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Unfamiliar formulation of Stokes Problem

  1. Mar 23, 2012 #1
    Hello, I'm trying out the escript python FEM software package which is so far rather impressive, if for no other reason than the developers have included a Stokes Flow solver. The problem I'm having, however, is that they have formulated the problem in a manner I have not encountered before, nor can seem to make it "work" in the manner I would expect it to. In particular, we have from from section 6.1 of the users manual:

    My basic problem is that I have not encountered what would normally be the Laplacian on the LHS of the above statement. That is, I would typically expect Stokes problem to be stated as
    [tex]\Delta v - \nabla p = f[/tex]
    which, components aside, does not seem to be an equivalent statement. Due to my application, the inclusion of the initial condition [itex]\sigma_{ij,j}[/itex] is unimportant, and conservation of mass ([itex]\nabla\cdot v=0[/itex]) is assumed in both cases.

    So, can anyone tell me what I'm doing wrong, or where I might find a derivation of the quoted formulation so that I can actually apply it?

    Thanks!
     
    Last edited: Mar 23, 2012
  2. jcsd
  3. Mar 23, 2012 #2
    See the explanation in section 1.5.
     
  4. Mar 23, 2012 #3
    Thanks, but I suppose I should clarify...

    The problem I'm having is not one of indices vs. operator, what I'm failing to see is how
    [tex]\nabla\cdot\left(\eta\left(\nabla v + \nabla^T v \right)\right)[/tex]
    is equivalent (in some sense?) to
    [tex]\eta\Delta v[/tex]
    That is, I'm assuming that they mean that [itex]\nabla^T[/itex] denotes the adjoint to [itex]\nabla[/itex], but even then that doesn't seem to add up...

    Cheers!
     
  5. Mar 23, 2012 #4
    [itex]\nabla^Tv[/itex] denotes the TRANSPOSE of [itex]\nabla v[/itex]

    If you sum them both and divide by 2, you get a symmetrical tensor called the "rate of stain tensor", let's call it ε

    For an incompressilble flow ([itex]\nabla · v = 0[/itex]) the law that relates the "viscous stress tensor σ" (I think this one is also called deviatoric stress tensor) to the "rate of strain tensor ε" is:

    σ= 2η·ε

    Now, in the equation of conservation of momentum, σ doesn't appear as such, but through its divergence. If you calculate its divergence (or just look it up, Navier-Poisson's Law), you get to the conclusion:

    [itex]\nabla · σ = - \nabla \times (η\nabla \times v)[/itex]

    Since η is constant you can get it out of the curl expression. Applying this property of operators you finally get to the laplacian of v

    [tex]\nabla \times \nabla \times \vec{v} = \nabla (\nabla \cdot \vec{v}) - \nabla^2 \vec{v} [/tex]

    Hope I could clarify!
     
  6. Mar 24, 2012 #5
    Brilliant, thank you!

    :-)
     
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