Diplomarbeit. Ausgefu hrt am Institut fu r. Analysis und Scientific Computing. der Technischen Universita t Wien. unter der Anleitung von

Transcription

1 Diplomarbeit Pressure Robust Discretizations for Navier Stokes Equations: Divergence-free Reconstruction for Taylor-Hood Elements and High Order Hybrid Discontinuous Galerkin Methods Ausgefu hrt am Institut fu r Analysis und Scientific Computing der Technischen Universita t Wien unter der Anleitung von Univ.Prof. Dipl.-Ing. Dr.techn. Joachim Scho berl durch Philip Lukas Lederer BSc Neuhauserstraße 5, 53 Salzburg September 7, 16

2

3 Abstract This thesis focuses on a well-known issue of discretization techniques for solving the incompressible Navier Stokes equations. Due to a weak treatment of the incompressibility constraint there are different disadvantages that appear, which can have a major impact on the convergence and physical behaviour of the solutions. First we approximate the equations with a well-known pair of elements and introduce an operator that creates a reconstruction into a proper space to fix the mentioned problems. Afterwards we use an H(div) conforming method that already handles the incompressibility constraint in a proper way. For a stable high order approximation an estimation for the saddlepoint structure of the Stokes equations is needed, known as the Ladyschenskaja-Babuška-Brezzi (LBB) condition. The independency of the estimation from the order of the polynomial degree is shown in this thesis. For that we introduce an H -stable extension that preserves polynomials. All operators and schemes are implemented based on the finite element library Netgen/NGSolve and tested with proper examples.

4

5 Acknowledgments I want to use this opportunity to express my gratitude to my supervisor Prof. Joachim Schöberl for supporting and improving this thesis. It was a very interesting and fascinating process of development over the last year and it helped me to enhance my knowledge of this topic. Discussions with him were always fruitful and lead me to finding solutions for arising problems. Next I want to thank Christoph Lehrenfeld for helping me to correct this work and always giving me useful and inspiring comments. Also I want to express my appreciation to my colleagues Christoph Wintersteiger, Matthias Hochsteger and Gerhard Kitzler. A lot of discussions and valuable remarks lead to an improvement of this thesis. Finally I want to thank my girlfriend Karoline and my family for supporting me during my studies.

6

7 Contents 1 Introduction 1 Navier Stokes equations 5.1 Description model Physical quantities Mass derivation Conservation of mass Conservation of momentum Stokes fluid Navier Stokes equations Stokes equations Discretization of the Stokes and the Navier Stokes problem Weak formulation Stokes problem Convective term of the Navier Stokes equations Analysis of the saddle point problem Abstract theory LBB-condition and the Brezzi theorem Analysis of the Stokes problem Approximation of the Stokes problem Basic results Error analysis Exact divergence-free Aubin Nitsche technique for the Stokes problem Finite elements for the Stokes problem Taylor-Hood element Discontinuous pressure Time discretization Basic definitions First order IMEX Second order IMEX Reconstruction operator for the Taylor-Hood element Basic definitions Vertex reconstruction Construction of R V h Error analysis

8 4.3 Element reconstruction Construction of R T h Error analysis Numerical examples Convergence rates for a smooth solution Stokes example with post processing Hagen-Poiseuille channel flow using the curl formulation of the Navier Stokes equations Two phase bubble Stokes example Hybrid discontinuous Galerkin method for the Navier Stokes equations Hybrid discontinuous Galerkin method for the Stokes equations Hybrid discontinuous Galerkin method Approximation of the Stokes equations High order discrete LBB-condition Local LBB-condition Global LBB-condition Numerical estimations Curl convection for the hybrid discontinuous Galerkin method Approximation error Numerical example H -extension Tangential extension E D Normal extension E N Correction theorem Proof of the H -continuous extension E Appendix Transport theorem of Reynold Jacobi Polynomials Sobolev Slobotezki space Finite Elements Interpolation spaces

9 1 Introduction Computational fluid dynamics, or as commonly abbreviated CFD, is a huge part in the field of numerical mathematics and engineering. For a long time, scientists and engineers have been trying to find a way to describe the motion of fluids and gases. Using the basic rules of physics, namely Newton s laws, one can derive different models for all kinds of fluids. This thesis considers the incompressible Navier Stokes equations which describe a flow of an incompressible fluid by the velocity u and the pressure p: u t ν u + (u )u + p = f in, div u = in, with suitable initial and boundary conditions. Due to the nonlinear structure and the incompressibility constraint this set of partial differential equations can not always be solved in an explicit form, and a discretization technique has to be used. The method considered in this thesis is a mixed finite element approximation, therefore the approaching derivations are treated in a weak sense. Although this works for a large amount of problems it may happen that the resulting solutions are affected with a large error and show a non physical behaviour. The occurring problems can be distinguished when we look at the error estimation of the mixed problem: u u h V inf u v h V + inf p q h Q. v h V h q h Q h We observe that the error of the velocity not only depends on the best approximation of the velocity, but also on the best approximation of the pressure. A pressure robust scheme, as it is mentioned in [Lin13], overcomes this dependency and also results in proper physical solutions. In this thesis we introduce pressure robust methods with an optimal convergence order. Outline of the thesis: In the first two chapters we focus on the derivation of the Navier Stokes equations and post some known results and approximation properties. We are going to distinguish the main causes for the resulting lack that we mentioned above and try to understand how we can solve this problem. In chapter three we introduce two versions of a reconstruction operator for the wellknown Taylor-Hood element. This operator corrects the divergence of the solution by equilibrating the error in a proper space. We show that this reconstruction does not affect the convergence order of the method, thus an optimal error estimation is provided. 1

10 1 Introduction We finish this chapter with some numerical examples, including steady and unsteady flows and a two phase Stokes example. In the fourth chapter we consider a discretization that was already introduced in a hybridised version of the method from Cockburn, Kanschat and Schötzau (see [CKS5]) by Christoph Lehrenfeld and Joachim Schöberl in [LS15], namely a high order hybrid discontinuous Galerkin ansatz. We also want to mention the work from Stenberg and Könnö, where a H(div) conforming ansatz was used for the Brinkman problem, see [KS11]. This mixed method fulfills the properties that lead to a proper physical description and an independent velocity error. Still, for a stable and optimal convergence it remains to show the independency of the LBB-constant β of the polynomial degree, so β β(k). For this proof we need an H -continuous extension that preserves polynomials. Under this assumption we show the k-robust LBB-condition and finish the chapter with a numerical example, namely an unsteady laminar flow around a cylinder. The aim of the last chapter is to show the existence of such an H -continuous extension. We split the result in three theorems. In the first step we show the existence of an extension for fixed values of the tangential gradient on the boundary. The second theorem is used to weakly correct the values of normal derivative under certain assumptions of the input. Finally we close the statement by showing that the error due to the assumptions of the second step is small enough. Implementation: All numerical examples were implemented and tested in the finite element library Netgen/NGSolve, see [Sch97] and [Sch14].

11 Notation: In this thesis we stick to the following notation: bounded subdomain of R or R 3 T quasi uniform mesh on h global element size of T ˆT reference element ω V vertex path ω T element path Π k (T ) polynomials of order k on T Π k (T ) element wise polynomials of order k Ih Πk standard nodal interpolator Ih BDMk BDM interpolator Π C h Clement operator Π F Fortin operator L projection on a Hilbert space Q P L Q P L (ω) R F P C R V h R T h L projection on constants on the domain ω linear mapping from the element to ˆT Piola transformation covariant transformation reconstruction operator defined by a vertex equilibration reconstruction operator defined by an element equilibration By a b we mean that there exists a constant c independent of a, b, k, h such that a cb 3

12

13 Navier Stokes equations In this chapter we derive the incompressible Navier Stokes equations which are a system of non linear partial differential equations and describe the motion of a fluid in space and time. For the derivation we concern conservation properties derived from physics. We proceed as in [SS14], [SA8] and [Bra15]..1 Description model A fluid can be described in two ways: i. Using a Lagrangian model, one considers a fluid as an amount of particles and describes them with their trajectories and their streamlines. Using this description, one can determine the position in space for every particle at each point in time. ii. Using an Eulerian model, one considers a fixed domain and describes the fluid by determining the velocity at each point in space for each point in time. In fluid dynamics the Eulerian model is more common (see [SS14]) and is also used for this thesis, but we want to remind that both descriptions can always be transformed into the other one.. Physical quantities In this thesis we assume a bounded fixed domain R d where d = or 3, with = Γ N Γ D, where Γ D = Γ in Γ wall is a Dirichlet boundary that describes either an inflow condition (u(x, t) = u in (x, t)) or a no-slip condition on walls (u(x, t) = ) and Γ N the Neumann boundary where we assume a do-nothing outflow. For the time we define an interval I T = [, T ] with T >. The physical quantities that appear are i. ρ(x, t) C 1 ( I T )...Density of the fluid ii. u(x, t) [C ( I T )] d...velocity of the fluid iii. p(x, t) C 1 ( I T )...Pressure of the fluid iv. f(x, t) [C ( I T )] d...volume force that acts on the fluid v. σ(x, t) [C 1 ( I T )] d d...stress tensor on the surface for all x and t I T. 5

14 Navier Stokes equations.3 Mass derivation Due to the two models for a fluid we have to consider two different derivations. Let b C 1 be an arbitrary function then we define i. b t...the change of time one observes at a fixed point x Db ii. Dt...The mass derivation which describes the change in time when one follows the particle. Using those two descriptions we observe that in a fluid with the velocity u we get the equivalence (see [SA8]).4 Conservation of mass Db Dt = b + (u )b. (.3.1) t Using the density ρ we can determine the mass of an arbitrary volume V (t) by m := ρ dx t I T. The conservation of mass implies that V (t) Dm Dt = D Dt V (t) ρ dx =. As V (t) is transported with velocity u in time, we have to be careful when we want to change the integral and the derivation. Using the mass derivation (.3.1), the Reynolds transport theorem (see appendix, 7.1.1) and V = V () we can show that D Dt V (t) ρ dx = V Dρ + ρ div u dx = Dt and thus, as V was arbitrary, we get the point wise equivalence ρ + div (ρu) =. t V ρ + div ρu dx =, t In this thesis we always assume a constant density in space and time, so we get the well-known incompressibility constraint for the velocity.5 Conservation of momentum div u =. (.4.1) The first law of classical mechanics (Newton s first law) states that the rate of change of the momentum of a body is balanced by the forces applied on this body: DP Dt = F. 6

15 .6 Stokes fluid Considering a fluid we can define the momentum of a body, thus a constant set of particles, by P = ρu dx. V (t) Using the stress tensor σ on the boundary and the volume force f as acting forces on the fluid we get D ρu dx = σ n ds + ρf dx. (.5.1) Dt V (t) V (t) V (t) Similar to before, due to the the properties of the mass derivation and the transport theorem of Reynolds (7.1.1), D ρu dx = ρ Du Dt V V Dt dx = div σ + ρf dx, (.5.) V and so, as V was arbitrary, we get the equations of the conservation of momentum.6 Stokes fluid ρ Du Dt = div σ + ρf. To close the set of equations we have to derive a dependency of the stress tensor σ of the velocity u and the pressure p. For this thesis we consider a Newtonian and Stokes fluid which has the properties ([Bra15]) i. σ = σ(ε(u)) with ε(u) = 1 ( u + ut ) ii. σ is homogeneous iii. σ is isotropic iv. If ε(u) = then σ ij = pδ ij. Using those properties we get or u t ρ Du Dt = µ u p + ρf, ν u + (u )u + p = f, (.6.1) where µ is the dynamic viscosity, ν = µ ρ is the kinematic viscosity and p = p ρ is a scaled pressure. Remark 1: From now on we always write p for the scaled pressure. 7

16 Navier Stokes equations.7 Navier Stokes equations The incompressibility constraint (.4.1) and the equations derived from the conservation of momentum (.6.1) are called the unsteady incompressible Navier Stokes equations: u t ν u + (u )u + p = f in div u = in u = u in on Γ in (.7.1) u = on Γ wall.8 Stokes equations A characteristic number for fluid dynamics is the Reynolds number defined by Re := UL ν, (.8.1) where U and L are a characteristic velocity and length. This number can be used to measure the ratio between inertia and friction forces. In the case of a steady flow ( u/ t = ) and a very small Reynolds number Re we can make an asymptotic expansion of the velocity to see that the convective term (u )u in the Navier Stokes equations (.7.1) vanishes ([Bra15]). We get the steady Stokes equations: ν u + p = f in div u = in u = u in on Γ in (.8.) u = on Γ wall 8

17 3 Discretization of the Stokes and the Navier Stokes problem In this chapter we present discretization techniques for the steady Stokes problem (.8.) and introduce a time discretization for the unsteady Navier Stokes equations (.7.1). We consider the weak formulation and analyze the problem to derive a condition that is necessary for a stable method. For the approximations we also study the properties of the error and identify a problem that arises using most of the standard methods. This leads us to the main aspects considered in chapter 4 and Weak formulation Stokes problem As the Stokes equations (.8.) can be seen as a set of partial differential equations for the velocity and the pressure we have two spaces for our solutions and test functions: V := [H 1 ()] d and Q := L = {q L () : q dx = }. (3.1.1) We multiply the first equation of (.8.) with a test function v V and the second equation with a test function q Q and integrate over to get ν uv dx + p v dx = fv dx div u q dx =. Next, assuming homogeneous boundary conditions u = on, we apply integration by part on the integrals of the first line ν uv dx + p v dx = ν u : v dx div v p dx, and so by defining two bilinear forms a(u, v) := b(v, q) := we get the variational formulation: ν u : v dx div v q dx, 9

18 3 Discretization of the Stokes and the Navier Stokes problem Problem 1: Find (u, p) V Q so that a(u, v) + b(v, p) = (f, v) L () v V b(u, q) = q Q (3.1.) Remark : To get a symmetric saddlepoint structure in the variational formulation the pressure p was scaled with a minus. Remark 3: The space Q was chosen as for v [H 1()]d and q H 1 () b(v, q) := div v q dx = v q dx, so the bilinear form b(v, q) does not change if we add a constant to q Convective term of the Navier Stokes equations To discretize the Navier Stokes equations we also have to find a weak formulation including the term (u u). For that we multiply the convective term with a test function v V and integrate over to get c(u, u, v) := (u )u v dx. By that we define the variational formulation Problem : Find (u, p) V Q so that a(u, v) + b(v, p) + c(u, u, v) = (f, v) L () v V b(u, q) = q Q. (3.1.3) A common approach to find a discretization of the Navier Stokes equations is to use this variational formulation as continuous basis, because the discretization properties of a convective term similar to (b )u, for an arbitrary but fixed function b, is well-known. A different approach is the following. We take a closer look on c(u, u, v): (u )u v dx = d i,j=1 u i ( x i u j )v j dx d = u i ( d u j )v j dx u i ( u i )v j dx x i,j=1 i x i,j=1 j }{{} d + u i ( u i )v j dx. x i,j=1 j }{{} A B 1

19 3.1 Weak formulation Using the product rule on one component we can write A as d i,j=1 u i ( u i )v j dx = x j u i ( u i )v j = 1 (u i ) v j, x j x j d i,j=1 1 (u i ) v j dx = 1 x j where u = u u. Next we integrate by part, to get 1 (u ) v dx = 1 (u ) div v dx + 1 (u ) v dx, (u )v n ds. For B we observe that in the case of i = j the terms vanish, and so by using the outer product (for d = 3) (u ) = u 1 x 1 u u 1 x u u 1 x 3 u x 1 u 3 x u 3 x 3 u 3 x 1 x x 3 and u 1 v 1 u 1 v u 1 v 3 v 1 u 1 v 1 u v 1 u 3 u v v u = u v 1 u v u v 3 v u 1 v u v u 3 u 3 v 1 u 3 v u 3 v 3 v 3 u 1 v 3 u v 3 u 3 u 1 v v 1 u u 1 v 3 v 1 u 3 = u v 1 v u 1 u v 3 v u 3, u 3 v 1 v 3 u 1 u 3 v v 3 u we can write B also as or d i,j=1 u i ( u j )v j u i ( u i )v j dx = (u ) : [u v v u] dx, x i x j (u ) : [u v v u] dx = ( u) (u v) dx. Using the identity a (b c) = c (a b) = (a b) c and u = curl u we get ( u) (u v) dx = (curl u u) v dx, and so (u )u v dx = u u) v dx + (curl 1 (u ) div v dx + 1 (u )v n ds. 11

20 3 Discretization of the Stokes and the Navier Stokes problem Defining c curl (u, u, v) := we get the variational formulation (curl u u) v dx, Problem 3: Find (u, p) V Q so that a(u, v) + b(v, p + 1/u ) + c curl (u, u, v) = (f, v) L () v V b(u, q) = q Q. (3.1.4) Remark 4: In this thesis we call (3.1.4) the curl formulation of the Navier Stokes equations. Remark 5: Due to b(v, p + 1/u ) the pressure p is scaled with 1/u. We call this scaled pressure p b := p + 1/u the Bernoulli pressure. Remark 6: As we defined the variational problem with homogeneous Dirichlet boundary conditions, the boundary integral (u )v n vanishes. When we want to solve a domain including Neumann boundaries, we have to include this integral on the right hand side of the variational formulation. 3. Analysis of the saddle point problem 3..1 Abstract theory A mixed variational formulation implies two Hilbert spaces V and Q, bilinear forms and continuous linear-forms a(u, v) : V V R, b(u, q) : V Q R f(v) : V R, g(q) : Q R. The problem is to find u V and p Q such that a(u, v) + b(v, p) = f(v) v V b(u, q) = g(q) q Q. One can also add up the two lines and define the bilinear form B(, ) : (V Q) (V Q) R by B((u, p), (v, q)) = a(u, v) + b(u, q) + b(v, p), (3..1) to write the mixed method as a single variational problem: Find (u, p) V Q : B((u, p), (v, q)) = f(v) + g(q) (v, q) V Q, (3..) see [Sch9]. 1

21 3. Analysis of the saddle point problem 3.. LBB-condition and the Brezzi theorem To guarantee a stable and unique solution for the mixed problem we introduce the theorem of Brezzi including the LBB-condition named after Olga Alexandrowna Ladyschenskaja, Ivo Babuška and Franco Brezzi. Theorem 3.1 (Brezzi s theorem). Assume that a(, ) and b(, ) are continuous bilinear forms a(u, v) α u V v V u, v V, b(v, q) β v V q Q v V, q Q. Assume a(.,.) is coercive on the kernel, i.e., a(u, u) α 1 u V u V with V := {v V : b(v, q) = q Q}, and the LBB-condition is fulfilled sup v V b(v, q) v V β 1 q Q q Q. (3..3) Then the mixed problem (3..) is uniquely solvable, and the solution fullfills the stability estimation u V + p Q c{ f V + g Q }, with the constant c depending on α 1, α, β 1, β. From the Brezzi theorem it also follows that the bilinear form B(, ) fulfills the inf-sup condition See [Sch9]. inf v,q V Q v,q sup u,p V Q u,p 3..3 Analysis of the Stokes problem B((u, p), (v, q)) ( v V + q Q )( u V + p Q ) β. We now want to use the Brezzi theorem for the Stokes problem. For the analysis we use the norms V := ν H 1 (), Q := 1 ν L (). 13

22 3 Discretization of the Stokes and the Navier Stokes problem We already defined the bilinear forms for the weak formulation (3.1.), so by using the Cauchy Schwarz inequality we see a(u, v) = ν u v dx C.S. u V v V, b(v, q) = div v q dx C.S. div v L () q L () v V q Q. On the kernel V := {v V : div v q dx = q Q} we observe coercivity by using the Poincare inequality (see appendix, theorem 7.18) a(u, u) α 1 u V, (3..4) where α 1 depends on the shape of. The LBB-condition was first shown by Nečas by proving an equivalent inequality for the dual operator of the divergence, namely the gradient. We observe that : L () [H 1()] d has a closed range and Nečas ([Neč67]) showed that By that ([Bra][154]) we get q L ( c() q H 1 () q Q. sup v V div v q dx v V β 1 q Q q Q. (3..5) Using Brezzi s theorem we find stable and unique solutions u and p of the steady Stokes problem. 3.3 Approximation of the Stokes problem In this section we want to take a closer look on the approximation of the saddle point problem. Note that the discrete version of the LBB-condition can not be derived from the continuous one (3..5) Basic results We define the finite-dimensional subspaces V h V and Q h Q. The h refers to a quasiuniform triangulation T (see appendix) which these approximation spaces are derived from. The discrete variational formulation is Problem 4: Find (u h, p h ) V h Q h so that a(u h, v h ) + b(v h, p h ) = (f, v h ) L () v h V h b(u h, q h ) = q h Q h. (3.3.1) 14

23 3.3 Approximation of the Stokes problem The continuity results of the bilinear forms follow from the infinite dimensional case and the coercivity on V h, := {v h V h : div v h q h dx = q h Q h } again can be derived from the Poincare inequality. The discrete LBB-condition sup v h V h div v h q h dx v h V β 1 q h Q q h Q h (3.3.) is the constraint that arises for the definitions of the approximation spaces. For example, it is not possible to use V h := [Π ()] and Q h := Π () (for d = ). In this thesis we use different couplings for V h and Q h. In section 3.4 we present two standard elements and in chapter 5 we use a hybrid discontinuous Galerkin ansatz introduced by Christoph Lehrenfeld and Joachim Schöberl ([LS15] and [Leh1]) Error analysis Due to an approximation we are always interested in the error u u h V and p p h Q. We show different results similar to [BF91]. Remark 7: In this section we always refer on the constants that appear in Brezzi s theorem 3.1 considered for the discrete Stokes problem (3.3.1). As mentioned α 1 depends on and α = 1, but also β = 1 as the scaling with ν is hidden in Q. Proposition 1. Let (u, p) be the solution of (3.1.) and (u h, p h ) be the solution of (3.3.1). Then we have ( u u h V 1 + α ) inf u v h α V + β inf p q h 1 v h V h, α Q. (3.3.3) 1 q h Q h Proof. Let v h be an arbitrary element of V h. As v h u h V h, we have (coercivity) α 1 v h u h V a(v h u h, v h u h ) = a(v h u, v h u h ) + a(u u h, v h u h ). For the second term we observe and so and a(u u h, v h u h ) = a(u, v h u h ) a(u h, v h u h ) = f(v h u h ) b(v h u h, p) f(v h u h ) b(v h u h, p h ) = b(v h u h, p p h ), α 1 v h u h V a(v h u, v h u h ) b(v h u h, p p h ) α v h u V v h u h V + β v h u h V p p h Q, v h u h V α α 1 u v h V + β α 1 p p h Q. 15

24 3 Discretization of the Stokes and the Navier Stokes problem Using the triangle inequality u u h V u v h V + v h u h V we get the result. Until now we only have an estimation on the manifold V h of discrete divergence-free test functions. For the error estimation on the full space V h we show Proposition. Let (u, p) be the solution of (3.1.) and (u h, p h ) be the solution of (3.3.1). Assume that β 1 is the constant of the discrete LBB-condition (3.3.). Then we have ( inf u v h V 1 + β ) v h V h, β 1 inf u v h V. v h V h Proof. Let w h be an arbitrary element of V h. First we observe that the discrete LBBcondition is equivalent to the existence of a solution of the variational problem Problem 5: For all g h Q h find u h V h so that b(u h, q h ) = (g h, q h ) L () q h Q h (3.3.4) u h V 1 β 1 g h Q. (3.3.5) We solve the equation b(r h, q h ) = (u w h, q h ) q h Q h, (3.3.6) and get r h V 1 β 1 u w h Q = 1 β 1 Next we define v h := r h + w h, and as b(v h, q h ) = b(r h, q h ) + b(w h, q h ) b(u w h, q h ) sup β u w h q h Q h q h Q β V. 1 = b(u, q h ) b(w h, q h ) + b(w h, q h ) =, the test function v h is an element of the kernel V h,. Finally we get ( u v h V = u w h r h V u w h V + r h V 1 + β ) u w h β V. 1 16

25 3.3 Approximation of the Stokes problem So all together we get Corollary 1. Let (u, p) be the solution of (3.1.) and (u h, p h ) be the solution of (3.3.1). Assume that β 1 is the constant of the discrete LBB-condition, then ( u u h V 1 + α ) ( 1 + β ) inf u v h α 1 β V + β inf p q h 1 v h V h α Q. 1 q h Q h (3.3.7) Two major aspects can be indicated looking at this error estimations. First of all, we observe that the error of the velocity approximation depends also on the solution of the pressure discretization, and secondly we notice that we can not guarantee a small error when the constant of the discrete LBB-condition tends to zero. Both problems can be eliminated by using proper spaces and a Fortin operator, respectively. Before we continue examining cases where we can improve the approximation error for the velocity we also bound the pressure error. Proposition 3. Let (u, p) be the solution of (3.1.) and (u h, p h ) be the solution of (3.3.1). Assume that β 1 is the constant of the discrete LBB-condition, then p p h Q (1 + β β 1 ) inf q h Q h p q h Q + α β 1 u u h V. Proof. For v = v h and q = q h subtract equation (3.3.1) from (3.1.) to get or a(u u h, v h ) + b(v h, p p h ) =, b(v h, q h p h ) = a(u u h, v h ) b(v h, p q h ). Using the discrete LBB-condition and the continuity of the bilinear forms we get and thus q h p h Q 1 β 1 b(v h, q h p h ) v h V = 1 β 1 a(u u h, v h ) b(v h, p q h ) v h V α β 1 u u h V + β β 1 p q h Q. p p h Q (1 + β β 1 ) inf q h Q h p q h Q + α β 1 u u h V. 17

26 3 Discretization of the Stokes and the Navier Stokes problem Discrete kernel subset property Analysing the proof of proposition 1 we see that the error of the pressure arises in the continuity estimation of b(v h u h, p p h ). We know that v h u h is an element of V h,, so the problem is that b(v h u h, p) does not have to be zero for every choice of V h and Q h. This is indeed the case when we have the discrete kernel subset property as in this case b(v h u h, p) =. V h, V, (3.3.8) Corollary. Let (u, p) be the solution of (3.1.) and (u h, p h ) be the solution of (3.3.1), and assume that V h, V h. Then we have ( u u h V 1 + α ) inf u v h α V. (3.3.9) 1 v h V h, Proof. Follows from proposition 1 and b(v h u h, p p h ) =. Fortin operator As mentioned above the case β 1 can cause a loss in precision and even a lack of convergence ([BF91][58]), so it is really important to study the behaviour of this constant. One approach is to use a Fortin operator Π F : V V h with the properties b(π F u u, q h ) = Π F u V c u V, q h Q h (3.3.1) with c c(h). If we have this properties we see that b(v h, q h ) b(π F v, q h ) b(v, q h ) sup sup v h V h v h V v V Π F = sup v V v V Π F v V sup v V b(v, q h ) c v V β 1 c q Q, and so the discrete LBB-condition follows from the continuous one. Using a Fortin operator also delivers an error estimation that is independent of the LBB constant β 1. Corollary 3. Let (u, p) be the solution of (3.1.) and (u h, p h ) be the solution of (3.3.1). Assume we have a Fortin operator Π F with the properties (3.3.1), then we get ( u u h V 1 + α ) u Π F u α V + β inf p q h 1 α Q. 1 q h Q h Proof. Follows from proposition 1 and the properties (3.3.1) of the operator Π F. 18

27 3.3 Approximation of the Stokes problem The construction of such a Fortin operator is normally not intuitively but is often constructed in the following way. First one constructs an operator Π F 1, which delivers a best approximation with a certain continuity. After that a second operator Π F provides local corrections to preserve the properties (3.3.1), so we have: Proposition 4. Assume Π F 1 : V V h and Π F : V V h such that Π F 1 u V c 1 u V, and for all u V b(π F u u, q h ) = q h Q h, Π F (u Π F 1 u) V c u V, then for Π F u := Π F (u ΠF 1 u) + ΠF 1 u properties (3.3.1) are fulfilled with c = c 1 + c. Proof. We observe b(π F u, q h ) = b(π F (u Π F 1 u), q h ) + b(π F 1 u, q h ) = b(u Π F 1 u, q h ) + b(π F 1 u, q h ) = b(u, q h ), and Π F u V Π F (u Π F 1 u) V + Π F 1 u V (c 1 + c ) u V Exact divergence-free In section we defined the weak formulation (3.3.1). Therefore, if one can solve the discrete problem, the velocity u h only fulfills div u h q h dx = q h Q h, which we call discrete divergence-free. In this section we want to analyze the problem under an even sharper assumption for the approximation. Assume we have the property that the divergence of a velocity test function v h is an element of the pressure space Q h, thus div V h Q h. (3.3.11) Then we have that from discrete divergence-free follows exact divergence-free, namely div v h q h dx = q h Q h div v h =, (3.3.1) 19

28 3 Discretization of the Stokes and the Navier Stokes problem and so also for the solution u h div u h =. We mention three advantages if this property is fulfilled. Remark 8: As div u L () a point evaluation is not legit, so div v h = is meant almost everywhere in. Remark 9: There are examples, which fulfill (3.3.11), e.g. the Scott Vogelius element (see [BF91]), but they are in general computationally expensive. In chapter 5 we use a recent approach that has the property (3.3.11) and is also cheap to compute. We also want to mention that one could use isogeometric methods from isogeometric analysis to fulfill this property. Discrete kernel subset property With (3.3.11) it is clear that we have V,h V as b(u h, q) = div u h q = q Q, }{{} = and so due to corollary we get a better approximation of the velocity. Energy losses Consider the unsteady Navier Stokes equations (.7.1) with density ρ = 1, ν = 1, no volume forces f and homogeneous Dirichlet boundary conditions. Due to the friction of the particles the kinetic energy should decrease in time, so d dt u L (). Using (.7.1) and integration by part we observe 1 d u dt L () = uu dx = u (ν u (u )u + p) dx = ν u : u (u )u u div u p dx. The convective term can then be written as u (u )u u dx = u i u dx = x i = 1 i=1,,3 u u dx = 1 i=1,,3 div u u dx, and, using the in compressibility constraint div u =, we get 1 d u dt L () = ν u : u dx. 1 u u i u dx x i

29 3.3 Approximation of the Stokes problem The question is if this is still valid for the approximation of u h : d u h dt L () = ν u h : u h dx 1 div u h u h dx div u h p h dx. }{{}}{{} = Due to u h / Q h, the convective part does not vanish as u h is only discrete divergence-free, but if (3.3.11) is fulfilled we also observe the proper physical behaviour d u h dt L () = ν u h : u h dx. Helmholtz decomposition The last point we want to mention is an algebraic property that arises from exact divergence-free functions and the resulting impact on the Stokes and Navier Stokes equations (see [Lin13]). First we observe that an arbitrary irrotational field φ and a divergence-free field ω with homogeneous boundary values are orthogonal with respect to the L scalar product as φ ω dx = φ div ω dx + φ ω n ds =. (3.3.13) Next we define the orthogonal complement of V due to the scalarproduct induced by the bilinear form a(, ) V := {v V : a(v, w) = w V }. By that and the orthogonality (3.3.13) we observe that the weak formulation of the Stokes equations (3.1.) splits into two parts. We get the variational formulation: Problem 6: and Find (u, p) V Q so that ν u : v dx = fv dx v V, (3.3.14) pv = fv v V. Assume we use a gradient field for the right hand side, namely f := φ, then we observe that ν u : v dx = f v dx = v V, and pv = f v dx v V, 1

30 3 Discretization of the Stokes and the Navier Stokes problem and thus the solution is (u =, p = φ), so the irrotational force was completely balanced by the pressure. When we want to approximate this we observe that in the case of div V h Q h the solution is given by (u h =, p h = PQ L h φ), where PQ L h is the L projection on Q h. This can be seen as for all v h V h we have div v h P }{{} Q L h φ dx = div v h φ = φv h, Q h and so (u h =, p h = P L Q h φ) solves (3.1.) with f = φ. In the case of using discrete divergence-free velocity fields this may not be fulfilled and we get a nonphysical velocity field u h (, ). Analyzing the curl formulation of the Navier Stokes equations (3.1.4) we recall the identity (u )u = ( u) u + 1 (u ), and the resulting pressure p + 1 u. Again using only discrete divergence-free test functions results in a bad approximation due to the appearing gradient term (u ). In chapter 4 we introduce a reconstruction operator to solve this problem and compare the results of discrete and exact divergence-free ansatz spaces in an unsteady Navier Stokes example Aubin Nitsche technique for the Stokes problem In this section we present a standard Aubin Nitsche technique to show that the convergence rate in the L norm of the velocity is one order higher than in the H 1 norm. For this we first define the dual problem of the Stokes problem using the bilinear form (3..1): Problem 7: Find (w, λ) V Q so that B((v, q), (w, λ)) = (f, v) L () v V, q Q. (3.3.15) For the Aubin Nitsche technique we have to assume a regularity property of the solution, which is presented in [KO76] on proper domains. Theorem 3.. Assume that the solution of the discrete Stokes problem (3.3.1) fulfills u u h H 1 () hk u H k+1 () and p p h L () hk p H k (), and that the dual problem fulfills an H /H 1 regularity, namely w H ( f L () and λ H 1 ( f L (), then we have the L estimation u u h L () hk+1 u H k+1 () + hk+1 p H k ().

31 3.4 Finite elements for the Stokes problem Proof. We first solve the dual problem with the error u u h as right hand side: B((v, q), (w, λ)) = (u u h, v) L () v V, q Q. Next we choose the test functions v := u u h and q := p p h, and get (u u h, u u h ) L () = B((u u h, p p h ), (w, λ)). Together with the Galerkin orthogonality of the Stokes problem (u u h ) w h + div w h (p p h ) = div (u u h )λ h =, for an w h V h and λ h Q h, and the standard interpolator Ih Π1 we get u u h L () = B((u u h, p p h ), (w I Π1 h and I Π h w, λ IΠ λ)). h (see appendix), Using the continuity of the bilinear form B (see Brezzi s theorem for the Stokes problem 3.1) we first see B((u u h, p p h ), (w Ih Π1 w, λ IΠ h λ)) u u h H 1 () w Ih Π1 w H 1 () + u u h H 1 () λ Ih Π λ L () + w Ih Π1 w H p p h L 1 () (), and so using the assumption of the discrete error and the error assumptions of I Πk h (k =, 1) we get u u h L () hk u H k+1 () h w H () + h k u H k+1 () h λ H 1 () + h w H () hk p H k 1 (). Together with the regularity estimates and after dividing one factor we have u u h L () hk+1 u H k+1 () + hk+1 p H k (). 3.4 Finite elements for the Stokes problem We have seen in the last sections that the choice of the approximation spaces has to fulfill the discrete LBB-condition (3.3.) to guarantee a stable and unique solution. In this chapter we introduce two different elements for the Stokes equations and show their approximation properties. Using the P P 1 Taylor-Hood element leads to a continuous pressure, but the test functions only preserve a discrete divergence-free property. How to eliminate this disadvantage is the subject of chapter 4. The P P element leads to a discontinuous pressure and an element wise conservation of the divergence, but has a worse error convergence rate. 3

32 3 Discretization of the Stokes and the Navier Stokes problem Taylor-Hood element Trying to solve the mixed formulation (3.3.1) we could make the choice of using the same polynomial degree for the approximation of the velocity and the pressure. This was quite famous when people started to approximate the Stokes equations as the results looked satisfying. Soon it came up that in this case the stability of the solution highly depends on the mesh as the kernel of the gradient can become big for some triangulations ([BF91][1]). This unpredictable behaviour of equal order interpolation methods for the velocity and the pressure was the impetus to find proper elements. An approach was given by Taylor and Hood ([HT73]) by using an approximation for the pressure of one degree lower than the velocity. We give a D example by analyzing the P P 1 element. For a given triangulation T we choose the spaces: V h := [Π (T )] [C ()] and Q h := Π 1 (T ) C (). Using a standard interpolation operator Ih V and a Clement operator (see appendix) ΠC h leads to v Ih V v H 1 () h v H 3 () and q Π C h q L () h p H (). Assuming enough regularity for the exact solutions u and p, we get the interpolation error of the solutions u h and p h of (3.3.1) for the Taylor-Hood element in the H 1 norm by and in the L norm (see 3.) u u h H 1 () + p p h L () h u H 3 () + h p H () u u h L () h3 u H 3 () + h 3 p H (), thus an optimal convergence order of the velocity and pressure error. For the discrete LBB-condition we refer to [GR86], but state the needed assumption on the mesh T : T has a set of interior nodes {V r } R r=1 such that {ω V r } R r=1 with ω Vr := T (3.4.1) T :v r T is a partition of Although the fact that the Taylor-Hood elements are quite popular due to an easy implementation and analysis, they do not have the property div V h Q h, and so only provide discrete divergence-free solutions Discontinuous pressure Instead of approximating the pressure with a continuous ansatz it is also possible to use a P P element, namely an approximation with polynomials of order two for the velocity and a piecewise constant approximation for the pressure, so V h := [Π (T )] [C ()] and Q h := Π (T ). 4

33 3.4 Finite elements for the Stokes problem By that the divergence-free condition can be written as div u h dx = u h n ds = T T, T T which reads as a conservation of mass on each triangle. Using a standard interpolation operator Ih V and the Clement operator (see appendix) ΠC h, we see that we lose one order of accuracy due to the poor approximation of the pressure v I V h v H 1 () h v H 3 () and q Π C h q L () h q H 1 (). To show the LBB-condition for the P P element, we construct a Fortin operator that fulfills (3.3.1) and proceed as in proposition 4. The idea of the construction of the second operator Π F is that b(u ΠF u, q h) = now reads as div (u Π F ( u) dx = u Π F u ) n ds =. Lemma 3.3. The choice T V h := [Π (T )] [C ()] and Q h := Π (T ) fulfills the discrete LBB-condition (3.3.). T Proof. Let Π F 1 be a Clement operator ΠC h. By that we get Π F 1 v V v V. v Π F 1 v Q h v V. Define Π F on all triangles in the following way (Π F v)(v i ) = i = 1,, 3 Π F v n ds = v n ds i, j = 1,, 3 i j, E ij E ij where V i are the vertices of the triangle. Note that Π F v V h, as the second condition can be fulfilled by a correct choice of (Π F u)(v ij) where V ij is the middle point of edge E ij. Then by construction we get div (v Π F v)q h dx = q h Q h, and using a scaling argument we get T Π F v V (T ) = Π F v V ˆv ( ˆT ) V ( ˆT ) 1 h v Q(T ) + v V (T ), 5

34 3 Discretization of the Stokes and the Navier Stokes problem where V (T ) and Q(T ) are with ν scaled H 1 (T ) and L (T ) norms, respectively. Next we observe that Π F (v Π F 1 v) V = Π F (v Π F 1 v) V (T ) T T 1 h (I Π F 1 )v) Q(T ) + (I Π F 1 )v) V (T ) v V. T T Defining Π F v := Π F 1 v + ΠF (v ΠF 1 v) and using proposition 4 the lemma is shown. 3.5 Time discretization In this section we want to discuss the discretization of the unsteady Navier Stokes equations (.7.1). For this we use an additive decomposition method, the IMEX scheme, which stands for an IMplicit EXplicit splitting method (see [ARS97]). The main idea is to handle the nonlinear convection term explicitly and use it as a force for the implicit scheme. The implicit part of the equation is the diffusion term and the incompressibility constraint. Due to that, we ensure that this constraint is fulfilled in each time step. For the ease, we use f := in this section Basic definitions For the time dependent discretization we use an approximation for the velocity u h and the pressure p h given by N u u h (x, t) := u i (t)ϕ i (x) and p h (x, t) := p i (t)ψ i (x), i=1 with u(t) := {u i (t)} Nu i=1, p(t) := {p i(t)} Np i=1 and {ϕ i} Nu i=1 as a basis for V h and {ψ i } Np a basis for Q h. By that we define the matrices M R Nu Nu M i,j := ϕ i ϕ j dx i, j = 1,..., N u, A R Nu Nu A i,j := ν ϕ i : ϕ j dx i, j = 1,..., N u, D R Nu Np D i,j := div ϕ i ψ j dx i = 1,..., N u and j = 1,..., N p, and for an arbitrary v := {v i } Nu i=1 and w := {w i} Nu C(v)w R Nu C i := (v h w h )ϕ i dx i=1 N p i=1 the vector N u N u with v h (x, t) := v i (t)ϕ i (x) and w h (x, t) := w i (t)ϕ i (x). i=1 i=1 i=1 as Using those definitions and the finite element discretization for the spatial domain (see (3.1. and (3.1.3)) we have the problem: 6

35 3.5 Time discretization γ γ 1 1 γ γ 1 γ γ γ γ 1 δ 1 δ δ 1 δ Table 3.1: Butcher tableaus for step, L-Stable IMEX scheme Problem 8: Find (u, p) R Nu+Np so that M u t + Au + Dp + C(u)u = in I T D T u = in I T u() = u (3.5.1) 3.5. First order IMEX Assume that the time interval I T is divided in equidistant steps =: t < t i < < t Nt := T, with length t = t i+1 t i for all i = 1,..., N t 1. We use the following approximation for the time derivation of u(t) u t := un+1 u n, t with u n := u(t n ), p n := p(t n ). Using an implicit Euler method for the stiffness A and the divergence constraint D and an explicit Euler method for the convection C leads to the system (M + ta)u n+1 + tdp n+1 = Mu n tc(u n )u n D T u n+1 =. Note that the convection term only appears on the right hand side of the system as the vector C(u n )u n can be calculated for each new time step Second order IMEX Instead of using first order Euler schemes for A, D and C we now use a diagonal Runge Kutta method. By that, the resulting IMEX scheme can be represented with the Butcher tableaus (3.1) with γ := and δ := 1 γ. The obtained systems are: (M + γ ta)u 1 + γ tdp 1 = Mu n γ tc(u n )u n D T u 1 = 7

36 3 Discretization of the Stokes and the Navier Stokes problem and (M + γ ta)u n+1 + γ tdp n+1 = Mu n t(1 γ)au 1 t ( (1 δ)c(u 1 )u 1 δc(u n )u n) D T u n+1 =. 8

37 4 Reconstruction operator for the Taylor-Hood element In the last chapter we introduced the P P 1 Taylor-Hood element using V h = [Π (T )] [C ()] and Q h = Π 1 (T ) C () as approximation spaces for the Stokes problem. The disadvantage of the element is that the velocity test functions preserve only a discrete divergence-free property div(v h ) q h dx = q h Q h. In this chapter we introduce two versions of reconstruction operators R h that cure this drawback. The first one deals with local problems on vertex patches and can handle the problem we observed due to the Helmholtz decomposition (3.3.3). The construction of the first operator is quite similar to an error estimator for the Maxwell equations introduced by Joachim Schöberl and Dietrich Braess, see [BS8]. The second one preserves also a proper approximation convergence and will fulfill enough regularity for an L error estimation using an Aubin-Nitsche duality argument. Remark 1: In this thesis we construct an operator for the P P 1 Taylor- Hood element, but we want to mention that the reconstruction is also possible for a bigger class of elements as for example the mini-element and high order Taylor-Hood elements. This is the topic of a paper in preparation with Joachim Schöberl, Alexander Linke and Christian Merdon. 4.1 Basic definitions For the reconstruction we define the spaces Σ h := BDM (T ) H(div)() and Qh := Π 1 (T )/R L (). (4.1.1) Note that Q h are piecewise polynomials of order one and do not have to be continuous over element edges. For the construction we use the L projection onto constants defined by P L (ω) R :L (ω) R u 1 u dx =: u ω ω ω 9

38 4 Reconstruction operator for the Taylor-Hood element for all ω, and define the vertex patch ω V by ω V := T and T ωv = {T : V T } T :V T for all vertices V T and the element patch ω T by ω T := T and T ωt = { T : T T } T :T T for all T T. On this patches we furthermore define the local spaces Σ h, (ω i ) := {τ h BDM (T ωi ) : τ h n = on ω i } and Qh (ω i ) = Π 1 (T ωi )/R, with i {T, V }. 4. Vertex reconstruction 4..1 Construction of R V h The main ides is to find a function σ h Σ h that fulfills div σ h = div u h exactly and then define the vertex reconstruction by R V h (u h) := u h σ h. To find σ h we solve local problems on all vertex patches ω V to get local solutions σ V. By summing up the local corrections we then have σ h. Problem 9: For a given div w h Π 1 (T ωv ), find (σh V, λ h) Σ h, (ω V ) Q h (ω V )/R so that σh V τ h dx + div τ h λ h dx = τ h Σ h, (ω V ) (4..1) ω V ω V div σh V ψ ω h dx = (φ V div w h )(ψ h ψ V h ) dx ψ h Q h (ω V )/R ω V ω V where φ V is the hat function corresponding to the vertex V. Remark 11: The right hand side of the second line of problem (4..1) is a valid linear form on the factorized space Q h (ω V )/R as it does not depend on the representative of one equivalence class. This is fulfilled as when we add a constant to ψ h it is eliminated on the right hand side as we subtract the mean value, and on the left side due to σh V = on ω V. 3

39 4. Vertex reconstruction Lemma 4.1. Equation (4..1) has a unique solution (σ V h, λ h) satisfying i. div σ V h = PL Q h (ω V ) ( (I P L (ω V ) R )φ V div w h ) on, (4..) where σ V h is trivially extended by onto. In particular, if then div σh V = PL Q (φ h (ω V ) V div w h ). ii. σh V L () h div w h L () (div w h, q h ) L () = q h Q h, Proof. First we show the existence and uniqueness of the saddlepoint problem (4..1). On the spaces Σ h (ω V ) and Q h (ω V ) we choose the norms and τ h Σh (ω V ) := τ h L (ω V ) + h div τ h L (ω V ), ψ h Qh (ω V ) := 1 h ψ h L (ω V ), and get a σ (σ h, τ h ) := σ h τ h dx C.S. σ h Σh (ω V ) τ h Σh (ω V ) ω V b σ (τ h, ψ h ) := div τ h ψ h dx C.S. τ h Σh (ω V ) ψ h Qh (ω V ). ω V For the coercivity on the kernel we point out that due to div Σ h, (ω V ) Q h (ω V )/R it follows that an element in the kernel fulfills an exact divergence-free property div τ h = and we get a σ (τ h, τ h ) = τ h τ h = τ h L (ω V ) = τ h Σ h (ω V ). ω V It remains to show the discrete LBB-condition to use Brezzi s theorem. We first show the LBB-condition on the reference patch ω V and then on ω V. It should be mentioned that there exist different reference patches due to the number of elements that belong to a vertex, but for each triangulation T there exist a finite number of reference patches. We use the BDM interpolator Ih BDM that provides (see appendix) b σ (I BDM h τ, ψ h ) = b σ (τ, ψ h ) ψ h Q h ( ω V ) and Ih BDM τ τ H H(div )( ωv ) 1 ( ω V ) τ [H 1 ( ω V )]. 31

40 4 Reconstruction operator for the Taylor-Hood element As [H 1 ( ω V )] H(div )( ω V ) we get for all ˆψh Q h ( ω V )/R sup τˆ h Σ h, ( ω V ) b σ (ˆτ h, ˆψ h ) ˆτ h H(div)( ωv ) b σ (Ih BDM ˆτ, sup ˆψ h ) ˆτ [H 1 ( ω V )] I BDM h ˆτ H(div)( ωv ) b σ (ˆτ, sup ˆψ h ) ˆτ [H 1 ( ω V )] ˆτ H 1 ( ω V ) Next we use the continuous LBB-condition (3..5) to get sup ˆτ h Σ h, ( ω V ) b σ (ˆτ h, ˆψ h ) ˆτ h H(div)( ωv ) ˆψ h L ( ω V ). To show the condition on ω V we choose for an arbitrary ψ h the functions ˆψ h = ψ h, and τ h := P( ˆτ h ), where P is the Piola transformation (see appendix). By that we get b σ (τ h, ψ h ) ω sup = sup V div τ h ψ h τ h Σ h, (ω V ) τ h Σh, τ h Σ h, (ω V ) τ h L (ω V ) + h div τ h L (ω V ) ω = sup V div ˆτ h ˆψh ˆτ Σ h, ( ω V ) ˆτ h L ( ω V ) + div ˆτ h L ( ω V ) ˆψ L h = 1 ( ω V ) h ψ h L (ω V ) = ψ h Qh (ω V ) Using Brezzi s theorem 3.1 the existence and uniqueness is proven. Next we observe that for ψ h R we get P L (ω V ) R ψ h = ψ h, and together with ω φ V div w h (ψ h ψ V h ) dx =, ω V and ω V div σ V h ψ h dx = ψ h ω V σ V h n dx =, it follows that the second line of problem (4..1) is also fulfilled for constants and thus we get div σh V ψ h dx = ω V ω φ V div w h (ψ h ψ V h ) dx ω V ψ h Q h (ω V ). Using the locality of the L projection on constants and an arbitrary q L () we see div σh V q dx = div σh ω V q dx = (φ V div w h )(I P L (ω V ) R )P L (q) dx Q V ω h (ω V ) V ( ) = (I P L (ω V ) R )(φ V div w h ) q dx, so the first equation is shown. If ω V P L Q h (ω V ) (div w h, q h ) L () = q h Q h, 3

41 4. Vertex reconstruction we get P L (ω V ) R (φ V div w h ) = 1 φ V div w h dx =, (4..3) ω V ω V as φ V Q h and therefore div σh V = PL Q (φ h (ω V ) V div w h ). The estimation for the norm is given due to stability estimation provided by Brezzi s theorem, thus σ V h L () = σ V h L (ω V ) P L Q h (ω V ) (φ V div w h ) Qh (ω V ) h div w h L (ω V ) h div w h L (). Now we can construct the reconstruction operator. Note that σh V a normal trace and by that we can define Σ h,(ω V ), so it has σ h := V T σ V h Σ h. Theorem 4.. The operator defined by R V h (w h) := w h σ h, fulfills i. If (div w h, q h ) L () = q h Q h, then (div R V h (w h), q h ) L () = q h Q h ii. w h R V h (w h) L () h div w h L () Proof. We observe that and div R V h (w h) = div w h div σ h = div w h div σh V V T = div w h ( ) P L (I P L (ω V ) Q h (ω V ) R )φ V div w h V T = div w h P L φ Q V div w h P L h Q h (ω V ) R V T }{{} =1 = (I P L Q h ) div w h L Q h, V T w h R V h (w h) L () = σ h L () σ V h L (ω V ) V T φ V div w h }{{} = P L (ω V ) V T h div w h L (ω V ) h div w h L (). 33

42 4 Reconstruction operator for the Taylor-Hood element 4.. Error analysis The reconstruction operator is used for the right hand side. Now we have the problem Problem 1: Find (u h, p h ) V h Q h so that a(u h, v h ) + b(v h, p h ) = (f, R V h (v h)) L () v h V h b(u h, q h ) = q h Q h. (4..4) Due to the Helmholtz decomposition, a discrete divergence-free velocity test function can result in a velocity u h (, ) when using a gradient field as force f = φ (see 3.3.3), but for the solution of the problem 4..4 we get a(u h, v h ) = ( φ, R V h (v h)) L () = (φ, div R V h }{{ (v h) ) } L () = v h V h, = and thus u h = (, ). For the error analysis we use the first lemma of Strang that provides an estimation for the velocity error (see [BF91][18]) and the error estimation from section (1) to get (f, w h ) u u h V inf u v h V + inf p q h Q + sup L (f, R V h (w h)) L. v h V h q h Q h w h V h w h V The last term can be estimated due to (f, w h ) L () (f, R V h (w h)) L () f L () wh R V h (w h) L () f L () h w h H 1 (). And so, using the error analysis for the P P 1 Taylor-Hood element (4..5) u u h V h u H 3 () + h p H () + h f L (). (4..6) For the pressure we also use the Strang lemma to get p p h V (f, w h ) inf p q h Q + u u h V + sup L (f, R V h (w h)) L. q h Q h w h V h w h V Similar to before we get for the Taylor-Hood element p p h L () h u H 3 () + h p H () + h f L (). (4..7) Here we see the problem. Due to the dominant h term, the error has a worse convergence rate than we would expect (O(h )). This is fixed in the next section. 34

43 4.3 Element reconstruction 4.3 Element reconstruction Construction of R T h Analogue to R V h we solve local problems to find a global reconstruction operator. The difference is that we use element patches (see section 4.1) instead and a smoothing operator on the test functions of the mixed problem. We define with the properties i. S Qh = id, ii. if q h R S( q h ) = q h, S : Q h Q h iii. the smoothing operator is quasi local, namely if q h = on ω T S( q h ) = on T. Using the smoothing operator for the right hand side we define the problem on ω T by Problem 11: For a given div w h Π 1 (T ωt ), find (σh T, λ h) Σ h, (ω T ) Q h (ω T )/R so that σh T τ h dx + div τ h λ h dx = τ h Σ h, (ω T ) (4.3.1) ω T ω T div σh T ψ h dx = div w h (ψ h S(ψ h )) dx ψ h Q h (ω T )/R. ω T T Lemma 4.3. Equation (4.3.1) has a unique solution (σh T, λ h) satisfying i. (div σh T, q h) L () = (div w h, q h S q h ) L (T ) q h Q h () ii. σh T H 1 () h div w h L () Proof. The existence and uniqueness is shown analogue as in lemma 4.1. Due to the second property of the smoothing operator S we have for a ψ h R div w h (ψ h Sψ h ) dx =, and due to the zero boundary values of σh T also T ω T div σ T h ψ h dx = ψ h ω T σ T h n dx =. By that the second line of (4.3.1) is also valid for constants and thus div σh T ψ h dx = div w h (ψ h S(ψ h )) dx ψ h Q h (ω T ). ω T T 35

44 4 Reconstruction operator for the Taylor-Hood element Using the quasi locality we can finally expand the equation on Q h. To see this we choose an arbitrary q h Q h and split it into q h = q h 1 + q h with supp( q1 h ) = /ω T and supp( q h ) = ω T. Then we get div σh T q h dx = div σh T q h 1 /ω T }{{} dx + div σh ω T q h dx = T = = div w h ( q h S( q h )) dx = div w h ( q h S( q h )) dx. T Finally we show the estimation in the dual norm. For that we choose an arbitrary g [Π (ω T )] and ψ Π 1 (ω T ) such that ψ = g and observe σh T g dx = σh T ψ dx = σh T nψ ds div σh T ψ dx ω T ω T ω T ω T = div w h (I S)ψ dx =, T }{{} = on T and so σ T h H 1 () = sup w [H 1 ()] = sup w [H 1 ()] = h σh T L (ω T ). σt h w dx w H 1 () = sup w [H 1 ()] σt h (I (ω T ) [PL R ] )w dx w H 1 () T ω T σ T h w dx w H 1 () Together with the estimation of Brezzi s theorem L (ω T ) h div w h L (ω T ), σ T h sup w [H 1 ()] σ T h L (ω T ) h w H 1 () w H 1 () we have σ T h H 1 ()) h div w h L (ω T ). Similar to before we can sum up all local reconstructions σ T h to define σ h := T T σ T h Σ h. Theorem 4.4. Let R T h w h := w h σ h then i. If (div w h, q h ) L () = q h Q h, then (div R T h (w h), q h ) L () = q h Q h. ii. w h R T h (w h) H 1 () h div w h L () 36

45 4.3 Element reconstruction Proof. For an arbitrary w h with (div w h, q h ) L () = q h Q h and a q h Q h we have: (div R T h (w h), q h ) L () = (div w h, q h ) L () T T (div σ T h, q h) L () = (div w h, q h ) L () T T (div w h, q h S q h ) L (T ) = (div w h, q h ) L () (div w h, q h ) L () + (div w h, S q h }{{} Q h ) L (). For the estimation in the dual norm we use the properties of each σh T 4.3, so for w h R T h w h = σ h we get σ h H 1 () = sup σ h w dx σh T w dx T T = sup w [H 1 ()] w [H 1 ()] = sup w [H 1 ()] sup w [H 1 ()] w H 1 () ω T σh T w dx T T w H 1 () T T h div w h L () = sup w [H 1 ()] w H 1 () T T σ T L h (ω T ) h w H 1 (ω T ) h w H 1 () T T shown in lemma ω T σh T (I (ω T ) PL R )w dx w H 1 () σh T L (ω T ) Remark 1: An example for the smoothing operator S from Π 1 (T ) Π 1 (T ) C () is given by averaging the values of the discontinuous linear polynomials in each vertex of T, also called Oswald-interpolator see [Osw93] Error analysis As for the first operator R V h we observe the same properties for the element reconstruction R T h with respect to the problems due to the Helmholtz decomposition (see section 4..). The main advantage can be seen in the error analysis. Again using Strang s lemma we get (f, w h ) u u h V inf u v h V + inf p q h Q + sup L (f, R V h (w h)) L v h V h q h Q h w h V h w h V p p h V (f, w h ) inf p q h Q + inf u u h V + sup L (f, R V h (w h)) L. q h Q h v h V h w h V h w h V The last term can be estimated using the dual norms (f, w h R T h (w h)) L () = w h R T h w h, f H 1 H 1 wh R T h (w h) H 1 () f H 1 () h w h H 1 () f H 1 (). 37

46 4 Reconstruction operator for the Taylor-Hood element Using the error analysis for the P P 1 Taylor-Hood element we get u u h V + p p h L () h u H 3 () + h p H () + h f H 1 (). We see that the error still has the convergence order O(h ), so the reconstruction operator had no impact on this property. For the L error we use an Aubin Nitsche technique as presented in 3. but due to the non conforming right hand side we can not proceed in the same way. As in section 3. we assume enough regularity of the dual problem and solve the problem with the error u u h as right hand side to get (u u h, u u h ) L () = B((u u h, p p h ), (w, λ)). Similar to the proof of theorem 3. we use the Galerkin orthogonality to subtract Ih Π1w and Ih Π λ, but due to the different right hand sides another term appears and we get u u h L () = B((u u h, p p h ), (w I Π1 h The first term is bounded similarly to theorem 3. B((u u h, p p h ), (w I Π1 h For the second term we use (f, I Π1 h w RT h (IΠ1 h w)) L () w, λ IΠλ)) (f, IΠ1 h w, λ IΠ h λ)) h u H 3 () h w H () Ih Π1 + h u H 3 () h λ H 1 () + h w H () h p H (). h w RT h (IΠ1 h w)). w RT h (IΠ1 h w) H 1 () f H 1 (), and theorem 4.4 Ih Π1 w RT h (IΠ1 h w) H 1 () h div I Π 1 h w L (). As the solution of the dual problem fulfills div w =, we can subtract this term to get Ih Π1 w RT h (IΠ1 h w) H div 1 () h I Π 1 h w div w L () and thus h I Π 1 h w w H 1 () h3 w H (), u u h L () h3 u H 3 () + h3 p H () + h3 f H 1 (). 4.4 Numerical examples In this section we present four different numerical examples using the reconstruction operators R V h and RT h. In the first example we compare the convergence orders of the standard Taylor-Hood element with and without the reconstruction for the right hand side. In the second example we solve a standard Stokes problem with non homogeneous 38

47 4.4 Numerical examples boundary conditions that induce a flow and use the operators to correct the divergence of the solution. In the third problem we concern a Hagen-Poiseuille channel flow by solving the unsteady Navier Stokes equations in time. The convection term is implemented in a curl formulation (see equation 3.1.4) which results in a bad approximation due to the Helmholtz decomposition (see 3.3.3) when we use discrete divergence-free test functions. Finally we look at a two phase Stokes flow where the pressure is discontinuous over the surface of a bubble Convergence rates for a smooth solution In the first example we want to examine the convergence rates of the pressure and the velocity error measured in the H 1 and the L norm. The domain is the unit square := (, 1) (, 1) and the exact velocity and pressure is given by u := curl ζ with ζ := x (x 1) y (y 1) p = x 3 + y 3 1. In figure 4.1 we can see the exact solutions. For the error observation we use different viscosities ν = 1, 1 3, 1 8 and either no reconstruction or R V h and RT h on the right hand side. (a) Velocity field and absolute value u (b) Pressure p Figure 4.1: Exact solutions of the first example 39

48 4 Reconstruction operator for the Taylor-Hood element h h 3 u u h H 1 () u u h L () p p h L () Number of dofs Figure 4.: Convergence rates of the P P 1 Taylor-Hood element for ν = 1 Classical Taylor -Hood In section we observed the classical convergence rates for the P P 1 Taylor-Hood element and u u h H 1 () + p p h L () h u H 3 () + h p H () u u h L () h3 u H 3 () + h 3 p H (). In figure 4. we observe the expected rates but we also distinguish a lack of accuracy that appears using a small viscosity, see figure 4.3. Although the convergence rate is still O(h ) the error gets really high. This can be seen when we look at the error analysis in section 3.3. as the Q norm was scaled by 1 ν and the V norm with ν and so u u h H 1 () inf u v h H v h V 1 () + 1 h ν inf p q h Q. q h Q h Note, that the pressure is induced by a gradient field (3x, 3y ), so an error in the approximation of the pressure results in an even bigger error for the velocity. Vertex reconstruction Now we use the vertex reconstruction R V h for the right hand side and again analyze the error. Due to the properties of the operator we expect a lower convergence rate, namely u u h H 1 () + p p h L () h u H 3 () + h p H () + h f L (), 4

49 4.4 Numerical examples 1 6 u u h H 1 () h ν = 1 ν = 1 3 ν = Number of dofs Figure 4.3: H 1 convergence rate for u of the P P 1 Taylor- Hood element for different viscosities so only a linear convergence. In figure 4.4 we see that the error of the velocity still has an O(h ) convergence, which is better than expected. The reason for this is not yet known, but still the convergence of the pressure error is less than compared to the Taylor-Hood element, namely O(h 3/ ). Anyway, the reconstruction helped in the dependency of ν as we can observe in figure 4.5, where we see no impact at all. Element reconstruction Finally we use the element reconstruction R T h for the right hand side. Again we observe the error and expect to see the same rates as for the Taylor-Hood element and u u h H 1 () + p p h L () h u H 3 () + h p H () + h f H 1 (), u u h L () + p p h L () h3 u H 3 () + h 3 p H () + h 3 f H 1 (), but no impact of the viscosity. In figure 4.6 we see, that compared to the vertex reconstruction, the pressure has also an O(h ) convergence rate and in figure 4.7 we can observe that changing the viscosity ν makes no difference. 41

50 4 Reconstruction operator for the Taylor-Hood element Number of dofs h h 3/ h h 3 u u h H 1 () u u h L () p p h L () Figure 4.4: Convergence rates of the P P 1 Taylor-Hood element for ν = 1 using R V h for the right hand side u u h H 1 () h ν = 1 ν = 1 3 ν = Number of dofs Figure 4.5: H 1 convergence rate for u of the P P 1 Taylor- Hood element for different viscosities using R V h for the right hand side 4

51 4.4 Numerical examples h h 3 u u h H 1 () u u h L () p p h L () Number of dofs Figure 4.6: Convergence rates of the P P 1 Taylor-Hood element for ν = 1 using R T h for the right hand side u u h H 1 () h ν = 1 ν = 1 3 ν = Number of dofs Figure 4.7: H 1 convergence rate for u of the P P 1 Taylor- Hood element for different viscosities using R T h for the right hand side 4.4. Stokes example with post processing In this example we use the reconstruction operator R T h as a post processing tool to generate a divergence-free velocity R T h (u h). The domain is the unit square := (, 1) 43

52 4 Reconstruction operator for the Taylor-Hood element (, 1) and we induce a flow by non homogeneous boundary conditions u = (, y(1 y)) on {} (, 1) u = (, ) on \ {} (, 1). The viscosity is chosen as ν = 1. In figure 4.8 the induced flow field u h and the pressure p h is shown, and in figure 4.9 we can observe an O(h) convergence of the error div u h L () although the error is really high, so the incompressibility constraint is not approximated very well. When we use the reconstruction the divergence is reduced to a value O(1e 16 ) for each refinement level. Due to the error analysis of the reconstruction operator, we also expect a good convergence in the H 1 and the L norm. (a) Velocity field and absolute value u h (b) Pressure p h Figure 4.8: Approximated solutions of the second example Hagen-Poiseuille channel flow using the curl formulation of the Navier Stokes equations Until now we only focused on the steady Stokes equations. In this example we want to analyze a Hagen-Poiseuille channel flow approximated by the unsteady Navier Stokes equations. We consider the domain = (, 1) (, ) with Dirichlet and Neumann boundaries Γ N := {1} (, ) and Γ D := \ Γ N, viscosity ν = 1 3 and together with c curl (u h, u h, v h ) = (curl u h u h ) v h v h V h 44

53 4.4 Numerical examples 1 1 h div u h L () Number of dofs Figure 4.9: div u h L () for the P P 1 Taylor-Hood element we solve the unsteady Navier Stokes equations using a first order IMEX scheme introduced in section 3.5. with timestep t =.1 and a steady Stokes solution as startvalue u(x, ) = u stokes. Using the boundary conditions the exact solution is given by: u = (1y( y), ) on {} (, ) u = (, ) on Γ D \ ({} (, ) Γ N ), u = (1y( y), ) and p =.x.1 (4.4.1) This example is chosen so that the convective term of the Navier Stokes equations is equal to zero Using the identity we get (u )u =. (u )u = curl u u + 1 u curl u u = 1 u. In this case the convective term is a gradient field and so due to the Helmholtz decomposition we expect a bad approximation using discrete divergence-free test functions like the P P 1 Taylor-Hood element preserves. To solve this problem we use the reconstruction operator in the convective term, namely c curl (u h, u h, v h ) = (curl u h R T h (u h)) R T h (v h) dx v h V h. 45

54 4 Reconstruction operator for the Taylor-Hood element We use the reconstruction operator on the one hand for the test function v h to get rid of the mentioned problem above and on the other hand on u h to make the term antisymmetric. In figure 4.1 we compare the two approximations at t =.5s. We see that the reconstruction operator has a major impact on the exactness of the solution. Remark 13: The curl formulation is chosen to present the properties of the extension operator. Although a standard convection formulation may result in a proper approximation for this example, there are cases where this is not valid anymore, so a curl formulation including the reconstruction can help. Figure 4.1: Absolute value u h of the approximation of the Hagen-Poiseuille flow at t =.5s. The upper picture is with the reconstruction R T h and the lower without Two phase bubble Stokes example The last example we give is a two phase Stokes flow. For this problem the exact solution is given by a discontinuous pressure that we approximate with continuous linear polynomials. Due to the coupling of the pressure and the velocity this bad approximation results in a non-physical behaviour of the velocity. We consider the domain = 1 with := {(x, y) : x + y.5} and 1 := ( 1, 1) ( 1, 1) \, 46

55 4.4 Numerical examples zero volume forces and introduce a force on the interface to solve: ν u + p = in i, i = 1, div u = in i, i = 1, [u] = on [ ν u n + p n] = f s on u = on. On the interface we use the surface tension force given by f s = τ κ n ds, where τ is a constant and κ is the mean curvature. In the case of a circle it is given by κ = 1 R, where R is the radius. For and τ = 1 this results in f s = n ds. With K := π 4 the exact velocity and pressure is given by u = (, ) in p =.5K in 1 p =.5K + in. We see that the pressure is discontinuous across the boundary and the velocity is constant zero. Using the standard Taylor-Hood element results in a bad approximation of the pressure and due to that an influence on the velocity so u h (, ). In figure 4.11 we see the solution for ν = 1e 3 for the Taylor-Hood element and in figure 4.1 using the reconstruction on the surface tension. In figure 4.13 we also plotted the pressure of the two solutions along the line from ( 1, ) to (1, ). One can see that the standard Taylor-Hood element produces oscillations in contrast to the reconstruction. 47

56 4 Reconstruction operator for the Taylor-Hood element (a) Absolute value u h (b) Pressure p h Figure 4.11: Approximated solutions of the fourth example with the standard Taylor- Hood element (a) Absolute value u h (b) Pressure p h Figure 4.1: Approximated solutions of the fourth example using the reconstruction operator on the surface tension 48

57 4.4 Numerical examples 1 p T H p R T h Number of dofs Figure 4.13: Pressure of the fourth example from ( 1, ) to (1, ) for the Taylor-Hood element with and without using the reconstruction operator on the surface tension In figure 4.14 we see the error of the standard Taylor-Hood element. The L error of pressure and the H 1 error of the velocity seem to converge only with order O(h 1/ ) and the L error of the velocity converges with a rate between O(h) and O(h 3/ ). Still, all errors are really high. In figure 4.15 we can see the effect of using the reconstruction operator R T h on the surface tension force. The L and H 1 error of the velocity is reduced close to zero and the L error seems to converge with order O(h 1/ ). The L pressure error (see figure 4.16) also seems to converge with order O(h 1/ ) h 1/ h h 3/ p p h L () u u h L () u u h H 1 () Number of dofs Figure 4.14: Error convergence of p h and u h for the two phase bubble flow without the reconstruction of the surface tension 49

58 4 Reconstruction operator for the Taylor-Hood element 1 1 h 1/ u u h L () u u h H 1 () Number of dofs Figure 4.15: Error convergence of u h for the two phase bubble flow with reconstruction of the surface tension 1 h 1/ p p h L () Number of dofs Figure 4.16: Error convergence of p h for the two phase bubble flow with reconstruction of the surface tension 5

59 5 Hybrid discontinuous Galerkin method for the Navier Stokes equations In this chapter we use a discretization method that was introduced by Joachim Schöberl and Christoph Lehrenfeld in [Leh1] and in [LS15] as a modified version of the method from Cockburn, Kanschat and Schötzau (see [CKS5]). In section 3.4 we already introduced a finite element pair with a discontinuous pressure that leads to an element wise divergence preserving velocity field but has a worse convergence order. The introduced method goes even further by using a hybridised discontinuous Galerkin approach for the velocity as well as a discontinuous pressure approximation. In the first section we present the method and show some existing results. We see that there already exists an h independent version of the discrete LBB-condition, so showing also the independence of the polynomial order k is the outcome of the second section. In the third section we introduce an implementation of the curl convection and finally show a numerical result in the last section. 5.1 Hybrid discontinuous Galerkin method for the Stokes equations Hybrid discontinuous Galerkin method We want to use a discontinuous Galerkin (DG) approach but want to construct a method that can generate exact divergence-free flow fields. This yields to an H(div)() conforming discretization, namely to normal continuous velocity components over edges of finite elements. Therefore we introduce the space W h := {u T h [Πk (T )] d : [u h n] E = E F}, where E is an element of the triangulation skeleton F and [[u h n] E is the jump on the common edge E of two elements T 1 and T [u h n] E = ( u h T1 u h T ) n1. For the pressure we use the space Q h := Π k 1 (T ). By that we get the property div (W H ) = Q h, see ([LS15]). As functions of W h do not belong to H 1 () anymore we use the interior penalty method introduced in [Arn8] to weakly enforce continuity of the velocity flow field but use a hybridised version to avoid a full coupling between neighbours. For this we add additional unknowns on the skeleton and define the facet space as F h := {u F h [Πk (F)] d : u F h n = }, 51

60 5 Hybrid discontinuous Galerkin method for the Navier Stokes equations for the tangential trace of the velocity and introduce the velocity space as V h := W h F h. Remark 14: The space F h was only introduced to implement a more efficient handling of linear systems, see [LS15], in fact it is essentially the same as the interior penalty DG formulation. Remark 15: For the construction of W h we used BDM k finite elements Remark 16: In this thesis we only consider the two dimensional case d =. By that we can define a tangential vector τ on each edge Approximation of the Stokes equations Using the interior penalty method [Arn8] and the notation u h = (u T h, uf h ), v h = (vh T, vf h ) and [vh τ ] := [(vt h vf h ) τ]τ and [uτ h ] := [(ut h uf h ) τ]τ we define the bilinear forms for the Stokes problem as in [LS15] by a HDG (u h, v h ) := ν u T h : vt h dx ν ut h T T T T n [vτ h ] ds ν vt h T n [uτ h ] ds + α T h [uτ h ][[vτ h ] ds, b HDG (u h, q h ) := div u T h q h dx. T T T Using the norm v h H 1 HDG () := T T v h L (T ) + k h [vτ h ] L ( T ) we can show for sufficiently large α continuity of the bilinear forms a HDG (, ) and b HDG (, ) and coercivity on the kernel. This was shown for an h-version already in [Leh1] where an inverse trace inequality (see [WH3]) was needed q h n L ( T ) k h q h L (T ) q h [Π k (T )]. We use this inequality also to see that v h HHDG 1 () is equivalent (on the discrete space W h ) to v h H 1 HDG () v h H 1 HDG () := T T v h L (T ) + k h [vτ h ] L ( T ) + h k v T h n L ( T ). (5.1.1) Also the discrete LBB-condition sup div v h q h v h V h v h H 1 HDG () β h q h L () q h Q h with β β(h) was proven in [Leh1]. In the next section we present a proof that we also have β β(k), and so an independence of the polynomial degree k. 5

61 5. High order discrete LBB-condition 5. High order discrete LBB-condition As mentioned in remark 14 the facet space F h was only introduced for an efficient handling of linear systems. Due to that fact we change to a DG formulation and notation in this section. Under the assumption that the polynomial degree k >, we want to show the global discrete LBB-condition for V h = BDM k (T ) and Q h = Π k 1 (T ), so sup v h V h div v h q h dx v h H 1 DG () β q h L () q h Q h (5..1) with v h HDG 1 () = T T v h L (T ) + E T k h [v h τ ] L (E) and β β(k). For this, we split the proof in two steps: i. First we prove the local LBB-condition on the reference element ˆT for test functions q Π k 1 ( ˆT ) L ( ˆT ). ii. We show the global LBB-condition using a transformation to the reference triangle. For the first step we use an extension that is continuous with respect to the H norm and preserves polynomials. Theorem 5.1. Assume a given function u [Π k ( ˆT )] where ˆT is the reference Element and ˆT u n dx =. Then there exists an operator E : [Πk ( ˆT )] Π k+1 ( ˆT ) so that for φ = E(u) we have curl φ n = u n on ˆT (5..) (u curl φ) τ L ( ˆT ) 1 k u H 1 ( ˆT ) (5..3) φ H ( ˆT ) u H 1 ( ˆT ) (5..4) The existence of such an extension is not trivial and is proven in the last chapter Local LBB-condition Theorem 5.. Let ˆT be the reference element with the vertices (, ), (, 1), (1, ). Then ˆT sup div v h q h dx β q h v h BDM k ( ˆT v ) h L H 1 DG, ( ˆT ( ˆT ) q h Π k 1 ( ˆT ) L ( ˆT ) (5..5) ) With BDM k ( ˆT ) = {v BDM k ( ˆT ) : v n = on ˆT } and v h HDG, 1 ( ˆT := ) v h L ( ˆT ) + k v h τ L (E), and β β(k). E ˆT 53

62 5 Hybrid discontinuous Galerkin method for the Navier Stokes equations Proof. Choose an arbitrary p Π k 1 ( ˆT ) L ( ˆT ). We define a Poincare-Operator Z a by Z a :Π k 1 ( ˆT ) [Π k ( ˆT )] p Z a (p)(x) := (x a) 1 with a ˆT and γ(t) := a + t(x a), and define Z by Z(p)(x) := θ(a)z a (p)(x) da, ˆT tp(γ(t)) dt, where θ C ( ˆT ) and ˆT θ dx = 1. By that we get div Z(p) = p and as the operator is also continuous with respect to the H 1 norm (see [CM8]) we have for u 1 := Z(p) u 1 H 1 ( ˆT ) p L ( ˆT ). (5..6) In figure 5.1 we can see an example for p = 1xy 1 and the corresponding velocity field Z a for a = (.,.) Figure 5.1: Pressure p, absolute value Z a and the velocity field Z a. The upper scale is for the pressure, the lower for Z a Next we use theorem 5.1 to find a function φ Π k+1 ( ˆT ) with curl φ n = u 1 n on ˆT (u 1 + curl φ) τ L ( ˆT ) 1 k u 1 H 1 ( ˆT ) (5..7) φ H ( ˆT ) u 1 H 1 ( ˆT ). Let u := curl φ and u := u 1 + u, then u ˆT n = (u 1 ˆT + u ˆT ) n = (u 1 ˆT u 1 ˆT ) n = div u = div u 1 + div u = p + div curl φ = p. 54

63 5. High order discrete LBB-condition In figure 5. we can see the vector field curl φ and the resulting velocity u for the example from above. Note that the normal trace of the result is equal to zero. (a) Absolute value of curl φ and u For the u we have the estimation (b) Left we see the field of curl φ and right the resulting field u Figure 5.: Result after the correction u = u 1 + curl φ u HDG, 1 ( ˆT ) u 1 HDG, 1 ( ˆT ) + u HDG, 1 ( ˆT ) u 1 HDG, 1 ( ˆT ) + u L ( ˆT + k u ) τ L }{{} (E) φ E ˆT H ( ˆT ) u 1 H 1 ( ˆT ) + k u 1 τ L (E) + k u τ L (E). E ˆT E ˆT 55

64 5 Hybrid discontinuous Galerkin method for the Navier Stokes equations Together with (u 1 + u ) τ L (E) 1 k u 1 H 1 ( ˆT ) and the triangle inequality we get u HDG, 1 ( ˆT ) u 1 H 1 ( ˆT ) p L ( ˆT ). (5..8) With that we see sup v h BDM k ( ˆT ) div v h p dx v h H 1 DG, ( ˆT ) =p {}}{ div u p dx u H 1 DG, ( ˆT ) = p L ( ˆT ) u H 1 DG, ( ˆT ) u H 1 DG, ( ˆT ) p L u H 1 DG, ( ˆT ) p L. (5..9) 5.. Global LBB-condition Theorem 5.3. Let k >, V h = BDM k (T ) and Q h = Π k 1 (T ). Then we have sup div v h q h dx β q h v h V h v h L () q h Q h (5..1) H 1 DG () with v h HDG 1 () = T T v h L (T ) + E T k h [v h τ ] L (E) and β β(k). Proof. To show the global LBB-condition we proceed in two steps. First we construct a low order function u h,1 and then we define a correction u h, using the local LBBcondition (5..5). Step ( 1: Assume an arbitrary p h Π k 1 (T ). First we construct a function u h,1 [Π (T )] [C ()] ) [H 1 ()] with div u h,1 T = ph T, where T p h T = T p h dx T T. To find u h,1 we solve a kind of Stokes problem on [H 1 ()] with a different right hand side for the divergence 56

65 5. High order discrete LBB-condition Problem 1: Find (u S, p S ) ([H 1 ()] L ) so that ν u : v dx + div p S q dx, = v V div u S q dx, = p T h q dx q Q Due to the LBB-condition for the continuous Stokes problem (3..5) this problem has an unique solution and we have div u S = p h T and u S H 1 () p h T L (). Next we use the Fortin Π F operator we introduced in section 3.4., to define u h,1 := Π F (u S ) [Π (T )] [C ()] Using the properties of Π F we have div u h,1 dx = div u S dx = p T h dx T T T T T H 1 () u S H 1 () T p h L (). u 1 h Step : Next we define p h := p h div u 1 h Πk 1 (T ) L (T ) T T. Now for all T T we look for a function u T h with div u T h = p h T u T h n = on T u T H h 1 DG (T ) p L h (T ). We do that by solving a problem on the reference element. For each element we define a reference right hand side ˆp := h p h T. Due to the local LBB-condition (5..5) we find a function û with div û = ˆp û n = on ˆT û H 1 DG, ( ˆT ) ˆp L ( ˆT ). Now we use the Piola transformation P (see appendix) to get the solution on T u T h := P(û) = 1 det(j ) J û, 57

66 5 Hybrid discontinuous Galerkin method for the Navier Stokes equations where J is the Jacobian matrix of the finite element mapping F : ˆT T. Due to the properties of P we get and as div u T h = 1 det(j ) }{{} =h u T h û H 1 DG, ( ˆT ) = û L ( ˆT ) + n = on ˆT E ˆT div û = 1 h ˆp = p h T k u h τ L (E) = h u T h L (T ) + k h u T h h τ ˆp L (E) L ( ˆT ) = h p h L (T ) E T }{{} h u T h H 1 DG (T ) we get u T h H 1 DG (T ) p h L (T ). Summing up the element wise solutions leads to u h, := u T h T T and we observe div u h, = p h. Using this correction u h := u h,1 + u h, we get div(u h ) = div(u h,1 ) + div(u h, ) = div(u h,1 ) + p h div(u h,1 ) = p h, and u h HDG 1 () u h,1 HDG }{{ 1 () + u h, HDG } 1 () T p h L () + u h, HDG 1 (T ) T T = u h,1 H 1 () p h L () + T T p h L (T ) p h L (). Finally we have sup v h V h div v hp h dx v h H 1 DG () div u hp h dx u h H 1 DG () = (p h) dx u h H 1 DG () u h H 1 DG () p h L () u h H 1 DG () p h L (). 58

67 5.3 Curl convection for the hybrid discontinuous Galerkin method 5..3 Numerical estimations In the previous section we showed that in the two dimensional case the discrete LBB constant is independent of the polynomial order k (on triangles). We also calculated the LBB constant numerically. The results are shown in table 5.1. As we can see the results are satisfying and also in the three dimensional case (for a tetrahedron) the independency seems to be fulfilled. k D/trig D/tet Table 5.1: Numerically calculated discrete LBB constant for the HDG method for and 3 dimensions 5.3 Curl convection for the hybrid discontinuous Galerkin method In this section we show an implementation of the curl convection c curl (u, u, v) for the HDG-method. For the time discretization we use the methods introduced in section 3.5, so the convection term only appears in an explicit form. The problem that appears is that for a function u H(div)(T ) the curl is only defined in a distributional sense. To see this choose an arbitrary function ϕ C () and the definition of the weak curl to formally get curl u, ϕ := u curl ϕ dx = u curl ϕ dx T T T = curl u ϕ dx + [u n] ϕ ds. T T T E F E Using this for functions w h, u h, v h W h and a standard DG upwind value û T h leads to c curl (w h, u h, v v ) = curl wh T (ut h vt h ) dx T T T + (ŵt h n ) (u T h vf h ) ds ( w T h n ) (u F h vt h ) ds. E T E It remains the implementation of the boundary term that leads to the Bernoulli pressure 1 (u )v n ds, in the case of Neumann boundaries. For the HDG method this leads to c bp (u h, v h ) := (u T h ) + (u F h ) (vh T n) ds. E Γ N E E 59

68 5 Hybrid discontinuous Galerkin method for the Navier Stokes equations Remark 17: The choice of the upwind value appears only for the facet values vh F in the curl convection. This leads to a stable discretization although this is not natural for the tangential components. Remark 18: We saw in section (4.4.3) that a reconstruction on exact divergence-free velocity functions improves the solution when using a curl formulation for the convective part of the Navier Stokes equations. We implemented the curl convection for the HDG method to show that due to the property div V h Q h, namely an exact divergence-free approximation, we can also use the curl formulation to get proper solutions. In [LS15] also the standard convection form was implemented. 5.4 Approximation error Due to the properties we are now able to show an optimal error convergence, namely an independent constant c c(k, h) in the error estimation. We start with the approximation error for the exact solutions (u, p) and the approximation u h and the L projection P L Q h p u u h H 1 HDG () + P L Q h p p h L (). Next we add and subtract a BDM interpolator to get P u u h H 1 HDG () + L L u Q h p p h I BDM k h u () H 1 HDG () + Ih BDMk H P u u h + L L 1 Q HDG () h p p h. () For the third and fourth term we use the inf-sup condition for the bilinear form B and the Galerkin orthogonality to get Ih BDMk H P u u h + L L 1 Q HDG () h p p h () 1 β sup v h,p h V h Q h v h,q h = 1 β sup v h,p h V h Q h v h,q h B((Ih BDMk u u h, PQ L h p p h ), (v h, q h )) v h H 1 HDG () + q h L () B((Ih BDMk u u, P L Q h p p), (v h, q h )) v h H 1 HDG () + q. h L () Next we use the definition of B and the commuting properties of the BDM interpolator and the L projection, namely to get b HDG (Ih BDMk u u, q h ) = b HDG (v h, PQ L h p p) =, I BDMk h u u h H 1 HDG () + P L Q h p p h L () 1 a HDG (Ih BDMk u u, v h ). β v h H 1 HDG () 6

69 5.5 Numerical example Due to the definition of a HDG we can bound it with the HDG norm, and use the equivalence (5.1.1) to get I a HDG (Ih BDMk u u, v h ) h BDMk u u v h H 1 H 1 HDG () HDG () v h H 1 HDG () v h H 1 HDG () Ih BDMk u u v h H 1 H 1 HDG () HDG () Ih BDMk v h H 1 HDG () u u, H 1 HDG () and so altogether P u u h H 1 HDG () + L L u Q h p p h c I BDM k u, () H 1 HDG () with c c(h, k). 5.5 Numerical example In the end we show an example using the curl formulation and the HDG-method to solve the Navier Stokes equations using a first order IMEX scheme The example we chose is the well-known benchmark of a laminar flow around a cylinder, see [ST96]. We simulate the D example with the inflow condition u(, y) = 6y(.41 y).41, which yields to a Reynolds number Re = 1 and induces an unsteady flow. In figure 5.3 one can see the solution at t = 3s and in figure 5.4 the drag and lift coefficient defined by c D = F D ρu D and c L = F L ρu D, h with F D = (ν u t Γ c n n y pn x ) ds and F L = (ν u t Γ c n n x + pn y ) ds where Γ c is the boundary of the cylinder, are plotted. For the simulation we use a mesh with 315 elements and chose the polynomial degree k = 3. 61

70 5 Hybrid discontinuous Galerkin method for the Navier Stokes equations (a) Velocity field and absolute value u h (b) Pressure p h Figure 5.3: Absolute value of the velocity u h and pressure p of a laminar flow around the cylinder 3 c D cl t Figure 5.4: Drag and lift coefficient for a laminar flow around a cylinder 6

71 6 H -extension In this section we prove the existence of an H -continuous operator E. Note that E enforces exact tangential values of the gradient on ˆT although the properties in (5..7) demand an exact normal trace. This can be done by switching from the operator to the rotated operator in 5.1. The proof is analogue, but the tangential and the normal values change their position. Theorem 6.1. Assume a given function u [Π k ( ˆT )] where ˆT is the reference element and ˆT u τ dx =. Then there exists an operator E : [Πk ( ˆT )] Π k+1 ( ˆT ) so that for φ = E(u) we have φ τ = u τ on ˆT (6..1) (u φ) n L ( ˆT ) 1 k u H 1 ( ˆT ) (6..) φ H ( ˆT ) u H 1 ( ˆT ). (6..3) Due to the different demands of the extension E(u), we show three theorems with different kinds of properties and use them afterwards to prove theorem 6.1. First we see that the tangential gradient of E(u) has the same values as the tangential trace of u. How to find a function that has this property is shown in the first theorem 6.3. We start with an extension that preserves a proper tangential gradient on the first edge. Afterwards we correct the error of the tangential gradients on the other two edges by defining two more extensions. The main challenge is the H -estimation where we use different techniques and the definition of the H 1/ seminorm. The final result E D (u) then fulfills E D (u) τ = u τ on ˆT. After the first step the normal gradient of E D (u) does not coincide with the normal trace of u at all. The second theorem 6.4 is used to correct the error u c = ( u E D (u) ) n. This can be done under certain assumptions, so we split the error u c in a good term u g c, where we can use theorem 6.4, and a bad term u b c. The constructed extension for the proof of the second theorem is also split in three parts correcting the values edge by edge. Again the H -estimation is a big challenge and uses similar techniques as in the proof of theorem 6.3. That the splitting of u c in two parts is a stable operation and how to handle u b c, is considered in the last theorem 6.5. We show that there exists a polynomial whose norm 63

72 6 H -extension is bounded in a proper way by the polynomial order k. For that we use a Lagrangian function to find a minimum under certain conditions. Finally we show the proof of 6.1 in section 6.4. Before we continue with the theorems 6.3, 6.4 and 6.5, we show a lemma which uses the technique of K-functionals. Lemma 6.. Let E 1 = (, 1) and assume a function a(s) C 1 (E 1 ) and u H 1/ (E 1 ). For (x, y) ˆT we define Then we have g(x, y) := 1 a(s)u(x + sy)ds. g H 1 ( ˆT ) u H 1/ (E 1 ). Proof. Due to g H 1 (T ) = g L (T ) + g L (T ) we show the proof in two steps. First we bound the L norm and then the H 1 seminorm. For an arbitrary y we define the horizontal line L y := {(t, y) : t [, 1 y]}. By that we get, using Cauchy Schwarzs inequality, 1 y 1 y ( 1 g L (L = y) g(x, y) dx = a(s)u(x + sy) ds) dx C.S. 1 y 1 a(s) ds 1 Using the substitution t = x + sy we get 1 y 1 u(x + sy) ds dx u(x + sy) ds dx = 1 y 1 y 1 1 y x+y x u(t) dt dx. u(x + sy) ds dx. In the next step we use Fubini s theorem to first change the order of the integration variables and then increase the integrated area, namely 1 y x+y u(t) dt dx = u(t) d(t, x) Fub. u(t) d(x, t), and so x g L (L y) 1 y 1 x 1 y x t x+y t u(t) dx dt = t y }{{} =y By that we can bound the L norm on ˆT 1 t 1 t y x t u(t) dt = u L (E 1 ) u H 1/ (E 1 ). 1 g L ( ˆT ) = g L (L dy y) u H 1/ (E 1 ). 64

73 To bound the H 1 seminorm we use the theory of interpolation spaces (see appendix 7.5). In the first step we make an estimation for the x derivation and then for the y derivation. Due to we get analogue as before g(x, y) = x g(x, y) x 1 L (L y) u (x + sy)a(s) ds, u L (E 1 ). (6..4) Next we observe that by using integration by part we can write the x derivation also as g(x, y) = x 1 = 1 y 1 u (x + sy) }{{} d ds u(x+sy) a(s) ds = 1 a (s)u(x + sy) ds + 1 (a(1)u(x + y) a()u(x)). y y }{{}}{{} =:A =:B For the first term we proceed as before to get For the second term we get A L (L y) 1 y u L (E 1 ). (6..5) B L (L = 1 1 y y) y (a(1)u(x + y) a()u(x)) dx 1 1 y y u(x + y) + u(x) dx, and together with the substitution t = x + y B L (L 1 1 y) y u(t) dt y y u(x) dx 1 y u L (E 1 ). (6..6) y Using estimation (6..4), (6..5) and (6..6) and the definition of the K-functional we have g(x, y) 1 x inf L u=u (L +u 1 y) y u L (E 1 ) + u 1 L (E 1 ) 1 K(y, u), y and so g(x, y) x L ( ˆT ) = 1 1 g(x, y) x For the y derivation we observe that = dy L (L y) 1 y K(y, u) dy g(x, y) = y 1 1 y K(y, u) dy = u H 1/ (E 1 ). u (x + sy)sa(s) ds. So by defining ã(s) = sa(s) we proceed similar to the x derivation to finish the proof. 65

74 6 H -extension 6.1 Tangential extension E D Theorem 6.3. Assume a given function u [Π k ( ˆT )] where ˆT is the reference element and ˆT u τ ds =. Then there exists an operator ED : [Π k ( ˆT )] Π k+1 ( ˆT ) so that for φ τ = E D (u) we have φ τ τ = u τ on ˆT (6.1.1) φ τ H ( ˆT ) u H 1 ( ˆT ). (6.1.) Proof. For the proof we split the extension E D in three edge extensions E E i 1, EE i,j and E E i 3. The main idea is to first extend the values of the lower edge and correct the values on the other two edges afterwards. All following examples are visualized in a three dimensional perspective, see figure 6.1. (, 1) (1, ) (, ) Figure 6.1: Three dimensional perspective of the reference element ˆT Step 1: The first extension E E i 1 has no further restrictions. For the ease we set E i = E 1 and define τ E1 as the tangential vector on E 1. For u τ = u τ E1 we set ψ(x) := x where c can be an arbitrary constant and u τ (s, ) ds + c, We observe that and so E E 1 1 (u)(x, y) = φ 1(x, y) := φ 1 x = φ 1 y = 1 1 φ 1 τ 1 E1 = φ 1 x ψ (x + sy) ds = ψ (x + sy)s ds = = E ψ(x + sy) ds. u τ (x + sy) ds u τ (x + sy)s ds (6.1.3) u τ (x) ds = u τ (x) = u τ. (6.1.4) 66

75 6.1 Tangential extension E D In figure 6. we can see the values and the partial derivation of the first extension φ 1 for u τ = 6x(1 x) 1. We now show Figure 6.: Extension φ 1, φ 1 x and φ 1 y for u τ = 6x(1 x) 1 φ 1 H ( ˆT ) = φ 1 L ( ˆT ) + φ 1 H 1 ( ˆT ) u H 1 ( ˆT ). For the L estimation we proceed as in the proof of lemma 6. 1 y φ 1 L (L = y) ( 1 1 ψ L (E 1 ) = ( ψ(x + sy) ds) dx C.S. x 1 y 1 u τ (s, ) ds) dx C.S. u τ L (E 1 ) u τ H 1/ (E 1 ) u H 1 ( ˆT ) ψ(x + sy) ds dx 1 x u τ (s, ) ds dx and so 1 φ 1 L ( ˆT ) = φ 1 L (L dy y) u H 1 ( ˆT ). To bound φ 1 H 1 ( ˆT ) get we use lemma 6. for each partial derivative (6.1.3), and so we φ 1 H 1 ( ˆT ) u τ H 1/ (E 1 ) u H 1 ( ˆT ), and φ 1 H ( ˆT ) u H 1 ( ˆT ). (6.1.5) Step : For E E i,j we have the restriction that E E i,j (u) τ Ej Ej =. 67

76 6 H -extension Again we set E i = E 1 and E j = E. For an arbitrary ũ we define and ψ (x) := x With that we define the extension E E 1, (ũ)(x, y) = φ (x, y) := u,τ := ũ τ E1, u,τ (s, ) ds ψ with ψ := 1 ψ (x + sy) ds or using integration by part for the second term y 1 x 1 we can write the extension as ψ (x + s(1 x)) ds = y 1 = y φ (x, y) = ψ (x + sy) ds + y }{{} On E we have y = 1 x and so =E E 1 1 (ũ)(x,y) φ E =, and due to a constant value on the edge E also On E 1 we have y = and so φ E1 τ E1 = ψ (x) x y 1 x 1 1 u,τ (s, ) ds. ψ (x + s(1 x)) ds, ψ (x + s(1 x))s ds u,τ (x + s(1 x))s ds, u,τ (x + s(1 x))s ds. } {{ } =:φ corr φ τ E E =. (6.1.6) = u,τ = ũ τ E1. (6.1.7) In figure 6.3 we can see the values and the partial derivations of the correction term φ corr and the resulting extension φ for the same example u,τ = 6x(1 x) 1. Note that the correction does not change the values and the derivation with respect to x on the edge E 1. 68

77 6.1 Tangential extension E D (a) Correction φ corr, φcorr x and φcorr y (b) Extension φ, φ and φ x y Figure 6.3: Values and partial derivatives of the correction φ corr and the extension φ for u,τ = 6x(1 x) 1 It remains to show the H estimation. For the first part E E 1 1 (ũ)(x, y) we showed the estimation in step 1, so we have E E 1 1 (ũ) H ũ ( ˆT ) H 1 ( ˆT ). It remains the estimation for φ corr. Again we split the norm in φ corr H ( ˆT ) = φcorr L ( ˆT ) + φcorr L ( ˆT ) + φ corr L ( ˆT ). 69

78 6 H -extension For the first term we get φ corr L ( ˆT ) = 1 = C.S. 1 x (φ corr 1 1 x x ( y ) dy dx x 1 u,τ (x + s(1 x))s ds) dy dx ( ) y ( 1 u,τ (t) dt) dy dx 1 x x }{{} x u,τ (t) dt dy dx u,τ L (E 1 ) ũ H 1 ( ˆT ). 1 1 x 1 u,τ (t) dt dy dx Next we calculate all partial derivatives up to order (always using integration by part for the u,τ terms) φ corr x 1 = y = y 1 x = y 1 x u,τ (x + s(1 x)) (1 s)s ds = }{{} = 1 1 x 1 1 d ds u,τ (x+s(1 x)) u,τ (x + s(1 x))(s 1) ds + u,τ (x + s(1 x))(1 s)s 1 }{{} = u,τ (x + s(1 x))(s 1) ds, φ corr y = 1 u,τ (x + s(1 x))s ds, φ corr x x = y (1 x) + y 1 x 1 y = (1 x) + y (1 x) 1 1 u,τ (x + s(1 x))(s 1) ds u,τ (x + s(1 x))(s 1)(1 s) ds 1 u,τ (x + s(1 x))(s 1) ds u,τ (x + s(1 x))(4s 3) ds + y (1 x) u,τ (x), φ corr x y = 1 1 x 1 u,τ (x + s(1 x))(s 1) ds, 7

79 6.1 Tangential extension E D and finally φ corr y y =. We start with the mixed second order derivative. First note that 1 by that we can substract the meanvalue s 1 ds =, to get u,τ (x,1) := 1 1 x 1 x u,τ (s) ds, φ corr x y = 1 1 x = 1 1 x 1 1 u,τ (x + s(1 x))(s 1) ds ( u,τ (x + s(1 x)) u,τ (x,1) ) (s 1) ds. And so using the Cauchy-Schwarz inequality φ corr 1 1 x ( φ corr ) x y = dy dx L ( ˆT ) x y 1 ( φ corr ) 1 x = dy dx x y C.S. = x 1 1 (1 x) 1 The inner integral can also be written as 1 x ( u,τ (t) u,τ (x,1) ) dt = 1 (1 x) ( u,τ (x + s(1 x)) u,τ (x,1) ) ds dx x ( u,τ (t) u,τ (x,1) ) dt dx. 1 1 x x (u,τ (t) u,τ (s)) dt ds, and so using Fubini s theorem φ corr x y L ( ˆT ) (1 x) 3 (u,τ (t) u,τ (s)) dt ds dx x x Fub. (u,τ (t) u,τ (s)) (1 x) 3 d(s, t, x). x s x t s,t min(s,t) 1 (1 x) 3 dx (u,τ (t) u,τ (s)) dt ds. 71

80 6 H -extension For the inner integral and w.l.o.g. assuming that s < t we get and so min(s,t) ( ) 1 (1 x) 3 dx = min(s, t) φ corr x y L ( ˆT ) 1 1 ( ) 1 ( ) 1, 1 s t s (u,τ (t) u,τ (s)) (t s) dt ds. Using the definition of the Sobolev Slobodeckij norm (see appendix) we get φ corr x y L ( ˆT ) 1 1 (u,τ (t) u,τ (s)) (t s) dt ds = u,τ H 1/ (E 1 ) ũ H 1 ( ˆT ). Next we look at the second derivative of φ corr y x x. Note that we have get φ corr x x 1 1 x x 1 1 or by adding and subtracting an integral φ corr x x 1 1 x x x u,τ (x + s(1 x))(s 1) ds u,τ (x + s(1 x))(4s 3) ds x u,τ (x), u,τ (x + s(1 x))(s 1) ds u,τ (x + s(1 x))(4s ) ds u,τ (x) u,τ (x + s(1 x)) ds. } {{ } =:A 1 x 1 and so we The first and the second term can be estimated in the same way as φ corr x y. We focus on the third term A. Again using the Cauchy Schwarz inequality and Fubini s theorem the L norm is bounded by A L ( ˆT ) = 1 = 1 Fub. t x 1 x 1 (1 x) 1 A dy dx C.S. 1 1 x 1 x (u,τ (x) u,τ (t)) 1 (u,τ (x) u,τ (t)) dt dx = (x t) d(x, t) = u,τ H 1/ (E 1 ) ũ H 1 ( ˆT ), (u,τ (x) u,τ (x + s(1 x)) ds dx 1 1 t x (u,τ (x) u,τ (t)) (1 x) d(x, t) (u,τ (x) u,τ (t)) (x t) dx dt 7

81 6.1 Tangential extension E D and by that the estimation for φ corr x x is given. It remains to bound the first order derivatives. The estimation for x derivation is similar to φ corr x y because y 1, and for the y derivation we observe as before φ corr 1 1 x ( 1 y = u,τ (x + s(1 x)s ds) dy dx L ( ˆT ) All together we have 1 (1 x) 1 u,τ L (E 1 ) ũ H 1 ( ˆT ). Step 3: For E E i 3 we have the restriction that u,τ (x + s(1 x) ds dx = 1 1 x u,τ (t) dt dx φ H ( ˆT ) ũ H 1 ( ˆT ). (6.1.8) E E i 3 (u) τ E j Ej = for j i. Again we set E i = E 1 and j =, 3. For an arbitrary ũ with ũ τ ds = and ũ E τ E = ũ E3 τ E3 = (6.1.9) we set and ˆT ψ 3 (x) := By that we define the last extension as E E 1 3 (ũ)(x, y) = φ 3(x, y) := 1 u 3,τ := ũ τ E1, x u 3,τ (s, ) ds. 1 ψ 3 (x + sy) ds y ψ 3 (x + s(1 x)) ds 1 x }{{} 1 y ψ 3 (s(x + y)) ds + y x + y }{{} =:φ corr, 3 =:φ corr,1 3 1 ψ 3 (s) ds. }{{} =:φ corr,3 3 Similar to the second extension we observe that on E due to y = 1 x and on E 3 due to x = the extension is constant, and so and φ 3 E τ E = φ 3 E3 τ E3 =, (6.1.1) φ 3 E1 τ E1 = ψ 3 = u 3,τ = ũ τ E1. (6.1.11) 73

82 6 H -extension In figure 6.4 we can see the values and the partial derivations of the correction φ corr, 3 and the resulting extension φ 3 for the same example u,τ = 6x(1 x) 1. Note that on the edge E 3 the partial derivation with respect to y of the correction φcorr, 3 y is the negative value of φ y in figure 6.3. This leads to φ 3 E3 τ E3 =. (a) Correction φ corr, 3, φcorr, 3 x and φcorr, 3 y (b) Extension φ 3, φ 3 and φ 3 x y Figure 6.4: Values and partial derivatives of the correction φ corr, 3 and the extension φ 3 for u,τ = 6x(1 x) 1 Again we have to show the H estimation. The first term of φ 3 can be written as E E 1 1 where we already have the estimation. The second term is the same as for E E 1, so it remains to show the estimations for φ corr, 3 and φ corr,3 3. Note that we have ψ 3 () =, 74

83 6.1 Tangential extension E D and by that we can write φ corr, 3 = y x + y = y = y and due to (6.1.9) also ψ 3 (1) =, so φ corr,3 3 = y 1 ψ 3 (s(x + y)) ds ψ 3(s(x + y))(s 1) ds y x + y ψ 3(s(x + y))(s 1) u 3,τ (s(x + y))(s 1) ds ψ 3 (x) ds = y 1 ψ 3(s)s ds + ψ 3 (s)s 1 1 = y u 3,τ (s)s ds. With similar techniques to the proofs for the estimation of E E 1 1 and E E 1 we can bound φ corr,3 ũ H ( ˆT ) H 1 ( ˆT ). 3 For φ corr, 3 we make the estimations part by part. Together with estimation follows from φ corr, 3 L ( ˆT ) C.S. 1 1 x 1 1 x 1 y u 3,τ (s(x + y)) ds dy dx y x + y x+y u 3,τ L (E 1 ) ũ H 1 ( ˆT ). y x+y u 3,τ (s(x + y)) ds dy dx 1 1 the L Next we calculate all partial derivatives using integration by part on all integrals where u 3,τ appears to get: φ corr, 3 x = y x + y 1 u 3,τ (s(x + y))(1 s) ds φ corr, 3 y = 1 u(s(x + y)(s 1) ds + y x + y 1 u 3,τ (s(x + y))(1 s) ds φ corr, 3 x x = y (x + y) + + y (x + y) y (x + y) u 3,τ (s(x + y))(1 s) ds u 3,τ (s(x + y))(4s ) ds u 3,τ (s(x + y)) u 3,τ (x + y) ds 75

84 6 H -extension φ corr, 3 x y x = (x + y) + + y (x + y) y (x + y) u 3,τ (s(x + y))(1 s) ds u 3,τ (s(x + y))(4s ) ds u 3,τ (s(x + y)) u 3,τ (x + y) ds Again we use that φ corr, 3 y y = 1 x + y x (x + y) y (x + y) y (x + y) x x+y 1 and 1 u 3,τ (s(x + y))(1 s) ds 1 1 u 3,τ (s(x + y))(1 s) ds u 3,τ (s(x + y))(4s ) ds u 3,τ (s(x + y)) u 3,τ (x + y) ds. y x+y 1 on ˆT. We start with the L norm of the first order x-derivation and proceed similar to the estimation of φ corr φ corr, 3 x L ( ˆT ) C.S. = = = 1 1 x x 1 1 x 1 x + y x y. ( u 3,τ (s(x + y)) u 3,τ (,x+y) ) ds dy dx 1 (x + y) x+y 1 1 x x+y x+y 1 1 x x+y x+y 1 1 x 1 1 as t s t x + y, and so φ corr, 3 x ( u 3,τ (t) u 3,τ (,x+y) ) dt dy dx x+y x+y (u 3,τ (t) u 3,τ (s)) dt ds dy dx (u 3,τ (t) u 3,τ (s)) (x + y) dt ds dy dx (u 3,τ (t) u 3,τ (s)) (t s) dt ds dy dx (u 3,τ (t) u 3,τ (s)) (t s) dt ds } {{ } = u 3,τ H 1/ (E 1 ) L ( ˆT ) u 3,τ H 1/ (E 1 ) ũ H 1 ( ˆT ). dy dx, We continue with the y derivation. First observe that the second integral is the same as the x derivation, so we have φ corr, 1 1 x 1 3 u 3,τ (s(x + y)) φ corr, 3 ds dy dx +. y x L ( ˆT ) L ( ˆT ) 76

85 6.1 Tangential extension E D Using Fubini s theorem and ζ = x + y we observe 1 1 x 1 u 3,τ (s(x + y)) ds dy dx = Fub. = 1 1 x 1 ζ 1 ζ 1 x + y ζ u 3,τ L (E 1 ), 1 u 3,τ (t) dt dy dx u 3,τ (t) dt dx dζ and so also φ corr, 3 y L ( ˆT ) ũ H 1 ( ˆT ). Finally we define two functions and ξ := 1 x + y 1 u 3,τ (s(x + y))(1 s) ds θ := 1 x + y 1 u 3,τ (s(x + y)) u 3,τ (x + y) ds, to bound the L norm of the second order derivation with the L norm of linear combinations of ξ and θ. For the estimation of ξ we proceed similarly for φcorr, 3 x but have to use Fubini s theorem due to the higher power of the x + y term. ξ L ( ˆT ) 1 Fub. 1 1 x x+y x+y s x y t x y t,s (u 3,τ (t) u 3,τ (s)) (x + y) 3 dt ds dy dx (u 3,τ (t) u 3,τ (s)) (x + y) 3 d(s, t, y) dx max(s x,t x) Without loss of generality let s < t to get and so 1 max(s x,t x) 1 (x + y) 3 dy (u 3,τ (t) u 3,τ (s)) dt ds dx. 1 1 dy = (x + y) 3 (x + y) max(s x,t x) t (t s), ξ L ( ˆT ) (u 3,τ (t) u 3,τ (s)) u 3,τ H 1/ (E 1 ) ũ H 1 ( ˆT ). (t s) dt ds dx. 77

86 6 H -extension For θ we get θ L ( ˆT ) 1 1 x 1 1 x 1 x + y 1 1 x x+y 1 1 (x + y) (u 3,τ (s(x + y)) u 3,τ (x + y)) dt dy dx x+y (u 3,τ (t) u 3,τ (x + y)) dt dy dx (u 3,τ (t) u 3,τ (x + y)) (x + y) dt dy dx. Together with t (x + y) t + x + y x + y for t x + y we get θ L ( ˆT ) 1 1 x x+y u 3,τ H 1/ (E 1 ) ũ H 1 ( ˆT ). (u 3,τ (t) u 3,τ (x + y)) (t (x + y)) dt dy dx As φ corr, 3 x x φ corr, 3 y y φ corr, 3 x y θ L ( ˆT L ( ˆT ) + ξ L ( ˆT ) ) θ L ( ˆT L ( ˆT ) + ξ L ( ˆT ) ) θ L ( ˆT ) + ξ L ( ˆT ), L ( ˆT ) we finally have Step 4: Now we can construct E D. Assume a given u with u τ ds =. φ 3 H ( ˆT ) ũ H 1 ( ˆT ). (6.1.1) ˆT First we define the linear transformations F and F 3 defined by the transformations of the vertices F : (1, ) (1, ) (, ) (, 1) and F 3 : (, ) (, ) (, 1) (1, ) (, 1) (, ) (1, ) (, 1) and the corresponding covariant transformations C, C 3 (see appendix) to define proper extensions from the other edges, so ( ) E E,1 (u)(x, y) = E E 1, (C u) (F (x, y)), 78

87 6.1 Tangential extension E D with and with E E,1 E (v) E 1, H ( ˆT (v), ) H ( ˆT ) ( ) E E 3 3 (u)(x, y) = E E 1 3 (C 3u) (F 3 (x, y)), E E 3 3 (v) H ( ˆT ) E E 1 3 (v) H ( ˆT ). By that we define φ 1 = E E 1 1 (u) φ = φ 1 + E E,1 (u φ 1 ) φ 3 = φ + E E 3 3 (u φ ), and set E D (u) = φ 3. Note that due to Green s theorem the surface integral over ˆT of the tangential component of φ is, and so also u φ τ ds =, and as ˆT u φ E1 τ E1 = u φ E τ E = assumption for E E 3 3 is fulfilled. Also we see that due to the construction of the extensions by integrals of u we get a function E D (u) Π k+1 ( ˆT ), and using the properties (6.1.6), (6.1.7) and (6.1.1), (6.1.11) we have E D (u) E1 τ E1 = φ E1 τ E1 + E E 3 3 (u φ ) τ E1 }{{ E1 } = = φ 1 E1 τ E1 + E E,1 (u φ 1 ) τ E1 }{{ E1 } = = φ 1 E1 τ E1 = u τ E1 E D (u) E τ E = φ E τ E + E E 3 3 (u φ ) τ E1 }{{ E1 } = = φ 1 E τ E + E E,1 (u φ 1 ) τ E E = φ 1 E τ E + (u φ 1 ) E τ E = u τ E 79

88 6 H -extension E D (u) E3 τ E3 = φ E3 τ E3 + E E 3 3 (u φ ) τ E3 E3 = φ E3 τ E3 + (u φ ) E3 τ E3 = u τ E3. The H estimation follows from (6.1.5), (6.1.8) and (6.1.1), so E D (u) E H ( ˆT ) φ H ( ˆT ) + E 3 3 (u φ ) H ( ˆT ) φ H ( ˆT ) + u φ H 1 ( ˆT ) φ H ( ˆT ) + u H 1 ( ˆT ) E φ 1 H ( ˆT ) + E,1 (u φ 1 ) + u H ( ˆT ) H 1 ( ˆT ) φ 1 H ( ˆT ) + u φ 1 H 1 ( ˆT ) + u H 1 ( ˆT ) φ 1 H ( ˆT ) + u H 1 ( ˆT ) u H 1 ( ˆT ) 6. Normal extension E N Theorem 6.4. Assume a given function u Π k (E 1 ) where E 1 = [, 1] is an edge of the reference element and has a zero of order two in the vertices. Then there exists an operator E N : Π k (E 1 ) Π k+1 ( ˆT ) so that for φ n = E N (u) we have φ n = on ˆT (6..1) φ n n = u on E 1 (6..) φ n n = on ˆT \ E 1 (6..3) φ n H ( ˆT ) u H 1/ (E 1). (6..4) Proof. The idea of the proof is similar to the proof of theorem 6.3. First we extend the values of the lower edge and correct the trace and the normal derivative afterwards. For the corrections we use cubic blending coefficients. Due to that it is not trivial that the corresponding extension still fulfills E N (u) Π k+1 ( ˆT ). Before we start with the proof we remind the reader of the definition of the H 1/ (E 1) norm u = u H 1/ (E 1) H 1/ (E 1 ) + u L (E 1 ) = 1 1 (u(x) u(y)) (x y) dx dy + 1 ( 1 1 x + 1 ) u (x) dx. x All following examples are visualized in a three dimensional perspective, see figure 6.1. Step 1: In the first step we define φ 1 (x, y) := y 1 a 1 (s)u(x + sy) ds, 8

89 6. Normal extension E N with a 1 (s) = 6s(1 s), and so 1 a 1 (s) ds = 1 and 1 For the gradient we use integration by part to get a 1(s)s ds = 1 and a() = a(1) =. and By that we observe φ 1 x = φ 1 y = 1 1 a 1(s)u(x sy) ds, a 1(s)su(x + ys) ds. φ 1 E1 = and 1 φ 1 n E1 E1 = φ 1 y = u a 1(s)s }{{} = 1 ds = u(x). In figure 6.5 we can observe the first extension φ 1 for the example u = 1x (1 x). Due to the constant value φ 1 = on the lower edge E 1 the derivation with respect to x is also zero. Figure 6.5: Values and partial derivatives of the extension φ 1 for u = 1x (1 x) Using lemma 6. we observe that φ 1 x u H 1/ H 1 ( ˆT (E 1 ) u H 1/ ) φ 1 y u H 1/ H 1 ( ˆT (E 1 ) u H 1/ ) (E 1) (E 1) 81

90 6 H -extension and due to we have φ 1 L ( ˆT ) = 1 1 x C.S. 1 1 x 1 1 x = y ( 1 a 1 (s)u(x + sy) ds) dy dx 1 y u(x + sy) ds dy dx y 1 x 1 1 x 1 1 x u(t) dt dy dx u(x + sy) ds dy dx u L (E 1 ) u H 1/ (E 1), φ 1 H ( ˆT ) u H 1/ (E 1). Step : In the next step we correct the values on the second edge E. For that we use two cubic polynomial blending functions, namely with the properties and define φ (x, y) = φ 1 (x, y) a a(s) = 3s s 3 and b(s) = s 3 s, a() = a () = a (1) = and a(1) = 1 (6..5) b() = b () = b(1) = and b (1) = 1. (6..6) ( ) ( ) y y φ 1 (x, 1 x) b (1 x) φ 1 (x, 1 x). 1 x 1 x y The first polynomial a corrects the values, the second one b corrects the derivative on the edge E. Due to the zero values on the lower edge, the extension from the first step φ 1 does not change on E 1. In figure 6.6 we can see a(s) and b(s) on (, 1) a(s) b(s) s Figure 6.6: a(s) and b(s) on (, 1) 8

91 6. Normal extension E N On the edges E 1 and E we now have φ E1 = φ 1 E1 a()φ 1 (x, 1 x) b()(1 x) φ 1 (x, 1 x) = y φ E = φ 1 E a(1)φ 1 (x, 1 x) b(1)(1 x) φ 1 (x, 1 x) y = φ 1 (x, 1 x) φ 1 (x, 1 x) =. In figure 6.7 we can observe the extension φ including the first correction for the example u = 1x (1 x). Note that the values and partial derivatives on the second edge E are corrected. Figure 6.7: Values (lower scale) and partial derivatives (upper scale) of the extension φ for u = 1x (1 x) Next we calculate the derivations using the chain and the product rule φ x (x, y) = φ 1 (x, y) x ( ) x ( y y 1 x )a φ 1 (x, 1 x) 1 x ( ) ( y φ1 a 1 x x (x, 1 x) φ ) (x, 1 x) y ( ) ( ) y y b φ1 (x, 1 x) x 1 x 1 x y ( ) ( y ) φ 1 b 1 x x x (x, 1 x) φ 1 (x, 1 x) x y φ y (x, y) = φ 1 y (x, y) y ( y 1 x )a ( ) y ( y y 1 x )b 1 x ( ) y φ 1 (x, 1 x) 1 x (1 x) φ 1 (x, 1 x). y 83

92 6 H -extension and observe φ y = φ 1 E1 y a () E1 1 x φ 1(x, 1 x) b () 1 x (1 x) φ 1 (x, 1 x) y = φ 1 y = u(x), E1 so φ n E1 E1 = u(x), and φ y = φ 1 E y a(1) E 1 x φ 1(x, 1 x) b (1) 1 x (1 x) φ 1 (x, 1 x) y = φ 1 y (x, 1 x) φ 1 (x, 1 x) =. y Due to the constant value of φ = on the edge E we have φ τ E E = φ x = φ E y = E and so φ n E E =. It remains to show the H estimation. For the first term of φ we already showed the estimation in the first step. For the rest we split φ into ( ) ( ) y y ψ := a φ 1 (x, 1 x) and ξ := b (1 x) φ 1 (x, 1 x). 1 x 1 x y We first show the estimation for ψ ψ(x, y) = y (3(1 x) y) (1 x) 1 u(x + s(1 x))a 1 (s) ds. For all estimations we use the Cauchy Schwarz inequality, Fubini s theorem and that y 1 x 1 on ˆT. By that we get ψ L ( ˆT ) = 1 1 x C.S. 1 1 x 1 1 x = 1 1 x 1 ψ dy dx 1 (3(1 x) y) u(x + s(1 x)) ds dy dx (3(1 x) y) 1 1 x x 1 x u(t) dt dy dx u(t) dt dy dx u L (E 1 ) u H 1/ (E 1). 84

93 6. Normal extension E N We continue with the first order derivatives. ψ y = 6 y ( 1 y ) 1 u(x + s(1 x))a 1 (s) ds 1 x 1 x ψ x = 6 y ( (1 x) 1 y ) 1 u(x + s(1 x))a 1 (s) ds 1 x y ( + (1 x) 3 y ) 1 x x (φ 1(x, 1 x)). y ( = 6 (1 x) 1 y ) 1 u(x + s(1 x))a 1 (s) ds 1 x y ( + (1 x) 3 y ) 1 (a 1 x 1(s) a 1(s)s)u(x + s(1 x)) ds For the y-derivation we have ψ y L ( ˆT ) C.S. = 1 1 x x 1 1 x 1 1 x 1 x u (x + s(1 x)) ds dy dx 1 x u(t) 1 t u(t) dt dy dx dt dy dx u L (E 1 ) u H 1/ (E 1). Due to the structure of the x-derivation we bound the norm similar to the y-derivation, to get ψ x L ( ˆT ) u H 1/ (E 1). It remains the estimation for the second order derivatives. ( ψ y y = y ) 1 u(x + s(1 x))a 1 (s) ds 1 x 1 x ( ψ y x = 1 y 1 x 18 y ) 1 (1 x) (1 x) u(x + s(1 x))a 1 (s) ds ( y + 6 (1 x) 1 y ) 1 (a 1 x 1(s) a 1(s)s)u(x + s(1 x)) ds 85

94 6 H -extension ( ψ x x = (1 x) 3 y (1 x) 4 y 3 y ( 1 y ) 1 (1 x) 3 (1 x) u(x + s(1 x))a 1 (s) ds ) 1 (a 1 x 1(s)s a 1(s))u(x + s(1 x)) ds ) ( 1 ) (a x 1(s) a 1(s)s)u(x + s(1 x)) ds y ( + (1 x) 3 y 1 x ( y = 18 (1 x) 4 y 3 ) 1 (1 x) 3 (1 x) u(x + s(1 x))a 1 (s) ds y ( + 6 (1 x) 3 1 y ) 1 (a 1 x 1(s)s a 1(s))u(x + s(1 x)) ds y ( + (1 x) 3 y ) 1 1 (36s 6s + 18)u(x + s(1 x)) ds 1 x 1 x y ( (1 x) 3 y ) 6 1 u(x) u(x + s(1 x)) ds. 1 x 1 x We start with the second derivative with respect to y. ψ C.S. 1 1 x 1 1 y y (1 x) u(x + s(1 x)) ds dy dx L ( ˆT ) = 1 x 1 x t 1 t = 1 (1 x) 1 x u(t) Fub. d(x, t) = (1 x) u(t) dt dx t 1 x t 1 (1 x) dx u(t) dt u(t) d(x, t) (1 x) t u(t) dt u H 1/ (E 1). The first term of the mixed derivation can be estimated as ψ L ( ˆT ) and the second term as ψ y y, so we also get L ( ˆT ) ψ y x u. H 1/ (E 1) L ( ˆT ) It remains the second order derivation of x. Again the first three terms can be estimated similar to before. It remains the last term ( y (1 x) 3 y ) 6 1 u(x) u(x + s(1 x)) ds. 1 x 1 x For this we refer to the estimation of in the proof of theorem 6.3 to bound φ corr x x L ( ˆT ) the L norm with u H 1/ (E 1 ). Altogether we have ψ x x L ( ˆT ) u H 1/ (E 1), 86

95 6. Normal extension E N and so ψ H ( ˆT ) u H 1/ (E 1). The estimations for ξ can be done in the same way by using the introduced techniques, and thus we get ψ H ( ˆT ) u H 1/ (E 1) and φ H ( ˆT ) u H 1/ (E 1). Step 3: For the last step we define and have E N (u)(x, y) = φ 3 (x, y) := φ (x, y) a ( ) y b x + y ( ) y + b x + y φ 3 E1 = φ E1 = φ 3 E = φ E a(y) φ (, 1) }{{} = ( ) y φ (, x + y) x + y (x + y) φ (, x + y) x (x + y) φ (, x + y), y b(1 x) φ (, 1) +b(1 x) φ (, 1) = } x {{}} x {{} = = φ 3 E3 = φ E3 a(1)φ (, y) = = φ (, y) φ (, y) =. In figure 6.8 we can observe the final extension φ 3 including the first and the second correction for the example u = 1x (1 x). Figure 6.8: Values (lower scale) and partial derivatives (upper scale) of the resulting extension φ 3 for u = 1x (1 x) 87