A finite element method for solving the Stokes-Darcy problem / eingereicht von: Manuela Redl

UNIVERSIT¨AT LINZ JOHANNES KEPLER

JKU

Technisch-Naturwissenschaftliche Fakult¨at

A Finite Element Method for Solving the

Stokes-Darcy Problem

MASTERARBEIT

zur Erlangung des akademischen Grades

Diplom-Ingenieurin

Industriemathematik

Manuela Redl

Institut f¨

ur Numerische Mathematik

Beurteilung:

A. Univ.-Prof. Dipl.-Ing. Dr. Walter Zulehner

The Stokes-Darcy problem appears in applications like modelling surface and ground-water ows or also in ltration processes. This thesis is about the coupling of a uid ow and a porous medium ow, where we consider a model consisting of a uid region where we assume a Stokes ow and a porous medium region, where we suppose Darcy's law. The two regions are separated by an interface and the uid can ow both ways across it. The Stokes and the Darcy equations must be coupled by suitable interface conditions in order to obtain a well posed problem.

We derive a weak mixed formulation for the Stokes-Darcy problem and make a complete analysis, where we show existence and uniqueness of a solution to the weak problem using Brezzi's theorem. Also other possible weak formulations are discussed. Then we analyse the corresponding discrete problem and make a nite element approx-imation with Taylor-Hood elements in the uid region and Raviart-Thomas elements in the porous medium region. The discretization leads us to a system of equations which we can solve.

Finally, numerical results are presented and the behaviour of the error is compared with the results obtained from theory.

Zusammenfassung

Das Stokes-Darcy Problem tritt in Anwendungen wie der Modellierung von Oberächen-und GrOberächen-undwasserströmen oder auch bei der Filtration auf. In dieser Masterarbeit geht es um die Koppelung von Fluidströmen und Strömungen durch ein poröses Medium, wobei wir ein Modell, bestehend aus einem Fluidbereich mit Stokeschem Fluss und einem porösen Bereich, in dem Darcy's Gesetz gilt, betrachten. Die zwei Gebiete sind durch ein Interface von einander getrennt, aber das Fluid kann von beiden Seiten aus hindurch ieÿen. Die Stokes und Darcy Gleichungen müssen durch passende Interface-bedingungen gekoppelt werden, sodass wir ein gut gestelltes Problem erhalten.

Wir leiten eine schwache Formulierung des Stokes-Darcy Problems her und erstellen eine vollständige Analyse, wobei wir Existenz und Eindeutigkeit einer Lösung des Problems mit Hilfe des Satzes von Brezzi zeigen. Zusätzlich werden weitere mögliche schwache Formulierungen behandelt. Anschlieÿend analysieren wir das dazugehörige diskrete Problem und machen eine Finite Elemente Approximierung mit Taylor-Hood Elementen im Fluidbereich und Raviart-Thomas Elemente im porösen Bereich. Durch die Diskretisierung erhalten wir ein Gleichungssystem, das wir lösen können.

Zum Schluss werden die numerischen Resultate präsentiert und das Verhalten des Fehlers wird mit den Ergebnissen aus der Theorie verglichen.

Acknowledgments

First of all, I want to express my gratitude to my supervisor Prof. Dr. Walter Zulehner for his guidance and the valuable time he took for numerous discussions and answering my questions.

I would like to give my deep thanks to my friends, my family and my boyfriend for their love and mental support. I am thankful for their great care, for their encourage-ment and that they are always there for me.

Last but not least I want to thank my helpful colleagues, who supported me during my studies. I appreciate the time we spent together.

Manuela Redl Linz, Juni 2015

Contents

1 Introduction and the model problem 1

2 Derivation of the weak formulation 5

2.1 Weak form of the Stokes equations . . . 5

2.2 Weak form of the Darcy equations . . . 8

2.3 Weak form of the Stokes-Darcy problem . . . 9

2.3.1 Inhomogeneous boundary conditions . . . 14

3 Analysis of the weak formulation 16 3.1 Further weak formulations of the Stokes-Darcy problem . . . 25

4 Discretization of the weak formulation 28 4.1 Analysis of the discrete problem . . . 30

4.2 Finite element approximation with Taylor-Hood and Raviart-Thomas elements . . . 35 4.2.1 Algebraic formulation . . . 37 4.2.2 Computational considerations . . . 48 5 Numerical results 54 6 Conclusion 58 Bibliography 59 iv

Chapter 1

Introduction and the model problem

In this introductory chapter, which is mainly based on Layton, Schieweck and Yotov [9], we present the Stokes-Darcy problem and give an overview how this thesis is structured.

The Stokes-Darcy problem deals with the coupling of a uid ow and a porous medium ow across an interface. More precisely, we consider two dierent regions, namely a uid and a porous medium region which are separated by an interface and where the incompressible uid can ow both ways across the interface. For instance the coupling of surface and groundwater ows can be modelled in such a way and questions like how substances and contaminants make their way from rivers and lakes to the groundwater can be treated. Another application where the Stokes-Darcy problem plays an important role is the ltration process for example in oil lters or also in arterial vessels, where blood is ltered through the vessel walls, see [5].

In this thesis we consider a model problem consisting of two dierent regions Ω1

and Ω2 (Ωj ⊂ Rd, d = 2 or 3), which are separated by an interface ΓI, see Figure

1.1. In Ω1 we assume a Stokes ow and in the porous medium region Ω2 we suppose

Darcy's law.

We use the following notations:

• Γj := ∂Ωj\ΓI, j = 1 or 2.

• nj is the unit outward normal vector of Ωj.

• uj : Ωj 7→ Rd is the velocity of the uid in Ωj.

• pj : Ωj 7→ R is the uid pressure in Ωj.

• The deformation rate tensor D is dened by D(u1) := 1₂(∇u1+∇uT1).

• T (u1, p1) :=−p1I + 2µD(u1) is the stress tensor. Here µ is the viscosity of the

uid.

Since we assume a Stokes ow in the region Ω1, (u1, p1) satises

{

− div T (u1, p1) = f1 in Ω1 (balance of momentum),

div u1 = 0 in Ω1 (conservation of mass). (1.1)

CHAPTER 1. INTRODUCTION AND THE MODEL PROBLEM 2

Ω1 = fluid region

Ω2 = porous medium region

Γ2 Γ1 Γ1 Γ1 Γ2 Γ2 ΓI n2 n1

Figure 1.1: The model problem. In the porous medium region Ω2, (u2, p2) satises

{

u2 = −K∇p2 in Ω2 (Darcy's law),

div u2 = f2 in Ω2 (balance of mass), (1.2)

where K is the rock permeability divided by the uid viscosity. Furthermore, let K be a symmetric and positive denite matrix in Rd×d.

We impose the following boundary conditions in our model problem:

u1 = 0 on Γ1 (no slip),

u2· n2 = 0 on Γ2 (no ow).

Furthermore, we couple problem (1.1) and (1.2) by appropriate interface conditions across ΓI:

• Conservation of mass across ΓI (ux continuity) is given by

u1· n1+ u2· n2 = 0 on ΓI. (1.3)

• Balance of normal forces on ΓI is expressed by

−⃗t(u1, p1)· n1 = p2 on ΓI, (1.4)

where ⃗t is the Cauchy stress vector (also called traction vector) satisfying

⃗t(u1, p1) = T (u1, p1)n1. Using the denition of the stress tensor T (u1, p1), (1.4)

is equivalent to

• Let τj, j = 1, . . . , d− 1, be an orthonormal system of tangent vectors on ΓI.

The condition on the tangential uid velocity on ΓI by Beavers and Joseph is

(u1− u2)· τj =

√ ˜

kj

µα (−⃗t(u1, p1))· τj, j = 1, . . . , d− 1, on ΓI, (1.6)

where ˜kj = τj · µKτj and α are assumed to be positive. This means that

the tangential slip velocity on ΓI is proportional to the shear stress along ΓI.

Since −⃗t(u1, p1)· τj = (−T (u1, u2)n1)· τj = p1In1· τj − 2µD(u1)n1· τj = −2µD(u1)n1· τj, condition (1.6) is equivalent to (u1− u2)· τj =− √ ˜ kj α 2D(u1)n1· τj, j = 1, . . . , d− 1, on ΓI. (1.7)

From observations it is known that the term u2·τj is much smaller than the other

appearing terms. This leads to the Beavers-Joseph-Saman condition which is obtained by neglecting the term u2· τj in (1.6):

u1· τj =

√ ˜

kj

µα (−⃗t(u1, p1))· τj, j = 1, . . . , d− 1, on ΓI. (1.8)

We will use the Beavers-Joseph-Saman condition because it is not clear whether the well-posedness of the problem is guaranteed if we use condition (1.6) by Beavers and Joseph. In Remark 2.4 we will explain where the diculties arise. This thesis is structured in the following way: In Chapter 2 we will derive the weak formulations for the Stokes equations and for the Darcy equations and use suitable interface conditions, which t in our problem setting, to couple the two problems. This leads to a weak mixed formulation of the Stokes-Darcy problem. In the end of this chapter also the case of inhomogeneous boundary conditions will be discussed.

Chapter 3 is about the analysis of the weak formulation. Existence and uniqueness of a solution to the problem will be proved using Brezzi's theorem and also other possible weak formulations and their analysis will be thematized.

Chapter 4 deals with the discretization of the weak problem. First we will analyse the discrete problem where we again verify the well-posedness of the problem. Then we will make a nite element approximation with Taylor-Hood elements in the uid region and Raviart-Thomas elements in the porous medium region. The coupling of the degrees of freedom on the interface will be explained in detail. We will end

CHAPTER 1. INTRODUCTION AND THE MODEL PROBLEM 4 up with an algebraic formulation in matrix form and will give some computational explanations.

In Chapter 5 numerical results will be presented and the behaviour of the error between the exact solution and the approximated solution will be compared with the theoretical results of Chapter 4.

Chapter 2

Derivation of the weak formulation

In this chapter, which is mainly based on Layton, Schieweck and Yotov [9], we develop two dierent weak formulations of the Stokes-Darcy problem. To this end we will rst derive the weak formulation for the uid region Ω1 and for the porous medium region

Ω2 separately and in the last section of this chapter we will couple these formulations

using the interface conditions.

2.1 Weak form of the Stokes equations

First of all, we consider the Stokes ow in the region Ω1. The balance of momentum

equation in (1.1) can be written as

−2µ div D(u1) +∇p1 = f1. (2.1)

As usual we derive the weak formulation by multiplying with a suciently smooth test function v1 and integrating over Ω1:

−2µ ∫ Ω1 div D(u1)· v1dx + ∫ Ω1 ∇p1· v1dx = ∫ Ω1 f1· v1dx. (2.2)

Using integration by parts, we have

−2µ ∫ Ω1 div D(u1)· v1dx =−2µ ∫ ∂Ω1 D(u1)n1· v1ds + 2µ ∫ Ω1 D(u1) :∇v1dx, (2.3) where A : B := ∑ i,j

AijBij denotes the Frobenius product of A and B. Since the

deformation rate tensor D(u1) is symmetric and AT : BT = A : B, we obtain

D(u1) :∇v1 = 1 2D(u1) :∇v1+ 1 2D(u1) T :∇v1 | {z } =D(u1):∇vT₁ = D(u1) : ( 1 2(∇v1+∇v T 1)) = D(u1) : D(v1). (2.4) 5

CHAPTER 2. DERIVATION OF THE WEAK FORMULATION 6 For the second integral in (2.2) we have

∫ Ω1 ∇p1 · v1dx = ∫ ∂Ω1 p1n1· v1ds− ∫ Ω1 p1div v1dx. (2.5) So we obtain −2µ ∫ ∂Ω1 D(u1)n1· v1ds + 2µ ∫ Ω1 D(u1) : D(v1) dx + ∫ ∂Ω1 p1n1· v1ds− ∫ Ω1 p1div v1dx = ∫ Ω1 f1· v1dx. (2.6)

The next step is to build in the boundary and interface conditions. Since the Dirichlet condition u1 = 0 on Γ1 can not be included in (2.6), we have to force

v1 = 0 on Γ1. Hence the integrals over Γ1 vanish. So it remains to consider the

integrals over the interface: ∫ ΓI (p1n1− 2µD(u1)n1)· v1ds = ∫ ΓI ((p_| 1I− 2µD(u_{z 1))_} =−T (u1,p1) n1)· v1ds = ∫ ΓI −⃗t · v1ds. (2.7)

The Cauchy stress vector ⃗t can also be written as

⃗t = (⃗t· n1)n1+

d−1

∑

j=1

(⃗t· τj)τj, for d = 2 or 3, (2.8)

because n1 together with τj, j = 1, d− 1 is an orthonormal system of Rd. Using (2.8)

and the interface conditions (1.4) and (1.8) we get that ∫ ΓI −⃗t · v1ds = ∫ ΓI ( −(⃗t · n1)n1+ d−1 ∑ j=1 −(⃗t · τj)τj ) · v1ds = ∫ ΓI p2n1 · v1ds + d−1 ∑ j=1 ∫ ΓI αµ √ ˜ kj (u1· τj)(τj· v1) ds. (2.9)

All in all we get the following weak form of the balance of momentum equation: 2µ ∫ Ω1 D(u1) : D(v1) dx + d−1 ∑ j=1 ∫ ΓI αµ √ ˜ kj (u1· τj)(v1· τj) ds − ∫ Ω1 p1div v1dx + ∫ ΓI p2v1 · n1ds = ∫ Ω1 f1· v1dx. (2.10)

Now we derive the weak formulation of the second equation in (1.1) by multiplying with a test function q1 and integrating over Ω1:

∫

Ω1

q1div u1dx = 0. (2.11)

To ensure that the integrals which appear in the weak formulations (2.10) and (2.11) are well dened, we have to choose appropriate function spaces. We need that v1, q1

and the rst weak derivatives of v1 are square integrable. The boundary condition

u1 = 0on Γ1 is an essential condition and has to be explicity enforced on our function

space.

Therefore we choose

X1 :={v1 ∈ (H1(Ω1))d: v1 = 0 on Γ1}, (2.12)

equipped with the norm ∥v1∥X1 := |v1|1,Ω1 := ∥∇v1∥0,Ω1 =

(∫ Ω1∇v1 :∇v1dx )1/2 for u1, v1 and let p1, q1 be in M1 := L2(Ω1). (2.13)

At this point we want to give a brief description of the spaces L2_{(Ω) and H}1_(Ω)

following [3]. L2_{(Ω) is the space of measurable functions p dened on Ω ⊂ R}d _which

are square-integrable, i.e., _∫

Ω

|p(x)|2_{dx <∞. (2.14)}

We identify p with q if p = q almost everywhere in Ω. The space L2_{(Ω), equipped with}

the inner product

(p, q)0,Ω = (p, q)L2_(Ω) :=

∫

Ω

pq dx, (2.15)

and the norm

∥p∥0,Ω:= (∫ Ω p2dx )1/2 (2.16) is a Hilbert space.

H1_{(Ω) is the space of functions u in L}2_{(Ω) which possess weak derivatives} ∂u ∂xi in

L2_{(Ω) for all i = 1, . . . , d , i.e.,}

∫ Ω ϕ∂u ∂xi dx =− ∫ Ω ∂ϕ ∂xi u dx for all ϕ ∈ C₀∞(Ω). (2.17)

C₀∞(Ω) denotes the space of innitely dierentiable functions with compact support. The scalar product on H1_{(Ω) is dened by}

(u, v)1,Ω := (u, v)0,Ω+ (∇u, ∇v)0,Ω (2.18)

and the corresponding norm and semi-norm by

∥u∥1,Ω:= ( ∥u∥2 0,Ω+∥∇u∥ 2 0,Ω )1/2

CHAPTER 2. DERIVATION OF THE WEAK FORMULATION 8 Note that H1_{(Ω) is a Hilbert space.}

To sum up, the weak formulation of the Stokes ow in Ω1 reads:

Find u1 in X1 and p1 in M1 such that

a1(u1, v1) + b1(v1, p1) +⟨p2, v1· n1⟩ΓI = (f1, v1)0,Ω1 for all v1 ∈ X1, (2.20)

b1(u1, q1) = 0 for all q1 ∈ M1, (2.21)

where the bilinear forms a1 and b1 are given by

a1(u1, v1) := 2µ ∫ Ω1 D(u1) : D(v1) dx + d−1 ∑ j=1 ∫ ΓI αµ √ ˜ kj (u1· τj)(v1· τj) ds, (2.22) b1(v1, p1) :=− ∫ Ω1 p1div v1dx (2.23) and ⟨p2, v1· n1⟩ΓI := ∫ ΓI p2v1· n1ds. (2.24)

2.2 Weak form of the Darcy equations

In the next step we derive the weak form of (1.2). In order to do this, we multiply Darcy's law by a suciently smooth test function v2. Then we integrate over the

porous medium region Ω2:

∫ Ω2 K−1u2· v2dx = ∫ Ω2 −∇p2· v2dx. (2.25)

Integration by parts yields ∫ Ω2 K−1u2· v2dx = ∫ Ω2 p2div v2dx− ∫ ΓI p2v2· n2ds− ∫ Γ2 p2v2· n2ds | {z } =0. .(2.26)

Since the no ow condition u2· n2 = 0 on Γ2 can not be directly incorporated in the

weak form (2.26), we have to restrict the function space for the test functions to a subspace whose elements fulll v2· n2 = 0 on Γ2.

The weak form of the mass balance in (1.2) is ∫ Ω2 q2div u2dx = ∫ Ω2 f2q2dx, (2.27)

To ensure that the weak form in (2.26) and (2.27), respectively, is well dened, we need that v2, div v2 and q2 are in L2(Ω2). Therefore we choose

M2 := L2(Ω2) (2.28)

for p2, q2 and

X2 :={v2 ∈ H(div; Ω2) :⟨v2· n2, ω⟩∂Ω2 = 0 for all ω ∈ H

1 0,ΓI(Ω2)}, (2.29) for u2, v2, where H_0,Γ1 I(Ω2) ={ω ∈ H 1_(Ω 2) : ω = 0 on ΓI}, (2.30) and H(div; Ω2) = {v2 ∈ (L2(Ω2))d: div v2 ∈ L2(Ω2)}, (2.31)

with the following norm

∥v2∥X2 :=∥v2∥H(div;Ω2) := ( ∥v2∥20,Ω2 +∥ div v2∥ 2 0,Ω2 )1/2 . (2.32)

Note that by div v2 we denote the weak divergence of v2 which is given by

∫ Ω2 div v2 ϕ dx =− ∫ Ω2 v2· ∇ϕ dx, for all ϕ in C0∞(Ω2), see [3].

We end up with the following mixed formulation for the porous medium region Ω2:

Find u2 in X2 and p2 in M2 such that

a2(u2, v2) + b2(v2, p2) +⟨p2, v2· n2⟩ΓI = 0 for all v2 ∈ X2 (2.33)

b2(u2, q2) = −(f2, q2)0,Ω2 for all q2 ∈ M2, (2.34)

where the bilinear forms a2 and b2 are given by

a2(u2, v2) := ∫ Ω2 K−1u2· v2dx (2.35) and b2(v2, p2) := − ∫ Ω2 p2div v2dx. (2.36)

2.3 Weak form of the Stokes-Darcy problem

The weak form of the Stokes-Darcy problem is now obtained by adding (2.33) to (2.20) and (2.34) to (2.21):

Find ui ∈ Xi and pi ∈ Mi, i = 1, 2,such that

a1(u1, v1) + a2(u2, v2) + b1(v1, p1) +

b2(v2, p2) +⟨p2, v1· n1⟩ΓI +⟨p2, v2· n2⟩ΓI = (f1, v1)0,Ω1, (2.37)

CHAPTER 2. DERIVATION OF THE WEAK FORMULATION 10 for all vi ∈ Xi and for all qi ∈ Mi, (i = 1, 2).

We introduce the following function spaces:

X := X1× X2 ={v = (v1, v2) : vi ∈ Xi} (2.39) and M := {q = (q1, q2) : qi ∈ Mi and ∫ Ω1 q1dx + ∫ Ω2 q2dx = 0} (2.40)

equipped with the norms

∥v∥X := (∥v1∥ 2 X1 +∥v2∥ 2 X2) 1/2 and ∥q∥M := (∥q1∥2M1 +∥q2∥ 2 M2) 1/2_.

We use the space M of functions with mean value zero over Ω to ensure uniqueness of the pressure pj in Ωj. A detailed explanation can be found in Remark 2.5.

Dening the bilinear forms

a(u, v) := a1(u1, v1) + a2(u2, v2) for u, v ∈ X, (2.41)

b(v, p) := b1(v1, p1) + b2(v2, p2) for v ∈ X and p in M, (2.42)

bI(v, µ) := ⟨v1 · n1+ v2· n2, µ⟩ΓI for v ∈ X and µ ∈ Λ, (2.43)

and the linear functionals

⟨F, v⟩ := (f1, v1)0,Ω1 and ⟨G, q⟩ := −(f2, q2)0,Ω2, (2.44)

we can simplify (2.37) and (2.38) to: Find u ∈ X, p ∈ M and λ ∈ Λ such that

a(u, v) + b(v, p) + bI(v, λ) = ⟨F, v⟩ for all v ∈ X, (2.45)

b(u, q) = ⟨G, q⟩ for all q ∈ M, (2.46)

with λ = p2|ΓI. The ux continuitiy equation (1.3) on ΓI can be written as

bI(u, µ) = 0 for all µ ∈ Λ. (2.47)

It remains to specify the space Λ such that the bilinear form bI(·, ·) is well dened. In

order to do this we need the following important trace theorems:

Theorem 2.1 (See [1, 10]). Let γu = u|∂Ω be the pointwise denition of the trace

or boundary values for smooth functions. Then the trace map γ can be extended for functions in H1_{(Ω), or more precisely, there exists a continuous map}

γ : H1(Ω)→ H1/2(∂Ω), (2.48) and a constant c > 0 such that

Here H1/2_{(∂Ω) := {u ∈ L}2_{(∂Ω) : |u|}

1/2,∂Ω < ∞} (see [1]), where the semi-norm

and norm is dened by

|u|1/2,∂Ω:= (∫ ∂Ω ∫ ∂Ω |u(x) − u(y)|2 |x − y|d dsxdxy )1/2 (2.50) and

∥u∥1/2,∂Ω := (∥u∥20,∂Ω +|u| 2 1/2,∂Ω)

1/2

. (2.51)

By H−1/2_{(∂Ω)(or equivalently by (H}1/2_(∂Ω))∗_{)we denote the dual space of H}1/2_(∂Ω)

which is the space of continuous linear functionals on H1/2_{(∂Ω)with the norm}

∥f∥−1/2,∂Ω := sup u∈H1/2_(∂Ω)

|f(u)| ∥u∥1/2,∂Ω

. (2.52)

Theorem 2.2 (See [10]). There exists a mapping

γn : H(div, Ω)→ H−1/2(∂Ω), (2.53)

such that γnu = u· n|∂Ω holds for smooth vector functions u. Furthermore, there exists

a constant c > 0 such that

∥γnu∥−1/2,∂Ω≤ c∥u∥H(div,Ω). (2.54)

Proof. We use Green's theorem which implies for smooth elds ϕ : Ω 7→ R and

u : Ω7→ Rd _∫ ∂Ω2 u· n ϕ ds | {z } γnu(ϕ):= = ∫ Ω div u ϕ dx + ∫ Ω u· ∇ϕ dx. (2.55)

Applying the Cauchy Schwarz inequality twice gives

|γnu(ϕ)| ≤ ∥ div u∥0,Ω∥ϕ∥0,Ω+∥u∥0,Ω∥∇ϕ∥0,Ω

≤ (∥ div u∥2 0,Ω+∥u∥ 2 0,Ω )1/2 ·(∥ϕ∥2 0,Ω+∥∇ϕ∥ 2 0,Ω )1/2 = ∥u∥H(div;Ω)∥ϕ∥1,Ω.

Since C∞_{(Ω)is dense in H}1_{(Ω) (see, e.g., [1]) and (C}∞_(Ω))d _{is also dense in H(div; Ω)}

(see [7]), we have that for u ∈ H(div, Ω), γnuis a bounded linear functional on H1(Ω),

i.e. γnu∈ H−1(Ω): ∥γnu∥−1,Ω:= sup ϕ∈H1_(Ω) |γnu(ϕ)| ∥ϕ∥1,Ω ≤ ∥u∥H(div;Ω). (2.56)

Thanks to the inverse trace theorem (see [1]) for a function ψ ∈ H1/2_{(∂Ω)there exists}

an extension ϕ ∈ H1_(Ω) s.t. ϕ|

∂Ω= ψ and

CHAPTER 2. DERIVATION OF THE WEAK FORMULATION 12 Hence, γnu = u· n on ∂Ω is in H−1/2(∂Ω): ∥γnu∥−1/2,∂Ω = sup ψ∈H1/2_(∂Ω) |γnu(ψ)| ∥ψ∥1/2,∂Ω ≤ sup ϕ∈H1_(Ω) Ctr|γnu(ϕ)| ∥ϕ∥1,Ω ≤ Ctr∥u∥H(div;Ω). (2.58)

Therefore, γn can be continuously extended to H(div; Ω).

Theorem 2.2 yields that vi· ni|∂Ωi ∈ H−1/2(∂Ωi)but for the bilinear form bI(·, ·) we

are interested for which space Λ the restriction of the functional vi· ni to the interface

ΓI is well dened.

To this end, we introduce the space H1/2

00 (ΓI)which is the completion of the smooth

functions on ∂Ω2 with compact support in ΓI with respect to the norm ∥ · ∥1/2,∂Ω2. It

holds that for any function µ in H1/2

00 (ΓI)the extension by zero on ∂Ωi gives a function

˜

µi ∈ H1/2(∂Ωi)with

∥˜µi∥1/2,∂Ωi ≤ C∥µ∥_H1/2

00 (ΓI), i = 1, 2. (2.59)

Note that we can dene

∥µ∥_H1/2

00 (ΓI):=∥˜µ2∥1/2,∂Ω2. (2.60)

Further we dene the extension E : H1/2

00 (ΓI)→ H0,Γ1 2(Ω2), µ 7→ Eµ := ϕ, where ϕ is

the weak solution of

   ∆ϕ = 0 in Ω2, ϕ = µ on ΓI, ϕ = 0 on Γ2. (2.61)

Eµ denotes the harmonic extension of µ, which is continuous, i.e.,

∥Eµ∥1,Ω2 ≤ C∥µ∥_H1/2 00 (ΓI). Choosing Λ := H₀₀1/2(ΓI) (⊂ L2(ΓI)) (2.62) yields ∥γnv2|ΓI∥Λ∗ = sup µ∈Λ |γnv2|ΓI(µ)| ∥µ∥Λ ≤ sup ϕ∈H1 0,Γ2(Ω2) |γnv2(ϕ)| ∥ϕ∥1,Ω2 ≤ sup ˜ ϕ∈H1_(Ω 2) |γnv2( ˜ϕ)| ∥˜ϕ∥1,Ω2 ≤ ∥v2∥H(div;Ω2), (2.63) i.e., v2· n2|ΓI ∈ (H 1/2 00 (ΓI))∗.

For i = 1 we know due to Theorem 2.1 that

vi· ni|ΓI ∈ L

2

(ΓI)≡ (L2(ΓI))∗ ⊆ (H001/2(ΓI))∗. (2.64)

Denition 2.3. We dene v2· n2|ΓI by

⟨v2· n2|ΓI, µ⟩ΓI :=⟨v2· n2, γEµ⟩∂Ω2, µ∈ H

1/2

00 (ΓI), (2.65)

where γ is the trace on ∂Ω2.

Remark 2.4. Remember the interface condition by Beavers and Joseph: (u1− u2)· τj =

√ ˜

kj

µα (−⃗t(u1, p1))· τj, j = 1, d− 1, on ΓI.

If we had used this condition instead of the Beavers-Joseph-Saman condition we would have got diculties because there does not exist a tangential trace of functions

u2 ∈ H(div, Ω2), that means u2 · τj|ΓI is not well dened. So we can not use the

condition by Beavers and Joseph in our setting.

In summary, the coupled Stokes-Darcy problem (1.1) + (1.2) with the interface conditions (1.3), (1.5) and (1.8) has the following weak formulation:

Find (u, p, λ) ∈ X × M × Λ such that 

 

a(u, v) + b(v, p) + bI(v, λ) = ⟨F, v⟩ for all v ∈ X,

b(u, q) = ⟨G, q⟩ for all q ∈ M,

bI(u, µ) = 0 for all µ ∈ Λ.

(2.66) This would be a possible weak formulation, however, for the analysis of the problem we derive another weak formulation using the space V, of functions in X, which fullls the ux continuity condition bI(v, µ) = 0for all µ ∈ Λ:

V :={v ∈ X : bI(v, µ) = 0 for all µ ∈ Λ}.

Therefore we get the following weak formulation: Find u ∈ V and p ∈ M such that

{

a(u, v) + b(v, p) = ⟨F, v⟩ for all v ∈ V,

b(u, q) = ⟨G, q⟩ for all q ∈ M. (2.67) Remark 2.5. Let c = (c, c) be a constant function in M1× M2. Then for v ∈ V

b(v,c) = − ∫ Ω1 c div v1dx− ∫ Ω2 c div v2dx =−c ∫ Ω div ˜v dx = −c ∫ ∂Ω ˜ v· n ds = −c ∫ Γ1 v1 |{z} = 0,on Γ1 ·n1ds− c ∫ Γ2 v2· n2 | {z } = 0,on Γ2 ds = 0.(2.68)

Here we used that v ∈ V implies ˜v ∈ H(div; Ω), with ˜v ∈ (L2_(Ω))d _{dened by}

˜

v|Ωj := vj, j = 1, 2 (see Lemma 3.6). Further we applied the Gauss theorem and

CHAPTER 2. DERIVATION OF THE WEAK FORMULATION 14 Equation (2.68) implies that if we have a solution (u, p) of the weak problem (2.67), then (u, p + c) is also a solution. To guarantee uniqueness we therefore dene

M := {q = (q1, q2) : qi ∈ Mi and

∫

Ω1q1dx +

∫

Ω2q2dx = 0} instead of M1 × M2.

However, we have to take into account that the weak formulation (2.67) has to be sat-ised for all test functions in M1 × M2. Since every function in M1× M2 can be

written as a function in M plus a constant, we have to verify that the weak formulation is also fullled for constant test functions c. This yields

b(u,c) = ⟨G, c⟩ for all c ∈ M1× M2, (2.69)

which is due to (2.68) equivalent to 0 =

∫

Ω2

f2dx. (2.70)

Hence, the function f2 has to fulll the solvability condition (2.70).

2.3.1 Inhomogeneous boundary conditions

If we consider instead of the no slip condition u1 = 0 and instead of the no ow

condition u2·n2 = 0the inhomogeneous conditions u1 = g1 on Γ1and u2·n2 = g2 on Γ2,

respectively, we obtain the following inhomogeneous problem: Find u ∈ Vg _{and p ∈ M such that}

{

a(u, v) + b(v, p) = ⟨F, v⟩ for all v ∈ V,

b(u, q) = ⟨G, q⟩ for all q ∈ M, (2.71) where Vg := {v = (v1, v2)∈ X1g1 × X g2 2 : bI(v, µ) = 0 for all µ ∈ Λ}, (2.72) Xg1 1 := {v1 ∈ (H1(Ω1))d : v1 = g1 on Γ1}, (2.73) Xg2 2 := {v2 ∈ H(div; Ω2) :⟨v2· n2, ω⟩∂Ω2 =⟨g2, ω⟩∂Ω2 (2.74) for all ω ∈ H1 0,ΓI(Ω2)}. Homogenization

We suppose that there exists ug = (ug1, ug2) ∈ V

g _{such that u}

g1|Γ1 = g1 and

ug2 · n2|Γ2 = g2. The ansatz u = ug + ω, ω ∈ V, leads us to the following

homo-geneous problem: Find ω ∈ V and p ∈ M such that {

a(ω, v) + b(v, p) = ⟨F, v⟩ − a(ug, v) =:⟨ ˜F , v⟩ for all v ∈ V,

b(ω, q) = ⟨G, q⟩ − b(ug, q) =:⟨ ˜G, q⟩ for all q ∈ M. (2.75)

Remark 2.6. Due to Remark 2.5 the solvability condition in the case of inhomoge-neous boundary conditions is of the following form:

Since for a function ug ∈ Vg it holds that ug ∈ H(div; Ω) (see Lemma 3.6) we

have b(ug, 1) =

∫

Ωdiv ugdx =

∫

∂Ωug · n ds and therefore the solvability condition is

equivalent to ∫ Ω2 f2dx + ∫ Γ1 g1· n1ds + ∫ Γ2 g2ds = 0. (2.77)

Chapter 3

Analysis of the weak formulation

Unless noted otherwise, the content of this chapter is mainly based on the paper of Layton, Schieweck and Yotov [9]. First we want to give a detailed analysis of the weak formulation (2.67). At the end of this chapter we will consider problem (2.66) and further weak formulations of the Stokes-Darcy problem.

A very important tool to verify existence and uniqueness of a solution to a weak mixed problem is the following theorem of Brezzi:

Theorem 3.1 (Brezzi). (See [4]) Let V and M be Hilbert spaces, F ∈ V∗ _{and G ∈ M}∗

be bounded linear functionals and let a : V × V −→ R and b : V × M −→ R be bilinear forms. Suppose that there exist constants α1, α2, β1, β2 > 0 such that

1. a(·, ·) is bounded on V , i.e,

|a(u, v)| ≤ α2∥u∥_V ∥v∥_V for all u, v ∈ V ,

2. b(·, ·) is bounded on V × M, i.e,

|b(v, q)| ≤ β2∥v∥V∥q∥M for all v ∈ V, q ∈ M,

3. a(·, ·) is coercive on Ker B, i.e.,

a(v, v)≥ α1∥v∥2V for all v ∈ W := Ker B = {v ∈ V : b(v, q) = 0 for all q ∈ M},

4. inf

0̸=q∈M_0̸=v∈Vsup

b(v, q) ∥v∥V∥q∥M

≥ β1 > 0.

Then there exists a unique solution to the weak problem: Find u ∈ V and p ∈ M, such that

{

a(u, v) + b(v, p) = ⟨F, v⟩ for all v ∈ V, b(u, q) = ⟨G, q⟩ for all q ∈ M. Moreover, we have ∥u∥V ≤ 1 α1 ∥F ∥V∗+ 1 β1 ( 1 + α2 α1 ) ∥G∥M∗ ∥p∥M ≤ 1 β1 ( 1 + α2 α1 ) ∥F ∥V∗+ α2 β2 1 ( 1 + α2 α1 ) ∥G∥M∗. 16

Proof. For a proof, see, e.g., [4].

Before we start discussing the assumptions of the theorem of Brezzi for the weak formulation (2.67), we will show some useful lemmas.

Lemma 3.2. The bilinear form bI(·, ·) is continuous on X × Λ, where X and Λ are

dened in (2.39) and (2.62), respectively. Proof. From (2.63) and (2.64) it follows that

∫

ΓI

vi· niµ ds≤ ∥vi· ni|ΓI∥Λ∗∥µ∥Λ≤ ∥vi∥H(div;Ωi)∥µ∥Λ. (3.1)

We know that ∥v2∥H(div;Ω2) =∥v2∥X2 ≤ ∥v∥X and ∥v1∥

2 H(div;Ω1)=∥v1∥ 2 0,Ω1+∥ div v1∥ 2 0,Ω1, whereby ∥ div v1∥20,Ω1 = ∫ Ω1 ( _d ∑ i=1 ∂v1i ∂xi )2 dx≤ ∫ Ω1 d d ∑ i=1 ( ∂v1i ∂xi )2 dx ≤ d∥∇v1∥20,Ω1 = d∥v1∥ 2 X1. (3.2) This implies ∥v1∥2H(div;Ω1) ≤ c 2 F∥∇v1∥20,Ω1 + d∥∇v1∥ 2 0,Ω1 = (c 2 F + d)∥v1∥2X1 ≤ C 2_∥v∥2 X, (3.3)

whereby we also used Friedrichs' inequality. All in all we obtain

bI(v, µ)≤ ∥v1∥H(div;Ω1)∥µ∥Λ+∥v2∥H(div;Ω2)∥µ∥Λ ≤ (C + 1)∥v∥X∥µ∥Λ.

Lemma 3.2 implies that V is a closed subspace of X since for every convergent sequence (vk)k∈N ∈ V the limit ¯v is also in V because the continuity of bI(·, ·) implies

that bI(¯v, µ) = 0for all µ ∈ Λ. Note that a closed subspace of a Hilbert space is again

a Hilbert space. Therefore V is a Hilbert space.

Lemma 3.3. For vi ∈ Xi∩ (H1(Ωi))d, i = 1, 2 it holds

∥vi∥Xi ≤ C1∥vi∥1,Ωi. (3.4)

Proof. First we consider the case i = 1, which is pretty simple:

∥v1∥2X1 =∥∇v1∥ 2 0,Ω1 ≤ ∥v1∥ 2 0,Ω1 +∥∇v1∥ 2 0,Ω1 =∥v1∥ 2 1,Ω1. (3.5)

In case of i = 2 we have, since v2 ∈ X2∩ (H1(Ω2))d and since the divergence operator

is a continuous operator (see (3.2)), that

∥v2∥2X2 =∥v2∥ 2 0,Ω2 +∥ div v2∥ 2 0,Ω2 ≤ ∥v2∥ 2 1,Ω2 + d∥v2∥ 2 1,Ω2 = C 2 1∥v2∥21,Ω2, with C1 := √ 1 + d.

CHAPTER 3. ANALYSIS OF THE WEAK FORMULATION 18 Lemma 3.4. For i = 1, 2 the bilinear forms ai(·, ·) and bi(·, ·) are bounded, i.e., there

exist C2 > 0 and C3 > 0 such that

|ai(ui, vi)| ≤ C2∥ui∥Xi∥vi∥Xi for all ui, vi ∈ Xi, (3.6)

|bi(vi, qi)| ≤ C3∥vi∥Xi∥qi∥Mi for all vi ∈ Xi, qi ∈ Mi. (3.7)

Proof. First, note that ∫ Ω1 |D(u1) : D(v1)| dx ≤ ∫ Ω1 ∥D(u1)∥F∥D(v1)∥F dx ≤ (∫ Ω1 ∥D(u1)∥2F dx )1/2(∫ Ω1 ∥D(v1)∥2Fdx )1/2 = (∫ Ω1 d ∑ i,j=1 1 4 ( ∂u1i ∂xj +∂u1j ∂xi )2 dx )1/2(∫ Ω1 d ∑ i,j=1 1 4 ( ∂v1i ∂xj + ∂v1j ∂xi )2 dx )1/2 ,

where we used the Cauchy-Schwarz inequality twice. Using the fact that 2∂u1i

∂xj ∂u1j ∂xi ≤ ( ∂u1i ∂xj) 2_{+ (}∂u1j ∂xi) 2 _{it follows} ∫ Ω1 d ∑ i,j=1 1 4 ( ∂u1i ∂xj + ∂u1j ∂xi )2 dx ≤ ∫ Ω1 d ∑ i,j=1 1 2 (( ∂u1i ∂xj )2 + ( ∂u1j ∂xi )2) dx = ∫ Ω1 d ∑ i,j=1 ( ∂u1i ∂xj )2 dx =|u1|21,Ω1. Hence, _∫ Ω1 |D(u1) : D(v1)| dx ≤ |u1|1,Ω1|v1|1,Ω1. (3.8)

Now we consider the second term of the bilinear form a1(·, ·).

∫ ΓI |(u1· τj)(v1· τj)| ds ≤ (∫ ΓI |u1· τj|2ds )1/2(∫ ΓI |v1· τj|2ds )1/2 ≤  ∫ ΓI ∥u1∥2∥τ_{| {z }}j∥2 =1 ds   1/2 ∫ ΓI ∥v1∥2∥τ_{| {z }}j∥2 =1 ds   1/2 = ∥u1∥0,ΓI∥v1∥0,ΓI ≤ C2 tr∥u1∥1,Ω1∥v1∥1,Ω1 ≤ C 2 tr(C 2 F + 1)|u1|1,Ω1|v1|1,Ω1 = C_tr2(C_F2 + 1)∥u1∥X1∥v1∥X1.

Here we made use of the Cauchy-Schwarz inequality, the trace theorem (see [3]) and the Friedrichs' inequality. Note that for x ∈ Rdwe denote by ∥x∥ the euclidean norm.

This and estimation (3.8) together with the triangular inequality implies |a1(u1, v1)| ≤ 2µ ∫ Ω1 |D(u1) : D(v1)| dx + d−1 ∑ j=1   √αµ˜_kj    ∫ ΓI |(u1· τj)(v1· τj)| ds ≤  2µ + (d − 1)αµC2 tr(C 2 F + 1) max j   √1_k˜_j      ∥u1∥X1∥v1∥X1.

Now we want to show that a2(·, ·) is bounded: Using that

|K−1_u

2· v2| ≤ λ_|max_{z(K−1_})

= 1

λmin(K)

∥u2∥∥v2∥ (3.9)

and with the Cauchy-Schwarz inequality we have

|a2(u2, v2)| =  ∫ Ω2 K−1u2· v2dx   ≤∫ Ω2 1 λmin(K) ∥u2∥ · ∥v2∥ dx ≤ 1 λmin(K) ∥u2∥0,Ω2∥v2∥0,Ω2 ≤ 1 λmin(K)

∥u2∥H(div;Ω2)∥v2∥H(div;Ω2)

= 1

λmin(K)∥u

2∥X2∥v2∥X2,

where λmin(K) is the minimal eigenvalue of K. Note that we obtain

ai(ui, vi)≤ C2∥ui∥Xi∥vi∥Xi for all ui, vi ∈ Xi,

with C2 = max{2µ + (d − 1)αµCtr2(CF2 + 1) max_j

 √1 ˜ kj  , 1 λmin(K)}.

In the next step we consider bi(·, ·) for i = 1, 2 :

|b1(v1, q1)| = | − ∫ Ω1 q1div v1dx| ≤ (∫ Ω1 q2₁dx )1/2(∫ Ω1 (div v1)2dx )1/2 = ∥q1∥M1∥ div v1∥0,Ω1 ≤ √d∥q1∥M1∥v1∥X1,

where we used the continuity of the divergence operator in the last step. For the bilinear form b2 we have

|b2(v2, q2)| = | − ∫ Ω2 q2div v2dx| ≤ (∫ Ω2 q2₂dx )1/2(∫ Ω2 (div v2)2dx )1/2 ≤ ∥q2∥M2∥v2∥H(div;Ω2) =∥q2∥M2∥v2∥X2.

So we showed, that bi(·, ·) is bounded on Xi× Mi and C3 = max{

√ d, 1}.

CHAPTER 3. ANALYSIS OF THE WEAK FORMULATION 20 In the following let W2 be a subspace of divergence free functions in X2:

W2 :={v2 ∈ X2 : div v2 = 0, a.e. x ∈ Ω2} ⊂ X2. (3.10)

Lemma 3.5. The bilinear form a1(·, ·) is coercive on X1, a2(·, ·) is coercive on W2

and therefore a(·, ·) is also on X1× W2 coercive:

a1(v1, v1) ≥ C4∥v1∥2X1 for all v1 ∈ X1, (3.11)

a2(v2, v2) ≥ C5∥v2∥2X2 for all v2 ∈ W2, (3.12)

a(v, v) ≥ min{C4, C5}∥v∥2X for all v ∈ X1× W2. (3.13)

Proof. For proving coercivity we need an important inequality called Korn inequality (see [3]): Let Ω ⊂ Rd_{, d ≥ 2, be an open and bounded set. Further let Γ ⊂ ∂Ω. Then}

there exists a positive constant CK such that

∫ Ω D(v) : D(v) dx≥ CK|v|21,Ω for all v ∈ (H 1 0,Γ(Ω)) d_{. (3.14)}

With the help of this inequality we are able to show

a1(v1, v1) = 2µ ∫ Ω1 D(v1) : D(v1) dx + d−1 ∑ j=1 αµ √ ˜ kj ∫ ΓI (v1· τj)2ds | {z } ≥0 ≥ 2µ ∫ Ω1 D(v1) : D(v1) dx ≥ 2µCK|v1|21,Ω1 = 2µCK∥v1∥ 2 X1. Using that K−1v2· v2 ≥ λ_|min(K_{z −1_}) =_λmax(K)1 ∥v2∥2 (3.15) we get a2(v2, v2) = ∫ Ω2 K−1v2· v2dx≥ 1 λmax(K) ∫ Ω2 ∥v2∥2dx = 1 λmax(K) ∥v2∥20,Ω2 = 1 λmax(K) ∥v2∥2H(div;Ω2) = 1 λmax(K) ∥v2∥2X2, for all v2 ∈ W2.

The inequalities (3.11) and (3.12) imply (3.13) because

a(v, v)≥ C4∥v1∥2X1 + C5∥v2∥

2

X2 ≥ min{C4, C5}∥v∥

2

X,

In order to prove coercivity of a(·, ·) on V × M, we need the following two lemmas, which you can also nd in the paper of Layton, Schieweck and Yotov [9]. The rst one states that a trace-continuous normal velocity has a well-dened divergence on the whole domain. Let Ω := interior(Ω1∪ Ω2). For a given v = (v1, v2) ∈ X, dene

˜

v ∈ (L2(Ω))dby ˜v|Ωj := vj and for q = (q1, q2)∈ M, dene ˜q ∈ L

2_{(Ω)by ˜q|}

Ωj := qj, j =

1, 2. As in [9] we write v and q instead of ˜v and ˜q to simplify notation. It should be clear from context what is meant.

Lemma 3.6. If v ∈ V , then v ∈ H(div; Ω). Proof. First we dene

g(x) = div vj(x) for x ∈ Ωj, j = 1, 2.

We want to show g = div v. Since vj ∈ H(div; Ωj), we can apply the divergence

theorem in each Ωj. So we have, for all ϕ ∈ C0∞(Ω),

∫ Ω v· ∇ϕ dx = ∫ Ω1 v1· ∇ϕ dx + ∫ Ω2 v2· ∇ϕ dx = − ∫ Ω1 (div v1)ϕ dx + ∫ ∂Ω1\ΓI ϕ v1· n1ds | {z } = 0. − ∫ Ω2 (div v2)ϕ dx + ∫ ∂Ω2\ΓI ϕ v2· n2ds | {z } = 0. + ∫ ΓI (v1· n1+ v2· n2)ϕ dx | {z } =0. . Since ϕ ∈ C∞

0 (Ω), the integrals over ∂Ωi\ΓI, i = 1, 2vanish. The last term is equal to

zero because ϕ ∈ C∞ 0 (Ω) implies ϕ|ΓI ∈ H 1/2 00 (ΓI). Therefore we obtain ∫ Ω v· ∇ϕ dx = − ∫ Ω g ϕ dx. (3.16)

We know that div vj ∈ L2(Ωj), thus g ∈ L2(Ω) and g is the weak L2 divergence of

v ∈ V.

We next dene the subspace W = Ker B ⊂ V,

W :={v ∈ V : b(v, q) = 0 for all q ∈ M}.

Lemma 3.7. The space W is a closed subspace of V and X. Moreover, v ∈ W implies div v = 0, a.e. x∈ Ω.

CHAPTER 3. ANALYSIS OF THE WEAK FORMULATION 22 Proof. For v ∈ W we know by Lemma (3.6) that v ∈ H(div; Ω), since W ⊂ V . Hence, for any q ∈ M b(v, q) =− ∫ Ω1 q1div v1dx− ∫ Ω2 q2div v2dx =− ∫ Ω q div v dx.

Further we know that div v ∈ L2_{(Ω) and using the divergence theorem we can show}

that div v has zero mean value over Ω: ∫ Ω div v dx = ∫ ∂Ω v · n ds = ∫ Γ1 v1 |{z} =0on Γ1. ·n1ds + ∫ Γ2 v2· n2 | {z } =0on Γ2. ds = 0.

This implies div v ∈ M. Setting q = div v yields

b(v, q) =−

∫

Ω

(div v)2dx = 0,

and thus, div v = 0.

Additionally W is a closed subspace of V because

b(v, q) = − ∫ Ω div v q dx≤ ∥ div v∥0,Ω∥q∥0,Ω = (∥ div v1∥20,Ω1 +∥ div v2∥ 2 0,Ω2 )1/2( ∥q1∥20,Ω1 +∥q2∥ 2 0,Ω2 )1/2 | {z } =∥q∥M. ≤ (d∥v1∥2X1 +∥v2∥ 2 X2∥ )1/2 ∥q∥M ≤ √ d∥v∥X∥q∥M, i.e., b(·, ·) is continuous on V × M.

For showing existence and uniqueness of a solution to (2.67) we will now examine the assumptions of Brezzi's theorem. Note that these assumptions, except the inf-sup condition, are rather easy to show.

1. F ∈ V∗ _{and G ∈ M}∗ _:

F is a bounded linear functional on V because |⟨F, v⟩| =  ∫ Ω1 f1v1dx   ≤ ∥f1∥0,Ω1∥v1∥0,Ω1 ≤ ∥f1∥0,Ω1CF|v1|1,Ω1 = ∥f1∥0,Ω1CF∥v1∥X1 ≤ Cf1∥v∥X, (3.17)

where we used the Cauchy-Schwarz inequality and Friedrichs' inequality. And for the linear functional G we have

|⟨G, q⟩| = ∫ Ω2 f2q2dx   ≤ ∥f2∥0,Ω2∥q2∥0,Ω2 ≤ Cf2∥q∥M. (3.18)

2. Boundedness of a(·, ·) on V × V :

From Lemma (3.4), the triangular inequality and Cauchy Schwarz inequality follows that for all u, v ∈ V ⊂ X

|a(u, v)| ≤ |a1(u1, v1)| + |a2(u2, v2)| ≤ C2(∥u1∥X1∥v1∥X1 +∥u2∥X2∥v2∥X2) = C2 ( ∥u1∥X1 ∥u2∥X2 ) · ( ∥v1∥X1 ∥v2∥X2 ) ≤ C2 ( ∥u1∥2X1 +∥u2∥ 2 X2 )1/2( ∥v1∥2X1 +∥v2∥ 2 X2 )1/2 = α2∥u∥X∥v∥X. 3. Boundedness of b(·, ·) on V × M: For all v ∈ V ⊂ X and all q ∈ M we have

|b(v, q)| ≤ |b1(v1, q1)| + |b2(v2, q2)| ≤ C3(∥v1∥X1∥q1∥M1 +∥v2∥X2∥q2∥M2) ≤ C3 ( ∥v1∥2X1 +∥v2∥ 2 X2 )1/2( ∥q1∥2M1 +∥q2∥ 2 M2 )1/2 = β2∥v∥X∥q∥M.

4. Coercivity of a(·, ·) on Ker B:

To exhibit coercivity of a(·, ·) on W = Ker B we have to nd an α1 > 0 such

that

a(v, v)≥ α1∥v∥2V for all v ∈ W.

Due to Lemma (3.7) div v = 0, since v ∈ Ker B. Hence, div v2 = 0, i.e. v2 ∈ W2,

and therefore coercivity follows from (3.13) with α1 = min{C4, C5}.

5. inf-sup condition on V × M:

Lemma 3.8. There exists a constant β1 such that

inf

0̸=q∈M_0̸=v∈Vsup

b(v, q) ∥v∥X∥q∥Q

≥ β1 > 0. (3.19)

Proof. (See [9]) Let q ∈ M\{0} be arbitrary but xed. If we could nd a v ∈ V satisfying b(v, q)≥ β1∥v∥X∥q∥M, (3.20) we would have sup 0̸=v∈V b(v, q) ∥v∥X∥q∥M ≥ _∥v∥b(v, q) X∥q∥M ≥ β1

and since q is arbitrary, (3.19) would follow. So it remains to construct such a

CHAPTER 3. ANALYSIS OF THE WEAK FORMULATION 24 For a given q = (q1, q2)∈ M, we dene ˜q(x) by ˜q|Ωi = qi. The function ˜q(x) has

mean value zero over Ω, thus ˜q ∈ L2 0(Ω).

The fact that the divergence operator div : H1

0(Ω)d−→ L20(Ω) is surjective (see

[7]), implies that there exists a ˜v ∈ H1

0(Ω)d satisfying div ˜v =−˜q in Ω (3.21) with ∥˜v∥1,Ω ≤ C∥˜q∥0,Ω. (3.22) So we have − ∫ Ω ˜ q div ˜v dx≥ C6∥˜v∥1,Ω∥˜q∥0,Ω. (3.23)

Given this ˜v, we dene v = (v1, v2) ∈ X by vi = ˜v|Ωi, (i = 1, 2). Since

˜

v ∈ (H1

0(Ω))d, we have v1|Γ1 = 0 and v2 · n2|Γ2 = 0. In addition,

v1|ΓI = v2|ΓI = ˜v|ΓI ∈ (H 1/2 00 (ΓI))d⊂ (L2(ΓI))d so that vi· ni ∈ L2(ΓI)and bI(v, µ) =⟨v1· n1+ v2· n2, µ⟩ΓI =⟨v1· n1− v2· n1, µ⟩ΓI = 0 for all µ ∈ L2_(Γ I). Therefore, v ∈ V.

Now we want to verify (3.20) for this v: We know

∥˜q∥0,Ω = (∫ Ω1 ˜ q2dx + ∫ Ω2 ˜ q2dx )1/2 = (∫ Ω1 q₁2dx + ∫ Ω2 q2₂dx )1/2 =∥q∥M,

and using Lemma 3.3 we can show

∥˜v∥2 1,Ω = ∥v1∥21,Ω1 +∥v2∥ 2 1,Ω2 ≥ 1 C2 1 (∥v1∥2X1 +∥v2∥ 2 X2) ≥ 1 C2 1 ∥v∥2 X. (3.24)

All in all we obtain

b(v, q) = − ∫ Ω1 div v1q1dx− ∫ Ω2 div v2q2dx =− ∫ Ω div ˜v ˜q dx ≥ C6∥˜v∥1,Ω∥˜q∥0,Ω ≥ C6 C1 ∥v∥X∥q∥M = β1∥v∥X∥q∥M.

So we have veried all assumptions of Brezzi's theorem and consequently we obtain the following

Remark 3.10. For the analysis of the other weak formulation (2.66) we refer the reader to [9] and [6]. Layton, Schwiewek and Yotov noted that if the porous medium is entirely enclosed within the uid region, i.e. ΓI = ∂Ω2, one can show the inf-sup

condition. Furthermore, they mentioned that in this case the unique weak solution to (2.67) is also the unique weak solution to (2.66) and the two formulations are equivalent.

3.1 Further weak formulations of the Stokes-Darcy

problem

In this section we want to show further weak fomulations of the Stokes-Darcy problem and give annotations on their well-posedness.

In [6], Galvis and Sarkis considered the following Stokes-Darcy equations    − div T (u1, p1) = f1 in Ω1, div u1 = 0 in Ω1, u1 = 0 on Γ1, (3.25)    u2 = −K∇p2+ f2 in Ω2, div u2 = 0 in Ω2, u2· n2 = 0 on Γ2, (3.26) and used the same interface conditions as treated in this thesis. They derived the following weak formulation:

Find (u, p, λ) ∈ X × M × Λ such that 

 

a(u, v) + b(v, p) + bΓI(v, λ) = l(v) for all v ∈ X,

b(u, q) = 0 for all q ∈ M,

bΓI(v, µ) = 0 for all µ ∈ Λ,

(3.27) where the bilinear forms a(·, ·), b(·, ·) are dened as in (2.41) and (2.42), respectively, and

l(v) := (f1, v1)0,Ω1 + (f2, v2)0,Ω2. (3.28)

The dierence to the weak formulation (2.66) lies in the choice of the space Λ and the denition of bΓI. They set Λ = H1/2_(Γ I) and dene bΓI : X × Λ → R by bΓI(v, µ) :=⟨v1· n1|ΓI, µ⟩ΓI +⟨v2· n2|ΓI, µ⟩ΓI. (3.29) Since v1· n2|ΓI ∈ L 2_(Γ

I), it is also a linear functional on H1/2(ΓI). Galvis and Sarkis

were able to show that functionals in H−1/2_(∂Ω

CHAPTER 3. ANALYSIS OF THE WEAK FORMULATION 26

∂Ω2\ΓI can be identied with functionals in H−1/2(ΓI) and this is why they dene

v2· n2|ΓI ∈ H −1/2_(Γ I) as ⟨v2· n2|ΓI, µ⟩ΓI :=⟨v2· n2, γ ˆEµ⟩∂Ω2, µ∈ H 1/2_(Γ I), (3.30)

where γ is the trace on ∂Ω2 and ˆE is dened in the next Lemma:

Lemma 3.11. (See [6]) There exists a continuous extension ˆ

E : H1/2_(Γ

I)→ H1(Ω2) : µ7→ ˆEµ := ϕ, where ϕ is the solution of the following weak

problem    −∆ϕ = 0 in Ω2, ϕ = µ on ΓI, ∂ϕ ∂n = 0 on Γ2. (3.31) Remark 3.12. Galvis and Sarkis showed in [6] that the weak problem (3.27) admits a unique solution.

Additionally, they noted that from the physical point of view the space H1/2_(Γ

I) is

the correct choice since the Lagrange multipliers are related to the Darcy pressure on the interface ΓI and the value of the Darcy pressure at ΓI∩∂(Ω1∪Ω2)is not prescribed

when ux boundary condition is imposed on the porous side exterior boundary Γ2.

Now we want to quote a further weak formulation, which has been analysed by Discacciati and Quarteroni in [5]. For that purpose we rewrite the rst two equations of (1.2) (Darcy's law and conservation of mass) as an elliptic equation, namely

− div(K∇p2) = f2. (3.32)

Using that u2 =−K∇p2 we get the following no ow condition

−K∇p2 · n2 = 0 on Γ2. (3.33)

In the following, let f2 = 0. Multiplying (3.32) with a suciently smooth test function

q2 in H1(Ω2) and using integration by parts yields

∫ Ω2 ∇q2 · K∇p2dx− ∫ ∂Ω2 q2K∇p2· n2ds = 0. (3.34)

Furthermore, we have that ∫ ∂Ω2 q2K∇p2· n2ds = ∫ Γ2 q2K_{| {z }}∇p2· n2 =0on Γ2. ds + ∫ ΓI q2 K_{| {z }}∇p2· n2 =u1·n1on ΓI. ds, (3.35)

Finally, we obtain the following weak form of (3.32): ∫ Ω2 ∇q2· K∇p2dx− ∫ ΓI q2u1· n1ds = 0 for all q2 ∈ H1(Ω2). (3.36)

For all u = (u1, p2)∈ Y := X1× H1(Ω2), v = (v1, q2)∈ Y and q1 ∈ M1 dene

a(u, v) := a1(u1, v1) + a2(p2, q2) +⟨p2, v1· n1⟩ΓI − ⟨q2, u1· n1⟩ΓI, (3.37)

b(u, q1) := b1(u1, q1), (3.38)

where a1(·, ·) and b1(·, ·) are dened as in (2.22) and (2.23), respectively, and

a2(p2, q2) :=

∫

Ω2

∇q2· K∇p2dx. (3.39)

Please note that the bilinear form a(u, v) is not symmetric.

All in all we get the following weak form of the Stokes-Darcy problem: Find u = (u1, p2)∈ Y, p1 ∈ M1 such that

{

a(u, v) + b(v, p1) = ⟨F, v⟩ for all v = (v1, q2)∈ Y,

b(u, q1) = 0 for all q1 ∈ M1, (3.40)

wherby the linear functional F is dened as in (2.44).

Discacciati and Quarteroni showed in [5] existence and uniqueness of a solution to problem (3.40), whereby they considered a model problem with more general boundary conditions. They split Γ1 into Γinf and Γf and Γ2 into Γbp and Γp and impose the

following boundary conditions:

u1 = uin on Γinf ,

u1 = 0 on Γf,

p2 = ϕp on Γp,

Chapter 4

Discretization of the weak formulation

This chapter, which is devoted to the discretization of the weak formulation, mainly follows the explanations by Layton, Schieweck and Yotov in [9] and is also based on Braess [3] and Bahriawati and Carstensen [2].

For simplicity we only consider the case d = 2, i.e., Ωj ⊂ R2 (j = 1, 2). We

introduce the following notation:

• Th

j is a mesh upon Ωj (j = 1, 2) with Ωj =∪K∈Th j K.

• K is a cell (e.g. a triangle). We assume that the cells are ane equivalent. • Eh(K) denotes the set of all edges E of the element K.

• Eh(ΓI)denotes the set of all element edges E which are part of the interface ΓI,

i.e. E ⊂ ΓI.

Further, we assume that the meshes Th

1 and T2h match at ΓI, which means that they

share the same edges on the interface.

For the discretization in the Stokes region we use nite element spaces

X₁h ⊂ X1 and M1h ⊂ M1, (4.1)

which are supposed to be inf-sup stable, i.e., there exists a constant β1 such that

inf q1∈M1h\{0} sup v1∈X1h\{0} b1(v1, q1) ∥v1∥X1∥q1∥M1 ≥ β1 > 0, (4.2)

and we assume that the discrete Korn inequality (D(v1), D(v1))0,Ω1 ≥ α1|v1| 2 1,Ω1 for all v1 ∈ X h 1 (4.3) is satised.

Note that under certain conditions one can show that the usual inf-sup condition for Xh

1 ∩ (H01(Ω1))d and M1h∩ L20(Ω1) implies that the spaces X1h and M1h full (4.2),

see [9].

Further we assume that the spaces Xh

1 and M1h contain at least polynomials of

degree r1 and r1 − 1, respectively, (r1 ≥ 1). We suppose that there exist (quasi)

interpolation operators I_Xh₁ : X1∩ (Hs(Ω1))d→ X1h and I h M1 : M1∩ H s (Ω1)→ M1h (4.4)

such that for all K ∈ Th

1 it holds that |v1− IXh1v1|m,K ≤ Ch s−m K |v1|s,δ(K), m = 0, 1, 1≤ s ≤ r1+ 1, (4.5) and ∥q1− IMh1q1∥0,K ≤ Ch s K|q1|s,δ(K), 0≤ s ≤ r1, (4.6)

where hK = diam(K) and δ(K) = K in most cases of usual interpolation operators

and in some cases of quasi interpolation operators the vicinity of K is meant by δ(K). In the Darcy region we use nite element spaces

X₂h ⊂ X2 and M2h ⊂ M2, (4.7)

where Xh

2 and M2h are assumed to include at least polynomials of degree r2 and l2,

respectively. Further we assume that

div(X₂h) = M₂h (4.8)

and that there exists an interpolation operator Ih X2 : (H

1_(Ω

2))d → X2h such that for

all v2 ∈ (H1(Ω2))d it holds that

(div I_Xh 2v2, q h 2)0,Ω2 = (div v2, q h 2)0,Ω2, q h 2 ∈ M h 2, (4.9) and ⟨Ih

X2v2 · n2, µ⟩E =⟨v2· n2, µ⟩E for all µ ∈ Pr2(E) and E ∈ Eh(ΓI). (4.10)

Note that by Pk we denote the space of polynomials of degree less or equal k, or more

precisely, Pk :={w(x, y) = ∑ i+j≤k i,j≥0 cijxiyj}. (4.11) For all vh 2 ∈ X2h we assume

vh₂ · n2|E ∈ Pr2(E) for all E ∈ Eh(K), K∈ T

h

2 . (4.12)

Suppose that Ih

M2 : M2 → M

h

2 is the L2-orthogonal projection such that for all

q2 ∈ M2 (I_Mh₂q2− q2, q2h)0,Ω2 = 0 for all q h 2 ∈ M h 2. (4.13)

CHAPTER 4. DISCRETIZATION OF THE WEAK FORMULATION 30 Further, the operators Ih

X2 and I

h

M2 are assumed to full for all K ∈ T

h 2 , ∥q2− IMh2q2∥0,K ≤ Ch s K|q2|s,K, 0≤ s ≤ l2+ 1, (4.14) |v2− IXh2v2|m,K ≤ Ch s−m K |v2|s,K, m ∈ {0, 1}, 1 ≤ s ≤ r2+ 1, (4.15) ∥ div(v2− IXh2v2)∥0,K ≤ Ch s K| div(v2)|s,K, 0≤ s ≤ l2+ 1. (4.16)

Now we dene the following nite element spaces for the Stokes-Darcy Problem:

Xh := X₁h× X₂h, Mh :={(q₁h, qh₂)∈ M₁h× M₂h : ∫ Ω1 q₁hdx + ∫ Ω2 qh₂dx = 0}. (4.17)

Λh :={µh ∈ L2(ΓI) : µh|E ∈ Pr2(E) for all E ∈ Eh(ΓI)}. (4.18)

Vh :={vh = (v₁h, v₂h)∈ Xh : bI(vh, µh) = 0 for all µh ∈ Λh}. (4.19)

Note that Λh _{̸⊂ Λ, since a function µ}h _{contained in L}2_(Γ

I) does not have to be zero

on ∂ΓI, and Vh ̸⊂ V.

Using this notation the discrete problem reads: Find (uh_{, p}h₎∈ Vh× Mh such that

{

a(uh, vh) + b(vh, ph) = ⟨F, vh⟩ for all vh ∈ Vh, (4.20) b(uh, qh) = ⟨G, qh⟩ for all qh ∈ Mh, (4.21)

is satised.

4.1 Analysis of the discrete problem

The analysis of the discrete problem (4.20) - (4.21) can be carried out using Brezzi's theorem. Note that the existence and uniqueness of a solution to problem (2.67) does not imply existence and uniqueness of a solution to the discrete problem.

The boundedness of a(·, ·) on Vh _{and b(·, ·) on V}h_{× M}h _{follows from the}

bound-edness of a(·, ·) on X and of b(·, ·) on X × M, respectively, since Vh ⊂ Xh ⊂ X and

Mh ⊂ M.

Coercivity of a(·, ·) on Ker Bh_:

For proving coercivity we have to show that there exists an α > 0 such that

a(vh, vh)≥ α∥vh∥2_X for all vh ∈ Wh, (4.22)

where

Wh := Ker Bh ={vh ∈ Vh : b(vh, qh) = 0 for all qh ∈ Mh}. (4.23) From Lemma 3.5 it follows that

a(vh, vh)≥ min{C4, C5}∥vh∥2X for all vh ∈ X1h× W2h, (4.24)

where Wh