumption and thus why I get 2*eta, not eta. Is this not true?:

    [itex]\eta(\nabla\mathbf{u}+\nabla\mathbf{u}^T)=2\eta\nabla\mathbf{u}[/itex]

    if so, what happend to the 2?

    What rules am I missing??
     
  5. Sep 9, 2014 #4

    olivermsun

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    In the original Eq. 1:
    $$\frac{\partial} {\partial x_j} \left[ \eta\left(\frac{\partial u_i}{\partial x_j}+\frac{\partial u_j}{\partial x_i}\right) + \ldots \right]$$ has the subscript ##j## both outside and inside the parentheses.
     
  6. Sep 9, 2014 #5
    The i-th component of the force (rewriting equation 1 with [itex]\lambda=0[/itex] ) is

    [itex]F_i=\eta\sum\limits_{j}{\frac{\partial^2{u_i}}{\partial^2{x_j}}}+\eta \frac{\partial}{\partial{x_i}}\sum\limits_{j}\frac{\partial{u_j}} {\partial{x_j}} [/itex].

    The second term is zero cause the divergence of u is zero hence all that is left is [itex]\eta\sum\limits_{j}{\frac{\partial^2{u_i}}{\partial^2{x_j}}}=\eta\nabla^2u_i[/itex].
     
    Last edited: Sep 9, 2014
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