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The Brinkman equations for steady flow of an incompressible fluid through rigid porous solid are:

$-\dfrac{\mu_0}{k}\mathbf{v} - \mathrm{grad}p + \mu_0 \Delta \mathbf{v} =0$ and $\mathrm{div}(\mathbf{v}) = 0$ in $\Omega$

where $k > 0$ is the permeability of the solid, $\mu_0 >0$ is the viscosity of the fluid, and $p,\mathbf{v}$ are the pressure and velocity fields.

Suppose that the domain is divided into two disjoint regions and $k$ assumes the values $k_0$ and $k_1$ on them.

My question: Is there a jump condition for this problem due to the discontinuity of $k$ ?

Ans: $\mathrm{jump}(\mathbf{t_n})=0$ where $\mathbf{t_n} := -p\mathbf{n} + \mu_0\nabla \mathbf{v}\;\mathbf{n}$

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There is a jump condition, but it is automatically satisfied by the solution (i.e., you don't need to do anything special). This is no different than solving $-\nabla \cdot k \nabla u = f$: there is a jump condition discussed in Jump condition for elliptic equation in standard finite element method but it is automatically satisfied.

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  • $\begingroup$ Could you please elaborate what will be the expression of the jump condition ? If there is a jump, I am unable to see how the term $\mu \Delta\mathbf{v}$ will make sense. I had asked the linked question too. $\endgroup$ – me10240 Aug 31 '14 at 22:26
  • $\begingroup$ The physically correct equation will have the term $\nabla \cdot \mu \nabla \mathbf v$ instead of the one you list. The jump condition will then be obvious. $\endgroup$ – Wolfgang Bangerth Aug 31 '14 at 23:59
  • $\begingroup$ I looked at the derivation of jump conditions from a continuum mechanics text, and realized that to get the correct form, I need to multiply throughout by $k$ first. Then, I obtained the jump condition as: $ \mathrm{jump}(k\boldsymbol{t_n}) = 0 $ where $\boldsymbol{t_n} = [- pI + \mu_0(\nabla \mathbf{v}]\mathbf{n} $, where $\mathbf{n}$ is the unit normal. Could you please confirm if this is correct ? If so I shall add it to the thread as an answer and mark the issue solved. $\endgroup$ – me10240 Sep 1 '14 at 1:38
  • $\begingroup$ Yes, that seems right. It's the flux/stress that's got the jump condition and that's exactly what you state here. $\endgroup$ – Wolfgang Bangerth Sep 1 '14 at 12:35
  • $\begingroup$ Based on your answer to another question of mine on jump conditions, I think my jump condition stated above is wrong. It should not have the $k$ factor. I am adding the correct answer to my original question. $\endgroup$ – me10240 Sep 2 '14 at 16:30

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