J OHA NNES KEPLER

UN IV ERS IT A T L INZNe t zw e r k f u r F o r s c h u n g , L e h r e u n d P r a x i s

Numerical Simulation of a Three-dimensional

Bingham Fluid Flow

Masterarbeit

zur Erlangung des akademischen Grades

Diplomingenieur

in der Studienrichtung

Industriemathematik

Angefertigt am Institut fur Numerische Mathematik

Betreuung:

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

Eingereicht von:

Philipp Laaber

Linz, Marz 2008

Johannes Kepler Universitat

A-4040 Linz · Altenbergerstraße 69 · Internet: http://www.uni-linz.ac.at · DVR 0093696

Abstract

Non-Newtonian fluids, especially Bingham fluids, are widely used in prac-tice. The numerical simulation of such flows involves many mathematicalproblems. In this thesis we present two numerical methods for computing aBingham fluid flow in a three-dimensional domain. Both methods are basedon an existing algorithm for a two-dimensional problem.

Starting from a classical formulation for several problems, we will derivemixed as well as dual dual formulations for two problems. We will proofexistence and uniqueness of a solution. The two-dimensional algorithm isextended in two different ways into three dimensions. The discretization ofboth problems is then done by finite elements. Existence, uniqueness as wellas error estimates are then shown. Numerical results for both realizationsconfirm the mathematical theory. At the end an example with practicalbackground is presented.

i

Contents

1 Introduction 1

2 Mathematical Model 3

2.1 Bingham fluid flow in a three-dimensional domain . . . . . . 32.2 Reduction on a 2D problem . . . . . . . . . . . . . . . . . . . 6

3 Variational formulations 10

3.1 Primal formulations . . . . . . . . . . . . . . . . . . . . . . . 103.1.1 The three-dimensional problem . . . . . . . . . . . . . 103.1.2 The two-dimensional case . . . . . . . . . . . . . . . . 16

3.2 Dual formulations . . . . . . . . . . . . . . . . . . . . . . . . 173.2.1 The three-dimensional problem . . . . . . . . . . . . . 173.2.2 The two-dimensional case . . . . . . . . . . . . . . . . 20

3.3 Existence and uniqueness of a solution . . . . . . . . . . . . . 243.3.1 Existence and uniqueness for the mixed problem . . . 253.3.2 Application to a fluid flow in a quadratic pipe . . . . . 33

4 A method for solving the weak systems 37

4.1 An Uzawa type method . . . . . . . . . . . . . . . . . . . . . 374.1.1 A time dependent approximation . . . . . . . . . . . . 374.1.2 A stabilized scheme . . . . . . . . . . . . . . . . . . . 43

4.2 Convergence results . . . . . . . . . . . . . . . . . . . . . . . 454.3 Application to the problem in 3D . . . . . . . . . . . . . . . . 48

4.3.1 The first approach . . . . . . . . . . . . . . . . . . . . 484.3.2 The second approach . . . . . . . . . . . . . . . . . . . 49

5 Approximation with Finite Elements 52

5.1 Discretization of the domain . . . . . . . . . . . . . . . . . . . 525.2 The discrete models . . . . . . . . . . . . . . . . . . . . . . . 54

5.2.1 The two-dimensional case . . . . . . . . . . . . . . . . 555.2.2 The three-dimensional case . . . . . . . . . . . . . . . 55

5.3 Existence and uniqueness for the discrete mixed problems . . 605.3.1 The two-dimensional problem . . . . . . . . . . . . . . 605.3.2 The three-dimensional problem . . . . . . . . . . . . . 60

ii

5.4 Convergence and discretization errors . . . . . . . . . . . . . 635.4.1 The two-dimensional problem . . . . . . . . . . . . . . 635.4.2 The three-dimensional case using piecewise linear func-

tions for p . . . . . . . . . . . . . . . . . . . . . . . . . 65

6 Numerical Results 67

6.1 Scalar problem in 2D . . . . . . . . . . . . . . . . . . . . . . . 676.1.1 Realization . . . . . . . . . . . . . . . . . . . . . . . . 676.1.2 Results . . . . . . . . . . . . . . . . . . . . . . . . . . 68

6.2 The three-dimensional problem . . . . . . . . . . . . . . . . . 736.2.1 Test . . . . . . . . . . . . . . . . . . . . . . . . . . . . 736.2.2 Comparison of the different algorithms . . . . . . . . . 80

6.3 An example related to a practical problem . . . . . . . . . . . 83

Bibliography 90

Eidesstattliche Erklarung 93

Curriculum Vitae 94

iii

Chapter 1

Introduction

The topic of this thesis was motivated by a cooperation with the companydTech Steyr. I am thankful for the support given by Mr. Peter Fischerand his employees, especially by Mr. Walter Hinterberger and Mr. SimonSchneiderbauer.

We will study the flow of a medium containing two phases. In some re-gions the medium acts like a fluid, in others the medium can be interpretedas a rigid one. Such non-Newtonian fluids are of high interest since theyappear in almost every environment. Typical examples are paints, ketchup,tooth paste, mud, lava, snow, blood and many more. The thesis is mainlybased on a paper from He and Glowinski ([18]). Therein an Uzawa-typealgorithm for the stationary flow of a Bingham fluid in a two-dimensionaldomain is developed. Our goal is to construct a similar algorithm for an ex-tended three-dimensional problem. The thesis is structured in the followingway:

In chapter 2 we derive equations for a non-Newtonian fluid with yieldstress. If a certain function of the stress passes a limit τc, the medium startsto flow, what means that there exists relations between the strain ratesand the stress. But if the yield limit is not reached, no strain-rates exist.Then the fluid acts like a rigid medium. A few constitutive laws, model-ing such a medium are presented. For our study we will choose a Binghamfluid. The classical formulation of such a fluid is then given. Using specialboundary conditions as well as a special domain, we will also consider atwo-dimensional problem.

In chapter 3 suitable weak formulations for the vector-valued problemin 3D, as well as for the scalar problem in 2D are given. We will see, thatthe usual mixed formulation involving a variational inequality of the secondkind, has a non-differentiable term to handle. In order to get rid of this term

1

we will derive a dual formulation for both cases. At the end of the chapterexistence and uniqueness for our problems are shown.

In chapter 4 we follow the ideas from He and Glowinski (see [18]). Anapproximation of the original problem by a time dependent one is suggested.It is shown that an additional but necessary stabilization leads to an algo-rithm which converges to the desired solution and which works well alsofor large values of τc. We then carry over the results to the vector-valuedthree-dimensional problem. For this we will use two different approaches.On the one hand we will interpret (u, p) as the primal variable and λ as thedual variable, on the other hand we will interpret u as the primal variableand (p, λ) as the dual variable. Both interpretations lead to two differentalgorithms for the vector-valued problem.

In chapter 5 we use Galerkin’s principle to discretize our problems withfinite elements. The two-dimensional problem is discretized in the sameway as it was done in [18], in order to compare our results. For the three-dimensional problem we will use a continuous approximation of the pressurefor the first approach and a discontinuous approximation for the secondapproach, respectively. Existence and uniqueness of a discrete solution isshown. The convergence of the discrete solution is then proven and errorestimates are developed for the velocity.

In the last chapter we will present our numerical results. First we willtest the convergence properties in the two-dimensional case for various valuesof τc and compare them with the results from [18]. Further we will test thethree-dimensional algorithm by choosing special parameters and prescribingthe pressure. The three-dimensional problem is then computed. Next, wewill compare our developed methods for the three-dimensional problem. Fi-nally we will use our algorithm for a problem with practical background. Incooperation with dTech Steyr, we will study the behavior of a snow cover.

2

Chapter 2

Mathematical Model

2.1 Bingham fluid flow in a three-dimensional do-

main

We want to study the stationary flow of an incompressible two-phase fluidwith yield limit. In this section we will derive equations describing thedesired flow. Our domain of interest should be an open bounded setΩ3D ⊂ R

3. As in fluid dynamics common, we will describe our fluid inEulerian coordinates. Using them, we can write down the equations ofmotion for the steady state case, the so called balance equations.

−divσ = f(x) (2.1)

with a volume forcef : Ω3D → R

3, x 7→ f(x)

and the stress tensor

σ : Ω3D → R3×3, u 7→ σ(u).

In (2.1), the operator on the left hand side is defined by

divσ :=

div (σ11, σ12, σ13)div (σ21, σ22, σ23)div (σ31, σ32, σ33)

.

Moreover for an incompressible fluid, it follows that the incompressibilitycondition

div u = 0

has to hold. These equations are valid for all sort of continua. Further itcan be shown (see [24]) that the stress tensor σ can be decomposed in aspherical part as well as a deviatoric part:

σ = −pI + τ.

In the case of a fluid the variable p denotes the hydrostatic pressure. Thetensor τ is called the deviator or the deviatoric stress.

3

Constitutive law

Up to now we have characterized an arbitrary fluid. In order to produce aclosed model we have to classify the properties of the fluid. In many fluidsthis is done by a given relationship between the deviatoric stress τ and thetensor of strain rates.

Definition. The tensor of strain-rates is defined by

Dij(u) =1

2

(∂ui

∂xj+∂uj

∂xi

)

for i = 1, 2, 3.

Definition. The second invariant of the tensor of strain-rates is defined,using the Einstein summation convention, by

DII =1

2DijDij

and the second invariant of the deviatoric stress is defined by

τII =1

2τijτij

respectively.

Now a first model would be, if the deviatoric stress only depends linearlyon the tensor of strain-rates. So

τ = µD(u).

We then call the fluid Newtonian. The constant µ is called the viscosity. Amore general class of fluids is the ones of generalized (or non-) Newtonianmodels, where τ is depending nonlinearly on u. For example the function µcan depend on the second invariant of the tensor of strain rates DII:

τ = µ(DII)D(u).

A class of models doing so, is the class of Herschel-Bulkley fluids. Thismodel is characterized by

τ = (2µDn−1II + τc/(DII)

1/2)D(u) ⇔ τ1/2II > τc

D(u) = 0 ⇔ τ1/2II ≤ τc,

(2.2)

where the variable τc is called the yield limit. So if a certain stress function

(here τ1/2II ) is under the given threshold τc, the medium acts like a rigid

body since there are no strain rates. But if the stress increases and the yieldlimit is passed, the medium starts to flow following the constitutive law (2.2).

4

There is now the question which kind of Herschel-Bulkley model wechoose. Here we are going to study the case n = 1. A fluid described bysuch a law is then called Bingham fluid. It is characterized by

τ = 2µD(u) + τcD(u)/(DII)1/2 ⇔ τ

1/2II > τc

D(u) = 0 ⇔ τ1/2II ≤ τc.

There are also variants of the Bingham model, for example a biviscous Bing-ham model in which two viscosities are considered (see [10]). But we do notwant to go in detail here and refer to e.g. [27] for more modified models,to [3] for a general characterization of constitutive laws and to [30] for acomparison of some models for a practical problem. Further for more flowproperties of Bingham fluids especially if the external loads vary, see [4].

Boundary Conditions

In order to complete our model, we need appropriate boundary conditions.For this we separate our boundary into three parts.

∂Ω3D := Γ3D = Γ3D

D ∪ Γ3D

N ∪ Γ3D

P

withΓ3D

D ∩ Γ3D

N = ∅, Γ3D

D ∩ Γ3D

P = ∅, Γ3D

P ∩ Γ3D

N = ∅.

Now several boundary conditions are possible. For instance a part of theboundary, let us say Γ3D

D , could defined by a wall. In this case we get a socalled no-slip condition, i.e.

u = 0 on Γ3D

D .

Further on another part the traction g(x), x ∈ Γ3D

N is known, so

σ · n = g(x) on Γ3D

N .

Often the domain Ω3D can be interpreted as a cylinder or prism with a largelength let us say in z-direction. Simulating such a domain can be done byspecial boundary conditions for u and p. We assume that

Ω3D = Ω × (0, L),

where Ω denotes the cross section and L the length of the prism. Let Γ3D

P

be the part on which special boundary conditions are needed, then

Γ3D

P := Ω × 0 ∪ Ω × L.

The conditions on u and p then read as

u|z=0= u|z=L

on Γ3D

P

5

andp|z=0

= p|z=L+ h(x) on Γ3D

P ,

respectively. The function h(·) is representing a jump on the interface.According to [29] we call these conditions periodic boundary conditions.Summarizing our results, a Bingham fluid flow (in a special domain Ω3D)can be modeled by the following

Problem 2.1 (Classical Formulation). Find

u ∈ C2(Ω3D) ∩ C(∂Ω3D), p ∈ C1(Ω3D)

such that∇p− div τ = f in Ω3D

τ = 2µD(u) + τcD(u)/(DII)1/2 ⇔ τ

1/2II > τc

D(u) = 0 ⇔ τ1/2II ≤ τc

div u = 0 in Ω3D

u = 0 on Γ3D

D

σn = g(x) on Γ3D

N

u|z=0= u|z=L

on Γ3D

P

p|z=0= p|z=L

+ i(x) on Γ3D

P .

2.2 Reduction on a 2D problem

Motivated by the work of He and Glowinski ([18]) we are going to set upboundary conditions and right hand side in such a way that we can re-duce the three-dimensional problem to a two-dimensional one. For this ourdomain Ω3D is assumed to be a prismatic domain with infinite length inz-direction. In order to simulate the flow along z-direction, we can take adomain with finite length and set appropriate periodic boundary conditions.So let

Ω3D = Ω × (0, 1)

with quadratic cross section Ω in the xy-plane. The boundary is split intotwo parts

Γ3D = Γ3D

D ∪ Γ3D

P ,

so Γ3D

N = ∅. On the one hand we consider a wall around the prism, so wehave a no-slip condition on it:

u = 0 on Γ3D

D ,

6

whereΓ3D

D = ∂Ω × (0, 1).

On the rest of the boundary

Γ3D

P := Ω × 0 ∪ Ω × 1

we set homogenous periodic boundary conditions for the velocity u, i.e.

u(x, y, 0) = u(x, y, 1) on Γ3D

P .

For the pressure p we assume a constant drop along the z-axis. Let c bethe drop in pressure per unit length, then we consider the following periodicboundary condition on p:

p(x, y, 0) = p(x, y, 1) + c on Γ3D

P .

We assume the flow to be laminar, which is automatically fulfilled if c is nottoo large (see [11]). Especially let f = 0. The model is then given by

Problem 2.2 (Classical Formulation with special boundary conditions).Find

u ∈ C2(Ω3D) ∩ C(∂Ω3D), p ∈ C1(Ω3D)

such that∇p− div τ = 0 in Ω3D

τ = 2µD(u) + τcD(u)/(DII)1/2 ⇔ τ

1/2II > τc

D(u) = 0 ⇔ τ1/2II ≤ τc

div u = 0 in Ω3D

u = 0 on Γ3D

D

u(x, y, 0) = u(x, y, 1) on Γ3D

P

p(x, y, 0) = p(x, y, 1) + c on Γ3D

P .

The following results can be found in [11].

Notation. For u = (u1, u2) we denote the absolute value by | · |, i.e.

|u| =√

u21 + u2

2.

Under the assumptions made on the flow, it follows that our velocityfield has the form (0, 0, u3(x, y)). So our strain-rate tensor can be writtenas

D(u) =

0 0 12

∂u3

∂x

0 0 12

∂u3

∂y12

∂u3

∂x12

∂u3

∂y 0

(2.3)

7

and therefore the deviatoric stress tensor τ has the special form

τ =

0 0 τ130 0 τ23τ31 τ32 0

,

where the entries τij only depend on x, y. If we consider now problem 2.2,the first three equations reduces to

∂p

∂x= 0 (2.4)

∂p

∂y= 0 (2.5)

∂p

∂z=

∂τ13∂x

+∂τ32∂y

. (2.6)

If we take a look at (2.6), the left hand side depends only on z because of(2.4) and (2.5), whereas the right hand side depends only on x, y. Thereforethe pressure must be of the form

p = kz, k ∈ R

and as we assumed boundary conditions for p, this leads to

p = −cz.

In the case τ1/2II ≤ τc it follows from (2.3) that D(u) = 0 reduces to

∇u3 = 0.

Because of the special form of the flow, the incompressibility condition isautomatically fulfilled. Now together with

D1/2II =

1

2|∇u3|

and (2.3) it follows

τ13 = µ∂u3

∂x+ τc|∇u3|−1∂u3

∂x,

τ23 = µ∂u3

∂y+ τc|∇u3|−1∂u3

∂y

for the case τ1/2II > τc. Now we can write down the classical formulation of

our reduced problem:

8

Problem 2.3 (Classical formulation for the two-dimensional case). Find

u ∈ C2(Ω) ∩ C(∂Ω)

such that−µ∆u− τc div

(∇u|∇u|

)

= c ⇔ τ1/2II > τc

∇u = 0 ⇔ τ1/2II ≤ τc

u = 0 on ∂Ω = Γ

9

Chapter 3

Variational formulations

We are now going to derive weak formulations for which we will have exis-tence and uniqueness of a solution. We will formulate variational problemsfor the 3D problem 2.2 as well as for the reduced 2D problem 2.3. Furtherwe will give dual formulations for both.

Throughout this thesis let V, V 3D, Q be Hilbert spaces with scalar prod-ucts (·, ·)V ,(·, ·)V 3D , (·, ·)Q inducing norms, ‖ · ‖V ,‖ · ‖V 3D and ‖ · ‖Q respec-tively. With V ∗,V 3D∗,Q∗ we will denote their dual spaces, where 〈·, ·〉 is theduality pairing. Moreover we will denote the L2- scalar product by (·, ·).Notation. We will denote the H1-norm by ‖ · ‖1 and the H1-semi normwith | · |1 respectively. The L2-norm will be denoted as ‖ · ‖0.

3.1 Primal formulations

3.1.1 The three-dimensional problem

For the sake of simplicity we will denote the vector valued solution u3D ofthe three-dimensional problem also by u. Note that in the next subsectionu is the scalar solution of the reduced two-dimensional problem. For thederivation of the variational formulation we used [20],[21],[11].

We choose arbitrary but fixed functions v ∈ V 3D

0 and q ∈ Q respectively,fulfilling

v = 0 on Γ3D

D ,

andv|P1

= v|P2on Γ3D

P .

Then v−u is a test function in V 3D

0 . The test spaces V 3D

0 and Q are specifiedlater. Next we introduce Ω3D

1 ,Ω3D

0 in such a way that

Ω3D = Ω3D

1 ∪ Ω3D

0 , Ω3D

1 ∩ Ω3D

0 = ∅, (3.1)

10

whereΩ3D

1 := x ∈ Ω3D∣∣D(u) 6= 0

andΩ3D

0 := x ∈ Ω3D∣∣D(u) = 0.

So Ω3D

0 is the region where τ1/2II ≤ τc, i.e. the fluid acts like a rigid medium.

Ω3D

1 is then the set where τc is passed, so where it acts like a fluid. Notethat either Ω3D

1 or Ω3D

0 can be empty. If we multiply now our equations byv − u ∈ V 3D

0 , q ∈ Q and integrate over the whole domain, we get for theincompressibility condition

∫

Ω3D

q div u dx = 0 ∀q ∈ Q. (3.2)

and for the balance equations∫

Ω3D

−divσ · (v − u) dx =

∫

Ω3D

f · (v − u) dx ∀v ∈ V 3D

0 . (3.3)

By Gauss’ theorem and the fact that σ is a symmetric tensor, the integralon the left hand side can be written as

∫

Ω3D

σ : D(v − u) dx−∫

∂Ω3D

(σn) · (v − u) ds. (3.4)

So (3.3) is equivalent to

−∫

Ω3D

p div(v − u) dx+

∫

Ω3D

τ : D(v − u) dx−∫

∂Ω3D

(σn) · (v − u) ds

=

∫

Ω3D

f · (v − u) dx ∀ v ∈ V 3D

0 , (3.5)

where we have used that the trace of D is just the divergence.

Now let us take a look on the second integral. According to the linearityof D(·) we have

∫

Ω3D

τ : D(v − u) dx =

∫

Ω3D

τ : D(v) dx−∫

Ω3D

τ : D(u) dx. (3.6)

From (3.1) we get∫

Ω3D

τ : D(v) dx =

∫

Ω3D1

τ : D(v) dx+

∫

Ω3D0

τ : D(v) dx.

Now in Ω3D

1 we can use the definition of the tensor τ so that we have∫

Ω3D1

τ : D(v) dx = 2µ

∫

Ω3D1

D(u) : D(v) dx+τc

∫

Ω3D1

D−1/2II (u)D(u) : D(v) dx.

11

With the well known Cauchy-inequality it follows that

D(u) : D(v) ≤ 2D1/2II (u)D

1/2II (v), (3.7)

we further have

τc

∫

Ω3D1

D−1/2II (u)D(u) : D(v) dx ≤ τc

∫

Ω3D1

2D1/2II (v) dx

so that∫

Ω3D1

τ : D(v) dx ≤ 2µ

∫

Ω3D1

D(u) : D(v) dx+ τc

∫

Ω3D1

2D1/2II (v) dx.

In Ω3D

0 we use the fact that τ1/2II ≤ τc. So applying inequality (3.7) once

more we have∫

Ω3D0

τ : D(v) dx ≤∫

Ω3D0

2τ1/2II D

1/2II (v) dx ≤ τc

∫

Ω3D0

2D1/2II (v) dx.

Moreover

2µ

∫

Ω3D1

D(u) : D(v) dx = 2µ

∫

Ω3D

D(u) : D(v) dx,

because of the definition of Ω3D

0 . So summarizing we get∫

Ω3D

τ : D(v) dx ≤ 2µ

∫

Ω3D

D(u) : D(v) dx+ τc

∫

Ω3D

2D1/2II (v) dx.

Using the same technique for the second integral in (3.6), we obtain∫

Ω3D

τ : D(u) dx ≤ 2µ

∫

Ω3D

D(u) : D(u) dx+ τc

∫

Ω3D

2D1/2II (u) dx.

Since Cauchy’s inequality is an equation if its arguments are the same, weeven have

∫

Ω3D

τ : D(u) dx = 2µ

∫

Ω3D

D(u) : D(u) dx+ τc

∫

Ω3D

2D1/2II (u) dx.

In the end we get

∫

Ω3D

τ : D(v − u) dx ≤ 2µ

∫

Ω3D

D(u) : D(v − u) dx

+ τc

∫

Ω3D

2D1/2II (v) dx− τc

∫

Ω3D

2D1/2II (u) dx. (3.8)

Now, let us take a look on the boundary integral in (3.4). There holds∫

Γ3D

(σn) · (v − u) ds =

∫

Γ3D

D

(σn) · (v − u)︸ ︷︷ ︸

=0

ds+

∫

Γ3D

P

(σn) · (v − u) ds.

12

Especially∫

Γ3D

P

(σn) · (v − u) ds =

∫

Γ3D

P

(pI)n · (v − u) ds+

∫

Γ3D

P

τn · (v − u) ds.

Using the identity n(x, y, 0) = −n(x, y, 1) as well as boundary conditions,we have∫

Γ3D

P

(pI)n · (v − u) ds =

∫

Ωpn|z=0

· (v − u)|z=0ds+

∫

Ωpn|z=1

· (v − u)|z=1ds

=

∫

Ω

(p|z=0

− p|z=1

)n|z=0

· (v − u)|z=0ds

=

∫

Ωcn|z=0

(v − u)|z=0ds (3.9)

and further∫

Γ3D

P

τn · (v − u) ds =

∫

Ω(τn)|z=0

· (v − u)|z=0ds+

∫

Ω(τn)|z=1

· (v − u)|z=1ds

= 0.

Hence, together with (3.5),(3.8) we have

2µ

∫

Ω3D

D(u) : D(v − u) dx−∫

Ω3D

p div (v − u) dx

+ τc

∫

Ω3D

2D1/2II (v) dx− τc

∫

Ω3D

2D1/2II (u) dx

≥∫

Ω3D

f · (v − u) dx+

∫

Ωcn|z=0

(v − u)|z=0ds. (3.10)

We define bilinear forms

a(·, ·) : V 3D

0 × V 3D

0 → R, a(u, v) =

∫

Ω3D

2D(u) : D(v) dx

b1(·, ·) : V 3D

0 ×Q→ R, b1(v, p) =

∫

Ω3D

p div v dx

as well as the continuous, convex but non-differentiable functional

j(·) : V → R, j(v) =

∫

Ω3D

2D1/2II (v) dx

and the linear, continuous operator

F ∈ V 3D

0∗, 〈F, v − u〉 =

∫

Ω3D

f · (v − u) dx+

∫

Ωcn|z=0

(v − u)|z=0ds.

13

System (3.10),(3.2) can then be formulated as

µa(u, v − u) + b1(v − u, p) + τcj(v) − τcj(u) ≥ 〈F, v − u〉 ∀v ∈ V 3D

0

b1(u, q) = 0 ∀q ∈ Q.(3.11)

We need that the first weak derivatives of u, v exist, and that they are inL2. Moreover p, q have to be in L2. Appropriate choices for our spaces aretherefore

V 3D = [H1(Ω3D)]3

V 3D

0 = v ∈ [H1(Ω3D)]3∣∣ v = 0 on Γ3D

D , v(x, y, 0) = v(x, y, 1)Q = L2(Ω3D)

So we have showed, that the balance equations imply a so called variationalinequality of the second kind.

Remark. Note that, although the constitutive law is dealing with two equiv-alences, the weak formulation (3.11) includes both cases: the one where the

medium acts like a fluid and the case where τ1/2II is under the given threshold

τc.

Now we show that from (3.11), we obtain back the balance equations aswell as the incompressibility condition. If D(u) 6= 0, then the functional j(·)is differentiable with directional derivative (using the chain rule)

〈j′(u), v〉 =d

dλj(u+ λv)|λ=0

=

∫

Ω3D

D−1/2II (u)D(u) : D(v) dx. (3.12)

Next, choosing test functions u + λv with λ > 0, the inequality in (3.11)equals

µa(u, v) + b1(v, p) + τc1

λ(j(u+ λv) − j(u)) ≥ 〈F, v〉,

after dividing the whole inequality by λ. Now letting λ→ 0 we get

µa(u, v) + b1(v, p) + τc〈j′(u), v〉 ≥ 〈F, v〉.

Taking now test functions ±v, it follows

µa(u, v) + b1(v, p) + τc〈j′(u), v〉 = 〈F, v〉.

Using (3.12), the application of Gauss’ theorem then yields

−∫

Ω3D

2µdiv (D(u)) · v dx+

∫

Ω3D

∇p · v dx

− τc

∫

Ω3D

div(

D−1/2II (u)D(u)

)

· v dx+

∫

∂Ω3D

(σn) · v ds

=

∫

Ω3D

f · v dx+

∫

Ωcn|z=0

v|z=0ds ∀v ∈ V 3D

0 . (3.13)

14

Since (3.13) is valid for every test function v, we can choose v such thatv = 0 on ∂Ω3D. So for D(u) 6= 0 this leads to

−div (2µD(u)) − div

(

τcD(u)

D1/2II (u)

)

+ ∇p = f. (3.14)

And therefore the deviatoric stress tensor has the form

τ = 2µD(u) + τcD−1/2II (u)D(u). (3.15)

In this case the tensor of strain rates is given by

D(u) =1

2µ(1 − τc

τ1/2II

)τ. (3.16)

Now from (3.15) it follows

τ1/2II = τc + 2µD

1/2II (u) > τc.

Expression (3.15) makes no sense if D(u) = 0. In this case τ1/2II ≤ τc has to

hold since otherwise we get from (3.16) that DII(u) > 0 in contradiction toD(u) = 0.

Now we know that (3.14) is valid, we get from (3.13)

∫

∂Ω3D

(σn) · v ds−∫

Ωcn|z=0

v|z=0ds = 0.

So it follows with the same argumentation used in (3.9) that

p(x, y, 0) = p(x, y, 1) + c on Γ3D

P .

Finally, the equation in (3.11) holds for every test function q from which theincompressibility condition

div u = 0 (3.17)

follows. Now from (3.14),(3.17) we obtain our classical formulation.

Summarizing our results, under additional integrability as well as differ-entiability conditions, we have seen that problem 2.3 is equivalent to:

Problem 3.1 (Mixed formulation). Find u ∈ V 3D

0 , p ∈ Q, such that

µa(u, v − u) + b1(v − u, p) + τcj(v) − τcj(u) ≥ 〈F, v − u〉 ∀v ∈ V 3D

0

b1(u, q) = 0 ∀q ∈ Q.

15

with the bilinear forms

a(·, ·) : V 3D

0 × V 3D

0 → R, a(u, v) =

∫

Ω3D

2D(u) : D(v) dx,

b1(·, ·) : V 3D

0 ×Q→ R, b1(v, p) =

∫

Ω3D

p div v dx,

the continuous, convex but non-differentiable functional

j(·) : V 3D

0 → R, j(v) =

∫

Ω3D

2D1/2II (v) dx

and the linear, continuous operator

F ∈ V 3D

0∗, 〈F, v − u〉 =

∫

Ω3D

f · (v − u) dx+

∫

Ω(cI)n|z=0

(v − u)|z=0ds.

The solutions u, p are searched in

V 3D

0 = v ∈ [H1(Ω3D)]3∣∣ v = 0 on Γ3D

D , v(x, y, 0) = v(x, y, 1)Q = L2(Ω3D).

3.1.2 The two-dimensional case

In the same way we can find an equivalent variational formulation for theproblem reduced to the cross section Ω. As we mentioned in the previoussubsection, the scalar solution of our problem will be also denoted by u.Analogously we get the following

Problem 3.2 (Primal formulation). Find u ∈ V0, such that

µa2(u, v − u) + τcj(v) − τcj(u) ≥ (c, v − u) ∀ v ∈ V0

with the bilinear form

a2(·, ·) : V0 × V0 → R, a2(u, v) =

∫

Ω∇u · ∇v dx

and the continuous, convex but non-differentiable functional

j(·) : V0 → R, j(v) =

∫

Ω|∇v| dx.

The spaces V, V0 are defined by

V = H1(Ω)

V0 = v ∈ V∣∣ v = 0 on Γ = H1

0 (Ω).

16

3.2 Dual formulations

To overcome the difficulties caused by the non-differentiable functional j(·),one usually does a regularization somewhere or, in our case, one derivesanother formulation in which the non-differentiability vanishes. The dualformulation given in the next subsections can be found in [11],[14].

3.2.1 The three-dimensional problem

Notation. The Frobenius norm of a tensor λ is given by

‖λ‖F := (λijλij)1/2.

Theorem 3.2.1. With the same notations as in the previous section prob-lem 3.1 is equivalent to the following dual dual formulation:

Problem 3.3 (Dual dual formulation). Find u ∈ V 3D

0 , p ∈ Q, λ ∈ Λ suchthat

µa(u, v − u) + b1(v − u, p) + τc√

2b2(v − u, λ) = 〈F, v − u〉 ∀v ∈ V 3D

0

b1(u, q) = 0 ∀q ∈ Qb2(u, η − λ) ≤ 0 ∀η ∈ Λ

where

b2(·, ·) : V 3D

0 × Λ → R, b2(v, λ) =

∫

Ω3D

λ : D(v) dx

and

V 3D

0 = v ∈ [H1(Ω3D)]3∣∣ v = 0 on Γ3D

D , v(x, y, 0) = v(x, y, 1)Q = L2(Ω3D)

Λ = λ ∈ [L2(Ω3D)]3×3∣∣λij = λji, ‖λ‖F ≤ 1, i, j = 1, 2, 3

Proof. The proof can be found in [11] for a more general case, or in [14] forthe two-dimensional case. First we show that

b2(u, η − λ) =

∫

Ω3D

(η − λ) : D(u) dx ≤ 0 (3.18)

is equivalent to

∫

Ω3D

λ : D(u) dx =

∫

Ω3D

‖D(u)‖F dx. (3.19)

17

”⇐ ”:From (3.19) we get that

∫

Ω3D

(η − λ) : D(u) dx =

∫

Ω3D

η : D(u) dx−∫

Ω3D

‖D(u)‖F dx

≤∫

Ω3D

‖η‖F︸ ︷︷ ︸

≤1

: ‖D(u)‖F dx−∫

Ω3D

‖D(u)‖F dx

≤ 0 ∀η ∈ Λ.

”⇒ ”:On the other hand if we choose η = D(u)/‖D(u)‖F ∈ Λ,

∫

Ω3D

(η − λ) : D(u) dx ≤ 0 ⇒∫

Ω3D

D(u) : D(u)

‖D(u)‖Fdx ≤

∫

Ω3D

λ : D(u) dx,

from which we get

∫

Ω3D

(D(u) : D(u))1/2 dx ≤∫

Ω3D

λ : D(u) dx.

Since we get from Cauchy’s inequality that

∫

Ω3D

λ : D(u) dx ≤∫

Ω3D

‖λ‖F · ‖D(u)‖F dx ≤∫

Ω3D

‖D(u)‖F dx,

(3.19) follows. So the equivalence is shown.

Now let us take arbitrary but fixed test functions v ∈ V 3D

0 , q ∈ Q. Fromproblem 3.3 it follows that

µa(u, v − u) + b1(v − u, p) + τcj(v) − τcj(u) − 〈F, v − u〉=τc

(

j(v) − j(u) −√

2(λ,D(v) −D(u)))

.

Since (3.19) holds, we get

j(u) =

∫

Ω3D

2

(1

2D(u) : D(u)

)1/2

dx =√

2(λ,D(u))

and together with

√2(λ,D(v)) ≤

√2

∫

Ω3D

‖λ‖F ‖D(v)‖F dx ≤ j(v)

we obtainτc

(

j(v) − j(u) −√

2(λ,D(v) −D(u)))

≥ 0.

18

Next we assume problem 3.1. With the setting

(Y, v) = µa(u, v) + b1(v, p) − 〈F, v〉

we can rewrite problem 3.1 as

(Y, v) + τcj(v) − ((Y, u) + τcj(u)) ≥ 0 ∀v ∈ V 3D

0 .

Now we choose test functions v = ±tw with t > 0 and w ∈ V 3D

0 . Then

t (±(Y,w) + τcj(w)) − ((Y, u) + τcj(u)) ≥ 0 ∀w ∈ V 3D

0 . (3.20)

From this, tending t to zero, it follows that

(Y, u) + τcj(u) ≤ 0. (3.21)

On the other hand, choosing test functions v = 1t (u−w) we get from (3.20)

that±(Y,w) ≥ τc(j(u) − j(u− w)) ≥ −τcj(w),

since the inequalityj(w − u) ≤ j(w) + j(u)

holds. In other words we get, that

|(Y,w)| ≤ τcj(w) ∀w ∈ V 3D

0 . (3.22)

From this automatically

(Y, u) + τcj(u) ≥ 0 (3.23)

follows. Because of (3.21),(3.23)

(Y, u) + τcj(u) = 0 (3.24)

must be true. Next we introduce the space

Φ = φ∣∣φ ∈ [L1(Ω3D)]3×3, φij = φji

equipped with the norm

‖φ‖Φ =

∫

Ω3D

‖φ‖F dx

as well as the mapping

π : V 3D

0 → Φ, w 7→ πw = D(w).

With these notations we can rewrite (3.22) as

|(Y,w)| ≤ τc√

2‖πw‖Φ.

19

Now, applying the theorem of Hahn-Banach yields that there exists

λ ∈ Φ∗ = φ∣∣φ ∈ [L∞(Ω3D)]3×3, φij = φji

such that(Y,w) = −τc

√2(λ,D(w)) (3.25)

and|(λ, πw)| ≤ ‖πw‖Φ. (3.26)

But (3.26) is equivalent to

‖λ‖Φ∗ = supπw∈Φ

|(λ, πw)|‖πw‖Φ

≤ 1

and (3.25) equals

µa(u,w − u) + b1(w − u, p) + τc√

2b2(w − u, λ) = 〈F,w − u〉 ∀w ∈ V 3D

0 .

Finally (3.24) yields∫

Ω3D

λ : D(u) dx =

∫

Ω3D

(D(u) : D(u))1/2 dx.

and therefore (3.18) follows.

Remark. As we have seen in the proof, it follows from the mixed problemnot only that λ is an element of [L2(Ω3D)]3×3. In fact it follows that λ is an[L∞(Ω3D)]3×3 function.

3.2.2 The two-dimensional case

It is clear that we can derive an equivalent dual formulation for the two-dimensional case (the proof can be found in [14]). Again u should be thesame variable for the 2D case. Then for the problem on the cross section wehave the following

Problem 3.4 (Dual formulation in the 2D case). Find u ∈ V0,w ∈W suchthat

µa2(u, v − u) + τcb(v − u,w) = (c, v − u) ∀v ∈ V0

b(u, φ− w) ≤ 0 ∀φ ∈W

with the bilinear form

b(·, ·) : V0 ×W → R, b(v, w) =

∫

Ωw · ∇v dx

and the convex set

W = w = (w1, w2) ∈ [L2(Ω)]2∣∣ |w| ≤ 1,

where |w| =√

w21 + w2

2.

20

Equivalence to the three-dimensional problem

We will show that the dual formulations in two, as well as in three dimensionsare equivalent if some extra settings hold. For this we will show that asolution (u,w) of the two-dimensional problem 3.4 can be extended to asolution of the three-dimensional problem 3.3.

Theorem 3.2.2. Let u3 ∈ V0 and w = (λ13, λ23) ∈ W be a solution of thetwo-dimensional problem 3.4, then

u = (0, 0, u3(x, y)), λ =1√2

0 0 λ13(x, y)

0 0 λ23(x, y)

λ31(x, y) λ32(x, y) 0

is a solution of the three-dimensional problem 3.3 in which f = 0 andp = −cz.Proof. First it is easy to see, that every test function v ∈ V 3D

0 can be writtenas

v(x, y, z) =

v1(x, y, z)v2(x, y, z)

v3(x, y, z) − v(x, y)

−

00

v(x, y)

=: v + v

where v is the mean value of v3 concerning the z-component,i.e.

v(x, y) =

∫

[0,1]v3(x, y, z) dz.

Such a decomposition is now chosen. We then have the following computa-tions:

a(u, v) =

=

∫

Ω×[0,1]2

0 0 12

∂u3

∂x

0 0 12

∂u3

∂y12

∂u3

∂x12

∂u3

∂y 0

: D(v) dx

=

∫

Ω×[0,1]21

2(∂u3

∂x(∂v1∂z

+∂v3∂x

) +∂u3

∂y(v2∂z

+∂v3∂y

)) dx

=

∫

Ω

∫

[0,1]

∂u3

∂x

∂v1∂z

+∂u3

∂x

∂v3∂x

− ∂u3

∂x

∂v

∂x+∂u3

∂y

∂v2∂z

+∂u3

∂y

∂v3∂y

− ∂u3

∂y

∂v

∂ydz dx

u3=u3(x,y)=

∫

Ω

∫

[0,1]

∂u3

∂x

∂v1∂z

+∂u3

∂y

∂v2∂z

dz dx

=

∫

Ω

∂u3

∂x

∫

[0,1]

∂v1∂z

dz dx +

∫

Ω

∂u3

∂y

∫

[0,1]

∂v2∂z

dz dx

=

∫

Ω

∂u3

∂x(v1(x, y, 1) − v1(x, y, 0)) dx +

∫

Ω

∂u3

∂y(v2(x, y, 1) − v2(x, y, 0)) dx

v∈V 3D0= 0,

21

and

a(u, v) =

∫

Ω×[0,1]2D(u) : D(v) dx

=

∫

Ω

∫

[0,1]

∂u3

∂x

∂v

∂x+∂u3

∂y

∂v

∂ydz dx

=

∫

Ω

∂u3

∂x

∂v

∂x+∂u3

∂y

∂v

∂ydx

=

∫

Ω∇u3 · ∇v dx

= a2(u3, v).

Moreover we have

b1(v, p) = −∫

Ω×[0,1]p div v dx

= −∫

Ω×[0,1]p div v dx

= −∫

Ω×[0,1]c∇z · v dx−

∫

∂Ω3D

pv · ndx

= −∫

Ω×[0,1]cv3 dx−

∫

∂Ω3D

pv · ndx.

Now, according to the periodic boundary conditions we have

b1(v, p) = −∫

Ωcv dx−

∫

Ωcv|z=0

· n|z=0dx.

For the second component of v we get

b1(v, p) = −∫

Ω×[0,1]cz∂v

∂zdx = 0.

Clearly there holdsb1(u, q) = 0.

Further

b2(v, λ) =1√2

∫

Ω×[0,1]λ13(

∂v1∂z

+∂v3∂x

− ∂v

∂x) + λ23(

∂v2∂z

+∂v3∂y

− ∂v

∂y) dx.

Since the statements∫ 1

0λ13(x, y)

∂v3∂x

dz = λ13∂

∂x

∫ 1

0v3 dz = λ13

∂v

∂x

and ∫ 1

0λ13

∂v1∂z

dz = λ13v1(x, y, 1) − λ13v1(x, y, 0) = 0

22

are true (and also for the second component of v), we get that

b2(v, λ) =1√2

∫

Ω

∫

[0,1]λ13

∂v1∂z

+ λ23∂v2∂z

dz dx

= 0.

For the other component of the test function we have

b2(v, λ) =1√2

∫

Ω

∫

[0,1]λ13

∂v

∂x+ λ23

∂v

∂ydz dx

=1√2

∫

Ωλ13

∂v

∂x+ λ23

∂v

∂ydx

=1√2

∫

Ωw · ∇v dx

=1√2b(v, w).

In the same manner we get

b2(u, µ− λ) =1√2b(u3, φ− w).

where φ = (φ13, φ23) ∈W and

µ =1√2

0 0 φ13(x, y)0 0 φ23(x, y)

φ31(x, y) φ32(x, y) 0

.

Now we use our assumption. The two-dimensional problem is satisfied andtherefore the following must hold:

µa2(u3, v) + τcb(v, w) = (c, v).

Now plugging in our results from above we get (note that f = 0)

µa(u, v) + b1(v, p) + τc√

2b2(v, λ) = 〈F, v〉

and from the inequalityb(u3, φ− w) ≤ 0,

we getb2(u, µ− λ) ≤ 0.

23

3.3 Existence and uniqueness of a solution

In this section we are going to list some results for mixed variational in-equalities in an abstract way. Later we will apply these theorems to ourproblems. The following assumptions as well as the theorems can be foundin [17]. Proofs and details are cited from [26],[13].

Let X,Y be Hilbert spaces equipped with scalar products (·, ·)X and(·, ·)Y respectively. Their norms, induced by the scalar products, are denotedby ‖ · ‖X and ‖ · ‖Y . For the dual spaces we will use the notation X⋆ andY ⋆. Further on, let 〈·, ·〉 be the duality pairing. We introduce

a(·, ·) : X ×X → R, b(·, ·) : X × Y → R, j(·) : X → R

f ∈ X⋆, g ∈ Y ⋆

then a variational inequality of the mixed type has the following abstractform:

Problem 3.5 (Abstract mixed formulation). Given f ∈ X⋆, g ∈ Y ⋆, find(u, p) ∈ X × Y such that

a(u, x− u) + b(x− u, p) + j(x) − j(u) ≥ 〈f, x− u〉 ∀x ∈ Xb(u, y) = 〈g, y〉 ∀ y ∈ Y.

We will consider the linear case, so a(·, ·), b(·, ·) are bilinear forms. Weassume that a(·, ·) as well as b(·, ·) are bounded, i.e. there exist constantsα, β1 > 0, such that

a(x1, x2) ≤ α‖x1‖X‖x2‖X ∀x1, x2 ∈ X

andb(x, y) ≤ β1‖x‖X‖y‖Y ∀x ∈ X, ∀ y ∈ Y.

By defining the linear operator

A : X → X⋆, 〈Ax, y〉 = a(x, y) ∀x ∈ X, ∀ y ∈ X

as well as the linear operator

B : X → Y ⋆, 〈Bx, y〉 = b(x, y) ∀x ∈ X, ∀ y ∈ Y

and its adjoint operator

B⋆ : Y → X⋆, 〈B⋆y, x〉 = b(x, y) ∀x ∈ X, ∀ y ∈ Y

the abstract form of a mixed formulation looks like the following way

24

Problem 3.6 (Abstract mixed formulation in operator form). Givenf ∈ X⋆, g ∈ Y ⋆, search (u, p) ∈ X × Y :

〈Au, x− u〉 + 〈B⋆p, x− u〉 + j(x) − j(u) ≥ 〈f, x− u〉 ∀x ∈ X〈Bu, y〉 = 〈g, y〉 ∀ y ∈ Y.

Definition. Let B be an operator which maps from X onto Y . The kernelof B is defined via

Ker B := x ∈ X : Bx = 0.

The operator A is assumed to be coercive on Ker B, i.e. there existsα0 > 0 such that

〈Ax, x〉 ≥ α0‖x‖2X , ∀x ∈ Ker B. (3.27)

For the bilinear form b(·, ·) we assume that the inf-sup condition holds: thereexists a constant β > 0, such that

infy∈Y

supx∈X

b(x, y)

‖x‖X‖y‖Y≥ β. (3.28)

Definition. A function f(·) is called lower semi continuous if for all a theset

x∣∣ f(x) ≤ a

is closed.

Definition. A function f(·) is called proper if it is not identically infinity,i.e.

f 6≡ ∞.

Finally we assume j(·) to be a non-negative, continuous, convex, properlower semi-continuous function.

Remark. It is important to know that we does not need differentiability forj(·).

3.3.1 Existence and uniqueness for the mixed problem

Definition. The polar set X0 ⊂ Y ⋆ of a subspace X ⊂ Y is defined by

X0 := y ∈ Y ⋆ : 〈y, x〉 = 0 ∀x ∈ X.

Definition. The orthogonal complement K⊥ of a set K ⊂ X is defined thefollowing way

K⊥ := x ∈ X : (x, y)X = 0 ∀y ∈ K.

25

Notation. WithIm A

we will denote the range (the image) of the operator A.

Theorem 3.3.1 (Closed Range Theorem). Let X,Y be Hilbert spaces,A : X → Y ⋆ a linear, continuous operator and let A⋆ : Y → X⋆ be the ad-joint operator, defined via 〈A⋆y, x〉 = 〈Ax, y〉. Then the following statementsare equivalent:

(a) Im A is closed,

(b) Im A⋆ is closed,

(c) Im A = (Ker A⋆)0,

(d) Im A⋆ = (Ker A)0.

Proof. The proof can be found in [33].

The following lemma gives us the possibility to use the inf-sup conditionin another way. It can be found in [13],[34].

Lemma 3.3.2. The three following properties are equivalent:

1. There exist a constant β > 0, such that

infy∈Y

supx∈X

b(x, y)

‖x‖X‖y‖Y≥ β. (3.29)

2. The operator B⋆ is an isomorphism from Y onto (Ker B)0 and

‖B⋆y‖X⋆ ≥ β‖y‖Y ∀ y ∈ Y. (3.30)

3. The operator B is an isomorphism from (Ker B)⊥ onto Y ⋆ and

‖Bx‖Y ⋆ ≥ β‖x‖X ∀x ∈ X. (3.31)

Proof. We show first (1) ⇔ (2):

Due to the definition of the dual operator B⋆, (3.29) is equivalent to

supx∈X

〈B⋆y, x〉‖x‖X

≥ β‖y‖Y

26

where the left hand side is the definition of the norm on X⋆, so that theinequality equals (3.30). From that, we get that B⋆ is injective. Furthermoreit also implies that the inverse of B⋆ is continuous. Using the continuity ofB⋆ as well as (3.30), it follows that Im B⋆ is a closed subspace of X⋆.Therefore it follows from the closed range theorem 3.3.1 that

Im B⋆ = (Ker B)0

what means that B⋆ is also surjective and therefore an isomorphism.

We prove (2) ⇔ (3):The equivalence of the isomorphisms is a direct consequence of theorem3.3.1. There rests the equivalence of the inequalities:We assume (3.30) and take an arbitrary x ∈ (Ker B)⊥. Then (x, .)X ∈(Ker B)0 and since we assumed (2), there exists y ∈ Y with B⋆y = (x, .)X .It follows then

‖Bx‖Y ⋆ ≥ b(x, y)

‖y‖Y=

‖x‖2X

‖y‖Y=

‖(x, .)X‖X⋆

‖y‖Y‖x‖X =

‖B⋆y‖X

‖y‖Y‖x‖X ≥ β‖x‖X .

We assume (3.31) and take an arbitrary y ∈ Y , then it follows that (y, .)Y ∈ Y ⋆.So there exists x ∈ (Ker B)⊥ such that Bx = (y, .)Y . It follows then

‖B⋆y‖X⋆ ≥ b(x, y)

‖x‖X=

‖y‖2Y

‖x‖X=

‖(y, .)X‖X⋆

‖x‖X‖y‖Y =

‖Bx‖Y

‖x‖X‖y‖Y ≥ β‖y‖Y .

According to Lemma 3.3.2, there exists a unique solution u0 ∈ (Ker B)⊥

ofBu0 = g in Y ⋆. (3.32)

We are now going to prove existence and uniqueness for the mixed prob-lem 3.6. For this we introduce a variational inequality of the second kindon the kernel Ker B, which provides a unique solution for the original prob-lem. Let u0 being the solution of (3.32). Then we introduce the followingauxiliary problem:

Problem 3.7 (Auxiliary problem on Ker B). Find w ∈ Ker B such that

〈Aw, x− w〉 + j(x+ u0) − j(w + u0) ≥ 〈f , x− w〉 ∀x ∈ Ker B.

with〈f , x− w〉 = 〈f, x− w〉 − 〈Au0, x− w〉.

Remark. If we can find a unique solution w ∈ Ker B of problem 3.7, thenu = u0 + w ∈ X is a unique solution of problem 3.6.

27

Since the functional x 7→ j(x+u0) is convex and continuous from Ker Bto R, f is bounded and we can apply the following theorem:

Theorem 3.3.3. Let A be a bounded, linear and coercive operator. Let f bea bounded functional. Further let j be a non-negative, continuous, convex,proper and lower-semi continuous functional. Then the following variationalinequality of the second kind has a unique solution:Find u ∈ X such that

a(u, x− u) + j(x) − j(u) ≥ 〈f, x− u〉 ∀x ∈ X. (3.33)

The proof requires the next lemma.

Lemma 3.3.4. Let A be a bounded, linear and coercive operator. Let f bea bounded functional. Further let j be a non-negative, continuous, convex,proper and lower-semi continuous functional. Let ρ > 0 then for an arbitrarybut fixed u ∈ X, the problem of finding y ∈ X such that

(y, x− y) + ρj(x)− ρj(y) ≥ (u, x− y) + ρ〈f, x− y〉 − ρa(u, x− y) ∀x ∈ X

has a unique solution.

Proof. For the detailed proof see [14] and the references therein. The mainidea is the following: the problem is a variational inequality of the secondkind with the special bilinear form (·, ·). Since the inner product is clearlysymmetric and positive definite, it follows the equivalence to a minimizationproblem: find y ∈ X such that

Jρ(u, y) ≤ Jρ(u, x) ∀x ∈ X

with

Jρ(u, x) =1

2‖x‖2

X + ρj(x) − (u, x) + ρa(u, x) − ρ〈f, x〉.

With the assumptions on j it follows then from results out of convex analysisthat the optimization problem has a unique solution.

We are now able to proof theorem 3.3.3. It can be found in [14].

Proof of theorem 3.3.3. First we will prove uniqueness. Let u1, u2 be twosolutions of the variational problem,i.e.

a(u1, x− u1) + j(x) − j(u1) ≥ 〈f, x− u1〉 ∀x ∈ X (3.34)

a(u2, x− u2) + j(x) − j(u2) ≥ 〈f, x− u2〉 ∀x ∈ X (3.35)

From the above inequalities it follows

j(ui) ≤ a(ui, x− ui) + j(x) − 〈f, x− ui〉 i = 1, 2 ∀x ∈ X (3.36)

28

Since j(·) is proper, what implies that j 6≡ +∞, there exists x0 ∈ X suchthat j(x0) < +∞, so we take x = x0 in (3.36) and get that j(ui) is finite fori = 1, 2. Therefore we can choose special test functions x = u2 in (3.34) aswell as x = u1 in (3.35). Adding these two inequalities yields

a(u1, u2 − u1) + a(u2, u1 − u2) ≥ 0 ⇔ a(u2 − u1, u2 − u1) ≤ 0.

Since a(·, ·) is bounded, we get

α0‖u2 − u1‖2 ≤ 0

what is equivalent tou2 = u1.

Now for proving existence of a solution we formulate an auxiliary problem.For the meanwhile let u ∈ X arbitrary, then we define z as the solution offinding z ∈ X such that

(z, x− z) + ρj(x) − ρj(z) ≥ (u, x− z) + ρ〈f, x− z〉 − ρa(u, x− z) ∀x ∈ X(3.37)

with ρ > 0. Now lemma 3.3.4 provides a unique solution.

Defining the mapping

Sρ : X → X, u 7→ Sρ(u) = z,

one can see that (3.33) is equivalent to finding a solution u of the fixed pointproblem

u = Sρ(u). (3.38)

So for showing existence it is sufficient to prove that Sρ is a contractionmapping. With the notation Sρ(u1) = z1, Sρ(u2) = z2 and the choicex = z2 and x = z1 respectively we get from (3.37)

(z1, z2 − z1) + ρj(z2)− ρj(z1) ≥ (u1, z2 − z1) + ρ〈f, z2 − z1〉 − ρa(u1, z2 − z1)

and

(z2, z1 − z2)+ ρj(z1)− ρj(z2) ≥ (u2, z1 − z2)+ ρ〈f, z1 − z2〉− ρa(u2, z1 − z2).

Adding these two inequalities leads to

‖z2 − z1‖2 ≤ (u2 − u1, z2 − z1) − ρa(u2 − u1, z2 − z1). (3.39)

Now we use the operator A ∈ L(X,X) to rewrite (3.39).

‖z2−z1‖2 ≤ ((I − ρA)(u2 − u1), z2 − z1) ≤ ‖I−ρA‖L(X,X)‖u2−u1‖‖z2−z1‖

29

or‖z2 − z1‖ ≤ ‖I − ρA‖L(X,X)‖u2 − u1‖.

We assume that ρ satisfies

0 < ρ <2α0

‖A‖2(3.40)

then

‖(I − ρA)x‖2 = ‖x‖2 − 2ρ(x,Ax) + ρ2‖Ax‖2

≤ ‖x‖2 − 2α0ρ‖x‖2 + ρ2‖A‖2‖x‖2

= (1 − 2α0ρ+ ρ2‖A‖2)‖x‖2

Since (3.40) holds, it follows that (1 − 2α0ρ+ ρ2‖A‖2) < 1 and therefore

‖I − ρA‖ < 1.

So it follows

‖Sρ(u2) − Sρ(u1)‖ = ‖z2 − z1‖ ≤ ‖I − ρA‖‖u2 − u1‖.

As the contraction property of Sρ is shown, it follows from Banach’s fixedpoint theorem that (3.38) and thus (3.33) has a unique solution.

So problem 3.7 has a unique solution w ∈ Ker B. To find a solution ofthe mixed problem, we are going to approximate problem 3.6 as well as 3.7by regularizing the nondifferentiable functional j. It can be shown ([26])that we can introduce a family of functionals jε : X → R parameterized byε ∈ (0, 1] with the following properties:

a) jε is convex and differentiable with Gateaux derivative

j′ε : X → X⋆. (3.41)

b)jε(x) → j(x) for ε→ 0, (3.42)

uniformly with respect to x ∈ X.

c)‖j′ε(x)‖X⋆ ≤ c, ∀x ∈ X, (3.43)

where the constant c should be independent of ε and x.

Our new problem then reads as

30

Problem 3.8 (Regularized abstract mixed formulation). Givenf ∈ X⋆, g ∈ Y ⋆, search (uε, pε) ∈ X × Y :

〈Auε, x− uε〉 + 〈B⋆pε, x− uε〉 + jε(x) − jε(uε) ≥ 〈f, x− uε〉 ∀x ∈ X〈Buε, y〉 = 〈g, y〉 ∀ y ∈ Y.

The auxiliary problem is replaced by

Problem 3.9 (Regularized auxiliary problem). Find wε ∈ Ker B such that

〈Awε, x− wε〉 + jε(x+ u0) − jε(wε + u0) ≥ 〈f , x− wε〉 ∀x ∈ Ker B.

with〈f , x− wε〉 = 〈f, x− wε〉 − 〈Au0, x− wε〉.

Applying theorem 3.3.3 leads to the existence of a unique solutionwε ∈ Ker B for the auxiliary problem 3.9. Now, as jε is differentiable, itfollows with the same technique used in section 3.1, that problem 3.9 isequivalent to the problem of finding uε = wε + u0 ∈ X such that

〈Auε, x〉 + 〈j′ε(uε), x〉 = 〈f, x〉 ∀x ∈ Ker B. (3.44)

In the same manner the inequality of problem 3.8 is equivalent to

〈Auε, x〉 + 〈j′ε(uε), x〉 + 〈B⋆pε, x〉 = 〈f, x〉 ∀x ∈ X.

We observe that f −Auε − j′ε(uε) ∈ (Ker B)0, since (3.44) holds. Thereforewe can apply lemma 3.3.2 and get a unique pε ∈ Y such that

B⋆pε = f −Auε − j′ε(uε). (3.45)

Thus we have shown

Lemma 3.3.5. Problem 3.8 has a unique solution (uε, pε) ∈ X × Y .

So regularizing problems 3.6, 3.7 leads to a unique solution. Now theonly thing to do is, to show that for ε → 0 the limits of our solutions solvethe original problems. We will see that uε converges strongly to u. In thecase of the multiplier pε this result is unfortunately not given. We will haveonly weak convergence of a subsequence.

First let us choose special test functions x = wε in problem 3.7 as wellas x = w in problem 3.9. Then adding this two inequalities leads to

〈Awε − Aw,w − wε〉 + jε(u) − j(u) + j(uε) − jε(uε) ≥ 0.

With uε − u = wε − w, (3.42) and the boundedness of A we get

α0‖w − wε‖2 ≤ jε(u) − j(u)

≤ C2ε

31

or equivalently‖w − wε‖ ≤

√

C2ε/α0.

Thus we end up with the next lemma.

Lemma 3.3.6. The solution wε of problem 3.9 converges strongly to thesolution w of problem 3.7.

Next we will to show that our sequences uε, pε are bounded indepen-dently from ε. The result above sure gives us the desired attribute for uε,since ‖uε‖ ≤

√

C/α0 + ‖u‖. For pε we use the inf-sup condition and get

β‖pε‖Y ≤ sup0 6=x∈X

|〈B⋆pε, x〉|‖x‖X

(3.45)

≤ sup0 6=x∈X

|〈f −Auε − j′ε(uε), x〉|‖x‖X

≤ ‖f‖X⋆ + ‖A‖L(X,X⋆)‖uε‖X + ‖j′ε(uε)‖X⋆

≤ ‖f‖X⋆ + ‖A‖L(X,X⋆)(√

C/α0 + ‖u‖X) + C3

where we used the boundedness of uε as well as the assumption on ‖j′ε(uε)‖X⋆ .So it follows (see [19]) that there exists a subsequence pε′ of pε which con-verges weakly to a limit p ∈ Y . Since we have uε → u for every subsequenceuε′ , there clearly holds uε′ → u.

Finally we have to show that the limits of our subsequences really solvethe original problem. Starting from our approximate problem 3.8 and usingthe second assumption on j(·) we get

〈B⋆pε′ , x〉 + 〈Auε′ , x〉 − 〈f, x− uε′〉 + jε(x)

≥ 〈B⋆pε′ , uε′〉 + 〈Auε′ , uε′〉 + jε(uε′)

≥ 〈B⋆pε′ , uε′〉 + 〈Auε′ , uε′〉 + j(uε).

With the assumptions on A,B⋆, the convergence of the subsequences, thecontinuity of j(·) we get, after taking the limit ε′ → 0 on both sides,

〈Au, x− u〉 + j(x) − j(u) + 〈B⋆p, x− u〉 ≥ 〈f, x− u〉.

Together with (3.32) we have

Theorem 3.3.7. Let a(·, ·), b(·, ·) be bounded bilinear forms. For b(·, ·)we assume that the inf-sup condition (3.28) holds. Let the operator A becoercive. Further let j(·) be a non-negative, continuous, convex, proper andlower semi-continuous functional. Its regularization family jε(·) is assumedto fulfill (3.41)-(3.43). Then there exists a solution (u, p) ∈ X × Y forproblem 3.6. Furthermore, u is unique.

32

By regularizing the original problems 3.6 and 3.7, we have seen that theauxiliary problem reduces to an equation what was essential for the existenceand uniqueness of the regularized mixed problem. So showing existence anduniqueness of the mixed formulation requires finding a suitable regularizingfamily jε with the wright assumptions.

In [17] one can see that the theorem also holds for more general assump-tions.

3.3.2 Application to a fluid flow in a quadratic pipe

We will now prove that our mixed problem fulfills all assumptions, such thatwe can apply the theorems from the previous subsection. From now on weconsider the case

Γ3D = Γ3D

D .

So our mixed problem, derived in the previous section has the form

µa(u, v − u) + b1(v − u, p) + τcj(v) − τcj(u) ≥ (f, v − u) ∀v ∈ V 3D

0

b1(u, q) = 0 ∀q ∈ Q.

So in the notation of the previous subsection, we choose

X = V 3D

0 , Y = Q, b(·, ·) = b1(·, ·)and A, j stay the same.

We know that if we can show all assumptions, then there exists a uniquesolution u for the two-dimensional case and a solution u, p for the three-dimensional case where u is unique but p not. The non-uniqueness of pcomes from the fact that the kernel of the operator B1 is non-trivial. Itcontains all constant functions. So if we demand another assumption onthe space Q we get a uniquely defined p too. In the literature usually onedemands that ∫

Ω3D

p dx = 0. (3.46)

So we seek p ∈ L20(Ω

3D), where

L20(Ω

3D) := v ∈ L2(Ω3D)∣∣

∫

Ω3D

v dx = 0.

From now on we will define Q := L20(Ω

3D).

We define the operator

A : V 3D

0 → V 3D

0⋆, 〈Au, v〉 = a(u, v) =

∫

Ω3D

2D(u) : D(v) dx, ∀ v ∈ V 3D

0

then A is linear. The boundedness of the bilinear form a(·, ·) is simple. Forthe coercitivity of the operator A on V 3D

0 , we can use the following lemma:

33

Lemma 3.3.8 (Korn’s inequality). Let Ω3D ⊂ R3 an open bounded set with

a Lipschitz continuous boundary. Then there exists a constantcK = cK(Ω3D) > 0 such that

∫

Ω3D

D(v) : D(v) dx ≥ c2K‖v‖21 ∀ v ∈ [H1

0 (Ω3D)]3.

Proof. Can be found in [11].

So the operator A is elliptic. By Cauchy’s inequality the bilinear formb1(·, ·) is bounded, so we define the bounded linear operator

B1 : V 3D

0 → Q⋆, 〈B1v, p〉 = b1(v, p) = −∫

Ω3D

p div v dx, ∀ p ∈ Q

as well as the adjoint operator

B⋆1 : Q→ V 3D

0⋆, 〈B⋆

1p, v〉 = b1(v, p) = −∫

Ω3D

pdiv v dx, ∀ v ∈ V 3D

0 .

Now we will show that the inf-sup condition holds. This condition for theoperator B1 = −div is proved in several papers. First of all we have thefollowing

Definition. For the dual space of H10 , H−1 its norm is defined by

‖p‖−1 := sup0 6=q∈H1

0(Ω3D)

(p, q)

‖q‖1.

The next lemma is our starting point for proving the inf sup condition.

Lemma 3.3.9 (Necas). Let Ω3D ⊂ R3 be an open bounded set with a Lips-

chitz continuous boundary. Then there exists a constant c > 0 such that forall p ∈ L2(Ω3D)

‖p‖0 ≤ c(‖p‖−1 + ‖∇p‖−1).

Proof. The proof is rather technical and can be found in [11]. For an alter-native proof see [5].

With it we have the following theorem.

Theorem 3.3.10. Let Ω3D ⊂ R3 an open bounded set with a Lipschitz

continuous boundary. Then there exists a constant c > 0 such that for allp ∈ L2

0(Ω3D)

‖p‖0 ≤ c‖∇p‖−1 (3.47)

holds.

The proof is cited from [34].

34

Proof. As the embedding [H10 (Ω3D)]3 → L2(Ω3D) is compact, its adjoint

embedding L2(Ω3D) → H−1(Ω3D) is also compact. Now we assume that(3.47) doesn’t hold. Then there exists a sequence (pk) ∈ L2

0(Ω3D) such

that ‖pk‖0 = 1 and ‖∇pk‖−1 → 0. According to the compact embeddingL2(Ω3D) → H−1(Ω3D) there exists a convergent subsequence(p′k) ∈ H−1(Ω3D). From the last lemma we then have, that (p′k) is Cauchyin L2(Ω3D) and therefore p′k → p in L2(Ω3D) where p ∈ L2

0(Ω3D).

Moreover the following must hold

∇p = limk→∞

∇p′k = 0

and therefore p has to be constant. As p ∈ L20(Ω

3D) it follows that p = 0.On the other hand this is inconsistent with ‖pk‖0 = 1.

Finally we need a regularization family jε(·) fulfilling (3.41)-(3.43). Ac-cording to [26] we define

jε(u) = τc

∫

Ω3D

√

D(u) : D(u) + ε2 dx.

With it, conditions (3.41) and (3.42) are clearly satisfied. To show (3.43),we need the derivative of jε(·). Similarly to (3.12) the derivative in directionv is given by

〈j′ε(u), v〉 = τc

∫

Ω3D

D(u) : D(v)√

D(u) : D(u) + ε2dx.

So we can estimate

|〈j′ε(u), v〉| ≤ τc

∫

Ω3D

2DII(u)1/2DII(v)

1/2

√

D(u) : D(u) + ε2dx

≤ τc

∫

Ω3D

√2DII(v)

1/2 dx

≤ 2τc‖v‖V 3D0

which implies‖j′ε(u)‖V 3D

0

⋆ ≤ c

with a constant c. In the same way we can prove the assumptions for thetwo-dimensional case. Since we have already showed the equivalence betweenthe mixed problem and the dual formulation, we end up with

Theorem 3.3.11. The dual dual formulation 3.3 and the dual formulation3.4 have a unique solution (u, p) ∈ V 3D

0 ×Q and u ∈ V0 respectively. More-over there exists a solution λ ∈ Λ and w ∈ W for problem 3.3 and problem3.4 respectively.

35

For the dual variables λ for the three-dimensional problem and w for thetwo-dimensional problem respectively, the kernel of the operators

B2 : V 3D

0 → Λ⋆, 〈B2u, λ〉 = b2(u, λ)

and for the two-dimensional case

B : V0 →W ⋆, 〈Bu,w〉 = b(u,w)

respectively, are not so easy to handle as it was for B1 (see also [16],[18]).For example looking at the two-dimensional problem, suppose that u is aunique solution and let w,w′ the dual solutions. Plugging in and subtractingyields

div(w′ − w) = 0. (3.48)

From functional analysis we have the decomposition

[L2(Ω)]3×3 = ∇[H10 (Ω)]3 ⊕ S0

withS0 := w ∈ [L2(Ω)]3×3

∣∣ divw = 0

so that we can decompose uniquely

w = w1 + w2, w1 ∈ ∇[H10 (Ω)]3, w2 ∈ S0.

Now we see from (3.48), that if u is a unique solution, then we have auniquely component of w in ∇H1

0 (Ω) but w2 and w′2 do not have to be the

same. The space S0 is much larger than the space of constant functions, sowe need more than a simple condition like (3.46) in order to get uniqueness.We will do this in the next chapter.

36

Chapter 4

A method for solving the

weak systems

Now we want to find a suitable method solving the variational inequalities.There are at least two possibilities to do that:

1. Regularizing the non-differentiable functional leads to an variationalequation which can then be solved with more common methods. Theanswers of how to regularize the functional and which methods areused, can be found in [14] and [16]. Similarly one can regularize theconstitutive law (viscosity regularization), see for example [12],[1],[2].

2. Using a dual formulation has the advantage not to handle a non-differentiable functional but has an extra variable to be solved. Meth-ods based an such formulations can be found in [15],[18],[25],[16] and [28].

In this chapter we are going to use dual formulations, developed in chapter3, combined with an Uzawa type method.

4.1 An Uzawa type method

4.1.1 A time dependent approximation

We use the ideas from [18]. Therein the developed method is applied tothe problem reduced onto the cross section of the prism which we alreadyderived in chapter 2 and chapter 3. Here we will list the results for the 2D-problem and then we will carry them over to the three-dimensional case.Starting point is the dual formulation 3.4. First we will show that we canreplace the inequality therein by an equation.

We introduce the orthogonal projection onto the set W ,

PW : L2(Ω) × L2(Ω) →W,

37

for the norm

‖µ‖L2(Ω)×L2(Ω) :=

(∫

Ω|µ|2 dx

)1/2

.

It is given by

PW(µ(x)) =µ(x)

sup (1, |µ(x)|) ∀µ ∈ L2(Ω) × L2(Ω).

whereW = w = (w1, w2) |w ∈ L2(Ω) × L2(Ω), |w| ≤ 1.

Lemma 4.1.1. The orthogonal projection PW fulfills the following property:

‖PW(µ) − PW(η)‖L2×L2 ≤ ‖µ− η‖L2×L2 . (4.1)

Proof. Since PW is the orthogonal projection, there holds for allw ∈ L2(Ω) × L2(Ω):

(w − PW(w), v − PW(w))L2×L2 ≤ 0 (4.2)

Let µ, η ∈ L2(Ω)×L2(Ω) arbitrary but fixed, then it follows from (4.2), withthe choice w = µ and v = PW(η), that

(PW(µ), PW(η) − PW(µ))L2×L2 ≥ (µ, PW(η) − PW(µ))L2×L2 . (4.3)

On the other hand, choosing w = η and v = PW(µ) leads to

(PW(η), PW(µ) − PW(η))L2×L2 ≥ (η, PW(µ) − PW(η))L2×L2 . (4.4)

Adding (4.3) and (4.4) now yields

(PW(µ) − PW(η), PW(η) − PW(µ))L2×L2 ≥ (µ− η, PW(η) − PW(µ))L2×L2 ,

what is equivalent to

‖PW(µ) − PW(η)‖2L2×L2 ≤ (µ− η, PW(µ) − PW(η))L2×L2

≤ ‖µ− η‖L2×L2 · ‖PW(µ) − PW(η)‖L2×L2 .

So we get‖PW(µ) − PW(η)‖L2×L2 ≤ ‖µ− η‖L2×L2 .

Notation. We will denote the norm ‖ · ‖L2×L2 also by ‖ · ‖0.

38

Now the inequality from problem 3.4,

〈Bu, φ− w〉 ≤ 0 ∀φ ∈W,

is clearly equivalent to

〈w + rBu− w, φ− w〉 ≤ 0 ∀φ ∈W (4.5)

for r > 0. And this is equivalent to

w = PW(w + rBu).

Therefore the dual formulation 3.4 is equivalent to the following problem:

Problem 4.1 (Dual formulation for the steady state case). Findu ∈ V0, w ∈W such that

µa2(u, v) + τcb(v, w) = (c, v) ∀ v ∈ V0

w = PW(w + rBu) (4.6)

with r > 0.

Now we could compute a solution (u,w) by the following natural algo-rithm:

Algorithm 4.1.2. Ifw0 ∈W

is given, then, for m ≥ 0, wm being known, we compute um and wm+1 by

µa2(um, v) = (c, v) − τcb(v, wm) ∀v ∈ V0

wm+1 = PW(wm + rBum)

Concerning the convergence we have the following result:

Theorem 4.1.3. Under the assumption

0 < r <2µ

τc(4.7)

we have for all w0 ∈W

limm→∞

um = u in V0 = H10 (Ω)

andlim

m→∞wm = w weakly in L2(Ω) × L2(Ω)

39

Proof. We denote um − u and wm − w by um and wm respectively. Since(u,w) solves problem 4.1 we have

µa2(um, v) = −τcb(v, wm) ∀v ∈ V0 (4.8)

and

‖wm+1‖0 = ‖Pw(wm + rBum)−PW(w− rBu)‖0

(4.1)

≤ ‖wm + rBum‖0. (4.9)

Using the identity

‖wm + rBum‖20 = ‖wm‖2

0 + 2r(Bum, wm) + r2‖Bum‖20,

(4.9) is equivalent to

‖wm‖20 − ‖wm+1‖2

0 ≥ −2rb(um, wm) − r2‖Bum‖20. (4.10)

Using test functions v = um in (4.8) leads to

b(um, wm) = − µ

τca2(um, um) = − µ

τc‖Bum‖2

0,

so that (4.10) is equal to

‖wm‖20 − ‖wm+1‖2

0 ≥ r

(2µ

τc− r

)

‖Bum‖20.

Now using (4.7), we get that the sequence ‖wm‖20 is decreasing. As the

sequence is bounded from below by zero, we get

limm→∞

(‖wm‖20 − ‖wm+1‖2

0) = 0

and thereforelim

m→∞‖Bum‖2

0 = limm→∞

‖∇um‖20 = 0.

Since theH10 -semi norm is equivalent to theH1-norm in V0, this is equivalent

tolim

m→∞‖um − u‖1 = 0.

For the convergence of the dual solution wm we refer to [18].

Remark. In each step of the algorithm we have to solve an equation of theform

µa2(um, v) = f,

what is the weak formulation of the Dirichlet-problem

−µ∆u = f in Ω

u = 0 on Γ.

So the initial problem of solving a variational inequality of the second kind,changed to the problem of solving a Poisson equation several times.

40

As we can see from the assumption, the algorithm deteriorates as theyield limit τc is getting larger and larger (under the assumption that everything else stays the same). So we are going to develop another method withbetter properties. The main idea in [18] is to approximate the solution u∞of

µa2(u∞, v − u∞) + τcj(v) − τcj(u∞) ≥ (c, v − u∞) ∀ v ∈ V0 (4.11)

by the time dependent problem

Problem 4.2 (Time dependent problem). Find u(t) ∈ L2loc(0,∞;H1

0 (Ω))such that

∫

Ω

∂

∂t∇u(t) · ∇(v − u(t)) dx+ µa2(u(t), v − u(t))

+τcj(v) − τcj(u(t)) ≥ (c, v − u(t)) ∀ v ∈ V0

(4.12)

u(0) = u0 (4.13)

where

L2loc(0,∞;H1

0 (Ω)) := v : [0,∞) → H10 (Ω) | ‖v‖L2([0,t],H1

0(Ω)) <∞, ∀t ∈ R

+

and

‖v‖L2([0,t],H10(Ω)) :=

(∫ t

0‖v(s)‖2

H10(Ω) ds

)1/2

.

Remark. We denoted the solution of (4.11) now by (u∞, w∞) in order toconfirm that it is a steady state solution.

The next theorem guarantees that the time-dependent solution convergesto the steady state solution.

Theorem 4.1.4. Let u(t) be the solution of problem 4.2 and let u∞ be thesolution of (4.11). Then there exists c ∈ R such that the estimate

‖u(t) − u∞‖1 ≤ c e−µt‖u0 − u∞‖1, ∀ t ≥ 0 (4.14)

holds.

Proof. We set u(t) − u∞ = u(t). If we choose test functions v = u∞ inproblem 4.2, v = u(t) in problem 4.1, respectively, we get, after adding thetwo inequalities,

〈 ∂∂t

∇u(t),∇(u(t) − u∞)〉 + µa2(u(t), u(t)) ≤ 0. (4.15)

Sinced

dt〈v(t), w(t)〉 = 〈∂v(t)

∂t, w(t)〉 + 〈v(t), ∂w(t)

∂t〉

41

and∂u∞∂t

= 0,

(4.15) is equivalent to

1

2

d

dt

∫

Ω|∇u(t)|2 dx ≤ −µ

∫

Ω|∇u(t)|2 dx.

After integrating this ordinary differential equation with respect to t we get∫

Ω|∇u(t)|2 dx ≤ e−2µt

∫

Ω|∇u(0)|2 dx

or in another notation, using the initial value (4.13),

|u(t) − u∞|21 ≤ e−2µt|u0 − u∞|21.Thanks to Friedrich’s inequality we then have

|u(t) − u∞|1 ≥ 1

c‖u(t) − u∞‖1.

Since|u0 − u∞|21 ≤ ‖u0 − u∞‖2

0

holds, we have showed (4.14).

For the time discretization of problem 4.2, the authors use a backwardEuler scheme as the following: Given

u0 = u0

a solution un+1 can be computed (for given un ) by

1

∆t

∫

Ω∇(un+1 − un) · ∇(v − un+1) dx+ µa2(u

n+1, v − un+1)

+ τcj(v) − τcj(un+1) ≥ (c, v − un+1) ∀ v ∈ V0

or equivalently

(1 + µ∆t)a2(un+1, v − un+1) + ∆tτcj(v) − ∆tτcj(u

n+1)

≥ a2(un, v) + ∆t(c, v − un+1) ∀ v ∈ V0, (4.16)

where ∆t denotes the time discretization step. In the same manner as we didit for (4.11), one can show that (4.16) is equivalent to the time discretizedmixed dual system:

Problem 4.3 (Time discretized dual dual system). Find u ∈ V0, w ∈ Wsuch that for all v ∈ V0

(1 + µ∆t)a2(un+1, v) + τc∆tb(v, w

n+1) = a2(un, v) + ∆t(c, v)

wn+1 = PW(wn+1 + rBun+1)

for given u0 = u0, w0 ∈W .

42

4.1.2 A stabilized scheme

Computing a solution of problem 4.3 now seems to be straight forward byusing a fixed point algorithm. Unfortunately there rests the problem ofbad convergence properties for large values of τc, everything else stayingthe same. So we will use a special regularization for problem 4.3 which re-solves the problem of bad convergence property and makes the dual variableunique. A so called (pseudo-) time disretized regularization term is added tothe projection such that our problem to be solved (after time discretizing)looks like the following:

Problem 4.4 (Regularized time discretized dual dual formulation). Givenu0 = u0 ∈ V0, w

0 ∈ W . For un, wn being known, find un+1 ∈ V0, wn+1 ∈ W

such that for all v ∈ V0

(1 + µ∆t)a2(un+1, v) + τc∆t b(v, w

n+1) = a2(un, v) + ∆t(c, v)

εwn+1 − wn

∆t+ wn+1 = PW(wn+1 + rBun+1) (4.17)

with ε > 0.

Remark. The regularization term guarantees that wn+1 ∈ W since (4.17)can be written as the convex combination

wn+1 =ε

ε+ ∆twn +

∆t

ε+ ∆tPW(wn+1 + rBun+1).

As we mentioned, the additional term on the left hand side causes thatthe dual variable is unique.

Theorem 4.1.5. Under the assumption

0 < r <2

τc(µ+

1

∆t) (4.18)

problem 4.4 has a unique solution (un+1, wn+1) ∈ V0 ×W for all (un, wn) ∈V0 ×W .

Proof. We define the map

Tn : W →W, Tn(φ) =ε

ε+ ∆twn +

∆t

ε+ ∆tPW(φ+ rBun+1

φ ),

where un+1φ is denoted as the solution of the Dirichlet-problem

(1 + µ∆t)a2(un+1φ , v) = a2(u

n, v)−∆tτcb(v, φ) + ∆t(c, v) ∀ v ∈ V0. (4.19)

We remark that for a fixed φ, uφ is a unique solution of our problem sincewe can apply the Lax-Milgram theorem to (4.19). Now we take arbitrary

43

φ1, φ2 ∈ W and denote φ1 − φ2 by φ. In the same manner we will denoteun+1

φ1− un+1

φ2by u. Since PW fulfills (4.1), it follows

‖Tn(φ1) − Tn(φ2)‖0 ≤ ∆t

ε+ ∆t‖φ+ rBu‖0. (4.20)

From the definition of the norm it follows

‖φ+ rBu‖20 = ‖φ‖2

0 + 2r(Bu, φ) + r2‖Bu‖20. (4.21)

Furthermore, from (4.19) we get (by subtracting the equations for un+1φ1

and

un+1φ2

, using test functions v = u)

(1 + µ∆t)a2(u, u) = −∆tτcb(u, φ) = −∆tτc(Bu, φ), (4.22)

so that we can combine (4.21),(4.22) and get out of (4.20)

‖Tn(φ1)−Tn(φ2)‖20 ≤

(∆t

ε+ ∆t

)2(

‖φ‖20 − 2r

1 + µ∆t

τc∆ta2(u, u) + r2‖Bu‖2

0

)

,

or equivalently

‖Tn(φ1) − Tn(φ2)‖20+r

[2

τc

(

µ+1

∆t

)

− r

](∆t

ε+ ∆t

)2

‖Bu‖20

≤(

∆t

ε+ ∆t

)2

‖φ‖20,

since

a2(u, u) =

∫

Ω|∇u|2 dx = ‖Bu‖2

0.

Now under the assumption (4.18) it follows that

‖Tn(φ1) − Tn(φ2)‖0 ≤ ∆t

ε+ ∆t‖φ1 − φ2‖0 ∀φ1, φ2 ∈W.

Since ∆tε+∆t < 1, Tn is a contraction and therefore has a unique fixed point

φ together with a unique solution uφ.

As we have shown that there exists a solution for the regularized problem,we can write down a natural fixed point iteration for solving the system.Remember that n denotes the time discretization step whereas m is nowintroduced to denote the inner iteration index.

Algorithm 4.1.6. Ifwn+1

0 = wn

then for m ≥ 0, wn+1m being known, we can compute un+1

m , wn+1m+1 by

(1 + µ∆t)a2(un+1m , v) = a2(u

n, v) − τc∆t b(v, wn+1m ) − ∆t(c, v) ∀ v ∈ V0

wn+1m+1 =

ε

ε+ ∆twn +

∆t

ε+ ∆tPW(wn+1

m + rBun+1m ).

44

4.2 Convergence results

In this section we will proof that (unm, w

nm+1) - computed by algorithm 4.1.6 -

converges to the desired solution (u∞, w∞), as well as that the regularizationresolves the problem of bad convergence properties for large values of τc.

Notation. We denote the measure of a set Ω by meas(Ω).

Theorem 4.2.1. Under the assumption

0 < r <2

τc

(

µ+1

∆t

)

(4.23)

there holds for all m ≥ 0

‖wn+1m − wn+1‖0 ≤

(∆t

ε+ ∆t

)m

‖wn+1 − wn‖0

≤(

∆t

ε+ ∆t

)m

2(meas(Ω))1/2. (4.24)

Further we obtain

‖un+1m − un+1‖1 ≤ τc

(µ+ 1/∆t)‖wn+1 − wn‖0

(∆t

ε+ ∆t

)m

≤ 2(meas(Ω))1/2 τc(µ+ 1/∆t)

(∆t

ε+ ∆t

)m

. (4.25)

Proof. Again we use the mapping

Tn : W →W, Tn(φ) =ε

ε+ ∆twn +

∆t

ε+ ∆tPW(φ+ rBun+1

φ ).

Then we have from the last theorem that Tn is a contraction with

‖Tn(φ1) − Tn(φ2)‖0 ≤ ∆t

ε+ ∆t‖φ1 − φ2‖0 ∀φ1, φ2 ∈W. (4.26)

From algorithm 4.1.6 we get that

wn+1m+1 = Tn(wn+1

m )

andwn+1 = Tn(wn+1).

The contraction property (4.26) then yields

‖wn+1m+1 − wn+1‖0 ≤

(∆t

ε+ ∆t

)

‖wn+1m − wn+1‖0 ∀m ≥ 0.

45

As we assumed wn+10 = wn we get, applying (4.26) further m times,

‖wn+1m − wn+1‖0 ≤

(∆t

ε+ ∆t

)m

‖wn+1 − wn‖0 ∀m ≥ 0. (4.27)

By definition of the convex set W , we get for an arbitrary w ∈W

‖w‖0 =

(∫

Ω|w|2 dx

)1/2

≤(∫

Ω1 dx

)1/2

= (meas(Ω))1/2. (4.28)

Combining (4.27) and (4.28) results in (4.24).To prove (4.25) we use the fact that un+1

m , un+1 are solutions of theDirichlet problem in algorithm 4.1.6. Therefore un+1

m − un+1 is a solution of

(1+µ∆t)a2(un+1m −un+1, v) = −τc∆t b(v, wn+1

m −wn+1) ∀ v ∈ V0, ∀m ≥ 0.

Choosing test functions v = un+1m − un+1 leads to

(1 + µ∆t)‖∇(un+1m − un+1)‖0 ≤ τc∆t‖wn+1

m − wn+1‖0 ∀m ≥ 0.

From (4.24) and Friedrich’s inequality we get (4.25).

So we have seen that the solution (unm, w

nm) of an inner iteration con-

verges to the desired limit (un+1, wn+1). Next we have to show the conver-gence of (un, wn) to the steady state solution (u∞, w∞). The next theoremguarantees the strong convergence of un as well as the weak convergence ofwn.

Theorem 4.2.2. Under the assumption

0 < r <2µ

τc(4.29)

we havelim

n→∞un = u∞ in V0 = H1

0 (Ω) (4.30)

as well aslim

n→∞wn = w∞ weakly in L2(Ω) × L2(Ω) (4.31)

for all (u0, w0) ∈ V0 ×W .

Proof. We only give a detailed proof for (4.30). For (4.31) we refer to [18]and the references therein. Again we will denote un+1 −u∞ and wn+1 −w∞

by un+1 and wn+1, respectively. As (u∞, w∞) is a solution of (4.1) and(un+1, wn+1) is a solution of problem 4.4 we then have

(1 + µ∆t)a2(un+1, v) = a2(u

n, v) − τc∆tb(v, wn+1) ∀ v ∈ V0, (4.32)

46

as well as (from (4.1))

∥∥∥∥εwn+1 − wn

∆t+ wn+1

∥∥∥∥

0

≤ ‖wn+1 + rBun+1‖0. (4.33)

In the same manner as in the proofs before, we now choose test functionsv = un+1 in (4.32) leading to

(1 + µ∆t)‖∇un+1‖20 = a2(u

n, un+1) − τc∆tb(un+1, wn+1). (4.34)

Further we have (remembering that ‖Bu‖20 = ‖∇u‖2

0)

‖wn+1 + rBun+1‖20 = ‖wn+1‖2

0 + 2rb(un+1, wn+1) + r2‖∇un+1‖20. (4.35)

Together with (4.34),(4.35) we then get from (4.33)

∥∥∥∥εwn+1 − wn

∆t+ wn+1

∥∥∥∥

2

0

≤ ‖wn+1‖20 + 2rb(un+1, wn+1) + r2‖∇un+1‖2

0

= ‖wn+1‖20 − 2r

1 + µ∆t

τc∆t‖∇un+1‖2

0

+2r

τc∆ta2(u

n, un+1) + r2‖∇un+1‖20.

(4.36)

Another formulation for the left hand side is

ε2∥∥∥∥

wn+1 − wn

∆t

∥∥∥∥

2

0

+ 2ε

∆t

∫

Ω(wn+1 − wn) · wn+1 dx+ ‖wn+1‖2

0,

so that we get from (4.36)

ε2∥∥∥∥

wn+1 − wn

∆t

∥∥∥∥

2

0

+ 2ε

∆t

∫

Ω(wn+1 − wn) · wn+1 dx

≤

− 2r1 + µ∆t

τc∆t‖∇un+1‖2

0 +2r

τc∆ta2(u

n, un+1) + r2‖∇un+1‖20. (4.37)

Moreover, using the inequality 2ab ≤ a2 + b2, we get

2

∫

Ω(wn+1 − wn) · wn+1 dx = 2‖wn+1‖2

0 − 2

∫

Ωwn · wn+1 dx

≤ 2‖wn+1‖20 − ‖wn‖2

0 − ‖wn+1‖20

= ‖wn+1‖20 − ‖wn‖2

0

as well as

2a2(un, un+1) = 2

∫

Ω∇un · ∇un+1 dx ≤ ‖∇un‖2

0 + ‖∇un+1‖20.

47

Summing up these two results we get from (4.37)

ε2∥∥∥∥

wn+1 − wn

∆t

∥∥∥∥

2

0

+ r

(2µ

τc− r

)

‖∇un+1‖20

≤ ε

∆t

(‖wn‖2

0 − ‖wn+1‖20

)+

r

τc∆t

(‖∇un‖2

0 − ‖∇un+1‖20

). (4.38)

From our assumption on r, (4.29), we get

0 < ε2∥∥∥∥

wn+1 − wn

∆t

∥∥∥∥

2

0

+ r

(2µ

τc− r

)

‖∇un+1‖20.

So it follows that the sequence (ε‖wn‖20 + (r/τc)‖∇un‖2

0)n is decreasing.Therefore the left hand side of (4.38) converges to zero, so it follows

limn→∞

∇un+1 = 0 in L2(Ω) × L2(Ω),

which is equivalent to (4.30).

4.3 Application to the problem in 3D

Our task is now to derive an analogous algorithm for the three-dimensionalproblem. For this we use two different ways treating the variables u, p, λ.A first choice would be, if we interpret (u, p) as the solution of a Stokesproblem, instead of u being the solution of a Dirichlet problem. This stepsounds like a natural one. Another possibility is that we interpret (p, λ)as one dual variable. This is also a logical generalization because in manyapplications p is already a dual variable.

4.3.1 The first approach

First we will look at problem 3.3:

Find u ∈ V 3D

0 , p ∈ Q, λ ∈ Λ such that

µa(u, v) + b1(v, p) + τc√

2b2(v, λ) = (f, v) ∀v ∈ V 3D

0

b1(u, q) = 0 ∀q ∈ Qb2(u, η − λ) ≤ 0 ∀η ∈ Λ.

Now with the same technique as in the 2D model, the system is equivalentto

Problem 4.5 (Dual dual formulation). Find u ∈ V 3D

0 , p ∈ Q, λ ∈ Λ suchthat

µa(u, v) + b1(v, p) + τc√

2b2(v, λ) = (f, v) ∀v ∈ V 3D

0

b1(u, q) = 0 ∀q ∈ Qλ = PΛ(λ+ rD(u))

48

with r > 0.The operator occurring on the right hand side is now the orthog-onal projection onto Λ given by:

PΛ : [L2(Ω3D)]3×3 → Λ, PΛ(µ)(x) =µ(x)

sup (1, ‖µ(x)‖F )∀µ ∈ [L2(Ω3D)]3×3.

A natural generalization of problem 4.4 looks then like

Problem 4.6 (Regularized time discretized dual dual formulation for 3D).Given u0 = u0 ∈ V 3D

0 , p0 ∈ Q,λ0 ∈ Λ. For un, pn, λn being known, findun+1 ∈ V 3D

0 , pn+1 ∈ Q,λn+1 ∈ Λ such that for all v ∈ V 3D

0 , q ∈ Q

(1 + µ∆t)a(un+1, v) + τc√

2∆t b2(v, wn+1) + ∆t b1(v, p

n+1) = a(un, v) + ∆t(f, v)

(pn+1, q) = (pn, q) + r(B1un+1, q)

ελn+1 − λn

∆t+ λn+1 = PΛ(λn+1 + rB2u

n+1)

with ε > 0.

Then the generalization of algorithm 4.1.6 can be written as

Algorithm 4.3.1. If

(u0, p0, λ0) ∈ V 3D

0 ×Q× Λ

is given, and ifpn+10 = pn, λn+1

0 = λn

then, for m ≥ 0, pn+1m , λn+1

m being known, we compute un+1m ,pn+1

m+1 and λn+1m+1

as follows:

(1 + µ∆t)a(un+1m , v) = a(un, v) − ∆tb1(v, p

n+1m ) − τc

√2∆tb2(v, λ

n+1m ) + (f, v)

pn+1m+1 = pn+1

m + r div un+1m

λn+1m+1 =

ε

ε+ ∆tλn +

∆t

ε+ ∆tPΛ(λn+1

m + rD(un+1m ))

where r, r > 0.

4.3.2 The second approach

In this case we introduce the following settings:

W = Q× Λ,

b : V 3D

0 ×W → R, b(v, w) = b(v, (w1, w2)) = b1(v, w1) + b2(v, w2)

49

with

b1 : V 3D

0 ×Q→ R, b1(v, p) =1

τc√

2b1(v, p),

as well as

PW : L2(Ω3D) × [L2(Ω3D)]3×3, (p, µ) 7→ (p, PΛ(µ))

and

B : V 3D

0 → [L2(Ω3D)]3×3 × [L2(Ω3D)]3×3, u 7→ Bu = (B1u,B2u)

with the linear operators B1, B2 defined by the bilinear forms b1(·, ·), b2(·, ·).Finally we introduce the norm for the product space W by

‖w‖W := max ‖w1‖0, ‖w2‖0.

Then our problem looks like the following:

Problem 4.7 (Regularized time discretized dual dual formulation 3D ).Find u∞ ∈ V 3D

0 , w∞ ∈ W = w = (w1, w2) |w1 ∈ Q, w2 ∈ Λ, ‖w2‖F ≤ 1such that

a(u∞, v) + τc√

2b(v, w∞) = (f, v) ∀ v ∈ V 3D

0

w∞ = PW(w∞ + rBu∞)

what is just the form of problem 4.1 with an extra factor√

2 in the firstline. So we can exactly apply the results from the previous section whichmeans that a solution of (4.39) can be computed the following way (again ndenotes the time discretization step whereas m denotes the inner iterationstep)

Algorithm 4.3.2. If

(u0, p0, λ0) ∈ V 3D

0 ×Q× Λ

is given, and ifpn+10 = pn, λn+1

0 = λn

then, for m ≥ 0, pn+1m , λn+1

m being known, we compute un+1m ,pn+1

m+1 and λn+1m+1

as follows:

(1 + µ∆t)a(un+1m , v) = a(un, v) − ∆tb1(v, p

n+1m ) − τc

√2∆tb2(v, λ

n+1m ) + (f, v)

pn+1m+1 =

ε

ε+ ∆tpn +

∆t

ε+ ∆t(pn+1

m + r div un+1m )

λn+1m+1 =

ε

ε+ ∆tλn +

∆t

ε+ ∆tPΛ(λn+1

m + rD(un+1m ))

wherer = r/τc

√2.

50

Remark. Applying the analysis leads to the same results except an extraconstant factor because of the factor

√2 in the dual formulation.

Remark. We want to mention that applying algorithms 4.3.1,4.3.2 leads tothe computation of the exact solution of a linear system in each step of thefixed-point iteration. In order to avoid this fact one can use an algorithmdealing with an approximate solution. For more details for such an inexactUzawa-type method, we refer to [7].

51

Chapter 5

Approximation with Finite

Elements

In this chapter we will use the finite element method for discretizing ourdeveloped dual systems in the three-dimensional case as well as in the two-dimensional one. We are going to discretize the generalizations for the three-dimensional problems in two different ways and then we will compare theconvergence properties in chapter 6. Definitions from the first section arecited from [9],[32],[31].

5.1 Discretization of the domain

For the three-dimensional case we consider a mesh T 3D

h of Ω3D, where

T 3D

h := T∣∣T is a tetrahedron in Ω3D.

The index h denotes the maximum of all diameters hT of all tetrahedronsT . The set of nodes is denoted by N 3D

h , the set of edges is denoted by E3D

h

and finally we denote the set of faces by F3D

h . We further distinguish theinner nodes from the ones lying on the boundary of the mesh,i.e.

N 3D

h = N 3D

h,Ω3D ∪N 3D

h,Γ3D

analogously for the edges

E3D

h = E3D

h,Ω3D ∪ E3D

h,Γ3D

as well, for the facesF3D

h = F3D

h,Ω3D ∪ F3D

h,Γ3D .

Analogously we can discretize the two-dimensional domain Ω by a mesh Th

withTh := T

∣∣T is a triangle in Ω.

52

The set of all nodes is here denoted by Nh, whereas the set of edges isdenoted by Eh. Moreover

Nh = Nh,Ω ∪Nh,Γ

andEh = Eh,Ω ∪ Eh,Γ.

We introduce the following definitions:

Definition. A mesh T 3D

h of a domain Ω is called an admissible triangulation,if

(R1) Ω =⋃

T∈T 3D

h

T

(R2) T1 ∩ T2 =

∅

a node

an edge

a face

for all T1 6= T2 ∈ T 3D

h ,

holds.

Definition. Let T 3D

h be a triangulation. For an element T ∈ T 3D

h let ρT

be the radius of the ball (circle in the two-dimensional case) with maximaldiameter BT . We call T 3D

h shape regular if the ratio of hT and ρT is boundedfrom above, independently from T and h, i.e. there exists κ > 0 such that

hT

ρT≤ κ ∀T ∈ Th.

Definition. A shape regular mesh Th fulfilling

∃κ > 0 :h

hT≤ κ, ∀T ∈ Th

is called quasi-uniform.

Our meshes are now assumed to be admissible and quasi-uniform. Theneighborhoods of a node x and a tetrahedron K are defined by

ωx :=⋃

T∈T 3D

h,x∈T

T

andωK :=

⋃

T∈T 3D

h,T∩K 6=0

T

respectively. So ωx contains all tetrahedrons which contain the node x,whereas ωK denotes the set which contains all tetrahedrons T which have acommon node, a common edge or a common face with the tetrahedron K.

53

5.2 The discrete models

We use Galerkin’s principle for the discretization with finite elements. Sothe infinite dimensional spaces V 3D

0 , Q,Λ and V0,W are replaced by finitedimensional subspaces V 3D

0,h, Qh,Λh and V0,h,Wh, respectively. Defining thesubspaces later, we first write down the discrete models for the mixed prob-lem 3.1

Problem 5.1 (Discrete mixed formulation). Find uh ∈ V 3D

0,h, ph ∈ Qh suchthat ∀ vh ∈ V 3D

0,h,∀ qh ∈ Qh

µa(uh, vh − uh) + b1(vh − uh, ph) + τcj(vh) − τcj(uh) ≥ (f, vh − uh)

b1(uh, qh) = 0

and for the problem reduced to the cross section 3.2

Problem 5.2 (Discrete primal formulation). Find uh ∈ V0,h such that∀ vh ∈ V0,h

µa2(uh, vh − uh) + τcj(vh) − τcj(uh) ≥ (c, vh − uh)

respectively. The dual formulations can then be written as

Problem 5.3 (Discrete dual dual formulation). Finduh ∈ V 3D

0,h, ph ∈ Qh, λh ∈ Λ such that ∀ vh ∈ V 3D

0,h,∀ qh ∈ Qh, ∀ ηh ∈ Λh

µa(uh, vh) + b1(vh, ph) + τc√

2 b2(vh, λh) = (f, vh)

b1(uh, qh) = 0

b2(uh, ηh − λh) ≤ 0

and in the 2D case

Problem 5.4 (Discrete dual formulation). Find uh ∈ V0,h, wh ∈ Wh suchthat ∀ vh ∈ V0,h, ∀φh ∈Wh

µa2(uh, vh) + τc√

2b(vh, wh) = (c, vh)

b2(uh, φh − wh) ≤ 0

respectively.

Lemma 5.2.1. The discrete problems 5.1 and 5.2 are equivalent to theirdiscrete dual formulations 5.3 and 5.4 respectively.

Proof. Can be done in the same way as we did it for the continuous case.

54

5.2.1 The two-dimensional case

In [18] the following finite dimensional subspaces are used for the approxi-mation of V, V0, L

2(Ω) × L2(Ω),W :

Vh = vh ∈ C0(Ωh)∣∣ vh|T ∈ P1, ∀T ∈ Th ⊂ V,

V0,h = vh ∈ Vh

∣∣ vh|Γh

= 0 ⊂ V0,

Lh = λh

∣∣λh ∈ L2(Ωh) × L2(Ωh), λh|T ∈ P0, ∀T ∈ Th ⊂ L2(Ωh) × L2(Ωh),

Wh = wh

∣∣wh ∈ Lh, |wh|T | ≤ 1∀T ∈ Th ⊂W.

Therein P0, P1 denote the spaces of polynomials in two variables of degree 0and 1 respectively. The finite element discretization of algorithm 4.1.6 looksthen as follows:

Algorithm 5.2.2 (Discrete 2D algorithm). If

wn+1h,0 = wn

h

then for m ≥ 0, wn+1h,m being known, we can compute un+1

h,m , wn+1h,m+1 by

(1 + µ∆t)a2(un+1h,m , vh) = a2(u

nh, vh) − τc∆t b(vh, w

n+1h,m ) − ∆t(c, vh) ∀ vh ∈ V0,h

wn+1h,m+1 =

ε

ε+ ∆twn

h +∆t

ε+ ∆tPWh

(wn+1h,m + rBun+1

h,m ).

5.2.2 The three-dimensional case

Continuous approximation of the pressure

Here we are going to discretize algorithm 4.3.1. We introduce the followingfinite dimensional subspaces for V 3D, V 3D

0 , Q, [L2(Ω3D)]3×3,Λ:

V 3D

h = vh ∈ [C0(Ω3D

h )]3∣∣ vh|T ∈ [P1]

3, ∀T ∈ T 3D

h ⊂ [H1(Ω3D)]3,

V 3D

0,h = vh ∈ Vh

∣∣ vh|Γ3D

h= 0 ⊂ V 3D

0 ,

Qh = qh ∈ C0(Ω3D

h )∣∣ qh|T ∈ P1, ∀T ∈ Th ⊂ Q,

Lh = λh

∣∣λh ∈ [L2(Ω3D

h )]3×3, λh|T ∈ [P0]3×3, ∀T ∈ T 3D

h ⊂ [L2(Ω3D

h )]3×3,

Λh = λh

∣∣λh ∈ Lh, ‖λh|T ‖F ≤ 1∀T ∈ T 3D

h ⊂ Λ.

55

The usage of linear ansatz functions for uh as well as for ph leads to anunstable system, i.e. the discrete inf-sup condition for b1(·, ·) is not fulfilled.By adding so called bubble functions the system becomes stable.A bubble function is defined by

bT (x) = λ1λ2λ3λ4

where the λi are the barycentric coordinates of x on T . With this definitionwe see that bT ∈ P4 vanishes on the boundary ∂T . We replace V 3D

0,h by

V3D

0,h = vh ∈ [C0(Ω3D)]3∣∣ vh|T = pT +βT bT , pT ∈ [P1]

3, βT ∈ R3 ∀T ∈ T 3D

h .

By adding bubble functions, we have to deal with more degrees of freedom.Luckily one can show that the bubble functions can be locally eliminatedbecause of their definition. As we use problem 5.3 for our implementation,we will show the elimination for this system. For uh ∈ ¯V 3D

0,h we make theansatz

uh = u1h + ub

h, u1h ∈ V 3D

0,h, ubh =

∑

T∈T 3D

h

βT bT ∈ [spanbT : T ∈ T 3D

h ]3.

We get, by using special test functions bT ei for i = 1, 2, 3 (where the eishould be the unit vectors in R

3) with a fixed T

µ

3∑

j=1

a(bT ej , bT ei)βT,j + µa(u1h, bT ei)

+ b1(bT ei, ph) +√

2τcb2(bT ei, λh) = (f, bT ei). (5.1)

Now we make some observations. As the bubble function bT is zero outsideof a tetrahedron T , it follows

a(u1h, bT ei) =

∫

T2D(u1

h) : D(bT ei) dx

= −∫

T2divD(u1

h) · bT ei dx︸ ︷︷ ︸

=0, u1

hlinear on T

+

∫

∂T2D(u1

h)n · bT ei ds︸ ︷︷ ︸

=0, bT =0 on ∂T

= 0

as well as

b2(bT ei, λh) =

∫

Tλh : D(bT ei) dx

= −∫

Tdivλh · bT ei dx

︸ ︷︷ ︸

=0, λh const. on T

+

∫

∂Tλhn · bT ei ds

︸ ︷︷ ︸

=0

= 0

56

and

b1(bT ei, ph) = −∫

Tph div bT ei dx

=

∫

T∇ph · bT ei dx−

∫

∂TphbT ei · nds

︸ ︷︷ ︸

=0

=

∫

T∇ph · bT ei dx.

So that we get from (5.1)

µ3∑

j=1

a(bT ej , bT ei)βT,j =

∫

T(f −∇ph) · bT ei dx. (5.2)

With the notations

γT =

∫

TbT dx, f

T=

1

γT

∫

TbT f dx, δT = µ(a(bT ej , bT ei))i,j∈1,2,3

we can write down the formula for the unknowns βT . From (5.2) we have

βT = δ−1T

(∫

TbT (f −∇ph) dx

)

= γT δ−1T (f

T −∇ph). (5.3)

One can easily check that δT is positive definite, so the inverse exists. Nowwe can plug in our result (5.3) into the second as well as the third equation(inequality resp.) of problem 5.3 and get

b1(uh, qh) = b1(u1h, qh) + b1(u

bh, qh)

= −∫

Ω3D

qh div u1h dx−

∑

T∈Th

∫

Tqh div bTβT dx

= −∫

Ω3D

qh div u1h dx+

∑

T∈Th

∫

T∇qh · bTβT dx−

∫

∂Tqhn · bTβT ds

︸ ︷︷ ︸

=0

(5.3)= −

∫

Ω3D

qh div u1h dx+

∑

T∈Th

α(T )

∫

T(δ−1

T (fT −∇ph)) · ∇qh dx

= 0

with

α(T ) =γ2

T

meas(T ).

57

Moreover we have

b2(uh, ηh − λh) = b2(u1h, ηh − λh) + b2(u

bh, ηh − λh)

= b2(u1h, ηh − λh) +

∑

T∈Th

∫

T(ηh − λh) : D(bTβT ) dx

= b2(u1h, ηh − λh) −

∑

T∈Th

∫

Tdiv (ηh − λh) · bTβT dx

︸ ︷︷ ︸

=0

−∫

∂T(ηh − λh)n · bTβT ds

︸ ︷︷ ︸

=0

= b2(u1h, ηh − λh) ≤ 0 ∀ ηh ∈ Λh.

Summarizing our results, the additional degrees of freedom coming from thebubble functions, can be eliminated and the new system to be solved hasthe following form:

Problem 5.5 (Discrete dual dual system with continuous pressure). Searchuh ∈ V 3D

0,h, ph ∈ Qh, λh ∈ Λh

µa(uh, vh) + b1(vh, ph) +√

2τc b2(vh, λh) = (f, vh) ∀ vh ∈ V 3D

0,h

b1(uh, qh) − ch(ph, qh) = 〈Gh, qh〉 ∀ qh ∈ Qh

b2(uh, ηh − λh) ≤ 0

with the mesh-dependent bilinear form

ch(ph, qh) =∑

T∈T 3D

h

α(T )

∫

T(δ−1

T ∇ph) · ∇qh dx

and the mesh-dependent linear functional

〈Gh, qh〉 =∑

T∈T 3D

h

α(T )

∫

T(δ−1

T fT) · ∇qh dx.

So our algorithm can be written as

Algorithm 5.2.3. If

(u0h, p

0h, λ

0h) ∈ V 3D

0,h ×Qh × Λh

is given, and ifpn+10,h = pn

h, λn+10,h = λn

h

58

then, for m ≥ 0, pn+1m,h , λ

n+1m,h being known, we compute un+1

m,h ,pn+1m+1,h and

λn+1m+1,h as follows:

(1 + µ∆t)a(un+1m,h , vh) = a(un

h, vh) − ∆tb1(vh, pn+1m,h ) − τc

√2∆tb2(vh, λ

n+1m,h ) + (f, vh)

(pn+1m+1,h, qh) = (pn+1

m,h , qh) + r(〈Gh, qh〉 − (B1un+1m,h , qh) + ch(ph, qh)

λn+1m+1,h =

ε

ε+ ∆tλn

h +∆t

ε+ ∆tPΛh

(λn+1m,h + rD(un+1

m,h ))

where r, r > 0.

Discontinuous approximation of the pressure

An alternative method would be if we treat p, λ as one variable. In thiscase we use for p the same ansatz functions as for λ. We choose the finitedimensional subspaces of V 3D

0 , Q, [L2(Ω3D)]3×3,Λ in the following way:

V 3D

h = vh ∈ [C0(Ω3D

h )]3∣∣ vh|T ∈ [P1]

3, ∀T ∈ T 3D

h ⊂ [H1(Ω3D)]3,

V 3D

0,h = vh ∈ V 3D

h

∣∣ vh|Γ3D

h= 0 ⊂ V 3D

0 ,

Qh = qh∣∣ qh|T ∈ P0, ∀T ∈ Th ⊂ Q,

Lh = λh

∣∣λh ∈ [L2(Ω3D

h )]3×3, λh|T ∈ [P0]3×3, ∀T ∈ T 3D

h ⊂ [L2(Ω3D

h )]3×3,

Λh = λh

∣∣λh ∈ Lh, ‖λh|T ‖F ≤ 1∀T ∈ T 3D

h ⊂ Λ.

So the discrete version of algorithm 4.3.2 looks like

Algorithm 5.2.4. If

(u0h, p

0h, λ

0h) ∈ V 3D

0,h ×Qh × Λh

is given, and ifpn+10,h = pn

h, λn+10,h = λn

h

then, for m ≥ 0, pn+1m,h , λ

n+1m,h being known, we compute un+1

m,h ,pn+1m+1,h and

λn+1m+1,h as follows:

(1 + µ∆t)a(un+1m,h , v) = a(un

h, v) − ∆tb1(v, pn+1m,h ) − τc

√2∆tb2(v, λ

n+1m,h ) + (f, v)

pn+1m+1,h =

ε

ε+ ∆tpn

h +∆t

ε+ ∆t(pn+1

m,h + r div un+1m,h )

λn+1m+1,h =

ε

ε+ ∆tλn

h +∆t

ε+ ∆tPΛ(λn+1

m,h + rD(un+1m,h ))

59

wherer = r/τc

√2.

5.3 Existence and uniqueness for the discrete mixed

problems

Here we want to show that our discretized problems also provide uniquesolutions.

5.3.1 The two-dimensional problem

For the two-dimensional problem we can apply the theorem cited from [14].

Theorem 5.3.1. The discrete problem 5.2 admits one and only one solu-tion.

Proof. The statement follows from the equivalence of the primal problem5.2 to minimizing

J(vh) :=1

2a2(vh, vh) − (c, vh) + j(vh)

which is shown in [14]. The functional J is continuous and strictly convexon Vh too and we have

lim‖vh‖→∞

J(vh) = +∞

since the discrete assumption

a2(vh, vh) ≥ α1‖vh‖21 ∀ vh ∈ Vh

is fulfilled. The uniqueness of the solution uh is then clear.

5.3.2 The three-dimensional problem

Similarly to chapter 2, we can guarantee the existence of a discrete solution,if the discrete assumptions are satisfied. So we have the following theoremwhich can be found in [17]

Theorem 5.3.2. Under the assumptions from theorem 3.3.7 with the ex-ception that the inf sup condition (3.28) is replaced by

infqh∈Qh

supvh∈V 3D

0,h

b1(vh, qh)

‖vh‖V 3D

0,h‖qh‖Qh

≥ β (5.4)

and the condition on coercivity (3.27) is replaced by the condition

〈Av, v〉 ≥ α0‖v‖2V 3D

h

, ∀ v ∈ Ker B1,h, (5.5)

the discrete problem 5.1 has a unique solution.

60

Since

Ker B1,h = vh

∣∣

∫

Ω3D

qh div vh dx = 0 for all qh ∈ Qh ⊂ V 3D

0,h,

condition (5.5) is fulfilled. For the discrete inf sup condition (5.4) we firstneed the following lemma, which can be found in [6],[34].

Lemma 5.3.3 (Fortin). We assume that there exists a surjective mappingΠh : V 3D → V 3D

h with

1. b1(Πhv, qh) = b1(v, qh) for all qh ∈ Qh as well as for all v ∈ V 3D and

2. ‖Πhv‖V 3D ≤ c‖v‖V 3D for all v ∈ V 3D with c > 0, independent of h.

If further the inf-sup condition for the spaces V 3D, Q holds, then the discretecounterpart follows.

Proof. There holds

β1‖qh‖Q ≤ sup0 6=v∈V 3D

b1(v, qh)

‖v‖V 3D

≤ c sup0 6=v∈V 3D

b1(Πhv, qh)

‖Πhv‖V 3D

≤ c sup0 6=vh∈V 3D

h

b1(vh, qh)

‖vh‖V 3D

and therefore

inf0 6=qh∈Qh

sup0 6=vh∈V 3D

h

b1(vh, qh)

‖vh‖V 3D‖qh‖Q≥ 1

cβ1.

Definition. The modified Clement interpolation operator is defined by

Rh : L1(Ω3D) → V 3D

0,h, Rhu 7→∑

x∈N 3D

0,Ω3D

(Pu)φx

with

P : L1(ω3D

x ) → R, u 7→ Pu =1

meas(ωx)

∫

ωx

u dx

and φx denotes the nodal basis function.

Remark. By standard theory ([24]) it follows the following error estimatefor the modified Clement interpolation operator

‖u−Rhu‖L2(T ) ≤ chT ‖u‖[H1(ωT )]3 . (5.6)

Fortin’s lemma now provides a method for proving the discrete inf-supcondition. With it we have (see [31])

Theorem 5.3.4. Assume that we have a shape regular triangularizationthen for the spaces ¯V 3D

h and Qh the discrete inf-sup condition is fulfilled.

61

Proof. We do the proof with the previous lemma. So let us define

Πhu := Rhu+∑

T∈T 3D

h

ψT

∫

T(u−Rhu)

(∫

TψT dx

)−1

dx

where

ψT = 256bT ∈ spanbT : T ∈ T 3D

h , ⇒ maxx∈T

ψT (x) = 1.

With the previous definitions it is clear that Πhu is a linear operator. Furtherwe have ∫

TΠhu dx =

∫

Tu dx (5.7)

and therefore∫

Ω3D

ph div Πhu dx = −∫

Ω3D

∇ph · Πhu dx

= −∑

T∈T 3D

h

∇ph

∫

TΠhu dx

(5.7)= −

∑

T∈T 3D

h

∇ph

∫

Tu dx

= −∫

Ω3D

∇ph · u dx

=

∫

3DΩph div u dx.

So the first condition from lemma 5.3.3 is fulfilled. Proving the secondassumption, we first need the following standard results (see [24]):

|ψT |[H1(T )]3 ≤ chn/2−1T (5.8)

and ∫

TψT dx ≥ chn

T . (5.9)

So it follows with (5.6),(5.8),(5.9)

|Πhu|[H1(T )]3 ≤ |Rhu|[H1(T )]3 + |ψT |[H1(T )]3

∣∣∣∣

∫

T(u−Rhu) dx

∣∣∣∣

∣∣∣∣

∫

TψT dx

∣∣∣∣

−1

≤ c1|u|[H1(ωT )]3 + c2h−n/2−1T meas (T )1/2‖u−Rhu‖L2(T )

≤ c3|u|[H1(ωT )]3 .

Summation over all elements T concludes the proof.

Remark. For the three-dimensional problem with piecewise constant ansatzfunctions, theorem 5.3.1 can be applied.

62

5.4 Convergence and discretization errors

Up to now we do not know if the unique solution derived from our discretizedsystems, is an approximation for the infinite-dimensional one. So we haveto prove that for h→ 0, the discrete solution converges. The next lemmataconcerning the convergence of the finite element method, as well as thetheorem for the two-dimensional case can be found in [14].

Further in this section we will give error estimates for the velocity u forthe two-dimensional problem as well as for the three-dimensional problemwith continuous pressure. The result for the two-dimensional problem canbe found in [14]. For the three-dimensional problem see [25].

5.4.1 The two-dimensional problem

Proving the convergence of the discrete solution requires the next lemma.

Lemma 5.4.1. ∀ v ∈ V0 = H10 (Ω) there exists vh ∈ V0,h such that vh → v

strongly in H10 (Ω)

Proof. We take an arbitrary v ∈ D(Ω), since the space of test functions

D(Ω) := v ∈ C∞(Ω)∣∣ ∃K ⊂ Ω compact = C∞

0 (Ω)

is dense in H10 (Ω). Let a triangle T ∈ Th be defined by its nodes M1,M2,M3,

we then introduce the affine linear function ΦT on T (the nodal basis) suchthat

ΦT (Mi) = v(Mi), for i = 1, 2, 3

We now define vh byvh = ΦT in T. (5.10)

It is shown in [8], using Taylor series expansion, that

|ΦT (x) − v(x)| ≤ c1(v)h ∀x ∈ T

and| gradΦT (x) − grad v(x)| ≤ c2(v)

√h ∀x ∈ T

where the constants c1(v), c2(v) depend on v. Together with (5.10) we getthat

vh → v in H10 (Ω).

Theorem 5.4.2. Let Th be an admissible and shape regular mesh, then thesolution uh of problem 5.2 converges strongly to the solution u of problem3.2 in H1

0 (Ω).

63

Proof. The soluion uh of problem 5.2 is given by

µa2(uh, vh − uh) + τcj(vh) − τcj(uh) ≥ (c, vh − uh) ∀ vh ∈ V0,h.

By choosing test functions vh = 0, we get

µa2(uh, uh) + τcj(uh) ≤ (c, uh).

Since j(uh) ≥ 0, it follows with Friedrich’s inequality

‖uh‖1 ≤ cF ‖c‖0 ∀h.

Now, from lemma 5.4.1 we get, for rh being the nodal interpolation operatoron V0,h, that

rhv ∈ V0,h ∀h, and rhv → v strongly in H10 (Ω).

As u is the solution of problem 3.2 and uh ∈ V0,h ⊂ V0, we can choose v = uh

and get

µa2(u, uh − u) + τcj(uh) − τcj(u) ≥ 〈c, uh − u〉 ∀h. (5.11)

In the same way with v = rhu ∈ V0,h ⊂ V0 in problem 5.2, we get

µa2(uh, rhu− u) + τcj(rhu) − τcj(uh) ≥ (c, rhu− uh). (5.12)

Adding (5.11) and (5.12) yields

µa2(uh − u, uh − u) ≤ a2(uh, rhu− u) + τc(j(rhu) − j(u)) + (c, u− rhu).

From this we get

‖uh − u‖21 ≤

(

cF ‖c‖0 + τc√

meas(Ω))

‖rhu− u‖1 + ‖c‖0‖rhu− u‖0.

From the proof of lemma 5.4.1 we have for the nodal interpolation operatorthe estimates

‖rhv − v‖1 ≤ c1(v)√h

and‖rhv − v‖0 ≤ c2(v)h.

Using the Galerkin orthogonality

a2(vh, rhv − v) = 0 ∀ v ∈ V, ∀ vh ∈ Vh,

we get that‖uh − u‖2

1 ≤ ατc√h+ βh

with α, β being independent of τc, h. So we have shown that (uh)h is stronglyconvergent in H1

0 (Ω).

64

Remark. Theorem 5.4.2 also provides an error estimate for the approximatesolution uh:

‖uh − u‖1 = O(4√h).

Under more restrictive assumptions, one can show that

‖uh − u‖1 = O(h| log h|1/2).

For this we refer to [14] and the references therein.

5.4.2 The three-dimensional case using piecewise linear func-

tions for p

The following results are from [25]. Looking at the three-dimensional prob-lem 5.5, we need the following standard assumptions:

a(u, v) ≤ α2‖u‖1‖v‖1

a(u, u) ≥ α1‖u‖21

b(v, q) ≤ β2‖v‖1‖q‖0

j(u) − j(v) ≤ β1‖u− v‖1

for all u, v ∈ V 3D

0 and for all p, q ∈ Q. These assumptions are all fulfilledas we have already shown in chapter 2. Further we need the followingassumptions on the mesh-dependent bilinear form ch(·, ·):

(C1) ch(p, q) is defined for any couple of functions p, q ∈ H1(Ω3D).

(C2) [·]h, defined by [qh]2h = ch(qh, qh) is a norm.

(C3) ∀ ph, qh ∈ Qh, ch(ph, qh) ≤ [ph]h[qh]h

(C4) There exists a positive constant independent of h, γ, and k > 0 suchthat

∀ vh ∈ V 3D

h ,∀ qh ∈ Qh, b(vh, qh) ≤ γ1

hk‖vh‖0[qh]h

(C5) There exists a positive constant independent of h, c, such that

∀ q ∈ H1(Ω3D), [q]h ≤ chk‖q‖1.

Conditions (C1),(C2),(C3),(C5) are easily checked. Showing (C4) requiresthe following lemma:

65

Lemma 5.4.3. Let the family of triangulations be quasi-uniform. Thenthere exists a constant c > 0, independent of h such that, for any uh ∈ V 3D

0,h

and qh ∈ Qh, we have

|b1(vh, qh)| ≤ c1

h‖vh‖0[qh]h.

Proof. First we deduce that α(T ) = O(h2T ) and therefore α(T ) ≥ cαh

2T . We

then have

[qh]2h =∑

T∈T 3D

h

α(T )

∫

T∇qh∇qh dx ≥ min

T∈T 3D

h

α(T )∑

T∈T 3D

h

∫

T∇qh · ∇qh dx

≥ cα(minhT

h2)2h2|qh|21

≥ cαcbh2|qh|21,

since the ratio hT /h is bounded away from zero by cb (quasi-uniform). Soon the one hand we have

|qh|1 ≤ c1

h[qh]h (5.13)

with c =√

1/cαcb. On the other hand we get

b1(vh, qh) = −∫

Ω3D

qh div vh dx =

∫

Ω3D

∇qh · vh dx

and, therefore, using Cauchy-Schwarz inequality together with (5.13),

b1(vh, qh) =

∫

Ω3D

∇qh · vh dx ≤ ‖vh‖0|qh|1 ≤ c1

h‖vh‖0[qh]h

follows.

With it we have an appropriate error estimate

Theorem 5.4.4. Let u and p a solution of problem 3.1 and uh,ph a solutionof

µa(uh, vh − uh) + τcj(vh) − τcj(uh) + b1(vh − uh, ph) ≥ (f, vh − uh)

b1(uh, qh) − ch(ph, qh) = 〈Gh, qh〉for all vh ∈ V 3D

0,h and for all qh ∈ Qh, what is the equivalent mixed formulation

to problem 5.5. Further we assume that u ∈ [H2(Ω3D)]3 and p ∈ H1(Ω3D).Then the following error bound holds:

‖u− uh‖1 ≤ ch1/2(|u|2 + |p|1)Proof. The technical proof can be found in [25].

Remark. The convergence of the discrete solution for the approximationwith constant functions follows from theorem 5.4.2.

Remark. The regularity condition u ∈ [H2(Ω3D)]3 is a result which can befound in [14].

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Chapter 6

Numerical Results

In this chapter we want to present our numerical results. For the scalartwo-dimensional problem, we tested our algorithm by comparing with theresults from [18]. Then we will test the three-dimensional algorithm usingspecial settings which lead to the results of the two-dimensional problem.At the end we will show results for the three-dimensional problem as wellas for a three-dimensional problem related to a practical application.

6.1 Scalar problem in 2D

6.1.1 Realization

For the two-dimensional problem we implemented algorithm 4.1.6 in C++.In each step of the fixed-point iteration we solved the arising linear system forun+1

h,m with a preconditioned CG-Solver. For preconditioning we used a simpleJacobi-preconditioner. The Dirichlet boundary conditions are implementedby a penalty term. For a better comparison, we used the same parametersas it was done in [18]:

ε = ∆t =1

µ, µ = 1.0, r =

µ

τc.

For the drop in pressure per length, we choose c = 10. Our yield limit τc ischosen to be 0.5, 1.5 and 2.5. To initialize algorithm 4.1.6 we have used

u0h = 0, w0

h = 0.

Concerning the stopping criterion, we used for the fixed-point iteration

‖wn+1h,m+1 − wn+1

h,m ‖0 ≤ 10−4 or m ≥ 5.

So the inner iteration is bounded by 5 steps. In the experiments it turnedout that this bound is enough. For the uzawa-like iteration (the outer one)we used, analogously to the related paper, that

‖∇un+1h −∇un

h‖0 ≤ 10−6.

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6.1.2 Results

In figure 6.1 we see the discretized quadratic cross section of our domain Ωusing triangles. We then used various values for the yield limit τc in orderto see properties of the solution vector u. If τc increases, then the rigidbehavior grows. One can see that for τc being large enough, no flow exists,since the domain of rigid behavior then captures the whole domain and dueto boundary conditions u = 0 follows. In figure 6.2 the rigid phase is verysmall, whereas in figure 6.3 we see that in the center a large region withD(u) = 0 is given.

Figure 6.1: The discretization of the domain Ω with triangles.

Concerning the convergence rate we see in figure 6.7 and figure 6.8 thatthe convergence rates for u and w slightly improve if we use finer meshes.Further it can be seen, that for a fixed mesh size the convergence gets worseif we increase the yield limit. This was already mentioned. Moreover we seethat adding the stabilization to the projection has the desired effect thatthe algorithm does not deteriorate (see figures 6.5 and 6.6). Finally in table6.4 uzawa-iteration numbers for various values of h and τc are given.

Our results are in agreement with them from [18].

68

Figure 6.2: Plot for the scalar velocity u with τc = 0.5. Discretization with16641 nodes. The rigid behavior of the flow can be seen in the center of thedomain (constant velocity).

Figure 6.3: Plot for the scalar velocity u with τc = 2.5. Discretization with16641 nodes. The rigid behavior of the flow can be seen on nearly the wholedomain.

69

Number of Uzawa-iterations

τc = 0.5 τc = 1.5 τc = 2.5 τc = 3.5

h = 1/32 : 1089 nodes 31 80 196 26h = 1/64 : 4225 nodes 29 86 158 26h = 1/128 : 16641 nodes 26 67 179 26

Figure 6.4: Number of outer iterations for various discretizations and variousvalues of τc. We see that the iteration numbers vary if τc increases, but staysnearly the same if h varies.

Figure 6.5: Visualization of ‖∇un+1h −∇un

h‖0 (y-axis) versus iteration num-bers (x-axis). Discretization with h = 1/128. The convergence rate getsworse if τc increases.

70

Figure 6.6: Visualization of ‖wn+1h −wn

h‖0 (y-axis) versus iteration numbers(x-axis). Discretization with h = 1/128. The convergence rate gets worse ifτc increases.

Figure 6.7: Visualization of ‖∇un+1h −∇un

h‖0 (y-axis) versus iteration num-bers for τ = 0.5. Discretization with various meshes. The convergence rateimproves for finer meshes.

71

Figure 6.8: Visualization of ‖wn+1h −wn

h‖0 (y-axis) versus iteration numbersfor τ = 0.5. Discretization with various meshes. The convergence rateimproves for finer meshes.

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6.2 The three-dimensional problem

6.2.1 Test

Similar to the previous section we will test our algorithm. Using the firstapproach, we choose our parameters in the same way as we did it in theorem3.2.2. So, under the assumptions

f = 0, c = 10.0, p = −cz, τc = 0.5,

and the choice

ε = ∆t =1

µ, µ = 1.0, r =

√2µ

τc, r =

r√2τc

=µ

τ2c

,

we will compare the results from the three-dimensional algorithm with theresults from the two-dimensional one. The error of u is again computed by‖∇un+1

h −∇unh‖0 and the error for λ is ‖wn+1

h −wnh‖0. The stopping criterion

is the same as in the 2D case:

‖wn+1h,m+1 − wn+1

h,m ‖0 ≤ 10−4 or m ≥ 5

for the inner iteration, and

‖∇un+1h −∇un

h‖0 ≤ 10−6

for the outer iteration. The initializations for u and λ are also the same asin the 2D case. For the implementation of the periodic boundary conditionswe refer to [29]. The discretization of Ω3D is done with tetrahedrons (seefigure 6.14). In figures 6.9 and 6.10 the test of the algorithm is presented.It can be seen that in each step of the algorithm, the error for u and λ inthe three-dimensional case coincides with the error in the two-dimensionalcase.

Next we are going to apply the same algorithm without prescribing p.All other parameters stay the same. To initialize the pressure we use

p0h = −4z2 − 6z.

The error for p is computed by

‖pn+1h − pn

h‖0.

In figures 6.11-6.13 the error between the solution u from algorithm 5.2.3and the two-dimensional solution, computed by algorithm 5.2.2, in the eu-clidean norm is plotted in the plane z = 0. One can see that there areminimal variations in some nodes. These variations makes sense, since the

73

Figure 6.9: Verification of the 3D algorithm for u in the case h = 1/8, τc =1.5. Visualization of ‖∇un+1

h −∇unh‖0 (y-axis) versus outer iteration number

(x-axis).

finite element space can not exactly approximate the pressure p = −cz.

The absolute value of the solution u as well as the pressure p can then beseen in figures 6.15 and 6.16 respectively. In figure 6.15 the rigid behavior ofthe flow is located in the center of the prism. This part is very small sincethe yield limit is τc = 0.5. In figure 6.16 the linear behavior of the pressurecan be seen.

We have seen that our algorithm in the three-dimensional case fur-nishes results which, on the one hand coincide with the results in the two-dimensional case if we prescribe the pressure, and on the other hand are agood approximation of the two-dimensional results if we does not prescribethe pressure. Further we have seen that the part of the domain Ω3D wherethe medium acts like a rigid body is located in the center. For the choiceτc = 0.5 the rigid behavior is not very dominant.

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Figure 6.10: Verification of the 3D algorithm for λ in the case h = 1/8, τc =1.5. Visualization of ‖wn+1

h − wnh‖0 (y-axis) versus outer iteration number.

Figure 6.11: Error in the first component u1 between the three-dimensionalsolution, using linear ansatz functions for p, and the two-dimensional solu-tion. Mesh size: h = 1/16, yield limit: τc = 0.5. The error is computed inthe euclidean norm.

75

Figure 6.12: Error in the second component u2 between the three-dimensional solution, using linear ansatz functions for p, and the two-dimensional solution. Mesh size: h = 1/16, yield limit: τc = 0.5. Theerror is computed in the euclidean norm.

76

Figure 6.13: Error in the third component u3 between the three-dimensionalsolution, using linear ansatz functions for p, and the two-dimensional solu-tion u. Mesh size: h = 1/16, yield limit: τc = 0.5. The error is computed inthe euclidean norm.

Figure 6.14: Discretization of the three-dimensional domain Ω3D with tetra-hedrons.

77

Figure 6.15: Visualization of the absolute value |u| of the three-dimensionalsolution, using linear ansatzfunctions for p. The discretization is done with4096 nodes. The part of the domain where the fluid acts like a rigid mediumcan be seen in the center of the prism.

78

Figure 6.16: Visualization of the hydrostatic pressure p of the three-dimensional problem using linear ansatzfunctions for p. The discretizationis done with 4096 nodes. The linear behavior of the pressure can be seen.

79

Figure 6.17: Comparison of the different algorithms. Visualization of‖∇un+1

h −∇unh‖0 (y-axis) for h = 1/16 and τc = 0.5 versus iteration numbers.

6.2.2 Comparison of the different algorithms

In chapter 4 we developed two different algorithms for the three-dimensionalproblem. There rests the question which one provides better convergenceproperties. In figures 6.17-6.20 the residual for the two algorithms as wellas for the two-dimensional algorithm is plotted. We see that first the stepfrom two into three dimensions is costly and second that the continuousapproximation of the pressure is leading to a better convergence rate. Table6.18 confirms this statement with iteration numbers for the two-dimensionalalgorithm, the three-dimensional algorithm using linear ansatz functions forp and the three-dimensional algorithm using constant ansatz functions.

80

Number of Uzawa-iterations for τc = 0.5

2D-algorithm 3D algorithm 3D algorithm5.2.2 5.2.3 5.2.4

64 nodes 19 125 100512 nodes 21 81 1884096 nodes 45 56 207

Figure 6.18: Number of outer iterations for various discretizations. Theyield limit is given by τc = 0.5.

Figure 6.19: Comparison of the different algorithms. Visualization of‖pn+1

h − pnh‖0 (y-axis) for h = 1/16 and τc = 0.5 versus iteration numbers.

81

Figure 6.20: Comparison of the different algorithms. Visualization of‖λn+1

h − λnh‖0 (y-axis) for h = 1/16 and τc = 0.5 versus iteration numbers.

82

6.3 An example related to a practical problem

At the end we wantto apply the three-dimensional algorithm to a problemwith practical background. In collaboration with dTech Steyr, we studiedthe behavior of a snow cover on a mountainside. Snow can be seen as amaterial with yield stress, so there are regions where the medium acts likea fluid as well as regions where the yield limit is not reached yet, i.e. wherethe medium acts like a rigid body. The latter are the more interesting, sincethese are the regions which eventually can slide down. Due to [22],[30] wedecided to use a Bingham model for simulating the flow on a snow coveredmountainside.

Our domain Ω is chosen to be a rectangular prism ([0, xmax]× [0, ymax]×[0, zmax]) with boundary Γ, under an angle of inclination Θ. We split the

Figure 6.21: Cross section in the xz-plane of the domain Ω rotated by Θwith rotation axis y.

boundary into four parts,

Γ = ΓD ∪ ΓN ∪ ΓPx ∪ ΓPy

whereΓD = [0, xmax] × [0, ymax] × 0

ΓN = [0, xmax] × [0, ymax] × zmaxΓPx = 0 × [0, ymax] × [0, zmax] ∪ xmax × [0, ymax] × [0, zmax]

83

ΓPy = [0, xmax] × 0 × [0, zmax] ∪ [0, xmax] × ymax × [0, zmax].

On the bottom of the snow cover,ΓD, the snow adheres to the mountainside,so we set a no-slip condition

u = 0 on ΓD.

On the top,ΓN , we demand that there are zero-tractions, i.e. this boundaryshould be strainless, so

σn = 0 on ΓN .

On the rest of the boundary, we want to simulate a very long (on ΓPx) aswell as very broad (on ΓPy) mountainside, so we set homogenous periodicboundary conditions:

u(0, y, z) = u(xmax, y, z) on ΓPx ,

u(x, 0, z) = u(x, ymax, z) on ΓPy .

Along x-direction the pressure decreases. The drop in pressure is directlyproportional to the difference in altitude ∆h of the snow cover. The propor-tionality factor is m ∗ g, where m denotes the mass of the snow cover and gthe acceleration of gravity. So

p(0, y, z) = p(xmax, y, z) +mg∆h on ΓPx ,

p(x, 0, z) = p(x, ymax, z) on ΓPy .

Our classical formulation then reads as follows:

Problem 6.1 (Classical Formulation for the snow cover). Find

u ∈ C2(Ω) ∩ C(∂Ω), p ∈ C1(Ω)

such that

∇p− div τ = f in Ω

τ = 2µD(u) + τcD(u)/(DII)1/2 ⇔ τ

1/2II > τc

D(u) = 0 ⇔ τ1/2II ≤ τc

div u = 0 in Ω

u = 0 on ΓD

σn = 0 on ΓN

u(0, y, z) = u(xmax, y, z) on ΓPx

p(0, y, z) = p(xmax, y, z) +mg∆h on ΓPx

u(x, 0, z) = u(x, ymax, z) on ΓPy

p(x, 0, z) = p(x, ymax, z) on ΓPy .

84

The volume force f is given by

f = (0, 0,−ρ ∗ g)

where ρ denotes the density.

For the numerical simulation we used the first approach developed inchapter 5. The regions where the snow acts like a rigid medium can be seenif the snow cover is high enough and the mountainside has a strong rise.Therefore we choose

zmax = 3m, Θ = 45.

The physical parameters for snow are given by

ρ = 400kg/m3, µ = 0.03kg ∗ sm

, τc = 750Pa

The mathematical parameters are chosen to be the same as there were forthe previous problem:

ε = ∆t =1

µ, r =

√2µ

τc, r =

µ

τ2c

.

Further we used the following stopping criterions for the outer iteration:

‖∇un+1h −∇un

h‖0 ≤ 10−5.

The stopping criterion for the inner iteration stays the same:

‖λn+1h,m+1 − λn+1

h,m ‖0 ≤ 10−4 or m ≥ 5.

In figure 6.22 the discretized domain is shown. Figure 6.23 then shows theplot for |u|. One can clearly see the regions where |u| = const., i.e. wherethe snow cover can slide down. Figures 6.24-6.26 show the error for u, p andλ. Using a discretization with 125 nodes, the algorithm converges after 249iterations.

85

Figure 6.22: Discretization of the domain with tetrahedrons.

Figure 6.23: Absolute Value of the velocity u. Discretization with 125 nodes.A large part of the domain involves the rigid behavior (constant velocity).

86

Figure 6.24: Visualization of ‖∇un+1h −∇un

h‖0 versus iteration number. Thealgorithm converges after 249 iterations.

Figure 6.25: Visualization of ‖pn+1h − pn

h‖0 versus iteration number.

87

Figure 6.26: Visualiztation of ‖λn+1h − λn

h‖0 versus iteration number.

88

Summary

The simulation of a Bingham fluid led to a mixed system involving a vari-ational inequality of the second kind. We have seen that the arising non-differentiable functional characterizes the difficulty of the problem. In orderto get rid of this we developed a dual dual formulation consisting of a sys-tem of equations and a projection. Although we did not have to treat theproblem of non-differentiability, there occurred the problem with the largekernel of the convex set Λ. For this reason the uniqueness of a dual variableis not given.

We presented two methods for solving the three-dimensional problem.The idea of using a time-dependent problem in order to approximate thesteady state problem, led to an uzawa-type method coupled with a fixed-point iteration together with a stabilization for the projection. The stabiliza-tion is necessary due to the non-uniqueness of the dual variable. Discretiza-tion with finite elements was done for both three-dimensional approaches.On the one hand we used a continuous approximation of the pressure stabi-lized with bubble functions on the other we chose a discontinuous approx-imation of the pressure without additional stabilization. We saw that wecan eliminate the additional degrees of freedom coming from the bubbles.In return we get an additional mesh dependent bilinear form. For both ap-proaches we derived existence and convergence results.

In the numerical results we have seen that the chosen approaches leadto an algorithm which works correctly. We have seen that the parameterrepresenting the yield limit causes higher iteration numbers if it increases.On the other hand it seemed that the iteration number decreases for finermeshes. Further we compared our results for the three-dimensional problem.It could be seen that the approach using the continuous pressure approxi-mation provides better convergence properties. Finally we have seen thatwe can apply the algorithm to a problem with practical background. Thebehavior of a snow cover on a mountainside was computed. If the angle ofinclination and the height of the snow cover are big enough, a domain inwhich the fluid acts like a rigid medium can be seen.

89

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Eidesstattliche Erklarung

Ich, Philipp Laaber, erklare an Eides statt, dass ich die vorliegende Diplo-marbeit selbststandig und ohne fremde Hilfe verfasst, andere als die angegebe-nen Quellen und Hilfsmittel nicht benutzt bzw. die wortlich oder sinngemaßentnommenen Stellen als solche kenntlich gemacht habe.

Linz, Marz 2008

Laaber Philipp

93

Curriculum Vitae

Name: Laaber PhilippNationality: AustriaDate of Birth: 12.11.1981Place of Birth: Steyr

Education:

1988-1992: Elementary school,Dietach

1992-2000: High School with an accent on natural sciences,Steyr

2001-2005: Baccalaureate studies in Technical Mathematics,Johannes Kepler University Linz

2005-2008: Master studies in Industrial Mathematics,Johannes Kepler University Linz

94