1. Introduction. In [21] a hierarchy of models for fluid flow through a porous medium was developed within the context of mixture theory. It was shown that the classical Darcy's model of porous medium flow is the simplest of this hierarchy, and some extensions and variations of it are proposed and justified. Supported by experimental evidence, the reasonableness of this hierarchy has been rigorously demonstrated in [21] .
One of the simplest models proposed in [21] is Darcy's model but with a porosity coefficient that depends on the pressure, that is
Here p is the pressure within the fluid, u is its velocity, and f represents an external force acting on the fluid. Reference [11] studied this model in the case when the porosity α is a bounded and smooth function, see also [1] . The authors of [11] also developed a heuristic analysis of the case of exponential dependence on pressure, that is
where α 0 and γ are positive parameters. The proposed formulation reduced this problem to the solution of two linear equations: a convection diffusion problem and a linear Darcy model, this formulation will be reviewed in section 3 below. The analysis of each of the resulting discrete linear problems has been investigated in [11] but the rigorous well posedness of this strategy remained an open problem. The first goal of this work is to attempt to fill this gap. Under some assumptions on the problem data we will show that the splitting strategy is meaningful and, thus, we will use it to define a solution to our problem. Reference [11] also proposed a discretization of this split formulation, and provided an error analysis for it, provided the discrete solution to the convection diffusion equation remained positive, see [11, (4.15) ]. This assumption, however, was not verified and this will be the second goal of this work. Our second goal is to close the gaps in the error analysis for the proposed numerical method.
Our presentation is organized as follows. In Section 2 we establish notation and recall some useful facts. Our problem of interest is presented in Section 3. The splitting formulation is introduced in Section 3.1, and its analysis is presented in Section 3.2. We introduce a suitable, realistic, and reasonable set of assumptions on the geometry and problem data which guarantee that the split formulation of our problem is well posed. the discretization is described and analyzed in Section 4. Finally, some numerical illustrations of the positivity that is at the heart of our analysis are presented in Section 5.
2. Notations and technical tools. Throughout the paper, Ω ⊂ R d with d ∈ {2, 3, 4}, is a bounded domain with Lipschitz boundary. The analysis can be extended to higher dimensions under suitable integrability and regularity of the data. Whenever X(D) is a normed space of functions over D, we indicate by · X its norm. X(D) ′ denotes the dual of X(D). If the spatial domain needs to be indicated, then we will denote it by · X(D) . For r ∈ [1, ∞), we denote the Banach space of Lebesgue r-
represents the space of essentially bounded measurable functions on Ω, with the usual norm. For k positive integer and 1 ≤ r ≤ ∞, W k,r (Ω) denotes the space of functions in L r (Ω) whose weak partial derivatives of order up to k are all in L r (Ω). With the
W s,r (Ω) is a Banach space with its natural norm v r W s,r = v r L r +|v| r W s,r . For r = 2, we set H s (Ω) = W s,2 (Ω) for s integer or s ∈ (0, 1). For vector-valued functions we use boldface and the spaces of these functions are denoted, for instance, by L r (Ω). We also need the space H(div, Ω) which is defined as
We use several facts about Sobolev spaces. The first is the trace property, namely, owing to the fact that the boundary of Ω is Lipschitz, if v ∈ H s (Ω) for s ∈ ( 1 2 , 1], then the trace of v on ∂Ω, which we denote simply by v| ∂Ω , belongs to the space H s−1/2 (∂Ω). Moreover, there is constant C with the estimate
We will also recall that vector fields in H(div, Ω) have a well defined trace of their normal component along the boundary of Ω. Namely, if n(x) is the outward normal vector at x ∈ ∂Ω, then for any v ∈ H(div, Ω) we have that v · n| ∂Ω is in H 1/2 (∂Ω) ′ and its action is defined via the divergence formula v · n, u| ∂Ω ∂Ω :=
In the event v · n| ∂Ω is Lebesgue integrable, then the duality pairing ·, · ∂Ω is a mere integration over ∂Ω.
The second property of Sobolev spaces that will be used frequently is the fact that they embed into function spaces of higher integrability. Precisely, from Sobolev embedding theorem, we have that for 0 < s ≤ 1,
if d = 2s, and any 2 * > 2 if d = 2s.
along with the estimate: there is a universal constant C = C(Ω, s) such that
Notice that if s = 1 and d ≤ 4, then 2 * ≥ 4.
3. The continuous problem. We begin by providing the exact formulation of our problem at hand. We assume that the boundary of Ω is divided into two pieces:
in Ω,
with the permeability function defined in (2) . While f and g are given data, the unknowns are the velocity u and pressure p of the fluid.
3.1. The splitting formulation. Let us now recall the transformation that allowed [11] to write this problem as two linear ones. This will be useful in identifying regularity requirements for the data f and g. Take the first equation in (3) and divide it by α. Incompressibility then implies that (4) div e −γp ∇p = div(e −γp f ).
Define q = e −γp and note that ∇q = −γe −γp ∇p, so that (4) can be rewritten as −∆q = γq div f + γf · ∇q.
In addition, since p = 0 on Γ w we have that q = 1 there. If we assume that f is sufficiently smooth to have a normal trace on Γ then taking a normal trace of the first equation in (3) we obtain
This, for the variable q means ∂ n q + γqf · n = α 0 γg.
In conclusion, for the variable q we have obtained the following convection diffusion problem
This motivates the following strategy to solve (3) . Fig. 1 . Geometry of the domain. The domain Ω is a bounded Lipschitz domain, where a finite number of strictly contained subdomains are removed. The "exterior" boundary is Γ, whereas the "interior" is Γw.
• Find q that solves (5).
• Define
• Find (U, P ) that solve
in Ω,
If (u, p) solves (3), and p ∈ L ∞ (Ω), thenα in (6) will be a function that is bounded from below and above by positive numbers. Equation (7) is now the classical linear Darcy's equation and its solution coincides with (u, p). The advantage of the splitting strategy is that it gives a meaning to a solution of a nonlinear system by transforming it into two linear systems when the set up leading to this transformation is applicable.
Analysis of the problem. We now provide an analysis of the splitting strategy. In order to do so, we will operate under the following assumptions: • Geometry. The domain Ω is constructed as follows: Figure 1 . Essentially we are working on annuli-type domains and Γ w is taken to be the inside boundary.
• Data regularity. We assume that the problem data satisfies the following conditions.
• Permeability. The parameters α 0 and γ are positive constants.
• Volume forcing. The volume forcing term f satisfies
, and δ > 0 if d = 4; and m > d.
• Boundary forcing. The boundary forcing term g satisfies
with m > d. • Sign conditions. We assume that the volume and boundary forcing terms satisfy the following sign conditions. • Volume forcing. The function f satisfies div f ≤ 0, in Ω, f · n ≥ 0, on Γ.
• Boundary forcing. The boundary forcing term g satisfies
We remark that the geometry assumption guarantees that the pieces of the boundary Γ and Γ w are well separated. In other words, there exists ǫ 0 > 0 such that for
We now introduce a subspace of functions in H 1 (Ω) whose trace vanish on Γ w :
Our notion of a solution to (3) is defined via the splitting strategy as follows. We immediately comment that, owing to the regularity conditions on f , the definition we give below makes sense.
Definition 1 (solution). We say that the triple (q, U,
for all φ ∈ H 1 w (Ω), and, withα defined in (6),
Let us now proceed to show that this formulation, under our imposed assumptions on the data, is well posed. We remark that the notion of solution given in Definition 1 assumes that for given a solution q of (8), the functionα is well defined almost everywhere in Ω and can be suitably used to solve the linear Darcy equation (9) . It is known thatα being a bounded and strictly positive function is sufficient to demonstrate that (9) is uniquely solvable, corresponding to appropriate data, see [8, Theorem 2.34 ]. As a consequence, the conditions we impose on the data must ensure that not only we have a unique solution to (8) but that it also gives aα that is bounded and strictly positive.
We begin by showing existence and uniqueness of solutions for subproblem (8). To simplify notation, we define the bilinear form
The subproblem (8) can now be rephrased, after the change of variables z = q − 1, as: given f , and g satisfying our conditions, find z ∈ H 1 w (Ω) such that
Well posedness of this problem is established in [11] via Lax-Milgram theorem under a smallness assumption on the gradient of f . The following proposition establishes well posedness for other classes of the data f , namely those that satisfy the regularity and sign conditions stated at the beginning of this subsection.
Proposition 2 (boundedness and coercivity). Let g ∈ H 1/2 00 (Γ) ′ , and f satisfy our regularity and sign conditions. Then problem (8) has a unique solution.
Proof. We will show that the equivalent problem (10) is well posed. Since f ∈ H(div, Ω), and g is in H
defines a bounded linear functional on H 1 w (Ω). This follows from the trace and embedding estimates discussed in Section 2. Thus, we only need to show that, under the given assumptions on f , the bilinear form A f , is bounded and coercive. Boundedness follows immediately from the regularity assumptions on f and its divergence. Indeed, since d ≤ 4, we have that 2 * ≥ 4. Consequently, Sobolev embedding shows that
Moreover, we have that
which together with Sobolev embedding shows the desired boundedness.
Let us now show coercivity, set φ = z ∈ H 1 w (Ω) on the left hand side of (10) to obtain
where, in the last step, we used the sign condition on the divergence of f and its normal trace on Γ, and that completes the proof.
We remark that after integration by parts on the left hand side of (10), we have that
It then follows that z is a weak solution of the mixed boundary value problem
Having shown that this problem always has a unique solution, we will make sure that (6) defines a suitable coefficient for (9) to make sense. The goal is to exploit the sign condition on g to guarantee that the solution q is bounded from below by a positive number. We begin by proving a regularity result for q.
Proposition 3 (regularity). Suppose that Ω satisfies our geometry assumptions, f and g satisfy our regularity assumptions, and f satisfies our sign conditions. Then, there exists ν ∈ (0, 1) such that q ∈ H 1 (Ω), the unique solution of (8) belongs to C 0,ν (Ω).
Proof. For q ∈ H 1 (Ω), the solution to (8) , the function z = q − 1 ∈ H 1 w (Ω) solves the mixed boundary value problem (11) . It follows from local Hölder regularity of solutions of elliptic equations; see, for example, [10, Theorem 8.24 ], that the regularity conditions on f , guarantee the existence of ν ∈ (0, 1) such that z ∈ C 0,ν loc (Ω), with the estimate that for any Ω ′ ⋐ Ω, then there exists a constant C = C(Ω ′ ) such that
In particular, since for any ǫ ∈ (0, ǫ 0 ), ∂(Γ + ǫ) ∩ Ω is in the interior of Ω, z is Hölder continuous on ∂(Γ + ǫ) ∩ Ω. As a consequence, since z = 0 on Γ w , we may apply the global Hölder regularity of solutions; see [10, Theorem 8.29 ] and the discussion following this result, to conclude that for any ǫ ∈ (0, ǫ 0 ), there exists ν ∈ (0, 1) (possibly different from the previous ν) such that z ∈ C 0,ν (Ω \ (Γ + ǫ)).
Next we study the regularity of z near Γ where a Robin-type boundary condition is imposed. Hölder regularity estimates for elliptic equations over Lipschitz domains with Neumann, and more generally with Robin, boundary conditions are obtained in the work of Nittka [19, 20] . Applying [20, Theorem 3.7] or [19, Proposition 3.6], we obtain that we can choose ǫ 0 > 0 such that a solution z to (11) is in C 0,ν ((Γ + ǫ 0 ) ∩ Ω) for some ν ∈ (0, 1) and for any ǫ ∈ (0, ǫ 0 ) with the estimate
Combining the above estimates we obtain that z ∈ C 0,ν (Ω) with the estimate that
which is what we needed to show.
With the regularity result of Proposition 3 at hand, we have that, in particular, q is continuous onΩ and so it is bounded. We must, additionally, show that this function is strictly positive inΩ, to be able to conclude thatα, defined in (6), is an admissible coefficient. This is the content of the following result.
Theorem 4 (positivity). In the setting of Proposition 3 assume, in addition, that g = 0 on Γ. Then, we have that the unique solution to problem (8) satisfies q(x) ≥ q 0 , ∀x ∈Ω for some q 0 > 0. As a consequenceα ∈ C(Ω) and there is α 0 > 0 such thatα(x) ≥ α 0 for all x ∈Ω.
Proof. We prove this in two steps. First, we show that the solution q is nonnegative and then applying a separate argument we show that q is strictly positive.
Step 1: In this step we show that q ≥ 0, using the argument developed by Chicco in [3] , adjusted to our setting, and which in turn uses Stampacchia's truncation method [22] . For k ∈ R, we begin by defining q k (x) = min{q(x) − k, 0}, and Ω(k) = {x ∈ Ω : q k < 0} k 0 = sup{k : q k ≡ 0 in Ω}.
We aim to show that q 0 ≥ 0. For that it suffices to show that k 0 ≥ 0. We argue by contradiction and assume that k 0 < 0. We prove first that (12) lim k→k0 |Ω(k)| = 0.
To do so, observe that from the definition of the set Ω(k), as ǫ ↓ 0 we have, for every x ∈ Ω, χ Ω(k0−ǫ) (x) → χ Ω(k0) (x). As a consequence, |Ω(k 0 )| = lim ǫ↓0 |Ω(k 0 − ǫ)| = 0. In addition, one can easily show that as ǫ ↓ 0,
where P = Ω(k 0 ) ∪ {x ∈ Ω : q = k 0 }. In this case, we have that lim ǫ↓0 |Ω(k 0 + ǫ)| = |{x ∈ Ω : q(x) = k 0 }|.
Assume that this limit is positive. Owing to the fact that |Ω(k 0 )| = 0 we have k 0 − q(x) ≤ 0 for every x ∈ Ω. Moreover, using that q ∈ H 1 (Ω) solves (8) and k 0 is a constant,
Where, to obtain the inequality, we used the sign condition on the divergence of f . The weak minimum principle of [3, Corollary 1] then implies that, either q(x) = k 0 in Ω or q(x) > k 0 almost everywhere in Ω. However, if q(x) = k 0 < 0 in Ω, we arrive at a contradiction, as q ∈ C(Ω) is positive (in fact q ≡ 1) on Γ w . On the other hand, if q(x) > k 0 almost everywhere, then |{x ∈ Ω : q(x) = k 0 }| = 0, which is again a contradiction. In conclusion, lim ǫ↓0 |Ω(k 0 + ǫ)| = 0 and (12) holds.
We now use (12) to deduce that for every η > 0, there exists k 1 ∈ (k 0 , 0) such that 0 < |Ω(k 1 )| < η and for all k ≤ k 1 q k (x) = 0 ∀x ∈ Ω \ Ω(k).
Note that for such k, we have q k ≤ 0 in Ω. Moreover, since q = 1 on Γ w and q is continuous, there is an open set O ⊂ Ω, such that Γ w ⊂ ∂O ∩ ∂Ω, where we have
for all x ∈Ō. Consequently, q k = 0 onŌ and this implies that q k ∈ H 1 w (Ω), which makes it a suitable test function in (8) . Therefore,
where we also used the sign conditions on the data. Now, the coercivity of A f , proved in Proposition 2, implies that
so that q k ≡ 0 onΩ (recall that q k ≡ 0 on Γ w ). However, we have arrived at a contradiction, since q k < 0 in Ω(k), which has positive measure. In conclusion, we must have that k 0 ≥ 0 and therefore q(x) ≥ 0, as we intended to show in this step.
Step 2. Applying the weak minimum principle in [22, Corollary 8.1] , the minimum of q, which could possibly be zero, can only be achieved at the boundary Γ. That is, on any set Ω ′ compactly contained in Ω we must have, inf{q(x) : x ∈ Ω ′ } > 0. Assume now that x 0 ∈ Γ is such that 0 = q(x 0 ) = inf{q(x) : x ∈Ω}. Since g = 0 in Γ, we can invoke the boundary Harnack inequality of [4, Theorem 3.1] to assert then that, for a sufficiently small ρ > 0,
which is a contradiction. Therefore q(x) > 0 inΩ, as we intended to show.
Remark 5 (sharpness and positivity under other conditions). The condition g = 0 may seem rather restrictive, as one may expect that g ≥ 0 may be sufficient. After all, under enough smoothness of the domain Ω and solution q, this is all is needed to conclude the strict positivity, say, via a boundary Hopf lemma; see [10, Lemma 3.4] or [9, Section 6.4.2]. However, the counterexamples of [15] , which are attributed to A. Castro, show that in the case that the boundary is merely Lipschitz, as it is our case of interest here, positivity may fail at "corner" points.
It is possible, nevertheless, to prove strict positivity under other assumptions: • In the case that Ω is convex, Γ w = ∅ (which does not satisfy our geometry assumptions), f · n ≥ f 0 > 0 and g ≥ 0; we can invoke the weak Harnack inequality of [15, Lemma 3.2] to conclude the strict positivity of q. Notice that the proof of Proposition 2 shows that, under the strict positivity assumption on f · n, we still have coercivity of A f . • In the case that Ω, f and g are sufficiently smooth to guarantee that q ∈ W 2,d (Ω), and g ≥ 0; we can invoke [18, Corollary 3.2 ] to arrive at the same conclusion.
Once we know thatα is bounded and strictly positive, the analysis of (9) is standard. We summarize the well posedness of the splitting strategy in the following result.
Theorem 6 (existence and uniqueness). Suppose that the domain Ω satisfies our geometry assumptions; and the data f and g satisfy our regularity, and sign conditions. Under these conditions, problem (3) has a unique solution in the sense of Definition 1.
Proof. Owing to Proposition 2 problem (8) has a unique solution which moreover, via Proposition 3, is continuous and strictly positive (Theorem 4). This implies that α, defined as in (6), is a bounded and strictly positive function. The conditions of [8, Theorem 2.34 ] are now satisfied and this implies that the linear Darcy equation (9) has a unique solution.
Having studied the continuous problem, we can proceed with its approximation. In addition to the conditions that guaranteed well posedness of the continuous problem we shall, to avoid unnecessary technicalities, assume that Ω is a polytope. This guarantees that Ω can be triangulated exactly. Given a conforming, and quasiuniform triangulation of Ω (see [6] for a definition of these notions) of size h > 0, we construct two finite element spaces W h ⊂ H 1 (Ω), and X h ⊂ L 2 (Ω). We also define M h = W h ∩ H 1 w (Ω). We assume that the family of pairs (X h , M h ) satisfies a discrete inf-sup condition: There is a constant β > 0 such that for all h > 0
Examples of suitable spaces can be readily found in the literature [6, 8, 12, 2] . To ensure positivity of discrete approximations to the variable q, we require that the Galerkin projection with respect to A f onto our space M h has almost optimal approximation properties in the max norm. In other words, if w ∈ H 1 (Ω) and w h ∈ W h are such that
Remark 7 (max norm estimates). The derivation of max norm error estimates for finite element approximations is a, rather technical, and underdeveloped subject. To our knowledge, most of the references that deal with this subject are only concerned with the Dirichlet problem for the Laplacian, and assume at least convexity of the domain; see for instance [17] for the Laplacian, and [7] for the Dirichlet problem for more general operators, but under an acuteness assumption on the triangulation. The only reference we are aware of that deals with mixed boundary conditions is [16] , where convexity is also assumed, and the differential operator is the Laplacian. While we admit that this is a weakness of our analysis, we shall proceed assuming that (14) holds. Another possible approach to obtain such an estimate is by deriving a discrete maximum principle, as it is detailed, for instance, in [5, Chapter III, Sections 20, 21] . This, however, imposes restrictions on the mesh.
We approximate the solution to (3) with the finite element spaces that we have just described. We will say that the triple (q h , u h , p h ) ∈ W h × X h × M h is a finite element approximation of the solution to (3), in the sense of Definition 1 if:
• We define
Our main goal now is to show that the discrete problem is well posed and to study its approximation properties.
Here we show that, under similar assumptions as for the continuous problem, the discrete problem (15)-(17) is uniformly well posed.
Theorem 8 (well posedness). In the setting of Theorem 6 or Remark 5 assume, in addition, that h is sufficiently small. Then problem (15)-(17) has a unique solution.
Proof. By conformity, coercivity of A f is inherited to M h so that problem (15) has a unique solution q h ∈ W h . Now, since q h ∈ W h is the Galerkin approximation of q ∈ H 1 (Ω), owing to (14) we have that
where I h denotes the Lagrange interpolant. Theorem 6 now implies that q ∈ C 0,ν (Ω) for some ν > 0 so that,
The previous reasoning shows that the coefficientα h , defined in (16), is a bounded and positive function uniformly in h. Therefore, since the discrete inf-sup condition (13) holds, we again invoke [8, Theorem 2.34] to conclude that problem (17) is uniformly well posed.
We now proceed with the error analysis of scheme (15)-(17).
Theorem 9 (error estimates). In the setting of Theorem 8 we have that, if h is sufficiently small,
where, in all estimates, the constants may depend on (q, U, P ) but are independent of h.
Proof. The estimate on q−q h is immediate. Let us focus on the estimates between (U, P ) and (u h , p h ). Setting v = v h ∈ X h and r = r h ∈ M h in (9) yields
Owing to the fact thatα h is uniformly bounded and positive, we can invoke discrete stability to conclude, from the previous identities, that Notice now that, for x ∈Ω,
which allows us to conclude.
The error estimate of the previous theorem can be combined with the regularity of Proposition 3 to obtain
We end by commenting that, if further regularity on q can be asserted, the pointwise estimate can be improved and a higher rate can be obtained.
Numerical illustrations of the error estimates we proved in Section 4.2 were presented in [11, Section 5] . Here then we confine ourselves to illustrating the positivity of the variable q, which is at the heart of the splitting formulation.
The computations were carried out with the help of the FreeFem++ package [14] . We used a piecewise linear discretization of the variable q. where r 2 = x 2 + y 2 , and κ > 0 is to be chosen. Notice that div f = κ 1 − 25 r 2 + (r − 5) 2 r < 0, in Ω, f · n ≥ 0, on Γ, so that this forcing fits within our theory. We consider two cases: 1. g = 0 and κ = 6 10 , which fits the framework of Theorem 4, 2. g = 1 10 with κ = 1 that does not. See, however, Remark 5. Table 1 shows the minimal value of q h as a function of the number of degrees of freedom. Notice that, as Theorem 8 shows, positivity can only be guaranteed for a sufficiently small mesh (sufficiently large number of degrees of freedom). After that, the minimum of q h seems to stabilize at a positive value. The problem data is γ = 2, α 0 = 4e γ and we set the problem data so that the exact solution is u = x x 2 + y 2 + y,
The minimal value of q h as a function of the number of degrees of freedom is illustrated in Table 2 . In this case, positivity is obtained for all values of h.