## 1. Flow through porous material : the Darcy model

We consider a closed bounded domain $\Omega\subset\mathbb R ^d$.

Whenever $\underline u,\underline v\in[L^2(\Omega)]^d$, we denote :

$(\underline u,\underline v)_0 = \int_\Omega\underline u\cdot\underline v$

If $\varphi,\psi\in L^2(\Omega)$, we define :

$(\varphi,\psi)_0=\int_\Omega\varphi\psi$

If moreover $A\subset\Omega$ has dimension $d-1$ or $d$, then we define the corresponding bilinear form :

$\langle\varphi,\psi\rangle_A=\int_A\varphi\psi$

### 1.1. Mixed Darcy problem

Find $\underline u:\Omega\to\mathbb R^d$ and $\varphi:\Omega\to\mathbb R$ such that :

$\left\{\begin{array}[c]{rll} \underline u +\underline{\underline\kappa}\nabla\varphi & =\underline f_1 & \text{in }\Omega\\ \nabla\cdot\underline u & = f_2 & \text{in }\Omega \end{array}\right.$

where $\underline f_1\in [L^2(\Omega)]^d$ is the source term and $f_2\in L^2(\Omega)$ is the incompressibility defect.

Here, $\underline{\underline\kappa}$ is a positive definite matrix (often assumed symmetric or even diagonal) representing the hydraulic conductivity whose coefficients are in $m s^{-1}$, and $\varphi=z+\frac p{\rho g}$ is the fluid charge or piezometric head expressed in $m$, where $g$ is the gravity acceleration, $\rho$ is the fluid density and $z$ denotes the height from a reference level.

Remark

We usually consider the case where $\underline{\underline\kappa}$ is symmetric, or even diagonal, and $\underline f_1\equiv0$. If moreover one wishes to consider incompressible flow, it suffices to set $f_2\equiv0$ as well.

### 1.2. Primal Darcy Problem

The mixed formulation yields :

$\underline u = -\underline{\underline\kappa}\nabla\varphi + \underline f_1\quad\text{ in }\Omega$

so writing $\nabla\cdot\underline u = f_2$ gives :

$- \nabla\cdot\underline{\underline\kappa}\nabla\varphi + \nabla\cdot\underline f_1 = f_2\quad\text{ in }\Omega$
Remark

In the sequel, we will focus on the case with no source term, i.e. $\underline f_1\equiv0$. For the sake of simplicity, we will denote $f=f_2$.

The primal form then becomes :

$-\nabla\cdot\underline{\underline\kappa}\nabla\varphi=f\quad\text{ in }\Omega$

Moreover, we suppose $\underline{\underline\kappa}$ is diagonal, and we denote $\kappa_i,i\in\{1,...,d\}$ its diagonal entries. We emphasize the fact that they are all positive constants.

### 1.3. Boundary conditions

We need to impose boundary conditions on $\partial\Omega$. As in the Stokes case, we will consider Dirichlet, Neumann and Robin type boundary conditions.

Let us decompose the boundary $\partial\Omega$ in two relatively open disjoint parts :

$\partial\Omega=\overline{\Gamma_{nat}}\cup\overline{\Gamma_{ess}}$
$\Gamma_{nat}\cap\Gamma_{ess}=\emptyset$

on which we impose the conditions :

$\left\{\begin{array}{rll} -(\underline{\underline\kappa}\nabla\varphi)\cdot\underline n & = u_{nat} & \text{ on }\Gamma_{nat}\\ \varphi & = \varphi_{ess} & \text{ on }\Gamma_{ess} \end{array}\right.$

Thus, the natural condition is a Neumann condition for $\varphi$ on which depends the variational formulation, and the essential condition is a Dirichlet condition for $\varphi$ on which depends the test function space.

### 1.4. Weak formulation

A nice functional space for this primal Darcy problem is the space of all the infinitely differentiable scalar functions on $\Omega$ vanishing close to the essential boundary : $V_0 = \left\{\psi\in C^\infty(\Omega)\quad|\quad\left.\psi\right|_{\Gamma_{ess}}=0\right\}$. Nevertheless, it is not complete. Our test function space will be its $H^1(\Omega)$ completion.

Definition

The test space for the primal Darcy problem is the space of function whose trace vanishes on the essential boundary :

$H_0^1(\Omega)\subset V=\overline{V_0}^{H^1(\Omega)}\subset H^1(\Omega)$

It inherits the $H^1(\Omega)$ norm.

Let us multiply the primal form by $\psi\in V$. The left hand side term becomes :

$\begin{array}{rl} \displaystyle\int_\Omega-(\nabla\cdot\underline{\underline\kappa}\nabla\varphi)\psi=&-\displaystyle\int_\Omega\nabla\cdot(\underline{\underline\kappa}(\nabla\varphi)\psi)+\int_\Omega(\underline{\underline\kappa}\nabla\varphi)\cdot\nabla\psi\\ =&\displaystyle-\int_{\Gamma_{nat}}\psi\underline{\underline\kappa}\nabla\varphi\cdot\underline n+\underline{\underline\kappa}\int_\Omega\nabla\varphi\cdot\nabla\psi\\ =&-\langle u_{nat},\psi\rangle_{\Gamma_{nat}}+(\underline{\underline\kappa}\nabla\varphi,\nabla\psi)_0 \end{array}$

while the right hand side term becomes $(f,\psi)_0$.

To impose the Dirichlet condition $\varphi=\varphi_{ess}$ on $\Gamma_{ess}$, one has to involve the trace operator, since the restriction of $\varphi$ is not properly defined anymore. By surjectivity of the trace operator, $\varphi_{ess}$ admits an extension (not unique) in $H^1(\Omega)$ we still denote $\varphi_{ess}$ for simplicity. The Dirichlet condition becomes :

$\varphi-\varphi_{ess}\in V$
Definition

Let $a:V\times V\to\mathbb R$ be a bilinear form and $\tilde f:V\to\mathbb R$ a linear form defined by :

$a(\varphi,\psi)=(\underline{\underline\kappa}\nabla\varphi,\nabla\psi)_0\quad\text{ ; }\quad\tilde f(\psi)=(f,\psi)_0-\langle u_{nat},\psi\rangle_{\Gamma_{nat}}$
Proposition

The weak formulation of the Darcy problem reads as follows :

$\left\{\begin{array}{ll} \text{find }\varphi\in H^1(\Omega)\text{ such that }\varphi-\varphi_{ess}\in V\text{ and}&\\ a(\varphi,\psi)=\tilde f(\psi)&\\ \end{array}\right.$
Theorem

The Darcy problem in weak form admits an unique solution $\varphi$.

Proof

This comes straightforwardly from Lax-Milgram theorem. It is sufficient to show that $a$ is bounded and coercive, and that $\tilde f$ is bounded.

• Thanks to Cauchy-Schwarz inequality, the boundedness of $a$ comes easily :

$|a(\varphi,\psi)|=|(\underline{\underline\kappa}\nabla\varphi,\nabla\psi)_0|\leqslant\max_{i\in\{1,...,d\}}\{\kappa_i\}|(\nabla\varphi,\nabla\psi)_0|\leqslant\max_{i\in\{1,...,d\}}\{\kappa_i\}\|\varphi\|_V\|\psi\|_V$
• The coercivity of $a$ is also easy to prove :

$a(\varphi,\varphi)=(\underline{\underline\kappa}\nabla\varphi,\nabla\varphi)_0\geqslant\min_{i\in\{1,...,d\}}\{\kappa_i\}(\varphi,\varphi)_V=\min_{i\in\{1,...,d\}}\{\kappa_i\}\|\varphi\|_V^2$
• Using the Cauchy-Schwarz inequality and the Poincaré inequality, we get the boundedness of $\tilde f$. We first apply the Riesz representation theorem to find $\tilde F\in L^2(\Omega)$ representing $\tilde f$.

$|\tilde f(\psi)|=|(\tilde F,\psi)_0|\leqslant\|\tilde F\|_0\|\psi\|_0\leqslant C\|\tilde F\|_0\|\psi\|_V$