1. Coupling Stokes and Darcy equations
1.1. Problem setting
We have : \(\underline n_f = -\underline n_p\text{ on }\Gamma\).
We aim to solve the Stokes equation in domain \(\Omega_f\) and the Darcy equation in domain \(\Omega_p\), subject to boundary conditions on \(\partial\Omega=(\partial\Omega_f\backslash\Gamma)\cup(\partial\Omega_p\backslash\Gamma)\) together with coupling relations along \(\Gamma\). We denote by \(\Gamma_f\text{ and }\Gamma_p\) the outer boundary of \(\Omega_f\text{ and }\Omega_p\) respectively.
We use the following notations :
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\(\underline u_f\text{ and }p_f\) are the fluid velocity and fluid pressure respectively within the Stokes domain \(\Omega_f\),
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\(\underline u_p\text{ and }\varphi\) are the fluid velocity and piezometric head respectively within the Darcy domain \(\Omega_p\),
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\(\underline{\underline D}(\underline u)\) is the symmetrized gradient defined by :
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\(\underline{\underline T}(\underline u,p)\) is the Cauchy stress tensor defined by :
1.2. Equations
Writing Stokes and Darcy problems on their respective domains yields the system :
1.3. Boundary conditions
The same essential/natural splitting as in the sole Darcy problem occurs here. One thus has to split the outer boundary \(\partial\Omega\) in four relatively open disjoint pieces. The subscripts \(f,p\) denote objects relative to the Stokes and The Darcy problem respectively, and the subscripts \(n,e\) denote respectively natural and essential boundary condition.
One can impose :
1.4. Coupling conditions
We impose the following coupling conditions along the interface \(\Gamma\).
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Continuity of the normal velocity, i.e. \(\underline u_f\cdot\underline n_f + \underline u_p\cdot\underline n_p = 0\), or :
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Continuity of the normal component of the Cauchy stress :
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Beavers-Joseph condition :
where \(\alpha\) is proportional to the square root of the permeability of \(\Omega_p\) and inversely proportional to the square root of the kinematic viscosity \(\nu\) and depends on the characteristic length of the pores in \(\Omega_p\).
It is common to neglect the tangential components of \(\underline u_p\), thus the Beavers-Joseph condition reduces to the Beavers-Joseph-Saffman condition :
See D. A. Nield, The Beavers-Joseph boundary condition and related matters : a historical and critical note, Transp. Porous Med. (2009) 78:537-540 and M. Hoffmann, The Navier-Stokes-Darcy problem, Master’s Thesis (2013). This identity comes from empirical observations.
1.5. Weak formulation
We basically just write independently the weak formulation for Darcy and Stokes problem, then make the coupling condition appear by substitution.
In the following, function restrictions on a boundary are to be understood in terms of trace operator.
From Stokes' equation, we have to take the divergence of the stress tensor, in particular of the deformation tensor, times a test function \(\underline v\in H_f\). We use the well-known identity :
where \(A:B=Tr(AB^T)\) denotes the generalized scalar product. Since \(D(\underline u)\) is symmetric, we can replace \(\nabla\underline v\) by \(D(\underline v)\), according to the following relations : \(Tr(AB^T)=Tr(BA^T)=Tr(BA)=Tr(AB)\) with \(A\) symmetric and any \(B\).
We define the following bilinear forms :
1.5.1. Stokes
Let \(\underline v\in H_f\). Multiplying and integrating gives :
We now focus on the left hand side term.
Notice that the \(\Gamma_{f,e}\) part vanishes since \(\left.\underline v\right|_{\Gamma_{f,e}}=0\). Writing \(\displaystyle\underline v=(\underline v\cdot\underline n_f)\underline n_f+\sum_{i=1}^{d-1}(\underline v\cdot\underline\tau_i)\underline\tau_i\) on the boundary gives :
allowing us to apply the coupling conditions :
so that the total weak Stokes equation is :
We may often use Saffmann’s approximation and a zero Neumann condition on the outer boundary, yielding :
1.5.2. Darcy
Since \(\underline u_p=\underline{\underline\kappa}\nabla\varphi\) and \(\nabla\cdot\underline u_p=0\), we rewrite Darcy’s equation :
Let us write its weak formulation :
so the weak Darcy is :
1.5.3. Sum up
Adding up these equations yields the weak formulation of this coupled Stokes-Darcy problem :