Rigorous justification of the effective boundary condition on a porous wall via homogenization

Abstract

Viscous flow through a reservoir with porous boundary is studied via asymptotic analysis and homogenization. Under the assumption of periodicity of the pores, the effective boundary condition is derived and rigorously justified. The velocity on the boundary satisfies a version of the Darcy law. The Darcy law for tangential component can also be seen as the Beavers–Joseph law.

Appendix A: The boundary layer problem

To start with, we consider the general problem

$$\begin{aligned}&-\mu \,\Delta _y \mathbf{B}+\nabla _y b =\mathbf{F}\,\,\,\text{ in }\,\,\, {\mathcal {G}}\,, \end{aligned}$$

(109)

$$\begin{aligned}&\text{ div}_y\, \mathbf{B}=G\,\,\,\text{ in }\,\,\, {\mathcal {G}}\,, \end{aligned}$$

(110)

$$\begin{aligned}&{\mathbf {T}} (\mathbf{B}, b){\mathbf {e}}_2={\mathbf {d}} \,\,\,\text{ on }\,\,\,\gamma ^N\,, \end{aligned}$$

(111)

$$\begin{aligned}&\mathbf{B} ={\mathbf {0}}\,\,\,\text{ on }\,\,\gamma ^D \end{aligned}$$

(112)

$$\begin{aligned}&({\mathbf {B}} , b ) \;\;\text{ is }~ 1\text{-periodic } \text{ in }\;y_1\;,\;\;\nabla _y \mathbf{B} \in L^2 ({\mathcal {G}})\,\,\;\; . \end{aligned}$$

(113)

By its weak solution, we mean $({\mathbf {B}}, b) \in {\mathcal {D}}({\mathcal {G}}) \times L^2_{loc} ({\mathcal {G}})$ such that

$$\begin{aligned} \text{ div}_y\, \mathbf{B}=G \end{aligned}$$

in the sense of distributions, and

$$\begin{aligned} \mu \int \limits _{{\mathcal {G}}} \nabla _y {\mathbf {B}}\;:\;\nabla _y {\mathbf {z}} = \int \limits _{{\mathcal {G}}}{\mathbf {F}}\,\cdot \,{\mathbf {z}} + \int \limits _{\gamma ^N} {\mathbf {z}}\cdot {\mathbf {d}}\;\;,\;\;\forall \;{\mathbf {z}} \in {{\mathcal {D}}}_0 ({\mathcal {G}}), \end{aligned}$$

(114)

where

$$\begin{aligned} {{\mathcal {D}}}({{\mathcal {G}}}) = \lbrace {\mathbf {V}}\in H^1_{loc}({\mathcal {G}})\;,\;{\mathbf {V}}={\mathbf {0}}\;\text{ on }\;\gamma ^D\;\;,\;\;\nabla _y {\mathbf {V}} \in L^2 ({\mathcal {G}})\rbrace . \end{aligned}$$

and

$$\begin{aligned} {{\mathcal {D}}}_0({{\mathcal {G}}}) = \lbrace {\mathbf {V}}\in H^1 ({{\mathcal {G}}})\;,\;{\mathbf {V}}={\mathbf {0}}\;\text{ on }\;\gamma ^D\;\;,\;\;\text{ div}_y{\mathbf {V}} =0 \rbrace . \end{aligned}$$

The pressure is then picked to satisfy

$$\begin{aligned} \nabla _y b = \mu \,\Delta _y \mathbf{B} + {\mathbf {F}} \end{aligned}$$

in the sense of distributions.

We prove the following result providing the existence and asymptotic behavior of the solution:

Theorem 3

Let ${\mathbf {d}} \in {\mathbb {R}}^2$ and let ${\mathbf {F}}\in L^2 ({\mathcal {G}})^2,\;G\in L^2 ({\mathcal {G}})$ be such that $e^{\alpha y_2} {\mathbf {F}}\in L^2 ({\mathcal {G}})^2 \;,\; e^{\alpha y_2} G \in L^2 ({\mathcal {G}})^2$, for some $\alpha >0$ and that for (a.e.) $s>0$

$$\begin{aligned} \langle {\mathbf {F}}\rangle = \int \limits _{\gamma (s)} {\mathbf {F}} (y_1 , s) \,\mathrm{d}y_1 =0\;\;,\;\langle G\rangle = \int \limits _{\gamma (s)} G (y_1 , s) \,\mathrm{d}y_1 =0\;\;, \end{aligned}$$

where $\gamma (s)={\mathcal {G}} \cap \{ y_2=s \}$. Then

• 1. There exists a weak solution $({\mathbf {B}} , b) \in {\mathcal {D}}({\mathcal {G}}) \times L^2_{loc} ({ {\mathcal {G}}})$ to the problem (109)–(113). Furthermore, ${\mathbf {B}}$ is unique while b is determined up to an additive constant.

• 2. There exist constants ${\mathbf {B}}^\infty \in {\mathbb {R}}^2$ and $b^\infty \in {\mathbb {R}}$ such that $e^{\tau y_2} ({\mathbf {B}} -{\mathbf {B}}^\infty ) \in H^1 ({\mathcal {G}})^2$, $e^{\tau y_2} (b -b^\infty ) \in L^2 ({\mathcal {G}})$ for some $\tau >0$.

• 3. The pressure b is determined up to a constant. Thus, we can choose $b^\infty =0$ . Then

  $$\begin{aligned} \langle b\rangle= & {} \int \limits _{\gamma (s)} b (y_1 , s) \,\mathrm{d}y_1 =0\;\;,\;\;\forall \;s>0\,, \end{aligned}$$

  (115)
  $$\begin{aligned} \langle B_1 \rangle= & {} \int \limits _{\gamma (s)} B_1 (y_1 , s) \,\mathrm{d}y_1 =B_1^\infty \;\;,\;\;\forall \;s>0\;, \end{aligned}$$

  (116)
  $$\begin{aligned} \langle B_2\rangle= & {} \int \limits _{\gamma (s)} B_2 (y_1 , s) \,\mathrm{d}y_1\equiv B_2^\infty =const.\;\;,\;\;\forall \;s>0\;\;. \end{aligned}$$

  (117)

• 4. If $G=0$ and ${\mathbf {F}}=0$ and $B^\infty _k\in {\mathbb {R}}$ are the constants from (117), $k=1,2$, then

  $$\begin{aligned} \mu \int \limits _{{\mathcal {G}}} \left| \nabla _y {\mathbf {B}} \right| ^2 = \int \limits _{\gamma ^N} {\mathbf {B}}\,\cdot \,{\mathbf {d}} ={\mathbf {B}}^\infty \cdot {\mathbf {d}} \;\;, \end{aligned}$$

  (118)

  so that

  $$\begin{aligned} {\mathbf {B}}^\infty \cdot {\mathbf {d}} >0\;\;, \end{aligned}$$

  where ${\mathbf {B}}^\infty = (B_1^\infty , B_2^\infty )$.

• 5. If $G=0$ and ${\mathbf {F}}=0$, then the linear operator ${\mathbf {L}}:{\mathbf {R}}^2 \rightarrow {\mathbf {R}}^2$, defined by ${\mathbf {L}}\, {\mathbf {d}}={\mathbf {B}}^\infty$, is symmetric and strictly positive (thus, it is invertible). Furthermore, its matrix in a canonic basis is diagonal.

• 6. Two problems (24)–(29) have unique solutions $({\mathbf {w}}^k , \pi ^k ,\tau _k) \in {\mathcal {D}}({\mathcal {G}}) \times L^2 ({\mathcal {G}}) \times {\mathbf {R}}$.

Proof

For the proof of the existence of the solution and its exponential decay, the reader should consult [20] or [30] (see also [21] or [25]) , with slight modification due to the boundary condition.

Let us confirm that the pressure is determined only up to a constant. It is not clear, on the first glance, since the boundary condition (109) involves the pressure. First of all, the pressure boundary condition should be understood only in the sense of the variational formulation (114). Secondly, since it is not a kinematic but a dynamic boundary condition, it is a kind of Neumann condition for the Navier-Stokes. Indeed, replacing b by $b +C$ in (114) produces an additional term $C\int \limits _{\gamma } z_2$, with $\gamma = \gamma ^N\cup \gamma ^D$, but (due to the incompressibility)

$$\begin{aligned} \int \limits _{\gamma } z_2 =0\;\;. \end{aligned}$$

To finish the proof of the assertion 3. and prove the assertion 4., we take the partial mean $\langle \;f\;\rangle = \int \limits _{-1/2}^{1/2} f(y_1 , y_2) \mathrm{d}y_1$ of (109) and (110). It gives

$$\begin{aligned} -\mu \frac{\mathrm{d}^2}{\mathrm{d}y_2^2} \langle B_1 \rangle= & {} 0\;\;\Rightarrow \;\;\langle B_1\rangle =\text{ const. }\;, \end{aligned}$$

(119)

$$\begin{aligned} -\mu \frac{\mathrm{d}^2}{\mathrm{d}y_2^2} \langle B_2 \rangle + \frac{\mathrm{d}}{\mathrm{d}y_2}\langle b \rangle= & {} 0\;\;\Rightarrow \;\;-\mu \frac{\mathrm{d}}{\mathrm{d}y_2} \langle B_2 \rangle + \langle b \rangle =\text{ const. }\,, \end{aligned}$$

(120)

$$\begin{aligned} \frac{\mathrm{d}}{\mathrm{d}y_2}\langle B_2\rangle= & {} 0\;\;\Rightarrow \;\;\langle B_2 \rangle =\text{ const. } \end{aligned}$$

(121)

Thus,

$$\begin{aligned} \langle b \rangle =\text{ const. } \equiv b^\infty \;\; \end{aligned}$$

(122)

and

$$\begin{aligned} \langle {\mathbf {B}} \rangle =\text{ const. } \equiv {\mathbf {B}}^\infty \;. \end{aligned}$$

(123)

As the pressure is determined up to a constant, we can take $b^\infty =0$. Using ${\mathbf {B}}$ as the test-function in (24) leads to

$$\begin{aligned} \mu \int \limits _{{\mathcal {G}}}\left| \nabla _y{\mathbf {B}}\right| ^2 -\int \limits _{{\mathcal {G}}}b\,\cdot \,G - \int \limits _{{\mathcal {G}}} {\mathbf {F}}\cdot {\mathbf {B}} +b^\infty \cdot B_2^\infty ={\mathbf {B}}^\infty \cdot {\mathbf {d}}\;\;. \end{aligned}$$

(124)

In particular, for $G=0$, ${\mathbf {F}}=0$ and $b^\infty =0$ it gives (118).

To prove the last assertion, first we easily verify that the operator is linear, due to the linearity of the Stokes system. Then, for any ${\mathbf {d}}\in {\mathbf {R}}^2$,

$$\begin{aligned} {\mathbf {d}}\cdot {\mathbf {L}}\,{\mathbf {d}} = {\mathbf {d}}\cdot {\mathbf {B}}^\infty >0\;. \end{aligned}$$

Strict positivity implies invertibility. Then we simply take ${\mathbf {d}}_k = {\mathbf {L}}^{-1} {\mathbf {e}}_k$. We still need to prove that ${\mathbf {d}}_k = \tau _k {\mathbf {e}}_k$. In other words, the matrix representation in canonic basis is diagonal. Let ${\mathbf {d}}_k = (d^1_k , d^2_k)$ , k=1,2. We look at the problem

$$\begin{aligned}&-\mu \,\Delta _y \mathbf{B}^k+\nabla _y b^k =\mathbf{0}\,\,\,\text{ in }\,\,\, {{\mathcal {G}}}\,, \end{aligned}$$

(125)

$$\begin{aligned}&\text{ div}_y\, \mathbf{B}^k=0\,\,\,\text{ in }\,\,\, {{\mathcal {G}}}\,, \end{aligned}$$

(126)

$$\begin{aligned}&{\mathbf {T}} (\mathbf{B}^k, b^k){\mathbf {e}}_2={\mathbf {d}}_k \,\,\,\text{ on }\,\,\,\gamma ^N\,, \end{aligned}$$

(127)

$$\begin{aligned}&\mathbf{B}^k ={\mathbf {0}}\,\,\,\text{ on }\,\,\gamma ^D\,, \end{aligned}$$

(128)

$$\begin{aligned}&({\mathbf {B}}^k , b^k ) \;\;\text{ is } \text{1-periodic } \text{ in }\;y_1\;,\;\;\nabla _y \mathbf{B}^k \in L^2 ({{\mathcal {G}}})\,. \end{aligned}$$

(129)

By definition of ${\mathbf {d}}_k$

$$\begin{aligned} \lim _{y_2\rightarrow +\infty } {\mathbf {B}}^k={\mathbf {e}}_k \; \end{aligned}$$

(130)

and, thus,

$$\begin{aligned} \langle {\mathbf {B}}^k \rangle (y_2) = \int \limits _{-1/2}^{1/2} {\mathbf {B}}^k (y_1 , y_2 ) \mathrm{d}y_1 = {\mathbf {e}}_k\;\;. \end{aligned}$$

We test the equation for ${\mathbf {B}}^1$ by ${\mathbf {B}}^2$ to get

$$\begin{aligned} \mu \int \limits _{{\mathcal {G}}} \nabla {\mathbf {B}}^1 \,\nabla {\mathbf {B}}^2 ={\mathbf {d}}_1\cdot \int \limits _{{\mathcal {G}}} {\mathbf {B}}^2 = {\mathbf {d}}_1\cdot {\mathbf {e}}_2\;\;. \end{aligned}$$

Testing the equation for ${\mathbf {B}}^2$ by ${\mathbf {B}}^1$ to get

$$\begin{aligned} \mu \int \limits _{{\mathcal {G}}} \nabla {\mathbf {B}}^2\, \nabla {\mathbf {B}}^1 ={\mathbf {d}}_2\cdot \int \limits _{{\mathcal {G}}} {\mathbf {B}}^1 = {\mathbf {d}}_2\cdot {\mathbf {e}}_1\;\;, \end{aligned}$$

proving that

$$\begin{aligned} d_2^1 = d_1^2 \;\;. \end{aligned}$$

At this point, we recall that, by definition, the domain is symmetric in the sense that $(y_1 , 0) \in \gamma ^N\;\Leftrightarrow \; (-y_1 , 0) \in \gamma ^N$, since $\gamma ^N =\langle -a , a \rangle \times \lbrace 0 \rbrace$. Then, it is easy to verify by direct computation that the functions

$$\begin{aligned} {\mathbf {W}}^1 (y_1 ,y_2)=B_1^1 (-y_1 , y_2) {\mathbf {e}}_1 -B_1^2 (-y_1 , y_2) {\mathbf {e}}_2 \;\;,\;\;\Pi ^1 (y_1 , y_2) = -b^1 (-y_1 , y_2) \end{aligned}$$

satisfy the Stokes system

$$\begin{aligned} -\mu \Delta {\mathbf {W}}^1 + \nabla \Pi ^1 ={\mathbf {0}}\;\;,\;\;\text{ div }\,{\mathbf {W}}^1=0\;\;\text{ in }\;\;{\mathcal {G}} \;\;. \end{aligned}$$

Also ${\mathbf {W}}^1={\mathbf {0}}$ on $\gamma ^D$ and

$$\begin{aligned} \lim _{y_2\rightarrow +\infty }{\mathbf {W}}^1 = {\mathbf {e}}_1 \;\;. \end{aligned}$$

Finally on $\gamma ^N$, we have:

$$\begin{aligned} -\mu \left( \frac{\partial W^1_1}{\partial y_2} + \frac{\partial W^1_2}{\partial y_1}\right) (y_1 ,0 )= & {} -\mu \left( \frac{\partial B^1_1}{\partial y_2} + \frac{\partial B^1_2}{\partial y_1}\right) (-y_1 ,0)=d^1_1\,,\\ -2\mu \frac{\partial W^1_2}{\partial y_2} (y_1 ,0) + \Pi ^1 (y_1 ,0)= & {} 2\mu \frac{\partial B^1_2}{\partial y_2} (-y_1 ,0)- b^1 (y_1 ,0) = -d^1_2\;\;. \end{aligned}$$

We know that the problem: find $({\mathbf {B}}^k , b^k,{\mathbf {d}}^k)$ such that (125)–(130), has a unique solution. Thus, ${\mathbf {W}}^1 ={\mathbf {B}}^1$ and ${\mathbf {d}}_1 = (d^1_1 , d^2_1) = (d^1_1 , - d^2_1)$. Thus, $d^2_1=0$. Since $d_2^1=d_1^2$, they are both equal zero and the matrix ${\mathbf {L}}$ is diagonal

$$\begin{aligned} {\mathbf {L}} = \left[ \begin{array}{rr} \tau _1&{}0\\ 0&{}\tau _2 \end{array} \right] \;\;, \end{aligned}$$

or, equivalently

$$\begin{aligned} {\mathbf {d}}_k = \tau _k {\mathbf {e}}_k \;\;. \end{aligned}$$

$\square$

For the boundary layer problems (24)–(28) and (35)–(40), we prove some more specific results:

Lemma 1

$$\begin{aligned}&\int \limits _{-1/2}^{1/2} {\mathbf {w}}^k ({\mathbf {y}}) \mathrm{d}y_1 ={\mathbf {e}}_k\;\;,\;\forall y_2\ge 0\,, \end{aligned}$$

(131)

$$\begin{aligned}&\int \limits _{-1/2}^{1/2}\pi ^k ({\mathbf {y}}) \mathrm{d}y_1 =0\;\;,\;\forall y_2\ge 0\,, \end{aligned}$$

(132)

$$\begin{aligned}&\tau _k= \mu \int \limits _{{\mathcal {G}}} \left| \nabla {\mathbf {w}}^k\right| ^2>0\;\;,\;k=1,2\;. \end{aligned}$$

(133)

Proof

From Theorem 3, we know that

$$\begin{aligned} \int \limits _{-1/2}^{1/2} {\mathbf {w}}^k ({\mathbf {y}}) \mathrm{d}y_1 =\text{ const. }\;\;. \end{aligned}$$

Since

$$\begin{aligned} \lim _{y_2\rightarrow \infty } {\mathbf {w}}^k ={\mathbf {e}}_k, \end{aligned}$$

we get the first claim (131). Thus, taking into account (29) and, again, Theorem 3 gives

$$\begin{aligned} \int \limits _{-1/2}^{1/2} \pi ^k ({\mathbf {y}}) \mathrm{d}y_1 =0\;\;\;,\;k=1,2\;. \end{aligned}$$

The last claim (118) from Theorem 3 implies

$$\begin{aligned} \mu \int \limits _{{\mathcal {G}}} \left| \nabla {\mathbf {w}}^k\right| ^2 = \tau _k \int \limits _{\gamma ^N} w_k^k = \tau _k\;\;\;. \end{aligned}$$

Thus, $\tau _k>0$. $\square$