Analysis of the Roughness Regimes for Micropolar Fluids via Homogenization

Abstract

We study the asymptotic behavior of micropolar fluid flows in a thin domain of thickness $\eta _\varepsilon$ with a periodic oscillating boundary with wavelength $\varepsilon$. We consider the limit when $\varepsilon$ tends to zero and, depending on the limit of the ratio of $\eta _\varepsilon /\varepsilon$, we prove the existence of three different regimes. In each regime, we derive a generalized Reynolds equation taking into account the microstructure of the roughness.

Appendix: Computation of the Coefficients of the Micropolar Reynolds Equation

In this appendix, we describe how to obtain the coefficient of the Reynolds equation

$$\begin{aligned} \mathrm{div}_{z'}\left( -{h^3(z')\over 1-N^2}\varPhi (h(z'),N,R_c)\nabla _{z'}\bar{p}(z') + b(z')\right) =0\quad \hbox {in }\omega , \end{aligned}$$

(104)

where $b(x')= {h^3(z')\over 1-N^2}\varPhi (h(z'),N,R_c)f'(z')$ and $\varPhi$ defined by (78), from the micropolar system posed in $\varOmega =\{(z',z_3)\in \mathbb {R}^2\times \mathbb {R}:\,z'\in \omega ,\ 0<z_3<h(z')\}$, given by

$$\begin{aligned}&\left\{ \begin{array}{ll}\displaystyle -\partial _{z_3}^2 {\bar{u}}_1+\partial _{z_1}{\bar{p}}(z')+2N^2\partial _{z_3} {\bar{w}}_2={\bar{f}}_1(z')&{}\ \,\mathrm{in}\,\varOmega ,\\ -R_c\partial _{z_3}^2 {\bar{w}}_2+4N^2 {\bar{w}}_2-2N^2\partial _{z_3}\bar{u}_1={\bar{g}}_2(z')&{}\ \,\mathrm{in}\,\varOmega , \end{array}\right. \end{aligned}$$

(105)

$$\begin{aligned}&\left\{ \begin{array}{l}\displaystyle -\partial _{z_3}^2 {\bar{u}}_2+\partial _{z_2}{\bar{p}}(z')-2N^2\partial _{z_3} {\bar{w}}_1={\bar{f}}_2(z')\ \,\mathrm{in}\,\varOmega ,\\ -R_c\partial _{z_3}^2 {\bar{w}}_1+4N^2 {\bar{w}}_1+2N^2\partial _{z_3}\bar{u}_2={\bar{g}}_1(z')\ \,\mathrm{in}\,\varOmega , \end{array}\right. \end{aligned}$$

(106)

together with

$$\begin{aligned} \partial _{z_1}\left( \int _0^{h(z')}{\bar{u}}_1(z',z_3)\,\mathrm{d}z_3\right) + \partial _{z_2}\left( \int _0^{h(z')}{\bar{u}}_2(z',z_3)\,\mathrm{d}z_3\right) =0\quad \,\mathrm{in}\,\omega , \end{aligned}$$

(107)

and boundary conditions

$$\begin{aligned} {\bar{u}}'(z',0)={\bar{u}}'(z',h(z'))={\bar{w}}'(z',0)={\bar{w}}'(z',h(z'))=0. \end{aligned}$$

(108)

We note that $({\bar{u}}_1, {\bar{w}}_2)$, with external forces $(\bar{f}',{\bar{g}}')$, and $({\bar{u}}_2, -{\bar{w}}_1)$, with external forces $({\bar{f}}',-{\bar{g}}')$, satisfy the same equations and boundary conditions. So we only describe the computation of $({\bar{u}}_1, \bar{w}_2)$.

First, from the first equation of (105) we have

$$\begin{aligned} \partial _{z_3}{\bar{u}}_1(z)=\left( \partial _{z_1}{\bar{p}}(z')-{\bar{f}}_1(z')\right) z_3+2N^2{\bar{w}}_2(z) + C(z'). \end{aligned}$$

(109)

Putting this into the second equation of (105), we have

$$\begin{aligned} \begin{array}{l} \partial _{z_3}^2{\bar{w}}_2(z)-{4N^2\over R_c}(1-N^2){\bar{w}}_2(z)\\ =-{2N^2\over R_c}\left( \partial _{z_1}{\bar{p}}(z')-\bar{f}_1(z')\right) z_3 -{1\over R_c} {\bar{g}}_2(z')+ {2N^2\over R_C} C(z'). \end{array} \end{aligned}$$

(110)

The solution is

$$\begin{aligned} \begin{array}{ll} {\bar{w}}_2(z)=&{}A(z')\cosh (kz_3)+B(z')\sinh (kz_3)\\ &{}+{1\over 2(1-N^2)}(\partial _{z_1}{\bar{p}}(z')-{\bar{f}}_1(z'))z_3\\ &{}+{1\over 2(1-N^2)}C(z') +{1\over 4N^2(1-N^2)}{\bar{g}}_2(z'), \end{array} \end{aligned}$$

(111)

where $k=\sqrt{{4N^2(1-N^2)\over R_c}}$ and A and B are unknowns functions.

Putting this solution into equation (110), we can write ${\bar{u}}_1$ as follows

$$\begin{aligned} \begin{aligned} {\bar{u}}_1(z)&= {z_3^2\over 2(1-N^2)}(\partial _{z_1}{\bar{p}}(z')-f_1(z'))\\&\quad +{2N^2\over k}(A(z')\sinh (kz_3)+B(z')\cosh (kz_3))\\&\quad +{z_3\over 1-N^2}C(z')+{z_3\over 2(1-N^2)}{\bar{g}}_2(z')+D(z'). \end{aligned} \end{aligned}$$

(112)

We rewrite C, D, as a function of A and B, using the boundary conditions. So, for ${\bar{u}}_1(z',0)={\bar{w}}_2(z',0)=0$, we, respectively, get

$$\begin{aligned}D(z')=-{2N^2\over k}B(z'),\quad C(z')=2(1-N^2)\left( -A(z')-{1\over 4N^2(1-N^2)}{\bar{g}}_2(z')\right) ,\end{aligned}$$

and so

$$\begin{aligned} \begin{aligned} {\bar{u}}_1(z)&={z_3^2\over 2(1-N^2)}(\partial _{z_1}{\bar{p}}(z')-{\bar{f}}_1(z'))+\left( {2N^2\over k}\sinh (kz_3)-2z_3\right) A(z')\\&\quad +{2N^2\over k}(\cosh (kz_3)-1)B(z')-{z_3\over 2N^2}{\bar{g}}_2(z'),\\ {\bar{w}}_2(z)&={z_3\over 2(1-N^2)}(\partial _{z_1}{\bar{p}}(z')-{\bar{f}}_1(z'))\\&\quad +(\cosh (kz_3)-1)A(z')+\sinh (k z_3)B(z'). \end{aligned} \end{aligned}$$

(113)

Using the boundary conditions ${\bar{u}}_1(z',h(z'))=\bar{w}_2(z',h(z'))=0$, we get the following system

$$\begin{aligned}Q\left( \begin{array}{c} A\\ B \end{array}\right) =-{h(z')\over 2(1-N^2)}(\partial _{z_1}{\bar{p}}(z')-{\bar{f}}_1(z'))\left( \begin{array}{c} h(z')\\ 1 \end{array}\right) + {\bar{g}}_2(z'){h(z')\over 2N^2} \left( \begin{array}{c} 1\\ 0 \end{array}\right) ,\end{aligned}$$

where Q is the matrix defined by

$$\begin{aligned}Q=\left( \begin{array}{cc} {2N^2\over k}\sinh (k h(z'))-2h(z') &{} {2N^2\over k}(\cosh (kh(z'))-1)\\ \cosh (kh(z'))-1&{} \sinh (kh(z')) \end{array}\right) .\end{aligned}$$

The solution of this system is given by

$$\begin{aligned} A(z')=-{h(z')\over 2(1-N^2)}(\partial _{z_1}{\bar{p}}(z')-{\bar{f}}_1(z'))A_1(z')+ {h(z')\over 2N^2}{\bar{g}}_2(z')A_2(z'),\\ B(z')=-{h(z')\over 2(1-N^2)}(\partial _{z_1}{\bar{p}}(z')-\bar{f}_1(z'))B_1(z')+ {h(z')\over 2N^2}{\bar{g}}_2(z')B_2(z'), \end{aligned}$$

where $A_1(z')$, $B_1(z')$ and $A_2(z')$, $B_2(z')$ are solution of

$$\begin{aligned}Q\left( \begin{array}{c} A_1\\ B_1 \end{array}\right) =\left( \begin{array}{c} h(z')\\ 1 \end{array}\right) \quad \hbox {and}\quad Q\left( \begin{array}{c} A_2\\ B_2 \end{array}\right) =\left( \begin{array}{c} 1\\ 0 \end{array}\right) .\end{aligned}$$

Calculating $A_i$, $B_i$ for $i = 1, 2$, we have

$$\begin{aligned}\begin{array}{l} \displaystyle A_1(z')=-{1\over 2},\\ \displaystyle A_2(z')={\sinh (kh(z'))\over -2h(z')\sinh (kh(z'))+{4N^2\over k}(\cosh (kh(z'))-1)},\\ \displaystyle B_1(z')={1\over 2}\coth \left( {kh(z')\over 2}\right) ,\\ \displaystyle B_2(z')={-(\cosh (kh(z')-1)\over -2h(z')\sinh (kh(z'))+{4N^2\over k}(\cosh (kh(z'))-1)}, \end{array}\end{aligned}$$

and then ${\bar{u}}_1$ and ${\bar{w}}_2$ are obtained by (113) as functions of ${\bar{p}}$, ${\bar{f}}_1$ and ${\bar{g}}_2$, by the following expressions

$$\begin{aligned} \begin{aligned} {\bar{u}}_1(z)&= \left[ {z_3^2\over 2(1-N^2)}+{1\over 4(1-N^2)}\left( {2N^2\over k}\sinh (kz_3)-2z_3\right) \right. \\&\quad \left. -{h(z')\over 2(1-N^2)}{N^2\over k}(\cosh (kh(z'))-1)\coth \left( {kh(z')\over 2}\right) \right] \left( \partial _{z_1}{\bar{p}}(z')-{\bar{f}}_1(z')\right) \\&\quad +\Big [-{z_3\over 2N^2}+{h(z')\over 2N^2}\Big (\left( {2N^2\over k}\sinh (kz_3)-2z_3\right) A_2\\&\quad +{2N^2\over k}(\cosh (kz_3)-1)B_2\Big )\Big ]{\bar{g}}_2(z'),\\ {\bar{w}}_2(z)&=\Big [{z_3\over 2(1-N^2)} +{h(z')\over 4(1-N^2)}\Big (\cosh (kz_3)-1\\&\quad -\coth \left( {kh(z')\over 2}\right) \sinh (kz_3)\Big )\Big ]\left( \partial _{z_1}{\bar{p}}(z')-{\bar{f}}_1(z')\right) \\&\quad +{h(z')\over 2N^2}\left[ \cosh (kz_3)A_2+\sinh (kz_3)B_2\right] \bar{g}_2(z'). \end{aligned} \end{aligned}$$

As it was pointed at the beginning, expressions for ${\bar{u}}_2,\bar{w}_1$ are obtained by using the expressions of ${\bar{u}}_2$, $\bar{w}_1$, and so we have

$$\begin{aligned} \begin{aligned} {\bar{u}}_2(z)&=\left[ {z_3^2\over 2(1-N^2)}+{1\over 4(1-N^2)}\left( {2N^2\over k}\sinh (kz_3)-2z_3\right) \right. \\&\quad \left. -{h(z')\over 2(1-N^2)}{N^2\over k}(\cosh (kh(z'))-1)\coth \left( {kh(z')\over 2}\right) \right] \left( \partial _{z_2}{\bar{p}}(z')-{\bar{f}}_2(z')\right) \\&\quad -\Big [-{z_3\over 2N^2}+{h(z')\over 2N^2}\left( \left( {2N^2\over k}\sinh (kz_3)-2z_3\right) A_2\right. \\&\quad \left. +{2N^2\over k}(\cosh (kz_3)-1)B_2\right) \Big ]{\bar{g}}_1(z'),\\ {\bar{w}}_1(z)&=-\Big [{z_3\over 2(1-N^2)} +{h(z')\over 4(1-N^2)}\Big (\cosh (kz_3)-1\\&\quad -\coth \left( {kh(z')\over 2}\right) \sinh (kz_3)\Big )\Big ]\left( \partial _{z_2}{\bar{p}}(z')-{\bar{f}}_2(z')\right) \\&\quad +{h(z')\over 2N^2}\left[ \cosh (kz_3)A_2+\sinh (kz_3)B_2\right] \bar{g}_1(z'). \end{aligned} \end{aligned}$$

We observe ${\bar{u}}'$ and ${\bar{w}}'$ can be rewritten as follows

$$\begin{aligned} \begin{aligned} {\bar{u}}'(z)&= \left[ {z_3^2\over 2(1-N^2)}+{1\over 4(1-N^2)}\left( {2N^2\over k}\sinh (kz_3)-2z_3\right) \right. \\&\quad \left. -{h(z')\over 2(1-N^2)}{N^2\over k}(\cosh (kh(z'))-1)\coth \left( {kh(z')\over 2}\right) \right] \left( \nabla _{z'}{\bar{p}}(z')-{\bar{f}}'(z')\right) \\&\quad -\Big [-{z_3\over 2N^2} +{h(z')\over 2N^2}\left( \left( {2N^2\over k}\sinh (kz_3)-2z_3\right) A_2\right. \\&\quad \left. +{2N^2\over k}(\cosh (kz_3)-1)B_2\right) \Big ](\bar{g}'(z'))^\perp , \end{aligned} \end{aligned}$$

(114)

and

$$\begin{aligned} \begin{aligned} {\bar{w}}'(z)&=\Big [{z_3\over 2(1-N^2)}+{h(z')\over 4(1-N^2)}\Big (\cosh (kz_3) -1\\&\quad -\coth \left( {kh(z')\over 2}\Big )\sinh (kz_3)\right) \Big ]\left( \nabla _{z'}{\bar{p}}(z')-{\bar{f}}'(z')\right) ^\perp \\&\quad +{h(z')\over 2N^2}\left[ \cosh (kz_3)A_2+\sinh (kz_3)B_2\right] \bar{g}'(z'). \end{aligned} \end{aligned}$$

(115)

Finally, integrating the expressions of ${\bar{u}}'$ and ${\bar{w}}'$ with respect to the variable $z_3$, it holds that

$$\begin{aligned} \begin{array}{l} \displaystyle \int _0^{h(z')}{\bar{u}}_j'(z',z_3)\,\mathrm{d}z_3=-{h^3(z')\over 1-N^2}\varPhi (h(z'),N,R_c)\left( \partial _{z_j}{\bar{p}}(z')-{\bar{f}}_j(z')\right) ,\\ \displaystyle \int _0^{h(z')}{\bar{w}}_j'(z',z_3)\,\mathrm{d}z_3=-{1\over 4N^3}\sqrt{{R_c\over 1-N^2}}\varPsi (h(y'),N,R_c){\bar{g}}_j(z'), \end{array}\end{aligned}$$

(116)

for $j=1,2$, with $\varPhi$ and $\varPsi$ defined by (78) and (79), respectively. Putting this in (107), we get the desired Reynolds equation (104).