# Unfamiliar formulation of Stokes Problem

1. Mar 23, 2012

### the.drizzle

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
$$\Delta v - \nabla p = f$$
which, components aside, does not seem to be an equivalent statement. Due to my application, the inclusion of the initial condition $\sigma_{ij,j}$ is unimportant, and conservation of mass ($\nabla\cdot v=0$) 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. Mar 23, 2012

### Hassan2

See the explanation in section 1.5.

3. Mar 23, 2012

### the.drizzle

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
$$\nabla\cdot\left(\eta\left(\nabla v + \nabla^T v \right)\right)$$
is equivalent (in some sense?) to
$$\eta\Delta v$$
That is, I'm assuming that they mean that $\nabla^T$ denotes the adjoint to $\nabla$, but even then that doesn't seem to add up...

Cheers!

4. Mar 23, 2012

### Alpha Floor

$\nabla^Tv$ denotes the TRANSPOSE of $\nabla v$

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 ($\nabla · v = 0$) 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:

$\nabla · σ = - \nabla \times (η\nabla \times v)$

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

$$\nabla \times \nabla \times \vec{v} = \nabla (\nabla \cdot \vec{v}) - \nabla^2 \vec{v}$$

Hope I could clarify!

5. Mar 24, 2012

### the.drizzle

Brilliant, thank you!

:-)